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 $A/B.$
+9519 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 \(\dfrac AB = \dfrac{2r^2}{\dfrac{3\sqrt 3}{4}R^2} = \dfrac{8}{3\sqrt3} \left(\dfrac{R}{r}\right)^{-2} = \dfrac{8\sqrt3}{9}\left(\dfrac{27}{32}\right) = \boxed{\dfrac{3\sqrt 3}{4}}\)
