AMC 10 · 2020 · #6
Grade 5 geometry-2d
Pick an answer.
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.
Pin down the mirrors
The grid size fixes both mirrors.
A non-square rectangle can only fold onto itself two ways — side to side and top to bottom.
A rectangle that is not a square can only fold onto itself two ways, side to side and top to bottom.
▸ Why?
A fold line meets what it folds square on and cuts it exactly in half, which pins the two mirrors.
▸ Why?
A reflection moves the rectangle without stretching, so a longer side can never land on a shorter one.
Write the mirrors as rules
Write both as coordinate rules.
Naming each square by its column and row lets you find its mirror by subtracting instead of by eyeballing.
5.G.A.1Visualize Spatial RelationshipsMirror the first square
The first square drags in four.
A square off both mirror lines has three partners it drags along, never fewer.
4.G.A.3Visualize Spatial RelationshipsMirror the second square
A square on an axis makes only two.
Sitting on a mirror line means the square is its own reflection, so it costs half as many partners.
4.G.A.3Visualize Spatial RelationshipsMirror the third square
The third also drags in four.
Different starting squares in different columns can never mirror onto each other.
4.G.A.3Draw A DiagramGet the lower bound
Removing the three already shaded gives the bound.
Count what symmetry forces, then subtract what is already there.
4.OA.A.3Extreme PrincipleCheck it is reachable
A real arrangement exists, so the answer is 7.
A set built by mirroring is automatically symmetric, so the lower bound is also a recipe.
4.G.A.3Extreme PrincipleThis 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.
- Pin down where the mirrors can be
- Write the two mirrors as rules
- Mirror the first shaded square
- Mirror the second shaded square
- Mirror the third shaded square
- Lower bound on the shading
- Check that 7 is reachable
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