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.
be a sequence of positive integers. Let 
be the number of 3-element subsequences 
with 
, such that 
and 
. Considering all such sequences 
, find the greatest value of .
, 
is the midpoint of 
. 
is an arbitrary point on 
. Let 
be the intersection of 
and 
. The line 
intersects with the circumcircle of 
, for the second time at 
. 
is the circumcenter of triangle 
. Prove that 
.
in the expansion of 
table. A cannon can fire at any 
cells of the board after that the tank will move to one of the adjacent cells (by side). Then the process is repeated. Find the smallest value of such that the cannon can definitely shoot the tank after some time.
is 
. Line 
meets line 
at point 
, and there is a point 
on 
such that 
. Line 
intersects at point 
. The perpendicular bisector of line segment intersects line 
at point 
, and line intersects 
at point 
.
is tangent to 
.
be the incenter of triangle 
, and let 
, 
, and 
be the circumcenters of triangles 
, 
, and 
, respectively. Prove that the circumcircles of triangles and 
are concentric.
be positive real numbers. Prove that 
.
, note 
., such that 
isn’t divisible by 
but 
is divisible by 
, 
, 
, the inequality![\[ \frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \leq \frac {1}{abc}
\]](http://latex.artofproblemsolving.com/f/a/9/fa964e29de91f0e2f3cb32e335a52d4521bc38fd.png)
for 
, we get 
, and since 
, 
, and 
. Plugging in 
, 
, and 
we get 
, 
, and 
. So 
is in the form 
, where 
is a polynomial (possibly with degree zero). Plugging in 
for we get 
, and testing other values we confirm this works, therefore 
. Therefore 
.
and 
and so 
Plugging this in, 
and so 
Using the 
relation again, we get 
and plugging in 
gives the result.
gives 
.
gives 
.
gives 
and .
gives .
, then 
, a contradiction as it implies 
is the zero polynomial. is a cubic polynomial with roots 
.
. 
.
and 
. We have 
and 
and so 
.
to get to get to get to get and .
for some polynomial 
. We plug in values of 
and 
to confirm that 
. Therefore the answer is 
combined with something in terms of having infinitely many roots.
is the only polynomial that satisfies the given equations; obviously, it works. First, it is easy to show via substitution that 
. Next, let 
for all 
. Plugging into the given equation, 
It follows that the polynomial 
is the zero polynomial, so 
is constant.
for some constant . Utilizing the second constraint, 
, so the answer is 
. Extract 
.
and 
give 
. Then plugging in 
gives 
. Since 
are zeroes of , we can write 
for some polynomial 
. Plugging this into the functional equation gives 
for all 
, so 
is a constant. Hence 
and 
, so 
. We then have that 
, so the requested sum is 
.
which are obvious roots. Indeed, suppose r is a different root. If it is positive then since P(r)=0 and r-1 is not 0, P(r+1) must be 0, hence inductively there are infinite roots, contradiction. On the other hand, if r is negative by "shifting the index" to get that (r-2)P(r)=(r+1)P(r-1), since P(r)=0 and r is not -1 then P(r-1) must be 0, implying again infinite roots. Hence P(x)=a(x-1)x(x+1), and using the given condition we solve it to get that a=2/3. Now plugging in x=7/2 yields 105/4 for a final answer of 
be a root of .
so 
or 
is also a root of .
so 
or 
is also a root of .
.
and 
so 
.
so the answer is 
is a polynomial such that 
and ![\[\frac{P(2x)}{P(x+1)}=8-\frac{56}{x+7}\]](http://latex.artofproblemsolving.com/1/a/b/1ab69a52821b33240a86fd8d9e64a4152262f3b8.png)
for all real for which both sides are defined. Find 
.
where 
is the degree of . Then, the coefficient of 
in 
is 
and the coefficient of in 
is 
, so 
.
gives , so let 
. Then, since is a root of the denominator, let 
. Then, since 
is a root of the denominator, let 
.
, so 
.
.
for every real 
.
,
.
,
.
,
.
,
for some polynomial 
.
,
.
is a constant polynomial. Back-substituting, we see that ![\[P(x)=(x-1)(x+1)Q(x)=(x-1)(x+1) \cdot xR(x) = c(x-1)x(x+1)\]](http://latex.artofproblemsolving.com/a/7/9/a792fa8a778493ac4a6f3dbd118f8995caec35e6.png)
for some constant 
.
and 
,
,
.
and ![\[P\left(\frac72\right) = \frac23 \cdot \frac52 \cdot \frac72 \cdot \frac92 = \frac{105}{4}.\]](http://latex.artofproblemsolving.com/f/e/6/fe6cfa5a35256c17654e66cffafd520c9dcdcaf6.png)
The answer is 
.
,
or 
.
if it is of degree 
,
.
,
,
,
and 
,
,
.
to get 
,
)
,
doesn't do anything, since 
term is 0 anyways. So we have by factoring 
.
always, which means 
The RHS vanishes, hence the LHS vanishes, i.e. 
The LHS vanishes, hence the RHS vanishes, i.e. 
The LHS vanishes, hence the RHS vanishes, i.e. 
for some polynomial 
Substituting this expression into the given identity, and cancelling like terms, we obtain 
for all 
Therefore, 
It follows that 
Hence, 
,
Then
,
,
while the second equation should be 
.
,
rather than 
,
instead of 
yields 
,
are roots. By the Factor theorem, we have 
.
gives 
.
.
.
,
.
for some polynomial 
which simplifies into 
.
.
.
.
,
.
and realize the polynomial is 
,
,
hence 
.
after some simplification, giving us 
as the answer.