[Unknown008]

Uh, jcaron2, the calculations are good, but the answer is not. Perhaps you've missed the part where I said that the answers ranged from 0 to 999, all positive integers in all the structured questions here.

Also, I don't understand why you picked and for the lengths of and , sorry :(

[jcaron2]

Uh, jcaron2, the calculations are good, but the answer is not. Perhaps you've missed the part where I said that the answers ranged from 0 to 999, all positive integers in all the structured questions here.

Also, I don't understand why you picked and for the lengths of and , sorry :(

LOL. Yeah, if all the answers are integers, I guess that one's got to be wrong.

Since all sides of the octahedron are equilateral triangles, all angles are 60 degrees (when viewed in their respective coordinate systems). In the calculation for d2, the coordinates of point A are (3,0). Thus, the coordinates of point F must be (3-f*cos 60,f*sin 60).

[Unknown008]

Oh, okay! I understood that.

Now, come to the part where you said that we can take the squares of d_1 and d_2 instead of the actual values to make things easier. I admit that squaring makes it easier, but I am not sure whether that keeps the real value...

I tried to find the derivative, directly, but the square root sure did make it difficult. I am presently with an equation with f^4...

[jcaron2]

Well, that's embarrassing...

The y-coordinate of the face center is , not . Silly mistake. Also, I noticed that I had written the x-coordinate as 3, rather than 0, but I didn't carry that mistake over into my calculation. I think that fixes it.

However, that being said, there was a much easier way to have done this in the first place (as you probably could have guessed). Given the constraints of the problem, that the shortest distance passes through point F on segment AB, we can just flatten the two equilateral triangles onto a single plane as shown in the figure below.

Now, calculating the distance is easy, especially since I've written the coordinates of the start and end points on the drawing. No need to even calculate f.







At least this time the answer's an integer!

[Unknown008]

Hey! Yes, that should be it! Strangely, I didn't find that mistake of yours either :(

Anyway, thanks jcaron2! :D

2 questions left! :)

[jcaron2]

It's ok, I think I can give you a greenie now, had spread the rep enough times... let's try.

30. Finally the last one!
A trapezium ABCD has AD and BC parallel and a point E is chosen on the base AD so that the line segments BE and CE divide the trapezium into three right-angled triangles. These three right-angled triangles are similar, but no two are congruent. In common units, all the triangles' sides lengths are integers. The length of AD is 2009. What is the length of BC?
~~~

Here, my drawing tells me that if the angle BEC is 90, and AB is a slant, then ABE must be 90 too. However, that means that AB and EC are parallel, and therefore the triangles ABE and BCE are congruent. However, if that were a rectangle, then, that would solve matters. Is that right?

Jerry, the drawing below is the only construction I can see that meets all the requirements. Since the triangles are similar, the angle is the same for all of them. Since angle ABC=90+ and angle BCD=90+ as well, and AD is parallel to BC, we know the trapezium is symmetrical. Thus AB=CD.

Now by similarity we can see that





The only numbers that add to 2009 and multiply out to a perfect square are



Thus, AB must be 980, AE must be 784, and DE must be 1225. The rest of the lengths can be found simply with the Pythagorean theorem.

Josh

[galactus]

Very good, jcaron. That is what I came up with too.

The first perfect squares I saw were, as in my previous post,

I did not draw out the trapezoid. Most of these problems involve some sort of observation.

[jcaron2]

Aw.. a pity.. Ok, I'll move on to the next one then. (that question was worth 8 marks)

28.
The country of Big Wally has a railway which runs in a a loop1080 km long. Three companies, A, B and C run trains on the track and plan to build stations. Company A will build three stations, equally spaced at 360 km intervals. Company B will build four stations at 270 km intervals and Company C will build five stations at 216 km intervals.

The government tells them to space their stations so that the longest distance between consecutive stations is as small as possible. What is this distance in kilometres?

Brute force in Matlab says the answer is 174.

[Unknown008]

Awesome Josh! :) I never thought that it was like that. Had to spread the rep :(

Now, how am I supposed to find 784 and 1225? I am not allowed any calculator in that 'exam'. Is there any sort of 'trick' that could help me?

And finally the last one, how exactly did you get that? The factors of 174 are 2, 3 and 29. I don't think you got that from the common factors nor multiples :confused:

[galactus]

I think he may have used an optimization algorithm in MatLab. Such as a Traveling

Salesman-type Problem.

If it is not too much trouble I would like to see the code. I have used Matlab for these as

Well, but am probably not as versed as Josh in its application.

174 does seem like a unlikely result. But, hey, who am I to argue with matlab:)

[Unknown008]

The weird thing in that is that those questions are intended for children of 11 and 12 years old! :eek: And without any sort of aid, only graph paper, compasses, ruler... wait, I'll take a look... where was I? Yes, scribbling paper. Which are not permitted: calculators, slide rules, log tables, maths stencils, mobile phones or other calculating aids. Oh, wait, it's not for 11 and 12 years old children, it's for "Australian School Years 11 and 12" :o

[jcaron2]

Awesome Josh! :) I never thought that it was like that. Had to spread the rep :(

Now, how am I supposed to find 784 and 1225? I am not allowed any calculator in that 'exam'. Is there any sort of 'trick' that could help me?

Notice that 2009 = 49*41. That you CAN do in your head. Since 49 is a perfect square, you can just solve the problem for 41 (The answer being 25 + 16, of course). Then, remultiply those numbers by 49 (and get 1225 and 784), and the results are still perfect squares which now add up to 2009.

And finally the last one, how exactly did you get that? The factors of 174 are 2, 3 and 29. I don't think you got that from the common factors nor multiples :confused:

Here's my code. I didn't do anything fancy like optimization or a golden search or anything like that. I just made a brute force nested for-loop that calculates the maximum distance for all possible combinations and reports back the best result.

A=0:360:1080;
B=0:270:1080;
C=0:216:1080;

mx = 1000; % Dummy value to start
for b=0:270
for c=0:216
d = max(diff(unique([A B+b C+c]))); % Finds the maximum distance between any two stations
if d<mx
bestc = c;
bestb = b;
mx = d; % If the value for this combination of b and c is smaller than the previous max, set the max to this value.
end
end
end

bestc
bestb
mx

--------------------

MATLAB'S OUTPUT:

bestc =

102


bestb =

6


mx =

174

In other words, the best case scenario is when you space a B station 6 km from an A, and space a C station 102 km from that same A.

[Unknown008]

Ok, I got the 2009 thingy! :D


And for the nested loop, how could I do that with only paper and pencil? :o

[Unknown008]

Hey guess what? I got the results! Not the solutions though :( These will come in later.

I've got 58 out of 135 (not quite a score to have, I know) but I'm in the 97th percentile!

I had a discussion with my math teacher. I didn't understand well why he did those, but he feels confident about it.

Ok, the highest common factor of the three distances is 18, right?

He then did 1080/18 giving 90.

There are 12 stations is total. 90/12 = 5.

Now, 18 * 5 = 90.

So, 90 km is the answer according to him and other friends...

[morgaine300]

I've got 58 out of 135 (not quite a score to have, I know) but I'm in the 97th percentile!

Congrats! If that's 97th percentile, then apparently people don't get very high scores. Obviously you have to consider that. :-)

[Unknown008]

Thanks morgaine! I know. This statistics was from my school though. There will be only one prize winner. One from form 3, that makes.. hm.. one 14/15 year old boy in the school (of course, he had a paper of different level than mine.

[morgaine300]

That 14/15 year old boy is just some smart alec - don't worry about him. :-)

You should be proud of what you have accomplished and not worry about always being the winner of everything. OK, it's a nice ego boost when you do something great, but in the long-run overall good accomplishments and doing your best are more important.

[/pep talk from much older person]

[Unknown008]

I'll try looking for other questions... I might have a similar one later on ;)

[Unknown008]

Ok guys! I got the answers now!!

You remember the parallelogram with 6 inscribed circles? Well, the answer was not E, that is 216. It was D. I wasn't so sure about what you were telling me galactus... and it seems that you overlooked something, which I think is an overestimate of the actual area.

For the others though, you were all right!
As for the question about the train stations, I still don't understand how to do it in the head... or on paper :(

[Unknown008]

Ok, got some more questions I'm stuck...

On my car, a particular brand of tyre lasts 40 000 kilometres on a front wheel or 60 000 kilometres on a rear wheel. By interchanging the front and rear tyres, the greatest distance, in kilometres, I can get from a set of four of these tyres is

(A) 52 000
(B) 50 000
(C) 48 000
(D) 40 000
(E) 44 000
~~~~~
I got 52 000. Is that right? I initially got 50 000 but since after exchanging the wheels, there was the front wheels not yet completely worn out, I thought that I may get some more distance, hence 52 000...