MATHCOUNTS AMC AIME USA(J)MO USAMO function calculus

I realize each year, some participants make ctdown that few anticipate, and others who were "good bets" fall short of their goals, but without naming names, I know of several competitiors this year who did "all the right things" effort wise -  some went to competitive math camps, many did as many practice problems as anyone else (I mean they did [b]all[/b] available problems at least once), were confident going into the competition, did extremely well at prior MATHCOUNTS competitions, etc, and amazingly (to many) either fell many places from prior Nats years, or simply fell way short of their expected goal.  And not to diminish the talents/work ethic of those who scored "better than anticipated" - but some of this year's top finishers at Nats had consistently lower scores ([i]much lower [/i]in some instances) when comparing their scores from both prior Mathcts competitions, as well as their scores like math SAT, AMC, AIME etc, to those who finished much lower, and well -- it seemed to happen more this year than in prior years - and I'm trying to get a "handle" on "Why"?

I know at least one of these "poor scorers" who aced the Target with all 8 right, but totally bombed on Sprint and they [i]still[/i] can't quite grasp what was so different about this yr's Sprint questions to those of the previous 20 years' tests.  Anyone have any theories about this?  Did all these "top kids" simply have a bad day all at the same time, or was there something about the questions that maybe they were more likely to "misread" or "overcomplicate" or ???????  (oh, I have to discount "actual test taking ability" as many of those that "fell short" had pwned many prior competitions).

This happens every year - some under-the-radar people do extremely well, and some who are expected to score well don't.  There's nothing really unusual about it...

But the "number of places" some people "fell" or "climbed" does seem unusual to me from just what I know of prior year's scores, participants, etc.  Of course, I don't know if anyone has ever done a statistical analysis of the exact participants in any given 3 - 6 year period (or whatever period would cover all possible "2 or 3-peaters") to see the exact pattern of what percentage of people either rose or fell from their prior year's placings -, and by how much, (oh, just focusing on Nats, as including States and Chapters would be unwieldy and include too many people) -- but my general observation was that there was much more variance this year, and in prior years there seemed to be a more logical assessment of who might end up in the top 12.  (of course, it's never an exact science - I'm just speaking "in general")
As I pointed out (as AMC scores etc. are often made public - especially the names of those with perfect scores), it seemed that many of the top kids at Nats didn't have the best scores from other tests overall - and it just seems -well, inconsistent - slightly "illogical" to me, just how that can be.  I just wonder if anyone thought this year's Nat's test somehow varied from prior years? And if so, how? Nature of problems? Way they were posed? ??? AIME15 - I don't know if you were at Nats this year or were just commenting "in general"?

Every single person in cd was a USAMO qualifier, so no one there was truly "under the radar".

Mewto55555, true, but there were [i]many[/i] USAMO qualifiers this year - I believe 25 or more, and of course, only 12 people could make it into ctdown. My point was that if you compared these (top 12) people's actual USAMO scores (with the exception of those few who clearly blew the averages out of the water on all accts!  :)  - plus this same group's AMC, AIME scores, etc, with non-top 12's, many might be surprised how many of those scoring "higher" on what most agree to be "higher level tests", (or who previously placed much higher at prior Nats than those in the ctdown), did not make ctdown, and I just found this interesting. And several participants I spoke to felt they did much "differently" (again, some better, some much worse) than on all the prior tests they practiced on.  Which indicates to me a potential "shift" in the style of questions .. or something.

OR .. maybe it just indicates that MATHCOUNTS success is not a good correlation or indicator of one's ability to ultimately succeed at much more complex, higher level math. Otherwise, wouldn't you expect all 25 USAMO'ers to make up the top 25 places?? (which of course, didn't occur)

[quote="mathlearner"]
OR .. maybe it just indicates that MATHCOUNTS success is not a good correlation or indicator of one's ability to ultimately succeed at much more complex, higher level math. Otherwise, wouldn't you expect all 25 USAMO'ers to make up the top 25 places?? (which of course, didn't occur)[/quote]

There are 2 things wrong with this.

1. MATHCOUNTS success is not a good correlation or indicator of one's ability to ultimately succeed at much more complex, higher level math. MATHCOUNTS doesn't require any actual creative thinking or unusual methods; mathcounts problems are all problems where if you have seen a certain, well-known, trick before, then the problem is trivial. It might take some time to solve it though, but I'm not sure that time required to solve indicates math skill.

2. Um, "wouldn't you expect all 25 USAMO'ers to make up the top 25 places??" is flawed. There are a lot of people at nats who shoudl have made usamo, but were unlucky on aime/amc.

Finally, I have to say, the real test is not a practice test. People perform differently on both. Example: CA team practices...

pythag011, mmm.. excellent points. I definitely agree with #1 - (and I think that's also quite reassuring to those "budding mathematicians" who may NOT have ever made Nats ctdown .. but I'm not so sure about #2.  I mean, I get your point, but not sure "luck" is as big a factor as some people like to think -- sure, you can get more or less problems you're familiar with, but I think most qualifiers qualify by more than a single problem, and thus, their mathematical acumen plays a much larger part. And also, focusing on the MATHCTS participants who "were" USAMO qualifiers - not those that [i]should have been [/i]indicates that (at least these)  participants were certainly "skilled enough" to make top 12 Nats, but somehow didn't ... :(

But the mystery question remains ... "[i]why[/i]"??

I've heard top coaches say the top 12 could differ if the test were given on a different day to the same group, but I wonder - is that variation more a function of the test, or just human factors? (ie: being tired, sick, etc.)  I'm guessing both.

I think this just shows how fine the line is between getting into the top 12 and "just missing".  One or two questions could make a difference of several places in the standings.  That is why I always hope for very hard tests so that there is  a very clear differentiation in both teams and individuals.

I thought all of the sprint tests this year were right on but the target rounds were too easy.  The team rounds were also a bit on the easy side making it difficult for any team to make up ground they had lost in the sprint or target.

The tests seem to run in cycles.  The two or three years before this year were on the hard side.  This year the cycle seemd to turn in the opposite direction.  I can usually tell by the school and chapter round what the difficulty of the state and national is going to be.

The team round was insanely easy (compared to previous years). Two people on our team had literally memorized #10.

As for this years test, I actually did one point better than my averages. Imo, this test was a lot more arithmetic/brute force/listing out intensive than previous ones, which suited me well, but perhaps not so much the others.

[quote="remy1140"]A lot of people in nats this year memorized #10 on team round. Considering how 8 teams got a 10/10 on team, this year was very easy, while last year, only one team got a perfect.

I think it's not only the Mathcounts tests that became much easier this year, but also some of the AMCs. In my opinion, the 2008 AMC8 was easier than the 2007 one, and there were more than 400 perfect scores this year, while last year, there were only about 100 perfect scores. I wonder if that somewhat has to do with how calculators are banned starting 2008.[/quote]

AMC 8 is only one AMC.

I do have to agree that the AMC 10s were easier, but the AMC 12s were in my opinion harder this year.

Was last year's team really that hard? Looking at the problems didn't give me that impression at ALL.

Uh, please excuse my ignorance but what do you mean when you say people "memorized problem #10 of the Nats team round? I mean, how do you "memorize" a problem before you are given it to solve? Do you mean this problem was seen on a prior test? If so, what year?

Basic derangement problems.  I'm not sure if it's appeared in MATHCOUNTS before, but I've definitely seen it in FTW (it asked for number of ways, not probability).

[quote="remy1140"]I saw almost the exact problem in Alcumus as well.[/quote]

Other than some silly brute force problems, all mathcounts problems are well known.

[quote="pythag011 should have"][quote="remy1140"]I saw almost the exact problem in Alcumus as well.[/quote]

Other than some silly brute force problems, most mathcounts problems are well known.[/quote] edited for truth

[quote="Ihatepie"][quote="pythag011"][quote="remy1140"]I saw almost the exact problem in Alcumus as well.[/quote]

Other than some silly brute force problems, most mathcounts problems are well known.[/quote] edited for truth[/quote]
Agreed. It says "pythag011 wrote: ..." but it isn't what pythag wrote. JK

Ok I edited it again.  :P Not the best grammar, but still.


geometry
ratio
geometric transformation
reflection
circumcircle
trigonometry
Euler

Let $ ABC$ be a triangle, $ P$ is an arbitrary point in the plane of triangle $ ABC$, let $ P_{1},P_{2},P_{3}$ be the reflections of $ P$ in the sides $ BC,CA,AB$, respectively. Dentoe by $ O$ the circumcenter, $ H$ the orthocenter of triangle $ ABC$, respectively. show that $ \frac {S_{\bigtriangleup HP_{1}P_{2}}}{S_{\bigtriangleup HP_{1}P_{3}}} \equal{} \frac {S_{\bigtriangleup OAB}}{S_{\bigtriangleup OAC}}$ (where $ S$ denotes the area of triangle $ XYZ$)

Let $ Q$ and $ R$ be a reflection point of $ P_{3}$ and $ P_{2}$ wrt $ BC$
$ H_{1}$ is a reflection of $ H$ wrt $ BC$. $ H_{2}$ is a reflection of $ H_{1}$ wrt $ AB$. $ H'$ is a reflection of $ H_{2}$ wrt $ BC$.
$ H_{3}$ is a reflection of $ H_{1}$ wrt $ AC$. $ H''$ is a reflection of $ H_{3}$ wrt $ BC$....-_-;;;
It is easy to show that $ H',H_{1},H''$ lie on a line.
$ HP_{1}P_{2}\equiv H_{1}PR$, $ HP_{1}P_{3} \equiv H_{1}PQ$....
$ PH_{1} \equal{} H'Q \equal{} H''R$...
$ \angle H'BH_{1}\equal{}\angle QBP\equal{}2\angle B$, $ \angle H''CH_{1}\equal{}\angle RCP\equal{}2\angle C$...
$ \frac {H'H_{1}}{H_{1}H''} \equal{} \frac {BH_{1}\sin B}{CH_{1}\sin C} \equal{} \frac {\sin 2B}{\sin 2C} \equal{} \frac {S_{AOC}}{S_{AOB}}$
So remain is to show that $ \frac {d(Q,PH_{1})}{d(R,PH_{1})} \equal{} \frac {H'H_{1}}{H_{1}H''}( \equal{} \frac {S_{AOC}}{S_{AOB}})$
But $ \frac {d(H',PH_{1})}{d(H'',PH_{1})} \equal{} \frac {H'H_{1}}{H_{1}H''} \equal{} \frac {S_{AOC}}{S_{AOB}}$
and $ d(H',PH_{1}) \minus{} d(Q,PH_{1}) \equal{} H'Q \sin \measuredangle (H'Q,PH_{1}) \equal{} H'Q \sin 2B \equal{} PH_{1}\sin 2B$,
$ d(H'',PH_{1}) \minus{} d(R,PH_{1}) \equal{} H''R\sin 2C \equal{} PH_{1}\sin 2C$
$ \rightarrow \frac {d(H',PH_{1}) \minus{} d(Q,PH_{1})}{d(H'',PH_{1}) \minus{} d(R,PH_{1})} \equal{} \frac {S_{AOC}}{S_{AOB}}$,
It's obvious...

I want to see your solution... this was a very curious problem for me. Although I satisfied with my solution, I think there are simpler solutions using easy technique...

Dear Soo Hong Lee
My solution is similar to yours, by trigonometry, we have $ S_{\bigtriangleup HP_{2}P_{3}} \equal{} \frac {R^2 \minus{} OP^2}{2}sin2A$. where $ R$ denotes the circumradius of triangle $ ABC$.
You are famous for your genius in China, we also like Han-sol Shin (username Leonhard Euler), all his geometrical problems are very interesting and pretty hard. Do you know him?

Who called me genius???? I'm just a lucky boy!! :blush:
BTW, of course, I know him. He is 1 year elder than me. But strange thing is that such a "monster" in geometry is in middle school grade in Korea Winter School in this year!!
I first met him in a web forum named "KAIST mathmatical problem solving group". That group have published a magazine "Math Letter", and I sometimes proposed easy problems include geometry(when I was young).
Then suddenly a strange person appeared in that forum, who used nickname "Leonhard Euler", and proposed numerous geometry problems hard enough...
I don't know how does he made such a curious and hard geometry problems, but It doesn't a accidental result, certainly.


group theory
abstract algebra
induction
superior algebra
superior algebra unsolved

If a group has only representations of dimension 1 and 2, what can I say about it? For example, a group that has representations of only dimension 1 will have to be abelian.

For the 1 and 2 dimensional case, I think the answer is that it's solvable, but I can't quite prove it...

This is an interesting question. Such groups are bound to have other nice properties; maybe we can even find a characterization, but for now I'll prove their solvability.

First, notice that the property of having irreducible complex representations of degree at most $2$ is preserved by passing to quotient groups and subgroups. For quotient groups this is clear: irreducible representations of $G/H$ give rise to irreducible representations of $G$ of the same degree after composing with the projection $G\to G/H$. On the other hand, let $H$ be any subgroup of $G$, and let $\rho$ be an irreducible representation of $H$. Also, let $\psi$ be an irreducible representation of $G$ which occurs as a summand in the representation $\mbox{Ind}^{G}_{H}\rho$ of $G$ induced from $\rho$. By [url=http://planetmath.org/encyclopedia/FrobeniusReciprocity.html]Frobenius recirpocity[/url], the multiplicity of $\psi$ in $\mbox{Ind}^{G}_{H}\rho$ is equal to the multiplicity of $\rho$ in the restriction $\mbox{Res}^{G}_{H}\psi$ of $\psi$ to $H$. We chose our representations such that this multiplicity is non-zero, so $\rho$ is a summand in the $1$ or $2$-dimensional representation $\mbox{Res}^{G}_{H}\psi$ of $H$. This proves that the degree of $\rho$ is at most $2$, as desired.

If the only $1$-dimensional representation is the trivial one, then $G$ has odd order (since the order is the sum of the squares of the degrees of the irreducible representations). The degree of an irreducible representation divides the order of $G$, so in this case there can be no irreducible representations of degree $2$. It follows that $G$ is trivial. This means that if $G$ is non-trivial, then it has non-trivial $1$-dimensional representations, so the derived subgroup $G'$ is proper. By induction on the order, $G'$ is solvable. Also, $G/G'$ is solvable, being abelian. Thus $G$ is solvable, and we're done.

P.S.

Notice also that all odd order subgroups of $G$ are abelian. Coupled with the solvability, this says quite a bit.


When was NEMO?  I think I was supposed to write questions, but I completely forgot about it.  Also, I don't think my middle school sent any teams; if it already happened, then someone probably should have posted a reminder of some type...

Ligon GT Magnet Middle School had two teams, and both teams took the top 2 places at NEMO. The winning team's captain was Nick Tobey.
1st place - Ligon
2nd place - Ligon
3rd place - Martin
4th place - Carnage
Aaah, I can't remember the rest
For individuals:
1st place - Calvin Deng; Ligon
2nd place - I think its Tejas Sundarasen...; Ligon
3rd place - Nick Tobey; Ligon
4th place - Ivan Kuznetsomething; Ligon
5th place - a girl from Martin I think
6th place - someone from Ligon
Not sure bout the other ones.
There weren't that many schools, about five maybe
LIGON DOMINATES  :D

Nemo was on... 1/5/08.

1. Calvin Deng, Ligon, 24
2. Allen Yang, Ligon, 23 (on individual tiebreaks; on tiebreaker round, 3rd)
3. Nick Tobey, Ligon, 23 (reversed with Allen on tiebreaker round)
4. Lisa Zheng, Martin, 16
5. Tejas Sundarasen, Ligon, 15
6. Alan Wu, Carnage, 15

so yeah

5th was also carnage
6th was carrington

only 6 teams

The 2nd place ligon team was winning until individual. They got 60/80 on application, while the 1st place ligon team got 57/80. Both Ligon teams got 8/10 on team. (Calvin missed #1 on team :|  )

so actually Allen Yang was 6th after tiebreakers because he got a 1 and everyone else got a 2 :wink:

I'm not trash-talking Calvin, eh? because... he did well, eh?　but missing an easy problem is not helpful.

[quote="not_trig"]so actually Allen Yang was 6th after tiebreakers because he got a 1 and everyone else got a 2 :wink:[/quote]

Heh. This is pretty amusing. Unfortunately, that is about all I have to offer on this matter.

You guys should make the tiebreakers easier, maybe.

[quote="fibointepi"]Calvin is my friend!!!  :spam: So don't trash talk about him! (By the way, he said some guy called Pranav said one-ninth)[/quote]

Somebody is their own friend.  :maybe:

Pranav said one-ninth... Whats that supposed to mean??

The answer was $ \frac{1}{12}$, but Pranav (a guy on my friend's team) said $ \frac{1}{9}$.  :cursing:

[quote="yoonzik"]so i was wrong. whatever. and who's trash talking calvin anyways?

and also, not_trig, one of the teachers said that the individual scores didnt count for the final team placements.[/quote]

Heh I don't completely know... Vitek and Vivek were being rather cryptic

To clear all this up:

To break ties on individuals for awards, we used a similar system to that employed by Mathcounts: we sort first by highest score, then by hardest problem answered correctly.  So, for example, if Jane and Jill both get 25 right, but Jane gets 1-25 right and Jill gets 1-24 and 26 right, then Jill is placed higher than Jane.

We used tiebreakers to determine plaques, I believe.  Scores were determined by a combination of team, individual, and application scores.  Team scores were multiplied by a scaling factor to make sure that they counted for an appropriate amount.

EDIT: Resultes were emailed out.  Maybe I'll post them later if I feel like it.

Oh ya Teki-Teki when are you going to post results on the NEMO website?

I was on the 2nd Place Team. Missed 6 in individual round!  :(

Welcome, Sanjay :lol:  You might want to check the Latex forum for help.

the results are already all done, i saw it on the enloe website

Sanjay, If you only missed 6, you must not have attempted all of the questions, or else you would have tied with me.

Sorry About that-

-I missed 15, mistype!

[quote="dnkywin"]Teki-Teki the results on the NEMO website still say that the top finishers were are Some Kid (Some School) and the winning teams were Some School. Please update.[/quote]

No, actually it says that our account has been suspended.  Dunno when you checked.  Hrmm... I am not in charge of the website either, so don't gripe to me.


modular arithmetic
function
algebra unsolved
algebra

Let $m$ be a positive integer and $\{a_n\}_{n\geq 0}$ be a sequence given by $a_0 = a \in \mathbb N$, and \[ a_{n+1} = \begin{cases} \displaystyle \frac{a_n}2 & \textrm { if } a_n \equiv 0 \pmod 2, \\ a_n + m & \textrm{ otherwise. } \end{cases}  \]
Find all values of $a$ such that the sequence is periodical (starting from the beginning).

I don't understand what's happening in this problem. All terms are clearly nonnegative integers and if the sequence is periodical from a certain rank, it is bounded and thus for large $n$ one can only have $a_{n+1}=\frac{a_n}{2}$ and $a_n$ even. Or, since we are dealing with integers, one should have $a_n=0$ for large $n$ and coming back the first term would be zero. I don't get something obvious (sorry if I'm making a huge mistake, I've been struggling with physics for more than 6 hours) or this isn't the statement of the real problem???

I might have misunderstood the problem wrong (I got it by telephone) ... other than that for this particular problem, your solution seems to be correct.

The problem statement is:

[i]Let $m$ be a positive integer. Find all positive integers $\alpha$ such that the sequence $\left\{ a_n \right\}^\infty_{n=0}$, defined as:[/i]

$a_0 = \alpha$
$a_{n+1} = \left\{ \begin{array}{ll} \frac{1}{2} a_n & a_n \equiv 0 \: (mod \: 2) \\ a_n + m & a_n \equiv 1 \: (mod \: 2) \end{array} \right.$

[i]for $n = 0, 1, 2, \ldots$, is periodic, that is, its terms repeat starting from: $a_0, a_1, a_2, \ldots, a_k$, for some $k$.[/i]

This is the problem which is changed the title of the one, Baltic Way Problem 5, 1997. ;)
[url=http://www.hh.schule.de/ifl/mathematik/baltic97.pdf]www.hh.schule.de/ifl/mathematik/baltic97.pdf[/url]

We can see almost same problems. Contest's problems are recycling, if so I am dissapointed. :(

Good problems are repeating! ;)

Oh, the problem from Baltic Way is way much easier... One just need to prove the sequence is bound and thats the end.

Quite similar problem in France 1996

http://www.imo.org.yu/othercomp/Fra/FraMO96.pdf

Quite spread problem.....

its also on mathematical olympiad challenges

[size=150][b]Problem 4.(BMO 2006)[/b][/size]

Solution:
If $n$ is even it is clear that no $m$ satisfies the problem.

Let $n$ be odd. Then the solutions are
$M=( {1,2,3,...,m}\cup {m+1,m+3,m+5,...,2m} )$

If $a=m$ the sequence is $m,2m,m,2m,....$
If $a=2m$ the sequence is $2m,m,2m,m....$

Let $S=M-{m,2m}$

Let's divide $S$ into subsets:

$P_1={1.2^0,1.2^1,....,1.2^{k_1}},$ where$m< 1.2^{k_1} <2m;\\ P_3={3.2^0,3.2^1,....,3.2^{k_3}},$where$m< 3.2^{k_3} <2m;\\ P_5={5.2^0,5.2^1,....,5.2^{k_5}},$ where$m< 5.2^{k_5} <2m;\\ ... P_{m-2}={(m-2).2^0,(m-2).2^1,....,(m-2).2^{k_{m-2}}}, where m< (m-2).2^{k_{m-2}} <2m;\\$
Then if $a_i\in P_t$ then $a_{i+1}$ is the previous element of $P_t$, if there is such, or $a_{i+1}=a_i + m$ which on the other hand is last element of $P_q$.

Let define the function $f(t)=q$ as the index of the set we will get in after we leave $P_t$.

If $f(t)=f(l)$ then $t+m=l+m$, and $t=l$ so $f$ i inection.

So we have $P{}_t{}_1\rightarrow P{}_t{}_2\rightarrow ... P{}_t{}_i\rightarrow P{}_t{}_1$.

Because at some moment we will be in one set twice, and if we take the first such set this must be $P{}_t{}_1$.

By this we prooved that if $a\in M$ then a is a solution and also we see that then $a_i\in M$ for all $i$ !!!

Let $a \notin M$. Then $a>m$ and consider $that a$ is solution.
If $a_i$ is odd then $a_i+1=a_i+m$ (this is even) and $a_i+m=2^k.a_{i+k}, a_{i+k}$- odd $\Rightarrow a_{i+k}=\frac{a_i +m}{2^k} \leq  \frac{a_i +m}{2} \Rightarrow (a_{i+k}-m) \leq  \frac 12 (a_i - m)$.

So, at some time we will have a member of the sequence smaller or equal than m, so we will have $a_j \in M$ for some $j$. Then all $a_J\in M (J>j)$, and because the sequence is periodic, $a_0 \in M$, contradiction.

Answer: $a\in {1,2,3,...,m}\cup {m+1,m+3,m+5,...,2m}$

[quote=n0vakovic]Let $A$ be the set of all solutions $a$. First, if $m$ is even there is no solution. So suppose $m$ is odd and WLOG suppose $a_0 = \min \{a_0, a_1, \dots, a_k \}$. Now $a_0$ is also odd and $a_0 \leq \frac{a_0+m}2$ so $a_0 \leq m$. If $a_i \leq m$ then $a_{i+1} = \frac{a_i} 2 \leq m$ or $a_{i+1} = a_i+m \leq 2m$ is even and $a_{i+2} = \frac{a_i+m}2 \leq m$, and we conclude that only numbers greater than $m$ in $A$ could be those even not greater than $2m$. Now we will prove that $A = \{1, 2, 3, \dots, m-1,m, m+1, m+3, \dots, 2m-2, 2m \}$. Consider oriented graph whose vertices are members of $A$ and edges from $x$ to $x/2$ if $x$ is even and from $x$ to $x+m$ if $x$ is odd. The out-degree of every vertice is 1, and since all $x \leq m$ has only in-edge from $2x$ and for all $x>m$ only in-edge is from $x-m$, the in-degree is also 1 for all vertices and that means the graph could be divided into cycles and that concludes the proof.[/quote]

Can $a_0 = 0$ ?


trigonometry
geometry
circumcircle
geometry solved

If a triangle $\triangle ABC$ is harmonic show that :

(1) $\Delta  = R \cdot \rho = R^2 \tan \omega $

(2) $\frac{1}{{\rho^2 }} = \frac{1}{{h_a^2 }} + \frac{1}{{h_b^2 }} + \frac{1}{{h_c^2 }}$

where $\Delta$ is the area, R is the circumradius, $ \rho$ is the radius of the second Lemoine circle,  $\omega$ is the Brocard angle, $h_a,h_b,h_c$ are the altitudes of triangle  $\triangle ABC$.

__________________________________________________________________________________________________________________________

[b]P.S.[/b] A triangle $\triangle ABC$ is said [i]harmonic[/i] if it satisfies one of the following equivalent conditions:

(a)  $a^2  + b^2  + c^2  = 4R^2 $

(b) $ \sin ^2 \alpha  + \sin ^2 \beta  + \sin ^2 \gamma  = 1$

where $a,b,c$ are the lengths of the sides $BC, CA, AB$ and $\alpha$ , $\beta$ , $\gamma$ are the angles with vertices $A,B,C$ .

Let's prove that $\Delta=R^2\tan{\omega}$.

By well-known theorem we have $\frac{a}{\sin{\alpha}}=\frac{b}{\sin{\beta}}=\frac{c}{\sin{\gamma}}=2R$. Then $\Delta=\frac{1}{2}ab\sin{\alpha}=2R^2\sin{\alpha}\sin{\beta}\sin{\gamma}$.

Let P be the first Brocard point. Since $PC=\frac{AC\sin{\angle{CAP}}}{\sin{\angle{APC}}}$ and $PC=\frac{BC\sin{\angle{CBP}}}{\sin{\angle{BPC}}}$ then $\frac{\sin{\omega}\sin{beta}}{\sin{\gamma}}=\frac{\sin{(\beta-\omega)}\sin{\alpha}}{\sin{\beta}}$. And since $\sin{(\beta-\omega)}=\sin{\beta}\cos{\omega}-\cos{\beta}\sin{\omega}$ we have $\cot{\omega}=\cot{\beta}+\frac{\sin{\beta}}{\sin{\alpha}\sin{\gamma}}$. But $\sin{\beta}=\sin{(\alpha+\gamma)}=\sin{\alpha}\cos{\gamma}+\sin{\gamma}\cos{\alpha}$. And finally we obtain $\cot{\omega}=\cot{\alpha}+\cot{\beta}+\cot{\gamma}$.

So, we have to prove that $\Delta=R^2\tan{\omega}$. By the previous equalities we have to prove $2\sin{\alpha}\sin{\beta}\sin{\gamma}(\cot{\alpha}+\cot{\beta}+\cot{\gamma})=1$.

It is well-known that $\cot{\alpha}+\cot{\beta}=\frac{\sin{\gamma}}{\sin{\alpha}\sin{\beta}}$, $\cot{\beta}+\cot{\gamma}=\frac{\sin{\alpha}}{\sin{\beta}\sin{\gamma}}$ and $\cot{\gamma}+\cot{\alpha}=\frac{\sin{\beta}}{\sin{\gamma}\sin{\alpha}}$.

Therefore $2\sin{\alpha}\sin{\beta}\sin{\gamma}(\cot{\alpha}+\cot{\beta}+\cot{\gamma})=\sin{\alpha}\sin{\beta}\sin{\gamma}((\cot{\alpha}+\cot{\beta})+(\cot{\beta}+\cot{\gamma})+(\cot{\gamma}+\cot{\alpha}))=\sin{\alpha}\sin{\beta}\sin{\gamma}(\cot{\alpha}+\cot{\beta})+\sin{\alpha}\sin{\beta}\sin{\gamma}(\cot{\beta}+\cot{\gamma})+\sin{\alpha}\sin{\beta}\sin{\gamma}(\cot{\gamma}+\cot{\alpha})=\sin^2{\alpha}+\sin^2{\beta}+\sin^2{\gamma}=1$ as desired.

P.S. I don't know what "second Lemoine circle" means. So I can't solve other parts of the problem.

[quote="Remike"]
P.S. I don't know what "second Lemoine circle" means. So I can't solve other parts of the problem.[/quote]

In following link some information are given on the second Lemoine circle

http://mathworld.wolfram.com/CosineCircle.html

Thanks for the solution of the first part of the problem !

Leon

[quote="Leon"]If a triangle $\triangle ABC$ is harmonic show that :

(1) $\Delta  = R \cdot \rho = R^2 \tan \omega $

(2) $\frac{1}{{\rho^2 }} = \frac{1}{{h_a^2 }} + \frac{1}{{h_b^2 }} + \frac{1}{{h_c^2 }}$

where $\Delta$ is the area, R is the circumradius, $ \rho$ is the radius of the second Lemoine circle,  $\omega$ is the Brocard angle, $h_a,h_b,h_c$ are the altitudes of triangle  $\triangle ABC$.[/quote]

This problem is straightforward.

At first, we will use the formula $\rho=R\tan\omega$ for the radius $\rho$ of the second Lemoine circle of triangle ABC.

It is well-known that the Brocard angle $\omega$ of triangle ABC satisfies $\cot\omega=\frac{a^2+b^2+c^2}{4\Delta}$. Taking the reciprocal of this equation, we get $\tan\omega=\frac{4\Delta}{a^2+b^2+c^2}$.

The area $\Delta$ of triangle ABC can be calculated using the formula $\Delta=\frac12 ah_a=\frac12 bh_b=\frac12 ch_c$. Thus, $h_a=\frac{2\Delta}{a}$, $h_b=\frac{2\Delta}{b}$, $h_c=\frac{2\Delta}{c}$.

Using these all formulas, we see that your equation (1):

$\Delta=R\cdot\rho$ is equivalent to
$\Delta=R\cdot R\tan\omega$ is equivalent to
$\Delta=R^2\tan\omega$ is equivalent to
$\Delta=R^2\frac{4\Delta}{a^2+b^2+c^2}$ is equivalent to
$1=\frac{4R^2}{a^2+b^2+c^2}$ is equivalent to
$a^2+b^2+c^2=4R^2$,

what is just the characteristic equation of a harmonic triangle.

Also, your equation (2):

$\frac{1}{\rho ^{2}}=\frac{1}{h_{a}^{2}}+\frac{1}{h_{b}^{2}}+\frac{1}{h_{c}^{2}}$ is equivalent to
$\frac{1}{\left( R\tan \omega \right) ^{2}}=\frac{1}{h_{a}^{2}}+\frac{1}{h_{b}^{2}}+\frac{1}{h_{c}^{2}}$ is equivalent to
$\frac{1}{\left( R\frac{4\Delta }{a^{2}+b^{2}+c^{2}}\right) ^{2}}=\frac{1}{\left( \frac{2\Delta }{a}\right) ^{2}}+\frac{1}{\left( \frac{2\Delta }{b}\right) ^{2}}+\frac{1}{\left( \frac{2\Delta }{c}\right) ^{2}}$ is equivalent to
$\frac{\left( a^{2}+b^{2}+c^{2}\right) ^{2}}{16\Delta ^{2}R^{2}}=\frac{a^{2}}{4\Delta ^{2}}+\frac{b^{2}}{4\Delta ^{2}}+\frac{c^{2}}{4\Delta ^{2}}$ is equivalent to
$\frac{\left( a^{2}+b^{2}+c^{2}\right) ^{2}}{4R^{2}}=a^{2}+b^{2}+c^{2}$ is equivalent to
$\frac{a^{2}+b^{2}+c^{2}}{4R^{2}}=1$ is equivalent to
$a^{2}+b^{2}+c^{2}=4R^{2}$,

what is again the characteristic equation of a harmonic triangle.

This solves your problem.

  Darij


Let $ a,b,c,x,y,z$ be real numbers such that $ x \plus{} y \plus{} z \equal{} 1$ . Prove that

$ ax \plus{} by \plus{} cz \plus{} 2 \sqrt {(xy \plus{} yz \plus{} xz)(ab \plus{} bc \plus{} ac)} \le {a \plus{} b \plus{} c}$


induction
combinatorics proposed
combinatorics

Suppose we have $n$ intervals of $\mathbb R$ which along any $k$ of them there are 2 which their intersection is non empty. Prove that there exist $k-1$ points in $\mathbb R$ which any of these subsets contain one of these points.

First modify the intervals as follows: when we encounter an interval which contains another one strictly, shrink the larger one to match the smaller one. This way, the main property of the collection of intervals is not altered, as can easily be seen, but when the process finishes, there won't be any intervals strictly containing other intervals in the collection.

Now let $I=[a,b]$ be one of the intervals which has the leftmost endpoint as its left endpoint. Any other such intervals will coincide with $I$. There are now two cases:

(a) There are no intervals in our set which cut $I$ along an interval $[c,b],\ c\in(a,b]$

In this situation, the collection of intervals we get by eliminating those which coincide with $I$ contains only intervals disjoint from $I$, and thus has the property that among any $k-1$ of them, some $2$ will cut each other. By induction on $k$, we can conclude that there are $k-2$ points such that each interval in this smaller set contains one of them, and adding an extra point which belongs to $I$, we get what we want.

(b) There is an interval $J=[c,d]$ with $c\in(a,b]$

Eliminate all intervals which coincide with $I$ in our collection, and replace them with the interval $[c,b]$. It's easy to see that because none of our intervals contains another one strictly, this new set of intervals still has the initial property (among any $k$ of them, $2$ will intersect each other). We can now perform the shrinking procedure in the beginning. The new intervals we get have their endpoints among those of the old ones, but there is at least one endpoint less (namely $a$). This means that we can proceed by induction on the number of endpoints.

I want to solve this problem:
but I cannot fo any further from here:
at first I will give  every interval , a vertics,and two vertics are connected iff their interval have somthing in common  (Their intersection be none empty)[I call this kind of graph , Intervally Graph].

then I proved two lemmas.
[color=darkblue]Lemma 1[/color].a cycle with more than 3 vertics cannot be Intervally Graph.
[color=darkblue]Lemma 2[/color]. if we have a $K_p$ Intervally Graph then their intervals have somthing in common.

can anyone help me to complete my solution.

let say$intervalA>{interval B}$if $A$ is completely righter than $B$.then refering to this poset the problem becomes MIRSKY theorem.

[quote="dashmiz"]then the problem becomes MIRSKY theorem.[/quote]

What is Mirsky theorem.

I created exactly the same problem:

We say that $I_{1}\geq I_{2}$ if they dont intersect and $I_{1}$ is to the right of $I_{2}$. This becomses a partially ordered set.  We are given that there is no antichain of lenght $k$, therefore , by Mirsky theorem (kindda Dilworth Dual), the Poset is the union of $k-1$ chains. Every chain is a set of intervals with a point in common so the problem is solved.

Here it is again, with the same proof :):

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=intervals&t=106694[/url]


I need help with the following geometry problems. Would be grateful if someone could give some kind of hints.

1. Let BE and CF be the altitudes of an acute triangle ABC, with E on AC and F on AB. Let O be the point of intersection of BE and CF. Take any line KL through O with K on AB and L on AC. Suppose M and N are located on BE and CF respectively, such that KM is perpendicular to BE and LN is perpendicular to CF. Prove that FM is parallel to EN.

2. In a triangle ABC, D is a point on BC such that AD is the internal bisector of angle A. Suppose (angle) B=2C and CD=AB. Prove that angle A=72 degrees.

Well, I've just solved the second one. Sometimes, looking at a problem after a small break is quite helpful.

I'll look at the first one now...

[quote="Pradhan P"]For the first question, consider the two cases where K lies in the interior of FB and K lying in the exterior of FB. Use the fact that ABC is acute to conclude similarly about L and EC. Now spot cyclic quadrilaterals and chase a few angles.[/quote]
Done... thanks a lot. Sometimes, problems are a lot easier than they seem.


[hide]
the probablity that it will  rain on day one is 2/3 then you don't want it to rain so it becomes 1/3 then 1/3
so 2/3*1/3*1/3=2/27[/hide]

[quote="G-UNIT"]  You guys are forgetting the events can be re-arranged 3 ways. [/quote]
i knew i messed up somehow. :)

Also, if you're wondering why there are 3 ways to arrange, it's because $\binom{3}{1}$ gives the number of ways to choose 1 day from 3 total.

In general, the formula for finding the number of days is $P^k(1-P)^{n-k}\cdot \binom{n}{k}$, where $n$ is the total number of days, $k$ is the number where it must rain (or whatever else), and $P$ is the probability that it rains on a given day.

[quote="Hamster1800"][quote="G-UNIT"]Thats wrong.

[hide="Answer"]

  The correct answer should be:

$\frac23 \cdot \frac13 \cdot \frac13\cdot 3=\boxed{\frac{6}{27}}$

  You guys are forgetting the events can be re-arranged 3 ways. [/hide][/quote]

That's wrong too :P

[hide="Correct Answer"]
$\frac 23 \cdot \frac 13 \cdot \frac 13 \cdot 3 = \boxed{\frac 29}$

SIMPLIFY
[/hide][/quote]

  brilliant!

[quote="236factorial"]The probability it will rain on any given day is $\frac{2}{3}$. What is the probability that it will only rain on one day out of three days?[/quote]
[hide]
The probability of it not raining is 1/3, so the answer is 3(2/3*1/3*1/3), since there are 3 days that it can rain. The answer is $\boxed{\frac{2}{9}}$[/hide]


Solve $ y^3\minus{}y^2\minus{}2y\plus{}1\equal{}0$ without $ cacdano$ technique

dont you mean the cardona technique?

and i have no idea what that is

Do you mean $ y^3 \plus{} y^2 \minus{} 2y \minus{} 1 \equal{} 0$?
If we let $ y \equal{} x \plus{} \frac1x$, and multiply the resulting equation by $ x^3$, don't we get $ x^6 \plus{} x^5 \plus{} \cdots \plus{} x \plus{} 1 \equal{} 0$, which has the primitive 7th roots of unity as solutions?

Nice, how did you come up with that?

It can be easily fixed by letting $ y \equal{} \minus{} x \minus{} \frac {1}{x}$ instead.

Haha, I was recently trying to find the 7th roots of unity in random ways...it's luck I guess. :P

Bah math154 was trying to find 7th roots of unity with me algebraiclly. Anyway, $ x\equal{}cis(\frac{2k\pi}{7})$ where k is an integer (and is not a multiple of 7). But where would you go from there?

Since $ y \equal{} x \plus{} \frac1x$, your answers would simply be $ y \equal{} \boxed{\operatorname{cis}\left(\frac {2k\pi}7\right) \plus{} \frac1{\operatorname{cis}\left(\frac {2k\pi}7\right)}}$.

[quote="t0rajir0u"]He means the [url=http://en.wikipedia.org/wiki/Cubic_equation#Cardano.27s_method]Cardano-Tartaglia method[/url].  Try doing it with Lagrange resolvents, then.[/quote]
Yes,sorry
But I found by computer $ cos (\frac{\pi }{7},\frac{3\pi }{7},\frac{5\pi }{7}$

How do you obtain the three solutions using primitive roots?

$ y \equal{} \frac{1}{x} \plus{} x$ doesn't seem to yield those solutions listed above.

Combining all previous posts:

We have $ y^3 \minus{} y^2 \minus{} 2y \plus{} 1 \equal{} 0$.

Substituting $ y \equal{} \minus{} x \minus{} \frac1x$, we find that

$ \minus{} \left(x^3 \plus{} \frac1{x^3} \plus{} 3x \plus{} \frac3x\right) \minus{} \left(x^2 \plus{} \frac1{x^2} \plus{} 2\right) \plus{} \left(2x \plus{} \frac2x\right) \plus{} 1 \equal{} 0$.

Simplifying, we get

$ \minus{} x^3 \minus{} x^2 \minus{} x \minus{} 1 \minus{} \frac1x \minus{} \frac1{x^2} \minus{} \frac1{x^3} \equal{} 0$, and multiplying through by $ \minus{} x^3$, we have

$ x^6 \plus{} x^5 \plus{} x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1 \equal{} 0$, which has roots

$ \operatorname{cis}\left(\frac {2\pi}7\right),\operatorname{cis}\left(\frac {4\pi}7\right),\operatorname{cis}\left(\frac {6\pi}7\right),\operatorname{cis}\left(\frac {8\pi}7\right),\operatorname{cis}\left(\frac {10\pi}7\right),\operatorname{cis}\left(\frac {12\pi}7\right)$.

Plugging back into our equation $ y \equal{} \minus{} x \minus{} \frac1x$, we have one root as

$ \minus{}\operatorname{cis}\left(\frac {2\pi}7\right)\minus{}\frac1{\operatorname{cis}\left(\frac {2\pi}7\right)}$

$ \equal{}\boxed{\minus{}2\cos\left(\frac{2\pi}7\right)}$.

The other two roots are $ \boxed{\minus{}2\cos\left(\frac{4\pi}7\right)}$ and $ \boxed{\minus{}2\cos\left(\frac{6\pi}7\right)}$.


How many forums (counting AoPS) do you belong to? Please list them, too. Social networks like facebook do not count.

uh... one? yep, that's AoPS. no wait... 2. 'cause my class made one too. cool i'm the first post other than the original! :D

Ive bene on a bunch, but Im only active on 3...well, not really on the other 2 anymore, now just AOPS

4:

AoPS
SmashBoards
Yugioh:The Abridged Series
studio12.co.cc (although it's basically a dead forum).

Three:

AoPS
CoD4Boards
Bungie

And why doesn't Facebook count?  :P

I belong to four:

AoPS
Chief Delphi
Game Maker Community
XKCD Forums

I don't belonD to any forum.

I belonG to AoPS, AoPS and nothing but AoPS (:()

I hadn't joined anything more, but I think that I might join mafiascum

1.This one
2.Math Help Forum
3. SOS Math
4. My Math Forum
5. Google Math

I've joined many and I've hosted a few.
Now I use AoPS only

I'm registred on many forums but I'm active on only two.
1. AOPS
2. is my own forum which i tried to make :)

Oh yeah, actually 5. I forgot Battle City. Too late to change my vote now, though.

I am registered in many but active in AoPS only.

[quote="7h3.D3m0n.117"]And why doesn't Facebook count?  :P[/quote]

I meant other than social networks...

I belong to tons of forums and am too lazy to list them all sorry. Probably nearing 25-30 now.

[quote="myyellowducky82"]i used to belong on a bunch but i gave them all up
But I'm a member of AoPS and Neopets--and even neopets doesn't let me on the forums, so that doesn't count.[/quote]
Neopets is sad. Do you actually believe that they actually used to say "Neopets will never advertise." Talk about walking the talk. Look at Neopets now.

oooooh yeah. definately. the last time i checked was half a year ago, but their NC mall is just--so---UUUUUUGGGGG!!!!! i mean honestly, you have to pay money for fake money! and they have GAMES that are advertisements!!!!!! (although they DO give you a lot of neopoints, though. :lol: )


I'm just starting preparing for the AMC 12 and the AIME next year. Where should I start? I've read Winning Solutions, should I read the "Art of Problem Solving", is the series that good? What else should I do/read. I'm extremely good at math I just need to learn more about this type of math if I want to do well.

Of course AoPS is a good place to start!  Just start doing problems.  If you find something you don't know, then research it.

[quote="scm007"]I'm extremely good at math I just need to learn more about this type of math if I want to do well.[/quote]

Be a little more humble.  It'll help, if for nothing more than you'll work harder.

I'm sure pretty much every single common poster on this board is "extremely good at math" relative to the other students at their schools. I'd be careful before calling myself "extremely good" at anything.

There are people are exceptionally good at math in this site..
I used to call myself pretty good at math in my school but in here, there are really a lot people who are better than me on many areas of math.  I don't know about AMC12 but you can try some problems I post in this forum except they're not gonna be that difficult because I don't know really hard questions.  For me, the best way to learn math is to produce the problem by yourself.  I make some problems by myself and some of them are pretty good.  For more problems and stuffs, visit Mathcounts Marathon in Mathcounts Forum or Problem Solving Marathon in this forum.  Both are posted by myself.  Hope this helps.


Homework?

I'm not sure this forum is the appropriate one :wink:

Pierre.

homework can't be this one.
actually,it's one of my friends give me.
and it is a question which is in Peking university test.It seems unusual.
anyway, i don't know the method.
who can post it?

Because $ 0<\equal{} \prod\frac{1}{9^n}<\equal{}\frac{1}{9^n}$for any$ n>\equal{}1$and lim $ \frac{1}{9^n}\equal{}0$ we can say that $ lim \prod\frac{1}{9^n}\equal{}0$

...
I know the reason why pierre said it is homework...
I made a mistake on inputting it...
The right problem is:
$ \boxed{\lim\prod(1+\frac{1}{9^n})}$
now, it looked better.-_-

[quote="pbornsztein"]Ok.

But, then, are you asking for the value of the limit?
The convergence is easy to prove (use that $ 1 \plus{} x \leq e^x$), but I'm note sure that the exact value of the limit is achievable :wink:

Pierre.[/quote]

Let $ a_m \equal{} \prod_{n \equal{} 0}^{m}{(1 \plus{} \frac {1}{9^n})}$. We have $ a_{m \plus{} 1} \equal{} a_{m}(1 \plus{} \frac {1}{9^{m\plus{}1}}) > a_m$.
$ a_m < \prod_{n \equal{} 0}^{\infty}{(1 \plus{} \frac {1}{9^n})}\leq e^{\sum_{n \equal{} 0}^{\infty}{\frac {1}{9^n}}} \equal{} e^{\frac {9}{8}}$.
So limit exist.

Yes, that's what I'm saying. But it only gives an upper bound for the exact value.
I'm quite sure that this value cannot be easily determined in the general case (when $ \frac 1 9$ is replaced by $ a$ with $ a \in ]0,1[$). Maybe for $ a \equal{} \frac 1 9$ is there a trick...

Pierre.

[quote="wxydx00"]the answer is:
$ A \equal{} lim\prod(1 \plus{} \frac {1}{9^n})$
  $ \equal{} lim\prod(1 \plus{} \frac {1}{3^{2n}})$
$ \frac {8}{9}A \equal{} lim(1 \minus{} \frac {1}{3^2})\prod(1 \plus{} \frac {1}{3^{2n}}) \equal{} 1$
  $ A \equal{} \frac {9}{8}$[/quote]

I'm affraid you are wrong : $ \lim_{m\rightarrow \plus{}\infty}(1 \minus{} \frac {1}{3^2})\prod_{n\equal{}1}^m(1 \plus{} \frac {1}{3^{2n}}) \neq 1$

$ \lim_{m\rightarrow \plus{}\infty}(1 \minus{} \frac {1}{3^2})\prod_{n\equal{}1}^m(1 \plus{} \frac {1}{3^{2n}})$ $ >(1 \minus{} \frac {1}{3^2})\prod_{n\equal{}1}^3(1 \plus{} \frac {1}{3^{2n}})\equal{}\frac{4788800}{4782969}>1$

I think you make a confusion with $ \lim_{m\rightarrow \plus{}\infty}(1 \minus{} \frac {1}{3^2})\prod_{n\equal{}1}^m(1 \plus{} \frac {1}{3^{2^n}})$ which indeed is $ \equal{}1$

Oh,yes..
So,where is the right method.
I make the$ \prod(1\plus{}\frac{1}{3^n})$to the$ (1\plus{}\frac{1}{3^2})(1\plus{}\frac{1}{3^4})(1\plus{}\frac{1}{3^8})...$


floor function
modular arithmetic
number theory proposed
number theory

There is a well known result:
For every integer $ n$ such that $ n \geq 3$:
\[ \sum_{k \equal{} 0}^{\lfloor \frac {n}{3} \rfloor} {n \choose {3k}} \equiv 0 \pmod{3}\]
Prove that for every prime number $ p$ and positive integer $ n$ such that $ n \geq p$ we have:
\[ \sum_{k \equal{} 0}^{\lfloor \frac {n}{p} \rfloor} {n \choose {pk}} \equiv 0 \pmod{p}\]
[hide]
[b]Hint.[/b]
Let
\[ P(x) \equal{} (1 \minus{} x)^n \equal{} \sum_{k \equal{} 0}^{n} ( \minus{} x)^{n \minus{} k} {n \choose {n \minus{} k}} \equal{} \sum_{k \equal{} 0}^{n} ( \minus{} 1)^k {n \choose k} x^k\]
Also let $ \varepsilon$ be the complex number satisfies:
\[ \varepsilon^{p \minus{} 1} \plus{} \varepsilon^{p \minus{} 2} \plus{} ... \plus{} \varepsilon \plus{} 1 \equal{} 0\]
Then:
\[ \sum_{k \equal{} 0}^{\lfloor \frac {n}{p} \rfloor} {n \choose {pk}} \equal{} \dfrac{P(1) \plus{} P(\varepsilon) \plus{} P(\varepsilon^2) \plus{} ... \plus{} P(\varepsilon^{p \minus{} 1})}{p}\]
Notice that $ P(1) \equal{} 0$ and $ P(\varepsilon^k) \equal{} (1 \minus{} \varepsilon)^k Q(\varepsilon)$ where $ Q$ is also an integer-coefficient polynomial.
[/hide]

It is not true for $ n \equal{} p$: $ \binom{p}{0} \plus{} \binom{p}{p} \equal{} 2\not \equal{} 0\mod p, p > 2$.
If it true for $ n$, then it is true for $ n \plus{} p$.
If $ p < n < 2p$, then it is true, for $ n \equal{} 2p$ $ 1 \plus{} C_{2p}^p \plus{} 1 \equal{} 4\mod p$ it is not true.
Therefore your statment is true for $ n > p, p\not |n$
Second method is $ \sum_i \binom{n}{mi} \equal{} \frac {1}{m} \sum_{k \equal{} 1}^m (1 \plus{} exp(\frac {2\pi i k}{m}))^n.$

Sorry, the problem has been mistyped. It should have been:
\[ \sum_{k\equal{}0}^{\lfloor \dfrac{n}{p} \rfloor} (\minus{}1)^k \binom{n}{kp} \equiv 0 \pmod{p}\]
Now, it is true for $ n \geq p$. The proof is hidden in my post at #1.

[quote="Mashimaru"]Sorry, the problem has been mistyped. It should have been:
\[ \sum_{k \equal{} 0}^{\lfloor \dfrac{n}{p} \rfloor} ( \minus{} 1)^k \binom{n}{kp} \equiv 0 \pmod{p}\]
Now, it is true for $ n \geq p$. The proof is hidden in my post at #1.[/quote]
It can be proven by applying Lucas theorem. Suppose $ n\equal{}pt\plus{}r$ where $ 0\leq r\leq p\minus{}1$ then we have:
\[ {{n}\choose{pk}} \equiv {{t}\choose{k}} (\mod p)\]
After that, we only need to notice that $ \sum_{i\equal{}0}^n (\minus{}1)^k {{n}\choose{k}}\equal{}0$


Here's a problem from my ARML homework that ive been working on but can't seem to do:

Let $z$ be a root of $x^5-1=0$ with $z \not = 1$

Compute the value of $\displaystyle z^{15}+z^{16}+z^{17}+...z^{50}$.

use the facts

1.  $x^5=1$
2.  $1+x+x^2+x^3+x^4$=0  (do you see why that's true?  if not, factor it has a geometric sequence, and use the fact that $z$ isn't 1.)

[hide]

Since 1 is clearly a root of this, factor it out:

such that  is a root of the second term.

You're trying to computer . Factor out :

Since , therefore  must equal 0 according to the original equation.

Therefore, you may simplify the second equation even more:

You can continue to factor out powers of  and repeating the process:

Going back to the original equation, .

Therefore, the expression is equal to .[/hide]


inequalities
calculus
limit
geometry
geometric transformation
reflection
calculus computations

Hi,
    I can't figure out why the delta- epilson proof proves that a certain limit exits :?: , conceptually it makes no sense. Even if we find an inequality for the delta.
for ex: |x-4|<=(1/3)Ɛ; how does this prove that a limit exits, conceptually and mathematically?

Thank you in advance :D [/code]

Historically, calculus grew and flourished pretty well during the nearly 200 years between Newton/Leibniz and Weierstrass, which might be some sort of hint about how seriously to take this in a calculus course ...

But anyway. The official definition of limit cannot be understood or comprehended piecemeal, or as an equation. It must be understood whole, as a logical flow, with all of the quantifiers and logical connectives ("for all" and "there exists" and "if ... then") in their proper places doing their proper work.

What you've written appears to be an attempt to deal with the concept piecemeal - there's no stated limit in sight (although I suppose we could guess what you might have meant), and the quantifiers and logical connectives are nowhere to be seen. As such, what you have said is incoherent. I can't reply to it directly for that reason.

Now, what is this Weierstrassian formality, this epsilon-delta stuff? It's not that it has something to do with limits - it [i]is[/i] the formal mathematical definition of limit.

Let's try to sneak up on this gradually. We say that $ \lim_{x\to a}f(x)\equal{}L;$ what does that mean?

First try: as $ x$ gets close to $ a,$ $ f(x)$ gets close to $ L.$

- That's much too squishy. What do we mean by close? Does $ x\plus{}10^{\minus{}10}$ tend to $ 0$ as $ x\to 0$ because $ 10^{\minus{}10}$ is "close" to zero? Well, no, that's not really what we mean (even if your graphing calculator can't tell the difference.) We need something that expresses what we really want - and here the word "arbitrary" comes to our rescue.

Second try: as $ x$ gets arbitrarily close to $ a,$ $ f(x)$ get arbitrarily close to $ L.$

- The problem with this is that the logical flow is unclear. Which "arbitrary" takes precedence? Upon further reflection, maybe we don't want both words to be the same.

Third try: we can make $ f(x)$ arbitrarily close to $ L$ by making $ x$ close enough to $ a$ (but not equal to $ a.$)

- Now we're getting somewhere. I like this definition. We can formalize it further by stating exactly what we mean by "arbitrarily close" and "close enough". We then arrive at the full formality:

$ \lim_{x\to a}f(x)\equal{}L$ iff for all $ \epsilon > 0$ (comment: no matter how small, arbitrarily small), there exists $ \delta > 0$ (comment: small enough, depending on how small $ \epsilon$ is) such that for all $ 0<|x\minus{}a|<\delta$ (for all $ x$ that close to $ a$ or closer), we have $ |f(x)\minus{}L|<\epsilon$ ($ f(x)$ is at least that close to $ L.$)

----

Now: why formalize the definition at all? The first thing to understand is that defining what we mean by limit, which is what we've just done, does not [i]find[/i] limits and never will find limits. It's not a computational tool. It's how we convince ourselves that what we've already found is correct. I should also mention that there is seldom any call to use this definition in a specific, concrete case (although such cases are often offered as pedagogical exercises). Instead, we use it to prove general theorems - general theorems like $ \lim_{x\to a}(f(x)\plus{}g(x))\equal{}\lim_{x\to a}f(x)\plus{}\lim_{x\to a}g(x),$ provided both limits on the right exist. (With similar theorems about mulitiplication, division, and composition.) And many other theorems.

We have to use the definition to prove theorems - why? Because if we're going to prove that a limit exists, and that it equals such and such a number, we have to know what it even means for a limit to exist and for it to equal something. We have to have the definition even to know what we're trying to prove.


geometry
calculus
derivative
3D geometry
sphere
topology
vector

I was wondering if there is a proof that, for example, the derivative of an expression for volume gives a formula for surface area (3 dimensional case).  Is there some way to prove this for higher dimensions as well?  I can see that it works with a sphere, but am not sure how to prove it for say an 8 dimensional object.  Can someone help me out on this?  Thank you very much.

This can only work for nice enough objects- note the unstated assumption that the figure depends only on some linear dimension.

In addition, we need that when this linear dimension is increased by $c$, we add a shell of uniform thickness $c$ to the figure. This thickness is measured perpendicular to the surface.

With that caveat, this does work for balls of any dimension, using the radius. It also works for cubes using half of the side length, and a few other figures.

"Expression for volume" is a little too general. The structure of what you're looking for is something like this:

Suppose $g: \mathbb{R}^n\mapsto\mathbb{R}$ satisfies certain regularity conditions (which I won't specify in complete detail, but we are going to want $\nabla g\ne0$ "most" of the time.) and suppose that for each $c,\,\{x: g(x)\le c\}$ is compact and hence has finite measure.

Let $V(y)$ equal the n-dimensional measure pf $\{x: g(x)\le y\}.$

Now consider the set $S_y=\{x: g(x)=y\}.$ As long as $\nabla g\ne 0$ on this set, it is a smooth compact $n-1$-dimensional manifold embedded in $\mathbb{R}^n.$ As such, it is possible to construct an $(n-1)$-dimensional measure on $S_y.$ We let $A(y)$ be this $n-1$-dimensional measure of the whole set $S_y.$

So, is the result you want? $A(y)=\frac{d}{dy}V(y).$ Not quite; to make that work, we would need that $|\nabla g|=1$ everywhere (namely, the figure is expanding at a constant rate in all directions.)

Any proof will have to pull out all of the heavy tools of vector calculus - the Implicit Function Theorem, Jacobians, and so on.

But note that we have to have the function $g$ in order for your question to even make sense.

Cute result: for the ball of radius $r$ in $\mathbb{R}^n,$ the $n$-dimensional measure of that ball is

$V(r)=\frac{\pi^{\frac n2}r^n}{\Gamma\left(\frac n2+1\right)}.$

The $(n-1)$-dimensional measure of the sphere of radius $r$ (the boundary of the ball) is

$A(r)=\frac{\pi^{\frac n2}r^{n-1}}{\Gamma\left(\frac n2\right)}.$

OK I understand that last part, provided those equations are known beforehand.  However, I did not know them until you posted them (for the n-dimensional ball).  How were those arrived at?  Thanks a lot for your help :)


No discussion on the 2005 AMC10A/12A should take place until Thursday, Feb. 3, at 12:00 EST, as stated [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25191]here[/url].

To be completely clear, discussion even includes writing about how you think you did, stating 'that was easy/hard', etc.  [b]To be safe, in any and all posts, pretend that the 2005 AMC 10A/12A test never existed.[/b]

If you have any questions about this rule, PM me or one of the administrators.

Sorry for the duplicate announcement.  Some people didn't read the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25191]first, more subtle one[/url] and I had to delete some threads.

What's "Source: ... and 'that w" mean? In particular the "that w" part.

He probably meant to say something like "That was easy/hard"

I wrote "This includes 'what score do you think you got?' and 'that was really hard'" in the source box so it would appear on the front screen.  It looks funny when you view the thread, but for someone who is scanning the forum, planning to post an early discussion of the AMCs, it will inform them quickly and stop them.

[quote="solafidefarms"]Very sorry that I did not read that. Will remember in future.[/quote]

That's okay -- I should have been more explicit about what constituted 'discussion' sooner.

Especially since the AMC Director is a member!

well, all AMC's would be sent in by like midnight tomorrow... so yeah... 12:00 am est on Thursday. why is there no : easy/hard evautaion? It is just opinion...

But still, that may affect performance.

 If your friend told you that the science test was really hard, would you worry and panic and get really nervous?

If you friend, however, told you it was hard as crazy, I bet you would be very worried. State of mind plays a huge role in test taking.

If discussion were to be allowed, it would be extremely difficult to rigorously define what constitutes an "acceptable" post. It's easier just to ban discussion about the AMC altogether until Thursday.

I can answer that, since I have no clue what was on the 2005 AMC tests.  If this is still too much explanation, the mods should feel free to delete this.

It's a theoretical concern but a legitimate one.  Suppose I tell you that the test was "very, very hard."  Well, this puts you on notice and may affect your strategy in several ways.  (1) You may believe (rightly) that there are many tricky questions on the test and that you have to be very careful, more so than usual.  (2) You may believe (rightly) that the last 5 problems are harder than most last-5 problems, and so you should skip those and focus on the first 20.  (3) You may believe (rightly) that there are many difficult questions, and some of them appear early in the test, so you know you need to skip around, even among the early questions, and try to find the easy stuff.

These things are all speculative, but telling someone that an exam is easy/hard can give them an overall idea what strategy they should pursue.  You tell me I'm going to get an easy exam, I'm going to play aggressive and go for 150.  You tell me it's hard, I'm going to play conservative and make really sure I get my 100 so I can move on to the next stage.  Is any of this stuff rock-solid?  No, but might as well dodge this theoretical concern and enforce a "code of silence" until a couple of days after.


Ben starts with the number $ 33$. If he squares this value, then adds $ 11$ and finally divides by $ 11$, what is the result?

Wow, that is a clever manipulation! Does this method work for almost all MATHCOUNTS/AMC 8-10 etc. problems?


I'm doing some math problems (i have a test tommorow) and am having trouble with unequations. I know the the unequal sign ('<, or >') changes in certain cases, but i don't know when. Sorry for my bad english, but i think you'll understand.

I think they change when you multiply an equation my a negative number.
5<-8x
-5/8>x
Is that what you are asking?

Regards,
Funcia

also when you divide by a negative number (duh!)

.....im in algebra 1 and i happen to be doing inequalities right now

btw, u call them inequalities
-jorian

it also switches when you take the reciprocal and they are both positive or both negative  :D


Να βρεθεί ο φυσικός αριθμός $ n$ αν ο αριθμός $ A\equal{}2^{17}\plus{}17\cdot2^{12}\plus{}2^{n}$ είναι τέλειο τετράγωνο ακεραίου αριθμού.

[b][color=red]Καλησπέρα μετά από καιρό! Καταρχήν εύχομαι σε όλους καλή σχολική και καλή ΟΛΥΜΠΙΑΚΗ χρονιά! Καλή επιτυχία στον ερχόμενο διαγωνισμό της ΕΜΕ και καλή σταδιοδρομία στους (πλέον) φοιτητές μας![/color] [/b]

Για την άσκηση είμαι σχεδόν σίγουρος ότι θα υπάρχει πιο σύντομος δρόμος με κάποια άλλη διαδικασία αλλά σας παραθέτω τις δικές μου σκέψεις:

Η παράσταση Α είναι ίση με $ 49\cdot 2^{12} + 2^n$. Θέλουμε λοιπόν να λύσουμε τη διοφαντική εξίσωση $ 49\cdot 2^{12} + 2^n = x^2$ για κάποιο ακέραιο $ x$ (ο οποίος προφανώς είναι μεγαλύτερος από $ 7\cdot 2^6$).

Ισοδύναμα $ (x - 7\cdot 2^6)(x + 7\cdot 2^6) = 2^n$. To $ n$ αποκλείεται να είναι ίσο με 0, άρα πρέπει να λύσουμε το σύστημα

$ \left\{\begin{array}{ll} x - 7\cdot 2^6 = 2^{n_1} , & \hbox{} \\
x + 7\cdot 2^6 = 2^{n_2} , & \hbox{.} \end{array} \right.$ με $ n_1,n_2 \geq 1$ και $ n_1 + n_2 = n$

Με αφαίρεση κατά μέλη των παραπάνω παίρνουμε $ 7\cdot 2^7 = 2^{n_2} - 2^{n_1}$. Εύκολα φαίνεται από την τελευταία, ότι $ n_1,n_2 \geq 7$ διότι διαφορετικά καταλήγουμε σε άτοπο [Τί γίνεται εάν τουλάχιστον ένας εκ των $ n_1, n_2$ (ή ακόμη χειρότερα [b]και οι δύο[/b]) είναι μικρότερος από 7 ?] Άρα καταλήγουμε στην εξίσωση

$ 7 = 2^{n_2 - 7} - 2^{n_1 - 7}$ η οποία είναι αδύνατη αν οι $ n_1,n_2$ είναι (και οι δύο) μεγαλύτεροι από 7. Άρα τουλάχιστον ένας είναι ίσος με 7 (αφού ούτε μικρότεροι από 7 μπορεί να είναι).

Αν $ n_2 = 7$ τότε παίρνουμε αδύνατη εξίσωση, ενώ αν $ n_1 = 7$, τότε πρέπει $ 8 = 2^{n_2 - 7}$ άρα $ n_2 = 10$, συνεπως

$ n = n_1 + n_2 \Rightarrow \boxed{n = 17}$

Ελπίζω να μη ξέχασα κάποια περίπτωση

Αλέξανδρος


A die is rolled.  What is the probability that the number rolled is greater than 2 and even?

...based on the other answers you have gotten on probability and dice rolling, what do you think it could be? Ask yourself the following questions and try to answer the problem:

-How many numbers are greater than 2?
-Of those, how many are even?

Is there a AND Probability formula I can use for AND and OR questions?

Let $P(x)$ be the probability of event x occurring, and let $P(y)$ be the probability of event y occurring.

Probability of event X AND event Y occurring is $P(x) \cdot P(y)$.

Probability of event X OR event Y occurring is $P(x)+P(y)-P(x)\cdot P(y)$

[quote="mysmartmouth"]
Probability of event X OR event Y occurring is $P(x)+P(y)-P(x)\cdot P(y)$[/quote]

Wait... isn't an OR event just $P(x)+P (y)$? (ie. mutually exclusive events)?

EDIT: yes I know this is unrelated to the problem at hand but...

Well if they are mutually exclusive, then it is what you said.  But if it is something like dice, then it is the formula I gave.

ok even numbers
2,4,6
greater than 2
so only 4,6 work

there are 6 sides on a standard die
2/6=$1/3$

[quote="mysmartmouth"]Probability of event X OR event Y occurring is $P(x)+P(y)-P(x)\cdot P(y)$[/quote]

Actually, in the general case this should be $P(x)+P(y)-P(x \text{ and }y)$.  For mutually exclusive events the last probability is $0$, for independent events it is $P(x) \cdot P(y)$, and in other cases it must be calculated separately.

In any case, moogra's approach is much simpler.

Yes, for some reason when I read it I was only thinking that OR events are mutually exclusive and disregarded when there is an intersection of the sets  :blush: . And yes I agree with moogra, that was the solution I was hinting at in my first post.

[hide="answer, I think"]Wouldnt the answer be 1/3? Since there are 3 even numbers, and two even numbers over 2, it would be 2/6=1/3.[/hide]


linear algebra
matrix
quadratics
inequalities
linear algebra unsolved

Prove or disprove:
For two real symmetric matrices $ A$ and $ B$, we know that diagonal elements of $ A$ majorize diagonal elements from $ B$ (assume that diagonal entries are sorted). Is it true that eigenvalues of $ A$ must majorize eigenvalues of $ B$ (again we sort both arrays of eigenvalues)?

No. For your result you need a bit more, namely that the induced quadratic form of $ A$ majorizes this one of $ B$.

[hide="Counterexample:"]$ A\equal{}\left(\begin{array}{cc}2&0\\0&2\end{array}\right)$, $ B\equal{}\left(\begin{array}{cc}1&2\\2&1\end{array}\right)$.[/hide]

By definition of majorization we have that traces of matrices $ A$ and $ B$ (the sum of diagonal entries) are equal and therefore your counterexample is not correct...

I assumed that you meant the following version: If $ \lambda_1\le\ldots\le\lambda_n$ and $ \mu_1\le\ldots\le\mu_n$ are the diagonal elements of $ A$ and $ B$, resp. with $ \lambda_i\ge\mu_i$.

What definition of "majorization" do you apply?

http://en.wikipedia.org/wiki/Majorization
Therefore, you have $ \lambda_1 \plus{} \ldots \plus{} \lambda_k \geqslant \mu_1 \plus{} \ldots \plus{} \mu_k$ for every $ 1 \leq k \leq n \minus{} 1$ and equality for $ k \equal{} n$.

It's false.
Let $ A,B$ be 2 symmetric real matrices s.t. $ trace(A)\equal{}trace(B)$. $ A,B$ are orthogonally similar to $ A',B'$ that have the same  constant diagonal (with eventually distinct changes of basis).
 Diagonal elements of $ A'$ majorize diagonal elements of $ B'$. Now assume the Alexillic's result is true. Then eigenvalues of $ A'$ majorize eigenvalues of $ B'$ and eigenvalues of $ A$ majorize eigenvalues of $ B$. The following obvious example gives the contradiction:
 $ A\equal{}\begin{pmatrix}0&0\\0&0\end{pmatrix},B\equal{}\begin{pmatrix}1&0\\0&\minus{}1\end{pmatrix}$.

I still do not understand :)
You have that diagonal elements of $ B$ majorize diagonal elements of $ A$, and the same holds for eigenvalues of matrices $ B$ and $ A$?

Hi Alexillic,  read once again my proof.
$ A',B'$ have the same diagonal $ [trace(A)/n,\cdots,trace(A)/n]$. I apply your false result to $ A',B'$ and not to $ A,B$.
If you don't understand my argument then consider this counterexample:
$ A\equal{}I_2,B\equal{}\begin{pmatrix}1&1\\1&1\end{pmatrix}$.


A pendulum 80 cm long is pulled aside, so that its bob is raised 20 cm from its lowest position and is tehn released. As the 50 g bob moves through its lowest position, (a) what is its speed and (b) what is the tension in the pendulum cord?

I know that $ T\minus{}mg\equal{}\frac{mv^2}{r}$, where $ T,m,v,r$ are tension, mass of the pendulum, velocity, and radius of the circle. Since $ T$ and $ v$ are what we're trying to find, I thought it would be easier to start with $ v$, as the book recommends.

Going through some trig, I found that the pendulum moved a total of 57.8 cm. But I don't know how to calculate the time it took, and I think I might be going about this in a wrong way in general.

Help?

To find its velocity, you use conservation of energy: $ E_{T}\equal{}U\equal{}mgh\equal{}(.05)(.2)(9.8)\equal{}.098J$.  At its lowest point, all of the energy is in the form of kinetic energy.  So $ .098\equal{}\frac{1}{2}mv^{2}\equal{}\frac{1}{2}(.05)v^{2}\rightarrow v\equal{}1.98 \mbox{m/s}$.


Let $ x^{4}\minus{}10x^{3}\plus{}ax^{2}\plus{}bx\plus{}130\equal{}0$ ($ a,b\in Z$) have all the roots being complex numbers which are not real. Moreover, all the roots are in the form $ p\plus{}qi$, where $ p,q\in Z$. Find the maximum value of $ a\plus{}b$.

We just have to factor $ 130$ into complex numbers or w/e. The rest is casework.


[quote="shobber"]I can solve problem 2 by using the conclution of problem 1, but I am not sure about my (and yours) solution to problem one.

If $f(x)$ is a second degree function, then $x_1 \geq x_2$ does not imply $f(x_1) \geq f(x_2)$...[/quote]

I basically showed that $a_n \ge n+2 \implies a_{n+1} \ge n+3$.
The thing is, $f(n+1)$ is a function of $f(n)$ [i] and [/i] $n$.  You could even, if you wanted to, show that \[ f(n+1) \ge f(n); f(n), f(n+1) \in \mathbb{N} \implies f(n+1) \ge f(n)+1 \] which completes the induction.

[quote="K81o7"]Part 1 can be solved by induction...
$a_{n+1}^2=a_n^2-na_n+1\ge (n+2)^2-n(n+2)+1=2n+5 \ge n+3$
[/quote]

A small imprecision there: You can't conclude $a_n^2-na_n\ge (n+2)^2-n(n+2)$, because of the minus sign. But you can patch it via $a_n(a_n-n)\ge (n+2)\cdot 2.$

[quote="shobber"]Let me say in another way:  After assuming $a_n \geq n+2$, why can you plug it into the equation $a_{n+1}=a_n^2-na_n+1$?[/quote]

Why not?
If $a_n \ge n+2$ then $a_{n+1}=a_n^2-na_n+1 = a_n(a_n-n)+1 \ge (n+2)(2)+1=2n+5$
The factored $a_n$ term is positive, so it's fine.

[quote="K81o7"][hide="1"]Part 1 can be solved by induction...
$a_{n+1}^2=a_n^2-na_n+1\ge (n+2)^2-n(n+2)+1=2n+5 \ge n+3$[/hide]

Part 2 seems to consist  of proving that the harmonic mean of $(1+a_n)$ is greater than or equal to $2n$...but there must be some cleverer way...[/quote]

I've tried some things and don't knows if it's okey or not; can somebody chek'it for me;

for $n\geq 1$   $a_n\geq 2n+1$ and for $n\geq 2 a_n\geq 2(n^2+1)$ then :
$\sum 1/(a_n+1)\leq  1/2\sum 1/(1+n^2)$
then :$\leq 1/2(1/2+\sum_{n\geq 2} (1/n-1/(n+1))$
then the result


symmetry
geometry
trigonometry
projective geometry
geometry proposed

Let $P$ and $Q$ is two point in the triangle $ABC$.$B_1\in{AP}\cap{CQ},B_2\in{AQ}\cap{CP}$ and $l_B$ is the line pase trough $B_1$ and $B_2$.$l_A$and$l_C$ are analogously defined.
Prove that $l_A$,$l_B$ and $l_C$ are concurrent. ;)

You should remove the '?' from the source field. It is definitly yours and

  yours alone.  Does anyone have problem with this?

  S Nastupaischim Novym Godom!

  Salute and thank you,

  M.T.

I writed '?',becaus I think that this problem is very nice,so sambody can get to it befor me :(  :D,
Ti znaesh ruski?
Ia toje!!:blush:
Tebia toge s nastupaishim novim godam jelayu vsevo xaroshava v nowm gadu!!!
Thank you.

[u]Nice concurrence ![/u] I shall give a equivalent enunciation, with the my notations:

$\blacksquare\ \bullet\parallel$Let $ABC$ be a triangle and two points $P$, $Q$ which belong to the inside of the triangle $ABC$. I note the following intersections: $X_a\in BQ\cap CP$, $Y_a\in BP\cap CQ$; $X_b\in CP\cap AQ$, $Y_b\in CQ\cap AP$; $X_c\in AP\cap BQ$, $Y_c\in AQ\cap BP$.
Prove that the lines $X_aY_a$, $X_bY_b$, $X_cY_c$ are concurrently.$\parallel\bullet\ \blacksquare$

[u]Demonstration.[/u] I note the intersections $U$, $V$, $W_x$, $W_y$ of the line $BC$ with the lines $AP$, $AQ$, $X_bX_c$, $Y_bY_c$ respectively. I will apply the Menelaus' theorem for the triangle $AUV$ to the transversals $\overline {BX_cQ}$, $\overline {BY_cP}$, $\overline {CX_bP}$, $\overline {CY_bQ}$, $\overline {X_bX_cW_x}$, $\overline {W_bY_cW_y}$:

$\frac{BU}{BV}\cdot \frac{QV}{QA}\cdot \frac{X_cA}{X_cU}=1$; $\frac{BV}{BU}\cdot \frac{Y_cA}{Y_cV}\cdot \frac{PU}{PA}=1$;

$\frac{CU}{CV}\cdot \frac{X_bV}{X_bA}\cdot \frac{PA}{PU}=1$; $\frac{CV}{CU}\cdot \frac{QA}{QV}\cdot \frac{Y_bU}{Y_bA}=1$;

$\frac{W_xV}{W_xU}\cdot \frac{X_cU}{X_cA}\cdot \frac{X_bA}{X_bV}=1$; $\frac{W_yU}{W_yV}\cdot \frac{Y_bA}{Y_bU}\cdot \frac{Y_cV}{Y_cA}=1$.

From the product of the above relations we obtain the relation $\frac{W_xU}{W_xV}=\frac{W_yU}{W_yV}$, i.e. $W_x\equiv W_y\equiv W$. Therefore, the lines $X_bX_c$, $Y_bY_c$, $BC$ are concurrently (in the point $W$). For the triangles $X_aX_bX_c$, $Y_aY_bY_c$ we observe that $C\in X_aX_b\cap Y_aY_b$, $B\in X_aX_c\cap Y_aY_c$, $W\in X_bX_c\cap Y_bY_c$ and $W\in BC$. From the Desarques' theorem results that the lines $X_aY_a$, $X_bY_b$, $X_cY_c$ are concurrently.

[u]Remark.[/u] There is a shorter prove using  the polar-line of a point w.r.t. a angle !

Nice solution Virgil,but if you know Papu's theorem,ther is wery short solution from it ;)(my solution :lol: )

Evidently, I know the Pappus's theorem, but I didn't think it. I hope yet that you like this solution. See that I made a little formal modification in the enunciation for the symmetry of the notations: $X_a\longleftrightarrow Y_a$.

[b]A very interesting remark.[/b]

$A\in X_aY_a\ \vee \ B\in X_bY_b\ \vee \ C\in X_cY_c\Longleftrightarrow W\in PQ$

You can apply the Desarques's theorem to the triangles, for example, $PX_bY_c$ and $QX_cY_b$, where $X_a\in PX_b\cap QX_c$, $Y_a\in PY_c\cap QY_b$, $A\in X_bY_c\cap X_cY_b$. Thus, $A\in X_aY_a$ $\Longleftrightarrow$ $PQ\cap X_bX_c\cap Y_bY_c\ne \emptyset.$

$\blacksquare\ P.S.\ 1^{\circ}$ Here is and the solution using the [u]Pappus' theorem:[/u]

$\{Y_b,P,X_c\}\subset AP$, $\{Y_c,Q,X_b\}\subset AQ\Longrightarrow$ the points $Y_a\in Y_bQ\cap PY_c$, $X_a\in PX_b\cap X_cQ$ and the intersection of the lines $Y_bX_b$, $X_cY_c$ are collinearly, i.e. the lines $X_aY_a$, $X_bY_b$, $X_cY_c$ are concurrently.

$\blacksquare\ P.S.\ 2^{\circ}$ [u][b]Please, dear Tiks, don't use the icons ![/b][/u]

I am a susceptible (impressionable) man.

[quote="Virgil Nicula"]$\blacksquare\ P.S.\ 1^{\circ}$ Here is and the solution using the [u]Pappus' theorem:[/u]

$\{Y_b,P,X_c\}\subset AP$, $\{Y_c,Q,X_b\}\subset AQ\Longrightarrow$ the points $Y_a\in Y_bQ\cap PY_c$, $X_a\in PX_b\cap X_cQ$ and the intersection of the lines $Y_bX_b$, $X_cY_c$ are collinearly, i.e. the lines $X_aY_a$, $X_bY_b$, $X_cY_c$ are concurrently.[/quote]
That also my solution.
[quote="Virgil Nicula"]$\blacksquare\ P.S.\ 2^{\circ}$ [u][b]Please, dear Tiks, don't use the icons ![/b][/u]

I am a susceptible (impressionable) man.[/quote]
No problem dear Virgil!.

Thanks.
Like you the remark when $W\in PQ$ ?
See a interesting problem at
http://www.mathlinks.ro/Forum/viewtopic.php?t=66939

[quote]Like you the remark when $W\in PQ$ ?[/quote]
Of cours I like it
[quote]See a interesting problem at
http://www.mathlinks.ro/Forum/viewtopic.php?t=66939[/quote]
I foude same solutions for this problem,but in all of them I use analitc geometry or trigonometry :(.


Let $a,b$, and $x$ be positive integers such that $x^{a+b}=a^b{b}$. Prove that $a=x$ and $b=x^{x}$.

I hope I made no mistakes:
[hide="Solution"]
We assume that $ x\ne a$.  Let $ a=\prod_{i=1}^{k} {p_i^{x_i}}$, $ b=\prod_{i=1}^k {p_i^{y_i}}$, and $ x=\prod_{i=1}^{k} {p_i^{z_i}}$.  Henceforth, we will ignore simply denote $ \prod_{i=1}^k$ as  $ \prod$ for simplicity.  Furthermore, when we refer to $ y_i$, we  do not mean $ y_i$ collectively, but for rather for a fixed $ y_i$ (or else otherwise noted.)  Then, \[ x^{a+b}=a^bb\iff \prod {p_i^{z_i\cdot (\prod {p_i^{y_i}}+\prod {p_i^{x_i}})}=\prod {p_i^{y_i+x_i\cdot \prod {p_i^{y_i}}}}\iff}\] \[ x_i\cdot \prod {p_i^{y_i}}+y_i=z_i\cdot (\prod {p_i^{x_i}}+\prod {p_i^{y_i}})\] for all $ 1\le i\le k$.  Now, if $ y_i<x_i$, then $ p_i^{y_i}|\prod {p_i^{x_i}}$, and $ p_i^{y_i}$ is a factor of all the other terms in the equation except $ y_i$.  This means that $ p_i^{y_i}|y_i$, which is only possible if $ y_i=0$ since $ y_i<p_i^{y_i}$.  If $ y_i=0$, then we can obtain that $ (x_i-z_i)\cdot \prod {p_i^{y_i}}=\prod {p_i^{x_i}}$.  Yet, $ p_i$ is not a factor of $ \prod {p_i^{y_i}}$ when it is a factor of $ \prod {p_i^{x_i}}$.  This means that $ p_i^{x_i}$ is a factor of $ x_i-z_i<x_i<p_i^{x_i}$, so $ x_i-z_i=0$, so $ \prod {p_i^{x_i}}=0$, which is a contradiction.  This means that $ x_i\le y_i$ for all $ 1\le i\le k$, so $ \prod {p_i^{x_i}}|\prod {p_i^{y_i}}$, meaning that $ \prod {p_i^{x_i}}|y_i$, so $ a|y_i$ for all $ 1\le i\le k$.  Letting $ y_i=ak_i$, we have that $ ak_i+x_ib=z_ia+z_ib$, so $ a(k_i-z_i)=b(z_i-x_i)$, so \[ \frac{k_i-z_i}{z_i-x_i}=\frac{b}{a}=\prod {p_i^{ak_i-x_i}}\] Now, since $ ak_i=y_i$, which is a multiple of $ a>x_i$, we have that $ ak_i>x_i$, meaning that $ \prod {p_i^{ak_i-x_i}}>ak_i-x_i\ge k_i-x_i$.  If $ k_i-z_i<0$, then if $ z_i-x_i>0$, we would have that the left hand side of the equation is negative when the other side is positive, giving us a contradiction.  Then, $ z_i-x_i<0$, so $ \frac{k_i-z_i}{z_i-x_i}=\frac{z_i-k_i}{x_i-z_i}<z_i-k_i+x_i-z_i<x_i$.  Yet, $ a\ge p_i^{x_i}\ge 2x_i$, meaning that \[ \prod {p_i^{ak_i-x_i}}>ak_i-x_i\ge a-x_i\ge x_i>\frac{k_i-z_i}{z_i-x_i}\] which is a contradiction.  Otherwise, we have that $ k_i-z_i, z_i-x_i>0$, so \[ \frac{k_i-z_i}{z_i-x_i}<k_i-z_i+z_i-x_i<\prod {p_i^{ak_i-x_i}}\] which is a contradiction, so we conclude that $ z_i=x_i$ for all $ 1\le i\le k$, so $ a=x$.  This is a contradiction, so we get that $ a=x$.  This means that $ x^{x+b}=x^b\cdot b$, so $ b=x^x$, as desired. [/hide]

[hide="what is wrong with this?"]
What??? $ x^b|a^bb$ thus $ x|a$ or some $ b$-th power divides $ b$.
We toss out second case as $ n^b\geq 2^b>b$
Thus $ x|a$. Suppose $ a\geq 2x$; then $ b\equal{}x^a(x/a)^b\geq x^a 2^b>b$.
Thus $ a\equal{}x$ and the result follows.
[/hide]

[quote="me@home"]
Thus $ x|a$. Suppose $ a\geq 2x$; then $ b \equal{} x^a(x/a)^b\geq x^a 2^b > b$.[/quote]
Sorry, I might be misunderstanding, but isn't $ \frac{x}{a}\le \frac {1}{2}$, so $ b\equal{}x^a\cdot (\frac{x}{a})^b\le x^a\cdot (\frac{1}{2})^b$?

[hide=Another solution]
[b]Step 1[/b] $x\mid a$.
For each prime $p$ we have $bv_p(x)<(a+b)v_p(x)=bv_p(a)+v_p(b)<b(v_p(a)+1)$, so $v_p(x)<v_p(a)+1$, hence $x\mid a$.
[b]Step 2[/b] $a\mid b$.
Write $a=cx$, then $x^{cx+b}=(cx)^bb$ reduces to $x^a=c^bb$.
For each prime $p\mid a$ we have $av_p(x)=bv_p(c)+v_p(b)$.
Let $v_p(a)=k\geq 1$, then $v_p(av_p(x))\geq k$, therefore $v_p(bv_p(c)+v_p(b))\geq k$.
If $v_p(b)=0$
- If $v_p(c)=0$, then $av_p(x)=0$, so $v_p(x)=0$, but this is impossible as $p\mid a\to p\mid x$
- Else, $v_p(v_p(c))=v_p(bv_p(c)+v_p(b))\geq k=v_p(a)\geq v_p(c)$, which is impossible.
If $0<v_p(b)<k$, then $v_p(v_p(b))\geq \min(v_p(bv_p(c)+v_p(b)),v_p(bv_p(c)))$.
However, since $v_p(v_p(b))<v_p(b)<k\leq v_p(bv_p(c)+v_p(b)$ and $v_p(v_p(b))<v_p(b)\leq v_p(bv_p(c))$, this is also impossible.
Therefore we have $v_p(b)\geq v_p(a)$ for all prime $p$, thus $a\mid b$.
[b]Step 3[/b] $a=x,b=x^x$.
For each prime $p$, from $a\mid b$ and $(a+b)v_p(x)=bv_p(a)+v_p(b)$, it follows that $a\mid v_p(b)$ for all prime $p$.
Therefore $b=d^a$ for some $d\in\mathbb{N}$. If $b=1$ it's obvious that $a=x=1$, so suppose that $b\geq 2$, so that $d\geq 2$.
Thus for each prime $p$, $(a+d^a)v_p(x)=d^av_p(a)+av_p(d)$.
This gives $v_p(x)=v_p(a)+\frac{av_p(d)-av_p(a)}{a+d^a}$.
Now, note that $av_p(a)-av_p(d)\leq av_p(a)<a^2<a+2^a<a+d^a$ for all $a$, so $v_p(x)>v_p(a)-1$.
However, from $x\mid a$, $v_p(x)\leq v_p(a)$, so $v_p(x)=v_p(a)$ for all prime $p$.
Therefore $a=x$, and thus $b=x^x$.
[/hide]


Several equal spherical planets are in outer space. On the surface of each planet is a set of points which is invisible from any of the remaining planets.
Prove that the sum of the areas of all these sets equals the surface area of one planet.

Please continue discussion at http://www.mathlinks.ro/Forum/viewtopic.php?t=6395 or http://www.mathlinks.ro/Forum/viewtopic.php?t=1465 .

  Darij


vector
algebra
polynomial
linear algebra
matrix
abstract algebra
calculus

Let $ K$ be a field,and $ F\equal{}K[\alpha]$ ba a extension field of $ F$,such that $ \alpha$ has minimal polynomial $ x^m\plus{}a_{m\minus{}1}x^{m\minus{}1}\plus{}\cdots\plus{}a_0\equal{}0$ in $ K[x]$,then a vector space $ V$ of dimension $ n$ of $ F$ can be regard as a vector space of $ K$ which has dimension $ mn$.
Then every homomorphism of $ V$ can be naturally maped to a matrix of $ M_{nm}(K)$,should this map be monmorphism?and the automorphism of $ V$ map to the invertible matrix?

Thank you for help.

The map is mono because if we fix a base of $ V$ over $ F$ such that the matrix representation of an endomorphism is $ A \equal{} A_0 \plus{} A_1 \alpha \plus{} \cdots \plus{} A_{m\minus{}1} \alpha^{m\minus{}1}$, then with respect to the natural base of $ V$ over $ K$ the first $ n$ columns would be exactly $ \left[A_0 A_1 \cdots A_{m\minus{}1}\right]^T$.

For automorphism part, I suspect that this correspondence (linear transformation of V over F -> linear transformation of V over K) is a homomorphism, which would automatically solve your problem, but currently I am not sure about how to show it.

thank you.
another question:
if there is $ m$ different automorphisms of $ F$ that fix $ K$,define it is $ t_1,t_2,...t_m$,then we have:

$ \det(A) \equal{} \prod_{i \equal{} 1}^m t_i(\det(f))$
Where $ f$ is the homorphism of $ V$ over $ F$,$ A$ is the correspondent homorphism of $ V$ over $ K$.

It is ture when $ n\equal{}1$(which is in algebraic number theory that seem not trivial)

Let's do this more systematically - wherever there is an assertion about matrices, it can't harm to translate it into a base-free linear maps form (of course, the exact opposite is equally useful).

What you have is the following situation:

$ F$ is a field extension of $ K$. You want it to be primitive, but this is pointless - it is enough that $ K$ is a commutative ring and $ F$ is a commutative $ K$-algebra which is free as an $ K$-module. Note that this free allows us to define a norm map $ \mathcal{N}: F\to K$ as follows: For every element $ f\in F$, the multiplication with $ f$ (as a map from $ F$ to $ F$) is an endomorphism of the free $ K$-module $ F$, thus has a determinant. This determinant is what we call the norm $ \mathcal{N}\left(f\right)$ of $ f$.

Now you have given a finite free $ F$-module $ V$. Of course, you can consider this $ V$ as a $ K$-module as well. Every endomorphism $ g$ of the $ F$-module $ V$ is also an endomorphism of the $ K$-module $ V$ (but not the other way round!), and thus has both a determinant $ \det_Fg\in F$ when seen as an endomorphism of the $ F$-module $ V$, and a determinant $ \det_Kg\in K$ when seen as an endomorphism of the $ F$-module $ V$.

The problem is now to show that $ \det_Kg \equal{} \mathcal{N}\left(\det_Fg\right)$ for any endomorphism $ g$ of the $ F$-module $ V$. In other words, we have to show that $ \det_K \equal{} \mathcal{N}\circ\det_F$ as maps from $ \mathrm{End}_FV$ to $ K$.

This $ \det_K \equal{} \mathcal{N}\circ\det_F$ formula has been posted in matrix form before:
http://www.mathlinks.ro/viewtopic.php?t=44950
http://www.mathlinks.ro/viewtopic.php?t=45040

The proof posted there was not the best, but a simpler one can be done by basically performing Gaussian elimination on the block matrix (treating the blocks $ A_{ij}$ as elements of a field - they are elements of a commutative ring which is generically an integral domain, and one can go into the quotient field of that domain). I must hurry back to IMO coordination, so sorry for any missing explanations and mistakes. Will try to detail this later.

  darij


geometry
circumcircle
analytic geometry
linear algebra
matrix
geometry proposed

find the center and radius of the circumcircle of triangle ABC with coordinates $ (\minus{}x,0), (x,0), (x_1,y_1)$

well i mean assume that the points are not collinear but otherwise obviously they exist
@lig could you explain?

What? Maybe there is a missing condition or something.

$ R \equal{} \frac {abc}{4A} \equal{} \frac {2x\cdot\sqrt {(x_1 \minus{} x)^2 \plus{} y_1^2}\cdot\sqrt {(x_1 \plus{} x)^2 \plus{} y_1^2}}{4|xy_1|} \equal{} \frac {\sqrt {(x_1 \minus{} x)^2 \plus{} y_1^2}\cdot\sqrt {(x_1 \plus{} x)^2 \plus{} y_1^2}}{2|y_1|}$

$ x,y_1\ne 0$...

[quote="mathemonster"]where (x_0,y_0) is the center right?[/quote]
Yes.
In Cartesian coordinates, the equation of a circle of radius $ r$ centered on $ (x_0,y_0)$ is $ (x\minus{}x_0)^2\plus{}(y\minus{}y_0)^2\equal{}r^2$.

[quote="Ligouras"]$ O(0,y) \Longrightarrow O(0, \frac {x_1^2 \plus{} y_1^2 \minus{} x^2}{2y_1})$

$ R \equal{} \sqrt {x^2 \plus{} y^2}$.[/quote]


algebra
polynomial
number theory unsolved
number theory

[quote="RobertuX"][quote="kiemkhach"]consider a sequence $(a_n)$ satisfies $a_0=1;a_1=9$
and $a_{n+1}=6a_n-a_{n-1}-4$

find $n$ such that $a_n$ is a perfect square[/quote]
Well,  $n=1$ and $n=2$ are two solutions, rightt????    :P  :P  :P but I suppose you want an $n>2$.
And in general

\[a_n=1+\sqrt 2\left((3+2\sqrt 2)^n-(3-2\sqrt 2)^n\right).\]

TCM[/quote]

$a_3=6a_2-a_1-4=54-1-4=49=7^2$???? :(
can you post your solution? :D I think ;n=1 and n=2 and n=3 are solutions??

[quote="RobertuX"][quote="RobertuX"]In general
\[a_n=1+\sqrt 2\left((3+2\sqrt 2)^n-(3-2\sqrt 2)^n\right).\]
[/quote]
Well, of course this 3 values $n=1,2,3$ are solutions, and using my expression $a_4$ is NOT a square.[/quote]
I don't understand! :( In general :$a_n=1+\sqrt{2}((3+\sqrt{2})^n-(3-\sqrt{2})^n)$
 how do you prove $a_n$ is not a square???($n>3$)

[quote="kiemkhach"]In general
\[a_n=1+\sqrt 2\left((3+2\sqrt 2)^n-(3-2\sqrt 2)^n\right).\]
how do you prove $a_n$ is not a square???($n>3$)[/quote]
Well, sorry for the last comment, by definition $a_{n+1}=6a_n-a_{n-1}-4$, this is so colled non homogeneous sequence, so first I get an equivalent homogeneous sequence, HOW? well, in this case the sum of coefficients of the homogeneous part is $1-6+1=-4$ so we need to consider $b_n=a_n-1$, in fact $b_n$ satisfies the relations:$b_0=0$, $b_1=8$ and $b_{n+1}-6b_n+b_{n-1}=0$ and to solve this kind of sequences you need to consider the charasteristic polynomial related to the sequence, in this case is $x^2-6x+1=0$ which roots are $\alpha=3-\sqrt 8$ and $\beta=3+\sqrt 8$, so $b_n=A\alpha^n+B\beta^n$, now we calculate $A$ and $B$ with the initial values, so $A+B=0$ and $\alpha A+\beta B=8$ then $A=-B=\sqrt 2$ so, you can write easily the expression of $a_n$. If you need for info please, read a book of Discrete Mathematics.

Sorry again,and best regards,


If $a,b,c$ are there side lengths of a triangle then
\[\frac{a^{2}b}{c}+\frac{b^{2}c}{a}+\frac{c^{2}a}{b}\ge a^{2}+b^{2}+c^{2}. \]

Ugly but quick solution:

Use Ravi substitution, and the inequality is equivalent to
\[x^{5}+y^{5}+z^{5}+x^{4}z+xy^{4}+yz^{4}\geq 2(x^{3}y^{2}+x^{2}z^{3}+y^{3}z^{2}) \]
Now use AM-GM to show that $x^{5}+xy^{4}\geq 2x^{3}y^{2}$ and similar for other variables, then sum up.


Given that $ 7x \plus{} 3y \equal{} 54$ and $ 3x \plus{} 7y \equal{} 46$, what is the value of $ x \plus{} y$?

We realize that if we add both equations up, we get $ 10x \plus{} 10y \equal{} 100$.  Perfect.  After dividing by $ 10$ we get $ \boxed{x \plus{} y \equal{} 10}$


ratio
number theory unsolved
number theory

$ a_{0}$=1,$ a_{n\plus{}1}$=$ a_{n}$+(cos133∏/355)/$ a_{n}$
find,with proof,the smallest n satisfying:$ a_{n}$ is an integer

any idea will be appreciated.
i wonder if it is suitable for the NT forum,or should I post it in algebra forum?

I think it's a trick question. Answer is $ n\equal{}0$.

Interesting problem.

Is 133/355 supposed to be 113/355?  That ratio seemed peculiar and since 355/113 is sometimes used as an approximation to Pi it made me think that was involved with the problem.

If not, is there any reason for the peculiar 133/355 ?

The difficulty in this problem appears to be in analyzing the non-linear recurrence relation. Nothing jumped out at me which made me think like Phil that n=0.

I did a quick numeric check though and the closest I found was at n=161218988 with a=11123.0000000001411...  But this closeness appears to be just because you're adding smaller and smaller numbers.

my bad,
the question is to find the smallest n>0


[quote="veezbo"]Is it true that 3-D geometry doesn't show up anymore at the AIME (or even below) level?[/quote]
Noway, I love 3-D geometry and they are quite easy. I remember seeing one on a recent AMC (200x)

No, there was one on last year's AMC 12B. (#20 I think?)

*sigh* Yes unfortunately, they are still alive and well. But they're not that common :)

Agh. It's just because I'm really bad at it and have to rederive some basic stuff because I haven't actually learned it at all.

I was hoping to spend more time just working on Euclidean Geometry because apparently while although I can consistently solve one problem above #10 and nearly every non-geometry problem under #10, I have still never actually solved an AIME-level geometry problem during the test.

Do you think that Intro to Geometry covers the 3-D geometry on AMC 10/12/AIME?

Definitely not the AIME. Probably the AMC 10 and most of AMC 12.

[quote="veezbo"]Is it true that 3-D geometry doesn't show up anymore at the AIME (or even below) level?[/quote]

No, e.g. see 2008 AIME I no. 15.  However, most of the 3-D geometry in AMC/AIME is often just 2-D geometry guised in 3-D, and therefore is usually easier than you think.

Are there any books you would recommend for mastering AIME/beggining Olympiad geometry? (not 3D, I mean in general)

There have been a discussion in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=296565]this[/url] link, especially towards the end. After reading the Intro to Geo book (or taking the course), Geometry Revisited, Problems in Plane Geometry, etc are recommended.

3-D does appear some on AMC/AIME but definitely not much so there's not much need worrying about it I guess

[quote="worthawholebean"]I wouldn't recommend a book - I'd recommend doing more problems.[/quote]

A book of problems, I think he means.

You really have to get the fundamentals down first. You can't just rush off and only do problems. And when you do, try hard, study the solution carefully, pinpoint your weak points, and try to eliminate stupid mistakes.


Let $ ABC$ is triangle, let $ AB\equal{}c, BC\equal{}a,CA\equal{}b$ and let $ M$ is a change point. Prove that : $ a.MA^2\plus{}b.MB^2\plus{}c.MC^2 \ge abc$.


Here is a challenge problem that I found really nice.

The product

\[(x+5)(x+10)(2x^2+3)(x^3+6x+16)^2(x+9)(x+4)^3(x+18)\]

is expanded.  In the resulting polynomial, how many of the coefficients are odd?

[quote="blahblahblah"]Ah, this reminds me of the course I took on polynomial rings.  But the problem doesn't REALLY use any of that junk, although the mapping I am using is of relevance to certain results on finite fields.   See spoiler for solution:

[hide]We will take the mapping $\varphi:\mathbb{Z}[x]\rightarrow (\mathbb{Z}/2)[x]$ and the result will follow quickly.  Let $p(x)$ be the original polynomial, then:

$p(x)=(x+5)(x+10)(2x^2+3)(x^3+6x+16)^2(x+9)(x+4)^3(x+18)$ and
$\varphi(p(x))=(x+1)(x)(1)(x^3)^2(x+1)(x)^3(x)=x^11(x+1)^2=x^{13}+x^{11}$.  These are the only 'surviving' terms, and hence the only terms with odd coefficients.

So, there are 2 odd coefficients, and maple agrees with me.

Using this method, it doesn't matter what we change the polynomials to; the answer is still very easy to find.
[/hide][/quote]
Excellent. But I wonder if there is another way of solving this without referencing to ring homomorphims...

In a naive sense, you can just take all coefficients $\bmod 2$ and realize that that accomplishes the same thing whether you do it to the factored version of the polynomial or the expanded version of the polynomial.

Certainly, since binary representations yield quickly those coefficients that are odd. This is more of an appropriate level topic in the Intermediate Forum  :)


What is the $ 100^{\text{th}}$ term of the sequence 2, 3, 5, 6, 7, 10, 11, $ \ldots$, which consists of all of the positive integers that are neither perfect squares nor perfect cubes?

Consider the sequence of the first $ 100$ positive integers.  $ 10$ of these are perfect squares, so we add the positive integers from $ 101$ to $ 110$ and get rid of the perfect squares.

Also, from $ 1$ to $ 110$ there are $ 4$ cubes, but $ 2$ of them are also perfect squares so we only need to get rid of $ 2$ cubes.

The new $ 100^\text{th}$ term is then $ \boxed{112}$


Ten people queqe before a tap to fill up their buckets. Each bucket requires a different time to fill. In what order should the people queqe up in order to minimize their combined waiting time? Do not just use pure logic to justify your answer.

In order to have the least waiting time in total, the people should line up in order from least time to fill in front to most time to fill in the back.  We know this because for any two people standing one in front of the other, switching their positions does not affect the waiting time of anyone else in the line.
The least total waiting time between those two people will be achieved by having the less-time-to fill man in front.  To show this, let us assume that one of them takes n time to fill and one takes x time to fill where x>n.
If x is in front of n, then their total waiting time will be x+(x+n)=2x+n.  If n is in front of x, then their total waiting time will be n+(n+x)=2n+x.  We know that 2x+n>2n+x.  Therefore, the waiting time will be minimized by putting the man with a bucket which fills more quickly in front.  One can now take any formation, and can undertake a series of these optimizations, and get to the shortest time being from least to greatest.
One can also see the total time as 10(first one in line)+9(second in line)+8(third in line)+...+(tenth in line).

Hehehe
[hide]
From least time to greatest, since the lightests buckets can get out of the way easier.[/hide]

[quote]
Hehehe

Click to reveal hidden content

From least time to greatest, since the lightests buckets can get out of the way easier.
[/quote]

Dangit not just pure logic. But you are right. So is K81o7 who gave the full solution.


inequalities
function
logarithms
inequalities proposed

Let $ \{a,b,c\}\subset[ k, \plus{} \infty)$ such that $ abc\geq1$ and
$ k$ is a real root of the equation $ x \plus{} \sqrt [3]x \plus{} 1 \equal{} 0.$ Prove that
\[ (a^2 \plus{} a \plus{} 1)(b^2 \plus{} b \plus{} 1)(c^2 \plus{} c \plus{} 1)\geq(a \plus{} 2)(b \plus{} 2)(c \plus{} 2)\]
I posted it here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=179021
but it still unsolved there.

I proved that $ (a^2\plus{}a\plus{}1)(b^2\plus{}b\plus{}1)(c^2\plus{}c\plus{}1)\geq \sqrt[3]{a^2b^2c^2}(a\plus{}2)(b\plus{}2)(c\plus{}2)$

To prove this just consider the function $ f(x)\equal{}\ln (a^2\plus{}a\plus{}1)\minus{}\frac{2}{3}\ln a\minus{}\ln (a\plus{}2)$

 :wink:

[quote="silouan"]I proved that $ (a^2 \plus{} a \plus{} 1)(b^2 \plus{} b \plus{} 1)(c^2 \plus{} c \plus{} 1)\geq \sqrt [3]{a^2b^2c^2}(a \plus{} 2)(b \plus{} 2)(c \plus{} 2)$

To prove this just consider the function $ f(x) \equal{} \ln (a^2 \plus{} a \plus{} 1) \minus{} \frac {2}{3}\ln a \minus{} \ln (a \plus{} 2)$

 :wink:[/quote]
But $ a$ can be negative. :wink:

I posted a proof only for positive reals (I forgot to write it )


A stone, which weighs 0.62 kg, is thrown at 5.5ms, its kinetic energy when at its maximum height above the surface of the ground is 2.3J. What is the height of the stone at this point, any equations and advice gratefully recieved.

Its initial kinetic energy is $\frac12 \cdot 0.62 \cdot 5.5^{2}=9.3775\textrm{J}$.

Assuming resistance is negligible, and that it was thrown from ground level, when it is at its highest point it has $9.3375-2.3=7.0375\textrm{J}$ of gravitational energy, and since $\textrm{GPE}=mgh$ we have $gh=\frac{7.0375}{0.62}=11.35$ and hence, taking $g=9.8ms^{-2}$
$h=1.16m$.

The two equation required are
kinetic energy=$\frac12mv^{2}$
gravitational energy=$mgh$


analytic geometry
geometry unsolved
geometry

Is it possible to find?
[img]http://img239.imageshack.us/img239/5181/dsfdfyj1.png[/img]

every posting good problems :)

thanks Shahzada

my figure can be easily made using Tales.
peharps. we have 3 constants, but in your problem we have only 2.

so. it's missing one value. without that we cannot going further.

but i'm not really sure about that. because $x+s = AB$ and $x+i = AC$

so, maybe it's possible to eliminate $x$, but i'm not confident of this.

i will try to find the equation of that line, tangent to the 2 circles in a easy way after.

c ya

thanks much yagaron.

If we change the size of circles such that the |MN| is same, CB and AB will change.It may missing something.If it is possible to solve there must be an equation included only MN,AC and AB not any angle.But i think it is wrong.what is your opinion?

we have 9 points. let's call it $A,B,C, ...$

so, $AB$ is a line $AC$ is another...

what i think is, if we have only 3 combinations of a pair of points it's possible to solve the problem

of course that equal combinations does not enter in the problem. like equal lengths.

because on my drawn all anothers points coordinates could be found only with those 3 mesures.

but find the equation of that line is a litle bit trouble.

with 2 random poinnts on the 2 circles we can make a line. so. we can make this line tangent to the 2 circles with some calculations.

or we can use the fact that the center of the circles and the tangential point have the a comon tangent.

or others things. using vectors...

but all solutions i made have a lot of calculations. i will find a simple one

What program do you use to make pictures?

http://www.geometryexpressions.com/downloads.php?PHPSESSID=14909a289ff7b8348e864ce42e14bda4

Geometric Expressions

you can try a free trial copy of this. :)

We can not find $x$.Let be $NA=y$.
$A,N,M$ are collinear.Since $\triangle CNA \sim \triangle MBA$, $\frac{CA}{MA}=\frac{NA}{BA}$.Then, $\frac{8}{6+y}=\frac{y}{x}$.$y$ must be given.


Let be given four circles $(O_1,R_1),(O_2,R_2),(O_3,R_3),(O,R)$ satisfying:
$(O_1,R_1),(O_2,R_2)$ tangent externally.
$(O_1,R_1),(O_3,R_3)$ tangent externally.
$(O_3,R_3),(O_2,R_2)$ tangent externally.
$(O,R)$ tangents to the three circles $(O_1),(O_2),(O_3)$ all internally.
Prove that:
$R_1+R_2+R_3\le 3(2\sqrt{3}-3)R$


Kris has an endless supply of 4-cent stamps and 7-cent stamps for mailing a package. What is the greatest amount of postage that cannot be affixed to a package using these stamps?

$ 4 \times 7 \minus{} 4 \minus{} 7 \equal{} 28 \minus{} 11 \equal{} \boxed{17}$.

you can immediately rule out all multiples of 4 and 7
so 4, 8, 12.... can't be it
neither can 7, 14, 21.....
but if you look closely 7 is 1 less than 8 or 3 more than 4
14 is 2 less than 16 or 2 more than 12
and 21 is 3 less than 24 or 1 more than 20
so after a certain amount no number will be unable to be synthesized by this combination
Kouichi Nakagawa showed us the theorem to solve this problem which is posted above but i will show it also below

4 X 7 - 4 - 7 = 17

since 4 X 7 gives you four 7s or seven 4s, by taking away one of each so that it can neither be 4 X 6 or 3 X 7 we come out with the largest number unable to be synthesized by any number of 4s or 7s

17!

[quote="pinkmuskrat"]Is this the Chicken McNugget Theorem? Does it at all relate to Simon's Factoring Trick? Thanks[/quote]
I have solved the problem that resembled this problem in a book of Japanese mathematics.
Chicken McNugget Theorem was introduced to the book as a theorem, but the name of the theorem was not on.
I found out the name of the theorem thanks to pinkmuskrat.

Here is a simple explanation.  It lacks elegance, but it helps organize one's view and is easily understood, completed, & remembered.

1 --  [b]NOT possible[/b] postage using a 4-cent or 7-cent stamp
2 --  [b]NOT possible[/b]
3 --  [b]NOT possible[/b]
4 --  Possible
5 --  [b]NOT possible[/b]
6 -- [b]NOT possible[/b]
7 --  Possible
8 -- Possible -- [i]Now we start asking the question; can we get this amount by adding 4 or 7 to a previous possibility?[/i]
9 -- [b]NOT possible[/b] -- Count back 4 or 7; both impossible postage amounts
10 -- [b]NOT possible[/b] -- Count back 4 or 7; both impossible amounts
11 -- Possible -- Count back 4;  since 7 is possible then 11 is possible
12 -- Possible -- Count back 4;  since 8 is possible then 12 is possible
13 -- [b]NOT possible[/b] -- Count back 4 or 7; both impossible
14 -- Possible
15 -- Possible
16 -- Possible
17 -- [b]NOT possible[/b] -- Count back 4 or 7; both impossible
18 -- Possible
19 -- Possible
20 -- Possible
21 -- Possible
[i]We need not check further.  17 is the greatest amount that our stamps cannot produce.
Adding combinations of 4-cent stamps to the previous four possibilities will produce all integer values 22 or greater.[/i]


inequalities
trigonometry
inequalities unsolved

ABC is acute  triangle. Show that
$  10(h_a+h_b+h_c)+36R \leq 9\sqrt{3}(a+b+c)  $

I do not believe this inequality is true. Can anyone find example to disprove this one ? Maybe A-> 90 and B=C -> 45

Hi everyone!

Dividing inequlity to $2R$ we obtain following inequality:
$10(\sin x\sin y+\sin y\sin z+\sin z\sin x)+18\leq 9\sqrt{3}(\sin x+\sin y+\sin z)$, where $x,y,z$ are angles of the triangle.

Let $(x,y,z)$ be optimal set for this inequality, i.e. RHS-LHS=min. Suppose $x>y$. Consider new triangle correcponding to set $(x-t,y+t,z)$. Then it is necessery $(RHS-LHS)'_{t=0}\geq 0$.
It easily follows that $10\sin(y-x)+(\cos y-\cos x)(9\sqrt{3}-10\sin z)\geq 0$ $\Leftrightarrow$ $10\cos\frac{x-y}{2}-\sin\frac{x+y}{2}(9\sqrt{3}-10\sin z)\leq 0$.
Since triangle is acute it implies that $x-y<z$; $x+y=\pi-z$, thus following inequality is true
$10\cos\frac{z}{2}-\sin\frac{z}{2}(9\sqrt{3}-10\sin z)\leq 0$.
But this is wrong inequality for $0\leq z\leq \pi/2$.
Indeed, $10\cos\frac{z}{2}-9\sqrt{3}\sin\frac{z}{2}\geq 0$ for $z\leq 2\arctan\frac{10}{9\sqrt{3}}$. If $z>2\arctan\frac{10}{9\sqrt{3}}$ then $\sin z>\frac{180\sqrt{3}}{343}$ $\Rightarrow$ $9\sqrt{3}-10\sin z<9$ and this implies $10\cos(z/2)>9\sin(z/2)>\sin(z/2)(9\sqrt{3}-10\sin z)$, since $\cos z\geq 1/\sqrt{2}\geq \sin(z/2)$.
Contradiction.

Hence in optimal set $(x,y,z)$ we have $x=y=z=\pi/3$.