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2023 AMC 10B Problem 10

2023 AMC 10B Problem 10AMC 10 Step by Step

Name: Date: September 13, 2026

Paths & Grids · difficulty 3 of 5 · about 4 min · position P1-10

You are playing a game. A $2$ $\times$ $1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3$ $\times$ $3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a square, after which you are told whether that square is covered by the hidden rectangle. What is the minimum number of turns you need to ensure that at least one of your guessed squares is covered by the rectangle?

A)
$3$
B)
$5$
C)
$4$
D)
$8$
E)
$6$

Answer

Key insight

Every domino contains an edge-middle square, so those four squares always work; three squares touch at most 4+3+3 of the 12 dominoes.

Solution

In the worst case every answer is "no" until we hit, so we need a set of squares that meets every possible placement of the rectangle.

There are $12$ placements: $6$ horizontal and $6$ vertical. Each placement pairs two adjacent squares, and adjacent squares are always a corner and an edge-middle square, or an edge-middle square and the center. So every placement contains one of the four edge-middle squares. Guessing those four squares guarantees a hit: $4$ turns suffice.

Three turns cannot suffice. Count how many placements contain a given square: a corner lies in $2$ , an edge-middle square in $3$ , the center in $4$ . Any three squares therefore meet at most $4 + 3 + 3 = 10$ placements, fewer than $12$ . Some placement avoids all three guesses, and the hidden rectangle could be exactly that one, so three "no" answers are always possible.

The answer is $\boxed{\textbf{(C)}\ 4}$ .

Why this works

"Guarantee a hit" questions are hitting-set problems: find a set of guesses meeting every possibility, then prove nothing smaller works. The construction comes from noticing which squares every placement must use; the lower bound comes from counting placements per square, an extremal count that bounds what three guesses can cover.

The trap

Guessing the center plus two corners (or just three squares) and not checking that some domino avoids them all.

Common mistakes

Guessing the center plus two corners (or just three squares) and not checking that some domino avoids them all.
Thinking adaptive play helps: a "no" answer only removes placements, and until a placement is forced the adversary can always say "no."

Techniques

Bound the quantity above/below or estimate to pin it down · Consider the largest/smallest element or boundary case