2021 JMPSC Invitationals Problems/Problem 14

Problem

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

Solution

[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm);  real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */  pen dotstyle = black; /* point style */  real xmin = -75.0614580781354, xmax = 144.30457756711317, ymin = -28.5847126201819, ymax = 97.17376695854563;  /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); pen qqwuqq = rgb(0,0.39215686274509803,0);   draw((0,79.2489157968718)--(0,0)--(31.036632190371098,0)--cycle, linewidth(1));  draw((0,27.518773386755257)--(0,0)--(31.036632190371098,0)--cycle, linewidth(1));  draw((0,27.518773386755257)--(0,0)--(31.036632190371098,0)--(17.56520990745875,34.39792061246379)--cycle, linewidth(1));  draw((3.017805668614108,0)--(3.0178056686141086,3.017805668614108)--(0,3.017805668614108)--(0,0)--cycle, linewidth(1) + ccqqqq);  draw(arc((0,27.518773386755257),4.267821705160478,-90,-41.56194864878782)--(0,27.518773386755257)--cycle, linewidth(1) + qqwuqq);  draw(arc((31.036632190371098,0),4.267821705160478,138.43805135121218,180)--(31.036632190371098,0)--cycle, linewidth(1) + qqwuqq);  draw(arc((17.56520990745875,34.39792061246379),4.267821705160478,-117.05101221915062,-68.61296086793845)--(17.56520990745875,34.39792061246379)--cycle, linewidth(1) + qqwuqq);  draw(arc((17.56520990745875,34.39792061246379),4.267821705160478,-158.61296086793843,-117.0510122191506)--(17.56520990745875,34.39792061246379)--cycle, linewidth(1) + qqwuqq);  draw((16.464718299532908,37.20791514527062)--(13.654723766726086,36.10742353734477)--(14.755215374651927,33.29742900453795)--(17.56520990745875,34.39792061246379)--cycle, linewidth(1) + ccqqqq);   /* draw figures */ draw((0,79.2489157968718)--(0,0), linewidth(1));  draw((0,0)--(31.036632190371098,0), linewidth(1));  draw((31.036632190371098,0)--(0,79.2489157968718), linewidth(1));  draw((0,27.518773386755257)--(0,0), linewidth(1));  draw((0,0)--(31.036632190371098,0), linewidth(1));  draw((31.036632190371098,0)--(0,27.518773386755257), linewidth(1));  draw(circle((15.51831609518555,13.75938669337763), 20.73978921320061), linewidth(1));  draw((0,27.518773386755257)--(0,0), linewidth(1));  draw((0,0)--(31.036632190371098,0), linewidth(1));  draw((31.036632190371098,0)--(17.56520990745875,34.39792061246379), linewidth(1));  draw((17.56520990745875,34.39792061246379)--(0,27.518773386755257), linewidth(1));  draw((17.56520990745875,34.39792061246379)--(0,0), linewidth(1));   /* dots and labels */ dot((0,0),linewidth(4pt) + dotstyle);  label("$A$", (-2.9352712609233205,-2.124218048187195), SW * labelscalefactor);  dot((0,27.518773386755257),dotstyle);  label("$B$", (-3.362053431439368,28.319576781957252), W * labelscalefactor);  dot((31.036632190371098,0),dotstyle);  label("$C$", (32.345388168403296,-1.9819573246818474), NE * labelscalefactor);  dot((0,79.2489157968718),dotstyle);  label("$D$", (-1.2281425788591294,81.52508737295736), NE * labelscalefactor);  dot((17.56520990745875,34.39792061246379),linewidth(4pt) + dotstyle);  label("$E$", (18.119315817868372,35.57487368072999), NE * labelscalefactor);  dot((9.672839652798576,18.94230539953703),linewidth(4pt) + dotstyle);  label("$F$", (9.299150960536716,20.779758436173815), N * labelscalefactor);  clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);   /* end of picture */ [/asy]

$\mathbf{Lemma}:$ We claim that $EF$ is the angle bisector of $\angle BEC$.


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


$\mathbf{End~Proof}$


By the angle bisector theorem and our $\mathbf{Lemma},$ \[\frac{FC}{BF}=\frac{EC}{BE} \qquad (1).\] We seek the lengths $EC$ and $BE$.


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


To find $BE$, we use the Pythagorean Theorem in $BED$. (We already found $\angle BEC=90^\circ$, which tells us that supplementary $\angle BED = 90^\circ$.) By the Pythagorean Theorem, $BE^2+DE^2=BD^2.$ We found that $DE=\frac{84}{13}$, and since we are given $BD=7$, we have \[BE=\sqrt{7^2-\left (\frac{84}{13}\right )^2}=\frac{35}{13} \qquad (3).\]


Our answer, by equation $(1)$, is $\frac{EC}{BE}$. From equation $(2),$ $EC=\frac{85}{13}$ and from equation $(3),$ $BE=\frac{35}{13}$. Therefore, our final answer is $\frac{\frac{85}{13}}{\frac{35}{13}}=\frac{17}{7} \implies \boxed{24}.$

~samrocksnature

Solution 2

Note $ABEC$ is cyclic, so $\angle BEC = \angle BED = 90^o$. By Power Of A Point we have $7(7+5)=DE \cdot DC \implies DE=\frac{84}{13}$, $EC=\frac{85}{13}$. Now, note \[\angle ABC = \angle BCA\]\[\qquad =\angle BEA\]\[= \angle AEC\]Therefore, by Angle Bisector Theorem, \[\frac{FC}{BF}=\frac{EC}{BE}=\frac{85}{35} \implies \frac{17}{7} \implies 17+7=\boxed{24}\]

~Geometry285

See also

  1. Other 2021 JMPSC Invitationals Problems
  2. 2021 JMPSC Invitationals Answer Key
  3. All JMPSC Problems and Solutions

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