AMC 10 · 2019 · #25

Grade 11 geometry-2d
centroid-2-to-1dilationequilateral-trianglelaw-of-cosinessine-area-formulaoptimization extreme-principleidentify-subproblemsconvert-to-algebra ↑ Prerequisites: centroid-2-to-1law-of-cosines
📏 Long solution 💡 4 insights
Problem
A convex quadrilateral has two fixed sides of lengths 2 and 6. Three of its vertex triples give triangles whose centroids happen to be the vertices of an equilateral triangle. Among all quadrilaterals satisfying this, find the largest possible area.

Pick an answer.

(A)
27
(B)
$16\sqrt3$
(C)
$12+10\sqrt3$
(D)
$9+12\sqrt3$
(E)
30
How to solve
Strategy Extreme Principle

The question asks for a maximum over a whole family of shapes, so Tool #14 (Extreme Principle) is the destination: get the area down to one expression in one variable, then push that variable to its best value and check the best value is legal. Getting there needs the other tools first. Tool #1 (Draw a Diagram) exposes the fact that all three named triangles contain the vertex C, so the three centroids can be measured from C. Tool #15 (Organize Information in More Ways) re-reads those three centroids as a shrunken copy of the midpoint triangle of △ ABD, which converts the strange equilateral condition into the clean statement "△ ABD is equilateral". Tool #7 (Identify Subproblems) cuts ABCD along the diagonal BD into two triangles whose areas can be handled separately. Tool #4 (Introduce a Variable) names the one remaining degree of freedom, the angle at C. Tool #13 (Convert to Algebra) turns both areas into functions of that angle and collapses the sum into a single sine wave, whose maximum is then read off.

1STEP 1

Find the shared vertex

All three contain one point.

CG₁ = 2/3 CX, CG₂ = 2/3 CY, CG₃ = 2/3 CZ
2STEP 2

Dilate from that point

Dilating carries the equilateral condition back.

△ G₁G₂G₃ → { dilation at C, ratio 3/2 } △ XYZ = midpoint triangle of △ ABD ⟹ AB = BD = DA = k
3STEP 3

Cut along the diagonal

Split the quadrilateral into two triangles.

[ABCD] = [ABD] + [BCD], [ABD] = √(3)/4k²
4STEP 4

Name the angle

Name the angle between the fixed sides.

k² = 2² + 6² - 2 · 2 · 6cosθ = 40 - 24cosθ ⟹ [ABD] = √(3)/4(40 - 24cosθ) = 10√(3) - 6√(3)cosθ
5STEP 5

Add the second triangle

The area becomes a function of that angle.

[BCD] = 1/2 · 2 · 6sinθ = 6sinθ ⟹ [ABCD] = 10√(3) + 6sinθ - 6√(3)cosθ
6STEP 6

Collapse into one sine wave

The two terms fold into one sine wave.

6sinθ - 6√(3)cosθ = 12(1/2sinθ - √(3)/2cosθ) = 12(sinθcos 60° - cosθsin 60°) = 12sin(θ - 60°)
7STEP 7

Push it to the maximum

The maximum is twelve plus ten root three.

[ABCD] = 10√(3) + 12sin(θ - 60°) ≤ 10√(3) + 12, equality at θ = 150° → 12 + 10√(3)
Answer
12+10√3
The five choices are close together numerically — 27, 16√3 ≈ 27.71, 12+10√3 ≈ 29.32, 9+12√3 ≈ 29.78, 30 — so a rough estimate cannot separate them and the amplitude has to be exact. It is: √(6² + (6√3)²) = 12 on the nose, which rules out anything above 10√3 + 12 and kills the two larger choices. Building the winning configuration confirms it. With θ = 150°, put B = (0,0) and D ≈ (7.796, 0); then C ≈ (1.846, -0.770) satisfies BC = 2, CD = 6, and A ≈ (3.898, 6.752) completes the equilateral triangle on the other side. Walking A → B → C → D turns the same way at all four corners, so the quadrilateral is convex, and the shoelace formula gives area ≈ 29.3205, matching 12 + 10√3 ≈ 29.3205. The centroid condition also passes a sanity test: it never mentions C's position, and indeed the conclusion AB = BD = DA says nothing about C — consistent with C being free to swing on the fixed sides 2 and 6.
💡Key takeaway

The equilateral centroid triangle is just a shrunken midpoint triangle, so it secretly says ABD is equilateral; after that the whole area is one sine wave in the angle at C, and its peak is 12 + 10√(3).

  • Every named triangle contains C
  • Scale up from C by 3/2
  • Cut along the diagonal BD
  • Name the angle at C
  • Add the second triangle
  • Collapse into one sine wave
  • Push the angle to the extreme