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?
Tomarken established part (i)
Note triangles AXY and DXY are congruent, as they are reflections.
Call CY=x by similar triangles BX=36/x
AY=DY=15x
AX=DX=1536/x
So using the similar triangles again
x/3 = (15x)/(1536/x)
Solving gives x=4.5
so
AY = 10.5
AX = 7
And using the law of cosines XY˛=85.75
XY = √85.75 ≈ 9.260129589

Posted by Jer
on 20140306 15:34:49 