GS. 3 If $P$ is an arbitrary point in the interior of the equilateral triangle $A B C$, find the probability that the area of $\triangle A B P$ is greater than each of the areas of $\triangle A C P$ and $\triangle B C P$.
$D, E, F$ be the mid-points of $A B, A C$ and $B C$ respectively.
The medians $C D, B E$ and $A F$ are concurrent at the centroid $G$.
It is easy to see that $\triangle C E G, \triangle C F G, \triangle A E G, \triangle A D G, \triangle B D G, \triangle B F G$ are congruent triangles having the same areas.
$P$ is any point inside the triangle $\Rightarrow P$ lies on or inside one of these six congruent triangles. As shown in the diagram, $P_{1}$ lies inside $\triangle B F G$. Let the feet of perpendiculars from $P_{1}$ to $A B$, $B C, C A$ be $K_{1}, L_{1}, H_{1}$ with lengths $x_{1}, y_{1}$ and $z_{1}$ respectively.
$P_{1} H_{1}$ and $A F$ meet at $R_{1}, P_{1} K_{1}$ intersects $B E$ at $M_{1}$, and $A F$ at $Q_{1}, L_{1} P_{1}$ produced meet $B E$ at $N_{1}$ By the properties on parallel lines, we can easily prove that $\Delta P_{1} M_{1} N_{1}$ and $\Delta P_{1} Q_{1} R_{1}$ are equilateral triangles. Let $P_{1} M_{1}=P_{1} N_{1}=N_{1} M_{1}=t, P_{1} Q_{1}=P_{1} R_{1}=Q_{1} R_{1}=s$
Let $H$ and $Y$ be the midpoints of $Q_{1} R_{1}$ and $N_{1} M_{1}$ respectively. $R_{1} H=0.5 s, Y M_{1}=0.5 t$
Let $U$ and $T$ be the feet of perpendiculars from $H$ to $A C$ and $A B$ respectively.
Let $X$ and $Z$ be the feet of perpendiculars from $Y$ to $B C$ and $A B$ respectively.
$U H=z_{1}-s+0.5 s \cos 60^{\circ}=z_{1}-0.75 s, Y Z=x_{1}-t+0.5 t \cos 60^{\circ}=x_{1}-0.75 t$
$H T=x_{1}+0.75 s, Y X=y_{1}+0.75 t$
It is easy to show that $\triangle A H U \cong \triangle A H T, \triangle B Y X \cong \triangle B Y Z$ (A.A.S.)
$U H=H T$ and $Y Z=Y X$ (corr. sides, $\cong \Delta ' s$ s) $\Rightarrow z_{1}-0.75 s=x_{1}+0.75 s, x_{1}-0.75 t=y_{1}+0.75 t$
$z_{1}=x_{1}+1.5 s, x_{1}=y_{1}+1.5 t \Rightarrow z_{1}>x_{1}>y_{1}$
$\therefore \frac{1}{2} A C \cdot z_{1}>\frac{1}{2} A B \cdot x_{1}>\frac{1}{2} B C \cdot y_{1} \Rightarrow$ area of $\triangle A C P_{1}>$ area of $\triangle A B P_{1}>$ area of $\triangle B C P_{1}$
If $P_{2}$ lies inside $\triangle B D G$, using a similar method, we can easily prove that
area of $\triangle A C P_{2}>$ area of $\triangle B C P_{2}>$ area of $\triangle A B P_{2}$.
If $P_{3}$ lies inside $\triangle A D G$, then area of $\triangle B C P_{3}>$ area of $\triangle A C P_{3}>$ area of $\triangle A B P_{3}$.
If $P_{4}$ lies inside $\triangle A E G$, then
area of $\triangle B C P_{4}>$ area of $\triangle A B P_{4}>$ area of $\triangle A C P_{4}$.
If $P_{5}$ lies inside $\triangle C E G$, then
area of $\triangle A B P_{5}>$ area of $\triangle B C P_{5}>$ area of $\triangle A C P_{5}$.
If $P_{6}$ lies inside $\triangle C F G$, then
area of $\triangle A B P_{6}>$ area of $\triangle A C P_{6}>$ area of $\triangle B C P_{6}$
In order that the area of $\triangle A B P$ is greater than each of the areas of $\triangle A C P$ and $\triangle B C P, P$ must lie inside $\triangle C E G$ or
$\triangle C F G$
Required probability
$=\frac{\text { Area of } \triangle C E G+\text { area of } \triangle C F G}{\text { Area of } \triangle A B C}=\frac{2}{6}=\frac{1}{3}$
Method 2 Suppose $P$ lies inside $\triangle B D G$.
Produce $A P, B P, C P$ to intersect $B C, C A, A B$ at $L, M, N$ respectively.
Let $S_{\triangle X Y Z}$ denotes the area of $\triangle X Y Z$.
$$
\begin{aligned}
\frac{S_{\triangle A P B}}{S_{\triangle A P C}} & =\frac{S_{\triangle A B M}-S_{\triangle B P M}}{S_{\triangle A C M}-S_{\triangle C P M}} \\
& =\frac{\frac{1}{2} B M \cdot A M \sin A M B-\frac{1}{2} B M \cdot P M \sin A M B}{\frac{1}{2} C M \cdot A M \sin A M C-\frac{1}{2} C M \cdot P M \sin A M C} \\
& =\frac{\frac{1}{2} B M \sin A M B \cdot(A M-P M)}{\frac{1}{2} C M \sin A M C \cdot(A M-P M)}=\frac{B M \sin A M B}{C M \sin A M C} \\
& =\frac{B M}{M C}\left(\because \sin A M B=\sin \left(180^{\circ}-A M C\right)=\sin A M C\right) \\
\frac{S_{\triangle A P B}}{S_{\triangle B P C}} & =\frac{S_{\triangle A B N}-S_{\triangle A P N}}{S_{\triangle B C N}-S_{\triangle C P N}} \\
& =\frac{\frac{1}{2} B N \cdot A N \sin A N B-\frac{1}{2} A N \cdot P N \sin A N B}{\frac{1}{2} B N \cdot N C \sin B N C-\frac{1}{2} N C \cdot P N \sin B N C} \\
& =\frac{\frac{1}{2} A N \sin A N B \cdot(B N-P N)}{\frac{1}{2} N C \sin B N C \cdot(B N-P N)}=\frac{A N \sin A N B}{N C \sin B N C} \\
& =\frac{A N}{N C}\left(\because \sin A N B=\sin \left(180^{\circ}-B N C\right)=\sin B N C\right)
\end{aligned}
$$

In order that the area of $\triangle A B P$ is greater than each of the areas of $\triangle A C P$ and $\triangle B C P$, $B M>M C$ and $A N>N C$
$\therefore P$ must lie inside $\triangle C E G$ or $\triangle C F G$
$$
\text { Required probability }=\frac{S_{\triangle C E G}+S_{\triangle C F G}}{S_{\triangle A B C}}=\frac{2}{6}=\frac{1}{3}
$$

Remark: The original question is
若 $P$ 是等邊三角形 $A B C$ 內部的隨意一點, 求 $\triangle A B P$ 的面積同時大於 $\triangle A C P$ 及 $\triangle B C P$ 的面積的概率。
If $P$ is an arbitrary point in the interior of the equilateral triangle $A B C$, find the probability that the area of $\triangle A B P$ is greater than both of the areas of $\triangle A C P$ and $\triangle B C P$.
There is a slight difference between the Chinese version and the English version.