Problem:
Let P be a point on the circle circumscribing square ABCD that satisfies PA⋅PC=56 and PB⋅PD=90. Find the area of ABCD.
Solution:
Ptolemy's theorem states that for cyclic quadrilateral WXYZ,WX⋅YZ+XY⋅WZ=WY⋅XZ. We may assume that P is between B and C. Let PA=a,PB=b,PC=c,PD=d, and AB=s. We have a2+c2=AC2=2s2, because AC is a diameter of the circle. Similarly, b2+d2=2s2. Therefore, (a+c)2=a2+c2+2ac=2s2+2(56)=2s2+112. Similarly, (b+d)2=2s2+180. By Ptolemy's Theorem on PCDA,as+cs=ds2, and therefore a+c=d2. By Ptolemy's on PBAD,bs+ds=as2, and therefore b+d=a2. By squaring both equations, we obtain
2d2=(a+c)2=2s2+1122a2=(b+d)2=2s2+180
Thus, a2=s2+90, and d2=s2+56. Plugging these values into a2+c2=b2+d2=2s2, we obtain c2=s2−90, and b2=s2−56. Now, we can solve using a and c (though using b and d yields the same solution for s).