Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 7"
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==Problem== | ==Problem== | ||
| + | [[Image:2006 CyMO-7.PNG|250px|right]] | ||
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| + | In the figure, <math>AB\Gamma</math> is an equilateral triangle and <math>A\Delta \perp B\Gamma</math>, <math>\Delta E\perp A\Gamma</math>, <math>EZ\perp B\Gamma</math>. If <math>EZ=\sqrt{3}</math>, then the length of the side of the triangle <math>AB\Gamma</math> is | ||
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| + | <math>\mathrm{(A)}\ \frac{3\sqrt{3}}{2}\qquad\mathrm{(B)}\ 8\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 3\qquad\mathrm{(E)}\ 9</math> | ||
==Solution== | ==Solution== | ||
| − | {{ | + | <math>\triangle EZ\Gamma</math> is a <math>30-60-90</math> [[right triangle]], so <math>Z\Gamma = \frac{\sqrt{3}}{\sqrt{3}} = 1</math>. Also <math>\angle ZE\Delta = 90 - 30 = 60^{\circ}</math>, so <math>\triangle ZE\Delta</math> also is a <math>30-60-90 \triangle</math>. |
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| + | Thus, <math>\Delta Z = \sqrt{3} \cdot \sqrt{3} = 3</math>. Adding, <math>\Delta Z + Z\Gamma = 4</math>, and a side of <math>\triangle AB\Gamma</math> is <math>2 \Delta \Gamma = 8\ \mathrm{(B)}</math>. | ||
==See also== | ==See also== | ||
{{CYMO box|year=2006|l=Lyceum|num-b=6|num-a=8}} | {{CYMO box|year=2006|l=Lyceum|num-b=6|num-a=8}} | ||
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| + | [[Category:Introductory Geometry Problems]] | ||
Latest revision as of 09:42, 27 April 2008
Problem
In the figure, 
is an equilateral triangle and 
, 
, 
. If 
, then the length of the side of the triangle is
Solution
is a 
right triangle, so 
. Also 
, so 
also is a 
.
Thus, 
. Adding, 
, and a side of 
is 
.
See also
| 2006 Cyprus MO, Lyceum (Problems) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 | ||
