AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be an acute triangle and let 
be the midpoint of 
. A circle 
passing through 
and meets the sides 
and 
at points 
and 
respectively. Let 
be the point such that 
is a parallelogram. Suppose that lies on the circumcircle of . Determine all possible values of 
.
be a triangle with circumcircle 
. Let 
and 
respectively denote the midpoints of the arcs and that do not contain the third vertex. Let 
denote the midpoint of arc 
(the arc including 
). Let 
be the incenter of . Let 
be the circle that is tangent to and internally tangent to at , and let 
be the circle that is tangent to and internally tangent to at . Show that the line 
, and the lines through the intersections of and , meet on .
be a circle with centre , and 
a convex quadrilateral such that each of the segments 
and 
is tangent to . Let be the circumcircle of the triangle 
. The extension of 
beyond meets at 
, and the extension of 
beyond 
meets at 
. The extensions of 
and 
beyond 
meet at 
and , respectively. Prove that ![\[A D+D T+T X+X A=C D+D Y+Y Z+Z C.\]](http://latex.artofproblemsolving.com/8/2/6/8261276053071cc6cc06dda824a883426bf1d4ac.png)
such that for every 
is satisfied? Explain the answer.
be a random point on the smaller arc of the circumcircle of square 
, and let 
be the intersection point of segments and 
. The feet of the tangents from point to the circumcircle of the triangle 
are and , where 
is the center of the square. Prove that points , , and lie on a single circle.
and 
such that
be an integer, and let 
be positive real numbers such that 
. Prove that![\[(1 + a_2)^2 (1 + a_3)^3 \dotsm (1 + a_n)^n > n^n.\]](http://latex.artofproblemsolving.com/7/7/f/77fb719e975225b59c64f1f3aa102c98a6c25f50.png)
colors, there exists a monochromatic regular hexagon.
is the smallest number such that if 
, for each coloring of the set 
with two colors there exists a monochromatic arithmetic progression of length 
. Prove that
.
points in plane. Prove that the number of isosceles triangles having their vertices among these points is 
. Find a configuration of points in plane such that the number of equilateral triangles with vertices among these points is 
.
![\[ a_n=2^n+3^n+6^n -1,\ n\geq 1. \]](http://latex.artofproblemsolving.com/6/4/9/6499d8f992f3e72765cab1f3afe15df99d01b12a.png)
and it is clear, since for any prime 
we have 
is divisible by 
by Fermat Small Theorem (
, while 
). For 
we have 
and so it is divisible by 6.
and it is clear, since for any prime we have is divisible by by Fermat Small Theorem (, while ). For we have and so it is divisible by 6.
because it doesn't seem ok (I mean 
would mean that 5 divides 15/2 from the definition of congruence).. So I wrote it like this 
and then we get 
which is obviously divisible by p (for p>=5), and beacuse 
and 
we see that there's no solution in prime numbers..
those residues coprime to are invertible, i.e. if 
and are coprime, then there exists 
such that 
modulo . So, if we write 
, we mean 
.
because it doesn't seem ok (I mean would mean that 5 divides 15/2 from the definition of congruence)..
! I think this is the easiest Number theory problem in IMO .
: 
is divisible by 
(It's proved easily by Fermat Little Theorem)
p>3
x=2^{p-2}, y=3^{p-2}, z=6^{p-2}
2x\equiv 3y\equiv 6z (\equiv 1) (\mod p)and
\gcd(p,2)=\gcd(p,3)=1
x\equiv 3z, y\equiv 2z (\mod p)
x+y+z\equiv 6z\equiv 1 (\mod p)or
p|a_{p-2}
.
, so the 4 solutions for the characteristic equality 
are 
by the explicit term, we get
and 
. is unique in both directions modulo any prime, 0 will appear again. Therefore 1 is the only solution.
is unique in both directions modulo any prime, 0 will appear again. Therefore 1 is the only solution.
!!!!!!!!! 
,
for all 
,
such that: 
and 
.
,
is a complex number so we are done
![\[2^n + 3^n + 6^n - 1 \equiv 0\pmod{p}\]](http://latex.artofproblemsolving.com/6/a/d/6ad29e27ec0af4da7cf507a4e6dd028d3b82747b.png)
Let 
.![\[\frac{1}{2} + \frac{1}{3} + \frac{1}{6} - 1 \equiv 0\pmod{p}\]](http://latex.artofproblemsolving.com/6/b/a/6ba086076d7eeff9592a9116904281db53aa0def.png)
so 
![\[2^{p-2}+3^{p-2}+6^{p-2}\equiv2^{-1}+3^{-1}+6^{-1}\equiv1\pmod{p}.\]](http://latex.artofproblemsolving.com/d/e/a/deabeef3910647848a5e1104e94d4999f44bcc38.png)
for all 
,![\[a_{p-2} \equiv \frac{1}{2}+\frac{1}{3}+\frac{1}{6}-1 \equiv 0\pmod{p}\]](http://latex.artofproblemsolving.com/5/2/3/523533c71a0b86d911bd502c4e4f364aeb4cb041.png)
for all 
.
divides 
and 
respectively. Thus, the only such 
contains all primes.
then we have 
which is true for any 
then we have 
which is true for all even 
.
.
to get 
.
,
,
.
is a power of 
which is not relatively prime to 
that divides 
,
so 
,
,
,
,
,
,
,
,
Note that 
which completes the problem.
which obviously works. It clearly suffices to show that 
.
,
modulo 
don't work. Now, suppose that 
Therefore 
![\[2^{p-2}+3^{p-2}+6^{p-2}-1\equiv \frac{1}{2}+\frac{1}{3}+\frac{1}{6}-1 \equiv 0 \pmod{p}\]](http://latex.artofproblemsolving.com/6/7/2/672f7522379f8e28a4f5f8d901177aa72acc5282.png)
Thus, 
for all primes 
satisfy this condition, but the cases where 
are messier. In particular any positive integer of the form 
does not satisfy this condition.
and 
By FLT, we can say 
where 
Because 
we can say 
so we are done.