Australian Maths Competition (4 Viewers)

[nothingbutlies]

Re: Australian Mathematics Competition

I am not quite sure what number question it was, but it was a 10 mark question. It was in the intermediate division. The question was on patterns, it showed pattern 1,2 and 3 then it said what would pattern 11 be. My answer was 66, and i am very confident on that. So can anybody help me out.
....
<>
....

Pattern 1^
----
.....<>
<>
.....<>
......
Pattern 2^
...
........<>
...<>
<>...<>
...<>
.......<>
...
Pattern 3^
What is pattern 11? That was the question im pretty sure, thank you in advanced for replys. <>=hexagon

I will try and describe how i worked it out. Pattern 1 has one hexagon, while pattern 2 has three hexagon. So if i add 2 + 1=3 i got the one from the hexagon in pattern 1 and the two from the pattern (2)<--....Pattern 3 if we add three hexagons to pattern 2 thats gives us 6 and it also gives us pattern 3. So basically the next hexagon pattern 4 i would add 4 hexagons to the previous pattern. So its basically 6 hexagons from pattern 3 so i did.....6+4+5+6+7+8+9+10+11=66

Sorry I don't really understand your working? The solution to the hexagon question was 231....

Did you misread the question? The question was not referring to the number of hexagons in pattern 11, but the number of matches/whatever required to make the pattern.

 

[maths94]

Re: Australian Mathematics Competition

Sorry I don't really understand your working? The solution to the hexagon question was 231....

Did you misread the question? The question was not referring to the number of hexagons in pattern 11, but the number of matches/whatever required to make the pattern.

nnumber if hexagons... or some shape.. also prove to me its 231

 

[3unitz]

Re: Australian Mathematics Competition

I have found a way to make the maximum distance 174 km. But i am still not happy with this result. Even though i do think that is the opitmal solution, I have done it in such an ad hoc manner that I can't be sure... I am sure that there is some far more elegant solution, if I could only think what it was... (sigh).

i also got this answer. my proof is below:

we can scale down each length by 3 (without loss of generality), giving us a total length of 360. this allows for a nice geometrical representation of distances as angles around a circle (360 degrees).

stations C can be represented as follows: View attachment 19015 (note that 72*3 = 216)

we now consider 2 cases for placing stations B. case 1 is where we have two Dmax's = 72 (this is when a station C is at the same location as a station B), and case 2 is where we have one Dmax = 72.

case 1:
set a station B at the same location as a station C, View attachment 19016

stations A must now be located to minimise both Dmax's: View attachment 19017

the best that can be done is Dmax = 60.

case 2:
case two is rather messy to draw on the comp, but here it is: View attachment 19018

now just equate the largest angles to minimise Dmax:

72 - Y - X = 48 + Y = 54 + X

which gives X = 4, Y = 10

so each angle is 58 degrees, which gives a minimum Dmax of 58*3 = 174

 

[Iruka]

Re: Australian Mathematics Competition

Yeah, that's basically what I did - you start with two Dmax's and then shrink them. But another interval on the circle opens up until you've got three Dmax's of 174 km each.

Anyway, it is good to see that you've sorted out the other cases.

 

[gurmies]

Re: Australian Mathematics Competition

i also got this answer. My proof is below:

We can scale down each length by 3 (without loss of generality), giving us a total length of 360. This allows for a nice geometrical representation of distances as angles around a circle (360 degrees).

Stations c can be represented as follows: View attachment 19015 (note that 72*3 = 216)

we now consider 2 cases for placing stations b. case 1 is where we have two dmax's = 72 (this is when a station c is at the same location as a station b), and case 2 is where we have one dmax = 72.

case 1:
set a station b at the same location as a station c, View attachment 19016

stations a must now be located to minimise both dmax's: View attachment 19017

the best that can be done is dmax = 60.

case 2:
case two is rather messy to draw on the comp, but here it is: View attachment 19018

now just equate the largest angles to minimise dmax:

72 - y - x = 48 + y = 54 + x

which gives x = 4, y = 10

so each angle is 58 degrees, which gives a minimum dmax of 58*3 = 174

<3

 

[F-LAU]

Re: Australian Mathematics Competition

the maths comp is the easiest ever. i always get distinctions and higher.

 

[ninetypercent]

Re: Australian Mathematics Competition

hopefully, the cut offs this year will be lower.

 

[gurmies]

Re: Australian Mathematics Competition

Anyone here confident that they got a medal?

 

[nothingbutlies]

Re: Australian Mathematics Competition

nnumber if hexagons... or some shape.. also prove to me its 231

oh ffs my firefox crashed halfway through my reply. stupid thing.

anyway um, are we sure we are talking about the same question here? The one I'm talking about definitely was asking for the number of lines in the pattern, not the number of hexagons.

So, they told us first pattern had 6 lines, second had 15 and the next had 27. This can be written as 6, 6+9, 6+9+12 etc.

Therefore the 11th one can be written as 6+9+12+...+36. Using sum of an arithmetic progression, the answer would be 11/2[12+(10*3)], which is 231.

If you haven't learnt arithmetic progressions etc yet, you could also solve the problem in a rather visual manner. The hexagons look rather like triangular numbers yes (well, they're 'stacked' in that manner). Taking the third pattern as an example, there are 6 hexagons, which is the third triangular number. Therefore there are 6*6 lines, since each hexagon has 6 sides. But at 3 'points', there are 3 sides which overlap and have already been counted as part of another hexagon. Now these three points are also in a triangular number shape. Therefore for the third shape number of lines would be [3rd triangular number x 6] - [2nd triangular number times 3]. So for 11th pattern, it would be [11th triangular number x 6] - [10th triangular number x 3] which is [66 x 6] - [55 x 3] which is 231.

Hope that helps

 

[ninetypercent]

Re: Australian Mathematics Competition

I'm confident that I will get a credit. I reckon I got around 60-70 points, but I'm not too sure whether that will get me a distinction. I'm praying that it will.

 

[Official]

Re: Australian Mathematics Competition

I'm confident that I will get a credit. I reckon I got around 60-70 points, but I'm not too sure whether that will get me a distinction. I'm praying that it will.

last two years, the prize cut off was only 77 for yr 11.

 

[addikaye03]

Re: Australian Mathematics Competition

Heres some random Q for you guys to try, i've got a solution for each that i will post up later if required:

3. Han stands at (-2,0) on a plane. Han's cake sits at (2,0) on the plane. Centred at (0,0) there is an impenetrable circle of radius 1. If Han is starving and infinitesimally thin, how far must he walk to reach his cake?

4. A restaurant offers 10% off the price of the bill that costs at least 10 dollars, 20% off the price of a bill that costs at least 20 dollars, and so on. Everything on the menu costs an integer number of dollars. What is the most you will pay?

6. An equalateral triangle has vertices at (0,0), (a,5) and (b,13). Find the area of the triangle.

I've got more after this if their a hit lol

 

[Timothy.Siu]

Re: Australian Mathematics Competition

Yeah, that's basically what I did - you start with two Dmax's and then shrink them. But another interval on the circle opens up until you've got three Dmax's of 174 km each.

Anyway, it is good to see that you've sorted out the other cases.

i intepreted the question differently.

doesn't it say the government tells them to space them so that the longest distance between the stations is as SHORT as possible?

or is this what u did.

 

[Iruka]

Re: Australian Mathematics Competition

Um, yeah, it is what we did, but I think you'd have to follow the conversation which went on over several days...

Basically, I started with two maximum intervals that were 180 km apart, and then I worked out that you could make both of them smaller, but at the expense of the second largest interval on the other side of the circle opening up. So to get the maximum interval as small as possible, we shrink the two big ones (and in so doing expand the second largest one) until there are three intervals all of 174 km.

 

[Timothy.Siu]

Re: Australian Mathematics Competition

ah okay, too complicated for me i guess

 

[Aerath]

Re: Australian Mathematics Competition

Um, yeah, it is what we did, but I think you'd have to follow the conversation which went on over several days...

Basically, I started with two maximum intervals that were 180 km apart, and then I worked out that you could make both of them smaller, but at the expense of the second largest interval on the other side of the circle opening up. So to get the maximum interval as small as possible, we shrink the two big ones (and in so doing expand the second largest one) until there are three intervals all of 174 km.

Damn, my guess was pretty close.
I guessed 178km.

 

[Revelrous]

Re: Australian Mathematics Competition

i reckon this exam was pretty easy tbh
i even forgot it was on until i turned up at school on the day looool

 

[Official]

Re: Australian Mathematics Competition

How long does it usually take for the results to come out?

 

[ninetypercent]

Re: Australian Mathematics Competition

I think they will be back by the end of term 3, since the Year 12's have to get their results back by then. maybe week 8?

 

[Aquawhite]

Re: Australian Mathematics Competition

They usually take their time with these results... Damn tests >_<