[/quote]
,
,
denote the number of its positive divisors (including 1 and itself). Determine all positive integers 
for which there exists a positive integer 
.
let 
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 
.
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 Davi-8191, Purple_Planet, Adventure10, Mango247, and 5 other users
Let 
be a circle and let 
be a line such that and have no common points. Further, let 
be a diameter of the circle ; assume that this diameter is perpendicular to the line , and the point 
is nearer to the line than the point 
. Let 
be an arbitrary point on the circle , different from the points and . Let 
be the point of intersection of the lines 
and . One of the two tangents from the point to the circle touches this circle at a point 
; hereby, we assume that the points and lie in the same halfplane with respect to the line . Denote by 
the point of intersection of the lines 
and . Let the line 
intersect the circle at a point 
, different from .
Prove that the reflection of the point in the line lies on the line 
.
be a circle and let 
be a line such that and have no common points. Further, let 
be a diameter of the circle ; assume that this diameter is perpendicular to the line , and the point 
is nearer to the line than the point 
. Let 
be an arbitrary point on the circle , different from the points and . Let 
be the point of intersection of the lines 
and . One of the two tangents from the point to the circle touches this circle at a point 
; hereby, we assume that the points and lie in the same halfplane with respect to the line . Denote by 
the point of intersection of the lines 
and . Let the line 
intersect the circle at a point 
, different from .Prove that the reflection of the point
in the line lies on the line 
.This post has been edited 7 times. Last edited by darij grinberg, Jun 1, 2006, 4:27 PM
.
(with directed segments!). Hence, 
,
hence 
.
and 
are similar hence 
,
.
.
,
intersect 
.
.
and 
,
transforms to circumcircle of 
since 
hence 
and obviously we obtain that 
and we are done.
. As far as I remember you have to apply Pascal at least twice. You're right, that the degenerate version plays an important role; furthermore you also have to consider a point at infinity once. Sorry for the lack of details, but I will try to reconstruct it this afternoon.
)
and 
.
are collinear. Thus 
.
are collinear. By the previous we get that 
is on 
.
are collinear. By the previous we get that 
is on 
,
,
,
intersects 
.
is tangent to circle 
are cocircle.
.
,
,
problem is proved. 
.
.
isosceles so 
,
.
intersect 
so 
, which implies 
be the intersection of 
and 
,
be the intersection of 
and 
.
,
,
,
are collinear, that is, we have that 
,
so that 
is directly similar to 
.
and 
,
maps 
to 
,
,
and 
are cyclic. But this means that 
lies on both 
,
.
and 
,
and 
.
.
,
.
across 
is tangent???
We claim that 
so 
Thus, 
so 
tangent to 
Also, 
so 
is cyclic. Then, 
which implies 
to get 
are collinear, so the intersection of 
lies on 
Then note that Pascal on 
gives that the intersection of 
as desired.
and 
.
we have that 
are collinear. Furthermore 
Hence 
.
.
,
are collinear, but since 
,
id est 
across 
be the intersection of 
and 
.
,
.
.
is isosceles. Because 
is a right triangle, 
.
is right and 
is right. Since 
is a right triangle, 
is isosceles with 
.
.
.
is cyclic. Therefore,
Therefore, 
to get 
.
finishes.
.
implies that 
form a line perpendicular to 
,
.
yields 
,
is tangent to 
harmonic. To finish, let 
,
which implies 
,
is the reflection of 
.
.
.
gives us
are collinear. Hence 
lies on 
.
and 
and 
and 
,
is just the line 
and since 
lies in 
and also note that 
so by La'Hire 
and all of those tell us that 
and by ortocenter config we get that 
is cyclic and has center 
,
cyclic so by miquel theorem in 
we get 
cyclic so by arcs we get 
so 
which means arcs 
are equal in 
and since 
are symetric w.r.t. 
lies on ![[asy]
size(10cm); draw(unitcircle); pair O = origin;
pair A = dir(180); pair B = dir(0); pair D = (1.6,0.7); pair C = 2*foot(O,A,D)-A; pair M = 1/conj(D); pair E = IP(M--(M+dir(M)*dir(-90)),unitcircle); pair F = extension(B,E,D,conj(D)); pair G = 2*foot(O,A,F)-A; pair G1= conj(G); pair X = extension(C,B,G,E); draw(F--X); draw(A--D--E--F--A--B); draw(G1--F, dotted); draw(G--X--C, lightred+dashed); dot(X);
dot("$A$", A, dir(180)); dot("$B$", B, dir(0)); dot("$D$", D, dir(0)); dot("$C$", C, 1.9*dir(85)); dot("$E$", E, dir(-90)*dir(M)); dot("$F$", F, dir(0)); dot("$G$", G, dir(100)); dot("$G'$", G1, dir(-90));
[/asy]](http://latex.artofproblemsolving.com/3/0/e/30e4a4e10bd948c7f64c3d4ccb51e1c01771697a.png)
forces 
to lie on 
at 
and 
therefore 
.
but 
,
.
is tangent to 
by Lemma 1.44 (Three Tangents) in EGMO, We get 
so we get 
therefore 
is cyclic.
tangent to circle we get 
.
and 
we get 
gives 
collinear, so 
.
,
,
.
be the intersection of 
;
be the intersection of 
;
then 
,
.
from Three Tangents Lemma, a fact I found that we will not use (though it does implies 
harmonic).
at 
;
is harmonic. Projecting,![\[ (AB;GU) \stackrel{E}{=} (SF;R, UE \cap d) \]](http://latex.artofproblemsolving.com/7/9/e/79ebba9fee0c18df8c5f27bd09696d59e9350709.png)
So it suffices to show that 
.
are on the unit circle, then 
.
must be real, and set it equal to its conjugate.
,
through the unit circle is 
.
and plug in 
and 
into the above equation.
for 
on the unit circle is 
.
.
into the above equation.
and 
for a positive real number 
greater than 
.
to avoid confusing it with point 
.
,
so![\[f = \frac{1 + e - ke}{1 - e}.\]](http://latex.artofproblemsolving.com/4/a/d/4adc6658da9d273a096cb6026491aa683e585678.png)
Thus,![\[g = \frac{f + 1}{\overline{f} + 1} = \frac{\frac{2 - ke}{1 - e}}{\frac{2e - k}{e - 1}} = \frac{ke - 2}{2e - k}.\]](http://latex.artofproblemsolving.com/c/f/4/cf49397003096ca24b25a0609d316ad3326f7215.png)
.
,
so![\[d = \frac{2e - e^{2}k}{1 - e^{2}}.\]](http://latex.artofproblemsolving.com/4/3/7/43712bbff49a8a7946cade4379b59204d07d5e3b.png)
![\[\frac{2ge}{g + e} = \frac{2e \cdot \frac{ke - 2}{2e - k}}{\frac{2e^{2} - 2}{2e - k}} = \frac{ke^{2} - 2e}{e^{2} - 1}\]](http://latex.artofproblemsolving.com/b/1/b/b1bb032d1ad24ada20eb771ee853cf9d2eb37504.png)
which is equal to the value of 
and 
,
less annoying. We have 
,![\[c = \frac{d + 1}{\overline{d} + 1} = \frac{2ge + g + e}{2 + g + e}.\]](http://latex.artofproblemsolving.com/a/f/a/afa8fd33423a67beb51e712c8bddb137aebbfd77.png)
,![\[f = \frac{-g(e + 1) - (e)(g - 1)}{-g - e} = \frac{2ge + g - e}{g + e}.\]](http://latex.artofproblemsolving.com/3/f/c/3fc82b17b11183f5229885e9627f7ed6e1e1d4b6.png)
are collinear. To do this, we want 
to be real. We first find that
We can also find![\[f - \overline{g} = \frac{2g^{2}e + g^{2} - eg - g - e}{g(g + e)}.\]](http://latex.artofproblemsolving.com/7/a/5/7a5c90fc7629461545fcd42b054bc02ce916c59c.png)
Thus,![\[\frac{c - f}{f - \overline{g}} = \frac{2g(e - g)(e + 1)}{(e + g + 2)(2g^{2}e + g^{2} - eg - g - e)}.\]](http://latex.artofproblemsolving.com/2/1/d/21deee438124dcab42018b7be75118915430c5ca.png)
We want to prove this is real, that is, that it's equal to its conjugate.![\[2g^{2}e^{2} + g^{2}e - e^{2}g - eg - e^{2} + 2g^{3}e + g^{3} - eg^{2} - g^{2} - eg + 4g^{2}e + 2g^{2} - 2eg - 2g - 2e.\]](http://latex.artofproblemsolving.com/d/7/3/d7371010831449d11c8a0b3e8c81b615c29a49d1.png)
Combining terms gives![\[2g^{2}e^{2} + 2g^{3}e + 4g^{2}e - e^{2}g + g^{3} - 4eg - e^{2} + g^{2} - 2e - 2g.\]](http://latex.artofproblemsolving.com/7/3/e/73e4153e930cb42e8321feacfbe64eec0e4a4297.png)
Its conjugate is equal to![\[\frac{2g + 2e + 4ge - g^{2} + e^{2} - 4g^{2}e - g^{3} + ge^{2} - 2g^{3}e - 2g^{2}e^{2}}{g^{3}e^{2}}.\]](http://latex.artofproblemsolving.com/c/0/3/c03b7c5001dae84baf8da6c8abef8cd1cfd4e9a2.png)
Conveniently, the numerator is the negative of the original! (Though we should expect this).
as desired.
possible (particularly, just looking at the expression for 
and 
are similar should inspire just finding the differences of points instead of doing determinant things.
.
so 
which is trivial due to symmethry of the new problem.
implies that 
and Pascals on 
implies that 
and the tangent to 
and we are done
