Around the convex quadrilateral $ABCD$, three rectangles are circumscribed .
It is known that two of these rectangles are squares. Is it true that the third one is necessarily a square?
(A rectangle is circumscribed around the quadrilateral $ABCD$ if there is one vertex $ABCD$ on each side of the rectangle).
This is quite a beautiful problem. Sketch of a solution:
The answer is yes.
First prove that AC is perpendicular to BD, with AC = BD. (This is nontrivial, but I'm not going to write up this part right now, someone else can fill it in if they want.)
Then the problem becomes: given a quadrilateral PQRS with perpendicular diagonals that are of equal length, and a rectangle circumscribed around it, must the rectangle a square? The answer is yes. Suppose the rectangle circumscribing it is JKLM, with P on JK, Q on KL, R on LM, S on MJ. Let PR, KL meet at X. (In my configuration, X and L are on different sides of JK). Let Q' be on MJ with QQ' parallel to ML. Let perpendicular from K to QQ' be Y. And let PR meet QS at Z. RYZQ is cyclic.
Angle RXL = angle YRP = angle SQQ' . So SQQ' is congruent to PRR', where R' is the projection of R onto JK.
Disclaimer: Not sure if this works for all configurations of PQRS in JKLM.