From an article today on the BBC news web site (http://news.bbc.co.uk/2/hi/uk_news/education/6589301.stm), 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 to the Royal Society of Chemistry's webpage: http://www.rsc.org/AboutUs/New ... ses/2007/ChineseMaths.asp

The contest closes at noon GMT on Friday, April 27, 2007.

I didn't understand question i because it uses a symbol I dont know, but the others are just a case of not being daunted. I cant be bothered (really not in the mood as covered in a different post) to get the actual answers, but basically if you're not scared off by little 1's under the id tags of the naming convention, then question ii and iii look pretty easy.

erm, i have fairly decent skills at maths. i know that you need to use trigonometry to do that question, but it uses alot of symbols that i dont know. Even if i did know, i wouldnt answer it, cos along with £500 i would get the biggest geek of the year award at school.

[edit] the bottom one is easy, though. I just need coffee or some sleep at the moment,[sarcasm] and i know that scotland is a small place, so anyone could read this and spread rumours. so im not going to answer them [/sarcasm][/edit]

The depressing thing is that any half decent A level student. The perpendicular sign isn't commonly used these days, but it should be known.

The more depressing thing is that I know I could've answered that about 2 years ago. Unfortunately 2 years of no real usage has caused me to forget the exact math and identities involved, and quite frankly reaching for the my notes would only depress me further this evening :'(

is it just me or do question 1 and 3 make little sense ?

Edit: Oops, TAA already pointed that out. The questions shouldn't be too hard for anyone with a decent A-level maths grade. It is easy to forget it after a year or two though I must say!

Quote from Shotglass :

is it just me or do question 1 and 3 make little sense ?

Makes sense to me..

Quote from the_angry_angel :

Makes sense to me..

I'm sure! Unlike..

Quote from the_angry_angel :

The depressing thing is that any half decent A level student.

Which, no matter how many times I read it, seems to feel lacking in some way.

if() {
;
}

Quote from the_angry_angel :

Makes sense to me..

in both questions the lines are skew and there is no angle between two skew lines

Acrobat ain't installed and for some reason won't install, and so I can't see the questions. Most importantly, I can barely solve a quadratic equation with a scientific calculator. When you say skew, do you mean parallel? If so, can the lines not be 180 degrees to each other?

Quote from SamH :

When you say skew, do you mean parallel? If so, can the lines not be 180 degrees to each other?

skew means neither parallel nor intersecting or in other words there is no common plane
like this:
http://mathworld.wolfram.com/i ... eps-gif/SkewLines_800.gif

the important bit is that they dont intersect and thus have no angle between them

Ahhh! Okies!

I'll just go back to doodling in the test paper margin then

Quote from SamH :

I'll just go back to doodling in the test paper margin then

how would you do that ? you cant even look at it without acrobat

edit: btw gsview works well for pdfs

Quote from Shotglass :

skew means neither parallel nor intersecting or in other words there is no common plane

Question one is to do with normals between 2 lines, and how they intersect. If a plane is a perpendicular bisector of another the dot product (I think - I honestly cant remember the actual identity and I cant be arsed to check) of the normals results in -1. Lines don't actually have to touch or cross to be perpendicular remember

Question three I'm a little hazy on. In some instances you can simply extend the lines, but they don't share a plane in this case, so that won't work. There's another method to prove this one, but for the life of me I can't remember the exact method. I do believe it's to do with normals (again) for some reason (since they're perpendicular you can make the normals.in a specific direction cross and measure the angle between that intersection, and that should mirror what the actual angle is between the 2 planes). Now I could be totally barking up the wrong tree there. My excuse being I've been drinking and had little sleep. Of course

I fully expect our local math nuts to come and hit me over the head with a cluebat and inform myself of the actual formulae.

The English test is ridiculously easy, that's GCSE level.

Probably could have answered the first one back when I was at college. Nowadays I'm only able to answer questions on car physics. Just how things change.

Quote from the_angry_angel :

Question one is to do with normals between 2 lines, and how they intersect.

where does it say any of that ?
i says prove that bd is perpendicular to a1c and those 2 lines are clearly skew

Quote :

Question three I'm a little hazy on, but if you extend the lines you should find they eventually intersect each other.

theres no way that those 2 would ever intersect

Quote from Shotglass :

where does it say any of that ?
i says prove that bd is perpendicular to a1c and those 2 lines are clearly skew

If you draw the prism top down, as if it were 2D, you should see that A1-C appears to be the perpendicular bisector of B-D. Incidentally I got it wrong, it's nothing to do with normals.

Consider it in a 2D form (from the top down), so that your standard cartesian representation of triangles works. You then find the "slope" of each triangle you can now see. The product of these 2 angles should then be -1 - which proves they are perpendicular, viewed from the top down plane. You can find this out from the details given about the prism.

Just because they're skewed in the drawn diagram doesn't mean that they aren't perpendicular from another perspective. Remember this is 3 dimensional, which results in a huge number of potential points of view.

Quote from Shotglass :

theres no way that those 2 would ever intersect

I reworded my response. Pretty sure I'm correct on the normals for this one though. I'll dig out my notes at lunch time if I feel I have the energy

I just noticed the english test! I have to say I don't remember doing that in my first year of maths, I think it must be for those who haven't done the maths A-level and are doing foundation stuff. I think you learn how to solve that kind of problem in year 8-9 (12-14 years).

Viewed from above, the quadrangle ABCD is a "kite" shape, with AC being the long axis, and BD the short axis. The triangles ABC and ADC are mirrored.

Using Pythagoras, the length of AC can be calculated: 4. Since the length of AB is 2, the angle between AD and AC is 60 degrees, and the angle between DC and AC is 30 degrees.

The length of AE is 1, EC is 3.

(i) BD is a normal vector for the plane A - A1 - C1 - C. Hence, BD is orthogonal to any vector in that plane.

(ii) View from the side, in the direction of the plane B - B1 - D1 - D. The requested angle is the same angle as the the angle between EA1 and EC1. The triangle E - A1 - C1 satisfies the Pythagorean equation, so it's 90 degrees.

(iii) Can in principle be solved by using the dot product: x.y = |x| * |y| * cos a. Choose B as origin, create 3D vectors for AD and BC1.

Quote from the_angry_angel :

Question three I'm a little hazy on. In some instances you can simply extend the lines, but they don't share a plane in this case, so that won't work. There's another method to prove this one, but for the life of me I can't remember the exact method. I do believe it's to do with normals (again) for some reason (since they're perpendicular you can make the normals.in a specific direction cross and measure the angle between that intersection, and that should mirror what the actual angle is between the 2 planes). Now I could be totally barking up the wrong tree there. My excuse being I've been drinking and had little sleep. Of course

sounds like a projection to me which is not asked in any way

Quote from the_angry_angel :

If you draw the prism top down, as if it were 2D, you should see that A1-C appears to be the perpendicular bisector of B-D. Incidentally I got it wrong, it's nothing to do with normals.

Consider it in a 2D form (from the top down), so that your standard cartesian representation of triangles works. You then find the "slope" of each triangle you can now see. The product of these 2 angles should then be -1 - which proves they are perpendicular, viewed from the top down plane. You can find this out from the details given about the prism.

Just because they're skewed in the drawn diagram doesn't mean that they aren't perpendicular from another perspective. Remember this is 3 dimensional, which results in a huge number of potential points of view.

now this time youre clearly talking about projections and the questions doesnt make clear that its about projections in any way
what it asks is if two lines are perpendicular and the angle between two lines is clearly defined as the smaller angle they form in their common plane at their intersection

from wiki:
An angle is the figure formed by two rays sharing a common endpoint, called the vertex of the angle.

or in other words you cant assign an angle to two skew lines unless you project them and for that youd need a clearly defined projection plane which is neither implied nor given anywhere in those questions

Quote from Shotglass :

from wiki:
An angle is the figure formed by two rays sharing a common endpoint, called the vertex of the angle.

For something to be perpendicular it doesn't need to share a common end point?

I appreciate I'm now well out of my field so I'm conceeding on this one. wsinda seems to be much more with it than me.

Quote from Shotglass :

theres no way that those 2 would ever intersect

Correct, but you can still define & calculate the angle between them. It's the same as the angle between the directional vectors of the two lines (and these intersect, by definition, in the origin). See also this Wikipedia page.

So I don't have to keep viewing it in a PDF file

And I have no idea where the equation starts :o

Quote from the_angry_angel :

For something to be perpendicular it doesn't need to share a common end point?

being perpendicular in repect to lines just means that the angle between them is 90° and to this the simple rule that the have to form an angle in the first place applies

Quote from wsinda :

Correct, but you can still define & calculate the angle between them. It's the same as the angle between the directional vectors of the two lines (and these intersect, by definition, in the origin). See also this Wikipedia page.

hm could be but for that AB would have to be the way vectors are denoted however in question 3 they clearly talk about lines with that exact same nomenclature
therefore the question is clearly about the (nonexistant) angle between those lines and not about the angle between their directional vectors

Quote from Shotglass :

therefore the question is clearly about the (nonexistant) angle between those lines and not about the angle between their directional vectors

That is the same, by definition. At least, that's what they made me believe when I studied maths at the university.