# Thread: Help on 2009 AIME II question 5.

1. ## Help on 2009 AIME II question 5.

5. Equilateral triangle is inscribed in circle , which has radius . Circle with radius is internally tangent to circle at one vertex of . Circles and , both with radius , are internally tangent to circle at the other two vertices of . Circles , , and are all externally tangent to circle , which has radius , where and are relatively prime positive integers. Find .
I called the radius of circle : $r$. I called the intersection between circle and : $x$. I called the intersection between circle and : $y$. I called the intersection between circle and : $Z$. The height of the triangle going through point x: $p$.The points $x$, $y$, and $z$ form equilateral triangle . I found line to be $6+r$. The height of the triangle would be $15$, because the radius of circle is 10, which would make the side length of the triangle $10\sqrt{3}$. Triangle EPY is a right triangle. So line segment EP would be $\sqrt{(4+r)^2 - 75}$ . So line segment $XP$ would be $(6+r) + \sqrt{(4+r)^2 - 75}$, which should equal 15. Solving for $r$, i would get 70/13.

A alternate solution I came up with was to use triangle $AEX$.I know angle $XAE$ is 60 degrees. So using the cosine law, I would get: $(r+4)^2 = (r-4)^2 + 10^2 - 2*10*(r-4)*(\cos 60)$ which also comes out to $r$ = 70/13.

However my answer is wrong. The correct answer is 27/5. Could someone tell me what I am doing wrong? Thank you.
*I am sorry if i posted this in the wrong section. This is mostly geometry, but there is a bit of trig at the end.

2. ## Re: Help on 2009 AIME II question 5.

Your calculation is not easy to follow. You define the points x, y and z but not the points B, C and E. Are we to assume that these are the centres of the various circles ? If so you should say so. Similarly, (if I'm reading this correctly), you define p to be the length of a line and then later it somehow becomes a point.
Anyway, having said that, I think that your mistake comes at the point where you calculate an expression for EP. It looks as if you are assuming that the line Ey passes through the centre of the circle C, but that is not the case, in other words, it's the 4+r that's wrong.

3. ## Re: Help on 2009 AIME II question 5.

Look at this picture. I hope it helps you solving the problem.

4. ## Re: Help on 2009 AIME II question 5.

Hi,
If you haven't yet solved the problem, the attachment shows a solution by assigning coordinates to the figure.