Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 8"
(solution)
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
<div style="float:right">
+
[[Image:2006 CyMO-8.PNG|250px|right]]
−
[[Image:2006 CyMO-8.PNG|250px]]
 
−
</div>
 
  
 
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.
 
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.
 
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
 
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
  
−
A. <math>\frac{2}{3}</math>
+
<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>
−
 
 
−
B. <math>\frac{1}{2}</math>
 
−
 
 
−
C. <math>\frac{3}{7}</math>
 
−
 
 
−
D. <math>\frac{3}{10}</math>
 
−
 
 
−
E. None of these
 
  
 
==Solution==
 
==Solution==
−
{{solution}}
+
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
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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_Cyprus_MO/Lyceum/Problem_8&oldid=27381"