A triangle $ABC$ is divided into four regions by $3$ lines parallel to $BC$ . The lines divide $AB$ into $4$ equal segments.

If the second largest region has area $225$ , find the area of $ABC$ .

A diagram if you need it:

Source:
**
AIMO 2017
**

The answer is 720.

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By the AAA (angle-angle-angle) similarity test, we can say that triangles $AB_{1}C_{1}, AB_{2}C_{2}, AB_{3}C_{3}, ABC$ are similar.

Since the lines divide $AB$ into 4 equal segments, the sides of the triangles are in the ratio $1:2:3:4$ (i.e. Side Ratio). Therefore their areas are in the ratio $1:4:9:16$ .

We know that region $B_{3}C_{3}C_{2}B_{2}$ has area $225$ . Thus if we let the area of triangle $AB_{1}C_{1} = x$ , then

$225 = AB_{3}C_{3} - AB_{2}B_{2} = 9x - 4x = 5x$

$x = \frac{225}{5} = 45$

Therefore the area of triangle $ABC$ is $16x = 16 \times 45 = 720$

The answer is

720.