2012 AMC 10A 真题与在线限时测验

计时

1:15:00

1.

Cagney 每 $20$ 秒给一个纸杯蛋糕涂糖霜，Lacey 每 $30$ 秒给一个纸杯蛋糕涂糖霜。两人一起工作， $5$ 分钟内可以给多少个纸杯蛋糕涂糖霜？

Cagney can frost a cupcake every $20$ seconds and Lacey can frost a cupcake every $30$ seconds. Working together, how many cupcakes can they frost in $5$ minutes?

$10$

$15$

$20$

$25$

$30$

答案：D

知识点：速率单位换算

难度评级：560

小提示：

先把两人的速度都换成每分钟能完成的个数。

Convert each frosting rate to cupcakes per minute

大提示：

把两个速度相加，再乘以 $5$ 分钟。

Add the two rates, then multiply by $5$ minutes

解答：

Cagney 每分钟完成 $60 \div 20 = 3$ 个，Lacey 每分钟完成 $60 \div 30 = 2$ 个。

$5$ 分钟内，他们能完成 $5(2 + 3) = 5 \cdot 5 = 25\text{。}$

所以正确答案是 D。

Cagney can make $60 \div 20 = 3$ cupcakes per minute, and Lacey can make $60 \div 30 = 2.$

In $5$ minutes, they can make $5(2 + 3) = 5 \cdot 5 = 25.$

Thus, D is the correct answer.

2.

一个边长为 $8$ 的正方形被切成两块全等的长方形。每个长方形的边长是多少？

A square with side length $8$ is cut in half, creating two congruent rectangles. What are the dimensions of one of these rectangles?

$2 \text{ 乘 } 4$

$2 \text{ by } 4$

$2 \text{ 乘 } 6$

$2 \text{ by } 6$

$2 \text{ 乘 } 8$

$2 \text{ by } 8$

$4 \text{ 乘 } 4$

$4 \text{ by } 4$

$4 \text{ 乘 } 8$

$4 \text{ by } 8$

答案：E

知识点：正方形（几何）矩形

难度评级：450

小提示：

每个长方形都有一条边仍然是 $8$ 。

One side of each rectangle stays $8$

大提示：

另一条边是正方形边长的一半。

The other side is half of the square side length

解答：

注意其中一条边仍与原正方形的边长相同。

因此一个尺寸是 $8$ ，另一个尺寸是原边长的一半，即 $4$ 。

所以正确答案是 E。

Note that the one of the sides remains the same as the original square.

This means that one dimension is $8.$ The other dimension is the original side cut in half, which is $4.$

Thus, E is the correct answer.

3.

一只小虫在数轴上爬行，起点是 $-2$ 。它先爬到 $-6$ ，然后掉头爬到 $5$ 。这只小虫一共爬了多少个单位？

A bug crawls along a number line, starting at $-2.$ It crawls to $-6,$ then turns around and crawls to $5.$ How many units does the bug crawl altogether?

$9$

$11$

$13$

$14$

$15$

答案：E

知识点：绝对值

难度评级：560

小提示：

加上从 $-2$ 到 $-6$ 的距离，以及从 $-6$ 到 $5$ 的距离。

Add the distance from $-2$ to $-6$ and from $-6$ to $5$

大提示：

数轴上的距离用绝对差来计算。

Distance on a number line uses absolute differences

解答：

它爬行的距离为 $|-2 - (-6)| = |4| = 4$ 即从 $-2$ 移动到 $-6$ 。接着它爬行的距离为 $|-6 - 5| = |-11| = 11$ 即从 $-6$ 移动到 $5$ 。

总距离为 $4 + 11 = 15\text{。}$

所以正确答案是 E。

It crawls a distance of $|-2 - (-6)| = |4| = 4$ when it moves from $-2$ to $-6.$ It then travels a distance of $|-6 - 5| = |-11| = 11$ as it moves from $-6$ to $5.$

The total distance is then $4 + 11 = 15.$

Thus, E is the correct answer.

4.

已知 $\angle ABC = 24^\circ$ ， $\angle ABD = 20^\circ$ 。 $\angle CBD$ 的最小可能度数是多少？

Let $\angle ABC = 24^\circ$ and $\angle ABD = 20^\circ .$ What is the smallest possible degree measure for $\angle CBD?$

$0$

$2$

$4$

$6$

$12$

答案：C

知识点：导角最优化

难度评级：770

小提示：

两个角共用射线 $BA$ 。

Both angles share ray $BA$

大提示：

当两个给定角尽可能重叠时，剩下的角最小。

The smallest angle occurs when the two given angles overlap as much as possible

解答：

注意两个角都共用射线 $AB$ 。为了使所求角最小，让 $D$ 位于 $A$ 和 $C$ 之间。

这样就有 $\begin{aligned} \angle CBD &= \angle ABC - \angle ABD \\ &= 4^{\circ} \end{aligned}\text{。}$

所以正确答案是 C。

Note that both of the angles share the ray $AB.$ To minimize the desired degree, we want $D$ to be between $A$ and $C.$

This would make $\begin{aligned} \angle CBD &= \angle ABC - \angle ABD \\ &= 4^{\circ}. \end{aligned}$

Thus, C is the correct answer.

5.

去年有 $100$ 只成年猫被送到 Smallville 动物收容所，其中一半是母猫。成年母猫中有一半各带着一窝小猫，每窝平均有 $4$ 只小猫。去年这个收容所一共收到多少只成年猫和小猫？

Last year $100$ adult cats, half of whom were female, were brought into the Smallville Animal Shelter. Half of the adult female cats were accompanied by a litter of kittens. The average number of kittens per litter was $4.$ What was the total number of cats and kittens received by the shelter last year?

$150$

$200$

$250$

$300$

$400$

答案：B

知识点：分数平均数

难度评级：870

小提示：

先求成年母猫的数量，再求带小猫的母猫数量。

Half the adult cats are female, and half of those have litters

大提示：

用带小猫的母猫数乘以平均每窝小猫数。

Multiply the number of litters by the average litter size

解答：

成年母猫有 $100 \div 2 = 50$ 只，其中带小猫的有 $50 \div 2 = 25$ 只。

因为平均每窝有 $4$ 只小猫，所以小猫总数为 $4 \cdot 25 = 100\text{。}$

因此成年猫和小猫的总数为 $100 + 100 = 200\text{。}$

所以正确答案是 B。

We have that there are $100 \div 2 = 50$ female cats. We then have that $50 \div 2 = 25$ cats that have kittens.

Since the average number of kittens per litter is $4,$ the total number of kittens is $4 \cdot 25 = 100.$

The total number of cats and kittens is then $100 + 100 = 200.$

Thus, B is the correct answer.

6.

两个正数的乘积为 $9$ 。其中一个数的倒数是另一个数倒数的 $4$ 倍。这两个数的和是多少？

The product of two positive numbers is $9.$ The reciprocal of one of these numbers is $4$ times the reciprocal of the other number. What is the sum of the two numbers?

$\dfrac{10}{3}$

$\dfrac{20}{3}$

$7$

$\dfrac{15}{2}$

$8$

答案：D

知识点：方程组分式方程

难度评级：1070

小提示：

设两个数为 $x$ 和 $y$ 。

Let the two numbers be $x$ and $y$

大提示：

使用 $xy=9$ ，并用倒数条件建立 $x$ 与 $y$ 的关系。

Use $xy=9$ and the reciprocal condition to relate $x$ and $y$

解答：

设两个数为 $x$ 和 $y$ ，满足 $xy = 9 \text{ 和 } \dfrac{1}{x} = \dfrac{4}{y}\text{。}$

由乘积条件与倒数条件得 $y = \dfrac{9}{x}$ $\dfrac{1}{x} = \dfrac{4x}{9}\text{。}$ $x = \dfrac{3}{2}\text{，}$ 其中 $x$ 为正。

于是另一个数为 $y = 9 \div \dfrac{3}{2} = 6\text{。}$

两数之和为 $6 + \dfrac{3}{2} = \dfrac{15}{2}\text{。}$

所以正确答案是 D。

Let the two numbers be $x$ and $y$ such that $xy = 9 \text{ and } \dfrac{1}{x} = \dfrac{4}{y}.$

We get that $y = \dfrac{9}{x}$ $\dfrac{1}{x} = \dfrac{4x}{9}.$ $x = \dfrac{3}{2},$ since $x$ is positive.

Then $y = 9 \div \dfrac{3}{2} = 6.$

The desired sum is then $6 + \dfrac{3}{2} = \dfrac{15}{2}.$

Thus, D is the correct answer.

7.

一个袋子里有若干弹珠，其中 $\dfrac{3}{5}$ 是蓝色的，其余是红色的。现在把红色弹珠的数量加倍，而蓝色弹珠数量不变。此时红色弹珠占全部弹珠的几分之几？

In a bag of marbles, $\dfrac{3}{5}$ of the marbles are blue and the rest are red. If the number of red marbles is doubled and the number of blue marbles stays the same, what fraction of the marbles will be red?

$\dfrac{2}{5}$

$\dfrac{3}{7}$

$\dfrac{4}{7}$

$\dfrac{3}{5}$

$\dfrac{4}{5}$

答案：C

知识点：比与比例分数

难度评级：870

小提示：

把原来的蓝色和红色数量看作比例 $3:2$ 。

Start with a $3:2$ ratio of blue to red marbles

大提示：

红色数量加倍后，蓝色与红色的比例变成 $3:4$ 。

Doubling the red marbles changes the ratio to $3:4$

解答：

不妨设袋中共有 $5$ 颗弹珠；因为只关心比例，所以可以这样假设。于是蓝色有 $3$ 颗，红色有 $2$ 颗。

红色数量加倍后有 $4$ 颗红弹珠，因此红色所占比例为 $\dfrac{4}{4 + 3} = \dfrac{4}{7}\text{。}$

所以正确答案是 C。

WLOG, let the number of marbles in the bag be $5.$ Since we only care about ratios, we can do this. Then there are $3$ blue marbles and $2$ red marbles.

Doubling the red marbles gives us $4$ of them. Then the fraction of red marbles is $\dfrac{4}{4 + 3} = \dfrac{4}{7}.$

Thus, C is the correct answer.

8.

三个非负整数两两相加得到的和分别为 $12$ ， $17$ 和 $19$ 。这三个数中的中间那个数是多少？

The sums of three whole numbers taken in pairs are $12,$ $17,$ and $19.$ What is the middle number?

$4$

$5$

$6$

$7$

$8$

答案：D

知识点：方程组

难度评级：1020

小提示：

把三个两两和相加。

Add the three pair sums

大提示：

三个两两和的总和是三个数总和的两倍。

The total of the three pair sums is twice the sum of the three numbers

解答：

设三个数为 $a, b, c$ ，且 $a \lt b \lt c$ 。由于三个和都不同，所以这三个数也互不相等。

题意给出 $a + b = 12, a + c = 17\text{，}$ 以及 $b + c = 19\text{。}$

把三式相加，得到 $2(a + b + c) = 48\text{。}$

因此 $a + b + c = 24\text{，}$ 再减去 $a + c = 17\text{，}$ 得到 $b = 7$ 。

所以正确答案是 D。

Let the three numbers be $a, b, c$ where $a \lt b \lt c.$ None of them are equal, since all three sums are different.

Then $a + b = 12, a + c = 17,$ and $b + c = 19.$

Adding all three equations together gives us $2(a + b + c) = 48.$

Then $a + b + c = 24,$ from which we can subtract $a + c = 17,$ to get $b = 7.$

Thus, D is the correct answer.

9.

有一对均匀的六面骰子。一枚骰子的六个面上只标有偶数，其中 $2$ 、 $4$ 和 $6$ 各出现两次。另一枚骰子的六个面上只标有奇数，其中 $1$ 、 $3$ 和 $5$ 各出现两次。同时掷这两枚骰子，顶面点数之和为 $7$ 的概率是多少？

A pair of six-sided fair dice are labeled so that one die has only even numbers (two each of $2,$ $4,$ and $6$ ), and the other die has only odd numbers (two each of $1,$ $3,$ and $5$ ). The pair of dice is rolled. What is the probability that the sum of the numbers on the tops of the two dice is $7?$

$\dfrac{1}{6}$

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

答案：D

知识点：骰子（概率）分类讨论

难度评级：960

小提示：

列出所有和为 $7$ 的偶奇数对。

List the even-odd pairs that sum to $7$

大提示：

每个列出的点数在对应骰子上都有两个面。

Each listed value appears on two faces of its die

解答：

和为 $7$ 的点数对是 $(2, 5), (4, 3), \text{ 和 } (6, 1)\text{。}$

有 $\dfrac{1}{3} \cdot \dfrac{1}{3} = \dfrac{1}{9}$ 的概率得到其中任意一个指定的点数对。

共有 $3$ 个这样的点数对，所以和为 $7$ 的总概率为 $3 \cdot \dfrac{1}{9} = \dfrac{1}{3}\text{。}$

所以正确答案是 D。

The pairs of numbers that sum to $7$ are $(2, 5), (4, 3), \text{ and } (6, 1).$

There is a $\dfrac{1}{3} \cdot \dfrac{1}{3} = \dfrac{1}{9}$ chance that we get any of these pairs.

There are $3$ pairs, which means that the total probability that the rolls sum to $7$ is $3 \cdot \dfrac{1}{9} = \dfrac{1}{3}.$

Thus, D is the correct answer.

10.

Mary 把一个圆分成 $12$ 个扇形。这些扇形的圆心角度数都是整数，并且构成一个等差数列。最小的扇形角的度数最小可能是多少？

Mary divides a circle into $12$ sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?

$5$

$6$

$8$

$10$

$12$

答案：C

知识点：等差数列丢番图方程最优化

难度评级：1420

小提示：

设等差数列为 $a,a+d,\ldots,a+11d$ 。

Let the arithmetic sequence be $a,a+d,\ldots,a+11d$

大提示：

使用 $12a+66d=360$ ，并在 $a$ 保持正数时让 $d$ 尽可能大。

Use $12a+66d=360$ and maximize $d$ while keeping $a$ positive

解答：

设最小扇形角为 $a$ ，等差数列的公差为 $d$ 。

于是 $\dfrac{12}{2}(a + a + 11d) = 12a + 66d$ 就是这个等差数列的和。

总和是 $360$ ，所以 $12a + 66d = 360$ $2a + 11d = 60\text{。}$

要使 $a$ 尽可能小，就要让 $d$ 尽可能大。若 $d = 5$ ，则 $a$ 不是整数；因此 $d = 4$ ，且 $a = 8$ 。

所以正确答案是 C。

Let $a$ be the smallest possible sector angle and $d$ be the difference in the arithmetic sequence.

Then we have that $\dfrac{12}{2}(a + a + 11d) = 12a + 66d$ is the sum of the arithmetic sequence.

We have that this sums to $360,$ so $12a + 66d = 360$ $2a + 11d = 60.$

We want to minimize $a,$ so we maximize $d.$ If $d = 5,$ then $a$ is not an integer, so $d = 4$ and $a = 8.$

Thus, C is the correct answer.

11.

两个圆外切，圆心分别为 $A$ 和 $B$ ，半径分别为 $5$ 和 $3$ 。一条公共外切线与射线 $AB$ 相交于点 $C$ 。求 $BC$ 的长度。

Externally tangent circles with centers at points $A$ and $B$ have radii of lengths $5$ and $3,$ respectively. A line externally tangent to both circles intersects ray $AB$ at point $C.$ What is $BC?$

$4$

$4.8$

$10.2$

$12$

$14.4$

答案：D

知识点：相切圆切线相似

难度评级：1420

小提示：

从两个圆心分别向公共切线作垂线。

Draw radii to the two tangent points

大提示：

以 $C$ 为顶点的两个直角三角形相似。

The two right triangles with vertex $C$ are similar

解答：

令 $x$ 为 $BC$ 。注意 $\triangle CEB$ 与 $\triangle CDA$ 由角角判定相似（切线与过切点的半径垂直）。

于是 $\dfrac{x}{3} = \dfrac{8 + x}{5}\text{。}$ 交叉相乘得到 $5x = 24 + 3x$ $x = 12\text{。}$

所以正确答案是 D。

Let $x$ be $BC.$ Note that $\triangle CEB$ and $\triangle CDA$ are similar due to angle-angle (tangent lines are perpendicular to radii).

Then $\dfrac{x}{3} = \dfrac{8 + x}{5}.$ Cross-multiplying gives us $5x = 24 + 3x$ $x = 12.$

Thus, D is the correct answer.

12.

根据闰年规则，年份能被 $400$ 整除就是闰年，例如 $2000$ 年；或者能被 $4$ 整除但不能被 $100$ 整除也是闰年，例如 $2012$ 年。Charles Dickens 诞辰 $200$ 周年的庆祝日是 $2012$ 年二月 $7$ 日，星期二。他出生那天是星期几？

A year is a leap year if and only if the year number is divisible by $400$ (such as $2000$ ) or is divisible by $4$ but not $100$ (such as $2012$ ). The $200$ th anniversary of the birth of novelist Charles Dickens was celebrated on February $7,$ $2012,$ a Tuesday. On what day of the week was Dickens born?

星期五

Friday

星期六

Saturday

星期日

Sunday

星期一

Monday

星期二

Tuesday

答案：A

知识点：日期与时间模运算

难度评级：1540

小提示：

按模 $7$ 计算星期几的偏移。

Count day shifts modulo $7$

大提示：

这 $200$ 年中要计入闰日，但 $1900$ 年不是闰年。

In $200$ years, include leap days but exclude $1900$

解答：

平年有 $365$ 天，向未来移动一年，星期几会向后移动一天，因为 $365 = 52 \cdot 7 + 1\text{。}$

闰年有额外的一天，所以星期几会向后移动两天。

在 $1812$ 年到 $2012$ 年之间，有五十个年份是 $4$ 的倍数，但要去掉 $1900$ 年，所以共有 $49$ 个闰年。

因此，往回推 $200$ 年需要往回移动 $200 + 49 = 249 = 35 \cdot 7 + 4$ 天，也就是星期几往回移动 $4$ 天。

星期二往前推四天是星期五，也就是 $1812$ 年二月 $7$ 日。

所以正确答案是 A。

On a typical year with $365$ days, moving one year into the future moves the day of the week forward one since $365 = 52 \cdot 7 + 1.$

On a leap year however, the day of the week gets moved forward twice since there is an extra day.

Fifty of the years between $2012$ and $1812$ are multiples of $4,$ but we have to discard $1900,$ which leaves us with $49$ leap years.

Therefore, moving back $200$ years means we have to go back $200 + 49 = 249 = 35 \cdot 7 + 4$ days, which means we move back $4$ days in the week.

This takes us back to Friday, which is the day that corresponds to February $7,$ $1812.$

Thus, A is the correct answer.

13.

数 $1$ 、 $2$ 、 $3$ 、 $4$ 和 $5$ 的迭代平均数按如下方式计算。把这五个数按某种顺序排列。先求前两个数的平均数，再求它与第三个数的平均数，然后求所得结果与第四个数的平均数，最后求所得结果与第五个数的平均数。用这个过程能得到的最大可能值与最小可能值之差是多少？

An iterative average of the numbers $1,$ $2,$ $3,$ $4,$ and $5$ is computed the following way. Arrange the five numbers in some order. Find the mean of the first two numbers, then find the mean of that with the third number, then the mean of that with the fourth number, and finally the mean of that with the fifth number. What is the difference between the largest and smallest possible values that can be obtained using this procedure?

$\dfrac{31}{16}$

$2$

$\dfrac{17}{8}$

$3$

$\dfrac{65}{16}$

答案：C

知识点：平均数代数变形最优化

难度评级：1600

小提示：

把最终的平均数写成一个加权和。

Write the final average as a weighted sum

大提示：

后放入的数权重更大。

Later entries receive larger weights

解答：

设排列顺序为 $a, b, c, d, e\text{。}$

则迭代平均数为 $\dfrac{\dfrac{\dfrac{\dfrac{a + b}{2} + c}{2} + d}{2} + e}{2}$ $= \dfrac{a + b + 2c + 4d + 8e}{16}\text{。}$

要使结果最小，应按 $5, 4, 3, 2, 1\text{，}$ 排列，得到 $\dfrac{5 + 4 + 6 + 8 + 8}{16} = \dfrac{31}{16}\text{。}$

要使结果最大，则反过来排列，得到 $\dfrac{1 + 2 + 6 + 16 + 40}{16} = \dfrac{65}{16}\text{。}$

两者之差为 $\dfrac{65}{16} - \dfrac{31}{16} = \dfrac{34}{16} = \dfrac{17}{8}\text{。}$

所以正确答案是 C。

Let the order of the numbers be $a, b, c, d, e.$

Then the iterative average is $\dfrac{\dfrac{\dfrac{\dfrac{a + b}{2} + c}{2} + d}{2} + e}{2}$ $= \dfrac{a + b + 2c + 4d + 8e}{16}.$

To minimize this, we make the order $5, 4, 3, 2, 1,$ which gives us a sum of $\dfrac{5 + 4 + 6 + 8 + 8}{16} = \dfrac{31}{16}.$

To maximize it, we have to reverse this order to get an average of $\dfrac{1 + 2 + 6 + 16 + 40}{16} = \dfrac{65}{16}.$

The difference between these is $\dfrac{65}{16} - \dfrac{31}{16} = \dfrac{34}{16} = \dfrac{17}{8}.$

Thus, C is the correct answer.

14.

Chubby 制作每边有 $31$ 个方格的非标准棋盘。棋盘的四个角都是黑格，并且每一行、每一列都红黑相间。这样的棋盘上有多少个黑格？

Chubby makes nonstandard checkerboards that have $31$ squares on each side. The checkerboards have a black square in every corner and alternate red and black squares along every row and column. How many black squares are there on such a checkerboard?

$480$

$481$

$482$

$483$

$484$

答案：B

知识点：基本计数奇偶性找规律

难度评级：1020

小提示：

每一行有 $16$ 个或 $15$ 个黑格。

Rows alternate between $16$ and $15$ black squares

大提示：

有 $16$ 行以黑格开始，另外 $15$ 行以白格开始。

There are $16$ rows of one type and $15$ of the other

解答：

共有 $15$ 行各含 $15$ 个黑格，另有 $16$ 行各含 $16$ 个黑格。

第一行有 $16$ 个黑格，其余各行按 $15$ 个与 $16$ 个交替。

因此共有 $16$ 行各含 $16$ 个黑格，其余各行含 $15$ 个黑格。

因此黑格总数为 $15^2 + 16^2 = 481\text{。}$

所以正确答案是 B。

Note that there are $15$ rows with $15$ black tiles and $16$ rows with $16$ black tiles.

This can be seen by observing that the first row has $16$ black tiles, and all the other rows alternate with $15$ and $16$ tiles.

Then, due to the alternating pattern, there will be a total of $16$ rows with $16$ tiles, and the other rows have $15$ tiles.

The total number of black squares is then $15^2 + 16^2 = 481.$

Thus, B is the correct answer.

15.

下图由三个单位正方形和两条线段组成。求 $\triangle ABC$ 的面积。

Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC?$

$\dfrac16$

$\dfrac15$

$\dfrac29$

$\dfrac13$

$\dfrac{\sqrt{2}}{4}$

答案：B

知识点：坐标几何三角形面积

难度评级：1480

小提示：

将 $A$ 和 $B$ 放在坐标轴上。

Place $A$ and $B$ on coordinate axes

大提示：

求出两条所画直线的交点。

Find the intersection of the two drawn lines

解答：

可以用坐标几何求两条直线的交点。

令 $A$ 为原点， $B = (1, 0)$ 。经过 $A$ 的直线斜率为 $-\dfrac{1}{2}$ ，方程为 $y = -\dfrac{1}{2}x\text{。}$ 经过 $B$ 的直线斜率为 $2$ ， $y$ 轴截距为 $-2$ ，所以方程为 $y = 2x - 2\text{。}$

联立两式，得到 $2x - 2 = -\dfrac{1}{2}x$ $x = \dfrac{4}{5}\text{。}$

因此点 $C$ 的 $y$ 坐标为 $-\dfrac{1}{2} \cdot \dfrac{4}{5} = -\dfrac{2}{5}\text{。}$

三角形 $ABC$ 的面积为 $\dfrac{1}{2} \cdot 1 \cdot \dfrac{2}{5} = \dfrac{1}{5}\text{。}$

所以正确答案是 B。

We can use coordinate geometry to figure out where the intersection of the two lines occurs.

Let $A$ be the origin and $B = (1, 0).$ Then the slope of the line through $A$ is $-\dfrac{1}{2},$ which makes the equation of the line $y = -\dfrac{1}{2}x.$ The slope of the line through $B$ is $2.$ The $y$ -intercept is $-2.$ This makes the equation of this line $y = 2x - 2.$

Equating the equations, we get $2x - 2 = -\dfrac{1}{2}x$ $x = \dfrac{4}{5}.$

This makes the $y$ -coordinate of $C$ $-\dfrac{1}{2} \cdot \dfrac{4}{5} = -\dfrac{2}{5}.$

The area of triangle $ABC$ is then $\dfrac{1}{2} \cdot 1 \cdot \dfrac{2}{5} = \dfrac{1}{5}.$

Thus, B is the correct answer.

16.

三名跑步者同时从一条 $500$ 米环形跑道的同一点出发，沿顺时针方向跑。他们的速度分别为每秒 $4.4$ ， $4.8$ 和 $5.0$ 米。三人第一次再次同时相遇时，他们已经跑了多少秒？

Three runners start running simultaneously from the same point on a $500$ -meter circular track. They each run clockwise around the course maintaining constant speeds of $4.4,$ $4.8,$ and $5.0$ meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?

$1{,}000$

$1{,}250$

$2{,}500$

$5{,}000$

$10{,}000$

答案：C

知识点：相对速度最小公倍数路程、速度与时间

难度评级：1600

小提示：

利用环形跑道上的相对速度。

Use relative speeds on the circular track

大提示：

求中等速度的跑者和最快的跑者何时都以整圈数领先最慢的跑者。

Find when the middle and fast runners lap the slow runner by whole laps

解答：

先求速度为 $4.8$ 米/秒的跑者追上最慢跑者所需的时间。

必须有 $4.8x - 4.4x = 500$ $x = 1250\text{，}$ 其中 $x$ 是较快跑者追上另一人的时间。

注意 $4.4 \cdot 1250 = 5500$ ，所以这两名跑者每次相遇都在起点。

现在要求最小的 $t$ ，使 $t$ 是 $1250$ 的倍数，并且最快的跑者也回到起点。

每 $1250$ 秒，最快的跑者跑 $1250 \cdot 5 = 6250$ 米；到 $2500$ 秒时，他跑了 $12500$ 米，正好是整数圈。

所以正确答案是 C。

Let us find the amount of time that it takes for the runner running at $4.8$ meters per second to lap the slowest person.

We must have that $4.8x - 4.4x = 500$ $x = 1250,$ where $x$ is the amount of time it takes for the faster runner to lap the other.

Note that $4.4 \cdot 1250 = 5500,$ which means that these two runners always intersect at the starting line.

We now have to find the least time, $t,$ such that $t$ is a multiple of $1250$ and the fastest runner ends up at the starting line.

Every $1250$ seconds, the fastest runner runs $1250 \cdot 5 = 6250$ meters. Then in $2500$ seconds, the fastest runner runs $12500$ meters, which is a whole number of laps.

Thus, C is the correct answer.

17.

整数 $a$ 和 $b$ 互质，满足 $a > b > 0$ ，且 $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}\text{。}$ 求 $a-b$ 。

Let $a$ and $b$ be relatively prime integers with $a > b > 0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}.$ What is $a-b?$

$1$

$2$

$3$

$4$

$5$

答案：C

知识点：因式分解分式方程二次方程

难度评级：1670

小提示：

分解 $a^3-b^3$ 。

Factor $a^3-b^3$

大提示：

方程会化为关于 $\frac{a}{b}$ 的二次方程。

The equation becomes a quadratic in $\frac{a}{b}$

解答：

先因式分解 $a^3 - b^3 = (a - b)(a^2 + ab + b^2)\text{。}$ 消去这个因子后得到 $\dfrac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \dfrac{73}{3}\text{。}$

交叉相乘并整理，得到 $70a^2 - 149ab + 70b^2 = 0\text{。}$ 因为 $b \neq 0$ ，可同除以 $b^2$ ，得到 $70\left(\dfrac{a}{b}\right)^2 - 149\dfrac{a}{b} + 70 = 0\text{。}$

用二次公式，并注意 $a \gt b$ ，可得 $\dfrac{a}{b} = \dfrac{10}{7}$ 。

由于 $a$ 和 $b$ 互质，所以 $a = 10$ ， $b = 7$ ，差为 $10 - 7 = 3$ 。

所以正确答案是 C。

Recall that we can factor $a^3 - b^3 = (a - b)(a^2 + ab + b^2).$ Canceling out this factor gives us that $\dfrac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \dfrac{73}{3}.$

Cross-multiplying and rearranging gives us $70a^2 - 149ab + 70b^2 = 0.$ Since $b \neq 0,$ we can divide through by $b^2$ to get $70\left(\dfrac{a}{b}\right)^2 - 149\dfrac{a}{b} + 70 = 0.$

Applying the quadratic formula and noting that $a \gt b$ gives us that $\dfrac{a}{b} = \dfrac{10}{7}.$

Since $a$ and $b$ are relatively prime, we have that $a = 10$ and $b = 7.$ Their difference is $10 - 7 = 3.$

Thus, C is the correct answer.

18.

图中的闭合曲线由 $9$ 段全等圆弧组成，每段圆弧长为 $\dfrac{2\pi}{3}$ ，每段对应圆的圆心都是边长为 $2$ 的正六边形的某个顶点。该闭合曲线围成的面积是多少？

The closed curve in the figure is made up of $9$ congruent circular arcs each of length $\dfrac{2\pi}{3},$ where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side $2.$ What is the area enclosed by the curve?

$2\pi+6$

$2\pi+4\sqrt{3}$

$3\pi+4$

$2\pi+3\sqrt{3}+2$

$\pi+6\sqrt{3}$

答案：E

知识点：扇形正多边形面积分割

难度评级：1930

小提示：

每段圆弧都是单位圆上的 $120^\circ$ 弧。

Each arc is a $120^\circ$ arc of a unit circle

大提示：

把这些扇形块围绕正六边形重新拼接。

Rearrange the sector pieces around the regular hexagon

解答：

每段圆弧的半径都是 $1$ ，长度都是 $\frac{2\pi}{3}$ ，所以对应的各个扇形面积都相同。如图所示，这些扇形块可以重新拼接：添加到正六边形上的部分合起来恰好构成一个单位圆。

正六边形由六个边长为 $2$ 的等边三角形拼成，所以它的面积为 $6\left(\dfrac{\sqrt3}{4}\cdot2^2\right)=6\sqrt3\text{。}$ 再加上那个单位圆，围成的总面积为 $\pi+6\sqrt3$ 。

所以正确答案是 E。

Each arc has radius $1$ and length $\frac{2\pi}{3},$ so the corresponding circular sectors all have the same area. As the diagram shows, the sector pieces may be rearranged: everything added to the regular hexagon combines to exactly one unit circle.

The regular hexagon is made of six equilateral triangles of side $2,$ so its area is $6\left(\dfrac{\sqrt3}{4}\cdot2^2\right)=6\sqrt3.$ Adding the unit circle gives a total enclosed area of $\pi+6\sqrt3.$

Thus, E is the correct answer.

19.

油漆工 Paula 和她的两名助手各自以固定但不同的速度刷漆。他们总是上午 $8:00$ 开始工作，并且三人每天午餐休息的时间相同。

星期一，三人一起刷完了一栋房子的 $50\%$ ，并在下午 $4:00$ 停工。星期二，Paula 不在，两名助手只刷完了房子的 $24\%$ ，并在下午 $2:12$ 停工。星期三，Paula 独自工作，到晚上 $7:12$ 完成整栋房子。

每天的午餐休息是多少分钟？

Paula the painter and her two helpers each paint at constant, but different, rates. They always start at $8:00$ AM, and all three always take the same amount of time to eat lunch.

On Monday the three of them painted $50\%$ of a house, quitting at $4:00$ PM. On Tuesday, when Paula wasn’t there, the two helpers painted only $24\%$ of the house and quit at $2:12$ PM. On Wednesday Paula worked by herself and finished the house by working until $7:12$ P.M.

How long, in minutes, was each day’s lunch break?

$30$

$36$

$42$

$48$

$60$

答案：D

知识点：速率方程组

难度评级：2060

小提示：

设午餐休息为 $m$ 分钟。

Let the lunch break be $m$ minutes

大提示：

使用星期一对应 $p+h$ ，星期二对应 $h$ ，星期三对应 $p$ 。

Use Monday for $p+h$ , Tuesday for $h$ , and Wednesday for $p$

解答：

设午餐休息为 $m$ 分钟，Paula 的工作速度为每分钟 $p$ 个百分点，两名助手的合计速度为每分钟 $h$ 个百分点。

星期一给出 $(p+h)(480-m)=50$ 。星期二给出 $h(372-m)=24$ 。星期三剩余工作量为百分之 $26$ ，所以 $p(672-m)=26$ 。

由后两式相加再减去第一式，得到 $108h-192p=0$ ，所以 $h=\dfrac{16}{9}p$ 。

代入星期一的方程得 $\dfrac{25}{9}p(480-m)=50$ ，而星期三方程为 $p(672-m)=26$ 。联立解得 $m=48$ 。

所以正确答案是 D。

Let the lunch break be $m$ minutes, Paula’s rate be $p$ percent per minute, and the helpers’ combined rate be $h$ percent per minute.

Monday gives $(p+h)(480-m)=50$ . Tuesday gives $h(372-m)=24$ . Since the remaining work on Wednesday was $26$ percent, Wednesday gives $p(672-m)=26$ .

Adding the Tuesday and Wednesday equations and subtracting the Monday equation gives $108h-192p=0$ , so $h=\dfrac{16}{9}p$ .

Substituting into Monday gives $\dfrac{25}{9}p(480-m)=50$ , while Wednesday gives $p(672-m)=26$ . Solving these two equations gives $m=48$ .

Thus, D is the correct answer.

20.

一个 $3 \times 3$ 的正方形被分成 $9$ 个单位小方格。每个小方格独立随机地染成白色或黑色，两种颜色的可能性相同。

然后把整个正方形绕中心顺时针旋转 $90^{\circ}$ ，并把每个位于原来黑格位置上的白格染成黑色，其余方格的颜色都保持不变。最终整个网格全为黑色的概率是多少？

A $3 \times 3$ square is partitioned into $9$ unit squares. Each unit square is painted either white or black with each color being equally likely, chosen independently and at random.

The square is then rotated $90^{\circ}$ clockwise about its center, and every white square in a position formerly occupied by a black square is painted black. The colors of all other squares are left unchanged. What is the probability the grid is now entirely black?

$\dfrac{49}{512}$

$\dfrac{7}{64}$

$\dfrac{121}{1024}$

$\dfrac{81}{512}$

$\dfrac{9}{32}$

答案：A

知识点：基本概率独立事件对称性分类讨论

难度评级：1980

小提示：

四个角在旋转下形成一个 $4$ 循环。

Corner squares form a $4$ -cycle under rotation

大提示：

对每个 $4$ 循环，数出不会让白格接到白色前驱的染色情况。

For each $4$ -cycle, count colorings with no white square receiving a white predecessor

解答：

中心格必须一开始就是黑色，贡献概率 $\dfrac12$ 。四个角在旋转下形成一个循环，四个边中点也形成另一个相同的循环。

对一个 $4$ 循环，除非有一个白格被旋转到原本也是白格的位置，否则最后四个位置都会变黑。成功的初始颜色序列为 $BBBB$ 、 $BBBW$ 的四种旋转，以及 $BWBW$ 的两种旋转，共 $7$ 种，占 $16$ 种。

边中点循环有相同的独立计数。因此概率为 $\dfrac12\left(\dfrac7{16}\right)^2=\dfrac{49}{512}$ 。

所以正确答案是 A。

The center square must initially be black, contributing probability $\dfrac12$ . The four corner squares form one cycle under the rotation, and the four edge-middle squares form another identical cycle.

For one $4$ -cycle, the final four positions are all black unless a white square is rotated into a position that was also white. The successful initial colorings are $BBBB$ , the four rotations of $BBBW$ , and the two rotations of $BWBW$ , for $7$ colorings out of $16$ .

The same count applies to the edge-middle cycle, independently. Therefore the probability is $\dfrac12\left(\dfrac7{16}\right)^2=\dfrac{49}{512}$ .

Thus, A is the correct answer.

21.

设点 $A=(0,0,0)$ 、 $B=(1,0,0)$ 、 $C=(0,2,0)$ 和 $D=(0,0,3)$ 。点 $E$ 、 $F$ 、 $G$ 和 $H$ 分别是线段 $\overline{BD}$ ， $\text{ } \overline{AB}$ ， $\text{ } \overline {AC}$ 和 $\overline{DC}$ 的中点。求四边形 $EFGH$ 的面积。

Let points $A=(0,0,0),$ $B=(1,0,0),$ $C=(0,2,0),$ and $D=(0,0,3).$ Points $E,$ $F,$ $G,$ and $H$ are midpoints of line segments $\overline{BD},$ $\text{ } \overline{AB},$ $\text{ } \overline {AC},$ and $\overline{DC}$ respectively. What is the area of $EFGH?$

$\sqrt{2}$

$\dfrac{2\sqrt{5}}{3}$

$\dfrac{3\sqrt{5}}{4}$

$\sqrt{3}$

$\dfrac{2\sqrt{7}}{3}$

答案：C

知识点：立体几何中点距离公式

难度评级：1730

小提示：

在三角形 $ABD$ 和 $ABC$ 中使用中位线。

Use midsegments in triangles $ABD$ and $ABC$

大提示：

这个四边形是长方形，边长可由 $AD$ 和 $BC$ 求出。

The quadrilateral is a rectangle with side lengths from $AD$ and $BC$

解答：

注意 $EF = \dfrac{1}{2}AD$ ，因为它是 $\triangle ABD$ 的中位线。同理， $HG = \dfrac{1}{2}AD$ ，且 $FG = \dfrac{1}{2}BC$ 。

又因为 $\overline{EF}$ 和 $\overline{HG}$ 垂直于 $xy$ 平面，所以它们垂直于 $\overline{FG}$ 和 $\overline{EH}$ 。

因此 $EFGH$ 是长方形，且 $EF = HG$ 。有 $EF = \dfrac{1}{2} \cdot 3 = \dfrac{3}{2}$ 。

还可得 $FG = \dfrac{1}{2} \sqrt{1^2 + 2^2} = \dfrac{\sqrt{5}}{2}\text{。}$

所以 $EFGH$ 的面积为 $EF \cdot FG = \dfrac{3}{2} \cdot \dfrac{\sqrt{5}}{2} = \dfrac{3\sqrt{5}}{4}\text{。}$

所以正确答案是 C。

Note that $EF = \dfrac{1}{2}AD$ since it is a midsegment of $\triangle ABD.$ Similarly, $HG = \dfrac{1}{2}AD$ and $FG = \dfrac{1}{2}BC.$

We also have that $\overline{EF}$ and $\overline{HG}$ are perpendicular to the $xy$ -plane, which means that they are perpendicular to $\overline{FG}$ and $\overline{EH}.$

This tells us that $EFGH$ is rectangle since $EF = HG.$ We have $EF = \dfrac{1}{2} \cdot 3 = \dfrac{3}{2}.$

We also have that $FG = \dfrac{1}{2} \sqrt{1^2 + 2^2} = \dfrac{\sqrt{5}}{2}.$

The area of $EFGH$ is then $EF \cdot FG = \dfrac{3}{2} \cdot \dfrac{\sqrt{5}}{2} = \dfrac{3\sqrt{5}}{4}.$

Thus, C is the correct answer.

22.

前 $m$ 个正奇数的和比前 $n$ 个正偶数的和多 $212$ 。所有可能的 $n$ 的和是多少？

The sum of the first $m$ positive odd integers is $212$ more than the sum of the first $n$ positive even integers. What is the sum of all possible values of $n?$

$255$

$256$

$257$

$258$

$259$

答案：A

知识点：求和平方差丢番图方程

难度评级：2260

小提示：

使用前 $m$ 个正奇数之和为 $m^2$ 。

Use that the first $m$ odd numbers sum to $m^2$

大提示：

把条件化为 $(2m+p)(2m-p)=847$ 。

Turn the condition into $(2m+p)(2m-p)=847$

解答：

前 $m$ 个正奇数之和为 $m^2$ ，前 $n$ 个正偶数之和为 $n(n+1)$ 。因此 $m^2=n(n+1)+212$ 。

把它看成关于 $n$ 的二次方程，判别式为 $1-4(212-m^2)=4m^2-847$ ，必须是一个奇平方数。设 $p^2=4m^2-847$ ，则 $(2m+p)(2m-p)=847$ 。

$847$ 的正因数配对为 $847\cdot1$ 、 $121\cdot7$ 和 $77\cdot11$ 。它们分别给出 $p=423,57,33$ 。

因为 $n=\dfrac{-1+p}{2}$ ，所以 $n$ 的可能值为 $211,28,16$ ，它们的和为 $255$ 。

所以正确答案是 A。

The first $m$ positive odd integers sum to $m^2$ , and the first $n$ positive even integers sum to $n(n+1)$ . Thus $m^2=n(n+1)+212$ .

As a quadratic in $n$ , this has discriminant $1-4(212-m^2)=4m^2-847$ , which must be an odd square. Let $p^2=4m^2-847$ . Then $(2m+p)(2m-p)=847$ .

The positive factor pairs of $847$ are $847\cdot1$ , $121\cdot7$ , and $77\cdot11$ . They give $p=423,57,33$ , respectively.

Because $n=\dfrac{-1+p}{2}$ , the possible values of $n$ are $211,28,16$ . Their sum is $255$ .

Thus, A is the correct answer.

23.

Adam、Benin、Chiang、Deshawn、Esther 和 Fiona 都有网络账号。在这六个人之间，有些但不是所有人互为好友；他们没有小组外的好友。若每个人的好友数都相同，则可能的好友关系图共有多少种？

Adam, Benin, Chiang, Deshawn, Esther, and Fiona have internet accounts. Some, but not all, of them are internet friends with each other, and none of them has an internet friend outside this group. Each of them has the same number of internet friends. In how many different ways can this happen?

$60$

$170$

$290$

$320$

$660$

答案：B

知识点：图论分类讨论双射

难度评级：2200

小提示：

按共同的好友数分类。

Case on the common number of friends

大提示：

好友图的补图会把度数 $1$ 与 $4$ 、度数 $2$ 与 $3$ 配对。

Use complements to pair the $1$ -friend and $4$ -friend cases, and the $2$ -friend and $3$ -friend cases

解答：

按每个人拥有的好友数分类。由于好友图既不是空图也不是完全图，这个数从 $1$ 到 $4$ 。

注意， $1$ 个好友和 $2$ 个好友的情况，分别通过取补图与 $4$ 个好友和 $3$ 个好友的情况对应，因为确定谁是好友也就确定了谁不是好友。

情况 $1$ ：每个人有 $1$ 个好友

这意味着 $6$ 个人必须分成 $3$ 对，每对中的两人互为好友。

第一个人的好友有 $5$ 种选择，剩下 $4$ 个人。

下一位未配对者的好友有 $3$ 种选择，剩余的 $2$ 个人则必须互为好友。

因此这种情况共有 $3 \cdot 5 = 15$ 种可能。

情况 $2$ ：每个人有 $2$ 个好友

这种情况有两种可能。第一种是分成两个三人组，每组三人彼此都是好友。

选择第一个三人组有 $\binom{6}{3} = 20$ 种方法。由于两个组可以互换，必须除以 $2$ ，得到 $20 \div 2 = 10$ 种配置。

第二种可能是好友关系形成一个 $6$ 环。

六个人沿环的每一种排列都给出这样的图。选择起点会使每个图被计算 $6$ 次，选择遍历方向又会被计算 $2$ 次，因此不同的 $6$ 环共有 $\frac{6!}{6\cdot2}=60$ 个。再加上 $10$ 个两三角形配置，本情况共有 $10+60=70$ 种配置。

所以总配置数为 $2(15 + 70) = 170\text{。}$

所以正确答案是 B。

We case on the value of friends that each person has. This value ranges from $1$ to $4$ , since the graph is neither empty nor complete.

Note that the cases for $1$ and $2$ friends correspond with the case for $4$ and $3$ friends, since choosing who are friends determines who are not friends.

Case $1:$ everyone has $1$ friend

This means that the $6$ people must split up into $3$ pairs where the people in each pair are friends.

There are $5$ choices for the friend for the first person. This leaves $4$ people remaining.

There are then $3$ choices for the friend of the next unpaired person. The remaining $2$ people are then forced to be friends.

Therefore, there are $3 \cdot 5 = 15$ possibilities for this case.

Case $2:$ everyone has $2$ friends

There are two possibilities for this case. There could be two triples where everyone in a triple is friends with each other.

For this possibility, there are $\binom{6}{3} = 20$ ways to choose the people in the first triple. We have to divide by $2$ since we can swap the pairs. This gives us $20 \div 2 = 10$ configurations.

The second possibility is that the friends form one $6$ -cycle.

Every ordering of the six people around a cycle gives such a graph. Each graph is counted $6$ times by the choice of starting person and $2$ times by the direction of traversal, so there are $\frac{6!}{6\cdot2}=60$ distinct $6$ -cycles. Together with the $10$ pairs of triangles, this case has $10+60=70$ configurations.

The total number of arrangements is then $2(15 + 70) = 170.$

Thus, B is the correct answer.

24.

正整数 $a$ 、 $b$ 和 $c$ 满足 $a\ge b\ge c$ ，且 $a^2-b^2-c^2+ab=2011$ 以及 $a^2+3b^2+3c^2-3ab-2ac-2bc$ $=-1997\text{。}$ 求 $a$ 。

Let $a,$ $b,$ and $c$ be positive integers with $a\ge b\ge c$ such that $a^2-b^2-c^2+ab=2011$ and $a^2+3b^2+3c^2-3ab-2ac-2bc$ $=-1997.$ What is $a?$

$249$

$250$

$251$

$252$

$253$

答案：E

知识点：代数变形丢番图方程分类讨论

难度评级：2350

小提示：

先把两个方程相加。

Add the two equations first

大提示：

三个平方差必须以某种顺序等于 $9,4,1$ 。

The three squared differences must be $9,4,1$ in some order

解答：

两个方程相加，得到 $2(a^2 + b^2 + c^2) - 2(ab + ac + bc)$ $= 14\text{。}$

把左边分组并因式分解，得到 $(a - b)^2 + (a - c)^2 + (b - c)^2$ $= 14\text{。}$

左边每一项都是非负整数的平方。三个平方数和为 $14$ 的唯一方式是 $9, 4$ 和 $1$ 。

在三组差中， $a - c$ 最大，所以 $a - c = 3$ 。

还不能确定另外两个差分别对应哪个平方。先试 $a - b = 1$ 和 $b - c = 2$ 。

把这些值代入第一个方程，得到 $a^2 - (a - 1)^2 - (a - 3)^2$ $+ a(a - 1) = 2011\text{。}$ 化简得 $7a = 2021$ 。由于 $2021$ 不能被 $7$ 整除，所以应有 $a - b = 2$ 、 $b - c = 1$ 。

在另一种情形下， $b=a-2$ ， $c=a-3$ 。第一个方程变为 $\begin{aligned} a^2-(a-2)^2-(a-3)^2\\ {}+a(a-2)&=2011 \end{aligned}\text{，}$ 即 $8a-13=2011$ 。因此 $a=253$ 。

所以正确答案是 E。

Adding together the equations gives us $2(a^2 + b^2 + c^2) - 2(ab + ac + bc)$ $= 14.$

We can group terms and factor this to get $(a - b)^2 + (a - c)^2 + (b - c)^2$ $= 14.$

Note that every term on the left hand side is a nonnegative square integer. The only triple of squares that add to $14$ is $9, 4,$ and $1.$

We have that $a - c$ is the biggest difference among the three pairs. Therefore, $a - c = 3.$

We cannot discern which of the other terms we can match with the other squares. Let us try $a - b = 1$ and $b - c = 2.$

Plugging in these values into the first equation gives us $a^2 - (a - 1)^2 - (a - 3)^2$ $+ a(a - 1) = 2011.$ Simplifying yields $7a = 2021.$ Since $2021$ is not divisible by $7,$ we have that $a - b = 2$ and $b - c = 1.$

In the other case, $b=a-2$ and $c=a-3.$ The first equation becomes $\begin{aligned} a^2-(a-2)^2-(a-3)^2\\ {}+a(a-2)&=2011, \end{aligned}$ or $8a-13=2011.$ Hence $a=253.$

Thus, E is the correct answer.

25.

实数 $x$ 、 $y$ 和 $z$ 独立且均匀地从区间 $[0,n]$ 中随机选取，其中 $n$ 是某个正整数。 $x$ 、 $y$ 和 $z$ 中任意两个数的距离都不小于 $1$ 的概率超过 $\dfrac{1}{2}$ 。满足条件的最小 $n$ 是多少？

Real numbers $x,$ $y,$ and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n.$ The probability that no two of $x,$ $y,$ and $z$ are within $1$ unit of each other is greater than $\dfrac{1}{2}.$ What is the smallest possible value of $n?$

$7$

$8$

$9$

$10$

$11$

答案：D

知识点：几何概率体积对称性

难度评级：2460

小提示：

先把三个随机数排序。

Order the three random numbers first

大提示：

从中间的数减去 $1$ ，从最大的数减去 $2$ 。

Subtract $1$ from the middle number and $2$ from the largest number

解答：

这个问题可以看作几何概率问题，把区间看成坐标轴上的范围。

不妨先考虑 $n \geq x \geq y \geq z \geq 0\text{。}$

满足这个顺序限制的点 $(x, y, z)$ 构成一个四面体。

这个四面体的高为 $n$ ，底面积为 $\dfrac{n^2}{2}$ ，所以体积为 $\dfrac{1}{3} \cdot \dfrac{n^2}{2} \cdot n = \dfrac{n^3}{6}\text{。}$

现在加入题目中的限制。需要找出满足下面条件的区域： $|x - y|, |x - z|, |y - z| \geq 1\text{。}$

根据已经规定的顺序，这些不等式可化为 $x - y \geq 1 \text{ 和 } y - z \geq 1\text{。}$

这两个限制形成如下图所示的另一个四面体。

注意，在新的四面体中，所有线性尺寸都减少了 $2$ 。因此高为 $n - 2$ ，底面积为 $\dfrac{(n - 2)^2}{2}$ 。

因此体积为 $\dfrac{1}{3} \cdot \dfrac{(n - 2)^2}{2} \cdot (n - 2)$ $= \dfrac{(n - 2)^3}{6}\text{。}$

所求概率为 $\dfrac{(n - 2)^3}{6} \div \dfrac{n^3}{6} = \dfrac{(n - 2)^3}{n^3}\text{。}$

逐一检查选项可知，使这个比例大于 $\dfrac{1}{2}$ 的最小值是 $10$ 。

所以正确答案是 D。

This problem lends itself to geometric probability since we can view the interval as a range on an axis.

WLOG, let $n \geq x \geq y \geq z \geq 0.$

Then we have that the points $(x, y, z)$ which satisfy this restriction form a tetrahedron.

The height of this tetrahedron is $n,$ and the base has an area of $\dfrac{n^2}{2}.$ This makes the volume $\dfrac{1}{3} \cdot \dfrac{n^2}{2} \cdot n = \dfrac{n^3}{6}.$

Now we have to apply the restrictions from the problem statement. We need to find the region where $|x - y|, |x - z|, |y - z| \geq 1.$

From our ordering condition that we imposed, these inequalities reduce to $x - y \geq 1 \text{ and } y - z \geq 1.$

These two restrictions form another tetrahedron as shown below.

Note that in the new tetrahedron, all the dimensions have been reduced by $2.$ This makes the height $n - 2$ and the base $\dfrac{(n - 2)^2}{2}.$

The volume is then $\dfrac{1}{3} \cdot \dfrac{(n - 2)^2}{2} \cdot (n - 2)$ $= \dfrac{(n - 2)^3}{6}.$

The desired probability is then $\dfrac{(n - 2)^3}{6} \div \dfrac{n^3}{6} = \dfrac{(n - 2)^3}{n^3}.$

Plugging in all the answer choices, we get that the smallest value such that this fraction is greater than $\dfrac{1}{2}$ is $10.$

Thus, D is the correct answer.