Olympiads/ Kangaroos

I did the Maclaurin Olympiad and thought it was slightly easier than expected.

I did the Maclaurin Olympiad and thought it was slightly easier than expected.

I found the first 5 questions very easy relative to other years. They took me ~1 hour 20 min for full proofs. Was unable to make any real progress on Q6

How d'yu do Q3??

How d'yu do Q3??

w.l.o.g assume it is a unit square (you can do x units instead). Let the length DX be y. You can find the distance from C to X in terms of y. Then, find the radius (you get sqrt(1+y^2)/2). Since the semicircle only touches AB, AB is a tangent, so the radius is perpendicular to it, so it is vertical. You can find the distance from O to AB since O is the midpoint of CX. I did this all using co-ordinates, which made it easier. You end up with sqrt(1+y^2)/2=1-(y/2) and that is easy to do. The result is that AX: DX is 1:3 (y=0.75)

I didn't manage to do Q3 or Q6

For Q5 I used the logic of if you take a 2x4 subsection of it then the top row contains 2 colours and then the row below contains the other 2 colours so the row below contains the other 2 colours (1st set of colours) so the row below contains the other 2 colours (the 2nd set of colours) because of this the top row and bottom row acts as if it was a magic square. So we can remove the middle 2 rows. Repeat the same logic to columns with a 4x2 subsection and then you only have the corners left in a magic square so must have all 4 colours

I'm guessing 10/9/0/6/10/0
For myself but seeing as it's my first time doing an Olympiad my guesses could be way off on the partial marks ones

I though Q6 was pretty easy but discussing with some Trinity Camp qualifiers, the problem in their consensus was that it was the hardest ever Maclaurin Olympiad problem

The idea behind the solution is to find a valid construction to generate solutions. If you are familiar with the proof there are infinitely many primitive pythagorean triples you may recognise the identity (m^2 + n^2)^2 = (2mn)^2 + (m^2 - n^2)^2 which holds for all integers m,n>0.

A very similar construction idea is required for Q6. What i did was notice that (k^n + 1)^{mn+1} = ((k^n + 1)^n)^m + (k(k^n + 1)^m)^n and hence for all integers k>0 we can generate infinitely many solutions. This construction I found quite quickly (perhaps due to intuition and luck) but i just tried letting y = x^n (an idea i got from looking at n=1,2,3) and it was fairly straight forwards from there to get x = k^n + 1, y = (k^n + 1)^n and z = k(k^n + 1)^m

Another construction (which is easier to find) involves the crucial observation that 2^{mn+1} = (2^n)^m + (2^m)^n. We can then multiply both sides by k^{mn(mn+1)} for some integer k>0 to get that k^{mn(mn+1)}2^{mn+1} = k^{mn(mn+1)}(2^n)^m + k^{mn(mn+1)}(2^m)^n which simplifies to (2*k^{mn})^{mn+1} = ((2^n)*(k^{n(mn+1)})^m + ((2^m)*(k^{m(mn+1)})^n which gives that x = 2*k^{mn}, y = (2^n)*(k^{n(mn+1)} and z = (2^m)*(k^{m(mn+1)}

I think the second construction is definitely easier to find with logic but the first is more concise. I think the problem was probably completely unapproachable if you had no familiarity with these construction ideas from problems since induction or recurrence relations (like for pell equations) as far as I can tell can't solve the problem

I'm curious to how other people solved Q5. I'm pretty sure that the correct approach involved a proof by contradiction in 2 cases: Case 1 = 2 corners on the same side have the same colour and Case 2 = 2 corners on the same diagonal have the same colour

I was fairly short on time for this question so I only got the first case done (in a kind of rushed but valid way)

I think, thinking about the problem now the easiest way to reach a contradiction is to first observe that we must have 4 squares of each colour since we can apply the pigeon-hole principal on the 4 non-overlapping 2x2 squares. Then in each case you can show if the 2 corners have the same colour then the grid will have 5 of one colour.

In the paper in the first case i made the observation that each square must have a different colour to each adjacent and diagonal square. Then I WLOG chose the colour of all the the squares on the side where the corners were the same colour e.g R B G R and then deduced that the 2 squares below the "R" and "G" must both be Y but this is a contradiction since they are adjacent. Then i ran out of time for case 2...