Triangle Solving – Problem 15: Find angle

In $\triangle ABC$, let the sides opposite $A,B,C$ be $a,b,c$. Suppose $$\sin^{2}A+\sin^{2}C=\sin^{2}B+\sin A\sin C.$$ (1) Find angle $B$. (2) If the area of $\triangle ABC$ is $\sqrt{3}$, find the minimum possible value of $a+c$, and determine the triangle's shape at equality.

Step 1. By the Law of Sines $a=2R\sin A$, $b=2R\sin B$, $c=2R\sin C$. The condition $\sin^{2}A+\sin^{2}C=\sin^{2}B+\sin A\sin C$ becomes $a^{2}+c^{2}=b^{2}+ac$.

Step 2. By the Law of Cosines, $b^{2}=a^{2}+c^{2}-2ac\cos B$. Substituting gives $2ac\cos B=ac$, so $\cos B=\frac12$ and $B=\frac{\pi}{3}$.

Step 3. For (2), the area is $S_{\triangle ABC}=\frac12 ac\sin B=\frac{\sqrt{3}}{4}ac$. Since $S_{\triangle ABC}=\sqrt{3}$, we get $ac=4$.

Step 4. By AM-GM, $a+c\ge 2\sqrt{ac}=4$, with equality when $a=c=2$.

Step 5. If $a=c$, then $A=C$. Together with $B=\frac{\pi}{3}$, we get $A=C=\frac{\pi}{3}$, so the triangle is equilateral.

(1) $B=\frac{\pi}{3}$. (2) The minimum of $a+c$ is $4$, attained when $\triangle ABC$ is equilateral.