AMC 10 · 2020 · #16

Grade 5 arithmetic

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

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

📏 Medium solution 💡 2 insights

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Problem

Bela and Jenn take turns placing real numbers on the closed interval [0, n], where n ＞ 4 is a fixed integer. Bela goes first and may pick anything in [0, n]. After that, each new pick must be more than 1 unit away from every previously chosen number. The first player who can't 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 $n>8$.

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Diagram): drawing the segment [0, n] instantly suggests a left-right symmetry around the midpoint $\frac{n}{2}$ . 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

Draw [0, n] on a number line and mark its midpoint m = $\frac{n}{2}$ ; the segment is symmetric about m, so each point x has a mirror partner n - x.

mirror(x) = n - x, x ∈ [0, n]

2STEP 2

Bela's plan: take the center m = $\frac{n}{2}$ first, then answer each of Jenn's picks x with its mirror n - x.

Bela's first pick =

\frac{n}{2}

; reply to Jenn's x is n - x

3STEP 3

The forbidden zone stays symmetric about m, so Jenn's new x always has its mirror n - x legal — the gap |n - 2x| ＞ 1 holds since x ≠ m.

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

4STEP 4

When Jenn can move, its mirror is legal for Bela, so Jenn is squeezed out first — Bela wins for every n ＞ 4, no parity split.

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

5STEP 5

Small cases: n = 5 → Bela picks 2.5, n = 6 → he picks 3; both times the mirror reply keeps Jenn losing, odd or even.

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

6STEP 6

Since Bela wins at n = 5 (odd) and n = 6 (even), every parity-based choice — (B), (C), (D), (E) — dies, leaving only (A).

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 10 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).

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