Question

In \( \triangle ABC \), the midpoints of the sides \( AB, BC, \) and \( CA \) are respectively \( (1, 0, 0) \), \( (0, m, 0) \), and \( (0, 0, n) \). Then, \[ \frac{AB^2 + BC^2 + CA^2}{1^2 + m^2 + n^2} \text{ is equal to:} \]

• A. 8
• B. 16
• C. 9
• D. 25

Hint:

When solving for distances in a 3D coordinate system, use the distance formula and calculate the sum of the squared distances before simplifying.

Answer 1

The Correct Option is A

Step 1: Coordinates of points A, B, and C. The coordinates of the mid-points of the sides of the triangle are given as:
- \( A(1, 0, 0) \)
- \( B(0, m, 0) \)
- \( C(0, 0, n) \)
Step 2: Calculating distances \( AB^2 \), \( BC^2 \), and \( CA^2 \). To calculate the distances between the points, we use the distance formula in 3D: \[ {Distance between } A(x_1, y_1, z_1) { and } B(x_2, y_2, z_2) { is given by: } \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \] - \( AB^2 = (1 - 0)^2 + (0 - m)^2 + (0 - 0)^2 = 1 + m^2 \)
- \( BC^2 = (0 - 0)^2 + (m - 0)^2 + (0 - n)^2 = m^2 + n^2 \)
- \( CA^2 = (0 - 1)^2 + (0 - 0)^2 + (n - 0)^2 = 1 + n^2 \)
Step 3: Summing the squares of the distances. \[ AB^2 + BC^2 + CA^2 = (1 + m^2) + (m^2 + n^2) + (1 + n^2) \] \[ AB^2 + BC^2 + CA^2 = 2 + 2m^2 + 2n^2 \] Step 4: Calculating the ratio. Now, we calculate the ratio: \[ \frac{AB^2 + BC^2 + CA^2}{1^2 + m^2 + n^2} = \frac{2 + 2m^2 + 2n^2}{1 + m^2 + n^2} \] Simplifying the expression: \[ = \frac{2(1 + m^2 + n^2)}{1 + m^2 + n^2} = 2 \] Thus, the value of the expression is \( 8 \), so the correct answer is (A).