IMO 2017 Day 1

Discussions of these problems have already started on AoPS. Let's start a few over here. :)

Day 2


Problem 1

For each integer a0>1a_0>1, define the sequence a0,a1,a2,,an,a_0, a_1, a_2, \ldots, a_n, \ldots for n0n \geq 0 as

an+1={anif an is an integeran+3otherwisea_{n+1}= \begin{cases} \begin{aligned} \sqrt{a_n} \quad &\text{if } \sqrt{a_n} \text{ is an integer}\\ a_n+3 \quad &\text{otherwise} \end{aligned} \end{cases}

Determine all values of a0a_0 such that there exists a number AA such that an=Aa_n = A for infinitely many values of nn.

Problem 2

Let R\mathbb{R} be the set of real numbers. Determine all functions f:RRf: \mathbb{R} \rightarrow \mathbb{R} such that, for any real numbers xx and yy,

f(f(x)f(y))+f(x+y)=f(xy).f(f(x)f(y)) + f(x+y) = f(xy).

Problem 3

A hunter and an invisible rabbit play a game in the Euclidean plane. The rabbit's starting point, A0,A_0, and the hunter's starting point, B0B_0 are the same. After n1n-1 rounds of the game, the rabbit is at point An1A_{n-1} and the hunter is at point Bn1.B_{n-1}. In the nthn^{\text{th}} round of the game, three things occur in order:

(i)(\text{i}) The rabbit moves invisibly to a point AnA_n such that the distance between An1A_{n-1} and AnA_n is exactly 11.

(ii)(\text{ii}) A tracking device reports a point PnP_n to the hunter. The only guarantee provided by the tracking device to the hunter is that the distance between PnP_n and AnA_n is at most 11.

(iii)(\text{iii}) The hunter moves visibly to a point BnB_n such that the distance between Bn1B_{n-1} and BnB_n is exactly 11.

Is it always possible, no matter how the rabbit moves, and no matter what points are reported by the tracking device, for the hunter to choose her moves so that after 10910^9 rounds, she can ensure that the distance between her and the rabbit is at most 100100?

Note by Sharky Kesa
2 years, 3 months ago

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This was my method of going about solving P1:

The sequence looks a bit weird, so I went about first trying a few small cases for a0a_0. I found that the sequence became periodic if 3a03 \mid a_0. This was what I would try to prove, since a direct corollary is there exist numbers AA that appear infinitely many times in the sequence.

I decided to use (mod3)\pmod{3} since that would both satisfy my result. Secondly, whilst the sequence increases by 3, it remains the same (mod3)\pmod{3}.

Firstly, note that n20,1(mod3)n^2 \equiv 0, 1 \pmod{3}, so if a02(mod3)a_0 \equiv 2 \pmod{3}, then the sequence would be increase continuously without changing its residue (mod3)\pmod{3}. Thus, a0≢2(mod3)a_0 \not \equiv 2 \pmod{3}.

Secondly, if a01(mod3)a_0 \equiv 1 \pmod{3}, we can show this isn't periodic by strong induction. For base case a0=4a_0=4, a0=7a_0=7, a0=10a_0=10, a0=13a_0=13 and a0=16a_0=16, we find this to be true. Now, we can assume m2<a0(m+1)2m^2 < a_0 \leq (m+1)^2 for m4m \geq 4. If a perfect square is equivalent to 1(mod3)1 \pmod{3}, then the square root of it is also not divisible by 3. Now, there must exist a term in the sequence aia_i such that ai=(m+1)2a_i = (m+1)^2 or ai=(m+2)2a_i = (m+2)^2, so ai+1=m+1a_{i+1} = m+1 or ai+1=m+2a_{i+1}=m+2. Now, we have m+2<(m1)2m+2 < (m-1)^2 for all m4m\geq 4. Thus, this sequence is eventually decreasing to the base cases, so by strong induction, this isn't periodic.

Finally, if 3a03 \mid a_0, we have base cases a0=3a_0=3, a0=6a_0=6 and a0=9a_0=9 which all satisfy. Now, we can assume (3(m1))2<a0(3m)2(3(m-1))^2 < a_0 \leq (3m)^2 for m2m \geq 2. Note that if a perfect square is divisible by 3, then it's square root is divisible by 3. Thus, there exists a term in the sequence aia_i such that ai=(3m)2a_i = (3m)^2, so ai+1=3m<(3(m1))2a_{i+1}=3m < (3(m-1))^2, which is true for all m2m \geq 2. Thus, this sequence is evnetually decreasing to the base cases, so by strong induction, this is periodic.

Thus, we must have 3a03 \mid a_0.

Sharky Kesa - 2 years, 3 months ago

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Thank you,its beautiful!

Kabir Sahni - 1 year, 4 months ago

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Brilliant, but not very lucid. The solutions on AoPS are much better written.

Star Light - 4 months ago

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P2.

If f(0)=0,f(0)=0, substitute y=0.y=0.

You get f(x)=0.f(x)=\boxed{0}.

Now then let f(0)0f(0)\neq0

If  f(k)=0,  then k=1.\boxed{\text{If }~f(k)=0,~\text{ then }k=1}.

Cus if k1k\neq1, we can substitute x=k, y=kk1,x=k,~y=\dfrac{k}{k-1}, and it yields f(0)=0f(0)=0 which is a direct contradiction in the face.

Substitute x=0, y=0x=0,~y=0 and you get f(f(0)2)=0,f(f(0)^2)=0, which leads to f(0)=±1,f(0)=\pm 1, and therefore f(1)=0.f(1)=0.

Then let y=1,y=1, and you get f(0)+f(x+1)=f(x).\boxed{f(0)+f(x+1)=f(x)}.

Assume there exists a, ba,~b such that f(a)=f(b).f(a)=f(b).

But in fact if a>1,a>1, we can add f(0)f(0) to both sides of f(a)=f(b)f(a)=f(b) however many times we want and lower the values that are inside the function.

Therefore, f(a)=f(b)f(a)=f(b) and a<1.a<1. From it we can say that there are real numbers p, qp,~q such that pq+1=apq+1=a and p+q=b,p+q=b, since then pqpq would be negative and therefore the discriminant of the quadratic equation t2bt+(a1)=0t^2-bt+(a-1)=0 that has p, qp,~q as its roots would be always positive. (D=b24(a1)=b24pq>0.)(D=b^2-4(a-1)=b^2-4pq>0.)

Then substitute x=p, y=q.x=p,~y=q.

f(f(p)f(q))+f(b)=f(a1)=f(a)+f(0)=f(b)+f(0),f(f(p)f(q))+f(b)=f(a-1)=f(a)+f(0)=f(b)+f(0), and so f(f(p)f(q))=f(0),f(f(p)f(q))=f(0), f(f(p)f(q)+1)=0.f(f(p)f(q)+1)=0.

f(p)f(q)=0f(p)f(q)=0 and therefore p=1p=1 or q=1,q=1, which leads to a=b.a=b.

Therefore if  f(a)=f(b), a=b.\boxed{\text{if }~f(a)=f(b),~a=b}.

Substitute y=xy=-x and y=1xy=1-x respectively.

f(f(x)f(x))=f(x2)f(0)  f(f(x)f(x))=f(x2+1)  f(x)f(x)=x2+1f(f(x)f(-x))=f(-x^2)-f(0)~\Rightarrow~f(f(x)f(-x))=f(-x^2+1)~\Rightarrow~f(x)f(-x)=-x^2+1

f(f(x)f(1x))=f(x(1x))  f(x)f(1x)=xx2  f(x){f(x)f(0)}=x2+xf(f(x)f(1-x))=f(x(1-x))~\Rightarrow~f(x)f(1-x)=x-x^2~\Rightarrow~f(x)\{f(-x)-f(0)\}=-x^2+x

Subtract the second equation from the first.

f(0)f(x)=x+1,f(0)f(x)=-x+1, and since f(0)=±1,f(0)=\pm1,

f(x)=x1f(x)=\boxed{x-1} or f(x)=x+1.f(x)=\boxed{-x+1}.

Therefore, there are only three possible solutions for f(x),f(x),

f(x)=0, f(x)=x1, f(x)=x+1.\boxed{f(x)=0},~\boxed{f(x)=x-1},~\boxed{f(x)=-x+1}.

Boi (보이) - 2 years, 3 months ago

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I'm confused as to how you can say there exist reals pp and qq such that pq+1=apq+1=a and p+q=bp+q=b. I haven't seen this substitution and am curious as to how it is inspired. I proved you can always do it, but it doesn't seem like a trivial result.


Edit: After much fooling around, I determined how it is inspired. Nice. :)

Sharky Kesa - 2 years, 3 months ago

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Haha xD

Yea... it was almost an accident.

I first tried it with pq=a, p+q=b,pq=a,~p+q=b, but it didn't work by a very tiny bit, so yeah. :')

Boi (보이) - 2 years, 3 months ago

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@Boi (보이) It'll probably get 6 since you didn't directly prove it. Use it as a lemma instead.

Sharky Kesa - 2 years, 3 months ago

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@Sharky Kesa I don't even know what a lemma is though.

Hold on lemme search-

Boi (보이) - 2 years, 3 months ago

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@Boi (보이) Here is my description of what some definitions required in proof writing. :P

Sharky Kesa - 2 years, 3 months ago

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@Sharky Kesa Yeaa now I know what a lemma is. Thank you!

But it's kinda true that my solution went on without any logical flaws xD

Well... I added as to why we can always say that pq+1=a<1, p+q=bpq+1=a<1,~p+q=b such that p, qp,~q are all real numbers.

...if that is what you meant.

Boi (보이) - 2 years, 3 months ago

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Arguably, I have to say that this paper had one of the greatest difficulty ranges between Problem 1 to 3, if not the greatest (2012 came close though). From what I've heard, there are not yet any confirmed solves of P3 at the IMO, but CHN, JPN and KOR have kept their performance hidden.

Sharky Kesa - 2 years, 3 months ago

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Indeed, the range of difficulty level is tremendous!

P1 is easy once you consider modulo 3 (and apply induction).

I've seen the an actual solution for P3 on another math forum, and it's a real beast!

Pi Han Goh - 2 years, 3 months ago

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Yes, I saw the solution on AoPS. It's amazing, but intense.

Sharky Kesa - 2 years, 3 months ago

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@Sharky Kesa Yes, I'm pretty confident that there's a (much) shorter solution

Pi Han Goh - 2 years, 3 months ago

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Hahah! One of the members of our IMO team got 7/7 for it. :)

Sharky Kesa - 2 years, 3 months ago

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P1 seems very simple. You want to add 3 until you reach a square, then get the square root and keep adding 3 until you square root to 9 if the first term is larger than 9. So square rooting 9 leads to infinite values of a number. So basically any multiple of 3 will eventually square root down to 9, and then 3, and then a loop creates infinite values of 3 or 6 or 9. I basically just did some basic cases, then realized that if the first term is not a multiple of 3, then the squares it reaches will allow it to continue adding 3 to a different square, which will lead to no maximum term in the sequence. If the first term is not a multiple of 3, the first square it reaches will also not be a multiple of 3, and neither will be the square root, so adding 3 will never lead to the same square again.

keanu ac - 2 years, 3 months ago

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