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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
\sqrt{a^2+b^2+2}+\sqrt{b^2+c^2+2 }+\sqrt{c^2+a^2+2}\ge 6
parmenides51   19
N an hour ago by NicoN9
Source: JBMO Shortlist 2017 A1
Let $a, b, c$ be positive real numbers such that $a + b + c + ab + bc + ca + abc = 7$. Prove
that $\sqrt{a^2 + b^2 + 2 }+\sqrt{b^2 + c^2 + 2 }+\sqrt{c^2 + a^2 + 2 } \ge 6$ .
19 replies
parmenides51
Jul 25, 2018
NicoN9
an hour ago
Inspired by Austria 2025
sqing   1
N an hour ago by sqing
Source: Own
Let $ a,b\geq 0 ,a,b\neq 1$ and $  a^2+b^2=1. $ Prove that$$   (a + b ) \left( \frac{a}{(b -1)^2} + \frac{b}{(a - 1)^2} \right) \geq 12+8\sqrt 2$$
1 reply
sqing
an hour ago
sqing
an hour ago
IMO Genre Predictions
ohiorizzler1434   51
N an hour ago by ethan2011
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
51 replies
1 viewing
ohiorizzler1434
May 3, 2025
ethan2011
an hour ago
Concurrency from isogonal Mittenpunkt configuration
MarkBcc168   17
N 3 hours ago by Ilikeminecraft
Source: Fake USAMO 2020 P3
Let $\triangle ABC$ be a scalene triangle with circumcenter $O$, incenter $I$, and incircle $\omega$. Let $\omega$ touch the sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ at points $D$, $E$, and $F$ respectively. Let $T$ be the projection of $D$ to $\overline{EF}$. The line $AT$ intersects the circumcircle of $\triangle ABC$ again at point $X\ne A$. The circumcircles of $\triangle AEX$ and $\triangle AFX$ intersect $\omega$ again at points $P\ne E$ and $Q\ne F$ respectively. Prove that the lines $EQ$, $FP$, and $OI$ are concurrent.

Proposed by MarkBcc168.
17 replies
MarkBcc168
Apr 28, 2020
Ilikeminecraft
3 hours ago
anyone who can help me this 2 problems?
auroracliang   2
N 3 hours ago 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
3 hours ago
What conic section is this? Is this even a conic section?
invincibleee   2
N 3 hours ago 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
3 hours ago
Spheres, ellipses, and cones
ReticulatedPython   0
3 hours ago
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
3 hours ago
0 replies
Looking for users and developers
derekli   13
N 3 hours ago 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
3 hours ago
trigonometric functions
VivaanKam   12
N 4 hours ago 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
4 hours ago
find number of elements in H
Darealzolt   1
N Yesterday at 6:47 PM by alexheinis
If \( H \) is the set of positive real solutions to the system
\[
x^3 + y^3 + z^3 = x + y + z
\]\[
x^2 + y^2 + z^2 = xyz
\]then find the number of elements in \( H \).
1 reply
Darealzolt
Yesterday at 1:50 AM
alexheinis
Yesterday at 6:47 PM
primes and perfect squares
Bummer12345   0
Yesterday at 5:08 PM
If $p$ and $q$ are primes, then can $2^p + 5^q + pq$ be a perfect square?
0 replies
Bummer12345
Yesterday at 5:08 PM
0 replies
simple trapezoid
gggzul   0
Yesterday at 4:44 PM
Let $ABCD$ be a trapezoid. By $x$ we denote the angle bisector of angle $X$ . Let $P=a\cap c$ and $Q=b\cap d$. Prove that $ABPQ$ is cyclic.
0 replies
gggzul
Yesterday at 4:44 PM
0 replies
geometry
JetFire008   0
Yesterday at 4:14 PM
Given four concyclic points. For each subset of three points take the incenter. Show that the four incentres form a rectangle.
0 replies
JetFire008
Yesterday at 4:14 PM
0 replies
Inequalities
sqing   11
N Yesterday at 3:02 PM by sqing
Let $a,b,c> 0$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1.$ Prove that
$$  (1-abc) (1-a)(1-b)(1-c)  \ge 208 $$$$ (1+abc) (1-a)(1-b)(1-c)  \le -224 $$$$(1+a^2b^2c^2) (1-a)(1-b)(1-c)  \le -5840 $$
11 replies
1 viewing
sqing
Jul 12, 2024
sqing
Yesterday at 3:02 PM
Integer
Fang-jh   13
N Apr 9, 2025 by alexanderhamilton124
Source: Chinese TST
Let $ n > 1$ be an integer, and $ n$ can divide $ 2^{\phi(n)} + 3^{\phi(n)} + \cdots + n^{\phi(n)},$ let $ p_{1},p_{2},\cdots,p_{k}$ be all distinct prime divisors of $ n$. Show that $ \frac {1}{p_{1}} + \frac {1}{p_{2}} + \cdots + \frac {1}{p_{k}} + \frac {1}{p_{1}p_{2}\cdots p_{k}}$ is an integer. ( where $ \phi(n)$ is defined as the number of positive integers $ \leq n$ that are relatively prime to $ n$.)
13 replies
Fang-jh
Apr 5, 2008
alexanderhamilton124
Apr 9, 2025
Source: Chinese TST
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Fang-jh
237 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ n > 1$ be an integer, and $ n$ can divide $ 2^{\phi(n)} + 3^{\phi(n)} + \cdots + n^{\phi(n)},$ let $ p_{1},p_{2},\cdots,p_{k}$ be all distinct prime divisors of $ n$. Show that $ \frac {1}{p_{1}} + \frac {1}{p_{2}} + \cdots + \frac {1}{p_{k}} + \frac {1}{p_{1}p_{2}\cdots p_{k}}$ is an integer. ( where $ \phi(n)$ is defined as the number of positive integers $ \leq n$ that are relatively prime to $ n$.)
This post has been edited 4 times. Last edited by Fang-jh, Apr 5, 2008, 10:00 AM
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Rust
5049 posts
#2 • 2 Y
Y by Adventure10, Mango247
Let $ n=3$, then $ \phi (n)=2$ and $ 2^2+3^2=13=p_1$ (k=1), therefoe $ \frac{1}{p_1}+\frac{1}{p_1}=\frac{2}{13}$ is not integer. :lol:
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zhaobin
2382 posts
#3 • 3 Y
Y by Adventure10, Derpy_Creeper, Mango247
Hello,But there is a condition:$ n$ can divide $ 2^{\phi(n)} + 3^{\phi(n)} + \cdots + n^{\phi(n)},$.
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TTsphn
1313 posts
#4 • 4 Y
Y by Adventure10, Mango247, Mango247, Mango247
Here is solution .
Let $ n = \prod _{i = 1}^r p_i^{a_i}$
We have property :
1.$ \varphi (p_i^{a_i})|\varphi (n)$ .
2. $ \varphi (n)\geq a_i,\forall i = 1...r$
Now consider problem :
$ T = \sum_{k = 2}^n k^{\varphi (n)} \equiv (\sum_{{j = 1}_{\gcd {j,p}} = 1}^nj) - 1(mod p_i^{a_i})$
We has $ n - \frac {n}{p}$ j such that $ \gcd(j,p) = 1$ so
$ T\equiv n - \frac {n}{p} - 1$
$ \Leftrightarrow \frac {n}{p} + 1 \equiv 0(\mod p^{a_i})$
It follow that $ a_i = 1,\forall i = 1,..,r$ and $ p_i|\frac {n}{p_i} + 1$
So $ \frac {1}{p_1} + ... + \frac {1}{p_1....p_r}$ is an integer.
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Erken
1363 posts
#5 • 2 Y
Y by Adventure10, Mango247
It is well-known that $ \phi(n)=n\left(1-\frac{1}{p_1}\right)\dots\left(1-\frac{1}{p_k}\right)$,

hence for all $ 1\leq i\leq k$,we have that $ p_i-1|\phi(n)$.
And there are exactly $ \left[\frac{n}{p_i}\right]$ numbers among $ 2,3\dots,n$,that are divisible by $ p_i$.
Also if $ a$ is not divsible by $ p_i$,then $ a^{\phi(n)}\equiv 1$ mod $ p$.So
$ S=2^{\phi(n)} + 3^{\phi(n)} + \cdots + n^{\phi(n)}\equiv n-1-\left[\frac{n}{p_i}\right]$.
But as we know $ p_i|n$ and $ n|S$,hence $ p_i|n-1-\left[\frac{n}{p_i}\right]$,or
$ \boxed{\left[\frac{n}{p_i}\right]\equiv -1}$ mod $ p_i$.
From boxed relation we conclude that $ p_i^2\not{|} n$.But this means that $ n=p_1p_2\dots p_k$,and
$ \prod_{j\neq i}p_j\equiv -1$ mod $ p_i$.
Since $ (p_i,p_j)=1$ for all $ i\neq j$,it is enough to show that

$ \prod_{j\neq i}p_j+\sum_{k\neq i}\left(\prod_{j\neq k}p_j\right)+1$ is divisible by $ p_i$,

which is clearly true,because as we've observed

$ \prod_{j\neq i}p_j\equiv -1$ mod $ p_i$
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cnyd
394 posts
#6 • 1 Y
Y by Adventure10
Here is my solution;

$ n = p_{1}^{\alpha_{1}}...p_{i}^{\alpha{i}}...p_{k}^{\alpha{k}}$

$ p_{i}|\sum_{j = 2}^{n} j^{\phi{n}}$

$ \implies$ $ \sum_{j = 2}^{n} j^{\phi{n}}\equiv n - 1 - p_{1}^{\alpha_{1}}.p_{i}^{\alpha_{i} - 1}...p_{k}^{\alpha{k}}$

if $ \alpha_{i}\geq 2$ $ \implies$ $ p_{i}|n - 1,p_{i}|n$ $ \implies$ $ p_{i}|1$ Contradiction!

$ \implies$ $ p_{i}||n$

$ \implies$ $ n = p_{1}...p_{k}$

and $ \frac {n}{p_{i}}\equiv - 1\pmod{p_{i}}$

$ \implies$ $ (\sum_{i = 1}^{k} \frac {1}{p_{i}}) + \frac {1}{p_{1}..p_{k}}$

$ p_{i}|\frac {n}{p_{i}} + 1$ $ \implies$ $ (\sum_{i = 1}^{k} \frac {1}{p_{i}}) + \frac {1}{p_{1}..p_{k}}\in\mathbb{Z}$
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subham1729
1479 posts
#7 • 2 Y
Y by Adventure10, Mango247
Suppose set of prime divisors of $n$ are $\{p_1,p_2,.....,p_k\}$.According to the given condition $p_i|n-1-\frac{n}{p_i}$ for all $1\leq i\leq k$.And so $p_i$ can divide $\frac{n}{p_i}$ for all $i$ and so $n$ must be a square free.So $p_i|1+\frac{n}{p_i}$ for all $i$ and now just multiplying those we're done.
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AnonymousBunny
339 posts
#8 • 2 Y
Y by Adventure10, Mango247
A bit too trivial for China TST.


Note that
\[\varphi (n) = \dfrac{n}{\displaystyle \prod_{i=1}^{k} p_i} (p_i-1) \equiv 0 \pmod{p_i-1} \quad \forall \ 1 \leq i \leq k.\]
Hence, we have that $a^{\phi (n)} \equiv 1 \pmod{p}$ for all $p \mid n.$ Then, note that
\begin{align*}
\displaystyle \sum_{j=1}^{k} j^{\varphi (n)} & = \displaystyle \sum_{\substack{2 \leq j \leq k \\ \gcd (j, p) = 1}} j^{\varphi (n)} + \displaystyle \sum_{\substack{2 \leq j \leq k \\ p \mid j}} j^{\varphi (n)} \\
& \equiv \displaystyle \sum_{\substack{2 \leq j \leq k \\ \gcd (j, p) = 1}} j^{\varphi (n)} \pmod{p }\\
& \equiv n - 1 - \dfrac{n}{p} \equiv -\dfrac{n}{p}-1 \pmod{p}. \end{align*}
Hence, we need $\dfrac{n}{p} + 1 \equiv 0 \pmod{p} \implies p \nmid \dfrac{n}{p}.$ Consequently, $v_p (n) \leq 1$ for all primes $p,$ so $n = p_1 p_2 \cdots p_k.$ Now, plugging $p_1, p_2, \cdots , p_k$ and multiplying these
\[{n= p_1 p_2 \cdots p_k \mid \displaystyle \prod_{i=1}^{k}  ( \dfrac{n}{p_i} + 1 ) 
\\ \implies p_1 p_2 \cdots p_k \mid \displaystyle \sum_{i=1}^{k} \displaystyle \prod_{\substack{1 \leq j \leq k \\ j \neq k}} p_i + 1 \ ^{(*)}\\
\implies  \displaystyle \sum_{i=1}^{k} ( \displaystyle \prod_{\substack{1 \leq j \leq k \\ j \neq k}} p_i ) \cdot \dfrac{1}{p_1 p_2 \cdots p_k} = \displaystyle \sum_{i=1}^{k} p_i^{-1} + ( \displaystyle \prod_{i =1}^{k} p_i} )^{-1} \in \mathbb{Z}. \quad \blacksquare\]

$^{(*)}$ This is because all other terms in the expansion of $\displaystyle \prod ( \dfrac{n}{p_i}+1 )$ are of the form $p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$ where the $a_i$'s are all positive integers.
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anantmudgal09
1980 posts
#9 • 2 Y
Y by Adventure10, Mango247
Indeed, too easy for China,
Anyway noting that $\phi(p^{\alpha i}) \mid \phi(n)$ we conclude that $\frac{n}{p_i}+1\equiv 0 \bmod p^{\alpha_ i}$ which kills it.
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Sprites
478 posts
#10
Y by
Since $\phi({p_i}^{\alpha_i})|\phi(N)$ for $N=\prod_{p_i}^{a_i}$ so $\frac{n}{p_i} \equiv -1 \pmod p_i$
So $p_i^2 \nmid n$ and hence $n = p_1 p_2 \cdots p_k$ and we are done since $ {n= p_1 p_2 \cdots p_k \mid \displaystyle \prod_{i=1}^{k} ( \dfrac{n}{p_i} + 1 )
\\ \implies p_1 p_2 \cdots p_k \mid \displaystyle \sum_{i=1}^{k} \displaystyle \prod_{\substack{1 \leq j \leq k \\ j \neq k}} p_i + 1 \ \\
\implies \displaystyle \sum_{i=1}^{k} ( \displaystyle \prod_{\substack{1 \leq j \leq k \\ j \neq k}} p_i ) \cdot \dfrac{1}{p_1 p_2 \cdots p_k} = \displaystyle \sum_{i=1}^{k} p_i^{-1} + ( \displaystyle \prod_{i =1}^{k} p_i} )^{-1} \in \mathbb{Z}. \quad $
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HamstPan38825
8859 posts
#11
Y by
This is just an easy grind.
Note that for each $p_i$ and $a$, either $p_i \mid a^{\phi(n)}$ or $p_i - 1 \mid a^{\phi(n)}$. Thus, the sum in the expression is congruent to the number of $a$ which are not multiples of $p_i$, which is equal to $n-1-\frac n{p_i}$ modulo $p_i$.

Thus $p_i \mid \frac n{p_i} + 1$, which implies that $n$ is squarefree, and also $$p_1p_2\cdots p_k \mid \sum_{i=1}^k \frac n{p_i} + 1,$$which implies the desired.
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Maximilian113
575 posts
#12
Y by
It suffices to show that for each prime $p_i|n,$ we have that $p_i | \prod_{1 \leq j \leq k, j \neq i} p_j + 1.$ By the condition, it follows that $\sum_{j=2}^{n} j^{\phi(n)} \equiv 0 \pmod{p_i}.$ Clearly for $j$ relatively prime to $p_i,$ $j^{\phi(n)} \equiv 1,$ while otherwise we get zero. Therefore, we have that $$n-1-\frac{n}{p_i} \equiv 0 \pmod{p_i}.$$If $p_i | \frac{n}{p_i},$ this yields $-1 \equiv 0,$ which is absurd. Therefore, $n$ is squarefree for all $p_i$ so $n=p_1p_2 \cdots p_k,$ and substituting yields the desired result. QED
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megarnie
5604 posts
#13
Y by
Claim: For any prime $p \mid n$, we have $n \equiv -p \pmod{p^2}$.
Proof: Fix some prime $p\mid n$. Note that $p - 1 \mid \phi(n)$, so $2^{\phi(n) } + 3^{\phi(n)} + \cdots + n^{\phi(n)}$ in modulo $p$ is just the number of positive integers in $[2,n]$ not divisible by $p$, which is $n - \frac np - 1$. This must be divisible by $p$, so $p\mid \frac np + 1 $, so $\frac np \equiv -1 \pmod p$. If we let $\frac np = pk - 1$, we have $n = p^2 k - p$, as desired. $\square$

This for one implies that $n$ is squarefree, so $n = p_1 p_2 \cdots p_k$. Now let $S$ be \[ \frac {1}{p_{1}} + \frac {1}{p_{2}} + \cdots + \frac {1}{p_{k}} + \frac {1}{p_{1}p_{2}\cdots p_{k}}\]
Suppose it wasn't an integer. Let $p$ be a prime so that $\nu_p(S) < 0$. This obviously means $p \in \{p_1, p_2, \ldots, p_k\}$. WLOG $p = p_1$. Then note that $\nu_p \left( S - \frac{1}{p_1} - \frac{1}{p_1 \cdot  p_2 \cdots p_k} \right) = \nu_p  \left( \frac{1}{p_2} + \frac{1}{p_3} + \cdots + \frac{1}{p_k} \right) 
 = 0$. Thus it suffices to show that $\nu_p \left( \frac{1}{p_1} + \frac{1}{p_1  \cdot p_2 \cdots p_k} \right) \ge 0$, which is equivalent to \[ \nu_p \left( \frac{1}{p} + \frac 1n \right) \ge 0\]
Note that $\frac 1p + \frac 1n = \frac{p + n}{p^2}$. Since $p^2 \mid p + n$, $\nu_p(p + n) \ge 2$, so \[\nu_p \left( \frac 1p + \frac 1n \right)  = \nu_p(p + n) - \nu_p(p^2) \ge 0,\]as desired.
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alexanderhamilton124
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#14 • 1 Y
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To prove that
\[
\frac{1}{p_1} + \frac{1}{p_2} + \cdots + \frac{1}{p_k} + \frac{1}{p_1 p_2 \cdots p_k}
\]is an integer, we can show that
\[
v_{p_i}\left(\frac{1}{p_1} + \frac{1}{p_2} + \cdots + \frac{1}{p_k} + \frac{1}{p_1 p_2 \cdots p_k}\right) \geq 0
\]for all \(1 \leq i \leq k\).

Note that
\[
1^{\phi(n)} + \cdots + n^{\phi(n)} = \frac{n}{p_i}(p_i - 1) \pmod{p_i} \implies -1 \equiv \frac{n}{p_i}(p_i - 1)
\]since
\[
2^{\phi(n)} + \cdots + n^{\phi(n)} \equiv 0 \pmod{p_i}.
\]This also means \(n\) is squarefree, and
\[
\frac{n}{p_i} \equiv -1 \pmod{p}.
\]
Note that
\[
v_{p_i}\left(\frac{1}{p_1} + \frac{1}{p_2} + \cdots + \frac{1}{p_k} + \frac{1}{p_1 p_2 \cdots p_k}\right) = 
v_{p_i}\left(\left(\frac{1}{p_i} + \frac{1}{p_1 \cdots p_k}\right) + \frac{1}{p_2} + \cdots\right).
\]
Note that
\[
v_{p_i}\left(\frac{1}{p_2} + \cdots\right) \geq 0
\]since no denominator is divisible by \(p_i\). Meanwhile,
\[
\frac{1}{p_i} + \frac{1}{p_1 \cdots p_k} = \frac{1}{p_i} + \frac{1}{n}, \quad \text{since } n \text{ is squarefree,}
\]which is equal to
\[
\frac{1}{p_i} \left(1 + \frac{p_i}{n}\right).
\]Note that
\[
\frac{p_i}{n} \equiv -1 \pmod{p_i},
\]so
\[
p_i \mid \left(1 + \frac{p_i}{n}\right),
\]hence
\[
v_{p_i}\left(\frac{1}{p_i} + \frac{1}{p_1 \cdots p_k}\right) \geq 0,
\]which means the \(v_{p_i}\) of the overall sum must be greater than or equal to 0 as well, so we're done.
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