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.
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Are you good at everything?!?!?!!
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As you know, one of the many things I'm NOT good at are inequalities ;)
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Sorry, let me rephrase that: "Except for classical inequalities, are you good at everything?!?!?!!" HAHAAHAH
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@Pi Han Goh – I need to study the Stolz lemma... the guy even had my first name ;)
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@Otto Bretscher – This is all that you need to know . Enjoy!
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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...
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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 ( n 2 n ) 1 / n much easily compared to Stirling's.
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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.
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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 6 0 + ( 4 ∗ 8 1 ∗ 6 0 ) = 9 0 .