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

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==Problem==
 
==Problem==
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[[Image:2006 CyMO-17.PNG|250px]]
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<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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A. <math>60^{\circ}</math>
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B. <math>50^{\circ}</math>
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C. <math>40^{\circ}</math>
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D. <math>45^{\circ}</math>
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E. <math>70^{\circ}</math>
  
 
==Solution==
 
==Solution==
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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==
 
{{CYMO box|year=2006|l=Lyceum|num-b=16|num-a=18}}
 
{{CYMO box|year=2006|l=Lyceum|num-b=16|num-a=18}}
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[[Category:Introductory Geometry Problems]]

Revision as of 16:43, 16 October 2007

Problem

2006 CyMO-17.PNG

$ABC$ is equilateral triangle of side $\alpha$ and $AD=BE=\frac{\alpha}{3}$. The measure of the angle $\ang CPE$ (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
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
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