# How to prove that this triangle is equilateral?

Question: If $\cos A +\cos B +\cos C=\frac{3}{2}$, prove that the triangle is equilateral.

My attempt: According to cosine rule, $\cos A=\frac{b^2+c^2-a^2}{2bc}$, $\cos B=\frac{c^2+a^2-b^2}{2ca}$ and $\cos C=\frac{a^2+b^2-c^2}{2ab}$.

$\cos A +\cos B +\cos C=\frac{3}{2}$

$\therefore\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}+\frac{a^2+b^2-c^2}{2ab}=\frac{3}{2}$

$\implies\frac{ab^2+ac^2-a^3+bc^2+ba^2-b^3+ca^2+cb^2-c^3}{2abc}=\frac{3}{2}$

I have tried simplifying the given equation using cosine rule but could not get far. Please help.

To prove that the triangle is equilateral, i have done the following: I have used the cosine rule to simplify the given equation, algebraic manipulation to bring it into the required form, triangle inequality to deduce that each term in the manipulated equation is non-negative and the fact that when the sum of non-negative terms is equal to zero, each term is equal to zero to set up separate equations to finally deduce that all three sides of the triangle are equal.

According to cosine rule, $\cos A=\frac{b^2+c^2-a^2}{2bc}$, $\cos B=\frac{c^2+a^2-b^2}{2ca}$ and $\cos C=\frac{a^2+b^2-c^2}{2ab}$.

$\cos A +\cos B +\cos C=\frac{3}{2}$

$\implies\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}+\frac{a^2+b^2-c^2}{2ab}=\frac{3}{2}$

$\implies\frac{ab^2+ac^2-a^3+bc^2+ba^2-b^3+ca^2+cb^2-c^3}{2abc}=\frac{3}{2}$

$\implies ab^2+ac^2-a^3+bc^2+ba^2-b^3+ca^2+cb^2-c^3=3abc$

$\implies ab^2+ac^2+bc^2+ba^2+ca^2+cb^2=a^3+b^3+c^3+3abc$

$\implies ab^2+ac^2+bc^2+ba^2+ca^2+cb^2-6abc=a^3+b^3+c^3-3abc$

$\implies ca^2-2abc+cb^2+ab^2-2abc+ac^2+bc^2-2abc+ba^2=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$\implies c(a^2-2ab+b^2)+a(b^2-2bc+c^2)+b(c^2-2ca+a^2)=\frac{1}{2}(a+b+c)(2a^2+2b^2+c^2-2ab-2bc-2ca)$

$\implies 2[c(a-b)^2+a(b-c)^2+b(c-a)^2]=(a+b+c)(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2)$

$\implies 2c(a-b)^2+2a(b-c)^2+2b(c-a)^2=(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]$

$\implies 2c(a-b)^2+2a(b-c)^2+2b(c-a)^2=(a+b+c)(a-b)^2+(a+b+c)(b-c)^2+(a+b+c)(c-a)^2$

$\implies 0=(a+b+c)(a-b)^2-2c(a-b)^2+(a+b+c)(b-c)^2-2a(b-c)^2+(a+b+c)(c-a)^2-2b(c-a)^2$

$\implies (a+b+c)(a-b)^2-2c(a-b)^2+(a+b+c)(b-c)^2-2a(b-c)^2+(a+b+c)(c-a)^2-2b(c-a)^2=0$

$\implies (a+b+c-2c)(a-b)^2+(a-2a+b+c)(b-c)^2+(a+b-2b+c)(c-a)^2=0$

$\implies (a+b-c)(a-b)^2+(-a+b+c)(b-c)^2+(a-b+c)(c-a)^2=0$

$\implies (a+b-c)(a-b)^2+(b+c-a)(b-c)^2+(c+a-b)(c-a)^2=0$

According to triangle inequality, $a+b>c, b+c>a$ and $c+a>b$.

$\therefore a+b-c>0, b+c-a>0$ and $c+a-b>0$.

When the sum of non-negative terms is equal to zero, each term$=0$.

$\therefore (a+b-c)(a-b)^2+(b+c-a)(b-c)^2+(c+a-b)(c-a)^2=0$

$\implies (a+b-c)(a-b)^2=0, (b+c-a)(b-c)^2=0$ and $(c+a-b)(c-a)^2=0$

$\implies a=b, b=c$ and $c=a$

$\implies a=b=c$

$\therefore$ The triangle is equilateral.

Or $$\sum_{cyc}(a^3-a^2b-a^2c+abc)=0.$$ Now, let $a\geq b\geq c$.

Thus, $$0=\sum_{cyc}(a^3-a^2b-a^2c+abc)=\sum_{cyc}(a^3-a^2b-ab^2+abc)=\sum_{cyc}a(a-b)(a-c)\geq$$ $$\geq a(a-b)(a-c)+b(b-a)(b-c)=(a-b)^2(a+b-c)\geq0,$$ which says that the equality occurs for $a=b=c$.

I Assume that all angles are acute. Now its wise to use concavity of $\cos$ to say (from Jensen's Inequality)

$$\cos(A)+\cos(B) + \cos(C) \ge 3\cos(\frac{A+B+C}{3}) = 3 \cos(\frac{\pi}{3})=\frac{3}{2}$$

equality occurs for $A=B=C = \pi / 3$.

I don't know how to handle obtuse triangles with this..,

A slick way is to combine Carnot's theorem with Euler's inequality $R\geq 2r$.
$$R\cos A+R\cos B + R\cos C$$ is the sum of the signed distances of the circumcenter from the triangle sides, and it equals $R+r$. Since $r<\frac{R}{2}$ unless $ABC$ is equilateral, $$R\cos A+R\cos B + R\cos C = \frac{3}{2}R$$ implies that $ABC$ is equilateral.