AMC 8 · 2015 · #2
Grade 6 geometry-2dProblem
Point O is the center of the regular octagon ABCDEFGH, and X is the midpoint of the side AB. What fraction of the area of the octagon is shaded?
Pick an answer.
AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A regular octagon $ABCDEFGH$ has center $O$, and $X$ is the midpoint of side $\overline{AB}$. The shaded region is the polygon $XBCDEO$ (it walks from $X$ along the octagon's boundary through $B, C, D, E$, then in to the center $O$, then back out to $X$). What fraction of the octagon's area is shaded?
Givens: Octagon $ABCDEFGH$ is regular (all sides and angles equal); $O$ is the center of the octagon; $X$ is the midpoint of side $\overline{AB}$; The shaded region is bounded by $X \to B \to C \to D \to E \to O \to X$; Answer choices: (A) $\tfrac{11}{32}$, (B) $\tfrac{3}{8}$, (C) $\tfrac{13}{32}$, (D) $\tfrac{7}{16}$, (E) $\tfrac{15}{32}$
Unknowns: The shaded area as a fraction of the whole octagon's area
Understand
Restated: A regular octagon $ABCDEFGH$ has center $O$, and $X$ is the midpoint of side $\overline{AB}$. The shaded region is the polygon $XBCDEO$ (it walks from $X$ along the octagon's boundary through $B, C, D, E$, then in to the center $O$, then back out to $X$). What fraction of the octagon's area is shaded?
Givens: Octagon $ABCDEFGH$ is regular (all sides and angles equal); $O$ is the center of the octagon; $X$ is the midpoint of side $\overline{AB}$; The shaded region is bounded by $X \to B \to C \to D \to E \to O \to X$; Answer choices: (A) $\tfrac{11}{32}$, (B) $\tfrac{3}{8}$, (C) $\tfrac{13}{32}$, (D) $\tfrac{7}{16}$, (E) $\tfrac{15}{32}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
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 ($\triangle OBC$, $\triangle OCD$, $\triangle ODE$) plus the half-triangle $\triangle OXB$. With those subproblems solved, the fraction is just simple addition.
Execute — Answer: D
3.G.A.2 Step 1 - Add spokes to the figure: draw segments from the center $O$ to each of the eight vertices $A, B, C, D, E, F, G, H$.
- Because the octagon is regular, these spokes cut it into $8$ congruent isosceles triangles ($\triangle OAB$, $\triangle OBC$, $\triangle OCD$, $\ldots$, $\triangle OHA$).
- Call the area of one such triangle $T$, so the whole octagon has area $8T$.
💡 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 $\tfrac{1}{8}$ of the whole.
6.G.A.1 Step 2 - Trace the shaded boundary $X \to B \to C \to D \to E \to O \to X$ and break it into pieces that match the spokes.
- Going from $B$ around to $E$ and back to $O$ covers exactly three central triangles: $\triangle OBC$, $\triangle OCD$, $\triangle ODE$.
- The only leftover is the little triangle $\triangle OXB$, which sits inside $\triangle OAB$.
💡 Tool #7: turn a weird $6$-sided region into a sum of pieces you already know — three standard $T$-triangles plus one extra.
4.NF.B.4 Step 3 - Find the area of $\triangle OXB$.
- Since $X$ is the midpoint of $\overline{AB}$, the segment $\overline{OX}$ is a median of $\triangle OAB$ from vertex $O$ to the opposite side.
- A median splits a triangle into two pieces of equal area, so $\triangle OXB$ is exactly half of $\triangle OAB$.
💡 Cutting one of the eight $T$-pieces in half gives $\tfrac{1}{2} T$ — a Grade 4 fraction-of-a-quantity calculation.
4.NF.B.4 Step 4 Add up the shaded pieces in terms of $T$.
💡 Three wholes plus a half is $3\tfrac{1}{2} = \tfrac{7}{2}$ — straightforward mixed-number addition.
5.NF.B.3 Step 5 - Divide the shaded area by the total octagon area.
- The $T$'s cancel because the answer is a pure ratio, leaving a fraction of small whole numbers.
💡 Dividing $\tfrac{7}{2}$ by $8$ is the Grade 5 interpretation of a fraction as a division: $\tfrac{7}{2} \div 8 = \tfrac{7}{16}$.
3.G.A.2 Add spokes to the figure: draw segments from the center $O$ to each of the eight 6.G.A.1 Trace the shaded boundary $X \to B \to C \to D \to E \to O \to X$ and break it i 4.NF.B.4 Find the area of $\triangle OXB$. Since $X$ is the midpoint of $\overline{AB}$, 4.NF.B.4 Add up the shaded pieces in terms of $T$. 5.NF.B.3 Divide the shaded area by the total octagon area. The $T$'s cancel because the a Review
Reasonableness: The shaded region covers a bit less than half the octagon — visually it spans $4$ of the $8$ pie slices, but one of them (the $\triangle OAB$ slice) is only half-shaded. That should give a fraction just under $\tfrac{4}{8} = \tfrac{1}{2}$. Our answer $\tfrac{7}{16} = 0.4375$ is exactly $\tfrac{1}{16}$ below $\tfrac{1}{2}$, matching the missing half-slice $\tfrac{1}{2} \cdot \tfrac{1}{8} = \tfrac{1}{16}$. Sanity check passes.
Alternative: Tool #3 (Eliminate Possibilities): each answer choice has denominator $16$ or $32$. Since the natural decomposition gives pieces of size $\tfrac{1}{8}$ (full slices) and $\tfrac{1}{16}$ (half slices), the answer must be a sum of sixteenths: only (D) $\tfrac{7}{16}$ fits without forcing a thirty-second. The other choices ($\tfrac{11}{32}$, $\tfrac{13}{32}$, $\tfrac{15}{32}$) cannot arise from this clean decomposition, and $\tfrac{3}{8} = \tfrac{6}{16}$ would mean exactly three slices, ignoring the half-slice $\triangle OXB$.
CCSS standards used (min grade 6)
3.G.A.2Partition shapes into parts with equal areas; express the area of each part as a unit fraction of the whole (Slicing the regular octagon into $8$ congruent triangles around the center, so each triangle is $\tfrac{1}{8}$ of the whole octagon.)4.NF.B.4Multiply a fraction by a whole number; apply to solve word problems (Computing $\tfrac{1}{2}$ of one central triangle to find $\text{Area}(\triangle OXB) = \tfrac{1}{2} T$, then summing $T + T + T + \tfrac{1}{2} T = \tfrac{7}{2} T$.)5.NF.B.3Interpret a fraction as division of the numerator by the denominator (Dividing the shaded area $\tfrac{7}{2} T$ by the total area $8T$ to get the ratio $\tfrac{7}{16}$.)6.G.A.1Find the area of polygons by composing into rectangles or decomposing into triangles (Decomposing the shaded region $XBCDEO$ into three full central triangles plus the half-triangle $\triangle OXB$.)
⭐ Once 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.
⭐ Once 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.
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