Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 11"
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==Problem== | ==Problem== | ||
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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 | ||
− | + | <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> | |
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− | B | ||
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− | C | ||
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− | D | ||
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− | E | ||
==Solution== | ==Solution== | ||
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<math>B=(0,4)</math> | <math>B=(0,4)</math> | ||
− | <math>\Gamma =(\frac{8}{3},\frac{4}{3})</math> | + | <math>\Gamma =\left(\frac{8}{3},\frac{4}{3}\right)</math> |
We find the area of triangles: | We find the area of triangles: |
Latest revision as of 10:37, 27 April 2008
Problem
The lines and intersect at the point . If the line intersects the axes and to the points and respectively, then the ratio of the area of the triangle to the area of the triangle equals
Solution
We find some coordinates:
We find the area of triangles:
See also
2006 Cyprus MO, Lyceum (Problems) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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