Leapfrog Investigation

This is an investigation task accessible to students as early as year 7, but requires vectors to be able to prove the result.

The investigation

Choose a starting point and three other points, called A, B, and C.

From the starting point, leapfrog over point A. By that, I mean land on the point P1 such that A is the midpoint of P1 and the starting point.

Then leapfrog from that point over B. Then C, A, B and C.

Where will you finish?

I encourage students to draw it out first, stressing the importance of accuracy. It makes for a good measuring and constructions with a ruler task for lower years. Below is an interactive version for the board. Move the points around, then use the "Jumps" slider to show the jumps.

For those that have studied vectors, a further challenge would be to ask them to prove the result.

Aditionally, does this hold true in 3D?

Want to send this to students and let them play around with it? Here's a link.

It bugs me when resources don't have solutions, so here's the proof!

The proof

Let points $A$, $B$ and $C$ have position vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$, respectively, treating the starting point as the origin without loss of generality.

After the first leapfrog jump, we have a position vector of $2\mathbf{a}$.

After the second leapfrog jump, we have a position vector of $2\mathbf{a}+2(\mathbf{b}-2\mathbf{a})=2\mathbf{b}-2\mathbf{a}$.

After the third leapfrog jump, we have a position vector of $2\mathbf{b}-2\mathbf{a}+2(\mathbf{c}-(2\mathbf{b}-2\mathbf{a}))=2\mathbf{a}-2\mathbf{b}+2\mathbf{c}$.

After the fourth: $2\mathbf{b}-2\mathbf{c}$.

After the fifth: $2\mathbf{c}$.

After the sixth: $2\mathbf{c}+2(\mathbf{c}-2\mathbf{c})=\mathbf{0}$.

The resultant vector is $\mathbf{0}$, so we finish where we started. Furthermore, we did not define the dimension of the vector, so this works in any number of dimensions (including 1D and 3D).