2021A AMC 12 真题（P版）

1 答案 B

第一项的指数是 \(1+2+3 = 6\)，所以第一项是 \(2^6 = 64\)。括号内的和是 \(2 + 4 + 8 = 14\)。因此原式的值为 \(64 - 14 = 50\)。

因此，正确答案是 B。

English

The exponent in the first term is \(1+2+3 = 6,\) so the first term is \(2^6 = 64.\) The parenthesized sum is \(2 + 4 + 8 = 14.\) Therefore the value is \(64 - 14 = 50.\)

Thus, the correct answer is B.

2 答案 D

因为 \(\sqrt{a^2+b^2}\) 从不为负，等式成立必须有 \(a + b \ge 0\)。两边平方得 \(a^2 + b^2 = (a+b)^2\) \(= a^2 + 2ab + b^2\)，化简为 \(2ab = 0\)，即 \(ab = 0\)。

反过来，如果 \(ab = 0\)，则 \(a^2 + b^2 = (a+b)^2\)；如果还满足 \(a + b \ge 0\)，那么 \(\sqrt{a^2+b^2} = |a+b| = a+b\)。所以这两个条件合在一起正好是所需条件。

因此，正确答案是 D。

English

Because \(\sqrt{a^2+b^2}\) is never negative, equality requires \(a + b \ge 0.\) Squaring both sides gives \(a^2 + b^2 = (a+b)^2\) \(= a^2 + 2ab + b^2,\) which simplifies to \(2ab = 0,\) i.e. \(ab = 0.\)

Conversely, if \(ab = 0\) then \(a^2 + b^2 = (a+b)^2,\) and if additionally \(a + b \ge 0\) then \(\sqrt{a^2+b^2} = |a+b| = a+b.\) So both conditions together are exactly what is needed.

Thus, the correct answer is D.

3 答案 D

较大的数末尾是 \(0\)，擦去这个数字会除以 \(10\) 得到较小的数。因此较大的数是较小数的 \(10\) 倍。设较小的数为 \(x\)，则两数之和为 \(x + 10x = 11x = 17{,}402\)，得 \(x = 1{,}582\)。

两个数是 \(1{,}582\) 和 \(15{,}820\)，它们的差为 \(15{,}820 - 1{,}582 = 14{,}238\)。

因此，正确答案是 D。

English

The larger number ends in \(0,\) and erasing that digit divides it by \(10\) to give the smaller number. So the larger number is \(10\) times the smaller. Writing the smaller number as \(x,\) the sum is \(x + 10x = 11x = 17{,}402,\) giving \(x = 1{,}582.\)

The two numbers are \(1{,}582\) and \(15{,}820,\) whose difference is \(15{,}820 - 1{,}582 = 14{,}238.\)

Thus, the correct answer is D.

4 答案 D

紫色的蛇不会减法，而任何不会减法的蛇也不会加法。所以每条紫色的蛇都不会加法。

每条开心的蛇都会加法。由于紫色的蛇不会加法，没有开心的蛇可能是紫色的；也就是说，开心的蛇不是紫色的。

因此，正确答案是 D。

English

A purple snake cannot subtract, and any snake that cannot subtract also cannot add. So every purple snake cannot add.

Every happy snake can add. Since purple snakes cannot add, no happy snake can be purple; that is, happy snakes are not purple.

Thus, the correct answer is D.

5 答案 E

设 \(n = \overline{ab}\) 为这个两位整数。则 \(1.\overline{ab} = 1 + \dfrac{n}{99}\)，而有限小数为 \(1.ab = 1 + \dfrac{n}{100}\)。正确乘积减去学生的乘积为 \[ \begin{aligned} &66\left(\frac{n}{99} - \frac{n}{100}\right) \\ &= 66 \cdot \frac{n}{9900} = \frac{n}{150} \end{aligned}\text{。}\]

令 \(\dfrac{n}{150} = 0.5\) 得到 \(n = 75\)。

因此，正确答案是 E。

English

Let \(n = \overline{ab}\) be the two-digit integer. Then \(1.\overline{ab} = 1 + \dfrac{n}{99}\) while the terminating value is \(1.ab = 1 + \dfrac{n}{100}.\) The correct product minus the student’s product is \[ \begin{aligned} &66\left(\frac{n}{99} - \frac{n}{100}\right) \\ &= 66 \cdot \frac{n}{9900} = \frac{n}{150}. \end{aligned} \]

Setting \(\dfrac{n}{150} = 0.5\) gives \(n = 75.\)

Thus, the correct answer is E.

6 答案 C

设红牌数为 \(r\)，总牌数为 \(t\)。由 \(\dfrac{r}{t} = \dfrac13\) 得 \(t = 3r\)。加入 \(4\) 张黑牌后，\(\dfrac{r}{t+4} = \dfrac14\)，所以 \(t + 4 = 4r\)。

代入 \(t = 3r\) 得 \(3r + 4 = 4r\)，因此 \(r = 4\)，\(t = 12\)。

因此，正确答案是 C。

English

Let \(r\) be the number of red cards and \(t\) the total. From \(\dfrac{r}{t} = \dfrac13\) we get \(t = 3r.\) After adding \(4\) black cards, \(\dfrac{r}{t+4} = \dfrac14,\) so \(t + 4 = 4r.\)

Substituting \(t = 3r\) gives \(3r + 4 = 4r,\) so \(r = 4\) and \(t = 12.\)

Thus, the correct answer is C.

7 答案 D

展开得 \[ \begin{aligned} &(xy-1)^2 + (x+y)^2 \\ &= x^2y^2 - 2xy + 1 + x^2 \\ &\quad {}+ 2xy + y^2 \\ &= x^2y^2 + x^2 + y^2 + 1\text{。} \end{aligned} \] 这可因式分解为 \((x^2 + 1)(y^2 + 1)\)。

每个因子都至少为 \(1\)，所以乘积至少为 \(1\)，且当 \(x = y = 0\) 时取到等号。

因此，正确答案是 D。

English

Expanding, \[ \begin{aligned} &(xy-1)^2 + (x+y)^2 \\ &= x^2y^2 - 2xy + 1 + x^2 \\ &\quad {}+ 2xy + y^2 \\ &= x^2y^2 + x^2 + y^2 + 1. \end{aligned} \] This factors as \((x^2 + 1)(y^2 + 1).\)

Each factor is at least \(1,\) so the product is at least \(1,\) with equality when \(x = y = 0.\)

Thus, the correct answer is D.

8 答案 C

模 \(2\) 计算，项 \(D_0, D_1, D_2, \ldots\) 的奇偶性为 \(E\)、\(E\)、\(O\)、\(O\)、\(O\)、\(E\)、\(O\)、\(E\)、\(E\)、\(O\)、\(O\)、\(O\)、\(E\)、\(O, \ldots\)，从 \(D_0\) 开始以 \(7\) 为周期重复，因为 \(D_7, D_8, D_9\) 的奇偶性与 \(D_0, D_1, D_2\) 相同，都是 \(E, E, O\)。

由于 \(2021 \equiv 5\)、\(2022 \equiv 6\)，且 \(2023 \equiv 0 \pmod 7\)，所求奇偶性分别与 \(D_5, D_6, D_0\) 相同，即 \(E, O, E\)。

因此，正确答案是 C。

English

Working modulo \(2,\) the terms \(D_0, D_1, D_2, \ldots\) have parities \(E,\) \(E,\) \(O,\) \(O,\) \(O,\) \(E,\) \(O,\) \(E,\) \(E,\) \(O,\) \(O,\) \(O,\) \(E,\) \(O, \ldots\) which repeat with period \(7\) starting from \(D_0\) (indeed \(D_7, D_8, D_9\) have the same parities \(E, E, O\) as \(D_0, D_1, D_2\)).

Since \(2021 \equiv 5,\) \(2022 \equiv 6,\) and \(2023 \equiv 0 \pmod 7,\) the parities match those of \(D_5, D_6, D_0,\) namely \(E, O, E.\)

Thus, the correct answer is C.

9 答案 C

因为 \(3 - 2 = 1\)，将乘积乘以 \(3 - 2\) 不改变它的值。于是 \[ (3-2)(3+2) = 3^2 - 2^2\text{，} \] 再乘以下一个因子 \((3^2 + 2^2)\) 得到 \(3^4 - 2^4\)，依此类推。每一步都会使指数加倍。

用完全部七个因子后，乘积逐层相消，化为 \(3^{128} - 2^{128}\)。

因此，正确答案是 C。

English

Since \(3 - 2 = 1,\) multiplying the product by \(3 - 2\) does not change it. Then \[ (3-2)(3+2) = 3^2 - 2^2, \] and multiplying by the next factor \((3^2 + 2^2)\) gives \(3^4 - 2^4,\) and so on. Each step doubles the exponent.

After using all seven factors, the product telescopes to \(3^{128} - 2^{128}.\)

Thus, the correct answer is C.

10 答案 E

每个圆锥中的液体形成一个与容器相似的小圆锥。设窄圆锥中液体的半径为 \(3\)、高度为 \(h_1\)，宽圆锥中液体的半径为 \(6\)、高度为 \(h_2\)。体积相等给出 \(\tfrac13\pi\cdot 9\cdot h_1 = \tfrac13\pi\cdot 36\cdot h_2\)，所以 \(h_1 = 4h_2\)。

投入弹珠后，每个圆锥中的体积都增加相同的 \(\Delta V = \tfrac43\pi\) 且二者起始体积都是 \(V\)。因为圆锥体积随高度的三次方缩放，新高度为 \(h\sqrt[3]{1 + \frac{\Delta V}{V}}\)，所以上升量为 \(h\left(\sqrt[3]{1 + \frac{\Delta V}{V}} - 1\right)\)。这个因子对两个圆锥相同，因此上升量之比为 \(h_1 : h_2 = 4 : 1\)。

因此，正确答案是 E。

English

The liquid in each cone forms a smaller cone similar to the container. Let the narrow liquid cone have radius \(3\) and height \(h_1,\) and the wide one radius \(6\) and height \(h_2.\) Equal volumes give \(\tfrac13\pi\cdot 9\cdot h_1 = \tfrac13\pi\cdot 36\cdot h_2,\) so \(h_1 = 4h_2.\)

Dropping the marble raises the volume by the same amount \(\Delta V = \tfrac43\pi\) in each cone, and both start with the same volume \(V.\) Because a cone’s volume scales as the cube of its height, the new height is \(h\sqrt[3]{1 + \frac{\Delta V}{V}},\) so each rise equals \(h\left(\sqrt[3]{1 + \frac{\Delta V}{V}} - 1\right).\) This factor is identical for the two cones, so the rises are in the ratio \(h_1 : h_2 = 4 : 1.\)

Thus, the correct answer is E.

11 答案 C

每次反射都可以把路径展开成一条直线。将起点 \((3, 5)\) 关于 \(y\)-轴反射到 \((-3, 5)\)，并将目标点 \((7, 5)\) 关于 \(x\)-轴反射到 \((7, -5)\)。总传播距离等于这两个像点之间的直线距离：\[ \begin{aligned} &\sqrt{(-3 - 7)^2 + (5 - (-5))^2} \\ &= \sqrt{100 + 100} = 10\sqrt{2} \end{aligned}\text{。}\]

因此，正确答案是 C。

English

Reflecting the path at each bounce turns it into a single straight segment. Reflect the start \((3, 5)\) across the \(y\)-axis to \((-3, 5),\) and reflect the target \((7, 5)\) across the \(x\)-axis to \((7, -5).\) The total travel distance equals the straight-line distance between these two images: \[ \begin{aligned} &\sqrt{(-3 - 7)^2 + (5 - (-5))^2} \\ &= \sqrt{100 + 100} = 10\sqrt{2}. \end{aligned} \]

Thus, the correct answer is C.

12 答案 A

由韦达定理，六个根的和为 \(10\)（即 \(z^5\) 系数的相反数），乘积为 \(16\)。因此每个根都是 \(2^{e_i}\) 的形式，并且 \(\sum e_i=4\)。因为 \(2^{e_i}\ge 1+e_i\)，六个根的和至少为 \(6+4=10\)。只有当每个 \(e_i\) 都是 \(0\) 或 \(1\) 时才取等号，所以六个根必为 \(2,2,2,2,1,1\)。

因此多项式为 \((z - 1)^2 (z - 2)^4\)。展开得 \[ \begin{aligned} &(z^2 - 2z + 1) \\ &\quad {}\cdot (z^4 - 8z^3 + 24z^2 - 32z + 16) \\ &= z^6 - 10z^5 + 41z^4 \\ &\quad {}- 88z^3 + 104z^2 \\ &\quad {}- 64z + 16 \end{aligned}\text{。}\]\(z^3\) 的系数是 \(B = -88\)。

因此，正确答案是 A。

English

By Vieta’s formulas the six roots sum to \(10\) (the negative of the \(z^5\) coefficient) and multiply to \(16.\) Thus every root is a power \(2^{e_i},\) with \(\sum e_i=4.\) Since \(2^{e_i}\ge 1+e_i,\) the roots have sum at least \(6+4=10.\) Equality holds only when every \(e_i\) is \(0\) or \(1,\) so the roots must be \(2,2,2,2,1,1.\)

So the polynomial is \((z - 1)^2 (z - 2)^4.\) Expanding, \[ \begin{aligned} &(z^2 - 2z + 1) \\ &\quad {}\cdot (z^4 - 8z^3 + 24z^2 - 32z + 16) \\ &= z^6 - 10z^5 + 41z^4 \\ &\quad {}- 88z^3 + 104z^2 \\ &\quad {}- 64z + 16. \end{aligned} \] The coefficient of \(z^3\) is \(B = -88.\)

Thus, the correct answer is A.

13 答案 B

每个列出的复数模都是 \(2\)，所以 \(z^5\) 的模是 \(32\)，其实部为 \(32\cos(5\theta)\)，其中 \(\theta\) 是 \(z\) 的辐角。五个辐角分别是 \(180^\circ\)，\(150^\circ\)，\(135^\circ\)，\(120^\circ\)，和 \(90^\circ\)。

乘以 \(5\) 后得到 \(900^\circ \equiv 180^\circ\)，\(750^\circ \equiv 30^\circ\)，\(675^\circ \equiv 315^\circ\)，\(600^\circ \equiv 240^\circ\)，和 \(450^\circ \equiv 90^\circ\)。其中最大的余弦是 \(\cos 30^\circ\)，来自 \(z = -\sqrt3 + i\)，实部为 \(16\sqrt3\)。

因此，正确答案是 B。

English

Each listed number has modulus \(2,\) so \(z^5\) has modulus \(32,\) and its real part is \(32\cos(5\theta),\) where \(\theta\) is the argument of \(z.\) The arguments are \(180^\circ,\) \(150^\circ,\) \(135^\circ,\) \(120^\circ,\) and \(90^\circ.\)

Multiplying by \(5\) gives \(900^\circ \equiv 180^\circ,\) \(750^\circ \equiv 30^\circ,\) \(675^\circ \equiv 315^\circ,\) \(600^\circ \equiv 240^\circ,\) and \(450^\circ \equiv 90^\circ.\) The largest cosine is \(\cos 30^\circ,\) from \(z = -\sqrt3 + i,\) giving real part \(16\sqrt3.\)

Thus, the correct answer is B.

14 答案 E

对第一个和，\(\log_{5^k} 3^{k^2} = \dfrac{k^2}{k}\log_5 3 = k\log_5 3\)，所以 \[ \begin{aligned} &\sum_{k=1}^{20} k\log_5 3 = \frac{20\cdot 21}{2}\log_5 3 \\ &= 210\log_5 3 \end{aligned}\text{。}\]

对第二个和，\(\log_{9^k} 25^k = \log_9 25 = \log_3 5\) 与 \(k\) 无关，所以和为 \(100\log_3 5\)。

由于 \(\log_5 3 \cdot \log_3 5 = 1\)，乘积为 \(210 \cdot 100 = 21{,}000\)。

因此，正确答案是 E。

English

For the first sum, \(\log_{5^k} 3^{k^2} = \dfrac{k^2}{k}\log_5 3 = k\log_5 3,\) so \[ \begin{aligned} &\sum_{k=1}^{20} k\log_5 3 = \frac{20\cdot 21}{2}\log_5 3 \\ &= 210\log_5 3. \end{aligned} \]

For the second sum, \(\log_{9^k} 25^k = \log_9 25 = \log_3 5,\) independent of \(k,\) so the sum is \(100\log_3 5.\)

Since \(\log_5 3 \cdot \log_3 5 = 1,\) the product is \(210 \cdot 100 = 21{,}000.\)

Thus, the correct answer is E.

15 答案 D

选 \(t\) 名男高音和 \(b\) 名男低音的权重为 \(\binom{6}{t}\binom{8}{b}\)。为了只保留 \(t - b \equiv 0 \pmod 4\)，使用 \(\omega = i\) 的单位根筛法：\[ \begin{aligned} &N + 1 \\ &= \frac14\sum_{j=0}^{3}(1 + i^{j})^6\,(1 + i^{-j})^8 \end{aligned}\text{。}\]

\(j = 0\) 项为 \(2^6\cdot 2^8 = 16384\)。\(j = 2\) 项含有因子 \((1 + i^2)^6 = 0\)。\(j = 1\) 和 \(j = 3\) 项分别为 \(-128i\) 和 \(128i\)，相互抵消。因此总和为 \(16384\)，且 \(\dfrac{16384}{4} = 4096\)。

这个计数包含空组，所以 \(N = 4096 - 1 = 4095\)，因此 \(N \equiv 95 \pmod{100}\)。

因此，正确答案是 D。

English

Choosing \(t\) tenors and \(b\) basses is weighted by \(\binom{6}{t}\binom{8}{b}.\) To keep only \(t - b \equiv 0 \pmod 4,\) apply a roots of unity filter with \(\omega = i:\) \[ \begin{aligned} &N + 1 \\ &= \frac14\sum_{j=0}^{3}(1 + i^{j})^6\,(1 + i^{-j})^8. \end{aligned} \]

The \(j = 0\) term is \(2^6\cdot 2^8 = 16384.\) The \(j = 2\) term has factor \((1 + i^2)^6 = 0.\) The \(j = 1\) and \(j = 3\) terms are \(-128i\) and \(128i,\) which cancel. So the sum is \(16384,\) and \(\dfrac{16384}{4} = 4096.\)

This count includes the empty group, so \(N = 4096 - 1 = 4095,\) and \(N \equiv 95 \pmod{100}.\)

Thus, the correct answer is D.

16 答案 C

数列共有 \(1 + 2 + \cdots + 200\) \(= \dfrac{200\cdot 201}{2}\) \(= 20100\) 项，所以中位数是第 \(10050\) 项和第 \(10051\) 项的平均数。

数值 \(n\) 占据的位置一直到 \(\dfrac{n(n+1)}{2}\)。因为 \(\dfrac{141\cdot 142}{2} = 10011\)，而 \(\dfrac{142\cdot 143}{2} = 10153\)，所以位置 \(10012\) 到 \(10153\) 全部等于 \(142\)。两个中间位置都落在这一段中，所以中位数是 \(142\)。

因此，正确答案是 C。

English

The list has \(1 + 2 + \cdots + 200\) \(= \dfrac{200\cdot 201}{2}\) \(= 20100\) terms, so the median is the average of the \(10050\)th and \(10051\)st terms.

The value \(n\) occupies positions up to \(\dfrac{n(n+1)}{2}.\) Since \(\dfrac{141\cdot 142}{2} = 10011\) and \(\dfrac{142\cdot 143}{2} = 10153,\) positions \(10012\) through \(10153\) all equal \(142.\) Both middle positions fall in this block, so the median is \(142.\)

Thus, the correct answer is C.

17 答案 D

令 \(D = (0,0)\)，\(B = (b, 0)\) 在一条轴上，\(A = (0, a)\) 在另一条轴上，使 \(AD \perp BD\)。因为 \(CD \parallel AB\)，可写成 \(C = t(b, -a)\)，其中 \(t\) 为某个实数。于是 \(CD = t\sqrt{a^2+b^2}\)，且 \(BC^2 = b^2(1-t)^2 + t^2a^2\)。由 \(BC = CD\) 得 \(t^2 = (1-t)^2\)，所以 \(t = \tfrac12\)。

因此 \(C = \left(\tfrac{b}{2}, -\tfrac{a}{2}\right)\)，且 \(CD = 43\) 给出 \(a^2 + b^2 = 4\cdot 43^2 = 7396\)。对角线 \(AC\) 与 \(BD\)（\(x\)-轴）交于 \(O = \left(\tfrac{b}{3}, 0\right)\)，而 \(P = \left(\tfrac{b}{2}, 0\right)\)。因此 \(OP = \tfrac{b}{6} = 11\)，所以 \(b = 66\)。

那么 \(a^2 = 7396 - 66^2 = 3040\)，所以 \(AD = a = \sqrt{3040} = 4\sqrt{190}\)。取 \(m = 4\)、\(n = 190\)，得 \(m + n = 194\)。

因此，正确答案是 D。

English

Place \(D = (0,0)\) with \(B = (b, 0)\) on one axis and \(A = (0, a)\) on the other, so that \(AD \perp BD.\) Since \(CD \parallel AB,\) write \(C = t(b, -a)\) for some \(t.\) Then \(CD = t\sqrt{a^2+b^2}\) and \(BC^2 = b^2(1-t)^2 + t^2a^2.\) Setting \(BC = CD\) gives \(t^2 = (1-t)^2,\) so \(t = \tfrac12.\)

Thus \(C = \left(\tfrac{b}{2}, -\tfrac{a}{2}\right),\) and \(CD = 43\) gives \(a^2 + b^2 = 4\cdot 43^2 = 7396.\) The diagonal \(AC\) meets \(BD\) (the \(x\)-axis) at \(O = \left(\tfrac{b}{3}, 0\right),\) while \(P = \left(\tfrac{b}{2}, 0\right).\) Hence \(OP = \tfrac{b}{6} = 11,\) so \(b = 66.\)

Then \(a^2 = 7396 - 66^2 = 3040,\) so \(AD = a = \sqrt{3040} = 4\sqrt{190}.\) With \(m = 4\) and \(n = 190,\) we get \(m + n = 194.\)

Thus, the correct answer is D.

18 答案 E

函数方程使 \(f\) 完全可加：若 \(x = \prod p^{e_p}\)，则 \(f(x) = \sum e_p\, f(p) = \sum e_p\, p\)，其中分母中的质数贡献负指数，因为 \(f(\frac{1}{p}) = -p\)。

逐项计算：\(f\!\left(\tfrac{17}{32}\right) = 17 - 5\cdot 2 = 7\)，\(f\!\left(\tfrac{11}{16}\right) = 11 - 4\cdot 2 = 3\)，\(f\!\left(\tfrac{7}{9}\right) = 7 - 2\cdot 3 = 1\)，\(f\!\left(\tfrac{7}{6}\right) = 7 - 2 - 3 = 2\)，而 \(f\!\left(\tfrac{25}{11}\right) = 2\cdot 5 - 11 = -1\)。只有最后一个为负。

因此，正确答案是 E。

English

The functional equation makes \(f\) completely additive: for \(x = \prod p^{e_p},\) we have \(f(x) = \sum e_p\, f(p) = \sum e_p\, p,\) where a prime in the denominator contributes a negative exponent (since \(f(\frac{1}{p}) = -p\)).

Evaluating: \(f\!\left(\tfrac{17}{32}\right) = 17 - 5\cdot 2 = 7,\) \(f\!\left(\tfrac{11}{16}\right) = 11 - 4\cdot 2 = 3,\) \(f\!\left(\tfrac{7}{9}\right) = 7 - 2\cdot 3 = 1,\) \(f\!\left(\tfrac{7}{6}\right) = 7 - 2 - 3 = 2,\) and \(f\!\left(\tfrac{25}{11}\right) = 2\cdot 5 - 11 = -1.\) Only the last is negative.

Thus, the correct answer is E.

19 答案 C

将右边写为 \(\cos\left(\tfrac{\pi}{2}\sin x\right) = \sin\left(\tfrac{\pi}{2} - \tfrac{\pi}{2}\sin x\right)\)。两个正弦相等要求 \[ \begin{aligned} &\frac{\pi}{2}\cos x \\ &= \frac{\pi}{2}(1 - \sin x) + 2\pi k \end{aligned} \] 或 \[ \begin{aligned} &\frac{\pi}{2}\cos x \\ &= \pi - \frac{\pi}{2}(1 - \sin x) + 2\pi k\text{。} \end{aligned} \]

第一种化为 \(\cos x + \sin x = 1 + 4k\)；因为 \(\cos x + \sin x \in [-\sqrt2, \sqrt2]\)，只有 \(k = 0\) 可行，得到 \(\cos x + \sin x = 1\)，在 \([0, \pi]\) 中的解为 \(x = 0\) 和 \(x = \tfrac{\pi}{2}\)。第二种化为 \(\cos x - \sin x = 1\)，在 \([0, \pi]\) 中唯一解为 \(x = 0\)。

不同的解是 \(x = 0\) 和 \(x = \tfrac{\pi}{2}\)，共 \(2\) 个。

因此，正确答案是 C。

English

Write the right side as \(\cos\left(\tfrac{\pi}{2}\sin x\right) = \sin\left(\tfrac{\pi}{2} - \tfrac{\pi}{2}\sin x\right).\) Equal sines require either \[ \begin{aligned} &\frac{\pi}{2}\cos x \\ &= \frac{\pi}{2}(1 - \sin x) + 2\pi k \end{aligned} \] or \[ \begin{aligned} &\frac{\pi}{2}\cos x \\ &= \pi - \frac{\pi}{2}(1 - \sin x) + 2\pi k. \end{aligned} \]

The first reduces to \(\cos x + \sin x = 1 + 4k;\) since \(\cos x + \sin x \in [-\sqrt2, \sqrt2],\) only \(k = 0\) works, giving \(\cos x + \sin x = 1,\) with solutions \(x = 0\) and \(x = \tfrac{\pi}{2}\) in \([0, \pi].\) The second reduces to \(\cos x - \sin x = 1,\) whose only solution in \([0, \pi]\) is \(x = 0.\)

The distinct solutions are \(x = 0\) and \(x = \tfrac{\pi}{2},\) for a total of \(2.\)

Thus, the correct answer is C.

20 答案 B

令 \(V = (0, 0)\)，焦点 \(F = (0, f)\)，准线为 \(y = -f\)，其中 \(f = FV\)。抛物线上的点 \(A = (x, y)\) 满足 \(x^2 = 4fy\)，且 \(AF = y + f = 20\)，所以 \(y = 20 - f\)。又 \(AV^2 = x^2 + y^2 = 4fy + y^2\) \(= 441\)。

代入 \(y = 20 - f\)： \[ \begin{aligned} &4f(20 - f) + (20 - f)^2 = 441 \\ &\;\Longrightarrow\; 3f^2 - 40f + 41 = 0\text{。} \end{aligned} \] 由 Vieta 公式，两个可能的 \(f\) 值之和为 \(\dfrac{40}{3}\)。

因此，正确答案是 B。

English

Let \(V = (0, 0),\) focus \(F = (0, f),\) and directrix \(y = -f,\) where \(f = FV.\) A point \(A = (x, y)\) on the parabola satisfies \(x^2 = 4fy\) and \(AF = y + f = 20,\) so \(y = 20 - f.\) Also \(AV^2 = x^2 + y^2 = 4fy + y^2\) \(= 441.\)

Substituting \(y = 20 - f:\) \[ \begin{aligned} &4f(20 - f) + (20 - f)^2 = 441 \\ &\;\Longrightarrow\; 3f^2 - 40f + 41 = 0. \end{aligned} \] By Vieta’s formulas, the sum of the two possible values of \(f\) is \(\dfrac{40}{3}.\)

Thus, the correct answer is B.

21 答案 A

根为 \(z = 1\)，\(z = -1 \pm i\sqrt3\)，和 \(z = -2 \pm i\sqrt2\)，对应点 \((1, 0)\)，\((-1, \pm\sqrt3)\)，和 \((-2, \pm\sqrt2)\)。由于关于 \(x\)-轴对称，椭圆具有形式 \(Ax^2 + Cy^2 + Dx + F = 0\)。

代入这些点得到 \(5x^2 + 6y^2 + 9x - 14 = 0\)。配方得 \[ 5\left(x + \tfrac{9}{10}\right)^2 + 6y^2 = \tfrac{361}{20}\text{，} \] 所以 \(a^2 = \tfrac{361}{100}\)（沿 \(x\) 方向），且 \(b^2 = \tfrac{361}{120}\)。于是 \(e^2 = 1 - \dfrac{b^2}{a^2} = 1 - \dfrac{100}{120} = \dfrac16\)，所以 \(e = \sqrt{\tfrac16}\)。

因此 \(m = 1\)，\(n = 6\)，得 \(m + n = 7\)。

因此，正确答案是 A。

English

The roots are \(z = 1,\) \(z = -1 \pm i\sqrt3,\) and \(z = -2 \pm i\sqrt2,\) giving the points \((1, 0),\) \((-1, \pm\sqrt3),\) and \((-2, \pm\sqrt2).\) By symmetry about the \(x\)-axis, the ellipse has the form \(Ax^2 + Cy^2 + Dx + F = 0.\)

Substituting the points yields \(5x^2 + 6y^2 + 9x - 14 = 0.\) Completing the square gives \[ 5\left(x + \tfrac{9}{10}\right)^2 + 6y^2 = \tfrac{361}{20}, \] so \(a^2 = \tfrac{361}{100}\) (along \(x\)) and \(b^2 = \tfrac{361}{120}.\) Then \(e^2 = 1 - \dfrac{b^2}{a^2} = 1 - \dfrac{100}{120} = \dfrac16,\) so \(e = \sqrt{\tfrac16}.\)

With \(m = 1\) and \(n = 6,\) we get \(m + n = 7.\)

Thus, the correct answer is A.

22 答案 D

设三个根为 \(r_1,r_2,r_3\)。由七次单位根可得 \(r_1+r_2+r_3=-\tfrac12\)。把这三个角都加倍只是把它们的余弦值重新排列，所以 \(\sum r_i^2=\tfrac32+\tfrac12\sum r_i=\tfrac54\)。于是 \(r_1r_2+r_1r_3+r_2r_3\) 等于 \(\tfrac12((\sum r_i)^2-\sum r_i^2)=-\tfrac12\)。

由积化和差公式得 \(4r_1r_2r_3=1+r_1+r_2+r_3=\tfrac12\)，所以 \(r_1r_2r_3=\tfrac18\)。因此它们的首一多项式为 \[ x^3 + \tfrac12 x^2 - \tfrac12 x - \tfrac18 = 0\text{。}\]

对比系数得 \(a = \tfrac12\)，\(b = -\tfrac12\)，\(c = -\tfrac18\)。因此 \(abc = \tfrac12\cdot\left(-\tfrac12\right)\cdot\left(-\tfrac18\right) = \tfrac{1}{32}\)。

因此，正确答案是 D。

English

Let the three roots be \(r_1,r_2,r_3.\) The seventh roots of unity give \(r_1+r_2+r_3=-\tfrac12.\) Doubling the three angles merely permutes their cosines, so \(\sum r_i^2=\tfrac32+\tfrac12\sum r_i=\tfrac54.\) Hence \(r_1r_2+r_1r_3+r_2r_3\) equals \(\tfrac12((\sum r_i)^2-\sum r_i^2)=-\tfrac12.\)

The product-to-sum identity gives \(4r_1r_2r_3=1+r_1+r_2+r_3=\tfrac12,\) so \(r_1r_2r_3=\tfrac18.\) Therefore their monic polynomial is \[ x^3 + \tfrac12 x^2 - \tfrac12 x - \tfrac18 = 0. \]

Matching coefficients, \(a = \tfrac12,\) \(b = -\tfrac12,\) \(c = -\tfrac18.\) Therefore \(abc = \tfrac12\cdot\left(-\tfrac12\right)\cdot\left(-\tfrac18\right) = \tfrac{1}{32}.\)

Thus, the correct answer is D.

23 答案 D

将方格分为中心格 \(C\)、边中格 \(E\) 和角格（吸收状态）。从 \(C\) 出发，每一步都会到达一个 \(E\) 格。从一个 \(E\) 格出发，四个相邻格中有两个是角格，一个是中心格，一个是另一个 \(E\) 格，所以 \(P(\text{角格}) = \tfrac12\)、\(P(\text{中心格}) = \tfrac14\)、\(P(E) = \tfrac14\)。

设 \(a_n\) 为从边格出发、在 \(n\) 步内到达角格的概率，且 \(c_n\) 为从中心出发的概率（第一步必到边格）。于是 \(a_0=c_0=0\)，并且 \(c_n=a_{n-1}\)。计算得 \(a_n=\tfrac12+\tfrac14c_{n-1}+\tfrac14a_{n-1}\)，且 \(a_1=\tfrac12\)，\(a_2=\tfrac12+\tfrac14\cdot\tfrac12=\tfrac58\)，\[ a_3 = \tfrac12 + \tfrac14\cdot\tfrac12 + \tfrac14\cdot\tfrac58 = \tfrac{25}{32}\text{。}\]

从中心开始且有四步可用时，概率等于 \(a_3 = \tfrac{25}{32}\)（第一步到达边格，剩下三步）。

因此，正确答案是 D。

English

Classify squares as center \(C,\) edge-middle \(E,\) or corner (absorbing). From \(C,\) every hop lands on an \(E\) square. From an \(E\) square, two of the four neighbors are corners, one is the center, and one is another \(E\) square, so \(P(\text{corner}) = \tfrac12,\) \(P(\text{center}) = \tfrac14,\) \(P(E) = \tfrac14.\)

Let \(a_n\) be the probability of reaching a corner within \(n\) hops starting from an edge square, and let \(c_n\) be the corresponding probability from the center. Set \(a_0=c_0=0.\) Then \(c_n=a_{n-1}\) and \(a_n=\tfrac12+\tfrac14c_{n-1}+\tfrac14a_{n-1}.\) Thus \(a_1=\tfrac12,\) \(a_2=\tfrac12+\tfrac14\cdot\tfrac12=\tfrac58,\) and \[ a_3 = \tfrac12 + \tfrac14\cdot\tfrac12 + \tfrac14\cdot\tfrac58 = \tfrac{25}{32}. \]

Starting from the center with four hops available, the probability equals \(a_3 = \tfrac{25}{32}\) (the first hop reaches an edge, leaving three hops).

Thus, the correct answer is D.

24 答案 D

在圆 \(\Omega\) 中，弦 \(QR\) 所对的圆周角 \(\angle QPR = 60^\circ\)，所以 \(QR = 2r\sin 60^\circ\)，于是 \(3\sqrt3 = r\sqrt3\)，得 \(r = 3\)。

设 \(A = (-7, 0)\)、\(B = (7, 0)\)，且 \(\Gamma: x^2 + y^2 = 49\)（上半圆）。因为 \(\Omega\) 在 \(P = (p, 0)\) 处与 \(AB\) 相切，其圆心为 \((p, 3)\)。两个圆方程相减得到直线 \(QR\)，而圆心 \((p,3)\) 到 \(QR\) 的距离必须等于 \(r\cos 60^\circ = \tfrac32\)。这给出 \((p^2 - 31)^2 = 9p^2 + 81\)，所以 \(p^2 = 16\)（根 \(p^2 = 55\) 会使 \(P\) 在 \(AB\) 外）。

当 \(p^2 = 16\) 时，点 \(P\) 到直线 \(QR\) 的距离是 \(\dfrac{49 - p^2}{\sqrt{4p^2 + 36}} = \dfrac{33}{10}\)。因此 \[ \begin{aligned} [\triangle PQR] &= \tfrac12\cdot QR\cdot d \\ &= \tfrac12\cdot 3\sqrt3\cdot\tfrac{33}{10} \\ &= \frac{99\sqrt3}{20}\text{。} \end{aligned} \] 所以 \(a = 99\)、\(b = 3\)、\(c = 20\)，且 \(a + b + c = 122\)。

因此，正确答案是 D。

English

In circle \(\Omega,\) the chord \(QR\) subtends the inscribed angle \(\angle QPR = 60^\circ,\) so \(QR = 2r\sin 60^\circ,\) giving \(3\sqrt3 = r\sqrt3,\) hence \(r = 3.\)

Place \(A = (-7, 0),\) \(B = (7, 0),\) with \(\Gamma: x^2 + y^2 = 49\) (upper half). Since \(\Omega\) is tangent to \(AB\) at \(P = (p, 0),\) its center is \((p, 3).\) Subtracting the two circle equations gives the line \(QR,\) and the distance from the center \((p,3)\) to \(QR\) must equal \(r\cos 60^\circ = \tfrac32.\) This yields \((p^2 - 31)^2 = 9p^2 + 81,\) so \(p^2 = 16\) (the root \(p^2 = 55\) places \(P\) outside \(AB\)).

With \(p^2 = 16,\) the distance from \(P\) to line \(QR\) is \(\dfrac{49 - p^2}{\sqrt{4p^2 + 36}} = \dfrac{33}{10}.\) Thus \[ \begin{aligned} [\triangle PQR] &= \tfrac12\cdot QR\cdot d \\ &= \tfrac12\cdot 3\sqrt3\cdot\tfrac{33}{10} \\ &= \frac{99\sqrt3}{20}. \end{aligned} \] So \(a = 99,\) \(b = 3,\) \(c = 20,\) and \(a + b + c = 122.\)

Thus, the correct answer is D.

25 答案 E

因为 \(f(n) = \dfrac{d(n)}{n^{\frac{1}{3}}}\) 是乘法函数，它的值可按质数幂分解为各个 \(p^e\ \|\ n\) 所对应的项 \(\dfrac{e + 1}{p^{\frac{e}{3}}}\) 的乘积。我们分别最大化每一项。

设 \(g_p(e)=\dfrac{e+1}{p^{\frac{e}{3}}}\)，则 \(\dfrac{g_p(e+1)}{g_p(e)}=\dfrac{e+2}{e+1}p^{-\frac{1}{3}}\)。这个比值随 \(e\) 递减，所以只要看它第一次小于 \(1\) 的位置，就能确定唯一的最大值。由此得到：对 \(p=2\) 取 \(e=3\)；对 \(p=3\) 取 \(e=2\)；对 \(p=5,7\) 取 \(e=1\)；而对每个质数 \(p\ge11\) 取 \(e=0\)。

因此 \(N = 2^3\cdot 3^2\cdot 5\cdot 7 = 2520\)，其数字和为 \(2 + 5 + 2 + 0 = 9\)。

因此，正确答案是 E。

English

Since \(f(n) = \dfrac{d(n)}{n^{\frac{1}{3}}}\) is multiplicative, its value factors over prime powers as a product of terms \(\dfrac{e + 1}{p^{\frac{e}{3}}}\) for each prime power \(p^e\ \|\ n.\) We maximize each term separately.

If \(g_p(e)=\dfrac{e+1}{p^{\frac{e}{3}}},\) then \(\dfrac{g_p(e+1)}{g_p(e)}=\dfrac{e+2}{e+1}p^{-\frac{1}{3}}.\) This ratio decreases with \(e,\) so checking where it first falls below \(1\) finds the unique maximum. It gives \(e=3\) for \(p=2,\) \(e=2\) for \(p=3,\) \(e=1\) for \(p=5,7,\) and \(e=0\) for every prime \(p\ge11.\)

Hence \(N = 2^3\cdot 3^2\cdot 5\cdot 7 = 2520,\) whose digit sum is \(2 + 5 + 2 + 0 = 9.\)

Thus, the correct answer is E.