.
and 
are points inside 
and 
are similar and quadrilaterals 
and 
are cyclic. Let lines 
and 
meet at 
and 
meet at 
.
are parallel.
meet the sides 
in 
respectively. Show that ![\[ \frac{AP}{PD} = \frac{AF}{FB} + \frac{AE}{EC}. \]](http://latex.artofproblemsolving.com/7/5/9/759482bace109eb619aaea692625f7854faa9fec.png)
.
represented by tokens to players 
to 
so that they all have the same number of tokens in their urns.
beats 
when, when 
.
What is 
satisfy 
.
and the incircle 
of 
at 
,
meets 
and intersects 
at 
meets 
at 
.
(different from 
and 
.
.
?
,
.
and 
.
,
and multiply to 
,
.
,
be an acute, scalene triangle, and let 
,
,
be the midpoints of 
,
,
,
intersect ray 
in points 
and 
respectively, and let lines 
and 
intersect in point 
,
,
denote the set of positive integers. Find all functions 
such that for positive integers 
![\[f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.\]](http://latex.artofproblemsolving.com/c/1/f/c1f8ffe04cfcc45b498f3931a4796d5c56dc04d0.png)
and 
are erected outside an acute triangle 
Suppose that ![\[\angle BC_1C+\angle CA_1A+\angle AB_1B=180^{\circ}.\]](http://latex.artofproblemsolving.com/3/3/7/3378f0c4409d14bc0afddf2616992c13f624be3b.png)
Prove that lines 
and 
are concurrent.
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 
.
lie on 
,
satisfy 
.
the external bisector?
.
.
,![\[
\frac{GB}{GC} \cdot \frac{QC}{QP} \cdot \frac{RP}{RB} = 1,
\]](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, 
,
.
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 
