2003 AMC 10B Problem 23

Attempt Problem 23 of the 2003 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 10B solutions, or check the answer key.

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23.

A regular octagon ABCDEFGHABCDEFGH has an area of one square unit. What is the area of the rectangle ABEF?ABEF?

1221 - \dfrac{\sqrt{2}}{2}

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

21\sqrt{2} - 1

12\dfrac{1}{2}

1+24\dfrac{1 + \sqrt{2}}{4}

Answer: D
Concepts:regular polygonarea decompositionsymmetry
Difficulty rating: 1660
Solution:

Let OO be the center. The octagon splits into 88 congruent triangles from O,O, so AOB\triangle AOB has area 18.\dfrac18.

Since OO is the midpoint of AE,AE, triangles AOBAOB and BOEBOE have equal area, so ABE\triangle ABE has area 14.\dfrac14. The rectangle ABEFABEF is twice this, namely 12.\dfrac12.

Thus, the correct answer is D.

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