(a) AG=32AM and GM=31AM. (b) The same argument applies cyclically to all three vertices in any dimension.
Worked solution
(a) m=(b+c)/2. Then
g−a=(b+c−2a)/3=32(m−a).
The scalar 2/3 lies between 0 and 1, so G lies on the segment AM. The remaining segment is 1/3 of AM, yielding the ratio.
(b) The expression for g is unchanged by permuting A, B and C. Repeating (a) for the midpoints of AC and AB shows G belongs to both remaining medians. Vector addition and scalar multiplication used here do not require z=0.
Equivalent justified methods accepted; respect any requested proof method.
Exact values unless specified. Sketches assessed by mathematical features, not artistic quality.
Part a: Write the midpoint vector.
[1 mark]
Part a: Show g−a=32(m−a).
[1 mark]
Part a: Infer segment membership and the ratio.
[1 mark]
Part b: Apply the symmetric argument to the other medians.
[1 mark]
Part b: Explain that the argument is independent of the plane orientation.
[1 mark]
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.