Tag Archives: coordinate proof

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.

octagon1

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..

octagon6

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.

QED

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

octagon8
Full solution here.

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

octagon9
Full second solution here.

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

octagon7
Full pdf and proof here.

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

octagon10
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.

Midpoints, midpoints, everywhere!

I didn’t encounter the Quadrilateral Midpoint Theorem (QMT) until I had been teaching a few years.  Following is a minor variation on my approach to the QMT this year plus a fun way I leveraged the result to introduce similarity.

In case you haven’t heard of it, the surprisingly lovely QMT says that if you connect, in order, the midpoints of the four sides of a quadrilateral–any quadrilateral–even if the quadrilateral is concave or if its sides cross–the resulting figure will always be a parallelogram.

Parallel1

Parallel2

Parallel3

This is a cool and easy property to explore on any dynamic geometry software package (GeoGebra, TI-Nspire, Cabri, …).

SKETCH OF THE TRADITIONAL PROOF:  The proof is often established through triangle similarity:  Whenever you connect the midpoints of two sides of a triangle, the resulting segment will be parallel to and half the length of the triangle’s third side.  Draw either diagonal in the quadrilateral to create two triangles.  Connecting the midpoints of the other two sides of each triangle creates two congruent parallel sides, so the quadrilateral connecting all four midpoints must be a parallelogram.

NEW APPROACH THIS YEAR:  I hadn’t yet led my class into similarity, but having just introduced coordinate proofs, I tried an approach I’d never used before.  I assigned a coordinate proof of the QMT.  I knew the traditional approach existed, but I wanted them to practice their new technique.  From a lab in December, they already knew the result of the QMT, but they hadn’t proved it.

PART I:  Let quadrilateral ABCD be defined by the points , A=(a,b), B=(c,d), C=(e,f),  and D=(g,h).  There are several ways to prove that the midpoints of ABCD are the vertices of a parallelogram.  Provide one such coordinate proof.

All groups quickly established the midpoints of the four sides:  AB_{mid}=\left( \frac{a+c}{2},\frac{b+d}{2} \right)BC_{mid}=\left( \frac{c+e}{2},\frac{d+f}{2} \right)CD_{mid}=\left( \frac{e+g}{2},\frac{f+h}{2} \right), and DA_{mid}=\left( \frac{g+a}{2},\frac{h+b}{2} \right).  From there, my students took three approaches to the final proof, each relying on a different sufficiency condition for parallelograms.

The most common was to show that opposite sides were parallel.  \displaystyle slope \left( AB_{mid} \text{ to } BC_{mid} \right) = \frac{\frac{a-e}{2}}{\frac{b-f}{2}}=\frac{a-e}{b-f} and \displaystyle slope \left( CD_{mid} \text{ to } DA_{mid} \right) =\frac{a-e}{b-f}, making those two midpoint segments parallel.  Likewise, \displaystyle slope \left( BC_{mid} \text{ to } CD_{mid} \right) = \displaystyle slope \left( DA_{mid} \text{ to } AB_{mid} \right) = \frac{c-g}{d-h}, proving the other opposite side pair also was parallel.  With both pairs of opposite sides parallel, the midpoint quadrilateral was necessarily a parallelogram.

I had two groups leverage the fact that the diagonals of parallelograms were mutually bisecting.    \displaystyle midpoint \left( AB_{mid} \text{ to } CD_{mid} \right) = \left( \frac{a+c+e+g}{4},\frac{b+d+f+h}{4}\right) = = midpoint \left( BC_{mid} \text{ to } DA_{mid} \right).  QED.

One student even proved that opposite sides were congruent.

While it was not readily available for my students this year, I can imagine allowing CAS for these manipulations if I use this activity in the future.

EXTENDING THE QMT TO SIMILARITY:  For the next stage, I asked my students to explains what happens when the QMT is applied to degenerate quadrilaterals.

PART II:  You could think of triangles as being degenerate quadrilaterals when two quadrilateral vertices coincide to make one side of the quadrilateral have side length 0.  Apply this to generic quadrilateral ABCD from above where points A and D coincide to create triangle BCD.  Use this to explain how the segment connecting the midpoints of any two sides of a triangle is related to the third side of the triangle.

I encourage you to construct this using a dynamic geometry package, but here’s the result.

Parallel4

 

Heres a brief video showing the quadrilateral going degenerate.

Notice the parallelogram still exists and forms two midpoint segments on the triangle (degenerate quadrilateral).  By parallelogram properties, each of these segments is parallel and congruent to the opposite side of the parallelogram, making them parallel to and half the length of the opposite side of the triangle.

CONCLUSION:  I think it critical to teach in a way that draws connections between ideas and units. This exercise made a lovely transition from quadrilaterals through coordinate proofs to the triangle midpoint theorem.