Another year, another IMO. Let's get cracking. :)

Let \(\Gamma\) be the circumcircle of acute triangle \(ABC\). Points \(D\) and \(E\) are on segments \(AB\) and \(AC\) respectively such that \(AD = AE\). The perpendicular bisectors of \(BD\) and \(CE\) intersect minor arcs \(AB\) and \(AC\) of \(\Gamma\) at points \(F\) and \(G\) respectively. Prove that lines \(DE\) and \(FG\) are either parallel or they are the same line.

Find all integers \(n \geq 3\) for which there exist real numbers \(a_1, a_2, \dots a_{n + 2}\) satisfying \(a_{n + 1} = a_1\), \(a_{n + 2} = a_2\) and

\[a_ia_{i + 1} + 1 = a_{i + 2}\]

For \(i = 1, 2, \dots, n\).

An *anti-Pascal* triangle is an equilateral triangular array of numbers such that, except for the numbers in the bottom row, each number is the absolute value of the difference of the two numbers immediately below it. For example, the following is an anti-Pascal triangle with four rows which contains every integer from \(1\) to \(10\):

\[4\] \[2\quad 6\] \[5\quad 7 \quad 1\] \[8\quad 3 \quad 10 \quad 9\]

Does there exist an anti-Pascal triangle with \(2018\) rows which contains every integer from \(1\) to \(1 + 2 + 3 + \dots + 2018\)?

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## Comments

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TopNewestProblem 3 We can observe from the nature of the anti-Pascal triangle that for 'n' rows, we have n + (n-1) + (n-2) +...+ 1 Now the Problem that remains to be solved is that whether all the natural numbers are distinct or not. This can be easily proved because we know that any natural number can be obtained as the difference of 'related' natural numbers. As we are free to chose the natural numbers at the base of the triangle, therefore there must be two or more same natural numbers. hence the repeating natural numbers now occupy the place of the other natural numbers.

thus, there exist an anti-Pascal triangle with 2018 rows which contains every integer from 1 to 1+2+3+...+2018.

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Unfortunately, that is incorrect reasoning. How do you justify, "This can be easily proved because we know that any natural number can be obtained as the difference of 'related' natural numbers. As we are free to chose the natural numbers at the base of the triangle, therefore there must be two or more same natural numbers. hence the repeating natural numbers now occupy the place of the other natural numbers." ?

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Mr. Sharky Kesa, what is the answer? My research tells me this was the hardest problem. Do you have or know someone who has the answer?

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IMOProblems/Problem_3No solution there :( At least give me a hint (or two).

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here (Scroll down to see it)

There is a solutionLog in to reply

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You can do better than that. Do not throw a bunch of statements without Justification.

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I do know there can not exist an anti-Pascal Triangle with 2018 rows which contains every integer from 1 to 2018*2019/2; But why? :´(

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My method to Q2:

Note that if we try the \(n=3\) case, we have \[ab+1=c\] \[bc+1=a\] \[ca+1=b\]

Subbing the first line into the second line gives us \(b(ab+1)+1=a\), so \(1+b = a(1-b^2)\). Thus, either \(b=-1\), or \(a=\frac{1}{1-b}\). If \(b=-1\), we have \(a+c=1\), \(ac=-2\), so \(a=-1\), \(c=2\). Thus, we have a possible sequence \(-1, -1, 2, -1, -1, 2, \ldots\), so all \(3 \mid n\) satisfy.

Now, note that \(a_i a_{i+1} + 1 = a_{i+2} \implies a_i a_{i+1} a_{i+2} + a_{i+2} = a_{i+2}^2\), and \(a_{i+1} a_{i+2} + 1 = a_{i+3} \implies a_i a_{i+1} a_{i+2} + a_i = a_i a_{i+3}\). Adding the first statement over \(i\) yields \(\sum a_i a_{i+1} a_{i+2} + \sum a_i = \sum a_i^2\). Using the second statement, we get \(\sum a_i a_{i+3} = \sum a_i^2\). However, by rearrangement inequality, \(\sum a_i a_{i+3} \leq \sum a_i^2\), so \(a_i = a_{i+3}\).

Thus, either \(a\) is periodic with a length of either \(1\) or \(3\). If period is \(1\), then we have \(a^2 + 1 = a\), which isn't solvable over the reals. Thus, the period is \(3\), which is satisfied by having \(a_i = a_{i+1} = -1\), \(a_{i+2} - 2\).

Therefore, all \(3 \mid n\) satisfy.

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@Sharky Kesa Wow.....my approach too was on similar tracks.......!!!! Btw, isn't this a fairly easy problem for the IMO??

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That's what I thought. Q2 was easier than Q1 in my opinion but looking at score predictions, a lot more people solved Q1 than Q2.

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I can´t find the solution :,(

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