For triangle ABC with sides $a,b,c$, $ \cos (A-B) = \frac{2 \sin A \sin B}{\sin C}$, and $\frac{2}{a} + \frac{1}{b} = 1$

Question

The question asks the minimal length of the perimeter of the triangle. I tried to use $\sin C = \sin(A+B)$, and move it to the other side, and get $\sin A \cos A + \sin B \cos B = 2 \sin A \sin B$, but I don't know how to continue from here, Thank you for helping.

Answer 1

We obtain: $$\cos(\alpha-\beta)\sin\gamma=\cos(\alpha-\beta)-\cos(\alpha+\beta)$$ or $$\cos\gamma+(1-\sin\gamma)\cos(\alpha-\beta)=0$$ or $$\cos^2\frac{\gamma}{2}-\sin^2\frac{\gamma}{2}+\left(\sin^2\frac{\gamma}{2}+\cos^2\frac{\gamma}{2}-2\sin\frac{\gamma}{2}\cos\frac{\gamma}{2}\right)\cos(\alpha-\beta)=0$$ or $$\left(\cos\frac{\gamma}{2}-\sin\frac{\gamma}{2}\right)\left(\cos\frac{\gamma}{2}+\sin\frac{\gamma}{2}+\left(\cos\frac{\gamma}{2}-\sin\frac{\gamma}{2}\right)\cos(\alpha-\beta)\right)=0$$ or $$\left(\cos\frac{\gamma}{2}-\sin\frac{\gamma}{2}\right)\left(\cos\frac{\gamma}{2}(1+\cos(\alpha-\beta))+\sin\frac{\gamma}{2}(1-\cos(\alpha-\beta))\right)=0,$$ which gives $$\cos\frac{\gamma}{2}-\sin\frac{\gamma}{2}=0,$$ $$\gamma=90^{\circ}$$ and $$a^2+b^2=c^2.$$ Can you end it now?

I got that the minimal value of a perimeter is equal to $10$.

Answer 2

The triangle looks a right triangle in which $A=60^o, B=30^o, C=90^o$ because:

$Cos (60-30)=\frac{2 Sin (60) Sin (30)}{Sin (90)}=\frac{\sqrt 3}{2}$

And also we have a point like D on AB from which two perpendiculars are dropped on CB=a and AC=b their ratio to heights a and b are $\frac{1}{b}$ and $\frac{2}{a}$ respectively such that we have:

$\frac{2}{a}+\frac{1}{b}+\frac{0}{h_c}=1$

Hence we have:

$b=\frac{c}{2}$

$a=\frac{c\sqrt 3}{2}$

Puting these in relation we get:

$a=\frac{3}{4(2+\sqrt 3}$

$b=\frac{\sqrt 3}{4(2+\sqrt 3)}$

$c=\frac{\sqrt 3}{2(2+\sqrt 3)}$

And the measure of primeter is:

$p=\frac{3(1+\sqrt 3}{4(2+\sqrt 3)}$