Following is a really fun problem Tom Reardon showed my department last May as he led us through some TI-Nspire CAS training. Following the introduction of the problem, I offer a mea culpa, a proof, and an extension.

**THE PROBLEM:**

Take any square and construct midpoints on all four sides.

Connect the four midpoints and four vertices to create a continuous 8-pointed star as shown below. The interior of the star is an octagon. Construct this yourself using your choice of dynamic geometry software and vary the size of the square.

Compare the areas of the external square and the internal octagon.

You should find that **t he area of the original square is always 6 times the area of the octagon.**

I thought that was pretty cool. Then I started to play.

**MINOR OBSERVATIONS:**

Using my Nspire, I measured the sides of the octagon and found it to be equilateral.

As an extension of Tom’s original problem statement, I wondered if the constant square:octagon ratio occurred in any other quadrilaterals. I found the external quadrilateral was also six times the area of the internal octagon for parallelograms, but not for any more general quadrilaterals. Tapping my understanding of the quadrilateral hierarchy, that means the property also holds for rectangles and rhombi.

**MEA CULPA:**

Math teachers always warn students to never, ever assume what they haven’t proven. Unfortunately, my initial exploration of this problem was significantly hampered by just such an assumption. I obviously know better (and was reminded afterwards that Tom actually had told us that the octagon was *not *equiangular–but like many students, I hadn’t listened). After creating the original octagon, measuring its sides and finding them all equivalent, I errantly assumed the octagon was regular. That isn’t true.

That false assumption created flaws in my proof and generalizations. I discovered my error when none of my proof attempts worked out, and I eventually threw everything out and started over. I knew better than to assume. But I persevered, discovered my error through back-tracking, and eventually overcame. That’s what I really hope my students learn.

**THE REAL PROOF:**

*Goal*: Prove that the area of the original square is always 6 times the area of the internal octagon.

Assume the side length of a given square is , making its area .

The octagon’s area obviously is more complicated. While it is not regular, the square’s symmetry guarantees that it can be decomposed into four congruent kites in two different ways. Kite *AFGH* below is one such kite.

Therefore, the area of the octagon is 4 times the area of *AFGH*. One way to express the area of any kite is , where and are the kite’s diagonals. If I can determine the lengths of and , then I will know the area of *AFGH* and thereby the ratio of the area of the square to the area of the octagon.

The diagonals of every kite are perpendicular, and the diagonal between a kite’s vertices connecting its non-congruent sides is bisected by the kite’s other diagonal. In terms of *AFGH*, that means is the perpendicular bisector of .

The square and octagon are concentric at point *A, *and point *E* is the midpoint of , so is isosceles with vertex *A*, and is the perpendicular bisector of .

That makes right triangles . Because , similarity gives . I know one side of the kite.

Let point *I* be the intersection of the diagonals of *AFGH*. is right isosceles, so is, too, with degrees. With , the Pythagorean Theorem gives . Point *I* is the midpoint of , so . One kite diagonal is accomplished.

Construct . Assuming degree angle measures, if , then and . Knowing two angles of gives the third: .

I need the length of the kite’s other diagonal, , and the Law of Sines gives

, or

.

Expanding using cofunction and angle sum identities gives

From right , I also know and . Therefore, , and the kite’s second diagonal is now known.

So, the octagon’s area is four times the kite’s area, or

Therefore, the ratio of the area of the square to the area of its octagon is

.

*QED*

**EXTENSIONS:**

This was so nice, I reasoned that it couldn’t be an isolated result.

I have extended and proved that the result is true for other modulo-3 stars like the 8-pointed star in the square for any *n*-gon. I’ll share that very soon in another post.

I proved the result above, but I wonder if it can be done without resorting to trigonometric identities. Everything else is simple geometry. I also wonder if there are other more elegant approaches.

Finally, I assume there are other constant ratios for other modulo stars inside larger *n*-gons, but I haven’t explored that idea. Anyone?

I have a proof for rectangle/octagon that does not use trig at all (property of centroid and similar triangles). I will check if it works for parallelogram.

We had this problem queued up for our blog, but you beat us to it. Our solution uses “area of a triangle” and not much more:

https://docs.google.com/document/d/1AvrWFOf1TeUhaJPFDJLH2BNCKu2QLSIsorjngJ6JoB0/edit

Yours is a MUCH lovelier response. Nice. My solution clearly betrays my serious algebraic bias…

I’m typesetting Tatiana’s response using triangles, similarity, and centroids.

The generalized solution is going to be nice, too.

Here’s my solution. http://gotmath.com/doc/octagon.pdf

Dave, your solution is almost too neat!

Dave, I should have replied long ago, but Simon is right … I love the beautiful symmetry of your solution. Thanks for the beautiful contribution.

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