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).