Triangle Solving – Problem 5: determine the range of

In a plane quadrilateral $ABCD$, suppose $|AB|\cdot|BC|=4$, $\angle ABC=\frac{2\pi}{3}$, and $\angle ADC=\frac{\pi}{3}$. When the length of the diagonal $AC$ is minimized, determine the range of $AD+DC$.

Step 1. In $\triangle ABC$, by the Law of Cosines $$AC^{2}=AB^{2}+BC^{2}-2\,AB\cdot BC\cos\angle ABC.$$ With $\angle ABC=\frac{2\pi}{3}$ and $AB\cdot BC=4$, we get $AC^{2}=AB^{2}+BC^{2}+4$.

Step 2. For fixed $AB\cdot BC=4$, we have $AB^{2}+BC^{2}\ge 2AB\cdot BC=8$, with equality when $AB=BC=2$. Hence $AC^{2}\ge 12$, so $AC_{\min}=2\sqrt{3}$.

Step 3. In $\triangle ACD$, $\angle ADC=\frac{\pi}{3}$. By the Law of Sines, $$\frac{AD}{\sin\angle ACD}=\frac{DC}{\sin\angle CAD}=\frac{AC}{\sin\angle ADC}=\frac{2\sqrt{3}}{\sin(\pi/3)}=4.$$ So $AD=4\sin\angle ACD$.

Step 4. Also $DC=4\sin\angle CAD=4\sin\left(\angle ACD+\frac{\pi}{3}\right)=2\sin\angle ACD+2\sqrt{3}\cos\angle ACD$.

Step 5. Therefore $$AD+DC=6\sin\angle ACD+2\sqrt{3}\cos\angle ACD=4\sqrt{3}\sin\left(\angle ACD+\frac{\pi}{6}\right).$$

Step 6. Since $\angle ACD\in(0,\pi-\angle ADC)=(0,\frac{2\pi}{3})$, we have $\angle ACD+\frac{\pi}{6}\in\left(\frac{\pi}{6},\frac{5\pi}{6}\right)$, so $\sin\left(\angle ACD+\frac{\pi}{6}\right)\in\left(\frac{1}{2},1\right]$.

Step 7. Hence $AD+DC\in(2\sqrt{3},4\sqrt{3}]$.

$$(2\sqrt{3},4\sqrt{3}]$$