MHB Problem of the Week #107 - April 14th, 2014

[Chris L T521]

Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Given a sphere of radius $r$, find the height of a pyramid of minimum volume whose base is square and whose base and triangular faces are all tangent to the sphere.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

No one answered this week's problem. I don't have a solution ready at this time due to moving my stuff back to Chicago and preparing for a trip to Hawaii (which is where I currently am), but I will post one as soon as I am able (i.e. sometime [early...hopefully] tomorrow).

UPDATE: Opalg submitted a nice solution to this problem late, and due to time constraints and lack of access to an internet connection, I'm going to use it as our solution for this problem. With all of that said, you can find Opalg's solution below.

[sp]

https://www.physicsforums.com/attachments/2361​

Let $h$ be the height of the pyramid, and $b$ the side of the square base. The diagram shows a cross-section through midpoints of two opposite sides of the base. With $\theta$ as shown in the diagram, $h = r(1+\csc\theta)$ and $\dfrac{\frac12b}h = \tan \theta$ so that $b = 2h\tan\theta$. The volume of the pyramid is $$V = \tfrac13b^2h = \tfrac43h^3\tan^2\theta = \tfrac43r^3(1+\csc\theta)^3\tan^2\theta.$$ Let $s = \sin\theta$. Then $1+\csc\theta = \frac{1+s}s$ and $\tan^2\theta = \frac{s^2}{1-s^2}$. So we can write $V$ in terms of $s$ as $$V = \tfrac43r^3\frac{s^2(1+s)^3}{s^2(1-s^2)} = \tfrac43r^3\frac{(1+s)^2}{s(1-s)}.$$ Then $\dfrac{dV}{ds} = \tfrac43r^3\dfrac{2(1+s)s(1-s) - (1+s)^2(1-2s)}{s^2(1-s)^2} = \tfrac43r^3\dfrac{(1+s)(3s-1)}{s^2(1-s)^2} = 0$ when $s = \frac13.$ So the minimum volume $\frac{32}3r^3$ occurs when $s = \frac13$ and $h = \frac{r(1+s)}s = 4r.$[/sp]