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; y=ax^2 +bx+c. 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, want to make an attempt? I'm sure you'll be interested too to know how to solve the problems I'll post ;)

Is this competition for fun, or for school? And how long before you know how you did?

As for the questions, if this is supposed to be the least difficult, I give up already. ;)

I did it yesterday, and have no idea when the results will come. It's a competition among the students of the same age as me, in the country. I know, that was actually the eighth question. I managed with the first seven. Or, you want to take a look at them too? ;)

I'm always willing to take a look at stuff. My math -- or at least my memory of it -- isn't near the level of the others here, so I just have to do stuff for my own fun.

I'm still trying to figure out what an "inverted" parabola is, as opposed to a, um, "normal" one. LOL. So is the curved part up or down?

LOL! Yup, my graphing program is not yet installed... I'll try get one online. It's a parabola with the coefficient of x^2 being negative (an inverted one, an upside down bowl ;))

Gotcha. If only I'd read this before the 2nd glass of wine. Now I'll have to think harder. :-)

So where's the first 7 questions?

Do you want me to post them too?

This took a lot of trial & error & thought since I don't know very many "rules" about parabolas. Like I had to spend quite a bit of time even finding an equation that would graph that way. (Can we do a linear one instead?) I did manage to figure out reasonably quickly that (a) can't be true.

I originally thought that C had to be greater than the absolute value of A, but finally figured out that isn't true. However, that I even had several equations where that was true eliminated (d). And (c) is kind of the reversal of that, cause as far as I can tell, B is 0 and makes no difference.

So I'm down to (b)... if I have everything else correct, this is it. We know A is negative and that C is positive. (We do know that, right? :o) If I'm right that B is 0, then it's answer (b). If you start with negative and subtract C, you're more negative. Hence, less than zero.

Yup, I think you're correct...

c) would be correct also if -a was greater than c. But since b) does not put such conditions, it must be b).

I took some time to realise that b=0 (which was quite obvious).

Ok, if you want the first 7 questions, here they are:

1. Find the value of (2009+9)-(2009-9)
a. 4000
b. 2018
c. 3982
d. 0
e. 18

Obviously, the answer is e;18

2. In the diagram(first attachment sorry for the bad quality of the image :o), x is equal to:
a. 140
b. 122
c. 80
d. 90
e. 98

Here the answer is e. 98

3. The graph of y=kx passes through (-2, -1). The value of k is
a. 2
b. -2
c. 4
d. \frac12
e. -\frac12

Of course the answer is d=1/2

4. The value of 0.6^{-2} is
a. -0.36
b. 0.036
c. 9/25
d. 25/9
e. 3.6

The answer is d. 25/9

5. (x - y) - 2(y - z) + 3(z - x) equals
a. - 2x - 3y + 5z
b. -2x - 3y - z
c. 4x + y - z
d. 4x + 3y - z
e. 2x + 3y - 5z

The answer is a. - 2x - 3y + 5z

6. On a string of beads, the largest bead is in the centre and the smallest beads are on the ends. He size of the beads increases from the ends to the centre as shown in the diagram (second attachment).

The smallest beads cost $1 each, the next smallest beads cost $2 each, the next smallest is $3 each and so on. How much change from $200 would there be for the beads on a string with 25 such beads?

a. $25
b. $31
c. $40
d. $52
e. $55

I got b. $31 I'm sure of it.

7. If a*b = a + \frac1b for every pair a, b of positive numbers, the value of 1*(2*3) is
a. 10/3
b. 10/7
c. 11/6
d. 9/2
e. 3/10

I got b. 10/7

The answers are in white. Just highlight them to see them.

Ok, next questions are easier.

9. in a school of 1000 students, 570 are girls. One-quarter of the students travel to school by bus and 313 boys do not go by bus. How many girls travel to go to school by bus?
a. 7
b. 63
c. 153
d. 180
e. 133

The answer is e. 133

10. A box in the dressing shed of a sporting team contains 6 green and 3 red caps. The probability that the first 2 caps taken at random from the box will be the same colour is
a. 1/2
b. 5/12
c. 2/3
d. 3/4
e. 2/9

The answer is a. 1/2

11. QRST is a square with T at (0,1) and S at (2, 0). Which of the following is an equation of the line through the origin which bisects the area of the square? (attachment 1)
a. y = x/2
b. y = x/3
c. y = (2x)/3
d. y = 2x
e. y = 3x

The answer is y = x/3

12. [I'm stuck at this one too :()
A rectangle PQRS (attachment 2)has PQ = 2x cm and PS = x cm. The diagonals PR and QS meet at T. X lies on RS so that QX divides the pentagon PQRST into two sections of equal area. The length, in centimetres of RX is
a. x/2
b. x
c. (5x)/4
d. (3x)/2
e. (3x)/4

I ruled out a, b and e because they are too short. I don't know for the rest...

#12:

The area of the pentagon PQRST is \frac{3x^{2}}{2}

Half of that is \frac{3x^{2}}{4}

If we let RX have length y, then the area of the triangle formed by QRX has area half the area of the pentagon which is \frac{xy}{2}=\frac{3x^{2}}{4}

Solving for y gives us y=RX=\frac{3x}{2}

Interesting mix.


1. Find the value of (2009+9)-(2009-9)

They just wanted you to have a success before delving into the real test? :)
(I really shouldn't comment. I have no idea what this competition is, or the age groups involved, etc.)



2. In the diagram(first attachment sorry for the bad quality of the image :o), x is equal to:

I don't remember too many geometry rules cause I really don't work with it -- I remember some basics, and sometimes I can figure stuff out by logic. But I have no clue where your answer is coming from. I just made up a rule cause I didn't know and screwed it up. :eek:


4. The value of 0.6^{-2} is

Well, that's easy. You pull out the calculator and put in (.6)(2nd)(y^x)(2)(+/-)(=). :D

Actually... I didn't cheat. I solved it manually. Really I did.


6. On a string of beads, the largest bead is in the centre and the smallest beads are on the ends. He size of the beads increases from the ends to the centre as shown in the diagram (second attachment).

I got b. $31 I'm sure of it.

You don't sound too sure. But it's right.


7. If a*b = a + \frac1b for every pair a, b of positive numbers, the value of 1*(2*3) is

I don't even get what they're saying. 1*(2*3) is 6. How does that relate to a & b? Seriously, I don't get it.


The answers are in white. Just highlight them to see them.

Nifty idea.

Except for (7) and not remembering triangles properties, I thought this was all pretty easy. (I tutor most of this stuff.) The parabola one almost seems out of place with this stuff. (I only solved it cause I don't have a time limit. :p) But I haven't looked at the other set yet. I'm supposed to be getting work done in my yard while it's still light.

As for 9 - 12... well, 9 was easy.

10 I got but it was the long way. I knew there were 72 ways it could be done, and then had to manually figure out that 36 of those would be the same color. I don't know how to do that any shortcut way. Might be one of those things that I actually know, but don't recognize it as being such. Probabilities can be like that.

11 & 12 are both getting beyond my memory. I have a feeling I could get 11 if I tried hard enough, but my mind isn't up for the challenge at the moment. And 12 - no clue. I don't even know what galactus is talking about. Way too many years since I did that kind of stuff.

(Notice I'm having trouble with shapes? :D)

Lol, what I'm giving answers, I'm 100&#37; sure of them, morgaine! :) It's OK, I gave the paper to my mum too, and she too was rusty!

Thanks galactus! I had actually found the answer some time after I posted the question. I'm kind of a little frustrated it was as easy as that... but I can do nothing, I was thinking it was some kind of 'logical' work like some questions. I should've done the calculations... :-/

And, the level is actually 'Senior Division', with 'Australian School Years 11 and 12', 'Time allowed: 75 minutes'.

Those aussie do have a high level of maths! Even being 17, I'm stuck at some of their questions!

Ok, the next one, 13, which I had problems as well.

13. The solution to the equation 5^x - 5^{x-2} = 120\sqrt5 is rational number of the form \frac ab where b is not equal to 0 an a and b are positive and have no common factors. What is the value of a+b?
a. 3
b. 5
c. 7
d. 9
e. 11

I started like this:

Let y = 5^x

So, we have: y - \frac {y}{25} = 120\sqrt5

y= \frac{3000\sqrt5}{24}

5^x= 125\sqrt5

5^x= 5^{3.5}

x= \frac 72

so a+b=9

Oh my... I have solved it... and saw my mistake... I'm weird, I'm really weird...

The mistake I did was 120*25=300, which of course is 3000. :o

14 How many points (x,y) on the circle x^2=y^2=50[/math are such that at least on eof the coordinates x, y is an integer?
a. 16
b. 30
c. 48
d. 60
e. 100

I got d. 60

15. An eyebrowis an arrangement of the numbers 1, 2, 3, 4 and 5 such that the second and forth numbers are each bigger than both their immediate neighbours. For example, (1, 3, 2, 5, 4) is an eyebrow an (1, 3, 4, 5, 2) is not. The number of eyebrows is:
a. 16
b. 12
c. 15
d. 24
e. 18

My answer: a. 16

16. The sum of the positive solutions to the equation [math](x^2-x)^2=18(x^2-x) - 72 is :
a. 5
b. 7
c. 8
d. 9
e. 18

I didn't know at first, but after the competition, I figured it out (another frustration) The answer is b. 7

17 On a clock face, what is the size, in degrees, of the acute angle between the line joining the 5 and the 9 and the line joining the 3 and the 8?
a. 15
b. 22.5
c. 30
d. 45
e. 60

The answer is d. 45

18. A positive fraction is added to its reciprocal. The sum is x/60 in lowest terms, where x is an integer. The number of possible values of x is
a. 1
b. 2
c. 3
d. 4
e. 5

I don't know how to do this.

I started with: \frac ab + \frac ba = \frac {a^2+b^2}{ab}

Then, equating, ab = 60 , a^2+b^2 = x

What I think now, is that one of the two must be a fraction, for a^2+b^2 > ab for all positive integers. Or I'm overlooking something...

18. A positive fraction is added to its reciprocal. The sum is x/60 in lowest terms, where x is an integer. The number of possible values of x is
a. 1
b. 2
c. 3
d. 4
e. 5

I don't know how to do this.

I started with: \frac ab + \frac ba = \frac {a^2+b^2}{ab}

Then, equating, ab = 60 , a^2+b^2 = x

What I think now, is that one of the two must be a fraction, for a^2+b^2 > ab for all positive integers. Or I'm overlooking something...

a^{2}+b^{2}=x, \;\ ab=60

The values of ab=60 that satisfy are:

15*4=60
20*3=60
10*6=20
30*2=60
12*5=60

15/4+4/15=241/60
10/6+6/10=34/15
30/2+2/30=226/15
12/5+5/12=169/60
20/3+3/20=409/60

3 of those have the required 60 as the denominator when in lowest terms.

Oh, so that's what I was overlooking... I was looking for proper fractions... Thanks galactus!

Ok, next (there are 30 in all):

19.

In triangle PQT, PQ = 10 cm, QT = 5 cm and angle PQT = 60 degrees. PW, PY and TQ are tangents to the circle with centre S at W, Y and V respectively. The radius of the crcle, in centimetres, is
a. \frac{5\sqrt3}{2+\sqrt3}

b. \frac{5(3-\sqrt3)}{2}

c. \frac{5}{1+\sqrt3}

d. \frac{5\sqrt3}{2}

e. \frac{25\sqrt3}{6}

I managed to get PT = 75^{0.5) [from cosine rule] and angle PTQ = 90 degrees [from sine rule]. That makes angle TPQ = 30 degrees, and angle YSW = 150 degrees. I'm stuck here.

19.

In triangle PQT, PQ = 10 cm, QT = 5 cm and angle PQT = 60 degrees. PW, PY and TQ are tangents to the circle with centre S at W, Y and V respectively. The radius of the crcle, in centimetres, is
a. \frac{5\sqrt3}{2+\sqrt3}

b. \frac{5(3-\sqrt3)}{2}

c. \frac{5}{1+\sqrt3}

d. \frac{5\sqrt3}{2}

e. \frac{25\sqrt3}{6}

I managed to get PT = 75^{0.5) [from cosine rule] and angle PTQ = 90 degrees [from sine rule]. That makes angle TPQ = 30 degrees, and angle YSW = 150 degrees. I'm stuck here.


R+Rtan(\frac{\pi}{6})=5

R=\frac{5(3-\sqrt{3})}{2}

See how I got that? I used some of the old surveyor trig. This problem is a lot how a curve in a road is laid out.

Is that some sort of formula? I can't get it. I see you took the 30 degrees of the place where the two longer tangents meet...

TV and TY are equal to the radius of the circle length.

If we look at angle VSW is equal to 60 degrees.

VQ=Rtan(\frac{\pi}{6})

And TV+VQ=5

R+Rtan(\frac{\pi}{6})=5 and solve for R.

See, we can find the tangent length, VQ, from the formula Rtan(\frac{I}{2})

where I is a central angle VSW, which is 60 degrees.

Oh, I get it finally! Wow, I didn't think I had to think like that lol! :D

Ok, I'll skip no. 20 for later, that was pretty a 'logical' question.

21.
A palindromic number is a 'symmetrical' number which reads the same forwards and backwards. For example, 55, 101 and 8668 are palindromic numbers. There are 90 four-digit palindromic numbers. How many of these four-digit palindromic numbers are divisible by 7?
a. 7
b. 9
c. 14
d. 18
e. 21

How am I supposed to find that in about 2 minutes? :confused: I would like to know the 'trick'

If I am thinking about this correctly, 1001 is palindromic and divisible by 7.

1001, 2002, 3003,.

If we had 770, we get the others.

1001, 1771, 2002, 2772, etc.

I believe there are 18 if we count all them.

Check me out on this on.

:eek: Terrific! Amazing! Thanks yet again galactus! :) I had to spread the rep... :(

For the next number. I'll post the pic tomorrow. The circles are touching each other so that if their centres are joined, three centres would form an equilateral triangle.

22. What is the area in square centimetres, of the parallelogram that would fit snugly around 6 circles, each of radius 3 cm, as shown in the diagram?
a. 108
b. 8(4+3\sqrt3)
c. 15(2+\sqrt3)
d. 12(9+5\sqrt3)
e. 216

I tried to get the area through \frac 12 (a)(b)sin\,c, then double that yo give (a)(b) sin\,c

The base is given by (3*4) + 2(extensions)

Extension = \frac{6}{\sqrt3}

Slant length = 6 + 2(extensions)

Area = (12+\frac{12}{\sqrt3})(6+\frac{12}{\sqrt3})sin 60

= 60\sqrt3 + 118

I can't seem to get my mistake.. :( I'll check in tomorrow, bedtime here.

EDIT: Ok here's the pic.

Break it up into various rectangles and they add to 216.

Shift the parallelogram to the right by 'pushing' on the top and transform it into a rectangle. Easier to envision that way.

Those aussie do have a high level of maths! Even being 17, I'm stuck at some of their questions!

Not sure it's any higher than what we've got here. I knew how to do a lot of this stuff by time I finished 10th grade, and the rest (at least what you've posted so far) by 11th when I got to higher algebra/trig. 12th I was taking analytic geometry (which at the time I thought was loads of fun - cough!) and some intro calculus (which I hated). Of course, not a high percentage of people were taking the courses I was - I was doing the "college prep." Which is kind of funny since the college I work at gives college credit for stuff I took in high school. My high school may have been a little over-ambitious.

Now as for that time limit... took me an hour to figure out #8. LOL.


13. The solution to the equation 5^x - 5^{x-2} = 120\sqrt5 is rational number of the form \frac ab where b is not equal to 0 an a and b are positive and have no common factors. What is the value of a+b?

This doesn't seem like enough info. Was this:


So, we have: y - \frac {y}{25} = 120\sqrt5

given with the problem?

EDIT: Never mind... That 5^x... equation in there was not showing up for me in your original post. That's darn weird.


15. An eyebrowis an arrangement of the numbers 1, 2, 3, 4 and 5 such that the second and forth numbers are each bigger than both their immediate neighbours. For example, (1, 3, 2, 5, 4) is an eyebrow an (1, 3, 4, 5, 2) is not. The number of eyebrows is:

Don't understand the way they word some of these. Do they mean the number of possible eyebrows you could make?


18. A positive fraction is added to its reciprocal. The sum is x/60 in lowest terms, where x is an integer. The number of possible values of x is

I made the mistake of paying attention to what you were doing and trying to follow it. I was about to attempt it like what galactus was doing, and then when I saw what you were doing I gave up and decided I didn't know how to do it. That's what I get. :p (I usually trust myself better than that.)

As for the rest, I've given up. They're getting into too many shapes. I don't even remember what an "acute" angle is. And circles - forget it. I can find the area and that's it. So this is starting to get way beyond my memory. A palindromic? I don't even remember ever knowing that. And as soon as I saw the word tengent I quit reaading. :D

And as soon as I saw the word tengent I quit reaading.

I apparently also quit knowing how to type... sheesh.

I assume 'tengent' is meant to be 'tangent'. A tangent isn't anything complicated. It's just a line that touches a circle or other curve at some point.

For instance, the tangent to the curve x^2 at x=1 is an example in the graph below.

Also, a palindromic number is just that. Same as a palindromic word. It's the same backwards as forwards.

1001, 2002, 3223, and so on

And Napoleon's famous line, "Able was I ere I saw Elba" is a palindrome. He was exiled to the island of Elba in case you wonder what that means.

I have noticed a lot of terms in math scare people because they think it's complicated, but it's not.

How about a 'rectangular parallelopiped'? That's just another name for a box.

As for high levels of math, check out a Chinese version of the same test. Then you can say, 'sheesh'.

Break it up into various rectangles and they add to 216.

Shift the parallelogram to the right by 'pushing' on the top and transform it into a rectangle. Easier to envision that way.

I didn't know that was possible... I mean, that when you 'push' the upper left vertex, making the new shape as a rectangle will keep the actual area.


Not sure it's any higher than what we've got here. I knew how to do a lot of this stuff by time I finished 10th grade, and the rest (at least what you've posted so far) by 11th when I got to higher algebra/trig. 12th I was taking analytic geometry (which at the time I thought was loads of fun - cough!) and some intro calculus (which I hated). Of course, not a high percentage of people were taking the courses I was - I was doing the "college prep." Which is kinda funny since the college I work at gives college credit for stuff I took in high school. My high school may have been a little over-ambitious.

Ok, let me rephrase my previous comment then: Mauritius has a so low level of Maths!


How about a 'rectangular parallelopiped'?. That's just another name for a box.

I have heard that somewhere. I think in my vectors class. That's a prism with a base in the shape of a parallelogram, right?

Ok, now for the number 23.

In 3009, King Warren of Australia suspects the Earls of Akaroa, Bairnsdale, Claremont, Darlingdust, Erina and Frankston are plotting a conspiracy against him. He questions each in private and they tell him:
Akaroa: Frankston is loyal but Erina is a traitor
Bairnsdale : Akaroa is loyal
Claremont : Frankston is loyal, but Bairnsdale is a traitor.
Darlingdust : Claremont is loyal but Bairnsdale is a traitor.
Erina: Darlingdust is a traitor.
Frankston: Akaroa is loyal.

Each traitor knows who the other traitors are, but will always give false information, accusing loyalists og being traitors and vice versa. Each loyalists tells the truth as he knows it, so his information can be trusted, but he may be wrong about those he claims to be loyal. How many traitors are there?
a. 1
b. 2
c. 3
d. 4
e. 5

I got d. 4 but my friend told me otherwise...

24.
Four circles of radius 1 cm are drawn with their centres at the four vertices of a square with side length 1 cm. The area, in centimetres, of the region overlapped by all four circles is
a. 2\sqrt3 - \pi
b. \pi - \sqrt2
c. 1+\frac13 - \sqrt3
d. \pi - 2\sqrt2
e. \frac {\pi - 3 -\sqrt3}{2}

i don't even know where to start this... :(

I figured #24. I think you have a typo on c. It should be \frac{\pi}{3}-\sqrt{3}+1. That is the solution.

If you do not have a drawing, make one with a ruler and a compass so that it is accurate.

You see a little square with bulging sides in the center of the original square. That is the area of overlap for the 4 circles.

It is kind of hard to explain what I did to arrive at the solution.

I found the intersection of two of the circles with centers (0,0) and (0,1):

x^{2}+y^{2}=1, \;\ x^{2}+(y-1)^{2}=1

They intersect at x=\frac{\sqrt{3}}{2}

Subtract half the width of the square, 1/2, and we have a side of a triangle we can use Pythagoras on.

\sqrt{(\frac{\sqrt{3}-1}{2})^{2}+(\frac{\sqrt{3}-1}{2})^{2}}=\frac{\sqrt{6}-\sqrt{2}}{2}. That is the side length of the 'bulgy' square. But we have to find the area of those bulges which are 4 circular segments.

These areas can be found from the formula for a circular segment and multiplying by 4:

2(\frac{\pi}{6}-sin(\frac{\pi}{6}))=\frac{{\pi}-3}{3}

The area of the square is (\frac{\sqrt{6}-\sqrt{2}}{2})^{2}=2-\sqrt{3}

Add them: 2-\sqrt{3}+(\frac{{\pi}-3}{3})=\fbox{\frac{\pi}{3}-\sqrt{3}+1}

There are many ways to tackle this.

I made this graph with my downloaded graphing utility. It is unconstrained, otherwise, it would look more like circles than ellipses, and the center region would look like a square with bulging sides. I also drew it with a compass and rule on paper.

I assume 'tengent' is meant to be 'tangent'. A tangent isn't anything complicated. It's just a line that touches a circle or other curve at some point.

Yes, tangent. Was having trouble typing. If I don't remember how to do anything with circles and curves, I'm not likely to remember how to do anything touching it. That was also the kind of thing I disliked in high school.


I have noticed a lot of terms in math scare people because they think it's complicated, but it's not.

I've never been scared by math or terms. I just don't remember a lot of stuff I learned over 30 years ago. There's a big difference. I tutor math - just not this stuff.


As for high levels of math, check out a Chinese version of the same test. Then you can say, 'sheesh'.

I was referring to the typos. I'm picky about that sort of thing and don't usually leave them. :-)

Oh yes, should be pi instead of 1 there :o. Thanks galactus!

And for the formula for the circular segment, I didn't know that...

Is it:

\frac 12 (\theta - sin\theta)

Where theta is the angle between the sides of the concerned sector?

Actually, it's \frac{1}{2}r^{2}({\theta}-sin{\theta})

Because the radius is 1, I just left it out instead of writing (1)^2.

Ok, Okay, that's why I asked, to confirm ;)

25. *That's about functions, \frac{x+6}{x} and doing ff(x), then fff(x) etc until f_n(x).

f_2 = \frac{7x+6}{x+6} and

f_3 = \frac{13x+42}{7x+6}

Let S be the complete set of real solutions of the equation f_n(x) = x. The number of elements in S is:
a.2
b. 2n
c. 2^n
d. 1
e. infinite*

Tying the math tags plus the codes is quite lengthy and I must go now. If you get confused, I'll post tomorrow. I think it's infinite though...

Ok, the exact question:

25.
Let f(x) = \frac{x+6}{x} and f_n(x) = f(f(...(f(x))...)) be the n-th fold composite of f. For example, f_2(x) = \frac{\frac{x+6}{x} +6}{\frac{x+6}{x}} = \frac{7x+6}{x+6} and f_3(x) = \frac{\frac{7x+6}{x+6} +6}{\frac{7x+6}{x+6}}.

Let S be the complete set of real solutions of the equation f_n(x) = x. The number of elements in S is:
a. 2
b. 2n
c. 2^n
d. 1
e. infinite

galactus, won't you take a look? :(

If I understand correctly, as we keep doing nested compositions, we get for example:

f_{4}(x)=\frac{\frac{13x+42}{7x+6}+6}{\frac{13x+42 }{7x+6}}=x

x=-2 \;\ and \;\ 3

The solutions remain the same on up the line.

For example, f_{7}(x)=x\Rightarrow \frac{463x+798}{133x+330}=x

x=-2 and 3.

And, it remains that way on up the line to f_{n}(x)=x

Generally, we can use the equation L=\frac{L+6}{L} and find that we have a quadratic with solutions

L=-2 and 3

So, there are 2 elements.

Assuming I am interpreting what they mean correctly. I think so.:)

Amazing! Then it was my english that wasn't that good. I failed to understand the 'what' they were asking. :(

Now come those 'structured' questions, which have answers in the range of 0 to 999.

26. The reciprocals of 4 positive integers add up to 19/20. Three of these integers are in the ratio 1:2:3. What is the sum of the four integers?

I don't know where to even start. :(

Check me out, but wouldn't they be
\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{30} = \frac{19}{20}?

The integers sum to 42.

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

I wonder how you got that... Trial and error I suppose? Darn, why ain't I able to 'see' the answers! :(

I immediately seen that 2,4,6 were in that ratio.

Then, I just solved

\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{x} = \frac{19}{20} for x.

That was it. I guess you could say I 'saw' it.;)

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 \sqrt{6^{2}+6^{2}}=6\sqrt{2}

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.

Check me out, but wouldn't they be
\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{30} = \frac{19}{20}?.

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.

216=2^{3}\cdot 3^{3}

270=\fbox{2} \cdot \fbox{3^{3}}\cdot 5

\;\ \;\ \;\ \;\ \nwarrow \;\ \nearrow
\;\ \;\ \;\ \;\ \;\ \;\ 54\;\

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

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... =/

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; y=ax^2 +bx+c. 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 y=m(x+n)(x-n)=mx^2-mn^2, where m and n are constants. Notice there is an x^2 term and a constant term, but no x 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

\frac{d}{dx}\left(ax^2+bx+c\right) = 2ax+b = 0 when x=0

2a(0)+b = 0

b=0

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/mathematics/amc-questions-384108-5.html#post1929361

And here: https://www.askmehelpdesk.com/mathematics/amc-questions-384108-4.html#post1927007

And here:https://www.askmehelpdesk.com/mathematics/amc-questions-384108-4.html#post1925448

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

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.

\sqrt{1960^{2}+441^{2}}=2009

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

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

23824
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

d_2=\sqrt{x_f^2+y_f^2}

d_2=\sqrt{(3-\frac{f}{2})^2+(f\frac{\sqrt{3}}{2})^2}

d_2=\sqrt{f^2-3f+9}

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 (3,3\frac{\sqrt{3}}{2}).

d_1=\sqrt{(3-\frac{f}{2})^2+(3\frac{\sqrt{3}}{2}-f\frac{\sqrt{3}}{2})^2}

d_1=\sqrt{f^2-\frac{15}{2}f+\frac{63}{4}}

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:

\frac{d}{df}(d_1^2+d_2^2) = 2f - \frac{15}{2} + 2f - 3 = 4f - \frac{21}{2}

f_{min}=\frac{21}{8}

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.

d_1=\frac{3\sqrt{21}}{8}

d_2=\frac{3\sqrt{57}}{8}

d^2 = (2d_1 + 2d_2)^2 = \frac{351+27\sqrt{133}}{8}

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

Josh

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.

\sqrt{1960^{2}+441^{2}}=2009

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 \frac{f}{2} and f\frac{\sqrt3}{2} for the lengths of d_1 and d_2, sorry :(

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 \frac{f}{2} and f\frac{\sqrt3}{2} for the lengths of d_1 and d_2, 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 \sqrt{3}, not \frac{3\sqrt{3}}{2}. 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.

(\frac{d}{2})^2=\frac{d^2}{4}=\left(3-\frac{-3}{2}\right)^2+\left(\sqrt{3}-\frac{3\sqrt{3}}{2} \right)^2

d^2=4\left( \frac{81}{4} + \frac{3}{4} \right)

d^2=84

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! :)

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 \alpha is the same for all of them. Since angle ABC=90+\alpha and angle BCD=90+\alpha as well, and AD is parallel to BC, we know the trapezium is symmetrical. Thus AB=CD.

Now by similarity we can see that

\frac{AE}{AB}=\frac{CD}{DE}=\frac{AB}{2009-AE}

AB^2=AE(2009-AE)

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

784*1225=980^2

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, \sqrt{1960^{2}+441^{2}}=2009

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

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

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.

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...

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...

Interchanging them at what point in time?

Well, there's no more information. I guess taking interchanging whenever we like.

Oh, I get it now. The maximum... i.e. if you change them at a time to achieve the maximum. Duh. Well, now I have to think again.

Lol! Take your time. A more toughie afterwards ;)

It's 48,000. I can tell you how I figured it out, but it was a lot of trial and error. When they get to 24,000 km switch them. I have no idea how to set it up in some equation to just get there. It took me a while to think through how it works, let alone how to make an equation.

The longer you leave them on, the more switching the back to the front will increase their mileage, but the front ones switched to the back will decrease. 24,000 is the point at which they both get exactly the same, If you switch them there, they each get another 24,000.

I had a feeling it was correct when they both came out to 48,000, the same number. But I did check switching at like 23,500 and 24,500 just to be sure.

What I actually did in my trial & error was thus:
Just using 20,000 miles as a starting point... er, kilometers... so how do you say mileage then? :confused:

The tires on the front at 20,000 will be halfway worn. If I move those to the back, they will have half their life left. But on the back, half the life is 30,000. For a total of 50,000. The ones on the back at 20,000 are 1/3 worn, leaving 2/3 life. Putting those on the front, 2/3 life is now 26,667 for a total of 46,667.

If we make it 30,000... That's .75 on the front, leaving .25 life. Move to the back with .25 life and that's 15,000 left, for a total of 30,000. (That one goes down as you increase the life before switching.) Then 30,000 on the back is .50 so switch to the front and you have .50 left and that's 20,000, for a total of 50,000. (That one increases as you go further cause it's taking advantage of the life on the back first.)

I just kind of tried a few intervals and plotted it on a graph and saw what happened. So I was able to home in closer. When I got to 25,000 I went to 26,000 and realized it got worse. So I went down to 24,000, and bingo, they both ended up with the same life at 48,000, which was the most I got.

So now it's up to you to make an equation for that. I can maximize something with a set of parameters, but have no clue how that's done when switching something off like that.

Oh, that's interesting. At 24,000, the front tires are .6 worn and the back .4 worn. The exact opposite proportion as the total km. That has to mean something, doesn't it?

Oh thanks! I found the ratio 2:3, I knew I had to make the front worn to some extent so that the ratio is inversed, but I somewhat messed up and came up with 2/3 worn at the front and 3/2 worn at the rear... I realise I had to use the total of ratios, 2/5 and 3/5 :o Thanks! :)

One I started, not yet done. The exams are taking my time...

Each face of a solid cube is divided into four as indicated in the diagram. Starting from vertex P , paths can be travelled to vertex Q along connected line segments. If each movement along the path takes one closer to Q, the number of possible paths from P to Q is
(A) 46
(B) 90
(C) 36
(D) 54
(E) 60

25939

Yeah, I started it too... then got stuck. I got an answer that isn't a choice, realized what I screwed up, and now can't figure out how to fix it. But I will get it -- it shouldn't be this difficult, darn it.

I have to get some more time to think about it too :p Cya I have to go now :)

Oh thanks! I found the ratio 2:3, I knew I had to make the front worn to some extent so that the ratio is inversed, but I somewhat messed up and came up with 2/3 worn at the front and 3/2 worn at the rear... I realise I had to use the total of ratios, 2/5 and 3/5 :o Thanks! :)

One I started, not yet done. The exams are taking my time...

Each face of a solid cube is divided into four as indicated in the diagram. Starting from vertex P , paths can be travelled to vertex Q along connected line segments. If each movement along the path takes one closer to Q, the number of possible paths from P to Q is
(A) 46
(B) 90
(C) 36
(D) 54
(E) 60

25939

\frac{6!}{2!2!2!}=\frac{720}{8}=90

Think of it as a 3-dimensional coordinate system. We have 2 in the x direction, 2 in the y direction, and 2 in the z direction. This is the number of ways we walk exactly 6 blocks from P to Q.

If I'm getting what you mean, this means that I have six vector directions:

\(1\\0\\0\), \(0\\1\\0\), \(0\\0\\1\)

All those twice. That makes 6! Now, each one is repeated twice, hence divided by 2!

Is that the "theory" behind it?

Yes, you can think of it that way. Think of x,y,z coordinates.

Ok, thanks! I'll try find more I'm stuck...