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Worked solution

Prove the three medians meet at the centroid

Self-Reading: A Special Point of a Triangle — The Centroid. Choose one vertex of a triangle as origin OO, with the other vertices A,BA,B.

A median joins a vertex to the midpoint of the opposite side; AMAM is the median from AA. Let GG be the intersection of the medians from AA and BB.

(i) Express the position vector of GG in terms of the position vectors of A,BA,B.

(ii) Determine AG:GMAG:GM.

(iii) Show that GG also lies on the median from OO.

CJC · Vectors Basic · Example 8 (Self-Reading)

(i) Position of G

Answer

OG→=13a+13b\overrightarrow{OG}=\tfrac13\mathbf a+\tfrac13\mathbf b.

Name the midpoint positions

ON→=12a,OM→=12b\overrightarrow{ON}=\frac12\mathbf a,\qquad\overrightarrow{OM}=\frac12\mathbf b

Let OA=a, OB=b. N is halfway along OA and M halfway along OB. The triangle is nondegenerate, so a,b are nonparallel.

Describe the median from A

OG→=OA→+AG→=a+λAM→=a+λ(OM→−OA→)=a+λ(12b−a)=(1−λ)a+λ2b.\begin{aligned}\overrightarrow{OG}&=\overrightarrow{OA}+\overrightarrow{AG}\\&=\mathbf a+\lambda\overrightarrow{AM}\\&=\mathbf a+\lambda(\overrightarrow{OM}-\overrightarrow{OA})\\&=\mathbf a+\lambda(\tfrac12\mathbf b-\mathbf a)\\&=(1-\lambda)\mathbf a+\frac\lambda2\mathbf b.\end{aligned}

A,G,M collinear gives AG=λ AM. Expand the displacement AM using its endpoint positions.

Describe the median from B

OG→=OB→+BG→=b+μBN→=b+μ(ON→−OB→)=b+μ(12a−b)=μ2a+(1−μ)b.\begin{aligned}\overrightarrow{OG}&=\overrightarrow{OB}+\overrightarrow{BG}\\&=\mathbf b+\mu\overrightarrow{BN}\\&=\mathbf b+\mu(\overrightarrow{ON}-\overrightarrow{OB})\\&=\mathbf b+\mu(\tfrac12\mathbf a-\mathbf b)\\&=\frac\mu2\mathbf a+(1-\mu)\mathbf b.\end{aligned}

B,G,N collinear gives BG=μ BN, a second expression for the same position.

Compare coefficients and solve

1−λ=μ2,λ2=1−μ,μ2+λ=1,μ+λ2=1,λ=μ=23,OG→=13a+13b.\begin{gathered}1-\lambda=\frac\mu2,\quad\frac\lambda2=1-\mu,\\\frac\mu2+\lambda=1,\quad\mu+\frac\lambda2=1,\\\lambda=\mu=\frac23,\qquad\overrightarrow{OG}=\frac13\mathbf a+\frac13\mathbf b.\end{gathered}

The independent coefficients give two simultaneous equations; λ=μ=2/3 yields the stated position.

How did this part go?

(ii) Median ratio

Answer

AG:GM=2:1AG:GM=2:1.

Use the fraction of the whole median

AG→=λAM→=23AM→⟹AG:GM=2:1\overrightarrow{AG}=\lambda\overrightarrow{AM}=\frac23\overrightarrow{AM}\quad\Longrightarrow\quad AG:GM=2:1

AG is two thirds of AM; GM is the remaining third.

How did this part go?

(iii) The third median

Answer

OG→=23OR→\overrightarrow{OG}=\tfrac23\overrightarrow{OR}, so GG lies on median OROR.

Use the midpoint of AB

OR→=12(a+b)=12a+12b\overrightarrow{OR}=\frac12(\mathbf a+\mathbf b)=\frac12\mathbf a+\frac12\mathbf b

R is the midpoint of AB, so OR is the median from O.

Recover the same point on OR

OG→=13a+13b=23(12a+12b)=23OR→\begin{aligned}&\overrightarrow{OG}\\ & =\frac13\mathbf a+\frac13\mathbf b\\ & =\frac23\left(\frac12\mathbf a+\frac12\mathbf b\right)\\ & =\frac23\overrightarrow{OR}\end{aligned}

The scalar 2/3 is between zero and one. Thus G lies on the median segment, not merely its infinite line.

How did this part go?

Compare your sketch

Expand diagram

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Left: triangle OAB with M midpoint of OB, N midpoint of OA, R midpoint of AB; medians AM, BN, OR meet at G. Right: O is relabelled C in triangle ABC. G divides each median from the vertex in ratio 2:1. Matching one-, two- and three-tick pairs identify the three side midpoints.
Left: triangle OAB with M midpoint of OB, N midpoint of OA, R midpoint of AB; medians AM, BN, OR meet at G. Right: O is relabelled C in triangle ABC. G divides each median from the vertex in ratio 2:1. Matching one-, two- and three-tick pairs identify the three side midpoints.
Left: triangle OAB with M midpoint of OB, N midpoint of OA, R midpoint of AB; medians AM, BN, OR meet at G. Right: O is relabelled C in triangle ABC. G divides each median from the vertex in ratio 2:1. Matching one-, two- and three-tick pairs identify the three side midpoints.

Solution notes

Both copies print the full two-page proof. Relabelling O as C shows that all three medians of any nondegenerate triangle ABC meet at its centroid, two thirds of the way from each vertex to the opposite midpoint. For a uniform triangular lamina in a uniform gravitational field, the centroid is also its centre of gravity; the source’s physical remark is qualified by these assumptions.