AMC 10 · 2020 · #14

Grade 5 logic

Archetype Pair-Sum Invariant └ Sub-archetype Forbidden-Complement Selection

symmetry-argumentinvariant-monovariantinterval-arithmetic symmetry-argumentpattern-recognition ↑ Prerequisites: symmetry-argumentparity

📏 Medium solution 💡 2 insights

📘 View easy version →

Problem

Two players take turns placing real numbers on the closed interval from zero to some fixed integer greater than four. The first player may pick anything. After that, each new pick must be more than one unit away from every number already chosen. The player who cannot move loses. Decide who wins with best play.

Pick an answer.

(A)

Bela will always win.

(B)

Jenn will always win.

(C)

Bela will win if and only if $n$ is odd.

(D)

Jenn will win if and only if $n$ is odd.

(E)

Jenn will win if and only if $n>8$.

How to solve

Strategy Draw a Diagram

Tool #1 (Diagram): drawing the segment [0, n] instantly suggests a left-right symmetry around the midpoint n/2. That symmetry is the key — if Bela starts at the center, every move by Jenn has a mirror image still available for Bela. Tool #9 (Easier Problem): test small concrete cases n = 5 and n = 6 on a number line to confirm the mirror strategy works no matter the parity of n. Tool #3 (Eliminate): the answer choices split on parity of n — once two small cases (one odd, one even) both go to Bela, choices (B), (C), (D), (E) all die and (A) remains.

1STEP 1

Define the mirror

Define a mirror map on the interval.

mirror(x) = n - x, x ∈ [0, n]

2STEP 2

Take the centre first

The first player takes the exact centre.

Bela's first pick = n/2; reply to Jenn's x is n - x

3STEP 3

Check the mirror move is always legal

The mirror move is always legal.

|(n - x) - x| = |n - 2x| ＞ 1 since x ≠ n/2

4STEP 4

The first player never gets stuck

If the opponent can move, so can the mirror.

Jenn moves → Bela has the mirror move → Bela never stuck first

5STEP 5

Check small cases

Small cases confirm it.

n = 5: pick 2.5; n = 6: pick 3

6STEP 6

Name the winner

The answer is that the first player always wins.

n = 5 → Bela; n = 6 → Bela → (A)

Answer

Bela will always win.

The mirror strategy is the classic symmetry argument for two-player games on a symmetric board, and it gives the first mover an automatic edge because they can claim the unique fixed point (the center). The answer (A) Bela always wins matches the intuition that moving first plus a symmetry to exploit is a forced win.

💡Key takeaway

This AMC 12 problem only needs Grade 5 number-line symmetry you already know — Bela picks the middle and then mirrors Jenn across that middle. Because the segment is symmetric, every legal move by Jenn has a legal mirror reply for Bela, so Jenn always runs out first. The answer is (A).