2006 Cyprus MO/Lyceum/Problem 7

Problem

2006 CyMO-7.PNG

In the figure, $AB\Gamma$ is an equilateral triangle and $A\Delta \perp B\Gamma$, $\Delta E\perp A\Gamma$, $EZ\perp B\Gamma$. If $EZ=\sqrt{3}$, then the length of the side of the triangle $AB\Gamma$ is

$\mathrm{(A)}\ \frac{3\sqrt{3}}{2}\qquad\mathrm{(B)}\ 8\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 3\qquad\mathrm{(E)}\ 9$

Solution

$\triangle EZ\Gamma$ is a $30-60-90$ right triangle, so $Z\Gamma = \frac{\sqrt{3}}{\sqrt{3}} = 1$. Also $\angle ZE\Delta = 90 - 30 = 60^{\circ}$, so $\triangle ZE\Delta$ also is a $30-60-90 \triangle$.

Thus, $\Delta Z = \sqrt{3} \cdot \sqrt{3} = 3$. Adding, $\Delta Z + Z\Gamma = 4$, and a side of $\triangle AB\Gamma$ is $2 \Delta \Gamma = 8\ \mathrm{(B)}$.

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