2020 AMC 12A Problem 14

Question

Regular octagon

ABCDEFGH

has area

n.

Let

m

be the area of quadrilateral

ACEG.

What is

\dfrac{m}{n}?

\dfrac{\sqrt{2}}{4}

\dfrac{\sqrt{2}}{2}

\dfrac{3}{4}

\dfrac{3\sqrt{2}}{5}

\dfrac{2\sqrt{2}}{3}

ACEG

is a square inscribed in the octagon

Place the octagon symmetrically and compute both areas; the four corner triangles make up the difference

The four vertices

A, C, E, G

form a square, since they are every other vertex of the regular octagon.

Taking a unit circumradius, the octagon’s area is

2\sqrt2

and the square

ACEG

has diagonal equal to the circle’s diameter, giving area

2.

The ratio is

\dfrac{2}{2\sqrt2} = \dfrac{1}{\sqrt2} = \dfrac{\sqrt2}{2}.

Thus, B is the correct answer.

Problem 14 in other years