Maclaurin olympiad

I'm currently preparing for the maclaurin olympiad for a while now, but i'm still struggling with a lot of questions. Any advice/useful resources??

Reply 1

mqb2766

Original post by yi123456

I'm currently preparing for the maclaurin olympiad for a while now, but i'm still struggling with a lot of questions. Any advice/useful resources??

Its in ~4 days? Reality is there isnt a lot you can do in that time, but are you strugging with the techniques, getting started/problem solving, ... or specific topics like geometry or ...?

Reply 2

yi123456

Original post by mqb2766

Its in ~4 days? Reality is there isnt a lot you can do in that time, but are you strugging with the techniques, getting started/problem solving, ... or specific topics like geometry or ...?

i guess getting started bc if i get some idea of how to do them, i could get somewhere.

Reply 3

mqb2766

Original post by yi123456

i guess getting started bc if i get some idea of how to do them, i could get somewhere.

It obviously depends on the question, so if you are/were stuck on some maybe post a couple to illustrate what you found difficult.

However, the basic ideas of problem solving so sketching the problem, simplifying it, substituting simple numbers, working backwards, guestimating, ... can all help you better understand the problem / get started. I like
https://www.worldscientific.com/worldscibooks/10.1142/9478#t=aboutBook
and the intro chapter (you can download) illustrates them for a simple handshaking problem.

Reply 4

chris8topheryu

How did everyone do on the Olympiad?

Reply 5

mqb2766

Original post by chris8topheryu

How did everyone do on the Olympiad?

Best to leave any discussion until tomorrow
https://ukmt.org.uk/intermediate-challenges/cayley-hamilton-maclaurin-olympiad-challenge

Reply 6

yi123456

Did anyone get question 6 cuz i had no clue how to start

And also question 4

Original post by yi123456

Did anyone get question 6 cuz i had no clue how to start

For 6, I began by drawing a bunch of hexagons and just trying it out with different numbers.
I thought that it was something to do with pairs of numbers adding up to N to prove that it works for all odd numbers except for 1.
Eg. For N = 7, you can have 6 + 1, 5 + 2, 4 + 3 and 7 itself.
This means that you can travel 7 hexagons in a direction, travel 7 more in a different direction, travel 7 hexagons back, and 7 to get back to the start.
For a number like 5, it's similar but with a more triangular route as there are only 2 pairs and 5 itself. This is simplifying it a bit as you have to go in descending order of distances so the bee would take a more convoluted route, that essentially boils down to the explanation above.

It also works for multiples of 4 where I think you can imagine it as a kind of number line (with a negative and a positive side) to represent a straight line of hexagons with 0 being the start. You can go to 4 at the start, travel 3 back to 1, travel 2 back to -1 and finally travel 1 to 0.
For other even numbers, it doesn't work but I didn't know quite how to prove it.

For question 4, I had no clue but I'd assume it has something to with niftily drawing some triangles.

Reply 9

breadmeatbread

for q4 i extended the straight parts of the road up and also drew the radii perpendicular to them from the centre of the sectors with radius R and x. then you get a kite with 120 90 90 and 60 degrees because a tangent and radius meet at 90. if you split it in half then you get a 30 60 90 triangle where the radius is opposite 30, so then the line opposite 90 is double that. you can then overlap the triangles for the x sector and R sector, and find that 2x is equal to 2R plus the distance between their centres, which is x minus r as both sectors with x and r meet on the inside bend. therefore 2x=2R+x-r so that x+r=2R

Reply 10

breadmeatbread

for q6 i said that each unit up or down changed the bees vertical position by 1/-1 and a unit left or right would change position 1/-1 horizontal and a half up or down. then i said that you need moves left and right to balance so there is 0 horizontal, which means half of the total moves spent displaces vertically. this means that if the total of the sum is a multiple of 3 it should be possible (unless N is 3 because 3 is greater than 2 which is a third of the sum 1+2+3). however it works for even sums from N down, as you can move a set number in one direction and then trace back exactly the same amount of steps. this means if the sum is divisible by 2 then the bee can still do it, so 3 is still valid.

then you use that n+n-1+...+1 is equal to n squared plus n all over 2, and you show for which n it is a multiple of 2 or 3 (if multiple of 2 number is 4k or 4k+3 and if of 3 then number can be 6k, 6k+2(exclude 2), 6k+3, 6k+5)

im not sure if theres any other possible values and my actual written solution was quite bad so i can imagine having missed out something important, hopefully this years boundaries are much lower

(edited 1 month ago)

Reply 11

SMaths

Original post by breadmeatbread

for q6 i said that each unit up or down changed the bees vertical position by 1/-1 and a unit left or right would change position 1/-1 horizontal and a half up or down. then i said that you need moves left and right to balance so there is 0 horizontal, which means half of the total moves spent displaces vertically. this means that if the total of the sum is a multiple of 3 it should be possible (unless N is 3 because 3 is greater than 2 which is a third of the sum 1+2+3). however it works for even sums from N down, as you can move a set number in one direction and then trace back exactly the same amount of steps. this means if the sum is divisible by 2 then the bee can still do it, so 3 is still valid.
then you use that n+n-1+...+1 is equal to n squared plus n all over 2, and you show for which n it is a multiple of 2 or 3 (if multiple of 2 number is 4k or 4k+3 and if of 3 then number can be 6k, 6k+2(exclude 2), 6k+3, 6k+5)
im not sure if theres any other possible values and my actual written solution was quite bad so i can imagine having missed out something important, hopefully this years boundaries are much lower