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: 
are concurrent. Let 
be the point of intersection. Let be the centre of 
. We can use Pascal on 
to show that 
are concurrent.
are concurrent iff is the polar of 
. Now is the polar of 
so 
are concurrent iff 
iff 
are concurrent. We can now use Pascal twice (find the hexagons yourself hahaha) to show that 
are collinear. It is now obvious due to the complete quadrilateral that 
. Take perspective through 
so 
. Using Pascal again (find the hexagons yourself) we get 
are collinear so we get another complete quadrilateral so 
. So 
; thus 
are concurrent.
and 
. Then these two points intersect at a point 
collinear with 
and (Notice the tangential quadrilateral with sides containing 
, 
, 
, 
). By Newton's theorem on the tangential quadrilateral containing the tangents through , , 
, , lines 
, 
, and 
are concurrent. 
, 
, and 
are collinear. Also, lies on by Newton's theorem, and thus, 
and 
. 
be the Pole of wrt , If a line through intersects at 
, then 
are on the same side of . Since 
(The polar of A,C), and we can use Menelaus to show that 
concur, hence 
. By properties of poles and polar 
. Hence 
lies on the line formed by 
, which is . lies on , and 
incircle
, then 
are collinear.
again, by the lemma 
or 
.
lies on the polar of , which is .
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
of a quadrilateral 
touches 
at 
,
on the segment 
inside 
meet 
respectively. Prove that 
are concurrent.
,
is a kite with 
and 
.