Solution to Problem #6
Al Zimmermann of White Plains NY, Philippe Fondanaiche of Paris (France),
and Robin Stokes of the University of New England (Australia) solved the
problem. The basic idea was to use Brahmagupta's formula for the area of a
cyclic quadrilateral:
A =
Sqrt[(s - a)(s - b)(s -
c)(s -
d)].
The resulting solutions were:
(4, 4, 4, 4) [a square]
(6, 6, 3, 3) [a kite]
(6, 3, 6, 3) [a rectangle]
(8, 5, 5, 2) [irregular]
(8, 5, 2, 5) [an isosceles trapezoid]
(14, 6, 5, 5) [irregular]
(14, 5, 6, 5) [an isosceles trapezoid]
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