2020 AMC 12B Problem 14

Question

Bela and Jenn play the following game on the closed interval

[0, n]

of the real number line, where

n

is a fixed integer greater than

4.

They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in the interval

[0, n].

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 will always win.

Jenn will always win.

Bela will win if and only if

n

is odd.

Jenn will win if and only if

n

is odd.

Jenn will win if and only if

n \gt 8.

Answer: A

Concepts:combinatorial game symmetry

Difficulty rating: 1500

Small hint:

Consider what happens if Bela plays exactly at the midpoint of the interval first

Big hint:

After claiming the center, Bela can mirror each of Jenn’s moves through the midpoint

Solution:

Bela first plays the midpoint

\tfrac{n}{2}.

This choice makes the configuration symmetric about the center of the interval.

Thereafter, whenever Jenn picks a number

x,

Bela responds with its mirror image

n - x.

Since Bela has already chosen

\tfrac n2,

Jenn’s legal move satisfies

\left|x-\tfrac n2\right|>1.

Therefore

|x-(n-x)|=2\left|x-\tfrac n2\right|>2,

so Bela’s response is far enough from Jenn’s new point. Symmetry shows that it is also far enough from every earlier point. Thus Bela always has a move whenever Jenn does, so Jenn is the first to be stuck. Bela always wins.

Thus, the correct answer is A.