2020 AMC 10B Problems/Problem 16
- The following problem is from both the 2020 AMC 10B #16 and 2020 AMC 12B #14, so both problems redirect to this page.
Contents
Problem
Bela and Jenn play the following game on the closed interval of the real number line, where is a fixed integer greater than . They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in the interval . Thereafter, the player whose turn it is chooses a real number that is more than one unit away from all numbers previously chosen by either player. A player unable to choose such a number loses. Using optimal strategy, which player will win the game?
Solution 1 (Symmetry)
If Bela selects the middle number in the range and then mirror whatever number Jenn selects, then if Jenn can select a number within the range, so can Bela. Jenn will always be the first person to run out of a number to choose, so the answer is
Solution 2 (Educated Guess)
First of all, realize that the parity of likely has no effect on the strategy, since Bela obviously wins when , and it's not hard to find a winning strategy for or (start with ). Similarly, there is no reason the strategy would change when
So we are left with and From here it is best to try out random numbers and try to find the strategy that will let Jenn win, but if you can't find it, realize that it is more likely the answer is since Bela has the first move and thus has more control.
Solution 3 (Simulation)
We can just play the game. We can draw a number line to and assume Bela and Jenn will only play the integer values. From now after playing one round of the game, Bela will win so answers and are eliminated. Now we want to test if Bela will win if is even so we can draw another number line this time up to . Now after playing the game, we can find that Bela won yet again so the answer is
-minor edits by jjbros
Video Solutions
https://youtu.be/3BvJeZU3T-M (for AMC 10) ~IceMatrix
https://youtu.be/0xgTR3UEqbQ (for AMC 12) ~IceMatrix
See Also
2020 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.