内切圆 + 角落 5×10 矩形

设正方形边长为 $s$，则内切圆半径 $r=s/2$，圆心在 $(s/2,s/2)$。

把矩形放在左下角，边平行于正方形边，则矩形远端角点坐标为 $(10,5)$（或 $(5,10)$，结果相同）。该角点恰好在圆上，因此

$$\left(\frac{s}{2}-10\right)^2+\left(\frac{s}{2}-5\right)^2=\left(\frac{s}{2}\right)^2.$$

令 $u=s/2$，得

$$(u-10)^2+(u-5)^2=u^2 \Rightarrow u^2-30u+125=0.$$

解得 $u=25$ 或 $u=5$，但需 $u>10$，所以 $u=25$。

因此 $s=2u=\boxed{50}$。

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英文解析

Let the side length of the square be $s$. Then the radius of the inscribed circle is $r=s/2$, and the center is at $(s/2,s/2)$.

Place the rectangle in the bottom-left corner with sides parallel to the square's sides. The coordinates of the far corner of the rectangle are $(10,5)$ (or $(5,10)$, yielding the same result). This corner lies exactly on the circle, therefore

$$\left(\frac{s}{2}-10\right)^2+\left(\frac{s}{2}-5\right)^2=\left(\frac{s}{2}\right)^2.$$

Let $u=s/2$, yielding

$$(u-10)^2+(u-5)^2=u^2 \Rightarrow u^2-30u+125=0.$$

Solving gives $u=25$ or $u=5$, but we require $u>10$, so $u=25$.

Thus $s=2u=\boxed{50}$.