inequality in a tetrahedron with incenter & orthocenter

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parmenides51

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parmenides51

#1 h Jul 20, 2018, 12:48 PM • 3 Y

Y by LiusTheo2014, Adventure10, Mango247

Let $OABC$ be a tetrahedron such that $OA \perp OB \perp OC \perp OA$ , $r$ be the radius of its inscribed sphere and $H$ be the orthocenter of triangle $ABC$ . Prove that $OH \le r(\sqrt3 +1)$

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mihaig

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mihaig

#2 h Jul 20, 2018, 3:00 PM • 4 Y

Y by LiusTheo2014, archimedes26, Adventure10, Mango247

Beautiful.

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arqady

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arqady

#3 h Jul 20, 2018, 4:18 PM • 4 Y

Y by LiusTheo2014, Pirkuliyev Rovsen, Adventure10, Mango247

For the collection.
The following problem was in my entrance exam to Moscow University in 1979.
Let $OABC$ be a tetrahedron such that $OA \perp OB$ , $OA \perp OC$ , $OB\perp OC$ and $H$ be an orthocenter of the triangle $ABC$ .
Prove that:
$OH\leq\frac{\sqrt[3]{HA\cdot HB\cdot HC}}{\sqrt2}.$

This post has been edited 2 times. Last edited by arqady, Jul 20, 2018, 4:24 PM

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mihaig

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mihaig

#4 h Jul 20, 2018, 5:10 PM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

For the collection.
The following problem was in my entrance exam to Moscow University in 1979.
Let $OABC$ be a tetrahedron such that $OA \perp OB$ , $OA \perp OC$ , $OB\perp OC$ and $H$ be an orthocenter of the triangle $ABC$ .
Prove that:
$OH\leq\frac{\sqrt[3]{HA\cdot HB\cdot HC}}{\sqrt2}.$

Respect!
Because I don't know,even in our modern times,how many people are able to solve this during an exam.
P.S. In order not to repeat myself,I chose an avant-garde proof for this problem,not as to main topic,a geometric one.

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WolfusA

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WolfusA

#5 h Aug 3, 2018, 3:31 PM • 2 Y

Y by Adventure10, Mango247

Let $OABC$ be a tetrahedron such that $OA \perp OB$ , $OA \perp OC$ , $OB\perp OC$ and $H$ be an orthocenter of the triangle $ABC$ . Then foot of $O$ on plane $ABC$ is orthocenter of triangle $ABC$ .
proof

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mihaig

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mihaig

#6 h Aug 4, 2018, 3:27 AM • 2 Y

Y by Adventure10, Mango247

WolfusA wrote:

Let $OABC$ be a tetrahedron such that $OA \perp OB$ , $OA \perp OC$ , $OB\perp OC$ and $H$ be an orthocenter of the triangle $ABC$ . Then foot of $O$ on plane $ABC$ is orthocenter of triangle $ABC$ .
proof

Wolfusica,you solved here an old dilemma known by any 8-th grade student.
But you didn't solve the proposed problem.

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mihaig

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mihaig

#7 h Aug 4, 2018, 3:38 AM • 2 Y

Y by Adventure10, Mango247

This is sharper:
https://leogiugiuc.wordpress.com/2018/07/24/accessible-geometry-problem-2/

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WolfusA

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WolfusA

#8 h Aug 6, 2018, 3:47 PM • 2 Y

Y by Adventure10, Mango247

mihaig wrote:

Wolfusica (!),you solved here an old dilemma known by any 8-th grade student.
But you didn't solve the proposed problem.

I posted it for the sake of completeness of solution. Your calculations are heavy based on this. Maybe 8-th grade student wants to know how to prove it?

mihaig wrote:

Respect!
Because I don't know,even in our modern times,how many people are able to solve this during an exam.

Referring to arqady's problem. I needed 10 minutes. Though you may say I wasn't on the exam. It's just "after easy calculations"
$2\sqrt2xyz\le\sqrt{x^2+y^2}\cdot\sqrt{z^2+y^2}\cdot\sqrt{x^2+z^2}$ where $AO=x,BO=y,CO=z$ .

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mihaig

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mihaig

#9 h Aug 6, 2018, 7:25 PM • 2 Y

Y by Adventure10, Mango247

Many palavras,less work.

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mihaig

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mihaig

#10 h Sep 12, 2018, 3:58 PM • 1 Y

Y by Adventure10

arqady wrote:

For the collection.
The following problem was in my entrance exam to Moscow University in 1979.
Let $OABC$ be a tetrahedron such that $OA \perp OB$ , $OA \perp OC$ , $OB\perp OC$ and $H$ be an orthocenter of the triangle $ABC$ .
Prove that:
$OH\leq\frac{\sqrt[3]{HA\cdot HB\cdot HC}}{\sqrt2}.$

Thank you,Mike (arqady)!
A true pearl.
Greetings!

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