In triangle $ABC$ , $A$ is the biggest angle and $C$ the smallest. If $A=2C$ and $a+c=2b$, find $a:b:c$

Question

In triangle $ABC$ , $A$ is the biggest angle and $C$ is the small angle. If $A=2C$ and $a+c=2b$, find $a:b:c$

I tried extending $AC$ and $AB$ to form an isosceles triangle with sides $a+c$ and $2b$, which generates a lot of other isosceles triangles, but I'm not really sure how to use them.

Answer 1

According to the sine rule,

$$ \frac ac =\frac{\sin A}{\sin C}=\frac{\sin 2C}{\sin C} =2\cos C$$

$$ \frac bc =\frac{\sin (A+C)}{\sin C}=\frac{\sin 3C}{\sin C}$$ $$=4\cos^2C-1= \frac{a^2}{c^2}-1$$

Plug the given $2b=a+c$ into the LHS,

$$\frac{2a^2}{c^2} - \frac ac -3 = 0$$

which yields the ratio,

$$ \frac ac = \frac 32$$

Again, use $2b= a+c$ to obtain,

$$a:b:c= 6:5:4$$

Answer 2

Hint:

$B=\pi-3C>0\implies3C<180^\circ$

$a+c=2b\implies\sin2C+\sin C=2\sin(\pi-3C)$

Use http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html

$2\sin3C/2\cos C/2=4\sin3C/2\cos3C/2$

$\sin\dfrac{3C}2(d-2(4d^3-3d))=0$ where $d=\cos\dfrac C2$

But $0<3C/2<90,\implies8d^2=7$

$\cos C=2(7/8)-1=\dfrac34$

Answer 3

Since we are only interested in side ratios, let $a=1$

$B = 180° - A - C = 180° - 3C$

Law of Sines:

$${1\over \sin(2C)} = {(1+c) \over 2 \sin(3C)} = {c \over \sin C}$$

$$1 = (1+c) - c = {2\sin(3C) \over \sin(2C)} - {\sin C \over \sin(2C)}$$ $$\sin(2C) = 2\sin(3C)-\sin(C)$$ $$2\sin C \cos C = 2(3\sin C - 4 \sin^3 C) - \sin C$$ $$2\cos C = 2(3 - 4(1-\cos^2c)) - 1 $$ $$8\cos^2 C - 2\cos C - 3 = 0$$ $$(4\cos C - 3)(2\cos C + 1) = 0$$ Since $C$ is the small angle, $\cos C = \Large{3 \over 4}$
$$c = {\sin C \over \sin(2C)} = {1 \over 2 \cos C} = {2\over3}$$

$$a:b:c = 1:{1+c\over2}:c = 1:{1\over2}\times{5\over3}:{2\over3} = 6:5:4$$