Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 7"

Revision as of 18:02, 19 October 2007

Problem

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

A. $\frac{3\sqrt{3}}{2}$

B. $8$

C. $4$

D. $3$

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 $2 \Delta \Gamma = 8\ \mathrm{(B)}$ .

@@ Line 10: / Line 10: @@
 B. <math>8</math>
-C.  <math>4</math>
+C. <math>4</math>
 D. <math>3</math>
@@ Line 17: / Line 17: @@
 ==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>. 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==
 {{CYMO box|year=2006|l=Lyceum|num-b=6|num-a=8}}
+[[Category:Introductory Geometry Problems]]

2006 Cyprus MO, Lyceum (Problems)
Preceded by Problem 6	Followed by Problem 8
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Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 7"

Revision as of 18:02, 19 October 2007

Problem

Solution

See also