View source for 2021 JMPSC Invitationals Problems/Problem 14

==Problem==
Let there be a <math>\triangle ACD</math> such that <math>AC=5</math>, <math>AD=12</math>, and <math>CD=13</math>, and let <math>B</math> be a point on <math>AD</math> such that <math>BD=7.</math> Let the circumcircle of <math>\triangle ABC</math> intersect hypotenuse <math>CD</math> at <math>E</math> and <math>C</math>. Let <math>AE</math> intersect <math>BC</math> at <math>F</math>. If the ratio <math>\tfrac{FC}{BF}</math> can be expressed as <math>\tfrac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime, find <math>m+n.</math>

==Solution==

<asy>
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import graph; size(10cm); 
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</asy>

<math>\mathbf{Lemma}:</math> We claim that <math>EF</math> is the angle bisector of <math>\angle BEC</math>. 


<math>\mathbf{Proof:}</math> Observe that <math>AB=5=AC</math>, which tells us that <math>\triangle ABC</math> is a <math>45-45-90</math> triangle. In cyclic quadrilateral <math>ABEC</math>, we have <cmath>\angle ABC = 45^\circ = \angle AEC</cmath> and <cmath>\angle BAC+\angle BEC=180^\circ \implies \angle BEC = 90^\circ.</cmath> Since <math>\angle BEA + \angle AEC = \angle BEC</math>, we have <math>\angle BEA =45^\circ = \angle AEC</math>. This means that <math>EA</math>, or equivalently <math>EF</math>, is an angle bisector of <math>\angle BEC</math> in <math>\triangle BEC</math>. 



<math>\mathbf{End~Proof}</math>



By the angle bisector theorem and our <math>\mathbf{Lemma},</math> <cmath>\frac{FC}{BF}=\frac{EC}{BE} \qquad (1).</cmath> We seek the lengths <math>EC</math> and <math>BE</math>.



To find <math>EC</math>, we can proceed by Power of a Point using point <math>D</math> on circle <math>(ABC)</math> to get <math>DE \cdot DC = DB \cdot DA.</math> Since <math>DC=13</math>, <math>DB = 7</math>, and <math>AD = 12</math>, we have <math>DE=\frac{84}{13}.</math> Since <math>CD=13</math>, we have <cmath>EC=CD-DE=\frac{85}{13} \qquad (2).</cmath> ~samrocksnature

==See also==
#[[2021 JMPSC Invitational Problems|Other 2021 JMPSC Invitational Problems]]
#[[2021 JMPSC Invitational Answer Key|2021 JMPSC Invitational Answer Key]]
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
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