geometric ineq

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0 replies

jlacosta

Apr 2, 2025

0 replies

external bisector in 2 angle

crocodilepradita 5

N 21 minutes ago by ihategeo_1969

Given a $\triangle ABC$ with incenter $I$ . Line $BI$ and $CI$ intersects $CA$ and $AB$ at $E$ and $F$ , respectively. Let $M$ and $N$ be the midpoints of $BI$ and $CI$ . Line $FM$ meets the external bisector of angle $B$ at $K$ and line $EN$ meets the external bisector of angle $C$ at $L$ . Prove that $K, B, L, C$ are concylic.

5 replies

crocodilepradita

Aug 22, 2024

ihategeo_1969

21 minutes ago

Interesting inequalities

sqing 6

N an hour ago by MathPerson12321

Source: Own

Let $a,b\geq 0$ and $a+b=2.$ Prove that
$ab( a^2+ b^2)^2 \leq \frac{128}{27}$ $ab( a^2-ab+ b^2)^2 \leq \frac{256}{81}$ $ab\sqrt{ab}( a^2+ b^2)^2 \leq \frac{1536}{343}\sqrt{\frac{6}{7}}$ $ab\sqrt{ab}( a^2-ab+ b^2)^2 \leq \frac{2048}{343\sqrt{7}}$

6 replies

1 viewing

sqing

2 hours ago

MathPerson12321

an hour ago

Difficult combinatorics problem about distinct sums under shifts

CBMaster 0

an hour ago

Source: Korea

Problem. Let $a_1, ..., a_n$ be the nonnegative integers in $\{0, 1, ..., m\}$ where $m=\left\lceil \frac{n^{2/3}}{4} \right\rceil$ . Define $A=\{a_i+a_j+(j-i)|1\leq i<j\leq n\}$ . Prove that $|A|\geq m$ .

Bonus problem (Open). Can we prove a tighter result than the one above? That is, is there a function $f(n)$ such that $f(n)=O(n^\alpha)$ where $\alpha>\frac{2}{3}$ , and the statement is still true when $m=f(n)$ ?
Or, is there a function $f(n)$ such that $f(n)\geq C \cdot n^{2/3}$ where $C>\frac{1}{4}$ , and the statement is still true when $m=f(n)$ ?.

0 replies

CBMaster

an hour ago

0 replies

4 variables with quadrilateral sides

mihaig 5

N an hour ago by mihaig

Source: VL

Let $a,b,c,d\geq0$ satisfying
$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}+\frac1{d+1}=2.$ Prove
$4\left(abc+abd+acd+bcd\right)\geq3\left(a+b+c+d\right)+4.$

5 replies

mihaig

Apr 25, 2025

mihaig

an hour ago

No more topics!

geometric ineq

ivanbart-15 2

N Sep 7, 2021 by Giulio2004

Source: Australia 1984

Let $\triangle ABC$ be a triangle and let the internal bisector of the angle $\angle BAC$ intersects the circumcircle again at $P$ . Similarly define $Q$ and $R$ . Prove that $AP+BQ+CR>AB+BC+CA$ .

2 replies

ivanbart-15

Aug 28, 2011

Giulio2004

Sep 7, 2021

geometric ineq

G H J

Reveal topic tags

geometry

circumcircle

geometry unsolved

G H BBookmark kLocked kLocked NReply

Source: Australia 1984

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ivanbart-15

296 posts

ivanbart-15

#1 h Aug 28, 2011, 9:21 PM • 2 Y

Y by Adventure10, Mango247

Let $\triangle ABC$ be a triangle and let the internal bisector of the angle $\angle BAC$ intersects the circumcircle again at $P$ . Similarly define $Q$ and $R$ . Prove that $AP+BQ+CR>AB+BC+CA$ .

This post has been edited 1 time. Last edited by ivanbart-15, Sep 24, 2011, 1:30 PM

Z K Y

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alpch

45 posts

alpch

#2 h Aug 28, 2011, 9:56 PM • 2 Y

Y by Adventure10, Mango247

Let $I$ is the intersections of internal bisector. If you write angles tou see that $IR=IA=IB$
in the $\triangle AIC$ ;
$CR=CI+IR=CI+IA>AC$ similary $CR>AC,AP>AB,BQ>BC$
the sum of this we get the solution

Z K Y

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Giulio2004

1 post

Giulio2004

#3 h Sep 7, 2021, 10:17 PM • 1 Y

Y by Mango247

Why IA=IB?

Z K Y

N Quick Reply