Question

If \( A(0,0,0),\ B(3,4,0),\ C(0,12,5) \) are the vertices of a triangle ABC, then the x-coordinate of its incenter is:

• A. \( \dfrac{25}{18 + 7\sqrt{2}} \)
• B. \( \dfrac{25}{26} \)
• C. \( \dfrac{39}{18 + 7\sqrt{2}} \)
• D. \( \dfrac{39}{26} \)

Hint:

To find the incenter of a triangle in 3D, use the weighted average of the coordinates of vertices, where the weights are the lengths of the opposite sides.

Answer 1

The Correct Option is C

Step 1: Use Incenter Formula in 3D 
The incenter of triangle \( ABC \) in 3D is: \[ I = \left( \frac{aA_x + bB_x + cC_x}{a + b + c},\ \frac{aA_y + bB_y + cC_y}{a + b + c},\ \frac{aA_z + bB_z + cC_z}{a + b + c} \right) \] Where \( a = BC,\ b = AC,\ c = AB \) are the side lengths opposite vertices \( A, B, C \) respectively. 
Step 2: Compute Side Lengths 
AB: \[ AB = \sqrt{(3-0)^2 + (4-0)^2 + (0-0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \] BC: \[ BC = \sqrt{(0-3)^2 + (12-4)^2 + (5-0)^2} = \sqrt{9 + 64 + 25} = \sqrt{98} \] AC: \[ AC = \sqrt{(0-0)^2 + (12-0)^2 + (5-0)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \] Step 3: Use Incenter Formula for x-coordinate 
Using \( A_x = 0,\ B_x = 3,\ C_x = 0 \) \[ x = \frac{aA_x + bB_x + cC_x}{a + b + c} = \frac{\sqrt{98} \cdot 0 + 13 \cdot 3 + 5 \cdot 0}{\sqrt{98} + 13 + 5} = \frac{39}{18 + 7\sqrt{2}} \] (since \( \sqrt{98} = 7\sqrt{2} \)) 
Therefore, the x-coordinate of the incenter is \( \boxed{\dfrac{39}{18 + 7\sqrt{2}}} \)