I called the radius of circle : $\displaystyle r$. I called the intersection between circle and : $\displaystyle x$. I called the intersection between circle and : $\displaystyle y$. I called the intersection between circle and : $\displaystyle Z$. The height of the triangle going through point x: $\displaystyle p$.The points $\displaystyle x$, $\displaystyle y$, and $\displaystyle z$ form equilateral triangle . I found line to be $\displaystyle 6+r$. The height of the triangle would be $\displaystyle 15$, because the radius of circle is 10, which would make the side length of the triangle $\displaystyle 10\sqrt{3}$. Triangle EPY is a right triangle. So line segment EP would be $\displaystyle \sqrt{(4+r)^2 - 75}$ . So line segment $\displaystyle XP$ would be $\displaystyle (6+r) + \sqrt{(4+r)^2 - 75}$, which should equal 15. Solving for $\displaystyle r$, i would get 70/13.5. 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 .

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