Tag Archives: geometry

From a Square to Ratios to a System of Equations

Here’s another ratio problem from @Five_Triangles, this time involving triangle areas bounded by a square.

Don’t read further until you’ve tried this for yourself. It’s a fun problem that, at least from my experience, doesn’t end up where or how I thought it would.


I see two big challenges here.

First, the missing location of point P is especially interesting, but is also likely to be quite vexing for many students. This led me to the first twist I found in the problem: the introduction of multiple variables and a coordinate system. Without some problem-solving experience, I don’t see that as an intuitive step for most middle school students. Please don’t interpret this as a knock on this problem, I’m simply agreeing with @Five_Triangle’s assessment that this problem is likely to be challenging for middle school students.

The second challenge I found emerged from the introduction the coordinate system: an underlying 2×2 system of equations. There are multiple ways to tackle a solution to a linear system, but this strikes me as yet another high hurdle for younger students.

Finally, I’m a bit surprised by my current brain block on multiple approaches for this problem. I suspect I’m blinded here by my algebraic bias in problem solving; surely there are approaches that don’t require this. I’d love to hear any other possibilities.


Because I was given properties of point P and not its location, the easiest approach I could see was to position the square on the xy-plane with point B at the origin, \overline{AB} along the y-axis, and \overline{BC} along the x-axis. That gave my point P coordinates (x,y) for some unknown values of x & y.


The helpful part of this orientation is that the x & y coordinates of P are automatically the altitudes of \Delta ABP and \Delta BCP, respectively. The altitudes of the other two triangles are determined through subtraction.


From here, I used the given ratios to establish one equation in terms of x & y.

\displaystyle \frac{\Delta ABP}{\Delta DAP} = \frac{\frac{1}{2}*12*x}{\frac{1}{2}*12*(12-y)} = \frac{3}{4}

Of course, since all four triangles have the same base lengths, the given area ratios are arithmetically equivalent to corresponding height ratios.  I used that to write a second equation.

\displaystyle \frac{\Delta BCP}{\Delta CDP} = \frac{y}{12-x} = \frac{1}{3}

Simplifying terms and clearing denominators leads to 4x=36-3y and 3y=12-x, respectively.

A VERY INTERESTING insight at this point is that there is an infinite number of locations within the square at which each ratio is true.  Specifically, the \Delta ABP : \Delta DAP = 3:4 ratio is true everywhere along the line 4x=36-3y.  This problem constrains us to only the points within the square with vertices (0,0), (12,0), (12,12), and (0,12), but setting that aside, anywhere along the line 4x=36-3y would satisfy the first constraint.  The same is true for the second line and constraint.

I think it would be very interesting for students to construct this on dynamic geometry software (e.g., GeoGebra or the TI-Nspire) and see the ratio remain constant everywhere along either line even though the triangle areas vary throughout.

Together, these lines form a 2×2 system of linear equations with the solution to both ratios being the intersection point of the two lines.  There are lots of ways to do this; I wonder how a typical 6th grader would tackle them. Assuming they have the algebraic expertise, I’d have work them by hand and confirm with a CAS.


The question asks for the area of \Delta ABP = \frac{1}{2}*12*x = 6*8 = 48.


Just two extensions this time.  Other suggestions are welcome.

  1. What’s the ratio of the area of \Delta BCP : \Delta DAP at the point P that satisfies both ratios??
    It’s not 1:4 as an errant student might think from an errant application of the transitive property to the given ratios.  Can you show that it’s actually 1:8?
  2. If a random point is chosen within the square, is that point more likely to satisfy the area ratio of \Delta ABP : \Delta DAP or the ratio of \Delta BCP : \Delta CDP?
    The first ratio is satisfied by the line 4x=36-3y which intersects the square on the segment between (9,0) and (0,12).  At the latter point, both triangles are degenerate with area 0.  The second ratio’s line intersects the square between (12,0) and (0,4).  As the first segment is longer (how would a middle schooler prove that?), it is more likely that a randomly chosen point would satisfy the \Delta ABP : \Delta DAP  ratio.  This would be a challenging probability problem, methinks.


What other possibilities do you see either for a solution to the original problem or an extension?

Value Process over Answers

Most of my thinking about teaching lately has been about the priceless, timeless value of process in problem solving over the ephemeral worth of answers.  While an answer to a problem puts a period at the end of a sentence, the beauty and worth of the sentence was the construction, word choice, and elegance employed in sharing the idea at the heart of the sentence.

Just as there are many ways to craft a sentence–from cumbersome plodding to poetic imagery–there are equally many ways to solve problems in mathematics.  Just as great writing reaches, explores, and surprises, great problem solving often starts with the solver not really knowing where the story will lead, taking different paths depending on the experience of the solver, and ending with even more questions.

I experienced that yesterday reading through tweets from one of my favorite middle and upper school problem sources, Five Triangles.  The valuable part of what follows is, in my opinion, the multiple paths I tried before settling on something productive.  My hope is that students learn the value in exploration, even when initially unproductive.

At the end of this post, I offer a few variations on the problem.

The Problem


Try this for yourself before reading further.  I’d LOVE to hear others’ approaches.

First Thoughts and Inherent Variability

My teaching career has been steeped in transformations, and I’ve been playing with origami lately, so my brain immediately translated the setup:

Fold vertex A of equilateral triangle ABC onto side BC.  Let segment DE be the resulting crease with endpoints on sides AB and AC with measurements as given above.

So DF is the folding image of AD and EF is the folding image of AE.  That is, ADFE is a kite and segment DE is a perpendicular bisector of (undrawn) segment AF.  That gave \Delta ADE \cong \Delta FDE .

I also knew that there were lots of possible locations for point F, even though this set-up chose the specific orientation defined by BF=3.

Lovely, but what could I do with all of that?

Trigonometry Solution Eventually Leads to Simpler Insights

Because FD=7, I knew AD=7.  Combining this with the given DB=8 gave AB=15, so now I knew the side of the original equilateral triangle and could quickly compute its perimeter or area if needed.  Because BF=3, I got FC=12.

At this point, I had thoughts of employing Heron’s Formula to connect the side lengths of a triangle with its area.  I let AE=x, making EF=x and EC=15-x.  With all of the sides of \Delta EFC defined, its perimeter was 27, and I could use Heron’s Formula to define its area:

Area(\Delta EFC) = \sqrt{13.5(1.5)(13.5-x)(x-1.5)}

But I didn’t know the exact area, so that was a dead end.

Since \Delta ABC is equilateral, m \angle C=60^{\circ} , I then thought about expressing the area using trigonometry.  With trig, the area of a triangle is half the product of any two sides multiplied by the sine of the contained angle.  That meant Area(\Delta EFC) = \frac{1}{2} \cdot 12 \cdot (15-x) \cdot sin(60^{\circ}) = 3(15-x) \sqrt3.

Now I had two expressions for the same area, so I could solve for x.

3\sqrt{3}(15-x) = \sqrt{13.5(1.5)(13.5-x)(x-1.5)}

Squaring both sides revealed a quadratic in x.  I could do this algebra, if necessary, but this was clearly a CAS moment.


I had two solutions, but this felt WAY too complicated.  Also, Five Triangles problems are generally accessible to middle school students.  The trigonometric form of a triangle’s area is not standard middle school fare.  There had to be an easier way.

A Quicker Ending

Thinking trig opened me up to angle measures.  If I let m \angle CEF = \theta, then m \angle EFC = 120^{\circ}-\theta, making m \angle DFB = \theta, and I suddenly had my simple breakthrough!  Because their angles were congruent, I knew \Delta CEF \sim \Delta BFD.

Because the triangles were similar, I could employ similarity ratios.


And that is one of the CAS solutions by a MUCH SIMPLER approach.

Extensions and Variations

Following are five variations on the original Five Triangles problem.  What other possible variations can you find?

1)  Why did the CAS give two solutions?  Because \Delta BDF had all three sides explicitly given, by SSS there should be only one solution.  So is the 13.0714 solution real or extraneous?  Can you prove your claim?  If that solution is extraneous, identify the moment when the solution became “real”.

2)  Eliminating the initial condition that BF=3 gives another possibility.  Using only the remaining information, how long is \overline{BF} ?

\Delta BDF now has SSA information, making it an ambiguous case situation.  Let BF=x and invoke the Law of Cosines.

7^2=x^2+8^2-2 \cdot x \cdot 8 cos(60^{\circ})

Giving the original BF=3 solution and a second possible answer:  BF=5.

3)  You could also stay with the original problem asking for AE.

From above, the solution for BF=3 is AE=10.5.  But if BF=5 from the ambiguous case, then FC=10 and the similarity ratio above becomes


4)  Under what conditions is \overline{DE} \parallel \overline{BC} ?

5)  Consider all possible locations of folding point A onto \overline{BC}.  What are all possible lengths of \overline{DE}?



Next Steps from a Triangle

Watching the news a couple mornings ago, an impossible triangle appeared on the screen.  Hopefully some readers might be able to turn some first ideas a colleague and I had into a great applied geometry lesson.  What follows are some teacher thoughts.  My colleagues and I hope to cultivate classes where students become curious enough to raise some of these questions themselves.



At first glance, the labeling seems off.  In Euclidean geometry, the Triangle Inequality says the sum of the lengths of any two sides of a triangle must exceed the length of the third side.  Unfortunately, the shorter two sides sum to 34 miles, so the longest side of 40 miles seems physically impossible.  Someone must have made a typo.  Right?

But to dismiss this as a simple typo would be to miss out on some spectacular mathematical conversations.  I’m also a big fan of taking problems or situations with prima facie flaws and trying to recover either the problem or some aspects of it (see two of previous posts here and here).


Without confirming any actual map distances, I first was drawn to the vagueness of the approximated side lengths.  Was it possible that this triangle was actually correct under some level of round-off adjustment?  Hopefully, students would try to determine the degree of rounding the graphic creator used.  Two sides are rounded to a multiple of 10, but the left side appears rounded to a nearest integer with two significant digits.  Assuming the image creator was consistent (is that reasonable?), that last side suggests the sides were rounded to the nearest integer.  That means the largest the left side could be would be 14.5 miles and the bottom side 20.5 miles.  Unfortunately, that means the third side can be no longer than 14.5+20.5=35 miles.  Still not enough to justify the 40 miles, but this does open one possible save.

But what if all three sides were measured to the nearest 10 instead of my assumed ones place?  In this case the sides would be approximately 10, 20, and 40.  Again, this looks bad at first, but a 10 could have been rounded from a 14.9, a 20 from a 24.9, making the third side a possible 14.9+24.9=39.8, completely justifying a third side of 40.    This wasn’t the given labeling, but it would have potentially saved the graphic’s legitimacy.


Is there another way the triangle might be correct?  Rarely do pre-collegiate geometry classes explore anything beyond Euclidean geometry.  One of my colleagues, Steve, proposed spherical geometry:

Does the fact that the earth is round play a part in these seemingly wrong values (it turns out “not really”… Although it’s not immediately clear, the only way to violate the triangle inequality in spherical geometry is to connect point the long way around the earth. And based on my admittedly poor geographical knowledge, I’m pretty sure that’s not the case here!)


Perhaps students eventually realize that the distances involved are especially small relative to the Earth’s surface, so they might conclude that the Euclidean geometry approximation in the graphic is likely fine.

Then again, why is the image drawn “as the crow flies”?  The difficult mountainous terrain in upstate New York make surface distances much longer than air distances between the same points.  Steve asked,

in the context of this problem (known location of escaped prisoners), why is the shortest distance between these points being shown? Wouldn’t the walking/driving distance by paths be more relevant?  (Unless the prisoners had access to a gyrocopter…)

The value of a Euclidean triangle drawn over mountainous terrain has become questionable, at best.


I suspect the triangle awkwardly tried to show the distances the escapees might have traveled.  Potentially interesting, but when searching for a missing person in the mountains–the police and news focus at the time of the graphic–you don’t walk the perimeter of the suspected zone, you have to explore the area inside.

A day later, I saw the search area around Malone, NY shown as a perfect circle.  (I wish I had grabbed that image, too.).  Around the same time, the news reported that the search area was 22 square miles.

  • Was the authorities’ 22 measure an approximation of a circle’s area, a polygon based on surface roads, or some other shape?
  • Going back to the idea of a spherical triangle, Steve hoped students would ask if they could “compute that from just knowing the side lengths? Is there a spherical Herons Formula?”
  • If the search area was a more complicated shape, could you determine its area through some sort of decomposition into simpler shapes?  Would spherical geometry change how you approach that question?  Steve wondered if any students would ask, “Could we compute that from just knowing the side lengths? Is there a spherical Herons Formula?
  • At one point near the end of the search, I hear there were about 1400 police officers in the immediate vicinity searching for the escapee.  If you were directing the search for a prison escapee or a lost hiker, how would you deploy those officers?  How long would it take them to explore the entire search zone?  How would the shape of the potential search zone affect your deployment plan?
  • If you spread out the searchers in an area, what is the probability that an escapee or missing person could avoid detection?  How would you compute such a probability?
  • Ultimately, I propose that Euclidean or spherical approximations seriously underestimated the actual surface area?  The dense mountainous terrain significantly complicated this search.  Could students extrapolate a given search area shape to different terrains?  How would the number of necessary searchers change with different terrains?
  • I think there are some lovely openings to fractal measures of surface roughness in the questions in the last bullet point.


Ultimately, we hope students would ask

  • What caused the graphic’s errors?  Based on analyses above and some Google mapping, we think “a liberal interpretation of the “approximately” label on each leg might actually be the culprit.”  What do the triangle inequality violations suggest about round-off errors or the use of significant digits?
  • The map appeared to be another iteration of a map used a few days earlier.  Is it possible that compounded rounding errors were partially to blame?
  • Surely the image’s designer new the triangle was an oversimplification of the reality.  Assuming so, why was this graphic used anyway?  Does it have any news value?  Could you design a more meaningful infographic?


Many thanks to Steve Earth for his multiple comments and thoughts that helped fill out this post.

Lovely or Tricky Triangle Question?

In addition to not being drawn to scale and asking for congruence anyway, I like this problem because it potentially forces some great class discussions.

One responder suggested using the Law of Sines (LoS) to establish an isosceles triangle.  My first thought was that was way more sophisticated than necessary and completely missed the fact that the given triangle information was SSA.

My initial gut reaction was this SSA setup was a “trick” ambiguous case scenario and no congruence was possible, but I couldn’t find a flaw in the LoS logic. After all, LoS fails when attempting to find obtuse angles, but the geometry at play here clearly makes angles B and C both acute.  That meant LoS should work, and this was actually a determinate SSA case, not ambiguous.  I was stuck in a potential contradiction.  I was also thinking with trigonometry–a far more potent tool than I suspected was necessary for this problem.

“Stuck” moments like this are GOLDEN for me in the classroom.  I could imagine two primary student situations here.  They either  1) got a quick “proof” without recognizing the potential ambiguity, or 2) didn’t have a clue how to proceed.  There are many reasons why a student might get stuck here, all of which are worth naming and addressing in a public forum.  How can we NAME and MOVE PAST situations that confuse us?  Perhaps more importantly, how often do we actually recognize when we’re in the middle of something that is potentially slipperier than it appears to be on the surface?


I read later that some invoked the angle bisector theorem, but I took a different path.  I’m fond of a property I asked my geometry classes to prove last year .

If any two of a triangle’s 1) angle bisector, 2) altitude, and 3) median coincide, prove that the remaining segment does, too, and that whenever this happens, the triangle will be isosceles with its vertex at the bisected angle.

Once I recognized that the angle bisector of angle BAC was also the median to side BC, I knew the triangle was isosceles.  The problem was solved without invoking any trigonometry or any similarity ratios.

Very nice problem with VERY RICH discussion potential.  Thanks for the tweet, Mr. Noble.

For more conversation on this, check out this Facebook conversation.

Circle and Square

Here’s another great geometry + algebra problem, posed by Megan Schmidt and pitched by Justin Aion to some students in his Geometry class.

Following is the problem as Justin posed it yesterday.


Justin described the efforts of three of his students’ on his his ‘blog.  Following is my more generalized approach.  Don’t read further if you want to solve this problem for yourself!


My first instinct in any case like this is build it in a dynamic geometry package and play.  Using my TI-Nspire, without loss of generality, I graphed a circle centered at the origin, constructed a tangent segment at the bottom of the circle centered on the y-axis, and then used that segment to construct a square.  I recognized that the locus of the upper right corners of all such squares would form a line.


That made it clear to me that for any circle, there was a unique square that intersected the circle three times as Megan had posed.

Seeing this and revealing its algebraic bias, my mind conceived an algebraic solution.  Assuming the radius of the circle is R, the equation of my circle is x^2+y^2=R^2 making the lower y-intercept of the circle (0,-R).  That made y=2x-R the locus line containing the upper right corner of the square.


To find generic coordinates of the upper right corner of the square in terms of R, I just needed to solve the system of equations containing the circle and the line.  That’s easy enough to compute by hand if you can handle quadratic algebra.  That manipulation is not relevant right now, so my Nspire CAS’s version is:


The output confirms the two intersections are (0,-R) and the unknown at \displaystyle \left( \frac{4R}{5} , \frac{3R}{5} \right).

Because of the horizontal symmetry of the square with respect to the y-axis, the system solution shows that the generic length of the side of the square is \displaystyle 2\left( \frac{4R}{5} \right) = \frac{8R}{5} .  The circle’s y-intercept at (0,-R) means the generic diameter of the circle is 2R.

Therefore, the generic ratio of the circle’s diameter to the square’s side length is

\displaystyle \frac{diameter}{side} = \frac{2R}{(8R)/5} = \frac{5}{4}.

And this is independent of the circle’s radius!  The diameter of the circle is always \frac{5}{4} of the square’s side.


For Megan’s particular case with a side length of 20, that gives a circle diameter of 25, confirming Justin’s students’ solution.

Does anyone have a different approach?  I’m happy to compile and share all I get.


While not necessary for the generalized solution, it was fun to see a 3-4-5 right triangle randomly appear in Quadrant 1.


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?