## Inscribed Rectangle

#### Problem

A rectangle and an equilateral triangle with equal area are inscribed in a circle with radius, 2.

Find the dimensions of the rectangle.

#### Solution

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, `x`^{2} + `y`^{2} = 16 [1].

Similarly with the triangle, 4`a`^{2} = `a`^{2} + 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 `x`^{2}`y`^{2} = 27 [2].

From [1] we get, `y`^{2} = 16 `x`^{2}.

Substituting this into [2] we get, `x`^{2}(16 `x`^{2}) = 27.

Leading to the quartic, `x`^{4} 16`x`^{2} + 27 = 0.

Solving this as a quadratic in `x`^{2} we obtain the roots, `x`^{2} = 837.

Using `y`^{2} = 16 `x`^{2} we observe that `y`^{2} = 16 (837).

That is, when `x`^{2} = 8 + 37, `y`^{2} = 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`?