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.
, the following inequality always holds:![\[
\frac{ab}{b+1} + \frac{bc}{c+1} + \frac{ca}{a+1} \geq \frac{3abc}{1 + abc}
\]](http://latex.artofproblemsolving.com/d/5/8/d582a3269c92041834c02153ee8e946a274e52bd.png)
and 
be positive integers. is a divisor of 
and is a divisor of 
.
., find all possible positive integer values of![\[
\frac{a}{b} + \frac{b}{c} + \frac{c}{a}.
\]](http://latex.artofproblemsolving.com/3/2/b/32b41635c159921e11245369e66185ec69b0d785.png)
be a triangle with 
and 
.
inside triangle satisfies the following conditions:![\[
XA^2 + BC^2 \leq XC^2 + AB^2 \leq XB^2 + AC^2.
\]](http://latex.artofproblemsolving.com/0/3/1/031aa58d7d3dabe636dfb4245ef2456a2a3d5df6.png)
Find the maximum value of 
such that![\[
\frac{XA}{AC} \geq m.
\]](http://latex.artofproblemsolving.com/c/5/e/c5ed2892b224aff3b2f3e20b4c9424707ff88690.png)
be a cyclic quadrilateral an let 
be a point on the side 
The diagonals 
meets the segments 
at 
The line through parallel to 
mmets the extension of the side 
beyond 
at 
The line through 
parallel to 
meets the extension of the side beyond at 
Prove that the circumcircles of the triangles 
and 
are tangent .
be an integer. Find all tuples of real numbers 
such that![\[
2a_1 - 3a_2,\ 2a_2 - 3a_3,\ \ldots,\ 2a_n - 3a_1
\]](http://latex.artofproblemsolving.com/c/0/9/c098bda64393b12049d319db49815947b2a0b7a4.png)
is a permutation of 
.
be non-negative real numbers such that 
Prove that
board. On each cell of the main diagonal there is a file. At each step, we may select some files and move them to the left, on their respective rows, by the same distance. What is the minimum number of necessary moves in order to put all files on the first column?
be the side lengths of a triangle. Prove that![\[\frac{a^3}{(a+b)^3}+\frac{b^3}{(b+c)^3}+\frac{c^3}{(c+a)^3} \leqslant \frac{9}{8}.\]](http://latex.artofproblemsolving.com/5/1/5/515cbd3cd69fee483312f90d4cb29d98ca761f41.png)
spheres centered at 
, with radius of 
and , which is remarked as 
and 
, respectively. Given 
points 
outside of (not including the surface of ).
as the number of sets 
such that the midpoint of 
lies entirely inside of .
Hint
to 
be positive integers such that 
and 
Determine the value of 
is on point 
such that 
and 
and 
is a point on 
such that 
and 
Given that 
calculate 
given that 
Hint
![\[f(x) =
\begin{cases}
x - 9, & \text{if } x > 100 \\
f(f(x + 10)), & \text{if } x \leq 100
\end{cases}\]](http://latex.artofproblemsolving.com/4/e/5/4e5b4486030a59e74df648819a1a59f82622b3b7.png)
.
be real numbers such that 
Find 
are on circle such that 
and 
Determine the path length from 
to 
formed by segment 
and arc 
such that the expression 
is also an integer.
such that 
is a multiple of 
and 
are real numbers such that 
and 
Calculate 
will follow the rules such that![\[
p_1 = (0,0), \quad p_{i+1} = (x_i + 1, y_i) \text{ or } (x_i, y_i + 1), \quad p_{10} = (4,5).
\]](http://latex.artofproblemsolving.com/6/0/3/6036ef2619ffb1e91db0ab310388217e472490f2.png)
How many sequences 
are possible such that 
is the only point with equal coordinates?
The probability that at least two people choose the same number can be written as 
Find 
on the positive integers using the rule that for 
For all prime , 
and for all other 
Find the smallest possible value of such that 
can be written as the sum of two distinct, non-negative integer powers of 
be the set of positive integers of such that 
for some other positive integer 
Find the only three-digit value of in 
be a positive integer and let be the integer that is formed by removing the first three digits from 
Find the value of with least value such that 
of triangle 
beyond to a point 
such that 
. Let be the midpoint of side . Let the bisector of 
intersect line 
at . Prove that 
.
and 
.
Doing this, we get that 
Now, we can solve this by factoring. By using RRT, we get that 
is a factor. Upon division, we get 
is the factored form. Now, we know that our only positive solutions are 
so our values of 
Summing these gives our answer of 
in or something like that idk my friend explained it pretty wonkily idk if he even solved it right but anyhow we press on. Transformations suck.
Furthermore, note that 
Thus, the answer is just 
so 
Cross multiplying and solving this quadratic, we get that our solution is 
Thus, 
We know from 
that 
so substituting this gives us 
Now, plugging this in to 
and 
gives us 
Solving these equations gives us 
Thus, by Vieta's, the sum of the roots is 
thus by median properties (the 2:1 ratio split thingy), we know that 
(For reference, I missed this part in-contest) Now, we can calculate 
and thus 
Now, by Pythagorean theorem, we know that the altitude from 
so our area is just 
Thus, simplifying and substituting 
we get that 
Simple guess and check gives us 
as our solutions. My dumb self forgot the former solution. Thus, 
giving us 
as our sum.
we get that 
Now if we try finding 
we will realize that 
this 
pattern continues until we once again reach 
thus 
Now return to the problem statement; 
but each inner function will just become 
hence the answer is 
Thus, simplifying, we get that 
or that 
Plugging this in to the first equation, we get that 
so this means that 
Now, 
thus 
and so by Special Right Triangles we do NOT in fact get 
but rather 
.
to be an integer. Plugging in factors of 
we see that there are 
integers that work.
Plugging in numbers gives us 
as our smallest value.
Furthermore, 
so 
meaning that 
.
and thus 
giving us 
Furthermore, we know that 
so our answer is just 
in the summation. Essentially, we have 
.
In other words, we want to find 
.
.
in such a polynomial is 
.
,
.
,
.
and 
and divide by 
.
and 
in our sum, subtract by 
.
term look kind of weird? Well, we actually notice 
,
.
,
,
is our answer!
,
terms. Drawing it out on the unit circle (which I heavily advise), it cycles between 
and 
.
,
,
.
,
for when 
terms! Anyways, the answer is 
.
so our coefficient is just 
Thus, our answer is just 
.
etc.
we get that it is 
Thus, we note that for an integer 
Since 
we know that all values of 
satisfy our conditions. Thus, our answer is just 
is our solution. For these Pell Equations, all solutions are of the form 
for 
Thus, we plug in 
giving us 
as our answer.
and let 
By our condition, we know that 
or that 
Furthermore, we know that 
does not work, so we try 
We know that 
by the fact that it must be divisible by a 
Noting that 
doesn't work, we try 
or that 
This, miraculously, works, and thus our value of 
