AoPS Academy
of non-negative real numbers such that the following three conditions are all satisfied:
;
;
of such that 
[/list]
is a permutation of 
, and they are both permutations of 
. Note that any list is a permutation of itself.
denote the sum of digits function. Does there exist 
such that![\[
n, s(n), \dots, s^{2024}(n)
\]](http://latex.artofproblemsolving.com/c/3/3/c33803a77912ed4dde5a5c0ae05934496f04707f.png)
are all distinct perfect squares?
having integer coefficients. On each move, Vasya pays him a ruble, and calls an integer 
of his choice, which has not yet been called by him. Petya has to reply with the number of distinct integer solutions of the equation 
. The game continues until Petya is forced to repeat an answer. What minimal amount of rubles must Vasya pay in order to win?
that satisfies:![\[
\sum_{i=1}^{n} a_{\left\lfloor \frac{n}{i} \right\rfloor} = n^k
\]](http://latex.artofproblemsolving.com/b/b/e/bbe7fe81e366fe390be5d17193d1581cc40dbece.png)
be a positive integer. Prove that for all integers 
, we have:![\[
\frac{c^{a_n} - c^{a_{n-1}}}{n} \in \mathbb{Z}
\]](http://latex.artofproblemsolving.com/0/4/4/04464c1a350e3f9a929ab51534018498680068fc.png)
, 
feet of altitude from 
to 
, 
diameter in 
, 
, 
and 
midpoint of 
.
by McLaurin we have 
so 
is tangent to 
in addition as 
we must have tangent to as well, also note that 
sre colinear due to the 
angles so as 
and the given colinearity we have that 
therefore 
are colinear and now we finish with 
therefore 
are colinear and now this means that since 
we in fact have 
as desired thus we are done 
.
then 
hence 
in the radical axis of 
and 
is also on the radical axis of and 
is the radical center of 
is on the radical axis of 
and which is 
where 
their intersections;
is diameter then 
i.e. on the polar of 
so is on the polar of 
is the polar of which leads to the desired result.
. Let 
, where 
and let 
Since 
, we will be done if we show that 
. Since 
from the Ceva/Menelaus picture, we obtain that 
Hence, 
Therefore, 
, so 
where we used Heron's formula. Thus, 
as desired.
. We wish to show that 
are collinear. and is the midpoint of 
, we obtain that 
.
and since 
, we obtain that 
. is the diameter in , thus 
, so are collinear, as desired.
, feet of altitude from to , diameter in , , and midpoint of . by McLaurin we have so is tangent to in addition as we must have tangent to as well, also note that sre colinear due to the angles so as and the given colinearity we have that therefore are colinear and now we finish with therefore are colinear and now this means that since we in fact have as desired thus we are done .
,obviously 
so well known X is the center of circle appolonius for points 
such that 
.By 2002 G7 
is tangent to 
and since is a chord tangent to incircle, 
is bisector of 
so 
so is on the circle appolonius.
(the line of centers). Since is the same as 
we are done
be a point on incircle such that 
tangent to incircle.Then 
. Since 
and midpoint of , from harmonic properties we get that , from 
we get that 
, so tangent to the 
. From IMO 2002 G7 we get that 
collinear, so 
. Let the 
excircle touches at 
.
and meet the incircle again at 
and ,
.
and 
.
i.e. 
are concyclic. This implies 
are collinear and 
.
, we are done.
at and be the antipode of .
, 
, and are collinear. Let 
intersect the incircle again at . By construction 
is harmonic so 
is tangent to the incircle. So 
as desired.
and 
are similar so triangles 
and 
are also similar finishing the problem.
, 
, 
, 
, inradius
and -exradius
is midpoint of , we obtain 
.
be a point which -excircle is tangent to 
![$$XD\cdot DA’=\frac{(a+c-b)(a+b-c)}{4}=(s-b)(s-c)=\frac{s(s-a)(s-b)(s-c)}{s(s-a)}=\frac{4[ABC]^2}{(a+b+c)(b+c-a)}=\frac{2[ABC]}{a+b+c}\cdot \frac{2[ABC]}{b+c-a}=rR$$](http://latex.artofproblemsolving.com/c/1/d/c1d252b9b3413deb8340b6ecb1232699d0a181f0.png)
So, 
is a diameter in 
we get 
and 
. Notice that ![\[-1=(P,D;X,\infty_{BC})\overset{I}{=}(Q,V;K,R)\]](http://latex.artofproblemsolving.com/1/1/6/11619d3d0e46d2b4061ec993ae8bf13dbf45eabc.png)
Now, it suffices to prove that , and 
are collinear. If we let meet again at 
, it suffices to then prove 
.![\[-1=(P,D;B,C)\overset{I}{=}(Q,V;B,C)\overset{I_a}{=}(I_aQ\cap BC,\infty_{BC};B,C)\]](http://latex.artofproblemsolving.com/1/3/b/13b308d0a8dc5ec23fe353012eb07e7f0864c6e8.png)
Hence , 
and are collinear and since 
, by butterfly theorem we get that , and 
are also collinear. Therefore ![\[(Q,V;K',R)\overset{D}{=}(U,I;I_a,S)\overset{R}{=}(M,\infty_{BC};D,T)=-1. \ \ \blacksquare\]](http://latex.artofproblemsolving.com/0/a/9/0a9028a4b2c26900500c99cbf67130ea639eb9b7.png)
.
. Since 
and is the midpoint of , from EGMO Lemma 9.17 we have 
. Therefore has equal power wrt and so 
collinear. But 
so 
is tangent to . Similarly, 
is tangent to , hence 
is the pole of wrt to . By La Hire’s Theorem, lies on the pole of wrt . Since 
is tangent to , lies on the pole of too, so the conclusion follows 
be the feet of the altitude from to , be the midpoint of 
,
be the antipode of . We know that 
are colinear. Because of homothety from , we have 
.
tangent to 
are colinear 
from two of this, 
We are done.
be a triangle with incenter and 
,
at 
.
meets 
and 
is by realicing 
and knowing what 
does) and then you can realice what was everything about, spotting harmonic boundles is extremely importsnt in oly
meet 
,
meet 
meet 
meet 
,
be the foot from 
meet 
since 
we get that 
so 
.