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| − | ==Problem==
| + | #REDIRECT [[2013 AMC 12B Problems/Problem 19]] |
| − | In triangle <math>ABC</math>, <math>AB = 13</math>, <math>BC = 14</math>, and <math>CA = 15</math>. Distinct points <math>D</math>, <math>E</math>, and <math>F</math> lie on segments <math>\overline{BC}</math>, <math>\overline{CA}</math>, and <math>\overline{DE}</math>, respectively, such that <math>\overline{AD} \perp \overline{BC}</math>, <math>\overline{DE} \perp \overline{AC}</math>, and <math>\overline{AF} \perp \overline{BF}</math>. The length of segment <math>\overline{DF}</math> can be written as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m + n</math>?
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| − | <math>\textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30</math>
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| − | ==Solution==
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| − | Since <math>\angle{AFB}=\angle{ADB}=90</math>, quadrilateral <math>ABDF</math> is cyclic. It follows that <math>\angle{ADE}=\angle{ABF}</math>. In addition, since <math>\angle{AFB}=\angle{AED}=90</math>, triangles <math>ABF</math> and <math>ADE</math> are similar. It follows that <math>AF=(13)(\frac{4}{5}), BF=(13)(\frac{3}{5})</math>. By Ptolemy, we have <math>13DF+(5)(13)(\frac{4}{5})=(12)(13)(\frac{3}{5})</math>. Cancelling <math>13</math>, the rest is easy. We obtain <math>DF=\frac{16}{5}\implies{16+5=21}\implies{\boxed{\textbf{(B)} 21}</math>
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| − | == See also ==
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| − | {{AMC10 box|year=2013|ab=B|num-b=22|num-a=24}}
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