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If you take a circle with a horizontal diameter and mark off any two points on the circumference above the diameter, then these two points together with the end points of the diameter form the vertices of a cyclic quadrilateral with the diameter as one of the sides. We refer to the quadrilaterals in question as diametric. In this note we consider diametric quadrilaterals having two equal sides. They can be reduced to the two forms (a, b, a, d) (an isosceles trapezoid) and (b, a, a, d) (a skewed kite). It is shown that these quadrilaterals are diametric if and only if (r, a, d) is a right triangle satisfying d > √2a where r = √(d2 − a2), in which case the area of either quadrilateral is (b+d)/4 √(d2 − b2), where b = d − 2a2/d. The integer case (Brahmaguta quadrilaterals) is then considered. It is shown that each primitive Pythagorean triple (t, u, v) with v > √2u determines a Brahmagupta diametric quadrilateral with two equal sides uv, two other sides v2 − 2u2, v2, and area ut3. Moreover every primitive diametric quadrilateral with integer sides, two of which are equal, and with integer diagonals, arises in this way.
The College Mathematics Journal is designed to enhance classroom learning and stimulate thinking regarding undergraduate mathematics. CMJ publishes articles, short Classroom Capsules, problems, solutions, media reviews and other pieces. All are aimed at the college mathematics curriculum with emphasis on topics taught in the first two years.