2020 AMC 12A Problem 14 — Solution | LIVE by Po-Shen Loh

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2020 AMC 12A Problem 14

Below is the professionally curated solution for Problem 14 of the 2020 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 12A solutions, or check the answer key.

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Concepts:regular polygon area ratio

Difficulty rating: 1690

14.

Regular octagon $ABCDEFGH$ has area $n.$ Let $m$ be the area of quadrilateral $ACEG.$ What is $\dfrac{m}{n}?$

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

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

$\dfrac{3}{4}$

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

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

Solution:

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.

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