Wednesday, January 27, 2010

Fox 236

There should be several purely geometric solutions.

1 comment:

  1. Let M be the midpoint of AB and M1 the midpoint of CD where ABCD is the trapezium whose height is h, A being the bottom left hand corner. Now Tr.CMD=Tr.CMM1+Tr.DMM1=(1/2)(MM1)h/2+(1/2)(MM1)h/2=(1/2)(MM1)h. But MM1=(1/2)(AD+BC). Hence, Tr, CMD =(1/2)(AD+BC)/2*(h)=(1/2)(A(ABCD)).