Find the radius of the sphere with respect to R and r

Question

The figure above shows a straight cone of base radius $R$ in which a sphere is inscribed that intersects the side of the cone in a circle of radius $r$. Point $C$ is the center of the base of the cone and tangent to the sphere. Points $A$ and $B$ are located on the same generatrix of the cone, where $A$ belongs to the circle of radius $r$ and $B$ to the circle of radius $R$.

Based on this hypothetical situation, judge the following items. The radius of the sphere is $\dfrac{R\sqrt r}{\sqrt{2R-r}}$

(S: True)

I try:

$CB = CA = R$

$\triangle VAO \sim \triangle VCB \sim \triangle VTA$

$\dfrac{VO+re}{VT}=\dfrac{r}{R}=\dfrac{VA}{VA+R}$

$\dfrac{VO}{VA+R}=\dfrac{re}{R}=\dfrac{VA}{VO+re}$

$\dfrac{r}{re}=\dfrac{VA}{VO}=\dfrac{VT}{VA}$

Answer 1

You've made a great start. As shown in the diagram below, draw from $A$ to meet $CB$ perpendicularly at $D$, and let $\lvert AD\rvert = h$.

Thus, $\lvert DB\rvert = R - r$. Using the Pythagorean theorem with $\triangle ADB$ gives that

$$\begin{equation}\begin{aligned} h^2 + (R - r)^2 & = R^2 \\ h^2 + R^2 - 2rR + r^2 & = R^2 \\ h^2 & = 2rR - r^2 \\ h & = \sqrt{r(2R - r)} \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Since $\triangle VTA \sim \triangle ADB$, then

$$\frac{\lvert VA\rvert}{\lvert VT\rvert} = \frac{R}{h} \tag{2}\label{eq2A}$$

Using the reciprocal of your final result, and \eqref{eq1A}, we get from \eqref{eq2A} that

$$\begin{equation}\begin{aligned} \frac{re}{r} & = \frac{R}{h} \\ re & = \frac{rR}{\sqrt{r(2R - r)}} \\ re & = \frac{R\sqrt{r}}{\sqrt{2R - r}} \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

Answer 2

From $\triangle VTA \sim \triangle VCB$,

$$\begin{align*} \frac{VA}{VA+R} &= \frac{r}{R}\\ VA\cdot R &= r(VA+R)\\ VA(R-r) &= rR\\ VA &= \frac{rR}{R-r} \end{align*}$$

From $\triangle VAO \sim \triangle VCB$,

$$\begin{align*} \frac{VO+re}{VA} &= \frac{R}{re}\\ \frac{VO}{VA} &= \frac{R}{re} - \frac{re}{VA}\\ &= \frac{R}{re} - \frac{re(R-r)}{rR}\\ \end{align*}$$

From $\triangle ATO \sim \triangle VAO$,

$$\begin{align*} \frac{re}{VO} &= \frac{r}{VA}\\ re &= r \frac{VO}{VA}\\ &= \frac{rR}{re} - \frac{re(R-r)}{R}\\ re\left(1+\frac{R-r}{R}\right) &= \frac{rR}{re}\\ (re)^2 &= \frac{rR^2}{2R-r}\\ re &= \frac{R\sqrt r}{\sqrt{2R-r}} \end{align*}$$