We want to arrange the integers $1-12$ in the small circles, squares, and hexagons of the diagram such that the sum of the 6 numbers along each large circle is 33, and the sum of the 4 numbers in each small circle, square, and hexagon is equal. In how many ways is this possible?

![](https://cdn.mathpix.com/cropped/2024_05_02_ff5e7f1fd30093f284e6g-1.jpg?height=599&width=645&top_left_y=297&top_left_x=729)
Let the numbers written in the small circles be $k_{1}, k_{2}, k_{3}, k_{4}$, the first two in the small circles intersected by the two large circles, and $k_{1}$ in the small circle at the center of the third large circle, the last two in the small circles intersected once by the large circle around $k_{1}$. Similarly, let the numbers written in the squares and hexagons be $n_{1}, n_{2}, n_{3}, n_{4}$, and $h_{1}, h_{2}, h_{3}, h_{4}$, respectively, as shown in Figure 1. According to the first requirement of the arrangement,

$$
\left(k_{1}+k_{2}+k_{3}+k_{4}\right)+\left(n_{1}+n_{2}+n_{3}+n_{4}\right)+\left(h_{1}+h_{2}+h_{3}+h_{4}\right)=78
$$

indeed, the sum of the first 12 natural numbers is 78. The sums in each of the parentheses must be equal according to the third requirement, so each sum is 26. Finally, due to the requirement for the large circles,

$$
\begin{aligned}
& \left(k_{1}+k_{2}\right)+\left(n_{1}+n_{2}\right)+\left(h_{3}+h_{4}\right)=33 \\
& \left(k_{1}+k_{2}\right)+\left(h_{1}+h_{2}\right)+\left(n_{3}+n_{4}\right)=33 \\
& \left(n_{1}+n_{2}\right)+\left(h_{1}+h_{2}\right)+\left(k_{3}+k_{4}\right)=33
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_ccc00b1f91502908701ag-1.jpg?height=536&width=637&top_left_y=716&top_left_x=684)

Figure 1

By adding the first two equations, considering the above, we get

$$
2\left(k_{1}+k_{2}\right)+2 \cdot 26=66, \quad k_{1}+k_{2}=7, \quad k_{3}+k_{4}=19
$$

Similarly, by adding the first and third, and the last two equations, it follows that the sum of the 1st and 2nd indexed numbers in the squares and hexagons is 7, and the sum of the 3rd and 4th indexed numbers is 19.

The sum of two different numbers chosen from the required numbers can be 7 and 19 in three different ways:
a)
b)

$$
\begin{array}{ccc}
\text { I. } & \text { II. } & \text { III. } \\
7=1+6 & =2+5= & 3+4 \\
19=7+12 & =8+11= & 9+10
\end{array}
$$

Therefore, the arrangement is possible, for example, as shown in Figure 2.

![](https://cdn.mathpix.com/cropped/2024_05_02_ccc00b1f91502908701ag-1.jpg?height=550&width=686&top_left_y=1793&top_left_x=706)

Any number from 1 to 6 can be written in the place of $k_{1}$; if this is chosen, $k_{2}$ must be $7 - k_{1}$. Any of the remaining 4 numbers can be chosen for $n_{1}$, and $n_{2}$ will then be $7 - n_{1}$, and finally, any of the remaining 2 numbers for $h_{1}$, and the other for $h_{2}$. Thus, the 6 letters can be filled with appropriate numbers in $6 \cdot 4 \cdot 2 = 48$ ways.

Similarly, it can be seen that no matter how we write the numbers at the intersection points of the large circles, the other 6 places can again be filled with the integers from 7 to 12 in 48 ways. The 12 spaces can thus be filled in $48 \cdot 48 = 2304$ ways.