such that 
.
round any of its endpoints. Is it possible that after several rotations the needle returns to initial position with the endpoints interchanged?
board, a nonnegative real number is written in such a way that the sum of the numbers in each row is equal to 
,
to be the largest value in row 
,
.
to be the largest value in column 
.
?
.
and 
be the feet of the altitudes from 
to 
and from 
to 
respectively.
and 
as the reflections of 
be the circumcircle of 
.
the second intersection of line 
with 
the intersection of ray 
with 
is the circumcenter of 
can be inscribed within a circle.
Y by Adventure10, Mango247
Two perpendicular chords of a circle, 
, which intersect at point 
, define on the circle four arcs with pairwise different length, with 
being the smallest of them. We draw the chords 
with 
and 
different from 
. If 
is the intersection point of 
and 
the intersection point of 
prove that 
.
, which intersect at point 
, define on the circle four arcs with pairwise different length, with 
being the smallest of them. We draw the chords 
with 
and 
different from 
. If 
is the intersection point of 
and 
the intersection point of 
prove that 
.This post has been edited 1 time. Last edited by parmenides51, Jun 21, 2022, 1:32 AM
.
cyclic.
.
?
.
and 
,![\[
\angle DCL = \angle DAM \quad \text{and} \quad \angle CDL = \angle CBN.
\]](http://latex.artofproblemsolving.com/a/8/8/a882fe91d1c6ad7677a472580efdcc06563c7645.png)
Thus, we obtain![\[
\angle DCL + \angle CDL = \angle DAM + \angle CBN.
\]](http://latex.artofproblemsolving.com/0/5/1/051482095e1ef8ec7dcf3571990cc869a53cff58.png)
Since 
,
is the point of intersection of 
and 
,
,![\[
\angle DCL + \angle CDL = \angle BZA + \angle CBN = 90^\circ,
\]](http://latex.artofproblemsolving.com/b/8/5/b8518c2473e8b17c8210a57906e142b0493e775d.png)
so![\[
\angle DLC = 90^\circ.
\]](http://latex.artofproblemsolving.com/b/1/0/b106adad4f673b5f0ab43c7805ff4db8aa35aa50.png)
be the point of intersection of 
and 
.
,
is a Simson line of the point 
with respect to the triangle 
.
and 
,![\[
\angle CPL = \angle CNL \quad \text{and} \quad \angle CNL = \angle CAD,
\]](http://latex.artofproblemsolving.com/1/7/d/17d438dc30c9823bfcdd740a96f0788a68a8f918.png)
so![\[
\angle CPL = \angle CAD.
\]](http://latex.artofproblemsolving.com/f/b/5/fb54bfbeea1144ae42e3aad1d18800b65b8d4e0a.png)
Since 
,![\[
\angle KTC = \angle ADC \quad \text{(1)}.
\]](http://latex.artofproblemsolving.com/9/9/6/996bd53868a148867f44eaf0b10b1e1602b429dd.png)
and 
are equal). Therefore,![\[
\angle ADC = \angle KNL \quad \text{(2)}.
\]](http://latex.artofproblemsolving.com/f/1/9/f195ff7af40182b7ab97e5a66c1d78b24ec44cc2.png)
