[/quote]
,
for which, for all 
,![\[
f(x) f(yf(x)) + y f(xy) = \frac{f\left(\frac{x}{y}\right)}{y} + \frac{f\left(\frac{y}{x}\right)}{x}
\]](http://latex.artofproblemsolving.com/9/0/c/90c180110402e1a32b70edb2b0a03a28727457d1.png)
such that 
.
.
be a convex quadrilateral with 
,
and 
.
and 
the reflections of 
in lines 
,
and 
meet the line 
at 
and 
,
and 
are tangent to each other.
.
.
.
.
and 
and 
,![\[ \angle ABS=\angle ACR=180^\circ-\angle BGC.\]](http://latex.artofproblemsolving.com/0/a/d/0ad01e29194358ea0d3e10cb74b0769d1bd3a656.png)
Prove that 
.
and 
for which![\[ \left \lfloor \frac{a^2}{b} \right \rfloor + \left \lfloor \frac{b^2}{a} \right \rfloor = \left \lfloor \frac{a^2 + b^2}{ab} \right \rfloor + ab.\]](http://latex.artofproblemsolving.com/c/3/6/c3682397d21443cb720e56251c260b984392a84a.png)
,
,
,
.
meets 
at 
.
be the reflection of 
.
meets 
and 
at 
meets 
.
,
,
form an arithmetic progression.
such that 
for all 
.
is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence 
,![\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]](http://latex.artofproblemsolving.com/1/7/5/1751a60482d729a36c71b77ac9c978e724f40da0.png)
right and 
,
.
we take two points 
so that 
and 
.
and 
containing inside 
and 
,
and 
,
and that 
perpendicular to the plane of the square. Show that the projections of 
,
,
and 
are coplanar if and only if ![$O\in [AC]\cup [BD]$](http://latex.artofproblemsolving.com/6/6/6/666df8f2afa87f15c79f9c96d9d1cea17a2a7af0.png)
.
.![\[
(2^x - 1)(5^x - 1) = y^n
\]](http://latex.artofproblemsolving.com/f/8/f/f8f013a96863bc5f39aa23edb98eed9aec54a17f.png)
have no positive integer solution 
.
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 
.
,
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 
