Date modified: Mon 2023-05-15 . 09:27 AM
Date created: Sat 2023-10-07 . 09:02 PM
# Ptolemy theorem with complex numbers. Ptolemy theorem states that given a cyclic quadrilateral (a quadrilateral inscribed in a circle), the product of the diagonals is equal to the sum of the product of the lengths of the opposing sides: ![[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 12.13.26.excalidraw.svg]] %%[[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 12.13.26.excalidraw.md|🖋 Edit in Excalidraw]], and the [[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 12.13.26.excalidraw.dark.svg|dark exported image]]%% Let us see if we can show this using complex numbers! ## The complex numbers statement of Ptolemy's theorem. Place this circle anywhere in the complex plane, and let $a,b,c,d$ be the four complex numbers describing the vertices of this cyclic quadrilateral. ![[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 12.19.31.excalidraw.svg]] %%[[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 12.19.31.excalidraw.md|🖋 Edit in Excalidraw]], and the [[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 12.19.31.excalidraw.dark.svg|dark exported image]]%% Then what we want to show is the equality $$|a-c||b-d| = |a-b||c-d| + |b-c||d-a|$$ We will next gather some ingredients to prove this. ## Ingredient 1: Cyclic quadrilaterals have opposing angles sum to $\pi$. We first need a simple result of cyclic quadrilaterals: **Lemma.** If $Q$ is a cyclic quadrilateral, then the sum of opposing angles is $\pi$. **Proof.** To see this, we consider the following two cases: Whether $Q$ contains the center or not: ![[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 15.58.02.excalidraw.svg]] %%[[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 15.58.02.excalidraw.md|🖋 Edit in Excalidraw]], and the [[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 15.58.02.excalidraw.dark.svg|dark exported image]]%% In the first case, connect the center to the vertices of $Q$, and notice this forms isosceles triangles: ![[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 16.01.27.excalidraw.svg]] %%[[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 16.01.27.excalidraw.md|🖋 Edit in Excalidraw]], and the [[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 16.01.27.excalidraw.dark.svg|dark exported image]]%% Since the interior angle sum of a quadrilateral is $2\pi$, which is $2a+2b+2c+2d=2\pi$, hence $(a+b)+(c+d) = \pi = (a+d)+(b+c)$. Done In the second case, again connect the center to the vertices of $Q$: ![[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 16.07.56.excalidraw.svg]] %%[[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 16.07.56.excalidraw.md|🖋 Edit in Excalidraw]], and the [[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 16.07.56.excalidraw.dark.svg|dark exported image]]%% Since $Q$ does not contain the center, we have a sliver of angle $x$ outside of the isosceles triangles on the ends. So we have $2a+2b+2c-2x = 2\pi$. In other words, $a+b+(c-x) = \pi = (a-x)+(b+c)$. Done. ## Ingredient 2: Angle between three complex numbers. Given three distinct complex numbers $x,y,z$, the angle from the segment formed by $x,y$ to the segment formed by $x,z$ is given by $$ arg \frac{x-z}{x-y} $$ ![[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 14.00.51.excalidraw.svg]]%%[[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 14.00.51.excalidraw.md|🖋 Edit in Excalidraw]], and the [[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 14.00.51.excalidraw.dark.svg|dark exported image]]%% ## Ingredient 3: When does triangle inequality hold for complex numbers? Given any two complex numbers $p,q$, we have triangle inequality $|p + q| \le |p| + |q|$. If one of them is zero, then equality surely holds. If neither are zero, then equality holds if they both have the same argument. Namely, they lie on the same ray emanating from the origin: ![[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 12.50.41.excalidraw.svg]] %%[[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 12.50.41.excalidraw.md|🖋 Edit in Excalidraw]], and the [[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 12.50.41.excalidraw.dark.svg|dark exported image]]%% In other words $p/q > 0$, a real positive number, that is, **Lemma.** Let $p,q$ be complex numbers, then $|p + q| = |p| + |q|$ if either (A) $p$ or $q$ is zero, or (B) $p,q\neq 0$ and $p/q > 0$. ## Proof of Ptolemy's theorem. First take our cyclic quadrilateral with vertices $a,b,c,d$. ![[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 14.13.33.excalidraw.svg]] %%[[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 14.13.33.excalidraw.md|🖋 Edit in Excalidraw]], and the [[0 inbox/---files/Ptolemy-theorem-with-complex-numbers 2023-05-14 14.13.33.excalidraw.dark.svg|dark exported image]]%% Then by ingredient 1, the opposing angles have sum $\theta + \phi = \pi$. Keeping the orientation to be positive, this means, by ingredient 2, $$ arg\frac{a-d}{a-b}+arg\frac{c-b}{c-d}=\pi $$ In other words the argument of the product is $\pi$: $$ arg\frac{a-d}{a-b}\frac{c-b}{c-d}=\pi $$ which means this product is negative: $$ \frac{a-d}{a-b}\frac{c-b}{c-d} < 0 $$ Equivalently, $$ \frac{a-d}{a-b}\frac{\color{blue} b-c}{c-d} > 0 $$ So by ingredient 3, we have equality case in triangle inequality: $$ \begin{align*} |(a-d)(b-c) + (a-b)(c-d)| \\ = |(a-d)(b-c)| + |(a-b)(c-d)| \end{align*} $$ Observe the RHS is just the sum of the product of the lengths of opposing sides. And the LHS simplifies to $$ \begin{align*} &|(a-d)(b-c) + (a-b)(c-d)| \\ &=|ab-ac-db+dc+ac-ad-bc+bd| \\ &=|ab +dc -ad-bc |\\ & = |a(b-d)-c(b-d)|\\ & = |a-c||b-d| \end{align*} $$ which is the product of the diagonals. Done! #geometry #complex-numbers