Let BAC be an arbitrary triangle with external squares ABIJ, BCKL, and CAMN. IBLP, KCNQ, and MAJR are parallelograms. Prove that PAQ, QBR, and RCP are isosceles right triangles.

Sorry, guys, this problem is really trivial.  AL is the "sum" of AB and BL and AQ is the "sum" of CQ and AC, each summand of AQ respectively equal and orthogonal to the summands of AL.  This follows from the easy fact that triangles PBL and CNQ are both congruent to ABC ("new theorems"?  Not.).  Hence AL is, of course, equal and orthogonal to AQ.

Posted by McWorter on 2005-10-25 19:55:10