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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TopNewestMy 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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