[/quote]
,
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)
with 
,
intersects the extension of 
at point 
.
is the midpoint of 
,
intersects the circumcircle of 
.
.
be positive real numbers such 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 
,
.
be reals such that 
P
,
,
is __
satisfying 
be a triangle and 
denote its inradius and ex-radii opposite to the vertices 
respectively. If 
and 
,
is obtuse
is acute
,
is semi perimeter
,![$ 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)
.
satisfy that: 
and 
.
, such that 
Prove that :
equality cases : ?
and 
Prove that
.
is in the interior of ![\[\angle PAD:\angle PBA:\angle DPA=1:2:3=\angle CBP:\angle BAP:\angle BPC\]](http://latex.artofproblemsolving.com/9/3/8/9388a91325842f1ec67a4693f0f10af46ed66c6d.png)
Prove that the following three lines meet in a point: the internal bisectors of angles 
and 
and the perpendicular bisector of segment 
.
Y by braveheartlong, Adventure10, Mango247
Let 
is centroid of 
triangle. Let 
is the perpendicular line with . Let 
is a point on . Let 
are midpoints of 
respectively. The line through point 
and perpendicular with meet 
at 
. The line through point 
and perpendicular with meet 
at 
. Let 
is a line through point and perpendicular with 
. Prove always pass a fixed point.
is centroid of 
triangle. Let 
is the perpendicular line with . Let 
is a point on . Let 
are midpoints of 
respectively. The line through point 
and perpendicular with meet 
at 
. The line through point 
and perpendicular with meet 
at 
. Let 
is a line through point and perpendicular with 
. Prove always pass a fixed point.Attachments:
This post has been edited 1 time. Last edited by thanhnam2902, Jan 29, 2008, 4:52 PM
are concurrent (*)
so
we have 
.
.
,
,
with 
fixed. midpoints 
,
are: 
and 
equation of 
eq 
then 
and 
slope of 
(Reducing terms and making the computations) is: 
then the slope of the perpendicular to 
then the equation of the perpendicular to 
the intersection of this line with the 
axis is: 
(point: 
)
(but on 
the line 
does not depend on 
Assume the proposition is true for the line 
itself. Let 
be the foot of normal from 
let 
be the perpendiculars from 
to 
let 
be the fixed point of 
and let a parallel to 
through 
at 
is a parallelogram, 
and 
Since 
is fixed, 
is also fixed. Therefore, it is sufficient to prove the proposition for 
cut perpendiculars to 
at points 
are midpoints of 
is a midline of the 
Since 
Parallels to 
through 
respectively, intersect at the reflection 
of 
in the perpendicular bisector of 
Define 3 points on the line at infinity: 
Consider the line 
and the line at infinity 
By Pappus theorem, the intersections 
are collinear.
be the foot of a normal from 
i.e., 
is the line supposedly passing through a fixed point. 
is cyclic because of the right angles 
hence 
Since 
are on the circumcircle of the 
and 
it follows that 
Let 
cut 
again at 
is right, 
is a diameter of 
so that 
is also right and 
of the triangle 
with respect to the sideline 
lies on the midperpendicular of the segment 
where 
is the orthocenter of 
and 
= 
where 
is the orthogonal projection of 
intersects the line segment 
such that 
and so, it is a fixed point.
be, the midpoints of the segments 
respectively and we denote as 
the orthogonal projections of 
respectively, on 
where 
is the 
are concurrent at one point so be it 
as the 
and 
and 
and 
we conclude that the points 
are collinear and 
intersects the line segment 
where 
as the fourth vertex of the rectangle 
is a fixed point on the line segment 
as the intersection point of the circumcircle of 
from its altitude 
linearly on 
,
denoting time. Observe that 
also move linearly on 
respectively. As a result of the above, for a particular point 
in the plane, the function 
is a quadratic in 
Hence, if it is zero for 
points, it is identically 
and we'd be done.
pass three a fixed point for some three choices of 
respectively to 
