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Number Theory Chain!
JetFire008   60
N 41 minutes ago by whwlqkd
I will post a question and someone has to answer it. Then they have to post a question and someone else will answer it and so on. We can only post questions related to Number Theory and each problem should be more difficult than the previous. Let's start!

Question 1
60 replies
JetFire008
Apr 7, 2025
whwlqkd
41 minutes ago
3 right-angled triangle area
NicoN9   1
N an hour ago by Mathzeus1024
Source: Japan Junior MO Preliminary 2020 P1
Right angled triangle $ABC$, and a square are drawn as shown below. Three numbers written below implies each of the area of shaded small right angled triangle. Find the value of $AB/AC$.

IMAGE
1 reply
NicoN9
Yesterday at 6:08 AM
Mathzeus1024
an hour ago
two sequences of positive integers and inequalities
rmtf1111   50
N an hour ago by math-olympiad-clown
Source: EGMO 2019 P5
Let $n\ge 2$ be an integer, and let $a_1, a_2, \cdots , a_n$ be positive integers. Show that there exist positive integers $b_1, b_2, \cdots, b_n$ satisfying the following three conditions:

$\text{(A)} \ a_i\le b_i$ for $i=1, 2, \cdots , n;$

$\text{(B)} \ $ the remainders of $b_1, b_2, \cdots, b_n$ on division by $n$ are pairwise different; and

$\text{(C)} \ $ $b_1+b_2+\cdots b_n \le n\left(\frac{n-1}{2}+\left\lfloor \frac{a_1+a_2+\cdots a_n}{n}\right \rfloor \right)$

(Here, $\lfloor x \rfloor$ denotes the integer part of real number $x$, that is, the largest integer that does not exceed $x$.)
50 replies
rmtf1111
Apr 10, 2019
math-olympiad-clown
an hour ago
inequalities
Tamako22   0
an hour ago
let $a,b,c> 1,\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}=1.$
prove that$$\sqrt{a}+\sqrt{b}+\sqrt{c}\ge \dfrac{2}{\sqrt{a}}+\dfrac{2}{\sqrt{b}}+\dfrac{2}{\sqrt{c}}$$
0 replies
Tamako22
an hour ago
0 replies
Problem 6
SlovEcience   2
N an hour ago by mashumaro
Given two points A and B on the unit circle. The tangents to the circle at A and B intersect at point P. Then:
\[ p = \frac{2ab}{a + b} \], \[ p, a, b \in \mathbb{C} \]
2 replies
SlovEcience
3 hours ago
mashumaro
an hour ago
A coincidence about triangles with common incenter
flower417477   3
N an hour ago by mashumaro
$\triangle ABC,\triangle ADE$ have the same incenter $I$.Prove that $BCDE$ is concyclic iff $BC,DE,AI$ is concurrent
3 replies
flower417477
Apr 30, 2025
mashumaro
an hour ago
this geo is scarier than the omega variant
AwesomeYRY   11
N an hour ago by LuminousWolverine
Source: TSTST 2021/6
Triangles $ABC$ and $DEF$ share circumcircle $\Omega$ and incircle $\omega$ so that points $A,F,B,D,C,$ and $E$ occur in this order along $\Omega$. Let $\Delta_A$ be the triangle formed by lines $AB,AC,$ and $EF,$ and define triangles $\Delta_B, \Delta_C, \ldots, \Delta_F$ similarly. Furthermore, let $\Omega_A$ and $\omega_A$ be the circumcircle and incircle of triangle $\Delta_A$, respectively, and define circles $\Omega_B, \omega_B, \ldots, \Omega_F, \omega_F$ similarly.

(a) Prove that the two common external tangents to circles $\Omega_A$ and $\Omega_D$ and the two common external tangents to $\omega_A$ and $\omega_D$ are either concurrent or pairwise parallel.

(b) Suppose that these four lines meet at point $T_A$, and define points $T_B$ and $T_C$ similarly. Prove that points $T_A,T_B$, and $T_C$ are collinear.

Nikolai Beluhov
11 replies
AwesomeYRY
Dec 13, 2021
LuminousWolverine
an hour ago
No function f on reals such that f(f(x))=x^2-2
N.T.TUAN   17
N an hour ago by Assassino9931
Source: VietNam TST 1990
Prove that there does not exist a function $f: \mathbb R\to\mathbb R$ such that $f(f(x))=x^2-2$ for all $x\in\mathbb R$.
17 replies
N.T.TUAN
Dec 31, 2006
Assassino9931
an hour ago
Hard diophant equation
MuradSafarli   4
N an hour ago by iniffur
Find all positive integers $x, y, z, t$ such that the equation

$$
2017^x + 6^y + 2^z = 2025^t
$$
is satisfied.
4 replies
MuradSafarli
Yesterday at 6:12 PM
iniffur
an hour ago
Geometry that "looks" hard
Pmshw   3
N 2 hours ago by Lemmas
Source: Iran 2nd round 2022 P6
we have an isogonal triangle $ABC$ such that $BC=AB$. take a random $P$ on the altitude from $B$ to $AC$.
The circle $(ABP)$ intersects $AC$ second time in $M$. Take $N$ such that it's on the segment $AC$ and $AM=NC$ and $M \neq N$.The second intersection of $NP$ and circle $(APB)$ is $X$ , ($X \neq P$) and the second intersection of $AB$ and circle $(APN)$ is $Y$ ,($Y \neq A$).The tangent from $A$ to the circle $(APN)$ intersects the altitude from $B$ at $Z$.
Prove that $CZ$ is tangent to circle $(PXY)$.
3 replies
Pmshw
May 9, 2022
Lemmas
2 hours ago
a