A rectangle and an equilateral triangle with equal area are inscribed in a circle with radius, 2.
Find the dimensions of the rectangle.
By drawing a diagonal in the rectangle, we form a diameter. In the case of the triangle, we use the geometrical result which states that the centre of an inscribed equilateral triangle is 1/3 the height of the triangle. Hence the radius is 2/3 the height of the triangle, and we deduce that the height of the triangle is 3.
Applying the Pythagorean theorem to the rectangle, x2 + y2 = 16 .
Similarly with the triangle, 4a2 = a2 + 9, so a = 3.
Hence the area of the triangle is 33.
As the areas are equal, the area of the rectangle, xy = 33,
so x2y2 = 27 .
From  we get, y2 = 16 x2.
Substituting this into  we get, x2(16 x2) = 27.
Leading to the quartic, x4 16x2 + 27 = 0.
Solving this as a quadratic in x2 we obtain the roots, x2 = 837.
Using y2 = 16 x2 we observe that y2 = 16 (837).
That is, when x2 = 8 + 37, y2 = 8 37, and vice versa.
Therefore the rectangle has dimensions, (8 + 37) by (8 37).
Can you generalise for a circle with radius, r?
Find the side length of the triangle if the rectangle measures, x by y.
What if the triangle has side length, a?