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 
.
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 .
, and equality holds if and only if the point R lies on the ray PT (the only exception is when 
; in this case, P = T, so the ray PT is undefined, and equality holds for all points R in the plane p, since we always have < RPQ = < TPQ = 90°)., it follows that 
; also, since the angles < RPQ and < TPQ are < 180°, their half-angles 
and 
are < 90°, so that from we can conclude that 
. Equality holds, as in the above, if and only if the point R lies on the ray PT (and, respectively, for all points R in the plane p if ).
with equality if and only if 
, i. e. if and only if < QRP = < PQR, i. e. if and only if triangle PQR is isosceles with base QR, i. e. if and only if PR = PQ, i. e. if and only if the point R lies on the sphere with center P and radius PQ.
(since )
(since ), and that we have used, i. e. if and only if the point R lies both on the ray PT (this condition should be ignored if ) and on the sphere with center P and radius PQ.
is maximum is the point of intersection of the ray PT with the sphere with center P and radius PQ (or, respectively, it can be any arbitrary point on the intersection of the plane p with the sphere with center P and radius PQ if ).
is the point we are looking for instead of 
. Let be the foot of on plane ![$[i][b]p[/b][/i]$](http://latex.artofproblemsolving.com/4/7/d/47d7c3c332aacb36a413eb82ed00f5bc90ad927d.png)
. Draw ray y that passes through . Let a = , b = 
, c = 
, x = 
, α = ∠
, β = ∠
and ![$[i]r[/i]$](http://latex.artofproblemsolving.com/a/7/e/a7e18c283a412e3621c0c615a6f2948c17258c7c.png)
= (
)/ = (a + x)/b. is on ray y with 
.
, and equality holds if and only if the point R lies on the ray PT (the only exception is when ; in this case, P = T, so the ray PT is undefined, and equality holds for all points R in the plane p, since we always have < RPQ = < TPQ = 90°).
. If T=P, 
. Also in both cases 
is fixed. So: is max, if and only if 
.
lies on line 
, when 
(I took 
as projection of onto the plane 
as others).
. Assume doesnt lie on . Take point 
on ray such that doesn't lie on segment and
. We can choose such because 
from triangle inequality on 
. 
So we get 
which is contradiction. So should lie on line .
and 
.
Where 
and 
.
.![\[ \frac{QP+PR}{QR} \]](http://latex.artofproblemsolving.com/b/8/e/b8ead012f276c34cfbd73d7507cfa1592728f80c.png)
is maximum.
