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Incongruency (Posted on 2007-03-23) Difficulty: 4 of 5
It is easy to divide a square into four congruent isosceles triangles, just draw both diagonals. But can you divide a square into four incongruent isosceles triangles?

See The Solution Submitted by Brian Smith    
Rating: 4.0000 (1 votes)

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Solution proof | Comment 3 of 4 |
(In reply to using geometer's sketchpad by Charlie)

Most of the construction is the same and the final picture is the same, but the proof is easier with a change to the final construction:

From the top left going clockwise, label the corners of the square A,B,C,D.

Draw a diagonal from A to C. The first isosceles triangle is in the upper right-- triangle ABC.

From A, lay off along the diagonal a length equal to one side of the square (call it length 1, for a unit square), and label the point E. The second isosceles triangle is AED.

The remaining length of the diagonal, EC, is sqrt(2) - 1. 

Bisect angle DAC, and label the intersection of the bisector with CD, F. And DAC is 45 degrees, with sine = 1/sqrt(2) and cosine = 1/sqrt(2). By half-angle formula tan(x/2) = (1 - cos x) / sin x. This comes out to sqrt(2) - 1.  By the symmetry of AEFD (which can be proved by making an additional point where the bisector cuts ED and showing congruent triangles), EF is also sqrt(2) -1. That shows that both triangles ECF and EFD are isosceles.

Edited on March 23, 2007, 3:10 pm
  Posted by Charlie on 2007-03-23 15:06:34

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