1997 AMC 8 Problem 15

Attempt Problem 15 of the 1997 AMC 8 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 1997 AMC 8 solutions, or check the answer key.

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15.

Each side of the large square in the figure is trisected (divided into three equal parts). The corners of an inscribed square are at these trisection points, as shown. The ratio of the area of the inscribed square to the area of the large square is
33\dfrac{\sqrt{3}}{3}
59\dfrac{5}{9}
23\dfrac{2}{3}
53\dfrac{\sqrt{5}}{3}
79\dfrac{7}{9}
Answer: B
Concepts:area ratioPythagorean Theorem
Small hint:
Use the Pythagorean theorem for the inscribed square side
Big hint:
Let the large square side be split into three equal parts
Solution:
Let 3x3x be the side length of the large square. Then we can find the side length of the inner square via (2x)2+x2=5x2=x5 \sqrt{(2x)^2 + x^2} = \sqrt{5x^2} = x\sqrt{5} from the Pythagorean Theorem.
The area of the larger square is (3x)2=9x2 (3x)^2 = 9x^2 and that of the inner square is (x5)2=5x2. (x\sqrt{5})^2 = 5x^2.
The ratio of the areas is then 59.\dfrac{5}{9}.
Thus, B is the correct answer.
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