Is there a way other than guess and check to figure it out?
Hi Ilikebugs,
Yes, you can solve this systematically. Let $X$ be the center of the circle which passes through points $M$, $O$, and $P$. Show that $\triangle MXO\cong \triangle OXN$. Argue that $m\angle XON = 30º$. Draw an altitude of $\triangle XON$ from vertex $X$ to side $ON$. Say $Y$ is the point of intersection of the altidtude with side $ON$. Then consider the right triangle $\triangle OXY$ and solve the hypotenuse $OX$, which is the radius you want.
Let's begin by constructing the line segment $\overline{MN}$. since $\overline{OM}=\overline{ON}=R$, we know $\triangle MON$ is isosceles, and so we can label $\angle OMN=\angle ONM=\theta$:
Given that the sum of the interior angles in a triangle is $180^{\circ}$, we may state:
$ \displaystyle 2\theta+60^{\circ}=180^{\circ}$
Solving this, we find:
$ \displaystyle \theta=60^{\circ}$
And so we know $\triangle MON$ is equilateral:
Next, let's decompose $\triangle MON$ into 3 congruent isosceles triangles sharing a central common vertex:
Now, using the formula for the area of a triangle, we may state:
$ \displaystyle 3\cdot\frac{1}{2}Rr\sin\left(30^{\circ}\right)=\frac{1}{2}R^2\sin\left(60^{\circ}\right)$
Can you now solve for $r$?
Well, what are $\sin\left(30^{\circ}\right)$ and $\sin\left(60^{\circ}\right)$?
Put away the calculator! This will help in the long run... Memorize the sine and cosine functions for the angles 0, 30, 45, 60, and 90 degrees. It will save you a lot of time and grief.
sin(30) = 1/2 and sin(60) = sqrt(3) / 2.
Put these into your equation for r. Can you solve it now?
-Dan