2021A AMC 10 真题（P版）

1 答案 D

\[ \begin{aligned} (2^2 - 2) - &(3^2 - 3) + (4^2 - 4) \\ &= 2 - 6 + 12 \\ &= 8 \end{aligned}\text{。}\]

所以正确答案是 D。

English

\[ \begin{aligned} (2^2 - 2) - &(3^2 - 3) + (4^2 - 4) \\ &= 2 - 6 + 12 \\ &= 8. \end{aligned} \]

Thus, D is the correct answer.

2 答案 C

设 Lara 的学校有 \(x\) 名学生，则 Portia 的学校有 \(3x\) 名学生。

因此 \[ \begin{aligned} 3x + x &= 2600 \\ x &= 650 \end{aligned}\text{。}\] 所以 \(3x = 1950\)。

所以正确答案是 C。

English

Let \(x\) be the number of students in Lara’s high school. Then Portia’s high school has \(3x\) students.

Therefore, \[ \begin{aligned} 3x + x &= 2600 \\ x &= 650. \end{aligned} \] Then \(3x = 1950.\)

Thus, C is the correct answer.

3 答案 D

设两个数为 \(x\) 和 \(y\)。不妨设 \(x\) 能被 \(10\) 整除，那么 \(x\) 的个位数字是 \(0\)。

擦去 \(x\) 的个位数字，相当于把它除以 \(10\)。题意还给出 \(\dfrac{x}{10} = y\)。

由两数之和可得 \[ \begin{aligned} \dfrac{x}{10} + x &= 17,402 \\ \dfrac{11x}{10} &= 17,402 \\ x &= 15,820 \end{aligned}\text{。}\] 因此两数之差为 \[ x - \dfrac{x}{10} = 14,238\text{。}\]

所以正确答案是 D。

English

Let \(x\) and \(y\) be the two numbers. WLOG, let \(x\) be divisible by \(10.\) Then the units digit of \(x\) is \(0.\)

If we erase the units digit, then we are essentially dividing \(x\) by \(10.\) The problem statement also gives us that \(\dfrac{x}{10} = y.\)

Therefore, \[ \begin{aligned} \dfrac{x}{10} + x &= 17,402 \\ \dfrac{11x}{10} &= 17,402 \\ x &= 15,820. \end{aligned} \] Then \[ x - \dfrac{x}{10} = 14,238. \]

Thus, D is the correct answer.

4 答案 D

每秒行驶的距离构成等差数列：\[ 5, 5 + 7, 5 + 2 \cdot 7, \ldots \]

等差数列求和的标准公式为 \[ n\cdot\dfrac{a_1+a_n}{2}\text{。}\] 已知项数为 \(30\)，首项为 \(5\)。末项为 \[5 + 29 \cdot 7 = 208\text{。}\]

把这些值代入公式，得到 \[ 30 \cdot \dfrac{5 + 208}{2} = 15 \cdot 213 = 3195\text{。}\]

所以正确答案是 D。

English

The distance travelled every second forms an arithmetic sequence: \[ 5, 5 + 7, 5 + 2 \cdot 7, \ldots \]

The standard arithmetic-sequence sum formula is \[ n\cdot\dfrac{a_1+a_n}{2}. \] We know the number of terms is \(30\) and the first term is \(5.\) The last term is \[5 + 29 \cdot 7 = 208.\]

Plugging these values into the expression yields \[ 30 \cdot \dfrac{5 + 208}{2} = 15 \cdot 213 = 3195. \]

Thus, D is the correct answer.

5 答案 B

全班所有人的总分为 \(8k\)。这 \(12\) 个分数的总分为 \(12 \cdot 14 = 168\)。

其余学生的总分为 \(8k - 168\)，人数为 \(k - 12\)。因此平均数为 \[ \dfrac{8k - 168}{k - 12}\text{。}\]

所以正确答案是 B。

English

The sum of the scores of everyone in the class is \(8k.\) The sum of the scores in the collection of \(12\) is \(12 \cdot 14 = 168.\)

This means that the sum of the scores of everyone not in the collection is \(8k - 168.\) There are also \(k - 12\) people not in the collection. Therefore, the average is \[ \dfrac{8k - 168}{k - 12}. \]

Thus, B is the correct answer.

6 答案 A

设到消防塔的全程距离为 \(2d\)，其中 \(d > 0\)。

于是 Chantal 徒步用了 \[ \dfrac{d}{4} + \dfrac{d}{2} + \dfrac{d}{3} = \dfrac{13d}{12} \] 小时。

Jean 在 \(\dfrac{13d}{12}\) 小时内走了 \(d\) 英里，所以他的速度为 \[ d \div \dfrac{13d}{12} = \dfrac{12}{13} \] 英里每小时。

所以正确答案是 A。

English

Let \(2d\) be the distance from the trailhead to the fire tower, where \(d > 0.\)

Then Chantal hiked for \[ \dfrac{d}{4} + \dfrac{d}{2} + \dfrac{d}{3} = \dfrac{13d}{12} \] hours.

If Jean travelled \(d\) miles in \(\dfrac{13d}{12}\) hours, then his speed was \[ d \div \dfrac{13d}{12} = \dfrac{12}{13} \] miles per hour.

Thus, A is the correct answer.

7 答案 D

紫色的蛇不会减法，而不会减法的蛇也不会加法，所以紫色的蛇不会加法。

另一方面，开心的蛇都会加法，所以开心的蛇不可能是紫色的。

所以正确答案是 D。

English

Note that the third condition ensures that purple snakes can’t add.

We also know that all happy snakes can add, which means that happy snakes can’t be purple as well.

Thus, D is the correct answer.

8 答案 E

设 \(N=10a+b\)，即由数字 \(a\) 和 \(b\) 组成的两位数。于是 \[1.\overline{ab}=1+\frac{N}{99}\] 而学生所用的有限小数是 \[1.ab=1+\frac{N}{100}\text{。}\]

正确的乘积比学生算出的乘积大 \(0.5\)，所以 \[ 66\left(\frac{N}{99}-\frac{N}{100}\right) =\frac{N}{150}=0.5\text{。}\] 因此 \(N=75\)。

所以正确答案是 E。

English

Let \(N=10a+b,\) the two-digit integer formed by the digits \(a\) and \(b.\) Then \[1.\overline{ab}=1+\frac{N}{99}\] while the terminating decimal the student used is \[1.ab=1+\frac{N}{100}.\]

The correct product exceeds the student’s product by \(0.5,\) so \[ 66\left(\frac{N}{99}-\frac{N}{100}\right) =\frac{N}{150}=0.5. \] Hence \(N=75.\)

Thus, E is the correct answer.

9 答案 D

展开得 \[ \begin{gathered} x^2y^2 - 2xy + 1 + x^2 + 2xy + y^2 \\ = x^2y^2 + x^2 + y^2 + 1 \end{gathered}\text{。}\] 每个平方项都非负，因此除 \(1\) 外的各项都取 \(0\) 时，总和取得最小值 \(1\)。

当 \(x = y = 0\) 时可以取到这个值。

所以正确答案是 D。

English

Expanding, we get \[ \begin{gathered} x^2y^2 - 2xy + 1 + x^2 + 2xy + y^2 \\ = x^2y^2 + x^2 + y^2 + 1. \end{gathered} \] Note that every square must be non-negative. Therefore, the minimum value is when all the terms except \(1\) are \(0,\) making the sum \(1.\)

This is attainable when \(x = y = 0.\)

Thus, D is the correct answer.

10 答案 C

把这个乘积乘以 \(3-2=1\)。反复应用平方差公式，得到 \[ \begin{aligned} (3-2)(3+2)&=3^2-2^2,\\ (3^2-2^2)(3^2+2^2)&=3^4-2^4 \end{aligned}\text{，}\] 同样的相消一直进行到最后一个因子。因此这个乘积等于 \(3^{128}-2^{128}\)。

所以正确答案是 C。

English

Multiply the product by \(3-2=1.\) Repeatedly applying the difference-of-squares identity gives \[ \begin{aligned} (3-2)(3+2)&=3^2-2^2,\\ (3^2-2^2)(3^2+2^2)&=3^4-2^4, \end{aligned} \] and the same cancellation continues through the final factor. Therefore the product is \(3^{128}-2^{128}.\)

Thus, C is the correct answer.

11 答案 E

按进制定义，把这个差写成 \(10\) 进制表达式：\[ 2b^3 + 2b + 1 - 2b^2 - 2b - 1 \] \[= 2b^3 - 2b^2 = 2b^2(b - 1)\text{。}\]

要使它能被 \(3\) 整除，\(b\) 或 \(b - 1\) 必须能被 \(3\) 整除。

唯一两个条件都不满足的选项是 \(8\)。

所以正确答案是 E。

English

We can express this expression in base \(10\) using the definition of bases: \[ 2b^3 + 2b + 1 - 2b^2 - 2b - 1 \]\[= 2b^3 - 2b^2 = 2b^2(b - 1). \]

For this to be divisible by \(3,\) either \(b\) or \(b - 1\) must be divisible by \(3.\)

The only answer choice that satisfies neither of these conditions is \(8.\)

Thus, E is the correct answer.

12 答案 E

设窄圆锥和宽圆锥的初始液面高度分别为 \(h_1\)、\(h_2\)。因为两者的液体体积相等，所以

\[\frac13\pi(3)^2h_1=\frac13\pi(6)^2h_2\text{，}\]

于是 \(h_1=4h_2\)。

投入相同的弹珠后，两个圆锥中液面以下的最终体积仍然相等，都等于原有液体体积加上一颗弹珠的体积。若新的液面半径分别为 \(3x\) 和 \(6y\)，由相似可得新的高度为 \(h_1x\) 和 \(h_2y\)。于是

\[\frac13\pi(3x)^2h_1x=\frac13\pi(6y)^2h_2y\text{。}\]

利用 \(h_1=4h_2\)，可化简为 \(x^3=y^3\)，所以 \(x=y\)。因此液面上升高度之比为

\[ \begin{aligned} h_1(x-1):h_2(y-1) &= h_1:h_2 \\ &= 4:1 \end{aligned}\text{。}\]

所以正确答案是 E。

English

Let the initial liquid heights in the narrow and wide cones be \(h_1\) and \(h_2.\) Since the liquid volumes are equal,

\[\frac13\pi(3)^2h_1=\frac13\pi(6)^2h_2,\]

so \(h_1=4h_2.\)

After the identical marbles are dropped in, each cone must contain the same final volume below the liquid surface: the original liquid volume plus the volume of one marble. If the new liquid-surface radii are \(3x\) and \(6y,\) similarity gives new heights \(h_1x\) and \(h_2y.\) Thus

\[\frac13\pi(3x)^2h_1x=\frac13\pi(6y)^2h_2y.\]

Using \(h_1=4h_2,\) this simplifies to \(x^3=y^3,\) so \(x=y.\) The rise ratio is therefore

\[ \begin{aligned} h_1(x-1):h_2(y-1) &= h_1:h_2 \\ &= 4:1. \end{aligned} \]

Thus, E is the correct answer.

13 答案 C

取 \(A=(0,0,0)\)、\(B=(2,0,0)\)、\(C=(0,3,0)\) 和 \(D=(0,0,4)\)。则

\[ \begin{aligned} BC &= \sqrt{2^2+3^2}=\sqrt{13}, \\ BD &= \sqrt{2^2+4^2}=2\sqrt5, \\ CD &= \sqrt{3^2+4^2}=5 \end{aligned}\text{，}\]

所以这个坐标模型与题目给出的所有棱长相符。三条从 \(A\) 出发的棱互相垂直，长度分别为 \(2,3,4\)，因此体积为

\[\frac16(2)(3)(4)=4\text{。}\]

所以正确答案是 C。

English

Place \(A=(0,0,0),\) \(B=(2,0,0),\) \(C=(0,3,0),\) and \(D=(0,0,4).\) Then

\[ \begin{aligned} BC &= \sqrt{2^2+3^2}=\sqrt{13}, \\ BD &= \sqrt{2^2+4^2}=2\sqrt5, \\ CD &= \sqrt{3^2+4^2}=5, \end{aligned} \]

so this coordinate model matches all the given edge lengths. The tetrahedron is a rectangular-corner tetrahedron with perpendicular edge lengths \(2,3,4\) from \(A,\) so its volume is

\[\frac16(2)(3)(4)=4.\]

Thus, C is the correct answer.

14 答案 A

由 Vieta 公式，六个根的和为 \(10\)，积为 \(16\)。因为积是 \(2\) 的幂，所以每个正整数根都是 \(2\) 的幂。把四个因子 \(2\) 分配给六个根，当其中四个根为 \(2\)、两个根为 \(1\) 时和最小，而这个和恰好已经等于 \(10\)。因此这些根是 \[1,1,2,2,2,2\text{。}\]

系数 \(B\) 是所有三个根乘积之和的相反数。按取零个、一个还是两个等于 \(1\) 的根来分类，得 \[ \begin{aligned} B&=-\left(\binom43 2^3+2\binom42 2^2\right.\\ &\qquad\left.+\binom41 2\right)\\ &=-88 \end{aligned}\text{。}\]

所以正确答案是 A。

English

By Vieta’s formulas, the six roots have sum \(10\) and product \(16.\) Because the product is a power of \(2,\) every positive integer root is a power of \(2.\) Distributing the four factors of \(2\) among six roots gives the least possible sum when four roots are \(2\) and two roots are \(1\); that sum is already \(10.\) Hence the roots are \[1,1,2,2,2,2.\]

The coefficient \(B\) is the negative of the sum of all products of three roots. Choosing zero, one, or two of the two roots equal to \(1\) gives \[ \begin{aligned} B&=-\left(\binom43 2^3+2\binom42 2^2\right.\\ &\qquad\left.+\binom41 2\right)\\ &=-88. \end{aligned} \]

Thus, A is the correct answer.

15 答案 C

令两式相等，得到 \[ Ax^2 + B = Cx^2 + D \] \[ x^2(A - C) = D - B \] \[ x^2 = \dfrac{D - B}{A - C} \geq 0\text{，}\]因为平方数非负。

因此 \(D - B\) 与 \(A - C\) 必须同号。

若先为 \((A, C)\) 和 \((B, D)\) 各选两个不同值，则有 \(2\) 种安排使两个差同号。

不过还要除以 \(2\)，因为两条曲线列出的顺序不区分。

所以所求选择数为 \[ \dfrac{1}{2} \binom{6}{2} \binom{4}{2} \cdot 2 = 90\text{。} \]

所以正确答案是 C。

English

Setting the equations equal to each other, we get \[ Ax^2 + B = Cx^2 + D \] \[ x^2(A - C) = D - B \] \[ x^2 = \dfrac{D - B}{A - C} \geq 0 \] since squares are non-negative.

This means \(D - B\) and \(A - C\) must both have the same sign.

If we choose two distinct values for \((A, C)\) and \((B, D),\) there are \(2\) ways to arrange them such that the numerator and denominator both have the same sign.

We have to divide by \(2,\) however, since the two curves are not considered distinct.

Therefore, the total number of tuples is \[ \dfrac{1}{2} \binom{6}{2} \binom{4}{2} \cdot 2 = 90. \]

Thus, C is the correct answer.

16 答案 C

这个数列共有 \[ \begin{aligned} 1+2+\cdots+200&=\frac{200\cdot201}{2}\\ &=20100 \end{aligned} \] 项，所以处于中间的两个位置是第 \(10050\) 项和第 \(10051\) 项。

到最后一个 \(141\) 为止共有 \[\frac{141\cdot142}{2}=10011\] 项，到最后一个 \(142\) 为止共有 \[\frac{142\cdot143}{2}=10153\] 项。因此中间的两项都等于 \(142\)，所以中位数是 \(142\)。

所以正确答案是 C。

English

The list contains \[ \begin{aligned} 1+2+\cdots+200&=\frac{200\cdot201}{2}\\ &=20100 \end{aligned} \] entries, so its two middle positions are \(10050\) and \(10051.\)

There are \[\frac{141\cdot142}{2}=10011\] entries through the last \(141,\) and \[\frac{142\cdot143}{2}=10153\] entries through the last \(142.\) Thus both middle entries are \(142,\) so the median is \(142.\)

Thus, C is the correct answer.

17 答案 D

因为 \(BC=CD\)，从 \(C\) 到 \(BD\) 的中线垂直于 \(BD\)，所以 \(\triangle BPC\) 是直角三角形。设 \(\angle DBC=\alpha\)。又因 \(AB\parallel CD\)，有 \(\angle ABD=\alpha\)，因此 \(\triangle BPC\sim\triangle BDA\)。

由于 \(P\) 是 \(BD\) 的中点，\(\frac{BD}{BP}=2\)。在相似关系中，\(BC\) 对应 \(AB\)，所以

\[ \begin{aligned} \frac{AB}{BC} &=2, \\ AB &=2\cdot43=86 \end{aligned}\text{。}\]

另外，\(\triangle ABO\sim\triangle CDO\)，所以

\[\frac{BO}{OD}=\frac{AB}{CD}=2\text{。}\]

设 \(BP=PD=t\)。由于 \(OP=11\)，且 \(P\) 是 \(BD\) 的中点，得到 \(BO=t+11\)、\(OD=t-11\)，所以

\[\frac{t+11}{t-11}=2\text{。}\]

解得 \(t=33\)，故 \(BD=66\)。最后，\(\triangle ABD\) 是直角三角形，所以

\[ \begin{aligned} AD &= \sqrt{AB^2-BD^2} \\ &= \sqrt{86^2-66^2} \\ &= 4\sqrt{190} \end{aligned}\text{。}\]

因此 \(m+n=4+190=194\)。

所以正确答案是 D。

English

Because \(BC=CD,\) the median from \(C\) to \(BD\) is perpendicular to \(BD.\) Thus \(\triangle BPC\) is a right triangle. Let \(\angle DBC=\alpha.\) Since \(AB\parallel CD,\) we also have \(\angle ABD=\alpha,\) so \(\triangle BPC\sim\triangle BDA.\)

Since \(P\) is the midpoint of \(BD,\) we have \(\frac{BD}{BP}=2.\) In the similarity, \(BC\) corresponds to \(AB,\) so

\[ \begin{aligned} \frac{AB}{BC} &=2, \\ AB &=2\cdot43=86. \end{aligned} \]

Also, \(\triangle ABO\sim\triangle CDO,\) so

\[\frac{BO}{OD}=\frac{AB}{CD}=2.\]

Since \(OP=11\) and \(P\) is the midpoint of \(BD,\) write \(BP=PD=t.\) Then \(BO=t+11\) and \(OD=t-11,\) so

\[\frac{t+11}{t-11}=2.\]

This gives \(t=33,\) hence \(BD=66.\) Finally, \(\triangle ABD\) is right, so

\[ \begin{aligned} AD &= \sqrt{AB^2-BD^2} \\ &= \sqrt{86^2-66^2} \\ &= 4\sqrt{190}. \end{aligned} \]

Thus \(m+n=4+190=194.\)

Thus, D is the correct answer.

18 答案 E

反复应用该函数方程可得：对每个质数 \(p\) 和正整数 \(e\)，都有 \(f(p^e)=ef(p)=ep\)。另外 \[f(a)=f\left(\frac ab\right)+f(b)\text{，}\] 所以 \(f(\frac{a}{b})=f(a)-f(b)\)。

用质因数分解逐项计算各个选项，得 \[ \begin{aligned} f(\frac{17}{32})&=17-5\cdot2=7,\\ f(\frac{11}{16})&=11-4\cdot2=3,\\ f(\frac{7}{9})&=7-2\cdot3=1,\\ f(\frac{7}{6})&=7-2-3=2,\\ f(\frac{25}{11})&=2\cdot5-11=-1 \end{aligned}\text{。}\] 只有最后一个值是负数。

所以正确答案是 E。

English

Repeated use of the functional equation gives \(f(p^e)=ef(p)=ep\) for every prime \(p\) and positive integer \(e.\) Also, \[f(a)=f\left(\frac ab\right)+f(b),\] so \(f(\frac{a}{b})=f(a)-f(b).\)

Evaluating the choices by prime factorization, \[ \begin{aligned} f(\frac{17}{32})&=17-5\cdot2=7,\\ f(\frac{11}{16})&=11-4\cdot2=3,\\ f(\frac{7}{9})&=7-2\cdot3=1,\\ f(\frac{7}{6})&=7-2-3=2,\\ f(\frac{25}{11})&=2\cdot5-11=-1. \end{aligned} \] Only the final value is negative.

Thus, E is the correct answer.

19 答案 E

按 \(x-y\) 和 \(x+y\) 的符号分类。例如在一种情况中，\(|x-y|=x-y\) 且 \(|x+y|=x+y\)，所以

\[ \begin{gathered} x^2+y^2=6x \\ \quad\Longrightarrow\quad (x-3)^2+y^2=9 \end{gathered}\text{。}\]

其余三种情况类似，得到半径为 \(3\)、圆心分别在 \((0,3)\)、\((-3,0)\) 和 \((0,-3)\) 的圆。相应的四段全等圆弧就构成了图像的边界。

围成区域由边长 \(6\) 的中心正方形和四个半径为 \(3\) 的半圆组成。正方形面积为 \(36\)，四个半圆的总面积等于两个半径为 \(3\) 的圆，即 \(18\pi\)。

因此面积为 \(36+18\pi\)，所以 \(m+n=36+18=54\)。

所以正确答案是 E。

English

Consider the four sign cases for \(x-y\) and \(x+y.\) In one case, for example, \(|x-y|=x-y\) and \(|x+y|=x+y,\) so

\[ \begin{gathered} x^2+y^2=6x \\ \quad\Longrightarrow\quad (x-3)^2+y^2=9. \end{gathered} \]

The other three cases similarly give circles of radius \(3\) centered at \((0,3),\) \((-3,0),\) and \((0,-3).\) The relevant arcs form the boundary shown by these four congruent circle pieces.

The region consists of a central square of side length \(6,\) together with four semicircles of radius \(3.\) The square contributes area \(36,\) and the four semicircles have the area of two full radius-\(3\) circles, namely \(18\pi.\)

Therefore the area is \(36+18\pi,\) so \(m+n=36+18=54.\)

Thus, E is the correct answer.

20 答案 D

一个排列有效，当且仅当相邻两项之间的四个比较符号交替出现。因此这些符号只能是升降升降或降升降升。

对升降升降型，最大的数 \(5\) 必须位于第 \(2\) 位或第 \(4\) 位。若它在第 \(2\) 位，设第 \(4\) 位上的数为 \(r\)。它的两个相邻数必须是小于 \(r\) 的两个不同的数，共有 \((r-1)(r-2)\) 种排法。对 \(r=1,2,3,4\) 求和，得到 \[0+0+2+6=8\] 个排列。由对称性，当 \(5\) 在第 \(4\) 位时另有 \(8\) 个，所以这种比较模式共有 \(16\) 个排列。

把每个数 \(x\) 替换为 \(6-x\)，可与降升降升型一一对应，所以另有 \(16\) 个。

总共有 \(16+16=32\) 个有效排列。

所以正确答案是 D。

English

A permutation is valid exactly when the four comparison signs between consecutive terms alternate. Thus the signs must be either up-down-up-down or down-up-down-up.

For the up-down-up-down pattern, the largest entry \(5\) must be in position \(2\) or position \(4.\) If it is in position \(2,\) let the entry in position \(4\) be \(r.\) Its two neighbors must be distinct numbers less than \(r,\) which can be ordered in \((r-1)(r-2)\) ways. Summing over \(r=1,2,3,4\) gives \[0+0+2+6=8\] permutations. By symmetry there are another \(8\) when \(5\) is in position \(4,\) for a total of \(16\) with this comparison pattern.

Replacing every entry \(x\) by \(6-x\) gives a bijection to the down-up-down-up permutations, so there are another \(16.\)

The total number of valid rearrangements is \(16+16=32.\)

Thus, D is the correct answer.

21 答案 C

设直线 \(AB,CD,EF\) 的交点形成三角形 \(PQR\)，直线 \(BC,DE,FA\) 的交点形成三角形 \(XYZ\)。因为六边形等角，这些外侧三角形都是等边三角形。

若等边三角形边长为 \(s\)，面积为 \(\frac{\sqrt3}{4}s^2\)。两个面积条件可写成：

\[ \begin{aligned} \frac{\sqrt3}{4}PQ^2 &=192\sqrt3, \\ \frac{\sqrt3}{4}YZ^2 &=324\sqrt3 \end{aligned}\text{。}\]

所以 \(PQ=16\sqrt3\)，\(YZ=36\)。为了说明这个周长关系，把六边形依次的边长记为 \(a,b,c,d,e,f\)。两条交替延长线截出的三角形的边长分别为 \(b+c+d\) 和 \(c+d+e\)，而六边形封闭又给出 \(a+f=c+d\)。因此它们的边长之和为 \[ \begin{aligned} &(b+c+d)\\ &\quad+(c+d+e)\\ &=b+e+2(c+d)\\ &=(a+b+c)+(d+e+f) \end{aligned}\text{，}\] 这正是六边形的周长。所以周长为

\[PQ+YZ=16\sqrt3+36\text{。}\]

因此 \(m+n+p=36+16+3=55\)。

所以正确答案是 C。

English

Let the intersections of lines \(AB,CD,EF\) form triangle \(PQR,\) and let the intersections of lines \(BC,DE,FA\) form triangle \(XYZ.\) Because the hexagon is equiangular, all these outer triangles are equilateral.

For an equilateral triangle with side length \(s,\) the area is \(\frac{\sqrt3}{4}s^2.\) Hence

\[ \begin{aligned} \frac{\sqrt3}{4}PQ^2 &=192\sqrt3, \\ \frac{\sqrt3}{4}YZ^2 &=324\sqrt3. \end{aligned} \]

So \(PQ=16\sqrt3\) and \(YZ=36.\) To justify the perimeter relation, write the consecutive hexagon side lengths as \(a,b,c,d,e,f.\) The two alternating-line triangles have side lengths \(b+c+d\) and \(c+d+e,\) while closure of the hexagon gives \(a+f=c+d.\) Hence their side-length sum is \[ \begin{aligned} &(b+c+d)\\ &\quad+(c+d+e)\\ &=b+e+2(c+d)\\ &=(a+b+c)+(d+e+f), \end{aligned} \] the hexagon’s perimeter. Therefore the perimeter is

\[PQ+YZ=16\sqrt3+36.\]

Thus \(m+n+p=36+16+3=55.\)

Thus, C is the correct answer.

22 答案 B

设借走的是第 \(a\) 张到第 \(b\) 张纸，并设 \(s=b-a+1\)。被借走的页码从 \(2a-1\) 到 \(2b\)，所以共有 \(2s\) 页，页码之和为 \(s(2a+2b-1)\)。

所有页码之和为 \(\frac{50\cdot51}{2}=1275\)。若剩余页码的平均数为 \(19\)，则 \[ \begin{aligned} 1275&-s(2a+2b-1)\\ &=19(50-2s),\\ s(2a+2b-39)&=325 \end{aligned}\text{。}\]

因为 \(b-a+1\) 是 \(325\) 的正因数，且不超过 \(25\)，所以它只能是 \(1,5,13,25\) 这四种可能。前两种会迫使 \(b>25\)，而 \(25\) 会借走所有纸张。因此唯一有效的可能是

\[ \begin{aligned} 2a+2b-39 &=25, \\ b-a+1 &=13 \end{aligned}\text{。}\]

所以 \(a+b=32\)，且 \(b-a=12\)，得到 \(a=10\)、\(b=22\)。因此借走了 \(13\) 张纸。

所以正确答案是 B。

English

Suppose the borrowed sheets are sheets \(a\) through \(b,\) and let \(s=b-a+1.\) The borrowed pages run from \(2a-1\) through \(2b,\) so there are \(2s\) borrowed pages and their sum is \(s(2a+2b-1).\)

The total sum of all page numbers is \(\frac{50\cdot51}{2}=1275.\) If the remaining pages have mean \(19,\) then \[ \begin{aligned} 1275&-s(2a+2b-1)\\ &=19(50-2s),\\ s(2a+2b-39)&=325. \end{aligned} \]

Because \(b-a+1\) is a positive divisor of \(325\) and is at most \(25,\) its only possibilities are \(1,5,13,25.\) The first two would force \(b>25,\) and \(25\) would remove every sheet. Thus the only valid possibility is

\[ \begin{aligned} 2a+2b-39 &=25, \\ b-a+1 &=13. \end{aligned} \]

Thus \(a+b=32\) and \(b-a=12,\) so \(a=10\) and \(b=22.\) Therefore \(13\) sheets were borrowed.

Thus, B is the correct answer.

23 答案 D

用 \(M\) 表示中心格，\(E\) 表示非角的边格，\(C\) 表示角格。Frieda 从 \(M\) 出发，第一次跳一定到 \(E\)。

从一个边格出发，到 \(C,E,M\) 的概率分别为 \(\frac12,\frac14,\frac14\)。从 \(M\) 出发下一步一定到 \(E\)。

现在列出四次以内首次到达角格的状态模式及其概率。

\[EC:\quad 1\cdot\frac12=\frac12\text{，}\]

\[EEC:\quad 1\cdot\frac14\cdot\frac12=\frac18\text{，}\]

\[EEEC:\quad 1\cdot\frac14\cdot\frac14\cdot\frac12=\frac1{32}\text{，}\]

\[EMEC:\quad 1\cdot\frac14\cdot1\cdot\frac12=\frac18\text{。}\]

把这些概率相加：

\[\frac12+\frac18+\frac1{32}+\frac18=\frac{25}{32}\text{。}\]

所以正确答案是 D。

English

Classify a square as \(M\) for the center, \(E\) for a non-corner edge square, and \(C\) for a corner. Frieda starts at \(M,\) and the first hop always takes her to an \(E.\)

From an edge square, the probabilities of moving to \(C,E,M\) are \(\frac12,\frac14,\frac14,\) respectively. From \(M,\) the next hop always goes to an \(E.\)

Now count the possible first-hit patterns within four hops:

\[EC:\quad 1\cdot\frac12=\frac12,\]

\[EEC:\quad 1\cdot\frac14\cdot\frac12=\frac18,\]

\[EEEC:\quad 1\cdot\frac14\cdot\frac14\cdot\frac12=\frac1{32},\]

\[EMEC:\quad 1\cdot\frac14\cdot1\cdot\frac12=\frac18.\]

Adding gives

\[\frac12+\frac18+\frac1{32}+\frac18=\frac{25}{32}.\]

Thus, D is the correct answer.

24 答案 D

注意，每个方程都会给出两条平行直线。

\[ (x + ay)^2 = 4a^2 \] 给出两条直线 \[ x + ay - 2a = 0 \] 和 \[ x + ay + 2a = 0\text{。} \] 这两条直线的斜率都是 \(-\dfrac{1}{a}\)。

类似地，\[ (ax-y)^2 = a^2 \] 给出直线 \[ ax - y - a = 0 \] 和 \[ ax - y + a = 0\text{。} \] 这些直线的斜率为 \(a\)。

两组直线互相垂直，因此围成一个矩形。

回忆两条平行线 \[ \begin{cases} Ax+By+C_1=0 \\ Ax+By+C_2=0 \end{cases} \] 之间的距离 \(d\) 为 \[ d = \dfrac{\mid C_2 - C_1 \mid}{\sqrt{A^2 + B^2}}\text{。} \]

用这个公式，第一组平行线之间的距离为 \[ \dfrac{4a}{\sqrt{a^2 + 1}}\text{。} \] 类似地，第二组平行线之间的距离为 \[ \dfrac{2a}{\sqrt{a^2 + 1}}\text{。} \]

这两个距离就是矩形的边长，相乘得面积 \[ \dfrac{8a^2}{a^2 + 1}\text{。} \]

所以正确答案是 D。

English

Note that each of the equations yields two parallel lines.

\[ (x + ay)^2 = 4a^2 \] results in the two lines \[ x + ay - 2a = 0 \] and \[ x + ay + 2a = 0. \] Both of these lines have a slope of \(-\dfrac{1}{a}.\)

Similarly, \[ (ax-y)^2 = a^2 \] results in the lines \[ ax - y - a = 0 \] and \[ ax - y + a = 0. \] These lines have slope \(a.\)

Note that each pair of lines is perpendicular to the other pair of lines. This shows that the equations form a rectangle.

Recall that the formula for the distance \(d\) between two parallel lines \[ \begin{cases} Ax+By+C_1=0 \\ Ax+By+C_2=0 \end{cases} \] is \[ d = \dfrac{\mid C_2 - C_1 \mid}{\sqrt{A^2 + B^2}}. \]

Using this formula, we get that the distance between the first pair of lines is \[ \dfrac{4a}{\sqrt{a^2 + 1}}. \] Similarly, the distance between the second pair of lines is \[ \dfrac{2a}{\sqrt{a^2 + 1}}. \]

These are the side lengths of the rectangle. Multiplying yields the area \[ \dfrac{8a^2}{a^2 + 1}. \]

Thus, D is the correct answer.

25 答案 E

中心格的颜色有 \(3\) 种选法。这种颜色的另外两枚筹码都不能放在边中格，因为这些格子与中心格相邻。因此它们只能占据四个角格中的两个，共有 \(\binom42=6\) 种选法。

无论这两个角格是哪一种情形——处于对角，还是处于同一边的两端——相邻条件都会把剩下两种颜色的位置唯一确定，至多相差二者的互换。因此对中心颜色及其两个角格的每一种选择，都恰有 \(2\) 种补全方式。总数为 \[3\binom42\cdot2=36\text{。}\]

所以正确答案是 E。

English

Choose the center color in \(3\) ways. Its other two chips cannot occupy any edge-middle square, because those squares are adjacent to the center. Thus they must occupy two of the four corners, which can be chosen in \(\binom42=6\) ways.

For either possible corner pattern—two opposite corners or two corners on the same side—the adjacency conditions force the remaining two colors up to interchanging them. Hence there are \(2\) completions for each choice of the center color and its two corners. The total is \[3\binom42\cdot2=36.\]

Thus, E is the correct answer.