IMO 2016 Day 1

I'm surprised nobody else has posted this. Yes, there are plenty of discussions in Art of Problem Solving, but why not here? And of course there are more problems.


Problem 1

Triangle BCFBCF has a right angle at BB. Let AA be the point on line CFCF such that FA=FBFA = FB and FF lies between AA and CC. Point DD is chosen so that DA=DCDA = DC and ACAC is the bisector of DAB\angle{DAB}. Point EE is chosen so that EA=EDEA = ED and ADAD is the bisector of EAC\angle{EAC}. Let MM be the midpoint of CFCF. Let XX be the point such that AMXEAMXE is a parallelogram. Prove that BDBD, FXFX, and MEME are concurrent.

Problem 2

Find all positive integers nn for which each cell of an n×nn \times n table can be filled with one of the letters I, M, and O in such a way that:

  • in each row and each column, one third of the entries are I, one third are M, and one third are O; and
  • in any diagonal, if the number of entries on the diagonal is a multiple of three, then one third of the entries are I, one third are M, and one third are O.

Note. The rows and columns of an n×nn \times n table are each labelled 11 to nn in a natural order. Thus each cell corresponds to a pair of positive integer (i,j)(i, j) with 1i,jn1 \le i,j \le n. For n>1n > 1, the table has 4n24n-2 diagonals of two types. A diagonal of first type consists all cells (i,j)(i, j) for which i+ji+j is a constant, and the diagonal of this second type consists all cells (i,j)(i, j) for which iji-j is constant.

Problem 3

Let P=A1A2AkP = A_1 A_2 \cdots A_k be a convex polygon in the plane. The vertices A1,A2,,AkA_1, A_2, \ldots, A_k have integral coordinates and lie on a circle. Let SS be the area of PP. An odd positive integer nn is given such that the squares of the side lengths of PP are integers divisible by nn. Prove that 2S2S is an integer divisible by nn.

Note by Ivan Koswara
3 years, 3 months ago

No vote yet
1 vote

  Easy Math Editor

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. 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 \( ... \) or \[ ... \] 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

I have an extension to Q1. (Almost got a complete solution for it, but i observed this, which follows):

Let this intersection point be labelled OO. Prove OO is the circumcentre of ΔXBE\Delta XBE.

Sharky Kesa - 3 years, 2 months ago

Log in to reply

Nice observation. If you complex trig bash it, the solution to your extension becomes very apparent. I'm currently typing up my solution to the original problem and can include your extension at the end.

Trevor Arashiro - 3 years, 2 months ago

Log in to reply

Nice, I have my own solution for it as well. Th essence for this extension is to just prove OE=OXOE=OX since OX=OBOX=OB is quite apparent.

Sharky Kesa - 3 years, 2 months ago

Log in to reply

Here's my synthetic solution. Let XFXF intersect BDBD at OO and EMEM at OO^{'}. We need to show that OO coincides with OO^{'}. A clear observation shows that B,C,F,D,XB,C,F,D,X are concyclic because {DCF}\angle\{DCF\}={DBF}\angle\{DBF\}. Moreover DD is the circumcenter of triangle CBACBA. A clear angle chase shows that {DXF}\angle \{DXF\}={DBF}\angle\{DBF\} showing OO as the circumcenter of triangle OBEOBE thus EOE{O^{'}}=BOBO Thus OO coincides with OO{'}.proved Your extension can be proved easily if you watch FF as the incenter of triangle DABDAB and watching XFADXFAD and CDEMCDEM as the parallelogram. Thanks proving OO as the circumcenter was really very helpful.

Rajyawardhan Singh - 1 year, 7 months ago

Log in to reply

Oh about it, u were talking to me.. OK I am also trying to solve it with another way...

Ayush Kumar - 1 year, 7 months ago

Log in to reply

Provide a solution using complex numbers.

Rajyawardhan Singh - 1 year, 7 months ago

Log in to reply

Well ur solution was really nice, I am also seeing it and trying to find a mistake..... 😂 😂😂😂😂😂😂😂😇😇😇😇😇😝😝😝😝😝😋

Ayush Kumar - 1 year, 7 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...