From an article today on the BBC news web site, the Royal Society of Chemistry in the UK is offering a £500 prize if you can solve a mathematics problem set as a pre-entrance qualification exam for prospective university students in China.
Here’s the link at the Royal Society of Chemistry’s webpage: http://www.rsc.org/AboutUs/News/PressReleases/2007/ChineseMaths.asp
The contest closes at noon GMT on Friday, April 27, 2007.
SOHCAHTOA
Holy shit, the first-year English test doesn’t even require hard square roots.
edit: and now the math nerd in me wants to try and solve the first one. sigh
I’d be the first to start chiming in on the substandard American math education. The lag girls start to show at age 12 is a travesty. This is the only culture I’ve seen with the concept of a nerd. However, there’s a lot more people in China and much fewer spots, so they have to be more competitive.
In Japan they bully and kill the stupid kids. Not the smart ones. Darwin for the win!
For contrast: this is the New York State Regents 2007 sample test. It’s been mandatory for all high school graduates to take (but i think passing rate is abominable)
… don’t you need Algeo to prove this? Like dot and cross products, and all that other stuff I have never seen since leaving high school?
Also, I got as far as AC = 4. I rule.
Arg. It would have to be geometry, my most hated of mathematical areas.
What does ABCD-A1B1C1D1 mean?
The questions don’t seem that hard, but the diagram is. I think that’s the joke. The difficulty doesn’t come from the math, but rather the horrible test designers.
It seems like a typo, but unsure. They might mean similarity, parallel, or equality.
It seems like fairly standard high school geometry, from what I remember, but a bit more overwhelming than usual in the number of lines involved. I think I could do it, but I’d need a refresher from my old notes first, and maybe not under test conditions.
Though I’m not even sure what the first question means, since BD and A1C don’t even intersect…
It’s the points that refer to the square prism in the diagram. ABCD is the base part of the object, and A1B1C1D1 is the top.
Lines don’t have to intersect to be perpendicular to each other in 3D space.
Edit: At least I don’t think so. Some Maths genius could probably tell us. hehe
I’m not a genius, but I’m pretty sure lines have to intersect to be perpendicular.
Lines are perpendicular if their dot product is 0, they don’t have to intersect.
I figured as much, but I don’t understand what the hyphen between them signifies.
This doesn’t seem that much beyond what I learned in 10th grade geometry, but I couldn’t do the problem now without boning up. I don’t even remember how to do a mathematical proof, for instance.
All of my math beyond arithmetic, and maybe some very elemental algebra, has atrophied.
I just figured it was some kind of geometric labeling convention, but I don’t know more beyond that.
Line AC = 4 : use pythagorean theorem on triangle ADC having been given AD and DC.
angle DAC = 60 degrees : soh
angle DCA = 30 : angles in triangles sum to 180 degrees
ADE = 30 degrees : AC perpedicular BD, we found angle DAC to be 60 degrees in #2, sum triangles 180 degrees
angle BDC = 60 degrees: given 90 degree angle ADC, step 4 found ADE=30 degrees
angle DEC = BEC = AEB = AED = 90 degrees : AC perpendicular DB
length AE = 1 : sin 30 =1/2 = O/H = AE/AD = AE/2
length EC = 3 : step #1 AC=4, AC-AE = 3 = EC
length DE = square root of 3 : pythagorean theorem on right triangle DEC, DC = 2(radical 3) given, EC =3 step #8
angle ABE = 30 degrees : step 7 AE = 1, given AB=2, soh, sin 30 degrees=1/2=AE/AB
angle EAB = 60 degrees : sum triangle angles = 180, step #10 gives 30 ABE= degrees
11.5. angle DAE= 60 degrees : sum triangle angles=180 , DAE=90, AED=90 step #6
angle BAD=120 degrees: sum of DAE and EAB (step 11)
length BE = sqrt of 3: in a 30-60-90 triangle side opposite 30=x, opposite 60=radical 3, opposite 90 = 2x, here x =1 = length AE step 7.
BC= 2 *sqrt of 3 : BE=sqrt of 3, EC = 3 step 8, pythagorean theorem
angle ACB = 30 degrees, angle ABC= 90, angle EBC=60: we have a triangle with sides in ratio x, radical 3, 2x, sides, therefore a 30-60-90 triangle,
(huh, the base is imperial star destroyer shaped)
angle DBC=60, : summing angles given and found in #12 and 15, all angles in 4 sided polygon sum to 360
QE = (3*sqrt3)/4 : A1AC forms a triangle, a line going straight up from E will intersect A1C, at a point i’ll call Q, forming proportional triangles, A1AC and DEC. using the two shorter sides on each we get ratios: sqrt3 over 4= QE/3 = A1A/AC = QE/EC. from givens and step #1 and 8.
hmmmmm…we have a BE = radical 3 * (4/4), step #13 and QE=radical 3 * (3/4). so we potentially have a 3:4:5 right triangle IF QB is radical 3 * (5/4)…
therefore muthafuckin’ parallel and since A1A is perpendicular, BED is perpendicular to the plane formed by AA1C1C. and therefore BD perpendicular to A1C.
(i) solved
try my hand at the other ones tomorrow.
Are they joking? Geometry was 8th grade. Why would you put it on a college entrance exam? What about calculus? Or that utility discipline, statistics?
Geometry in the 8th grade? Maybe some basic stuff, but the real stuff is generally sophomore in HS. And most high school kids don’t take calculus until college, if at all. I had calc in senior year due to being in the nerd classes, but most kids get trig.
I don’t even remember how to do a mathematical proof, for instance.
I don’t either, for the most part. I do remember a lot of geometry, trig and linear algebra, though, since I use them all the time on the job.
We did basic calculus in our last two years of highschool - derivatives and integrals from first principles to using the rules. I think we even did some basic differential calculus.