Tetrahedron inside a hemisphere of radius 1

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Tetrahedron inside a hemisphere of radius 1

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Source: Problem 6, Polish NO 1988

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1198 posts

Megus

#1 h Oct 16, 2005, 2:18 PM • 2 Y

Y by Adventure10, Mango247

Find the largest possible volume for a tetrahedron which lies inside a hemisphere of radius $1$ .

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#2 h Apr 26, 2011, 3:56 PM • 1 Y

Y by Adventure10

Are you sure it's a hemisphere and not sphere? It's a lot more complicated with a hemisphere than a sphere.

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#3 h Jul 12, 2011, 1:42 PM • 2 Y

Y by Adventure10, Mango247

I have found $V_{\max}=\frac{\sqrt{3}}{4}$ , is it correct?

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#4 h Jul 12, 2011, 3:35 PM • 1 Y

Y by Adventure10

tuan119 wrote:

I have found $V_{\max}=\frac{\sqrt{3}}{4}$ , is it correct?

$V_{\max}=\frac{\sqrt{3}}{4}$ is correct.

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#5 h Jul 13, 2011, 2:22 AM • 2 Y

Y by Adventure10, Mango247

//cdn.artofproblemsolving.com/images/cf7d72977021302816e6b30bdeef52e01fcff82c.jpg

We have: $V=\frac{1}{3}S_{EFD}.d(C;(EFD));$
We use: $a=2R\sin E; b=2R\sin F; c=2R\sin D$ and $S_{EFD}=\frac{abc}{4R}=\frac{abc}{4}; \implies a=b=c=\sqrt3; h=1$
$V_{tetrahedron} \le V_{C.DEF}$ , where $C, D, E, F \in$ hemisphere $(H)$ .
$\implies V_{\max} =\frac{\sqrt3}{4}$

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#6 h Jul 13, 2011, 3:23 AM • 2 Y

Y by Adventure10, Mango247

To find the max. volume is very easy but to eliminate other possible configurations is difficult.

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#7 h Nov 7, 2011, 6:58 PM • 2 Y

Y by Adventure10, Mango247

It's obvious to see the configuration where the volume attains its maximum, but to solve the problem, one has to show the way to eliminate other configuration which is not easy.

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#8 h Apr 11, 2018, 10:12 AM • 2 Y

Y by Adventure10, Mango247

Problem is nice. We want to take the biggest value of height and area of base. Height is at most $R=1$ and base, which is triangle has an area at most $\frac{3r^2\sqrt{3}}{4}$ where $r$ is a radius of circle circumscribed on base. We know that $r\le R=1$ . So $V\le \frac{1}{3}\frac{3r^2\sqrt3}{4}\cdot R\le \frac{\sqrt3}{4}$
We see that $CDEF$ in post $5$ satisfies conditions if $CDE$ is an equilateral triangle so the answer is $\frac{\sqrt3}{4}$

This post has been edited 3 times. Last edited by WolfusA, Apr 11, 2018, 10:14 AM

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Smita

#9 h Apr 11, 2018, 11:20 AM • 2 Y

Y by Adventure10, Mango247

tuan119 wrote:

http://img38.picoodle.com/i57c/duytuan119/id_bb4_u9n46.jpg

We have: $V=\frac{1}{3}S_{EFD}.d(C;(EFD));$
We use: $a=2R\sin E; b=2R\sin F; c=2R\sin D$ and $S_{EFD}=\frac{abc}{4R}=\frac{abc}{4}; \implies a=b=c=\sqrt3; h=1$
$V_{tetrahedron} \le V_{C.DEF}$ , where $C, D, E, F \in$ hemisphere $(H)$ .
$\implies V_{\max} =\frac{\sqrt3}{4}$

but how can u prove that it is the maximum volume????

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#10 h Apr 11, 2018, 12:30 PM • 2 Y

Y by Adventure10, Mango247

I have proved this...

WolfusA wrote:

Height is at most $R=1$ and base, which is triangle has an area at most $\frac{3r^2\sqrt{3}}{4}$ where $r$ is a radius of circle circumscribed on base. We know that $r\le R=1$ . So $V\le \frac{1}{3}\frac{3r^2\sqrt3}{4}\cdot R\le \frac{\sqrt3}{4}$

EDIT: I see, the question was addressed to tuan119. You're right. That's no proof.

This post has been edited 1 time. Last edited by WolfusA, Apr 11, 2018, 12:31 PM

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