Subfield: Geometry


Question: In the diagram, eleven circles of four different radius 1, each circle labelled $X$ has radius 2, the circle labelled $Y$ has radius 4 , and the circle labelled $Z$ has radius $r$. Each of the circles labelled $W$ or $X$ is tangent to three other circles. The circle labelled $Y$ is tangent to all ten of the other circles. The circle labelled $Z$ is tangent to three other circles. Determine positive integers $s$ and $t$ for which $r=\frac{s}{t}$.


![](https://cdn.mathpix.com/cropped/2023_12_21_dff3b889a8199883de5cg-1.jpg?height=346&width=536&top_left_y=857&top_left_x=1250)

Solution:
We label the centres of the outer circles, starting with the circle labelled $Z$ and proceeding clockwise, as $A, B, C, D, E, F, G, H, J$, and $K$, and the centre of the circle labelled $Y$ as $L$.

![](https://cdn.mathpix.com/cropped/2023_12_21_c26842f50f756ae98092g-1.jpg?height=613&width=954&top_left_y=1103&top_left_x=691)

Join $L$ to each of $A, B, C, D, E, F, G, H, J$, and $K$. Join $A$ to $B, B$ to $C, C$ to $D, D$ to $E, E$ to $F, F$ to $G, G$ to $H, H$ to $J, J$ to $K$, and $K$ to $A$.

When two circles are tangent, the distance between their centres equals the sum of their radii.

Thus,

$$
\begin{array}{r}
B C=C D=D E=E F=F G=G H=H J=J K=2+1=3 \\
B L=D L=F L=H L=K L=2+4=6 \\
C L=E L=G L=J L=1+4=5 \\
A B=A K=r+2 \\
A L=r+4
\end{array}
$$

By side-side-side congruence, the following triangles are congruent:

$$
\triangle B L C, \triangle D L C, \triangle D L E, \triangle F L E, \triangle F L G, \triangle H L G, \triangle H L J, \triangle K L J
$$

Similarly, $\triangle A L B$ and $\triangle A L K$ are congruent by side-side-side.

Let $\angle A L B=\theta$ and let $\angle B L C=\alpha$.



By congruent triangles, $\angle A L K=\theta$ and

$$
\angle B L C=\angle D L C=\angle D L E=\angle F L E=\angle F L G=\angle H L G=\angle H L J=\angle K L J=\alpha
$$

The angles around $L$ add to $360^{\circ}$ and so $2 \theta+8 \alpha=360^{\circ}$ which gives $\theta+4 \alpha=180^{\circ}$ and so $\theta=180^{\circ}-4 \alpha$.

Since $\theta=180^{\circ}-4 \alpha$, then $\cos \theta=\cos \left(180^{\circ}-4 \alpha\right)=-\cos 4 \alpha$.

Consider $\triangle A L B$ and $\triangle B L C$.

![](https://cdn.mathpix.com/cropped/2023_12_21_b77e2de70eb587b4b2bfg-1.jpg?height=288&width=464&top_left_y=577&top_left_x=928)

By the cosine law in $\triangle A L B$,

$$
\begin{aligned}
A B^{2} & =A L^{2}+B L^{2}-2 \cdot A L \cdot B L \cdot \cos \theta \\
(r+2)^{2} & =(r+4)^{2}+6^{2}-2(r+4)(6) \cos \theta \\
12(r+4) \cos \theta & =r^{2}+8 r+16+36-r^{2}-4 r-4 \\
\cos \theta & =\frac{4 r+48}{12(r+4)} \\
\cos \theta & =\frac{r+12}{3 r+12}
\end{aligned}
$$

By the cosine law in $\triangle B L C$,

$$
\begin{aligned}
B C^{2} & =B L^{2}+C L^{2}-2 \cdot B L \cdot C L \cdot \cos \alpha \\
3^{2} & =6^{2}+5^{2}-2(6)(5) \cos \alpha \\
60 \cos \alpha & =36+25-9 \\
\cos \alpha & =\frac{52}{60} \\
\cos \alpha & =\frac{13}{15}
\end{aligned}
$$

Since $\cos \alpha=\frac{13}{15}$, then

$$
\begin{aligned}
\cos 2 \alpha & =2 \cos ^{2} \alpha-1 \\
& =2 \cdot \frac{169}{225}-1 \\
& =\frac{338}{225}-\frac{225}{225} \\
& =\frac{113}{225}
\end{aligned}
$$



and

$$
\begin{aligned}
\cos 4 \alpha & =2 \cos ^{2} 2 \alpha-1 \\
& =2 \cdot \frac{113^{2}}{225^{2}}-1 \\
& =\frac{25538}{50625}-\frac{50625}{50625} \\
& =-\frac{25087}{50625}
\end{aligned}
$$

Finally,

$$
\begin{aligned}
\cos \theta & =-\cos 4 \alpha \\
\frac{r+12}{3 r+12} & =\frac{25087}{50625} \\
\frac{r+12}{r+4} & =\frac{25087}{16875} \\
\frac{(r+4)+8}{r+4} & =\frac{25087}{16875} \\
1+\frac{8}{r+4} & =\frac{25087}{16875} \\
\frac{8}{r+4} & =\frac{8212}{16875} \\
\frac{2}{r+4} & =\frac{2053}{16875} \\
\frac{r+4}{2} & =\frac{16875}{2053} \\
r+4 & =\frac{33750}{2053} \\
r & =\frac{25538}{2053}
\end{aligned}
$$

Therefore, the positive integers $s=25538$ and $t=2053$ satisfy the required conditions.

Final answer: $25538$,$2053$