Evdokimov M.A.

16 cards with integers from 1 to 16 are laid face down in a $4 \times 4$ table so that cards with consecutive numbers are adjacent (touching by a side). What is the minimum number of cards that need to be flipped simultaneously to definitely determine the location of all numbers (regardless of how the cards are arranged)?
Evaluation. Let's number the cells as shown in Figure 1.

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

Fig. 1

![](https://cdn.mathpix.com/cropped/2024_05_06_afad11de590b368f242eg-08.jpg?height=412&width=237&top_left_y=1249&top_left_x=908)

Fig. 2

![](https://cdn.mathpix.com/cropped/2024_05_06_afad11de590b368f242eg-08.jpg?height=398&width=235&top_left_y=1253&top_left_x=1249)

Fig. 3

Notice that one of the cells numbered 1 must be open, otherwise the red and blue ways of filling the table in Figure 2 would be indistinguishable. One of the cells numbered 2 must also be open, otherwise the red and blue ways of filling the table in Figure 3 would be indistinguishable.

Similarly, at least one of the cells numbered 3, 4, 5, 6, 7, 8 must be open, meaning that at least 8 cards must be open.

Example. We will prove that by seeing the numbers in the second and third columns, we can restore the numbers in the first and fourth columns. Notice that in the black cells of the chessboard coloring, all numbers have the same parity, and in the white cells, the other parity. By seeing the second and third columns, we understand the parity of the number in each cell.

From the open cells, we highlight those for which the number written in the cell does not have all neighboring numbers open. From each such cell, we draw an edge to the only unopened neighboring cell and uniquely restore the number in it.

Notice that if an edge leads to a corner, we will restore the number in it. If no edge leads to a corner cell, then it contains the extreme number, i.e., 1 or 16, and since we know the parity of the number in each cell, we can also restore the number in the corner. Thus, the numbers in the corners are definitely restored.

If there are corner cells for which not all neighboring numbers are open, we draw an edge from each such corner to the unopened neighboring cell and uniquely restore the number in it.

The numbers in the non-corner cells of the first and fourth columns may remain un-restored. Consider any of them. No edge leads to it from the neighboring column or from the corner, so in this cell, there is definitely an extreme number (since it has at most one neighboring cell with a number). By parity, it is easy to determine which extreme number should be there.

Thus, we have restored the numbers in all cells.

## [ Lengths of sides, heights, medians, and bisectors ] Problem $\underline{67255}$ Topics: [ Triangle inequality (other) $]$

Difficulty: $3+$

Grades: 8,9

Author: Bakayev E.V.

In triangle ABC, medians $\mathrm{BK}$ and $\mathrm{CN}$ intersect at point M. What is the maximum number of sides of quadrilateral ANMK that can have a length of 1?

## Solution

For example, any triangle where $\mathrm{AB}=\mathrm{AC}=2$ works. Let's prove that there cannot be three equal sides.

![](https://cdn.mathpix.com/cropped/2024_05_06_afad11de590b368f242eg-09.jpg?height=443&width=798&top_left_y=705&top_left_x=1)

Method 1. Suppose that at least three sides of quadrilateral ANMK are equal to 1. There are only two fundamentally different cases. 1) $\mathrm{AN}=\mathrm{NM}=\mathrm{MK}=1$. Then $\mathrm{NB}=1, \mathrm{MB}=2$, so $\mathrm{MN}+\mathrm{NB}=\mathrm{MB}$. 2) $\mathrm{AN}=\mathrm{NM}=\mathrm{MK}=1$. Then $\mathrm{AC}=2, \mathrm{NC}=3$, so $\mathrm{NA}+\mathrm{AC}=\mathrm{NC}$. In both cases, a contradiction with the triangle inequality is obtained.

Method 2. If more than two sides of the quadrilateral are equal to 1, then either $\mathrm{AK}=\mathrm{NA}$ or $\mathrm{KM}=\mathrm{MN}$. In the first case, triangle ABC is isosceles. In the second case, $\mathrm{BK}=\mathrm{CN}=3$, so in this case, triangle ABC is also isosceles. From this, it follows that $\mathrm{AK}=\mathrm{KM}=\mathrm{MN}=\mathrm{NA}$, meaning that $\mathrm{AKMN}$ is a rhombus. Contradiction, since lines AK and NM are not parallel.

Method 3. Let $\mathrm{L}$ be the midpoint of side $\mathrm{BC}$. If $\mathrm{NM}=\mathrm{NA}=\mathrm{NB}$, then triangle AMB is a right triangle. Triangle LMK is similar to triangle AMB with a coefficient of 0.5, so MK = LM, and therefore the hypotenuse of triangle AMK is greater than the hypotenuse of triangle LMK, i.e., $\mathrm{AK}>\mathrm{LK}=\mathrm{NA}$. Thus, in quadrilateral ANMK, there are no three sides equal to each other.

The case when KM = KA is similar, and if $KM \neq KA$ and $NM \neq NA$, then among the segments NM, NA, KM, KA, there are also no three equal to each other.

## Answer

2 sides.