Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 8"

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==Problem==
 
==Problem==
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[[Image:2006 CyMO-8.PNG|250px|right]]
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In the figure <math>AB\Gamma \Delta E</math> is a regular 5-sided polygon and <math>Z</math>, <math>H</math>, <math>\Theta</math>, <math>I</math>, <math>K</math> are the points of intersections of the extensions of the sides.
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If the area of the "star" <math>AHB\Theta \Gamma I\Delta KEZA</math> is 1, then the area of the shaded quadrilateral <math>A\Gamma IZ</math> is
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<math>\mathrm{(A)}\ \frac{2}{3}\qquad\mathrm{(B)}\ \frac{1}{2}\qquad\mathrm{(C)}\ \frac{3}{7}\qquad\mathrm{(D)}\ \frac{3}{10}\qquad\mathrm{(E)}\ \text{None of these}</math>
  
 
==Solution==
 
==Solution==
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In the quadrilateral <math>A\Gamma IZ</math>, we have three isosceles triangles <math>A\Gamma\Delta</math>, <math>AZE</math>, and <math>\Gamma \Delta I</math>. Those are congruent to each other, as well as <math>HAB</math>, <math>B\Gamma\Theta</math>, and <math>EK\Delta</math>. Also, <math>AE\Delta</math> is congruent to <math>AB\Gamma</math>. Thus we have two figures of equal area: <math>A\Gamma IZ</math> and a combination of two figures: <math>HB\Theta\Gamma A</math> and <math>EK\Delta</math>. Since the area of the whole star is 1, the area of <math>AZI\Gamma</math> is <math>\frac{1}{2}\mathrm{(B)}</math>.
  
 
==See also==
 
==See also==
 
{{CYMO box|year=2006|l=Lyceum|num-b=7|num-a=9}}
 
{{CYMO box|year=2006|l=Lyceum|num-b=7|num-a=9}}

Latest revision as of 08:18, 12 August 2008

Problem

2006 CyMO-8.PNG

In the figure $AB\Gamma \Delta E$ is a regular 5-sided polygon and $Z$, $H$, $\Theta$, $I$, $K$ are the points of intersections of the extensions of the sides. If the area of the "star" $AHB\Theta \Gamma I\Delta KEZA$ is 1, then the area of the shaded quadrilateral $A\Gamma IZ$ is

$\mathrm{(A)}\ \frac{2}{3}\qquad\mathrm{(B)}\ \frac{1}{2}\qquad\mathrm{(C)}\ \frac{3}{7}\qquad\mathrm{(D)}\ \frac{3}{10}\qquad\mathrm{(E)}\ \text{None of these}$

Solution

In the quadrilateral $A\Gamma IZ$, we have three isosceles triangles $A\Gamma\Delta$, $AZE$, and $\Gamma \Delta I$. Those are congruent to each other, as well as $HAB$, $B\Gamma\Theta$, and $EK\Delta$. Also, $AE\Delta$ is congruent to $AB\Gamma$. Thus we have two figures of equal area: $A\Gamma IZ$ and a combination of two figures: $HB\Theta\Gamma A$ and $EK\Delta$. Since the area of the whole star is 1, the area of $AZI\Gamma$ is $\frac{1}{2}\mathrm{(B)}$.

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 7
Followed by
Problem 9
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