[ Auxiliary similar triangles ]

Given an equilateral triangle $A B C$. Point $K$ is the midpoint of side $A B$, point $M$ lies on side $B C$, and $B M: M C=1: 3$. A point $P$ is chosen on side $A C$ such that the perimeter of triangle $P K M$ is the smallest possible. In what ratio does point $P$ divide side $A C$?
Since side KM of triangle $ABC$ is fixed, it is necessary to find a point $P$ on side $AC$ such that the sum $KP + PM$ is the smallest possible. Let $M'$ be the point symmetric to point $M$ with respect to line $AC$. According to problem 55557, the intersection point of segment $KM'$ with side $AC$ is the desired point $P$. Let the side of triangle $ABC$ be $4a$. Then $CM = 3a$ and $AK = 2a$. Point $M'$ is symmetric to point $M$ with respect to line $AC$, so

$CM' = CM = 3a$, and since $\angle ACM' = \angle C = 60^\circ = \angle KAP$, triangle $AKP$ is similar to triangle $CM'P$ by two angles. Therefore,

$AP : PC = AK : CM' = 2 : 3$.

## Answer

$2 : 3$.

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