2020 AMC 10B Problem 16

P16-20~3 minPrint

Games & Processes Exploit symmetry to reduce work or pair up objects

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?

Show solution without answering

Key insight

Bela takes the exact middle first, then answers each of Jenn's numbers with its mirror image across the middle. That reply is always allowed, so Jenn gets stuck first.

Solution

Try an example first. Let $n = 10$ . Bela picks $5$ , the middle. Say Jenn picks $2$ . Bela answers with $8$ , which is $2$ reflected across $5$ (since $10 - 2 = 8$ ). Say Jenn picks $9.5$ . Bela answers with $0.5$ . Whatever Jenn does on one side, Bela copies on the other side. This suggests a strategy.

Bela's strategy.

On his first turn, Bela picks the middle number $\tfrac n2$ .
Whenever Jenn picks a number $x$ , Bela picks its mirror image $n - x$ . It is just as far from the middle as $x$ , but on the other side.

Why Bela's answer is always allowed. After each of Bela's turns the chosen numbers are balanced: for every chosen number on one side of the middle, its mirror image has been chosen on the other side.

Jenn's $x$ is more than $1$ away from every chosen number. Reflecting the whole picture across the middle, $n - x$ is more than $1$ away from the mirror images of those numbers, and those mirror images are exactly the same set of chosen numbers. So $n - x$ is more than $1$ away from every number chosen before Jenn's turn.
It is also more than $1$ away from Jenn's own new number $x$ . Jenn's $x$ must be more than $1$ away from the middle $\tfrac n2$ , and $x$ and $n - x$ are on opposite sides of the middle at equal distances. So they are more than $2$ apart.

Why Bela wins. Chosen numbers are more than $1$ apart, so only finitely many fit in $[0, n]$ and the game must end. Bela can always answer Jenn's move, so Bela is never the one who gets stuck. The player who first cannot move must be Jenn. None of this depends on whether $n$ is odd or even.

The answer is $\boxed{\textbf{(A)}\ \text{Bela will always win.}}$

Why this works

When a game board has a center, the first player can often win by taking the center and then copying every opponent move on the opposite side. The proof has two checks, and both matter: the copied move is allowed, and it does not clash with the move the opponent just made. The answer choices about odd and even $n$ are there to distract you.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Bela starts by choosing the midpoint $\tfrac n2$ . From then on, whenever Jenn chooses a number $x$ , Bela answers with its mirror image $n - x$ .

Why this is always legal: the set of chosen numbers is symmetric about $\tfrac n2$ after each of Bela's turns. If Jenn's $x$ is more than $1$ from every chosen number, then $n - x$ is more than $1$ from every chosen number as well (distances are preserved by reflection). And $n - x \ne x$ , since $x = \tfrac n2$ is already taken and any legal $x$ is more than $1$ away from it; also $n - x$ is not within $1$ of $x$ , because $|x - (n-x)| = 2|x - \tfrac n2| > 2$ .

So Bela always has a move after Jenn does. The game ends (choices are spaced more than $1$ apart on a bounded interval, so only finitely many fit), and the player who first cannot move is necessarily Jenn.

The answer is $\boxed{\textbf{(A)}\ \text{Bela will always win.}}$

Review in the AoPS books

Intermediate Counting & Probability, Chapter 13: Events With States (§13.4)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)
Introduction to Counting & Probability, Chapter 3: Correcting for Overcounting (§3.4)