LOL, yup.

26.
We say a number is ascending if its digits are strictly increasing. For example, 189 and 3468 are ascending while 142 and 466 are not. For which ascending 3-digit number n (between 100 and 999) is 6n also ascending?
~~~
Sigh, just got it... 578.

27.
A regular octahedron has edges of length 6 cm. If d cm is the shortest distance from the centre of one face measured around the surface of the octahedron, what is the value of d^2?

EDIT: My drawing is not to scale.

I am not quite getting this question. The center of a face? When they say "around the

surface" of the octahedron, do they mean one much traverse each side in the shortest

distance? I wonder. I see you have 'start' and 'end'.

The straight line distance from the 'start' to the 'end' would be

Square this and get 72. I doubt if that's what they are getting at though.

This question also bugs me... Could it be from the centre of a face, to the centre of the opposite face? Then, around the surface is like making the octahedron a solid shape, and say for example 'what minimum length of thread can you use to link the two centres?' That way, the problem would seem to be 'tougher'.

Also, the 'start' and the 'end' are not on the edge, but on a face. I think that on the drawing, the dot is found on the left 'down-slanting' face and the right dot on the right 'up-slanting' face...

I am not going to tackle this one because I am unsure of what they even mean.

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 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?

Perhaps the gcd's.

To visualize, draw a straight line and scale the various stations. You'll see. Then see if it matches what you calculate using the hints above.

Quote:

---

Originally Posted by galactus

Check me out, but wouldn't they be
?.

The integers sum to 42.

2, 4, and 6 are in the ratio 1:2:3.

---

Darn - I was almost there and then gave up. :rolleyes:

I still can't get it. After the drawing, 216 km was the greatest distance. So, I moved all the stations for C so that the 216 is now halved, that is 108 km. However, since all the other stations of the same company have to move as well, more 'space' is created, and the new longest is slightly less than 216 km. I don't know how to find accurately that distance... :(

The LCM of the distances give 1080 km. That means all fit in the loop. And I'm stuck there.

GCD? I don't know that... is that greatest common denominator?

Yes, Greatest Common Divisor or Denominator.

gcd(270,360)=90

gcd(216,360)=72

gcd(270,216)=54

The smallest one is 54. That is the shortest longest distance. That is what I was getting at.

270-216=54. There is no smaller gcd, so this is it.

Ok, you mean, that the answer to that question is 54?

What you posted was LCM, so got me confused.

Also, I didn't know about gcd until now. What I knew was HCF, highest common factor. Seems that they call it differently in different countries.

Yes, sorry, a typo. I meant the GCD instead of LCM. We know the LCM is 1080 as given.

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?

Let me think about that railway problem some more. It may very well be 108.

There are 12 stations, so 1080/12=90. One would think the distance could not be less than 90.

Well, from my drawing, the longest distance is slightly less than 216/2 = 108

Quote:

---

Originally Posted by me

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?

---

That doesn't work. Through trig, it includes that the side EC is a fraction... =/

Quote:

---

Originally Posted by Unknown008

Ok, I'll have a bunch of challenging questions from the AMC (Australian Mathematical Competition) I did today. I'll post one at a time so as not to confuse the posters and myself. The questions I suppose will be of ascending difficulty, those which I wasn't able to solve.

1. There's a given equation; . There was a sketch along, that of an inverted parabola, which had a positive y-intercept and the turning point was on the y-axis.

Which is true?
a) a + b + c = 0
b) a + b - c < 0
c) -a + b - c > 0
d) a + b + c < 0
e) There is not enough information.

I ruled out a) and d), since there is a solution other than 0 when putting x = 1.
The others, I'm at a lost.

Thanks for replying :)

Survivorboi, wanna make an attempt? I'm sure you'll be interested too to know how to solve the problems I'll post ;)

---

As you already pointed out, clearly a < 0 since the parabola points down, and clearly c > 0 since the y-intercept is positive.

That just leaves b. Since the vertex of the parabola is on the y-axis (where x = 0), b has to be 0. This can be shown in several ways. Here are a couple:

1) The two x-intercepts are equidistant from the origin. In other words, if one intercept is at x=n, the other is automatically at x=-n. Thus, the equation for the parabola is , where m and n are constants. Notice there is an term and a constant term, but no term. Thus b=0 in the equation of this parabola.

2) The slope (derivative) of the parabola is zero at the vertex. This coincides with x=0, since the vertex is on the y-axis. So

when





In any event, if we know a<0, b=0, and c>0, the only statement from above that is definitely true is that a + b - c < 0.

Hey, thanks jcaron2! That was a really good explanation!

If you could take a look here: https://www.askmehelpdesk.com/mathem...ml#post1929361

And here: https://www.askmehelpdesk.com/mathem...ml#post1927007

And here:https://www.askmehelpdesk.com/mathem...ml#post1925448

These are the last 3 questions where I haven't got the answers and working yet. :o

Quote:

---

Originally Posted by Unknown008

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?

---

I have not looked at this to any length, but I noticed the 2009. Normally, with these problems there is some observation to make. Two integers that lead to 2009 are 1960 and 441.



This may have something to do with it. Just a thought.

BC may be 1960. Check and see if you wish.

Attachment 23824

Quote:

---

Originally Posted by Unknown008

27.
A regular octahedron has edges of length 6 cm. If d cm is the shortest distance from the centre of one face measured around the surface of the octahedron, what is the value of d^2?

---

All right, I'll take a crack at this one. Sorry my last post was way behind the times. Somehow I overlooked the fact that there were five additional pages of posts after the first one. :-)

Judging by the picture, the question has got to mean from the center of one face to the center of the face on the opposite side of the octahedron. So let's label some of the vertices so you can more easily follow along with my logic (or lack thereof?). See the attached drawing.

First of all, because of symmetry we know that the shortest distance will pass through the point E which is the midpoint of AD. Also, the total distance, d, will be 2d1 + 2d2 by my drawing. (It's not very clear, but d1 represents the length of the dotted segment from the center of the upper right face to F). We know the coordinates of all of the labeled points except F. We only know that F will lie somewhere on AB and will be such that d is minimized. Let's use f for the length of segment AF. Now we can write out d1 and d2 in terms of f.

Let's start with d2. We'll use a coordinate system where ABD is in the XY plane, and AD is along the X axis, and EB is along the Y axis. This means E is at the origin. The distance d2 can be written as







If we use an analogous coordinate system for triangle ABC (where the origin would be at the midpoint of AC), we can calculate d1. First, it's helpful to realize that the center of ABC will be at .





In order to minimize d we need to minimize the sum of the distances d1 and d2. However, since all of these lengths are positive numbers, we can make the problem much simpler by realizing that we accomplish the same thing by minimizing the sum of the squares of those distances. That gets rid of the radicals for us and makes the derivatives trivial.

So to minimize with respect to f, we simply take the derivative and set it equal to zero:





Now we simply need to plug this value of f back into the equations for d1 and d2, add them up (twice each for the full distance d), and square it to get the final answer. I'll skip most of the simple algebra. See if your answers agree with mine.







Seems like kind of a wacky answer. Maybe I made an algebra error somewhere? Let me know if you get a different one.

Josh

Quote:

---

Originally Posted by galactus

I have not looked at this to any length, but I noticed the 2009. Normally, with these problems there is some observation to make. Two integers that lead to 2009 are 1960 and 441.



This may have something to do with it. Just a thought.

BC may be 1960. Check and see if you wish.

---

Thanks for the reply, that was a good catch, these Pythagorean triple... I tried that. I'm not sure of it, but if I have a triangle, and make another triangle inside the triangle with edges on the sides of the larger triangle, and each side of the smaller triangle have to be parallel to one side of the larger triangle, I end up with a smaller 'inverted' triangle, which have sides equal to half the sides of the larger triangle. However, that cannot be the solution, since the answers are integers, and not above 999. :(

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 :(

Quote:

---

Originally Posted by 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 :(

---

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).

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...

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!

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

Anyway, thanks jcaron2! :D

2 questions left! :)

Quote:

---

Originally Posted by Unknown008

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

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.

Quote:

---

Originally Posted by Unknown008

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.

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:

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:)

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

Quote:

---

Originally Posted by 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?

---

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.

Quote:

---

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.

Ok, I got the 2009 thingy! :D


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

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...

Quote:

---

Originally Posted by Unknown008

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. :-)

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.

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]

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

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 :(

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...