# 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
2 years, 11 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
• Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

## Comments

Sort by:

Top Newest

Problem 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.

- 2 years, 10 months ago

Log in to reply

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." ?

- 2 years, 10 months ago

Log in to reply

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?

- 2 years, 10 months ago

Log in to reply

I do know of the solution, but it is by someone else (that is why I haven't posted it here). Have a look at the Art of Problem Solving website to get a solution.

- 2 years, 10 months ago

Log in to reply

https://artofproblemsolving.com/wiki/index.php?title=2018IMOProblems/Problem_3

No solution there :( At least give me a hint (or two).

- 2 years, 10 months ago

Log in to reply

There is a solution here (Scroll down to see it)

- 2 years, 10 months ago

Log in to reply

Thank you, very much!!!!!!

- 2 years, 10 months ago

Log in to reply

Can you please clear me, why there is many people complaining about the originality of Problem 3? I saw the statistics... Come on! All nations on earth were humiliated by problem 3.

- 2 years, 10 months ago

Log in to reply

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

- 2 years, 10 months ago

Log in to reply

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? :´(

- 2 years, 10 months ago

Log in to reply

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.

- 2 years, 11 months ago

Log in to reply

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

- 2 years, 11 months ago

Log in to reply

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.

- 2 years, 11 months ago

Log in to reply

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??

- 2 years, 11 months ago

Log in to reply

:| 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).

- 2 years, 11 months ago

Log in to reply

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......!!

- 2 years, 11 months ago

Log in to reply

Can somebody tell me please the anwser of Q3; the "Anti-Pascal Triangle Array"?? PLease ;)

I can´t find the solution :,(

- 2 years, 10 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...