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Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 8"
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<math>\angle BPE=75+60^\circ=180^\circ-2\angle PBE\implies \angle PBE=\frac{45}{2}</math>. <math>\angle ABE=\angle ABP+\angle PBE=45+\frac{45}{2}^\circ=\frac{135}{2}^\circ\implies m+n=\boxed{137}</math>.
 
<math>\angle BPE=75+60^\circ=180^\circ-2\angle PBE\implies \angle PBE=\frac{45}{2}</math>. <math>\angle ABE=\angle ABP+\angle PBE=45+\frac{45}{2}^\circ=\frac{135}{2}^\circ\implies m+n=\boxed{137}</math>.
  
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*[[Mock AIME 1 2006-2007/Problem 7 | Previous Problem]]
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*[[Mock AIME 1 2006-2007 Problems/Problem 7 | Previous Problem]]
  
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*[[Mock AIME 1 2006-2007 Problems/Problem 9 | Next Problem]]
  
 
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*[[Mock AIME 1 2006-2007]]

Revision as of 14:52, 3 April 2012

Problem

Let $ABCDE$ be a convex pentagon with $AB\sqrt{2}=BC=CD=DE$, $\angle ABC=150^\circ$, $\angle BCD=75^\circ$, 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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