AMC 10 · 2017 · #19

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length
equilateral-trianglelaw-of-cosinessimilar-figuressymmetry-argument symmetry-argumentidentify-subproblems ↑ Prerequisites: equilateral-triangle
📏 Long solution 💡 3 insights
Problem
Start with an equilateral triangle ABC. Slide past B along line AB to a point B' with BB' = 3*AB. In the same rotating sense, slide past C along line BC to C' with CC' = 3*BC, and past A along line CA to A' with AA' = 3*CA. Compare the area of the outer triangle A'B'C' to the area of ABC.

Pick an answer.

(A)
9:1
(B)
16:1
(C)
25:1
(D)
36:1
(E)
37:1

AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

Draw the figure and pick a convenient side length, AB = 1. The three extensions are built the same way, so rotating the picture 120 degrees about the center maps it onto itself: the outer triangle A'B'C' must be equilateral too. That collapses the whole problem into one subproblem, finding a single outer side, since area for similar triangles scales as the square of the side. One outer side, say B'C', sits inside a triangle whose two sides and included angle are easy to read off.

1STEP 1

Draw it and use the symmetry

Set AB = 1. A 120-degree turn maps the figure onto itself, so the outer triangle A'B'C' is equilateral — one side settles it.

AB=BC=CA=1, △ A'B'C' is equilateral
2STEP 2

Two sides of triangle B'BC'

In triangle B'BC' at corner B, one side is BB' = 3; the other runs along collinear B, C, C', so BC' = 1 + 3 = 4.

BB'=3, BC'=BC+CC'=1+3=4
3STEP 3

The angle between those two sides

B' sits opposite A across B, so ray BB' reverses ray BA; swinging to BC' gives angle B'BC' = 180 - 60 = 120 degrees.

∠ B'BC' = 180° - 60° = 120°
4STEP 4

Find B'C' with a right triangle

Drop a perpendicular from C' to AB' at D: BD = 2, DC' = sqrt(12), B'D = 5, so Pythagoras gives B'C'² = 25 + 12 = 37, B'C' = sqrt(37).

B'C'² = B'D² + DC'² = 5² + 12 = 37, B'C'=√(37)
5STEP 5

Square the side ratio

Both are equilateral, so similar; area scales as the side ratio squared: (sqrt(37)/1)² = 37, giving area ratio 37 : 1, choice (E).

([△ A'B'C'])/([△ ABC]) = (√(37)/1)² = 37
Answer
37:1
Each outer side stretches well past the original, so the outer triangle should dwarf ABC and a large ratio is expected. The four wrong choices 9, 16, 25, 36 are all perfect squares, which would happen only if the outer side came out a whole number; here the 120-degree corner forces B'C'² = 37, a non-square, so the side sqrt(37) is irrational and 37:1 is the natural fit. A quick law-of-cosines check on the same triangle agrees: 3² + 4² - 2(3)(4)cos120 = 9 + 16 + 12 = 37.
💡Key takeaway

When a shape is grown by a length factor, its area grows by that factor squared, so finding one outer side settles the whole area ratio.

  • Draw it and use the symmetry
  • Two sides of triangle B'BC'
  • The angle between those two sides
  • Find B'C' with a right triangle
  • Square the side ratio

More like this

Same archetype — closest grade level first.