Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 8"
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==Problem== | ==Problem== | ||
− | Let <math>ABCDE</math> be a convex pentagon with <math>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>. | + | 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>. |
==Solution== | ==Solution== | ||
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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>. | ||
− | *[[Mock AIME 1 2006-2007/Problem 7 | Previous Problem]] | + | *[[Mock AIME 1 2006-2007 Problems/Problem 7 | Previous Problem]] |
− | *[[Mock AIME 1 2006-2007/Problem 9 | Next Problem]] | + | *[[Mock AIME 1 2006-2007 Problems/Problem 9 | Next Problem]] |
*[[Mock AIME 1 2006-2007]] | *[[Mock AIME 1 2006-2007]] |
Latest revision as of 14:54, 21 August 2020
Problem
Let be a convex pentagon with
,
,
, and
. If
where
and
are relatively prime positive integers, find
.
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]](http://latex.artofproblemsolving.com/7/5/a/75a1a47b166889a6586a8ba9f1ef40758ff387ae.png)
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.
.
.