Cyclic quadrilateral ABCD is circumscribed by a unit circle. Let E, F be the orthocenters of ΔBCD, ΔACD respectively.
If ABEF is a square then find the area of the quadrilateral.
There seem to be too many degrees of freedom to get a unique area.
Playing around with Geometer's sketchpad:
If A and B are very close, the triangles are nearly the same isosceles right triangle and the quadrilateral is nearly this same triangle. The area is then very close to 1.
Actually, there are two solutions here: one with the square inside the circle and one with the square partly outside the circle. In the first case, the area is just under 1, in the second its just over 1.
If A and B are separated by 90 degrees of arc, the square becomes inscribed in the circle and so the area is 2.
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Posted by Jer
on 2018-11-10 22:56:35 |
Ill defined problem?