Embracing tetrahedra

Geometry Level 4

Consider a regular tetrahedron with a volume of 60. Reflect this tetrahedron about its center to obtain a second tetrahedron. Find the volume of the union of these two tetrahedra.


The answer is 90.

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1 solution

Otto Bretscher
Oct 24, 2015

The union of the two tetrahedra is the disjoint union of one of them (the big red one in the attached figure, say) with four smaller regular tetrahedra (colored yellow in the figure). The small yellow tetrahedra have half the height (and therefore an eighth of the volume) of the big red tetrahedron. Thus the volume of the union is 60 + ( 4 1 8 60 ) = 90 60+\left(4*\frac{1}{8}*60\right)=\boxed{90} .

Are you good at everything?!?!?!!

Pi Han Goh - 5 years, 7 months ago

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As you know, one of the many things I'm NOT good at are inequalities ;)

Otto Bretscher - 5 years, 7 months ago

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Sorry, let me rephrase that: "Except for classical inequalities, are you good at everything?!?!?!!" HAHAAHAH

Pi Han Goh - 5 years, 7 months ago

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@Pi Han Goh I need to study the Stolz lemma... the guy even had my first name ;)

Otto Bretscher - 5 years, 7 months ago

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@Pi Han Goh Thanks for the effort, but they say you can't teach an old dog new tricks... somehow this things looks too contrived and "forced" to me, kind of like "classical inequalities" ;) I'm sure it's an "art form", though...

Otto Bretscher - 5 years, 7 months ago

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@Otto Bretscher Man you're stubborn! Haha! It simplifies most limits much easily. It's especially useful when don't have to apply the Stirling's approximation which is a tedious task to simplify.

See this . Brian's one is the standard approach which requires plenty of simplification, whereas mine is as close to a one-line solution.

You can simplify limits like ( 2 n n ) 1 / n \dbinom{2n}n^{1/n} much easily compared to Stirling's.

Pi Han Goh - 5 years, 7 months ago

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@Pi Han Goh I did that problem with a Riemann sum as well... that is one of the techniques that seem natural to me.

Otto Bretscher - 5 years, 7 months ago

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