2018 AIME I Problems/Problem 8
Contents
Problem
Let be an equiangular hexagon such that , and . Denote by the diameter of the largest circle that fits inside the hexagon. Find .
Video Solution by Punxsutawney Phil
https://www.youtube.com/watch?v=oc-cDRIEzoo
Video Solution by Walt S
https://www.youtube.com/watch?v=wGP9bjkdh1M
Solution 1
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 .
-expiLnCalc
Solution 2
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
~novus677
See Also
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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