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 real numbers. For 
, let![\[\{a_n^{(i)}\}_{n \geq 0}\]](http://latex.artofproblemsolving.com/4/c/7/4c74470e7f3812966ba17e6fca535f085dced237.png)
be an infinite real sequence satisfying![\[a_0^{(i)} = 0.\]](http://latex.artofproblemsolving.com/0/f/6/0f6a0e940e48c2bf62fc12579f2396fb86d7b896.png)
It is known that:![\[a_1^{(1)}, a_1^{(2)}, \cdots, a_1^{(2025)}\]](http://latex.artofproblemsolving.com/7/9/5/795b375a183f2c553294be354f67826951092b84.png)
are not all zero.
and any , the following holds:![\[p_i \cdot a_n^{(i+1)} = a_{n-1}^{(i)} + a_n^{(i)} + a_{n+1}^{(i)},\]](http://latex.artofproblemsolving.com/1/a/c/1ac323cfc66546d8172f307b020d955b086d99f8.png)
![\[\{a_n^{(2026)}\}\]](http://latex.artofproblemsolving.com/b/6/e/b6e825cc8d5a1a8b2a56cf20ecde79c4c51ba9a6.png)
satisfies![\[a_n^{(2026)} = a_n^{(1)}, \, n = 0, 1, 2, \cdots.\]](http://latex.artofproblemsolving.com/c/d/f/cdf1e2f03ae1c343f34a017157886e4b6fca2d21.png)
such that for infinitely many positive integers 
,![\[\max \left\{ |a_n^{(1)}|, |a_n^{(2)}|, \cdots, |a_n^{(2025)}|\right\} \geq r.\]](http://latex.artofproblemsolving.com/2/5/5/255ba07c42ee84b59da2d3f1959a95d81c1c6b99.png)
for which there exists an infinite subset 
of the positive integers such that: for any pairwise distinct positive integers 
, the sum 
and the product 
are both square-free.
be a finite set of real numbers, 
be a real number, and 
be 2025 non-zero real numbers. Define![\[A =
\left\{
(x_1, x_2, \cdots, x_{2025}) : x_1, x_2, \cdots, x_{2025} \in X \text{ and } \sum_{i=1}^{2025} \lambda_i x_i = d
\right\},\]](http://latex.artofproblemsolving.com/2/e/1/2e18b6d25bfaa3e3d52f7efe5e4cba7a9cb87db9.png)
![\[B =
\left\{
(x_1, x_2, \cdots, x_{2024}) : x_1, x_2, \cdots, x_{2024} \in X \text{ and } \sum_{i=1}^{2024} (-1)^i x_i = 0
\right\},\]](http://latex.artofproblemsolving.com/4/1/f/41f6b126cc3cf24757b37147906c767d49a0fdd9.png)
![\[C =
\left\{
(x_1, x_2, \cdots, x_{2026}) : x_1, x_2, \cdots, x_{2026} \in X \text{ and } \sum_{i=1}^{2026} (-1)^i x_i = 0
\right\}.\]](http://latex.artofproblemsolving.com/d/e/6/de6586b88004e3f5aa7774525b00ee975c072ff9.png)
Show that 
.
and 
satisfies the following recursion formulas: 
, 
and 
. Show that 
, for 
.
be a finite, nonempty set of points in the plane such that, for every point 
in , there exist points 
in (distinct from ) such that 
. What is the smallest possible number of points in ? and are such that, for every positive integer 
,![\[ a_n > 0,\qquad\ b_n>0,\qquad\ a_{n+1}=a_n+\dfrac{1}{b_n},\qquad\ b_{n+1} = b_n+\dfrac{1}{a_n}. \]](http://latex.artofproblemsolving.com/4/f/c/4fc9fd1cb31a25378268f20e019e639667e52910.png)
Prove that 
.
and centroid 
. 
is the reflection of over , similarily define 
and 
. over , it is the same as reflecting over 
over and over , 
over and over .?
be an isosceles triangle with 
, and let 
be a point inside side 
such that 
. Let 
and 
be two points inside sides 
and 
, respectively, such that 
. Let the perpendicular bisector of 
meet line segment 
at 
, and let the circumcircles of triangles and 
meet again at point 
, different from ., , are collinear. Prove that 
.
with 
with the property that![\[\varphi(a^m-1)|a^{\varphi(m)}-1 ?\]](http://latex.artofproblemsolving.com/e/0/7/e07ae15e0267f4725151c919b47aaac7b950351d.png)
such that![\[(a^2-b^2)(b^2-c^2) = abc.\]](http://latex.artofproblemsolving.com/7/1/8/7188261c40a45f5e15533100432d36abbf515efd.png)
. If this is still too hard assume in 
that 
is a prime.
![$P \in \mathbb R[x,y]$](http://latex.artofproblemsolving.com/8/a/0/8a0576299e976c51607959fb68c7c742b2a49c6a.png)
with ![$P \not\in \mathbb R[x] \cup \mathbb R[y]$](http://latex.artofproblemsolving.com/a/f/7/af71ba8c643d5bd8a2928fa6714ccd91713bd641.png)
and 
homeomorphismes of 
, 
is an homoemorphisme too.
, 
and 
. The interior bisector of the angle 
intersects the side at the point 
. Calculate the length of the line segment 
.
, 
, 
. If the altitude to the side whose length is divide the side into two line segments prove that their difference is 4. such that
, 
and 
. Prove that 
.
.
so let be 
.
.
such that 
. Find the minimum value of the following expression:![\[E(a,b)=3\sqrt{1+2a^2}+2\sqrt{40+9b^2}.\]](http://latex.artofproblemsolving.com/a/3/9/a397af42f0436880e28ca7c846e287041c0160e3.png)
be real numbers such that .Prove that![\[ 3\sqrt{1+2a^2}+2\sqrt{40+9b^2}\geq5\sqrt{11}.\]](http://latex.artofproblemsolving.com/6/1/e/61ed1561359910f241c62f3e0c68c311ceee887c.png)
Let 
Prove that![\[ \frac{(a+b)^2}{\sqrt{a^2-1}+ \sqrt{b^2-4}} \geq 6.\]](http://latex.artofproblemsolving.com/9/2/1/92111918033f09a397fa44494991988eebc65128.png)
Let 
. Prove that: 
. Note that 
is continuous, since the terms inside the square roots are always strictly greater than 
. We obtain 
. Now we can easily see that 
, so there can be at most one root of 
. This, by inspection is at 
, which produces a value of 
, which is indeed the minimum.
and prove 
obviously, and also 
.
![$f:(0,1] \rightarrow R, \ f(x) = 3x^2-2x$](http://latex.artofproblemsolving.com/2/1/2/212096ffe462ef536c15850b145ef028a4a3fd00.png)
.
.![$x \in (0;\frac{1}{3}]$](http://latex.artofproblemsolving.com/6/0/0/60067b6e4c73d94a4de7090458dc81d694bcc8bd.png)
the funtion is decreasing, and for ![$x \in [\frac{1}{3},1]$](http://latex.artofproblemsolving.com/7/4/f/74f38da682f70e0d5edaa90301aff61ff3b3692b.png)
the function is increasing. So minimum value of the function is 
So we get:
.