Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 11"

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==Problem==
 
==Problem==
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[[Image:2006 CyMO-11.PNG|250px|right]]
[[Image:2006 CyMO-11.PNG|250px]]
 
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The lines <math>(\epsilon):x-2y=0</math> and <math>(\delta):x+y=4</math> intersect at the point <math>\Gamma</math>. If the line <math>(\delta)</math> intersects the axes <math>Ox</math> and <math>Oy</math> to the points <math>A</math> and <math>B</math> respectively, then the ratio of the area of the triangle <math>OA\Gamma</math> to the area of the triangle <math>OB\Gamma</math> equals
 
The lines <math>(\epsilon):x-2y=0</math> and <math>(\delta):x+y=4</math> intersect at the point <math>\Gamma</math>. If the line <math>(\delta)</math> intersects the axes <math>Ox</math> and <math>Oy</math> to the points <math>A</math> and <math>B</math> respectively, then the ratio of the area of the triangle <math>OA\Gamma</math> to the area of the triangle <math>OB\Gamma</math> equals
  
A. <math>\frac{1}{3}</math>
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<math>\mathrm{(A)}\ \frac{1}{3}\qquad\mathrm{(B)}\ \frac{2}{3}\qquad\mathrm{(C)}\ \frac{3}{5}\qquad\mathrm{(D)}\ \frac{1}{2}\qquad\mathrm{(E)}\ \frac{4}{9}</math>
 
 
B. <math>\frac{2}{3}</math>
 
 
 
C. <math>\frac{3}{5}</math>
 
 
 
D. <math>\frac{1}{2}</math>
 
 
 
E. <math>\frac{4}{9}</math>
 
  
 
==Solution==
 
==Solution==

Latest revision as of 10:37, 27 April 2008

Problem

2006 CyMO-11.PNG

The lines $(\epsilon):x-2y=0$ and $(\delta):x+y=4$ intersect at the point $\Gamma$. If the line $(\delta)$ intersects the axes $Ox$ and $Oy$ to the points $A$ and $B$ respectively, then the ratio of the area of the triangle $OA\Gamma$ to the area of the triangle $OB\Gamma$ equals

$\mathrm{(A)}\ \frac{1}{3}\qquad\mathrm{(B)}\ \frac{2}{3}\qquad\mathrm{(C)}\ \frac{3}{5}\qquad\mathrm{(D)}\ \frac{1}{2}\qquad\mathrm{(E)}\ \frac{4}{9}$

Solution

We find some coordinates:

$O=(0,0)$

$A=(4,0)$

$B=(0,4)$

$\Gamma =\left(\frac{8}{3},\frac{4}{3}\right)$

We find the area of triangles:

$[OAB]=8$

$[OA\Gamma]=\frac{\frac{4}{3}*4}{2}=\frac{8}{3}$

$[OB\Gamma]=[OAB]-[OA\Gamma]=\frac{16}{3}$

$\frac{[OA\Gamma]}{[OB\Gamma]}=\frac{1}{2} \Rightarrow \mathrm {(D)}$

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 10
Followed by
Problem 12
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