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.

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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