## Task 3 - 260613

The connecting roads of three places $A, B, C$ form a triangle. In the middle of the road from $B$ to $C$ lies another place $D$. The distance from $A$ via $B$ to $C$ is $25 \mathrm{~km}$, from $B$ via $C$ to $A$ is $27 \mathrm{~km}$, and from $C$ via $A$ to $B$ is $28 \mathrm{~km}$.

A pioneer group hikes on the mentioned roads from place $A$ to place $D$ via the shortest route.

a) Through which of the two places $B$ or $C$ does the pioneer group pass?

Justify your decision!

b) How much time does she save compared to the longer of the two possible routes from $A$ to $D$ if she travels $4 \mathrm{~km}$ per hour?

c) How long is the total distance she has traveled?
a) Because $A B+B C=25 \mathrm{~km}$ and $A C+B C=27 \mathrm{~km}$, $A C$ is $2 \mathrm{~km}$ longer than $A B$. Therefore, because $C D=B D$, a hike from $A$ over $C$ to $D$ is $2 \mathrm{~km}$ longer than a hike from $A$ over $B$ to $D$.

Since the pioneer group chose the shorter of the two routes, they walk via the location $B$.

b) According to the considerations in a), the pioneer group saves $2 \mathrm{~km}$ on the shorter route from $A$ to $D$ compared to the longer route. If they cover $4 \mathrm{~km}$ per hour, they need half an hour for $2 \mathrm{~km}$. Thus, they save $\frac{1}{2}$ hour on a hike from $A$ to $D$ on the shorter route compared to the longer one.

c) If one were to walk from $A$ over $B$ to $C$, then from $C$ over $A$ to $B$, and then from $B$ over $C$ to $A$, one would walk twice the perimeter of the triangle $A B C$ and, because $25+28+27=80$, cover a total of $80 \mathrm{~km}$.

Therefore, because $80: 2=40$, the perimeter of the triangle $A B C$ is $40 \mathrm{~km}$. Subtracting $A C+B C=27 \mathrm{~km}$ from it leaves $A B=40 \mathrm{~km}-27 \mathrm{~km}=13 \mathrm{~km}$.

Subtracting $A C+A B=28 \mathrm{~km}$ from the perimeter leaves $B C=40 \mathrm{~km}-28 \mathrm{~km}=12 \mathrm{~km}$. Therefore, $B D=12 \mathrm{~km}: 2=6 \mathrm{~km}$. The desired route thus measures $A B+B D=13 \mathrm{~km}+6 \mathrm{~km}=19 \mathrm{~km}$.