be a triangle with circumcircle 
and let 
be the 
-
be the intersection points of 
intersects the tangent line at 
at a point 
.
.
be reals such that 
. Prove that
Let 
. Prove that
and 
Prove that
be real numbers such that 
Prove that
be non-negative real numbers such that 
.![\[
P = \frac{a}{a^2 + 2} + \frac{b}{b^2 + 2} + \frac{c}{c^2 + 2}.
\]](http://latex.artofproblemsolving.com/d/b/e/dbea7db769c4996fc081ee75837e177b298cffcb.png)
.
+ 
+ 
Find the maximum positive integer 
such that for every subset 
with n elements, there always exist two elements a, b in T such that:
Find 
Y by khan.academy, cubres, Mathdreams
Consider a triangle 
and some point 
on 
. The perpendicular from to 
intersects the circumcircle of 
at 
and the perpendicular from to 
intersects the circumcircle of 
at 
. Show that the line 
does not depend on the choice of .
(Shining Sun, USA)
and some point 
on 
. The perpendicular from to 
intersects the circumcircle of 
at 
and the perpendicular from to 
intersects the circumcircle of 
at 
. Show that the line 
does not depend on the choice of .(Shining Sun, USA)
This post has been edited 6 times. Last edited by Bata325, Apr 13, 2025, 2:54 AM
Reason: italics
Reason: italics
denote the circumcenter of 
.
.
denote the foot of the altitude from ![\[\angle{BAO} = \frac{180^{\circ} - \angle{AOB}}{2} = \frac{180^{\circ} - 2\angle{C}}{2} = 90^{\circ} - \angle{C} = \angle{HXC} = 180^{\circ} - \angle{BXH} = \angle{BAQ}.\]](http://latex.artofproblemsolving.com/e/2/d/e2d9375826c14c5e7197102e8c1d14f2dad41068.png)
Hence, the points 
are collinear. Similarly 
are all collinear.
be the midpoint of 
hence 
are the centres of the circles 
respectively
and 
such that 
and 
and 
are the perpendicular bisectors of the radial axes 
and 
hence 
and 
so 
and 
and the required claim is proved
![\[\angle{BAO} = \frac{180^{\circ} - \angle{AOB}}{2} = \frac{180^{\circ} - \angle{C}}{2} = 90^{\circ} - \angle{C} = \angle{HXC} = 180^{\circ} - \angle{BXH} = \angle{BAQ}.\]](http://latex.artofproblemsolving.com/7/d/9/7d972f3079ea48f9b57a8f171777f9dabada44e0.png)
Hence, the points 
hence 
where ![\[\measuredangle BAQ = \measuredangle CXQ = 90 + \measuredangle BCA\]](http://latex.artofproblemsolving.com/9/4/2/9429375d4f0c28c6a6939e228bc971645d77e8ec.png)
which implies that ![\[\measuredangle CAQ = \measuredangle CXP = 90 + \measuredangle CBA\]](http://latex.artofproblemsolving.com/0/0/a/00ab53c14582f60a8519bd3daef06f3048629879.png)
which implies that 
