2011 AMC 12A 详解 | LIVE by Po-Shen Loh

1.

一个手机套餐每月费用为 $\$20$ 外加每条短信 $5$ ¢，以及超过 $30$ 小时后每分钟 $10$ ¢。Michelle 一月份发送了 $100$ 条短信，并通话 $30.5$ 小时。她需要支付多少钱？

A cell phone plan costs $\$20$ each month, plus $5$ ¢ per text message sent, plus $10$ ¢ for each minute used over $30$ hours. In January Michelle sent $100$ text messages and talked for $30.5$ hours. How much did she have to pay?

$\$24.00$

$\$24.50$

$\$25.50$

$\$28.00$

$\$30.00$

知识点：钱币单位换算

难度评级：840

小提示：

$30.5$ 小时比 $30$ 小时套餐多出 $30$ 分钟

$30.5$ hours is $30$ minutes over the $30$ -hour allowance

大提示：

把基本费用、短信 $5$ ¢ 乘以条数、以及超时分钟数乘以 $10$ ¢ 相加

Add the base fee, $5$ ¢ times the texts, and $10$ ¢ times the overage minutes

解答：

短信费用为 $100 \cdot 5 = 500$ 美分 $= \$5$ 。她比 $30$ 小时套餐多通话了 $30$ 分钟，因此超时费用为 $30 \cdot 10 = 300$ 美分 $= \$3$ 。

总费用为 $\$20 + \$5 + \$3 = \$28$ 。

因此，正确答案是 D。

The text charge is $100 \cdot 5 = 500$ cents $= \$5.$ She talked $30$ minutes past the $30$ -hour allowance, so the overage is $30 \cdot 10 = 300$ cents $= \$3.$

The total is $\$20 + \$5 + \$3 = \$28.$

Thus, the correct answer is D.

2.

有 $5$ 枚硬币按图示平放在桌上。按从上到下的顺序，这些硬币应怎样排列？

There are $5$ coins placed flat on a table according to the figure. What is the order of the coins from top to bottom?

$(C, A, E, D, B)$

$(C, A, D, E, B)$

$(C, D, E, A, B)$

$(C, E, A, D, B)$

$(C, E, D, A, B)$

知识点：逻辑推理

难度评级：840

小提示：

轮廓完整、没有被打断的硬币在最上面

A coin whose outline is a complete, unbroken circle lies on top

大提示：

在每个重叠处，弧线没有被打断的那枚硬币在上方

At each overlap, the coin whose arc is drawn without interruption is the higher one

解答：

硬币 $C$ 画成一个完整且没有中断的圆，所以没有东西盖住它，它在最上面。

继续观察其余重叠处，未被遮住的弧线显示每枚硬币在下一枚硬币上方： $C$ 盖住 $E$ ， $E$ 盖住 $D$ ， $D$ 盖住 $B$ 而 $A$ 盖住 $B$ ，同时在其他硬币下方。因此从上到下的顺序是 $(C, E, D, A, B)$ 。

因此，正确答案是 E。

Coin $C$ is drawn as a complete, unbroken circle, so nothing covers it and it lies on top.

Reading the remaining overlaps, each coin’s uncovered arc shows it sits above the next: $C$ covers $E,$ $E$ covers $D,$ $D$ covers $B,$ and $A$ covers $B$ while lying under the others. This gives the top-to-bottom order $(C, E, D, A, B).$

Thus, the correct answer is E.

3.

一个小洗发水瓶可装 $35$ 毫升洗发水，而一个大瓶可装 $500$ 毫升洗发水。Jasmine 想购买最少数量的小瓶，足以完全装满一个大瓶。她必须买多少瓶？

A small bottle of shampoo can hold $35$ milliliters of shampoo, whereas a large bottle can hold $500$ milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?

$11$

$12$

$13$

$14$

$15$

知识点：取整函数估算

难度评级：880

小提示：

计算 $500$ 除以 $35$

Divide $500$ by $35$

大提示：

因为 $14$ 瓶不够 $500$ ，要把商向上取整

Since $14$ bottles fall short of $500,$ round the quotient up

解答：

十四瓶可装 $14 \cdot 35 = 490$ 毫升，这还不够。十五瓶可装 $15 \cdot 35 = 525$ 毫升，已经足够。

所以 Jasmine 需要 $15$ 瓶。

因此，正确答案是 E。

Fourteen bottles hold $14 \cdot 35 = 490$ milliliters, which is not enough. Fifteen bottles hold $15 \cdot 35 = 525$ milliliters, which suffices.

So Jasmine needs $15$ bottles.

Thus, the correct answer is E.

4.

在一所小学，三年级、四年级、五年级学生每天平均分别跑 $12$ ， $15$ ，和 $10$ 分钟。三年级学生人数是四年级的两倍，四年级学生人数是五年级的两倍。这些学生每天平均跑多少分钟？

At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of $12,$ $15,$ and $10$ minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students?

$12$

$\dfrac{37}{3}$

$\dfrac{88}{7}$

$13$

$14$

知识点：加权平均数比与比例

难度评级：1040

小提示：

设五年级学生人数为 $1$ 则四年级为 $2$ ，三年级为 $4$

Let the number of fifth graders be $1,$ so fourth graders is $2$ and third graders is $4$

大提示：

平均数为 $\dfrac{4 \cdot 12 + 2 \cdot 15 + 1 \cdot 10}{4 + 2 + 1}$

The average is $\dfrac{4 \cdot 12 + 2 \cdot 15 + 1 \cdot 10}{4 + 2 + 1}$

解答：

三、四、五年级的人数比为 $4 : 2 : 1$ 。加权平均数为 $\begin{aligned} \dfrac{4\cdot12+2\cdot15+10}{7} \\ = \dfrac{48+30+10}{7} \\ = \dfrac{88}{7} \end{aligned}\text{。}$

因此，正确答案是 C。

Take the grade sizes in the ratio $4 : 2 : 1$ for third, fourth, and fifth grades. The weighted average is $\begin{aligned} \dfrac{4\cdot12+2\cdot15+10}{7} \\ = \dfrac{48+30+10}{7} \\ = \dfrac{88}{7}. \end{aligned}$

Thus, the correct answer is C.

5.

去年夏天，Town Lake 上生活的鸟中 $30\%$ 是鹅， $25\%$ 是天鹅， $10\%$ 是鹭， $35\%$ 是鸭。在不是天鹅的鸟中，鹅占百分之几？

Last summer $30\%$ of the birds living on Town Lake were geese, $25\%$ were swans, $10\%$ were herons, and $35\%$ were ducks. What percent of the birds that were not swans were geese?

$20$

$30$

$40$

$50$

$60$

知识点：百分数条件概率

难度评级：990

小提示：

不是天鹅的鸟占总数的 $75\%$

The non-swan birds make up $75\%$ of the total

大提示：

把 $\dfrac{30}{75}$ 化成百分数

Compute $\dfrac{30}{75}$ as a percent

解答：

不是天鹅的鸟占总数的 $100\% - 25\% = 75\%$ ，而鹅占总数的 $30\%$ 。所求比例为 $\dfrac{30}{75} = \dfrac{2}{5} = 40\%\text{。}$

因此，正确答案是 C。

The birds that are not swans make up $100\% - 25\% = 75\%$ of the total, and geese are $30\%$ of the total. The requested fraction is $\dfrac{30}{75} = \dfrac{2}{5} = 40\%.$

Thus, the correct answer is C.

6.

一个篮球队的队员投进了一些三分球、一些两分球和一些一分罚球。他们由两分球得到的分数与由三分球得到的分数一样多。他们罚中的次数比投中的两分球次数多一。球队总得分为 $61$ 分。他们罚中了多少球？

The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team’s total score was $61$ points. How many free throws did they make?

$13$

$14$

$15$

$16$

$17$

知识点：方程组换元法

难度评级：1170

小提示：

设投中的两分球个数为 $a$ ；两分球和三分球得到的分数各为 $2a$

Let $a$ be the number of two-point shots; the points from two-point and three-point shots are each $2a$

大提示：

罚球个数为 $a + 1$ ，所以总分为 $2a + 2a + (a + 1)$

The free throws number $a + 1,$ so the total is $2a + 2a + (a + 1)$

解答：

设投中的两分球个数为 $a$ 。两分球得到 $2a$ 分，三分球也得到同样的 $2a$ 分。罚球个数为 $a + 1$ ，得到 $a + 1$ 分。

总分为 $\begin{aligned} 2a + 2a + (a + 1) \\ = 5a + 1 = 61 \end{aligned}\text{，}$ 所以 $a = 12$ ，罚球个数为 $a + 1 = 13$ 。

因此，正确答案是 A。

Let $a$ be the number of two-point shots. The two-point shots score $2a$ points, and the three-point shots score the same $2a$ points. The free throws number $a + 1$ and score $a + 1$ points.

The total is $\begin{aligned} 2a + 2a + (a + 1) \\ = 5a + 1 = 61, \end{aligned}$ so $a = 12$ and the free throws number $a + 1 = 13.$

Thus, the correct answer is A.

7.

Ms. Demeanor 班上 $30$ 名学生中的多数人在学校书店买了铅笔。这些学生每人买了相同数量的铅笔，且这个数量大于 $1$ 。每支铅笔的价格（单位：美分）大于每名学生买的铅笔数量，所有铅笔总费用为 $\$17.71$ 。每支铅笔的价格是多少美分？

A majority of the $30$ students in Ms. Demeanor’s class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than $1.$ The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was $\$17.71.$ What was the cost of a pencil in cents?

$7$

$11$

$17$

$23$

$77$

知识点：质因数分解整除性极限情形界定

难度评级：1370

小提示：

分解 $1771 = 7 \cdot 11 \cdot 23$ ，并令（学生数）（每人铅笔数）（单价） $= 1771$

Factor $1771 = 7 \cdot 11 \cdot 23$ and let (students)(pencils)(cost) $= 1771$

大提示：

多数意味着超过 $15$ 名学生，所以学生数为 $23$

A majority means more than $15$ students, so the number of students is $23$

解答：

总费用为 $1771$ 美分，且 $1771 = 7 \cdot 11 \cdot 23$ 。写成（学生数）（每人铅笔数）（每支铅笔价格） $= 1771$ ，学生数必须是该数的一个因数，并且是 $30$ 人中的多数，即超过 $15$ 。唯一这样的因数是 $23$ 。

因此（铅笔数）（单价） $= 77 = 7 \cdot 11$ ，且单价 $\gt$ 铅笔数 $\gt 1$ ，所以每人买 $7$ 支铅笔，每支 $11$ 美分。

因此，正确答案是 B。

Total cents is $1771 = 7 \cdot 11 \cdot 23.$ Writing (students)(pencils each)(cost per pencil) $= 1771,$ the number of students is a divisor of $1771$ that is a majority of $30,$ hence more than $15.$ The only such divisor is $23.$

Then (pencils)(cost) $= 77 = 7 \cdot 11$ with cost $\gt$ pencils $\gt 1,$ forcing $7$ pencils at $11$ cents each.

Thus, the correct answer is B.

8.

在八项数列 $A$ ， $B$ ， $C$ ， $D$ ， $E$ ， $F$ ， $G$ ， $H$ 中， $C$ 的值为 $5$ ，且任意三个连续项的和为 $30$ 。求 $A + H$ ？

In the eight-term sequence $A,$ $B,$ $C,$ $D,$ $E,$ $F,$ $G,$ $H,$ the value of $C$ is $5$ and the sum of any three consecutive terms is $30.$ What is $A + H?$

$17$

$18$

$25$

$26$

$43$

知识点：找规律代数变形

难度评级：1190

小提示：

连续三项的和都相等，会迫使数列每三项重复一次

Consecutive triples having equal sums forces the sequence to repeat every three terms

大提示：

因此 $H$ 等于 $B$ ，所以 $A + H$ 与 $A + B$ 相同

So $H$ equals $B,$ which means $A + H$ is the same as $A + B$

解答：

因为 $A + B + C = B + C + D = 30$ ，所以 $D = A$ ；同理可知，这个数列以 $3$ 为周期重复。因此第八项 $H$ 等于 $B$ 。

由 $A + B + C = 30$ 且 $C = 5$ ，得 $A + H = A + B = 30 - 5 = 25$ 。

因此，正确答案是 C。

Since $A + B + C = B + C + D = 30,$ we get $D = A,$ and likewise the sequence repeats with period $3.$ Thus $H,$ the eighth term, equals $B.$

From $A + B + C = 30$ and $C = 5,$ we have $A + H = A + B = 30 - 5 = 25.$

Thus, the correct answer is C.

9.

在一个双胞胎和三胞胎大会上，有 $9$ 组双胞胎和 $6$ 组三胞胎，且都来自不同家庭。每个双胞胎成员都与除自己兄弟姐妹外的所有双胞胎成员握手，并与一半的三胞胎成员握手。每个三胞胎成员都与除自己兄弟姐妹外的所有三胞胎成员握手，并与一半的双胞胎成员握手。一共发生了多少次握手？

At a twins and triplets convention, there were $9$ sets of twins and $6$ sets of triplets, all from different families. Each twin shook hands with all the twins except his/her sibling and with half the triplets. Each triplet shook hands with all the triplets except his/her siblings and with half the twins. How many handshakes took place?

$324$

$441$

$630$

$648$

$882$

知识点：数对计数分类讨论

难度评级：1440

小提示：

有 $18$ 名双胞胎成员和 $18$ 名三胞胎成员；分别数双胞胎之间、三胞胎之间、以及双胞胎和三胞胎之间的握手

There are $18$ twins and $18$ triplets; count twin-twin, triplet-triplet, and twin-triplet handshakes separately

大提示：

每个双胞胎成员与 $16$ 名双胞胎成员握手，每个三胞胎成员与 $15$ 名三胞胎成员握手；这两部分总数都要除以 $2$

Each twin greets $16$ twins and each triplet greets $15$ triplets; divide those two totals by $2$

解答：

共有 $18$ 名双胞胎成员和 $18$ 名三胞胎成员。

双胞胎之间的握手：每个双胞胎成员与 $18 - 2 = 16$ 名其他双胞胎成员握手，得到 $\dfrac{18 \cdot 16}{2} = 144$ 。

三胞胎之间的握手：每个三胞胎成员与 $18 - 3 = 15$ 名其他三胞胎成员握手，得到 $\dfrac{18 \cdot 15}{2} = 135$ 。

双胞胎和三胞胎之间的握手：每个双胞胎成员与 $18$ 名三胞胎成员中的一半握手，得到 $18 \cdot 9 = 162$ （每次这种握手只计一次）。

总数为 $144 + 135 + 162 = 441$ 。

因此，正确答案是 B。

There are $18$ twins and $18$ triplets.

Twin-twin handshakes: each twin shakes $18 - 2 = 16$ other twins, giving $\dfrac{18 \cdot 16}{2} = 144.$

Triplet-triplet handshakes: each triplet shakes $18 - 3 = 15$ other triplets, giving $\dfrac{18 \cdot 15}{2} = 135.$

Twin-triplet handshakes: each twin shakes half the $18$ triplets, giving $18 \cdot 9 = 162$ (each such handshake counted once).

The total is $144 + 135 + 162 = 441.$

Thus, the correct answer is B.

10.

一对标准 $6$ 面公平骰子掷一次。掷出的点数和决定一个圆的直径。圆的面积的数值小于该圆周长的数值的概率是多少？

A pair of standard $6$ -sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle’s circumference?

$\dfrac{1}{36}$

$\dfrac{1}{12}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{5}{18}$

知识点：骰子（概率）圆不等式

难度评级：1370

小提示：

若直径为 $d$ ，面积为 $\dfrac{\pi d^2}{4}$ ，周长为 $\pi d$

With diameter $d,$ the area is $\dfrac{\pi d^2}{4}$ and the circumference is $\pi d$

大提示：

不等式 $\dfrac{\pi d^2}{4} \lt \pi d$ 化简为 $d \lt 4$

The inequality $\dfrac{\pi d^2}{4} \lt \pi d$ reduces to $d \lt 4$

解答：

若直径为 $d$ ，面积 $\lt$ 周长意味着 $\dfrac{\pi d^2}{4} \lt \pi d$ ，即 $d \lt 4$ 。因为 $d \ge 2$ ，所以点数和必须是 $2$ 或 $3$ 。

点数和为 $2$ 的概率是 $\dfrac{1}{36}$ ，点数和为 $3$ 的概率是 $\dfrac{2}{36}$ ，合计 $\dfrac{3}{36} = \dfrac{1}{12}$ 。

因此，正确答案是 B。

For diameter $d,$ area $\lt$ circumference means $\dfrac{\pi d^2}{4} \lt \pi d,$ i.e. $d \lt 4.$ Since $d \ge 2,$ this needs a sum of $2$ or $3.$

A sum of $2$ has probability $\dfrac{1}{36}$ and a sum of $3$ has probability $\dfrac{2}{36},$ totaling $\dfrac{3}{36} = \dfrac{1}{12}.$

Thus, the correct answer is B.

11.

圆 $A$ 、 $B$ 和 $C$ 的半径都为 $1$ 。圆 $A$ 和圆 $B$ 共有一个切点。圆 $C$ 与 $\overline{AB}$ 的中点相切。位于圆 $C$ 内部、但在圆 $A$ 和圆 $B$ 外部的面积是多少？

Circles $A,$ $B,$ and $C$ each have radius $1.$ Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}.$ What is the area inside circle $C$ but outside circle $A$ and circle $B?$

$3 - \dfrac{\pi}{2}$

$\dfrac{\pi}{2}$

$2$

$\dfrac{3\pi}{4}$

$1 + \dfrac{\pi}{2}$

知识点：圆面积相切圆面积分割

难度评级：1540

小提示：

设圆心为 $A(-1,0)$ ， $B(1,0)$ ，和 $C(0,1)$ ；从 $C$ 到 $A, B$ 的距离都是 $\sqrt2$

Place the centers at $A(-1,0),$ $B(1,0),$ and $C(0,1);$ the distance from $C$ to each of $A, B$ is $\sqrt2$

大提示：

所求面积是圆 $C$ 的面积减去圆 $C$ 分别与圆 $A$ 和圆 $B$ 的两个透镜形重叠区域

The wanted area is the area of $C$ minus the two lens-shaped overlaps of $C$ with $A$ and with $B$

解答：

设 $A = (-1, 0)$ ， $B = (1, 0)$ ，则它们的切点是原点，也就是 $\overline{AB}$ 的中点。因此 $C = (0, 1)$ ，因为圆 $C$ 经过原点。

从 $C$ 到 $A$ （以及到 $B$ ）的距离为 $\sqrt2$ 。两个圆心距离为 $\sqrt2$ 的单位圆重叠成一个透镜形，其面积为 $\begin{aligned} 2\cos^{-1}\!\left(\tfrac{\sqrt2}{2}\right) \\ {}- \tfrac{\sqrt2}{2}\sqrt{4 - 2} \\ = 2 \cdot \tfrac{\pi}{4} - 1 = \tfrac{\pi}{2} - 1 \end{aligned}\text{。}$

圆 $A$ 和圆 $B$ 只在原点相交，所以两个透镜形区域不重叠。所求面积为 $\pi - 2\left(\tfrac{\pi}{2} - 1\right) = 2\text{。}$

因此，正确答案是 C。

Place $A = (-1, 0),$ $B = (1, 0),$ so their tangency point is the origin, the midpoint of $\overline{AB}.$ Then $C = (0, 1),$ since $C$ passes through the origin.

The distance from $C$ to $A$ (and to $B$ ) is $\sqrt2.$ Two unit circles whose centers are $\sqrt2$ apart overlap in a lens of area $\begin{aligned} 2\cos^{-1}\!\left(\tfrac{\sqrt2}{2}\right) \\ {}- \tfrac{\sqrt2}{2}\sqrt{4 - 2} \\ = 2 \cdot \tfrac{\pi}{4} - 1 = \tfrac{\pi}{2} - 1. \end{aligned}$

Circles $A$ and $B$ meet only at the origin, so the two lenses do not overlap. The wanted area is $\pi - 2\left(\tfrac{\pi}{2} - 1\right) = 2.$

Thus, the correct answer is C.

12.

一艘机动船和一只木筏都从河上的码头 $A$ 出发，向下游前进。木筏以河水水流速度漂流。机动船相对于河水保持恒定速度。机动船到达下游码头 $B$ 后，立即掉头向上游返回。离开码头 $A$ $9$ 小时后，它在河上与木筏相遇。机动船从 $A$ 到 $B$ 花了多少小时？

A power boat and a raft both left dock $A$ on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock $B$ downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river $9$ hours after leaving dock $A.$ How many hours did it take the power boat to go from $A$ to $B?$

$3$

$3.5$

$4$

$4.5$

$5$

知识点：相对速度路程、速度与时间对称性

难度评级：1580

小提示：

在水流参考系中思考，此时木筏静止不动

Work in the frame of the water, in which the raft stays put

大提示：

相对于河水，船往返速度相同，所以去程和回程用时相等

Relative to the water the boat travels at the same speed going out and coming back, so it takes equal times each way

解答：

全部相对于河水来测量。在这个参考系中，木筏静止在船出发的位置，而船相对于河水以恒定速度 $v$ 运动，下游和上游方向都是如此。

船离开木筏，向外行驶一段时间，再以相同的相对速度返回木筏，所以去程和回程用时相等。因此到 $B$ 的下行路程花了 $9$ 的一半，即 $4.5$ 小时。

因此，正确答案是 D。

Measure everything relative to the water. In that frame the raft is stationary at the point where the boat started, and the boat moves at its constant speed $v$ relative to the water, both downstream and upstream.

The boat leaves the raft, travels away for some time, then returns to it at the same relative speed, so it spends equal times going and returning. Hence the outbound leg to $B$ takes half of $9,$ which is $4.5$ hours.

Thus, the correct answer is D.

13.

三角形 $ABC$ 的边长为 $AB = 12$ ， $BC = 24$ ，且 $AC = 18$ 。过 $\triangle ABC$ 内心且平行于 $\overline{BC}$ 的直线分别交 $\overline{AB}$ 于 $M$ ，交 $\overline{AC}$ 于 $N$ 。 $\triangle AMN$ 的周长是多少？

Triangle $ABC$ has side-lengths $AB = 12,$ $BC = 24,$ and $AC = 18.$ The line through the incenter of $\triangle ABC$ parallel to $\overline{BC}$ intersects $\overline{AB}$ at $M$ and $\overline{AC}$ at $N.$ What is the perimeter of $\triangle AMN?$

$27$

$30$

$33$

$36$

$42$

知识点：内切圆、内心与内切圆半径平行线等腰三角形

难度评级：1600

小提示：

设 $I$ 为内心。因为 $MN \parallel BC$ ，所以 $\angle MIB$ 等于 $\angle IBC$

Let $I$ be the incenter. Since $MN \parallel BC,$ the angle $\angle MIB$ equals $\angle IBC$

大提示：

这使得 $\triangle MBI$ 为等腰三角形，且 $MB = MI$ ，同理 $NC = NI$

That makes $\triangle MBI$ isosceles with $MB = MI,$ and similarly $NC = NI$

解答：

设 $I$ 为内心。因为 $\overline{BI}$ 平分 $\angle B$ ，且 $MN \parallel BC$ ，由内错角可得 $\angle MIB = \angle IBC = \angle MBI$ ，所以 $\triangle MBI$ 是等腰三角形，且 $MB = MI$ 。同理 $NC = NI$ 。

因此 $\triangle AMN$ 的周长为 $\begin{gathered} AM + MN + NA \\ = AM + (MI + IN) + NA \\ = AM + MB + NC + NA \\ = AB + AC = 12 + 18 = 30 \end{gathered}\text{。}$

因此，正确答案是 B。

Let $I$ be the incenter. Because $\overline{BI}$ bisects $\angle B$ and $MN \parallel BC,$ alternate angles give $\angle MIB = \angle IBC = \angle MBI,$ so $\triangle MBI$ is isosceles with $MB = MI.$ Similarly $NC = NI.$

Therefore the perimeter of $\triangle AMN$ is $\begin{gathered} AM + MN + NA \\ = AM + (MI + IN) + NA \\ = AM + MB + NC + NA \\ = AB + AC = 12 + 18 = 30. \end{gathered}$

Thus, the correct answer is B.

14.

假设 $a$ 和 $b$ 是独立随机选取的一位正整数。点 $(a, b)$ 位于抛物线 $y = ax^2 - bx$ 上方的概率是多少？

Suppose $a$ and $b$ are single-digit positive integers chosen independently and at random. What is the probability that the point $(a, b)$ lies above the parabola $y = ax^2 - bx?$

$\dfrac{11}{81}$

$\dfrac{13}{81}$

$\dfrac{5}{27}$

$\dfrac{17}{81}$

$\dfrac{19}{81}$

知识点：抛物线基本概率分类讨论

难度评级：1690

小提示：

点在抛物线上方当且仅当 $b \gt a \cdot a^2 - b \cdot a$ ，即 $b(a + 1) \gt a^3$

The point is above the parabola when $b \gt a \cdot a^2 - b \cdot a,$ i.e. $b(a + 1) \gt a^3$

大提示：

对每个 $a$ ，统计 $\{1, \ldots, 9\}$ 中可行的 $b$ ；只有 $a = 1, 2, 3$ 可能有解

Count valid $b$ in $\{1, \ldots, 9\}$ for each $a;$ only $a = 1, 2, 3$ allow any

解答：

代入 $x = a$ 、 $y = b$ ，点在抛物线上方当且仅当 $b \gt a^3 - ab$ ，即 $b(a + 1) \gt a^3$ 。

当 $a = 1$ ： $b \gt \tfrac12$ ，所有 $9$ 个值都可行。当 $a = 2$ ： $b \gt \tfrac83$ ，所以 $b \ge 3$ 有 $7$ 个。当 $a = 3$ ： $b \gt \tfrac{27}{4} = 6.75$ ，所以 $b \ge 7$ 有 $3$ 个。当 $a \ge 4$ 时，没有 $b \le 9$ 可行。

总数为 $9 + 7 + 3 = 19$ ，共 $81$ 种情况，所以概率为 $\dfrac{19}{81}$ 。

因此，正确答案是 E。

Substituting $x = a,$ $y = b,$ the point is above the parabola when $b \gt a^3 - ab,$ i.e. $b(a + 1) \gt a^3.$

For $a = 1:$ $b \gt \tfrac12,$ all $9$ values work. For $a = 2:$ $b \gt \tfrac83,$ so $b \ge 3,$ giving $7.$ For $a = 3:$ $b \gt \tfrac{27}{4} = 6.75,$ so $b \ge 7,$ giving $3.$ For $a \ge 4,$ no $b \le 9$ works.

The count is $9 + 7 + 3 = 19$ out of $81,$ so the probability is $\dfrac{19}{81}.$

Thus, the correct answer is E.

15.

一个半径为 $2$ 的半球的圆形底面放在一个高为 $6$ 的正方形棱锥的底面上。该半球与棱锥的另外四个面相切。棱锥底面边长是多少？

The circular base of a hemisphere of radius $2$ rests on the base of a square pyramid of height $6.$ The hemisphere is tangent to the other four faces of the pyramid. What is the edge-length of the base of the pyramid?

$3\sqrt{2}$

$\dfrac{13}{3}$

$4\sqrt{2}$

$6$

$\dfrac{13}{2}$

知识点：立体几何棱锥距离公式

难度评级：1870

小提示：

取通过顶点以及两条相对底边中点的截面

Take the cross-section through the apex and the midpoints of two opposite base edges

大提示：

在这个截面中，斜边直线从 $\left(\tfrac{s}{2}, 0\right)$ 到 $(0, 6)$ ，且它到中心的距离为 $2$

In that cross-section, the slant line runs from $\left(\tfrac{s}{2}, 0\right)$ to $(0, 6),$ and its distance from the center is $2$

解答：

设底面边长为 $s$ ，底面中心在原点，顶点高度为 $6$ 。用通过顶点和两条相对底边中点的竖直平面截取。侧面在截面中表现为从 $\left(\tfrac{s}{2}, 0\right)$ 到 $(0, 6)$ 的直线。

这条直线是 $\tfrac{2}{s}x + \tfrac16 y = 1$ 。半球与该面相切，所以原点到这条直线的距离等于半径 $2$ ： $\dfrac{1}{\sqrt{\tfrac{4}{s^2} + \tfrac{1}{36}}} = 2\text{。}$

因此 $\tfrac{4}{s^2} + \tfrac{1}{36} = \tfrac14$ ，所以 $\tfrac{4}{s^2} = \tfrac{2}{9}$ ， $s^2 = 18$ ，得 $s = 3\sqrt2$ 。

因此，正确答案是 A。

Let the base have side $s,$ centered at the origin, with apex at height $6.$ Cut with the vertical plane through the apex and the midpoints of two opposite base edges. The slant face appears as the line from $\left(\tfrac{s}{2}, 0\right)$ to $(0, 6).$

This line is $\tfrac{2}{s}x + \tfrac16 y = 1.$ The hemisphere is tangent to the face, so the distance from the origin to this line is the radius $2:$ $\dfrac{1}{\sqrt{\tfrac{4}{s^2} + \tfrac{1}{36}}} = 2.$

Then $\tfrac{4}{s^2} + \tfrac{1}{36} = \tfrac14,$ so $\tfrac{4}{s^2} = \tfrac{2}{9}$ and $s^2 = 18,$ giving $s = 3\sqrt2.$

Thus, the correct answer is A.

16.

凸五边形 $ABCDE$ 的每个顶点都要指定一种颜色。有 $6$ 种颜色可选，并且每条对角线的两个端点必须颜色不同。共有多少种不同的涂色方法？

Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?

$2520$

$2880$

$3120$

$3250$

$3750$

知识点：图论容斥原理

难度评级：1820

小提示：

五条对角线 $AC, CE, EB, BD, DA$ 在这些顶点之间形成一个 $5$ -环

The five diagonals $AC, CE, EB, BD, DA$ form a single $5$ -cycle among the vertices

大提示：

长度为 $n$ 的环用 $k$ 种颜色作正常涂色的数量是 $(k-1)^n + (-1)^n (k-1)$

The number of proper colorings of a cycle of length $n$ with $k$ colors is $(k-1)^n + (-1)^n (k-1)$

解答：

对角线按顺序连接顶点 $A - C - E - B - D - A$ ，形成一个 $5$ -环。题目条件正是要求这个环被正常涂色。

长度为 $n$ 的环用 $k$ 种颜色作正常涂色的数量为 $(k-1)^n + (-1)^n (k-1)$ 。取 $n = 5$ 且 $k = 6$ ， $\begin{gathered} 5^5 + (-1)^5 \cdot 5 \\ = 3125 - 5 = 3120 \end{gathered}\text{。}$

因此，正确答案是 C。

The diagonals connect the vertices in the order $A - C - E - B - D - A,$ which is a $5$ -cycle. The condition is exactly that this cycle is properly colored.

The number of proper $k$ -colorings of a cycle of length $n$ is $(k-1)^n + (-1)^n (k-1).$ With $n = 5$ and $k = 6,$ $\begin{gathered} 5^5 + (-1)^5 \cdot 5 \\ = 3125 - 5 = 3120. \end{gathered}$

Thus, the correct answer is C.

17.

半径为 $1$ ， $2$ ，和 $3$ 的三个圆两两外切。由这些切点确定的三角形面积是多少？

Circles with radii $1,$ $2,$ and $3$ are mutually externally tangent. What is the area of the triangle determined by the points of tangency?

$\dfrac{3}{5}$

$\dfrac{4}{5}$

$1$

$\dfrac{6}{5}$

$\dfrac{4}{3}$

知识点：相切圆坐标几何鞋带公式

难度评级：1920

小提示：

圆心形成一个边长为 $1+2$ ， $2+3$ ， $1+3$ ，即 $3$ - $4$ - $5$ 的直角三角形

The centers form a triangle with sides $1+2,$ $2+3,$ $1+3,$ i.e. a $3$ - $4$ - $5$ right triangle

大提示：

把直角放在半径为 $1$ 的圆心处，并在连接圆心的线段上定位各个切点

Put the right angle at the center of the radius- $1$ circle and locate each tangency point on the connecting segment

解答：

圆心之间的距离等于半径和： $3$ ， $4$ ，和 $5$ ，形成一个直角三角形，直角在半径为 $1$ 的圆心处。把该圆心放在 $(0,0)$ ，半径为 $2$ 的圆心放在 $(3,0)$ ，半径为 $3$ 的圆心放在 $(0,4)$ 。

切点在线段上，距离等于相应半径： $(1, 0)$ ， $(0, 1)$ ，以及斜边上的 $(3,0) + 2 \cdot \tfrac{(-3,4)}{5} = \left(\tfrac95, \tfrac85\right)$ 。

由鞋带公式，面积为 $\begin{gathered} \tfrac12\left| 1\left(1 - \tfrac85\right) + 0 + \tfrac95(0 - 1) \right| \\ = \tfrac12 \cdot \tfrac{12}{5} = \tfrac65 \end{gathered}\text{。}$

因此，正确答案是 D。

The centers are separated by the sums of radii: $3,$ $4,$ and $5,$ a right triangle with the right angle at the radius- $1$ center. Place that center at $(0,0),$ the radius- $2$ center at $(3,0),$ and the radius- $3$ center at $(0,4).$

The tangency points lie on the segments at distances equal to the radii: $(1, 0),$ $(0, 1),$ and on the hypotenuse at $(3,0) + 2 \cdot \tfrac{(-3,4)}{5} = \left(\tfrac95, \tfrac85\right).$

By the shoelace formula the area is $\begin{gathered} \tfrac12\left| 1\left(1 - \tfrac85\right) + 0 + \tfrac95(0 - 1) \right| \\ = \tfrac12 \cdot \tfrac{12}{5} = \tfrac65. \end{gathered}$

Thus, the correct answer is D.

18.

假设 $|x + y| + |x - y| = 2$ 。 $x^2 - 6x + y^2$ 的最大可能值是多少？

Suppose that $|x + y| + |x - y| = 2.$ What is the maximum possible value of $x^2 - 6x + y^2?$

$5$

$6$

$7$

$8$

$9$

知识点：绝对值最优化正方形（几何）

难度评级：1840

小提示：

方程 $|x+y| + |x-y| = 2$ 表示 $\max(|x|, |y|) = 1$ ，即正方形 $[-1,1]^2$ 的边界

The equation $|x+y| + |x-y| = 2$ means $\max(|x|, |y|) = 1,$ the boundary of the square $[-1,1]^2$

大提示：

在该正方形上，当 $x = -1$ 且 $y = \pm 1$ 时， $x^2 - 6x + y^2$ 最大

On that square, $x^2 - 6x + y^2$ is largest when $x = -1$ and $y = \pm 1$

解答：

恒等式 $|x+y| + |x-y|$ $= 2\max(|x|, |y|)$ 将条件变成 $\max(|x|, |y|) = 1$ ，即满足 $|x| \le 1$ 且 $|y| \le 1$ 的正方形边界。

在这个区域上， $x^2 - 6x + y^2$ 会随着 $x$ 变小以及 $y^2$ 变大而增大，所以最大值在 $x = -1$ 、 $y = \pm 1$ 处取得： $1 + 6 + 1 = 8\text{。}$

因此，正确答案是 D。

The identity $|x+y| + |x-y|$ $= 2\max(|x|, |y|)$ turns the condition into $\max(|x|, |y|) = 1,$ the boundary of the square with $|x| \le 1$ and $|y| \le 1.$

On this region $x^2 - 6x + y^2$ increases as $x$ decreases and as $y^2$ increases, so the maximum is at $x = -1,$ $y = \pm 1:$ $1 + 6 + 1 = 8.$

Thus, the correct answer is D.

19.

在一个有 $N$ 名选手的比赛中，被授予精英身份的选手人数等于 $2^{1 + \lfloor \log_2 (N - 1) \rfloor} - N\text{。}$ 假设有 $19$ 名选手被授予精英身份。 $N$ 的两个最小可能值之和是多少？

注： $\lfloor x \rfloor$ 是小于或等于 $x$ 的最大整数。

At a competition with $N$ players, the number of players given elite status is equal to $2^{1 + \lfloor \log_2 (N - 1) \rfloor} - N.$ Suppose that $19$ players are given elite status. What is the sum of the two smallest possible values of $N?$

Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x.$

$38$

$90$

$154$

$406$

$1024$

知识点：取整函数 2的幂分类讨论

难度评级：2150

小提示：

令 $m = \lfloor \log_2 (N-1) \rfloor$ ，则 $N = 2^{m+1} - 19$

Let $m = \lfloor \log_2 (N-1) \rfloor,$ so $N = 2^{m+1} - 19$

大提示：

为了使这个 $m$ 一致，需有 $2^m \le N - 1$ ，这迫使 $2^m \ge 20$

For that $m$ to be consistent, $2^m \le N - 1,$ which forces $2^m \ge 20$

解答：

令 $m = \lfloor \log_2 (N - 1) \rfloor$ ，则精英人数为 $2^{m+1} - N = 19$ ，得 $N = 2^{m+1} - 19$ 。

一致性要求 $2^m \le N - 1 = 2^{m+1} - 20$ ，即 $2^m \ge 20$ ，所以 $m \ge 5$ 。

两个最小选择是 $m = 5$ ，给出 $N = 64 - 19 = 45$ ，以及 $m = 6$ ，给出 $N = 128 - 19 = 109$ 。它们的和为 $154$ 。

因此，正确答案是 C。

Let $m = \lfloor \log_2 (N - 1) \rfloor,$ so the elite count is $2^{m+1} - N = 19,$ giving $N = 2^{m+1} - 19.$

Consistency requires $2^m \le N - 1 = 2^{m+1} - 20,$ i.e. $2^m \ge 20,$ so $m \ge 5.$

The two smallest choices are $m = 5$ giving $N = 64 - 19 = 45,$ and $m = 6$ giving $N = 128 - 19 = 109.$ Their sum is $154.$

Thus, the correct answer is C.

20.

令 $f(x) = ax^2 + bx + c$ ，其中 $a$ 、 $b$ 和 $c$ 是整数。假设 $f(1) = 0$ 、 $50 \lt f(7) \lt 60$ 、 $70 \lt f(8) \lt 80$ ，并且对某个整数 $k$ ，有 $5000k \lt f(100) \lt 5000(k+1)$ 。求 $k$ 。

Let $f(x) = ax^2 + bx + c,$ where $a,$ $b,$ and $c$ are integers. Suppose that $f(1) = 0,$ $50 \lt f(7) \lt 60,$ $70 \lt f(8) \lt 80,$ and $5000k \lt f(100) \lt 5000(k+1)$ for some integer $k.$ What is $k?$

$1$

$2$

$3$

$4$

$5$

知识点：二次方程方程组极限情形界定

难度评级：2030

小提示：

由 $f(1) = 0$ ，得 $c = -a - b$ ，所以 $f(7) = 6(8a + b)$ 且 $f(8) = 7(9a + b)$

From $f(1) = 0,$ $c = -a - b,$ so $f(7) = 6(8a + b)$ and $f(8) = 7(9a + b)$

大提示：

这些范围迫使 $8a + b = 9$ 且 $9a + b = 11$

The bounds force $8a + b = 9$ and $9a + b = 11$

解答：

因为 $f(1) = a + b + c = 0$ ，所以 $c = -a - b$ 。因此 $f(7) = 48a + 6b = 6(8a + b)\text{，}$ $f(8) = 63a + 7b = 7(9a + b)\text{。}$

由 $50 \lt 6(8a + b) \lt 60$ 得 $8a + b = 9$ ，由 $70 \lt 7(9a + b) \lt 80$ 得 $9a + b = 11$ 。相减得 $a = 2$ ，进而 $b = -7$ ， $c = 5$ 。

所以 $f(100) = 20000 - 700 + 5$ $= 19305$ ，它满足 $5000 \cdot 3 \lt 19305 \lt 5000 \cdot 4$ ，因此 $k = 3$ 。

因此，正确答案是 C。

Since $f(1) = a + b + c = 0,$ we have $c = -a - b.$ Then $f(7) = 48a + 6b = 6(8a + b),$ $f(8) = 63a + 7b = 7(9a + b).$

From $50 \lt 6(8a + b) \lt 60$ we get $8a + b = 9,$ and from $70 \lt 7(9a + b) \lt 80$ we get $9a + b = 11.$ Subtracting, $a = 2,$ then $b = -7$ and $c = 5.$

So $f(100) = 20000 - 700 + 5$ $= 19305,$ which lies in $5000 \cdot 3 \lt 19305 \lt 5000 \cdot 4,$ giving $k = 3.$

Thus, the correct answer is C.

21.

令 $f_1(x) = \sqrt{1 - x}$ ；对整数 $n \ge 2$ ，令 $f_n(x) = f_{n-1}\!\left(\sqrt{n^2 - x}\right)$ 。若 $N$ 是使 $f_n$ 的定义域非空的最大 $n$ 值，且 $f_N$ 的定义域为 $\{c\}$ ，求 $N + c$ 。

Let $f_1(x) = \sqrt{1 - x},$ and for integers $n \ge 2,$ let $f_n(x) = f_{n-1}\!\left(\sqrt{n^2 - x}\right).$ If $N$ is the largest value of $n$ for which the domain of $f_n$ is nonempty, the domain of $f_N$ is $\{c\}.$ What is $N + c?$

$-226$

$-144$

$-20$

$20$

$144$

知识点：函数根式递推

难度评级：2270

小提示：

从内向外构造定义域： $f_n$ 要求 $\sqrt{n^2 - x}$ 落在 $f_{n-1}$ 的定义域中

Build the domains outward: $f_n$ needs $\sqrt{n^2 - x}$ to lie in the domain of $f_{n-1}$

大提示：

前三个定义域为 $(-\infty,1]$ 、 $[3,4]$ 和 $[-7,0]$ ；继续往下推，同时记住平方根非负

The first three domains are $(-\infty,1],$ $[3,4],$ and $[-7,0];$ continue while remembering that a square root is nonnegative

解答：

每一步都要求 $\sqrt{n^2 - x}$ 落在 $f_{n-1}$ 的定义域中。追踪定义域：

$f_1: (-\infty, 1]$ 。 $f_2: \sqrt{4 - x} \in (-\infty, 1] \Rightarrow [3, 4]$ 。 $f_3: \sqrt{9 - x} \in [3, 4] \Rightarrow [-7, 0]$ 。 $f_4: \sqrt{16 - x} \in [-7, 0] \Rightarrow \{16\}$ （只有值 $0$ 可能）。 $f_5: \sqrt{25 - x} = 16 \Rightarrow \{-231\}$ 。

对 $f_6$ ，我们需要 $\sqrt{36 - x} = -231$ ，这是不可能的，所以定义域为空。因此 $N = 5$ ， $c = -231$ ，且 $N + c = -226$ 。

因此，正确答案是 A。

Each step requires $\sqrt{n^2 - x}$ to lie in the domain of $f_{n-1}.$ Tracking the domains:

$f_1: (-\infty, 1].$ $f_2: \sqrt{4 - x} \in (-\infty, 1] \Rightarrow [3, 4].$ $f_3: \sqrt{9 - x} \in [3, 4] \Rightarrow [-7, 0].$ $f_4: \sqrt{16 - x} \in [-7, 0] \Rightarrow \{16\}$ (only the value $0$ is possible). $f_5: \sqrt{25 - x} = 16 \Rightarrow \{-231\}.$

For $f_6$ we would need $\sqrt{36 - x} = -231,$ impossible, so the domain is empty. Hence $N = 5,$ $c = -231,$ and $N + c = -226.$

Thus, the correct answer is A.

22.

设 $R$ 为一个正方形区域， $n \ge 4$ 为整数。若从 $R$ 内部一点 $X$ 发出 $n$ 条射线，可以把 $R$ 分成 $n$ 个面积相等的三角形，则称 $X$ 为 $n$ -射线分割点。有多少个点是 $100$ -射线分割点但不是 $60$ -射线分割点？

Let $R$ be a square region and $n \ge 4$ an integer. A point $X$ in the interior of $R$ is called $n$ -ray partitional if there are $n$ rays emanating from $X$ that divide $R$ into $n$ triangles of equal area. How many points are $100$ -ray partitional but not $60$ -ray partitional?

$1500$

$1560$

$2320$

$2480$

$2500$

知识点：面积格点补集计数

难度评级：2460

小提示：

$n$ -射线分割点形成网格：当 $n = 2m$ ，它们是内部点 $\left(\tfrac{i}{m}, \tfrac{j}{m}\right)$ ，构成 $(m-1) \times (m-1)$ 个点

The $n$ -ray partitional points form a grid: for $n = 2m,$ they are the interior points $\left(\tfrac{i}{m}, \tfrac{j}{m}\right),$ an $(m-1) \times (m-1)$ array

大提示：

一个点同时是 $100$ -射线和 $60$ -射线分割点，当且仅当它是 $20$ -射线分割点

A point is both $100$ - and $60$ -ray partitional exactly when it is $20$ -ray partitional

解答：

将正方形缩放为 $[0,1]^2$ ，并令 $X=(x,y)$ 。射线必须包括通向四个顶点的射线。每个小三角形的面积都是 $\frac{1}{n}$ 。以底边各段为底的三角形面积之和为 $\frac{y}{2}$ ，所以这样的三角形有 $\frac{ny}{2}$ 个。同理，沿上、左、右三边的个数分别是 $\frac{n(1-y)}{2}$ ， $\frac{nx}{2}$ ，和 $\frac{n(1-x)}{2}$ 。

这四个数都必须是正整数。因此 $n$ 是偶数，并且 $\begin{gathered} X=\left(\dfrac{2i}{n},\dfrac{2j}{n}\right), \\ 1\le i,j\le\dfrac n2-1 \end{gathered}\text{。}$ 反过来，按上述数量把每条边等分，并将分点连接到 $X$ ，就会得到 $n$ 个等面积三角形。因此这些恰好是分割点。

当 $n=100$ 时，这些点为 $(\frac{i}{50},\frac{j}{50})$ ，其中 $1\le i,j\le49$ ，共 $49^2=2401$ 个。这样的点同时也是 $60$ 射线分割点，当且仅当对某些整数 $c,d$ ，有 $\frac{i}{50}=\frac{c}{30}$ 且 $\frac{j}{50}=\frac{d}{30}$ 。因此 $i$ 和 $j$ 都必须是 $5$ 的倍数。每个坐标有 $9$ 种选择，所以重合的点有 $9^2=81$ 个。

所以所求数量为 $2401 - 81 = 2320$ 。

因此，正确答案是 C。

Scale the square to $[0,1]^2$ and write $X=(x,y).$ The rays must include those through the four vertices. Every small triangle has area $\frac{1}{n}.$ The triangles whose bases partition the bottom side together have area $\frac{y}{2},$ so their number is $\frac{ny}{2}.$ Similarly, the numbers along the top, left, and right sides are $\frac{n(1-y)}{2},$ $\frac{nx}{2},$ and $\frac{n(1-x)}{2}.$

These four numbers must be positive integers. Hence $n$ is even and $\begin{gathered} X=\left(\dfrac{2i}{n},\dfrac{2j}{n}\right), \\ 1\le i,j\le\dfrac n2-1. \end{gathered}$ Conversely, partitioning each side into the indicated number of equal segments and joining the division points to $X$ produces $n$ equal-area triangles. Thus these are exactly the partitional points.

For $n=100,$ the points are $(\frac{i}{50},\frac{j}{50})$ with $1\le i,j\le49,$ giving $49^2=2401.$ Such a point is also $60$ -ray partitional exactly when $\frac{i}{50}=\frac{c}{30}$ and $\frac{j}{50}=\frac{d}{30}$ for integers $c,d.$ Thus $i$ and $j$ must both be multiples of $5.$ There are $9$ choices for each, so the overlap has $9^2=81$ points.

So the count is $2401 - 81 = 2320.$

Thus, the correct answer is C.

23.

令 $f(z) = \dfrac{z + a}{z + b}$ ，且 $g(z) = f(f(z))$ ，其中 $a$ 和 $b$ 是复数。假设 $|a| = 1$ ，并且对所有使 $g(g(z))$ 有定义的 $z$ ，都有 $g(g(z)) = z$ 。 $|b|$ 的最大可能值与最小可能值之差是多少？

Let $f(z) = \dfrac{z + a}{z + b}$ and $g(z) = f(f(z)),$ where $a$ and $b$ are complex numbers. Suppose that $|a| = 1$ and $g(g(z)) = z$ for all $z$ for which $g(g(z))$ is defined. What is the difference between the largest and smallest possible values of $|b|?$

$0$

$\sqrt{2} - 1$

$\sqrt{3} - 1$

$1$

$2$

知识点：复数矩阵单位根

难度评级：2560

小提示：

用矩阵 $\begin{pmatrix} 1 & a \\ 1 & b \end{pmatrix}$ 表示 $f$ ；则 $g(g(z)) = z$ 意味着 $f^4$ 是恒等变换

Represent $f$ by the matrix $\begin{pmatrix} 1 & a \\ 1 & b \end{pmatrix};$ then $g(g(z)) = z$ means $f^4$ is the identity

大提示：

阶为 $4$ 的情形迫使 $(\operatorname{tr})^2 = 2\det$ ，从而得到 $b^2 = -(1 + 2a)$

The order- $4$ case forces $(\operatorname{tr})^2 = 2\det,$ which gives $b^2 = -(1 + 2a)$

解答：

直接复合可得 $g(z)=\dfrac{Az+B}{Cz+D}\text{，}$ 其中 $A=1+a$ ， $B=a(1+b)$ ， $C=1+b$ ，且 $D=a+b^2$ 。

矩阵 $\begin{pmatrix}A&B\\C&D\end{pmatrix}$ 表示变换 $g$ 。要使 $g\circ g$ 为恒等变换，它的平方必须是标量矩阵。比较非对角元和两个对角元，得到两种可能：或者 $B=C=0$ 且 $A=D$ ，或者 $A+D=0$ 。第一种给出 $b=-1$ （其中 $a\ne-1$ ）；第二种给出 $b^2=-(1+2a)\text{。}$

在第二种情形中， $|b|^2=|1+2a|$ 。当 $a$ 绕单位圆变化时， $|1+2a|$ 的取值从 $1$ 到 $3$ ，所以 $1\le|b|\le\sqrt3$ 。两个端点都能取得：分别取 $(a,b)=(-1,1)$ 和 $(a,b)=(1,i\sqrt3)$ 。单独的情形 $b=-1$ 也满足 $|b|=1$ 。因此所求差为 $\sqrt3-1$ 。

因此，正确答案是 C。

Direct composition gives $g(z)=\dfrac{Az+B}{Cz+D},$ where $A=1+a,$ $B=a(1+b),$ $C=1+b,$ and $D=a+b^2.$

The matrix $\begin{pmatrix}A&B\\C&D\end{pmatrix}$ represents $g.$ For $g\circ g$ to be the identity, its square must be scalar. Comparing the off-diagonal entries and the two diagonal entries gives two possibilities: either $B=C=0$ and $A=D,$ or $A+D=0.$ The first gives $b=-1$ (with $a\ne-1$ ); the second gives $b^2=-(1+2a).$

In the second case, $|b|^2=|1+2a|.$ As $a$ runs around the unit circle, $|1+2a|$ ranges from $1$ to $3,$ so $1\le|b|\le\sqrt3.$ Both endpoints occur: take $(a,b)=(-1,1)$ and $(a,b)=(1,i\sqrt3).$ The separate case $b=-1$ also has $|b|=1.$ Therefore the requested difference is $\sqrt3-1.$

Thus, the correct answer is C.

24.

考虑所有满足 $AB = 14$ 、 $BC = 9$ 、 $CD = 7$ 且 $DA = 12$ 的四边形 $ABCD$ 。在这样的四边形内部或边界上能放入的最大圆的半径是多少？

Consider all quadrilaterals $ABCD$ such that $AB = 14,$ $BC = 9,$ $CD = 7,$ and $DA = 12.$ What is the radius of the largest possible circle that fits inside or on the boundary of such a quadrilateral?

$\sqrt{15}$

$\sqrt{21}$

$2\sqrt{6}$

$5$

$2\sqrt{7}$

知识点：内切圆、内心与内切圆半径婆罗摩笈多公式圆内接四边形最优化

难度评级：2460

小提示：

若半径为 $r$ 的圆能放入，从圆心向四边分割可知该四边形的面积至少为 $21r$

If a circle of radius $r$ fits, splitting from its center shows the quadrilateral’s area is at least $21r$

大提示：

布雷特施奈德不等式用圆内接情形给出面积上界；取等的四边形也有内切圆，因为两组对边之和相等

Bretschneider’s inequality bounds the area by the cyclic case; the equality case is also tangential because opposite side sums are equal

解答：

设以 $X$ 为圆心、半径为 $r$ 的圆能放入其中一个四边形。如果 $h_1,h_2,h_3,h_4$ 是 $X$ 到四条边所在直线的距离，那么每个 $h_i\ge r$ 。将四边形分成四个三角形，得到 $\begin{aligned} K&=\dfrac12(14h_1+9h_2 \\ &\qquad+7h_3+12h_4) \\ &\ge21r \end{aligned}\text{。}$

布雷特施奈德不等式表明，给定这些边长的四边形面积不超过圆内接情形： $\begin{aligned} K^2&\le(21-14)(21-9) \\ &\qquad\cdot(21-7)(21-12) \\ &=7\cdot12\cdot14\cdot9, \\ K&\le42\sqrt6 \end{aligned}\text{。}$

因此 $r\le \frac{K}{21}\le2\sqrt6$ 。等号可以达到：具有这些边长的圆内接四边形也有内切圆，因为 $14+7=9+12$ ，其内切圆半径为 $\frac{K}{21}=2\sqrt6$ 。

因此，正确答案是 C。

Suppose a circle of radius $r$ centered at $X$ fits in one of the quadrilaterals. If $h_1,h_2,h_3,h_4$ are the distances from $X$ to the four side lines, then each $h_i\ge r.$ Splitting the quadrilateral into four triangles gives $\begin{aligned} K&=\dfrac12(14h_1+9h_2 \\ &\qquad+7h_3+12h_4) \\ &\ge21r. \end{aligned}$

Bretschneider’s inequality bounds the area of any quadrilateral with these sides by the cyclic case: $\begin{aligned} K^2&\le(21-14)(21-9) \\ &\qquad\cdot(21-7)(21-12) \\ &=7\cdot12\cdot14\cdot9, \\ K&\le42\sqrt6. \end{aligned}$

Therefore $r\le \frac{K}{21}\le2\sqrt6.$ Equality is attainable: the cyclic quadrilateral with these sides is also tangential because $14+7=9+12,$ and its incircle has radius $\frac{K}{21}=2\sqrt6.$

Thus, the correct answer is C.

25.

三角形 $ABC$ 满足 $\angle BAC = 60^\circ$ ， $\angle CBA \le 90^\circ$ ， $BC = 1$ ，且 $AC \ge AB$ 。设 $H$ ， $I$ ，和 $O$ 分别为 $\triangle ABC$ 的垂心、内心和外心。假设五边形 $BCOIH$ 的面积达到最大可能值。求 $\angle CBA$ ？

Triangle $ABC$ has $\angle BAC = 60^\circ,$ $\angle CBA \le 90^\circ,$ $BC = 1,$ and $AC \ge AB.$ Let $H,$ $I,$ and $O$ be the orthocenter, incenter, and circumcenter of $\triangle ABC,$ respectively. Assume that the area of the pentagon $BCOIH$ is the maximum possible. What is $\angle CBA?$

$60^\circ$

$72^\circ$

$75^\circ$

$80^\circ$

$90^\circ$

知识点：外接圆、外心与外接圆半径最优化三角学

难度评级：2840

小提示：

因为 $\angle BAC = 60^\circ$ ，所以点 $B, C, O, I, H$ 都在同一个圆上

Because $\angle BAC = 60^\circ,$ the points $B, C, O, I, H$ all lie on one circle

大提示：

在这个圆上，要使四边形 $BOIH$ 面积最大，应使从 $O$ 到 $B$ 的三段连续小弧相等

On that circle, maximize the quadrilateral $BOIH$ by making the three consecutive subarcs from $O$ to $B$ equal

解答：

令 $B=\angle CBA$ ，并令 $C=\angle BCA=120^\circ-B$ 。因为 $AC\ge AB$ ，所以 $B\ge C$ ，从而 $60^\circ\le B\le90^\circ$ 。标准角度公式给出 $\begin{aligned} \angle BOC&=2\angle A=120^\circ, \\ \angle BHC&=180^\circ-\angle A=120^\circ, \\ \angle BIC&=90^\circ+\dfrac{\angle A}{2}=120^\circ \end{aligned}\text{。}$ 因此 $B,C,O,I,H$ 位于同一个圆上。

此外， $BC=1$ 且 $A=60^\circ$ 确定外接圆半径 $OB=OC=\frac{1}{\sqrt3}$ ，所以 $\triangle BCO$ 以及经过 $B,C,O$ 的圆都是固定的。在 $C$ 处追角可得 $\begin{aligned} \angle OCI&=30^\circ-\dfrac C2, \\ \angle ICH&=30^\circ-\dfrac C2 \end{aligned}\text{。}$ 因此相应的弦满足 $OI=IH$ 。

五边形面积等于固定面积 $[BCO]$ 加上 $[BOIH]$ 。当两个点分割固定弧 $OB$ 时，圆内接四边形在三段连续小弧相等时面积最大（等价地，使它们的正弦和最大）。因此在最大值处有 $OI=IH=HB$ 。

在 $\triangle BOC$ 中， $\angle OCB=30^\circ$ 。等弦使 $\angle OCI=\angle ICH$ ，且 $\angle ICH=\angle HCB$ ，所以这些角都等于 $10^\circ$ 。因此 $30^\circ-\tfrac C2=10^\circ$ ，得 $C=40^\circ$ 和 $B=80^\circ$ 。

因此，正确答案是 D。

Write $B=\angle CBA$ and $C=\angle BCA=120^\circ-B.$ Since $AC\ge AB,$ we have $B\ge C,$ so $60^\circ\le B\le90^\circ.$ The standard angle formulas give $\begin{aligned} \angle BOC&=2\angle A=120^\circ, \\ \angle BHC&=180^\circ-\angle A=120^\circ, \\ \angle BIC&=90^\circ+\dfrac{\angle A}{2}=120^\circ. \end{aligned}$ Hence $B,C,O,I,H$ lie on one circle.

Also $BC=1$ and $A=60^\circ$ fix the circumradius $OB=OC=\frac{1}{\sqrt3},$ so $\triangle BCO$ and the circle through $B,C,O$ are fixed. Angle chasing at $C$ gives $\begin{aligned} \angle OCI&=30^\circ-\dfrac C2, \\ \angle ICH&=30^\circ-\dfrac C2. \end{aligned}$ Thus the corresponding chords satisfy $OI=IH.$

The pentagon’s area is the fixed area $[BCO]$ plus $[BOIH].$ For two points dividing a fixed arc $OB,$ an inscribed quadrilateral has greatest area when its three consecutive subarcs are equal (equivalently, maximize the sum of their sines). Hence at the maximum $OI=IH=HB.$

In $\triangle BOC,$ $\angle OCB=30^\circ.$ Equal chords make $\angle OCI=\angle ICH,$ $\angle ICH=\angle HCB,$ and each of these angles is $10^\circ.$ Therefore $30^\circ-\tfrac C2=10^\circ,$ so $C=40^\circ$ and $B=80^\circ.$

Thus, the correct answer is D.