2023A AMC 10 真题（P版）

1 答案 E

她们相向骑行，所以速度相加。两人以每小时 \(18 + 12 = 30\) 英里的速度缩短这 \(45\) 英里的距离，因此经过 \(\frac{45}{30} = 1.5\) 小时相遇。Alice 从 \(A\) 出发，所以这时她骑了 \(18 \cdot 1.5 = 27\) 英里。因此，正确答案是 E。

English

They ride toward each other, so their speeds add. That closes the \(45\)-mile gap at \(18 + 12 = 30\) mph, and they meet after \(\frac{45}{30} = 1.5\) hours. Alice starts at \(A,\) so by then she’s gone \(18 \cdot 1.5 = 27\) miles. Thus, E is the correct answer.

2 答案 A

设披萨重量为 \(p\)。一杯橙子片重 \(\frac14\) 磅，所以天平两边满足 \(\frac13 p + \frac72 \cdot \frac14 = \frac34 p + \frac12 \cdot \frac14\)，也就是 \(\frac13 p + \frac78 = \frac34 p + \frac18\)。整理披萨项得到 \(\frac68 = \left(\frac34 - \frac13\right)p = \frac{5}{12}p\)。因此 \(p = \frac34 \cdot \frac{12}{5} = \frac95 = 1\frac45\)。所以答案是 A。

English

Let \(p\) be the pizza’s weight. A cup of orange slices is \(\frac14\) pound, so the two sides balance as \(\frac13 p + \frac72 \cdot \frac14 = \frac34 p + \frac12 \cdot \frac14,\) that is \(\frac13 p + \frac78 = \frac34 p + \frac18.\) Collect the pizza terms: \(\frac68 = \left(\frac34 - \frac13\right)p = \frac{5}{12}p.\) So \(p = \frac34 \cdot \frac{12}{5} = \frac95 = 1\frac45.\) Therefore, the answer is A.

3 答案 A

如果一个完全平方数能被 \(5\) 整除，它就能被 \(25\) 整除，所以它形如 \((5k)^2 = 25k^2\)。我们需要 \(25k^2 \lt 2023\)，即 \(k^2 \lt 80.9\)。这允许 \(k = 1, 2, \ldots, 8\)，共有 \(8\) 个平方数。因此，正确答案是 A。

English

If a perfect square is divisible by \(5,\) it’s divisible by \(25,\) so it looks like \((5k)^2 = 25k^2.\) We need \(25k^2 \lt 2023,\) i.e. \(k^2 \lt 80.9.\) That allows \(k = 1, 2, \ldots, 8,\) which is \(8\) squares. Thus, A is the correct answer.

4 答案 D

在任意四边形中，每条边都短于另外三条边之和。设最长边为 \(s\)。其余边长之和为 \(26 - s\)，所以 \(s \lt 26 - s\)，得到 \(s \lt 13\)，因此 \(s \le 12\)。能否达到 \(12\) 呢？边长 \(4, 12, 9, 1\) 可行，因为 \(12 \lt 4 + 9 + 1\)。所以最大长度为 \(12\)。因此，答案是 D。

English

In any quadrilateral each side is shorter than the sum of the other three. Call the longest side \(s.\) The rest sum to \(26 - s,\) so \(s \lt 26 - s,\) which gives \(s \lt 13\) and hence \(s \le 12.\) Can we hit \(12?\) The sides \(4, 12, 9, 1\) work, since \(12 \lt 4 + 9 + 1.\) So the greatest length is \(12.\) Therefore, the answer is D.

5 答案 E

把所有数分解成质因数：\(8^5 \cdot 5^{10} \cdot 15^5\) \(= 2^{15} \cdot 5^{10} \cdot 3^5 \cdot 5^5\) \(= 2^{15} \cdot 5^{15} \cdot 3^5\) \(= 10^{15} \cdot 243\)。这就是 \(243\) 后面接 \(15\) 个零，所以共有 \(3 + 15 = 18\) 位数字。因此，正确答案是 E。

English

Factor everything into primes. \(8^5 \cdot 5^{10} \cdot 15^5\) \(= 2^{15} \cdot 5^{10} \cdot 3^5 \cdot 5^5\) \(= 2^{15} \cdot 5^{15} \cdot 3^5\) \(= 10^{15} \cdot 243.\) That’s \(243\) followed by \(15\) zeros, so it has \(3 + 15 = 18\) digits. Thus, E is the correct answer.

6 答案 D

按关联次数计数。每条边在 \(2\) 个面上，所以六个面的值之和是所有边值总和的 \(2\) 倍。每个顶点在 \(3\) 条边上，所以所有边值总和是顶点值总和的 \(3\) 倍。连起来，立方体的值为 \(2 \cdot 3 \cdot 21 = 126\)。因此，答案是 D。

English

Count by incidences. Each edge lies on \(2\) faces, so the six face values together are \(2\) times the total of all edge values. Each vertex lies on \(3\) edges, so the total edge value is \(3\) times the vertex sum. Chaining these, the cube’s value is \(2 \cdot 3 \cdot 21 = 126.\) Therefore, the answer is D.

7 答案 B

累加和只能通过开头几次掷骰恰好到达 \(3\)，且这些方式互斥：单独掷出 \(3\)（概率 \(\frac16\)），再是 \(1,2\) 和 \(2,1\)（各为 \(\frac1{36}\)），以及 \(1,1,1\)（概率 \(\frac1{216}\)）。相加得 \(\frac{36}{216} + \frac{6}{216} + \frac{6}{216} + \frac{1}{216} = \frac{49}{216}\)。因此，正确答案是 B。

English

The total can only reach exactly \(3\) through the opening rolls, and these ways are disjoint: \(3\) alone (probability \(\frac16\)), then \(1,2\) and \(2,1\) (each \(\frac1{36}\)), and \(1,1,1\) (probability \(\frac1{216}\)). Add them up: \(\frac{36}{216} + \frac{6}{216} + \frac{6}{216} + \frac{1}{216} = \frac{49}{216}.\) Thus, B is the correct answer.

8 答案 D

Breadus 读数是经过 \((110, 0)\) 和 \((350, 100)\) 的华氏温度线性函数，所以 \(B = \frac{100}{350 - 110}(F - 110)\) \(= \frac{5}{12}(F - 110)\)。代入 \(F = 200\)：\(B = \frac{5}{12} \cdot 90 = 37.5\)。因此，答案是 D。

English

The Breadus reading is linear in Fahrenheit through \((110, 0)\) and \((350, 100),\) so \(B = \frac{100}{350 - 110}(F - 110)\) \(= \frac{5}{12}(F - 110).\) Plug in \(F = 200:\) \(B = \frac{5}{12} \cdot 90 = 37.5.\) Therefore, the answer is D.

9 答案 E

年份 \(2023\) 已经给出两个 \(2\)（偶数次）、一个 \(0\) 和一个 \(3\)。因此，为了让每个数字总体都出现偶数次，月份 \(MM\) 和日期 \(DD\) 的四个数字必须再提供一个 \(0\)、一个 \(3\)，然后另外两个数字相同，同时还要让 \(2\) 的个数保持偶数。逐一检查合法的月日组合，得到 \(01/13\)、\(01/31\)、\(02/23\)、\(03/11\)、\(03/22\)、\(10/13\)、\(10/31\)、\(11/03\) 和 \(11/30\)，恰好有 \(9\) 个日期。因此，正确答案是 E。

English

The year \(2023\) already gives two \(2\)s (even), one \(0\), and one \(3\). So to make every digit occur an even number of times, the four digits of \(MM\) and \(DD\) must supply one more \(0\), one more \(3\), and two equal digits, while keeping the number of \(2\)s even. Checking the legal dates gives \(01/13,\) \(01/31,\) \(02/23,\) \(03/11,\) \(03/22,\) \(10/13,\) \(10/31,\) \(11/03,\) and \(11/30,\) exactly \(9\) dates. Thus, E is the correct answer.

10 答案 D

设当前 \(n\) 次小测的平均分为 \(m\)。再得一次 \(11\) 分后平均分为 \(m + 1\)：\(\frac{mn + 11}{n + 1} = m + 1\)，整理得 \(m + n = 10\)。再得三次 \(11\) 分后平均分为 \(m + 2\)：\(\frac{mn + 33}{n + 3} = m + 2\)，即 \(3m + 2n = 27\)。解这组方程得 \(m = 7\)。因此，答案是 D。

English

Let \(m\) be the current mean over \(n\) quizzes. One more \(11\) makes the mean \(m + 1:\) \(\frac{mn + 11}{n + 1} = m + 1,\) which tidies up to \(m + n = 10.\) Three more \(11\)s make it \(m + 2:\) \(\frac{mn + 33}{n + 3} = m + 2,\) i.e. \(3m + 2n = 27.\) Solve the pair and \(m = 7.\) Therefore, the answer is D.

11 答案 C

每个角上的直角三角形有直角边 \(a\) 和 \(b\)。外层正方形的一条边给出 \(a+b=\sqrt3\)，内接正方形的一条边给出 \(a^2+b^2=2\)。相减求积：\(2ab=(a+b)^2-(a^2+b^2)\) \(=3-2\) \(=1\)，所以 \(ab=\tfrac12\)。于是 \(a\) 和 \(b\) 是 \(t^2-\sqrt3\,t+\tfrac12=0\) 的根，即 \(\tfrac{\sqrt3\pm1}{2}\)。较短边与较长边之比为 \(\dfrac{\sqrt3-1}{\sqrt3+1}\) \(=\dfrac{(\sqrt3-1)^2}{2}\) \(=2-\sqrt3\)。因此，正确答案是 C。

English

Each corner right triangle has legs \(a\) and \(b.\) A side of the outer square gives \(a+b=\sqrt3,\) and a side of the inscribed square gives \(a^2+b^2=2.\) Subtract to find the product: \(2ab=(a+b)^2-(a^2+b^2)\) \(=3-2\) \(=1,\) so \(ab=\tfrac12.\) Then \(a\) and \(b\) are the roots of \(t^2-\sqrt3\,t+\tfrac12=0,\) namely \(\tfrac{\sqrt3\pm1}{2}.\) The ratio of the smaller leg to the larger is \(\dfrac{\sqrt3-1}{\sqrt3+1}\) \(=\dfrac{(\sqrt3-1)^2}{2}\) \(=2-\sqrt3.\) Thus, C is the correct answer.

12 答案 B

倒序 \(N\) 后，它的个位是 \(N\) 的首位。若倒序数能被 \(5\) 整除，则该数字为 \(0\) 或 \(5\)。三位数不能以 \(0\) 开头，所以 \(N\) 以 \(5\) 开头，也就是 \(500 \le N \le 599\)（倒序数总会以 \(5\) 结尾）。现在只需数其中 \(7\) 的倍数：从 \(7 \cdot 72 = 504\) 到 \(7 \cdot 85 = 595\)，共有 \(14\) 个数。因此，答案是 B。

English

When we reverse \(N,\) its last digit is the first digit of \(N.\) For the reversal to be divisible by \(5,\) that digit is \(0\) or \(5.\) A three-digit number can’t start with \(0,\) so \(N\) starts with \(5,\) meaning \(500 \le N \le 599\) (and the reversal ends in \(5,\) always fine). Now just count multiples of \(7\) here: from \(7 \cdot 72 = 504\) to \(7 \cdot 85 = 595,\) that’s \(14\) numbers. Therefore, the answer is B.

13 答案 C

设 \(A\) 为 Abdul，\(C\) 为 Chiang，且 \(AC = 48\)，\(B\) 为 Bharat，满足 \(\angle B = 60^\circ\)。所有以 \(60^\circ\) 角看见 \(AC\) 的点都在同一圆弧上，所以所有可行的 \(B\) 都在一个圆上，其中弦 \(AC\) 所对圆周角为 \(60^\circ\)。正弦定理给出该圆直径为 \(\frac{AC}{\sin 60^\circ} = \frac{48}{\frac{\sqrt3}{2}} = 32\sqrt3\)。现在 \(AB\) 是一条弦，而弦最长时为直径。所以 \(AB = 32\sqrt3\)，\(AB^2 = 1024 \cdot 3 = 3072\)。因此，正确答案是 C。

English

Let \(A\) be Abdul, \(C\) be Chiang with \(AC = 48,\) and \(B\) be Bharat with \(\angle B = 60^\circ.\) Every point seeing \(AC\) at \(60^\circ\) lies on one circular arc, so all valid \(B\) sit on a circle where chord \(AC\) subtends \(60^\circ.\) The law of sines gives its diameter, \(\frac{AC}{\sin 60^\circ} = \frac{48}{\frac{\sqrt3}{2}} = 32\sqrt3.\) Now \(AB\) is a chord, and a chord is longest when it’s a diameter. So \(AB = 32\sqrt3\) and \(AB^2 = 1024 \cdot 3 = 3072.\) Thus, C is the correct answer.

14 答案 B

一个数 \(n \le 100\) 只有在 \(n\) 能被 \(11\) 整除时，才可能有能被 \(11\) 整除的因数，所以 \(n \in \{11, 22, \ldots, 99\}\)。写成 \(n = 11m\)，其中 \(m \le 9\)。这里 \(m\) 不是 \(11\) 的倍数，所以 \(d(11m) = 2\,d(m)\)，而能被 \(11\) 整除的因数恰好是 \(11d\) 这种形式，共有 \(d(m)\) 个。因此对于每个这样的 \(n\)，概率都是 \(\frac{d(m)}{2\,d(m)} = \frac12\)。对全部 \(100\) 个起始数字取平均，概率为 \(\frac{1}{100}\sum_{m=1}^{9}\frac12 = \frac{9}{200}\)。因此，答案是 B。

English

A number \(n \le 100\) can only have a divisor divisible by \(11\) when \(n\) is divisible by \(11,\) so \(n \in \{11, 22, \ldots, 99\}.\) Write \(n = 11m\) with \(m \le 9.\) Here \(m\) is not a multiple of \(11,\) so \(d(11m) = 2\,d(m),\) and the divisors that are multiples of \(11\) are exactly the \(d(m)\) numbers \(11d.\) That makes the chance \(\frac{d(m)}{2\,d(m)} = \frac12\) for each such \(n.\) Averaging over all \(100\) starting numbers, the probability is \(\frac{1}{100}\sum_{m=1}^{9}\frac12 = \frac{9}{200}.\) Therefore, the answer is B.

15 答案 E

半径为 \(r\) 的圆面积为 \(\pi r^2\)。因此半径 \(2k\) 与 \(2k-1\) 之间的阴影环形区域面积为 \(\pi\big((2k)^2 - (2k-1)^2\big)\) \(= (4k-1)\pi\)。有 \(2n\) 个圆时，阴影总面积为 \(\pi\sum_{k=1}^{n}(4k-1) = \pi(2n^2 + n)\)。我们需要 \(2n^2 + n \ge 2023\)。当 \(n = 31\) 时为 \(1953\)，当 \(n = 32\) 时为 \(2080\)。所以 \(n = 32\)，也就是 \(2n = 64\) 个圆。因此，正确答案是 E。

English

A circle of radius \(r\) has area \(\pi r^2.\) So the shaded ring between radius \(2k\) and \(2k-1\) has area \(\pi\big((2k)^2 - (2k-1)^2\big)\) \(= (4k-1)\pi.\) With \(2n\) circles the shaded total is \(\pi\sum_{k=1}^{n}(4k-1) = \pi(2n^2 + n).\) We want \(2n^2 + n \ge 2023.\) At \(n = 31\) it’s \(1953,\) at \(n = 32\) it’s \(2080.\) So \(n = 32,\) which means \(2n = 64\) circles. Thus, E is the correct answer.

16 答案 B

设有 \(L\) 名左手选手和 \(2L\) 名右手选手，所以共有 \(3L\) 名选手，比赛场数为 \(\binom{3L}{2}\)。每场比赛有一名获胜者，且左手胜场是右手胜场的 \(1.4\) 倍，所以胜场按 \(7 : 5\) 分配，总场数必须是 \(12\) 的倍数。左手选手至多赢下所有至少有一名左手选手参加的比赛，共 \(\binom{L}{2} + 2L^2\) 场。因此 \[ \frac{7}{12}\binom{3L}{2} \leq \binom{L}{2} + 2L^2\text{，} \] 化简得 \(L \leq 3\)。当 \(L = 1\) 或 \(L = 2\) 时，总场数不是 \(12\) 的倍数；当 \(L = 3\) 时，总场数为 \(\binom{9}{2} = 36\)。这种结果确实可以实现：左手选手赢下对阵右手选手的全部 \(18\) 场比赛和左手选手之间的全部 \(3\) 场比赛，而右手选手赢下彼此之间的全部 \(15\) 场比赛。这样双方的胜场数分别为 \(21\) 和 \(15\)。因此总共进行了 \(36\) 场比赛，答案是 B。

English

Say there are \(L\) left-handers and \(2L\) right-handers, so \(3L\) players and \(\binom{3L}{2}\) games. Every game has one winner, and left wins are \(1.4\) times right wins, so the wins split \(7 : 5\) and the total must be a multiple of \(12\). Left-handers can win at most all games involving at least one left-hander, namely \(\binom{L}{2} + 2L^2\). Hence \[ \frac{7}{12}\binom{3L}{2} \leq \binom{L}{2} + 2L^2, \] which gives \(L \leq 3\). For \(L = 1\) and \(L = 2\), the total is not divisible by \(12\). For \(L = 3\), there are \(\binom{9}{2} = 36\) games. This is attainable if the left-handers win all \(18\) cross-group games and all \(3\) games among themselves, while the right-handers win their \(15\) internal games. The win totals are \(21\) and \(15\), so the answer is \(36\). Therefore, the answer is B.

17 答案 A

令 \(A = (0,0)\)、\(B = (30,0)\)、\(C = (30,28)\)、\(D = (0,28)\)，且 \(P = (30, p)\) 在 \(BC\) 上，\(Q = (30 - q, 28)\) 在 \(CD\) 上。三个直角三角形给出 \(AP = \sqrt{30^2 + p^2}\)、\(QA = \sqrt{28^2 + (30 - q)^2}\)、\(PQ = \sqrt{(28 - p)^2 + q^2}\)，并且三者都必须是整数。当 \(0 \leq p \leq 28\) 时，将等式写成 \((AP-p)(AP+p)=900\)，可知只有 \(p=0\) 和 \(p=16\) 两种可能，对应 \(AP=30\) 和 \(AP=34\)。同理，令 \(x=30-q\)，则 \((QA-x)(QA+x)=784\)，且 \(0 \leq x \leq 30\)，所以 \(x=0\) 或 \(x=21\)。将这四种组合代入 \(PQ\) 的公式检验，只有 \(p=16\)、\(x=21\) 可行。因此 \(q=9\)、\(QA=35\)，并且 \(PQ=\sqrt{12^2+9^2}=15\)。所以 \(\triangle APQ\) 的周长为 \(34+15+35=84\)，正确答案是 A。

English

Set \(A = (0,0)\), \(B = (30,0)\), \(C = (30,28)\), \(D = (0,28)\), with \(P = (30, p)\) on \(BC\) and \(Q = (30 - q, 28)\) on \(CD\). The three right triangles give \(AP = \sqrt{30^2 + p^2}\), \(QA = \sqrt{28^2 + (30 - q)^2}\), and \(PQ = \sqrt{(28 - p)^2 + q^2}\). For \(0 \leq p \leq 28\), the equation \((AP-p)(AP+p)=900\) gives only \(p=0\) and \(p=16\), with \(AP=30\) and \(AP=34\). Similarly, setting \(x=30-q\), the equation \((QA-x)(QA+x)=784\) with \(0 \leq x \leq 30\) gives \(x=0\) or \(x=21\). Testing these four combinations in the formula for \(PQ\), only \(p=16\), \(x=21\) works. Thus \(q=9\), \(QA=35\), and \(PQ=\sqrt{12^2+9^2}=15\). The perimeter of \(\triangle APQ\) is \(34+15+35=84\). Thus, A is the correct answer.

18 答案 D

每个菱形有 \(4\) 条边，每条棱由 \(2\) 个面共享，所以 \(E = \frac{12 \cdot 4}{2} = 24\)。已知 \(F = 12\)，由欧拉公式得 \(V = 2 - F + E = 14\)。设 \(x\) 个顶点有 \(3\) 条棱相交，其余 \(14 - x\) 个顶点有 \(4\) 条棱相交。度数之和等于棱数的两倍：\(3x + 4(14 - x) = 2E = 48\)，所以 \(x = 8\)。因此，答案是 D。

English

Each rhombus has \(4\) edges, and every edge is shared by \(2\) faces, so \(E = \frac{12 \cdot 4}{2} = 24.\) With \(F = 12,\) Euler’s formula gives \(V = 2 - F + E = 14.\) Suppose \(x\) vertices have \(3\) edges and the other \(14 - x\) have \(4.\) The degrees sum to twice the edge count: \(3x + 4(14 - x) = 2E = 48,\) so \(x = 8.\) Therefore, the answer is D.

19 答案 E

旋转使中心到每个点及其像的距离相等。因此 \(P\) 到 \(A\) 和 \(A'\) 等距，也到 \(B\) 和 \(B'\) 等距，所以它是两条垂直平分线的交点。从 \((3,3)\) 到 \((4,3)\) 的线段 \(BB'\) 的垂直平分线是 \(x = 3.5\)。从 \((1,2)\) 到 \((3,1)\) 的线段 \(AA'\) 的垂直平分线是 \(2x - y = 2.5\)。于是 \(y = 2(3.5) - 2.5 = 4.5\)，所以 \(P = (3.5, 4.5)\)，\(|r - s| = |3.5 - 4.5| = 1\)。因此，正确答案是 E。

English

A rotation keeps its center equidistant from each point and its image. So \(P\) is equidistant from \(A\) and \(A',\) and from \(B\) and \(B',\) which puts it at the intersection of two perpendicular bisectors. The bisector of \(BB'\) from \((3,3)\) to \((4,3)\) is \(x = 3.5.\) The bisector of \(AA'\) from \((1,2)\) to \((3,1)\) is \(2x - y = 2.5.\) Then \(y = 2(3.5) - 2.5 = 4.5,\) so \(P = (3.5, 4.5)\) and \(|r - s| = |3.5 - 4.5| = 1.\) Thus, E is the correct answer.

20 答案 D

按行把格子标为 \(a\)、\(b\)、\(c\)、\(d\)、\(e\)、\(f\)、\(g\)、\(h\)、\(i\)。左上角方块 \(a, b, d, e\) 是四种颜色的一个排列，所以有 \(4! = 24\) 种。方块 \(\{b, c, e, f\}\) 也必须包含四种颜色，而 \(b, e\) 已确定，所以 \(\{c, f\}\) 是剩下两种颜色的某种顺序：\(2\) 种。对 \(\{g, h\}\) 同理，它们是除 \(d, e\) 外的两种颜色，另有 \(2\) 种。最后 \(i\) 被迫取 \(\{e, f, h\}\) 中缺少的颜色，并且只有在 \(f \ne h\) 时才可行。在 \(2 \cdot 2 = 4\) 种顺序组合中，恰有一种满足 \(f = h\)，所以 \(3\) 种保留下来。总数为 \(24 \cdot 3 = 72\)。因此，答案是 D。

English

Label the cells row by row as \(a,\) \(b,\) \(c;\) \(d,\) \(e,\) \(f;\) \(g,\) \(h,\) \(i.\) The top-left block \(a, b, d, e\) is a permutation of the four colors, so \(4! = 24\) ways. The block \(\{b, c, e, f\}\) is also all four colors, and \(b, e\) are fixed, so \(\{c, f\}\) is the remaining two in some order: \(2\) ways. Same story for \(\{g, h\},\) the two colors apart from \(d, e,\) another \(2\) ways. That leaves \(i,\) forced to whatever color is missing from \(\{e, f, h\},\) and that only works when \(f \ne h.\) Of the \(2 \cdot 2 = 4\) order combinations, exactly one has \(f = h,\) so \(3\) survive. The total is \(24 \cdot 3 = 72.\) Therefore, the answer is D.

21 答案 D

把每个条件翻译成函数值：\(P(1) = 1\)、\(P(0) = 0\)、\(P(9) = 0\)、\(P(4) = 0\)。所以 \(0, 4, 9\) 是根。三次多项式可以吗？带这些根的首一三次多项式有 \(P(1) = (1)(-3)(-8) = 24 \ne 1\)，所以不行。次数最小的首一多项式是 \(4\) 次：\(P(x) = x(x - 4)(x - 9)(x - c)\)。现在 \(P(1) = (1)(-3)(-8)(1 - c)\) \(= 24(1 - c)\) \(= 1\)，所以 \(1 - c = \frac{1}{24}\)，且 \(c = \frac{23}{24}\)。这就是唯一的非整数根，因此 \(m + n = 23 + 24 = 47\)。因此，正确答案是 D。

English

Translate each condition into a value: \(P(1) = 1,\) \(P(0) = 0,\) \(P(9) = 0,\) and \(P(4) = 0.\) So \(0, 4, 9\) are roots. Could a cubic do it? A monic cubic with those roots has \(P(1) = (1)(-3)(-8) = 24 \ne 1,\) so no. The least-degree monic polynomial is degree \(4:\) \(P(x) = x(x - 4)(x - 9)(x - c).\) Now \(P(1) = (1)(-3)(-8)(1 - c)\) \(= 24(1 - c)\) \(= 1,\) so \(1 - c = \frac{1}{24}\) and \(c = \frac{23}{24}.\) That’s the lone non-integer root, so \(m + n = 23 + 24 = 47.\) Thus, D is the correct answer.

22 答案 D

把 \(C_1, C_2\) 的圆心放在 \(\left(\pm\frac14, 0\right)\)。由对称性，位于两圆内部的最大圆以原点为圆心，半径为 \(r_3\)，其中 \(1 - r_3 = \frac14\)，所以 \(r_3 = \frac34\)。设 \(C_4\) 圆心为 \((0, y)\)，半径为 \(r\)。与 \(C_1\) 内切给出 \(\sqrt{\frac1{16} + y^2} = 1 - r\)，与 \(C_3\) 外切给出 \(y = \frac34 + r\)。把第二个方程代入第一个：\(\frac1{16} + \left(\frac34 + r\right)^2 = (1 - r)^2\)。化简得 \(\frac72 r = \frac38\)，所以 \(r = \frac{3}{28}\)。因此，答案是 D。

English

Put the centers of \(C_1, C_2\) at \(\left(\pm\frac14, 0\right).\) By symmetry the largest circle inside both sits at the origin with radius \(r_3,\) where \(1 - r_3 = \frac14,\) so \(r_3 = \frac34.\) Let \(C_4\) be centered at \((0, y)\) with radius \(r.\) Internal tangency to \(C_1\) gives \(\sqrt{\frac1{16} + y^2} = 1 - r,\) and external tangency to \(C_3\) gives \(y = \frac34 + r.\) Substitute the second into the first: \(\frac1{16} + \left(\frac34 + r\right)^2 = (1 - r)^2.\) This collapses to \(\frac72 r = \frac38,\) so \(r = \frac{3}{28}.\) Therefore, the answer is D.

23 答案 C

相差 \(20\) 的互补因数为 \(b\) 和 \(b + 20\)，乘积为 \(N\)，所以 \(N = b^2 + 20b\)，且 \(N + 100 = (b + 10)^2\)。相差 \(23\) 的一对给出 \(4N + 529 = (2d + 23)^2\)。设 \(N + 100 = k^2\)。则 \(4k^2 + 129 = m^2\)，所以 \((m - 2k)(m + 2k)\) \(= 129\) \(= 3 \cdot 43\)。正因数对 \(3,43\) 给出 \(k=10\)，从而得到不合要求的 \(N=0\)。因数对 \(1,129\) 给出 \(m = 65\)、\(k = 32\)，于是 \(N = 32^2 - 100 = 924\)。检验：\(924 = 22 \cdot 42 = 21 \cdot 44\)，数字和为 \(9 + 2 + 4 = 15\)。因此，正确答案是 C。

English

Complementary divisors differing by \(20\) are \(b\) and \(b + 20\) with product \(N\), so \(N = b^2 + 20b\) and \(N + 100 = (b + 10)^2\). A pair differing by \(23\) gives \(4N + 529 = (2d + 23)^2\). Set \(N + 100 = k^2\). Then \(4k^2 + 129 = m^2\), so \((m - 2k)(m + 2k)\) \(= 129\) \(= 3 \cdot 43\). The positive factor pair \(3,43\) gives \(k=10\), hence the inadmissible value \(N=0\). The pair \(1,129\) gives \(m = 65\), \(k = 32\), hence \(N = 32^2 - 100 = 924\). Check it: \(924 = 22 \cdot 42 = 21 \cdot 44\), and the digit sum is \(9 + 2 + 4 = 15\). Thus, C is the correct answer.

24 答案 C

设 \(d=\tfrac37\)。把与框架某条固定边相接的那些积木的斜边延长。由于所有相关的角都是 \(60^\circ\)，这些延长线在一端围成一个边长为 \(1\) 的正三角形，在另一端围成一个边长为 \(1-d\) 的正三角形。于是这条框架边被分成长度依次为 \(d,1,1\) 和 \(1-d\) 的四段，所以它的长度是 \(d+1+1+(1-d)=3\)。边长为 \(t\) 的正六边形面积为 \(\tfrac{3\sqrt3}{2}t^2\)。因此未被占据的面积等于边长为 \(3\) 的框架面积减去六个单位积木的面积：\(\tfrac{3\sqrt3}{2}\cdot 3^2\) \(- 6\cdot\tfrac{3\sqrt3}{2}\) \(= \tfrac{27\sqrt3}{2} - 9\sqrt3\) \(= \tfrac{9\sqrt3}{2}\)。因此，答案是 C。

English

Let \(d=\tfrac37.\) Extend the slanted edges of the blocks that meet a fixed side of the frame. Because all the relevant angles are \(60^\circ,\) the extensions form an equilateral triangle of side \(1\) at one end and an equilateral triangle of side \(1-d\) at the other. Thus that frame side is partitioned into lengths \(d,1,1,\) and \(1-d,\) so its length is \(d+1+1+(1-d)=3.\) A regular hexagon of side \(t\) has area \(\tfrac{3\sqrt3}{2}t^2.\) Therefore the uncovered area is the area of the side-\(3\) frame minus the areas of the six unit blocks: \(\tfrac{3\sqrt3}{2}\cdot 3^2\) \(- 6\cdot\tfrac{3\sqrt3}{2}\) \(= \tfrac{27\sqrt3}{2} - 9\sqrt3\) \(= \tfrac{9\sqrt3}{2}.\) Therefore, the answer is C.

25 答案 A

固定 \(R\)。在其他 \(11\) 个顶点中，有 \(5\) 个到该固定顶点的距离为 \(1\)，\(5\) 个距离为 \(2\)，还有 \(1\) 个（对顶点）距离为 \(3\)。从这 \(11\) 个顶点中有序选出不同的 \(Q, S\)，共有 \(11 \cdot 10 = 110\) 对。满足 \(d(R,Q) = d(R,S)\) 的对数为 \(5\cdot4 + 5\cdot4 + 1\cdot0 = 40\)，所以 \(P(\text{相等}) = \frac{40}{110} = \frac{4}{11}\)。由 \(Q\) 与 \(S\) 的对称性，\(\gt\) 和 \(\lt\) 的情况平分剩余概率，所以 \(P(d(Q,R) \gt d(R,S))\) \(= \frac{1 - \frac4{11}}{2}\) \(= \frac{7}{22}\)。因此，正确答案是 A。

English

Fix \(R.\) Of the other \(11\) vertices, \(5\) sit at distance \(1,\) \(5\) at distance \(2,\) and \(1\) (the opposite vertex) at distance \(3.\) Pick ordered distinct \(Q, S\) from these \(11:\) that’s \(11 \cdot 10 = 110\) pairs. The ones with \(d(R,Q) = d(R,S)\) number \(5\cdot4 + 5\cdot4 + 1\cdot0 = 40,\) so \(P(\text{equal}) = \frac{40}{110} = \frac{4}{11}.\) By the symmetry between \(Q\) and \(S,\) the \(\gt\) and \(\lt\) cases split the rest evenly, so \(P(d(Q,R) \gt d(R,S))\) \(= \frac{1 - \frac4{11}}{2}\) \(= \frac{7}{22}.\) Thus, A is the correct answer.