A certain square and a certain equilateral triangle have the same perimeter. The square and triangle are inscribed in circles, as shown below. If $A$ is the area of the circle containing the square, and $B$ is the area of the circle containing the triangle, then find $A/B.$

Guest Apr 14, 2020

#1**+1 **

For the square one, construct two radii to two consectuive vertices.

Let O be the center, and A, B be the vertices.

Let r be the radii of the circle on the left.

\(\angle AOB = 90^\circ\)

By Pythagoras theorem,

\(r^2 + r^2 = AB^2\\ AB = \sqrt2 r\)

Perimeter of square = \(4 \sqrt2 r\)

Do the same thing for the circle on the right, but notice that this time \(\angle AOB = 120^\circ\).

Let R be the radii of the circle on the right.

\(AB^2 = R^2 + R^2 - 2(R)(R)\cos 120^\circ\\ AB = \sqrt3R\)

Perimeter of triangle = \(3\sqrt 3 R\)

If the perimeters are the same, then

\(3\sqrt 3 R = 4\sqrt2 r\\ \dfrac{R}{r} = \dfrac{4\sqrt6}{9} --- (1)\)

We proceed to calculate the areas.

Using the notations in the problem,

\(A = \left(\sqrt2 r\right)^2 = 2r^2\)

\(B = \dfrac{\sqrt3}{4} \left(\sqrt3R\right)^2 = \dfrac{3\sqrt3}{4}R^2\)

By (1), we get \(\dfrac AB = \dfrac{2r^2}{\dfrac{3\sqrt 3}{4}R^2} = \dfrac{8}{3\sqrt3} \left(\dfrac{R}{r}\right)^{-2} = \dfrac{8\sqrt3}{9}\left(\dfrac{27}{32}\right) = \boxed{\dfrac{3\sqrt 3}{4}}\)

.MaxWong Apr 14, 2020