Problem 6

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Problem 6

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Polish Math Olympiad 2025 Finals P6

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

71 posts

blug

#1 h Apr 4, 2025, 12:17 PM

Y by

A strictly decreasing function $f:(0, \infty)\Rightarrow (0, \infty)$ attaining all positive values and positive numbers $a_1\ne b_1$ are given. Numbers $a_2, b_2, a_3, b_3, ...$ satisfy
$a_{n+1}=a_n+f(b_n),\;\;\;\;\;\;\;b_{n+1}=b_n+f(a_n)$ for every $n\geq 1$ . Prove that there exists a positive integer $n$ satisfying $|a_n-b_n| >2025$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

185 posts

#2 h Apr 5, 2025, 9:58 AM

Y by

Obviously $f$ is continious. WLOG $a_1>b_1$ . First we prove by induction that $a_n>b_n$ and $d_n=a_n-b_n$ is increasing:
$d_{n+1}=d_n+f(b_n)-f(a_n)$ . By induction hypothesis $b_n<a_n$ and thus $f(b_n)>f(a_n)$ which implies $d_{n+1}>d_n$ . Also, $d_1>0$ and $d$ was increasing, thus $d_{n+1}>0$ which means $a_{n+1}>b_{n+1}$ .
Great, now suppose $d_n<2025$ for all $n$ , then it is bounded from above and hence has a limit, say $l$ . Then $\sum_{n=1} f(b_n)-f(a_n) < \infty$ Now we prove that $\lim_{n \to \infty} b_n = \infty$ Suppose it isn't, then $b_n$ has a limit, say $k$ . Then $\lim_{n \to \infty} a_n = k+l$ And so $\lim_{n \to \infty} f(b_n)-f(a_n) = f(k)-f(k+l)>0$ But $d_{n+1}-d_n=f(b_n)-f(a_n)$ , and so $d_n$ goes to infinity, a contradiction. Thus, $b_n$ goes to infinity, and so does $a_n$ . As $a_{n+1}-a_n=f(b_n)$ , we know that $\sum_{n=1} f(b_n)$ diverges.
Now we are ready to finish. Choose any small $\varepsilon$ . Take $N$ to be a big positive integer, and choose $n$ to be large enough that $f(a_n) < \frac{l-\varepsilon}{N}$ and $d_n>l-\varepsilon$ . Then $b_{n+N}=b_n+f(a_n)+f(a_{n+1})+\ldots+f(a_{n+N}) \leq b_n+Nf(a_n) \leq b_n+l-\varepsilon < a_n$ Great, now split the sum $\sum_{k=n}^{\infty} f(b_k)-f(a_k)$ into $N$ sums depending on $k\mod N$ . We will consider one of them because the other are the same. $f(b_n)-f(a_n)+f(b_{n+N}) - f(a_{n+N} + \ldots + f(b_{n+uN})-f(a_{n+uN}) \geq f(b_n) - f(a_{n+uN})$ because $b_{t+N} < a_t$ for $t \geq n$ . Thus, when $u$ goes to infinity we get $\sum_{k=n+pN} f(b_k)-f(a_k) \geq f(b_n)$ Other sums can be considered in the same way. Thus, we get $\sum_{k=n}^{\infty} f(b_k)-f(a_k) \geq f(b_n)+f(b_{n+1})+\ldots+f(b_{n+N-1})$ But now remember how we were choosing $n$ and $N$ . We could take $n$ to be large enough that $d_n > l-\varepsilon$ and $N$ to be the floor of $\frac{l-\varepsilon}{f(a_n)}$ , and so $N \geq \frac{l-\varepsilon}{f(a_n)}-1$ . Then we can bound the sum above $\sum_{k=n}^{\infty} f(b_k)-f(a_k) \geq f(b_n)+f(b_{n+1})+\ldots+f(b_{n+N-1}) \geq Nf(a_n) \geq l-\varepsilon-f(a_n) > l-\varepsilon_2=const>0$ where $\varepsilon_2$ is a small number bigger than $\varepsilon$ . And so the series $\sum_{n=1} f(b_n)-f(a_n)$ diverges, which is a contradiction. Thus there exists a positive integer $n$ such that $d_n>2025$ .

This post has been edited 1 time. Last edited by atdaotlohbh, Apr 5, 2025, 10:32 AM

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a