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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
2-var inequality
sqing   8
N a few seconds ago by sqing
Source: Own
Let $ a,b\geq 0    $. Prove that
$$ \frac{a }{a^2+2b^2+1}+ \frac{b }{b^2+2a^2+1}\leq \frac{1}{\sqrt{3}} $$$$   \frac{a }{2a^2+ b^2+2ab+1}+ \frac{b }{2b^2+ a^2+2ab+1}  \leq \frac{1}{\sqrt{5}} $$$$ \frac{a }{2a^2+ b^2+ ab+1}+ \frac{b }{2b^2+ a^2+ ab+1} \leq \frac{1}{2} $$$$\frac{a }{a^2+2b^2+2ab+1}+ \frac{b }{b^2+2a^2+2ab+1}\leq \frac{1}{2} $$
8 replies
1 viewing
sqing
Today at 1:39 AM
sqing
a few seconds ago
Number Theory
fasttrust_12-mn   7
N 3 minutes ago by KTYC
Source: Pan African Mathematics Olympiad P1
Find all positive intgers $a,b$ and $c$ such that $\frac{a+b}{a+c}=\frac{b+c}{b+a}$ and $ab+bc+ca$ is a prime number
7 replies
fasttrust_12-mn
Aug 15, 2024
KTYC
3 minutes ago
Solution
KTYC   0
4 minutes ago





(a + b)/(a + c) = (b + c)/(b + a)


Cross-multiply,

(a + b)² = (a + c)(b + c)

(a + b)² = ab + bc + ac + c²

ab + bc + ac = (a + b)² - c²

ab + bc + ac = (a + b - c)(a + b + c)
—-Eqn(1)

—————————-

Since a, b, c ∈ ℕ,

(a + b + c) ∈ ℕ

(ab + bc + ac) ∈ ℕ


From Eqn(1),

(a + b - c) ∈ ℕ

—————————-

Since (ab + bc + ac) is prime,

The only positive factors of
(ab + bc + ac) are
1 and (ab + bc + ac)


Since c ∈ ℕ,

(a + b - c) < (a + b + c)


Hence,


a + b - c = 1
—-Eqn(2)

a + b + c = ab + bc + ac
—-Eqn(3)


From Eqn(2),

a + b = c + 1
—-Eqn(4)


From Eqn(3),

a + b + c = c(a + b) + ac
—-Eqn(5)


Sub. Eqn(4) into Eqn(5):

c + 1 + c = c(c + 1) + ac

2c + 1 = c² + c + ac

c² + c + ac - 2c - 1 = 0

c² + ac - c - 1 = 0

c² + c(a - 1) - 1 = 0
—-Eqn(6)

——————————

Eqn(6) is a quadratic equation about c


Hence,

Discriminant of Eqn(6)
= (a - 1)² - 4(1)(-1)
= (a - 1)² + 4


Since c ∈ ℕ,
Discriminant of Eqn(6) is a perfect square


Hence,

[(a - 1)² + 4] is a perfect square


Let
(a - 1)² + 4 = k²
where k ∈ ℕ


Hence,

k² - (a - 1)² = 4

[k - (a - 1)][k + (a - 1)] = 4

(k - a + 1)(k + a - 1) = 4
—-Eqn(7)

——————————

Since a, k ∈ ℕ,

a ≥ 1

k ≥ 1


Hence,

(k + a - 1) ≥ (1 + 1 - 1)

(k + a - 1) ≥ 1


Hence,

(k + a - 1) ∈ ℕ


4 ∈ ℕ


Hence,

From Eqn(7),

(k - a + 1) ∈ ℕ


Hence,

4 is a factor of two positive integers in Eqn(7)

——————————

(k - a + 1) + (k + a - 1)
= 2k,
which is even


Hence,

(k - a + 1) and (k + a - 1) have the same parity


Since 4 is even,
(k - a + 1) and (k + a - 1) are even

——————————

Express 4 as a product of two even positive integers,

4 = 2 x 2


Hence,

k - a + 1 = 2
—-Eqn(8)

k + a - 1 = 2
—-Eqn(9)


From Eqn(8) and Eqn(9),

k - a + 1 = k + a - 1

-a + 1 = a - 1

-a - a = -1 - 1

-2a = -2

a = 1
—-Eqn(10)


Sub. Eqn(10) into Eqn(6):

c² + c(1 - 1) - 1 = 0

c² = 1


Hence,

c = 1 —-Eqn(11)
Or
c = -1 (rej. as c > 0)


Sub. Eqn(10) and Eqn(11) into Eqn(4):

1 + b = 1 + 1

b = 1

——————————————————

Therefore,


Ans:

(a, b, c)
= (1, 1, 1)
0 replies
KTYC
4 minutes ago
0 replies
Inequality
MathsII-enjoy   4
N 9 minutes ago by ehuseyinyigit
A interesting problem generalized :-D
4 replies
MathsII-enjoy
Saturday at 1:59 PM
ehuseyinyigit
9 minutes ago
A Collection of Good Problems from my end
SomeonecoolLovesMaths   6
N an hour ago by SomeonecoolLovesMaths
This is a collection of good problems and my respective attempts to solve them. I would like to encourage everyone to post their solutions to these problems, if any. This will not only help others verify theirs but also perhaps bring forward a different approach to the problem. I will constantly try to update the pool of questions.

The difficulty level of these questions vary from AMC 10 to AIME. (Although the main pool of questions were prepared as a mock test for IOQM over the years)

Problem 1

Problem 2

Problem 3
6 replies
SomeonecoolLovesMaths
Yesterday at 8:16 AM
SomeonecoolLovesMaths
an hour ago
parallelogram in a tetrahedron
vanstraelen   0
4 hours ago
Given a tetrahedron $ABCD$ and a plane $\mu$, parallel with the edges $AC$ and $BD$.
$AB \cap \mu=P$.
a) Prove: the intersection of the tetrahedron with the plane is a parallelogram.
b) If $\left|AC\right|=14,\left|BD\right|=7$ and $\frac{\left|PA\right|}{\left|PB\right|}=\frac{3}{4}$,
calculates the lenghts of the sides of this parallelogram.
0 replies
vanstraelen
4 hours ago
0 replies
Polynomial
kellyelliee   0
Today at 3:57 AM
Let the polynomial $f(x)=x^2+ax+b$, where $a,b$ integers and $k$ is a positive integer. Suppose that the integers
$m,n,p$ satisfy: $f(m), f(n), f(p)$ are divisible by k. Prove that:
$(m-n)(n-p)(p-m)$ is divisible by k
0 replies
kellyelliee
Today at 3:57 AM
0 replies
Arithmetic Series and Common Differences
4everwise   6
N Today at 2:12 AM by epl1
For each positive integer $k$, let $S_k$ denote the increasing arithmetic sequence of integers whose first term is $1$ and whose common difference is $k$. For example, $S_3$ is the sequence $1,4,7,10,...$. For how many values of $k$ does $S_k$ contain the term $2005$?
6 replies
4everwise
Nov 10, 2005
epl1
Today at 2:12 AM
find number of elements in H
Darealzolt   0
Today at 1:50 AM
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 \( H \).
0 replies
Darealzolt
Today at 1:50 AM
0 replies
old problem from an open contest
Darealzolt   0
Today at 1:41 AM
Given that $a, b \in \mathbb{R}$ satisfy
\[
a + \frac{1}{a + 2015} = b - 4030 + \frac{1}{b - 2015}
\]and $|a - b| > 5000$. Determine the value of
\[
\frac{ab}{2015} - a + b.
\]
0 replies
Darealzolt
Today at 1:41 AM
0 replies
f_n(x)=\sum sin(nx)/n
Urumqi   6
N Today at 1:04 AM by Urumqi
$F_n(x)=\sum_{k=1}^{n}\frac{\sin (kx)}{k}$, prove that for all $x \in (0,\pi), F_n(x)>0$.

Thanks.
6 replies
Urumqi
Yesterday at 2:13 AM
Urumqi
Today at 1:04 AM
Looking for users and developers
derekli   9
N Today at 12:57 AM 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
9 replies
derekli
Yesterday at 12:57 AM
musicalpenguin
Today at 12:57 AM
Regular tetrahedron
vanstraelen   6
N Yesterday at 11:36 PM by Math-lover1
Given the points $O(0,0,0),A(1,0,0),B(\frac{1}{2},\frac{\sqrt{3}}{2},0)$
a) Determine the point $C$, above the xy-plane, such that the pyramid $OABC$ is a regular tetrahedron.
b) Calculate the volume.
c) Calculate the radius of the inscribed sphere and the radius of the circumscribed sphere.
6 replies
vanstraelen
Yesterday at 3:23 PM
Math-lover1
Yesterday at 11:36 PM
How many pairs
Ecrin_eren   5
N Yesterday at 10:19 PM by imbadatmath1233


Let n be a natural number and p be a prime number. How many different pairs (n, p) satisfy the equation:

p + 2^p + 3 = n^2 ?



5 replies
Ecrin_eren
May 2, 2025
imbadatmath1233
Yesterday at 10:19 PM
Four-variable FE mod n
TheUltimate123   2
N Apr 19, 2025 by cosmicgenius
Source: PRELMO 2023/3 (http://tinyurl.com/PRELMO)
Let \(n\) be a positive integer, and let \(\mathbb Z/n\mathbb Z\) denote the integers modulo \(n\). Determine the number of functions \(f:(\mathbb Z/n\mathbb Z)^4\to\mathbb Z/n\mathbb Z\) satisfying \begin{align*}     &f(a,b,c,d)+f(a+b,c,d,e)+f(a,b,c+d,e)\\     &=f(b,c,d,e)+f(a,b+c,d,e)+f(a,b,c,d+e). \end{align*}for all \(a,b,c,d,e\in\mathbb Z/n\mathbb Z\).
2 replies
TheUltimate123
Jul 11, 2023
cosmicgenius
Apr 19, 2025
Four-variable FE mod n
G H J
G H BBookmark kLocked kLocked NReply
Source: PRELMO 2023/3 (http://tinyurl.com/PRELMO)
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TheUltimate123
1740 posts
#1 • 1 Y
Y by MS_asdfgzxcvb
Let \(n\) be a positive integer, and let \(\mathbb Z/n\mathbb Z\) denote the integers modulo \(n\). Determine the number of functions \(f:(\mathbb Z/n\mathbb Z)^4\to\mathbb Z/n\mathbb Z\) satisfying \begin{align*}     &f(a,b,c,d)+f(a+b,c,d,e)+f(a,b,c+d,e)\\     &=f(b,c,d,e)+f(a,b+c,d,e)+f(a,b,c,d+e). \end{align*}for all \(a,b,c,d,e\in\mathbb Z/n\mathbb Z\).
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jasperE3
11291 posts
#2
Y by
bump, unsolved$~~~~~~~~~$
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cosmicgenius
1488 posts
#3 • 1 Y
Y by yofro
Scam problem; it turns out this is a standard exercise in group cohomology. Rename $n$ to $m$. The answer is $m^{m^3-m^2+m}$. Write $G = \mathbb Z / m \mathbb Z$.

For any ring $R$ (commutative with unity) two $R$-modules $A$ and $B$, given any projective resolution of $A$,
\[ \cdots \stackrel{d_3}{\longrightarrow} P_2 \stackrel{d_2}\longrightarrow P_1 \stackrel{d_1}\longrightarrow P_0 \stackrel{\varepsilon}\longrightarrow A \longrightarrow 0, \]we can consider the corresponding cochain complex given by applying the contravariant functor $\operatorname{Hom}_R (-, B)$ and ignoring $A$:
\[ 0 \stackrel{d^0}\longrightarrow \operatorname{Hom}_R (P_0, B) \stackrel{d^1}\longrightarrow \operatorname{Hom}_R (P_1, B) \stackrel{d^2}\longrightarrow \operatorname{Hom}_R (P_2, B) \stackrel{d^3}\longrightarrow \cdots. \]It is a basic fact of homological algebra that the cohomology $\ker d^{n+1} / \operatorname{im} d^n$ of this cochain complex does not depend on the choice of projective resolution; we denote this $R$-module as $\operatorname {Ext} _{R}^{n}(A,B)$.

To see why this is useful, it turns out that considering $\mathbb Z$ as a $\mathbb Z[G]$-module where $G$ acts trivially, there is a famous free resolution called the bar resolution
\[ \cdots \stackrel{d_3}{\longrightarrow} F_2 \stackrel{d_2}\longrightarrow F_1 \stackrel{d_1}\longrightarrow F_0 \stackrel{\varepsilon}\longrightarrow \mathbb Z \longrightarrow 0, \]where $F_n$ is the $\mathbb Z[G]$ module with group structure $\mathbb Z[G]^{\otimes (n+1)}$ and module structure on simple tensors given by $g \cdot (g_0 \otimes g_1 \otimes \dots \otimes g_n) = (gg_0) \otimes g_1 \otimes \dots \otimes g_n$, $\varepsilon$ is the augmentation map (taking each $g \in G$ to $1$), and each $d_n$ is given by extending the map
\begin{align*}
d_n(g_0 \otimes g_1 \otimes \dots \otimes g_n) &= \sum_{i=0}^{n-1} (-1)^i g_0 \otimes \dots \otimes g_i g_{i+1} \otimes \dots \otimes g_n \\
&\quad\;\;+ (-1)^n g_0 \otimes \dots \otimes g_{n-1}. 
\end{align*}Now consider the $\mathbb Z[G]$-module $B := \mathbb Z / m\mathbb Z$ where $G$ acts trivially (the fact that this is the same group as $G$ is a coincidence), and apply the contravariant functor $\operatorname{Hom}_{\mathbb Z[G]} (-, B)$ to get a cochain complex,
\[ 0 \stackrel{d^0}\longrightarrow \operatorname{Hom}_{\mathbb Z[G]} (F_0, B) \stackrel{d^1}\longrightarrow\operatorname{Hom}_{\mathbb Z[G]} (F_1, B) \stackrel{d^2}\longrightarrow \operatorname{Hom}_{\mathbb Z[G]} (F_2, B) \stackrel{d^3}\longrightarrow \cdots. \]Considering each $F_n$ as a free $\mathbb Z[G]$-module with basis $1 \otimes g_1 \otimes \dots \otimes g_n$ for $g_1, \dots, g_n \in G$, the group structure of $\operatorname{Hom}_{\mathbb Z[G]} (F_n, B)$ is isomorphic to abelian group of functions $G^n \to B$ under pointwise addition, and the map $d^n$ is given by the map
\begin{align*}
d^n(f) (g_1, \dots, g_n) &= g_1 \cdot f(g_2, \dots, g_n) + \sum_{i=1}^{n-1} (-1)^i f(g_1, \dots, g_i g_{i+1}, \dots, g_n) \\
&\quad\;\;+ (-1)^n f(g_1, \dots, g_{n-1}).
\end{align*}And voilà, since $g_1$ acts trivially, we wish to find the size of $\ker d^5$. Now using the definition of $\operatorname{Ext}$ and the first isomorphism theorem, we get
\begin{align*}
\# \ker d^5 &= \# \operatorname{Ext}_{\mathbb Z[G]}^4 (\mathbb Z, B) \cdot \#\operatorname{im} d^4 \\
&= \# \operatorname{Ext}_{\mathbb Z[G]}^4 (\mathbb Z, B) \cdot \frac{\# \operatorname{dom} d^4}{\#\ker d^4} \\
&\hdots \\
&= \prod_{n=0}^4 \left(\# \operatorname{Ext}_{\mathbb Z[G]}^n (\mathbb Z, B) \cdot \#\operatorname{dom} d^n\right)^{(-1)^{4-n}}.
\end{align*}Of course, for $n \ge 1$, we have $\operatorname{dom} d^n = \operatorname{Hom}_{\mathbb Z[G]} (F_{n-1}, B)$ which has $m^{m^{n-1}}$ elements, and $\operatorname{dom} d^0 = 0$, so it suffices to compute $\# \operatorname{Ext}_{\mathbb Z[G]}^n (\mathbb Z, B)$. To do this, we'll consider a different projective resolution of the $\mathbb Z[G]$-module $\mathbb Z$. Specifically, if $\sigma$ is a generator of $G$, then we can check that $N := 1+\sigma+\sigma^2+\dots+\sigma^{m-1}$ satisfies $N(1-\sigma) = 0$, so we have the free resolution
\[ \cdots \stackrel{\times (1 - \sigma)}{\longrightarrow} \mathbb Z[G] \stackrel{\times N}\longrightarrow \mathbb Z[G] \stackrel{\times (1 - \sigma)}\longrightarrow \mathbb Z[G] \stackrel{\varepsilon}\longrightarrow \mathbb Z \longrightarrow 0, \]where again $\varepsilon$ takes any $g \in G$ to $1$ (See the example on Dummit and Foote p. 801). This has corresponding chain complex under $\operatorname{Hom}_{\mathbb Z[G]} (-, B)$,
\[ 0 \longrightarrow B \stackrel{0}{\longrightarrow} B \stackrel{0}\longrightarrow B \stackrel{0}\longrightarrow \cdots, \](every morphism is $0$) under the identification $\operatorname{Hom}_{\mathbb Z[G]} (\mathbb Z[G], B) \cong B$ since $\sigma$ acts as $1$. Thus, $\# \operatorname{Ext}_{\mathbb Z[G]}^n (\mathbb Z, B) = |B|$ for all $n$, and we have
\[ \# \ker d^5 = m^{m^3 - m^2 + m^1 - m^0 + 0} \cdot m^{1 - 1 + 1 - 1 + 1} = m^{m^3 - m^2 + m}, \]where the first term is the contribution from $\operatorname{dom}$ and the second term is the contribution from $\operatorname{Ext}$, as desired.
This post has been edited 2 times. Last edited by cosmicgenius, Apr 19, 2025, 7:05 PM
Reason: typos v2
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