[/quote]
,
satisfy 
.
and the incircle 
of 
at 
,
meets 
at 
and intersects 
at 
,
meets 
at 
.
(different from 
)
and 
.
.
are being replaced with 
.
will appear on the board.
,
,
,
lie on a semicircle 
in this order with diameter 
,
and 
lie on segments 
respectively such that 
and 
.
passes through 
.
,
are concurrent.
such that 
is a divisor of 
.
such that for all positive real numbers 
,![\[
f^2(x) \geq f(x+y)\left(f(x) + y\right).
\]](http://latex.artofproblemsolving.com/7/1/f/71fb0d51492ee4c3b2bf4049e8224882fec16020.png)
.
the equation: 
,
.
lie on segments 
respectively. Points 
are such that 
and 
are equlateral, 
lie on opposite sides of 
,
lie on the opposite siedes of 
is a parallelogram. Take 
on the line 
so that 
and 
.
.
and 
be two points lying on the circle 
such that 
and 
are perpendicular and parallel to bisector of 
,
and 
bisect two arcs 
.
be given positive integer. Suppose that there is only one positive integer 
which satisfies the equation
for some positive integers 
.
be positive real numbers such that 
.
knowing that the denominators are positive real numbers.
such that 
,
.
with integer coefficients such that for any positive integer 
,
is positive and 
Y by Adventure10, Mango247
In 
ABC, points D, M lie on side BC and AB respectively, point P lies on segment AD. Line DM intersects segments BP, AC (extended part), PC (extended part) at E, F and N respectively. Show that if DE=DF, then DM=DN.
ABC, points D, M lie on side BC and AB respectively, point P lies on segment AD. Line DM intersects segments BP, AC (extended part), PC (extended part) at E, F and N respectively. Show that if DE=DF, then DM=DN.This post has been edited 1 time. Last edited by jred, Jun 29, 2013, 10:59 AM
???
???
and 
to intersect 
,
,
through 
,
Then, 
