This is CTC's first ever tournament! Double elimination, number of people will be everyone who has signed up by 6 pm forum time, Friday the 15th.

People eligible to sign up (and their brand new CTC ratings!):

Skaldjfhrulez - 1200
Violin321 -1200
Kl2836 -1200
Mathpassion101-1200
Xoangieexo -1200
Lotsofmath -1200
Yoh2241 -1200
AIME15 -1200
Jongao -1200
Dojo -1200
Pacman2812-1200
Vamathletes -1200
Splatyango -1200
Mathblitz -1200
Weird888 -1200
Dragonlaird7 -1200
Dannyhamtx -1200
Kookamunga -1200
Nikeballa96 -1200
Pi guy -1200
27r -1200
Yclarinet -1200
Chenhsi -1200
Mind Storm-1200

People who have signed up:

Jain_harshi-1200
Myyellowducky82-1200
Jjx1-1200
Pinkpiglet-1200
Pi_dimension-1200
Guineapiggies-1200
Isabella2296-1200
Mewto55555-1200
RomanianGenius-1200
Vallon22-1200
Superllamel-1200
Cherry_pi-1200
Dragon96-1200
Textangle-1200
FantasyLover-1200
Ernie-1200
Aime_is_hard-1200

I am joining. Finally there is a tourney :lol:

Seeding is based on your rating. Seeing as everyone is tied, I will draw names out of a hat. Or a bowl. Or something roughly hemispherically shape.

Can we use guest accounts for this tournament?

i join xD

[quote]The message is too small. Please make the message longer before submitting.[/quote]

Mehh.  Screw that.

I think I should know this, but I don't.

When are we gonna start?

What is this with people not reading the first post? Next friday!

56- superllamel VS vallon22 (superllamel wins by coinflip)
57- mewto55555 VS vallon22
58- isabella2296 VS ernie (ernie defeats isabella2296, 3-0)
59- mewto55555/vallon22 VS ernie
60- RomanianGenius VS superllamel
61- W59 VS RomanianGenius/superllamel

sorry izzy. congrats on 5th!
[hide="seeding"]1. Ernie
2. Pinkpiglet
3. Mewto55555
4. Guineapiggies
5. Violin321
6. Myyellowducky82
7. Pi-dimension
8. Kl2836
9. Lotsofmath
10. Mathnerd711
11. Aime_is_hard
12. Bisbmard9
13. Isabella2296
14. Jain_harshi
15. FantasyLover
16. Aime15
17. Jjx1
18. Vallon22
19. Dragon96
20. Textangle
21. NIkeballa96
22. Johnnybeard
23. Emilylambkin
24. Dragonlaird7
25. Discrete_math
26. Splatyango
27. Superllamel
28. RomanianGenius
29. Cherry_pi
30. Dannyhamtx
31. BYE
32. BYE[/hide]
[hide="games"]

[b]

[color=purple]1-ERNIE V BYE (ernie recieves a bye)[/color][color=green]
2-AIME15 V JJX1 (aime15 wins by forfeit)
3-KL2836 V DISCRETE_MATH (kl2836 wins 3-0)
4-DRAGONLAIRD7 V LOTSOFMATH (lotsofmath wins by forfeit)
5-VIOLIN321 V ROMANIANGENIUS (romaniangenius wins 3-0)
6-BISBMARD9 V NIKEBALLA96 (bisbmard9 wins 3-1)
7-TEXTANGLE V ISABELLA2296 (isabella2296 wins 3-0)
8-GUINEAPIGGIES V CHERRY_PI(guineapiggies leads 2-1, cherry_pi forfeits)
9-MEWTO55555 V DANNYHAMTX (mewto55555 wins 3-0)
10-JAIN_HARSHI V DRAGON96 (dragon96 wins 3-0)[/color][color=purple]
11-AIME_IS_HARD V JOHNNYBEARD(Aime_is_hard wins by coinflip)[/color][color=green]
12-SUPERLLAMEL V MYYELLOWDUCKY82 (superllamel wins by forfeit)[/color][color=purple]
13-SPLATYANGO V PI-DIMENSION(splatyango wins by coinflip)
14-EMILYLAMBKIN V MATHNERD711(mathnerd711 wins by lack of activity)[/color][color=green]
15-FANTASYLOVER V VALLON22 (vallon22 wins 3-2)[/color][color=purple]
16-PINKPIGLET V BYE (pinkpiglet receives a bye)

17-BYE V JJX1[/color][color=green]
18-DISCRETE_MATH V DRAGONLAIRD7 (discrete_math wins by forfeit)[/color][color=purple]
19-VIOLIN321 V NIKEBALLA96 (tied at 2-2,violin321 wins by coinflip)[/color][color=green]
20-TEXTANGLE V CHERRY_PI(textangle wins 3-0)
21-DANNYHAMTX V JAIN_HARSHI(dannyhamtx wins 3-0)
22-JOHNNYBEARD V MYYELLOWDUCKY82 (johnnybeard wins by forfeit!)[/color][color=purple]
23-PI-DIMENSION V EMILYLAMBKIN(pi-dimension wins by lack of activity)
24-FANTASYLOVER V BYE[/color][color=green]
25-ERNIE V AIME15(ernie wins 3-1)
26-KL2836 V LOTSOFMATH(kl2836 wins 3-0)
27-ROMANIANGENIUS V BISBMARD9(romaniangenius wins 3-0)
28-ISABELLA2296 V GUINEAPIGGIES(isabella2296 wins 3-0)
29-MEWTO55555 V DRAGON96(mewto55555 wins by forfeit)[/color][color=purple]
30-AIME_IS_HARD V SUPERLLAMEL(superllamel wins by lack of activity)
31-SPLATYANGO V MATHNERD711(mathnerd711 wins by lack of activity)[/color][color=green]
32-VALLON22 V PINKPIGLET (vallon22 wins 3-1)

33-JJX1 V PINKPIGLET(pinkpiglet wins 3-0)[/color][color=purple]
34-DISCRETE_MATH V SPLATYANGO(splatyango wins by coinflip)
35-VIOLIN321 V AIME_IS_HARD(aime_is_hard wins by lack of activity)
36-TEXTANGLE V DRAGON96(dragon96 wins by lack of activity)[/color][color=green]
37-DANNYHAMTX V GUINEAPIGGIES(dannyhamtx wins by forfeit)[/color][color=purple]
38-JOHNNYBEARD V BISBMARD9(bisbamrd9 wins by lack of activity)[/color][color=green]
39-PI-DIMENSION V LOTSOFMATH(lotsofmath wins 3-0)
40-FANTASYLOVER V AIME15(fantasylover wins by forfeit)[/color][color=purple]

41-PINKPIGLET V SPLATYANGO(pinkpiglet wins by coinflip)
42-AIME_IS_HARD V DRAGON96(dragon96 wins by coinflip)[/color][color=green]
43-DANNYHAMTX V BISBMARD9(dannyhamtx wins 3-1)
44-LOTSOFMATH V FANTASYLOVER (fantasylover wins 3-0)

45-ERNIE V KL2836(ernie wins 3-1)
46-ROMANIANGENIUS V ISABELLA2296(romaniangenius wins 3-2)
47-MEWTO55555 V SUPERLLAMEL(superllamel wins 3-0)
48-MATHNERD711 V VALLON22(vallon22 wins 3-2)[/color][color=purple]

49-PINKPIGLET V MATHNERD711(mathnerd711 wins by coinflip)[/color][color=green]
50-DRAGON96 V MEWTO55555(mewto55555 wins 3-0)
51-DANNYHAMTX V ISABELLA2296(isabella2296 wins 3-0)
52-FANTASYLOVER V KL2836(kl2836 wins 3-1)

53-MATHNERD711 V MEWTO55555(mewto55555 wins 3-0)[/color][color=purple]
54-ISABELLA2296 V KL2836(isabella2296 wins by coinflip)[/color][color=green]
55-ERNIE V ROMANIANGENIUS(romaniangenius wins 3-0)[/color][color=purple]
56-SUPERLLAMEL V VALLON22(superllamel wins by coinflip)[/color][color=orange]

57-MEWTO55555 V VALLON22[/color][color=green]
58-ISABELLA2296 V ERNIE(ernie wins 3-0)[/color][color=red]

59-W57 V ERNIE
60-ROMANIANGENIUS V SUPERLLAMEL

61-W59VL60

62-W60VW61 (W60 MUST LOSE TWICE TO LOSE)

[/b][/color][/hide]
[hide="PLACING"]

1. W62
2. L62
3. L61
4. L59
5. L57
5. Isabella2296
7. Mathnerd711
7. Kl2836
9. Pinkpiglet
9. Dragon96
9. Dannyhamtx
9. FantasyLover
13. Splatyango
13. Aime_is_hard
13. Bisbmard9
13. Lotsofmath
17. Jjx1
17. Discrete_math
17. Violin321
17. Textangle
17. Guineapiggies
17. Johnnybeard
17. Pi-Dimension
17. Aime15
25. Bye
25. Dragonlaird7
25. Nikeballa96
25. Cherry_pi
25. Jain_harshi
25. Myyellowducky82
25. Emilylambkin
25. Bye[/hide][/quote]

[hide="Ratings"]ernie- 1258
AIME15- 1171
jjx1- 1150
kl2836- 1248
Discrete_Math- 1200
lotsofmath- 1198
dragonlaird7- 1150
violin321- 1175
RomanianGenius- 1298
bisbmard9- 1177
nikeballa96- 1175
Textangle- 1200
isabella2296- 1225
guineapiggies- 1175
cherry_pi- 1150
mewto55555- 1266
dannyhamtx- 1221
jain_harshi- 1150
dragon96- 1177
AIME_is_hard- 1200
johnnybeard- 1223
superllamel- 1252
myyellowducky82- 1152
splatyango- 1200
PI-Dimension- 1175
emilylambkin- 1200
mathnerd711- 1158
FantasyLover- 1204
vallon22- 1270
pinkpiglet- 1202[/hide]

no i didnt, mewto

Wow ernie. Now I feel ad for putting that off until now  :P  You didn't count purples right? Hm whatever. And by the way if vallon has not responded to my prod pm by this time a week from now, he forfeits. Just a heads-up.

Vallon has responded, but has not set up a time. IF he doesn't set up a time (or at least give me a list of possible times) by the end of [b]Tuesday September 30[/b] he is out.

ouch
that hurts =)
what about the game: 60-ROMANIANGENIUS V SUPERLLAMEL?

edit: mewto pwns vallon 3-0
go 5th place!

As we enter the Final 4, the competition heats up. Can RG upset the out-of-practice superllamel sending him into the losers bracket? Mewto or Ernie, who will win? My predictions:

1. Mewto55555(will he take down ernie,rg, AND super?)
2. Superllamel (defeated by mew in the finals)
3. RG (crushed by mew in the semis)
4. Ernie (loses this round in a close match topped with an incredible comeback by mew in Game 5)

To the final 4, good luck, and may the force be with you!  :yoda:
[hide="seeding"][b]1. Ernie[/b]
2. Pinkpiglet[b]
3. Mewto55555[/b]
4. Guineapiggies
5. Violin321
6. Myyellowducky82
7. Pi-dimension
8. Kl2836
9. Lotsofmath
10. Mathnerd711
11. Aime_is_hard
12. Bisbmard9
13. Isabella2296
14. Jain_harshi
15. FantasyLover
16. Aime15
17. Jjx1
18. Vallon22
19. Dragon96
20. Textangle
21. NIkeballa96
22. Johnnybeard
23. Emilylambkin
24. Dragonlaird7
25. Discrete_math
26. Splatyango[b]
27. Superllamel
28. RomanianGenius[/b]
29. Cherry_pi
30. Dannyhamtx
31. BYE
32. BYE[/hide]
[hide="games"]

[b]

[color=purple]1-ERNIE V BYE (ernie recieves a bye)[/color][color=green]
2-AIME15 V JJX1 (aime15 wins by forfeit)
3-KL2836 V DISCRETE_MATH (kl2836 wins 3-0)
4-DRAGONLAIRD7 V LOTSOFMATH (lotsofmath wins by forfeit)
5-VIOLIN321 V ROMANIANGENIUS (romaniangenius wins 3-0)
6-BISBMARD9 V NIKEBALLA96 (bisbmard9 wins 3-1)
7-TEXTANGLE V ISABELLA2296 (isabella2296 wins 3-0)
8-GUINEAPIGGIES V CHERRY_PI(guineapiggies leads 2-1, cherry_pi forfeits)
9-MEWTO55555 V DANNYHAMTX (mewto55555 wins 3-0)
10-JAIN_HARSHI V DRAGON96 (dragon96 wins 3-0)[/color][color=purple]
11-AIME_IS_HARD V JOHNNYBEARD(Aime_is_hard wins by coinflip)[/color][color=green]
12-SUPERLLAMEL V MYYELLOWDUCKY82 (superllamel wins by forfeit)[/color][color=purple]
13-SPLATYANGO V PI-DIMENSION(splatyango wins by coinflip)
14-EMILYLAMBKIN V MATHNERD711(mathnerd711 wins by lack of activity)[/color][color=green]
15-FANTASYLOVER V VALLON22 (vallon22 wins 3-2)[/color][color=purple]
16-PINKPIGLET V BYE (pinkpiglet receives a bye)

17-BYE V JJX1[/color][color=green]
18-DISCRETE_MATH V DRAGONLAIRD7 (discrete_math wins by forfeit)[/color][color=purple]
19-VIOLIN321 V NIKEBALLA96 (tied at 2-2,violin321 wins by coinflip)[/color][color=green]
20-TEXTANGLE V CHERRY_PI(textangle wins 3-0)
21-DANNYHAMTX V JAIN_HARSHI(dannyhamtx wins 3-0)
22-JOHNNYBEARD V MYYELLOWDUCKY82 (johnnybeard wins by forfeit!)[/color][color=purple]
23-PI-DIMENSION V EMILYLAMBKIN(pi-dimension wins by lack of activity)
24-FANTASYLOVER V BYE[/color][color=green]
25-ERNIE V AIME15(ernie wins 3-1)
26-KL2836 V LOTSOFMATH(kl2836 wins 3-0)
27-ROMANIANGENIUS V BISBMARD9(romaniangenius wins 3-0)
28-ISABELLA2296 V GUINEAPIGGIES(isabella2296 wins 3-0)
29-MEWTO55555 V DRAGON96(mewto55555 wins by forfeit)[/color][color=purple]
30-AIME_IS_HARD V SUPERLLAMEL(superllamel wins by lack of activity)
31-SPLATYANGO V MATHNERD711(mathnerd711 wins by lack of activity)[/color][color=green]
32-VALLON22 V PINKPIGLET (vallon22 wins 3-1)

33-JJX1 V PINKPIGLET(pinkpiglet wins 3-0)[/color][color=purple]
34-DISCRETE_MATH V SPLATYANGO(splatyango wins by coinflip)
35-VIOLIN321 V AIME_IS_HARD(aime_is_hard wins by lack of activity)
36-TEXTANGLE V DRAGON96(dragon96 wins by lack of activity)[/color][color=green]
37-DANNYHAMTX V GUINEAPIGGIES(dannyhamtx wins by forfeit)[/color][color=purple]
38-JOHNNYBEARD V BISBMARD9(bisbamrd9 wins by lack of activity)[/color][color=green]
39-PI-DIMENSION V LOTSOFMATH(lotsofmath wins 3-0)
40-FANTASYLOVER V AIME15(fantasylover wins by forfeit)[/color][color=purple]

41-PINKPIGLET V SPLATYANGO(pinkpiglet wins by coinflip)
42-AIME_IS_HARD V DRAGON96(dragon96 wins by coinflip)[/color][color=green]
43-DANNYHAMTX V BISBMARD9(dannyhamtx wins 3-1)
44-LOTSOFMATH V FANTASYLOVER (fantasylover wins 3-0)

45-ERNIE V KL2836(ernie wins 3-1)
46-ROMANIANGENIUS V ISABELLA2296(romaniangenius wins 3-2)
47-MEWTO55555 V SUPERLLAMEL(superllamel wins 3-0)
48-MATHNERD711 V VALLON22(vallon22 wins 3-2)[/color][color=purple]

49-PINKPIGLET V MATHNERD711(mathnerd711 wins by coinflip)[/color][color=green]
50-DRAGON96 V MEWTO55555(mewto55555 wins 3-0)
51-DANNYHAMTX V ISABELLA2296(isabella2296 wins 3-0)
52-FANTASYLOVER V KL2836(kl2836 wins 3-1)

53-MATHNERD711 V MEWTO55555(mewto55555 wins 3-0)[/color][color=purple]
54-ISABELLA2296 V KL2836(isabella2296 wins by coinflip)[/color][color=green]
55-ERNIE V ROMANIANGENIUS(romaniangenius wins 3-0)[/color][color=purple]
56-SUPERLLAMEL V VALLON22(superllamel wins by coinflip)[/color][color=green]

57-MEWTO55555 V VALLON22(mewto55555 wins 3-0)
58-ISABELLA2296 V ERNIE(ernie wins 3-0)[/color][color=orange]

59-MEWTO55555 V ERNIE
60-ROMANIANGENIUS V SUPERLLAMEL[/color][color=red]

61-W59VL60

62-W60VW61 (W60 MUST LOSE TWICE TO LOSE)

[/b][/color][/hide]
[hide="PLACING"]

1. W62
2. L62
3. L61
4. L59
5. Vallon22
5. Isabella2296
7. Mathnerd711
7. Kl2836
9. Pinkpiglet
9. Dragon96
9. Dannyhamtx
9. FantasyLover
13. Splatyango
13. Aime_is_hard
13. Bisbmard9
13. Lotsofmath
17. Jjx1
17. Discrete_math
17. Violin321
17. Textangle
17. Guineapiggies
17. Johnnybeard
17. Pi-Dimension
17. Aime15
25. Bye
25. Dragonlaird7
25. Nikeballa96
25. Cherry_pi
25. Jain_harshi
25. Myyellowducky82
25. Emilylambkin
25. Bye[/hide][/quote][/quote]

[hide="seeding"][b]1. Ernie[/b]
2. Pinkpiglet[b]
3. Mewto55555[/b]
4. Guineapiggies
5. Violin321
6. Myyellowducky82
7. Pi-dimension
8. Kl2836
9. Lotsofmath
10. Mathnerd711
11. Aime_is_hard
12. Bisbmard9
13. Isabella2296
14. Jain_harshi
15. FantasyLover
16. Aime15
17. Jjx1
18. Vallon22
19. Dragon96
20. Textangle
21. NIkeballa96
22. Johnnybeard
23. Emilylambkin
24. Dragonlaird7
25. Discrete_math
26. Splatyango[b]
27. Superllamel
28. RomanianGenius[/b]
29. Cherry_pi
30. Dannyhamtx
31. BYE
32. BYE[/hide]
[hide="games"]

[b]

[color=purple]1-ERNIE V BYE (ernie recieves a bye)[/color][color=green]
2-AIME15 V JJX1 (aime15 wins by forfeit)
3-KL2836 V DISCRETE_MATH (kl2836 wins 3-0)
4-DRAGONLAIRD7 V LOTSOFMATH (lotsofmath wins by forfeit)
5-VIOLIN321 V ROMANIANGENIUS (romaniangenius wins 3-0)
6-BISBMARD9 V NIKEBALLA96 (bisbmard9 wins 3-1)
7-TEXTANGLE V ISABELLA2296 (isabella2296 wins 3-0)
8-GUINEAPIGGIES V CHERRY_PI(guineapiggies leads 2-1, cherry_pi forfeits)
9-MEWTO55555 V DANNYHAMTX (mewto55555 wins 3-0)
10-JAIN_HARSHI V DRAGON96 (dragon96 wins 3-0)[/color][color=purple]
11-AIME_IS_HARD V JOHNNYBEARD(Aime_is_hard wins by coinflip)[/color][color=green]
12-SUPERLLAMEL V MYYELLOWDUCKY82 (superllamel wins by forfeit)[/color][color=purple]
13-SPLATYANGO V PI-DIMENSION(splatyango wins by coinflip)
14-EMILYLAMBKIN V MATHNERD711(mathnerd711 wins by lack of activity)[/color][color=green]
15-FANTASYLOVER V VALLON22 (vallon22 wins 3-2)[/color][color=purple]
16-PINKPIGLET V BYE (pinkpiglet receives a bye)

17-BYE V JJX1[/color][color=green]
18-DISCRETE_MATH V DRAGONLAIRD7 (discrete_math wins by forfeit)[/color][color=purple]
19-VIOLIN321 V NIKEBALLA96 (tied at 2-2,violin321 wins by coinflip)[/color][color=green]
20-TEXTANGLE V CHERRY_PI(textangle wins 3-0)
21-DANNYHAMTX V JAIN_HARSHI(dannyhamtx wins 3-0)
22-JOHNNYBEARD V MYYELLOWDUCKY82 (johnnybeard wins by forfeit!)[/color][color=purple]
23-PI-DIMENSION V EMILYLAMBKIN(pi-dimension wins by lack of activity)
24-FANTASYLOVER V BYE[/color][color=green]
25-ERNIE V AIME15(ernie wins 3-1)
26-KL2836 V LOTSOFMATH(kl2836 wins 3-0)
27-ROMANIANGENIUS V BISBMARD9(romaniangenius wins 3-0)
28-ISABELLA2296 V GUINEAPIGGIES(isabella2296 wins 3-0)
29-MEWTO55555 V DRAGON96(mewto55555 wins by forfeit)[/color][color=purple]
30-AIME_IS_HARD V SUPERLLAMEL(superllamel wins by lack of activity)
31-SPLATYANGO V MATHNERD711(mathnerd711 wins by lack of activity)[/color][color=green]
32-VALLON22 V PINKPIGLET (vallon22 wins 3-1)

33-JJX1 V PINKPIGLET(pinkpiglet wins 3-0)[/color][color=purple]
34-DISCRETE_MATH V SPLATYANGO(splatyango wins by coinflip)
35-VIOLIN321 V AIME_IS_HARD(aime_is_hard wins by lack of activity)
36-TEXTANGLE V DRAGON96(dragon96 wins by lack of activity)[/color][color=green]
37-DANNYHAMTX V GUINEAPIGGIES(dannyhamtx wins by forfeit)[/color][color=purple]
38-JOHNNYBEARD V BISBMARD9(bisbamrd9 wins by lack of activity)[/color][color=green]
39-PI-DIMENSION V LOTSOFMATH(lotsofmath wins 3-0)
40-FANTASYLOVER V AIME15(fantasylover wins by forfeit)[/color][color=purple]

41-PINKPIGLET V SPLATYANGO(pinkpiglet wins by coinflip)
42-AIME_IS_HARD V DRAGON96(dragon96 wins by coinflip)[/color][color=green]
43-DANNYHAMTX V BISBMARD9(dannyhamtx wins 3-1)
44-LOTSOFMATH V FANTASYLOVER (fantasylover wins 3-0)

45-ERNIE V KL2836(ernie wins 3-1)
46-ROMANIANGENIUS V ISABELLA2296(romaniangenius wins 3-2)
47-MEWTO55555 V SUPERLLAMEL(superllamel wins 3-0)
48-MATHNERD711 V VALLON22(vallon22 wins 3-2)[/color][color=purple]

49-PINKPIGLET V MATHNERD711(mathnerd711 wins by coinflip)[/color][color=green]
50-DRAGON96 V MEWTO55555(mewto55555 wins 3-0)
51-DANNYHAMTX V ISABELLA2296(isabella2296 wins 3-0)
52-FANTASYLOVER V KL2836(kl2836 wins 3-1)

53-MATHNERD711 V MEWTO55555(mewto55555 wins 3-0)[/color][color=purple]
54-ISABELLA2296 V KL2836(isabella2296 wins by coinflip)[/color][color=green]
55-ERNIE V ROMANIANGENIUS(romaniangenius wins 3-0)[/color][color=purple]
56-SUPERLLAMEL V VALLON22(superllamel wins by coinflip)[/color][color=green]

57-MEWTO55555 V VALLON22(mewto55555 wins 3-0)
58-ISABELLA2296 V ERNIE(ernie wins 3-0)

59-MEWTO55555 V ERNIE(mewto55555 wins 3-2)[/color][color=orange]
60-ROMANIANGENIUS V SUPERLLAMEL[/color][color=red]

61-MEWTO55555 V L60

62-W60VW61 (W60 MUST LOSE TWICE TO LOSE)

[/b][/color][/hide]
[hide="PLACING"]

1. W62
2. L62
3. L61
4. Ernie
5. Vallon22
5. Isabella2296
7. Mathnerd711
7. Kl2836
9. Pinkpiglet
9. Dragon96
9. Dannyhamtx
9. FantasyLover
13. Splatyango
13. Aime_is_hard
13. Bisbmard9
13. Lotsofmath
17. Jjx1
17. Discrete_math
17. Violin321
17. Textangle
17. Guineapiggies
17. Johnnybeard
17. Pi-Dimension
17. Aime15
25. Bye
25. Dragonlaird7
25. Nikeballa96
25. Cherry_pi
25. Jain_harshi
25. Myyellowducky82
25. Emilylambkin
25. Bye[/hide][/quote][/quote][/quote]

[hide="seeding"][b]1. Ernie[/b]
2. Pinkpiglet[b]
3. Mewto55555[/b]
4. Guineapiggies
5. Violin321
6. Myyellowducky82
7. Pi-dimension
8. Kl2836
9. Lotsofmath
10. Mathnerd711
11. Aime_is_hard
12. Bisbmard9
13. Isabella2296
14. Jain_harshi
15. FantasyLover
16. Aime15
17. Jjx1
18. Vallon22
19. Dragon96
20. Textangle
21. NIkeballa96
22. Johnnybeard
23. Emilylambkin
24. Dragonlaird7
25. Discrete_math
26. Splatyango[b]
27. Superllamel
28. RomanianGenius[/b]
29. Cherry_pi
30. Dannyhamtx
31. BYE
32. BYE[/hide]
[hide="games"]

[b]

[color=purple]1-ERNIE V BYE (ernie recieves a bye)[/color][color=green]
2-AIME15 V JJX1 (aime15 wins by forfeit)
3-KL2836 V DISCRETE_MATH (kl2836 wins 3-0)
4-DRAGONLAIRD7 V LOTSOFMATH (lotsofmath wins by forfeit)
5-VIOLIN321 V ROMANIANGENIUS (romaniangenius wins 3-0)
6-BISBMARD9 V NIKEBALLA96 (bisbmard9 wins 3-1)
7-TEXTANGLE V ISABELLA2296 (isabella2296 wins 3-0)
8-GUINEAPIGGIES V CHERRY_PI(guineapiggies leads 2-1, cherry_pi forfeits)
9-MEWTO55555 V DANNYHAMTX (mewto55555 wins 3-0)
10-JAIN_HARSHI V DRAGON96 (dragon96 wins 3-0)[/color][color=purple]
11-AIME_IS_HARD V JOHNNYBEARD(Aime_is_hard wins by coinflip)[/color][color=green]
12-SUPERLLAMEL V MYYELLOWDUCKY82 (superllamel wins by forfeit)[/color][color=purple]
13-SPLATYANGO V PI-DIMENSION(splatyango wins by coinflip)
14-EMILYLAMBKIN V MATHNERD711(mathnerd711 wins by lack of activity)[/color][color=green]
15-FANTASYLOVER V VALLON22 (vallon22 wins 3-2)[/color][color=purple]
16-PINKPIGLET V BYE (pinkpiglet receives a bye)

17-BYE V JJX1[/color][color=green]
18-DISCRETE_MATH V DRAGONLAIRD7 (discrete_math wins by forfeit)[/color][color=purple]
19-VIOLIN321 V NIKEBALLA96 (tied at 2-2,violin321 wins by coinflip)[/color][color=green]
20-TEXTANGLE V CHERRY_PI(textangle wins 3-0)
21-DANNYHAMTX V JAIN_HARSHI(dannyhamtx wins 3-0)
22-JOHNNYBEARD V MYYELLOWDUCKY82 (johnnybeard wins by forfeit!)[/color][color=purple]
23-PI-DIMENSION V EMILYLAMBKIN(pi-dimension wins by lack of activity)
24-FANTASYLOVER V BYE[/color][color=green]
25-ERNIE V AIME15(ernie wins 3-1)
26-KL2836 V LOTSOFMATH(kl2836 wins 3-0)
27-ROMANIANGENIUS V BISBMARD9(romaniangenius wins 3-0)
28-ISABELLA2296 V GUINEAPIGGIES(isabella2296 wins 3-0)
29-MEWTO55555 V DRAGON96(mewto55555 wins by forfeit)[/color][color=purple]
30-AIME_IS_HARD V SUPERLLAMEL(superllamel wins by lack of activity)
31-SPLATYANGO V MATHNERD711(mathnerd711 wins by lack of activity)[/color][color=green]
32-VALLON22 V PINKPIGLET (vallon22 wins 3-1)

33-JJX1 V PINKPIGLET(pinkpiglet wins 3-0)[/color][color=purple]
34-DISCRETE_MATH V SPLATYANGO(splatyango wins by coinflip)
35-VIOLIN321 V AIME_IS_HARD(aime_is_hard wins by lack of activity)
36-TEXTANGLE V DRAGON96(dragon96 wins by lack of activity)[/color][color=green]
37-DANNYHAMTX V GUINEAPIGGIES(dannyhamtx wins by forfeit)[/color][color=purple]
38-JOHNNYBEARD V BISBMARD9(bisbamrd9 wins by lack of activity)[/color][color=green]
39-PI-DIMENSION V LOTSOFMATH(lotsofmath wins 3-0)
40-FANTASYLOVER V AIME15(fantasylover wins by forfeit)[/color][color=purple]

41-PINKPIGLET V SPLATYANGO(pinkpiglet wins by coinflip)
42-AIME_IS_HARD V DRAGON96(dragon96 wins by coinflip)[/color][color=green]
43-DANNYHAMTX V BISBMARD9(dannyhamtx wins 3-1)
44-LOTSOFMATH V FANTASYLOVER (fantasylover wins 3-0)

45-ERNIE V KL2836(ernie wins 3-1)
46-ROMANIANGENIUS V ISABELLA2296(romaniangenius wins 3-2)
47-MEWTO55555 V SUPERLLAMEL(superllamel wins 3-0)
48-MATHNERD711 V VALLON22(vallon22 wins 3-2)[/color][color=purple]

49-PINKPIGLET V MATHNERD711(mathnerd711 wins by coinflip)[/color][color=green]
50-DRAGON96 V MEWTO55555(mewto55555 wins 3-0)
51-DANNYHAMTX V ISABELLA2296(isabella2296 wins 3-0)
52-FANTASYLOVER V KL2836(kl2836 wins 3-1)

53-MATHNERD711 V MEWTO55555(mewto55555 wins 3-0)[/color][color=purple]
54-ISABELLA2296 V KL2836(isabella2296 wins by coinflip)[/color][color=green]
55-ERNIE V ROMANIANGENIUS(romaniangenius wins 3-0)[/color][color=purple]
56-SUPERLLAMEL V VALLON22(superllamel wins by coinflip)[/color][color=green]

57-MEWTO55555 V VALLON22(mewto55555 wins 3-0)
58-ISABELLA2296 V ERNIE(ernie wins 3-0)

59-MEWTO55555 V ERNIE(mewto55555 wins 3-2)[/color][color=purple]
60-ROMANIANGENIUS V SUPERLLAMEL(romaniangenius wins by forfeit)

61-MEWTO55555 V SUPERLLAMEL(mewto55555 wins by forfeit)[/color][color=orange]

62-ROMANIANGENIUS V MEWTO55555 (W60 MUST LOSE TWICE TO LOSE)

[/b][/color][/hide]
[hide="PLACING"]

1. W62
2. L62
3. Superllamel
4. Ernie
5. Vallon22
5. Isabella2296
7. Mathnerd711
7. Kl2836
9. Pinkpiglet
9. Dragon96
9. Dannyhamtx
9. FantasyLover
13. Splatyango
13. Aime_is_hard
13. Bisbmard9
13. Lotsofmath
17. Jjx1
17. Discrete_math
17. Violin321
17. Textangle
17. Guineapiggies
17. Johnnybeard
17. Pi-Dimension
17. Aime15
25. Bye
25. Dragonlaird7
25. Nikeballa96
25. Cherry_pi
25. Jain_harshi
25. Myyellowducky82
25. Emilylambkin
25. Bye[/hide]

SUperllamel forfeits to rg, forfeits to me, the drops out of ctc. Rg and i will play as soon as I fix my computer.

oh so superllamel, 27r, Fantasy Lover and I all drop out of CTC.

Yeah, I'm dropping out of CTC also.

I'll bet 100000 dollars that mewto wins.

im gone from ctc though i was never truely in

Anyway, Mewto55555 wins Round 1 3-2 (4-0,4-1,3-4,1-4,4-3)

Romaniangenius wins round 2 3-1(0-4,4-2,4-3,4-3)

CONGRATULATIONS ROMANIANGENIUS OUR NEW CHAMPION!  :first:

Also, good job to myself  :winner_second: and superllamel  :winner_third: for their top 3 finishes.

[hide="seeding"]1. Ernie
2. Pinkpiglet
3. Mewto55555
4. Guineapiggies
5. Violin321
6. Myyellowducky82
7. Pi-dimension
8. Kl2836
9. Lotsofmath
10. Mathnerd711
11. Aime_is_hard
12. Bisbmard9
13. Isabella2296
14. Jain_harshi
15. FantasyLover
16. Aime15
17. Jjx1
18. Vallon22
19. Dragon96
20. Textangle
21. NIkeballa96
22. Johnnybeard
23. Emilylambkin
24. Dragonlaird7
25. Discrete_math
26. Splatyango
27. Superllamel
28. RomanianGenius
29. Cherry_pi
30. Dannyhamtx
31. BYE
32. BYE[/hide]
[hide="games"]

[b]

[color=purple]1-ERNIE V BYE (ernie recieves a bye)[/color][color=green]
2-AIME15 V JJX1 (aime15 wins by forfeit)
3-KL2836 V DISCRETE_MATH (kl2836 wins 3-0)
4-DRAGONLAIRD7 V LOTSOFMATH (lotsofmath wins by forfeit)
5-VIOLIN321 V ROMANIANGENIUS (romaniangenius wins 3-0)
6-BISBMARD9 V NIKEBALLA96 (bisbmard9 wins 3-1)
7-TEXTANGLE V ISABELLA2296 (isabella2296 wins 3-0)
8-GUINEAPIGGIES V CHERRY_PI(guineapiggies leads 2-1, cherry_pi forfeits)
9-MEWTO55555 V DANNYHAMTX (mewto55555 wins 3-0)
10-JAIN_HARSHI V DRAGON96 (dragon96 wins 3-0)[/color][color=purple]
11-AIME_IS_HARD V JOHNNYBEARD(Aime_is_hard wins by coinflip)[/color][color=green]
12-SUPERLLAMEL V MYYELLOWDUCKY82 (superllamel wins by forfeit)[/color][color=purple]
13-SPLATYANGO V PI-DIMENSION(splatyango wins by coinflip)
14-EMILYLAMBKIN V MATHNERD711(mathnerd711 wins by lack of activity)[/color][color=green]
15-FANTASYLOVER V VALLON22 (vallon22 wins 3-2)[/color][color=purple]
16-PINKPIGLET V BYE (pinkpiglet receives a bye)

17-BYE V JJX1[/color][color=green]
18-DISCRETE_MATH V DRAGONLAIRD7 (discrete_math wins by forfeit)[/color][color=purple]
19-VIOLIN321 V NIKEBALLA96 (tied at 2-2,violin321 wins by coinflip)[/color][color=green]
20-TEXTANGLE V CHERRY_PI(textangle wins 3-0)
21-DANNYHAMTX V JAIN_HARSHI(dannyhamtx wins 3-0)
22-JOHNNYBEARD V MYYELLOWDUCKY82 (johnnybeard wins by forfeit!)[/color][color=purple]
23-PI-DIMENSION V EMILYLAMBKIN(pi-dimension wins by lack of activity)
24-FANTASYLOVER V BYE[/color][color=green]
25-ERNIE V AIME15(ernie wins 3-1)
26-KL2836 V LOTSOFMATH(kl2836 wins 3-0)
27-ROMANIANGENIUS V BISBMARD9(romaniangenius wins 3-0)
28-ISABELLA2296 V GUINEAPIGGIES(isabella2296 wins 3-0)
29-MEWTO55555 V DRAGON96(mewto55555 wins by forfeit)[/color][color=purple]
30-AIME_IS_HARD V SUPERLLAMEL(superllamel wins by lack of activity)
31-SPLATYANGO V MATHNERD711(mathnerd711 wins by lack of activity)[/color][color=green]
32-VALLON22 V PINKPIGLET (vallon22 wins 3-1)

33-JJX1 V PINKPIGLET(pinkpiglet wins 3-0)[/color][color=purple]
34-DISCRETE_MATH V SPLATYANGO(splatyango wins by coinflip)
35-VIOLIN321 V AIME_IS_HARD(aime_is_hard wins by lack of activity)
36-TEXTANGLE V DRAGON96(dragon96 wins by lack of activity)[/color][color=green]
37-DANNYHAMTX V GUINEAPIGGIES(dannyhamtx wins by forfeit)[/color][color=purple]
38-JOHNNYBEARD V BISBMARD9(bisbamrd9 wins by lack of activity)[/color][color=green]
39-PI-DIMENSION V LOTSOFMATH(lotsofmath wins 3-0)
40-FANTASYLOVER V AIME15(fantasylover wins by forfeit)[/color][color=purple]

41-PINKPIGLET V SPLATYANGO(pinkpiglet wins by coinflip)
42-AIME_IS_HARD V DRAGON96(dragon96 wins by coinflip)[/color][color=green]
43-DANNYHAMTX V BISBMARD9(dannyhamtx wins 3-1)
44-LOTSOFMATH V FANTASYLOVER (fantasylover wins 3-0)

45-ERNIE V KL2836(ernie wins 3-1)
46-ROMANIANGENIUS V ISABELLA2296(romaniangenius wins 3-2)
47-MEWTO55555 V SUPERLLAMEL(superllamel wins 3-0)
48-MATHNERD711 V VALLON22(vallon22 wins 3-2)[/color][color=purple]

49-PINKPIGLET V MATHNERD711(mathnerd711 wins by coinflip)[/color][color=green]
50-DRAGON96 V MEWTO55555(mewto55555 wins 3-0)
51-DANNYHAMTX V ISABELLA2296(isabella2296 wins 3-0)
52-FANTASYLOVER V KL2836(kl2836 wins 3-1)

53-MATHNERD711 V MEWTO55555(mewto55555 wins 3-0)[/color][color=purple]
54-ISABELLA2296 V KL2836(isabella2296 wins by coinflip)[/color][color=green]
55-ERNIE V ROMANIANGENIUS(romaniangenius wins 3-0)[/color][color=purple]
56-SUPERLLAMEL V VALLON22(superllamel wins by coinflip)[/color][color=green]

57-MEWTO55555 V VALLON22(mewto55555 wins 3-0)
58-ISABELLA2296 V ERNIE(ernie wins 3-0)

59-MEWTO55555 V ERNIE(mewto55555 wins 3-2)[/color][color=purple]
60-ROMANIANGENIUS V SUPERLLAMEL(romaniangenius wins by forfeit)

61-MEWTO55555 V SUPERLLAMEL(mewto55555 wins by forfeit)[/color][color=green]

62-ROMANIANGENIUS V MEWTO55555  (mewto55555 wins 3-2)
63-ROMANIANGENIUS V MEWTO55555  (romaniangenius wins 3-1)

[/b][/color][/hide]

CONGRATULATIONS TO ALL WHO PARTICIPATED IN THIS TOURNAMENT! THE FINAL RESULTS ARE AS FOLLOWS:

1. RomanianGenius
2. Mewto55555
3. Superllamel
4. Ernie
5. Vallon22
5. Isabella2296
7. Mathnerd711
7. Kl2836
9. Pinkpiglet
9. Dragon96
9. Dannyhamtx
9. FantasyLover
13. Splatyango
13. Aime_is_hard
13. Bisbmard9
13. Lotsofmath
17. Jjx1
17. Discrete_math
17. Violin321
17. Textangle
17. Guineapiggies
17. Johnnybeard
17. Pi-Dimension
17. Aime15
25. Dragonlaird7
25. Nikeballa96
25. Cherry_pi
25. Jain_harshi
25. Myyellowducky82
25. Emilylambkin

[hide="For Ernie"]Ernie, is there anyway you could tally up the ratings? My computer is messed up. The final round counts as two games (62and 63) If you can't I'll do it when I fix my comp.[/hide]

[hide="For Izzy"]Please lock this.[/hide]

aw, you coulda asked me to do it =)

anywayz, locked


search
pigeonhole principle
irrational number
real analysis
real analysis unsolved

Suppose q is an irrational number,prove that the set {Z+Z*q} is dense in R.
Here Z means the set of integers,R means the real numbers field,+ and *means as usual.
thank you l[/b]

This essentially amounts to proving that the set $ [nx(mod1) | n\in Z]$ is dense in the unit interval for all irrational $ x$ since given a real $ y$ choose an integer multiple of $ x$ that has fractional part arbitrarily close to $ y$ then add the appropriate integer to fill in the gap.
For a proof of this density result type "dense mod 1" into the search thing at the top of the page and then look at the thread started by Kuba - Grobber gives a proof.
If you want to try yourself then use a sort of pigeonhole-ish argument.


Hi Im doing a project for my stats class.
I basically need to ask 60 ppl their age and a random number guessed by them between 1 and 100.
I am too shy to ask people in real life, so if you guys can come through for me that would be highly appreciated.
Please post your age and a random number between 1 and 100.

If you feel its not secure to post your age on this forum please send me a private message with your age and the random number.

Thanks to everyone in advance.

13 years, I choose 37, since it's the most commonly chosen number between 1 and 100.

[i]Age[/i]: 62 [i]Favorite Number[/i]: 3
[hide="Huh?"](This is called response bias. Just did this to warn you :wink: I'm rlly 14.)[/hide]

Age: 13
Number: 42.  No jk, 77, cause there were 2 times in a row where that was the number the teacher was thinking of.

age 11 7 [hide] because it is my lucky number[/hide]

13, birthday was about a week ago.

so, 1

$ 13\frac{24}{73}$ is my age, 42 is my number.

Dude, no. I'm pming my age. lol
and ..number

Age 17, Number 35

Curiosity, is it have to be integer? :D

I'm $ \minus{}5\pi e^{i\pi}$, and I pick $ \sqrt{163}\plus{} \zeta(3)$

Hey guys, see the OP's post time ... most likely his project has already been due.

[color=red][size=84][PhireKaLk6781:]Locked. Oops, I never realized this was a revival. You're allowed to laugh at me for that. xD[/size][/color]


The total number of coins with three friends A, B and C is 10. Also, the sum of the
reciprocals of the number of coins with the three mentioned friends is 1. If the number of coins
with each of the three mentioned friends is an integer, then what can be the difference between the
number of coins with A and B?

[hide="hint 1"]There aren't very many possible values for the number of coins each friend has.  Try to get some bounds on these.[/hide]

[hide="hint 2"]Because their reciprocals sum to 1, each number must be greater than 1.  What happens if all are greater than 2?[/hide]

[hide="solution"]Let $ a$ be the number of coins $ A$ has; define $ b$ and $ c$ similarly.

Then, we have $ a \plus{} b \plus{} c \equal{} 10$.  Due to the constraints of the problem, $ a,b,c\geq2$.  Assume that some person does not have 2 coins (i.e. that $ a,b,c > 2$).  We know that at least one of $ a$, $ b$, and $ c$ is greater than $ 3$.  Thus, $ \frac1a \plus{} \frac1b \plus{} \frac1c < \frac13 \plus{} \frac13 \plus{} \frac13 \equal{} 1$, a contradiction.

WLOG, let $ a \equal{} 2$.  Then, we have a system of two equations with two unknowns:

$ b \plus{} c \equal{} 8$  and  $ \frac1b \plus{} \frac1c \equal{} \frac12$

Substituting, we have:

$ b \equal{} 8 \minus{} c\Rightarrow \frac1{8 \minus{} c} \plus{} \frac1c \equal{} \frac12$

$ 2(8 \minus{} c \plus{} c) \equal{} c(8 \minus{} c)$

$ c^2 \minus{} 8c \plus{} 16 \equal{} 0\Rightarrow c \equal{} 4$

It follows that the friends have 2, 4, and 4 coins, in some order.  Therefore, the only possible values of the difference between the number of coins of $ A$ and $ B$ are 2 and 0.[/hide]

This is probably on the upper limit of difficulty of problems for HSB...


[quote="Ulanbek_Kyzylorda KTL"]Let a,b,c>=0 and a^4+b^4+c^4=<2(a^2b^2+b^2c^2+c^2a^2). Prove that
                                           a^2+b^2+c^2=<2(ab+bc+ca).[/quote]
Because $ \sqrt a\plus{}\sqrt b\equal{}\sqrt{a\plus{}b\plus{}2\sqrt{ab}}\geq\sqrt c.$ :wink:

$ a^4\plus{}b^4\plus{}c^4\leq 2(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)$
$ \Longleftrightarrow (a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(a\minus{}b\plus{}c)(\minus{}a\plus{}b\plus{}c)\geq0$
$ \Longleftrightarrow a,b,c$ are left of side of triangle put that $ a\equal{}x\plus{}y$ , $ b\equal{}y\plus{}z$ and $ c\equal{}z\plus{}x$ with $ x,y,z\geq 0$
your inequality is:
$ (x\plus{}y)^2\plus{}(y\plus{}z)^2\plus{}(z\plus{}x)^2\leq 2[(x\plus{}y)(y\plus{}z)\plus{}(y\plus{}z)(z\plus{}x)\plus{}(z\plus{}x)(x\plus{}y)]$
$ \Leftrightarrow xy\plus{}yz\plus{}zx\geq 0$
obvious!


I got 30 total last year on State
20 on Sprint
10 on Target
I got 11th
That was before i got specialized AOPS topic bokss

Notes:
I missed so many of the beginning problems
I only missed two on the last ten
4th place got 35
I could've broke any tiebreakers
I'm pretty sure i missed number 1 on sprint
I missed 3 questions on the target because
1. I couldn't understand the graph quick enough and I didn't feel like reading carefully
2. I forgot to minus 1 from my answer
3. I forgot to use complementary probability
Sprint tired me out literally
My hand was hurting at the end of the sprint round

I did it- but at home 2 days ago :P

23 on Sprint.
7 on Target

23+7(2) = 37

I got a 42  (28 + 14) but that was probably because I did it at home for practice.  If I was there at the actual competition my score probably would have been about 20 points less because of nerves.

I've only done the sprint since we go over the target round at school (our team made it to state for once) and I got 18.  I need to read more carefully and not rush the arithmetic.  I could have had a 23 if I'd made no careless errors.  Nerves don't really affect me, with the wooden walls around me I forget I'm at a compeition.

32/18 :oops: /14 last yr, in KS.
1st written was 35
2nd was 34
3rd-7th were 32. I got 7th  :roll: , and cd up to 4th.

I would have taken 1st in Kansas then! Oh well I got first in WI anyway :P  37 (25,12)

Oh for last years test I actually did pretty well 26 sprint 16 target 26+16=42(that was practice at home) but on the real thing I do horrible. Plus when I did that test again like 5 months later I got a 28 on sprint.

second with 25/14. Apparently, 64-36 ISN'T 26...

I assume someone who got an 8 who is trying to appear smarter than they actually are.

27/12=39.  Lets just say I wasn't performing very well on target -_-.

That was barely good enough for 3rd in michigan, considering 7th was 38.

[quote="perfect628"]Apparently, 64-36 ISN'T 26...[/quote]

What are you talking about?? Anyway I do really bad on the real thing so practice scores dont even matter that much.

[quote="perfect628"]On the target problem about the sides of the right triangle, and lengths 6 and 8, I did $\sqrt{8^{2}-6^{2}}$ to find the third side. I did 64-36 in my head, and somehow ended up with 26.[/quote]

Oh okay.

I got 16 on sprint, 4 on target, making a total score of 24. i sucked so bad last year... i got a higher score in states when i was in [i]6th[/i] grade...

I got 25/16, good enough for first last year.

[quote="bpms"][quote="jhredsox"]who put a 46?????????????

yes this deserves a double post!

jorian[/quote]
A liar.
Last year the highest score was a 44 by both Nathan Benjamin and Karan Batra[/quote]

lol, there were 2!!! :rotfl:

who voted 45?
Actually I was mistaken, it is possible for someone to have gotten a 46, just extremely unlikely (the report covered most of the states and most of the others were mentioned on AoPS)

[quote="jhredsox"]i think you did better than that, bpms.

like a 39 maybe.

jorian[/quote]
I wish, that would've put me at 2nd.

last year, I was pretty bad...

23  :blush:   :(

but that was BEFORE I found the AoPS book and took the online lessons!


calculus integration logarithms function limit derivative

I tried finding a definite log integral on my ti-89, but it returned undef (probably because it was from 0 to something greater than 1). Does anyone know a way to find an approximate value? It doesnt have to be on an 89. Anything would be much appreciated.

An approximate value is infinity (actually -infinity since it's under the axis). If you go from 0 to anything greater than 0 it will give you infinity (undefined). It might help to take into consideration what the indefinite integral is, that $\int \ln(x) dx = x \ln(x) - 1$. And of course when you plug in 0 to ln you're going to get undefined.

Also, calculus problems should be posted in the calculus sections, not in pre-olympiad.

sorry, by log integral, i meant $li(x)=\displaystyle\int_0^x\displaystyle\frac{dt}{\ln t}$.
The only things i know about it are that its an upper bound for $\pi(n)$, but that seems like a bad way to calculate it.

Also, I dont know how to move this, so if a moderator/admin or someone who does know how could move this it would be much appreciated.

[quote="gopherhole112"]sorry, by log integral, i meant $li(x)=\displaystyle\int_0^x\displaystyle\frac{dt}{\ln t}$.
The only things i know about it are that its an upper bound for $\pi(n)$, but that seems like a bad way to calculate it.
[/quote]
Actually, it's not an upper bound for the prime counting function (although it is for all practical purposes).  Littlewood proved that the plots of $\pi(x)$ and $\text{li}(x)$ cross infinitely many times.
More information is at [url]http://mathworld.wolfram.com/LogarithmicIntegral.html[/url].

Ah, I was not familiar with the term log integral, I assumed it just meant the integral of a logarithm.

What!?! Where's the beautiful logic of mathematics? If you ask me, the log integral should the integral of a log!

Hokkage, I believe that you made a couple of mistakes:
$\int\ln(x)dx=x\ln(x)-x+C$
Also, just because the log function itself decreases without bound near 0, does [u]not[/u] mean that its integral is infinite (see [url]http://mathworld.wolfram.com/ImproperIntegral.html[/url]).  As a matter of fact, $\displaystyle\lim_{x\rightarrow\ 0^+}(x\ln(x)-x)=\lim_{x\rightarrow\ 0^+}\frac{\ln(x)}{1/x}-0=\lim_{x\rightarrow\ 0^+}\frac{1/x}{-1/x^2}=\lim_{x\rightarrow\ 0^+}-x=0$.
So, $\displaystyle\int_0^x\ln(t)dt=x\ln(x)-x-(0)$.

Draperp, yes I did make a mistake in that integral (my calculus teacher would kill me). I forgot the formula and I didn't feel like doing parts so I guessed and checked, and I think I took the derivative wrong  :oops: on that guess.

And I forgot about impropers! Geeze. Summer is doing horrible things to my calculus. Thanks for pointing those out, maybe I won't go so fast next time.

[quote="draperp"]Danbert, $li(x)=\displaystyle\int_{\mu}^x\frac{dt}{\ln(t)}$ for $x>1$.
($\mu=1.4513692...$)[/quote]

That amounts to the same thing I wrote, except that my integral is improper.
$\int_0^\mu\frac{dt}{\ln(t)}=0$

And "$x>1$" should be "$x>\mu$".

[quote="Danbert"]And "$x>1$" should be "$x>\mu$".[/quote]
No, $\int_a^b$ is still okay even when $b<a$.
(And I meant to address my previous answer Gopherhole112. Sorry.)


I know there isn't much time left before ARML, but I am almost finished a mock ARML Individual round.  I should be done by tonight, if anyone is interested in taking it.

Oh, there will certainly be people who are interested, myself included.  Just take a look at the number of replies to Altheman's announcement of his mock ARML.

Yeah I'd probably do it, if I have spare time...

wahoo i will consider doing it...if you want somebody to check over...i can do some of that too

I'll probably do it; I need the extra practice.


Why is the area of the trapezoid one fouth of the rectangle?

http://www.artofproblemsolving.com/Wiki/index.php/1987_AIME_Problems/Problem_6

"Rectangle ABCD is divided into four parts of equal area by five segments as shown in the figure..."


Let $T$ be a real number satisfying the property: For any nonnegative real numbers $a,b,c,d,e$ with sum equal to $1$, it is possible to arrange them around a circle such that the products of any two neighboring numbers are no greater than $T$.  Determine the minimum value of $T$.

Order the numbers as $a \leq b \leq c \leq d \leq e$. Now, if $a=b=0$ and $c=d=e=1/3$, there are two positive adjacent numbers, so that the product $1/9$ is present. Thus $T \geq 1/9$.

Now, to show that this $T$ always works: put your numbers in the order $a, e, b, c, d$ so that the products are $ae, eb, bc, cd, da$. Now $be \geq ae, cd \geq bc, cd \geq da$ so that the greatest of them is either $be$ or $cd$. If $cd>1/9$, then by AM-GM $c+d>2/3$; also $d^2 \geq cd>1/9$, so $d>1/3$, then $e>1/3$. Adding up gives $e+c+d>1$, absurd. On the other hand, if $be>1/9$ then $(b+c+d)e \geq 3be>1/3>1/4$, so that, by AM-GM, $b+c+d+e>1$, absurd.


Let $ P$ be the set of all points in $ R^n$ with rational coordinates.For points $ A,B\in P$ ,one can move from $ A$ to $ B$ if $ AB \equal{} 1$.Prove that: every point in $ P$ cab reached from any other point in $ P$ by a finite sequence of moves if only if $ n\geq 5$

any ideas
I have not found an idea which to slove problem?
can anyone help me?


LaTeX
function
algebra
floor function
number theory unsolved
number theory

prove that n_c_7 -[n/7] is always divided by 7.
p.s-i dont have latex .you may have some problem in understanding the notaions.i hope you can understand n_c_7 and the greatest integer function is denoted of n/7  by[n/7].
please help.

It is a direct corollary of Lucas's Theorem.

please prove it.it'll be morehelpfull if it is written in LATEX.


More specifically, anyone have the will and ability to sort those USAMO scores by score?

thanks

[quote="Xuan768"]I wrote a tight C++ program related to problem number 5 on the USAMO, and it appears that I must have gotten at least one point for it.  :lol:[/quote]
computers are excellent at guessing

Sort of.

You really should have gotten into computer science, Mark.


algebra
polynomial
algorithm
induction
combinatorics theorems
combinatorics

Hello, I'm a italian student that I'm studing the AKS algorithm avaible on address http://www.math.princeton.edu/~annals/issues/2004/Sept2004/Agrawal.pdf.

In this paper is wrote :

"So there exist at least $l+1$ distinct polynomials of degree one in $G$. Therefore, there exists at least $\binom{t+l}{t-1}$ distinct polynomials of degree < $t$ in $G$.

The question is : Why? help somebody me?

Thanks and excuse me for my english!

For $t = 1$, it counts the polynomial $f(x) = 1$.  For $t = 2$, it counts that polynomial and the $l+1$ polynomials of degree 1.  For $t \geq 3$, it counts the polynomials which can be made by multiplying up to $t-1$ of our linear polynomials together.

In other words, there are $t+l \choose t-1$ ways to choose up to $t-1$ objects from among $l+1$, if repitition is allowed but order isn't important.

Excuse me, but I have understood the theorem in this way:

The number of polynomial of degree less than $t$, is :

$\sum_{i=0}^{t-1}\binom{l+1+i-1}{i}$ (from combination with ripetition).

Now, I prove that $\sum_{i=0}^{t-1}\binom{l+i}{i}\geq \binom{t+l}{t-1}$ for induction on $t$

case base : $t=1$ then $\binom{l}{0}\geq \binom{l+1}{0}$
inductive reasoning : Let true the theorem for $t_{0}-1$ and prove for $t_{0}$

$\sum_{i=0}^{t_{0}}\binom{l+i}{i}= \sum_{i=0}^{t_{0}-1}\binom{l+i}{i}+\binom{l+t_{0}}{t_{0}}\geq \binom{t_{0}+l}{t_{0}-1}+\binom{t_{0}+l}{t_{0}}= \binom{t_{0}+1+l}{t_{0}}.$

Is correct ?


a long time ago, I read the Art and Craft of Problem Solving by Zeitz, and I remember they had a problem:

Prove that any positive integer n can can be written as a difference of two subsets of distinct powers of 3.

And if I remember correctly, he gave a proof using generating functions...does anybody know this proof?

Basically, prove by induction on n that
$\prod_{k=0}^n (x^{3^k}+1+x^{-3^k})=x^t+x^{t-1}+\cdots+x^{-t}$,
where $t=3^0+3^1+\cdots+3^n$. In the limiting case, the LHS gives the generating function for expressing a number as a sum or difference of different powers of 3, and the RHS has each power of x exactly once. So, there is exactly one way to express each integer (not restricted to positive integers) in this way.

IMO, a very nice proof!

[quote="zabelman"]IMO, a very nice proof![/quote]Indeed!  :cool:

that is cool, although i dont think you need induction because of the nice identity:

$\displaystyle (x^{3^n}+1+x^{3^{-n}}) = \frac{x^{3^{n+1}}-1}{(x^{3^n}-1)(x^{3^n})}$

then you can cancel nicely in the numerator and denominator.

Could it also be proved by giving an algorithm to convert $n$ from base 3 to a sum or difference of powers of 3? Basically take

$n=a_ka_{k-1}...a_1$ (base 3), where $a_i$ is 0, 1 or 2 for $i=1, 2, ..., n$, then take the smallest $i$ such that $a_i=2$ or $a_i=3$, substract 3 to give $b_i=-1$ or $b_i=0$ and add 1 to $a_{i+1}$. If $a_{i+1}=0$ and the new digit is $b_{i+1}=1$ you repeat for the next $i$, if $a_{i+1}=1$ or $2$, you do the same for digit $i+1$ and so on. Clearly all the digits to the right of $i+1$ are the ones we want and the algorithm must end (or else $k$ would be infinite), so it should work. ;)


function
analytic geometry
algebra unsolved
algebra

[quote="treegoner"]Let $f$ be a strictly increasing function from $R$ to $R$ such that
                 $ f(a+b-ab) = f(a) + f(b) - f(a)f(b)$ for every $a,b$
   Find $f$.[/quote]
$f(x)=x$ ??  ;)

[quote="treegoner"]Let $f$ be a strictly increasing function from $R$ to $R$ such that
                 $ f(a+b-ab) = f(a) + f(b) - f(a)f(b)$ for every $a,b$
   Find $f$.[/quote]

It is a disguised form of $h(AB) = h(A)h(B)$, with coordinates centered at 1 rather than 0.

Consider the operation $a * b = a + b - ab$ (multiplication, written relative to 1)
and let $g(x) = 1 - f(x)$.

$ g(a * b) = g(a) g(b) $.

Now let $A = 1-a$ and $B = 1-b$.  Then  $g (1 - AB) = g(1 - A) g(1-B)$.
Let $h(x) = g(1-x) = 1 - f(1-x)$.
$h(AB) = h(A)h(B)$

We have $h(x) = sgn(x) |x|^a$   for some $a > 0$ and this gives a family of
solutions for $f(x)$.


parameterization
integration
limit
probability and stats

Let X be an exponential random variable with parameter c. Find the expectation(if it exists) of 1/X.

$ E(1/X)\equal{}\int_0^{\infty}\frac1x\cdot\frac1ce^{\minus{}x/c}\,dx\equal{}\plus{}\infty$ since the integrand behaves like a constant times $ \frac1x$ as $ x\to 0.$

In general: suppose $ X$ is a nonnegative random variable with density $ f(x)$ such that $ \lim_{x\to0\plus{}}f(x)>0.$ Then $ E(1/X)\equal{}\plus{}\infty.$

[quote="Kent Merryfield"]In general: suppose $ X$ is a nonnegative random variable with density $ f(x)$ such that $ \lim_{x\to0 \plus{} }f(x) > 0.$ Then $ E(1/X) \equal{} \plus{} \infty.$[/quote]

It's really interesting and surprising to know. I m interested in knowing the proof. :roll:

replace $ \frac{1}{c} \exp(\frac{\minus{}x}{c})$ by $ f(x)$ in the proof of Kent Merryfield.

[quote="limsup"]It's really interesting and surprising to know. I m interested in knowing the proof. :roll:[/quote]

That smiley indicates sarcasm, just so you know.

so take my answer as  second order sarcasm ...


I use texniccenter, but MathType is fine.

You can copy/paste LaTeX images into Word, but I normally use Equation Editor.  My school has MathType, though.

[quote="thatin"]I think that Math Type is better than LaTeX... it is like Windows and DOS :P[/quote]

Probably whatever thing you like most is best for you, but certainly latex and math type is not like windows and dos. Windows is a simpler, more intuitive environment which lets you do basicly the same things DOS did. Math Type is a simpler, more intuitive environment [i]for writing math equations[/i] which in the proccess of simplifyng loses a great amount of the possibilities latex offers you. Everything you can do wiht Math Type, if I remember correctly because I don't use it anymore, is well given with a few buttons. Understand you can do a lot of things with a few buttons and most of the time it is enough, but sometimes you want truly specific stuff. Well the thing with latex is that probably someone has wanted to do it before and there are gurus who write specific packages in order to be able to do exactly that. And in the extremely outlandish situation in which there is no package you can always make it yourself if you're in the mood to learn how to do it.  Also latex is not simply for writing equations but a whole typesetting system for building entire documents, from letters to books, which I'm guessing math type is not.
But if math type serves you well that's what you should use. Nevertheless, mathematicians and physicists are switching to latex, I think. At work, it can be hard to overrun this tendencies.

Ursula

ok! windows and unix then :)
I must admit that I have never needed specific stuff.

[quote="lingomaniac88"]I usually create math documents using Microsoft Word.  What do you think of this?[/quote]

I do too, occasionally. It depends on what the content is - I'd go with LaTeX on equation-heavy things. But the thing is, it's just so much easier than opening LaTeX. I mean most of the time I'd be done before I even had a heading done for the LaTeX article.

But I'd definitely recommend you learn LaTeX if you don't know it already. It's very handy for documents that you'd like to look good, where...you know... Sum[x = 0 => k][x^2] just wouldn't do.

You just have to make the head and all this stuff once, simple save some starting document somewhere.
I'm even writing normal letters with $\LaTeX$...

i created a header, it takes like fifteen secs to create a header then

You should try the new Office 2007.  Granted, Excel 2007 is horrid in terms of usability, they've added equation support to Word.

Do you think it would be worthwhile for me to get the LaTeX typesetting stuff for my computer and use that as well?


geometry
analytic geometry
geometry proposed

Given a traingle $ ABC$ with area $ 1$. Let $ L,\ M,\ N$ be the points on the sides $ BC,\ CA,\ AB$ respectively and let $ P,\ Q,\ R$ be the points on the sides $ LM,\ MN,\ NL$ respectively. Find the maximum value of the following expression.

\[ \sqrt [3]{\triangle{CRL}} \plus{} \sqrt [3]{\triangle{APM}} \plus{} \sqrt [3]{\triangle{BQN}} \plus{} \sqrt [3]{\triangle{BLP}} \plus{} \sqrt [3]{\triangle{CMQ}} \plus{} \sqrt [3]{\triangle{ANR}}
\]

Note that $ \triangle{XYZ}$ denotes the area of the triangle $ XYZ$.

I have get pretty expressions for the six areas with areal coordinates, but I wonder the role of the cubic roots in this problem.
Some particular cases of the triangles LMN, PQR have appear in the old literature, but I have not found nothing so general like this.

[hide]Notice $ \Delta CRL=\frac{LC}{BC}\Delta CRB=\frac{RL}{NL}\frac{LC}{BC}\Delta CNB=\frac{NB}{AB}\frac{RL}{NL}\frac{LC}{BC}$ since $ \Delta ABC=1$.  We similarly find for the other triangles

$ \Delta CMQ=\frac{NA}{AB}\frac{QM}{MN}\frac{MC}{AC}$

$ \Delta APM=\frac{LC}{BC}\frac{PM}{LM}\frac{AM}{AC}$

$ \Delta ANR=\frac{LB}{BC}\frac{NR}{NL}\frac{AN}{AB}$

$ \Delta BLP=\frac{MC}{AC}\frac{PL}{ML}\frac{BL}{BC}$

$ \Delta BQN=\frac{AM}{AC}\frac{NQ}{MN}\frac{NB}{AB}$

Notice by AM-GM that

$ \sqrt[3]{\Delta CRL}=\sqrt[3]{\frac{NB}{AB}\frac{RL}{NL}\frac{LC}{BC}}\le\frac{\frac{NB}{AB}+\frac{RL}{NL}+\frac{LC}{BC}}{3}$

We find similar maximums for the other triangles.  Add them up and we have

$ \sqrt[3]{\Delta CRL}+\sqrt[3]{\Delta CMQ}+\sqrt[3]{\Delta APM}+\sqrt[3]{\Delta ANR}+\sqrt[3]{\Delta BLP}+\sqrt[3]{\Delta BQN}\le \frac{1}{3}(\frac{NB}{AB}+\frac{RL}{NL}+\frac{LC}{BC}+\frac{NA}{AB}+\frac{QM}{MN}+\frac{MC}{AC}+\frac{LC}{BC}+\frac{PM}{LM}+\frac{AM}{AC}+\frac{LB}{BC}+\frac{NR}{NL}+\frac{AN}{AB}+\frac{MC}{AC}+\frac{PL}{ML}+\frac{BL}{BC}+\frac{AM}{AC}+\frac{NQ}{MN}+\frac{NB}{AB})$

A little simplifying gives that the right hand side is simply $ 3$.  So the maximum value is $ 3$ which is attained when $ MNL$ is the midpoint triangle of $ ABC$ and $ PQR$ is the midpoint triangle of $ MNL$.
[/hide]

Very nice solution, cosinator. Congratulations.

The expressions of the areas of six triangles using areal coordinates are the same as in the solution of cosinator, with a weird notation...that of cosinator is much better.

[quote="cosinator"]Notice $ \Delta CRL = \frac {LC}{BC}\Delta CRB = \frac {RL}{NL}\frac {LC}{BC}\Delta CNB = \frac {NB}{AB}\frac {RL}{NL}\frac {LC}{BC}$ since $ \Delta ABC = 1$.  We similarly find for the other triangles

$ \Delta CMQ = \frac {NA}{AB}\frac {QM}{MN}\frac {MC}{AC}$

$ \Delta APM = \frac {LC}{BC}\frac {PM}{LM}\frac {AM}{AC}$

$ \Delta ANR = \frac {LB}{BC}\frac {NR}{NL}\frac {AN}{AB}$

$ \Delta BLP = \frac {MC}{AC}\frac {PL}{ML}\frac {BL}{BC}$

$ \Delta BQN = \frac {AM}{AC}\frac {NQ}{MN}\frac {NB}{AB}$

Notice by AM-GM that

$ \sqrt [3]{\Delta CRL} = \sqrt [3]{\frac {NB}{AB}\frac {RL}{NL}\frac {LC}{BC}}\le\frac {\frac {NB}{AB} + \frac {RL}{NL} + \frac {LC}{BC}}{3}$

We find similar maximums for the other triangles.  Add them up and we have

$ \sqrt [3]{\Delta CRL} + \sqrt [3]{\Delta CMQ} + \sqrt [3]{\Delta APM} + \sqrt [3]{\Delta ANR} + \sqrt [3]{\Delta BLP} + \sqrt [3]{\Delta BQN}\le \frac {1}{3}(\frac {NB}{AB} + \frac {RL}{NL} + \frac {LC}{BC} + \frac {NA}{AB} + \frac {QM}{MN} + \frac {MC}{AC} + \frac {LC}{BC} + \frac {PM}{LM} + \frac {AM}{AC} + \frac {LB}{BC} + \frac {NR}{NL} + \frac {AN}{AB} + \frac {MC}{AC} + \frac {PL}{ML} + \frac {BL}{BC} + \frac {AM}{AC} + \frac {NQ}{MN} + \frac {NB}{AB})$

A little simplifying gives that the right hand side is simply $ 3$.  So the maximum value is $ 3$ which is attained when $ MNL$ is the midpoint triangle of $ ABC$ and $ PQR$ is the midpoint triangle of $ MNL$.
[/quote]

Splendid! I have nothing to add more than this.

kunny


number theory unsolved
number theory
109

Prove that if $ a^2\plus{}ac\minus{}c^2\equal{}b^2\plus{}bd\minus{}d^2$, for the integers $ a>b>c>d>0$, then $ ab\plus{}cd$ is a composite number.


trigonometry
limit
real analysis
real analysis unsolved

Let $ (a_n)_{n\geq 0}$ and $ (b_n)_{n\geq 0}$ two sequences such that :

$ a_0>0$, $ a_{n\plus{}1}\equal{}a_n\cdot e^{\minus{}a_n}$, $ \forall n\geq 0$

$ b_0 \in (0,1)$, $ b_{n\plus{}1}\equal{}b_n \cdot \cos \sqrt{b_n}$, $ \forall n\geq 0$.

Compute $ \lim\limits_{n\to\infty}\frac{a_n}{b_n}$.

Note that both $ a_n$ and $ b_n$ approaches to zero. Then by Stolz's theorem,

\begin{eqnarray*}
\lim_{n\to\infty} \frac{a_n}{b_n}
& = & \lim_{n\to\infty} \frac{1 / b_n}{1 / a_n}\\
& = & \lim_{n\to\infty} \frac{(1/b_{n+1}) - (1/b_n)}{(1/a_{n+1}) - (1/a_n)}\\
& = & \lim_{n\to\infty} \frac{\frac{\sec(\sqrt{b_n}) - 1}{b_n}}{\frac{e^{a
_n} - 1}{a_n}}\\
& = & \frac{1}{2}\\
\end{eqnarray*}

[quote="sos440"]\begin{eqnarray*} & = & \lim_{n\to\infty} \frac {\frac {\sec(\sqrt {b_n}) - 1}{b_n}}{\frac {e^{a _n} - 1}{a_n}} \\
& = & \frac {1}{2} \\
\end{eqnarray*}[/quote]How do you conclude that limit is $ \frac {1}{2}$?

Denominator clearly converges to $ 1$. Also, note that $ (\sec \sqrt{x})' \equal{} \frac{\sin\sqrt{x}}{2\sqrt{x}} \sec^2 \sqrt{x}$, which converges to $ \frac{1}{2}$ as $ x \to 0^\plus{}$. Thus the limit is $ \frac{1}{2}$.

[quote="JoeBlow"]
How do you conclude that limit is $ \frac {1}{2}$?[/quote]

At the numerator we have to compute $ \lim\limits_{n\to\infty} \frac {1 \minus{} \cos \sqrt {b_n}}{b_n}$.But using the relation $ 1 \minus{} \cos \alpha \equal{} 2\sin ^2 \frac {\alpha}{2}$ and the well-known $ \lim\limits_{x\to 0} \frac {\sin x}{x} \equal{} 1$ we reach the same result.


The sum of the last four digits of Marian's phone number is 30. How many such four-digit sequences are there?

An alternative way to solve this is using... generating functions!

In general, if we are looking for the solutions to any $ a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} \cdots \plus{} a_n \equal{} k$ we can look for the coefficient of $ x^k$ in the product of the generating functions of $ a_1, a_2, a_3, a_4, \cdots, a_n$.

We are looking for $ a \plus{} b \plus{} c \plus{} d \equal{} 30$. Each of a, b, c, d can be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Thus their generating function is $ (x^0 \plus{} x^1 \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} x^5 \plus{} x^6 \plus{} x^7 \plus{} x^8 \plus{} x^9)$.

So the problem is equivalent to finding the coefficient of $ x^{30}$ in $ (1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} x^5 \plus{} x^6 \plus{} x^7 \plus{} x^8 \plus{} x^9)^4$. This is equal to $ 84$.

[quote="Kouichi Nakagawa"]{9, 9, 9, 3} : 4
{9, 9, 8, 4} : 12
{9, 9, 7, 5} : 12
{9, 9, 6, 6} : 6
{9, 8, 8, 5} : 12
{9, 8, 7, 6} : 24
{9, 7, 7, 7} : 4
{8, 8, 8, 6} : 4
{8, 8, 7, 7} : 6

$ 4 \plus{} 12 \plus{} 12 \plus{} 6 \plus{} 12 \plus{} 24 \plus{} 4 \plus{} 4 \plus{} 6 \equal{} \boxed{84}$[/quote]

This is what I did too (though I missed the ones that start with 8 on my first attempt).  But can anyone comment on the intuition behind why this works out to $ \dbinom{9}{3}$?

What's a 'distribution'? (the topic that this question is in) :?:

[quote=math154]No guys, let's be tricky.

$ a \plus{} b \plus{} c \plus{} d \equal{} 30\implies(9 \minus{} a) \plus{} (9 \minus{} b) \plus{} (9 \minus{} c) \plus{} (9 \minus{} d) \equal{} 6$

$ \implies\binom{6 \plus{} 4 \minus{} 1}{4 \minus{} 1} \equal{} \binom93 \equal{} \boxed{84}$.[/quote]

Nice trick!

Commenting so I can remember to jot this tricky thinking down later. :P

[quote=math154]No guys, let's be tricky.

$ a \plus{} b \plus{} c \plus{} d \equal{} 30\implies(9 \minus{} a) \plus{} (9 \minus{} b) \plus{} (9 \minus{} c) \plus{} (9 \minus{} d) \equal{} 6$

$ \implies\binom{6 \plus{} 4 \minus{} 1}{4 \minus{} 1} \equal{} \binom93 \equal{} \boxed{84}$.[/quote]

why does this work?

I'm not sure either.
Does it have anything to do with the thing about $n$ candies and $k$ kids (ways to distribute) is $\binom{n + k - 1}{k-1}$??

[quote=Samchooo][quote=mathmaster010][quote=Samchooo]The solutions are from 2009. I doubt the people that posted the solutions are going to see your post.[/quote]

it does not matter[/quote]

actually, it does because there isn't a point of posting[quote=ojaswupadhyay]Please hide your solutions.[/quote]

anymore.[/quote]

okay

Other people can explain the solution, the original poster of the solution isn't the only one who understands it


Let $X(t)$ be a solution of the equation $X'=f(X)$,where $f: R\to R$.Show that $X$ is monotonuous.


function
vector
algebra
functional equation
algebra proposed

A calculating ruler is a ruler for doing algebric calculations. This ruler has three arms, two of them are sationary and one can move freely right and left. Each of arms is gradient. Gradation of each arm depends on the algebric operation ruler does. For eaxample the ruler below is designed for multiplying two numbers. Gradations are logarithmic.
[img]http://aycu05.webshots.com/image/5604/2000468517162383885_rs.jpg[/img]
For working with ruler, (e.g for calculating $x.y$) we must move the middle arm that the arrow at the beginning of its gradation locate above the $x$ in the lower arm. We find $y$ in the middle arm, and we will read the number on the upper arm. The number written on the ruler is the answer.
1) Design a ruler for calculating $x^{y}$. Grade first arm ($x$) and ($y$) from 1 to 10.
2) Find all rulers that do the multiplication in the interval $[1,10]$.
3) Prove that there is not a ruler for calculating $x^{2}+xy+y^{2}$, that its first and second arm are grade from 0 to 10.

If we want to calculate the value of the monotonically increasing with respect to each variable function $t(x,y)$ when $x,y\in [a,b]$ with such a rule we need to find three monotonic increasing funtions $f,g,h$, so that $f(x)+g(y)=h(t(x,y))$, where the value each one returns is the distance in the corresponding sub-rule between the origin and the tick signalling the If we want to calculate the value of the monotonically increasing with respect to each variable function $t(x,y)$ when $x,y\in [a,b]$ with such a rule we need to find three monotonic increasing funtions $f,g,h$, so that $f(x)+g(y)=h(t(x,y))$, where the value each one returns is the distance in the corresponding sub-rule between the origin and the tick signalling the number we introduced as argument. Without loss of generality we can suppose $f(a)=g(a)=h(t(a,a))=0$.
1) We can take  $f(x)=h(c)=ln(lnx)$, $g(y)=lny$, so that $f(x)+g(y)=ln(ln(x))+lny=ln(yln(x))=ln(ln(x^{y}))=h(x^{y})$. I don't know if there is something more to be said.
2) We have $f(x)+g(y)=h(xy)$, $x,y\in [1,10]$. By the assumption made at the beginnign we can let $f(1)=g(1)=h(1)=0$. This leads by alternatively letting $x=1$ and $y=1$, to  $f=g=h$,  and hence Cauchy's functional equation $f(x)+f(y)=f(xy)$ which in this case, being $f(x)$ monotonic, has only solution $f(x)=c (lnx)$ for arbitrary constant $c$; here $c>0$. [ To see this result is just necessary to make the change $g(x)=f(e^{x})$ in orther to get the first Cauchy f.e.: $g(x)+g(y)=g(x+y)$ ].
3) By alternatively letting $x=0$, $y=0$ we get that $f(x)=g(x)=h(x^{2})$, so we should have $h(x^{2})+h(y^{2})=h(x^{2}+xy+y^{2})$.
On one hand making the change $x=y=\sqrt{z}$, leads to $2h(z)=h(3z)$. On the other $x=\frac{y}{\sqrt{3}}=\sqrt z$ leads to $h(z)+h(3z)=h((4+\sqrt{3})z) \Rightarrow 3h(z)=h((4+\sqrt{3})z)$. Calling $q(x)=ln (h(e^{x}))$ (also monotonically increasing), and letting $S$ the set of vectors such that $(z,w)\in S\leftrightarrow q(z)=w$ we get from the previous two equations: $(v\in S \leftrightarrow v+a\in S) \wedge (v\in S \leftrightarrow v+b\in S)$ where $a=(ln3,ln2)\;\;b=(ln(4+\sqrt{3}),ln3)$. But $a$ and $b$ are linearly indipendent, so $\forall N\in \mathbb{R}^{+},\;\exists (a,b) \in (S\cap [0,ln 10]\times \mathbb{R})/b>N$ which is clearly impossible. (I know this last part should be written more in detail; sorry.)

Being so long, there must be many mistakes.


invariant
combinatorics proposed
combinatorics

let $\alpha\geq 0$ be a given real. let $0\leq p_i\leq 1$, $i=1,...,2n$ (actually, we could replace $2n$ by an odd number, but that makes the problem statement a bit more complicated).
prove that there exists $0\leq x\leq 1$ s.t.
\[\sum_{i=1}^{2n} \frac 1{|x-p_i|^{\alpha}}\leq 2(4n)^{\alpha}\sum_{i=1}^{n} \frac 1{(2i-1)^{\alpha}}\].

the original problem treated the case of $\alpha=1$ and gave a slightly worse bound.

nobody wants to solve it(or is it just too hard?)? you have to try it since the solution i found (after maybe 2 days) is just too beautiful.

Actually I don't know if anyone made the effort to read the formula :). Could you please post the original IMO training problem. I think it's reasonable to give others the chance to start with the beginning :). Thanks.

Edit: Thanks :)

Okay, the original problem is IMO training 2005, pre-Rostock homework set (I) problem 6.2:

[i]Given n real numbers $p_1$, $p_2$, ..., $p_n$ such that $0\leq p_i\leq 1$ for all i = 1, 2, ..., n.
Prove the existence of a real number x with $0\leq x\leq 1$ satisfying

$\sum_{i=1}^n\frac{1}{\left|x-p_i\right|}\leq 8\cdot n\cdot\left(1+\frac13+\frac15+...+\frac{1}{2n-1}\right)$.[/i]

Actually, this is a weaker version of the particular case of Peter's problem above for $\alpha=1$.

By the way, I am thinking about whether it would make sense to post all the homework problems so far (18 problems) on ML. Actually, many of them are boring or trivial (this one, of course, not).

  darij

actually, the bound of the original problem for $n$ numbers is the same as my bound for $2n$ numbers(i don't know if that's just accidentally), therefore the original problem isn't really my problem for $\alpha=1$.

ok, seems like no one is interested in that problem, therefore i'll post the solution.

assume all points are distinct - the rest is done via taking the limit (or appropriately changing some slight details of the proof).

place a balloon at each of the points $p_i$. then blow up all balloons at the same speed. it's allowed for the balloons to leave the interval $[0; 1]$ and they may change their midpoints - only at the beginning they're centered at the points $p_i$. if two balloons touch each other, the point where they touch is invariant but they are still getting bigger, if three balloons are in a row, the first touching the second touching the third, the midpoint of the second is invariant and the two outer ones are pushed away by the second one which gets larger and so on - so all in all they are acting the way one conjectures them to do.

now, after some time, all points of the interval $[0; 1]$ are covered. in that moment, stop to blow the balloons. let $x$ be (one of the) last point(s) which are uncovered up to the moment everything out of the interval $[0; 1]$ is covered. we have two observations:

1) if $r$ is the radius of all the balloons at the end, then the midpoints of all the neighboring balloons have distance $2r$. furthermore, since $[0; 1]$ is covered, $r\geq \frac 1{4n}$. $x$ is some point where the balloons touch. all in all, something like that:
0----x---1
OOOOOOO

2) if we replace the points $p_i$ by the midpoints of the balloons, the sum of the distances (to the corresponding powers) gets at most bigger - one easily sees that by looking at what happens if we do this instantly(use e.g. karamata).

therefore we only have to evaluate the sum we get in the end, but that's pretty easy. we may assume that $r=\frac 1{4n}$ since for larger $r$ the sum gets smaller and we may assume that $x$ is exactly in the middle of all the balloons. this gives the desired bound.

btw, this shows that there is one $x$ for all $\alpha$.

[quote="Pascual2005"]didnt get it[/quote]
Me too.Can you,peter,write it in a more slightly way

This problem and the "baloon idea" remembers me about a nice problem Lhvietbao told me: If $A_1,A_2,\ldots,A_n$ are points in the plane then the set of points $P$ that satisfy $|PA_1PA_2 \ldots PA_n|\leq 1$ can be covered by a set of disks with sum of radii at most $C$, where $C$ is some constant (independent on $n$). I was told $C=e$ is good, but I myself did manage a bigger bound. Also I think the assertion holds for any dimension. You can try this one, Peter.

I have an idea similar to Peter's, but worse for this problem.

For any point $A_i, 1\leq i \leq 2n$ consider segment $S_i$ with length $\frac 1{4n}$ and midpoint $A_i$. Associate with $S_i$ the point $A_i$. then, perform the following step while possible:

(*)If there is some collection of segments $S_1,S_2,\ldots S_k$ with total length $\frac k{4n}$(so $k$ associated points) and they can be covered by a single segment with legth $\frac k{4n}$, then replace them by this new segment $S$, and associate all the ponits previously associated to $S_1,S_2,\ldots, S_k$ to $S$.

It's easy to see that each segment will have length $\frac{k}{4n}$ for some $k$ and will have exactly $k$ associated points. Clearly each segment contains all the points associated to it and and each $A_i$ is associated to exactly one segment (it may belong to more segments, but is associated just to one).

Now for each segment $S$ consider the "fat" segment $S'$ sharing the midpoint with $S$ but twice bigger. The total length of all fat segments is $1$, so some point $X$ is not covered by them.

Now let's estimate $\prod XA_i$. Firstly, let's see how many $A_i$ can be at distance at most $r$ from $X$. Suppose $S_1,S_2,\ldots,S_k$ with lengths $\frac{r_1}{4n}, \frac{r_2}{4n}, \ldots \frac{r_k}{4n}$. As the fat segment $S_i'$ doesn't contain $X$, it means that $X$ is at distance at least $\frac{r_k}{8n}$ from the segment $S_i$. Therefore $\frac{r_k}{8n}\leq r$ and so the segment centered at $x$ and with length $6r$ contains $S_i$. Since we cannot perform (*) for $S_1,S_2,\ldots,S_k$ we deduce that $\frac{r_1}{4n}+\ldots+\frac{r_k}{4n}<6r$ so $r_1+r_2+\ldots+r_n<24rn$ so there are less than $24rn$ points at distance at most $r$ from $x$.

Therefore, at most one point can have distance no more than $\frac 1{24n}$ from $X$, at most two $\frac{2}{24n}$ and so on. Hence the sum seeked in problem is at most $\sum_{i=1}^n (\frac{24n}{i})^{\alpha}=(24n)^{\alpha}\sum_{i=1}^n \frac 1{i^\alpha}$. We can actually replace $24$ by $\frac{2}{(\sqrt 2-1)^2}=11.65$(and this is the best constant for my method), but this is still way too big compared to Peter's bound(however better than the original one).

I guess I understood Peter's idea (very nice!), but I doubt I can explain it better than him.

Can you generalize the problem for larger dimensions?


$X$ consists of many pairwise disjoint Lebesgue-measurable sets. Every set in $X$ has a positive measure.

Show that $X$ is countable.

Two things: when you say "$X$ is countable", you must mean that there are countably many sets in the union that makes up $X$, not that the whole set is countable. Second: we must assume that the context you implied by "Lebesgue measurable" was  in $\mathbb{R}$ (or $\mathbb{R}^n$, or at the very least some $\sigma$-finite measure space.)

Let's say we're working on $\mathbb{R}$ and $X=\cup X_{\alpha}.$  We start by assuming that each $X_{\alpha}$ is contained in $[-n,n]$. (We make that assumption by intersecting each set with that interval, and discarding those whose intersections have measure zero.) Now, how many of the $X_{\alpha}$ have measure that exceeds $\frac1k$? Since their disjoint union must fit into a set of finite measure (the interval $[-n,n]$), only finitely many. Taking the union over $k$ leaves us with countably many sets. Now remove the restriction to the interval $[-n,n]$ by taking the union over $n$. That leaves us with a countable union of countable collections of sets, hence a countable collection.


Given a set of 1999 closed unit intervals on the real line, S is the set of points in exactly an odd number of such intervals.

Show $|S| \geq 1$, where we understand $|S|$ to be the length of the finite union of all it's disjoint intervals.

I solved it, its good Q

Please note that this is problem from the May version of Olymon, which is not due until June 15:
[url=http://www.math.toronto.edu/~barbeau/olymon57.pdf]www.math.toronto.edu/~barbeau/olymon57.pdf[/url]

It seems like May Olymon is too easy this month; in one day I already solved 6/7


[quote="xzlbq"][size=200]Let a,b,c be real positive numbers  and a+b+c=1, Prove  that

$ 5 \geq \frac {a \plus{} 1}{b \plus{} 2} \plus{} \frac {b \plus{} 2}{c \plus{} 3} \plus{} \frac {c \plus{} 3}{a \plus{} 1} \geq \frac {19}{6}$

BQ[/size][/quote]

try  $ a \equal{} \frac {x}{x \plus{} y \plus{} z},b \equal{} \frac {y}{x \plus{} y \plus{} z},c \equal{} \frac {z}{x \plus{} y \plus{} z}$

$ x,y,z > 0$

$ \frac {a \plus{} 1}{b \plus{} 2} \plus{} \frac {b \plus{} 2}{c \plus{} 3} \plus{} \frac {c \plus{} 3}{a \plus{} 1} \geq \frac {19}{6}$

$ \Longleftrightarrow169yzx \plus{} 88z^3 \plus{} 63y^3 \plus{} 38zx^2 \plus{} 120z^2x \plus{} 244yz^2 \plus{} 6x^2y \plus{} 75y^2x \plus{} 213y^2z\geq 0$

$ 5 \geq \frac {a \plus{} 1}{b \plus{} 2} \plus{} \frac {b \plus{} 2}{c \plus{} 3} \plus{} \frac {c \plus{} 3}{a \plus{} 1}$

$ \Longleftrightarrow  91yzx\plus{}22x^3\plus{}6y^3\plus{}56zx^2\plus{}35z^2x\plus{}7yz^2\plus{}65x^2y\plus{}48y^2x\plus{}14y^2z\geq 0$

[size=200]Let $ a,b,c,k_1,k_2,k_3$be real positive numbers  and $ a\plus{}b\plus{}c\equal{}1$, research

$ f(k_1,k_2,k_3) \geq \frac {a \plus{} 1}{b \plus{} 2} \plus{} \frac {b \plus{} 2}{c \plus{} 3} \plus{} \frac {c \plus{} 3}{a \plus{} 1} \geq g(k_1,k_2,k_3)$

 [/size]
 BQ

[size=200]if $ a\plus{}b\plus{}c\equal{}1$,then:
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \geq  \frac{1\minus{}a}{1\minus{}b}\plus{}\frac{1\minus{}b}{1\minus{}c}\plus{}\frac{1\minus{}c}{1\minus{}a}$[/size]
BQ

Isn't this spamming?
Isn't he/she should get banned or something his/her history is full of spamming.

[quote="xzlbq"][size=200]Let a,b,c be real positive numbers  and a+b+c=1, Prove  that

$ 5 \geq \frac {a \plus{} 1}{b \plus{} 2} \plus{} \frac {b \plus{} 2}{c \plus{} 3} \plus{} \frac {c \plus{} 3}{a \plus{} 1} \geq \frac {19}{6}$

BQ[/size][/quote]due to the conditions, we have $ c \plus{} 3\ge b \plus{} 2\ge a \plus{} 1\ge1$. So the minimum is attained for $ c \equal{} 0$ and the maximum for $ a \equal{} 0$. For both cases, the problem is then reduced to a 2 variable problem. The same argument again shows: minimum for $ (1,0,0)$ and the maxmum for $ (0,0,1)$.

EDIT: just saw
[quote="xzlbq"][quote="hedeng123"][quote="xzlbq"][size=200]Let $ a,b,c,k_1,k_2,k_3$be real positive numbers  and $ a \plus{} b \plus{} c \equal{} 1$, research

$ f(k_1,k_2,k_3) \geq \frac {a \plus{} 1}{b \plus{} 2} \plus{} \frac {b \plus{} 2}{c \plus{} 3} \plus{} \frac {c \plus{} 3}{a \plus{} 1} \geq g(k_1,k_2,k_3)$

 [/size]
 BQ[/quote]

what ??[/quote]

[size=200]Let $ a,b,c,k_1,k_2,k_3$be real positive numbers  and $ a \plus{} b \plus{} c \equal{} 1$, research

$ f(k_1,k_2,k_3) \geq \frac {a \plus{} k_1}{b \plus{} k_2} \plus{} \frac {b \plus{} k_2}{c \plus{} k_3} \plus{} \frac {c \plus{} k_3}{a \plus{} k_1} \geq g(k_1,k_2,k_3)$

 [/size]
 BQ[/quote]

I suppose you mean: give the minimum and the maximum of $ \frac {a \plus{} k_1}{b \plus{} k_2} \plus{} \frac {b \plus{} k_2}{c \plus{} k_3} \plus{} \frac {c \plus{} k_3}{a \plus{} k_1}$ in terms of $ k_1,k_2,k_3$. Why not say it like that :?:

[size=200]Let$ x_1,x_2,x_3,x_4$ be real positive numbers  and$ x_1\plus{}x_2\plus{}x_3\plus{}x_4\equal{}1$, Prove  that

$ \frac {x_1}{x_2} \plus{} \frac {x_2}{x_3} \plus{} \frac {x_3}{x_4} \plus{} \frac {x_4}{x_1} \geq \frac {1 \minus{} x_1}{1 \minus{} x_2} \plus{} \frac {1 \minus{} x_2}{1 \minus{} x_3} \plus{} \frac {1 \minus{} x_3}{1 \minus{} x_4} \plus{} \frac {1 \minus{} x_4}{1 \minus{} x_1}$

BQ[/size]

[quote="xzlbq"][size=200]if $ a \plus{} b \plus{} c \equal{} 1$,then:
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq \frac {1 \minus{} a}{1 \minus{} b} \plus{} \frac {1 \minus{} b}{1 \minus{} c} \plus{} \frac {1 \minus{} c}{1 \minus{} a}$[/size]
BQ[/quote]

this was an old problem I remember...

[quote="xzlbq"][size=200]if $ a \plus{} b \plus{} c \equal{} 1$,then:
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq \frac {1 \minus{} a}{1 \minus{} b} \plus{} \frac {1 \minus{} b}{1 \minus{} c} \plus{} \frac {1 \minus{} c}{1 \minus{} a}$[/size]
BQ[/quote]

[quote="xzlbq"][size=200]Let$ x_1,x_2,x_3,x_4$ be real positive numbers  and$ x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \equal{} 1$, Prove  that

$ \frac {x_1}{x_2} \plus{} \frac {x_2}{x_3} \plus{} \frac {x_3}{x_4} \plus{} \frac {x_4}{x_1} \geq \frac {1 \minus{} x_1}{1 \minus{} x_2} \plus{} \frac {1 \minus{} x_2}{1 \minus{} x_3} \plus{} \frac {1 \minus{} x_3}{1 \minus{} x_4} \plus{} \frac {1 \minus{} x_4}{1 \minus{} x_1}$

BQ[/size][/quote]

Couldn't you start a new thread for these?  :mad:

[quote="xzlbq"][size=200]Let$ x_1,x_2,x_3,x_4$ be real positive numbers  and$ x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \equal{} 1$, Prove  that

$ \frac {x_1}{x_2} \plus{} \frac {x_2}{x_3} \plus{} \frac {x_3}{x_4} \plus{} \frac {x_4}{x_1} \geq \frac {1 \minus{} x_1}{1 \minus{} x_2} \plus{} \frac {1 \minus{} x_2}{1 \minus{} x_3} \plus{} \frac {1 \minus{} x_3}{1 \minus{} x_4} \plus{} \frac {1 \minus{} x_4}{1 \minus{} x_1}$

BQ[/size][/quote]

tsds 1 time. :lol:


function
algebra
domain
calculus
derivative
real analysis
real analysis unsolved

1) Give an example to show that a sequence of functions may be uniformly continuous, pointwise equicontinuous, but not uniformly equicontinuous, when their domain is noncompact?

2) Give an example of a sequence of smooth equicontinuous functions fn: [a,b] -> R (real line)  whose derivatives are unbounded.

3) If f: R->R is continuous and the sequence fn(x) = f(nx) is equicontinuous, what can be said about f???

Thanks in advance..

1) what is " pointwise equicontinuous "?
3) what we need to say about $f$? :)

The term "pointwise equicontinuous" is not very common. If I understand it right, it means that we have a family of functions defined on some domain $D$ such that for every $\varepsilon>0$ and every $x\in D$, there exists $\delta=\delta(\varepsilon,x)>0$ such that $|y-x|<\delta$ implies that $|f(y)-f(x)|<\varepsilon$ for all $f$ in our family.

As to the second question, "What can be said about $f$?" is just a politically correct form of "Find all such $f$!" :lol:

So there must be $f=const$ in 3) i think.
Suppose $x < y$ and $f(x)\neq f(y)$. Then $|f(x)-f(y)|=2\varepsilon >0$. Then $\exists \delta >0$ such that
$|x_1-x_2|<\delta \forall \; n \; \Rightarrow |f_n(x_1)-f_n(x_2)|<\varepsilon.$ Let $n>\frac{|x-y|}{\delta}$. Then $\left|\frac{x}{n} - \frac{y}{n} \right|< \delta$ and  we must have $|f_n(x/n)-f_n(y/n)|<\varepsilon$. But  \[ 2\varepsilon = |f(x)-f(y)| = |f_n(x/n) - f_n(y/n)| < \varepsilon \] Contradiction...
One more remark : we don't need condition that $f \in C({\mathbb R})$... :)


geometry
perimeter
calculus
integration
rotation
geometric transformation
reflection

is it just me or is this one not that hard?
[hide="I dont know why anyone would need a calculator"]
5-5-5, 6-5-4, 7-5-3, 6-6-3, 7-6-2, 7-7-1, 7-4-4

7 triangles
[/hide]

[quote="DanK"]Would you need to count scalene triangles twice?  Because I don't think these are the same:

[img]http://i9.photobucket.com/albums/a70/littlekasper/triangles.jpg[/img][/quote]

Those are congruent though.

[quote="boo"][quote="mathgeniuse^ln(x)"]In how many non-congruent triangles of perimeter 15 do all the sides have an integral length?

No Calculator[/quote]

[hide]5-5-5
5-6-4
4-7-4
5-7-3
3-6-6
6-2-7
1-7-7

9 ways.[/hide][/quote]

can you count?
that's 7

[quote="boo"][quote="mathgeniuse^ln(x)"]In how many non-congruent triangles of perimeter 15 do all the sides have an integral length?

No Calculator[/quote]

[hide]5-5-5
5-6-4
4-7-4
5-7-3
3-6-6
6-2-7
1-7-7

9 ways.[/hide][/quote]

I think you mean 7 right.

Eitherway, that is the correct answer we were looking for.

[quote="DanK"]Are they congruent?  You can never rotate them so the sides are arranged the same way.  Doesn't that constitute a different shape?  And since congruent figures have identical size and shape, that would make them not congruent?[/quote] You can reflect them though. As long as it doesn't require transformations.

As long as you rotate, reflect, or translate, that is okay.


geometry
AoPS Books
AMC 10
number theory
prime factorization
\/closed

what exactly does that mean...are you referring to strategies not included in those books but at the same level or a higher level like olympiad?

[quote="rrusczyk"][quote="PenguinIntegral"]What classes would cover material that is not covered in AoPS 1 or 2. Which ones contain strategies that are not contained in 1 or 2?[/quote]

The intro classes have considerable overlap with the AoPS books, but go into greater detail with an even higher focus on strategy.  The non-intro classes contain a considerable amount of material that is not in the AoPS books.[/quote]

So what would your reccomend for someone(me) who wants to get the most for their dollar? Are the intro courses worth it for someone with both Aops books(Just got the second one and the solutions for the first, still as brilliant as ever!)

[quote="rrusczyk"][quote="ffdbzathf"]you should try doing the "are you ready" pretests and "do you need this" post-tests to the classes to see where you stand.[/quote]

This is a good recommendation.

As for what best benefit for your dollar, it depends on what type of student you are.  If you are the type who gets a ton out of 1-1 interaction and being able to bring your own questions to class, the Independent Study is your best bet.  If you have mastered the Intro AoPS book, much of the Intro classes will be review for you (but certainly not all).

What areas are you weak/strong in?[/quote]
I don't know, really. A few days ago, I went through the first fifteen questions of an amc 10, and I learned alot from the solutions of the three I missed.

I  remember learning two things. The first was that  square has all the exponents of its prime factorization even(I promptly smacked myself on the head and exclaimed "duh!") and that recognizing factorizations is a very useful tool. I haven't found much use for the factorizations yet, but thinking of squares as prime factorizations led me to some unique solutions in the "Using the integers" sections of vol 1 of AoPS.
(This was also before I got the solution manual, yesterday)

I also spend generous amounts of time (>2 hours) thinking about hard problems. The technique used to solve it is usually simple when I look up the answer, and because I remeber some peticularitys of the problems, I sometimes use that method on other problems with the same structure.

I like doing problems more than I like reading about how to do problems. If a new technique is introduced in a solution, I feel like I know it better because I have worked with the problem a lot and noted some things about it. That makes things stick better than reading a chapter of AoPS for me. I usually forget a lot.

[quote="rrusczyk"]If you are trying to choose a subject class, I'd fall back on ff's suggestion - use the pre & post tests as a guide.

Your approach to learning (problems, problems, problems) is, of course, the most effective.  Reading only primes the mind for discovery - you have to get in and think about problems to really understand the tools.[/quote]
Thanks for the complements on my learning method ;)

I really have to talk with my parents on this one.

Hmm, I'm not sure I understand the merits on spending >2 hrs. on an AoPs/Contest problem. Usually I stop after 30-40 minutes if I'm not making progress. And when I look at the solution I usually find that I was thinking entirely in the wrong direction and probably wouldn't have thought of the solution anyway. This way I can gain some insight and move through problems moderately quickly, yet still only have to look up the solution to 1-2 of the earlier problems in each chapter of AoPs vol.2. But then again, I may be rushing for fear of the upcoming AIME...

Generally, I strongly agree with EFuzzy's approach to training - put a shot clock on yourself when working problems.  Make it longer for harder problems.  After that time, read the solution and learn from it.  Then do the problem again in a couple weeks.


Find all real functions $f: \mathbb R\to \mathbb R$ satisfying
\[f\left(x^{2}+y\right)=2f(x)+f(y^{2}),\]
for all $x$ and $y$ in the domain.

did u mean this $f(x^{2}+y)=2f(x)+f(y^{2})$ :huh:

sorry, is =
anyway, thank for try helping.


calculus
integration
function
logarithms
limit
calculus computations

A little help?

For what value of a does the integral

$\int^{\infty}_{1}(\frac{ax}{x^{2}+1}-\frac{1}{2x})dx$

converge?  Evaluate the corresponding integral(s).  (Give the answer as a function of a.)

Thanks.

The integrand is $\frac{(2a-1)x^{2}-1}{2x(x^{2}+1)}$. If $2a-1\ne 0$, this can be compared (by Limit Comparison test) to $1/x$, hence the integral diverges. If $2a-1=0$, the integral is convergent by comparison to $1/x^{3}$.

The antiderivative is $\frac{a}{2}\log(x^{2}+1)-\frac{1}{2}\log x$. When $a=1/2$, this simplifies to $\frac14\log\frac{x^{2}+1}{x^{2}}$, which $\to 0$ as $x\to \infty$. The integral is $-\frac14\log 2$.

[quote]Give the answer as a function of a. [/quote]
 :what?:

[quote="datawerd"]I don't think we've learned those techniques in my class yet.
This is my answer so far, perhaps someone could check it for me.  And tell me if I can even do it this way.

$\int^{\infty}_{1}(\frac{ax}{x^{2}+1}-\frac{1}{2x})dx=[\frac{a}{2}\ln{x^{2}+1}-\ln{2x}]^{\infty}_{1}\\\\ =\lim_{b\rightarrow\infty}[(\frac{a}{2}\ln{b^{2}+1}-\ln{2b})-(\frac{a}{2}\ln{1^{2}+1}-\ln{2(1)})]\\\\ =\lim_{b\rightarrow\infty}[(\frac{a}{2}\ln{b^{2}+1}-\ln{2b})-(\frac{a-2}{2}\ln{2})]$

So after I found this, I evaluated the function for several values of a, and found that when $1<a<1$, the integral diverges.[/quote]
It's wrong. You dont right determin integral. there can't be
$\int^{\infty}_{1}(\frac{ax}{x^{2}+1}-\frac{1}{2x})dx=[\frac{a}{2}\ln{x^{2}+\huge{1}}-\ln{2x}]^{\infty}_{1}\\\\$$

[quote="datawerd"]$\int^{\infty}_{1}(\frac{ax}{x^{2}+1}-\frac{1}{2x})dx=\lim_{b\rightarrow\infty}[\frac{a}{2}\ln({x^{2}+1})-\ln({2x})]^{b}_{1}\\\\ =\lim_{b\rightarrow\infty}[\frac{a}{2}\ln{(b^{2}+1)}-\ln{(2b)}-\frac{a}{2}\ln{2}+\ln{2}]$

Is this right?  How do I show what values of $a$ makes the integral divergent.[/quote]
Ok: it's not right $\int\frac{1}{2x}dx=ln(2x)$
you must write this: $\int\frac{1}{2x}dx$=$\frac{1}{2}\int\frac{1}{x}dx$=$\frac{lnx}{2}$


Show that if a,b,c are positive, then
$ \frac {(a \plus{} b)^3}{3a^2 \plus{} 3ab \plus{} b^2} \plus{} \frac {(b \plus{} c)^3}{3b^2 \plus{} 3bc \plus{} c^2} \plus{} \frac {(c \plus{} a)^3}{3c^2 \plus{} 3ca \plus{} a^2} \ge \frac {8}{7} (a \plus{} b \plus{} c)$

follows directly from:
$ \frac {a^3}{3a^2 \plus{} 3ab \plus{} b^2}\geq \frac {12a \minus{} 5b}{49}$

I don't think that's right.
We have $ 3a^2b\plus{}3ab^2 \le 3(a^3\plus{}b^3)$
And if you add what you said cyclicly, you get exactly the right side.
But $ 3a^2b\plus{}3ab^2 \le 3(a^3\plus{}b^3)$, not $ \ge$.

I don't understand your comment.
If you sum cyclically the above inequality you get $ \sum_{cyc}\frac{a^3}{3a^2\plus{}3ab\plus{}b^2}\geq \frac17 (a\plus{}b\plus{}c)$
if we add $ a\plus{}b\plus{}c$ to both sides and then regroup $ \frac{a^3}{3a^2\plus{}3ab\plus{}b^2}\plus{}b\equal{}\frac{(a\plus{}b)^3}{3a^2\plus{}3ab\plus{}b^2}$ we get your inequality.


Given is a decreasing sequence $ x_{n}$ such that for each natural number $ n$ the following inequality holds:

$ x_{1}\plus{}2x_{4}\plus{}...\plus{}nx_{n^{2}}\leq\ 1$

Prove the inequality:

$ x_{1}\plus{}x_{2}\plus{}...\plus{}x_{n}\leq\ 3$.

We can see that $ x_{1}\leq1$ and $ x_{4}\leq0$. Since $ x_{n}$ is decreasing we have $ x_{n}\leq0$ for $ n\geq4$ and $ x_{2}\leq1$ and $ x_{3}\leq1$.
From all that we have $ x_{1}\plus{}x_{2}\plus{}...\plus{}x_{n}\leq3$.

Sorry but I don't understand how do you conclude that $ x_{4}\leq 0$ :?:

Oh, I'm wrong, I made a terrible mistake. It doesn't have to be $ x_{4}\leq0$.

If $ x_{k}<0$ for $ k>n$ we can change $ x_{i}\equal{}0$ for $ i>n$.
Let $ k\equal{}[\sqrt n]$, then obviosly $ x_{1}\plus{}x_{2}\plus{}...\plus{}x_{n}\le\sum_{i\equal{}1}^{k}(2i\plus{}1)x_{i^{2}}\le 3sum_{i\equal{}1}^{k}ix_{i^{2}}\le 3.$


[u]Private message[/u] me your answers to see how well you did! [b][color=red](Please [u]don't[/u] post your answers here.)[/color][/b]

1. What is the maximum possible time allowance per question in a FTW game? (1 point)

2. What is the maximum number of players in a normal game? (1 points)

3. At what ratings do you receive the second and fourth rating badges, respectively? (4 points)

4. An FTW normal game is Ranking-scored, has 10 questions, with 35 seconds per question. Find the difference between the maximum and minimum amount of time from when the "Start game" button is clicked to when the results screen is shown, assuming no lags or glitches. Show all your work. (12 points)

5. Without glitches, how many seconds after time is up can you choose to "Give up"? Round your answer to the nearest tenth. Note that the answer may be zero or negative. (3 points)

6. What is the most common answer to an FTW question? (2 points)

7. When are you allowed to reveal someone else's real name and not have it be considered a ToS infringement? (2 points)

8. Let $ \frac {x}{y}$ be the ratio in lowest integer terms of the length of a LanguageFilter ban to the length of a FloodingFilter ban. Find $ x^3 \minus{} y^3$. (3 points)

9. What causes a Guest to have a five-digit or four-digit number? (3 points)

10. How can a Guest play more games than the two that are allowed per day? (2 points)

11. Who was the first moderator (not administrator) of FTW chat? (2 points)

12. The 10-day period from March 21 to March 30 saw the joining to AoPS of four users that would eventually prove to be very influential to FTW, two of which joined on the same day. Name these four users. (8 points)

13. The FTW Leaderboard's clock is set in what time zone? (2 points)

14. In the year 2008, one user won over twice as many normal games and the user who placed second in normal games won. Name these two respective users (6 points), and for the nerdy bonus points (6 bonus points), how many games did each of them win?

15. Name the first five users to pass 2000 rating, in order (Hint: One of the five is banned.). (10 points)

[Freebee, hopefully]. Who programmed FTW? (5 points)

Describe the following FTW glitches. (3 points each)
16. 8x
17. Flashies
18. Second Lobby
19. Trapped Game
20. Fall 2009 Rating Save Glitch

Who do the following nicknames refer to? Provide the full formal username. (2 points each)
21. Lonny
22. Eli
23. Levlev
24. Ding
25. Mabbitt
26. Ylxy
27. Rizzy

Attribute these FTW chat quotes. (1 point each)
28. "I never said you never said [that] I never said I did."
29. "Not really, it went +2 +2 +2 -12 +2 +3."
30. "Sicily is a god dang multi like me?"
31. "Special ways are izzy...i mean easy."
32. "I'm going to have fun in the shower with my friends."

In about a week, I will post a scoreboard up with the best scores! Yes, partial credit is given for some questions.
[color=red]
A few corrections (you can resubmit for questions with problems):

#12: in 2008
#20: the [i]second[/i] rating glitch[/color]

Again, just in case:
[u]Private message[/u] me your answers to see how well you did! [b][color=red](Please [u]don't[/u] post your answers here.)[/color][/b]

[quote="phirekalk"]
15. Name the first five users to pass 2000 rating, in order (Hint: One of the five is banned.). (10 points) [/quote]actually, there are 2 banned.

I can't believe that there was no multis section in this, considering how 2/3 of these posts are from multis. And even more unbelievable is that r15s11z55y89w21 was the one who posted this quiz, and still no multi part.

A multis section would be a leak of classified intelligence and a breach of ethics. And dragon96, 2/3 (now 3/4) of the posts have users with legal multis, but all have some type of multi.

It is the end of Day 1 of the presentation of this test, and six people have submitted!

[u]Leaderboard[/u]
1. dragon96 - 63
2. ilovepink - 37
3. rizikin - 36
4. k00lperson - 35
5(tie). agentcx - 34
5(tie). vahalla - 34

Yay, I got a 53! Also, for the first five people who reached 2000, I think there is only 1 multi, since it is only the first five people.

i'm amazed.
That
pinky beat me
only
by
2 points
heh

Final Scoreboard:

1. dragon96 - 63
2. xxrxxhxx - 53
3. ilovepink - 37
4. rizikin - 36
5. k00lperson - 35
6(tie). agentcx - 34
6(tie). vahalla - 34

Answers!

1. What is the maximum possible time allowance per question in a FTW game? (1 point)
[b]120 seconds[/b]

2. What is the maximum number of players in a normal game? (1 points)
[b]20[/b]

3. At what ratings do you receive the second and fourth rating badges, respectively? (4 points) [b]1350 and 1750[/b]

4. An FTW normal game is Ranking-scored, has 10 questions, with 35 seconds per question. Find the difference between the maximum and minimum amount of time from when the "Start game" button is clicked to when the results screen is shown, assuming no lags or glitches. Show all your work. (12 points)
[b]346[/b]; [i]an explanation is on my [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=328532]blog[/url][/i]

5. Without glitches, how many seconds after time is up can you choose to "Give up"? Round your answer to the nearest tenth. Note that the answer may be zero or negative. (3 points)
[b]1 second[/b]; [i]you can give up any time before "second 0" is over[/i]

6. What is the most common answer to an FTW question? (2 points)
[b]4[/b]; [i]on harder problems, however, the answer is more likely to be 6, 8, 15, or 28, so guess those instead of 4[/i]

7. When are you allowed to reveal someone else's real name and not have it be considered a ToS infringement? (2 points)
[b]When it's part of their username[/b]

8. Let $ \frac {x}{y}$ be the ratio in lowest integer terms of the length of a LanguageFilter ban to the length of a FloodingFilter ban. Find $ x^3 \minus{} y^3$. (3 points)
[b]19[/b]; [i]a LanguageFilter ban is 15 minutes while a FloodingFilter ban is 10 minutes, these are in the ratio 3:2, and 3³-2³=27-8=19[/i]

9. What causes a Guest to have a five-digit or four-digit number? (3 points)
[b]Guest was created from a proxy[/b] OR [b]Guest was created while a "second lobby" formed[/b]

10. How can a Guest play more games than the two that are allowed per day? (2 points)
[b]Click on a game icon, wait for someone else to enter, join button automatically appears[/b]

11. Who was the first moderator (not administrator) of FTW chat? (2 points)
[b]iin77[/b]; [i]now inactive because he graduated from high school, note that levans is an administrator[/i]

12. The 10-day period from March 21 to March 30 saw the joining to AoPS of four users that would eventually prove to be very influential to FTW, two of which joined on the same day. Name these four users. (8 points)
[b]isabella2296, hurdler, AIME15, jjx1[/b]

13. The FTW Leaderboard's clock is set in what time zone? (2 points)
[b]Eastern Time[/b]

14. In the year 2008, one user won over twice as many normal games and the user who placed second in normal games won. Name these two respective users (6 points), and for the nerdy bonus points (6 bonus points), how many games did each of them win?
[b]isabella2296 - 1097 games won[/b]
[b]iin77 - 540 games won[/b]

15. Name the first five users to pass 2000 rating, in order (Hint: One of the five is banned.). (10 points)
[b]AIME15, Mewto55555, Ttocs45, mousy, Jongy[/b]; [i]note that Jongy had username "Jongao" previously[/i]

[Freebee, hopefully]. Who programmed FTW? (5 points)
[b]levans[/b]

Describe the following FTW glitches. (3 points each)
16. 8x
[b]In FTW2, at the start, one of the bugs were that inputting one line will result in eight lines of your text appearing. [/b]
17. Flashies
[b]The game screen flashes, often with multiple questions, making it hard to read the question or submit an answer. [/b]; [i]neoflashies, which replaced flashies, occur less frequently but are also more detrimental to your game score, as you can't submit at all if your computer contracts neoflashies[/i]
18. Second Lobby
[b]Upon exiting a game, you find yourself in a different lobby that eventually everyone else also moves to upon refreshing. [/b]; [i]any game attempted to be made during the formation of a second lobby gets marked "undefined"[/i]
19. Trapped Game
[b]Game fails to close properly, and someone else joins causing the game to error and not be able to proceed to the final scoreboard. [/b]; [i]occurs much less frequently now, but an interesting rise in cases occurred last week[/i]
20. Fall 2009 Rating Save Glitch
[b]Leaving a losing game and quickly finishing another game with yourself saves your rating from the loss that would occur in that game.[/b]

Who do the following nicknames refer to? Provide the full formal username. (2 points each)
21. Lonny [b]vallon22[/b]
22. Eli [b]AIME15[/b]
23. Levlev [b]levans[/b]
24. Ding [b]^_^[/b]
25. Mabbitt [b]1=2[/b]
26. Ylxy [b]PhireKaLk6781[/b]
27. Rizzy [b]rizikin[/b]

Attribute these FTW chat quotes. (1 point each)
28. "I never said you never said [that] I never said I did." [b]isabella2296[/b]
29. "Not really, it went +2 +2 +2 -12 +2 +3." [b]ilovepink[/b]
30. "Sicily is a god dang multi like me?" [b]policecap[/b]
31. "Special ways are izzy...i mean easy." [b]Mewto55555[/b]
32. "I'm going to have fun in the shower with my friends." [b]AIME15[/b]

I don't remember saying #28 :o

I also must say that I feel affronted I was not a part of the nickname section. :(

Correction on #31:

The sequence went:

mewmew: special ways are izzy
mewmew: i mean easy
mewmew: lol i said izzy

It was a joke.
[**:**] PhireKaLK6781: Why doesn't anyone call me ylxy?
(later)
[**:**] aleph0: Hi ylxy
[**:**] PhireKaLK6781: LOL

(or something like that)

I would like to point out that iin77 was not a moderator.  He was deemed by levans as a muter.

[quote="isabella2296"]I don't remember saying #28 :o[/quote]

[16:22] isabella2296: I never said you never said thta I never said I did.
[16:22] Mewto55555: ah well
[16:22] isabella2296: pwnt!
[16:22] Mewto55555: gj!
[16:22] Mewto55555: you are very intelligent
[16:23] Mewto55555: idk if anyone does

[quote="isabella2296"]I also must say that I feel affronted I was not a part of the nickname section. :( [/quote]

I already have a freebee question.

[quote="isabella2296"]Correction on #31:

The sequence went:

mewmew: special ways are izzy
mewmew: i mean easy
mewmew: lol i said izzy[/quote]

That's why I put the ellipses.

#20: I told you, this was discovered in Spring 2009...additionally, you described it incorrectly...:wink:
#32: I don't recall saying this....

@AIME15: I do remember you saying that once.

[quote="$ LaTeX$"]For #10, you could also just refresh and get a different guest number.  :) [/quote]

That would be a different guest.

[quote="$ LaTeX$"][quote="PhireKaLk6781"]Final Scoreboard:
10. How can a Guest play more games than the two that are allowed per day? (2 points)
[b]Click on a game icon, wait for someone else to enter, join button automatically appears[/b]

20. Fall 2009 Rating Save Glitch
[b]Leaving a losing game and quickly finishing another game with yourself saves your rating from the loss that would occur in that game.[/b]
[/quote]

For #10, you could also just refresh and get a different guest number.  :)

For #20, at first I thought you were talking about username changes. Because you could just refresh your page, and your rating would go back to what it was before your username was changed.[/quote]
i used that to get over 9000 rating.
actually, i didn't, but i should've.

Lonny?
ok, i guess =)

and BTW, when did you start planning to do this?

[quote="PhireKaLk6781"][quote="vallon22"]and BTW, when did you start planning to do this?[/quote]

I don't understand what you mean. Elaborate?[/quote]

I think he meant when you started to create the quiz.

Edit: I meant "I think he meant when you to think about creating a knowledge quiz on FTW."

[quote="$ LaTeX$"][quote="PhireKaLk6781"][quote="vallon22"]and BTW, when did you start planning to do this?[/quote]

I don't understand what you mean. Elaborate?[/quote]

I think he meant when you started to create the quiz.[/quote]

I'm quite sure he didn't mean that, given the time that something is posted is stamped on each post.

i mean when did you start planning to have a quiz like this
you have questions about 2008 leaderboard, so you've either been planning since 2008, or have an excellent memory for this like this

Oh, I came up with the idea of the quiz the day I made it. As for the 2008 Leaderboard, it's because I took a PrintScreen of that leaderboard for memories.  :D

lol Smart idea to print screen the leaderboard, I miss the leaderboard in the "good old days" when Mewto55555 had a 1500 and he would make fun of me and say "hey, mousy, why don't you play me?  if you lose you only lose 2 points!"  lol

*sniff* one of the five was banned off the leaderboard.  Teehee sorry.

I woulda gotten a 58 if I took the quiz.  Nice Test.