AMC 8 · 2015 · #2

Grade 6 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

area-trianglesfraction-arithmeticline-symmetry identify-subproblemsarea-difference ↑ Prerequisites: fraction-arithmeticarea-triangles

📏 Medium solution 💡 3 insights 📊 Diagram

📘 View easy version →

Problem

A regular octagon ABCDEFGH has center O, and X is the midpoint of side 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?

Pick an answer.

(A)

$\frac{11}{32}$

(B)

$\frac{3}{8}$

(C)

$\frac{13}{32}$

(D)

$\frac{7}{16}$

(E)

$\frac{15}{32}$

AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

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.

1STEP 1

Draw a spoke from O to every vertex: the octagon splits into 8 congruent triangles of area T, so its total area is 8T.

Area(octagon) = 8T

2STEP 2

Match the shaded boundary to the spokes: it covers three whole slices — △ OBC, △ OCD, △ ODE — plus the small leftover triangle △ OXB.

Shaded = △ OBC + △ OCD + △ ODE + △ OXB

3STEP 3

X is the midpoint of AB, so OX is a median of △ OAB and halves its area: △ OXB has area $\frac{1}{2}$ T.

Area(△ OXB) =

\frac{1}{2}

· Area(△ OAB) =

\frac{1}{2}

4STEP 4

Add the shaded pieces in terms of T: T + T + T + $\frac{1}{2}$ T = $\frac{7}{2}$ T.

Shaded = T + T + T +

\frac{1}{2}

T =

\frac{7}{2}

5STEP 5

Divide shaded by total; the T's cancel: $\frac{\frac{7}{2}T}{8T}$ = $\frac{7}{16}$ , choice (D).

\frac{Shaded}{Octagon}

\frac{\frac{7}{2}T}{8T}

\frac{7}{16}

→ (D)

Answer

\frac{7}{16}

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 △ OAB slice) is only half-shaded. That should give a fraction just under

\frac{4}{8}

\frac{1}{2}

. Our answer

\frac{7}{16}

= 0.4375 is exactly

\frac{1}{16}

below

\frac{1}{2}

, matching the missing half-slice

\frac{1}{2}

\frac{1}{8}

\frac{1}{16}

. Sanity check passes.

💡Key takeaway

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.

More like this