A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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 \((a + b + c + d)(h_a + h_b + h_c + h_d) \geq kV.\)

  • This Question is Closed
  1. thomas5267
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So h are the altitude of the triangles which make up the tetrahedron?

  2. zmudz
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @thomas5267 from what I understand of the problem, yes

  3. beginnersmind
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1441666807027:dw| I think it's like this.

  4. beginnersmind
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I apologize for the terrible art.

  5. beginnersmind
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    What are we looking at?

  6. beginnersmind
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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

  7. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1441667234670:dw|

  8. beginnersmind
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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.

  9. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So \(h_n= \frac{V}{n}\) where \(n\) = face side?

  10. beginnersmind
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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.

  11. beginnersmind
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  12. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh alright

  13. beginnersmind
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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.

  14. beginnersmind
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  15. zmudz
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oops, meant to say maximum. I'm pretty sure there is a maximum

  16. beginnersmind
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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|

  17. beginnersmind
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |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.

  18. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.