Question

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

$\dfrac{\sqrt{10} + 5}{10}$

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

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

$\dfrac{2\sqrt{19} - 4}{5}$

$\dfrac{9 - \sqrt{17}}{5}$

Solution:

By symmetry let $A = (a, b)$ with $a \gt 0$ and $B = (-a, b).$ The square sits on the tangent point $(0,1),$ so its horizontal width is $2a$ and its height is $b - 1.$

Since these are equal, $2a = b - 1,$ giving $b = 2a + 1.$

Substituting into $a^2 + b^2 = 4$ yields $5a^2 + 4a - 3 = 0.$ The positive root is $a = \dfrac{\sqrt{19} - 2}{5},$ so the side length is $2a = \dfrac{2\sqrt{19} - 4}{5}.$

Thus, the correct answer is D.

2012 AMC 12A Problem 12

Solution:

Problem 12 in Other Years