# Prove that given a triangle satisfying $8\prod \sin\frac{A}{2}=\prod \cos(A-B)$ then that triangle is equilateral.

Prove that given a triangle $ABC$ satisfying $$8 \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} = \cos(A-B)\cos(B-C)\cos(C-A)$$ then that triangle is equilateral.

• Suggest you re-word the question to something like "prove that given a triangle satisfying (your equation) then that triangle is equilateral." In your question you already have said it holds for equilaterals, but it seems you want it to hold only for equilaterals. – coffeemath Jul 26 '15 at 6:28
• @LzuTao, Hi Lzu. I did not receive any notification. I just accidently found this question. – Arashium Jul 26 '15 at 7:05
• Just adding that, you have two equations with three variables. Equilateral triangle might not be the only solution. – Arashium Jul 26 '15 at 7:26
• Have a look at the infinite solutions: link – Arashium Jul 26 '15 at 7:50
• Guess: expand the RHS? What's the rule for cosines, cos(A+B) = cosAcosB - sinAsinB? – BCLC Jul 26 '15 at 17:36

Let $p(a,b,c)$ be the left side of your equation, and $q(a,b,c)$ be the right side, using lower case for your $A,B,C$ for no real reason. The equation is then saying that $h(a,b,c)=0$ where $h(a,b,c)=p(a,b,c)-q(a,b,c).$
Now if $a+b<\pi$ we can get a legal triangle with angles $a,b,\pi-a-b.$ Define $w(a,b)=h(a,b,\pi-a-b),$ then if your equation were to hold only for equilateral triangles, we could have $w(a,b)=0$ only for $a=b=\pi/3.$ However for example $w(0.1,0.2)=-0.7648<0$ while $w(1.1,0.2)=0.3647>0.$ This means from the intermediate value theorem there is some $c$ between $0.1$ and $1.1$ for which $w(c,0.2)=0.$ However this $c$ is not $\pi/3$ [or rather more simply, $0.2 \neq \pi/3$] so that there are triangles not equilateral which satisfy your equation.