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Problem 5
SlovEcience   3
N 2 hours ago by GioOrnikapa
Let \( n > 3 \) be an odd integer. Prove that there exists a prime number \( p \) such that
\[ p \mid 2^{\varphi(n)} - 1 \quad \text{but} \quad p \nmid n. \]
3 replies
SlovEcience
Today at 1:15 PM
GioOrnikapa
2 hours ago
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IMO ShortList 2002, geometry problem 2
orl   27
N 3 hours ago by ZZzzyy
Source: IMO ShortList 2002, geometry problem 2
Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \]
27 replies
orl
Sep 28, 2004
ZZzzyy
3 hours ago
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FE based on (x+1)(y+1)
CrazyInMath   4
N 3 hours ago by jasperE3
Source: 2023 CK Summer MSG I-A
Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that \[f(xf(y)+f(x+y)+1)=(y+1)f(x+1)\]holds for all $x,y\in\mathbb{R}$.

Proposed by owoovo.shih and CrazyInMath
4 replies
CrazyInMath
Aug 14, 2023
jasperE3
3 hours ago
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Multiplicative polynomial exactly 2025 times
Assassino9931   1
N 3 hours ago by sami1618
Source: Bulgaria Balkan MO TST 2025
Does there exist a polynomial $P$ on one variable with real coefficients such that the equation $P(xy) = P(x)P(y)$ has exactly $2025$ ordered pairs $(x,y)$ as solutions?
1 reply
Assassino9931
Yesterday at 10:14 PM
sami1618
3 hours ago
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Geometric inequality problem
mathlover1231   1
N 3 hours ago by Double07
Given an acute triangle ABC, where H and O are the orthocenter and circumcenter, respectively. Point K is the midpoint of segment AH, and ℓ is a line through O. Points P and Q are the projections of B and C onto ℓ. Prove that KP + KQ ≥BC
1 reply
mathlover1231
5 hours ago
Double07
3 hours ago
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i love mordell
MR.1   1
N 3 hours ago by MR.1
Source: own
find all pairs of $(m,n)$ such that $n^2-79=m^3$
1 reply
MR.1
3 hours ago
MR.1
3 hours ago
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MM 2201 (Symmetric Inequality with Weird Sharp Case)
kgator   1
N 4 hours ago by CHESSR1DER
Source: Mathematics Magazine Volume 97 (2024), Issue 4: https://doi.org/10.1080/0025570X.2024.2393998
2201. Proposed by Leonard Giugiuc, Drobeta-Turnu Severin, Romania. Find all real numbers $K$ such that
$$a^2 + b^2 + c^2 - 3 \geq K(a + b + c - 3)$$for all nonnegative real numbers $a$, $b$, and $c$ with $abc \leq 1$.
1 reply
kgator
4 hours ago
CHESSR1DER
4 hours ago
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Interesting Inequality Problem
Omerking   0
4 hours ago
Let $a,b,c$ be three non-negative real numbers satisfying $a+b+c+abc=4.$
Prove that
$$\frac{a}{a^{2}+1}+\frac{b}{b^{2}+1}+\frac{c}{c^{2}+1} \leq\frac{6}{13-3ab-3bc-3ca}$$
0 replies
Omerking
4 hours ago
0 replies
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[SEIF Q1] FE on x^3+xy...( ͡° ͜ʖ ͡°)
EmilXM   18
N 4 hours ago by jasperE3
Source: SEIF 2022
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that any real numbers $x$ and $y$ satisfy
$$x^3+f(x)f(y)=f(f(x^3)+f(xy)).$$Proposed by EmilXM
18 replies
EmilXM
Mar 12, 2022
jasperE3
4 hours ago
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RGB chessboard
BR1F1SZ   1
N 4 hours ago by alfonsoramires
Source: 2025 Argentina TST P3
A $100 \times 100$ board has some of its cells coloured red, blue, or green. Each cell is coloured with at most one colour, and some cells may remain uncoloured. Additionally, there is at least one cell of each colour. Two coloured cells are said to be friends if they have different colours and lie in the same row or in the same column. The following conditions are satisfied:
[list=i]
[*]Each coloured cell has exactly three friends.
[*]All three friends of any given coloured cell lie in the same row or in the same column.
[/list]
Determine the maximum number of cells that can be coloured on the board.
1 reply
BR1F1SZ
Tuesday at 11:15 PM
alfonsoramires
4 hours ago
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interesting way to derive the quadratic formula
Soupboy0   2
N Mar 28, 2025 by Mathematicalprodigy37
If you have the quadratic $ax^2+bx+c$, call the roots $r$ and $s$ with $r \ge s$. Then by Vieta's, $r+s = \frac{-b}{a}$ and $rs = \frac{c}{a}$. If we can find $r+s$, we can find $r-s$. Note that $(r-s)^2=r^2-2rs+s^2$ and $(r+s)^2=r^2+2rs+s^2$. Therefore, $(r+s)^2-4rs=(r-s)^2$, and $r-s=\sqrt{(r+s)^2-4rs} = \sqrt{\frac{-b}{a^2}^2-\frac{4c}{a}}=\sqrt{\frac{b^2-4ac}{a^2}}=\frac{\sqrt{b^2-4ac}}{a}$. Now all we have to do is solve the following system:
$r+s=\frac{-b}{a}$ and $r-s=\frac{\sqrt{b^2-4ac}}{a}$. Solving for $r$ by elimination, we get $2r = \frac{-b+\sqrt{b^2-4ac}}{a}$, so ${r=\frac{-b+\sqrt{b^2-4ac}}{2a}}$, which is the quadratic formula. Substituting this into our first equation, we get $\frac{-b+\sqrt{b^2-4ac}}{2a}+s=\frac{-b}{a}$, and solving for $s$ yields $s=\frac{-b-\sqrt{b^2-4ac}}{2a}$,
2 replies
Soupboy0
Mar 28, 2025
Mathematicalprodigy37
Mar 28, 2025
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interesting way to derive the quadratic formula
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Reveal topic tags
quadratics
algebra
quadratic formula
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Soupboy0
311 posts
Soupboy0
#1 h  Mar 28, 2025, 3:33 PM • 2 Y
Y by Mathematicalprodigy37, mithu542
If you have the quadratic $ax^2+bx+c$, call the roots $r$ and $s$ with $r \ge s$. Then by Vieta's, $r+s = \frac{-b}{a}$ and $rs = \frac{c}{a}$. If we can find $r+s$, we can find $r-s$. Note that $(r-s)^2=r^2-2rs+s^2$ and $(r+s)^2=r^2+2rs+s^2$. Therefore, $(r+s)^2-4rs=(r-s)^2$, and $r-s=\sqrt{(r+s)^2-4rs} = \sqrt{\frac{-b}{a^2}^2-\frac{4c}{a}}=\sqrt{\frac{b^2-4ac}{a^2}}=\frac{\sqrt{b^2-4ac}}{a}$. Now all we have to do is solve the following system:
$r+s=\frac{-b}{a}$ and $r-s=\frac{\sqrt{b^2-4ac}}{a}$. Solving for $r$ by elimination, we get $2r = \frac{-b+\sqrt{b^2-4ac}}{a}$, so ${r=\frac{-b+\sqrt{b^2-4ac}}{2a}}$, which is the quadratic formula. Substituting this into our first equation, we get $\frac{-b+\sqrt{b^2-4ac}}{2a}+s=\frac{-b}{a}$, and solving for $s$ yields $s=\frac{-b-\sqrt{b^2-4ac}}{2a}$,
This post has been edited 1 time. Last edited by Soupboy0, Mar 28, 2025, 3:33 PM
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Owinner
53 posts
Owinner
#2 h  Mar 28, 2025, 3:48 PM
Y by
nice ive never seen it proved that way
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Mathematicalprodigy37
17 posts
Mathematicalprodigy37
#3 h  Mar 28, 2025, 3:54 PM
Y by
nice use of Vieta's formulas! :clap:
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