# UK's RSC maths challenge

1. Apr 24, 2007

### KTC

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

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...

Last edited by a moderator: May 2, 2017
2. Apr 25, 2007

### 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.

3. Apr 25, 2007

### 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 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.

Last edited: Apr 25, 2007
4. Apr 25, 2007

### Office_Shredder

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

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

5. Apr 25, 2007

### Edgardo

*quickly runs to the computer and writes an e-mail to the Royal Society of Chemistry in hope to get the £500 *

Last edited: Apr 25, 2007
6. Apr 25, 2007

### Office_Shredder

Staff Emeritus
Now if only you had parts a and c

7. Apr 25, 2007

### 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

8. Apr 25, 2007

### disregardthat

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

9. Apr 25, 2007

### KTC

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

10. Apr 25, 2007

### disregardthat

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

11. Apr 25, 2007

### 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

12. Apr 26, 2007

### 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?

13. Apr 26, 2007

### Alkatran

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

14. Apr 26, 2007

### disregardthat

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

15. Apr 27, 2007

### uart

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 doesnt 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 dont 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.

Last edited: Apr 27, 2007
16. Apr 29, 2007

### disregardthat

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

17. Apr 29, 2007

### 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.

18. Apr 29, 2007

### Moo Of Doom

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.

19. May 18, 2007