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

(Problem)
(Solution)
Line 17: Line 17:
  
 
==Solution==
 
==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>.
+
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 \Gamma PE = 60^{\circ} \ \mathrm{(A)}</math>.
  
 
==See also==
 
==See also==

Revision as of 22:05, 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 $\ang \Gamma PE = 60^{\circ} \ \mathrm{(A)}$ (Error compiling LaTeX. ! Undefined control sequence.).

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Invalid username
Login to AoPS