AMC 10 · 2018 · #12

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
coordinate-geometryarea-circlessimilar-figures convert-to-algebra ↑ Prerequisites: coordinate-geometryarea-circles
📏 Medium solution 💡 2 insights
Problem
AB is a diameter of a circle with AB=24, so the circle has center O (the midpoint of AB) and radius 12. A point C moves around the circle (never landing on A or B). For each position of C we mark the centroid of △ ABC — the average of its three vertices. Those centroids sweep out their own closed curve (a circle with two points missing). Find the area enclosed by that curve, rounded to the nearest whole number.

Pick an answer.

(A)
25
(B)
38
(C)
50
(D)
63
(E)
75

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Convert to Algebra

Tool #13 (Convert to Algebra): a moving point is hard to track in words, but if we drop the figure onto coordinates and apply the centroid formula, the answer falls out as plain arithmetic. Tool #1 (Draw a Diagram): putting the center O at the origin with A=(-12,0) and B=(12,0) makes A and B symmetric, so their contribution to the centroid collapses to zero. Tool #5 (Look for a Pattern): the centroid formula reveals that every centroid is exactly its point C shrunk to one-third of the way toward O — a single shrinking rule that turns the big circle into a smaller circle of the same shape, so we never have to track C point by point.

1STEP 1

Drop the figure onto coordinates

Center O at the origin, so A=(-12,0), B=(12,0), and C=(x,y) rides the circle x²+y²=144 — A and B are mirror images.

O=(0,0), A=(-12,0), B=(12,0), C=(x,y) with x²+y²=144
2STEP 2

Apply the centroid formula

The centroid averages the three vertices; A and B's coordinates cancel (-12+12=0, 0+0=0), leaving G=(x/3, y/3).

G=((-12+12+x)/3, (0+0+y)/3)=(x/3, y/3)
3STEP 3

Read the formula as a shrinking rule

Reading G=(x/3,y/3): each C is scaled by 1/3 toward O. That single rule is a dilation about O with factor 1/3, valid for every C.

G=1/3 C (dilation about O with factor 1/3)
4STEP 4

Shrink the whole circle

A dilation maps a circle to a circle, scaling lengths by 1/3, so G traces a circle of radius 4 about O; C=A,B just drops two points.

OG=1/3 OC=1/3· 12 = 4 → centroids form a circle of radius 4
5STEP 5

Take the area of the small circle

The traced curve is a circle of radius 4, so its area is π·4²=16π, rounded to the nearest whole number.

Area=π· 4² = 16π ≈ 50.27 ≈ 50 → (C)
Answer
50
The centroid circle has radius 4, exactly one-third of the original radius 12, so its area should be (1/3)²=1/9 of the big circle's area π· 12² = 144π. Indeed 1/9· 144π = 16π ≈ 50.3, matching. The value 16π rounds to 50, choice (C). The other choices correspond to wrong radii: 25≈π r² needs r≈ 2.8, 75 needs r≈ 4.9 — neither matches the clean factor-1/3 shrink, so they are ruled out.
💡Key takeaway

The centroid is always point C pulled one-third of the way toward the center, so it traces a circle of radius 4 — and π·4²=16π≈ 50.

  • Drop the figure onto coordinates
  • Apply the centroid formula
  • Read the formula as a shrinking rule
  • Shrink the whole circle
  • Take the area of the small circle

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