for all x, y 
0, where 
is the prime number.
satisfying
for all integers 
.
,
,
means 
applied 
,
means 
applied 
Y by
In triangle 
, let the incircle 
touch sides 
at 
, respectively. Let 
lie on the line through 
perpendicular to 
. Let 
be the intersections of 
with 
, respectively. Let 
be the projections of 
onto line . Let 
be the second intersections of 
with the incircle . Let 
be the intersection of 
and 
. Prove that the line 
bisects segment 
.
, let the incircle 
touch sides 
at 
, respectively. Let 
lie on the line through 
perpendicular to 
. Let 
be the intersections of 
with 
, respectively. Let 
be the projections of 
onto line . Let 
be the second intersections of 
with the incircle . Let 
be the intersection of 
and 
. Prove that the line 
bisects segment 
.Attachments:
cut 
at 
.
,
,
is the external bisector of angle 
,
,
.
.
bisects 
.
and 
lie on 
,
satisfy 
.
the external bisector?
.
.
with collinear points 
,![\[
\frac{GB}{GC} \cdot \frac{QC}{QP} \cdot \frac{RP}{RB} = 1,
\]](http://latex.artofproblemsolving.com/7/0/6/70690bba0146b6777345ccda1d3f43066e67b757.png)
which is equivalent (as a consequence of Thales' Theorem) to![\[
\frac{DB}{DC} \cdot \frac{LC}{LD} \cdot \frac{KD}{KB} = 1,
\]](http://latex.artofproblemsolving.com/a/3/5/a35d5dfe4548b70b8d19700dd562c92e72679d3f.png)
or![\[
\frac{LD}{KD} = \frac{DB}{KB} \cdot \frac{LC}{DC} = \frac{DP}{RK} \cdot \frac{QL}{DP} = \frac{QL}{RK}.
\]](http://latex.artofproblemsolving.com/6/3/1/631acbd0e91460ca454803bffcf14a4ee7e81b3f.png)
From this, the two right triangles 
and 
are similar. As a consequence, 
,
.
be the midpoint of 
and 
,![\[
\frac{\sin\angle QDT}{\sin\angle RDT} \cdot \frac{\sin\angle LND}{\sin\angle LNM} \cdot \frac{\sin\angle KMN}{\sin\angle KMD} = \frac{DR}{DQ} \cdot \frac{\sin\angle LND}{\sin\angle NLD} \cdot \frac{\sin\angle MKD}{\sin\angle KMD} = \frac{DR}{DQ} \cdot \frac{DL}{DN} \cdot \frac{DM}{DK} = 1.
\]](http://latex.artofproblemsolving.com/9/9/0/990bfcd59b2291dcd8874162f8765b1d5d98a2b1.png)
Thus, the lines 
