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.

(*) Difficulty level: University