AMC 10 · 2002 · #17
Grade 8 geometry-2dPick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The triangle sits inside a shape whose exact corners are hard to picture from the labels alone, so the load-bearing move is tool #1 (Draw a Diagram): pin the octagon down in coordinates so every vertex has a known address. The clean way to get those coordinates is tool #17 (Visualize Spatial Relationships) — see the regular octagon as a square with four corners sliced off. Once A, D, and G have coordinates, tool #7 (Identify Subproblems) turns the area into two easy pieces: a horizontal base and a vertical height.
Slice the octagon from a square
Cut four right isosceles corners off a square. Each cut's hypotenuse is an octagon side of 2, so ℓ² + ℓ² = 2² and each leg is √2.
A regular octagon is a square with equal right-triangle corners shaved off, and each shaved side is that triangle's hypotenuse.
A regular octagon is a square with equal right-triangle corners shaved off.
▸ Why?
The octagon is exactly the square minus those four corner pieces, so the areas and lengths add back up.
▸ Why?
Each shaved side is that corner triangle's hypotenuse, so its length comes from the two equal legs.
Put every vertex in coordinates
The square's side is 2 + 2√2, so labelling around from the bottom-left gives A=(√2, 0), D=(2√2+2, √2+2), G=(0, √2+2).
Once the octagon is a square-minus-corners, each vertex address is just built from the side length 2 and the leg √2.
6.G.A.3Draw A DiagramRead the base and height
D and G share the height √2+2, so DG is a flat base of length 2√2+2, and A sits y=0, a drop of √2+2 below it.
Two points at the same height make a flat base, and the third point's vertical distance to it is the height.
6.G.A.3Identify SubproblemsMultiply out the area
Area is half of base times height: (2√2+2)(√2+2) = 8 + 6√2, and half of that is 4 + 3√2 — choice (C).
Half base times height still works even when the lengths carry a √2 — just expand carefully.
6.G.A.1Identify SubproblemsSee the regular octagon as a square with its corners shaved off, drop it into coordinates, then any triangle's area is just half base times height.
- Slice the octagon from a square
- Put every vertex in coordinates
- Read the base and height
- Multiply out the area
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