2021B AMC 10 真题（P版）

1 答案 D

从 \(-9\) 到 \(9\) 的每个整数都满足条件，共有 \(9-(-9)+1 = 19\) 个。

所以正确答案是 D。

English

Every integer from \(-9\) to \(9,\) inclusive, works. This yields \(9-(-9)+1 = 19\) solutions.

Thus, the correct answer is D .

2 答案 D

因为 \(\sqrt{u^2}=|u|\)，所以原式等于 \[|3-2\sqrt3|+|3+2\sqrt3|\text{。}\] 又因为 \(2\sqrt3>3\)，所以它化为 \[(2\sqrt3-3)+(3+2\sqrt3)=4\sqrt3\text{。}\]

所以正确答案是 D。

English

Because \(\sqrt{u^2}=|u|,\) the expression equals \[|3-2\sqrt3|+|3+2\sqrt3|.\] Since \(2\sqrt3>3,\) this becomes \[(2\sqrt3-3)+(3+2\sqrt3)=4\sqrt3.\]

Thus, the correct answer is D .

3 答案 C

设低年级学生有 \(j\) 人，高年级学生有 \(s\) 人。那么 \(j+s = 28\)，且 \(0.25j = 0.1s\)。这说明 \(2.5j = s\)，所以 \(3.5j = 28\)。因此 \(j=8\)。

所以正确答案是 C。

English

Let the number of juniors be \(j\) and the number of seniors be \(s.\) Then, \(j+s = 28\) and \(0.25j = 0.1s.\) This means \(2.5j = s,\) so \(3.5j = 28.\) This makes \(j=8.\)

Thus, the correct answer is C .

4 答案 B

蓝蓝配对用掉 \(2\cdot 23 = 46\) 名穿蓝色衬衫的学生。因此还有 \(57-46=11\) 名穿蓝色衬衫的学生在混合配对中，也就恰有 \(11\) 名穿黄色衬衫的学生在混合配对中。

剩下只有 \(64\) 名穿黄色衬衫的学生与另一名穿黄色衬衫的学生配对，所以黄黄配对数为 \(\dfrac{64}2 = 32\)。

所以答案是 B。

English

There are \(2\cdot 23 = 46\) students with blue shirts that are in a pair with just blue shirts. This means there are \(57-46=11\) students in blue shirts who are paired with someone wearing a yellow shirt, meaning exactly \(11\) people wearing yellow shirts are paired with someone wearing a blue shirt.

This leaves just \(64\) students wearing a yellow shirt who are paired with someone else wearing a yellow shirt. This yields \(\dfrac{64}2 = 32 \) pairs.

Thus, the answer is B .

5 答案 B

乘积为 \(30\) 的一位正整数对只有 \(5,6\)。乘积为 \(24\) 的一位数因数对是 \(3,8\) 和 \(4,6\)。因为四个年龄互不相同，若取 \(4,6\) 就会重复年龄 \(6\)，所以另一对必须是 \(3,8\)。因此四个年龄是 \(3,5,6,8\)，它们的和为 \(22\)。

所以答案是 B。

English

The only pair of single-digit positive integers with product \(30\) is \(5,6.\) The single-digit factor pairs of \(24\) are \(3,8\) and \(4,6.\) Because all four ages are distinct, the pair \(4,6\) would repeat the age \(6,\) so the other pair must be \(3,8.\) Thus the ages are \(3,5,6,8,\) whose sum is \(22.\)

Thus, the answer is B .

6 答案 C

设上午班有 \(3x\) 人，那么下午班有 \(4x\) 人。

上午班总分为 \(84\cdot 3x = 252x\)，下午班总分为 \(70\cdot 4x = 280x\)。因此总分为 \(532x\)，总人数为 \(7x\)。

所以所有学生的平均分为 \(\dfrac{532x}{7x} = 76\)。

所以正确答案是 C。

English

Let the number of people in the first class be \(3x.\) This means the number of people in the second class is \(4x.\)

Thus, the sum of the scores of the first class is \(84\cdot 3x = 252x\) and the sum of the scores for the people in the second class is \(70\cdot 4x = 280x.\) This means the total sum is \(532x,\) with \(7x\) people.

Therefore, the average of all the students is \( \dfrac{532x}{7x} = 76.\)

Thus, the correct answer is C .

7 答案 D

在 \(\ell\) 的同一侧，所有在 \(A\) 处相切的圆互相包含。若套在一起的圆半径为 \(r_1>r_2>\cdots\)，则恰好在一个圆内的面积为 \(\pi(r_1^2-r_2^2)\)；更小圆内的点同时位于至少三个圆内，不符合“恰好一个”的条件。

要使面积最大，把半径 \(7\) 的圆单独放在一侧，把半径 \(5,3,1\) 的圆放在另一侧。这给出

\[ \begin{aligned} &7^2\pi+(5^2-3^2)\pi \\ &=49\pi+16\pi=65\pi \end{aligned}\text{。}\]

所以答案是 D。

English

On one side of \(\ell,\) circles tangent at \(A\) are nested. For nested circles with radii \(r_1>r_2>\cdots,\) the points inside exactly one of those circles have area \(\pi(r_1^2-r_2^2)\) if there are at least two circles; a third smaller nested circle does not count because its points are inside three circles, not exactly one.

To maximize the area, put the circle of radius \(7\) alone on one side, and put the circles of radii \(5,3,1\) on the other side. This gives

\[ \begin{aligned} &7^2\pi+(5^2-3^2)\pi \\ &=49\pi+16\pi=65\pi. \end{aligned} \]

Thus, the answer is D .

8 答案 A

在外层 \(15\times15\) 的一圈中，最上面一行包含 \(211,212,\ldots,225\)，因此第二行中正好在 \(211\) 下方的数是 \(210\)。

这是第二行的最大数。内层 \(13\times13\) 螺旋的右上角是 \(13^2=169\)。在整个方格第二行中，内层这一行的数从 \(157\) 到 \(169\)，因此最小数是 \(157\)。

所求和为 \(210+157=367\)。

所以答案是 A。

English

In the outer \(15\times15\) ring, the top row contains \(211,212,\ldots,225,\) so the number just below \(211\) in the second row is \(210.\) This is the greatest number in the second row.

The inner \(13\times13\) spiral has \(13^2=169\) in its upper-right corner. In the second row of the full grid, the inner-ring entries run from \(157\) to \(169.\) Thus the least entry in that row is \(157.\)

The required sum is \(210+157=367.\)

Thus, the answer is A .

9 答案 D

倒推变换。点 \((-6,3)\) 关于 \(y=-x\) 反射后得到 \((-3,6)\)。

接着把 \((-3,6)\) 绕 \((1,5)\) 顺时针旋转 \(90^\circ\)，以撤销原来的逆时针旋转。相对于 \((1,5)\)，该点为 \((-4,1)\)。顺时针旋转四分之一圈得到 \((1,4)\)，再平移回去得到 \((2,9)\)。

因此 \(a=2\)、\(b=9\)，所以 \(b-a=7\)。

所以答案是 D。

English

Work backward. Reflecting \((-6,3)\) across \(y=-x\) gives \((-3,6).\)

Now undo the \(90^\circ\) counterclockwise rotation by rotating \((-3,6)\) clockwise about \((1,5).\) Relative to \((1,5),\) the point is \((-4,1).\) A clockwise quarter-turn sends this to \((1,4),\) and translating back gives \((2,9).\)

Thus \(a=2,\) \(b=9,\) and \(b-a=7.\)

Thus, the answer is D .

10 答案 A

因为水是从一个容器倒入另一个容器，所以体积保持不变。圆锥中水的体积为 \(\frac{r^2h \pi}3 = \frac{12^2\cdot 18 \pi}3 = 864 \pi\)。

圆柱中水的体积为 \(r^2h \pi =24^2h \pi = 576h \pi\)。于是 \(576 h \pi = 864 \pi\)，所以 \(h = 1.5\)。

所以答案是 A。

English

The volumes must be the same since the water is poured from one to another. The volume of the cone is \(\frac{r^2h \pi}3 = \frac{12^2\cdot 18 \pi}3 = 864 \pi.\)

The volume of the cylinder is \(r^2h \pi =24^2h \pi = 576h \pi.\) This makes \(576 h \pi = 864 \pi,\) so \(h = 1.5.\)

Thus, the answer is A .

11 答案 D

假设切成 \(l\times w\) 的网格。内部小块数为 \((l-2)(w-2)\)，总块数为 \(lw\)。因为内部小块数等于周边小块数，所以内部小块占总块数的一半：

\[(l-2)(w-2)=\frac{lw}{2}\text{。}\]

展开得到 \(lw-4l-4w+8=0\)，整理为：

\[(l-4)(w-4)=8\text{。}\]

\(8\) 的正因数对给出 \((l,w)=(5,12)\) 或 \((6,8)\)，顺序可交换。它们分别产生 \(60\) 或 \(48\) 块，所以最大可能数量是 \(60\)。

所以答案是 D。

English

Suppose the cuts make an \(l\times w\) grid of pieces. The number of interior pieces is \((l-2)(w-2),\) and the total number of pieces is \(lw.\) Since the number of interior pieces equals the number of perimeter pieces, the interior pieces make up half the total:

\[(l-2)(w-2)=\frac{lw}{2}.\]

Multiplying out gives \(lw-4l-4w+8=0,\) or

\[(l-4)(w-4)=8.\]

The positive factor pairs of \(8\) give \((l,w)=(5,12)\) or \((6,8),\) up to order. These produce \(60\) or \(48\) pieces, respectively, so the greatest possible number is \(60.\)

Thus, the answer is D .

12 答案 C

分解质因数得 \[N =2^3\cdot 3^5\cdot 5\cdot 7\cdot 17^2\text{。}\]

若 \(x\) 是 \(N\) 的一个奇因数，则 \(2x,4x,8x\) 都是 \(N\) 的偶因数，它们的和为 \(14x\)。把所有奇因数都这样配对，偶因数之和就是奇因数之和的 \(14\) 倍。因此所求比为 \(1:14\)。

所以答案是 C。

English

Using prime factorization, we get \[N =2^3\cdot 3^5\cdot 5\cdot 7\cdot 17^2.\]

If we have an odd divisor \(x\) of \(N,\) then \(2x,4x,8x\) are divisors of \(N,\) which has a combined sum of \(14x.\) If we take the sum of every odd divisor, then the even divisors must have a sum which is \(14\) times the sum of the odd divisors. Therefore, the requested ratio is \(1:14.\)

Thus, the answer is C .

13 答案 B

第一个条件表示 \[3n^2 + 2n+d = 263\text{。}\]

类似地，第二个条件表示 \[3n^2 +2n +4 \] \[= 6^3 +6^2 +6d+1 \] \[= 253 + 6d\text{。}\]

两式相减得 \[4-d = 6d-10\] \[7d = 14\] \[d=2\text{。}\] 因此 \(3n^2 + 2n + 2 = 263\)，所以 \(n(3n+2) = 261\)。

这给出 \(n=9\)。因此 \(n+d = 11\)。

所以答案是 B。

English

The first statement means \[3n^2 + 2n+d = 263.\]

Similarly, the second statement means \[3n^2 +2n +4 \]\[= 6^3 +6^2 +6d+1 \]\[= 253 + 6d.\]

Subtracting these shows us that \[4-d = 6d-10\] \[7d = 14\] \[d=2.\] Therefore, \(3n^2 + 2n + 2 = 263,\) so \(n(3n+2) = 261.\)

This implies \(n=9.\) Therefore, \(n+d = 11.\)

Thus, the answer is B .

14 答案 B

两条长度为 \(38\) 的弦到圆心的距离相等。它们不能位于两条最外侧的直线上：否则圆心会在中间直线上，该线所截的弦会是直径，长度大于 \(38\)，而不是给定的 \(34\)。因此两条相等弦位于相邻直线上，圆心在两线中间。设这个半间距为 \(d\)。则每条长 \(38\) 的弦到圆心的距离是 \(d\)，长 \(34\) 的弦到圆心的距离是 \(3d\)。

若圆的半径为 \(r\)，则

\[r^2=19^2+d^2=17^2+(3d)^2\text{。}\]

因此 \(19^2-17^2=8d^2\)，所以 \(72=8d^2\)，得到 \(d=3\)。相邻平行线的距离是 \(2d=6\)。

所以答案是 B。

English

The two chords of length \(38\) are equally far from the center of the circle. They cannot lie on the two outer lines: then the center would lie on the middle line, whose chord would be a diameter longer than \(38,\) not the given \(34.\) Therefore, the equal chords lie on adjacent lines, with the center halfway between them. Let that half-distance be \(d.\) Then each \(38\)-chord is distance \(d\) from the center, and the \(34\)-chord is distance \(3d\) from the center.

If the circle has radius \(r,\) then

\[r^2=19^2+d^2=17^2+(3d)^2.\]

Thus \(19^2-17^2=8d^2,\) so \(72=8d^2,\) and \(d=3.\) The distance between adjacent parallel lines is \(2d=6.\)

Thus, the answer is B .

15 答案 B

由 \(x+\frac{1}{x} = \sqrt{5}\)，两边平方得 \[x^2 + 2 + \frac 1{x^2} = 5\] \[x^2 + \frac 1{x^2} = 3\text{。}\] 再平方一次，得 \[x^4 + 2 + \frac 1{x^4} = 9 \]\[x^4 -7 + \frac 1{x^4} = 0\text{。}\] 两边乘以 \(x^7\)，得到 \[x^{11}-7x^{7}+x^3=0\text{。}\]

所以答案是 B。

English

Since \(x+\frac{1}{x} = \sqrt{5},\) squaring yields \[x^2 + 2 + \frac 1{x^2} = 5\] \[x^2 + \frac 1{x^2} = 3.\] Squaring again yields \[x^4 + 2 + \frac 1{x^4} = 9 \]\[x^4 -7 + \frac 1{x^4} = 0.\] Multiplying by \(x^7\) yields \[x^{11}-7x^{7}+x^3=0.\]

Thus, the answer is B .

16 答案 C

能被 \(15\) 整除的数，个位必须是 \(0\) 或 \(5\)。若上坡整数的个位是 \(0\)，则这个数只能是 \(0\)，不是正整数。因此只需考虑个位为 \(5\) 的数。

接着要求这个上坡整数是 \(3\) 的倍数。其余数字只能从 \(\{1,2,3,4\}\) 中选，所选集合的数字和除以 \(3\) 必须余 \(1\)。数字 \(3\) 取或不取不影响余数，所以先在不含 \(3\) 的情况下计数再乘以 \(2\)。满足条件的子集只有 \(3\) 个，即 \(\{1\}, \{4\}\) 和 \(\{1,2,4\}\)。因此总共有 \(6\) 个子集。

所以正确答案是 C。

English

If a number is divisible by \(15,\) it has a units digit of \(0\) or \(5.\) If the units digit is \(0\) and the digits are strictly increasing, then the number is \(0,\) which isn’t positive. Therefore, we can just look at numbers with a units digit of \(5.\)

Next, we need to find uphill integers that are a multiple of \(3.\) This means the other digits are a subset of \(\{1,2,3,4\}.\) Taking the sum of the set must have a remainder of \(1\) when divided by \(3.\) Also, having or taking out \(3\) wouldn’t affect the remainder, so we can take the number of subsets without a \(3\) and multiply it by \(2.\) There are only \(3\) such subsets, namely \(\{1\}, \{4\}, \) and \(\{1,2,4\}.\) Thus, there are \(6\) total subsets.

Thus, the correct answer is C .

17 答案 C

Oscar 的得分为 \(4\)，所以他的牌必须是 \(1\) 和 \(3\)。在剩余牌中，和为 \(7\) 的唯一一对是 \(2\) 和 \(5\)，所以它们是 Aditi 的牌。

未使用的牌是 \(4,6,7,8,9,10\)。Tyrone 的得分 \(16\) 只能来自 \(6+10\) 或 \(7+9\)。若 Tyrone 拿到 \(7\) 和 \(9\)，剩余的 \(4,6,8,10\) 无法配成 Ravon 的得分 \(11\)。因此 Tyrone 拿到 \(6\) 和 \(10\)。

剩余的牌是 \(4,7,8,9\)。Ravon 必须拿 \(4\) 和 \(7\)，得分才为 \(11\)，留下 \(8\) 和 \(9\) 给 Kim，得分为 \(17\)。所以 Ravon 拿到了 \(4\) 号牌。

所以答案是 C。

English

Oscar’s score of \(4\) forces his cards to be \(1\) and \(3.\) From the remaining cards, the only pair with sum \(7\) is \(2\) and \(5,\) so those are Aditi’s cards.

The unused cards are now \(4,6,7,8,9,10.\) Tyrone’s score of \(16\) can come only from \(6+10\) or \(7+9.\) If Tyrone had \(7\) and \(9,\) then the remaining cards \(4,6,8,10\) could not be paired to give Ravon’s score \(11.\) Thus Tyrone has \(6\) and \(10.\)

The remaining cards are \(4,7,8,9.\) Ravon must have \(4\) and \(7\) to score \(11,\) leaving \(8\) and \(9\) for Kim’s score of \(17.\) Therefore, Ravon was given card \(4.\)

Thus, the answer is C .

18 答案 C

第一个不同的点数是偶数的概率为 \(\frac 36\)。

在已经出现一个偶数后，第二个不同点数仍为偶数的概率为 \(\frac 25\)。

在已经出现两个不同偶数后，第三个不同点数仍为偶数的概率为 \(\frac 14\)。

因此所求概率为 \[\dfrac{3\cdot 2\cdot 1}{6\cdot 5\cdot 4} = \dfrac 1{20}\text{。}\]

所以答案是 C。

English

The probability that the first number is even is \(\frac 36.\)

The probability that the second distinct number is even is \(\frac 25.\)

The probability that the third distinct number is even is \(\frac 14.\)

The combined probability is \[\dfrac{3\cdot 2\cdot 1}{6\cdot 5\cdot 4} = \dfrac 1{20}.\]

Thus, the answer is C .

19 答案 D

设所有整数之和为 \(s\)，最大数为 \(g\)，最小数为 \(l\)，集合 \(S\) 的大小为 \(n\)。

两个含 \(n-1\) 个元素的集合的平均数给出 \[ \begin{aligned} s-l&=40(n-1),\\ s-g&=32(n-1) \end{aligned}\text{。}\] 两式相减并利用 \(g-l=72\)，得 \(72=8(n-1)\)，所以 \(n=10\)。

同时去掉最大数和最小数后，还剩 \(8\) 个数，其和为 \(8\cdot35=280\)。另一方面，只去掉最小数时剩下的和为 \(9\cdot40=360\)，所以 \(g=360-280=80\)。于是 \(l=80-72=8\)，原来的总和为 \(s=280+80+8=368\)。

平均值为 \(\frac{s}{n}=\frac{368}{10}=36.8\)。

所以答案是 D。

English

Let the sum of all the integers be \(s,\) the greatest number be \(g,\) the least number be \(l,\) and the size of \(S\) be \(n.\)

The two averages of sets with \(n-1\) elements give \[ \begin{aligned} s-l&=40(n-1),\\ s-g&=32(n-1). \end{aligned} \] Subtracting and using \(g-l=72\) gives \(72=8(n-1),\) so \(n=10.\)

Removing both extremes leaves \(8\) numbers with sum \(8\cdot35=280.\) On the other hand, removing only the least number leaves a sum of \(9\cdot40=360,\) so \(g=360-280=80.\) Hence \(l=80-72=8,\) and the original sum is \(s=280+80+8=368.\)

The required average is \(\frac{s}{n}=\frac{368}{10}=36.8.\)

Thus, the answer is D .

20 答案 D

设 \(F\) 为与 \(A,B\) 和 \(C\) 都相连但未标字母的点。因为所有画出的线段长度都是 \(2\)，三角形 \(ABF\) 和 \(CBF\) 是位于 \(BF\) 两侧的等边三角形。因此 \(\angle ABC=120^\circ\)，所以

\[[ABC]=\frac12\cdot2\cdot2\sin120^\circ=\sqrt3\text{。}\]

另一侧同理可得 \([ADE]=\sqrt3\)。此外，在 \(\triangle ABC\) 中应用余弦定理，得 \(AC^2=12\)，同理 \(AD^2=12\)。因此等腰三角形 \(ACD\) 到底边 \(CD=2\) 的高为

\[\sqrt{(\sqrt{12})^2-1^2}=\sqrt{11}\text{。}\]

所以 \([ACD]=\frac12\cdot2\cdot\sqrt{11}=\sqrt{11}\)。五边形的总面积为

\[\sqrt{12}+\sqrt{11}\text{，}\]

所以 \(m+n=12+11=23\)。

所以答案是 D。

English

Let \(F\) be the unlabeled point joined to \(A,B,\) and \(C.\) Because all the drawn segments have length \(2,\) triangles \(ABF\) and \(CBF\) are equilateral and lie on opposite sides of \(BF.\) Hence \(\angle ABC=120^\circ,\) so

\[[ABC]=\frac12\cdot2\cdot2\sin120^\circ=\sqrt3.\]

The same reasoning on the other side gives \([ADE]=\sqrt3.\) Also, the Law of Cosines in \(\triangle ABC\) gives \(AC^2=12,\) and similarly \(AD^2=12.\) Thus the altitude of isosceles triangle \(ACD\) to its base \(CD=2\) is

\[\sqrt{(\sqrt{12})^2-1^2}=\sqrt{11}.\]

Therefore \([ACD]=\frac12\cdot2\cdot\sqrt{11}=\sqrt{11}.\) The pentagon’s total area is

\[\sqrt{12}+\sqrt{11},\]

so \(m+n=12+11=23.\)

Thus, the answer is D .

21 答案 A

取坐标 \(A=(0,1)\)、\(B=(0,0)\)、\(C=(1,0)\)、\(D=(1,1)\)。因为 \(C'D=\frac13\)，所以 \(C'=(\frac23,1)\)，从而 \(AC'=\frac23\)。

折叠把 \(C\) 反射到 \(C'\)，所以边 \(BC\) 的像是过 \(C'\) 和 \(E\) 的直线。把 \(B=(0,0)\) 关于 \(CC'\) 的垂直平分线反射，得到 \((-\frac{2}{15},\frac25)\)。过该点与 \(C'\) 的直线与 \(AB\) 相交于 \(E=(0,\frac12)\)。

因此 \(AE=\frac12\)，并且

\[EC'=\sqrt{\left(\frac23\right)^2+\left(\frac12\right)^2}=\frac56\text{。}\]

\(\triangle AEC'\) 的周长为

\[\frac12+\frac23+\frac56=2\text{。}\]

所以答案是 A。

English

Use coordinates with \(A=(0,1),\) \(B=(0,0),\) \(C=(1,0),\) and \(D=(1,1).\) Since \(C'D=\frac13,\) we have \(C'=(\frac23,1),\) so \(AC'=\frac23.\)

The fold reflects \(C\) to \(C',\) so the image of side \(BC\) is the line through \(C'\) and \(E.\) Reflecting \(B=(0,0)\) across the perpendicular bisector of \(CC'\) gives \((-\frac{2}{15},\frac25).\) The line through this point and \(C'\) meets \(AB\) at \(E=(0,\frac12).\)

Thus \(AE=\frac12,\) and

\[EC'=\sqrt{\left(\frac23\right)^2+\left(\frac12\right)^2}=\frac56.\]

The perimeter of \(\triangle AEC'\) is

\[\frac12+\frac23+\frac56=2.\]

Thus, the answer is A .

22 答案 D

固定 Ang 的摆放方式，并用 Ang 放入的颜色标记每个盒子。Ben 和 Jasmin 各自选择五种颜色的一个排列，因此共有 \((5!)^2\) 个等可能的摆放对。

若指定 \(k\) 个盒子都收到三块同色积木，则 Ben 和 Jasmin 在这 \(k\) 个盒子中都必须与 Ang 的颜色相同。这可以用 \(((5-k)!)^2\) 种方式完成。由容斥原理，成功的摆放对数为

\[ \begin{aligned} &\binom51(4!)^2-\binom52(3!)^2 \\ &\quad {}+\binom53(2!)^2-\binom54(1!)^2 \\ &\quad {}+\binom55(0!)^2 \end{aligned}\text{。}\]

上式等于

\[2880-360+40-5+1=2556\text{。}\]

这等于

\[\frac{2556}{(5!)^2}=\frac{2556}{14400}=\frac{71}{400}\text{。}\]

因此 \(m+n=71+400=471\)。

所以答案是 D。

English

Fix Ang’s placement and label each box by the color Ang put in it. Ben and Jasmin each choose a permutation of the five colors, so there are \((5!)^2\) equally likely pairs of placements.

For a specified set of \(k\) boxes to receive three blocks of the same color, both Ben and Jasmin must match Ang in those \(k\) boxes. This can happen in \(((5-k)!)^2\) ways. By inclusion-exclusion, the number of successful placement pairs is

\[ \begin{aligned} &\binom51(4!)^2-\binom52(3!)^2 \\ &\quad {}+\binom53(2!)^2-\binom54(1!)^2 \\ &\quad {}+\binom55(0!)^2. \end{aligned} \]

This equals

\[2880-360+40-5+1=2556.\]

Therefore the probability is

\[\frac{2556}{(5!)^2}=\frac{2556}{14400}=\frac{71}{400}.\]

Thus \(m+n=71+400=471.\)

Thus, the answer is D .

23 答案 C

硬币半径为 \(\frac12\)，所以它的圆心均匀分布在一个 \(7\times7\) 的正方形内，该区域面积为 \(49\)。

对每个角上的阴影三角形，能覆盖到它的圆心位置是在允许区域内距离该三角形不超过 \(\frac12\) 的点集。每个角对应一个高为 \(\frac{1+\sqrt2}{2}\) 的等腰直角三角形，所以面积为

\[\left(\frac{1+\sqrt2}{2}\right)^2=\frac{3+2\sqrt2}{4}\text{。}\]

四个角合计贡献 \(3+2\sqrt2\)。

中心阴影菱形是一个边长为 \(2\sqrt2\) 的正方形。把它向外扩张距离 \(\frac12\)，除了菱形本身面积 \(8\)，还增加总面积 \(4\sqrt2\) 的四个矩形，以及合起来面积为 \(\frac\pi4\) 的四个四分之一圆。因此中心部分贡献

\[8+4\sqrt2+\frac\pi4\text{。}\]

有利面积为

\[ \begin{aligned} &3+2\sqrt2+8+4\sqrt2+\frac\pi4 \\ &=11+6\sqrt2+\frac\pi4 \end{aligned}\text{。}\]

概率为

\[ \begin{aligned} &\frac{11+6\sqrt2+\frac\pi4}{49} \\ &=\frac{44+24\sqrt2+\pi}{196} \end{aligned}\text{。}\]

所以 \(a+b=44+24=68\)。

所以答案是 C。

English

The coin has radius \(\frac12,\) so its center is uniformly distributed over a \(7\times7\) square of area \(49.\)

A shaded corner triangle contributes the set of center positions within distance \(\frac12\) of that triangle, inside the allowed center square. For each corner this is a right isosceles triangle whose altitude is \(\frac{1+\sqrt2}{2},\) so its area is

\[\left(\frac{1+\sqrt2}{2}\right)^2=\frac{3+2\sqrt2}{4}.\]

All four corners contribute \(3+2\sqrt2.\)

The center shaded diamond is a square of side \(2\sqrt2.\) Expanding it by distance \(\frac12\) adds four rectangles of total area \(4\sqrt2\) and four quarter-circles of total area \(\frac\pi4,\) in addition to the diamond’s area \(8.\) Thus the center contribution is

\[8+4\sqrt2+\frac\pi4.\]

The favorable area is

\[ \begin{aligned} &3+2\sqrt2+8+4\sqrt2+\frac\pi4 \\ &=11+6\sqrt2+\frac\pi4. \end{aligned} \]

The probability is

\[ \begin{aligned} &\frac{11+6\sqrt2+\frac\pi4}{49} \\ &=\frac{44+24\sqrt2+\pi}{196}. \end{aligned} \]

So \(a+b=44+24=68.\)

Thus, the answer is C .

24 答案 B

对长度为 \(n\) 的单独一堵墙，根据所有可能操作计算它的 Sprague-Grundy 值。这里用到的各个墙长的值为

\[ \begin{aligned} &g(1)=1,\quad g(2)=2, \\ &g(3)=3,\quad g(4)=1, \\ &g(5)=4,\quad g(6)=3 \end{aligned}\text{。}\]

多堵墙的局面在这些值按位异或为 \(0\) 时，正好是轮到行动者的必败局面。逐一计算选项：

\[(6,1,1): 3\oplus1\oplus1=3\text{，}\]

\[(6,2,1): 3\oplus2\oplus1=0\text{，}\]

\[(6,2,2): 3\oplus2\oplus2=3\text{，}\]

\[(6,3,1): 3\oplus3\oplus1=1\text{，}\]

\[(6,3,2): 3\oplus3\oplus2=2\text{。}\]

只有 \((6,2,1)\) 是轮到行动者的必败局面，所以 Beth 恰好在这个初始配置下有必胜策略。

所以答案是 B。

English

For a single wall of length \(n,\) compute its Sprague-Grundy value from the possible moves. For the wall lengths needed here, the values are

\[ \begin{aligned} &g(1)=1,\quad g(2)=2, \\ &g(3)=3,\quad g(4)=1, \\ &g(5)=4,\quad g(6)=3. \end{aligned} \]

For several walls, the position is losing for the player to move exactly when the xor of the wall values is \(0.\) Evaluating the choices gives

\[(6,1,1): 3\oplus1\oplus1=3,\]

\[(6,2,1): 3\oplus2\oplus1=0,\]

\[(6,2,2): 3\oplus2\oplus2=3,\]

\[(6,3,1): 3\oplus3\oplus1=1,\]

\[(6,3,2): 3\oplus3\oplus2=2.\]

Only \((6,2,1)\) is losing for the player to move, so Beth has a guaranteed win exactly for that starting configuration.

Thus, the answer is B .

25 答案 E

对固定斜率 \(m\)，集合 \(S\) 中位于 \(y=mx\) 上或下方的点数为

\[\sum_{x=1}^{30}\lfloor mx\rfloor\text{，}\]

这适用于答案附近的斜率。

当 \(m=\frac23\) 时，把 \(x=3k+1,3k+2,3k+3\) 分组，其中 \(k=0,1,\ldots,9\)。于是

\[\lfloor \frac{2x}{3}\rfloor=2k,\ 2k+1, \ 2k+2\text{，}\]

每个区块上的和为 \(6k+3\)。因此总数为

\[\sum_{k=0}^9(6k+3)=270+30=300\text{。}\]

若 \(m<\frac23\)，十个比值 \(\frac{y}{x}=\frac{2}{3}\) 的点不再被计入，所以点数小于 \(300\)。因此区间下端为 \(\frac23\)。

下一个大于 \(\frac23\) 的可能比值 \(\frac{y}{x}\)，在 \(1\le x,y\le30\) 下可按 \(x\) 模 \(3\) 检查。最佳候选为

\[\frac{19}{28},\qquad \frac{20}{29},\qquad \frac{21}{30}=\frac{7}{10}\text{，}\]

其中最小的是 \(\frac{19}{28}\)。因此区间长度为

\[\frac{19}{28}-\frac23=\frac1{84}\text{。}\]

所以 \(a+b=1+84=85\)。

所以答案是 E。

English

For a fixed slope \(m,\) the number of points in \(S\) on or below \(y=mx\) is

\[\sum_{x=1}^{30}\lfloor mx\rfloor,\]

for the slopes near the answer.

At \(m=\frac23,\) grouping \(x=3k+1,3k+2,3k+3\) for \(k=0,1,\ldots,9\) gives

\[\lfloor \frac{2x}{3}\rfloor=2k,\ 2k+1, \ 2k+2,\]

whose sum over each block is \(6k+3.\) Thus the total is

\[\sum_{k=0}^9(6k+3)=270+30=300.\]

If \(m<\frac23,\) the ten points with ratios \(\frac{y}{x}=\frac{2}{3}\) are no longer counted, so the count is less than \(300.\) Therefore the lower end is \(\frac23.\)

The next possible ratio \(\frac{y}{x}\) greater than \(\frac23,\) with \(1\le x,y\le30,\) is minimized by checking \(x\) modulo \(3.\) The best candidates are

\[\frac{19}{28},\qquad \frac{20}{29},\qquad \frac{21}{30}=\frac{7}{10},\]

and the smallest is \(\frac{19}{28}.\) Hence the interval length is

\[\frac{19}{28}-\frac23=\frac1{84}.\]

Thus \(a+b=1+84=85.\)

Thus, the answer is E .