AMC 10 · 2020 · #6
Easy mode Grade 5The picture below shows a grid of small squares, 5 across and 4 down. Three of the squares are already shaded.
You may shade more squares, but you may not take away any shading. When you are finished, the picture should have two lines of symmetry — lines you could fold the picture along so that the two halves land exactly on top of each other.
What is the smallest number of extra squares you need to shade?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A grid of unit squares $5$ columns wide and $4$ rows tall has $3$ squares shaded. Shade more unit squares so the whole figure ends up with two lines of symmetry, and find the smallest number of extra squares that can do it.
Givens: The figure is a $5$-wide by $4$-tall grid of unit squares ($20$ squares in all); Label each square by (column, row): columns $1$–$5$ left to right, rows $1$–$4$ bottom to top; The three already-shaded squares are (column $2$, row $4$), (column $3$, row $2$), and (column $5$, row $1$); Choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $8$
Unknowns: The least number of additional unit squares that must be shaded
Understand
Restated: A grid of unit squares $5$ columns wide and $4$ rows tall has $3$ squares shaded. Shade more unit squares so the whole figure ends up with two lines of symmetry, and find the smallest number of extra squares that can do it.
Givens: The figure is a $5$-wide by $4$-tall grid of unit squares ($20$ squares in all); Label each square by (column, row): columns $1$–$5$ left to right, rows $1$–$4$ bottom to top; The three already-shaded squares are (column $2$, row $4$), (column $3$, row $2$), and (column $5$, row $1$); Choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $8$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #17 Visualize Spatial Relationships, #14 Extreme Principle, #3 Eliminate Possibilities
Draw the grid and mark the two candidate mirror lines on it (tool #1) — the outline of the figure is a non-square rectangle, so the mirror lines are forced before any shading is chosen. Then use tool #17: mirror each shaded square across each line and see which new squares are dragged in. Tool #14 handles the word 'least' properly — first prove a lower bound (every mirror image is unavoidable), then exhibit a shading that hits exactly that bound, so the minimum is achieved and not merely approached. Tool #3 matches the count to a choice.
Execute — Answer: D
4.G.A.3 Step 1 Pin down where the mirrors can be
- The outline of the whole figure is a rectangle $5$ units wide and $4$ units tall.
- Any line of symmetry of the finished figure has to be a line of symmetry of that outline too.
- A rectangle that is not a square has exactly two: the vertical line down the middle and the horizontal line across the middle.
- So the two lines are decided before we shade anything.
💡 A non-square rectangle can only fold onto itself two ways — side to side and top to bottom.
5.G.A.1 Step 2 Write the two mirrors as rules
- Under the vertical mirror, column $1$ swaps with column $5$ and column $2$ swaps with column $4$, while column $3$ stays put.
- Under the horizontal mirror, row $1$ swaps with row $4$ and row $2$ swaps with row $3$.
- That turns 'mirror this square' into arithmetic on the labels.
💡 Naming each square by its column and row lets you find its mirror by subtracting instead of by eyeballing.
4.G.A.3 Step 3 Mirror the first shaded square
- Take the shaded square at (column $2$, row $4$).
- If the figure is symmetric across both lines, then its vertical mirror, its horizontal mirror, and the mirror of the mirror must all be shaded as well.
- Neither of its center's coordinates sits on a mirror line, so all four squares in this family are different.
💡 A square off both mirror lines has three partners it drags along, never fewer.
4.G.A.3 Step 4 Mirror the second shaded square
- The shaded square at (column $3$, row $2$) sits in the middle column, so the vertical mirror sends it to itself.
- Only the horizontal mirror produces something new.
- This family has just $2$ squares.
💡 Sitting on a mirror line means the square is its own reflection, so it costs half as many partners.
4.G.A.3 Step 5 Mirror the third shaded square
- The shaded square at (column $5$, row $1$) is a corner square, off both mirror lines, so it drags in three more corner squares.
- Listing all three families side by side shows no square is counted twice.
💡 Different starting squares in different columns can never mirror onto each other.
4.OA.A.3 Step 6 Lower bound on the shading
- Every square in all three families is forced to be shaded.
- The three families are disjoint, so the finished figure needs at least $4 + 2 + 4 = 10$ shaded squares.
- Three of those are already shaded, so at least $7$ more are needed.
💡 Count what symmetry forces, then subtract what is already there.
4.G.A.3 Step 7 Check that $7$ is reachable
- Shading exactly those $10$ squares and nothing else really works: the set of $10$ was built by closing the three starting squares under both mirrors, so reflecting it in either line sends the set back onto itself.
- That figure has both lines of symmetry, so $7$ is not just a floor — it is achieved, and it is the answer (D).
💡 A set built by mirroring is automatically symmetric, so the lower bound is also a recipe.
4.G.A.3 The outline of the whole figure is a rectangle $5$ units wide and $4$ units tall 5.G.A.1 Under the vertical mirror, column $1$ swaps with column $5$ and column $2$ swaps 4.G.A.3 Take the shaded square at (column $2$, row $4$). If the figure is symmetric acro 4.G.A.3 The shaded square at (column $3$, row $2$) sits in the middle column, so the ver 4.G.A.3 The shaded square at (column $5$, row $1$) is a corner square, off both mirror l 4.OA.A.3 Every square in all three families is forced to be shaded. The three families ar 4.G.A.3 Shading exactly those $10$ squares and nothing else really works: the set of $10 Review
Reasonableness: Sanity-check the final picture directly. Row $1$ gets columns $1,2,4,5$ shaded and row $4$ gets the same four — those two rows swap under the horizontal mirror and each reads the same left-to-right as right-to-left, so both mirrors are fine there. Rows $2$ and $3$ each get only column $3$, the middle column, which is fixed by the vertical mirror, and the two rows swap under the horizontal mirror. Total shaded is $4 + 1 + 1 + 4 = 10$, so $10 - 3 = 7$ extra squares. Choice (A) $4$ is what you would get by mirroring only once instead of across both lines, and (E) $8$ is what you get by forgetting that the middle-column square needs only one partner.
Alternative: Tool #16 (Change Focus): instead of tracking squares, track the four quarter-blocks the two mirrors cut the grid into, plus the middle column. Fold the grid along both mirrors so all four quarters land on top of each other. Each shaded square lands somewhere in the folded quarter, and a symmetric figure is exactly one that is a full unfolding of some set of folded cells. The two corner-ish squares land on two different folded cells (cost $4$ each when unfolded) and the middle-column square lands on a cell of the fold line (cost $2$ when unfolded), giving $4 + 4 + 2 = 10$ again.
CCSS standards used (min grade 5)
4.G.A.3Recognize a line of symmetry for a two-dimensional figure (Ruling the two mirror lines of the $5 \times 4$ rectangle, mirroring each shaded square, and checking the final figure really is symmetric.)5.G.A.1Use a pair of perpendicular number lines forming a coordinate system (Naming each unit square by its (column, row) pair so reflection becomes the arithmetic rules $c \mapsto 6-c$ and $r \mapsto 5-r$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Adding the three family sizes $4 + 2 + 4 = 10$ and subtracting the $3$ already-shaded squares to get $7$.)
⭐ This AMC 12 problem needs only Grade 4–5 symmetry: each shaded square drags in its mirror images — four of them if it is off both mirror lines, but only two if it sits on one — so $4 + 2 + 4 = 10$ squares must be shaded and $7$ still need shading.
⭐ This AMC 12 problem needs only Grade 4–5 symmetry: each shaded square drags in its mirror images — four of them if it is off both mirror lines, but only two if it sits on one — so $4 + 2 + 4 = 10$ squares must be shaded and $7$ still need shading.
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