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Combinatoric
spiderman0   1
N 4 hours ago by MathBot101101
Let $ S = \{1, 2, 3, \ldots, 2024\}.$ Find the maximum positive integer $n \geq 2$ such that for every subset $T \subset S$ with n elements, there always exist two elements a, b in T such that:

$|\sqrt{a} - \sqrt{b}| < \frac{1}{2} \sqrt{a - b}$
1 reply
spiderman0
Yesterday at 7:46 AM
MathBot101101
4 hours ago
Combinatorial proof
MathBot101101   10
N 5 hours ago by MathBot101101
Is there a way to prove
\frac{1}{(1+1)!}+\frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}=1-\frac{1}{{n+1)!}
without induction and using only combinatorial arguments?

Induction proof wasn't quite as pleasing for me.
10 replies
MathBot101101
Apr 20, 2025
MathBot101101
5 hours ago
Simiplifying a Complicated Expression
phiReKaLk6781   6
N 5 hours ago by lbh_qys
Simplify: $ \frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-a)(b-c)}+\frac{c^3}{(c-a)(c-b)}$
6 replies
phiReKaLk6781
Mar 15, 2010
lbh_qys
5 hours ago
Geometry Angle Chasing
Sid-darth-vater   2
N Yesterday at 10:21 PM by Sid-darth-vater
Is there a way to do this without drawing obscure auxiliary lines? (the auxiliary lines might not be obscure I might just be calling them obscure)

For example I tried rotating triangle MBC 80 degrees around point C (so the BC line segment would now lie on segment AC) but I couldn't get any results. Any help would be appreciated!
2 replies
Sid-darth-vater
Monday at 11:50 PM
Sid-darth-vater
Yesterday at 10:21 PM
Absolute value
Silverfalcon   8
N Yesterday at 7:46 PM by zhoujef000
This problem seemed to be too obvious.. And I think I"m wrong.. :D

Problem:

Consider the sequence $x_0, x_1, x_2,...x_{2000}$ of integers satisfying

\[x_0 = 0, |x_n| = |x_{n-1} + 1|\]

for $n = 1,2,...2000$.

Find the minimum value of the expression $|x_1 + x_2 + ... x_{2000}|$.

My idea

Pretty sure I'm wrong but where did I go wrong?
8 replies
Silverfalcon
Jun 27, 2005
zhoujef000
Yesterday at 7:46 PM
Tetrahedrons and spheres
ReticulatedPython   3
N Yesterday at 7:26 PM by vanstraelen
Let $OABC$ be a tetrahedron such that $\angle{AOB}=\angle{AOC}=\angle{BOC}=90^\circ.$ A sphere of radius $r$ is circumscribed about tetrahedron $OABC.$ Given that $OA=a$, $OB=b$, and $OC=c$, prove that $$r^2+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{9\sqrt[3]{4}}{4}$$with equality at $a=b=c=\sqrt[3]{2}.$
3 replies
ReticulatedPython
Monday at 6:39 PM
vanstraelen
Yesterday at 7:26 PM
Σ to ∞
phiReKaLk6781   3
N Yesterday at 6:12 PM by Maxklark
Evaluate: $ \sum\limits_{k=1}^\infty \frac{1}{k\sqrt{k+2}+(k+2)\sqrt{k}}$
3 replies
phiReKaLk6781
Mar 20, 2010
Maxklark
Yesterday at 6:12 PM
Geometric inequality
ReticulatedPython   0
Yesterday at 5:12 PM
Let $A$ and $B$ be points on a plane such that $AB=n$, where $n$ is a positive integer. Let $S$ be the set of all points $P$ such that $\frac{AP^2+BP^2}{(AP)(BP)}=c$, where $c$ is a real number. The path that $S$ traces is continuous, and the value of $c$ is minimized. Prove that $c$ is rational for all positive integers $n.$
0 replies
ReticulatedPython
Yesterday at 5:12 PM
0 replies
Inequalities
sqing   27
N Yesterday at 3:51 PM by Jackson0423
Let $   a,b    $ be reals such that $  a^2+ab+b^2 =3$ . Prove that
$$ \frac{4}{ 3}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{11}{4 }$$$$ \frac{13}{ 4}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{2}{3 }$$$$ \frac{3}{ 2}\geq  \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq -\frac{17}{6 }$$$$ \frac{19}{ 6}\geq  \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2}$$Let $   a,b    $ be reals such that $  a^2-ab+b^2 =1 $ . Prove that
$$ \frac{3}{ 2}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }+ab \geq \frac{4}{15 }$$$$ \frac{14}{ 15}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }-ab \geq -\frac{1}{2 }$$$$ \frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq \frac{13}{42 }$$$$ \frac{41}{ 42}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2 }$$
27 replies
sqing
Apr 16, 2025
Jackson0423
Yesterday at 3:51 PM
Problem of the Week--The Sleeping Beauty Problem
FiestyTiger82   1
N Yesterday at 3:24 PM by martianrunner
Put your answers here and discuss!
The Problem
1 reply
FiestyTiger82
Yesterday at 2:30 PM
martianrunner
Yesterday at 3:24 PM
a