Find the ratio of the volume of a cone circumscribed around a regular triangular pyramid

Find the ratio of the volume of a cone circumscribed around a regular triangular pyramid to the volume of a cone inscribed in this pyramid.

We find the height h of an equilateral triangle lying at the base of the pyramid (the sides are equal to a):

h = √ (a ^ 2 – (a / 2) ^ 2) = a√3 / 2.

The radius of the circumscribed circle R is equal to 2/3 of the height of the triangle at the base:

R = (h / 3) * 2 = ((a√3 / 2) / 3) * 2 = a√3 / 3.

The radius of the inscribed circle r is equal to 1/3 of the height of the triangle at the base:

r = h / 3 = (a√3 / 2) / 3 = a√3 / 6.

Base areas of cones:

SR = nR ^ 2 = n * (a√3 / 3) ^ 2 = n * (a ^ 2 * 3) / 9 = (n * a ^ 2) / 3.

Sr = nr ^ 2 = n * (a√3 / 6) ^ 2 = n * a ^ 2 * 3/36 = (n * a ^ 2) / 12.

The volumes of cones of the same height are related as the areas of their bases:

VR / Vr = SR / Sr = ((n * a ^ 2) / 3) / ((n * a ^ 2) / 12) = 12/3 = 4.

Answer. The volume of the described cone is 4 times the volume of the inscribed cone.