Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 11"
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==Solution== | ==Solution== | ||
| − | {{ | + | We find some coordinates: |
| + | |||
| + | <math>O=(0,0)</math> | ||
| + | |||
| + | <math>A=(4,0)</math> | ||
| + | |||
| + | <math>B=(0,4)</math> | ||
| + | |||
| + | <math>\Gamma =(\frac{8}{3},\frac{4}{3})</math> | ||
| + | |||
| + | We find the area of triangles: | ||
| + | |||
| + | <math>[OAB]=8</math> | ||
| + | |||
| + | <math>[AO\Gamma]=\frac{\frac{4}{3}*4}{2}=\frac[8}{3}</math> | ||
| + | |||
| + | <math>[BO\Gamma]=[OAB]-[AO\Gamma]=\frac{16}{3}</math> | ||
| + | |||
| + | <math>\frac{[AO\Gamma]}{[BO\Gamma]}=\frac{1}{2} \Rightarrow \mathrm {(D)}</math> | ||
==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}} | ||
Revision as of 22:48, 19 October 2007
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
A. 
B. 
C. 
D. 
E. 
Solution
We find some coordinates:
We find the area of triangles:
![$[OAB]=8$](http://latex.artofproblemsolving.com/a/d/c/adc725ea15c033d507efaff92d9aaa5d05fb2021.png)
$[AO\Gamma]=\frac{\frac{4}{3}*4}{2}=\frac[8}{3}$ (Error compiling LaTeX. Unknown error_msg)
![$[BO\Gamma]=[OAB]-[AO\Gamma]=\frac{16}{3}$](http://latex.artofproblemsolving.com/d/7/4/d74b87a2523c72e6ca195ee8d001361469d30a9b.png)
![$\frac{[AO\Gamma]}{[BO\Gamma]}=\frac{1}{2} \Rightarrow \mathrm {(D)}$](http://latex.artofproblemsolving.com/d/d/e/dde52bf0f31dbc212a2d21eb53847df077e3d976.png)
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 | ||
