Sphere Radius Calculation for Points A, B, C

This problem asks us to find the radius of a sphere, named \(S\). The sphere \(S\) must be the one with the smallest possible radius that passes through three specific points in three-dimensional space: \(A(1, 0, 0)\), \(B(0, 1, 0)\), and \(C(0, 0, 1)\).

Analyzing Points A(1,0,0), B(0,1,0), C(0,0,1)

Let's start by examining the given points:

• Point A is located at \((1, 0, 0)\) on the x-axis.
• Point B is located at \((0, 1, 0)\) on the y-axis.
• Point C is located at \((0, 0, 1)\) on the z-axis.

To understand the geometric relationship between these points, we calculate the distances between each pair:

• Distance between A and B (\(AB\)):

  \(d(A, B) = \sqrt{(1-0)^2 + (0-1)^2 + (0-0)^2} = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{1 + 1} = \sqrt{2}\)

• Distance between B and C (\(BC\)):

  \(d(B, C) = \sqrt{(0-0)^2 + (1-0)^2 + (0-1)^2} = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}\)

• Distance between A and C (\(AC\)):

  \(d(A, C) = \sqrt{(1-0)^2 + (0-0)^2 + (0-1)^2} = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}\)

Since all three distances are equal (\(AB = BC = AC = \sqrt{2}\)), the points \(A\), \(B\), and \(C\) form an equilateral triangle.

Determining the Smallest Sphere Through Triangle ABC

The problem specifies the "smallest radius sphere" passing through these three points. For three non-collinear points, the smallest sphere that contains them is their circumsphere. The center of this sphere is the circumcenter of the triangle formed by the points, and its radius is the circumradius of that triangle.

Finding the Sphere Center (Circumcenter)

Let the center of the sphere be \((h, k, l)\) and its radius be \(R\). The distance from the center to each point \(A\), \(B\), and \(C\) must be equal to \(R\). Using the distance formula:

• Distance from \((h,k,l)\) to \(A(1,0,0)\):

  \((h-1)^2 + (k-0)^2 + (l-0)^2 = R^2 \implies (h-1)^2 + k^2 + l^2 = R^2 \quad (1)\)

• Distance from \((h,k,l)\) to \(B(0,1,0)\):

  \((h-0)^2 + (k-1)^2 + (l-0)^2 = R^2 \implies h^2 + (k-1)^2 + l^2 = R^2 \quad (2)\)

• Distance from \((h,k,l)\) to \(C(0,0,1)\):

  \((h-0)^2 + (k-0)^2 + (l-1)^2 = R^2 \implies h^2 + k^2 + (l-1)^2 = R^2 \quad (3)\)

Now, we equate the expressions for \(R^2\):

Equating (1) and (2):

\((h-1)^2 + k^2 + l^2 = h^2 + (k-1)^2 + l^2\)

\(h^2 - 2h + 1 + k^2 = h^2 + k^2 - 2k + 1\)

\(-2h = -2k \implies h = k\)

Equating (2) and (3):

\(h^2 + (k-1)^2 + l^2 = h^2 + k^2 + (l-1)^2\)

\(k^2 - 2k + 1 + l^2 = k^2 + l^2 - 2l + 1\)

\(-2k = -2l \implies k = l\)

This tells us that the center of the sphere must have equal coordinates, i.e., the center is of the form \((h, h, h)\).

The points \(A\), \(B\), and \(C\) lie on the plane defined by the equation \(x + y + z = 1\). The circumcenter of the triangle \(ABC\) must lie on this plane. Since the center of our sphere is \((h, h, h)\), it must satisfy this plane equation:

\(h + h + h = 1\)

\(3h = 1 \implies h = \frac{1}{3}\)

Thus, the center of the sphere \(S\) is \(O' = (\frac{1}{3}, \frac{1}{3}, \frac{1}{3})\).

Calculating the Sphere Radius

Now that we have the center \(O'(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})\), we can find the radius \(R\) by calculating the distance from the center to any of the given points, for example, point \(A(1, 0, 0)\):

\(R^2 = \left(\frac{1}{3} - 1\right)^2 + \left(\frac{1}{3} - 0\right)^2 + \left(\frac{1}{3} - 0\right)^2\)

\(R^2 = \left(-\frac{2}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2\)

\(R^2 = \frac{4}{9} + \frac{1}{9} + \frac{1}{9}\)

\(R^2 = \frac{6}{9} = \frac{2}{3}\)

Taking the square root to find the radius:

\(R = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}\)

Alternative Method: Equilateral Triangle Properties

We identified that triangle \(ABC\) is equilateral with a side length \(a = \sqrt{2}\).

The radius of the circumcircle (\(r_{circ}\)) of an equilateral triangle with side length \(a\) can be calculated using the formula:

\(r_{circ} = \frac{a}{\sqrt{3}}\)

Substituting the side length \(a = \sqrt{2}\):

\(r_{circ} = \frac{\sqrt{2}}{\sqrt{3}}\)

For the smallest radius sphere passing through the vertices of triangle \(ABC\), this triangle must lie on a great circle of the sphere. Therefore, the radius of the sphere \(S\) is equal to the circumradius of triangle \(ABC\).

\(R = r_{circ} = \frac{\sqrt{2}}{\sqrt{3}}\)

Summary of Sphere Radius Result

Both calculation methods consistently show that the radius of the sphere \(S\) is \(\frac{\sqrt{2}}{\sqrt{3}}\).