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.

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)

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…

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.

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)…

imagine a plane in the shape of a square perpendicular to A1AC with points Aclone and A1clone at the same height as A and A1 respectively. if it existed it, AAclone would be perpendicular to AC. so EAB=60 degrees (step 11), therefore AcloneAB would be 30. IF AAClone was parallel to EB, then their angles would be the same in the crooks formed by the angles AcloneAB and ABE. ABE=30 degrees (step 10)

therefore muthafuckin’ parallel and since A1A is perpendicular, BED is perpendicular to the plane formed by AA1C1C. and therefore BD perpendicular to A1C.

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.