Three thin metal rods of lengths $8, 15, 17 \mathrm{~cm}$ were welded to form a triangle, and a solid sphere with a radius of $5 \mathrm{~cm}$ was placed on the horizontally held triangle frame. In what ratio does the plane of the triangle divide the volume of the sphere?
![](https://cdn.mathpix.com/cropped/2024_05_02_d250ce3d6c6511b020eeg-1.jpg?height=499&width=566&top_left_y=119&top_left_x=774)

Solution. Let's choose cm as the unit. Since \(8^{2} + 15^{2} = 17^{2}\), the triangle is a right triangle, with an area of \(t = \frac{8 \cdot 15}{2} = 60\), and a perimeter of \(k = 40\). Based on these, the radius of the inscribed circle is \(r = \frac{2t}{k} = 3\). Therefore, the plane \(S\) of the triangle intersects a sphere with radius \(R = 5\) at a circle with this radius. According to the Pythagorean theorem, the center of the sphere is \(d = 4\) units away from the plane \(S\). The height of the spherical cap cut by \(S\) is thus \(h = R - d = 1\), and the volume of the spherical cap is:

\[
V_{1} = \frac{h(3r^{2} + h^{2})\pi}{6} = \frac{28\pi}{6}
\]

while the volume of the entire sphere is:

\[
V = \frac{4R^{3}\pi}{3} = \frac{1000\pi}{6}
\]

The desired ratio is therefore: \(\frac{V_{1}}{V - V_{1}} = \frac{28}{972} = 0.03\).