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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</math></center> | </math></center> | ||
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{[BDE]}{[ABC]} - \frac{[CEF]}{[ABC]} |
− | \ &= 1 - p(1-r) | + | \ &= 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>. | ||
+ | |||
+ | === Solution 2 === | ||
+ | |||
+ | By the barycentric area formula, our desired ratio is equal to | ||
+ | <cmath>\begin{align*} | ||
+ | \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-(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{061}</math> | ||
+ | |||
+ | === Solution 3 (Informal) === | ||
+ | |||
+ | Since the only conditions are that <math>p + q + r = \frac{2}{3}</math> and <math>p^2 + q^2 + r^2 = \frac{2}{5}</math>, we can simply let one of the variables be equal to 0. In this case, let <math>p = 0</math>. Then, <math>q + r = \frac{2}{3}</math> and <math>q^2 + r^2</math> = <math>\frac{2}{5}</math>. Note that the ratio between the area of <math>DEF</math> and <math>ABC</math> is equivalent to <math>(1-q)(1-r)</math>. Solving this system of equations, we get <math>q = \frac{1}{3} \pm \sqrt{\frac{4}{45}}</math>, and <math>r = \frac{1}{3} \mp \sqrt{\frac{4}{45}}</math>. Plugging back into <math>(1-q)(1-r)</math>, we get <math>\frac{16}{45}</math>, so the answer is <math>\boxed{061}</math> | ||
+ | |||
+ | === Note === | ||
+ | |||
+ | Because the givens in the problem statement are all regarding the ratios of the sides, the side lengths of triangle <math>ABC</math>, namely <math>13, 15, 17</math>, are actually not necessary to solve the problem. This is clearly demonstrated in all of the above solutions, as the side lengths are not used at all. | ||
== See also == | == See also == |
Latest revision as of 19:52, 28 June 2024
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
Solution 3 (Informal)
Since the only conditions are that and , we can simply let one of the variables be equal to 0. In this case, let . Then, and = . Note that the ratio between the area of and is equivalent to . Solving this system of equations, we get , and . Plugging back into , we get , so the answer is
Note
Because the givens in the problem statement are all regarding the ratios of the sides, the side lengths of triangle , namely , are actually not necessary to solve the problem. This is clearly demonstrated in all of the above solutions, as the side lengths are not used at all.
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 |
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