Yes, forks. And that they predate chopsticks. I mean, what the frigate?!!??
So, to distract myself from this distressing thought, here's a cute little fact: draw three circles \(C_a,C_b,C_c\) so that any two of them are tangent, then there exist precisely two circles \(C_d,C_{d'}\) tangent to all three circles.
Even cuter little fact: label the respective radii of \(C_a,C_b,C_c,C_d\) by the letters \(a,b,c,d\), then \begin{align}2(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2})=(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})^2.\end{align} A less cute fact: the above identity also holds if we replace \(d\) with \(d'\), i.e.: the radius of \(C_{d'}\); I was just too lazy to figure out how to say this nicely. Oh, and I also forgot to mention that if one of these circles encircles the other three, then we think of its radius as being a negative number.
For a much more eloquent description of this little gem, consider the following poem by the Nobel prize winner Frederick Soddy, published in Nature (one of the most respected journals out there), June 1936:
The Kiss Precise
For pairs of lips to kiss maybe
Involves no trigonometry.
'Tis not so when four circles kiss
Each one the other three.
To bring this off the four must be
As three in one or one in three.
If one in three, beyond a doubt
Each gets three kisses from without.
If three in one, then is that one
Thrice kissed internally.
Four circles to the kissing come.
The smaller are the benter.
The bend is just the inverse of
The distance from the center.
Though their intrigue left Euclid dumb
There's now no need for rule of thumb.
Since zero bend's a dead straight line
And concave bends have minus sign,
The sum of the squares of all four bends
Is half the square of their sum.
To spy out spherical affairs
An ocular surveyor
Might find the task laborious,
The sphere is much the gayer,
And now besides the pair of pairs
A fifth sphere in the kissing shares.
Yet, signs and zero as before,
For each to kiss the other four
The square of the sum of all five bends
Is thrice the sum of their squares.
Involves no trigonometry.
'Tis not so when four circles kiss
Each one the other three.
To bring this off the four must be
As three in one or one in three.
If one in three, beyond a doubt
Each gets three kisses from without.
If three in one, then is that one
Thrice kissed internally.
Four circles to the kissing come.
The smaller are the benter.
The bend is just the inverse of
The distance from the center.
Though their intrigue left Euclid dumb
There's now no need for rule of thumb.
Since zero bend's a dead straight line
And concave bends have minus sign,
The sum of the squares of all four bends
Is half the square of their sum.
To spy out spherical affairs
An ocular surveyor
Might find the task laborious,
The sphere is much the gayer,
And now besides the pair of pairs
A fifth sphere in the kissing shares.
Yet, signs and zero as before,
For each to kiss the other four
The square of the sum of all five bends
Is thrice the sum of their squares.
So, peeps usually call this result Descartes' Circle Theorem, even though it's possible (I dun-actually-no) that this has been studied since the ancient Greeks. So, I've read somewhere that heaps of peeps have rediscovered this result over the years, and that there are a fair few proofs of it (including one by Newton!). I think that I know two, three ways to prove this result (but not its generalizations - which are, I think...also due to Soddy?).
- Mathod (**) 1: Coordinate bash - just write this up in \(\mathbb{R}^2\) and bash it out. But this is probably gonna be pretty nasty.
- Mathod 2: Inversions - Mobius transformations on the complex line \(\mathbb{C}\) take circles to circles, and so too do reflections (such as complex conjugation) on \(\mathbb{C}\), so the map\[\varphi:z\rightarrow\frac{r^2}{\bar{z}-\bar{w}}+w\] is a special kinda map that flips the inside of a circle of radius \(r\) based at \(w\) to the outside and the outside in so that circles are still preserved. And we can use maps of this form (called inversions) to get our circles in some standard form and then bash some stuff in this standard form. Still, nasty.
- Mathod 3: Heron's formula - I actually found a proof for this in a paper by Coxeter, who has a heap of stuff named after him in maths. He himself yoinked this proof from another paper by a Mr. Philip Beecroft in one of the most deliciously named journals ever: "The Lady's and Gentleman's Diary for the year of our Lord 1842, being the second after Bissextile, designed principally for the amusement and instruction of Students in Mathematics: comprising many useful and entertaining particulars, interesting to all persons engaged in that delightful pursuit".
Okay, so let's just try to briefly walk through how this proof works. I guess I should probably link the article that I pulled this from, for those who're interested. Coxeter's a very good writer, so it's actually pretty readable if you're willing to Google a few definitions. The first idea is to notice that given a configuration of four pairwise tangent circles \(C_a,C_b,C_c,C_d\), the three tangent points of any three such circles will lie on a unique circle. Therefore, we get four new circles \(C_\alpha,C_\beta,C_\gamma,C_\delta\) respectively meeting the tangents points of \[\{C_b,C_c,C_d\},\{C_a,C_c,C_d\},\{C_a,C_b,C_d\},\{C_a,C_b,C_c\}.\]
Psst: I was gonna draw a picture of this, but it got too messy. =(
Well, I'm not going to justify those facts, so let's just do some high school geometry instead and prove the darn result!
The first thing that we should notice is that, by construction, the circle \(C_{\delta}\) is circumscribed by the triangle with vertices placed at the centers of \(C_a,C_b\) and \(C_c\). Moreover, since the normal (line) to a tangent line of a circle goes through its center, we may split triangle \(\triangle ABC\) up into six little right-angled triangles each of height \(\delta\).
Invoking Heron's formula, we get that: \begin{align}\delta^2(a+b+c)^2=abc(a+b+c),\end{align}
which is a pretty sweet lookin' shabang. Now let's do something similar with the area of triangle \(\triangle BCD\) with its vertices centered at the circles \(C_b, C_c\) and \(C_d\). We see that this area - which we can compute by Heron's formula in terms of \(b,c\) and \(d\), may also be obtained by subtracting the area of the four smaller triangles from the two larger ones as per the following picture (you'll notice that my circles are pretty out of scale with respect to each other when compared with the first pic =P).
Therefore, we obtain in this case: \begin{align}\alpha^2(b+c+d)^2&=bcd(b+c+d).\end{align}
Which looks pretty similar to our previous formula. And at this point we just ply a few symmetry arguments, wlogs and a tiny bit of algebraic tweaking to obtain the following set of algebraic thingies: \begin{align} \alpha^{-2}&=b^{-1}c^{-1}+b^{-1}d^{-1}+c^{-1}d^{-1}\\ a^{-2}&=\beta^{-1}\gamma^{-1}+\beta^{-1}\delta^{-1}+\gamma^{-1}\delta^{-1}\\ \beta^{-2}&=a^{-1}c^{-1}+a^{-1}d^{-1}+c^{-1}d^{-1}\\ b^{-2}&=\alpha^{-1}\gamma^{-1}+\alpha^{-1}\delta^{-1}+\gamma^{-1}\delta^{-1}\\ \gamma^{-2}&=a^{-1}b^{-1}+a^{-1}d^{-1}+b^{-1}d^{-1}\\ c^{-2}&=\alpha^{-1}\beta^{-1}+\alpha^{-1}\delta^{-1}+\gamma^{-1}\delta^{-1}\\ \delta^{-2}&=a^{-1}b^{-1}+a^{-1}c^{-1}+b^{-1}c^{-1}\\ d^{-2}&=\alpha^{-1}\beta^{-1}+\alpha^{-1}\gamma^{-1}+\beta^{-1}\gamma^{-1}.\end{align} Well, we've got all of these cross-terms and stuff on the right hand side, so why not look at an algebraic expression which might generate such cross-terms (***): \begin{align}(a^{-1}+b^{-1}+c^{-1}+d^{-1})^2&=a^{-2}+b^{-2}+c^{-2}+d^{-2}+2\text{( all the cross terms! )}\\&=a^{-2}+b^{-2}+c^{-2}+d^{-2}+\alpha^{-2}+\beta^{-2}+\gamma^{-2}+\delta^{-2}\\&=(\alpha^{-1}+\beta^{-1}+\gamma^{-1}+\delta^{-1})^2.\end{align} So we see that: \begin{align}a^{-1}+b^{-1}+c^{-1}+d^{-1}=\alpha^{-1}+\beta^{-1}+\gamma^{-1}+\delta^{-1}.\end{align} Now playing the same game, but with fewer cross-terms, we obtain: \begin{align}-a^{-2}+b^{-2}+c^{-2}+d^{-2}&=2\alpha^{-1}(\alpha^{-1}+\beta^{-1}+\gamma^{-1}+\delta^{-1})\\&=2\alpha^{-1}(a^{-1}+b^{-1}+c^{-1}+d^{-1}).\end{align} Completing the square with the LHS and cancelling stuff, we find that:\begin{align}-a^{-1}+b^{-1}+c^{-1}+d^{-1}=2\alpha^{-1}.\end{align} At this point my brain's all like: come on maaaaaan, there's gots to be some monodromy goin' on in dis! Bah! I can't quite figure it out yet...(makes squinty eyes) one day, my friend, one day....
Anyhoo, so there are three other similar equations to this last guy, and if we just square both sides for each of these four equations, expand, add them together, then all the cross-terms cancel and we're left with: \begin{align}a^{-2}+b^{-2}+c^{-2}+d^{-2}=\alpha^{-2}+\beta^{-2}+\gamma^{-2}+\delta^{-2}.\end{align} So adorable.
Anyhooo, putting the fact that the reciprocal sums and the reciprocal square sums of \(\{a,b,c,d\}\) and \(\{\alpha,\beta,\gamma,\delta\}\) are equal, we finally get what we came for: \begin{align}2(a^{-2}+b^{-2}+c^{-2}+d^{-2})&=a^{-2}+b^{-2}+c^{-2}+d^{-2}+\alpha^{-2}+\beta^{-2}+\gamma^{-2}+\delta^{-2}\\&=(a^{-1}+b^{-1}+c^{-1}+d^{-1})^2.\end{align}
Claps chalk off hands and walks off.
*: according to something I saw on TV and Wikipedia, not the forks page, but the Chinese inventions page linked in the next line
**: I kept trying to type "method" but typed "mathod" instead. So I figured that I might as well stick to it, maths+method=mathod. =P
***: I'll admit that this is pretty weak motivation, even for me. =)
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