AMC 10 · 2012 · #14

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Line Intersection Vertices then Area

equilateral-trianglearea-trianglesthirty-sixty-ninety-triangle identify-subproblems ↑ Prerequisites: equilateral-triangle

📏 Medium solution 💡 3 insights

Problem

A square has side length 2√(3). Two equilateral triangles sit inside it: one has its base on the bottom side of the square, the other has its base on the top side. The region where the two triangles overlap is a rhombus. Find the area of that rhombus.

Pick an answer.

(A)

$\frac{3}{2}$

(B)

$\sqrt 3$

(C)

$2\sqrt 2 - 1$

(D)

$8\sqrt 3 - 12$

(E)

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

AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The shapes are all defined by position, so a coordinate diagram pins down every point exactly. Once the square and both triangles have coordinates, the rhombus is just the overlap, and its four corners are points I can compute. I break the work into small pieces: find the triangle height, find the corners, then combine the two diagonals into the area.

1STEP 1

Place the square on a grid

Set the square from (0,0) to (2√(3),2√(3)). The bottom triangle points up, the top one points down, and each base is 2√(3).

square side = 2√(3)

2STEP 2

Find each triangle's height

Cut it in half: height = √(3)/2 times s, so with s = 2√(3) the height is 3, putting the apexes at (√(3),3) and (√(3),2√(3)-3).

h = √(3)/2 · 2√(3) = 3

3STEP 3

Locate the two side corners

The left edges y=√(3)x and y=-√(3)x+2√(3) meet at (1,√(3)); by symmetry the right corner is (2√(3)-1,√(3)).

√(3)x = -√(3)x + 2√(3) → x = 1

4STEP 4

Measure the two diagonals

Subtract the shared coordinates: the vertical diagonal is 6-2√(3) and the horizontal one is 2√(3)-2.

d₁ = 6-2√(3), d₂ = 2√(3)-2

5STEP 5

Combine into the area

Half the product of the diagonals: 1/2(6-2√(3))(2√(3)-2) = 1/2(16√(3)-24) = 8√(3)-12, choice (D).

1/2(6-2√(3))(2√(3)-2) = 1/2(16√(3)-24) = 8√(3)-12

Answer

8√3 - 12

Numerically 8√(3)-12 ≈ 1.86. The whole square has area (2√(3))² = 12, and the small overlap rhombus is a thin sliver near the center, so an area under 2 is sensible. The diagonals 6-2√(3)≈ 2.54 and 2√(3)-2≈ 1.46 are both positive and fit inside the square, confirming the corners are placed correctly.

💡Key takeaway

Put the figure on a grid, find the four corners of the overlap, and a rhombus's area is just half the product of its two diagonals.

Place the square on a grid
Find each triangle's height
Locate the two side corners
Measure the two diagonals
Combine into the area

Place the square on a grid

Find each triangle's height

Locate the two side corners

Measure the two diagonals

Combine into the area

More like this