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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:

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

We find the area of triangles:

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

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

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

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by Problem 10	Followed by Problem 12
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 "http://www.artofproblemsolving.com/Wiki/index.php/2006_Cyprus_MO/Lyceum/Problem_11"

Category: Introductory Geometry Problems

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