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

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The lines <math>(\epsilon):x-2y=0</math> and <math>(\delta):x+y=4</math> intersect at the point <math>C</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>OAC</math> to the area of the triangle <math>OBC</math> equals
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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
  
 
A. <math>\frac{1}{3}</math>
 
A. <math>\frac{1}{3}</math>

Revision as of 21:56, 17 October 2007

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

A. $\frac{1}{3}$

B. $\frac{2}{3}$

C. $\frac{3}{5}$

D. $\frac{1}{2}$

E. $\frac{4}{9}$

Solution

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See also

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