Squares and Octagons, A compilation

My last post detailed my much-too-long trigonometric proof of why the octagon formed by connecting the midpoints and vertices of the edges of a square into an 8-pointed star is always 1/6 of the area of the original square.


My proof used trigonometry, and responses to the post on Twitter  and on my ‘blog showed many cool variations.  Dave Radcliffe thought it would be cool to have a compilation of all of the different approaches.  I offer that here in the order they were shared with me.

Method 1:  My use of trigonometry in a square.  See my original post.

Method 2:  Using medians in a rectangle from Tatiana Yudovina, a colleague at Hawken School.

Below, the Area(axb rectangle) = ab = 16 blue triangles, and
Area(octagon) = 4 blue triangles – 2 red deltas..


Now look at the two green, similar triangles.  They are similar with ratio 1/2, making

Area(red delta) = \displaystyle \frac{b}{4} \cdot \frac{a}{6} = \frac{ab}{24}, and

Area(blue triangle) = \displaystyle \frac{1}{16} ab

So, Area(octagon) = \displaystyle 2 \frac{ab}{24}-4\frac {ab}{16}=\frac{1}{6}ab.


Method 3:  Using differences in triangle areas in a square (but easily extended to rectangles)from @Five_Triangles (‘blog here).

Full solution here.

Method 4:  Very clever shorter solution using triangle area similarity in a square also from @Five_Triangles (‘blog here).

Full second solution here.

Method 5:  Great option Using dilated kitesfrom Dave Radcliffe posting as @daveinstpaul.

Full pdf and proof here.

Method 6:  Use fact that triangle medians trisect each other from Mike Lawler posting as @mikeandallie.

Tweet of solution here.

Method 7:  Use a coordinate proof on a specific square from Steve Ingrassia, a colleague at Hawken School.  Not a quick proof like some of the geometric solutions, but it’s definitely different than the others.

If students know the formula for finding the area of any polygon using its coordinates, then they can prove this result very simply with nothing more than simple algebra 1 techniques.   No trig is required.

The area of polygon with vertices (in either clockwise or counterclockwise order, starting at any vertex) of (x_1, y_1), (x_2, y_2), …, (x_n, y_n) is

\displaystyle Area = \left| \frac{(x_1y_2-x_2y_1)+(x_2y_3-x_3y_2)+...+(x_{n-1}y_n-x_ny_{n-1})}{2} \right|

Use a 2×2 square situated with vertices at (0,0), (0,2), (2,2), and (2,0).  Construct segments connecting each vertex with the midpoints of the sides of the square, and find the equations of the associated lines.

  • L1 (connecting (0,0) and (2,1):    y = x/2
  • L2 (connecting (0,0) and (1,2):   y=2x
  • L3 (connecting (0,1) and (2,0):  y= -x/2 + 1
  • L4 (connecting (0,1) and (2,2):  y= x/2 + 1
  • L5 (connecting (0,2) and (1,0):  y = -2x + 2
  • L6 (connecting (0,2) and (2,1):  y= -x/2 + 2
  • L7 (connecting (1,2) and (2,0):  y = -2x + 4
  • L8 (connecting (2,2) and (1,0):  y = 2x – 2

The 8 vertices of the octagon come at pairwise intersections of some of these lines, which can be found with simple substitution:

  • Vertex 1 is at the intersection of L1 and L3:   (1, 1/2)
  • Vertex 2 is at the intersection of L3 and L5:  (2/3, 2/3)
  • Vertex 3 is at the intersection of L2 and L5:  (1/2, 1)
  • Vertex 4 is at the intersection of L2 and L4:  (2/3, 4/3)
  • Vertex 5 is at the intersection of L4 and L6:  (1, 3/2)
  • Vertex 6 is at the intersection of L6 and L7:  (4/3, 4/3)
  • Vertex 7 is at the intersection of L7 and L8:  (3/2, 1)
  • Vertex 8 is at the intersection of L1 and L8:  (4/3, 2/3)

Using the coordinates of these 8 vertices in the formula for the area of the octagon, gives

\displaystyle \frac{ \left| 1/3 +1/3+0+(-1/3)+(-2/3)+(-1/3)+0 \right|}{2} = \frac{2}{3}

 Since the area of the original square was 4, the area of the octagon is exactly 1/6th of the area of the square.

Thanks, everyone, for your contributions.


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