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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
inequality thing
BinariouslyRandom   0
4 minutes ago
Source: Philippine MO 2025 P5
Find the largest real constant $k$ for which the inequality \[ (a^2+3)(b^2+3)(c^2+3)(d^2+3) + k(a-1)(b-1)(c-1)(d-1) \ge 0 \]holds for all real numbers $a$, $b$, $c$, and $d$.
0 replies
BinariouslyRandom
4 minutes ago
0 replies
Sharygin 2025 CR P12
Gengar_in_Galar   8
N 17 minutes ago by Kappa_Beta_725
Source: Sharygin 2025
Circles $\omega_{1}$ and $\omega_{2}$ are given. Let $M$ be the midpoint of the segment joining their centers, $X$, $Y$ be arbitrary points on $\omega_{1}$, $\omega_{2}$ respectively such that $MX=MY$. Find the locus of the midpoints of segments $XY$.
Proposed by: L Shatunov
8 replies
1 viewing
Gengar_in_Galar
Mar 10, 2025
Kappa_Beta_725
17 minutes ago
Sharygin 2025 CR P17
Gengar_in_Galar   6
N 21 minutes ago by Kappa_Beta_725
Source: Sharygin 2025
Let $O$, $I$ be the circumcenter and the incenter of an acute-angled scalene triangle $ABC$; $D$, $E$, $F$ be the touching points of its excircle with the side $BC$ and the extensions of $AC$, $AB$ respectively. Prove that if the orthocenter of the triangle $DEF$ lies on the circumcircle of $ABC$, then it is symmetric to the midpoint of the arc $BC$ with respect to $OI$.
Proposed by: P.Puchkov,E.Utkin
6 replies
Gengar_in_Galar
Mar 10, 2025
Kappa_Beta_725
21 minutes ago
Sharygin 2025 CR P21
Gengar_in_Galar   4
N 29 minutes ago by Kappa_Beta_725
Source: Sharygin 2025
Let $P$ be a point inside a quadrilateral $ABCD$ such that $\angle APB+\angle CPD=180^{\circ}$. Points $P_{a}$, $P_{b}$, $P_{c},$ $P_{d}$ are isogonally conjugated to $P$ with respect to the triangles $BCD$, $CDA$, $DAB$, $ABC$ respectively. Prove that the diagonals of the quadrilaterals $ABCD$ and $P_{a}P_{b}P_{c}P_{d}$ concur.
Proposed by: G.Galyapin
4 replies
Gengar_in_Galar
Mar 10, 2025
Kappa_Beta_725
29 minutes ago
helpppppppp me
stupid_boiii   1
N 2 hours ago by vanstraelen
Given triangle ABC. The tangent at ? to the circumcircle(ABC) intersects line BC at point T. Points D,E satisfy AD=BD, AE=CE, and ∠CBD=∠BCE<90 ∘ . Prove that D,E,T are collinear.
1 reply
stupid_boiii
Yesterday at 4:22 AM
vanstraelen
2 hours ago
Algebra Polynomials
Foxellar   2
N 3 hours ago by Foxellar
The real root of the polynomial \( p(x) = 8x^3 - 3x^2 - 3x - 1 \) can be written in the form
\[
\frac{\sqrt[3]{a} + \sqrt[3]{b} + 1}{c},
\]where \( a, b, \) and \( c \) are positive integers. Find the value of \( a + b + c \).
2 replies
Foxellar
4 hours ago
Foxellar
3 hours ago
geometry
luckvoltia.112   0
Today at 2:29 AM
Let \( \triangle ABC \) be an acute triangle with \( AB < AC \), and its vertices lie on the circle \( (O) \). Let \( AD \) be the altitude from vertex \( A \). Let \( E \) and \( F \) be the feet of the perpendiculars from \( D \) to the lines \( AB \) and \( AC \), respectively. Let \( EF \) intersect the circle \( (O) \) again at points \( P \) and \( Q \) such that \( E \) lies between \( Q \) and \( F \). Let the lines \( AD \) and \( EF \) intersect at point \( G \). Let \( I \) be the midpoint of segment \( AD \). Let \( AO \) intersect line \( BC \) at point \( K \).
a) Prove that \( AP = AQ = AD \).
b) Prove that line \( OI \) is parallel to line \( KG \).
c)Let \( H \) be the orthocenter of triangle \( ABC \), and let \( M \) be the midpoint of segment \( BC \). $S$ is the center (HBC). Let point \( T \) lie on line \( DS \) such that ray \( KD \) is the angle bisector of \( \angle GKT \). Prove that lines \( AD \) and \( MT \) intersect at a point lying on circle \( (O) \).
0 replies
luckvoltia.112
Today at 2:29 AM
0 replies
Great Geometry with Squares on sides of triangles
SomeonecoolLovesMaths   2
N Today at 1:50 AM by happypi31415
Three squares are drawn on the sides of triangle \(ABC\) (i.e., the square on \(AB\) has \(AB\) as one of its sides and lies outside \(ABC\)). Show that the lines drawn from the vertices \(A\), \(B\), and \(C\) to the centers of the opposite squares are concurrent.

IMAGE
2 replies
SomeonecoolLovesMaths
Yesterday at 9:44 PM
happypi31415
Today at 1:50 AM
A suspcious assumption
NamelyOrange   2
N Today at 1:30 AM by maromex
Let $a,b,c,d$ be positive integers. Maximize $\max(a,b,c,d)$ if $a+b+c+d=a^2-b^2+c^2-d^2=2012$.
2 replies
NamelyOrange
Yesterday at 1:53 PM
maromex
Today at 1:30 AM
n is divisible by 5
spiralman   1
N Yesterday at 8:42 PM by KSH31415
n is an integer. There are n integers such that they are larger or equal to 1, and less or equal to 6. Sum of them is larger or equal to 4n, while sum of their square is less or equal to 22n. Prove n is divisible by 5.
1 reply
spiralman
Wednesday at 7:38 PM
KSH31415
Yesterday at 8:42 PM
Monochromatic Triangle
FireBreathers   1
N Yesterday at 8:08 PM by KSH31415
We are given in points in a plane and we connect some of them so that 10n^2 + 1 segments are drawn. We color these segments in 2 colors. Prove that we can find a monochromatic triangle.
1 reply
FireBreathers
Yesterday at 2:28 PM
KSH31415
Yesterday at 8:08 PM
how difficult are these problems
rajukaju   1
N Yesterday at 7:28 PM by Shan3t
I can solve only the first 4 problems of the last general round of the HMMT competition: https://hmmt-archive.s3.amazonaws.com/tournaments/2024/nov/gen/problems.pdf

As a prediction, would this mean I am good enough to qualify for AIME? How does the difficulty compare?

1 reply
rajukaju
Yesterday at 6:43 PM
Shan3t
Yesterday at 7:28 PM
Maximum value of function (with two variables)
Saucepan_man02   1
N Yesterday at 1:39 PM by Saucepan_man02
If $f(\theta) = \min(|2x-7|+|x-4|+|x-2 -\sin \theta|)$, where $x, \theta \in \mathbb R$, then maximum value of $f(\theta)$.
1 reply
Saucepan_man02
Yesterday at 1:25 PM
Saucepan_man02
Yesterday at 1:39 PM
It is given that $M=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{23}=\frac{n}{23!},
Vulch   3
N Yesterday at 11:58 AM by mohabstudent1
It is given that $M=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{23}=\frac{n}{23!},$ where $n$ is a natural number.What is the remainder when $n$ is divided by $13?$
3 replies
Vulch
Apr 9, 2025
mohabstudent1
Yesterday at 11:58 AM
Very easy number theory
darij grinberg   102
N Apr 29, 2025 by ND_
Source: IMO Shortlist 2000, N1, 6th Kolmogorov Cup, 1-8 December 2002, 1st round, 1st league,
Determine all positive integers $ n\geq 2$ that satisfy the following condition: for all $ a$ and $ b$ relatively prime to $ n$ we have \[a \equiv b \pmod n\qquad\text{if and only if}\qquad ab\equiv 1 \pmod n.\]
102 replies
darij grinberg
Aug 6, 2004
ND_
Apr 29, 2025
Very easy number theory
G H J
Source: IMO Shortlist 2000, N1, 6th Kolmogorov Cup, 1-8 December 2002, 1st round, 1st league,
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Fibonacci_11235
45 posts
#92
Y by
Why is the problem equivalent to $(A)$ $x^2 \equiv 1 \pmod n$ for all $\gcd(x,n) = 1$?
We are given that $a \equiv b \pmod n \Leftrightarrow ab \equiv 1 \pmod n$ for $a$ and $b$ relatively prime to $n$.
But we used $a \equiv b \pmod n \Rightarrow ab \equiv 1 \pmod n$ to get $(A)$?
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ryanbear
1056 posts
#93
Y by
Only if case:
From the first $12$ values, get that $2,3,4,6,8$, and $12$ work.
Then, if $n$ does not divide $2$, then $a=b=2$ results in $4 \equiv 1 \pmod n$, which is false.
Similarly, if $n$ does not divide $3$, then $a=b=3$ results in $9 \equiv 1 \pmod n$, which is false.
So $n$ has to divide $6$
Get that $24$ also works but $18$ does not.
If $n \ge 25$, then if $n$ does not divide $5$, then $a=b=5$ results in $25 \equiv 1 \pmod n$, which is false.
So $n$ has to divide $30$.
Get that $30$ does not work and for $n \ge 60$, $a=7$ results in it not working
So $n$ has to divide $420$, but then $a=11$ does not work, so it has to divide $420*11$, and then it repeats, so it does not work since $p_1*p_2*...*p_n > (p_{n+1})^2$ for large $n$ where $p_n$ is the nth prime. This can be proved by induction because it is true for $n=4$ and if $n$ works, then the LHS is multiplied by $p_{n+1}$, but the RHS is multiplied by $(\frac{p_{n+2}}{p_{n+1}})^2 \ge 2^2 = 4$, so the LHS will remain larger. So the only numbers that work are $\boxed{2,3,4,6,8,12,24}$
Testing, all of those satisfy the if case.
This post has been edited 1 time. Last edited by ryanbear, Mar 11, 2024, 6:54 PM
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Cusofay
85 posts
#94
Y by
It is easy to see that the condition is equivalent to $a^2 \equiv 1 \pmod{n}$ for any $a$ with $gcd(a,n)=1$. Set $n=2^xk$ where $k$ is odd. Then $gcd(n,k+2)=1$ thus from the condition $$(k+2)^2\equiv 1\pmod{n} \Rightarrow 4 \equiv 1 \pmod{k} $$Hence $k\in\{1,3\}$ so $5^2\equiv 1 \pmod{n}$. Some case work gives us $n\in \{2,3,4,6,8,12,24  \}$

$$\mathbb{Q.E.D.}$$
This post has been edited 2 times. Last edited by Cusofay, Mar 30, 2024, 12:36 PM
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blueprimes
356 posts
#95
Y by
I think I have a slightly different solution than those presented above.

We claim the answer is $n = 2, 3, 4, 6, 8, 12, 24$. All can be checked to work.

Now first, we will show that $n$ can only have $2$ or $3$ as prime factors. For the sake of contradiction, assume some prime $p_1 \ge 5$ divides $n$, and consider the prime factorization $n = p_1^{e_1} p_2^{e_2} \dots p_k^{e_k}$ where $p_i$ are primes and all $e_i$ are positive integers. Let $g$ be a primitive root of $p_1$. Then by Chinese Remainder Theorem, we can choose some residue $h \pmod{n}$ where
$$h \equiv g \pmod{p_1^{e_1}}, h \equiv 1 \pmod{p_2^{e_2}}, \dots, h \equiv 1 \pmod{p_k^{e_k}}.$$Clearly the order of $h$ in $\pmod{n}$ is $p - 1 > 2$. Then we can choose $a = h, b = h^{p - 2}$ which must be distinct and multiply to $1 \pmod{n}$. Contradiction. Then the only prime factors of $n$ is $2$ or $3$.

Now we can look at $a = 5, b = 5$, and we must have $25 \equiv 1 \pmod{n}$ which implies $n$ must be a factor of $24$ besides $1$. This implies the earlier mentioned solution set, and we are done.
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Krave37
46 posts
#96
Y by
Proving that $a^2 \equiv 1 \pmod{n}$ is enough for both way.

Let us think about the case when $n = p$ where $p$ is a prime.

For a prime $(a+1)(a-1) \equiv 0 \pmod{p}$, this means either $p$ divides $a+1$ or $a-1$ or it can be both.

This we can observe as $a$ is the list of all numbers co prime to $n$, that $p = 2,3$ are the only primes satisfying this.

Now, we can see that for any general $n$, if $n$ has factors other than powers of $2$ and $3$, there is a problem.

Therefore, $n = 2^x.3^y$.
To such a number the smallest coprime will be 5, so $5^2 -1 \equiv 0 \pmod{2^x.3^y}$ holds,

This means that $2^x.3^y\leq 24$

By checking we get the solutions, 2,4,8,6,12,24
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Sedro
5851 posts
#97
Y by
The only $n\ge 2$ that work are those satisfying $n\mid 24$. It is straightforward to check that these $n$ have the desired property; we now prove they are the only possible $n$.

Suppose $n$ has the property that if $\gcd(a,n)=\gcd(b,n)=1$ and $a\equiv b \pmod{n}$, then $ab\equiv 1\pmod{n}$. This is equivalent to saying that all $a$ coprime to $n$ have order $2$ modulo $n$. By CRT, if $p \mid n$, then $a$ must have at most order $2$ modulo $p$. Note that if $p>3$, then $p$ has a primitive root with order greater than $2$, so no prime greater than $3$ can divide $n$. Also observe that by CRT again, if $p^k \mid \mid n$, then $a$ must have at most order $2$ modulo $p$. For an integer $r\ge 4$, we have $3^2 < 2^r$, so $3$ has an order greater than $2$ modulo $2^r$, and for an integer $s\ge 2$, we have $2^2 < 3^s$, so $2$ has an order greater than $2$ modulo $3^r$. Thus, $v_2(n)\le 3$ and $v_3(n)\le 1$, which implies $n\mid 24$. $\blacksquare$
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a_0a
34 posts
#98
Y by
Solution
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mahmudlusenan
24 posts
#99
Y by
just try to prove that $a^2 \equiv 1 \pmod{n}$ and $b^2 \equiv 1 \pmod{n}$
so $n\mid a^2-1$ and $n\mid b^2-1$ for all a,b $(a,n)=(b,n)=1$
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ezpotd
1286 posts
#100
Y by
We only consider $n$ prime power, as all other $n$ must be composed of prime powers that work.

Now observe for odd prime powers greater than $4$, take $a = b = 2$, then $ab = 4$ which is not $1$ mod $n$. This leaves us to check $3$, which obviously works, and even prime powers. Observe $16$ fails since $16 \mid 5 \cdot 13 - 1$, so we check $8$. We check that $a^2 \equiv 1$ mod $8$ for all odd $8$, by injectivity of multiplication we can check the only if part, so $8$ works.

By CRT, if two relatively prime things work, so do their product, and we know that the possible prime power divisors of $n$ are $2,3,4,8$, so the answer is just the divisors of $24$.
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pie854
243 posts
#103
Y by
Note that this implies $a^2\equiv 1\pmod n$ for all $a$ coprime to $n$. Due to primitive roots, it follows that $n=2^k$ or $n=3\cdot 2^k$. It's easy to check such $n$ works when $k\le 3$ and doesn't work when $k>3$.
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Maximilian113
575 posts
#104
Y by
Note that $a \equiv b \implies a^2-1 \equiv 0 \pmod n.$ Therefore, for each prime power divisor of $n,$ $p_i^{e_i},$ $$a \equiv \pm 1 \pmod{p_i^{e_i}}$$if and only if $a$ is relatively prime to $n.$ Therefore, $p_i = 2, 3.$ Testing gives $2, 4, 8$ and $3,$ so the answers are just the divisors of $24.$
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DakshAggarwalRedsurgance
10 posts
#105
Y by
let a no. be good if for all a,b relatively prime to n, we have
a = b(mod n) implies ab = 1 (mod n ).
If some no. satisfies the inverse as well than call it perfect.

Note that some no. is good only if all the factors of n are good.

Claim 1 - If some prime p divides n (a perfect no.) than so do all the primes less than equal to p. for all p > 3
proof- you can use the fact that if p does not divide n that p^2 - 1 must be divisibe by n.
note 5 is not a good no.
so good no. are multilples of 2 and/or 3.
using 5^2 - 1 is divisible by perfect no. we are done.
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Turtwig113
128 posts
#106
Y by
Take any prime $p \geq 5$ dividing $n$. Then $ab \equiv 1 \pmod{p}$ must be true if $a \equiv b \pmod{p}$, and $a, b$ are relatively prime to $p$. Now by Chinese Remainder Theorem we get for any $0 < k < p$ that there exist $a, b \equiv k \pmod{p}$ that are relatively prime to $n$, yielding $p=2$ or $p=3$. Therefore $n=2^x3^y$ for some $x, y$, and by plugging $a=5, b=5$ we get $2^m3^n \mid 24$. Checking all the divisors of 24 yields that all of them work as possible $n$, therefore the answers are $\boxed{1, 2, 3, 4, 6, 8, 12, 24}$.
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sharknavy75
703 posts
#107
Y by
If $n \in \mathbb{P}$ and $n \not \equiv 0 \pmod 2$ and $n \not \equiv 0 \pmod 3,$ then $2\cdot \frac{n+1}{2} \equiv 1 \pmod n.$ By Chinese Remainder Theorem, $n = 2^x3^y$ where $x, y \in \mathbb{N}_0.$ We can say $n \le 24$ by using $a \equiv b \equiv 5 \pmod n.$ We can easily verify the solutions $n$ satisfy the condition $n \not = 1$ and $n \mid 24.$
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ND_
53 posts
#108
Y by
We get that $a^2 \equiv 1~ (mod~p)$ for every prime $p$ dividing $n$ and $gcd(a,n)=1$. Let $a$ be a primitive root of $p$, then $2 \mid p-1 = ord_p(a) \Rightarrow n=2^l3^m$.
Since $5 \nmid n$, $5^2 \equiv 1$ (mod n) $\Rightarrow n \mid 24$
$n=24$ works, so the answer is $n=2,3,8,3,6,12,24$
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