What was the outcome of the UK's RSC maths challenge?

[KTC]

http://news.bbc.co.uk/1/hi/education/6589301.stm"

The sad thing is I don't think I can do it even if I try. Nevermind first year, 3/4 worth of a Maths degree...

 

[Alkatran]

This is very startling, but I do see a few problems. For one, the questions are from different classes of tests. Assuming the diagnostic question isn't cherry-picked to be simple, its still what you would give to struggling students to see if they even have a chance. Entrance exams, on the other hand, are meant to filter out the best and brightest. Also, because China has so many more people, they can afford to cut away a larger portion of applicants with harder tests.

I'm confident I can solve the entrance question, but it might take me an hour.

 

[uart]

Yeah I also had the feeling that they probably cherry picked the easiest UK question and the hardest Chinese question there, just to prove their point.

I'm confident I can solve the entrance question, but it might take me an hour.

I know the feeling, I'm not particularly good at visualizing 3D problems. It's like I know I can do it by tediously finding vertice coordinates and finding equations of planes and normals etc, but you know I just couldn't be bothered. :)

BTW. After staring at it for a little while I could find enough symmetry to solve part b) easily enough without any "brute force". The answer to that part is 90 degrees.

 

[Office_Shredder]

http://news.bbc.co.uk/1/hi/education/6588695.stm

Within the Uk the emphasis had shifted over the years. Where once there was "a huge amount of geometry and algebra", students these days spent more time on statistics, probability and data interpretation.

"I recall admitting and then teaching some very bright and well-taught Chinese students while a tutor in Oxford," he said.

"At school these students had done lots of geometry, algebra and all kinds of puzzle-solving.

"But I found myself having to teach introductory calculus to some of them. Some had just not done it."

Similarly, in the United States, many good students might not learn calculus until they got to college.

"So I could set a test for university entrants in China (or the US) which many British sixth form maths students could do, based on some calculus, which could make a similarly unbalanced media story in the Chinese papers," he said.

Basically, half the problem is they chose a math subject that's not particularly focused on in the UK

 

[Edgardo]

BTW. After staring at it for a little while I could find enough symmetry to solve part b) easily enough without any "brute force". The answer to that part is 90 degrees.

Thanks for the answer, lol...
*quickly runs to the computer and writes an e-mail to the Royal Society of Chemistry in hope to get the £500 *

 

[Office_Shredder]

Thanks for the answer, lol...
*quickly runs to the computer and writes an e-mail to the Royal Society of Chemistry*

Now if only you had parts a and c

 

[uart]

Haha, Ok if anyone wants to try for the money then the answer to part c) is 0.68472 radians (to 5 dp) and the answer to part a) is "by inspection". :p

 

[disregardthat]

How can the line $$B D$$ lie on the line $$A_1 C$$ They does not intersect..

 

[KTC]

The question didn't say $$BD$$ lie on the line $$A_{1}C$$. It said prove that it's $$\perp$$.

 

[disregardthat]

Doesn't that sign mean that a line is 90degrees on another line like the little image looks like? What does $$\perp$$ mean?

 

[KTC]

Yes, perpendicular. But you're in 3D.

What you get if you project the line $$A_{1}C$$ onto the plane $$ABCD$$ is on? Hence uart of "by inspection". ;) :D

 

[disregardthat]

I don't understand, what does $$B D \perp A_1 C$$ mean here excactly.
You mentioned the plane $$ABCD$$ But how does that fit in here?

 

[Alkatran]

I don't understand, what does $$B D \perp A_1 C$$ mean here excactly.
You mentioned the plane $$ABCD$$ But how does that fit in here?

Two lines P + t*Q and R + t*S are perpendicular if and only if Q . S = 0.

 

[disregardthat]

I have to read some about vectors. Very amusing signature by the way

 

[uart]

I don't understand, what does $$B D \perp A_1 C$$ mean here excactly.
You mentioned the plane $$ABCD$$ But how does that fit in here?

Jarle, in two dimensions the only instance in which two lines never intersect is if they are parallel. In three dimenions however it often happens that two lines never intersect, even when they are not parallel.

Just becuse two lines don't intersect it doesn't mean that the angle between them isn't still well defined. For a really simple example think of the 2D case of two parallel lines, they don't interesect but I'm sure you'll agree that the angle between them is well defined as zero. In general you can always translate a vector (slide it left-right, up-down etc) any way you wish, and provided you don't do any rotation, it will still point in the same direction. So if it helps you to visualize it you can just translate the lines until they do intersect and then consider the angle between the translated lines.

 

[disregardthat]

But wouldn't the angle between those two lines change from what point of view you wish to measure it?

 

[Gib Z]

Does the angle (0) between two parallel lines in 2 dimensions change from different points of view?

To tell you the truth, i was just being a smartass trying to say something that could be correct. If its not, I think as long as we use the same point of view for all calculations, things are fine.

 

[Moo Of Doom]

But wouldn't the angle between those two lines change from what point of view you wish to measure it?

What you're doing there, is linearly projecting the lines onto a plane - an operation that is known NOT to preserve angles. Angles are measured in the space the vectors are in (a 3-dimensional space in this case), not by projecting to 2 dimensions along some arbitrary viewpoint kernel.

To find the cosine of the angle between two vectors in 3-dimensions, simply take their dot product, and divide by the product of their norms. Then take the arccos of the result to get the angle.

 

[Edgardo]

Winner of the challenge

Here's a report about the winner of the challenge:
http://www.rsc.org/AboutUs/News/PressReleases/2007/CompWinner.asp