Mock Geometry AIME 2011 Problems/Problem 7

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Problem

In trapezoid $ABCD,$ $AB||CD,$ and $AB\perp BC.$ There is a point $P$ on side $AD$ such that the circumcircle of triangle $BPC$ is tangent to $AD.$ If $AB=3, AD=78, CD=75,$ $CP-BP$ can be expressed in the form $\frac{a\sqrt{b}} {c},$ where $a,b,c$ are positive integers and $a,c$ are relatively prime. Find $a+b+c.$

Solution

Let $H$ be the foot of the perpendicular from $A$ to $CD$. Then $DH=75-3=72$. By the Pythagorean Theorem on $\Delta AHD$, $78^2=AH^2+72^2$. Solving yields $AH=30$, and so $BC=AH=30$.


Let $E$ be the intersection of lines $AD$ and $BC$. $\Delta EBA \sim \Delta ECD$ by $AA$ similarity. Then $\frac{EB}{EC}=\frac{AB}{CD}$, or $\frac{EB}{30+EB}=\frac{3}{75}$, and so $EB = \frac{5}{4}$. Also, $\frac{EA}{ED}=\frac{AB}{CD}$, or $\frac{EA}{78+EA}=\frac{3}{75}$, and so $EA = \frac{13}{4}$. Therefore, $\cos{\angle E} = \frac{EB}{EA}=\frac{5}{13}$. By the Power of a Point of $E$, $EP^2=EB*EC=\frac{5}{4}*\frac{125}{4}$. Then $EP=\frac{25}{4}$.


From the Law of Cosines of $\Delta EPC$, $CP^2=(\frac{25}{4})^2 + (\frac{125}{4})^2-2(\frac{25}{4})(\frac{125}{4})(\frac{5}{13})$. Solving yields $CP=\frac{75\sqrt{26}}{13}$. Similarly, from the Law of Cosines of $\Delta EPB$, $BP^2= (\frac{5}{4})^2 + (\frac{25}{4})^2-2(\frac{5}{4})(\frac{25}{4})(\frac{5}{13})$. Solving yields $BP=\frac{15\sqrt{26}}{13}$.


Then $CP-BP= \frac{75\sqrt{26}}{13} - \frac{15\sqrt{26}}{13} = \frac{60\sqrt{26}}{13}$, and so $a+b+c=60+26+13=\boxed{099}$.

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