AMC 10 · 2003 · #16

Grade 8 geometry-2d
Archetype Geometric Probability by Area └ Sub-archetype Area-Proportional Region Probability
area-trianglesgeometric-probabilityequilateral-triangle symmetry-argumentcomplementary-counting ↑ Prerequisites: area-triangles
📏 Long solution 💡 2 insights
Problem
A point P is dropped at random inside an equilateral triangle ABC, with every spot equally likely. Joining P to the three vertices cuts the triangle into three smaller triangles. Find the probability that the triangle on side AB has a bigger area than both of the other two.

Pick an answer.

(A)
$\frac{1}{6}$
(B)
$\frac{1}{4}$
(C)
$\frac{1}{3}$
(D)
$\frac{1}{2}$
(E)
$\frac{2}{3}$
How to solve
Strategy Change Focus / Count the Complement

Measuring the winning region head-on looks like a chore. Tool #16 (Change Focus) shifts the question twice and both shifts do real work. First shift: the three triangles have equal bases, so comparing areas is the same as comparing how far P is from each side — the question becomes about distances, not areas. Second shift: instead of measuring the winning region, notice that the three winning regions are copies of each other, so the answer follows from the fact that they fill the triangle. Tool #1 (Draw a Diagram) marks the boundaries between the three regions. Tool #17 (Visualize Spatial Relationships) supplies the 120° turn that carries one region exactly onto the next — this is what makes "copies of each other" a proof and not a feeling. Tool #7 (Identify Subproblems) separates the two jobs: describe the regions, then compare their sizes.

1STEP 1

Equal bases turn area into distance

All three bases are equal, so the common factor cancels and the biggest triangle is the one P is farthest from.

[ABP]=1/2s d(P,AB), [BCP]=1/2s d(P,BC), [ACP]=1/2s d(P,CA)
2STEP 2

Name the three winning regions

Name the three winning regions; ties lie on lines, which carry zero probability.

P(△ ABP biggest)=[W_AB]/[ABC]
3STEP 3

Draw the boundaries and see the zone

The angle bisectors cut off a corner region, which is exactly the winning zone for that side.

W_AB=quadrilateral O–M_BC–C–M_CA
4STEP 4

A third of a turn swaps the zones

A one-third turn about the centre carries each zone onto the next, so they are congruent.

rotate 120°: W_AB↦ W_BC↦ W_CA↦ W_AB
5STEP 5

Three equal shares of the whole

Three equal zones fill the triangle, so the probability is 1/3, choice (C).

3[W_AB]=[ABC] → P=[W_AB]/[ABC]=1/3 → (C)
Answer
1/3
Measure the zone directly with coordinates as a check. Put A=(0,0), B=(1,0), C=(1/2,√(3)/2). The center is O=(1/2,√(3)/6), the midpoint of BC is (3/4,√(3)/4), and the midpoint of CA is (1/4,√(3)/4). The shoelace formula on that quadrilateral gives area √(3)/12, while [ABC]=√(3)/4, and (√(3)/12)/(√(3)/4)=1/3 — matching. The size is also visually sensible: the corner zone is a noticeable chunk, clearly more than the sliver 1/6 would suggest and clearly less than half the triangle, which rules out (A), (D), and (E).
💡Key takeaway

Equal bases mean the biggest triangle is simply the one whose side P is farthest from, and a 120° turn shuffles the three 'farthest from' zones into each other, so each zone must be exactly one third of the triangle.

  • Equal bases turn area into distance
  • Name the three winning regions
  • Draw the boundaries and see the zone
  • A third of a turn swaps the zones
  • Three equal shares of the whole

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