Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 17"
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I like pie (talk | contribs) (Standardized answer choices; minor edits to solution) |
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==Problem== | ==Problem== | ||
− | + | [[Image:2006 CyMO-17.PNG|250px|right]] | |
− | [[Image:2006 CyMO-17.PNG|250px]] | ||
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− | <math>AB\Gamma</math> is | + | <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 |
− | + | <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> | |
− | + | ==Solution== | |
+ | Label point <math>F</math> on <math>A\Gamma</math> such that <math>\Gamma F=\frac{\alpha}{3}</math>. | ||
− | + | 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>. | |
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==See also== | ==See also== |
Revision as of 13:41, 26 April 2008
Problem
is equilateral triangle of side and . The measure of the angle $\ang \Gamma PE$ (Error compiling LaTeX. ! Undefined control sequence.) is
Solution
Label point on such that .
By symmetry we see that the triangle in the middle is equilateral, so the measure of is , and the answer is .
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 |