incenter of ABC is orthocenter of triangles of excenters

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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

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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!

incenter of ABC is orthocenter of triangles of excenters

parmenides51 3

N Oct 17, 2024 by AshAuktober

Source: Indonesia INAMO Shortlist 2009 G10 https://artofproblemsolving.com/community/c1101409_

Given a triangle $ABC$ with incenter $I$ . It is known that $E_A$ is center of the ex-circle tangent to $BC$ . Likewise, $E_B$ and $E_C$ are the centers of the ex-circles tangent to $AC$ and $AB$ , respectively. Prove that $I$ is the orthocenter of the triangle $E_AE_BE_C$ .

3 replies

parmenides51

Dec 10, 2021

AshAuktober

Oct 17, 2024

incenter of ABC is orthocenter of triangles of excenters

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Source: Indonesia INAMO Shortlist 2009 G10 https://artofproblemsolving.com/community/c1101409_

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parmenides51

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parmenides51

#1 h Dec 10, 2021, 7:38 PM

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MathLuis

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MathLuis

#2 h Dec 10, 2021, 7:49 PM

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Apply I-E lemma 3 times and ur done

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removablesingularity

564 posts

removablesingularity

#3 h Nov 17, 2023, 3:54 AM

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Prove $AE_A$ perpendicular with $E_B E_C$ via $\angle E_B A C = \frac{1}{2} \left(\pi - \angle A\right) = 90 - \frac{\angle A}{2}$ and $\angle CAE_A = \frac{A}{2}$ , hence $\angle E_BAE_A = 90 - \frac{\angle A}{2} + \frac{\angle A}{2} = 90$ .

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AshAuktober

987 posts

AshAuktober

#4 h Oct 17, 2024, 6:51 AM

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This is... trivial?
Notice that $ABC$ is the orthic triangle of $E_AE_BE_C$ by angle chase so we're done.

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