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2012 AMC 10B Problem 14

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2012 AMC 10B Problem 14AMC 10 Step by Step
Name:                             Date: September 13, 2026
Quadrilaterals & Polygon Areas · difficulty 3 of 5 · about 5 min · position P11-15

Two equilateral triangles are contained in a square whose side length is 232\sqrt 3 . The bases of these triangles are the opposite sides of the square, and their intersection is a rhombus. What is the area of the rhombus?

  1. A)

    32\frac{3}{2}

  2. B)

    3\sqrt 3

  3. C)

    22−12\sqrt 2 - 1

  4. D)

    83−128\sqrt 3 - 12

  5. E)

    433\frac{4\sqrt 3}{3}

Answer

D

Key insight

Each triangle has height 3, so the apexes are 6 - 2√3 apart; the side lines cross at height √3, making the other diagonal 2√3 - 2.

Solution

Put the square at (0,0)(0,0), (23,0)(2\sqrt3, 0), (23,23)(2\sqrt3, 2\sqrt3), (0,23)(0, 2\sqrt3). An equilateral triangle with side 232\sqrt3 has height 32⋅23=3\frac{\sqrt3}{2} \cdot 2\sqrt3 = 3, which is less than 23≈3.462\sqrt3 \approx 3.46, so each triangle fits inside the square.

The lower triangle has its base on the bottom side and apex (3,3)(\sqrt3, 3); the upper triangle has its base on the top side and apex (3,23−3)(\sqrt3, 2\sqrt3 - 3). These two apexes are the top and bottom vertices of the rhombus, so the vertical diagonal is
3−(23−3)=6−23.3 - (2\sqrt3 - 3) = 6 - 2\sqrt3.

The left vertex is where the left sides cross. The lower triangle's left side is y=3 xy = \sqrt3\,x (slope tan⁡60∘\tan 60^\circ); the upper triangle's left side is y=23−3 xy = 2\sqrt3 - \sqrt3\,x. Setting them equal gives x=1x = 1, y=3y = \sqrt3. By symmetry the right vertex is at x=23−1x = 2\sqrt3 - 1, so the horizontal diagonal is 23−22\sqrt3 - 2.

The rhombus area is half the product of the diagonals:
12(6−23)(23−2)=12(123−12−12+43)=12(163−24)=83−12.\frac12 (6 - 2\sqrt3)(2\sqrt3 - 2) = \frac12\left(12\sqrt3 - 12 - 12 + 4\sqrt3\right) = \frac12(16\sqrt3 - 24) = 8\sqrt3 - 12.

The answer is (D) 83−12\boxed{\textbf{(D)}\ 8\sqrt3 - 12}.

Why this works

The figure is symmetric about both the vertical and horizontal midlines of the square, so the rhombus's diagonals lie on those midlines and only two points need computing: an apex and one crossing of the slanted sides. Coordinates turn "where do these sides meet" into solving two linear equations, and the diagonal formula avoids any angle work.

Alternative approach

The rhombus's sides lie along the triangles' sides, so its angles are 60∘60^\circ and 120∘120^\circ; a 60∘60^\circ rhombus is two equilateral triangles whose side equals the short diagonal s=23−2s = 2\sqrt3 - 2. Area =2⋅34s2=32(16−83)=83−12= 2 \cdot \frac{\sqrt3}{4} s^2 = \frac{\sqrt3}{2}(16 - 8\sqrt3) = 8\sqrt3 - 12.

The trap

Assuming each apex reaches the opposite side of the square (it does not, since the height 3 is less than 2√3), or dropping the 1/2 in the rhombus area formula.

Common mistakes

  • Assuming each apex reaches the opposite side of the square (it does not, since the height 3 is less than 2√3), or dropping the 1/2 in the rhombus area formula.
  • Using slope 11 or 13\frac{1}{\sqrt3} for the triangle sides instead of 3\sqrt3, which misplaces the side vertices.

Techniques

Place the figure on coordinates and compute · Exploit symmetry to reduce work or pair up objects