Consider the triangle ABC as shown in the diagram. Show that if ∠B = 2 ∠A then b2=a2+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=u2,     b=uv,     c=v2−u2.


First you have to prove the identity b2=a2+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 ∆ABX is an isosceles triangle. The angle of this triangle at B is 180° − 2α, and hence the angles at A and X are α. Therefore ∆XAC and ∆ABC are similar; hence
 a
b
 =  BC
AC
 =  AC
XC
 =  b
a+c
 ,
and thus b2=a2+ac.

Method 2 By the Cosine Rule:
b2
= a2 + c2 −2ac cos2α
a2
= b2 + c2 −2bccosα
Subtracting the equations, using the double angle formula and rearranging them a bit we get :
2b2 − 2a2 = 2c(bcosα− acos2α+a).    (1)

By the Sine Rule:
 a
sinα
=  b
sin2α
 a
sinα
=  b
2sinαcosα
cosα
=  b
2a
 .     (2)

Combining (1) and (2):
2b2 − 2a2
= 2c(  b2
2a
 −  2ab2
4a2
 + a)
= 2c(  b2
2a
 −  b2
2a
 + a)
b2 − a2
= ca
b2
= a2 + ac.

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)=b2, and as a and a+c are coprime, a and a+c must be perfect squares. Let a=u2, a+c=v2 so that
a=u2,     b=uv,     c=v2−u2.

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.