AMC
USA(J)MO
USAMO
AIME
MATHCOUNTS
AMC 10
AMC 12

[hide="USAMO and Nationals..."]
[hide="USAMO Qualifiers going"](State, Rank last year, AIME Score, Rank at state this year)

- Nathan Benjamin (Indiana)      (6th)           (8 AIME)     (Won)
- Sachin Shinde (Indiana)           (Didn't go) (13 AIME)    (4th)
- Neal Wu (Lousiana)                  (1st)             (9 AIME)     (Won)
- Anand Oza (Maryland)             (56th)          (9 AIME)     (3rd)
- Neil Gurram (Michigan)           (?)                 (9 AIME)    (2nd)
- Matthew Vengalil (Michigan)  (Didn't go)  (8 AIME)    (Won)
- Andrew Ardito (New York)        (9th)           (8 AIME)     (Won)
- Daniel Li (Virginia)                   (47th)         (10 AIME)   (Won)
- Sam Keller (Washington)         (19th)          (8 AIME)     (Won)
[/hide]
[hide="USAMO Qualifiers that were 7th graders there last year"](State, Rank last year AIME score, Rank at state if there)

- Ian Osborn (New Jersey)  (21st)  (9 AIME)  (6th)
- Jeffrey Chan (Texas)        (30th)  (8 AIME) (9th)
[/hide]
? = Unknown[/hide]
[hide="Returning People"](State, Rank Last Year)

[hide="2nd Year Eighth Graders"]
Steven          Hwang         (Alabama)
Jonathan      Aldrich         (Arkansas)
Karan            Batra           (Arkansas)
Kevin            Yang           (California)                     (17th)
Scott            McClary        (Colorado)
Tommy         Gill              (Delaware)
Robert          Knetsche      (Defense Department)
Benjamin      Horkley        (DC)                              (35th)
Philip             Bauer         (Florida)                         (26th)
Arjun            Puranik         (Illinois)                       (32nd)
Nathan         Benjamin     (Indiana)                      (Made Countdown)
Ryan             Gao             (Maine)
Anand          Oza              (Maryland)                     (56th)
Neil               Gurram        (Michigan)
Christopher   Mueller        (Minnesota)
George          Yu                (Missouri)
Aidan            Shackleton  (Montana)
BJ                 Juneeau      (Nevada)
Jack              Ingalis          (New Mexico)
Eric               McCabe       (North Carolina)           (13th)
Ashvin          Srivatsa        (Ohio)
Matt               Tanzer        (Pennsylvania)
Xinran            Liu              (Tennessee)
Kristian         Fennessy      (Virgin Islands)
Brian             Hamrick       (Virginia)                        (44th)
Daneil            Li                (Virginia)                        (47th)
Sam              Keller           (Washington)                (19th)
Colin              Kemp          (West Virgnia)
Iris                 Xu                (Wisconsin)
David             Wang           (Wyoming)
[/hide]
[hide="3rd Year Eighth Graders"]
Neal              Wu                (Louisiana)                     (Won)
Lucas            Hubbard        (Maine)
Alec              Patterson       (Montana)
Daesun         Yim               (New Jersey)
Andrew         Ardito            (New York)                    (Made Countdown)
[/hide]
[hide="2nd Year Seventh Graders"]
Jonathan      Horton           (Arizona)
Ben Alan       White            (Tennessee)
Kevin            Chen             (Texas)                             (29th)
Kyle              Stankowski    (Wisconsin)
[/hide][/hide]
[hide="State Cut-Offs"]
(CDR is counted if official)
California: 41
Texas: 40
Illinois: 39
Indiana: 39
New Jersey: 39
Washington: 39
Massachusetts: 38
Michigan: 38
Minnesota: 38
Ohio: 37
Virginia: 37
New York: 36+
Connecticut: 35
Florida: 35
Maine: 35-
Kansas: 32
Arkansas: 31
Tennessee: 31
Delaware: 30
Utah: ~30
North Dakota: 24
Virgin Islands: ~15

Colorado 10th: 32
Oklahoma 7th: 28

~ = Around, + = May be higher, - = May be lower
[/hide]
[hide="Nationals Grade Distribution"]
6th Graders: 6
7th Graders: 57
8th Graders: 165[/hide]
Please add information to the missing information of topics we've already covered, or add suggestions for new information!

Edits: Sam Keller placed 19th, Brian Hamrick placed 44th, MathCounts has it missing here: http://www.mathcounts.org/webarticles/articlefiles/536-Returning_Students_06.pdf Ohio, Masschusettes, New Jersey, Minnesota cutoffs, Ian Osborn's rank, Mixed up 2 people with the same name's states, jeez sorry!

[quote="easyas3.14159..."]Did you sit at your computer and hunt down all this stuff? :P[/quote]

Zenith Xan and I gathered USAMO a couple weeks ago. I added the state cutoffs 1 by 1 over the past couple of weeks as I saw them, the Nationals Grade Distribution was done by Xantos C. Guin and Zenith Xan double checked after the State Department people were added. I just typed the Returning People a couple of minutes before posting this topic. Klebian helped also.

[quote="lotrgreengrapes7926"]Nice goal, Ignite. :lol: This must have taken a long time. My name is there once.
Yay, this is my 50th post.  :D[/quote]

You are Ryan Gao or Lucas Hubbard  :ninja: I am going to say Ryan Gao because you got a 36 and first place was 39 or 38.

[quote="lotrgreengrapes7926"][quote="Ignite168"]
[quote="lotrgreengrapes7926"]Nice goal, Ignite. :lol: This must have taken a long time. My name is there once.
Yay, this is my 50th post.  :D[/quote]

You are Ryan Gao or Lucas Hubbard  :ninja: I am going to say Ryan Gao because you got a 36 and first place was 39 or 38.[/quote]

Yep. Plus, Lucas has an AoPS name, 10SNE1. I'm the one who posted that stupidly easy question saying it was hard.   :oops:  :D[/quote]

And I'm the one who can't imagine a folded paper right, anyway, do you know Maine's cutoff? We know it can't be higher than 36.

[quote="Ignite168"]
And I'm the one who can't imagine a folded paper right, anyway, do you know Maine's cutoff? We know it can't be higher than 36.[/quote]

No, sorry. I don't know. It's probably 34 or 35, hopefully not lower.

[quote="Phelpedo"][quote="Ignite168"][hide="USAMO and Nationals..."]
[hide="USAMO Qualifiers going"](State, Rank last year, AIME Score, Rank at state this year)

- Nathan Benjamin (Indiana)      (6th)           (8 AIME)     (Won)
- Sachin Shinde (Indiana)           (Didn't go) (13 AIME)    (4th)
- Neal Wu (Lousiana)                  (1st)             (9 AIME)     (Won)
- Anand Oza (Maryland)             (56th)          (9 AIME)     (3rd)
- Neil Gurram (Michigan)           (?)                 (9 AIME)    (2nd)
- Matthew Vengalil (Michigan)  (Didn't go)  (8 AIME)    (Won)
- Andrew Ardito (New York)        (9th)           (8 AIME)     (Won)
- Daniel Li (Virginia)                   (47th)         (10 AIME)   (Won)
- Sam Keller (Washington)         (19th)          (8 AIME)     (Won)
[/hide]
[hide="USAMO Qualifiers that were 7th graders there last year"](State, Rank last year AIME score, Rank at state if there)

- Ian Osborn (New Jersey)  (21st)  (9 AIME)  (?)
- Jeffrey Chan (Texas)        (30th)  (8 AIME) (9th)
[/hide]
? = Unknown[/hide]
[hide="Returning People"](State, Rank Last Year)

[hide="2nd Year Eighth Graders"]
Steven          Hwang         (Alabama)
Jonathan      Aldrich         (Arizona)
Karan            Batra           (Arkansas)
Kevin            Yang           (California)                     (17th)
Scott            McClary        (Colorado)
Tommy         Gill              (Delaware)
Robert          Knetsche      (Defense Department)
Benjamin      Horkley        (DC)                              (35th)
Philip             Bauer         (Florida)                         (26th)
Arjun            Puranik         (Illinois)                       (32nd)
Nathan         Benjamin     (Indiana)                      (Made Countdown)
Ryan             Gao             (Maine)
Anand          Oza              (Maryland)                     (56th)
Neil               Gurram        (Michigan)
Christopher   Mueller        (Minnesota)
George          Yu                (Missouri)
Aidan            Shackleton  (Montana)
BJ                 Juneeau      (Nevada)
Jack              Ingalis          (New Mexico)
Eric               McCabe       (North Carolina)           (13th)
Ashvin          Srivatsa        (Ohio)
Matt               Tanzer        (Pennsylvania)
Xinran            Liu              (Tennessee)
Kristian         Fennessy      (Virgin Islands)
Brian             Hamrick       (Virginia)
Daneil            Li                (Virginia)                        (47th)
Sam              Keller           (Washington)                (19th)
Colin              Kemp          (West Virgnia)
Iris                 Xu                (Wisconsin)
David             Wang           (Wyoming)
[/hide]
[hide="3rd Year Eighth Graders"]
Neal              Wu                (Louisiana)                     (Won)
Lucas            Hubbard        (Maine)
Alec              Patterson       (Montana)
Daesun         Yim               (New Jersey)
Andrew         Ardito            (New York)                    (Made Countdown)
[/hide]
[hide="2nd Year Seventh Graders"]
Jonathan      Horton           (Arkansas)
Ben Alan       White            (Tennessee)
Kevin            Chen             (Texas)                             (29th)
Kyle              Stankowski    (Wisconsin)
[/hide][/hide]
[hide="State Cut-Offs"]
(CDR is counted if official)
California: 41
Texas: 40
Illinois: 39
Indiana: ~39
Michigan: 38
Washington: 38
Massachusetts: 37+
Ohio: 37
Virginia: 37
New York: 36+
Connecticut: 35
Florida: 35
Maine: 35-
Kansas: 32
Arkansas: 31
Tennessee: 31
Utah: ~30
North Dakota: 24
Delaware: ~20

Colorado 10th: 32
Oklahoma 7th: 28

~ = Around, + = May be higher, - = May be lower
[/hide]
[hide="Nationals Grade Distribution"]
6th Graders: 6
7th Graders: 57
8th Graders: 165[/hide]
Please add information to the missing information of topics we've already covered, or add suggestions for new information!

Edit 1: Sam Keller placed 19th, sorry about that, apparently, Mathcounts.org forgot to put it in their .pdf as well here: http://www.mathcounts.org/webarticles/articlefiles/536-Returning_Students_06.pdf Thanks samath!

Edit 2: Cutoff for Ohio was 37, thanks ehehheehee![/quote]

Are you sure it was Sachin Schinde, not Nathan Benjamin, that got the 13 on the AIME?[/quote]
Your naiveness is astounding :ninja:. Sorry, but the AIME and mathcounts are COMPLETELY different. Doing good in mathcounts doesn't guarantee that you do good on the AIME and vice versa. Sachin Schinde got a 13 on the AIME and he only made it because fanatic got UNlucky.
P.S. Neil Gurram got 59th or 69th last year at nats.
EDIT: TYpo meant to say fanatic got unlucky

NJ's cutoff was at least 35 - 8th place got a 35. If I remember correctly NJ's cutoff was 39.

Confirmation: written cutoff from NJ was 39.

Yes, I am sure it was Sachin with the 13 on the AIME... grumble grumble.

Massachussetts cutoff was 38.
I got 4th place with a 38.
Please change it.

[quote="kyyuanmathcount"][quote="xxm0rg07hxx"]I definitely don't remember seeing a Jonathan Horton at any of our practices.  Also, I've done about all of my team rounds with a Jonathan Aldrich.  Your list is messed up and you have no life.[/quote]

You're in charge of making the list of AoPSers with a life.[/quote]

That's not that hard  :roll:

[quote="pgpatel"][quote="Phelpedo"][quote="Ignite168"][hide="USAMO and Nationals..."]
[hide="USAMO Qualifiers going"](State, Rank last year, AIME Score, Rank at state this year)

- Nathan Benjamin (Indiana)      (6th)           (8 AIME)     (Won)
- Sachin Shinde (Indiana)           (Didn't go) (13 AIME)    (4th)
- Neal Wu (Lousiana)                  (1st)             (9 AIME)     (Won)
- Anand Oza (Maryland)             (56th)          (9 AIME)     (3rd)
- Neil Gurram (Michigan)           (?)                 (9 AIME)    (2nd)
- Matthew Vengalil (Michigan)  (Didn't go)  (8 AIME)    (Won)
- Andrew Ardito (New York)        (9th)           (8 AIME)     (Won)
- Daniel Li (Virginia)                   (47th)         (10 AIME)   (Won)
- Sam Keller (Washington)         (19th)          (8 AIME)     (Won)
[/hide]
[hide="USAMO Qualifiers that were 7th graders there last year"](State, Rank last year AIME score, Rank at state if there)

- Ian Osborn (New Jersey)  (21st)  (9 AIME)  (?)
- Jeffrey Chan (Texas)        (30th)  (8 AIME) (9th)
[/hide]
? = Unknown[/hide]
[hide="Returning People"](State, Rank Last Year)

[hide="2nd Year Eighth Graders"]
Steven          Hwang         (Alabama)
Jonathan      Aldrich         (Arizona)
Karan            Batra           (Arkansas)
Kevin            Yang           (California)                     (17th)
Scott            McClary        (Colorado)
Tommy         Gill              (Delaware)
Robert          Knetsche      (Defense Department)
Benjamin      Horkley        (DC)                              (35th)
Philip             Bauer         (Florida)                         (26th)
Arjun            Puranik         (Illinois)                       (32nd)
Nathan         Benjamin     (Indiana)                      (Made Countdown)
Ryan             Gao             (Maine)
Anand          Oza              (Maryland)                     (56th)
Neil               Gurram        (Michigan)
Christopher   Mueller        (Minnesota)
George          Yu                (Missouri)
Aidan            Shackleton  (Montana)
BJ                 Juneeau      (Nevada)
Jack              Ingalis          (New Mexico)
Eric               McCabe       (North Carolina)           (13th)
Ashvin          Srivatsa        (Ohio)
Matt               Tanzer        (Pennsylvania)
Xinran            Liu              (Tennessee)
Kristian         Fennessy      (Virgin Islands)
Brian             Hamrick       (Virginia)
Daneil            Li                (Virginia)                        (47th)
Sam              Keller           (Washington)                (19th)
Colin              Kemp          (West Virgnia)
Iris                 Xu                (Wisconsin)
David             Wang           (Wyoming)
[/hide]
[hide="3rd Year Eighth Graders"]
Neal              Wu                (Louisiana)                     (Won)
Lucas            Hubbard        (Maine)
Alec              Patterson       (Montana)
Daesun         Yim               (New Jersey)
Andrew         Ardito            (New York)                    (Made Countdown)
[/hide]
[hide="2nd Year Seventh Graders"]
Jonathan      Horton           (Arkansas)
Ben Alan       White            (Tennessee)
Kevin            Chen             (Texas)                             (29th)
Kyle              Stankowski    (Wisconsin)
[/hide][/hide]
[hide="State Cut-Offs"]
(CDR is counted if official)
California: 41
Texas: 40
Illinois: 39
Indiana: ~39
Michigan: 38
Washington: 38
Massachusetts: 37+
Ohio: 37
Virginia: 37
New York: 36+
Connecticut: 35
Florida: 35
Maine: 35-
Kansas: 32
Arkansas: 31
Tennessee: 31
Utah: ~30
North Dakota: 24
Delaware: ~20

Colorado 10th: 32
Oklahoma 7th: 28

~ = Around, + = May be higher, - = May be lower
[/hide]
[hide="Nationals Grade Distribution"]
6th Graders: 6
7th Graders: 57
8th Graders: 165[/hide]
Please add information to the missing information of topics we've already covered, or add suggestions for new information!

Edit 1: Sam Keller placed 19th, sorry about that, apparently, Mathcounts.org forgot to put it in their .pdf as well here: http://www.mathcounts.org/webarticles/articlefiles/536-Returning_Students_06.pdf Thanks samath!

Edit 2: Cutoff for Ohio was 37, thanks ehehheehee![/quote]

Are you sure it was Sachin Schinde, not Nathan Benjamin, that got the 13 on the AIME?[/quote]
Your naiveness is astounding :ninja:. Sorry, but the AIME and mathcounts are COMPLETELY different. Doing good in mathcounts doesn't guarantee that you do good on the AIME and vice versa. Sachin Schinde got a 13 on the AIME and he only made it because fanatic got UNlucky.
[/quote]
Being a USAMO participant, I happen to know a thing or two about the AIME myself, thank you very much. The Benjamins have proven themselves to be quite adept at both the AIME and Mathcounts (David having scored a 12 last year), so it was not such an extraordinary thought that Nathan, as opposed to a relative unknown, might have gotten the 13. If you knew a little bit yourself, you might realize that, while success in MathCounts doesn't necessitate success in the AMC, the two are often related.

[quote="Phelpedo"][/quote][quote="pgpatel"][quote="Phelpedo"][quote="Ignite168"][hide="USAMO and Nationals..."]
[hide="USAMO Qualifiers going"](State, Rank last year, AIME Score, Rank at state this year)

- Nathan Benjamin (Indiana)      (6th)           (8 AIME)     (Won)
- Sachin Shinde (Indiana)           (Didn't go) (13 AIME)    (4th)
- Neal Wu (Lousiana)                  (1st)             (9 AIME)     (Won)
- Anand Oza (Maryland)             (56th)          (9 AIME)     (3rd)
- Neil Gurram (Michigan)           (?)                 (9 AIME)    (2nd)
- Matthew Vengalil (Michigan)  (Didn't go)  (8 AIME)    (Won)
- Andrew Ardito (New York)        (9th)           (8 AIME)     (Won)
- Daniel Li (Virginia)                   (47th)         (10 AIME)   (Won)
- Sam Keller (Washington)         (19th)          (8 AIME)     (Won)
[/hide]
[hide="USAMO Qualifiers that were 7th graders there last year"](State, Rank last year AIME score, Rank at state if there)

- Ian Osborn (New Jersey)  (21st)  (9 AIME)  (?)
- Jeffrey Chan (Texas)        (30th)  (8 AIME) (9th)
[/hide]
? = Unknown[/hide]
[hide="Returning People"](State, Rank Last Year)

[hide="2nd Year Eighth Graders"]
Steven          Hwang         (Alabama)
Jonathan      Aldrich         (Arizona)
Karan            Batra           (Arkansas)
Kevin            Yang           (California)                     (17th)
Scott            McClary        (Colorado)
Tommy         Gill              (Delaware)
Robert          Knetsche      (Defense Department)
Benjamin      Horkley        (DC)                              (35th)
Philip             Bauer         (Florida)                         (26th)
Arjun            Puranik         (Illinois)                       (32nd)
Nathan         Benjamin     (Indiana)                      (Made Countdown)
Ryan             Gao             (Maine)
Anand          Oza              (Maryland)                     (56th)
Neil               Gurram        (Michigan)
Christopher   Mueller        (Minnesota)
George          Yu                (Missouri)
Aidan            Shackleton  (Montana)
BJ                 Juneeau      (Nevada)
Jack              Ingalis          (New Mexico)
Eric               McCabe       (North Carolina)           (13th)
Ashvin          Srivatsa        (Ohio)
Matt               Tanzer        (Pennsylvania)
Xinran            Liu              (Tennessee)
Kristian         Fennessy      (Virgin Islands)
Brian             Hamrick       (Virginia)
Daneil            Li                (Virginia)                        (47th)
Sam              Keller           (Washington)                (19th)
Colin              Kemp          (West Virgnia)
Iris                 Xu                (Wisconsin)
David             Wang           (Wyoming)
[/hide]
[hide="3rd Year Eighth Graders"]
Neal              Wu                (Louisiana)                     (Won)
Lucas            Hubbard        (Maine)
Alec              Patterson       (Montana)
Daesun         Yim               (New Jersey)
Andrew         Ardito            (New York)                    (Made Countdown)
[/hide]
[hide="2nd Year Seventh Graders"]
Jonathan      Horton           (Arkansas)
Ben Alan       White            (Tennessee)
Kevin            Chen             (Texas)                             (29th)
Kyle              Stankowski    (Wisconsin)
[/hide][/hide]
[hide="State Cut-Offs"]
(CDR is counted if official)
California: 41
Texas: 40
Illinois: 39
Indiana: ~39
Michigan: 38
Washington: 38
Massachusetts: 37+
Ohio: 37
Virginia: 37
New York: 36+
Connecticut: 35
Florida: 35
Maine: 35-
Kansas: 32
Arkansas: 31
Tennessee: 31
Utah: ~30
North Dakota: 24
Delaware: ~20

Colorado 10th: 32
Oklahoma 7th: 28

~ = Around, + = May be higher, - = May be lower
[/hide]
[hide="Nationals Grade Distribution"]
6th Graders: 6
7th Graders: 57
8th Graders: 165[/hide]
Please add information to the missing information of topics we've already covered, or add suggestions for new information!

Edit 1: Sam Keller placed 19th, sorry about that, apparently, Mathcounts.org forgot to put it in their .pdf as well here: http://www.mathcounts.org/webarticles/articlefiles/536-Returning_Students_06.pdf Thanks samath!

Edit 2: Cutoff for Ohio was 37, thanks ehehheehee![/quote]

Are you sure it was Sachin Schinde, not Nathan Benjamin, that got the 13 on the AIME?[/quote]
Your naiveness is astounding :ninja:. Sorry, but the AIME and mathcounts are COMPLETELY different. Doing good in mathcounts doesn't guarantee that you do good on the AIME and vice versa. Sachin Schinde got a 13 on the AIME and he only made it because fanatic got UNlucky.
[/quote][quote="Phelpedo"]
Being a USAMO participant, I happen to know a thing or two about the AIME myself, thank you very much. The Benjamins have proven themselves to be quite adept at both the AIME and Mathcounts (David having scored a 12 last year), so it was not such an extraordinary thought that Nathan, as opposed to a relative unknown, might have gotten the 13. If you knew a little bit yourself, you might realize that, while success in MathCounts doesn't necessitate success in the AMC, the two are often related.[/quote] Sorry  :blush: . I didn't mean to be rude I, thought that you immediately connected Nathan Benjamin to the 13 on the AIME because he is a Mathcounts genius. Although he did get a perfect on the AMC10 last year. David got a perfect on the AMC12 this year.

[quote="Hamster1800"][quote="Ignite168"][quote="Hamster1800"]I'm a returning mathcounts person who ranked in the top 57 last year and you don't have that. MathCounts also forgot to put it on their pdf. I'll leave it up to you to figure out who I am, what grade I'm in, and where I placed. :D[/quote]

44th place Brian Hamrick ;)[/quote]

Dang you're fast...only took you 1 hour and 36 minutes :o[/quote]

I already knew that before you posted ;)

I got unlucky with the last millisecond switch thing but yes Sachin is a much better mathematician and he probably deserved it over me. However, I'm probably better at Mathcounts.

[quote="pgpatel"]David got a perfect on the AMC12 this year.[/quote]

That's like the least of his achievements :D

[quote="Hamster1800"][quote="pgpatel"]:rotfl: . By the way Hamster1800 got second at state for the second year in a row.[/quote]

Yes, yes, as i stated [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=475084&highlight=#475084]here[/url]. Anyway, my goal is to do better than last year (24/46 => 44th place).[/quote]

Congratulations, Brian!

[quote="binonunquineist"][quote="asdf4"]Heres whats ironic.
I got 1st last year at state, but this year  i got 2nd.  :(
Oh well, at least im still going to nats.[/quote]

I'll tell you something thats ironic. I got a better ranking at nationals than in chapter (4th in chapter, 3rd at nationals).[/quote]

Nice Job PEOPLE!!!

Why must you be in 7th grade like I?

[quote="pianoforte"]there are two people named David Wang (kansas, wyoming)....[/quote]
My teammate's name also is David Huang (pronounced David Wang)

There were so many people named David...it was fun watching my David teamates groan every time they heard "David"

There were 12 davids and 7 kevins at nationals. Interesting.

Sorry Ignite168 for yet another correction, but the cutoff for WA was 39. I happen to know because I was the cutoff...

[quote="pgpatel"]Where would a 17 (11/3) have gotten?[/quote]

probably 180ish

A score of 17 was a ten way tie between 183 through 192.  No way to tell where you would have been in the tie breaker without knowing someone who got a 17.


Let $p, q$ be distinct prime numbers. Find all integer solutions to: \[\frac{p}{x}+\frac{q}{y}= 1\]

[hide="Solution"]
\[\frac{p}{x}+\frac{q}{y}=1\\\implies py+qx=xy\\\implies (x-p)(y-q)=pq.\]
The completion of this solution is trivial.
[/hide]


Does any one have any advice on how to prepare for the KCATM?

I already posted this before you.  Please don't just repeat the same thing what another person has posted.

yo deltaunix, what school do u go to I think i might know u

Anyone know the KS mathcounts results for this year?

i dont know the results for everyone but im guessing they will tell in the banquet

[quote=aastha.sharma02]Probably one of the hundreds of Asians there with glasses[/quote]

In 2008 I mean....


Let $ x,$ $ y$ and $ z$ are positive numbers such that $ x \plus{} y \plus{} z \equal{} 1.$ Prove that:
1) \[ \left(2 \plus{} \frac {x}{y^{2} \plus{} yz \plus{} z^{2}}\right)\left(2 \plus{} \frac {y}{z^{2} \plus{} zx \plus{} x^{2}}\right)\left(2 \plus{} \frac {z}{x^{2} \plus{} xy \plus{} y^{2}}\right)\ge27\]
2) for all $ 0\leq \alpha\leq1 \plus{} \sqrt5$ the following inequality holds:
\[ \left(2 \plus{} \frac {\alpha x}{y^{2} \plus{} yz \plus{} z^{2}}\right)\left(2 \plus{} \frac {\alpha y}{z^{2} \plus{} zx \plus{} x^{2}}\right)\left(2 \plus{} \frac {\alpha z}{x^{2} \plus{} xy \plus{} y^{2}}\right)\ge(2 \plus{} \alpha)^{3}\]


analytic geometry
combinatorics proposed
combinatorics

Given a set of lines in the plane with no three meeting at a point, form a graph G whose vertices are the intersections of the lines, with two vertices adjacent if they appear consecutively on one of the lines. Prove that the chromatic number of G is at most 3.

I assume that the set of lines is finite, no? Anyway, it doesn't matter because we can deduce the infinite case from the finite one by compacity argument.

Since the number of lines is finite, there is a finite number of intersection points. Thus, we may choose an orthonormal system of ccordinates such that these intersection points are $M_1, \cdots , M_k$, with $M_i(xi,y_i)$ and $x_1 < x_2 < \cdots < x_k$.
Now we 3-color the $M_i$'s in the increasing order of their respective subscripts, as follows :
Color $M_1$ with color $c_1$ and $M_2$ with color $c_2$.
Assume that, for $i \in \{1, \cdots , k-1 \}$, the points $M_1, \cdots, M_i$ have been properly colored with no more than three colors.
Then, note that $M_{i+1}$ has at most 4 neighbours (since no three lines meet at $M_{i+1}$), and among them at most two have an $x$-coordinate smaller than $x_{i+1}$. Thus, give any non-used color (by its already colored neighbours) to $M_{i+1}$. This produces a proper 3-coloring of the intersection points and we are done.

Note that if there is more than two lines, there is a triangle so that the chromatic number is indeed 3.

Pierre.


Corn costs 99 cents per pound, and beans cost 45 cents per pound. If Shea buys
24 total pounds of corn and beans, and it costs $ \$18.09$, how many pounds of corn did
Shea buy? Express your answer as a decimal to the nearest tenth.

Why doesn't anyone do this question?  It's not too hard  :(

Let $ C$ be the pounds of corn Shea buys, and let $ B$ be the pounds of beans Shea buys.  We start off by writing equations:

$ 99C \plus{} 45B \equal{} 1,809$     [first equation]
$ C \plus{} B \equal{} 1,080$     [second equation]

Now, we use elimination, so we multiply the second equation by $ 45$:

$ 99C \plus{} 45B \equal{} 1,809$     [first equation]
$ 45C \plus{} 45B \equal{} 1,080$     [second equation]

We subtract the second equation from the first equation here:

$ 54C \equal{} 729$

Then, we divide and cancel.

$ C \equal{} \frac {729}{54} \equal{} \frac {27}{2} \equal{} 13.5$

So, to the nearest tenth, $ \boxed{13.5}$ pounds is our final answer.


Hey guys, new to the community. I'm currently a 12th grade student from Toronto and I'm in need of a bit of advice.

First a little background about myself: I've always been into math since I was young, learning stuff a few grades ahead right from grade 1. In middle school, I aced every local contest. Something happened in gr 9, and that was, i never really progressed in math besides the school curriculum (which never really taught me more than i already knew). Here i find myself 3 years later, 1 year shy of going to university (engineering) and i'm just an average student in my class.

What i need advice with: after browsing this site a bit, i've come to realize that math is vast outside of the highschool curriculum. I'm not sure where exactly to focus and what would be most beneficial for an engineering student. To clarify: i'm asking what areas of math should i focus on to excel in EE. I have until next September to bring myself up to speed. All responses are appreciated :D

Electrical engineering (if I've got that right) is still a fairly broad subject, but I believe you'll need to be fairly proficient in calculus, particularly the theory of differential equations, and probably with a year of linear algebra thrown in.  If you're restricting yourself to mathematics that will help you as an engineer, it'll be the kind of classes you'll need to take in college anyway, so what you could do is look up course requirements for an EE major at several colleges you're looking at.


Just so you know....
Laps ARE back and forth in swimming....

[quote="KingSmasher3"]As always, the Perfect score on the AMC8, is desired by more and more people.
I would like to start an AMC8 marathon to train people to get a perfect score!  :)
Also, when you posted a problem, put a number to represent the hardness of the problem. (1=easy, 5=hard)
Also, one may choose whether or not to post the multiple choice.  :wink:

[hide="New Problem (3)"]George swims 5 laps around a rectangular swimming pool. He then swims 7 back-and-forths along the length f the swimming pool.
George then notices that the total distance he swam for both excericises turned out to be the same!
What the ratio of the length to width of the swimming pool?  [/hide]

I hope this works out smoothly! Have fun![/quote]

[hide="Solution"]We construct the equation $ 10(l \plus{} w) \equal{} 14l$, and we get $ 4l \equal{} 10w$.

Therefore, the ratio of the length to the width of the swimming pool is 5/2.[/hide]

[hide="New problem"]What is the area of a hexagon with side length 3?[/hide]

[hide]Ans is $ 27\sqrt3/2$
because when we split it into six equaliteral triangles we find the area of one which is 9sqrt3 /4 and then multiply by six.[/hide]

[hide]What is the units digit of $ {2008}^{2008}$[/hide]

[hide="Solution"]We look at the first few powers of 8 until we find a pattern - $ 8, 4, 2, 6, \cdots$.  2008 is a multiple of 4, so the units digit is 6.
[/hide]
[hide="New problem"]What is the general term of this sequence:
$ 625, 125, 5, 1, \dfrac{1}{5} \cdots$.[/hide]

Guys, we already have plenty of marathons...see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=288069]Middle School Marathon[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194579]MATHCOUNTS Marathon[/url].


Can anyone tell me where I can find a material at olympiad-level on affine transformations? I first saw it in Kiran's notes, but this part is very superficial.

Have you tried google?
http://www.google.com/search?q=affine+transformations


2 squares are painted in blue and 2 squares are painted in red on a $ 3\times 3$ board in such a way that two square with same color is neither at same row nor at same column. In how many different ways can these four squares be painted?

$\textbf{(A)}\ 198  \qquad\textbf{(B)}\ 288  \qquad\textbf{(C)}\ 396  \qquad\textbf{(D)}\ 576  \qquad\textbf{(E)}\ 792$

Red squares can be put in $ 9\cdot 4\equal{}36$ ways. Then by simple counting of cases for blue squares, we find that there are $ 22$ allowable positions. Hence $ N\equal{}36\cdot 22\equal{}792$

[quote="Farenhajt"]Red squares can be put in $ 9\cdot 4 \equal{} 36$ ways.[/quote]  No: you must divide by 2 for the two possible orders you can add them in.  Similarly, the 22 is a double-count.

Right you are  :blush: ... so it's $ 198$


Ok on page 44 the authors provided a list of valid manipulations for modular arithmetic and multiplication. It stated (sorry no LaTeX) that [b]a[/b] [b]is congruent to [/b][b]b mod m[/b] and [b]p is congruent to q mod m[/b] and that [b](a+b) is congruent to (p+q) (mod m). [/b]I'm a little confused, i thought it was the other way around, that  [b](a+q) is congruent to (b+p)(mod m)[/b] at least according to their previous example. I tried their way out with [b]6 is congruent to 1 mod 5[/b] and [b]7 is congruent to 2 mod 5[/b] now 7 should, at least according to their formula, be congruent to 9 mod 5 which is reduced down to [b]2 is congruent to 4 mod 5[/b], which doesn't make sense, am i missing something? The other way makes more sense: [b]13 is congruent to 3 mod 5.[/b]

sorry it's not reduced down to that (poor choice of words), those are just the respective remainders in their smallest integer form.


[quote="ADN"]$ a,b,c\in R^{\plus{}}$, prove that:
$ \sqrt{\frac{a}{a\plus{}b}}\plus{}\sqrt{\frac{b}{b\plus{}c}}\plus{}\sqrt{\frac{c}{c\plus{}a}}\leq\frac{3\sqrt{2}}{2}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17549


is there a method to experimentally determine the half-life of elements particularly of those with high value of half life?
Thanks in advance

[quote="JRav"]There is an equation for half-life, which I can't remember at this time.  Clearly, for elements like U-238, there have to be alternative methods but I assume that by monitoring decay rate, they can come up with a pretty general idea.  If you analyze a half-life curve it is logarithmic (although upside-down in nature), and the equation does contain the natural logarithm function.[/quote]

Just elaborating on things here. Radioactive decay is a first order process. Say we are considering the decay of $ N$ then:

$ \frac {dN}{dt} \equal{} \minus{} kN$ where $ k$ is a constant. Then solving this:

$ \int^{N}_{N_0} \frac {dN}{N} \equal{} \int^{t}_{0} \minus{} k.dt$

Hence: $ \ln{\frac {N}{N_0}} \equal{} \minus{} kt$ then rearranging this we get $ N \equal{} N_0e^{ \minus{} kt}$. Hence we can actually predict the number of particles present in a sample over a period of time given the the right initial conditions.

Currently we cannot predict whether a given atom is going to disintegrate next or not though.

When $ \frac{N}{N_0} \equal{} \frac{1}{2} \equal{} e^{\minus{}kt}$ then $ t \equal{} \frac{1}{k} \ln{2}$ is the half-life.


You have to compile it yourself from the source code on the Asymptote website.


new user and i want to know how to access answer keys which are sent from other members using pm's

thanks!

[quote="007math"]You can't access another member's private message box.

Mods: Move to questions forum.[/quote]

I think he/she meant opening it not hacking into someone's account...


[quote="abstraction"]I'm currently studying computer science, but have gained a certain affinity for mathematics over the years. I want delve into pure math, but I'm not sure where to start. Can anyone point me towards any books to get me on my way?[/quote]

First you need to make sure you know how to write proofs (very important). Make sure you spend sufficient time on a book that introduces you to mathematical logic and proof writing. There are several good ones to choose from, but "How to Prove It" is a good choice.

Then you can look at calculus at a rigorous level with Apostol volumes I and II. I havn't actually worked through the entire books, but I here that they're pretty good. If you've had a little more math you can get into Spivak's Calculus on Manifolds.

See the FAQ post.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11946

[quote="abstraction"]I think (or hope) all CS curriculum teaches their students how to write, understand and critique proofs. Theoretical Computer Science is based around such things.[/quote]

Yeah I think that most computer scientists learn how to write proofs in discrete math courses, but I could be mistaken.


Hello gents,

Μία σχετικά εύκολη και όμορφη άσκηση στην ανάλυση είναι να βρεθεί (αν υπάρχει) το άπειρο ρίζικο: $ \sqrt{2\plus{}\sqrt{2\plus{}\sqrt{2\plus{}\cdots}}}$.

Όσοι δεν το έχετε δει, αξίζει να το προσπαθήσετε. Όσοι το έχετε δει, τί μπορείτε να πείτε για την παράσταση:
$ \sqrt{1\plus{}\sqrt{2\plus{}\sqrt{3\plus{}\cdots}}}$ :?:

Cheerio,

Durandal 1707

[quote="Durandal"] Όσοι το έχετε δει, τί μπορείτε να πείτε για την παράσταση:
$ \sqrt {1 \plus{} \sqrt {2 \plus{} \sqrt {3 \plus{} \cdots}}}$ :?:
[/quote]

Χρησιμοποιώντας $ \sqrt {n \plus{} \sqrt {n \plus{} 1} \plus{} 1} < \sqrt {n} \plus{} 1$, μπορώ να δείξω ότι η παράσταση συγκλίνει και είναι το πολύ 2. Είμαι σχεδόν σίγουρος όμως πως το όριο πρέπει να είναι μικρότερο.

EDIT: Ισχύει και η ανισότητα $ \sqrt {n \plus{} \sqrt {n \plus{} 1} \plus{} a} < \sqrt {n} \plus{} a$ όπου $ a \equal{} \frac{\minus{}1 \plus{} \sqrt{1 \plus{} 4\sqrt{2}}}{2} \equal{} 0.79...$ που δίνει ότι η παράσταση είναι το πολύ 1+α.

Dem, για τη σύγκλιση το έβαλα ;)

O λόγος είναι ένα cute (και όχι exceedingly hard) θεωρηματάκι που πέτυχα χτες:

Αν $ r > 1$, το όριο $ \lim_{n\to\infty} \sqrt [r]{a_1 \plus{} \sqrt [r]{a_2 \plus{} \sqrt [r]{\cdots \plus{} \sqrt [r]{a_n}}}}$ υπάρχει αν και μόνο αν η ακολουθία $ \sqrt [r^n]{a_n}$ είναι φραγμένη.

Cheerio,

Durandal 1707

[b]Edit[/b]: Interestingly enough, το θεώρημα επεκτείνεται (με λίγο διαφορετική όμως διατύπωση) και για τα continued fractions της περίπτωσης $ r \equal{} \minus{} 1$.

[quote="mathsforfun!"]Για το δευτερο οριο μπορει καποιος να βρει που παει? Dur το εχεις υπολογισει ?Εχω αρχισει και εκνευριζομαι.. :mad: :wallbash_red:[/quote]

Guys, δεν έχω ιδέα αν υπολογίζεται το όριο, μόνο για τη σύγκλιση το έβαλα! Θα δοκιμάσω αργότερα να δω μήπως βγάλω κάτι, αλλά δεν πιστεύω ότι υπάρχει τίποτα...

Cheerio,

Durandal 1707

[quote="Durandal"]Dem, για τη σύγκλιση το έβαλα ;)[/quote]

Ok. Νόμιζα πως γνώριζες το όριο. Θα προσπαθήσω να αποδείξω το θεώρημα αργότερα.

[quote="Extremal"]http://www.mathlinks.ro/viewtopic.php?search_id=348189509&t=152113
see post #5[/quote]

That's not the same, that's Ramanujan's identity...

Cheerio,

Durandal 1707

if I understand right, you are asking  to find limit of the expression
[quote="Durandal"]...

Αν $ r > 1$, το όριο $ \lim_{n\to\infty} \sqrt [r]{a_1 + \sqrt [r]{a_2 + \sqrt [r]{\cdots + \sqrt [r]{a_n}}}}$
...
[/quote]
 :)
here http://www.mathlinks.ro/viewtopic.php?search_id=348189509&t=152113 I also was talking  about generalisation  fo such problems  see my #5 post :
[quote="Extremal"]...
(you can also generelaze it more if you have sequance :
$ x_{n} = (a_{1} + b_{1}(a_{2} + ... + b_{n - 1}(a_{n})^{\frac {1}{p}})^{\frac {1}{p}})^\frac {1}{p}$ )
...

[/quote]

in your case $ p = r$,  $ b_{j} = 1$ for all integer $ j$  thus we get your particular case.
Now I try to explain how to solve this one :

[quote="Extremal"]...
(you can also generelaze it more if you have sequance :
$ x_{n} = (a_{1} + b_{1}(a_{2} + ... + b_{n - 1}(a_{n})^{\frac {1}{p}})^{\frac {1}{p}})^\frac {1}{p}$ )
...

[/quote]
where all $ a_{j},b_{j} \geq 0$
suppose $ x_{n} \to C < + \infty$
 then note that :
$ x_{n} \leq x_{n + 1}$ or in other words sequence $ x_{n}$ is monotone increasing, because
$ x_{n} = \sqrt [p]{a_{1} + b_{1}\sqrt [p]{a_{2} + .... + b_{n - 1}\sqrt [p]{a_{n}}}}$
 and
$ x_{n + 1} = \sqrt [p]{a_{1} + b_{1}\sqrt [p]{a_{2} + .... + b_{n - 1}\sqrt [p]{a_{n} + b_{n}\sqrt [p]{a_{n + 1}}}}}$ thus
$ x_{n} \leq x_{n + 1}$  is equivalent to prove :
\[ \sqrt [p]{a_{1} + b_{1}\sqrt [p]{a_{2} + .... + b_{n - 1}\sqrt [p]{a_{n}}}} \leq \sqrt [p]{a_{1} + b_{1}\sqrt [p]{a_{2} + .... + b_{n - 1}\sqrt [p]{a_{n} + b_{n}\sqrt [p]{a_{n + 1}}}}}
\]
after all  simplify you get :
$ 0 \leq b_{n} \sqrt [p]{a_{n + 1}}$ and this is true since all $ a_{j},b_{j} \geq 0$
thus we have
$ x_{n} \leq x_{n + 1}$ hence
 $ x_{n} \leq \lim _{n \to \infty} x_{n} = C < + \infty$
or $ x_{n} \leq C$ for all $ n$
note that
$ x_{n} = \sqrt [p]{a_{1} + b_{1}\sqrt [p]{a_{2} + b_{2}\sqrt [p]{a_{3} .... + b_{n - 1}\sqrt [p]{a_{n}}}}}\geq \sqrt [p]{0 + b_{1}\sqrt [p]{a_{2} + b_{2}\sqrt [p]{a_{3} .... + b_{n - 1}\sqrt [p]{a_{n}}}}} \geq \sqrt [p]{0 + b_{1}\sqrt [p]{0 + b_{2}\sqrt [p]{a_{3} .... + b_{n - 1}\sqrt [p]{a_{n}}}}} \geq \sqrt [p]{0 + b_{1}\sqrt [p]{0 + b_{2}\sqrt [p]{0 + .... + b_{n - 1}\sqrt [p]{a_{n}}}}} \geq .... \geq \sqrt [p]{b_{1}\sqrt [p]{b_{2}\sqrt [p]{.... b_{n - 1}\sqrt [p]{a_{n}}}}}$
thus we have :
$ \sqrt [p]{b_{1}\sqrt [p]{b_{2}\sqrt [p]{.... b_{n - 1}\sqrt [p]{a_{n}}}}} \leq x_{n} \leq C < + \infty$ for all $ n$.
after taking $ \ln$ of both sides inequality we have:
$ \left| \left( \sum_{1 \leq k < n}\frac {\ln b_{k}}{p^{k}} \right) + \frac {\ln a_{n}}{p^{n}} \right| \leq |\ln C | < + \infty$
for all $ n$.  thus sequence $ \left\{ c_{n} = \left| \left( \sum_{1 \leq k < n}\frac {\ln b_{k}}{p^{k}} \right) + \frac {\ln a_{n}}{p^{n}} \right| \right\}_{n = 0}^{\infty}$ is bounded if $ \lim_{n \to \infty} x_{n} = C < \infty$
OK.
Now I will prove : if sequence $ c_{n}$ is bounded for all $ n$ or $ \sup _{n \in \mathbb{N}} c_{n} < + \infty$ then  $ \lim_{n \to \infty} x_{n}$ exist  and is finite.
Let's $ \sup _{n \in \mathbb{N}} c_{n} = C < + \infty$
it's same as :
$ \sqrt [p]{b_{1}\sqrt [p]{b_{2}\sqrt [p]{.... b_{n - 1}\sqrt [p]{a_{n}}}}} \leq e^{C}$
or
$ a_{n} \leq e^{Cp^{n}}b_{1}^{ - p^{n - 1}}b_{2}^{ - p^{n - 2}}...b_{n - 1}^{ - p}$ for all $ n \in \mathbb{N}$
thus
\[{{{ x_{n} = \sqrt [p]{a_{1} + b_{1}\sqrt [p]{a_{2} + b_{2}\sqrt [p]{a_{3} + .... + b_{n - 1}\sqrt [p]{a_{n}}}}} \leq \sqrt [p]{e^{Cp} + b_{1}\sqrt [p]{a_{2} + b_{2}\sqrt [p]{a_{3} + .... + b_{n - 1}\sqrt [p]{a_{n}}}}}} \leq \sqrt [p]{e^{Cp} + b_{1}\sqrt [p]{e^{Cp^{2}}b_{1}^{ - p} + b_{2}\sqrt [p]{a_{3} + .... + b_{n - 1}\sqrt [p]{a_{n}}}}}} \leq \sqrt [p]{e^{Cp} + b_{1}\sqrt [p]{e^{Cp^{2}}b_{1}^{ - p} + b_{2}\sqrt [p]{e^{Cp^{3}}b_{1}^{ - p^{2}}b_{2}^{ - p} + .... + b_{n - 1}\sqrt [p]{a_{n}}}}}} \leq .... \leq \sqrt [p]{e^{Cp} + b_{1}\sqrt [p]{e^{Cp^{2}}b_{1}^{ - p} + b_{2}\sqrt [p]{e^{Cp^{3}}b_{1}^{ - p^{2}}b_{2}^{ - p} + .... + b_{n - 1}\sqrt [p]{e^{Cp^{n}}b_{1}^{ - p^{n - 1}}b_{2}^{ - p^{n - 2}}...b_{n - 1}^{ - p}}}}} = e^{C} \underbrace{\sqrt [p]{1 + \sqrt [p]{1 + \sqrt [p]{ + .... + \sqrt [p]{1}}}}}_{n}
\]
here is simple exercises : try to show that :
\[ \lim_{n \to \infty} \underbrace{\sqrt [p]{1 + \sqrt [p]{1 + \sqrt [p]{ + .... + \sqrt [p]{1}}}}}_{n} = A < + \infty
\]
thus we get :
$ x_{n} < e^{C}A < + \infty$ or $ x_{n}$ is bounded, but also monotone increasing , therefore  $ \lim_{n \to \infty} x_{n}$ exist and is finite.
OK.

thus we get :
in your particular case limit exist and is finite  [b] if and only if [/b] $ \sup_{n \in \mathbb{N}} |\frac {\ln a_{n}}{p^{n}} | < + \infty$

Now we need to find limit  of $ x_{n}$ hereI just repeat my steps in http://www.mathlinks.ro/viewtopic.php?search_id=1040848384&t=152113
suppose there exist  sequence $ \beta_{n} \geq 0$ for all $ n$ and  such as :
$ \beta ^{p}_{n} = a_{n} + b_{n} \beta_{n + 1}$

note that $ \beta_{1} = \sqrt [p]{a_{1} + b_{1}\sqrt [p]{a_{2} + .... + b_{n - 1}\beta_{n}}}$
and also :
$ x_{n} \leq y_{n} = \sqrt [p]{a_{1} + b_{1}\sqrt [p]{a_{2} + .... + b_{n - 1}\beta_{n}}} = \beta_{1}$
thus for all $ n \in \mathbb{N}$ we have $ x_{n} \leq \beta_{1}$

Now note that this inequality is true :
$ \sqrt [p]{a + bC} \leq \sqrt [p]{C} \sqrt [p]{a + b}$ for  ($ a,b \geq 0$ and $ C \geq 1$).
and for $ C = \frac {\beta_{n}}{\sqrt [p]{a_{n}}}$ inequality  give as
$ y_{n} \leq x_{n} \sqrt [p^{n}]{\frac {\beta_{n}^{p}}{a_{n}}}$ thus we obtain :
\[ \beta_{1}\sqrt [p^{n}]{\frac {a_{n}}{\beta_{n}^{p}}} \leq x_{n} \leq \beta_{1}
\]
thus $ x_{n} \to \beta_{1}$ provided $ p^{ - n} \ln \frac {a_{n}}{\beta_{n}^{p}} \to 0$ as $ n \to \infty$


AMC AIME quadratics MATHCOUNTS Vieta search

I, unfortunately, don't have the money to buy the AIME solutions, so I'm asking for help here.

(This is from the 1983 AIME)

3. What is the product of the real roots of the equation

x^2 + 18x + 30 = 2*sqrt(x^2 + 18x + 45)

I'm totally clueless as to how I should approach this problem. If I remove the square root by squaring both sides, the degree of the equation becomes 4 and there's no way I can solve that.

Perhaps I can factor both quadratic equations and work from there?

(-18 +- sqrt(324 - 4*30))/2

The problem with that is that sqrt(204) ends up being a non-integer answer. As you may know, the AIME only has integer answers.

Thanks!

Here's a hint:

Let v=sqrt(x^2 + 18x + 45) so v^2 = x^2 +18x + 45.  Keeping in mind that v can't be negative, your equation becomes v^2 - 15 = 2v.

That looks like it will be a lot easier to solve now.

Thanks. That should have been an easy problem.

J/w...I got 20 for that problem.  Tell me if I'm right or wrong.  Yea I pretty much did the substitution method for that problem.

By a mathcounts natl. alumni from 2003.  59th place oh yeah!   :)

let $\sqrt(x^2+18x+45) = a$ so then $a^2-15=2a$ so then we solve for $a$ getting ... 5 (or -4, but that's extraneous) then because $\sqrt(x^2+18x+45)=5)$ we get $x^2+18x+45=25$ subtracting 25 from both sides we use Vieta's to find that $x_1x_2 = 20$

[quote="hrithikguy"][hide]Let $y = x^2 +18x+30$
$y = 2 \sqrt{y + 15}$
$y^2 = 4y + 60$
$y^2 - 4y - 60 = 0 \Rightarrow (y+6)(y-10) = 0 \Rightarrow y = -6, 10$

However, $y > 0$, so $y = 10$

$x^2 + 18x + 20 = 0$

By Vieta's the answer is $\boxed {20}$[/hide][/quote]

Shouldn't the answer be 30 because $0=y=x^2 +18x+30= 2 \sqrt{y + 15}$

[quote="AlphaMath1"]It is never stated that $y=0$. In fact $y\ne 0$.[/quote]

Isn't that when we want the root ,then f(x)=0 is the root equation?

[quote="ejiant"]I, unfortunately, don't have the money to buy the AIME solutions, so I'm asking for help here.[/quote]
You can find solutions on this website. You could click on the link 'Contests', then go to USA, then AIME, and then you can find solutions from there. You could also click on 'Resources', then go to the AoPS Wiki, and then search AIME, look for a link that says past problems and solutions and then also find the solutions that way.


function
inequalities solved
inequalities

Let x, y, z be positive reals satisfying
      (x+y+z)^3 = 32xyz
Find the maximum and minimum of
     P = (x^4+y^4+z^4)/(x+y+z)^4

========

not the solution but ideas only :D
let x+y+z=1==> xyz=1/32
then we have to find the extrema of
2B 2 -4B + 9/8
WHERE B=xy+yz+zx
but damn it ,i find it damn tough
ne1 here who has got different ideas :(

My almost solution (since I didn't have the courage to take to the end all those huge computations) is the followimg:
   So, we have to find the values of  \sum x 2 , since then it's easy. Put x=a/4, y=b/4 and z=c/2. We find that abc=1 and a+b+2c=4. We must find the values of a 2 +b 2 +4c 2 . But a 2 +b 2 +4c 2 =(4-2c) 2 -2/c+4c 2 . Hence we must find all values of 4c 2 -8c-1/c when c is a positive such that there are positive reals a,b such that a+b=4-2c, ab=1/c. This comes down to c+1/ \sqrt c \leq 2. It folloes that c belongs to [(3- \sqrt 5)/2,1] . So we must study the function f(x)=4x 2 - 8x-1/x on this interval, which is messy.
   I'm sure Namdung has a very smart solution to this problem, since mine is...

hey harazi
i came xactly the same stuff yesterday as u wrote,
but would u call that solution messy ,i mean it seems pretty natural unless,of course u r not having sumthing divine in ur mind :D


function
limit
induction
inequalities
real analysis
real analysis unsolved

Hello !
Let $ f$ a  function defined on $ \mathbb{R}$ , $ a$ and $ b$ on $ \mathbb{R}$ such that :  $ 0<a\leq 1$ and $ 0<|b|<1$
We have : $ \lim_{x\rightarrow 0}f(x)\equal{}0$ and $ \lim_{x\rightarrow 0}\frac{f(x)\minus{}af(bx)}{x}\equal{}0$
Prove that : $ \lim_{x\rightarrow 0}\frac{f(x)}{x}\equal{}0$

If $ \displaystyle\mathop{\lim}\limits_{x\to0}\frac {f(x)}{x}$ exists, then, because $ |b|\leq1$,  $ \displaystyle\mathop{\lim}\limits_{x\to0}\frac {f(bx)}{bx}$ exists and also  $ \displaystyle\mathop{\lim}\limits_{x\to0}\frac {f(bx)}{bx} \equal{} \mathop{\lim}\limits_{x\to0}\frac {f(x)}{x}$.
Then
$ 0 \equal{} \displaystyle\mathop{\lim}\limits_{x\to0}\frac {f(x) \minus{} af(bx)}{x} \equal{} \mathop{\lim}\limits_{x\to0}\frac {f(x)}{x} \minus{} ab\mathop{\lim}\limits_{x\to0}\frac {f(bx)}{bx} \equal{} \mathop{\lim}\limits_{x\to0}\frac {f(x)}{x} \minus{} ab\mathop{\lim}\limits_{x\to0}\frac {f(x)}{x} \equal{} (1 \minus{} ab)\,\mathop{\lim}\limits_{x\to0}\frac {f(x)}{x}$.
Because $ |1 \minus{} ab| \leq 1$, we have $ \displaystyle\mathop{\lim}\limits_{x\to0}\frac {f(x)}{x} \equal{} 0$.
NOTICE: I did not use that $ \mathop{\lim}\limits_{x\to0}f(x) \equal{} 0$, but I assume that $ \displaystyle\mathop{\lim}\limits_{x\to0}\frac {f(x)}{x}$ exists. So I hope it helps somehow.

$ \lim_{x\rightarrow 0}\frac {f(x) \minus{} af(bx)}{x} \equal{} 0$
Let $ n\in \mathbb{N}$
Substitute for $ x$, $ b^nx$ and multiply by $ (ab)^n$
$ \lim_{x\rightarrow 0}\frac {a^nf(b^nx) \minus{} a^{n \plus{} 1}f(b^{n \plus{} 1}x)}{x} \equal{} 0$.
Then (telescopic sum)
$ \lim_{x\rightarrow 0}\frac {f(x) \minus{} a^{n \plus{} 1}f(b^{n \plus{} 1}x)}{x} \equal{} \lim_{x\rightarrow 0}\sum_{k \equal{} 0}^n \frac {a^kf(b^kx) \minus{} a^{k \plus{} 1}f(b^{k \plus{} 1}x)}{x} \equal{} 0$.
Let $ \epsilon > 0$.
Now $ \lim_{n\to \infty} a^{n \plus{} 1}f(b^{n \plus{} 1}x) \equal{} 0$, so there is an $ N\in\mathbb{N}$ so that for $ n\ge N$ and for $ x \in (\minus{}1,1)$: $ |a^{n \plus{} 1}f(b^{n \plus{} 1}x)| < \frac {\epsilon}{2}$
There is a $ \delta>0$ so that for $ |x|<\delta$ and for chosen $ n>N$ we have $ |\frac {f(x) \minus{} a^{n \plus{} 1}f(b^{n \plus{} 1}x)}{x}| <\frac{\epsilon}{2}$
and
$ |\frac {f(x)}{x}| \le |\frac {f(x)}{x}| \minus{} |\frac {a^{n \plus{} 1}f(b^{n \plus{} 1}x)}{x}| \plus{} \frac {\epsilon}{2} \le |\frac {f(x) \minus{} a^{n \plus{} 1}f(b^{n \plus{} 1}x)}{x}| \plus{} \frac {\epsilon}{2} < \epsilon$

[quote="grigkost"]NOTICE: I did not use that $ \mathop{\lim}\limits_{x\to0}f(x) \equal{} 0$, but I assume that $ \displaystyle\mathop{\lim}\limits_{x\to0}\frac {f(x)}{x}$ exists. So I hope it helps somehow.[/quote]
By assuming that $ \lim_{x\to0}\frac {f(x)}{x} \equal{} 0$ you're making an assumption that is much stronger than $ f(x)\to0.$

And you need at least $ \lim_{x\to0} f(x) \equal{} 0$ or else the problem is false. Just try letting $ f(x)$ equal a nonzero constant to see that.

We can use that :
$ \LARGE(\forall \epsilon>0)(\exists \alpha >0)(\forall x \in\mathbb{R}^{*}): 0<|x|<\alpha\Rightarrow\left|\frac{f(x)}{x}\right|<\epsilon$

No you can't use that. That's the statement that $ \lim_{x\to0}\frac{f(x)}{x}\equal{}0.$ In other words, it's the very thing you're trying to prove.

Mr Kent Merryfield:
have you other lanes to offer us
or the proof of  $ \LARGE(\forall \epsilon>0)(\exists \alpha >0)(\forall x \in\mathbb{R}^{*}): 0<|x|<\alpha\Rightarrow\left|\frac{f(x)}{x}\right|<\epsilon$

Let $ \epsilon > 0$ be given. Then We have:

1)   $ (\exists \alpha_{1} > 0)(\forall x \in\mathbb{R}^{*}): 0 < |x| < \alpha_{1}\Rightarrow|f(x)| < \frac {\epsilon}{2}$ (it is true for every $ \epsilon > 0$ so it must be true for this one.
Also since |b|<1, we have $ |bx|<|x|< \alpha_{1}\Rightarrow|f(bx)| <\frac{\epsilon}{2}$ (am I right here!).

2)   $ (\exists \alpha_{2} > 0)(\forall x \in\mathbb{R}^{*}): 0 < |x| < \alpha_{2}\Rightarrow|\frac {f(x) - af(bx)}{x}| < \frac {\epsilon}{2}.$

We must find an $ \alpha > 0: |x| < \alpha \Rightarrow|\frac {f(x)}{x}| < \epsilon.$

We know:
$ ||\frac {f(x)}{x}| - |\frac {af(bx)}{x}|| < |\frac {f(x)}{x} - \frac {af(bx)}{x}| < \frac {\epsilon}{2}\Rightarrow - \frac {\epsilon}{2} < |\frac {f(x)}{x}| < \frac {\epsilon}{2} + |a\frac {f(bx)}{x}|$
$ < \frac {\epsilon}{2} + |\frac {a}{x}|\frac {\epsilon}{2} < \frac {\epsilon}{2}(1 + |\frac {a}{\alpha_{1}(1 - |b|)}|) < \epsilon$ when $ \alpha_{1} = \frac {(1 - |b|)}{a}$ and $ \alpha = max(\alpha_{1}, \alpha_{2})$.

I am not sure if I am making sense here.

[quote="akech"]$ < \frac {\epsilon}{2} \plus{} \left|\frac {a}{x}\right|\frac {\epsilon}{2} < \ldots < \epsilon$ when $ \alpha_{1} \equal{} \frac {(1 \minus{} |b|)}{a}$ and $ \alpha \equal{} max(\alpha_{1}, \alpha_{2})$.

I am not sure if I am making sense here.[/quote]Hmm, unfortunately not, since you simply conclude that $ \frac{1}{x}$ is bounded around 0 - which does not seem to be very true to me. ;)

[quote="*210*"]Let $ f$ a  function defined on $ \mathbb{R}$ , $ a$ and $ b$ on $ \mathbb{R}$ such that :  $ 0 \& lt;a\leq 1$ and $ 0 \& lt;|b| \& lt;1$
We have : $ \lim_{x\rightarrow 0}f(x) = 0$ and $ \lim_{x\rightarrow 0}\frac {f(x) - af(bx)}{x} = 0$
Prove that : $ \lim_{x\rightarrow 0}\frac {f(x)}{x} = 0$[/quote][hide="Sketch of proof"]By induction prove that for $ g(x): =f(x)-af(bx)$ we have\[ \sum_{m=0}^{n-1}a^mg(b^mx)=f(x)-a^{n}f(b^{n}x)\] yielding\[ f(x)=\sum_{m\ge 0}a^mg(b^mx)\](pointwise limit). Given $ \varepsilon>0$, there is some $ \delta>0$ such that $ |g(x)|<\varepsilon|x|$ for every $ 0<|x|<\delta$. For these $ x$ it holds $ 0<|bx|\le|x|<\delta$ such that \[ |f(x)|\le\varepsilon|x|\sum_{m\ge 0}q^m=\frac{\varepsilon}{1-q}|x|\] with $ q: =|ab|<1$. Because $ q$ is independent of $ \varepsilon$, the claim follows.[/hide]

[quote="Third Edition"][hide=Sketch of proof][/hide][/quote]
What was wrong with my proof?

[quote="PeterP"]What was wrong with my proof?[/quote]
I must admit that I wasn't aware of it before.  :oops:  Now that I quickly went through it, what are the arguments for this inequality?

[quote="PeterP"]$ |\frac {f(x)}{x}| \le |\frac {f(x)}{x}| \minus{} |\frac {a^{n \plus{} 1}f(b^{n \plus{} 1}x)}{x}| \plus{} \frac {\epsilon}{2}$ [/quote]
In other words, how do you derive this with what you have prepared already: $ |a^{n \plus{} 1}f(b^{n \plus{} 1}x)|<\frac{\varepsilon}{2}|x|$?

Maybe I am missing something obvious here?


When the music stopped the five friends returned to there table and amused themselves by asking each other riddles. How many can you get?

The boy in red asked his first.
[b]Frank[/b]:Lask week I turned off the light in my bad room and managed to get to bed before the room was dark. If the bed was 10 feet from the light switch, how did I do that?

The boy in blue said:
[b]Henry[/b]:Whenever my cousin comes to visit me at the apartment she always gets off five floors too soon and walks up the rest of the way. Can you tell me why?

The girl in red said.
[b]Jane[/b]: One night my uncle was reading an exciting book when his wife turned out the light. Even though the room was pitch-dark he went right on reading. How could he do that?

The girl in green said.
[b]Mabel[/b]: This morning one of my ear-rings fell into my coffee. Even though my cup was full, the ring didn't get wet. How come?

The girl in blue asked the last riddle.
[b]Laura[/b]:Yesterday, my father was caught in the rain withouta het or umbrella. There was nothing over his hat and his clothes was got soaked. But not a hairon his head got wet? Why was that?

[hide="Frank"]It happened during the daytime, when the sun was out.[/hide]

[hide="Henry"]Henry's cousin is too short to reach the elevator button of Henry's floor.[/hide]

[hide="Jane"]Jane's uncle was reading an electronic book on a computer's lighted screen.[/hide]

[hide="Mabel"]The cup was filled with coffee beans, not liquid coffee.[/hide]

[hide="Laura"]Laura's father is bald.[/hide]

I got the same answers for Frank and Laura as cf249, but for Jane I got

[hide] Her uncle is blind, he is reading braille with his fingers.[/hide]


Every time I try to enter my grade into my USAMTS profile, it keeps going back to "none selected"...

What is going on here?

try switching your browser?
ex) mozilla firefox or google chrome or safari or IE or opera or Netscape Navigator?


conics
parabola
geometry
circumcircle
perpendicular bisector
angle bisector
geometry solved

A and B are two points on a parabola whose focus is S.

SA meets the line through B parallel to the axis of the parabola at U.

SB meets the line through A parallel to the  axis of the parabola at V.

Prove that a circle can be drawn to touch AU, UB,  BV and VA.

Nice problem!

We will denote by d(P; g) the distance from a given point P to a given line g.

Let w be the directrix of our parabola. The lines AU and BV are parallel to the axis of the parabola, and thus perpendicular to its directrix w. Let these lines AU and BV meet the directrix w at the points X and Y. Then, since $AU\perp w$ and $BV\perp w$, the points X and Y are the orthogonal projections of the points A and B on the line w, and thus, AX = d(A; w) and BY = d(B; w).

Let T be the circumcenter of the triangle SXY. Then, clearly, this circumcenter T must lie on the perpendicular bisectors of the sides SX, SY and XY of the triangle SXY.

But it is well-known that a parabola is the locus of all points equidistant from the focus and the directrix; thus, the point A, lying on our parabola, must be equidistant from the focus S and the directrix w of our parabola. In other words, AS = d(A; w). Since AX = d(A; w), we obtain AS = AX. In other words, the triangle SAX is an isosceles triangle with the base SX. Now recall the fact that the perpendicular bisector of the base of an isosceles triangle coincides with the angle bisector of the angle at the apex; thus, in the isosceles triangle SAX, the perpendicular bisector of the base SX coincides with the angle bisector of the angle SAX. Our former result that the point T lies on the perpendicular bisector of the segment SX can be now rewritten as follows: The point T lies on the angle bisector of the angle SAX. Hence, the point T is equidistant from the lines AS and AX; in other words, d(T; AS) = d(T; AX). Since the line AS coincides with the line VA, and the line AX coincides with the line AU, we thus can state that d(T; VA) = d(T; AU).

Similarly, we can prove d(T; UB) = d(T; BV).

Finally, since the lines AU and BV are both perpendicular to the line w and meet this line at the points X and Y, these lines AU and BV are just the perpendiculars to the line w at the points X and Y, respectively. Thus, the perpendicular bisector of the segment XY is the center line between the lines AU and BV. Any point on this perpendicular bisector must be thus equidistant from the lines AU and BV. Consequently, since the point T lies on the perpendicular bisector of the segment XY, it is equidistant from the lines AU and BV. In other words, d(T; AU) = d(T; BV). Together with d(T; VA) = d(T; AU) and d(T; UB) = d(T; BV), this yields d(T; VA) = d(T; AU) = d(T; BV) = d(T; UB). Thus, the point T is equidistant from the four lines AU, UB, BV and VA, and consequently, there exists a circle with center T which touches all these four lines AU, UB, BV and VA.

Problem solved.

  Darij


[quote="Klaas"]Find all pairs of prime numbers (p,q) for which number (p+1)^q is a square of natural number.

 :roll:[/quote]

Either $ q$ is even, and so $ q\equal{}2$, either $ p\plus{}1$ is a perfect square, and so $ p\equal{}(u\minus{}1)(u\plus{}1)$ and so $ u\equal{}2$ (since $ p$ is prime).

So two families of solutions :
$ (p,2)$ for any prime $ p$
$ (3,q)$ for any prime $ q$

But this seems not to be an olympiad problem.  :?:

[hide="list"][list]
(p,2)
(3,q)[/list][/hide]

@pco: Olympiad problems go in olympiad forum, usually.

Can you find all (p,q) such that $ (p\plus{}1)^q$ is a perfect 7th power?(Except q=7)

[quote="1=2"][hide="list"][list]
(p,2)
(3,q)[/list][/hide]

@pco: Olympiad problems go in olympiad forum, usually.

Can you find all (p,q) such that $ (p \plus{} 1)^q$ is a perfect 7th power?(Except q=7)[/quote]

If $ (p \plus{} 1)^q$ is a perfect 7th power, $ q\ne7$, therefore, $ p \plus{} 1 \equal{} m^7$

Hence $ m^7 \minus{} 1 \equal{} (m \minus{} 1)(m^6 \plus{} m^5 \plus{} \dots \plus{} 1) \equal{} p$

Since $ p \in \mathbb{P}$, either $ m \equal{} 2$ or $ p \equal{} 2$ ($ p\equal{}2$ yields no solutions)

Case $ m \equal{} 2$

$ p \equal{} 2^6 \plus{} 2^5 \plus{} ...$

This is a GP, $ a \equal{} 1, r \equal{} 2, n \equal{} 7$

Therefore $ p \equal{} \frac {1(2^7 \minus{} 1)}{2 \minus{} 1} \equal{} 127$

Hence $ (p,q) \equal{} (127,q)$

[quote="1=2"]@pco: Olympiad problems go in olympiad forum, usually.[/quote]
But easy ones like this better go to Intermediate or Pre-Olympiad.


algebra
polynomial
number theory proposed
number theory

Find all polynomial $ P(x)\in R[x]$ such that for all $ r\in Q$ then exist $ d\in Q$ such that $ f(d)\equal{}r$
Note :We knew that $ P(x)\equal{}ax\plus{}b$ where $ a,b\in Q$
But how many solution for this problem ?

Any pair $ (a,b)$ with $ a \neq 0$ gives a solution.
Please state your question more clearly.

[quote="TTsphn"]Find all polynomial $ P(x)\in R[x]$ such that for all $ r\in Q$ then exist $ d\in Q$ such that $ f(d) \equal{} r$
Note :We knew that $ P(x) \equal{} ax \plus{} b$ where $ a,b\in Q$
But how many solution for this problem ?[/quote]
This is the unique solution,except the case when $ P$ is constant.


AMC
USA(J)MO
USAMO
search
number theory unsolved
number theory

is there exist any set of integer numberslike X,inwichfor each N ,the equality of (a+2b=N)

has exactly one answer for a,b in X?

Yes, this is an old USAMO question. Use the search button ...

or consider the "base $-4$ representation".


Find all functions $ f: \mathbb{R}\to \mathbb{R}$ such that:
$ f\left(f\left(x\right) \plus{} y\right) \equal{} 2x \plus{} f\left(f\left(y\right) \minus{} x\right)$ for any two reals $ x$ and $ y$.

you can find this shortlist problem here:[url]http://www.kalva.demon.co.uk/short/soln/sh02a1.html[/url]

This is my first stab:

$x=0, y=x \Rightarrow f(f(0)+x)=f(f(x))$ (1)
$y=0 \Rightarrow f(f(x))= 2x+f(f(0)-x)$ (2)

Let f(0)=c and from (1) and (2) get $f(c+x) = 2x+f(c-x)$.
Define $g(x)=f(c+x) \Rightarrow g(x)-x = g(-x)-(-x)$.
Let $h(x)=g(x)-x$. The last equation shows that $h(x)$ can be any even function. Thus,
$f(x)=x-c+h(x-c)$ where h is an arbitrary even function.

I still need to determine c if possible. Is my work correct so far? I've checked my answer for the trivial case of h(x)=0.

I know, Lordwings, that I haven't yet checked your favorite possibility.

[quote="mathisfun1"]I know, Lordwings, that I haven't yet checked your favorite possibility.[/quote]
I wonder why you told that.
Guilty conscience? :rotfl:
[quote="mathisfun1"]The last equation shows that $h(x)$ can be any even function.[/quote]
It only shows that $h(x)$ must be an even function (it can't be a solution that is not an even function).
Solving functional equations is like an elimination game, but at the end you must prove that all your final candidates are really solutions.

$f(f(x)+y)=2x+f(f(y)-x)=2x+2y+f(f(-x)-y)=2y+f(f(-y)+x)$.
Therefore (y=f(-x) give) $f(f(x)+f(-x))=2g(x)+3c$, were $c=f(0),g(x)=f(x)-x-c, \ g(0)=0$.
It give $g(x)=g(-x)$. Let $z=x+y-c$ then $f(f(x)+y)=f(g(x)+z)=g(x)+z+c+g(g(x)+z))$ and $2z+f(f(-x)-y)=2z+f(g(-x)-z)=2z+g(-x)-z+c+g(g(-x)-z)$.
Because $g(x)=g(-x)$, it give $g(g(x)+z)=g(g(x)-z)$ for all x,z.
If exist b, suth that $a=g(b)\not =0$, then $g(2a)=0$.
We have 2 symmetric property:
$g(x)=g(-x),g(0)=0$, and $g(a+x)=g(a-x)$. Therefore $g(2a+x)=g(a-(a+x))=g(-x)=g(x)$.
If $g(x)$ is continiusly, we have for all t: $t\in (0,a)$ exist x $g(x)=t$. Therefore $g(x+2t)=g(x) \ \forall x\in R,t\in (0,a)$. It is impossible.
Therefore if f(x) is continiosly, then $f(x)=x+c$.
I think this equation had infinetely many noncontiniosly solutions.


quadratics
modular arithmetic
algorithm
number theory unsolved
number theory

Shown below is exercise 10.19.3 in Uspensky and Heaslett's Elementary Number Theory. Chapter 10 introduces Legendre and Jacobi symbols, proves the quadratic reciprocity law, and provides attack methods for quadratic congruences of prime moduli (among other things).  The problem is presented verbatim and then interpreted, to facilitate snaring the authors' insights.  My flawed approach, which might be repairable, is also included.  Request help, thanks.

Verbatim:
Devise a method for solving the congruence $ x^2 \equiv a\pmod{p}$ if the prime $ p \equiv 1\pmod{8}$ and one quadratic nonresidue of $ p$ is known.  Solve the congruence $ x^2 \plus{} 7\equiv 0\pmod{281}$ by using the fact that 3N281...

Answer:  $ x\equiv \pm 67\pmod{281}.$

Interpretation:
The book was written in the 1930's, which predates computers.  It references, for example, "A Photo Electric Number Sieve", so computer/calculator use in the (intended) method should be "light."  Just before presenting this problem, the chapter concludes with sections 18 and 19.  Section 18 discusses solving quadratic congruences for prime moduli congruent to 3,5, or 7 (mod 8).  In this situation, one of $ \frac{p\minus{}1}{2}, \frac{p\minus{}1}{4}$ will (critically) be odd. Section 19 converts $ x^2 \equiv a\pmod{p}$ to $ a \plus{} py \equal{} x^2$ and then attacks "excluding" _virtually_ all values of $ y$ between 0 and $ \frac{p\minus{}1}{4}.$

The exclusion method is presented in a separate post-chapter exercise.  Based on that + the verbatim problem, I suspect that the authors' intend subtly altering the method of section 18.

My flawed approach:
Solve $ x^2 \equiv a\pmod{\text{prime}\; p \equal{} 8n\plus{}1},$ given that $ g$ is a quadratic nonresidue of $ p$:
Let $ e$ denote $ g^{\minus{}1}a\pmod{p}$ and let $ \alpha$ denote the index of $ g$ with respect to $ e \Rightarrow$
[since $ g$ is a nonresidue]  $ \alpha$ is odd and $ a \equiv e^{\alpha \plus{} 1}\pmod{p} \Rightarrow
x \equiv \pm e^{(\alpha \plus{} 1)/2}.$

In particular, with $ p,a,g \equal{} 281, \minus{}7, 3,$ respectively,
$ e \equal{} 94(\minus{}7) \equiv 185\pmod{281}$ and $ 185^{(41)} \equiv 3\pmod{281} \Rightarrow
x \equiv \pm 185^{(21)} \equiv \pm 67\pmod{281}.$

This approach's flaw is that it is unproven/undetermined whether $ g$ is in the orbit of $ e.$  I also tried unsuccessfully to use the following:
Suppose $ (p\minus{}1) \equal{} 2^{\beta \plus{} 3}\times (\text{odd}\; r)$ and $ g$ belongs to the exponent $ \gamma.$  Then $ 2^{\beta \plus{} 3}$ divides $ \gamma.$

In another math forum, someone referred me to the link below, which seems to be directly on point.

http://en.wikipedia.org/wiki/Shanks-Tonelli_algorithm

See also p10-11 (of 14 pages) of the link below.  This portion of the pdf proves that each iteration of the Shanks-Tonelli Algorithm moves "closer" to the root of $ x^2 \equiv a\pmod{p}.$

http://www.math.vt.edu/people/brown/doc/sqrts.pdf


Solve the equation: $\left(\frac{x}{x+2}\right)^{2}=3x^{2}-6x-3$


geometry
geometric transformation
rotation
combinatorics unsolved
combinatorics

I saw in Engel's Problem-Solving Strategies some proof to prove F.L.T (Fermat Little Theorem).

The proof is about the number of distinct necklaces that can be done with $ n$ pearls that may be colored with $ a$ colors such that not all the pearls in the necklace are the same color. Since there are $ n$ places to put the pearls and $ a$ options for every place, there are $ a^n$ necklaces. Discarding the $ a$ necklaces with all pearls in one color, we get $ a^n \minus{} a$ necklaces such that not all the pearls are the same color. Since there are $ n$ rotations of the necklace, the number of distinct necklaces is $ (a^n \minus{} a)/n$ which is an integer because of its interpretation. Then $ a^n$ is congruent to $ a$ modulo $ n$. But... WHERE DID WE ASSUME THAT $ n$ WAS PRIME?? Isn't it an integer even if $ n$ is composite?

[quote="Pul de Algodoncito"]there are $ n$ rotations of the necklace[/quote] This is not true for $ n$ composite -- consider, for example, the necklace RBRB.

[quote="pythag011"]Fermat's Little Thereom is true only for prime n.[/quote]

Yes, this is exactly the question the OP was addressing...why the proof he posted doesn't work for composite n.


I didn't take rounds 1 or 2. Can i still take round 3?

Yep  :D

[quote="usamts.org"]Round 3 is now available!
If you did not participate in Round 1 or Round 2, you can still participate in Round 3.[/quote]


function
inequalities proposed
inequalities

f:R - R/    x, y  are  real numbers,  Find all  functions such that
f[ x + f ( xy ) ] = f( x ) + x f( y )

I think you should post this in the Algebra section, even if it does involve some inequalities.

must  i  choose  section?? i  think it depends  on me. i noticed  it. if  you can try solve  it   :D


algebra
polynomial
real analysis
real analysis unsolved

Find the radius convergence of $\sum \frac{z^{n}}{\sin(n\pi x)}$ where x an irrational number

It heavily depends on how well $x$ is approximable by rationals. For almost all $x$ the speed is at most polynomial, so the answer is $1$, but you can easily find $x$ for which it is any given number in $[0,1)$ as well. So, what are we expected to answer here? :?


AMC
AMC 8
AMC 10
AIME
algebra
polynomial
counting

Solve a problem, PM me the solution, and get the next question. Simple.

This tournament consists of 20 questions:
#1-5 are approximately Sprint Level
#6-8 are approximately Target Level
#9-11 are approximately Team Level
#12-14 are approximately AMC8-10 Level
#15-17 are approximately AMC10-12 Level
#18-20 are approximately AIME level

If you get a problem wrong you have to wait 1 hour before submitting another answer.
Also, once someone reaches at or above Team Level, those three levels behind will be eliminated from the competition.

To join, just PM me the answer to question 1.

Question 1:
Evaluate the following:
$ \frac{{1\over2}\plus{}{1\over3}}{{1\over6}}$

Current Leaderboard:
Question 3: mewto55555, high_flyer10
Question 2: Twin Prime Conjecture, 007math
Question 1: everyone else

Question 9 must be reached in order to eliminate everyone not currently playing.

Current Leaderboard:
Question 5: mewto55555
Question 4: Twin Prime Conjecture, high_flyer10
Question 3: no one
Question 2: 007math, Discrete_Math
Question 1: everyone else

Question 9 must be reached in order to eliminate everyone not currently playing.

Current Leaderboard:
Question 6: Twin Prime Conjecture, mewto55555
Question 5: no one
Question 4: high_flyer10
Question 3: no one
Question 2: 007math, Discrete_Math, AIME15, isabella2296, mz94, FantasyLover
Question 1: everyone else

Question 9 must be reached in order to eliminate everyone not currently playing.

Current Leaderboard:
Question 7: mewto55555
Question 6: Twin Prime Conjecture
Question 5: no one
Question 4: high_flyer10, FantasyLover, arb95, mz94
Question 3: mathemagician1729, AIME15
Question 2: 007math, Discrete_Math, isabella2296, ghjk, xpmath
Question 1: everyone else

Question 9 must be reached in order to eliminate everyone not currently playing.

Please put some marking for the question number.

Re: Question 1 for question 3 is not very helpful for compiling the leaderboard.

If this leaderboard is incorrect please correct me. It will appear on the next leaderboard.

Current Leaderboard:
Question 9: Twin Prime Conjecture
Question 8:
Question 7: mewto55555
Question 6: AIME15
Question 5:
Question 4: high_flyer10, FantasyLover, arb95, mz94
Question 3: mathemagician1729, suma_milli
Question 2: 007math, Discrete_Math, isabella2296, ghjk, xpmath, AIME_is_hard

Question 1 Submissions - CLOSED!!!

Current Leaderboard:
Question 10: Twin Prime Conjecture
Question 9:
Question 8:
Question 7: mewto55555
Question 6: AIME15, 007math
Question 5:
Question 4: high_flyer10, FantasyLover, arb95, mz94
Question 3: mathemagician1729, suma_milli, isabella2296, AIME_is_hard
Question 2: Discrete_Math, ghjk, xpmath

Once TPC reaches Question 12, all those people before Question 5 will lose! Hurry!

Also, by that time questions 2 to 5 will be released for public interest.

Current Leaderboard:
Question 10: Twin Prime Conjecture
Question 9:
Question 8:
Question 7: mewto55555
Question 6: AIME15, 007math, FantasyLover
Question 5:
Question 4: high_flyer10, arb95, mz94
Question 3: mathemagician1729, suma_milli, isabella2296, AIME_is_hard
Question 2: Discrete_Math, ghjk, xpmath

Once TPC reaches Question 12, all those people before Question 5 will lose! Hurry!!!

You can edit posts you know too..instead of clogging up topic...

Current Leaderboard:
Question 10: Twin Prime Conjecture
Question 9:
Question 8:
Question 7: mewto55555, FantasyLover
Question 6: AIME15, 007math, arb95, mathemagician1729, ghjk, mz94
Question 5:
Question 4: high_flyer10
Question 3: suma_milli, isabella2296, AIME_is_hard, lotsofmath
Question 2: Discrete_Math, xpmath
Remember, Question 1 signups are closed.

[hide="As of 5:35 Pacific:"]Current Leaderboard:
Question 10: Twin Prime Conjecture
Question 9:
Question 8:
Question 7: mewto55555, FantasyLover, mathemagician1729, ghjk
Question 6: AIME15, 007math, arb95, mz94
Question 5:
Question 4: high_flyer10
Question 3: suma_milli, isabella2296, AIME_is_hard, lotsofmath
Question 2: Discrete_Math, xpmath
[/hide]

[quote="007math"][quote="james4l"][b]Note for Question 7: You need to find only ONE VALID PAIR. You must also show that it works.[/b]

Current Leaderboard:
Question 12: Twin Prime Conjecture
Question 11:
Question 10:
Question 9: FantasyLover
Question 8:
Question 7: mewto55555, arb95, mathemagician1729, mz94
Question 6: AIME15, 007math, ghjk

EDIT: TPC has reached question 12! high_flyer10, suma_milli, isabella2296, AIME_is_hard, lotsofmath, Discrete_Math, xpmath are now out...

[hide="Questions 2-5 for public interest"]Question 2
Determine the maximum value for the following polynomial: -4x^2+3x+2.

Answer: 41/16 (2 9/16 or 2.5625 are also accepted)

Question 3
Given 3 boxes: red, green, and blue, and 6 indistinguishable balls: 3 red, 2 green, and 1 blue, find the number of combinations such that no ball goes into a box of the same color.

Answer: 24

Question 4
How many squares divisible by 20 are there below 1000000?

Answer: 100

Question 5
Given that 5x = 3n+16 and that n,x are integers, what is x mod 3?

Answer: 2
[/hide][/quote]

Wait a second..I am on Q7!![/quote]
I'm on Q7 either!
By the way, do we have to show our work for this question?

[b]Note for Question 7: You need to find only ONE VALID PAIR. You must also show that it works.[/b]

Current Leaderboard:
LEG ONE
Question 20:
Question 19:
Question 18 [eliminate q12 and before]:
Question 17:
Question 16:
Question 15 [eliminate q8 and before]:
Question 14:
[b]Leader - Question 13: FantasyLover[/b]
Question 12: Twin Prime Conjecture
Question 11:
Question 10: arb95, mathemagician1729, mewto55555
Question 9:
Question 8:
Question 7: mz94,  007math, ghjk
Question 6: AIME15

Once somebody reaches question 15 those before question 9 will be out!

[hide="Questions 2-5 for public interest"]Question 2
Determine the maximum value for the following polynomial: -4x^2+3x+2.

Answer: 41/16 (2 9/16 or 2.5625 are also accepted)

Question 3
Given 3 boxes: red, green, and blue, and 6 indistinguishable balls: 3 red, 2 green, and 1 blue, find the number of combinations such that no ball goes into a box of the same color.

Answer: 24

Question 4
How many squares divisible by 20 are there below 1000000?

Answer: 100

Question 5
Given that 5x = 3n+16 and that n,x are integers, what is x mod 3?

Answer: 2
[/hide][/quote]

no, since the least penalty you can get is penalty 1...

This is a tiebreak round for:
1) pythag011 - [35]
6) xpmath - [35]
4) AIME15 - [30]
5) Bugi - [30]
9) dragon96 - [25]

SUBMIT!

Also, these seem strangely difficult for MATHCOUNTS...:)

Well I did say that this was proof-based...

Ok, change of plan:

Your tiebreak round will be of a 5-problem round 1-ish type contest.

Problems: Question 1 is worth 3, then 4, then 5, then 6, then 7.

No chance of redoing problems, justification required.

Top 3 to reach 50 points advance.

1. In right triangle ABC, B is the right angle. Find the sum of the squares of the distances of the angle trisector trisecting B, given the sides of the triangle are 1, rt(3), and 2.

2. You have a cone of height 6 and diameter 6. Find the elliptical crossection formed when a plane, cutting the cone into two pieces of equal volume, intersects the cone.

3. Given a bag with twenty numbers (1~20), find the expected value of the sum of the first 5 picked numbers (without replacement).

4. Determine the number of permutations of the word "HEPTAGON" such that at least one letter is still in its original position.

5. Find the probability such that pythag011 qualifies after tiebreaker given that everyone writes down a random answer to each question, but due to the 1337ness of AoPSers, each random answer has 50% of being correct!

Ahem...?

This is a tiebreak round for:
1) pythag011 - [35]
6) xpmath - [35]
4) AIME15 - [30]
5) Bugi - [30]
9) dragon96 - [25]

[quote="dragon96"]Wait, why are there proofs in Middle School Tournaments? And they appear to be insanely difficult to an average 6-8th grader.[/quote]

Right, but I don't think anyone participating is an average 6-8th grader...

AIME15 was the one who said he was an average 6-8 th grader (SO NOT TRUE!!! if anyone's *below average* thats me!!). :juggle:  :surf:  :surf:

[quote="james4l"]BTW, due date = March 14.

Clarification on #2: the plane must intersect the base on the circle (the base perimeter)
Clarification on #5: Assume pythag011 and xpmath have 35 pts, AIME15 and Bugi have 30 pts, and dragon96 has 25 pts. Then find the probability that pythag011 ranks among the first three people. Point values (3,4,5,6,7) apply.

*cough* pythag011 *cough* you are not *cough* an average 7th *cough* grader *cough*[/quote]

Yes I am...

So how does the qualification for the amazing race thing work?

You need 50pts to reach the next round...


Let $ ABCD$ a rectangle. Prove that for any point $ P$

$ \overline{PA}^2\plus{}\overline{PC}^2\equal{}\overline{PB}^2\plus{}\overline{PD}^2$.

[quote="LoboSolitario"]Let $ ABCD$ a rectangle. Prove that for any point $ P$

$ \overline{PA}^2 \plus{} \overline{PC}^2 \equal{} \overline{PB}^2 \plus{} \overline{PD}^2$.[/quote]

$ (x\minus{}a)^2\plus{}(y\minus{}1)^2\plus{}(x\plus{}a)^2\plus{}(x\plus{}1)^2\equal{}(x\minus{}a)^2\plus{}(y\plus{}1)^2\plus{}(x\plus{}a)^2\plus{}(y\plus{}1)^2$ ;)

[quote="LoboSolitario"]What means $ x, a$ and $ y$?[/quote]
$ let A(a,b) B(a,d) C(c,d) D(c,b) P(x,y)$
$ (a\minus{}x)^2\plus{}(b\minus{}y)^2\plus{}(c\minus{}x)^2\plus{}(d\minus{}y)^2\equal{}(a\minus{}x)^2\plus{}(d\minus{}y)^2\plus{}(c\minus{}x)^2\plus{}(b\minus{}y)^2$

Thanks Chris. I was just wondering for a solution without coordinates, but i think there's not an easy one (not like this  :D).

[quote="LoboSolitario"]What means $ x, a$ and $ y$?[/quote]

\begin{align*}
P&=(x,y)\\
A&=(a,1)\\
B&=(a,-1)\\
C&=(-a,-1)\\
D&=(-a,1)
\end{align*}

Draw the perpendicular from $ P$ to one of the sides of the rectangle and use pithagoras.

 :)

This is known as the [[British Flag Theorem]]. If P is on the border or on the outside, it is an extension of the BFT.

mmm..... bacon fettuccinni tomato...


my school doesn't have AMC 8 test,  Any1 knows the site in San Ramon, CA area ?

Thanks

If you don't mind the drive, you can (usually) take it at Harvest Park Middle School in Pleasanton.

Thanks,  I can have the test there even I am not the student of that school ?  Do u know who I can contact with ?

Contact Mr. Randy Lomas - his email is [url=http://206.110.20.121/faculty/teachers/files/page14_1.htm]here[/url].

I remember a student last year took the AMC 8 even though he didn't attend the school.

I'm a home schooler.  I'v been taking the AMC 8 at different random places for the last 3 years, but I have two more years still to take it.  Can't wait till november I think I might actually do really good, thanks to AoPS.


MATHCOUNTS
algebra
binomial theorem
Pascal\u0027s Triangle

You are given the binomial $x+y$. Raise it to the 10th power.

Find:

1. The coefficient of the $x^{4}y^{6}$ term.

2. The sum of the coefficients.

3. Try to find an extension to this problem.

[size=75][color=red]Please keep topics in the MATHCOUNTS Forum to a MATHCOUNTS level.  The Binomial Theorem and the Binomial Expansion are not MATHCOUNTS-level topics.  Thread moved from MATHCOUNTS to High School Basics.  -- moderator rcv[/color][/size]

[quote="Arvind_sn"]You are given the binomial $x+y$. Raise it to the 10th power.

Find:

1. The coefficient of the $x^{4}y^{6}$ term.

2. The sum of the coefficients.

3. Try to find an extension to this problem.[/quote]
Isn't this just the binomial theorem?
[hide="anyways"]
$(x+y)^{k}=\sum^{k}_{n=0}\binom{k}{n}x^{n}y^{k-n}$
So the coefficient of the $x^{4}y^{6}=\binom{10}{4}=210$.
Now plug in x=1, y=1.
$2^{10}=\sum^{10}_{k=0}\binom{10}{k}$, so the sum of the coefficients is 1024.[/hide]

[hide]
1. The coefficient of the $x^{4}y^{6}$ term is equal to $\binom{10}{6}=210$

2. Obvious Lemma:
Given 1 1 is the 1st row of the pascal triangle (which is basically a table of binomial expansion values), the sum of the values in the nth row is $2^{n}$
So the sum of the coefficients is $2^{1}0=1024$

3. Meh, I'm too lazy to think up extensions.
[/hide]

Extension: what if it was $x+y+z$ raised to the 10th power?

Well there are 12C2=66 terms..but I dont know how to add the coefficiants.

If its (a+b+c)^10 sum of coefficiants, I know there are 66 terms but I dont know how to add.

I'm going to take an educated guess.
Since sum in a row in 2-d pascal's triangle (for binomials) =$2^{n}$
I'm guessing sum in a "row" for 3-d pascal's triangle (for trinomials)=$3^{n}$
So my guess is $59049$

For each $(a+b+c)$, there are0 3 choices: a,b, or c, which gives $3^{n}$ as the sum of coefficients in $(a+b+c)^{n}$

[url]http://en.wikipedia.org/wiki/Multinomial_theorem[/url] has a good generalization

For more information on the answers for the original problem, see
[url]http://www.artofproblemsolving.com/Wiki/index.php/Pascal%27s_Triangle[/url]
and
[url]http://www.artofproblemsolving.com/Wiki/index.php/Binomial_theorem[/url].

[quote="CircleSquared"]Yay!

Lemma:
The sum of the coefficients of an expanded $n$-nomial to the $m$th power is $n^{m}$[/quote]

Just a suggestion:
It's good that you come up with these patterns and what not but don't just sit on them, actually try to prove them because it's pretty useless in the long run if you don't know why they work

[quote="CircleSquared"]Umm I'm not really good at proofs of that kind...  I just remember the lemma  :oops:
[hide="hint for anyone who wants to try"]
Each set of n-nomials can be represented by an n-dimensional Pascal's triangle.
[/hide][/quote]

[hide="Better yet, instead of trying to sum the rows of a multidimensional Pascal triangle..."] Consider the n-nomial $(a_{1}+a_{2}+a_{3}+\cdots+a_{n})^{m}$, where $a_{i}$ are different variables. Now expand it out. If we let $a_{1}=a_{2}=\cdots=a_{n}=1$, then each of the $a_{i}$ parts of each term will become 1, which won't change the coefficient of that term. But after we get rid of all the $a_{i}$ parts, all we have left is a sum of coefficients. So the sum of the coefficients can be found by setting $a_{i}=1$ for all $0<i\leq n$. But we know that this is equal to $(1+1+1...1)^{m}=n^{m}$. Thus the sum of the coefficients of an $n$-nomial raised to the $m$th power is $n^{m}$.[/hide]

Now then, binomial expansions usually belong in high school basics...

Billy


hallo,
this is an easy one. find all functions $f: R\to R$ such that:
(1) for all $x,y \in R$ we have \[f(xy)=f(x)f(y)\] (2) for all $x \in R-\{0\}$ we have \[f(1+x)=f(1)+f(x)(1+\frac{2}{x}).\] :)

Solution for both conditions at once:

Putting $y=1$ into (1) we find either $f(x)\equiv 0$, which satisfies (2) as well and therefore is a solution, or $f(1)=1$ for $f(x)\not\equiv 0$. Hence

$f(1+x)=1+f(x)\left(1+{2\over x}\right)$

Putting $A=f(-1)$, from (1) we get $f(-x)=Af(x)$, therefore

$f(1-x^{2})=1+f(-x^{2})\left(1-{2\over x^{2}}\right)=1+Af^{2}(x)\left(1-{2\over x^{2}}\right)\qquad(*)$

and also

\begin{eqnarray*}f(1-x^{2}) &=& f(1-x)f(1+x)\\ &=& \left(1+f(-x)\left(1-{2\over x}\right)\right)\left(1+f(x)\left(1+{2\over x}\right)\right)\\ &=& 1+Af(x)\left(1-{2\over x}\right)+f(x)\left(1+{2\over x}\right)+Af^{2}(x)\left(1-{4\over x^{2}}\right)\qquad(**)\end{eqnarray*}

Equating $(*)$ and $(**)$ we get

$A\left(1-{2\over x}\right)+1+{2\over x}=Af(x){2\over x^{2}}\iff f(x)=\frac{A(x^{2}-2x)+x^{2}+2x}{2A}$

Putting $x=-1$ we get $A=\frac{3A-1}{2A}\iff A=1\lor A={1\over 2}$

Putting $x=1$ we get $1=\frac{-A+3}{2A}\iff A=1$

Therefore $A=1$ which gives $f(x)=x^{2}$, and that function obviously satisfies both initial conditions. Hence all the solutions are $f(x)=0$ and $f(x)=x^{2}$.

[quote="silouan"]You posted two different problems or it is one problem with two conitions ?[/quote]
hallo,
it is one problem with two conditions  :)


Do you want to have a plaque from the AMC for the highest AMC 10B scroe in NC?

You better take the AMC10 B this coming Wednesday. You do not have many strong competitors this time.

heh I'm taking 12B after failing 10A

and there is Enloe taking B's...

I'm taking the 10B in that city called AMCVILLE, I mean, it's just awesome.

I failed 10A, and hopefully I won't fail 10B?


ratio
geometry
similar triangles
geometry proposed

[img]http://img182.imageshack.us/img182/6407/colinealdq1.png[/img]

$\overline{AQ}$, $\overline{BP}$ and $\overline{CR}$ are external bisectors from angles $A$, $B$ and $C$. Prove that $\measuredangle \ BQP+\measuredangle \ BQR=180^\circ$

I believe an application of menelaus' theorem is in order, the ratio AR:RB can be determined by similar triangles once you draw a parallel through R to AC, in terms of the sides of the triangle.

[i am not using directed distances...]

let the parallel through R to AC intersect BC at X, then we use the angle bisector/parallel line to get $\angle XRC=\angle XCR$, so $XC=XR$,

now since $\triangle ABC\sim\triangle RBX$, we have

$\frac{c}{RB}=\frac{a}{RX-a}=\frac{b}{RX}=\frac{b-a}{a}$
(where last step is using proportionality rules)

hence, $RB=\frac{ac}{b-a}$, then $\frac{AR}{RB}=\frac{ c+\frac{ac}{b-a}}{\frac{ ac}{b-a}}=\frac{bc-ac+ac}{ac}=\frac{b}{a}$

it is evident that $\frac{AR}{RB}*\frac{BQ}{QC}*\frac{CP}{PA}=\frac{b}{a}*\frac{c}{b}*\frac{b}{a}=1$

so P, Q, R are collinear by menelaus' theorem

That's a idea...

C'mon, solve it  :ninja:


modular arithmetic
AMC
AIME
number theory

For how many ordered pairs of positive integers $ (x,y)$ is $ x \plus{} 2y \equal{} 100$?

$ \textbf{(A)}\  33 \qquad \textbf{(B)}\  49 \qquad \textbf{(C)}\  50 \qquad \textbf{(D)}\  99 \qquad \textbf{(E)}\ 100$

[hide="Solution"]B,

Notice that for odd numbers x, this equation doesn't work. Also, for each number y from 1 to 50, there is an x so that x+2y=100. Thus, the solutions are y=1,2,3,...,49 and there are 49 solutions.

y=0 and x=0 are excluded.[/hide]

[hide]First, put the equation in y equals form:
y = -1/2x + 100
Now, we know all even numbers work until 50, when y = 0, but since y has to be a positive integer it is actually 49, B (right?)[/hide]

[hide]Neither x or y can be 0, so y can be any number from 1 to 49. [b]B[/b][/hide]

I really don't understand these graph type questions much. :blush:

[hide="more formal modular arithmetic solution"]
$x+2y=100$
Taking it mod 2, we get
$x\equiv 0\pmod{2}$
So $x$ is even

Then do what 4everwise did after that[/hide]

I love how 4everwise posts a problem, waits 5 minutes, no one posts a solution, then he posts one himself.

[quote="13375P34K43V312"]

I love how 4everwise posts a problem, waits 5 minutes, no one posts a solution, then he posts one himself.[/quote]

I think he might be posting these for the resource section because of the format and when you post a problem into the resource section it asks for a post number for the solution to the problem...

Yeah that's what I used to do too when I was posting AIMEs and everyone would be like "huh?? why do you post the solution like 5 minutes after you post the problem????"

X and Y have to both be positive and a positive + a positive can NEVER equal a negative number it just can't.
And Don't revive old topics it is annoying and can kill your post rating.

[quote="TennisVictor"]I don't understand why x and y cannot equal 0?[/quote]

Obviously $ 0$ is not positive. But the question stipulated

[quote="4everwise"] ordered pairs of[b] positive [/b]integers $ (x,y)$[/quote]

Does that answer your question?  :)

[quote="monkeygirl13"]And Don't revive old topics it is annoying and can kill your post rating.[/quote]

Actually, it is ok to revive old topics if you have something to add to the discussion, particularly problems that appear in the resources section of the website since people will continue to view those topics long after their initial creation.


Let $\triangle ABC$ be a triangle. Denote $a=BC,\ \  b=CA,\ \  c=AB,\ \  \alpha = \angle A,\ \  \beta = \angle B, \ \  \gamma = \angle C$. Show that if
\[ b\cos \left( {\frac{{\alpha  - \gamma }}{2}} \right) - 3c\cos \left( {\frac{{\alpha  + \gamma }}{2}} \right) = 0\]
then $a=2c$.

\[
\begin{array}{l}
 b\cos \left( {\frac{{A - C}}{2}} \right) = 3c\cos \left( {\frac{{A + C}}{2}} \right) \\
 \sin B\cos \left( {\frac{{A - C}}{2}} \right) = 3\sin C\cos \left( {\frac{{A + C}}{2}} \right) \\
 2\cos \frac{B}{2}\cos \left( {\frac{{A - C}}{2}} \right) = 3\sin C \\
 2\sin \left( {\frac{{A + C}}{2}} \right)\cos \left( {\frac{{A - C}}{2}} \right) = 3\sin C \\
 \sin A + \sin C = 3\sin C \\
 \sin A = 2\sin C \\
 a = 2c \\
 \end{array}
\]


Well, whaddaya know, it's begun. Confirmations received from everyone.
Here is a list of everyone playing in this game.

Ars
Petervdd
Kheg_k
Slamchess
Sunny
Supernova
Pianoforte
FibFreak
S2
Dts
Wobster109
GoBraves
Ladykn1ght
Lasgnaman
Melissa
Ignite168

Here are the roles:

4 Mafia
3 Citizens
2 Angels
1 Inspector
1 Vigilante
1 Prosecutor
1 Fairy
1 Priest
1 Devil
1 Mason

And the explanation.
[hide]Mafia make one kill each night. You may confer in the first day, after that, you may ONLY talk in the night.

Angels decide on ONE person to save, if he is targeted by mafia or vigilante, he will be saved.

Inspector chooses on person to inspect each night. If said person is mafia, miller, or devil, inspector will be told he has made a successful inspection. Otherwise, unsuccessful.

Vigilante is innocent. He makes a night kill each night.
The Fairy is innocent. He has the power to automatically removing the power of vigilante for that night.
Vigilante kill may be stopped by angel, even if fairy does not remove his powers.

Prosecutor inspects one person each night. If said person is vigilante or miller, prosecutor will be told that he has made a successful inspection.

Priest inspects one person each night. If said person is angel, inspector, or devil, (not miller) he will be told that he has made a successful inspection. Otherwise, unsuccessful.

Devil is the same, except for angel, inspector, or priest. (He is guilty, by the way)

One Citizen will be a miller, meaning he will show up as guilty to Inspector and prosecutor. However, he is an innocent.

The Mason has the power of induction. At one night he chooses, he may induct any one person. After seeing a confirmation on his induction, he will be allowed to converse with the inductee. The mason will be told the inductee's other role, and the inductee will be told who the mason is. The mason and the inductee may confer through PM the night of induction, the first day of induction, and every night after that.

(Note, the only true guilties in this game are the mafia and the devil)[/hide]

If they are any questions, ask them within the next 20 hours. If I do not receive any questions before then, I will send out the role PMs, and the game will begin. It will be day.
Same with any remarks. If you don't like my roles, say something. I want your opinion.

Are mafia not allowed to talk at all during the day? Or just not allowed to pm each other during the day?

I have no preference on the mason issue.

Mafia may not PM in the day. They may talk.

I would add a person, and put a mason in the game.  I personally like having masons around.

Sorry to post in this topic, but then wouldn't everybody know who that mason is as it's the one that's added?

No. Roles have not been sent yet. Hey, ignite, since you're here, would you like to play in this game?

EDIT: OK, since I let ignite in this game (due to activity and quick reply), I think we're set. I'll assign roles, and send them in a few minutes. If you have any specific questions about your role, feel free to PM me.

I would definitely like to play in this game if I could but if I'm near the bottom of the signup list, I think you should let someone near the top of the list play. :)

OK people.

ROLE PM's Have been sent!

If you have a specific question about your role, PM me. If you think it's general enough, ask on here.

(Make sure, if you have a role, that you have read about it. There may be some changes that I have made, so even if its standard, just check in my explanations.)

IT IS DAY.

Let's hope this doesn't take 2 weeks like the old games I've been reading :P

Also.
IMPORTANT:
This is my policy for voting.

[hide]An accusation does not count as a vote.
Neither does a seconding.

You MAY NOT vote until a defense has been given, or until 24 hours after the accusation has been seconded.
If you are going out of town, or may be unavailable, please post so. I accidentally cleared my inbox without realizing people had written when they would be gone.

Once you have voted, you may ONLY change your vote within 24 hours of your vote. And no retractions. Vote changing ONLY.

Also, if a vote is stalling, I will give clearance to change your vote.

You may NOT change your vote because someone responded to an accusation after you voted, unless many people are favoring said person to get off the block.

As long as an accusation has not been seconding, any number of accusations can be made.
An accusation can be retracted at any time, as long as it has not been seconded.
You may not accuse more than one person at one time. You may second someone with an accusation standing, however, your accusation will become inactive.
[/hide]
16 people.

9 to lynch.

I don't think there are any inactives yet at least because everyone has had to confirm  pretty recently.

Edit: I break my promise not to accuse or second :P

Just something I don't get... vigilante and prosecutor are both innocent and thus at citizen's side? Then what exactly is the use of a prosecutor?

Well, the prosecutor can see if someone who roleclaimed vigilante is lying, I guess. Or maybe help find the miller.

[quote="lasagnaman"]sooo.... random lynch, or inactive?[/quote]

well... i suppose we can start with lasagnaman due to eagerness... but i wont accuse yet, it is too early...

[quote="Peter VDD"]We do not need klebian. We just all post our roles, and we will know a group of 2 people of which one is devil, and the game is over.

And a lucky guess of saving dts... I also expected dts to be angel, as I expected the mafia to kill him. He was the brain behind it all. ;)[/quote]

*sigh*

All right, since it's clear to me that the Mafia really don't care whether they win or lose...

I'm the [color=red]Devil[/color]. :diablo:

And I laughed for a minute straight when the Angels saved me.

[quote="Peter VDD"]Lol... the devil was the person helping us most in catching the mafia? :D[/quote]

I wasn't [i]trying[/i] to help, but kheg_k and wobster109 kind of made themselves obvious... so I figured, they're going to get lynched anyway, so why not make me the one who lynches them?  Then, they'll never think I'm guilty.

And it worked, didn't it?  Did anyone think I was guilty?

I did as I was priest found you as one of the angels,  the devil, or inspector. I figured out that Ignite was inspector and that you kept tried to lynch me so that is why I suspected you. Yeah but when the angels saved you for a while I did not suspect you but later started suspecting you. But you already figured that out. And Fibfreak's comment about the angels being Ignite and me being angels kind of made me bielive that she was an angel once I saw that you wanted to kill me.  :D

Hmm. Well, as far as I can tell, game over.

CITIZENS WIN

I will edit this post with all the stastistics I can. Role players, please help me here.

Ars- Priest
Petervdd- Prosecutor
Kheg_k- Mafia
Slamchess- Citizen
Sunny- Mafia
Supernova- Fairy
Pianoforte- Mafia
FibFreak- Angel
S2- Mason
Dts- Devil
Wobster109- Mafia
GoBraves- Citizen (miller)
Ladykn1ght- Angel
Lasgnaman- Citizen
Melissa- Vigilante
Ignite168- Inspector

This is the PMs I received:

Mafia: S2, dts, ---, ---

Inspector: Ars, slamchess, perfectnumber, melissa

Priest: dts, sunny, ignite, peter

Fairy: no, no, no, yes

Prosecutor: lasagnaman, wobster, perfectnumber, ladyknight

Devil: ignite, supernova, peter, ars

Angel: ladyknight, dts, ---, ignite

Vigilante: no one, ---, petervdd, pianoforte

If you have any more questions, I can probably answer them.

Umm dts submitted a mafia kill?

Wow, dts fooled me, even though I thought he was suspicious (I remember saying If you weren't saved from mafia, I would accuse you two or three times), I wasn't thinking of the devil of all  :blush:

Yay citizens!

[quote="dts"]I'm the [color=red]Devil[/color]. :diablo:

And I laughed for a minute straight when the Angels saved me.[/quote]
that is [b]really[/b] funny!  :rotfl:

yeah, it's too bad i was doing this thing called *not paying attention* when the mafia were supposed to be killing someone...anyway, next game will be better

[quote="Peter VDD"]Umm... wait a second. Am I correct that ars spotted the devil on turn 1 and did not tell us? :o[/quote]

Well, Priest is only told "Successful" or "Unsuccessful" for inspections, so all Ars knew was that I was either Inspector, Angel, or Devil.

On with Game 18.

Nice game guys.  Though we lost!   :(

I think in general...this game did move slightly faster than the other ones.  I'm definitely keeping the "must respond in __ hours" rule in the next game I mod.

[quote="Melissa"]Aaargh! I missed the end of the game. Actually I could not decide whether to kill pianoforte or perfectnumber628... now I see it wouldn't have made much of a difference :D  Yay for innocents![/quote]

Actually, it would have made a HUGE difference. Ignite knew that perfectnumber was mafia, but no one knew that pianoforte was mafia as well.


Would you consider logarithms advanced math?

Why don't you simply go through the advanced middle school and non-advanced high school forums thread by thread and note down the interesting problems?

[quote="Rep123max"]Lately a lot of people at my school have been asking me about math. I think I can get them really interested in math, but I am in need of some "cool" problems. The thing is, I need these problems to be relatively simple, nothing above AMC-10 level. So does anyone have some "cool" problems or really interesting ones that do not require advanced math? Thanks.[/quote]

Just search in Internet: there are a lot of problems there to see!!!!

There are also a lot of books out there, for instance "Hard-to-solve math puzzles" is a collection of 101 problems (and solutions) that I like. It seems a bit heavy on the problems that require guess-and-check, but most of the problems are pretty cool.


integration
real analysis
real analysis unsolved

Let $ f(x) \equal{} \frac {1}{3}(x^2 \minus{} 2)^2$

Find the arc length of the graph $ f(x)$ for $ 2 \leqslant x \leqslant 4$

[quote="kunny"]$ \int_2^4 \sqrt {1 \plus{} f'(x)^2}\ dx$[/quote]

I know that formula but i don't know how to integrate after this ,you just have to try that.  :|

hello, i think it can only be done by a numerical way.
Sonnhard.