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

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<math>AB\Gamma</math> is [[equilateral triangle]] of side <math>\alpha</math> and <math>AD=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>\ang \Gamma PE</math> is
  
 
A. <math>60^{\circ}</math>
 
A. <math>60^{\circ}</math>

Revision as of 21:59, 17 October 2007

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 $\ang \Gamma PE$ (Error compiling LaTeX. ! Undefined control sequence.) is

A. $60^{\circ}$

B. $50^{\circ}$

C. $40^{\circ}$

D. $45^{\circ}$

E. $70^{\circ}$

Solution

Draw a segment $BF$ such that $BF = \frac{\alpha}{3}$. By symmetry we see the triangle in the middle is equilateral, so the measure of $\ang CPE = 60^{\circ} \ \mathrm{(A)}$ (Error compiling LaTeX. ! Undefined control sequence.).

See also

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