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.
. 
amount of people stand on coordinates 
where 
. Every person got a water cup and two people are considered to be neighbour if the distance between them is 
. At the first minute, the person standing on coordinates 
got litres of water, and the other 
people's water cup is empty. Every minute, two neighbouring people are chosen that does not have the same amount of water in their water cups, and they equalize the amount of water in their water cups. will not have more than 
litres of water.
be an integer. You have two 
boards. Each board contains the numbers to inclusive, one number per square, arbitrarily arranged on each board. A move consists of exchanging two rows or two columns on the first board (no moves can be made on the second board). Show that it is possible to make a sequence of moves such that for all 
and 
, the number that is in the 
row and 
column of the first board is different from the number that is in the row and column of the second board.
be a convex pentagon such that![\[ \angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE.
\]](http://latex.artofproblemsolving.com/d/3/d/d3d9a82f4318190298a8f008d417a364e03f1fca.png)
The diagonals 
and 
meet at 
. Prove that the line 
bisects the side 
.
, with 
, 
is the incenter, 
is the intersection of 
-excircle and 
. Point 
lies on the external angle bisector of 
such that and lieas on the same side of the line 
and 
. Point 
lies on such that 
. Circle 
is tangent to 
and 
at 
, circle 
is tangent to and 
at 
and both circles pass through the inside of triangle . if 
is the Midpoint od the arc , which does not contain , prove that lies on the radical axis of and .
be a sequence of positive real numbers, and 
be a positive integer, such that![\[a_n = \max \{ a_k + a_{n-k} \mid 1 \leq k \leq n-1 \} \ \textrm{ for all } \ n > s.\]](http://latex.artofproblemsolving.com/f/1/2/f12c9c5e0f154549118a33a2f359329333f432a7.png)
and 
, such that![\[a_n = a_{\ell} + a_{n - \ell} \ \textrm{ for all } \ n \geq N.\]](http://latex.artofproblemsolving.com/b/0/b/b0b2bd4507621302eeee27ae50ed9c0afd72c6ef.png)
and on a given circle meet at point . Let be a point on segment and let 
meet the circle again at . The line through 
parallel to intersects at 
. Prove that 
is parallel to if and only if 
.
, and 
be positive real numbers, such that 
. Prove the inequality![\[\frac{x^3}{2 + x} + \frac{y^3}{2 + y} + \frac{z^3}{2 + z} \ge 1.\]](http://latex.artofproblemsolving.com/b/e/5/be5819a67c3cd78f2dea35fdccf48688c720ce3c.png)
When does the equality hold?
of positive integers such that ![\[ k!=(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}). \]](http://latex.artofproblemsolving.com/3/5/3/353bbb15bac205053c26208af8cac7c32a296f55.png)
Proposed by Gabriel Chicas Reyes, El Salvador
![\[\frac{\operatorname{lcm}(x, y) + \operatorname{lcm}(y, z)}{\operatorname{lcm}(x, z)}\]](http://latex.artofproblemsolving.com/1/e/8/1e8b3e2d9a4d0917ec622449b058702700542f22.png)
for positive integers 
, 
, ?
be prime, let be a positive integer, show that ![\[ \gcd\left({p - 1 \choose n - 1}, {p + 1 \choose n}, {p \choose n + 1}\right) = \gcd\left({p \choose n - 1}, {p - 1 \choose n}, {p + 1 \choose n + 1}\right). \]](http://latex.artofproblemsolving.com/9/d/4/9d409ce186320241eb358b5fbbb8a5a66120899c.png)
be a right triangle with a right angle in and a circumcenter 
. On the sides and , the points 
and lie in such a way that 
. Let and 
be the projection of and on , respectively. Prove that 
is half as long as .
of real numbers that satisfy the system of equations![\[ \begin{cases}x^3 = 3x-12y+50, \\ y^3 = 12y+3z-2, \\ z^3 = 27z + 27x. \end{cases}\]](http://latex.artofproblemsolving.com/d/4/d/d4d6c485b4653577e732ff235448bb779914a576.png)
then multiply all of the equations,
.
. Substituting them, we get the following equations:![\[ \begin{cases}a^3-3a=50-4b, \\ b^3-3b=\frac{9c-2}8, \\ c^3-3c=a. \end{cases}\]](http://latex.artofproblemsolving.com/a/1/9/a19ab62ed146a7761cf5f004cb55d3ed05ecde31.png)
Note that the function 
is a cubic polynomial function. 
has two solutions, namely 
.
. If 
then 
and if 
, then 
. If 
then 
.
then 
and if 
then 
.
, then 
, giving 
. If , then 
, which means 
. If then 
, contradiction.
, then 
, giving 
. If , then 
, which means 
. If then , contradiction.
is forced, giving 
and 
, which clearly work.
is the only solution.
.![\[(x, y, z) = (\alpha + 2, \beta + 4, \gamma + 6)\]](http://latex.artofproblemsolving.com/7/c/f/7cf0f39abd7d2a24b930c04cb64bee8709eb9504.png)
![\[(\alpha + 2)^3 = 3\alpha - 12\beta + 8\]](http://latex.artofproblemsolving.com/8/c/6/8c62e932e8e9496b78fb6a3f9d8b4febdb27e9b5.png)
![\[(\beta + 4)^3 = 12\beta + 3\gamma + 64\]](http://latex.artofproblemsolving.com/6/f/b/6fb540c563317ed5e6f39777f0e987f4da1413c1.png)
![\[(\gamma + 6)^3 = 27\alpha + 27\gamma + 216\]](http://latex.artofproblemsolving.com/3/c/e/3ce31b2e4c33f0b38b184d030c5bf951b843c7ff.png)
![\[\implies\]](http://latex.artofproblemsolving.com/c/3/0/c30ac6e41a136bd85e681667b55d9a3c825ee00e.png)
![\[\alpha(\alpha +3)^2 = -12\beta\]](http://latex.artofproblemsolving.com/f/f/5/ff57086005e680583660cd538c57b045386440bc.png)
![\[\beta(\beta + 6)^2 = 3\gamma\]](http://latex.artofproblemsolving.com/8/0/a/80ae7ff7bb124e175315ce8e9483e8bd88f0c25e.png)
![\[\gamma(\gamma + 9)^2 = 27\alpha\]](http://latex.artofproblemsolving.com/c/6/d/c6d7ca9a2db52675f269bf445181d4e774dc902f.png)
![\[\alpha\beta\gamma(\alpha +3)^2(\beta + 6)^2(\gamma + 9)^2 = -972\alpha\beta\gamma\]](http://latex.artofproblemsolving.com/2/4/f/24fa185f9a62545d5196015c7c035f5194c1a8a0.png)
FTSOC, assume that 
.
results in![\[(\alpha +3)^2(\beta + 6)^2(\gamma + 9)^2\ = -972\]](http://latex.artofproblemsolving.com/3/f/3/3f316a19d805b53ac059182b3ede1662bd4d0d6a.png)
which is a contradiction due to the trivial inequality.
. Plugging 
into 
, 
, or 
, we can clearly see that it makes the rest of the variables , so our answer for 
is .
. Motivated by this answer, we notice our equations can be rearranged to be
, 
, or 
to avoid a trivial inequality contradiction. However, any one of the these imply the other two, which finishes. 
.
Multiplying we have,
Clearly if all of the factors 
, 
and 
are nonzero we have no solutions. However if we have 
, 
or 
we find the claimed solution so we are done.
This can clearly be shown to work just by plugging in.
and 
After plugging in and reducing the equation, we get the following systems of equations:
If any of 
are equal to 
then all of them must be simply by plugging in, giving us the solution claimed at the beginning of our solution. Henceforth assume that are nonnegative.![\[
-\frac{p}{12q} = \left(\frac{1}{p+3}\right)^2,
\]](http://latex.artofproblemsolving.com/a/3/4/a3473d77c997fce634847ec787644ec992ef817d.png)
so![\[
-\frac{81p}{q} = 972 \cdot \left(\frac{1}{p+3}\right)^2.
\]](http://latex.artofproblemsolving.com/2/6/c/26c9412bb67c0ec5a06851bfa7ec5f838d50956b.png)
Meanwhile, the second and third equations multiply to![\[
qr((q+6)(r+9))^2 = 81pr,
\]](http://latex.artofproblemsolving.com/d/1/5/d1591e876da0291923bc1dc6e5b289b315b5613e.png)
which, since 
rearranges to![\[
\frac{81p}{q} = ((q+6)(r+9))^2.
\]](http://latex.artofproblemsolving.com/2/1/7/2172e9ebb4934b62e9520402f78defebd57099d4.png)
Adding our two equations together, we get![\[
972 \cdot \left(\frac{1}{p+3}\right)^2 + ((q+6)(r+9))^2 = 0.
\]](http://latex.artofproblemsolving.com/9/c/b/9cbb335217865cfd3ae938c5df63c25a67e718ff.png)
Since the square of a real number is nonnegative, all of the squares in the above expression are 
In particular, 
which is impossible.
and so the only solution to our system of equations is 
as claimed.
is the only one. Now note that the equations are equivalent to:
It is easy to see that , , or are independently sufficient enough to yield as a solution, so assume 
. Then multiplying the three equations and dividing by 
gives
which is clearly impossible. So is the only solution and we are finished.
, which clearly works.
.
Now assume none of 
are zero, since if one is then clearly all are.
In particular, if 
denotes the sign of 
, 
a clear contradiction. Therefore the only working solution is 
. 
, 
, 
![\[ \begin{cases}a^3+6a^2+9a=-12b ...(1) , \\ b^3+12b^2+36b=3c ...(2) , \\ c^3+18c^2+81c=27a ...(3) \end{cases}\]](http://latex.artofproblemsolving.com/d/e/9/de93ce1bf37ee5ffea2803f03dae3597fec5936b.png)
we will prove it by a contradiction.
and 
and 
we get:
.
and 
and 
we get:
.
from 
we find that 
and finally from we find that 
. 
Then our equations become
However, observe that for each one we can complete the square:
From the second and third equations, 
and 
have the same sign, while 
and have the same sign. However, from the first equation and have opposite signs, therefore they are both Hence 
too. This yields as our only solution.
, to which we motivate the substituitions 
. Then the equations become:
or 
. If all are nonzero, then we multiply all of these equations, cancel 
, and arrive at a contradiction. If any one is zero, the rest all are so we are done.
,
,
.![\[\sum_{i=1}^{n}\frac{1}{1+x_{i}}\cdot \sum_{i=1}^{n}x_{i}\leq \sum_{i=1}^{n}\frac{x_{i}^{k+1}}{1+x_{i}}\cdot \sum_{i=1}^{n}\frac{1}{x_{i}^{k}}\]](http://latex.artofproblemsolving.com/2/6/4/264d110e2a40c7f787c650aa8b8dc30af5d28263.png)
,
,
so from 2 we have 
so x=2,y=4,z=6
,
so from 2 we have 
and take 
from the last equation into the first one, so that it turns into
and 
.
,
and 
.
and 
.
,
,
,
.
(Or 
,
.
,
.
or 
,
.
to make more of the coefficients match. Then we get 
,
.
pretty much jumps out. We note that
,
,
,
.
equals 2, then they all do. If not, multiplication yields that
,
,
and 
(this is motivated by observing 
)
Note 
and 
.
.
,
which is why I attempted the bounding in the first place.
,
,
.
which gave 
the function 
is increasing so we have 
by the first equation. Also the third equation yields that, from 
.
.
.
.
works so 
had to be all integers. I tried substituting floors and what not to prove that this was the case, but it turned out too messy. I proceeded to try to just guess integer solutions, starting from 
,
but tried dumb things first like bounding on just 
.
as otherwise we get ![\[ \begin{cases} -12(y-4)=(x-2)(x-1)^2, \\ 3(z-6)=(y-4)(y+2)^2, \\ 27(x-2)=(z-6)(z+3)^2. \end{cases}\]](http://latex.artofproblemsolving.com/e/8/e/e8ec743edbabcb686307be6aec1da3d6cbe5b19f.png)
Taking the product and cancelling the common (non-zero) factor 
we see that both sides have an opposite sign unless 
.
a contradiction! Similar arguments apply for the other two factors.
,
and similar statements with all the inequality directions reversed.
so 
It is easy to check that none of 
or 
or 
produce valid solutions. Therefore, 
.
.
,
,
.
are of opposite sign, 
are of the same sign, and 
are of the same sign. The only way this can happen is when 
.
Multiplying yields ![\[(x + 1)^2 (y + 2)^2 (z + 3)^2 (x - 2)(y - 4)(z - 6) = -972(x - 2)(y - 4)(z - 6).\]](http://latex.artofproblemsolving.com/9/7/d/97dbb06c306f58ba081db70a77d870e9635b7158.png)
Now if ![\[(x + 1)^2 (y + 2)^2 (z + 3)^2 = -972,\]](http://latex.artofproblemsolving.com/2/d/1/2d15d85201dc01a6b7fc50a61536aa826c23442b.png)
a contradiction.
We can check this works.
The equations become 
Now it is easy to see that if one of 
is zero then all are. But multiplying all together gives 
impossible if 
Thus 
so 
the equations becomes:
Now we have 
Which implies 
.
by looking at signs. The singular solution