点到三边距离和的期望

对任意内点 $X$，令 $d_a,d_b,d_c$ 分别为 $X$ 到边长为 $a,b,c$ 的三边的垂距，则三角形面积可分解为三个小三角形面积之和：

$$A=\frac12 a d_a+\frac12 b d_b+\frac12 c d_c.$$

对均匀随机点，点到某一条边的距离 $d_a$ 与“对边小三角形面积”成正比，而面积比的期望等于 $1/3$，因此

$$\mathbb{E}[d_a]=\frac{2A}{3a}=\frac{h_a}{3},$$

其中 $h_a$ 为对边 $a$ 的高。对 $b,c$ 同理，所以

$$\mathbb{E}[d_a+d_b+d_c]=\frac{h_a+h_b+h_c}{3}.$$

本题 45-60-75 为直角三角形（比例 3-4-5），面积

$$A=\frac12\cdot 45\cdot 60=1350.$$

对应三条高：

$$h_{45}=\frac{2A}{45}=60,\quad h_{60}=\frac{2A}{60}=45,\quad h_{75}=\frac{2A}{75}=36.$$

因此

$$\mathbb{E}[d_a+d_b+d_c]=\frac{60+45+36}{3}=\boxed{47}.$$

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英文解析

For any interior point $X$, let $d_a, d_b, d_c$be the perpendicular distances from $X$to the three sides of length $a, b, c$, respectively. Then the area of the triangle can be decomposed into the sum of the areas of the three smaller triangles:

$$A=\frac12 a d_a+\frac12 b d_b+\frac12 c d_c.$$

For a uniformly random point, the distance $d_a$ to a particular side is proportional to the "area of the opposite small triangle," and the expected value of the ratio of areas equals $1/3$. Therefore,

$$\mathbb{E}[d_a]=\frac{2A}{3a}=\frac{h_a}{3},$$

where $h_a$ is the altitude to side $a$. Similarly for $b$and $c$, so

$$\mathbb{E}[d_a+d_b+d_c]=\frac{h_a+h_b+h_c}{3}.$$

In this problem, the triangle with sides $45-60-75$ is a right triangle (ratio $3-4-5$), with area

$$A=\frac12\cdot 45\cdot 60=1350.$$

The corresponding altitudes are:

$$h_{45}=\frac{2A}{45}=60,\quad h_{60}=\frac{2A}{60}=45,\quad h_{75}=\frac{2A}{75}=36.$$

Thus,

$$\mathbb{E}[d_a+d_b+d_c]=\frac{60+45+36}{3}=\boxed{47}.$$