A smooth curve inside the equilateral triangle

Calculus Level 4

An equilateral triangle (blue) has a smooth curve drawn inside it (red). The smooth curve is actually made up of three parabolic segments each connecting the midpoint of one side to the midpoint of an adjacent side. Each parabolic segment is tangent to the triangle sides at the midpoints of the sides. Find the ratio of the area enclosed by the red curve to the area of the equilateral triangle.

\dfrac{2}{3}

\dfrac{3}{4}

\dfrac{4}{5}

\dfrac{5}{6}

1 solution

David Vreken
Jun 4, 2021

Label the diagram as follows:

$\triangle AFE$ is an Archimedes Triangle of the bottom left parabola, and since this parabola divides the area of an Archimedes triangle in ratio $2:1$ , if the area of the region bounded by $AF$ , $AE$ , and the bottom left parabola is $k$ , then the area of the region bounded by $EF$ and the bottom left parabola is $2k$ .

By symmetry, we can then label the area of each region as follows:

Therefore, the ratio of the areas is $\cfrac{A_{\text{red enc.}}}{A_{\triangle ABC}} = \cfrac{3 \cdot 2k + 3k}{3 \cdot k + 3 \cdot 2k + 3k} = \cfrac{9k}{12k} = \boxed{\cfrac{3}{4}}$ .

A smooth curve inside the equilateral triangle

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