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2012 AMC 12A Problem 12

Problem 12 of 25IntermediateAlgebraGeometry

A square region ABCDABCD is externally tangent to the circle with equation x2+y2=1x^2 + y^2 = 1 at the point (0,1)(0, 1) on the side CD.CD. Vertices AA and BB are on the circle with equation x2+y2=4.x^2 + y^2 = 4. What is the side length of this square?

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Solution

By symmetry let A=(a,b)A = (a, b) with a>0a \gt 0 and B=(−a,b).B = (-a, b). The square sits on the tangent point (0,1),(0,1), so its horizontal width is 2a2a and its height is b−1.b - 1. Since these are equal, 2a=b−1,2a = b - 1, giving b=2a+1.b = 2a + 1. Substituting into a2+b2=4a^2 + b^2 = 4 yields 5a2+4a−3=0.5a^2 + 4a - 3 = 0. The positive root is a=19−25,a = \dfrac{\sqrt{19} - 2}{5}, so the side length is 2a=219−45.2a = \dfrac{2\sqrt{19} - 4}{5}. Thus, the correct answer is D.

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Concepts: coordinate geometry · quadratic · symmetry

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