# IMO 2018 Day 1

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

Day 2

## Problem 1

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.

## Problem 2

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 $$i = 1, 2, \dots, n$$.

## Problem 3

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 $$1 + 2 + 3 + \dots + 2018$$?

Note by Sharky Kesa
1 week, 2 days ago

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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.

- 1 week, 2 days ago

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

- 1 week, 2 days ago

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.

- 1 week, 2 days ago

Wait Seriously??? Well maybe Q1 was easier......I can't judge since my geometry is pretty weak......!! Btw, are you planning on being selected for the IMO??

- 1 week, 2 days ago

:| I've been planning on that for the last few years, without success. I'm still hopeful for next year though (it will be my last year).

- 1 week, 2 days ago

Well, if that's the case, then ALL THE BEST!!!!! Do well and just never loose hope!!!! Over here, I got to know abt IMO only last year.......and being in class 11th right now, I can try for it in this and the next year......!!

- 1 week, 2 days ago