[/quote]
,
such that ![\[2f(m)f(n)-f(n-m)-1\]](http://latex.artofproblemsolving.com/0/f/7/0f77b5addbf8f5e603eb1b6923cdac795cd132e9.png)
is a perfect square for all integer 
,
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 
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 
Y by son7, itslumi, Mango247, idkk
Let 
be the set of all lines in the plane and let 
be a function that assigns to each line 
a point 
on 
. Suppose that for any point 
, and for any three lines 
passing through , the points 
, and lie on a circle.
Prove that there is a unique point
such that 
for any line passing through .
Australia
be the set of all lines in the plane and let 
be a function that assigns to each line 
a point 
on 
. Suppose that for any point 
, and for any three lines 
passing through , the points 
, and lie on a circle.Prove that there is a unique point
such that 
for any line passing through .Australia
This post has been edited 2 times. Last edited by elitza, May 22, 2023, 11:39 AM
-
.
collinear and 
,
,
,
can't be cyclic. 
.
such that 
.
,
work, so the number of pairs is at least 
.
,![\[n(n-1-k)\le nk,\]](http://latex.artofproblemsolving.com/5/4/2/542a64ab3c769d72d8f1775a666646f054501039.png)
or 
,
be a circle of nonzero radius such that 
for each 
(exists by the problem condition).
on 
-
through 
.
more lines through 
such that 
,
.
such that 
since 
,
has at most two intersections with 
.
such that 
.
-
.
,
,
.
,
,
,
,
,
concur at some 
such that 
,
,
(this exists since 
,
.
is also 
.
and 
.
far enough away from these 
points such that it is on none of 
,
circumcircles. We'll try to show that 
.
as 
,
.
and 
,
,
or 
,
or 
.
It is easy to manually verify that these Boolean conditions actually imply that three of 
lie on 
,
through 
,
,
.
,
cannot be both 
![\[f(\ell) = \text{the unique point } X \in \ell \text{ satisfying } \measuredangle(\overline{PX}, \ell) = \theta\]](http://latex.artofproblemsolving.com/2/6/0/26020dcd8b11b53916db668398ae422e167675a5.png)
for some fixed pair 
,
define 
such that 
and 
are disjoint.
,
such that![\[\measuredangle(\overline{PX}, \ell) = \measuredangle(\overline{PD}, \overline{BC}) = \measuredangle(\overline{PE}, \overline{CA}) = \measuredangle(\overline{PF}, \overline{AB}) = \theta.\]](http://latex.artofproblemsolving.com/c/f/b/cfbdd5571a54776d324a79b6d158be1641f8a279.png)
![[asy]
size(300);
pair A = dir(120); dot("$A$", A, dir(A));
pair B = dir(210); dot("$B$", B, dir(B));
pair C = dir(-30); dot("$C$", C, dir(C));
pair D = 3/4*B + 1/4*C; dot("$D$", D, dir(D));
pair E = 3/7*C + 4/7*A; dot("$E$", E, dir(E));
pair F = 2/5*A + 3/5*B; dot("$F$", F, dir(F));
pair P = intersectionpoints(circumcircle(A,E,F), circumcircle(B,D,F))[1]; dot("$P$", P, dir(P));
guide extendedLine(pair P1, pair P2) {
return 0.2 * (P1 - P2) + P1 -- 0.2 * (P2 - P1) + P2;
}
draw(extendedLine(A, B) ^^ extendedLine(B, C) ^^ extendedLine(C, A));
draw(circumcircle(A,E,F) ^^ circumcircle(B,D,F) ^^ circumcircle(C,D,E));
pair L1 = dir(30);
pair L2 = 1.2*dir(-60);
draw(extendedLine(L1, L2), dashed);
pair Z = L2; dot("$Z$", Z, dir(Z));
draw(Z--A^^Z--B^^Z--C, dotted);
label("$\ell$", L1, dir(L1));
pair BZ = intersectionpoints(circumcircle(B,D,F), B--Z)[1]; dot("$B_Z$", BZ, dir(BZ));
pair CZ = intersectionpoints(circumcircle(C,D,E), C--Z)[0]; dot("$C_Z$", CZ, dir(CZ));
[/asy]](http://latex.artofproblemsolving.com/3/2/4/32458a4590585e0dd1e6b74092d4c055072b932b.png)
along with cyclic variants, and ![\[Z, f(\ell), f(\overline{AZ}), f(\overline{BZ}), f(\overline{CZ})\]](http://latex.artofproblemsolving.com/4/4/a/44adf28a008744dac4898d380b1b948d870d4239.png)
lie on a common circle 
.![\[\lvert\{A, B, C\} \cap \{f(\overline{AZ}), f(\overline{BZ}), f(\overline{CZ})\}\rvert \leq 1;\]](http://latex.artofproblemsolving.com/8/f/0/8f04299737d1ee918b96f5d1e0226c1eab2b285e.png)
otherwise, 
and 
.
,
by a short angle chase.
or 
,
are pairwise distinct.
be the point of concurrence of 
.
are concyclic, and 
.
is the second intersection of 
and 
by considering 
and 
.
?
passes through 
,
,
.
a point 
Note that for every noncollinear points 
such that 
circles 
concur by Miquel theorem.
.
we may suppose 
therefore 
has common point 
Consider arbitrary point 
By the same reason we may suppose that 
thus 
satisfies the condition, so take arbitrary line 
passing through 
If 
we obtain 
In the other case take two different cases 
of 
We get 
satisfying the condition we conclude 
contradiction.
such that:
fix a circle in such that for all 
( 
) , 
is on that circle.
:
to be that 
circle and define 
to be a tangent to 
passing through 
,
be a point on 
and 
intersect in two points!
witch is impossible, because 
in two points 
)
so by the problems statement: 
witch is a contradiction to 
such that: 
:
call such points 
.
such that 
and 
are both 
gives a contradiction. So there exists a unique 
be a foot of a altitude from 
,
.
and define 
similarly. Then by miquel, the circles 
all concur at a point.
.
.
,
must be cyclic, impossible since 
,
passing through 
(say), that is not one of the sides, I claim 
unless 
.
.
with the circles.
and 
as well, then 
,
and 
.
is fixed, we must have 
also to be fixed, or that 
to be fixed. But this happens for finitely many 
.
.
.
must be concyclic, impossible as 
are collinear.
for any line passing through a vertex is the second intersection with the circles. In particular, this actually implies that the circle corresponding to any point not on one of the three sides of the three tangents, passes through 
.
not lying on one of the six lines mentioned above, and also not equal to 
.
.
for all lines 
also satisfies this property. But then 
equals both 
contradiction. So we must actually have that 
for any line 
,
,
onto 
.
.
Otherwise, pick 
are distinct. Let 
be the result of 
to 
,
,
,
.
by substituting 
and 
,
for the points on 
[if there are multiple pick any 
] .
and 
this forces 
. and such that 
is tangent to 
this implies 
and such that 
are not collinear and then 
and 
contradiction ! .
.
is infinite .
at 
is another line through 
say 
obviously 
.
and pick any point 
in 
but not 
[if didnt exist just 
for 
and 
for 
where 
,
using 
in the following algorithm,
[if exists] , observe for a suitable choice of 
, 
, 
will intersect 
at some point 
such that 
are not concyclic .
is basically 
, say the tangent to 
is 
now say 
lie on circle. and then 
now play with 
from that POV u easilty get contradiction.
is not recursive by basic NT and chasing down the angles between 
radius] and repeat similar process as what we did before.
, pick two lines 
from 
and 
from 
, such that if 
intersects at 
and 
and 
then 
never lies on 
and 
do not lie on a circle but now observe the line 
observe 
are concyclic and 
are as well but that implies 
