Consider the triangle $\mathrm{ABC}$ as shown in the diagram. Show that if $\angle B=2\angle A$ then $b^2=a^2+\mathrm{ac}$. 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:

${\displaystyle a=u^2,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}b=\mathrm{uv},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}c=v^2-u^2.}$

First you have to prove the identity $b^2=a^2+\mathrm{ac}$ 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 $\Delta \mathrm{ABX}$ 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 $\Delta \mathrm{XAC}$ and $\Delta \mathrm{ABC}$ are similar; hence

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

and thus $b^2=a^2+\mathrm{ac}$.

Method 2 By the Cosine Rule:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{b^2}& =a^2+c^2-2\mathrm{ac}\cos 2\alpha \\ \multicolumn{1}{c}{a^2}& =b^2+c^2-2\mathrm{bc}\cos \alpha \end{array}}$

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

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

By the Sine Rule:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{\frac{a}{\sin \alpha }}& =\frac{b}{\sin 2\alpha }\\ \multicolumn{1}{c}{\frac{a}{\sin \alpha }}& =\frac{b}{2\sin \alpha \cos \alpha }\\ \multicolumn{1}{c}{\cos \alpha }& =\frac{b}{2a}.\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(2)\end{array}}$

Combining (1) and (2):

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{2b^2-2a^2}& =2c(\frac{b^2}{2a}-\frac{2{\mathrm{ab}}^2}{4a^2}+a)\\ \multicolumn{1}{c}{}& =2c(\frac{b^2}{2a}-\frac{b^2}{2a}+a)\\ \multicolumn{1}{c}{b^2-a^2}& =\mathrm{ca}\\ \multicolumn{1}{c}{b^2}& =a^2+\mathrm{ac}.\end{array}}$

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

${\displaystyle a=u^2,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}b=\mathrm{uv},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}c=v^2-u^2.}$

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:

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.