Triangle Solving – Problem 10: Find the range of

An acute triangle $ABC$ is inscribed in the unit circle (so its circumradius is $1$). Let the sides opposite $A,B,C$ be $a,b,c$. Suppose $$a^{2}+b^{2}-c^{2}=4a^{2}\cos A-2ac\cos B.$$ Find the range of $\frac{ac}{b}$.

A. $(2\sqrt{3},3\sqrt{3})$ B. $(\sqrt{3},3\sqrt{3})$ C. $(\frac{\sqrt{3}}{2},2\sqrt{3})$ D. $(\frac{\sqrt{3}}{2},\sqrt{3})$

Step 1. By the Law of Cosines, $a^{2}+b^{2}-c^{2}=2ab\cos C$. The given equation becomes $2ab\cos C=4a^{2}\cos A-2ac\cos B$.

Step 2. Dividing by $2a>0$ gives $b\cos C=2a\cos A-c\cos B$.

Step 3. Since the circumradius is $R=1$, the Law of Sines yields $a=2\sin A$, $b=2\sin B$, $c=2\sin C$. Substituting into Step 2 gives $\sin B\cos C=2\sin A\cos A-\sin C\cos B$.

Step 4. Rearranging, $\sin B\cos C+\sin C\cos B=2\sin A\cos A$. The left-hand side is $\sin(B+C)=\sin(\pi-A)=\sin A$. Thus $\sin A=2\sin A\cos A$, so $\cos A=\frac12$ and $A=\frac{\pi}{3}$.

Step 5. Therefore $a=2\sin A=\sqrt{3}$, and $$\frac{ac}{b}=\sqrt{3}\cdot\frac{\sin C}{\sin B}.$$

Step 6. Since $B+C=\pi-A=\frac{2\pi}{3}$, we have $C=\frac{2\pi}{3}-B$, so $$\frac{\sin C}{\sin B}=\frac{\sin\left(\frac{2\pi}{3}-B\right)}{\sin B}=\frac{\frac{\sqrt{3}}{2}\cos B+\frac12\sin B}{\sin B}=\frac{\sqrt{3}}{2}\cot B+\frac12.$$

Step 7. Hence $$\frac{ac}{b}=\frac{3}{2}\cot B+\frac{\sqrt{3}}{2}.$$

Step 8. Because the triangle is acute and $B+C=\frac{2\pi}{3}$, we have $B\in\left(\frac{\pi}{6},\frac{\pi}{2}\right)$, so $\cot B\in(0,\sqrt{3})$.

Step 9. Therefore $\frac{ac}{b}\in\left(\frac{\sqrt{3}}{2},2\sqrt{3}\right)$, so the correct choice is C.

C