Difference between revisions of "2020 AMC 12B Problems/Problem 14"
Argonauts16 (talk | contribs) (→Solution) |
Argonauts16 (talk | contribs) (→Problem 14) |
||
Line 1: | Line 1: | ||
− | ==Problem | + | ==Problem== |
Bela and Jenn play the following game on the closed interval <math>[0, n]</math> of the real number line, where <math>n</math> is a fixed integer greater than <math>4</math>. They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in the interval <math>[0, n]</math>. 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? | Bela and Jenn play the following game on the closed interval <math>[0, n]</math> of the real number line, where <math>n</math> is a fixed integer greater than <math>4</math>. They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in the interval <math>[0, n]</math>. 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? |
Revision as of 20:44, 7 February 2020
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
We can see that if Bela chooses , she splits the line into two halves. After this, she can simply mirror Jenn's moves, and because she now goes after Jenn, Bela will always win. Thus, our answer is
See Also
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by 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.