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Q&A

Given a triangle with squares on two sides, the line segments joining the centres of the squares to the midpoint of the third side are equal and perpendicular

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I am reading Tristan Needham's Visual Complex Analysis (2012 reprint, OUP), and in $\S$1.III.3: Geometry, the author gives a geometric proof of the following fact, shown in Figure [12b] on page 16:

Figure[12a] shows an arbitrary quadrilateral with squares constructed outward on each of the sides and line segments joining the centres of the opposite squares. Figure [12b] shows an arbitrary triangle with squares constructed outward on two sides and line segments joining the centres of the squares to the midpoint of the remaining side of the triangle. The centres of the squares are labelled $p$ and $m$, and the midpoint of the side is labelled $m$.

Image taken from Visual Complex Analysis, Tristan Needham, page 16.

Here squares have been constructed on two sides of an arbitrary triangle, and, as the picture suggests, the line-segments from their centres to the midpoint $m$ of the remaining side are perpendicular and of equal length.

Visual Complex Analysis, Tristan Needham, page 17.

To this end, the author shows the following fact. Let $\mathcal{T}_v$ be a translation of the plane by $v$, that is, $\mathcal{T}_v(z) = z + v$ for all $z \in \mathbb{C}$. Let $\mathcal{R}_a^\theta$ be a rotation of the plane about $a$ by an angle $\theta$ in the counter-clockwise direction. One can show that $\mathcal{R}_a^\theta(z) = e^{\iota\theta}z+k$ where $k = a(1-e^{\iota\theta})$. Then we have the following:

Let $\mathcal{M} = \mathcal{R}_{a_n}^{\theta_n} \circ \dotsb \circ \mathcal{R} _{a_2}^{\theta_2} \circ \mathcal{R} _{a_1}^{\theta_1}$ be the composition of $n$ rotations, and let $\Theta = \theta_1 + \theta_2 + \dotsb + \theta_n$ be the total amount of rotation. In general, $\mathcal{M} = \mathcal{R}_c^\Theta$ (for some $c$), but if $\Theta$ is a multiple of $2\pi$ then $\mathcal{M} = \mathcal{T}_v$ for some $v$.

Visual Complex Analysis, Tristan Needham, page 19.

Now, the author asks us to consider $\mathcal{M} = \mathcal{R}_m^\pi \circ \mathcal{R}_p^{(\pi/2)} \circ \mathcal{R}_s^{(\pi/2)}$. Since $(\pi/2) + (\pi/2) + \pi = 2\pi$, the above result says that $\mathcal{M} = \mathcal{T}_v$ for some $v$. To find $v$, it suffices to compute $\mathcal{M}(z)$ for any one point.

Next, the author says, "Clearly, $\mathcal{M}(k) = k$", but this is not so clear to me. I presume that $k = a(1 - e^{\iota\theta})$ for some appropriate $a$ and $\theta$, since the letter $k$ is not used for any other purpose in this section. But, it is not clear to me which $a$ and $\theta$ I am supposed to take here since $\mathcal{M}$ is a composition of three rotations.

Nevertheless, I tried just taking the origin to see if it is fixed under $\mathcal{M}$. I get

$$ \begin{gather} \mathcal{R}_s^{(\pi/2)}(0) = s(1 - \iota),\\ \mathcal{R}_p^{(\pi/2)}\bigl(s(1-\iota)\bigr) = \iota s (1 - \iota) + p(1 - \iota) = (s + p) + \iota (s - p),\\ \mathcal{R}_m^\pi\bigl( (s + p) + \iota (s - p) \bigr) = -(s + p) - \iota (s - p) + 2m. \end{gather} $$

How can I easily see that the last expression is equal to $0$ (presuming my calculations are correct)?

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4 comments

Out of curiosity, once he shows $\mathcal M$ is the identity, how does he complete the proof? msh210‭ 3 months ago

@msh210 It goes like this: "Thus $$\mathcal{R}_p^{(\pi/2)} \circ \mathcal{R}_s^{(\pi/2)} = (\mathcal{R}_m^\pi)^{-1} \circ \mathcal{M} = \mathcal{R}_m^\pi.$$ If we define $s' = \mathcal{R}_m^\pi(s)$ then $m$ is the midpoint of $ss'$. But, on the other hand, $$s' = \Bigl( \mathcal{R}_p^{(\pi/2)} \circ \mathcal{R}_s^{(\pi/2)} \Bigr)(s) = \mathcal{R}_p^{(\pi/2)}(s).$$ Thus the triangle $sps'$ is isosceles and has a right angle at $p$, so $sm$ and $pm$ are perpendicular and of equal length. Done." Bahudari Ragam‭ 3 months ago

@msh210 I usually add the remainder of any proofs at the end of my questions (or I used to, on SE), but I somehow forgot to do so here. Thanks for asking about it! Bahudari Ragam‭ 3 months ago

Thank you for the comment! msh210‭ 3 months ago

1 answer

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Your calculations look correct to me, but I think this is perhaps meant to be seen geometrically (apologies if this is already obvious to you and you were looking for a non-geometric explanation).

Let the vertices of the triangle be $x$, $y$, $z$, starting at the vertex shared by the two squares, and labeling clockwise. Consider point $z$ as it is transformed by $\mathcal{M}$. A quarter-turn around $s$ must take $z$ to $x$; it's just a rotation of the square. Likewise, a quarter-turn around $p$ must take $x$ to $y$, and a half-turn around $m$ must take $y$ to $z$. So $\mathcal{M}$ takes $z$ to itself. By the quoted result, $\mathcal{M}$ is a translation, and therefore it must be the identity transformation.

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3 comments

+1 This makes perfect sense. If this is what the author also meant, then I'm not sure why he phrased it as $\mathcal{M}(k) = k$, though. Bahudari Ragam‭ 3 months ago

I suspect $k$ is an idiosyncratic or just unfortunate choice of independent variable. It looks like the author is just saying that $\mathcal{M}$ is the identity function, like $f(x) = x$—at least, I don't see any reason to say anything more nuanced than that about $\mathcal{M}$, given what's discussed here. r~~‭ 3 months ago

Okay, now this is actually a bit embarrassing for me. I just had to turn the page and find out in Figure [15a] that the point $k$ is indeed the point you call $z$. So, if you'll excuse me, I shall go drown myself now. Bahudari Ragam‭ 3 months ago

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