2001 AIME I Problems/Problem 9

Problem

In triangle $ABC$ , $AB=13$ , $BC=15$ and $CA=17$ . Point $D$ is on $\overline{AB}$ , $E$ is on $\overline{BC}$ , and $F$ is on $\overline{CA}$ . Let $AD=p\cdot AB$ , $BE=q\cdot BC$ , and $CF=r\cdot CA$ , where $p$ , $q$ , and $r$ are positive and satisfy $p+q+r=2/3$ and $p^2+q^2+r^2=2/5$ . The ratio of the area of triangle $DEF$ to the area of triangle $ABC$ can be written in the form $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution

Solution 1

$[asy] /* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */ real p = 0.5, q = 0.1, r = 0.05; /* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */ pointpen = black; pathpen = linewidth(0.7) + black; pair A=(0,0),B=(13,0),C=IP(CR(A,17),CR(B,15)), D=A+p*(B-A), E=B+q*(C-B), F=C+r*(A-C); D(D(MP("A",A))--D(MP("B",B))--D(MP("C",C,N))--cycle); D(D(MP("D",D))--D(MP("E",E,NE))--D(MP("F",F,NW))--cycle); [/asy]$

We let $[\ldots]$ denote area; then the desired value is

$\frac mn = \frac{[DEF]}{[ABC]} = \frac{[ABC] - [ADF] - [BDE] - [CEF]}{[ABC]}$

Using the formula for the area of a triangle $\frac{1}{2}ab\sin C$ , we find that

$\frac{[ADF]}{[ABC]} = \frac{\frac 12 \cdot p \cdot AB \cdot (1-r) \cdot AC \cdot \sin \angle CAB}{\frac 12 \cdot AB \cdot AC \cdot \sin \angle CAB} = p(1-r)$

and similarly that $\frac{[BDE]}{[ABC]} = q(1-p)$ and $\frac{[CEF]}{[ABC]} = r(1-q)$ . Thus, we wish to find $\begin{align*}\frac{[DEF]}{[ABC]} &= 1 - \frac{[ADF]}{[ABC]} - \frac{[BDE]}{[ABC]} - \frac{[CEF]}{[ABC]} \\ &= 1 - p(1-r) - q(1-p) - r(1-q)\\ &= (pq + qr + rp) - (p + q + r) + 1 \end{align*}$ We know that $p + q + r = \frac 23$ , and also that $(p+q+r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + rp) \Longleftrightarrow pq + qr + rp = \frac{\left(\frac 23\right)^2 - \frac 25}{2} = \frac{1}{45}$ . Substituting, the answer is $\frac 1{45} - \frac 23 + 1 = \frac{16}{45}$ , and $m+n = \boxed{061}$ .

Solution 2

By the barycentric area formula, our desired ratio is equal to $\begin{align*} \begin{vmatrix} 1-p & p & 0 \\ 0 & 1-q & q \\ r & 0 & 1-r \notag \end{vmatrix} &=1-p-q-r+pq+qr+pr\\ &=1-(p+q+r)+\frac{(p+q+r)^2-(p^2+q^2+r^2)}{2}\\ &=1-\frac{2}{3}+\frac{\frac{4}{9}-\frac{2}{5}}{2}\\ &=\frac{16}{45}, \end{align*}$ so the answer is $\boxed{061}$

Solution 3 (Informal)

Since the only conditions are that $p + q + r = \frac{2}{3}$ and $p^2 + q^2 + r^2 = \frac{2}{5}$ , we can simply let one of the variables be equal to 0. In this case, let $p = 0$ . Then, $q + r = \frac{2}{3}$ and $q^2 + r^2$ = $\frac{2}{5}$ . Note that the ratio between the area of $DEF$ and $ABC$ is equivalent to $(1-q)(1-r)$ . Solving this system of equations, we get $q = \frac{1}{3} \pm \sqrt{\frac{4}{45}}$ , and $r = \frac{1}{3} \mp \sqrt{\frac{4}{45}}$ . Plugging back into $(1-q)(1-r)$ , we get $\frac{16}{45}$ , so the answer is $\boxed{061}$

Note

Because the givens in the problem statement are all regarding the ratios of the sides, the side lengths of triangle $ABC$ , namely $13, 15, 17$ , 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.

2001 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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