# Inscribed Circles in Triangles

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?

• Do you mean "altitude" instead of "perpendicular bisector"? A perpendicular bisector doesn't always separate a triangle into two triangles, but an altitude does. – Blue Aug 21 '14 at 12:18
• Yeah, sorry, i meant altitude, not perpendicular bisector – user152574 Aug 21 '14 at 12:19
• Do you understand the question, or do you want me to explain it again. – user152574 Aug 21 '14 at 12:19
• 3rd row from below: I think you meant to say "$2^{2014}$ circles". – Ludolila Aug 21 '14 at 12:21
• Yeah, sorry about that – user152574 Aug 21 '14 at 12:22

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.
• After the first attitude, you have $2$ right triangles. Now if you divide a right triangle into $2$ smaller triangles by a attitude (to the hypotenuse), then each sub-triangle is similar to the big one. I don't see the answer being a half unless the original triangle is special. – Quang Hoang Aug 24 '14 at 10:36