AMC 10 · 2013 · #11

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

equilateral-triangleperimetersystems-of-equations convert-to-algebraidentify-subproblems ↑ Prerequisites: equilateral-triangle

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Two parallel cuts split an equilateral triangle into three pieces of equal perimeter. Find the total length of the cuts.

Pick an answer.

(A)

(B)

$\dfrac{3}{2}$

(C)

$\dfrac{21}{13}$

(D)

$\dfrac{13}{8}$

(E)

$\dfrac{5}{3}$

How to solve

Strategy Introduce a Variable

The entire picture is pinned down by two numbers: how far down the side the first cut sits, and how far down the second sits. Name those two lengths and every side of every piece becomes a short expression in them, so 'all three perimeters are equal' turns into two linear equations in two unknowns. One thing needs care. Solving the equations only shows what the lengths would have to be if such a picture exists; it does not show that one does. So the found values get put back into the picture at the end, to confirm the cuts land in the required order and that the three perimeters really do come out equal.

1STEP 1

Each cut makes a small equilateral triangle

Each cut makes a smaller equilateral triangle.

∠ ADE = ∠ ABC = 60°, ∠ AED = ∠ ACB = 60° → DE = AD = AE

2STEP 2

Name the two cut depths

Two letters describe everything.

a = AD = AE = DE, b = AF = AG = FG, DE + FG = a + b

3STEP 3

Write the three perimeters

All three perimeters write out cleanly.

P_ADE = 3a, P_DFGE = a + b + 2(b-a) = 3b - a, P_FBCG = b + 1 + 2(1-b) = 3 - b

4STEP 4

Turn 'all equal' into two equations

Equal perimeters give two equations.

4a = 3b, b = 3 - 3a → 4a = 9 - 9a → a = 9/13, b = 12/13

5STEP 5

Check the picture really exists

The picture really exists.

0 ＜ 9/13 ＜ 12/13 ＜ 1, 3a = 3b - a = 3 - b = 27/13

6STEP 6

Add the cuts, then recount a different way

The cuts total 21/13, choice (C).

3·27/13 = 3 + 2(DE+FG) → 2(DE+FG) = 81/13 - 3 = 42/13 → DE + FG = 21/13

Answer

21/13

The two cut lengths came out as 9/13 and 12/13. Both lie strictly between 0 and 1 and in the order the labelled picture demands, and all three pieces share the perimeter 27/13, so the configuration in the problem really does exist and is unique. The size is sensible: each cut is shorter than BC = 1, so DE + FG ＜ 2, and 21/13 ≈ 1.615 sits in that range while still exceeding 1. The five choices are packed close together (1, 1.5, ≈ 1.615, 1.625, ≈ 1.667), so a rough decimal cannot separate them; the exact arithmetic is what settles it, and the denominator 13 born from 13a = 9 appears in only one choice.

💡Key takeaway

A cut parallel to the base slices off a smaller copy of the same triangle, so naming how far down each cut sits turns 'same perimeter' into two easy equations.

Each cut makes a small equilateral triangle
Name the two cut depths
Write the three perimeters
Turn 'all equal' into two equations
Check the picture really exists
Add the cuts, then recount a different way