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.
+2 w
where 
is a positive integer and 
is a prime.(Don't get mad at me,I've used the search function and did not see a correct and complete solution anywhere.)
such that 
divides 
for all positive integers 
and 
with 
.
, are there infinitely many positive integers 
such that 
is a perfect square? (Here 
denotes the integer part of the real number 
?
be an acute triangle with 
, Let 
be the intouch triangle with 
,
,
,, let 
be the intersecttion of the perpendicular from 
to 
with 
, and 
.
and 
are concylic
the set 
can be partitioned into triples in such a way that the numbers from each triple are the lengths of the sides of some obtuse triangle. be a triangle with circumcircle 
, circumcenter 
, and orthocenter 
. Assume that 
and that 
. Let 
and 
be the midpoints of sides and 
, respectively, and let 
and 
be the feet of the altitudes from 
and 
in 
, respectively. Let 
be the intersection of line 
with the tangent line to at 
. Let 
be the intersection point, other than , of with the circumcircle of 
. Let 
be the intersection of lines 
and . Prove that 
.
be the foot of perpendicular from to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle . A circle 
with centre 
passes through and , and it intersects sides and at and 
respectively. Let be the foot of altitude from to 
, and let be the midpoint of . Prove that the circumcentre of triangle 
is equidistant from and .
is a cyclic quadrilateral inscribed in a circle of radius one unit. If 
, prove that is a square.
, 
, 
, 
satisfy 
and 
. Prove that there exists a permutation 
of 
such that 
.
be a cyclic quadrilateral. A circle centered at passes through and and meets lines 
and again at points and (distinct from 
). Let denote the orthocenter of triangle 
Prove that if lines 
are concurrent, then triangle and 
are similar.
such that 
, where 
are non-negative reals such that 
.
.
be the given sequence and let 
.
and 
for all 
.
terms.
is the answer, just take 16 real numbers satisfying
.
and 
for 
. Thus we found all sequences with the given properties.
terms exists. Consider the matrix![\[\begin{bmatrix} x_1 & x_2 & x_3 & x_4 & x_5 & x_6 & x_7 & x_8 & x_9 & x_{10} \ \ \ x_{11} \\ x_2 & x_3 & x_4 & x_5 & x_6 & x_7 & x_8 & x_9 & x_{10} & x_{11} \ \ \ x_{12} \\ x_3 & x_4 & x_5 & x_6 & x_7 & x_8 & x_9 & x_{10} & x_{11} & x_{12} \ \ \ x_{13} \\ x_4 & x_5 & x_6 & x_7 & x_8 & x_9 & x_{10} & x_{11} & x_{12} & x_{13} \ \ \ x_{14} \\ x_5 & x_6 & x_7 & x_8 & x_9 & x_{10} & x_{11} & x_{12} & x_{13} & x_{14} \ \ \ x_{15} \\ x_6 & x_7 & x_8 & x_9 & x_{10} & x_{11} & x_{12} & x_{13} & x_{14} & x_{15} \ \ \ x_{16} \\ x_7 & x_8 & x_9 & x_{10} & x_{11} & x_{12} & x_{13} & x_{14} & x_{15} & x_{16} \ \ \ x_{17} \end{bmatrix}.\]](http://latex.artofproblemsolving.com/6/c/5/6c529101af135c3f0589e2c954664d9967cf001f.png)
works 
are any two distinct positive integers such that none of them divides the other, and if a sequence has the property that the sum of any consecutive terms is always positive and the sum of any 
consecutive terms is negative then the maximal length of this sequence is 
where 
. One can find a construction without too much difficulty.-sum the sum of consecutive terms. If we suppose, for contradiction, that the sequence has terms, then every 
-sum can be written as an 
-sum with a 
-sum subtracted, and must hence be positive. Similarly, every 
-sum can be written as a -sum with a -sum subtracted, and must therefore be negative. And, finally, every term can be written as a -sum with a -sum subtracted, and must therefore be positive again. However, if every term is positive, then it's impossible for every -sum to be negative, and we've reached a contradiction.
![\[ 5,5,-13,5,5,5,-13,5,5,-13,5,5,5,-13,5,5. \]](http://latex.artofproblemsolving.com/5/c/5/5c549f6c5c9138ff7540fab045cef2fb23cf40f7.png)
But how do you find such an example?
, meaning that if we can make it so that we get exactly two negative numbers of pretty large magnitude in each block of length and in each block of length we would be done by making the other numbers all equal to some positive number. It's not so obvious to pick the numbers 
and 
, but maybe this is easier?![\[20, 20, -51, 20, 20, 20, -51, 20, 20, -51, 20, 20, 20, -51, 20, 20.\]](http://latex.artofproblemsolving.com/f/8/1/f815c033f368c8adb22c2f14e3c3930db334866f.png)
reaches from the first 
to right before the last 
.
. For 
, we can consider any subset of 
.
, then suppose 
. Then, 
and 
. Therefore,
which is clearly not possible. So, any sequence of real numbers has seven consecutive terms which sum to a negative value and has eleven consecutive terms which sum to a positive value. Thus, the maximum number of terms in the sequence is 
.
then take the first 17 numbers 
and write them as follows.
, and 
can be constructed (posted in other posts above).
