2023B AMC 10 真题（P版）

1 答案 C

总共有 \(3 + \frac{1}{3} = \frac{10}{3}\) 杯橙汁。平均分到四个杯子中，每杯为 \(\frac{10}{3} \div 4 = \frac{5}{6}\)。因此每个满杯需要倒出 \(1 - \frac{5}{6} = \frac{1}{6}\) 杯。所以正确答案是 C。

English

There’s \(3 + \frac{1}{3} = \frac{10}{3}\) glasses of juice in all. Split four ways, each glass ends up with \(\frac{10}{3} \div 4 = \frac{5}{6}.\) So a full glass has to pour out \(1 - \frac{5}{6} = \frac{1}{6}.\) Thus, C is the correct answer.

2 答案 B

设原价为 \(P\)。折扣和税后，Carlos 需要支付 \(0.8P \times 1.075 = 0.86P\)。他能负担当且仅当 \(0.86P \le 43\)，所以 \(P \le 50\)。因此最贵的原价为 \(\$50\)。所以正确答案是 B。

English

Let \(P\) be the original price. After the discount and tax, Carlos pays \(0.8P \times 1.075 = 0.86P.\) He can afford it when \(0.86P \le 43,\) which means \(P \le 50.\) So the priciest shoes he can swing start at \(\$50.\) Therefore, the answer is B.

3 答案 D

直角三角形的斜边是外接圆的直径。因此圆 \(A\) 的直径为 \(5\)，圆 \(B\) 的直径为 \(13\)。面积之比等于直径平方之比：\(\left(\frac{5}{13}\right)^2 = \frac{25}{169}\)。所以正确答案是 D。

English

The hypotenuse of a right triangle is a diameter of the circle around it. So circle \(A\) has diameter \(5\) and circle \(B\) has diameter \(13.\) Areas scale as the square of that, giving \(\left(\frac{5}{13}\right)^2 = \frac{25}{169}.\) Thus, D is the correct answer.

4 答案 C

先换算成厘米。宽度为 \(6.5\) 毫米 \(= 0.65\) 厘米，长度为 \(25\) 米 \(= 2500\) 厘米。面积为 \(0.65 \times 2500 = 1625\) 平方厘米。所以正确答案是 C。

English

Put everything in centimeters first. The strip is \(6.5\) mm \(= 0.65\) cm wide and \(25\) m \(= 2500\) cm long. Its area is \(0.65 \times 2500 = 1625\) square centimeters. Therefore, the answer is C.

5 答案 A

设原数之和为 \(S\)，数的个数为 \(n\)。Maddy 每个数加 \(3\)，总和为 \(S + 3n = 45\)。Lara 把每个数变为原来的三倍，所以 \(3S = 45\)，得 \(S = 15\)。因此 \(15 + 3n = 45\)，所以 \(n = 10\)。所以正确答案是 A。

English

Let \(S\) be the original sum and \(n\) the number of entries. Maddy adds \(3\) to each, so her total is \(S + 3n = 45.\) Lara triples each, so hers is \(3S = 45,\) giving \(S = 15.\) Then \(15 + 3n = 45,\) so \(n = 10.\) Thus, A is the correct answer.

6 答案 E

序列从 \(1, 3, 4, 7, 11, 18, \ldots\) 可见奇偶性为奇、奇、偶，并以周期 \(3\) 重复。因此 \(L_n\) 为偶数当且仅当 \(n\) 能被 \(3\) 整除。在 \(1 \le n \le 2023\) 中，\(3\) 的倍数有 \(\lfloor \frac{2023}{3} \rfloor = 674\) 个。所以正确答案是 E。

English

Track the parities: \(1, 3, 4, 7, 11, 18, \ldots\) run odd, odd, even, then repeat with period \(3.\) So \(L_n\) is even exactly when \(n\) is divisible by \(3.\) Among \(1 \le n \le 2023,\) that’s \(\lfloor \frac{2023}{3} \rfloor = 674\) multiples of \(3.\) Therefore, the answer is E.

7 答案 B

设 \(O\) 为共同中心。旋转把 \(A\) 送到 \(E\)，所以 \(OA = OE\)，且 \(\angle AOE = 20^\circ\)。三角形 \(OAE\) 是等腰三角形，底角为 \(\angle OAE = \frac{180^\circ - 20^\circ}{2} = 80^\circ\)。正方形对角线 \(AC\) 平分 \(A\) 点的直角，所以 \(\angle OAB = 45^\circ\)。两角相减，得到 \(\angle EAB = 80^\circ - 45^\circ = 35^\circ\)。所以正确答案是 B。

English

Let \(O\) be the shared center. The rotation carries \(A\) to \(E,\) so \(OA = OE\) and \(\angle AOE = 20^\circ.\) That makes triangle \(OAE\) isosceles, with base angles \(\angle OAE = \frac{180^\circ - 20^\circ}{2} = 80^\circ.\) The diagonal \(AC\) splits the right angle at \(A,\) so \(\angle OAB = 45^\circ.\) Subtracting, \(\angle EAB = 80^\circ - 45^\circ = 35^\circ.\) Thus, B is the correct answer.

8 答案 A

只看个位数。\(2\) 的幂个位数以 \(2, 4, 8, 6\) 为周期 \(4\)，而 \(2023 \equiv 3 \pmod 4\)，所以 \(2022^{2023}\) 个位数为 \(8\)。\(3\) 的幂个位数按 \(3, 9, 7, 1\) 循环，同样每四项重复一次；又因为 \(2022 \equiv 2 \pmod 4\)，所以 \(2023^{2022}\) 个位数为 \(9\)。相加 \(8 + 9 = 17\)，个位数为 \(7\)。所以正确答案是 A。

English

Only the units digits matter. Powers of \(2\) cycle \(2, 4, 8, 6\) with period \(4,\) and \(2023 \equiv 3 \pmod 4,\) so \(2022^{2023}\) ends in \(8.\) Powers of \(3\) cycle \(3, 9, 7, 1,\) and \(2022 \equiv 2 \pmod 4,\) so \(2023^{2022}\) ends in \(9.\) Add them: \(8 + 9 = 17,\) so the units digit is \(7.\) Therefore, the answer is A.

9 答案 B

连续平方数 \(k^2\) 和 \((k+1)^2\) 的差为 \((k+1)^2 - k^2 = 2k + 1\)。需要 \(2k + 1 \le 2023\)，所以 \(k \le 1011\)。因此 \(k\) 可取 \(1, 2, \ldots, 1011\)，共有 \(1011\) 对。所以正确答案是 B。

English

Consecutive squares \(k^2\) and \((k+1)^2\) differ by \((k+1)^2 - k^2 = 2k + 1.\) We need \(2k + 1 \le 2023,\) which gives \(k \le 1011.\) So \(k\) runs \(1, 2, \ldots, 1011,\) for \(1011\) pairs. Thus, B is the correct answer.

10 答案 C

假设每次猜测都没有命中。那么多米诺完全位于未猜的小方格中，所以未猜的小方格中必须含有两个相邻格。要保证命中，需要让未猜格中没有任何两个相邻。把方格黑白相间地染色。两两不相邻的小方格最多有 \(5\) 个，即四个角和中心。因此至少要猜 \(9 - 5 = 4\) 个格子。\(4\) 个也足够：猜四个边中点。每个隐藏矩形都会覆盖一个边中点，所以必定命中。正确答案是 C。

English

Suppose every guess misses. Then the domino lies entirely on unguessed squares, so those squares include two adjacent cells. To force a hit, we need the unguessed squares to have no two adjacent. Color the grid like a checkerboard. The biggest set of pairwise non-adjacent squares has \(5\) cells, one color’s worth: the four corners and the center. So we must guess at least \(9 - 5 = 4\) squares. And \(4\) is enough: guess the four edge midpoints, since every domino covers one square of each color, hence one edge midpoint. Therefore, the answer is C.

11 答案 B

设 \(a, b, c \ge 1\) 分别为 \(\$20, \$50, \$100\) 纸币张数。则 \(20a + 50b + 100c = 800\)，化为 \(2a + 5b + 10c = 80\)。因为 \(2a\) 和 \(10c\) 都是偶数，\(5b\) 也必须为偶数，所以 \(b = 2t\)。此时 \(a = 40 - 5t - 5c \ge 1\) 条件等价于 \(t + c \le 7\)，其中 \(t, c \ge 1\)。这样的正整数对数量为 \(1 + 2 + \cdots + 6 = 21\)。所以正确答案是 B。

English

Let \(a, b, c \ge 1\) count the \(\$20, \$50, \$100\) bills. Then \(20a + 50b + 100c = 800,\) which divides down to \(2a + 5b + 10c = 80.\) Both \(2a\) and \(10c\) are even, so \(5b\) is too, forcing \(b = 2t.\) Now \(a = 40 - 5t - 5c \ge 1\) means \(t + c \le 7.\) With \(t, c \ge 1,\) the pairs number \(1 + 2 + \cdots + 6 = 21.\) Thus, B is the correct answer.

12 答案 C

当 \(x \gt 10\) 时，每个因子 \((x - i)^i\) 都为正，所以 \(P(x) \gt 0\)。向左移动时，穿过 \(x = i\) 只有在 \(i\) 为奇数时才变号，即 \(i = 9, 7, 5, 3, 1\)。因此十一个区间从右到左的符号依次为 \(+, +, -, -, +, +, -, -, +, +, -\)，其中六个区间上为正。所以正确答案是 C。

English

For \(x \gt 10,\) every factor \((x - i)^i\) is positive, so \(P(x) \gt 0.\) Now move left. Crossing \(x = i\) flips the sign only when \(i\) is odd, that is at \(i = 9, 7, 5, 3, 1.\) So the eleven intervals, right to left, carry signs \(+, +, -, -, +, +, -, -, +, +, -.\) Six are positive. Therefore, the answer is C.

13 答案 B

令 \(u = |x|\)、\(v = |y|\)。则 \(|u - 1| + |v - 1| \le 1\) 是以 \((1, 1)\) 为中心、两条对角线长都为 \(2\) 的菱形，面积为 \(2\)，且它完全位于 \(u, v \ge 0\) 中。映射 \((x, y) \mapsto (|x|, |y|)\) 对 \(u, v \gt 0\) 的点是四对一，所以原区域面积为 \(4 \times 2 = 8\)。所以正确答案是 B。

English

Substitute \(u = |x|,\) \(v = |y|.\) Then \(|u - 1| + |v - 1| \le 1\) is a diamond centered at \((1, 1)\) with diagonals of length \(2,\) so it has area \(2,\) and it sits entirely in \(u, v \ge 0.\) The map \((x, y) \mapsto (|x|, |y|)\) is four-to-one over \(u, v \gt 0,\) so the full region has area \(4 \times 2 = 8.\) Thus, B is the correct answer.

14 答案 C

若 \(m = 0\)，方程迫使 \(n^2 = 0\)，得到 \((0, 0)\)。否则两者都非零。不妨假设 \(|m| \le |n|\)。则 \(m^2 n^2 = m^2 + mn + n^2 \le 3n^2\)，所以 \(m^2 \le 3\)，即 \(m = \pm 1\)。当 \(m = 1\) 时，\(1 + n + n^2 = n^2\)，得 \(n = -1\)。当 \(m = -1\) 时，得 \(n = 1\)。加上 \((0,0), (1,-1), (-1,1)\)，共有三个有序对。所以正确答案是 C。

English

If \(m = 0,\) the equation forces \(n^2 = 0,\) giving \((0, 0).\) Otherwise both are nonzero; assume \(|m| \le |n|.\) Then \(m^2 n^2 = m^2 + mn + n^2 \le 3n^2,\) so \(m^2 \le 3\) and \(m = \pm 1.\) Take \(m = 1:\) \(1 + n + n^2 = n^2\) gives \(n = -1.\) Take \(m = -1:\) \(n = 1.\) That leaves \((0,0), (1,-1), (-1,1),\) three in all. Therefore, the answer is C.

15 答案 C

将乘积分组为 \((2!\,3!)(4!\,5!)\cdots(14!\,15!)\cdot 16!\)。因为 \((2k)!(2k+1)!\) \(= \bigl((2k)!\bigr)^2(2k+1)\)，每一对都是平方数乘以一个奇数。这些奇数 \(3, 5, 7, 9, 11, 13, 15\) 的乘积为 \(3^4 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13\)，平方自由部分为 \(7 \cdot 11 \cdot 13\)。同时 \(16! = 2^{15} 3^6 5^3 7^2 \cdot 11 \cdot 13\) 的平方自由部分为 \(2 \cdot 5 \cdot 11 \cdot 13\)。两者相乘后，整个乘积的平方自由部分为 \(2 \cdot 5 \cdot 7\)。因此最小的 \(m\) 是 \(2 \cdot 5 \cdot 7 = 70\)。所以正确答案是 C。

English

Group the product as \((2!\,3!)(4!\,5!)\cdots(14!\,15!)\cdot 16!.\) Since \((2k)!(2k+1)!\) \(= \bigl((2k)!\bigr)^2(2k+1),\) each pair is a perfect square times an odd number. Those odd numbers \(3, 5, 7, 9, 11, 13, 15\) multiply to \(3^4 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13,\) with squarefree part \(7 \cdot 11 \cdot 13.\) And \(16! = 2^{15} 3^6 5^3 7^2 \cdot 11 \cdot 13\) has squarefree part \(2 \cdot 5 \cdot 11 \cdot 13.\) Multiply the two: the squarefree part of the whole thing is \(2 \cdot 5 \cdot 7.\) That’s the smallest \(m,\) namely \(2 \cdot 5 \cdot 7 = 70.\) Thus, C is the correct answer.

16 答案 E

一个“升数”就是选择至少 \(2\) 个数字并按递增顺序写出。数字 \(0\) 不能出现，因为它不能在首位，也不能跟在更小的数字后面。因此数字来自 \(\{1, \ldots, 9\}\)，所以 \(U = 2^9 - 1 - 9 = 502\)。一个“降数”可以以 \(0\) 结尾，所以可从 \(\{0, \ldots, 9\}\) 中选择 \(\ge 2\) 个数字，得到 \(D = 2^{10} - 1 - 10 = 1013\)。因此 \(|U - D| = |502 - 1013| = 511\)。所以正确答案是 E。

English

An upno is just a choice of at least \(2\) digits, written in increasing order. A \(0\) can never appear: it can’t lead and can’t follow a smaller digit. So the digits come from \(\{1, \ldots, 9\},\) giving \(U = 2^9 - 1 - 9 = 502.\) A downno can end in \(0,\) so its digits are any subset of \(\{0, \ldots, 9\}\) of size \(\ge 2,\) giving \(D = 2^{10} - 1 - 10 = 1013.\) So \(|U - D| = |502 - 1013| = 511.\) Therefore, the answer is E.

17 答案 D

\(12\) 条边给出 \(4(a + b + c) = 13\)，所以 \(a + b + c = \frac{13}{4}\)。\(6\) 个面给出 \(2(ab + bc + ca) = \frac{11}{2}\)，所以 \(ab + bc + ca = \frac{11}{4}\)。空间对角线长度为 \[ \begin{gathered} \sqrt{a^2 + b^2 + c^2} \\ = \small \sqrt{(a+b+c)^2 - 2(ab+bc+ca)} \\ = \sqrt{\frac{169}{16} - \frac{11}{2}} \\ = \sqrt{\frac{81}{16}} = \frac{9}{4}\text{。} \end{gathered} \] 所以正确答案是 D。

English

The \(12\) edges give \(4(a + b + c) = 13,\) so \(a + b + c = \frac{13}{4}.\) The \(6\) faces give \(2(ab + bc + ca) = \frac{11}{2},\) so \(ab + bc + ca = \frac{11}{4}.\) The space diagonal is \[ \begin{gathered} \sqrt{a^2 + b^2 + c^2} \\ = \small \sqrt{(a+b+c)^2 - 2(ab+bc+ca)} \\ = \sqrt{\frac{169}{16} - \frac{11}{2}} \\ = \sqrt{\frac{81}{16}} = \frac{9}{4}. \end{gathered} \] Thus, D is the correct answer.

18 答案 E

清除分母得 \(c = 15a + 14b\)。对 \(210 = 2 \cdot 3 \cdot 5 \cdot 7\) 的质因数分别取模：\(c \equiv a \pmod 2\)，\(c \equiv 2b \pmod 3\)，\(c \equiv 4b \pmod 5\)，\(c \equiv a \pmod 7\)。因此 \(\gcd(c, 210) = 1\) 当且仅当 \(a\) 既不是 \(2\) 的倍数也不是 \(7\) 的倍数，且 \(b\) 既不是 \(3\) 的倍数也不是 \(5\) 的倍数，这正是 \(\gcd(a, 14) = 1\) 且 \(\gcd(b, 15) = 1\)。所以 III 为真，也使 II 为真。I 不一定为真。例如 \(a = 1, b = 3\)，则 \(\gcd(a, 14) = 1\)，但 \(c = 57\) 能被 \(3\) 整除。因此只有 II 和 III 为真，正确答案是 E。

English

Clear denominators to get \(c = 15a + 14b.\) Reduce modulo the primes of \(210 = 2 \cdot 3 \cdot 5 \cdot 7:\) \(c \equiv a \pmod 2,\) \(c \equiv 2b \pmod 3,\) \(c \equiv 4b \pmod 5,\) and \(c \equiv a \pmod 7.\) So \(\gcd(c, 210) = 1\) iff \(a\) is divisible by neither \(2\) nor \(7\) and \(b\) is divisible by neither \(3\) nor \(5,\) which is exactly \(\gcd(a, 14) = 1\) and \(\gcd(b, 15) = 1.\) That settles III, and it makes II true since the “and” implies the “or.” Statement I fails, though: take \(a = 1, b = 3.\) Then \(\gcd(a, 14) = 1,\) yet \(c = 57\) is divisible by \(3.\) So only II and III hold. Therefore, the answer is E.

19 答案 B

四个方向由对称性相同。假设她向东跳。她跳出正方形当且仅当原来的 \(x\) 坐标加跳跃距离 \(d\) 超过 \(6\)。固定 \(d\) 时，\(x\) 在 \([0, 6]\) 上均匀分布，所以横坐标大于 \(6 - d\) 的概率为 \(\frac{d}{6}\)。现在对 \(d\) 在 \([0, 1]\) 上取平均，得到 \(\frac{1}{6} \cdot \frac{1}{2} = \frac{1}{12}\)。所以正确答案是 B。

English

The four directions behave the same by symmetry, so say she hops east. She lands outside exactly when her \(x\)-coordinate plus the hop distance \(d\) tops \(6.\) Fix \(d.\) Her \(x\)-coordinate is uniform on \([0, 6],\) so it beats \(6 - d\) with probability \(\frac{d}{6}.\) Now average over \(d\) uniform on \([0, 1]:\) \(\frac{1}{6} \cdot \frac{1}{2} = \frac{1}{12}.\) Thus, B is the correct answer.

20 答案 A

这条曲线由四段全等的半圆弧组成。若弧所在圆的半径为 \(r\)，则总长度是单段半圆弧长 \(\pi r\) 的 \(4\) 倍。四段弧相交于四个点，这四点构成一个内接于球面大圆的正方形；每段弧的直径就是该正方形的一条边。球半径为 \(2\)，所以正方形边长为 \(2\sqrt2\)，从而 \(r = \sqrt2\)。（也可以这样验证：小圆所在平面到球心的距离为 \(\frac{2}{\sqrt2} = \sqrt2\)，所以小圆半径为 \(\sqrt{2^2 - (\sqrt2)^2} = \sqrt2\)。）因此总长度为 \(4 \cdot \pi\sqrt2 = \pi\sqrt{32}\)，所以 \(n = 32\)。正确答案是 A。

English

The curve is four congruent semicircular arcs, so its length is \(4\) times one semicircle, \(\pi r,\) where \(r\) is the arc radius. The arcs meet at four points that form a square inscribed in a great circle of the radius-\(2\) sphere, and each arc’s diameter is a side of that square, a chord of length \(2\sqrt2.\) So \(r = \sqrt2.\) (Check it another way: the small circle sits in a plane at distance \(\frac{2}{\sqrt2} = \sqrt2\) from the center, giving radius \(\sqrt{2^2 - (\sqrt2)^2} = \sqrt2.\)) The total length is \(4 \cdot \pi\sqrt2 = \pi\sqrt{32},\) so \(n = 32.\) Therefore, the answer is A.

21 答案 E

所有 \(3^{2023}\) 种放法等可能。用符号筛计数每个盒子都是奇数的放法：\(\frac{1}{8}\sum_{s \in \{\pm 1\}^3}(s_1 s_2 s_3)\) \(\cdot (s_1 + s_2 + s_3)^{2023}\)。当 \(s = (1,1,1)\) 和 \(s = (-1,-1,-1)\) 时，两项都等于 \(3^{2023}\)。对于其余六种符号选择，括号内的和为 \(1\) 或 \(-1\)，而整项都等于 \(-1\)，所以这六项合计为 \(-6\)。因此计数为 \(\frac{2 \cdot 3^{2023} - 6}{8} = \frac{3^{2023} - 3}{4}\)。除以总数，概率为 \(\frac{3^{2023} - 3}{4 \cdot 3^{2023}} = \frac{1}{4} - \frac{1}{4 \cdot 3^{2022}}\)，略小于 \(\frac{1}{4}\)。所以正确答案是 E。

English

All \(3^{2023}\) assignments are equally likely. A sign filter counts the ones with every bin odd: \(\frac{1}{8}\sum_{s \in \{\pm 1\}^3}(s_1 s_2 s_3)\) \(\cdot (s_1 + s_2 + s_3)^{2023}.\) For \(s = (1,1,1)\) and \(s = (-1,-1,-1),\) both terms equal \(3^{2023}.\) For each other sign choice, the sum in parentheses is \(1\) or \(-1,\) and the full term equals \(-1,\) so these six terms total \(-6.\) Thus the count is \(\frac{2 \cdot 3^{2023} - 6}{8} = \frac{3^{2023} - 3}{4}.\) Dividing, the probability is \(\frac{3^{2023} - 3}{4 \cdot 3^{2023}} = \frac{1}{4} - \frac{1}{4 \cdot 3^{2022}},\) a hair under \(\frac{1}{4}.\) Thus, E is the correct answer.

22 答案 B

令 \(n = \lfloor x \rfloor\)。方程 \(n^2 - 3x + 2 = 0\) 给出 \(x = \frac{n^2 + 2}{3}\)。为了与这个定义一致，需要 \(n \le \frac{n^2 + 2}{3} \lt n + 1\)。左边不等式等价于 \(n^2 - 3n + 2 \ge 0\)，对整数 \(n\) 均成立。右边不等式等价于 \(n^2 - 3n - 1 \lt 0\)，只对 \(n \in \{0, 1, 2, 3\}\) 成立。这些值给出 \(x = \frac{2}{3}, 1, 2, \frac{11}{3}\)，所以有 \(4\) 个不同的值。正确答案是 B。

English

Set \(n = \lfloor x \rfloor.\) Then \(n^2 - 3x + 2 = 0\) gives \(x = \frac{n^2 + 2}{3}.\) For this to be consistent we need \(n \le \frac{n^2 + 2}{3} \lt n + 1.\) The left side, \(n^2 - 3n + 2 \ge 0,\) holds for every integer \(n.\) The right side, \(n^2 - 3n - 1 \lt 0,\) holds only for \(n \in \{0, 1, 2, 3\}.\) Those give \(x = \frac{2}{3}, 1, 2, \frac{11}{3},\) so there are \(4\) distinct values. Therefore, the answer is B.

23 答案 B

真实和为 \(S = na + \frac{n(n-1)}{2}d\)。因为一项相差 \(1\)，写下的总和满足 \(222 = S \pm 1\)，所以 \(S = 221\) 或 \(223\)。又因为 \(2S = n\bigl(2a + (n-1)d\bigr)\)，所以 \(n\) 整除 \(2S\)。由 \(a \ge 1\)、\(d \ge 2\) 还可得 \(2S \ge 2n^2\)，即 \(n^2 \le S\)。当 \(S = 223\) 时，\(446\) 的因数中没有位于 \(3\) 到 \(\sqrt{223}\) 之间的数。当 \(S = 221 = 13 \cdot 17\) 时，\(442\) 的因数中只有 \(n = 13\) 落在这个范围内。于是 \(2a + 12d = 34\)，即 \(a + 6d = 17\)。由于 \(a\)、\(d\) 是正整数且 \(d \gt 1\)，只能有 \(a = 5\)、\(d = 2\)。因此 \(a + d + n = 5 + 2 + 13 = 20\)，正确答案是 B。

English

The true sum is \(S = na + \frac{n(n-1)}{2}d.\) Since one term is off by \(1,\) the written total satisfies \(222 = S \pm 1,\) so \(S = 221\) or \(223.\) Also \(2S = n\bigl(2a + (n-1)d\bigr),\) so \(n\) divides \(2S.\) Since \(a \ge 1\) and \(d \ge 2,\) we have \(2S \ge 2n^2,\) hence \(n^2 \le S.\) For \(S = 223,\) no divisor of \(446\) lies between \(3\) and \(\sqrt{223}.\) For \(S = 221 = 13 \cdot 17,\) the only divisor of \(442\) in this range is \(n = 13.\) Thus \(2a + 12d = 34,\) or \(a + 6d = 17.\) Since \(a\) and \(d\) are positive integers with \(d \gt 1,\) we get \(a = 5,\) \(d = 2.\) Then \(a + d + n = 5 + 2 + 13 = 20.\) Thus, B is the correct answer.

24 答案 E

固定 \(w\)。随着 \(u, v\) 在 \([0, 1]^2\) 中变化，点 \((2u - 3w,\ v + 4w)\) 填满一个 \(2 \times 1\) 的轴对齐矩形，其左下角为 \((-3w, 4w)\)。当 \(w\) 从 \(0\) 到 \(1\) 变化时，这个矩形沿向量 \((-3,4)\) 平移，该向量长度为 \(5\)。扫过的区域是一个中心对称的六边形。它的三组对边的长度分别为 \(2, 1\) 和 \(5\)，分别来自矩形的两条边以及平移线段。因此它的周长为 \(2(2+1+5)=16\)。所以，答案是 E。

English

Fix \(w.\) As \(u, v\) sweep \([0, 1]^2,\) the point \((2u - 3w,\ v + 4w)\) fills a \(2 \times 1\) axis-aligned rectangle with lower-left corner \((-3w, 4w).\) As \(w\) runs from \(0\) to \(1,\) this rectangle slides along the vector \((-3,4),\) whose length is \(5.\) The swept region is a centrally symmetric hexagon. Its opposite pairs of sides have lengths \(2, 1,\) and \(5,\) inherited from the two sides of the rectangle and the sliding segment. Therefore its perimeter is \(2(2+1+5)=16.\) Therefore, the answer is E.

25 答案 B

设原正五边形外接圆半径为 \(R\)。将一个顶点折到中心时，折痕是中心到该顶点线段的垂直平分线，距离中心 \(\tfrac{R}{2}\)。五条折痕围成新的正五边形，其内切半径为 \(\tfrac{R}{2}\)。原正五边形内切半径为 \(R\cos 36^\circ\)，所以新旧正五边形相似比为 \(\frac{1}{2\cos 36^\circ}\)，面积缩放因子为 \(\left(\tfrac{1}{2\cos 36^\circ}\right)^2\)。代入 \(\cos 36^\circ = \tfrac{1+\sqrt5}{4}\)：\(\tfrac{2}{3+\sqrt5} = \tfrac{3-\sqrt5}{2}\)。因此新面积为 \((\sqrt5+1)\cdot\tfrac{3-\sqrt5}{2} = \sqrt{5} - 1\)。所以正确答案是 B。

English

Let the original pentagon have circumradius \(R.\) Folding a vertex to the center creases along the perpendicular bisector of the center-to-vertex segment, a line at distance \(\tfrac{R}{2}\) from the center. Those five creases bound the new regular pentagon, whose apothem is \(\tfrac{R}{2}\) (the original apothem was \(R\cos 36^\circ\)). So the new pentagon is similar with ratio \(\frac{1}{2\cos 36^\circ},\) and its area is the old area times \(\left(\tfrac{1}{2\cos 36^\circ}\right)^2.\) Plug in \(\cos 36^\circ = \tfrac{1+\sqrt5}{4}:\) the factor becomes \(\tfrac{2}{3+\sqrt5} = \tfrac{3-\sqrt5}{2}.\) So the new area is \((\sqrt5+1)\cdot\tfrac{3-\sqrt5}{2} = \sqrt{5} - 1.\) Thus, B is the correct answer.