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| − | ==Problem==
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| − | Circle <math>\omega</math> has chord <math>AB</math> of length <math>18</math>. Point <math>X</math> lies on chord <math>AB</math> such that <math>AX = 4.</math> Circle <math>\omega_1</math> with radius <math>r_1</math> and <math>\omega_2</math> with radius <math>r_2</math> lie on two different sides of <math>AB.</math> Both <math>\omega_1</math> and <math>\omega_2</math> are tangent to <math>AB</math> at <math>X</math> and <math>\omega.</math> If the sum of the maximum and minimum values of <math>r_1r_2</math> is <math>\frac{m}{n},</math> find <math>m+n</math>.
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| − | ==Solution==
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| − | Let <math>r</math> be the radius of <math>\omega</math> and let <math>C</math> be the midpoint of <math>AB</math> and let <math>OC = x.</math> Note that <math>r^2 - x^2 = 81</math>. WLOG assume that <math>r_2\geq r_1.</math>
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| − | Since <math>AX = 4</math> and <math>AB = 18,</math> we have <math>XC = \frac{AB}{2}-AX = 5.</math>
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| − | By the Pythagorean Theorem, we have
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| − | <cmath>(O_1X+CO)^2+(XC)^2 = (OO_1)^2</cmath>
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| − | <cmath>(O_2X-CO)^2+(XC)^2 = (OO_2)^2.</cmath>
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| − | which is the same as
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| − | <cmath>(r_1+x)^2+25 = (r-r_1)^2</cmath>
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| − | <cmath>(r_2-x)^2+25 =(r-r_2)^2.</cmath>
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| − | \
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| − | Solving for <math>r_1</math> and <math>r_2,</math> we have that
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| − | <cmath>r_1 = \frac{r^2-x^2-25}{2(r+x)}</cmath>
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| − | <cmath>r_2 = \frac{r^2-x^2-25}{2(r-x)}.</cmath>
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| − | Thus,
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| − | <cmath>
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| − | r_1r_2 = \frac{((r^2-x^2)-5^2)^2}{4(r^2-x^2)} = \frac{784}{81},
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| − | </cmath>
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| − | meaning that the minimum and maximum value of <math>r_1r_2</math> are both <math>\frac{784}{81}</math> so the answer is <math>\boxed{1649}.</math>
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