Elementary Techniques used in the IMO (International Mathematical Olympiad) - Invariants

This is continuation from Post 8. For readers who have read this, please do NOT post your solutions in the other post, instead post it in the comments below in THIS post. Thank you.


Hints \LARGE \text{Hints}

  1. Well, since we cannot spam this directly, consider working backwards. Clearly, Tom needs to ensure that the buckets have a1,,a51 a_1,\ldots,a_5\le 1 or else the bucket would overflow on Jerry's turn. Suppose then that Jerry adds w1,,w5 w_1,\ldots,w_5 to the buckets, where w1++w5=1 w_1+\cdots+w_5=1 . So if we think about it further, we can see that if there are 2 2 nonadjacent buckets (we are trying to make Jerry win now), Jerry wins if we have ai+wi>1 a_i + w_i > 1 and ai+2+wi+2>1 a_{i+2} + w_{i+2} > 1 . Tom needs to prevent Jerry from setting (wi,wi+2)=(1+ϵai,1+ϵai+2) (w_i,w_{i+2})=(1+\epsilon-a_i,1+\epsilon-a_{i+2}) , so all he does is to ensure that 2+2ϵaiai+2>1=w1++w5 2+2\epsilon-a_i-a_{i+2}>1=w_1+\cdots+w_5 for all ϵ>0ai+ai+21 \epsilon>0 \rightarrow a_i+a_{i+2}\le 1 . Now it is just your job to prove that these invariants hold at every step.

  2. There isn't a really detailed hint for this question. Remark first that: If we start with a point on the convex hull of S \mathcal{S} and a line \ell that is “tangent” to the convex hull then we will only iterate over the points from the convex hull. The key motivation behind the solution is the fact that in rotations there are some really important angles such as π,π2 \pi, \frac{\pi}{2} . Also think, if we draw a line \ell , then the plane is split into two sections and there is one point only on \ell , how do we know that we can iterate? Maybe we can see what happens if we swap the two sections? Do you see the invariance?

  3. Clearly we need to construct a monovariant. Think: Is 20132013 2013^{2013} purposely there to put you off? Try some small cases. It becomes apparent that we need an alternating odd even structure. Prove that this is possible. Try casework if everything else fails. We need a monovariance that looks something like: max{a,b,c}g(a,b,c) max\{ a,b,c \} - g(a,b,c) Here we define g(a,b,c)=# of a,b,c that are equal to max{a,b,c} g(a,b,c) = \text{\# of} \space a,b,c \space \text{that are equal to} \space \max \{ a,b,c \} . The alternating structure suggests strongly to use 3max{a,b,c}g(a,b,c) 3 max\{ a,b,c \} - g(a,b,c) . Use induction to prove that this invariance holds a every step.

Note by Anqi Li
5 years, 9 months ago

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