The first time I recall encountering this problem was about 10-15 years ago when I was reading David Wells’ *Curious and Interesting Geometry*. Not having taught geometry for several years, I’d forgotten about it until I encountered it again last week in Paul Lockhart’s *Measurement*.

My statement of the problem:

Draw any quadrilateral with non-intersecting sides. Place a point at the midpoint of each side. Connect the four midpoints in clockwise order. 1) What shape does the resulting quadrilateral

alwaysassume? 2) How does the area of the new quadrilateral compare to the area of the original quadrilateral? Of course, you need to prove thy claims.

Draw several different quadrilaterals in your investigation. So long as the sides don’t overlap, nothing else matters. The pattern will emerge. I love the stunning and unexpected emergence of order.

Don’t read any further until you’ve played with this for yourself. The joy of mathematical discovery is worth it! Above all, give yourself and your students lots of time to explore. Don’t be too quick to offer suggestions.

**PARTIAL SOLUTION ALERT! DON’T READ FURTHER UNTIL YOU’VE PLAYED WITH THE QUADRILATERAL PROBLEM ABOVE!**

I suggest using TI-nSpire, Geogebra, Geometer’s Sketchpad, or some other dynamic geometry software to model this. Using my TI, I was able to quickly explore an entire spectrum of results. Following are two representative images.

No matter what type of non-overlapping quadrilateral you draw, a parallelogram *always* seems to emerge. I thought a while through various approaches to discover an elegant way to prove this, and in the process discovered the area solution. Find your own proof before reading further.

**FINAL SOLUTION ALERT! DON’T READ FURTHER UNTIL YOU HAVE YOUR OWN SOLUTION.**

My great insight happened when I imagined a diagonal drawn in a quadrilateral, splitting the original into two triangles. That insight reminded me of a cool triangle property: In any triangle, if you connect the midpoints of two sides, the resulting segment is parallel to and half the length of the third side.

In any quadrilateral ABCD, let W, X, Y, and Z be the respective midpoints of segments AB, BC, CD, and AD. Draw diagonal AC of ABCD, creating triangles ABC and ACD.

By the triangle property noted above, segments WX and ZY are each parallel to and half the length of WXYZ. That is sufficient to establish that WXYZ is a parallelogram. QED.

Two triangles sharing the same base doesn’t seem like a condition imposing lots of order, but that is just enough to lay the inevitable conditions for creating a highly structured parallelogram. The emergence of a highly ordered parallelogram from a seemingly random quadrilateral was inevitable! As one of my students said, “Math works.”

**To establish the area condition**, I offer a proof without words.

I’d love to hear how any of you approach the problem. I’ll post any responses.

Chris,

This is awesome. What did you use to produce that video? Geogebra?

It could be done in Geogebra, but I used my TI-nSpire software just because it happened to already be active on my computer. I knew I wanted to rotate two of the triangles and translate the others. The fun part was figuring out how to make that happen slowly enough for others to follow. If you’re interested, I could send you the .tns file or recreate it in Geogebra and send you that.