AMC 10 · 2013 · #16

Grade 8 geometry-2d

Archetype Arithmetic Expression Decomposition └ Sub-archetype Multi-Segment Path or Count Sum

angle-sum-polygonisosceles-triangletrigonometric-ratiosperimeter identify-subproblemsinvariant-monovariantextreme-principle ↑ Prerequisites: angle-sum-polygonisosceles-triangle

📏 Long solution 💡 3 insights

Problem

An equal-angled pentagon of fixed total side length is extended into a star. Find the spread of the star's outline length.

Pick an answer.

(A)

(B)

$\frac{1}{2}$

(C)

$\frac{\sqrt{5}-1}{2}$

(D)

$\frac{\sqrt{5}+1}{2}$

(E)

$\sqrt{5}$

How to solve

Strategy Identify Subproblems

The star is not one complicated object; it is five triangular points, one standing on each side of the pentagon. Splitting it that way is the whole game. Equiangular fixes every angle in sight, so all five points have identical angles, and identical angles force identical shape. Then each point's outline length is a fixed multiple of the side it stands on, and adding the five pieces turns the total into that multiple times the pentagon's perimeter. Name the five side lengths so the cancellation is visible rather than guessed. Two things still need proof and not assertion: that the star exists at all for every such pentagon, and that the family of pentagons really contains more than one shape, since otherwise 'maximum minus minimum' would be an empty question.

1STEP 1

Fix every angle in the picture

Equal angles fix every angle in the picture.

((5-2) × 180°)/5 = 108°, 180° - 108° = 72°

2STEP 2

Show the star always exists

The star always exists.

72° + 72° = 144° ＜ 180° → the two extensions meet

3STEP 3

All five points are one shape

All five points are the same shape.

180° - 72° - 72° = 36°; each point is 72°-72°-36° → edge = k · base

4STEP 4

Add the ten outline pieces

So the outline is a fixed multiple of the total.

s = Σ_i=1⁵ 2k x_i = 2k (x₁+x₂+x₃+x₄+x₅) = 2k · 1 = 2k

5STEP 5

Push the shape to its extremes

Even extreme shapes give the same length.

(x₁,…,x₅) = (3/10, (7-√(5))/40, (7+√(5))/40, (7+√(5))/40, (7-√(5))/40) → s = 2k again

6STEP 6

A constant has zero spread

A constant has spread 0, choice (A).

s_max - s_min = 2k - 2k = 0

Answer

Pin the constant to be sure it is a real number and not a fiction. In a 72°-72°-36° triangle, bisecting a base angle cuts off a smaller triangle with the same three angles, and comparing the two similar triangles gives the ratio k = (1+√(5))/2, the golden ratio. So s = 2k = 1 + √(5) ≈ 3.236 for every equiangular convex pentagon of perimeter 1. Checking directly: the regular pentagon gives 1+√(5), and so does the lopsided example from Step 5, computed independently. The answer choices confirm the trap: 1+√(5) itself is not offered, but (D) (√(5)+1)/2 = k and (E) √(5) sit right next to it, waiting for anyone who computes the star's perimeter and then forgets that the question asked for its range. Since the perimeter never varies, that range is a single point and the difference is 0. Worth noting how little this needs: the whole argument runs on the pentagon angle sum and AA similarity, so a student who has finished grade-8 geometry can close it. The exact value of k is a bonus check, not a requirement.

💡Key takeaway

All five star points have the same angles, so each point's edges are the same fixed multiple of the side it stands on, and the star's perimeter can only depend on the pentagon's perimeter.

Fix every angle in the picture
Show the star always exists
All five points are one shape
Add the ten outline pieces
Push the shape to its extremes
A constant has zero spread