Math Photo: Hotel Hexagons

I like the multiple takes on hexagons here, both in size and orientation.  Not bad for random hotel wall art!  This was in Cambridge, though, so maybe that explains it.

Varignon’s Theorem

Varignon’s Theorem is one of my favorite results in elementary geometry:  connect the adjacent midpoints of the four sides of any quadrilateral, and a parallelogram is formed!  It is a magical result that defies expectations, and it’s so much fun to play around with, explore, and extend.

Steven Strogatz shared his favorite proof of Varignon’s Theorem on Twitter yesterday, and so I felt compelled to share mine.  This is a standard proof of Varignon, but it is so clean and elegant:  it is an immediate consequence of the Triangle Midsegment Theorem and the transitivity of parallelism.

Strogatz’s vector proof is beautiful and efficient, but the power of transitivity really shines in this elementary geometric proof.

I created a a simple Desmos demonstration to explore Varignon’s Theorem.  And like all compelling mathematical results, there are so many fascinating follow-up questions to ask!

Related Posts

Regents Recap — June 2016: Still Not a Trig Function

I don’t know exactly why, but fake graphs on Regents exams really offend me.  Take a look at this “sine” curve from the June, 2016 Algebra 2 Trig exam.

Looking at this graph makes me uneasy.  It’s just so … pointy.  Here’s an actual sine graph, courtesy of Desmos.

Now this fake sine curve isn’t nearly as bad as these two half-ellipses put together, but I just don’t understand why we can’t have nice graphs on these exams.  It only took me a few minutes to put this together in Desmos.  Let’s invest a little time in mathematical fidelity.

Related Posts

Regents Recap — June 2016: Algebra is Hard

In my ongoing analysis of New York State’s math Regents exams, the most discouraging issues I encounter are the mathematical errors.  Consider this two-point problem from the June 2016 Common Core Algebra 2 exam.

There are no issues with this simple algebraic identity, but there is a serious issue with how this problem was graded.

With each exam a set of model responses are published by the state.  These serve as exemplar student work and inform graders of what correct and incorrect responses look like.  Here is a model response for this problem.

Sadly, according to the official scoring guidelines, this perfectly valid and 100% correct response earns only one point out of two.

The purported reason for the penalty is that “The student made an error by not manipulating expressions independently in an algebraic proof”.  It’s unclear what, if anything,  this means, but there is no requirement that expressions be manipulated independently in an algebraic proof.  This is an artificial criticism.

I suspect the complaint has to do with multiplying both sides of the equation by some quantity.  I have occasionally heard teachers argue that, when proving an identity, you can’t multiply both sides of an equation by the same thing.  Their reasons vary, but the most common explanation is that in doing so you are assuming the sides are equal, which is what you are trying to prove.

This is faulty mathematics.  For the most part, there is no issue with multiplying both sides of a purported identity by the same quantity:  if the original equation is true, the new equation will be true, and if the original equation is false, the new equation will be false.  In general, the equations are logically equivalent, that is, true and false under exactly the same circumstances.

For example, consider the true equation 4 = 4 and the false equation 4 = 5.  Notice that

4 = 4   and   3*4 = 3*4

are both true, and

4 = 5   and   3*4 = 3*5

are both false.  Multiplying both sides by 3 does not change the truth value of either equation.

Now, I said for the most part because there is one particular situation in which multiplying both sides of an equation by the the same quantity can be problematic.  Multiplying both sides of an equation by zero can turn a false equation into a true equation.  For example

3 = 2

is clearly false, but

0*3 = 0*2

is true.

So, the full mathematical story is that the statements a = b and ka = kb are logically equivalent if $k \neq 0$.  That is to say, multiplying both sides of an equation by the same quantity will preserve its truth value if you aren’t multiplying by zero.

$\frac{x^3 + 9}{x^3 + 8 } = 1 + \frac{1}{x^3+8}$

$(x^3 + 8)(\frac{x^3 + 9}{x^3 + 8 }) = (x^3 + 8)(1 + \frac{1}{x^3+8})$

Here, both sides of the original equation have been multiplied by $(x^3 + 8)$.  As long as $x^3 + 8 \neq 0$, these two equations are logically equivalent, and so proving that the latter equation is an identity is equivalent to proving that the original equation is an identity.

Could $x^3 + 8$ equal 0?  Generally speaking, yes.  But conveniently, the exam authors went to the trouble to tell us that $x \neq -2$, which means $x^3 + 8 \neq 0$.  Therefore, the mathematics of the model response is perfectly valid.  It should earn full credit.

The error in these official grading instructions demonstrates a serious lack of mathematical understanding on the part of those who produce these exams.  Errors like this, and this, and this, should give pause to those who automatically assume that exams like these are valid measurements of mathematical knowledge and ability.

And there are serious and real consequences here.  Not only are student losing points for perfectly valid mathematical work, official state documents are demonstrating incorrect mathematics to classroom teachers.  Without correction, this erroneous mathematics will likely be passed along to students.

While there can be legitimate disagreement and debate about the extent and use of testing in education, I think we can all agree that the tests themselves should not undermine the teaching and learning of content.  And that’s exactly what errors like this do.

Related Posts

Regents Recap — June 2016: How Much Should This Be Worth?

The following problem appeared on the June 2016 Common Core Algebra 2 Regents exam.

This is a straightforward and reasonable problem.  What’s unreasonable is that it is only worth two points.

The student here is asked to construct a representation of a mathematical object with six specific properties:  it must be a cosine curve;  it must be a single cycle; it must have amplitude 3; it must have period $\pi / 2$; it must have midline y = -1; and it must pass through the point (0,2).

That seems like a lot to ask for in a two-point problem, but the real trouble comes from the grading guidelines.

According to the official scoring rubric, a response earns one point if “One graphing error is made”.  Failure to satisfy any one of the six conditions would constitute a graphing error.  So a graph that satisfied five of the six required properties would earn one point out of two.  That means a response that is 83% correct earns 50% credit.

It gets worse.  According to the general Regents scoring guidelines, a combination of two graphing errors on a single problem results in a two-point deduction.  That means a graph with four of the six required properties, and thus two graphing errors, will earn zero points on this problem.  A response that is 66% correct earns 0% credit!

The decision to make this six-component problem worth two points creates a situation where students are unfairly and inconsistently evaluated.  It makes me wonder if those in charge of these exams actually considered the scoring consequences of their decision, especially since there are two obvious and simple fixes:  reduce the requirements of the problem, or increase its point value.

This is another example of how tests that are typically considered objective are significantly impacted by arbitrary technical decisions made by those creating them.

Related Posts