Yesterday, God Plays Dice posed a scalene triangle puzzle:
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: x2 + y2 - xy = z2
Plug stuff in and simplify like crazy,
z(a-b) = (a2 + b2 + 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 (m2 - n2 , 2mn, m2 + n2 ) 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 = b2 + 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 (m2 - n2, 2mn, m2 + n2)
Scalene-60 is (m2 - n2, m2 + mn + n2, m2 + 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!