 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Diagonal mystery (Posted on 2019-05-04) 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?

 No Solution Yet Submitted by Danish Ahmed Khan No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Solution Comment 2 of 2 | 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.

 Posted by Brian Smith on 2019-05-05 15:41:20 Please log in:

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