Isosceles Triangles in a Quadrilateral

In this post, I’ll discuss two issues. First, I’ll look at a problem taken from a major textbook, and explain why the solution is wrong. Then, I’ll discuss why this particular problem bothers me in the greater context of mathematics education.

First, the problem. This is a question from Pearson’s Common Core Geometry supplemental materials (Chapter 6-3, Practice K): “A classmate drew a quadrilateral with two diagonals. This divided the figure into four isosceles triangles. Is the quadrilateral a parallelogram? Use a drawing to justify your answer.”

The answer in the book is that, yes, it is a parallelogram. In reality, though, it is possible to construct a quadrilateral that satisfies the description but which is not a parallelogram.

Solution

I will be using this diagram to explain my answer.quadrilateralsegments

The reasoning in the book is that, if the four triangles are all isosceles, then \(g \cong h \cong e \cong f\), and therefore \(ABCD\) is a rectangle (and hence a parallelogram). Based on the hypothesis, the conclusion is true, but the hypothesis itself is false.

The hypothesis is based on the faulty assumption that the four sides of the quadrilateral represent the base sides of the isosceles triangles. That’s not necessarily the case, and it’s not specified in the problem.

Let us start instead with the assumption that \(a \cong g\) and that \(a \not\cong h\). That gives us one isosceles triangle; how could we construct the rest?

That gives us three options for \(\triangle AED\): \(g \cong d\), \(g \cong f\), or \(d \cong f\). If \(g \cong d\), then \(a \cong g \cong d\), so \(\odot A\) with radius \(|a|\) would have three radii touching \(\overleftrightarrow{BD}\), which is impossible. So \(g \not\cong d\).

However, there are still two possibilities: Either \(g \cong f\) or \(d \cong f\). Both of these would create a second isosceles triangle.

Working this way, it is not too difficult to create one possibility: \(a \cong g \cong f \cong c\) and \(e \cong h\). All four triangles are isosceles, and it is not necessarily a rectangle. I have not exhaustively analyzed all possibilities, but the existence of a single counterexample is enough to disprove a conjecture (in this case, that all such shapes are parallelograms).

quadrilateralsegments2I created this in GeoGebra using two sliders, \(s_1\) and \(s_2\), for the congruent lengths so I could explore the patterns. Point E (blue in the diagram) is at the origin. Points A and C are at coordinates \((s_1, 0)\) and \((-s_2, 0)\), respectively.

For the triangles on the side (where the side of the quadrilateral is one of the congruent sides of the triangle), we can use the Law of Cosines to calculate the angle at E: \(s_1^2 = s_2^2 + s_1^2 – 2s_1s_2\cos{\theta}\) \(\Rightarrow \cos{\theta} = \frac{s_2}{2s_1}\). We can use this angle to create a right triangle with a hypotenuse of \(2s_1\) and sides of \(s_2\) and \(\sqrt{4s_1^2 – s_2^2}\), so point B is at \((s_2^2 / (2s_1), s_2 \sqrt{4s_1 – s_2^2} / (2s_1))\) and point D is at \(((-s_2) / 2, (-\sqrt{4s_1^2 – s_2^2}) / 2)\).

As shown in the diagram, the non-congruent sides of the quadrilateral will be parallel, meaning that this is an isosceles trapezoid if \(s_1 \not\cong s_2\) (and a rectangle otherwise). Rectangles and isosceles trapezoids share the property that their diagonals are congruent, but there are other quadrilaterals that share this property. However, we can be more specific: A quadrilateral is a rectangle or an isosceles trapezoid if and only if its diagonals intersect to form two pairs of consecutive congruent segments.

I have not yet confirmed whether this is the only situation in which the four triangles will all be isosceles. Furthermore, while it’s true for all rectangles (since diagonals of rectangles are congruent and bisect each other, and hence form four isosceles triangles regardless), it’s only true for the subset of isosceles trapezoids described above.

Edit:

Emmanouil Lamprakis points out that a kite with a diagonal equal to the lengths of the sides of a shared vertex contains at least four isosceles triangles, albeit not the four smallest ones. Given kite ABCD with diagonal intersection at E, where AB = AC = AD, triangles ABC, ACD, BCD, and ABD will all be isosceles. When angle BAC has a measure of 45°, for that matter, triangles DEA and BEA will also be isosceles. I think the most obvious interpretation of the original problem is that “divided… into” refers to the four small triangles (in this case, ABE, BCE, CDE, and DAE), but lack of rigorous clarity is another valid complaint about this sort of question. Because of that, I would consider a student who found the kite solution to be correct, and I would rewrite the question for proper clarity.

Mistakes in Textbooks

So the answer in the textbook, which states without caveat that the shape formed is a parallelogram, is incorrect.

Textbooks are written at least in part by humans, and humans make mistakes. As both a teacher and a student, I don’t think I’ve ever dealt with a mathematics textbook that didn’t have some errors here or there. I’ve caught other errors in this textbook, for that matter; they have been obvious matters, such as mislabeled points. Students generally expect books to be perfect, and so these errors tend to make them question the book’s credibility, but with the number of questions and answers in a mathematics book, it is nigh on impossible to filter each and every one out.

However, pedagogically, errors like this one trouble me more than the obvious ones. The average student will not see the isosceles trapezoid solution; I myself didn’t see it directly, but had to build the construction in GeoGebra and notice the pattern. For that student, a single problem like this will be answered, and they will go on without any earth-shattering change in perspective.

The advanced student, meanwhile, will feel like there’s something a bit off about the solution. Perhaps they’ll reason to the isosceles trapezoid, or perhaps they’ll just feel a Disturbance in the Force. Either way, though, they won’t be satisfied with the answer. They’ll question the book. And some teachers will respond by digging in, insisting that the book is right. They may even reject the student’s counterexample outright.

I had math teachers like that: It says so in the book, so that’s the correct answer. I also had math teachers who allowed for the possibility that the book was wrong, so long as I could prove it. The latter are the math teachers we need, but how many of those are there?

A recent government report says that as many as 1 in 8 high school math teachers are neither certified in nor in possession of a major in mathematics. Naturally, people can be competent mathematicians without being holding college certification; my point, though, is that there are many, many math teachers who aren’t trained in higher levels of mathematics. I was recently reacquainted with a colleague who remembered our two-year-old conversation about high school math teachers not needing to be mathematicians; our lack of consensus about what constitutes a “mathematician” notwithstanding, I’ve seen ample anecdotal evidence that there are many math teachers who struggle with concepts even a step or two beyond what it is they’re expected to teach.

It has been a long-term struggle getting math majors into public school classrooms. Too many high school math teachers are mediocre at math, and too many elementary teachers communicate their own discomfort with math to their students (at a recent PD, I overheard some middle school teachers laughing about their lack of understanding of factorials–it took three of them to even remember the word; one of them said, “It’s a good thing we don’t have to teach high school”). Many are the rants by serious mathematicians about how acceptable it is to say, “Welp, LOL, I just don’t get math”, while we as a culture would openly mock someone who says, “Welp, LOL, I just don’t know how to read”; it’s that much more winceworthy when it comes from math teachers.

Because of this, though, it’s that much more important that textbooks avoid situations where a clever student and a not-as-clever teacher argue about these details. There’s a very real possibility that a student with a strong grasp of the material will see the solution, and a teacher with poorer understanding will scoff or even punish the student for that. I know because it happened to me in my youth, and as far as I can tell, the math competence of math teachers has just gotten worse in the interim: Some of my students have PTSD-levels of anxiety about so much as experimenting with mathematics, because they’re convinced that my veneer of encouragement is just a facade and that I’m waiting to pounce with mockery, as teachers in their past have done.

Textbooks, in other words, are a quicker point in the pipeline for correction: There are less than a dozen major titles, as opposed to the thousands of teachers in the country. We need to make sure these deep-level conceptual errors in textbooks are minimized, if not eliminated.

 

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