All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Shared Vertex (Posted on 2013-09-25)

Let squares ABCD and AB'C'D' ( both labeled counter-
clockwise ) share vertex A.

Prove that the midpoints of line segments BD, B'D,
B'D', and BD' are the vertices of a square.

 See The Solution Submitted by Bractals No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Coordinatized Solution Comment 1 of 1
Place the square with coordinates
A=(0,0)
B=(1,0)
C=(1,1)
D=(0,1)

rotate by angle x so that
A=(0,0)
B'=(cosx,sinx)
C'=(cosx-sinx,cosx+sinx)
D'=(-sinx,cosx)

call the midpoints:
of BD=(1/2,1/2) = E
of B'D=(cos(x)/2,(1+sin(x))/2) = F
of B'D'=((cos(x)-sin(x))/2,(cos(x)+sin(x))/2) = G
of BD'=((1-sin(x))/2,cos(x)/2) = H

To show a quadrilateral is a square we can show its consecutive sides are perpendicular and its diagonals are perpendicular.

Slope EF = Slope GH = sin(x)/(cos(x)-1)
Slope FG = Slope HE = (1-cos(x),sin(x))

Slope EG = (cos(x)+sin(x)-1)/(cos(x)-sin(x)-1)
Slope FH = (-cos(x)+sin(x)+1)/(cos(x)+sin(x)-1)

QED

 Posted by Jer on 2013-09-27 10:16:15

 Search: Search body:
Forums (0)