Anybody up for a challenge?

[f1.redbaron]

Most of us are either engineering students or have a degree in engineering, so solving a simple grade 12 math question shouldn't pose a problem. Yeah, right

Solve it, and send it for a chance to win 500 Pounds.

http://news.bbc.co.uk/2/hi/uk_news/educ ... tm#chinese

[Tom]

I can do the English ones easily:
i, 5m
ii, 6m^2
iii, 36.87degrees

I'm definetly going to show my math teacher the Chinese one, its all just vectors I think and its probably not as hard as it looks.

[joseff]

Tom is correct: doing everything with vectors is the easiest.
without vectors you could do i & ii easily though:

i) By definition of a square prism, ABCD is orthogonal to AA'C'C and therefore BD is perpendicular to A'C. QED.

ii) Consider ABCD: given AD=2, DC=2*(3^0.5) AD perpendicular to DC, therefore by Pythagoras' Theorem AC=4.
AC perpendicular to BD therefore AC * DE = AD * DC
DE = 3^.5
Consider the rectangle AA'C'C, which is orthogonal to both A'BD and BC'D and therefore the angle between both planes is the angle between A'E and EC'
By Pythagoras' Theorem:
A'E = 2, EC' = 2*(3^0.5) and therefore A'EC' is a right angle.
The angle between A'BD and BC'D is 90deg.

iii) we'll have to do this vector-style.
Frame of reference:
i // EC, j // EB, k // AA'
therefore the vectors: AD = (i -3^0.5j 0) and BC' (3i -3^0.5j 3^0.5k)
let the containing angle = a, therefore:
cos a = (AD.BC') / (|AD| * |BC'|)
cos a = 6 / (2 * 15^0.5)
a = arc cos (6 / (2 * 15^0.5))

And at this point I'm just too lazy to run the number through the calculator

Edited to insert i, j, k

[Saribro]

I can do the English ones easily:
iii, 36.87degrees

And you started out so well :p.
They asked for tan(theta) in fractional form, aka 3/4, not the angle itself :).

In any case, that English test is beyond silly, the Chinese test seems a similar level to what we got as entrance exam for engineering.

[wazojugs]

I can do the English ones easily:
i, 5m
ii, 6m^2
iii, 36.87degrees

I'm definetly going to show my math teacher the Chinese one, its all just vectors I think and its probably not as hard as it looks.

Did part 3 have to given as a fraction? if so is it 3/5 giving 0.6 then the inv sine of that gives 36.86 degrees.

Angle (^) = (5/sin 90) x (3/sin ^)
cross multiplying gives 3 x sin 90/ 5 therefore 3 x 1/5 or 3/5 = 0.6

[Tom]

You're completely right, I'm using the technical studies way, of course in maths it would most likely be expressed as a surd or radian. This is something i have to work on for my exams, reading the question properly, in truth I just looked at the first 3 words and thought 'bah, this is too easy.'

I was using the calaulator from the start menu so its probably completely wrong anyway.

[wazojugs]

tom, always read the question. i used to sit and read the whole exam paper questions to get the brain going and then start. if you go in straight away you can miss read the question. best thing i was told.

[wazojugs]

looking back at the amount of math exams i have done it worries me that it took a while for the brain to kick in doing these

[f1.redbaron]

I can do the English ones easily:
i, 5m
ii, 6m^2
iii, 36.87degrees

I'm definetly going to show my math teacher the Chinese one, its all just vectors I think and its probably not as hard as it looks.

Did part 3 have to given as a fraction? if so is it 3/5 giving 0.6 then the inv sine of that gives 36.86 degrees.

Angle (^) = (5/sin 90) x (3/sin ^)
cross multiplying gives 3 x sin 90/ 5 therefore 3 x 1/5 or 3/5 = 0.6

Wait...now you have me confused. They are looking for the fraction of tan theta. Tan theta is found by dividing opposite (3) by adjacent (4), meaning that tan theta = 3/4. So Tom's value of ~37 degrees is correct as arctan (0.75) ~ 37 degrees. Why would you use sine?

----------
Joseff, I'd submit your answer if I were you. Maybe you score 500 pounds.

[joseff]

Joseff, I'd submit your answer if I were you. Maybe you score 500 pounds.

It's been slashdotted. Now every geek in the known world will be submitting answers

[Ciro Pabón]

... and all the geeks at Slashdot will be wrong, I think.

It seems to me that your demonstration for point 1 is not complete.

The fact that the bottom face of the prism ABCD is perpendicular to the plane AA1C1C, which is true for any "square prism" (right prism), does not imply that the diagonal AB on this bottom face is also perpendicular to this plane.

You have to use the condition given about AC perpendicular to BD. If this is not true, then AC1 wouldn't be perpendicular to BD.

f1.redbaron is right.

[joseff]

It seems to me that your demonstration for point 1 is not complete.

You're right. The correct answer for (i) is:
AC perpendicular to BD
AA'C'C perpendicular to ABCD by definition of square prism
therefore BD perpendicular to AA'C'C
therefore BD perpendicular to A'C
QED

Man, after all these years writing "QED" still gives me tingles

[f1.redbaron]

I would like to correct one thing. It is, of course, possible to use sine to find the answer, but it would make your life needlessly difficult.
Since tan = sin/cos, and we have found sine to be 3/5, cosine can be found in the similar manner, i.e. it would be 4/5 = 0.8.

So dividing 0.6 by 0.8 we get 0.75, which is the answer we're looking for.

------

About the first problem. Are any of you guys having difficulties converting a 2D picture into 3D in your mind. If the picture is clear enough (say like those ones in Mechanics books), I have no problem with them. But if they are like the ones from BBC's site, or similar ones that I had come across on the exams, I spend a long time trying to visualise the forces (or whatever is in the problem). Ironically, after that, the problems become a joke.

I know it sounds like a stupid question, but does anybody have any tips on how to approach those problems?

[checkered]

F1redbaron, I have sometimes

encountered situations where the one devising the task has evidently first drawn a clear picture of the situation and then decided that it'd make his "clever" question far too easy to solve. Ergo, they draw a poorer representation or add something that has little to do with the problem and it becomes harder. This, of course, serves no pedagogical purpose and most likely is an unnecessary diversion for the students. But teachers and professors alike are imperfect beings, too.

It helps to bear in mind that usually the subject matter being tested is fairly limited. If the test/exam questions seem unrelated and more diverse than during coursework, the correct starting point indeed is to first make the connection and not be baffled by the paraphernalia. In the end, it is about the subject. Simplify, for example in the "chinese" problem, it pays to draw it by the given figures by yourself and not get stuck with what you got presented. (Edit: try an axonometric representation, for example)

Do what you can, don't preoccupy yourself with things you can't do. (Edit: Ever seen one of those "funniest home videos" shows (which they rarely are)? There's quite often a clip of a young kid on his/her bike driving into a mailbox on a driveway without any other obstructions around. I'm pretty certain that's an accurate graphic representation of how people who are somehow overwhelmed by math feel. It's not a failure of perception, logic, capability or action - it's about getting these to work concertedly.)

Testing, of course, isn't comparable to actually doing research, innovation and such.

The comparison between the two questions presented by the UKRSC is quite haphazard anyway; both of the whole tests certainly must feature both easier and harder tasks and we don't know the balance between those. The most striking difference between the two, for me, isn't the level of ease or difficulty - knowledge is knowledge.

But the English one is very mechanical and demands only the application of prelearned (preprogrammed) tools that need not be understood in a very fundamental way while the Chinese one has an element of proving something. And that goes into the area of logic, which is at the base of any mathematical application. This approach, IMHO, is worth extra investment if, in the end, we want people thinking for themselves.

[wazojugs]

I can do the English ones easily:
i, 5m
ii, 6m^2
iii, 36.87degrees

I'm definetly going to show my math teacher the Chinese one, its all just vectors I think and its probably not as hard as it looks.

Did part 3 have to given as a fraction? if so is it 3/5 giving 0.6 then the inv sine of that gives 36.86 degrees.

Angle (^) = (5/sin 90) x (3/sin ^)
cross multiplying gives 3 x sin 90/ 5 therefore 3 x 1/5 or 3/5 = 0.6

Wait...now you have me confused. They are looking for the fraction of tan theta. Tan theta is found by dividing opposite (3) by adjacent (4), meaning that tan theta = 3/4. So Tom's value of ~37 degrees is correct as arctan (0.75) ~ 37 degrees. Why would you use sine?


easy i am using the sin rule which works out the angle or side of any triangle (right angled or not)

A/sin a = B/sin b = C/ sinc

alot easier formula to use