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hard problem
Cobedangiu   1
N 17 minutes ago by Cobedangiu
problem
1 reply
+2 w
Cobedangiu
an hour ago
Cobedangiu
17 minutes ago
Gheorghe Țițeica 2025 Grade 8 P3
AndreiVila   1
N 25 minutes ago by sunken rock
Source: Gheorghe Țițeica 2025
Two regular pentagons $ABCDE$ and $AEKPL$ are given in space, such that $\angle DAC = 60^{\circ}$. Let $M$, $N$ and $S$ be the midpoints of $AE$, $CD$ and $EK$. Prove that:
[list=a]
[*] $\triangle NMS$ is a right triangle;
[*] planes $(ACK)$ and $(BAL)$ are perpendicular.
[/list]
Ukraine Olympiad
1 reply
AndreiVila
Yesterday at 9:00 PM
sunken rock
25 minutes ago
Geometry Problem in Taiwan TST
chengbilly   3
N 29 minutes ago by Hakurei_Reimu
Source: 2025 Taiwan TST Round 2 Independent Study 2-G
Given a triangle $ABC$ with circumcircle $\Gamma$, and two arbitrary points $X, Y$ on $\Gamma$. Let $D$, $E$, $F$ be points on lines $BC$, $CA$, $AB$, respectively, such that $AD$, $BE$, and $CF$ concur at a point $P$. Let $U$ be a point on line $BC$ such that $X$, $Y$, $D$, $U$ are concyclic. Similarly, let $V$ be a point on line $CA$ such that $X$, $Y$, $E$, $V$ are concyclic, and let $W$ be a point on line $AB$ such that $X$, $Y$, $F$, $W$ are concyclic. Prove that $AU$, $BV$, $CW$ concur at a single point.

Proposed by chengbilly
3 replies
chengbilly
Mar 27, 2025
Hakurei_Reimu
29 minutes ago
Gheorghe Țițeica 2025 Grade 9 P4
AndreiVila   1
N 34 minutes ago by AlgebraKing
Source: Gheorghe Țițeica 2025
For all $n\in\mathbb{N}$, we denote by $s(n)$ the sum of its digits. Find all integers $k\geq 2$ such that there exist $a,b\in\mathbb{N}$ with $$s(n^3+an+b)\equiv s(n)\pmod k,$$for all $n\in\mathbb{N}^*$.
1 reply
1 viewing
AndreiVila
Yesterday at 9:19 PM
AlgebraKing
34 minutes ago
(2^n + 1)/n^2 is an integer (IMO 1990 Problem 3)
orl   104
N 37 minutes ago by John_Mgr
Source: IMO 1990, Day 1, Problem 3, IMO ShortList 1990, Problem 23 (ROM 5)
Determine all integers $ n > 1$ such that
\[ \frac {2^n + 1}{n^2}
\]is an integer.
104 replies
orl
Nov 11, 2005
John_Mgr
37 minutes ago
Something nice
KhuongTrang   25
N 39 minutes ago by Zuyong
Source: own
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$
25 replies
KhuongTrang
Nov 1, 2023
Zuyong
39 minutes ago
(a²-b²)(b²-c²) = abc
straight   1
N 40 minutes ago by straight
Find all triples of positive integers $(a,b,c)$ such that

\[(a^2-b^2)(b^2-c^2) = abc.\]
If you can't solve this, assume $gcd(a,c) = 1$. If this is still too hard assume in $a \ge b \ge c$ that $b-c$ is a prime.
1 reply
+1 w
straight
Mar 24, 2025
straight
40 minutes ago
Equality occurs in strange points
arqady   10
N 40 minutes ago by sqing
Let $a$, $b$ and $c$ are non-negatives such that $a+b+c\neq0$. Prove that:
1) \[\sum_{cyc}\sqrt{a^2+b^2}\leq\frac{11(a^2+b^2+c^2)+7(ab+ac+bc)}{3\sqrt2(a+b+c)}\]
2)\[\sum_{cyc}\sqrt{a^2+7ab+b^2}\leq\frac{2(a^2+b^2+c^2)+7(ab+ac+bc)}{a+b+c}\]
3)\[\sum_{cyc}\sqrt{4a^2+ab+4b^2}\leq\frac{5(a^2+b^2+c^2)+4(ab+ac+bc)}{a+b+c}\]
10 replies
arqady
Aug 28, 2011
sqing
40 minutes ago
A number theory problem from the British Math Olympiad
Rainbow1971   3
N an hour ago by pooh123
Source: British Math Olympiad, 2006/2007, round 1, problem 6
I am a little surprised to find that I am (so far) unable to solve this little problem:

[quote]Let $n$ be an integer. Show that, if $2 + 2 \sqrt{1+12n^2}$ is an integer, then it is a perfect square.[/quote]

I set $k := \sqrt{1+12n^2}$. If $2 + 2 \sqrt{1+12n^2}$ is an integer, then $k (=\sqrt{1+12n^2})$ is at least rational, so that $1 + 12n^2$ must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs $(n, k)$ for which this happens are $(0,1), (2,7), (28,97)$ and $(390, 1351)$, and that, for every such pair $(n,k)$, the "next" such pair can be calculated as
$$
\begin{bmatrix}
7 & 2 \\
24 & 7 
\end{bmatrix} \begin{bmatrix}
n \\
k 
\end{bmatrix}
.$$The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?

Thank you.




3 replies
Rainbow1971
Yesterday at 8:39 PM
pooh123
an hour ago
Tricky NT
Hip1zzzil   1
N an hour ago by pooh123
Source: 2024 Korea
find the minimum value of positive integer which when multiplied by 2024, its digits are only 1 or 0.
1 reply
Hip1zzzil
Yesterday at 11:10 PM
pooh123
an hour ago
An interesting inequality
JK1603JK   4
N an hour ago by Nguyenhuyen_AG
Source: unknown
Let a,b,c>=0 and a^2+b^2+c^2+abc=4 then prove \frac{1}{a+b+2}+\frac{1}{b+c+2}+\frac{1}{c+a+2} \le \frac{6-(a+b+c)}{4}
When does equality occur?
4 replies
JK1603JK
Mar 21, 2025
Nguyenhuyen_AG
an hour ago
Complex number inequalities
jhz   2
N Mar 27, 2025 by flower417477
Source: 2025 CTST P18
Find the smallest real number $M$ such that there exist four complex numbers $a,b,c,d$ with $|a|=|b|=|c|=|d|=1$, and for any complex number $z$, if $|z| = 1$, then\[|az^3+bz^2+cz+d|\le M.\]
2 replies
jhz
Mar 26, 2025
flower417477
Mar 27, 2025
Complex number inequalities
G H J
G H BBookmark kLocked kLocked NReply
Source: 2025 CTST P18
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jhz
10 posts
#1 • 2 Y
Y by MS_asdfgzxcvb, sky.mty
Find the smallest real number $M$ such that there exist four complex numbers $a,b,c,d$ with $|a|=|b|=|c|=|d|=1$, and for any complex number $z$, if $|z| = 1$, then\[|az^3+bz^2+cz+d|\le M.\]
This post has been edited 1 time. Last edited by jhz, Mar 26, 2025, 12:28 AM
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MS_asdfgzxcvb
56 posts
#2
Y by
jhz wrote:
Find the smallest real number $M$ such that there exist four complex numbers $a,b,c,d$ with $|a|=|b|=|c|=|d|=1$, and for any complex number $z$, if $|z| = 1$, then\[|az^3+bz^2+cz+d|\le M.\]
WLOG \(c,d=1\). By ROUF we can easily get \(M>\sqrt 6\). This inequality is strict. The best upper bound I get is \(M\le \sqrt {6.117}\), for \(a=e^{-0.852i}, b=e^{1.8105i}.\)
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flower417477
358 posts
#4 • 1 Y
Y by MS_asdfgzxcvb
The answer should be $2(\sqrt{5}-1)$
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