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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>.
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| − | ==Solutions==
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| − | ==Solution Diagram==
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| − | [asy]
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| − | draw((0,0)--(12,20.78)--(24,0)--cycle);
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| − | draw((1,1.73)--(2,0));
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| − | draw((9,15.59)--(15,15.59));
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| − | draw((14,0)--(19,8.66));
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| − | label("<math>A</math>",(9,15.59),NW);
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| − | label("<math>B</math>",(15,15.59),NE);
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| − | label("<math>C</math>",(19,8.66),NE);
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| − | label("<math>D</math>",(14,0),S);
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| − | label("<math>E</math>",(2,0),S);
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| − | label("<math>F</math>",(1,1.73),NW);
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| − | pair O;
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| − | O=(11.25,7.36);
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| − | dot(O);
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| − | label("<math>O</math>",O,SW);
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| − | draw(Circle(O,6.06));
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| − | [/asy]
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| − | asymptote code for a picture
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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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