Points A, B, C, D are lying on a square of integer side length such that each point divides the corresponding side into integer segments.
If AC=2√26 and BD=2√29, what is the maximum area of the quadrilateral ABCD?
Let the square be WXYZ with A on WX, B on XY, C on YZ, and D on ZW. Let s=WX=XY=YZ=ZW, a=WA, b=XB, c=YC, and d=ZD.
Then the area of ABCD can be expressed as s^2 - s*(a+c)/2 - s*(b+d)/2 + (a+c)*(b+d)/2.
Note that the area formula is dependent only on s and the sums a+c and b+d. Then the exact position of segments AC and BD do not affect the area as long as their slopes relative to the square are consistent.
AC=2*sqrt(26)=sqrt(104) and BD=2*sqrt(29)=sqrt(116). Then to satisfy the integer constraints in the problem, each of 104 and 116 needs to be a sum of two squares with a square in common. Then there is one solution: 104=10^2+2^2 and 116=10^2+4^2.
The sides of the square must be s=10. AC then implies that a+c equals either 8 or 12 and BD implies that b+d equals either 6 or 14.
Then the possible areas of ABCD are:
10^2 - 10*8/2 - 10*6/2 + 8*6/2 = 54
10^2 - 10*8/2 - 10*14/2 + 8*14/2 = 46
10^2 - 10*12/2 - 10*6/2 + 12*6/2 = 46
10^2 - 10*12/2 - 10*14/2 + 12*14/2 = 54
Then the maximum area of ABCD is 54.
Solution
