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Problem 904

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Pythagorean Angle

Given a right-angled triangle with integer sides, the smaller angle formed by the two medians drawn on the the two perpendicular sides is denoted by $\theta$.

Let $f(\alpha ,L)$ denote the sum of the sides of the right-angled triangle minimizing the absolute difference between $\theta$ and $\alpha$ among all right-angled triangles with integer sides and hypotenuse not exceeding $L$.

If more than one triangle attains the minimum value, the triangle with the maximum area is chosen. All angles in this problem are measured in degrees.

For example, $f(30,{10}^2)=198$ and $f(10,{10}^6)=1600158$.

Define $F(N,L)=\sum _{n=1}^{N}f(\sqrt[3]{n},L)$.

You are given $F(10,{10}^6)=16684370$.

Find $F(45000,{10}^{10})$.

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毕达哥拉斯角

给定一个各边长为整数的直角三角形，在其两条相互垂直的边上各画一条中线，这两条中线交叉所形成的较小角记为$\theta$。

令$f(\alpha ,L)$表示，在所有各边长为整数且斜边不超过$L$的直角三角形中，使$\theta$与$\alpha$的绝对差值最小的三角形的各边长之和。

如果最小绝对差值对应多个三角形，则选择其中面积最大的一个。本题中的所有角度均以度数表示。

例如，$f(30,{10}^2)=198$，$f(10,{10}^6)=1600158$。

定义$F(N,L)=\sum _{n=1}^{N}f(\sqrt[3]{n},L)$。

已知$F(10,{10}^6)=16684370$。

求$F(45000,{10}^{10})$。

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