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| − | ==Problem==
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| − | In a circle centered at <math>O</math> with radius <math>7,</math> we have non-intersecting chords <math>AB</math> and <math>DC.</math> <math>O</math> is outisde of quadrilateral <math>ABCD</math> and <math>AB<CD.</math> Let <math>X = AO\cup CD</math> and <math>Y = BO\cup CD.</math> Suppose that <math>XO+YO = 7</math>. If <math>YC-DX=2</math> and <math>XY = 3</math>, then <math>AB = \frac{a\sqrt{b}}{c}</math> for <math>\gcd(a,c) = 1</math> and squareless <math>b.</math> Find <math>a+b+c.</math>
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| − | ==Solution==
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| − | Let <math>CX = x+2, DY=x</math>.
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| − | Then, by power of the point we have that
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| − | <cmath>(x+2)(x+3) = 49 - YO^2 </cmath>
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| − | <cmath>(x+5)x = 49 - XO^2</cmath>
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| − | and subtracting gives that <math>XO^2 - YO^2 = 6</math>.
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| − | Since we know that <math>XO + YO = 7</math>, dividing gives
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| − | that <math>XO - YO = \frac{6}{7}</math> so <math>XO = \frac{55}{14}</math>
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| − | and <math>YO = \frac{43}{14}</math>.
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| − | Then, by law of cosines, it follows that
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| − | <cmath>
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| − | OY^2 + OX^2 - 2 \cdot OX \cdot OY \cos\angle YOX = YX^2
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| − | </cmath>
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| − | which implies that <math>\cos\angle YOX = \frac{311}{473}</math>.
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| − | Then, <math>AB^2 = OA^2 + OB^2 - 2 \cdot OA \cdot OB \cos\angle YOX</math>
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| − | which implies that <math>AB = \frac{126 \sqrt{473}}{473}</math> so the
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| − | answer is then <math>\boxed{1072}</math>.
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