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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, x2 + y2 = 16   [1].

Similarly with the triangle, 4a2 = a2 + 9, so a = radical3.

Hence the area of the triangle is 3radical3.

As the areas are equal, the area of the rectangle, xy = 3radical3,
so x2y2 = 27   [2].

From [1] we get, y2 = 16 minus x2.

Substituting this into [2] we get, x2(16 minus x2) = 27.

Leading to the quartic, x4 minus 16x2 + 27 = 0.

Solving this as a quadratic in x2 we obtain the roots, x2 = 8plus or minusradical37.

Using y2 = 16 minus x2 we observe that y2 = 16 minus (8plus or minusradical37).
That is, when x2 = 8 + radical37, y2 = 8 minus radical37, and vice versa.

Therefore the rectangle has dimensions, radical(8 + radical37) by radical(8 minus radical37).


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?

Problem ID: 63 (Jan 2002)     Difficulty: 3 Star

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