2018 AMC 12B Problems/Problem 25
Circles , , and each have radius and are placed in the plane so that each circle is externally tangent to the other two. Points , , and lie on , , and respectively such that and line is tangent to for each , where . See the figure below. The area of can be written in the form for positive integers and . What is ?
Let be the center of circle for , and let be the intersection of lines and . Because , it follows that is a triangle. Let ; then and . The Law of Cosines in gives which simplifies to . The positive solution is . Then , and the required area is The requested sum is .
Let and be the centers of and respectively and draw , , and . Note than and are both right. Furthermore, since is equilateral, and . Mark as the base of the altitude from to . By special right triangles, and . since and , we can find find . Thus, . This makes . This makes the answer .
Let be the center of circle for . Let be the centroid of , which also happens to be the centroid of . Because and , . is the height of , thus is .
Applying cosine law on , one finds that . Multiplying by to solve for the height of , one gets . Simply multiplying by and then calculating the equilateral triangle's area, one would get the final result of .
This makes the answer .
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