ABCD is a square with point E inside the square.
Line segments AE, BE, and DE are drawn.
The length of AE is 8 and angle AEB is a right angle.
What is the area of triangle AED?
I solved it by assuming that there was a unique answer:
If this is solvable (we have been given enough information) then WLOG we can assume E is at the center of the square so that all four angles at the center between A and B, B and C, C and D, and D and E are right angles. The triangle in question is a right triangle with legs equal to 8, and has an area of 8^2 / 2 = 32.
The same is true when E is coincident with B, which is of course on the semicircle connecting A with B that forms the locus of E when AEB is a right angle.
When is or is approaching A, this can only be a limit as E approaches A, with the requirement for a larger and larger square with a longer and longer but narrower and narrower triangle. I haven't done the calculus, but the assurance that the answer doesn't change, combined with the one or two solutions, gives confidence 32 is correct.

Posted by Charlie
on 20230329 21:52:20 