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2006 Cyprus MO/Lyceum/Problem 11

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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\...

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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