and 
.
lie on segment 
lie on segment 
,
and 
.
be the circumcenter of triangle 
,
the orthocenter of triangle 
the point of intersection of the lines 
and 
.
consecutive even numbers, such that the square of their product divides the sum of their squares.
be a positive integer. Find all real solutions 
to the system:![\[a_1^2 + a_1 - 1 = a_2\]](http://latex.artofproblemsolving.com/a/4/e/a4e26b83c929b82800576bac11e12de0c3aa1cf6.png)
![\[ a_2^2 + a_2 - 1 = a_3\]](http://latex.artofproblemsolving.com/9/9/5/995e8eec93828b82901b0639463f907f5ff95d08.png)
![\[\hspace*{3.3em} \vdots \]](http://latex.artofproblemsolving.com/e/e/1/ee15e1fefe2ad550d3c9b1c51207eaf66fbfa41f.png)
![\[a_{n}^2 + a_n - 1 = a_1\]](http://latex.artofproblemsolving.com/7/c/7/7c727d1f2044c1bf7c80981e1074bb3cde807b8f.png)
and 
are the feet of the altitudes through 
and 
respectively. The incenters of the triangles 
and 
are 
and 
respectively; the circumcenters of the triangles 
and 
are 
and 
respectively. Prove that 
and 
are parallel.
be a function differentiable at 0 with 
.
,
,
,
.
and its absolute and relative error, known that its absolute error is equal or lower than 
If G has a normal subgroup of order 
then prove that 
is abelian without using Sylow Theorems.
![\[
\int_0^1 \int_{x^2}^x (x^2 + y^2)^{-1/2} \, dy \, dx
\]](http://latex.artofproblemsolving.com/e/a/9/ea9f3a093b1d4c0fa39e5b6d42afa2f71dace1e0.png)
a prime number, with 
and 
,
is a group?
which verify relation![\[
\left( f'(x) \right)^2 + f''(x) \leq 0, \forall x \in \mathbb{R}.
\]](http://latex.artofproblemsolving.com/3/a/2/3a235e8be4c61c32f376999e61d64973db21dc75.png)
is a finite set, let 
of subsets of 
if 
for each 
,
for each 
.
are there? Prove your answer.
be a continuous function such that 
for all 
.
for 
.
Y by Adventure10, Mango247
Let 
be an acute triangle and 
its circumcircle. Let 
be the line tangent to at 
. Let 
and 
be the intersections of the circumference with center 
and radius 
with lines and 
, respectively. Prove the orthocenter of lies on line 
.
be an acute triangle and 
its circumcircle. Let 
be the line tangent to at 
. Let 
and 
be the intersections of the circumference with center 
and radius 
with lines and 
, respectively. Prove the orthocenter of lies on line 
.
be the altitude of triangle 
.
.
.
or 
is midpoint of 
.
.
is isosceles triangle with 
.![\[\angle H'EA= \frac{ \angle ABD}{2}= 90^{\circ}- \angle DAB= \angle BAO.\]](http://latex.artofproblemsolving.com/5/5/d/55d2d8b3e9836bd465aa61467b571a5f33602594.png)
.
.
or 
.
and 
.
.
are between 
.
.
.
is the perpendicular bisector of ![\[\angle AEH=\angle HAC=90^\circ-\angle C=90^\circ-\angle DAB=\frac{\angle DBA}{2}=\angle AED\]](http://latex.artofproblemsolving.com/1/a/e/1aeaad5754b366b5c6ccc2fd97bdc06bad18ede3.png)
Since 
,
and 
lie on the same side of 
,
are collinear.
.
.
is cyclic 
.
.
which means that 
are collinear as desired.
,
,
are concyclic. I claim that 
and then 
.
and 
and 
.
,
and hence 
.
,
and it follows that 
be projections of 
and let 
be midpoint of 
.
is cyclic.
and so 
are collinear and so since 
respectively, 
is cyclic.
.
being cyclic, we also know that 
.
are collinear
be the unit circle, and 
and 
the reflections of 
.
and 
are parallelograms. By the reflection formula, we have 
and 
.
and 
.
.![\[q=\frac{h-e}{h-d}=\frac{a+b+c-a-b+a^2/b}{a+b+c-b-c+ac/b}=\frac{ c+a^2/b}{a+ac/b}=\frac{bc+a^2}{ab+ac}=\frac{1/a^2+1/bc}{1/ac+1/ab}=\overline q,\]](http://latex.artofproblemsolving.com/f/0/4/f04fd8491fc28f6c4593ce72d3580b62dffb8e34.png)
so 
are collinear as desired. 
etc.
Thus, if 
is the foot from 
Since 
we have 
hence done.
are concyclic, since 
. It implies 
be 
such that they lie on the unit circle, where 
the circumcenter of 
be 
We have that 
and 
Hence, 
To compute 
we can drop an altitude from 
Thus, we get that 
We can now compute the orthocenter by shifting 
We get that the orthocenter of 
is 
and hence the orthocenter of 
Clearly, 
We see that
and hence, 
are collinear.
and let 
denote the foot of the perpendicular from 
Also, let 
denote the foot of the perpendicular from 
Note that 
and 
and 
respectively. Firstly, by the complex foot formula, we have 
from which it follows that the complex number that represents 
Since the antipode of 
we can view 
as the foot of the perpendicular from 
to the line connecting 
and 
Hence, applying the complex foot formula again, we get 
Therefore, since 
we get
from which it follows that the complex number representing 
Now, since it is well-known that the orthocenter is 
we wish to show that 
and 
are collinear. Note that
Then, we have
concluding the proof. 
