# 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 *) (* * 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. *) LSectionArea = 1 - Pi / 4; (* The indefinite integral of (1 - sqrt(2x - x^2)) dx. *) TheIntegral[x_] := Block[{t = x - 1}, t - (Sqrt[x * (2 - x)] * t + ArcSin[t]) / 2] ConcaveTriangleArea[i_] := Block[{ slope = 1 / i, a = slope^2 + 1, b = -2 * (slope + 1), c = 1, x = (2 * c) / (-b + Sqrt[b^2 - 4 * a * c])}, (x^2 * slope / 2) + (TheIntegral[1] - TheIntegral[x])] i = 1; While[ConcaveTriangleArea[i] / LSectionArea >= 1 / 1000, i++] i