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Graph of polynomials
Ecrin_eren   0
15 minutes ago
Sure! Here's the same translation without LaTeX:


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The graph of the quadratic polynomial with real coefficients y = px^2 + qx + r, called G1, intersects the graph of the polynomial y = x^2, called G2, at points A and B. The lines tangent to G2 at points A and B intersect at point C. It is known that point C lies on G1. What is the value of p?

0 replies
Ecrin_eren
15 minutes ago
0 replies
Find the value
sqing   9
N 19 minutes ago by sqing
Source: 2025 Tsinghua University
Let $A= \lim_{n\to\infty}\tan^n\left(\frac{\pi}{4}+\frac{1}{n}\right)  . $ Find the value of $[100A] .$
9 replies
sqing
5 hours ago
sqing
19 minutes ago
Dear Sqing: So Many Inequalities...
hashtagmath   30
N 26 minutes ago by sqing
I have noticed thousands upon thousands of inequalities that you have posted to HSO and was wondering where you get the inspiration, imagination, and even the validation that such inequalities are true? Also, what do you find particularly appealing and important about specifically inequalities rather than other branches of mathematics? Thank you :)
30 replies
hashtagmath
Oct 30, 2024
sqing
26 minutes ago
Advanced topics in Inequalities
va2010   14
N 38 minutes ago by sqing
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!
14 replies
1 viewing
va2010
Mar 7, 2015
sqing
38 minutes ago
The point F lies on the line OI in triangle ABC
WakeUp   13
N 41 minutes ago by Nari_Tom
Source: All-Russian Olympiad 2012 Grade 10 Day 2
The point $E$ is the midpoint of the segment connecting the orthocentre of the scalene triangle $ABC$ and the point $A$. The incircle of triangle $ABC$ incircle is tangent to $AB$ and $AC$ at points $C'$ and $B'$ respectively. Prove that point $F$, the point symmetric to point $E$ with respect to line $B'C'$, lies on the line that passes through both the circumcentre and the incentre of triangle $ABC$.
13 replies
WakeUp
May 31, 2012
Nari_Tom
41 minutes ago
VOLUNTEERING OPPORTUNITY OPEN TO HIGH/MIDDLE SCHOOLERS
im_space_cadet   0
an hour ago
Hi everyone!
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Thanks,
im_space_cadet
0 replies
im_space_cadet
an hour ago
0 replies
(2^n -1)!! -1 is divided by 2^n
parmenides51   5
N an hour ago by parmenides51
Source: 2023 Grand Duchy of Lithuania, MC p4 (Baltic Way TST)
Note that $k\ge 1$ for an odd natural number $$k! ! = k \cdot (k - 2)  \cdot  ...  \cdot  1.$$Prove that $2^n$ divides $(2^n -1)!! -1$ for all $n \ge 3$.
5 replies
parmenides51
Mar 23, 2024
parmenides51
an hour ago
Transforming a grid to another
Severus   3
N an hour ago by Project_Donkey_into_M4
Source: STEMS 2021 Cat B P5
Sheldon was really annoying Leonard. So to keep him quiet, Leonard decided to do something. He gave Sheldon the following grid

$\begin{tabular}{|c|c|c|c|c|c|}
\hline
1 & 1 & 1 & 1 & 1 & 0\\ 
\hline
1 & 1 & 1 & 1 & 0 & 0\\ 
\hline
1 & 1 & 1 & 0 & 0 & 0\\ 
\hline
1 & 1 & 0 & 0 & 0 & 1\\ 
\hline
1 & 0 & 0 & 0 & 1 & 0\\
\hline
0 & 0 & 0 & 1 & 0 & 0\\
\hline
\end{tabular}$

and asked him to transform it to the new grid below

$\begin{tabular}{|c|c|c|c|c|c|}
\hline
1 & 2 & 18 &24 &28 &30\\
\hline
21 & 3 & 4 &16 &22 &26\\
\hline
23 &19 & 5 & 6 &14 &20\\
\hline
32 &25 &17 & 7 & 8 &12\\
\hline
33 &34 &27 &15 & 9 &10\\
\hline
35 &31 &36 &29 &13 &11\\
\hline
\end{tabular}$

by only applying the following algorithm:

$\bullet$ At each step, Sheldon must choose either two rows or two columns.

$\bullet$ For two columns $c_1, c_2$, if $a,b$ are entries in $c_1, c_2$ respectively, then we say that $a$ and $b$ are corresponding if they belong to the same row. Similarly we define corresponding entries of two rows. So for Sheldon's choice, if two corresponding entries have the same parity, he should do nothing to them, but if they have different parities, he should add 1 to both of them.

Leonard hoped this would keep Sheldon occupied for some time, but Sheldon immediately said, "But this is impossible!". Was Sheldon right? Justify.
3 replies
1 viewing
Severus
Jan 24, 2021
Project_Donkey_into_M4
an hour ago
Inspired by Bet667
sqing   1
N an hour ago by sqing
Source: Own
Let $ x,y\ge 0 $ such that $k(x+y)=1+xy. $ Prove that$$x+k^2y+\frac{1}{x}+\frac{k^2}{y} \geq \frac{k^2(k+1)^2+(k-1)^2}{k}$$Where $ k\in N^+.$
Let $ x,y\ge 0 $ such that $2(x+y)=1+xy. $ Prove that$$x+4y+\frac{1}{x}+\frac{4}{y} \geq \frac{37}{2}$$
1 reply
sqing
Today at 3:10 AM
sqing
an hour ago
FE inequality from Iran
mojyla222   2
N an hour ago by sami1618
Source: Iran 2025 second round P5
Find all functions $f:\mathbb{R}^+ \to \mathbb{R}$ such that for all $x,y,z>0$
$$
3(x^3+y^3+z^3)\geq f(x+y+z)\cdot f(xy+yz+xz) \geq (x+y+z)(xy+yz+xz).
$$
2 replies
mojyla222
Yesterday at 9:20 AM
sami1618
an hour ago
Not homogenous inequality
Nguyenhuyen_AG   0
an hour ago
Let $a,b,c$ are positive real numbers. Prove that
\[\frac{1}{(2a+1)(2b+1)}+\frac{1}{(2b+1)(2c+1)}+\frac{1}{(2c+1)(2a+1)} \geqslant \frac{3}{3+2(ab+bc+ca)}.\]
0 replies
Nguyenhuyen_AG
an hour ago
0 replies
Calculating combinatorial numbers
lgx57   5
N Mar 30, 2025 by generatingFraction
Try to simplify this expression:

$$\sum_{i=1}^n \sum_{j=1}^i C_{n}^i C_{n}^j$$
5 replies
lgx57
Mar 30, 2025
generatingFraction
Mar 30, 2025
Calculating combinatorial numbers
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lgx57
26 posts
#1
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Try to simplify this expression:

$$\sum_{i=1}^n \sum_{j=1}^i C_{n}^i C_{n}^j$$
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alexheinis
10548 posts
#2 • 1 Y
Y by MihaiT
We have $s=\sum_{k=1}^n\sum_{l=1}^k {n\choose k}{n\choose l}=\sum_{l=1}^n \sum_{k=l}^n {n\choose k}{n\choose l}=\sum_{k=1}^n \sum_{l=k}^n {n\choose k}{n\choose l}$.
Hence $2s=\sum_{k=1}^n {n\choose k}^2+\sum_{k=1}^n \sum_{l=1}^n {n\choose k}{n\choose l}={{2n}\choose n}-1+(2^n-1)^2$.

@generatingFraction: welcome to AOPS. To use latex just put the formulas within $-signs. It might not work for new members but it will afterwards.
This post has been edited 2 times. Last edited by alexheinis, Mar 30, 2025, 9:34 AM
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lgx57
26 posts
#3
Y by
alexheinis wrote:
We have $s=\sum_{k=1}^n\sum_{l=1}^k {n\choose k}{n\choose l}=\sum_{l=1}^n \sum_{k=l}^n {n\choose k}{n\choose l}=\sum_{k=1}^n \sum_{l=k}^n {n\choose k}{n\choose l}$.
Hence $2s=\sum_{k=1}^n {n\choose k}^2+\sum_{k=1}^n \sum_{l=1}^n {n\choose k}{n\choose l}={{2n}\choose n}-1+(2^n-1)^2$.

Why is $\sum_{k=1}^n {n\choose k}^2={{2n} \choose n}-1$?
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generatingFraction
5 posts
#4
Y by
You have 2n balls, calculate the numbers to choose n balls. It is the sum of choosing k in the first n balls and choosing n - k in the last n balls.
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generatingFraction
5 posts
#5
Y by
Also you can use generating function.
[x ^ k] (1 + x) ^ n = \binom{ n }{ k }
So \sum_{k = 1} ^ n \binom{ n }{ k } * \binom{ n }{ n - k } = [x ^ n] (1 + x) ^ {2n} = \binom{ 2n }{ n }
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generatingFraction
5 posts
#6
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By the way, how do your guys edit math formula ? The system showed new users can't send image.
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