AMC 10 · 2015 · #22

Grade 8 arithmetic
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
regular-pentagongolden-ratioisosceles-trianglesimilar-triangles symmetry-argument ↑ Prerequisites: isosceles-triangleregular-pentagon
📏 Long solution 💡 4 insights 📊 Diagram
Problem
Regular pentagon ABCDE has all five diagonals drawn, creating a five-pointed star and a small regular pentagon FGHIJ in the middle. Given AG = 1, find FG + JH + CD.

Pick an answer.

(A)
3
(B)
$12-4\sqrt5$
(C)
$\dfrac{5+2\sqrt5}{3}$
(D)
$1+\sqrt5$
(E)
$\dfrac{11+11\sqrt5}{10}$

AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

The whole problem lives in one picture, so Tool #1 (Draw a Diagram) drives it: mark which points sit on which diagonal. The key sighting is that FG, JH, and CD are the bases of three triangles AFG, AJH, ADC that all share the 36° corner at A — they are scaled copies of one '36-72-72' golden triangle. Tool #4 (Introduce a Variable) pins down the one ratio every golden triangle obeys, base = leg · √5−12\frac{√5-1}{2}, which is where √5 enters. Tool #7 (Identify Subproblems) then turns the single sum into three short length computations, and Tool #5 (Look for a Pattern) climbs from the smallest triangle to the largest by adding one inner-pentagon side at a time.

1STEP 1

Spot three golden triangles

The two diagonals from A split its 108° corner into three 36° wedges, so AFG, AJH, ADC are three nested 36-72-72 golden triangles.

∠ FAG = ∠ JAH = ∠ DAC = 36° → △ AFG ∼ △ AJH ∼ △ ADC
2STEP 2

Find the golden base-to-leg ratio

Bisecting a 72° base angle carves out a smaller copy, so b²+b-1=0 and base = leg · √5−12\frac{√5-1}{2} in every golden triangle.

b² + b - 1 = 0 → b = √5−12\frac{√5-1}{2} (b ＞ 0)
3STEP 3

Smallest triangle from AG = 1

AG = 1 is the leg of the smallest triangle AFG, so its base is FG = √5−12\frac{√5-1}{2}.

FG = √5−12\frac{√5-1}{2}
4STEP 4

Climb to the middle triangle

GH = FG = √5−12\frac{√5-1}{2}, so the middle leg is AH = 1 + √5−12\frac{√5-1}{2} = 1+√52\frac{1+√5}{2} and its base is JH = 1.

AH = 1 + √5−12\frac{√5-1}{2} = 1+√52\frac{1+√5}{2}, JH = 1+√52\frac{1+√5}{2} · √5−12\frac{√5-1}{2} = 1
5STEP 5

Climb to the outer triangle

By the mirror symmetry swapping A ⇔ C, the far piece HC = AG = 1, so AC = 1+√52\frac{1+√5}{2} + 1 = 3+√52\frac{3+√5}{2} and CD = 1+√52\frac{1+√5}{2}.

AC = 1+√52\frac{1+√5}{2} + 1 = 3+√52\frac{3+√5}{2}, CD = 3+√52\frac{3+√5}{2} · √5−12\frac{√5-1}{2} = 1+√52\frac{1+√5}{2}
6STEP 6

Add the three lengths

Add the three bases: FG = <spanclass="hl−ask">√5</span>−12\frac{<span class="hl-ask">√5</span>-1}{2} and CD = 1+√52\frac{1+√5}{2} combine to √5, plus the middle JH = 1, giving the total → (D).

FG + JH + CD = √5−12\frac{√5-1}{2} + 1 + 1+√52\frac{1+√5}{2} = √5 + 1 = 1 + √5 → (D)
Answer
1+√5
Check numerically: FG = √5−12\frac{√5-1}{2} ≈ 0.618, JH = 1, CD = 1+√52\frac{1+√5}{2} ≈ 1.618, so the sum is about 3.236, matching 1+√5 ≈ 3.236. The five choices in decimals are (A) 3, (B) 12-4√5 ≈ 3.06, (C) 5+2√53\frac{5+2√5}{3} ≈ 3.16, (D) ≈ 3.236, (E) 11+11√510\frac{11+11√5}{10} ≈ 3.56; only (D) fits. It is also reassuring that JH = 1 = AG and that CD is the largest piece (it is the side of the big pentagon), as the picture shows.
💡Key takeaway

Inside a regular pentagon every triangle made by the diagonals is the same golden triangle at a different size, so find one base-to-leg ratio and just scale it up.

  • Spot three golden triangles
  • Find the golden base-to-leg ratio
  • Smallest triangle from AG = 1
  • Climb to the middle triangle
  • Climb to the outer triangle
  • Add the three lengths

More like this

Same archetype — closest grade level first.