AoPS Academy
+2 w
[/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 the orthocenter of acute triangle 
, let 
be the foot of the altitude from 
to 
, and let 
be the reflection of across 
. Suppose that the circumcircle of triangle 
intersects line at two distinct points 
and 
. Prove that is the midpoint of 
.
+2 w
+1 w
lamps 
arranged in a circle in that order. At any given time, each lamp is either on or off. Every second, each lamp undergoes a change according to the following rule:
, if 
have the same state in the previous second, then is off right now. (Indices taken mod .) is on right now.
which is on. Prove that for infinitely many integers all the lamps will be off eventually, after a finite amount of time.
and 
, let us consider the equation
[list=a]
and 
, find the least positive integer 
satisfying this equation. and , there exist infinitely many positive integers satisfying this equation.
denotes the greatest common divisor of positive integers 
and .)
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 if and only if the following condition holds: distinct from 
and 
, there exists 
such[/center]
is directly similar to either 
or 
.[/center]
is directly similar to 
if there exists a sequence of rotations, translations, and dilations sending 
to , 
to , and 
to 
.
for which there exist real numbers 
satisfying 
, 
and
for 
.
such that 
holds for all positive rational numbers 
.
and 
. Prove that
Equality holds when 
Equality holds when 
and are prime-related if 
or 
for some prime 
. Find all positive integers , such that has at least three divisors, and all the divisors can be arranged without repetition in a circle so that any two adjacent divisors are prime-related.
and are included as divisors.
is chosen inside a convex quadrilateral 
. Could it happen that
, and 
, a triangle can be formed. Could it happen that from segments of lengths 
a right-angled triangle can be formed? for which it is possible to arrange all divisors of that are greater than 1 in a circle so that no two adjacent divisors are relatively prime.
. Prove that for every nonnegative integer , there is an odd number and a parabolic triangle with vertices at three distinct points with integer coordinates with area 
.
. Prove that for every nonnegative integer , there is an odd number and a parabolic triangle with vertices at three distinct points with integer coordinates with area .
and 
. This reduces the area to the following.![$A = \frac{1}{2}(2^{2n+1})(2^{2n+1}+1)^{4} = [2^{n}(2^{2n+1}+1)^{2}]^{2}$](http://latex.artofproblemsolving.com/7/d/2/7d2fcd982bd4dd4ed41b4bd92c44bc7ca87d5f49.png)
.
, and let 
, and 
.
. (After bashing)
set the vertices at 
,
,
.
set the vertices at 
,
,
.
is odd for all integers 
,
,
.
,
,
and 
.
instead of 
because it looks nicer 
which as area 1. For 
,
.
from now on.
By Shoelace, the area is 
When 
,
WLOG 
,
all perfect squares, so we need a Pythagorean triple. Amusingly, we can do the following:
with 
Then, assign 
as 
,
,
so the area would be 
so 
Therefore, we just need to show that for any positive integer 
of 
as one of the sides, and making the other sides 
which will both be odd for 
then
then 
then 
then 
?
,
is 
.
for each 
.
.
,
.
,
but multiply both 
.
,
,
and 
.
.
,
,
,
and 
we get 
.
in the first case, 
in the second case, and 
in the last case, we are done. (Whether we are given 
,
or 
.
,
,
with 
,
can be![\[(4^n, 4^{2n-1}+1, 4^{2n-1}-1).\]](http://latex.artofproblemsolving.com/2/e/e/2eeb37232b12e3b362324c6bbff9cf6a4e0f335d.png)
[for 
,
as the vertices]
![\[a(b^2-a^2)=4^n\cdot m^2\]](http://latex.artofproblemsolving.com/2/5/0/2507fc28e3413202d65a73cfbb1f67ed5cb82a07.png)
by Shoelace (actually this isn't necessary at all lol just use normal base x height). Let 
for “convenience”. So 
,
.
.
and 
,
,
.
holds, so we are done. 
has area 
.
and 
have areas 
and 
,
is a complete residue class modulo 3, hence, for given 
is a perfect square and 
,
.
Let the difference 
be equivalent to 
and let 
.
.
and 
for some odd integer 
.
.
,
.
a solution to 
It is well known that more solutions 
can be generated by raising 
to the power of a nonnegative integer, including the 
t
.
to the power of a power of 2. When we have a solution 
,
gives us the solution pair 
Note that this increases the 2-adic valuation of 
by one, and as 
,
.
,
,
,
.![$[XYZ]= \lvert 2a^3- 2ab^2\rvert$](http://latex.artofproblemsolving.com/d/c/4/dc411657bfd0dfcd22728b8615a4629afeb8342c.png)
![$[XYZ]=2ab^2-2a^3$](http://latex.artofproblemsolving.com/d/6/b/d6b1daa3e571c7abea6b4af8177629f6c76b58d5.png)
Set 
and calculating, we get 
.
.
,
,
.
,
and the parabolic triangle with vertices 
has area
,
and the parabolic triangle with vertices 
has area
,
and the parabolic triangle with vertices 
has area
![\[Area = \dfrac{1}{2} \left|(ab^2 + bc^2 + ca^2) - (ba^2 + cb^2 + ac^2) \right|\]](http://latex.artofproblemsolving.com/0/5/c/05c774fe2270ec3e4ed8da0ad522f9193590213f.png)
which is equal to :- 
and 
= 
,
and 
= 
and so on is given by the determinant![\[\frac 12 \begin{vmatrix} x_1 & x_1^2 & 1 \\ x_2 & x_2^2 & 1 \\ x_3 & x_3^2 & 1 \end{vmatrix} = \frac 12(x_1-x_2)(x_2-x_3)(x_3-x_1).\]](http://latex.artofproblemsolving.com/d/e/f/def1dd5fd136c90809b6988b8f54b25a2e5be059.png)
Now take 
and 
,![\[\frac 12 (x_1-x_2)(x_2-x_3)(x_3-x_1) = m^2\left(\frac{m^2-2}2\right)^2 \left(\frac{m^2+2}2\right)^2.\]](http://latex.artofproblemsolving.com/a/1/9/a194504dcf92097bd934b761986d6893303b602e.png)
By varying 
,
.
![\[\left\{ \begin{array}{ll}
P=(2^{2n}(2^{2n+1}+1), 2^{4n}(2^{2n+1}+1)^2) \\
Q=(-2^{2n}(2^{2n+1}+1), 2^{4n}(2^{2n+1}+1)^2) \\
R=((2^{2n}+1)(2^{2n+1}+1), (2^{2n}+1)^2(2^{2n+1}+1)^2)
\end{array}\right. \]](http://latex.artofproblemsolving.com/1/6/7/16776fe8fee02bb87481fec488c2701d5ee6a57b.png)
to get ![$[PQR]=(2^n(2^{2n+1}+1))^2$](http://latex.artofproblemsolving.com/b/a/f/baf385b8652ca50fb4f8856209a16132dd0c982f.png)
,