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If any officer can move more than $\delta$, then that officer can simply chase down the fugitive. Thus, I propose modifying the question to allow each officer to move at most distance $\delta$, but they all pull from a given pool of movement of size $c\delta$. Then we can ask what values of $c$ guarantee the capture of the fugitive. We can squeeze $c$ from two directions:

1. How small can $c$ be such that officers can always catch the fugitive?
2. How large can $c$ be such that the fugitive can still escape (for suitable choice of $\delta$)?

I claim that any constant $c>2$ works for direction 1, and we only need three officers to reach this bound. To see this, pick an equilateral triangle containing the fugitive, and place one of the three officers on each of the three edges of the equilateral triangle - specifically, on the orthogonal projection of the fugitive's location to that edge.

After each fugitive's move, the officers can move until they are each still at the orthogonal projection of the fugitive's location to their edge. This takes $$\delta ( | \cos \theta| + |\cos (\theta+ 2\pi)| + \cos (\theta + 4\pi)| ) \leq 2 \delta $$ movement, with the inequality because $\cos \theta + \cos(\theta + 2\pi/3) + \cos (\theta + 4\pi/3) =0$ so if one is positive and two are negative, the positive term is at most $1$ and the sum of the two negative terms are at most $1$, and similarly with one negative and two positive. Equality is attained for $\theta = 0 , \pi/3, 2\pi/3, \pi, 4\pi/3, 5\pi/3$, i.e. for moves parallel to an edge.

Because we keep the officers on the orthogonal projections, the fugitive can never escape this triangle, as then the fugitive's orthogonal projection would equal their location so their path would cross an officer. Since $c>2$, we have a little bit of extra movement, which we can use to shrink the triangle each turn.

The race is still on to improve this constant!

If any officer can move more than $\delta$, then that officer can simply chase down the fugitive. Thus, I propose modifying the question to allow each officer to move at most distance $\delta$, but they all pull from a given pool of movement of size $c\delta$. Then we can ask what values of $c$ guarantee the capture of the fugitive. We can squeeze $c$ from two directions:

1. An lower bound $l$ for the officers such that they can always capture the fugitive if $c\gt l$.
2. An upper bound $u$ for the fugitive such that they can escape if $c\lt u$.

I claim that $l=2$ works, and we only need three officers to reach this bound. To see this, pick an equilateral triangle containing the fugitive, and place one of the three officers on each of the three edges of the equilateral triangle - specifically, on the orthogonal projection of the fugitive's location to that edge.

After each fugitive's move, the officers can move until they are each still at the orthogonal projection of the fugitive's location to their edge. This takes $$\delta ( | \cos \theta| + |\cos (\theta+ 2\pi)| + \cos (\theta + 4\pi)| ) \leq 2 \delta $$ movement, with the inequality because $\cos \theta + \cos(\theta + 2\pi/3) + \cos (\theta + 4\pi/3) =0$ so if one is positive and two are negative, the positive term is at most $1$ and the sum of the two negative terms are at most $1$, and similarly with one negative and two positive. Equality is attained for $\theta = 0 , \pi/3, 2\pi/3, \pi, 4\pi/3, 5\pi/3$, i.e. for moves parallel to an edge.

Because we keep the officers on the orthogonal projections, the fugitive can never escape this triangle, as then the fugitive's orthogonal projection would equal their location so their path would cross an officer. Since $c>2$, we have a little bit of extra movement, which we can use to shrink the triangle each turn.

The race is still on to improve this constant!

If any officer can move more than $\delta$, then that officer can simply chase down the fugitive. Thus, I propose modifying the question to ask: If each officer can move at most distance $\delta$, but they all pull from a given pool of movement of size $c\delta$, then how small can $c$ be such that they can always catch the fugitive?

I claim that any constant $c>2$ works, and we only need three officers to reach this bound. To see this, pick an equilateral triangle containing the fugitive, and place one of the three officers on each of the three edges of the equilateral triangle - specifically, on the orthogonal projection of the fugitive's location to that edge.

After each fugitive's move, the officers can move until they are each still at the orthogonal projection of the fugitive's location to their edge. This takes $$\delta ( | \cos \theta| + |\cos (\theta+ 2\pi)| + \cos (\theta + 4\pi)| ) \leq 2 \delta $$ movement, with the inequality because $\cos \theta + \cos(\theta + 2\pi/3) + \cos (\theta + 4\pi/3) =0$ so if one is positive and two are negative, the positive term is at most $1$ and the sum of the two negative terms are at most $1$, and similarly with one negative and two positive. Equality is attained for $\theta = 0 , \pi/3, 2\pi/3, \pi, 4\pi/3, 5\pi/3$, i.e. for moves parallel to an edge.

Because we keep the officers on the orthogonal projections, the fugitive can never escape this triangle, as then the fugitive's orthogonal projection would equal their location so their path would cross an officer. Since $c>2$, we have a little bit of extra movement, which we can use to shrink the triangle each turn.

The race is still on to improve this constant!

If any officer can move more than $\delta$, then that officer can simply chase down the fugitive. Thus, I propose modifying the question to allow each officer to move at most distance $\delta$, but they all pull from a given pool of movement of size $c\delta$. Then we can ask what values of $c$ guarantee the capture of the fugitive. We can squeeze $c$ from two directions:

1. How small can $c$ be such that officers can always catch the fugitive?
2. How large can $c$ be such that the fugitive can still escape (for suitable choice of $\delta$)?

I claim that any constant $c>2$ works for direction 1, and we only need three officers to reach this bound. To see this, pick an equilateral triangle containing the fugitive, and place one of the three officers on each of the three edges of the equilateral triangle - specifically, on the orthogonal projection of the fugitive's location to that edge.

After each fugitive's move, the officers can move until they are each still at the orthogonal projection of the fugitive's location to their edge. This takes $$\delta ( | \cos \theta| + |\cos (\theta+ 2\pi)| + \cos (\theta + 4\pi)| ) \leq 2 \delta $$ movement, with the inequality because $\cos \theta + \cos(\theta + 2\pi/3) + \cos (\theta + 4\pi/3) =0$ so if one is positive and two are negative, the positive term is at most $1$ and the sum of the two negative terms are at most $1$, and similarly with one negative and two positive. Equality is attained for $\theta = 0 , \pi/3, 2\pi/3, \pi, 4\pi/3, 5\pi/3$, i.e. for moves parallel to an edge.

Because we keep the officers on the orthogonal projections, the fugitive can never escape this triangle, as then the fugitive's orthogonal projection would equal their location so their path would cross an officer. Since $c>2$, we have a little bit of extra movement, which we can use to shrink the triangle each turn.

The race is still on to improve this constant!

If any officer can move more than $\delta$, then that officer can simply chase down the fugitive. Thus, I propose modifying the question to ask: If each officer can move at most distance $\delta$, but they all pull from a given pool of movement of size $c\delta$, then how small can $c$ be such that they can always catch the fugitive?

I claim that $c\leq 2\sqrt{2}$, and we only need four officers to reach this bound. To see this, orient the playing field and place each officer at the four cardinal directions, with the partner at the opposite side. Think of them as living on a rectangle. Using their joint speed, they can always remain on that rectangle and be opposite their partner, with the fugitive directly between them and their partner. If they ever have any extra movement, they can squeeze in the rectangle.

This shows that any constant $c>2\sqrt{2}$ works, but what about $2\sqrt{2}$ exactly? In that case, the only time that there is no extra speed that can be used to squeeze in is when the fugitive goes perfectly diagonal. In that case the officers can lag just a little behind the fugitive, and use that extra lag to squeeze in a bit. The fugitive has to start backtracking when he gets too close to a corner (where the two close officers could just capture him directly), and the officers once again get some extra speed to use.

Now, the race is on to improve this constant!

If any officer can move more than $\delta$, then that officer can simply chase down the fugitive. Thus, I propose modifying the question to ask: If each officer can move at most distance $\delta$, but they all pull from a given pool of movement of size $c\delta$, then how small can $c$ be such that they can always catch the fugitive?

I claim that any constant $c>2$ works, and we only need three officers to reach this bound. To see this, pick an equilateral triangle containing the fugitive, and place one of the three officers on each of the three edges of the equilateral triangle - specifically, on the orthogonal projection of the fugitive's location to that edge.

After each fugitive's move, the officers can move until they are each still at the orthogonal projection of the fugitive's location to their edge. This takes $$\delta ( | \cos \theta| + |\cos (\theta+ 2\pi)| + \cos (\theta + 4\pi)| ) \leq 2 \delta $$ movement, with the inequality because $\cos \theta + \cos(\theta + 2\pi/3) + \cos (\theta + 4\pi/3) =0$ so if one is positive and two are negative, the positive term is at most $1$ and the sum of the two negative terms are at most $1$, and similarly with one negative and two positive. Equality is attained for $\theta = 0 , \pi/3, 2\pi/3, \pi, 4\pi/3, 5\pi/3$, i.e. for moves parallel to an edge.

Because we keep the officers on the orthogonal projections, the fugitive can never escape this triangle, as then the fugitive's orthogonal projection would equal their location so their path would cross an officer. Since $c>2$, we have a little bit of extra movement, which we can use to shrink the triangle each turn.

The race is still on to improve this constant!