[/quote]
,
be real numbers such that 
and 
Prove that
people arranged in a circle, and 
coins are distributed among them, where each person has at least 
coins. Each person is then assigned a random index number in 
such that no two people have the same number. Then every minute, if 
is the number of minutes passed, the person with index number congruent to 
be a convex cyclic quadrilateral with circumcircle 
.
produced beyond 
meet 
produced beyond 
,
.
be a line through 
and 
at 
and 
respectively, so that 
(respectively 
)
)
and 
are points on the arc 
of 
so that 
are tangent to 
are collinear.
.
Where 
.![$\sqrt[3]{ab}+ \sqrt[3]{cd} < \sqrt[3]{(a+b+c)(b+c+d)}$](http://latex.artofproblemsolving.com/0/5/0/0502ba65f0545b3af84c41b2549d6256550b473b.png)
.
,
,
be real positive numbers. Show that ![\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a+b}{b+c}+\frac{b+c}{a+b}+1\]](http://latex.artofproblemsolving.com/0/9/5/0959464bbb9b821a11df292f9e6addded7326371.png)
is almost squarefree if there exists a prime number 
such that 
and 
is squarefree. Prove that for any almost squarefree positive integer 
is an integer.
be a natural number. At a mathematical olympiad training camp the same 
such that 
for all 
?
have the same incenter 
.
is concyclic iff 
is concurrent
, such that for all 
:![\[ f(x+f(y))=\dfrac{f(x)}{1+f(xy)}\]](http://latex.artofproblemsolving.com/1/4/1/1418fad9fd00b31c38bfbc5657e2a322d66ef450.png)
Y by Adventure10, megarnie, GeoKing
Suppose 
is a connected graph with 
edges. Prove that it is possible to label the edges 
in such a way that at each vertex which belongs to two or more edges, the greatest common divisor of the integers labeling those edges is equal to 1.
Note: Graph-Definition. A graph consists of a set of points, called vertices, together with a set of edges joining certain pairs of distinct vertices. Each pair of vertices
belongs to at most one edge. The graph 
is connected if for each pair of distinct vertices 
there is some sequence of vertices 
such that each pair 
is joined by an edge of 
.
is a connected graph with 
edges. Prove that it is possible to label the edges 
in such a way that at each vertex which belongs to two or more edges, the greatest common divisor of the integers labeling those edges is equal to 1.Note: Graph-Definition. A graph consists of a set of points, called vertices, together with a set of edges joining certain pairs of distinct vertices. Each pair of vertices
belongs to at most one edge. The graph 
is connected if for each pair of distinct vertices 
there is some sequence of vertices 
such that each pair 
is joined by an edge of 
.This post has been edited 1 time. Last edited by orl, Aug 15, 2008, 1:51 PM
(mario pickavet, bronze '91)
be the vertices involved in the path. If 
,
be another vertex. We can find a path 
,
,
are not among the 
'
,
will be equal to 
vertices have been labelled, label the unique edge 
Otherwise, label the edges 
Since 
we are done.
s.t. 
are connected with each other and 
as 
as 
and 
respectively and 
as 4. Look at the vertex 
!?
such that 
(WLOG 
)
,
in succession. Then, each vertex is labeled either with 2 consecutive integers or with 
where we add edges to 
between pairs of odd-degree vertices. Then, 
such that each walk starts and ends at odd-degree vertices and each odd-degree is the endpoint of exactly one walk.
,![$[1, k_1]$](http://latex.artofproblemsolving.com/d/1/1/d119cc6926863efd3a3bafc64a8a81c175be6c2c.png)
in order, the second one with ![$[k_1+1, k_1+k_2]$](http://latex.artofproblemsolving.com/c/e/b/ceb9dee824e705b391db60eddfc1d78a53b533d9.png)
,![$[k_1+\cdots+k_{m-1}+1, k_1+\cdots+k_m]$](http://latex.artofproblemsolving.com/3/7/c/37c0ed81a5f949ec1189960c87b28beef99e05bb.png)
.
of edge labels equal to one. 
and 
to those first and last edges and of course the may not be coprime.
be the vertices of odd degree of 
,
in order. Otherwise, we will add the edges 
Let this modified graph be 
Note that this may result in the graph not being simple anymore, but that's fine since the problem still holds. Now, observe that 
from the cycle would partition the cycle into walks from 
to 
,
to 
,
,
to 
This makes it such that none of the walks can be cycles since the 
'
?
,
of 
.
where 
has an edge not in 
.
and replacing the edge 
with 
.
(with base case 
)
,
,
.
.
for which 
,
,
,
for which 
,
,
.
with 
with 
,
with 
.
for 
differ by 
.
edges, and is obviously still connected, unless it is empty. If it is empty then 
,
,
,
so that the condition is satisfied. Now, consider these labelings in conjunction with our labelings of the other edges. We know that the labelings are fine for 
since they do not change based upon the labelings of the induced graph since they are only adjacent to other 
'
.
,
for which 
,
.
in 
in a graph). Let 
be a cycle in 
.
with 
,
be the smallest such index for which 
,
be the second-smallest such index for which 
.
.
,
by minimality of 
with 
with 
with 
,
(these vertices have degree 
edges. Thus, by the inductive hypothesis we can label the edges 
so that the condition is satisfied for the reduced graph. Now, consider the entire graph. Then, for each vertex 
,
and 
,
be the vertices in the cycle. Then, label 
with 
,
with 
.
,
to get a connected graph on 
edges, so we can label the edges 
so that it works on the reduced graph. Now, it is not hard to see that it works in the original graph -- for all vertices which are not a 
,
(so it is 
)
.
for which 
,
.
with 
with 
,
with 
.
,
,
,
edges, so by the inductive hypothesis we can label it and then this labeling works for 
'
and 
all work since their neighboring edges have labels that are adjacent numbers (and so they have gcd 
consecutively. By maximality of the walk, every vertex in the walk (including the endpoints) is now valid; now, consider the subgraph 
.
,
.
be the spanning tree of 
is still connected, so by the induction hypothesis, we can label the edges of 
.
,
so labeling the remaining edges in arbitrary order will be satisfying the condition. Thus we're done.
degree.
.
.
is connected to goes from degree 1 to degree 2 when you include 
edges in this tree, we have the first 
