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 a positive integer and let 
be distinct integers in the set 
such that divides 
for 
. Prove that does not divide 
:![\[
2 \left( 5^x + 6^x - 3^x \right) = 7^x + 9^x.
\]](http://latex.artofproblemsolving.com/f/b/6/fb6d5ebb4f448d8eeb73cce3288fea3a56d7f528.png)
, 
be the orthocenter of it and 
be any point on the side 
. The points 
are on the segments 
, respectively, such that the points 
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
. Prove that the points 
and lie on the same line.
of non-negative real numbers such that the following three conditions are all satisfied:
;
;
of such that 
[/list]
is a permutation of 
, and they are both permutations of 
. Note that any list is a permutation of itself.
denote the sum of digits function. Does there exist 
such that![\[
n, s(n), \dots, s^{2024}(n)
\]](http://latex.artofproblemsolving.com/c/3/3/c33803a77912ed4dde5a5c0ae05934496f04707f.png)
are all distinct perfect squares?
having integer coefficients. On each move, Vasya pays him a ruble, and calls an integer 
of his choice, which has not yet been called by him. Petya has to reply with the number of distinct integer solutions of the equation 
. The game continues until Petya is forced to repeat an answer. What minimal amount of rubles must Vasya pay in order to win?
+1 w
that satisfies:![\[
\sum_{i=1}^{n} a_{\left\lfloor \frac{n}{i} \right\rfloor} = n^k
\]](http://latex.artofproblemsolving.com/b/b/e/bbe7fe81e366fe390be5d17193d1581cc40dbece.png)
be a positive integer. Prove that for all integers 
, we have:![\[
\frac{c^{a_n} - c^{a_{n-1}}}{n} \in \mathbb{Z}
\]](http://latex.artofproblemsolving.com/0/4/4/04464c1a350e3f9a929ab51534018498680068fc.png)
be real positive numbers such that: 
. Prove that: 
satisfying :![\[n^{2p}=m^2+n^2+p+1\]](http://latex.artofproblemsolving.com/7/4/e/74ee592cb8b3f3a3996ea34f40b78fde9d3672fc.png)
and are integers and 
is a prime number.
be a positive integer and let 
be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of around the circle such that the product of any two neighbors is of the form 
for some positive integer .
be integer numbers 
satisfy that:
. Prove that: are even
is a perfect square number can’t be any power 
of a positive integer number
be a triangle. Draw circles 
, 
, and 
such that is tangent to 
and 
, and and are defined similarly. Let 
be the insimilicenter of and . Define 
and 
similarly. Prove that 
, 
, and 
are concurrent.
be the exsimilicenter of 
. Note 
. Define 
similarly. The Monge D'Alembert theorem gives that 
, 
, 
, 
are collinear. But then 
and 
are perspective from line , so by Desargue's theorem they are perspective from a point and the result follows.
for exsimilicentre 
respectively. Define 
, 
, 
.
, we get these tuples of points are colinear: 
, 
, 
, 
.
are perspecive axially. By Desargue's, they're perspective centrally as desired.
be a triangle. Draw circles , , and such that is tangent to and , and and are defined similarly. Let be the insimilicenter of and . Define and similarly. Prove that , , and are concurrent.
and are perspective , where 
is the center of and similarly others.
.Define 
analogously. Then as 
are the in-similicenters of of 
and 
respectively, it follows that 
is the ex-similicenter of 
. Similarly
and 
are the ex-similicenters of 
and 
respectively 
. Hence by Desargue's theorem the lemma follows.
is the common tangent to 
. Consequently, 
and 
. Another application of Desargue's theorem on and finishes the problem 
be a triangle. Draw circles , , and such that is tangent to and , and and are defined similarly. Let be the insimilicenter of and . Define and similarly. Prove that , , and are concurrent.
as the Exsimilicenters of 
. So, by Theorem we get that 
. But lies on 
respectively 
. But By Monge D-Albert Circle Theorem we get that 
. So, 
are perspective pairs of triangle. So, By Desargues Theorem we get that 
are concurrent.
for 
, and let 
be the insimilicenter of and 
. Then, the lines 
, 
, 
are concurrent. be 
for each 
. Let 
. Note then that ![\[I_{ij} = \frac{w_i}{w_i+w_j} O_i + \frac{w_j}{w_i+w_j} O_j.\]](http://latex.artofproblemsolving.com/8/b/7/8b79c54037c813a5e649d994ca44c27fb6b3a570.png)
Note that ![\[\frac{\sum_{i=1}^4 w_iI_i}{\sum_{i=1}^4 w_i} = \frac{w_1+w_2}{w_1+w_2+w_3+w_4}I_{12}+\frac{w_3+w_4}{w_1+w_2+w_3+w_4}I_{34},\]](http://latex.artofproblemsolving.com/6/f/3/6f3852f7918537d90cacdbdef34b1889df666fcb.png)
so ![\[\frac{\sum_{i=1}^4 w_iI_i}{\sum_{i=1}^4 w_i}\in I_{12}I_{34}.\]](http://latex.artofproblemsolving.com/a/2/a/a2a1f5a13fa7bf22b03a3adad240a6e771462964.png)
The point on the left side is symmetric in , so we have the desired result. , , , and the incircle but with negative radius.
and 
.
such that:
be the exsimilicenter of 
and 
By Monge, 
and 
are collinear. Hence, 