Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 8"
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− | + | ==Problem== | |
+ | Let <math>ABCDE</math> be a convex pentagon with <math>\frac{AB}{\sqrt{2}}=BC=CD=DE</math>, <math>\angle ABC=150^\circ</math>, <math>\angle BCD=75^\circ</math>, and <math>\angle CDE=165^\circ</math>. If <math>\angle ABE=\frac{m}{n}^\circ</math> where <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>m+n</math>. | ||
− | [[Mock AIME 1 2006-2007]] | + | ==Solution== |
+ | <center><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></center> | ||
+ | |||
+ | Let <math>P</math> be a point in <math>ABCDE</math> such that <math>\angle ABP=\angle BAP=45^\circ</math>. We see that <math>\angle CBP=115^\circ</math> and thus <math>BP||CD</math>. Since <math>BP=BC=CD</math>, we have that <math>BCDP</math> is a rhombus. Therefore <math>\angle CDP=115^\circ</math> so <math>\angle PDE=60^\circ</math>. Since <math>PD=CD=DE</math> we have that <math>\triangle PDE</math> is equilateral. | ||
+ | |||
+ | <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>. | ||
+ | |||
+ | *[[Mock AIME 1 2006-2007 Problems/Problem 7 | Previous Problem]] | ||
+ | |||
+ | *[[Mock AIME 1 2006-2007 Problems/Problem 9 | Next Problem]] | ||
+ | |||
+ | *[[Mock AIME 1 2006-2007]] |
Latest revision as of 13:54, 21 August 2020
Problem
Let be a convex pentagon with , , , and . If where and are relatively prime positive integers, find .
Solution
Let be a point in such that . We see that and thus . Since , we have that is a rhombus. Therefore so . Since we have that is equilateral.
. .