2018 AIME I Problems/Problem 8

Let $ABCDEF$ be an equiangular hexagon such that $AB=6, BC=8, CD=10$ , and $DE=12$ . Denote $d$ the diameter of the largest circle that fits inside the hexagon. Find $d^2$ .

Solution 1

- cooljoseph

First of all, draw a good diagram! This is always the key to solving any geometry problem. Once you draw it, realize that $EF=2, FA=16$ . Why? Because since the hexagon is equiangular, we can put an equilateral triangle around it, with side length $6+8+10=24$ . Then, if you drew it to scale, notice that the "widest" this circle can be according to $AF, CD$ is $7\sqrt{3}$ . And it will be obvious that the sides won't be inside the circle, so our answer is $\boxed{147}$ .

-expiLnCalc

Solution 2

Like solution 1, draw out the large equilateral triangle with side length $24$ . Let the tangent point of the circle at $\overline{CD}$ be G and the tangent point of the circle at $\overline{AF}$ be H. Clearly, GH is the diameter of our circle, and is also perpendicular to $\overline{CD}$ and $\overline{AF}$ .

The equilateral triangle of side length $10$ is similar to our large equilateral triangle of $24$ . And the height of the former equilateral triangle is $\sqrt{10^2-5^2}=5\sqrt{3}$ . By our similarity condition, $\frac{10}{24}=\frac{5\sqrt{3}}{d+5\sqrt{3}}$

Solving this equation gives $d=7\sqrt{3}$ , and $d^2=\boxed{147}$

~novus 677

2018 AIME I (Problems • Answer Key • Resources)
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2018 AIME I Problems/Problem 8

Solution 1

Solution 2

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