[/quote]
,
and 
.
Let 
.
and 
such that 
.
are positive isn't necessary.
such that 
is a perfect square.
and 
be non-zero integers such that 
Prove that ![\[(a_1^2-1)(a_2^2-1)\cdots(a_n^2-1)\]](http://latex.artofproblemsolving.com/b/f/f/bffd77e507fc4ddc8368c8239c523a4df40448c4.png)
is a perfect square.
be positive real numbers.Prove that![$$\sqrt[3]{ab}+\sqrt[3]{cd}\leq\sqrt[3]{(a+b+c)(c+d+a)}.$$](http://latex.artofproblemsolving.com/b/7/9/b796d0f9a2d9430bf5efb83b901427b13386f478.png)
When equality holds?
be an integer. Ion draws a regular 
with incentre 
and excentres 
and 
,
and 
respectively. The incircle touches 
and 
at 
and 
respectively. Prove that the circles 
and 
have a common point other than 
safisty: 
and 
,![$\sqrt{ad^3 g^4} +\sqrt[3]{be^3 h^4} + \sqrt[8]{cf^3 k^4}$](http://latex.artofproblemsolving.com/b/f/a/bfa17e8bb4cc05a84ed6a757f506d6d0ed2b0081.png)
segments 
are parallel. Angle bisectors of 
and 
meet at 
.
and 
meet at 
.
is cyclic
be positive real numbers such that 
and 
,
and 
lie on segments 
,
and 
intersect at point 
.
and 
intersect at a second point 
intersect at a second point 
lies on segment 
such that 
.
and 
are similar.
Y by Tawan, centslordm, megarnie, thedragon01, Adventure10, Mango247
The edges of 
are each labeled with 
or 
such that any triangle has sum of labels at least 
Determine the minimum possible average of all 
labels.
(Here is defined as the complete graph on 2017 vertices, with an edge between every pair of vertices.)
Proposed by Michael Ma
are each labeled with 
or 
such that any triangle has sum of labels at least 
Determine the minimum possible average of all 
labels.(Here
is defined as the complete graph on 2017 vertices, with an edge between every pair of vertices.)Proposed by Michael Ma
be the set of vertices, 
let 
be the set of edges of 
.
be the subgraph of 
.
which associates each vertex with the maximum degree of vertices connected to it.
which associates each vertex with the average of the degrees of vertices connected to it. Obviously 
so if we show that 
we are done (remember that in this case degrees are defined on 
is the set of ordered couples of connected vertices in 
so if 
is the set of non ordered couples of connected vertices in 
)
be a function that associates each vertex to the number of edges in 
which starts from it. Clearly 
Besides note that 
(remember that 
is defined on 
such that 
and such that its degree in 
.
as the set of vertices which are connected to 
(obviously we are considering each edge as a couple of vertices) in order to respect the hypothesis that any triangol has sum of labels at least 
.
Hence, since 
is integer, it must be at least 
.
then the minimum average of labels of edges in 
Calling 
the edges of 
except for edges connecting 
and 
for 
which will be labelled with 
.
is minimum. In such labelling, consider edges 
so 
are both labelled 
from 
to 
with 
labelled 
for 
,![$$T= \frac{1}{ \binom{2017}{2}} \left[ 2 \cdot \left( \binom{2017}{2}-1008 \right)+1008 \right]= 2- \frac{1008}{\binom{2017}{2}}=2-\frac{1}{2017}.$$](http://latex.artofproblemsolving.com/0/8/f/08ff561da316a0a21e1662ab6272ec275bbdb925.png)
Construction: Consider 
vertices 
with 
are labelled 
where 
edges is still minimal.
,
,
,
.
there are 
,
.
)
edges labelled 
of its edges is minimally 
.
,
is 
.
:
is labeled 
and 
contribute a sum of at least 
.![\[ \binom{2m-1}{2} \left( 2 - \frac{1}{2m-1} \right) + (8m-4) + 1 = \binom{2m+1}{2} \left( 2 - \frac{1}{2m+1} \right) \]](http://latex.artofproblemsolving.com/2/5/5/2556873cf7ad148055f651874584545cf82f3626.png)
as desired.
with 
and 
,
with 
vertices, one can generate one on 
vertices by two vertices 
with weight 
for even 
for odd 
is also 
,
case of equality case. I have no clue where I should adjust the problem to.
-
edges. We call a pair of vertices 
fat if there exists a vertex 
different fat pairs.
-
) graph on 
edges, but we will start by adding first all the non-adjacent edges(call them initial edges), which are at most 
at number. It is clear that any edge we will add from this point(call those edges - babies) will be adjacent to at least one initial edge. Everytime we add a baby, let's say 
and color it in red, if it isn't already painted in red, where 
is an initial edge, if there are more segments that we can choose, we pick one randomly. Suppose that at a certain point, after adding a baby , let's say 
for which 
is an initial edge and 
is a baby, but this means that segment 
is also already painted in red, which means that there is an initial edge different from 
.
edges, which implies that the minimum sum of the labels that we can achieve is 
,
.![\[ \binom{2m-1}{2} \left( 2 - \frac{1}{2m-1} \right) + 8m + 1 = \binom{2m+1}{2} \left( 2 - \frac{1}{2m+1} \right) \]](http://latex.artofproblemsolving.com/d/2/e/d2e357d1e06a4ec274d9fb70ece7b6c0ef6fcc56.png)
as desired.
since there are only 
vertices when we delete the edge. Then, we compute![\[ \binom{2m-1}{2} \left( 2 - \frac{1}{2m-1} \right) + 8m - \mathbf{3} = \binom{2m+1}{2} \left( 2 - \frac{1}{2m+1} \right), \]](http://latex.artofproblemsolving.com/f/c/7/fc7bb5c4c5b23ebd0482b1ee36dfd9620be0782a.png)
so fortunately the solution works out. This is a funny case of two wrongs luckily (or unluckily depending on how you look at it) making the solution not fundamentally wrong.
and they have exactly one common vertice. In other words: take any vertice and if it has at least one edge 
or 
(as there's no 
l
.
labeled with ones and the rest with twos.
,
,
labels in a 
.
is trivial to verify.
,
.
labeled 
incident to it that is also a 
must be labeled 
is ![\[\frac{\left(2-\frac{1}{n-2}\right)\binom{n-2}{2}+2\cdot(n-2)+1}{\binom{n}{2}}=2-1/n,\]](http://latex.artofproblemsolving.com/9/c/d/9cdff5faf8a534e45940776f13f84e518c359033.png)
which completes the induction and the proof.
.
,
,
,
which fits. Now suppose that 
.
subgraph of the 
,
,
this means that the average for 
,
Construction is 
the labels of 
and 
sum to 
So just remove 
decreasing 
Repeat until no edges with label 
and 
.
.
be a simple graph. For each 
,
to be the average degree of the neighbors of 
Proof of the lemma:
Hence the lemma. 
.
and 
,
.
Which gives us 
,
.
