be a set of integers with the following properties:
.
and 
,
.
,
is composite, then all positive divisors of 
such that 
is an integer for every positive integer 
be the orthocenter of acute triangle 
,
be the foot of the altitude from 
to 
,
.
intersects line 
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 
and 
,
such[/center]
is directly similar to either 
.
is directly similar to 
if there exists a sequence of rotations, translations, and dilations sending 
to 
to 
to 
.
and 
is a point not on the line and that 
and 
Find the area of 
be a positive integer, and let 
be nonnegative integers such that 
Prove that![\[
\sum_{i=0}^n i\binom{a_i}{2}\le\frac{1}{2}\binom{a_0+a_1+\dots+a_n}{2}.
\]](http://latex.artofproblemsolving.com/c/4/9/c49d0ccfef69402d05dc44d721172498c605cf7e.png)
Note: 
for all nonnegative integers 
.
be a polynomial of degree 
such that the polynomial 
divides 
is zero. Prove that 
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
.
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 
as free variables. Let 
.
and 
we find,
Okay that wasn't too bad. Now similarly we can compute,
Note that 
,
and 
are just reflections over the real axis of 
,
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.
there is an involution swapping 
,
and 
.
,
is the reflection of ![\[a_1=\overline{a} , b_1=\overline{b} \text{ and } c_1 = \overline{c}\]](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 
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 
must be the same!
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.
