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.
satisfies 
. If 
, find the maximum possible value of 
.
+1 w
and a constant 
, it is given that 
and 
are also friendly polynomials. Prove that 
.
lying outside the circle 
, two tangents are drawn touching at points 
and 
. A point 
is chosen on the segment 
. Let points 
and 
be the midpoints of segments 
and 
respectively. The circumcircle of triangle 
intersects again at point 
(
). Prove that the line 
passes through the centroid of triangle 
.
be an acute-angled triangle with 
and let 
be the foot of the
-angle bisector on 
. The reflections of lines 
and 
in line meet and at points
and 
respectively. A line through meets and at 
and 
respectively such that and 
while lies strictly between 
and . Prove that the circumcircles of
and 
are tangent to each other.
with circumcircle 
. Point in such that 
is the angle bisector of 
. Circle with center and radius 
intersects and at and respectively, 
. Let be the midpoint of arc in that didn't have . Prove that 
angle bisector of 
if and only if 
.
be real numbers satisfying
. Prove that:
be an acute triangle with 
, and let be the projection from onto . Let be a point on the extension of 
past such that 
. Let be on the perpendicular bisector of 
such that and are on the same side of and![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of across and be 
and 
, respectively. Let 
and 
such that 
. Let 
and 
intersect the circumcircles of 
and 
at 
and , respectively. Let 
and 
intersect 
and 
at and . Let 
intersect at . Prove that 
.
be an acute triangle inscribed in a circle 
with orthocenter and altitude . The line passing through perpendicular to 
intersects at . The perpendicular bisector of intersects 
at . Let 
intersect at . Let be the reflection of across 
. The circumcircle of triangle 
intersects at different from . Prove that 
and 
intersect on the circumcircle of triangle 
. be an equilateral triangle with side lenght is 
.Let ![$D \in [AB]$](http://latex.artofproblemsolving.com/0/7/9/07914568284f3822cf15730a6a0e868f0bd79996.png)
is a point. Perpendiculars from to ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
and ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
intersects with and at points and respectively. Perpendiculars from and to ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
intersects with at points 
and 
. Prove that![$$[E_1F_1]=\frac{3}{4}$$](http://latex.artofproblemsolving.com/3/e/f/3efdf53cb3ff4b2a7fa60691f5aac9b3eeecf00c.png)
with incircle . Let 
be the point on such that the circumcircle of 
is tangent to and let the -excircle be tangent to at 
. Prove that the tangent from to and the tangent from to (distinct from ) meet on the line parallel to and passing through .
, let be the foot of perpendicular from to and also suppose that and 
are excenters oposite to the side 
in triangles 
and 
. If is the point that incircle touches , prove that 
are concyclic.
, let be the foot of perpendicular from to and also suppose that and are excenters oposite to the side in triangles and . If is the point that incircle touches , prove that are concyclic.
. Now note 
and 
now also we had 
. Where 
. Now so we've 
and that implies 
now so we've 
and certainly that implies 
and so done.
can be any point on . For this, if we denote 
as the projections of 
onto , then we have to prove that 
. Moreover, assume without loss of generality that 
. Then the problem is equivalent to proving that(after a few simplifications)![\[4xybc\sin B \sin C = ((c+t)^2-x^2)((b+t)^2-y^2)\]](http://latex.artofproblemsolving.com/d/a/e/dae5731d020f09be0238f2ed5a2abdd1562d5133.png)
![\[bc\sin B \sin C = t^2(1+\cos\alpha)(1-\cos\alpha)\]](http://latex.artofproblemsolving.com/b/f/e/bfea3f8ed677ea40b72ea5ff510cef5b54863293.png)
. This is obvious by sine rule in triangles 
.
and got the following result:, is any point on and the -excircle touches at . Also, are the -excentres of triangle 
respectively. Then 
is cyclic.
is actually the excircle tangency... You have to be kidding me.
are the tangent points of 
with . We get
.
so done.
be projection of and onto line . Then , by properties of 
. We see that,
and 
. Now, 
. Now using Pythagoras theorem, one see that 
. Similarly, 
. Thus we have 
Thus 
and easy to see that 
. So is a cyclic quadrilateral. and 
is its diameter. So we are done! 
.
be the midpoint of 
and 
be the projection of 
onto 
then 
thus 
has the same power wrt 
let its center be 
then 
thus is the intersection of the perpindacular bisectors of
and
and we have is on the perpindacular bisector of 
thus 
is the intouch of the triangle instead
and onto be and respectively and are the and excenters of 
and 
respectively
concyclic
as and as . Let be the foot from onto 
and 
the foot from . Now note that it suffices to show that 
Note that
so
which is symmetric in 
and 
. Hence done!
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and be the foot of altitude from and respectively be the midpoint of 
, and let be the midpoint of , let be the midpoint of 
becomes the midline in trapezoid 
.
is a right angle, so 
has diameter 
. Hence, it suffices to prove and 
be the feet of the altitudes from and to , respectively. Hence, we are able to write
![\[DR^2+DS^2 = PR^2+PS^2\]](http://latex.artofproblemsolving.com/9/4/9/9490fa6765676630932c95ad01e44bb03e5a3291.png)
![\[\iff DS^2-PS^2 = PR^2-DR^2.\]](http://latex.artofproblemsolving.com/c/5/a/c5a01f900e1d954ad389dbb613cb5626891717ef.png)
lie on in that order. Notice that is the definition of the -excenter of . This means that
is the -excenter of , so
lies on . Relabel point of the problem as 
. Let be incenter of and 
be orthogonal projections of on . Suppose that is midpoint of 
. Since 
we have 
. To prove 
we need to show that lies on perpendicular of 
. But also lies on perpendicular of 
then it suffice to prove that 
. Indeed, we have 
. So it means that lies on perpendicular of . Hence 
or 
so let the point on be 
and the A-intouch point be 
are the excenters of 
respectively. Let be the foot from to 
and the foot from 
Then note that we simply need to show 
by Pythagorean theorem. Let 
Then because 
and 
we are done by lengths
and 
.
Where 
be feet from 
to 
and since 
then 
so 
so 
therefore 
? Please cite the theorems you used, thank you .
lie on 
and the result follows.
?
be the semiperimeter of 
,
,
,
.
,
.
be the feet of altitudes from 
to 
and 
.
,