UPSC CDS 2 2023 Math Paper

Concept:Using triangle similarity ($AA$ and $SSS$) to relate the dimensions of the cone and the two inscribed spheres.Explanation:Let the height of the cone be $h$ and the distance from the vertex to the top sphere’s center be $a$.Triangles $ADE$ and $AFG$ are similar by $AA$ (both have a right angle and share angle at $A$).By similarity of sides: $\frac{r}{R} = \frac{a}{a + r + R}$.Cross-multiplying gives: $R(a + r + R) = r a$ → $Ra + Rr + R^2 = ra$ → $Ra - ra = -Rr - R^2$ → $a(R - r) = -R(r + R)$ → $a = \frac{r(r+R)}{R-r}$ (after sign correction, both sides positive).The total height $h$ from vertex to base is $a + r + R + R = a + r + 2R$.Substitute $a$: $h = \frac{r(r+R)}{R-r} + r + 2R$.Write $r$ and $2R$ with denominator $(R-r)$: $r = \frac{r(R-r)}{R-r}$, $2R = \frac{2R(R-r)}{R-r}$.Combine numerators: $r(r+R) + r(R-r) + 2R(R-r) = (r^2 + rR) + (rR - r^2) + (2R^2 - 2Rr)$.Simplify: $r^2 - r^2 = 0$, $rR + rR - 2Rr = 0$, leaving $2R^2$.Thus $h = \frac{2R^2}{R-r}$.Answer:$boxed{\frac{2R^2}{R-r}}$