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AMC 8 · 2005 · #3

Easy mode Grade 5
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Problem

Picture a big square ABCDABCD split into a 4×44\times 4 grid of small squares. A few of the small squares are already colored black.

We want the diagonal line from BB to DD to be a line of symmetry. That means if you fold the picture along BD‾\overline{BD}, every black square should land on another black square.

To make this work, we may need to color a few more small squares black. What is the smallest number of extra small squares we have to color?

Pick an answer.

(A)
1
(B)
2
(C)
3
(D)
4
(E)
5
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Toolkit + CCSS Solution

Understand

Restated: A $4 \times 4$ grid sits inside square $ABCD$, and five small squares are already shaded black. Find the smallest number of additional small squares that must be shaded so the diagonal $\overline{BD}$ becomes a line of symmetry of the black pattern.

Givens: Square $ABCD$ is divided into a $4 \times 4$ grid of small squares; Five small squares are already shaded black (top-left corner, two in the top row near $B$, one on the right edge, and one on the bottom row near $D$); We want $\overline{BD}$ to be a line of symmetry for the shaded pattern; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$

Unknowns: The minimum number of additional small squares that must be shaded black

Understand

Restated: A $4 \times 4$ grid sits inside square $ABCD$, and five small squares are already shaded black. Find the smallest number of additional small squares that must be shaded so the diagonal $\overline{BD}$ becomes a line of symmetry of the black pattern.

Givens: Square $ABCD$ is divided into a $4 \times 4$ grid of small squares; Five small squares are already shaded black (top-left corner, two in the top row near $B$, one on the right edge, and one on the bottom row near $D$); We want $\overline{BD}$ to be a line of symmetry for the shaded pattern; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$

Plan

Primary tool: #11 Find an Invariant

Secondary: #1 Draw a Diagram, #2 Make a Systematic List

A line of symmetry is exactly the statement that the black pattern is invariant under reflection across $\overline{BD}$ — tool #11. The reflection across $\overline{BD}$ swaps each square with its mirror partner, so the unknowns aren't numbers but pairings. Tool #1 (Draw a Diagram) lets us mark every black square right on the grid, and tool #2 (Make a Systematic List) walks through each black square one at a time to record its mirror image. Any mirror image that isn't already black is a square we must add — counting those new squares gives the minimum.

Execute — Answer: D

#1 Draw a Diagram 5.G.A.1 Step 1
  • Label the grid with coordinates so we can describe each small square.
  • Put $D$ at the origin so columns and rows run from $0$ to $3$.
  • Then the diagonal $\overline{BD}$ becomes the line $y = x$, and reflecting a square at column $c$, row $r$ across that line lands it at column $r$, row $c$ — just swap the two numbers.
$$\text{Reflection across } \overline{BD}: \; (c, r) \longleftrightarrow (r, c)$$

💡 Setting up perpendicular number lines for the grid is the Grade 5 "coordinate system" move, and it turns the mirror image into the easy rule "swap the two coordinates."

#2 Make a Systematic List 5.G.A.2 Step 2

List the five already-black squares by their $(\text{column}, \text{row})$ coordinates by reading them off the figure.

$$\text{Black squares} = \{(0,3),\; (1,0),\; (2,3),\; (3,1),\; (3,3)\}$$

💡 Naming each black square with its coordinates is the Grade 5 "plot a point" step — now we can talk about them precisely.

#2 Make a Systematic List 4.G.A.3 Step 3
  • For each black square, write down its mirror image across $\overline{BD}$ by swapping the two coordinates.
  • The square $(3,3)$ sits on the diagonal, so its mirror image is itself.
$$(0,3) \to (3,0), \;\; (1,0) \to (0,1), \;\; (2,3) \to (3,2), \;\; (3,1) \to (1,3), \;\; (3,3) \to (3,3)$$

💡 Symmetry across a line is the Grade 4 "line of symmetry" idea: every point has a partner on the other side, except points right on the line.

#11 Find an Invariant 4.G.A.3 Step 4
  • Check each mirror image against the original list.
  • If it is already black, the symmetry condition is satisfied for that pair.
  • If it is not, that mirror square must be added.
$$(3,0), (0,1), (3,2), (1,3) \notin \text{Black squares}; \;\; (3,3) \in \text{Black squares}$$

💡 The invariant is "every black square's twin is also black." Spot every place that invariant currently fails — those are the squares we have to fix.

#11 Find an Invariant 4.G.A.3 Step 5
  • Count the new squares we must shade.
  • Four mirror images $(3,0), (0,1), (3,2), (1,3)$ are missing, so we must shade exactly those four.
  • Adding them is enough: each new square's mirror image is already on the original list, so no further additions are needed.
$$\text{Minimum squares to add} = 4 \;\Rightarrow\; \textbf{(D)}$$

💡 Once every black square is paired with a black twin, the invariant holds and the diagonal really is a line of symmetry.

[1] #1 5.G.A.1 Label the grid with coordinates so we can describe each small square. Put $D$ at
[2] #2 5.G.A.2 List the five already-black squares by their $(\text{column}, \text{row})$ coord
[3] #2 4.G.A.3 For each black square, write down its mirror image across $\overline{BD}$ by swa
[4] #11 4.G.A.3 Check each mirror image against the original list. If it is already black, the s
[5] #11 4.G.A.3 Count the new squares we must shade. Four mirror images $(3,0), (0,1), (3,2), (1

Review

Reasonableness: Sanity-check by pairing up all $9$ black squares after the fix: $\{(0,3),(3,0)\}$, $\{(1,0),(0,1)\}$, $\{(2,3),(3,2)\}$, $\{(3,1),(1,3)\}$, and the self-mirror $(3,3)$. Every pair is a mirror image across $y=x$, so the diagonal $\overline{BD}$ is indeed a line of symmetry. Could we do it with fewer? No — each of the four original off-diagonal black squares had a missing partner, and a single new square can fix at most one missing partner, so we need at least $4$. That rules out (A), (B), (C), and (E).

Alternative: Tool #1 (Draw a Diagram) on its own: fold the picture along the diagonal $\overline{BD}$ in your head (or on paper). Each of the five black squares lands on the spot where its twin should be. Mark every landing spot that is white — there are exactly four of them: the reflections of the top-left, top-row pair, right-edge, and bottom-row blacks. Coloring those four white spots makes the folded picture match itself, giving answer (D).

CCSS standards used (min grade 5)

  • 4.G.A.3 Recognize a line of symmetry for a two-dimensional figure (Treating $\overline{BD}$ as a line of symmetry: every black square must have a black mirror image, and squares on the line are their own mirror image.)
  • 5.G.A.1 Use a pair of perpendicular number lines forming a coordinate system (Setting up column-row coordinates on the $4 \times 4$ grid so the diagonal becomes $y = x$ and reflection becomes the "swap coordinates" rule.)
  • 5.G.A.2 Represent real-world and mathematical problems by graphing points (Recording each black square as an ordered pair $(c, r)$ and locating its reflected partner $(r, c)$ on the same grid.)

⭐ A line of symmetry just means every black square has a black mirror twin. Pair up the black squares, find the lonely ones, and color in their twins — four were missing, so the answer is (D).

⭐ A line of symmetry just means every black square has a black mirror twin. Pair up the black squares, find the lonely ones, and color in their twins — four were missing, so the answer is (D).

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