Let $A B C$ be any triangle, and choose points $A^{\prime}, B^{\prime}$, and $C^{\prime}$ independently and uniformly on the sides $B C, C A$, and $A B$, respectively. For a point $Z$ in the plane, let $p(Z)$ denote the probability that the triangle bounded by the lines $A A^{\prime}$, $B B^{\prime}$, and $C C^{\prime}$ contains $Z$. Determine the interior point $Z$ of the triangle $A B C$ for which $p(Z)$ is maximized.
Let $Z_{A}, Z_{B}$, and $Z_{C}$ be the intersection points of the lines $A Z, B Z$, and $C Z$ with the opposite sides of triangle $A B C$. Denote

$$
\alpha:=\frac{B Z_{A}}{B C}, \quad \beta:=\frac{C Z_{B}}{C A} \quad \text { and } \quad \gamma:=\frac{A Z_{C}}{A B}
$$

the ratios in which these points divide the respective sides. Then

$$
\frac{Z_{A} C}{B C}=\frac{B C-B Z_{A}}{B C}=1-\frac{B Z_{A}}{B C}=1-\alpha
$$

and similarly

$$
\frac{Z_{B} A}{C A}=1-\beta, \quad \text { and } \quad \frac{Z_{C} B}{A B}=1-\gamma
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_fdd652518331dd71eff3g-1.jpg?height=431&width=553&top_left_y=674&top_left_x=775)

According to Ceva's theorem,

$$
\begin{aligned}
1 & =\frac{B Z_{A}}{Z_{A} C} \cdot \frac{C Z_{B}}{Z_{B} A} \cdot \frac{A Z_{C}}{Z_{C} B}=\frac{B Z_{A}}{B C} \cdot \frac{B C}{Z_{A} C} \cdot \frac{C Z_{B}}{C A} \cdot \frac{C A}{Z_{B} A} \cdot \frac{A Z_{C}}{A B} \cdot \frac{A B}{Z_{C} B}= \\
& =\frac{\alpha \beta \gamma}{(1-\alpha)(1-\beta)(1-\gamma)}
\end{aligned}
$$

follows, hence

$$
\alpha \beta \gamma=(1-\alpha)(1-\beta)(1-\gamma)
$$

It is easily seen that the (possibly degenerate) triangle bounded by the lines $A A^{\prime}, B B^{\prime}$, and $C C^{\prime}$ contains the point $Z$ if and only if either $B A^{\prime} \leq B Z_{A}$ and $C B^{\prime} \leq C Z_{B}$, and $A C^{\prime} \leq A Z_{C}$ hold; or if $B A^{\prime} \geq B Z_{A}, C B^{\prime} \geq C Z_{B}$, and $A C^{\prime} \geq A Z_{C}$ hold. This means that

$$
p(Z)=\alpha \beta \gamma+(1-\alpha)(1-\beta)(1-\gamma)=2 \alpha \beta \gamma=2(1-\alpha)(1-\beta)(1-\gamma)
$$

the latter equalities hold due to (1). From the arithmetic and geometric mean inequality,

$$
\begin{aligned}
& \sqrt[3]{\frac{p(Z)}{2}}=\sqrt[3]{\alpha \beta \gamma} \leq \frac{\alpha+\beta+\gamma}{3}, \quad \text { and } \\
& \sqrt[3]{\frac{p(Z)}{2}}=\sqrt[3]{(1-\alpha)(1-\beta)(1-\gamma)} \leq \frac{3-\alpha-\beta-\gamma}{3}
\end{aligned}
$$

follows, hence

$$
2 \sqrt[3]{\frac{p(Z)}{2}} \leq \frac{\alpha+\beta+\gamma}{3}+\frac{3-\alpha-\beta-\gamma}{3}=1
$$

therefore $p(Z) \leq \frac{1}{4}$ follows. Equality can hold if and only if both arithmetic and geometric mean inequalities hold with equality, i.e., if $\alpha=\beta=\gamma$ holds. In this case, however, (1) implies $\alpha=\beta=\gamma=\frac{1}{2}$, i.e., $Z$ is the centroid of triangle $A B C$. It is easily verified that for the centroid, all the above estimates hold with equality. This proves that the answer to the problem is the centroid.

Remark. The problem can be easily solved using barycentric coordinates. Let $\alpha, \beta, \gamma$ be the barycentric coordinates of $Z$, i.e., $\vec{Z}=\alpha \vec{A}+\beta \vec{B}+\gamma \vec{C}$, where $\vec{X}$ denotes the position vector of point $X$ and $\alpha+\beta+\gamma=1$. Then the expression

$$
p(Z)=2 \cdot \frac{\alpha \beta \gamma}{(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)}
$$

needs to be maximized. From the arithmetic and geometric mean inequality,

$$
(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha) \geq 2 \sqrt{\alpha \beta} \cdot 2 \sqrt{\beta \gamma} \cdot 2 \sqrt{\gamma \alpha}=8 \alpha \beta \gamma
$$

follows, hence

$$
p(Z)=2 \cdot \frac{\alpha \beta \gamma}{(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)} \leq 2 \cdot \frac{\alpha \beta \gamma}{8 \alpha \beta \gamma}=\frac{1}{4}
$$

follows, with equality holding only if $\alpha=\beta=\gamma$. The required probability is therefore maximal when $\alpha=\beta=\gamma=\frac{1}{3}$, i.e., at the centroid.