2024 AIME I 真题与在线限时测验

计时

3:00:00

1.

Aya 每天早晨都会步行 $9$ 千米，然后在咖啡店停留。若她以每小时 $s$ 千米的恒定速度行走，包括在咖啡店停留的 $t$ 分钟在内，全程共用 $4$ 小时。若她以每小时 $s + 2$ 千米的速度行走，包括同样的 $t$ 分钟在内，全程共用 $2$ 小时 $24$ 分钟。假设 Aya 以每小时 $s + \frac{1}{2}$ 千米的速度行走。求包括在咖啡店停留的 $t$ 分钟在内，全程需要的分钟数。

Every morning Aya goes for a $9$ -kilometer-long walk and stops at a coffee shop afterwards. When she walks at a constant speed of $s$ kilometers per hour, the walk takes her $4$ hours, including $t$ minutes spent in the coffee shop. When she walks $s + 2$ kilometers per hour, the walk takes her $2$ hours and $24$ minutes, including $t$ minutes spent in the coffee shop. Suppose Aya walks at $s + \frac{1}{2}$ kilometers per hour. Find the number of minutes the walk takes her, including the $t$ minutes spent in the coffee shop.

答案

答案：204

知识点：路程、速度与时间二次方程

难度评级：1890

小提示：

把两次行程写成方程： $\frac{9}{s} + \frac{t}{60} = 4$ 和 $\frac{9}{s+2} + \frac{t}{60} = \frac{12}{5}$

Convert both trips into equations: $\frac{9}{s} + \frac{t}{60} = 4$ and $\frac{9}{s+2} + \frac{t}{60} = \frac{12}{5}$

大提示：

两式相减可消去 $t$ ，并得到 $s(s+2) = \frac{45}{4}$

Subtracting the two equations eliminates $t$ and gives $s(s+2) = \frac{45}{4}$

解答：

以小时为单位，两种情形给出 $\frac{9}{s} + \frac{t}{60} = 4$ 和 $\frac{9}{s+2} + \frac{t}{60} = \frac{12}{5}\text{。}$ 两式相减得 $\frac{9}{s} - \frac{9}{s+2} = \frac{8}{5}$ ，所以 $\frac{18}{s(s+2)} = \frac{8}{5}$ ，从而 $s(s+2) = \frac{45}{4}$ 。方程 $s^2 + 2s - \frac{45}{4} = 0$ 的正根为 $s = \frac{5}{2}$ 。

因此 $\frac{t}{60} = 4 - \frac{9}{\frac{5}{2}} = \frac{2}{5}$ ，所以 $t = 24$ 分钟。以 $s + \frac{1}{2} = 3$ 千米每小时行走时，步行本身需要 $\frac{9}{3} = 3$ 小时，所以总时间为 $180 + 24 = 204$ 分钟。

Measuring time in hours, the two scenarios say $\frac{9}{s} + \frac{t}{60} = 4$ and $\frac{9}{s+2} + \frac{t}{60} = \frac{12}{5}.$ Subtracting, $\frac{9}{s} - \frac{9}{s+2} = \frac{8}{5},$ so $\frac{18}{s(s+2)} = \frac{8}{5},$ giving $s(s+2) = \frac{45}{4}.$ The positive root of $s^2 + 2s - \frac{45}{4} = 0$ is $s = \frac{5}{2}.$

Then $\frac{t}{60} = 4 - \frac{9}{\frac{5}{2}} = \frac{2}{5},$ so $t = 24$ minutes. Walking at $s + \frac{1}{2} = 3$ kilometers per hour takes $\frac{9}{3} = 3$ hours, so the total is $180 + 24 = 204$ minutes.

2.

存在实数 $x$ 和 $y$ ，且二者都大于 $1$ ，使得 $\log_x\left(y^x\right) = \log_y\left(x^{4y}\right) = 10$ 。求 $xy$ 。

There exist real numbers $x$ and $y,$ both greater than $1,$ such that $\log_x\left(y^x\right) = \log_y\left(x^{4y}\right) = 10.$ Find $xy.$

答案

答案：25

知识点：对数代数变形

难度评级：2070

小提示：

把指数提到前面： $x \log_x y = 10$ 且 $4y \log_y x = 10$

Bring the exponents down: $x \log_x y = 10$ and $4y \log_y x = 10$

大提示：

将两个方程相乘；因为 $\log_x y \cdot \log_y x = 1$ ，对数项会消失

Multiply the two equations: since $\log_x y \cdot \log_y x = 1,$ the logarithms disappear

解答：

将指数从对数中提出来，条件变为 $x \log_x y = 10$ 和 $4y \log_y x = 10\text{。}$ 两式相乘，并使用 $\log_x y \cdot \log_y x = 1$ ，得到 $4xy = 100$ ，所以 $xy = 25$ 。

这样的 $x$ 和 $y$ 的确存在。令 $y = \frac{25}{x}$ 。连续函数 $x\log_x(\frac{25}{x})$ 在 $x \to 1^+$ 时趋于无穷大，而在 $x = 2$ 时小于 $10$ 。因此它在某个 $1 \lt x \lt 2$ 处等于 $10$ ，此时 $y \gt 1$ 。

Pulling the exponents out of the logarithms, the conditions become $x \log_x y = 10$ and $4y \log_y x = 10.$ Multiplying these equations and using $\log_x y \cdot \log_y x = 1$ gives $4xy = 100,$ so $xy = 25.$

Such $x$ and $y$ do exist. Set $y = \frac{25}{x}.$ The continuous function $x\log_x(\frac{25}{x})$ tends to infinity as $x \to 1^+,$ while at $x = 2$ it is less than $10.$ Hence it equals $10$ for some $1 \lt x \lt 2,$ where $y \gt 1.$

3.

Alice 和 Bob 玩如下游戏。他们面前有一堆 $n$ 个筹码。两人轮流操作，Alice 先手。每一轮，当前玩家从筹码堆中取走 $1$ 个或 $4$ 个筹码。取走最后一个筹码的玩家获胜。求不超过 $2024$ 的正整数 $n$ 的个数，使得无论 Alice 如何行动，Bob 都有保证获胜的策略。

Alice and Bob play the following game. A stack of $n$ tokens lies before them. The players take turns with Alice going first. On each turn, the player removes $1$ token or $4$ tokens from the stack. The player who removes the last token wins. Find the number of positive integers $n$ less than or equal to $2024$ such that there is a strategy that guarantees that Bob wins, regardless of Alice’s moves.

答案

答案：809

知识点：组合游戏模运算小情形

难度评级：2110

小提示：

先找出小的 $n$ 中哪些是轮到行动者必败的位置：当 $n - 1$ 和 $n - 4$ 都是必胜位置时， $n$ 正好是必败位置

Work out which small $n$ are losses for the player about to move: $n$ is a loss exactly when both $n - 1$ and $n - 4$ are wins

大提示：

必败位置以 $5$ 为周期重复：它们是 $n \equiv 0$ 或 $2 \pmod 5$ 。Bob 正好在这些 $n$ 上获胜

The losing positions repeat with period $5:$ they are $n \equiv 0$ or $2 \pmod 5.$ Bob wins exactly for those $n.$

解答：

若轮到行动的玩家在最优游戏下会输，就称 $n$ 为必败位置。我们断言必败位置正是 $n \equiv 0$ 或 $2 \pmod 5$ 。从这样的 $n$ 出发，取走 $1$ 个或 $4$ 个筹码会留下 $n \equiv 4, 1, 3 \pmod 5$ ，不会再次留下 $0$ 或 $2$ 。而从任何 $n \equiv 1, 3, 4 \pmod 5$ 出发，一步就能到达 $\equiv 0$ 或 $2 \pmod 5$ 的位置（分别取走 $1$ 、 $1$ 、 $4$ 个筹码）。由于 $n = 0$ 对轮到行动的玩家是败局，归纳可确认这个模式。

Bob 获胜当且仅当 $n$ 对 Alice 是必败位置。在 $1 \le n \le 2024$ 中， $5$ 的倍数有 $404$ 个，满足 $n \equiv 2 \pmod 5$ 的数有 $405$ 个（从 $2$ 到 $2022$ ），总共为 $404 + 405 = 809$ 。

Call $n$ a losing position if the player about to move loses with best play. We claim the losing positions are exactly $n \equiv 0$ or $2 \pmod 5.$ From such an $n,$ removing $1$ or $4$ tokens leaves $n \equiv 4, 1, 3 \pmod 5$ — never again $0$ or $2$ — while from any $n \equiv 1, 3, 4 \pmod 5$ one move reaches a position $\equiv 0$ or $2 \pmod 5$ (remove $1,$ $1,$ $4$ tokens respectively). Since $n = 0$ is a loss for the player to move, induction confirms the pattern.

Bob wins exactly when $n$ is a losing position for Alice. Among $1 \le n \le 2024$ there are $404$ multiples of $5$ and $405$ values $n \equiv 2 \pmod 5$ (from $2$ to $2022$ ), for a total of $404 + 405 = 809.$

4.

Jen 参加抽奖，从 $S = \{1, 2, 3, \ldots, 9, 10\}$ 中选择 $4$ 个不同元素。随后从 $S$ 中随机抽出四个元素。如果 Jen 选的数中至少有两个被抽中，她就中奖；如果她选的四个数全部被抽中，她就中头奖。在 Jen 已经中奖的条件下，她中头奖的概率为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

Jen enters a lottery by selecting $4$ distinct elements of $S = \{1, 2, 3, \ldots, 9, 10\}.$ Then four elements of $S$ are drawn at random. Jen wins a prize if at least two of her numbers were drawn, and wins the grand prize if all four of her numbers were drawn. The probability that Jen wins the grand prize given that Jen wins a prize is $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：116

知识点：条件概率组合

难度评级：2230

小提示：

中头奖必然意味着中奖，所以条件概率是 $\frac{P(\text{头奖})}{P(\text{中奖})}$ 。

Winning the grand prize forces winning a prize, so the conditional probability is $\frac{P(\text{grand})}{P(\text{prize})}$

大提示：

统计抽出的 $4$ 个数与 Jen 的彩票恰好重合 $2$ 、 $3$ 、或 $4$ 个的情况；所有四数组合等可能

Count draws that share exactly $2,$ $3,$ or $4$ numbers with Jen’s ticket — every draw of $4$ numbers is equally likely

解答：

所有 $\binom{10}{4} = 210$ 种抽法等可能。与 Jen 的彩票恰好有 $k$ 个数重合的抽法数为 $\binom{4}{k}\binom{6}{4-k}$ ，所以中奖的抽法数为 $\begin{aligned} &\binom{4}{2}\binom{6}{2} + \binom{4}{3}\binom{6}{1} \\ &\quad {}+ \binom{4}{4}\binom{6}{0} \\ &= 90 + 24 + 1 = 115 \end{aligned}\text{，}$ 其中正好 $1$ 种抽法中头奖。

因为中头奖必然中奖，条件概率为 $\frac{\frac{1}{210}}{\frac{115}{210}} = \frac{1}{115}$ ，所以 $m + n = 1 + 115 = 116$ 。

All $\binom{10}{4} = 210$ draws are equally likely. The number of draws sharing exactly $k$ numbers with Jen’s ticket is $\binom{4}{k}\binom{6}{4-k},$ so the number winning a prize is $\begin{aligned} &\binom{4}{2}\binom{6}{2} + \binom{4}{3}\binom{6}{1} \\ &\quad {}+ \binom{4}{4}\binom{6}{0} \\ &= 90 + 24 + 1 = 115, \end{aligned}$ and exactly $1$ of these wins the grand prize.

Since the grand prize implies a prize, the conditional probability is $\frac{\frac{1}{210}}{\frac{115}{210}} = \frac{1}{115},$ so $m + n = 1 + 115 = 116.$

5.

矩形 $ABCD$ 的边长为 $AB = 107$ 和 $BC = 16$ ，矩形 $EFGH$ 的边长为 $EF = 184$ 和 $FG = 17$ 。点 $D$ 、 $E$ 、 $C$ 、 $F$ 按这个顺序位于直线 $DF$ 上，且 $A$ 与 $H$ 位于直线 $DF$ 的两侧，如图所示。点 $A$ 、 $D$ 、 $H$ 、 $G$ 共圆。求 $CE$ 。

Rectangle $ABCD$ has dimensions $AB = 107$ and $BC = 16,$ and rectangle $EFGH$ has dimensions $EF = 184$ and $FG = 17.$ Points $D,$ $E,$ $C,$ and $F$ lie on line $DF$ in that order, and $A$ and $H$ lie on opposite sides of line $DF,$ as shown. Points $A,$ $D,$ $H,$ and $G$ lie on a common circle. Find $CE.$

答案

答案：104

知识点：坐标几何圆垂直平分线

难度评级：2390

小提示：

将直线 $DF$ 放在 $x$ 轴上，并令 $D$ 为原点；用 $DE$ 表示 $A$ 、 $D$ 、 $H$ 、 $G$ 的坐标

Put line $DF$ on the $x$ -axis with $D$ at the origin, and write the coordinates of $A,$ $D,$ $H,$ and $G$ in terms of $DE$

大提示：

圆心位于 $y = -8$ （ $\overline{AD}$ 的垂直平分线）上，也位于 $\overline{HG}$ 的垂直平分线上；令它到 $D$ 和 $H$ 的距离相等

The circle’s center lies on $y = -8$ (bisector of $\overline{AD}$ ) and on the vertical bisector of $\overline{HG};$ equate its distances to $D$ and $H$

解答：

将直线 $DF$ 放在 $x$ 轴上，令 $D = (0, 0)$ 且 $C = (107, 0)$ ，于是 $A = (0, -16)$ 。设 $DE = e$ 。那么 $E = (e, 0)$ 、 $F = (e + 184, 0)$ ，第二个矩形位于直线上方： $H = (e, 17)$ 、 $G = (e + 184, 17)$ 。

经过 $A$ 、 $D$ 、 $H$ 、 $G$ 的圆的圆心，既在竖直线段 $\overline{AD}$ 的垂直平分线 $y = -8$ 上，也在水平线段 $\overline{HG}$ 的垂直平分线 $x = e + 92$ 上。令圆心到 $D$ 和到 $H$ 的距离平方相等， $\begin{aligned} &(e + 92)^2 + 8^2 \\ &= 92^2 + 25^2 = 9089 \end{aligned}\text{，}$ 所以 $(e + 92)^2 = 9025$ ，且 $e + 92 = 95$ ，得到 $e = 3$ 。

因此 $CE = DC - DE$ $= 107 - 3 = 104$ 。

Put line $DF$ on the $x$ -axis with $D = (0, 0)$ and $C = (107, 0),$ so $A = (0, -16).$ Let $DE = e.$ Then $E = (e, 0),$ $F = (e + 184, 0),$ and the second rectangle sits above the line: $H = (e, 17)$ and $G = (e + 184, 17).$

The center of the circle through $A,$ $D,$ $H,$ $G$ lies on the perpendicular bisector of the vertical segment $\overline{AD},$ the line $y = -8,$ and on the perpendicular bisector of the horizontal segment $\overline{HG},$ the line $x = e + 92.$ Equating the center’s squared distances to $D$ and to $H,$ $\begin{aligned} &(e + 92)^2 + 8^2 \\ &= 92^2 + 25^2 = 9089, \end{aligned}$ so $(e + 92)^2 = 9025$ and $e + 92 = 95,$ giving $e = 3.$

Therefore $CE = DC - DE$ $= 107 - 3 = 104.$

6.

考虑在一个 $8 \times 8$ 方格中，从左下角沿网格线走到右上角、长度为 $16$ 的路径。求这样的路径中，恰好改变方向四次的路径数，如下图中的例子所示。

Consider the paths of length $16$ that follow the lines from the lower left corner to the upper right corner on an $8 \times 8$ grid. Find the number of such paths that change direction exactly four times, as in the examples shown below.

答案

答案：294

知识点：格路分拆与有序分拆

难度评级：2340

小提示：

恰好转向四次的路径由五段交替向右和向上的直线段组成

A path with exactly four turns consists of five straight runs that alternate between rightward and upward

大提示：

若路径从向右开始，则三段向右和两段向上的长度都是正整数，并且各自总和为 $8$ ；先数这些组成，再将结果乘二

If the path starts rightward, the three rightward runs and two upward runs are positive integers summing to $8$ each — count the compositions, then double

解答：

恰好改变方向四次的路径由五段最长的直线段组成，方向在向右与向上之间交替。若第一段向右，模式为 $R, U, R, U, R$ ：三段向右的正长度总和为 $8$ ，两段向上的正长度总和为 $8$ 。这样的组成数分别为 $\binom{7}{2} = 21$ 和 $\binom{7}{1} = 7$ ，给出 $21 \cdot 7 = 147$ 条路径。

从向上开始的路径由对称性也有 $147$ 条。总数为 $147 + 147 = 294$ 。

A path that changes direction exactly four times consists of five maximal straight runs, alternating between rightward and upward moves. If the first run is rightward, the pattern is $R, U, R, U, R:$ three rightward runs with positive lengths summing to $8,$ and two upward runs with positive lengths summing to $8.$ The counts of such compositions are $\binom{7}{2} = 21$ and $\binom{7}{1} = 7,$ giving $21 \cdot 7 = 147$ paths.

Paths starting upward are counted symmetrically, another $147.$ The total is $147 + 147 = 294.$

7.

求表达式 $(75 + 117\mathrm{i})z + \frac{96 + 144\mathrm{i}}{z}$ 实部的最大可能值，其中 $z$ 是满足 $|z| = 4$ 的复数。这里 $\mathrm{i} = \sqrt{-1}$ 。

Find the largest possible real part of $(75 + 117\mathrm{i})z + \frac{96 + 144\mathrm{i}}{z}$ where $z$ is a complex number with $|z| = 4.$ Here $\mathrm{i} = \sqrt{-1}.$

答案

答案：540

知识点：复数三角恒等式最优化

难度评级：2410

小提示：

写作 $z = 4(\cos\theta + \mathrm{i}\sin\theta)$ ，并分别取每一项的实部

Write $z = 4(\cos\theta + \mathrm{i}\sin\theta)$ and take the real part of each term

大提示：

你会得到形如 $a\cos\theta + b\sin\theta$ 的表达式，其最大值为 $\sqrt{a^2 + b^2}$

You get an expression of the form $a\cos\theta + b\sin\theta,$ whose maximum is $\sqrt{a^2 + b^2}$

解答：

写 $z = 4(\cos\theta + \mathrm{i}\sin\theta)$ ，所以 $\frac{1}{z} = \frac{1}{4}(\cos\theta - \mathrm{i}\sin\theta)$ 。 $(75 + 117\mathrm{i})z$ 的实部为 $4(75\cos\theta - 117\sin\theta)$ $= 300\cos\theta - 468\sin\theta$ ，而 $(96 + 144\mathrm{i}) \cdot \frac{1}{4}(\cos\theta - \mathrm{i}\sin\theta)$ 的实部为 $24\cos\theta + 36\sin\theta$ 。

总实部为 $324\cos\theta - 432\sin\theta$ ，它关于 $\theta$ 的最大值是 $\begin{aligned} &\sqrt{324^2 + 432^2} \\ &= 108\sqrt{3^2 + 4^2} \\ &= 108 \cdot 5 = 540 \end{aligned}\text{。}$

Write $z = 4(\cos\theta + \mathrm{i}\sin\theta),$ so $\frac{1}{z} = \frac{1}{4}(\cos\theta - \mathrm{i}\sin\theta).$ The real part of $(75 + 117\mathrm{i})z$ is $4(75\cos\theta - 117\sin\theta)$ $= 300\cos\theta - 468\sin\theta,$ and the real part of $(96 + 144\mathrm{i}) \cdot \frac{1}{4}(\cos\theta - \mathrm{i}\sin\theta)$ is $24\cos\theta + 36\sin\theta.$

The total real part is $324\cos\theta - 432\sin\theta,$ whose maximum over $\theta$ is $\begin{aligned} &\sqrt{324^2 + 432^2} \\ &= 108\sqrt{3^2 + 4^2} \\ &= 108 \cdot 5 = 540. \end{aligned}$

8.

可以放置八个半径为 $34$ 的圆，使它们都与 $\triangle ABC$ 的 $\overline{BC}$ 相切，并且这些圆依次两两相切，第一个圆与 $\overline{AB}$ 相切，最后一个圆与 $\overline{AC}$ 相切，如图所示。类似地，也可以按同样方式放置 $2024$ 个半径为 $1$ 的圆，使它们都与 $\overline{BC}$ 相切。 $\triangle ABC$ 的内切圆半径可表示为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

Eight circles of radius $34$ can be placed tangent to $\overline{BC}$ of $\triangle ABC$ so that the circles are sequentially tangent to each other, with the first circle being tangent to $\overline{AB}$ and the last circle being tangent to $\overline{AC},$ as shown. Similarly, $2024$ circles of radius $1$ can be placed tangent to $\overline{BC}$ in the same manner. The inradius of $\triangle ABC$ can be expressed as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：197

知识点：相切圆内切圆、内心与内切圆半径三角学

难度评级：2560

小提示：

圆心位于 $BC$ 上方高度 $\rho$ 处，相邻圆心相距 $2\rho$ ，两端圆心到 $B$ 和 $C$ 的距离分别为 $\rho\cot\frac{B}{2}$ 与 $\rho\cot\frac{C}{2}$

The centers sit at height $\rho$ above $BC,$ spaced $2\rho$ apart, and the end centers lie at distances $\rho\cot\frac{B}{2}$ and $\rho\cot\frac{C}{2}$ from $B$ and $C$

大提示：

两条圆链测得的是同一条 $BC$ ，这会确定 $\cot\frac{B}{2} + \cot\frac{C}{2}$ ；内切圆满足 $BC = r\left(\cot\frac{B}{2} + \cot\frac{C}{2}\right)$

Both chains measure the same $BC,$ which determines $\cot\frac{B}{2} + \cot\frac{C}{2};$ the incircle satisfies $BC = r\left(\cot\frac{B}{2} + \cot\frac{C}{2}\right)$

解答：

对一条由 $n$ 个半径为 $\rho$ 的圆组成、且都与 $\overline{BC}$ 相切的圆链，圆心位于高度 $\rho$ 处，相邻圆心相距 $2\rho$ 。第一个圆与 $\overline{AB}$ 和 $\overline{BC}$ 相切，所以它的圆心在从 $B$ 出发的角平分线上，到 $B$ 的水平距离为 $\rho\cot\frac{B}{2}$ ；类似地，最后一个圆心到 $C$ 的距离为 $\rho\cot\frac{C}{2}$ 。因此令 $k = \cot\frac{B}{2} + \cot\frac{C}{2}$ ，有 $BC = \rho k + 2\rho(n - 1)\text{。}$

两条圆链给出 $34k + 34 \cdot 14 = BC$ $= k + 2 \cdot 2023$ ，所以 $33k = 3570$ ， $k = \frac{1190}{11}$ ，从而 $BC = k + 4046 = \frac{45696}{11}$ 。

内切圆就是由一个半径为 $r$ 的圆组成的圆链： $BC = rk$ 。因此 $r = \frac{BC}{k} = \frac{45696}{1190} = \frac{192}{5}\text{，}$ 所以 $m + n = 192 + 5 = 197$ 。

For a chain of $n$ circles of radius $\rho$ tangent to $\overline{BC},$ the centers lie at height $\rho$ with consecutive centers $2\rho$ apart. The first circle is tangent to $\overline{AB}$ and $\overline{BC},$ so its center lies on the bisector from $B,$ at horizontal distance $\rho\cot\frac{B}{2}$ from $B;$ similarly the last center is $\rho\cot\frac{C}{2}$ from $C.$ Hence with $k = \cot\frac{B}{2} + \cot\frac{C}{2},$ $BC = \rho k + 2\rho(n - 1).$

The two chains give $34k + 34 \cdot 14 = BC$ $= k + 2 \cdot 2023,$ so $33k = 3570$ and $k = \frac{1190}{11},$ whence $BC = k + 4046 = \frac{45696}{11}.$

The incircle is a chain of one circle of radius $r:$ $BC = rk.$ Therefore $r = \frac{BC}{k} = \frac{45696}{1190} = \frac{192}{5},$ and $m + n = 192 + 5 = 197.$

9.

设 $A$ 、 $B$ 、 $C$ 、 $D$ 是双曲线 $\frac{x^2}{20} - \frac{y^2}{24} = 1$ 上的点，使得 $ABCD$ 是一个菱形，且它的对角线在原点相交。求一个最大数，使它对所有这样的菱形 $ABCD$ 都小于 $BD^2$ 。

Let $A,$ $B,$ $C,$ and $D$ be points on the hyperbola $\frac{x^2}{20} - \frac{y^2}{24} = 1$ such that $ABCD$ is a rhombus whose diagonals intersect at the origin. Find the largest number less than $BD^2$ for all rhombuses $ABCD.$

答案

答案：480

知识点：双曲线菱形最优化

难度评级：2710

小提示：

菱形的对角线是经过原点的互相垂直的直线；若 $B = (x, y)$ 且 $y = mx$ ，则 $BD^2 = 4(x^2 + y^2)$

The diagonals are perpendicular lines through the origin; if $B = (x, y)$ with $y = mx,$ then $BD^2 = 4(x^2 + y^2)$

大提示：

另一条垂直的对角线也必须与双曲线相交，所以两个斜率的平方都严格介于 $\frac{5}{6}$ 与 $\frac{6}{5}$ 之间；研究 $BD^2$ 随斜率的变化

The perpendicular diagonal must also meet the hyperbola, so both slopes squared lie strictly between $\frac{5}{6}$ and $\frac{6}{5};$ study $BD^2$ as the slope varies

解答：

菱形的对角线互相垂直且互相平分，所以 $C = -A$ 、 $D = -B$ ，并且 $OA \perp OB$ 。设直线 $BD$ 的斜率为 $m$ ，于是 $B = (x, mx)$ ，并满足 $x^2\left(\frac{1}{20} - \frac{m^2}{24}\right) = 1\text{，}$ 即 $x^2 = \frac{120}{6 - 5m^2}\text{，}$ 这要求 $m^2 \lt \frac{6}{5}$ 。于是 $BD^2 = 4(x^2 + m^2x^2)$ $= \frac{480(1 + m^2)}{6 - 5m^2}$ 。直线 $AC$ 的斜率为 $-\frac{1}{m}$ ，所以它与双曲线相交仅当 $\frac{1}{m^2} \lt \frac{6}{5}$ ，也就是 $m^2 \gt \frac{5}{6}$ 。

在区间 $\frac{5}{6} \lt m^2 \lt \frac{6}{5}$ 上，量 $\frac{480(1 + m^2)}{6 - 5m^2}$ 随 $m^2$ 严格递增：当 $m^2 \to \frac{5}{6}$ 时，它趋近于 $480 \cdot \frac{\frac{11}{6}}{\frac{11}{6}} = 480$ ，而当 $m^2 \to \frac{6}{5}$ 时，它无界增长。因此 $BD^2$ 正好取到 $(480, \infty)$ 中的值，并且永不等于 $480$ 。

对每一个这样的菱形都小于 $BD^2$ 的最大数因此是 $480$ 。

The diagonals of a rhombus are perpendicular bisectors of each other, so $C = -A,$ $D = -B,$ and $OA \perp OB.$ Let line $BD$ have slope $m,$ so $B = (x, mx)$ with $x^2\left(\frac{1}{20} - \frac{m^2}{24}\right) = 1,$ i.e. $x^2 = \frac{120}{6 - 5m^2},$ which requires $m^2 \lt \frac{6}{5}.$ Then $BD^2 = 4(x^2 + m^2x^2)$ $= \frac{480(1 + m^2)}{6 - 5m^2}.$ Line $AC$ has slope $-\frac{1}{m},$ so it meets the hyperbola only when $\frac{1}{m^2} \lt \frac{6}{5},$ that is $m^2 \gt \frac{5}{6}.$

On the interval $\frac{5}{6} \lt m^2 \lt \frac{6}{5},$ the quantity $\frac{480(1 + m^2)}{6 - 5m^2}$ is strictly increasing in $m^2:$ as $m^2 \to \frac{5}{6}$ it tends to $480 \cdot \frac{\frac{11}{6}}{\frac{11}{6}} = 480,$ and as $m^2 \to \frac{6}{5}$ it grows without bound. Hence $BD^2$ takes exactly the values in $(480, \infty)$ and never equals $480.$

The largest number that is less than $BD^2$ for every such rhombus is therefore $480.$

10.

设 $\triangle ABC$ 的边长为 $AB = 5$ 、 $BC = 9$ 、 $CA = 10$ 。在 $\triangle ABC$ 的外接圆上，过 $B$ 和 $C$ 作切线，两条切线交于点 $D$ ，且 $\overline{AD}$ 与外接圆交于 $P \ne A$ 。长度 $AP$ 等于 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质整数。求 $m + n$ 。

Let $\triangle ABC$ have side lengths $AB = 5,$ $BC = 9,$ and $CA = 10.$ The tangents to the circumcircle of $\triangle ABC$ at $B$ and $C$ intersect at point $D,$ and $\overline{AD}$ intersects the circumcircle at $P \ne A.$ The length of $AP$ is equal to $\frac{m}{n},$ where $m$ and $n$ are relatively prime integers. Find $m + n.$

答案

答案：113

知识点：圆幂切线余弦定理

难度评级：2920

小提示：

由切线-弦定理， $\angle DBC = \angle DCB = \angle A$ ，所以 $DB = \frac{\frac{BC}{2}}{\cos A}$

By the tangent-chord angle, $\angle DBC = \angle DCB = \angle A,$ so $DB = \frac{\frac{BC}{2}}{\cos A}$

大提示：

点的幂给出 $DP \cdot DA = DB^2$ ，所以 $AP = DA - \frac{DB^2}{DA}$ ；用三角形 $ABD$ 中的余弦定理求 $DA$

Power of the point gives $DP \cdot DA = DB^2,$ so $AP = DA - \frac{DB^2}{DA};$ compute $DA$ from the law of cosines in triangle $ABD$

解答：

由切线-弦定理， $\angle DBC = \angle DCB = \angle A$ ，所以三角形 $DBC$ 是等腰三角形，且 $DB = \frac{\frac{BC}{2}}{\cos A}$ 。余弦定理给出 $\cos A = \frac{10^2 + 5^2 - 9^2}{2 \cdot 10 \cdot 5} = \frac{11}{25}$ 和 $\cos B = \frac{9^2 + 5^2 - 10^2}{2 \cdot 9 \cdot 5} = \frac{1}{15}$ ，所以 $DB = \frac{\frac{9}{2}}{\frac{11}{25}} = \frac{225}{22}$ 。

因为 $D$ 位于 $BC$ 与 $A$ 相对的一侧， $\angle ABD = A + B$ ，且 $\begin{aligned} &\cos(A + B) \\ &= \frac{11}{25} \cdot \frac{1}{15} \\ &\quad {}- \frac{6\sqrt{14}}{25} \cdot \frac{4\sqrt{14}}{15} \\ &= \frac{11 - 336}{375} \\ &= -\frac{13}{15} \end{aligned}\text{。}$ 因此在三角形 $ABD$ 中使用余弦定理，得到 $\begin{aligned} &DA^2 = 5^2 + \left(\tfrac{225}{22}\right)^2 \\ &\quad {}+ 2 \cdot 5 \cdot \tfrac{225}{22} \cdot \tfrac{13}{15} \\ &= \frac{105625}{484} \end{aligned}\text{，}$ $DA = \frac{325}{22}\text{。}$

点 $D$ 的幂给出 $DP \cdot DA = DB^2$ ，所以 $\begin{aligned} &AP = DA - \frac{DB^2}{DA} \\ &= \frac{DA^2 - DB^2}{DA} \\ &= \frac{\frac{105625 - 50625}{484}}{\frac{325}{22}} \\ &= \frac{55000}{22 \cdot 325} \\ &= \frac{100}{13} \end{aligned}\text{，}$ 因此 $m + n = 100 + 13 = 113$ 。

By the tangent-chord angle, $\angle DBC = \angle DCB = \angle A,$ so triangle $DBC$ is isosceles with $DB = \frac{\frac{BC}{2}}{\cos A}.$ The law of cosines gives $\cos A = \frac{10^2 + 5^2 - 9^2}{2 \cdot 10 \cdot 5} = \frac{11}{25}$ and $\cos B = \frac{9^2 + 5^2 - 10^2}{2 \cdot 9 \cdot 5} = \frac{1}{15},$ so $DB = \frac{\frac{9}{2}}{\frac{11}{25}} = \frac{225}{22}.$

Since $D$ lies on the opposite side of $BC$ from $A,$ $\angle ABD = A + B,$ and $\begin{aligned} &\cos(A + B) \\ &= \frac{11}{25} \cdot \frac{1}{15} \\ &\quad {}- \frac{6\sqrt{14}}{25} \cdot \frac{4\sqrt{14}}{15} \\ &= \frac{11 - 336}{375} \\ &= -\frac{13}{15}. \end{aligned}$ The law of cosines in triangle $ABD$ then gives $\begin{aligned} &DA^2 = 5^2 + \left(\tfrac{225}{22}\right)^2 \\ &\quad {}+ 2 \cdot 5 \cdot \tfrac{225}{22} \cdot \tfrac{13}{15} \\ &= \frac{105625}{484}, \end{aligned}$ $DA = \frac{325}{22}.$

The power of $D$ gives $DP \cdot DA = DB^2,$ so $\begin{aligned} &AP = DA - \frac{DB^2}{DA} \\ &= \frac{DA^2 - DB^2}{DA} \\ &= \frac{\frac{105625 - 50625}{484}}{\frac{325}{22}} \\ &= \frac{55000}{22 \cdot 325} \\ &= \frac{100}{13}, \end{aligned}$ and $m + n = 100 + 13 = 113.$

11.

正八边形的每个顶点独立地以相同概率染成红色或蓝色。若这个八边形可以通过某次旋转，使得所有蓝色顶点最终落到原本是红色顶点的位置上，则称该染色满足条件。这个概率为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

Each vertex of a regular octagon is independently colored either red or blue with equal probability. The probability that the octagon can then be rotated so that all of the blue vertices end up at positions where there had been red vertices is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：371

知识点：基本概率双重计数分类讨论

难度评级：2990

小提示：

若有 $b$ 个蓝色顶点，则七个非零旋转的重合数 $|B \cap (B + k)|$ 之和为 $b^2 - b$ ，所以所有 $b \le 3$ 的染色都满足条件

With $b$ blue vertices, the overlaps $|B \cap (B + k)|$ summed over the seven nonzero rotations equal $b^2 - b,$ so every coloring with $b \le 3$ works

大提示：

当 $b = 4$ 时，某个旋转必须把蓝色集合恰好送到它的补集；分别对 $k$ 为奇数、 $k \equiv 2 \pmod 4$ 、以及 $k = 4$ 计数

For $b = 4$ a rotation must carry the blue set exactly onto its complement; count those sets separately for $k$ odd, $k \equiv 2 \pmod 4,$ and $k = 4$

解答：

将顶点标为 $0, \ldots, 7$ ，并令 $B$ 为蓝色顶点集合， $b = |B|$ 。旋转 $k$ 满足条件，当且仅当 $(B + k) \cap B = \varnothing$ 。由于 $B + k$ 必须放入 $8 - b$ 个红色位置中，必有 $b \le 4$ 。把 $|B \cap (B + k)|$ 对所有八个旋转求和，会把 $(i, j) \in B \times B$ 的所有有序对各计一次（通过 $k = i - j$ ），总数为 $b^2$ ，而 $k = 0$ 项贡献 $b$ 。所以当 $b \le 3$ 时，七个非零旋转的重合总数仅为 $b^2 - b \le 6$ ，必有某个旋转没有重合：所有 $b \le 3$ 的 $1 + 8 + 28 + 56 = 93$ 种染色都成功。

当 $b = 4$ 时，不相交会迫使 $B + k$ 正好等于 $B$ 的补集。若 $k$ 为奇数，循环 $0, k, 2k, \ldots$ 会遍历全部顶点，并且必须在 $B$ 与其补集之间交替，所以 $B$ 是偶数顶点集合或奇数顶点集合：共有 $2$ 个集合。若 $k \equiv 2 \pmod 4$ ，则 $B$ 在两个 $4$ -循环 $\{0, 2, 4, 6\}$ 和 $\{1, 3, 5, 7\}$ 中各取一对相对顶点：共有 $2 \cdot 2 = 4$ 个集合，例如 $\{0, 1, 4, 5\}$ 。若 $k = 4$ ，则 $B$ 从每对 $\{i, i + 4\}$ 中恰取一个顶点：共有 $2^4 = 16$ 个集合。前两类都包含某对相对顶点的两个成员，而第三类从不这样；偶数或奇数顶点集合又把它们的相对顶点对都取自同一个 $4$ -循环，所以三类互不重叠，总共有 $2 + 4 + 16 = 22$ 个集合。

因此在 $2^8 = 256$ 种染色中，有 $93 + 22 = 115$ 种满足条件，概率为 $\frac{115}{256}$ ，所以 $m + n = 115 + 256 = 371$ 。

Label the vertices $0, \ldots, 7$ and let $B$ be the blue set, $b = |B|.$ Rotation by $k$ works exactly when $(B + k) \cap B = \varnothing.$ Since $B + k$ must fit inside the $8 - b$ red positions, $b \le 4.$ Summing $|B \cap (B + k)|$ over all eight rotations counts all pairs $(i, j) \in B \times B$ once (via $k = i - j$ ), a total of $b^2,$ and the $k = 0$ term contributes $b.$ So for $b \le 3$ the seven nonzero rotations share only $b^2 - b \le 6$ overlaps, and some rotation has none: all $1 + 8 + 28 + 56 = 93$ colorings with $b \le 3$ succeed.

For $b = 4,$ disjointness forces $B + k$ to be exactly the complement of $B.$ If $k$ is odd, the cycle $0, k, 2k, \ldots$ visits all vertices and must alternate between $B$ and its complement, so $B$ is the evens or the odds: $2$ sets. If $k \equiv 2 \pmod 4,$ then $B$ meets each of the $4$ -cycles $\{0, 2, 4, 6\}$ and $\{1, 3, 5, 7\}$ in an antipodal pair: $2 \cdot 2 = 4$ sets, such as $\{0, 1, 4, 5\}.$ If $k = 4,$ then $B$ contains exactly one of each pair $\{i, i + 4\}:$ $2^4 = 16$ sets. The first two families contain both members of some antipodal pair while the third never does, and the evens/odds take both their antipodal pairs from one $4$ -cycle, so the three families are disjoint: $2 + 4 + 16 = 22$ sets.

In total $93 + 22 = 115$ of the $2^8 = 256$ colorings work, so the probability is $\frac{115}{256}$ and $m + n = 115 + 256 = 371.$

12.

定义 $f(x) = \left||x| - \tfrac{1}{2}\right|$ 和 $g(x) = \left||x| - \tfrac{1}{4}\right|$ 。求下列两条图像的交点个数： $y = 4g(f(\sin(2\pi x)))$ 与 $x = 4g(f(\cos(3\pi y)))\text{。}$

Define $f(x) = \left||x| - \tfrac{1}{2}\right|$ and $g(x) = \left||x| - \tfrac{1}{4}\right|.$ Find the number of intersections of the graphs of $y = 4g(f(\sin(2\pi x)))$ and $x = 4g(f(\cos(3\pi y))).$

答案

答案：385

知识点：函数绝对值交点计数

难度评级：3160

小提示：

当 $u = |\sin 2\pi x|$ 从 $0$ 变到 $1$ 时，值 $4||u - \frac{1}{2}| - \frac{1}{4}|$ 在 $1$ 与 $0$ 之间折返四次；数出每条图像的单调弧

As $u = |\sin 2\pi x|$ runs from $0$ to $1,$ the value $4||u - \frac{1}{2}| - \frac{1}{4}|$ zigzags between $1$ and $0$ four times; count the monotone arcs of each graph

大提示：

一条图像的每条自下而上的弧，与另一条图像的每条自左而右的弧恰好相交一次；然后检查两条图像都经过的角点 $(1, 1)$

Each bottom-to-top arc of one graph crosses each left-to-right arc of the other exactly once — then examine the corner $(1, 1),$ which both graphs pass through

解答：

两个右端表达式的值都在 $[0, 1]$ 中，所以所有交点都在单位正方形内。在那里可用 $\varphi(u) = 4\left||u - \tfrac{1}{2}| - \tfrac{1}{4}\right|$ 写出两条曲线：第一条为 $y = \varphi(|\sin 2\pi x|)$ ，第二条为 $x = \varphi(|\cos 3\pi y|)$ 。当 $u$ 从 $0$ 增加到 $1$ 时， $\varphi(u)$ 分段线性地按 $1 \to 0 \to 1 \to 0 \to 1$ 变化，拐点在 $u = \frac{1}{4}, \frac{1}{2}, \frac{3}{4}$ 。对 $x \in [0, 1]$ ， $|\sin 2\pi x|$ 单调扫过 $[0, 1]$ 共 $4$ 次，所以第一条图像由 $4 \cdot 4 = 16$ 条单调弧组成，每条弧都在一个窄竖条内从 $0 \le y \le 1$ 的全范围上升或下降。类似地，当 $y \in [0, 1]$ 时， $|\cos 3\pi y|$ 单调扫过 $[0, 1]$ 共 $6$ 次，所以第二条图像由 $24$ 条单调弧组成，每条弧都在一个窄横条内跨过 $0 \le x \le 1$ 的全范围。

取两条图像各一条弧，分别位于竖条 $[a, b]$ 和横条 $[c, d]$ 中。在矩形 $[a, b] \times [c, d]$ 内，第一条弧连接底边和顶边，第二条弧连接左边和右边，并且二者都单调，因此这两条弧恰好相交一次。这给出 $16 \cdot 24 = 384$ 个交点。

角点 $(1, 1)$ 处还藏着一个额外交点，它在两条图像上： $\varphi(|\sin 2\pi|) = \varphi(0) = 1$ ，且 $\varphi(|\cos 3\pi|) = \varphi(1) = 1$ 。在它附近，第一条图像满足 $y \approx 1 - 8\pi(1 - x)$ ，而第二条图像满足 $x \approx 1 - 18\pi^2(1 - y)^2$ ，所以最后两条弧除了已经计入的横截相交外，还在共同端点 $(1, 1)$ 相遇。总数为 $384 + 1 = 385$ 。

Both right-hand sides take values in $[0, 1],$ so every intersection lies in the unit square, and there we may write both curves using $\varphi(u) = 4\left||u - \tfrac{1}{2}| - \tfrac{1}{4}\right|:$ the first is $y = \varphi(|\sin 2\pi x|)$ and the second is $x = \varphi(|\cos 3\pi y|).$ As $u$ increases from $0$ to $1,$ $\varphi(u)$ runs linearly $1 \to 0 \to 1 \to 0 \to 1$ with corners at $u = \frac{1}{4}, \frac{1}{2}, \frac{3}{4}.$ For $x \in [0, 1],$ $|\sin 2\pi x|$ sweeps $[0, 1]$ monotonically $4$ times, so the first graph consists of $4 \cdot 4 = 16$ monotone arcs, each climbing or descending through the full range $0 \le y \le 1$ within a narrow vertical strip. Likewise $|\cos 3\pi y|$ sweeps $[0, 1]$ monotonically $6$ times for $y \in [0, 1],$ so the second graph consists of $24$ monotone arcs, each crossing the full range $0 \le x \le 1$ within a narrow horizontal strip.

Take one arc of each graph, living in the vertical strip $[a, b]$ and the horizontal strip $[c, d].$ Inside the rectangle $[a, b] \times [c, d],$ the first arc joins the bottom edge to the top edge and the second joins the left edge to the right edge, and each is monotone, so the two arcs cross exactly once. This yields $16 \cdot 24 = 384$ intersection points.

One further point hides at the corner $(1, 1),$ which lies on both graphs: $\varphi(|\sin 2\pi|) = \varphi(0) = 1$ and $\varphi(|\cos 3\pi|) = \varphi(1) = 1.$ Near it the first graph is $y \approx 1 - 8\pi(1 - x)$ while the second satisfies $x \approx 1 - 18\pi^2(1 - y)^2,$ so the two final arcs meet at their shared endpoint $(1, 1)$ in addition to the transversal crossing already counted. The total is $384 + 1 = 385.$

13.

设 $p$ 是最小的质数，使得存在整数 $n$ ，令 $n^4 + 1$ 能被 $p^2$ 整除。求最小的正整数 $m$ ，使得 $m^4 + 1$ 能被 $p^2$ 整除。

Let $p$ be the least prime number for which there exists an integer $n$ such that $n^4 + 1$ is divisible by $p^2.$ Find the least positive integer $m$ such that $m^4 + 1$ is divisible by $p^2.$

答案

答案：110

知识点：乘法阶模运算质数

难度评级：3160

小提示：

若 $n^4 + 1$ 能被 $p$ 整除，则 $n$ 在模 $p$ 下的阶为 $8$ ，所以 $p \equiv 1 \pmod 8$ ；测试最小的这类质数

If $n^4 + 1$ is divisible by $p$ then $n$ has order $8$ modulo $p,$ so $p \equiv 1 \pmod 8;$ test the smallest such prime

大提示：

模 $17$ 下 $-1$ 的四次方根是 $\pm 2$ 和 $\pm 8$ ；用一次 Hensel 提升把每个根提升为模 $289$ 的解，再比较大小

The fourth roots of $-1$ modulo $17$ are $\pm 2$ and $\pm 8;$ lift each to a solution modulo $289$ with a Hensel step and compare

解答：

若 $n^4 + 1$ 能被 $p$ 整除，则 $n^8 \equiv 1$ 且 $n^4 \equiv -1 \pmod p$ ，所以 $n$ 在模 $p$ 下的阶为 $8$ ，从而 $p - 1$ 是 $8$ 的倍数（且 $p = 2$ 不行，因为 $n^4 + 1 \equiv 2 \pmod 4$ ）。最小的满足 $p \equiv 1 \pmod 8$ 的质数是 $17$ ，而且确实有 $2^4 = 16 \equiv -1 \pmod{17}$ 。由于导数 $4n^3$ 在这样的 $n$ 处不被 $17$ 整除，每个根都能提升为模 $17^2 = 289$ 的根，所以 $p = 17$ 。

模 $17$ 下 $-1$ 的四次方根是 $\pm 2$ 和 $\pm 8$ 。为了提升 $n = 8$ ，设 $n = 8 + 17t$ ：模 $289$ 下， $\begin{aligned} &n^4 + 1 \equiv 8^4 + 1 \\ &\quad {}+ 4 \cdot 8^3 \cdot 17t \\ &= 17(241 + 2048t) \end{aligned}\text{，}$ 所以需要 $241 + 2048t \equiv 3 + 8t$ $\equiv 0 \pmod{17}$ ，得到 $t \equiv 6$ ，从而 $n \equiv 8 + 102 = 110 \pmod{289}$ 。

同样计算可将 $2$ 、 $15$ 、 $9$ 分别提升到 $155$ 、 $134$ 、 $179$ ，所以最小正整数 $m$ 为 $110$ 。确实， $110^4 + 1 = 146410001$ $= 289 \cdot 506609$ 。

If $n^4 + 1$ is divisible by $p,$ then $n^8 \equiv 1$ and $n^4 \equiv -1 \pmod p,$ so $n$ has order $8$ modulo $p$ and $p - 1$ is divisible by $8$ (and $p = 2$ fails since $n^4 + 1 \equiv 2 \pmod 4$ ). The smallest prime $p \equiv 1 \pmod 8$ is $17,$ and indeed $2^4 = 16 \equiv -1 \pmod{17}.$ Because the derivative $4n^3$ is not divisible by $17$ at such an $n,$ each root lifts to a root modulo $17^2 = 289,$ so $p = 17.$

The fourth roots of $-1$ modulo $17$ are $\pm 2$ and $\pm 8.$ To lift $n = 8,$ set $n = 8 + 17t:$ modulo $289,$ $\begin{aligned} &n^4 + 1 \equiv 8^4 + 1 \\ &\quad {}+ 4 \cdot 8^3 \cdot 17t \\ &= 17(241 + 2048t), \end{aligned}$ so we need $241 + 2048t \equiv 3 + 8t$ $\equiv 0 \pmod{17},$ giving $t \equiv 6$ and $n \equiv 8 + 102 = 110 \pmod{289}.$

The same computation lifts $2,$ $15,$ and $9$ to $155,$ $134,$ and $179$ respectively, so the least positive $m$ is $110.$ Indeed $110^4 + 1 = 146410001$ $= 289 \cdot 506609.$

14.

设 $ABCD$ 是一个四面体，满足 $AB = CD = \sqrt{41}$ 、 $AC = BD = \sqrt{80}$ ，以及 $BC = AD = \sqrt{89}$ 。四面体内部存在一点 $I$ ，使得 $I$ 到四个面的距离都相等。这个距离可写成 $\frac{m\sqrt{n}}{p}$ 的形式，其中 $m$ 、 $n$ 、 $p$ 是正整数， $m$ 与 $p$ 互质，且 $n$ 不被任何质数的平方整除。求 $m + n + p$ 。

Let $ABCD$ be a tetrahedron such that $AB = CD = \sqrt{41},$ $AC = BD = \sqrt{80},$ and $BC = AD = \sqrt{89}.$ There exists a point $I$ inside the tetrahedron such that the distances from $I$ to each of the faces of the tetrahedron are all equal. This distance can be written in the form $\frac{m\sqrt{n}}{p},$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime. Find $m + n + p.$

答案

答案：104

知识点：立体几何体积海伦公式

难度评级：3270

小提示：

相对棱相等使这个四面体可以嵌入一个长方体中，其三个面上对角线分别为 $\sqrt{41}$ 、 $\sqrt{80}$ 、 $\sqrt{89}$

Equal opposite edges let you embed the tetrahedron in a rectangular box whose face diagonals are $\sqrt{41},$ $\sqrt{80},$ and $\sqrt{89}$

大提示：

这个长方体的尺寸是 $4 \times 5 \times 8$ 。等距点是内切球球心，并且 $r = \frac{3V}{S}$ ，四个面全等。

The box is $4 \times 5 \times 8.$ The equidistant point is the insphere center, and $r = \frac{3V}{S}$ with all four faces congruent.

解答：

相对棱相等的四面体可以嵌入一个长方体中，六条棱成为长方体面上的对角线。若长方体尺寸为 $a \times b \times c$ ，则 $a^2 + b^2 = 41$ 、 $a^2 + c^2 = 80$ 、且 $b^2 + c^2 = 89$ 。相加得 $a^2 + b^2 + c^2 = 105$ ，所以 $(a, b, c) = (4, 5, 8)$ 。从长方体中去掉四个体积为 $\frac{abc}{6}$ 的角上四面体，剩下的体积为 $\begin{aligned} &V = abc - 4 \cdot \frac{abc}{6} \\ &= \frac{abc}{3} = \frac{160}{3} \end{aligned}\text{。}$

四个面都是边长为 $\sqrt{41}$ 、 $\sqrt{80}$ 、 $\sqrt{89}$ 的全等三角形。用 Heron 公式的形式 $16F^2 = 2(a^2b^2 + b^2c^2 + c^2a^2)$ $- (a^4 + b^4 + c^4)$ ，并代入边长平方 $41, 80, 89$ ，得到 $16F^2 = 28098 - 16002 = 12096$ ，所以 $F = \sqrt{756} = 6\sqrt{21}$ 。

到四个面距离相等的点是内切球球心，将四面体分解为四个以各面为底的棱锥，得到 $V = \frac{1}{3} r \cdot 4F$ 。因此 $\begin{aligned} &r = \frac{3V}{4F} = \frac{160}{24\sqrt{21}} \\ &= \frac{20}{3\sqrt{21}} = \frac{20\sqrt{21}}{63} \end{aligned}\text{，}$ 所以 $m + n + p = 20 + 21 + 63$ $= 104$ 。

A tetrahedron with equal opposite edges embeds in a rectangular box with the six edges as face diagonals. If the box has dimensions $a \times b \times c,$ then $a^2 + b^2 = 41,$ $a^2 + c^2 = 80,$ and $b^2 + c^2 = 89.$ Adding gives $a^2 + b^2 + c^2 = 105,$ so $(a, b, c) = (4, 5, 8).$ The box minus four corner tetrahedra of volume $\frac{abc}{6}$ each leaves $\begin{aligned} &V = abc - 4 \cdot \frac{abc}{6} \\ &= \frac{abc}{3} = \frac{160}{3}. \end{aligned}$

All four faces are congruent triangles with sides $\sqrt{41},$ $\sqrt{80},$ $\sqrt{89}.$ By Heron’s formula in the form $16F^2 = 2(a^2b^2 + b^2c^2 + c^2a^2)$ $- (a^4 + b^4 + c^4)$ applied to the squared sides $41, 80, 89,$ we get $16F^2 = 28098 - 16002 = 12096,$ so $F = \sqrt{756} = 6\sqrt{21}.$

The point equidistant from all four faces is the insphere center, and decomposing the tetrahedron into four pyramids over the faces gives $V = \frac{1}{3} r \cdot 4F.$ Hence $\begin{aligned} &r = \frac{3V}{4F} = \frac{160}{24\sqrt{21}} \\ &= \frac{20}{3\sqrt{21}} = \frac{20\sqrt{21}}{63}, \end{aligned}$ and $m + n + p = 20 + 21 + 63$ $= 104.$

15.

设 $\mathcal{B}$ 是所有表面积为 $54$ 、体积为 $23$ 的长方体的集合。设 $r$ 为能容纳 $\mathcal{B}$ 中每一个长方体的最小球的半径。 $r^2$ 的值可写为 $\frac{p}{q}$ ，其中 $p$ 和 $q$ 是互质正整数。求 $p + q$ 。

Let $\mathcal{B}$ be the set of rectangular boxes with surface area $54$ and volume $23.$ Let $r$ be the radius of the smallest sphere that can contain each of the rectangular boxes that are elements of $\mathcal{B}.$ The value of $r^2$ can be written as $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

答案

答案：721

知识点：韦达定理长方体最优化

难度评级：3370

小提示：

条件给出 $ab + bc + ca = 27$ 和 $abc = 23$ ，球必须容纳空间对角线，所以要最大化 $a^2 + b^2 + c^2$

The constraints say $ab + bc + ca = 27$ and $abc = 23,$ and the sphere must contain the space diagonal, so maximize $a^2 + b^2 + c^2$

大提示：

$a^2 + b^2 + c^2 = (a + b + c)^2 - 54$ ，而 $a + b + c$ 的极值发生在两个维度相等时；解出该共同值满足的三次方程

$a^2 + b^2 + c^2 = (a + b + c)^2 - 54,$ and the extreme of $a + b + c$ occurs when two dimensions are equal — solve the resulting cubic in that common value

解答：

对尺寸为 $a, b, c$ 的长方体，条件为 $2(ab + bc + ca) = 54$ 和 $abc = 23$ ，所以 $ab + bc + ca = 27$ 。能容纳一个长方体的最小球以该长方体的空间对角线为直径，所以 $\begin{aligned} &r^2 = \max_{\mathcal{B}} \frac{a^2 + b^2 + c^2}{4} \\ &= \max_{\mathcal{B}} \frac{(a + b + c)^2 - 54}{4} \end{aligned}\text{。}$

当 $ab + bc + ca$ 和 $abc$ 固定时， $s = a + b + c$ 在一个区间内变化；在端点处，三次多项式 $t^3 - st^2 + 27t - 23$ 有重根，也就是两个维度相同。令 $b = c$ ：则 $2ab + b^2 = 27$ 且 $ab^2 = 23$ ，消去 $a$ 得 $\frac{b(27 - b^2)}{2} = 23$ ，即 $b^3 - 27b + 46 = 0$ ，其因式分解为 $(b - 2)(b^2 + 2b - 23) = 0$ 。根为 $b = 2$ 和 $b = 2\sqrt{6} - 1$ 。

当 $b = 2$ 时， $a = \frac{23}{4}$ ，且 $s = \frac{23}{4} + 4 = \frac{39}{4} = 9.75$ ；当 $b = 2\sqrt{6} - 1$ 时， $s \approx 9.31$ ，较小。所以 $a^2 + b^2 + c^2$ 的最大值为 $\left(\frac{39}{4}\right)^2 - 54 = \frac{657}{16}$ ，得到 $r^2 = \frac{657}{64}$ ，因此 $p + q = 657 + 64 = 721$ 。

For a box with dimensions $a, b, c,$ the conditions are $2(ab + bc + ca) = 54$ and $abc = 23,$ so $ab + bc + ca = 27.$ The smallest sphere containing a box has the box’s space diagonal as a diameter, so $\begin{aligned} &r^2 = \max_{\mathcal{B}} \frac{a^2 + b^2 + c^2}{4} \\ &= \max_{\mathcal{B}} \frac{(a + b + c)^2 - 54}{4}. \end{aligned}$

With $ab + bc + ca$ and $abc$ fixed, $s = a + b + c$ ranges over an interval, and at an endpoint the cubic $t^3 - st^2 + 27t - 23$ has a double root, meaning two dimensions coincide. Setting $b = c:$ $2ab + b^2 = 27$ and $ab^2 = 23,$ so eliminating $a$ gives $\frac{b(27 - b^2)}{2} = 23,$ i.e. $b^3 - 27b + 46 = 0,$ which factors as $(b - 2)(b^2 + 2b - 23) = 0.$ The roots are $b = 2$ and $b = 2\sqrt{6} - 1.$

For $b = 2,$ $a = \frac{23}{4}$ and $s = \frac{23}{4} + 4 = \frac{39}{4} = 9.75;$ for $b = 2\sqrt{6} - 1,$ $s \approx 9.31$ is smaller. So the maximum of $a^2 + b^2 + c^2$ is $\left(\frac{39}{4}\right)^2 - 54 = \frac{657}{16},$ giving $r^2 = \frac{657}{64}$ and $p + q = 657 + 64 = 721.$