AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
denote the set of positive integers. Consider a function 
. For any 
we write 
. Suppose that 
has the following two properties:, then 
;
is finite.
is periodic.
, find all functions where 
for all 
,
he may move to any of the points 
, 
, 
, 
and 
. Show that if he starts at 
he can never get to 
.
be the prime numbers listed in increasing order, and let 
be a real number between 0 and 1. For positive integer 
, define![\[ x_k = \begin{cases} 0 & \mbox{if} \; x_{k-1} = 0, \\[.1in] {\displaystyle \left\{ \frac{p_k}{x_{k-1}} \right\}} & \mbox{if} \; x_{k-1} \neq 0, \end{cases} \]](http://latex.artofproblemsolving.com/e/1/7/e17b3fc5962fba938dee877dac266c3cd024a6e3.png)
denotes the fractional part of . (The fractional part of is given by 
where 
is the greatest integer less than or equal to .) Find, with proof, all satisfying 
for which the sequence 
eventually becomes 0.
of positive integers, Ali proceed the following algorythm: In the i-th step he markes all rational numbers in the interval ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
which have denominator equal to 
. Then he write down the number 
equal to the length of the smallest interval in which both two ends of that is a marked number. Find all sequences with 
and such that for all 
we have
be an acute triangle (
) with circumcircle 
. Point 
is chosen on arc 
, not containing 
, so that, letting 
intersect at 
, one has 
. Points 
and 
lie on lines 
and 
respectively, such that 
is a parallelogram. Point 
is chosen on arc , not containing 
, such that 
. Line 
intersects 
at 
, and line 
intersects 
at 
. Line 
intersects , and 
at points 
, 
and 
respectively.
.
such that for all 
we have:
and 
be prime numbers. Prove that if 
is a perfect square, then 
is also a perfect square.
-term sequence 
in which each term is either 0 or 1 is called a binary sequence of length . Let 
be the number of binary sequences of length containing no three consecutive terms equal to 0, 1, 0 in that order. Let 
be the number of binary sequences of length that contain no four consecutive terms equal to 0, 0, 1, 1 or 1, 1, 0, 0 in that order. Prove that 
for all positive integers .
points are given on a circle. On a player's turn, that player must draw a triangle using three unused points as vertices, without creating any crossing edges. The first player who cannot move loses. If Elmo's clone goes first and players alternate turns, who wins? (Your answer may be in terms of .) greater than 10. Then, she multiplies all the digits of and obtains an odd number. Find all possible values of the units digit of .
cups labeled 
, where the 
-th cup has capacity liters. In total, there are liters of water distributed among these cups such that each cup contains an integer amount of water. In each step, we may transfer water from one cup to another. The process continues until either the source cup becomes empty or the destination cup becomes full.
Prove that from any configuration where each cup contains an integer amount of water, it is possible to reach a configuration in which each cup contains exactly 1 liter of water in at most 
steps.
Prove that in at most 
steps, one can go from any configuration with integer water amounts to any other configuration with the same property.
satisfy 
. Prove that
When does the equality occur?
where 
and 
.
![[asy]size(10cm);
import markers;
pair X1=(0,2),X2=(0,-2),Y1=(sqrt(7),1),Y2=(-sqrt(7),1),Z1=(sqrt(7),-1),Z2=(-sqrt(7),-1);
fill(X1--Y1--Z1--cycle^^X1--Y1--Z2--cycle,cyan);draw(X1--Y1--Z1--cycle^^X1--Y1--Z2--cycle,blue);
draw(anglemark(Y1,Z2,X1,10)^^anglemark(Z1,X1,Y1,10),blue);
draw(anglemark(X1,Y1,Z2,10,12)^^anglemark(Y1,Z1,X1,10,12),blue);
D(MP("X_1\;(0,2)",X1,N)--MP("Y_1\;(\sqrt{7},1)",Y1,NE)--MP("Z_1\;(\sqrt{7},-1)",Z1,SE)--MP("X_2\;(0,-2)",X2,S)--MP("Z_2\;(-\sqrt{7},-1)",Z2,SW)--MP("Y_2\;(-\sqrt{7},1)",Y2,NW)--X1);
label("$\triangle X_1Y_1Z_1\sim\triangle Z_2X_1Y_1$",(0,-3));
dot(X1^^X2^^Y1^^Y2^^Z1^^Z2);
[/asy]](http://latex.artofproblemsolving.com/7/a/0/7a0905fae32e34c8667c745d10a5f5e9d6735020.png)
where and .
(though slightly annoying, I'll follow the notation of the diagram from the official solution). Now let 
be the point such that 
is a parallelogram. From this we already know that 
.
such that 
( on the same side of 
with respect to 
). In other words, we consider the spiral similarity with center 
that maps to , and define as the image of 
via this transformation. Now, by definition we get that 
is similar to our original triangle . We shall now prove that 
is similar to the original triangle as well. More precisely, we'll show that 
.
, where in the last step we have used that 
, due to the similarity. So it is enough to prove that 
. be the lengths of the sides opposite to 
in our original triangle. Now 
, hence the ratio of the similarity is 
. Therefore, 
, and we have 
, as wanted.
are similar, and this holds without any extra requirements on the initial triangle .
is constructed such that 
is a parallelogram. Notice this implies that 
is a parallelogram as well, because it has a pair of opposite sides that are parallel and of equal length.
(respective sides are equal), and also 
(two halves of a parallelogram). So we have 6 similar triangles now.
and 
are congruent (two halves of a parallelogram). So if happens to be similar to the initial triangle , then all 8 triangles 
are similar and we are done.
if and only if 
in the starting triangle. Provided that this is the case, we further need 
, that is 
. Now, 
, and because of the similarity we have that 
, hence 
. So we need 
, and if we set 
we have 
, which is the same to saying that 
and 
for some 
. 
AND 
??
in the plane such that all of the triangles 
?
