MIT college AMC USA(J)MO USAMO Harvard MATHCOUNTS

Do you think that if you make USAMO you have a great chance to make MIT? I mean, do they really care? (apart from GPA, SAT scores, etc...)

they care more if you make it to the IMO :D

You should ask Joe Laurendi (joml88) about this.  He was a USAMO participant as a junior (he won't be one as a senior), and he was recently accepted into MIT.  Ask him, or he'll find this topic soon enough anyway!

[quote="mysmartmouth"]  Ask him, or he'll find this topic soon enough anyway![/quote]

 :D

Yes, I think it improved [i]my[/i] chances of getting in greatly.  They definitely care.

Of course, you need to have other qualities to complement your math achievements.  For example, I had a high GPA, decent SAT's, a few extracurriculars that I was very involved with, and living in SC probably helped me out too (geographic diversity thing).

Of course, I have to add the caveat that nothing you can do will absolutely ensure you of getting into any particular college.

i just find it very amusing that they have room for AMC/AIME scores on the MIT app- these are definately considered if you took them :) making USAMO definately wouldn't hurt  :yup:

Making USAMO is nice for MIT. I think if you legitimately make a MOP (top 30) you are pretty much guaranteed to get into MIT (provided you don't commit any felonies or flunk out or something), which is what I like about the admissions process.

As for Harvard, you can be a two time IMO medalist by the end of your junior year (Anders Kaseorg) and still get rejected, or get 41/42 at the IMO and 600/600 at the IOI and still not get in (Eric Price) ... granted Eric didn't get those things on his college app, but he wasn't exactly slouching before senior year either...

Sometimes you get the impression that being a "balanced individual" (and drawing this conclusion solely from a paper application) is slightly overweighted when some absolutely amazingly talented people don't get one of 1600 accepts...

[quote="JBL"][quote="darktreb"]Making USAMO is nice for MIT. I think if you legitimately make a MOP (top 30) you are pretty much guaranteed to get into MIT (provided you don't commit any felonies or flunk out or something), which is what I like about the admissions process. [/quote]

This isn't true, and I'm really tired of the "silver bullet" approach to college admissions.  [b]There is no silver bullet.[/b]  More importantly, every single question in the "does such-and-such college like such-and-such attribute" falls into a category in which you have people who don't really know answering other people who don't really know.  Use some common sense: imagine you were an admissions officer.  What do you think you would think if you looked at someone's application?  He got into USAMO -- that tells me he's a good test-taker (probably reflected by SAT scores), likes math.  Okay, good.  If you were a college, would you accept someone just because of that?  No.  Would that improve your opinion of someone?  Yes, of course.[/quote]

How many MOPpers do you know that got rejected from MIT?

If I read someone's application and saw they made the USAMO, I would not merely say they are a good test taker.  I know how amazing of an accomplishment it is and the hard work required in the [i]years[/i] of practice it takes to make USAMO.  Likewise, the admissions officers at MIT seem to be quite knowledgeable about USAMO as well.

OK, it almost goes without saying that you cannot determine if someone will be admitted to [i]any[/i] college based on just one aspect of the application.  However, those who qualify for USAMO/MOP usually have other things going for them as well.  And I don't believe this is a coincidence.  In all honesty, to get at that USAMO level one must be intelligent and very hard working, at the least.  They love to learn and thirst for knowledge.  And, of course, they are also good test takers, as you already pointed out.  Thus, it is reasonable to expect that, on average, MOP qualifiers will also have grades, standardized test scores, extracurriculars, and so forth to stand a major shot of getting into MIT.

My point is that I don't think it's awful advice to say that making MOP will almost guarantee you a spot at MIT.  That ever so trite (in my opinion) disclaimer that no one aspect of a person's college application will decide their admission decision should also be added.

[quote="joml88"]My point is that I don't think it's awful advice to say that making MOP will almost guarantee you a spot at MIT.[/quote]

You were doing fine until this point.  Then you switched your causations around again!  Someone who goes to MOP is reasonably likely to get into MIT, [i]but not primarily because she went to MOP[/i].  Rather, the type of person who goes to MOP is the type of person who is reasonably likely to get into MIT, [i]even if she doesn't go to MOP[/i].  As a result, saying to someone that MIT accepts a lot of MOPpers seems to be totally unhelpful in terms of their own process of college planning.  I think as a general rule advice that says, "If you are one of the top 30 in the country at X, you will be selected by college Y" is unhelpful for exactly the same reasons.

Ihave a friend who got Gold in the IMO and 2 silvers and he was rejected in MIT ... he's international though

[quote="JBL"]I think as a general rule advice that says, "If you are one of the top 30 in the country at X, you will be selected by college Y" is unhelpful for exactly the same reasons.[/quote]
I guess it (the advice) might be unhelpful to some, but it can provide motivation and/or a goal to shoot for.  I mean, being extremely motivated makes it soo much easier to work hard and [i]become[/i] the type of person who is reasonably likely to get into MIT.  I, for one, definitely changed a lot during my journey to make MOP, and those changes wouldn't have occured if I didn't have all that motivation from seeing MOP as an extremely worthwhile goal.  I think the whole point of the advice is to provide something to shoot for -- it certainly isn't impossible to [i]become[/i] the type of person who is likely to get into MIT -- you just need something to fuel those changes.

Yeah I'm definitely not an authority on the subject matter... it was just my 2 cents :)

I don't think getting to MOP is a bad thing, and I think people who strive to reach that level (in math or any other subject) are admirable for putting so much effort into improving themselves, whether or not they actually achieve their goal.  But I don't think anyone is going to be [i]more[/i] resolute, admirable or successful because doing well is going to get them into MIT.  Moreover, since only a very small number of people get to be in the top 30 of anything, making that the target strikes me as somewhat cruel.  Even if MOP were a sure-thing into MIT (which it's not), telling people indiscriminantly that the way to get into MIT is via MOP is bad advice for the vast majority of them, whose own personal paths to MIT will probably look quite different.

I'm already a sophmore ... do you think I still have a good shot at making USAMO/MOP/IMO?   :(

[quote="JBL"][quote="darktreb"]Making USAMO is nice for MIT. I think if you legitimately make a MOP (top 30) you are pretty much guaranteed to get into MIT (provided you don't commit any felonies or flunk out or something), which is what I like about the admissions process. [/quote]

This isn't true, and I'm really tired of the "silver bullet" approach to college admissions.  [b]There is no silver bullet.[/b]  More importantly, every single question in the "does such-and-such college like such-and-such attribute" falls into a category in which you have people who don't really know answering other people who don't really know.  Use some common sense: imagine you were an admissions officer.  What do you think you would think if you looked at someone's application?  He got into USAMO -- that tells me he's a good test-taker (probably reflected by SAT scores), likes math.  Okay, good.  If you were a college, would you accept someone just because of that?  No.  Would that improve your opinion of someone?  Yes, of course.[/quote]

Actually, in all fairness, I think you are dead wrong about the "no silver bullet" concept. I have never heard of someone who made a 30 person MOP, put it on their college app, and then got rejected from MIT. I was even told at my first 30 person MOP, by multiple people, that my MIT and Caltech admissions were virtually guaranteed. This is under the assumption that everyone who goes to a 30 person MOP is generally a pretty good student (very reasonable) and does well overall in school (very reasonable) and is not a criminal (very reasonable).

In various cases, I would assert that MOP was indeed the REASON why some people got into MIT. If you take MOP off of some candidates' applications and replace it with a bunch of school extracurriculars (schol bowl, usamo qualifier, some assorted regional/state level stuff), you are left with a weaker candidate. In fact, I have a very good friend who attended Exeter (a superior school to mine), was a 5 time USAMO qualifier (like myself), had similar scores on standardized tests, had similar grades to mine (a little better maybe), and all in all had an extremely similar app to mine. The only major difference was that I had a 30 person MOP on my app (and just one actually). I was accepted, and he was waitlisted.

On the other hand, for international students, it's VERY TOUGH regardless of what you have done. That's just the way the system works. One of the TAs for my Abstract Algebra 2 class right now scored a perfect 42 at IMO 2001 for China.

The only "reason" saying MIT accepts a lot of MOPpers would be unhelpful is because MOP is not something you can guarantee yourself to get into, regardless of your effort. It is difficult and takes a combination of talent, effort, and the right circumstances. That's one of the reasons why it is weighted highly.

[quote="darktreb"]This is under the assumption that everyone who goes to a 30 person MOP is not a criminal (very reasonable). [/quote]

whoa this is quite a revelation

in an effort to be slightly germane, I would add, "and also don't you think squabbling over whether making MOP is a silver bullet is missing the point? sheesh. do the best you can do and see what happens ... ? " but for some reason people don't seem to like that

[quote="MysticTerminator"][quote="darktreb"]This is under the assumption that everyone who goes to a 30 person MOP is not a criminal (very reasonable). [/quote]

whoa this is quite a revelation[/quote]

 :rotfl:

[quote="MysticTerminator"]in an effort to be slightly germane, I would add, "and also don't you think squabbling over whether making MOP is a silver bullet is missing the point? sheesh. do the best you can do and see what happens ... ? " but for some reason people don't seem to like that[/quote]

Yep, people like to look for the magic bullet, rather than living life in recognition of the probabilistic nature of most planning. I of course would like MIT to acknowledge participation on this forum as the silver bullet for admission   ;)  but MIT has its own way of doing things, and if MIT passes over a particular applicant with outstanding credentials (as it has done in my state in recent years), the applicant simply has to go to another college and thrive there. Amazingly, many people who have never gone to MIT still manage to have happy, productive lives.   :o

[quote="joml88"]

How many MOPpers do you know that got rejected from MIT?

If I read someone's application and saw they made the USAMO, I would not merely say they are a good test taker.  I know how amazing of an accomplishment it is and the hard work required in the [i]years[/i] of practice it takes to make USAMO.  Likewise, the admissions officers at MIT seem to be quite knowledgeable about USAMO as well.

OK, it almost goes without saying that you cannot determine if someone will be admitted to [i]any[/i] college based on just one aspect of the application.  However, those who qualify for USAMO/MOP usually have other things going for them as well.  And I don't believe this is a coincidence.  In all honesty, to get at that USAMO level one must be intelligent and very hard working, at the least.  They love to learn and thirst for knowledge.  And, of course, they are also good test takers, as you already pointed out.  Thus, it is reasonable to expect that, on average, MOP qualifiers will also have grades, standardized test scores, extracurriculars, and so forth to stand a major shot of getting into MIT.

My point is that I don't think it's awful advice to say that making MOP will almost guarantee you a spot at MIT.  That ever so trite (in my opinion) disclaimer that no one aspect of a person's college application will decide their admission decision should also be added.[/quote]

I'd have to agree with JBL, there are definitely other things colleges would look for aside from USAMO and MOP, and the two definitely do not mean you automatically get in.

Take this example, a guy from our school:

Anonymous*
13th Place - Nationals MathCounts
3-Time USAMO Qualifier
Black MOPper
Perfect score on PSAT, SAT, ACT

MIT: Rejected
The reason being, his grades weren't the best.

He applied again later on as a college freshman I believe. He was rejected again.
He is probably one of the smartest people I know, and look what happened.

Personally, I know of 10 math students who live in my state and were accepted into MIT this year. Not a single one ever made the USAMO.

Another thing, making the USAMO isn't all that difficult. Making MOP is a different story, but it will not get you anywhere by itself. As JBL said, it is more likely that MOPpers are motivated and thus excel in all areas (beyond math).

*EDIT: privacy it is.

You should at least try to preserve some privacy for said person.

Yes, here on AoPS we use screennames partly to preserve privacy, and I pretend that I don't know who JBL is and so forth. If we mention examples, let's be sure to anonymize the examples enough that the actual person is not easily identifiable.

[quote]

Another thing, making the USAMO isn't all that difficult. Making MOP is a different story, but it will not get you anywhere by itself. As JBL said, it is more likely that MOPpers are motivated and thus excel in all areas (beyond math).

[/quote]

After all, correlation implies causation  :D
Just kidding, I do agree.

[quote="xxazurewrathxx"]

I'd have to agree with JBL, there are definitely other things colleges would look for aside from USAMO and MOP, and the two definitely do not mean you automatically get in.

Take this example, a guy from our school:

Anonymous*
13th Place - Nationals MathCounts
3-Time USAMO Qualifier
Black MOPper
Perfect score on PSAT, SAT, ACT

MIT: Rejected
[/quote]

Well maybe that person made Black MOP in his/her senior year.

[quote="Karth"]What is a black MOPper?[/quote]

They put people into groups at MOP based on their skill level.  Black MOP is the highest group (so in this group are the 6 IMO contestants, the USAMO winners, and probably the HM winners...).  (the groups are named like karate belts, I believe)

If I'm still a sophmore, will I still have time to become a Black MOPper? I'll work as hard as possible (even with SATs, etc. but SAT and school will come first above all).  :)

[quote="joml88"][quote="Karth"]I'm already a sophmore... do i still have much time? Do i still have a good shot at making USAMO/MOP/IMO?[/quote]

[quote]If I'm still a sophmore, will I still have time to become a Black MOPper? I'll work as hard as possible (even with SATs, etc. but SAT and school will come first above all). [/quote]

This type of question has to be the most popular on the forums... However, it's one of the most annoying  :|   We can't tell you how good a shot you have of usamo/mop/whatever.  We can try to tell you how you might want to go about preparing.  But its really [i]you[/i] you must do the work and [i]you[/i] who will be taking the test and things.[/quote]

Okay... so how can I go about preparing? :|

[quote="Karth"]

Okay... so how can I go about preparing? :|[/quote]

Eh playing a large number of video games is the best way to prepare :lol:

But seriously just do problems.


Is there a general way to determine whether a $ m\times n$ board has a closed knight's tour?

Are you actually asking this because you want to know the answer or because you are looking for a proof of the result?  Wikipedia gives an answer, but not a full proof and I would guess that a proof would be highly nontrivial.


geometry
geometric transformation
rotation

15 people sit around a table. When they sit down, they did not notice that a name tag is in front of each seat, and they found that no any name tag and the person sitting there match each other. Prove that after some rotation of the sitting order, at least two people will match the name tag where they sit

Interesting problem. I first found it here:
[url]http://mathcentral.uregina.ca/mp/previous2006/sept06sol.php[/url]
(the web states that this is problem number 6 on the 1975 Canadian Mathematical Olympiad).


inequalities
number theory unsolved
number theory

Let $ q_i,p$ be distinct primes, $ k,a$ integers $ \ge 2$

Be $ \prod_{i \equal{} 1}^{k} q_i \equal{} p^a \plus{} 2$

Show that

$ \prod_{i \equal{} 1}^{k} (q_i \minus{} 1) < p^a \minus{} p^{a \minus{} 1} \plus{} 2$

I don't know if I understood this right, but

$ 31 \cdot 43 \equal{} 1333 \equal{} 11^3 \plus{} 2$

But $ 30 \cdot 42 \equal{} 1260 > 1212 \equal{} 11^3 \minus{} 11^2 \plus{} 2$

Is there a condition that $ k \geq a$ ?

maybe this is true , i mean we must have $ k \ge a$.
let's suppose that we have this condition then we can easily prove what is asked,
just see that we have
$ \prod_{i \equal{} 1}^{k}q_{i} \equal{} p^{a} \plus{} 2$
which implies that
$ \prod_{i \equal{} 1}^{k}(q_{i} \minus{} 1) < p^{a} \minus{} p^{a \minus{} 1} \plus{} 2 \Longleftrightarrow \prod_{i \equal{} 1}^{k}(1 \minus{} \frac {1}{q_{i}}).(p^a \plus{} 2) < p^{a} \minus{} p^{a \minus{} 1} \plus{} 2 \Longleftrightarrow \prod_{i \equal{} 1}^{k}(1 \minus{} \frac {1}{q_{i}}) < 1 \minus{} \frac {p^{a \minus{} 1}}{p^a \plus{} 2},1 \minus{} \frac {1}{p} < 1 \minus{} \frac {p^{a \minus{} 1}}{p^a \plus{} 2}$

so we have only to prove that $ \prod_{i \equal{} 1}^{k}(1 \minus{} \frac {1}{q_{i}}) < 1 \minus{} \frac {1}{p}$.
but just see that if there exists $ 1 \le j \le k$ such that $ q_j < p$ then we'll have
$ \prod_{i \equal{} 1}^{k}(1 \minus{} \frac {1}{q_{i}}) < 1 \minus{} \frac {1}{q_j} < 1 \minus{} \frac {1}{p}$
and we are done so all primes $ q_i$ must be greater than $ p$ which implies that
$ p^k < \prod_{i \equal{} 1}^{k}q_{i} \equal{} p^{a} \plus{} 2$
so we must have $ k < a$ which is a clear contradiction with the assumption $ k \ge a$.

Yes you are true I forgot this condition.
Thank you...

Different taste of this inequality / equation

Let $ q_i,p$ be distinct primes, $ k,a$ integers $ \ge 2$

Be $ \prod_{i \equal{} 1}^{k} q_i \equal{} p^a \plus{} 2$

Show that

$ \prod_{i \equal{} 1}^{k} (q_i \minus{} 1) \equal{}  p^a \minus{} p^{a \minus{} 1} \plus{} 2$
can only be true if k=a=1


quadratics
algebra
polynomial
superior algebra
superior algebra solved

Prove that the rank and signature of the following real quadratic form
\[\sum_{i=1}^n \sum_{j=1}^n (\lambda i j + i + j)x_i x_j \quad (n>1)\]
has no relation with $\lambda$ (i mean they are constant).

Let $a_{i,j}=\lambda\cdot ij+i+j$.
Let $A=(a_{i,j})_{i,j=\overline{1,n}}$.

Let $U\in M_{n\times 2}(\mathbb{R})$ and $V\in M_{2\times n}(\mathbb{R})$ defined by:
$u_{i,j}=\lambda\cdot ij+i+j\ \forall i=\overline{1,n}\ j=\overline{1,2}$ and
$v_{1,i}=2-i$ and $v_{2,i}=i-1$ for all $i=\overline{1,n}$.

It is easy to see that $A=U\cdot V$.

Let $P_A$ denote the characteristic polynomial of A.
It is known that $P_A(X)=P_{V\cdot U}(X)\cdot X^{n-2}$.

After computations we get $\det(V\cdot U)=-\frac{n^2(n^2-1)}{12}$.
So $\det(V\cdot U)<0$. Notice that it is independent of $\lambda$.
Since A is symmetric, all its eigenvalues are real. So the eigenvalues of $V\cdot U$ are real. Since their product is negative, one is positive and the other negative.
So A has one positive and one negative eigenvalue and 0 as eigenvalue of multiplicity n-2.
Thus the quadratic form has rank 2 and signature 1 for all $\lambda$.


Find all functions $f : \mathbb{Q}^{+} \rightarrow \mathbb{Q}^{+}$, where $\mathbb{Q}^{+}$ is the set of positive rationals, such that $f(x+1) = f(x) + 1$ and $f(x^3) = f(x)^3$ for all $x$.

It's easy to show that $f(x+k) = f(x) + k$
Lets take $p,q \in \mathbb{N}$, $(p,q)=1$

$f((\dfrac{p}{q} + q^2))^3) = f(\dfrac{p}{q} + q^2) = (f(\dfrac{p}{q}) + q^2)^3 =$
$= f(\dfrac{p}{q})^3 + 3(f(\dfrac{p}{q}))^2 + 3f(\dfrac{p}{q})q^4 + q^6$

On the other hand:
$f((\dfrac{p}{q} + q^2))^3) = f((\dfrac{p}{q})^3) + 3p^2 + 3pq^3 + q^6 =$
$= (f(\dfrac{p}{q}))^3 + 3p^2 + 3pq^3 + q^6$
So we have got:
$(f(\dfrac{p}{q}))^2 \cdot q^2 + f(\dfrac{p}{q}) \cdot q^4 = p^2 + pq^3$
Finally:
$q \cdot f(\dfrac{p}{q}) - p \ 0$
$f(\dfrac{p}{q}) = \dfrac{p}{q}$

I've seen it before, but dunno where ;o

In 'Kolko matematyczne dla olimpijczykow' by H. Pawlowski :)


Let $ a;b;c$ be positive real numbers.Prove that
$ a^{b\plus{}c}\plus{}b^{c\plus{}a}\plus{}c^{a\plus{}b} \geq 1$
When equality occurs?

[quote="pro_math"]Let $ a;b;c$ be positive real numbers.Prove that
$ a^{b \plus{} c} \plus{} b^{c \plus{} a} \plus{} c^{a \plus{} b} \geq 1$
When equality occurs?[/quote]
This is a problem in Y&Y of VietNam
so , please lock topic !


inequalities
function
calculus
blogs
advanced fields
advanced fields unsolved

Let $ U\in R^n$ be open, bounded with smooth boundary. Then there exists a constant such that for ever $ u\in H_{0}^{1}(U)$,
\[ \|u\|_{L^2(U)}\leq c \|Du\|_{L^2(U)}.\]

Was this assigned as a homework problem without hints?  That's kind of tough.

First, prove it for compactly supported smooth functions and then use density.

To prove it for compactly supported functions, the key is to use the Fundamental Theorem of Calculus to bound sup u in terms of |grad u|, and integrate and stuff.  If you need more help than that, you can look it up in pretty much any grad level elliptic PDE book.  (Specifically, try Evans.)

[quote="yenlee"]
To prove it for compactly supported functions, the key is to use the Fundamental Theorem of Calculus to bound sup u in terms of |grad u|, and integrate and stuff.[/quote]
Or just use Fourier transform and Plansherel. (You will need to notice that $ \widehat u$ cannot be concentrated near the origin in the $ L^2$-sense, of course).

you can read the [url=http://terrytao.wordpress.com/2009/04/30/245c-notes-4-sobolev-spaces/]blog[/url] of Tao.

Or just study Evan's book(page 261-266...)  :lol:


[quote="lsrith"][b][i] :?: PLease, thank you first! I cannot find out the IMO 2008 web page in english. I can find in Spain. Unfortunately, I cannot understand. Can you help get through to the one in English? [/i][/b][/quote]

I think it says that currently the url is just reserved, no data is available about the olympiad.

@quangchuc, please don't say that . All teams always have occasion  :(

[quote="Erken"][quote="quangchuc"]I think Viet Nam will win! Do you think so?[/quote]
I think it will be China,but it is too early to talk about it...,and i also think that we must talk about it in another specialize for this theme topic :wink:[/quote]

Moroco may be  :P

Why not?

[quote="kdano"]I think it says that currently the url is just reserved, no data is available about the olympiad.[/quote]
The IMO'2008 web-page has finally come alive:

   http://www.imo-2008.es/index.html

They even list the names of the problem selection committee (with Kuczma, Kós, Savchev).
This could cause all kinds of security issues.

Thanks to all of you who join this IMO2008 discussion. I finally find out the web page. :)

what do you think about morroco's level in maths? The truth please.(i want answers please :) )

I know sure that there is a graet potential there, but they do not do much about it, if you know what i mean.

you're right mohamed,but how to improve this potential??? :?:

please who knows how they choose the contestants who will participate in the imo

hi rachid ( 7ta nta mn l bled ??  :lol: ) ,

after  evry phase in the  national olympiad , they call the best contestants in every academy to rabat , to participate in some short and weak training program , durig this they pass some tests , and the best 6 sudents in the last test are going to IMO  :P

[quote="anas"]yes Morocco has potential , but Moroccan contestants are not well trained  :( , besides the Moroccan trainers are not involved in math competitions , and they just recycle problems  :rotfl: , finally Morocco is ranked 124/125 in education level  :!:  :o  , so evry moroccan who get  silver or even a bronze medal in IMO , must be a genuise  :D[/quote]

That ranking says almost nothing about how good the mathematicians are of a particuraly country, see netherlands for example.

And about moroccan getting silver medals, you should check 10 years ago how the results were then ....
The olympiad comittee sucks as the government !

HI anas(ana mn marrakech)
i understand from what you said that only the contestants who participated in the imo who has the chance to participate a second time, so the other students from other schools haven't the chance to participate?.

hello again rachid!

I see that you didn't understand, anyway see here:

[url]http://www.men.gov.ma/dajc/notes/notesannee/%20notes%202007/note%201272007.doc[/url]
 :wink:

Hallo guys, what's up???

I wish Greece will take a good place in this year's IMO.


If $ 2^{20}\minus{}2^{19}\equal{}2^x$ , what is the value of $ x$ ?

since $ 2^{19} \times 2 \equal{} 2^{20}$ that means that $ 2^{19} \plus{} 2^{19} \equal{} 2^{20}$
so we get our answer....$ 19$

If one notices that $ 2^n + 2^n=2(2^n)=2^{n+1}$ then the problem is over rather quickly... if one doesn't notice, they should head for the fact that for a>0, $ a^x=a^y \iff x=y$:

Start:  $ 2^{20}-2^{19}=2^x$
Rewrite both sides as a product of a power of 2:
$ 2^{19}(2-1)=2^x$
$ 2^{19}=2^x$, so $ x=19$.

A minutely harder example problem:

$ 4^8=2^{y+3}$
$ (2^2)^8=2^{y+3}$
$ 2^{16}=2^{y+3}$
$ 16=y+3$
$ y=13$


Από τον διαγωνισμό IMC
Nα δείξετε ότι $ tan(sinx)>sin(tanx)$ για κάθε $ x$ στο $ (0,\pi/2)$

[quote="r_boris"]Από τον διαγωνισμό IMC
Nα δείξετε ότι $ tan(sinx) > sin(tanx)$ για κάθε $ x$ στο $ (0,\pi/2)$[/quote]
Λυση


Do you guys love to help out other people doing their math homework? Because I just loving doing that provided it is either challenging (which almost always isn't) or advanced enough. I do it because I like math and explaining it. And also because I feel that if you teach something you get to know it much better.

[quote="archimedes1"]If it's regular classroom math, then no.

[quote]
"First, try bringing $ x$ over to the right side of the equation, to set it equal to zero."
"Huh?"
"Subtract $ x$ from both sides."
"Oh... okay...  Now what?"
"Now, what does this equation look like, that you've seen before?"
"Huh?"
"What type of equation is this, that you know how to solve?"
"Um...  an algebra equation?"
"It's a quadratic equation.    How do you solve a quadratic equation?"
"Huh?"
[/quote]

I'm usually horrible at explaining things to others in math class, because none of them are motivated to learn.  They don't care about what they're doing, they just expect me to do it for them.  I don't like having teachers delegate me to help out the other students when I'm done with the assignment.

On the other hand, I love writing solutions to competiton-level problems, and explaining things in a Math Club - type environment.[/quote]
Same here, I find it really hard to explain classroom math... i usually get asekd in advisory/study hall or enrichment
while I'm not the one who people ask for help in my math class. The second smartest person is, he seems to be able to explain it; actually no, he probably just ends up showing "how" to do it. The other students always asks him to compare answers.

For an old discussion about this,

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=162901

Yeah, I actually find explaining harder topics easier, because the only people who listen want to, not because it's required. And true, quite noticeably obvious, most people who are pretty good don't like it. They see math as a subject where you just memorize formulas and plug them in.

As a college professor, I've had the conversation archimedes1 describes (and believe me, it is possible to have that conversation in a calculus class). My metaphorical description for such a conversation is "pushing a piece of string." Of course, this is my job. I also get plenty of the other kind of conversation.

That's nothing. I once had to teach a kid with attention deficit disorder how to memorize the multiplication of any number from 0 to 10. It's frustrating to teach a kid how to do math when he can't even add 5 and 7. I have, however, tutored a kid who caught on very quickly when I gave hints or guided him through a question. That was certainly a fun and rewarding experience. So, if you can find motivated students who just have a little trouble with applying certain concepts, it can be of great joy to help them towards mathematical triumphs.

This is weird; I was just about to sign up to be a tutor in math at my high school. I've heard lots of complaints that the people just don't show up or waste your time.

My experience with homework helping is most of the people just want the answers and don't really care about the method in which they derive that answer. (Of course they change their mind when test time comes around). Therefore, I only help people who want to actually learn the concept (strangely they are mostly girls  in my math class).

Honestly, for most of my friends, I'm much happier doing their math homework and having them do my chem or something, but for those who are actually interested in learning, yeah, sure I'll explain. Having a captive audience helps.


I have a table that has 5 rows and 3 columns. The leftmost column has three characters, the rightmost column has three characters, and the middle column of each row is usually three or more lines long. How do I put vertical spacing between each row (around the \hline)? (I tried \\[1in] but it doesn't work because the middle column takes up more vertical space than the rightmost column does.)

I'm not exactly sure what you want (an example would help) but you can increase the distance between the rows with \renewcommand{\arraystretch}{[i]factor[/i]} where [i]factor[/i] is a multiplying factor. The current value is 1 and a value of 1.5 would increase the spacing by 50%, so it's a matter of trying different values.

You can do it for the first row by doing something like \hline \\[-2ex] (add a new row and a negative vertical space) and for the last row \\[1ex] \hline. But, your first post seems to indicate that this doesn't work so could you give a simple example of the sort of table you want?

I was using tabular, and not array.
I think the \\[-2ex] doesn't work because the middle column is extra long.

[code]\begin{center}
\begin{tabular}{ p{0.3in}  p{4.9in} | p{1in} }

asdf & Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text & asdf \\

\hline

asdf & Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text & asdf \\

\hline

etc.....[/code]

[quote="123456789"]I was using tabular, and not array.[/quote]Everything I said applies to both array and tabular. They behave the same.
[quote="123456789"]
I think the \\[-2ex] doesn't work because the middle column is extra long.[/quote]I'm not sure how the width of a column has any effect on the vertical spacing. \\[-2ex] works fine, and \renewcommand{\arraystretch}{2} (before \begin{tabular}) is even better (but can't be shown here), but the middle column has no relevance to either. I am clearly not understanding exactly what the problem is.
Here for example is the \\[[i]n[/i]ex] method, ridiculously exaggerated so it shows up:
$ \begin{tabular}{ p{0.3in}  p{4.9in} | p{1in} }
\hline
&&\\[-0.5ex]
asdf &  Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text & asdf \\
\hline
asdf & Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text & asdf \\[6ex]\hline
\end{tabular}$

But for the spacing above an \hline, you had to use [6ex] because you accounted for the extra lines in the middle column. However, each row's middle column will take up a different number of rows. How do I have latex automatically account for that?

No, I used 6ex because I wanted to overdo it. The number of lines in the middle column does not affect what you use. The vertical spacing is done [i]after[/i] the text is written and will [i]always[/i] be the same no matter how many lines of text there are. Let's use real values of -2ex and 1ex:

$ \begin{tabular}{ p{0.3in} p{4.9in} | p{1in} } \hline & & \\
[ \minus{} 2ex] asdf & Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text & asdf \\
\hline asdf & Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text & asdf \\
[1ex]\hline \end{tabular}$

Now lets add more lines but the same vertical spacing and you get:

$ \begin{tabular}{ p{0.3in} p{4.9in} | p{1in} } \hline & & \\
[ \minus{} 2ex] asdf & Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text & asdf \\
\hline asdf & Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text & asdf \\
[1ex]\hline \end{tabular}$

You get exactly the same vertical spacing between the text and the lines.

If you want this effect for every row, not just the first and last rows, then use \renewcommand{\arraystretch}{factor}

But stevem, look at the example below. I used [3ex] in both cases, but the result is different:

\begin{tabular}{ p{0.3in} p{4.9in} | p{1in} }
asdf & Lots of text Lots of text Lots of text  & asdf \\[3ex]
\hline asdf & Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text & asdf \\
[3ex]\hline \end{tabular}

[quote="123456789"]I don't need more spacing between rows of text. I just want spacing between the last row of a cell and the \hline beneath it, and spacing between the first row of a cell and the \hline above it.[/quote]
I've been trying to follow this as it doesn't make much sense since a cell is one of the little rectangles and a row is a whole line. Thus [i]last row of a cell[/i] makes no sense.
So I wonder. Do you want the text in the first and third (asdf in your example) columns to be vertically centered? If so, then use the array package (also works for tabular) which has, instead of p{0.3in}, the option m{0.3in}. This will vertically centre the text in a cell.

No, I want the horizontal alignment just as it is in the previous example, just with breathing room between the \hlines.

In that case use \renewcommand{\arraystretch}{2} (or whatever factor you want) which will give the same breathing room between the lines for each row.

But what if I want to keep the spacing between lines equal? I don't want extra space between every line, only between final lines and the \hline, and first lines and the \hline preceding them.

Whatever, I think I'll just stick with no space. Thanks anyway.

I know you've given up but I think I have the perfect solution for you now that I understand what you want:)
The booktabs package allows you to add vertical space with the command \addlinespace after a whole row, rather than adding vertical space in just the last column.
So try:
[code]\usepackage{booktabs}
...
\begin{tabular}{ p{0.3in} p{4.9in}  p{1in} }
asdf & Lots of text Lots of text Lots of text & asdf \\
\hline asdf & Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text Lots of text  & asdf \\ \addlinespace[-1ex]
 \hline
\end{tabular}[/code]
I'm reasonably sure this is what you want. The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=booktabs]booktabs package[/url] does lots of other things to improve the layout of tables and has lots of advice, but you can ignore or use them as you please.

Thanks. When I get a chance, I'll try it.

Well, it kind of works. I get that extra line of space, but then the vertical dividers between the columns gets cut off before it reaches the next \hline.

I've decided I don't really need this extra space anymore. Thanks for your help.


Prove that $ \displaystyle\sum\limits_{n\equal{}1}^k {F_n}^2\equal{}{F_k}{F_{k\plus{}1}}$.

And the prize for the "best" title goes to...

Do never again use such unreadable pseudo-LaTeX as title. And the source line is not for mirroring the title!

Try defining arbitrary sums with induction  :wink:
And you still could have changed the title  :roll:

[hide="Combinatorially"]Shift the indices so that $ F_0 \equal{} F_1 \equal{} 1$ instead and $ F_n$ counts the number of ways to tile a $ 1 \times n$ rectangle. We want $ \sum_{n\equal{}0}^{k} F_n^2 \equal{} F_k F_{k\plus{}1}$. How many ways to tile a $ 1 \times k$ and $ 1 \times (k\plus{}1)$ rectangle with squares and dominoes?
Way 1: $ F_k F_{k\plus{}1}$
Way 2: Align the rectangles and condition on which of the $ k\plus{}1$ positions has the last square. Suppose it is position $ n$, $ 0 \leq n \leq k$. Since the parity of the length of the two rectangles is different and we need an even number of positions past the last square, which rectangle the square is in is uniquely determined by the position. All of the spaces past the square are dominoes. There are two $ 1 \times n$ rectangles left to tile, for which there are $ F_n^2$ ways. Summing all these cases up gives $ \sum_{n\equal{}0}^{k} F_n^2$.[/hide]


Hey, who's going to Nationals 2007?
-Kevin Hu, IL

fanatic ia going from IN, but i don't know if he/she wants his/her name spoken.

I'm not going to nationals this year. I'm retiring from old age :D

[quote="Fanatic"]Hey Kevin. I know you're going as 1st from Illinois :p

Yes I'm going.

Other AoPSer's that are confirmed:

Daine
L33tSp34kF3v3r
ksstan
footballrocks
1=2

... of course, and more.[/quote]

Don't forget me!   :D

dude no you're not. all you ever do is lie.

Great. Fifty bazillion AoPSers are doing nationals. I'm gonna be #229. Just watch me.

so that's at least 10 people who say their going. Impressive, since by my count only 20 states have had State yet. that's one out of 8 people who are on this forum.

Johnz DEFINITELY Lied.

Well, maybe he didn't. but that means i've met with him several times.

I'm surprised I haven't been tripped and had my leg broken [as he wished upon me in a pm] yet.

LOL johnz told me when he first joined that he was in 9th grade.

Next year I'll be in 7th grade, AND I'm moving to Kentucky (if all plans hold) because my dad thinks that there is less of a chance my sister will run away. Put the two and two together, and I might be going to Nationals next year!

I have a Kevin Chen in one of my classes and another Kevin Chen in a study hall :lol:

Kevin Chens are taking over!!!

hihihihihi and myself are both going to nationals, to add to the list
even though i is not really that goodlier of a smart fellow! ok that sentence was just for fun

[quote="SplashD"]Who's on the Florida team?[/quote]
1st: Liam Bressler
2nd: Anthony Grebe
3rd: Preet Patel
4th: Andrew Harn

I think I misspelled Andrew's last name.

Kentucky Team:
Evan Miller
Some kid from Owensboro
Michael Druggan
Chris Cai

[quote="gdtea"]Kevin is a popular name of math genuine.[/quote]

Kevin is my middle name lol  :rotfl:

I'm going to nationals, which is what inspired me to remember this website :)

gah....that was me. I forgot to log off my dad's account.

I took 3rd in Washington State, so I'm headed to Nationals.

(Seems like nobody's posted here for a while, but I just created this account, so...)

[quote="ra5249"]yay for kevins! XD
wats california team?[/quote]

This thread seems kinda dead, but I'll answer anyway.

CA team in order of score:
Nathan Pinsker
Patrick Yang
Albert Wu
Daichi Tsuna

I'm from Oklahoma, the retard state.
I'm going, and this is my team:

1.  Nikhal Mutalik
2  Anson Hwang
3  Me
4  Joey Copp

Since Oklahoma's so retarded, I doubt anybody's heard of us, cause our abilities are only average  ( for azns ) :)

"average" for azns? lol. I'd say something politically incorrect right now, but...  :roll:


trigonometry
geometry
ratio
geometric transformation
reflection
invariant
trapezoid

Let $ M$ and $ N$ be points on the side $ BC$ of triangle $ ABC$, with the point $ M$ lying on the segment $ BN$, such that $ BM \equal{} CN$. Let $ P$ and $ Q$ be points on the segments $ AN$ and $ AM$, respectively, such that $ \measuredangle PMC \equal{}\measuredangle MAB$ and $ \measuredangle QNB \equal{}\measuredangle NAC$. Prove that $ \measuredangle QBC \equal{}\measuredangle PCB$.

Suppose circle $PMC$ and line $AN$ intersect at $S$. Also circle $BQN$ intersect with $AM$ at $T$. If we prove $M$, $N$ , $S$ and $T$ are on a circle then problem is done. You see $\triangle ACS$ and $\triangle TBA$ are similiar.
\[ \frac{AS}{AT}=\frac{AC}{BT}=\frac{AC}{CN}\cdot\frac{BM}{BT}=\frac{\sin \angle ANM}{\sin \angle CAS}\cdot\frac{\sin \angle BTM}{\sin \angle AMN}=\frac{\sin \angle ANM}{\sin \angle AMN}=\frac{AM}{AN} \]
So $AS\cdot AN=AM\cdot AT$. Therefore $N$, $M$, $S$, $T$ are on a circle.

I am trying to prove it with Area Ratios, but stucked...
This problem is equal to $\frac{\triangle BQN}{\triangle CPM}=\frac{BQ}{CP}$. Anyone has any ideas?

It definitely cannot be solved by area ratio only, because an attempt via trigo shows that you have to use the formulae like $\sin(x+y)$, $\sin x+\sin y$, etc. This implies simply handling the ratios [u]must not[/u] work. One has to deal with something like concyclicity, similarity, etc.

The area ratio of two triangles with $\angle{B}=\angle{B'}$ is $\frac{\triangle{ABC}}{\triangle{A'B'C'}}=\frac{BA\cdot BC}{B'A'\cdot B'C'}$. I am trying to do it in that way.

Let's go for a nonstandard proof:

The reflection with respect to the perpendicular bisector of the segment BC maps the point B to the point C and vice versa; also, it maps the point M to the point N and vice versa (in fact, the points M and N lie on the segment BC and satisfy BM = CN; this means that they are symmetric to each other with respect to the midpoint of the segment BC, and thus with respect to the perpendicular bisector of the segment BC as well). Let D be the image of the point A under this reflection, and let R be the image of the point P under this reflection. Altogether, the reflection with respect to the perpendicular bisector of the segment BC maps the points B, C, M, N, A, P to the points C, B, N, M, D, R, respectively. Hence, as the point P lies on the segment AN, its image R lies on the segment DM, and we have < RNB = < PMC, < MDB = < NAC and < RBC = < PCB (since reflections are congruence transformations and keep angles invariant - non-directed angles at least). Since the points N and D are the reflections of the points M and A in the perpendicular bisector of the segment BC, the quadrilateral AMND is an isosceles trapezoid; thus, it has a circumcircle.

Let the line NQ meet the line BD at a point U, and let the line NR meet the line AB at a point V. Since < UNM = < QNB = < NAC = < MDB = < UDM, the points U, M, N, D lie on one circle, i. e. the point U lies on the circumcircle of the quadrilateral AMND. Since < VNM = < RNB = < PMC = < MAB = < VAM, the points V, M, N, A lie on one circle, i. e. the point V also lies on the circumcircle of the quadrilateral AMND. Altogether, we see that both points U and V lie on the circumcircle of the quadrilateral AMND, so the hexagon AMDUNV is cyclic; thus, according to the Pascal theorem, the points $AM\cap UN$, $MD\cap NV$ and $DU\cap VA$ are collinear. Since $AM\cap UN=Q$, $MD\cap NV=R$ and $DU\cap VA=B$, this means that the points Q, R and B are collinear. Consequently, < QBC = < RBC. Since < RBC = < PCB, this yields < QBC = < PCB, and the problem is solved.

  Darij

Shobber i could not understand what do you mean , can you explain.


$ ABC$ is an equilateral triangle. $ D$ is a point inside $ \triangle ABC$ such that $ AD \equal{} 8$, $ BD \equal{} 13$, and $ \angle ADC \equal{} 120^\circ$. What is the length of $ DC$?

$\textbf{(A)}\ 12  \qquad\textbf{(B)}\ 13  \qquad\textbf{(C)}\ 14  \qquad\textbf{(D)}\ 15  \qquad\textbf{(E)}\ 16$


Let $ k$ be a positive integer. Find all positive integers $ n$ such that $ 3^k$ divides $ 2^n\minus{}1$.

this can be done using the very nice fact that $ 2$ is a primitive root of $ 3^n$ for evry positive integer n.
This can be proved using the fact that $ 2^{phi(3^n)} \equal{} 1 \plus{} 3^{n}(mod3^{n \plus{} 1})$ where phi is the euler's totient or
Using Exponent Lifting

Hence this holds iff $ phi(3^k)$ divides $ n$

Im quite sure my solution is wrong(im very new to NT :blush: )
Now,
$ 3$ divides $ 2^{n}\plus{}1$ for odd n
so, $ 3$ does not divide $ 2^{n}\minus{}1$ for odd n
so, $ 3^{k}\equal{}(2^{n1}\plus{}1)(2^{n1}\minus{}1)$
so, $ 3$ divides $ 2^{n1}\plus{}1$ and n1 is odd
and $ 3$ divides $ 2^{n1}\minus{}1$ and n1 is odd - contradiction
so no solutions


set theory
combinatorics unsolved
combinatorics

There are arbitrary 7 points in the plane. Circles are drawn through every 4 possible concyclic points. Find the maximum number of circles that can be drawn.

I think that the maximum number of circles is $ 5$(just guess)
Let $ ABCD$ is a cyclic quadrilateral.$ AB\cap CD\equal{}F,AD \cap BC\equal{}E$.
The circle $ (ABE)$ cuts the circle $ (ADF)$ at $ M$.We have that $ MEAB,MDAF,ABCD,MCDE,MACF$ are cyclic quadrilaterals.
Is that right?

The maximum number is $ \boxed{6}$ . Consider an equilateral triangle, the midpoint of its edges and its center. (see below).

[url=http://img260.imageshack.us/i/china1990tst.png/][img]http://img260.imageshack.us/img260/4910/china1990tst.png[/img][/url]

We'll will prove that the required number of circles is at most $ 7$. Let transpose our problem into set theory terms.

Choose $ A \equal{} \{[7]\} \equal{} \{1,2,3,4,5,6,7\},$ and let $ M_{1}, M_{2}, \dots, M_{k} \subset A,$ with
\[ |M_i| \equal{} 4 \ \wedge\ |M_i\cap M_j|\leq 2, \ \forall\ i,j\in A, \ i\neq j\]
Then $ k\leq 7.$

[b][u]Proof.[/b][/u] Define $ C_i: \equal{} \{j\in \{[k]\} \ : \ i\in M_j \}, \ \forall\ i\in A$. By counting in two ways we get that:
\[ 2\cdot\dbinom{k}{2}\geq \displaystyle\sum_{i < j}|M_i\cap M_j| \equal{} \displaystyle\sum_{i \equal{} 1}^7 \dbinom{|C_i|}{2} \equal{} \frac {1}{2}\left (\displaystyle\sum_{i \equal{} 1}^7|C_i|^2 \minus{} \displaystyle\sum_{i \equal{} 1}^7|C_i|\right )\stackrel{\mathrm{CBS}}{\geq}\frac {1}{2}\left [\frac {1}{7}\left (\displaystyle\sum_{i \equal{} 1}^7 |C_i|\right )^2 \minus{} \displaystyle\sum_{i \equal{} 1}^7|C_i|\right] \equal{} \frac {8}{7}k^2 \minus{} 2k\]
So  $ \frac {8}{7}k^2 \minus{} 2k\leq k^2 \minus{} k\ \Longrightarrow\ \frac {1}{7}k^2\leq k\ \Longrightarrow\ \boxed{k\leq 7}$

We used above that $ 4k \equal{} \displaystyle\sum_{i \equal{} 1}^k |M_i| \equal{} \displaystyle\sum_{i \equal{} 1}^7 |C_i|$. As an afferent observation, I found that $ k\equal{}7$ is reachable.

[i][hide="Example"][/i]

\[ M_1\equal{}\{1,2,3,4\}, \ M_2\equal{}\{1,2,5,6\}, \ M_3\equal{}\{1,3,5,7\}, \ M_4\equal{}\{1,4,6,7\}, \ M_5\equal{}\{2,3,6,7\}, \ M_6\equal{}\{2,4,5,7\}, \ M_7\equal{}\{3,4,5,6\}\][/hide]

Suppose that we can draw $7$ circles through our points. Then (using what i wrote above) through any point pass exactly $4$ circles. Consider a point $P$ and the $4$ circles passing through it. Note that these circles intersect each other in $6$ points. Making an inversion with pole $P$ and random power our figure maps into a complete quadrangle. It's easy to check that we cannot draw more than $2$ circles using these points. (or else there will be a circle that contains $3$ collinear points, absurd). Hence the contradiction...

This is similar to Mathcamp Admissions this year lol.

The answer is 6. For construction, consider the 7 points $A(0,0), B(1,1), C(1, 2), D(0,3), E(2, 1), F(2,2), G(3,0)$. Then invert about the circle with center $(-5,0)$ and radius $1$. Then the 6 circles are $A'B'C'D', A'E'F'D', A'B'E'G', A'C'F'G', B'C'E'F', D'C'E'G'$, which can be verified by noticing the original points are either cyclic are collinear.

Now to prove that 6 is the maximum. I claim that each point has at most 4 circles going through it. Assume 5 circles are going through a point $P$. Inverting about $P$, we have 6 points and 5 lines with 3 points on each line, which is impossible. Hence each point has at most 4 circles going through it.

Note that if all points had at most 3 circles going through it, we would have less than 6 circles, so assume there exists a point $Q$ with 4 circles going through it. Invert about $Q$. The 4 circles form a complete quadrilateral. In the 6 points that form the complete quadrilateral, there are at most 2 cyclic quadrilaterals. Hence at most 6 circles can be drawn.


integration
calculus
real analysis
real analysis unsolved

suppose $ f_1,f_2, ...$ are such that $ \int_{\Bbb R^n}f_k \equal{} 1$ and $ 0 \leq f_k \leq 1/k$ for all $ k$. Prove that $ \int_{\Bbb R^n}\sup_{k}f \equal{} \infty$

Since the limit of the integrals is not the same as the integral of the limit, a certain well-known theorem must fail to apply to this sequence.


The equation $ 4x\plus{}6y\equal{}128$ has many solutions. If $ x$ and $ y$ are positive integers, what value of $ x$ is associated with the smallest positive integer value of $ y\minus{}x$?

We are given that $ 4x \plus{} 6y \equal{} 128\implies2x \plus{} 3y \equal{} 64$, so taking both sides $ \bmod2$, we have $ y \equal{} 0\pmod2\implies y \equal{} 2m$ for some integer $ m\geq1$ (since $ y > 0$).

Plugging this back into our original equation, we have $ 2x \equal{} 64 \minus{} 3y \equal{} 64 \minus{} 6m\implies x \equal{} 32 \minus{} 3m$, and

$ y \minus{} x \equal{} 2m \minus{} (32 \minus{} 3m) \equal{} 5m \minus{} 32$.

The smallest positive value of $ 5m\minus{}32$ occurs when $ m\equal{}7\implies5m\minus{}32\equal{}3$, which works when $ (x,y)\equal{}(\boxed{11},14)$.

2x+3y=64, y-x=z
3y-3x=3z
64=3z+5x, as z is minimal, that means x is maximum possible such that 64-5x>0, from which x=11, and z=3.

cant y=2 and x=29 which would give u -27????


How many triangles are in this figure?\\
(Some triangles may overlap other triangles.)
[asy]/* AMC8 1998 #4P */
size(72, 24);
pair a=(0,0), b=(36,0), c=(72,0), d=(54, 9), e=(24,24);
draw(e--a--c--e--b--d);[/asy]

Actually, I count six triangles.

You might have missed the biggest triangle.

I see 5 as well (yes, counting the big one.)

The big one, the three little ones, and then two of the little ones at the right make 5.

Whoops, I think I may have counted the triangle on the far left twice.

Sorry, my bad. :blush:

[asy]/* AMC8 1998 #4P */
size(72, 24);
pair a=(0,0), b=(36,0), c=(72,0), d=(54, 9), e=(24,24);
draw(e--a--c--e--b--d);[/asy]

I got 5 as well. My method for these problems is organizing the larger triangles by how many little triangles in each.

Triangles consisting of:
1 little triangle- 3
2 little triangles- 1
3 little triangles- 1

This would be for a count of $ \boxed{5}$ in total.


the following image will automatically give the answer

Since the circles are tangent, that means the radius of one of the circles makes a 90 degree angle at the point of tangence. Since the radius of the other circle makes the same angle, the points are conlinear.
(This isn't really a proper proof but I hope you get the idea).

let the tangent line to the circles by $l$ and $X$ is on line $l$ distinct from $M$, and $A$, $B$ are the centers of the circles

by definition of tangency, $\angle XMA=\angle XMB=90$, then

$\angle XMA+\angle XMB=180=\angle AMB$

therefore $\angle AMB$ is a straight angle or a line and $A$, $M$, and $B$ are collinear

[quote="nxxx"]Circles $\omega_{1}$ and $\omega_{2}$ are tangent at point M. Prove that M and centers of this circles are collinear.[/quote]
[hide]
Draw Circle 1.
Draw a radius of this circle, and draw the Tangent Line to that point. Then draw another radius and its Circle.

The TL is 90 degrees to each radius, thus the Radii are parallel. But since they are connected, it is collinear? Something to that effect.

[/hide]


Let $ n\ge 3$ be an integer. Show that there exist odd positive integers $ x$ and $ y$ with $ 2^n \equal{} 7x^2 \plus{} y^2$.

This is in Engel; one way to find a recursion is to note that $ 2^3 \equal{} (1)^2 \plus{} 7(1)^2 \equal{} N(1 \plus{} \sqrt { \minus{} 7})$, so $ 2^n\cdot2^3 \equal{} N(y \plus{} x\sqrt { \minus{} 7})N(1 \plus{} \sqrt { \minus{} 7}) \equal{} N(y \minus{} 7x \plus{} (x \plus{} y)\sqrt { \minus{} 7})$, and then $ 2^{n \plus{} 1} \equal{} N\left(\frac {y \minus{} 7x}{2} \plus{} \frac {x \plus{} y}{2}\sqrt { \minus{} 7}\right)$. Similarly, $ 2^3 \equal{} N(1 \minus{} \sqrt { \minus{} 7})$ so $ 2^n\cdot2^3 \equal{} N(y \plus{} x\sqrt { \minus{} 7})N(1 \minus{} \sqrt { \minus{} 7}) \equal{} N(y \plus{} 7x \plus{} (x \minus{} y)\sqrt { \minus{} 7})$ and $ 2^{n \plus{} 1} \equal{} N\left(\frac {y \plus{} 7x}{2} \plus{} \frac {x \minus{} y}{2}\sqrt { \minus{} 7}\right)$. It's easy to show that exactly one of these is an odd ordered pair.

Another way is to show
$ (x^2 \plus{} 7y^2)(a^2 \plus{} 7b^2) \equal{} 7(ay \plus{} xb)^2 \plus{} (ax \minus{} 7by)^2$
Then prove for n=3,4,5

EDIT: correction from below post

Nice. By the way, RobRoobiks, the last term should $ (ax\minus{}7by)^{2}$.

Sorry but what does $ N(1\plus{}\sqrt{\minus{}7})$ mean? I mean that $ N$ before the brackets.  :blush:

@ Rob Roobiks can you explain how $ (x^{2}\plus{}7y^{2})(a^{2}\plus{}7b^{2})\equal{}7(ay\plus{}xb)^{2}\plus{}(ax\minus{}7ab)^{2}$ proves the statement?

I get that you got it from Lagrange but how do you prove that it can always be expressed as a power of 2?

I don't get it lol. =(

This seems to me to be the basis for a proof by strong induction along the following lines:

For $ m > n$, $ 2^{n} \equal{} x^{2} \plus{} 7y^{2}$ and $ 2^{m \minus{} n} \equal{} a^{2} \plus{} 7b^{2}$, gives $ 2^{n} \equal{} (x^{2} \plus{} 7y^{2})(a^{2} \plus{} 7b^{2}) \equal{} (ax \plus{} 7by)^{2} \plus{} 7(ay \plus{} bx)^{2}$

[By the way, I think "N" stands for "norm".]

Thank you AndrewTom. Ill edit it right away


function
algebra
system of equations
algebra unsolved

Given an integer $n  \ge2,$solve in real numbers the system of equations
max{$1, x_{1}$} = $x_{2}$
max{$2, x_{2}$} = $2x_{3}$
· · · · · · · · · · · ·
max{$n, x_{n}$} = $nx_{1}$.

We'll use two obvious properties of $\max$ function:

1. If $c>0$, then $c\max\{a,b\}=\max\{ca,cb\}$

2. $\max\{\max\{a,b\},c\}=\max\{a,b,c\}$

Multiply the third equation by $2!$, the fourth by $3!$, the fifth by $4!$ etc... the last by $(n-1)!$:

\begin{eqnarray*}\max\{1!,x_{1}\}&=& x_{2}\\ \max\{2!,x_{2}\}&=& 2!x_{3}\\ \max\{3!,2!x_{3}\}&=& 3!x_{4}\\ \max\{4!,3!x_{4}\}&=& 4!x_{5}\\ &\vdots& \\ \max\{(n-1)!,(n-2)!x_{n-1}\}&=& (n-1)!x_{n-1}\\ \max\{n!,(n-1)!x_{n}\}&=& n!x_{1}\end{eqnarray*}

Each right-hand side appears on the left-hand side of the next equation, hence the property 2 allows us to perform "chain agglomeration" of the equations:

$\max\{n!,(n-1)!,\dots,2!,1!,x_{1}\}=n!x_{1}$

From there we get either $n!x_{1}=n!$ or $n!x_{1}=x_{1}$. However, the second possibility yields $x_{1}=0$, which doesn't satisfy the above equation. Therefore $n!x_{1}=n!\iff x_{1}=1$, and inductively we easily show $x_{1}=x_{2}=\dots=x_{n}=1$, which is the unique solution.


algebra
polynomial
calculus
integration
function
Support
complex numbers

Let $p$ a complex non-constant polynomial and $n(z)$ the number of distinct roots of the equation $ P(x)=z$. What is the value of the sum of the numbers $n-n(z)$ where n is the degree of p and z runs over the set of complex numbers?

Let $z_1,...,z_k$ be distinct roots of $P'(z)=0$. Then if $z\neq P(z_i)$ for all $i$ then $P(x)-z=0$ have all roots of degree 1, so $n(z)=n$ and $n-n(z)=0$. On the other hand if $z=P(z_i)$ then $P(z)-z=0$ has a root $z_i$ of degree $n_i+1$, where $n_i$ is a degree of $z_i$ in $P'(z)=0$. Since $n_1+...+n_k=n-1$ we conclude that Grobber's answer is correct.  :?

I hope it is correct.

That's exactly what I did. ;) We need a bit more work, because we might have $P(z_i)=P(z_j)$, for example, but it comes down to what you did here.

I know what you mean, but first of all, you need an interpretation on the term "sum over all complex numbers" to make this a proper problem.

We don't need this, since the summands are 0 for all $z$ except finite number.

It does make sense as written, Buffalo, at least after the fact. Let $f(z)=n-n(z)$ be defined on $\mathbb{C}.$ We can consider $\mathbb{C}$ to be a measure space with counting measure (each finite set has measure equal to its cardinality). What we're asking for is the integral of $f(z)$ with respect to that measure. Of course, a function only has a chance of being integrable if its support is countable, and an integer-valued function like $f$ can only be integrable if its support is finite. The argument that Myth used contains within itself the proof that $f(z)=0$ for all but finitely many $z$'s, hence it is integrable with respect to counting measure.[/quote]

Hmm... Is that clear explanation?  :lol:  :lol:  :lol:


schur 's inequaliti for n number is true or fales

[quote]WTF is "Schur inequality for n numbers"? Search for "Schur general*" to find lots of different generalizations of Schur, some of them applying to n numbers, but there is no direct extension of Schur to n numbers...

darij[/quote]
I don't know WTK

http://www.mathlinks.ro/Forum/viewtopic.php?&t=89757
contains the "Zaimi-Marku" in post 5 for example which is a form of schur in n variables. :P

[quote]http://www.mathlinks.ro/Forum/viewtopic.php?&t=89757
contains the "Zaimi-Marku" in post 5 for example which is a form of schur in n variables. [/quote]
but in this topic no one prove  :(

maybe i'll post the proof but not now cuz i'm bizy (most probably in january :wink: )


Let $ ABC$ be an acute triangle with orthocenter $ H$ and let $ X$ be an arbitrary point in its plane. The circle with diameter $ HX$ intersects the lines $ AH$ and $ AX$ at $ A_{1}$ and $ A_{2}$, respectively. Similarly, define $ B_{1}$, $ B_{2}$, $ C_{1}$, $ C_{2}$. Prove that the lines $ A_{1}A_{2}$, $ B_{1}B_{2}$, $ C_{1}C_{2}$ are concurrent.

[hide][i]Remark[/i]. The triangle obviously doesn't need to be acute.[/hide]

See the equivalence with http://www.mathlinks.ro/Forum/viewtopic.php?t=126138 ?

  darij

Points $ A$, $ X$, $ A_{1}$,$ A_{2}$,$ B_{1}$,$ B_{2}$,$ C_{1}$,$ C_{2}$ are cyclic.
From trigonometric form of Ceva theorem it is sufficiantly to prove that
$ \frac{B_{1}A_{2}}{A_{2}C_{1}}\frac{C_{1}B_{2}}{B_{2}A_{1}}\frac{A_{1}C_{2}}{C_{2}B_{1}}\equal{}1$
Let $ D$, $ E$, $ F$ are feet of the altitudes from vertices $ A$, $ B$, $ C$ respectively.
Points $ A$, $ H$, $ A_{2}$, $ E$ are cyclic.
$ \angle B_{1}HA_{2}\equal{}\angle EHA_{2}$ ( or $ \angle B_{1}HA_{2}\equal{}180\minus{}\angle EHA_{2}$ )
$ \angle EHA_{2}\equal{}\angle EAA_{2}$ ( or $ \angle EHA_{2}\equal{}180\minus{}\angle EAA_{2}$ )
$ \angle EAA_{2}\equal{}\angle CAX$ ( or $ \angle EAA_{2}\equal{}180\minus{}\angle CAX$ )
$ sin\angle B_{1}HA_{2}\equal{}sin\angle CAX$
The same conclusion we can get using directed angles.
Similarly $ sin\angle C_{1}HA_{2}\equal{}sin\angle BAX$, $ sin\angle C_{1}HB_{2}\equal{}sin\angle ABX$, $ sin\angle A_{1}HB_{2}\equal{}sin\angle CBX$,
$ sin\angle A_{1}HC_{2}\equal{}sin\angle BCX$, $ sin\angle B_{1}HC_{2}\equal{}sin\angle ACX$
But from trigonometric form of Ceva theorem $ \frac{sin\angle CAX}{sin\angle BAX}\frac{sin\angle ABX}{sin\angle CBX}\frac{sin\angle BCX}{sin\angle ACX}\equal{}1$

Just for sake of completeness, note that this problem has been posted in the same form here: [url]http://www.mathlinks.ro/viewtopic.php?t=80025[/url].

[quote="Shishkin"]Points $ A$, $ X$, $ A_{1}$,$ A_{2}$,$ B_{1}$,$ B_{2}$,$ C_{1}$,$ C_{2}$ are cyclic.
From trigonometric form of Ceva theorem it is sufficiantly to prove that
$ \frac {B_{1}A_{2}}{A_{2}C_{1}}\frac {C_{1}B_{2}}{B_{2}A_{1}}\frac {A_{1}C_{2}}{C_{2}B_{1}} \equal{} 1$
[/quote]

i dun understand..why Ceva can be used in this way? I can't find out which triangle Ceva theorem is applied to it...and why sin C_1HB_2=sin ABX??

plz help. :)

[quote="Shishkin"]Points $ A$, $ X$, $ A_{1}$,$ A_{2}$,$ B_{1}$,$ B_{2}$,$ C_{1}$,$ C_{2}$ are cyclic.
From trigonometric form of Ceva theorem it is sufficiantly to prove that
$ \frac {B_{1}A_{2}}{A_{2}C_{1}}\frac {C_{1}B_{2}}{B_{2}A_{1}}\frac {A_{1}C_{2}}{C_{2}B_{1}} \equal{} 1$
[/quote]

yeah , I , don't understand too

1)the triangle$ABC$and$A_1B_1C_1$are similar.
2)$A_2A_1B_1=XAC$and similar othe angle:)
3)because that $XA,XB,XC$are concurrent in triangle $ABC$ than $A_1A_2,B_1B_2,C_1C_2$are concurrent in traingle $A_1B_1C_1$:)
im so lazy to write compleat solution :oops: if have question ask it :wink:

My Solution :
Let feet of perpendiculars of $A,B,C$ be respectively $A',B',C'$
We've $\angle {B_1A_1C_1}=\angle {B'HC}=\angle {C'HB}=A$ and same for $B$ and $C$.
Which implies $A_1B_1C_1$ is similar to $ABC$.
Now $\angle {B_1A_1A_2}=\angle{B_1HA_2}=\angle {XAC}$ same for $B,C$ also.
So $A_1A_2$ correspondence to the Isogonal of $AX$. Same for $B,C$.
Now since $AX,BX,CX$ are concurrent hence their Isogonals are also concurrent. So done  :wink:


In the square $ ABCD$, $ M, \, N$ are the midpoints of $ AB, \, AD$, $ \{ Q \}= CM \cap BN$, $ \{ O \}= AC \cap BD$, $\{ P \}= DE \cap BC$, $E \in QC$.

Prove that:

(a) $ CM \perp BN$;

(b) $ DQ = AB$;

(c) $ P$ is the midpoint of $ BC$ if and only if $ E$ is the midpoint of $ QC$;

(d) $ N,O,P$ are collinear if and only if $ E$ is the midpoint of $ QC$.

[i]Ion Trandafir[/i]

[quote="Andreas"][hide="(a)"]By LAL $\Delta BAN \equiv \Delta CBM \Rightarrow \widehat{MBQ}= \widehat{BCM}\Rightarrow \widehat{MQB}= \widehat{MBC}= 90^{\circ}$.[/hide]

[hide="(b)"]$DCQN$ is cyclic. Hence $\widehat{NQD}= \widehat{NCD}$. Since $N$ is midpoint of $AD$, then $\widehat{NQD}= \widehat{BCM}\Rightarrow \widehat{DCQ}= \widehat{DQC}\Rightarrow DQ = DC$.[/hide]

[hide="(c)"]$PB = PC \Leftrightarrow \widehat{PDC}= \widehat{PCE}\Leftrightarrow \widehat{PEC}= 90^{\circ}$, iff $DE$ is the altitude of an isosceles triangle. Hence iff $CE = EQ$.[/hide]

[hide="(d)"]$NO$ is parallel to $DC$. So $N, O, P$ are collinear iff $NP$ is parallel to $DC$, iff $ND = PC$, and by (c) iff $CE = EQ$.[/hide][/quote]

Nice solutions, but what do you mean in part (a) by LAL?

[quote="lotrgreengrapes7926"][quote="computer_nuke"]Nice solutions, but what do you mean in part (a) by LAL?[/quote]I don't know exactly what it stands for, maybe leg-angle-leg, but it's some form of triangle congurency. It's also called SAS, side-angle-side.[/quote]

O I see, but nice proof anyway =)


A piece of graph paper is folded once so that $ (0,2)$ is matched with $ (4,0)$ and $ (7,3)$ is matched with $ (m,n)$. Find $ m \plus{} n$.

$ \textbf{(A)}\ 6.7\qquad
\textbf{(B)}\ 6.8\qquad
\textbf{(C)}\ 6.9\qquad
\textbf{(D)}\ 7.0\qquad
\textbf{(E)}\ 8.0$

[hide="Explanation"]Actually, at first, there is a trivial error in your solution: The slope of the line through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $\frac{y_2-y_1}{x_2-x_1}$ and not $\frac{x_2-x_1}{y_2-y_1}$. Hence, your slope should be not $\frac {4-2}{0-1} = \frac {2}{-1} = -2$, but $\frac {0-1}{4-2} = \frac {-1}{2} = -\frac12$.

But this is not all. Actually, $\frac {0-1}{4-2} = \frac {-1}{2} = -\frac12$ is not the slope of the folding line (i. e. the line along which the paper is folded). It is just the slope of the line through the points (0, 2) and (4, 0) (and through their midpoint (2, 1)). Now you need the slope of the folding line. This folding line is clearly [i]perpendicular[/i] to the line through the points (0, 2) and (4, 0) (in fact, since the folding matches the point (0, 2) with the point (4, 0), these points must be symmetric to each other with respect to the folding line, so the line joining these points is perpendicular to the folding line). Now remember that if a line l has slope k, then every line perpendicular to l has slope $-\frac{1}{k}$. Since the line through the points (0, 2) and (4, 0) has slope $-\frac12$, the folding line, being perpendicular to this line, has slope $-\frac{1}{-\frac12}=-\left(-2\right)=2$.[/hide]

I hope this was of some help (unfortunately, I don't exactly know which facts that I have used are known to you).

  Darij

Thank you, darij!

I got it now.

[hide="Understanding_this_problem"]After reading the darij's explanation, I found where I was misleading. So, I found the right slope and used that and the midpoint of (m,n) and (7,3) which is expressed as $(\frac {n+3}{2}, \frac {m+7}{2})$.

Then I can set them into two equation:

$\frac {n-3}{m-7} = \frac {-1}{2}$

$\frac {n+3}{2} = 2*\frac {m+7}{2} - 3$

Solving from this, I get $m=\frac {3}{5}$ and $n=\frac {31}{5}$. Add m+n, we'll get 6.8.

Thanks again![/hide]


I assume you're referring to coordinate systems?  Cartesian coordinates are the ones you already know: $ (x,y,z)$.  It is a rectangular system.  We also have polar, which is for circular symmetry, cylindrical, which is just 3D polar, and spherical, which is for spherical symmetry.

It's your most familiar coordinate grid system, like where you have $ (0,0)$, $ (2, 3)$, $ (\minus{}5, 10)$, etc...
http://en.wikipedia.org/wiki/Cartesian_coordinate_system

As someone already pointed out on another of your threads, we're not doing the problem for you.  That doesn't help you at all.  Sorry.

[quote="TKelectronic"]1)  3e^(j*π/4)
2)  1/e^j
3)  e^(j*π/4) + 2e^(-j*π/4)

I would really appreciate your help on exersise above.
Any suggestions-entries are most welcome.[/quote]
fixed so it looks better
$ 1. 3e^{\frac {j \pi}{4}}$
$ 2. \frac {1}{e^j}$
$ 3. e^{\frac {j \pi}{4}} \plus{} 2e^{\frac { \minus{} j \pi}{4}}$

Use $ e^{j\theta}\equal{}\cos\theta\plus{}j\sin\theta$, where $ \theta$ is a real number.

(@moogra: Those are $ \pi$s, not $ n$s.)


What is the smallest positive integer that is a multiple of 8, 12, and 20 and is larger than the least common multiple of 24, 40 and 60?

$ 120$ is the least common multiple of $ 24$, $ 40$, and $ 60$. Multiples of $ 8$, $ 12$, and $ 20$ have to be divisible by $ 120$. Thus, the answer would be the smallest whole number multiple of $ 120$, excluding $ 120$ itself, which is $ 120\times2\equal{}\fbox{240}$.


Como esta cerca el examen de la APMO ...
les sugiero revisen los examenes anteriores para tener una referencia del nivel de estos examenes ..
aki edite las preguntas del año pasado ...

Saludos,

[color=blue][size=150]XVII APMO[/size][/color]
[color=blue] Date: March, 2005.[/color]

$\blacktriangleright$ [color=blue]Problema 1.[/color]
Probar que para cada n\'umero real irracional $a$, existe los
n'umeros reales irracionales $b$ y $b'$ tal que $a+b$ y $ab'$ son
ambos son racionales mientras $ab$ y $a+b'$ ambos son
irracionales.

$\blacktriangleright$ [color=blue]Problema 2.[/color]
Sean $a,b$ y $c$ n\'umeros reales positivos tal que $abc = 8$.
Probar que:
$\frac{a^2}{\sqrt{(1+a^3)(1+b^3)}} + \frac{b^2}{\sqrt{(1+b^3)(1+c^3)}} + \frac{c^2}{\sqrt{(1+c^3)(1+a^3)}} \geq \frac{4}{3}.$

$\blacktriangleright$ [color=blue]Problema 3.[/color]
Probar que existe un tri\'angulo el cual puede ser cortado en 2005
triangulos congruentes.

$\blacktriangleright$ [color=blue]Problema 4.[/color]
En una ciudad mediana, existen $n \times n$ casas indexadas por
$(i,j)$ para $1\leq i,j \leq n$, con $(1,1)$ como la casa en la
esquina superior izquierda, donde $i$ y $j$ son los indices de la
fila y la columna respectivamente. En el momento 0, un incendio se
inicia en la casa indexada por $(1,c)$ donde $c \leq n/2$. Durante
cada periodo subsecuente $[t,t+1]$, los bomberos defienden una
casa la cual a\'un no esta en llamas mientras el fuego se extiende
a todos los $\emph{vecinos}$ indefensos de cada casa que estuvo en
llamas en el tiempo $t$. Una vez que una casa es defendida, esta
permanece asi todo el tiempo. El proceso termina cuando el fuego
no puede extenderse mas. A lo mas, cuantas casas pueden ser
salvadas por los bomberos?
Nota: Una casa indexada por $(i,j)$ es un \emph{vecino} de una
casa indexada por $(k,l)$, si $\mid i-k \mid + \mid j-l \mid = 1.$

$\blacktriangleright$ [color=blue]Problema 5.[/color]
En un triangulo $ABC$, los puntos $M$ y $N$ estan en los lados
$AB$ y $AC$, respectivamente, tal que $MB=BC=CN$. Sean $R$ y $r$
el radio de la circunferencia circunscrita e inscrita del
tri\'angulo $ABC$, respectivamente. Expresar la razon
$\frac{MN}{BC}$ en terminos de $R$ y $r$.

$\text{\LaTeX}{}ed$ by Carlos Bravo 2006 - MathLinks AoPS
$www.artofproblemsolving.com$

Aki ..
la version en pdf .. del examen

XVII APMO !!!--

CARLOS BRAVO ;)
LIma - Peru


linear algebra
matrix
vector
inequalities
triangle inequality
least common multiple

Let $ A\equal{}[a_{ij}]_{n\times n}$ be a matrix with nonnegative entries such that
                  $ \sum_{i\equal{}1}^{n}\sum_{j\equal{}1}^{n}a_{ij}\equal{}n$.
a) Prove that $ |\det A|\leq 1$.
b) If $ |\det A|\equal{} 1$ and $ \alpha\in \mathbb{C}$ is an arbitrary eigenvalue of $ A$, show that $ | \alpha |\equal{}1$.
( we call $ \alpha\in \mathbb{C}$ an eigenvalue of $ A$ if there exists a nonzero vector $ x\in \mathbb{C}^{n}$ such that  $ Ax\equal{}\alpha x$.)

You can see a solution have in IMC Longlist !

Yes, IMC 2005 longlist problem 60. I proved the first part on http://www.mathlinks.ro/Forum/viewtopic.php?t=216071 . Now here is a better second part: Prove that if $ \left|\det A\right|\equal{}1$, then $ A$ is a permutation matrix.

  darij

By Hadamard's inequality we have $|det(A)| \leq || C_{1}||...|| C_{n}||$ (column vectors). We apply AM-GM plus the triangle inequality and use the non-negativity of the entries to get $(||C_{1}||...||C_{n}||)^{1/n} \leq \frac{||C_{1}||+...+||C_{n}||}{n}\leq \frac{\sum_{i=1}^{n}\sum_{j=1}^{n}a_{ij}}{n}=1$.

To prove the second part, note that equality holds if and only if only one entry in each column is non-zero and no row contains two non-zero entries.
Then, if the non-zero entries are $a_{1},...,a_{n} > 0$ their product is $1$ and their sum $n$, thus by AM-GM they are all equal to $1$. Thus, $A$ is a permutation matrix so all its eigenvalues have magnitude $1$ (they are roots of unity, as for each permutation matrix $P$ there is a positive number $k$ such that $P^{k}=I$, just pick the least common multiple of the orders of all the cycles in the permutation and subtract 1, thus we have $\lambda^k=1$ for any eigenvalue).


Hi!!
find all functions f:IR--->IR such that:
f(x+y)=f(x)+f(y)+xy

Let g(x)=f(x)+x^2/2  then Original condition becomes g(x+y)=g(x)+g(y)
So g(x) is of the form c^x where c is a constant. But setting x and y =0 in the original condition we get g(x)=0
  f(x)=c^x-x^2/2 (except when x=0) or if I make some mistakes ...

[quote] mathematikos said:
hi stephencheng!!
I think you have to put g(x)=f(x)-x²/2 no??[/quote]

You are right. Sorry for making such a foolish mistake.


Suppose that $ \left(u_n\right)$ is a sequence of real numbers satisfying $ u_{n \plus{} 2} \equal{} 2u_{n \plus{} 1} \plus{} u_{n}$, and that $ u_3 \equal{} 9$ and $ u_6 \equal{} 128$. What is $ u_5$?

$ \textbf{(A)}\ 40 \qquad \textbf{(B)}\ 53 \qquad \textbf{(C)}\ 68 \qquad \textbf{(D)}\ 88 \qquad \textbf{(E)}\ 104$

[hide="Solution"]
So with 3 numbers in a sequence, you know that the first plus 2 * the second = the third.
We have _ _ 9 _ _ 128
Try plugging a number for the first spot.
Putting in a 1, you get 1 + 2(u second spot) = 9
8 = 2 u
second u is a 4
slowly work up till you get the fifth, and check with the sixth(128)
Our attempt = 1 4 9 _ _ 128
4 +2(9) = 22
9 + 2(22) = 53
[color=red]CHECK: 22 + 2(53) = 128[/color]
So the answer is 53(B)!
[/hide]

[hide="more latexified"]well $ 128 \equal{} 2u_5 \plus{} u_4$ and $ u_5 \equal{} 2u_4 \plus{} 9$ so $ 128 \equal{} 4u_4 \plus{} 18 \plus{} u_4$ and $ 5u_4 \equal{} 110$ therefore $ u_4 \equal{} 22$ so $ 2\times 22 \plus{} 9 \equal{} 53 \longrightarrow \fbox{\mbox{B}}$ [/hide]

[hide]
Kind of the same thing with the plug-in method, but I tried to do as little work as possible (i.e. not having to solve for $ u_4$ heh)

First off,

$ 2u_{n \plus{} 1} \equal{} u_{n \plus{} 2} \minus{} u_n$

here we go:

$ 128 \equal{} u_{4 \plus{} 2} \equal{} 2u_{4 \plus{} 1} \plus{} u_4$

$ 128 \equal{} 2u_5 \plus{} \frac {1}{2}(2u_{3 \plus{} 1})$

$ 128 \equal{} 2u_5 \plus{} \frac {1}{2}(u_{3 \plus{} 2} \minus{} u_3)$

$ 256 \equal{} 4u_5 \plus{} u_5 \minus{} u_3$

$ 256 \equal{} 5u_5 \minus{} 9$

$ 265 \equal{} 5u_5$

$ u_5 \equal{} \boxed{53 \equal{} B}$[/hide]

Sorry to revive this so much later, but I used an easier way.  In problems like these there is about a 99% chance that all the numbers in the sequence will be integers, so...

u5 = 2u4 + 9

Any integer multiplied by 2 becomes even, then if you add an odd number such as 9 you get an odd number.  In the answers, the only odd number is B)53.  After that all you need to do is check.


Ummm... is this supposed to be right?

[quote]AoPS Classes Math Jam Monday, September 15, at 7:30 PM ET/4:30 PM PT.[/quote]

As said on the top...

I'll assume it is Wednesday, but I just want to make sure...

Actually, I think they meant Monday September 13.  Good catch.  I think Wednesday September 15 is another Math Jam, on the Putnam exam.


In a statinoaery store, pencils have one price and pens have another price.  Two pencils and three pens cost 78 cents.  But three pencils and two pens cos 72 cents.  How much does one pencil cost?
[hide="hint"]set up a system of equations[/hide]

[hide] im too lazy to show my work, the answer is 12 cents [/hide]

[hide]let c be pencil and p be pen.
2c + 3p = 78
3c + 2p = 72

solving, which everyone should know how to do, c is 12 cents. [/hide]


calculus
integration
derivative
limit
real analysis
real analysis unsolved

Let $a\in \mathbb R$ and let $f: [a,\infty) \rightarrow \mathbb R$ be Riemann integrable . Suppose that $\int_{a}^{\infty}f(x)dx$ converges, $f$ is differentiable, has a Riemann integrable derivative and $\int_{a}^{\infty}f'(x)dx$ converges.

Given the above, prove that $\lim_{x\rightarrow \infty}f(x) =0$.

Trying to be very, very thorough...

By the Fundamental Theorem of Calculus, $f(x)=f(a)+\int_{a}^{x}f(t)\,dt.$

But one of the hypotheses of the problem is that $\lim_{x\to\infty}\int_{a}^{x}f(t)\,dt=\int_{a}^{\infty}f(t)\,dt$ exists.

Hence, $\lim_{x\to\infty}f(x)=f(a)+\int_{a}^{\infty}f(t)\,dt$ exists.

Suppose $\lim_{x\to\infty}f(x)=L\ne0.$ Suppose, for the sake of the argument, that $L>0.$

There exists $N$ such that if $x>N,$ then $f(x)>\frac{L}2.$ For $b>N,$

$\int_{a}^{b}f(x)\,dx>\int_{a}^{N}f(x)\,dx+\int_{N}^{b}\frac{L}2\,dx= \int_{a}^{N}f(x)\,dx+\frac{L(b-N)}2=C+\frac{L}2\,b.$

In the limit as $b\to\infty,$ this goes to $+\infty,$ and it follows that $\int_{a}^{\infty}f(x)\,dx=\infty.$

Similarly, if $L<0,$ we would have $\int_{a}^{\infty}f(x)\,dx=-\infty.$

Thus the hypothesis that $\int_{a}^{\infty}f(x)\,dx$ converges rules out the possibility of $L\ne 0.$ We conclude that $L=\lim_{x\to\infty}f(x)=0.$

----

Note: if you removed the hypothesis that the improper integral of $f'$ converges, the theorem would be false. Who can produce a counterexample?


algebra
polynomial
system of equations
algebra unsolved

Solve in real numbers $ x,y,z$ the system of equations:
$ x^5\equal{}5y^3\minus{}4z$
$ y^5\equal{}5z^3\minus{}4x$
$ z^5\equal{}5x^3\minus{}4y$

With Maple.
We obtain $ x(x\minus{}1)(x\plus{}1)(x\minus{}2)(x\plus{}2)p_{120}(x)\equal{}0$ where $ p_{120}$ is a polynomial of degree $ 5^3\minus{}5\equal{}120$ whose all coefficients are positive. Then this polynomial has no real roots. After it's easy: the solutions are $ x\equal{}y\equal{}z\in\{0,1,\minus{}1,2,\minus{}2\}$.

A precision:
$ p_{120}$ has no real roots because it's an EVEN polynomial whose all the coefficients are positive.

hello, what version of Maple do you use?
Sonnhard.

Hi Sonnhard,
I use the Maple's command "with(Groebner)"; you find "grosso modo" the same version (it must contain the software FGb due to JC Faugere) in Maple 10-13.


limit
logarithms
calculus
calculus computations

n is positive integer.
\[ \lim_{n\to \infty} \frac{(n\plus{}1)^3 \ln (n\plus{}1)}{n^3} \minus{} \ln n \equal{}0.\]

The limit writes as $ \lim_{n \to \infty} \left ( \ln(n\plus{}1) \minus{} \ln n \plus{} \frac {3\ln(n\plus{}1)} {n} \plus{} \frac {3\ln(n\plus{}1)} {n^2} \plus{} \frac {\ln(n\plus{}1)} {n^3} \right )$ $ \equal{} \lim_{n \to \infty} \ln(1 \plus{} \frac {1} {n}) \equal{} 0$.

So try now to compute $ \lim_{n \to \infty} \left ( \frac {(n\plus{}1)^4\ln(n\plus{}1)} {n^3} \minus{} (n\plus{}4)\ln n \right )$, which is slightly more interesting.

[quote="mavropnevma"]The limit writes as $ \lim_{n \to \infty} \left ( \ln(n \plus{} 1) \minus{} \ln n \plus{} \frac {3\ln(n \plus{} 1)} {n} \plus{} \frac {3\ln(n \plus{} 1)} {n^2} \plus{} \frac {\ln(n \plus{} 1)} {n^3} \right )$ $ \equal{} \lim_{n \to \infty} \ln(1 \plus{} \frac {1} {n}) \equal{} 0$.

So try now to compute $ \lim_{n \to \infty} \left ( \frac {(n \plus{} 1)^4\ln(n \plus{} 1)} {n^3} \minus{} (n \plus{} 4)\ln n \right )$, which is slightly more interesting.[/quote]

use the same method, your problem's answer is 1.

thanks


What is the leftmost digit of $2006^{2006}$.

P.S. u can use your calculators


Two polynomials $ f(x) \equal{} a_{100}x^{100} \plus{} a_{99}x^{99} \plus{} \ldots \plus{} a_1x \plus{} a_0$ and $ f(x) \equal{} b_{100}x^{100} \plus{} b_{99}x^{99} \plus{} \ldots \plus{} b_1x \plus{} b_0$ of degree 100 differ by a permutation of coefficients. It is known that $ a_i\not \equal{} b_i$ for $ i \equal{} 0,1,\ldots,100$
Is it possible that $ f(x)\ge g(x)$  for all real numbers x?

[quote="SimonM"]Is this just the rearrangement inequality?[/quote]
When $ 0 < x < 1$, $ x^{100} < x^{99} < \ldots < 1$, and when $ x > 1$, $ x^{100} > x^{99} > \ldots > 1$. So I don't think it is.

we take $ h(x)\equal{}f(x)\minus{}g(x)\equal{}\sum_{k\equal{}0}^{100}(a_k\minus{}b_k)x^k$
we suppose that $ \forall x\in\mathbb{R}: \ h(x)\ge 0$, we have $ h(1)\equal{}\sum_{k\equal{}0}^{100}(a_k\minus{}b_k)\equal{}0$ so $ (x\minus{}1)^2|h(x)$
so $ \sum_{k\equal{}1}^{100}k(a_k\minus{}b_k)\equal{}h'(1)\equal{}0$ gives $ \sum_{k\equal{}1}^{100}ka_k\equal{}\sum_{k\equal{}1}^{100}kb_k$
i continue tomorow :arrow:


calculus
derivative
trigonometry
function
calculus computations

I can't figure this out...maybe I'm missing something....

what is $dx/d{\theta}$ for $x=2cos(3\theta)$

in my book it says $dx/d{\theta}=-6sin(3\theta)$... but i'm not sure where the 6 came from?

Thanks

nevermind guys...i just realized it's because of the chain rule....forgot bout that


don't answer!; it's from current Polish MO

[quote="inny"]Triangle $ ABC$ ($ \sphericalangle ACB > 90$) is inscribed in circle S. D is a point where $ CS$ cuts $ AB$. $ |AC| \plus{} |BC| \equal{} 2|CS|$. Proof that circles inscribed in triangles $ ADC$ and $ BDC$ have the same radius.

P.S. Sorry for my English..[/quote]
A problem of a Vietnamese Magazine of Mathematics.


geometry
geometric transformation
reflection
articles
LaTeX

Suddenly (as in just today) TeXnicCenter won't compile my document. When I click "build output", it just freezes. The freezing isn't bad--I can click Ctrl-Alt-Delete and immediately close the window with Task Manager. The output file doesn't reflect changes I make. This is the first time I've ever had a problem like this with TeXnicCenter, so I really don't know where to look for reasons. I'm using Windows 2000 if that's any help. Sorry if I'm not being detailed enough...I don't know what details I should include and what I shouldn't.

I need to turn in the Intermediate Counting/Probability midterm tomorrow, so this is precisely the worst time for this to happen, so I need help ASAP.

Without knowing more details, the only thing I can suggest is first rerunning the Profile Wizard (by selecting Define Output Profiles from the Build menu, and then clicking on the Wizard button), and if that doesn't help, you can try manually fiddling with the Profiles and/or as a last resort just reinstalling Texnic Center from scratch.

Even reinstalling didn't work :( Thanks for the help anyway...any other ideas? When I accidentally had it do a DVI file, it didn't freeze, just had a lot of errors because I had things like pdfpagewidth in the document. Should I edit that stuff out and start doing everything with a DVI file? I really would prefer pdf, but...

It sounds like there might be something wrong with the document you're trying to compile. Copy and paste this code into a file and try to compile: [code]\documentclass{article}
\begin{document}
Hello!
\end{document}[/code] If that works, then it's probably the document that is messed up, and you just have to debug it. If it doesn't work, then there's probably something wrong with your LaTeX compiler, and you might have to reinstall that. (TeXnicCenter is merely a shell, i.e. a convenient interface to use MikTeX to build your files; MikTeX is the brains behind it that actually compiles the documents. Reinstalling TeXnicCenter won't fix any problems you may have with MikTeX.)
Good luck!

It just started working again. I have no idea what happened there, but o-kay. Thanks, people.


geometry
geometric transformation
reflection
geometry solved

Three circles $(O_1),(O_2),(O_3)$ concurent at $O$.
The line $\Delta$ pass throught $O$ and intersects $(O_1),(O_2),(O_3)$ at $A_1,A_2,A_3$ respectively. Three lines $\Delta_1, \Delta_2, \Delta_3$ pass throught $A_1,A_2,A_3$ and perpendicular to $O_2O_3,O_3O_1,O_1O_2$ respectively. Prove that  $\Delta_1, \Delta_2, \Delta_3$ concurent.

We will use a lemma from http://cut-the-knot.com/Curriculum/Geometry/Orthopole.shtml :

[color=blue][b]Lemma 1.[/b] Let ABC be a triangle and g any line. If X, Y, Z are the feet of the perpendiculars from the points A, B, C to the line g, then the perpendiculars to the lines BC, CA, AB through the points X, Y, Z are concurrent.[/color]

Now, in your problem, let $D_1$, $D_2$, $D_3$ be the antipodes of the point O on the circles $\left(O_1\right)$, $\left(O_2\right)$, $\left(O_3\right)$, respectively; in other words, let $D_1$, $D_2$, $D_3$ be the reflections of the point O in the points $O_1$, $O_2$, $O_3$. Then, the points $O_2$ and $O_3$ are the midpoints of the segments $OD_2$ and $OD_3$, respectively, and therefore, $O_2O_3\parallel D_2D_3$. Similarly, $O_3O_1\parallel D_3D_1$ and $O_1O_2\parallel D_1D_2$.

On the other hand, since the point $D_1$ is the antipode of the point O on the circle $\left(O_1\right)$, the segment $OD_1$ is a diameter of this circle, and since the point $A_1$ lies on this circle, we thus have $\measuredangle OA_1D_1=90^{\circ}$. Hence, the point $A_1$ is the foot of the perpendicular from the point $D_1$ to the line $\Delta$. Similarly, the points $A_2$ and $A_3$ are the feet of the perpendiculars from the points $D_2$ and $D_3$ to the line $\Delta$. Thus, by Lemma 1, the perpendiculars to the lines $D_2D_3$, $D_3D_1$, $D_1D_2$ through the points $A_1$, $A_2$, $A_3$ are concurrent. But these perpendiculars to the lines $D_2D_3$, $D_3D_1$, $D_1D_2$ are, of course, perpendiculars to the lines $O_2O_3$, $O_3O_1$, $O_1O_2$ (since $O_2O_3\parallel D_2D_3$, $O_3O_1\parallel D_3D_1$ and $O_1O_2\parallel D_1D_2$). And thus we can state that the perpendiculars to the lines $O_2O_3$, $O_3O_1$, $O_1O_2$ through the points $A_1$, $A_2$, $A_3$ are concurrent. This solves the problem.

  darij

I have used the same Lamma to solve this problem. However, my proof to the Lemma is really monstrous... :(

Oh, I forgot to ask what is the defintion of antipode?
Does it refer to the reflection of a point wrt to another point?

If P is a point lying on a circle k, then the [i]antipode[/i] of the point P on the circle k is the reflection of P in the center of k. In other words, the [i]antipode[/i] of the point P on the circle k is the point P' such that the segment PP' is a diameter of the circle k.

  Darij


Find the best $ k$ such that the following inequality holds for positive reals $ a,b,c$ :

$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\minus{}3\geq k(\frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\minus{}\frac{3}{2})$

Link to the Greek forum

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=207043

[quote="silouan"]Find the best $ k$ such that the following inequality holds for positive reals $ a,b,c$ :

$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} 3\geq k(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \minus{} \frac {3}{2})$[/quote]

Here is the solution I have submitted to the Mathlinks contest. :)
Let $ a\equal{}t^3,b\equal{}t,c\equal{}1$ $ (t>0)$, then the inequality becomes
\[ \frac{(t\minus{}1)^2(t^3\plus{}3t^2\plus{}2t\plus{}1)}{t^3} \ge k\cdot \frac{(t\minus{}1)^2(2t^6\plus{}2t^5\plus{}3t^4\plus{}2t^3\plus{}2t^2\plus{}t\plus{}2)}{2t(t^2\plus{}1)(t^3\plus{}1)}\]
\[ \Leftrightarrow k \le \frac{2(t^3\plus{}3t^2\plus{}2t\plus{}1)(t^2\plus{}1)(t^3\plus{}1)}{2t^6\plus{}2t^5\plus{}3t^4\plus{}2t^3\plus{}2t^2\plus{}t\plus{}2}\]
Now, let $ t \rightarrow \plus{}\infty$, we get $ k \le 1$. We will show that it is the value which we find, that is
\[ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b} \plus{}\frac{3}{2}\]
\[ \Leftrightarrow \sum \frac{ca}{b(b\plus{}c)} \ge \frac{3}{2}\]
By the Cauchy Schwarz Inequality, we get
\[ LHS \ge \frac{(ab\plus{}bc\plus{}ca)^2}{2abc(a\plus{}b\plus{}c)} \ge \frac{3}{2}\]
We have done.
In conclusion
\[ k_{\max} \equal{}1\]
:)


how do you guys factor high order polynomials and see symmetries in polynomials? do you guys memorize identities or do you just see  patterns like that?

I mean for instance, can you guys tell immediately the coefficients of polynomial just by looking at symmetries and stuffs like that, and what exactly are the thought process?

for example, the coefficients of
$a(a-b)(b-c)+b(b-c)(c-a)+c(c-a)(a-b)$ ?

You asked a question about factoring and then gave an example about expansion...

As far as factoring goes, most of the factoring tricks I can think of are basically just elegant applications of what is sometimes referred to as the Factor Theorem:

[b]Lemma:[/b]  Given a polynomial $f(x)$, if $f(a) = 0$ then $x-a$ divides $f(x)$.

actually, I meant like if you are given the expansion of somethings that are supposedly
$a(a-b)(b-c)+b(b-c)(c-a)+c(c-a)(b-a)$

how would you figure out that the expression is,
$a(a-b)(b-c)+b(b-c)(c-a)+c(c-a)(b-a)$


Prove that for all $n$,
\[\left(\frac{n}{e}\right)^{n}< n! < e \left( \frac{n}{2}\right)^{n}\]

Second:

Take $ln$ of both sides:

\[\sum_{i=1}^{n}{\ln{i}}< 1+n.\ln{\frac{n}{2}}\]

But by Jensen:

\[\sum_{i=1}^{n}{\ln{i}}\leq n.\ln{\frac{n+1}{2}}\]

So we are done if we can prove that

\[n.\ln{\frac{n+1}{2}}< 1+n.\ln{\frac{n}{2}}\]

or

\[(\frac{n+1}{2})^{n}< (\frac{n}{2})^{n}.e\]

Which reduces to
\[e > (1+\frac{1}{n})^{n}\]

evident by definition of $e$.


geometry
circumcircle
search
geometry proposed
U sviti matematyky

Let $ P$ be arbitrary point inside the triangle $ ABC,$ $ \omega_A,$ $ \omega_B$ and $ \omega_C$ be the circumcircles of triangles $ BPC,$ $ APC$ and $ APB$ respectively. Denote by $ X,Y,Z$ the intersection points of straight lines $ AP,$ $ BP,$ $ CP$ with circles $ \omega_A,$ $ \omega_B,$ $ \omega_C$ respectively ($ X,Y,Z\ne P$). Prove that $ \frac{AP}{AX}\plus{}\frac{BP}{BY}\plus{}\frac{CP}{CZ}\equal{}1.$

This problem can be solved easily with [b]Ptoleme theorem[/b].


Given a triangle $ ABC$ inscribed in a circle $ (O)$, $ M$ and $ N$ are the midpoints of the small arcs $ BC$ and $ AC$ respectively. Take points $ D$ and $ E$ on the rays $ BA$ and $ AB$ respectively such that $ AD\equal{}AC$  and  $ BE\equal{}BC$ (both points are at the exterior of the segment $ AB$). Prove that $ DN$ and $ EM$ intersect at $ K$ which lies on $ (O)$.

Nick Rapanos

$ AM$ intersect $ BC$ at $ X$, $ BN$ intersect $ AC$ at $ Y$. The triangle $ ABX$ is similar to $ AMC$ so $ AM*AX\equal{}bc$. It is well known that $ AY\equal{}\frac{bc}{a\plus{}c}$ so $ bc\equal{}AY*AE\equal{}AX*AM$ so triangle $ AEM$ is similar to $ AXY$. If ME intersect the circle $ (O)$ at $ T$(if $ T\equiv C$ then $ \angle BCM\equal{}\frac{A}{2}\equal{}\angle CEB$ so $ A\equal{}B$ and similar we can prove that $ D,N,C$ are collinear )then $ \angle (MT,MA)\equal{}\angle (TC,CA)\equal{}\angle (XY,YA)$ so $ TC//XY$. Similar we can prove that the line $ DN$ and the line from $ C$ parallel to $ XY$ intersect on the circle $ (O)$. So $ DN$ and $ EM$ intersect on the circle $ (O)$

It is enough to prove that $ \angle CMK \equal{} \angle CNK.$

But it is easy to show that $ \angle DCN \equal{} \angle ECM \equal{} \frac {\angle B}{2} \minus{} \frac {\angle A}{2}$ $ ,(1)$

Also, from the similar right triangles $ \bigtriangleup A'AC,$ $ \bigtriangleup M'MC$ and $ \bigtriangleup B'BC,$ $ \bigtriangleup N'NC,$ it is easy to show that $ \frac {CM}{CN} \equal{} \frac {CE}{CD}$ $ ,(2)$

$ ($ where $ A',\ B'\ M',\ N',$ are the orthogonal projections on $ CD,\ CE,\ BC,\ AC,$ respectively $ ).$

From $ (1),$ $ (2),$ we conclude that the triangles $ \bigtriangleup CEM,$ $ \bigtriangleup CDN$ are similar and so, we have $ \angle MEC \equal{} \angle NDC$ $ ,(3)$

From $ (1),$ $ (3)$ $ \Longrightarrow$ $ \angle CMK \equal{} \angle CNK$ and the proof is completed.

Kostas Vittas.


Prove that in triangle ABC, Ra = K / (s-a), where Ra is the exradius of excircle opposite vertex A, K is the area, s the semiperimeter, and a the length of side BC.

There's no need for double post, your question is answered.

http://www.mathlinks.ro/viewtopic.php?t=292329


We condole with victims on Haiti's Earthquake.

Japanese Communities Moderator

kunny

Yes.

I heard the earthquake had a magnitude of 7.0.

That ~7 magnitude, as a shallow-focus strike-slip earthquake, very close to the city of Port-au-Prince. Shallow-focus is worse from a damage point of view than deep, strike-slip is worse from a direct-damage point of few than other mechanisms (although at least it shouldn't have caused any tsunamis), and Haiti does not, in general, have earthquake-resistant construction techniques. All told, this will probably turn out to be the single worst natural disaster in the Western Hemisphere is a quarter of a century, and maybe longer. And Haiti may be the one country in the hemisphere least capable of coping with such a disaster. They're going to need a lot of help.

Indeed, Haiti is the only country in the Western Hemisphere with a "Very Poor" (<\$2000) PC GDP.

It's amazing how close to the capital it was...15 km...that's 9 miles.

Before I left to the Philippines,
I had to do a group project about a French-speaking-country.
I did my report on Haiti.
I'm Korea for a layover at the moment...
and in my hotel room,
I was watching BBC,
and that came up.
I then got really worried and scared.
I know so much about that country,
and it's falling into pieces in front of me.
I really want to help. The pictures of the result of the quake is just heartbreaking.
~7.0 magnitude...
that just scares me more.

For a country that never had much at all, to have a 7.0 magnitude earthquake is devastating. Some officals have estimated a death toll of up to 500,000. It's sad. I am glad that so many people are trying to help. On the Today show they mentioned that Iceland, USA, Canada, France and China have all sent supplies and/or help.

[quote="bluecarneal"]This is where I would've least expected a quake, especially one that large.[/quote]
Ask what happened to the once-thriving city of [url=http://en.wikipedia.org/wiki/1692_Jamaica_earthquake]Port Royal, Jamaica.[/url]

In plate tectonic terms, the Caribbean Sea is on a separate plate from North America, with the Caribbean moving relatively eastward and North America relatively westward. To the east, in the island arc of the Lesser Antilles, the plate boundary is a thrust fault and the islands themselves are primarily volcanic. Along the Greater Antilles (Cuba, Hispaniola, Puerto Rico) to the north, this is a transform boundary and hence subject to strike-slip faulting. Apparently, the main plate-boundary fault lies off the north shore of Hispaniola (the island that includes both Haiti and the Dominican Republic), but there are other strike-slip faults further south taking up some of the motion, with some sources talking about "microplates" in between the Caribbean and North America. Both the 1692 Port Royal earthquake and the 2010 Port-au-Prince earthquake took place on such faults.

It is true that Central America and southern and western Mexico are more tectonically active than the Greater Antilles, but that does not mean the latter region is inactive.

A comparison of magnitude: the 1989 Loma Prieta earthquake in California was a shallow, partly strike-slip earthquake of just about the same magnitude as this one. It hit a rich first-world area with earthquake experience and rigorous building codes. The urban area closest to the epicenter was Santa Cruz, and Santa Cruz was further away from the epicenter than Port-au-Prince was in this one. Even so, a historic shopping area in downtown Santa Cruz was completely destroyed, dozens of motorists died when a freeway collapsed in Oakland, 50 miles away, and the Bay Bridge was closed for months.

PhireKaLk6781 is right; it was a terrible piece of luck to have this strike so very close to a large city.

There just are some parts of world that are like "Natural Disaster Capitals" and are places where a lot of them happen. Best example in USA: Charleston, SC.

I agree with all of you. It's very odd. The epicenter was in Port-au-Prince, I was surprised it didn't cause a tidal wave. Very sad,, so many lives lost in one a couple tremors.....

Have you heard what our neighborhood right wing nut Pat has been saying about the earthquake? : http://www.huffingtonpost.com/2010/01/13/pat-robertson-haiti-curse_n_422099.html

Sad. Just sad. He also claimed it was a blessing in disguise so the Haitian people could build newer and shinier buildings.

On a totally different note, i'e been thinking about the effect this disaster has had on the global community. How all powers are banding together and channeling their resources to aid Haiti, how petty differences are put aside. Which really begs the question - Is there any way other than a tragedy to so uniformly and powerfully band people together?

[quote="PhireKaLk6781"][quote="bluecarneal"]Hurrricane, hurricane, tropical storm, hurricane, then earthquake?
This is where I would've least expected a quake, especially one that large.[/quote]

7.0, although tremendous, isn't colossal, considering that there is an average of one 8.0+ per year. But Haiti as such a poor country can't afford much disaster-resistance design, such that any natural disaster is magnified many orders of magnitude in Haiti. Comparison: Hurricane Andrew only killed about 200 people when it rampaged through Florida with a Category 5 landfall, Hurricane Jeanne barely scraped Haiti with minimal Category 1 status and killed over 3000 people.

Haiti does lie on a major plate boundary: North American--Caribbean, so it is an earthquake danger zone.[/quote]

We filled our average of 8+ magnitude earthquakes this year, and it is only march.