[{"id":"math-line-circle-001","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":1,"stem":"Find the slope of the line passing through $A(1, 2)$ and $B(3, 6)$.","stemZh":"求过 $A(1, 2)、B(3, 6)$ 两点的直线的斜率。","options":[{"id":"A","text":"1/2"},{"id":"B","text":"2"},{"id":"C","text":"−2"},{"id":"D","text":"4"}],"answer":"B","explanation":"Using $k = (y_{2} − y_{1})/(x_{2} − x_{1})$, we get $k = (6 − 2)/(3 − 1) = 4/2 = 2$. Option A is the result of swapping numerator and denominator.","explanationZh":"由斜率公式 $k = (y_{2} − y_{1})/(x_{2} − x_{1})，得 k = (6 − 2)/(3 − 1) = 4/2 = 2。A$ 项是把分子分母颠倒的结果。","tags":["easy","slope","two-point-form"],"status":"published"},{"id":"math-line-circle-002","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":1,"stem":"Find the slope of the line passing through $P(2, 5)$ and $Q(2, −1)$.","stemZh":"求过 $P(2, 5)、Q(2, −1)$ 两点的直线的斜率。","options":[{"id":"A","text":"0"},{"id":"B","text":"4"},{"id":"C","text":"The slope does not exist"},{"id":"D","text":"−3"}],"answer":"C","explanation":"Both points have $x = 2$, so the line is vertical $(x = 2). A$ vertical line has no $slope: k = (y_{2} − y_{1})/(x_{2} − x_{1})$ would require division by 0. Its angle of inclination is 90°, but the slope is undefined.","explanationZh":"两点横坐标都是 $2，直线为 x = 2，垂直于 x$ 轴。此时 $x_{2} − x_{1} = 0，斜率不存在（倾斜角为 90°，但斜率无定义）。$","tags":["easy","slope","vertical-line"],"status":"published"},{"id":"math-line-circle-003","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":1,"stem":"Find the distance between the points $A(1, 2)$ and $B(4, 6)$.","stemZh":"$求 A(1, 2) 与 B(4, 6)$ 两点间的距离。","options":[{"id":"A","text":"5"},{"id":"B","text":"7"},{"id":"C","text":"$\\sqrt{13}$"},{"id":"D","text":"25"}],"answer":"A","explanation":"$|AB| = \\sqrt{(4 − 1)^{2} + (6 − 2)^{2}} = \\sqrt{9 + 16} = \\sqrt{25} = 5$. Option D is the value inside the square root before taking the root.","explanationZh":"$|AB| = \\sqrt{(4 − 1)^{2} + (6 − 2)^{2}} = \\sqrt{9 + 16} = \\sqrt{25} = 5。D$ 项是忘记开平方的结果。","tags":["easy","distance-formula"],"status":"published"},{"id":"math-line-circle-004","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":2,"stem":"Find the midpoint of the segment joining $A(−2, 3)$ and $B(4, −1)$.","stemZh":"$求 A(−2, 3) 与 B(4, −1)$ 连线的中点坐标。","options":[{"id":"A","text":"$(2, 2)$"},{"id":"B","text":"$(−6, −3)$"},{"id":"C","text":"$(1, −1)$"},{"id":"D","text":"$(1, 1)$"}],"answer":"D","explanation":"Midpoint $= ((x_{1} + x_{2})/2, (y_{1} + y_{2})/2) = ((−2 + 4)/2, (3 + (−1))/2) = (1, 1)$. Option A adds without halving; option B multiplies instead of dividing.","explanationZh":"中点坐标 $= ((x_{1} + x_{2})/2, (y_{1} + y_{2})/2) = ((−2 + 4)/2, (3 − 1)/2) = (1, 1)。A$ 项忘记除以 2，B 项错误地相乘。","tags":["easy","midpoint"],"status":"published"},{"id":"math-line-circle-005","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":1,"stem":"What is the angle of inclination of the line $y =$ x?","stemZh":"直线 $y = x$ 的倾斜角是多少？","options":[{"id":"A","text":"30°"},{"id":"B","text":"45°"},{"id":"C","text":"60°"},{"id":"D","text":"90°"}],"answer":"B","explanation":"The slope is $k = 1$ and $k = \\tan θ$ with $0° ≤ θ < 180°$, so $\\tan θ = 1$ gives $θ = 45°$.","explanationZh":"斜率 $k = 1，又 k = \\tan θ（0° ≤ θ < 180°），由 \\tan θ = 1 得 θ = 45°。$","tags":["easy","angle-of-inclination"],"status":"published"},{"id":"math-line-circle-006","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":1,"stem":"Write the equation of the line with slope 2 and y-intercept 3.","stemZh":"写出斜率为 2、在 y 轴上截距为 3 的直线方程。","options":[{"id":"A","text":"$y = 3x + 2$"},{"id":"B","text":"$y = 2x − 3$"},{"id":"C","text":"$y = 2x + 3$"},{"id":"D","text":"$2x + 3y = 1$"}],"answer":"C","explanation":"In slope-intercept form $y =$ kx + b, k is the slope and b is the y-intercept, so $y = 2x + 3$. Option A swaps the two numbers.","explanationZh":"斜截式 $y =$ kx + b 中 k 为斜率、b 为 y 轴截距，故方程为 $y = 2x + 3。A$ 项把两个数交换了位置。","tags":["easy","slope-intercept-form","intercept"],"status":"published"},{"id":"math-line-circle-007","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":2,"stem":"Write the standard equation of the circle with centre $(1, −2)$ and radius 3.","stemZh":"写出圆心为 $(1, −2)、半径为 3$ 的圆的标准方程。","options":[{"id":"A","text":"$(x + 1)^{2} + (y − 2)^{2} = 9$"},{"id":"B","text":"$(x − 1)^{2} + (y + 2)^{2} = 3$"},{"id":"C","text":"$(x − 1)^{2} + (y − 2)^{2} = 9$"},{"id":"D","text":"$(x − 1)^{2} + (y + 2)^{2} = 9$"}],"answer":"D","explanation":"The standard form is $(x − a)^{2} + (y − b)^{2} = r^{2}$ with centre $(a, b)$. Here $(x − 1)^{2} + (y − (−2))^{2} = 3^{2}$, i.e. $(x − 1)^{2} + (y + 2)^{2} = 9$. Note the right-hand side is $r^{2} = 9$, not $r = 3$.","explanationZh":"标准方程为 $(x − a)^{2} + (y − b)^{2} = r^{2}，其中 (a, b)$ 为圆心。代入得 $(x − 1)^{2} + (y + 2)^{2} = 3^{2} = 9，注意右边是 r^{2}$ 而不是 r。","tags":["easy","circle-equation","standard-equation"],"status":"published"},{"id":"math-line-circle-008","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":1,"stem":"Where does the point $P(3, 4)$ lie relative to the circle $x^{2} + y^{2} = 25$?","stemZh":"$点 P(3, 4)$ 与圆 $x^{2} + y^{2} = 25$ 的位置关系是？","options":[{"id":"A","text":"On the circle"},{"id":"B","text":"Inside the circle"},{"id":"C","text":"Outside the circle"},{"id":"D","text":"At the centre"}],"answer":"A","explanation":"Substitute the point into the left $side: 3^{2} + 4^{2} = 9 + 16 = 25$, which equals $r^{2}$. Hence P lies exactly on the circle.","explanationZh":"$把点代入左边：3^{2} + 4^{2} = 9 + 16 = 25 = r^{2}，所以点 P$ 在圆上。","tags":["easy","point-and-circle"],"status":"published"},{"id":"math-line-circle-009","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":2,"stem":"Which of the following lines is parallel to $y = 3x + 1$?","stemZh":"下列哪条直线与 $y = 3x + 1$ 平行？","options":[{"id":"A","text":"$y = −3x + 1$"},{"id":"B","text":"$y = (1/3)x + 1$"},{"id":"C","text":"$y = 3x − 5$"},{"id":"D","text":"$x + 3y = 1$"}],"answer":"C","explanation":"Two non-vertical lines are parallel when their slopes are equal. The given line has slope 3, and $y = 3x − 5$ also has slope $3 (with a$ different intercept, so the lines are distinct). Option B has the negative reciprocal slope and option A the negative slope.","explanationZh":"两直线平行（不重合）的条件是斜率相等。已知直线斜率为 $3，y = 3x − 5$ 斜率也是 3 且截距不同，故平行。B 项是负倒数斜率，A 项是相反数。","tags":["easy","parallel-lines","slope"],"status":"published"},{"id":"math-line-circle-010","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":2,"stem":"Which of the following lines is perpendicular to $y = 2x + 1$?","stemZh":"下列哪条直线与 $y = 2x + 1$ 垂直？","options":[{"id":"A","text":"$y = 2x − 1$"},{"id":"B","text":"$y = −(1/2)x + 4$"},{"id":"C","text":"$y = (1/2)x + 4$"},{"id":"D","text":"$y = −2x + 4$"}],"answer":"B","explanation":"For two lines with slopes $k_{1}$ and $k_{2}$, perpendicular means $k_{1}k_{2} = −1$. Here $k_{1} = 2$, so $k_{2} = −1/2$, giving $y = −(1/2)x + 4$. Option D uses the negative slope instead of the negative reciprocal.","explanationZh":"两直线垂直的条件是 $k_{1}k_{2} = −1。已知 k_{1} = 2，故 k_{2} = −1/2，即 y = −(1/2)x + 4。D$ 项错误地取了相反数而不是负倒数。","tags":["easy","perpendicular","slope"],"status":"published"},{"id":"math-line-circle-011","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":3,"stem":"Find the distance from the point $P(1, 2)$ to the line $3x + 4y − 5 = 0$.","stemZh":"求点 $P(1, 2)$ 到直线 $3x + 4y − 5 = 0$ 的距离。","options":[{"id":"A","text":"6/5"},{"id":"B","text":"6"},{"id":"C","text":"1"},{"id":"D","text":"5/6"}],"answer":"A","explanation":"$d = |Ax_0 + By_0 + C| / \\sqrt{A^{2} + B^{2}} = |3(1) + 4(2) − 5| / \\sqrt{9 + 16} = |3 + 8 − 5|/5 = 6/5$. Option B forgets to divide by $\\sqrt{A^{2} + B^{2}} = 5$, and option D divides the wrong way round.","explanationZh":"$d = |Ax_0 + By_0 + C| / \\sqrt{A^{2} + B^{2}} = |3 + 8 − 5| / \\sqrt{9 + 16} = 6/5。B$ 项忘记除以 $\\sqrt{A^{2} + B^{2}} = 5，D$ 项分子分母颠倒。","tags":["medium","point-to-line-distance"],"status":"published"},{"id":"math-line-circle-012","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":3,"stem":"Find the point of intersection of the lines $2x + y = 5$ and $x − y = 1$.","stemZh":"求直线 $2x + y = 5 与 x − y = 1$ 的交点坐标。","options":[{"id":"A","text":"$(1, 2)$"},{"id":"B","text":"$(2, −1)$"},{"id":"C","text":"$(3, 1)$"},{"id":"D","text":"$(2, 1)$"}],"answer":"D","explanation":"Adding the two equations gives $3x = 6$, so $x = 2$; substituting into $x − y = 1$ gives $y = 1$. The intersection point is $(2, 1)$, which satisfies both equations.","explanationZh":"两式相加得 $3x = 6，x = 2；代入 x − y = 1 得 y = 1。交点为 (2, 1)，代入两方程均成立。$","tags":["medium","intersection"],"status":"published"},{"id":"math-line-circle-013","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":3,"stem":"Find the equation of the line through $(−1, 3)$ that is parallel to $2x − y + 4 = 0$.","stemZh":"求过点 $(−1, 3)$ 且与直线 $2x − y + 4 = 0$ 平行的直线方程。","options":[{"id":"A","text":"$2x − y − 5 = 0$"},{"id":"B","text":"$2x − y + 5 = 0$"},{"id":"C","text":"$x + 2y − 5 = 0$"},{"id":"D","text":"$2x + y + 5 = 0$"}],"answer":"B","explanation":"The line $2x − y + 4 = 0$ has slope 2, so the required line has slope 2 as $well: y − 3 = 2(x + 1)$, i.e. $y = 2x + 5$, or $2x − y + 5 = 0. Check: 2(−1) − 3 + 5 = 0$.","explanationZh":"直线 $2x − y + 4 = 0$ 的斜率为 $2，所求直线斜率也为 2：y − 3 = 2(x + 1)，即 2x − y + 5 = 0。验证：2(−1) − 3 + 5 = 0。$","tags":["medium","parallel-lines","point-slope-form"],"status":"published"},{"id":"math-line-circle-014","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":3,"stem":"Find the centre and radius of the circle $x^{2} + y^{2} − 4x + 6y − 3 = 0$.","stemZh":"求圆 $x^{2} + y^{2} − 4x + 6y − 3 = 0$ 的圆心和半径。","options":[{"id":"A","text":"centre $(−2, 3)$, radius 4"},{"id":"B","text":"centre $(2, −3)$, radius 16"},{"id":"C","text":"centre $(2, −3)$, radius 4"},{"id":"D","text":"centre $(2, 3)$, radius 4"}],"answer":"C","explanation":"Complete the $square: (x^{2} − 4x + 4) + (y^{2} + 6y + 9) = 3 + 4 + 9$, so $(x − 2)^{2} + (y + 3)^{2} = 16$. The centre is $(2, −3)$ and $r = \\sqrt{16} = 4$. Option B mistakes $r^{2} = 16$ for the radius.","explanationZh":"$配方：(x^{2} − 4x + 4) + (y^{2} + 6y + 9) = 3 + 4 + 9，即 (x − 2)^{2} + (y + 3)^{2} = 16。圆心为 (2, −3)，半径 r = \\sqrt{16} = 4。B$ 项把 $r^{2} = 16$ 当成了半径。","tags":["medium","completing-the-square","general-form"],"status":"published"},{"id":"math-line-circle-015","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":3,"stem":"For the line $3x + 4y + c = 0$ to be tangent to the circle $x^{2} + y^{2} = 25$, what must $|c|$ be?","stemZh":"若直线 $3x + 4y + c = 0$ 与圆 $x^{2} + y^{2} = 25$ 相切，则 $|c|$ 等于多少？","options":[{"id":"A","text":"5"},{"id":"B","text":"−25"},{"id":"C","text":"125"},{"id":"D","text":"25"}],"answer":"D","explanation":"A line is tangent to a circle when the distance from the centre to the line equals the $radius: |c|/\\sqrt{3^{2} + 4^{2}} = 5$, so $|c|/5 = 5$ and $|c| = 25 (c = 25$ or $c = −25)$. Option A is the radius itself; option B cannot be an absolute value.","explanationZh":"$相切条件是圆心到直线的距离等于半径：|c|/\\sqrt{3^{2} + 4^{2}} = 5，即 |c|/5 = 5，得 |c| = 25（c = \\pm25）。A$ 项是半径本身；B 项绝对值不可能为负。","tags":["medium","tangent-condition","point-to-line-distance"],"status":"published"},{"id":"math-line-circle-016","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":3,"stem":"Under which condition does $x^{2} + y^{2} +$ Dx + Ey $+ F = 0$ represent a (real) circle?","stemZh":"方程 $x^{2} + y^{2} +$ Dx + Ey $+ F = 0$ 表示圆的充要条件是？","options":[{"id":"A","text":"$D^{2} + E^{2} − 4F > 0$"},{"id":"B","text":"$D^{2} + E^{2} − 4F < 0$"},{"id":"C","text":"$D^{2} + E^{2} − 4F = 0$"},{"id":"D","text":"$D + E − F > 0$"}],"answer":"A","explanation":"Completing the square gives $(x + D/2)^{2} + (y + E/2)^{2} = (D^{2} + E^{2} − 4F)/4$. This is a real circle exactly when the right side is positive, i.e. $D^{2} + E^{2} − 4F > 0$. If it equals 0 the equation is a single point, and if it is negative no real points satisfy it.","explanationZh":"配方得 $(x + D/2)^{2} + (y + E/2)^{2} = (D^{2} + E^{2} − 4F)/4。当 D^{2} + E^{2} − 4F > 0$ 时表示实圆；等于 0 时表示一个点；小于 0 时不表示任何实图形。","tags":["medium","general-form","circle-equation"],"status":"published"},{"id":"math-line-circle-017","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":3,"stem":"Find the equation of the line through the intercepts $(0, 4)$ and $(2, 0)$.","stemZh":"求过 $(0, 4) 和 (2, 0)$ 两点的直线方程。","options":[{"id":"A","text":"$x + 2y = 4$"},{"id":"B","text":"$2x + y = 4$"},{"id":"C","text":"$2x − y = 4$"},{"id":"D","text":"$x − 2y = 4$"}],"answer":"B","explanation":"The slope is $(0 − 4)/(2 − 0) = −2$, so $y = −2x + 4$, i.e. $2x + y = 4$. Equivalently use the intercept form $x/2 + y/4 = 1. Check: (0, 4)$ gives $4 = 4$ and $(2, 0)$ gives $4 = 4$.","explanationZh":"斜率 $k = (0 − 4)/(2 − 0) = −2，故 y = −2x + 4，即 2x + y = 4。也可用截距式 x/2 + y/4 = 1。代入两点均成立。$","tags":["medium","intercept-form","two-point-form"],"status":"published"},{"id":"math-line-circle-018","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":3,"stem":"A line has slope −1. What is its angle of inclination?","stemZh":"一条直线的斜率为 −1，则它的倾斜角为多少？","options":[{"id":"A","text":"45°"},{"id":"B","text":"−45°"},{"id":"C","text":"135°"},{"id":"D","text":"90°"}],"answer":"C","explanation":"$k = \\tan θ$ with $0° ≤ θ < 180°$. Since $\\tan θ = −1 < 0, θ$ is $obtuse: θ = 180° − 45° = 135°. A$ negative angle such as −45° is not allowed for an angle of inclination.","explanationZh":"$由 k = \\tan θ 且 0° ≤ θ < 180°，\\tan θ = −1 < 0$ 说明 $θ 为钝角：θ = 180° − 45° = 135°。倾斜角的取值范围不允许 −45°。$","tags":["medium","angle-of-inclination"],"status":"published"},{"id":"math-line-circle-019","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":3,"stem":"Find the distance between the parallel lines $3x + 4y + 1 = 0$ and $3x + 4y − 9 = 0$.","stemZh":"求两条平行直线 $3x + 4y + 1 = 0 与 3x + 4y − 9 = 0$ 之间的距离。","options":[{"id":"A","text":"1"},{"id":"B","text":"5"},{"id":"C","text":"4"},{"id":"D","text":"2"}],"answer":"D","explanation":"For parallel lines Ax + By $+ C_1 = 0$ and Ax + By $+ C_2 = 0, d = |C_1 − C_2| / \\sqrt{A^{2} + B^{2}} = |1 − (−9)| / \\sqrt{9 + 16} = 10/5 = 2$. Equivalently take any point on one line and use the point-to-line distance formula.","explanationZh":"两平行线 Ax + By $+ C_1 = 0 与$ Ax + By $+ C_2 = 0$ 的距离为 $d = |C_1 − C_2| / \\sqrt{A^{2} + B^{2}} = |1 − (−9)|/5 = 10/5 = 2。也可在其中一条上取一点代入点到直线距离公式。$","tags":["medium","parallel-lines","point-to-line-distance"],"status":"published"},{"id":"math-line-circle-020","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":3,"stem":"Where does the point $P(1, 1)$ lie relative to the circle $(x − 2)^{2} + (y − 3)^{2} = 4$?","stemZh":"$点 P(1, 1)$ 与圆 $(x − 2)^{2} + (y − 3)^{2} = 4$ 的位置关系是？","options":[{"id":"A","text":"Outside the circle"},{"id":"B","text":"On the circle"},{"id":"C","text":"Inside the circle"},{"id":"D","text":"At the centre"}],"answer":"A","explanation":"The centre is $C(2, 3)$ and $r = 2. |PC|^{2} = (1 − 2)^{2} + (1 − 3)^{2} = 1 + 4 = 5$, so $|PC| = \\sqrt{5} > 2 =$ r. Hence P lies outside the circle.","explanationZh":"圆心 $C(2, 3)，半径 r = 2。|PC|^{2} = (1 − 2)^{2} + (1 − 3)^{2} = 5，故 |PC| = \\sqrt{5} > 2，点在圆外。$","tags":["medium","point-and-circle"],"status":"published"},{"id":"math-line-circle-021","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":4,"stem":"Which of the following equations does NOT represent a circle?","stemZh":"下列方程中不表示圆的是？","options":[{"id":"A","text":"$x^{2} + y^{2} − 2x + 4y + 1 = 0$"},{"id":"B","text":"$x^{2} + y^{2} − 2x + 4y + 5 = 0$"},{"id":"C","text":"$x^{2} + y^{2} − 2x + 4y − 1 = 0$"},{"id":"D","text":"$x^{2} + y^{2} − 2x + 4y + 4 = 0$"}],"answer":"B","explanation":"All four have $D = −2, E = 4$, so $r^{2} = (D^{2} + E^{2} − 4F)/4 = (4 + 16 − 4F)/4 = 5 −$ F. For $B, F = 5$ gives $r^{2} = 0: (x − 1)^{2} + (y + 2)^{2} = 0$, which is the single point $(1, −2)$, not a circle. The others give $r^{2} = 4, 6$ and 1 respectively.","explanationZh":"四式的 $D = −2、E = 4，故 r^{2} = (D^{2} + E^{2} − 4F)/4 = 5 −$ F。B $项 F = 5 时 r^{2} = 0，即 (x − 1)^{2} + (y + 2)^{2} = 0，只表示点 (1, −2)，不是圆。其余各项 r^{2}$ 分别为 4、6、1。","tags":["hard","general-form","circle-equation"],"status":"published"},{"id":"math-line-circle-022","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":4,"stem":"Find the distance from $P(1, −1)$ to the line passing through $A(2, 1)$ and $B(−1, 3)$.","stemZh":"求点 $P(1, −1)$ 到过 $A(2, 1)、B(−1, 3)$ 两点的直线的距离。","options":[{"id":"A","text":"$8\\sqrt{13}/13$"},{"id":"B","text":"$4\\sqrt{13}/13$"},{"id":"C","text":"8/13"},{"id":"D","text":"$2\\sqrt{13}/13$"}],"answer":"A","explanation":"Slope of $AB: k = (3 − 1)/(−1 − 2) = −2/3$. Using point-slope form at $A: y − 1 = (−2/3)(x − 2) → 2x + 3y − 7 = 0$. Then $d = |2(1) + 3(−1) − 7| / \\sqrt{4 + 9} = 8/\\sqrt{13} = 8\\sqrt{13}/13$.","explanationZh":"AB 的斜率 $k = (3 − 1)/(−1 − 2) = −2/3，由点斜式得 y − 1 = (−2/3)(x − 2)，即 2x + 3y − 7 = 0。故 d = |2 − 3 − 7| / \\sqrt{4 + 9} = 8/\\sqrt{13} = 8\\sqrt{13}/13。$","tags":["hard","point-to-line-distance","two-point-form"],"status":"published"},{"id":"math-line-circle-023","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":4,"stem":"For what value of m does $x^{2} + y^{2} + 4x − 2y + m = 0$ represent a circle of radius 3?","stemZh":"当 m 为何值时，方程 $x^{2} + y^{2} + 4x − 2y + m = 0$ 表示半径为 3 的圆？","options":[{"id":"A","text":"−5"},{"id":"B","text":"4"},{"id":"C","text":"−4"},{"id":"D","text":"0"}],"answer":"C","explanation":"Complete the $square: (x + 2)^{2} + (y − 1)^{2} = 5 −$ m. Setting $r^{2} = 9$ gives $5 − m = 9$, so $m = −4$. The sign is the trap: a negative constant term makes the right-hand side larger.","explanationZh":"配方得 $(x + 2)^{2} + (y − 1)^{2} = 5 −$ m。令 $r^{2} = 9，即 5 − m = 9，得 m = −4。注意符号：常数项为负时右边反而更大。$","tags":["hard","completing-the-square","general-form"],"status":"published"},{"id":"math-line-circle-024","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":4,"stem":"Find the equation of the tangent to the circle $x^{2} + y^{2} = 13$ at the point $(2, 3)$.","stemZh":"求圆 $x^{2} + y^{2} = 13$ 在点 $(2, 3)$ 处的切线方程。","options":[{"id":"A","text":"$2x + 3y = 13$"},{"id":"B","text":"$3x + 2y = 13$"},{"id":"C","text":"$2x − 3y = 13$"},{"id":"D","text":"$2x + 3y = −13$"}],"answer":"A","explanation":"First check that $(2, 3)$ is on the $circle: 2^{2} + 3^{2} = 13$. For $x^{2} + y^{2} = r^{2}$ the tangent at $(x_{0}, y_{0})$ is $x_{0}x + y_{0}y = r^{2}$, giving $2x + 3y = 13$. The radius to $(2, 3)$ has slope 3/2, and the tangent's slope −2/3 is its negative reciprocal.","explanationZh":"先验证 $(2, 3) 在圆上：2^{2} + 3^{2} = 13。圆 x^{2} + y^{2} = r^{2} 在 (x_{0}, y_{0})$ 处的切线为 $x_{0}x + y_{0}y = r^{2}，即 2x + 3y = 13。半径斜率为 3/2，切线斜率 −2/3$ 恰为其负倒数。","tags":["hard","tangent-condition","circle-equation"],"status":"published"},{"id":"math-line-circle-025","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":5,"stem":"How many tangent lines can be drawn from the point $P(3, 4)$ to the circle $x^{2} + y^{2} = 4$?","stemZh":"过点 $P(3, 4)$ 可作圆 $x^{2} + y^{2} = 4$ 的几条切线？","options":[{"id":"A","text":"0"},{"id":"B","text":"1"},{"id":"C","text":"2"},{"id":"D","text":"Infinitely many"}],"answer":"C","explanation":"The centre is $O(0, 0)$ and $r = 2. |OP| = \\sqrt{9 + 16} = 5 > 2$, so P lies outside the circle, and exactly two tangents can be drawn from it. (One tangent if P is on the circle, none if P is inside.)","explanationZh":"圆心 $O(0, 0)，半径 r = 2。|OP| = \\sqrt{9 + 16} = 5 > 2，点在圆外，故可作两条切线。（点在圆上只能作一条，点在圆内则没有切线。）$","tags":["hard","tangent-condition","point-and-circle"],"status":"published"},{"id":"math-line-circle-026","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":5,"stem":"The lines kx $+ y + 1 = 0$ and x + ky $+ 2 = 0$ are perpendicular. Find k.","stemZh":"若直线 kx $+ y + 1 = 0 与 x +$ ky $+ 2 = 0$ 垂直，求 k。","options":[{"id":"A","text":"−1"},{"id":"B","text":"1"},{"id":"C","text":"2"},{"id":"D","text":"0"}],"answer":"D","explanation":"For general-form lines $A_{1}x + B_{1}y + C_{1} = 0$ and $A_{2}x + B_{2}y + C_{2} = 0$, perpendicular means $A_{1}A_{2} + B_{1}B_{2} = 0$. Here $k·1 + 1·k = 2k = 0$, so $k = 0$. Indeed the lines become $y = −1 (horizontal)$ and $x = −2 (vertical)$. Using $k_{1}k_{2} = −1$ fails here because one line is vertical and has no slope.","explanationZh":"一般式下两直线垂直的条件为 $A_{1}A_{2} + B_{1}B_{2} = 0，即 k·1 + 1·k = 2k = 0，得 k = 0。此时两直线为 y = −1（水平）与 x = −2（竖直）。若用 k_{1}k_{2} = −1$ 会失效，因为其中一条直线斜率不存在。","tags":["hard","perpendicular","general-form"],"status":"published"},{"id":"math-line-circle-027","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":5,"stem":"Find the point on the line $2x + y − 4 = 0$ that is closest to the origin.","stemZh":"求直线 $2x + y − 4 = 0$ 上到原点距离最小的点。","options":[{"id":"A","text":"$(4/5, 8/5)$"},{"id":"B","text":"$(8/5, 4/5)$"},{"id":"C","text":"$(2, 0)$"},{"id":"D","text":"$(0, 4)$"}],"answer":"B","explanation":"The closest point is the foot of the perpendicular from O. The given line has slope −2, so the perpendicular through O has slope $1/2: y = x/2$, i.e. $x = 2y$. Substituting into $2x + y = 4$ gives $5y = 4$, so $y = 4/5$ and $x = 8/5$. The point is $(8/5, 4/5)$.","explanationZh":"最近点是原点到该直线的垂足。已知直线斜率为 $−2，过原点的垂线斜率为 1/2，即 y = x/2（x = 2y）。代入 2x + y = 4 得 5y = 4，y = 4/5，x = 8/5，即 (8/5, 4/5)。$","tags":["hard","perpendicular","foot-of-perpendicular"],"status":"published"},{"id":"math-line-circle-028","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":5,"stem":"A circle passes through $A(1, 1), B(3, 1)$ and $C(1, 3)$. Find its centre.","stemZh":"一圆过 $A(1, 1)、B(3, 1)、C(1, 3)$ 三点，求圆心坐标。","options":[{"id":"A","text":"$(2, 1)$"},{"id":"B","text":"$(1, 2)$"},{"id":"C","text":"$(3, 3)$"},{"id":"D","text":"$(2, 2)$"}],"answer":"D","explanation":"The centre lies on the perpendicular bisector of every chord. Chord AB is horizontal from $(1, 1)$ to $(3, 1)$, so its perpendicular bisector is $x = 2$. Chord AC is vertical from $(1, 1)$ to $(1, 3)$, so its perpendicular bisector is $y = 2$. The centre is $(2, 2)$, with $r = \\sqrt{2}$.","explanationZh":"圆心在任意弦的垂直平分线上。弦 AB 水平，其中垂线为 $x = 2；弦$ AC 竖直，其中垂线为 $y = 2。故圆心为 (2, 2)，半径 r = \\sqrt{2}。$","tags":["hard","circle-equation","perpendicular"],"status":"published"},{"id":"math-line-circle-029","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":4,"stem":"Find the area enclosed by the circle $x^{2} + y^{2} − 6x − 8y = 0$.","stemZh":"求圆 $x^{2} + y^{2} − 6x − 8y = 0$ 的面积。","options":[{"id":"A","text":"$5π$"},{"id":"B","text":"$10π$"},{"id":"C","text":"$25π$"},{"id":"D","text":"$100π$"}],"answer":"C","explanation":"Completing the $square: (x − 3)^{2} + (y − 4)^{2} = 9 + 16 = 25$, so $r = 5 (note F = 0\\text{ here, so }r^{2} = (D^{2} + E^{2})/4)$. The area is $πr^{2} = 25π$.","explanationZh":"配方得 $(x − 3)^{2} + (y − 4)^{2} = 9 + 16 = 25，故 r = 5（此处 F = 0，r^{2} = (D^{2} + E^{2})/4）。面积 S = πr^{2} = 25π。$","tags":["hard","completing-the-square","general-form"],"status":"published"},{"id":"math-line-circle-030","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"line-circle","type":"single-choice","difficulty":5,"stem":"The circle $(x − 1)^{2} + (y − 2)^{2} = 4$ has two tangents parallel to $3x + 4y = 0$. One of them is $3x + 4y − 1 = 0$. Find the other.","stemZh":"$圆 (x − 1)^{2} + (y − 2)^{2} = 4$ 有两条与 $3x + 4y = 0$ 平行的切线，其中一条为 $3x + 4y − 1 = 0，求另一条。$","options":[{"id":"A","text":"$3x + 4y − 21 = 0$"},{"id":"B","text":"$3x + 4y + 21 = 0$"},{"id":"C","text":"$3x + 4y − 11 = 0$"},{"id":"D","text":"$3x + 4y + 1 = 0$"}],"answer":"A","explanation":"Write a parallel tangent as $3x + 4y − c = 0$. Tangency requires the distance from the centre $(1, 2)$ to equal $r = 2: |3(1) + 4(2) − c|/5 = 2 → |11 − c| = 10 → c = 1$ or $c = 21$. So the other tangent is $3x + 4y − 21 = 0$.","explanationZh":"设平行切线为 $3x + 4y − c = 0。相切要求圆心 (1, 2)$ 到直线的距离等于 $r = 2：|3 + 8 − c|/5 = 2，即 |11 − c| = 10，解得 c = 1 或 c = 21。故另一条切线为 3x + 4y − 21 = 0。$","tags":["hard","tangent-condition","parallel-lines"],"status":"published"},{"id":"math-conics-001","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":1,"stem":"For the ellipse $x^{2}/25 + y^{2}/16 = 1$, which axis contains the major axis?","stemZh":"椭圆 $x^{2}/25 + y^{2}/16 = 1$ 的长轴在哪个轴上？","options":[{"id":"A","text":"The x-axis"},{"id":"B","text":"The y-axis"},{"id":"C","text":"The line $y = x$"},{"id":"D","text":"Neither axis"}],"answer":"A","explanation":"In $x^{2}/a^{2} + y^{2}/b^{2} = 1$ with $a > b > 0$, the larger denominator is $a^{2}$. Here $25 > 16$, so $a^{2} = 25$ sits under $x^{2}$, and the major axis lies on the x-axis with length $2a = 10$.","explanationZh":"标准方程 $x^{2}/a^{2} + y^{2}/b^{2} = 1（a > b > 0）中，较大的分母是 a^{2}。此处 25 > 16，a^{2} = 25 在 x^{2}$ 下方，故长轴在 x 轴上，长轴长 $2a = 10。$","tags":["easy","ellipse","major-axis"],"status":"published"},{"id":"math-conics-002","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":1,"stem":"For the ellipse $x^{2}/25 + y^{2}/16 = 1$, find the focal half-distance c.","stemZh":"椭圆 $x^{2}/25 + y^{2}/16 = 1$ 的半焦距 c 为多少？","options":[{"id":"A","text":"9"},{"id":"B","text":"3"},{"id":"C","text":"41"},{"id":"D","text":"$\\sqrt{41}$"}],"answer":"B","explanation":"For an $ellipse, c^{2} = a^{2} − b^{2} = 25 − 16 = 9$, so $c = 3 (the$ foci are at $(\\pm3, 0))$. Option A is $c^{2}$, and $\\sqrt{41}$ would come from wrongly adding as in a hyperbola.","explanationZh":"椭圆中 $c^{2} = a^{2} − b^{2} = 25 − 16 = 9，故 c = 3（焦点为 (\\pm3, 0)）。A$ 项是 $c^{2}；\\sqrt{41}$ 是误用双曲线的 $c^{2} = a^{2} + b^{2}$ 得出的。","tags":["easy","ellipse","focal-distance"],"status":"published"},{"id":"math-conics-003","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":2,"stem":"Find the eccentricity of the ellipse $x^{2}/25 + y^{2}/16 = 1$.","stemZh":"求椭圆 $x^{2}/25 + y^{2}/16 = 1$ 的离心率。","options":[{"id":"A","text":"3/5"},{"id":"B","text":"4/5"},{"id":"C","text":"5/3"},{"id":"D","text":"3/4"}],"answer":"A","explanation":"$a = 5, b = 4$ and $c^{2} = a^{2} − b^{2} = 9$, so $c = 3$. Hence $e = c/a = 3/5$. Option B is b/a and option C is a/c, neither of which is the eccentricity.","explanationZh":"$a = 5，b = 4，c^{2} = a^{2} − b^{2} = 9，c = 3，故 e = c/a = 3/5。B$ 项是 b/a，C 项是 a/c，都不是离心率。","tags":["easy","ellipse","eccentricity"],"status":"published"},{"id":"math-conics-004","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":2,"stem":"For the hyperbola $x^{2}/9 − y^{2}/16 = 1$, find the focal half-distance c.","stemZh":"双曲线 $x^{2}/9 − y^{2}/16 = 1$ 的半焦距 c 为多少？","options":[{"id":"A","text":"$\\sqrt{7}$"},{"id":"B","text":"7"},{"id":"C","text":"5"},{"id":"D","text":"25"}],"answer":"C","explanation":"For $a hyperbola, c^{2} = a^{2} + b^{2} = 9 + 16 = 25$, so $c = 5$ and the foci are $(\\pm5, 0)$. Note the relation differs from the ellipse: here $a^{2}$ and $b^{2}$ are added, not subtracted.","explanationZh":"双曲线中 $c^{2} = a^{2} + b^{2} = 9 + 16 = 25，故 c = 5，焦点为 (\\pm5, 0)。注意与椭圆不同：这里是 a^{2} + b^{2}$ 而不是相减。","tags":["easy","hyperbola","focal-distance"],"status":"published"},{"id":"math-conics-005","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":1,"stem":"Find the equations of the asymptotes of the hyperbola $x^{2}/9 − y^{2}/16 = 1$.","stemZh":"求双曲线 $x^{2}/9 − y^{2}/16 = 1$ 的渐近线方程。","options":[{"id":"A","text":"$y = \\pm(3/4)x$"},{"id":"B","text":"$y = \\pm(9/16)x$"},{"id":"C","text":"$y = \\pm(16/9)x$"},{"id":"D","text":"$y = \\pm(4/3)x$"}],"answer":"D","explanation":"For $x^{2}/a^{2} − y^{2}/b^{2} = 1$ the asymptotes are $y = \\pm(b/a)x$. Here $a = 3$ and $b = 4$, so $y = \\pm(4/3)x$. Option A inverts the ratio b/a into a/b.","explanationZh":"双曲线 $x^{2}/a^{2} − y^{2}/b^{2} = 1$ 的渐近线为 $y = \\pm(b/a)x。此处 a = 3、b = 4，故为 y = \\pm(4/3)x。A$ 项把 b/a 错成了 a/b。","tags":["easy","hyperbola","asymptote"],"status":"published"},{"id":"math-conics-006","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":1,"stem":"Find the focus of the parabola $y^{2} = 8x$.","stemZh":"求抛物线 $y^{2} = 8x$ 的焦点坐标。","options":[{"id":"A","text":"$(4, 0)$"},{"id":"B","text":"$(2, 0)$"},{"id":"C","text":"$(0, 2)$"},{"id":"D","text":"$(−2, 0)$"}],"answer":"B","explanation":"Comparing $y^{2} = 8x$ with $y^{2} = 2px$ gives $2p = 8$, so $p = 4$. The focus is $(p/2, 0) = (2, 0)$ and the directrix is $x = −2$. Option A uses p itself instead of p/2.","explanationZh":"$与 y^{2} = 2px$ 比较得 $2p = 8，p = 4。焦点为 (p/2, 0) = (2, 0)，准线为 x = −2。A$ 项误把 p 当成了焦点横坐标（应为 p/2）。","tags":["easy","parabola","focus"],"status":"published"},{"id":"math-conics-007","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":2,"stem":"Find the directrix of the parabola $y^{2} = 8x$.","stemZh":"求抛物线 $y^{2} = 8x$ 的准线方程。","options":[{"id":"A","text":"$x = 2$"},{"id":"B","text":"$x = −4$"},{"id":"C","text":"$x = −2$"},{"id":"D","text":"$y = −2$"}],"answer":"C","explanation":"$y^{2} = 2px$ with $2p = 8$ gives $p = 4$, so the directrix is $x = −p/2 = −2$. The focus $(2, 0)$ and the directrix $x = −2$ are symmetric about the vertex (the origin).","explanationZh":"$由 y^{2} = 2px 得 2p = 8，p = 4，故准线为 x = −p/2 = −2。焦点 (2, 0)$ 与准线 $x = −2$ 关于顶点（原点）对称。","tags":["easy","parabola","directrix"],"status":"published"},{"id":"math-conics-008","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":1,"stem":"Which of the following equations represents a hyperbola?","stemZh":"下列方程中表示双曲线的是？","options":[{"id":"A","text":"$x^{2}/4 − y^{2}/9 = 1$"},{"id":"B","text":"$x^{2}/4 + y^{2}/9 = 1$"},{"id":"C","text":"$y^{2} = 4x$"},{"id":"D","text":"$x^{2} + y^{2} = 4$"}],"answer":"A","explanation":"A minus sign between the two squared terms gives a hyperbola. Option B (plus sign) is an ellipse, option C is a parabola (only one variable is $squared)$, and option D is a circle.","explanationZh":"两个平方项之间为减号时表示双曲线。B 项为加号，是椭圆；C 项只有一个变量平方，是抛物线；D 项是圆。","tags":["easy","hyperbola","standard-equation"],"status":"published"},{"id":"math-conics-009","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":2,"stem":"For the ellipse $x^{2}/25 + y^{2}/16 = 1$, what is the length of the major axis?","stemZh":"椭圆 $x^{2}/25 + y^{2}/16 = 1$ 的长轴长为多少？","options":[{"id":"A","text":"5"},{"id":"B","text":"8"},{"id":"C","text":"16"},{"id":"D","text":"10"}],"answer":"D","explanation":"$a^{2} = 25$, so $a = 5$ and the major axis has length $2a = 10$. The minor axis has length $2b = 8$, which is option B — a common confusion.","explanationZh":"$a^{2} = 25，a = 5，长轴长为 2a = 10。短轴长为 2b = 8（即 B$ 项），这是常见的混淆点。","tags":["easy","ellipse","major-axis"],"status":"published"},{"id":"math-conics-010","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":1,"stem":"In which direction does the parabola $y^{2} = 2px (p > 0)$ open?","stemZh":"抛物线 $y^{2} = 2px（p > 0）的开口方向是？$","options":[{"id":"A","text":"Upward"},{"id":"B","text":"To the right"},{"id":"C","text":"To the left"},{"id":"D","text":"Downward"}],"answer":"B","explanation":"Since $y^{2} ≥ 0$ and $p > 0$, we need $x ≥ 0$, so the curve opens towards the positive x-direction, i.e. to the right. Its axis of symmetry is the x-axis and its focus is $(p/2, 0)$.","explanationZh":"$由 y^{2} ≥ 0 且 p > 0$ 可知 $x ≥ 0，故开口朝 x$ 轴正方向，即向右。对称轴为 x 轴，焦点为 $(p/2, 0)。$","tags":["easy","parabola","axis-of-symmetry"],"status":"published"},{"id":"math-conics-011","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":3,"stem":"An ellipse centred at the origin has its major axis on the x-axis with $a = 5$ and $c = 3$. Find its standard equation.","stemZh":"中心在原点、长轴在 x 轴上的椭圆，a $= 5，c = 3，求其标准方程。$","options":[{"id":"A","text":"$x^{2}/25 + y^{2}/9 = 1$"},{"id":"B","text":"$x^{2}/25 + y^{2}/34 = 1$"},{"id":"C","text":"$x^{2}/25 + y^{2}/16 = 1$"},{"id":"D","text":"$x^{2}/5 + y^{2}/4 = 1$"}],"answer":"C","explanation":"For an ellipse $c^{2} = a^{2} − b^{2}$, so $b^{2} = a^{2} − c^{2} = 25 − 9 = 16$. The equation is $x^{2}/25 + y^{2}/16 = 1$. Option A puts $c^{2}$ under $y^{2}$, and option D uses a and b instead of $a^{2}$ and $b^{2}$ in the denominators.","explanationZh":"椭圆中 $c^{2} = a^{2} − b^{2}，故 b^{2} = a^{2} − c^{2} = 25 − 9 = 16，方程为 x^{2}/25 + y^{2}/16 = 1。A$ 项把 $c^{2}$ 放到了 $y^{2}$ 下面；D 项分母用了 a、b 而非 $a^{2}、b^{2}。$","tags":["medium","ellipse","standard-equation"],"status":"published"},{"id":"math-conics-012","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":3,"stem":"Find the eccentricity of the hyperbola $x^{2}/9 − y^{2}/16 = 1$.","stemZh":"求双曲线 $x^{2}/9 − y^{2}/16 = 1$ 的离心率。","options":[{"id":"A","text":"3/5"},{"id":"B","text":"5/3"},{"id":"C","text":"4/3"},{"id":"D","text":"25/9"}],"answer":"B","explanation":"$c^{2} = a^{2} + b^{2} = 9 + 16 = 25$, so $c = 5$ and $a = 3$, giving $e = c/a = 5/3$. For a hyperbola $e > 1 always$, which immediately rules out option A.","explanationZh":"$c^{2} = a^{2} + b^{2} = 9 + 16 = 25，c = 5，a = 3，故 e = c/a = 5/3。双曲线的离心率恒大于 1，据此可直接排除 A$ 项。","tags":["medium","hyperbola","eccentricity"],"status":"published"},{"id":"math-conics-013","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":3,"stem":"A hyperbola centred at the origin has its real axis on the x-axis with $a = 3$ and $c = 5$. Find its standard equation.","stemZh":"中心在原点、实轴在 x 轴上的双曲线，a $= 3，c = 5，求其标准方程。$","options":[{"id":"A","text":"$x^{2}/9 − y^{2}/25 = 1$"},{"id":"B","text":"$x^{2}/9 + y^{2}/16 = 1$"},{"id":"C","text":"$x^{2}/25 − y^{2}/9 = 1$"},{"id":"D","text":"$x^{2}/9 − y^{2}/16 = 1$"}],"answer":"D","explanation":"For a hyperbola $c^{2} = a^{2} + b^{2}$, so $b^{2} = c^{2} − a^{2} = 25 − 9 = 16$. The equation is $x^{2}/9 − y^{2}/16 = 1$. Notice $b^{2}$ comes from subtracting $a^{2}$ from $c^{2}$, the opposite of the ellipse rule.","explanationZh":"双曲线中 $c^{2} = a^{2} + b^{2}，故 b^{2} = c^{2} − a^{2} = 25 − 9 = 16，方程为 x^{2}/9 − y^{2}/16 = 1。注意这里 b^{2} = c^{2} − a^{2}，与椭圆的 b^{2} = a^{2} − c^{2}$ 正好相反。","tags":["medium","hyperbola","standard-equation"],"status":"published"},{"id":"math-conics-014","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":3,"stem":"Find the foci of the ellipse $x^{2}/16 + y^{2}/25 = 1$.","stemZh":"求椭圆 $x^{2}/16 + y^{2}/25 = 1$ 的焦点坐标。","options":[{"id":"A","text":"$(\\pm3, 0)$"},{"id":"B","text":"$(0, \\pm5)$"},{"id":"C","text":"$(0, \\pm3)$"},{"id":"D","text":"$(\\pm5, 0)$"}],"answer":"C","explanation":"Here $25 > 16$, so $a^{2} = 25$ is under $y^{2}$ and the major axis lies on the y-axis. Then $c^{2} = a^{2} − b^{2} = 25 − 16 = 9, c = 3$, and the foci are $(0, \\pm3)$. Option A has the right value of c but puts the foci on the wrong axis.","explanationZh":"$因 25 > 16，a^{2} = 25 在 y^{2}$ 下方，长轴在 $y 轴上。c^{2} = a^{2} − b^{2} = 25 − 16 = 9，c = 3，焦点为 (0, \\pm3)。A 项 c$ 值正确但焦点放错了坐标轴。","tags":["medium","ellipse","major-axis"],"status":"published"},{"id":"math-conics-015","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":3,"stem":"For the parabola $y^{2} = 12x$, find the distance from the focus to the directrix.","stemZh":"抛物线 $y^{2} = 12x$ 的焦点到准线的距离为多少？","options":[{"id":"A","text":"3"},{"id":"B","text":"12"},{"id":"C","text":"4"},{"id":"D","text":"6"}],"answer":"D","explanation":"$2p = 12$ gives $p = 6$. The focus is $(3, 0)$ and the directrix is $x = −3$, so the distance between them is $p = 6$. Option A is $p/2 (the$ focal distance from the vertex) and option B is 2p.","explanationZh":"$由 2p = 12 得 p = 6。焦点为 (3, 0)，准线为 x = −3，二者距离为 p = 6。A$ 项是 p/2（焦点到顶点的距离），B 项是 2p。","tags":["medium","parabola","directrix"],"status":"published"},{"id":"math-conics-016","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":3,"stem":"P is a point on the parabola $y^{2} = 8x$ with x-coordinate 2, and F is its focus. Find $|PF|$.","stemZh":"点 P 在抛物线 $y^{2} = 8x$ 上，横坐标为 2，F 为焦点，求 $|PF|。$","options":[{"id":"A","text":"2"},{"id":"B","text":"4"},{"id":"C","text":"6"},{"id":"D","text":"8"}],"answer":"B","explanation":"By the definition of $a parabola, |PF|$ equals the distance from P to the directrix $x = −2$, which is $2 − (−2) = 4. Directly: y^{2} = 16$ gives $P = (2, 4)$ or $(2, −4)$, and the distance to $F(2, 0)$ is 4.","explanationZh":"$由抛物线定义，|PF|$ 等于 P 到准线 $x = −2$ 的距离，即 $2 − (−2) = 4。直接计算：y^{2} = 16，P 为 (2, 4) 或 (2, −4)，到 F(2, 0)$ 的距离也是 4。","tags":["medium","parabola","focus"],"status":"published"},{"id":"math-conics-017","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":3,"stem":"An ellipse has $a = 5$ and $b = 3$. Find the distance between its two foci.","stemZh":"椭圆的 $a = 5，b = 3，求两焦点间的距离。$","options":[{"id":"A","text":"8"},{"id":"B","text":"4"},{"id":"C","text":"16"},{"id":"D","text":"10"}],"answer":"A","explanation":"$c^{2} = a^{2} − b^{2} = 25 − 9 = 16$, so $c = 4$. The distance between the foci (the focal distance) is $2c = 8$, not $c = 4$. Option D is 2a, the length of the major axis.","explanationZh":"$c^{2} = a^{2} − b^{2} = 25 − 9 = 16，c = 4。两焦点间的距离（焦距）为 2c = 8，而不是 c = 4。D 项 2a = 10$ 是长轴长。","tags":["medium","ellipse","focal-distance"],"status":"published"},{"id":"math-conics-018","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":3,"stem":"For the hyperbola $x^{2}/4 − y^{2}/9 = 1$, what is the length of the real axis?","stemZh":"双曲线 $x^{2}/4 − y^{2}/9 = 1$ 的实轴长为多少？","options":[{"id":"A","text":"2"},{"id":"B","text":"6"},{"id":"C","text":"9"},{"id":"D","text":"4"}],"answer":"D","explanation":"$a^{2} = 4$ gives $a = 2$, so the real axis has length $2a = 4$. The imaginary axis has length $2b = 6 (option B)$. Do not confuse the axis length with a itself.","explanationZh":"$a^{2} = 4，a = 2，故实轴长为 2a = 4。虚轴长为 2b = 6（B$ 项）。注意区分轴长与半轴长 a。","tags":["medium","hyperbola","real-axis"],"status":"published"},{"id":"math-conics-019","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":3,"stem":"Which of the following points lies on the ellipse $x^{2}/25 + y^{2}/16 = 1$?","stemZh":"下列哪个点在椭圆 $x^{2}/25 + y^{2}/16 = 1$ 上？","options":[{"id":"A","text":"$(3, 4)$"},{"id":"B","text":"$(5, 0)$"},{"id":"C","text":"$(4, 3)$"},{"id":"D","text":"$(2, 2)$"}],"answer":"B","explanation":"Substituting $(5, 0)$ gives $25/25 + 0 = 1$, so it lies on the ellipse (it is a vertex). Checking the $others: (3, 4)$ gives $9/25 + 1 = 34/25 ≠ 1, (4, 3)$ gives $16/25 + 9/16 ≠ 1$, and $(2, 2)$ gives $4/25 + 4/16 ≠ 1$.","explanationZh":"代入 $(5, 0)：25/25 + 0 = 1，该点在椭圆上（是长轴端点）。其余：(3, 4) 得 9/25 + 1 = 34/25 ≠ 1；(4, 3) 得 16/25 + 9/16 ≠ 1；(2, 2) 得 4/25 + 4/16 ≠ 1。$","tags":["medium","ellipse","standard-equation"],"status":"published"},{"id":"math-conics-020","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":3,"stem":"Which of the following could be the eccentricity of an ellipse?","stemZh":"下列哪个数可能是椭圆的离心率？","options":[{"id":"A","text":"3/5"},{"id":"B","text":"1"},{"id":"C","text":"5/3"},{"id":"D","text":"4/3"}],"answer":"A","explanation":"For an ellipse $e = c/a$ with $0 < c < a$, so $0 < e < 1$. Only 3/5 lies in this $range: e = 1$ corresponds to a parabola and $e > 1 (options C\\text{ and }D)$ to a hyperbola.","explanationZh":"椭圆的 $e = c/a，且 0 < c <$ a，故 $0 < e < 1。只有 3/5$ 在此范围内：e $= 1$ 是抛物线，e $> 1（C、D$ 两项）是双曲线。","tags":["medium","eccentricity","ellipse"],"status":"published"},{"id":"math-conics-021","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":5,"stem":"The equation $x^{2}/m + y^{2}/4 = 1$ represents an ellipse with eccentricity 1/2. Find all possible values of m.","stemZh":"方程 $x^{2}/m + y^{2}/4 = 1$ 表示离心率为 1/2 的椭圆，求 m 的所有可能值。","options":[{"id":"A","text":"3 only"},{"id":"B","text":"16/3 only"},{"id":"C","text":"3 or 16/3"},{"id":"D","text":"4"}],"answer":"C","explanation":"Two cases. If $m > 4$ the major axis is on the $x-axis: e^{2} = (m − 4)/m = 1/4 → 4m − 16 = m → m = 16/3$. If $0 < m < 4$ the major axis is on the $y-axis: e^{2} = (4 − m)/4 = 1/4 → m = 3$. Both are valid, so $m = 3$ or $m = 16/3$.","explanationZh":"分两种情况：若 $m > 4，长轴在 x 轴上，e^{2} = (m − 4)/m = 1/4，得 m = 16/3；若 0 < m < 4，长轴在 y 轴上，e^{2} = (4 − m)/4 = 1/4，得 m = 3。两者都成立，故 m = 3 或 16/3。$","tags":["hard","ellipse","eccentricity"],"status":"published"},{"id":"math-conics-022","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":4,"stem":"The hyperbola $x^{2}/9 − y^{2}/m = 1$ has asymptotes $y = \\pm(2/3)x$. Find m.","stemZh":"双曲线 $x^{2}/9 − y^{2}/m = 1$ 的渐近线为 $y = \\pm(2/3)x，求$ m。","options":[{"id":"A","text":"6"},{"id":"B","text":"4"},{"id":"C","text":"2"},{"id":"D","text":"36"}],"answer":"B","explanation":"Here $a^{2} = 9$ so $a = 3$, and $b^{2} =$ m. The asymptotes are $y = \\pm(b/a)x = \\pm(\\sqrt{m}/3)x$. Setting $\\sqrt{m}/3 = 2/3$ gives $\\sqrt{m} = 2$, so $m = 4$. Option C is the value of b rather than $b^{2}$.","explanationZh":"此处 $a^{2} = 9，a = 3，b^{2} =$ m。渐近线为 $y = \\pm(b/a)x = \\pm(\\sqrt{m}/3)x。令 \\sqrt{m}/3 = 2/3 得 \\sqrt{m} = 2，故 m = 4。C$ 项是 b 而不是 $b^{2}。$","tags":["hard","hyperbola","asymptote"],"status":"published"},{"id":"math-conics-023","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":4,"stem":"Find the equation of the parabola with focus $(0, 2)$ and directrix $y = −2$.","stemZh":"求焦点为 $(0, 2)、准线为 y = −2$ 的抛物线方程。","options":[{"id":"A","text":"$y^{2} = 8x$"},{"id":"B","text":"$x^{2} = −8y$"},{"id":"C","text":"$x^{2} = 4y$"},{"id":"D","text":"$x^{2} = 8y$"}],"answer":"D","explanation":"The vertex is midway between the focus and the directrix, i.e. the origin, and the parabola opens upward along the y-axis. The distance from the focus to the directrix is $p = 4$, and for $x^{2} = 2py$ we get $x^{2} = 8y (check:\\text{ focus }(0, p/2) = (0, 2))$.","explanationZh":"顶点在焦点与准线中间，即原点；开口沿 y 轴向上。焦点到准线的距离 $p = 4，标准方程 x^{2} = 2py 即 x^{2} = 8y（验证：焦点 (0, p/2) = (0, 2)）。$","tags":["hard","parabola","directrix"],"status":"published"},{"id":"math-conics-024","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":5,"stem":"P is a point on the right branch of the hyperbola $x^{2}/9 − y^{2}/16 = 1$. If $|PF_{2}| = 10$, where $F_{2}$ is the right focus, find $|PF_{1}|$ for the left focus $F_{1}$.","stemZh":"点 P 在双曲线 $x^{2}/9 − y^{2}/16 = 1 的右支上，F_{2}$ 为右焦点，若 $|PF_{2}| = 10，求 |PF_{1}|。$","options":[{"id":"A","text":"16"},{"id":"B","text":"4"},{"id":"C","text":"6"},{"id":"D","text":"14"}],"answer":"A","explanation":"By the definition of $a hyperbola, ||PF_{1}| − |PF_{2}|| = 2a = 6$. On the right branch P is closer to $F_{2}$, so $|PF_{1}| − |PF_{2}| = 6$ and $|PF_{1}| = 10 + 6 = 16$. Option B is the value obtained on the left branch, and option C is 2a itself.","explanationZh":"由双曲线定义 $||PF_{1}| − |PF_{2}|| = 2a = 6。P$ 在右支上时离 $F_{2}$ 更近，故 $|PF_{1}| − |PF_{2}| = 6，|PF_{1}| = 10 + 6 = 16。B$ 项是左支上的结果，C 项是 2a 本身。","tags":["hard","hyperbola","focal-distance"],"status":"published"},{"id":"math-conics-025","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":4,"stem":"P is a point on the ellipse $x^{2}/25 + y^{2}/16 = 1$ with foci $F_{1}$ and $F_{2}$. Find the perimeter of triangle $PF_{1}F_{2}$.","stemZh":"点 P 在椭圆 $x^{2}/25 + y^{2}/16 = 1 上，F_{1}、F_{2}$ 为焦点，求 $△PF_{1}F_{2}$ 的周长。","options":[{"id":"A","text":"10"},{"id":"B","text":"6"},{"id":"C","text":"16"},{"id":"D","text":"12"}],"answer":"C","explanation":"By the definition of an $ellipse, |PF_{1}| + |PF_{2}| = 2a = 10$. Also $c^{2} = 25 − 16 = 9$ so $c = 3$ and $|F_{1}F_{2}| = 2c = 6$. The perimeter is $10 + 6 = 16$.","explanationZh":"由椭圆定义 $|PF_{1}| + |PF_{2}| = 2a = 10；又 c^{2} = 25 − 16 = 9，c = 3，|F_{1}F_{2}| = 2c = 6。故周长为 10 + 6 = 16。$","tags":["hard","ellipse","focal-distance"],"status":"published"},{"id":"math-conics-026","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":5,"stem":"Find the equation of the tangent to the parabola $y^{2} = 4x$ at the point $(1, 2)$.","stemZh":"求抛物线 $y^{2} = 4x$ 在点 $(1, 2)$ 处的切线方程。","options":[{"id":"A","text":"$y = x + 1$"},{"id":"B","text":"$y = x − 1$"},{"id":"C","text":"$y =$ −x + 3"},{"id":"D","text":"$y = 2x$"}],"answer":"A","explanation":"First $check: 2^{2} = 4(1)$, so $(1, 2)$ is on the parabola. Substituting $y = x + 1$ into $y^{2} = 4x$ gives $(x + 1)^{2} = 4x → x^{2} − 2x + 1 = 0 → (x − 1)^{2} = 0, a$ double root, so the line touches the curve at $x = 1. (Equivalently$, for $y^{2} = 2px$ the tangent at $(x_{0}, y_{0})$ is $y_{0}y = p(x + x_{0}).)$","explanationZh":"先验证 $2^{2} = 4 \\times 1，(1, 2)$ 在抛物线上。把 $y = x + 1$ 代入 $y^{2} = 4x 得 (x + 1)^{2} = 4x，即 (x − 1)^{2} = 0，有二重根，说明直线与曲线相切于 x = 1。（也可用切线公式 y_{0}y = p(x + x_{0})。）$","tags":["hard","parabola","tangent"],"status":"published"},{"id":"math-conics-027","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":5,"stem":"A hyperbola $x^{2}/a^{2} − y^{2}/b^{2} = 1 (a > 0, b > 0)$ has eccentricity 2. Find the acute angle between its two asymptotes.","stemZh":"双曲线 $x^{2}/a^{2} − y^{2}/b^{2} = 1（a > 0，b > 0）的离心率为 2，求两条渐近线的锐夹角。$","options":[{"id":"A","text":"30°"},{"id":"B","text":"90°"},{"id":"C","text":"120°"},{"id":"D","text":"60°"}],"answer":"D","explanation":"$e = c/a = 2$ gives $c = 2a$, so $b^{2} = c^{2} − a^{2} = 3a^{2}$ and $b/a = \\sqrt{3}$. The asymptotes $y = \\pm\\sqrt{3}x$ make angles of $\\pm60°$ with the x-axis, so the acute angle between them is $60° (the$ supplementary angle is $120°)$.","explanationZh":"$e = c/a = 2，即 c = 2a，故 b^{2} = c^{2} − a^{2} = 3a^{2}，b/a = \\sqrt{3}。渐近线 y = \\pm\\sqrt{3}x 与 x$ 轴夹角为 $\\pm60°，故两渐近线的锐夹角为 60°（另一角为 120°）。$","tags":["hard","hyperbola","asymptote"],"status":"published"},{"id":"math-conics-028","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":4,"stem":"The ellipse $x^{2}/25 + y^{2}/16 = 1$ and the hyperbola $x^{2}/4 − y^{2}/m = 1$ have the same foci. Find m.","stemZh":"椭圆 $x^{2}/25 + y^{2}/16 = 1$ 与双曲线 $x^{2}/4 − y^{2}/m = 1$ 有相同的焦点，求 m。","options":[{"id":"A","text":"4"},{"id":"B","text":"9"},{"id":"C","text":"5"},{"id":"D","text":"13"}],"answer":"C","explanation":"For the $ellipse, c^{2} = 25 − 16 = 9$, so the foci are $(\\pm3, 0)$. For the $hyperbola, c^{2} = a^{2} + b^{2} = 4 + m$, and this must equal 9, so $m = 5$. The key is that the two conics use different relations linking a, b and c.","explanationZh":"椭圆中 $c^{2} = 25 − 16 = 9，焦点为 (\\pm3, 0)。双曲线中 c^{2} = a^{2} + b^{2} = 4 +$ m，令其等于 $9 得 m = 5。关键在于椭圆与双曲线中$ a、b、c 的关系式不同。","tags":["hard","hyperbola","focal-distance"],"status":"published"},{"id":"math-conics-029","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":4,"stem":"For which values of k does the equation $x^{2}/k + y^{2}/4 = 1$ represent a hyperbola?","stemZh":"当 k 取何值时，方程 $x^{2}/k + y^{2}/4 = 1$ 表示双曲线？","options":[{"id":"A","text":"$k > 0$"},{"id":"B","text":"$k < 0$"},{"id":"C","text":"$k > 4$"},{"id":"D","text":"$0 < k < 4$"}],"answer":"B","explanation":"A hyperbola needs the two squared terms to have opposite signs. Since the coefficient of $y^{2}$ is positive $(4 > 0)$, we need $1/k < 0$, i.e. $k < 0$; the equation then reads $y^{2}/4 − x^{2}/(−k) = 1$. For $k > 0$ it is an ellipse.","explanationZh":"$双曲线要求两个平方项系数异号。y^{2} 的系数为正（4 > 0），故需 1/k < 0，即 k < 0，此时方程化为 y^{2}/4 − x^{2}/(−k) = 1。k > 0$ 时为椭圆。","tags":["hard","hyperbola","standard-equation"],"status":"published"},{"id":"math-conics-030","subject":"math","moduleId":"geometry-algebra","unitId":"analytic-geometry","topicId":"conics","type":"single-choice","difficulty":5,"stem":"A parabola has its vertex at the origin, the x-axis as its axis of symmetry, opens to the right, and passes through $(2, 4)$. Find the distance from its focus to its directrix.","stemZh":"抛物线顶点在原点，对称轴为 x 轴，开口向右，且过点 $(2, 4)，求其焦点到准线的距离。$","options":[{"id":"A","text":"2"},{"id":"B","text":"8"},{"id":"C","text":"16"},{"id":"D","text":"4"}],"answer":"D","explanation":"The form is $y^{2} = 2px$ with $p > 0$. Substituting $(2, 4): 16 = 2p(2) = 4p$, so $p = 4$. The focus is $(2, 0)$, the directrix is $x = −2$, and the distance between them is $p = 4$. Option B is 2p and option A is p/2.","explanationZh":"设方程为 $y^{2} = 2px（p > 0）。代入 (2, 4)：16 = 2p \\times 2 = 4p，得 p = 4。焦点为 (2, 0)，准线为 x = −2，二者距离为 p = 4。B$ 项是 2p，A 项是 p/2。","tags":["hard","parabola","directrix"],"status":"published"},{"id":"math-vectors-001","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":1,"stem":"Let $\\overrightarrow{a} = (3, −4)$. Find $|\\overrightarrow{a}|$.","stemZh":"已知 $\\overrightarrow{a} = (3, −4)，求 |\\overrightarrow{a}|。$","options":[{"id":"A","text":"5"},{"id":"B","text":"7"},{"id":"C","text":"1"},{"id":"D","text":"$\\sqrt{7}$"}],"answer":"A","explanation":"$|\\overrightarrow{a}| = \\sqrt{3^{2} + (−4)^{2}} = \\sqrt{25} = 5.$","explanationZh":"$|\\overrightarrow{a}| = \\sqrt{3^{2} + (−4)^{2}} = \\sqrt{25} = 5。}$","tags":["easy","magnitude"],"status":"published"},{"id":"math-vectors-002","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":1,"stem":"Let $\\overrightarrow{a} = (1, 2)$ and $\\overrightarrow{b} = (3, −1)$. Find $\\overrightarrow{a} + \\overrightarrow{b}$.","stemZh":"已知 $\\overrightarrow{a} = (1, 2)，\\overrightarrow{b} = (3, −1)，求 \\overrightarrow{a} + \\overrightarrow{b}。$","options":[{"id":"A","text":"$(4, 3)$"},{"id":"B","text":"$(4, 1)$"},{"id":"C","text":"$(2, 1)$"},{"id":"D","text":"$(−2, 3)$"}],"answer":"B","explanation":"Add $coordinates: \\overrightarrow{a} + \\overrightarrow{b} = (1 + 3, 2 + (−1)) = (4, 1)$. Option D is $\\overrightarrow{a} − \\overrightarrow{b}$.","explanationZh":"$坐标相加：\\overrightarrow{a} + \\overrightarrow{b} = (1 + 3, 2 + (−1)) = (4, 1)。D$ 选项是 $\\overrightarrow{a} − \\overrightarrow{b}。$","tags":["easy","addition"],"status":"published"},{"id":"math-vectors-003","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":1,"stem":"Which statement about plane vectors is correct?","stemZh":"下列关于平面向量的说法正确的是？","options":[{"id":"A","text":"$|\\overrightarrow{a}| = |\\overrightarrow{b}|$ implies $\\overrightarrow{a} = \\overrightarrow{b}$"},{"id":"B","text":"$\\overrightarrow{a} ∥ \\overrightarrow{b}$ implies $|\\overrightarrow{a}| = |\\overrightarrow{b}|$"},{"id":"C","text":"Equal vectors have equal magnitude and the same direction"},{"id":"D","text":"Two vectors with the same direction are equal"}],"answer":"C","explanation":"Two vectors are equal exactly when they have the same magnitude AND the same direction. Equal magnitude alone (A) or equal direction alone (D) is not enough; parallel vectors (B) need not have equal length.","explanationZh":"两向量相等当且仅当大小相等且方向相同。仅有大小相等（A）或仅有方向相同（D）都不够；平行向量（B）长度不一定相等。","tags":["easy","concept"],"status":"published"},{"id":"math-vectors-004","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":1,"stem":"Which statement about the zero vector $\\overrightarrow{0}$ is correct?","stemZh":"关于零向量 $\\overrightarrow{0}，下列说法正确的是？$","options":[{"id":"A","text":"The zero vector has magnitude 1"},{"id":"B","text":"The zero vector is perpendicular to every vector"},{"id":"C","text":"$|\\overrightarrow{0}|$ is not defined"},{"id":"D","text":"$|\\overrightarrow{0}| = 0$"}],"answer":"D","explanation":"The zero vector has length 0, so $|\\overrightarrow{0}| = 0$; its direction is regarded as arbitrary.","explanationZh":"零向量的长度为 $0，即 |\\overrightarrow{0}| = 0，其方向规定为任意的。$","tags":["easy","zero-vector"],"status":"published"},{"id":"math-vectors-005","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":2,"stem":"Let $\\overrightarrow{a} = (2, −3)$ and $\\overrightarrow{b} = (−1, 5)$. Find $\\overrightarrow{a} − \\overrightarrow{b}$.","stemZh":"已知 $\\overrightarrow{a} = (2, −3)，\\overrightarrow{b} = (−1, 5)，求 \\overrightarrow{a} − \\overrightarrow{b}。$","options":[{"id":"A","text":"$(3, −8)$"},{"id":"B","text":"$(3, 8)$"},{"id":"C","text":"$(1, −8)$"},{"id":"D","text":"$(−3, −8)$"}],"answer":"A","explanation":"$\\overrightarrow{a} − \\overrightarrow{b} = (2 − (−1), −3 − 5) = (3, −8)$. Watch the double signs.","explanationZh":"$\\overrightarrow{a} − \\overrightarrow{b} = (2 − (−1), −3 − 5) = (3, −8)，注意符号。$","tags":["easy","subtraction"],"status":"published"},{"id":"math-vectors-006","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":2,"stem":"Let $\\overrightarrow{a} = (−2, 4)$. Find $−\\overrightarrow{2a}$.","stemZh":"已知 $\\overrightarrow{a} = (−2, 4)，求 −\\overrightarrow{2a}。$","options":[{"id":"A","text":"$(4, 8)$"},{"id":"B","text":"$(4, −8)$"},{"id":"C","text":"$(−4, 8)$"},{"id":"D","text":"$(−4, −8)$"}],"answer":"B","explanation":"Multiply both coordinates by $−2: −\\overrightarrow{2a} = (−2·(−2), −2·4) = (4, −8)$.","explanationZh":"两个坐标都乘 $−2：−\\overrightarrow{2a} = (−2·(−2), −2·4) = (4, −8)。$","tags":["easy","scalar-multiplication"],"status":"published"},{"id":"math-vectors-007","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":2,"stem":"Let $\\overrightarrow{a} = (2, m)$ and $\\overrightarrow{b} = (4, 6)$. If $\\overrightarrow{a} ∥ \\overrightarrow{b}$, find m.","stemZh":"已知 $\\overrightarrow{a} = (2, m)，\\overrightarrow{b} = (4, 6)，若 \\overrightarrow{a} ∥ \\overrightarrow{b}，求$ m。","options":[{"id":"A","text":"−3"},{"id":"B","text":"12"},{"id":"C","text":"3"},{"id":"D","text":"4"}],"answer":"C","explanation":"For $\\overrightarrow{a} = (x_{1}, y_{1})$ and $\\overrightarrow{b} = (x_{2}, y_{2}), \\overrightarrow{a} ∥ \\overrightarrow{b} ⇔ x_{1}y_{2} − x_{2}y_{1} = 0$, so $2·6 − 4m = 0$ and $m = 3$.","explanationZh":"$\\overrightarrow{a} = (x_{1}, y_{1})，\\overrightarrow{b} = (x_{2}, y_{2})，\\overrightarrow{a} ∥ \\overrightarrow{b} ⇔ x_{1}y_{2} − x_{2}y_{1} = 0，故 2·6 − 4m = 0，m = 3。$","tags":["easy","collinear"],"status":"published"},{"id":"math-vectors-008","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":2,"stem":"Let $\\overrightarrow{a} = (1, 2)$ and $\\overrightarrow{b} = (2, −1)$. Find $\\overrightarrow{a} · \\overrightarrow{b}$.","stemZh":"已知 $\\overrightarrow{a} = (1, 2)，\\overrightarrow{b} = (2, −1)，求 \\overrightarrow{a} · \\overrightarrow{b}。$","options":[{"id":"A","text":"4"},{"id":"B","text":"3"},{"id":"C","text":"1"},{"id":"D","text":"0"}],"answer":"D","explanation":"$\\overrightarrow{a} · \\overrightarrow{b} = 1·2 + 2·(−1) = 2 − 2 = 0$, so $\\overrightarrow{a} ⟂ \\overrightarrow{b}$.","explanationZh":"$\\overrightarrow{a} · \\overrightarrow{b} = 1·2 + 2·(−1) = 0，因此 \\overrightarrow{a} ⟂ \\overrightarrow{b}。$","tags":["easy","dot-product"],"status":"published"},{"id":"math-vectors-009","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":1,"stem":"Given $A(1, 2)$ and $B(4, 6)$, find the coordinate representation of $\\overrightarrow{AB}$.","stemZh":"已知 $A(1, 2)，B(4, 6)，求 \\overrightarrow{AB}$ 的坐标。","options":[{"id":"A","text":"$(3, 4)$"},{"id":"B","text":"$(−3, −4)$"},{"id":"C","text":"$(5, 8)$"},{"id":"D","text":"$(1, 2)$"}],"answer":"A","explanation":"$\\overrightarrow{AB} = (x_B − x_A, y_B − y_A) = (4 − 1, 6 − 2) = (3, 4)$. Option B is $\\overrightarrow{BA}$.","explanationZh":"$\\overrightarrow{AB} = (x_B − x_A, y_B − y_A) = (4 − 1, 6 − 2) = (3, 4)。B$ 选项是 $\\overrightarrow{BA}。$","tags":["easy","coordinates"],"status":"published"},{"id":"math-vectors-010","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":2,"stem":"Let $\\overrightarrow{a} = (3, 4)$. Find the unit vector in the same direction as $\\overrightarrow{a}$.","stemZh":"已知 $\\overrightarrow{a} = (3, 4)，求与 \\overrightarrow{a}$ 同方向的单位向量。","options":[{"id":"A","text":"$(3, 4)$"},{"id":"B","text":"$(3/5, 4/5)$"},{"id":"C","text":"$(4/5, 3/5)$"},{"id":"D","text":"$(1, 0)$"}],"answer":"B","explanation":"$|\\overrightarrow{a}| = 5$, so the unit vector is $\\overrightarrow{a}/|\\overrightarrow{a}| = (3/5, 4/5)$.","explanationZh":"$|\\overrightarrow{a}| = 5，故单位向量为 \\overrightarrow{a}/|\\overrightarrow{a}| = (3/5, 4/5)。$","tags":["easy","unit-vector"],"status":"published"},{"id":"math-vectors-011","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":3,"stem":"Let $|\\overrightarrow{a}| = 2, |\\overrightarrow{b}| = 1$ and $\\overrightarrow{a} · \\overrightarrow{b} = 1$. Find the angle $θ$ between $\\overrightarrow{a}$ and $\\overrightarrow{b}$.","stemZh":"已知 $|\\overrightarrow{a}| = 2，|\\overrightarrow{b}| = 1，\\overrightarrow{a} · \\overrightarrow{b} = 1，求 \\overrightarrow{a} 与 \\overrightarrow{b}$ 的夹角 $θ。$","options":[{"id":"A","text":"30°"},{"id":"B","text":"45°"},{"id":"C","text":"60°"},{"id":"D","text":"120°"}],"answer":"C","explanation":"$\\cos θ = (\\overrightarrow{a} · \\overrightarrow{b})/(|\\overrightarrow{a}||\\overrightarrow{b}|) = 1/(2·1) = 1/2$, so $θ = 60°. A$ negative dot product would give 120°.","explanationZh":"$\\cos θ = (\\overrightarrow{a} · \\overrightarrow{b})/(|\\overrightarrow{a}||\\overrightarrow{b}|) = 1/(2·1) = 1/2，故 θ = 60°。若数量积为负才得 120°。$","tags":["medium","angle","dot-product"],"status":"published"},{"id":"math-vectors-012","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":3,"stem":"Let $\\overrightarrow{a} = (1, 2)$ and $\\overrightarrow{b} = (−2, 1)$. Find the angle between $\\overrightarrow{a}$ and $\\overrightarrow{b}$.","stemZh":"已知 $\\overrightarrow{a} = (1, 2)，\\overrightarrow{b} = (−2, 1)，求 \\overrightarrow{a} 与 \\overrightarrow{b}$ 的夹角。","options":[{"id":"A","text":"0°"},{"id":"B","text":"30°"},{"id":"C","text":"60°"},{"id":"D","text":"90°"}],"answer":"D","explanation":"$\\overrightarrow{a} · \\overrightarrow{b} = 1·(−2) + 2·1 = 0$, so $\\cos θ = 0$ and the vectors are $perpendicular: θ = 90°$.","explanationZh":"$\\overrightarrow{a} · \\overrightarrow{b} = 1·(−2) + 2·1 = 0，故 \\cos θ = 0，两向量垂直，夹角为 90°。$","tags":["medium","perpendicular","dot-product"],"status":"published"},{"id":"math-vectors-013","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":3,"stem":"Given $A(1, 1), B(3, 5)$ and $C(4, 7)$, which statement is correct?","stemZh":"已知 $A(1, 1)，B(3, 5)，C(4, 7)，下列结论正确的是？$","options":[{"id":"A","text":"A, B, C are collinear"},{"id":"B","text":"$|\\overrightarrow{AB}| = |\\overrightarrow{BC}|$"},{"id":"C","text":"B is the midpoint of AC"},{"id":"D","text":"$\\overrightarrow{AB} · \\overrightarrow{AC} = 0$"}],"answer":"A","explanation":"$\\overrightarrow{AB} = (2, 4)$ and $\\overrightarrow{AC} = (3, 6)$. Since $2·6 − 3·4 = 0$, the vectors are parallel, and they share the point A, so A, B, C are collinear. Also $|\\overrightarrow{AB}| = \\sqrt{20} ≠ \\sqrt{5} = |\\overrightarrow{BC}|$.","explanationZh":"$\\overrightarrow{AB} = (2, 4)，\\overrightarrow{AC} = (3, 6)，2·6 − 3·4 = 0，两向量平行且共点$ A，故 A、B、C 三点共线；且 $|\\overrightarrow{AB}| = \\sqrt{20} ≠ \\sqrt{5} = |\\overrightarrow{BC}|。$","tags":["medium","collinearity"],"status":"published"},{"id":"math-vectors-014","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":3,"stem":"Let $\\overrightarrow{a} = (2, 1)$ and $\\overrightarrow{b} = (1, λ)$. If $\\overrightarrow{a} ⟂ \\overrightarrow{b}$, find $λ$.","stemZh":"已知 $\\overrightarrow{a} = (2, 1)，\\overrightarrow{b} = (1, λ)，若 \\overrightarrow{a} ⟂ \\overrightarrow{b}，求 λ。$","options":[{"id":"A","text":"2"},{"id":"B","text":"−2"},{"id":"C","text":"1/2"},{"id":"D","text":"−1/2"}],"answer":"B","explanation":"$\\overrightarrow{a} ⟂ \\overrightarrow{b} ⇔ \\overrightarrow{a} · \\overrightarrow{b} = 2·1 + 1·λ = 0$, so $λ = −2 (the\\text{ sign is the }trap)$.","explanationZh":"$\\overrightarrow{a} ⟂ \\overrightarrow{b} ⇔ \\overrightarrow{a} · \\overrightarrow{b} = 2·1 + 1·λ = 0，故 λ = −2，注意符号。$","tags":["medium","perpendicular","dot-product"],"status":"published"},{"id":"math-vectors-015","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":3,"stem":"In triangle ABC, M is the midpoint of BC. Express $\\overrightarrow{AM}$ in terms of $\\overrightarrow{AB}$ and $\\overrightarrow{AC}$.","stemZh":"在 △ABC 中，M 为 BC 的中点，用 $\\overrightarrow{AB}、\\overrightarrow{AC}$ 表示 $\\overrightarrow{AM}。$","options":[{"id":"A","text":"$\\overrightarrow{AB} + \\overrightarrow{AC}$"},{"id":"B","text":"$(1/2)(\\overrightarrow{AB} − \\overrightarrow{AC})$"},{"id":"C","text":"$(1/2)(\\overrightarrow{AB} + \\overrightarrow{AC})$"},{"id":"D","text":"$\\overrightarrow{AB} − \\overrightarrow{AC}$"}],"answer":"C","explanation":"By the parallelogram law, if D completes the parallelogram ABDC then the diagonal $\\overrightarrow{AD} = \\overrightarrow{AB} + \\overrightarrow{AC}$ passes through the midpoint M of BC, so $\\overrightarrow{AM} = (1/2)(\\overrightarrow{AB} + \\overrightarrow{AC})$.","explanationZh":"$由平行四边形法则，\\overrightarrow{AD} = \\overrightarrow{AB} + \\overrightarrow{AC}，其中$ AD 过 BC 中点 M，故 $\\overrightarrow{AM} = (1/2)(\\overrightarrow{AB} + \\overrightarrow{AC})。$","tags":["medium","midpoint","addition"],"status":"published"},{"id":"math-vectors-016","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":3,"stem":"Let $\\overrightarrow{a} = (1, −2)$ and $\\overrightarrow{b} = (3, 4)$. Find $|\\overrightarrow{2a} − \\overrightarrow{b}|$.","stemZh":"已知 $\\overrightarrow{a} = (1, −2)，\\overrightarrow{b} = (3, 4)，求 |\\overrightarrow{2a} − \\overrightarrow{b}|。$","options":[{"id":"A","text":"$\\sqrt{5}$"},{"id":"B","text":"5"},{"id":"C","text":"$\\sqrt{89}$"},{"id":"D","text":"$\\sqrt{65}$"}],"answer":"D","explanation":"$\\overrightarrow{2a} − \\overrightarrow{b} = (2, −4) − (3, 4) = (−1, −8)$, so $|\\overrightarrow{2a} − \\overrightarrow{b}| = \\sqrt{1 + 64} = \\sqrt{65}.$","explanationZh":"$\\overrightarrow{2a} − \\overrightarrow{b} = (2, −4) − (3, 4) = (−1, −8)，故 |\\overrightarrow{2a} − \\overrightarrow{b}| = \\sqrt{1 + 64} = \\sqrt{65}。}$","tags":["medium","magnitude","scalar-multiplication"],"status":"published"},{"id":"math-vectors-017","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":3,"stem":"Which of the following is true for all plane vectors $\\overrightarrow{a}$ and $\\overrightarrow{b}$?","stemZh":"对任意平面向量 $\\overrightarrow{a}、\\overrightarrow{b}，下列恒成立的式子是？$","options":[{"id":"A","text":"$|\\overrightarrow{a} + \\overrightarrow{b}| ≤ |\\overrightarrow{a}| + |\\overrightarrow{b}|$"},{"id":"B","text":"$|\\overrightarrow{a} + \\overrightarrow{b}| = |\\overrightarrow{a}| + |\\overrightarrow{b}|$"},{"id":"C","text":"$|\\overrightarrow{a} · \\overrightarrow{b}| = |\\overrightarrow{a}||\\overrightarrow{b}|$"},{"id":"D","text":"$\\overrightarrow{a} · \\overrightarrow{b} = 0$ implies $\\overrightarrow{a} = \\overrightarrow{0}$ or $\\overrightarrow{b} = \\overrightarrow{0}$"}],"answer":"A","explanation":"The triangle inequality $|\\overrightarrow{a} + \\overrightarrow{b}| ≤ |\\overrightarrow{a}| + |\\overrightarrow{b}|$ always holds. Equality in B needs the same $direction; |\\overrightarrow{a} · \\overrightarrow{b}| = |\\overrightarrow{a}||\\overrightarrow{b}||\\cos θ| ≤ |\\overrightarrow{a}||\\overrightarrow{b}|$; and $\\overrightarrow{a} · \\overrightarrow{b} = 0$ also happens for perpendicular nonzero vectors.","explanationZh":"三角不等式 $|\\overrightarrow{a} + \\overrightarrow{b}| ≤ |\\overrightarrow{a}| + |\\overrightarrow{b}|$ 恒成立。B $取等号需同向；|\\overrightarrow{a} · \\overrightarrow{b}| = |\\overrightarrow{a}||\\overrightarrow{b}||\\cos θ| ≤ |\\overrightarrow{a}||\\overrightarrow{b}|；\\overrightarrow{a} · \\overrightarrow{b} = 0$ 也可能是两非零向量垂直。","tags":["medium","properties"],"status":"published"},{"id":"math-vectors-018","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":3,"stem":"Let $\\overrightarrow{a} = (3, 4)$. Find the vector of magnitude 10 that has the same direction as $\\overrightarrow{a}$.","stemZh":"已知 $\\overrightarrow{a} = (3, 4)，求与 \\overrightarrow{a}$ 同向且模为 10 的向量。","options":[{"id":"A","text":"$(30, 40)$"},{"id":"B","text":"$(6, 8)$"},{"id":"C","text":"$(6/5, 8/5)$"},{"id":"D","text":"$(3, 4)$"}],"answer":"B","explanation":"$|\\overrightarrow{a}| = 5$, so the required vector is $(10/5)\\overrightarrow{a} = \\overrightarrow{2a} = (6, 8)$. Using $\\overrightarrow{10a}$ gives the wrong length 50.","explanationZh":"$|\\overrightarrow{a}| = 5，所求向量为 (10/5)\\overrightarrow{a} = \\overrightarrow{2a} = (6, 8)。若直接取 \\overrightarrow{10a}，模长为 50，错误。$","tags":["medium","unit-vector","scalar-multiplication"],"status":"published"},{"id":"math-vectors-019","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":3,"stem":"In parallelogram $ABCD, \\overrightarrow{AB} = \\overrightarrow{a}$ and $\\overrightarrow{AD} = \\overrightarrow{b}$. Find $\\overrightarrow{BD}$.","stemZh":"在平行四边形 ABCD $中，\\overrightarrow{AB} = \\overrightarrow{a}，\\overrightarrow{AD} = \\overrightarrow{b}，求 \\overrightarrow{BD}。$","options":[{"id":"A","text":"$\\overrightarrow{a} + \\overrightarrow{b}$"},{"id":"B","text":"$\\overrightarrow{a} − \\overrightarrow{b}$"},{"id":"C","text":"$\\overrightarrow{b} − \\overrightarrow{a}$"},{"id":"D","text":"$−\\overrightarrow{a} − \\overrightarrow{b}$"}],"answer":"C","explanation":"$\\overrightarrow{BD} = \\overrightarrow{AD} − \\overrightarrow{AB} = \\overrightarrow{b} − \\overrightarrow{a}$. The sum $\\overrightarrow{a} + \\overrightarrow{b}$ is the other diagonal $\\overrightarrow{AC}$.","explanationZh":"$\\overrightarrow{BD} = \\overrightarrow{AD} − \\overrightarrow{AB} = \\overrightarrow{b} − \\overrightarrow{a}；\\overrightarrow{a} + \\overrightarrow{b}$ 是另一条对角线 $\\overrightarrow{AC}。$","tags":["medium","parallelogram","subtraction"],"status":"published"},{"id":"math-vectors-020","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":3,"stem":"Let $\\overrightarrow{a} = (1, 2)$ and $\\overrightarrow{b} = (−3, 4)$. Find $\\cos θ$, where $θ$ is the angle between $\\overrightarrow{a}$ and $\\overrightarrow{b}$.","stemZh":"已知 $\\overrightarrow{a} = (1, 2)，\\overrightarrow{b} = (−3, 4)，求 \\overrightarrow{a} 与 \\overrightarrow{b}$ 夹角 $θ$ 的余弦。","options":[{"id":"A","text":"1/5"},{"id":"B","text":"2/5"},{"id":"C","text":"$\\sqrt{5}$"},{"id":"D","text":"$\\sqrt{5}/5$"}],"answer":"D","explanation":"$\\overrightarrow{a} · \\overrightarrow{b} = 1·(−3) + 2·4 = 5, |\\overrightarrow{a}| = \\sqrt{5}, |\\overrightarrow{b}| = 5$, so $\\cos θ = 5/(5\\sqrt{5}) = 1/\\sqrt{5} = \\sqrt{5}/5$.","explanationZh":"$\\overrightarrow{a} · \\overrightarrow{b} = 1·(−3) + 2·4 = 5，|\\overrightarrow{a}| = \\sqrt{5}，|\\overrightarrow{b}| = 5，故 \\cos θ = 5/(5\\sqrt{5}) = \\sqrt{5}/5。$","tags":["medium","angle","dot-product"],"status":"published"},{"id":"math-vectors-021","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":4,"stem":"Let $\\overrightarrow{a} = (k, 1)$ and $\\overrightarrow{b} = (4, k)$. Find all real k for which $\\overrightarrow{a} ∥ \\overrightarrow{b}$.","stemZh":"已知 $\\overrightarrow{a} = (k, 1)，\\overrightarrow{b} = (4, k)，求使 \\overrightarrow{a} ∥ \\overrightarrow{b}$ 的所有实数 k。","options":[{"id":"A","text":"$\\pm2$"},{"id":"B","text":"2"},{"id":"C","text":"−2"},{"id":"D","text":"4"}],"answer":"A","explanation":"$x_{1}y_{2} − x_{2}y_{1} = k·k − 4·1 = k^{2} − 4 = 0$ gives $k = 2$ or $k = −2$. Taking only $k = 2$ misses the opposite-direction case.","explanationZh":"$x_{1}y_{2} − x_{2}y_{1} = k·k − 4·1 = k^{2} − 4 = 0，得 k = 2 或 k = −2；只取 k = 2$ 会漏掉反向平行的情形。","tags":["hard","collinear"],"status":"published"},{"id":"math-vectors-022","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":4,"stem":"For nonzero vectors $\\overrightarrow{a}$ and $\\overrightarrow{b}, \\overrightarrow{a} · \\overrightarrow{b} < 0$. What can be concluded about the angle $θ$ between them?","stemZh":"非零向量 $\\overrightarrow{a}、\\overrightarrow{b}$ 满足 $\\overrightarrow{a} · \\overrightarrow{b} < 0，则它们的夹角 θ$ 满足？","options":[{"id":"A","text":"$θ$ is acute $(0° ≤ θ < 90°)$"},{"id":"B","text":"$θ$ is obtuse $(90° < θ ≤ 180°)$"},{"id":"C","text":"$θ = 90°$"},{"id":"D","text":"$\\overrightarrow{a}$ and $\\overrightarrow{b}$ have opposite directions"}],"answer":"B","explanation":"$\\overrightarrow{a} · \\overrightarrow{b} = |\\overrightarrow{a}||\\overrightarrow{b}|\\cos θ$ and $|\\overrightarrow{a}||\\overrightarrow{b}| > 0$, so $\\overrightarrow{a} · \\overrightarrow{b} < 0$ means $\\cos θ < 0$, i.e. $90° < θ ≤ 180°$: an obtuse angle. Opposite directions $(θ = 180°)$ is only one special case.","explanationZh":"$\\overrightarrow{a} · \\overrightarrow{b} = |\\overrightarrow{a}||\\overrightarrow{b}|\\cos θ，且 |\\overrightarrow{a}||\\overrightarrow{b}| > 0，故 \\overrightarrow{a} · \\overrightarrow{b} < 0$ 等价于 $\\cos θ < 0，即 90° < θ ≤ 180°，夹角为钝角；反向（θ = 180°）只是其中一个特例。$","tags":["hard","dot-product","angle"],"status":"published"},{"id":"math-vectors-023","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":4,"stem":"Which statement about plane vectors is correct?","stemZh":"下列关于平面向量的命题正确的是？","options":[{"id":"A","text":"If $\\overrightarrow{a} ∥ \\overrightarrow{b}$ and $\\overrightarrow{b} ∥ \\overrightarrow{c}$, then $\\overrightarrow{a} ∥ \\overrightarrow{c}$"},{"id":"B","text":"If $\\overrightarrow{a} · \\overrightarrow{b} = \\overrightarrow{a} · \\overrightarrow{c}$, then $\\overrightarrow{b} = \\overrightarrow{c}$"},{"id":"C","text":"The zero vector $\\overrightarrow{0}$ is parallel to every vector"},{"id":"D","text":"If $|\\overrightarrow{a} + \\overrightarrow{b}| = |\\overrightarrow{a} − \\overrightarrow{b}|$, then $\\overrightarrow{a} = \\overrightarrow{b}$"}],"answer":"C","explanation":"By definition $\\overrightarrow{0}$ is parallel to every vector. A fails when $\\overrightarrow{b} = \\overrightarrow{0} (then \\overrightarrow{a}\\text{ and }\\overrightarrow{c}\\text{ may be }arbitrary); B$ fails because $\\overrightarrow{a} · (\\overrightarrow{b} − \\overrightarrow{c}) = 0$ only gives perpendicularity or $\\overrightarrow{a} = \\overrightarrow{0}; D$ gives $\\overrightarrow{a} · \\overrightarrow{b} = 0$, i.e. $\\overrightarrow{a} ⟂ \\overrightarrow{b}$.","explanationZh":"规定零向量与任意向量平行。A $在 \\overrightarrow{b} = \\overrightarrow{0}$ 时不成立；B $中 \\overrightarrow{a} · (\\overrightarrow{b} − \\overrightarrow{c}) = 0$ 只能说明垂直或 $\\overrightarrow{a} = \\overrightarrow{0}；D$ 实际推出 $\\overrightarrow{a} · \\overrightarrow{b} = 0，即 \\overrightarrow{a} ⟂ \\overrightarrow{b}。$","tags":["hard","zero-vector","properties"],"status":"published"},{"id":"math-vectors-024","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":4,"stem":"Let $|\\overrightarrow{a}| = 1, |\\overrightarrow{b}| = 2$ and $\\overrightarrow{a} ⟂ \\overrightarrow{b}$. Find $|\\overrightarrow{a} − \\overrightarrow{2b}|$.","stemZh":"已知 $|\\overrightarrow{a}| = 1，|\\overrightarrow{b}| = 2，且 \\overrightarrow{a} ⟂ \\overrightarrow{b}，求 |\\overrightarrow{a} − \\overrightarrow{2b}|。$","options":[{"id":"A","text":"3"},{"id":"B","text":"$\\sqrt{5}$"},{"id":"C","text":"5"},{"id":"D","text":"$\\sqrt{17}$"}],"answer":"D","explanation":"$|\\overrightarrow{a} − \\overrightarrow{2b}|^{2} = |\\overrightarrow{a}|^{2} − \\overrightarrow{4a} · \\overrightarrow{b} + 4|\\overrightarrow{b}|^{2} = 1 − 0 + 16 = 17$, so $|\\overrightarrow{a} − \\overrightarrow{2b}| = \\sqrt{17}$. Note that the factor 2 must be squared to 4.","explanationZh":"$|\\overrightarrow{a} − \\overrightarrow{2b}|^{2} = |\\overrightarrow{a}|^{2} − \\overrightarrow{4a} · \\overrightarrow{b} + 4|\\overrightarrow{b}|^{2} = 1 − 0 + 16 = 17，故 |\\overrightarrow{a} − \\overrightarrow{2b}| = \\sqrt{17}；注意系数 2$ 平方后为 4。","tags":["hard","magnitude","dot-product"],"status":"published"},{"id":"math-vectors-025","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":4,"stem":"The points $A(1, 2), B(2, 5)$ and $C(4, k)$ are collinear. Find k.","stemZh":"已知 $A(1, 2)，B(2, 5)，C(4, k)$ 三点共线，求 k。","options":[{"id":"A","text":"11"},{"id":"B","text":"9"},{"id":"C","text":"7"},{"id":"D","text":"−7"}],"answer":"A","explanation":"$\\overrightarrow{AB} = (1, 3)$ and $\\overrightarrow{AC} = (3, k − 2)$. Collinearity gives $1·(k − 2) − 3·3 = 0$, so $k − 2 = 9$ and $k = 11$.","explanationZh":"$\\overrightarrow{AB} = (1, 3)，\\overrightarrow{AC} = (3, k − 2)，共线得 1·(k − 2) − 3·3 = 0，即 k − 2 = 9，k = 11。$","tags":["hard","collinearity"],"status":"published"},{"id":"math-vectors-026","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":5,"stem":"In triangle ABC, a point P satisfies $\\overrightarrow{AP} = (1/3)\\overrightarrow{AB} + (2/3)\\overrightarrow{AC}$. Which statement is true?","stemZh":"在 △ABC 中，点 P 满足 $\\overrightarrow{AP} = (1/3)\\overrightarrow{AB} + (2/3)\\overrightarrow{AC}，则下列说法正确的是？$","options":[{"id":"A","text":"P is the midpoint of BC"},{"id":"B","text":"P lies on BC and BP : PC $= 2 : 1$"},{"id":"C","text":"P lies on AB"},{"id":"D","text":"P is the centroid of triangle ABC"}],"answer":"B","explanation":"The coefficients add to 1, so P lies on line BC. $\\overrightarrow{BP} = \\overrightarrow{AP} − \\overrightarrow{AB} = (2/3)(\\overrightarrow{AC} − \\overrightarrow{AB}) = (2/3)\\overrightarrow{BC}$, hence BP : PC $= 2 : 1$. The centroid would be $(1/3)(\\overrightarrow{AB} + \\overrightarrow{AC})$.","explanationZh":"系数之和为 1，故 P 在直线 BC $上。\\overrightarrow{BP} = \\overrightarrow{AP} − \\overrightarrow{AB} = (2/3)(\\overrightarrow{AC} − \\overrightarrow{AB}) = (2/3)\\overrightarrow{BC}，所以$ BP : PC $= 2 : 1；重心对应的是 (1/3)(\\overrightarrow{AB} + \\overrightarrow{AC})。$","tags":["hard","collinearity","geometry-application"],"status":"published"},{"id":"math-vectors-027","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":5,"stem":"Nonzero vectors $\\overrightarrow{a}$ and $\\overrightarrow{b}$ satisfy $|\\overrightarrow{a}| = |\\overrightarrow{b}| = |\\overrightarrow{a} − \\overrightarrow{b}|$. Find the angle between $\\overrightarrow{a}$ and $\\overrightarrow{b}$.","stemZh":"非零向量 $\\overrightarrow{a}、\\overrightarrow{b}$ 满足 $|\\overrightarrow{a}| = |\\overrightarrow{b}| = |\\overrightarrow{a} − \\overrightarrow{b}|，求 \\overrightarrow{a} 与 \\overrightarrow{b}$ 的夹角。","options":[{"id":"A","text":"30°"},{"id":"B","text":"45°"},{"id":"C","text":"60°"},{"id":"D","text":"120°"}],"answer":"C","explanation":"Let $|\\overrightarrow{a}| = |\\overrightarrow{b}| =$ r. Squaring $|\\overrightarrow{a} − \\overrightarrow{b}| = r$ gives $r^{2} + r^{2} − 2r^{2}\\cos θ = r^{2}$, so $\\cos θ = 1/2$ and $θ = 60° (the$ three vectors form an equilateral triangle).","explanationZh":"$设 |\\overrightarrow{a}| = |\\overrightarrow{b}| =$ r，由 $|\\overrightarrow{a} − \\overrightarrow{b}| = r$ 平方得 $r^{2} + r^{2} − 2r^{2}\\cos θ = r^{2}，故 \\cos θ = 1/2，θ = 60°（三边相等构成等边三角形）。$","tags":["hard","angle","magnitude"],"status":"published"},{"id":"math-vectors-028","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":4,"stem":"Let $\\overrightarrow{a} = (2, 1)$ and $\\overrightarrow{b} = (1, 3)$. Find the projection vector of $\\overrightarrow{a}$ in the direction of $\\overrightarrow{b}$.","stemZh":"已知 $\\overrightarrow{a} = (2, 1)，\\overrightarrow{b} = (1, 3)，求 \\overrightarrow{a} 在 \\overrightarrow{b}$ 方向上的投影向量。","options":[{"id":"A","text":"$(1, 3)$"},{"id":"B","text":"$(5, 15)$"},{"id":"C","text":"$(1/2, 3)$"},{"id":"D","text":"$(1/2, 3/2)$"}],"answer":"D","explanation":"The projection is $((\\overrightarrow{a} · \\overrightarrow{b})/|\\overrightarrow{b}|^{2}) \\overrightarrow{b} = (5/10)(1, 3) = (1/2, 3/2)$. Forgetting to divide by $|\\overrightarrow{b}|^{2} = 10$ gives $(5, 15)$.","explanationZh":"投影向量为 $((\\overrightarrow{a} · \\overrightarrow{b})/|\\overrightarrow{b}|^{2}) \\overrightarrow{b} = (5/10)(1, 3) = (1/2, 3/2)；若漏除以 |\\overrightarrow{b}|^{2} = 10$ 会得到 $(5, 15)。$","tags":["hard","projection","dot-product"],"status":"published"},{"id":"math-vectors-029","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":5,"stem":"Let $|\\overrightarrow{a}| = 2, |\\overrightarrow{b}| = 3$ and $(\\overrightarrow{a} + \\overrightarrow{b}) · (\\overrightarrow{a} − \\overrightarrow{2b}) = −17$. Find the angle between $\\overrightarrow{a}$ and $\\overrightarrow{b}$.","stemZh":"已知 $|\\overrightarrow{a}| = 2，|\\overrightarrow{b}| = 3，(\\overrightarrow{a} + \\overrightarrow{b}) · (\\overrightarrow{a} − \\overrightarrow{2b}) = −17，求 \\overrightarrow{a} 与 \\overrightarrow{b}$ 的夹角。","options":[{"id":"A","text":"60°"},{"id":"B","text":"30°"},{"id":"C","text":"120°"},{"id":"D","text":"90°"}],"answer":"A","explanation":"$Expand: |\\overrightarrow{a}|^{2} − \\overrightarrow{a} · \\overrightarrow{b} − 2|\\overrightarrow{b}|^{2} = 4 − \\overrightarrow{a} · \\overrightarrow{b} − 18 = −14 − \\overrightarrow{a} · \\overrightarrow{b} = −17$, so $\\overrightarrow{a} · \\overrightarrow{b} = 3$. Then $\\cos θ = 3/(2·3) = 1/2$ and $θ = 60°$.","explanationZh":"展开得 $|\\overrightarrow{a}|^{2} − \\overrightarrow{a} · \\overrightarrow{b} − 2|\\overrightarrow{b}|^{2} = 4 − \\overrightarrow{a} · \\overrightarrow{b} − 18 = −14 − \\overrightarrow{a} · \\overrightarrow{b} = −17，故 \\overrightarrow{a} · \\overrightarrow{b} = 3，\\cos θ = 3/(2·3) = 1/2，θ = 60°。$","tags":["hard","dot-product","angle"],"status":"published"},{"id":"math-vectors-030","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"vectors","type":"single-choice","difficulty":5,"stem":"Nonzero vectors $\\overrightarrow{a}$ and $\\overrightarrow{b}$ satisfy $|\\overrightarrow{a} + \\overrightarrow{b}| = |\\overrightarrow{a}| + |\\overrightarrow{b}|$. What must be true?","stemZh":"非零向量 $\\overrightarrow{a}、\\overrightarrow{b}$ 满足 $|\\overrightarrow{a} + \\overrightarrow{b}| = |\\overrightarrow{a}| + |\\overrightarrow{b}|，则必有？$","options":[{"id":"A","text":"$\\overrightarrow{a} ⟂ \\overrightarrow{b}$"},{"id":"B","text":"$\\overrightarrow{a}$ and $\\overrightarrow{b}$ have the same direction"},{"id":"C","text":"$|\\overrightarrow{a}| = |\\overrightarrow{b}|$"},{"id":"D","text":"$\\overrightarrow{a} = \\overrightarrow{b}$"}],"answer":"B","explanation":"Equality holds in the triangle inequality exactly when the angle is 0°, i.e. the vectors have the same direction. Equal length (C) or even equality (D) is not required.","explanationZh":"三角不等式取等号的条件是夹角为 0°，即两向量同向；并不要求长度相等（C），更不要求两向量相等（D）。","tags":["hard","properties","magnitude"],"status":"published"},{"id":"math-complex-001","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":1,"stem":"Let $z = 3 − 4i$. Find the real part of z.","stemZh":"已知 $z = 3 − 4i，求 z$ 的实部。","options":[{"id":"A","text":"3"},{"id":"B","text":"−4"},{"id":"C","text":"4"},{"id":"D","text":"−3"}],"answer":"A","explanation":"For $z = a +$ bi the real part is $a = 3$ and the imaginary part is $b = −4$.","explanationZh":"$z = a +$ bi 的实部为 $a = 3，虚部为 b = −4。$","tags":["easy","real-imaginary-parts"],"status":"published"},{"id":"math-complex-002","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":1,"stem":"What is $i^{2}$ equal to?","stemZh":"$i^{2}$ 等于多少？","options":[{"id":"A","text":"1"},{"id":"B","text":"−1"},{"id":"C","text":"i"},{"id":"D","text":"−i"}],"answer":"B","explanation":"By definition i is the imaginary unit with $i^{2} = −1$.","explanationZh":"虚数单位 i 满足 $i^{2} = −1。$","tags":["easy","powers-of-i"],"status":"published"},{"id":"math-complex-003","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":1,"stem":"Let $z = 2 + 3i$. Find the conjugate $\\bar{z}$.","stemZh":"已知 $z = 2 + 3i，求其共轭复数 \\bar{z}。$","options":[{"id":"A","text":"−2 + 3i"},{"id":"B","text":"2 + 3i"},{"id":"C","text":"2 − 3i"},{"id":"D","text":"−2 − 3i"}],"answer":"C","explanation":"The conjugate keeps the real part and flips the sign of the imaginary $part: \\bar{z} = 2 − 3i$.","explanationZh":"$共轭复数实部不变、虚部变号：\\bar{z} = 2 − 3i。$","tags":["easy","conjugate"],"status":"published"},{"id":"math-complex-004","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":1,"stem":"Let $z = 3 + 4i$. Find $|z|$.","stemZh":"已知 $z = 3 + 4i，求 |z|。$","options":[{"id":"A","text":"7"},{"id":"B","text":"1"},{"id":"C","text":"25"},{"id":"D","text":"5"}],"answer":"D","explanation":"$|z| = \\sqrt{a^{2} + b^{2}} = \\sqrt{9 + 16} = 5$. Note that 25 is $|z|^{2}$, not $|z|$.","explanationZh":"$|z| = \\sqrt{a^{2} + b^{2}} = \\sqrt{9 + 16} = 5；注意 25 是 |z|^{2}，不是 |z|。$","tags":["easy","modulus"],"status":"published"},{"id":"math-complex-005","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":2,"stem":"Compute $(2 + 3i) + (1 − 4i)$.","stemZh":"计算 $(2 + 3i) + (1 − 4i)。$","options":[{"id":"A","text":"3 − i"},{"id":"B","text":"3 + i"},{"id":"C","text":"1 + 7i"},{"id":"D","text":"3 − 7i"}],"answer":"A","explanation":"Add real and imaginary parts $separately: (2 + 1) + (3 − 4)i = 3 −$ i.","explanationZh":"$实部与虚部分别相加：(2 + 1) + (3 − 4)i = 3 −$ i。","tags":["easy","addition"],"status":"published"},{"id":"math-complex-006","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":2,"stem":"Compute $(2 + 3i) − (1 − 4i)$.","stemZh":"计算 $(2 + 3i) − (1 − 4i)。$","options":[{"id":"A","text":"1 − i"},{"id":"B","text":"1 + 7i"},{"id":"C","text":"3 − i"},{"id":"D","text":"1 − 7i"}],"answer":"B","explanation":"$(2 − 1) + (3 − (−4))i = 1 + 7i$; the imaginary part picks up a plus sign.","explanationZh":"$(2 − 1) + (3 − (−4))i = 1 + 7i，虚部要注意变号。$","tags":["easy","subtraction"],"status":"published"},{"id":"math-complex-007","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":2,"stem":"Compute $i^{5}$.","stemZh":"计算 $i^{5}。$","options":[{"id":"A","text":"1"},{"id":"B","text":"−1"},{"id":"C","text":"i"},{"id":"D","text":"−i"}],"answer":"C","explanation":"$i^{4} = 1$, so $i^{5} = i^{4} · i =$ i.","explanationZh":"$i^{4} = 1，故 i^{5} = i^{4} · i =$ i。","tags":["easy","powers-of-i"],"status":"published"},{"id":"math-complex-008","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":2,"stem":"Compute $(1 + i)(1 − i)$.","stemZh":"计算 $(1 + i)(1 − i)。$","options":[{"id":"A","text":"0"},{"id":"B","text":"1"},{"id":"C","text":"$\\sqrt{2}$"},{"id":"D","text":"2"}],"answer":"D","explanation":"$(1 + i)(1 − i) = 1 − i^{2} = 1 − (−1) = 2$. Note $\\sqrt{2}$ is $|1 + i|$, not the product.","explanationZh":"$(1 + i)(1 − i) = 1 − i^{2} = 1 + 1 = 2；\\sqrt{2} 是 |1 + i|，不是乘积。$","tags":["easy","multiplication"],"status":"published"},{"id":"math-complex-009","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":1,"stem":"Let $z = a +$ bi with $a, b ∈ ℝ$. When is z a real number?","stemZh":"$设 z = a + bi（a, b ∈ ℝ），z$ 为实数的条件是？","options":[{"id":"A","text":"$b = 0$"},{"id":"B","text":"$a = 0$"},{"id":"C","text":"$a = b = 0$"},{"id":"D","text":"$a > 0$"}],"answer":"A","explanation":"z is real exactly when its imaginary part $b = 0; a = 0 (with b ≠ 0)$ gives a pure imaginary number.","explanationZh":"z 为实数当且仅当虚部 $b = 0；a = 0（且 b ≠ 0）时 z$ 为纯虚数。","tags":["easy","real-imaginary-parts"],"status":"published"},{"id":"math-complex-010","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":2,"stem":"Let $z = −2 + 5i$. Find the imaginary part of z.","stemZh":"已知 $z = −2 + 5i，求 z$ 的虚部。","options":[{"id":"A","text":"5i"},{"id":"B","text":"5"},{"id":"C","text":"−2"},{"id":"D","text":"−2i"}],"answer":"B","explanation":"The imaginary part is the real coefficient $b = 5$, not $5i. (The$ real part is $−2.)$","explanationZh":"虚部是实数 $b = 5，而不是 5i（实部为 −2）。$","tags":["easy","real-imaginary-parts"],"status":"published"},{"id":"math-complex-011","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":3,"stem":"Let $z = (m − 1) + (m + 2)i$ with m real. For which m is z a pure imaginary number?","stemZh":"$设 z = (m − 1) + (m + 2)i（m ∈ ℝ），z$ 为纯虚数时 $m = ？$","options":[{"id":"A","text":"−2"},{"id":"B","text":"2"},{"id":"C","text":"1"},{"id":"D","text":"0"}],"answer":"C","explanation":"A pure imaginary number needs real part 0 and imaginary part $nonzero: m − 1 = 0$ gives $m = 1$, and then $m + 2 = 3 ≠ 0$. The value $m = −2$ makes the imaginary part 0, so $z = −3$ is real.","explanationZh":"纯虚数要求实部为 0 且虚部不为 $0：m − 1 = 0 得 m = 1，此时 m + 2 = 3 ≠ 0；m = −2$ 时虚部为 $0，z = −3$ 是实数。","tags":["medium","pure-imaginary"],"status":"published"},{"id":"math-complex-012","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":3,"stem":"Compute $(3 + 2i)/(1 − i)$.","stemZh":"计算 $(3 + 2i)/(1 − i)。$","options":[{"id":"A","text":"1 + 5i"},{"id":"B","text":"$1/2 − (5/2)i$"},{"id":"C","text":"$5/2 + (1/2)i$"},{"id":"D","text":"$1/2 + (5/2)i$"}],"answer":"D","explanation":"Multiply by the $conjugate: (3 + 2i)(1 + i)/((1 − i)(1 + i)) = (1 + 5i)/2 = 1/2 + (5/2)i$. The denominator is 2, not 1.","explanationZh":"$分子分母同乘共轭：(3 + 2i)(1 + i)/((1 − i)(1 + i)) = (1 + 5i)/2 = 1/2 + (5/2)i；分母是 2$ 不能漏除。","tags":["medium","division"],"status":"published"},{"id":"math-complex-013","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":3,"stem":"Let $z = (1 + i)^{3}$. Find $|z|$.","stemZh":"已知 $z = (1 + i)^{3}，求 |z|。$","options":[{"id":"A","text":"$2\\sqrt{2}$"},{"id":"B","text":"2"},{"id":"C","text":"$\\sqrt{2}$"},{"id":"D","text":"4"}],"answer":"A","explanation":"$|z| = |1 + i|^{3} = (\\sqrt{2})^{3} = 2\\sqrt{2}$. Direct $check: (1 + i)^{3} = 2i(1 + i) = −2 + 2i$, and $|−2 + 2i| = \\sqrt{8} = 2\\sqrt{2}$.","explanationZh":"$|z| = |1 + i|^{3} = (\\sqrt{2})^{3} = 2\\sqrt{2}；也可直接算：(1 + i)^{3} = 2i(1 + i) = −2 + 2i，模为 \\sqrt{8} = 2\\sqrt{2}。$","tags":["medium","modulus"],"status":"published"},{"id":"math-complex-014","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":3,"stem":"Let $z = 2 +$ i. Compute $z · \\bar{z}$.","stemZh":"已知 $z = 2 +$ i，计算 $z · \\bar{z}。$","options":[{"id":"A","text":"3 + 4i"},{"id":"B","text":"5"},{"id":"C","text":"$\\sqrt{5}$"},{"id":"D","text":"3"}],"answer":"B","explanation":"$z · \\bar{z} = (2 + i)(2 − i) = 4 + 1 = 5 = |z|^{2}$, always a real number. Note 3 + 4i is $z^{2}$, not $z · \\bar{z}$.","explanationZh":"$z · \\bar{z} = (2 + i)(2 − i) = 5 = |z|^{2}，结果一定是实数；3 + 4i 是 z^{2}，不是 z · \\bar{z}。$","tags":["medium","modulus","conjugate"],"status":"published"},{"id":"math-complex-015","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":3,"stem":"Compute $(1 + i)^{4}$.","stemZh":"计算 $(1 + i)^{4}。$","options":[{"id":"A","text":"4"},{"id":"B","text":"4i"},{"id":"C","text":"−4"},{"id":"D","text":"−4i"}],"answer":"C","explanation":"$(1 + i)^{2} = 2i$, so $(1 + i)^{4} = (2i)^{2} = 4i^{2} = −4$.","explanationZh":"$(1 + i)^{2} = 2i，故 (1 + i)^{4} = (2i)^{2} = 4i^{2} = −4。$","tags":["medium","powers-of-i","multiplication"],"status":"published"},{"id":"math-complex-016","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":3,"stem":"Let x, y be real and $(x + yi) + (2 − 3i) = 5 +$ i. Find $(x, y)$.","stemZh":"$设 x, y ∈ ℝ 且 (x + yi) + (2 − 3i) = 5 +$ i，求 $(x, y)。$","options":[{"id":"A","text":"$(3, −4)$"},{"id":"B","text":"$(−3, 4)$"},{"id":"C","text":"$(7, −2)$"},{"id":"D","text":"$(3, 4)$"}],"answer":"D","explanation":"Equality of complex numbers means equal real and equal imaginary $parts: x + 2 = 5$ and $y − 3 = 1$, so $x = 3, y = 4$.","explanationZh":"复数相等即实部、虚部分别相等：x $+ 2 = 5，y − 3 = 1，得 x = 3，y = 4。$","tags":["medium","equality"],"status":"published"},{"id":"math-complex-017","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":3,"stem":"Let $z = (m^{2} − 1) + (m − 1)i$ with m real. For which m is z a real number?","stemZh":"$设 z = (m^{2} − 1) + (m − 1)i（m ∈ ℝ），z$ 为实数时 $m = ？$","options":[{"id":"A","text":"1"},{"id":"B","text":"−1"},{"id":"C","text":"$\\pm1$"},{"id":"D","text":"0"}],"answer":"A","explanation":"The imaginary part must $vanish: m − 1 = 0$, so $m = 1 (then z = 0,\\text{ which is }real)$. For $m = −1$ the real part is 0 but the imaginary part is −2, so $z = −2i$ is not real.","explanationZh":"虚部为 $0：m − 1 = 0，得 m = 1（此时 z = 0$ 是实数）。m $= −1$ 时实部为 0 但虚部为 $−2，z = −2i$ 不是实数。","tags":["medium","real-imaginary-parts"],"status":"published"},{"id":"math-complex-018","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":3,"stem":"Compute $i + i^{2} + i^{3} + i^{4}$.","stemZh":"计算 $i + i^{2} + i^{3} + i^{4}。$","options":[{"id":"A","text":"1"},{"id":"B","text":"0"},{"id":"C","text":"i"},{"id":"D","text":"−1"}],"answer":"B","explanation":"$i + (−1) + (−i) + 1 = 0$; any four consecutive powers of i add up to 0.","explanationZh":"$i + (−1) + (−i) + 1 = 0，任意连续四个 i$ 的幂之和都为 0。","tags":["medium","powers-of-i"],"status":"published"},{"id":"math-complex-019","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":3,"stem":"Compute $(2 − i)^{2}$.","stemZh":"计算 $(2 − i)^{2}。$","options":[{"id":"A","text":"5"},{"id":"B","text":"3 + 4i"},{"id":"C","text":"3 − 4i"},{"id":"D","text":"5 − 4i"}],"answer":"C","explanation":"$(2 − i)^{2} = 4 − 4i + i^{2} = 4 − 4i − 1 = 3 − 4i$. Using $i^{2} = 1$ instead of −1 gives 5 − 4i.","explanationZh":"$(2 − i)^{2} = 4 − 4i + i^{2} = 4 − 4i − 1 = 3 − 4i；若误用 i^{2} = 1$ 会得到 5 − 4i。","tags":["medium","multiplication"],"status":"published"},{"id":"math-complex-020","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":3,"stem":"Let $z = 3 +$ bi with b real and $b < 0$. If $|z| = 5$, find b.","stemZh":"$设 z = 3 +$ bi（b $∈ ℝ，b < 0），若 |z| = 5，求$ b。","options":[{"id":"A","text":"4"},{"id":"B","text":"16"},{"id":"C","text":"−16"},{"id":"D","text":"−4"}],"answer":"D","explanation":"$9 + b^{2} = 25$ gives $b^{2} = 16$, so $b = \\pm4$; the condition $b < 0$ selects $b = −4$.","explanationZh":"$9 + b^{2} = 25 得 b^{2} = 16，b = \\pm4；由 b < 0 取 b = −4。$","tags":["medium","modulus"],"status":"published"},{"id":"math-complex-021","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":4,"stem":"Let $z = (m^{2} − m − 2) + (m^{2} − 1)i$ with m real. For which m is $z = 0$?","stemZh":"$设 z = (m^{2} − m − 2) + (m^{2} − 1)i（m ∈ ℝ），z = 0 时 m = ？$","options":[{"id":"A","text":"−1"},{"id":"B","text":"1"},{"id":"C","text":"2"},{"id":"D","text":"−2"}],"answer":"A","explanation":"$z = 0$ needs both parts $zero: m^{2} − m − 2 = 0$ gives $m = 2$ or $m = −1$, and $m^{2} − 1 = 0$ gives $m = \\pm1$. The common value is $m = −1$.","explanationZh":"$z = 0$ 需实部、虚部同时为 $0：m^{2} − m − 2 = 0 得 m = 2 或 −1，m^{2} − 1 = 0 得 m = \\pm1，公共解为 m = −1。$","tags":["hard","equality","real-imaginary-parts"],"status":"published"},{"id":"math-complex-022","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":4,"stem":"A complex number z satisfies $z · \\bar{z} = 25$ and $z + \\bar{z} = 6$. Find z.","stemZh":"复数 z 满足 $z · \\bar{z} = 25，z + \\bar{z} = 6，求$ z。","options":[{"id":"A","text":"3 + 4i"},{"id":"B","text":"$3 \\pm 4i$"},{"id":"C","text":"$4 \\pm 3i$"},{"id":"D","text":"$\\pm5$"}],"answer":"B","explanation":"Write $z = a +$ bi. Then $z + \\bar{z} = 2a = 6$, so $a = 3$, and $z · \\bar{z} = a^{2} + b^{2} = 25$ gives $b^{2} = 16, b = \\pm4$. Both $z = 3 + 4i$ and $z = 3 − 4i$ work.","explanationZh":"$设 z = a +$ bi，则 $z + \\bar{z} = 2a = 6，a = 3；z · \\bar{z} = a^{2} + b^{2} = 25 得 b^{2} = 16，b = \\pm4，故 z = 3 \\pm 4i。$","tags":["hard","modulus","conjugate"],"status":"published"},{"id":"math-complex-023","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":4,"stem":"Compute $i^{1} + i^{2} + i^{3} + … + i^{2024}$.","stemZh":"计算 $i^{1} + i^{2} + i^{3} + … + i^{2024}。$","options":[{"id":"A","text":"1"},{"id":"B","text":"i"},{"id":"C","text":"0"},{"id":"D","text":"−1"}],"answer":"C","explanation":"Powers of i repeat with period 4 and each block $i + i^{2} + i^{3} + i^{4} = i − 1 − i + 1 = 0$. Since $2024 = 4 \\times 506$, the sum is 0.","explanationZh":"i 的幂以 4 为周期，每四项之和 $i + i^{2} + i^{3} + i^{4} = 0；2024 = 4 \\times 506，故总和为 0。$","tags":["hard","powers-of-i"],"status":"published"},{"id":"math-complex-024","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":4,"stem":"Let $z = (1 + i)/(1 − i)$. Compute $z^{2024}$.","stemZh":"$设 z = (1 + i)/(1 − i)，计算 z^{2024}。$","options":[{"id":"A","text":"i"},{"id":"B","text":"−1"},{"id":"C","text":"−i"},{"id":"D","text":"1"}],"answer":"D","explanation":"$z = (1 + i)^{2}/((1 − i)(1 + i)) = 2i/2 = i$, so $z^{2024} = i^{2024} = (i^{4})^{506} = 1$.","explanationZh":"$z = (1 + i)^{2}/((1 − i)(1 + i)) = 2i/2 =$ i，故 $z^{2024} = i^{2024} = (i^{4})^{506} = 1。$","tags":["hard","division","powers-of-i"],"status":"published"},{"id":"math-complex-025","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":4,"stem":"Let $z = a +$ bi with $a, b ∈ ℝ$. Which expression always equals $|z|^{2}$?","stemZh":"$设 z = a + bi（a, b ∈ ℝ），下列哪个式子恒等于 |z|^{2}？$","options":[{"id":"A","text":"$z · \\bar{z}$"},{"id":"B","text":"$z^{2}$"},{"id":"C","text":"$a^{2} − b^{2}$"},{"id":"D","text":"$(a + b)^{2}$"}],"answer":"A","explanation":"$z · \\bar{z} = (a + bi)(a − bi) = a^{2} + b^{2} = |z|^{2}$, always a nonnegative real number. $z^{2} = a^{2} − b^{2} + 2abi$ is generally not real.","explanationZh":"$z · \\bar{z} = (a + bi)(a − bi) = a^{2} + b^{2} = |z|^{2}，恒为非负实数；而 z^{2} = a^{2} − b^{2} + 2abi$ 一般不是实数。","tags":["hard","modulus","conjugate"],"status":"published"},{"id":"math-complex-026","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":5,"stem":"Let $z = (m^{2} − 4) + (m^{2} − 3m + 2)i$ with m real. For which m is z a pure imaginary number?","stemZh":"$设 z = (m^{2} − 4) + (m^{2} − 3m + 2)i（m ∈ ℝ），z$ 为纯虚数时 $m = ？$","options":[{"id":"A","text":"2"},{"id":"B","text":"−2"},{"id":"C","text":"$\\pm2$"},{"id":"D","text":"1"}],"answer":"B","explanation":"Real part 0 gives $m^{2} − 4 = 0$, so $m = \\pm2$. Check the imaginary part: at $m = 2$ it is $4 − 6 + 2 = 0$, giving $z = 0 (not$ pure imaginary); at $m = −2$ it is $4 + 6 + 2 = 12 ≠ 0$, giving $z = 12i$.","explanationZh":"实部为 $0 得 m^{2} − 4 = 0，m = \\pm2。再检验虚部：m = 2$ 时虚部 $4 − 6 + 2 = 0，z = 0$ 不是纯虚数；m $= −2$ 时虚部 $4 + 6 + 2 = 12 ≠ 0，z = 12i。$","tags":["hard","pure-imaginary"],"status":"published"},{"id":"math-complex-027","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":5,"stem":"Let $z = (a^{2} − 1) + 2ai$ with a real. Find the minimum value of $|z|$.","stemZh":"$设 z = (a^{2} − 1) + 2ai（a ∈ ℝ），求 |z|$ 的最小值。","options":[{"id":"A","text":"0"},{"id":"B","text":"1/2"},{"id":"C","text":"1"},{"id":"D","text":"2"}],"answer":"C","explanation":"$|z|^{2} = (a^{2} − 1)^{2} + 4a^{2} = a^{4} + 2a^{2} + 1 = (a^{2} + 1)^{2}$, so $|z| = a^{2} + 1 ≥ 1$, with equality at $a = 0 (then z = −1)$.","explanationZh":"$|z|^{2} = (a^{2} − 1)^{2} + 4a^{2} = a^{4} + 2a^{2} + 1 = (a^{2} + 1)^{2}，故 |z| = a^{2} + 1 ≥ 1，当 a = 0$ 时取等号（此时 $z = −1）。$","tags":["hard","modulus"],"status":"published"},{"id":"math-complex-028","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":4,"stem":"Let $z_{1} = 3 + 4i$ and $z_{2} = 4 − 3i$. Find $|z_{1} · z_{2}|$.","stemZh":"已知 $z_{1} = 3 + 4i，z_{2} = 4 − 3i，求 |z_{1} · z_{2}|。$","options":[{"id":"A","text":"7"},{"id":"B","text":"1"},{"id":"C","text":"5"},{"id":"D","text":"25"}],"answer":"D","explanation":"$|z_1 · z_2| = |z_1| · |z_2| = 5 · 5 = 25$. Direct $check: z_1z_2 = (3 + 4i)(4 − 3i) = 24 + 7i$ and $|24 + 7i| = \\sqrt{576 + 49} = 25$.","explanationZh":"$|z_1 · z_2| = |z_1| · |z_2| = 5 · 5 = 25；直接计算 z_1z_2 = 24 + 7i，|24 + 7i| = \\sqrt{576 + 49} = 25。$","tags":["hard","multiplication","modulus"],"status":"published"},{"id":"math-complex-029","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":5,"stem":"Let $z = 1 +$ i. Compute $z + z^{2} + z^{3} + z^{4}$.","stemZh":"$设 z = 1 +$ i，计算 $z + z^{2} + z^{3} + z^{4}。$","options":[{"id":"A","text":"−5 + 5i"},{"id":"B","text":"5 + 5i"},{"id":"C","text":"−5 − 5i"},{"id":"D","text":"5 − 5i"}],"answer":"A","explanation":"$z^{2} = 2i, z^{3} = 2i(1 + i) = −2 + 2i, z^{4} = (z^{2})^{2} = −4. Summing: (1 + 0 − 2 − 4) + (1 + 2 + 2 + 0)i = −5 + 5i$.","explanationZh":"$z^{2} = 2i，z^{3} = 2i(1 + i) = −2 + 2i，z^{4} = (z^{2})^{2} = −4，相加得 (1 + 0 − 2 − 4) + (1 + 2 + 2 + 0)i = −5 + 5i。$","tags":["hard","powers","addition"],"status":"published"},{"id":"math-complex-030","subject":"math","moduleId":"geometry-algebra","unitId":"vectors-complex","topicId":"complex","type":"single-choice","difficulty":5,"stem":"Let $z = a +$ bi with $a, b ∈ ℝ$. If $z^{2}$ is a real number, what must hold?","stemZh":"$设 z = a + bi（a, b ∈ ℝ），若 z^{2}$ 是实数，则必有？","options":[{"id":"A","text":"$a = 0$"},{"id":"B","text":"$a = 0$ or $b = 0$"},{"id":"C","text":"$b = 0$"},{"id":"D","text":"$a = b = 0$"}],"answer":"B","explanation":"$z^{2} = (a^{2} − b^{2}) + 2abi$ is real exactly when $2ab = 0$, i.e. $a = 0$ or $b = 0$. Either condition alone is sufficient, so the answer must allow both.","explanationZh":"$z^{2} = (a^{2} − b^{2}) + 2abi$ 为实数当且仅当 $2ab = 0，即 a = 0 或 b = 0；二者都足以保证结论，故答案须包含两种情况。$","tags":["hard","real-imaginary-parts","multiplication"],"status":"published"},{"id":"math-space-001","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":1,"stem":"A cube has edge length 2. Its volume is","stemZh":"棱长为 2 的正方体，其体积为","options":[{"id":"A","text":"4"},{"id":"B","text":"6"},{"id":"C","text":"8"},{"id":"D","text":"12"}],"answer":"C","explanation":"Volume of $a cube: V = a^{3}$. With $a = 2$ we get $V = 2^{3} = 8. 6a^{2}$ would be the surface area, 12a is the total edge length.","explanationZh":"正方体体积公式 $V = a^{3}，取 a = 2 得 V = 2^{3} = 8。6a^{2} 是表面积，12a$ 是棱长总和。","tags":["easy","cube","volume"],"status":"published"},{"id":"math-space-002","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":1,"stem":"The volume of a cube is 27. Its edge length is","stemZh":"正方体的体积为 27，其棱长为","options":[{"id":"A","text":"3"},{"id":"B","text":"9"},{"id":"C","text":"4"},{"id":"D","text":"$\\sqrt{3}$"}],"answer":"A","explanation":"$V = a^{3} = 27$, so $a = \\sqrt[3]{}27 = 3$. The cube root is required — do not divide by $3 (that$ would give $9)$.","explanationZh":"$由 V = a^{3} = 27 得 a = \\sqrt[3]{}27 = 3。这里要开三次方，不能除以 3（否则得 9）。$","tags":["easy","cube","volume"],"status":"published"},{"id":"math-space-003","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":1,"stem":"A sphere has radius 3. Its volume is","stemZh":"半径为 3 的球，其体积为","options":[{"id":"A","text":"$36π$"},{"id":"B","text":"$12π$"},{"id":"C","text":"$108π$"},{"id":"D","text":"$9π$"}],"answer":"A","explanation":"$V = (4/3)πr^{3} = (4/3)π·27 = 36π$. Note $36π = 4π·9 = 4πr^{2}$ is the surface area only if you forget the factor r/3; the correct volume keeps the 4/3 factor.","explanationZh":"球体积公式 $V = (4/3)πr^{3}，取 r = 3 得 V = (4/3)π·27 = 36π。注意不要漏掉系数 4/3。$","tags":["easy","sphere","volume"],"status":"published"},{"id":"math-space-004","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":1,"stem":"A sphere has radius 2. Its surface area is","stemZh":"半径为 2 的球，其表面积为","options":[{"id":"A","text":"$4π$"},{"id":"B","text":"$16π$"},{"id":"C","text":"$8π$"},{"id":"D","text":"$32π$"}],"answer":"B","explanation":"$S = 4πr^{2} = 4π·4 = 16π$. Squaring the radius comes $first: 8π$ would come from using 2r, and $32π$ from using $r^{3}$.","explanationZh":"球表面积公式 $S = 4πr^{2}，取 r = 2 得 S = 4π·4 = 16π。要先把半径平方，2r 与 r^{3}$ 的用法都是错的。","tags":["easy","sphere","surface-area"],"status":"published"},{"id":"math-space-005","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":1,"stem":"A cube has edge length a. Its surface area is","stemZh":"棱长为 a 的正方体，其表面积为","options":[{"id":"A","text":"$a^{2}$"},{"id":"B","text":"$4a^{2}$"},{"id":"C","text":"$12a^{2}$"},{"id":"D","text":"$6a^{2}$"}],"answer":"D","explanation":"A cube has 6 congruent square faces, each of area $a^{2}$, so $S = 6a^{2}. a^{2}$ is one face only and $12a^{2}$ double-counts.","explanationZh":"正方体有 6 个全等的正方形面，每个面积为 $a^{2}，故 S = 6a^{2}。a^{2}$ 只是一个面的面积。","tags":["easy","cube","surface-area"],"status":"published"},{"id":"math-space-006","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":2,"stem":"A rectangular box measures 3 by 4 by 5. Its volume is","stemZh":"长、宽、高分别为 3、4、5 的长方体，其体积为","options":[{"id":"A","text":"12"},{"id":"B","text":"20"},{"id":"C","text":"94"},{"id":"D","text":"60"}],"answer":"D","explanation":"$V = 3·4·5 = 60$. The value $94 = 2(3·4 + 4·5 + 3·5)$ is the surface area, not the volume.","explanationZh":"长方体体积 $V = 3·4·5 = 60。94 = 2(3·4 + 4·5 + 3·5)$ 是表面积，不是体积。","tags":["easy","cuboid","volume"],"status":"published"},{"id":"math-space-007","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":2,"stem":"A cube has edge length 1. Its space diagonal has length","stemZh":"棱长为 1 的正方体，其体对角线长为","options":[{"id":"A","text":"1"},{"id":"B","text":"$\\sqrt{3}$"},{"id":"C","text":"$\\sqrt{2}$"},{"id":"D","text":"2"}],"answer":"B","explanation":"The space diagonal is $\\sqrt{1^{2} + 1^{2} + 1^{2}} = \\sqrt{3}. \\sqrt{2}$ is a face diagonal, obtained by using only two dimensions.","explanationZh":"体对角线长 $= \\sqrt{1^{2} + 1^{2} + 1^{2}} = \\sqrt{3}。\\sqrt{2}$ 是面对角线，只用了两个方向的长度。","tags":["easy","cube","space-diagonal"],"status":"published"},{"id":"math-space-008","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":2,"stem":"In the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$, which pair of lines is skew (neither parallel nor intersecting)?","stemZh":"在正方体 $ABCD-A_{1}B_{1}C_{1}D_{1}$ 中，哪一对直线是异面直线（既不平行也不相交）？","options":[{"id":"A","text":"AB and CD"},{"id":"B","text":"AB and $A_{1}B_{1}$"},{"id":"C","text":"AB and AD"},{"id":"D","text":"AB and $CC_{1}$"}],"answer":"D","explanation":"AB lies in the bottom face while the vertical edge $CC_{1}$ passes through C, which is not on AB; the two lines are neither parallel nor intersecting, so they are skew. AB ∥ CD, AB $∥ A_{1}B_{1}$, and AB meets AD at A.","explanationZh":"AB 在底面内，而竖棱 $CC_{1}$ 经过不在线 AB 上的点 C，两直线既不平行也不相交，因此是异面直线。AB ∥ CD，AB $∥ A_{1}B_{1}，AB 与$ AD 相交于 A。","tags":["easy","skew-lines","line-position"],"status":"published"},{"id":"math-space-009","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":2,"stem":"A cylinder has base radius 1 and height 2. Its lateral (side) surface area is","stemZh":"底面半径为 1、高为 2 的圆柱，其侧面积为","options":[{"id":"A","text":"$4π$"},{"id":"B","text":"$2π$"},{"id":"C","text":"$6π$"},{"id":"D","text":"$8π$"}],"answer":"A","explanation":"Lateral area $= 2πrh = 2π·1·2 = 4π$. The value $2π$ is the volume $πr^{2}h$, and $8π$ comes from replacing r by the diameter 2.","explanationZh":"圆柱侧面积 $S = 2πrh = 2π·1·2 = 4π。2π$ 是体积 $πr^{2}h，8π$ 是误把直径当半径代入的结果。","tags":["easy","cylinder","surface-area"],"status":"published"},{"id":"math-space-010","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":2,"stem":"A solid stands on a horizontal table. Its front view is a rectangle, its side view is a rectangle, and its top view is a circle. The solid is","stemZh":"一个几何体放在水平桌面上，它的正视图是矩形，侧视图是矩形，俯视图是圆。这个几何体是","options":[{"id":"A","text":"a cone"},{"id":"B","text":"a sphere"},{"id":"C","text":"a cylinder"},{"id":"D","text":"a cube"}],"answer":"C","explanation":"Only a cylinder gives two rectangular side views and a circular top view. A sphere would show circles in all three views, a cone shows a triangle from the side, and a cube never shows a circle.","explanationZh":"只有圆柱的正视图、侧视图都是矩形而俯视图是圆。球的三视图都是圆，圆锥的侧视是三角形，正方体的视图不会出现圆。","tags":["easy","three-view","cylinder"],"status":"published"},{"id":"math-space-011","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":3,"stem":"In the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$ of edge 1, the angle between the skew lines $AB_{1}$ and $BC_{1}$ is","stemZh":"在棱长为 1 的正方体 $ABCD-A_{1}B_{1}C_{1}D_{1}$ 中，异面直线 $AB_{1} 与 BC_{1}$ 所成的角是","options":[{"id":"A","text":"30°"},{"id":"B","text":"45°"},{"id":"C","text":"60°"},{"id":"D","text":"90°"}],"answer":"C","explanation":"Translate $BC_1$ to $AD_1 (they$ are parallel). Then the required angle is $∠B_1AD_1$, and $AB_1, AD_1, B_1D_1$ are all face diagonals of length $\\sqrt{2}$, so triangle $AB_1D_1$ is equilateral and the angle is 60°.","explanationZh":"$将 BC_1$ 平移到 $AD_1（二者平行），所求角即为 ∠B_1AD_1。AB_1、AD_1、B_1D_1$ 都是长为 $\\sqrt{2}$ 的面对角线，故 $△AB_1D_1$ 是等边三角形，夹角为 60°。","tags":["medium","skew-lines","angle"],"status":"published"},{"id":"math-space-012","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":3,"stem":"In the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$ of edge 1, the angle between the line $A_{1}D$ and the base plane ABCD is","stemZh":"在棱长为 1 的正方体 $ABCD-A_{1}B_{1}C_{1}D_{1}$ 中，直线 $A_{1}D$ 与底面 ABCD 所成的角是","options":[{"id":"A","text":"30°"},{"id":"B","text":"60°"},{"id":"C","text":"90°"},{"id":"D","text":"45°"}],"answer":"D","explanation":"$A_{1}$ projects to A on the base plane, so AD is the projection of $A_{1}D$ and the required angle is $∠A_{1}DA$. Since $AA_{1} =$ AD $= 1, \\tan∠A_{1}DA = AA_{1}/AD = 1$, giving 45°.","explanationZh":"$A_{1}$ 在底面上的射影是 A，故 AD $是 A_{1}D$ 的射影，所求角为 $∠A_{1}DA。因为 AA_{1} =$ AD $= 1，\\tan∠A_{1}DA = AA_{1}/AD = 1，所以夹角为 45°。$","tags":["medium","line-plane-angle","cube"],"status":"published"},{"id":"math-space-013","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":3,"stem":"To conclude that a line l is perpendicular to a plane $α$, it is enough to know that l is perpendicular to","stemZh":"要判定直线 l 垂直于平面 $α，只需知道 l$ 垂直于","options":[{"id":"A","text":"one line lying in $α$"},{"id":"B","text":"two lines lying in $α$"},{"id":"C","text":"two intersecting lines lying in $α$"},{"id":"D","text":"infinitely many lines lying in $α$"}],"answer":"C","explanation":"The criterion requires two INTERSECTING lines in the plane. Two parallel lines are not enough (a line can be perpendicular to a whole family of parallel lines without being perpendicular to the $plane)$, and neither is a single line.","explanationZh":"判定定理要求直线垂直于平面内两条相交直线。只垂直于两条平行线是不够的（直线可以垂直于平面内一族平行线而不垂直于平面），只垂直于一条直线更不行。","tags":["medium","line-plane-perpendicular","criterion"],"status":"published"},{"id":"math-space-014","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":3,"stem":"A cone has base radius 3 and height 4. Its volume is","stemZh":"底面半径为 3、高为 4 的圆锥，其体积为","options":[{"id":"A","text":"$12π$"},{"id":"B","text":"$36π$"},{"id":"C","text":"$15π$"},{"id":"D","text":"$24π$"}],"answer":"A","explanation":"$V = (1/3)πr^{2}h = (1/3)π·9·4 = 12π$. Dropping the factor 1/3 gives $36π; 15π = πrl$ is the lateral area (the slant height is $l = \\sqrt{3^{2}+4^{2}} = 5)$.","explanationZh":"$V = (1/3)πr^{2}h = (1/3)π·9·4 = 12π。漏掉系数 1/3$ 会得到 $36π；15π = πrl$ 是侧面积（母线 $l = \\sqrt{3^{2}+4^{2}} = 5）。$","tags":["medium","cone","volume"],"status":"published"},{"id":"math-space-015","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":3,"stem":"A cone has base radius 3 and slant height 5. Its total surface area is","stemZh":"底面半径为 3、母线长为 5 的圆锥，其全面积（表面积）为","options":[{"id":"A","text":"$15π$"},{"id":"B","text":"$24π$"},{"id":"C","text":"$9π$"},{"id":"D","text":"$30π$"}],"answer":"B","explanation":"Total surface area $=$ lateral area + base area $= πrl + πr^{2} = π·3·5 + π·9 = 15π + 9π = 24π$. The value $15π$ is the lateral area alone and $9π$ is the base alone.","explanationZh":"表面积 $=$ 侧面积 + 底面积 $= πrl + πr^{2} = π·3·5 + π·9 = 15π + 9π = 24π。15π 只是侧面积，9π$ 只是底面积。","tags":["medium","cone","surface-area"],"status":"published"},{"id":"math-space-016","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":3,"stem":"A cube has edge length 2. The volume of the sphere inscribed in the cube (touching all six faces) is","stemZh":"棱长为 2 的正方体，其内切球（与六个面都相切）的体积为","options":[{"id":"A","text":"$4π/3$"},{"id":"B","text":"$4π$"},{"id":"C","text":"$32π/3$"},{"id":"D","text":"$4\\sqrt{3}π$"}],"answer":"A","explanation":"The inscribed sphere has diameter equal to the edge, so $r = 1$ and $V = (4/3)πr^{3} = 4π/3$. Here $4π$ is the surface $area, 32π/3$ comes from using $r = 2$, and $4\\sqrt{3}π$ is the circumscribed sphere $(R = \\sqrt{3})$.","explanationZh":"内切球的直径等于棱长，故 $r = 1，V = (4/3)πr^{3} = 4π/3。4π 是表面积，32π/3$ 是误取 $r = 2，4\\sqrt{3}π$ 是外接球（R $= \\sqrt{3}）的体积。$","tags":["medium","sphere","inscribed-sphere"],"status":"published"},{"id":"math-space-017","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":3,"stem":"A rectangular box measures 1 by 2 by 2. Its space diagonal has length","stemZh":"长、宽、高分别为 1、2、2 的长方体，其体对角线长为","options":[{"id":"A","text":"3"},{"id":"B","text":"$\\sqrt{5}$"},{"id":"C","text":"$\\sqrt{6}$"},{"id":"D","text":"$2\\sqrt{2}$"}],"answer":"A","explanation":"Space diagonal $= \\sqrt{1^{2} + 2^{2} + 2^{2}} = \\sqrt{9} = 3. \\sqrt{5}$ and $2\\sqrt{2} = \\sqrt{8}$ come from omitting one $dimension, \\sqrt{6}$ from using 1, 1, 2.","explanationZh":"体对角线长 $= \\sqrt{1^{2} + 2^{2} + 2^{2}} = \\sqrt{9} = 3。\\sqrt{5} 与 2\\sqrt{2} = \\sqrt{8} 都是漏掉一个方向的结果，\\sqrt{6}$ 是误用 1、1、2 计算的。","tags":["medium","cuboid","space-diagonal"],"status":"published"},{"id":"math-space-018","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":3,"stem":"Which condition guarantees that plane $α$ is parallel to plane $β$?","stemZh":"下列哪个条件可以保证平面 $α$ 平行于平面 $β？$","options":[{"id":"A","text":"Two lines in $α$ are parallel to $β$"},{"id":"B","text":"Two intersecting lines in $α$ are parallel to $β$"},{"id":"C","text":"One line in $α$ is parallel to one line in $β$"},{"id":"D","text":"Both $α$ and $β$ are perpendicular to a third plane $γ$"}],"answer":"B","explanation":"The criterion needs two INTERSECTING lines of one plane parallel to the other plane. Two parallel lines in $α$ are not enough, and two planes both perpendicular to $γ$ may still intersect each other.","explanationZh":"判定定理要求一个平面内有两条相交直线都平行于另一个平面。两条平行线不够；两个平面都垂直于第三个平面时，它们仍可能相交。","tags":["medium","plane-plane-parallel","criterion"],"status":"published"},{"id":"math-space-019","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":3,"stem":"An ant sits at vertex A of a cube of edge 1 and walks on the surface to the opposite vertex $C_{1}$. The shortest possible distance is","stemZh":"一只蚂蚁在棱长为 1 的正方体的顶点 A 处，沿表面爬到相对的顶点 $C_{1}，最短路程为$","options":[{"id":"A","text":"$\\sqrt{3}$"},{"id":"B","text":"3"},{"id":"C","text":"$1 + \\sqrt{2}$"},{"id":"D","text":"$\\sqrt{5}$"}],"answer":"D","explanation":"Unfold two adjacent faces into a 2 by 1 rectangle; the straight segment then has length $\\sqrt{2^{2} + 1^{2}} = \\sqrt{5}$. The value $\\sqrt{3}$ is the straight space diagonal (which leaves the surface) and $1 + \\sqrt{2}$ is a longer edge-plus-diagonal route.","explanationZh":"把相邻两个面展开成 $2\\times1$ 的矩形，最短路线长为 $\\sqrt{2^{2} + 1^{2}} = \\sqrt{5}。\\sqrt{3} 是体对角线（离开了表面），1 + \\sqrt{2}$ 是先走棱再走面对角线的较长路线。","tags":["medium","net","shortest-path"],"status":"published"},{"id":"math-space-020","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":3,"stem":"In the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$, the plane through $A_{1}, B, C$ and $D_{1}$ makes with the base plane ABCD a dihedral angle of","stemZh":"在正方体 $ABCD-A_{1}B_{1}C_{1}D_{1}$ 中，过 $A_{1}、B、C、D_{1}$ 的平面与底面 ABCD 所成的二面角为","options":[{"id":"A","text":"30°"},{"id":"B","text":"45°"},{"id":"C","text":"60°"},{"id":"D","text":"90°"}],"answer":"B","explanation":"The two planes meet along BC. Cut by the plane through $A, B, A_{1}, B_{1}$: the section of the base is BA and the section of the oblique plane is $BA_{1}$, and $∠ABA_{1} = 45°$ because $AA_{1} =$ AB. So the dihedral angle is 45°.","explanationZh":"两平面交于 BC。用过 $A、B、A_{1}、B_{1}$ 的平面去截：底面截线为 BA，斜面截线为 $BA_{1}。由于 AA_{1} = AB，∠ABA_{1} = 45°，故二面角为 45°。$","tags":["medium","dihedral-angle","cube"],"status":"published"},{"id":"math-space-021","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":4,"stem":"A cube has edge length 2. Find the surface area of the sphere that passes through all eight vertices of the cube.","stemZh":"棱长为 2 的正方体，求经过它八个顶点的球的表面积。","options":[{"id":"A","text":"$12π$"},{"id":"B","text":"$4π$"},{"id":"C","text":"$48π$"},{"id":"D","text":"$16π$"}],"answer":"A","explanation":"The circumscribed sphere has diameter equal to the space diagonal $2\\sqrt{3}$, so $R = \\sqrt{3}$ and $S = 4πR^{2} = 4π·3 = 12π$. Here $4π$ is the inscribed sphere $(r = 1), 48π$ uses the full diagonal as the radius, and $16π$ uses $R = 2$.","explanationZh":"外接球的直径等于体对角线 $2\\sqrt{3}，故 R = \\sqrt{3}，S = 4πR^{2} = 4π·3 = 12π。4π$ 是内切球（r $= 1），48π 是把整条体对角线当半径，16π$ 是误取 $R = 2。$","tags":["hard","sphere","circumscribed-sphere"],"status":"published"},{"id":"math-space-022","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":5,"stem":"In the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$ of edge 1, the distance from A to the plane $A_{1}BD$ is","stemZh":"在棱长为 1 的正方体 $ABCD-A_{1}B_{1}C_{1}D_{1}$ 中，点 A 到平面 $A_{1}BD$ 的距离为","options":[{"id":"A","text":"$\\sqrt{3}$"},{"id":"B","text":"$\\sqrt{3}/2$"},{"id":"C","text":"1/2"},{"id":"D","text":"$\\sqrt{3}/3$"}],"answer":"D","explanation":"Use the plane $x + y + z = 1 (it$ contains $A_1, B$ and D). The distance from $A(0,0,0)$ to it is $|0 − 1|/\\sqrt{1^{2}+1^{2}+1^{2}} = 1/\\sqrt{3} = \\sqrt{3}/3. Equivalently$, compare volumes of the tetrahedron $A-A_1BD: (1/3)·(\\sqrt{3}/2)·d = 1/6$.","explanationZh":"平面 $A_1BD$ 的方程为 $x + y + z = 1，点 A(0,0,0)$ 到它的距离 $d = |0 − 1|/\\sqrt{1^{2}+1^{2}+1^{2}} = 1/\\sqrt{3} = \\sqrt{3}/3。也可用四面体 A-A_1BD 的体积法：(1/3)·(\\sqrt{3}/2)·d = 1/6。$","tags":["hard","point-to-plane-distance","cube"],"status":"published"},{"id":"math-space-023","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":4,"stem":"A cone has base radius 4 and height 3. Its lateral surface area is","stemZh":"底面半径为 4、高为 3 的圆锥，其侧面积为","options":[{"id":"A","text":"$12π$"},{"id":"B","text":"$20π$"},{"id":"C","text":"$16π$"},{"id":"D","text":"$25π$"}],"answer":"B","explanation":"The slant height is $l = \\sqrt{r^{2} + h^{2}} = \\sqrt{16 + 9} = 5$, so the lateral area is $πrl = π·4·5 = 20π$. Using the height 3 instead of the slant height gives $12π; 16π$ is the base area and $25π = πl^{2}$.","explanationZh":"母线 $l = \\sqrt{r^{2} + h^{2}} = \\sqrt{16 + 9} = 5，故侧面积 S = πrl = π·4·5 = 20π。把高 3$ 当成母线会得到 $12π；16π 是底面积，25π = πl^{2}。$","tags":["hard","cone","surface-area"],"status":"published"},{"id":"math-space-024","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":5,"stem":"An ant crawls on the surface of a rectangular box measuring 1 by 2 by 2 from one vertex to the opposite vertex. The shortest possible distance is","stemZh":"一只蚂蚁沿长、宽、高分别为 1、2、2 的长方体表面，从一个顶点爬到相对的顶点，最短路程为","options":[{"id":"A","text":"3"},{"id":"B","text":"$\\sqrt{13}$"},{"id":"C","text":"$2\\sqrt{2}$"},{"id":"D","text":"$\\sqrt{17}$"}],"answer":"B","explanation":"Unfolding gives a rectangle whose sides are $(1 + 2)$ and 2, so the distance is $\\sqrt{3^{2} + 2^{2}} = \\sqrt{13}$. The other $unfolding, (2 + 2)$ by 1, gives $\\sqrt{17}$, which is longer; 3 is the straight space diagonal, which leaves the surface.","explanationZh":"展开后得到边长为 $(1 + 2) 与 2$ 的矩形，最短路程为 $\\sqrt{3^{2} + 2^{2}} = \\sqrt{13}。另一种展开（2 + 2）与 1$ 给出 $\\sqrt{17}，更长；3$ 是体对角线，离开了表面。","tags":["hard","net","shortest-path"],"status":"published"},{"id":"math-space-025","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":4,"stem":"A cube is inscribed in a sphere of radius $1 (all$ eight vertices of the cube lie on the sphere). The volume of the cube is","stemZh":"一个正方体内接于半径为 1 的球（正方体八个顶点都在球面上），正方体的体积为","options":[{"id":"A","text":"$8\\sqrt{3}/9$"},{"id":"B","text":"8"},{"id":"C","text":"$2\\sqrt{3}/3$"},{"id":"D","text":"8/9"}],"answer":"A","explanation":"The space diagonal of the cube equals the diameter 2, so $a\\sqrt{3} = 2$ and $a = 2\\sqrt{3}/3$. Then $V = a^{3} = 8·3\\sqrt{3}/27 = 8\\sqrt{3}/9$. The value 8 uses the diameter as the edge, while $2\\sqrt{3}/3$ is the edge length itself, not the volume.","explanationZh":"正方体的体对角线等于球的直径 $2，即 a\\sqrt{3} = 2，得 a = 2\\sqrt{3}/3。于是 V = a^{3} = 8·3\\sqrt{3}/27 = 8\\sqrt{3}/9。取 8 是把直径当棱长，2\\sqrt{3}/3$ 是棱长本身而非体积。","tags":["hard","cube","inscribed-sphere"],"status":"published"},{"id":"math-space-026","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":4,"stem":"In the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$, which of the following statements is FALSE?","stemZh":"在正方体 $ABCD-A_{1}B_{1}C_{1}D_{1}$ 中，下列哪个结论是错误的？","options":[{"id":"A","text":"AB ⊥ plane $ADD_{1}A_{1}$"},{"id":"B","text":"$A_{1}A ⊥$ plane ABCD"},{"id":"C","text":"AB ⊥ plane ABCD"},{"id":"D","text":"AC ⊥ plane $BB_{1}D_{1}D$"}],"answer":"C","explanation":"AB lies IN the base plane ABCD, so it cannot be perpendicular to it. The others hold: AB ⊥ AD and AB $⊥ AA_{1} (two\\text{ intersecting lines of plane }ADD_{1}A_{1}); AA_{1}$ is an edge perpendicular to the base; and AC ⊥ BD together with AC $⊥ BB_{1}$ gives AC ⊥ plane $BB_{1}D_{1}D$.","explanationZh":"AB 本身在底面 ABCD 内，不可能垂直于该平面。其余都成立：AB ⊥ AD 且 AB $⊥ AA_{1}（平面 ADD_{1}A_{1} 内两条相交直线）；AA_{1}$ 是垂直于底面的棱；由 AC ⊥ BD 且 AC $⊥ BB_{1} 得$ AC ⊥ 平面 $BB_{1}D_{1}D。$","tags":["hard","line-plane-perpendicular","cube"],"status":"published"},{"id":"math-space-027","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":5,"stem":"In the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$, which line is parallel to the plane $A_{1}BD$?","stemZh":"在正方体 $ABCD-A_{1}B_{1}C_{1}D_{1}$ 中，哪条直线平行于平面 $A_{1}BD？$","options":[{"id":"A","text":"AB"},{"id":"B","text":"$C_{1}D_{1}$"},{"id":"C","text":"$CD_{1}$"},{"id":"D","text":"$A_{1}C_{1}$"}],"answer":"C","explanation":"With $A(0,0,0), B(1,0,0), D(0,1,0), A_{1}(0,0,1)$ the plane $A_{1}BD$ is $x + y + z = 1$, with normal $(1,1,1)$. The direction of $CD_{1}$ is $(−1,0,1)$, whose dot product with the normal is 0, and C does not lie on the plane, so $CD_{1} ∥$ plane $A_{1}BD$. The other three lines each meet the plane $(AB\\text{ at B, }A_{1}C_{1}\\text{ at }A_{1},\\text{ and the line }C_{1}D_{1}\\text{ meets the extended }plane)$.","explanationZh":"$设 A(0,0,0)、B(1,0,0)、D(0,1,0)、A_{1}(0,0,1)，则平面 A_{1}BD 为 x + y + z = 1，法向量为 (1,1,1)。CD_{1}$ 的方向向量为 $(−1,0,1)，与法向量数量积为 0，且 C$ 不在该平面内，故 $CD_{1} ∥$ 平面 $A_{1}BD。其余三条直线都与该平面相交（AB$ 交于 $B，A_{1}C_{1}$ 交于 $A_{1}，C_{1}D_{1}$ 所在直线与平面相交）。","tags":["hard","line-plane-parallel","cube"],"status":"published"},{"id":"math-space-028","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":5,"stem":"A sphere is inscribed in a right circular cylinder, touching both bases and the lateral surface. The ratio V(sphere) : V(cylinder) is","stemZh":"一个球内切于一个直圆柱（与两个底面和侧面都相切），则 V(球) : V(圆柱) 等于","options":[{"id":"A","text":"1 : 2"},{"id":"B","text":"3 : 4"},{"id":"C","text":"2 : 3"},{"id":"D","text":"1 : 3"}],"answer":"C","explanation":"The cylinder has radius r and height 2r. So $V(sphere) = (4/3)πr^{3}$ and $V(cylinder) = πr^{2}·2r = 2πr^{3}$, giving the ratio $(4/3) : 2 = 2 : 3$.","explanationZh":"圆柱底面半径为 r、高为 $2r，故 V(球) = (4/3)πr^{3}，V(圆柱) = πr^{2}·2r = 2πr^{3}，二者之比为 (4/3) : 2 = 2 : 3。$","tags":["hard","sphere","cylinder","volume-ratio"],"status":"published"},{"id":"math-space-029","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":4,"stem":"The lateral surface of a cylinder with base radius 1 and height 2 is cut along a generator and flattened. The resulting rectangle measures","stemZh":"把底面半径为 1、高为 2 的圆柱沿一条母线剪开并展平，所得矩形的两边长为","options":[{"id":"A","text":"$π$ and 2"},{"id":"B","text":"$2π$ and 2"},{"id":"C","text":"$2π$ and 4"},{"id":"D","text":"$4π$ and 2"}],"answer":"B","explanation":"One side of the rectangle is the height 2, the other is the circumference of the $base, 2πr = 2π$. Using $πr$ would give $π$, and using $2π(2r) = 4π$ would give a strip that is twice as wide as the actual lateral surface.","explanationZh":"矩形的一边是高 $2，另一边是底面周长 2πr = 2π。用 πr$ 会得到 $π，用 2π(2r) = 4π$ 则把侧面宽度算成了实际的两倍。","tags":["hard","net","cylinder"],"status":"published"},{"id":"math-space-030","subject":"math","moduleId":"geometry-algebra","unitId":"spatial-geometry","topicId":"space","type":"single-choice","difficulty":5,"stem":"In the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$ of edge 1, the tetrahedron with vertices $A_{1}, B, C_{1}$ and D has volume","stemZh":"在棱长为 1 的正方体 $ABCD-A_{1}B_{1}C_{1}D_{1}$ 中，以 $A_{1}、B、C_{1}、D$ 为顶点的四面体的体积为","options":[{"id":"A","text":"1/6"},{"id":"B","text":"2/3"},{"id":"C","text":"1/2"},{"id":"D","text":"1/3"}],"answer":"D","explanation":"Its six edges are face diagonals of length $\\sqrt{2}$, so it is a regular tetrahedron, obtained from the cube by removing four congruent corner tetrahedra $(A-A_1BD, B-A_1BC_1, C_1-BC_1D, D-A_1C_1D)$, each of volume 1/6. Hence $V = 1 − 4·(1/6) = 1/3$.","explanationZh":"该四面体的六条棱都是长为 $\\sqrt{2} 的面对角线，是正四面体。它由正方体挖去四个全等的角上四面体（A-A_1BD、B-A_1BC_1、C_1-BC_1D、D-A_1C_1D）得到，每个体积为 1/6，故 V = 1 − 4·(1/6) = 1/3。$","tags":["hard","tetrahedron","volume","cube"],"status":"published"}]

Geometry & Algebra