Difference between revisions of "2001 AIME I Problems/Problem 9"
Line 32: | Line 32: | ||
By the barycentric area formula, our desired ratio is equal to | By the barycentric area formula, our desired ratio is equal to | ||
− | <cmath>\ | + | <cmath>\begin{align*} |
\begin{vmatrix} | \begin{vmatrix} | ||
1-p & p & 0 \ | 1-p & p & 0 \ | ||
0 & 1-q & q \ | 0 & 1-q & q \ | ||
r & 0 & 1-r \notag | r & 0 & 1-r \notag | ||
− | \end{vmatrix} =1-p-q-r+pq+qr+pr=1-(p+q+r)+\frac{(p+q+r)^2-( | + | \end{vmatrix} &=1-p-q-r+pq+qr+pr\ |
+ | &=1-(p+q+r)+\frac{(p+q+r)^2-(p^2+q^2+r^2)}{2}\ | ||
+ | &=1-\frac{2}{3}+\frac{\frac{4}{9}-\frac{2}{5}}{2}\ | ||
+ | &=\frac{16}{45} | ||
+ | \end{align*},</cmath> so the answer is <math>\boxed{61.}</math> | ||
== See also == | == See also == | ||
{{AIME box|year=2001|n=I|num-b=8|num-a=10}} | {{AIME box|year=2001|n=I|num-b=8|num-a=10}} |
Revision as of 21:05, 27 May 2016
Contents
[hide]Problem
In triangle , , and . Point is on , is on , and is on . Let , , and , where , , and are positive and satisfy and . The ratio of the area of triangle to the area of triangle can be written in the form , where and are relatively prime positive integers. Find .
Solution
Solution 1
We let denote area; then the desired value is
Using the formula for the area of a triangle , we find that
and similarly that and . Thus, we wish to find We know that , and also that . Substituting, the answer is , and .
Solution 2
By the barycentric area formula, our desired ratio is equal to so the answer is
See also
2001 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.