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2012 AMC 10B Problem 14

P11-15~5 minPrint
Quadrilaterals & Polygon AreasTriangles: Area & PythagoreanPlace the figure on coordinates and computeExploit symmetry to reduce work or pair up objects

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?

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Key insight

Each triangle is 3 tall, so the two tips are 6 - 2√3 apart. The rhombus has a 60-degree angle at each tip, so it is two equilateral triangles.

Solution

1. First find how tall each triangle is. An equilateral triangle with side ss has height 32s\frac{\sqrt3}{2}s. With s=23s = 2\sqrt3 the height is 32⋅23=3\frac{\sqrt3}{2} \cdot 2\sqrt3 = 3.
2. The square is 23≈3.462\sqrt3 \approx 3.46 tall, so neither tip reaches the far side.
3. Measure heights from the bottom of the square. The bottom triangle's tip is at height 33. The top triangle hangs down 33 from the top, so its tip is at height 23−3≈0.462\sqrt3 - 3 \approx 0.46.
4. These two tips are the top and bottom corners of the rhombus. The distance between them (the rhombus's vertical diagonal) is
3−(23−3)=6−23.3 - (2\sqrt3 - 3) = 6 - 2\sqrt3.
5. Now look at the top corner. The two rhombus sides that meet there lie along the two slanted sides of the bottom triangle, so the angle there is 60∘60^\circ.
6. Cut the rhombus along its horizontal diagonal. The top piece has two equal sides (all sides of a rhombus are equal) with a 60∘60^\circ angle between them, so it is an equilateral triangle. The bottom piece is the same.
7. The diagonals of a rhombus cut each other in half, so each equilateral piece has height 12(6−23)=3−3\frac12(6 - 2\sqrt3) = 3 - \sqrt3.
8. For an equilateral triangle, side =23×= \dfrac{2}{\sqrt3} \times height. So each side is
2(3−3)3=63−2=23−2.\frac{2(3 - \sqrt3)}{\sqrt3} = \frac{6}{\sqrt3} - 2 = 2\sqrt3 - 2.
9. The rhombus is two triangles, each with area 12×side×height\frac12 \times \text{side} \times \text{height}, so its area is side times height:
(23−2)(3−3)=63−6−6+23=83−12.(2\sqrt3 - 2)(3 - \sqrt3) = 6\sqrt3 - 6 - 6 + 2\sqrt3 = 8\sqrt3 - 12.

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

Why this works

The picture is symmetric left-to-right and top-to-bottom, so the rhombus's corners are easy to locate: two of them are the triangles' tips. The angles of the rhombus come straight from the 60∘60^\circ angles of the equilateral triangles, and a rhombus with a 60∘60^\circ angle is always two equilateral triangles glued together. Checking whether the tips reach the far side (they don't) is the first thing to do.

Alternative approach

Use "area of a rhombus = half the product of its diagonals." The vertical diagonal is 6−236 - 2\sqrt3 from step 4. The horizontal diagonal is the shared side of the two equilateral pieces, which is 23−22\sqrt3 - 2. So the area is
12(6−23)(23−2)=12(123−12−12+43)=12(163−24)=83−12.\tfrac12(6 - 2\sqrt3)(2\sqrt3 - 2) = \tfrac12(12\sqrt3 - 12 - 12 + 4\sqrt3) = \tfrac12(16\sqrt3 - 24) = 8\sqrt3 - 12.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.