Wednesday, May 11, 2011

Ceva using vectors--special case

Some time ago we looked at Ceva's theorem (post). I'm starting on a book about Vector Calculus, and saw this question early in Chapter 1.

Using vectors, prove that the lines from each vertex of a triangle to the midpoint of the opposite side cross at a single point. The picture is as shown below:

We solve this by constructing parametric equations for the midpoint lines. The first one starts at O and travels along the vector a + (b-a)/2, which is the diagonal of the parallelogram formed by a and b. The equation is:

u[a + (b-a)/2]

where u is the parameter. Similarly, construct the line extending from A to its opposing side. It starts from A and travels along the vector b/2 - a. The equation is:

a + v(b/2 - a)

At the point where the vectors cross, these are equal:

u[a + (b-a)/2] = a + v(b/2 - a)

I had a little trouble at this point, and I must confess I peeked at the answer. We can solve this by considering that both the a and the b terms must balance. (Even though a isn't perpendicular to b, the part of b which is perpendicular to a is a constant fraction k of the whole).

This leads to:

½ k ub = ½ k vb
u = v = ⅔

ua - ½ ua = a - va
½ u + v = 1
u = v = ⅔

Given this suggested solution u = v = ⅔ , we can easily verify that

u[a + (b-a)/2]
⅔[a + (b-a)/2]
= ⅔ a + ⅓ b - ⅓ a = ⅓ (a + b)

a + v(b/2 - a)
a + ⅔(b/2 - a)
= a + ⅓ b -⅔ a = ⅓ (a + b)

Now consider the third side. The midpoint line starts from B and travels along the vector a/2 - b. The equation is:

b + w(a/2 - b)

We observe that for w = ⅔ this becomes:

b + w(a/2 - b)
b + ⅔(a/2 - b)
b + ⅓ a - ⅔ b = ⅓ (a + b)

Thus, all three lines intersect at the same point.

I haven't got an extension to the general case just yet (i.e. any point, not just the midpoint).

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