Mock Geometry AIME 2011 Problems/Problem 7

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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Mock Geometry AIME 2011 Problems/Problem 7

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