AMC 10 · 2016 · #17

Grade 10 geometry-2d
coordinate-geometryequilateral-trianglecentroid-2-to-1rotation-isometry coordinate-geometrysymmetry-argument ↑ Prerequisites: equilateral-trianglepythagorean-theorem
📏 Long solution 💡 4 insights
Problem
Equilateral triangles on a square's sides have centres that form another square. Find the area ratio.

Pick an answer.

(A)
1
(B)
$\frac{2+\sqrt{3}}{3}$
(C)
$\sqrt{2}$
(D)
$\frac{\sqrt{2}+\sqrt{3}}{2}$
(E)
$\sqrt{3}$
How to solve
Strategy Draw a Diagram

The phrase "center of an equilateral triangle" is a position, not a number, so tool #1 (Draw a Diagram) with coordinate axes is the move that makes it computable: put the square on the grid and every center becomes an ordered pair. Tool #9 (Solve an Easier Related Problem) fixes the side length at 1, which is free because the question asks for a ratio and scaling the whole picture scales both areas equally. Tool #7 (Identify Subproblems) splits the work into three small, independent questions — how tall is an equilateral triangle, where inside it is the center, and how do you get the area of a square from its diagonal. Tool #17 (Visualize Spatial Relationships) supplies the fact that a quarter turn about the square's center carries the whole picture onto itself, which both proves EFGH is a square and hands over its center for free. Tool #4 (Introduce a Variable) names the half-diagonal R so the final area is one clean formula, and tool #3 (Eliminate Possibilities) confirms the closed form against the five decimal values at the end.

1STEP 1

Fix the side length at 1

A ratio lets the side be fixed at one.

[EFGH]/[ABCD] is unchanged by scaling, so set AB=BC=CD=DA=1
2STEP 2

Put the square on the grid

A grid places every corner.

A=(0,1), B=(1,1), C=(1,0), D=(0,0), O=(1/2,1/2)
3STEP 3

Height of an equilateral triangle

The triangle's height follows at once.

h²+(1/2)²=1² → h=√(3)/2; apex=(1/2, 1+√(3)/2)
4STEP 4

Locate the center E

That locates one centre.

E=(1/2, 1+√(3)/6) since 1/3·√(3)/2=√(3)/6
5STEP 5

Spin the picture a quarter turn

A quarter turn gives the other three.

E=(1/2, 1+√(3)/6), F=(1+√(3)/6, 1/2), G=(1/2, -√(3)/6), H=(-√(3)/6, 1/2)
6STEP 6

Area from the half-diagonal

The half-diagonal gives the area.

R=(3+√(3))/6, R²=(12+6√(3))/36=(2+√(3))/6, [EFGH]=2R²=(2+√(3))/3
7STEP 7

Cross-check and take the ratio

A side check confirms choice (B).

EF²=2(1/2+√(3)/6)²=(2+√(3))/3; [EFGH]/[ABCD]=((2+√(3))/3)/1=(2+√(3))/3 → (B)
Answer
(2+√(3))/3
Numerically √(3)≈ 1.7321, so √(3)/6≈ 0.2887 and R≈ 0.7887, giving [EFGH]=2R²≈ 1.2440 against [ABCD]=1. The five choices are 1, ≈ 1.2440, ≈ 1.4142, ≈ 1.5731, ≈ 1.7321, so only (B) is in range and no two choices are close enough to confuse. The size is also believable in advance: square ABCD has half-diagonal √(2)/2≈ 0.7071 while EFGH has half-diagonal ≈ 0.7887, so EFGH should be a bit larger, ruling out (A) 1; and the ratio of two squares' areas is the square of the ratio of their half-diagonals, (0.7887/0.7071)²≈ 1.244, matching. Coordinates confirm the shape is genuinely a square: all four sides come out equal at √((2+√(3))/3)≈ 1.1154 and both diagonals equal 2R≈ 1.5774, and 1.1154²+1.1154²=1.5774² confirms the right angles. One trap worth naming: the center of an equilateral triangle is 1/3 of the height above the base, not 1/2 — using 1/2 would give R=1/2+√(3)/4 and a ratio near 1.44, close to (C) but not equal to any choice, so a careless reader would not even be rescued by the answer list.
💡Key takeaway

The center of an equilateral triangle sits one third of the way up from its base, so each corner of EFGH pokes out √(3)/6 past a side of the unit square — and that alone fixes the area ratio at (2+√(3))/3.

  • Fix the side length at 1
  • Put the square on the grid
  • Height of an equilateral triangle
  • Locate the center E
  • Spin the picture a quarter turn
  • Area from the half-diagonal
  • Cross-check and take the ratio