kitchen table math, the sequel: international math test for future teachers

Thursday, April 15, 2010

international math test for future teachers

America’s future math teachers, on average, earned a C on a new test comparing their skills with their counterparts in 15 other countries, significantly outscoring college students in the Philippines and Chile but placing far below those in educationally advanced nations like Singapore and Taiwan.

The researchers who led the math study in this country, to be released in Washington on Thursday, judged the results acceptable if not encouraging for America’s future elementary teachers. But they called them disturbing for American students heading to careers in middle schools, who were outscored by students in Germany, Poland, the Russian Federation, Singapore, Switzerland and Taiwan.

On average, 80 percent to 100 percent of the future middle school teachers from the highest-achieving countries took advanced courses like linear algebra and calculus, while only 50 percent to 60 percent of their counterparts in the United States took those courses, the study said.

U.S. Falls Short in Measure of Future Math Teachers
by Sam Dillon
Here's the report: The Preparation Gap: Teacher Education for Middle School Mathematics in Six Countries (pdf file)

NCTM response
“There are so many people who bash our teachers’ math knowledge that to be honest these results are better than what a lot of people might expect,” said Hank Kepner, professor of mathematics education at the University of Wisconsin, Milwaukee, who is president of the National Council of Teachers of Mathematics. “We show up pretty well here, right in the middle of the pack.”
NCTM motto: Good enough is good enough.

I could do this problem, which is a good thing, given that I've completed ALEKS geometry.

On the other hand, I had to think about it -- and I don't remember having learned that the angle bisector also bisects the opposite side in a parallelogram.


Anonymous said...

Re: "And I don't remember having learned that the angle bisector also bisects the opposite side in a parallelogram"

I doesn't have to --- consider a square.

Michael Weiss said...

This just happens to be a parallelogram in which the acute angles measure 60°, and in which the side lengths are in a 2:1 ratio. If either of those conditions fails to hold the angle bisectors don't bisect the opposite side.

Katharine Beals said...

I'm proud to report that my autistic 13 1/2 year old figured got all three correct answers in about 5 minutes.

Michael Weiss said...

Oops, let me correct my statement above.

The angles don't matter. In any parallelogram whose sides are in a 2:1 ratio, the bisectors of two angles joined by one of the long side will meet at the midpoint of the opposite side. And conversely.

...also, they will be perpendicular.

Anonymous said...

Indeed, the bisectors will be perpendicular wherever they meet.

Catherine Johnson said...

now I'm confused

Catherine Johnson said...

Katharine - wow!

Anonymous said...

The ed system can make excuses for poor student performance. After all, they have to accept everyone, they come from all backgrounds, etc. But what's the excuse for poor teacher performance?


GoogleMaster said...

Don't get sucked into the misleading coincidence that the angle bisector also bisects the opposite side in this case. As several posters pointed out, this is usually not the case.

When all else fails, draw the picture and label everything you are given. Then label everything else you can figure out. This one uses a 60-60-60 triangle, a 30-60-90 triangle, and properties of parallelograms and isosceles triangles.

ABCD is a parallelogram, and BAD = 60 deg. Opposite angles in a parallelogram are equal, so DCB also = 60 deg.

Angles in a quadrilateral add up to 360 deg, and again opposite angles in a parallelogram are equal, so ADC = CBA = 120 deg.

AM is the angle bisector of BAD, so by the definition of angle bisector, BAM = MAD = 30 deg.

BM is the angle bisector of ABC, so CBM = MBA = 60 deg.

Angles in a triangle sum to 180 deg, so MCB + CBM + BMC = 180. But MCB = DCB, so 60 + 60 + BMC = 180, so BMC = 60 deg.

Angles in a triangle sum to 180 deg, so AMB + MBA + BAM = 180. AMB + 60 + 30 = 180, so AMB = 90 deg.

Angles in a triangle sum to 180 deg, so ADM + DMA + MAD = 180. But ADM = ADC, so 120 + DMA + 30 = 180, so DMA = 30 deg.

MAD-ADM-DMA is a 30-120-30 (isosceles) triangle. In an isosceles triangle, the sides opposite the equal angles are equal. So DM = AD.

CBM-BMC-MCB is a 60-60-60 (equilateral) triangle. In an equilateral triangle, all sides are equal, so MC = CB = BM.

AMB-MBA-BAM is a 90-60-30 triangle. In this special right triangle, the sides are in the ratio 1:2:sqrt(3). So AB = 2(BM).

Opposite sides of a parallelogram are equal, so AB = DM + MC.

AB = DM + MC, but AB = 2(BM) and MC = BM, so DM + BM = 2(BM), so DM = BM. So now we have DM = AD = BM = MC = CB, and AB = 2(BM).

The perimeter of the parallelogram is AD + DM + MC + CB + AB = BM + BM + BM + BM + 2(BM) = 6(BM). But we are given that the perimeter is 6 cm, so BM = 1 cm = DM = AD = MC = CB, and AB = 2 cm. AM = sqrt(3)(BM) = sqrt(3) cm.

AB = 2 cm
AM = sqrt(3) cm
BM = 1 cm

le radical galoisien said...

it's a simple case of solving for all the angles. which then makes you realise that AD = BC = DM = MC.

So thus AB is 2 cm and then a simple application of pythagora's theorem.

use sin and cos relationships to check.

In this case, sin BAM = sin(30 deg) = BM / AB = 1/2 (yay)