## Problem 3.

Let $A B C$ be a triangle and $D, E$, and $F$ be points located on the segments $A C, B A$, and $C B$ respectively, such that the segments $A F, B D, C E$ concur at a point $P$ inside the triangle. We know that $B P=6, P D=6, P C=9, P E=3$, and $A F=20$. Find the area of triangle $A B C$.
Solution.
![](https://cdn.mathpix.com/cropped/2024_04_26_40ba6606ee42bc08e8e3g-2.jpg?height=886&width=1164&top_left_y=702&top_left_x=388)

We use several times that the areas of two triangles with the same base are proportional to the heights and that the areas of two triangles with the same height are proportional to the bases.

The letters $x, y, z, \ldots$ in the left figure indicate the areas of the small triangles where they are located. We denote by $S$ the total area of the triangle $ABC$ requested. We have

$$
\frac{u+v}{S}=\frac{PD}{BD}=\frac{1}{2}, \quad \frac{x+y}{S}=\frac{EP}{EC}=\frac{1}{4}
$$

It also holds that

$$
\frac{PF}{AF}=\frac{PF}{20}=\frac{z+t}{S}=\frac{S-(x+y)-(u+v)}{S}=1-\frac{1}{2}-\frac{1}{4}
$$

from which $PF=5$ and $AP=15$.

By comparing the areas of the triangles $ABD$ and $BDC$ and of $PDA$ and $PDC$, we get

$$
\frac{CD}{AD}=\frac{z+t+u}{x+y+v}=\frac{u}{v}=\frac{z+t}{x+y}=\frac{1/4 S}{1/4}=1, \text{ so } CD=AD \text{ and } u=v; \text{ hence } S=4u
$$

Now we draw a parallel $DG$ to $AF$ (right figure). The triangles $CAP$ and $CDG$ are similar with a ratio of $1/2$. Therefore, $DG=\frac{1}{2} AP=\frac{15}{2}$ and $PG=GC=\frac{9}{2}$.

Since $PD^2 + PG^2 = DG^2$, the triangle $DPG$ is a right triangle with a right angle at $P$ as indicated in the figure. Its area is $u/2=27/2$, so $u=27$ and $S=4u=108$.