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Grid combo with tilings
a_507_bc   7
N 36 minutes ago by john0512
Source: All-Russian MO 2023 Final stage 10.6
A square grid $100 \times 100$ is tiled in two ways - only with dominoes and only with squares $2 \times 2$. What is the least number of dominoes that are entirely inside some square $2 \times 2$?
7 replies
a_507_bc
Apr 23, 2023
john0512
36 minutes ago
sqrt(2)<=|1+z|+|1+z^2|<=4
SuiePaprude   3
N 38 minutes ago by alpha31415
let z be a complex number with |z|=1 show that sqrt2 <=|1+z|+|1+z^2|<=4
3 replies
SuiePaprude
Jan 23, 2025
alpha31415
38 minutes ago
Simple but hard
Lukariman   5
N 41 minutes ago by Giant_PT
Given triangle ABC. Outside the triangle, construct rectangles ACDE and BCFG with equal areas. Let M be the midpoint of DF. Prove that CM passes through the center of the circle circumscribing triangle ABC.
5 replies
Lukariman
Today at 2:47 AM
Giant_PT
41 minutes ago
inequalities
Ducksohappi   0
43 minutes ago
let a,b,c be non-negative numbers such that ab+bc+ca>0. Prove:
$ \sum_{cyc} \frac{b+c}{2a^2+bc}\ge \frac{6}{a+b+c}$
P/s: I have analysed:$ S_a=\frac{b^2+c^2+3bc-ab-ac}{(2b^2+ac)(2c^2+2ab)}$, similarly to $S_b, S_c$, by SOS
0 replies
Ducksohappi
43 minutes ago
0 replies
bulgarian concurrency, parallelograms and midpoints related
parmenides51   7
N an hour ago by Ilikeminecraft
Source: Bulgaria NMO 2015 p5
In a triangle $\triangle ABC$ points $L, P$ and $Q$ lie on the segments $AB, AC$ and $BC$, respectively, and are such that $PCQL$ is a parallelogram. The circle with center the midpoint $M$ of the segment $AB$ and radius $CM$ and the circle of diameter $CL$ intersect for the second time at the point $T$. Prove that the lines $AQ, BP$ and $LT$ intersect in a point.
7 replies
parmenides51
May 28, 2019
Ilikeminecraft
an hour ago
Interesting inequalities
sqing   3
N an hour ago by sqing
Source: Own
Let $a,b,c \geq 0 $ and $ab+bc+ca- abc =3.$ Show that
$$a+k(b+c)\geq 2\sqrt{3 k}$$Where $ k\geq 1. $
Let $a,b,c \geq 0 $ and $2(ab+bc+ca)- abc =31.$ Show that
$$a+k(b+c)\geq \sqrt{62k}$$Where $ k\geq 1. $
3 replies
sqing
Today at 4:34 AM
sqing
an hour ago
Fast-growing sequence is always an integer
MarkBcc168   18
N an hour ago by Thapakazi
Source: Fake USAMO 2020 P1, Fake USAJMO 2020 P3
Suppose that $C$ is a positive integer and $a_1,a_2,a_3,\hdots$ is an infinite sequence of positive integers satisfying
$$a_{n+1}=\sqrt{a_n^3-Ca_n}$$for all positive integers $n$. Prove that this sequence must be eventually constant, i.e. there exists a positive integer $N$ such that $a_N = a_{N+1} = a_{N+2} = \hdots$.

Proposed by YRNG-BCC168.
18 replies
MarkBcc168
Apr 28, 2020
Thapakazi
an hour ago
AC bisects BE, BC = DE, CD//BE, <BAC = <DAE, AB/BD=AE/ED
parmenides51   3
N an hour ago by pku
Source: China Northern MO 2012 p7 CNMO
As shown in figure , in the pentagon $ABCDE$, $BC = DE$, $CD \parallel BE$, $AB>AE$. If $\angle BAC = \angle DAE$ and $\frac{AB}{BD}=\frac{AE}{ED}$. Prove that $AC$ bisects the line segment $BE$.
IMAGE
3 replies
parmenides51
Oct 28, 2022
pku
an hour ago
Angle Relationships in Triangles
steven_zhang123   3
N an hour ago by Bergo1305
In $\triangle ABC$, $AB > AC$. The internal angle bisector of $\angle BAC$ and the external angle bisector of $\angle BAC$ intersect the ray $BC$ at points $D$ and $E$, respectively. Given that $CE - CD = 2AC$, prove that $\angle ACB = 2\angle ABC$.
3 replies
steven_zhang123
Wednesday at 11:09 PM
Bergo1305
an hour ago
D1032 : A general result on polynomial 2
Dattier   4
N an hour ago by alpha31415
Source: les dattes à Dattier
Let $P \in \mathbb Q[x,y]$ with $\max(\deg_x(P),\deg_y(P)) \leq d$ and $\forall (a,b) \in \mathbb Z^2 \cap [0,d]^2, P(a,b) \in \mathbb Z$.

Is it true that $\forall (a,b) \in\mathbb Z^2, P(a,b) \in \mathbb Z$?
4 replies
Dattier
Wednesday at 5:19 PM
alpha31415
an hour ago
Non-negative real variables inequality
KhuongTrang   2
N Apr 29, 2025 by NguyenVanHoa29
Source: own
Problem. Let $a,b,c\ge 0: ab+bc+ca>0.$ Prove that$$\color{blue}{\frac{\left(2ab+ca+cb\right)^{2}}{a^{2}+4ab+b^{2}}+\frac{\left(2bc+ab+ac\right)^{2}}{b^{2}+4bc+c^{2}}+\frac{\left(2ca+bc+ba\right)^{2}}{c^{2}+4ca+a^{2}}\ge \frac{8(ab+bc+ca)}{3}.}$$
2 replies
KhuongTrang
Apr 24, 2025
NguyenVanHoa29
Apr 29, 2025
Non-negative real variables inequality
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Source: own
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KhuongTrang
731 posts
#1 • 4 Y
Y by math90, NguyenVanHoa29, Zuyong, TNKT
Problem. Let $a,b,c\ge 0: ab+bc+ca>0.$ Prove that$$\color{blue}{\frac{\left(2ab+ca+cb\right)^{2}}{a^{2}+4ab+b^{2}}+\frac{\left(2bc+ab+ac\right)^{2}}{b^{2}+4bc+c^{2}}+\frac{\left(2ca+bc+ba\right)^{2}}{c^{2}+4ca+a^{2}}\ge \frac{8(ab+bc+ca)}{3}.}$$
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Quantum-Phantom
275 posts
#2
Y by
Are there any easier methods?

After multiplying both sides by \(\prod\limits_{\rm cyc}\left(a^2+4ab+b^2\right)\), we need to show that
\[\frac13\sum_{\rm cyc}a^2b^2(a+3b)(3a+b)(a-b)^2+abc\cdot f(a,b,c)\ge0,\]where $f(a,b,c)$ is a fifth degree polynomial:
\[\sum_{\rm cyc}\left(2a^5+\frac83a^4b+\frac83ab^4+\frac{22}3a^2b^2c-\frac{14}3a^3b^2-\frac{14}3a^2b^3-\frac{16}3a^2b^2c\right).\]By the $uvw$ method, it is not hard to show that $f(a,b,c)\ge0$ is true.

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NguyenVanHoa29
9 posts
#3 • 1 Y
Y by arqady
I think it is a concave function according to w^3 and the rest is easy checking.
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