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Geometry Problem
Itoz   2
N an hour ago by Itoz
Source: Own
Given $\triangle ABC$. Let the perpendicular line from $A$ to $BC$ meets $BC,\odot(ABC)$ at points $S,K$, respectively, and the foot from $B$ to $AC$ is $L$. $\odot (AKL)$ intersects line $AB$ at $T(\neq A)$, $\odot(AST)$ intersects line $AC$ at $M(\neq A)$, and lines $TM,CK$ intersect at $N$.

Prove that $\odot(CNM)$ is tangent to $\odot (BST)$.
2 replies
2 viewing
Itoz
Yesterday at 11:49 AM
Itoz
an hour ago
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Why is the old one deleted?
EeEeRUT   11
N an hour ago by Mathgloggers
Source: EGMO 2025 P1
For a positive integer $N$, let $c_1 < c_2 < \cdots < c_m$ be all positive integers smaller than $N$ that are coprime to $N$. Find all $N \geqslant 3$ such that $$\gcd( N, c_i + c_{i+1}) \neq 1$$for all $1 \leqslant i \leqslant m-1$

Here $\gcd(a, b)$ is the largest positive integer that divides both $a$ and $b$. Integers $a$ and $b$ are coprime if $\gcd(a, b) = 1$.

Proposed by Paulius Aleknavičius, Lithuania
11 replies
EeEeRUT
Apr 16, 2025
Mathgloggers
an hour ago
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Congruence related perimeter
egxa   2
N 2 hours ago by LoloChen
Source: All Russian 2025 9.8 and 10.8
On the sides of triangle \( ABC \), points \( D_1, D_2, E_1, E_2, F_1, F_2 \) are chosen such that when going around the triangle, the points occur in the order \( A, F_1, F_2, B, D_1, D_2, C, E_1, E_2 \). It is given that
\[ AD_1 = AD_2 = BE_1 = BE_2 = CF_1 = CF_2. \]Prove that the perimeters of the triangles formed by the triplets \( AD_1, BE_1, CF_1 \) and \( AD_2, BE_2, CF_2 \) are equal.
2 replies
egxa
Yesterday at 5:08 PM
LoloChen
2 hours ago
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number theory
Levieee   7
N 2 hours ago by g0USinsane777
Idk where it went wrong, marks was deducted for this solution
$\textbf{Question}$
Show that for a fixed pair of distinct positive integers \( a \) and \( b \), there cannot exist infinitely many \( n \in \mathbb{Z} \) such that
\[ \sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}. \]
$\textbf{Solution}$

Let
\[ x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}. \]
Then,
\[ x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}. \]So:
\[ x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}. \]
Therefore,
\[ \sqrt{(n + a)(n + b)} \in \mathbb{N}. \]
Let
\[ (n + a)(n + b) = k^2. \]Assume \( n + a \neq n + b \). Then we have:
\[ n + a \mid k \quad \text{and} \quad k \mid n + b, \]or it could also be that \( k \mid n + a \quad \text{and} \quad n + b \mid k \).

Without loss of generality, we take the first case:
\[ (n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b. \]
Thus,
\[ k_1 k_2 = \frac{n + b}{n + a}. \]
Since \( k_1 k_2 \in \mathbb{N} \), we have:
\[ k_1 k_2 = 1 + \frac{b - a}{n + a}. \]
For infinitely many \( n \), \( \frac{b - a}{n + a} \) must be an integer, which is not possible.

Therefore, there cannot be infinitely many such \( n \).
7 replies
Levieee
Yesterday at 7:46 PM
g0USinsane777
2 hours ago
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inequalities proplem
Cobedangiu   4
N 2 hours ago by Mathzeus1024
$x,y\in R^+$ and $x+y-2\sqrt{x}-\sqrt{y}=0$. Find min A (and prove):
$A=\sqrt{\dfrac{5}{x+1}}+\dfrac{16}{5x^2y}$
4 replies
Cobedangiu
Yesterday at 11:01 AM
Mathzeus1024
2 hours ago
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3 var inquality
sqing   0
2 hours ago
Source: Own
Let $ a,b,c $ be reals such that $ a+b+c=0 $ and $ abc\geq \frac{1}{\sqrt{2}} . $ Prove that
$$ a^2+b^2+c^2\geq 3$$Let $ a,b,c $ be reals such that $ a+2b+c=0 $ and $ abc\geq \frac{1}{\sqrt{2}} . $ Prove that
$$ a^2+b^2+c^2\geq \frac{3}{ \sqrt[3]{2}}$$$$ a^2+2b^2+c^2\geq 2\sqrt[3]{4} $$
0 replies
sqing
2 hours ago
0 replies
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Combinatorics
TUAN2k8   0
2 hours ago
A sequence of integers $a_1,a_2,...,a_k$ is call $k-balanced$ if it satisfies the following properties:
$i) a_i \neq a_j$ and $a_i+a_j \neq 0$ for all indices $i \neq j$.
$ii) \sum_{i=1}^{k} a_i=0$.
Find the smallest integer $k$ for which: Every $k-balanced$ sequence, there always exist two terms whose diffence is not less than $n$. (where $n$ is given positive integer)
0 replies
TUAN2k8
2 hours ago
0 replies
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pqr/uvw convert
Nguyenhuyen_AG   4
N 2 hours ago by SunnyEvan
Source: https://github.com/nguyenhuyenag/pqr_convert
Hi everyone,
As we know, the pqr/uvw method is a powerful and useful tool for proving inequalities. However, transforming an expression $f(a,b,c)$ into $f(p,q,r)$ or $f(u,v,w)$ can sometimes be quite complex. That's why I’ve written a program to assist with this process.
I hope you’ll find it helpful!

Download: pqr_convert

Screenshot:
IMAGE
IMAGE
4 replies
Nguyenhuyen_AG
Today at 3:39 AM
SunnyEvan
2 hours ago
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A nice lemma about incircle and his internal tangent
manlio   0
2 hours ago
Have you a nice proof for this lemma?
Thnak you very much
0 replies
manlio
2 hours ago
0 replies
J
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Nice problem about a trapezoid
manlio   0
2 hours ago
Have you a nice solution for this problem?
Thank you very much
0 replies
manlio
2 hours ago
0 replies
J
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IHC 10 Q25: Eight countries participated in a football tournament
xytan0585   0
2 hours ago
Source: International Hope Cup Mathematics Invitational Regional Competition IHC10
Eight countries sent teams to participate in a football tournament, with the Argentine and Brazilian teams being the strongest, while the remaining six teams are similar strength. The probability of the Argentine and Brazilian teams winning against the other six teams is both $\frac{2}{3}$. The tournament adopts an elimination system, and the winner advances to the next round. What is the probability that the Argentine team will meet the Brazilian team in the entire tournament?

$A$. $\frac{1}{4}$

$B$. $\frac{1}{3}$

$C$. $\frac{23}{63}$

$D$. $\frac{217}{567}$

$E$. $\frac{334}{567}$
0 replies
xytan0585
2 hours ago
0 replies
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square root problem that involves geometry
kjhgyuio   8
N Apr 6, 2025 by mathprodigy2011
If x is a nonnegative real number , find the minimum value of √x^2+4 + √x^2 -24x +153

8 replies
kjhgyuio
Apr 5, 2025
mathprodigy2011
Apr 6, 2025
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square root problem that involves geometry
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kjhgyuio
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kjhgyuio
#1 h  Apr 5, 2025, 3:56 AM • 1 Y
Y by PikaPika999
If x is a nonnegative real number , find the minimum value of √x^2+4 + √x^2 -24x +153
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kjhgyuio
45 posts
kjhgyuio
#2 h  Apr 5, 2025, 3:57 AM • 1 Y
Y by PikaPika999
Here’s a hint: Try completing the square
This post has been edited 1 time. Last edited by kjhgyuio, Apr 5, 2025, 3:57 AM
Reason: nil
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ND_
44 posts
ND_
#3 h  Apr 5, 2025, 5:14 AM • 1 Y
Y by PikaPika999
kjhgyuio wrote:
If x is a nonnegative real number , find the minimum value of $\sqrt{x^2+4} + \sqrt{x^2 -24x +153}$
Let \( P \) and \( Q \) be \( (0,2) \) and \( (12,3) \) and \( X \) be a point on the x axis, then our required expression is \( XP + XQ \), which is minimized when \( X \) lies on \( P'Q \), where \( P' = (0,-2) \). So the minimum value is \( \sqrt{12^2 + (3 - (-2))^2} = 13 \).
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kjhgyuio
45 posts
kjhgyuio
#4 h  Apr 5, 2025, 9:07 AM
Y by
ND_ wrote:
kjhgyuio wrote:
If x is a nonnegative real number , find the minimum value of $\sqrt{x^2+4} + \sqrt{x^2 -24x +153}$
Let \( P \) and \( Q \) be \( (0,2) \) and \( (12,3) \) and \( X \) be a point on the x axis, then our required expression is \( XP + XQ \), which is minimized when \( X \) lies on \( P'Q \), where \( P' = (0,-2) \). So the minimum value is \( \sqrt{12^2 + (3 - (-2))^2} = 13 \).

Another method : complete the square to find √x^2-24x+153 =√(x-12)^2+9 let 2 and x be the legs of a right angles triangle .then ,based on pythagoras.c^2=√x^2+4 similarly for √x^2-24x+153 let (x-2) and 3 be the legs of a right angled triangle C^2 =√(x-2)^2+9 find that the smallest configuration of x leads to making a big triangle with hypotenuse √x^2+4 + √x^2 -24x +153 and the two legs have lengths 5 and 12.Using pythagoras again 5^2+12^2=169 √169 =13
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ehuseyinyigit
810 posts
ehuseyinyigit
#5 h  Apr 5, 2025, 9:22 AM
Y by
Released!! That Minkoski problem being shared per a week have been released.
$$\sqrt{x^2+4}+\sqrt{(x-12)^2+9}\geq \sqrt{12^2+5^2}=13$$
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kjhgyuio
45 posts
kjhgyuio
#6 h  Apr 5, 2025, 9:46 AM
Y by
ehuseyinyigit wrote:
Released!! That Minkoski problem being shared per a week have been released.
$$\sqrt{x^2+4}+\sqrt{(x-12)^2+9}\geq \sqrt{12^2+5^2}=13$$

?
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Primeniyazidayi
71 posts
Primeniyazidayi
#8 h  Apr 5, 2025, 11:39 AM
Y by
kjhgyuio wrote:
ehuseyinyigit wrote:
Released!! That Minkoski problem being shared per a week have been released.
$$\sqrt{x^2+4}+\sqrt{(x-12)^2+9}\geq \sqrt{12^2+5^2}=13$$

?

Here is the Minkowski inequality:
If $x_1,\cdots,x_n,y_1,\cdots,y_n$ are real numbers and $p$ is a random positive real number such that $1 < p < \infty$,then $$\sqrt[p]{\sum_{i=1}^n \left| x_i + y_i \right|^p} \le \sqrt[p]{\left| \sum_{i=1}^n x_i \right|^p} + \sqrt[p]{\left| \sum_{i=1}^n y_i \right|^p}$$
This post has been edited 1 time. Last edited by Primeniyazidayi, Apr 5, 2025, 12:44 PM
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kjhgyuio
45 posts
kjhgyuio
#9 h  Apr 6, 2025, 12:23 AM
Y by
Primeniyazidayi wrote:
kjhgyuio wrote:
ehuseyinyigit wrote:
Released!! That Minkoski problem being shared per a week have been released.
$$\sqrt{x^2+4}+\sqrt{(x-12)^2+9}\geq \sqrt{12^2+5^2}=13$$

?

Here is the Minkowski inequality:
If $x_1,\cdots,x_n,y_1,\cdots,y_n$ are real numbers and $p$ is a random positive real number such that $1 < p < \infty$,then $$\sqrt[p]{\sum_{i=1}^n \left| x_i + y_i \right|^p} \le \sqrt[p]{\left| \sum_{i=1}^n x_i \right|^p} + \sqrt[p]{\left| \sum_{i=1}^n y_i \right|^p}$$
ok thanks
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mathprodigy2011
313 posts
mathprodigy2011
#10 h  Apr 6, 2025, 1:21 AM
Y by
taking the derivative is probably much easier if you don't see the geometric intuition
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