What is the smallest perimeter of an integer sided right triangle one of whose angles is less than one degree?

A quick attempt led me to a triangle (115, 6612, 6613) with perimeter 13340, its smallest angle is arctan(115/6612) < 1°. I am not quite sure it is the required triangle, but I promise to look deeper to this interesting problem later tomorrow.

* * * * * *

Coming back after a double-check of my yesterday's attempt to post a quick solution, I found Fred's answer confirming the result. Thanks Fred!

Initially I recalled Euclid's formulas for Pythagorean triangles:

a = m² - n², b = 2mn, c = m² + n² (m > n - natural)

If m and n are sufficiently large, but very close, the leg a will be much shorter than b, in order to have a/b < tan 1°, or denoting the ratio r = m/n:

(m² - n²) / (2mn) = ((m/n)² - 1)/ (2(m/n)) = (r² - 1)/ (2r) < tan 1°,

or r² - 2r*tan 1° - 1 < 0, or 0 < r < tan 1° + √(tan² 1° + 1) = 1.017607...

Taking m and n differing by 1, the quick calculation showed

58/57 < 1.017607... < 57/56, so m = 58, n = 57,

a = 115, b = 6612, c = 6613 and perimeter 13340.

If m - n > 1, the inequalities above require larger integers m and n and lead to triangles with greater perimeter.

Duke's answer is the right one.

Because to minimize the perimeter with a given (approximate) shape,

in this case, with one very small angle,

you can't do better than to have the two longest sides differ by 1.

And all such IRTs are of the form:

a = 2n + 1

b = ½(a² - 1) = 2n² + 2n = n(a + 1)

c = ½(a² + 1) = 2n² + 2n + 1 = b + 1

where n is any positive integer.

The perimeter is then:

p = a + b + c = 4n² + 6n + 2 = 2(2n + 1)(n + 1)

So now just look for a case where b/a is close to tan89º = 57.28996..., and because

b/a = n + n/(2n + 1) = n + ½ - O(1/n)

the choice for n is

n = 57

And that gives Duke's triangle,

(a,b,c) = (115, 6612, 6613)

p = 13340