# nayuki/Project-Euler-solutions

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 # # Solution to Project Euler problem 587 # Copyright (c) Project Nayuki. All rights reserved. # # https://www.nayuki.io/page/project-euler-solutions # https://github.com/nayuki/Project-Euler-solutions # import itertools, math # Start by defining the coordinate system in a convenient way. The position and scale of the diagram don't # matter because we only care about the ratio of areas, not the absolute areas. So, let the bottom left # of the diagram be the origin (x = 0, y = 0), and let each circle to have a radius of 1. # # The leftmost circle is centered at (1, 1), and its equation is (x - 1)^2 + (y - 1)^2 = 1. # The diagonal line has slope = s = 1 / n (for any positive n), and the line's equation is y = s * x. # From basic geometry, the area of the blue L-section is 1 - pi / 4. # # Let's find the x-coordinate where the diagonal line intersects the first circle. # Take the equation of the circle and substitute y = s * x for the line: # # (x - 1)^2 + (s*x - 1)^2 = 1. # (x^2 - 2x + 1) + (s^2 x^2 - 2s*x + 1) = 1. # (1 + s^2)x^2 + (-2 - 2s)x + 1 = 0. # # We can apply the quadratic formula with a = 1 + s^2, b = -2 - 2s, c = 1. There are two solutions for x, # and we only want the smaller value. Thus, let X = (-b - sqrt(b^2 - 4ac)) / (2a). Or equivalently # with more numerical stability (using the Citardauq formula), X = (2c) / (-b + sqrt(b^2 - 4ac)). # # The orange concave triangle can be divided into two parts by a vertical line: # # - The left part is a proper triangle, whose area is easily seen as x * y / 2 = X^2 * s / 2. # # - The right part is the region between the circle and the baseline. Let's re-express # the circle's equation in terms of y, and only keep the lower semicircle: # # (x - 1)^2 + (y - 1)^2 = 1. # (y - 1)^2 = 1 - (x - 1)^2. # y - 1 = -sqrt(1 - (x - 1)^2). # y = 1 - sqrt(1 - (x - 1)^2). # y = 1 - sqrt(1 - (x^2 - 2x + 1)). # y = 1 - sqrt(2x - x^2). # # Now, the indefinite integral of f(x) = 1 - sqrt(2x - x^2) with respect to x # is F(x) = (x - 1) - [sqrt(2x - x^2) * (x - 1) + asin(x - 1)] / 2. # Finding this integral is not obvious, but verifying it is a fairly straightforward # mechanical procedure involving differentiation and simplification. # # The area of the right part is the integral of f(x) for x from X to 1, because the start is # the x-coordinate where line meets the circle, and the end is where the circle meets the baseline. # Hence the area is equal to F(1) - F(X). # # All in all, for any given n, the area of the orange concave triangle is X^2 * s / 2 + F(1) - F(X). # The rest of the algorithm is a brute-force search with n = 1, 2, 3, ... until the ratio condition is met. # # Additional notes: # - Intuitively, as n increases and the slope gets smaller, the area of the orange concave triangle should strictly # decrease. This statement is in fact true, but proving it involves a big pile of differentiation and algebra. # 0. We need to show that X (which is the x-coordinate of the line-circle intersection) increases with n. # We'd differentiate X with respect to n, and get an expression that is always positive for any positive n. # 1. Because X increases with n, the area of the right part, with its always-positive integrand, must decrease. # 2. As for the left part, we'd differentiate X^2 * s / 2 with respect to n, and get a huge messy formula. # It turns out this formula is negative for all n > 1. Hence the area of this triangle also decreases with n. # After we prove that increasing n leads to decreasing orange area, we could use # binary search to find the minimum value of n needed to meet the ratio requirement. # - The use of floating-point arithmetic, for basic arithmetic operations (+ - * /) and irrational functions (sqrt, # asin) alike, is inherently difficult or impossible to prove the correctness of. Furthermore, the algorithms # for irrational functions are hard to understand and beyond the scope of this problem, and the error bounds for # all operations are difficult to reason about. # It should be possible to solve this particular problem using only integer arithmetic in a provably correct way. # The basic idea would be to round the result of each operation both down and up to an integer fraction, # keep track of pessimistic intervals that are guaranteed to contain the true value, accept a comparison only # if the intervals don't overlap, and recompute everything at a higher precision if a comparison is inconclusive. # Note: Because it doesn't seem easy to compute pi and asin(), it might be better to # approximate integrals directly using the Darboux definition of lower and upper sums. def compute(): # The indefinite integral of (1 - sqrt(2x - x^2)) dx. def integral(x): t = x - 1.0 return t - (math.sqrt(x * (2.0 - x)) * t + math.asin(t)) / 2.0 lsectionarea = 1.0 - math.pi / 4.0 for i in itertools.count(1): slope = 1.0 / i a = slope**2 + 1.0 b = -2.0 * (slope + 1.0) c = 1.0 x = (2.0 * c) / (-b + math.sqrt(b * b - 4 * a * c)) concavetrianglearea = (x**2 * slope / 2) + (integral(1.0) - integral(x)) if concavetrianglearea / lsectionarea < 0.001: return str(i) if __name__ == "__main__": print(compute())