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.
with circumcenter 
and orthocenter 
, let 
be an arbitrary point on the circumcircle of triangle such that does not lie on line 
and that line 
is not parallel to line 
. Let 
be the point on the circumcircle of triangle such that 
is perpendicular to , and let 
be the point on line 
such that 
. Let 
and 
be the points on the circumcircle of triangle such that 
is a diameter, and 
and 
are parallel. Let 
be the midpoint of 
.
and 
are perpendicular.
and 
are perpendicular.
with 
, a tangent to the circumcircle of at point 
intersects the extension of 
at point 
. 
is the midpoint of 
, and 
intersects the circumcircle of at 
. Prove that 
.
be positive real numbers such that 
. Prove that![\[ \frac{a_{1} a_{2} \cdots a_{n} \left[ 1 - (a_{1} + a_{2} + \cdots + a_{n}) \right] }{(a_{1} + a_{2} + \cdots + a_{n})( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})} \leq \frac{1}{ n^{n+1}}. \]](http://latex.artofproblemsolving.com/6/5/5/65558feba82e32266d3d3cbdfea85e079483403f.png)
with integer coefficients such that for all positive number 
and prime 
satisfying 
, we have 
.
, where 
, then the number of integers in the range of 
is __
be a triangle and 
denote its inradius and ex-radii opposite to the vertices 
respectively. If 
and 
, then which of the following is/are true?
is obtuse
is acute
, where 
is semi perimeter
, where is semi perimeter
satisfy that: 
and 
. Find the maximum: 
such that it is possible to divide a triangle in smaller triangles such that there are not three collinear vertices and such that each vertex belongs to the same number of segments.
. is a good value, and consider the planar graph given by the configuration (that is whose vertices are the vertices of the triangles and the edges are the sides of the triangles). Since no three vertices in the configuration are collinear, we deduce that each of the faces of this graph are triangular (I omit the not bounded one) and even are the triangles themselves. Thus there are 
faces.
edges, and each of these edges belong to exactly 
triangles except for the three of the initial triangles. It follows that there are exactly 
edges. has to be odd.
be the common degree of the vertices, and 
be the number of the vertices. Since the sum of the degree is twice the number of edges, we deduce that 
.
(I still omit the not bounded face), we deduce that 
. (1) is odd, let 
where 
.
which has to be an integer.
is even, then 
so that 
has to divide 
, and is odd and greater than 
. Thus, 
and 
. is a solution. It suffices to consider the initial triangle itself.
is odd. Then, the relation (1) becomes 
where 
. It follows that 
divides where 
.
which in turn leads to 
.
is a solution. Just choose an interior point of the initial triangle and join it to the vertices.
is also a solution, as given by the well-known Morley's configuration (since here we do not care about the equilateral triangle, you just have to choose three interior point such that among the vertices there is no three collinear ones).
...
vertices each with degree 
for a total of 
edges and 
triangles. and choose interior points, say 
in it which are the vertices of a convex hexagon, so that 
is joined with 
, 
is joined with 
and 
is joined with 
(thus choose and label carefully these points...). Now, choose interior points of this hexagon, say 
, and darw edges 
and 
and 
and 
. We are done...
.
representation. Suppose that all the vertices have degree . Hence, 
This implies that 
by Euler's relation, we know that 
, so if we let 
, we have that 
, so 
, which implies that , as claimed. For 
the example is trivial. For , we can take the Morley configuration as said above. For , we can consider the following example:
prove that![$\frac{a}{b} + \sqrt {\frac{b}{c}} + \sqrt[3]{{\frac{c}{a}}} > \frac{5}{2}$](http://latex.artofproblemsolving.com/6/a/6/6a657cb647e279b0745e3189c3f9d54aa25bf976.png)
be reals such that 
P
satisfying 
![$ a,b\in[\frac{1}{2},1] $](http://latex.artofproblemsolving.com/0/d/2/0d24f742e1f7db553b3a8795c061c796f07a897f.png)
.
Let ![$ a,b\in[\frac{1}{2},2] $](http://latex.artofproblemsolving.com/a/8/d/a8d4c8205bac0e34129f582a8f255355eda12e0b.png)
.
and 
Prove that
