Duke Math Meet

Sun Nov 01, 2009 6:17 pm, by dysfunctionalequations

I went to this competition.

Code:
Power Round:
1. YAY
2. YAY
3. YAY
4. YAY
5. YAY
6. YAY
7. YAY
8. YAY
9. YAY

Team Round:
1. DARN
2. YAY
3. YAY
4. YAY
5. YAY
6. YAY
7. YAY
8. DARN
9. YAY
10. YAY

Individual Round:
1. YAY
2. YAY
3. YAY
4. YAY
5. YAY
6. YAY
7. DARN
8. DARN
9. DARN
10. YAY

Relay Round:
1. DARN
2. Semi-YAY


Yes.

Also Benjamin Gunby is very tricky.

Trivialize this entry

5 Comments have been made

whoa darn how did you do #10 on individual

Click to reveal hidden content
Okay so it actually turns into one sequence a_1=2, a_2=7, a_3=8, a_{n+3}=\frac{a_{n+2}+a_{n+1}+a_n}{3} and you're trying to find the limit of this sequence. So then you realize that if you only have two terms a_1=x, a_2 = y, a_{n+2}=\frac{a_n+a_{n+1}}{2} then the answer is \frac{x+2y}{3}. Then you think "HEY IT WOULD BE PRETTY TRICKY IF THE ANSWER WERE \frac{1\cdot2 + 2\cdot 7 + 3\cdot 8}{6}." This is admittedly very cheap and ridiculous, but it is what I did during the competition.

An actual legitimate way to do this is possibly to actually solve the recurrence relation - I'm not actually sure how messy that turns out to be, though.

hey that is quite tricky

Okay wait

actual way to do this
So let f(a,b,c) denote the limit of the sequence a_1=a, a_2=b, a_3=c, a_n=\frac{a_{n-1}+a_{n-2}+a_{n-3}}{3}. Then we have the properties

(i) f(a,b,c)+f(x,y,z)=f(a+x, b+y, c+z)
(ii) f(a,a,a)=a
(iii) f(a,b,c)=f\left(b,c,\frac{a+b+c}{3}\right).

So from the first property we get that f(x,y,z)=f(x,0,0)+f(0,y,0)+f(0,0,z)=c_1 x+c_2 y+c_3 z for some constants c_1, c_2, c_3 for all rational x,y,z. From the third property, we get c_2=2c_1, c_3=3c_1. Therefore, f(x,y,z)=c_1(x+2y+3z). c_1 must be \frac{1}{6} by the second property. YAY

Blah

Darn I ran out of time in Individual 10 while bashing the recurrence...
Our team was 6th overall, 0.5 behind AAST B
  • Posted Wed Nov 11, 2009 11:08 am, by bpgbcg

actually, quite trivial...

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  • Hm it looks like geo is my second best olympiad subject, yet still far from algebra.

    By gauss1181, on Sun Oct 11, 2009 11:58 am

  • your polynomial on Mewmew's blog definitely has real, distinct roots

    By stevenmeow, on Mon Oct 05, 2009 6:12 pm

  • Darn. Maybe I should post more geometry here.

    Meaning, REAL geometry.

    This is unfortunate.

    By dysfunctionalequations, on Thu Oct 01, 2009 3:37 am

  • Darn I looked through 6 pages for geo, but NO THERE IS NO GEO HERE WHAT (besides that one IBMO, which I got bored and wrote a solution for hmm). Okay that 2005 IMO thing is like, a theorem with a name which totally should not be on the IMO.

    By serialk11r, on Wed Sep 30, 2009 7:34 pm

  • ‮ I can type backwards. I can write \LaTeX code backwards.

    By dysfunctionalequations, on Wed Sep 30, 2009 2:04 pm

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