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2015 AMC 8

Problem 2

Problem 2 · 2015 AMC 8 Medium
Geometry & Measurement area-fractionsymmetry
Figure for AMC 8 2015 Problem 2
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Answer: D — 7/16.
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Hint 1 of 2
Don't try to measure the shaded shape directly. The center O is begging you to slice: draw a spoke from O to each vertex and the octagon falls into 8 identical wedges.
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Hint 2 of 2
Once everything is in equal pieces, the area question becomes a counting question — how many of the 8 wedges (and which half-wedges) are shaded? (A center point in a regular polygon almost always means 'cut into equal slices.')
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Approach: cut the octagon into 8 equal wedges from the center, then count
  1. Draw a spoke from O to every vertex. A regular octagon splits into 8 congruent triangular wedges, so each wedge is exactly 1/8 of the area — now area is just bookkeeping.
  2. The shaded region is three whole wedges (OBC, OCD, ODE) plus ▵OXB. Since X is the midpoint of AB, OX cuts wedge OAB into two equal halves, so ▵OXB is half a wedge.
  3. Shaded = 3½ wedges out of 8 = (7/2)/8 = 7/16.
  4. Why this transfers: a midpoint that splits a triangle off the same base line splits its area in half too (same height, half the base). That's how the half-wedge appears without any computation.
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