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:

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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Out of curiosity, once he shows $\mathcal M$ is the identity, how does he complete the proof? msh210‭ 24 days 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." Gourimanohari Ragam‭ 24 days 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! Gourimanohari Ragam‭ 24 days ago

Thank you for the comment! msh210‭ 24 days ago

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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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+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. Gourimanohari Ragam‭ 27 days 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~~‭ 27 days 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. Gourimanohari Ragam‭ 27 days ago