Triangle Solving – Problem 13: Find the radius of circle

Ptolemy's theorem states that for a cyclic quadrilateral, the product of the diagonals equals the sum of the products of opposite sides. Let $ABCD$ be a cyclic quadrilateral inscribed in circle $O$. Suppose $AC=\sqrt{3}\,BD$, $\angle ADC=2\angle BAD$, and $$AB\cdot CD+BC\cdot AD=4\sqrt{3}.$$ Find the radius of circle $O$.

A. 4 B. 2 C. $\sqrt{3}$ D. $2\sqrt{3}$

Step 1. By Ptolemy's theorem, $AC\cdot BD=AB\cdot CD+BC\cdot AD=4\sqrt{3}$.

Step 2. Since $AC=\sqrt{3}\,BD$, we get $\sqrt{3}\,BD^{2}=4\sqrt{3}$, hence $BD=2$.

Step 3. Let the circle radius be $R$. Using the chord formula (equivalently, the Law of Sines in the corresponding triangles), $$\frac{AC}{\sin\angle ADC}=\frac{BD}{\sin\angle BAD}=2R.$$

Step 4. From $AC=\sqrt{3}\,BD$, it follows that $\sin\angle ADC=\sqrt{3}\sin\angle BAD$.

Step 5. Using $\angle ADC=2\angle BAD$, we have $$\sin(2\angle BAD)=\sqrt{3}\sin\angle BAD\Rightarrow 2\sin\angle BAD\cos\angle BAD=\sqrt{3}\sin\angle BAD.$$

Step 6. Since $0<\angle BAD<\pi$, $\sin\angle BAD>0$, so $\cos\angle BAD=\frac{\sqrt{3}}{2}$ and $\sin\angle BAD=\frac12$.

Step 7. Therefore $2R=\frac{BD}{\sin\angle BAD}=\frac{2}{1/2}=4$, so $R=2$. Hence the correct choice is B.

B