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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
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0 replies
jlacosta
Mar 2, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
P(a+1)=1 for every root a
MTA_2024   1
N 11 minutes ago by pco
Find all real polynomials $P$ of degree $K$ having $k$ distinct real roots ($k \in \mathbb N$), such that for every root $a$ : $P(a+1)=1$.
1 reply
MTA_2024
23 minutes ago
pco
11 minutes ago
Functional equation
socrates   30
N 36 minutes ago by NicoN9
Source: Baltic Way 2014, Problem 4
Find all functions $f$ defined on all real numbers and taking real values such that \[f(f(y)) + f(x - y) = f(xf(y) - x),\] for all real numbers $x, y.$
30 replies
socrates
Nov 11, 2014
NicoN9
36 minutes ago
Algebra Functions
pear333   1
N an hour ago by whwlqkd
Let $P(z)=z-1/z$. Prove that there does not exist a pair of rational numbers $x,y$ such that $P(x)+P(y)=4$.
1 reply
pear333
Today at 12:20 AM
whwlqkd
an hour ago
Polynomial equation with rational numbers
Miquel-point   2
N an hour ago by Assassino9931
Source: Romanian TST 1979 day 2 P1
Determine the polynomial $P\in \mathbb{R}[x]$ for which there exists $n\in \mathbb{Z}_{>0}$ such that for all $x\in \mathbb{Q}$ we have: \[P\left(x+\frac1n\right)+P\left(x-\frac1n\right)=2P(x).\]
Dumitru Bușneag
2 replies
Miquel-point
Apr 15, 2023
Assassino9931
an hour ago
Area problem
MTA_2024   0
an hour ago
Let $\omega$ be a circle inscribed inside a rhombus $ABCD$. Let $P$ and $Q$ be variable points on $AB$ and $AD$ respectively, such as $PQ$ is always the tangent line to $\omega$.
Prove that for any position of $P$ and $Q$ the area of triangle $\triangle CPQ$ is the same.
0 replies
MTA_2024
an hour ago
0 replies
Geometry
srnjbr   0
an hour ago
In triangle ABC, D is the leg of the altitude from A. l is a variable line passing through D. E and F are points on l such that AEB=AFC=90. Find the locus of the midpoint of the line segment EF.
0 replies
srnjbr
an hour ago
0 replies
Geometry
srnjbr   0
an hour ago
in triangle abc, l is the leg of bisector a, d is the image of c on line al, and e is the image of l on line ab. take f as the intersection of de and bc. show that af is perpendicular to bc
0 replies
srnjbr
an hour ago
0 replies
<QBC =<PCB if BM = CN, <PMC = <MAB, <QNB = < NAC
parmenides51   1
N 2 hours ago by dotscom26
Source: 2005 Estonia IMO Training Test p2
On the side BC of triangle $ABC$, the points $M$ and $N$ are taken such that the point $M$ lies between the points $B$ and $N$, and $| BM | = | CN |$. On segments $AN$ and $AM$, points $P$ and $Q$ are taken so that $\angle PMC = \angle  MAB$ and $\angle QNB = \angle NAC$. Prove that $\angle QBC = \angle PCB$.
1 reply
parmenides51
Sep 24, 2020
dotscom26
2 hours ago
Bosnia and Herzegovina EGMO TST 2017 Problem 2
gobathegreat   2
N 2 hours ago by anvarbek0813
Source: Bosnia and Herzegovina EGMO Team Selection Test 2017
It is given triangle $ABC$ and points $P$ and $Q$ on sides $AB$ and $AC$, respectively, such that $PQ\mid\mid BC$. Let $X$ and $Y$ be intersection points of lines $BQ$ and $CP$ with circumcircle $k$ of triangle $APQ$, and $D$ and $E$ intersection points of lines $AX$ and $AY$ with side $BC$. If $2\cdot DE=BC$, prove that circle $k$ contains intersection point of angle bisector of $\angle BAC$ with $BC$
2 replies
gobathegreat
Sep 19, 2018
anvarbek0813
2 hours ago
Another NT FE
nukelauncher   58
N 2 hours ago by andrewthenerd
Source: ISL 2019 N4
Find all functions $f:\mathbb Z_{>0}\to \mathbb Z_{>0}$ such that $a+f(b)$ divides $a^2+bf(a)$ for all positive integers $a$ and $b$ with $a+b>2019$.
58 replies
nukelauncher
Sep 22, 2020
andrewthenerd
2 hours ago
Easiest Functional Equation
NCbutAN   7
N 2 hours ago by InftyByond
Source: Random book
Find all functions $f: \mathbb R \to \mathbb R$ such that $$f(yf(x)+f(xy))=(x+f(x))f(y)$$Follows for all reals $x,y$.
7 replies
NCbutAN
Mar 2, 2025
InftyByond
2 hours ago
Lower bound for integer relatively prime to n
62861   23
N 3 hours ago by YaoAOPS
Source: USA Winter Team Selection Test #1 for IMO 2018, Problem 1
Let $n \ge 2$ be a positive integer, and let $\sigma(n)$ denote the sum of the positive divisors of $n$. Prove that the $n^{\text{th}}$ smallest positive integer relatively prime to $n$ is at least $\sigma(n)$, and determine for which $n$ equality holds.

Proposed by Ashwin Sah
23 replies
62861
Dec 11, 2017
YaoAOPS
3 hours ago
Number of quadratic residues mod $p$ up to $x$
Gauler   0
3 hours ago
Call an integer $n$ to be $y$-smooth if all of its prime factors are $\le y$. Let $\Psi(x,y)$ denote the number of $y$-smooth integers upto $x$. Let $y = y(p)$ be the least quadratic non-residue mod $p$. Show that there are at least $\Psi(x,y)$ quadratic residues mod $p$ up to $x$.
0 replies
Gauler
3 hours ago
0 replies
Property of Arithmetic function
Gauler   0
3 hours ago
If $f$ is a non-negative arithmetic function and
$$F(\sigma)=\sum_{n=1}^\infty \frac{f(n)}{n^\sigma}$$is convergent for some $0<\sigma<1$, then prove that
$$\sum_{n\le x} f(n) + x\sum_{n>x} \frac{f(n)}{n} \le x^\sigma F(\sigma).$$
0 replies
Gauler
3 hours ago
0 replies
Wait wasn&#039;t it the reciprocal in the paper?
Supercali   7
N Yesterday at 11:23 AM by kes0716
Source: India TST 2023 Day 2 P1
Let $\mathbb{Z}_{\ge 0}$ be the set of non-negative integers and $\mathbb{R}^+$ be the set of positive real numbers. Let $f: \mathbb{Z}_{\ge 0}^2 \rightarrow \mathbb{R}^+$ be a function such that $f(0, k) = 2^k$ and $f(k, 0) = 1$ for all integers $k \ge 0$, and $$f(m, n) = \frac{2f(m-1, n) \cdot f(m, n-1)}{f(m-1, n)+f(m, n-1)}$$for all integers $m, n \ge 1$. Prove that $f(99, 99)<1.99$.

Proposed by Navilarekallu Tejaswi
7 replies
Supercali
Jul 9, 2023
kes0716
Yesterday at 11:23 AM
Wait wasn&#039;t it the reciprocal in the paper?
G H J
G H BBookmark kLocked kLocked NReply
Source: India TST 2023 Day 2 P1
The post below has been deleted. Click to close.
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Supercali
1260 posts
#1 • 1 Y
Y by MS_asdfgzxcvb
Let $\mathbb{Z}_{\ge 0}$ be the set of non-negative integers and $\mathbb{R}^+$ be the set of positive real numbers. Let $f: \mathbb{Z}_{\ge 0}^2 \rightarrow \mathbb{R}^+$ be a function such that $f(0, k) = 2^k$ and $f(k, 0) = 1$ for all integers $k \ge 0$, and $$f(m, n) = \frac{2f(m-1, n) \cdot f(m, n-1)}{f(m-1, n)+f(m, n-1)}$$for all integers $m, n \ge 1$. Prove that $f(99, 99)<1.99$.

Proposed by Navilarekallu Tejaswi
This post has been edited 4 times. Last edited by Supercali, Jul 11, 2023, 12:23 PM
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AnonymousBunny
339 posts
#3
Y by
Let $g(m,n)=1/f(m,n)$. Consider a random walk starting from $(m+1,n)$ where at each step we go left or down 1 unit with equal probability. Then, $g(m,n)$ is the probability that we touch the line $y=0$ before we touch the line $x=0$.

So we want to estimate the following quantity: start a random walk from $(n+1,n)$. What is the probability we hit $y=0$ before $x=0$? Let $E$ denote this event.

Let $P_i=(x_i,y_i)$ be the $i$th point in the step and let $v$ be the largest $i$ such that $|y_i-x_i|=1$. Condition on $v=k$; we want to estimate $P(x_k>y_k|v=k)$. I claim this is $\geq 1/2+1/(2n-1)$

Consider a set of lattice paths where from each $(x,y)$ we go to either $(x+1,y+1)$ or $(x+1,y-1)$.
Let $A$ be the number of such length $k$ paths from $(0,1)$ to (k,1) which touch the line $y=0$, $B$ be the number of paths from $(0,1)$ to $(k,1)$ which don't touch the line $y=0$ and let $C$ be the number of paths from $(0,1)$ to $(k,-1)$. By standard bijection arguments, the conditional probability we want to compute is $(A+B)/(A+B+C)$. Moreover, by reflection argument, $B=C$, and by using Catalan numbers, $A=(A+B/((k-1)/2)$. Since $k \leq 2n$, we get that $A/(B+C) \geq 1/2+ 1/(2n-1)$.

Now simply note that $E$ is true iff $x_v>y_v$ where $v$ is defined as above. Since for each conditioning of $v$ the conditional probability is $\geq 1/2+ 1/(2n-1)$ we get that $g(n,n) \geq 1/2+ 1/(2n-1)$.

In fact this inequality must be strict because the conditional probability is strictly larger for small values of $v$ (which have a nonzero probability).
This post has been edited 1 time. Last edited by AnonymousBunny, Jul 11, 2023, 11:57 AM
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falcon311
100 posts
#4
Y by
in the real test it was typod so $2^k$ became $\frac{1}{2^k}$
The typo made it super easy...
Everyone except for 1 person solved it.
This post has been edited 1 time. Last edited by falcon311, Jul 13, 2023, 4:00 AM
Reason: missed some info
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Rijul saini
904 posts
#5 • 2 Y
Y by Ankoganit, MS_asdfgzxcvb
Let $g(m,n) = \frac{1}{f(m,n)}$. Then $g(0,k) = \frac{1}{2^k}$, $g(k,0) = 1$ for all integers $k \ge 0$, and $$2 \cdot g(m,n) = g(m-1,n) + g(m, n-1)$$for all integers $m,n \ge 1$.

Let $G(x,y) = \sum_{m,n \ge 0} g(m,n) x^m y^n$ be the generating function of $g(m,n)$.
Claim 1: $\color{blue}{G(x,y) = \frac{1}{(2-x-y)} + \frac{1}{(1-x)(2-x-y)}}$
Proof: Multiplying both sides of $2 \cdot g(m,n) = g(m-1,n) + g(m, n-1)$ by $x^my^n$ and summing over all $m,n \ge 1$, we get
\begin{align*}
\sum_{m,n \ge 1} 2 \cdot g(m,n) x^m y^n &= \sum_{m,n \ge 1} g(m-1,n) x^m y^n +  \sum_{m,n \ge 1} g(m, n-1)x^m y^n \\
2 \left(G - \frac{1}{1-x} - \frac{2}{2-y} + 1 \right) &= x \left(G - \frac{1}{1-x} \right) +  y \left(G - \frac{2}{2-y} \right)\\
(2-x-y)G &= \frac{2-x}{1-x} = 1 + \frac{1}{1-x}\\
G(x,y) &= \frac{1}{(2-x-y)} + \frac{1}{(1-x)(2-x-y)}
\end{align*}
Claim 2: $\color{blue}{g(m,n) = \frac{1}{2^{m+n+1}} \binom{m+n}{n} + \sum_{k =0}^{m} \binom{n+k}{n} \frac{1}{2^{n+k+1}} }$.
Proof: Taking coefficient of $x^my^n$ in the generating function from Claim 1.

Claim 3: $\color{blue}{\sum_{k=0}^{n} \binom{n+k}{n} \frac{1}{2^{n+k+1}} = \frac 12}$.
Proof: This seems fairly well known. There are three proofs I know of this. The easiest is to simply prove it by induction on $n$. The second to consider the situation of flipping a fair coin repeatedly until having $n+1$ heads, and then seeing if the $(n+1)^{th}$ head appears before the $(n+1)^{th}$ tail. For fun, here's a third solution using generating functions:

\begin{align*}
\sum_{k=0}^{n} \binom{n+k}{n} \frac{1}{2^k} &= \sum_{k=0}^{n} [x^k] \left(1+\frac{x}{2} \right)^{n+k} \\
&=\sum_{k=0}^{n} [x^n] x^{n-k}\left(1+\frac x2 \right)^{n+k} \stackrel{k \mapsto n-k}{=} [x^n] \sum_{k=0}^n x^k \left(1+ \frac x2 \right)^{2n-k} \\
& = [x^n] \left(1+ \frac x2 \right)^{2n}  \frac{1 - \left(\frac{x}{1+ \frac x2} \right)^{n+1} }{1 - \left(\frac{x}{1+ \frac x2} \right)} \\
& = [x^n] \left(1+ \frac x2 \right)^{n}  \frac{\left(1+\frac x2 \right)^{n+1} - x^{n+1}}{1 -  \frac x2} \\
& = [x^n] \frac{\left(1+\frac x2 \right)^{2n+1}}{1 -  \frac x2} - \underbrace{[x^n] x^{n+1} \cdot \left(1+ \frac x2 \right)^{n}  \frac{1}{1 -  \frac x2} }_{=0} \\
& \stackrel{y = \frac x2}{=} \frac{1}{2^n} [y^n] \frac{(1+y)^{2n+1}}{1-y} \\
& =\frac{1}{2^n} \sum_{k=0}^{n} \binom{2n+1}{k} = \frac{1}{2^n} \cdot 2^{2n} = \boxed{2^n}.
\end{align*}
Claim 4: $\color{blue}{g(n,n) > \frac{1}{2 - \frac{1}{n+1}} }$, and therefore $\color{blue}{g(99,99) > \frac{1}{1.99} }$.
Proof: By Claim 2, $g(n,n) = \frac{1}{2^{2n+1}} \binom{2n}{n} + \sum_{k =0}^{n} \binom{n+k}{n} \frac{1}{2^{n+k+1}} =  \frac{1}{2^{2n+1}} \binom{2n}{n} + \frac 12$, where the last equality is by Claim 3.
Now, $\frac{1}{2 - \frac{1}{n+1} } = \frac{n+1}{2n+1} = \frac 12 + \frac{1}{2(2n+1)}.$ Thus, it suffices to observe that $\binom{2n}{n} > \frac{1}{2n+1} \cdot 2^{2n}$, which is true because it is the largest number among the $(2n+1)$ numbers $\binom{2n}{k}$ for $0 \le k \le 2n$ and their sum is $2^{2n}$.
This post has been edited 2 times. Last edited by Rijul saini, Nov 23, 2023, 4:30 AM
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anantmudgal09
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#6 • 4 Y
Y by starchan, math_comb01, everythingpi3141592, MS_asdfgzxcvb
The solution that follows uses a pretty fun main idea: instead of computing the recursion by going down all the way to the $x$ or $y$ axes, we instead go down all the way to the line $y=-x$ which has binary values, and simply compute the number of paths going to each point there. I have not seen this trick used elsewhere, but it felt really cool :)


Define $a_{m, n}=\frac{1}{f(m, n)}$ for all $m, n \ge 0$, this is well-defined as $f$ takes only positive values. Then $a_{0, k}=2^{-k}$ and $a_{k, 0}=1$ for all $k \ge 0$, and $a_{m, n}=\frac{1}{2}(a_{m, n-1}+a_{m-1, n})$ for all $m, n \ge 1$.

Extending $a_{m, n}=\frac{1}{2}(a_{m, n-1}+a_{m-1, n})$ gives $a_{1, -1}=1, a_{2, -1}=1, \dots, a_{k, -1}=1$ for all $k \ge 1$. Extending downwards in similar fashion, $a_{m, n}=1$ for all $m>0, n<0$ with $m+n \ge 0$. Similarly, since $\frac{1}{2^{k+1}}=\frac{1}{2}\left(\frac{1}{2^k}+0\right)$, we conclude that $a_{m, n}=0$ for all $m<0, n>0$ with $m+n<0$. Thus, the diagonal element $a_{k, -k}$ is $0$ if $k<0$ and $1$ if $k \ge 0$.

Consider now the computation for $a_{n, n}$: expanding the recursion till we reach the diagonal line $y=-x$, gives it as a linear combination of the diagonal elements. The weight of each $a_{x, -x}$ is given by adding $\frac{1}{2^{\text{path length}}}$ over all paths from $(n, n)$ to $(x, -x)$ that go either down or left in each step. The length of all such paths if $(n-x)+(n-(-x))=2n$, and the number of such paths is $\binom{2n}{n+x}$. Thus, $$a_{n, n}=\frac{1}{2^{2n}} \cdot \left(\sum_{k=0}^{2n} a_{k-n, n-k} \cdot \binom{2n}{k}\right)=\frac{1}{2^{2n}} \cdot \left(\sum_{k=n}^{2n} \binom{2n}{k} \right)=\frac{1}{2^{2n}} \left(\frac{1}{2}\left(2^{2n}+\binom{2n}{n}\right)\right).$$
By the well-known fact that $\binom{2n}{n} \ge \binom{2n}{k}$ for any $k$, and $2^{2n}=\sum_{r=0}^{2n} \binom{2n}{k} < (2n+1)\binom{2n}{n}$ for $n>1$, we get that $a_{n, n} > \frac{1}{2}\left(1+\frac{1}{2n+1}\right)=\frac{n+1}{2n+1}$, so $f(n, n) < 2-\frac{1}{n+1}$ for all $n>1$. So, $f(99, 99) < 1.99$.
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Pyramix
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Let $g(n)=\frac1{f(n)}$ for every $n$. Then, we have
\[g(m,n)=\frac{g(m-1,n)+g(m,n-1)}2\]and $g(0,k)=\frac1{2^k}$, $g(k,0)=1$.
We need to show that $g(n,n)>\frac{n+1}{2n+1}$ for every $n$. Here's a standard solution of the same, using generating functions. (We explicitly find $g(n,n)$ in terms of $n$.)

Let $G(x,y)=\sum_{m,n\geq0}g(m,n)x^my^n$
Then, we have that $G(x,y)-\left(\frac{x+y}2\right)G(x,y)=\frac12+\frac1{2(1-x)}$ as most terms just cancel and we're left with geometric series.
So, we get $G(x,y)=\frac1{2-x-y}+\frac1{(2-x-y)(1-x)}$
Now, coefficient of $x^my^n$ in $G(x,y)$ is $\frac1{2^{m+n+1}}\binom{m+n}{n}+\frac1{2^{m+n+1}}\sum_{t=0}^{m}2^{m-t}\binom{n+t}{n}$.
So, we have
\[g(n,n)=\frac1{2^{2n+1}}\binom{2n}n+\frac1{2^{2n+1}}\sum_{t=0}^{n}2^{n-t}\binom{n+t}n\]It is well-known that $\sum_{t=0}^{n}\frac1{2^{n+t+1}}\binom{n+t}n$ is actually just $\frac12$.
Here's a proof using generating functions:
Let $H(x)=\sum_{t=0}^{n}x^{n-t}\left(1+\frac x2\right)^{n+t}$ We just want the coefficient of $x^n$ from this and to show it is $2^n$.
Note that $H(x)=\sum_{t=0}^{n}x^t\left(1+\frac x2\right)^{2n-t}=\left(1+\frac x2\right)^{2n}\left(\frac{1-\left(\frac{x}{1+\frac x2}\right)^{n+1}}{1-\left(\frac{x}{1+\frac x2}\right)}\right)=\left(1+\frac x2\right)^n\frac{\left(1+\frac x2\right)^{n+1}-x^{n+1}}{1-\frac x2}$
and, as usual, ignore the large powers.
So, it is same as finding coefficient of $x^n$ from $\left(1+\frac x2\right)^{2n+1}\left(1+\frac x2+\left(\frac x2\right)^2+\cdots\right)$, but that is simply $\sum_{i=0}^{n}\binom{2n+1}i2^{-n}=2^n$, as required.

So, we have $g(n,n)=\frac12+\frac1{2^{2n+1}}\binom{2n}n$. We had to show that $g(n,n)>\frac{n+1}{2n+1}$. It suffices to show that $\binom{2n}n>\frac{2^{2n}}{2n+1}$. But that is true, because $\binom{2n}n>\binom{2n}{n-1}>\cdots>\binom{2n}0$ and the sum of these numbers is more than $2^{2n}$. We're done. $\blacksquare$
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kes0716
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AnonymousBunny wrote:
Let $g(m,n)=1/f(m,n)$. Consider a random walk starting from $(m+1,n)$ where at each step we go left or down 1 unit with equal probability. Then, $g(m,n)$ is the probability that we touch the line $y=0$ before we touch the line $x=0$.

So we want to estimate the following quantity: start a random walk from $(n+1,n)$. What is the probability we hit $y=0$ before $x=0$? Let $E$ denote this event.

Let $P_i=(x_i,y_i)$ be the $i$th point in the step and let $v$ be the largest $i$ such that $|y_i-x_i|=1$. Condition on $v=k$; we want to estimate $P(x_k>y_k|v=k)$. I claim this is $\geq 1/2+1/(2n-1)$

Consider a set of lattice paths where from each $(x,y)$ we go to either $(x+1,y+1)$ or $(x+1,y-1)$.
Let $A$ be the number of such length $k$ paths from $(0,1)$ to (k,1) which touch the line $y=0$, $B$ be the number of paths from $(0,1)$ to $(k,1)$ which don't touch the line $y=0$ and let $C$ be the number of paths from $(0,1)$ to $(k,-1)$. By standard bijection arguments, the conditional probability we want to compute is $(A+B)/(A+B+C)$. Moreover, by reflection argument, $B=C$, and by using Catalan numbers, $A=(A+B/((k-1)/2)$. Since $k \leq 2n$, we get that $A/(B+C) \geq 1/2+ 1/(2n-1)$.

Now simply note that $E$ is true iff $x_v>y_v$ where $v$ is defined as above. Since for each conditioning of $v$ the conditional probability is $\geq 1/2+ 1/(2n-1)$ we get that $g(n,n) \geq 1/2+ 1/(2n-1)$.

In fact this inequality must be strict because the conditional probability is strictly larger for small values of $v$ (which have a nonzero probability).

I think the reflection argument shows $A=C$, not $B=C$.
Furthermore, for computing $A/B$, my calculation gave $(A+B)/C = (k+2)/k$. Hence $A:B:C = k : 2 : k$, and the probability is $\frac{k+2}{2k+2} \geq \frac{1}{2} + \frac{1}{2n+1}$, which is weaker than $\frac{1}{2} + \frac{1}{2n-1}$ you claimed but still suffices because that value is $\frac{100}{199} = 1.99^{-1}$. If you have spare time please calculate again and edit post or reply to check out who is right :)
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kes0716
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#9
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Surprised that everyone's solution uses relatively hard concepts like random walk or generating functions.
Actually it is straight forward induction (on $m+n$) that
$$\frac{1}{f(m,n)} =\frac{1}{2^{m+n}}\sum_{i=0}^n\binom{m+n}{i} $$(it holds for $m=0$ or $n=0$ by definition, and assuming true for $m+n-1$ the inductive step follows from $\sum_{i=0}^{n-1}\binom{m+n-1}{i}+\sum_{i=0}^n\binom{m+n-1}{i}=\sum_{i=0}^n\binom{m+n}{i}$.)
Now using the fact that $\binom{198}{99}$ is maximum among $\binom{198}{r}$,
$$\frac{1}{f(99,99)}=\frac{1}{2^{198}}\sum_{i=0}^{99}\binom{198}{i}=\frac{1}{2}(1+\frac{1}{2^{198}}\binom{198}{99})>\frac{1}{2}(1+\frac{1}{199})=\frac{100}{199}$$
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