Problem:
Circles Ο1β and Ο2β intersect at two points P and Q, and their common tangent line closer to P intersects Ο1β and Ο2β at points A and B, respectively. The line parallel to AB that passes through P intersects Ο1β and Ο2β 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 mnβ, where m and n are positive integers and n is not divisible by the square of any prime. Find m+n.
Solution:
Let O1β and O2β be the centers of circles Ο1β and Ο2β, respectively. Let lines XY and AO1β intersect at point C, and lines XY and BO2β intersect at point D.
Since AB is tangent to circle Ο1β, we know O1βAβ₯AB. Because XYβ₯AB, it follows that O1βAβ₯XY. Also, since X and P lie on Ο1β, the segment O1βA is the perpendicular bisector of XY. Thus, PC=2PXβ=5.
Similarly, PD=2PYβ=7, so CD=CP+PD=5+7=12.
Since O1βAβ₯CD, O1βAβ₯AB, O2βBβ₯CD, and O2βBβ₯AB, quadrilateral ABDC is a rectangle. Hence, AB=CD=12.
Let PQ and AB intersect at point M. Since AB is symmetric with respect to radical axis PQ, then M is the midpoint of AB, so AM=2ABβ=6.
Now apply the Power of a Point Theorem in circle Ο1β with tangent MA and secant MPQ: MA2=MPβ MQ. Solving gives 36=MPβ MQ, and from the problem, we find MP=4.
Now consider right trapezoid AMPC. Using the Pythagorean Theorem: AC=MP2β(AMβCP)2β=42β(6β5)2β=16β1β=15β.
To find the area of trapezoid XABY, we compute: [XABY]=21β(AB+XY)β AC=21β(12+24)β 15β=1815β.