Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 11"
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==Problem== | ==Problem== | ||
− | + | [[Image:2006 CyMO-11.PNG|250px|right]] | |
− | [[Image:2006 CyMO-11.PNG|250px]] | ||
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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: | ||
Line 39: | Line 29: | ||
==See also== | ==See also== | ||
{{CYMO box|year=2006|l=Lyceum|num-b=10|num-a=12}} | {{CYMO box|year=2006|l=Lyceum|num-b=10|num-a=12}} | ||
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+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 09:37, 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 | |
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 |