Mock AIME 1 2006-2007 Problems/Problem 8

Problem

Let $ABCDE$ be a convex pentagon with $\frac{AB}{\sqrt{2}}=BC=CD=DE$ , $\angle ABC=150^\circ$ , $\angle BCD=75^\circ$ , and $\angle CDE=165^\circ$ . If $\angle ABE=\frac{m}{n}^\circ$ where $m$ and $n$ are relatively prime positive integers, find $m+n$ .

Solution

$[asy]defaultpen(fontsize(8)); pair A=expi(pi*5/12)+expi(0)+expi(pi/2), B=expi(pi*5/12), C=(0,0), D=expi(0), E=expi(0)+expi(pi/12), P=expi(pi*5/12)+expi(0); draw(A--B--C--D--E--A);draw(B--P--E--B);draw(D--P--A); label("A",A,(1,0));label("B",B,(-1,0));label("C",C,(-1,0));label("D",D,(1,-1)); label("E",E,(1,0));label("P",P,(1,0)); dot(A^^B^^C^^D^^E^^P);[/asy]$

Let $P$ be a point in $ABCDE$ such that $\angle ABP=\angle BAP=45^\circ$ . We see that $\angle CBP=115^\circ$ and thus $BP||CD$ . Since $BP=BC=CD$ , we have that $BCDP$ is a rhombus. Therefore $\angle CDP=115^\circ$ so $\angle PDE=60^\circ$ . Since $PD=CD=DE$ we have that $\triangle PDE$ is equilateral.

$\angle BPE=75+60^\circ=180^\circ-2\angle PBE\implies \angle PBE=\frac{45}{2}$ . $\angle ABE=\angle ABP+\angle PBE=45+\frac{45}{2}^\circ=\frac{135}{2}^\circ\implies m+n=\boxed{137}$ .

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Mock AIME 1 2006-2007 Problems/Problem 8

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