6. The game of hockey is popular all over the world. Much in the game depends on the goalkeeper. To hone the skills of goalkeepers and ensure a training process that does not depend on other players, a puck launcher was created. The machine can be set to eject pucks at a specified frequency, speed, and angle.

Let the machine be installed on the ice at a distance of $l=12 \text{ m}$ from the goal. The shots are made in a plane perpendicular to the ice surface and the goal line, with a certain fixed initial speed of the puck $V_{0}$ and at various angles $\alpha$ to the ice surface. Let the point of ejection be the origin of the coordinate system. The x-axis is directed perpendicular to the center line of the hockey rink towards the goal. The y-axis is directed upwards, perpendicular to the ice surface. The goalkeeper has a trap for pucks, depicted by the point $M$. The trajectory of the puck, while in the air and considered as a material point, in the specified coordinate system over time $t$ is described by the equations: $\left\{\begin{array}{l}x=V_{0} t \cos \alpha, \\ y=V_{0} t \sin \alpha-\frac{g t^{2}}{2} .\end{array}\right.$ Determine the range of possible values of the square of the initial ejection speed of the puck, for each of which the puck hits the trap and the maximum possible height of the trap at the moments of catching the puck does not exceed 1 m. For simplification of calculations, assume that the acceleration due to gravity $g=10 \text{ m} / \text{s}^{2}$.
Solution. Express time from the first equation of the system and substitute into the second

$y(x)=\frac{V_{0} x}{V_{0} \cos \alpha} \sin \alpha-\frac{g\left(\frac{x}{V_{0} \cos \alpha}\right)^{2}}{2}=x \operatorname{tg} \alpha-\frac{g}{2}\left(\frac{x}{V_{0} \cos \alpha}\right)^{2} \Rightarrow$ $y(x)=x \operatorname{tg} \alpha-\frac{g}{2} \frac{(x)^{2}}{V_{0}^{2}}\left(1+\operatorname{tg}^{2} \alpha\right) \Rightarrow h=l \operatorname{tg} \alpha-\frac{g}{2} \frac{(l)^{2}}{V_{0}^{2}}\left(1+\operatorname{tg}^{2} \alpha\right)$.

To ensure the puck hits the trap at height $h$, the condition must be met $\frac{g}{2} \frac{l^{2}}{V_{0}^{2}}\left(1+\operatorname{tg}^{2} \alpha\right)-l \operatorname{tg} \alpha+h=0 \Rightarrow \frac{g}{2} \frac{l^{2}}{V_{0}^{2}} \operatorname{tg}^{2} \alpha-l \operatorname{tg} \alpha+h+\frac{g}{2} \frac{l^{2}}{V_{0}^{2}}=0$,

$D=(l)^{2}-4 \frac{g}{2} \frac{l^{2}}{V_{0}^{2}}\left(h+\frac{g}{2} \frac{l^{2}}{V_{0}^{2}}\right)=0 \Rightarrow h_{\max }=\frac{V_{0}^{2}}{2 g}-\frac{g}{2} \frac{l^{2}}{V_{0}^{2}}$.

In this case $\operatorname{tg} \alpha=\frac{l}{g l^{2} / V_{0}^{2}}=\frac{V_{0}^{2}}{g l}$.

This can also be obtained by completing the square,

$$
h=l \operatorname{tg} \alpha-\frac{g}{2} \frac{l^{2}}{V_{0}^{2}}\left(1+\operatorname{tg}^{2} \alpha\right)=-\frac{g}{2} \frac{l^{2}}{V_{0}^{2}}\left(\operatorname{tg} \alpha-\frac{V_{0}^{2}}{g l}\right)^{2}+\frac{V_{0}^{2}}{2 g}-\frac{g}{2} \frac{l^{2}}{V_{0}^{2}} \leq \frac{V_{0}^{2}}{2 g}-\frac{g}{2} \frac{l^{2}}{V_{0}^{2}}
$$

Determine the range of the square of the velocity if $l=12$ m and $0 \leq h_{\max } \leq 1$ :

$$
0 \leq h=\frac{V_{0}^{2}}{2 g}-\frac{g}{2} \frac{144}{V_{0}^{2}} \leq 1 \Rightarrow 0 \leq \frac{V_{0}^{4}-g^{2} \cdot 144}{2 g V_{0}^{2}} \leq 1 \Rightarrow\left\{\begin{array}{l}
V_{0}^{4} \geq 144 g^{2} \\
V_{0}^{4}-2 g V_{0}^{2}-g^{2} \cdot 144 \leq 0
\end{array}\right.
$$

$\left\{\begin{array}{l}V_{0}^{4} \geq 144 g^{2} \\ V_{0}^{4}-2 g V_{0}^{2}-g^{2} \cdot 144 \leq 0\end{array} \Rightarrow\left\{\begin{array}{l}V_{0}^{2} \geq 12 g \\ \left(V_{0}^{2}-g\right)^{2}-g^{2} \cdot 145 \leq 0\end{array} \Rightarrow\right.\right.$

$\left\{\begin{array}{l}V_{0}^{2} \geq 12 g \\ \left(\left(V_{0}^{2}-g\right)-g \cdot \sqrt{145}\right)(\underbrace{\left(V_{0}^{2}-g\right)+g \cdot \sqrt{145}}_{>0}) \leq 0\end{array} \Rightarrow\left\{\begin{array}{l}V_{0}^{2} \geq 12 g \\ V_{0}^{2} \leq g+g \cdot \sqrt{145}\end{array} \Rightarrow 12 g \leq V_{0}^{2} \leq g+g \cdot \sqrt{145}\right.\right.$

$120 \leq V_{0}^{2} \leq 10+10 \cdot \sqrt{145}$

Answer: $120 \leq V_{0}^{2} \leq 10+10 \cdot \sqrt{145}$ or approximately [120; 130]

# Criteria for evaluating the olympiad work

Profile: MathematicsSubject: Mathematics

Class: 11, variants 1 and 2

Task 1 (maximum score 12)

| Criterion (select one criterion) | Score |
| :---: | :---: |
| No point listed below is completed, and/or the correct answer is simply recorded. |  |
| It is proven that the polynomial $P(t)$ has three distinct real roots, or a connection between the coefficients of this polynomial and its roots is established. | 3 |
| It is proven that the polynomial $P(t)$ has three distinct real roots, a connection between the coefficients of this polynomial and its roots is established. The conditions that allow finding the coefficients of the polynomial $R(t)$ are correctly recorded. | 6 |
| A computational error is made in the solution of the problem | 9 |
| Provided | 12 |

Task 2 (maximum score 16)

| Criterion (select one criterion) | Score |
| :--- | :---: |
| No point listed below is completed, and/or the correct answer is simply recorded. | 0 |
| All three connection schemes are listed. The number of combinations is correctly calculated for one of the schemes. | 4 |
| All three connection schemes are listed. The number of combinations is correctly calculated for two of the schemes. | 8 |
| All three connection schemes are listed. The number of combinations is correctly calculated for all three schemes. | 12 |
| The final probability is