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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
jlacosta
May 1, 2025
0 replies
Inspired by Austria 2025
sqing   4
N 3 minutes ago by Tkn
Source: Own
Let $ a,b\geq 0 ,a,b\neq 1$ and $  a^2+b^2=1. $ Prove that$$   (a + b ) \left( \frac{a}{(b -1)^2} + \frac{b}{(a - 1)^2} \right) \geq 12+8\sqrt 2$$
4 replies
sqing
Today at 2:01 AM
Tkn
3 minutes ago
Erasing the difference of two numbers
BR1F1SZ   1
N 15 minutes ago by BR1F1SZ
Source: Austria National MO Part 1 Problem 3
Consider the following game for a positive integer $n$. Initially, the numbers $1, 2, \ldots, n$ are written on a board. In each move, two numbers are selected such that their difference is also present on the board. This difference is then erased from the board. (For example, if the numbers $3,6,11$ and $17$ are on the board, then $3$ can be erased as $6 - 3=3$, or $6$ as $17 - 11=6$, or $11$ as $17 - 6=11$.)

For which values of $n$ is it possible to end with only one number remaining on the board?

(Michael Reitmeir)
1 reply
BR1F1SZ
Yesterday at 9:48 PM
BR1F1SZ
15 minutes ago
3-var inequality
sqing   1
N 28 minutes ago by Natrium
Source: Own
Let $ a,b,c\geq 0 ,a+b+c =1. $ Prove that
$$\frac{ab}{2c+1} +\frac{bc}{2a+1} +\frac{ca}{2b+1}+\frac{27}{20} abc\leq \frac{1}{4} $$
1 reply
sqing
May 3, 2025
Natrium
28 minutes ago
Geo metry
TUAN2k8   0
41 minutes ago
Help me plss!
Given an acute triangle $ABC$. Points $D$ and $E$ lie on segments $AB$ and $AC$, respectively. Lines $BD$ and $CE$ intersect at point $F$. The circumcircles of triangles $BDF$ and $CEF$ intersect at a second point $P$. The circumcircles of triangles $ABC$ and $ADE$ intersect at a second point $Q$. Point $K$ lies on segment $AP$ such that $KQ \perp AQ$. Prove that triangles $\triangle BKD$ and $\triangle CKE$ are similar.
0 replies
TUAN2k8
41 minutes ago
0 replies
geometry
JetFire008   1
N Today at 4:23 AM by ohiorizzler1434
Given four concyclic points. For each subset of three points take the incenter. Show that the four incentres form a rectangle.
1 reply
JetFire008
Yesterday at 4:14 PM
ohiorizzler1434
Today at 4:23 AM
A pentagon inscribed in a circle of radius √2
tom-nowy   2
N Today at 4:20 AM by ohiorizzler1434
Can a pentagon with all rational side lengths be inscribed in a circle of radius $\sqrt{2}$ ?
2 replies
tom-nowy
Today at 2:37 AM
ohiorizzler1434
Today at 4:20 AM
Inequalities
sqing   8
N Today at 3:12 AM by sqing
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. $
$$ a+b+3c^3\geq\sqrt{\frac{2} {3}}$$$$ a+b+2c^4\geq \frac{5} {8}$$
8 replies
sqing
Sunday at 12:46 PM
sqing
Today at 3:12 AM
trapezoid
Darealzolt   0
Today at 2:03 AM
Let \(ABCD\) be a trapezoid such that \(A, B, C, D\) lie on a circle with center \(O\), and side \(AB\) is parallel to side \(CD\). Diagonals \(AC\) and \(BD\) intersect at point \(M\), and \(\angle AMD = 60^\circ\). It is given that \(MO = 10\). It is also known that the difference in length between \(AB\) and \(CD\) can be expressed in the form \(m\sqrt{n}\), where \(m\) and \(n\) are positive integers and \(n\) is square-free. Compute the value of \(m + n\).
0 replies
Darealzolt
Today at 2:03 AM
0 replies
Inequalities
sqing   2
N Today at 1:47 AM by sqing
Let $ a,b,c\geq 0 ,   2a +ab + 12a bc \geq 8. $ Prove that
$$  a+  (b+c)(a+1)+\frac{4}{5}  bc \geq 4$$$$  a+  (b+c)(a+0.9996)+ 0.77  bc \geq 4$$
2 replies
sqing
May 4, 2025
sqing
Today at 1:47 AM
anyone who can help me this 2 problems?
auroracliang   2
N Yesterday at 11:51 PM by ReticulatedPython
1. Let r be the radius of the largest circle which is tangent to the parabola y=x^2 at x=0 and which lies entirely on or inside (that is, above) the parabola, find r.

2. Counting number n has the following property,: if we take any 50 different numbers from 1,2,3, ... n, there always are two numbers with the difference of 7. what is the largest value among all value of n?


thanks a lot
2 replies
auroracliang
Nov 3, 2024
ReticulatedPython
Yesterday at 11:51 PM
What conic section is this? Is this even a conic section?
invincibleee   2
N Yesterday at 11:48 PM by ReticulatedPython
IMAGE

The points in this are given by
P = (sin2A, sin4A)∀A [0,2π]
Is this a conic section? what is this?
2 replies
invincibleee
Nov 15, 2024
ReticulatedPython
Yesterday at 11:48 PM
Spheres, ellipses, and cones
ReticulatedPython   0
Yesterday at 11:38 PM
A sphere is inscribed inside a cone with base radius $1$ and height $2.$ Another sphere of radius $r$ is internally tangent to the lateral surface of the cone, but does not intersect the larger inscribed sphere. A plane is tangent to both of these spheres, and passes through the inside of the cone. The intersection of the plane and the cone forms an ellipse. Find the maximum area of this ellipse.
0 replies
ReticulatedPython
Yesterday at 11:38 PM
0 replies
Looking for users and developers
derekli   13
N Yesterday at 11:31 PM by DreamineYT
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
13 replies
derekli
May 4, 2025
DreamineYT
Yesterday at 11:31 PM
trigonometric functions
VivaanKam   12
N Yesterday at 11:06 PM by aok
Hi could someone explain the basic trigonometric functions to me like sin, cos, tan etc.
Thank you!
12 replies
VivaanKam
Apr 29, 2025
aok
Yesterday at 11:06 PM
Easy IMO 2023 NT
799786   133
N Apr 24, 2025 by Maximilian113
Source: IMO 2023 P1
Determine all composite integers $n>1$ that satisfy the following property: if $d_1$, $d_2$, $\ldots$, $d_k$ are all the positive divisors of $n$ with $1 = d_1 < d_2 < \cdots < d_k = n$, then $d_i$ divides $d_{i+1} + d_{i+2}$ for every $1 \leq i \leq k - 2$.
133 replies
799786
Jul 8, 2023
Maximilian113
Apr 24, 2025
Easy IMO 2023 NT
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 2023 P1
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Ywgh1
139 posts
#135 • 1 Y
Y by radian_51
IMO 2023 p1

We claim that the answer is $n=p^k$, where $p$ is a prime.

Let $p_1$ and $p_2$ be the first two primes of $n$. Now let $s$ be the largest number, such that $p_1^s < p_2$

So we have that
\[\frac{n}{p_2},\frac{n}{p_1^s},\frac{n}{p_1^{s-1}}\]are consecutive divisors of $n$.
Hence we must have that.
$$\frac{n}{p_2}|(\frac{n}{p_1^s}+\frac{n}{p_1^{s-1}}) \implies p_1^s|p_2(p_1+1)$$Which is a contradiction, hence $n$ has less than 2 primes.
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ezpotd
1263 posts
#137 • 1 Y
Y by radian_51
I claim the answer is only prime powers. It is easy to see that these work. To see nothing else works, assume for the sake of contradiction some $n$ with more than $1$ prime power worked. Then take the second smallest prime divisor, let it be $q$, and the largest divisor less than it, $p^{k}$, where $k$ can be $1$. We would then have $\frac nq \mid \frac{n}{p^{k - 1}} + \frac{n}{p^k}$, which is obviously not true by taking $\nu_p$ on both sides, the left side has $\nu_p(\frac nq) = \nu_p(n)$, right hand side has $\nu_p({\frac{n}{p^{k - 1}} + \frac{n}{p^k}}) = \nu_p({\frac{n}{p^k}})$, so the left side cannot divide the right.
This post has been edited 1 time. Last edited by ezpotd, Sep 30, 2024, 1:00 AM
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lelouchvigeo
181 posts
#139 • 1 Y
Y by S_14159
Nothing new
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cursed_tangent1434
620 posts
#140
Y by
Trivial NT on the IMO. Finally found some time to actually post my solution from the contest.

We claim that the answer is all positive integers of the form $p^m$ for some prime $p$ and $m\ge 2$. These clearly all work since for all $1\le i \le m+1$ , $d_i=p^{i-1}$. Thus,
\[d_i=p_{i-1}\mid p_i + p_{i+1}=d_{i+1}+d_{i+2}\]quite clearly.

Now, we show that no other composite $n$ work. Say $n$ has atleast two distinct primes divisors, of which the smallest two are $p<q$. Note that we have $d_i = p^{i-1}$ for all $1\le i \le r$ for some $r \ge 2$ and $d_{r+1}=q$ as the divisors are arranged in increasing order. Now, if $n$ satisfies the desired condition we must have,
\begin{align*}
d_{m-r} &\mid d_{m-r+1} + d_{m-r+2}\\
\frac{n}{d_{r+1}} & \mid \frac{n}{d_r} + \frac{n}{d_{r-1}}\\
\frac{n}{q} & \mid \frac{n}{p^{r-1}} + \frac{n}{p^{r-2}}\\
p^{r-1} & \mid q + pq\\
p &\mid q+pq\\
p& \mid q
\end{align*}which is a very clear contradiction since $q>p$ are both primes. Thus, there are no such $n$ of this form, which finishes the proof.
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EVKV
70 posts
#141
Y by
Clear answer is n =p^c where p is a prime
Assume that n can have more than 1 prime factor and assume some prime q also divides n
Well you get that p|q contradiction.
Too easy for P1 tho
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maths_enthusiast_0001
133 posts
#142 • 1 Y
Y by radian_51
My first NT after decades..... :D
IMO 2023 P1 wrote:
Determine all composite integers $n>1$ that satisfy the following property: if $d_1$, $d_2$, $\ldots$, $d_k$ are all the positive divisors of $n$ with $1 = d_1 < d_2 < \cdots < d_k = n$, then $d_i$ divides $d_{i+1} + d_{i+2}$ for every $1 \leq i \leq k - 2$.
All numbers of the form $p^l$ where $p$ is a prime and $l>1$ work evidently. We now show these are the only ones.
Let $m$ denote the number of divisors of $n$. Then we have $d_{i}d_{m+1-i}=n$. Thus,
$$d_{i} \mid d_{i+1}+d_{i+2}$$$$\implies \frac{n}{d_{m-i+1}} \mid \frac{n}{d_{m-i}}+\frac{n}{d_{m-i-1}}$$$$\implies d_{m-i}d_{m-i-1} \mid d_{m-i+1}(d_{m-i}+d_{m-i-1})$$$$\implies \boxed{d_{i}d_{i+1} \mid d_{i+2}(d_{i}+d_{i+1})}  [\because i \mapsto m-i-1]$$Now $i=1$ gives, $d_{1}d_{2} \mid d_{3}(d_{1}+d_{2}) \implies d_{2} \mid d_{3}$ since $d_{1}=1$. Also, $d_{2}$ is a prime number (say $p$). If $d_{3}$ has any prime factor $q$(say) other than $p$ then $d_{1}<q<d_{3};q \neq d_{2}$ which is a blatant contradiction. Thus $\boxed{d_{3}=p^{2}}$ as it can not be $p^e$ where $e \geq 3$ otherwise $d_{2} < p^{2} < d_{3}$ again, a contradiction. Now by an easy strong induction and similar argument we can conclude that $d_{i+1}=p^{i}$. Thus $n=p^{l}$ where $l>1$ and $p$ is a prime number. $\blacksquare$ ($\mathcal{QED}$)
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Rohit-2006
238 posts
#143 • 1 Y
Y by radian_51
Suppose $n$ has a prime divisors $\geq 2$.

Say the least ones are $p$ and $q$ and $p$ be the minimum.

Let the multiplicity of $p$ be $m$.

At the $(k+1)$-th step, $k \leq m$:

$p^{k-1} < q$, so $q$ can write

$p^{k-1} \mid p^k + q \quad ( \rightarrow \leftarrow)$.

Hence, $n = p^t$, $t \in \mathbb{N}$ and $p$ is prime.
This post has been edited 2 times. Last edited by Rohit-2006, Jan 17, 2025, 10:34 AM
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iStud
268 posts
#144 • 1 Y
Y by radian_51
Note that since $d_{k-2}\mid d_{k-1}+d_k$ and $d_{k-2}\mid n=d_k$, then $d_{k-2}\mid d_{k-1}$. Recall that $d_{k-2}=\frac{n}{d_3}$ and $d_{k-1}=\frac{n}{d_2}$, so $\frac{\frac{n}{d_2}}{\frac{n}{d_3}}\in\mathbb{N}\Longleftrightarrow\frac{d_3}{d_2}\in\mathbb{N}\Longleftrightarrow d_2\mid d_3$. But we also have $d_2\mid d_3+d_4$, so we must have $d_2\mid d_4\Longleftrightarrow\frac{\frac{n}{d_2}}{\frac{n}{d_4}}\in\mathbb{N}\Longleftrightarrow\frac{d_{k-1}}{d_{k-3}}\in\mathbb{N}\Longleftrightarrow d_{k-3}\mid d_{k-1}$. Again, we know that $d_{k-3}\mid d_{k-2}+d_{k-1}$ so we must have $d_{k-3}\mid d_{k-2}$. Repeating the process, eventually we'll end with $1=d_1\mid d_2\mid d_3\mid\dots\mid d_{k-1}\mid d_k=n$. Notice that for any natural number $n$, the three largest divisors of $n$ are $n,\frac{n}{p},\frac{n}{q}$ for prime $2\le p<q$ or $n,\frac{n}{p},\frac{n}{p^2}$ for prime $p\ge 2$. If the three largest divisors of $n$ are $n,\frac{n}{p},\frac{n}{q}$, then $\frac{\frac{n}{p}}{\frac{n}{q}}\in\mathbb{N}\Longleftrightarrow\frac{q}{p}\in\mathbb{Z}\Longleftrightarrow p\mid q$, contradiction. So the three largest divisors of $n$ are $n,\frac{n}{p},\frac{n}{p^2}$ for prime $p\ge 2$. In the similar spirit, we end up showing that $n=p_{k-1}$, which indeed works for any prime $p\ge 2$. Done. $\blacksquare$
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cubres
119 posts
#145
Y by
Storage - grinding IMO problems
This post has been edited 1 time. Last edited by cubres, Feb 3, 2025, 9:44 PM
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megahertz13
3183 posts
#146
Y by
The answer is $n=p^k$ for a prime number $p$ and a positive integer $k$. These clearly work, and we prove that all other $n$ fail. Assume the contrary, and let the smallest prime factors of $n$ be $a$ and $b$.

Case 1: $a < b < a^2$. Then we have the smallest three divisors are $1, a, b$, so the largest three are $\frac{n}{b}, \frac{n}{a}, n$. However, we need $$\frac{n+\frac{n}{a}}{\frac{n}{b}}=\frac{1+\frac{1}{a}}{\frac{1}{b}}=b+\frac{b}{a}$$to be an integer, but $b/a$ is clearly not an integer. There are no cases here.

Case 2: $a<a^2<\dots<a^k<b$ for $k>1$. We know that $$\frac{b+a^k}{a^{k-1}}=a+\frac{b}{a^{k-1}}$$must be an integer, but this is clearly impossible.

This concludes the problem.

Note: I almost messed up and disregarded Case 1.
This post has been edited 1 time. Last edited by megahertz13, Feb 11, 2025, 10:46 PM
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Ilikeminecraft
616 posts
#147
Y by
I claim $n$ is a perfect power. Assume not. Let $p < q$ be the smallest two prime divisors of $n.$ Let $\frac nq,\frac n{p^c}$ be the two consecutive divisors. Hence, we get $\frac nq \mid \frac n{p^c} + \frac n{p^{c - 1}},$ so $p\mid p + q,$ or $p \mid q.$ All perfect powers work.
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ray66
35 posts
#148
Y by
Let $p$ and $q$ be the smallest two prime factors of $n$. Then $\frac{n}{q} | n\frac{1+p}{p}$ so $\frac{q(p+1)}{p} \in \mathbb{Z}$. But that means $p | q$, impossible. So $n=p^k$ and only has 1 distinct prime factor. All $n$ of this form work for $k \ge 2$.
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Jupiterballs
47 posts
#149
Y by
Let the smallest prime dividing $n$ be $p$
Then $d_{k-1} = \frac{n}{p}$
Which gives that $d_{k-2} \mid d_{k-1} + d_{k} = \frac{n}{p} + n = \frac{n}{p}(p+1)$

Now, we form cases:-
Case-1) $d_{k-2} \mid \frac{n}{p}$
implying that $d_{k-2} = \frac{n}{p^2}$ (easy to see why)

Case-2) $v_p (d_{k-2}) = v_p (n)$, or $d_{k-2}$ does not divide $n$
Which means that we need another prime $p$
Then this implies that $gcd(p, p+1) = p$

Which is absurd, therefore $d_{k-2} = \frac{n}{p^2}$
Reiterating this gives us that $d_1 = \frac{n}{p^{k-1}}$
or $n = p^{k-1}$, which works clearly
q.e.d
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anudeep
172 posts
#150
Y by
We claim $n$ of the form $p^{\alpha}$ with $p$ being a prime and $\alpha\in\mathbb{Z}_{\ge 2}$ are the only solutions (easy to check why these work).
Now we shall show that these are the the only solutions. Suppose this is not true. Let $p,q$ be the smallest two distinct prime divisors of $n$ with $p<q$. Let $j$ be the largest integer such that $p^j\lvert n$ and $p^j<q$ (assume $j>1$ as the case $n=pq$ is trivial). One may notice that $d_j=p^{j-1}$, $d_{j+1}=p^j$ and $d_{j+2}=q$. By the given condition it must be that,
$$p^{j-1}\lvert (p^j+q).$$From the above we say $p^{j-1}\lvert q$, which is ridiculous and hence the claim. $\square$
This post has been edited 2 times. Last edited by anudeep, Apr 20, 2025, 6:30 AM
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Maximilian113
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\begin{align*}
d_{k-2} | d_{k-1}+d_k, d_{k-2} | d_k \implies d_{k-2} | d_{k-1} \implies d_2 | d_3. \\
d_2 | d_3, d_2 | d_3+d_4 \implies d_2 | d_4 \implies d_{k-3} | d_{k-1}. \\
d_{k-3} | d_{k-2}+d_{k-1} \implies d_{k-3} | d_{k-2}.
\end{align*}Hence applying this argument in general yields $d_{x} | d_{x+1} \implies d_{x-1} | d_x.$ So $d_1 | d_2 | d_3 | \cdots | d_{k-1} | d_k.$ So if some prime $p | n$ where $p$ is minimal we have $d_1=1, d_2=p \implies p | d_i \, \forall i \geq 2.$ So there is no other prime $q \neq p$ dividing $n.$

Hence $n=p^k$ for some prime $p$ and positive integer $k > 1.$ It is easy to see that this works.
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