# Sequence of right triangles

Consider the right triangle \(\triangle_1 \) with vertices at \((0,0) \), \((0,1) \), and \((1,1). \) Note that the angle at the origin is \(\frac{\pi}{4}. \) Use the hypotenuse of \(\triangle_1 \) as one of the legs of another right triangle, \(\triangle_2, \) with an angle of \(\frac{\pi}{8} \) at the origin, and whose second leg extends towards the \(x \)–axis. Continue constructing adjacent right triangles \(\triangle_n \) in the same fashion, with one leg supported on the hypothenuse of the previous right triangle \(\triangle_{n-1}, \) with angle \(\frac{\pi}{2^{n+1}} \) at the origin, and with a second leg extending towards the \(x \)–axis.

Note that, since the angles at the origin form the series \(\sum_{n=2}^\infty \frac{\pi}{2^n}, \) (that add up to \(\frac{\pi}{2} \)) the limit of the previous process is an actual segment on the \(x \)–axis.

- What is the length of this segment?
- Find an equation in polar coordinates \((\rho,\theta) \) of a curve passing through all the vertices of the resulting triangles, except possibly \((0,0). \)
- Find an equation for that curve in rectangular coordinates.

Let us denote \( h_n \) the hypotenuse of triangle \( \triangle_n. \) By construction we have \( h_1=\sqrt{2} \) and, for each consecutive steps, the definition of cosine in our setup gives us that

We may solve the recurrence with the aid of the formula of sine of double angles: \( \sin 2\alpha = 2 \sin \alpha \cos \alpha. \) In our case, setting \( \alpha = \frac{\pi}{2^{n+1}} \), we have

or equivalently,

which is the same as

We use this identity to obtain the answer to the first two questions. For instance, an expression in polar coordinates of a function that goes through all the vertices of the triangles could be given by

And in that case, the length of the segment at the limit is given by

I leave the third part of the puzzle an an exercise to the reader.

## Miscellaneous

This is the \( \LaTeX \) code used with `tikz`

to produce the diagram above. Note the power of the `plot`

command, that allows us to obtain a very accurate graph of any function.

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\begin{tikzpicture}[scale=2]
\draw[very thick,->] (-0.49,0) -- (1.99,0) node[right]{ $x$ };
\draw[fill=red] (0,0) -- (0,1) -- (1,1) -- cycle;
\draw[fill=red!60!white] (0,0) -- (22.5:1.5307) -- (1,1) -- cycle;
\draw[fill=red!30!white] (0,0) -- (11.25:1.5607) -- (22.5:1.5307) -- cycle;
\draw[fill=red!10!white] (0,0) -- (5.625:1.56827) -- (11.25:1.5607) -- cycle;
\draw[domain=0.001:3.141,smooth,variable=\t,very thick,blue] %
plot ({\t r}:1.570796*{sin(\t r)}/\t);
\draw (90:0.2) arc (90:45:0.2);
\draw (45:0.3) arc (45:22.5:0.3);
\draw (22.5:0.4) arc (22.5:11.25:0.4);
\draw (11.25:0.5) arc (11.25:5.625:0.5);
\draw[very thick,->] (0,-0.49) -- (0,1.49) node[above]{ $y$ };
\draw[fill=white] (0,1) circle (0.5pt);
\draw[fill=white] (1,1) circle (0.5pt);
\draw[fill=white] (22.5:1.5307) circle (0.5pt);
\draw[fill=white] (11.25:1.5607) circle (0.5pt);
\draw[fill=white] (5.625:1.56827) circle (0.5pt);
\end{tikzpicture}