# nayuki/Project-Euler-solutions

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 # # Solution to Project Euler problem 94 # Copyright (c) Project Nayuki. All rights reserved. # # https://www.nayuki.io/page/project-euler-solutions # https://github.com/nayuki/Project-Euler-solutions # import eulerlib, fractions, itertools # Consider an arbitrary almost equilateral triangle with side lengths (c, c, c +/- 1). # Split it down the middle to get a right triangle, and label the new sides. # /\ /| # c / \ c c / | b # / \ --> / | # -------- ----- # c +/- 1 a # Note that a = (c +/- 1) / 2, and a^2 + b^2 = c^2 (Pythagorean theorem). # # We know that c is an integer. The area of the original triangle is a*b, # which is an integer by definition from the problem statement. # - If a is an integer, then b is an integer (so that a*b is an integer), # thus (a,b,c) is a Pythagorean triple. # - Otherwise a is an integer plus a half, then b must be even, # but a^2 + b^2 is not an integer, which contradicts c being an integer. # # Conversely, consider an arbitrary Pythagorean triple (a,b,c). # If 2a = c +/- 1, then we can form an almost equilateral triangle: # /|\ # c / | \ c # / | \ # --------- # 2a # For this to happen, the Pythagorean triple must be primitive. Because if not, # then a = 0 mod k and c = 0 mod k for some k > 1, which means 2a = 0 mod k which # cannot equal c +/- 1 = +/- 1 mod k. So we only need to generate primitive triples. # # Pythagorean triples theorem: # Every primitive Pythagorean triple with a odd and b even can be expressed as # a = st, b = (s^2-t^2)/2, c = (s^2+t^2)/2, where s > t > 0 are coprime odd integers. def compute(): LIMIT = 10**9 ans = 0 # What search range do we need? # c = (s^2+t^2)/2. Perimeter = p = 3c +/- 1 = 3/2 (s^2+t^2) +/- 1 <= LIMIT. # We need to keep the smaller perimeter within limit for # the search to be meaningful, so 3/2 (s^2+t^2) - 1 <= LIMIT. # With t < s, we have that s^2+t^2 < 2s^2, so 3/2 (s^2+t^2) - 1 < 3s^2 - 1. # Therefore it is sufficient to ensure that 3s^2 - 1 <= LIMIT, i.e. s^2 <= (LIMIT+1)/3. for s in itertools.count(1, 2): if s * s > (LIMIT + 1) // 3: break for t in range(s - 2, 0, -2): if fractions.gcd(s, t) == 1: a = s * t b = (s * s - t * t) // 2 c = (s * s + t * t) // 2 if a * 2 == c - 1: p = c * 3 - 1 if p <= LIMIT: ans += p if a * 2 == c + 1: p = c * 3 + 1 if p <= LIMIT: ans += p # Swap the roles of a and b and try the same tests # Note that a != b, since otherwise c = a * sqrt(2) would be irrational if b * 2 == c - 1: p = c * 3 - 1 if p <= LIMIT: ans += p if b * 2 == c + 1: p = c * 3 + 1 if p <= LIMIT: ans += p return str(ans) if __name__ == "__main__": print(compute())