topology
MIT
college
ARML
Ross Mathematics Program
summer program
PROMYS

Who's all applying? I for one just found out about SIMUW a few days ago and am working on the proofs as quickly as possible.  I've never attended a math event such as this and am quite excited at the possibility.

And previous attendee's thoughts and experiences?

I went last year and I think I'll try to apply again this year. It was a pretty cool program and I learned a fair amount while there. What I though they did nicely was provide for a wide range of students. There were a number of students that didn't have much math experience outside the classroom while some of us knew a fair amount, but they found a way of challenging everyone most of the time without blowing too many people away.

Its not classroom math, we learn lots of stuff (topology, abstract algebra, number theory) that you don't usually see.

There's lots of free time and we don't take the math extremely seriosly. If you're a genius or really obsessed with math this might not be for you. Otherwise, I had a great time, and met a lot of great people.

Thanks for the summary, I'm neither a genius (unfortunately =p) nor obsessed with math, just merely interested in learning about more abstract facets of mathematics.  Hopefully I see you there.

i defenately recomend it. :)

jefferey said everything else.

I'm trying to work through my math problems and i found it extremely hard.
so far i've only "solved" TWO questions and the rest are just 'looking' too complicated.

I'm not sure if they would accept those who love math, but are not extremely smart at it.

Plus, how many question minimum do you have to solve in order to have a "chance" in being accepted?

A quick glance at the questions tells me that they're harder than last year, when all the problems were pretty easy (I don't have an immediate solution for a couple of these). From what I know, two problems should give you a reasonable shot. Admissions isn't super competitive.

Yeah, I only solved 6 out of the 8.  You should have a good chance, especially since you're a girl (at least I'm guessing that based on your name =p)

Good luck and I hope to see you there.

Looks like decisions are coming out (I got my acceptance yesterday, a couple of days after I sent it in). I'm not completely sure if I'm going to go though.

Who else is thinking of going?

I got accepted and I'm definately going.

I'm accepted. Waiting on RSI right now. If I don't get into RSI, I'll be attending SIMUW.

I'm coming to simuw for sure (from Vancouver) and so is this girl from my school.

Hi everybody!

I also just got accepted and I'll be coming from Oregon. This program looks like it'll be fun; I look forward to meeting you all there.

Akira

Hey, I'm from Oregon too, just got accepted. You know what's sad, after spending some good couple hours on solving them all, I was looking over AoPS volume 2 the other day and found out that the basic proofs of the concepts I was using were all in there (that I had come up with on my own)... that's what i hate about math, you forget random things easily if you don't use them.

paladin8, what's RSI? Can you still apply for that? (probably not...)

[quote="Uh oh!"]Second, I can teach you all of the number theory that you'd learn at PROMYS during SIMUW if you really want to learn that stuff, since I'll be a TAC at SIMUW and I did the Ross problem sets which PROMYS is based on.[/quote]

Haha, do you remember all of that? I know I don't :P.

[quote="Uh oh!"]
First, SIMUW is free, and you'll get lots of free stuff. PROMYS costs money, plus the cost of transportation. Second, I can teach you all of the number theory that you'd learn at PROMYS during SIMUW if you really want to learn that stuff, since I'll be a TAC at SIMUW and I did the Ross problem sets which PROMYS is based on. Third, SIMUW really tries to create a sense of family and community which is what makes it such an amazing place. The math is good and all, but it's the other experiences that make SIMUW such an unparalleled program.

I could go on with many more reasons, but I doubt you'd want to read that much, but if you need more reasons, feel free to ask.
[quote]

What are the other reasons?
How do SIMUW kids compare with Ross kids in math?

[quote="Uh oh!"]

And Sam, you better come.[/quote]

I'm thinking about it

To Jeffrey, I do remember pretty much all of it since it was only last summer and I'm taking a number theory course right now at Yale that went over a lot of the important stuff, although in not nearly the same depth.

Some more reasons to go to SIMUW:
1. You'll meet people from around the Pacific Northwest who you'll see at countless competitions. It's nice to develop that sort of network and to actually be able to keep in touch with the friends that you'll make rather than only being able to talk to people online since they live on the other side of the country.
2. SIMUW encourages people to have fun, as well as do math. The Saturday excursions are great and there's plenty of mandatory fun that's actually fun, from Mafia to trips to the beach to poker tournaments.
3. SIMUW will expose you to a lot more different types of math than other camps. Rather than just learning about number theory, for example, there are six different lectures, and you still go pretty far in depth to each one. I'm particularly excited for knot theory this year since it was one of my favorites.
4. You get to spend the summer in Seattle where the weather does not get unbearably hot and humid like so many other places in the country. It's really nice and sunny, but it never feels muggy like you're swimming through the air.
5. Most people do a lot better on the AMC and AIME after going to SIMUW.
6. Although it might not seem like such a good thing to some people, the fact that, for the most part, your parents will be able to come and pick you up if you start feeling homesick is really, really nice, especially if it's going to be the first time you're spending a significant time on your own.
7. You'll get to meet Amir and me, two of the coolest people in the world.

As to the mathematical ability of SIMUW kids versus Ross kids, it just depends on who ends up going. I think it's fair to say that Amir and I were the top two mathematicians in the state in our year (we were the only seniors to qualify for the USAMO last year) and we both went to SIMUW after our junior years. Both of us went to Ross this past summer since we couldn't go back to SIMUW, but Amir decided to leave early since he kinda hated it there, but I stayed and I finished more problem sets than anybody else there. That being said, there have been some amazing mathematicians to come out of Ross since it's been going on for so long. One of the junior counselors at Ross left for two weeks in the middle to go participate at the International Physics Olympiad and came back with a gold medal. Go figure.

In general, while the Ross kids probably were better at math than the SIMUW kids, I don't think that it makes that much of a difference. It depends more on you, your ability, and what you want to get out of the camp.

Again, if you want more reasons to go to SIMUW, just ask, although it would be nice if you asked questions about things in particular rather than asking for more reasons in general since there's so much stuff to talk about it's hard to know where to start.

Are you implying that Amir and I aren't insanely smart in nearly every type of math?  :P

Not that I want to create a debate between SIMUW and PROMYS since I don't actually think anybody has gone to both programs, one of the cool things about SIMUW is that we have courses in those cool topics like topology and elliptic curves. When I went, we had a course on elliptic curves and knot theory which had lots of topology in it. You can check out what cool courses we'll have this summer at the SIMUW website.

Again, I'd like to say that if you're interested in number theory in particular, I'd be more than happy to spend some time over the summer going over it since it's my favorite area of math right now. Number theory pretty much rocks my socks off and I agree that it really helps you think about math. If you're really hardcore, I can show you the Ross problem sets and you can work on those which will most definitely keep you occupied for the six weeks.

And SIMUW has lots of game playing as well. I'm sure Amir will force people to play mafia with him and I'm always up for some DDR or some set. SIMUW actually has a stash of games that you can check out from us TACs and you're bound to find people to play with if you don't feel like doing math. Also, we'll have other cool things like playing some volleyball or ultimate, going to go get ice cream from the Mix, and seeing movies as random things to do whenver people feel the impulse, in addition to all of the planned stuff like going to get the new Harry Potter book at midnight.

[quote="Uh oh!"]I think it's fair to say that Amir and I were the top two mathematicians in the state in our year[/quote]
false statement; not me, just you.
[quote="Uh oh!"]You'll get to meet Amir and me, two of the coolest people in the world.
[/quote]
a very true statement. :)
[quote="Uh oh!"]Are you implying that Amir and I aren't insanely smart in nearly every type of math?  :P

[/quote]

of course not, nobody dares to declare such thing! :P
[quote="Uh oh!"]
 I'm sure Amir will force people to play mafia with him [/quote]

oh i will!!

those who do get accepted to simuw should definitely go, especially if you're still young and have other years to go to other mathcamps (or possibly simuw again!). friendships that you make at simuw are quite valuable  (I became friends with edward during simuw2005, and we have been close friends since.)
to make it short, just go! you wont regret it, i guarantee!

Hello everyone,
My name is Alysia, and I'm going to be a TAC for SIMUW 2007.  I'm from southern CA, and this will also be my frist experience with SIMUW; I'm so excited to be involved with this program! I look forward to meeting all of you in a couple months ! :lol:

does someone know when we should be receiving more info about simuw (and actual letters in the mail)?

if i recall correctly, they usually send out information packets; and forms that need to be filled out and other things that you guys need to do in mid may.

[quote="paladin8"][quote="Uh oh!"]To Jeffrey, I do remember pretty much all of it since it was only last summer and I'm taking a number theory course right now at Yale that went over a lot of the important stuff, although in not nearly the same depth.[/quote]

I see. Good stuff.

For those interested in hearing from a PROMYS alum (you can ask t0rajir0u, too), I'd say that PROMYS is good for one of two reasons: (1) to stimulate your interest in NT and other types of math, or (2) to become godlike at NT. And a nice bonus is having tons of fun and meeting some very intelligent individuals.

Since I went after my freshman year, my math abilities were pretty limited, but I still got a ton out of PROMYS. Working on the problem sets with other people was really fun, and I basically learned all the NT I know from there. I went from knowing next to nothing about it to... well, where I am today, minus some new stuff.

That said, I believe that if you have a strong math background, you could get even more out of PROMYS. And there isn't just NT, either, there are advanced seminars if you want to do them and be really hardcore (topics are Geometry and Symmetry, Modular Forms, Abstract Algebra... not for the faint of heart, of course). Plus the counselors are like insanely smart in every type of math, so they give mini-courses in cool stuff like topology and elliptic curves.

There's SO MUCH to learn at PROMYS, so if that's what you're into, PROMYS is definitely a good choice. As for life there, it is great too; I wouldn't bet that Towers dorm is nicer than the UW dorms, but that's ok. Playing a variety of games, Mafia in particular, through the night, literally from 9 PM to like 6 AM, is very important. Basically, PROMYS is awesome.[/quote]

But is it really worth it for only an hour of instruction a day (I heard this from someone, so I may be wrong).

If PROMYS is like Ross, you only go to lecture for one hour each weekday, plus seminar three times a week for an hour each time, and then there are various other lectures you can go to, but none of them are required. But with Ross, you're expected to do math pretty much every other waking moment, so you end up doing math all the time and it's not unusual to see people staying up all night doing math. At SIMUW, on the other hand, you're in class for a lot longer (about 5 hours each weekday, Wednesdays are special), but it's not all lectures since you spend classtime working with other people on problems that the professors give out. Outside of class, you can do math if you want to, but it's not at all necessary. Obviously since it's math camp, lots of people choose to do math in the free time, but it's not forced on you like it is at Ross (depending on how rebellious you are, of course. By the end of Ross, most people were hiding in the business school library playing computer games). There is some forced mandatory fun at SIMUW, but I think it's actually fun, and I'll make sure that any fun that I make mandatory will be enjoyable to people.

From my understanding of SIMUW, it is supposed to be comprised of math that is challenging, but more importantly math that student's [b]enjoy [/b]doing.  I feel that this program places a higher emphasis in sparking student's interest, and enhancing the overall learning experience rather than making a super intense mathematical program where you do math all day everyday.  I have been a counselor for programs where the math portion was literally "Math Bootcamp" (math basically 24/7) and the experience those students had with math was probably the most negative expereince that they've ever had (and also damaging to their mathematical careers)   Now I'm not saying that the programs mentioned are Math bootcamps by any means, especially since I have no experience with any of them.  I think that in making your decision about which camp to chose, pick the one where you will have the best overall experience.
From what I've read, SIMUW seems to have the best balance between math and fun.  Having a balance betweent the two is essential because without fun, yes you'll learn but you're most likely not going to enjoy what you're learning. One of my professors always says "If you don't enjoy math, why do you do it? what's the point?"

At SIMUW you're guaranteed to have awesome educational and enjoyable experience!!!!

and remember:
doing math + having fun while doing math + having more fun + edward's mandatory fun= a down right good time!!! :P

See you in a little over a month!!!  :lol:
-Alysia

hey the full schedule's up!!

or maybe it's been there and i just didn't see it for a while...

it actually got updated in the past few days...(there is still some course info's missing though).


parameterization
real analysis
real analysis theorems

hi,
i have a question thats killing me...
i'm required to solve this ODE :

    y"(x)-3y'(x)-2y(x)=sinh(x)

using the Variation of parameters Method.

The solution was EXTREMELY long,involving many sinh and cosh terms.I've tried countless times,just cannot get to that answer.

Urgent help is needed please! thanks.

Write $\sinh x=\frac12e^{x}-\frac12e^{-x}.$ Then solve two separate problems, once with $\frac12e^{x}$ as the right hand side; the other with $-\frac12e^{-x}$ as the right hand side.


geometry
inradius
function
congruent triangles

Let $ I(r)$ denote the number of non-congruent triangles with integer side length and inradius $ r$.  For example, $ I(1) \equal{} 1$ and $ I(2) \equal{} 5$.  Find a simple function $ F(r)$ such that $ I(r) \equal{} O(F(r))$.  How about $ I(r) \equal{} \Theta(F(r))$?

(Such a function is not given in the OEIS.)

What do you mean by $ O(F(r))$ and $ \Theta(F(r))$?


Functional Analysis
advanced fields
advanced fields theorems

Let $ \mathcal{H}$ be a Hilbert space and let $ \mathcal{M}$ be a closed subspace. Define $ \mathcal{M}^{\perp}\equal{}\{x\in\mathcal{H}: (x\cdot a)\equal{}0\forall a\in\mathcal{M}\}$. Then, every $ z\in\mathcal{H}$ can be written $ z\equal{}x\plus{}y$ where $ x\in\mathcal{M},y\in\mathcal{M}^{\perp}$

What is wrong with this proof?
Let $ a\in\mathcal{H},x\in\mathcal{M}$. Observe that $ y\equal{}a\minus{}(a\cdot x)\frac{x}{\|x\|^2}\in\mathcal{M}^{\perp}$. As $ \mathcal{M}$ is a closed subspace, $ y'\equal{}(a\cdot x)\frac{x}{\|x\|^2}\in\mathcal{M}$ and hence $ a\equal{}y\plus{}y'$

Why?
$ (y\cdot x)\equal{}(a\cdot x)\minus{}(a\cdot x)\frac{(x\cdot x)}{\|x\|^2}\equal{}0$

yes, now look at the definition of $ M^{\perp}$.

you only solved the problem in case of $ dim(M)\equal{}1$.


geometry
angle bisector
geometry proposed

[b]Theorem 1[/b]. Let $ ABC$ be a given triangle. Let $ d_{A_{1}}$, $ d_{A_{2}}$ be two lines passing through the vertex $ A$, so that they are symmetric to each other with respect to the internal angle bisector of $ \angle{BAC}$ (in other words, the lines $ d_{A_{1}}$, $ d_{A_{2}}$ are two isogonals wrt. $ \angle{BAC}$). In the same manner, take two isogonals $ d_{B_{1}}$, $ d_{B_{2}}$  with respect to $ \angle{ABC}$ and two isogonals  $ d_{C_{1}}$, $ d_{C_{2}}$ with respect to $ \angle{BCA}$. Prove that the hexagon determined by these $ 6$ lines has concurrent diagonals.

[b]Theorem 2[/b]. Let $ ABC$ be a given triangle. Let $ X_{1}$, $ X_{2}$ be two points lying on the side $ BC$, such that they are symmetric to each other with respect to the midpoint of $ BC$ (in other words, the points $ X_{1}$ and $ X_{2}$ are isotomic with respect to the segment $ BC$). Now, let $ Y_{1}$, $ Y_{2}$ be two isotomic points wrt. $ CA$, and let $ Z_{1}$, $ Z_{2}$ be two isotomic points wrt. $ AB$. Prove that the hexagon bounded by the lines $ AX_{1}$, $ AX_{2}$, $ BY_{1}$, $ BY_{2}$, $ CZ_{1}$, $ CZ_{2}$ has concurrent diagonals.

This problem is well known ( at least to me ) in its more general formulation :

[b][size=100][color=DarkBlue]GENERAL PROBLEM. Let be the pairs of points $ D,\ D'$ and $ E,\ E'$ and $ F,\ F'$ on the side-segments $ BC,\ AC,\ AB$ respectively, of a given triangle $ \bigtriangleup ABC,$ isotomic comjugates with respect to the corresponded side-segment $ ($ $ D,\ E,\ F,$ adjoining points to $ B,\ C,\ A,$ respectively $ ).$ We denote the points $ A'\equiv BE'\cap CF,\ B'\equiv AD\cap CF',\ C'\equiv AD'\cap BE$ and $ A''\equiv BE\cap CF',\ B''\equiv AD'\cap CF,\ C''\equiv AD\cap BE'.$ Prove that :

(a) – The line segments $ AA',\ BB',\ CC',$ are concurrent at one point so be it $ M.$
(b) – The line segments $ AA'',\ BB'',\ CC'',$ are concurrent at one point so be it $ N.$
(c) – The line segments $ A'A'',\ BB'',\ CC'',$ are concurrent at one point so be it $ P.$
(d) – The points $ M,\ N,\ P,$ are collinear.[/color][/size][/b]

$ \bullet$ This problem is also true when the above pairs of points, are the traces on $ BC,\ AC,\ AB,$ of arbitrary isogonal lines through their opposite vertices $ A,\ B,\ C$ respectively, of $ \bigtriangleup ABC.$

$ \bullet$ It is also true for some other kinds of pairs of points on $ BC,\ AC,\ AB.$

I have mentioned these results in [b][size=100]Hyacinthos Forum[/size][/b], two years ago and I will post here later more details about their proofs.

Kostas Vittas.

[quote="vittasko"][b][size=100][color=DarkBlue]GENERAL PROBLEM. Let be the pairs of points $ D,\ D'$ and $ E,\ E'$ and $ F,\ F'$ on the side-segments $ BC,\ AC,\ AB$ respectively, of a given triangle $ \bigtriangleup ABC,$ isotomic comjugates with respect to the corresponded side-segment $ ($ $ D,\ E,\ F,$ adjoining points to $ B,\ C,\ A,$ respectively $ ).$ We denote the points $ A'\equiv BE'\cap CF,\ B'\equiv AD\cap CF',\ C'\equiv AD'\cap BE$ and $ A''\equiv BE\cap CF',\ B''\equiv AD'\cap CF,\ C''\equiv AD\cap BE'.$ Prove that :

(a) – The line segments $ AA',\ BB',\ CC',$ are concurrent at one point so be it $ M.$
(b) – The line segments $ AA'',\ BB'',\ CC'',$ are concurrent at one point so be it $ N.$
(c) – The line segments $ A'A'',\ BB'',\ CC'',$ are concurrent at one point so be it $ P.$
(d) – The points $ M,\ N,\ P,$ are collinear.[/color][/size][/b][/quote]
[b][size=100]PROOF.[/size][/b] - Let be the points $ X\equiv BC\cap AA',\ Y\equiv AC\cap BB',\ Z\equiv AB\cap CC'$

and it is enough to prove that $ \frac {XB}{XC}\cdot \frac {YC}{YB}\cdot \frac {ZA}{ZB} \equal{} 1$

Applying the [b][size=100]Ceva theorem[/size][/b], for the concurrent cevians at point $ A',$

we have that $ \frac {XB}{XC}\cdot \frac {E'C}{E'A}\cdot \frac {FA}{FB} \equal{} 1$ $ \Longrightarrow$ $ \frac {XB}{XC} \equal{} \frac {E'A}{E'C}\cdot \frac {FB}{FA}$ $ ,(1)$

By the same way, for the concurrency cevians at points $ B',\ C',$

we have that $ \frac {YC}{YA} \equal{} \frac {F'B}{F'A}\cdot \frac {DC}{DB}$ $ ,(2)$ and $ \frac {ZA}{ZB} \equal{} \frac {D'C}{D'B}\cdot \frac {EA}{AC}$ $ ,(3)$

From $ (1),$ $ (2),$ $ (3)$ $ \Longrightarrow$ $ \frac {XB}{XC}\cdot \frac {YC}{YB}\cdot \frac {ZA}{ZB} \equal{} 1$ $ ,(4)$

From $ (4),$ we conclude that the line segments $ AA',\ BB',\ CC',$ are concurrent at one point so be it $ M$ and the proof of the part [b][size=100](a)[/size][/b] of the problem is completed.

$ \bullet$ Because of now, the concurrency of $ AA',\ BB',\ CC',$ based on the [b][size=100]Second theorem of concurrent segments[/size][/b], we conclude the results of [b][size=100](b)[/size][/b], [b][size=100](c)[/size][/b], [b][size=100](d)[/size][/b], as the problem states and we are done.

[b][size=100]REMARK.[/size][/b] - The points $ M\equiv AA'\cap BB'\cap CC'$ and $ N\equiv AA''\cap BB''\equiv CC'',$ are [b][size=100]isotomic conjugates[/size][/b] with respect to the given triangle $ \bigtriangleup ABC,$ as well.

[b][size=100]COMMENT.[/size][/b] - The concept of the solution of this kind of problems, is that we need to prove directly, only the part [b][size=100](a)[/size][/b], or the part [b][size=100](b)[/size][/b]. The other results, as alrerady been shown, may be concluded based on the [b][size=100]Second theorem of concurrent segments.[/size][/b]

Kostas Vittas.

PS. I will post here later the proofs of the part [b][size=100](a)[/size][/b], for some other kinds of pairs of points $ D,\ D'$ and $ E,\ E'$ and $ F,\ F',$ on $ BC,\ AC,\ AB$ respectively, as mentioned in the post #2#.

[quote="¬[ƒ(Gabriel)³²¹º]¼"]for the isogonal lines we can prove a) and b) with the Jacobi theorem and then c) and d) with the Second theorem of concurrent segments[/quote]
Exactly, dear [b][size=100]Gabriel[/size][/b]. For the case of the Isogonals, the easy proof by [b][size=100]Jacobi theorem[/size][/b], of the part [b][size=100](a)[/size][/b], or of the part [b][size=100](b)[/size][/b], is the first step of the solution.

$ \bullet$ The same ( it is enough only to prove directly, the part [b][size=100](a)[/size][/b], or the part [b][size=100](b)[/size][/b] ), for the case of the pairs of points $ D,\ D'$ and $ E,\ E'$ and $ F,\ F',$ on the side-segments $ BC,\ AC,\ AB$ rerspectively, as their intersection points from an arbitrary circle ( easy to prove ).

Kostas Vittas.

Dear Kostas, thank you for your interest. I kinda had the same proof (by making use of Jacobi). I am also aware of the more general version of the theorem (it is called the Lamoen Cross-Triangle Theorem - according to Darij). See here: [url]http://www.cip.ifi.lmu.de/~grinberg/LamoenCross.zip[/url]. However, I liked these two particular cases.

Thank you dear [b][size=100]Cosmin[/size][/b]  for the reference about the [b][size=100]Lamoen cross triangle theorem[/size][/b] – according to [b][size=100]Darij[/size][/b].

I see that it is exactly what I say as the [b][size=100]Second theorem of concurrent segments[/size][/b].

I also think ( I will check it more carefully ) that the proof I have in mind about it, based on the [b][size=100]Desarques theorem[/size][/b], is the same as the given one by [b][size=100]darij[/size][/b] and I didn’t any think about his extended results, as for the harmonic conjugations.

Finally, when we think some result in geometry as our discovery, [b][size=100]Darij[/size][/b] probably has already preceded about it.  :(

But never mind,  :) I am glad because I have enjoyed this nice theorem, in trying to solve some difficult problems, when I was confused of so many  lines in the schema and so, it is easier to distinguish three concurrent segments, instead of the relative perspective triangles.

The [b][size=100]Desmic theorem[/size][/b], as an alternative name mentioned by [b][size=100]Darij[/size][/b], is more attractive to me.

Thank you and best regards, Kostas Vittas.


symmetry
number theory proposed
number theory

Solve in the positive integers the equation
$xy(x^{2}+y^{2})=2z^{4}$

You can assume $(x,y)=1$ and $2^{c}\parallel x$. Make transformation $x \to 2^{c}x$ so that $2 \not | x$. Then our equation rewrites as:

$2^{c-1}xy(2^{2c}x^{2}+y^{2})=z^{4}$. Hence $4|c-1$ and $x,y$ must be 4th powers. Hence we are to solve $4m^{8}+n^{8}=k^{4}$, $(n,m)=1$, $2\not | mn$. Using Sophie Germain's identity we have:

$(n^{4}+2m^{4}-2m^{2}n^{2})(n^{4}+2m^{4}+2m^{2}n^{2})=k^{4}$ and as:

$(n^{4}+2m^{4}-2m^{2}n^{2},n^{4}+2m^{4}+2m^{2}n^{2})=1$ both must be fourth powers. Hence

$(n^{2}+m^{2})^{2}+(m^{2})^{2}=(a^{2})^{2}$
$(n^{2}-m^{2})^{2}+(m^{2})^{2}=(b^{2})^{2}$ now to get solutions or decide that there is none (I haven't got enough free will to check it, I leave it as an exercise  :P ) we treat them both as pythagorean equations so we are able to characterize solutions (I mean the only solutions to $a^{2}+b^{2}=c^{2}$ are $a=k(m^{2}-n^{2})$, $b=2kmn$, $c=k(m^{2}+n^{2})$)

Why do we have to have 2|x to start with? Can we not have x,y both odd and $2|x^{2}+y^{2}$, or does your solution cover this possibility in a way I haven't appreciated?

You're right - I haven't noticed that and for now I don't know how to fix it.

You missed many solutions .For example .Each triple
(x,y,z)=(t,t,t)  with t every integer is a solution  :wink:

[quote="juna"]...hence $xy=2$, $x^{2}+y^{2}=z^{4}$ or $xy=z^{4}$, $x^{2}+y^{2}=2$.[/quote]

Why¿

@silouan: no, they are covered by the coprime tupel $(1,1,1)$.

[quote]You missed many solutions .For example .Each triple
(x,y,z)=(t,t,t) with t every integer is a solution [/quote]
Yes, I missed them.
ZetaX: can assume $(xy,z)=1$. It demonstrates that the coprime solutions no.

Yours is shorter¿
- this is not the shortest-proof-competition.
- the other proof is in my oppinion good to read (and seems correct; the "known facts" there were posted before, the latter one follows even from the first one), whereas yours has a lot of gaps to fill.

For example:
Why can you assume $\gcd(xy,z)=1$¿

[quote="juna"]I thought what is it apparent things
If $(xy,z)=d$ then
let $x=dx_{1}$, $y=dy_{1}$, $z=dz_{1}$ then[/quote]
This can only be done in integers if $(x,y,z)=d$, $(xy,z)=d$ doesn't suffice.

[quote="juna"]let $(y,z)=1$, $x=dx_{1}$, $z=dz_{1}$ then[/quote]
Why can you assume $(y,z)=1$¿

In total, I find your post near to impossible to read. It's possible that it is correct, but I fail to understand what you do and don't want to guess at every point how it should be or how it should be understood.

[quote="juna"][quote]Why can you assume $(y,z)=1$¿[/quote]
In view of a symmetry $x,y$[/quote]
What symmetry¿ Or what condition you want to use¿

But these two things are not the only problems I had...
And now it's already as long as libra_gold's solution...

I do not understand that to you not clearly. We consider all cases of common divisors $x,y,z$
case#1 $x=dx_{1}$, $y=dy_{1}$, $z=dz_{1}$
case#2 $x=dx_{1}$, $(y,z)=1$, $z=dz_{1}$ we may given $(x,z)=1$, $y=dy_{1}$, $z=dz_{1}$ - x,y - symmetry in this case
case#3 $x=d_{1}x_{1}$, $y=d_{2}y_{1}$, $z=d_{1}d_{2}z_{1}$
[quote]And now it's already as long as libra_gold's solution...[/quote]
Let's aggregate parts and we shall see...
[quote]But these two things are not the only problems I had...[/quote]
I am ready to consider these problems

I just want to ask you one question.

Do you mean #If $(x,y)=1$ then $(xy,z)=1$#?

No. If $(x,y)=1$ then can be $(x,z)\neq 1$, $(y,z)\neq 1$ hence $(xy,z)\neq 1$. Look through my posts more closely.

One line solution.
Write the equation in the form $(x+y)^{4}-(x-y)^{4}=(2z)^{4}$  :)


[quote="epitomy01"]Let $ a, b, c$ be positive real numbers. Prove that:
$ \displaystyle{\frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}} \plus{} a \plus{} b \plus{} c \geq \displaystyle\frac {2(a \plus{} b \plus{} c)^3}{3(ab \plus{} bc \plus{} ca)}$

Note: This problem is from Issue 1 2008, and deadline is past (16th March i think), so I believe I can post it here.[/quote]
The solution is the following
first, note that (it is easy)
\[ \sum \frac{a^2}{b} \ge \frac{(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)}{ab\plus{}bc\plus{}ca}\]
It suffices to prove
\[ \frac{(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)}{ab\plus{}bc\plus{}ca} \plus{}a\plus{}b\plus{}c \ge \frac {2(a \plus{} b \plus{} c)^3}{3(ab \plus{} bc \plus{} ca)}\]
that is
\[ a^2\plus{}b^2\plus{}c^2 \ge ab\plus{}bc\plus{}ca\]
which is true. :)

[quote="epitomy01"]Let $ a, b, c$ be positive real numbers. Prove that:
$ \displaystyle{\frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}} \plus{} a \plus{} b \plus{} c \geq \displaystyle\frac {2(a \plus{} b \plus{} c)^3}{3(ab \plus{} bc \plus{} ca)}$[/quote]

Use s.o.s method

[quote="manlio"][quote="can_hang2007"]first, note that (it is easy)
\[ \sum \frac {a^2}{b} \ge \frac {(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca}
\]
[/quote]

Can you please show your proof? For me it is not easy  :wink:

Thank you very much.[/quote]
Just expanding my friend. :)
The inequality is equivalent to
\[ (\sum ab^3)(\sum ab) \ge abc(\sum a)(\sum a^2)\]
\[ \Leftrightarrow \sum a^2b^4\plus{}\sum ab^4c\plus{}\sum a^2b^3c \ge \sum a^4bc \plus{}\sum a^3b^2c \plus{}\sum a^2b^3c\]
\[ \Leftrightarrow \sum a^2b^4 \ge \sum a^3b^2c\]
Setting $ x\equal{}ab^2,y\equal{}bc^2,z\equal{}ca^2$, then it becomes
\[ x^2\plus{}y^2\plus{}z^2 \ge xy\plus{}yz\plus{}zx\]
which is true by the Am-GM. :)

Thank you very much.

I tried to prove it by applying Cauchy-Schwarz al lhs but I failed  :wink:


inequalities
rearrangement inequality
algebra unsolved
algebra

sorry for my mistakes for 1...
1)Let {a_1,a_2,a_3,a_4,a_5}={1,2,3,4,5}.Find the maximum and minumum value of a_1+2a_2+3a_3+4a_4+5a_5=?

2)For positive integers x,y,z
   x+y+z=90
   Let x/n=y/n+1=z/n+2 for n element of positive integer.How many positive (x,y,z) integers can be found that satisfy these conditions?

3)A is a set that all elements are smaller than 102.A does not include the sum of any elements.What can be the maximum number of element of A?

4)Find (x,y) integers that satisfy the equation

            2x+5y=xy-1

Hello, ssupermath!

Quote:4) Find (x,y) integers that satisfy the equation: [hide]

Solve for y:  y = (2x + 1)/(x - 5) = 2 + 11/(x - 5)

Since y is an integer, (x - 5) must divide 11.

The only divisors of 11 are: -11, -1, 1, 11.

Hence: x - 5 = {-11,-1, 1, 11}.

And:  x = {-6,  4,  6,  16}.

The corresponding y-values are: y = {1, -9, 13, 3}.

Answers: (x,y) = (-6,1), (4,-9), (6,13), (16,3).[/hide]

[quote]1)Let {a_1,a_2,a_3,a_4,a_5}={1,2,3,4,5}.Find the maximum and minumum value of a_1+2a_2+3a_3+4a_4+5a_5=? [/quote]

From the rearrangement inequality, the maximum is
1*1 + 2*2 + 3*3 + 4*4 + 5*5 = 65

and the minimum

1*5 + 2*4 + 3*3 + 4*2 + 5*1 = 35


Which two integers yeild the same result whether the larger of them is divided by the other or they are multiplied together.

well, we have lots of solutions:
1. the larger number is 0, and the other is any negative integer
2. the larger number is 1 and the smaller is -1
3. the larger number is any integer bigger than 1 and the smaller number is eather 1 or -1.

This was far too easy. So, what's the catch?

edit number 2. You can have any number greater than -1 and -1

example: -0.5 and -1

$\frac{-0.5}{-1} = 0.5$
$(-0.5)(-1) = 0.5$

yes, but 0.5 is not an integer and 0 is already accounted for

oh right forgot that they had to be integers  :blush:

anyway, if we were to move to the real numbers then you'd have to change it :D


modular arithmetic
LaTeX
calculus
calculus computations

How does one prove that $ im(z)\equal{}.5$ in the following equation?

$ icosh(z)\equal{}sinh(z)$

Well, that doesn't seem to be the answer. I get that $ \text{Im}\,z\equiv\frac{\pi}4\pmod{\pi}.$

Of course, I could also be misreading your bad $ \text{\LaTeX}.$

Here's the procedure: $ \cosh z=\frac{e^{z}+e^{-z}}2$ and $ \sinh z=\frac{e^{z}-e^{-z}}2,$ so

$ i\,\frac{e^{z}+e^{-z}}2=\frac{e^{z}-e^{-z}}2$

$ ie^z+ie^{-z}=e^z-e^{-z}$

$ (1-i)e^z=(1+i)e^{-z}$

$ e^{2z}=\frac{1+i}{1-i}=i.$

Since $ i=e^{i\pi/2}$ we get $ \text{Re}\,z=0$ and $ 2\,\text{Im}\,z=\frac{\pi}2+2\pi k$ for some $ k\in\mathbb{Z}.$

What was bad about my latex?

Anyways, I made a typo, the equation should have read:

$ icosh( \pi z) \equal{} sinh( \pi z)$

I did the calculations a couple days ago, and I think my answer was z = r + .5i, r ∈ ℝ, but how do I prove this?

Use \cosh and \sinh, please - don't omit the backslash. Look at the code in my previous post to see how. And I wasn't 100% sure what you meant by $ im(z)$ - again, check my code for a suggestion about that.

But all that change does in divide my answers by $ \pi$. We now have that the answers to this problem are $ z\equal{}i\left(\frac14\plus{}k\right)$ for $ k\in\mathbb{Z}.$


function
parameterization
algebra unsolved
algebra

Let $z \in S$ be a fixed parameter. The parameter $k \in \mathbb{N}$ is also fixed and $i \in \mathbb{N}$ runs in the sum from $1$ to $k$. The variable $x \in S$ is the argument of the function or in other words: $x \mapsto f(x)$ where $f: S \mapsto S$. The function f satisfies: $\sum^k_{i=1} f^i(x) = (z + k) \cdot x + z \cdot k$   Find all such functions $f$. Please consider the cases $S \in \mathbb{N}, \mathbb{Z}, \mathbb{Q}, \mathbb{R}$ or $S \in \mathbb{C}.$

[hide="An example"]

An example to see which parameters are fixed in the problem: $k = 3, z = 0, \in \mathbb{N}: f(f(f(x))) + f(f(x)) + f(x) = 3x.$

[/hide]

Is there known answer, or it is like an open question? :maybe:


Find all integer numbers $a$, $b$ and $c$ such that $\frac{1}{2}(a+b)(b+c)(c+a)+(a+b+c)^3=1-abc$.

The given equation is equivalent to $(2a+b+c)(a+2b+c)(a+b+2c) = 2$...

Pierre.


greatest common divisor
number theory
least common multiple
calculus
integration

why is the GCF of A and B mutliplied by the LCM of A and B equal to AB?

[hide="Proof"]I'll argue that the LCM is equal to the product of the two numbers divided by their GCF (this is an equivalent statement). Assume this is not the case. Well, the product of the two numbers divided by their GCF must still have a factor of the GCF present. Let $ g$ be the GCF for convenience, then for some integers $ m$ and $ n$, $ a \equal{} gm$ and $ b \equal{} gn$. Therefore, $ ab / g$ is still integral and can be written as either $ bm$ or $ an$, so both $ a$ and $ b$ divide the given quantity. Now, assume furthermore that there exists a smaller integer $ k$ which can be divided by both $ a$ and $ b$. This implies that both $ m$ and $ n$ share a common factor. However, this implies that $ g$ was not the GCD of $ a$ and $ b$, for if $ g$ were the GCD, then it would have included this common factor. Contradiction, so the LCM is equal to the product of the two numbers divided by their GCF, so the product of the LCM and GCF is the product of the two given integers.

[/hide]

[hide="Alternatively"]
Let $ A\equal{}A'\cdot d$ and $ B\equal{}B'\cdot d$ where $ \gcd(A',B')\equal{}1$. From where we can get that $ \gcd(A,B)\equal{}d$ and $ lcm(A,B)\equal{}A'\cdot B'\cdot d$. Multiply these two together to get,
\[ \gcd(A,B) \cdot lcm(A,B) \equal{} A'\cdot B'\cdot d\cdot d \equal{} (A'\cdot d)(B'\cdot d) \equal{} A\cdot B\]
And thus proven.
[/hide]

$ a \plus{} b \equal{} \text{min}(a, b) \plus{} \text{max}(a, b)$.


How many sevens are written in the number $ 4444^{4444}$ ??

What I did was put $ 4444^{4444}$ into Mathematica, copy the output, paste it into a word processor and "Find and Replace"d all the 7's with spaces.  It said it replaced 1605 occurrences, so this is the answer.  I don't think there is any number-theoretic way to do this.

Edit: http://www.wolframalpha.com/input/?i=DigitCount[4444%5E4444%2C+10%2C+7] says that it is in fact 1605.


limit
inequalities unsolved
inequalities

[quote="N.T.TUAN"]Not exists max and min :P[/quote]
Why can you say that?

[quote="chien than"]Let $x;y$ be real numbers.Find min and max of:
$A=x^{3}-9xy+y^{3}$[/quote]
Fix $y$. Then $\lim_{x\to \pm\infty}A=\pm\infty$.


wrong destination for the link.
the question that goes with the link i pressed was

"Five coins look the same, but one is a counterfeit coin with a different weight than each of the four genuine coins. Using a balance scale, what is the least number of weighings needed to ensure that, in every case the counterfeit coin is found and is shown to be heavier or lighter?"

Well, the link seems to be fixed now.

If you count the number of toothpicks in the diagram provided, you can see a pattern:
stage 1: 3 toothpicks
stage 2: 5 toothpicks
stage 3: 7 toothpicks

There are actually 2 very similar ways to see this pattern:

Stage n: the nth odd number
Stage n: 3n-(n-1) or 2n+1<--which is essentially how you calculate the nth odd number

So just plug in 15: 2n+1=30+1= Ans: 31


Let $ a,b\in \mathbb{R}$ . Show that :

\[ \frac{a}{\sqrt{a^2\plus{}b^2}}\plus{}\frac{b}{\sqrt{9a^2\plus{}b^2}}\plus{}\frac{2ab}{\sqrt{(a^2\plus{}b^2)(9a^2\plus{}b^2)}}\le \frac{3}{2}\]

This problem had been posted many times before.
We solve this by trigonometric method.
Let $ cosx\equal{}\frac{|a|}{\sqrt{a^2\plus{}b^2}}$, $ cos y\equal{}\frac{|b|}{\sqrt{9a^2\plus{}b^2}}$
Then use the lemma that if $ x\plus{}y\plus{}z\equal{}0$,then $ cosx\plus{}cosy\plus{}cosz \leq \frac{3}{2}$

[quote="Pain rinnegan"]Let $ a,b\in \mathbb{R}$ . Show that :
\[ \frac {a}{\sqrt {a^2 \plus{} b^2}} \plus{} \frac {b}{\sqrt {9a^2 \plus{} b^2}} \plus{} \frac {2ab}{\sqrt {(a^2 \plus{} b^2)(9a^2 \plus{} b^2)}}\le \frac {3}{2}\]
[/quote]

I've seen the problem some days before,I've solved it using only AM-GM.Note that the inequality is equivalent to the following:

$ \frac{1}{ \sqrt{1\plus{}3x} } \plus{} \frac{1}{ \sqrt{1\plus{}3y} } \plus{} \frac{2}{ \sqrt{ (1\plus{}3x)(1\plus{}3y) } } \leq  \frac{3}{2}$

With $ x,y >0$ and $ xy\equal{}1$

Now,you're free to use AM-GM easily  :wink:  :)


The children sitting around a circle are playing the game as follows. At first the teacher gives each child an even number of candies (bigger than $ 0$, may be equal, maybe different). A certain child gives half of his candies to his neighbor on the right. Then the child who has just received candies does the same if he has an even number of candies, otherwise he gets one candy from the teacher and then does the job; and so on. Prove that after several steps there will be a child who will be able, giving the teacher half of his candies, to make the numbers of candies of all the children equal.

Just a remark: it is very similar to http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=2747 but in that thread all children pass candy simultaneously.


USAMTS
algorithm
algebra
system of equations

Let's say you had a 4-by-4 system of equations. Do you have to show every step of how you explicitly solved it, or can you just say you used a calculator to solve it and you got these answers, etc...?

You can use Mathematica or some other program and just show that the system solver actually gives you the correct answer.

[quote="Math Geek"][...] and show that the system solver actually gives you the correct answer.[/quote]

So all you have to do is retrieve the answers and then plug it in to the equations and check that they work?

But I guess you have to cite your calculations.

[quote="worthawholebean"]As long as you prove that your solution works, you're fine. It may be more interesting mathematically to show how you found it, but that is not strictly necessary.[/quote]
That's where the citation part comes into play, especially when you use well-known proofs from mathematical journals or encyclopedias like Wolfram MathWorld.

But what I'm solving here is just a mundane 4-by-4 system of equations (with, as expected, 4 variables). Blah. It is not interesting.

So do I state that I used my calculator's solver, then plug the values the calculator spits out back into the four equations and show that they are all satisfied?

Hey, Math Geek, are you doing usamts this year as well? That's great.

To make this post not a spam, I will give my suggestion to the OP: if it is an ordinary system of equations which can be solved using some elementary algorithms, then you are probably fine if you don't show the calculations. That said, I do not guarantee it would earn all points.

[quote="Math Geek"]Hmm, yongyi, i don't think telling people that you're using a 4x4 system of equations is very wise, cuz it's sort of a giveaway for people to approach any of their problems using that method.  Then again, enough people have viewed this for me to accurately say that the damage has been done. :(

HINT: Be more vague next time.[/quote]

Let me just say this. My question was not a giveaway. Read my pm for details.

If you are talking about a linear system of 4 equations in 4 variables, you are probably fine not to show your steps.  If you are talking about a nonlinear system of 4 equations, I would suggest you show your work.

on the USAMTS website it says that u can use programs as long as you show that the solver satisfies the problem. So you dont have to technically solve it by hand even if its nonlinear.


ratio
geometry
3D geometry
sphere
integration
calculus
calculus computations

Math Day at the Beach is our contest for high schools in Southern California. The following question was on our team round, and only 4 of the 32 teams got the correct answer.

A plane intersects a sphere of radius $R$, with the distance from the plane to the center of the sphere being $x$. The plane cuts a cap-shaped piece off from the sphere. A cone is inscribed in this cap in such a way that the circular base of the sphere cap is the base of the cone. What is the limit as $x$ tends to $R$ from below of the volume of the cone divided by the volume of the spherical cap?

[hide="My cat when it`s in pain... in other words, the solution."]
Let $r$ be the radius of the small circle. Then, $V_\text{cone}= \frac{\pi r^{2}h}3 = \frac{\pi \left( R^{2}-x^{2}\right) \left( R-x \right)}3$.
We have that $V_\text{cap}= \pi \int_{x}^{R}\left( R^{2}-t^{2}\right) \, dt$, which, after some computations, should give $\frac{\pi \left( 2 R^{3}+x^{3}-3 R^{2}x \right)}3 = \frac{\pi \left( x-R \right)^{2}\left( x+2R \right)}3$.
The limit should be $\frac23$.
[/hide]

I don't understand what is hard in this problem?  :) usual problem from usual book.  :wink:

The "usual book" that I have has a two-dimensional analogue of this problem in a section that would be covered in a first-semester calculus course, but way out near the end of a list of problems, where the hardest problems and "special challenges" lie, and I don't see the three dimensional version anywhere in that book.

The problem had essentially three steps. The first step was decoding the verbally presented information, forming a mental picture, and deciding what needed to be done. The second step - the easiest one - was computing the volumes. The third step was doing the necessary algebraic work to find the limit.

I would characterize what perfect_radio posted as "streamlined" or "efficient." In particular, he has presented the volumes already usefully factored. My conjecture is that many teams had volumes computed but hadn't yet reached that kind of useful factorization when time ran out on them. (Note that you need to find two factors of $(R-x)$ in both numerator and denominator.) Algebraic manipulation is an art, and doing algebraic manipulation efficiently is not a skill that everyone has.

No, this wasn't deep analysis, it was "merely" computational calculus. But not all computations are easy or immediate. I personally was not surprised that the problem proved difficult. This was one of 8 questions which teams of six students had to answer in 30 minutes, total. In many cases, not everyone on the team would have known any calculus at all; I was banking on the assumption that everyone had someone in a calculus course.

[quote="Kent Merryfield"]This was one of 8 questions which teams of six students had to answer in 30 minutes, total.[/quote]
This is the key piece of information. Although the problem is "usual", it does take time to do all the steps and 30 minutes pass pretty quickly.

[hide="Answer."]
[url=http://en.wikipedia.org/wiki/Pablo_Francisco]Pablo Ridson Francisco[/url] is an American stand-up comedian born in Tucson, Arizona now living in Los Angeles.
[/hide]

I remember that problem. I didn't do it, since I was working on the complex number problem (which I wasn't able to solve...). At home, I found that it was pretty doable, but the rest of my team couldn't figure it out... :P

I don't really understand what it means by a "cap-shaped piece."  Can someone please elaborate?


Let $a,b,c >0$ real numbers...

Prove that:
\[ \frac{a}{b+2c} + \frac{b}{c+2a} +\frac{c}{a+2b} \geq  1. \]

Carlos Bravo :)
LIMA-PERU

It's just Cauchy-Schwartz
$\frac{a}{b+2c} + \frac{b}{c+2a} +\frac{c}{a+2b} = \frac{a^2}{ab+2ac} + \frac{b^2}{bc+2ba} +\frac{c^2}{ca+2cb} \geq \frac{(\sum a)^2}{3\sum ab}\geq 1$

Another solution for this inequality is to to subsitute $m=2c+b, n=2a+c$ and $p=2b+a$.
 Then $a=\frac{p+4n-2m}{9}$, $b=\frac{m+4p-2n}{9}$ and $c=\frac{n+4m-2p}{9}.$
 So the given inequality reduces to $\frac{p}{m}+\frac{m}{n}+\frac{n}{p}+4(\frac{n}{m}+\frac{p}{n}+\frac{m}{p})\geq 15.$
  But, this last inequality follows immediately from AM-GM. ;)


This is kind of a technical question, but for the activities list on college applications, if there was an activity during the summer between..say, your junior and senior year, would you list it under 11th or 12th grade?  Or both?  Or does it really matter?
Also, on the summer activities, do summer activities during the summer between your 8th and 9th grade years count?  And if so, do you list them as 9th grade?  Then that would just make everything else one grade higher, so i guess they are related questions. thanks.

The application forms I have seen distinguish between school year activities and summer activities, but perhaps you are looking at a different form from the ones I have seen. Is there a downloadable copy of the form you are puzzled about somewhere, so that I could take a look at it too?

yea. i was wondering the same thing. On any app that says lists activities (summer camps included) and then put the grade(s) which the activity was done. This is on most apps, like Stanford, MIT, the common app (i think).
yea. does anyone know. its probably somewhere on collegeconfidential.

On the [url=http://app.commonapp.org/index.cfm?APP=AppOnline&ACT=Display&DSP=Forms]Common Application[/url], I would check the box for the grade you have already attained. That is, if you did it in the summer between tenth grade and eleventh grade, check tenth grade. And don't worry about it too much, as I doubt this is a "deal-breaker" in a college application.

Good luck in your applications.

[quote="Geehoon"]I did what tokenadult suggested.
you're not in 12th grade yet until school year of senior starts.[/quote]

that's how my son did it too. But he included summer after 8th grade because he attended a high school program.


i will start with a phrase. read it aloud to yourself, and choose a phrase that sounds similiar to the one above. every phrase will be a phrase that sounds similiar to the phrase in the above post but is slightly changed. a few posts wont show a significant change, but over time, you will end up with something completely different.

I will begin:

I just got rickrolld because I deserved it.

next person changes that phrase to something sounding similiar.

This is a lot like Genetic Mutation, which I think just died.


[quote="Davron"]Let ABCD be a convex, and let I and J be the incentres of trinagles ABC and DBC, respectively. The line IJ intersects the lines AB and DC at the points E and F, respectively. Given that the lines AB and DC intersect at P, and PE=PF, prove that the points A,B,C,D lie on a circle.

davron L.F.[/quote]

Let's note
$\angle IJC+\angle IBC=\angle PFE+\angle JCF +\angle IBC=(90^\circ-\frac{\angle BPC}{2})+(90^\circ-\frac{\angle PCB}{2})+(90^\circ-\frac{\angle PBC}{2})=(270^\circ-\frac{\angle CPB+ \angle PCB+ \angle PBC}{2}) = 180^\circ$
So  $BCJI$ is inscribed in a circle.
Thus $\angle CJB =\angle CIB \Rightarrow \angle CDB = \angle CAB \Rightarrow$
$ABCD$ is inscribed in a circle.

V:)

Please explain how can you say that :
$<IJC=<PFE+<JCF$ thanks.

davron

[quote="Davron"]Please explain how can you say that :
$<IJC=<PFE+<JCF$[/quote]

In fact, by the sum of angles in triangle JCF, we have $180^{\circ}-\measuredangle CJF=\measuredangle CFJ+\measuredangle JCF$; in other words, $\measuredangle IJC=\measuredangle PFE+\measuredangle JCF$.

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=58690 for this problem.

  Darij

ahhh now i got | stupidly, wrongly took the extensions of BA and CD lol :(

davron

\[ \underline {\mathrm {Very\ nice\ and\ simple\ problem\ !\ Here\ is\ a\ clear\ proof.}}  \]  :huuh: We suppose that the line $BC$ separates the points $P$ and $A\ .$ Therefore,
$PE=PF\Longleftrightarrow m(\widehat {BEF})=m(\widehat {CFE})=\phi\Longleftrightarrow 2\phi +B+C=360^{\circ} \Longleftrightarrow$
$\boxed {180^{\circ}-\phi=\frac{B+C}{2}}\Longleftrightarrow m(\widehat {BIE})=180^{\circ}-\phi-\frac B2=\frac C2=m(\widehat {BCJ})\Longleftrightarrow$
$\widehat {BIE}\equiv \widehat {BCJ}\Longleftrightarrow$ [u][b]The quadrilateral $BIJC$ is cyclically[/u][/b] $\Longleftrightarrow$
$\widehat {BIC}\equiv\widehat {BJC}\Longleftrightarrow 90^{\circ}+\frac 12\cdot m(\widehat {BAC})=90^{\circ}+\frac 12\cdot m(\widehat {BDC})\Longleftrightarrow$
$\widehat {BAC}\equiv\widehat {BDC}\Longleftrightarrow$ [b][u]The quadrilateral $ABCD$ is cyclically.[/u][/b]


search
geometry
geometric transformation

Does anybody know any Physics forum that is similar to this one? i've researched some and found a few forums such as the advanced physics forum, but none of them close to this one..

Which forum are you referring to when you say "this one"? Do you mean [url=http://www.physicsforums.com/]www.physicsforums.com[/url]?

Try doing an internet search for "physics forum". I'm sure some results will come up. You could also try the sci.physics usenet newsgroup.

He's asking

Mathematics : AoPS Forum :: Physics : ???

That's a perfect translation TZF, I couldn't have put it better :)

Does such a forum exist? I've searched and haven't found and none of you guys know any so guess there isn't.. thanks anyways


What are you talking about???

Could you clarify a little?

I think this is a reference to the Black Eyed Peas song "My Hump"...at least I think it's something like that.

So its asking the next line in the song?  I don't really know that song.

If it's asking the next line, I don't think that any of them are exactly right.

um, i don't get this topic.  I get that this is a song but theres no point to this topic.  My answer to the poll is the 1st option, by the way. :)  :)

I'm asking you a question, and you answer.  It's as simple as that.

It isn't so simple when you don't know how to answer.

I'm get get get get get you drunk....

  lol  :D

  The song's ok, but I find the "mix your milk with my cocoa" part especially disturbing...

For some reason, this reminds me of bananas.

this topic needs to get locked or something.

Yea this is so pointless and stupid it should get locked.

[quote="G-UNIT"]I'm get get get get get you drunk....

[/quote]

I believe there were only [b]4[/b] "get" 's

I thought it was "Imma get get get get you drunk..." or something like that.

It gets stuck in my head  :|

[quote="daveed"]Oh my gosh!  On wednesday, I was at programming team, and was working on a problem when, in a burst of creativity, decided to name all my variables, classes, and methods as lines from that song.

And then at the end, the teacher had to read my solution

... :blush:[/quote]
ROFLMAO  :rotfl:
Definitely interesting class names.

my hump, my hump, my hump, my lovely little lumps, check it out.
I drive these brothers crazy, i do it on the daily, they treat me really nicely, they buy me all these ices.

Have you all been brainwashed or something????? :o


I had a question about the rules. It is can we submit multiple proofs of certain parts of problems if we are unsure about a proof and we make it one solution? i.e. can i say here is a proof of blah blah blah. an alternate proof is blab blab blab.  if so, will we be counted off for having one of the two incorrect.  i realize that this would not be logical for a non-proof problem (you cannot submit multiple final results and expect to get it right) and may be viewed as kind of "guessing at the proof" but in its defense, if it shows the same result, isn't it just similar to showing your work, or what you tried to do to solve the problem, which isn't necessarily bad?

thanks

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=163379

Read this.

Personally, I would think it would be best if you looked over your work and selected the proof that looked the most correct.


There must be a faster way than trial and error:

There exists a unique pair $ (x,y)$ integers so $ x^2\plus{}84x\plus{}2008\equal{}y^2$ find $ x\plus{}y$

Hint please?

Subtract $ y^2$ and view this as a quadratic in $ x$. How can you exploit the fact that $ x$ and $ y$ are integers?

Using TachyonPulse's hint I get four possibilities for $ x, y$ integers, (but only one if they are restricted to positive integers):

[hide=" Solution"]

$ x^{2} \plus{} 84x \plus{} 2008 \minus{} y^{2} \plus{} 0$

$ x \equal{} \frac { \minus{} 84 \pm \sqrt {7056 \minus{} 4(2008 \minus{} y^{2})}}{2}$

$ \equal{} \minus{} 42 \pm \sqrt {y^{2} \minus{} 244}$

So we need $ y^{2} \minus{} 244$ to be a square, say $ z^{2}$

$ (y \minus{} z)(y \plus{} z) \equal{} 244$

Considering the factors of $ 244$, we get a number of pairs of simultaneous equations of which the only two pairs which work (make $ x, y$ integers and $ y^{2} \minus{} 244$ a square) are

$ y \minus{} z \equal{} 2$ and $ y \plus{} z \equal{} 122$, giving $ y \equal{} 62, z \equal{} 60$  and $ y \minus{} z \equal{} \minus{} 2$ and $ y \plus{} z \equal{} \minus{} 122$, giving $ y \equal{} \minus{} 62, z \equal{} \minus{} 60$

$ x \equal{} \minus{} 42 \pm 60 \equal{} 18$ or $ \minus{} 102$. This leads to four solutions:

$ (x, y) \equal{} (18, 62), (18, \minus{} 62), ( \minus{} 102, 62)$ and $ ( \minus{} 102, \minus{} 62)$

and so $ x \plus{} y \equal{} 80, \minus{} 44, \minus{} 40$ or $ \minus{} 164$.

[/hide]

P.S. kunny has given a path to the solutions of both your problems in response to your post of problem 8. It shows a neater solution than this and confirms that $ x$ and $ y$ have to be positive integers, so the only correct answer is $ x\plus{}y\equal{}80$

Yes, I meant positive integers.

Thank you guys

already discissed at

[url]http://www.mathlinks.ro/resources.php?c=182&cid=45&year=2008&p=1075124[/url]


1.)  Solve  2x – 5(4 – x)  =  4(2x – 6) – 2

  2.) solve  2x/5 + 8 = x/2 - 4

These problems have taken me for a loop how do i solve them  :huh:  :?

[hide="2"]
Hmm... I'm not sure if this works. Tell me if it's wrong.
$\frac{2x}{5}+8=\frac{x}{2}-4
\\$
First, I think you multiply the whole equation by 10.
$\frac{20x}{5}+80=\frac{10x}{2}-40\\
\\
4x-80=5x-40\\
\\
-40=x$[/hide]

you are right until you got to the -40 part
see you must remember that what you do the left you must do to the right so

                4x - 80 = 5x - 40
                  i take a short cut to this step
                  if you are good with sign numbres you can do this
                 4x - 5x = 80 -40
                          x = 40 see what i did when i moved the -80 to the other side of the = sign it became + 80 - 40= + 40
and the + 5x became - 5x. thank you so much for getting me this far i was going to throw the labtop out the window!

This would've been a nice problem in the equation solving contest  :D  Except it's a little easy for round 3.

[quote="236factorial"][hide="number2"]

$\frac{2x}{5}+8=\frac{x}{2}-4$

$\frac{-x}{10}=-12$

$x=120$[/hide][/quote]

OHH! Blah, I knew that. :blush:  Thanks for showing that.  :D


Princeton college MATHCOUNTS MIT \/closed

Hey. Poll.

Do you think we should have Math Jams more often?

[quote="solafidefarms"]What should AoPS have more math jams on?[/quote]

Lots of stuff! Maybe a math jam specific to specific contests and stuff like that. More math problem solving jams and college jams.

But I really want more math jams for specific contests, especially the newer ones like Princeton Math Competition.

[quote="lotrgreengrapes7926"]We should ask Mildorf to hold an iTest Math Jam. :D[/quote]

...if iTest didn't make money off of selling solutions

I guess we could hold an informational math jam about the Princeton Math Contest, if enough people wanted it. But since there are no past tests yet, we can't really hold a PUMaC-themed problem solving math jam. :P

[quote="kyyuanmathcount"]I noticed there aren't any MATHCOUNTS prep or countdown mathjams.  Those were really fun, would somebody consider hosting one?[/quote]

Yes. I completely agree. I would really appreciate those math jams too. They were very useful as well as fun. I hope you bring them back.

[quote="MithsApprentice"]It was mentioned at the last MIT Math Jam that there would potentially be more informational ones from other college admissions officers. Anyone know how that's coming along?[/quote]We have one from Harvey Mudd on October 24.

Will there be any mathcounts countdown ones? Also, can there be mathjams for mc after county state and nats like there is for AMC 10/12 where the admins go over requested problems from the competition?

[quote="bpms"]Will there be any mathcounts countdown ones?[/quote]Possibly in 2007, as we get closer to the MC state/national dates.

[quote]Also, can there be mathjams for mc after county state and nats like there is for AMC 10/12 where the admins go over requested problems from the competition?[/quote]We do not have permission to use current-year MC problems for something like this.

[quote="neelnal"][quote="mysmartmouth"]How does the classroom work as far as resource drain/availability?

It would be nice if in the future, individual AoPS users could schedule the classroom to use for a math jam.  Kind of like the Mock Countdowns people hold via AOL Instant Messenger.  Would that ever be possible?[/quote]

That would be cool, but I can see why it could also lead to chaos.  :maybe:[/quote]

how? he said people can schedule it which I think means you can sign up for time slots.

[quote="SplashD"][quote="bpms"][quote="neelnal"]oh ok. Well, I thought that people would just flood the admins with requests to have their own jam.[/quote]
People would... How can they make sure only certain people enter the jam?[/quote]

well maybe the person who signed up for the time slot could be the mod and would be able to kick out anybody who isn't supposed to be there.[/quote]
Well obviously they would otherwise the incorrect and correct answers would show up giving no benefit over AIM.


Hey, ppl from the AoPS

      I have a therom that is ture but need prove,

          thx everyone :P

Wrong forum, send this one to Intermediate Topics. This same problem has been posted already many times, I'll post the links after I search for them.

EDIT: Ok I found one [5 minutes of search]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=130109[/url]


In a narrow alley of width $w$ a ladder of length $a$ is placed with its foot at point $P$ between the walls. Resting against one wall at $Q$, the distance $k$ above the ground makes a $45^\circ$ angle with the ground. Resting against the other wall at $R$, a distance $h$ above the ground, the ladder makes a $75^\circ$ angle with the ground. The width $w$ is equal to

$\textbf {(A) } a \qquad \textbf {(B) } RQ \qquad \textbf {(C) } k \qquad \textbf {(D) } \frac{h+k}{2} \qquad \textbf {(E) } h$

[hide="hint"]Form a square so that the answer is easier to find[/hide]

[hide="hint"]The triangle in the center is equilateral (why?)[/hide]

[quote="Hokkage"][hide="Answer"]A) a. The angle between the two ground angles (45 and 75) is 60, because it's a straight line and thus 180 degrees. Since the the triangle formed has lengths a, a, and w, the two other angles must be the same and add up to 120, thus making it equilateral. Therefore w = a.[/hide][/quote]Not there yet. Think square.

[hide="Finished I think"]Ah, what I got isn't the horizontal, but the other side of the triangle. Since you have congruent triangles (the angle above the intersection of the top side of the triangle with k is 45) then the length is h. Sorry if this answer is unclear. E) h.[/hide]

Since no one posted a solution yet...
[hide="Solution"]
Connect the tops of the two ladders to form a length of $a$, then apply pythagorean theorem: $a^2=w^2+(h-k)^2$. We can also apply pythagorean theorem on the $75$ degree triangle: $(w-k)^2+h^2=a^2$.

Thus $(w-k)^2+h^2=w^2+(h-k)^2$. Simplifying yields $w=h$.

[quote=4everwise] Resting against one wall at $Q$, the distance $k$ above the ground makes a $45^\circ$ angle with the ground.
[/quote]
I culdn't understand this. Could you,please, "translate" it?


Pe laturile $AB$ si $AC$ ale triunghiului ascutitunghic ABC se construiesc in exterior triunghiurile asemenea $ABM$ si $ACN$ astfel incat $m(\angle M)=m(\angle N)=90^\circ$ si $m(\angle MBA)=m(\angle NCA)=m(\angle BAC)$. Din mijlocul  $A_{1}$ al laturii $BC$ se duce perpendiculara $A_{1}A'$ pe $MN$. Analog obtinem dreptele $B_{1}B'$ si $C_{1}C'$. Sa se arate ca $A_{1}A'$, $B_{1}B'$ si  $C_{1}C'$ sunt concurente.

cred ca am o solutie destul de urata...
[hide="solutie"]
pentru inceput, voi "lamuri" pozitia punctelor $M$,$N$ etc...
fie $\triangle UVW$ determinat de paralela prin $A$ la $BC$, paralela prin $B$ la $AC$ si paralela prin $C$ la $AB$. (a.i. $U$ se opune lui $A$, etc). deci $\triangle ABC$ este triunghi median in $\triangle UVW$.
din conditiile problemei, rezulta ca $M\in[UW]$ si $N\in[UV]$. mai mult, $AM\bot UW$, deci $AM$ este linie mijlocie in triunghiul $WVV'$, unde $V'$ este piciorul inaltimii din $V$ pe $UW$. asadar $M$ este mijlocul lui $[WV']$. analog celelalte puncte sunt mijloacele segmentelor determinate de piciorul unei inaltimi si unul din varfuri.
acum, voi arata ca dreptele din enunt sunt chiar mediatoarele segmentelor $[MN]$, etc...
cum $A_{1}$ este mijlocul lui $[BC]$, rezulta ca $A_{1}\in [AU]$ (deoarece $ABUC$ este paralelogram), ba chiar $A_{1}$ este mijlocul lui $[AU]$. cum triunghiurile $\triangle AMU$ si $\triangle ANU$ sunt dreptunghice (in $M,N$), rezulta ca $MA_{1}$ si $NA_{1}$ sunt mediane, deci $MA_{1}=NA_{1}=AA_{1}$. deci perpendiculara din $A_{1}$ pe $MN$ este chiar mediatoarea acestui segment.
acum mai ramane sa arat ca aceste mediatoare sunt concurente.
pentru aceasta, voi reformula problema.
fie $\triangle ABC$ un triunghi, $A',B',C'$ picioarele inaltimilor. punctele $X,Y,Z,T,U,V$ sunt mijloacele segmentelor $[BA'],[A'C],[CB'],[B'A],[AC'],[C'B]$. atunci mediatoarele segmentelor $[XT],[YU],[ZV]$ sunt concurente.
demonstratie :
fie $S$ intersectia dintre mediatoarele segmentelor $[XT]$ si $[YU]$. ideea va fi sa exprim faptul ca $SX=ST$ si $SY=SU$, cu formula medianei. (este mai convenabil sa ridic la patrat).
$4SX^{2}=2(SB^{2}+SA'^{2})-BA'^{2}$.
$4ST^{2}=2(SA^{2}+SB'^{2})-AB'^{2}$.
$4SY^{2}=2(SC^{2}+SA'^{2})-A'C^{2}$.
$4SU^{2}=2(SA^{2}+SC'^{2})-AC'^{2}$.
adunand prima cu a patra si a doua cu a treia (tinand cont de egalitatile de mai sus), obtin:
$2(SB^{2}+SA'^{2}+SA^{2}+SC'^{2})-BA'^{2}-AC'^{2}=2(SA^{2}+SB'^{2}+SC^{2}+SA'^{2})-AB'^{2}-A'C^{2}$.
deoarece perpendicularele ridicate in $A',B',C'$ sunt concurente (in $H$), din lema lui carnot se obtine ca $BA'^{2}+AC'^{2}+CB'^{2}=B'A^{2}+A'C^{2}+C'B^{2}$.
deci, grupand favorabil termenii mai sus, se obtine
$2(SB^{2}+SC'^{2}-SB'^{2}-SC'^{2})=C'B^{2}-CB'^{2}$, adica
$2(SB^{2}+SC'^{2})-BC'^{2}=2(SB'^{2}+SC^{2})-B'C^{2}$.
insa acestea din urma sunt chiar lungimi de mediane, mai precis, membrul stang este $4SV^{2}$ si membrul drept este $4SZ^{2}$.
deci $SV=SZ$, adica $S$ se afla si pe mediatoarea lui $[VZ]$ (ceea ce aveam de demonstrat).
[/hide]

Si eu am o demonstratie.  :D [hide]Ca si in prima demonstratie se arata ca dreapta $A_{1}A'$ este mediatoarea lui $MN$. Se pot considera mijloacele laturilor triunghiului initial si pe urma se obtine o congruenta de triunghiuri. Fie $B_{2}$ si $C_{2}$ picioarele perpendicularelor din $B$ si $C$, iar $S$ mijlocul segmentului $B_{2}C_{2}$. Se demonstreaza vectorial ca $ASA_{1}A'$ este paralelogram si deci dreapta suport a laturii $A_{1}A'$ este simediana in triunghiul median triunghiului $ABC$. Scriind si analoagele rezulta concluzia.[/hide]


Is it possible to prove problem #5 on the USAMO using induction? It wasn't on the AMC website's list of possible proofs, but is it possible that it's still valid? Also, how much do the graders take off if you miss an answer by a little bit? (specifically, i said 2k^3+3k^2+k instead of 2k^3+3k^2+3k because i started with k^2 instead of k^2+1). Thanks for reading!

Please use the links in the [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=27&year=2006]Resource pages[/url] to discuss the problems.  For example, problem 5 can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=490682#p490682]here[/url]

Well I personally proved 5 without induction.

You can work backwards and consider the reverse functions.

Hey i proved it by induction, too. Look at the 2 paths ofthe frog and look at its lust jump in the longer path on a number j<2^i.
j+2^(m index j +1)>2^i, and at this point, u crush with induction ;)

hahaha! that's what i did! except i didn't finish the full part of the induction but i proved that part at least...except i didn't think i'd get anywhere so i abandoned it  :(


modular arithmetic
Additive Number Theory

In how many ways can $2^{n}$ be expressed as the sum of four squares of natural numbers?

For $ n\ge3$ we have $ x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2 \equal{} 2^n$, and since $ x^2\equiv0,1,4\pmod8$, we know that $ 2|\gcd(x,y,z,t)$.
So cancelling a factor $ 4$ from both sides we get at least one of $ x,y,z,y$ odd, and by the above this implies we need $ 0 \plus{} 0 \plus{} 0 \plus{} 1 \equal{} 1$ or $ 1 \plus{} 1 \plus{} 1 \plus{} 1 \equal{} 4$ modulo 8, but the only way to have $ 2^n\equiv1,4\pmod8$ is $ n \equal{} 1,2$ respectively.
The first has no solutions, the latter only $ 1^2 \plus{} 1^2 \plus{} 1^2 \plus{} 1^2 \equal{} 2^2$, so the final answer is: there are no solutions for $ n$ odd, and for $ n \equal{} 2k$ there is a unique solution $ \left(2^{k \minus{} 1}\right) \plus{} \left(2^{k \minus{} 1}\right) \plus{} \left(2^{k \minus{} 1}\right) \plus{} \left(2^{k \minus{} 1}\right) \equal{} 2^{2k}$.

[size=75][color=red][mod team: solution rewritten readably][/color][/size]


Substitute $x=a+2000$ to get

$\sqrt{{a\over 23}+1}+\sqrt{{a\over 22}+1}=\sqrt{{a\over 1977}+1}+\sqrt{{a\over 1978}+1}$

$\sqrt{{a\over 23}+1}-\sqrt{{a\over 1978}+1}+\sqrt{{a\over 22}+1}-\sqrt{{a\over 1977}+1}=0$

$\frac{{a\over 23}-{a\over 1978}}{\sqrt{{a\over 23}+1}+\sqrt{{a\over 1978}+1}}+\frac{{a\over 22}-{a\over 1977}}{\sqrt{{a\over 22}+1}+\sqrt{{a\over 1977}+1}}=0$

\begin{eqnarray*}a\left(\frac{1}{23\cdot1978\left(\sqrt{{a\over 23}+1}+\sqrt{{a\over 1978}+1}\right)}+\frac{1}{22\cdot1977\left(\sqrt{{a\over 22}+1}+\sqrt{{a\over 1977}+1}\right)}\right)=0\end{eqnarray*}

Since the expression in the parentheses is clearly positive, the only solution is $a=0\iff x=2000$.


geometry
circumcircle
incenter
geometric transformation
reflection
Euler
Mafia

In a triangle $ABC$ two lines, symmetric to $AC$ respectively with respect to $AB$ and $BC$, intersect in $K$. Prove that points $B$, $K$ and $O$ are collinear (where $O$ is the circumcenter of triangle $ABC$).

[quote="Megus"]symmetric to $AC$ respectively with respect to $AB$ and $BC$[/quote]

What does this mean?

First let's assume that $B$ is acute.

Because the lines are symmetric with respect to $AC$, so $AB$, $CB$ are the exterior angle bisectors of $\angle{CAK}$ and $\angle{ACK}$. Thus $B$ is the excenter of $\triangle{AKC}$. Use the angle relations of the excenter, we know that $\angle{ABC}=90^o-\frac{1}{2} \angle{AKC}$, from which we can get that $\angle{AKC}=180^o-2\angle{ABC}$. Since $O$ is the circumcentre, so $\angle{AOC}=2\angle{B}$, thus $\angle{AKC}+\angle{AOC}=180^o$, hence $A,K,C,O$ lies on the same circle. But since $OC=OA$, so $OK$ is th angle bisector of $\angle{AKC}$. Because $KB$ is also the angle bisector of $\angle{AKC}$, therefore $K,O,B$ lies on the same line.

If $B$ is obtuse, then $B$ becomes the incenter of $\angle{AKC}$. The proof is basically the same.

exercise: give a two-lines proof based on the fact that a point has an isogonal conjugate with respect to a triangle (theorem 1 from http://www.mathlinks.ro/Forum/viewtopic.php?t=18472 post #2).

another exercise, maybe a bit harder: prove that the line joining the point K to the reflection of B in CA is parallel to the euler line of triangle ABC.

  darij (and the other pentacrats who are playing mafia right now)


analytic geometry
graphing lines
slope
calculus

If L is a line that intersects the curve with equation $ x^2\plus{}xy\minus{}4y\equal{}5$ at (1,a) and (3,b), then the slope of the line that is perpendicular to L is?

Oh how I wish I could use calculus in this thread...  No matter.
First, you're going to need those missing y-coordinates:
$ 1\plus{}y\minus{}4y\equal{}5\Rightarrow y(1\minus{}4)\equal{}4\Rightarrow y\equal{}\minus{}\frac{4}{3}$ so that is $ a$
$ 9\plus{}3y\minus{}4y\equal{}5\Rightarrow y(3\minus{}4)\equal{}\minus{}4\Rightarrow y\equal{}4$ so $ b\equal{}4$.
So now we have our two points, $ (1,\minus{}\frac{4}{3})$ and $ (3,4)$.  So now just find the slope between the two:
$ \frac{4\plus{}\frac{4}{3}}{3\minus{}1}\equal{}\frac{8}{3}$.
So then the slope of the perpendicular is just the negative reciprocal: $ \minus{}\frac{3}{8}$.


[quote="indianamath"]what score are you expecting this year?

i expect either a perfect or getting 1 wrong. i'm taking the algebra 1 test.[/quote]

ditto, except for the one wrong part...

I hopefully dont lose to some *cough* certain people.

[quote="undefined117"]hoping for a 32/32 on the Comprehensive, though it's probably not gonna happen.

realistically, i [u]always[/u] make one or two mistakes on a test even when i'm SURE i aced it.[/quote]

and comprehensive is going to be hard to get into top 5% (last year you needed 30/32), especially with all the good math people (like all the USAMO qualifiers and stuff)-not that people taking it arent good-are taking the teest

yeah, but it only goes back to 2003 though.. or is that when they started?

and i don't like how they use frames on that page..

EDIT: NVM about the 2003 thing i said above hee

i need to read more carefully.

[quote="Art of Owna"][quote="indianamath"]what score are you expecting this year?

i expect either a perfect or getting 1 wrong. i'm taking the algebra 1 test.[/quote]

Did you get highest last year in state?  And did you go to that awards ceremony?[/quote]

i tied for first.

this is next week or on the 21st..or what..i just signed up i dont exactly know when it is.

21st.(yeah i pmed you. but maybe this question will come up again.)

[quote="Art of Owna"][quote="mwpl11"][quote="mathcountsfreak"]Yeah, I'm taking algebra 1 but I'm not expecting a perfect or anything because I took a practice ICTM today and got 36 (it was last years alg 1) but MAYBE I'll do better in the competition.

EDIT: 50th post![/quote]

can you get the practice tests on the internet? or on the website?[/quote]
ya you just type into google ictm indiana math ___ where ___ is the year..it even has the 1999 tests[/quote]

grrrr...i cant find them. it just has news articles. could someone give me a link cuz i like can't read....

lol wtf i have to get up at 7:15 tomorrow..and get to the high school by 7:45..

[quote="undefined117"]haha.. ours starts at 9:30 a.m., so i have plenty of time.. gonna do like 10 rounds on the arithmetic page, and yeah.

this'll be one of the few days when i'll actually sleep before 10 p.m. :D[/quote]

I think ours starts at 9:30 too but I live like 30 mins away from the place and I think our math teachers just wants us to have an early start so we arent late.  (Im not exactly sure..since I really didnt read the information about it though)

nah i'll go to bed at 11. see you there!! Udayan!! Good luck tomorow ppl.

hmm i have to start at 9 but i have to be there by 8:45... :( Oh well thats like sleeping in compared to when i usually wake up...lol

Oh, and when will we be able to discuss scores and questions and stuff? because tomorrow's the only day to take it, so we should be able to dicuss like sunday or so...right?

dude! sweet.... i'm taking geometry sexiness... and all you guys are talking algebra I and comprehensive, so i can't own u ^_- lol

lol that woudnt let me take comprehensive because I was still in algebra II in school.

when are we allowed to discuss the problems?

i've checked my entire test and i think i got 30/32

and TWO lucky guesses :D

Oh we already know our scores....I dont know if you can discuss though..

Are you allowed to discuss the test?

If you arent are you allowed to discuss place and score?

And I wish I took Comprehensive because I heard it was much more "interesting" (not saying how) this year than previous years (not that I dont like my algebra II score)

Yeah, "interesting" it was. I felt as if this year's test were designed more to confuse! A lot of the answers were "none of the above" or something along those lines. Does anyone know how this year's comprehensive test, written by Rose Hulman, compares to Rose-Hulmans annual HS math contest?


Let  - 1< ax^2+bx+c<1      for all x in l-1;1l

prove that for all x in l-1;1l  we find  -1<cx^2+bx+a<1

[url=http://www.mathlinks.ro/viewtopic.php?t=145147&search_id=1228108607]Take a look here[/url]


$a_{ij}$ $(i=1,...,m; j=1,...,n)$ are non-negative real numbers such that $0 \le a_{i1} \le ... \le a_{in}$ $(i=1,...,m)$. Let $a'_{i1},...,a'_{in}$ be a permutation of $a_{i1},...,a_{in}$. Then
\[\displaystyle\sum_{j=1}^{n}(\displaystyle\prod_{i=1}^{m}a_{ij}) \ge \displaystyle\sum_{j=1}^{n}(\displaystyle\prod_{i=1}^{m}a'_{ij})\]
and
\[\displaystyle\prod_{j=1}^{n}(\displaystyle\sum_{i=1}^{m}a_{ij}) \le \displaystyle\prod_{j=1}^{n}(\displaystyle\sum_{i=1}^{m}a'_{ij}).\]
I saw it from a chinese article, its name is 微微???等?. I am not sure whether it is a well-known inequality. Can anyone show me a proof and its english name? and when does the equality hold?  :)

I posted the first inequality [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24822]here[/url], but no one gave a proof.

And shouldn't the second inequality be reversed, in light of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=17280]this[/url] thread?

Oh! you are right, Billzhao. I have edited it. :)


function
limit
Support
algebra
polynomial
inequalities
real analysis

Hi,

I have a question.
let $ b > 1$ and $ r,s,t \in \mathbb{C}$, and $ E \equal{} \{ z\in \mathbb{C}, r \plus{} s e^z \plus{} t e^{bz} \equal{} 0 \}$, consider a continuous function $ f : \mathbb{R}\rightarrow \mathbb{C}$ which is uniform limit on every compact of a sequence of linear combinaison of $ x\mapsto e^{zx}$ with $ z\in E$.

I suppose that $ f$ is not $ 0$ and $ \lim_{ \plus{} \infty} f \equal{} 0$.

Does $ E$ contain $ z$ such that $ \Re(z) < 0$.

The formula $ r f(x) \plus{} sf(x \plus{} 1) \plus{} tf(x \plus{} b) \equal{} 0$ may help, I hope.

Thanks

the function $ f$ is called a mean periodic function

[quote="holomorf"]Does $ E$ contain $ z$ such that $ \Re(z) < 0$.
[/quote]
Obviously not if $ |r|>|s|\plus{}|t|$.

of course I agree.

But I know nothing about (r,s,t). I just know the behaviour of f.

Note that if $ \lim\limits_{x\rightarrow \plus{}\infty} e^{zx}\equal{}0$ if and only if $ \Re(z)<0$, that's why I hope that such $ z$ exists.
The difficulty is that uniform compact limit is weaker than uniform limit on R.

If you're interested, there's a non trivial theorem which says that if a continuous function f is such that $ f \star \mu\equal{}0$ where $ \mu$ is a Radon measure with compact support, so $ f$ is compact uniform limit of sequence of linear combinaison $ x\mapsto e^{zx}P(x)$ where $ P$ is polynomial and $ e^{zx}P(x) \star \mu\equal{}0$. (it's the theorem of mean-periodic functions)

In my issue, $ \mu$ is just $ r\delta_0\plus{}s\delta_1\plus{}t\delta_\beta$. And $ e^{zx} \star \mu \equal{}0\Leftrightarrow r\plus{}se^{z}\plus{}te^{\beta z}\equal{}0$.
And I'm asking if the behaviour at infinity gives us some data on the z which appears.

Ah, sorry, I somehow misread the post entirely. The answer is "yes". I have no time to post the details right now but I'll try to do it later today. The argument is somewhat involved but completely standard. The first observation is that zeroes are well-separated in the imaginary direction. This allows us to split each approximating polynomial $ g$ into the sum of $ g_k$ where each $ g_k$ contains at most 3 terms and the exponents in different $ g_k$ are well-separated. The next step is to prove the inequality $ \sum_k\|g_k\|^2_{L^2(a-T,a+T)}\le
C|\g\|_{L^2(a-2T,a+2T)}^2$ for all $ a>0$ and sufficiently large $ T>0$. Then we prove the estimate $ \|g_k\|^2_{L^2(-T,T)}\le C\sum_{m=1}^3\|g_k\|^2_{L^2(am-T,am+T)}$ for all $ a>0$ under the assumption that all real parts are non-negative, and finally we show that $ \|g\|_{L^2(-T/2,T/2)}^2\le C \sum_k\|g_k\|^2_{L^2(a-T,a+T)}$. The final estimate is
\[ \|g\|_{L^2(-T/2,T/2)}^2\le C \sum_{m=1}^3\|g\|^2_{L^2(am-2T,am+2T)}\]
which obviously finishes the story by first passing from $ g$ to $ f$ and then passing to the limit as $ a\to+\infty$.