Bunuel wrote:

Within equilateral triangle A, circle B is inscribed; within circle B, square C is inscribed; within square C, circle D is inscribed; within circle D, square E is inscribed; within square E, circle F is inscribed; finally, within circle F, equilateral triangle G is inscribed.

What is the ratio of the area of triangle A to the area of triangle G?

(A) 4 √2 : 1

(B) 8 : 1

(C) 8 √2 : 1

(D) 16 : 1

(E) 16 √2 : 1

Let's start with equilateral triangle G.

Triangle G is inscribed within circle F or circle F is the circumcircle of triangle G.

circumradius of a circumcircle of a triangle is given by \(\frac{abc}{4 \triangle}\) , where abc is the product of sides of a triangle and \(\triangle\) is the area of triangle.

let the side of triangle G be \(x\)

then the area of triangle G will be \(\frac{\sqrt3}{4}\) * \(x^2\)

then the circumradius of triangle G or radius of circle F will be \(\frac{x^3}{4* \frac{\sqrt3}{4} * x^2}\) = \(\frac{x}{\sqrt3}\) = \(\frac{\sqrt3x}{3}\)

Circle F is inscribed within square E, so the side of square will be equal to diameter of circle F = \(2\frac{\sqrt3x}{3}\)

The diagonal of a square is given by \(\sqrt2\) * side of square. Diagonal of square E will be \(2*\sqrt{2}*\frac{\sqrt3x}{3}\) = \(2\frac{\sqrt6x}{3}\)

This diagonal will be equal to diameter of circle D which circumscribes square E and the diameter of circle D will be equal to the side of square C in which circle D is inscribed.

So the side of square C will be \(2\frac{\sqrt6x}{3}\)

Diagonal of square C will be \(\sqrt2\)*\(2*\frac{\sqrt6x}{3}\) = \(4\frac{\sqrt3x}{3}\)

Diagonal of square C will be equal to diameter of Circle B

The radius of circle circle B will be \(2\frac{\sqrt3x}{3}\) ....(1)

Circle B is the incircle of triangle A.

Inradius of a circle is given by \(\frac{\triangle}{S}\) where \(\triangle\) is the area of triangle and S is semi perimeter = 1/2(sum of sides of triangle.)

Let y be the side of traingle A,

the area of triangle A is \(\frac{\sqrt3}{4}\) * \(y^2\)

and semi perimeter is \(3y/2\)

Inradius or radius of circle B will be \(\frac{\frac{\sqrt3}{4}y^2}{\frac{3y}{2}}\) = \(\frac{\sqrt3}{6}y\) ....(2)

(1) = (2) =radius of circle B

or \(2\frac{\sqrt3x}{3}\) = \(\frac{\sqrt3}{6}y\)

or 4x = y

or \(\frac{y}{x}=4\)

Ratio of area of triangle A and area of triangle G will be equal to ratio of square of their sides.

or \(\frac{y^2}{x^2}=4^2=16\)

So the ratio will be \(16:1\)

Answer:- D