[/quote]
,
be a triangle with circumcircle 
and let 
be the 
-
and 
be the intersection points of 
and 
be the projections of 
intersects the tangent line at 
at a point 
.
.
and 
: Prove that :
board and made 
moves so that it visited all squares and returned to the starting square. At every moment, the distance from the center of the square the king was on to the center of the board was calculated. A move is called 
if this distance becomes smaller after the move. Find the maximum possible number of pleasant moves. (The chess king moves to a square adjacent either by side or by corner.)
is lovely if there exist a positive integer 
and (not necessarily distinct) positive integers 
,
,
,
such that 
and 
for 
.
,
be an odd prime. Prove that:![\[\displaystyle\sum_{k=1}^{p-1}k^{2p-1} \equiv \frac{p(p+1)}{2} \pmod{p^2}\]](http://latex.artofproblemsolving.com/a/5/5/a55e75d4d9769ceec3a3ce359aac676c7c7875fd.png)
,
,
,
'![\[n=a_1+a_2+\dots+a_m,\]](http://latex.artofproblemsolving.com/6/c/2/6c2a979a5ff6c40bf2d7152c8c32088fc36f848b.png)
where 
,
,
,
.
![\[
\begin{cases}
x - y z = 1,\\[2pt]
y - z x = 2,\\[2pt]
z - x y = 4.
\end{cases}
\]](http://latex.artofproblemsolving.com/b/2/d/b2d94e49491a71f86109103f985fe3f238cf4d95.png)
be a function such that
for all 
.
.
such that the equality 
holds for all pairs of real numbers 
.
is an obvious solution. Now, let 
.
or 
.
.
and 
is a square of an integer for all nonnegative integers 
.
is defined as follows: 
if 
and 
if 
.
,
be all positive integers smaller than 
such that 
for all 
is the largest positive integer that divides both 
and 
.
.
such that 
.
is a benefactor of 
because 
.
benefactors.
denotes the least natural such that
Find all naturals 
such that 
.
such that the distance between 
to the hunter. The only guarantee provided by the tracking device to the hunter is that the distance between 
such that the distance between 
and 
Y by anantmudgal09, bobcats4thewin, laegolas, Ankoganit, mira74, nmd27082001, JasperL, kk108, rkm0959, mathwiz0803, Lam.DL.01, Tawan, don2001, Ultroid999OCPN, integrated_JRC, Amir Hossein, Davi-8191, W.Sun, gyluo, Wizard_32, Varuneshwara, tintin21, mijail, The_Maitreyo1, OlympusHero, Shashank-, DarthVader2618, tenebrine, franzliszt, megarnie, samrocksnature, centslordm, tigerzhang, rayfish, fidgetboss_4000, TheHimMan, Lamboreghini, JG666, ImSh95, rohan.sp, fuzimiao2013, aidan0626, EpicBird08, LoloChen, Adventure10, Mango247, Sedro, KhaiMathAddict
A hunter and an invisible rabbit play a game in the Euclidean plane. The rabbit's starting point, 
and the hunter's starting point, 
are the same. After 
rounds of the game, the rabbit is at point 
and the hunter is at point 
In the 
round of the game, three things occur in order:
rounds, she can ensure that the distance between her and the rabbit is at most 
Proposed by Gerhard Woeginger, Austria
and the hunter's starting point, 
are the same. After 
rounds of the game, the rabbit is at point 
and the hunter is at point 
In the 
round of the game, three things occur in order:- The rabbit moves invisibly to a point
such that the distance between and is exactly 
- A tracking device reports a point
to the hunter. The only guarantee provided by the tracking device to the hunter is that the distance between and is at most
- The hunter moves visibly to a point
such that the distance between 
and is exactly
rounds, she can ensure that the distance between her and the rabbit is at most 
Proposed by Gerhard Woeginger, Austria
This post has been edited 6 times. Last edited by djmathman, Jun 16, 2020, 4:13 AM
Reason: problem author
Reason: problem author
This only works for 
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 
and 
(such a case gives an equality in the bound above (unless I made a mistake)). Note that this is not always possible.
is already big (~10) and it seems like tracing back more than one point 
then we have literally no info, so we may get 
-
-
below the top of the 
if the rabbit can be anywhere in an 
.
.
.
to 
.
to 
rounds. With some algebra (rabbits are good at multiplying), we can show that 
,
in at most 
.
to 
in 
rounds. Thus we can get from 
to 
rounds, which is much less than 
,
in fact, and we are done.
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.
such that 
,
is the projection of 
,
.
.![[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]](http://latex.artofproblemsolving.com/f/5/0/f503e827c41ec47146606e22b46ff7c0f0d7ad33.png)
.
-
,
.
.
Let 
above. Check that ![\[A'B'^2\ge x^2+\frac{2(m-x)}{m+\sqrt{m^2-1}}\ge x^2+\frac{m-x}m\ge x^2+\frac1{101},\]](http://latex.artofproblemsolving.com/2/3/3/233bf5e3cd7f0ad7ef7f65a262f20a141d2ec146.png)
as desired. 
times; that is, ![\[A_NB_N^2\ge\left\lfloor\frac{10^9}{101}\right\rfloor\cdot\frac1{101}\ge\frac{10^9-100}{101^2}>100^2,\]](http://latex.artofproblemsolving.com/b/0/2/b02bd74bc871a57c4b9709a5555af152badfb823.png)
so 
,
in at most 
moves. For convenience, denote 
moves, such that the angle from the ray 
to the line is at most 
.
centered about 
,
unit above 
,
by using this process to increase by 
and then leaking his location to start again. As the distance squared increase by at least 
each time, this takes at most 
moves.
away from the hunter if they happen to choose antipodal points, so we need at least 
.
moves, it is possible for the distance between hunter and rabbit to increase to at least 
.
.
.
and 
.
as the rabbit moves. This is clearly possible because as the rabbit moves on a straight line from 
turns, when the rabbit reaches point 
where 
.
.
,![\[PH^2 = 5k^2 - 2k\sqrt{4k^2 - 1}\]](http://latex.artofproblemsolving.com/a/2/b/a2bd990cd9d02584f5d541b2c5ee98d001ea7eb6.png)
since 
so therefore, we actually have 
.![\[4k^2 - 2k\sqrt{4k^2 - 1} \geq \frac12\]](http://latex.artofproblemsolving.com/c/8/e/c8e9f38385e113450c75b71bdd9e24ecc697a36d.png)
which can be easily proven by expanding, hence 
as desired. 
.
.
,
moves to increase the square of the distance by 
,
,
.
turns, the hunter cannot guarantee being within distance 
from the rabbit.
and the rabbit at point 
be line 
.
.
,
.
times, we can extend the distance 
be distance between hunter
)
)
.
then the rabbit can keeping moving in the same straight line as the hunter and so will never be caught.Hence assume 
for all 
.
be a point 
be point such that 
and 
.
ed);
by 
every 
or 
then note that the rightmost that the hunter can go is 
then if the rabbit chose 
the distance ![[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]](http://latex.artofproblemsolving.com/b/f/f/bff9a9c43432aa5bd484fbec63a637d8db318312.png)
wherein the rabbit ends up either at 
.
hence proving the claim 
in 
rounds so the rabbit wins and we are done
,
and the hunter is at point 
such that the distance between 
moves, the hunter cannot guarantee being within distance 
away from the rabbit. Let 
and 
be the locations of the rabbit and the hunter, respectively, after move 
.
be the 
-
and 
.
on move 
is the point such that 
and 
moves, we claim that 
,
over 
,
,
for all 
.
as 
,
.
be the reflection of 
over 
,
.
so 
.
.
,
and 
is increasing over ![$[0^{\circ},90^{\circ}]$](http://latex.artofproblemsolving.com/7/2/b/72bc44e818e4fa902174b0aa4eaff595769b40fa.png)
.
,
,
to line 
.
or 
is at least 
or 
is at least
We claim that this is at least 
.
Substituting this into the equation, we find that
Since 
,
,
,
turns, the rabbit can increase the square of the distance by at least 
.
turns, the rabbit can increase the square of the distance to at least 
,
.
moves, the hunter cannot guarantee being within distance 
with 
such that 
and 
.
that the hunter can now go to, clearly it is optimal for 
and 
.![\[IA^2=1+IN^2=1+(\sqrt{RA^2-1}-RI)^2\]](http://latex.artofproblemsolving.com/8/9/e/89e84c82034d176f625378fa53fe78b2bdb9053a.png)
![\[IA^2=1+(\sqrt{k^2-1})-(k-d))^2\]](http://latex.artofproblemsolving.com/d/9/5/d95bb297d950f208f32cb03cce95e779d63e3dcc.png)
![\[IA^2\geq 1+((k-\frac{1}{k})-(k-d))^2=1+(\frac{(d-1)^2}{k^2})\]](http://latex.artofproblemsolving.com/3/4/7/3479ba88dc160f5048cee8989aa0fbce629c1fef.png)
if 
,
which it is.
keeping the number of moves to 
at most.
moves in such a way that the angle formed by 
.
)
turn of the rabbit, the tracker can always ping on the intersection of 
turn.
and the rabbit is at the point 
a point on the circle having radius of 
is the point 
units above 
and then Law of Cosines yields the folloiwing
Hence, the rabbit can get from distance 
by using the process to increase by 
and then revealing it's location to start all over again. As the squared distance increases by a minimum of 
every time, this takes at most of ![\[ \frac{(2n+1)(n+3)}{3} \cdot (n+2)\]](http://latex.artofproblemsolving.com/5/a/4/5a4938b67a75801194a3a414f8dfd8146e740395.png)
moves.
from the hunter if the hunter chooses antipodal points, hence we need a minimum of ![\[1+ \sum_{n = 1}^{100}\frac{(2n+1)(n+3)(n+2)}{3} < 100 + \sum_{n = 2}^{100} n^3 = 25502559 < 10^9\]](http://latex.artofproblemsolving.com/1/f/2/1f22468199845ac5cc2471520b04411727298319.png)
hence the rabbit wins, which finishes the problem.
turns, let the rabbit goes to either ![[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]](http://latex.artofproblemsolving.com/1/7/a/17a1637b73705bc901a4f1f84b46b4a4080841da.png)
times we get that in at most 
and the hunter's starting point, 
are the same. After 
rounds of the game, the phantom is at point 
and the hunter is at point 
In the 
-
such that there exist points 
satisfying 
such that 
rounds, he can ensure that the distance between him and the phantom is at most 
?
are the successive positions of the rabbit. Note that in this formulation we explicitly allow the points 
to be different at each round. Whenever we want to calculate where the phantom might go, given its previous positions 
we search for possible points 
that satisfy the given condition. You may not agree with this, you may say - Why? The positions 
are already selected, because they represent the positions the rabbit has already been to! We only need to choose the last position 
of the rabbit!
- no one knows, but they meet all the conditions!
as the shadow-position of the corresponding point 
Assume at some moment the hunter is at a point 
and the phantom - at a point 
at distance 
Suppose the phantom has revealed that its last shadow-position 
coincides with 
Further, the phantoms steps at points 
which lie on the ray 
(see fig 1) with 
and this distance is a bit less than 
with center 
so that 
is the projection of 
and 
on the line 
The smaller the angle 
the closer 
is to 
Note that for the last position 
that lies on the arc 
can serve as its shadow-point (fig. 1). Let us estimate the difference between the distances 
and 
at the right hand side can be made as small as we want providing 
which means the phantom's lead after these 
rounds is eventually a bit smaller than 
That is, 
At this moment the phantom looks at the hunter. If he is in the half plane below the line 
to 
and reveals its last shadow-position as being 
Otherwise, the phantom jumps to 
and correspondingly reveals its last shadow-position being 
The distance between the hunter and the phantom now is 
and since we can make 
as small as we want, the distance 
is close to 
Thus
between 
hunter at 
and the last shadow-position at 
Moreover, we increased the distance 
Repeating this action several times we can increase the distance 
from 
to 
Since the last sum can be made as large as we want, it follows the phantom can escape from the hunter.
.
Now, let us put 
which is satisfied when 
In this case, back to 
it yields
So, starting from a distance 
after 
rounds. We repeat this procedure while 
This means that after every at most 
rounds, we increase the distance by 
Thus, after at most 
steps, this distance certainly exceeds 
which is a LOCAL change by 1/2 every time (if you only consider distance squared, so we'll just get distance squared is 
)
moves has >100 distance already.
which is the reflection of 
.
be on 
.
or 
will be greater than 
,
.
and along ![\[\sqrt{((d-k)+\sqrt{k^2-1})^2+1}=\sqrt{d^2+2(k-d)(k-\sqrt{k^2-1})}=\sqrt{d^2+\frac{2(k-d)}{k+\sqrt{k^2-1}}}\]](http://latex.artofproblemsolving.com/e/4/e/e4e2651d37d261ae2820ba4f85cc3840241fb83b.png)
After the first move, the hunter cannot guarantee being less than distance of 
.
.![\[\sqrt{d^2+\frac{1}{2}}\]](http://latex.artofproblemsolving.com/f/a/1/fa1f2a625ae89b959d47fe6d2a737375b7d4b307.png)
of the rabbit. Thus, after 
moves, the hunter cannot guarantee being within distance![\[\sqrt{d^2+10000}>100\]](http://latex.artofproblemsolving.com/e/e/4/ee4d949bccd1ea2ebe43e2224e7402c6fe988d0d.png)
which is a contradiction! Thus, the hunter cannot guarantee.
