Square ABCD has side length 24.
Point P is inside the square so that
AP = DP = 13.
What is the length of PC?
Triangle APD is isosceles. Its line of symmetry intersects AD at X and BC at Y.
PXD is a right triangle so XD=12, XP=5, and YP=19.
PYC is also a right triangle so PC^2=19^2+12^2=505
PC=sqrt(505)
Incidentally, if you consider P' on the outside of the square, YP=29 and PC=sqrt(985). Both 505 and 985 are semiprimes.
Edited on January 14, 2018, 12:15 pm

Posted by Jer
on 20180113 15:15:16 