Mock AIME 6 2006-2007 Problems/Problem 9

Problem

$ABC$ is a triangle with integer side lengths. Extend $\overline{AC}$ beyond $C$ to point $D$ such that $CD=120$ . Similarly, extend $\overline{CB}$ beyond $B$ to point $E$ such that $BE=112$ and $\overline{BA}$ beyond $A$ to point $F$ such that $AF=104$ . If triangles $CBD$ , $BAE$ , and $ACF$ all have the same area, what is the minimum possible area of triangle $ABC$ ?

Solution

Let "a", "b", and "c", be the lengths of sides $AB$ , $BC$ and $CA$ respectively.

Let "h_a", "h_b", and "h_c", be the heights of $\Delta ABC$ from sides $AB$ , $BC$ and $CA$ respectively.

Since the areas of triangles $CBD$ , $BAE$ , and $ACF$ are equal, then,

$104h_a=112h_b=120h_c$

Therefore,

$\frac{h_a}{h_b}=\frac{112}{104}=\frac{14}{13}$ and $\frac{h_a}{h_c}=\frac{120}{104}=\frac{15}{13}$

Since the area of $\Delta ABC$ is half any base times it's height, then:

$ah_a=bh_b=ch_c$

Therefore,

$b=\frac{h_a}{h_b}a=\frac{14}{13}a$ and $c=\frac{h_a}{h_c}a=\frac{15}{13}a$

Since "a", "b", and "c", and $13$ is a prime number, then the minimum integer value that $a$ can have in order for $b$ and $c$ to also be integer is $13$