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 .
.
for 
means that 
.
. Then, we wish to prove that 
. By AM-GM, 
so it suffices to prove that 
, which is true because 
.
, we just need to prove that 
implies that 
. Expanding gives 
as desired.
, the assertion is true. We will now prove that the assertion is true for .
such that 
, 
, 
don't form the side lengths of a triangle, then 
and 
by AM-GM so 
as desired.
. By AM-GM and our claim, 
as desired.
are equal to 
, 
, 
where , are positive reals and is nonnegative. By Cauchy-Schwarz, we have
Now
by AM-GM. To reach contradiction it is now sufficient to show 
which is true. We are done.
. Then ![\[t_1 + t_2 + t_3 = \frac25 t_1 + \frac35 t_1 + (t_2 + t_3) \geq \frac25 t_1 + \frac85(t_2 + t_3)\]](http://latex.artofproblemsolving.com/0/a/c/0acd6d3ef414f23102f5794365916ac04db3c1c9.png)
and ![\[\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} \geq \frac{1}{t_1} + \frac{4}{t_2 + t_3}\]](http://latex.artofproblemsolving.com/a/7/2/a7235fbb9869141c281b54ebbd1937a89bad445e.png)
by Cauchy-Schwarz, so we have ![\[(t_1 + t_2 + \dots + t_n)\left(\frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n}\right)\]](http://latex.artofproblemsolving.com/7/6/6/76643350b294b5322309ee8358dca57f236afcb5.png)
![\[ \geq \left(\frac25 t_1 + \frac85(t_2 + t_3) + t_4 + \dots + t_n\right)\left(\frac{1}{t_1} + \frac{4}{t_2 + t_3} + \frac{1}{t_4} + \dots + \frac{1}{t_n}\right)\]](http://latex.artofproblemsolving.com/5/6/b/56be74ad5093a6bdb0b8c517fedd46f1846faee2.png)
![\[ \overset{\text{C-S}}{\geq} \left(\sqrt{\frac25} + \sqrt{\frac{32}{5}} + (n - 3)\right)^2 = \left(n + \sqrt{10} - 3\right)^2.\]](http://latex.artofproblemsolving.com/7/c/0/7c0c0364f68030796905b430e213165b3fd94381.png)
But 
is strictly increasing, so 
, a contradiction. 
and 
? I originally tried using 
's, which was very close but broke slightly for 
. But if we use arbitrary coefficients and 
, it suffices to maximize the function ![\[f(x) = \sqrt x + 2\sqrt{2 - x},\]](http://latex.artofproblemsolving.com/c/3/7/c37b71145f5b1d1c15cfcfa44cbe556c1213ebbe.png)
which by C-S or similar methods is greatest when 
. This fits perfectly when .
. Assume to the contrary that there exists some tuple 
such that the inequality is satisfied and 
. We show that if 
we can form a similar tuple with 
elements. Of course, just remove 
, the decrease on the left is 
, the decrease on the right is 
by AM-GM, so the inequality is preserved. Now we can just reduce to the case, where we MUST have 
. Since 
are increasing for 
, if the inequality holds true for 
, it must hold for 
. Now plugging this value of gives 
, which is NEVER true by AM-GM, so the original inequality can never hold in the conditions described.
WLOG we can assume is the largest. Thus we need to show 
. If we rewrite as a quadratic function in respect to (considering and 
as parameters) we have:
We however know:
.WLOG we can assume 
. We know that 
. We will first show that 
. This follows from the fact:
From this it directly follows that 
, implying that 
are the sides of a triangle for 
. We are now left to show that 
from which the desired will follow (because 
are the smallest). For simplicity let 
. FTSOC let 
for some 
. We know that by Cauchy-Schwarz 
. We will now show that 
.
From here it follows that:
![\[\left(t_1 + t_2 + \dots + t_n\right)
\left( \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n} \right) = n+\sum_{i, j \in \{1, 2, \dots, n\}, i\neq j} \frac{t_i}{t_j} +\frac{t_j}{t_i}\]](http://latex.artofproblemsolving.com/6/0/3/603a241e4653fd3df03efa30e822979ccf517f55.png)
Thus we get if for any 
, 
and 
that 
we get a contradiction,
that 
, thus I will
, we get that proving that is equivalent to proving![\[\frac{1}{a}+\frac{1}{b}-\frac{a+b}{c(c+k)}\geq 0\]](http://latex.artofproblemsolving.com/d/6/3/d6347eedbee72e919170ae49d15d63dc6872409c.png)
When and is a positive real. Thus the above is equivalent to:![\[\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\geq 0\]](http://latex.artofproblemsolving.com/e/6/e/e6e5ffd6f245a148efe013b7f83b0e5e68252740.png)
![\[\frac{b+a}{ab}-\frac{1}{a+b}\]](http://latex.artofproblemsolving.com/4/9/7/497c7d2bc149121cfb147ceba6006cffeb0b1315.png)
The above is minimised when 
, thus its equivalent to:![\[\frac{2a}{a^2}-\frac{1}{2a}\geq 0\]](http://latex.artofproblemsolving.com/3/d/9/3d9bcb384a9e7159c50d6748733c5ecf1e01c2f4.png)
Which is clearly true.
, and FTSOC let 
. The condition then gives by Cauchy that 
where the summation is taken over all ordered pairs in . Now notice that 
has no integer solutions for 
, so we just need to prove that 
by AMGM. Also, 
by Cauchy so this reduces into 
which is obvious.
that if there exists at least one triple 
that are not the sides of a triangle, then 
For 
we must show that if 
cannot be the sides of a triangle, then 
WLOG, assume that 
then we have that 
Also, because our inequality is homogenous, assume that 
Letting 
it suffices to show that 
But by AM-GM, 
so it suffices to show that 
But, 
by 
applications of AM-GM at the end. Therefore, our base case is proven.
Now, assume that for 
our proposition holds. Consider the sequence 
of positive real numbers. Then if there is at least one triple, say WLOG 
which are not the sides of a triangle then we have by our inductive hypothesis that
Therefore, our induction is complete. QED which are not the sides of a triangle, and 
But by our induction above, we have that 
a contradiction. Therefore, all are the sides of a triangle. QED
such that 
Then, we have that:
Thus, all that is left is to prove that 
Expanding, we have that 
Now, differentiate w.r.t. 
We get 
We claim that 
However, this is clearly true because 
while 
Thus, 
is increasing as 
increases. Thus, we can assume that 
This clearly implies that 
Hence, we are done.
.![\[\sqrt{a+\frac{9}{b+2c}}+\sqrt{b+\frac{9}{c+2a}}+\sqrt{c+\frac{9}{a+2b}}+\frac{2(ab+bc+ca)}{9}\ge\frac{20}{3}\]](http://latex.artofproblemsolving.com/b/3/2/b32e0bcba967ca847e878ac186c3168cd6c5fc66.png)
,![\[n^2 + 1 > \left( t_1 + t_2 + \cdots + t_n \right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \cdots + \frac{1}{t_n} \right).\]](http://latex.artofproblemsolving.com/1/a/b/1ab410ebc17d9f6c38a3e7ae0f6efa955ffddfd3.png)
Show that 
.
.
,
,
.
are the three sides of some triangle for all 
.
terms is at least 
,
and the inequality will be violated.
.
when 
,
,
and 
verify 
then any three of them are sides of a triangle. I think I've managed to solve this problem and if my solution is correct, then it's a really hard and nice problem.
.
![\[ (t_1+ \cdots + t_n) ( \frac{1}{t_1} + \cdots + \frac{1}{t_n}) = n^2 + \sum_{i<j} (\sqrt{\frac{t_i}{t_j}} - \sqrt{\frac{t_j}{t_i}} )^2 \]](http://latex.artofproblemsolving.com/8/2/e/82e1a00585a2b5adbd6c59119f4bf55c41b04f5e.png)
then![\[{{ (\sqrt{a/b} - \sqrt{b/a})^2 + (\sqrt{a/c} - \sqrt{c/a})^2 =
\frac{ (a-b)^2}{ab} } + \frac{ (a-c)^2}{ac} }
\geq \frac{(a-b)c}{ab} + \frac{(a-c)b}{ac}
= c/b + b/c -(b+c)/a \geq 2 - 1 = 1 \]](http://latex.artofproblemsolving.com/3/c/1/3c1c7545ca6ebf41b70af49f67869215ec2039d5.png)
,
.
.
Thus, we want
if 
Since 
,
which is clearly a contradiction. Thus, for all 
,
and 
are side lengths of a triangle which implies that 
,
and 
are as well. Swapping the indices we get the conclusion for all 
,
,
and 
.
.
and do a second degree equation in 
.
,
so 
which is a contradiction from Cauchy.
Mainly for storage.
.
.
,
and 
.
.
.![$(s+u+v)(\frac{1}{s} + \frac{1}{u} + \frac{1}{v}) \ge (s+u+v)(\frac{4}{s+u} + \frac{1}{v}) = 5 + \frac{4v}{s+u} + \frac{s+u}{v} \ge 5 + 5 \cdot \sqrt[5]{\frac{v^3}{(s+u)^3}} \ge 10$](http://latex.artofproblemsolving.com/3/2/2/322de8dea08b6057f70ea16e1396449a2cc2af43.png)
when 
.
We rephrase the problem as such:
the number 
is written. A table is "good" if all of the numbers in it sum up to a number strictly less than 
.
for all 
.
written term by term:![\[
\begin{array}{c|ccccccc}
& \frac{1}{t_1}
& \frac{1}{t_2}
& \frac{1}{t_3}
& \frac{1}{t_4}
& \frac{1}{t_5}
& \cdots
& \frac{1}{t_n} \\
\hline
t_1 & & & & & & & \\
t_2 & & & & & & & \\
t_3 & & & & & & & \\
t_4 & & & & & & & \\
t_5 & & & & & & & \\
\vdots & & & & & & & \\
t_n & & & & & & & \\
\end{array}
\]](http://latex.artofproblemsolving.com/2/2/c/22c8c239fc3d7d4edd456b3ca64c7d274d2fa718.png)
Let us now start filling in the table with numbers - we immediately notice that the cells along the main diagonal all have the number 
inside of them, so we can immediately fill these ones up:![\[
\begin{array}{c|ccccccc}
& \frac{1}{t_1}
& \frac{1}{t_2}
& \frac{1}{t_3}
& \frac{1}{t_4}
& \frac{1}{t_5}
& \cdots
& \frac{1}{t_n} \\
\hline
t_1 &1 &a &b &c &d & &v \\
t_2 &a' &1 &g &h &i & &w \\
t_3 &b' &g' &1 &l &m & &x \\
t_4 &c' &h' &l' &1 &p & &y \\
t_5 &d' &i' &m' &p' &1 & &z \\
\vdots & & & & & &\ddots & \\
t_n &v' &w' &x' &y' &z' & &1 \\
\end{array}
\]](http://latex.artofproblemsolving.com/b/a/0/ba08e85c9ac36fc7a91ac7b4c59b0528aa25db6f.png)
Now we observe the following
etc.)
for all real 
.
with cell 
)
,
cells (two by two symmetric in regards to the main diagonal) with a sum of the numbers in them at least 
such that 
- assuming that such exists would invariably lead to a contradiction, thus completing the problem.
.
(note that if there exists a triple that does not form a triangle, then the triple 
also does not form one) and we have thus reduced the problem to proving that
Since the expression on the left is scalable, we can assume that 
,
and 
.
From here, we use the useful lemma 
(can be derived - pun intended - from the second derivative of the LHS function and proving that it is reached for 
) to get the inequality we need to prove to be simply
Since this function is increasing for 
(this can be verified by expressing the difference 
)
at 
,
cells with a sum of their numbers at least 
,