[/quote]
,
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 
to the side 
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 
,
and 
.
is prime, 
with 
and 
be an acute triangle with orthocenter 
and circumcenter 
.
and 
be points on the line segments 
and 
respectively such that 
is a parallelogram. Prove that 
.
that satisfy 
for all 
and 
satisfies the following recursion formulas: 
,
.
and 
be points on segment 
denote the circle passing through 
denote the circle passing through 
)
.
Y by Adventure10, Mango247
A circle 
with radius 
is given. A collection 
of triangles is called good, if the following conditions hold:
such that, for each positive integer 
, there exists a good collection of triangles, each of perimeter greater than .
with radius 
is given. A collection 
of triangles is called good, if the following conditions hold:
- each triangle from is inscribed in ;
- no two triangles from have a common interior point.
such that, for each positive integer 
, there exists a good collection of triangles, each of perimeter greater than .
.
so, there must exist 
such that 
.
be the height from 
.
and 
.
.
.
.
which can be less enough to make 
.
.
.
,
.
and 
.
So, 
satisfy the condition.
satisfy the problem too and we are done.
.
(and hence 
)
be a large integer. Construct a 
-
.
,
,
.![[asy] size(6cm); draw(unitcircle); pair A = dir(170); pair B = dir(165); pair C = dir(0); dot("$A_{k-1}$", A, dir(180)); dot("$A_k$", B, dir(150)); dot("$A_M$", C, dir(C)); filldraw(A--B--C--cycle, rgb(.8,.8,.8), black); dot("$A_0$", dir(180), dir(225)); [/asy]](http://latex.artofproblemsolving.com/a/6/5/a65af27638e25c59ad9bee4c561e2b3dc85b87cb.png)
t
)
where we have used Taylor's theorem in the form 
and 
.
,
for all 
.
.
,
,
for some 
.
.
,
,
be the side lengths and 
the semiperimeter. We have 
and 
,
,
by Heron's formula. 
such that 
. Length of each side of a triangle in 
,so each side must have length greater than or equal to 
Hence number of triangles will be bounded . Contradiction.
(Just take one coinciding with two sides). Let positive integer 
. Consider a rectangle inscribed in the unit circle with one side 
.
.
and 
due to the choice of 
due to the choice of 
![$t\in (0;4].$](http://latex.artofproblemsolving.com/a/b/a/aba911eb07f0e688bc46232e4eac28457113cf0e.png)
Let 
denotes the perimeter of 
Let 
be a cemicircle of 
Pick 
by the triangle inequality 
Next, for 
because of 
we can choose take 
on arc 
of 
Thus the collection of 
for 
is good.
triangles, each of perimeter greater than 
Pick an arbitrary triangle 
the shortest side has length at least 
Using formula connecting area of triangle and it's circumradius we obtain 
Since triangles of collection are disjoint, it follows that 
and hence the conclusion.
.
so 
so if ![\[A=\frac{abc}{4R}=\frac{abc}{4}\ge \frac{(t-4)^3}{4}\]](http://latex.artofproblemsolving.com/9/3/2/932ab6d252aec8d831f3638ba00dba06c38f0b41.png)
which means that enough triangles will cover the circle so there can't be infinite. If 
then we first start by noting that any right triangle works, so let 
be any point on the circle and let it be close to 
and 
.
be the perimeter minus 
and we get 
and 
so 
so we're done.
.
,
we get that the triangles will have an area that is bounded below therefore if we pick 
.
be a diameter. Pick 
.
between 
which obviously exists. Now realize that 
and 
so summing gives 
.
![$\boxed{(0,4]}$](http://latex.artofproblemsolving.com/8/7/7/8773f43378aa982d2f85cc80179f002e58d50291.png)
,
is a set of all points on the remaining semicircle
be such a point 
on arc 
,
(if there are multiple such points, choose the closest to 
.
.
.
,
.
,
triangles can be in the collection
,
,
.
and area 
.
is impossible.
.
.
.
.
twice and ![\[ 2 \bigl(\sin \theta + 2 \cos \tfrac\theta2\bigr) = 4 + 2 \theta + O(\theta^2) > 4 + \varepsilon. \]](http://latex.artofproblemsolving.com/4/9/3/493600d8ce116ccedfc9fd17156c7df04a37804c.png)
This means 
,
.
.
triangles, which is finite, contradiction. 
is possible.
,
,
.
,![\[ 2 (\sin \varepsilon + \cos (k-1)\varepsilon + \cos k\varepsilon) = 4 + 2 \varepsilon - O_k(\varepsilon^2) = 4 + 2 \varepsilon - O_n(\varepsilon^2), \]](http://latex.artofproblemsolving.com/e/b/0/eb0bebf6c7ee41861f320d69b7edbc3628ea305f.png)
so picking a sufficiently small 
points 
such that the arc 
for each 
,
for each 
by a suitable choice of 
,
.![\[\sin \varepsilon + \cos\left(\frac{n\varepsilon}2\right) + \cos\left(\frac{(n-2)\varepsilon}2\right) > 2.\]](http://latex.artofproblemsolving.com/8/7/7/877fc0b08f81d2c11031c978cccefae9915ea49e.png)
Let 
and observe 
.![\[f'(x) = \cos x - \frac nx \sin \left(\frac{nx}2\right) - \frac{n-2}2\sin\left(\frac{nx}2\right) \geq 0\]](http://latex.artofproblemsolving.com/3/a/6/3a6c590aad6486f27ed6b275a78d1d71911e27e8.png)
on some interval ![$[0, \varepsilon_0]$](http://latex.artofproblemsolving.com/2/6/b/26b4270f20fc5f566c305bfab96cfc2ee9927704.png)
.
,
,
.
.![\[[\Delta] = \frac 12 S_1 S_2 \sin \theta > \frac 12\varepsilon^2 \cdot \frac{\varepsilon}{2R} = \frac 14 \varepsilon^3.\]](http://latex.artofproblemsolving.com/9/8/6/98616002cb9a836a54d533b5caca96e5538b853d.png)
Picking 
yields a contradiction. 
has perimeter less than 
,
