The Pythagorean Diagram problem

My Pythagorean problem was very popular. 33600 Triangles -- Mark Thompson, Michael Reid, Patrick Hamlyn. Bill Daly -- 5376 Triangles. 1344 Triangles -- Joe Logic, Al Zimmermann, Livio Zucca. Only Joseph DeVincentis sent me the construction below, with 336 triangles (1/2, 2/3, 5/6). Can anyone prove this is minimal, or find a construction with fewer congruent triangles? Definitely not a trivial problem.

**material added august 24th**

Roger Phillips found a 4-piece dissection of the Pythagorean Figure into a 7x8 rectangle. On the topic of dissections, a French site by Alain Rousseau has beautiful animations of famous dissections.

**material added august 28th**

The Pythagorean Figure in 56 triangles by Huang and Reid.

The 13 didoms -- the ways to put two 1-2 triangles together.

Livio Zucca's didom doggie

For those that want to try generalizing the problem (how many
triangles for the 5-12-13 figure?), two identities may be useful (n>m)
**Tan[ 1/2 ArcTan[(2 m n)/(n ^{2} - m^{2})] ] = m/n**

For integers m and n, {n

Michael Reid has found a dissection with a different triangle.

Brendan Owen and Roel Huisman both found that
two sets of didoms + triangles **can** cover the Pythagorean figure.
I've been playing with the pieces, and think they are a lot of fun.
So, I'm making 16 of them as packages.

Brendan Owen solved one of my idle questions:
Can two sets of didoms make a dodecagon? The answer is **YES**!
For the extraordinary solution, again visit
Livio
Zucca's site.

Roel Huismen shows that a 7x22 rectangle can be made with the DSP set and two squares.

Rectangle by Roel Huisman

Wei-Hwa Huang found that the 7-24-25 triangle can be outlined with
1-2 triangles (see above).

The mathematical recreations community has voted that the interesting
new polyform will be call **polydoms**. Polyformist Andrew
Clarke coined the term. A few weeks of work has revealed a lot
about these pieces. Two sets of didoms can make a dodecagon.
Add 4 doms, and you can make the 3-4-5 triangle figure. Roel Huisman
surprised me by using the same set to make a rectangle.

Roger Phillips studied the general problem:
`Haven't got as far as trying any tilings yet, but I can prove that
for many (including the 5-12-13) it's not possible.`

`First, a definition: for the right triangle (n^2 - m^2), 2nm, (n^2+m^2)
let's call the n,m right triangle the "root" triangle. Thus for every complex
number z = n + im with integer coefficients n > m > 0, z represents the
root triangle forming an angle A with the positive real axis, and z^2 =
(n^2 - m^2) + i2nm represents the larger triangle, whose angle is 2A.`

`The area of the large triangle is nm(n^2 - m^2) = 2(n^2 - m^2)(area
of root triangle), so the root area always divides the large triangle area.
To tile the Pythagorean figure with the root triangle, we need its area
to divide the three squares, which have a combined area of 2(n^2 + m^2)^2.
So we need the following to be an integer:`
` 2(n^2 + m^2)^2/(nm/2)`
` = 4(n^4 + 2n^2m^2 + m^4)/(nm)`
` = 8nm + 4(n^4 + m^4)/(nm)`
`So we need nm to divide`
` 4(n^4 + m^4)`
` = 4(n - m)^4 + 4(n^3m - 6n^2m^2 + 4nm^3)`
`Since nm divides the second term, we need nm to divide 4(n - m)^4.`

`For the 3-4-5 triangle, (n,m) = (2,1) and 2 does divide 4. But this
is the only triangle whose root is of the form (m + 1, m) (ie 3-4-5, 5-12-13,
7-24-25, ...) which can be tiled by its root triangle, as for these we
need m(m + 1) to divide 4, which only has an integer solution for m = 1.`
`For triangles with root triangles of the form (m + 2, m) (ie 8-6-10,
12-16-20, 16-30-34, ...) we need m(m + 2) to divide 64, which (noting that
all factors of 64 are powers of 2) only has a solution when m = 2.`
`For n - m = 3, we need m(m + 3) to divide 324, which has solutions
for m = 1 and 3.`
`--`
`Roger Phillips`