2022 AIME II Problems/Problem 15
Contents
Problem
Two externally tangent circles and have centers and , respectively. A third circle passing through and intersects at and and at and , as shown. Suppose that , , , and is a convex hexagon. Find the area of this hexagon.
Solution 1
First observe that and . Let points and be the reflections of and , respectively, about the perpendicular bisector of . Then quadrilaterals and are congruent, so hexagons and have the same area. Furthermore, triangles and are congruent, so and quadrilateral is an isosceles trapezoid. Next, remark that , so quadrilateral is also an isosceles trapezoid; in turn, , and similarly . Thus, Ptolmey's theorem on yields , whence . Let . The Law of Cosines on triangle yields and hence . Thus the distance between bases and is (in fact, is a triangle with a triangle removed), which implies the area of is .
Now let and ; the tangency of circles and implies . Furthermore, angles and are opposite angles in cyclic quadrilateral , which implies the measure of angle is . Therefore, the Law of Cosines applied to triangle yields
Thus , and so the area of triangle is .
Thus, the area of hexagon is .
~djmathman
Solution 2
Denote by the center of . Denote by the radius of .
We have , , , , , are all on circle .
Denote . Denote . Denote .
Because and are on circles and , is a perpendicular bisector of . Hence, .
Because and are on circles and , is a perpendicular bisector of . Hence, .
In ,
Hence,
In ,
Hence,
In ,
Hence,
Taking , we get . Thus, .
Taking these into (1), we get . Hence,
Hence, .
In ,
In , by applying the law of sines, we get
Because circles and are externally tangent, is on circle , is on circle ,
Thus, .
Now, we compute and .
Recall and . Thus, .
We also have
Thus,
Therefore,
~Steven Chen (www.professorchenedu.com)
See Also
2022 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.