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

Revision as of 00:38, 14 August 2025

Problem

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

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 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 $d^2=\boxed{147}$

~novus677

@@ Line 7: / Line 7: @@
 ==Video Solution by Walt S==
 https://www.youtube.com/watch?v=wGP9bjkdh1M
-==Solution 1==
-[[image:2018_AIME_I-8.png|center|500px]]
-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>.
--expiLnCalc
 ==Solution 2==

2018 AIME I (Problems • Answer Key • Resources)
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