Difference between revisions of "2001 AIME I Problems/Problem 9"
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== Solution == | == Solution == | ||
+ | |||
+ | === Solution 1 === | ||
<center><asy> | <center><asy> | ||
/* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */ | /* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */ | ||
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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>. | ||
+ | === Solution 2 === | ||
+ | |||
+ | By the barycentric area formula, our desired ratio is equal to | ||
+ | <cmath>\det A= | ||
+ | \begin{vmatrix} | ||
+ | 1-p & p & 0 \ | ||
+ | 0 & 1-q & q \ | ||
+ | r & 0 & 1-r \notag | ||
+ | \end{vmatrix} =1-p-q-r+pq+qr+pr=1-(p+q+r)+\frac{(p+q+r)^2-(pq+qr-pr)}{2}=1-\frac{2}{3}+\frac{\frac{4}{9}-\frac{2}{5}}{2}=\frac{16}{45},</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 18:25, 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.