[/quote]
,
, s.t :
be real numbers so that 
and 
Show that
,
be the orthocenter of 
.
at 
at 
.
be the midpoint of 
,
meet 
.
at 
.
is tangent to 
,
by ![\[\displaystyle f(x) = \left\{ \begin{array}{lr} x+\frac 12 & \text{if}\ \ x < \frac 12\\ x^2 & \text{if}\ \ x \ge \frac 12 \end{array} \right.\]](http://latex.artofproblemsolving.com/c/b/d/cbd9b412893774283a438296cdeb05b80cc878bf.png)
Let 
and 
be two real numbers such that 
.
and 
by 
,
,
for 
.
such that ![\[(a_n - a_{n-1})(b_n-b_{n-1})<0.\]](http://latex.artofproblemsolving.com/a/5/c/a5c8844d595bb1186d4804beadd69a3bc1d40ea3.png)
,
- point 
is tangent to line segment 
.
is tangent to line 
.
and circumcentre 
.
cross 
be the mid-arc points of 
,
cross 
at 
cross 
at 
.
and 
are concurrent on the external bisector of 
.
,
and 
to 
and 
,
and 
meet 
is tangent to the circumcircle.
and a point 
outside the circle such that 
. The perpendicular bisector of 
at 
and 
.The segment 
intersects 
at 
intersects 
at 
.The rays 
intersect 
respectively. Prove that the triangles 
and 
are equilateral triangles and the lines , 
are concurrent.
such that for any real numbers 
and 
,![\[
f\left(x^2+2y f(x)\right) + (f(y))^2 \leq (f(x+y))^2
\]](http://latex.artofproblemsolving.com/8/4/3/8437449cd3f8f0da24d17a60b55dcb563e6b5655.png)
of positive integers satsifying 
.
penguin and each one is numbered using numbers 
.
is called 
if 
where 
is 
prime divisor of 
(
where 
is number of 
)
Y by Adventure10
In a triangle 
, let 
and 
be the midpoints of 
and 
, respectively, and let 
be the foot of the altitude through 
. Show that the line 
, the angle bisector of 
and the circumcircle of 
pass through a common point.
Alternate version: In a triangle, let and be the midpoints of and , respectively. The line and the angle bisector of meet at 
. Show that 
is a right angle.
, let 
and 
be the midpoints of 
and 
, respectively, and let 
be the foot of the altitude through 
. Show that the line 
, the angle bisector of 
and the circumcircle of 
pass through a common point.Alternate version: In a triangle
, let and be the midpoints of and , respectively. The line and the angle bisector of meet at 
. Show that 
is a right angle.
.
.
thus the result.
