Intriguing ornament

Geometry Level 4

At the home of a mathematician friend I noticed an unusual ornament the other day that consists of a sphere within a concentric cube. It would have been inappropriate to take a picture under the circumstances, but , fortunately, I found a somewhat similar picture online (the ornament comes in more pleasing colours, of course). The host explained that the ornament is skillfully designed so that exactly half of the surface area of the sphere is exposed. Find the ratio R a \frac{R}{a} between the radius R R of the sphere and the side length a a of the cube!


The answer is 0.6.

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2 solutions

Steven Chase
Dec 25, 2018

The surface area of a spherical cap is 2 π R h 2 \pi R \, h , where R R is the radius of the sphere and h h is the height of the cap (which, in this case, is the height of the portion protruding from the cube). Six of these spherical cap areas combine to equal half the sphere surface area.

12 π R h = 2 π R 2 R = 6 h h = R 6 12 \pi \, R \, h = 2 \pi R^2 \\ R = 6 h \\ h = \frac{R}{6}

The cube side length is then:

a = 2 R 2 h = 2 R R 3 = 5 R 3 R a = 3 5 = 0.6 a = 2 R - 2 h = 2R - \frac{R}{3} = \frac{5 R}{3} \\ \frac{R}{a} = \frac{3}{5} = 0.6

Yes, exactly! Thank you! This was a gentle one, in the spirit of the season ;)

Otto Bretscher - 2 years, 5 months ago

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Indeed, thanks, and Merry Christmas! I have posted an easy festive one in the emag section as well.

Steven Chase - 2 years, 5 months ago

A rational answer is a happy ending. (I was wondering why you didn't make exposed sphere equal exposed cube. I'm sure you considered it but the exact solution is immensely ugly.)

Jeremy Galvagni - 2 years, 5 months ago

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That sounds fun. Will you post it?

Steven Chase - 2 years, 5 months ago

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@Steven Chase You asked for it. You need to use the quadratic formula to find R a = π + π 2 + 15 π 6 π 0.234 \frac{R}{a} = \frac{-\pi +\sqrt{\pi^{2}+15\pi }}{6\pi} \approx 0.234

Jeremy Galvagni - 2 years, 5 months ago

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@Jeremy Galvagni Oh, actually I was hoping you would post it as a problem

Steven Chase - 2 years, 5 months ago

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@Steven Chase Too similar to Otto's and not so nice. My brain is crackling though. I may come up with something worth sharing.

Jeremy Galvagni - 2 years, 5 months ago

Yes, indeed, as a Swiss I like exact answers, not just rational but terminating whenever possible ;)

Otto Bretscher - 2 years, 5 months ago
Huan Bui
Dec 25, 2018

@Steven Chase has given a comprehensive solution. However, since I always like to see how things are derived from first principles (in this case it is the spherical cap surface area formula), here is my go at the problem.

The sphere's exposed area is: 6 ϕ = 0 2 π θ = 0 arccos ( a / 2 R ) R 2 sin θ d θ d ϕ = 1 2 4 π R 2 6 \int_{\phi=0}^{2\pi}\int_{\theta = 0}^{\arccos{(a/2R)}}R^2\sin{\theta}\,d\theta\,d\phi = \frac{1}{2}4\pi R^2 . This implies 6 ( 1 a 2 R ) = 1 6\left(1-\frac{a}{2R}\right) = 1 , i.e., R a = 3 5 = 0.6 \frac{R}{a} = \frac{3}{5} = \boxed{0.6} .

Yes, Comrade, nicely done!

The formula that @Steven Chase uses is actually quite interesting conceptually, and well worth knowing: It works for a "cap" of a sphere as well as for a slice. It is a consequence of the fact, known to Archimedes, that the "natural" projection of a circumscribed cylinder onto a sphere has a scaling factor of 1 (as I made you guys prove in Problem 4 on the Final exam). Fernando told me that Archimedes considered this to be his greatest mathematical discovery: He was so proud of it that he had a sphere and a circumscribed cylinder chiseled on his tomb (as Cicero reports).

Otto Bretscher - 2 years, 5 months ago

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The formula was worth knowing indeed! I actually had it on my "cheatsheet'' for the Final Exam - as a result I was able to check my answer for Problem 4. I figured that the formula is so simple that I might as well know it instead of spending time deriving it with "heavy calculus machinery.'' I don't know what you think but for me the formula never feels very intuitive...

Huan Bui - 2 years, 5 months ago

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Yes, Comrade, this actually does become quite intuitive if you meditate over it long enough (as most things do, in maths and otherwise!). Let me give you some hints on how I think about it, without giving it all away. It's some simple but beautiful math, well worth thinking about. As in many cases (think inverse square field!) it's a case of two countervailing tendencies cancelling each other out perfectly.

Let S ( z ) S(z) be the surface area of a sphere of radius R R up to height z z , the lower cap if you will. The main point will be to explain why d S d z = 2 π R \frac{dS}{dz}=2\pi R , a constant. The formulas for a cap or, more generally, a slice of a sphere are immediate corollaries. Let r 2 = x 2 + y 2 r^2=x^2+y^2 as usual, and d s 2 = d r 2 + d z 2 ds^2=dr^2+dz^2 . Note that d S = 2 π r d s dS=2\pi r ds (If you wish to draw a figure, work in the r z rz -plane.) Now we have two similar triangles, one involving R , z , r R,z,r and the other d r , d z , d s dr,dz,ds ; this is just an expression of the fact that a tangent to the circle is perpendicular to the radius at that point. It follows that r R = d z d s \frac{r}{R}=\frac{dz}{ds} so that d S = 2 π r d s = 2 π R d z dS=2\pi rds=2\pi Rdz , as claimed.

As we go up, r r decreases and d s d z \frac{ds}{dz} increases, and those tendencies are inversely proportional and cancel each other out. To put it in more geometrical terms: As you go up, the rings around the z z -axis shrink in diameter, but they become "flatter" and thus, wider.

Otto Bretscher - 2 years, 5 months ago

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