MATHCOUNTS AMC USA(J)MO USAMO HMMT percent

Doesn't everyone think that sixteen people should be included in the Countdown Round at Nationals? I mean, with twelve people, it's pretty obvious that the people in charge want to give an advantage in the Countdown Round to those who did better in the written round, but I believe it's an obscenely big advantage, and shouldn't even be in existence at all. The Countdown Round is supposed to be a completely different part of the competition, the written round shouldn't impact the result of the countdown round [u]too[/u] much.

Take this year for example. I mean, more people thought that Bobby Shen, Richard Ni, Michael Wu, and Carl (he should have known to read, we would have finished second, but that's a different story...) would make countdown than Michael Zhao, the two Minnesota kids, and Paul Davey. Not to say that they were not deserving, but still, it would be better if sixteen had made it.

By the way, according to Carl, he claims that if he had come in 9th (which was quite possible with ties), he might have made the finals, and he's actually right because he would have a significantly easier path to beat people, and he's pretty good (not as good as Kevin Chen, Allen, and the New Jersey kid).

Does anyone have thoughts on this matter?

I honestly think that MATHCOUNTS should stick with what they have.  If you took that test, then I think you'd agree that a great advantage should be given to those few who manage to all but conquer it.  MATHCOUNTS has been around for some 20-odd years now, and I think they should stay put on the Countdown Round format.  If I had it my way, I think all 228 should be in the C-Down, and then we play the tourney from there, based upon your rank on the Written, of course.  That way, the real "Quick Draw" would win the whole thing, not just the insanely smart kids that do well on the Written (most of the time, these two are one in the same)  So, I think that MATHCOUNTS should just play the Countdown as they do now.

so you're saying the 1 seed will have less chance if you place him/her against the 16 seed first round?

i see a lot of bias here though...

[hide="unrelevant information"]Just commenting here, you can interpret the meaning yourself.  :wink:

The four people you listed who had a 35 and who you think should have made countdown all took first place at states. The four other people who had come second to those same four in their respective states all had a respectable showing in the top 12. Coincidence?

Furthermore HALF of the top 12 took second place at states. [/hide]

The truth is, there's always going to be a cutoff and there's always going to be people just below the cutoff. You could make the same argument pertaining to extending the size of each state team for the people who got fifth(Fanatic will probably comment on this). This is totally besides the point but if you're really good like Kevin Chen and have an off day and make a couple of mistakes, you can still take third place written. In my opinion, the MN and NE people are actually rather good at countdown. If you expand countdown to 16, what's stopping you from expanding it to 32 or 64 or 228? It is what it is. Word.

Actually, if you'd ask me, I'd prefer the old system with 10 people in CD and it being a ladder format... that way winning written has more rewards. I think a lot of the skill goes into written, and there's much more like involved in the 7 problems you get for cd...

and yeah i would've love the 16 for cd and 5 per state thingy, but not gonna happen.

By the way, Nathan Salo got third at States and was in the Countdown Round.

You know while we're at it why don't we make 10 tests and take the best score, change it to 5 people from each state, and put all 285 people in a ladder style countdown round.

People have off days.  It's a fact of life.  You can't change it.  Deal with it.

I agree that the ladder style is better, though.  But bracket-style equals TV, and TV equals money.

I think SplashD's argument here is the best: Just like with the USAMO, there will always be a cutoff, and there will always be people right below it that people think should have made it.  With Mathcounts, in particular, I think there's a tendency to be inconsistent, where you might show a lot of potential in 7th grade or through other contests, but then not make the countdown.

The reason I think giving first-round byes in the countdown is a good idea is that the countdown round winner is the one who gets the title of "national champion".  If this is the case, the written round winners should get a leg up in the countdown so that the written round serves a purpose in the individual competition other than just picking out the ones who get to do countdown.  In fact, I'd like to see more bias towards the written winners, not less, though the ladder format is a bit much.

Ideally, I would also like to see a countdown for all 228 kids.  If that were the case, though, I'd start everyone on equal footing and name a "written round winner" and a "countdown round winner".  Unfortunately, most years that would result in a split "national championship".

By the way, state results mean little in terms of your chances at nationals.  My two years, I finished third and second at state, and both times I was the top student from Minnesota at nats.  One year, the guy who won state didn't even make the top 200.  This year, of the 12 countdown students, 4 finished first at state, 5 finished second, and 3 finished third.

[quote="ra5249"]Following on Sly Si's idea, would it work to make a "written" winner, a "countdown" winner, and then make those two winners face against each other?[/quote]

But would they "face off" in a written test or a countdown test?

Oh you know what would be sick? If like we divided everyone into 12 pools of 19 then did a round robin in each pool.

Then two people from each pool advance along with 8 wild cards then 32 people do a bracket

except that would take forever

[quote="footballrocks41237"]You know while we're at it why don't we make 10 tests and take the best score, change it to 5 people from each state, and put all 285 people in a ladder style countdown round.[\quote]

do you realize how long that would take?

[quote="indianamath"][quote="footballrocks41237"]You know while we're at it why don't we make 10 tests and take the best score, change it to 5 people from each state, and put all 285 people in a ladder style countdown round.[/quote]

do you realize how long that would take?[/quote]

He was being sarcastic, and mocking suggestions to ensure a 'true' winner that some people are making.

[quote="df_calculus"]Take this year for example. I mean, more people thought that Bobby Shen, Richard Ni, Michael Wu, and Carl (he should have known to read, we would have finished second, but that's a different story...) would make countdown than Michael Zhao, the two Minnesota kids, and Paul Davey. Not to say that they were not deserving, but still, it would be better if sixteen had made it.[/quote]

Hey. Michael Zhao was on my team last year :mad: :D

i thknk they should just make 2. they already work hard enough generating 1.

I think the current system is fine. It's right to reward those who do well on the written round. if you're really good and you do bad on the written, you can still win the whole thing.

[quote="mathwhiz23"]such a faceoff would not be possible. there would be no valid test to determine a national champion. the only solution is to have two national champions or a "written round winner" and a "countdown round winner."

too bad that might not happen either because its bad for publicity and tv purposes to have two winners. looks like the formats not changing any time soon.[/quote]

I think you're exactly right. They need a definite person they can point to and say "this is the winner in 2007".

Aside: In Mathleague (formerly GPML), they separate sprint and target rounds, as well as grade levels (9th through 12th), as well as small and large schools. This makes the awards ceremonies long and drawn out. Just my personal thoughts...

Personally, I think the system is good as it is. It's exciting to watch the Countdown round, especially the final rounds where $\$$2,000 is literally at stake (well, it's scholarships, but it's all the same). There's no need to change the cutoff.

And by the way, I've had enough bad experience with cutoffs. I got a 99 on the AMC-12. The cutoff was 100.

Woot! 1337th post!!!

[quote="SplashD"][quote="Fanatic"]Actually, I would like the idea of taking like 3 sprint rounds and 3 targets, and posting the results after each one. If I found out I was in 14th after the 1st, I would be going ballistic!![/quote]

Riddle: You know what would be even better?

[hide="Answer"]If MATHCOUNTS had a Guts round (for those of you who don't know what this is, look up HMMT) in addition to or instead of the countdown round. (I knew the Guts round was the funnest part of HMMT for me). True, it'd be hard to manage, though I think it's much more exciting to watch than countdown. I admit this isn't practical but real-time results are better than announcers yelling out people's names so fast that haven't came to the realization yet that they have actually been called. [/hide][/quote]

OH DUDE MASSACHUSETTS WOULD OWN OWN OWN OWN THAT UP
But actually, you're riht, countdown doesn't teach very much...maybe working quickly is important in real life but working quickly in teams is much more realistic.

[quote="mathnerd314"][quote="guyinPA"]I actually Like sly si's (hope I spelled right) idea of having a separate Countdown and Written Champion!  Anyone out there in Mathcounts listening??

 (If a student is THAT good, they could potentially win both) -- but I believe that they require two separate sets of skills - one focused on quickness and one on longer analysis - in real life, does an engineer or scientist REALLY need to solve the problem in literally Seconds, or do most great ideas require diligence, thought, and extended effort?[/quote]

By your logic, we should abandon Countdown entirely.[/quote]

I suggested that in a subtle way in my previous post...that would not be a bad idea at all.

We could always have a countdown winner and a written winner and have them face off in the [b]OCTAGON[/b]

[img]http://blogs.chron.com/fighting/ufc47_liddell_vs_ortiz-215.jpg[/img]

No, really though...  Have a sort of a written speed test....you know 60 questions 30 minutes sprint difficulty....  Pit the countdown winner against written.  This way you need both speed and uhh written skill....

[quote="rd5493"]that's a good idea, except that 19 isn't quite a round number, and how do you have a round robin w/ 19 people...

maybe MATHCOUNTS should build off of this concept...[/quote]

I'm surprised no one has said anything about this yet. You can have a round robin with any number of people because in a round robin everyone faces everyone, then the person with the highest score wins. If there's a tie they normally have some sort of tie breaking procedure. If two people each beat the same number of people the one that won out of the two of them gets the higher spot. It gets a little complicated when 3 or 4 people tie, but you could say whoever took the least number of questions to beat the same number of opponents wins.

As for the countdown round, I say get rid of it completely, or only for cases when there is a tie in the scores.

zserf - if you're suggesting get rid of countdown - what is round robin for?? Written?

And for footballrocks.. I think your suggestion is still closer to a countdown where speed wins out - why? Because I know Many people who given the full 45 seconds or more, or even 30 seconds COULD answer most countdown questions 100% correctly, but just can't do it in THREE or even 10 seconds -- compare these people to the other 90 + percent of the population that couldn't get the countdown questions right even if given unlimited time (ok, not the nat competitors - I'm talking in general) - but you see, the written round - which is also timed, by the way, gives a slightly better indication of who can do a problem right, and not "just faster" -- but I wouldn't mind if there was only a written round, however, I think they do the countdown round more for the benefit of the spectators, parents, audience, - don't ya think?

Hey, can someone start a poll whether they'd rather Keep or get rid of state and nats countdown round? not sure how to make a poll on here :(

[quote="guyinPA"]zserf - if you're suggesting get rid of countdown - what is round robin for?? Written?
[/quote]

Someone mentioned it earlier, then someone else said you couldn't do a round robin with nineteen people, so I was informing them you could.

[quote="footballrocks41237"]We could always have a countdown winner and a written winner and have them face off in the [b]OCTAGON[/b][/quote]

I think chessboxing would be better

[img]http://www.nanoblog.com/images/chessboxing.jpg[/img]

http://en.wikipedia.org/wiki/Chessboxing

[quote="guyinPA"]To Indiana math who wrote
"[i]I think the current system is fine. It's right to reward those who do well on the written round. if you're really good and you do bad on the written, you can still win the whole thing[/i]." (sorry, tried to do the quote thing and didn't come out right - had to edit)

First, it sounds like you are contradicting yourself. How is the person who did well on the written rewarded if someone beats him or her in the ctdown round?
How is that fair? I mean how/why is that fair to the person who DIDN'T do "bad" in the written round?  Basically, are you saying speed or performance in a countdown round should have more weighting than the written? Would you feel differently if you did great at the written round but did lousy on ctdown? Just wondered... Maybe I just didn't understand what you were trying to say?  :?:[/quote]

And let me ask you: what do you consider to be "fair"? If they do what you apparently want and get rid of the countdown round or whatever, there will be people complaining too. Would you feel differently if you did great at the countdown round but did lousy on written? Just wondered... Maybe I just didn't understand what you were trying to say.

The current system is fine. It's fair in the sense that if someone tries hard enough, they can do well. And unless you can convince me that your idea of "fair" is actually fair to everyone, that's the only definition I care about.

[quote="13375P34K43V312"]Oh you know what would be sick? If like we divided everyone into 12 pools of 19 then did a round robin in each pool.[/quote]

I would lose instantly.

What would be really cool:

Announcer: "OK!!! 10 year olds in that room, 11 year olds in that room, 12 year olds in that room, 13 year olds in that room, and 14 year olds in that room. Let me know if you don't know where to go."

Then they have a mini competition for each room, then the winners in each room face off in countdown round.

I think they should do it how they used to many years ago and use the Masters Round for the top 10 or 12 to determine the National Champion.

guys guys guys the round robin thing was me kidding, w/e. but if we had infinite time, i would be up for a 228 person round robin tournament

but that masters round idea...why did they ever get rid of that? that's actually like a really good idea. screw countdown, they RUTHLESSLY DEPRIVED ME of my opportunity to own.

[quote="SplashD"][quote="Fanatic"]Actually, I would like the idea of taking like 3 sprint rounds and 3 targets, and posting the results after each one. If I found out I was in 14th after the 1st, I would be going ballistic!![/quote]

Riddle: You know what would be even better?

[hide="Answer"]If MATHCOUNTS had a Guts round (for those of you who don't know what this is, look up HMMT) in addition to or instead of the countdown round. (I knew the Guts round was the funnest part of HMMT for me). True, it'd be hard to manage, though I think it's much more exciting to watch than countdown. I admit this isn't practical but real-time results are better than announcers yelling out people's names so fast that haven't came to the realization yet that they have actually been called. [/hide][/quote]
strangest thing, I had a dream 2 weeks ago about me being in a guts round at nats in the same room as written with the same style.

I think that in my dream, China had a team somehow, and they pwned up, getting first.

the way people are making the contest now, is pretty unfair in some places. the old way seemed much better, and there are even better ways, but they just won't use them!

I don't really like the countdown round much. Some people have some advantages, such as reading faster among others...
I wish that they go back to taking the top 10 for Masters

Well, they could always of course take the top two in written and top two in countdown and pitch them together in the masters round to determine national champion :)

someone might have mentioned this before

masters used to be way shorter as well. u got like 10(i think) minutes for a 7 minute presentation, so it would go faster.


The integer 72 is the first of three consecutive integers 72, 73, and 74, that can each be expressed as the sum of the squares of two positive integers. The integers 72, 288, and 800 are the first three members of an infinite increasing sequence of integers with the above property. Find a function that generates the sequence and give the next three members.

[quote="Peter VDD"]Have I misunderstood the question or is this trivial? $f(n)=2(n+1)^2(n+2)^2$ is such a function? (and there are infinitely much)[/quote]

Why trivial??

[quote="Peter VDD"]I can't remember why I said it was trivial, but it is a valid sequence anyhow:

\[ f(n)=(n^2+3n+2)^2+(n^2+3n+2)^2  \] \[ f(n)+1=(n^2+4n+3)^2+(n^2+2n)^2  \] \[ f(n)+2=(n^2+3n+3)^2+(n^2+3n+1)^2  \]

I guess I then saw a trick to derive this in some seconds, but some trial-and-error works fine here, and the actual sequence is nearly given in the problem.[/quote]

All right!


First notice that these numbers are conveniently close to each other, which indicates to us that we can use some sort of factoring trick. In fact, we can rewrite this as $ (723)(723) \minus{} (723 \minus{} 1)(723 \plus{} 1) \implies 723^2 \minus{} 723^2 \minus{} 1^2 \equal{} \boxed{\minus{}1}$.


calculus
derivative
limit
integration
real analysis
function
real analysis unsolved

We have $ f: [0,\infty)\rightarrow\Re$ with first derivative where

$ \lim_{x\rightarrow\infty}(f(x) \plus{} f'(x)) \equal{} 0$.

Then prove that:

$ \lim_{x\rightarrow\infty}f(x) \equal{} 0$.

Let $ g(x) \equal{} e^xf(x).$ Then $ \lim_{x\to\infty}e^{ \minus{} x}g'(x) \equal{} \lim_{x\to\infty}f(x) \plus{} f'(x) \equal{} 0.$

Given $ \epsilon > 0,$ find $ A$ such that for $ x > A,$ $ |e^{ \minus{} x}g'(x)| < \epsilon,$ which we can rewrite as $ |g'(x)| < \epsilon e^x.$ Then, for $ x > A,$

$ g(x) \equal{} g(A) \plus{} \int_A^xg'(t)\,dt.$

$ |g(x)| < |g(A)| \plus{} \epsilon\int_A^xe^t\,dt$

$ |g(x)| < |g(A)| \plus{} \epsilon e^x.$ Divide by $ e^x$ to get

$ |f(x)| < |g(A)|e^{ \minus{} x} \plus{} \epsilon.$

Take the limit - well, the limsup - to get that $ \limsup_{x\to\infty}|f(x)|\le\epsilon.$

The is sufficient to say that $ \limsup_{x\to\infty}|f(x)| \equal{} 0,$ which implies that $ \lim_{x\to0}f(x) \equal{} 0.$

A similar version: assume $ f$ to be a continuous function on $ [0, \infty)$ such that\[ \mathop {\lim }\limits_{x \to \infty } \left( {f\left( x \right) \plus{} \int_0^x {f\left( t \right)dt} } \right)
\]
exists as a finite limit. Then \[ \mathop {\lim }\limits_{x \to \infty } f\left( x \right) \equal{} 0.
\]


In Skinien there 2009 towns where each of them is connected with exactly 1004 other town by a highway. Prove that starting in an arbitrary town one can make a round trip along the highways such that each town is passed exactly once and finally one returns to its starting point.

Well, should Dirac's Theorem be used somehow?

Let me know if this is correct :maybe:

Let the largest cycle in the graph be $ C_1$.  With the remaining vertices, take the largest cycle to be $ C_2$.  Continue like this defining $ C_3,C_4,...,C_k$.  Let $ s \equal{} |C_k|$.  It is clear that $ |C_1|\ge|C_2|\ge\cdots\ge|C_k|$.  Additionally, if $ 2009 > |C_1|$, it follows that $ 1004\ge s$.  Thus, every vertex of $ C_k$ is connected to another vertex outside of $ C_k$.

It is clear that for every vertex $ v\in C_k$, $ v$ is connected to at most $ \frac {1}{2}|C_i|$ vertices of $ C_i$ for $ i < k$, since otherwise $ v$ would be connected to two adjacent vertices and thus the maximality of $ C_i$ would be compromised.  Consider the spacing (number of vertices) between vertices connected to $ v$ in a cycle $ C_i$ for $ i\ne k$.  We shall show that the spacing is strictly less than $ s \plus{} 1$.  Suppose otherwise.  Then, in some cycle, $ C_i$, $ v$ is connected to at most $ \frac {1}{2}(|C_i| \minus{} s)$.  Otherwise, the maximum number of connected vertices in $ C_j$ ($ j\ne i,k$) is $ \frac {1}{2}|C_j|$.  It follows that the number of vertices outside of $ C_k$ connected to $ v$ is at most

$ \minus{} \frac {1}{2}s \plus{} \frac {1}{2}\left(C_1 \plus{} C_2 \plus{} \cdots \plus{} C_{k \minus{} 1}\right)$

$ \equal{}\minus{} \frac {1}{2}s \plus{} \frac {1}{2}\left(2009 \minus{} s\right)$

$ \equal{}1004.5 \minus{} s$

Since $ v$ can be connected to at most $ s \minus{} 1$ vertices of $ C_k$ it follows that $ deg(v)\le1003.5$, a contradiction.

Now we consider two consecutive vertices $ v_1,v_2\in C_k$.  Consider the vertices in $ C_1$ to which they are connected.  Let $ a\in C_1$ to which $ v_1$ is connected.  If $ v_2$ is connected to some vertex withing a spacing of $ s \minus{} 1$ of $ a$ then we may join the two cycles $ C_1$ and $ C_k$ to create a cycle larger than $ C_1$.  Otherwise, the closest vertices $ b,c\in C_1$ on either side of $ a$ (in the cycle $ C_1$) connected to $ v_2$ are a spacing of $ s$ away from $ a$.  But then the spacing between the vertices $ b,c$ is at least $ s\plus{}1$ (as long as $ |C_1|>s$; the case $ |C_1|\equal{}s$ is trivial) which is a contradiction to the fact that all spacings of vertices in $ C_1$ connected to $ v_2$ are at most $ s$.

What is $ s \equal{} 1$?  Then you can't consider the two consecutive vertices in $ C_k$ anymore.

BTW, note that the problem is [b]false[/b] if "exactly $ 1004$" is replaced by "at least $ 1004$" (divide the cities into two classes of size $ 1004$ and $ 1005$, and connect each city to all the cities in the other class), so somewhere in your argument you are going to have to use that the degrees are exactly $ 1004$.

Hmm...

Well if $ s \equal{} 1$, then the spacing on each cycle other than $ C_k$ is exactly $ 2$.  Then it follows that in order to satisfy the equality, all $ C_i$ have an even number of vertices.  More specifically, $ |C_{k \minus{} 1}|\ge2$.  Then, if $ C_{k \minus{} 1}$ is not connected to $ C_k$, we may use the same argument as before.  Otherwise, suppose $ C_{k \minus{} 1} \equal{} \{v_1,v_2,...,v_n\}$, where the $ v_i$ are in order.  Also let $ C_k \equal{} \{v\}$ and suppose $ v$ is connected to $ v_1$.  Then, considering the set $ C \equal{} C_k\cup C_{k \minus{} 1}$, we may follow the same argument as before by taking $ v$ and $ v_n$ as the two vertices.  Then we may create a larger cycle by joining $ v,v_1,v_2,...,v_n$ in that order.

How's that?

And, kevinatcausa, I'm not quite sure of your example.  Wouldn't the class of $ 1004$ have degree $ 1003$?

Each city is connected to all of the cities in the [b]other[/b] class, so the vertices in the class of size $ 1004$ all have degree $ 1005$, and vice versa.

If you work through your argument with that example, what happens is that $ |C_1|\equal{}2008$, and $ |C_2|\equal{}1$.    The vertex $ v$ in $ C_2$ is connected to every alternate vertex in $ C_1$,  and can't be added to the new cycle.   Note that in this case the spacing is equal to $ s\plus{}1$, not strictly less, and $ v$ is connected to $ \frac{1}{2}|C_i|$, not at most $ \frac{1}{2}(|C_i|\minus{}s).$

Right.  I see.

The problems arise when $ k \equal{} 2$ (for $ s \equal{} 1$ and when the 'spacing' is of no use - when $ v$ is connected to only one vertex of $ C_1$).  This can be easily fixed though by analyzing the case by itself, and this utilizes the equality of all degrees.

Let me know if there are any other problems, but I think this is it.


Let $OABC$ be a trihedral angle such that  \[ \angle BOC = \alpha, \quad \angle COA = \beta, \quad \angle AOB = \gamma , \quad \alpha + \beta + \gamma = \pi . \] For any interior point $P$ of the trihedral angle let $P_1$, $P_2$ and $P_3$ be the projections of $P$ on the three faces. Prove that $OP \geq PP_1+PP_2+PP_3$.

[i]Constantin Cocea[/i]

[img]http://i19.photobucket.com/albums/b200/mariusdamian/pbg.png[/img]
$P\in (A_1 B_1 C_1)$
It follows $OP\geq d(O;(A_1 B_1C_1))=h.$
But $S_{OA_1 B_1}=S_{OB_1 C_1}=S_{OC_1A_1}=S_{A_1 B_1 C_1}$
and $V_{OA_1 B_1 C_1}=V_{OPA_1 B_1}+V_{OPB_1 C_1}+V_{OPC_1 A_1}.$
It results $OP\geq h=PP_1+PP_2+PP_3$
and hence the conclusion.

My opinion near to before solution:
A plane is constructed which passes through point P and cuts OA, OB at D and E respectively and also cuts CO at  Z such that: OD=EZ, DE=OZ, OE=DZ which holds since :
 $\alpha  + \beta  + \gamma  = \pi  = \mathop D\limits^ \wedge   + \mathop E\limits^ \wedge   + \mathop Z\limits^ \wedge  .$
Hence
$\mathop {ODE}\limits^\vartriangle   = \mathop {OEZ}\limits^\vartriangle   = \mathop {ODZ}\limits^\vartriangle   = \mathop {DEZ}\limits^\vartriangle  ,$
$OP \geqslant OO^' $
with O΄ the projection of O on plane DEZ and
$V_{P.ODE}  + V_{P.OZE}  + V_{P.ODZ}  = V_{O.ZDE}  \Rightarrow ......End.$

S.E.Louridas


geometry
analytic geometry
geometry proposed

Let ABC a triangle and P a point in its interior. Lines PA,PB,PC intesct BC,CA,AB at D,E,F, respectvely. Prove that [PAF] + [PBD] + [PCE] = [ABC]/2

I really don't think this is true in general, man. Think about what happens if $P$ tends to a point on $BC$ different from the midpoint of $BC$. $[PBD],[PCE]\to 0$, while $[PAF]$ tends to an area which is not half of $[ABC]$.

This problem is a A2 of USA TST 2004.
I no find a solution  :(

Actually, the equation $\left[PAF\right]+\left[PBD\right]+\left[PCE\right]=\frac{\left[ABC\right]}{2}$ holds if and only if the point P lies on (at least) one of the three medians of the triangle ABC or on the line at infinity. This can be easily verified by a calculation using barycentric coordinates; I am not sure whether I will have an opportunity to find a synthetic proof in the next weeks.

  darij

Let $s_1$ be the area of $\triangle{APE}$ ,$s_2$ the area of $\triangle{CPE}$, and so on cloclwise until $s_6$ the area of $\triangle{APF}$. Then we easily get that:
1.$s_1s_3s_5=s_2s_4s_6$
2. $\frac{s_1}{s_2}=\frac{s_1+s_6+s_5}{s_2+s_3+s_4}$ and three similar relations.
Using 1. and 2. it's not very hard to prove that $(s_1-s_2)(s_3-s_4)(s_5-s_6)=0$.


I'm looking for a good piano concerto to jump into this summer, I want to do the Concerto competition at my high school next year as a freshman. For a quick backround of my experience...

I've played for a little over 9 years, under the same teacher (except she's moving away in two weeks, I found a new one), in recent years when I've become more serious, I've focused mostly on Beethoven sonatas, I've done four in the last four years, including, most recently, op. 31, no. 2 (the tempest) and op. 109. The last two years, I've poked around at Romanticism and Contemporary, I've worked on the Liszt Sonnetto del Petrarca for about one and a half years, playing it four or five in various competitions and recitals, and picked up a big one by Messiaen about 6-7 months ago. I haven't done a whole lot with Baroque, I picked up Bach's Italian Concerto (it's not really a concerto) in fourth grade, and played it in around 10-15 competitions and recititals, up until the end of seventh grade, and did a little bit of Bach this year (Sinfonia from Partita no. 2).

Any ideas?

Given your former reperatoire, I would suggest:

Beethoven 2 or 4, since you've probably got a good musical feel for his expressions in your sonata experience, Schumann in A minor or, if you want something more contemporary,
Grieg in A minor or Ravel (these are a bit more difficult, technically and musically).

Mendelssohn's Capriccio Italienne might be worth looking at as well.

Hmm... I think most people find Beethoven 4 and 5 to be rather difficult concertos, 1-3 are a bit easier, and probably more appropriate (3rd one is really cool).

For a romantic concerto, I would say Schumann is really nice, but the 3rd movement is quite tricky. But if this is your first romantic concerto, I think the Grieg, either of the two Mendelssohn, or Saint-Saens 2 are more popular choices.

If this is your first concerto period and you don't have much experience playing in ensembles, it might be better to a Classical era one that's relatively easy for you, from a technical viewpoint. That way you have a bit of room for when you actually play the concerto with others.

beethoven concerto n° 3 in c-moll is beautiful and it isn't much difficult


Let a,b,c>0 such that (2b-a)(2c-b)(2a-c)(a+b+c)^3>27(abc)^2. Prove that max{a,b,c}>4min{a,b,c}

This is clearly stupid. Do not try it until I write the correct statement.

Sorry, the corect statement is:
  For all a,b,c>0 such that max{a,b,c}<2min{a,b,c} we have 27(abc)^2>=(2b-a)(2c-b)(2a-c)(a+b+c)^3.


ARML geometry function algebra polynomial calculus AMC

I made a Mock ARML (individual round) that my team did at practice today. I figured I might as well post it here. Anyways, obviously, if you're on the Georgia ARML team, don't take it, since you have seen the problems already.

Otherwise,
Spend [b]ten minutes[/b] on each pair.
Do [b]not[/b] use calculators.
PM me your answers to me by the end of the day on [b]Wednesday[/b] even if you did not solve too many. This is admittedly [b]way more difficult[/b] than an actual ARML: the high score at practice today was a three - and many of the people taking the test scored sixes or sevens on the real individual round last year. Don't focus on getting a valid estimation of what your score will be on Saturday - focus on learning. Unofficially work on some problems after the ten minutes is over until you solve them.
[b]Read[/b] the problems very carefully. There were apparently countless reading mistakes at practice today.
Have fun and good luck!

[hide="Pair One"]
1. $ ABCD$ is a convex quadrilateral such that $ |AB| \equal{} 5$, $ |BC| \equal{} 17$, $ |CD| \equal{} 7$, and $ |DA| \equal{} 25$. Given that $ m\angle{ABC} \plus{} m\angle{BCD} \equal{} 270^{\circ}$, find the area of $ ABCD$.

2. Given that the sum of all positive integers with exactly two proper divisors, each of which is less than $ 30$, is $ 2397$, find the sum of all positive integers with exactly three proper divisors, each of which is less than $ 30$ (a proper divisor of $ n$ is a positive integer that divides but is not equal to $ n$).[/hide]
[hide="Pair Two"]
3. A variation of Pascal's triangle is constructed by writing the numbers $ 2$ and $ 3$ in the top row and writing each subsequent term as the sum of the two terms above it. Find the fifth term from the left in the thirteenth row.
\[ \begin{tabular}{ccccccccccc} & & & & & & & & & & \\
& & & & 2 & & 3 & & & & \\
& & & 2 & & 5 & & 3 & & & \\
& & 2 & & 7 & & 8 & & 3 & & \\
& 2 & & 9 & & 15 & & 11 & & 3 & \\
2 & & 11 & & 24 & & 26 & & 14 & & 3 \\
\end{tabular}
\]
4. Equilateral triangle $ ABC$ has a side length of $ 7$. A ball begins at vertex $ A$, rolls through the interior of the triangle, bounces off side $ BC$, and settles at point P. Given that $ |BP| \equal{} 3$ and $ |CP| \equal{} 5$, find the total distance that the ball travels.
[/hide]
[hide="Pair Three"]
5. Al is thinking of a function, $ f(x)$. He reveals to Bob that the function is a polynomial of the form $ f(x) \equal{} ax^8 \plus{} bx^5 \plus{} cx^2 \plus{} dx \plus{} e$, where $ a$, $ b$, $ c$, $ d$, and $ e$ are [b]complex[/b] number coefficients. Bob wishes to determine the value of $ d$. For any [b]complex[/b] number $ x$ that Bob asks about, Al will tell him the value of $ f(x)$. At least how many values of $ x$ must Bob ask about in order to definitively determine the value of $ d$.

6. John has a pile of $ 63$ blocks. On top of the pile is one block. Below this block are two smaller blocks. Below each of these two blocks are two even smaller blocks. Below each of these blocks are two still smaller blocks, and so on until the last row, which contains $ 32$ blocks. John removes blocks one at a time, removing only blocks that currently have no blocks on top of them. Find the number of ways in which John can remove exactly seven blocks.
[/hide]
[hide="Pair Four"]
7. Let $ f(x)$ equal the number of zeroes to the right of the rightmost non-zero digit in the decimal form of $ x!$, and let $ n \equal{} \frac {5^{2008} \plus{} 2}{3}$. Given that $ f(n)$ can be written as $ \frac {5^{a} \minus{} b}{c}$, where $ b$ and $ c$ are relatively prime positive integers, $ b$ is less than $ 10^5$, and $ c$ is less than $ 10^2$, find $ a \plus{} b \plus{} c$.

8. Given that $ \sum_{i \equal{} 0}^{n}a_ia_{n \minus{} i} \equal{} 1$ and $ a_n > 0$ for all non-negative integers $ n$, evaluate $ \sum_{j \equal{} 0}^{\infty}\frac {a_j}{2^j}$.
[/hide]

[quote="jmerry"]I'll dispute the appropriateness of #8: it's a calculus question, and ARML questions should never require calculus.[/quote]
It is not.  There is a simple solution with no calculus.  In fact, I don't see how you would use calculus to solve it.  Could you maybe PM me your calculus solution, if it's nice?  I'm curious.

Also, Eric, are you going to make the slight change to number 8?

Yes; because Altheman pointed out an ambiguity in number eight, I am going to make a slight change. jmerry, I will PM you my solution to number eight in a couple minutes. I also made a slight change on #2 - I accidentally used the terms "proper divisor" and "proper factor" interchangeably, but they are not interchangeable - nevertheless, I think it was fairly clear what the problem was asking.

Thanks to everyone who participated. Discussion of the problems is now permitted.

[hide="Results and Statistics"]
Altheman and tjhance tied for the highest score out of eight participants - both of them solved five problems correctly. No one submitted a correct answer to number six except for worthawholebean, who solved it after running out of time. Each participant solved at least one problem. No one other than Altheman and tjhance solved more than three.

The number of people to solve each problem correctly out of eight participants:
#1 - 6
#2 - 1 (E^(pi*i)=-1)
#3 - 4
#4 - 5
#5 - 3
#6 - 0
#7 - 1 (tjhance)
#8 - 2

Average Score - 2.75[/hide]
I might make some more tests next year. Perhaps I will have more luck with a mock AIME since some of my problems seem to be rather time consuming.

[hide="#6"]
Obviously, the pile is $ 6$ layers. Each time we remove a brick, we add one more choice for the next one - $ 1$ choice for the first, $ 2$ for the second, etc. So if we remove the restriction of $ 6$ layers, we have $ 7!$ choices. However, we must now subtract each selection of bricks which tries to grab from the $ 7$th layer. To select one of these, we must always take a brick from the lowest layer possible, and there are always two possible choices. This means there are $ 2^6$ ways to try to remove a brick from the $ 7$th layer or $ \boxed{7!\minus{}2^6\equal{}4976}$ ways to remove only bricks from the first $ 6$.
[/hide]

Ohhhhhh

For some reason, I did not think that the order in which you remove the blocks matters. I assumed it was only significant which blocks you took. This wrong assumption made this a bit harder...

Is number 3 105? If not, why?

Sorry for the necropost.

[quote="tanuagg13"]If not, why?[/quote]  You seem a little confused about how this whole idea of "presenting a solution" is supposed to work ;)

Before this gets locked, I need to ask a few questions.

1.) I don't see how to do this without un-clever bash.
2.) Also easy, except way too bashy to be true.

Solutions for as many problems as possible would be appreciated. Sorry for being so demanding. I just didn't do so well (3 -_-) so I would love to see clever solutions to these, thank you! :)

For number 1, [hide]extend edges AB and CD until they meet.  The resulting angle is a right angle.  Subtracting the areas of two right triangles gives the area of the quadrilateral; subtracting the equations that you get from the Pythagorean Theorem gives a closely related expression.  Not too bashy (though obviously not at ARML number 1 difficulty, either).[/hide]

For number 2, [hide]you are given $2^2+3^2+5^2+7^2+11^2+\ldots+29^2$ and asked to compute the value of $(2^3+3^3+5^3)+\sum_{p,q}pq$ where the sum is over pairs of primes smaller than 30.  You can get this sum from the given by summing up the primes less than 30 and squaring.[/hide]

For number 3, [hide]the form of the object means we should expect the result to be just a sum of multiples of shifted copied of Pascal's triangle.  Indeed, subtracting off two copies of Pascal's triangle quickly reveals that the $k$th entry in the $n$th row ($k$ starting from 0) is $2\binom nk + \binom{n-1}{k-1}$, and this is easy to prove with induction.  Now plug in 13 for $n$ and 4 for $k$ to get the answer (which is much larger than 105).[/hide]

For number 4, [hide]it helps to know or remember that the 3-5-7 triangle has largest angle 120 degrees.  Of course proving this once you guess it is very easy.[/hide]

[quote="bdejean"]5.

[hide="Short sorta solution"]Ask about two additive inverses and subtract them.  Now we have a linear equation in terms of $b$ and $d$.

Ask about another pair to get a second linear equation in terms of $b$ and $d$.

Solve.

So this is 4.  I don't [b]know[/b] that you can't do it in less, but I suspect so.[/hide][/quote]
We can do better.  [hide]In particular, most of the exponents are the same mod 3, so we can use third roots of unity: let $\omega$ be a primitive third root of unity.  Bob asks the values at 1, $\omega$ and $\omega^2$, getting as output values $(a + b + c) + d + e$, $(a + b + c)\omega^2 + d\omega + e$ and $(a + b + c)\omega + d \omega^2 + e$.  This provides a system of three linear equations in the three unknowns $a + b + c$, $d$ and $e$, with a unique solution (e.g., because the matrix of coefficients has nonzero determinant), so we can compute $d$.  Now you must argue that for any two input values there are at least two different values of $d$ that work; this is not hard to show by twiddling $e$.[/hide]

For number 8, [hide]ignoring issues of convergence, compute $\left(\sum_{j \geq 0} \frac{a_j}{2^j}\right)^2 = \sum_{n \geq 0} \frac{\sum_{j = 0}^n a_j a_{n - j}}{2^n}$.  Thus our desired sum has square 2; since (if it converges) it is clearly positive, the answer must be $\sqrt{2}$.  Now of course jmerry's objection is relevant: to really call this a solution, we must justify why the series in question converges.[/hide]

all problems have been posted in separate threads in HSM  , I just created the related post collectionw

[url=https://artofproblemsolving.com/community/c3731776_2008_mock_arml_2_individual]here[/url]

enjoy  / start solving

more USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more ARML mocks [url=https://artofproblemsolving.com/community/c2442143_arml_mocks]here[/url],
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


Let $ a,b,c$ be positive real number such that $ abc\equal{}1$. Prove that:
\[ \frac{b\plus{}c}{\sqrt{a}} \plus{} \frac{c\plus{}a}{\sqrt{b}} \plus{} \frac{a\plus{}b}{\sqrt{c}} \geq \sqrt{a} \plus{} \sqrt{b} \plus{} \sqrt{c} \plus{} 3\]

[hide="Solution"]Let $ \sqrt {a} \equal{} x$, $ \sqrt {b} \equal{} y$, $ \sqrt {c} \equal{} z$, so $ xyz\equal{}1$ then we have to show
$ \frac {x^2 \plus{} y^2}{z} \plus{} \frac {y^2 \plus{} z^2}{x} \plus{} \frac {z^2 \plus{} x^2}{y} \ge x \plus{} y \plus{} z \plus{} 3$
But $ \frac {x^2}{z} \plus{} \frac {z^2}{x} \ge \frac {(x \plus{} z)^2}{z \plus{} x} (Cauchy) \equal{} x \plus{} z$
So $ LHS \ge 2(x \plus{} y \plus{} z)$, but obviously $ x \plus{} y \plus{} z \ge 3$ by AM-GM[/hide]

[hide="Not nice solution"]Let $ a\equal{}x^6,b\equal{}y^6,c\equal{}z^6$. Since $ \frac{1}{\sqrt{a}}\equal{}\sqrt{bc}$, we have $ LHS\equal{}\sum_{sym}x^9y^3$, and we have $ RHS\equal{}x^3\plus{}y^3\plus{}z^3\plus{}3\equal{}x^6y^3z^3\plus{}x^3y^6z^3\plus{}x^3y^3z^6\plus{}3x^4y^4z^4$. Now use Miurhead.[/hide]

[quote="varunrocks"]Did the first guy to post the solution just use T2's lemma.[/quote]
Yes, Titu's lemma is another form of Cauchy.

[hide="Is It Right??"]Can be written as
\[ \frac {\sqrt {a}(b \plus{} c)}{a} \plus{} \frac {\sqrt {b}(c \plus{} a)}{b} \plus{} \frac {\sqrt {c}(a \plus{} b)}{c}\geq\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \plus{} 3\]
Using chebycheff we get $ \frac {\sqrt {a}(b \plus{} c)}{a} \plus{} \frac {\sqrt {b}(c \plus{} a)}{b} \plus{} \frac {\sqrt {c}(a \plus{} b)}{c}\geq\frac {\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}}{3}(\sum_{cyc}\frac {(b \plus{} c)}{a})$
But $ \sum_{cyc}\frac {(b \plus{} c)}{a}\ge 6$ put this value there and we are left to prove that $ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}\ge3$ which is true by $ A.M\ge G.M$[/hide]

[hide]$ \sum \frac{b\plus{}c}{\sqrt{a}}\geq 2\sum_{cyc}\sqrt{\frac{bc}{a}}\equal{}\sum_{cyc}\left(\sqrt{\frac{bc}{a}}\plus{}\sqrt{\frac{ca}{b}}\right)$
$ \geq 2\sum \sqrt{a}\geq \sum \sqrt{a}\plus{}3\sqrt[6]{abc}\equal{}\sum \sqrt{a}\plus{}3$
I think it is correct . [/hide]


What are all values of $x$ for which the series $\sum_{n=1}^{\infty}\frac{(x-2)^{n}}{n3^{n}}$ converges?

The ratio test will give you the radius of convergence. The endpoints need separate treatment; one converges while the other diverges.

The ratio test is totally overkill. Just look at the terms and determine the radius of convergence by definition (i.e., for what x are the terms bounded?).

HI:YOU CAN use  From the ratio test or n-th root test that you can find them in any Calculus book.
                                                                            Happy new year :maybe:


The diameter of a circle is divided into $ n$ equal parts. On each part a semicircle is construced. As $ n$ becomes very large, the sum of the lengths of the arcs of the semicircles approaches a length:

$ \textbf{(A)}$ equal to the semi-circumference of the original circle
$ \textbf{(B)}$ equal to the diameter of the original circle
$ \textbf{(C)}$ greater than the diameter but less than the semi-circumeference of the original circle
$ \textbf{(D)}$ that is infinite
$ \textbf{(E)}$ greater than the semi-circumference but finite

[hide="Click for solution"]
For each semi-circle the diameter is $ \frac{2r}{n}$ and the length of its arc is $ \frac{\pi r}{n}$.  The sum of $ n$ such arcs is $ \frac{n \cdot \pi r}{n}\equal{}\pi r\equal{}$ semi-circumference, or $ \boxed{\textbf{(A)}}$.
[/hide]


I found this problem:

[i]"Find all the possible positive integers x,y which solve:

7^x-4^x=3
and all positive integers x,y which solve:

7^x-3^x=4"[/i]
I was told to use MOD, but what is it?
Thank you very much!

P.S. Could somebody explain me how to use LaTex in this forum? Thank you, I'm still a newbie here!

[quote="davi24"]P.S. Could somebody explain me how to use LaTex in this forum? Thank you, I'm still a newbie here![/quote]

Click "LaTex" on the bar thing at the left. To use it in the forums, just type a dollar sign, the latex, and another dollar sign. For example, to see the code for this : $7^x-4^x=3$, just put your mouse over it.  :)

\[7^x-4^x=3\]
\[7^x-3^x=4\]

Subtract the equations, to get

\[-3^x+4^x=1 \iff 4^x=1+3^x\]

Look at $3^x \mod {10}$ - It's a repeating pattern: {3,9,7,1}.
Look at $4^x \mod {10}$ - It's a repeating pattern: {4,6,4,6}.

In which case are these values 1 unit apart? When $x=1$. Indeed, $1+3^1=4^1$.

Thus the answer is $x=1$.

$x=1$ is one solution. Now prove that function in always increasing/decreasing!

$4^x$ grows at a faster rate than $3^x$. Does that count as a proof?

the word increasing/decreasing is screaming calculus at me... which is something we're not supposed to use rite?

There are two separate complications here:  chess64, you seem to be treating those as a pair of simultaneous equations when in fact they seem to be two seperate problems.  Second, the question asks about two variables, x and y, but then both questions only have an x.  This indicates to me that there is a typo and that seamus's suggestion will become irrelevant as a result.

This does not look like a question designed for modular arithmetic.  Indeed, we note that in the calculuation chess64 did, those modular congruences also work when x = 5.  However, needless to say this is not in fact a solution.

To understand that something is increasing most certainly does not require the use of calculus -- to quantify how a smooth curve increases instantaneously is a question of calculus, but it is not at all necessary to use that here.

really sorry... I meant
7^x-4^y=3
7^x-3^y=4
Thank you for all you answers anyway!

[quote="davi24"]Now I try with LaTeX...

$7^x-4^y=3$

$7^x-3^y=4$[/quote]
just one question: both of the equations at the same time, or each seperatly?  :|
(btw, nice job on the latex :P )

Each seperately. I think solutions are only (1,1) and (1,1), but I don't know how to demonstrate.
Thank you  :lol:

1) $7^x-4^y=3$
    $7^x=3+4^y$
     It's pretty obvious then that $x=y$ and they both are $1$

2) $7^x-3^y=4$
    $7^x=4+3^y$
     Same as before $x=y$ which equals $1$

"pretty obvious"?  :?
can you prove that there is only one solution? ;)

Wait, you're right...unless I use logs. But they should be the only integer answers.

It's obvious that, if you prove x=y, than the equation is really easy, cause
$7^x-4^x=3$ has x=y=1 as a solution, and since $7^x$ grows more rapidly than $4^x$ it's the only one. But how do you prove x=y?

By the way, only an idea...
I think I've found what a MOD is...
So, let's see $7^x$ MOD 8;
for x=1, it's 7.
The growth $7^(^x^+^1^)-7^x$ is $6*7^x$.
So, if $7^x$ MOD 8 is 7, then 7*6=42 and the rest is 2.
But 7+2=9 MOD 8 = 1.
Now let's see next growth.
$7^x$ MOD 8 = 1, we said.
So 1+6*1=7.
Or, which is the same, $7^x$ mod 8 is 7,1,7,1,... (sorry for my english).
Now let's see that $4^y$ MOD 8 is zero if y>1. So the only case in which the difference of the rests can be three is x=y=1. Is this (stupid) prove valid? Thank you!

I think it's right cause u can prove the second equation in the same way by using mod 9.
Thank you all!

can you prove the second one with $\mod 9$?  :?

$7^x\equiv 1,4$ or $7 \mod 9$
and $3^y\equiv 0 \mod 9$ if $y>1$
so you have still infinitly many cases where the equation becomes $4\mod 9 \equiv 4\mod9$  :|

the first one seems okay though ;)

You're right. I could solve it by using mod 9 and then using mod 13 (which is ugly cause it gives a lot of different results but I was able to exclude all the other solutions)

Notice that x must be in the form 2+3k to give 4 as rest with mod 9.
If you see $7^2^+{}^3{}^k$ mod 13, you get 10,11,3,2; if you see $3^y$ mod 13 you get 3,9,1. None of the differences (10-3, 10-9,...) can be 4, so we can exclude any other solution besides 1,1


inequalities
function
inequalities proposed

Let $x,y,z>0$ and $xyz=1$. Find minimum of the expression:
$\large \frac{x}{y^{5}+2}+\frac{y}{z^{5}+2}+\frac{z}{x^{5}+2}$

Vasc and anyone, can you help me? :blush:

the expression can get as big as we want so I guess you're asking for the minimum of the expression...

oh, I sorry because I mistaken   :blush:
I will fix it :D
This problem very difficult and I hope who help me :)
Can EV theorem prove it , Vasc?

It was also posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=107246.

I don't think so because this problem with number $5$. It very difficult  :D

The fact is that, problem with number $4$ is unsolved now. So does anyone have a solution? It's very famous in Viet Nam, by a myserious person  :P

I agree idea of hungkhtn :D I think this inequality is very very difficult.
In mathnfriend forum has a solution by funcion of Bùi Việt Anh with number $5$. But that solution only solve in the case $a\ge b\ge c$
Anyone can solve it?  :)


Find x,y,z є (a ,∞) a>0  with
3xyz+a^2*(x+y+z)=2a(xy+yz+zx)

[hide="Really Messy"]
Call $ x \equal{} a \plus{} m$, $ y \equal{} a \plus{} n$, and $ z \equal{} a \plus{} p$.  We have that
\[ 3xyz \plus{} a^2(x \plus{} y \plus{} z) \equal{} 3(a^3 \plus{} a^2(m \plus{} n \plus{} p) \plus{} a(mn \plus{} np \plus{} mp) \plus{} mnp) \plus{} a^2(3a \plus{} m \plus{} n \plus{} p) \equal{} 6a^3 \plus{} 4a^2(m \plus{} n \plus{} p) \plus{} 3a(mn \plus{} np \plus{} mp) \plus{} 3mnp
\]
Now, we have that
\[ 2a(xy \plus{} yz \plus{} xz) \equal{} 2a(3a^2 \plus{} (mn \plus{} mp \plus{} np) \plus{} a(2n \plus{} 2m \plus{} 2p)) \equal{} 6a^3 \plus{} 4a^2(m \plus{} n \plus{} p) \plus{} 2a(mn \plus{} mp \plus{} np)
\]

\[ 6a^3 \plus{} 4a^2(m \plus{} n \plus{} p) \plus{} 2a(mn \plus{} np \plus{} mp) \equal{} 6a^3 \plus{} 4a^2(m \plus{} n \plus{} p) \plus{} 3a(mn \plus{} np \plus{} mp) \plus{} mnp
\]

\[ a(mn \plus{} np \plus{} mp) \plus{} mnp \equal{} 0
\]
Since a>0 and m, n, and p are positive, we have that mnp and mn+np+mp are both 0.  Thus, we have that two of m, n, and p are 0.  Without loss of generality, call m=n=0.  Thus, x=a=y and z>a.  [/hide]

A bit less messy if we
[hide="do this"]divide the whole equation by $ a^3$ and plug $ {x\over a}\equal{}X\plus{}1, {y\over a}\equal{}Y\plus{}1, {z\over a}\equal{}Z\plus{}1$, with $ X,Y,Z\geqslant 0$[/hide]


function
trigonometry
induction
algebra proposed
algebra

Determine all continuous functions $ f: [\minus{}1;1] \to \mathbb R$ such that \[ f(2x^2\minus{}1)\equal{}2x \cdot f(x), \quad \forall x \in [\minus{}1,1].\]

[quote="thuyanh158"]determine all continuous function :$ [ \minus{} 1;1] \to R$ such that : $ f(2x^2 \minus{} 1) \equal{} 2x.f(x)$.  $ \forall x \in [ \minus{} 1,1]$
 :roll:[/quote]

Let $ P(x)$ be the assertion $ f(2x^2\minus{}1)\equal{}2xf(x)$
Let $ Q(t)$ be the assertion $ P(\cos(t))$ : $ f(\cos(2t))\equal{}2\cos(t)f(\cos(t))$

Comparing $ P(x)$ and $ P(\minus{}x)$, we get that $ f(\minus{}x)\equal{}\minus{}f(x)$ $ \forall x\neq 0$ and so, with continuity, $ f(0)\equal{}0$

So $ f(\cos(\frac{(2k\plus{}1)\pi}2))\equal{}0$

Then $ Q(\frac{(2k\plus{}1)\pi}4)$ $ \implies$ $ f(\cos(\frac{(2k\plus{}1)\pi}2))\equal{}2\cos(\frac{(2k\plus{}1)\pi}4)f(\cos(\frac{(2k\plus{}1)\pi}4))$ and so $ f(\cos(\frac{(2k\plus{}1)\pi}4))\equal{}0$

And an immediate induction gives $ f(\cos(\frac{(2k\plus{}1)\pi}{2^n}))\equal{}0$ $ \forall k,n\in\mathbb N$

And continuity implies then $ f(x)\equal{}0$ $ \forall x$

thank you very much for your useful post. it's very kind of you  :roll:


find the function f:R ->R such that:
f( (x+y) / (x-y) ) =( f(x)+f(y) ) / ( f(x)-f(y) ) [/hide][/code]

$ f(x) \equal{} \minus{}x$ surely does not work.   And it should be needless to say that answers (even correct answers) to such problems without solutions are worthless.

[hide="A Start"]
[b]Finding f(1):[/b]

Let $ y \equal{} 0$. Then, $ f(1) \equal{} \frac {f(x) \plus{} f(0)}{f(x) \minus{} f(0)} \iff [f(1) \minus{} 1]f(x) \equal{} [f(1) \plus{} 1] f(0)$

Hence, either $ f(1) \minus{} 1 \equal{} 0 \implies f(1) \equal{} 1$ or:

$ f(x) \equal{} \frac {(f(1) \plus{} 1) f(0)}{f(1) \minus{} 1}$, implying that it is a constant.

However, from the equation as stated, $ f(x)$ cannot be a constant, since the denominator of the right hand side would then always be undefined.

Hence, $ f(1) \equal{} 1$.

[b]Invariant:[/b]
Note that $ \frac {x \plus{} y}{x \minus{} y} \equal{} \frac {ax \plus{} ay}{ax \minus{} ay}$

Hence, $ \frac {f(x) \plus{} f(y)}{f(x) \minus{} f(y)} \equal{} \frac {f(ax) \plus{} f(ay)}{f(ax) \minus{} f(ay)}$ for some third parameter $ a$.

[b]Another Relationship:[/b]

Consider the case when $ x \equal{} 1$.

$ \frac {f(1) \plus{} f(y)}{f(1) \minus{} f(y)} \equal{} \frac {f(a) \plus{} f(ay)}{f(a) \minus{} f(ay)}$

$ 1 \plus{} \frac {2f(y)}{f(1) \minus{} f(y)} \equal{} 1 \plus{} \frac {2f(ay)}{f(a) \minus{} f(ay)}$

$ \frac {f(y)}{f(1) \minus{} f(y)} \equal{} \frac {f(ay)}{f(a) \minus{} f(ay)}$

$ f(y)f(a) \minus{} f(y)f(ay) \equal{} f(1)f(ay) \minus{} f(y)f(ay)$

$ f(y)f(a) \equal{} f(1)f(ay) \iff \boxed{f(ay) \equal{} f(a)f(y)}$
[/hide]

f(x)=x is result.
if you want to solve that problem , you must solve this function:
f:R ->R such that:
f(x+y)=f(x)+f(y)
f(x.y)=f(x).f(y)
but it also very easy for you

I hope you will fun with this problem
f:R ->R
f(a.x+b)=c.f(x)


[quote="Remike"]The sequences $\{a_n\}_{n=1}^{\infty}$ and $\{b_n\}_{n=1}^{\infty}$ are said to be [i]strongly distinct[/i], if there are no $i,j$ such that $a_i=b_j$. The sequence $\{a_n\}_{n=1}^{\infty}$ is said to be [i]stronger[/i] than the sequence $\{b_n\}_{n=1}^{\infty}$ if for every natural $n$ the product $a_1 \cdot \ldots \cdot a_n$ is divisible by the sum $b_1+\ldots+b_n$. Prove or disprove the existance of pairwise strongly distint sequences $\{x_n\}_{n=1}^{\infty}$, $\{y_n\}_{n=1}^{\infty}$ and $\{z_n\}_{n=1}^{\infty}$ such that $\{x_n\}$ is stronger than $\{y_n\}$, $\{y_n\}$ is stronger that $\{z_n\}$, and $\{z_n\}$ is stronger than $\{x_n\}$.[/quote]
$\{x_n\}$ is stronger than $\{y_n\}$ implies $y_1\mid x_1$.
$\{y_n\}$ is stronger than $\{z_n\}$ implies $z_1\mid y_1$.
$\{z_n\}$ is stronger than $\{x_n\}$ implies $x_1\mid z_1$.
Then $|x_1|=|y_1|=|z_1|$ that contradicts to the fact that the sequences $\{x_n\}_{n=1}^{\infty}$, $\{y_n\}_{n=1}^{\infty}$ and $\{z_n\}_{n=1}^{\infty}$ are pairwise strongly distint.


A survey of 2000 shoppers showed that 1400 shop at M-Mart, 850 shop at Glen Valley Market and 390 shop at both stores. What percent of the shoppers do not shop at either store?

First, we find how many people shop.  We do $ 1400\plus{}850\minus{}390\equal{}1860$.  This is called PIE, and you can find info on it [url=http://www.artofproblemsolving.com/Wiki/index.php/PIE]here[/url].  We then get that $ 2000\minus{}1860\equal{}140$ shop at neither.  To find the percent, we do $ \frac{140}{2000}\equal{}\frac{7}{100}\equal{}\boxed{7\%}$


thats what mr.Linker told me.

is it because of SAT?
we lost one of our brilliant teammates...

by the way,
whos Tao Zuoyu Stuyvesant and how good is he compare to hooyoung?
is there anybody who knows him?

Can it really be because of the SAT's??? ARML does count as an "unexpected event" according to their website, so a makeup is offered...Mr. Linker even sent the email that will be offered June 18th at Stuyvesant...So what can be the reason...and who will be his replacement if this is true???

I agree, but as far as I know, it is going to be Tao Zuoyu from Stuyvesant.

Zuoyu Tao, you mean.

Do we really have nobody else? He might indeed be the 6th senior in the Stuy Math Team, but that's 6th of . . . what, 8? He comes after Peak, Yuriy, Yitian, Mark and Ilya, only one of which is on the A team. No offense to anyone, but I don't understand why he is being chosen.

wow, sho's on AoPS?!

I cant say anything since I dont know him personally.
was he on NYSML this year? what team?

[SELF-CENSORED]
N team, according to the list I have.

I assume you were being facetious.  Otherwise, you really ought not to say something like that about AoPS on AoPS.

I won't be able to come to ARML because I will be going to training camp for the USACO (http://www.usaco.org).

wow, hooyoung is going to USACO camp and MOP! Congrats and hope you enjoy both :)


algebra
polynomial
vector
invariant
number theory
relatively prime
superior algebra

Can anyone help me out with this my professor said this in class before break but I'm not sure why this is true

For a finite vector space $ V$ with proper subspace $ W$ let $ T: V\to V$ be a linear transformation with invariant subspace $ W$.  Let $ h$ be the minimal polynomial of $ T$, $ h_{1}$ be the minimal polynomial of $ T$ restricted to $ W$ and $ h_{2}$ be the minimal polynomial of the linear transformation $ T$ induces on $ V/W$ then

$ h_{1}|h$ and $ h_{2}|h$
and if $ h_{1}$ and $ h_{2}$ are relatively prime then $ h\equal{}h_{1}h_{2}$

[quote="stuyguy89"]Can anyone help me out with this my professor said this in class before break but I'm not sure why this is true

For a finite vector space $ V$ with proper subspace $ W$ let $ T: V\to V$ be a linear transformation with invariant subspace $ W$.  Let $ h$ be the minimal polynomial of $ T$, $ h_{1}$ be the minimal polynomial of $ T$ restricted to $ W$ and $ h_{2}$ be the minimal polynomial of the linear transformation $ T$ induces on $ V/W$ then

$ h_{1}|h$ and $ h_{2}|h$
and if $ h_{1}$ and $ h_{2}$ are relatively prime then $ h \equal{} h_{1}h_{2}$[/quote]

h(T)=0 on V, so h(T)=0 on W.  The minimal polynomial divides all such polynomials, so h1 divides h.  Similarly, h(T)=0 on V, so h(T)=0 on V/W, so h2 divides h.  Now h3(T)=h1(T)*h2(T) is 0 on W since it is divisible by h1.  Also h1(T)*h2(T) is 0 on V/W since it is divisible by h2.  If v is any element of V, then h2(T)(v+W) = 0+W, so h2(T)(v) is in W.  But h1(T)(w) = 0 for all w in W, including w=h2(T)(v).  Hence h3(T)(v)=0.

So h is divisible by h1 and h2, and h divides h3=h1*h2.


Let triangle $ ABC$; $ I,O$ are incentre,circumcentre respectively. Denote that $ A',B',C'$ lie on segment $ BC,AC,AB$ for wich :$ I{A'} \bot BC,I{B'} \bot AC,I{C'} \bot AB$. $ d$ throw $ A'$ such that $ d \bot B'C' \equal{} H$.$ A'' \in d$ such that $ HA' \equal{} HA''$. Prove that $ BC,IO,AA''$ are concurrent

[img]http://www.picresize.com/images/rsz_1lg06.jpg[/img]


find x and y  if
$x+\sqrt{y}=5$
$y+\sqrt{x}=3$

by induction i made

a complet solution is looong

[hide]
Let $x=a^{2}$ and $y=b^{2}$, where $a,b$ nonnegative. Then the equations become
\[a^{2}+b=5,\ b^{2}+a=3\]
The first equation is $b=5-a^{2}$, so substitute into the second:
\[(5-a^{2})^{2}+a=3\]
\[0=a^{4}-10a^{2}+a+22=(a-2)(a^{3}+2a^{2}-6a-11)\]
The case $a=2$ gives $x=4,\ y=1$. Now we put bounds on the three roots of the cubic; let $P(t)=t^{3}+2t^{2}-6t-11$. Then
\[P(-3)=-2,\ P(-2)=1,\ P(0)=-11,\ P(9/4)=-191/64,\ P(3)=16\]
so the three roots lie in between -3 and -2, between -2 and 0, and between 9/4 and 3. However, $a$ and $b$ nonnegative and $a^{2}+b=5$ requires $0\leq a\leq \sqrt{5}<9/4$, so the cubic only has extraneous roots and $\boxed{x=4,\ y=1}$ is the only solution.
[/hide]

[quote="icx"]A complete solution would be great, Thales. :maybe:

We must have $(x,y) \in \mathbb{Z}^{2}_{+}$. The smallest perfect squares less or equal to 5 are 0, 1 and 4. We see that only $x=4$ and $y=1$ suit.  :D[/quote]

Why must this be so? It appears correct but it seems impossible to prove...
Is it just a misunderstanding of the problem?

http://www.me@homepage

mister irc made the same think i do and said i am wrong

hauhua

he need to read a calc book

[quote="chess64"][quote="scorpius119"][hide]...let $P(t)=t^{3}+2t^{2}-6t-11$. Then
\[P(-3)=-2,\ P(-2)=1,\ P(0)=-11,\ P(9/4)=-191/64,\ P(3)=16 \]
so the three roots lie in between -3 and -2, between -2 and 0, and between 9/4 and 3.[/hide][/quote]

What motivated you to find $P(9/4)$?[/quote]

I wanted to show that the third root was greater than $\sqrt{5}$. I could have just plugged in $\sqrt{5}$ (actually, that would have been easier), but rationals are generally easier to work with. This suggested trying $p/q$ where $p^{2}-5q^{2}=1$ (so that $p/q$ is near $\sqrt{5}$ and greater than it) and $9/4$ is the simplest of them.


trigonometry
geometry
circumcircle
perpendicular bisector
angle bisector
geometry unsolved

Let $ABC$ be a triangle with $AC > AB$.  Let $P$ be the intersection point of the perpendicular bisector of $BC$ and the internal angle bisector of $\angle{A}$.  Construct points $X$ on $AB$ (extended) and $Y$ on $AC$ such that $PX$ is perpendicular to $AB$ and $PY$ is perpendicular to $AC$.  Let $Z$ be the intersection point of $XY$ and $BC$.
Determine the value of $\frac{BZ}{ZC}$.

The triangle $\triangle AXY$ is isosceles with $AX = AY$, hence $\angle BXZ \equiv \angle AXY = \angle AYX = 180^\circ - \angle CYX$. Using the sine theorem for the triangles $\triangle BXZ,\ \triangle CYZ,$

$\frac{BX}{BZ} = \frac{\sin \angle{BZX}}{\sin \angle{BXZ}} = \frac{\sin \angle{CZY}}{\sin \angle{CYZ}} = \frac{CY}{CZ}$

$\frac{BZ}{CZ} = \frac{BX}{CY} = \frac{AX - AB}{AC - AY}$

P is the midpoint of the arc BC of the circumcircle (O) of the triangle $\triangle ABC$ opposite to the vertex A. Let the circle (P) with radius PB = PC meet the lines AB, AC at B', C', respectively. Since the triangles $\triangle BPB', \triangle CPC'$ are both isosceles, X is the midpoint of BB' and Y is the midpoint of CC'. Since the circle (P) is centered on the bisector AP of the angle $\angle A$, we have $AC' = AB = c$ and $AB' = AC = b$. Therefore, $AX = AY = \frac{b + c}{2}$. It follows that

$\frac{BZ}{CZ} = \frac{\frac{b + c}{2} - c}{b - \frac{b + c}{2}} = 1$

It can just be observed that $XY$ is the [i]Simson's Line [/i]of point $P$. Thus it contains the midpoint of $BC$.

[hide="Motivation/Hints"]
[hide="Hint 0"]Always draw a good diagram! Guess the value of $\frac{BZ}{ZC}$. If the answer is not obvious, draw it again![/hide]
[hide="Hint 1"]Once you have a good diagram, like that of happiface, we see that the problem will involve many cyclic quadrilaterals. What is true of $P$? [hide="Answer"]Show that $P$ lies on the circumcircle of $\triangle ABC$. This fact is useful in many problems.[/hide][/hide]
[hide="Hint 2"]What does $AXPY$ look like? [hide="Answer"]Prove that $AXPY$ is a kite.[/hide][/hide]
[hide="Hint 3"]What's so special about $AXPY$ being a kite?[hide="Answer"]Call $Q$ the intersection point of $AP$ and $XY$. It's well-known that $XQ=QY$ and $\angle PQX=\angle PQY=90^\circ$.[/hide][/hide]
[hide="Hint 4"]We need facts about $B,Z,C$. How do we transform parts of $AXPY$ to get $B,Z,C$? [hide="Answer"]You can think of this as rotating and dilating $X,Q,Y$ about $P$.[/hide][/hide]
[hide="Hint 5"]What is something between $B,C,P$ and $X,Y,P$? [hide="Answer"]Prove that $\angle PBC=\angle PXY$.[/hide][/hide]
[hide="Hint 6"]How do we make this useful? Don't worry, I was stuck trying to use $\angle PBC=\angle PXY$ directly for a long time. [hide="Answer"]The hint above is actually supposed to be prove that $\angle PBZ=\angle PXZ$, but failure is good for learning.[/hide][/hide]
[hide="Hint 7"]What does this tell us? [hide="Answer"]$BXPZ$ is cyclic.[/hide][/hide]
[hide="Hint 8"]We want $BZ=ZC$, so how do we use the fact that $BXPZ$ is cyclic? [hide="Answer"]$\angle BZP=90^\circ$.[/hide][/hide]
[hide="Hint 9"]You're almost finished! If you made enough progress independent of the hints, you definitely have the ability to piece together the solution![/hide]
[/hide]
[hide="Solution without Simson Line/Trig"]
Since the circumcircle of $\triangle ABC$ meets the angle bisector of $\angle BAC$ at the midpoint of $\widehat{BC}$ and meets the perpendicular bisector of $BC$ at the midpoint of $\widehat{BC}$, we can alternately define $P$ as the midpoint of $\widehat{BC}$ on the circumcircle of $\triangle ABC$.

Since $AP$ is the angle bisector of $\angle BAC=\angle XAY$, so $\angle XAP=\angle PAY$. We know that $\angle AXP=90^\circ=\angle AYP$ and $AP=AP$, so by AAS congruence, $\triangle AXP\cong\triangle AYP$. Then $AXPY$ is a kite. Let $XY$ meet $AP$ at $Q$. It is well-known that $XQ=QY$, and $\angle XQP=\angle PQY=90^\circ$.

We claim that $\angle PBZ=\angle PXZ$. $Z$ lies on $BC$, so $\angle PBZ=\angle PBC$. Since $ABPC$ is cyclic, $\angle PBC=\angle PAC$. But $Y$ lies on $AC$, so $\angle PAC=\angle PAY$. Since $\angle AXP=90^\circ=\angle AXY$, so $AXPY$ is cyclic, so $\angle PAY=\angle PXY$. Since $Z$ lies on $XY$, $\angle PXY=\angle PXZ$, and combining equalities gives us the desired result.

Since $\angle PBZ=\angle PXZ$, so $BXPZ$ is cyclic. Then $\angle BZP=180-\angle BXP=180-\angle AXP=90^\circ$. Analogously, $\angle CZP=90^\circ$. Since $P$ is the midpoint of $\widehat{BC}$, $\angle PBZ=\angle PBC=\angle PCB=\angle PCZ$. Also, $PZ=PZ$, so AAS congruence tells us that $\triangle BZP\cong\triangle CZP$, so $BZ=ZC$.
[/hide]

[hide=mental math solution]
Let the bisector of $\angle A$ hit $\odot(ABC)$ at $P'$. Then $P'$ lies on the perpendicular bisector of $BC$, since $P'B=P'C$ as $P'$ is the arc midpoint of $\overarc{BC}$. So $P' = P$ and $P \in \odot(ABC)$.

Now $XY$ is the Simpson line of $P$ wrt $ABC$ so $PZ \perp BC, Z \in BC$. $P$ lies on the perpendicular bisector of $BC$ so $Z$ is the midpoint of $BC$ and $BZ/ZC=\boxed{1}$.
[/hide]

I don't know whether my proof is right or not or it can be done by this way:

It is well known that, $ABCD$ is cyclic and $BP=PC$.
$\angle AXP=\angle AYP=90^\circ$. So $AXPY$ is cyclic and $XP=YP$.
$\angle XBP=\angle ACP=\angle XCP$.
$\angle CYP=\angle AXP=\angle BXP$.
So, $\triangle BPX=\triangle CPY$
so, $\angle BPX=\angle CPY$
Let us denote the midpoint of $BC$ by $M$.
We join $X$,$M$ ;$Y$,$M$.
As $PM$ is perpendicular to $BC$, $BMPX$ and $CYMP$ cyclic.
So, $\angle BPX=\angle BMX=\angle CPY=\angle CMY$.
Then $X,Y,M$ collinear and $XY$ intersect $BC$ at $M=Z$,
and $BZ=CZ$.


Two getting-started-level problems from the challenges my Algebra teacher hands out sometimes. Don't worry about them being homework...they actually are, but I've done them already, and they're not near difficult enough that I'd ask anyone for help  :lol:

There is a 15-by-25 foot barn. A goat is tied by a rope to one 25-foot side of the barn, 10 feet from the nearest corner. Find a formula or set of formulas to describe the area that the goat can graze for any rope length r up to 40 feet.

Triangle ABC has sides AB=10 and AC=19. D is on AB and E is on AC such that AD=6 and AE=9. If the area of triangle ABC is 38, find the area of triangle ADE.

#1[hide]

If rope is shorter than 10 ft: ((pi)(r 2))/(2)
If rope is between 10 and 15 ft: ((pi)(3r 2-20r+100))/(4)
If rope is between 15 and 25 ft: ((pi)(4r 2 -50r+325))/(4)
If rope is between 25 and 30 ft: ((pi)(5r 2  -100r +950))/(4)
If rope is between 30 and 40 ft: ((pi)(6r 2  -160r +1850))/(4)

[/hide]

After tons of editting I think i'm finally done lol. Whether right or wrong, that was fun :P

2
[hide]Can use a certain formula that says that (AD*AE)/(AC*AB) is equal to the ratio of the triangle formed by ADE to triangle ABC. This comes out 54/190 = 27/95.
-edit- But the area of the triangle is (27/95) * 38 =10.8.[/hide]

Tarquin is right on #2. Yours, the first two you're right, the others I'm too lazy to check for sure. Probably right though.


trigonometry
SFFT
special factorizations
complex numbers

[quote="nutz_for2.718281828"][quote="mathgeniuse^ln(x)"][quote="vmsvl"]$4\cos{15x}\cos{5x}\cos{3x}\cos{x}+\cos{15x}\cos{5x}+\cos{3x}\cos{x}=0$[/quote]

[hide]Maybe we could call, $\cos{15x}\cos{5x}=a$, and $\cos{3x}\cos{x}=b$.

So we get $4ab+a+b=0$.

And I don't know what to do.[/hide][/quote]

[hide="continuing from where you left off..."]

$4ab+a+b=0$

$4ab+a+b+\frac{1}{4}=\frac{1}{4}$

$(2a+\frac{1}{2})(2b+\frac{1}{2})=\frac{1}{4}$

Maybe you can put a in terms of b, or vice versa...[/hide][/quote]

But you only have 1 equation...

[quote="mathgeniuse^ln(x)"][quote="bpms"][quote="nutz_for2.718281828"]You know what a and b are, just use the addition formula too much and get a in terms of b...[/quote]
You don't have to apply simon's favorite factoring trick at every given oppurtunity :roll:[/quote]

We know, but it was that it seemed promising.[/quote]
Simon's Favorite Factoring trick is only good when you are looking for integers, the chance that cos15xcos5x is an integer is practically epsilon.

True, True, and the yeah the chances will be small.

partial solution
[hide]

using product-to-sum formulas, the original equation become

$(\cos 20x+\cos 10x)(\cos 4x+\cos 2x)+\frac{1}{2}(\cos 10x+\cos 20x)+\frac{1}{2}(\cos 4x+\cos 2x)=0$

$\frac{1}{2}(\cos 24x+\cos 22x+\cos 20x+...+\cos 2x) = 0$
this sum can be found using complex numbers ...

[/hide]

[quote="mathgeniuse^ln(x)"][quote="vmsvl"]$4\cos{15x}\cos{5x}\cos{3x}\cos{x}+\cos{15x}\cos{5x}+\cos{3x}\cos{x}=0$[/quote]

Where does this question come from?[/quote]
From diendantoanhoc.net

[quote="mathgeniuse^ln(x)"][quote="vmsvl"]$4\cos{15x}\cos{5x}\cos{3x}\cos{x}+\cos{15x}\cos{5x}+\cos{3x}\cos{x}=0$[/quote]

Where does this question come from?[/quote]
From mathnfriend.net

[quote="nutz_for2.718281828"]All right! That's it!

I'm going into brute force!!!!

[hide="more stuff"]Where I left off,

$\left(2a+\frac{1}{2}\right)\left(2b+\frac{1}{2}\right)=\frac{1}{4}$

So $\cos{15x}\cos{5x}=a$ and $\cos{3x}\cos{x}=b$

We now want the formula for $\cos{5x}$ when we know $\cos{x}$

$\cos{2x}=\cos^{2}{x}-\sin^{2}{x}$

$\sin{2x}=2\cos{x}\sin{x}$

$\cos{3x}=\cos{2x}\cos{x}-\sin{2x}\sin{x}=\cos^{3}{x}-\sin^{2}{x}\cos{x}-2\cos{x}\sin^{2}{x}=\cos^{3}{x}-3\cos{x}\sin^{3}{x}$

$\sin{3x}=\sin{x}\cos^{2}{x}-\sin^{3}{x}+2\sin^{2}{x}\cos{x}$

$\cos{5x}=\cos{(3x+2x)}=\cos{3x}\cos{2x}-\sin{2x}\sin{3x}=(\cos^{3}{x}-3\cos{x}\sin^{3}{x})(\cos^{2}{x}-\sin^{2}{x})-2\sin{x}\cos{x}(\sin{x}\cos^{2}{x}-\sin^{3}{x}+2\sin^{2}{x}\cos{x})=something$ $big$[/hide]

You can take over.... :rotfl:[/quote]
You can just use DeMoivre's theorem to figure out what $\cos 5x$ is :|

sorry, I'm a semi-beginner and semi-intermediate... I'm stuck........

[quote="jcmtnez"]Here it is the solution:

Lets call
$cos(5x) = a$
$cos(x) = b$

Then:

$4\cos(15x)\cos(5x)\cos(3x)\cos(x) = \cos(x)\cos(5x) = ab$

$\cos(15x)\cos(5x) = \frac{\cos(x)}{2}= \frac{a}{2}$

$\cos(3x)\cos(x) = \frac{\cos(x)}{2}= \frac{b}{2}$

From these identities we get the following:

$2ab+a+b=0$

Now, lets call $cos(x) = t$

and we'll get:

$a = (5t-20t^{3}+16^{5})$
$b= t$

This one variable equation follows by substitution:

$32t^{6}+16t^{5}-40t^{4}-20t^{3}+10t^{2}+6t = 0$

Clearly $t=0$, therefore $cos(x)=0$, and $x = 90+180k$ which are the solutions to the equation.

Check it out with the calculator is amazing how simple it is the solution, I didnt expect it.

If you want to know were did I get the identities and the final equation look at this web page, it is a great resource:

http://mcraefamily.com/MathHelp/GeometryTrigEquivCos3xEtcB.htm[/quote]

I'm sorry to say this, but this is really really wrong  :maybe:

The three first equations are wrong...do you even know how to use the addition formula? And it's not because $t=0$ is a root, that there aren't others. In fact, as $32t^{5}+16t^{4}-40t^{3}-20t^{2}+10t+6 = 0$, we have at least one other real root...

One can obtain [hide="this"]\[\begin{aligned}0 & = 4\cos 15x\cos 5x\cos 3x\cos x+\cos 15x\cos 5x+\cos 3x\cos x \\ & = \frac{1}{2}\sum_{i=1}^{12}\cos 2ix = \frac{1}{2}\cos 13x\sum_{i=1}^{5}\cos (2i+1)x = \frac{1}{2}\cos 13x\cos 6x\left(\cos 5x+\cos 3x+\cos x\right) \\ & = \frac{1}{2}\cos 13x\cos 6x\cos 3x\left(2\cos 2x+1\right)\end{aligned}\][/hide]It's easy to continue from there.


LaTeX
ARML
HMMT
Gauss
AMC
AIME
Math Prize for Girls

\[ \begin{tabular}{|c|c|}\hline \multicolumn{2}{|c|}{\textbf{\large Complete SC Memberlist}} \\
\hline N{a}me & AoPS SN \\
\hline \multicolumn{2}{|c|}{\textbf{Moderators}} \\
\hline Joe & joml88 \\
\hline Sean & mysmartmouth \\
\hline \multicolumn{2}{|c|}{\textbf{Teachers/Coaches}} \\
\hline Mr. Rybarczyk & prybarczyk \\
\hline Mr. Neal & gauss202 \\
\hline \multicolumn{2}{|c|}{\textbf{Members}} \\
\hline Brian & bbowers \\
\hline Junda & chen52808 \\
\hline Ben & bencruisin \\
\hline Adam & chamoo \\
\hline David & furious \\
\hline David & DummAddict \\
\hline Asif & arkmastermind \\
\hline Jim & scmathman13 \\
\hline Joey & miraculouspostmaster \\
\hline Carmen & your\_{c}hinese\_{b}uddy \\
\hline Harry & origamiidiot \\
\hline Ella & SoltzElla \\
\hline Jaydeep & srimani \\
\hline Ian & apeman5291 \\
\hline Shobhit & Iversonfan2005 \\
\hline Bella & dark\_{s}unburst \\
\hline Rono & rs9999 \\
\hline David & drummerdude413 \\
\hline Ankit & DanielHStennis08 \\
\hline Anne & avcamethyst \\
\hline Ravi & Anaxerzia \\
\hline \end{tabular}
\]

\[ \begin{tabular}{|c|c|}\hline \multicolumn{2}{|c|}{\textbf{\large Honorary NC Members}} \\
\hline Nam{e} & AoPS SN \\
\hline John & jb05 \\
\hline Arnav & MysticTerminator \\
\hline \end{tabular}
\]
Please note that this is a partial list.  Please add yourself to the list by replying.  Feel free to correct any mistakes as well.  If you would like your name removed, please contact me immediately.

LAST NAMES REMOVED PER REQUEST.

[quote="mysmartmouth"]OK, I fixed that.  And who is Chen again?  I think Mr. R mentioned who he was, but I forgot by now.[/quote]

Junda

[quote="jb05"]I feel very left out here, with the doubled effect of my name not on the list and my state not included in the title of this forum.[/quote]

lol...sorry about that, it's just that only like one person from NC has been active on the Carolinas forum, so we just made a separate forum for SC. But I'm sure if you ask Valentin or Mcrawford, they'll make a new forum for NC. Either way, we don't want NCers hanging around here...jk

Oh, and BTW, MY NAME'S "Iversonfan2005" NOT "iversonfan2005"!!!!!!!!!! :mad:  :mad:

I like $\LaTeX$ tables :D  Oh, for some reason the $\LaTeX$ code didn't stay formatted and got compressed...I'm only noting that because I imagine Sean will be editing it and it would probably be hard to figure it out with it all scruched up and stuff...Sean, if you need any help with that, let me know.  Btw, don't try putting the header "Name" in the first column.  For some reason $\LaTeX$ doesn't like that and will give you an error message.

Oh yeah, I added AIM sn's to the table.  Some of the SC All state team members have been getting together in chatrooms on AIM.  I highly recommend getting an AIM account as well.  You'll be informed more quickly of events and you'll get to know your fellow teammates.  We can also get together and discuss problems if people are in the mood.

[quote="your_chinese_buddy"]um....am i supposed to be on the list if i'm on the state math team??  :huh:[/quote]
you will be once you tell us who you are. we would like to get everyone to introduce themselves in the "introduction" stickie.

[quote="your_chinese_buddy"]um....am i supposed to be on the list if i'm on the state math team??  :huh:[/quote]

Yes.  Please introduce yourself [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20914]here[/url]

[quote="mysmartmouth"]OK, Joe.  I hate latex.  You're officially in charge of the memberlist.  Lol.  And please fix it; I've butchered it pretty badly.  Also, there's a buch of new people to add.  Look around.[/quote]

You didn't butcher it.  You just need to use \_ for underscores since just putting _ will make it a subscript.  Also, you should have put the \hline after the line, not before it.  Other than that you it was good. ;)

[quote="furious"]mathguy was an old account of mine. i think joe remembers the situation with that. lol. anyways, you can take it off the list.[/quote]

Can you explain?  I'd like to hear this story!   :rotfl:

[quote="your_chinese_buddy"]um....am i supposed to be on the list if i'm on the state math team??  :huh:[/quote]

no, some people on that list are not even on the all state math team

if you're a member of the SC AOPS forum, you can be added onto the list ;)

Joe:
1) I successfully added two new members! Yay! I can work Latex tables! Please maintain this list in alphabetical order.

2) Can we reformat the table such that the Moniker & AoPS SN & AIM SN line is ABOVE the Moderators heading? Or does that not work/look bad?

Everyone Else:
1) Please post missing info here; Joe or I will then add it to the chart. If you are new, please post your information in the Introduction thread.

2) If you do not have an [url=http://www.aim.com/]AOL Instant Messenger[/url] (AIM) screename, I encourage you to get one. They are free here. They are excellent for instant communication among team members, and I have worked with Joe to solve many problems through AIM. Also, you do not need to download anything to use AIM! You can now use AIM from Internet Explorer or Firefox through [url=http://www.aim.com/get_aim/express/aim_expr.adp]AIM Express[/url]. If you already have an AIM screename, please list it here, or PM me, so I can add it to the chart.

Thanks!

Woah, where'd the post go?

I think sean sincerely enjoys deleting arnav's posts. </obvious>

It's a pity, that one was truly a work of art though few would have gotten it :'(

oh, thanks sean

We have, however, determined that Arnav is exempted from spam laws (see "dude", NC forum)

I'm Anne.  can you add me to the list? i just registered.

Let it be known from here on out that Gray is a member of this forum.

This is Annie.
Why did we revive this...?

New people are joining AoPS!

Are you doing Math Prize for Girls next month? If so, HMMT problems are good ones to use for practice!

@gauss oh alright (:

Yes I am (this month actually)! And thanks! Thats a good idea, I wasn't sure where to go after doing the 3 previous tests :b

AIME problems are probably even better if you can get old ones. Send me a message if you need some.


$A, B, C$ are angles of a triangle. Prove:

(i)$sin A + sin B + sin C \leq \frac{3\sqrt{3}}{2}$

(ii)$csc A + csc B + csc C \geq 2\sqrt{3}$

(iii)$(csc A)(csc B)(csc C) \geq \frac{8\sqrt{3}}{9}$

(1) $\sin{x}$ is concave within $[0, \pi]$ thus $\sum \sin{A} \leq 3 \sin{\frac{\pi}{3}}=\frac{3\sqrt{3}}{2}$.

(2) By AM-HM we have $\sum \csc{A} \geq \frac{9}{\sum \sin{A}} \geq 2\sqrt{3}$.

(3) $\prod \sin{A} \leq \left( \frac{\sum \sin{A}}{3} \right)^3 \leq \frac{3\sqrt{3}}{8}$. Therefore we get: $\prod \csc{A} \geq \frac{8\sqrt{3}}{9}$.


[hide="1"]
2+2+2=6, which is a multiple of 3, so 1/3??[/hide]
[hide="2"]
4/18, add 5:
7/14, you need to add 5[/hide]
[hide="3"]
1/2*1/2*1/2*1/2
h      h     t      t
It is the same thing for 3 and 4 heads,
$3\cdot\frac{1}{4}^4$ So the answer is:
$\frac{3}{16}$[/hide]

(1) $\frac{1}{3}$
(2) $6$
 once again, math has given us easy yet fun problems :)

[quote="math92"]1) What is the probablity that a multiple of 2 is also a multiple of 3?

2) A bag contains 8 white marbles, 6 red marbles, and 4 blue marbles.  HOw many blue marbles must be added to the bag so that the probablitity of choosing a blue marble is: $\frac{1}{2}$

3) Matt will flip a coin 4 times.  What is the probablity that $\frac{1}{2}$ or more of the flips will be Heads?[/quote]

The correct answer  for number 3 is 11/16. The reason is as follows. If you enumerate the sequences in which EXACTLY 2 heads occuring:

HHTT, HTHT, THHT, TTHH, THTH, THHT   (total of 6).

Similarly, the sequences in which EXACTLY 3 heads occuring:

HHHT, HHTH, HTHH, THHH (total of 4).

Finally following is the sequence in which EXACTLY 4 heads occuring:

HHHH (total of 1).

So there is a total of 6+4+1 = 11 sequences in which atleast two or more heads are occuring. Since there are a total of 16 possibilities, the probability of atleast two or more heads occuring is 11/16.

[quote="math92"]according to my answers, pascal got 2 and 3 wrong, and jhollenbeck got 2 wrong.  For number 2 you guys added 8, 6, and 4 wrong.  8+6+4=18 not 14. :)[/quote]

Actually, they forgot to consider the blue marbles in the total.

Yes, your right. I'll edit it.
[hide="2"]
4/18=2/9
So, you need to add 5.[/hide]
For 3, I don't get why I was wrong   :?

[quote="pascal_1623"]
For 3, I don't get why I was wrong   :?[/quote]
look at mthiyaga's solution, it explains the answer very well.

[quote="pascal_1623"]For 3, I don't get why I was wrong   :?[/quote]

For the case of HHTT, you stated the probablity of that happening was 1/16, which is correct. But, as mthiyaga illustrates, you can throw in the order HTHT or HTTH, and each will also have a 1/16 possibility. since there are 4 choose 2 = 6 ways to arrange two heads and two tails, you must multiplied by 6 to achieve 6/16=3/8, which is the probability of throwing exactly two heads. Others follow a similar pattern.


http://www.artofproblemsolving.com/Forum/weblog.php?w=1092

note there are bunch of links on the top that says "Category: PHP" which simply redirects to the blog again.

looks like it escaped from the category box.

In gauss1181's too. The general life is on top.


function
real analysis
real analysis unsolved

Source: Paper
what is the invers of the function
$f(x)=2^{\frac{1}{x}}+3^{\frac{1}{x}}+5^{\frac{1}{x}}$
 :?:

There is almost certainly no closed form answer.

Yes,you can't find the inverse.At least I can't find it and I don't think there is a general formula for finding something like this.Who knows? :?  :huh:  :?  :?


Έστω $ ABC$ τρίγωνο με $ \widehat{A}\equal{}45$     $ \widehat{C}\equal{}30$ και έστω $ AM$ η διάμεσος. Να βρεθεί η γωνία $ \widehat{MAC}$

Could you translate the context in English, Γιώργος? :)

Thank you very much, cretanman.  :)

I have just solved the problem in 3 minutes, but I wonder if this solution can be called a synthetic soultion.

Take a point $ H$ such that $ BH$ is perpendicular to $ AC.$ Let $ AH \equal{} 1,$ we have $ BH \equal{} 1,\ AB \equal{} \sqrt {2},\ CH \equal{} \sqrt {3}$ and $ BM \equal{} CM \equal{} 1.$

Pappus' theorem gives $ AB^2 \plus{} AC^2 \equal{} 2(AM^2 \plus{} BM^2)\Longleftrightarrow (\sqrt {2})^2 \plus{} (1 \plus{} \sqrt {3})^2 \equal{} 2(AM^2 \plus{} 1^2)$, yielding $ AM \equal{} \sqrt {2 \plus{} \sqrt {3}}.$

Then use sine law, we have $ \frac {CM}{\sin \angle{MAC}} \equal{} \frac {AM}{\sin \angle{ACM}}\Longleftrightarrow \frac {1}{\sin \angle{MAC}} \equal{} \frac {\sqrt {2 \plus{} \sqrt {3}}}{\sin 30^\circ}$, yielding $ \sin \angle{MAC} \equal{} \frac {1}{2\sqrt {2 \plus{} \sqrt {3}}} \equal{} \frac {\sqrt {2 \minus{} \sqrt {3}}}{2} \equal{} \frac {\sqrt {6} \minus{} \sqrt {2}}{4}$
$ \therefore \angle{MAC} \equal{} 15^\circ.$

[quote="stergiu"][quote="cretanman"]Let $ ABC$ a triangle with $ \widehat{A} \equal{} 45^{\circ}$, $ \widehat{C} \equal{} 30^{\circ}$ and $ AM$ is the median. ($ M$ is the midpoint of $ BC$). Find the angle $ \widehat{MAC}$.[/quote]

 [color=blue][i]This is  a nice and simple problem. In 1996 I found it an old issue of Gazeta Matematica.I gave the solution to my friend  Sotiris Abrazis , but he lost it. Recently he asked me to solve the problem again :)  It took me half an hour and more to solve it again(synthetically). Any way ,  Juniors must try for a while with pure synthetic geometry to find a solution. I do not think that seniors will have trouble with it ![/i][/color]

Babis[/quote]

Κύριε Μπάμπη από τον κ Σωτήρη Αμπράζη πήρα αυτή την άσκηση διότι όπως αναφέρατε έχασε τη λύση. :lol:

Μία συνθετική λύση με την ύλη της Α΄λυκείου.


Adrian has drawn a circle in the xy-plane whose radius is a positive integer at most 2008. The origin lies somewhere inside the circle. You are allowed to ask him questions of the form “Is the point $ (x, y)$ inside your circle?” After each question he will answer truthfully “yes” or “no”. Show that it is always possible to deduce the radius of the circle after at most sixty questions. [i][Note: Any point which lies exactly on the circle may be considered to lie inside the circle.][/i]

[hide="hint"]ask questions (0,y) ( in order to find one point outside (0,y'+1) and another inside (0,y')) similarly with (o,-y) and (x,0)[/hide]

Observe that $ 4016<4096\equal{}2^{12}$.
Ask the question1: " Is $ (0,4016)$ within the circle?"
                      q2: " Is $ (0,2008)$ within the circle?"
Depending on the answer,
                      q3: " Is $ (0,1004)\quad or\quad (0,3012)$ within the circle?"
With $ \leq 12$ questions we can determine the co ordinate $ (0,r)\in circle$.
Similarly we can determine the co ordinate $ (0,\minus{}l)\in circle$,$ l>0$
Centre of the circle lies on the line $ x\equal{}\frac{r\plus{}l}{2}$

Similarly atmost another 24 questions determine the line perpendicular to y axis passing through the centre and hence the co ordinates of the centre.Atmost another 12 questions settle the radius.
A problem on same vien : http://www.mathlinks.ro/viewtopic.php?t=181161

[quote="srikanth"]Atmost another 12 questions settle the radius.
 [/quote]

How??  :o


Find all the functions $ f$ continous in $ x_0\in\mathbb{R}$ for wich
$ f(x\plus{}y)\equal{}f(x)\plus{}f(y)\minus{}af(x)f(y),\forall x,y\in\mathbb{R} .$ $ a$ const.

[quote="cckek"]Find all the functions $ f$ continous in $ x_0\in\mathbb{R}$ for wich
$ f(x \plus{} y) \equal{} f(x) \plus{} f(y) \minus{} af(x)f(y),\forall x,y\in\mathbb{R} .$ $ a$ const.[/quote]

If $ a\equal{}0$, we have just Cauchy's equation and then the continuous solutions are $ f(x)\equal{}cx$

If $ a\neq 0$, let $ h(x)\equal{}1\minus{}af(x)$. we get $ h(x\plus{}y)\equal{}h(x)h(y)$ with two kind of continuous solutions :
$ h(x)\equal{}0$
$ h(x)\equal{}e^{cx}$

And so the solutions for $ f(x)$:
$ f(x)\equal{}\frac{1}{a}$

$ f(x)\equal{}\frac{1\minus{}e^{cx}}{a}$

Yes, that is the solution. Congrat. Nice :D


What are the details on AIME qualification? How will teachers be notified? When will they be notified? etc...

I believe your school is sent a sheet that has the scores of all participants and based on it, you'll know who qualified.

I mean for USAMTS to AIME qualification.

We have not yet coordinated with the AMC how this will work.  We will notify you when we have this information.


induction
geometry
geometric transformation
reflection
rectangle

Does anyone know a formula for the sum of the 1st n numbers?
almost like an arithmetic factorial.

hello, we have
$ \sum_{i\equal{}1}^{n}i\equal{}\frac{n(n\plus{}1)}{2}$.
Sonnhard.

[quote="yildobekir"]$ 1 \plus{} 2 \plus{} 3 \plus{} \ldots \plus{} n \equal{} \frac {n(n \minus{} 1)}{2}$[/quote]

Its $ \dfrac {n(n\plus{}1)}{2}$.  NOT $ \frac {n(n \minus{} 1)}{2}$.

One can prove that $ 1 \plus{} 2 \plus{} \dots \plus{} n \equal{} \frac{n(n\plus{}1)}{2}$ is true using induction, I believe.

[hide="Induction"] First of all, we shall start off with a base case (or whatever its called). Our base case is $ n\equal{}1$. Clearly, if we plug $ n\equal{}1$ into the formula, the equation is true. Next, we make the inductive hypothesis that it works for some $ n$: that is, for some positive $ n$, we assume $ 1\plus{}2\plus{}3\plus{}\dots\equal{}n\equal{}\frac{n(n\plus{}1)}{2}$ is true. We will now show that it works for $ n\plus{}1$. Evaluating,

$ 1\plus{}2\plus{}3\plus{}\dots\plus{}n\plus{}n\plus{}1\equal{}(1\plus{}2\plus{}3\plus{}\dots\plus{}n)\plus{}(n\plus{}1)\equal{}\frac{n(n\plus{}1)}{2}\plus{}(n\plus{}1)$

$ \equal{}\frac{n(n\plus{}1)}{2}\plus{}\frac{2(n\plus{}1)}{2}\equal{}\frac{(n\plus{}1)(n\plus{}2)}{2}$.

Hence, the induction is proven for all positive integers $ n$.[/hide]

If you know the proof by induction, you have to know the formula. But if you know to derive it, even if you don't know the formula, you can derive it amidst any exam in the rough page if you know the idea.

Let

$ S=1+2+3+\cdots+n$

$ S=n+(n-1)+(n-2)+\cdots+1$

Adding the above equations, vertically, we have $ 2S=(1+n)+(2+n-1)+\cdots(n+1) \Longrightarrow 2S=(n+1)+(n+1)+\cdots(n+1)$ where $ (n+1)$ repeats $ n$ times

And so, $ 2S=n(n+1) \Longrightarrow S=\frac{n(n+1)}{2}$ and we are done. :wink:

Consider one dot. Consider 2 dots below it as follows:

.
..

Continue in this fashion.

.
..
...

etc.

We can reflect this (place one dot at the end of the last line, etc.) to get a rectangle $ n$ by $ n\plus{}1$.

Thus, we have $ \frac{n(n\plus{}1)}{2}$.

See the picture below.

 [geogebra]c3d6ef4fb2bffe7ec699f1be79c990ad46d9c733[/geogebra]

The shown example is 1+2+3+4, but this is general and can be extended.