Question

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

Solution:

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 $MQ = MP + 5,$ $36 = MP(MP + 5), \qquad MP = 4.$

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 Problem 9

Solution:

Problem 9 in Other Years