In the diagram, the circles with centres $B$ and $D$ have radii 1 and $r$, respectively, and are tangent at $C$. The line through $A$ and $D$ passes through $B$. The line through $A$ and $S$ is tangent to the circles with centres $B$ and $D$ at $P$ and $Q$, respectively. The line through $A$ and $T$ is also tangent to both circles. Line segment $S T$ is perpendicular to $A D$ at $C$ and is tangent to both circles at $C$.

![](https://cdn.mathpix.com/cropped/2024_04_17_bceb232fd49f7f0368beg-4.jpg?height=331&width=656&top_left_y=507&top_left_x=710)

(a) There is a value of $r$ for which $A S=S T=A T$. Determine this value of $r$.

(b) There is a value of $r$ for which $D Q=Q P$. Determine this value of $r$.

(c) A third circle, with centre $O$, passes through $A, S$ and $T$, and intersects the circle with centre $D$ at points $V$ and $W$. There is a value of $r$ for which $O V$ is perpendicular to $D V$. Determine this value of $r$.

![](https://cdn.mathpix.com/cropped/2024_04_17_bceb232fd49f7f0368beg-6.jpg?height=189&width=208&top_left_y=242&top_left_x=1593)
(a) Since $A S=S T=A T$, then $\triangle A S T$ is equilateral.

This means that $\angle T A S=\angle A S T=\angle A T S=60^{\circ}$.

Join $B$ to $P, B$ to $S, D$ to $Q$ and $D$ to $S$.

![](https://cdn.mathpix.com/cropped/2024_04_30_2d11aa317d3213e95c24g-07.jpg?height=345&width=569&top_left_y=337&top_left_x=884)

Since $A S$ is tangent to the circle with centre $B$ at $P$, then $B P$ is perpendicular to $P S$. Since $B P$ and $B C$ are radii of the circle with centre $B$, then $B P=B C=1$.

Consider $\triangle S B P$ and $\triangle S B C$.

Each is right-angled (at $P$ and $C$ ), they have a common hypotenuse $B S$, and equal side lengths $(B P=B C)$.

This means that $\triangle S B P$ and $\triangle S B C$ are congruent.

Thus, $\angle P S B=\angle C S B=\frac{1}{2} \angle A S T=30^{\circ}$.

This means that $\triangle S B C$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, and so $S C=\sqrt{3} B C=\sqrt{3}$.

Since $\angle C S Q=180^{\circ}-\angle C S P=180^{\circ}-60^{\circ}=120^{\circ}$, then using a similar argument we can see that $\triangle D S C$ is also a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

This means that $C D=\sqrt{3} S C=\sqrt{3} \cdot \sqrt{3}=3$.

Since $C D$ is a radius of the circle with centre $D$, then $r=C D=3$.

(b) Solution 1

From the given information, $D Q=Q P=r$.

Again, join $B$ to $P, B$ to $S, D$ to $Q$, and $D$ to $S$.

As in (a), $\triangle S B P$ and $\triangle S B C$ are congruent which means that $S P=S C$.

Using a similar argument, $\triangle S D C$ is congruent to $\triangle S D Q$.

This means that $S C=S Q$.

Since $S P=S C$ and $S C=S Q$, then $S P=S Q$.

Since $Q P=r$, then $S P=S Q=\frac{1}{2} r$.

Suppose that $\angle P S C=2 \theta$.

Since $\triangle S B P$ and $\triangle S B C$ are congruent, then $\angle P S B=\angle C S B=\frac{1}{2} \angle P S C=\theta$.

Since $\angle Q S C=180^{\circ}-\angle P S C=180^{\circ}-2 \theta$, then $\angle Q S D=\angle C S D=\frac{1}{2} \angle Q S C=90^{\circ}-\theta$.

Since $\triangle S D Q$ is right-angled at $Q$, then $\angle S D Q=90^{\circ}-\angle Q S D=\theta$.

This means that $\triangle S B P$ is similar to $\triangle D S Q$.

Therefore, $\frac{S P}{B P}=\frac{D Q}{S Q}$ and so $\frac{\frac{1}{2} r}{1}=\frac{r}{\frac{1}{2} r}=2$, which gives $\frac{1}{2} r=2$ and so $r=4$.

Solution 2

From the given information, $D Q=Q P=r$.

Join $B$ to $P$ and $D$ to $Q$. As in (a), $B P$ and $D Q$ are perpendicular to $P Q$. Join $B$ to $F$ on $Q D$ so that $B F$ is perpendicular to $Q D$.

![](https://cdn.mathpix.com/cropped/2024_04_30_2d11aa317d3213e95c24g-08.jpg?height=417&width=610&top_left_y=388&top_left_x=866)

This means that $\triangle B F D$ is right-angled at $F$.

Also, since $B P Q F$ has three right angles, then it must have four right angles and so is a rectangle.

Thus, $B F=P Q=r$ and $Q F=P B=1$.

Since $Q D=r$, then $F D=r-1$.

Also, $B D=B C+C D=1+r$.

Using the Pythagorean Theorem in $\triangle B F D$, we obtain the following equivalent equations:

$$
\begin{aligned}
B F^{2}+F D^{2} & =B D^{2} \\
r^{2}+(r-1)^{2} & =(r+1)^{2} \\
r^{2}+r^{2}-2 r+1 & =r^{2}+2 r+1 \\
r^{2} & =4 r
\end{aligned}
$$

Since $r \neq 0$, then it must be the case that $r=4$.
(c) As in Solution 1 to (b), $\triangle S B P$ is similar to $\triangle D S Q$ and $S P=S Q$. Therefore, $\frac{S P}{B P}=\frac{D Q}{S Q}$ or $\frac{S P}{1}=\frac{r}{S P}$ which gives $S P^{2}=r$ and so $S P=\sqrt{r}$.

Thus, $S P=S Q=S C=\sqrt{r}$.

Next, $\triangle A P B$ is similar to $\triangle A Q D$ (common angle at $A$, right angle).

Therefore, $\frac{A B}{B P}=\frac{A D}{D Q}$ and so $\frac{A B}{1}=\frac{A B+B D}{r}$ and so $A B=\frac{A B+1+r}{r}$.

Re-arranging gives $r A B=A B+1+r$ and so $(r-1) A B=r+1$ and so $A B=\frac{r+1}{r-1}$.

This means that $A C=A B+B C=A B+1=\frac{r+1}{r-1}+1=\frac{(r+1)+(r-1)}{r-1}=\frac{2 r}{r-1}$.

Next, draw the circle with centre $O$ that passes through $A, S$ and $T$ and through point $V$ on the circle with centre $D$ so that $O V$ is perpendicular to $D V$.

![](https://cdn.mathpix.com/cropped/2024_04_30_2d11aa317d3213e95c24g-09.jpg?height=564&width=745&top_left_y=889&top_left_x=793)

Let the radius of this circle be $R$. Note that $O S=A O=R$.

Consider $\triangle O S C$.

This triangle is right-angled at $C$.

Using the Pythagorean Theorem, we obtain the following equivalent equations:

$$
\begin{aligned}
O S^{2} & =O C^{2}+S C^{2} \\
R^{2} & =(A C-A O)^{2}+S C^{2} \\
R^{2} & =(A C-R)^{2}+S C^{2} \\
R^{2} & =A C^{2}-2 R \cdot A C+R^{2}+S C^{2} \\
2 R \cdot A C & =A C^{2}+S C^{2} \\
R & =\frac{A C}{2}+\frac{S C^{2}}{2 A C} \\
R & =\frac{2 r}{2(r-1)}+\frac{(\sqrt{r})^{2}}{4 r /(r-1)} \\
R & =\frac{r}{r-1}+\frac{r-1}{4}
\end{aligned}
$$

Since $O V$ is perpendicular to $D V$, then $\triangle O V D$ is right-angled at $V$.

Using the Pythagorean Theorem, noting that $O V=R$ and $D V=r$, we obtain the following equivalent equations:

$$
\begin{aligned}
O V^{2}+D V^{2} & =O D^{2} \\
R^{2}+r^{2} & =(O C+C D)^{2} \\
R^{2}+r^{2} & =(A C-A O+C D)^{2} \\
R^{2}+r^{2} & =\left(\frac{2 r}{r-1}-R+r\right)^{2} \\
R^{2}+r^{2} & =\left(\frac{2 r+r(r-1)}{r-1}-R\right)^{2} \\
R^{2}+r^{2} & =\left(\frac{r^{2}+r}{r-1}-R\right)^{2} \\
R^{2}+r^{2} & =\left(\frac{r^{2}+r}{r-1}\right)^{2}-2 R\left(\frac{r^{2}+r}{r-1}\right)+R^{2} \\
2 R\left(\frac{r^{2}+r}{r-1}\right) & =\left(\frac{r^{2}+r}{r-1}\right)^{2}-r^{2} \\
2 R\left(\frac{r(r+1)}{r-1}\right) & =\frac{r^{2}(r+1)^{2}}{(r-1)^{2}}-r^{2} \\
2 R & =\frac{r-1}{r(r+1)} \cdot \frac{r^{2}(r+1)^{2}}{(r-1)^{2}}-\frac{r-1}{r(r+1)} \cdot r^{2} \\
2 R & =\frac{r(r+1)}{r-1}-\frac{r(r-1)}{r+1}
\end{aligned}
$$

Since $R=\frac{r}{r-1}+\frac{r-1}{4}$, we obtain:

$$
\frac{2 r}{r-1}+\frac{r-1}{2}=\frac{r(r+1)}{r-1}-\frac{r(r-1)}{r+1}
$$

Multiplying both sides by $2(r+1)(r-1)$, expanding, simplifying, and factoring, we obtain the following equivalent equations:

$$
\begin{aligned}
4 r(r+1)+(r-1)^{2}(r+1) & =2 r(r+1)^{2}-2 r(r-1)^{2} \\
\left(4 r^{2}+4 r\right)+(r-1)\left(r^{2}-1\right) & =2 r\left((r+1)^{2}-(r-1)^{2}\right) \\
\left(4 r^{2}+4 r\right)+\left(r^{3}-r^{2}-r+1\right) & =2 r\left(\left(r^{2}+2 r+1\right)-\left(r^{2}-2 r+1\right)\right) \\
\left(4 r^{2}+4 r\right)+\left(r^{3}-r^{2}-r+1\right) & =2 r(4 r) \\
r^{3}-5 r^{2}+3 r+1 & =0 \\
(r-1)\left(r^{2}-4 r-1\right) & =0
\end{aligned}
$$

Now $r \neq 1$. (If $r=1$, the circles would be the same size and the two common tangents would be parallel.)

Therefore, $r \neq 1$ which means that $r^{2}-4 r-1=0$.

By the quadratic formula,

$$
r=\frac{4 \pm \sqrt{(-4)^{2}-4(1)(-1)}}{2}=\frac{4 \pm \sqrt{20}}{2}=2 \pm \sqrt{5}
$$

Since $r>1$, then $r=2+\sqrt{5}$.