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?
Try it yourself first — the explanation is most useful after you’ve attempted it.
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Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: The playing field is the interval $[0, n]$ with $n$ a fixed integer and $n > 4$; Bela moves first; players alternate; Every new pick must be at distance $> 1$ from every previously chosen number; A player who cannot pick a legal number loses; Choices: (A) Bela always wins, (B) Jenn always wins, (C) Bela iff $n$ odd, (D) Jenn iff $n$ odd, (E) Jenn iff $n > 8$
Unknowns: Which player has a forced win, and under what condition on $n$
Understand
Restated: 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.
Givens: The playing field is the interval $[0, n]$ with $n$ a fixed integer and $n > 4$; Bela moves first; players alternate; Every new pick must be at distance $> 1$ from every previously chosen number; A player who cannot pick a legal number loses; Choices: (A) Bela always wins, (B) Jenn always wins, (C) Bela iff $n$ odd, (D) Jenn iff $n$ odd, (E) Jenn iff $n > 8$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #9 Solve an Easier Related Problem, #3 Eliminate Possibilities
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.
Execute — Answer: A
#1 Draw a Diagram 4.G.A.3Step 1
Draw the segment $[0, n]$ on a number line and mark its midpoint $m = n/2$.
The interval is left-right symmetric about $m$: a point $x$ in $[0, n]$ has a mirror partner $n - x$, also in $[0, n]$, at the same distance from $m$.
$$\text{mirror}(x) = n - x,\quad x \in [0, n]$$
💡 A segment has a line of symmetry through its middle — every point has a partner.
#1 Draw a Diagram 4.G.A.3Step 2
Bela's strategy: on turn 1 he picks $m = n/2$ (the exact center).
After that, whenever Jenn picks some $x$, Bela picks its mirror $n - x$.
$$\text{Bela's first pick} = n/2;\quad \text{reply to Jenn's } x \text{ is } n - x$$
💡 Start at the symmetry center, then copy the opponent across the middle.
#1 Draw a Diagram 5.G.A.1Step 3
Why the mirror move is always legal for Bela.
The set of forbidden points (within distance $1$ of any prior pick) is itself symmetric about $m$ after every Bela move: the center pick is its own mirror, and each Jenn pick $x$ is paired with Bela's previous reply $n - x$.
So if Jenn just placed $x$ legally, the mirror $n - x$ is at least $1$ unit from every earlier pick (because the earlier picks form a symmetric set) and at distance $|n - x - x| = |n - 2x|$ from $x$ — and this distance is $> 1$ as long as $x \ne m$, which is exactly the case since the center $m$ was Bela's first pick and is already forbidden to Jenn.
$$|(n - x) - x| = |n - 2x| > 1 \text{ since } x \ne n/2$$
💡 Symmetric forbidden zone in $\Rightarrow$ symmetric forbidden zone out — Bela's mirror move stays legal.
#1 Draw a Diagram 4.G.A.3Step 4
Conclusion: whenever Jenn has a legal move, Bela has its mirror as his legal move.
So Bela is never the first to run out — Jenn must lose first.
Bela wins for every $n > 4$, no parity condition.
$$\text{Jenn moves} \Rightarrow \text{Bela has the mirror move} \Rightarrow \text{Bela never stuck first}$$
💡 If you can always copy, you never run out before your opponent.
#9 Solve an Easier Related Problem 5.G.A.1Step 5
Sanity-check the strategy on small cases with Tool #9 (Easier Problem).
For $n = 5$: Bela picks $2.5$.
The remaining legal zone is $[0, 1.5) \cup (3.5, 5]$.
Whatever Jenn picks there, $5 - x$ is in the other piece and stays legal until Jenn is squeezed out first.
💡 One odd win and one even win knock out every parity-dependent choice.
[1]
#1 4.G.A.3Draw the segment $[0, n]$ on a number line and mark its midpoint $m = n/2$. The
[2]
#1 4.G.A.3Bela's strategy: on turn 1 he picks $m = n/2$ (the exact center). After that, wh
[3]
#1 5.G.A.1Why the mirror move is always legal for Bela. The set of forbidden points (withi
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#1 4.G.A.3Conclusion: whenever Jenn has a legal move, Bela has its mirror as his legal mov
[5]
#9 5.G.A.1Sanity-check the strategy on small cases with Tool #9 (Easier Problem). For $n =
[6]
#3 2.OA.C.3Eliminate the other choices (Tool #3). Since Bela wins for $n = 5$ (odd) and $n
Review
Reasonableness: 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.
Alternative: Tool #9 (Easier Problem) alone: simulate the game with only integer picks on $n = 5, 6, 7, 8$ using a number line (Tool #1). In every case Bela wins by starting at the middle integer (or near it) and mirroring. Confirming four small cases with no parity exception again points to choice (A).
CCSS standards used (min grade 5)
2.OA.C.3 Determine whether a group of objects has an odd or even number (Checking both an odd and an even value of $n$ to rule out parity-conditional choices.)
4.G.A.3 Recognize a line of symmetry for a two-dimensional figure (Identifying the segment $[0, n]$ as symmetric about its midpoint $n/2$ and building Bela's mirror strategy.)
5.G.A.1 Use a pair of perpendicular number lines forming a coordinate system (Reasoning about distances on the number line $[0, n]$ to verify the mirror move stays more than $1$ unit from every prior pick.)
⭐ 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 $\textbf{(A)}$.
⭐ 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 $\textbf{(A)}$.
