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be a positive integer. Paul has a 
rectangular strip consisting of 
unit squares, where the 
square is labelled with 
for all 
. He wishes to cut the strip into several pieces, where each piece consists of a number of consecutive unit squares, and then translate (without rotating or flipping) the pieces to obtain an 
square satisfying the following property: if the unit square in the row and 
column is labelled with 
, then 
is divisible by 
.
+1 w cities in a country, where 
is an integer. Some pairs of cities are connected by direct (two-way) flights. For two cities 
and 
we define:
A 
between and as a sequence of distinct cities 
, 
, such that there are direct flights between 
and 
for every 
;
A 
between and as a path between and such that no other path between and has more cities;
A 
between and as a path between and such that no other path between and has fewer cities. and in the country, there exist a long path and a short path between them that have no cities in common (except and ). Let 
be the total number of pairs of cities in the country that are connected by direct flights. In terms of , find all possible values 
be integer numbers 
satisfy that:
. Prove that: are even
is a perfect square number
can’t be any power 
of a positive integer number
.
touches the sides CA and AB of the triangle ABC, its center must lie on the angle bisector of the angle between these two sides, i. e. on the angle bisector of the angle CAB. On the other hand, since this semicircle has its base on the line BC, its center must lie on the line BC. Thus, the center of the semicircle with radius is the point of intersection of the angle bisector of the angle CAB with the line BC, i. e. the point X. Similarly, the centers of the semicircles with radii 
and 
are the points Y and Z.. The lines CA and AB touch both of these circles; thus, they are the external common tangents of these two circles; consequently, the point of intersection A of these lines CA and AB is the external center of similitude of these two circles. Hence, there exists a homothety with center A which maps the circle with center X and radius to the incircle of triangle ABC.
(in fact, it maps the circle with center X and radius to the incircle of triangle ABC with radius r), but on the other hand, the ratio of this homothety is 
(in fact, it has the center A, and it maps the circle with center X and radius to the incircle of triangle ABC with center I). Hence, 
. Similarly, 
and 
. Hence, the equation becomes 
, so that 
, and we are done.
, we have that
such that 
such that 
, and the bound is sharp (equality is obtained for the degenerate triangle with sidelengths 0, 1, 1).. Division by r transforms this into 
., we have 
. Now, 
(this is a well-known fact; it was proven, e. g., in http://www.mathlinks.ro/Forum/viewtopic.php?t=50582 post #3, second solution, in the equivalent form 
). Thus, 
, so that 
. Hence, 
. Similarly, 
and 
. Hence, the inequality in question, , becomes 
, or, equivalently, 
. But this is a well-known triangle inequality and was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=42896 . Proof complete.
on the side 
of the triangle 
to the lines 
and 
are 
and 
, where 
are the altitudes of the triangle , for some 
.we search for which 
that give us 
, so the equal distances (that are the same of ) are 
.
show that![\[\frac{a_1}{a_1+a_2}+\frac{a_2}{a_2+a_3}+\cdots +\frac{a_n}{a_n+a_1}\ge\frac{a_1+a_2+\cdots +a_n}{a_1+a_2+\cdots +a_n-(n-2)\sqrt[n]{a_1a_2\cdots a_n}}>1.\]](http://latex.artofproblemsolving.com/c/e/2/ce286b2083f9fcbb7a976b511eef0a223a464a77.png)
is the radius of the incircle of ![\[ \frac{2}{r} = \frac{1}{r_a} + \frac{1}{r_b} + \frac{1}{r_c}. \]](http://latex.artofproblemsolving.com/c/d/e/cde2afd5a6e644c42d1a2aba859cbb4c8ad08123.png)
is the bisector of 
,
,
is the center
,![\[ r_a=AD\cdot\sin\frac{\hat A} 2=\frac{bc}{b+c}\sin\hat A=\frac{2sr}{b+c} \]](http://latex.artofproblemsolving.com/0/9/b/09bd4cd586515daf33f87be3938ab3c6b2a64e33.png)
.![\[ \frac 1{r_a}+\frac 1{r_b}+\frac 1{r_c}=\frac{b+c}{2sr}+\frac{c+a}{2sr}+\frac{a+b}{2sr}=\frac{4s}{2sr}=\frac 2 r \]](http://latex.artofproblemsolving.com/f/0/f/f0f8c80511f44c5cf70bf9b384644b3418153460.png)
denote the area ![$S=[ABC]$](http://latex.artofproblemsolving.com/b/3/a/b3ae3d445111e4dd28be75922309d3270079368c.png)
and the semiperimeter
.
of the point 
is circumscribed to the circle 
,
and
,
(the ray 
is the bisector of the angle 
)![$2pr=2S=[ABA'C]=p(ABA'C)\cdot \rho_a=(b+c)\rho_a\Longrightarrow \frac{1}{\rho_a}=\frac{b+c}{2pr}$](http://latex.artofproblemsolving.com/1/1/5/11553886155e869be06b6886dc4fb18753c3edcb.png)
a.s.o.
