`(0,0,0), (a,0,0), (0,b,0) and (0,0,c)` are four distinct points. What are the coordinates of the point which is equidistant from the four points?

Options

A. `((a+b+c)/3,(a+b+c)/3,(a+b+c))`
B. `(a,b,c)`
C. `(a/2,b/2,c/2)`
D. `(a/3,b/3,c/3)`

Answer

To find the coordinates of the point that is equidistant from the four given points \((0,0,0)\), \((a,0,0)\), \((0,b,0)\), and \((0,0,c)\), we can follow these steps:

### Step-by-Step Solution:

1. **Define the Points**:
Let the four points be:
- \( O(0, 0, 0) \)
- \( A(a, 0, 0) \)
- \( B(0, b, 0) \)
- \( C(0, 0, c) \)

We need to find a point \( P(x, y, z) \) that is equidistant from all these points.

2. **Distance from Point O**:
The distance \( PO \) from point \( P \) to point \( O \) is given by the distance formula:
\[ PO = \sqrt{(x - 0)^2 + (y - 0)^2 + (z - 0)^2} = \sqrt{x^2 + y^2 + z^2} \]

3. **Distance from Point A**:
The distance \( PA \) from point \( P \) to point \( A \) is:
\[ PA = \sqrt{(x - a)^2 + (y - 0)^2 + (z - 0)^2} = \sqrt{(x - a)^2 + y^2 + z^2} \]

4. **Distance from Point B**:
The distance \( PB \) from point \( P \) to point \( B \) is:
\[ PB = \sqrt{(x - 0)^2 + (y - b)^2 + (z - 0)^2} = \sqrt{x^2 + (y - b)^2 + z^2} \]

5. **Distance from Point C**:
The distance \( PC \) from point \( P \) to point \( C \) is:
\[ PC = \sqrt{(x - 0)^2 + (y - 0)^2 + (z - c)^2} = \sqrt{x^2 + y^2 + (z - c)^2} \]

6. **Set Distances Equal**:
Since \( P \) is equidistant from all four points, we can set the distances equal to each other. Start with \( PO = PA \):
\[ \sqrt{x^2 + y^2 + z^2} = \sqrt{(x - a)^2 + y^2 + z^2} \]
Squaring both sides gives:
\[ x^2 + y^2 + z^2 = (x - a)^2 + y^2 + z^2 \]
Simplifying this, we find:
\[ x^2 = (x - a)^2 \]
Expanding the right side:
\[ x^2 = x^2 - 2ax + a^2 \]
This simplifies to:
\[ 0 = -2ax + a^2 \implies 2ax = a^2 \implies x = \frac{a}{2} \]

7. **Repeat for Other Distances**:
Now, set \( PO = PB \):
\[ \sqrt{x^2 + y^2 + z^2} = \sqrt{x^2 + (y - b)^2 + z^2} \]
Squaring both sides gives:
\[ x^2 + y^2 + z^2 = x^2 + (y - b)^2 + z^2 \]
This simplifies to:
\[ y^2 = (y - b)^2 \]
Expanding gives:
\[ y^2 = y^2 - 2by + b^2 \implies 0 = -2by + b^2 \implies 2by = b^2 \implies y = \frac{b}{2} \]

Finally, set \( PO = PC \):
\[ \sqrt{x^2 + y^2 + z^2} = \sqrt{x^2 + y^2 + (z - c)^2} \]
Squaring gives:
\[ x^2 + y^2 + z^2 = x^2 + y^2 + (z - c)^2 \]
This simplifies to:
\[ z^2 = (z - c)^2 \]
Expanding gives:
\[ z^2 = z^2 - 2cz + c^2 \implies 0 = -2cz + c^2 \implies 2cz = c^2 \implies z = \frac{c}{2} \]

8. **Final Coordinates**:
Thus, the coordinates of the point \( P \) that is equidistant from the four points are:
\[ P\left(\frac{a}{2}, \frac{b}{2}, \frac{c}{2}\right) \]