Kalinin A.

Anya and Borya, whose speeds are constant but not necessarily the same, simultaneously set out from villages A and B towards each other. If Anya had set out 30 minutes earlier, they would have met 2 km closer to village B. If Borya had set out 30 minutes earlier, the meeting would have taken place closer to village A. By how much?
Let Kolya (half an hour after Borya) and Tolya (half an hour before Borya) also leave from B at the same speed, and from A - Tanya (half an hour before Anya).

First method. Anya meets Tolya, Borya, and Kolya sequentially at equal time intervals. Therefore, the distances between the adjacent meeting points are also equal. It is clear that Anya's meeting with Kolya happened half an hour later, and at the same point where Tanya met Borya, as mentioned in the condition. Thus, the distances between the adjacent meeting points of Anya with the three boys are 2 km, and, in particular, the meeting point of Anya with Tolya is 2 km closer to A than the meeting point of Anya with Borya.

Second method. Consider the graphs of the movements of Anya, Borya, Tanya, and Tolya. These are two pairs of parallel lines, so the points of their intersections are the vertices of a parallelogram. Two opposite vertices have the same ordinate (the meeting point of Anya with Borya coincides with the meeting point of Tanya with Tolya), which means that this point is the arithmetic mean of the ordinates of the other two vertices. Therefore, the other two meeting points lie on opposite sides but at the same distance of 2 km from this point.

## Answer

2 km.

## [ Rectangles and Squares. Properties and Criteria ] Problem 65570 Topics: [ Ratios of Linear Elements of Similar Triangles ] [ Criteria for Equality of Right Triangles ] Difficulty: $3+$ Grades: $8,9,10,11$ Author: Akoyan A.V.  Given a square $A B C D$, $M$ and $N$ are the midpoints of sides $B C$ and $A D$. On the extension of diagonal $A C$ beyond point $A$, point $K$ is taken. Segment $K M$ intersects side $A B$ at point $L$. Prove that angles $K N A$ and $L N A$ are equal.

## Solution

Let line $A B$ intersect segment $K N$ at point $T$. Note that segment $K C$ intersects segment $M N$ at its midpoint. Since segment $L T$ is parallel to $M N$, it cuts off a similar triangle $L K T$ from triangle $M K N$, and therefore $K C$ intersects $L T$ at its midpoint as well. Consequently, right triangles $A N T$ and $A N L$ are equal by two legs. Therefore, $\angle L N A = \angle T N A = \angle K N A$.