Is it possible to cut a square into seven isosceles right triangles, no two of which are congruent?
Yes, it is possible.
Draw a square 5sqrt2 units on each side. I labeled mine ABCD starting with the upper left corner and going clockwise.
Draw the diagonal AC. Draw point E on DC such that DE = 2sqrt2 units long and EC = 3sqrt2 units long. Draw point F on AC such that EF is perpendicular to DC. EF is 3sqrt2 units long.
Draw point G on AD such that AFG is a right triangle with <F = 90 degrees.
Draw point H on FG such that FHE is a right triangle with <H = 90 degrees.
Draw point I on HE such that DIE is a right triangle with <I = 90 degrees. Draw GI.
Triangles in descending order of size:
ABC: AB=BC=5sqrt2. AC=10.
CEF: CE=EF=3sqrt2. CF=6.
AFG: AF=FG=4. AG=4sqrt2.
EHF: EH=HF=3. EF=3sqrt2.
DIE: DI=IE=2. DE=2sqrt2.
DGI: DG=GI=sqrt2. DI=2.
GHI: GH=HI=1. GI=sqrt2.
changed strange boxes and a's with squigly lines to 'sqrt'.
Edited on February 1, 2005, 2:15 am
Posted by Dustin
on 2005-02-01 02:13:25