[/quote]
,
be the set of integers, and let 
be a function. Prove that there are infinitely many integers 
such that the function 
defined by 
is not bijective.
,
such that 
.
and 
respectively be the incentre, circumcentre and circumcircle of triangle 
.
are chosen on 
,
is the foot of the altitude from 
to 
.
are similarly defined, prove that lines 
and 
concurr on 
.
and 
be tangency points of the incircle of 
,
and 
respectively. Let the circle with centre 
passing through 
and 
.
,
and 
,
.
and 
be feet of perpendiculars from 
and 
respectively. Prove that 
,
and 
are concurrent.
that is not isosceles at 
,
be its incircle, which is tangent to 
at 
,
and 
intersect the line passing through 
at 
and 
,
and perpendicular to 
intersect 
and 
at 
and 
,
are collinear and that 
are concurrent.
positive integer numbers such that the least common multiple (LCM) of all these numbers is not a perfect square. Mykhailo consecutively hides one of these numbers and writes down the LCM of the remaining 
numbers that are not hidden. What is the maximum number of the 
satisfying: 
.
min
is a prime number and 
are positive integers. Prove that there exist infinitely 
are divisible by 
if 
if 
primes, and let 
be the sum of the divisors of ![\[\left|\sum_{d|n}\frac{\mu(d)\sigma(d)}{d}\right| \geq \frac{1}{n}, \]](http://latex.artofproblemsolving.com/8/e/b/8eb88de7cd7ea8e2853dbc80cae24bdc6b77f9be.png)
be a sequance of natural integers. Proof that, any number in the sequance can be written only as 
for any 
and not necessarily distinct primes 
for 
.
are same in some case.
![\[2^{xy}f(xy-1)+2^{x+y+1}f(x)f(y)=4xy-2,\forall x,y\in\mathbb{R}\]](http://latex.artofproblemsolving.com/d/1/d/d1de7196fd3ba60732c14fa2b156ec880b4829d0.png)
has no integer solutions
Y by Davi-8191, jhu08, megarnie, mathematicsy, mathlearner2357, rayfish, Adventure10, Mango247, and 6 other users
Let 
be a point in the plane of 
, and 
a line passing through . Let 
be the points where the reflections of lines 
with respect to intersect lines 
respectively. Prove that are collinear.
be a point in the plane of 
, and 
a line passing through . Let 
be the points where the reflections of lines 
with respect to intersect lines 
respectively. Prove that are collinear.This post has been edited 1 time. Last edited by tc1729, Apr 25, 2012, 9:56 PM
.
in the expression left it unchanged, and since I was running out of time, said that you could show the same for switching 
.
is the same, and they are collinear.
are cyclic. Darn.
are isogonal with respect t
.
are isogonal with respect t
.
are isogonal with respect t
.
.
.
and 
we find,
Okay that wasn't too bad. Now similarly we can compute,
Note that 
and 
,
Then it suffices to show that,
Now we expand by minors to find,
and we're done.
![\[
\begin{vmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{vmatrix} = a_{11} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33}\end{vmatrix}- a_{12} \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33}\end{vmatrix} + a_{13} \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32}\end{vmatrix}
\]](http://latex.artofproblemsolving.com/4/3/0/4303e1e80bfa2af32ac91a5488ea0584d1d33141.png)
matrices, by considering the alternate sum and difference of the smaller 
matrices. If you have EGMO, it's mentioned in the addendum on linear algebra.
by reverse reconstruction as the intersection of 
there is an involution swapping 
,
and 
.
,
![\[a_1=\overline{a} , b_1=\overline{b} \text{ and } c_1 = \overline{c}\]](http://latex.artofproblemsolving.com/e/6/d/e6d2fd41b1b34208b22eef98a66cdbb0d209cf39.png)
Now, note that since 
,
Similarly, we obtain that,![\[b' = \frac{\overline{b}(\overline{c}a-c\overline{a})}{\overline{b}(\overline{a}-\overline{c})+b(c-a)}\]](http://latex.artofproblemsolving.com/b/a/c/bac638de120d8e65baa8fa567f028cd0d413d2e3.png)
and![\[c' = \frac{\overline{c}(\overline{b}a-b\overline{a})}{\overline{c}(\overline{b}-\overline{a})+c(a-b)}\]](http://latex.artofproblemsolving.com/6/1/0/6102ed5282407b52ab9b31e68250a888b8bf333f.png)
Now, let 
.
We then let, 
, 
and 
.
,
and 
.
one can do a quick check and verify that 
,
,
,
and 
.![\[\overline{E}=\frac{R(aAQ-bBP)}{Q(aAR-cCP)}\]](http://latex.artofproblemsolving.com/c/7/c/c7c01765226e0b556ddfdde912b4918d26af110d.png)
we wish to have 
.
So, it suffices to show that 
which is quite clear since,
as desired. So, it is clear that 
.
thus 
in quadrilateral 
thus 
Similarly 
tangent as well and now it is easy to check these three must be the same line.
.
,
are symmetric wrt 
.
,
,
,
and 
and 
are similar
and thus 
are cyclic
is the simson line of 
We want to prove that 
.
by reflection, we have that 
by reflection, we have that 
.
.
,
are isogonal conjugates in 
and quadrilateral 
,
are equal iff. 
are isogonal conjugates. This can also be approached with the theory of perfect isogonal six-point sets; 
is a perfect isogonal six-point set).
,
.
make up a complete quadrilateral with intersections in 
move P on 
Then our 
desired points each have 
so we need to check 
cases so that they are collinear.
t
in the original problem we have the lines through 
that are parralel to 
so we need to have 4 cases again. These cases are where 
for 
goes now theire reflection about a point in infinity just lies on the line at infinity , doing a back transformation we het the concurrency changes to collinearity and we are done
,
,
.
is defined to be 
,
with respect to 
and we are done.
