Kick it

This problem can appear particularly daunting at first glance, and offers a number of tempting pseudo-solutions that are, unfortunately, false. Before we can begin, we need to make some reductions, as there are simply too many variables!

Let $G$ be the goalie's starting position, and $B$ the ball's starting position in all the following paragraphs.

---

Observation 1: The goalie's optimal path is a straight line.

More formally, if there exists a set of paths the goalie can take which enable him to intercept the ball, then at least one of them must be a straight line. This is intuitively proven since the ball's path is deterministic after the kick at time $t = 0$ ; If the goalie can take a path $P$ to point $X$ , and intercept the ball at point $X$ at time $t_x$ , then the goalie can also take a straight line path to point $X$ and reach it at time $t \leq t_x$ , and thus still intercept the ball. So, we only need to worry about the goalie moving in straight lines.

---

Observation 2: The angle between the scoring point, $B$ , and $G$ should be maximized.

I haven't found an elegant way to prove this, so I wholeheartedly welcome proofs easier on the eyes than mine!

More formally, for a trajectory of the ball from point $B$ to a point $X$ on the goal line, we should maximize $\angle GBX$ in order to minimize $S$ . We will prove the contrapositive: Decreasing $\angle GBX$ requires us to increase $S$ in order to avoid the goalie.

Suppose that, for a given angle $\theta = \angle GBX$ , the ball must be kicked at infimum speed $S$ in order to avoid the goalie and score. That is, for any speed $S' < S$ , at that same angle, the goalie would intercept the ball. Let $I(x)$ be the set of points at which the goalie can intercept the ball for kicking speed $S -x$ at angle $\theta$ , and let $L$ be the limit of $I(\epsilon )$ as $\epsilon \rightarrow 0$ . It turns out that $L$ is a single point, and not a set of points, but we don't need to prove this to continue.

Cast a ray from $B$ to some point $P \in L$ , and draw a circle, $C$ , centered at $G$ , with radius $|GP|$ . Note that $\theta = \angle GBX = \angle GBP$ . Now rotate $P$ around $B$ towards $G$ to form $P'$ , and note that $\angle GBP > \angle GBP'$ . Additionally, note that $P'$ is on the interior of $C$ , regardless of the magnitude of the rotation. Since the goalie was capable of intercepting the ball at any point in $C$ by the time he's able to reach $P$ , the goalie is able to reach $P'$ in even less time. Thus, at this new angle, the speed of the ball must be increased in order to escape the reach of the goalie.

---

These observations are enough to boil down the problem to a single inequality. It is important to note, however, that avoiding the goalie's reach entirely, which we have been reasoning about, and avoiding the goalie's reach just long enough to cross the goal line , are two different things numerically (Consider, for example, if the goal were 7300 meters wide). However, for this problem, as we'll see below, it turns out that these are the same thing.

Since we're skeptical and lazy, let's shoot at each goal post, and determine which one yields the better infimum speed to pass the goalie. Define $d(s,t)$ to be the distance between $G$ and the ball at time $t$ , given velocity $s$ of the ball. This function is easily defined by Pythagoras' for each trajectory. Using Observation 1, we then consider:

$d(s,t) \leq 10 t$

For precisely $s$ and $t$ where the above inequality holds, the goalie is able to intercept the ball. Therefore, we are looking for $s$ where $t$ has no solution (the goalie can't make the interception).

---

For the left goal post:

$d(s,t) = \sqrt{(0.5 - (2 - \frac{s * t * 5.65}{ \sqrt{10^2 + 5.65^2}}))^2 + (8 - \frac{s * t * 10}{ \sqrt{10^2 + 5.65^2}})^2}$

The inequality thus yields at least one solution for $t$ precisely when $s \leq \frac{5 \sqrt{13983785}}{604} \approx 30.956$ .

For the right goal post:

$d(s,t) = \sqrt{(0.5 - (2 + \frac{s * t * 1.65}{ \sqrt{10^2 + 1.65^2}}))^2 + (8 - \frac{s * t * 10}{ \sqrt{10^2 + 1.65^2}})^2}$

The inequality yields at least one solution for $t$ precisely when $s \leq \frac{5 \sqrt{1088585}}{564} \approx 29.2534$ .

Thus, shooting for the right goal post is optimal, and allows us to kick the ball at the slowest possible speed to get past the goalie and score. Multiplying by 99, we get $99S \approx 2896.1$ , and so we answer 2896 .

---

For shooting at the right goal post, the point $P$ as mentioned in the proof of Observation 2 exists at time $t \approx 0.2961$ . This means the goalie makes the catch near the point $(3.410, 8.546)$ for $S$ just below the observed infimum, which is well ahead of the goal line, so we have not overlooked the scenario described earlier.