Geometry problem

Let $a,b,c,d \in R$ , such that $a+b+c+d=4 .$ Prove that :
$a^4+b^4+c^4+d^4+3 \geq \frac{7}{4}(a^3+b^3+c^3+d^3)$ $a^4+b^4+c^4+d^4+ \frac{252}{25} \geq \frac{88}{25}(a^3+b^3+c^3+d^3)$ equality cases : ?

2 replies

SunnyEvan

6 hours ago

ektorasmiliotis

an hour ago

Inspired by JK1603JK

sqing 6

N an hour ago by sqing

Source: Own

Let $a,b,c\geq 0$ and $ab+bc+ca=1.$ Prove that $\frac{abc-2}{abc-1}\ge \frac{4(a^2b+b^2c+c^2a)}{a^3b+b^3c+c^3a+1}$

6 replies

sqing

Today at 3:31 AM

sqing

an hour ago

Geometry problem

kjhgyuio 1

N 2 hours ago by Mathzeus1024

Source: smo

In trapezium ABCD,AD is parallel to BC and points E and F are midpoints of AB and DC respectively. If
Area of AEFD/Area of EBCF =√3 + 1/3-√3 and the area of triangle ABD is √3 .find the area of trapezium ABCD

1 reply

kjhgyuio

Apr 1, 2025

Mathzeus1024

2 hours ago

D1018 : Can you do that ?

Dattier 1

N 2 hours ago by Dattier

Source: les dattes à Dattier

We can find $A,B,C$ , such that $\gcd(A,B)=\gcd(C,A)=\gcd(A,2)=1$ and $\forall n \in \mathbb N^*, (C^n \times B \mod A) \mod 2=0$ .

For example :

$C=20$
$A=47650065401584409637777147310342834508082136874940478469495402430677786194142956609253842997905945723173497630499054266092849839$

$B=238877301561986449355077953728734922992395532218802882582141073061059783672634737309722816649187007910722185635031285098751698$

Can you find $A,B,C$ such that $A>3$ is prime, $C,B \in (\mathbb Z/A\mathbb Z)^*$ with $o(C)=(A-1)/2$ and $\forall n \in \mathbb N^*, (C^n \times B \mod A) \mod 2=0$ ?

1 reply

Dattier

Mar 24, 2025

Dattier

2 hours ago

2025 Caucasus MO Juniors P3

BR1F1SZ 1

N 2 hours ago by FarrukhKhayitboyev

Source: Caucasus MO

Let $K$ be a positive integer. Egor has $100$ cards with the number “ $2$ ” written on them, and $100$ cards with the number “ $3$ ” written on them. Egor wants to paint each card red or blue so that no subset of cards of the same color has the sum of the numbers equal to $K$ . Find the greatest $K$ such that Egor will not be able to paint the cards in such a way.

1 reply

BR1F1SZ

Mar 26, 2025

FarrukhKhayitboyev

2 hours ago

1 area = 2025 points

giangtruong13 0

2 hours ago

In a plane give a set $H$ that has 8097 distinct points with area of a triangle that has 3 points belong to $H$ all $\leq 1$ . Prove that there exists a triangle $G$ that has the area $\leq 1$ contains at least 2025 points that belong to $H$ ( each of that 2025 points can be inside the triangle or lie on the edge of triangle $G$ )X

0 replies

giangtruong13

2 hours ago

0 replies

Burak0609

Burak0609 0

2 hours ago

So if $2 \nmid n\implies$ $2d_2+d_4+d_5=d_7$ is even it's contradiction. I mean $2 \mid n and d_2=2$ .
If $3\mid n \implies d_3=3$ and $(d_6+d_7)^2=n+1,3d_6d_7=n \implies d_6^2-d_6d_7+d_7^2=1$ ,we can see the only solution is $d_6=d_7=1$ and it is contradiction.
If $4 \mid n d_3=4$ and $(d_6+d_7)^2-4d_6d_7=1 \implies d_7=d_6+1$ . So $n=4d_6(d_6+1)$ .İt means $8 \mid n$ .
If $d_6=8 n=4.8.9=288$ but $3 \nmid n$ .İt is contradiction.
If $d_5=8$ we have 2 option. Firstly $d_4=5 \implies 2d_2+d_4+d_5=17=d_7 d_6=16$ but $10 \mid n$ is contradiction. Secondly $d_4=7 \implies d_7=2.2+7+8=19 and d_6=18$ but $3 \nmid n$ is contradiction. I mean $d_4=8 \implies d_7=d_5+12, n=4(d_5+11)(d_5+12) and d_5 \mid n=4(d_5+11)(d_5+12)$ . So $d_5 \mid 4.11.12 \implies d_5 \mid 16.11$ . If $d_5=16 d_6=27$ but $3 \nmid n$ is contradiction. I mean $d_5=11,d_6=22,d_7=23$ . The only solution is $n=2024$ .

0 replies

Burak0609

2 hours ago

0 replies

Good Partitions

va2010 25

N 3 hours ago by lelouchvigeo

Source: 2015 ISL C3

For a finite set $A$ of positive integers, a partition of $A$ into two disjoint nonempty subsets $A_1$ and $A_2$ is $\textit{good}$ if the least common multiple of the elements in $A_1$ is equal to the greatest common divisor of the elements in $A_2$ . Determine the minimum value of $n$ such that there exists a set of $n$ positive integers with exactly $2015$ good partitions.

25 replies

va2010

Jul 7, 2016

lelouchvigeo

3 hours ago

An inequality on triangles sides

nAalniaOMliO 7

N 4 hours ago by navier3072

Source: Belarusian National Olympiad 2025

Numbers $a,b,c$ are lengths of sides of some triangle. Prove the inequality $\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \geq \frac{a+b}{2c}+\frac{b+c}{2a}+\frac{c+a}{2b}$

7 replies

nAalniaOMliO

Mar 28, 2025

navier3072

4 hours ago

D is incenter

Layaliya 3

N 4 hours ago by rong2020

Source: From my friend in Indonesia

Given an acute triangle $ABC$ where $AB > AC$ . Point $O$ is the circumcenter of triangle $ABC$ , and $P$ is the projection of point $A$ onto line $BC$ . The midpoints of $BC$ , $CA$ , and $AB$ are $D$ , $E$ , and $F$ , respectively. The line $AO$ intersects $DE$ and $DF$ at points $Q$ and $R$ , respectively. Prove that $D$ is the incenter of triangle $PQR$ .

3 replies

Layaliya

Yesterday at 11:03 AM

rong2020

4 hours ago

Geometry problem

Mnjr 3

N Sep 11, 2016 by jayme

In the isosceles triangle $ABC$ ( $AC=BC$ ) point $O$ is the circumcenter, the $I$ incenter, and $D$ lies on $BC$ so that lines $OD$ and $BI$ are perpendicular. Prove that $ID$ and $AC$ are parallel.

3 replies

Mnjr

Sep 7, 2016

jayme

Sep 11, 2016

Geometry problem

G H J

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Mnjr

34 posts

Mnjr

#1 h Sep 7, 2016, 3:58 PM • 2 Y

Y by rezareza14, Adventure10

In the isosceles triangle $ABC$ ( $AC=BC$ ) point $O$ is the circumcenter, the $I$ incenter, and $D$ lies on $BC$ so that lines $OD$ and $BI$ are perpendicular. Prove that $ID$ and $AC$ are parallel.

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Mnjr

34 posts

Mnjr

#2 h Sep 7, 2016, 5:22 PM • 2 Y

Y by Adventure10, Mango247

any solution?

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