2006 Cyprus MO/Lyceum/Problem 11

Problem

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:

$[OAB]=8$

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

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

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

2006 Cyprus MO, Lyceum (Problems)
Preceded by Problem 10	Followed by Problem 12
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2006 Cyprus MO/Lyceum/Problem 11

Problem

Solution

See also