The New York Times just posted an article regarding an international assessment of middle school mathematics teachers.

U.S. Falls Short in Measure of Future Math Teachers

Fair enough: my job next year is to address this problem. That’s not what bothers me about the article.

What bothers me is the sample problem:

On the figure, *ABCD* is a parallelogram, angle *BAD* = 60º, *AM* and *BM* are angle bisectors of angles *BAD* and *ABC* respectively. If the perimeter of *ABCD* is 6 cm, find the sides of triangle *ABM*.

(Try to solve it yourself before moving on.)

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**Issue #1**: I (and likely many people doing the problem) used a meta-argument on the perimeter. Implicitly the figure is divided into what looks like 6 equal parts, so with a perimeter of 6 cm, probably each part is 1 cm. At the very least this step comes across as artificially manufactured test trickery rather than real mathematics.

**Issue #2**: I realize it is “standard”, but do people really need the properties of the 30-60-90 triangle? This strikes me as “legacy knowledge” from back when trigonometry was inconvenient to do. (The article says nothing about calculator use, but I suspect they were not allowed.)

These aren’t significant or serious issues, but they do make me worry about the quality of the other questions on the test.

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Aaron K, on April 16, 2010 at 9:46 am said:Why does one need a ‘meta argument?’

BAM is a 30 degree angle; MBA and CBM are 60 degree angles. The angle (not drawn) bisecting CBM is perpendicular to CM, which tells us CMB is 60 degrees. Therefore triangle CMB is equilateral, so BM and CB are the same length. Call this X.

From the 30-60-90 identity, AB’s length is twice BM (and therefore twice CB). Call this 2x

2X + 2*2*X = 6

6X = 6

X = 1

Again, using the 30-60-90 identity, AB is 2, BM is 1, and AM is root 3.

Jason Dyer, on April 16, 2010 at 9:53 am said:No, one does not need a meta-argument. The point is one can make one in the first place, that is, “game” the test.

Aaron K, on April 16, 2010 at 11:43 am said:Ah, now I get it. At least (a) the figure was not drawn to scale and (b) it wasn’t a multiple-choice question.

Kate Nowak, on April 16, 2010 at 10:04 am said:FWIW I didn’t assume M was the midpoint of DC, I just labeled angles and made inferences from there.

I probably overemphasize 30-60-90 and 45-45-90 in Geometry, because I know fluency with them will make Trig easier for kids and save them time on the SAT.

I don’t really want to spend so much time on it. It’s not that useful. But my effectiveness is judged primarily on test results. The alignment between what we are nominally doing with kids (21st c. skills, etc, etc) and how we’re evaluated (% students achieving mastery on a multiple choice test where a computer is not allowed, etc) is imperceptible.

So I guess “Is this a good question?” and “Is this a good test?” depend on what we expect math teachers to be doing with kids and preparing them for.

Jason Dyer, on April 16, 2010 at 10:34 am said:Granted re: the SAT.

I’m not sure it makes Trig easier. It’s like teaching FOIL for multiplying binomials. Sure, it makes that specific case faster, but then students fall down when they need to generalize.

LSquared, on April 16, 2010 at 11:45 am said:Am I the only one who used the isosceles triangle? After you find out that MCB is equilateral, and AMB is 30-60-90, it’s only a short step to ADM is 30-120-30, and therefore isosceles. That’s where I got the length ratios–I use need the 30-60-90 triangle for length AM. I’d hope that someone who might be teaching geometry would be flexible enough to figure this out.

ruby_omnibus, on April 16, 2010 at 5:52 pm said:I forget the ratio length of 30-60-90 and have to resort to Pythagoras’ theorem.

As we know that △AMB is a right triangle, △ADM is an isosceles triangle, and△ MBC is an equilateral triangle, I need to know the ratio of line.

I divide the parallelogram into two parallelograms with a line drawn from point M to the line AB. Let me name the intersection as Q.

Inside the new parallelogram MQBC, line MB bisects it. As we know △MBC is equilateral, it is easy to prove △MBQ is equilateral too and thus MQ=QB.

Inside the right △AMB, it is divided by MQ. As we know △MBQ is equilateral, it is easy to prove △MAQ is isosceles and MQ=AQ. (or we can deduce the same fact from parallelogram MDAQ)

We get that MQ=AQ=BQ, and MQ=MB (equilateral triangle), MB=BC (equilateral triangle). We can deduce two facts. First, the property of parallelogram and AQ=BQ, we get BM=MC. Q and M is the mid-points of line DC and AB respectively. Second, BC=BQ=AQ, its implies that AB=2 BC.

With perimeter of 6 cm, AB = 2 cm, BC =1cm, DC = 1cm and AD = 1cm. Hence, MB = 1cm.

With △AMB is a right triangle. From Pythagoras’ theorem, we know AM² + MB² = AB².

Let the length of AM is x.

x² + 1² = 2²

x² + 1 = 4

x² = 3

x = √ 3

The answer is AM = √ 3 cm, MB = 1 cm, AB = 2 cm.

Mr. W, on April 16, 2010 at 9:24 pm said:I went with the same exact method as Aaron K did.

I had to make sure M wasn’t the midpoint, although most of my students would probably have guessed the sides because of the reasons giving of thinking M is the midpoint and then dividing it to get 1& 1 and then check to make sure it works.

Only one way to find out though…opening quiz question on Monday!!!

jd2718, on April 17, 2010 at 7:08 am said:I’m annoyed that M is the midpoint, is lettered “M” and is guessably the midpoint – unwarranted guesses should not lead to correct answers.

And about “good question” – nah. It’s ok. Traditional. Ignore the guessing part, they have to know what an angle bisector is, the 30-60-90 relationship, they have to notice two 60s, know about equilateral triangles, reason a bit about the ratio of the sides, they can skip the isosceles distraction…

It’s just a series of facts and recognitions that have a rough order to them.

What do you think of this, instead:

On the figure, ABCD is a parallelogram, AM and BM are angle bisectors of angles BAD and ABC respectively. M is the midpoint of DMC. Find the measure of angle BAD.

Jonathan

Jason Dyer, on April 17, 2010 at 8:10 am said:On the figure, ABCD is a parallelogram, AM and BM are angle bisectors of angles BAD and ABC respectively. M is the midpoint of DMC. Find the measure of angle BAD.Much better!

Paralelogramo e triângulo: um problema modelo de um teste para futuros professores « problemas | teoremas, on April 17, 2010 at 2:06 pm said:[…] Trigonometria | Tags: Matemática, Matemática-secundário, Problema O blogue The Number Warrior publicou recentemente um post no qual, citando o artigo de Sam Dillon U.S. Falls Short in Measure of […]

jd2718, on April 18, 2010 at 8:57 am said:I think my problem is under-determined. We need the ratio of the lengths of the sides to determine the angle, or vv.

aaronthill, on April 19, 2010 at 4:47 pm said:It is under-determined: Let ABCD be a rectangle with the length of AB twice the length of BC. This also satisfies the criteria stipulated, but has a different measure for BAD.

Actually saying “AM and BM are angle bisectors of angles BAD and ABC respectively” implies (as long as ABCD is a parallelogram) that M is the midpoint of CD.

Takis Konstantopoulos, on April 26, 2010 at 4:51 am said:Although the question could have been phrased in a better way, I don’t find anything fundamentally wrong with it.

Jason Dyer, on April 26, 2010 at 7:08 am said:I’m not sure “phrasing” is it, unless you can come up with some clever way to remove the meta-argument through rephrasing. It is an issue with a.) the convienent perimeter of 6 closely matching what looks like 6 equal line segments b.) the use of the letter M as a midpoint (even though one is supposed to prove it is so) and c.) the problem is solvable in about 20 seconds by guess-and-check when realizing the above two facts.

Now, I wouldn’t call this a terrible-horrible question, but it was picked as the exemplar to show off in the New York Times. If this is what the questions are like in general, that means the test as a whole can be gamed, which means as a statistical measure, it is measuring ability to take tests more than mathematical ability. Hence my umbrage.

Takis Konstantopoulos, on April 26, 2010 at 8:16 am said:I can see your point. In other words, it’s as if the question suggests the answer.

Newspapers are not good in giving a good picture about mathematics education. Your comment reminds me of a silly “math” quiz posted on the BBC site some time ago.