1. A mathematician left point A for point B. After some time, a physicist also left point A for point B. Catching up with the mathematician after 20 km, the physicist, without stopping, continued to point B and turned back. They met again 20 km from B. Then each continued in their respective directions. Upon reaching points A and B respectively, they turned and walked towards each other again. How many kilometers from point A will they meet for the third time, if the distance between points A and B is 100 km?
# Answer: 45.

Solution. From the first meeting to the second meeting, the mathematician walked a total of 100 - 20 - 20 = 60 km, while the physicist walked -100 - 20 + 20 = 100 km. It is clear from this that the ratio of their speeds is 6:10 or 3:5. From the second to the third meeting, they will walk together 100 + 100 = 200 km, during which the mathematician will walk \(\frac{3}{8}\), and the physicist \(\frac{5}{8}\) of this distance, meaning the mathematician will walk 75 km. Thus, he will first walk the remaining 20 km to point B, and then, turning back, he will walk another 75 - 20 = 55 km. Therefore, he will be 100 - 55 = 45 km from point A.

## Criteria.

If the solution is incorrect - 0 points.

If the system of equations is correctly set up - 3 points.

If there is a computational error - 5 points.

If the solution is correct - 7 points.