A twenty-filler, a ten-filler, and a one-crown coin are placed next to each other so that each touches the other two. What is the area of the triangle that can be circumscribed around the coins? $\left(r_{1}=21 \mathrm{~mm}, r_{2}=19 \mathrm{~mm}, r_{3}=23 \mathrm{~mm}\right)$.
Let $ABC$ be the sought triangle, and let $O_1O_2O_3$ be the triangle determined by the centers of the given circles, and let $D, D_1, E, E_1, F$, and $F_1$ be the points of tangency. The sought area is then

$$
T=ADO_1D_1 + BEO_2E_1 + CFO_3F_1 + DEO_2O_1 + E_1F_1O_3O_2 + FD_1O_1O_3 + O_1O_2O_3
$$

But

$$
ADO_1D_1 = 2ADO_1 = r_1 \cdot AD = r_1^2 \operatorname{ctg} \frac{A}{2}
$$

similarly

$$
BEO_2E_1 = r_2^2 \operatorname{ctg} \frac{B}{2} \text{ and } CFO_3F_1 = r_3^2 \operatorname{ctg} \frac{C}{2} \text{. }
$$

Furthermore,

$$
DEO_2O_1 = \frac{r_1 + r_2}{2} \sqrt{(r_1 + r_2)^2 - (r_1 - r_2)^2} = (r_1 + r_2) \sqrt{r_1 r_2}
$$

likewise

$$
E_1F_1O_3O_2 = (r_2 + r_3) \sqrt{(r_2 r_3)}(r_1 + r_2) \sqrt{r_1 r_2} \quad \text{ and } \quad FD_1O_1O_3 = (r_3 + r_1) \sqrt{(r_1 r_3)}
$$

and finally

$$
O_1O_2O_3 = \sqrt{(r_1 + r_2 + r_3) r_1 r_2 r_3}
$$

thus

$$
\begin{gathered}
T = r_1^2 \operatorname{ctg} \frac{A}{2} + r_2^2 \operatorname{ctg} \frac{B}{2} + r_3^2 \operatorname{ctg} \frac{C}{2} + \\
+ (r_1 + r_2) \sqrt{r_1 r_2} + (r_2 + r_3) \sqrt{r_2 r_3} + (r_3 + r_1) \sqrt{r_1 r_3} + \sqrt{(r_1 + r_2 + r_3) r_1 r_2 r_3}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_d0fb2173137a4e244d73g-1.jpg?height=401&width=593&top_left_y=1114&top_left_x=755)

Now let's determine one of the angles of the $ABC$ triangle, for example, angle $A$. Drop perpendiculars from $O_1$ to $EO_2$ and $FO_3$, thus obtaining points $G$ and $H$.

Then

$$
A \angle = O_2O_1O_3 \angle + O_3O_1H \angle - O_2O_1G \angle,
$$

where the angles on the right-hand side are provided by the formulas

$$
\begin{gathered}
\sin \frac{O_2O_1O_3}{2} = \sqrt{\frac{r_2 r_3}{(r_1 + r_2)(r_1 + r_3)}} \\
\sin O_3O_1H = \frac{r_3 - r_1}{r_3 + r_1} \quad \text{ and } \quad \sin O_2O_1G = \frac{r_1 - r_2}{r_1 + r_2}
\end{gathered}
$$

The angles $B$ and $C$ can also be determined in the same manner.

Substituting the given numerical values, we get

$$
T = 1467.5 \text{ mm}^2
$$

(Kornél Riesz, Budapest.)

The problem was also solved by: Deutsch I., Enyedi B., Haar A., Pivnyik I., Prékopa D., Riesz K., Riesz M.