Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 17"

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[[Image:2006 CyMO-17.PNG|250px|right]]
 
[[Image:2006 CyMO-17.PNG|250px|right]]
  
<math>AB\Gamma</math> is equilateral triangle of side <math>\alpha</math> and <math>A\Delta=BE=\frac{\alpha}{3}</math>. The measure of the angle <math>\ang \Gamma PE</math> is
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<math>AB\Gamma</math> is equilateral triangle of side <math>\alpha</math> and <math>A\Delta=BE=\frac{\alpha}{3}</math>. The measure of the angle <math>\angle\Gamma PE</math> is
  
 
<math>\mathrm{(A)}\ 60^\circ\qquad\mathrm{(B)}\ 50^\circ\qquad\mathrm{(C)}\ 40^\circ\qquad\mathrm{(D)}\ 45^\circ\qquad\mathrm{(E)}\ 70^\circ</math>
 
<math>\mathrm{(A)}\ 60^\circ\qquad\mathrm{(B)}\ 50^\circ\qquad\mathrm{(C)}\ 40^\circ\qquad\mathrm{(D)}\ 45^\circ\qquad\mathrm{(E)}\ 70^\circ</math>

Latest revision as of 22:00, 30 November 2015

Problem

2006 CyMO-17.PNG

$AB\Gamma$ is equilateral triangle of side $\alpha$ and $A\Delta=BE=\frac{\alpha}{3}$. The measure of the angle $\angle\Gamma PE$ is

$\mathrm{(A)}\ 60^\circ\qquad\mathrm{(B)}\ 50^\circ\qquad\mathrm{(C)}\ 40^\circ\qquad\mathrm{(D)}\ 45^\circ\qquad\mathrm{(E)}\ 70^\circ$

Solution

Label point $F$ on $A\Gamma$ such that $\Gamma F=\frac{\alpha}{3}$.

By symmetry we see that the triangle in the middle is equilateral, so the measure of $\angle\Gamma PE$ is $60^{\circ}$, and the answer is $\mathrm{(A)}$.

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 16
Followed by
Problem 18
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