# Problem 3. Maximum 14 points

Settlements $A, B$, and $C$ are connected by straight roads. The distance from settlement $A$ to the road connecting settlements $B$ and $C$ is 100 km, and the sum of the distances from settlement $B$ to the road connecting $A$ and $C$, and from settlement $C$ to the road connecting $A$ and $B$ is 300 km. It is known that settlement $D$ is equidistant from the roads connecting settlements $A, B, C$ and lies within the area bounded by these roads. Any resident of all settlements spends no more than 1 liter of fuel for every 10 km of road. What is the maximum amount of fuel that would be needed by a motorist who needs to get from settlement $D$ to any of the roads connecting the other settlements?

#
# Solution

The settlements form a triangle $\mathrm{ABC}$, and point $\mathrm{D}$, being equidistant from the sides of the triangle, is the incenter of the triangle (i.e., the center of the inscribed circle). Note that the fuel consumption will be maximal when the distance from point $\mathrm{D}$ to the sides of triangle $\mathrm{ABC}$ is maximal, or, equivalently, when the radius $\mathrm{r}$ of the inscribed circle in the triangle is maximal under the conditions of the problem. The conditions of the problem can be easily reformulated as follows (see the figure):

![](https://cdn.mathpix.com/cropped/2024_05_06_521382fedbd8d60cdabcg-04.jpg?height=768&width=857&top_left_y=1158&top_left_x=377)

$$
\text { ha=100, hb+hc=300, }
$$

where $\mathrm{ha}, \mathrm{hb}, \mathrm{hc}$ are the heights of triangle ABC. It is not difficult to establish that the following equality holds:

$$
1 \mathrm{ha} + 1 \mathrm{hb} + 1 \mathrm{hc} = 1 \mathrm{r} \text {. }
$$

Indeed, let $S$ be the area of the triangle, then

$$
\text { S=12aha=12bhb=12chc=pr. }
$$

From this, we have:

$$
\text { ha=2Sa, hb=2Sb, hc=2Sc, r=Sp. }
$$

And we get:

$$
1 \mathrm{ha} + 1 \mathrm{hb} + 1 \mathrm{hc} = \mathrm{a} / 2 \mathrm{~S} + \mathrm{b} / 2 \mathrm{~S} + \mathrm{c} / 2 \mathrm{~S} = (\mathrm{a} + \mathrm{b} + \mathrm{c}) / 2 \mathrm{~S} = \mathrm{p} / \mathrm{S} = 1 \mathrm{r} \text {. }
$$

From the condition, it follows that

$$
1 \mathrm{r} = 1 / 100 + 1 / \mathrm{hb} + 1 / \mathrm{hc} = 1 / 100 + 1 / \mathrm{hb} + 1 / (300 - \mathrm{hb}) = 1 / 100 + 300 / (\mathrm{hb} (300 - \mathrm{hb})).
$$

The radius $\mathrm{r}$ is maximal when the sum on the right side of the equation is minimal. This is equivalent to the maximality of the expression

$$
\mathrm{hb} / (300 - \mathrm{hb}).
$$

This is a quadratic function and its maximum is achieved at its vertex, i.e., as it is easy to understand, when $\mathrm{hb} = 150$. Therefore, the maximum value of the radius of the inscribed circle is found from the equation

$$
1 \mathrm{r} = 1 / 100 + 300 / (150 * (300 - 150)) = 1 / 100 + 1 / 75 = 7 / 300 \Leftrightarrow \mathrm{r} = 300 / 7 \text { km. }
$$

From this, the maximum fuel consumption is:

$$
\text { r10 = 30 / 7 = 307 liters. }
$$

Answer: 307 liters.