HMMT November

Let $\angle C A P=\alpha$, By the Law of Sines, $\frac{\sqrt{3}}{\sin 2 \alpha}=\frac{1}{\sin \alpha}$ which rearranges to $\cos \alpha=\frac{\sqrt{3}}{2} \Rightarrow \alpha=\frac{\pi}{6}$. This implies that $\angle B A C=\frac{\pi}{2}$. By the Pythagorean Theorem, $2 A B^{2}=(\sqrt{3}+1)^{2}$, so $A B^{2}=2+\sqrt{3}$. Applying Stewart's Theorem, it follows that $A P^{2}=\frac{(\sqrt{3}+1)(2+\sqrt{3})}{\sqrt{3}+1}-\sqrt{3} \Rightarrow A P=\sqrt{2}$.