This is a “Power Round”-style problem set exploring how facts in Euclidean geometry can be “extended” to complex numbers.
Ptolemy’s theorem says that for a convex cyclic quadrilateral $ABCD$, $$AB\cdot CD + AD\cdot BC = AC\cdot BD$$ For convenience, we will try to ignore the ordering of the vertices $A,B,C,D$ along the circle. Then, instead we will have that $$AB\cdot CD \pm AC\cdot BD \pm AD \cdot BC = 0$$ for some choice of signs.
Try to check this condition for a bunch of quadrilaterals on the unit circle (by hand).
Show that this condition is equivalent to some polynomial in $AB^2,AC^2,AD^2, BC^2, BD^2, CD^2$ being 0. That is, find a polynomial $P$ such that the above equation is true for some choice of $\pm$-signs if and only if $$P(AB^2,AC^2,AD^2, BC^2, BD^2, CD^2) = 0$$ for a polynomial $P$. (Possible hint: how might you turn $a\pm b = 0$ into a polynomial $P(a,b)=0$?)
Set $A=(1,0), B=(0,1), C=(-1,0), D = (i,\sqrt{2})$. These “lie on the same circle” with equation $x^2+y^2=1$ (even though point $D$ doesn’t quite make sense…). Verify that $P=0$ above. (Hint: pick a choice of sign that makes the simpler version work.)
This is amazing! Could it really be that Ptolemy’s theorem also holds for $\mathbb C$-points?
(Hint: we may assume $P$ looks like $yP_1(x) + P_2(x)$. Then $(1-x^2)P_1(x)^2 - P_2(x)^2 = 0$ is true for all $x\in [-1,1]$, so this is true as polynomials. But this cannot possibly happen!)
In particular, $P(x,y) = 0$ must hold for any $(x,y)\in \mathbb C^2$ where $x^2+y^2=1$.
Interpretation: $P(x,y)=0$ is some property of a single point on a circle. Then if it is true for all points on the real circle, then it is true for all points on the complex circle.
The natural next step we might want to do is as follows: suppose that for all pairs of points $(x_1,y_1), (x_2,y_2)$ on a circle, $P(x_1,y_1,x_2,y_2) = 0$. Now we hope that there exists polynomials $Q_1,Q_2$ such that $$P = (x_1^2+y_1^2-1) Q_1 + (x_2^2+y_2^2-1)Q_2$$
This seems more difficult to do with this method, so instead we’ll try something else.
One way to get all points $(x,y)$ on a circle (specifically, the unit circle) is to set $(x,y) = (\cos \theta, \sin\theta)$, and to let $\theta$ range from $0$ to $2\pi$. But $\sin$ and $\cos$ are weird, so instead we bring in $$(x,y) = \left(\frac{e^{i\theta} + e^{-i \theta}}2, \frac{e^{i\theta} - e^{-i\theta}}{2i}\right)$$ Substituting $t=e^{i\theta}$ (and noticing that $t^{-1} = e^{-i\theta}$): $$(x,y) = \left(\frac{t}{2} + \frac{1}{2t}, \frac{t}{2i} - \frac{1}{2it}\right)$$
Now we try proving the earlier claim again.
Given a polynomial $\tilde P$, suppose $\tilde P(t)/t^n = 0$ for all complex $t$ with unit length (i.e. $|t| = 1$). Show that $P(t)/t^n = 0$ for all complex $t$.
Show that for any complex $x',y'$ with $(x')^2 + (y')^2 = 1$, there exists a $t'\in \mathbb C$ such that $$(x',y') = \left(\frac{t'}{2} + \frac{1}{2t'}, \frac{t'}{2i} - \frac{1}{2it'}\right)$$ (Hint: the expressions $x'\pm iy'$ will look very clean.)
Conclude that Ptolemy’s theorem holds for points with coordinates in $\mathbb C$.
(Possible approaches: Fundamental theorem of algebra or Fourier)
blah generalize to multiple $t_i$’s, go back and prove complex Ptolemy.
For two circles $\Gamma_1,\Gamma_2$ with radii $r_1,r_2$ resp. centered at $(x_1,y_1),(x_2,y_2)$, define $$d(\Gamma_1,\Gamma_2) = (x_1-x_2)^2 + (y_1 - y_2)^2 - (r_1-r_2)^2$$
If $r_1=0$, show that this is precisely the power of $(x_1,y_1)$ with respect to $\Gamma_2$.
Fix $\Gamma_1$. Describe all $\Gamma_2$ such that $d(\Gamma_1,\Gamma_2)$ is constant.
Maybe a slight warning here is in order. Give three circles for which $d(\Gamma_1,\Gamma_2) = d(\Gamma_2,\Gamma_3) = 0$ but $d(\Gamma_1,\Gamma_3)\neq 0$.
I want (?) this to be a theorem:
Theorem. (Better Casey) On $\mathbb C^3$, define $$d(x,y)^2 = (x_1-y_1)^2 + (x_2 - y_2)^2 + (x_3-y_3)^2$$ Suppose $d(O,A)= d(O,B) = d(O,C) = d(O,D) = 0$. Then $$d(A,B)d(C,D) \pm d(A,C)d(B,D) \pm d(A,D)d(B,C) = 0$$
I originally drafted this to be a candidiate for a future SMT Power Round. I chose Ptolemy’s theorem because it seemed like the easiest nontrivial geometry “identity” that uses just metric distance, but I think maybe having better examples in general could be nice.
I also tried giving a talk about this but it didn’t pan out too well either. Still some distance away from really being able to communicate the excitement that you can just “extend” geometry to the complex numbers and get away with it most of the time.