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Peculiar pentagon (Posted on 2012-11-15) Difficulty: 3 of 5
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?

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 3.0000 (1 votes)

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Solution General Proof Comment 1 of 1
It turns out |AE| = 4|XY| for any 5 points.  They don't have to form a convex pentagon.

Let Ax be the x-coordinate of A etc.  So that A = (Ax,Ay) etc.

The length of AE = sqrt((Ax-Ex) + (Ay-Ey))

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((Xx-Yx)+(Xy-Yy))
= sqrt(((Ax-Ex)/4) + ((Ay-Ey)/4))
= sqrt((Ax-Ex) + (Ay-Ey))/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 2012-11-15 11:09:03
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