Lessons taught; Lessons learnt

I was browsing through the Nrich website yesterday, and found this interesting problem:

If you are given the coordinates of the midpoints of the edges of a pentagon, can you find the coordinates of the vertices of the pentagon?

This was, apparently, originally given as a question in an Oxford extrance paper from 1926!

If you follow the link, you’ll see that the Nrich site gets you to investigate this problem with pentagons, triangles, and quadrilaterals, using the excellent free dynamic geometry app Geogebra. I use Geogebra regularly in my teaching, so I’m happy to see it appearing in well-known sites like Nrich.

I’ll let you think about this problem by yourself (Nrich gives the hint that you should think about simultaneous equations), and will focus on just one aspect of it in this post…

Midpoints of a Quadrilateral

It’s very easy to convince yourself by messing around with the dynamic geometry that it’s always possible to find a pentagon or a triangle with any given midpoints, but that quadrilaterals are more tricky. Indeed, the default configuration of midpoints presented to you seems to be impossible.

Why is this? Perhaps there is a special property which needs to be satisfied by the midpoints of quadrilaterals. A great way to look at this is to set up the situation in a dynamic geometry package, and see if anything interesting suggests itself for further investigation:

Geogebra applet will be here if you enable Java.

(Source: midpoints_of_a_quadrilateral)

It appears that the midpoints always form a parallelogram, and that this holds even when we don’t have a ‘proper’ quadrilateral, but a bow-tie shape.

How could we prove this? One route is to use vector arithmetic.

Using Vectors

Let’s call the four points of the quadrilateral P₁, P₂, P₃, and P₄, and the four midpoints M₁, M₂, M₃, and M₄.

A sample quadrilateral with midpoints

We have

M₁ = ½ ( P₁ + P₂ )
M₂ = ½ ( P₂ + P₃ )
M₃ = ½ ( P₃ + P₄ )
M₄ = ½ ( P₄ + P₁ )

Now, we want to show that the quadrilateral formed by the midpoints is a parallelogram. We can do this if we show that the vector that takes us from M₁ to M₂ (M₂ – M₁) is the same as the vector that takes us from M₄ to M₃ (M₃ – M₄), and that the vector that takes us from M₂ to M₃ is the same as the vector that takes us from M₁ to M₄. Both of these come from the observation that

M₁ + M₃ = ½ ( P₁ + P₂ + P₃ + P₄ ) = M₂ + M₄,

from which we can conclude that M₂ – M₁ = M₃ – M₄ and M₃ – M₂ = M₄ – M₁. M₁M₂M₃M₄ is therefore a parallelogram.

Alternative Proofs and Further Questions

This seems like the type of property which could be proved in many different ways. While I like using vectors, the traditional way to prove something like this would be to use synthetic (Euclidean) geometry. I’ve never been particularly great at this, but if someone has a simple proof in this style, I’d love to see it.

This also raises several further questions:

• When is the parallelogram a rectangle?
• When is the parallelogram a square?
• What happens for hexagons?

Next in this series

Midpoints (2): Pentagons

Related Posts (automatically generated)

1. Midpoints (3): A return to quadrilaterals
2. Midpoints (2): Pentagons
3. MA Conference: A circular NRICHing activity