Problem 11-2. Mr. $A$ stood at the point with coordinates $(0,0)$ for an hour. During the same hour, moving uniformly and in a straight line, Mr. $B$ traveled from the point $(22,0)$ to the point $(2,20)$. During this same hour, Miss $C$, also moving uniformly and in a straight line, traveled from the point $(30,4)$ to the point $(0,24)$. How many times during the observation period did the area of triangle $A B C$ take on integer values? The initial and final moments are included.
Answer: 53.

Solution. Informally speaking, the problem in this task is not to find a path of calculations leading to the answer; but to find a path to the answer that passes through not too many intermediate calculations. Let's show how to do this.

As is known, the area of a triangle formed by vectors $\overrightarrow{\left(x_{1}, y_{1}\right)}$ and $\overrightarrow{\left(x_{2}, y_{2}\right)}$ is $\left|\frac{1}{2}\left(x_{1} y_{2}-x_{2} y_{1}\right)\right|$. If the point $\overrightarrow{(x, y)}$ moves uniformly and in a straight line, then its coordinates depend on time $t$ as $x=x_{0}+x^{\prime} t$, $y=y_{0}+y^{\prime} t$. Then the value $f(t)=\frac{1}{2}\left(x_{1} y_{2}-x_{2} y_{1}\right)$ as a function of $t$ is a polynomial of the second degree. Let's choose the time axis so that zero is in the middle of the observation interval, and the initial and final moments have coordinates -1 and 1, respectively. Let $f(t)=a x^{2}+b x+c$. The coordinates of points $B$ and $C$ in the middle of the observation interval are easily calculated, these are $(12,10)$ and $(15,14)$, respectively. Thus, $f(-1)=\frac{1}{2}(22 \cdot 4-30 \cdot 0)=$ 44, $f(0)=\frac{1}{2}(12 \cdot 14-15 \cdot 10)=9, f(1)=\frac{1}{2}(2 \cdot 24-0 \cdot 20)=24$. From which
$c=f(0)=9, b=\frac{f(1)-f(-1)}{2}=-10, a=\frac{f(1)+f(-1)-2 f(0)}{2}=25$. The minimum of the quadratic trinomial is reached at the point $\frac{-b}{2 a}=\frac{1}{5} ; f\left(\frac{1}{5}\right)=8$. Thus, we see that the minimum of the expression $\frac{1}{2}\left(x_{1} y_{2}-x_{2} y_{1}\right)$ fell within the observation interval, moreover, it is positive, that is, the modulus is equal to the expression under the modulus. In total, the area first decreased from 44 to 8, then increased from 8 to 24, thus taking integer values $1+(44-8)+(24-8)=53$ times.

| Rating | Score | Content of the criterion |
| :---: | :---: | :--- |
| + | 20 | Full solution |
| .+ | 17 | Only an arithmetic error. Ignoring the second passage of the same integer area. |
| $+/ 2$ | 16 | The formula for $S(\mathrm{t})\left(100 t^{2}-120 t+44\right.$ or  $44-2 t+t^{2} / 36$ or $\left.t^{2} / 4-6 t+44\right)$, or $2 S$ is correctly found, further  reasoning is absent |
| $-/+$ | 10 | The idea of continuity, giving only a lower bound  (the values at the initial and final moments are found,  the difference between them is calculated) or an arithmetic  error in calculating the formula for $S(t)$ |
| - | 0 | The solution is completely incorrect/only the answer |