2012 AMC 12A Problem 12

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}

Use the symmetry across the

y

-axis to write

A = (a, b)

and

B = (-a, b)

The side length is

2a

horizontally and

b - 1

vertically, so

b = 2a + 1

; substitute into

a^2 + b^2 = 4

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.

Problem 12 in other years