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zmudz
 one year ago
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
\((a + b + c + d)(h_a + h_b + h_c + h_d) \geq kV.\)
zmudz
 one year ago
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 \((a + b + c + d)(h_a + h_b + h_c + h_d) \geq kV.\)

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thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0So h are the altitude of the triangles which make up the tetrahedron?

zmudz
 one year ago
Best ResponseYou've already chosen the best response.0@thomas5267 from what I understand of the problem, yes

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441666807027:dw I think it's like this.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2I apologize for the terrible art.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2What are we looking at?

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2@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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441667234670:dw

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2Anyway, 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So \(h_n= \frac{V}{n}\) where \(n\) = face side?

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2Anyway, 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
 one year ago
Best ResponseYou've already chosen the best response.2@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.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2Correction: \[V = \frac{1}{3}a*h_a\] \[h_a =\frac{3V}{a} \] \[(a + b + c + d)(h_a + h_b + h_c + h_d) \geq kV.\] \[(a + b + c + d)(\frac{3V}{a} + \frac{3V}{b} + \frac{3V}{c} + \frac{3V}{d} ) \geq kV.\] \[3(a + b + c + d)(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} ) \geq k.\] 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
 one year ago
Best ResponseYou've already chosen the best response.2E.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
 one year ago
Best ResponseYou've already chosen the best response.0oops, meant to say maximum. I'm pretty sure there is a maximum

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2Actually, 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
 one year ago
Best ResponseYou've already chosen the best response.2dw: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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