.
be the feet of the altitudes from 
to 
respectively. Let 
be the circle with diameter 
.
be the intersection of 
and 
be the orthocenter of 
be the intersection of 
and 
be the lines through 
tangent to 
respectively. Let 
be the intersection of 
and 
.
and 
.
be the line through 
and let 
be the reflection of 
is tangent to 
.
,
,
terms is divisible by 
be a positive integer divisible by 3, and let 
be non-negative real numbers satisfying 
.![\[
M_n = \min_{1 \le i \le n} \{x_i x_{3i}\},
\]](http://latex.artofproblemsolving.com/d/b/b/dbbf0f6bf20f506d6cb63357b06ec6c52bebed10.png)
where the indices are taken modulo 
.
be odd natural number and 
be pairwise distinct numbers. Prove that someone can divide the difference of these number into two sets with equal sum.
)
board is filled with some nonnegative integer. Two numbers in the filling are said to be adjacent if their cells share a common side. (Note that two numbers in cells that share only a corner are not adjacent). The filling is called a garden if it satisfies the following two conditions:
or 
.
and 
be positive real numbers such that ![\(a+b+c=4\sqrt[3]{abc}\)](http://latex.artofproblemsolving.com/2/8/2/28214a1814762b71858db1543fae1a50212eaafa.png)
.![\[2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.\]](http://latex.artofproblemsolving.com/7/0/4/7043a170ac5535465a5a1b4daccb1f1de3043280.png)
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 
