Given a right triangle with integer sides, show that there is another integer-sided scalene triangle with a 60 degree angle and 1.5 times the perimeter of the given triangle.

Isabel posted the solution, but it involved getting several examples and making conjectures from them about the scalene-60 triple. Since I dislike the production of conjectures from data, and strongly prefer derivations from first principles, here is my detailed solution:

Consider scalene-60 triangles, there is one angle bigger than 60 degrees and one smaller. (Total is 180)

Say the side opposite 60 is z.

Let the smaller one be x=z-a, and the bigger one be y=z+b where both a and b are positive.

From triangle inequalities, you get

Then using cosine rule: x

^{2}+ y

^{2}- xy = z

^{2}

Plug stuff in and simplify like crazy,

z(a-b) = (a

^{2}+ b

^{2}+ ab)

If a=b, then this becomes a=b=0, the equilateral triangle.

As the RHS is strictly non-zero, a>b.

Simplifying, z = (a-b) + 3ab/(a-b)

(z-a, z, z+b) generates all such tuples! The only condition on a,b is that (a-b) must divide 3ab.

Now (a-b) divides a if and only if (a-b) divides b.

(Proof: x divides a and x divides a-b, therefore x must divide b, now set x = a-b)

So we can scale a and b by (a-b) and get another triple.

In other words, it is enough to consider the two cases: a-b = 1, and a-b = 3.

Actually, we have parametrized all possible scalene-60 triples:

With a-b=3, we get one sequence and another with a-b=1

As an aside, this is an interesting picture:

If you draw a 60 degree angle and mark off 8 on one arm, use that endpoint to draw an arc of radius 7, it will intersect the other arm at 2 places -- at 3 and at 5. The angles correspond to arccos(1/7) and 180-arccos(1/7)

Coming back to the puzzle, all right triangles can be parametrized as (m

^{2}- n

^{2}, 2mn, m

^{2}+ n

^{2}) or a scalar multiple thereof, where m and n are co-prime and have opposite parity.

The perimeter is 2m(m+n).

1.5 times perimeter is 3m(m+n)

Can we map this to the scalene-60s?

Let us see. Try the first sequence: a-b=3

gives us z = b

^{2}+ 3b + 3 for triangle (z-a, z, z-b)

with perimeter 3z-3 = 3(z-1) = 3(b+2)(b+1)

Looks similar to what we want!

Now it is just a matter of fiddling with the definitions of (m,n) and (z,a,b) to get the common parametric form:

Right triple is (m

^{2}- n

^{2}, 2mn, m

^{2}+ n

^{2})

Scalene-60 is (m

^{2}- n

^{2}, m

^{2}+ mn + n

^{2}, m

^{2}+ 2mn)

[Substitute b=(m-n)/n; and scale the triple (z,a,b) by n]

We are home and have a sequence of scalene-60 triangles left over!

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