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| − | ==Problem==
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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 by <math>d</math> the diameter of the largest circle that fits inside the hexagon. Find <math>d^2</math>.
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| − | ==Solution 1==
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| − | [[image:2018_AIME_I-8.png|center|500px]]
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| − | - cooljoseph
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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
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| − | ==Solution 2==
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| − | Like solution 1, draw out the large equilateral triangle with side length <math>24</math>. Let the tangent point of the circle at <math>\overline{CD}</math> be G and the tangent point of the circle at <math>\overline{AF}</math> be H. Clearly, GH is the diameter of our circle, and is also perpendicular to <math>\overline{CD}</math> and <math>\overline{AF}</math>.
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| − | The equilateral triangle of side length <math>10</math> is similar to our large equilateral triangle of <math>24</math>. And the height of the former equilateral triangle is <math>\sqrt{10^2-5^2}=5\sqrt{3}</math>. By our similarity condition,
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| − | <math>\frac{10}{24}=\frac{5\sqrt{3}}{d+5\sqrt{3}}</math>
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| − | Solving this equation gives <math>d=7\sqrt{3}</math>, and <math>d^2=\boxed{147}</math>
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| − | ~novus677
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| − | ==See Also==
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| − | {{AIME box|year=2018|n=I|num-b=7|num-a=9}}
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| − | {{MAA Notice}}
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