The answer to this problem must be negative, otherwise it would be boring.

Any idea about who proposed this problem?

tarzanjunior wrote: Any idea about who proposed this problem? Maybe It'll sound weird but why do I think thus was proposed from Taiwan or Hong Kong :/ Purely speculating here.

Too much hearsay. Moderator, please delete.

The best estimate I got so far is $d_{n+1}\le\sqrt{\left(\sqrt{d_n^2-1}-1\right)^2+1}+1.$ This only works for $n$ less than approximately 650000 which is a lot smaller than $10^9.$ This makes me think of winning strategy for the rabbit. I believe that if such a strategy exists, then we will have our tracking device always lying so that $\angle B_{n-1}P_nA_n = 90^\circ$ and $P_nA_n=1$ (such a case gives an equality in the bound above (unless I made a mistake)). Note that this is not always possible. Important note: I got bound by considering only the latest info ($P_n$) from the tracking device, so there is still a chance of the existence of a more advanced winning strategy for the hunter.

I think the answer is yes. Because if the answer is no, then you just need to prove that the rabbit can escape the hunter when it's making every step to be farthest from the hunter. Since we can now assume that hunter knows that rabbit is going away from him every step it wouldn't be that trivky to prove that he can/can't do what problem is stating. Therefore the answer must be positive.

did anyone solve this on the contest?

You know, I tried to model the situation described in my above post starting when $d_n=A_nB_n$ is already big (~10) and it seems like tracing back more than one point $P_n$ doesn't really help us, it decreases the possible area of rabbit just by several percents, which is not really helpful. Therefore we are left with two possibilities: either few starting points $P_n$ are crucial in determining the position of a rabbit throughout the game, or rabbit can escape in this scenario. Note that if $P_1=B_0$ then we have literally no info, so we may get $d_1$ as big as 2 (if the hunter and rabbit choose opposite directions), so we already start to lose some info from not the latest $P_n$. Oh, wait, I got another thought: the only way for us to be able to precisely tell which are the possible rabbit's positions is for all $P_n$ to lie on some straight line.

It would be quite amusing if there were to be an inversion solution to this, especially given its geometric flavor

The answer is no. WLOG let the rabbit control the tracking device. We define an $r$-muffin to be the region enclosed by a lower semicircle of radius 1 and an arc of a circle of radius $r$ placed above the semicircle. Call the top of the muffin to be the midpoint of the arc of radius $r$, and the center of the muffin to be the center of its semicircle. In particular, the unit circle is a 1-muffin. If the rabbit can be anywhere in an $r$-muffin (with the tracking device on the center of the muffin), in the next round the rabbit can be anywhere in an $(r+1)$-muffin with the top just 1 higher (and placing the tracking device on the center of the muffin) (see diagram below). Furthermore, if the hunter is a distance $d$ below the top of the $r$-muffin (that is, the hunter is below the top of the muffin and the distance from the hunter to the line through the top and tangent to the muffin is $d$), then in the next round the hunter will be at least a distance $d$ below the top of the $(r+1)$-muffin. We say the game is in state $(r,d)$ if the rabbit can be anywhere in an $r$-muffin and the hunter is at least distance $d$ below the top of the muffin. Then in 1 round, the rabbit can make the game change from state $(r,d)$ to $(r+1,d)$. Another game changer is the following: if the rabbit can be anywhere in an $r$-muffin, let $A$ be the leftmost point of the muffin, then in the next round the rabbit can be anywhere in the unit circle centered at $A$ (which is a 1-muffin). Now the top of the 1-muffin can be anywhere, so we pick the top to be the point furthest from the hunter (and rotate so the top is really the top). WLOG the hunter is to the right of the muffin, then with some geometry, the game state changed from $(r,d)$ to $\left(1,\sqrt{(d-r+\sqrt{r^2-1})^2+1}\right)$. This is only beneficial if $r>d$. So we can change the state from $(2d,d)$ to $\left(1,\sqrt{(-d+\sqrt{4d^2-1})^2+1}\right)$. Thus we can change the game state from $(1,d)$ to $\left(1,\sqrt{(-d+\sqrt{4d^2-1})^2+1}\right)$ in at most $2d+1$ rounds. With some algebra (rabbits are good at multiplying), we can show that $\sqrt{(-d+\sqrt{4d^2-1})^2+1}>\sqrt{d^2+\frac12}$, so we can change the game state from $(1,d)$ to $\left(1,\sqrt{d^2+\frac12}\right)$ in at most $2d+1$ rounds. Our aim is the state $(1,100)$. We can change the state from $(1,\sqrt{s})$ to $\left(1,\sqrt{s+\frac12}\right)$ in $2\sqrt{s}+1$ rounds. Thus we can get from $(1,1)$ to $(1,100)$ in at most $4(\sqrt{2}+\sqrt{3}+\cdots+\sqrt{10000})+20000\sim 4000000$ rounds, which is much less than $10^9$. Now we just need to get to $(1,1)$. In the first round, place the tracking device at $A_0$, then the rabbit can be anywhere on the unit circle. In the second round, place the tracking device diametrically opposite the hunter, then we will be in state $(1,2)$ in fact, and we are done.

Someone give oneplusone a trophy for solving this monster-problem! Congrats!

@oneplusone are you at IMO 2017. If so, did you solve this on contest because that would be really impressive.

oneplusone graduated a while ago.

One plus one good dolution

We claim that the hunter cannot guarantee being distance at most $100$ from the rabbit after $10^9$ rounds. On the first step, have the rabbit move in some direction, and assume the tracking device returns the original starting point. Then in the worst case the hunter will be at a distance $2$ from the rabbit. Let $R \ge 2$, and let $N$ be the least integer which is at least $R+1$. We claim that if the rabbit is a distance exactly $R$ from the hunter at a certain time, in $N$ steps, the hunter cannot ensure that the distance between her and the rabbit is less than distance $R + 0.01 R^{-2}$. First, the rabbit will reveal his position. WLOG the hunter is at $(-R,0)$ and the rabbit is at $(0,0)$. Let $X = \sqrt{N^2 - 0.25}$. In particular, $N-X = \frac{0.25}{N+X} \ge \frac{0.25}{2N}.$ Now the rabbit will flip a coin and either go to the point $(X, 0.5)$ or $(X, -0.5)$ in the next $N$ steps, and the tracking device will travel along the x-axis. From the hunter's perspective, after $N$ steps she has no idea whether the rabbit is at $(X,0.5)$ or $(X,-0.5)$, since these situations are symmetric. Furthermore, after $N$ steps, the hunter is at a point $P$ whose x-coordinate is at most $N-R$. If $P$ is above the x-axis, then her distance to $(X,-0.5)$ is at least \begin{align*} \sqrt{(X-(N-R))^2+0.25} &= \sqrt{X^2 - 2X(N-R) + (N-R)^2 + 0.25} \\ &= \sqrt{N^2 - 2X(N-R) + (N-R)^2}\\ &= \sqrt{R^2 + (N-R)(2N-2X)}\\ &\ge \sqrt{R^2 + (2N-2X)} \\ &\ge \sqrt{R^2 + \frac{1}{4N}} \\ &\ge \sqrt{R^2 + \frac{1}{4(R+2)}} \\ &\ge R + 0.05 R^{-2}. \end{align*}If $P$ is below the x-axis, the same inequalities follow for $(X, 0.5)$. Therefore with some probability, the rabbit will be at least $R + 0.05 R^{-2}$ away from the hunter. Therefore if the rabbit starts from a distance of $R$ and repeats this process $20(R+1)^2$ times, he will be able to escape to a distance of at least $R+1$ in at most $20(R+1)^3$ steps, and so after at most $7 \times 10^8$ steps, the rabbit will be able to distance himself from the hunter by more than $100$. For the rest of the steps, the rabbit can just run directly away from the hunter and escape to safety! EDIT: As pointed out by MathStudent2002, what I had here originally actually exceeded the $10^9$ threshhold, but this was fixed by changing the $0.01$ to $0.05$. Thanks you to oneplusone, who points out that if we pick $N \approx 2R$ instead, then we can get a distance increase of $R^{-1}$ instead in $2R$ steps.

Not quite. You need to repeat the process $100R^2$ times but each process is $R+1$ steps, so you need $100R^3$ steps to go from $R$ to $R+1$ and that will exceed $10^9$. Can be fixed by picking $N\sim 2R$ then $R$ will increase by of order $R^{-1}$ instead of $R^{-2}$.

Note that ksun's solution can also be optimized by going to $(X,\pm 1)$ instead, he solved the problem for the tracker reporting a disk of diameter $1$. This makes getting it $<10^9$ easier. In fact, if we instead let the tracking machine have radius $\frac{1}{19}$, we can do some optimizations to ksun's argument (setting $N\approx 2R$, not using the approximation of $0.05R^{-1}$, etc.). This gets very close to the $10^9$ bound. These also give a rabbit strategy of approximately $2.7$ million steps.

It is interesting in ksun's solution that the distance between hunter and rabbit initially shrinks before it grows.

Very difficult to solve in competition!! And oneplusone has given a beautiful solution!!!!!!

Reef334 wrote: Claim: the hunter's best strategy should be to move directly towards the tracker. Of course, no! All the posts above based on that assumption are..., ok, at leats do not understand well what should be proved. The best strategy for the hunter is to move directly towards the real position of the rabbit!! Suppose she has a secret Oracle, who somehow knows everything, including the rabbit's affairs. So, the hunter calls the Oracle and he tells her where exactly the animal is. But even so, the hunter will lose! Because the rabbit will cheat, but in such a way no one can prove it and there is a positive probability, rabbit plays decently. So, the hunter and the rabbit know they both cheating, but the hunter cannot prove anything without revealing herself of doing the same. Seriously, this problem is a philosophical one. (every one can make that easy calculations). There are two possible answers - yes or no. And before starting of thinking of the best strategies, one should clear up what exactly every answer means, and what type of arguments should be used in either case. I've written some thoughts in the same spirit here.

The answer is no. Consider the following process, occurring over $m$ rounds. For convenience, we will refer to this as an $m$-move. The rabbit moves from point $A$ to the point $A'$ such that $AA'=m$, one unit each round, and if $P'$ is the projection of $A'$ onto line $AB$, then $A'P'=1$. During each round, let the dog report the projection of the rabbit's location onto line $AB$. This is always possible since the distance from the rabbit to line $AB$ never exceeds $1$. Because the hunter doesn't know which half-plane the rabbit is on, he must move along $\overline{AB}$. Otherwise, it is possible that the rabbit is on the other side of the line as he moves, and he moves farther away from the rabbit. For convenience, assume the hunter is magically able to deduce the exact location of the rabbit at the end of the process. [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair B, Bp, A, Ap, P; B=(0,0); Bp=(2,0); A=(5,0); Ap=(5,0)+2*dir(20); P=foot(Ap,B,A); draw(B--P); draw(Bp--Ap--P,dashed); draw(B--Bp,linewidth(1.5)); draw(A--Ap,linewidth(1.5)); dot("$B$",B,S); dot("$B'$",Bp,S); dot("$A$",A,S); dot("$A'$",Ap,NE); dot("$P'$",P,SE); [/asy][/asy] Claim: Suppose that at some point in time, the rabbit is at point $A$ and the hunter is at point $B$, and furthermore $AB=x$. After a $101$-move, if the rabbit moves to $A'$ and the hunter moves to $B'$, then the rabbit can guarentee that $\textstyle A'B'\ge\sqrt{x^2+1/101}$. Proof. Recall that $A'P'=1$, whence $B'P'=x+\sqrt{m^2-1}-m$. It follows that \begin{align*} A'B'^2&\ge1+\left(x+\sqrt{m^2-1}-m\right)^2\\ &=1+x^2+(m^2-1)+m^2+2x\sqrt{m^2-1}-2m\sqrt{m^2-1}-2xm\\ &=x^2+2m^2-2xm+2x\sqrt{m^2-1}-2m\sqrt{m^2-1}\\ &=x^2+2(m-x)\left(m-\sqrt{m^2-1}\right). \end{align*}Let $m=101$ above. Check that \[A'B'^2\ge x^2+\frac{2(m-x)}{m+\sqrt{m^2-1}}\ge x^2+\frac{m-x}m\ge x^2+\frac1{101},\]as desired. $\blacksquare$ Finally, we can perform a $101$-move $\lfloor10^9/101\rfloor$ times; that is, \[A_NB_N^2\ge\left\lfloor\frac{10^9}{101}\right\rfloor\cdot\frac1{101}\ge\frac{10^9-100}{101^2}>100^2,\]so $A_NB_N>100$, as desired.

The answer is no. Suppose at some point that the rabbit is $d$ away from the hunter, and the hunter, who is at $B$, knows exactly where the rabbit is, which is point $A$. Then, we claim that we can increase the square of the distance by $O\left(\frac{1}{d}\right)$ in at most $\lceil d\rceil+1$ moves. For convenience, denote $d_0=\lceil d\rceil$ To do this, the rabbit moves in a straight line for $d_0+1$ moves, such that the angle from the ray $\overrightarrow{BA}$ to the line is at most $2\sin^{-1}\frac{1}{2d_0}$. As the chord determined by the intersections of this line and $\overrightarrow{BA}$ on the circle of radius $r\le d_0$ centered about $A$ always has length less than 1, on the rth turn of the rabbit, the tracker can always ping on the intersection of $\overrightarrow{BA}$ with the circle of radius $r$. Thus, even if the hunter were able to deduce the rabbit's strategy, he would have no way of narrowing down which line the rabbit was moving along until the $d_0+1$st turn. Now, after $d_0+1$ turns, the hunter is somewhere within the circle of radius $d_0+1$ centered about $B$, let's say $B'$, while the rabbit is at $A'$, a point on the circle of radius $d_0+1$ centered about $A$. If $B'$ is on one side of $BA$, we suppose that the rabbit actually decided to go on the line with angle $2\sin^{-1}\frac{1}{2d_0}$ from $BA$ on the other side of $BA$. Then, if $X$ is the point $d_0-d+1$ unit above $A$ on $\overrightarrow{BA}$, we have $XA'\ge B'A'$, and by Law of Cosines, $$A'X=(d_0+1)^2+(d_0-d+1)^2-2(d_0+1)(d_0-d+1)\cos^{-1}\left(2\sin^{-1}\frac{1}{2d_0+2}\right)=d^2+(d_0+1)(d_0-d+1)\frac{4(d_0+1)^2-1}{(d_0+1)^4}$$$$\ge d^2+\frac{3}{d_0+1}$$ So, the rabbit can get from distance $n$ to $n+1$ by using this process to increase by $\frac{3}{d+1}$ and then leaking his location to start again. As the distance squared increase by at least $\frac{3}{n+3}$ each time, this takes at most $\frac{(2n+1)(n+3)}{3}\cdot (n+2)$ moves. Now, note that at the beginning the rabbit can immediately get a distance $2$ away from the hunter if they happen to choose antipodal points, so we need at least $$1+\sum_{n=2}^{100}\left\lceil\frac{(2n+1)(n+3)(n+2)}{3}\right\rceil<100+\sum_{n=2}^{100}n^3=25502559<10^9$$

We start the proof with a claim. Claim: Suppose that at some time, the distance between hunter and rabbit is $d \geq 1$. Then, in $2\lceil d\rceil$ moves, it is possible for the distance between hunter and rabbit to increase to at least $\sqrt{d^2 + \tfrac12}$. Proof: Suppose that initially, the hunter is at $A$ and the rabbit is at $B$, where $\overline{AB} = d$. Let $k = \lceil d\rceil > d$. Suppose that the rabbit moves in a straight line toward some point $P$ closer to $B$ than $A$ such that $\delta(P, \overrightarrow{AB}) = 1$ and $\overline{PB} = 2k$. We may assume a sub-optimal case for the hunter, that the tracker always reports some position on ray $\overrightarrow{AB}$ as the rabbit moves. This is clearly possible because as the rabbit moves on a straight line from $B$ to $P$, its distance to $\overrightarrow{AB}$ is strictly less than $1$ since $\delta(P, \overrightarrow{AB}) = 1$. Hence, since the hunter only has access to the tracker, in this case, the hunter moves at a pace of $1$ unit per turn along the line formed by $A$ and $B$ while starting at $A$. This goes on for $2k$ turns, when the rabbit reaches point $P$ and the hunter is at some point $H \neq A$ where $HB = 2k - d > k$. Now consider triangle $PBH$. Note that if we let $\theta = \angle PBH$, we can find\[PH^2 = 5k^2 - 2k\sqrt{4k^2 - 1}\]since $\cos{\theta} = \tfrac{\sqrt{4k^2 - 1}}{2k}$ so therefore, we actually have $PH > \sqrt{5k^2 - 2k\sqrt{4k^2 - 1}}$. This expression is not very nice, so in fact we can make the argument that\[4k^2 - 2k\sqrt{4k^2 - 1} \geq \frac12\]which can be easily proven by expanding, hence $PH > \sqrt{k^2 + \frac12} \geq \sqrt{d^2 + \frac12}$ as desired. $\square$ After the first turn, it is clearly possible for $AB = 2$. Suppose that under the strategy outlined by the claim above, after $10^9$ moves, the distance $AB \leq 100$. Since the distance never exceeds $100$, it never takes over $200$ moves to increase the square of the distance by $\tfrac12$, hence the square of the distance must increase by at least $\tfrac{10^9}{200} \cdot \tfrac12 > 10^6$, which is a clear contradiction due to size. We are done. $\blacksquare$

The answer is $\boxed{\text{no}}$. The main claim to this problem is the following: Claim: Let the distance between the hunter and the rabbit be $d$. After $\lfloor 500d \rfloor$ turns, the hunter cannot guarantee being within distance $\sqrt{d^2+\frac{1}{2}}$ from the rabbit. Proof: Let the hunter initially be at point $H$ and the rabbit at point $R$. Let $\ell$ be line $HR$. Let $A$ and $B$ be reflections across $\ell$ such that the projection from $A$ and $B$ to $\ell$ has length $1$ and $RA = RB = \lfloor 500d \rfloor$. Let the rabbit move across line $RA$, and suppose that the tracking device reports a point $P$ on $\ell$ every time. (We can safely assume this happens because we are asked if, regardless of where the tracking device reports, the hunter can still win). Note that if this happens, the hunter has no way of knowing which side of $\ell$ the rabbit is on, so if the hunter wants to guarantee the win, she must move along line $\ell$. After $\lfloor 500d \rfloor$ moves, the rabbit will arrive at point $A$. Suppose that the hunter is at point $X$ on line $\ell$, where $HX = \lfloor 500d \rfloor$. Now this is a simple computation using the Pythagorean Theorem, with details left to the interested reader. $\square$ Note that applying this claim $2 \cdot 10^4$ times, we can extend the distance $d$ to $100$ in under $10^9$ moves, as desired.

fattypiggy123 wrote: 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 $n^{\text{th}}$ round of the game, three things occur in order: The rabbit moves invisibly to a point $A_n$ such that the distance between $A_{n-1}$ and $A_n$ is exactly $1.$ 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 $1.$ 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 $100?$ Proposed by Gerhard Woeginger, Austria The answer is NO the hunter cannot ensure the required conditions. Let $d_n$ be distance between hunter($H_n$) and rabbit($R_n$) are $n$ rounds.Let $\epsilon=0.0025$.If at any point $d_n\geq 100$ then the rabbit can keeping moving in the same straight line as the hunter and so will never be caught.Hence assume $d_n<100$ for all $n\leq 10^9$.Let $H'$ be a point $200$ units from $H_n$ and $Y_1,Y_2$ be point such that $Y_1Z=1=Y_2Z$ and $R_nZ=200$ . Keep in mind that $$(400\epsilon-2\epsilon d_n)>(400\cdot 0.0025-2\cdot 0.0025\cdot 100)=\tfrac{1}{2}$$ed); Claim : The rabbit has a way of increasing $d_n^2$ by $\tfrac{1}{2}$ every $200$ rounds. Proof : Now if the rabbit goes either to $R_n\to Y_1$ or $R_n\to Y_2$ then note that the rightmost that the hunter can go is $H'$ and if he doesnt go to this point then he will definitely end up to the left of $H'$.If he went above $H_nH'$ then if the rabbit chose $Y_2$ the distance $d_n$ between would increase. [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair B, Bp, A, Ap, P,Ap2; B=(0,0); Bp=(2,0); A=(5,0); Ap=(5,0)+2*dir(20); Ap2=(5,0)-2*dir(160); P=foot(Ap,B,A); draw(B--P); draw(Bp--Ap--P,dashed); draw(Bp--Ap2--P,dashed); draw(B--Bp,linewidth(1.5)); draw(A--Ap,linewidth(1.5)); draw(A--Ap2,linewidth(1.5)); dot("$H_n$",B,S); dot("$R_n$",Bp,S); dot("$H'$",A,S); dot("$Y_1$",Ap,NE); dot("$Y_2$",Ap2,SE); dot("$Z$",P,SE); [/asy][/asy] Therefore the best bet for the hunter is to go from $H_n\to H'$ wherein the rabbit ends up either at $Y_1/Y_2$.Hence \begin{align*}d_{n+1}^2&=1+(H'Z)^2 \\ &= 1+(d_n-(200-R_nZ)^2 \\ &\approx 1+(d_n-\epsilon)^2 \\ &= d_n^2+(1+\epsilon^2-2\epsilon d_n) \\ &\approx d_n^2+(400\epsilon-2\epsilon d_n)\\ &> d_n^2+\tfrac{1}{2}\end{align*}hence proving the claim .$\square$ Back to the original problem by our claim the rabbit can increase $d_n^2$ by $\tfrac{1}{2}$ in every $200$ rounds,Hence $d_n^2$ reaches $100^2$ in $<2\cdot 10^4\cdot 200=4\cdot 10^6<10^9$ rounds so the rabbit wins and we are done.$\blacksquare$

dame dame

We present instead the claim that after $4d$ moves, the hunter cannot guarantee being within distance $\sqrt{d^2+\frac{1}{2}}$ of the rabbit. Let the hunter start at $H$ and the rabbit start at $R$. Assume that the rabbit reveals its location R, which renders all previous information about it unnecessary to solving the problem. The rabbit can choose $2$ points $A,B$ with $A$ above and $B$ below line $RH$ such that $AB=2$ and $k=RA=RB$. Let the midpoint between $A,B$ be $N$ Now, the rabbit will go to either $X$ or $Y$ which will take $k$ hops, and he will report the location of point $P_n$ on the line $RH$. For all points $I$ that the hunter can now go to, clearly it is optimal for $HI=k$ and $I$ is on line $RH$, so $IA=IB$. Because of this, you can now get: \[IA^2=1+IN^2=1+(\sqrt{RA^2-1}-RI)^2\]\[IA^2=1+(\sqrt{k^2-1})-(k-d))^2\]\[IA^2\geq 1+((k-\frac{1}{k})-(k-d))^2=1+(\frac{(d-1)^2}{k^2})\] So for this problem it remains to show that this is greater than $d^2+0.5$ if $k=4d$, and this is true if $d\geq1$ which it is.

Can someone check this please? The answer is $\boxed{\text{no}}$. Assume that at some point the rabbit is $d$ units away from the hunter and the hunter is aware of the rabbit's decision, which let us label as point $A$. We now make the following claim: Claim: The square of the distance can be increased by $O \left( \frac{1}{d}\right)$ keeping the number of moves to $\lceil d \rceil + 1$ at most. To achieve this, the rabbit moves in a straight line for $\lceil d\rceil +1$ moves in such a way that the angle formed by $\overrightarrow{BA}$ to the line does not exceed $2 \sin^{-1} \frac{1}{2 \lceil d\rceil}$. Since the chord that is formed by the intersections of the this line and $\overrightarrow{BA}$ on the circle having a radius $p$ ($p \le \lceil d \rceil$) and a center $A$ always has length that does not exceed $1$. Hence, on the $r^{\text{th}}$ turn of the rabbit, the tracker can always ping on the intersection of $\overrightarrow{BA}$ with the circle of radius $p$. Therefore, even if the hunter was able to deduce the rabbit's strategy, there is no way for him to find out the line along which the rabbit is moving until the $\lceil d \rceil + 1^{\text{st}}$ turn. After the $\lceil d \rceil + 1^{\text{st}}$ turn, the hunter is in an arbitrary position inside the circle of radius $\lceil d \rceil + 1$ that is centered about the point $B$. Let's say the hunter is at the point $B_1$ and the rabbit is at the point $A_1$ a point on the circle having radius of $\lceil d \rceil + 1$ having a center $A$. If the point $B_1$ is on one side of $BA$, let's assume that the rabbit is moving along the line with an angle of $2 \sin^{-1} \frac{1}{2 \lceil d\rceil}$ from $BA$ on the other side if $BA$. Then let's say if $P_1$ is the point $\lceil d\rceil - d + 1$ units above $A$ on the line $\overrightarrow{BA}$ we get that $P_1A_1 \ge B_1A_1$ and then Law of Cosines yields the folloiwing \begin{align*} A_1P_1 &= (+1)^2+(\lceil d \rceil-d+1)^2-2(\lceil d \rceil+1)(\lceil d \rceil-d+1)\cos^{-1}\left(2\sin^{-1}\frac{1}{2\lceil d \rceil+2}\right) \\ &= d^2+(\lceil d \rceil+1)(\lceil d \rceil-d+1)\frac{4(\lceil d \rceil+1)^2-1}{(\lceil d \rceil+1)^4} \\ &\ge \frac{3}{\lceil d \rceil + 1} \end{align*}Hence, the rabbit can get from distance $k$ to $k+1$ by using the process to increase by $\frac{3}{(d+1)}$ and then revealing it's location to start all over again. As the squared distance increases by a minimum of $\frac{3}{(n+3)}$ every time, this takes at most of \[ \frac{(2n+1)(n+3)}{3} \cdot (n+2)\]moves. Observe that at the beginning, the rabbit can get away by a distance of $2m$ from the hunter if the hunter chooses antipodal points, hence we need a minimum of \[1+ \sum_{n = 1}^{100}\frac{(2n+1)(n+3)(n+2)}{3} < 100 + \sum_{n = 2}^{100} n^3 = 25502559 < 10^9\]hence the rabbit wins, which finishes the problem.

Thanks to L567 for explaining me the motive of this problem! Claim - Let the distance between the rabbit and the hunter be some non zero number $d$ at some instant. The hunter cannot guarantee being within $\sqrt{d^2+\frac{1}{2}}$ of the rabbit after $\lfloor 500d \rfloor$ moves. Proof - At an instant, let the hunter be at $A$ and the rabbit is at $B$. After $n=\lfloor 500d \rfloor$ turns, let the rabbit goes to either $X$ or $Y$ such that $X$ and $Y$ are symmetric while the hunter goes to a point $A'$ which lies on the line $AB$. [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair B, Bp, A, Ap, P,Ap2; B=(0,0); Bp=(3,0); A=(6,0); Ap=(6,0)+2*dir(40); Ap2=(6,0)-2*dir(140); P=foot(Ap,B,A); draw(B--P); draw(Bp--Ap--P,dashed); draw(Bp--Ap2--P,dashed); draw(B--Bp,linewidth(1.5)); draw(A--Ap,linewidth(1.5)); draw(A--Ap2,linewidth(1.5)); dot("$A$",B,S); dot("$B$",Bp,S); dot("$A'$",A,S); dot("$X$",Ap,NE); dot("$Y$",Ap2,SE); dot("$M$",P,SE); [/asy][/asy] $A'X=A'Y=\sqrt{(\sqrt{n^2-1}-(n-d))^2+1}=\sqrt{d^2+2(n-d)(n-\sqrt{n^2-1})}=\sqrt{d^2+\frac{2(n-d)}{n+\sqrt{n^2-1}}}\geq \sqrt{d^2+\frac{n-d}{n}}>\sqrt{d^2+\frac{1}{2}}$ Now, applying this claim $2\cdot 10^4$ times we get that in at most $10^9$ moves, the distance between rabbit and the hunter is more than $100$. So, the answer is $\boxed{\text{No}}$

Could I pose a question for all those guys that argue like: "the rabbit goes to this or that direction and the hunter goes in another direction (in between)..." . What if the hunter cheats and mounts another device on the rabbit, but this time a real one that shows its exact position. So, what if the hunter goes straight after the rabbit? I mean, leave alone the real position of the rabbit. It doesn't matter what direction it heads on. It may sit in a chair and manipulating the pointing device. What matters is only the position this device reveals. The real position of the rabbit does not exist, until the moment he claims - I am here - and if it's consistent with the rules, that is, if he could present a list of moves, not contradicting with the rules, that lead to this imaginary position, then it's a proof that the rabbit's location is exactly there.

The point of this post is to show another wording of the problem. I am sure that if this had been the text given at 2017 IMO, the results could have been much, much better. I will not do the exact calculations, but rather just show that the rabbit may escape. (Same problem, different wording) A hunter and a phantom play a game in the Euclidean plane. The phantom's starting point, $ P_0,$ and the hunter's starting point, $ B_0$ are the same. After $ n-1$ rounds of the game, the phantom is at point $ P_{n-1}$ and the hunter is at point $ B_{n-1}.$ In the $ n$-th round of the game, two things occur in order: (i) The phantom moves visibly to a point $ P_n$ such that there exist points $ A_0,A_1,A_2,\dots,A_n$ satisfying $ A_0\equiv P_0\,,\, |A_{j-1}A_j|=1\,,\,|P_jA_j|\le 1, j=1,2,\dots,n.$ (ii) The hunter moves visibly to a point $ B_n$ such that $ |B_{n-1}B_n|=1.$ Is it always possible, no matter how the phantom moves, for the hunter to choose his moves so that after $ 10^9$ rounds, he can ensure that the distance between him and the phantom is at most $ 100$? Let's first see that this is the same problem, although there is no rabbit in it. The rabbit is hidden behind the words. In fact, the points $ A_0=P_0, A_1,A_2,\dots$ are the successive positions of the rabbit. Note that in this formulation we explicitly allow the points $ A_1,A_2,\dots, A_n$ to be different at each round. Whenever we want to calculate where the phantom might go, given its previous positions $ P_0,P_1,\dots,P_{n-1},$ we search for possible points $ A_j, j=1,2,\dots,n$ that satisfy the given condition. You may not agree with this, you may say - Why? The positions $ A_1,A_2,\dots,A_{n-1}$ are already selected, because they represent the positions the rabbit has already been to! We only need to choose the last position $ A_n$ of the rabbit! I would disagree with you! Remember that the rabbit is invisible, no one can see it. So, if at, say, $ n$-th position, the hunter complains that the rabbit is cheating, he has no reasons - there is a possibility the rabbit is a fair player. Indeed, there is a probability the rabbit's positions were indeed $ A_1,A_2,\dots,A_n$ - no one knows, but they meet all the conditions! We further refer to the point $ A_i$ as the shadow-position of the corresponding point $ P_i.$ Assume at some moment the hunter is at a point $ B$ and the phantom - at a point $ P$ at distance $ d:=|BP|.$ Suppose the phantom has revealed that its last shadow-position $ A$ coincides with $ P.$ Further, the phantoms steps at points $ P_1,P_2,\dots,P_m$ which lie on the ray $ BP$ (see fig 1) with $ PP_1=P_1P_2=\dots=P_{m-1}P_m$ and this distance is a bit less than $ 1.$ We construct arcs $ A'_iA''_i$ with center $ P$ and radius $ PA'_i=PA''_i=i, i=1,2,\dots,m$ so that $ P_i$ is the projection of $ A'_i$ and $ A''_i$ on the line $ BP.$ The smaller the angle $ \angle A'_1PA''_2,$ the closer $ P_iP_{i+1}$ is to $ 1.$ Suppose for the final arc we have $ A'_mP_m=A''_mP_m=1.$ Note that for the last position $P_m$ of the phantom, any point $ A_m$ that lies on the arc $ A'_mA''_m$ can serve as its shadow-point (fig. 1). Let us estimate the difference between the distances $ |PA'_m|$ and $ |PP_m|.$ \begin{align*} |PA'_m|-|PP_m|&=|PA'_m|-\sqrt{|PA'_m|^2-1}\\ &=m-\sqrt{m^2-1} =\frac{1}{\sqrt{m^2-1}+m}\\ &\le \frac{1}{m}=\varepsilon. \end{align*} Note that $\varepsilon$ at the right hand side can be made as small as we want providing $m$ is large enough. The distance traveled by the hunter is equal to $ PA'_m=m,$ which means the phantom's lead after these $ m$ rounds is eventually a bit smaller than $ d=|BP|.$ That is, $|B'P_m|\ge d-\varepsilon.$ At this moment the phantom looks at the hunter. If he is in the half plane below the line $ BP$ the phantom jumps from $ P_m$ to $ P'= A'_m$ and reveals its last shadow-position as being $ A'=A'_m.$ Otherwise, the phantom jumps to $ P'= A''_m$ and correspondingly reveals its last shadow-position being $ A'=A''_m.$ The distance between the hunter and the phantom now is $ |B'P'|\ge \sqrt{(d-\varepsilon)^2+1},$ and since we can make $ \varepsilon$ as small as we want, the distance $ |B'P'|$ is close to $ \sqrt{d^2+1}.$ Thus $$ \displaystyle |B'P'|-d \approx \sqrt{d^2+1}-d=\frac{1}{\sqrt{d^2+1}+d}>\frac{1}{3d}\qquad(1).$$ To recap, we started with a distance $ d$ between $ P$ and $ B$ and the phantom last shadow-position being $ P$ and ended up with phantom at a point $ P',$ hunter at $ B',$ and the last shadow-position at $ P'.$ Moreover, we increased the distance $ BP$ from $ d$ to $ \displaystyle d+\frac{1}{3d}.$ Repeating this action several times we can increase the distance $ PB$ from $ 1$ to $ \displaystyle \sum_{i=1}^k\frac{1}{3i}.$ Since the last sum can be made as large as we want, it follows the phantom can escape from the hunter. In order to get full score on this problem it remains only to make routine estimates and establish that after $10^9$ rounds the phantom makes the distance greater than 100 units. Look again at $(1)$. The precise estimate is \begin{align*} |B'P'|-d &\ge \sqrt{(d-\varepsilon)^2+1}-d=\frac{(d-\varepsilon)^2+1=d^2}{\sqrt{(d-\varepsilon)^2+1}+d}\\ &=\frac{1+\varepsilon^2-2d\varepsilon}{\sqrt{(d-\varepsilon)^2+1}+d}\qquad(2). \end{align*}Now, let us put $\frac{1}{m}=\varepsilon\le \frac{1}{4d}$ which is satisfied when $m\ge 4d.$ In this case, back to $(2),$ it yields $$|B'P'|-d\ge \frac{1}{2}\cdot \frac{1}{\sqrt{(d-\varepsilon)^2+1}+d}\ge \frac{1}{6d}.$$So, starting from a distance $d$ between $B$ and $P$ we increased this distance to $d+\frac{1}{6d}$ after $m=\left\lceil 4d\right\rceil$ rounds. We repeat this procedure while $d<100.$ This means that after every at most $\left\lceil 4d\right\rceil\le 400$ rounds, we increase the distance by $\frac{1}{6d}>1/600.$ Thus, after at most $400\cdot 600\cdot 100=24\cdot 10^6$ steps, this distance certainly exceeds $100$. Remark: More details in my blog.

yayayayayayyayayayayayayayayayayayayayayayayayayyayayayayayayay Solved with hint 19% from OTIS (for fun, i added 10 but i wonder who added 19\% and 39\% after i asked for help but tried to play it off as if they didn't want to help :thonk:) The key is that even if the rabbit HAD revealed its location from point B a distance d away from point A the hunter, draw the line AB, oriented horizontally. Then the rabbit can move n times diagonally up right to B', s.t. if C is the foot from B' to AB, B'C=1. Indeed, since the hunter doesn't know which side of the line the rabbit's on, to minimize the risk (this is because he needs to GUARANTEE), he just moves n directly right to some point A'. Now compute $$A'B=n-d\Rightarrow A'C=\sqrt{n^2-1}-(n-d)\Rightarrow A'B'^2= 1 + \left(\sqrt{n^2-1}-(n-d)\right)^2$$$$\ge 1 + \left(n-\tfrac1n) - (n-d) \right)^2= 1 + (d-1/n)^2>d^2 + \frac12\forall n \ge 4d,$$which is a LOCAL change by 1/2 every time (if you only consider distance squared, so we'll just get distance squared is $>100^2$). Simply taking n=400 suffices (this means it takes 400 moves to increase by 1/2): $400 \cdot 2 \cdot 100^2 < 10^9$ moves has >100 distance already.

Suppose the hunter is at $B_{n-1}$ while the rabbit has just moved to $A_n$. If the tracking device reports $P_n$, then consider $A'_n$ which is the reflection of $A_n$ across $B_{n-1}P_n$. Let $B'_n$ be on $B_{n-1}P_n$ such that $B_{n-1}B'_n=1$. If the hunter goes to a point $B_n$ that strays from the direction of $B_{n-1}P_n$ then either $B_nA_n$ or $B_nA'_n$ will be greater than $B'_nA_n$, which is equal to $B'_nA'_n$. Since the hunter cannot determine whether the rabbit is at $A_n$ or $A'_n$, she guarantees the least distance if she goes directly towards $P_n$. We can use this fact to continuously lead the hunter along a straight line $\ell$, initially containing both the hunter and the rabbit, while the rabbit strays further and further away $\ell$. As long as our tracker continues to report points on $\ell$, the hunter must go along $\ell$. Let $k$ be any number. Suppose in a single move, we move the rabbit away from $\ell$ by $\tfrac{\sqrt{k^2-1}}{k}$ and along $\ell$, but away from the hunter. Any time over the first $k$ moves, the rabbit's distance to $\ell$ is at most $1$ so the tracking device can always find a point on $\ell$ to report. Let the distance to the hunter initially be $d$. Then, after $k$ moves of this sort, the distance will be \[\sqrt{((d-k)+\sqrt{k^2-1})^2+1}=\sqrt{d^2+2(k-d)(k-\sqrt{k^2-1})}=\sqrt{d^2+\frac{2(k-d)}{k+\sqrt{k^2-1}}}\]After the first move, the hunter cannot guarantee being less than distance of $2$. After $k$ moves, the hunter cannot guarantee being within a distance of $\sqrt{d^2+\tfrac{k-d}{k}}$. It is evident that if the distance ever becomes greater than $100$ then the rabbit can simply run directly away from the hunter and the hunter will never be able to catch up. Therefore, if the hunter wants to ensure she is at most $100$ away from the rabbit at the end, she must ensure that she is always within $100$ away. Suppose she can do this. Then we can set $k=200$. After $200$ moves, the hunter cannot guarantee being within distance \[\sqrt{d^2+\frac{1}{2}}\]of the rabbit. Thus, after $4000000=4\cdot 10^6$ moves, the hunter cannot guarantee being within distance \[\sqrt{d^2+10000}>100\]which is a contradiction! Thus, the hunter cannot guarantee.