Question

圆 $\omega_1$ 和 $\omega_2$ 相交于两点 $P$ 和 $Q$ ，更靠近 $P$ 的公切线分别在点 $A$ 与点 $B$ 处与 $\omega_1$ 和 $\omega_2$ 相切。过 $P$ 且平行于 $\overline{AB}$ 的直线第二次分别交 $\omega_1$ 和 $\omega_2$ 于点 $X$ 与 $Y$ ，已知 $PX = 10$ 、 $PY = 14$ 、 $PQ = 5$ 。则梯形 $XABY$ 的面积为 $m\sqrt{n}$ ，其中 $m$ 和 $n$ 是正整数，且 $n$ 不被任何素数的平方整除。求 $m + n$ 。

Circles $\omega_1$ and $\omega_2$ intersect at two points $P$ and $Q,$ and their common tangent line closer to $P$ intersects $\omega_1$ and $\omega_2$ at points $A$ and $B,$ respectively. The line parallel to $\overline{AB}$ that passes through $P$ intersects $\omega_1$ and $\omega_2$ for the second time at points $X$ and $Y,$ respectively. Suppose $PX = 10,$ $PY = 14,$ and $PQ = 5.$ Then the area of trapezoid $XABY$ is $m\sqrt{n},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n.$

答案

答案：33

知识点：圆幂根轴梯形

难度评级：2920

解答：

因为 $\omega_1$ 在 $A$ 处的切线平行于弦 $XP$ ，点 $A$ 是弧 $XP$ 的中点，所以从 $A$ 向直线 $XY$ 作垂线会落在 $\overline{XP}$ 的中点；同理，从 $B$ 作垂线会落在 $\overline{PY}$ 的中点。由于 $X$ 和 $Y$ 位于 $P$ 的两侧，梯形的两条平行边为 $XY = 10 + 14 = 24$ 和 $AB = \frac{10}{2} + \frac{14}{2} = 12$ 。

直线 $PQ$ 是根轴，所以它与切线的交点 $M$ 满足 $MA^2 = MP \cdot MQ = MB^2$ ： $M$ 是 $\overline{AB}$ 的中点，并且 $MA = 6$ 、 $MQ = MP + 5$ ，于是 $\begin{gathered} 36 = MP(MP + 5), \\ MP = 4. \end{gathered}$

沿 $AB$ 建立坐标： $A$ 和 $B$ 的垂足分别是 $XP$ 与 $PY$ 的中点，所以 $P$ 距第一个垂足 $5$ 个单位，而 $M$ 距 $A$ 为 $6$ 个单位。因此 $M$ 与 $P$ 的水平偏移为 $6 - 5 = 1$ ，梯形的高 $h$ 满足 $h^2 = MP^2 - 1 = 15$ 。面积为 $\frac{24 + 12}{2}\sqrt{15} = 18\sqrt{15},$ 所以 $m + n = 18 + 15 = 33$ 。

Since the tangent to $\omega_1$ at $A$ is parallel to the chord $XP,$ the point $A$ is the midpoint of arc $XP,$ so the perpendicular from $A$ to line $XY$ lands at the midpoint of $\overline{XP};$ similarly the perpendicular from $B$ lands at the midpoint of $\overline{PY}.$ As $X$ and $Y$ are on opposite sides of $P,$ the parallel sides of the trapezoid are $XY = 10 + 14 = 24$ and $AB = \frac{10}{2} + \frac{14}{2} = 12.$

Line $PQ$ is the radical axis, so its intersection $M$ with the tangent line satisfies $MA^2 = MP \cdot MQ = MB^2:$ $M$ is the midpoint of $\overline{AB},$ and with $MA = 6$ and $MQ = MP + 5,$ $\begin{gathered} 36 = MP(MP + 5), \\ MP = 4. \end{gathered}$

Set up coordinates along $AB:$ the feet of $A$ and $B$ are the midpoints of $XP$ and $PY,$ so $P$ lies $5$ units from the first foot, while $M$ lies $6$ units from $A.$ Hence the horizontal offset between $M$ and $P$ is $6 - 5 = 1,$ and the height $h$ of the trapezoid satisfies $h^2 = MP^2 - 1 = 15.$ The area is $\frac{24 + 12}{2}\sqrt{15} = 18\sqrt{15},$ so $m + n = 18 + 15 = 33.$

2023 AIME II 第 9 题

答案：33

难度评级：2920

解答：

其他年份的第 9 题