2018 AIME I Problems/Problem 8
Let be an equiangular hexagon such that , and . Denote by the diameter of the largest circle that fits inside the hexagon. Find .
First of all, draw a good diagram! This is always the key to solving any geometry problem. Once you draw it, realize that . Why? Because since the hexagon is equiangular, we can put an equilateral triangle around it, with side length . Then, if you drew it to scale, notice that the "widest" this circle can be according to is . And it will be obvious that the sides won't be inside the circle, so our answer is .
Like solution 1, draw out the large equilateral triangle with side length . Let the tangent point of the circle at be G and the tangent point of the circle at be H. Clearly, GH is the diameter of our circle, and is also perpendicular to and .
The equilateral triangle of side length is similar to our large equilateral triangle of . And the height of the former equilateral triangle is . By our similarity condition,
Solving this equation gives , and
This is because the altitude of our equilateral triangle with side length is perpendicular to the tangent line to the circle, which implies they are all degrees (two degree angles from altitude, two degree angles from tangent lines). This allows us to calculate further. Tilt your head degrees clockwise if you can't see what is being done. ~IronicNinja
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