### Triangles with multiple angles (2)

With this characterization of triangles in which an angle is twice another we can, for instance, prove that there is exactly one of these triangles so that the lengths of the sides are consecutive integers^{1}: the one with \(a = 4,b = 6\) or \(c = 5\). Or we can even determine general expressions for the sine or the cosine of a double-angle.

By similar process, using our knowledge for \(n=2\), we may find an explicit relation between the lengths of the sides of a triangle with one angle equal to three times another.

By construction, in the triangle \(\triangle ABQ\) we have \(\angle B = 2 \angle A\), and, therefore, \(|AQ|^{2}- |BQ|^{2} = |AB||BQ|\), that is,
\[\left(b-\frac{a^{2}}{b}\right)^{2}-\left(\frac{ac}{b}\right)^{2}=\left(\frac{ac}{b}\right)c\]
which is equivalent to \((b^{2} - a^{2})^{2} - ac^{2}(a + b) = 0\) and, canceling \(a + b\), also to\[(b^{2} - a^{2} + c^{2})(b - a) - bc^{2} = 0.\]
To prove the reciprocal, let us go back to reference [1]. Euler states that if for a triangle \(\triangle ABC\), with corresponding sides \(a\), \(b\), \(c\), the equality \((b^{2} - a^{2} + c^{2})(b - a) - bc^{2} = 0\) holdes, then \(\angle B = 3\angle A\).
The picture used by Euler to illustrate this statement, which involves a circle with center on vertex \(C\) and its intersection points with side \(AC\), is only possible if \(b>a\). Obviously this is a necessary condition for the angle on \(B\) to be three times the angle on \(A\), but this order among \(a\) and \(b\) is not a consequence of hypothesis \((b^{2}-a^{2}+c^{2})(b-a)-bc^{2} = 0\).
This equality only guarantees that, if \(b < a\), then we must have \(b^{2} -a^{2} +c^{2} < 0\), and, consequentely, \(c < a\).

For instance, if, \(a = 3\), \(c = 2\) and \(b\) is the solution of the equation \[(b^{2} - 5)(b - 3) = 4b\]
in \(\left]1,\sqrt{5}\right[\) then \(a\), \(b\) and \(c\) are the lengths of the sides of a triangle \(\triangle ABC\), opposite to vertices \(A\), \(B\), \(C\), respectively, that satisfies condition \((b^{2} - a^{2} + c^{2})(b-a) - bc^{2} = 0\) but \(\angle B\neq3\angle A\).^{2}

However, if we admit that for the triangle \(\triangle ABC\), with sides \(a\), \(b\), \(c\), we have \[(b^{2} - a^{2} + c^{2})(b-a) - bc^{2} = 0 \mbox{ and } b>a\] then, we may fix on the side \(AC\), a point \(Q\) such that the angle \(\angle QBC\) equals to \(\angle A\). This way, the triangle \(\triangle BQC\) is similar to a \(\triangle ABC\), with sides \(|BQ|=\frac{ac}{b},|CQ|=\frac{a^{2}}{b}\) and, therefore \(|AQ|=b-\frac{a^{2}}{b}\). Besides that, since, by hypothesis, we have \[\left(b-\frac{a^{2}}{b}\right)^{2}-\left(\frac{ac}{b}\right)^{2}=\left(\frac{ac}{b}\right)c\] that is, \(|AQ|^{2} - |BQ|^{2} = |AB||CQ|\), we conclude, from the \(n = 2\) case, that for the original triangle \(\triangle ABQ\), the angle \(\angle ABQ\) equals \(2 \angle QAB\). From this we proved that, for the triangle \(\triangle ABC\), We must have \(\angle B = 3\angle A\).

^{1}Problem proposed, in 1968, for the International Mathematical Olympiad.

^{2}In this triangle, there isn’t even any angle equal to three times another.