Sample Question 12:

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?

\(\textbf{(A)}\ \frac{\sqrt{10}+5}{10}\qquad\textbf{(B)}\ \frac{2\sqrt{5}}{5}\qquad\textbf{(C)}\ \frac{2\sqrt{2}}{3}\qquad\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}\qquad\textbf{(E)}\ \frac{9-\sqrt{17}}{5}\)

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Answer Keys

Question 12: D

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Solutions

Question 12
Step 1: Note that the circles have radii of 1 and 2 respectively. Draw a triangle based on the figure given in the problem with one of its vertices at the center of the circles (O), one at the base of the square (E), and one at the top-left vertex of the square (A). The sides of the triangle are expressed in terms of \(s\) (the length of the side of the square).

Step 2: Since \(AO\) is the radius of the larger circle, which is 2, we can write the Pythagorean Theorem as:
\(\left( \frac{s}{2} \right) ^2 + (s+1)^2 = 2^2\)

Step 3: Simplify this to get \((1/4)s^2 + s^2 + 2s + 1 = 4\), then rearrange to give \((5/4) s^2 + 2s - 3 = 0\), then multiply by 4 to obtain the quadratic \(5s^2 + 8s - 12 = 0\).

Step 4: Solve this quadratic equation using the quadratic formula. The solution for the length of the side of the square is
\(s = \frac{-8+\sqrt{8^2-4(5)(-12)}}{10} = \frac{-8+\sqrt{304}}{10} = \frac{-8+4\sqrt{19}}{10} = \frac{2\sqrt{19}-4}{5}\)

So, the answer is \((\textbf{D)}\ \frac{2\sqrt{19}-4}{5}\).