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Intermediate Counting
RenheMiResembleRice   4
N 5 hours ago by Apple_maths60
A coin is flipped, a 6-sided die numbered 1 through 6 is rolled, and a 10-sided die numbered 0
through 9 is rolled. What is the probability that the coin comes up heads and the sum of the
numbers that show on the dice is 8?
4 replies
RenheMiResembleRice
Yesterday at 7:46 AM
Apple_maths60
5 hours ago
Inequalities
sqing   2
N Yesterday at 2:33 PM by DAVROS
Let $a,b$ be real numbers such that $ a^2+b^2+a^3 +b^3=4   . $ Prove that
$$a+b \leq 2$$Let $a,b$ be real numbers such that $a+b + a^2+b^2+a^3 +b^3=6 . $ Prove that
$$a+b \leq 2$$
2 replies
sqing
Saturday at 1:10 PM
DAVROS
Yesterday at 2:33 PM
Might be the first equation marathon
steven_zhang123   33
N Yesterday at 2:15 PM by eric201291
As far as I know, it seems that no one on HSM has organized an equation marathon before. Click to reveal hidden textSo why not give it a try? Click to reveal hidden text Let's start one!
Some basic rules need to be clarified:
$\cdot$ If a problem has not been solved within $5$ days, then others are eligible to post a new probkem.
$\cdot$ Not only simple one-variable equations, but also systems of equations are allowed.
$\cdot$ The difficulty of these equations should be no less than that of typical quadratic one-variable equations. If the problem involves higher degrees or more variables, please ensure that the problem is solvable (i.e., has a definite solution, rather than an approximate one).
$\cdot$ Please indicate the domain of the solution to the equation (e.g., solve in $\mathbb{R}$, solve in $\mathbb{C}$).
Here's an simple yet fun problem, hope you enjoy it :P :
P1
33 replies
steven_zhang123
Jan 20, 2025
eric201291
Yesterday at 2:15 PM
Inequalities
hn111009   6
N Yesterday at 1:26 PM by Arbelos777
Let $a,b,c>0$ satisfied $a^2+b^2+c^2=9.$ Find the minimum of $$P=\dfrac{a}{bc}+\dfrac{2b}{ca}+\dfrac{5c}{ab}.$$
6 replies
hn111009
Yesterday at 1:25 AM
Arbelos777
Yesterday at 1:26 PM
Congruence
Ecrin_eren   2
N Yesterday at 8:42 AM by Ecrin_eren
Find the number of integer pairs (x, y) satisfying the congruence equation:

3y² + 3x²y + y³ ≡ 3x² (mod 41)

for 0 ≤ x, y < 41.

2 replies
Ecrin_eren
Apr 3, 2025
Ecrin_eren
Yesterday at 8:42 AM
Olympiad
sasu1ke   3
N Yesterday at 1:00 AM by sasu1ke
IMAGE
3 replies
sasu1ke
Saturday at 11:52 PM
sasu1ke
Yesterday at 1:00 AM
How to judge a number is prime or not?
mingzhehu   1
N Saturday at 11:14 PM by scrabbler94
A=(10X1+1)(10X+1),X1,X∈N+
B=(10 X1+3)(10X+7),X∈N,X1∈N
C=(10 X1+9)(10X+9), X∈N,X1∈N
D=(10 X1+1)(10X+3), X1∈N+,X∈N
E=(10 X1+7)(10X+9),X∈N,X1∈N
F=(10 X1+1)(10X+7),X1∈N+,X∈N
G=(10 X1+3)(10X+9),X∈N,X1∈N
H=(10 X1+1)10X+9),X1∈N+,X∈N
I=(10 X1+3)(10X+3),X1∈N,X∈N
J=( 10X1+7)(10X+7),X∈N,X1∈N

For any natural number P∈{P=10N+1,n∈N},make P=A or B or C
If P can make the roots of function group(ABC) without any root group completely made up of integer, P will be a prime
For any natural number P∈{P=10N+3,n∈N},make P=D or E
If P can make the roots of function group(DE) without any root group completely made up
of integer, P will be a prime
For any natural number P∈{P=10N+7,n∈N},make P=F or G
If P can make the roots of function group(FG) without any root group completely made up
of integer, P will be a prime
For any natural number P∈{P=10N+9,n∈N},make P=H or I or J
If P can make the roots of function group(GIJ) without any root group completely made up
of integer, P will be a prime
1 reply
mingzhehu
Saturday at 2:45 PM
scrabbler94
Saturday at 11:14 PM
inequality
revol_ufiaw   3
N Saturday at 2:55 PM by MS_asdfgzxcvb
Prove that that for any real $x \ge 0$ and natural number $n$,
$$x^n (n+1)^{n+1} \le n^n (x+1)^{n+1}.$$
3 replies
revol_ufiaw
Saturday at 2:05 PM
MS_asdfgzxcvb
Saturday at 2:55 PM
What is an isogonal conjugate and why is it useful?
EaZ_Shadow   6
N Saturday at 2:40 PM by maxamc
What is an isogonal conjugate and why is it useful? People use them in Olympiad geometry proofs but I don’t understand why and what is the purpose, as it complicates me because of me not understanding it.
6 replies
EaZ_Shadow
Dec 28, 2024
maxamc
Saturday at 2:40 PM
Any nice way to do this?
NamelyOrange   3
N Saturday at 2:00 PM by pooh123
Source: Taichung P.S.1 math program tryouts

How many ordered pairs $(a,b,c)\in\mathbb{N}^3$ are there such that $c=ab$ and $1\le a\le b\le c\le60$?
3 replies
NamelyOrange
Apr 2, 2025
pooh123
Saturday at 2:00 PM
China Western Mathematical Olympiad 2015 ,Problem 3
sqing   8
N Feb 5, 2025 by shanelin-sigma
Source: China Yinchuan Aug 16, 2015
Let the integer $n \ge 2$ , and $x_1,x_2,\cdots,x_n $ be positive real numbers such that $\sum_{i=1}^nx_i=1$ .Prove that$$\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(\sum_{1\le i<j\le n} x_ix_j\right)\le \frac{n}{2}.$$
8 replies
sqing
Aug 16, 2015
shanelin-sigma
Feb 5, 2025
China Western Mathematical Olympiad 2015 ,Problem 3
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G H BBookmark kLocked kLocked NReply
Source: China Yinchuan Aug 16, 2015
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sqing
41446 posts
#1 • 1 Y
Y by Adventure10
Let the integer $n \ge 2$ , and $x_1,x_2,\cdots,x_n $ be positive real numbers such that $\sum_{i=1}^nx_i=1$ .Prove that$$\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(\sum_{1\le i<j\le n} x_ix_j\right)\le \frac{n}{2}.$$
This post has been edited 1 time. Last edited by sqing, Aug 16, 2015, 11:25 AM
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MariusStanean
655 posts
#2 • 4 Y
Y by rkm0959, zsgvivo, Adventure10, Mango247
Suppose that $x_1\le x_2\le\ldots \le x_n$ and with Cebasev inequality we have
$$2\cdot LHS=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(1-\sum_{i=1}^n x_i^2\right)=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(\sum_{i=1}^n x_i(1-x_i)\right)\le n\sum_{i=1}^n x_i=n$$
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sqing
41446 posts
#3 • 1 Y
Y by Adventure10
Very nice.
Thanks.
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zsgvivo
28 posts
#4 • 2 Y
Y by Adventure10, Mango247
MariusStanean wrote:
Suppose that $x_1\le x_2\le\ldots \le x_n$ and with Cebasev inequality we have
$$2\cdot LHS=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(1-\sum_{i=1}^n x_i^2\right)=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(\sum_{i=1}^n x_i(1-x_i)\right)\le n\sum_{i=1}^n x_i=n$$

what is Cebasev inequality
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sqing
41446 posts
#5 • 2 Y
Y by zsgvivo, Adventure10
zsgvivo wrote:
MariusStanean wrote:
Suppose that $x_1\le x_2\le\ldots \le x_n$ and with Cebasev inequality we have
$$2\cdot LHS=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(1-\sum_{i=1}^n x_i^2\right)=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(\sum_{i=1}^n x_i(1-x_i)\right)\le n\sum_{i=1}^n x_i=n$$

what is Cebasev inequality
https://en.wikipedia.org/wiki/Chebyshev%27s_sum_inequality
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zsgvivo
28 posts
#6 • 1 Y
Y by Adventure10
sqing wrote:
zsgvivo wrote:
MariusStanean wrote:
Suppose that $x_1\le x_2\le\ldots \le x_n$ and with Cebasev inequality we have
$$2\cdot LHS=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(1-\sum_{i=1}^n x_i^2\right)=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(\sum_{i=1}^n x_i(1-x_i)\right)\le n\sum_{i=1}^n x_i=n$$

what is Cebasev inequality
https://en.wikipedia.org/wiki/Chebyshev%27s_sum_inequality

thanks!
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bel.jad5
3750 posts
#7 • 2 Y
Y by Adventure10, Mango247
@fair enogh...thanks
This post has been edited 1 time. Last edited by bel.jad5, Jul 24, 2019, 1:03 PM
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WolfusA
1900 posts
#8 • 2 Y
Y by Adventure10, Mango247
For $a+b<1, a,b\ge 0$ holds $\frac{1}{1-a}\le\frac{1}{1-b}\iff a\le b$ and $a(1-a)\le b(1-b)\iff a-b\le (a-b)(a+b)\iff (1-a-b)(a-b)\le0\iff a\le b$ which proves sequences $\left(\frac{1}{1-x_i}\right)$, $\left(x_i(1-x_i)\right)$ are non-decreasing for $0<x_1\le x_2\le\ldots \le x_n, x_1+x_2+...+x_n=1$ so we can use Chebyshev
This post has been edited 1 time. Last edited by WolfusA, Jul 24, 2019, 11:57 AM
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shanelin-sigma
150 posts
#12
Y by
MariusStanean wrote:
Suppose that $x_1\le x_2\le\ldots \le x_n$ and with Cebasev inequality we have
$$2\cdot LHS=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(1-\sum_{i=1}^n x_i^2\right)=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(\sum_{i=1}^n x_i(1-x_i)\right)\le n\sum_{i=1}^n x_i=n$$

But $x_i(1-x_i)$ isn’t always increase as $x_i$ increase
Should we check the case if $x_n> \frac 12$?
This post has been edited 1 time. Last edited by shanelin-sigma, Feb 5, 2025, 12:29 AM
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