[/quote]
,
by 
different methods. The same inequality can be used twice but in different ways, dm me if you have any question about the rules.
,
be the 
-
and 
are placed on line 
such that 
is between 
,
and 
,
be the 
,
be the reflection of the orthocenter of triangle 
,
denote the number of its positive divisors (including 1 and itself). Determine all positive integers 
for which there exists a positive integer 
.
let 
be the foot of the altitude from 
and 
,
,
be the incenter, 
and 
the other intersection points of the circle 
with the lines 
and 
,
.
such that![\[(a^2-b^2)(b^2-c^2) = abc.\]](http://latex.artofproblemsolving.com/7/1/8/7188261c40a45f5e15533100432d36abbf515efd.png)
.
that 
is a prime.
is divided into rectangles with two cells. Two rectangles are called adjacent if they share a segment of length 1 or 2. In each rectangle the amount of adjacent with it rectangles is written.
is an integer, then it is a perfect square.[/quote]
.
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
,
,
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?
interesting if 
of 
and 
is also a divisor of 
Y by megarnie, Rounak_iitr, aidan0626, lpieleanu
Alice the architect and Bob the builder play a game. First, Alice chooses two points 
and 
in the plane and a subset 
of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each other, and no three cities he places may be collinear. Finally, roads are constructed between the cities as follows: for each pair 
of cities, they are connected with a road along the line segment 
if and only if the following condition holds:
Note:
is directly similar to 
if there exists a sequence of rotations, translations, and dilations sending 
to 
, 
to 
, and 
to 
.
and 
in the plane and a subset 
of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each other, and no three cities he places may be collinear. Finally, roads are constructed between the cities as follows: for each pair 
of cities, they are connected with a road along the line segment 
if and only if the following condition holds:
For every city 
distinct from 
and 
, there exists 
such
distinct from 
and 
, there exists 
such
that 
is directly similar to either 
or 
.
Alice wins the game if (i) the resulting roads allow for travel between any pair of cities via a finite sequence of roads and (ii) no two roads cross. Otherwise, Bob wins. Determine, with proof, which player has a winning strategy.
is directly similar to either 
or 
.
Note:
is directly similar to 
if there exists a sequence of rotations, translations, and dilations sending 
to 
, 
to 
, and 
to 
.
to be the exterior of the circle with diameter 
.
.
are connected by a path. The base case 
is vacuously true. Now given two cities 
with distance 
,
,
we have 
,
.
,
if and only if the disk with diameter 
contains no other points of 
and 
be two points in different connected components. Since 
is not an edge, there exists a third point 
inside the disk with diameter 
.
to find a new point 
,![[asy]
size(6cm); pair p = (-1,0); pair q = (1,0); pair r = (0.4,0.7); draw(unitcircle, blue); draw(circle(midpoint(p--r), abs(p-r)/2), red); pair s = (-0.3,0.6); draw(p--q, dashed+blue); draw(p--r, dashed+red); draw(s--r, dashed); dot("$p$", p, dir(p)); dot("$q$", q, dir(q)); dot("$r$", r, dir(r)); dot("$s$", s, dir(s));
picture X; real eps = 0.05; draw(X, (-eps,-eps)--(eps,eps), black+1.4); draw(X, (eps,-eps)--(-eps,eps), black+1.4); add(X); add(shift(midpoint(p--r))*X); add(shift(midpoint(s--r))*X); label("$\delta_1$", (0,0), 2*dir(45), deepgreen); label("$\delta_2$", midpoint(p--r), 2*dir(-10), deepgreen); label("$\delta_3$", midpoint(s--r), 2*dir(100), deepgreen);[/asy]](http://latex.artofproblemsolving.com/8/5/e/85e4167367d3fd13fe8d676d5c49d49e7565bfe6.png)
,
,
,![\[ \delta_i^2 \le \delta_{i-1}^2 - 1 \]](http://latex.artofproblemsolving.com/7/6/c/76c8152140356ed4aa66e64bfcc3329b50da1d1a.png)
and this eventually generates a contradiction for large 
,
.
and 
meet, meaning 
is a convex quadrilateral. WLOG assume 
(each quadrilateral has an angle at least 
)
contains 
,
iff there is no 
such that 
.![\[ \delta_1 > \delta_2 > \dots \]](http://latex.artofproblemsolving.com/5/8/3/583027b67431f834ac74af0822dc76ecc7f10503.png)
instead, and since there are finitely many distances one arrives at a contradiction.
and 
for all 
,
A cartoon of the graph is shown below. ![[asy]size(8cm);
dotfactor *= 1.5; pair A1 = (0,0); pair A2 = (2,0); pair A3 = (4,0); pair A4 = (6,0); pair B1 = (1,3**0.5); pair B2 = (3,3**0.5); pair B3 = (5,3**0.5); pair B4 = (7,3**0.5); dot("$A_1$", A1, dir(-90), blue); dot("$A_2$", A2, dir(-90), blue); dot("$A_3$", A3, dir(-90), blue); dot("$B_1$", B1, dir(90), blue); dot("$B_2$", B2, dir(90), blue); dot("$B_3$", B3, dir(90), blue); label("$\dots$", A4, blue); label("$\dots$", B4, blue); draw("$2.08$", A1--A2, dir(-90), blue, Margins); draw("$2.04$", A2--A3, dir(-90), blue, Margins); draw(A3--A4, blue, Margins); draw("$2.06$", B1--B2, dir(90), blue, Margins); draw("$2.02$", B2--B3, dir(90), blue, Margins); draw(B3--B4, blue, Margins); draw(rotate(60)*"$2.09$", B1--A1, dotted); draw(rotate(-60)*"$2.07$", B1--A2, dotted); draw(rotate(60)*"$2.05$", B2--A2, dotted); draw(rotate(-60)*"$2.03$", B2--A3, dotted); draw(rotate(60)*"$2.01$", B3--A3, dotted); [/asy]](http://latex.artofproblemsolving.com/6/6/3/6630cf97c32731f13f58153d57fea98288452ff7.png)
In that case, 
and 
will be disconnected from each other: none of the edges 
or 
are formed. In this case the relative neighbor graph consists of the edges 
and 
.
![\[ AB \text{ connected} \iff \forall C \in \mathcal T, \angle ACB< 90^\circ.\]](http://latex.artofproblemsolving.com/9/7/f/97fbd4b36a8078b3c8675d188d9f34d43a88ef4e.png)
Here 
is Bob's cities.
and 
cross. If they do, then quadrilateral 
is clearly convex. So one of its angles is at least 90 degrees, but that means the diagonal it subtends cannot be connected.
are connected. Indeed we start with the base case 
.
and isn't connected by a road, then there exists some 
with 
,![\[ PR^2 + QR^2 \le PQ^2 < 2. \]](http://latex.artofproblemsolving.com/4/d/a/4da3b80d0d0655b3d49c844c88c30b1054bc311a.png)
But 
and 
,
are connected. Then if 
isn't connected directly, there must exist a point ![\[ PR^2 + QR^2 \le PQ^2 < n, \]](http://latex.artofproblemsolving.com/e/6/5/e65f16b2f4d31235ef7785f9c49d58e8d27ca280.png)
which as 
and 
are greater than 
and 
.
which is connected to 
.
.
I think.
,
intersect, then 
is convex but 
so its angles cannot add up to 
.
,
.
but each 
so done.
.
be the complement of 
.
such that 
is a cyclic harmonic, at most one lies in 
,
that is, interpolate between the 
decrease enough to obtain a contradiction (using the notation of Evan's solution).
.
that meet, then 
is a convex quadrilateral but internal sum of angles is strictly less than 
,
not connected directly by a road then it means there exists a city 
which is a contradiction, now suppose it was true for 
,
like this:
wasn't connected by a road then then there exists a city 
and thus 
(similar for the 2nd) then the 
are connected thus so if 
.
.
,
,
,
are all less than 
which is a clear contradiction.
then any point in the circle is distance less than one to both 
is true, then consider 
.![\[AC^2=AB^2-BC^2+2AB\cdot BC\cos(\angle ACB)<AB^2-1=k-1\]](http://latex.artofproblemsolving.com/b/e/0/be054bd1e1e094c61b84053c3feddc54fa39d9a6.png)
so 
,
such that 
.
,
because 
iff 
other cities 
.![\[\angle ACB, \angle ADB, \angle CAD, \angle CBD < 90^{\circ}\]](http://latex.artofproblemsolving.com/c/2/a/c2ab97d750fcb98b7628c3f1d20b408a2c4035aa.png)
so cannot add to 
.
city 
.
.
,![\[AB^2 \ge AC^2 + BC^2\]](http://latex.artofproblemsolving.com/7/5/9/759f3b8f0be9289989cf8d86dc3033d8f77d57ec.png)
,
,
.
,
which is not allowed.
,![\[AC^2, BC^2 \le AB^2 - 1 \le k\]](http://latex.artofproblemsolving.com/4/d/4/4d418bb31bf4b01cea92c77e803d5d76b822c81a.png)
true).
so done. Alice wins.
