正方形 ABCD 的边长为 4 。点 P 和 Q 分别在 \overlineAD 和 \overlineCD 上，且 AP = (8)/(5) 、 DQ = (10)/(3) 。一条路径从 P 到 Q 的线段开始，之后在正方形 ABCD 的边上反射继续前进（入射角等于反射角），如下图所示。如果路径碰到正方形的顶点，就在那里终止；否则它将永远继续。这条路径会在哪个顶点终止？

(E) 路径会永远继续下去。

2025 AMC 10B 真题与在线限时测验

计时

1:15:00

1.

一袋 $350$ 克的咖啡豆说明上写着：正确冲泡一大杯手冲咖啡需要 $20$ 克咖啡豆。用这袋咖啡豆最多可以正确冲泡多少大杯咖啡？

The instructions on a $350$ -gram bag of coffee beans say that proper brewing of a large mug of pour-over coffee requires $20$ grams of coffee beans. What is the greatest number of properly brewed large mugs of coffee that can be made from the coffee beans in that bag?

$16$

$17$

$18$

$19$

$20$

答案：B

知识点：取整函数

难度评级：860

小提示：

每杯需要 $20$ 克，所以把咖啡豆总质量除以 $20$ 。

Each mug uses $20$ grams, so divide the total mass by $20$

大提示：

不完整的一杯不能算一杯，所以要把 $\frac{350}{20}$ 向下取整。

A partial mug does not count, so round $\frac{350}{20}$ down to a whole number

解答：

每杯需要 $20$ 克，因此 $\frac{350}{20} = 17.5$ 。半杯不能算作一杯，所以向下取整，最多能冲泡 $17$ 大杯，正确答案是 B。

Each mug needs $20$ grams, so we divide: $\frac{350}{20} = 17.5.$ A half mug isn’t a mug, so round down. That leaves $17.$ Thus, B is the correct answer.

2.

Jerry 写下前 $2025$ 个正平方数的个位数字： $1$ ， $4$ ， $9$ ， $6$ ， $5$ ， $6$ ， $\ldots$ 。他写下的所有数字之和是多少？

Jerry wrote down the ones digit of each of the first $2025$ positive squares: $1,$ $4,$ $9,$ $6,$ $5,$ $6,$ $\ldots$ What is the sum of all the numbers Jerry wrote down?

$9025$

$9070$

$9090$

$9115$

$9160$

答案：D

知识点：个位数字完全平方数找规律

难度评级：990

小提示：

$n^2$ 的个位数字只由 $n$ 的个位数字决定，所以这个列表每 $10$ 项循环一次。

The ones digit of $n^2$ depends only on the ones digit of $n,$ so the list repeats every $10$ terms

大提示：

一个完整循环 $1, 4, 9, 6, 5, 6, 9, 4, 1, 0$ 的和是 $45$ ；共有 $202$ 个完整循环，还剩 $5$ 项。

One full block $1, 4, 9, 6, 5, 6, 9, 4, 1, 0$ sums to $45;$ there are $202$ full blocks and $5$ leftover terms

解答：

$n^2$ 的个位数字只取决于 $n$ 的个位数字，因此序列每 $10$ 项循环： $1, 4, 9, 6, 5, 6, 9, 4, 1, 0$ 。一个循环的和为 $45$ 。又 $2025 = 202 \cdot 10 + 5$ ，所以共有 $202$ 个完整循环，再加上前五项，总和为 $202 \cdot 45$ $+ (1 + 4 + 9 + 6 + 5)$ $= 9090$ $+ 25 = 9115$ 。因此正确答案是 D。

The ones digit of $n^2$ depends only on the ones digit of $n,$ so the list repeats every $10$ terms: $1, 4, 9, 6, 5, 6, 9, 4, 1, 0.$ One block sums to $45.$ Now $2025 = 202 \cdot 10 + 5,$ so we get $202$ full blocks plus the first five terms: $202 \cdot 45$ $+ (1 + 4 + 9 + 6 + 5)$ $= 9090$ $+ 25 = 9115.$ Therefore, the answer is D.

3.

一个类似帕斯卡三角形的三角形，第一行为 $10$ ，第二行为 $10$ 后接 $1$ 。在之后的每一行中，第一个数是 $10$ ，最后一个数是 $1$ ，并且和标准帕斯卡三角形一样，行中每个其他数都是它正上方两个数之和。前四行如下图所示。

第 $11$ 行所有数之和的各位数字之和是多少？

A Pascal-like triangle has $10$ as the top row and $10$ followed by $1$ as the second row. In each subsequent row the first number is $10,$ the last number is $1,$ and, as in the standard Pascal’s Triangle, each other number in the row is the sum of the two numbers directly above it. The first four rows are shown below.

What is the sum of the digits of the sum of the numbers in the $11$ th row?

$11$

$13$

$14$

$16$

$17$

答案：D

知识点：杨辉三角递推等比数列

难度评级：1270

小提示：

只追踪每一行的和。每个内部数都会被它下方的两个数各计入一次。

Track only the row sums. Each interior entry is counted by both of the two entries below it

大提示：

证明 $S_n = 2 S_{n-1}$ （当 $n \ge 3$ ），且 $S_2 = 11$ ，再计算 $S_{11}$ 。

Show $S_n = 2 S_{n-1}$ for $n \ge 3$ with $S_2 = 11,$ then compute $S_{11}$

解答：

设 $S_n$ 为第 $n$ 行的和。第 $n-1$ 行的每个数都会贡献到下一行的两个位置，而固定的边界数 $10$ 和 $1$ 正好补上边缘缺少的项。因此 $S_n = 2 S_{n-1}$ （当 $n \ge 3$ ）。由 $S_2 = 11$ ，得到 $S_n = 11 \cdot 2^{n-2}$ ，所以 $S_{11} = 11 \cdot 2^9 = 5632$ 。其各位数字之和为 $5 + 6 + 3 + 2 = 16$ 。因此正确答案是 D。

Let $S_n$ be the sum of row $n.$ Each entry in row $n-1$ feeds the two entries just below it, and the fixed border numbers $10$ and $1$ exactly make up for the terms lost at the edges. So $S_n = 2 S_{n-1}$ for $n \ge 3.$ With $S_2 = 11,$ this gives $S_n = 11 \cdot 2^{n-2},$ so $S_{11} = 11 \cdot 2^9 = 5632.$ Its digits sum to $5 + 6 + 3 + 2 = 16.$ Thus, D is the correct answer.

4.

七进位两位数 $\underline{a}\,\underline{b}$ 的值等于九进位两位数 $\underline{b}\,\underline{a}$ 的值。 $a + b$ 是多少？

The value of the two-digit number $\underline{a}\,\underline{b}$ in base seven equals the value of the two-digit number $\underline{b}\,\underline{a}$ in base nine. What is $a + b?$

$7$

$9$

$10$

$11$

$14$

答案：A

知识点：进制位值丢番图方程

难度评级：1130

小提示：

按位值计算，七进位的 $\underline{a}\,\underline{b}$ 是 $7a + b$ ，九进位的 $\underline{b}\,\underline{a}$ 是 $9b + a$ 。

By place value, $\underline{a}\,\underline{b}$ in base seven is $7a + b,$ and $\underline{b}\,\underline{a}$ in base nine is $9b + a$

大提示：

方程化简为 $3a = 4b$ ；再找出 $a$ 是合法七进位数字的解。

The equation reduces to $3a = 4b;$ find digits with $a$ a valid base-seven digit

解答：

按位值，七进制的 $\underline{a}\,\underline{b}$ 是 $7a + b$ ，九进制的 $\underline{b}\,\underline{a}$ 是 $9b + a$ 。令两者相等： $7a + b = 9b + a$ ，所以 $6a = 8b$ ，即 $3a = 4b$ 。因此 $a = 4t$ ，且 $b = 3t$ 。因为它们是两位数， $t \ne 0$ ；又因为 $a$ 是七进制数字，所以 $t \le 1$ 。因此 $a = 4$ 、 $b = 3$ ，所以 $a + b = 7$ 。所以答案是 A。

By place value, $\underline{a}\,\underline{b}$ in base seven is $7a + b,$ and $\underline{b}\,\underline{a}$ in base nine is $9b + a.$ Set them equal: $7a + b = 9b + a,$ so $6a = 8b,$ that is $3a = 4b.$ Thus $a = 4t$ and $b = 3t.$ Because these are two-digit numerals, $t \ne 0;$ because $a$ is a base-seven digit, $t \le 1.$ Hence $a = 4$ and $b = 3,$ so $a + b = 7.$ Therefore, the answer is A.

5.

在 $\triangle ABC$ 中， $AB = 10$ 、 $AC = 18$ 、 $\angle B = 130^\circ$ 。设 $O$ 是经过 $A$ 、 $B$ 、 $C$ 三点的圆的圆心。求 $\angle CAO$ 的度数。

In $\triangle ABC,$ $AB = 10,$ $AC = 18,$ and $\angle B = 130^\circ.$ Let $O$ be the center of the circle containing points $A,$ $B,$ and $C.$ What is the degree measure of $\angle CAO?$

$20$

$30$

$40$

$50$

$60$

答案：C

知识点：外接圆、外心与外接圆半径圆周角等腰三角形

难度评级：1310

小提示：

$O$ 是外心，所以 $OA = OB = OC$ 。把圆心角 $\angle AOC$ 与圆周角 $\angle B$ 联系起来。

$O$ is the circumcenter, so $OA = OB = OC.$ Relate the central angle $\angle AOC$ to the inscribed angle $\angle B$

大提示：

因为 $\angle B$ 是钝角，所以 $\angle AOC = 360^\circ - 2\angle B$ ；再使用等腰三角形 $OAC$ 。

Because $\angle B$ is obtuse, $\angle AOC = 360^\circ - 2\angle B;$ then use the isosceles triangle $OAC$

解答：

因为 $O$ 是外心， $OA = OB = OC$ 。圆周角 $\angle B = 130^\circ$ 对应弧 $AC$ ，且由于 $B$ 是钝角，对应的小圆心角为 $\angle AOC = 360^\circ$ $- 2 \cdot 130^\circ = 100^\circ$ 。三角形 $OAC$ 是等腰三角形，所以 $\angle CAO = \tfrac{180^\circ - 100^\circ}{2} = 40^\circ$ 。边长 $AB$ 和 $AC$ 实际上不需要用到。因此正确答案是 C。

Since $O$ is the circumcenter, $OA = OB = OC.$ The inscribed angle $\angle B = 130^\circ$ subtends arc $AC,$ and because $B$ is obtuse, the central angle is $\angle AOC = 360^\circ$ $- 2 \cdot 130^\circ = 100^\circ.$ Triangle $OAC$ is isosceles, so $\angle CAO = \tfrac{180^\circ - 100^\circ}{2} = 40^\circ.$ (The lengths $AB$ and $AC$ never enter.) Thus, C is the correct answer.

6.

直线 $y = \tfrac{1}{3}x + 1$ 把由 $0 \le x \le 2$ 和 $0 \le y \le 2$ 定义的正方形区域分成上、下两部分。直线 $x = a$ 又把下方区域分成面积相等的两部分。若 $a$ 可以写成 $\sqrt{s} - t$ ，其中 $s$ 和 $t$ 是正整数，求 $s + t$ 。

The line $y = \tfrac{1}{3}x + 1$ divides the square region defined by $0 \le x \le 2$ and $0 \le y \le 2$ into an upper region and a lower region. The line $x = a$ divides the lower region into two regions of equal area. Then $a$ can be written as $\sqrt{s} - t,$ where $s$ and $t$ are positive integers. What is $s + t?$

$18$

$19$

$20$

$21$

$22$

答案：C

知识点：梯形面积二次方程

难度评级：1410

小提示：

$x = a$ 左侧的下方区域是一个梯形，竖直边长为 $1$ 和 $\tfrac{a}{3} + 1$ ，宽为 $a$ 。

The lower region left of $x = a$ is a trapezoid with vertical sides $1$ and $\tfrac{a}{3} + 1$ and width $a$

大提示：

下方总面积为 $\tfrac{8}{3}$ ，所以令 $a + \tfrac{a^2}{6}$ 等于其一半 $\tfrac{4}{3}$ ，再解这个二次方程。

The total lower area is $\tfrac{8}{3},$ so set $a + \tfrac{a^2}{6}$ equal to half of it, $\tfrac{4}{3},$ and solve the quadratic

解答：

下方区域的面积为 $\int_0^2\left(\tfrac{x}{3} + 1\right)dx = \tfrac{2}{3} + 2 = \tfrac{8}{3}$ 。 $0 \le x \le a$ 的部分是面积为 $a + \tfrac{a^2}{6}$ 的梯形。它应等于总面积的一半，即 $\tfrac{4}{3}$ ，所以 $a^2 + 6a - 8 = 0$ ，从而 $a = -3 + \sqrt{17}$ 。因此 $s = 17$ ， $t = 3$ ， $s + t = 20$ 。因此正确答案是 C。

The lower region has area $\int_0^2\left(\tfrac{x}{3} + 1\right)dx = \tfrac{2}{3} + 2 = \tfrac{8}{3}.$ The slice with $0 \le x \le a$ is a trapezoid of area $a + \tfrac{a^2}{6}.$ We want that to be half the total, namely $\tfrac{4}{3},$ so $a^2 + 6a - 8 = 0$ and $a = -3 + \sqrt{17}.$ Then $s = 17,$ $t = 3,$ and $s + t = 20.$ Therefore, the answer is C.

7.

Frances 站在一扇上锁的门正南方 $15$ 米处，这扇门位于一条东西走向的栅栏上。栅栏正后方有一盒巧克力，位于上锁的门以东 $x$ 米处。另有一扇未上锁的门位于巧克力盒以东 $9$ 米处，还有一扇未上锁的门位于上锁的门以西 $8$ 米处。Frances 可以走向一扇未上锁的门，穿过它，再走向巧克力盒。碰巧无论经由哪一扇未上锁的门，她走的总距离都相同。求 $x$ ？

Frances stands $15$ meters directly south of a locked gate in a fence that runs east-west. Immediately behind the fence is a box of chocolates, located $x$ meters east of the locked gate. An unlocked gate lies $9$ meters east of the box, and another unlocked gate lies $8$ meters west of the locked gate. Frances can reach the box by walking toward an unlocked gate, passing through it, and walking toward the box. It happens that the total distance Frances would travel would be the same via either unlocked gate. What is the value of $x?$

$3\tfrac{2}{7}$

$3\tfrac{3}{7}$

$3\tfrac{4}{7}$

$3\tfrac{5}{7}$

$3\tfrac{6}{7}$

答案：C

知识点：坐标几何勾股定理根式

难度评级：1500

小提示：

把上锁的门放在原点，Frances 放在 $(0, -15)$ ；每条路线都是先直线走到门，再沿栅栏走到盒子。

Put the locked gate at the origin and Frances at $(0, -15);$ each route is a straight walk to a gate plus a walk along the fence to the box

大提示：

西侧路线总长为 $17 + (x + 8)$ ；令它等于 $\sqrt{(x + 9)^2 + 225} + 9$ ，再平方。

The west route totals $17 + (x + 8);$ set it equal to $\sqrt{(x + 9)^2 + 225} + 9$ and square

解答：

把栅栏放在 $x$ -轴上，上锁的门在原点，Frances 在 $(0, -15)$ 。则盒子在 $(x, 0)$ ，东门在 $(x + 9, 0)$ ，西门在 $(-8, 0)$ 。东侧路线为 $\sqrt{(x + 9)^2 + 15^2} + 9$ ；西侧路线为 $\sqrt{8^2 + 15^2}$ $+ (x + 8) = 17 + x + 8$ 。令二者相等，得 $\sqrt{(x + 9)^2 + 225} = x + 16$ 。平方并化简： $18x + 306 = 32x + 256$ ，所以 $x = \tfrac{50}{14} = \tfrac{25}{7} = 3\tfrac{4}{7}$ 。因此正确答案是 C。

Put the fence on the $x$ -axis, the locked gate at the origin, and Frances at $(0, -15).$ Then the box is at $(x, 0),$ the east gate at $(x + 9, 0),$ and the west gate at $(-8, 0).$ The east route is $\sqrt{(x + 9)^2 + 15^2} + 9;$ the west route is $\sqrt{8^2 + 15^2}$ $+ (x + 8) = 17 + x + 8.$ Set them equal: $\sqrt{(x + 9)^2 + 225} = x + 16.$ Square and simplify to get $18x + 306 = 32x + 256,$ so $x = \tfrac{50}{14} = \tfrac{25}{7} = 3\tfrac{4}{7}.$ Thus, C is the correct answer.

8.

Emmy 对 Max 说：“我今天订了 $36$ 件数学俱乐部运动衫。”Max 问：“每件多少钱？”Emmy 回答：“我给你一个提示。总价是 $\$\underline{A}\,\underline{B}\,\underline{B}.\underline{B}\,\underline{A}$ ，其中 $A$ 和 $B$ 是数字，且 $A \ne 0$ 。”停顿片刻后，Max 说：“这个价格不错。”求 $A + B$ 。

Emmy says to Max, “I ordered $36$ math club sweatshirts today.” Max asks, “How much did each shirt cost?” Emmy responds, “I’ll give you a hint. The total cost was $\$\underline{A}\,\underline{B}\,\underline{B}.\underline{B}\,\underline{A},$ where $A$ and $B$ are digits and $A \ne 0.$ ” After a pause, Max says, “That was a good price.” What is $A + B?$

$7$

$8$

$11$

$14$

$15$

答案：C

知识点：整除性模运算位值

难度评级：1350

小提示：

把总价写成美分： $\underline{A}\,\underline{B}\,\underline{B}.\underline{B}\,\underline{A}$ 是 $10001A + 1110B$ 美分，并且必须能被 $36$ 整除。

Write the total in cents: $\underline{A}\,\underline{B}\,\underline{B}.\underline{B}\,\underline{A}$ is $10001A + 1110B$ cents, and it must be divisible by $36$

大提示：

把条件模 $36$ 化简为 $7A + 6B \equiv 0$ ；检查数字即可找到唯一的 $A, B$ 。

Reduce the condition modulo $36$ to $7A + 6B \equiv 0;$ test digits to find the unique $A, B$

解答：

以美分为单位，总价是 $10000A + 1000B + 100B$ $+ 10B + A$ $= 10001A + 1110B$ 。它要平均分给 $36$ 件运动衫，所以能被 $36$ 整除。因为 $10001 \equiv 29$ 且 $1110 \equiv 30 \pmod{36}$ ，所以需要 $29A + 30B \equiv 0$ ，也就是 $7A + 6B \equiv 0 \pmod{36}$ 。在 $A \ne 0$ 的数字解中，唯一可行的是 $A = 6, B = 5$ ，因为 $7 \cdot 6 + 6 \cdot 5 = 72$ 。这对应 $\$655.56$ ，也就是每件 $\$18.21$ ，所以 $A + B = 11$ 。因此正确答案是 C。

In cents the total is $10000A + 1000B + 100B$ $+ 10B + A$ $= 10001A + 1110B.$ Split evenly among $36$ shirts, so it’s divisible by $36.$ Now $10001 \equiv 29$ and $1110 \equiv 30 \pmod{36},$ so we need $29A + 30B \equiv 0,$ which reduces to $7A + 6B \equiv 0 \pmod{36}.$ The only digit solution with $A \ne 0$ is $A = 6, B = 5,$ since $7 \cdot 6 + 6 \cdot 5 = 72.$ That’s $\$655.56,$ or $\$18.21$ a shirt, so $A + B = 11.$ Therefore, the answer is C.

9.

有多少个整数有序三元组 $(x, y, z)$ 满足下面的不等式组？

$-x - y - z \le -2$ $-x + y + z \le 2$ $x - y + z \le 2$ $x + y - z \le 2$

How many ordered triples of integers $(x, y, z)$ satisfy the following system of inequalities?

$-x - y - z \le -2$ $-x + y + z \le 2$ $x - y + z \le 2$ $x + y - z \le 2$

$4$

$8$

$11$

$15$

$17$

答案：C

知识点：换元法奇偶性系统列举

难度评级：1560

小提示：

后三个不等式关于 $x, y, z$ 对称；令 $p = -x + y + z$ ， $q = x - y + z$ ， $r = x + y - z$ 。

The last three inequalities are symmetric in $x, y, z;$ let $p = -x + y + z,$ $q = x - y + z,$ $r = x + y - z$

大提示：

此时 $p, q, r \le 2$ ，三者同奇偶，且 $p + q + r = x + y + z \ge 2$ ；数这样的三元组。

Then $p, q, r \le 2$ with equal parity and $p + q + r = x + y + z \ge 2;$ count such triples

解答：

令 $p = -x + y + z$ ， $q = x - y + z$ ， $r = x + y - z$ 。后三个不等式给出 $p, q, r \le 2$ ，第一个不等式给出 $x + y + z \ge 2$ ，并且 $p + q + r = x + y + z$ 。又因为 $x = \tfrac{q + r}{2}$ 等，所以 $p, q, r$ 必须同奇偶。现在数每项 $\le 2$ 、同奇偶、总和在 $[2, 6]$ 中的三元组。偶数情形有 $(2,2,2)$ 、 $(2,2,0)$ 的排列、 $(2,0,0)$ 的排列，以及 $(2,2,-2)$ 的排列，共 $10$ 个。奇数情形只有 $(1,1,1)$ 。总共 $11$ 个，并且每个都对应唯一的 $(x, y, z)$ 。因此正确答案是 C。

Let $p = -x + y + z,$ $q = x - y + z,$ $r = x + y - z.$ The last three inequalities say $p, q, r \le 2,$ the first says $x + y + z \ge 2,$ and $p + q + r = x + y + z.$ Since $x = \tfrac{q + r}{2}$ and so on, $p, q, r$ must all share the same parity. Now count triples with each part $\le 2,$ equal parity, and sum in $[2, 6].$ The even ones are $(2,2,2),$ the permutations of $(2,2,0),$ of $(2,0,0),$ and of $(2,2,-2),$ giving $10.$ The only odd one is $(1,1,1).$ That’s $11$ in all, and each yields a unique $(x, y, z).$ Thus, C is the correct answer.

10.

设 $f(n) = n^3 - 5n^2 + 2n + 8$ ， $g(n) = n^3 - 6n^2 + 5n + 12$ 。使 $\dfrac{f(n)}{g(n)}$ 也为整数的所有整数 $n$ 之和是多少？

Let $f(n) = n^3 - 5n^2 + 2n + 8,$ and let $g(n) = n^3 - 6n^2 + 5n + 12.$ What is the sum of all integer values of $n$ for which $\dfrac{f(n)}{g(n)}$ is also an integer?

$2$

$3$

$4$

$5$

$6$

答案：A

知识点：多项式因式分解整除性

难度评级：1510

小提示：

两个三次多项式都有根 $n = -1$ ；把它们分解因式并约去公因式。

Both cubics have $n = -1$ as a root; factor each and cancel common factors

大提示：

比值化简为 $\dfrac{n - 2}{n - 3} = 1 + \dfrac{1}{n - 3}$ ；它为整数只可能在 $n - 3$ 整除 $1$ 时发生。

The ratio simplifies to $\dfrac{n - 2}{n - 3} = 1 + \dfrac{1}{n - 3};$ it is an integer only when $n - 3$ divides $1$

解答：

分解因式： $f(n) = (n + 1)(n - 2)(n - 4)$ ， $g(n) = (n + 1)(n - 3)(n - 4)$ 。避开 $n \in \{-1, 3, 4\}$ 这些使 $g$ 为零或出现 $\tfrac{0}{0}$ 的值后，可约去公因式： $\dfrac{f(n)}{g(n)} = \dfrac{n - 2}{n - 3} = 1 + \dfrac{1}{n - 3}$ 。这只有在 $n - 3 = \pm 1$ 时才是整数，所以 $n = 2$ 或 $n = 4$ 。但 $n = 4$ 使 $g$ 为零，因此只有 $n = 2$ ，和为 $2$ 。因此正确答案是 A。

Factor both cubics: $f(n) = (n + 1)(n - 2)(n - 4)$ and $g(n) = (n + 1)(n - 3)(n - 4).$ Away from $n \in \{-1, 3, 4\},$ where $g$ vanishes or the ratio is $\tfrac{0}{0},$ the common factors cancel and $\dfrac{f(n)}{g(n)} = \dfrac{n - 2}{n - 3} = 1 + \dfrac{1}{n - 3}.$ That’s an integer only when $n - 3 = \pm 1,$ so $n = 2$ or $n = 4.$ But $n = 4$ kills $g,$ so only $n = 2$ survives, and the sum is $2.$ Therefore, the answer is A.

11.

星期一， $6$ 名学生同时来到辅导中心，每人随机分配给当天值班的 $6$ 名导师之一。星期二，同样的 $6$ 名学生又来了，同样的 $6$ 名导师也在值班，学生再次随机分配给导师。恰有 $2$ 名学生两天都见到同一位导师的概率是多少？

On Monday, $6$ students went to the tutoring center at the same time, and each one was randomly assigned to one of the $6$ tutors on duty. On Tuesday, the same $6$ students showed up, the same $6$ tutors were on duty, and the students were again randomly assigned to the tutors. What is the probability that exactly $2$ students met with the same tutor both Monday and Tuesday?

$\dfrac{1}{16}$

$\dfrac{3}{16}$

$\dfrac{1}{4}$

$\dfrac{3}{8}$

$\dfrac{1}{2}$

答案：B

知识点：错位排列排列基本概率

难度评级：1590

小提示：

每天的分配都是一个排列；比较两天就等同于看一个随机排列，而“同一位导师”对应不动点。

Each day’s assignment is a permutation; comparing the two days is like a single random permutation, and “same tutor” means a fixed point

大提示：

先用 $\binom{6}{2}$ 种方式选择那 $2$ 名匹配的学生，再把其余 $4$ 名错排（ $D_4 = 9$ ）；最后除以 $6!$ 。

Choose the $2$ matched students in $\binom{6}{2}$ ways and derange the other $4$ (there are $D_4 = 9$ derangements); divide by $6!$

解答：

每天的分配都是把 $6$ 名学生分给 $6$ 名导师的一个排列。比较两天时，保持同一导师的学生数就是 $\tau = \pi_{\text{Tue}}^{-1}\pi_{\text{Mon}}$ 的不动点个数，而这个置换是 $6$ 个元素上的均匀随机排列。要恰有 $2$ 个不动点，先选出这 $2$ 个，有 $\binom{6}{2}$ 种；其余 $4$ 个要错排，有 $D_4 = 9$ 种。因此概率为 $\dfrac{\binom{6}{2} D_4}{6!} = \dfrac{15 \cdot 9}{720} = \dfrac{3}{16}$ 。所以正确答案是 B。

Each day’s assignment is a permutation of the $6$ students among the $6$ tutors. Comparing the two days, the number who keep the same tutor is the number of fixed points of $\tau = \pi_{\text{Tue}}^{-1}\pi_{\text{Mon}},$ itself a uniformly random permutation of $6$ elements. We want exactly $2$ fixed points, so choose those $2$ in $\binom{6}{2}$ ways and derange the other $4,$ where $D_4 = 9.$ The probability is $\dfrac{\binom{6}{2} D_4}{6!} = \dfrac{15 \cdot 9}{720} = \dfrac{3}{16}.$ Thus, B is the correct answer.

12.

下图显示了一个等边三角形、一个有 $60^\circ$ 角的菱形和一个正六边形。每个图形中都放有一些两两相切的全等圆盘。分别令 $T$ 、 $R$ 、 $H$ 表示每种情况下圆盘总面积与外部多边形面积之比。

下列哪一项正确？

The figure below shows an equilateral triangle, a rhombus with a $60^\circ$ angle, and a regular hexagon, each of them containing some mutually tangent congruent disks. Let $T,$ $R,$ and $H,$ respectively, denote the ratio in each case of the total area of the disks to the area of the enclosing polygon.

Which of the following is true?

$T = H = R$

$H \lt R = T$

$H = R \lt T$

$H \lt R \lt T$

$H \lt T \lt R$

答案：C

知识点：相切圆面积比正多边形

难度评级：1710

小提示：

若一个圆盘与夹角为 $\theta$ 的两条边相切，则圆心到顶点的距离为 $\dfrac{r}{\sin(\frac{\theta}{2})}$ 。

For a disk tangent to two sides meeting at angle $\theta,$ its center is at distance $\dfrac{r}{\sin(\frac{\theta}{2})}$ from the vertex

大提示：

利用相切条件把每种圆盘半径 $r$ 用多边形的尺寸表示，再计算（圆盘面积）/（多边形面积）。

Use the tangency conditions to write each disk radius $r$ in terms of the polygon’s size, then compute (disk area) / (polygon area) in each case

解答：

设三角形边长为 $s$ 。三个半径为 $r$ 的圆盘给出 $s = 2r(1 + \sqrt3)$ ，所以 $T = \dfrac{3\pi r^2}{(\frac{\sqrt3}{4})s^2}$ $= \dfrac{(2\sqrt3 - 3)\pi}{2} \approx 0.73$ 。对于边长为 $a$ 的菱形，两个圆盘位于长对角线上，满足 $a\sqrt3 = 6r$ ，所以 $r = \tfrac{a}{2\sqrt3}$ ，且 $R = \dfrac{2\pi r^2}{(\frac{a^2\sqrt3}{2})} = \dfrac{\pi\sqrt3}{9} \approx 0.60$ 。对于边长为 $a$ 的正六边形，六个圆盘中每个都与一条边在中点处相切，同样有 $r = \tfrac{a}{2\sqrt3}$ ，且 $H = \dfrac{6\pi r^2}{(\frac{3\sqrt3}{2})a^2}$ $= \dfrac{\pi\sqrt3}{9} \approx 0.60$ 。因此 $H = R \lt T$ ，正确答案是 C。

Take the triangle with side $s.$ Three disks of radius $r$ give $s = 2r(1 + \sqrt3),$ so $T = \dfrac{3\pi r^2}{(\frac{\sqrt3}{4})s^2}$ $= \dfrac{(2\sqrt3 - 3)\pi}{2} \approx 0.73.$ For the rhombus with side $a,$ the two disks sit on the long diagonal $a\sqrt3 = 6r,$ so $r = \tfrac{a}{2\sqrt3}$ and $R = \dfrac{2\pi r^2}{(\frac{a^2\sqrt3}{2})} = \dfrac{\pi\sqrt3}{9} \approx 0.60.$ For the hexagon with side $a,$ each of the six disks touches a side at its midpoint, again giving $r = \tfrac{a}{2\sqrt3}$ and $H = \dfrac{6\pi r^2}{(\frac{3\sqrt3}{2})a^2}$ $= \dfrac{\pi\sqrt3}{9} \approx 0.60.$ So $H = R \lt T.$ Therefore, the answer is C.

13.

一个 $30$ - $60$ - $90^\circ$ 直角三角形斜边上的高，被到最短边的中线分成长度为 $x \lt y$ 的两段。求 $\dfrac{x}{x + y}$ 。

The altitude to the hypotenuse of a $30$ - $60$ - $90^\circ$ right triangle is divided into two segments of lengths $x \lt y$ by the median to the shortest side of the triangle. What is the ratio $\dfrac{x}{x + y}?$

$\dfrac{3}{7}$

$\dfrac{\sqrt3}{4}$

$\dfrac{4}{9}$

$\dfrac{5}{11}$

$\dfrac{4\sqrt3}{15}$

答案：A

知识点：特殊直角三角形高线中线（几何）

难度评级：1660

小提示：

把直角放在原点，两条直角边沿坐标轴（短边为 $1$ ，长边为 $\sqrt3$ ）；求高的垂足和中线方程。

Set the right angle at the origin with legs along the axes (short leg $1,$ long leg $\sqrt3$ ); find the foot of the altitude and the median line

大提示：

求这条高与到短边的中线的交点，再取靠近端点的那段与整条高的比值。

Intersect the altitude with the median to the short side, then take the ratio of the near segment to the whole altitude

解答：

令直角顶点 $C = (0,0)$ ，短边 $CB = 1$ ，其中 $B = (1, 0)$ ，长边 $CA = \sqrt3$ ，其中 $A = (0, \sqrt3)$ 。从 $C$ 到斜边 $AB$ 的高的垂足为 $H = \left(\tfrac34, \tfrac{\sqrt3}{4}\right)$ ，且这条高在直线 $x = \sqrt3\,y$ 上。从 $A$ 到 $CB$ 中点 $\left(\tfrac12, 0\right)$ 的中线与这条高相交于 $\left(\tfrac37, \tfrac{\sqrt3}{7}\right)$ 。这把 $CH$ （长度 $\tfrac{\sqrt3}{2}$ ）分成 $\tfrac{4\sqrt3}{14}$ 和 $\tfrac{3\sqrt3}{14}$ ，所以 $x = \tfrac{3\sqrt3}{14}$ ，于是 $\dfrac{x}{x + y} = \dfrac{\frac{3\sqrt3}{14}}{\frac{\sqrt3}{2}} = \dfrac{3}{7}$ 。因此正确答案是 A。

Place the right angle at $C = (0,0),$ the short leg $CB = 1$ with $B = (1, 0),$ and the long leg $CA = \sqrt3$ with $A = (0, \sqrt3).$ The altitude from $C$ to hypotenuse $AB$ has foot $H = \left(\tfrac34, \tfrac{\sqrt3}{4}\right)$ and runs along $x = \sqrt3\,y.$ The median from $A$ to the midpoint $\left(\tfrac12, 0\right)$ of $CB$ meets that altitude at $\left(\tfrac37, \tfrac{\sqrt3}{7}\right).$ This cuts $CH$ (length $\tfrac{\sqrt3}{2}$ ) into $\tfrac{4\sqrt3}{14}$ and $\tfrac{3\sqrt3}{14},$ so $x = \tfrac{3\sqrt3}{14}$ and $\dfrac{x}{x + y} = \dfrac{\frac{3\sqrt3}{14}}{\frac{\sqrt3}{2}} = \dfrac{3}{7}.$ Thus, A is the correct answer.

14.

九名运动员参加篮球队选拔，且没有两人身高相同。他们依次从一个袋子中随机抽取腕带，不放回；袋中有 $3$ 条蓝色、 $3$ 条红色、 $3$ 条绿色腕带。他们被分成蓝组、红组和绿组。每组最高的成员被指定为该组队长。三名队长正好是最高的三名运动员的概率是多少？

Nine athletes, no two of whom are the same height, try out for the basketball team. One at a time, they draw a wristband at random, without replacement, from a bag containing $3$ blue bands, $3$ red bands, and $3$ green bands. They are divided into a blue group, a red group, and a green group. The tallest member of each group is named the group captain. What is the probability that the group captains are the three tallest athletes?

$\dfrac{2}{9}$

$\dfrac{2}{7}$

$\dfrac{9}{28}$

$\dfrac{1}{3}$

$\dfrac{3}{8}$

答案：C

知识点：基本概率条件概率

难度评级：1500

小提示：

队长是最高的三人，当且仅当这三人分到三个不同的组。

The captains are the three tallest exactly when the three tallest end up in three different groups

大提示：

依次把前三高的人分到九个位置中：后两人每次进入新组的概率分别是 $\tfrac{6}{8}$ 和 $\tfrac{3}{7}$ 。

Assign the top three to slots one at a time: the probabilities that each new one lands in a fresh group are $\tfrac{6}{8}$ then $\tfrac{3}{7}$

解答：

三名队长正好是最高的三人，当且仅当这三人进入三个不同的组；这样每人都是自己组中最高的。把这三人依次放入 $9$ 个位置（每组 $3$ 个）中。第二高的人避开第一高所在组的概率是 $\tfrac{6}{8}$ ，第三高的人避开前两人所在组的概率是 $\tfrac{3}{7}$ 。所以所求概率为 $\tfrac{6}{8} \cdot \tfrac{3}{7} = \tfrac{9}{28}$ 。因此正确答案是 C。

The captains are the three tallest exactly when those three land in three different groups, since each is then the tallest of its own group. Drop them into the $9$ slots one at a time ( $3$ per group). The second tallest misses the first’s group with probability $\tfrac{6}{8},$ and the third misses both with probability $\tfrac{3}{7}.$ So the probability is $\tfrac{6}{8} \cdot \tfrac{3}{7} = \tfrac{9}{28}.$ Therefore, the answer is C.

15.

级数

$\sum_{k=1}^{\infty} \frac{1}{k^3 + 6k^2 + 8k}$

可以表示为 $\dfrac{a}{b}$ ，其中 $a$ 和 $b$ 是互质的正整数。求 $a + b$ 。

The sum

$\sum_{k=1}^{\infty} \frac{1}{k^3 + 6k^2 + 8k}$

can be expressed as $\dfrac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. What is $a + b?$

$89$

$97$

$102$

$107$

$129$

答案：D

知识点：部分分式裂项相消

难度评级：1600

小提示：

把分母分解为 $k(k + 2)(k + 4)$ ，然后使用部分分式。

Factor the denominator as $k(k + 2)(k + 4)$ and use partial fractions

大提示：

各项 $\tfrac{\frac{1}{8}}{k} - \tfrac{\frac{1}{4}}{k + 2} + \tfrac{\frac{1}{8}}{k + 4}$ 会裂项相消；只有 $k = 1, 2, 3, 4$ 留下余项。

The pieces $\tfrac{\frac{1}{8}}{k} - \tfrac{\frac{1}{4}}{k + 2} + \tfrac{\frac{1}{8}}{k + 4}$ telescope; only $k = 1, 2, 3, 4$ leave a remainder

解答：

分解 $k^3 + 6k^2 + 8k = k(k + 2)(k + 4)$ ，并作部分分式分解： $\dfrac{1}{k(k + 2)(k + 4)} = \dfrac{\frac{1}{8}}{k}$ $- \dfrac{\frac{1}{4}}{k + 2}$ $+ \dfrac{\frac{1}{8}}{k + 4}$ 。对所有 $k$ 求和时， $\tfrac1n$ 在 $n \ge 5$ 的系数都会相消，所以只剩前几项： $\tfrac18\left(1 + \tfrac12 - \tfrac13 - \tfrac14\right)$ $= \tfrac18 \cdot \tfrac{11}{12} = \tfrac{11}{96}$ 。因此 $a + b = 11 + 96 = 107$ ，正确答案是 D。

Factor $k^3 + 6k^2 + 8k = k(k + 2)(k + 4),$ then split into partial fractions: $\dfrac{1}{k(k + 2)(k + 4)} = \dfrac{\frac{1}{8}}{k}$ $- \dfrac{\frac{1}{4}}{k + 2}$ $+ \dfrac{\frac{1}{8}}{k + 4}.$ Summing over all $k,$ the coefficient of $\tfrac1n$ cancels for $n \ge 5,$ so only the first few terms survive: $\tfrac18\left(1 + \tfrac12 - \tfrac13 - \tfrac14\right)$ $= \tfrac18 \cdot \tfrac{11}{12} = \tfrac{11}{96}.$ So $a + b = 11 + 96 = 107.$ Thus, D is the correct answer.

16.

一个圆被分成 $6$ 个大小互不相同的扇形。接着把其中 $2$ 个扇形涂红、 $2$ 个涂绿、 $2$ 个涂蓝，并要求相邻的两个扇形颜色不同。下图展示了一种涂色方式。

一共有多少种不同的涂色方式？

A circle has been divided into $6$ sectors of different sizes. Then $2$ of the sectors are painted red, $2$ painted green, and $2$ painted blue so that no two neighboring sectors are painted the same color. One such coloring is shown below.

How many different colorings are possible?

$12$

$16$

$18$

$24$

$28$

答案：D

知识点：图论补集计数分类讨论

难度评级：1800

小提示：

大小不同的扇形形成一个固定的 $6$ 个位置的环；需要用每种颜色恰好两次来做合法涂色。

The unequal sectors form a fixed cycle of $6$ positions; you need proper colorings using each color exactly twice

大提示：

一个 $6$ -环共有 $2^6 + 2 = 66$ 种合法的 $3$ 色涂法；去掉不是每种颜色恰好出现两次的情形。

A $6$ -cycle has $2^6 + 2 = 66$ proper $3$ -colorings; remove those that do not use each color exactly twice

解答：

六个大小不同的扇形构成一个由 $6$ 个可区分位置组成的固定环，所以要计算 $6$ 环的合法 $3$ 色涂色，并要求每种颜色恰好使用两次。合法 $3$ 色涂色共有 $(3-1)^6 + (3-1) = 66$ 种。其中有 $6$ 种只使用两种颜色：缺少的颜色有 $3$ 种选择，其余两种颜色有 $2$ 种交替方式。另有 $36$ 种的颜色计数为 $(3,2,1)$ ：出现三次的颜色有 $3$ 种选择，它所占的交替位置组有 $2$ 种选择，出现两次的颜色有 $2$ 种选择，再从其余 $3$ 个位置中选择它的 $2$ 个位置，有 $3$ 种方法。剩下的 $66 - 6 - 36 = 24$ 种恰好每种颜色使用两次。所以答案是 D。

The six unequal sectors form a fixed cycle of $6$ distinguishable positions, so we want proper $3$ -colorings of a $6$ -cycle that use each color exactly twice. There are $(3-1)^6 + (3-1) = 66$ proper $3$ -colorings altogether. Of these, $6$ use only two colors: choose the missing color in $3$ ways, then alternate the other two colors in $2$ ways. Another $36$ have color counts $(3,2,1)$ : choose the color used three times in $3$ ways, choose one of the $2$ alternating sets of positions for it, choose the color used twice in $2$ ways, and choose its $2$ positions among the other $3$ in $3$ ways. The remaining $66 - 6 - 36 = 24$ use each color exactly twice. Therefore, the answer is D.

17.

考虑一个由 $n$ 个正整数组成的递减序列 $x_1 \gt x_2 \gt x_3 \gt \cdots \gt x_n$ ，满足以下条件。序列前 $3$ 项的平均数是 $2025$ 。对所有 $4 \le k \le n$ ，序列前 $k$ 项的平均数比前 $k - 1$ 项的平均数小 $1$ 。

$n$ 的最大可能值是多少？

Consider a decreasing sequence of $n$ positive integers $x_1 \gt x_2 \gt x_3 \gt \cdots \gt x_n$ that satisfies the following conditions. The average of the first $3$ terms in the sequence is $2025.$ For all $4 \le k \le n,$ the average of the first $k$ terms in the sequence is $1$ less than the average of the first $k - 1$ terms in the sequence.

What is the greatest possible value of $n?$

$1013$

$1014$

$1016$

$2016$

$2025$

答案：B

知识点：平均数求和极限情形界定

难度评级：1910

小提示：

令 $A_k$ 为前 $k$ 项的平均数；条件给出当 $k \ge 3$ 时 $A_k = 2028 - k$ 。

Let $A_k$ be the average of the first $k$ terms; the conditions give $A_k = 2028 - k$ for $k \ge 3$

大提示：

则当 $k \ge 4$ 时， $x_k = k A_k$ $- (k-1)A_{k-1} = 2029$ $- 2k$ ；要求它始终是正整数。

Then $x_k = k A_k$ $- (k-1)A_{k-1} = 2029$ $- 2k$ for $k \ge 4;$ require it to stay a positive integer

解答：

令 $A_k$ 为前 $k$ 项的平均数。则 $A_3 = 2025$ ，且 $A_k = A_{k-1} - 1$ （当 $k \ge 4$ ），所以 $A_k = 2028 - k$ 。部分和为 $S_k = k(2028 - k)$ ，因此当 $k \ge 4$ 时各项为 $x_k = S_k - S_{k-1} = 2029 - 2k$ ，也就是 $x_4 = 2021, x_5 = 2019, \ldots$ 。只要 $2029 - 2k \gt 0$ ，即 $k \le 1014$ ，这些项就保持为正，此时 $x_{1014} = 1$ 。这个上界可以达到：取 $(x_1,x_2,x_3) = (2030,2023,2022)$ ，它是递减的，和为 $6075$ ，并且各项都大于 $x_4 = 2021$ 。所以最大可能值为 $n = 1014$ ，因此正确答案是 B。

Let $A_k$ be the average of the first $k$ terms. Then $A_3 = 2025$ and $A_k = A_{k-1} - 1$ for $k \ge 4,$ so $A_k = 2028 - k.$ The partial sum is $S_k = k(2028 - k),$ and for $k \ge 4$ the terms are $x_k = S_k - S_{k-1} = 2029 - 2k,$ namely $x_4 = 2021, x_5 = 2019, \ldots.$ These stay positive as long as $2029 - 2k \gt 0,$ that is $k \le 1014,$ with $x_{1014} = 1.$ This bound is attainable: take $(x_1,x_2,x_3) = (2030,2023,2022),$ which is decreasing, sums to $6075,$ and lies above $x_4 = 2021.$ Thus the greatest possible value is $n = 1014,$ so B is the correct answer.

18.

求和 $\lfloor\sqrt{1}\rfloor + \lfloor\sqrt{2}\rfloor + \lfloor\sqrt{3}\rfloor$ $+ \cdots + \lfloor\sqrt{2024}\rfloor$ $+ \lfloor\sqrt{2025}\rfloor$ 的个位数字是多少？（记 $\lfloor x \rfloor$ 为小于或等于 $x$ 的最大整数。）

What is the ones digit of the sum $\lfloor\sqrt{1}\rfloor + \lfloor\sqrt{2}\rfloor + \lfloor\sqrt{3}\rfloor$ $+ \cdots + \lfloor\sqrt{2024}\rfloor$ $+ \lfloor\sqrt{2025}\rfloor?$ (Recall that $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x.$ )

$1$

$2$

$3$

$5$

$8$

答案：D

知识点：取整函数前n项平方和

难度评级：1730

小提示：

当 $\lfloor\sqrt{n}\rfloor = m$ 时，对应的 $2m + 1$ 个数满足 $m^2 \le n \le (m + 1)^2 - 1$ 。

$\lfloor\sqrt{n}\rfloor = m$ for the $2m + 1$ values $m^2 \le n \le (m + 1)^2 - 1$

大提示：

因为 $\sqrt{2025} = 45$ ，对 $m = 1$ 到 $44$ 求和 $m(2m + 1)$ ，再加上最后一项 $45$ 。

Since $\sqrt{2025} = 45,$ sum $m(2m + 1)$ for $m = 1$ to $44$ and add the final term $45$

解答：

对每个 $m$ ，当 $\lfloor\sqrt{n}\rfloor = m$ 时，对应的 $2m + 1$ 个整数满足 $m^2 \le n \le (m + 1)^2 - 1$ 。因为 $\sqrt{2025} = 45$ ，所以 $1 \le m \le 44$ 的项贡献 $\sum_{m=1}^{44} m(2m + 1)$ ，再加上 $n = 2025$ 对应的 $45$ 。这个和为 $\sum_{m=1}^{44}(2m^2 + m)$ $= 2 \cdot \tfrac{44 \cdot 45 \cdot 89}{6}$ $+ \tfrac{44 \cdot 45}{2} = 58740$ $+ 990 = 59730$ ，所以总和是 $59775$ ，个位数字为 $5$ 。因此正确答案是 D。

For each $m,$ $\lfloor\sqrt{n}\rfloor = m$ on the $2m + 1$ integers $m^2 \le n \le (m + 1)^2 - 1.$ Since $\sqrt{2025} = 45,$ the terms with $1 \le m \le 44$ contribute $\sum_{m=1}^{44} m(2m + 1),$ and $n = 2025$ tacks on $45.$ That sum is $\sum_{m=1}^{44}(2m^2 + m)$ $= 2 \cdot \tfrac{44 \cdot 45 \cdot 89}{6}$ $+ \tfrac{44 \cdot 45}{2} = 58740$ $+ 990 = 59730,$ so the total is $59775.$ Its ones digit is $5.$ Therefore, the answer is D.

19.

一个容器底部是 $1 \times 1$ 的正方形，顶部开口是 $3 \times 3$ 的正方形，四个侧面是全等的梯形，如图所示。从空容器开始，一根以恒定速率出水的水管用 $35$ 分钟把容器装到梯形侧面的中线高度。

还需要多少分钟才能把容器剩余部分装满？

A container has a $1 \times 1$ square bottom, a $3 \times 3$ open square top, and four congruent trapezoidal sides, as shown. Starting when the container is empty, a hose that runs water at a constant rate takes $35$ minutes to fill the container up to the midline of the trapezoids.

How many more minutes will it take to fill the remainder of the container?

$70$

$85$

$90$

$95$

$105$

答案：D

知识点：长度、面积与体积的缩放关系体积相似

难度评级：1660

小提示：

这个容器是一个金字塔的一部分；从顶点到边长为 $w$ 的截面为止的体积按 $w^3$ 缩放。

The container is a piece of a pyramid; the volume from the apex up to side length $w$ scales as $w^3$

大提示：

体积分成 $2^3 - 1^3 = 7$ 份（到中线，边长 $2$ ）和 $3^3 - 2^3 = 19$ 份（剩余部分）；用 $35$ 分钟按比例换算。

Volumes split as $2^3 - 1^3 = 7$ (up to the midline, side $2$ ) and $3^3 - 2^3 = 19$ (remainder); scale the $35$ minutes

解答：

这个容器是一个正方台：高度比例为 $t$ 的水平截面边长为 $1 + 2t$ 。把侧面向上延长到顶点，则到边长为 $w$ 的截面为止的体积按 $w^3$ 缩放。因此整个容器有 $3^3 - 1^3 = 26$ 份，到中线（边长 $2$ ）的部分有 $2^3 - 1^3 = 7$ 份，剩余部分有 $3^3 - 2^3 = 19$ 份。 $7$ 份用时 $35$ 分钟，所以每份 $5$ 分钟。剩下 $19$ 份需要 $95$ 分钟。因此正确答案是 D。

The container is a square frustum: a horizontal slice at height fraction $t$ has side $1 + 2t.$ Extend the sides up to their apex, and the volume out to where the side length is $w$ scales as $w^3.$ So the whole container is $3^3 - 1^3 = 26$ parts, the piece up to the midline (side $2$ ) is $2^3 - 1^3 = 7$ parts, and the rest is $3^3 - 2^3 = 19$ parts. Those $7$ parts take $35$ minutes, so each part is $5$ minutes. The remaining $19$ parts take $95$ minutes. Thus, D is the correct answer.

20.

四个全等的半圆内接于边长为 $1$ 的正方形中，使得它们的直径在正方形的边上，每条直径的一个端点位于正方形的一个顶点，并且相邻半圆互相相切。一个以正方形中心为圆心的小圆与四个半圆都相切，如下图所示。

小圆的直径可以写成 $(\sqrt{a} + b)(\sqrt{c} + d)$ ，其中 $a$ ， $b$ ， $c$ 和 $d$ 是整数。求 $a + b + c + d$ 。

Four congruent semicircles are inscribed in a square of side length $1$ so that their diameters are on the sides of the square, one endpoint of each diameter is at a vertex of the square, and adjacent semicircles are tangent to each other. A small circle centered at the center of the square is tangent to each of the four semicircles, as shown below.

The diameter of the small circle can be written as $(\sqrt{a} + b)(\sqrt{c} + d),$ where $a,$ $b,$ $c,$ and $d$ are integers. What is $a + b + c + d?$

$3$

$5$

$8$

$9$

$11$

答案：A

知识点：相切圆坐标几何根式

难度评级：1930

小提示：

先由相邻两个半圆相切求出半圆半径 $\rho$ ，它们的直径分别从相邻顶点开始。

First find the semicircle radius $\rho$ from the tangency of two adjacent semicircles whose diameters start at neighboring vertices

大提示：

小圆半径等于正方形中心到某个半圆圆心的距离减去 $\rho$ ；再把 $2t$ 分解成两个二项式的乘积。

The small radius equals (distance from the square’s center to a semicircle’s center) minus $\rho;$ then factor $2t$ into two binomials

解答：

设每个半圆半径为 $\rho$ ，圆心可取如 $(\rho, 0)$ 和 $(1, \rho)$ 。相邻半圆相切，所以这两个圆心距离为 $2\rho$ ： $(1 - \rho)^2 + \rho^2 = 4\rho^2$ 。这给出 $2\rho^2 + 2\rho - 1 = 0$ ，所以 $\rho = \tfrac{\sqrt3 - 1}{2}$ 。小圆半径为 $t$ ，圆心在 $\left(\tfrac12, \tfrac12\right)$ 。它与一个半圆相切时，到该半圆圆心的距离等于 $\rho + t$ 。这个距离是 $\sqrt{2 - \sqrt3}$ ，所以 $t = \sqrt{2 - \sqrt3} - \rho$ 。直径为 $2t = \sqrt6 - \sqrt2$ $- \sqrt3 + 1 = (\sqrt3 - 1)(\sqrt2 - 1)$ 。因此 $a + b + c + d = 3$ $+ (-1) + 2 + (-1) = 3$ ，正确答案是 A。

Let each semicircle have radius $\rho,$ with centers like $(\rho, 0)$ and $(1, \rho).$ Adjacent semicircles are tangent, so these centers are $2\rho$ apart: $(1 - \rho)^2 + \rho^2 = 4\rho^2.$ This gives $2\rho^2 + 2\rho - 1 = 0,$ so $\rho = \tfrac{\sqrt3 - 1}{2}.$ The small circle of radius $t$ sits at $\left(\tfrac12, \tfrac12\right),$ and it’s tangent to a semicircle when its distance to that center equals $\rho + t.$ That distance is $\sqrt{2 - \sqrt3},$ so $t = \sqrt{2 - \sqrt3} - \rho,$ and the diameter is $2t = \sqrt6 - \sqrt2$ $- \sqrt3 + 1 = (\sqrt3 - 1)(\sqrt2 - 1).$ So $a + b + c + d = 3$ $+ (-1) + 2 + (-1) = 3.$ Therefore, the answer is A.

21.

一个 $3 \times 3$ 方格中的 $9$ 个小正方形要被涂成红、蓝、黄三色，要求每个红色小方格至少与一个蓝色小方格共边，每个蓝色小方格至少与一个黄色小方格共边，每个黄色小方格至少与一个红色小方格共边。可以通过旋转和/或反射互相得到的涂色视为相同。共有多少种不同的涂色？

Each of the $9$ squares in a $3 \times 3$ grid is to be colored red, blue, or yellow in such a way that each red square shares an edge with at least one blue square, each blue square shares an edge with at least one yellow square, and each yellow square shares an edge with at least one red square. Colorings that can be obtained from one another by rotations and/or reflections are to be considered the same. How many different colorings are possible?

$3$

$9$

$12$

$18$

$27$

答案：C

知识点：伯恩赛德引理分类讨论

难度评级：2100

小提示：

这些规则形成一个有方向的循环：红色需要蓝色邻居，蓝色需要黄色邻居，黄色需要红色邻居。

The rules form a directed cycle: red needs a blue neighbor, blue needs a yellow, yellow needs a red

大提示：

先数固定带标号的三阶方格上所有合法涂色，再按正方形的 $8$ 个旋转和反射对称分组。

Count all valid colorings of the fixed labeled grid, then group them by the $8$ rotations and reflections of the square

解答：

先计算各位置可区分的方格涂色。固定中心方格为红色，并按循环顺序列出四个边中点的颜色。在旋转或反射意义下，边中点的唯一可能模式是 $YRBR$ 和 $YRBY$ 。第一种模式有 $4$ 种放置方式，以及 $3$ 个可能的循环角落字符串 $BRYB, BYRB$ ，和 $BYYB$ ；第二种模式有 $8$ 种放置方式，以及 $2$ 个可能的角落字符串 $BYBB$ 和 $BYRB$ 。所以红色中心时共有 $4 \cdot 3 + 8 \cdot 2 = 28$ 种涂色。中心有 $3$ 种颜色选择，所以带标号涂色共有 $84$ 种。

现在应用伯恩赛德引理。恒等变换固定全部 $84$ 种涂色。没有非恒等旋转能固定合法涂色。水平轴和竖直轴的两个反射各固定 $6$ 种涂色，而两个对角线反射均不固定任何涂色。因此在旋转和反射意义下，涂色数为 $\dfrac{84 + 6 + 6}{8} = 12$ 。所以正确答案是 C。

First count colorings of a grid whose positions are distinguished. Fix the center square as red and list the four edge-middle colors cyclically. Up to a rotation or reflection, the only possible edge patterns are $YRBR$ and $YRBY.$ The first has $4$ placements and $3$ possible cyclic corner strings, $BRYB, BYRB,$ and $BYYB;$ the second has $8$ placements and $2$ possible corner strings, $BYBB$ and $BYRB.$ Thus there are $4 \cdot 3 + 8 \cdot 2 = 28$ colorings with a red center. The center has $3$ possible colors, so there are $84$ labeled colorings.

Now apply Burnside’s lemma. The identity fixes all $84$ colorings. No nonidentity rotation fixes a valid coloring. Each of the two reflections across a horizontal or vertical axis fixes $6$ colorings, while each diagonal reflection fixes none. Therefore, the number of colorings up to rotations and reflections is $\dfrac{84 + 6 + 6}{8} = 12.$ Thus, C is the correct answer.

22.

随机选择一个七位正整数。已知它的各位数字之和为 $61$ ，求它能被 $11$ 整除的概率。

A seven-digit positive integer is chosen at random. What is the probability that the number is divisible by $11,$ given that the sum of its digits is $61?$

$\dfrac{3}{14}$

$\dfrac{3}{11}$

$\dfrac{2}{7}$

$\dfrac{4}{11}$

$\dfrac{3}{7}$

答案：A

知识点：条件概率整除性隔板法

难度评级：2040

小提示：

数字和 $61$ 比最大值 $63$ 少 $2$ ，所以所有数字都是 $9$ ，只是总共少掉 $2$ 。

A digit sum of $61$ is $2$ below the maximum $63,$ so the digits are all $9$ except a total shortfall of $2$

大提示：

用交错数字和判断能否被 $11$ 整除；把它写成奇数位亏缺与偶数位亏缺的关系。

For divisibility by $11$ use the alternating digit sum; express it through the shortfalls on the odd- versus even-position digits

解答：

数字和 $61 = 63 - 2$ ，表示七个数字全为 $9$ 后总亏缺为 $2$ ，这样的数有 $\binom{2 + 6}{6} = 28$ 个。要能被 $11$ 整除，需要 $O - E \equiv 0 \pmod{11}$ ，其中 $O$ 是 $4$ 个奇数位数字之和， $E$ 是 $3$ 个偶数位数字之和。设亏缺为 $d_O + d_E = 2$ 。则 $O - E = 9 - d_O$ $+ d_E = 11 - 2 d_O$ ，只有当 $d_O = 0$ 时才是 $11$ 的倍数。因此全部亏缺 $2$ 都在 $3$ 个偶数位上，有 $\binom{2 + 2}{2} = 6$ 种。概率为 $\tfrac{6}{28} = \tfrac{3}{14}$ 。因此正确答案是 A。

A digit sum of $61 = 63 - 2$ means all seven digits are $9$ except for a total deficit of $2,$ which gives $\binom{2 + 6}{6} = 28$ numbers. For divisibility by $11$ we need $O - E \equiv 0 \pmod{11},$ where $O$ sums the $4$ odd-position digits and $E$ the $3$ even ones. Write the deficits as $d_O + d_E = 2.$ Then $O - E = 9 - d_O$ $+ d_E = 11 - 2 d_O,$ a multiple of $11$ only when $d_O = 0.$ So all of the deficit $2$ falls on the $3$ even positions, giving $\binom{2 + 2}{2} = 6$ ways. The probability is $\tfrac{6}{28} = \tfrac{3}{14}.$ Therefore, the answer is A.

23.

一个由小正方形组成的矩形网格有 $141$ 行和 $91$ 列。每个小正方形里有放两个数字的空间。Horace 和 Vera 都把从 $1$ 到 $141 \times 91 = 12{,}831$ 的数字填入网格。Horace 按行填写：他把 $1$ 到 $91$ 依次从左到右填入第 $1$ 行，把 $92$ 到 $182$ 依次从左到右填入第 $2$ 行，并如此继续到第 $141$ 行。Vera 按列填写：她把 $1$ 到 $141$ 依次从上到下填入第 $1$ 列，再把 $142$ 到 $282$ 依次从上到下填入第 $2$ 列，并如此继续到第 $91$ 列。有多少个小正方形中两人写下了相同的数字？

A rectangular grid of squares has $141$ rows and $91$ columns. Each square has room for two numbers. Horace and Vera each fill in the grid by putting the numbers from $1$ through $141 \times 91 = 12{,}831$ into the squares. Horace fills the grid horizontally: he puts $1$ through $91$ in order from left to right into row $1,$ puts $92$ through $182$ into row $2$ in order from left to right, and continues similarly through row $141.$ Vera fills the grid vertically: she puts $1$ through $141$ in order from top to bottom into column $1,$ then $142$ through $282$ into column $2$ in order from top to bottom, and continues similarly through column $91.$ How many squares get two copies of the same number?

$7$

$10$

$11$

$12$

$19$

答案：C

知识点：丢番图方程模运算

难度评级：2300

小提示：

在第 $i$ 行第 $j$ 列，Horace 写 $91(i - 1) + j$ ，Vera 写 $141(j - 1) + i$ ；令二者相等。

At row $i,$ column $j,$ Horace writes $91(i - 1) + j$ and Vera writes $141(j - 1) + i;$ set them equal

大提示：

方程化简为 $9i - 14j = -5$ ；找出哪些列 $j$ 会使 $i$ 是范围内的整数。

The equation reduces to $9i - 14j = -5;$ find the columns $j$ that make $i$ an integer in range

解答：

在第 $i$ 行第 $j$ 列，Horace 写的是 $91(i - 1) + j$ ，Vera 写的是 $141(j - 1) + i$ 。令二者相等并化简，得到 $9i - 14j = -5$ ，所以 $i = \tfrac{14j - 5}{9}$ ，恰好在 $j \equiv 1 \pmod 9$ 时为整数。对于 $j = 1, 10, 19, \ldots, 91$ ，共有 $11$ 个值，且对应的 $i$ 为 $1, 15, 29, \ldots, 141$ ，都在范围内。因此有 $11$ 个小正方形匹配，正确答案是 C。

At row $i,$ column $j,$ Horace writes $91(i - 1) + j$ and Vera writes $141(j - 1) + i.$ Set them equal and simplify to get $9i - 14j = -5,$ so $i = \tfrac{14j - 5}{9},$ an integer exactly when $j \equiv 1 \pmod 9.$ For $j = 1, 10, 19, \ldots, 91,$ that’s $11$ values, and $i$ runs $1, 15, 29, \ldots, 141,$ all within range. So $11$ squares match. Thus, C is the correct answer.

24.

一只青蛙按如下规则在数轴上跳跃。它从 $0$ 开始。如果它在 $0$ ，那么它以概率 $\tfrac12$ 移动到 $1$ ，并以概率 $\tfrac12$ 消失。对于 $n = 1$ ， $2$ 或 $3$ ，如果它在 $n$ ，那么它以概率 $\tfrac14$ 移动到 $n + 1$ ，以概率 $\tfrac14$ 移动到 $n - 1$ ，并以概率 $\tfrac12$ 消失。

青蛙到达 $4$ 的概率是多少？

A frog hops along the number line according to the following rules. It starts at $0.$ If it is at $0,$ then it moves to $1$ with probability $\tfrac12$ and it disappears with probability $\tfrac12.$ For $n = 1,$ $2,$ or $3,$ if it is at $n,$ then it moves to $n + 1$ with probability $\tfrac14,$ it moves to $n - 1$ with probability $\tfrac14,$ and it disappears with probability $\tfrac12.$

What is the probability that the frog reaches $4?$

$\dfrac{1}{101}$

$\dfrac{1}{100}$

$\dfrac{1}{99}$

$\dfrac{1}{98}$

$\dfrac{1}{97}$

答案：E

知识点：随机游走递推概率方程组

难度评级：2170

小提示：

令 $p_n$ 为从位置 $n$ 出发最终到达 $4$ 的概率，并取 $p_4 = 1$ ；对每个状态列一个方程。

Let $p_n$ be the probability of ever reaching $4$ starting from $n,$ with $p_4 = 1;$ set up one equation per state

大提示：

使用 $p_0 = \tfrac12 p_1$ ，并对 $n = 1, 2, 3$ 使用 $p_n = \tfrac14 p_{n-1} + \tfrac14 p_{n+1}$ ，然后解这个方程组。

Use $p_0 = \tfrac12 p_1$ and $p_n = \tfrac14 p_{n-1} + \tfrac14 p_{n+1}$ for $n = 1, 2, 3,$ then solve the system

解答：

令 $p_n$ 为从位置 $n$ 出发到达 $4$ 的概率，且 $p_4 = 1$ 。规则给出 $p_0 = \tfrac12 p_1$ 、 $p_1 = \tfrac14 p_0 + \tfrac14 p_2$ 、 $p_2 = \tfrac14 p_1 + \tfrac14 p_3$ 、 $p_3 = \tfrac14 p_2 + \tfrac14$ 。向上求解可得 $p_1 = \tfrac27 p_2$ 和 $p_2 = \tfrac{7}{26} p_3$ 。继续代回得到 $p_3 = \tfrac{26}{97}$ 、 $p_2 = \tfrac{7}{97}$ 、 $p_1 = \tfrac{2}{97}$ ，最后 $p_0 = \tfrac{1}{97}$ 。因此正确答案是 E。

Let $p_n$ be the probability of reaching $4$ from position $n,$ with $p_4 = 1.$ The rules give $p_0 = \tfrac12 p_1,$ $p_1 = \tfrac14 p_0 + \tfrac14 p_2,$ $p_2 = \tfrac14 p_1 + \tfrac14 p_3,$ and $p_3 = \tfrac14 p_2 + \tfrac14.$ Work upward: $p_1 = \tfrac27 p_2$ and $p_2 = \tfrac{7}{26} p_3.$ These unwind to $p_3 = \tfrac{26}{97},$ $p_2 = \tfrac{7}{97},$ $p_1 = \tfrac{2}{97},$ and finally $p_0 = \tfrac{1}{97}.$ Therefore, the answer is E.

25.

正方形 $ABCD$ 的边长为 $4$ 。点 $P$ 和 $Q$ 分别在 $\overline{AD}$ 和 $\overline{CD}$ 上，且 $AP = \tfrac{8}{5}$ 、 $DQ = \tfrac{10}{3}$ 。一条路径从 $P$ 到 $Q$ 的线段开始，之后在正方形 $ABCD$ 的边上反射继续前进（入射角等于反射角），如下图所示。如果路径碰到正方形的顶点，就在那里终止；否则它将永远继续。

这条路径会在哪个顶点终止？

Square $ABCD$ has sides of length $4.$ Points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{CD},$ respectively, with $AP = \tfrac{8}{5}$ and $DQ = \tfrac{10}{3}.$ A path begins along the line segment from $P$ to $Q$ and continues by reflecting against the sides of $ABCD$ (with congruent incoming and outgoing angles), as shown in the figure. If the path hits a vertex of the square, then it terminates there; otherwise it continues forever.

At which vertex does the path terminate?

$A$

$B$

$C$

$D$

路径会永远继续下去。

The path continues forever.

答案：B

知识点：反射（几何）奇偶性最大公约数坐标几何

难度评级：2520

小提示：

处理反射路径最方便的方法是“展开”：在一张由反射正方形组成的网格中沿直线前进。

Reflecting billiard paths is easiest by “unfolding”: follow a straight line through a grid of reflected copies of the square

大提示：

找出这条直线第一次经过的网格顶点 $(4a, 4b)$ ； $a$ 和 $b$ 的奇偶性决定它对应原正方形的哪个顶点。

Find the first grid corner $(4a, 4b)$ the line hits; the parities of $a$ and $b$ determine which actual vertex it is

解答：

令 $A = (0,0)$ 、 $B = (4,0)$ 、 $C = (4,4)$ 、 $D = (0,4)$ ，则 $P = \left(0, \tfrac85\right)$ 、 $Q = \left(\tfrac{10}{3}, 4\right)$ 。初始方向为 $\left(\tfrac{10}{3}, \tfrac{12}{5}\right) \parallel (25, 18)$ 。把台球路径展开成由反射正方形拼成的网格，并从 $P$ 出发沿直线前进。在网格顶点 $(4a,4b)$ 处，由横坐标得 $t = \tfrac{4a}{25}$ 。于是纵坐标要求 $10 + 18a = 25b$ ，即 $a \equiv 5 \pmod{25}$ 。第一个正的可能值是 $a = 5$ ，由此得 $b = 4$ ，展开后的顶点为 $(20,16)$ 。横向穿过 $5$ 个小正方形（奇数），说明它落在 $x = 4$ 这条边上；纵向穿过 $4$ 个小正方形（偶数），说明它落在 $y = 0$ 上。这个顶点就是 $(4,0) = B$ 。因此正确答案是 B。

Place $A = (0,0),$ $B = (4,0),$ $C = (4,4),$ $D = (0,4),$ so $P = \left(0, \tfrac85\right)$ and $Q = \left(\tfrac{10}{3}, 4\right).$ The initial direction is $\left(\tfrac{10}{3}, \tfrac{12}{5}\right) \parallel (25, 18).$ Unfold the billiard into a grid of reflected copies and follow the straight line from $P.$ At a grid corner $(4a,4b),$ the horizontal coordinate gives $t = \tfrac{4a}{25}.$ The vertical coordinate then requires $10 + 18a = 25b,$ so $a \equiv 5 \pmod{25}.$ The first positive possibility is $a = 5,$ giving $b = 4$ and the unfolded corner $(20,16).$ Crossing $5$ cells across (odd) puts it on the side $x = 4,$ and $4$ cells up (even) puts it on $y = 0.$ That’s vertex $(4,0) = B.$ Thus, B is the correct answer.