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2017 AMC 12B Problem 8

Below is the professionally curated solution for Problem 8 of the 2017 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 12B solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:quadraticPythagorean Theoremratio and proportion

Difficulty rating: 1440

8.

The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side to the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?

3−12\dfrac{\sqrt{3} - 1}{2}

12\dfrac{1}{2}

5−12\dfrac{\sqrt{5} - 1}{2}

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

6−12\dfrac{\sqrt{6} - 1}{2}

Solution:

Let xx and yy be the short and long sides, so the diagonal is x2+y2\sqrt{x^2 + y^2} and x2y2=y2x2+y2.\dfrac{x^2}{y^2} = \dfrac{y^2}{x^2 + y^2}. Writing r=x2y2,r = \dfrac{x^2}{y^2}, the right side is y2x2+y2=1r+1,\dfrac{y^2}{x^2 + y^2} = \dfrac{1}{r + 1}, so r=1r+1,r = \dfrac{1}{r+1}, giving r2+r−1=0.r^2 + r - 1 = 0. The positive root is r=5−12.r = \dfrac{\sqrt{5} - 1}{2}.

Thus, the correct answer is C.

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