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

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==Problem==
 
==Problem==
<div style="float:right">
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[[Image:2006 CyMO-17.PNG|250px|right]]
[[Image:2006 CyMO-17.PNG|250px]]
 
</div>
 
  
<math>ABC</math> is [[equilateral triangle]] of side <math>\alpha</math> and <math>AD=BE=\frac{\alpha}{3}</math>. The measure of the angle <math>\ang CPE</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
  
A. <math>60^{\circ}</math>
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<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>
  
B. <math>50^{\circ}</math>
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==Solution==
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Label point <math>F</math> on <math>A\Gamma</math> such that <math>\Gamma F=\frac{\alpha}{3}</math>.
  
C. <math>40^{\circ}</math>
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By symmetry we see that the triangle in the middle is equilateral, so the measure of <math>\angle\Gamma PE</math> is <math>60^{\circ}</math>, and the answer is <math>\mathrm{(A)}</math>.
 
 
D. <math>45^{\circ}</math>
 
 
 
E. <math>70^{\circ}</math>
 
 
 
==Solution==
 
Draw a segment <math>BF</math> such that <math>BF = \frac{\alpha}{3}</math>. By symmetry we see the triangle in the middle is equilateral, so the measure of <math>\ang CPE = 60^{\circ} \ \mathrm{(A)}</math>.
 
  
 
==See also==
 
==See also==

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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