2022A AMC 10 真题（P版）

1 答案 D

这个连分数可以逐层化简如下：

\[\begin{aligned} 3 +& \dfrac{1}{3 + \dfrac{1}{3 + \dfrac{1}{3}}} \\ &= 3 + \dfrac{1}{3 + \dfrac{1}{\dfrac{10}{3}}} \\ &= 3 + \dfrac{1}{3 + \dfrac{3}{10}} \\ &= 3 + \dfrac{1}{\dfrac{33}{10}} \\ &= 3 + \dfrac{10}{33} \\ &= \dfrac{109}{33} \end{aligned}\text{。}\]

所以正确答案是 D。

English

We can simplify this expression as follows:

\[\begin{aligned} 3 +& \dfrac{1}{3 + \dfrac{1}{3 + \dfrac{1}{3}}} \\ &= 3 + \dfrac{1}{3 + \dfrac{1}{\dfrac{10}{3}}} \\ &= 3 + \dfrac{1}{3 + \dfrac{3}{10}} \\ &= 3 + \dfrac{1}{\dfrac{33}{10}} \\ &= 3 + \dfrac{10}{33} \\ &= \dfrac{109}{33}. \end{aligned}\]

Thus, D is the correct answer.

2 答案 B

我们可以列一个比例来解决这个问题：

\[ \dfrac{15}{57} = \dfrac{x}{27}\text{。}\]

交叉相乘，得到 \[ x = \dfrac{15}{57} \cdot 27 = \dfrac{135}{19} \approx 7\text{。}\]

所以正确答案是 B。

English

We can set up a proportion to solve this problem:

\[ \dfrac{15}{57} = \dfrac{x}{27}. \]

Cross multiplying, we get \[ x = \dfrac{15}{57} \cdot 27 = \dfrac{135}{19} \approx 7. \]

Thus, B is the correct answer.

3 答案 E

设这三个数分别为 \(x, y\) 和 \(z\)。题目中的条件给出下面的关系：

\[\begin{aligned} x+y+z&=96 &&\text{(1)} \\ x&=6z &&\text{(2)} \\ z&=y-40 &&\text{(3)} \end{aligned}\text{。}\]

把 \((3)\) 变形，得到 \(y = z + 40\)。再把这个新等式和 \((2)\) 代入 \((1)\)，得到 \[ 6z + z + 40 + z = 96 \] \[ 8z + 40 = 96 \] \[ 8z = 56 \Rightarrow z = 7\text{。}\]

由此得到 \[ x = 6 \cdot z = 6 \cdot 7 = 42 \] 以及 \[ y = z + 40 = 7 + 40 = 47\text{。}\]

因此 \(y - x = 47 - 42 = 5\)。

所以正确答案是 E。

English

Let \(x, y,\) and \(z\) be the three numbers. The conditions from the problem give us the following relations:

\[\begin{aligned} x+y+z&=96 &&\text{(1)} \\ x&=6z &&\text{(2)} \\ z&=y-40 &&\text{(3)}. \end{aligned}\]

Rearranging \((3),\) we get \(y = z + 40.\) Plugging this new equation and \((2)\) into \((1),\) we get \[ 6z + z + 40 + z = 96 \] \[ 8z + 40 = 96 \] \[ 8z = 56 \Rightarrow z = 7. \]

From this, we get that \[ x = 6 \cdot z = 6 \cdot 7 = 42 \] and \[ y = z + 40 = 7 + 40 = 47. \]

Therefore, \(y - x = 47 - 42 = 5.\)

Thus, E is the correct answer.

4 答案 E

每加仑行驶 \(x\) 英里，等价于每英里消耗 \(\frac{1}{x}\) 加仑。

因为 \(1\) 英里等于 \(\frac{1}{m}\) 千米，所以每千米耗油量为 \(\frac{m}{x}\) 加仑。

每加仑为 \(l\) 升，因此每千米消耗 \(\frac{lm}{x}\) 升；每 \(100\) 千米消耗 \(\frac{100lm}{x}\) 升。

所以正确答案是 E。

English

A car that gets \(x\) miles per gallon uses \(\frac{1}{x}\) gallons per mile.

Since \(1\) mile is \(\frac{1}{m}\) kilometers, this is \(\frac{m}{x}\) gallons per kilometer.

Multiplying by \(l\) liters per gallon gives \(\frac{lm}{x}\) liters per kilometer, so for \(100\) kilometers the fuel used is \(\frac{100lm}{x}\) liters.

Thus, E is the correct answer.

5 答案 C

参考图形：

由 \(AP = QC = s\)，可知 \(PB = BQ\)。因此 \(\triangle PBQ\) 是以 \(PQ=s\) 为斜边的等腰直角三角形。由勾股定理，\(PB = \dfrac{s}{\sqrt{2}}\)。

我们还知道 \[ 1 = AB = AP + PB = s + \dfrac{s}{\sqrt{2}}\text{。} \]

整理可得 \[ 1 = (1 + \dfrac{1}{\sqrt{2}})s \] 因而 \[ s = \dfrac{1}{1 + \dfrac{1}{\sqrt{2}}} = \dfrac{\sqrt{2}}{\sqrt{2} + 1}\text{。} \]

将这个分式有理化，得到

\[ \dfrac{\sqrt{2}}{\sqrt{2} + 1} \cdot \dfrac{\sqrt{2} - 1}{\sqrt{2} - 1} = 2 - \sqrt{2}\text{。} \]

所以正确答案是 C。

English

Consider the diagram:

Since \(AP = QC = s,\) we know that \(PB = BQ.\) This shows that \(\triangle PBQ\) is an isosceles right triangle with hypotenuse \(PQ=s.\) Using the Pythagorean theorem, we get that \(PB = \dfrac{s}{\sqrt{2}}.\)

We also know that \[ 1 = AB = AP + PB = s + \dfrac{s}{\sqrt{2}}. \]

This equation simplifies to \[ 1 = (1 + \dfrac{1}{\sqrt{2}})s \] Which implies that \[ s = \dfrac{1}{1 + \dfrac{1}{\sqrt{2}}} = \dfrac{\sqrt{2}}{\sqrt{2} + 1}. \]

We can rationalize this fraction to get

\[ \dfrac{\sqrt{2}}{\sqrt{2} + 1} \cdot \dfrac{\sqrt{2} - 1}{\sqrt{2} - 1} = 2 - \sqrt{2}. \]

Thus, C is the correct answer.

6 答案 A

根据平方根定义，\[\sqrt{(a - 1)^2} = |a - 1|\text{。}\]

因为 \(a < 0\)，所以 \(a - 1 < 0\)，从而 \(|a - 1| = 1 - a\)。

原式化为 \[ |a - 2 - (1 - a)| = |2a - 3|\text{。} \] 又因为 \(a < 0\)，所以 \(2a - 3 < 0\)。因此 \(|2a - 3| = 3 - 2a\)。

所以正确答案是 A。

English

By the definition of square root, we get that \[\sqrt{(a - 1)^2} = |a - 1|.\]

Since \(a < 0,\) we get that \(a - 1 < 0,\) which means that \(|a - 1| = 1 - a.\)

The whole expression therefore simplifies to \[ |a - 2 - (1 - a)| = |2a - 3|. \] Since \(a < 0,\) we know that \(2a - 3 < 0.\) This means that \(|2a - 3| = 3 - 2a.\)

Thus, A is the correct answer.

7 答案 B

分解质因数：\(18=2\cdot3^2\)，\(45=3^2\cdot5\)，\(180=2^2\cdot3^2\cdot5\)。

最小公倍数条件要求 \(n\) 中质因数 \(2\) 的指数为 \(2\)，质因数 \(5\) 的指数为 \(1\)，质因数 \(3\) 的指数至多为 \(2\)，其他质因数的指数均为 \(0\)。最大公因数条件 \(\gcd(n,45)=15=3\cdot5\) 进一步要求质因数 \(3\) 在 \(n\) 中的指数恰为 \(1\)。

因此 \(n=2^2\cdot3\cdot5=60\)，其各位数字之和为 \(6\)。

所以正确答案是 B。

English

Prime factorize \(18=2\cdot3^2,\) \(45=3^2\cdot5,\) and \(180=2^2\cdot3^2\cdot5.\)

The lcm condition forces the exponent of \(2\) in \(n\) to be \(2,\) the exponent of \(5\) to be \(1,\) the exponent of \(3\) to be at most \(2,\) and every other prime exponent to be \(0.\) The gcd condition \(\gcd(n,45)=15=3\cdot5\) further forces the exponent of \(3\) in \(n\) to be exactly \(1.\)

Therefore \(n=2^2\cdot3\cdot5=60,\) and the sum of its digits is \(6.\)

Thus, B is the correct answer.

8 答案 D

这 \(6\) 个数的平均数为 \[ \dfrac{1 + 7 + \cdots + X}{6} = \dfrac{20 + X}{6}\text{。}\]

这个值可以等于数据集中的任一项，因此分别讨论。

\[ \dfrac{20 + X}{6} = 1 \iff X = -14 \]

\[ \dfrac{20 + X}{6} = 7 \iff X = 22 \]

\[ \dfrac{20 + X}{6} = 5 \iff X = 10 \]

\[ \dfrac{20 + X}{6} = 2 \iff X = -8 \]

\[ \dfrac{20 + X}{6} = X \iff X = 4 \]

所有正的 \(X\) 值之和为 \(36\)。

所以正确答案是 D。

English

The average of the \(6\) numbers is \[ \dfrac{1 + 7 + \cdots + X}{6} = \dfrac{20 + X}{6}. \]

This value can equal any of the terms in the set, so we can case on what it equals.

\[ \dfrac{20 + X}{6} = 1 \iff X = -14 \]

\[ \dfrac{20 + X}{6} = 7 \iff X = 22 \]

\[ \dfrac{20 + X}{6} = 5 \iff X = 10 \]

\[ \dfrac{20 + X}{6} = 2 \iff X = -8 \]

\[ \dfrac{20 + X}{6} = X \iff X = 4 \]

Adding up all the positive values for \(X,\) we get \(36.\)

Thus, D is the correct answer.

9 答案 D

左下区域有 \(5\) 种颜色可选。左上区域与它相邻，因此有 \(4\) 种选择。中间下方区域同时与前两个区域相邻，因此有 \(3\) 种选择。

右上区域也与前面两个区域相邻，因此有 \(3\) 种选择。最后，右下区域同样有 \(3\) 种选择。

把这些选择相乘，得到 \(5\cdot4\cdot3\cdot3\cdot3=540\) 种涂色方法。

所以正确答案是 D。

English

There are \(5\) choices for the color of the bottom left rectangle. This forces there to be \(4\) choices for the top left rectangle. The middle bottom rectangle touches both of the previous ones, so there are \(3\) color options for this rectangle.

The rectangle in the top right is also limited to \(3\) colors since it touches the two previous rectangles. Finally, the rectangle in the bottom right also has \(3\) color options.

Multiplying these together, we get \(5\cdot4\cdot3\cdot3\cdot3=540\) total colorings.

Thus, D is the correct answer.

10 答案 E

如图，将原长方形的宽和高分别记为 \(a\) 和 \(b\)。于是有 \[a^2 + b^2 = 64\] 和 \[(a - 2)^2 + (b - 2)^2 = 32\text{。}\]

后一式化简为 \[ a^2 + b^2 - 4a - 4b + 4 + 4 = 32\text{，} \] 也就是 \[ 72 - 4(a + b) = 32\text{。} \] 因此 \[ a + b = 10\text{。} \]

将它平方，得到 \[ a^2 + b^2 + 2ab = 100\text{，} \] 从而 \[ 2ab = 36\text{，} \] 因此面积 \((ab)\) 为 \(18\)。

所以正确答案是 E。

English

We can label \(a\) and \(b\) as the width and height as in the diagram. Then we get that \[a^2 + b^2 = 64\] and \[(a - 2)^2 + (b - 2)^2 = 32.\]

The latter expression simplifies to \[ a^2 + b^2 - 4a - 4b + 4 + 4 = 32, \] which is the same as \[ 72 - 4(a + b) = 32. \] From this we get \[ a + b = 10. \]

Squaring this, we get \[ a^2 + b^2 + 2ab = 100, \] which gets us that \[ 2ab = 36, \] which means that the area \((ab)\) is \(18.\)

Thus, E is the correct answer.

11 答案 C

可将 \(4096\) 写成 \(2^{12}\)，所以 \(\dfrac{1}{4096} = 2^{-12}\)。令两个给定表达式相等，得到 \[ 2^m \cdot 2^{-6} = 2 \cdot 2^{\frac{-12}{m}}\text{。} \] 比较指数得 \[ m - 6 = 1 + \dfrac{-12}{m}\text{。} \]

两边乘以 \(m\)，得到 \[ m^2 - 6m = m - 12 \] 因而 \[ m^2 - 7m + 12 = 0 \]\[(m-4)(m-3)\]\[m=4,~m=3\]

因此，所有解的和为 \(7\)。

所以正确答案是 C。

English

We can rewrite \(4096\) as \(2^{12},\) so \(\dfrac{1}{4096} = 2^{-12}.\) Then if we equate the given expressions, we get \[ 2^m \cdot 2^{-6} = 2 \cdot 2^{\frac{-12}{m}}. \] Equating the exponents, we get \[ m - 6 = 1 + \dfrac{-12}{m}. \]

Multiplying by \(m,\) we get \[ m^2 - 6m = m - 12 \] and so \[ m^2 - 7m + 12 = 0 \]\[(m-4)(m-3)\]\[m=4,~m=3\]

Therefore, we can see that the sum of the solutions is \(7.\)

Thus, C is the correct answer.

12 答案 A

对于第一个问题，总是说真话者会回答“是”，总是说谎者也会回答“是”，第一次选择说谎的交替者同样会回答“是”；第一次选择说真话的交替者则回答“否”。记相应人数后可写成 \[ 22 = t + l + a_l\text{。}\]

对于第二个问题，总是说谎者会回答“是”；第一次选择说谎的交替者这轮必须说真话，也会回答“是”。其余两类回答“否”，所以 \[ 15 = l + a_l\text{。}\]

由此得到 \(t = 7\)。校长第一轮只给一直说真话的孩子糖果，因此总共只发给他们 \(7\) 块糖。

所以正确答案是 A。

English

For the first question, the truth-tellers will respond yes, the liars will respond yes, and the alternaters who decided to lie first will say yes. The alternaters who decide to tell the truth first will say no. Denote this as \[ 22 = t + l + a_l. \]

For the second questions, the liars will respond yes, and the alternaters who decided to lie first will say yes (they are forced to tell the truth for this question). The truth-tellers will respond no, and the alternaters who told the truth first would lie this round, responding no. Denote this as \[ 15 = l + a_l. \]

From this, we get that \(t = 7.\) The principal only gives candy to children who always tell the truth in the first round, therefore only giving them \(7\) candies total.

Thus, A is the correct answer.

13 答案 C

参考下图：

设 \(Y\) 为 \(\overline{BD}\) 与 \(\overline{AC}\) 的交点。由角平分线定理，\(AB:AC=BP:PC=2:3\)，所以设 \(AB=2x\)、\(AC=3x\)。

关于角平分线 \(\overline{AP}\) 的反射把射线 \(AB\) 映到射线 \(AC\)。因为 \(BY\perp AP\)，它把 \(B\) 映到 \(Y\)。所以 \(AY=AB=2x\)，从而 \(YC=AC-AY=x\)。

因为 \(AD\parallel BC\)，且 \(B,Y,D\) 共线、\(A,Y,C\) 共线，所以 \(\triangle BYC\sim\triangle DYA\)。因此 \[\frac{AD}{BC}=\frac{AY}{YC}=2\text{。}\]

最后，\(BC=BP+PC=5\)，所以 \(AD=2BC=10\)。

所以正确答案是 C。

English

Consider the following diagram:

Let \(Y\) be the intersection of \(\overline{BD}\) and \(\overline{AC}.\) By the Angle Bisector Theorem, \(AB:AC=BP:PC=2:3,\) so write \(AB=2x\) and \(AC=3x.\)

Reflection across the angle bisector \(\overline{AP}\) sends ray \(AB\) to ray \(AC.\) Because \(BY\perp AP,\) it sends \(B\) to \(Y.\) Thus \(AY=AB=2x,\) and hence \(YC=AC-AY=x.\)

Since \(AD\parallel BC,\) with \(B,Y,D\) collinear and \(A,Y,C\) collinear, we have \(\triangle BYC\sim\triangle DYA.\) Therefore \[\frac{AD}{BC}=\frac{AY}{YC}=2.\]

Finally, \(BC=BP+PC=5,\) so \(AD=2BC=10.\)

Thus, C is the correct answer.

14 答案 E

数 \(8\) 到 \(14\) 之间不能互相配对，所以它们必须分别与 \(1\) 到 \(7\) 中的一个数配对。特别地，\(7\) 必须与 \(14\) 配对，因为没有其他可用的数至少是 \(7\) 的两倍。

再看其余各数的搭档。\(8\) 和 \(9\) 可与 \(1-4\) 中任一数配对；\(10\) 和 \(11\) 可与 \(1-5\) 中任一数配对；\(12\) 和 \(13\) 可与 \(1-6\) 中任一数配对。

\(8\) 有 \(4\) 种选择，随后 \(9\) 只剩 \(3\) 种，因为 \(8\) 已占用一个数。\(10\) 仍有 \(3\) 种选择，因为虽已占用 \(2\) 个选择，但它多出一个可选数 \((5)\)。接着 \(11\) 有 \(2\) 种，\(12\) 有 \(2\) 种，\(13\) 只有 \(1\) 种。

相乘得到 \[ 4 \cdot 3 \cdot 3 \cdot 2 \cdot 2 = 144\text{。}\]

所以正确答案是 E。

English

The numbers from \(8\) through \(14\) cannot be paired with one another, so they must be paired with the numbers from \(1\) through \(7.\) In particular, \(7\) must be paired with \(14,\) since no other available number is at least twice \(7.\)

Now let’s look at what the other numbers can pair with. \(8\) and \(9\) can pair with any number \(1-4.\) \(10\) and \(11\) can pair with any number \(1-5,\) and \(12\) and \(13\) can pair with any number \(1-6.\)

\(8\) can pair with \(4\) numbers, but then \(9\) only has \(3\) options since \(8\) took one. \(10\) then has \(3\) options, since \(2\) choices are taken, but it has one more to choose from \((5).\) \(11\) then has \(2\) options, \(12\) has \(2\) options, and \(13\) only has \(1.\)

Multiplying these together yields \[ 4 \cdot 3 \cdot 3 \cdot 2 \cdot 2 = 144. \]

Thus, E is the correct answer.

15 答案 D

注意到 \(7^2 + 24^2\) 与 \(15^2 + 20^2\) 相等。这迫使 \(AC = 25\)，否则 \(\angle B\) 与 \(\angle D\) 会同时为锐角或同时为钝角，与它们的和为 \(180^{\circ}\) 矛盾。

又因为 \(\angle B\) 是直角，所以 \(AC\) 是圆的直径。圆的面积为 \(\dfrac{625}{4} \pi\)。

四边形的面积等于两个三角形的面积之和，即 \[ \dfrac{1}{2}(7 \cdot 24 + 20 \cdot 15) = \]\[ 84 + 150 = 234\text{。}\]

圆内而四边形外的面积为 \[ \dfrac{625}{4} \pi - 234 = \dfrac{625 \pi - 936}{4}\text{。}\]

因此 \[ a + b + c = 625 + 936 + 4 \]\[ = 1565\text{。}\]

所以正确答案是 D。

English

Notice that \(7^2 + 24^2\) and \(15^2 + 20^2\) are both the same. This forces \(AC = 25\) since otherwise \(\angle B\) and \(\angle D\) would both be acute or obtuse, violating the fact that their sum is \(180^{\circ}.\)

Also since \(\angle B\) is right, we know that \(AC\) is the diameter of the circle. The area of the circle is then \(\dfrac{625}{4} \pi.\)

To find the area of the quadrilateral, we can find the area of each of the triangles, which is \[ \dfrac{1}{2}(7 \cdot 24 + 20 \cdot 15) = \] \[ 84 + 150 = 234. \]

To find the area outside the quadrilateral, we subtract to get \[ \dfrac{625}{4} \pi - 234 = \dfrac{625 \pi - 936}{4}. \]

Therefore, \[ a + b + c = 625 + 936 + 4 \] \[ = 1565. \]

Thus, D is the correct answer.

16 答案 D

设原长方体的三个尺寸为 \(h, l\) 和 \(w\)。新长方体的体积为 \[ (h + 2)(l + 2)(w + 2)\text{。} \] 展开得 \[ hlw + 2(hl + hw + lw) \] \[+ 4(l + h + w) + 8\text{。} \] 再由韦达定理，\[ hlw = -\dfrac{D}{A} = \dfrac{3}{5}\text{，} \] \[ hl + hw + lw = \dfrac{C}{A} = \dfrac{29}{10}\text{，} \] 且 \[ l + h + w = -\dfrac{B}{A} = \dfrac{39}{10}\text{。} \]

把这些值代入展开式，得到 \[ \dfrac{3}{5} + 2 \cdot \dfrac{29}{10} + 4 \cdot \dfrac{39}{10} + 8 = 30\text{。} \]

所以正确答案是 D。

English

Let \(h, l,\) and \(w\) be the dimensions of the old box. Then the volume of the new box is \[ (h + 2)(l + 2)(w + 2). \] Expanding, we get \[ hlw + 2(hl + hw + lw) \]\[+ 4(l + h + w) + 8. \] We can use Vieta’s formulas to find the terms in this expression. We get that \[ hlw = -\dfrac{D}{A} = \dfrac{3}{5}, \] \[ hl + hw + lw = \dfrac{C}{A} = \dfrac{29}{10}, \] and \[ l + h + w = -\dfrac{B}{A} = \dfrac{39}{10}. \]

Plugging these values into the expression, we get \[ \dfrac{3}{5} + 2 \cdot \dfrac{29}{10} + 4 \cdot \dfrac{39}{10} + 8 = 30. \]

Thus, D is the correct answer.

17 答案 D

这些循环小数满足 \[ \begin{aligned} 0.\overline{\underline{a}\underline{b}\underline{c}}&=\frac{100a+10b+c}{999},\\ 0.\overline a&=\frac a9,\quad 0.\overline b=\frac b9,\\ 0.\overline c&=\frac c9 \end{aligned}\text{。}\]

代入后两边同乘 \(999\)，得到 \[100a+10b+c=37(a+b+c)\text{，}\] 也就是 \[7a=3b+4c\text{。}\]

对每个 \(a\in\{1,2,\ldots,9\}\)，在这个线性方程中逐一检验非零数字 \(b,c\)，得到 \[ \begin{gathered} (1,1,1),(2,2,2),(3,3,3),\\ (4,4,4),(4,8,1),(5,1,8),\\ (5,5,5),(5,9,2),(6,2,9),\\ (6,6,6),(7,7,7),(8,8,8),\\ (9,9,9) \end{gathered}\text{。}\] 因此这样的整数共有 \(13\) 个。

所以正确答案是 D。

English

The repeating decimals satisfy \[ \begin{aligned} 0.\overline{\underline{a}\underline{b}\underline{c}}&=\frac{100a+10b+c}{999},\\ 0.\overline a&=\frac a9,\quad 0.\overline b=\frac b9,\\ 0.\overline c&=\frac c9. \end{aligned} \]

Substitution and multiplication by \(999\) give \[100a+10b+c=37(a+b+c),\] or \[7a=3b+4c.\]

For each \(a\in\{1,2,\ldots,9\},\) checking the nonzero digits \(b,c\) in this linear equation gives \[ \begin{gathered} (1,1,1),(2,2,2),(3,3,3),\\ (4,4,4),(4,8,1),(5,1,8),\\ (5,5,5),(5,9,2),(6,2,9),\\ (6,6,6),(7,7,7),(8,8,8),\\ (9,9,9). \end{gathered} \] Thus there are \(13\) integers.

Thus, D is the correct solution.

18 答案 A

因为题目涉及角度和反射，使用极坐标会更方便。

设极坐标为 \((r, \theta)\)。逆时针旋转 \(k\) 度后变为 \((r, \theta + k^{\circ})\)，再反射后变为 \((r, 180^{\circ} - \theta - k^{\circ})\)。

因此 \[ T_k(r, \theta) = (r, 180^{\circ} - \theta - k^{\circ})\text{。}\]

由此可见 \[ T_{k + 1}(T_k(r, \theta)) = \] \[T_{k + 1}(r, 180^{\circ} - \theta - k^{\circ}) = \] \[(r, \theta - 1^{\circ})\text{。}\]

现在分析点 \((1, 0^{\circ})\) 的变化。

经过 \(T_1\) 后，得到 \((1, 179^{\circ})\)。

经过 \(T_2\) 后，得到 \((1, -1^{\circ})\)。

经过 \(T_3\) 后，得到 \((1, 178^{\circ})\)。

经过 \(T_4\) 后，得到 \((1, -2^{\circ})\)。

\[\vdots\]

经过 \(T_{2m - 1}\) 后，得到 \((1, 180^{\circ} - m^{\circ})\)。

经过 \(T_{2m}\) 后，得到 \((1, -m^{\circ})\)。

由此可见，角度第一次回到 \(0^{\circ}\) 是在 \(T_{2(180)-1}=T_{359}\) 之后。因此 \(n=359\)。

所以正确答案是 A。

English

Since we are working with angles and reflections, working with polar coordinates would make this problem easier to deal with.

Let \((r, \theta)\) be a polar coordinate. Rotating this by \(k\) degrees counterclockwise maps the point to \((r, \theta + k^{\circ})\) and then reflecting it maps it to \((r, 180^{\circ} - \theta - k^{\circ}).\)

Therefore, we have that \[ T_k(r, \theta) = (r, 180^{\circ} - \theta - k^{\circ}). \]

From this, we can see that \[ T_{k + 1}(T_k(r, \theta)) = \]\[T_{k + 1}(r, 180^{\circ} - \theta - k^{\circ}) = \]\[(r, \theta - 1^{\circ}). \]

Now, let’s analyze what happens to the point \((1, 0^{\circ}).\)

After \(T_1,\) we get \((1, 179^{\circ}).\)

After \(T_2,\) we get \((1, -1^{\circ}).\)

After \(T_3,\) we get \((1, 178^{\circ}).\)

After \(T_4,\) we get \((1, -2^{\circ}).\)

\[\vdots\]

After \(T_{2m - 1},\) we get \((1, 180^{\circ} - m^{\circ}).\)

After \(T_{2m},\) we get \((1, -m^{\circ}).\)

From this, we can see that the first time the angle is back to \(0^{\circ}\) is after \(T_{2(180)-1}=T_{359}.\) Therefore \(n=359.\)

Thus, A is the correct answer.

19 答案 C

将调和和乘以 \(L_{17}\)，得到 \(h=\sum_{i=1}^{17}\frac{L_{17}}{i}\)。

对 \(1\le i\le16\)，项 \(\frac{L_{17}}{i}\) 仍能被 \(17\) 整除，所以这些项贡献 \(0\pmod{17}\)。

因此 \(h\equiv \frac{L_{17}}{17}\pmod{17}\)。最小公倍数 \(L_{17}\) 含有素数幂因子 \(16,9,5,7,11,13,17\)，所以 \[ \begin{aligned} \frac{L_{17}}{17} &\equiv 16\cdot9\cdot5\cdot7\cdot11\cdot13 \\ &\pmod{17} \end{aligned}\text{。}\]

模 \(17\) 化简后，这就是 \((-1)\cdot9\cdot5\cdot7\cdot11\cdot13\) \(\equiv5\pmod{17}\)。

所以正确答案是 C。

English

Multiplying the harmonic sum by \(L_{17},\) we get \(h=\sum_{i=1}^{17}\frac{L_{17}}{i}.\)

For \(1\le i\le16,\) the term \(\frac{L_{17}}{i}\) is still divisible by \(17,\) so these terms contribute \(0\pmod{17}.\)

Thus \(h\equiv \frac{L_{17}}{17}\pmod{17}.\) The least common multiple \(L_{17}\) contains the prime-power factors \(16,9,5,7,11,13,17,\) so \[ \begin{aligned} \frac{L_{17}}{17} &\equiv 16\cdot9\cdot5\cdot7\cdot11\cdot13 \\ &\pmod{17}. \end{aligned} \]

Reducing modulo \(17,\) this is \((-1)\cdot9\cdot5\cdot7\cdot11\cdot13\) \(\equiv5\pmod{17}.\)

Thus, C is the correct answer.

20 答案 E

设等差数列为 \[ a, a + d, a + 2d, a + 3d\text{，}\]等比数列为 \[ b, br, br^2, br^3\text{。}\]

则 \[ a + b = 57 \tag*{(1)}\text{，}\]\[ a + d + br = 60 \tag*{(2)}\text{，}\]且 \[ a + 2d + br^2 = 91 \tag*{(3)}\text{。}\]

用 \((2)\) 减去 \((1)\)，再用 \((3)\) 减去 \((2)\)，得到 \[ d + b(r - 1) = 3 \] 和 \[ d + br(r - 1) = 31\text{。}\]

两式相减，得 \[ b(r - 1)^2 = 28\text{。}\]

令 \(t=b(r-1)=br-b\)，它是整数。则 \(t^2=28b\)，所以对某个非零整数 \(u\)，有 \(t=14u\)、\(b=7u^2\)。因为 \(a=57-b>0\)，必须有 \(u\in\{-2,-1,1,2\}\)。

现在 \(r=1+\frac{t}{b}=1+\frac{2}{u}\)。当 \(u=-1\) 或 \(u=-2\) 时，分别得到 \(r=-1\) 和 \(r=0\)，不能形成正数等比数列。若 \(u=2\)，则 \(b=28, r=2, a=29\)，且 \(d=3-t=-25\)，使等差数列第三项为负。

因此 \(u=1\)，所以 \(b = 7, r = 3, a = 50\)，且 \(d = -11\)。等差数列为 \[ 50,39,28,17\text{，}\]等比数列为 \[ 7,21,63,189\text{。}\]

所求答案为 \(17 + 189 = 206\)。

所以正确答案是 E。

English

Let the arithmetic sequence be \[ a, a + d, a + 2d, a + 3d \] and the geometric sequence be \[ b, br, br^2, br^3. \]

Then \[ a + b = 57 \tag*{(1)}, \] \[ a + d + br = 60 \tag*{(2)}, \] and \[ a + 2d + br^2 = 91 \tag*{(3)}. \]

Subtracting \((1)\) from \((2)\) and \((2)\) from \((3),\) we get \[ d + b(r - 1) = 3 \] and \[ d + br(r - 1) = 31. \]

Subtracting these, we get \[ b(r - 1)^2 = 28. \]

Let \(t=b(r-1)=br-b,\) which is an integer. Then \(t^2=28b,\) so \(t=14u\) and \(b=7u^2\) for some nonzero integer \(u.\) Because \(a=57-b>0,\) we must have \(u\in\{-2,-1,1,2\}.\)

Now \(r=1+\frac{t}{b}=1+\frac{2}{u}.\) The cases \(u=-1\) and \(u=-2\) give \(r=-1\) and \(r=0,\) respectively, so they cannot produce a positive geometric sequence. If \(u=2,\) then \(b=28, r=2, a=29,\) and \(d=3-t=-25,\) making the third arithmetic term negative.

Therefore \(u=1,\) so \(b = 7, r = 3, a = 50,\) and \(d = -11.\) The arithmetic sequence is \[ 50,39,28,17, \] and the geometric sequence is \[ 7,21,63,189. \]

The desired answer is \(17 + 189 = 206.\)

Thus, E is the correct answer.

21 答案 B

从正上方观察碗口。把底部正方形放在 \((\pm\tfrac12,\pm\tfrac12,0)\)。正六边形中，与所接边相邻的一条边，其平行于该边和垂直于该边的分量分别为 \(\tfrac12\) 与 \(\tfrac{\sqrt3}{2}\)。

在底部正方形的一个顶点处，相邻两个六边形的对应边重合。比较它们的水平分量可知，在六边形平面内，垂直于所接边的单位向量的水平分量为 \(\frac{1}{\sqrt3}\)。正六边形的对边与所接边相距 \(\sqrt3\)，所以水平向外位移为 \(1\)。

因此，碗口的四条单位长度边都位于底部正方形对应边之外一个单位处。其俯视图是一个 \(3\) 乘 \(3\) 的正方形，去掉四个直角边为 \(1\) 的等腰直角三角形：

面积为 \[3^2-4\left(\frac12\cdot1^2\right)=7\text{。}\]

所以正确答案是 B。

English

View the rim from directly above. Place the bottom square at \((\pm\tfrac12,\pm\tfrac12,0).\) In a regular hexagon, an edge adjacent to the attached side has components \(\tfrac12\) parallel and \(\tfrac{\sqrt3}{2}\) perpendicular to that side.

At a corner of the bottom square, the corresponding edges of two adjacent hexagons coincide. Comparing their horizontal components shows that the horizontal component of a unit vector perpendicular to an attached side within its hexagon is \(\frac{1}{\sqrt3}.\) The opposite side of a regular hexagon is \(\sqrt3\) units from the attached side, so its horizontal outward displacement is \(1.\)

Consequently, the four unit-length sides of the rim lie one unit beyond the four sides of the bottom square. Its top view is therefore a \(3\)-by-\(3\) square with four isosceles right corner triangles of leg \(1\) removed:

Its area is \[3^2-4\left(\frac12\cdot1^2\right)=7.\]

Thus, B is the correct answer.

22 答案 D

设第一次扫描拿到 \(n\) 张卡片，其中 \(1 \leq n \leq 12\)。

选择这 \(n\) 张卡片所占的位置后，剩余卡片的位置也被确定，因为两组卡片都必须按递增顺序排列。

共有 \(\binom{13}{n}\) 种方法选择这 \(n\) 张卡片的位置；但若这 \(n\) 张卡片恰好位于最前面，则所有卡片都会在第一次扫描中拿完。

因此对固定的 \(n\)，共有 \(\binom{13}{n} - 1\) 种排列。

对 \(1\le n\le12\) 求和，得到 \[ \begin{gathered} \sum_{n=1}^{12}\left(\binom{13}{n}-1\right) \\ = (2^{13}-2)-12 \\ = 8178 \end{gathered}\text{。}\]

所以正确答案是 D。

English

Let \(n\) be the number of cards picked up on the first pass, where \(1 \leq n \leq 12.\)

If we choose the spaces that the \(n\) cards occupy, the positions of the remaining cards are determined since they must be placed in order.

There are \(\binom{13}{n}\) ways to choose where the \(n\) cards go, but if the \(n\) cards are placed at the very beginning, then all the cards will be picked up on the first pass.

Therefore, for a given \(n\) there are \(\binom{13}{n} - 1\) ways to arrange the cards.

Summing over \(1\le n\le12,\) we get \[ \begin{gathered} \sum_{n=1}^{12}\left(\binom{13}{n}-1\right) \\ = (2^{13}-2)-12 \\ = 8178. \end{gathered} \]

Thus, D is the correct answer.

23 答案 B

令 \(P'\) 为 \(P\) 关于 \(\overline{BC}\) 的垂直平分线的反射点。

这样得到两个等腰梯形 \(CBPP'\) 和 \(DAPP'\)。

由反射对称可得 \[\begin{gathered} P'A = PD = 4 \\ P'D = PA = 1 \\ P'C = PB = 2 \\ P'B = PC = 3 \end{gathered}\text{。}\]

对两个圆内接梯形使用托勒密定理，知道对角线乘积等于两组对边乘积之和。因此 \[\begin{gathered} PP' \cdot AD + 1 = 16 \\ PP' \cdot BC + 4 = 9 \end{gathered}\text{。}\]

于是 \(PP' \cdot AD = 15\)，且 \(PP' \cdot BC = 5\)。两式相除得到 \(\dfrac{BC}{AD} = \dfrac{1}{3}\)。

所以正确答案是 B。

English

Let \(P'\) be the reflection of \(P\) across the perpendicular bisector of \(\overline{BC}.\)

This forms two new isosceles trapezoids: \(CBPP'\) and \(DAPP'.\)

Therefore, we get \[\begin{gathered} P'A = PD = 4 \\ P'D = PA = 1 \\ P'C = PB = 2 \\ P'B = PC = 3. \end{gathered}\]

Using Ptolemy’s theorem, we know that the product of the diagonals is equal to the sum of the products of the opposite sides. Therefore: \[\begin{gathered} PP' \cdot AD + 1 = 16 \\ PP' \cdot BC + 4 = 9. \end{gathered}\]

This gets us \(PP' \cdot AD = 15\) and \(PP' \cdot BC = 5.\) Dividing these two equations yields \(\dfrac{BC}{AD} = \dfrac{1}{3}.\)

Thus, B is the correct answer.

24 答案 E

把这五个数字依次看作五辆车的首选车位。车位编号为 \(0,1,2,3,4\)，每辆车若首选车位空着就停在那里，否则停在右边第一个空位。若把首选车位按非递减顺序排列为 \(b_1\le b_2\le\cdots\le b_5\)，则所有车都能停下，当且仅当 \[b_i\le i-1\qquad(1\le i\le5)\text{。}\]这些不等式正是题目中的条件。

为了计数，加入第六个车位，把 \(0,1,\ldots,5\) 号车位排列在一个圆上。对任意一个 \(6^5\) 种首选字符串，五辆车都能停下，且恰好留下一个空位。把所有首选位置都旋转一格，空位也随之旋转。因此每个由六个字符串组成的轨道中，每一种可能的空位恰好出现一次。

所以恰有 \(\frac{6^5}{6}=6^4=1296\) 个圆形首选字符串留下 \(5\) 号车位为空。这样的字符串中没有车首选 \(5\) 号车位；紧接着这个空位把圆切开，就恰好得到车位 \(0\) 到 \(4\) 上的成功停车序列。因此所求字符串数为 \(1296\)。

所以正确答案是 E。

English

Regard the five digits, in order, as the preferred parking spaces of five cars. Spaces are numbered \(0,1,2,3,4,\) and each car takes its preferred space if possible, or else the first empty space to its right. If the preferences sorted into nondecreasing order are \(b_1\le b_2\le\cdots\le b_5,\) all cars park exactly when \[b_i\le i-1\qquad(1\le i\le5).\] These inequalities are precisely the conditions in the problem.

To count such preference strings, add a sixth space and arrange spaces \(0,1,\ldots,5\) in a circle. For any of the \(6^5\) preference strings, all five cars park and exactly one space remains empty. Rotating every preference by one position rotates the empty space as well. Thus each orbit of six preference strings has each possible empty space exactly once.

Therefore exactly \(\frac{6^5}{6}=6^4=1296\) circular preference strings leave space \(5\) empty. No car in such a string prefers space \(5,\) and cutting the circle immediately after that empty space gives exactly a successful parking sequence on spaces \(0\) through \(4.\) Hence the desired number of strings is \(1296.\)

Thus, E is the correct answer.

25 答案 B

令 \(r\) 为 \(R\) 的每条边上的格点数；类似地，令 \(s\) 对应 \(S\)，令 \(t\) 对应 \(T\)。注意，一个矩形中的格点数等于沿其宽方向的格点数与沿其长方向的格点数之积。

第一个条件给出 \[ r^2 = \dfrac{9}{4} \cdot s^2 \]\[r = \dfrac{3}{2} \cdot s \tag*{(1)} \]

\(R \cup S\) 中的格点数等于两个区域各自格点数之和，但在 \(y\) 轴上有重叠，也就是正方形 \(S\) 与该轴相接之处。

因此第二个条件给出 \[ t^2 = \dfrac{1}{4}(r^2 + s^2 - s) \] \[ t^2 = \dfrac{1}{4}(\dfrac{9}{4} \cdot s^2 + s^2 - s) \] \[ t^2 = \dfrac{1}{4} \cdot \dfrac{13s^2 - 4s}{4} \] \[ 16t^2 = s(13s - 4)\text{。}\] 由 \((1)\) 可知 \(s\) 是 \(2\) 的倍数。把 \(s\) 换成 \(2j\)，得到 \[ 16t^2 = 2j(26j - 4) \] \[ 4t^2 = j(13j - 2)\text{。}\] 为了使乘积能被 \(4\) 整除，\(j\) 必须被 \(2\) 整除。再把 \(j\) 换成 \(2k\)，得到 \[ 4t^2 = 2k(26k - 2) \] \[ t^2 = k(13k - 1) \tag*{(2)} \]

令 \(x\) 为矩形 \(S \cap T\) 底边上的格点数，令 \(y\) 为矩形 \(R \cap T\) 底边上的格点数。

这样，\(S \cap T\) 中的格点数为 \(xt\)，而 \(R \cap T\) 中的格点数为 \(yt\)。

第三个条件给出 \[ \dfrac{xt}{s^2} = 27 \cdot \dfrac{yt}{r^2} \] \[ \dfrac{x}{s^2} = 27 \cdot \dfrac{y}{\dfrac{9}{4} s^2} \] \[ x = 12y\text{。}\]

又已知 \(t = x + y - 1\)（其中已计入重叠部分），于是 \[ t = 13y - 1 \tag*{(3)} \]

由 \((3)\) 可得 \[ \begin{gathered} t \equiv -1 \pmod{13}, \\ t^2 \equiv 1 \pmod{13} \end{gathered}\text{。}\]

然而由 \((2)\) 可得 \[ t^2 \equiv -k \pmod{13} \] \[ k \equiv -1 \pmod{13}\text{。}\]

同时由 \((2)\) 可知 \(k\) 是平方数，因为它与 \(13k - 1\) 互质，而它们的乘积是平方数。

因此 \(k\) 必须是满足 \(k\equiv-1\pmod{13}\) 的完全平方数。较小的正平方数 \(1,4,9,16\) 都不满足这一同余条件，而 \(k=25\) 满足，所以 \(25\) 是最小可能值。

由这个 \(k\) 值可得 \(j = 2 \cdot 25 = 50\)、\(s = 2 \cdot 50 = 100\)，以及 \(r = \dfrac{3}{2} \cdot 100 = 150\)。还可算得 \[ t^2 = 25(13 \cdot 25 - 1) = 25 \cdot 324 \] \[ t = 5 \cdot 18 = 90\text{。}\]因此 \[ r + s + t = 340\text{。}\]不过题目问的是边长之和。每个正方形的边长都比每边的格点数少 \(1\)，三个正方形共要减去 \(3\)。

这个值可以达到：由方程 \((3)\) 得 \(y=7\)，从而 \(x=84\)。因为 \(x\le s=100\)、\(y\le r=150\)，每边有 \(90\) 个格点的正方形 \(T\) 可以跨过 \(y\) 轴形成所需的重叠，而且它的上方两个顶点位于 \(R\cup S\) 中。

因此所求答案为 \(340 - 3 = 337\)。

所以正确答案是 B。

English

Let \(r\) be the number of lattice points on the side length of \(R.\) Similarly define \(s\) for \(S\) and \(t\) for \(T.\) Note that the number of lattice points in a rectangle is the product of the number of lattice points along its width and the number of lattice points along its length.

The first condition gives us that \[ r^2 = \dfrac{9}{4} \cdot s^2 \]\[r = \dfrac{3}{2} \cdot s \tag*{(1)} \]

The number of lattice points in \(R \cup S\) is the sum of the lattice points in each of the regions, but there is overlap along the \(y\)-axis where \(S\) touches it.

The second condition, therefore, yields \[ t^2 = \dfrac{1}{4}(r^2 + s^2 - s) \] \[ t^2 = \dfrac{1}{4}(\dfrac{9}{4} \cdot s^2 + s^2 - s) \] \[ t^2 = \dfrac{1}{4} \cdot \dfrac{13s^2 - 4s}{4} \] \[ 16t^2 = s(13s - 4). \] From \((1),\) we get that \(s\) is a multiple of \(2.\) We can substitute \(s\) with \(2j\) to get \[ 16t^2 = 2j(26j - 4) \] \[ 4t^2 = j(13j - 2). \] For the product to be divisible by \(4,\) \(j\) must be divisible by \(2.\) We can again substitute \(j\) with \(2k\) to get \[ 4t^2 = 2k(26k - 2) \] \[ t^2 = k(13k - 1) \tag*{(2)} \]

Let \(x\) be the number of lattice points along the bottom of the rectangle formed by \(S \cap T\) and \(y\) be the number of lattice points along the bottom of the rectangle formed by \(R \cap T.\)

Using these variables, we get that the number of lattice points in \(S \cap T\) is \(xt\) and in \(R \cap T\) is \(yt.\)

The third condition gives us that \[ \dfrac{xt}{s^2} = 27 \cdot \dfrac{yt}{r^2} \] \[ \dfrac{x}{s^2} = 27 \cdot \dfrac{y}{\dfrac{9}{4} s^2} \] \[ x = 12y. \]

We also know that \(t = x + y - 1\) (accounting for overlap), and this yields \[ t = 13y - 1 \tag*{(3)} \]

\((3)\) gives us that \[ \begin{gathered} t \equiv -1 \pmod{13}, \\ t^2 \equiv 1 \pmod{13}. \end{gathered} \]

However, by \((2),\) we get that \[ t^2 \equiv -k \pmod{13} \] \[ k \equiv -1 \pmod{13}. \]

By \((2),\) we also get that \(k\) is a perfect square since it is relatively prime to \(13k - 1,\) and they must multiply to a perfect square.

Thus \(k\) must be a perfect square satisfying \(k\equiv-1\pmod{13}.\) The smaller positive squares \(1,4,9,16\) do not satisfy this congruence, while \(k=25\) does, so \(25\) is the least possible value.

From this value of \(k,\) we get that \(j = 2 \cdot 25 = 50,\) \(s = 2 \cdot 50 = 100,\) and \(r = \dfrac{3}{2} \cdot 100 = 150.\) We can also find that \[ t^2 = 25(13 \cdot 25 - 1) = 25 \cdot 324 \] \[ t = 5 \cdot 18 = 90. \] Therefore, \[ r + s + t = 340. \] The question, however, asked for the sum of the side lengths. The side lengths of the squares are \(1\) less than the number of lattice points on the side, so we have to subtract \(3.\)

This value is attainable: equation \((3)\) gives \(y=7\) and hence \(x=84.\) Since \(x\le s=100\) and \(y\le r=150,\) a square \(T\) with \(90\) lattice points per side can straddle the \(y\)-axis with the required overlaps, and its top vertices lie in \(R\cup S.\)

Therefore, the desired answer is \(340 - 3 = 337.\)

Thus, B is the correct answer.