Question

Let $S_1$ and $S_2$ be two circles drawn inside a unit square ABCD, touching each other externally. Further, the circle $S_1$ touches the sides AD and DC; and the circle $S_2$ touches the sides AB and BC. If the area of $S_2$ is twice the area of $S_1$, then the radius of $S_1$ is

• A. $3\sqrt{2} - 4$
• B. $3 - 2\sqrt{2}$
• C. $2 - \sqrt{2}$
• D. $\frac{3}{\sqrt{2}} - 2$

Hint:

Using the diagonal of the square is a great way to verify: the total diagonal of the unit square is $\sqrt{2}$.
The diagonal is also equal to $r_1\sqrt{2} + r_1 + r_2 + r_2\sqrt{2}$, which directly yields $(r_1+r_2)(1+\sqrt{2}) = \sqrt{2}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
Two circles are inscribed in opposite corners of a unit square such that they touch the adjacent sides of the square and touch each other externally.
We need to find the radius of the smaller circle $S_1$ using the geometric relations of the square.

Step 2: Key Formula or Approach:
Let the vertices of the unit square be positioned on a Cartesian coordinate plane:
- Let $D = (0, 0)$, $A = (0, 1)$, $B = (1, 1)$, and $C = (1, 0)$.
- The distance between the centers of two externally touching circles is $d = r_1 + r_2$.

Step 3: Detailed Explanation:

• Let $r_1$ be the radius of $S_1$ and $r_2$ be the radius of $S_2$.
- $S_1$ touches $AD$ ($x = 0$) and $DC$ ($y = 0$). Its center is:
\[ C_1 = (r_1, r_1) \]
- $S_2$ touches $AB$ ($y = 1$) and $BC$ ($x = 1$). Its center is:
\[ C_2 = (1 - r_2, 1 - r_2) \]

• Since the two circles touch each other externally, the distance between their centers is $r_1 + r_2$:
\[ C_1C_2 = \sqrt{(1 - r_2 - r_1)^2 + (1 - r_2 - r_1)^2} = \sqrt{2}(1 - r_1 - r_2) \]
Set this equal to $r_1 + r_2$:
\[ \sqrt{2}(1 - (r_1 + r_2)) = r_1 + r_2 \]
\[ \sqrt{2} = (r_1 + r_2)(\sqrt{2} + 1) \implies r_1 + r_2 = \frac{\sqrt{2}}{\sqrt{2} + 1} = 2 - \sqrt{2} \]

• We are given that the area of $S_2$ is twice the area of $S_1$:
\[ \pi r_2^2 = 2\pi r_1^2 \implies r_2 = \sqrt{2} r_1 \]

• Substitute $r_2 = \sqrt{2} r_1$ into our sum equation:
\[ r_1 + \sqrt{2} r_1 = 2 - \sqrt{2} \]
\[ r_1(1 + \sqrt{2}) = 2 - \sqrt{2} \]

• Solve for $r_1$:
\[ r_1 = \frac{2 - \sqrt{2}}{\sqrt{2} + 1} = \frac{\sqrt{2}(\sqrt{2} - 1)}{\sqrt{2} + 1} \]
Multiply numerator and denominator by $(\sqrt{2} - 1)$:
\[ r_1 = \sqrt{2}(\sqrt{2} - 1)^2 = \sqrt{2}(2 - 2\sqrt{2} + 1) \]
\[ r_1 = \sqrt{2}(3 - 2\sqrt{2}) = 3\sqrt{2} - 4 \]



Step 4: Final Answer:
The radius of $S_1$ is $3\sqrt{2} - 4$.