Tokarev S.I.

Can the natural numbers from 1 to 81 be written in the cells of a $9 \times 9$ table so that the sum of the numbers in each $3 \times 3$ square is the same?
Table $T$, depicted in the left figure, contains each of the numbers $1,2, \ldots, 9$ nine times and has the property that the sum of the numbers in each $3 \times 3$ square is 36.

| 1 | 4 | 7 | 1 | 4 | 7 | 1 | 4 | 7 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 2 | 5 | 8 | 2 | 5 | 8 | 2 | 5 | 8 |
| 3 | 6 | 9 | 3 | 6 | 9 | 3 | 6 | 9 |
| 7 | 1 | 4 | 7 | 1 | 4 | 7 | 1 | 4 |
| 8 | 2 | 5 | 8 | 2 | 5 | 8 | 2 | 5 |
| 9 | 3 | 6 | 9 | 3 | 6 | 9 | 3 | 6 |
| 4 | 7 | 1 | 4 | 7 | 1 | 4 | 7 | 1 |
| 5 | 8 | 2 | 5 | 8 | 2 | 5 | 8 | 2 |
| 6 | 9 | 3 | 6 | 9 | 3 | 6 | 9 | 3 |

| 7 | 8 | 9 | 4 | 5 | 6 | 1 | 2 | 3 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 4 | 5 | 6 | 1 | 2 | 3 | 7 | 8 | 9 |
| 1 | 2 | 3 | 7 | 8 | 9 | 4 | 5 | 6 |
| 7 | 8 | 9 | 4 | 5 | 6 | 1 | 2 | 3 |
| 4 | 5 | 6 | 1 | 2 | 3 | 7 | 8 | 9 |
| 1 | 2 | 3 | 7 | 8 | 9 | 4 | 5 | 6 |
| 7 | 8 | 9 | 4 | 5 | 6 | 1 | 2 | 3 |
| 4 | 5 | 6 | 1 | 2 | 3 | 7 | 8 | 9 |
| 1 | 2 | 3 | 7 | 8 | 9 | 4 | 5 | 6 |

It is clear that the table $Q$, obtained by rotating it by $90^{\circ}$ (right figure), also has the specified properties. Notice now that the same numbers in table $T$ correspond to different numbers in table $Q$, so there is a one-to-one correspondence between the cells of the $9 \times 9$ table and pairs $(a, b)$, where $a, b=1,2, \ldots, 9$. If we now take a new $9 \times 9$ table and in its cell corresponding to the pair $(a, b)$, write the number $9a + b - 9$, then each of the numbers from 1 to 81 will appear in this table exactly once, and the sum of the numbers in each $3 \times 3$ square will be the same (it is equal to $9 \cdot 45 + 45 - 9 \cdot 9 = 369$, see the figure).

| 7 | 35 | 63 | 4 | 32 | 60 | 1 | 29 | 57 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 13 | 41 | 69 | 10 | 38 | 66 | 16 | 44 | 72 |
| 19 | 47 | 75 | 25 | 53 | 81 | 27 | 50 | 78 |
| 61 | 8 | 36 | 58 | 5 | 33 | 55 | 2 | 30 |
| 67 | 14 | 42 | 64 | 11 | 39 | 70 | 17 | 45 |
| 73 | 20 | 48 | 79 | 26 | 54 | 76 | 23 | 51 |
| 34 | 62 | 9 | 31 | 59 | 6 | 28 | 56 | 3 |
| 40 | 68 | 15 | 37 | 65 | 12 | 43 | 71 | 18 |
| 46 | 74 | 21 | 52 | 80 | 22 | 49 | 77 | 24 |

## Answer

![](https://cdn.mathpix.com/cropped/2024_05_06_0bf3a2f278e52693111ag-12.jpg?height=226&width=1475&top_left_y=150&top_left_x=11)

Author: Dvoryaninov S.

A point spotlight located at vertex $B$ of an equilateral triangle $ABC$ illuminates an angle $\alpha$. Find all such values of $\alpha$, not exceeding $60^{\circ}$, such that for any position of the spotlight, when the illuminated angle is entirely within angle $ABC$, a triangle can be formed from the illuminated and two unilluminated segments of side $AC$.

## Solution

Let $\alpha \leq 30^{\circ}$. Take points $K$ and $L$ on the base $AC$ such that $\angle KBC = \angle LBA = \alpha$; $T$ is the midpoint of segment $AC$ (right figure). Take a point $R$ on the base such that $KT = RC$. Let $M$ lie at the intersection of segments $RC$ and $LC$. Take a point $N$ such that $\angle NBM = \alpha$. Note that $N$ lies on segment $AK$.

We have $NM > KM = KT + TM = CR + TM > CM + TM = TC = \frac{1}{2}AC$, hence $AN + MC < MN$.

Therefore, a triangle cannot be formed from these segments.

The only remaining possibility is $\alpha = 30^{\circ}$.

Let $MN$ be the illuminated segment, $\angle MBN = \alpha = 30^{\circ}$ (left figure). Clearly, $BN$ is not perpendicular to $AC$. (Otherwise, point $M$ would coincide with $A$ and, thus, the beam would not be inside the triangle.)
![](https://cdn.mathpix.com/cropped/2024_05_06_0bf3a2f278e52693111ag-12.jpg?height=330&width=736&top_left_y=1498&top_left_x=662)

Reflecting point $C$ over line $BN$, we get point $D$, which does not lie on $AC$ (otherwise, $BN \perp AC$) (right figure).

Triangle $BDC$ is isosceles, so $\angle DBN = \angle CBN = x$ and $NC = DN$. Clearly, $\angle MBN = \angle NBM - \angle NBD = 30^{\circ} - x$,

$\angle ABM = \angle ABC - \angle MBN - \angle NBC = 60^{\circ} - 30^{\circ} - x = 30^{\circ} - x$.

Thus, $ABD$ is an isosceles triangle, $BM$ is its bisector, and hence $AM = MD$. In triangle $DMN$, the sides are equal to segments $AM$, $MN$, and $NC$, so a triangle can be formed from them.

## Answer

$\alpha = 30^{\circ}$.