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 positive integer. A board with a format 
is divided in equal squares.Determine all integers ≥3 such that the board can be covered in 
(or 
) pieces so that there is exactly one empty square in each row and each column.
.The Incircle touches sides 
and 
in 
and 
,respectivly. Lines CI and EF intersect at 
. The point 
≠ is on the line AI so that 
=
.If 
is the foot of the altitude from 
in triangle 
,prove that points , and are colinear.
be an acute scalene triangle and let 
be a point in its interior. Let 
, 
, 
be projections of onto triangle sides 
, 
, , respectively. Find the locus of points such that 
, 
, 
are concurrent and 
.
and 
such that 
+ 
+ 2 is a perfect square.
for which
is a prime number
such that there exists a sequence of positive real numbers 
, 
, with the property that for every it holds that:![\[
a_1 + \cdots + a_{n+1} < \alpha \cdot a_n.
\]](http://latex.artofproblemsolving.com/9/b/7/9b73deb46032844b324288b02470ff0623ac0ff1.png)
Proposed by Pavle Martinović
, let 
be the point which minimizes the sum of squares of distances from the sides of the triangle. Let 
the projections of on the sides of the triangle . Show that is the barycenter of 
.
be the set of all positive real numbers. Find all strictly monotone (increasing or decreasing) functions 
such that there exists a two-variable polynomial 
with real coefficients satisfying
for all 
.
.
. Let 
be the midpoint of the arc 
not containing 
and define 
similarly. Show that the orthocenter of 
is the incenter of .
, which clearly work.
denote the assertion. Clearly 
gives 
. Then from 
we see 
implying is even. Thus can be written as 
for some polynomial 
. work. It is to show that these work from just plugging.
fails. To see this we consider the leading coefficients of the polynomial. Note that we want the following to fail,
This is equivalent to showing that the leading coefficients of the following fail,
Setting 
and 
we see we need,
However it is easy to see the LHS is much greater than the RHS for 
, so it is impossible for this equation to hold. Thus we are done.
gives 
so then we can plug in 
to get that is even.
. Taking 
gives 
. Solving, we get 
.
we find that .
we find that 
, so must be even.
, 
and 
.
, and we obtain![\[ P(5k) + P(3k) + P(8k) = 2P(7k) \]](http://latex.artofproblemsolving.com/0/1/d/01dbee029b82151f3a309cf5af8f0b93de642a67.png)
Now, suppose that 
has degree . Then we obtain![\[ 5^n + 3^n + 8^n = 2\cdot 7^n \]](http://latex.artofproblemsolving.com/4/8/9/48968f6a13dbd3e538b437453c2935ede8599de3.png)
From Bernoulli we see that 
. Thus, 
. Taking modulo 6, we see that 
.
, as must be even.
for some reals 
and 
. To see how 
works, note that![\[ (a-b)^4 + (b-c)^4 + (c-a)^4 - 2(a+b+c)^4 = -6(ab+bc+ca)(ab+bc+ca+a^2+b^2+c^2) \]](http://latex.artofproblemsolving.com/5/5/b/55b5bb18d2789e2c80dfec95e4cb4c6687ef1716.png)
and 
works as![\[ (a-b)^2 + (b-c)^2 + (c-a)^2 - 2(a+b+c)^2 = -6(ab+bc+ca) \]](http://latex.artofproblemsolving.com/2/7/b/27bcd8e83e3c3a7fa6d78263a92599eef5beff71.png)
so indeed all polynomials of that form work. 
. Let's play with the coefficients of the polynomial, that is, if there is some 
on the left in the expression, where say is an expression of 
, then we should have the same thing on the right.
behaves: 
$$\Rightarrow \a_0=0.$$
behaves: 
is not always 
, so 
.
behaves: 
can essentially be anything, and we leave it that way.
further, but I'm so lazy, a lot of numbers appear with it, and something tells me that equality does not always happen, I thought. Then, it would be worthwhile to come up with some universal values for so that the condition 
would be satisfied at the same time, and so that the coefficients could be conveniently estimated.
looks like this: 
It is very inconvenient to disclose this whole thing with large values of 
, then let's find convenient values for 
Then, let's look at 
, I thought it would be quite convenient if 
and the value would be identical to 
for any . With the same logic, I made it so that 
, then let's see what happens to 
:
We get two roots, I see that the first one is more convenient and I take it. Let's look at our coefficient under these values: 
Let's say is not , then 
With 
and odd, the two sides will have different signs, and with 
and even, the growth will be faster on the left than on the right (it can be proved by induction that there will always be more on the left with .) Works for 
, but this does not mean that this will always be the case. If 
we have checked. Check 
, use the facts that 
, 
and make sure that the equality is identical.
and must be zero, and these two can take any value.
works where 
, 
.
denote the assertion 
.
.
. So 
.
.![\[ P\left(\frac{5c}{6}\right) + P\left(\frac{c}{2}\right) + P\left(\frac{4c}{3}\right) = 2P\left(\frac{7c}{6}\right). \]](http://latex.artofproblemsolving.com/f/f/a/ffac9e9e5f2e7743e7b1af924d8b9e8edbff4800.png)
term, we get that,![\[ a_i\left(\frac{5}{6}\right)^{2i} + a_i\left(\frac{1}{2}\right)^{2i} + a_i\left(\frac{4}{3}\right)^{2i} = 2\left(\frac{7}{6}\right)^{2i}. \]](http://latex.artofproblemsolving.com/a/3/5/a35651a4e49833d4c916728421f9a0c3d564e0a0.png)
.
, then 
. Now note that 
for all 
.
. Checking gives that 
and 
both work. So we get our desired form and we are done. 
for real numbers 
works and we want to prove its uniqueness. and 
are solutions, 
is a solution for all real numbers 
so it's enough to find all solutions of the form 
.
and 
gives us that is even
, then this obviously doesn't hold.
, then 
, which works.
, then 
, which also works.
, then 
which is a contradiction.
, then 
contradiction.
for any real number 
, which clearly works., we find that . Then, plugging 
, we have that is even.
, but this isn't possible since 
implies over reals. The next best thing would be to consolidate terms of the left hand side: 
Plugging 
gives us that 
However, we also have 
We will only use 
which when plugged into 
we get 
We will equate coefficients of 
on both sides, noting that is even. If the coefficient of 
in is 
, then we have 
Thus, or each nonnegative integer , either 
or 
This is only true for 
and , as we can test that doesn't work, and 
for 
Thus, only the and coefficients are allowed to be nonzero, thus done.
, makes it so that all terms are of the form 
This is very nice because this means that the coefficent of this only depends on the coefficient of , allowing us to easily equate coefficients. By only setting "homogeneous" constraints, like 
that I used, one can prevent the input of the polynomial from having a constant shift. In fact, 
is really the only substitution you need to solve this problem.
for real numbers works and we want to prove its uniqueness. and are solutions, is a solution for all real numbers so it's enough to find all solutions of the form . and gives us that is even, then this obviously doesn't hold., then , which works., then , which also works., then which is a contradiction., then contradiction. suffices
for real numbers 
, then it is true that the linear combination of any two solutions is still a solution, but in this case there are no solutions of the form even though there does exist solutions.
is not the only basis for real polynomials, and there are plenty of other choices, so closure under addition like this does not imply that only checking these suffices
I wish you good luck in the future!
where 
. This polynomial indeed satisfies the problem.
and 
for all 
. So we can write 
where 
. Now we put 
into the equation and after clearing the denominator we get 
for all , where 
We can check that 
for all . And for 
we can prove via the multinomial theorem that 
.
, which will finish the problem. Suppose ftsoc that 
and let 
. Since 
, it follows that the coefficient of 
in 
is some nonzero number times 
. Thus we must have 
, which is the desired contradiction.
be the set of solutions. We prove that 
so is an even function. is a 
-vector space
both work.
to get 
Hence, if a polynomial works, then we require 
where 
However, simply note that 
for 
. It can be shown (painfully) that 
works, thus 
works.
works, thus works, we shall show that this is the only solution.
and 
is even
, we immediately obtain![\[f(a) + f(0) + f(-a) = 2f(a) \Longrightarrow f(a) = f(-a) + f(0)\]](http://latex.artofproblemsolving.com/2/b/a/2ba188c9c1c48089b129b1a84eda8de0300f436a.png)
Since the constant terms of both sides of the equation have to be equal, it follows that , thus is even. ![\[f(a-b) + f(b-c) + f(c-a) = 2f(a+b+c)\]](http://latex.artofproblemsolving.com/9/0/9/909c138a61d0d1f6859d7822fb7893b8db2f62ac.png)
if 
works, then 
, 
, etc, also has to work. Thus by claim 
, we simply need to show that 
fails for 
. Let us first show that 
fails fails for 
to get rid of the algebraic condition, thus our equation becomes![\[{(a-b)}^{2k} + {\left(b + \frac{ab}{a+b}\right)}^{2k} + {\left(\frac{-ab}{a+b} - a\right)}^{2k} = 2{\left(a+b-\frac{ab}{a+b}\right)}^{2k}\]](http://latex.artofproblemsolving.com/c/a/a/caa05030c62f1aba4337d3d4ac6f2a777652f548.png)
We can multiply both sides by 
to obtain![\[{(a^2-b^2)}^{2k} + {(b^2 + 2ab)}^{2k} + {(a^2 + 2ab)}^{2k} = 2{(a^2 + ab + b^2)}^{2k}\]](http://latex.artofproblemsolving.com/7/3/4/73441f6405e2b94cce970f61398dd199daf37a79.png)
Now we can substitute 
and 
which gives![\[3^{2k} + 5^{2k} + 8^{2k} = 2 \cdot 7^{2k} \Longrightarrow {\left(\frac{3}{7}\right)}^{2k} + {\left(\frac{5}{7}\right)}^{2k} + {\left(\frac{8}{7}\right)}^{2k} = 2\]](http://latex.artofproblemsolving.com/1/2/2/1228a1e5cf8f1bf0aa746fa9e853e48f0dcb71cf.png)
Since 
by the binomial theorem, we are done. works, then , , etc, also has to work. works for all 
with 
however there exists some 
such that 
, but 
fails.
denote the largest value such that 
works, but 
fails., it follows that if and 
then 
works, thus we can just get rid of these terms and only look at the function
, we also preemptively define 
denote some triple of values such that 
and for which![\[{(p-q)}^{k} + {(q-r)}^{k} + {(r-q)}^{k} \neq 2{(p+q+r)}^{k}\]](http://latex.artofproblemsolving.com/3/8/0/380f8e875791accd07888e501e00e269fe98bde3.png)
such a triple clearly exists since we assumed that doesn't fulfill the functional equation.
, since we have assumed that 
works, it follows that![\[g(p-q) + g(q-r) + g(r-p) = 2g(p+q+r)\]](http://latex.artofproblemsolving.com/6/4/0/640cd8076426eaf2a8a94c65cd989b08b4a34aaf.png)
Thus![\[a_{k} \cdot \delta = 2h(p+q+r) - h(p-q) - h(q-r) - h(r-p)\]](http://latex.artofproblemsolving.com/2/5/e/25e8535ee9a5d52bc6cbb7c5dbf1805712c18067.png)
Notice that if the functional equation holds for some , it also holds for some 
where 
is an arbitrary constant. Thus by scaling the variables by we have![\[a_{k} \cdot \delta \cdot c^{k} = 2h(cp+cq+cr) - h(cp-cq) - h(cq-cr) - h(cr-cp)\]](http://latex.artofproblemsolving.com/4/b/7/4b73951e14e1aca190bda2fe72897b2dde744943.png)
Since 
is a degree 
polynomial at best, it follows that the right hand side grows by at most 
, whereas the left hand side grows by 
., the left hand side cannot possibly equal the right hand side. Thus we are done.
. Then, if 
,
works since 
, and there are no other solutions. is even and of degree 4, let 
. Plugging in, 
.
.
for any real 
, which "clearly" works. gives , then 
gives 
. So 
gives![\[f(2a-1)+f(a^2-2a)+f(a^2-1)=2f(a^2-a+1).\]](http://latex.artofproblemsolving.com/e/e/f/eef54fefa664c737db4d625d07c54bd9f11227c4.png)
Suppose that the leading term of 
is 
, then 
. Then by comparing the coefficient of 
on both sides, we get![\[4k(2k-1)-2k=2(2k+k(2k-1))\]](http://latex.artofproblemsolving.com/a/3/0/a3032d39fbad69cd86d0b8bd6c3ddec149f7929e.png)
which eventually reduces to 
, impossible. 
.![\[\sqrt{a+\frac{9}{b+2c}}+\sqrt{b+\frac{9}{c+2a}}+\sqrt{c+\frac{9}{a+2b}}+\frac{2(ab+bc+ca)}{9}\ge\frac{20}{3}\]](http://latex.artofproblemsolving.com/b/3/2/b32e0bcba967ca847e878ac186c3168cd6c5fc66.png)
we have the following relations![\[ f(a-b) + f(b-c) + f(c-a) = 2f(a+b+c). \]](http://latex.artofproblemsolving.com/4/4/9/4491a53f9c6f96c18e42eb984b8e83566140bfdb.png)
is in the image of f (I mean, for any such a,b,c there exists a number 
such that that number equals f(g(a,b,c))).
,
whenever 
divides 
.
and 
if 
are linearly independent, so each coefficient 
must be zero.
or 
,
,
.
.
.
,
for all even 
.
for all 
.
(so that 
)
.
satisfying the functional equation
for all 
.
.
,
or letting 
,
.
appears with multiplicity at least 
on the 
,
as well.
,
,
so that 
(since 
are the only possibly solutions. A quick check shows that these solutions are indeed valid.
.
,
.