Problem 471

Problem 471

Triangle inscribed in ellipse

The triangle ΔABC is inscribed in an ellipse with equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, 0 < 2b < a, a and b integers.

Let r(a,b) be the radius of the incircle of ΔABC when the incircle has center (2b, 0) and A has coordinates $(\frac{a}{2}, \frac{\sqrt{3}}{2}b)$.

For example, r(3,1) = ½, r(6,2) = 1, r(12,3) = 2.

Let $G(n)=\sum_{a=3}^{n}\sum_{b=1}^{\lfloor \frac{a-1}{2} \rfloor}r(a,b)$

You are given G(10) = 20.59722222, G(100) = 19223.60980 (rounded to 10 significant digits).

Find G(1011).

Give your answer in scientific notation rounded to 10 significant digits. Use a lowercase e to separate mantissa and exponent.

For G(10) the answer would have been 2.059722222e1.


三角形ΔABC内接于椭圆$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,其中0 < 2b < a,a和b均为整数。

当三角形ΔABC的内接圆圆心是(2b, 0),且A点坐标是$(\frac{a}{2}, \frac{\sqrt{3}}{2}b)$时,记r(a,b)是三角形ΔABC内接圆的半径。

例如,r(3,1) = ½,r(6,2) = 1,r(12,3) = 2。

记$G(n)=\sum_{a=3}^{n}\sum_{b=1}^{\lfloor \frac{a-1}{2} \rfloor}r(a,b)$

已知G(10) = 20.59722222,G(100) = 19223.60980(保留10位有效数字)