2022 AIME II Problems/Problem 15
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)
Solution 3
Let points and be the reflections of and respectively, about the perpendicular bisector of We establish the equality of the arcs and conclude that the corresponding chords are equal Similarly
Ptolemy's theorem on yields The area of the trapezoid is equal to the area of an isosceles triangle with sides and
The height of this triangle is The area of is
Denote hence
Semiperimeter of is
The distance from the vertex to the tangent points of the inscribed circle of the triangle is equal
The radius of the inscribed circle is
The area of triangle is
The hexagon has the same area as hexagon
The area of hexagon is equal to the sum of the area of the trapezoid and the areas of two equal triangles and so the area of the hexagon is
vladimir.shelomovskii@gmail.com, vvsss
Solution 4
Let circle 's radius be , then the radius of circle is . Based on Brahmagupta's Formula,
the hexagon's Area .
Now we only need to find the .
Connect and , and , and let be the point of intersection between and circle , based on the " 2 Non-Congruent Triangles of 'SSA' Scenario " , we can immediately see and therefore get an equation from the "Power of A Point Theorem:
(1).
Similarly,
(2).
We can also get two other equations about these 4 segments from Ptolemy's Theorem:
(3)
(4)
Multiply equations (1) and (2), and equations (3) and (4) respectively, we will get a very simple and nice equation of :
,
then:
.
This result is good enough for us to find the hexagon's area, which:
.
eJMaSc
Video Solution
~MathProblemSolvingSkills.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.