Difference between revisions of "2018 AIME I Problems/Problem 8"

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Let <math>ABCDEF</math> be an equiangular hexagon such that <math>AB=6, BC=8, CD=10</math>, and <math>DE=12</math>. Denote <math>d</math> the diameter of the largest circle that fits inside the hexagon. Find <math>d^2</math>.
 
Let <math>ABCDEF</math> be an equiangular hexagon such that <math>AB=6, BC=8, CD=10</math>, and <math>DE=12</math>. Denote <math>d</math> the diameter of the largest circle that fits inside the hexagon. Find <math>d^2</math>.
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==Solutions==
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==Solution Diagram==
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First of all, draw a good diagram! This is always the key to solving any geometry problem. Once you draw it, realize that <math>EF=2, FA=16</math>. Why? Because since the hexagon is equiangular, we can put an equilateral triangle around it, with side length <math>6+8+10=24</math>. Then, if you drew it to scale, notice that the "widest" this circle can be according to <math>AF, CD</math> is <math>7\sqrt{3}</math>.  And it will be obvious that the sides won't be inside the circle, so our answer is <math>\boxed{147}</math>.
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-expiLnCalc

Revision as of 17:03, 7 March 2018

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 $d^2$.

Solutions

Solution Diagram

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