centroid construction, ruler only, using ABC, circumcircle, incenter

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0 replies

jlacosta

Apr 2, 2025

0 replies

ARMO 2025

Oksutok 0

5 minutes ago

All-Russian MO 2025 Day 1

0 replies

Oksutok

5 minutes ago

0 replies

general form

pennypc123456789 1

N 7 minutes ago by sqing

If $a,b,c$ are positive real numbers, $k \ge 3$ then
$\frac{a + b}{a + kb + c} + \dfrac{b + c}{b + kc + a}+\dfrac{c + a}{c + ka + b} \geq \dfrac{6}{k+2}$

1 reply

pennypc123456789

3 hours ago

sqing

7 minutes ago

JBMO Shortlist 2023 A4

Orestis_Lignos 6

N 9 minutes ago by Novmath

Source: JBMO Shortlist 2023, A4

Let $a,b,c,d$ be positive real numbers with $abcd=1$ . Prove that

$\sqrt{\frac{a}{b+c+d^2+a^3}}+\sqrt{\frac{b}{c+d+a^2+b^3}}+\sqrt{\frac{c}{d+a+b^2+c^3}}+\sqrt{\frac{d}{a+b+c^2+d^3}} \leq 2$

6 replies

Orestis_Lignos

Jun 28, 2024

Novmath

9 minutes ago

2 var inquality

sqing 5

N 21 minutes ago by sqing

Source: Own

Let $a,b\geq 0$ and $a+ b+2ab=4 .$ Prove that
$3(a+b-2)(2 - ab) \ge (a-1)(b-1)(a-b)$ $9 (a+b-2)(3 - 2ab) \ge 2\sqrt 5(a-1)(b-1)(a-b)$ $9(a+b-2)(6 - 5ab) \ge2\sqrt {14} (a-1)(b-1)(a-b)$

5 replies

sqing

Today at 1:50 AM

sqing

21 minutes ago

No more topics!

centroid construction, ruler only, using ABC, circumcircle, incenter

parmenides51 1

N Jun 23, 2021 by ppanther

Source: 2018 Olympiad of Rusanovsky Lyceum in Kyiv, Ukraine , IX_X Team p8

Given a triangle $ABC$ , a circle circumcribed around it and the center of the inscribed circle. Without using a compass, use a ruler (without divisions) to construct the centroid of the triangle $ABC$ .

(G. Filippovsky)

1 reply

parmenides51

Jun 23, 2021

ppanther

Jun 23, 2021

centroid construction, ruler only, using ABC, circumcircle, incenter

G H J

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Source: 2018 Olympiad of Rusanovsky Lyceum in Kyiv, Ukraine , IX_X Team p8

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parmenides51

30629 posts

parmenides51

#1 h Jun 23, 2021, 7:02 PM • 2 Y

Y by Mango247, UH16

This post has been edited 2 times. Last edited by parmenides51, Jul 4, 2021, 11:55 PM
Reason: latex

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ppanther

160 posts

ppanther

#2 h Jun 23, 2021, 7:29 PM • 1 Y

Y by UH16

Denote by $I$ the incenter. Intersect rays $AI,BI,CI$ with the circumcircle at the arc midpoints $D,E,F$ , respectively. Construct $X = DF \cap BI$ and $Y = DE \cap CI$ .

Then $XY \parallel BC$ , since $\angle IXY = \angle EFI = \angle IBC$ , as $I$ is the orthocenter of $\triangle DEF$ .

Finally, if $L = BY \cap CX$ , then $IL \cap BC$ is the midpoint of $BC$ by Ceva.

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