# 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?

Thanks in advance.

• 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

## 1 Answer

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.

• Thanks for your answer but could you please explain how the 2 triangles are similar to each other. – user152574 Aug 24 '14 at 10:07
• Also, does the answer happen to be half. – user152574 Aug 24 '14 at 10:07
• 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
• I have tried do this question by using the similar triangles however i still not able to solve it. Could you please explain further how to do his question. Thanks. – user152574 Aug 27 '14 at 12:47
• could you please elaborate on how to do this question. Thanks – user152574 Aug 30 '14 at 12:06