[/quote]
,
has a unique solution in integers with 
and 
.
as 
fir 
.
for all 
.
and orthocenter 
.
intersect 
respectively. Lines through 
intersects 
at 
respectively. Point 
is a parallelogram. Prove that lines 
and 
are concurrent at a point on 
.
is a cyclic quadrilateral inscribed in a circle of radius one unit. If 
,
are positive real numbers.Prove that
be a function taking in positive integers and outputting nonnegative integers, defined as follows:
is the number of positive integers 
with 
such that the equation 
has an integer solution 
.
such tha
(Adit Aggarwal, India.)
be positive reals such that 
.
be reals such that 
Prove that
.
of triangle 
meets 
at 
be a point on the (internal) angle bisector of 
such that 
.
meet line 
and 
.
is isosceles.
such that, for all positive integers 
and 
and 
Show that
with center 
.
the incenter of triangle 
the contact points of the incircle with sides 
of triangle 
the foot of the perpendicular line from 
.
passes from the antidiametric point 
in the circle 
is a diametre of the circle 
Y by
Let 
be a 90-degree triangle with hypotenuse 
. Let 
and 
distinct points on segment and 
be the foot of the perpendicular from to 
and to 
, respectively. 
and 
intersect at 
.
Lines
and intersect at 
and lines 
and intersect at 
.
Prove that
if and only if 
.
be a 90-degree triangle with hypotenuse 
. Let 
and 
distinct points on segment and 
be the foot of the perpendicular from to 
and to 
, respectively. 
and 
intersect at 
.Lines
and intersect at 
and lines 
and intersect at 
.Prove that
if and only if 
.This post has been edited 1 time. Last edited by KAME06, Nov 4, 2024, 8:26 PM
and 
.
.
if and only if 
are homothetic with centre 
,
.
circumcentre of 
.
by problem conditions. So we have the fact 
,
and 
because 
,
then 
because 
.
.
means that 
are symmilar back to front of part 1. So we´re done.
