Difference between revisions of "2001 AIME I Problems/Problem 9"
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and similarly that <math>\frac{[BDE]}{[ABC]} = q(1-p)</math> and <math>\frac{[CEF]}{[ABC]} = r(1-q)</math>. Thus, we wish to find | and similarly that <math>\frac{[BDE]}{[ABC]} = q(1-p)</math> and <math>\frac{[CEF]}{[ABC]} = r(1-q)</math>. Thus, we wish to find | ||
− | < | + | <cmath>\begin{align*}\frac{[DEF]}{[ABC]} &= 1 - \frac{[ADF]}{[ABC]} - \frac{[DEF]}{[BDE]} - \frac{[CEF]}{[ABC]} |
− | \\ &= 1 - p(1-r) + q(1-p) + r(1-q)\\ &= (pq + qr + rp) - (p + q + r) + 1 \end{align}</ | + | \\ &= 1 - p(1-r) + q(1-p) + r(1-q)\\ &= (pq + qr + rp) - (p + q + r) + 1 \end{align*}</cmath> |
We know that <math>p + q + r = \frac 23</math>, and also that <math>(p+q+r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + rp) \Longleftrightarrow pq + qr + rp = \frac{\left(\frac 23\right)^2 - \frac 25}{2} = \frac{1}{45}</math>. Substituting, the answer is <math>\frac 1{45} - \frac 23 + 1 = \frac{16}{45}</math>, and <math>m+n = \boxed{061}</math>. | We know that <math>p + q + r = \frac 23</math>, and also that <math>(p+q+r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + rp) \Longleftrightarrow pq + qr + rp = \frac{\left(\frac 23\right)^2 - \frac 25}{2} = \frac{1}{45}</math>. Substituting, the answer is <math>\frac 1{45} - \frac 23 + 1 = \frac{16}{45}</math>, and <math>m+n = \boxed{061}</math>. | ||
Revision as of 16:30, 13 March 2015
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
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 .
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.