Triangle Solving – Problem 4: Multiple choice (select all that apply)

Multiple choice (select all that apply). In $\triangle ABC$, let the sides opposite $A,B,C$ be $a,b,c$. Suppose $a^{2}-b^{2}-c^{2}+bc=0$. Which statements are true?

A. $A=\frac{\pi}{3}$ B. If $a=\sqrt{3}$ and $\cos B=\frac{4}{5}$, then $c=\frac{4\sqrt{3}-3}{5}$ C. If $a=2$, then the maximum possible area of $\triangle ABC$ is $\sqrt{3}$ D. If $b\sin C=\sin C+\sqrt{3}\cos C$, then $c=2$

Step 1. From $a^{2}-b^{2}-c^{2}+bc=0$ we get $b^{2}+c^{2}-a^{2}=bc$. By the Law of Cosines, $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}=\frac{1}{2}$, so $A=\frac{\pi}{3}$. Thus A is true.

Step 2. For B: with $A=\frac{\pi}{3}$ we have $\sin A=\frac{\sqrt{3}}{2}$. The Law of Sines gives $\frac{a}{\sin A}=2R$, so $2R=\frac{\sqrt{3}}{\sqrt{3}/2}=2$. Hence $b=2R\sin B=2\sin B$. Since $\cos B=\frac{4}{5}$, $\sin B=\frac{3}{5}$ and $b=\frac{6}{5}$. Substituting into $b^{2}+c^{2}-bc=a^{2}=3$ gives $c=\frac{3+4\sqrt{3}}{5}$, so B is false.

Step 3. For C: since $a^{2}=b^{2}+c^{2}-bc$, we have $b^{2}+c^{2}=4+bc\ge 2bc$, so $bc\le 4$, with equality at $b=c=2$. The area is $S=\frac12 bc\sin A=\frac{\sqrt{3}}{4}bc\le \sqrt{3}$. Thus C is true.

Step 4. For D: by the Law of Sines, $b\sin C=c\sin B$. The condition gives $c\sin B=\sin C+\sqrt{3}\cos C=2\sin\left(C+\frac{\pi}{3}\right)=2\sin(C+A)=2\sin B$. Since $\sin B>0$, we get $c=2$. Thus D is true.

Step 5. Therefore, the correct choices are A, C, and D.

ACD