Two circles are placed in an equilateral triangle as shown in the figure. What…

Let the side of the equilateral triangle be a. Inradius (radius of the larger circle) r = a/(2√3) = a√3/6.

• Triangle altitude h = a√3/2. The distance from the apex to the center of the larger circle is h − r = a√3/2 − a√3/6 = a√3/3.

• The common tangent through the top of the larger circle lies a distance r below the apex, so the small top triangle has altitude r. Therefore the linear scale between the small triangle and the whole triangle is r/h = (a√3/6)/(a√3/2) = 1/3.

• Hence the radius of the smaller circle is 1/3 of the larger radius: r_small = r/3 = (a√3/6)/3 = a√3/18 = a/(6√3).

• Area of the smaller circle = π r_small^2 = π (a/(6√3))^2 = π a^2 /108.

• Area of the equilateral triangle = (√3/4) a^2.

• Ratio (smaller circle) : (triangle) = (π a^2 /108) : (√3/4 a^2) = π/(27√3).

Therefore the required ratio is π : 27√3.