Prove that addition for matrices is associative.

Is this too easy?

If it's too easy, also prove that the addiction of $n$ matrices, rather than just three, is associative. :D

If you think of addition as a binary operation, there will be many ways to associate the $n$ terms with parentheses. But then it's way past getting started level.

[quote="AntonioMainenti"]If it's too easy, also prove that the addiction of $n$ matrices, rather than just three, is associative. :D

[/quote]

That's what I want, and I'm not addicted to matrices  :rotfl:

Just look at how the addition of matrices is defined.

To add matrices, add the "items" in the corresponding places. So adding matrices is, in fact, the same as adding real numbers, and since addition of real numbers is associative, the addition of matrices must be associative as well.

I know it's not very rigorous, I could perfectly rewrite it in a more mathematical way, but of course we're in "Getting Started" here so I don't think that is a good idea.

That's how i proved (well, a little more detailed but this is the internet), I just wanted to see if there was a more formal way.

No, I don't think a more formal way exists. After all we're proving this straight from the definition. I don't think another "simple" proof can be found, but I don't think we should bother since the proof is obvious and easy enough :)

For some reason my textbook didn't contain answers to these proofs, and there were like 10 of these similar ones. Of course, if we could prove 1+1=2, that would be fun  :P


Find the quotient and the remainder when $ 3x^{4} \minus{} x^{3} \minus{} 3x^{2} \minus{} 14x \minus{} 8$ is divided by $ x^{2} \plus{} x \plus{} 2$.

[I have quotient $ \equal{} 3x^{2} \minus{} 4x \minus{} 5$ and remainder $ \equal{} \minus{} x \plus{} 2$.]

Thanks. You're right: I left out a $ 3$ and have edited it in.

Couldn't you just do long division?  Although that would be very hard to $ %Error. "LATEX" is a bad command.
$

Synthetic division works only for linear divisors, so it's only a practical method when dealing with divisors that can be factored into linear factors. Unfortunately, $ x^2 \plus{} x \plus{} 2$ has imaginary roots, which are very hard to work with.


I got into Brown this year and was waitlisted at Stanford. Brown has binding decisions so I am deciding whether to accept at Brown or decline and wait for Stanford waitlist decisions.

I hope someone can compare both colleges in terms of:

1) Math and Comp. Sci.
2) Campus life
3) Undergrad Research Opportunities

I know little about Brown, but I can give you a quick overview of Stanford for your three points of interest.

1) The math program is strong -- there is a good honors freshman sequence and plenty of good professors around. I don't know that it especially stands out among the top math programs, but it's definitely up there. As for computer science, it would be tough to find a department better than Stanford's. From my experience the courses have been excellent and very thorough. Many of the professors are famous for their research and teach really awesome classes in their fields. A lot of thought is put into the introductory computer science courses as well. You can expect a lot from a Stanford CS graduate.

2) The main benefit of campus life at Stanford relative to other academically strong schools is how laid back people are. Of course people take classes seriously, but it's California and you can't help but feel a little less stressful. Of course having never attended another university I can't really compare -- this is just my impression. I really like the environment here overall, though. There are plenty of events to attend and awesome people to meet. I can give you more details if you have specific questions; campus life as a whole is a broad topic.

3) This is probably one of Stanford's strongest points as a research institution. It is not difficult to find research opportunities as an undergraduate. Although I have not done any research yet, several of my friends are doing research over the summer with professors that are at the forefront of their fields. So not only will you be able to do real research, you'll get to work with both amazing peers and mentors who really know what they're doing. I'm guessing you'd be able to find similar research opportunities at other universities, but I think at Stanford research is much more readily available and advertised to undergraduates.

From what I've heard about Brown, its main strength as far as math is in applied math and its pure math department is lacking.  I'm afraid I can't give you specifics, though.

[quote]Brown has binding decisions[/quote]

What do you mean?

What I hear from my grad-school buddies is that Brown is very strong in the areas in which it has professors, but is quite small (so those areas are not all-encompassing).  I don't think this should matter too much to an undergrad, though.

You should also probably look at the wait list statistics, and see how many people usually get off of Stanford's waitlist.

Not on topic but, thanks tokenadult, I was wondering how that works.  :)

[quote="tokenadult"]If the original poster had a regular round admission from Brown, he should accept that offer of admission by May 1st by submitting a deposit, and if he is interested in Stanford still he should indicate a willingness to be on the waiting list. Those two actions are consistent. If he had an early decision offer of admission to Brown, he should have withdrawn his application to Stanford long ago.

It is routine for applicants who are put on the waiting lists of highly desired colleges to first accept (by May 1st) an offer of admission from some other college, possibly including paying a deposit. If offered admission from the waiting list, you can accept the offer of admission from your preferred college, but may forfeit your deposit--a small price to pay if you really like the waiting list college a lot better. This advice is general, and this situation happens hundreds of times a year.[/quote]

Yes, I do have a regular round of admission from Brown.
My problem is that Brown does not ask for any deposit, but once you accept admission to Brown, you are committed to attend Brown and you must withdraw all other pending applications. (That is what I meant by Binding decisions).

So either I accept to Brown and forgo Stanford or I decline Brown's offer for admission and wait for Stanford to accept students from the waiting list.

Anyway, thanks for you help and advice!


I think a lot of MC problems have shortcuts, and it would be nice to know them.  Anyone have any to share?  I'll start, I guess.

You know those grid problems where you have to count how many ways there are of going from one corner to the other going only in two directions?  Well, the old-fashioned way (at least for me) is to count the number of ways to get to each intersection and keep adding.  It's easier, though, to say that there are m+n choose n ways for a m by n grid (Can you see why?).

Any other ideas?  It's true that MC can be as shallow as guess and check, but not always...

Well, your question is way too broad to answer here, but if you have some specific questions, feel free to ask. Here's my Sprint Round strategy from last year:

1. If a problem looks like it'll take a while or be brute force, skip it and come back to it if you have time.
2. Try to have gotten to the end of the test by 25:00, even if you skipped a problem or two on the way.
3. Around 35:00, go back and read the last sentence of every problem. Most stupid mistakes (wrong answer forms, incorrect units, ...) can be caught this way.
4. 39:00 - guess.

Probably a little late, but oh well. Also, MC is not usually as shallow as guess and check - there's almost always a faster way.

-Adam Hesterberg

The question is like a bug is traveling from one corner of a m*n grid to the opposite corner how many way can he do it.
The answer is (m+n)Cn. Reason, no matter how you move, you need to move up m times and move right n times. Which we can transfer into a a sequence problem. Imagine a conquence consist m Us and n Rs. What we wanted to find out corresponds to total number of way of arranging it, or (m+n)Cn


inequalities
geometry
perimeter
search
trigonometry
geometry proposed

Given a triangle $ ABC$.Let $ AM,BN,CP$ be the medians of the triangle $ ABC$. Suppose that$ AM,BN,CP$ intersect the circumircle of the triangle$ ABC$ at $ D,E,F$ respectively .Prove that: $ DE\plus{}EF\plus{}FD \ge AB\plus{}BC\plus{}CA$ :)

[quote="quykhtn-qa1"]Given a triangle $ ABC$.Let $ AM,BN,CP$ be the medians of the triangle $ ABC$. Suppose that$ AM,BN,CP$ intersect the circumircle of the triangle$ ABC$ at $ D,E,F$ respectively .Prove that: $ DE \plus{} EF \plus{} FD \ge AB \plus{} BC \plus{} CA$ :)[/quote]
NO ONE INTERESTED IN THIS INEQUALITY??? :maybe:

More than perimeter they are area and inradii see [url=http://www.mathlinks.ro/viewtopic.php?search_id=482334692&t=85211]here[/url], using lemma in the last post $ (A',B',C') > > (A,B,C)$ we can see $ \sum\sin A'\ge\sum\sin A$ this mean $ p(A'B'C')\ge p(ABC)$ or in your post $ p(DEF)\ge p(ABC)$.


Let $ a, b, c > 0$. Prove that $ 3x\plus{}2y\plus{}4z \ge \sqrt{xy}\plus{}3\sqrt{yz}\plus{}5\sqrt{zx}$.

[hide="Solution"] $ \frac{x \plus{} y}{2} \plus{} \frac{3y \plus{} 3z}{2} \plus{} \frac{5x \plus{} 5z}{2} \ge \sqrt{xy} \plus{} 3 \sqrt{yz} \plus{} 5 \sqrt{zx}$ by AM-GM. [/hide]

[quote="t0rajir0u"][hide="Solution"] $ \frac {x \plus{} y}{2} \plus{} \frac {3y \plus{} 3z}{2} \plus{} \frac {5x \plus{} 5z}{2} \ge \sqrt {xy} \plus{} 3 \sqrt {yz} \plus{} 5 \sqrt {zx}$ by AM-GM. [/hide][/quote]

Yes. It's true. Thank you.

[b]The Arithmetic Mean-Geometric Mean Inequality:[/b]  For positive reals $ r_1, r_2, ... r_n$,

$ \frac{r_1 \plus{} r_2 \plus{} ... \plus{} r_n}{n} \ge \sqrt[n]{r_1 r_2 ... r_n}$.

[b]For two variables:[/b]  $ \frac{x \plus{} y}{2} \ge \sqrt{xy}$.

[quote="t0rajir0u"][b]The Arithmetic Mean-Geometric Mean Inequality:[/b]  For positive reals $ r_1, r_2, ... r_n$,

$ \frac {r_1 \plus{} r_2 \plus{} ... \plus{} r_n}{n} \ge \sqrt [n]{r_1 r_2 ... r_n}$.

[b]For two variables:[/b]  $ \frac {x \plus{} y}{2} \ge \sqrt {xy}$.[/quote]

Som one say it is [color=darkblue][b]Cauchy innequaliti.[/b][/color]

[b]Cauchy[/b], short for [b]Cauchy-Schwartz Inequality[/b],

states that

$ a_1^2\plus{}a_2^2\plus{}a_3^2\plus{}\hdots\plus{}a_n^2)(b_1^2\plus{}b_2^2\plus{}\hdots\plus{}b_n^2) \geq (a_1b_1\plus{}a_2b_2\plus{}\hdots\plus{}a_nb_n)^2$.

Equality occors iff either all $ a_i$ are $ 0$ or there is

a constant $ t$ such that $ b_i\equal{}ta_i$ for all $ i$.

The case the all the $ b_i$ are $ 0$ is covered when $ t\equal{}0$.

Well, some authors [b]do[/b] call AG ineq Cauchy's inequality. That is probably because Cauchy was the first to prove it in its general form.
It is sometimes a cause of misunderstanding, as most authors call Cauchy-Schwarz-Bunjakowsky inequality just "Cauchy's".

Title in ru.wikipedia:

[quote]Неравенство Коши (неравенство о среднем арифметическом и среднем геометрическом)[/quote]

or, translated

[quote]Cauchy's inequality (inequality between arithmetical and geometrical means)[/quote]

Is there a particular reason we were told that a,b,and c are all greater than 0???

That makes sense, especially since square roots are involved...


as you can see here [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=224442[/url], there is this new program, i made this topic so people can test geometric figures and maybe post geometry problems....
 [geogebra]23e4a1f8ec12aec8dddef211d5d5015d24a8f906[/geogebra]

Yeah... geogebra is amazing.

Do you know how to set up coordinate axes on it?  I know that I can just draw two lines, but in order to do that, I need the first line to be parallel (and then I can use the altitude feature to draw the second line) to the sides of my computer screen.  How can I do that?  Also, how come it won't let me name any of the lines I draw with 'y'?  I can call it 'a' or 'line', but it won't let me label it as 'y'.


calculus
integration
function
trigonometry
complex analysis
real analysis
real analysis unsolved

What is answer for $\int {Tan(Tanx)} dx$?
This problem is making rounds in two other math fora claiming this is a very easy problem, whereas my claim is that it is not possible to find by usual methods.

Please post your comment with solution.

You are correct. There is no closed form. You can check on http://integrals.wolfram.com if you wish. It is a useful tool for verifying this sort of question.

its quite simple..

I = int{ tan(tanx) dx}

let tanx =y

therefore
dy/dx = sec^2 x
=> dy/dx = 1+tan^2 x
=> dx = dy/(1+tan^2x) = dy/(1+y^2)

therefore

I = int{(tany/(1+y^2))dy}

you can solve this by using integration by parts.. with tany as the first function and 1/(1+y^2) as the second...

aditya

sorry mistake.. tany would be the second function according to ILATE (inverse, log, arithmatic, trigo and exp)

i think you did not read my answer correctly..

i did not say...

tan(tanx) = tan^2 x

i assumed tanx =y that makes tan(tanx) = tany

aditya :rotfl:

The integral you transform it to is not possible to find an antiderivative to.

Thinking of definite integrals:

Note that $\tan(\tan x)$ has infinitely many singularities in each period of the function. Thinking of this in complex analysis terms, we have infinitely many simple poles in the interval $\left(0,\frac{\pi}2\right)$ and an essential singularity at $\frac{\pi}2.$

The function would not be integrable over any interval whose closure contains any of these singularities. The nearest singularity to zero would be at $\arctan\left(\frac{\pi}2\right).$


calculus
integration
number theory
algebra unsolved
algebra

(1) for what integral value of x such that the expression
(x^2) +13x +44 is a perfect square.
(x^2) means x square.

u should post this in the number theory forum.
anyway, u have onlyto find all the integer solutions of
$ \minus{}7\equal{}t^2\minus{}(2y)^2$ which is easy


geometric series
AoPS Books
geometric sequence

[hide] Set the sum equal to S. Multiplying by 3, we get $3S=1+\frac{2}{3}+\frac{3}{9}+...$. Then we subtract the original equation $S=\frac{1}{3}+\frac{2}{9}+\frac{3}{27}+...$, to get $2S=1+\frac{1}{3}+\frac{1}{9}+...$. Using the formula for infinite geometric series, we get $\frac{1}{1-\frac{1}{3}}=\frac{1}{\frac{2}{3}}=\frac{3}{2}$. So $2S=\frac{3}{2} \longrightarrow S=\boxed{\frac{3}{4}}$. :D  [/hide]

[hide]We can express this as a bunch of individual infinite series (the vertical columns):

+(1/3)
+(1/9)+(1/9)
+(1/27)+(1/27)+(1/27)
...

The sum of an infinite series is $\frac{a}{1-r}$, and all of the infinite series above differ only in the first term, and the first terms makes up a new infinite series. So the sum is

$\frac{\frac{1/3}{1-1/3}}{1-1/3}=(\frac{1}{3})(\frac{3}{2})(\frac{3}{2})=\boxed{\frac{3}{4}}$.[/hide]

[quote="numbergeek"]Arithmetico-Gemoteric series. :)[/quote]

Yes but who remembers the general formula.I usually do the question like the others have and the derivation is basically like that as well.

[quote="numbergeek"]Arithmetico-Gemoteric series. :)[/quote]

Where can I find out more about these types of series? Are they in the AoPS books?

[hide]This is basically an infinite number of geometric sequences in one sequence.

We take note that the sum is basically split like this

$\frac{1}{3}$

$\frac{1}{9}+\frac{1}{9}$

...

So we addup the columns to get

$\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\frac{1}{54}+...$

and this equals

$\frac{3}{4}$[/hide]

In case anyone wants more questions on arithmetico-geometric series-

1)$1+(1+d)\frac{1}{5}+(1+2d)\frac{1}{5^2}+(1+3d)\frac{1}{5^3}+\cdots=\frac{15}{8}$.Find $d$.

2)Generalise the sum as well as $nth$ term of an arithmetico geometric sequence

3)Find $1+5^2x+9^2x^2+13^2x^3+\cdots \infty$,where $|x|<1$

I'd be glad to help if anyone wants the methods.


$f(x) = |x-\left\lfloor x \right\rfloor-\frac{1}{2}|$

What is the period of $f(x)$?

$A.$ $1$
$B.$ $2$
$C.$ $3$
$D.$ $4$
$E.$ $f(x)$ is not a period function

[hide]$f(x)=|x-\left\lfloor x \right\rfloor-\frac{1}{2}|$
$f(x+1)=|x+1-\left\lfloor x+1 \right\rfloor-\frac{1}{2}|=|x-\left\lfloor x \right\rfloor-\frac{1}{2}|$

$f(x)=f(x+1)$ Period is 1 $\boxed{A}$[/hide]


Hey guys,

I'm planning to hold a MathCounts practice for the kids from Dickerson on Wednesday afternoon and was wondering is any of the locals might be available to talk to them. Specifically, I'm looking for someone that has gone to Nationals. (Santosh, Miles, Harrison, someone else?) This is a really talented group. Two of them were the 2 that tied for the Middle School championship at Rockdale and Claus (the 7th grader) believes that he scored a 105 on the AMC12B!) It's quite possible that there may be 3 or 4 potential ARML members (down the road).

I'm planning to run the practice from about 3PM until about 5PM at Walton.

Thanks,
Tom

This is the sort of thing I would love to do and feel like I should be doing, but unfortunately, that's a really awkward time for me.  Specifically, this Wednesday, I will be on a bus for New York at that time. :(   I would also like to help the kids at Mabry (my old middle school).  I wonder if maybe there could be small regional MathCounts practices for interested people in the future, perhaps?

—Miles

I'd be happy to host a regional practice. I'm committed to Dodgen and Dickerson of course, but would like to see all of Cobb County grow. My intent is to host an ARML style practice. Basically working through an entire competition. I hope that this won't be the last practice that we have before state and you would be more than welcome at a future practice. I don't remember where Mabry placed at the chapter competition. (Or if they attended.) I'm pretty sure that Hightower Trail is the other Cobb County team going to state.

Fulton


The triangle ABC has sides AB=3, BC=4, AC=5. The inscribed circle is tangent to AB in C’, BC in A’, and AC in B’. What is the ratio between the areas of the triangles A’B’C’ and ABC ?

How'd you get that?  I got this:
[hide]First, the big triangle is a right triangle.  Then inscribe the circle and label the parts of the sides of the big triangle that are cut by the circle (they are tangents to the circle) x, x, 3-x, 3-x, 4-x, 4-x.  You'll get that the hypotenuse =4-x+3-x, and it also=5, so x=1.  Then with that, you can get the coordinates of the little triangle to be (0,1) (1,0) (1.5,2).  Use the formula A=.5|determinant of vertices| to get an area of 5/4.  That makes the ratio 5/4*6=5/24.[/hide]

I agree with Danbert.

[hide]so AC'=AB', CB'=CA', BC'=BA'

BA'+AC'=4,  AC'+C'B=3,  AB'+B'C=5.

Solve we have BA'=1, A'C=3, AC'=2.

Use [AB'C']=2(2)(sinBAC)/2,  [BA'C']=1(1)/2, [CA'B']=3(3)(sinACB)/2
since sinBAC=4/5,  sinACB=3/5. Sub it back in you get
 [AB'C']+[BA'C']+[CA'B']=24/5, [A'B'C']=[ABC] - 24/5=6/5. Divide.[/hide]

[quote="Wumbate"]coordinates of the little triangle to be (0,1) (1,0) (1.5,2)[/quote]

The third coordinate is not true.

I find the answer of the small triangle to be [hide]6 - 1/2 - 27/10 - 8/5 = 6/5, which makes the ratio 1/5 as Danbert said.[/hide]


geometry
trapezoid
perpendicular bisector

[color=darkblue]Let $ABCD$ be a trapezoid for which $AB\parallel CD$ and $A=D$.
Denote the middlepoint $M$ of the side $[AD]$.
Define the point $P$ so that $PA\perp MB$ and $PD\perp MC$.
Prove that $MP\perp BC$.

[b]Indication.[/b] $X\in PA\cap MB\ ,\ Y\in PD\cap MC\Longrightarrow$ the points $X,Y,B,C$ are conciclycally.

[b][u]Remark.[/u][/b] Maybe there are more proofs for this nice and easy (in my opinion) problem.
I have at least three proofs (syntetically, metrically and analytically).

[b]See[/b][/color] http://www.mathlinks.ro/Forum/viewtopic.php?t=100104

[hide="Hint 1 (for syntetical method)."] $AM^{2}=MX\cdot MB\ ,\ DM^{2}=MY\cdot MC\ ,\ AM=DM\Longrightarrow$
$MX\cdot MB=MY\cdot MC\Longrightarrow$ the points $X,Y,B,C$ are conciclycally.[/hide]
[hide="Hint 2 (for metrical method)."] [b][u]Lemma.[/u][/b] $AB\perp CD\Longleftrightarrow AC^{2}-AD^{2}=BC^{2}-BD^{2}\ .$[/hide]

Sorry Shobber, $A=m(\widehat{BAD})$ and $D=m(\widehat{ADC})\ .$ Here is a particular case of the problem from http://www.mathlinks.ro/Forum/viewtopic.php?t=100104

In the problem posted in: http://www.mathlinks.ro/Forum/viewtopic.php?t=100104

Make $B_{1},A, B_{2}$ be collinear

[quote="Virgil Nicula"][quote="shobber"]It seems that now you are doing a research on this kind of diagram, Virgil!

Since $\angle{A}=\angle{D}$, so easily we can get $PA=PD$. But because the locus of the point with equal distance to A and D is the perpendicular bisector of AD, thus P is on the line and thus $PM \perp BC$.[/quote]No, $PA\ne PD$ ! Read with more attention.[/quote]
 :D I nearly did $PA=PD$ in my solution. Good thing I didn't post it.

[color=darkblue][b][u]A nice equivalent enunciation.[/u][/b] Given are a  triangle $ABC$ and a line $d$ for which $A\in d$.
Define the points: $M\in d$, $BM\perp d$; $N\in d$ $CN\perp d$; the point $P$ so that $PM\perp AB$ and $PN\perp AC$.
Prove that $AP\perp BC$ if and only if $AM=AN$.[/color]


topology
advanced fields
advanced fields unsolved

If $ G$ is a topological group and $ H\triangleleft G$, but $ H$ is not closed in $ G$, is it possible that $ G/H$ with the standard quotient topology is not a topological group? I know that closed gets you $ T_2$, but do you lose $ T_1$ sometimes if $ H$ is not closed?

OK, let's try an example: $ \mathbb{R}/\mathbb{Q}$. The quotient has the indiscrete topology- only two open sets.

hm? of course $ G/H$ is a topological group. it is $ T_1$ iff it is $ T_2$ iff $ H$ is closed.

Ah, that may be the problem. In my definition of topological group it is required to be $ T_1$, so by those biconditionals, $ H$ not closed implies $ G/H$ not $ T_1$ and thus not a topological group. Without requiring $ T_1$, I knew that the quotient would still satisfy the topological group properties. Thanks.


apply the intermediate  value property of continuous functions to show  that the given equation has a solution in the given interval.

X^3 =5 on (1,2)

First you should notice three things:
(1) the function $f(x)=x^{5}$ is continuous;
(2) $f(1)<5$;
(3) $f(2)>5$.

Given (1-3), the intermediate value theorem yields that $f(x)=5$ for some $x$ between $1$ and $2$.


move the sums, its trivial

|i+1|>|i|

square root of 2 to the power of |i| is postive

[quote="Tomaths"]Its trivial

|i+1|>|i|

[/quote]
$i=-1$.Its not trivial at all.


Bonjour tout le monde! :lol:
Je ne peux pas parler le francais tres bien, alors je suis desolee si vous ne pouvez pas bien comprendre ce que j'ecris mais j'ai besoin de votre aide...
Je suis etudiante de dixieme annee dans une ecole canadienne et pour quelque annee maintenant j'etudiais le francais. Pour le projet de l'unite deux "souvenirs d'enfance," on a du (en groupes de six) ecrire un livre d'enfants. Le mienne est 6 pages mais si on ajoute les photos et si on met le font a 14 c'est 30 pages.
De toute facon...peut quelq'un m'aider a verifier ce que j'ai ecrit?
Assez de paroles...svp le projet est pour le 25 janvier mais je dois aller a l'emprimer devant cette date! Et je dois le memoriser aussi...Y a-t-il quelq'un qui peut m'aider?? Peut on m'envoyer un pm et je vais vous envoyer mon adresse d'email.
Merci,
ANI

P.S. La plupart de gens pensent que les canadiens savent parler le francais tres bien comme l'anglais et le francais sont les langues officieles, mais confiez-moi, la plupart des canadiens DETESTENT le francais...au Toronto au moins... :?

Peut-on effacer le derniere message parce que j'ai recu l'aide dont j'avais besoin.
Merci


1. If there were 10 birds on a tree branch, and a hunter shot two thirds of half the birds, how many were remaining?

2. If a shape had four sides and two corners, would it be considered a polygon?

3. $\frac{x}{7}=1$ what is x? lol i dont know why i put this here $\LaTeX$

4. Bob walked into a building and sat down in a chair. A nurse knocked him out, and he was in the waiting room when he came around. He was bandaged and walked away without saying anything. Why?

[quote="ckck"][quote=".:Logic:."][hide]
1) None, some would be dead and fall off the Branch.
The others would fly away due to the gunshots.
2) No
3) 7
4) He had to get surgery. He was knocked out by anesthetics.
Then surgery was done, and when he came around, he was bandaged from the surgery (assume some incision was made) and was in the waiting room, ready to go. :)
[/hide][/quote]
[hide="Better one for 3(i think)"]X is a letter, and the 24th letter of the alphabet.[/hide][/quote]

[hide]Logic, for number 1 and 2 you are right. Ckck for number 3 you are right. No one has solved number 4[/hide][/hide]

[quote="krums3"]4.  By 'knocked out' of the chair, he means he got up to let the nun sit down.[/quote]

Nope  :D  This is a hard one

Original poster,

your first one doesn't make sense for consideration because
(2/3) of (1/2) = (1/3).  And you can't hit 1/3 of 10 birds; you must hit a whole number of birds.

It turns out, they are NOT "easy," so you can drop that word.  (Leave it to the readers to decide.)  You contradicted yourself with your post of:

"Nope  :D This is a hard one"

ckck,

for the 3rd question, you didn't answer the question.

It asked about a lower case ex.  Your response was about a capital ex.  If the poster above you can't get it right, you shouldn't be allowed to get it right either.

Also if the basis of this question is that the "x" asked about is different than the ex in italics in the equation, the original poster's technicality is as much as mine pointed out in the paragraph above this one, IMO.


http://mathworld.wolfram.com/CoinProblem.html they say An explicit solution is known for n=3, where can I find it?

I don't know the exact definition of 'explicit solution', but I was under the impression that there were quick algorithms for n=3, but no closed form. I could be wrong though. I can't seem to find a link outlining the algorithm either.

I found it on CiteSeer:
[url=http://citeseer.ist.psu.edu/cache/papers/cs/11744/http:zSzzSzwww.cs.cmu.eduzSzafszSzcs.cmu.eduzSzuserzSzkannanzSzpubliczSzwwwzSz.zSzPaperszSz.zSzfrobenius.pdf/kannan89solution.pdf]Solution to the Frobenus Problem and Generalizations.[/url]


Let $ \mathsf{Q}$ be a $ (2n \plus{} 1) \times (2n \plus{} 1)$ orthogonal matrix $ (n \in \mathbb{N})$ with $ \det \mathsf{Q} \equal{} 1$. Show that $ \mathsf{Q}$ has a unit eigenvalue. What is a geometric interpretation of this result for a $ 3 \times 3$ matrix?

$ Q$ can be diagonalized over the complex numbers, and all its eigenvalues have modulus $ 1$. if $ Q$ has real entries, the complex eigenvalues come in conjugate pairs whose products are all 1. thus, since $ |Q|\equal{}1$, the product of the real eigenvalues is $ 1$, and since the dimension of $ Q$ is odd, at least one of them must be $ 1$.

Hence $ Q$ fixes at least one line. The geometric interpretation in the $ 3\times 3$ case is that that line is the axis of rotation.

(Real orthogonal matrices with determinant $ 1$ are rotations.)

[quote="pleurestique"]$ Q$ can be diagonalized over the complex numbers, and all its eigenvalues have modulus $ 1$.[/quote]

Sorry, but how do we know this?

By Schur's Lemma, any complex matrix can be unitarily triagularized. That is, there exists $ P$ with $ P^*P\equal{}I$ such that $ P*QP\equal{}T,$ where $ T$ is upper triangular. (I'm using $ A^*$ to mean the Hermitian adjoint, or conjugate transpose, of $ A.$) Then $ Q\equal{}PTP^*$ and $ QQ^T\equal{}QQ^*\equal{}PTP^*PT^*P^*\equal{}PTT^*P^*\equal{}I,$ from which it follows that $ TT^*\equal{}I.$

But the inverse of any upper triangular matrix is upper triangular, which means that $ T^*$ is both upper triangular and lower triangular, which means that $ T$ is diagonal.

Let $ T\equal{}\begin{bmatrix}\lambda_1&&&\\&\lambda_2&&\\&&\ddots&\\&&&\lambda_n\end{bmatrix}.$

From the fact that $ TT^*\equal{}I,$ we get that $ \lambda_j\overline{\lambda_j}\equal{}1$ for each $ j.$

So $ Q$ can diagonalized over the complexes, and $ |\lambda_j|\equal{}1$ for each eigenvalue $ \lambda_j.$

More generally, any normal matrix is unitarily similar to the diagonal matrix of its eigenvalues ($ A$ is a normal matrix iff $ AA^* \equal{} A^*A$).

[quote="Kent Merryfield"]By Schur's Lemma, any complex matrix can be unitarily triagularized. That is, there exists $ P$ with $ P^*P \equal{} I$ such that $ P*QP \equal{} T,$ where $ T$ is upper triangular. (I'm using $ A^*$ to mean the Hermitian adjoint, or conjugate transpose, of $ A.$) Then $ Q \equal{} PTP^*$ and $ QQ^T \equal{} QQ^* \equal{} PTP^*PT^*P^* \equal{} PTT^*P^* \equal{} I,$ from which it follows that $ TT^* \equal{} I.$
[/quote]

I learnt something new here, Schur's Lemma is not yet covered in my course.

Thanks Prof Merryfield and Carcul!


I didn't know where to post this, since it doesn't seem to be what I call a geometry problem. Anyway, here it goes:

In the plane lines $\ell_i,\ i\in\overline{1,n}$ are given, together with a point $P$. We project $P$ on $\ell_1$, then we project this point on $\ell_2$ and so on indefinitely (when we reach $\ell_n$ we go bak to $\ell_1$). Show that the set of points we have obtained is bounded.

The easiest way is to use complex numbers.
Let the points we numbered $P_1, \cdots P_n \cdots$ with affices
 $z_1, \cdots z_n \cdots$.
The we can see that $z_{i+1}=w z_i+ \alpha$ where $|w|<1$ for $w, \alpha$ depending on the line we project.
The if we tale $M=max\{|z|, \frac{\|alpha\|}{1-|w|}\}$ we can see that if $|z_i|<M$ then $|z_{i+1}<M$ hence the points are bounded

Thanks! I did something very similar, but in more detail. I chose an origin on each line and then worked with an upper bound for the distances from the various projections on a line to the origin on that line, but I didn't think it could be made as compact as you have made it.


Generalize the sum $ \displaystyle\sum_{i\equal{}1}^{n}S(i)$, where $ S(k)$ denotes the sum of the digits of $ k$.

Very interesting problem.
[hide="An attempt."]
Let the number of digits in $ n$ be $ m\equal{}1\plus{}\left\lfloor \log{n} \right\rfloor$.

$ \sum_{i\equal{}1}^{n}S(i)\equal{}\left( 45\sum_{j\equal{}0}^m \left\lfloor \frac{n}{10^{i\plus{}1}} \right\rfloor \right) \plus{} \left( \sum_{j\equal{}0}^m \left\lfloor \frac{n}{10^i} \right\rfloor(\mod 10\ \ ) \right)$
[/hide]


abstract algebra
group theory
superior algebra
superior algebra unsolved

(1) A ring $ R\neq 0,1\in R$ is a skew field if and only if every ringt $ R$-module is free.

(2) Show that an abelian group which has no maximal subgroup is divisible.

(3)For an abelian group $ A$ the torsion subgroup T(A) is defiend by

    $ T(A)=\{a|a\in A,\exists z\in Z[z\neq 0,az=0]\}$

    $ A$ is divisible$ \Rightarrow T(A)$ is a direct summand in $ A$.

(2)and (3)is easy
who can prove (1)?prove it please,thanks


hello,

let u_n the last nonnul digit of (n!) .Prove that u_n is
not periodic.

bye.

I'm pretty sure it was solved before here (at least posted :)).


When her three grandchildren came to visit, Grandma decided to give each of them a portion of her change, all of which was quarters. She gave the first grandchild half of all the quarters, plus half a quarter. She gave the second grandchild half of the quarters that were left, plus half a quarter. After that, she gave the third grandchild half the quarters left, plus half a quarter. By the time the grandmother had given the third grandchild his quarters, she had no quarters left. How many quarters did the grandmother have originally?

[hide="hint"][hide="are you sure"][hide="think yourself"]Grandma didn't damage any quarters. What does that tell you about the number of quarters she originally had?[/hide][/hide][/hide]

[hide]1/2 left
times 2 gives you 1
+1/2
1.5
times 2 gives 3
+.5
3.5
7=ans.[/hide]


I wouldn't be able to get this one even if my life depended on it : )
In a circle of radius 10, M is a point on chord PQ such that PM = 5 and MQ = 10.  Chords AB and CD are drawn through M and points X and Y are the respective points of intersection of chords AD and BC with chord PQ.  Given that XM = 3, find MY.  Express your answer as a rational number in lowest terms.

here i'll draw a picture if that helps  :)

[quote="gnosis"]Maybe if u would be more accurate in stating the facts of the problem we could help but from what u sad if AD and BC pass by M which is on PQ then which are X and Y??[/quote]

No, he stated it correctly.  [b]AB[/b] and [b]CD[/b] are the ones that pass through PQ at M, and it's [b]AD[/b] and [b]BC[/b] that pass through at X and Y.

Now, this is one thing that I keep seeing all over the place, and it's really been bugging me that for some reason, I've never heard a thing about what it is.  What exactly is power of a point?

Power of a point is a series of very similar geometry theorums, I'm too lazy to right about them, so instead I'll provide a link. [url]http://mathworld.wolfram.com/CirclePower.html[/url]

[url=http://www.cut-the-knot.org/pythagoras/PPower.shtml]http://www.cut-the-knot.org/pythagoras/PPower.shtml[/url]

this might be easier to understand...

Aaah, now I get it!  Thank you for the links.

I had some of the stuff in that with the angles and such in geometry last year, but we never went over the power of a point theorem that i can remember.

So, using that I got $MY=6$.  Is that correct?

sorry I get the power point theorem but I still don't understand how to get the answer if you don't know AM and CM... or is something else involved?  How is the radius involved?
Thanks for the help!

Well, doesn't power of a point mean that all the triangles in the diagram are similar?  That way, the ratio of $XM: MY$ should be $1: 2$, just like the ratio of $PM: MQ$.  Or did I do something stupid again?

[quote="DanK"]Well, doesn't power of a point mean that all the triangles in the diagram are similar?[/quote]
I don't think so...
It is true that some of the triangles are similar, though.
(For instance, $\triangle AMD\sim\triangle CMB$)


I have a question about a physics problem that I am trying to solve using calculus:

"A projectile is fired at a rate of 60 m/sec, ignoring air friction, what is the maximum possible distance away (on the x-axis) you can be if you want to projectile to go down a "chimney" with dimensions 10 m x 1.8 m (h x w). In other words, it has to be 10 m in the air 1.8 m before hitting to ground."

Let me know if you have an answer or method of solving. I have gotten several equations, but I always have one too many variables for the number of equations I have.

Thanks,
Brett

D/m*m so ,

            10m*1.8m=180m      then,

       180/60=3m/s     so it is 3m/s  away from X-axis
 :D  :D


Let p_0(x) = x^3 + 313x^2 - 77x - 8, and p_n(x) = p_{n-1}(x-n). What is the coefficient of x in p_20(x)?

[quote="jin thynj"]Let $ p_0(x) \equal{} x^3 \plus{} 313x^2 \minus{} 77x \minus{} 8$, and $ p_n(x) \equal{} p_{n\minus{}1}(x\minus{}n)$. What is the coefficient of $ x$ in $ p_{20}(x)$?[/quote]

We first note that for $ n>0$, $ p_n(x)\equal{}p_0\left(x\minus{}\dfrac{n\cdot(n\plus{}1)}{2}\right)$

$ p_{20}(x)\equal{}p_{19}(x\minus{}20)\equal{}p_{18}(x\minus{}39)\cdots \equal{}p_0(x\minus{}(20\plus{}19\plus{}\cdots \plus{}1))$

So $ p_{20}(x)\equal{}p_0(x\minus{}210)\equal{}(x\minus{}210)^3 \plus{} 313(x\minus{}210)^2 \minus{} 77(x\minus{}210) \minus{} 8$

Taking each coefficient of $ x$ in each term we have $ (3\cdot (\minus{}210)^2)\plus{}(313\cdot (2)(\minus{}210))\minus{}77\equal{}\boxed{763}$

Please use Contests (or Resources on MathLinks Skin) to see the answers to previous years' AIME questions.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=461986#461986

Locked.


factor the following:
$12x^2 \cdot y^3 \cdot z +18x^3 \cdot y^2 \cdot z^2-24xy^4 \cdot z^3$

$\displaystyle 12x^2 \cdot y^3 \cdot z +18x^3 \cdot y^2 \cdot z^2-24xy^4 \cdot z^3$

$= 6x{y}^2z \cdot (2xy + 3{x}^2z - 4{y}^2{z}^2)$

i dont think it simplifies further than that

i agree.  I'm too lazy to use Latex[/hide]


Yes, i`m not sure about it. I want to know when it´s right. :lol:

[quote="probability1.01"]Yeah, I see counterexamples.[/quote]

   Where are the counterexamples?

[quote="zhaobin"]are you sure about it[/quote]

   YES, now i am!!! :)

yeah :)
but I think it is another form of hoider inequality.  ;)
if I rememeber correctly

[quote="Lucas Molina"]Prove that $ \frac {(\sum_{i = 1}^{n}a_i)^k}{(\sum_{i = 1}^{n}b_i)^p}\leq{\sum_{i = 1}^{n}\frac {a_i^k}{b_i^p}}$, for $ a_i,b_i,k,p > 0$.[/quote]

  Proof for all $ k > p$:
   By Jensen, let $ f(x) = x^k$, then

$ f(\frac {\sum_{i = 1}^{n}\frac {a_i}{b_i^{\frac {p}{k}}}}{n})\leq\frac {\sum_{i = 1}^{n}\frac {a_i^k}{b_i^p}}{n}$. Then we have that

$ \frac {\sum_{i = 1}^{n}\frac {a_i}{b_i^{\frac {p}{k}}}}{n^k}\leq\frac {\sum_{i = 1}^{n}\frac {a_i^k}{\sqrt [k]{b_i^p}}}{n}$ =>

$ \frac {\sum_{i = 1}^{n}\frac {a_i}{\sqrt [k]{b_i^p}}}{n^{k - 1}}\leq\sum_{i = 1}^{n}\frac {a_i^k}{b_i^p}$

  Then , we have to prove that $ \frac {(\sum_{i = 1}^{n}a_i)^k}{(\sum_{i = 1}^{n}b_i)^p}\leq{\frac {(\sum_{i = 1}^{n}\frac {a_i}{\sqrt [k]{b_i^p}})}{n^{k - 1}}}$

   Then we need to prove that $ \frac {\sum_{i = 1}^{n}a_i}{(\sum_{i = 1}^{n}b_i)^{\frac {p}{k}}}\leq{\frac {\sum_{i = 1}^{n}\frac {a_i}{\sqrt [k]{b_i^p}}}{n^{\frac {k - 1}{k}}}}$.

      See that $ \frac {\sum_{i = 1}^{n}a_i}{\sum_{i = 1}^{n}\sqrt [k]{b_i^p}}\leq{\sum_{i = 1}^{n}\frac {a_i}{\sqrt [k]{b_i^p}}}$, then we going to prove that

$ (\sum_{i = 1}^{n}b_i)^{\frac {p}{k}}\leq{n^{\frac {k - 1}{k}}.\sum_{i = 1}^{n}\sqrt [k]{b_i^p}}$.   It`s proved by Jensen , then we have that $ 2k\geq{p + 1}$ => $ k > p$.

I'm sure there is something wrong with you.
but I don't know where.
let $a_i=1,b_i=1$
$LHS=n^(k-p),RHS=n$,if k>p+1 will be not true.
let $a_i=a,b_i=1$,
I forget the conterexample for $k<p+1$,
and  when $k=p+1$,it is Radon's inequality.

indeed,Radon's inequality is equal to holder inequality.
hope you can get what I said,
and then prove what I said :D

Correct statement (i think):

\hline
H$\ddot{o}$lder version 2: for $u, v \in N$, $u \geq v + 1$, $\sum_{i=1}^n \frac{x_i^u}{y_i^v} \geq \frac{ (\sum_{i=1}^n x_i)^u }{ n^{u-v-1}(\sum_{i=1}^n y_i)^v}$ with equality when $x_i/y_i$ are equal for all i.
\\
\hline


The altitudes though the vertices A, B, C of an acute triangle ABC meet the opposite sides at D, E, F. The line through D parallel to EF meets the lines AC and AB at Q and R, respectively. The line EF meets BC atP. Prove that the cicumcircle of the triangle PQR passes through the midpoint of BC.

Dear Mathlinkers,
this problem has been posted and resolved on this site, but I don't know where.
I have two references:
O.M.  IRAN  (1998)
Hubei Math Contest (1994).
Sincerely
Jean-Louis

This is a problem of IMO shorlist 1997. I have seen it twice in Mathlinks.
[url=http://www.mathlinks.ro/viewtopic.php?t=219812]Circumcircle of triangle PQR passes through midpoint of BC[/url]


search
inequalities unsolved
inequalities

let a,b,c be positive numbers such that $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}. \plus{} \frac {1}{d} \equal{} 4$. . . prove that
$ \frac {\sqrt [3]{a^3 \plus{} b^3}}{2} \plus{} \frac {\sqrt [3]{b^3 \plus{} c^3}}{2} \plus{} \frac {\sqrt [3]{d^3 \plus{} c^3}}{2} \plus{} \frac {\sqrt [3]{a^3 \plus{} d^3}}{2} \plus{} \leq 2(a \plus{} b \plus{} c \plus{} d) \minus{} 4$

Posted before, please search on site or National box.

Here is the link:
http://www.mathlinks.ro/viewtopic.php?p=770259#p770259

[quote="stephen38"]Can anyone tell me how to obtain this:
[quote="TomciO"]My solution from the contest:
Lemma. For positive reals $ x, y$ we have:
$ \sqrt [3]{\frac {x^{3} \plus{} y^{3}}{2}}\leq \frac {x^{2} \plus{} y^{2}}{x \plus{} y}$
[/quote]
I'm so curious to know it.Thanks. :)[/quote]
It is well-known  :wink:


calculus
derivative
trigonometry
real analysis
real analysis unsolved

Someone can tell me is this correct ?
$ f\equal{}x^{2}y\minus{}cos(z)\equal{}0$
$ y \cdot \frac{\partial z}{\partial y}\minus{}x \cdot  \frac{\partial z}{\partial x}\equal{}y \cdot (\minus{}\frac{x^2}{sinz})\minus{}x \cdot
(\minus{}\frac{2xy}{sinz})\equal{}\frac{x^{2}y}{sinz}$

I am sorry that I don't know exactly what you intended.
$ \frac {\partial f}{\partial x}  \cdot \frac {df}{dx} \plus{} \frac {\partial f}{\partial y}  \cdot \frac {df}{dy} \plus{} \frac {\partial f}{\partial z}  \cdot \frac {df}{dz}\equal{}0$     :?:
Would you give me more explanation ?


geometry
geometric transformation
homothety
circumcircle

Three circles $  K, L , M $ tangent externally each other at points $  A , B . C $. Circles  $ K , L$   are tangent at  $A $ and circles  $L , M  $ at  $ B $ . If the line $ AC $ meets circle $ M $ at  $D$  , prove that  $AB$   $\perp BD $

[u]Babis[/u]

  [i]( Based on Darij Grinberg's solution for an other difficult problem) [/i]

very easy.

in triangle ABC we have :arcAC/2+arcAB/2+arcBA/2+arc(BC+CD)/2+arcBC/2=180

but by homothecy its clear that arcAC=arcCD

then we have :AB+BC+AC=90

angle ABD=arcAB/2+arc(BC+CD)/2=arcAB/2+arcBC/2+arcAC/2=180/2=90

I encourage all of you to try for a simpler solution, without  <<...hompthecy >> , something that is unknow to the most ... preolympiad 'ers.Use very  elementary theorems.You ' ll need to drow some auxiliary but very basic segments!!!

   Babis

Because the circumcircle of the triangle $ABC$ is the incircle of the triangle $UVW$, where $U,\ V,\ W$ are the centers of the circles $k,\ l,\ m$ respectively, I offer a equivalent enunciation of this problem:

"The incircle of the triangle $ABC$ touchs the its sides in the points $D\in BC,\ E\in CA,\ F\in AB$. I note the second intersection $L$ between the line $DF$ with the circle $w=C(C,CD)$. Prove that $EL\perp EF.\"$

[b]A solution.[/b]

$m(\widehat {EFL})=m(\widehat {EDL})=\frac 12 (A+B),\ m(\widehat {FLE})=m(\widehat {DLE})=\frac 12 m(\widehat {DCE})$ $=\frac 12 C\Longrightarrow$

$m(\widehat {LEF})=180^{\circ} -\frac 12 (A+B+C)=90^{\circ}\Longrightarrow EL\perp EF.$

Perfect !

This is exactly my solution ,  too!

 Babis

in my solution , u can see( arc AC=arc CD),without using homothecy.

i said homothecy there to make it clear . ;)


geometry
rhombus
incenter
geometric transformation
homothety
similar triangles
geometry unsolved

I am sorry
I cannot figure out...
I have found on wikipedia what "Pascal theorem" is
Is the conclusion of this theorem to the hexagon $ TTD_2D_3D_1D_4$ the following, that the three intersections: $ TT, D_3D_1$, $ TD_2, D_1D_4$ and $ D_2D_3, D_4T$ are on the same line(on the projective plane)?

[quote="Number1"]Let $ O$ be incenter, $ D_1,D_2,D_3,D_4$ toucing point with $ BC,CD,EF,AB$.
Let $ X$ and $ Y$ be polar points for $ TF$ and $ AE$.

By Pascal theorem for degenerate hexagon $ TTD_2D_3D_1D_4$
we get $ X,Y,O$ are collinear.

Then $ AE\parallel{}TF$[/quote]

[quote="LaurT"]Let ABCD be a rhombus. A tangent line to its incircle intersects BC at E and CD at F. The incircle touches AD at T. Prove that AE and TF are parallel.[/quote]

Probably this will help:

 I have seen  a solution(elementary) in Crux ( issues  6 , page 351 ) of this year , using homothecy.

 Babis

[quote="LaurT"]Let ABCD be a rhombus. A tangent line to its incircle intersects BC at E and CD at F. The incircle touches AD at T. Prove that AE and TF are parallel.[/quote]
Dear [b]stergiu[/b], is this the solution you were speaking of?
[hide="Solution"]
Apply Brianchon's theorem to $ ATDFEB$, so $ BD$, $ AF$, and $ ET$ concur.  Furthermore, we have that $ AB$ and $ FD$ are parallel as are $ BE$ and $ TD$.  Combining this with the concurrence, we have that a homothety maps triangle $ ABE$ to triangle $ FDT$ and thus $ FT$ and $ AE$ are parallel.  [/hide]

Let us to present an elementary approach.

We denote the point $ P\equiv BD\cap ET$ and because of $ BC\parallel AD,$ we have that $ \frac {BE}{DT} \equal{} \frac {PE}{PT} \equal{} \frac {PB}{PD}$ $ ,(1)$

Let $ T'$ be, the tangency point of the incircle $ (O)$ of $ ABCD,$ to its side-segment $ CD.$

It is easy to prove ( by an angle changing in the configuration of the triangle $ \bigtriangleup CEF,$ with the circle $ (O)$ as its $ C$-excircke ) that $ \angle BOE \equal{} \angle DFO$ and so, from the similar triangles $ \bigtriangleup BEO,\ \bigtriangleup DOF,$ we have $ \frac {BE}{DO} \equal{} \frac {BO}{DF}$ $ \Longrightarrow$ $ (BE)\cdot (DF) \equal{} (DO)^{2}$

$ \Longrightarrow$ $ (BE)\cdot (DF) \equal{} (DT')\cdot (DC)$ $ ,(2)$ $ ($ because of $ (DO)^{2} \equal{} (DT')\cdot (DC)$ from the right triangle $ \bigtriangleup OCD$ $ ).$

From $ (2)$ $ \Longrightarrow$ $ \frac {BE}{DT'} \equal{} \frac {DC}{DF}$ $ \Longrightarrow$ $ \frac {BE}{DT} \equal{} \frac {AB}{DF}$ $ ,(3)$

From $ (1),$ $ (3)$ $ \Longrightarrow$ $ \frac {AB}{DF} \equal{} \frac {PB}{PD}$ $ ,(4)$

From $ (4)$ and because OF $ AB\parallel DF,$ we conclude the collinearity of the points $ A,\ P,\ F$ and so, we have that $ \frac {PB}{PD} \equal{} \frac {PA}{PF}$ $ ,(5)$

Hence, from $ (1),$ $ (5)$ $ \Longrightarrow$ $ \frac {PE}{PT} \equal{} \frac {PA}{PF}$ $ \Longrightarrow$ $ AE\parallel TF$ and the proof is completed.

Kostas Vittas.

Hi everyone.
I know that Vittasko,Number 1, Quatto Master and Babis have given brilliant proofs for this problem. However I was just wondering if an inversive solution is feasible here. Since we know that in triangle $ ABE$ and $ TFD$ we have $ AB$ parallel to $ DF$ and $ BE$ parallel to $ TD$ when we invert the whole figure with respect to the incircle of the rhombus we would get a family of circles passing through $ O$(the center of the incircle of the rhombus.) many of which are tangent circles in pairs. Now we can use this to prove that under the inversion lines $ AE$ and $ TF$ are transformed into tangent circles the point of tangency being $ O$ the center of incircle of the rhombus, since this would mean that lines $ AE$ and $ TF$ are parallel.
Could anyone help me with a rigorous proof using this approach or inversion in general.

Thanks in advance.

Hi everyone,
Can someone please help me or otherwise give an inversive solution to this problem.

Here is my solution:
Let U,V,W,M be the points of tangency of the circle to BC, CD, AB, and EF, O the center of the circle with radius r, let $ WB \equal{} BU \equal{} VD \equal{} DT \equal{} x$, and let $ \angle WOB \equal{} \angle BOU \equal{} \alpha$. Then, we have to show $ \triangle ABE$ and $ \triangle TFD$ are similar, or $ \frac {x}{x \plus{} VF} \equal{} \frac {x \plus{} UE}{x \plus{} AW}$, or $ x \cdot AW \equal{} x(VF \plus{} UE) \plus{} VF \cdot UE$. But note that $ \triangle AOB$ is a right triangle with altitude OW, so $ x \cdot AW \equal{} r^2$. So we have to show $ r^2 \equal{} x(VF \plus{} UE) \plus{} VF \cdot UE$.
Note that $ \angle WOV \equal{} 180$, so $ \angle UOV \equal{} 180 \minus{} 2 \alpha$. But $ \angle UOE \equal{} \angle EOM$ and $ \angle MOF \equal{} \angle FOV$, so $ \angle EOF \equal{} (1/2) \angle UOV \equal{} 90 \minus{} \alpha$. Thus, $ \frac {r}{x} \equal{} cot \alpha \equal{} tan(90 \minus{} \alpha) \equal{} tan EOF \equal{} tan(EOM \plus{} MOF) \equal{} \frac {\frac {UE}{r} \plus{} \frac {VF}{r}}{1 \minus{} \frac {UE \cdot VF}{r^2}}$. But
$ \frac {r}{x} \equal{} \frac {\frac {UE}{r} \plus{} \frac {VF}{r}}{1 \minus{} \frac {UE \cdot VF}{r^2}}$ is equivalent to what we wanted to show.


given that $a^{2}+b^{2}=ab$, find $\frac{a^{5}}{b^{5}}$

wait a minute...I don't think I'm doing anything wrong, but plugging the answer back into the equation gives an incorrect eequation ($2a^{2}=-a^{2}???$)  :blush:
I started with $a^{3}+b^{3}=0$, then divided by $(a+b)$ to get $a^{2}+b^{2}-ab=0$, which can be rearranged into the original equation.  this means that $\frac{a^{5}}{b^{5}}=\frac{a^{3}}{b^{3}}=-1$
[b]WHAT AM I DOING WRONG?????[/b]

Unless I am wrong:

$\frac{a^{5}}{b^{5}}= \frac{1}{2}\pm \frac{\sqrt{3}}{2}i$

ok I figured out what I did wrong  :D If you want, I will leave it as a challenge to try and figure out by yourself

if you just want to know,
[hide]going through my solution backwards is fine, but dividing by $(a-b)$ is equal to dividing by zero when $a=-b$.  This invalidates the equation[/hide]

Isn't it when $a=b$?

[hide]Shouldn't the answer be $1,-1$?[/hide]


When not on the moving sidewalk, Kelsey can walk the length of the sidewalk in 3 minutes. If she stands on the sidewalk as it moves, she can travel the length in 2 minutes. If Kelsey walks on the sidewalk as it moves, how many minutes will it take her to travel the same distance? Assume she always walks at the same speed, and express your answer as a decimal to the nearest tenth.

[hide="answer"]k = speed of Kelsey
s = speed of sidewalk
l = length of sidewalk
$k = \frac{l}{3}$
$s = \frac{l}{2}$
$k+s = \frac{l}{3} + \frac{l}{2}$
$k+s = \frac{5l}{6} = \frac{l}{\frac{6}{5}}$
So she can walk down the sidewalk in 6/5 minutes, or [color=green]1.2 minutes[/color][/hide]

[hide]
When walking not on the sidewalk, she goes $\frac 13$ in a minute.
When standing, or the speed of the sidewalk is $\frac 12$ per minute.
Thus she can do $\frac 13 + \frac 12= \frac 56$ per minute when walking on the sidewalk moving.
It then takes her $\frac 65$ minutes = $1.2$[/hide]

I agree with those answers.
A moving sidewalk is just an escalator but flat.
I was on one at the airport. They move really slow.

While waiting for our flight from Detroit to Atlanta, I wanted to know how long it would take me to walk backwards on the moving sidewalk without actually doing it. So I used the same process being the genius penguin I am. Then I decided to check my answer by being an idiot. My teacher said that you would never need to do this kind of problems in real life.

they could at least say 'moving flat escalator' in the question, i'm like that would be pretty weird if you were standing on the street and the sidewalk started moving. :)

[quote="ritchjp"]It would if you lived in LA. Of course everything would move with it.[/quote]

Haha that's where I live.  :rotfl:


algebra
polynomial
AoPS Books
abstract algebra

Hi,

I've recently gotten into problem solving as a hobby.  Its really fun (though I'm kinda late to the game, junior in high school  :(  ).  Anyways, I encountered this problem and found the answer (C), but I kinda brute forced it.  Does anyone know of a more, how should I put this, elegant way of obtaining a solution?

10. Which of the following divides $ n^3 \plus{} 11n$ for all positive integers $ n$?
A. 5  B. 7  C. 6  D. 4  E. 9

Thanks!

I'm not sure what you define as brute-forcing because I just played around with mods and parity to arrive at a solution. First after factoring the expression I obtained $ n(n^2 \plus{} 11)$. If n is even then the expression always divisible by 2 because the n on the outside is divisible by 2. If n is odd then $ n^2 \plus{} 11$ is always divisible by 2. Then I considered the residues of $ n     modulo 3$. By plugging in the possible residues of 0,1,and 2 modulo 3 into $ n$ and their respective squares into $ n^2$ I found by considering each of the three cases that the expression is always divisible 3, which gives C as an answer. I don't think there is any more elegant way than factoring and using modular arithmetic(which really didn't take me too long). But I could be wrong.

Plug in $ n \equal{} 1$ to eliminate A, B, and E, then $ n \equal{} 2$ to eliminate D. :D

[hide="Simple strategy"] Test out a few values to see what the biggest number [i]could[/i] be.  Once you think you've got it, try to prove it.  Then you know it can't be a bigger number. [/hide]
[hide="Complicated strategy"] A polynomial of degree $ k$ has at most $ k$ roots $ \bmod p$ for $ p$ prime.  You might not know what this means, but essentially it means that since the expression you're given has degree $ 3$, the only primes involved in the answer are $ 2$ or $ 3$. [/hide]

[quote] Simple strategy
Test out a few values to see what the biggest number could be. Once you think you've got it, try to prove it. Then you know it can't be a bigger number.

Complicated strategy
A polynomial of degree k has at most k roots \bmod p for p prime. You might not know what this means, but essentially it means that since the expression you're given has degree 3, the only primes involved in the answer are 2 or 3. [/quote]

Thanks!

@Brut3Forc3 and phi-unit
By brute forcing, I meant what Brut3Forc3 said (haha how ironic  :rotfl:  ).  I was just wondering if there was some way to do it like t0rajir0u posted.  I like to see the principles behind solutions, I know that plugging in numbers will give me the same answers, but this is just cooler  :P .  I don't really know how the complicated solution works but sounds interesting.  See, I'm planning to get into problem solving as best as I can but as I said, I am late to the game.  How much will the AoPS books help with problems like this?  I've borrowed Vol. 1 and Vol. 2 from a friend who bought them but never used them.  Do they cover stuff like this?  If not, what books do?

Thanks again!

@t0rajir0u:

I considered your strategy, it's pretty interesting. My interpretation for this is: we can always find a congruent polynomial of the original one $ \bmod p$, suppose it's $ (n \minus{} x_1)(n \minus{} x_2)...(n \minus{} x_k)$, which can only take at most $ k$ distinct roots $ \bmod p$, i.e. be a multiple of $ p$. Therefore, to make it divisible by $ p$ for all $ n$, we must have $ p\leq k$. Is this correct?

Essentially.  The theorem is called [url=http://en.wikipedia.org/wiki/Lagrange%27s_theorem_%28number_theory%29]Lagrange's theorem[/url].


function
LaTeX
parameterization
trigonometry
calculus
derivative
calculus computations

Suppose a smooth curve in $ \mathbb{R}^2$ passes through the origin. Let $ s$ denote arclength as measured along the curve, with $ s > 0$ to the right of the origin and $ s < 0$ to the left.

Fix a constant $ k > 0.$ Suppose that the curve satisfies the property that $ y \equal{} ks^2.$

Can you find a more conventional representation (such as parametric form) for this curve? And does the curve have some familiar name?

As a possible intermediate step: Show that there is a particular number (which depends on $ k$) such that the arclength of the entire curve cannot exceed this number. In other words, this is a curve of finite length and bounded extent.

The total length of the curve is 1/2k.

y=ks^2 gives s as a function of y. Differentiating with respect to x, we get an equation in dy/dx

The equation is,

{dy/dx}^2 = 1/{1/4ky - 1}.  This puts a bound on y: y<1/4k. this means, s^2 < 1/4k^2. i.e., s<1/2k.

the above equation is solved by the substituting, y= 1/{t^2 + 4k}.

The final solution turns out to be,

  x= sqr.root{y/4k - y^2 }  -  (1/4k)tan^-1 {sqr.root(1/4ky - 1)}  + pi/8k

I am not able to get this in the form, y=f(x).

One small comment: we actually have $ \minus{} \frac1{2k}\le s\le\frac1{2k},$ with the curve extending on both sides of the origin, so the total length of the curve is $ \frac1k.$

It's possible to get a nicer looking form than that. Not $ y \equal{} f(x)$ but rather a pair of parametric equations, $ x \equal{} x(t),y \equal{} y(t).$

Please learn a bit of [[LaTeX]].  (See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]here[/url].)

We see that taking $ y$ as a function of $ s$, $ y'(s) \equal{} 2k s$, so $ |2ks| \leq 1$.  This suggests the parametrization $ s \equal{} \frac {\sin t}{2k}$, so $ y \equal{} \frac {\sin^2 t}{4k}$ and $ \frac {dy}{dt} \equal{} \frac {\sin t \cos t}{2k}$.  We have (taking all derivatives w.r.t. $ t$) that $ s'(t)^2 \equal{} y'(t)^2 \plus{} x'(t)^2$, so $ \frac {\cos^2 t}{4k^2} \equal{} \frac {\sin^2 t \cos^2 t}{4k^2} \plus{} x'(t)^2$, and this gives $ x'(t)^2 \equal{} \frac {\cos^4 t}{4k^2}$.  Then (because I don't want to think about signs right now) we conclude from this that $ x'(t) \equal{} \frac {\cos^2 t}{2k}$, and from here we can integrate to get the formula for $ x$.  A few details to be patched up, but is this the sort of thing you were looking for?

(Admission: I actually didn't think of the parametrization for $ s$ in the way I described in the second sentence, but rather because I started by parametrizing by arclength, which gives $ y'(s) \equal{} 2k s$ and so $ x'(s) \equal{} \sqrt {1 \minus{} 4k^2 s^2}$.  This spits out an arcsine when you integrate, and [i]that's[/i] what got me to the substitution I used above.)

It is a pretty famous curve. It got quite a bit of attention throughout the 17th century.

substituting cot^2 t = 1/4ky - 1 in the solution mentioned in my previous post, the parametric equations turn out to be:
         x = (1-cos t)/8k; y = (t+ sin t)/8k

[quote="bharathhm"] x = (1-cos t)/8k; y = (t+ sin t)/8k[/quote]
Now, this is starting to look familiar. Does anyone recognize the curve from that description?

This is the cycloid oriented so its cusps point upward.

Athough the cycloid has many uses and appears in many problems, expressing it in exactly the form I started with here emphasizes the tautochrone property. (Had jmerry not already posted that particular link, I was planning to do so myself.)

What is the motion of a bead sliding on this wire without friction? Gravitational potential energy is proportional to $ y$ and $ y$ is in turn proportional to $ s^2.$ This is precisely the setup needed to have $ \frac{d^2s}{dt^2}\equal{}\minus{}cs,$ which is the equation of simple harmonic motion. And in simple harmonic motion, frequency is independent of amplitude, which is simply another way to state the tautochrone property.


hi,

i am currently doing some coursework of which requires me to need to find out what it means, can you help. the table below is what i have for information.

[img]http://www.redihot.com/images/english_03.gif[/img]

thanks alot, pleae click it to see it better.

thanks alot for not helping, although i found out, the 100 represents the average so the mirror is less meaning it is less popular than the average newspaper

Errr, if you wanted help, youd really have to give more information. What the hell are we supposed to be getting the mean of exactly?

just so you know, what you found is stated on the top left corner where it says

"index: 100=UK national average"

that's about all i see that you're gonna get from the table, unless if there's some more information you have, i.e specifically what you're supposed to do.


How many triangles $ ABC$ are there if radius of the inscribed circle of the triangle $ ABC$ is $ 1$ and all sides are integer?

Let $ BC\equal{}a,AB\equal{}c,AC\equal{}b$ where $ a\leq{b}\leq{c}$ are integers. If $ r\equal{}1$ then $ a,b,c$ satisfy $ (b\plus{}c\minus{}a)(a\plus{}b\minus{}c)(a\plus{}c\minus{}b)\equal{}4(a\plus{}b\plus{}c)$.
Moreover $ a\leq{12}$.
Proof: $ 4(a\plus{}b\plus{}c)\leq{1}2c$ and $ (b\plus{}c\minus{}a)(a\plus{}b\minus{}c)(a\plus{}c\minus{}b)\geq{ac}$. Thus $ ac\leq{12c}$ and $ a\leq{12}$.
For $ a\leq{b}\leq{c}\leq{3000}$ there is only one solution: $ 3,4,5$.

Another note is that when this occurs(inradius as 1)

K=s  where s is the semiperimeter of the triangle

Moreover $ c\leq{b}\plus{}12$.
Proof: $ c\leq{a}\plus{}b\leq{b}\plus{}12$.
For $ a\leq{b}\leq{c}\leq{1}0^5$ there is only one solution: $ 3,4,5$.

[quote="loup blanc"]Let $ BC \equal{} a,AB \equal{} c,AC \equal{} b$ where $ a\leq{b}\leq{c}$ are integers. If $ r \equal{} 1$ then $ a,b,c$ satisfy $ (b \plus{} c \minus{} a)(a \plus{} b \minus{} c)(a \plus{} c \minus{} b) \equal{} 4(a \plus{} b \plus{} c)$.
Moreover $ a\leq{12}$.
Proof: $ 4(a \plus{} b \plus{} c)\leq{1}2c$ and $ (b \plus{} c \minus{} a)(a \plus{} b \minus{} c)(a \plus{} c \minus{} b)\geq{ac}$. Thus $ ac\leq{12c}$ and $ a\leq{12}$.
For $ a\leq{b}\leq{c}\leq{3000}$ there is only one solution: $ 3,4,5$.[/quote]
Just let $ x \equal{} b \plus{} c \minus{} a,y \equal{} a \plus{} b \minus{} c,z \equal{} a \plus{} c \minus{} b$ then $ xyz \equal{} 4(x \plus{} y \plus{} z)$.
Wlog $ x \le y \le z$. If $ x \ge 4$ then: $ 4(yz \minus{} 1) \le x(yz \minus{} 1) \equal{} 4(y \plus{} z) \iff yz \minus{} 1 \le y \plus{} z \iff (y \minus{} 1)(z \minus{} 1) \le 2$ which is impossible. So $ x \in \{1,2,3\}$.
Since $ a \plus{} b \minus{} c \equiv a \minus{} b \plus{} c \equiv \minus{} a \plus{} b \plus{} c \bmod 2$ we see $ x \equiv y \equiv z \bmod 2$. Since $ 4 \mid xyz$ we see $ 2 \mid x,y,z$ and therefore $ x \equal{} 2$. Inserting:
$ 2yz \equal{} 4(y \plus{} z \plus{} 2) \iff yz \equal{} 2(y \plus{} z \plus{} 2) \iff (y \minus{} 2)(z \minus{} 2) \equal{} 8$, and therefore $ y \equal{} 4,z \equal{} 6$, which in turn gives $ a \equal{} 5,c \equal{} 4,b \equal{} 3$. So the only triangle is the $ 3,4,5$ triangle.

Hi Mathias,
you have not $ x(yz\minus{}1)\equal{}4(y\plus{}z)$.

[quote="loup blanc"]Hi Mathias,
you have not $ x(yz \minus{} 1) \equal{} 4(y \plus{} z)$.[/quote]
I'm sorry for my silly mistake! It should be $ x(yz\minus{}4) \equal{} 4(y\plus{}z)$. The inequality then becomes $ (y\minus{}1)(z\minus{}1) \le 5$ which is impossible too :)

[quote="Mathias_DK"][quote="loup blanc"]Hi Mathias,
you have not $ x(yz \minus{} 1) \equal{} 4(y \plus{} z)$.[/quote]
I'm sorry for my silly mistake! It should be $ x(yz \minus{} 4) \equal{} 4(y \plus{} z)$. The inequality then becomes $ (y \minus{} 1)(z \minus{} 1) \le 5$ which is impossible too :)[/quote]
[quote]Couldnt understand this. $ 4(yz \minus{} 1) \le x(yz \minus{} 4)$[/quote]

Hi ertanrock, reread posts #6,#7.
$ 4(yz\minus{}4)\leq{x}(yz\minus{}4)\equal{}4(y\plus{}z)$ implies $ (y\minus{}1)(z\minus{}1)\leq{5}$ that is impossible because $ 4\leq{y}\leq{z}$.

[quote="loup blanc"]Hi ertanrock, reread posts #6,#7.
$ 4(yz \minus{} 4)\leq{x}(yz \minus{} 4) \equal{} 4(y \plus{} z)$ implies $ (y \minus{} 1)(z \minus{} 1)\leq{5}$ that is impossible because $ 4\leq{y}\leq{z}$.[/quote]

ok now. Got it.
I thought like that. $ 4(yz \minus{} 1)\leq{x}(yz \minus{} 4)$


Heron triangles are triangles with integer sides and integer area.They are Primitive if their sides taken together have no common factors(not necessarily pairwise)
1) Prove there are infinitely many primitive Heron Triangles two of whose sides have a common factor of 5.
(example:(9,65,70),)
2)determine a primitive heron triangle two of whose sides have a common factor of 65
3)prove or disprove the existence of a primitive heron triangle each of whose sides is a square? :D

[quote="anorithdialga"]any one has a solution?[/quote]

yes,   anyone has a solution

coincidentally, i m reading the same only,ill respond in 4 long days!!
edit:wastage of post i think

Ok this is the first one:
[hide="2"]Every right triangle is a heron triangle, so their sum or difference should be also a heron triangle, thus forming acute and obtuse heron triangles. Here, we have to get two right triangles with sides multiple of 65. A right triangle with hypotenuse 65 gives options as (63,16,65) and (33,56,65). Choosing any one and creating an isosceles triangle with it does the job.[/hide]


$ \frac{a}{b}\equal{}\sqrt{2}\plus{}1$. Then the following expression value is ?

$ \frac{a^{3}\plus{}b^{3}}{a^{3}\minus{}a^{2}.b\plus{}a.b^{2}}$

$ a^3 \plus{} b^3 \equal{} (a \plus{} b)(a^2 \minus{} ab \plus{} b^2)$

$ a^3 \minus{} a^2b \plus{} ab^2 \equal{} a^3 \minus{} ab(a \minus{} b) \equal{} a\left(a^2 \minus{} ab \plus{} b^2\right)$

$ \frac {a^{3} \plus{} b^{3}}{a^{3} \minus{} a^{2}.b \plus{} a.b^{2}} \equal{} \frac {(a \plus{} b)(a^2 \minus{} ab \plus{} b^2)}{a\left(a^2 \minus{} ab \plus{} b^2\right)}$

$ \equal{} \frac {a \plus{} b}{a}$

$ \equal{} b\left(\frac {\frac {a}{b} \plus{} 1}{a}\right)$

$ \equal{} \left(\frac {\frac {a}{b} \plus{} 1}{\frac {a}{b}}\right)$

$ \equal{} \frac {\sqrt {2} \plus{} 2}{\sqrt {2} \plus{} 1}$

Thanks, but both of the answers are correct ?

[quote="luiseduardo"]$ \frac {a}{b} \equal{} \sqrt {2} \plus{} 1$. Then the following expression value is ?

$ \frac {a^{3} \plus{} b^{3}}{a^{3} \minus{} a^{2}.b \plus{} a.b^{2}}$[/quote]

Here is my solution:

$ \frac {a^{3} \plus{} b^{3}}{a^{3} \minus{} a^{2}.b \plus{} a.b^{2}}\equal{}\frac{(a\plus{}b)(a^2\minus{}ab\plus{}b^2)}{a(a^2\minus{}ab\plus{}b^2)}\equal{}1\plus{}\frac{b}{a}$

$ \equal{}1\plus{}\frac{1}{\sqrt{2}\plus{}1}\equal{}1\plus{}\sqrt{2}\minus{}1\equal{}\boxed{\sqrt{2}}$.


geometry
perimeter
geometric transformation
dilation
trigonometry
number theory
relatively prime

The perimeter of triangle $APM$ is $152,$ and the angle $PAM$ is a right angle. A circle of radius $19$ with center $O$ on $\overline{AP}$ is drawn so that it is tangent to $\overline{AM}$ and $\overline{PM}.$ Given that $OP=m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

is the answer:

[hide]

$\frac{m}{n}= \frac{7}{8}\implies m+n = 15$?

where is this from?

[/hide]

this problem was more: "can you simplify algebra" rather than any cool geometry...

anyway, denote $OP=x$, $PT=y$, dialate by 1/19, denote the circle tangent to $PM$ at $T$, so $OT=OA=1$, and $p=8$

$\triangle OTP\sim\triangle MAP$, so

$\frac{1}{MA}=\frac{x}{MP}=\frac{y}{x+1}=\frac{1+x+y}{8}$

so $(7-x)y=(x+1)^{2}$, and $1+y^{2}=x^{2}$, so $(x-7)^{2}[(x-1)(x+1)]=(x+1)^{4}$, since this is easily expanded, and solved, we obtain $x=\frac{5}{3}$, then undo the dialation, so $OP=\frac{95}{3}\implies m+n=\boxed{098}$

[quote="Altheman"]this problem was more: "can you simplify algebra" rather than any cool geometry...
[/quote]
[color=red]A cool problem [/color]:D
 [hide]
Let restate the problem. Will just change the letters and will dilate by 19 (see Altheman solution). Triangle $ABC$ is right ($A=90$) with perimeter of 8. Circle with center $D$ on $AC$ touches $AB$ and $BC$ and has radius 1. Find $DC$.
 Construct $GF$ parallel to $AC$ and tangent to the circle.
Let $AB=x$. The perimeter of $\Delta BFG=BF+FE+EG+BG=BF+FH+GA+BG=2x$ as $\Delta BFG \sim \Delta BAC$ we have $\frac{BG}{BA}=\frac{2x}{8}$
$\frac{x-1}{x}=\frac{2x}{8}\Rightarrow x^{2}-4x+4=0$ or $x=AB=2$.
$\tan{DAB}=\frac{DA}{BA}=\frac12$. As $CD$ is angle bisector $\tan{ABC}= \frac43$ (by double angle formula).
So $\frac43=\frac{AC}{AB}$, $AC=\frac83$, $DC=\frac53$ [/hide]

excuse me, but what does the p mean in your solution?
denote the circle tangent to PM at T, so OT=OA=1, and p=8

In Altheman's solution, $ p$ represents the [i]perimeter[/i] of $ \triangle APM$.

[hide=Solution]We see that the circle can only be tangent at $A.$ Suppose it it tangent at $X$ instead. Then $OX \perp AM,$ but $OA\perp AM.$ Thus, $OA \parallel OX.$ But they meet at $O,$ so this is a contradiction. Therefore, circle $O$ is tangent to $AM$ at $A,$ and $OA=19.$

Call the point where $O$ meets $PM,$ $Y.$ Note that $OY=19.$ Then, let $x=AM=MY$ ($AM=MY$ because they are tangents from $M$), and $OP=a.$ By Pythagoras, $PY = \sqrt{a^2-19^2}.$

$\triangle POY \sim \triangle PMA$ by AA similarity because they share $\angle APM$ and $\angle PAM = \angle PYO = 90.$ Hence, $AP/AM=PY/OY,$ or $\frac{a+19}{x}=\frac{\sqrt{a^2-19^2}}{19}.$ Solving for $x,$ we have $x=\frac{19a+19^2}{\sqrt{a^2-19^2}}.$

We know that the perimeter is $152,$ which means that $a+19+2x+\sqrt{a^2-19^2}=152.$ Substituting our value for $x,$$$a+19+\frac{38a+2\cdot19^2}{\sqrt{a^2-19^2}}+\sqrt{a^2-19^2}=152.$$Multiplying by $\sqrt{a^2-19^2},$$$(a+19)\sqrt{a^2-19^2}+38a+2\cdot 19^2+a^2-19^2=152\sqrt{a^2-19^2}.$$Moving the square roots to the LHS and everything else to the LHS and factoring,$$(a+19-152)\sqrt{a^2-19^2}=-(a+19)^2.$$Squaring and using difference of squares,$$(a-7 \cdot 19)^2 (a+19)(a-19)=(a+19)(a+19)^3.$$Canceling and expanding (and moving to one side),$$9a^2-570a+19^2\cdot 50 =0.$$Using the quadratic formula, the only solution is $\frac{95}{3}\implies \boxed{098}.$ [/hide]

[hide=Solution]Scale down by $19$; then the radius is $1$ and the perimeter is $8$. Let $T$ be the tangency point between the circle and $\overline{PM}$. Let $x=OP$. Since $\overline{OM}$ bisects $\angle PMA$, we can use the angle-bisector theorem to get $AM=k$ and $PM=kx$. Additionally, $TM=AM=k$.

Because the perimeter is $8$, we have $kx+k+x+1=8 \implies k(x+1)=7-x \implies k = \frac{7-x}{x+1}$.

Since $PT=kx-k$, we use the Pythagorean Theorem on $\triangle PTO$ to get:
$$(kx-k)^2+1=x^2$$
$$k^2(x-1)^2 = x^2-1$$
$$k^2(x-1) = x+1$$
$$\left(\frac{7-x}{x+1}\right)^2(x-1)=x+1$$
$$(x-7)^2(x-1)=(x+1)^3$$
$$x^3-15x^2+63x-49=x^3+3x^2+3x+1$$
$$18x^2-60x+50=0$$
$$9x^2-30x+25=0$$
$$(3x-5)^2=0$$
$$x=\frac53$$
Scaling up, we get the answer $\frac{95}{3}\implies\boxed{098}$.

Scale down by $19$, and reflect $M$ across $AP$ to $M'$ so that said circle is the incircle of $\triangle MM'P$. Let $(p,m):=(PA,MA)$.
[list=1]
[*] The perimeter condition becomes $p+m+\sqrt{p^2+m^2}=8$. [/*]
[*] The radius condition becomes, by $A=rs$,
\[pm=1(AM+PM)=\text{Perimeter}(\triangle APM)-AP=8-p.\] [/*]
[/list]
Equation (1) becomes
\begin{align*}
p^2+m^2&=(8-p-m)^2\\
&=(p+m)^2-16(p+m)+64\\
\implies32&=8(p+m)-pm.
\end{align*}
Substituting $pm=8-p$ and $m=\tfrac{8-p}{p}$ from Equation (2),
\begin{align*}
8(p+\tfrac{8-p}{p})-(8-p)&=32\\
8(p^2+(8-p))-p(8-p)&=32(8-p)\\
9p^2-48p+64&=0.
\end{align*}
So $p=\tfrac{8}{3}$ or $OP=\tfrac{5}{3}$, giving $\tfrac{95}{3}$.


[quote="mr. math"]In a room, each 22 liters of air weighs 29 grams. Each liter is 100 cubic centimeters. How much does a cubic meter of air in that room weigh? Express your answer to the nearest whole gram.[/quote]
Each liter by the way, it 1000 cubic centimeters since 1 cubic centimeter is 1 milliliter.
[hide] $1 m^{3}=(100)^{3}cm^{3}=1000000 cm^{3}$.  We also know that 1 Liter=1000 cc (cc is cubic centimeters), so 1 cubic meter is (1000)(1000) cc=1000 liters.  Thus, since each liter weighs 29 grams, so 1000 liters weighs 29 kilograms, which is our answer. [/hide]


Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ,$ find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}.$

[quote="1234567890"]I got that exact same answer, but I didn't know AIME answers can be 0, or can they? :)[/quote]

Yeah, they can. It's 000-999. 1000 choice MC test? :P

the question ask for "the least integer that is GREATER than......", not "not less than", so -1 can greater than -1, the answer is 0

The question is amazingly easy (like 10 sec) if you know that the answer can ONLY be -1, 0 or 1 (AIME only 000 or 001), no matter what inteter power z is taken to and no matter the angle degree (in this situation its 0)...

there is a generalized formula in AoPS vol 2 for these types of problems

[quote=Elemennop][hide]

Let , then , so , so , so then the given equation evaluates to , and the answer is .

[/hide][/quote]
huh?

w0W that revive.

Probably displaying issues from so old. That's why it looks so clipped and stutter-y.

[quote=JesusFreak197][hide="Answer"]We have $z=e^{i\theta}$, so $e^{i\theta}+\frac{1}{e^{i\theta}}=\frac{\cos \theta}{\cos ^2\theta +\sin ^2\theta}+\cos \theta +i\sin \theta - \frac{i\sin \theta}{\cos ^2\theta +\sin ^2\theta}=2\cos \theta$. Therefore, $\theta =3^\circ=\frac{\pi}{60}$. From there, $z^{2000}+\frac{1}{z^{2000}}=e^{\frac{100\pi (i)}{3}}+e^{\frac{-100\pi (i)}{3}}=2\cos \frac{4\pi}{3}=-1$, so our answer is $\boxed{0}$.[/hide][/quote]

Isn't Euler identity excluded from most content sets?

I just realized I bumped a 9 year old topic oops

I think this is much nicer to look at.

[hide]
Let $z=r(\cos\theta+i\sin\theta).$ The crucial claim is that $r=1$ (and the rest of the problem follows pretty easily from here).

Proof: Notice that $z+\frac{1}{z}=r(\cos\theta+i\sin\theta)+\frac{1}{r}(\cos\theta+i\sin\theta)=\cos\theta(r+\frac{1}{r})+i\sin\theta(r-\frac{1}{r}).$ Since $2\cos 3^{\circ}$ has no imaginary part, $r-\frac{1}{r}=0\to r=\frac{1}{r}\to r^2=1\to r=1$ (as $r$ is a positive number).

Now the rest of the problem follows.

Plugging $r=1$ into the equation we see that $2\cos\theta=2\cos 3^{\circ},$ so $z=cis 3^{\circ}.$

Thus $z^{2000}+z^{-2000}=cis(3\cdot 2000)+cis(3\cdot -2000)=-1,$ so the answer is $0.$
[/hide]

Whaaat I didn’t know $0$ could be an answer!

$z + \frac{1}{z} = 2 \cos 3^\circ \implies z^2 - (2 \cos 3 )z + 1 = 0.$ So, $z = \frac{2\cos 3 \pm \sqrt{4\cos^2 3 - 4}}{2} = \cos 3 + i\sin 3.$ By De Moivre's Theorem, $z^{2000} = \cos 6000^\circ + i\sin 6000^\circ = -\frac{1}{2} - \frac{\sqrt{3}}{2}i$ Thus, $z^{2000}+\frac{1}{z^{2000}} = 2\cos 240^\circ = -1.$ Therefore the answer is the least integer greater than $-1,$ which is $\boxed{0}.$

Let's denote $z = e^{i \theta}$. Then we have $e^{i\theta} + e^{-i\theta} = 2 \cos 3^\circ$, and want to find $e^{i2000\theta} + e^{-i2000\theta}$. From the first equation, we have $\cos \theta + i \sin \theta + \cos (-\theta) + i \sin (-\theta) = 2 \cos \theta = 2 \cos 3^\circ$, so $\cos \theta = \cos 3^\circ \implies \theta = 3^\circ$. We want to find $\cos 2000\theta + i \sin 2000\theta + \cos (-2000\theta) + i \sin (-2000\theta) = 2\cos(2000\theta) = 2 \cos(6000^\circ)$. This evaluates to $-1$, so the answer is $\boxed{0}$.


http://en.wikipedia.org/wiki/Convex_hull
http://mathworld.wolfram.com/ConvexHull.html

:)

Thanks..... :) ......Just by the way , are there any proofs for that expression on mathworld ??  :?


Suppose that metric space (X,d) is connected. Prove that for every two points a,b $ \in$ X, and for every positive number $ \delta$ exists finite sequence of points $ c_1, ... , c_n \in$ X which satisfy conditions: $ a \equal{} c_1, b \equal{} c_n$, and for every k $ \in$ {1,...,n}: $ d(c_k,c_{k \plus{} 1})\le \delta$

Let $ S(a)$ denote the set of points $ b$ reachable in this way. Let $ T(a)$ denote the set of points not reachable.

Then $ S(a)$ is open, since each $ b \in S(a)$ has a $ \delta$ neighborhood that is also reachable. $ T(a)$ is also open. For if $ d \in T(a)$ is not reachable
in finitely many $ \delta$-steps, then nothing in a $ \delta$-neighborhood of $ d$ can be reachable either. Otherwise we may extend a path to such a point to $ d$.

Then $ X \equal{} S(a) \cup T(a)$, while $ S(a) \cap T(a) \equal{} \emptyset$ and each set $ S(a), T(a)$ is open. Therefore either $ S(a) \equal{} \emptyset$ or $ T(a) \equal{} \emptyset$. Since $ a \in S(a)$, we conclude $ T(a) \equal{} \emptyset$ and so $ S(a) \equal{} X$.


I think the date is between May 6 and May 9!!    :P

May 7th to May 10th, as found [url=http://mathcounts.org/Page.aspx?pid=294]here[/url].

If you require some extra ...uh... "motivation," look at the [url=http://www.swandolphin.com/home.html]hotel[/url]. Don't spend all your time looking at it though; go practice.

[quote="xpmath"]If you require some extra ...uh... "motivation," look at the [url=http://www.swandolphin.com/home.html]hotel[/url]. [/quote]

Where did you read that? I didn't see that on the mathcounts website... I guess I am still in awe.

At [url=http://mathcounts.org/Page.aspx?pid=1336]this page[/url], it's stated that "The 2009 Raytheon MATHCOUNTS National Competition will be held at Walt Disney World's Swan and Dolphin Resort in Orlando, Florida!"

Searching "Swan and Dolphin Resort" on Google gives the website.


Given integers a>b>0 let

x= \sqrt a+ \sqrt b             y= \sqrt a- \sqrt b

If a-b is twice an odd integer prove that both x and y are irrational.

Sorry for my typing ; naturally x and y are distinct numbers

x= \sqrt a+ \sqrt b

y= \sqrt a- \sqrt b

Suppose x is rational. Then xy=a-b is rational, so y is rational.
Then 2 \sqrt a=x+y is rational, so   \sqrt a si rational.
Similarly  \sqrt b is rational.
Since a,b are integers and  \sqrt a, \sqrt b are rational, it follows that a and b are squares.
Let a=A 2 , b= B2.
So A 2 -B 2 =4k+2, which is impossible mod 4 (2 divides A-B iff 2 divides A+B).


For sufficiently large $A, n$, can it be said that $A^n + 1$ is always squarefree?

Here is a topic concerning a closely related problem.
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30001]http://www.mathlinks.ro/Forum/viewtopic.php?t=30001[/url]

Suppose that for all sufficently large $A$, and $n$, $A^n + 1$ is squarefree.

Since for all $A,n$ [b]sufficently large[/b] $A^n + 1$ is squarefree, we can take $A$ to be even and not divisible by $3$ (we just take an $A$ being a bit larger ;)). Hence $A^n + 1 \equiv 1 ( \mbox{mod } 2 )$ for all $n$. Take $n = 2 k$, for some positive integer $k$. Then $A^n + 1 \equiv A^{2 k} + 1 \equiv 2 ( \mbox{mod } 3 )$. This means that we can take sufficently large $A, n$  such that $2 \nmid A^n + 1$ and $3 \nmid A^n + 1$ and $A^n + 1$ squarefree. Hence $A^n + 1 = \prod_{j = 1}^m p_j$, with $p_j \neq 2, 3$ for all $j = 1, \ldots, m$. Hence $A^n + 1 \equiv \pm 1 (\mbox{mod } 6 )$, but $A^n + 1 \equiv A^{2 k} + 1 \equiv 2 ( \mbox{mod } 6 )$. Contradiction.

If $A$ is given, then remark that if there is some positive integer $k_0$ such that $A^{k_0} + 1 = a \cdot b^2$ is not squarefree and $A^n + 1$ is squarefree for all sufficently large $n$ then $A^{k_0 \cdot n} + 1$ is also squarefree for a sufficently large $n$. But $( A^{k_0 \cdot n} + 1 ; A^{k_0} + 1 ) = A^{k_0} + 1$. Hence $a \cdot b^2 = A^{k_0} + 1| A^{k_0 \cdot n} + 1$ and $A^{k_0 \cdot n} + 1$ is not squarefree. Contradiction.

Also I doubt there is some $A$ s.t for all $n$, $A^n + 1$ is squarefree. Hence that would imply that there is no integer $A$ s.t for all sufficently large $n$, $A^n + 1$ is squarefree.

I hope it is clear.

Just take any $A \equiv -1 \mod 4$ and any odd $n$ and you get big numbers such that $4|A^n+1$...


Prove that the perpendicular drawn from the point $ (4,1)$ on the join of $ (2,\minus{}1)$ and $ (6,5)$ divides it in the ratio of $ 5: 13$ . I got its answer $ 5: 8$. I think there is some problem in calculation.So please show your solution with calculation. Thanks a lot.

hello, let $ A(2;\minus{}1)$, $ B(6;5)$ and $ C(4;1)$ then the line say $ g$ through $ A,B$ is $ y\equal{}3/2x\minus{}4$, the line which is perpendicular to $ g$ is $ h$: $ y\equal{}\minus{}2/3x\plus{}11/3$ hence the intersection point $ D(46/13;17/13)$ so we get
$ \overline{AD}\equal{}10/13\sqrt{13}$
$ \overline{DB}\equal{}16/13\sqrt{13}$
and we get
$ \frac{\overline{AD}}{\overline{DB}}\equal{}5/8$.
Sonnhard.

Thanks. I got the same.It means that it is written wrong in the book.Thanks again. :-)


Let $ a,b,c > 0$, prove that  $ \sum_{cyc}\frac {a^2}b\ge3\sqrt [6]{\frac{a^6 \plus{} b^6 \plus{} c^6}3}$

[quote="akai"]Let $ a,b,c > 0$, prove that  $ \sum_{cyc}\frac {a^2}b\ge3\sqrt [6]{\frac {a^6 \plus{} b^6 \plus{} c^6}3}$[/quote]

Please help me, Can_hang2007, arqady, dduclam, ... or nother one ??? Thanks !

[quote="akai"]Let $ a,b,c > 0$, prove that  $ \sum_{cyc}\frac {a^2}b\ge3\sqrt [6]{\frac {a^6 \plus{} b^6 \plus{} c^6}3}$[/quote]
See my hint here: http://www.mathlinks.ro/viewtopic.php?t=184602 :)

We have a more weak inequality be
$ \sum_{cyc}\frac {a^2}b\ge\sum\sqrt [6]{\frac {a^6 \plus{} b^6 }2}$