A piece of paper has the precise shape of an equilateral triangle ABC which is creased along a line XY, with X on AB and Y on AC. so that A falls on some point D on BC.

(i) Are the triangles XBD and DCY always similar?
If so, prove it. If not, give a counterexample.

(ii) If AB = 15 and BD = 3, what is the length of the crease XY?

Geometry was never my strong suit but I think this works:

(i) Angles XBD, XDY, and DCY are all 60 degrees.  Let angle ADY = n.  Then angles AYD and BDX = (120 - n).  So triangles XBD and DCY are always similar.

(ii) Note AB = BC.  So if AB = 15 and BD = 3, then DC = 12 and CY = 3.  Also angle DCY is 60 degrees, so we can use the law of cosines to compute the length of DY.  We also note that DY = DX and angle XDY is 60 degrees, so XY = DY (triangle XDY is equilateral).  Plugging into the formula gives XY = sqrt(117) ~= 10.82.

 Edit:  I think part i is ok but my answer to part ii is incorrect.  I will revise it shortly.  

Edited on March 5, 2014, 1:29 pm

Edited on March 5, 2014, 1:32 pm

Posted by tomarken on 2014-03-05 13:23:18