## The Shipwrecked Surfer

16 Oct

I had saved this task from the August 2006 Mathematics Teacher to use at some point. One of the authors has a link to the article/paper here.

The problem:

A surfer, shipwrecked on an island in the shape of an equilateral triangle, wants to build a hut so that the sum of its distances to the three beaches is minimal. Where should the hut be located? (See Figure 1.)

I like how the problem is worded; students first have to make sense of the problem, and most have to persevere in solving it. I also like that students have to think about what we mean when we say the distance from a point (hut) to a line (beach).

Students started their work on paper. Many students chose the centroid as the point of interest. Others chose randomly. They used rulers and compasses to test their conjectures. One student figured out pretty quickly that it didn’t matter where you put the point. It doesn’t matter where you put the point? No – the sum of the distances from the point to the sides is the same no matter where you put the point.

We moved to our dynamic geometry software. We placed a point inside the triangle. We constructed the perpendicular from the point to each side. We measured the length of each segment on the perpendicular from the point to the side. We calculated the sum of the lengths. We moved the point around inside the triangle and verified that the sum is the same no matter where we put the point.

So we can put the point anywhere. Now what? What’s the significance of placing the point anywhere? What’s the significance of the sum of the distances?

What’s significant about the current location of the point of interest for this student?

Someone figured out that the sum of the distances equals the length of the altitude. Is that important? The altitude is the length of the segment perpendicular from any vertex to the opposite side.

Look what happens if we redefine our point of interest on the altitude. What type of polygon is IBXC? Compare the value of XI to the sum of CI and BI.

Somewhere around here the bell rang.

And that was it. We haven’t revisited the task, although a few students reflected on the task as part of their problem solving points for this quarter:

“The addition of the length to the sides from any random point in a triangle is equal to the altitude because the altitude is exactly that. If you put a point at the top corner of a triangle and add the distance from there to all three sides you get the altitude. The altitude is just adding all the distances to the sides at a certain point.”

“The entire reason the point inside the triangle was equidistance from the sides was because the triangle was equilateral. I’m not sure if this is more complicated than it seems, but if all of the angles are the same, all the sides are the same, and any way you rotate and turn the triangle it still looks the same, then a point inside the triangle is going to be equidistance from the sides because every point on the plane is equidistance. Think about it this way: If I move my point .2 cm to the left, then the distance from the side on its left is shortened by .2 cm, but the distance from the side on right is lengthened by.2 cm. Up, down, left, right, every distance’s total is equal.”

One of the best parts about teaching is that we get “do-overs”. Not always this year and with these students…but definitely next year with new students. How could I make this conversation more meaningful – and get to the mathematics sooner? I could pose the task the class period before. Students can do their paper work on the task outside of class – and come into class ready to share their conjectures. That would get us to the technology sooner – and to the deep exploration for which the technology provides the opportunity.

And so the journey continues…