Here’s another brute-force number theory puzzle, in the spirit of the Russian puzzle I looked at a couple weeks ago. From what would Martin Gardner tweet: “Two triangles share one angle but have six different side lengths (all whole numbers between 1 and 9). Find those lengths.”

Do we expect such triangles to exist? Well, clearly there are lots of different-sized right triangles, which share an angle. To take the smallest possible example, 3-4-5 and 6-8-10 are both the sides of right triangles, and they share an angle. But sadly, 10 is too large…

This is a bit tricky to think about, because the angles of an integer-sided triangle aren’t going to be anything nice. So instead of looking at the angles themselves let’s look at their cosines. Given a triangle with sides having lengths (a, b, c), the angle C, opposite the side of length c satisfies

which we can rearrange to give

This is just the law of cosines from “high school trigonometry”, which is a generalization of the Pythagorean theorem to non-right-angled triangles.

So now the search for a solution is just bookkeeping. Let’s look at all the possible triplets (a, b, c) of distinct integers that could be the sides of a triangle. For the sake of keeping things canonical, let’s force . The triangle inequality says that , or , so we get . Furthermore we can see that *a* must be at least 4, since each of those four inequalities is strict. So we loop over all those possible triples and store them in a Python dictionary according to the cosines of their angles. The following code will ask for an upper bound on the side lengths and carry out the search.

#Import some useful data structures. from collections import defaultdict from sets import Set #Greatest common divisor (by the Euclidean algorithm). For reducing fractions. def gcd(a,b): while b: a, b = b, a%b return a #Takes as input the three sides of a triangle; #Outputs red_num, red_den such that red_num/red_den is the cosine of the angle opposite the first side given #as a fraction in lowest terms def cosine_law(c, a, b): raw_num = (-c**2 + a**2 + b**2) raw_den = (2*a*b) g = gcd(raw_num, raw_den) red_num = raw_num/g red_den = raw_den/g return red_num, red_den out = defaultdict(set) maxn = int(raw_input("Largest side to allow? ")) + 1 #Loop over possible triples a, b, c with a > b > c > a-b #(i. e. in decreasing order and satisfying the triangle inequality) #and add the cosines of their angles to the dictionary out for a in range(4, maxn): for b in range(-int(-a/2), a): for c in range(a-b+1, b): if gcd(gcd(a,b),c) == 1: cosA = cosine_law(a,b,c) cosB = cosine_law(b,a,c) cosC = cosine_law(c,a,b) #store with vertex first, then larger of two second out[cosA].add((a,b,c)) out[cosB].add((b,a,c)) out[cosC].add((c,a,b)) #Look through out for pairs of triangles which share an angle #and have all three sides different. numsols = 0 for key in out: if len(out[key]) >= 2: triangles = list(out[key]) for i in triangles: for j in triangles: sides = Set([i[0], i[1], i[2], j[0], j[1], j[2]]) if len(sides) == 6 and i[0] >= j[0]: #print key, i, j numsols = numsols+1

And if we input 9 when it asks for that largest side the output is precisely

`(11, 16) (6, 8, 7) (3, 4, 2)`

The 2-tuple is the rational number 11/16, the cosine of the angle that these triangles have in common. The triplets represent triangles with an angle , with the side opposite that angle listed first. So the angle opposite the side of length 6 in a 6-7-8 triangle and the angle opposite the side of length 3 in a 2-3-4 triangle both have cosine $11/16$; they are both angles of $\cos^{-1} 11/16$, or about 46.6 degrees.

This leads naturally to the question: are such pairs of triangles rare? The answer is “not at all”. There are 30 solutions to this problem with all sides under 15, 951 with all sides under 30, and 81,014 with all sides under 100.

In fact, heuristically, we expect lots of such solutions. The number of different integer triangles with all their sides less than n is on the order of n^{3}. From the law of cosines, the angles of these triangle will have cosines that are rational numbers with numerator and denominator on the order of n^{2}; there are n^{4} such different numbers. We can wave our hands and say the magic words birthday problem; since the number of possible cosines is much greater than the *square root* of the number of angles we need to find cosines for, if everything happens “at random” there ought to be lots of collisions — and indeed there are.

Reblogged this on lava kafle kathmandu nepal <a href="https://plus.google.com/102726194262702292606" rel="publisher">Google+</a>.

Typo: a > b + c is degenerate. =)

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