[Geometry]
Problem: Three thin metal rods of lengths 9,12 and 15 are welded together to form a right-angled triangle, which is held in a horizontal position. A solid sphere of radius 5 rests in the triangle so that it is tangent to each of the three sides. Assuming that the thickness of the rods can be neglected, how high above the plane of the triangle is the top of the sphere?

Solution: Consider the cross-section of the sphere in the plane defined by the triangle. This crosssection will be a circle, since any cross-section of a sphere is a circle. This circle will be tangent to the three sides of the triangle, ie. will be the inscribed circle (or incircle) of the triangle. Let the centre of this circle be $O$, and its radius be $r$. We calculate the value of $r$.

<img_3449>

Join $O$ to the three points of tangency, $P, Q, R$, and to the three vertices $A, B, C$. Then $O P, O Q$ and $O R$ (radii) will form right angles with the three sides of the triangle. Consider the three triangles $\triangle A O B$, $\triangle B O C$ and $\triangle C O A$. Each of these triangles has a height of $r$ and they have bases 15, 9 and 12, respectively. Since the area of $\triangle A B C$ is equal to the sum of the areas of $\triangle A O B, \triangle B O C$, and $\triangle C O A$, So comparing areas,

$$
\begin{aligned}
\frac{1}{2}(9)(12) & =\frac{1}{2}(9)(r)+\frac{1}{2}(12)(r)+\frac{1}{2}(15)(r) \\
54 & =\frac{1}{2} r(9+12+15) \\
r & =3
\end{aligned}
$$

<img_3572>

Now join the centre of the cross-sectional circle to the centre of the sphere and let this distance be $h$. Now, the line joining the centre of the circle to the centre of the sphere will be perpendicular to the plane of the triangle, so we can form a right-angled triangle by joining the centre of the sphere to any point on the circumference of the cross-sectional circle. By Pythagoras,

$$
\begin{aligned}
h^{2}+r^{2} & =25 \\
h & =4
\end{aligned}
$$

This tells us that the top of the sphere is 9 units above the plane of the triangle, since the top of the sphere is 5 units above the centre of the sphere.

Final answer: 5