Competition · AMC preparation · step 4 of 4
AMC 8 · 2015 · #2
Grade 6 geometry-2d
Pick an answer.
AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The figure already provides a diagram, but the key move (Tool #1) is to add to it: draw every spoke from the center O to each vertex. That single addition slices the octagon into 8 congruent triangles, turning an awkward 6-sided shaded region into a sum of standard pieces. Tool #7 (Identify Subproblems) then splits the shaded region XBCDEO into pieces that exactly match those slices — three full central triangles (△ OBC, △ OCD, △ ODE) plus the half-triangle △ OXB. With those subproblems solved, the fraction is just simple addition.
Draw spokes from the center
Draw a spoke from O to every vertex: the octagon splits into 8 congruent triangles of area T, so its total area is 8T.
Partitioning a regular shape into equal pieces around its center is the Grade 3 idea of splitting a shape into equal parts, then naming each part as a unit fraction 1/8 of the whole.
Drawing a spoke from the center O to each vertex cuts the regular octagon into eight triangles of equal area, so the whole octagon's area is exactly eight times one triangle's area T.
▸ Why?
The eight triangles all have the same area, so the octagon is eight copies of one triangle's area rather than a mix of different sizes.
▸ Why?
Each triangle can be laid perfectly onto the next by turning the whole octagon one-eighth of a full turn about its center O, because a regular octagon has all sides and all angles equal and so lands back exactly on itself after that turn.
▸ Why?
That turn is a rigid motion, and a rigid motion sets a shape down onto its image without any stretching, so a triangle and the one it lands on keep the same lengths and angles and therefore the same area.
▸ Why?
The octagon's whole area is the sum of these eight triangles, because the spokes split it into exactly these pieces with no gaps and no overlaps.
Break the shaded region apart
Match the shaded boundary to the spokes: it covers three whole slices — △ OBC, △ OCD, △ ODE — plus the small leftover triangle △ OXB.
Tool #7: turn a weird 6-sided region into a sum of pieces you already know — three standard T-triangles plus one extra.
6.G.A.1Identify SubproblemsFind the half-triangle piece
X is the midpoint of AB, so OX is a median of △ OAB and halves its area: △ OXB has area T.
Cutting one of the eight T-pieces in half gives 1/2 T — a Grade 4 fraction-of-a-quantity calculation.
4.NF.B.4Identify SubproblemsAdd the shaded pieces
Add the shaded pieces in terms of T: T + T + T + T = T.
Three wholes plus a half is 3 1/2 = 7/2 — straightforward mixed-number addition.
4.NF.B.4Identify SubproblemsDivide by the octagon area
Divide shaded by total; the T's cancel: = , choice (D).
Dividing 7/2 by 8 is the Grade 5 interpretation of a fraction as a division: 7/2 ÷ 8 = 7/16.
5.NF.B.3Identify SubproblemsOnce you draw all the spokes from the center, this AMC 8 problem becomes a Grade 5 "count the pie slices" job — three whole slices plus one half, out of eight.
- Draw spokes from the center
- Break the shaded region apart
- Find the half-triangle piece
- Add the shaded pieces
- Divide by the octagon area
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