This question appeared in this year's UNSW Maths competition. It was question 5b and it was the only question that i couldn't do. Sorry if my explanation is bad as it is complicated to understand without a diagram.
In part a) you are given that the radius of a inscribed circle in any triangle is: $\displaystyle \frac{2s}{a+b+c}$ where $s$ is the area of the triangle and $a, b$ and $c$ are the length of the sides of the triangle.
So the question is:
If there is a triangle, the altitude is drawn so that there is $2$ triangles now. Now, the altitude is drawn again for each of the $2$ triangles, so that there is $4$ triangles. This process is down $2014$ times in total, so there is $2^{2014}$ triangles in the original triangle. Now, a circle is inscribed in each triangle so that there are $2^{2014}$ circles.
What is the fraction of the area of the circle over the area of the original triangle?
Thanks in advance.
Hint: After the first attitude, the triangles are divided into two groups containing of similar (right) triangles. The ratio of the areas of the corresponding circles to the triangles are the same for all triangles in a group.
Note: If the original triangle happens to be a right triangle as @Blue mention, there's no need to divide the $2^{2014}$ triangles into two groups.