# 1985 AIME Problems/Problem 6

## Problem

As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$.

## Solution 1

Let the interior point be $P$, let the points on $\overline{BC}$, $\overline{CA}$ and $\overline{AB}$ be $D$, $E$ and $F$, respectively. Let $x$ be the area of $\triangle APE$ and $y$ be the area of $\triangle CPD$. Note that $\triangle APF$ and $\triangle BPF$ share the same altitude from $P$, so the ratio of their areas is the same as the ratio of their bases. Similarly, $\triangle ACF$ and $\triangle BCF$ share the same altitude from $C$, so the ratio of their areas is the same as the ratio of their bases. Moreover, the two pairs of bases are actually the same, and thus in the same ratio. As a result, we have: $\frac{40}{30} = \frac{124 + x}{65 + y}$ or equivalently $372 + 3x = 260 + 4y$ and so $4y = 3x+ 112$.

Applying identical reasoning to the triangles with bases $\overline{CD}$ and $\overline{BD}$, we get $\frac{y}{35} = \frac{x+y+84}{105}$ so that $3y = x + y + 84$ and $2y = x + 84$. Substituting from this equation into the previous one gives $x = 56$, from which we get $y = 70$ and so the area of $\triangle ABC$ is $56 + 40 + 30 + 35 + 70 + 84 = \Rightarrow \boxed{315}$.

## Solution 2

This problem can be done using mass points. Assign B a weight of 1 and realize that many of the triangles have the same altitude. After continuously using the formulas that state (The sum of the two weights) = (The middle weight), and (The weight $\times$ side) = (Other weight) $\times$ (The other side), the problem yields the answer $\boxed{315}$

## Solution 3

Let the interior point be $P$ and let the points on $\overline{BC}$, $\overline{CA}$ and $\overline{AB}$ be $D$, $E$ and $F$, respectively. Also, let $[APE]=x,[CPD]=y.$ Then notice that by Ceva's, $\frac{FB\cdot DC\cdot EA}{DB\cdot CE\cdot AF}=1.$ However, we can deduce $\frac{FB}{AF}=\frac{3}{4}$ from the fact that $[AFP]$ and $[BPF]$ share the same height. Similarly, $x=\frac{84CE}{EA}$ and $y=\frac{35DC}{BD}.$ Plug and chug and you get $xy=84\cdot 35\cdot \frac{3}{4}=2205.$ Then notice by the same height reasoning, $\frac{84}{x}=\frac{119+y}{x+70}.$ Clear the fractions and combine like terms to get $35x=5880-xy.$ We know $xy=2205$ so subtraction yields $35x=3675,$ or $x=105.$ Plugging this in to our previous ratio statement yields $\frac{84}{105}=\frac{4}{5}=\frac{119+y}{175},$ so $y=21.$ Basic addition gives us $105+84+21+35+30+40=\boxed{315}.$

-dchen

## Solution 4

Let the interior point be $P$ and let the points on $\overline{BC}$, $\overline{CA}$ and $\overline{AB}$ be $D$, $E$ and $F$, respectively. Then the cevians $AD,BF,CE$ are concurrent, so we can use Ceva's Theorem, letting $\frac{BD}{DC}=\frac{a}{b}$ and $\frac{CF}{FA}=\frac{c}{d}$. Notice that $\frac{AE}{EB}=\frac{[\Delta APE]}{[\Delta EPB]}=\frac43.$ $$\frac{4}{3}\cdot \frac{a}{b}\cdot \frac{c}{d}=1\implies \frac{d}{c}=\frac{a}{b}\cdot \frac{4}{3}.$$

We know that $[\Delta CPD]=35\cdot \frac ba$ and $[\Delta APF]=84\cdot \frac dc,$ so $$[\Delta ABC] = 84+84\cdot \frac dc + 35\cdot \frac ba + 40+30+35 = \left(1+\frac ba\right)(40+30+35).$$ We will now solve for $\frac ba$:

$$84+84\cdot \frac ab\cdot \frac 43 + 35\cdot \frac ba = \frac ba\cdot 105.$$ $$12+16\cdot \frac ab + 5\cdot \frac ba = \frac ba\cdot 15$$ $$10\left(\frac ba\right)^2-12\cdot \frac ba-16=0$$ Factoring this gives $\left(\frac ba-2\right)\left(10\cdot \frac ba - 8\right)=0,$ so the area of $\triangle ABC$ is $$\left(1+\frac ba\right)(40+30+35)=3\cdot 105=\boxed{315.}$$

~ RubixMaster21