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

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==Problem==
 
==Problem==
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<div style="float:right">
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[[Image:2006 CyMO-7.PNG|250px|right]]
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[[Image:2006 CyMO-7.PNG|250px]]
 
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</div>
 
  
 
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
 
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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A. <math>\frac{3\sqrt{3}}{2}</math>
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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>
  
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B. <math>8</math>
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==Solution==
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<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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C. <math>4</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>.
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D. <math>3</math>
 
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E. <math>9</math>
 
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==Solution==
 
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<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==
 
==See also==

Latest revision as of 09:42, 27 April 2008

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
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