Fiddle Eagles, Fiddle!

You’re betting on the final two football games of the season for your home team, the Fiddle-delphia Eagles. Thanks to the wonders of time travel, you happen to know that the Eagles will win one game and lose the other. Unfortunately, you can’t remember in which order they do so. Maybe the Eagles win the first game and lose the second, or maybe they lose the first and win the second.

You have $\$100$, of which you can bet any amount (including fractions of pennies) that the Eagles will win. So if you bet $x$ dollars on the first game and the Eagles win, you’ll have $100+x$ dollars to bet on the second game. But if the Eagles lose that first game, you’re left with $100−x$ dollars to bet on the second game.

You want to implement a betting strategy that guarantees you’ll have as much money as possible after both games. If you did so, then after the two games how much money would you be guaranteed to have?

In this case, where you know there are only two outcomes for the final two games (WL or LW), then you know with perfect clarity what the outcome of the final game is once you observe the outcome of the first game. Now here's where we get into some semantics, what exactly does "you're betting ... for your home team" mean? If you are just betting for either the Figgles as they are affectionately known or their opponents, then you can end up with a guaranteed $\$200$ by betting nothing on the first game, then doubling up on the second game with the sure lock. However, that wouldn't be a very fun answer, so let's instead assume that we can only bet on the Figgles to win and that the question involves deciding how much to optimally bet for the first game and the second game.

Assume that we bet $x \in [0,100]$ on the first game, observe outcome $o \in \{W, L\}$ and then bet $y \in [0, U(x,o)]$ where $$U(x,o) = \begin{cases} 100+x, &\text{if $o = W$;}\\ 100-x, &\text{if $o = L$}\end{cases}$$ on the second game. Since we know that if $o = W,$ then the Figgles will lose the second game and vice versa, we see that the final money balance will be $$V(x, y, o) = \begin{cases} 100 + x - y, & \text{if $o = W$;}\\ 100 - x + y & \text{if $o = L$}\end{cases}$$ then the problem becomes $$\max_{x, y} \min \{ V(x,y, W), V(x,y, L) \}.$$ Since this is multistage, we should first optimize the choice of $y$ in each case. In particular $\hat{y}(W) = 0,$ since if the Figgles won the first game they will lose the second and $\hat{y}(L) = 100 -x,$ since you might as well double up everything on the sure win knowing that the Figgles lost the first game.

So we can reduce the problem by introducing $$\hat{V}(x,o) = V(x, \hat{y}(o), o) = \begin{cases} 100 + x, &\text{if $o = W$;}\\ 2 ( 100 - x ), &\text{if $o = L$}\end{cases}$$ and then solving for the optimal strategy we get an ending money balance of $$\max_{x \in [0,100]} \min \{ \hat{V}(x, W), \hat{V}(x, L) \} = \max_{x \in [100]} \min \{ 100 -x, 200 - 2x \} = 133.\bar{3}$$ when originally wagering $\hat{x} = 33.\bar{3}$ on the first game.

Seeing the forest for the cylinders

You find yourself amidst what appears to be an infinite grid of trees. For the purposes of this puzzle, let’s suppose each tree is a perfect cylinder. You are at the point $(0, 0),$ but there’s a tree with a radius of $r_1 = 0.25$ units (and diameter $0.5$ units) centered at every other point in the plane with integer coordinates.

Of course, you can’t actually see infinitely many trees. Most of them are obscured by the trees immediately around you. As you look around in all directions, how many distinct trees are you able to see?

Let's set up the generic problem, with some unknown radius $r \gt 0,$ then we can try to solve each of the pieces of this fiddle, er, riddle.

Assume that I am sitting at the origin and I look along the line $y=m x.$ If we assume for the time being that each of these perfectly cylindrical trees were see through, then I could see a tree centered at $(a,b) \in \mathbb{Z}^2$ if and only if the set $$\{ (x,y) \mid y = mx, (x-a)^2 + (y-b)^2 = r^2 \} \ne \emptyset.$$ Simplifying a bit, I could see any tree centered at $(a,b)$ provided that $$(x-a)^2 + (mx - b)^2 -r^2 = (1+m^2) x^2 -2( a + bm) x + (a^2 + b^2 - r^2) = 0$$ has a solution, that is, if and only if the discriminant \begin{align*}0 \leq \Delta( m \mid a, b, r ) &= 4(a + bm)^2 - 4(1 + m^2) (a^2 + b^2 - r^2) \\ &= 4 ( -(b^2 - r^2) + 2 ab m - (a^2 - r^2) m^2 ).\end{align*} Rearranging a bit, and solving for $m$ we see that these translucent trees would be visible if and only if $$\frac{ab - r \sqrt{(a+b)^2 - r^2}}{a^2 - r^2} \leq m \leq \frac{ab + r \sqrt{(a+b)^2 - r^2}}{a^2 - r^2}.$$ However, since we don't have translucent cylindrcal trees, if we are looking along $y = mx$ then we could only see the two trees in $$(\hat{a}(m), \hat{b}(m)) \in \arg\min \left\{ |a| + |b| \left|\right. \frac{ab - r \sqrt{(a+b)^2 - r^2}}{a^2 - r^2} \leq m \leq \frac{ab + r \sqrt{(a+b)^2 - r^2}}{a^2 - r^2} \right\}.$$ Due to all of the symmetry of this problem, only method is to allow $m$ to slowly increase from $m = 0$ to $m = 1,$ and then rely on symmetry (since there is nothing uniquely special about the trees in the lower half of the first quadrant) and inclusion/exclusion (since some of the trees will be double counted) to answer either the question of how many tress do we see or which trees is furthest away.

So in particular, we see that for $m \in [0, \frac{r}{\sqrt{1-r^2}}],$ I would be looking at the tree at $(1,0).$ Similarly, for any $m \in [\frac{1-r\sqrt{2-r^2}}{1-r^2}, 1],$ I would be looking at the tree centered at $(1,1).$ So in particular for $r = 0.25,$ we have filled in the set $$ m \in [0, \frac{\sqrt{15}}{15}] \sqcup [\frac{16-\sqrt{31}}{15}, 1]$$ covered by $(1,0)$ and $(1,1).$ The next smallest tree (based on the $1$-norm) is centered at $(2,1),$ which can be seen for any $m \in [ \frac{32-\sqrt{79}}{63}, \frac{32 + \sqrt{79}}{63} ].$ At this point we still have only filled a disjoint union of intervals with $(1,0),$ $(1,1)$ and $(2,1),$ so filling in the remaining gaps, in the translucent tree problem we would be able to see the tree centered at $(3,1)$ for any $m \in [\frac{48-\sqrt{159}}{143}, \frac{48+\sqrt{159}}{143} ],$ but since $\frac{48 + \sqrt{159}}{143} \gt \frac{32 - \sqrt{79}}{63} \gt \frac{\sqrt{15}}{15} \gt \frac{48-\sqrt{159}}{143}$, I can only see the tree $(3,1)$ for $m \in (\frac{\sqrt{15}}{15}, \frac{32 - \sqrt{79}}{63}).$ Similarly, I can see $(3,2)$ for any $m \in (\frac{48 + \sqrt{159}}{143}, \frac{16-\sqrt{31}}{15}),$ so in totality there are 8 trees in the lower half of the first quadrant that can be seen, that is, $$T(m) = \begin{cases} (1,0), & m \in [0, \frac{\sqrt{15}}{15}]; \\ (3, 1), & m \in ( \frac{\sqrt{15}}{15}, \frac{48 - \sqrt{159}}{143}); \\ (2, 1), & m \in [ \frac{48 - \sqrt{159}}{143}, \frac{48 + \sqrt{159}}{143} ];\\ (3,2), & m \in (\frac{48 + \sqrt{159}}{143}, \frac{16-\sqrt{31}}{15}); \\ (1,1), & m \in [\frac{16 - \sqrt{31}}{15}, 1].\end{cases}$$

So since there are 5 in the lower half of the first quadrant, we should multiply by 8 to get to 40 trees in the whole plane. However, this would double count the eight points $(\pm 1, 0),$ $(0, \pm 1),$ $(\pm 1, \pm 1),$ so we need to subtract 8 from the total, leaving $32$ trees of diameter $r=0.25$ can be seen from the origin.