Let ABCDE be a convex pentagon in which the sides have length 1, 2, 3, 4, and 5, but not necessarily in that order. Let F, G, H, I be the midpoints of AB, BC, CD, DE, respectively. Then let X be the midpoint of FH and Y the midpoint of GI, and suppose that the length XY is an integer. How long is AE?
It turns out AE = 4XY for any 5 points. They don't have to form a convex pentagon.
Let Ax be the xcoordinate of A etc. So that A = (Ax,Ay) etc.
The length of AE = sqrt((AxEx)² + (AyEy)²)
Applying all the midpoints
F = ((Ax+Bx)/2) , (Ay+By)/2)
G = ((Bx+Cx)/2) , (By+Cy)/2)
H = ((Cx+Dx)/2) , (Cy+Dy)/2)
I = ((Dx+Ey)/2) , (Dy+Ey)/2)
and then
X = ((Ax+Bx+Cx+Dx)/4 , (Ay+By+Cy+Dy)/4)
Y = ((Bx+Cx+Dx+Ex)/4 , (By+Cy+Dy+Ey)/4)
The length XY = sqrt((XxYx)²+(XyYy)²)
= sqrt(((AxEx)/4)² + ((AyEy)/4)²)
= sqrt((AxEx)² + (AyEy)²)/4
Which is precisely 1/4 of AE.
So if XY is to be an integer, AE = 4 so that XY = 1.

Posted by Jer
on 20121115 11:09:03 