2010B AMC 10 真题（P版）

1 答案 C

第一项为 \[\begin{aligned}100(100 - 3) &= 100 \cdot 97 \\&= 9700\end{aligned}\text{，}\]而第二项为 \[\begin{aligned}100 \cdot 100 - 3 &= 10000 - 3\\ &= 9997\end{aligned}\text{。}\]

因此差为 \[ 9700 - 9997 = - 297\text{。}\]

所以正确答案是 C。

English

We have that \[ \begin{aligned}100(100 - 3) &= 100 \cdot 97 \\&= 9700\end{aligned} \] and \[ \begin{aligned}100 \cdot 100 - 3 &= 10000 - 3\\ &= 9997.\end{aligned} \]

As such, their difference is: \[ 9700 - 9997 = - 297. \]

Thus, C is the correct answer.

2 答案 C

\(45\) 分钟等于 \[\dfrac{45}{60} = \dfrac{3}{4}\] 小时。第二场会议用时为 \[2 \cdot \dfrac{3}{4} = \dfrac{3}{2}\] 小时，所以两场会议共用 \[\dfrac{3}{4} + \dfrac{3}{2} = \dfrac{9}{4}\] 小时。这占整个工作日的 \[100 \% \cdot \dfrac{\frac{9}{4}}{9} = 100 \% \cdot \dfrac{1}{4} = 25 \%\text{。}\]

所以正确答案是 C。

English

Note that \(45\) minutes is \[ \dfrac{45}{60} = \dfrac{3}{4} \] hours. The second meeting is then \[ 2 \cdot \dfrac{3}{4} = \dfrac{3}{2} \] hours. Both meetings take a total of \[ \dfrac{3}{4} + \dfrac{3}{2} = \dfrac{9}{4} \] hours then. The percent of Makayla’s work day spent attending meetings is \[ 100 \% \cdot \dfrac{\frac{9}{4}}{9} = 100 \% \cdot \dfrac{1}{4} = 25 \%. \]

Thus, C is the correct answer.

3 答案 C

为了尽量延迟出现同色一双，最多可以先取出每种颜色各一只袜子。

一共有 \(4\) 种颜色，因此取出四只时仍可能没有同色一双。

再取一只就必然与其中一种颜色配成一双，所以需要 \(4 + 1 = 5\) 只。

所以正确答案是 C。

English

To maximize the number of socks, we want to grab as many single socks as possible before getting a pair.

There are \(4\) colors, which means that we can draw one sock of each color before drawing a pair.

This means that it takes at least \(4 + 1 = 5\) socks to be drawn before a pair is guaranteed.

Thus, C is the correct answer.

4 答案 C

依次计算得 \[\heartsuit(1) = \dfrac{1 + 1^2}{2} = 1\text{，}\]\[\heartsuit(2) = \dfrac{2 + 2^2}{2} = 3\text{，}\]以及 \[\heartsuit(3) = \dfrac{3 + 3^2}{2} = 6\text{。}\]

因此总和为 \[ 1 + 3 + 6 = 10\text{。}\]

所以正确答案是 C。

English

We have \[ \heartsuit(1) = \dfrac{1 + 1^2}{2} = 1, \] \[ \heartsuit(2) = \dfrac{2 + 2^2}{2} = 3, \] and \[ \heartsuit(3) = \dfrac{3 + 3^2}{2} = 6. \]

Then \[ 1 + 3 + 6 = 10. \]

Thus, C is the correct answer.

5 答案 B

注意 \(31\) 除以 \(7\) 时余数为 \(3\)。

因此，如果这个月从星期六、星期日、星期二或星期三开始，星期一和星期三的天数就不相等。

所以这个月只能从星期一、星期四或星期五开始。

所以正确答案是 B。

English

Note that \(31\) days leaves a remainder of \(3\) when divided by \(7.\)

This means that if the month starts on a Saturday, Sunday, Tuesday, or Wednesday, there will be an uneven number of Mondays and Wednesdays.

Then the month can only start on a Monday, Thursday, or Friday.

Thus, B is the correct answer.

6 答案 B

参考下图：

由于直径的两端与圆心在同一直线上，\[ \angle AOC = 180^{\circ} - 50^{\circ} = 130^{\circ}\text{。}\]

\(\triangle AOC\) 是等腰三角形，所以 \[ \angle CAO = \dfrac{180^{\circ} - 130^{\circ}}{2} = 25^{\circ}\text{。}\]

因为 \(\angle CAB = \angle CAO\)，所以 \(\angle CAB = 25^{\circ}\)。

所以正确答案是 B。

English

Consider the following diagram:

We have that \[ \angle AOC = 180^{\circ} - 50^{\circ} = 130^{\circ}. \]

Since \(\triangle AOC\) is isosceles, we have that \[ \angle CAO = \dfrac{180^{\circ} - 130^{\circ}}{2} = 25^{\circ}. \]

Since \(\angle CAB = \angle CAO,\) we have that \(\angle CAB = 25^{\circ}.\)

Thus, B is the correct answer.

7 答案 D

为求三角形的面积，可以向长为 \(12\) 的边作高。

于是得到一个直角三角形，它的一条直角边为 \(12 \div 2 = 6\)，斜边为 \(10\)。

另一条直角边的长度为 \[ \sqrt{10^2 - 6^2} = \sqrt{64} = 8\text{。}\]

三角形面积为 \[ \dfrac{8 \cdot 12}{2} = 48\text{。}\]

于是长方形的长为 \(48 \div 4 = 12\)。它的周长为 \[ 2(4 + 12) = 2 \cdot 16 = 32\text{。}\]

所以正确答案是 D。

English

To find the area of the triangle, we can drop the altitude to the side of length \(12.\)

Then we have a right triangle with one leg \(12 \div 2 = 6\) and hypotenuse \(10.\)

The other leg has length \[ \sqrt{10^2 - 6^2} = \sqrt{64} = 8. \]

The area of the triangle is then \[ \dfrac{8 \cdot 12}{2} = 48. \]

The length of the rectangle is then \(48 \div 4 = 12.\) Its perimeter is \[ 2(4 + 12) = 2 \cdot 16 = 32. \]

Thus, D is the correct answer.

8 答案 E

\(x\) 必须同时整除 \(48\) 和 \(64\)，因此必须整除它们的最大公因数。

\(48\) 和 \(64\) 的最大公因数是 \(16\)，所以 \(x\) 必须整除 \(16\)。

\(16\) 有 \(5\) 个正因数，即 \(2\) 的不超过 \(16\) 的各个幂。

所以正确答案是 E。

English

Note that \(x\) must divide both \(48\) and \(64,\) which means that it must divide their greatest common divisor.

The greatest common divisor of \(48\) and \(64\) is \(16,\) which means \(x\) divides \(16.\)

\(16\) has \(5\) factors, namely all the powers of \(2\) up to \(16.\)

Thus, E is the correct answer.

9 答案 D

忽略括号时，Larry 会得到 \[ 1 - 2 - 3 - 4 + e = e - 8\text{。}\]

按括号正确计算，\[\begin{aligned} &1 - (2 -(3 - (4 + e)))\\ & = 1 - (2 - (-1 - e))\\ &= 1 - (3 + e) \\ &=-2 - e\end{aligned}\text{。}\]

两者相等，所以 \[ e - 8 = -2 - e \] \[ e = 3\text{。}\]

所以正确答案是 D。

English

Ignoring the parentheses, Larry would get \[ 1 - 2 - 3 - 4 + e = e - 8. \]

Evaluating with the parentheses, one would get \[\begin{aligned} &1 - (2 -(3 - (4 + e)))\\ & = 1 - (2 - (-1 - e))\\ &= 1 - (3 + e) \\ &=-2 - e.\end{aligned}\]

Both of these values are the same, so \[ e - 8 = -2 - e \]\[ e = 3. \]

Thus, D is the correct answer.

10 答案 C

\(40\) 分钟是 \(\frac{2}{3}\) 小时。设 Shelby 在晴天中骑了 \(h\) 小时。

则雨中骑了 \(\frac{2}{3} - h\) 小时，所以总路程为 \[30h + 20\left(\dfrac{2}{3} - h\right) = 10h + \dfrac{40}{3}\] 英里。这个值等于 \(16\)，因此 \[10h + \dfrac{40}{3} = 16\] \[h = \dfrac{4}{15}\text{。}\]

因此雨中骑行时间为 \[60\left(\dfrac{2}{3} - \dfrac{4}{15}\right) = 60 \cdot \dfrac{2}{5} = 24\] 分钟。

所以正确答案是 C。

English

Note that \(40\) minutes is \(\frac{2}{3}\) hours. Let Shelby drive \(h\) hours in the sun.

Then she drives \(\frac{2}{3} - h\) hours in the rain, which means she travels a total of \[ 30h + 20\left(\dfrac{2}{3} - h\right) = 10h + \dfrac{40}{3} \] miles. We have this equals \(16,\) so \[ 10h + \dfrac{40}{3} = 16 \]\[ h = \dfrac{4}{15}. \]

Then Shelby drives \[ 60\left(\dfrac{2}{3} - \dfrac{4}{15}\right) = 60 \cdot \dfrac{2}{5} = 24 \] minutes in the rain.

Thus, C is the correct answer.

11 答案 A

设商品标价为 \(p\)。优惠券 A 节省 \(0.15p\)，优惠券 B 节省 \(\$ 30\)。

优惠券 C 节省 \[ 0.25(p - 100) = 0.25p - 25\text{。}\]

因此必须满足 \[0.15p \ge 30\] \[p \geq 200\] 以及 \[0.15p \ge 0.25p - 25\] \[250 \geq p\text{。}\]

这给出 \(x = 200\)、\(y = 250\)，所以 \(y - x = 50\)。

所以正确答案是 A。

English

Let \(p\) be the price of the item. Then coupon A saves \(0.15p.\) Coupon B saves \(\$ 30.\)

Coupon C will save \[ 0.25(p - 100) = 0.25p - 25. \]

We must have that \[ 0.15p \ge 30 \]\[ p \geq 200 \] and \[ 0.15p \ge 0.25p - 25 \]\[ 250 \geq p. \]

This shows that \(x = 200\) and \(y = 250.\) Therefore \(y - x = 50.\)

Thus, A is the correct answer.

12 答案 D

为了最小化 \(x\)，尽量让学生保持原答案。

由于回答“是”的比例从百分之五十增到百分之七十，至少有 \(70 - 50 = 20\) 个百分点所对应的学生必须改变答案。

为了最大化 \(x\)，可以让所有原来回答“否”的学生都改成“是”，但最终仍需要百分之七十回答“是”。

这要求原来回答“是”的学生中有 \(70 - 50 = 20\) 个百分点仍回答“是”，所以原来回答“是”的百分之五十中，最多有 \(50 - 20 = 30\) 个百分点可以改变。

因此最多有 \(50 + 30 = 80\) 个百分点的学生改变答案。最大值与最小值之差为 \(80 - 20 = 60\)。

所以正确答案是 D。

English

To minimize \(x,\) we want to have as many kids as possible maintain their answer.

We then need at least \(70 - 50 = 20\) percent of the students to change their answer.

To maximize \(x,\) we can have everybody that answered no change their answer, but some who answered yes must stay yes.

We need \(70 - 50 = 20\) percent of the people who said yes to stay yes, which means only \(50 - 20 = 30\) percent can switch.

This makes the maximum percent of students that can switch \(50 + 30 = 80\) percent. The difference is then \(80 - 20 = 60.\)

Thus, D is the correct answer.

13 答案 C

先处理外层绝对值。有两种可能：\[x = 2x - |60 - 2x|\] 或 \[x = -2x + |60 - 2x|\text{。}\]

这两个方程分别化为 \[x = |60 - 2x|\] 和 \[3x = |60 - 2x|\text{。}\]

每个方程都有 \(2\) 种情况。第一个方程给出 \[x = 60 - 2x\] 和 \[x = 2x - 60\text{。}\]

分别求解得 \[3x = 60\] \[x = 20\] 以及 \[-x = -60\] \[x = 60\text{。}\]

第二个方程给出 \[3x = 60 - 2x\] 和 \[3x = 2x - 60\text{。}\]

再次分别求解，得 \[5x = 60\] \[x = 12\] 和 \(x = -60\)。由于原方程右边非负，\(x\) 不能为负数。

所有解之和为 \[ 20 + 60 + 12 = 92\text{。}\]

所以正确答案是 C。

English

We first take care of the outer absolute value. We have either \[ x = 2x - |60 - 2x| \] or \[ x = -2x + |60 - 2x|. \]

These simplify to \[ x = |60 - 2x| \] and \[ 3x = |60 - 2x|. \]

We have \(2\) cases for each equation. For the first one, we have \[x = 60 - 2x\] and \[x = 2x - 60.\]

Solving both gives us \[ 3x = 60 \]\[ x = 20 \] and \[ -x = -60 \]\[ x = 60. \]

For the other equation, we have \[3x = 60 - 2x\] and \[3x = 2x - 60.\]

Again solving both, we have \[ 5x = 60 \]\[ x = 12 \] and \(x = -60.\) Note that \(x\) cannot be negative since the right side of the original equation is nonnegative.

Adding up all the solutions gives us \[ 20 + 60 + 12 = 92. \]

Thus, C is the correct answer.

14 答案 B

前 \(n\) 个正整数的和为 \(\dfrac{n(n + 1)}{2}\)。

因此有 \[\dfrac{\frac{99 \cdot 100}{2} + x}{100} = 100x\text{，}\]化简得 \[99 \cdot 50 = (100^2 - 1)x\] \[= 101 \cdot 99x\text{，}\]所以 \(x = \dfrac{50}{101}\)。

所以正确答案是 B。

English

Recall that the sum of the first \(n\) integers is \(\dfrac{n(n + 1)}{2}.\)

Then, we have that \[ \dfrac{\frac{99 \cdot 100}{2} + x}{100} = 100x, \] which simplifies to \[ 99 \cdot 50 = (100^2 - 1)x\]\[ = 101 \cdot 99x, \] by difference of squares. Dividing gives us \(x = \dfrac{50}{101}.\)

Thus, B is the correct answer.

15 答案 C

设 Jesse 答对 \(x\) 题，答错 \(y\) 题。

因此必须同时满足 \[4x-y=99\] 和 \[x+y\leq50\text{。}\]

我们有 \[y = 4x - 99\]\[5x - 99 \leq 50\text{。}\]

整理得 \(x \leq 29.8\)。因为 \(x\) 是整数，所以最大值为 \(29\)。

所以正确答案是 C。

English

Let \(x\) be the number of questions Jesse answered correctly and \(y\) be the number he answered incorrectly.

Then \[4x-y=99\] and \[x+y\leq50.\]

We have that \[ y = 4x - 99 \]\[ 5x - 99 \leq 50. \]

Rearranging and simplifying tells us that \(x \leq 29.8.\) Since \(x\) is an integer, its maximum value is \(29.\)

Thus, C is the correct answer.

16 答案 B

从 \(O\) 向 \(\overline{AB}\) 作高，垂足为 \(X\)。

由边长关系，\(\triangle OBX\) 是 \(30-60-90\) 特殊直角三角形。

扇形 \(AOB\) 的面积为 \[ \dfrac{1}{6} \cdot \pi \cdot \left(\dfrac{\sqrt3}{3}\right)^2 = \dfrac{\pi}{18}\text{。}\]

\(\triangle AOB\) 的面积为 \[ \dfrac{1}{2} \cdot 2 \cdot \dfrac{\sqrt3}{6} \cdot \dfrac{1}{2} = \dfrac{\sqrt3}{12}\text{。}\]

因此该边外、圆内的一个圆弓形面积为 \[ \dfrac{\pi}{18} - \dfrac{\sqrt3}{12}\text{。}\]

这样的区域有 \(4\) 个，总面积为 \[ 4\left(\dfrac{\pi}{18} - \dfrac{\sqrt3}{12}\right) = \dfrac{2\pi}{9} - \dfrac{\sqrt3}{3}\text{。}\]

所以正确答案是 B。

English

Drop the altitude from \(O\) to \(\overline{AB}\) at \(X.\)

Looking at the side lengths, we see that \(\triangle OBX\) is a \(30-60-90\) triangle.

This means that the area of sector \(AOB\) is \[ \dfrac{1}{6} \cdot \pi \cdot \left(\dfrac{\sqrt3}{3}\right)^2 = \dfrac{\pi}{18}. \]

We also have that the area of \(\triangle AOB\) is \[ \dfrac{1}{2} \cdot 2 \cdot \dfrac{\sqrt3}{6} \cdot \dfrac{1}{2} = \dfrac{\sqrt3}{12}. \]

The area of the sector outside the square and inside the circle is then \[ \dfrac{\pi}{18} - \dfrac{\sqrt3}{12}. \]

There are \(4\) of these regions, which gives us a total area of \[ 4\left(\dfrac{\pi}{18} - \dfrac{\sqrt3}{12}\right) = \dfrac{2\pi}{9} - \dfrac{\sqrt3}{3}. \]

Thus, B is the correct answer.

17 答案 B

设有 \(n\) 所学校，则比赛中共有 \(3n\) 名参赛者。

因为中位数是某一名学生的分数，\(3n\) 是奇数，所以 \(n\) 是奇数。Carla 排第 \(64\) 要求 \(3n\ge64\)，因此奇数 \(n\) 至少为 \(23\)。

Andrea 的中位名次是 \(\dfrac{3n+1}{2}\)。因为她是队内最高分，所以名次在 Beth 之前，故 \(\dfrac{3n+1}{2}\lt37\)。这给出 \(3n\lt73\)，所以奇数 \(n\) 至多为 \(23\)。

因此共有 \(23\) 所学校。

所以正确答案是 B。

English

Let there be \(n\) schools. Then there are \(3n\) participants in the contest.

Because the median is one student’s score, \(3n\) is odd, so \(n\) is odd. Carla’s \(64\)th place requires \(3n\ge64,\) and hence the odd integer \(n\) is at least \(23.\)

Andrea’s median place is \(\dfrac{3n+1}{2}.\) Because she had the highest score on her team, she finished ahead of Beth, so \(\dfrac{3n+1}{2}\lt37.\) This gives \(3n\lt73,\) and therefore the odd integer \(n\) is at most \(23.\)

Thus there are \(23\) schools.

Thus, B is the correct answer.

18 答案 E

注意 \[abc + ab + a = a(bc + b + 1)\text{。}\]因此若 \(a\) 能被 \(3\) 整除，整个表达式也能被它整除。

因为 \(2010\) 能被 \(3\) 整除，\(a\) 能被 \(3\) 整除的概率为 \(\frac{1}{3}\)。

现在考虑 \(a\) 不能被 \(3\) 整除的情况。要使表达式能被 \(3\) 整除，必须有 \(bc + b + 1\) 能被 \(3\) 整除。

这意味着 \[ bc + b = b(c + 1) \equiv 2 \pmod{3}\text{。}\]

唯一可能是一个因子为 \(2\)（模 \(3\)），另一个因子为 \(1\)（模 \(3\)）。

两种次序中的每一种，两个因子各有 \(\frac{1}{3}\) 的概率得到所需余数，因此概率为 \[\dfrac{1}{3} \cdot \dfrac{1}{3} = \dfrac{1}{9}\text{。}\]

共有两种次序，所以这种情况的概率为 \(\dfrac{2}{9}\)。

总概率为 \[ \dfrac{1}{3} \cdot 1 + \dfrac{2}{3} \cdot \dfrac{2}{9} = \dfrac{13}{27}\text{。}\]

所以正确答案是 E。

English

Note that \[ abc + ab + a = a(bc + b + 1). \] This means that if \(a\) is divisible by \(3,\) the whole expression is as well.

Since \(2010\) is divisible by \(3,\) we have that \(a\) is divisible by \(3\) with probability \(\frac{1}{3}.\)

Now consider \(a\) not divisible by \(3.\) For the expression to be divisible by \(3,\) we must have that \(bc + b + 1\) is divisible by \(3.\)

This means that \[ bc + b = b(c + 1) \equiv 2 \pmod{3}. \]

The only possibility for this is that one of the factors is \(2\) mod \(3\) and the other is \(1\) mod \(3.\)

For each of the two cases, there is a \(\frac{1}{3}\) chance that each of the factors is the desired modulus, for a probability of \[\dfrac{1}{3} \cdot \dfrac{1}{3} = \dfrac{1}{9}.\]

There are two cases, which means that this happens with a \(\dfrac{2}{9}\) probability.

The total probability is then \[ \dfrac{1}{3} \cdot 1 + \dfrac{2}{3} \cdot \dfrac{2}{9} = \dfrac{13}{27}. \]

Thus, E is the correct answer.

19 答案 B

参考下图：

由圆面积公式可知 \(BO = \sqrt{156}\)，因为它是半径。

延长 \(\overline{AO}\) 与 \(\overline{BC}\) 交于 \(X\)。设 \(s\) 为 \(\triangle ABC\) 的边长。

于是 \[BX = \dfrac{s}{2}\] 且 \[AX = \dfrac{s\sqrt3}{2}\text{。}\]

\(\triangle OXB\) 是直角三角形，所以 \[ (\sqrt{156})^2 = \left(\dfrac{s}{2}\right)^2\] \[ + \left(\dfrac{s\sqrt3}{2} + 4\sqrt3\right)^2\text{。}\]

化简得 \[ 156 = s^2 + 12s + 48 \] \[ s^2 + 12s - 108 = 0 \] \[ (s - 6)(s + 18) = 0\text{。}\]

由于 \(s\) 为正，\(s = 6\)。

所以正确答案是 B。

English

Consider the following diagram:

Using the formula for the area of a circle, we have that \(BO = \sqrt{156}\) since it is a radius.

Extend \(\overline{AO}\) to intersect \(\overline{BC}\) at \(X.\) Let \(s\) be the side length of \(\triangle ABC.\)

Then we have that \[ BX = \dfrac{s}{2} \] and \[ AX = \dfrac{s\sqrt3}{2}. \]

We have that \(\triangle OXB\) is right, which means that we can apply the Pythagorean Theorem. This gives us \[ (\sqrt{156})^2 = \left(\dfrac{s}{2}\right)^2\]\[ + \left(\dfrac{s\sqrt3}{2} + 4\sqrt3\right)^2. \]

Simplifying, we get \[ 156 = s^2 + 12s + 48 \] \[ s^2 + 12s - 108 = 0 \] \[ (s - 6)(s + 18) = 0. \]

Since \(s\) is positive, we must have that \(s = 6.\)

Thus, B is the correct answer.

20 答案 D

参考下图：

设正六边形边长为 \(1\)。较小圆内切于一个边长为 \(1\) 的等边三角形。

这个等边三角形的内切圆半径为 \(\dfrac{\sqrt3}{6}\)。面积为 \[ \pi \cdot \left(\dfrac{\sqrt3}{6}\right)^2 = \dfrac{\pi}{12}\text{。}\]

设较大圆圆心为 \(O\)，从 \(O\) 向 \(\overline{GH}\) 作垂线，垂足为 \(J\)。并连接 \(\overline{OG}\)。

\(\triangle OJG\) 是直角三角形，且由 \(\angle HGI = 60^{\circ}\) 可知，\(\triangle OJG\) 是 \(30-60-90\) 三角形。

设 \(OJ = r\)，则 \(OG = 2r\)。又 \(OG\) 等于正六边形的高、等边三角形的高与圆半径之和。

于是 \[ OG = \dfrac{\sqrt3}{2} + \sqrt3 + r\text{。}\]

代入 \(OG\)，得 \[2r = \dfrac{\sqrt3}{2} + \sqrt3 + r\text{。}\]化简得 \[r = \dfrac{3\sqrt3}{2}\text{。}\]

较大圆面积为 \[ \pi \cdot \left(\dfrac{3\sqrt3}{2}\right)^2 = \dfrac{27}{4}\pi\text{。}\]

所求比值为 \[ \dfrac{\frac{27\pi}{4}}{\frac{\pi}{12}} = 81\text{。}\]

所以正确答案是 D。

English

Consider the following diagram:

Assume the regular hexagon has side length \(1.\) The smaller circle is inscribed in an equilateral triangle of side length \(1.\)

The inradius of this equilateral triangle is \(\dfrac{\sqrt3}{6}.\) The area of the circle is then \[ \pi \cdot \left(\dfrac{\sqrt3}{6}\right)^2 = \dfrac{\pi}{12}. \]

Let \(O\) be the center of the larger circle. Drop the perpendicular from \(O\) to \(\overline{GH}\) at \(J.\) Draw \(\overline{OG}.\)

We have that \(\triangle OJG\) is right. Since \(\angle HGI = 60^{\circ},\) we also have that \(\triangle OJG\) is a \(30-60-90\) triangle.

Let \(OJ = r.\) Then \(OG = 2r.\) We also have that \(OG\) is the sum of the height of the hexagon, equilateral triangle, and radius of the circle.

Then \[ OG = \dfrac{\sqrt3}{2} + \sqrt3 + r. \]

Substituting in \(OG,\) we get \[ 2r = \dfrac{\sqrt3}{2} + \sqrt3 + r. \] Simplifying gives us \[ r = \dfrac{3\sqrt3}{2}. \]

The area of the larger circle is then \[ \pi \cdot \left(\dfrac{3\sqrt3}{2}\right)^2 = \dfrac{27}{4}\pi. \]

The desired ratio is then \[ \dfrac{\frac{27\pi}{4}}{\frac{\pi}{12}} = 81. \]

Thus, D is the correct answer.

21 答案 E

任意一个 \(4\) 位数都可写成 \(abcd\)，展开后为 \[10^3a + 10^2b + 10c + d\text{。}\]对回文数，\(a = d\) 且 \(b = c\)，所以可化为 \[1001a + 110b\text{。}\]因为 \(1001\) 能被 \(7\) 整除，所以还需要 \(110b\) 能被 \(7\) 整除。

只有 \(b\) 为 \(0\) 或 \(7\) 时才可能成立，因为 \(110\) 不能被 \(7\) 整除。

有 \(9\) 种 \(a\) 和 \(2\) 种 \(b\)，符合条件的回文数共有 \(9 \cdot 2 = 18\) 个。

四位回文数总数为 \(9 \cdot 10\)，因为千位有 \(9\) 种选择、百位有 \(10\) 种选择。

所求概率为 \[ \dfrac{18}{9 \cdot 10} = \dfrac{1}{5}\text{。}\]

所以正确答案是 E。

English

Note that we can express any \(4\) digit number as \(abcd.\) This can be expressed in long form as \[ 10^3a + 10^2b + 10c + d. \] Since in a palindrome, we have that \(a = d\) and \(b = c.\) We can simplify this to get \[ 1001a + 110b. \] Note that \(1001\) is divisible by \(7.\) This means that \(110b\) must also be divisible by \(7.\)

The only way for this to happen is if \(b\) is \(0\) or \(7\) since \(110\) is not divisible by \(7.\)

There are \(9\) options for \(a\) and \(2\) options for \(b,\) for a total of \(9 \cdot 2 = 18\) palindromes.

The total number of palindromes is \(9 \cdot 10\) since there are \(9\) options for the thousands digit and \(10\) options for the hundreds digit.

The desired probability is then \[ \dfrac{18}{9 \cdot 10} = \dfrac{1}{5}. \]

Thus, E is the correct answer.

22 答案 C

可以用补集计数。无限制地分配糖果的方法总数为 \[ 3^7 = 2187\text{。}\]

要数无效安排，需要考虑红袋或蓝袋为空的情况。

若红袋为空，则每块糖只有 \(2\) 个袋子可选，共 \[2^7 = 128\] 种安排。蓝袋为空也同样有 \(128\) 种。

这两类有一个重叠情况：红袋和蓝袋都为空。因此有效分法的数量为 \[ 2187 - (128 + 128 - 1) = 1932\text{。}\]

所以正确答案是 C。

English

We can count this with complementary counting. The total number of ways to distribute the candies with no restrictions is \[ 3^7 = 2187. \]

To find the number of invalid arrangements, we have to count the number of ways where either the red or blue bag is empty.

For the case where the red bag is empty, each candy has \(2\) options for the bag that goes into. There are then \[ 2^7 = 128 \] arrangements for this case. Similarly, there are \(128\) arrangements for the case where the blue bag is empty.

There is an overlap of one case where both bags are empty. The final answer is then \[ 2187 - (128 + 128 - 1) = 1932. \]

Thus, C is the correct answer.

23 答案 D

设 \(a_{ij}\) 表示第 \(i\) 行、第 \(j\) 列的数。递增条件迫使 \(a_{11}=1\)、\(a_{33}=9\)，而 \(a_{22}\) 只能是 \(4\)、\(5\) 或 \(6\)。

若 \(a_{22}=4\)，则 \(\{a_{12},a_{21}\}=\{2,3\}\)。从 \(5\)、\(6\)、\(7\) 和 \(8\) 中选两个放入左下方的一对位置 \(\{a_{31},a_{32}\}\)；另外两个放入右上方的一对位置 \(\{a_{13},a_{23}\}\)。每一对都只有一种递增排列。因此共有 \(2\binom42=12\) 个数阵：左上角周围的 \(2\)、\(3\) 有两种次序，左下方的一对有 \(\binom42\) 种选择。把各数反向并旋转数阵，同样有 \(12\) 个中心为 \(6\) 的数阵。

若 \(a_{22}=5\)，选择位置 \(a_{12}\)、\(a_{13}\) 和 \(a_{23}\) 中的三个数。它们可以是 \(\{2,3,4,6,7,8\}\) 的任意三元子集，但不能是 \(\{2,3,4\}\) 或 \(\{6,7,8\}\)；因为这两种选择会把三个小数或三个大数全放在中心的一侧，从而违反与中心数的必要大小关系。其余每一种选择都唯一确定剩余各数及其递增次序。因此共有 \(\binom63-2=18\) 个数阵。

总数为 \(12+18+12=42\)。

所以正确答案是 D。

English

Let \(a_{ij}\) be the entry in row \(i\) and column \(j.\) The increasing conditions force \(a_{11}=1,\) \(a_{33}=9,\) and \(a_{22}\) to be \(4,\) \(5,\) or \(6.\)

If \(a_{22}=4,\) then \(\{a_{12},a_{21}\}=\{2,3\}.\) Choose which two of \(5,\) \(6,\) \(7,\) and \(8\) go in the bottom-left pair \(\{a_{31},a_{32}\};\) the other two go in the top-right pair \(\{a_{13},a_{23}\}.\) Each pair then has only one increasing order. There are \(2\binom42=12\) arrays: two orders for \(2,\) \(3\) around the upper-left corner and \(\binom42\) choices for the bottom-left pair. By reversing the digits and rotating the array, there are also \(12\) arrays with center \(6.\)

If \(a_{22}=5,\) choose the three entries in positions \(a_{12},\) \(a_{13},\) and \(a_{23}.\) They can be any three of \(\{2,3,4,6,7,8\}\) except \(\{2,3,4\}\) or \(\{6,7,8\};\) those two choices would put all three small or all three large entries on one side and violate a required comparison with the center. Every other choice uniquely determines the remaining entries and their increasing orders. This gives \(\binom63-2=18\) arrays.

The total is \(12+18+12=42.\)

Thus, D is the correct answer.

24 答案 E

设两队第一节得分均为 \(a\)。设 Raiders 的公比为 \(r\)，Wildcats 的公差为 \(d\)。

将 \(r=\frac{m}{n}\) 写成最简分数。由于 Raiders 四节得分都是整数，\(a=n^3A\)，其中 \(A\) 为正整数，且总分为 \(A(n^3+n^2m+nm^2+m^3)\le100\)，所以只需检查 \(\dfrac{3}{2},2,3,4\)。

对 \(r=\dfrac{3}{2}\) 或 \(r=4\)，Raiders 的可行总分无法使 Wildcats 总分成为 \(4a+6d\) 的形式。对 \(r=3\)，方程 \(40a=4a+6d+1\) 模 \(6\) 不可能。

对 \(r=2\)，方程为 \(15a=4a+6d+1\)，所以 \(11a=6d+1\)。总分不超过 \(100\) 的唯一正整数解是 \(a=5,d=9\)。

上半场总分为 \(5+10+5+14=34\)。

所以正确答案是 E。

English

Let the first-quarter score for each team be \(a.\) Let the Raiders have common ratio \(r\) and let the Wildcats have common difference \(d.\)

Write \(r=\frac{m}{n}\) in lowest terms. Since the Raiders’ four quarter scores are integers, \(a=n^3A\) for some positive integer \(A\), and their total is \(A(n^3+n^2m+nm^2+m^3)\le100\). Thus the only possible ratios are \(\dfrac{3}{2},2,3,4\).

For \(r=\dfrac{3}{2}\) or \(r=4\), the only possible Raiders totals give Wildcats totals that are not of the form \(4a+6d\). For \(r=3\), the equation \(40a=4a+6d+1\) is impossible modulo \(6\).

For \(r=2\), the equation is \(15a=4a+6d+1\), so \(11a=6d+1\). The only positive solution with total at most \(100\) is \(a=5,d=9\).

The first-half total is \(5+10+5+14=34.\)

Thus, E is the correct answer.

25 答案 B

因为 \(1\)、\(3\)、\(5\) 和 \(7\) 是 \(P(x)-a\) 的根，可写成 \(P(x)-a=(x-1)(x-3)\) \(\cdot(x-5)(x-7)Q(x)\)，其中 \(Q(x)\) 也有整数系数。

将 \(2\)、\(4\)、\(6\) 和 \(8\) 代入 \(x\)，得到 \(-2a=-15Q(2)\) \(=9Q(4)\) \(=-15Q(6)\) \(=105Q(8)\)。因此 \(a\) 必须是 \(\operatorname{lcm}(15,9,105)=315\) 的倍数。

这个下界可以达到：取 \(Q(x)=42\) \(+(x-2)(x-6)(60-8x)\)，并定义 \(P(x)=315\) \(+(x-1)(x-3)\) \(\cdot(x-5)(x-7)Q(x)\)。该多项式有整数系数并满足所有要求。

所以正确答案是 B。

English

Because \(1,\) \(3,\) \(5,\) and \(7\) are roots of \(P(x)-a\), write \(P(x)-a=(x-1)(x-3)\) \(\cdot(x-5)(x-7)Q(x)\), where \(Q(x)\) has integer coefficients.

Substituting \(2,\) \(4,\) \(6,\) and \(8\) for \(x\) gives \(-2a=-15Q(2)\) \(=9Q(4)\) \(=-15Q(6)\) \(=105Q(8)\). Hence \(a\) must be a multiple of \(\operatorname{lcm}(15,9,105)=315\).

This lower bound is attainable: take \(Q(x)=42\) \(+(x-2)(x-6)(60-8x)\) and define \(P(x)=315\) \(+(x-1)(x-3)\) \(\cdot(x-5)(x-7)Q(x)\). This polynomial has integer coefficients and satisfies the required values.

Thus, B is the correct answer.