The Ratio Theorem

A point in a given ratio

Let A and B have position vectors a and b. "P divides AB in the ratio 2 : 1" means AP : PB = 2 : 1: cut AB into three equal parts, and P sits two parts from A and one part from B.

With a = (7, 1) and b = (1, 7), P is the point (3, 5).

Walking there

Go from O to A, then two thirds of the way along AB. AB = b − a, so OP $=a+\left(\frac{2}{3}\right)\left(b-a\right)$= a + (2/3)(b − a).

Collect the a terms and the b terms: OP $=\left(\frac{1}{3}\right)a+\left(\frac{2}{3}\right)b=\frac{a+2b}{3}$= (1/3)a + (2/3)b = (a + 2b)/3. With the numbers, $\frac{a+2b}{3}=\frac{7+2,1+14}{3}=\frac{9,15}{3}=\left(3,5\right)$(a + 2b)/3 = (7 + 2, 1 + 14)/3 = (9, 15)/3 = (3, 5).

Notice which end gets which weight. b gets 2, which is the part from A to P, and a gets 1, the part from P to B. Each end is weighted by the part on the far side of P, so the nearer end gets the larger share.

The ratio theorem

The same walk with AP : PB $=\lambda$= λ : $\mu$μ gives OP $=a+\left(\frac{\lambda }{\lambda +\mu }\right)\left(b-a\right)=\frac{\mu a+\lambda b}{\lambda +\mu }$= a + (λ/(λ + μ))(b − a) = (μa + λb)/(λ + μ). This is the ratio theorem.

The two weights, $\frac{\mu }{\lambda +\mu }$μ/(λ + μ) and $\frac{\lambda }{\lambda +\mu }$λ/(λ + μ), add to 1. That is why the sum is divided by $\lambda +\mu$λ + μ. With $\lambda =\mu =1$λ = μ = 1 the theorem gives $\frac{a+b}{2}$(a + b)/2, the midpoint.

The centroid

The centroid G of triangle ABC lies on each median, two thirds of the way from the vertex. The median from A goes to the midpoint M of BC, where OM $=\frac{b+c}{2}$= (b + c)/2.

G divides AM in the ratio 2 : 1, so OG $=\frac{1\times a+\frac{2\times \left(b+c\right)}{2}}{3}=\frac{a+b+c}{3}$= (1 × a + 2 × (b + c)/2)/3 = (a + b + c)/3. The other two medians give the same point, which is why the three medians meet.

The usual slips

Swapping the weights. Writing $\frac{\lambda a+\mu b}{\lambda +\mu }$(λa + μb)/(λ + μ) gives the point dividing AB in the ratio $\mu$μ : $\lambda$λ, nearer the other end.

Forgetting to divide by $\lambda +\mu$λ + μ. $\mu a+\lambda b$μa + λb on its own is $\left(\lambda +\mu \right)\times$(λ + μ) × OP, a point $\lambda +\mu$λ + μ times as far from O as P.

Reading $\lambda$λ : $\mu$μ as a part against the whole. In 2 : 1, AP is $\frac{2}{3}$2/3 of AB, not $\frac{2}{1}$2/1.