4. Students could present another solution to this problem. In such a case, it is important to have a well-argued and clear proof that the point they found is indeed the one sought. Accordingly, the maximum number of points for finding (constructing) and justifying it, depending on the degree of argumentation, is up to 11 points. Calculating its coordinates and the length of segment BA - up to 3 points.

## Task 2 (14 points)

A lawyer and an economist received an odd number of coins of the same denomination (more than four). They can divide the coins between themselves in only one of three ways, each of which is performed in three steps, pursuing their own personal interest (each wants to get more coins):

Step 1: The lawyer divides all the coins into two arbitrary parts, each containing at least two coins.

Step 2: The economist divides each of these parts again into two parts, each containing at least one coin.

The methods differ only in the third step:

Step 3:

Method 1: The economist takes the largest and the smallest parts.

Method 2: The economist takes both middle parts.

Method 3: The economist chooses either Method 1 or Method 2, but in exchange for the right to choose, he gives one coin to the lawyer.

The economist won in the division. Which method did he choose and why?
# Solution:

The economist chose method 1. Let "E" represent the economist, and "L" represent the lawyer. Suppose L divides the coins into parts with a and b coins in each, such that a < b. Then E, pursuing their own interest, divides b into two parts - 1 coin and b-1 coins. These parts become the largest and smallest, respectively. That is, with the first method, E takes the larger part. In this case, it is clear that $a=n, b=n+1$, where $n$ is the number of coins.

The second method of division would give the economist only $\mathrm{n}$ coins. Indeed, $\mathrm{a}=2, \mathrm{~B}=2 \mathrm{n}-1$ on the first step, then the best division on the second step is $2=1+1, 2n-1=n+n-1$.

The third method, when dividing on the first step $\mathrm{a}=\mathrm{n}, \mathrm{b}=\mathrm{n}+1$, gives the option to choose the part with n +1 coins, which, according to the division rules, must be given away.

It should be noted that the "struggle" is essentially over one coin in all three cases.

## Criteria: