For a cube, find the ratio of the surface area of the sphere inscribed in the cube to the surface area of the sphere circumscribed about the cube.

Introduction / Context:
This problem compares two spheres related to the same cube. One sphere is inscribed in the cube, touching each face, and the other is circumscribed about the cube, passing through all eight vertices. The question asks for the ratio of their surface areas. Understanding the relationships between these radii and the cube side is a useful exercise in three dimensional geometry.

Given Data / Assumptions:

• Consider a cube with side length a.
• Inscribed sphere touches all six faces of the cube.
• Circumscribed sphere passes through all eight vertices of the cube.
• Surface area of a sphere with radius r is 4 * π * r^2.

Concept / Approach:
For the inscribed sphere, the diameter equals the side length of the cube because it just fits between opposite faces. For the circumscribed sphere, the diameter equals the space diagonal of the cube. By expressing both radii in terms of a, we can compute the ratio of their surface areas. Constants like 4 and π will cancel in the ratio, leaving a simple relationship.

Step-by-Step Solution:
Step 1: Radius of inscribed sphere. Diameter = side of cube = a. So radius r_in = a / 2. Step 2: Radius of circumscribed sphere. Diameter = space diagonal of cube = a * sqrt(3). So radius r_out = (a * sqrt(3)) / 2. Step 3: Surface area formulas. Surface area of inscribed sphere S_in = 4 * π * (a / 2)^2 = 4 * π * a^2 / 4 = π * a^2. Surface area of circumscribed sphere S_out = 4 * π * (a * sqrt(3) / 2)^2. Compute radius squared: (a * sqrt(3) / 2)^2 = 3 * a^2 / 4. So S_out = 4 * π * (3 * a^2 / 4) = 3 * π * a^2. Step 4: Ratio of surface areas. S_in : S_out = π * a^2 : 3 * π * a^2 = 1 : 3.

Verification / Alternative check:
We can check numerically by taking a convenient side length, say a = 2 units. Then r_in = 1, r_out = sqrt(3). Surface areas become S_in = 4 * π * 1^2 = 4π, and S_out = 4 * π * (sqrt(3))^2 = 4 * π * 3 = 12π. The ratio is 4π : 12π = 1 : 3, confirming our algebraic result.

Why Other Options Are Wrong:
1 : 2, 1 : 4 and 1 : 5 correspond to incorrect relationships between the radii and ignore the exact geometric connection through the space diagonal.
2 : 3 suggests the circumscribed sphere is only 1.5 times the surface area of the inscribed sphere, which contradicts the explicit calculation showing a factor of 3.

Common Pitfalls:
Students sometimes confuse face diagonal with space diagonal, or misapply Pythagoras theorem. The face diagonal is a * sqrt(2), whereas the space diagonal is a * sqrt(3). Another frequent error is to compare radii directly instead of areas; radius ratio r_in : r_out is 1 : sqrt(3), but for areas the ratio involves the square, leading to 1 : 3. Remember that surface area depends on the square of the radius, not the radius itself.

Final Answer:
The required ratio of surface areas is 1 : 3.