.
such that 
.
be an increasing sequence of positive integer such that 
.
there exists 
such that 
be natural numbers such that![\[
\frac{a^n - 1}{(a - 1)^n + 1}
\]](http://latex.artofproblemsolving.com/6/2/7/627f1d5f55e7594f2e6dcf5d541e796b7a5bfae2.png)
is a natural number.
is odd.
be a prime divisor of ![\[
a^{(q - 1)/2} \equiv 1 \pmod{q}.
\]](http://latex.artofproblemsolving.com/b/2/1/b2135d8c5526e577722718e6d02bcc52559b0555.png)
3. Prove that if a is prime and 
,![\[
2^{(a - 1)/2} \equiv 1 \pmod{a}.
\]](http://latex.artofproblemsolving.com/2/d/c/2dc6cc1fe27cc72e72d63b13cf07939e19051245.png)
be a positive real number. Find all functions 
such that 
and![\[
f(af(x) - f(y) + y) \leq x + f(y) - y, \quad \forall x, y \in \mathbb{R}^+.
\]](http://latex.artofproblemsolving.com/e/2/9/e296b71ad8b8bd977e2bb2a3630a5bc020b8e03a.png)
Y by buzzychaoz, Adventure10, Mango247
Let the circumcenter of triangle 
be 
. 
is the projection of 
onto 
. The extension of 
intersects the circumcircle of 
at 
. The projections of onto 
are 
, and 
is the circumcentre of triangle 
. Define 
similarly.
Prove:
are concurrent
be 
. 
is the projection of 
onto 
. The extension of 
intersects the circumcircle of 
at 
. The projections of onto 
are 
, and 
is the circumcentre of triangle 
. Define 
similarly.Prove:
are concurrent
is the orthocenter. By simple angle chasing, we get that 
by a simple angle chase. So 
.
.
.
and 
.
.
which is of course going to cyclically cancel out when we use Trig-Ceva.
are similar and coefficient of similarity is 
.
.
;
be the intersections of 
and 
.
we see 
.
must be isosceles since 
by simple angle chasing, and since 
then 
is the midpoint of 
.
is midpoint of 
and so 
is a parallelogram. So 
and 
are heights of triangle 
and 
and so 
.
lies within 
and that 
have complex coordinates 
respectively. We easily find that the circumcenter of 
has coordinates 
so we have that 
satisfies 
Moreover because 
we have that 
Solving these two equations we easily find that 
Also, using the well-known formula for projections of points on to side lengths, we find that 
and 
and 
Now it is trivial to verify that 
so 
is a parallelogram.
since 
is clearly cyclic. Similarly, 
and therefore 
Now that we have all the angles in 
an easy angle chase yields 
But since 
are isogonal conjuagates with respect to 
are vertices of the pedal triangle of 
it is well-known that 
so we have that 
as desired.
,
.
is a parallelogram .
are symmetry WRT the midpoint of 
,
is parallel to 
( 
the circumcenter of 
is the midpoint of 
) 
.
concur at the orthocenter of 
.
,
be the reflection of 
.
.
and 
swap under isogonal conjugation about 
.
is a pair of isogonal conjugates. Let 
be the projections of 
.
are concyclic. Note that 
hence 
.
under homothety centered at 
and we're done. 
concur at the orthocenter of 
I have name orthocenter of 
as 
Kindly adjust 
be the 
a
.
and 
homothety sending 
with the same homothety 
circumcenter of 
circumcenter of 
be 
(Don't forget to see that 
is a parallelogram!) 
and 
are isogonal. Notice that 
so 
Also, 
by AA so 
by SAS. Thus, 
as 
Hence, 
and 
and Trig Ceva on triangle 
,
,
,
,
,
,
is the angle bisector of 
,
is the angle bisector of 
.
,
.
.
,
be the midpoint of 
passes through 
,
.
,
,
taking 
.
,
at 
to 
to 
.
,
,
,
.
.
,
and
.
,
,
.
,
as required.
,
,
.
,
,
and 
intersect at the orthocentre of 
,
.
The result follows by multiplying cyclically and trig Ceva.
Hence, 
Thus, 
Similarly, we can say that 
is a parallelogram
(i.e the intersection point of 
and 
be the circumcenter of 
.
and 
passes through the midpoint of 
.
