A hunter and an invisible rabbit play a game in the Euclidean plane. The rabbit's starting point, $A_0,$ and the hunter's starting point, $B_0$ are the same. After $n-1$ rounds of the game, the rabbit is at point $A_{n-1}$ and the hunter is at point $B_{n-1}.$ In the $n^{ ext{th}}$ round of the game, three things occur in order:
i. The rabbit moves invisibly to a point $A_n$ such that the distance between $A_{n-1}$ and $A_n$ is exactly $1.$
ii. A tracking device reports a point $P_n$ to the hunter. The only guarantee provided by the tracking device to the hunter is that the distance between $P_n$ and $A_n$ is at most $1.$
iii. The hunter moves visibly to a point $B_n$ such that the distance between $B_{n-1}$ and $B_n$ is exactly $1.$
Is it always possible, no matter how the rabbit moves, and no matter what points are reported by the tracking device, for the hunter to choose her moves so that after $10^9$ rounds, she can ensure that the distance between her and the rabbit is at most $100?$

Question

A hunter and an invisible rabbit play a game in the Euclidean plane. The rabbit's starting point, $A_0,$ and the hunter's starting point, $B_0$ are the same. After $n-1$ rounds of the game, the rabbit is at point $A_{n-1}$ and the hunter is at point $B_{n-1}.$ In the $n^{	ext{th}}$ round of the game, three things occur in order:
i. The rabbit moves invisibly to a point $A_n$ such that the distance between $A_{n-1}$ and $A_n$ is exactly $1.$
ii. A tracking device reports a point $P_n$ to the hunter. The only guarantee provided by the tracking device to the hunter is that the distance between $P_n$ and $A_n$ is at most $1.$
iii. The hunter moves visibly to a point $B_n$ such that the distance between $B_{n-1}$ and $B_n$ is exactly $1.$
Is it always possible, no matter how the rabbit moves, and no matter what points are reported by the tracking device, for the hunter to choose her moves so that after $10^9$ rounds, she can ensure that the distance between her and the rabbit is at most $100?$

ChooseAUsename · Answer

trivial

OronSH · Answer

but can you prove it

ChooseAUsename · Answer

Yes

OronSH · Answer

what is your proof

ChooseAUsename · Answer

bump again

OronSH · Answer

where is your proof $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ bruh imagien dollar signs don't work for latex

OronSH · Answer

how do i type inline latex bruh

OronSH · Answer

bruh \$\frac12+\frac13=\frac56\$ why does this not work

ChooseAUsename · Answer

Consider a positive integer $n>d$, to be chosen later. Let the hunter start at $B$ and the rabbit at $A$, as shown. Let $\ell$ denote line $AB$. Now, we may assume the rabbit reveals its location $A$, so that all previous information becomes irrelevant. The rabbit chooses two points $X$ and $Y$ symmetric about $l$ such that $XY=2$ and $AX=AY=n$, as shown. The rabbit can then hop to either $X$ or $Y$, pinging the point $P_n$ on the $\ell$ each time. This takes $n$ hops. Now among all points $H$ the hunter can go to, $\min \max\{HX,HY\}$ is clearly minimized with $H\in\ell$ by symmetry. So the hunter moves to a point $H$ such that $BH=n$ as well. In that case the new distance is $HX=HY$. We now compute \begin{align*}HX^2&=1+HM^2=1+(\sqrt{AX^2+1}-AH)^2\&=1+(\sqrt{n^2-1}-(n-d))^2\&\ge1+\left(\left(n-\frac1n\right)-(n-d)\right)^2\&=1+(d-1/n)^2\end{align*} which exceeds $d^2+\frac12$ whenever $n\ge4d$.$\qquad\square$

OronSH · Answer

oh wait does $this work$ o it's \(

OronSH · Answer

@chooseausename wrote:

Consider a positive integer $n>d$, to be chosen later. Let the hunter start at $B$ and the rabbit at $A$, as shown. Let $\ell$ denote line $AB$. Now, we may assume the rabbit reveals its location $A$, so that all previous information becomes irrelevant. The rabbit chooses two points $X$ and $Y$ symmetric about $l$ such that $XY=2$ and $AX=AY=n$, as shown. The rabbit can then hop to either $X$ or $Y$, pinging the point $P_n$ on the $\ell$ each time. This takes $n$ hops. Now among all points $H$ the hunter can go to, $\min \max\{HX,HY\}$ is clearly minimized with $H\in\ell$ by symmetry. So the hunter moves to a point $H$ such that $BH=n$ as well. In that case the new distance is $HX=HY$. We now compute $$\begin{align*}HX^2&=1+HM^2=1+(\sqrt{AX^2+1}-AH)^2\&=1+(\sqrt{n^2-1}-(n-d))^2\&\ge1+\left(\left(n-\frac1n\right)-(n-d)\right)^2\&=1+(d-1/n)^2\end{align*}$$ which exceeds $d^2+\frac12$ whenever $n\ge4d$.$\qquad\square$

bruh sobad you didn't even copy evan chen's solution properly

OronSH · Answer

you didn't even finish the problem

Fortish · Answer

@oronsh wrote:

you didn't even finish the problem

He can guide you to the answer but he can't give it directly.

Fortish · Answer

@oronsh wrote:

oh wait does $this work$ o it's \(

please remain on topic

OronSH · Answer

ok but where's evan's diagram

ChooseAUsename · Answer

attachment

Extrinix · Answer

At least source your inputs, please. Evan Chen - IMO Notes

OronSH · Answer

bruh not even asymptote so bad

OronSH · Answer

@extrinix wrote:

At least source your inputs, please. Evan Chen - IMO Notes

ok

Extrinix · Answer

@oronsh wrote:

@extrinix wrote:

At least source your inputs, please. Evan Chen - IMO Notes

ok

Talking to @chooseausename

OronSH · Answer

ok

OronSH · Answer

@chooseausename wrote:

ok

ChooseAUsename · Answer

@extrinix wrote:

@oronsh wrote:

@extrinix wrote:

At least source your inputs, please. Evan Chen - IMO Notes

ok

Talking to @chooseausename

ok

ChooseAUsename · Answer

attachment

Fortish · Answer

Please reread the rules or if you didn't read the rules: https://questioncove.com/rules

Fortish · Answer

@thisgirlpretty are you able to lock this question, ppl keep spamming.

Fortish · Answer

@vocaloid can you lock.

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