[/quote]
,
, such that 
Prove that :
equality cases : ?
of an equilateral triangle 
.
extended such that 
is on 
extended such that 
.
is the intersection of 
and 
.
?
where 
are real numbers, all 
not equal to 
.
s.t 
satisfying the condition![\[
2f(n)^2 = n^2 + g(n)^2 \quad \text{for all } n \in \mathbb{N}^*.
\]](http://latex.artofproblemsolving.com/6/a/1/6a112ea3fe44273fdd4dc445e9bd7387d365925f.png)
Prove that if 
for all 
,
has infinitely many fixed points.
be a diameter of a circle 
,
be the tangent at 
.
be a fixed, positive real, and consider all pairs of points 
and 
lying on 
.
and 
intersect 
and 
,
pass through a common point.
is described on a circle 
be 
respectively. A circle with a center 
,
.
,
.
,
and lines 
,
are concurrent.
satisfying![\[\begin{aligned}
\begin{cases}
a+b+c+d=0,\\
a^2+b^2+c^2+d^2=12,\\
abcd=-3.\\
\end{cases}
\end{aligned}\]](http://latex.artofproblemsolving.com/8/b/1/8b1923e0303c0aa79332ee4540ee1885a9cf3f08.png)
and 
,
intersects line 
and 
.
is tangent to 
,
,
are the sides of a quadrilateral inscribed in a given circle. Prove that the product 
acquires its maximum when the quadrilateral is a square.
satisfy that 
.
Y by Rounak_iitr, ItsBesi
Four points 
, 
, 
and 
lie on a circle 
such that 
. The tangent line to at point intersects the tangent line to at and the line 
at 
and 
. The circle and the circumcircle of triangle 
intersect again at 
. Prove that 
.
Proposed by Mahdi Etesamifard
, 
, 
and 
lie on a circle 
such that 
. The tangent line to at point intersects the tangent line to at and the line 
at 
and 
. The circle and the circumcircle of triangle 
intersect again at 
. Prove that 
.Proposed by Mahdi Etesamifard
This post has been edited 1 time. Last edited by Tafi_ak, Dec 23, 2022, 9:52 AM
![\[ \angle KAB=\angle BAC=\angle CAD=\angle BCK=\angle BDC=\angle BDA=\angle BMA=\angle BCA=\angle DCL=\angle DLC \]](http://latex.artofproblemsolving.com/6/3/d/63d5711e8856843cc1136522c4a9c8add9cf52f6.png)
That gives 
,
are collinear. Also notice that 
is a kite because of 
,
,
.
bisects 
.
as the antipode of 
.
.
and 
.
,
gives 
.
,
Now note points 
are collinear on the perpendicular bisector of segment 
.
![[asy]
size(200);
pair O=(0,0),B=dir(117),C=2*foot(B,O,dir(90))-B,A=2*foot(C,O,B)-C,D=2*foot(B,O,C)-B,K=2*A*C/(A+C),L=2*foot(C,A,D)-A,N=-B;
fill(B--A--C--cycle,green+white+white);
fill(C--D--L--cycle,green+white+white);
draw(unitcircle,royalblue);
draw(circumcircle(K,A,L),purple);
draw(A--K--L,brown);
draw(B--C^^A--L,orange);
draw(N--C,red);
draw(N--K);
draw(A--N--L,cyan);
draw(B--A--C--D,magenta);
dot("$A$",A,dir(A));
dot("$B$",B,dir(160));
dot("$C$",C,dir(C));
dot("$D$",D,dir(-40));
dot("$K$",K,dir(K));
dot("$L$",L,dir(L));
dot("$N$",N,dir(N));
markscalefactor=0.015;
draw(anglemark(A,K,B));
markscalefactor=0.018;
draw(anglemark(N,K,C));
[/asy]](http://latex.artofproblemsolving.com/5/d/8/5d8cb9f8d5451d0334ca907fbbb5f1d692b4fb62.png)
and 
we have:
It follows
Thus 
.
,
is perpendicular bisector of segment 
.
Look at 
now. Note
particularly 
.
of 
not containing 
we have that 
.
.
and 
and as 
is isosceles we have that 
is perpendicular bisector of 
,
,
thus 
.
to be 
.
,
are collinear.
which meets 
,
and hence 
are collinear. Hence 
so 
and let 
be the antipode of 
is cyclic :
.
and 
,
and 
.
.
and
Hence
Therefore 
.
are collinear, so 
,
,
,
so 
,
.
thus 
.
.
,
.
So it suffices to show 
thus 
is a parallelogram 
and we're done.
and 
is tangent to 
but 
then 
We have 
by 
cyclic and 
by 
cyclic. Then we obtain 
that means 
and 
are tangents to 
because 
.
,
,
,
that not contains 
