7. Let the incircle of equilateral triangle $ABC$ have a radius of 2, with the center at $I$. If point $P$ satisfies $PI=1$, then the maximum value of the ratio of the areas of $\triangle APB$ to $\triangle APC$ is $\qquad$ .
Answer: $\frac{3+\sqrt{5}}{2}$.
Solution: From $P I=1$, we know that point $P$ lies on the unit circle $K$ with center $I$.

Let $\angle B A P=\alpha$. Take a point $P_{0}$ on circle $K$ such that $\alpha$ reaches its maximum value $\alpha_{0}$, at which point $P_{0}$ should fall within $\angle I A C$, and $A P_{0}$ is the tangent point of circle $K$. Since $0<\alpha \leq \alpha_{0}<\frac{\pi}{3}$, we have
$$
\frac{S_{\triangle A P B}}{S_{\triangle A P C}}=\frac{\frac{1}{2} A P \cdot A B \cdot \sin \alpha}{\frac{1}{2} A P \cdot A C \cdot \sin \left(\frac{\pi}{3}-\alpha\right)}=\frac{\sin \alpha}{\sin \left(\frac{\pi}{3}-\alpha\right)} \leq \frac{\sin \alpha_{0}}{\sin \left(\frac{\pi}{3}-\alpha_{0}\right)}=\frac{\sin \left(\frac{\pi}{6}+\theta\right)}{\sin \left(\frac{\pi}{6}-\theta\right)},
$$

where $\theta=\alpha_{0}-\frac{\pi}{6}=\angle I A P_{0}$.
From $\angle A P_{0} I=\frac{\pi}{2}$, we know that $\sin \theta=\frac{I P_{0}}{A I}=\frac{1}{2 r}=\frac{1}{4}$, thus $\cot \theta=\sqrt{15}$, so
$$
\frac{\sin \left(\frac{\pi}{6}+\theta\right)}{\sin \left(\frac{\pi}{6}-\theta\right)}=\frac{\frac{1}{2} \cos \theta+\frac{\sqrt{3}}{2} \sin \theta}{\frac{1}{2} \cos \theta-\frac{\sqrt{3}}{2} \sin \theta}=\frac{\cot \theta+\sqrt{3}}{\cot \theta-\sqrt{3}}=\frac{\sqrt{15}+\sqrt{3}}{\sqrt{15}-\sqrt{3}}=\frac{3+\sqrt{5}}{2} .
$$

According to (1) and (2), when $P=P_{0}$, the maximum value of $\frac{S_{\triangle A P B}}{S_{\triangle A P C}}$ is $\frac{3+\sqrt{5}}{2}$.