Circles in a Triangle 02

Focus first on the top circle. Let $S$ be the point of tangency to the right side of the triangle and $T$ the point of tangency with the big circle. Also, let $X$ be the point of tangency of the big circle with the base of the triangle.

Now since $\Delta AMP$ is a right triangle, $\angle PAM$ is $30^{\circ}$ and $PM = r = 4$ , we have that $AP = 4\csc(30^{\circ}) = 8$ , and thus $AT = 8 + 4 = 12$ .

Next, since $O$ lies $\frac{2}{3}$ the way along the median from $A$ , we must have that $AT = TO = OX = 12$ , implying that the median length, which is also the height of the equilateral triangle, is $3*12 = 36$ .

The side of the triangle is then

$a = 36\csc(60^{\circ}) = 36*(\frac{2\sqrt{3}}{3}) = 24\sqrt{3}$ ,

and so $\dfrac{a}{\sqrt{3}} = \boxed{24}$ .

I just realized that the measure of the radius of P, Q and R is one-third of that of O, thus taking radius O as 4 3=12.. Since the radius of the biggest circle is one-third of the height of the triangle, we have the height as 12 3=36.. Multiplying by 2 and then dividing by square root of 3 will yield the side of the equilateral triangle=24*(square root of 3).. Upon getting the answer, as we divide square root of three to the result, it yields the answer 24..

We have a=2(sqrt3)R so we need to find 2R. But r=1/3R => R=12 => 2R=a/(sqrt3)=24