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.
such that for all reals 
Proposed by Aritra12, India
let 
be the number whose 
-th digit after the decimal point is the 
-th digit after the decimal point of 
. Show that if is rational then so is 
.
be reals such that 
and 
Prove that
Let be reals such that and 
Prove thatLet be reals such that and 
Prove that
and 
Prove that
Let and 
Prove that
, 
, and 
be the midpoints of sides BC, CA, and AB, respectively. Point 
is located on the ray 
so that 
is similar to 
. Points 
and 
on rays 
and 
, respectively, are defined similarly. Prove that lines 
, 
, and 
are concurrent, i.e. these three lines intersect at a point.
and a constant 
, it is given that 
and 
are also friendly polynomials. Prove that 
.
be a triangle. Distinct points 
, 
, 
lie on sides 
, 
, and 
, respectively, such that quadrilaterals 
and 
are cyclic. Line 
meets the circumcircle of 
again at 
. Let 
denote the reflection of across . Show that lies on the circumcircle of 
. be an acute triangle with circumcenter 
such that 
, let be the intersection of the external bisector of 
with , and let be a point in the interior of such that 
is similar to 
. Show that 
.
be the set of real numbers. Find all functions 
that satisfy the following condition. Here, 
is the function obtained by composing 
times, that is, 
, 
be an odd prime number and 
be the set of positive integers. Suppose that a function 
satisfies the following properties:
.
for any pair of relatively prime positive integers 
not both equal to 1;
for any pair of relatively prime positive integers .
with circumcenter 
. Let 
be a point on 
such that 
. Denote 
the point on satisfying 
. If 
is the projection of onto , show that 
passes through the orthocenter of .
and for all 
, 
Determine the smallest 
such that 
.
, so 
. Summing from 
to 
, we get![\[n!a_n=1!a_1+\sum_{k=1}^{n-1}k!\implies a_n=\frac{2025+\sum\limits_{k=1}^{n-1}k!}{n!}.\]](http://latex.artofproblemsolving.com/b/d/9/bd99b2f350454dbab326def02e1bfde29c059bec.png)
Clearly 
, 
. For 
we have![\[a_n>\frac{(n-1)!+(n-2)!}{n!}=\frac1n+\frac1{(n-1)n}=\frac1{n-1}\ge\frac1{2025},\]](http://latex.artofproblemsolving.com/7/5/2/752e750aa744d1cf297f7df5489b22eaeed1f84b.png)
so it does not work.
, we have![\[\sum_{k=1}^{n-3}k!<(n-2)!\sum_{k=1}^{n-3}\frac1{2^{n-2-k}}<(n-2)!,\]](http://latex.artofproblemsolving.com/9/d/7/9d7a50882ea2f0901d4884de88202b2d8a676140.png)
so
So the answer is 
.
So we want to find the smallest such that 
. Obviously 
. Then the rest is just routine checking (i included it in the script) to get 
.
: is the smallest number such that 
,
: is the smallest number such that ,![\[
\begin{aligned}
&\frac{1}{2025} < \frac{1}{2026} \\[10pt]
&\frac{a_{k-1} + 1}{k} > \frac{\frac{1}{2025} + 1}{k} > \frac{\frac{1}{2026} + 1}{k} = \frac{2027}{2026k} \\[10pt]
&\frac{2027}{2026k} < \frac{1}{2025} \\[10pt]
&2025 \cdot \frac{2027}{2026k} = 2,025.999 < k \\[10pt]
&k = 2026.
\end{aligned}
\]](http://latex.artofproblemsolving.com/9/a/7/9a7b6ad626ab3b6c69b7c7827befd34df0c2b63d.png)
and 
imply that either 
or 
.
implies that 
.
also implies that 
implies that for any 
,
or 
.
,
and 
imply that 
for all 
.
.
such that 
,
gives 
for all such 
.
then gives 
and thus 
.
which means that 
.
for all 
for some 
,
which is impossible. Thus 
in this case.
and 
.
Prove 
Find maximum 
