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The Villarceau circles are things whose existence is surprising. To find radii of Villarceau circles, I stupidly went through a bit of trigonometry and got a much simpler result than I expected, and then realized there was a glaringly obvious way to do it that I hadn't thought of.

In the $xyz$-space imagine a circle of radius $r>0$ in the $xz$-plane, whose center is at a distance $R>r$ from the $z$-axis, and revolve it about the $z$-axis, getting a torus embedded in $\mathbb R^3$. The intersection of that surface with the $xz$-plane is two circles not crossing each other. A line $\ell_1$ touches one of those circles on one side and another on the other side, and that line is in a plane parallel to the $y$-axis, and the intersection of that plane with the torus is the union of two Villarceau circles. So I thought: let's draw a line $\ell_2$ touching both circles on the same side, and the other line $\ell_3$ touching both circles on the same side, and the distance from the intersection of $\ell_1$ with $\ell_2$ to the intersection of $\ell_1$ with $\ell_3$ is the diameter of the Villarceau circle. So I thought: first, the distance from the center to the point of tangency of $\ell_1$ with one circle, is $$\sqrt{R^2-r^2}.\tag{1}$$ Add to that the distance the distance from that point of tangency to the point of intersection of $\ell_1$ with the nearest of those two parallel $\ell$s, and that distance is $$ \frac{r^2}{R+\sqrt{R^2 - r^2}}.\tag{2} $$ So the sum of $(1)$ and $(2)$ involves finding a common denominator, doing some routine cancelations, getting $$ \frac{R\left(R+\sqrt{R^2-r^2}\,\right)}{R+\sqrt{R^2-r^2}} $$ and one more cancellation gives you $R$. Then I realized that the obvious way to see that the radius is $R$ doesn't involve doing any of that.

But I recognized that bit of trivial algebra from a routine calculus problem: $$ \frac{d}{dx} \log\left(x+\sqrt{x^2-1}\,\right). $$ Going through the usual algorithms, you get this down to $$ \frac{x+\sqrt{x^2-1}}{\sqrt{x^2-1}\left(x+\sqrt{x^2-1}\,\right)} $$ and again to one last cancelation to get $\dfrac{1}{\sqrt{x^2-1}}$.

SO MY QUESTION IS: What particular commonality between these two problems causes this same bit of algebra to occur in both places?

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Psst. The (algebra) tag is no longer being used. – user2468 Oct 14 '12 at 23:38
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@JenniferDylan : Would you suggest some other tag, or do you prefer to delete it? (And apparently it is still being used, since, at least for now, this question bears it.) – Michael Hardy Oct 15 '12 at 2:01
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+1 for using the algebra tag and taking a stand on the fiasco by which algebra-meaninglesslabels are used as the permanent replacement. – zyx Oct 19 '12 at 17:46
    
I think algebra-precalculus is close enough - algebraic manipulations with various expression fall under this tag nicely. But I did not retag, since I don't want to go into retagging war. – Martin Sleziak Oct 23 '12 at 7:58
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@zyx Meta is a better place for such discussions - and you can find several discussions about this tag there. I just wanted to say that what you call fiasco was done based on community consensus. – Martin Sleziak Oct 23 '12 at 8:00

Well, if you may not have realized, $\frac{\ln(x\pm\sqrt{x^2-1})}{i}+\frac{\pi}2+2\pi n=\arccos(x)$. The derivative for $\arccos(x) \, dx$, as you may already know, is $-\frac1{\sqrt{1-x^2}}$

This relates the derivative of the logarithm you presented with trigonometry/geometry.

A derivation of the $\arccos(x)$ can be found here.

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Wow, I can't believe this question is this old... – Simple Art Dec 17 '15 at 22:39
    
ok, Remind me on Monday to see if I can see where you're going with this${}\,\ldots\qquad{}$ – Michael Hardy Dec 17 '15 at 23:07
    
LoL, Monday? I believe the points of tangency you speak of are related to trigonometry, which upon differentiating produce these equations. – Simple Art Dec 17 '15 at 23:10
    
I may be somewhat busy until Monday....... – Michael Hardy Dec 17 '15 at 23:16
    
That is ok with me. – Simple Art Dec 17 '15 at 23:22

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