Competition · AMC preparation · step 4 of 4
AMC 8 · 2012 · #23
Grade 7 geometry-2dPick an answer.
AMC 8 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #1 (Diagram) is the natural first move for a 2D geometry problem with no figure given — drawing the hexagon reveals a key fact: a regular hexagon is exactly 6 small equilateral triangles joined at the center. That turns the hexagon's area into a sum we can control. Tool #7 (Subproblems) splits the work into two clean pieces: (a) compare the side of one small triangle to the side of the big triangle, and (b) use the side ratio to get the area of one small triangle, then multiply by 6. Tool #9 (Easier Problem) covers the side-to-area scaling: equilateral triangles are all similar, so doubling the side multiplies the area by 2² = 4 — no square-root-of-3 formula needed.
Match the perimeters
Equal perimeters give 3 s_t = 6 s_h, so s_t = 2 s_h — the triangle's side is twice the hexagon's.
Writing one equation from the equal-perimeter condition and solving for s_t is the Grade 6 one-step-equation move.
6.EE.B.7Identify SubproblemsSplit the hexagon into 6 triangles
Cut the hexagon from its center into 6 congruent equilateral triangles of side s_h, so its area is six small triangles.
Decomposing a polygon into triangles to find its area is the Grade 6 "area by composition/decomposition" standard.
6.G.A.1Draw A DiagramCompare the two triangle sizes
Both triangles are similar with side ratio 2, so areas scale by 2² — the big triangle is 4 times a small one.
The "side scales by k → area scales by k²" rule is the Grade 7 scale-drawing fact, used here without needing the √(3)/4s² formula.
The large equilateral triangle covers exactly 4 times the area of one small equilateral triangle, because its side (2s) is twice the small triangle's side (s).
▸ Why?
Both triangles are equilateral, so every angle in each measures 60° and their angles match; triangles with equal angles are similar, and for similar triangles the area grows as the square of the side ratio. The sides are in the ratio 2s : s = 2, so the areas are in the ratio 2² = 4.
Find the small triangle's area
The big triangle's area 4 is 4 × a small one, so each small triangle has area 1.
Solving 4 · A_small = 4 for A_small is a Grade 6 one-step equation.
6.EE.B.7Identify SubproblemsMultiply by 6 for the hexagon
Six small triangles of area 1 fill the hexagon, so its area is 6 — choice (C).
Adding up the 6 congruent pieces of the decomposed hexagon gives its total area directly.
6.G.A.1Draw A DiagramCut the hexagon into 6 small equilateral triangles — the big triangle has the same shape but doubled side, so its area is 4 times one small triangle. That makes each small triangle area 1, and 6 × 1 = 6.
- Match the perimeters
- Split the hexagon into 6 triangles
- Compare the two triangle sizes
- Find the small triangle's area
- Multiply by 6 for the hexagon
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