Triangles with multiple angles (*)

Introduction

Given a plane triangle $$\triangle ABC$$, the equality between the angles $$\angle B=\angle A$$ is equivalent to a simple relation between opposite sides to these angles, $$|AC| = |BC|$$. What if $$\angle B = n \angle A$$ for some natural number $$n$$?

This question was considered by Euler in 1765 in a 35 pages paper [1], which is cataloged in the Eneström index as E324. Euler begins by studying some particular cases, from $$n=1$$ to $$n=5$$, and realized the existence of a pattern in the equations obtained. However he only presents the cases for $$n$$ up to $$13$$ and doesn’t present a complete proof of the equivalences that he claims to be true. Therefore, it is worth returning to this problem.

Given a triangle $$\triangle ABC$$, with vertices $$A,B,C$$, let $$a$$, $$b$$ and $$c$$ be the lengths of the opposite sides to $$A$$,$$B$$ and $$C$$, respectively. Let us recall how can we deduce geometrically that the equality $$\angle B = \angle A$$ is equivalent to $$a=b$$.

First suppose $$\angle B = \angle A$$ and consider the line segment $$CP$$ perpendicular to the line $$AB$$. This way we have two similar right triangles, $$\triangle APC$$ and $$\triangle BPC$$, with angles $$\angle A,\frac{\pi}{2}$$ and $$\frac{\pi}{2} - \angle A$$. Besides that, these triangles have a common side $$CP$$, which is corresponding by similarity. So, these triangles are congruent and $$a=b$$. Conversely, if $$a=b$$, let $$M$$ be the midpoint of $$AB$$ (figure 1). The triangles $$\triangle AMC$$ and $$\triangle BMC$$ are congruent, so, in particular, $$\angle B = \angle A$$.

Now let us consider the triangles $$\triangle ABC$$ such that $$\angle B = 2 \angle A$$. If the bissector of the angle $$\angle B$$ intersects $$AC$$ at the point $$Q$$, then the triangles $$\triangle ABC$$ and $$\triangle BQC$$ are similar. By Thales’ Theorem, we have

$\frac{|AC|}{|BC|}=\frac{|AB|}{|BQ|}=\frac{|BC|}{|CQ|}$ that is,$\frac{b}{a}=\frac{c}{|BQ|}=\frac{a}{|CQ|}$ and, consequently, $$|BQ|=\frac{ac}{b}$$ and $$|CQ|=\frac{a^{2}}{b}$$. Besides that, by construction, we have $$|AC| = |AQ| + |CQ|$$, so, $$|AQ|=b-\frac{a^{2}}{b}$$. Since in the triangle $$\triangle ABQ$$, we have $$\angle B = \angle A$$, we also have $$|AQ| = |BQ|$$. Replacing the above values in this equality, we obtain $$b-\frac{a^{2}}{b}=\frac{ac}{b}$$ which is equivalent to $$b^{2}-a^{2} = ac$$. Conversely, suppose that for the triangle $$\triangle ABC$$, with corresponding sides $$a$$, $$b$$ and $$c$$, the equality $$b^{2}-a^{2} = ac$$ holds. Then $$b>a$$ and we may fix a point $$Q$$ on the side $$AC$$ such that the angle $$\angle QBC$$ is equal to $$\angle A$$. This way we created a triangle $$\triangle BQC$$ similar to $$\triangle ABC$$, and, as we saw before, $|BQ|=\frac{ac}{b},|CQ|=\frac{a^{2}}{b} \mbox{ and } |AQ|=b-\frac{a^{2}}{b}.$Since by hypothesis $$b-\frac{a^{2}}{b}=\frac{ac}{b}$$, that is, $$|AQ| = |BQ|$$, we know that the triangle $$\triangle AQB$$ is isosceles and $$\angle ABQ = \angle A$$. It follows that, in the triangle $$\triangle ABC$$, we have $$\angle B = 2 \angle A$$.

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Translated for Atractor by a CMUC team, from its original version in Portuguese. Atractor is grateful for this cooperation.