AMC 10 · 2010 · #19

Grade 8 geometry-2d

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

law-of-cosinesequilateral-trianglearea-trianglesvieta-formulasquadratic-equations identify-subproblems ↑ Prerequisites: law-of-cosines

📏 Long solution 💡 4 insights

Problem

An equiangular hexagon ABCDEF (every interior angle 120°) has sides alternating AB=CD=EF=1 and BC=DE=FA=r. The triangle ACE joining every other vertex has area equal to 70% of the hexagon's area. Find the sum of all values of r that make this true.

Pick an answer.

(A)

$\frac{4\sqrt{3}}{3}$

(B)

$\frac{10}{3}$

(C)

(D)

$\frac{17}{4}$

(E)

AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The area equation hides two separate area computations, so split the work (Tool #7 Identify Subproblems): first find the area of △ ACE, then find the area of the whole hexagon, both in terms of r. Drawing the figure (Tool #1) reveals two structural gifts — the three corner triangles ABC, CDE, EFA are congruent, which forces △ ACE to be equilateral, and the hexagon is a big equilateral triangle with three unit corners sliced off. Since everything is already labelled with the variable r (Tool #4), setting [△ ACE] = 0.7 · [hexagon] collapses to a single quadratic in r, and the question only asks for the sum of its roots.

1STEP 1

Triangle ACE is equilateral

Corner triangles ABC, CDE, EFA each put sides 1 and r around a 120° angle, so SAS gives AC = CE = EA: triangle ACE is equilateral.

△ ABC ≅ △ CDE ≅ △ EFA (SAS) → AC = CE = EA

2STEP 2

Find side AC

In triangle ABC drop a perpendicular from A: the 30-60-90 piece and Pythagoras give AC² = r² + r + 1.

AC² = (√(3)/2)² + (r + 1/2)² = 3/4 + r² + r + 1/4 = r² + r + 1

3STEP 3

Area of triangle ACE

An equilateral triangle's area is $\frac{\sqrt{3}}{4}$ times its side squared, so triangle ACE carries the factor r² + r + 1.

[△ ACE] = √(3)/4 (r² + r + 1)

4STEP 4

Area of the hexagon

Extending the short sides makes the hexagon a big equilateral triangle of side r+2 minus three unit triangles: factor r² + 4r + 1.

[hex] = √(3)/4(r+2)² - 3·√(3)/4(1)² = √(3)/4[(r+2)² - 3] = √(3)/4 (r² + 4r + 1)

5STEP 5

Set up and simplify the equation

Impose the 70% condition: the shared $\frac{\sqrt{3}}{4}$ cancels and the geometry collapses to r² - 6r + 1 = 0.

r² + r + 1 = 0.7 (r² + 4r + 1) → 0.3r² - 1.8r + 0.3 = 0 → r² - 6r + 1 = 0

6STEP 6

Sum of all valid r

The discriminant 36 - 4 is positive and both roots are positive, so by Vieta the two values of r add to 6.

r = (6 ± √(32))/2 = 3 ± 2√(2); (3 + 2√(2)) + (3 - 2√(2)) = 6 → (E) 6

Answer

Sanity-check one root. Take r = 3 + 2√(2) ≈ 5.828. Then r² + r + 1 ≈ 33.97 + 5.83 + 1 = 40.80 and r² + 4r + 1 ≈ 33.97 + 23.31 + 1 = 58.28; the ratio 40.80/58.28 ≈ 0.700, exactly the required 70%. The other root r = 3 - 2√(2) ≈ 0.172 is a genuinely different but still positive hexagon, so two values of r really do exist and both must be counted. Their sum is 6, matching (E). The answer is a plain integer, consistent with a competition asking for a 'sum of all possible values'.

💡Key takeaway

Write both areas in terms of r: the inner triangle is equilateral with side² = r²+r+1, and the hexagon is a big triangle minus three unit corners with area factor r²+4r+1. The 70% rule turns into r² - 6r + 1 = 0, and the sum of its two roots is (E) 6.

Triangle ACE is equilateral
Find side AC
Area of triangle ACE
Area of the hexagon
Set up and simplify the equation
Sum of all valid r