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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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0 replies
1 viewing
jlacosta
May 1, 2025
0 replies
foldina a rectangle paper 3 times
parmenides51   1
N 23 minutes ago by TheBaiano
Source: 2023 May Olympiad L2 p4
Matías has a rectangular sheet of paper $ABCD$, with $AB<AD$.Initially, he folds the sheet along a straight line $AE$, where $E$ is a point on the side $DC$ , so that vertex $D$ is located on side $BC$, as shown in the figure. Then folds the sheet again along a straight line $AF$, where $F$ is a point on side $BC$, so that vertex $B$ lies on the line $AE$; and finally folds the sheet along the line $EF$. Matías observed that the vertices $B$ and $C$ were located on the same point of segment $AE$ after making the folds. Calculate the measure of the angle $\angle DAE$.
IMAGE
1 reply
parmenides51
Mar 24, 2024
TheBaiano
23 minutes ago
Divisibility NT
reni_wee   0
25 minutes ago
Source: Japan 1996, ONTCP
Let $m,n$ be relatively prime positive integers. Calculate $gcd(5^m+7^m, 5^n+7^n).$
0 replies
reni_wee
25 minutes ago
0 replies
Modular arithmetic at mod n
electrovector   3
N an hour ago by Primeniyazidayi
Source: 2021 Turkey JBMO TST P6
Integers $a_1, a_2, \dots a_n$ are different at $\text{mod n}$. If $a_1, a_2-a_1, a_3-a_2, \dots a_n-a_{n-1}$ are also different at $\text{mod n}$, we call the ordered $n$-tuple $(a_1, a_2, \dots a_n)$ lucky. For which positive integers $n$, one can find a lucky $n$-tuple?
3 replies
electrovector
May 24, 2021
Primeniyazidayi
an hour ago
Largest Divisor
4everwise   19
N 2 hours ago by reni_wee
What is that largest positive integer $n$ for which $n^3+100$ is divisible by $n+10$?
19 replies
4everwise
Dec 22, 2005
reni_wee
2 hours ago
Sequences problem
BBNoDollar   3
N 2 hours ago by BBNoDollar
Source: Mathematical Gazette Contest
Determine the general term of the sequence of non-zero natural numbers (a_n)n≥1, with the property that gcd(a_m, a_n, a_p) = gcd(m^2 ,n^2 ,p^2), for any distinct non-zero natural numbers m, n, p.

⁡Note that gcd(a,b,c) denotes the greatest common divisor of the natural numbers a,b,c .
3 replies
BBNoDollar
Yesterday at 5:53 PM
BBNoDollar
2 hours ago
Can anyone solve this binomial identity
sasu1ke   0
3 hours ago


\[
\sum_{0 \leq k \leq l} (l - k) \binom{m}{k} \binom{q + k}{n}
= \binom{l + q + 1}{m + n + 1},
\]\[
\text{integers } l, m \geq 0,\quad \text{integers } n \geq q \geq 0.
\]
0 replies
sasu1ke
3 hours ago
0 replies
[ABCD] = n [CDE], areas in trapezoid - IOQM 2020-21 p1
parmenides51   3
N 3 hours ago by iamahana008
Let $ABCD$ be a trapezium in which $AB \parallel CD$ and $AB = 3CD$. Let $E$ be then midpoint of the diagonal $BD$. If $[ABCD] = n \times  [CDE]$, what is the value of $n$?

(Here $[t]$ denotes the area of the geometrical figure$ t$.)
3 replies
parmenides51
Jan 18, 2021
iamahana008
3 hours ago
Looking for users and developers
derekli   5
N 5 hours ago by musicalpenguin
Guys I've been working on a web app that lets you grind high school lvl math. There's AMCs, AIME, BMT, HMMT, SMT etc. Also, it's infinite practice so you can keep grinding without worrying about finding new problems. Please consider helping me out by testing and also consider joining our developer team! :P :blush:

Link: https://stellarlearning.app/competitive
5 replies
derekli
Today at 12:57 AM
musicalpenguin
5 hours ago
Sequences and GCD problem
BBNoDollar   0
6 hours ago
Determine the general term of the sequence of non-zero natural numbers (a_n)n≥1, with the property that gcd(a_m, a_n, a_p) = gcd(m^2 ,n^2 ,p^2), for any distinct non-zero natural numbers m, n, p.

⁡Note that gcd(a,b,c) denotes the greatest common divisor of the natural numbers a,b,c .
0 replies
BBNoDollar
6 hours ago
0 replies
Sum of digits is 18
Ecrin_eren   13
N 6 hours ago by NamelyOrange
How many 5 digit numbers are there such that sum of its digits is 18
13 replies
Ecrin_eren
Yesterday at 1:10 PM
NamelyOrange
6 hours ago
Inequalities
sqing   1
N 6 hours ago by DAVROS
Let $ a,b,c>0 $ and $ a+b\leq 16abc. $ Prove that
$$ a+b+kc^3\geq\sqrt[4]{\frac{4k} {27}}$$$$ a+b+kc^4\geq\frac{5} {8}\sqrt[5]{\frac{k} {2}}$$Where $ k>0. $
$$ a+b+3c^3\geq\sqrt{\frac{2} {3}}$$$$ a+b+2c^4\geq \frac{5} {8}$$
1 reply
sqing
Today at 12:46 PM
DAVROS
6 hours ago
an algebra problem
Asyrafr09   2
N Today at 12:03 PM by pooh123
Determine all real number($x,y,z$) that satisfy
$$x=1+\sqrt{y-z^2}$$$$y=1+\sqrt{z-x^2}$$$$z=1+\sqrt{x-y^2}$$
2 replies
Asyrafr09
Today at 10:05 AM
pooh123
Today at 12:03 PM
Inequalities
sqing   1
N Today at 11:51 AM by sqing
Let $ a,b,c\geq 0 ,   2a +ab + 12a bc \geq 8. $ Prove that
$$  a+  (b+c)(a+1)+\frac{4}{5}  bc \geq 4$$$$  a+  (b+c)(a+0.9996)+ 0.77  bc \geq 4$$
1 reply
sqing
Today at 5:23 AM
sqing
Today at 11:51 AM
When to look at solutions - pre calc
omerrob13   1
N Today at 10:51 AM by abartoha
Hey all.
I am doing the precalc book, and unfortunately, im getting into the habit of looking in the solutions quite fast on a problem I did not able to make any progress on.
My goal is mainly to develop problem solving and reasonning skills.

I divide the problems in AOPS to 2:

- Challenge problems at the end of the of each chapter.
- The problems that teach you the material itself, and the problems at the end of each section (1.1,1.2, etc...)

For non challenging problems, It takes around 20 mins of me not be able to solve a problem, and look at the solutions for it

Is it too little?
My goal is mainly to develop problem solving and reasoning skills.
I'm not sure if it's too little time to bring to a regular problem, or its ok to give 20 mins to a problem and continue if making no progress.
1 reply
omerrob13
Today at 9:36 AM
abartoha
Today at 10:51 AM
From Recursion to Inequality
mojyla222   2
N Apr 20, 2025 by missionjoshi.65
Source: IDMC 2025 P2
$\{a_n\}_{n\geq 1}$ is a sequence of real numbers with $a_1=1,\;a_2 =2$ such that for all $n\geq 1$
$$a_{n+2}=\dfrac{a_{n+1}^{2}}{1+a_{n}}+a_{n+1}.$$Prove that

$$\dfrac{1}{1+a_{1}+a_{2}}+\dfrac{1}{1+a_{2}+a_{3}}+\cdots + \dfrac{1}{1+a_{1403}+a_{1404}}>\dfrac{2^{1403}-1}{2^{1404}}.$$
Proposed by Mojtaba Zare
2 replies
mojyla222
Apr 20, 2025
missionjoshi.65
Apr 20, 2025
From Recursion to Inequality
G H J
G H BBookmark kLocked kLocked NReply
Source: IDMC 2025 P2
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mojyla222
103 posts
#1
Y by
$\{a_n\}_{n\geq 1}$ is a sequence of real numbers with $a_1=1,\;a_2 =2$ such that for all $n\geq 1$
$$a_{n+2}=\dfrac{a_{n+1}^{2}}{1+a_{n}}+a_{n+1}.$$Prove that

$$\dfrac{1}{1+a_{1}+a_{2}}+\dfrac{1}{1+a_{2}+a_{3}}+\cdots + \dfrac{1}{1+a_{1403}+a_{1404}}>\dfrac{2^{1403}-1}{2^{1404}}.$$
Proposed by Mojtaba Zare
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RagvaloD
4913 posts
#2 • 1 Y
Y by Calc_1
First let show $a_{n+1} \geq 2a_{n}$
for $n=1$ it is true
$\frac{a_{n+1}}{a_n}=\frac{a_n}{1+a_{n-1}}+1 \geq \frac{2a_{n-1}}{1+a_{n-1}}+1 \geq 2$
$a_1=1,a_2=2,a_3=4,a_4=\frac{28}{3}>8=2^3$ and so $a_{n+1}>2^n$ for $n \geq 3$

$\frac{1}{1+a_n}=\frac{a_{n+2}-a_{n+1}}{a_{n+1}^2}$
$1+a_n+a_{n+1}=\frac{a_{n+1}^2}{a_{n+2}-a_{n+1}}+a_{n+1}=\frac{a_{n+1}a_{n+2}}{a_{n+2}-a_{n+1}}$
$\frac{1}{1+a_n+a_{n+1}}=\frac{1}{a_{n+1}}-\frac{1}{a_{n+2}}$

So $\dfrac{1}{1+a_{1}+a_{2}}+\dfrac{1}{1+a_{2}+a_{3}}+\cdots + \dfrac{1}{1+a_{1403}+a_{1404}}=\frac{1}{a_2}-\frac{1}{a_3}+\frac{1}{a_3}-\frac{1}{a_4}+...+\frac{1}{a_{1404}}-\frac{1}{a_{1405}}=\frac{1}{a_2}-\frac{1}{a_{1405}}>\frac{1}{2}-\frac{1}{2^{1404}}=\frac{2^{1403}-1}{2^{1404}}$
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missionjoshi.65
2 posts
#3 • 1 Y
Y by khan.academy
I have a similar solution as above.
\[
a_{n+2} = a_{n+1} \left( \frac{a_{n+1}}{1+a_n} + 1 \right) = a_{n+1} \left( \frac{a_{n+1} + 1 + a_n}{1+a_n} \right)
\]
Dividing by \(a_{n+1}\), we get:

\[
\frac{a_{n+2}}{a_{n+1}} = \frac{1+a_n+a_{n+1}}{1+a_n}
\]
Rearranging this gives:

\[
1+a_n+a_{n+1} = (1+a_n)\frac{a_{n+2}}{a_{n+1}}
\]
\[
\frac{1}{1+a_n+a_{n+1}} = \frac{1}{(1+a_n)\frac{a_{n+2}}{a_{n+1}}} = \frac{a_{n+1}}{(1+a_n)a_{n+2}} = \frac{a_{n+2}-a_{n+1}}{a_{n+1}a_{n+2}} = \frac{1}{a_{n+1}} - \frac{1}{a_{n+2}}
\]
So the given inequality turns into
\[
\sum_{k=1}^{1403} \frac{1}{1+a_k+a_{k+1}} = \sum_{k=1}^{1403} \left( \frac{1}{a_{k+1}} - \frac{1}{a_{k+2}} \right) = \frac{1}{2} - \frac{1}{a_{1405}} > \frac{1}{2} - \frac{1}{2^{1404}}
\]
Hence it suffices to show that \(a_{1405} > 2^{1404}\)

Now we proceed by induction. Let \(P(n)\) denote the induction statement.

\(P(n): a_{n+1} > 2^{n}\) for \(n \geq 3\). It is easy to see that the base cases satisfy this for \(n=3\) and \(n=4\).

Now, suppose it is true for \(n = k\), then we have

\[
a_{k+1} > 2^{k}
\]
Now, consider \(P(k+1)\):
\begin{align*}
a_{k+2} &= \frac{a^2_{k+1}}{1+a_{k}} + a_{k+1} \\
&> \frac{a_{k+1}^2}{a_{k+1}} + a_{k+1} \\
&> 2^{k+1}
\end{align*}
The second last inequality follows from \(a_{k+1} - a_k > 1\) which is trivial from induction.

Thus, \(a_{n+1} > 2^n \quad \forall \; n \geq 3\) which implies \(a_{1405} > 2^{1404}\) and our proof is completed.
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