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