Double angle triples

Consider the triangle $ABC$ as shown in the diagram. Show that if $\angle B = 2 \angle A$ then $b^2=a^2+ac$. Find integer solutions of this equation (for example, $a=4$, $b=6$ and $c=5$) and hence find examples of triangles with sides of integer lengths and one angle twice another.

This problem was solved by Yatir Halevi, of Maccabim-Reut High School, Israel. He found a general parametric formula for triangles where one angle is twice another that gives the lengths of the sides $a$, $b$ and $c$ (all integers) of such triangles. The angle opposite side $b$ is double the angle opposite side $a$. Choosing different values of the parameters $u$ and $v$ you get triangles given by:

$$a=u^2,b=uv,c=v^2-u^2.$$

Image

First you have to prove the identity $b^2=a^2+ac$ for such triangles. It can be proved using the similar triangles in the diagram or alternatively from the Sine and Cosine Rules.

Method 1

By construction $\Delta ABX$ is an isosceles triangle. The angle of this triangle at $B$ is $180^\circ - 2\alpha$, and hence the angles at $A$ and $X$ are $\alpha$. Therefore $\Delta XAC$ and $\Delta ABC$ are similar; hence

$$\frac{a}{b}=\frac{BC}{AC}=\frac{AC}{XC}=\frac{b}{a+c},$$

and thus $b^2=a^2+ac$.

Method 2

By the Cosine Rule:

$$\begin{array}{rl}{b}^2& =a^2+c^2-2ac\cos 2\alpha \\ a^2& =b^2+c^2-2bc\cos \alpha \end{array}$$

Subtracting the equations, using the double angle formula and rearranging them a bit we get :

$$2b^2-2a^2=2c(b\cos \alpha -a{\cos }^2\alpha +a).(1)$$

By the Sine Rule:

$$\begin{array}{rl}\frac{a}{\sin \alpha }& =\frac{b}{\sin 2\alpha }\\ \frac{a}{\sin \alpha }& =\frac{b}{2\sin \alpha \cos \alpha }\\ \cos \alpha & =\frac{b}{2a}.(2)\end{array}$$

Combining (1) and (2):

$$\begin{array}{rl}2b^2-2a^2& =2c(\frac{b^2}{2a}-\frac{2a{b}^2}{4a^2}+a)\\ & =2c(\frac{b^2}{2a}-\frac{b^2}{2a}+a)\\ b^2-a^2& =ca\\ b^2& =a^2+ac.\end{array}$$

To find integer solutions, we may assume that $a$, $b$ and $c$ have no common factors. Then $a$ and $a+c$ have no common factors (for if $p$ divides $a$ and $a+c$ then it divides $c$, and hence also $b$). As $a(a+c)=b^2$, and as $a$ and $a+c$ are coprime, $a$ and $a+c$ must be perfect squares. Let $a=u^2$, $a+c=v^2$ so that

$$a=u^2,b=uv,c=v^2-u^2.$$

Note that not every solution gives rise to a triangle (for all three triangle inequalities must be satisfied); for example, $u=1$ and $v=4$ gives $(a,b,c) = (1,4,15)$ and this does not give a triangle. The following are a few examples of values which do give triangles:

u	v				a	b	c
2	3				4	6	5
3	4				9	12	7
3	5				9	15	16
4	5				16	20	9

Note: it can be proved that $a$, $b$ and $c$ give a triangle if and only if $v > u > v/2$; see Math. Gazette , Vol. 412, June 1976, p.130.