zmudz
  • zmudz
Let \(a, b, c,\) and \(d\) be the areas of the triangular faces of a tetrahedron, and let \(h_a, h_b, h_c,\) and \(h_d\) be the corresponding altitudes of the tetrahedron. If \(V\) denotes the volume of the tetrahedron, find the maximum \(k\) such that \((a + b + c + d)(h_a + h_b + h_c + h_d) \geq kV.\)
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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thomas5267
  • thomas5267
So h are the altitude of the triangles which make up the tetrahedron?
zmudz
  • zmudz
@thomas5267 from what I understand of the problem, yes
beginnersmind
  • beginnersmind
|dw:1441666807027:dw| I think it's like this.

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beginnersmind
  • beginnersmind
I apologize for the terrible art.
beginnersmind
  • beginnersmind
What are we looking at?
beginnersmind
  • beginnersmind
@thomas5267 "So h are the altitude of the triangles which make up the tetrahedron?" No, they are the altitudes of the tetrahedron. As in V=(1/3)a*h_a
Jhannybean
  • Jhannybean
|dw:1441667234670:dw|
beginnersmind
  • beginnersmind
Anyway, I'd start with \((a + b + c + d)(h_a + h_b + h_c + h_d) \geq kV.\) \((a + b + c + d)(\frac{V}{a} + \frac{V}{b} + \frac{V}{c} + \frac{V}{d} ) \geq 3kV.\) \((a + b + c + d)(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} ) \geq 3k.\) Which seem like should be some sort of inequality between arithmetic and harmonic means.
Jhannybean
  • Jhannybean
So \(h_n= \frac{V}{n}\) where \(n\) = face side?
beginnersmind
  • beginnersmind
Anyway, we know that the arthmetic mean is greater than equal than the harmonic mean, so using some algebra we get that k is less then or equal to 1/3.
beginnersmind
  • beginnersmind
@Jhannybean I used the volume formula for the tetrahedron \[V = \frac{1}{3}a*h_a \] Where a is the area of a face and h_a is the corresponding altitude.
Jhannybean
  • Jhannybean
Oh alright
beginnersmind
  • beginnersmind
Correction: \[V = \frac{1}{3}a*h_a\] \[h_a =\frac{3V}{a} \] \[(a + b + c + d)(h_a + h_b + h_c + h_d) \geq kV.\] \[(a + b + c + d)(\frac{3V}{a} + \frac{3V}{b} + \frac{3V}{c} + \frac{3V}{d} ) \geq kV.\] \[3(a + b + c + d)(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} ) \geq k.\] The inequality between arithmetic and harmonic means states that: \[\frac{a_1+a_2+a_3+a_4}{4} \geq \frac{4}{\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4} }\] or \[\frac{a_1+a_2+a_3+a_4}{16} \geq \frac{1}{\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4} }\] So \[\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4} \geq \frac{1}{16}\frac{1}{a_1+a_2+a_3+a_4}\] At this point I'm actually confused. It seems like k has a minimum, not a maximum.
beginnersmind
  • beginnersmind
E.g. substituting tells us that the LHS of the original expression is always greater or equal to 3/16, with equality happening if a=b=c=d
zmudz
  • zmudz
oops, meant to say maximum. I'm pretty sure there is a maximum
beginnersmind
  • beginnersmind
Actually, if you think about it geometrically, it's clear that there's no maximum. You can construct a tetrahedron like this: |dw:1441674391911:dw|
beginnersmind
  • beginnersmind
|dw:1441674489410:dw| You can see that the areas of ABC (d) and ABD (c) are fairly large. Also, the altitude h_a is pretty long. So when you look at the product on the left hand side the terms c*h_a and d*h_a are large. However the volume V is quite small. So the constant k is large. Specifically we can make k arbitrarily large by "closing" the two large triangles.

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