Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 17"
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==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 | + | 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 21:05, 17 October 2007
Problem
is equilateral triangle of side 
and 
. The measure of the angle $\ang \Gamma PE$ (Error compiling LaTeX. Unknown error_msg) is
A. 
B. 
C. 
D. 
E. 
Solution
Draw a segment 
such that 
. 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. Unknown error_msg).
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 | ||
