Tag Archives: square

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.

Squares and Octagons

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.


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 the 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.


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.


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.


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 2x, making its area 4x^2.

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 \frac{1}{2}D_1\cdot D_2, where D_1 and D_2 are the kite’s diagonals. If I can determine the lengths of \overline{AG} and \overline {FH}, 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 \overline{AG} is the perpendicular bisector of \overline{FH}.

The square and octagon are concentric at point A, and point E is the midpoint of \overline{BC}, so \Delta BAC is isosceles with vertex A, and \overline{AE} is the perpendicular bisector of \overline{BC}.

That makes right triangles \Delta BEF \sim \Delta BCD.  Because \displaystyle BE=\frac{1}{2} BC, similarity gives \displaystyle AF=FE=\frac{1}{2} DC=\frac{x}{2}.  I know one side of the kite.

Let point I be the intersection of the diagonals of AFGH.  \Delta BEA is right isosceles, so \Delta AIF is, too, with m\angle{IAF}=45 degrees.  With \displaystyle AF=\frac{x}{2}, the Pythagorean Theorem gives \displaystyle IF=\frac{x}{2\sqrt{2}}.  Point I is the midpoint of \overline{FH}, so \displaystyle FH=\frac{x}{\sqrt{2}}.  One kite diagonal is accomplished.


Construct \overline{JF} \parallel \overline{BC}.  Assuming degree angle measures, if m\angle{FBC}=m\angle{FCB}=\theta, then m\angle{GFJ}=\theta and m\angle{AFG}=90-\theta.  Knowing two angles of \Delta AGF gives the third:  m\angle{AGF}=45+\theta.


 I need the length of the kite’s other diagonal, \overline{AG}, and the Law of Sines gives

\displaystyle \frac{AG}{sin(90-\theta )}=\frac{\frac{x}{2}}{sin(45+\theta )}, or

\displaystyle AG=\frac{x \cdot sin(90-\theta )}{2sin(45+\theta )}.

Expanding using cofunction and angle sum identities gives

\displaystyle AG=\frac{x \cdot sin(90-\theta )}{2sin(45+\theta )}=\frac{x \cdot cos(\theta )}{2 \cdot \left( sin(45)cos(\theta ) +cos(45)sin( \theta) \right)}=\frac{x \cdot cos(\theta )}{\sqrt{2} \cdot \left( cos(\theta ) +sin( \theta) \right)}

From right \Delta BCD, I also know \displaystyle sin(\theta )=\frac{1}{\sqrt{5}} and \displaystyle cos(\theta)=\frac{2}{\sqrt{5}}.  Therefore, \displaystyle AG=\frac{x\sqrt{2}}{3}, and the kite’s second diagonal is now known.

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

\displaystyle 4\left( \frac{1}{2} D_1 \cdot D_2 \right) = 2FH \cdot AG = 2 \cdot \frac{x}{\sqrt{2}} \cdot \frac{x\sqrt{2}}{3} = \frac{2}{3}x^2

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

\displaystyle \frac{area_{square}}{area_{octagon}} = \frac{4x^2}{\frac{2}{3}x^2}=6.



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?

Two Squares, Two Triangles, and some Circles

Here’s another fun twist on another fun problem from the Five Triangles ‘blog.  A month ago, this was posted.

What I find cool about so many of the Five Triangles problems is that most permit multiple solutions.  I also like that several Five Triangles problems initially appear to not have enough information.  This one is no different until you consider the implications of the squares.

I’ve identified three unique ways to approach this problem.  I’d love to hear if any of you see any others.  Here are my solutions in the order I saw them.  The third is the shortest, but all offer unique insights.

Method 1: Law of Cosines

This solution goes far beyond the intended middle school focus of the problem, but it is what I saw first.  Sometimes, knowing more gives you additional insights.

Because DEF is a line and EF is a diagonal of a square, I know m\angle CEF=45^{\circ}, and therefore m\angle CED=135^{\circ}.  \Delta CEF is a 45-45-90 triangle with hypotenuse 6, so its leg, CE has measure \frac{6}{\sqrt{2}}=3\sqrt{2}.  Knowing two sides and an angle in \Delta DEC means I could apply the Law of Cosines.

DC^2 = 4^2 + (3\sqrt{2})^2 - 2\cdot (3\sqrt{2}) \cdot \cos(135^{\circ})=58

Because I’m looking for the area of ABCD,  and that is equivalent to DC^2, I don’t need to solve for the length of DC to know the area I seek is 58.

Method 2: Use Technology

I doubt many would want to solve using this approach, but if you don’t see (or know) trigonometry, you could build a solution from scratch if you are fluent with dynamic geometry software (GeoGebra, TI-Nspire, GSP).  My comfort with this made finding the solution via construction pretty straight-forward.

  1. Construct segment EF with fixed length 6.
  2. Build square CEGF with diagonal EF.  (This can be done several ways.  I was in a transformations mood, so I rotated EF 90^{\circ} to get the other endpoints.)
  3. Draw line EF  and then circle with radius 4 through point E.
  4. Mark point D as the intersection of circle and line EF outside CEGF .
  5. Draw a segment through points and C.  (The square of the length of CD is the answer, but I decided to go one more step.)
  6. Construct square ABCD with sides congruent to CD.  (Again, there are several ways to do this.  I left my construction marks visible in my construction below.)
  7. Compute the area of ABCD.

Here is my final GeoGebra construction.


Method 3: The Pythagorean Theorem

Sometimes, changing a problem can make it much easier to solve.

As soon as I saw the problem, I forwarded it to some colleagues at my school.  Tatiana wrote back with a quick solution.  In the original image, draw diagonal, CG, of square CEGF. Because the diagonals of a square perpendicularly bisect each other, that creates right \Delta DHC with legs 3 and 7.  That means the square of the hypotenuse of \Delta DHC (and therefore the area of the square) can be found via the Pythagorean Theorem.

DC^2 = 7^2+3^2 = 58

Method 4: Coordinate Geometry

OK, I said three solutions, and perhaps this approach is completely redundant given the Pythagorean Theorem in the last approach, but you could also find a solution using coordinate geometry.

Because the diagonals of a square are perpendicular, you could construct ECFG with its center at the origin.  I placed point C at (0,3) and point E at (3,0).  That means point D is at (7,0), making the solution to the problem the square of the length of the segment from (0,3) to (7,0).  Obviously, that can be done with the Pythagorean Theorem, but in the image below, I computed number i in the upper left corner of this GeoGebra window as the square of the length of that segment.



Area 10 Squares – Proof & Additional Musings

Additional musings on the problem of Area 10 Squares:

Thanks, again to Dave Gale‘s inspirations and comments on my initial post. For some initial clarifications, what I was asking in Question 3 was whether these square areas ultimately can all be found after a certain undetermined point, thereby creating a largest area that could not be drawn on a square grid. I’m now convinced that the answer to this is a resounding NO–there is no area after which all integral square areas can be constructed using square grid paper. This is because there is no largest un-constructable area (proof below). This opens a new question.

Question 4:
Is there some type of 2-dimensional grid paper which does allow the construction of all square areas?

The 3-dimensional version of this question has been asked previously, and this year in the College Math Journal, Rick Parris of Exeter has “proved that if a cube has all of its vertices in then the edge length is an integer.”

Dave’s proposition above about determining whether an area 112 (or any other) can be made is very interesting. (BTW, 112 cannot be made.) I don’t have any thoughts at present about how to approach the feasibility of a random area. As a result of my searches, I still suspect (but haven’t proven) that non-perfect square multiples of 3 that aren’t multiples of pre-existing squares seem to be completely absent. This feels like a number theory question … not my area of expertise.

Whether or not you decide to read the following proof for why there are an infinite number of impossible-to-draw square areas using square grids, I think one more very interesting question is now raised.

Question 5:
Like the prime numbers, there is an infinite number of impossible-to-draw square areas. Is there a pattern to these impossible areas? (Remember that the pattern of the primes is one of the great unanswered questions in all of mathematics.)

My proof does not feel the most elegant to me. But I do like how it proves the infinite nature of these numbers without ever looking at the numbers themselves. It works by showing that there are far more integers than there are ways to arrange them on a square grid, basically establishing that there is simply not enough room for all of the integers forcing some to be impossible. I don’t know the formal mathematics name for this principle, but I think of it as a reverse Pigeonhole Principle. Rather than having more pigeons than holes (guaranteeing duplication), in this case, the number of holes (numbers available to be found) grows faster than the number of pigeons (the areas of squares that can actually be determined on a square grid), guaranteeing that there will always be open holes (areas of squares that cannot be determined on using a square grid).

This exploration and proof far exceeds most (all?) textbooks, but the individual steps require nothing more than the ability to write an equation for an exponential function and find the sum a finite arithmetic sequence. The mathematics used here is clearly within the realm of what high school students CAN do. So will we allow them to explore, discover, and prove mathematics outside our formal curricula? I’m not saying that students should do THIS problem (although they should be encouraged in this direction if interested), but they must be encouraged to do something real to them.

Now on to a proof for why there must be an infinite number of impossible-to-draw square areas on a square grid.

This chart shows all possible areas that can be formed on a square grid. The level 0 squares are the horizontal squares discussed earlier. It is lower left-upper right symmetric (as noted on Dave’s ‘blog), so only the upper triangle is shown.

From this, the following can be counted.
Level 1 – Areas 1-9: 6 of 9 possibilities found (yellow)
Level 2 – Areas 10-99: 40 of 90 possibilities found (orange)
Level 3 – Areas 100-999: 342 of 900 possibilities found (blue)

The percentage of possible numbers appears to be declining and is always less than the possible number of areas. But a scant handful of data points does not always definitively describe a pattern.

Determining the total number of possible areas:
Level 1 has 9 single-digit areas. Level 2 has 90 two-digit areas, and Level 3 has 900 three-digit areas. By this pattern, Level M has M-digit areas. This is the number of holes that need to be filled by the squares we can find on the square grid.

Determining an upper bound for the number of areas that can be accommodated on a square grid:
Notice that if a horizontally-oriented square has area of Level M, then every tilted square in its column has area AT LEAST of Level M. Also, the last column that contains any Level M areas is column where floor is the floor function.

In the chart, Column 1 contains 2 areas, and every Column N contains exactly (N+1) areas. The total number of areas represented for Columns 1 through N is an arithmetic sequence, so an upper bound for the number of distinct square areas represented in Columns 1 to N (assuming no duplication, which of course there is) is .

The last column that contains any Level M areas has column number . Assuming all of the entries in the data chart up to column are Level M (another overestimate if is not an integer), then there are

maximum area values to fill the Level M area holes. This is an extreme over-estimate as it ignores the fact that this chart also contains all square areas from Level 1 through Level (M-1), and it also contains a few squares which can be determined multiple ways (e.g., area 25 squares).

Both of these are dominated by base-10 exponential functions, but the number of areas to be found has a coefficient of 9 and the number of squares that can be found has coefficient 1/2. Further, the number of squares that can be found is decreased by an exponential function of base , accounting in part for the decreasing percentage of found areas noted in the data chart. That is, the number of possible areas grows faster than the number of areas that actually can be created on square grid paper.

While this proof does not say WHICH areas are possible (a great source for further questions and investigation!), it does show that the number of areas of squares impossible to find using a square grid grows without bound. Therefore, there is no largest area possible.