IMO 2018 Day 1

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

Day 2


Problem 1

Let Γ\Gamma be the circumcircle of acute triangle ABCABC. Points DD and EE are on segments ABAB and ACAC respectively such that AD=AEAD = AE. The perpendicular bisectors of BDBD and CECE intersect minor arcs ABAB and ACAC of Γ\Gamma at points FF and GG respectively. Prove that lines DEDE and FGFG are either parallel or they are the same line.

Problem 2

Find all integers n3n \geq 3 for which there exist real numbers a1,a2,an+2a_1, a_2, \dots a_{n + 2} satisfying an+1=a1a_{n + 1} = a_1, an+2=a2a_{n + 2} = a_2 and

aiai+1+1=ai+2a_ia_{i + 1} + 1 = a_{i + 2}

For i=1,2,,ni = 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 11 to 1010:

44 262\quad 6 5715\quad 7 \quad 1 831098\quad 3 \quad 10 \quad 9

Does there exist an anti-Pascal triangle with 20182018 rows which contains every integer from 11 to 1+2+3++20181 + 2 + 3 + \dots + 2018?

Note by Sharky Kesa
1 year, 1 month ago

No vote yet
1 vote

</code>...<code></code> ... <code>.">   Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. 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 1

paragraph 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 </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 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.

Lalit Choudhary - 1 year 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." ?

Sharky Kesa - 1 year 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?

Xizlon Tho - 1 year ago

Log in to reply

@Xizlon Tho 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.

Sharky Kesa - 1 year ago

Log in to reply

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

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

Xizlon Tho - 1 year ago

Log in to reply

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

Sharky Kesa - 1 year ago

Log in to reply

@Sharky Kesa Thank you, very much!!!!!!

Xizlon Tho - 1 year ago

Log in to reply

@Sharky Kesa 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.

Xizlon Tho - 1 year ago

Log in to reply

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

Xizlon Tho - 1 year 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? :´(

Xizlon Tho - 1 year ago

Log in to reply

My method to Q2:

Note that if we try the n=3n=3 case, we have ab+1=cab+1=c bc+1=abc+1=a ca+1=bca+1=b

Subbing the first line into the second line gives us b(ab+1)+1=ab(ab+1)+1=a, so 1+b=a(1b2)1+b = a(1-b^2). Thus, either b=1b=-1, or a=11ba=\frac{1}{1-b}. If b=1b=-1, we have a+c=1a+c=1, ac=2ac=-2, so a=1a=-1, c=2c=2. Thus, we have a possible sequence 1,1,2,1,1,2,-1, -1, 2, -1, -1, 2, \ldots, so all 3n3 \mid n satisfy.

Now, note that aiai+1+1=ai+2    aiai+1ai+2+ai+2=ai+22a_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 ai+1ai+2+1=ai+3    aiai+1ai+2+ai=aiai+3a_{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 ii yields aiai+1ai+2+ai=ai2\sum a_i a_{i+1} a_{i+2} + \sum a_i = \sum a_i^2. Using the second statement, we get aiai+3=ai2\sum a_i a_{i+3} = \sum a_i^2. However, by rearrangement inequality, aiai+3ai2\sum a_i a_{i+3} \leq \sum a_i^2, so ai=ai+3a_i = a_{i+3}.

Thus, either aa is periodic with a length of either 11 or 33. If period is 11, then we have a2+1=aa^2 + 1 = a, which isn't solvable over the reals. Thus, the period is 33, which is satisfied by having ai=ai+1=1a_i = a_{i+1} = -1, ai+22a_{i+2} - 2.

Therefore, all 3n3 \mid n satisfy.

Sharky Kesa - 1 year, 1 month 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??

Aaghaz Mahajan - 1 year, 1 month 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.

Sharky Kesa - 1 year, 1 month ago

Log in to reply

@Sharky Kesa 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??

Aaghaz Mahajan - 1 year, 1 month ago

Log in to reply

@Aaghaz Mahajan :| 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).

Sharky Kesa - 1 year, 1 month ago

Log in to reply

@Sharky Kesa 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......!!

Aaghaz Mahajan - 1 year, 1 month ago

Log in to reply

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

I can´t find the solution :,(

Xizlon Tho - 1 year ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...