I was just wondering if any of you are in the Florida State Competition? I am, and so is my team from Trinity Preparatory School, Winter Park, Orange County.

Oooh I will be at Orlando. I got first at the Big Bend Chapter. I think I got lucky (sort of not really). Our team didn't make it cuz we were really stupido. Last year our team got 1st in state. And the year before that. So our coach is disappointed. Now onword to prepare for state.

What school do you go to? I go to Trinity Prep, of Winter Park, Seminole County.

I go to Fairview Middle School

Leon County   (Big Bend Chapter)

People, quit arguing about "I could beat you any other time" - "NO YOU CAN'T!". If you want to argue with each other, do it over PM.

I'm going to states this year.
I go to Bay Point Middle in Pinellas County.
I really want to get to countdown because last year as a sixth grader i got 22nd place.


only $n=2$
for $n>2$ put $\frac{1}{n}<a_{1}<1$ and the rest of them are $1$.


geometry
3D geometry
number theory proposed
number theory

Prove that every integer number can be represented as a sum of 5 cubes.

[i](6 points)[/i]

Too many work! ;) Try to do it in one line. :)

Let x natural and let $a$ such that $6$ divides $x-a$. Then if $ x-a^3=6b$ we have $ x=(b+1)^3+(b-1)^3+(-b)^3+(-b)^3+a^3$.It is more than one line. :(


Let a, b, c, d be natural numbers that satisfy $ b^2\plus{}1\equal{}ac$ and $ c^2\plus{}1\equal{}bd$. Prove that $ a\plus{}c\equal{}3b, b\plus{}d\equal{}3c$

[hide="Might help.."]$ b^2 \plus{} 1 \equal{} ac$
$ c^2 \plus{} 1 \equal{} bd$
_______________(-)

$ b^2 \minus{} c^2 \equal{} ac \minus{} bd$

$ b^2 \plus{} bd \equal{} c^2 \plus{} ac$

$ b(b \plus{} d) \equal{} c(c \plus{} a)$

Plugging in $ a\plus{}c \equal{} 3b$ and $ b\plus{}d \equal{} 3c$,

$ b(3c) \equal{} c(3b)$

$ 3bc \equal{} 3bc$

Now I don't think this proves it but maybe it helps a bit. [/hide]

Note $ \frac{b^2 \plus{} c^2 \plus{} 1}{bc} \in \mathbb{Z}$, eg $ b^2 \minus{} kbc \plus{} c^2 \plus{} 1 \equal{} 0 \quad (*)$ [note that $ \frac{a\plus{}c}{b} \equal{} \frac{b\plus{}d}{c} \equal{} k$]. As $ b^2 \plus{} c^2 \ge 2bc$ it follows that $ k > 2$, and its well-known the equation has solutions only for $ k\equal{}3$ (see [url=http://en.wikipedia.org/wiki/Markov_number]here[/url]). This can be shown by Vieta jumping, namely if $ (b,c)$ satisfies $ (*)$ with $ b>c$, then considering $ (*)$ as a quadratic in $ b$ and using vieta's shows $ \frac{c^2\plus{}1}{b}$ is also a root, which is strictly less than $ b$ (unless $ b \equal{} c \equal{} 1$). Repeating eventually gives $ (1,1)$, from which we easily find $ k\equal{}3$.

See also: [url=http://www.mathlinks.ro/viewtopic.php?t=150373]PEN A5[/url]
[hide="Index"]infinite descent, vieta jumping[/hide]


2 tangent circles have radius 4 and 9. a common tangent is drawn. a circle is mutually tangent to the common tangent and both circles. find the radius of this circle.

This problem looks very familiar to me but I can't remember where from.  Draw triangles and rectangles and you should eventually get it.

By Pythag, the distance between the foot of atitudes from the bigger circle centers to the line is 12. Call the radius of the smaller circle x. The distances between the foot of the altitude from the small circle center to the line and that of the other two altitudes is $ \sqrt {(4 \plus{} x)^2 \minus{} (4 \minus{} x)^2}$ and $ \sqrt {(9 \plus{} x)^2 \minus{} (9 \minus{} x)^2}$, again by Pythag, and they must add up to 12. Solving for x gives 36/25 as the answer.

Basically same question (slightly different with values, etc..)

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1034891#1034891

 :)


Why is $e^{iy}=\cos y+i\sin y$?  :blush:

[hide]Look at the corresponding Taylor Series.[/hide]

[hide]Let $z=\cos x+i\sin x$.  Then we have \[dz=-\sin x+i\cos x=iz\implies\frac{dz}{z}=i\implies\ln z=ix\implies z=e^{ix},\] which is what we wanted to show.
[/hide]

If you don't know any calculus, assume the following equalities and see if you can prove it yourself.

$e^{x}= \sum_{n=0}^\infty \frac{x^{n}}{n!}$

$\cos{x}= \sum_{n=0}^\infty \frac{(-1)^{n}x^{2n}}{(2n)!}$

$\sin{x}= \sum_{n=0}^\infty \frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$

Now of course your next question should be why are those equations true, but... then you would need to get in to the calculus  :wink:

That way (Taylor Series) and the method I posted above are the only two ways of proving this I have seen.  Surely there must be a non-calculus proof of this;  does anyone know of such a proof?

The one that I know (or have seen, rather) is the Taylor Series one.


AMC
AIME
trigonometry
trig identities
Law of Cosines

a circle radius r is inscribed in ABCD. It touches AB at P and CD at Q. AP=19, PB = 26, CQ =37 QD = 23. Find r.

[hide]
Let $O$ be the center of the incircle. Draw the radii of the circle to the points of tangency, and lines connecting the center to the vertices of the quadrilateral. Call the measure of $\angle BAD$ and $\angle BCD$ $A$ and $C$, respectively. From the right triangles formed by the lines drawn in:

$\cos \displaystyle \left(\frac{A}{2} \right)=\frac{19}{\sqrt{19^2+r^2}}$
$\cos \displaystyle \left(\frac{C}{2} \right)=\frac{37}{\sqrt{37^2+r^2}}$

$\cos A=\frac{19^2-r^2}{19^2+r^2}$

$\cos C=\frac{37^2-r^2}{37^2+r^2}$

By Law of Cosines,

$BD^2=AB^2+AD^2-2(AB)(AD)(\cos A)$

and

$BD^2=BC^2+DC^2-2(BC)(DC)(\cos C)$

Set the two RHS equal to each other, and substitute in the expressions for $\cos A$ and $\cos C$.

$AB^2+AD^2-2(AB)(AD)(\cos A)=BC^2+DC^2-2(BC)(DC)(\cos C)$
$45^2+42^2-2(45)(42)\cos A=63^2+60^2-2(63)(60)(\cos C)$

There's more algebra, which I won't type out, but you eventually get to:

$2r^4-1294r^2=0$
$r=\boxed{\sqrt{647}}$
[/hide]


Putting $ x\equal{}y\equal{}z\equal{}1$, we get $ 7^3\geq64\times4^3\iff7\geq16$, which is false. So the inequality is not true.

[quote="155488"]sorry $ (12x^{2} \plus{} (y \plus{} z)^{2})(12y^{2} \plus{} (x \plus{} z)^{2})(12z^{2} \plus{} (x \plus{} y)^{2})\geq 64(x \plus{} y)^{2}(y \plus{} z)^{2}(z \plus{} x)^{2}$[/quote]
It's my problem.I posted it ,last year. can_hang2007 have a nice solution for it.
See here http://www.mathlinks.ro/viewtopic.php?t=321144


Does there exist a measurable space s.t. the set of measurable subsets is countably infinite?

(sorry if you solve it in a few seconds :))

I think this is impossible.

Suppose that all single-point sets are measurable. I think that if this fails, we can replace points with measurable sets of minimal size. (? - I'm not sure of this. This may be a flaw in the argument.)

In order to have infinitely many measurable subsets, the set must be infinite, so it must contain a one-to-one copy of $\mathbb{N}.$ Restrict our attention to that set. Since the collection of measurable subsets is closed under countable unions, and since single-point sets are assumed to be measurable, then every subset of $\mathbb{N},$ being a countable union of singletons, is measurable. And there are $2^{\aleph_0}$ of those.

The thing is that I forgot the exact details of my proof, but it definitely went along these lines. I didn't use that minimality thing. It's enough to find infinitely many disjoint sets in our space, and then we can work with those.

And where is a puzzle? :(
These are absolutely straightforward arguments. :?

It's not my faule that this is what they named that web page :). Besides, I did apologize for it in advance :). I just thought it was a nice fact to ponder, even if it is so easy.

Here's how to clean up Kent's argument: consider a maximal family of disjoint measurable sets $F$. Every measurable set is contained in $\bigcup F$ and for $A$ measurable, $B\in F$, the intersection of $A$ and $B$ is either empty or $B$. If $F$ is finite there are finitely many measurable sets, and if $F$ is infinite there are uncountably many countable unions of sets in $F$.

This is precisely what I did, as I recall.


Let $ f$ and $ g$ be two continuous and bijective functions defined from $ \mathbb R$  to $\mathbb  R$ such that $$ f(g^{ - 1}(x)) + g(f^{ -1}(x)) = 2x$$ for all reals $x$ and there exists a real number $ c$ such that $ f(c) = g(c).$
Prove that $ f(x) = g(x)$ for every real number $ x$.

[quote="tchebytchev"]let $ f$ and $ g$ be two continious and bijectives functions defined from $ R$  to $ R$ such that $ f(g^{ \minus{} 1}(x)) \plus{} g(f^{ \minus{} 1}(x)) \equal{} 2x$ and there exist a real nuber $ c$ such that $ f(c) \equal{} g(c).$
prove that $ f(x) \equal{} g(x)$ for every real number $ x$.[/quote]

Taking $ h(x)\equal{}f(g^{\minus{}1}(x))$ and $ u\equal{}g(c)$, this problem is equivalent to :
let $ h$ be a continuous bijective function defined from $ R$  to $ R$ such that $ h(x) \plus{} h^{ \minus{} 1}(x) \equal{} 2x$ and there exists a real number $ u$ such that $ h(u)\equal{}u.$
prove that $ h(x) \equal{} x$ for every real number $ x$.

Let $ x_0$ such that $ h(x_0)\minus{}x_0\equal{}a\neq 0$. We have $ h(x_0)\equal{}x_0\plus{}a$ and, using $ h(x) \plus{} h^{ \minus{} 1}(x) \equal{} 2x$, we get $ h(x_0\plus{}na)\equal{}(n\plus{}1)a$ $ \forall n\in\mathbb{Z}$.
More, $ h(x)$, since bijective and continuous, is strictly monotonous, and so strictly increasing (since $ h(x_0\plus{}na)\equal{}(n\plus{}1)a$) .

If $ a>0$, let $ p\in\mathbb{Z}$ such that $ x_0\plus{}pa\leq u<x_0\plus{}(p\plus{}1)a$. Since $ h(x)$ is strictly increasing, we have $ h(u)\geq h(x_0\plus{}pa)\equal{}x_0\plus{}(p\plus{}1)a>u$, and so $ h(u)>u$, which is impossible since we know that $ h(u)\equal{}u$

If $ a<0$, let $ p\in\mathbb{Z}$ such that $ x_0\plus{}(p\plus{}1)a\leq u<x_0\plus{}pa$. Since $ h(x)$ is strictly increasing, we have $ h(u)< h(x_0\plus{}pa)\equal{}x_0\plus{}(p\plus{}1)a\leq u$,  and so $ h(u)<u$, which is impossible since we know that $ h(u)\equal{}u$

So it does not exist any $ x_0$ such that $ h(x_0)\minus{}x_0\equal{}a\neq 0$. And so $ \boxed{h(x)\equal{}x}$ $ \forall x$

Q.E.D.

i think there is a mistake;$ h(x_0\plus{}na)\equal{}(n\plus{}1)a$ is not true ,because for $ n\equal{}0$ we have $ f(x_0)\equal{}a$

[quote="drapt@"]i think there is a mistake;$ h(x_0 \plus{} na) \equal{} (n \plus{} 1)a$ is not true ,because for $ n \equal{} 0$ we have $ f(x_0) \equal{} a$[/quote]

Yes, this is just a copy mistake : $ h(x_0 \plus{} na) \equal{} x_0\plus{}(n \plus{} 1)a$
Same correction on the next line.


search
inequalities
inequalities unsolved

Of Jack Garfulkel. In Hungkhtn's book, problem 2.56 in Chapter 2, page 148 (for once who had this book). You can search this forum because it was once discussed here,

Good Luck,

If I don't remember wrongly then Garfunkel inequality is
$\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}\geq\frac{5}{4}\sqrt{x+y+z}$
And a notation for you or some one I don't know who is hungkhtn or his absurd book I have never known about therefore don't recall it in my topic, in this day I listen too much about that absurd book. If you want to show me books please show the more famous books than some unknow books.
Let me show some classic books that if you read you will know more about world of inequalities
1/Inequalities  by G. H. Hardy (Author), J. E. Littlewood (Author), G. Pólya (Author)
2/Analytic Inequalities  by D. S. Mitrinovic (Author), P. M. Vaic (Author)
3/Classical and New Inequalities in Analysis (Mathematics and its Applications)  by Dragoslav S. Mitrinovic, J.E. Pecaric, and A.M. Fink

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=arqady&t=56071

Search with title "Arqady".

To Skywalker. Because your problem was posted in Mathlinker so I cited a book involving it. Hungkhtn's book is not famous all over the world but very famous in Vietnam and furthermore, Hung is my brother, so I know exactly this problem is on his book. I must affirm that it is not absurd book (please don't say that) because it is best seller in Vietnamese. Sorry if it annoy you. Because many problems of yours are so old, searching to them are a hard work and citing a book is easier, so I choose citing a book. That's the reason.

[quote="GacKiem"] ...because it is best seller in Vietnamese.[/quote]
 :D Do you know the best seller book of the world it is Bible  :D, I think you should learn inequality in Bible becasue it is more best-selling than your book.

[quote="GacKiem"]Because many problems of yours are so old, searching to them are a hard work and citing a book is easier, so I choose citing a book. That's the reason.[/quote]
I don't say mine is new, but in the forum if you want to reply you must send the link,or if you want to cite, you should show the famous book, don't cite from a nameless book.

Stop discussing this matter here! This book is of Hungkhtn, not mine. I only like him. I also cited the link to show a satisfied solution.  :mad:

For you, if you want, a nice problem proposed by Kien. Don't say bad thing about the book you don't know.  :mad:

Let $a,b,c$ be non-negative real numbers. Prove that
\[\sqrt{\frac{a^{3}}{a^{2}+ab+b^{2}}}+\sqrt{\frac{b^{3}}{b^{2}+bc+c^{2}}}+\sqrt{\frac{c^{3}}{c^{2}+ca+a^{2}}}\ge \frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{3}}.\]

The underlying problem is that the Internet is everyone's and no one's at all. Anyone here can have more than one ID with a password and say things as if they are quite different people. Actually they might be just a single one. So it is understandable if someone wants to say beautifully about something again and again. Day by day it has come to be an "excellent" marketing strategy.

You shouldn't be angry with me,I really don't know that book (and I belive all people in the world except a small number of Vietnamese are as me  :D  :D )
However I have to say thank you about your link with nice solution of Peter Scholze!
I will think about your problem!
I don't know what pvthuan said but I see Vietnamese prefer spam to post solution  :D

[quote="me@home"][quote="skywalker"]Let $x,y,z>0$ prove that
\[\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}\geq\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{2}}\]
[/quote]
By AmGm
\[\frac{x}{\sqrt{y+z}}\geq 2\sqrt{x}\frac{x}{x+y+z}\]
Thus
\[\sum\frac{x}{\sqrt{y+z}}\geq \sum 2\sqrt{x}\frac{x}{x+y+z}\geq \frac{2}{3}\sum \sqrt{x}\]
The last by rearrangement[/quote]

I think that you are wrong!

$\frac{2}{3}\sum \sqrt{x}\leq \frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{2}}$ Because $\frac{2}{3}\leq \frac{1}{\sqrt 2}$ :wink:

Do you solve my similar problem I posted above, skywalker?

Yes! Very good Job! Yours are the same as mine.  :lol:


[quote="pavel kozlov"]Let $\alpha,\beta$ be different roots of the equation:
$a\cos{x}+b\sin{x}=c$.
Prove, that $\cos^2{\frac{\alpha-\beta}{2}}=\frac{c^2}{a^2+b^2}$.[/quote]
Let $\frac{a}{\sqrt{a^2+b^2}} = cosd$,$\frac{b}{\sqrt{a^2+b^2}} = sind$
Then the equation is equivalent to
$cos(d-x) = \frac{c}{\sqrt{a^2+b^2}}$
So $d-\alpha =\beta -d +2k\pi$ ($k \in Z$)
$\to$ $d- \alpha = \frac{\alpha-\beta}{2} +k\pi$
$\to$ ${\cos^2{\frac{\alpha-\beta}{2}}=\frac{c^2}{a^2+b^2}}$


If $a,b,c$ are pozitive real numbers, solve:

$\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}\geq \frac{27}{2(a+b+c)^{2}}$

you can use http://en.wikipedia.org/wiki/Holder%27s_inequality  :wink:

[quote="N.T.TUAN"]This is other solution           [url=http://www.kalva.demon.co.uk/jbalkan/soln/jsol024.html]6th Balkan 2002/4[/url][/quote]

Junior Balkan rather.

@mods: Please merge this topic with http://www.mathlinks.ro/Forum/viewtopic.php?t=1045 .

  darij


AMC AIME blogs probability USA(J)MO USAMO geometry

I'm officially going to be writing another Mock AIME, hopefully posting it up in a week or so (shooting for this weekend, we'll see if I have time... otherwise it might be over Thanksgiving break too unless people have objections to taking a mock AMC and AIME in the same four-day weekend). I'll probably post it, give you about 3 days to submit answers, grade them, and then post results (maybe solutions or you guys can). It will be self-timed; we're all honest people here, right? :D

Same rules as a regular AIME, I'll probably post it up in this same thread.

And if someone wants to inform me as to which number Mock AIME this is, that'd be great.

it would be the first for the 2006 i believe.

i will participate as well.

Oh, should we say here that we're intending on participating? I shouldn't have any problem as long as the time isn't set, so I'll do it as well.

OK, yeah it'll 2006 Mock AIME 1, and I'm planning on doing it probably Sunday, 12 AM EST (midnight before Sunday) to Wednesday, 12 AM EST (midnight after Tuesday)... or Friday, 12 AM EST (midnight before Friday) to Monday, 12 AM EST (midnight after Sunday).

So... depending on when I finish the test, you can probably be sure it'll be one of those timeframes. Or within an hour of each. I will give precisely 72 hours after posting it for you to submit solutions. Please put them in the form:

1. Answer
2. Answer
etc.

That'll make grading much easier. Anyway, I hope to write a better test than last time, which had a few errors and some problems that were too easy/quick. So if you want you can post (why, I don't know), and now I should get back to writing that test.

Hmm so are you making this more difficult than the real AIME?

I'll probably be able to do it...unless something comes in the way.

I'll do it!  :D

Well, if he wants solutions... you might want to ask some people to send some solutions.

[quote="JesusFreak197"]Wait... by Answer, do you mean just the integer solution from 0 to 999, or can we send you our solution to the problem, as well?[/quote]

Umm, I'd rather have just the answer, for grading purposes. There will be a discussion thread made if anyone wants to post solutions (I doubt I'll have time to make an official one).

And to eryaman, I don't think it will be much harder than a real one, for one thing I can't write problems that are too hard. But difficulty level is always hard to judge when you're the problem writer so you'll find out soon enough.

[quote="FMako"]I'm interested.  In the discussed time region, should we pm you the answers in the discussed format?[/quote]

Yes, please.

[quote="furious"]I'm in
and it's for this weekend correct?[/quote]

Yup, I'm about 2/3 the way done now, and I'll revise a bit... so probably Friday-Sunday, the 19th to the 21st, all day on all three days (i will be gone at a state competition friday and saturday, but i won't PM/post grades until after the deadline anyway).

I agree with #9, but #7 seemed easy enough :huh: I was running out of time so I decided not to finish 9 anyway.

I solved 14 problems, sort of guessed on #13( I took about 2 hour 20 minutes).

BTW I hope the brackets in #14 does not mean greatest integer function.

My opinion of the test:
1-6 were easy, and I think I got them all.
7 was also straightforward, but I didn't realize a certain number was not prime, so I missed it.
8 was straightforward, but I may have messed up in counting.
9 looked simple, but I got intimidated by multiplying, so I didn't solve it.
In 10, I forgot the formulas, and couldn't remember
11 was hard, but I managed to get it
12 is an often used problem, but I think I made an arithmetic erroe.
13-15 I didn't get.
S

I liked it, although it was a bit easy. High median score at least, 13+ was rather difficult.

Geometry was fun on this one, just the kind I like... easy.   The analytic questions were nice.  13-15.... looked wayyyy too intimidating.  There were a few too many easy questions in the first 10.

[quote="1234567890"]In 10, I forgot the formulas, and couldn't remember[/quote]

Same. I learned them in Algebra II, but I forgot them now. :wallbash:

There has been at least one correct solution to everyone problem now EXCEPT #15. I'm fairly sure my answer is correct, but I'm getting a bit worried since no one has provided the same answer yet: for those curious here's a diagram (I think it was written clearly... and yes I suppose the extensions were not necessary but I think it made it more clear for some people, perhaps poorly worded on my part).

[quote="beta"]BTW I hope the brackets in #14 does not mean greatest integer function.[/quote]

No, it doesn't. I'd LaTeX the floors in anyway.

[quote="JesusFreak197"]It didn't say that the extension of $AF$ met the extension of $CD$, it just said that it met $CD$. That's where I got confused. I thought it might mean that, but I wasn't sure, and I don't know if I could have gotten it anyway.[/quote]

I see. The extensions weren't necessary at all, though, since a line is defined to be infinite. I would have drawn a diagram, but I have issues putting images into a PDF file.

[quote="Altheman"]i see that your construction works, but what about this one?[/quote]

Umm, $\tan{\angle BDE}$ does not quite equal $3$ in that... I'm fairly sure that if you account for that piece of information, $G$ has to lie beyond $C$.

oh ok, i see now...that is pretty difficult to see though

ahhhhhhh, solved for IH in 15 lol
and included 10 as a digit in #2.....

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

[url=https://artofproblemsolving.com/community/c3745703_2006_jeffrey_wang_1st_mock_aime]here[/url]

enjoy  / start solving

USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more AIME mocks [url=https://artofproblemsolving.com/community/c2439872_aime_mocks]here[/url]
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)

PS. Answer key is [url=https://artofproblemsolving.com/community/c5h62491p373622]here[/url]


calculus
Ross Mathematics Program
derivative
integration
vector
function
geometry

My school does not have AP Calculus BC, and I want to study on my own for the BC exam. We are using the Addison-Wesley textbook called "Calculus: A Complete Course" 2nd edition by Ross L. Finney, et al.

We will have covered chapters 1 to 7 by the end of the course to prepare for the AB exam. These are:

1. Prerequisites for Calculus
2. Limits and Continuity
3. Derivatives
4. Applications of Derivatives
5. The Definite Integral
6. Differential Equations and Mathematical Modeling
7. Applications of Definite Integrals

The next few chapters are:

8. L'Hopital's Rule, Improper Integrals, and Partial Fractions
9. Infinite Series
10. Parametric, Vector, and Polar Functions
11. Vectors and Analytic Geometry in Space
12. Vector-Valued Functions and Motion in Space
13. Multivariable Functions and Their Derivatives
14. Multiple Integrals
15. Integration in Vector Fields

Which of chapters 8 to 15 do I need to study to be ready for the BC exam?

We are currently on chapter 5 at school. Do each of these topics build on top of one another? Or, can I start studying, say, chapters 8, 9, and 10 right now without reading 6 and 7?

Based on what I have read about the BC course description, it seems to me that I only have to do chapters 8, 9, and 10. However, I am still not sure when I can start any of these chapters, i.e. I don't know if they are independent of the intervening chapters or not.

Any recommendations would be greatly appreciated because my school is starting to register for AP exams this week.

The BC exam covers all of those topics up through Chapter 10. But if you're only up to Chapter 5 right now, I would strongly advise you to take the AB rather than the BC exam in May. You really have a long road to go to get ready for the BC Exam from there. The topics in chapters 5, 6, and 7 are absolutely essential to everything in calculus - and they take some time to work through.

No matter which test you take though, my advice to anyone getting ready for the AP Calculus exam is to get a really good Test Prep book and work through it relentlessly until the exam. (I prefer the one by Barrons, but there are other good ones out there). Using those books as your primary source is better than a regular textbook because they cut right to the chase of what you need to know for the exam, and they have the exact kinds of questions that will be on it. Then you can use your textbook as a backup to explain the things that you don't understand from the Test Prep Book. (Note: This is not the teacher in me talking - who would like all students to understand why all the things in calculus work; it's the pragmatist in me - who realizes that the AP Calulus Exam is problem-solving oriented).

While I'm not saying it's impossible to get ready for the BC exam from where you are right now, I think it would be very risky to try it. Just getting through the rest of the topics for the AB exam and prepping for it could keep you busy until May.

[quote="mathnovice"]. . . . We are using the Addison-Wesley textbook called "Calculus: A Complete Course" 2nd edition by Ross L. Finney, et al.

We will have covered chapters 1 to 7 by the end of the course to prepare for the AB exam.

[snip]

The next few chapters are:

8. L'Hopital's Rule, Improper Integrals, and Partial Fractions
9. Infinite Series
10. Parametric, Vector, and Polar Functions
11. Vectors and Analytic Geometry in Space
12. Vector-Valued Functions and Motion in Space
13. Multivariable Functions and Their Derivatives
14. Multiple Integrals
15. Integration in Vector Fields

We are currently on chapter 5 at school. Do each of these topics build on top of one another? Or, can I start studying, say, chapters 8, 9, and 10 right now without reading 6 and 7?

Based on what I have read about the BC course description, it seems to me that I only have to do chapters 8, 9, and 10. Any recommendations would be greatly appreciated because my school is starting to register for AP exams this week.[/quote]

The College Board [url=http://www.collegeboard.com/student/testing/ap/sub_calbc.html]AP calculus BC Web page[/url]  includes a link to a complete course description for the BC course, which would indicate what topics are covered. My educated guess is that your educated guess is correct about which chapters of the Finney textbook are covered by the BC exam. The front matter of the textbook must include some statement by the author about how independent the chapters are, but you could always read ahead anyhow to see how much you understand and how many exercises you can do.

I would expect your friendly public library would have a certain number of cram books about the BC syllabus that you could check out and use for test prep.

Hope this helps!

I'm doing the same thing you're doing, and I think I have the same textbook. I've found that the last few chapters in the book are studiable without knowledge of the other sections. This is especially true with chapter 9, which you probably already have a fair bit of knowledge in if you do a fair amount of contests. The others are studiable if you focus on the basic concepts, not the mechanics of working the problems. You will have to wait until you learn u-substitution, etc, until you can work all the problems. Good luck on the studying...

You can take an AP Exam without the course at your school?

Oh by the way I have the same textbook and I sort of browse through another Calculus textbook (calculus 8th edition by varberg). The best way to prepare is to go through an AP CALC BC Study book lke the one Barrons offers or College Board AP stiff and read it. go read in your textbook the parts that you are not familiar with when you read the book.

Chapter 12 isn't on the exam.

[quote="rlee"]Buy the AP Cliff's Calculus AB/BC book to study.

(Shameless plug for my high school Calc BC teacher who wrote it :)).

But seriously, it is a good book. Onto BC material--I always felt that the BC material was very short. With some dedication, one could learn  chapters 6 to 10 in a month without any big problems. Chapters 13 to 15 are definitely not on the test. You definitely need to know 1-10. I hope by "infinite series" you mean all of that Taylor/Maclaurin polynomial stuff too. You also need to know Chapter 12, which would be motion in space.

If you do decide to prep for Calc BC, I would advise saving Chapter 9 for last. It is most certainly, by a long shot, the most difficult set of concepts in calculus in my opinion to understand and apply. After understanding just what the definite integral is, all other concepts should come easily that are tied with it. L'Hopital's rule is easy to learn. Polar and parametric are a little tricky but aren't too bad.[/quote]

By the way...what do you have to know about vector functions on the test. It is the one gap in my knowledge right now...

Yeah, me too. I've never been taught about vector and parametric functions.

Whoa-hoh. Parametric and vector functions are DEFINITELY tested on the AP exam. If you don't know them, you'd better know them by May.

will there be any of the six long problems that require a deep understanding of parametric, vector, or polar functions?

That depends on how they write the test.  It is certainly possible.  If you don't know, it's probably very advisable that you go get a review book specifically for the exam and look at it and see.  (For those who don't like spending money on things like that, either your public library could have some, or you could do my personal favorite and go sit in a bookstore and read it there.)

[quote="mathnovice"]Does anyone know a source for past AP exam multiple choice or short response questions?[/quote]

I don't know about mult choice, but you can buy a book with 15 years worth of free response with solutions at:
http://www.moems.org/calc.htm

You can also buy from the College Board each year's multiple choice and free response tests. However each year costs about $15 (It includes the scoring guide on the free-response) and most AP CALC teachers have them so just ask them. I have the Free-Response by George Lechner that was mentioned above and I have recent free-response solutions from websites (try google).

What is a monotonic function?

This is exercise 27 from lesson 8.1. Evaluate lim x --> 0+ (lnx - ln sin x). Can you use L'Hopital's rule on this? There is no way to change it into a fraction.

That's not right chruchill. You don't know what the relative rates at which they tend to infinity, so adding them together gives you an indeterminate form. For example, what would you say about lim x->:inf: ((x+1) - (x))? What about lim x->:inf: (x2 - x)? Or lim x->:inf: (:sqrt:x - x)?

But for this problem, consider what you know about natural logs - specifically, when you add or subtract them.

Churchill, your answer of 0 happens to be correct, but your method is fallacious.  You can't do arithmetic with infinity like that.  Otherwise, you could just as well argue that the limit (as x approaches 0 from above) of ln(x2) - ln(sin x) is also 0.  But that limit is actually -:inf:.

Here is a correct approach for solving the original problem.  Note that ln(x) - ln(sin x) equals ln(x / sin x) for all x between 0 and :pi:.  So all you have to do is find the limit of x / sin(x) as x approaches 0.  That's a well-known limit, or you can find it via L'Hopital's rule.

I just noticed that gauss beat me to the punch.  Oh well.

You can't [i]really[/i] find the limit of sin(x)/x as x-->0 with L'Hpital's rule. How do you propose to find the derivative of sin(x) without evaluating that limit?

I understand that you can get it of the form ln (x/sin x), but you can't use L'Hopital's rule on it. Is there no analytic method of evaluating the limit?

Since most people used the squeeze theorem (cos(t) :le: sin(t)/t :le: 1/cos(t) ) to prove that sin(x)/x=1 you can always flip the whole thing

It is safe to say that

lim x-> :inf: ln(sinx/x) = ln(lim x-> :inf: sinx/x)

So just apply L'Hopital's Rule like that.

Since when did they include vector calculus on the AP Test. I've never heard of this...

It's not hardcore vector calculus. Just position vectors, velocity vectors, acceleration vectors, and basic stuff like that.

[i]Moderator's note: This reply was edited to avoid horizontal scrolling caused by an overly long hyperlink. No substantive change was made in the content of the helpful post below.[/i]

The [url=http://www.collegeboard.com/student/testing/ap/calculus_bc/topic.html]Calculus BC Topic Outline[/url] from collegeboard.com.

Hey, why is the thread scrolling horizontally again on a full-screen window viewing this forum on a 1024 by 768 pixel monitor? Are there mathematical expressions here that are taking up whole lines and then some? Could someone please edit posts to see if we can get everything back into the same horizontal spacing that the rest of the forum has?

Just checking. If the first person to post the horizontally scrolling message doesn't edit it, then I'll have to try to edit it myself.

(After edit:) The problem was fixed; it was a result of a long URL inserted into a message without the use of the URL command attached to shorter link text.

Click the quote button to see how I did it. [url=http://www.artofproblemsolving.com]This[/url] is the AoPS homepage.

[quote="churchilljrhigh"]I think it was the hyperlink! Can someone teach me how to edit hyperlinks so they are like a few words, but when clicked they lead to the website?[/quote]

Thanks, Churchilljrhigh, for figuring out which post was at issue.

As Complex Zeta mentioned, you can make any passage of text in one of your messages into a hyperlink by writing a little bit of link text, then selecting that text while composing your message and using the URL control above your message composition form field. That puts URL codes (opening and closing URL codes) on both sides of your link text. Then you add an equal sign to the first URL code, just inside the closing bracket, and paste in the URL of your choice. Or you can put in all those codes by hand typing if you like.

Here is an [url=http://www.collegeboard.com/student/testing/ap/calculus_bc/topic.html]example link[/url] to show what it looks like, pointing to the College Board page that Gauss202 so kindly shared.

Thanks for spotting the message that was causing the scrolling, and for your participation on this forum.

There's an excellent resource for web-link-shrinking:  [url]http://www.tinyurl.com[/url] will take any url and convert it into a short one of 30 digits or less.  The shorter link (which will look something like http://tinyurl.com/ys44s  (don't click on that -- I just made it up and am not responsible for where it leads)) is permanent and nice to use.  And free.

According to the BC syllabus these aren't on the exam, but could someone just confirm for me?

1. relative rates of growth (i.e. order, little-oh, big-oh notation)?

2. limit comparison test/ direct comparison test

3. using integral tables, trigonmetric substitutions

You definitely have to know how to take some integrals especially some basic trig ones like int(sec(x),x) and int(csc(x),x). If you can do those like right now instantly without hesistation I think you are pretty much trig-ready.

1. No, O-notation isn't on it.
2. I am not *sure* that limit and direct comparison tests are on it, but I would definitely know them. How can you do series without those tests? I'd definitely learn them as well as all the other divergence/convergence tests.
3. No, you don't need to know how to use a table of integrals. By trig-substitution, if you mean manipulating trig using power-reducing rules or trig identities, then yes, you need to know them.

If you mean trig sub like let t = sin :theta: , then yes, I think you need to know that too :). At least I do.


Let $ x_1,x_2,...,x_2010$ be the positive real number satisfying $ (x_1\plus{}x_2\plus{}...\plus{}x_{2010})^2$≥$ 2009(x_1^4\plus{}x_2^4\plus{}...\plus{}x_{2010}^4)$.Prove that $ x_i,x_j,x_k$ are 3 sides of a triangle.(0<i<j<k≤n)

[quote="SUPERMAN2"]Let $ x_1,x_2,...,x_2010$ be the positive real number satisfying $ (x_1 \plus{} x_2 \plus{} ... \plus{} x_{2010})^2$≥$ 2009(x_1^4 \plus{} x_2^4 \plus{} ... \plus{} x_{2010}^4)$.Prove that $ x_i,x_j,x_k$ are 3 sides of a triangle.(0<i<j<k≤n)[/quote]
It is China MO 1999:wink:


1. A regular n-star is the union of n equal line segments $P_{1}P_{2}, P_{2}P_{3}, ... , P_{n}P_{1}$ in the plane such that the angles at $P_{i}$ are all equal and the path $P_{1}P_{2}...P_{n}P_{1}$ turns counterclockwise through an angle less than 180 degrees at each vertex. There are no regular 3-stars, 4-stars or 6-stars, but there are two non-similar regular 7-stars. How many non-similar regular 1000-stars are there?

Could someone explain the solution to this one?

Also, I'm not sure what they mean by "the path $P_{1}P_{2}...P_{n}P_{1}$ turns counterclockwise through an angle less than 180 degrees at each vertex"...

Imagine n-vertices $A_1, A_2, A_3, ..., A_n$ or a regular n-gon labeled counter-clocklwise. Regular n-gons themselves are not considered regular n-stars, so the consecutive vertices $P_1-P_2-P_3-...-P_n- P_1$ of an n-star can be

$A_1-A_3-A_5-...-A_{n-1}-A_1$
$A_1-A_4-A_7-...-A_{n-2}-A_1$
$A_1-A_5-A_9-...-A_{n-3}-A_1$
etc.

For example, the 2 non-similar regular 7-stars are $A_1-A_3-A_5-A_7-A_2-A_4-A_6-A_1$ and $A_1-A_4-A_7-A_3-A_6-A_2-A_5-A_1$. If you try this with 8-, 9, stars, you quickly find that the step must be more then 1 (no regular polygons), not a factor of n (you would run in cycles, never exhausting all available vertices) and less than ${\frac n 2}$ (stars with larger steps would be similar to those with steps less than ${\frac n 2}$). Find how many numbers m are there between $1 < m < \frac n 2$, which do not have a common factor with $n = 1000$. Since $1000 = 2^3 5^3$, you have to take numbers 1 to 500,  elliminate 1, all even numbers, all numbers ending with the digit 5, and no other numbers. There are 4 numbers per decade (ending with 1, 3, 7, or 9) not having a common factor with 1000, take away the initial one:

$N = \left( \frac{1000}{2} : 10 \right) \cdot 4 - 1 = 50 \cdot 4 - 1 = 199$ non-similar regular 1000-stars

[quote="probability1.01"]Hint: this problem has to do with the number of numbers less than 1000 relatively prime to 1000.  However, by their definition, I don't see why a regular n-gon doesn't count as a star...  :?[/quote]

The original problem actually stated that "each of the n line segments intersects at least one of the other line segments at a point other than an endpoint".  This was a shortened version.

yetti: thanks for the solution--I understand it now.

But I still don't know what they mean by "the path $P_{1}P_{2}...P_{n}P_{1}$ turns counterclockwise through an angle less than 180 degrees at each vertex"...

You can consider the segments $P_iP_{i+s\ mod\ n}$ oriented - they have size (all equal) and direction (from $P_i$ to $P_{i+s\ mod\ n}$). Likewise, angles can be oriented, being positive, when oriented counter-clockwise, and negative, when oriented clockwise. Take the oriented line $P_iP_{i+s\ mod\ n}$, which makes an oriented angle $\alpha_i$ with some fixed line in the plane (say, the x-axis, which itself is oriented) and add some counter-clockwise oriented (i.e., positive) angle $\phi < 180^o$, fixed for a given n-star. If the sum of the 2 oriented angles exceed $360^o$, subract $360^o$ from the sum. Each oriented segment (or line) of the star will form the same oriented angle $\phi$ with the 2 adjacent oriented segments.


Let $A_1$ be the center of the square inscribed in acute triangle $ABC$ with two vertices of the square on side $BC$.  Thus one of the two remaining vertices of the square is on side $AB$ and the other is on $AC$. Points $B_1,\ C_1$ are defined in a similar way for inscribed squares with two vertices on sides $AC$ and $AB$, respectively. Prove that lines $AA_1,\ BB_1,\ CC_1$ are concurrent.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Here is my solution:

Let X, Y, Z be the centers of the squares BCTU, CAVW, ABSR constructed on the segments BC, CA, AB and pointing into the exterior of triangle ABC. The square inscribed in triangle ABC with two vertices on BC, one on AB and one on AC is homothetic to the square constructed on the segment BC into the exterior of triangle ABC, the center of homothety being A; hence, the centers $A_1$ and X of these two squares are collinear with A, i. e. the line $AA_1$ coincides with the line AX. Similarly, the line $BB_1$ coincides with BY, and the line $CC_1$ coincides with CZ. Hence, instead of proving that the lines $AA_1$, $BB_1$, $CC_1$ concur, it will be enough to show that the lines AX, BY, CZ concur.

Well, why do the lines AX, BY, CZ concur? There are several proofs of this fact; here is the most elementary of them: I will show that the lines AX, BY, CZ are perpendicular to the lines YZ, ZX, XY, respectively; then, it will follow that AX, BY, CZ are the altitudes of triangle XYZ and hence concur.

Why are the lines AX, BY, CZ perpendicular to YZ, ZX, XY?

Well, the spiral dilatation with center B, rotation angle 45 and dilatation factor $\sqrt2$ maps C to T and Z to A. Hence, the segment TA is the image of the segment CZ, so that

  < (CZ; TA) = 45

(where < (k; l) denotes the directed angle between two lines k and l).

The spiral dilatation with center C, rotation angle 45 and dilatation factor $\frac{1}{\sqrt2}$ maps T to X and A to Y. Hence, the segment XY is the image of the segment TA, so that

  < (TA; XY) = 45,

and < (CZ; XY) = < (CZ; TA) + < (TA; XY) = 45 + 45 = 90, so that CZ is perpendicular to XY. Similarly, AX is perpendicular to YZ and BY is perpendicular to ZX, and we are done.

[By the way, the same approach using spiral dilatations shows that AX = YZ, BY = ZX and CZ = XY.]

  Darij

Let $D,E,F$ be the intersections of $AA_1,BB_1,CC_1$ with $BC,CA,AB$ respectively. The homothety which takes the square with one side on $BC$ to the external square having $BC$ as a side has $A$ as its center, so the center of the large square also lies on $AA_1$, so the problem is reduced to the following:

Erect right isosceles triangles $BCM,CAN,ABP$ (the right angles are at $M,N,P$) outside $ABC$. Show that $AM,BN,CP$ are concurrent. this is, of course, Vecten's point. Here's a proof of the concurrence:

Let $X$ be the midpoint of $CA$. The point $M$ is obtained from $P$ by the composition of two spiral similarities, one (call it $S_1$) centered at $A$, of angle $\frac \pi 4$ and ratio $\sqrt 2$, the other one (call it $S_2$) centered at $C$, of angle $\frac \pi 4$ and ratio $\frac {\sqrt 2}2$. The ratios cancel out, so the result is a rotation of anle $\frac \pi 2$, and the center is the fixed point of the transformation, which is clearly $X$ (just apply $S_1$ and then $S_2$ to $X$ to see that it's fixed).

This means that $AM$ is the image of $NP$ through a rotation of angle $\frac \pi 2$, so, in particular, $AM\perp NP$. The concurrence we want to prove is thus simply the concurrence of the altitudes of $MNP$.

[Edit: Darij posted his message while I was typing mine, and I'd hate to have typed it for nothing, so I'm posting it :)]

To solve this problem also it is possible to use Ceva's theorem in trigonometrical form.

Let the square with center $ A_1$ have vertices $ M$ and $ N$ lying on $ AB$ and $ AC$ respectively, let $ AA_1$ meet $ BC$ at $ X$ and $ MN$ at $ S$, then
$ \frac {BX}{XC} \equal{} \frac {MS}{SN} \equal{} \frac {[AMA_1]}{[ANA_1]} \equal{} \frac {\frac {1}{2}AM \cdot MA_1 \cdot sin AMA_1}{\frac {1}{2} AN \cdot NA_1 \cdot sin ANA_1} \equal{} \frac {AM}{AN} \cdot \frac {sin(B \plus{} 45)}{sin(C \plus{} 45)} \equal{} \frac {AB}{AC} \cdot \frac {sin(B \plus{} 45)}{sin(C \plus{} 45)}$,
so by Ceva the three lines will concur, as everything cancels.

A solution by using the [b][i]trigonometric form of Ceva's theorem.[/i][/b]

[hide="Solution"]

Let $~$ $\square{KLMN}$ $~$ be the square with center $~$ $A_1$ $~$ such that $~$ $K \in \overline{AB}\; \wedge\; N \in \overline{AC}$ $~$.

$~$ $A_1$ $~$ is the center of $~$ $\square{KLMN}\; \Longrightarrow\; \boxed{ \overline{A_1K} = \overline{A_1N} }\; \wedge\; \boxed{ \angle{A_1KN} = 45^\circ = \angle{A_1NK} }\quad \cdots\, (1)$

$\square{KLMN}$ $~$ is a square $~$ $\Longrightarrow\; KN \parallel LM\; \Longrightarrow\; KN \parallel BC\; \Downarrow$

$\Longrightarrow\; \boxed{ \angle{AKN} = \angle{ABC} = \angle{B} }\; \wedge\; \boxed{ \angle{ANK} = \angle{ACB} = \angle{C} }\quad \cdots\, (2)$

In $~$ $\triangle{AKA_1}\; \wedge\; \triangle{ANA_1}\; \stackrel{ \text{Law of sines} }{\Downarrow}$

$\Longrightarrow\; \frac { \sin \angle{KAA_1} } { \sin \angle{A_1KA} } = \frac { \overline{A_1K} } { \overline{A_1A} } \stackrel { (1) } {=} \frac { \overline{A_1N} } { \overline{A_1A} } = \frac { \sin \angle{NAA_1} } { \sin \angle{A_1NA} }\; \Downarrow$

$\Longrightarrow\; \frac { \sin \angle{KAA_1} } { \sin \angle{NAA_1} } = \frac { \sin \angle{A_1KA} } { \sin \angle{A_1NA} } \stackrel { (1)\; \wedge\; (2) } {=} \frac { \sin \angle{ B + 45^\circ } } { \sin \angle{ C + 45^\circ } }\; \Downarrow$

$\Longrightarrow\;  \boxed{ \frac { \sin \angle{BAA_1} } { \sin \angle{CAA_1} } = \frac { \sin \angle{ B + 45^\circ } } { \sin \angle{ C + 45^\circ } } }\quad \cdots\, (3)$

Similarly $~$ $\boxed{ \frac { \sin \angle{CBB_1} } { \sin \angle{ABB_1} } = \frac { \sin \angle{ C + 45^\circ } } { \sin \angle{ A + 45^\circ } } }\; \wedge\; \boxed{ \frac { \sin \angle{ACC_1} } { \sin \angle{BCC_1} } = \frac { \sin \angle{ A + 45^\circ } } { \sin \angle{ B + 45^\circ } } }\quad \cdots\, (4)$

$ \stackrel { (3)\; \wedge\; (4) } { \Longrightarrow }\; \boxed { \frac { \sin \angle{BAA_1} } { \sin \angle{CAA_1} } \cdot \frac { \sin \angle{ACC_1} } { \sin \angle{BCC_1} } \cdot \frac { \sin \angle{CBB_1} } { \sin \angle{ABB_1} } = } $

$ = \frac { \sin \angle{ B + 45^\circ } } { \sin \angle{ C + 45^\circ } } \cdot \frac { \sin \angle{ A + 45^\circ } } { \sin \angle{ B + 45^\circ } } \cdot \frac { \sin \angle{ C + 45^\circ } } { \sin \angle{ A + 45^\circ } } = $

$ \boxed{ = 1 } \; \stackrel { \text{trigonometric form of Ceva's Theorem} } { \Downarrow}$

Lines $~$ $AA_1,\, BB_1\; \wedge\; CC_1$ are concurrent. $~$ $\blacksquare$

[/hide]

Hello, I have a particularly nice solution!

[hide="Solution"]
We first make the same observation [b]darij grinberg[/b] did about constructing the squares on the exterior. Using the same letters for the points:

Now, we use coordinates to prove the desired result.

WLOG let $A$ be $(0,0)$, $B$ be $(1,0)$, and $C$ be $(m,n)$, with $m,n \ge 0$

We first find $X$. Rotate $C$ 90 degrees clockwise around $B$ to get $(n+1,1-m)$. The midpoint of this point and $C$ will be $X$, so $X$ lies at $\left(\frac{m+n+1}{2}, \frac{1+m-n}{2}\right)$.

Similarly, rotating $C$ 90 degrees counter-clockwise around $A$ yields $(-n,m) \implies Y$ is at $\left(\frac{m-n}{2}, \frac{m+n}{2}\right)$

Clearly, $Z$ is at $\left(\frac{1}{2},-\frac{1}{2}\right)$.

Now, let us find the intersection $P$ of $AX$ and $BY$ and show it lies on $CZ$.

The equation of $AX$ is $y=\frac{1-m+n}{1+m+n}x$ and of $BY$ is $y=\frac{m+n}{m-n-2}(x-1)$.

They intersect when $\frac{1-m+n}{1+m+n}x=\frac{m+n}{m-n-2}(x-1)$

or $\left(\frac{m+n}{m-n-2}-\frac{1-m+n}{1+m+n}\right)x=\frac{m+n}{m-n-2}$

Multiplying through by $(m-n-2)(1+m+n)$, we have:

$[(m+n)(m+n+1)-(1-m+n)(m-n-2)]x=(m+n)(m+n+1)$

For convenience, let $(m+n)(m+n+1)-(1-m+n)(m-n-2)=k$

so the x-coordinate of $P$ is $\frac{(m+n)(m+n+1)}{k}$

and since the equation of $AX$ is $y=\frac{1-m+n}{1+m+n}x$, the y-coordinate is $\frac{(m+n)(1-m+n)}{k}$.

Meanwhile, the equation of $CZ$ is merely $y=\frac{2n+1}{2m-1}\left(x-\frac{1}{2}\right)-\frac{1}{2}$

It is sufficient to show that the coordinates of $P$ satisfy this equation, or that:

$\frac{(m+n)(1-m+n)}{k}=\frac{2n+1}{2m-1}\left(\frac{(m+n)(m+n+1)}{k}-\frac{1}{2}\right)-\frac{1}{2}$

Multiplying through by $2k(2m-1)$ and moving the $\frac{1}{2}$ over, we must show:

$2(m+n)(1-m+n)(2m-1)+k(2m-1)=2(2n+1)(m+n)(m+n+1)-(2n+1)k$

Moving the $(2n+1)k$ over this becomes

$2(m+n)(1-m+n)(2m-1)+k(2m+2n)=2(2n+1)(m+n)(m+n+1)$

Dividing through by $2(m+n)$ and subbing back in $k$, we must show that:

$(1-m+n)(2m-1)+(m+n)(m+n+1)-(1-m+n)(m-n-2)=(2n+1)(m+n+1)$

or that

$(1-m+n)[(2m-1)-(m-n-2)]=(m+n+1)[(2n+1)-(m+n)]$

which is equivalent to

$(1-m+n)(m+n+1)=(m+n+1)(1-m+n)$

which is clearly true. Thus, the intersection $P$ of $AX$ and $BY$ lies on $CZ$, so $AA_1$, $BB_1$, and $CC_1$ are concurrent.[/hide]

Dear Mathlinkers,
following the nice idea of Darij, the point of intersection is the first Vecten point.
You can see
http://perso.orange.fr/jl.ayme  vol. 5  Vecten p. 25
Sincerely
Jean-Louis

Let the ray $AA_1$ meet $BC$ at $X$. Since $AX$ passes through the center of the square, we have $VX = q$, $WX = p$. By similar triangles $BX/XC = p/q$. Hence $BX/XC = (BX + p)/(XC + q) = BW/VC = (BV + s)/(CW + s)$, where $s$ is the side of the square. But $BV = s \cot B, CW = s \cot C$, so $BX/XC = (\cot B + 1)/(\cot C + 1)$. The desired result immediately follows by Ceva's Theorem.  $\Box$

[u]See[/u] [b]PP12[/b] ([u]an easy extension[/u]) from [url=http://www.artofproblemsolving.com/blog/56429][color=darkred][b]here[/b][/color][/url].

[hide="Solution"]
Denote $A,B,C$ as the measures of angles $\angle{BAC},\angle{ABC},\angle{ACB}$ respectively.

Let $D$ and $E$ be the vertices of the square with center $A_1$ that touches $AB$ and $AC$ respectively. Since $DE||BC$, $\angle{ADE}=B$ and $\angle{AED}=C$. Note that $\angle{EDA_1}=\angle{DEA_1}=45$, so $\angle{ADA_1}=\angle{ADE}+45=B+45$ and $\angle{AEA_1}=\angle{AED}+45=C+45$. By Law of Sines on $\triangle{ADA_1}$ and $\triangle{AEA_1}$, we get
\[ \frac{A_1D}{\sin\angle{A_1AB}}=\frac{AA_1}{\sin\angle{ADA_1}}\text{ and }\frac{A_1E}{\sin\angle{A_1AC}}=\frac{AA_1}{\sin\angle{AEA_1}}.\]
But $A_1D=A_1E$, so dividing the two equations produces
\[ \frac{\sin\angle{A_1AB}}{\sin\angle{A_1AC}}=\frac{\sin(B+45)}{\sin(C+45)}.\]
By the same logic,
\[ \frac{\sin\angle{B_1BC}}{\sin\angle{B_1BA}}=\frac{\sin(C+45)}{\sin(A+45)}\text{ and }\frac{\sin\angle{C_1CA}}{\sin\angle{C_1CB}}=\frac{\sin(A+45)}{\sin(B+45)}\]
so
\[ \frac{\sin\angle{A_1AB}}{\sin\angle{A_1AC}}\frac{\sin\angle{B_1BC}}{\sin\angle{B_1BA}}\frac{\sin\angle{C_1CA}}{\sin\angle{C_1CB}}=\frac{\sin(B+45)}{\sin(C+45)}\frac{\sin(C+45)}{\sin(A+45)}\frac{\sin(A+45)}{\sin(B+45)}=1\]
Thus, by Trig Ceva, $AA_1,BB_1,CC_1$ are concurrent. // [/hide]

As [b]Darij & Grobber[/b] pointed out, is more convenient to prove $AM, BN, CP$ concurent, where $M, N, P$ are the centers of the squares erected outwardly onto the sides of the triangle.
Let $X=AM\cap BC, Y=BN\cap AC, Z=CP\cap AB$.
Triangles $\Delta ABM,\Delta ACM$ have the same base, hence the ratio of their areas is $\frac{S_{ABM}}{S_{ACM}}=\frac{BX}{CX}$, but $2S_{ABM}=AB\cdot BM\cdot sin(B+45^\circ), 2S_{ACM}=AC\cdot CM\cdot sin(C+45^\circ)$; with $BM=CM$ we get $\frac{S_{ABM}}{S_{ACM}}=\frac{AB\cdot sin(B+45^\circ)}{AC\cdot sin(C+45^\circ)}=\frac{BX}{CX}$ and other two similar relations for $Y,Z$ which, multiplied side by side will give the required $\frac{BX}{CX}\cdot\frac{CY}{AY}\cdot\frac{AZ}{BZ}=1$.

However one cannot compare this solution with[b] Darij's[/b]!

Best regards,
sunken rock

We consider the square on side $BC$ and let its intersections with $AB$ and $AC$ be $X$ and $Y$ respectively.Let $A_1X=A_1Y=x$(circumradius of the square).It is also easy to note that $XY \parallel BC$.Thus $\angle{AXA_1}=45+B$ and $\angle{AYA_1}=45+C$.Thus applying the [b]sine rule[/b] in $\triangle{XAA_1}$ and $\triangle{YAA_1}$ respectively we get

$\frac{x}{sin{\angle{BAA_1}}}=\frac{AA_1}{sin(45+B)}$...(1)

$\frac{x}{sin{\angle{CAA_1}}}=\frac{AA_1}{sin(45+C)}$...(2)

Dividig (2) by (1) we get

$\frac{sin{\angle{BAA_1}}}{sin{\angle{CAA_1}}}=\frac{sin(45+B)}{sin(45+C)}$

Similarly considering the squares on the other sides we get

$\frac{sin{\angle{CBB_1}}}{sin{\angle{ABB_1}}}=\frac{sin(45+C)}{sin(45+A)}$ and

$\frac{sin{\angle{ACC_1}}}{sin{\angle{BCC_1}}}=\frac{sin(45+A)}{sin(45+B)}$

Multiplying these expressions we get

$\frac{sin{\angle{BAA_1}}}{sin{\angle{CAA_1}}} \dot \frac{sin{\angle{CBB_1}}}{sin{\angle{ABB_1}}} \dot \frac{sin
{\angle{ACC_1}}}{sin{\angle{BCC_1}}}=1$

so by converse of Ceva's theorem we are done!!

Easy Solution: Let the square with center $A_1$ touch $AC$ at $M$ and $AB$ at $L$. Using the Law of Sines on $\Delta AA_1K$, we find that$\frac{\sin{\angle A_1AK}}{A_1K}=\frac{\sin{(C+45)}}{AA_1}$. Similarly, we find that $\frac{\sin{\angle A_1AL}}{A_1L}=\frac{\sin{(B+45)}}{AA_1}$ in $\Delta AA_1L$. Dividing the two equations, we find that $\frac{\sin{\angle A_1AK}}{\sin {\angle A_1AL}}=\frac{\sin{(C+45)}}{\sin{(B+45)}}$. Multiplying the similar expressions for $B$ and $C$ yields the result using Trig Ceva. $\Box$

Let $X$ be the center of the square that has side $BC$ such that the square is on the opposite side of $BC$ as $A$. Define $Y$ and $Z$ similarly. Then by homothety the problem is equivalent to proving that $AX,BY,CZ$ concur. However we are done by Jacobi's Theorem as all desired angles are equal to $45$.

[hide=Solution]Let $WXYZ$ be an inscribed square in $ABC$ such that $W$ lies on $AB,$ $X$ lies on $AC,$ and $Y,Z$ lie on $BC,$ such that $WXYZ$ is in clockwise orientation. From the Law of Sines on triangle $AA_1X$ and $AA_1W,$ we find that $$\frac{\sin{\angle{XAA_1}}}{\sin{\angle{AXA_1}}}=\frac{A_1X}{AA_1}\text{ and }\frac{\sin{\angle{WAA_1}}}{\sin{\angle{AWA_1}}}=\frac{A_1W}{AA_1}.$$ Since $A_1$ is the center of the square, we must have $A_1W=A_1X.$ Dividing the two relations, we find that $$\frac{\sin{\angle{XAA_1}}}{\sin{\angle{WAA_1}}}=\frac{\sin{\angle{AXA_1}}}{\sin{\angle{AWA_1}}}\cdot\frac{\cancel{A_1X}}{\cancel{A_1W}}\cdot\frac{\cancel{AA_1}}{\cancel{AA_1}}=\frac{\sin{\angle{AXA_1}}}{\sin{\angle{AWA_1}}}.$$ Since $X$ lies on $AC$ and $W$ lies on $AB,$ we must have
\begin{align*}
\angle{AXA_1}=180-\angle{A_1XC}&\Rightarrow\sin{\angle{AXA_1}}=\sin{\angle{A_1XC}}\text{ and }\\
\angle{AWA_1}=180-\angle{A_1WB}&\Rightarrow\sin{\angle{AWA_1}}=\sin{\angle{A_1WB}}.
\end{align*}
Now, from the Law of Sines on triangle $XZC$ and $WYB,$ we find that
\begin{align*}
\frac{\sin{\angle{ZXC}}}{ZC}=\frac{\sin{\angle{XCZ}}}{XZ}&\text{ and }\frac{\sin{\angle{BWZ}}}{BY}=\frac{\sin{\angle{WBY}}}{WY}\\
\sin{\angle{A_1XC}}=\frac{ZC\cdot \sin{\angle{ACB}}}{XZ}&\text{ and }\sin{\angle{A_1WB}}=\frac{BY\cdot\sin{\angle{ABC}}}{WY}.
\end{align*}
Dividing the two relationships, we find that
\begin{align*}
\frac{\sin{\angle{A_1XC}}}{\sin{\angle{A_1WB}}}&=\frac{ZC\cdot WY}{XZ\cdot BY}\cdot\frac{\sin{\angle{ACB}}}{\sin{\angle{ABC}}}\tag{1}.
\end{align*}
Now, since $XZ$ and $WY$ are both diagonals of the square, we must have $XZ=WY.$ Also, $$ZC=YC+ZY=\frac{XY}{\tan{\angle{C}}}+XY.$$ Similarly, $$BY=BZ+ZY=\frac{WZ}{\tan{\angle{B}}}+WZ.$$ Thus, substituting into (1), we have
\begin{align*}
\frac{\sin{\angle{A_1XC}}}{\sin{\angle{A_1WB}}}&=\frac{ZC\cdot\cancel{WY}}{\cancel{XZ}\cdot BY}\cdot\frac{\sin{\angle{ACB}}}{\sin{\angle{ABC}}}\\
&=\frac{XY\left(\frac{1}{\tan{\angle{C}}}+1\right)}{WZ\left(\frac{1}{\tan{\angle{B}}}+1\right)}\cdot\frac{\sin{\angle{ACB}}}{\sin{\angle{ABC}}}\\
&=\frac{\cancel{XY}\left(\frac{\cos{\angle{C}}}{\sin{\angle{C}}}+1\right)}{\cancel{WZ}\left(\frac{\cos{\angle{B}}}{\sin{\angle{B}}}+1\right)}\cdot\frac{\sin{\angle{C}}}{\sin{\angle{B}}}\\
&=\frac{\cos{\angle{C}}+\sin{\angle{C}}}{\cos{\angle{B}}+\sin{\angle{B}}}.
\end{align*}
Therefore,
\begin{align*}
\frac{\sin{\angle{CAA_1}}}{\sin{\angle{BAA_1}}}\cdot\frac{\sin{\angle{ABB_1}}}{\sin{\angle{CBB_1}}}\cdot\frac{\sin{\angle{BCC_1}}}{\sin{\angle{ACC_1}}}&=\frac{\cos{\angle{C}}+\sin{\angle{C}}}{\cos{\angle{B}}+\sin{\angle{B}}}\cdot\frac{\cos{\angle{A}}+\sin{\angle{A}}}{\cos{\angle{C}}+\sin{\angle{C}}}\cdot\frac{\cos{\angle{B}}+\sin{\angle{B}}}{\cos{\angle{A}}+\sin{\angle{A}}}\\
&=1.
\end{align*}
By the converse of Ceva's Theorem, we may conclude that $AA_1,BB_1,$ and $CC_1$ are concurren, as desired.[/hide]

[hide=Storage]
Taking a homothety of $2$ at $A$ maps $A'$ to a point $D$ where $D$ is the center of the square with side $BC$ outside circle. As such, $AA'$ is the same line $AD$. Define $BE$ equal to $BB'$ and $CF$ equivalent to $CC'$ similarily. Thus, we only need to show that $AD, BE, CF$ all concur. Then, by Law of Sines, $$\frac{\sin \angle BAD}{\sin \angle DAC} = \frac{\frac{\sin \angle BAD}{\sin \angle DB}}{\frac{\sin \angle DAC}{DC}} = \frac{\frac{\sin \angle ABD}{AD}}{\frac{\sin \angle ACD}{AD}} = \frac{\sin \angle ABD}{\sin \angle ACD} = \frac{\sin(\angle B + 45^{\circ})}{\sin(\angle C + 45^{\circ})}$$ As such, by trignometric Ceva's \begin{align*}\frac{\sin \angle ABE}{\sin \angle EBC} \cdot \frac{\sin \angle CAD}{\sin \angle DAB} \cdot \frac{\sin \angle BCF}{\sin \angle FCA} &= \\ \frac{\sin(\angle A + 45^{\circ})}{\sin(\angle C + 45^{\circ})} \cdot \frac{\sin(\angle C + 45^{\circ})}{\sin(\angle B + 45^{\circ})} \cdot \frac{\sin(\angle B + 45^{\circ})}{\sin(\angle A + 45^{\circ})} &= 1 \end{align*}

[/hide]

We proceed using Trig Ceva (ch-rig chay-vuh).

Let the square with center $A_1$ be $DEFG$, where $D, E$ are on $BC$ and $F, G$ are on $AC$ and $AB$, respectively.  By the Law of Sines on $\triangle AGA_1$, $$\frac{\sin \angle BAA_1}{A_1G} = \frac{\sin\angle AGA_1}{AA_1}$$and similarly on $\triangle AFA_1$, $$\frac{\sin\angle CAA_1}{A_1F} = \frac{\sin\angle AFA_1}{AA_1}.$$Dividing the first equation by the second and simplifying, $$\color{blue} \frac{\sin\angle BAA_1}{\sin\angle CAA_1} = \frac{\sin\angle AGA_1}{\sin\angle AFA_1} = \frac{\sin (45^{\circ} + B)}{\sin(45^{\circ}+C)}.$$We have similar expressions for the other three ratios: $\frac{\sin\angle CBB_1}{\sin\angle ABB_1} = \frac{\sin (45^{\circ} + C)}{\sin(45^{\circ} + A)}$ and $\frac{\sin\angle ACC_1}{\sin\angle BCC_1} = \frac{\sin(45^{\circ} + A)}{\sin(45^{\circ} + B}$.  Multiplying these together results in $1$, and so we are done. $\square$

Consider a triangle $\triangle X_AY_AZ_A$ where $\angle X_A=135-\angle A$, $\angle Y_A=\angle A$, $\angle Z_A=45$, and the distance from $X_A$ to $Y_AZ_A$ is $1$. Let $a=Y_AZ_A$. Define $b, c$ similarly.

Let $A_2$ be the point where $AA_1$ meets $BC$. Then, there is a homothety from $A$ that sends $A_1$ to $A_2$ and makes the square into a new square with a vertex $P$ on $AB$ and a vertex $Q$ on $AC$ with $PQ\parallel BC$. So, $\angle A_2PQ=\angle A_2QP=45$. So, $\angle A_2PB=\angle QPB-\angle QPA_2=135-\angle B$. So, $\triangle PBA_2\sim\triangle X_BY_BZ_B$. Similarly, $\triangle QCA_2\sim \triangle X_CY_CZ_C$. Since the distance from $P$ to $BA_2$ is the same as the distance from $Q$ to $CA_2$, $\frac{BA_2}{CA_2}=\frac{b}{c}$.

By the same reasoning, $\frac{AC_2}{BC_2}=\frac{a}{b}$ and $\frac{CB_2}{AB_2}=\frac{c}{a}$. Multiplying, we get $\frac{b}{c}\cdot\frac{a}{b}\cdot\frac{c}{a}=1$. So, by Ceva's theorem, the three lines are concurrent.

ISL Marabot solve

Let, the point (of the square symmetric to $A$) on $AB$ and $AC$ are $A_B,A_C$. And define $B_C,B_A,C_A,C_B$ similarly.
Now, by using sine law on $\triangle AA_BA_1, \triangle AA_CA_1$ we get,
$$\dfrac{\sin {\angle A_BAA_1}}{\sin{\angle A_CAA_1}} = \dfrac{\sin{\angle B+45^\circ}}{\sin{\angle C+45^\circ}}$$
As the same, $$\dfrac{\sin {\angle B_CBB_1}}{\sin{\angle B_ABB_1}} = \dfrac{\sin{\angle C+45^\circ}}{\sin{\angle A+45^\circ}}$$
$$\dfrac{\sin {\angle C_ACC_1}}{\sin{\angle C_BCC_1}} = \dfrac{\sin{\angle A+45^\circ}}{\sin{\angle B+45^\circ}}$$
Now, by multiplying these we get,
$$\dfrac{\sin {\angle A_BAA_1}}{\sin{\angle A_CAA_1}}\times \dfrac{\sin {\angle B_CBB_1}}{\sin{\angle B_ABB_1}} \times \dfrac{\sin {\angle C_ACC_1}}{\sin{\angle C_BCC_1}} =1$$
So, $AA_1,BB_1,CC_1$ are concurrent by inverse of trig-ceva.

Let the square involved in constructing $A_1$ be $PQSR$ where $P$ is on $AB$, $Q$ is on $AC$, and $R,S$ are on $BC$. Triangles $APQ$ and $ABC$ are clearly similar. If we let the height from $A$ to $BC$ be $h$ and let the length of $BC$ be $a$, the side length of the square, $s$, satisfies $\frac{s}{a}=\frac{h-s}{h}$ which solves to $s=\frac{ah}{a+h}$.
We now employ barycentric coordinates with reference triangle $ABC$. We know that $P=(\frac{a}{a+h},\frac{h}{a+h},0)$ and $Q=(\frac{a}{a+h},0,\frac{h}{a+h})$. By EFFT, if $R=(0,r,1-r)$, then $$\overrightarrow{PR}=(\frac{a}{a+h},\frac{h}{a+h}-r,r-1)$$ and $$PR\perp BC\rightarrow 0=a^2(r-1-(\frac{h}{a+h}-r))+(\frac{a}{a+h})(c^2-b^2).$$ This equation solves to
\begin{align*}
r&=\frac{a^2(-1-\frac{h}{a+h})+(\frac{a}{a+h})(c^2-b^2)}{-2a^2} \\
&=\frac{a^2(-(a+h)-h)+(a)(c^2-b^2)}{-2a^2(a+h)} \\
&=\frac{c^2-b^2-a^2-2S}{-2a^2-2S} \\
&=\frac{2S+a^2+b^2-c^2}{2a^2+2S} \\
&=\frac{S+S_C}{a^2+S}.
\end{align*}
Therefore,
\begin{align*}
A_1&=\left(\frac{a}{2(a+h)},\frac{S+S_C}{2(a^2+S)},\frac{h}{2(a+h)}+\frac12-\frac{S+S_C}{2(a^2+S)}\right) \\
&=\left(\frac{a}{2(a+h)},\frac{S+S_C}{2a(a+h)},\frac{h}{2(a+h)}+\frac{a(a+h)}{2a(a+h)}-\frac{S+S_C}{2a(a+h)}\right) \\
&=\left(a^2:S+S_C:ah+a(a+h)-S-S_C\right) \\
&=\left(S_B+S_C:S+S_C:S+S_B\right).
\end{align*}
We claim that the wanted concurrency point is $(\frac{1}{S+S_A}:\frac{1}{S+S_B}:\frac{1}{S+S_C}).$ We see that line $AA_1$ is $-\frac{v}{w}=\frac{z}{y}=\frac{S+S_B}{S+S_C},$ which our concurrency point satisfies. It is easy to show that the concurrency point is also on the other two lines by symmetry. QED.

Let $A_2$ be the center of the square erected on $BC$ outside $ABC$. So there's a homothety at $A$ mapping the square centered at $A_1$ to the one centered at $A_2$, so we focus only on ray $AA_2$.
[asy]size(10cm);
pair B = (-1 , 1); pair C = (1 , 1); pair E = (1 , -1); pair F = (-1 , -1); pair P = (-0.7 , 1); pair Q = (-0.1 , 1); pair R = (-0.1 , 1.6); pair S = (-0.7 , 1.6);
pair A = extension(B, S, C, R); pair A_1 = midpoint(P--R); pair A_2 = midpoint(B--E); draw(A--A_2, red); draw(B--A_2--C, blue); draw(E--A--F, grey);
filldraw(P--Q--R--S--cycle, invisible, blue); filldraw(B--C--E--F--cycle, invisible, blue);
filldraw(A--B--C--cycle, invisible, red);
dot("$B$", B, dir(180)); dot("$C$", C, dir(0)); dot(E); dot(F); dot(P); dot(Q); dot(R); dot(S); dot("$A$", A, dir(A)); dot("$A_1$", A_1, dir(225)); dot("$A_2$", A_2, dir(270));
/* -----------------------------------------------------------------+ |                 TSQX: by CJ Quines and Evan Chen                  | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ B 180 = (-1,1) C 0 = (1,1) E .= (1,-1) F .= (-1,-1) P .= (-0.7,1) Q .= (-0.1,1) R .= (-0.1,1.6) S .= (-0.7,1.6) A = extension B S C R A_1 225 = midpoint P--R A_2 270 = midpoint B--E A--A_2 / red B--A_2--C / blue E--A--F / grey P--Q--R--S--cycle / 0.1 cyan / blue B--C--E--F--cycle / 0.1 cyan / blue A--B--C--cycle / 0.1 lightred / red */
[/asy]
By law of sines on $\triangle BAA_2$ and $\triangle CAA_2$, \[ \frac{\sin \angle BAA_2}{\sin \angle CAA_2} = \frac{BA_2 \cdot \sin (\angle B + 45^{\circ})}{AA_2} \div \frac{CA_2 \cdot \sin (\angle C + 45^{\circ})}{AA_2} = \frac{\sin(\angle B + 45^{\circ})}{\sin(\angle C + 45^{\circ})} \] so multiplying cyclically via trig ceva finishes.

@2above
Here is a more efficient bash:

There exists a homothety on $A$ mapping square $A_1$ with a new square externally off $BC$. Let's call the center of this square $A_2$. Similarly define $B_2$ and $C_2$. It suffices to show $AA_2, BB_2, CC_2$ concurrent. Now we can use bary. By Conway's formula, we can compute $A_2$ as $(-a^2:S_C+S:S_B+S)$, and can compute $B_2$ and $C_2$ in a similar fashion. They meet at $$((S_B + S)(S_C + S) : (S_A + S)(S_C + S) : (S_B + S)(S_A + S)),$$ which is the Outer Vecten Point. $\blacksquare$

[hide=.]Solved with a hint to "find a homothety"[/hide]

Image from StefanS

By law of sines on $\triangle{AKA_1}$ and $\triangle{ANA_1}$, we get that $\frac{\sin{\angle{KAA_1}}}{KA_1}=\frac{\sin{\angle{AKA_1}}}{AA_1}$ and $\frac{\sin{\angle{NAA_1}}}{NA_1}=\frac{\sin{\angle{ANA_1}}}{AA_1}$. Denote $\angle{KAA_1}$ as $\alpha_1$ and $\angle{NAA_1}$ as $\alpha_2$. We have $\frac{\sin{\angle{B+45}}}{KA_1}=\frac{\sin{\angle{\alpha_1}}}{AA_1}$ and $\frac{\sin{\angle{C+45}}}{NA_1}=\frac{\sin{\angle{\alpha_2}}}{AA_1}$ respectively. Now divide the first equation by the second to get $\frac{\sin{\angle{B+45}}}{\sin{\angle{C+45}}}=\frac{\sin{\angle{\alpha_1}}}{\sin{\angle{\alpha_2}}}$.

If we extend the orange line to meet $BC$ at $M$, we need to find $\frac{BM}{MC}$ and use Ceva's Theorem to prove concurrency. By ratio lemma, $\frac{BM}{MC}=\frac{AB*\sin{\alpha_1}}{AC*\sin{\alpha_2}}=\frac{\sin{\angle{B+45}}}{\sin{\angle{C+45}}}$ from which we found earlier. Doing the same with $B$ and $C$ as we did with $A$, we can use trig ceva to finish. $\blacksquare{}$

Taking a homothety at $A$ gives that $A_2$ lies on $AA_1$ where $A_2$ is the center of the square with side $BC$ outside the triangle.

It remains to show that $AA_2$, $BB_2$, and $CC_2$ are collinear.

Note that
\[
    \frac{\sin BAA_2}{\sin CAA_2} = \frac{\sin BAA_2 \slash BA_2}{\sin CAA_2 \slash CA_2} = \frac{\sin ABA_2 \slash AA_2}{\sin ACA_2 \slash AA_2} = \frac{\sin(B + 45^\circ)}{\sin(C + 45^\circ)}
\]

The result follows by Trignometric Ceva then.

Take a homothety sending the square to the outer square with side $BC$. The result follows by Jacobi's theorem / Trig Ceva.

Take a homothety that sends the square to the square erected off side $BC$ and similarly for the other ones. Call the new centers $A_2$, $B_2$, and $C_2$. Then the concurrency is equivalent to \[ 1=\prod \frac{\sin \angle BA_2A}{\sin \angle AA_2C} = \prod \frac{\sin \angle ABA_2 \cdot \frac{BA_2}{AA_2}}{\sin \angle ACA_2 \cdot \frac{CA_2}{AA_2}} = \prod \frac{\sin (45+B)}{\sin (45+C)} = 1,\] and we are done.

Huh… I thought I posted on this topic before, but apparently I haven’t. It suffices to take a homothety that brings each of the squares outside (which works since the triangle formed by two vertices with a gertice of the triangle is similar to the original), then the centers of those form 45 degree angles, which by Jacobi’s concur (or by trig ceva).

lul

Homothety at $A$ sends the square to square outside $ABC$ with side $BC$, and similarly for other vertices. Now Jacobi's theorem finishes.


Let $ A, B\in C^{m\times n}$, show that if $ A^\minus{}\plus{}B^\minus{}\equal{}A^\minus{}(A\plus{}B)B^\minus{}$, where $ A^\minus{}\in A\{1\},B^\minus{}\in B\{1\}$, then $ rank(A)\equal{}rank(B)$.
I know  $ A^\minus{}\plus{}B^\minus{}\equal{}A^\minus{}(A\plus{}B)B^\minus{}$implies $ rank(A^\minus{})\equal{}rank(B^\minus{})$.

Pre and post multiply
 $ A^{\minus{}}\plus{}B^{\minus{}}\equal{}A^{\minus{}}(A\plus{}B)B^{\minus{}}$ by $ A$, we have $ A\equal{}AA^\minus{}BB^\minus{}A$,
 therefore, $ rank(B)\geq rank(A)$. Pre and post multiply
 $ A^{\minus{}}\plus{}B^{\minus{}}\equal{}A^{\minus{}}(A\plus{}B)B^{\minus{}}$ by $ B$, we have $ B\equal{}BA^\minus{}AB^\minus{}B$,
 therefore, $ rank(A)\geq rank(B)$. Hence, $ rank(A)\equal{}rank(B)$.


Let $a,b,x,y,z$ be positive real numbers. Prove that \[\frac{x}{ay+bz}+\frac{y}{az+bx}+\frac{z}{ax+by}\ge \frac{3}{a+b}.\]

[quote="chess64"]Let $a,b,x,y,z$ be positive real numbers. Prove that
\[\frac{x}{ay+bz}+\frac{y}{az+bx}+\frac{z}{ax+by}\ge \frac{3}{a+b}. \]
[/quote]

[hide]By Cauchy,

\[\frac{x}{ay+bz}+\frac{y}{az+bx}+\frac{z}{ax+by}\ge \frac{(x+y+z)^{2}}{(a+b)(xy+yz+zx)}\]

and

\[\frac{(x+y+z)^{2}}{(a+b)(xy+yz+zx)}\ge \frac{3}{a+b}\Leftrightarrow \sum (x-y)^{2}\ge 0 \][/hide]

[hide]
Since $x(ay+bz)+y(az+bx)+z(ax+by)$ factors nicely into $(a+b)(xy+yz+zx)$, Cauchy gives us
\[(a+b)(xy+yz+zx)\left( \sum_\text{cyc}\frac{x}{ay+bz}\right) =\left( \sum_\text{cyc}x(ay+bz) \right) \left( \sum_\text{cyc}\frac{x}{ay+bz}\right) \]

\[\geq (x+y+z)^{2}\]

\[\sum_\text{cyc}\frac{x}{ay+bz}\geq \frac{(x+y+z)^{2}}{(a+b)(xy+yz+zx)}\]
Then $(x+y+z)^{2}\geq 3(xy+yz+zx)$, so the right hand side is $\geq \frac{3}{a+b}$.
[/hide]

[size=59]edit: beaten...[/size]


algebra
polynomial
pigeonhole principle
algebra unsolved

$k$ and $n$ are positive integers that are greater than $1$. $N$ is the set of positive integers. $A_1, A_2, \cdots A_k$ are pairwise not-intersecting subsets of $N$ and $A_1 \cup A_2 \cup \cdots \cup A_k = N$.

Prove that for some $i \in \{ 1,2,\cdots,k \}$, there exsits infinity many non-factorable n-th degree polynomials so that coefficients of one polynomial are pairwise distinct and all the coeficients are in $A_i$.

[quote="shobber"] $A_1, A_2, \cdots A_k$ are [a partition of the set of positive integers].  Prove that for some $i \in \{ 1,2,\cdots,k \}$, there exsits infinity many non-factorable n-th degree polynomials so that coefficients of one polynomial are pairwise distinct and all the coeficients are in $A_i$.[/quote]

This is immediate if you have some quantitative version of the statement that "among monic degree $n$ polynomials the set of irreducible (non-factorizable) polynomials has density 1".   One of the sets, say $A_i$, has positive upper density larger than $c > 0$.  The set of polynomials with coefficients in $A_i$ has upper density at least $c^n$ in the set of degree $n$ polynomials; in particular, it must intersect the set of irreducible polynomials.

Well, that's nice, but I didn't know this density fact, so I solved it using Eisenstein's irreducibility criterion (EIC).

EIC: If we have polynomial $f\in Z[x]$
$f(x)=a_{n}x^{n}+...+a_{1}x+a_{0}$
and prime $p$ such that $p/a_{i}$ for $i=0,...,n-1$
$p$ does not divide $a_{n}$ and
$p^{2}$ does not divide $a_{0}$, then
$f$ is irreducible.

Solution to the problem:
Let $p$ be a prime. Let $N_{1}= N-pN$. $(pN =\{ p,2p,... \})$
Consider all $j$ such that $A_{j}\cap pN$ is not finite set. Lets call the name of the set of such j's $J$.
We have infinite many numbers divisible by $p$, but not divisible by $p^{2}$. So exist $K$, such that $K$ is subset of$J$ and if $k\in K$ then exist numbers in $A_{k}$ divisible by $p$, but not divisible by $p^{2}$.
If $k\in K$ and $A_{k}\cap N_{1}\neq \emptyset$ then we do the following:
	1.take $a_{n}$ be in $A_{k}\cap N_{1}$.(fixed)
	2.take $a_{0}$ be in $A_{k}$ and $a_{0}$is divisible by $p$, but not divisible by $p^{2}$.(fixed)
	3.take $a_{i}$ (i=1,...,n-1) be in $A_{k}$, $a_{i}$ is divisible by $p$and $a_{i}\neq a_{0}$, and all $a_{i}$ different.

Then $f$ is irreducible! And there can be taken infinite number of such polynomials. Then the problem is solved!
So for all $k\in K$, $A_{k}\cap N_{1}= \emptyset$

Now take $p_{2}\neq p$ and $N_{2}= N_{1}-p_{2}N_{1}$. By similar arguement there are another sets $A_{l}$ such that $A_{l}$ is subset of $N_{1}$ and $A_{l}\cap N_{2}= \emptyset$.
Building $N_{3}, N_{4}$ and so on we reach moment when there are no $A_{i}$ left, contradiction!

P.S. Sorry for the poor writing, I tried to make it as clear as I can.

Some parts of the proof are not clear to me so I'll try to rewrite it.
We can assume that all subsets are nonempty. Let $ p_1<p_2<...$ be prime numbers. Let $ T_i\equal{}\mathbb{N} \backslash p_{i}\mathbb{N}$ and $ K_i\equal{}\{ap| a\in T_i\}$. We first consider $ K_1$, which is clearly infinite set, hence there exists [b]nonempty[/b] family of subsets $ F_1$ such that $ A\in F_1 \Rightarrow |A\cap K_1|\equal{}\infty$ (by this I mean that intersection of these two sets is infinite). If we can find $ A\in F_1$ such that $ |A\cap T_1|\neq 0$ then if $ a\in A$ and $ a\in T_1$ we can take $ a_n\equal{}a$ and other coefficients in the set $ K_1$ to get infinietly many irreducible polynomials (using Eisenstein Criterion), hence we can assume, that $ A\in F_1\Rightarrow |A\cap T_1|\equal{}0$.

By the same argument we get nonempty families of subsets $ F_i$ such that $ A\in F_i\Rightarrow |A\cap T_i|\equal{}0$, so for every $ i>0$ there exists subset $ A_{t(i)}$ such that $ |A_{t(i)}\cap T_i|\equal{}0$. Using Pigeonhole Principle there is a subset (WLOG) $ A_1$ and numbers $ i_1, i_2,...$ such that $ |A_1 \cap T_{i_m}|\equal{}0$ for $ m\equal{}1,2,...$. We assumed $ A_1$ is nonempty so we can find $ x\in A$. Let $ l\in \mathbb{N}$ such that $ p_{i_l}>x$. Then $ |A_1 \cap T_{i_l}|\equal{}0$ and since $ x<p_{i_l}\Rightarrow x\perp p$ we have $ x\in T_{i_l}$ and $ x$ is not an element of $ A_1$, contradiction.

Clearly one set $A_i$ contains an infinite subset $S$ of primes. Consider polynomials of the form $P(x)= \sum_{i=0}^{n} a_ix^i $ with $a_n$, $a_{n-1}$, $...$, $a_1$ chosen arbitrarily from $S$ and $a_0$ chosen such that $|a_0| > |a_1| + ... + |a_n|$. Its clear there are infitely many such polynomials. We claim they are irreducible. First, suppose $P(x)$ has a complex root $z$ such that $|z| \le 1$. Then

$  |a_0| > |a_1| + ... + |a_n| \ge  |a_1||z| + ... + |a_n||z^n| \ge |a_1z^1 + ... + a_nz^n| = |a_0|  $

Which is a contradiction. It follows all roots of $P(x)$ have modulus greater than $1$. Suppose $P(x)=f(x)g(x)$ with $f$, $g$ integer polynomials of degree at least $1$. Then $a_0=|P(0)|=|f(0)||g(0)|$. Since $a_0$ is prime, it follows one of $|f(0)|$, $|g(0)|$ is $1$, WLOG $|f(0)|$. But then the modulus of the product of the roots of $f$ at most $1$, hence $f$ has a root of modulus at most $1$, so $P$ does too, a contradiction.

i've got the same solution as malcolm :D

Alternatively, we can use Perron's irreducibility criterion after noting that one of the sets is infinite.

What's Perron's irreducibillity criterion?


[b]1st pre-TST 2004, 1 December 2003[/b]

problem 1: http://mathlinks.ro/viewtopic.php?t=5752
problem 2: http://mathlinks.ro/viewtopic.php?t=5751
problem 3: http://mathlinks.ro/viewtopic.php?t=4357

[b]2nd pre-TST 2004, 8 December 2003[/b]

problem 4: http://mathlinks.ro/viewtopic.php?t=5753 (auxiliary problem in post #7)
problem 5: http://mathlinks.ro/viewtopic.php?t=5758
problem 6: http://mathlinks.ro/viewtopic.php?t=5642

[b]1st TST 2004, 14 February 2004[/b]

problem 1: http://mathlinks.ro/viewtopic.php?p=19085
problem 2: http://mathlinks.ro/viewtopic.php?p=19086
problem 3: http://mathlinks.ro/viewtopic.php?p=19087

[b]2nd TST 2004, 17 February 2004[/b]

problem 1: http://mathlinks.ro/viewtopic.php?p=19088
problem 2: http://mathlinks.ro/viewtopic.php?p=19089
problem 3: http://mathlinks.ro/viewtopic.php?p=19090

[b]3rd TST 2004, 14 March 2004[/b]

problem 1: http://mathlinks.ro/viewtopic.php?p=19092
problem 2: http://mathlinks.ro/viewtopic.php?p=14710
problem 3: http://mathlinks.ro/viewtopic.php?p=19094

[b]4rd TST 2004, 4 April 2004[/b]

problem 1: http://mathlinks.ro/viewtopic.php?p=19095
problem 2: http://mathlinks.ro/viewtopic.php?p=18550
problem 3: http://mathlinks.ro/viewtopic.php?p=19129

[b]5th TST 2004, 25 April 2004[/b]

problem 1: http://mathlinks.ro/viewtopic.php?t=5438
problem 2: http://mathlinks.ro/viewtopic.php?t=5439
problem 3: http://mathlinks.ro/viewtopic.php?t=5440
problem 4: http://mathlinks.ro/viewtopic.php?t=5441
problem 5 (intended instead of 2): http://mathlinks.ro/viewtopic.php?t=5442
problem 6 (intended instead of 3): http://mathlinks.ro/viewtopic.php?t=5443

[b]6th TST 2004, 23 May 2004[/b]

problem 1: http://mathlinks.ro/viewtopic.php?p=100351
problem 2: http://mathlinks.ro/viewtopic.php?p=19960
problem 3: http://mathlinks.ro/viewtopic.php?p=100355

[b]7th TST 2004, 24 May 2004[/b]

problem 1: http://mathlinks.ro/viewtopic.php?p=20086
problem 2: http://mathlinks.ro/viewtopic.php?p=20087
problem 3: http://mathlinks.ro/viewtopic.php?p=20088

  Darij

hi,

if it is not too much trouble... would you be so kind of making the problems into a pdf file?

thx a lot,
psykyk

Hmm... maybe I will, but after the last two TSTs.

  Darij

When will the last two Tsts for German MO 2004 be held?
Oh, and again please post all the problems in pdf please...
Thx a lot

  Palosh alias Downito

________________________
"Our days are never coming back... " (Highway Song -> System of a down)

hey no problem then... since you posted them i thought you had a tex copy yourself... if that's not the case i can do many copy-pastes myself.

psykyk


Hello friends,

I am stuck with a problem as is the eigenvalue decompositon of an invertible matrix square matrix is unique?

I mean let $A$ and $B$ be two square invertible matrices. Is it possible for them to have same eigenvalues? Can there be any number of square invertible  matrices with same eigenvalues?

Thank you

srihari

It may help to think about triangular matrices.

You might also want to think about the diagonalization problem.


calculus
integration
number theory
prime numbers
number theory unsolved

Solve an equation $\sigma (p^{a})=\sigma (q^{b})$.
Where $p>q$ are positive prime numbers and $a,b$ are positive integers.
$\sigma (n)$ is the sum of the positive integral divisors of $n$.

[hide]$\sigma\left(p^{a}\right)=p+p^{2}+\cdots+p^{a}=\frac{p\left(1-p^{a}\right)}{1-p}$

$\sigma\left(q^{b}\right)=q+q^{2}+\cdots+q^{b}=\frac{q\left(1-q^{b}\right)}{1-q}$

$\frac{p\left(1-p^{a}\right)}{1-p}=\frac{q\left(1-q^{b}\right)}{1-q}\implies p^{a}\equiv 1\bmod q\implies q^{b}\equiv 1\bmod p$

Since $\gcd\left(p,q\right)=1$, one solution will come from $a+1=q$ and $b+1=p$.

Any ideas on how to solve it from here?[/hide]

Yeah, that's right...
But both $1$ will be eliminated at the process...
That doesn't change the result...

No one can solve it really?
May be it unable to solve...


geometry
3D geometry
tetrahedron
probability
trigonometry
trig identities
Law of Cosines

4 points $O$, $A$, $B$, and $C$ form a tetrahedron.

Given: $\angle AOB=\angle AOC=60^\circ$, $\angle BOC=90^\circ$, $OB=OC=1$, $OA=2$

Draw a perpendicular line from $O$ to $\triangle ABC$, and let $H$ be the intersection at $\triangle ABC$. Find the length of $OH$.

I dont really understand

if $\angle AOB=\angle AOC=90^0$ then , $ABC$ would become a triangle where $O$ lies on $BC$ which I dont think that $A,B,C,O$ would form a tetrahedron :?

Sorry, it was a typo :blush:

Edited.
($\angle AOB=AOC=60^\circ$, not $90^\circ$)

[quote="probability1.01"]Is the answer $\sqrt{\frac{2}{5}}$?[/quote]
 :yup:

Post your solution :)

[quote="frt"][quote="shyong"]well another thing is if $\angle AOB=\angle AOC=60^o$ then it follows $\angle BOC=120^o$ which contradict with the question $\angle BOC=90^o$..... :?[/quote]
Why? $\angle AOB=\angle AOC=60^\circ$ does not necessarily mean $\angle BOC=120^\circ$...

For example, think about a regular tetrahedron... $\angle AOB=\angle AOC=\angle BOC=60^\circ$, right?[/quote]

I think i got what you mean . I have intepret the question wrongly by thinking that it is in 2D plane  :blush:

Hm if probability doesn't post one, I'll do it :D
Actually made a cut-out of paper for this tetrahedron, turns out the perpendicular isn't actually on ABC.

OK here's my solution:

Let the intersection of the plane containing O, A, and H and ABC be D.
I'm gonna solve for OD, and the cos of ODH. Since the triangle ODH is right, we'll have OH :D.

Since AOB is 60, with the 2 sides coming out of it being 2 and 1, the triangle is a 30-60-90 triangle. Same for AOC. Thus, AB and AC are both sqrt(3)

Since BOC is 90, and the sides coming out of it are 1, the third side is sqrt(2)

The sides of ABC are sqrt(3), sqrt(3), and sqrt(2).

Since the tetrahedron is semetric with respect to the above plane, D is the midpoint of BC. We get AD= sqrt(5/2).

OD=sqrt(2)/2, because D is the midpoint of BC, and OBC is a 45-45-90 triangle with unit length legs.

It is already given that AO=2.

So, the triangle ADO has sides, 2, sqrt(5/2), and sqrt(2)/2.
The Law of Cosines gives that the cosine of ADO is -1/sqrt(5). ODH is 180-ADO, so its cosine is 1/sqrt(5).

So, ODH is a right triangle where the cos of ODH is 1/sqrt(5), and the hypotenuse is sqrt(2)/2. We get that the OH= sqrt(2/5).

If nothing else :

[hide]Let  X be midpoint  of BC, Now OBC is right-angle, so BC= $\sqrt 2$ and  OX = $(1/(\sqrt 2)) $
Similarly Look at triangle OBA (OBA is 90 degree here, as  angle BOA is 60 deg.  with  adj. side as 1 &2) so
so AB= $\sqrt 3$  and thus  AX = $\sqrt { 3- 1/2} = \sqrt {5/2} $
All you see now is triange  AXO, you  know all the sides,  so you know cos of angle XAO , all you have to multiply its sin with OA(=2) .
[/hide]

[hide="OR"]Let $O(0,0,0), C(1,0,0), B(0,1,0) and A(1,1, \sqrt 2)$ be the points. Volume of the tetrahedron can be calculated as:
$(1/6)(Area \ of  \ base)(Height)$  and thus height can be calculated [/hide]


Let $ \pi_1: \mathbb{R} \times \mathbb{R}, (x, y) \mapsto \pi_1(x, y)\equal{}x$, let $ X$ be the subspace $ (\{0\}\times \mathbb{R})\cup (\mathbb{R}\times \{0\}) \subseteq \mathbb{R} \times \mathbb{R}$

Prove $ f\equal{}\pi_1|_X$ is a closed map but not an open map.

I'm assuming you are using the standard topology on $ \mathbb{R}$ and $ \mathbb{R}\times \mathbb{R}$, and the subspace topology on $ X$.

Not an Open Map: Let $ U \equal{} \{(x,y)\in \mathbb{R}^2: y > 0\}$. $ U$ is open in $ \mathbb{R}\times \mathbb{R}$, so $ V \equal{} U\cap X \equal{} \{(0,y): y > 0\}$ is open in $ X$. But then $ \pi _1(V) \equal{} \{0\}$ is not open in $ \mathbb{R}$, so $ \pi _1$ is not an open map.

Closed Map: Let $ C$ be a closed set in $ X$. Let $ C_1 \equal{} C\cap \mathbb{R}\times \{0\}$ and $ C_2 \equal{} C\cap \{0\}\times \mathbb{R}$. Clearly $ C \equal{} C_1\cup C_2$, so that $ \pi _1(C) \equal{} \pi _1(C_1\cup C_2) \equal{} \pi _1(C_1)\cup \pi _1(C_2) \equal{} C_1\cup \{0\}$. Now, there is a closed set $ B\subset \mathbb{R}\times \mathbb{R}$ s.t. $ B$ is closed and $ B\cap X \equal{} C$. Since $ X$ is closed in $ \mathbb{R}\times \mathbb{R}$, $ C$ is closed in $ \mathbb{R}\times \mathbb{R}$. The standard topology on $ \mathbb{R}$ is simply the subspace topology of $ \mathbb{R}\times \{0\}$ in $ \mathbb{R}\times \mathbb{R}$ with the standard topology. Thus, $ C_1$ is closed in $ \mathbb{R}$ because $ C_1\equal{}C\cap \mathbb{R}\times \{0\}\Rightarrow \pi _1(C)\equal{}C_1\cup \{0\}$ is closed, so that $ \pi _1$ is a closed mapping.


calculus
integration
inequalities
function
logarithms
real analysis
real analysis solved

Sure, but I'll leave out all the details.

The key is observing that we have equality if $f(x)=\frac{e^x}{e-1}$. Define $g(x)=\frac{f(x)}{\frac{e^x}{e-1}}$. We have $\int^1_0 e^x(g(x)-1)dx=0$ (you can easily check this) and we have to prove something which is equivalent to $\int^1_0 e^x(\ln g(x))dx\le 0$ (you can easily check this too :D). It follows from the fact that $\ln(g(x))\le g(x)-1$, because for any real $t$, $\ln(t+1)\le t$ (we apply it for $t=g(x)-1$).

There is no doubt about your solution, grobber. :)

[quote="grobber"]Sure, but I'll leave out all the details.  ;)

It follows from the fact that $\ln(g(x))\le g(x)-1$[/quote]

This is the key to solve this problem.

kunny

[quote="kunny"]The continuous function $f(x)$ satisfies $f(x)>0$ at $[0,1]$ and $\int^1_0
f(x)dx=1$.
Prove that $\int^1_0 e^x\ln f(x)dx\leqq1-(e-1)\ln (e-1)$[/quote]
greeting from uec.ac.jp again :).

For 2 density functions of two probality P, Q: p(x), q(x). We have:

[tex] D(P||Q) = \int \limits_0^1 p(x)\log \frac{p(x)}{q(x)}\,dx \geq 0[/tex]
We define [tex] 0\log 0[/tex] as 0 and suppose that [tex] \int_0^1 f(x), \int_0^1 g(x)[/tex] and the intergral above are defined. D(P||Q) is called the divergence of P and Q. In also have:

For any convex function [tex]\phi(x)[/tex] positive and define in [0,1]. We have:

[tex] D(P||Q) = \int \limits_0^1 p(x) \phi \left( \frac{p(x)}{q(x)}\right)\,dx \geq 0[/tex]
In kunny's problem, just replace [tex]e^x[/tex] by its normalized form [tex] \frac{e^x}{\int_0^1 e^x\,dx}[/tex] we get the answer straghtforward.

Thank you very much,vuhung. :)

uec.a.c.jp? Oh,Really?

kunny


Let $x_1,x_2,...,x_k$ be $k$ distinct real numbers with non zero sum.
Prove that there are numbers: $n_1,n_2,...,n_k\in Z$ satisfying
$\sum_i n_ix_i>0$ and $\sum_i n_{\pi(i)}x_i<0$ for all permutative $\pi$ of $(1,2,..,k)$

I was a bit suprise when I saw your post Grobber. It is incredible to believe that you can find out my problem existed in the great forum before. But after seeing that topic I think I know the reason :D :).


How many n-digit numbers are there in which the sum of the digits is 9n - 8 ?

It's coefficient which is near exponent 9n-8 in polynomial:
$(1+x+x^2+\ldots+x^9)^n$
But I don't know whether it leads to nice and short solution :D

I propose you a different solution. I’ve found it on the Komal. There other solutions that use the combinations.

As each digit of such a number is at least 1, the numbers can be obtained by reducing a few digits of the number consisting of all nines so that the total reduction is 8. Therefore each number in question is assigned to a unique sequence of n natural numbers adding up to 8.

The task is to determine the number of ways to express 8 as the sum of n non-negative integers. Sums of the same terms in a different order are considered different.

A sum of this kind can be modelled by dividing a stick of length 8 into parts of unit length and making n-1 cuts along the divisions. It is allowed to make more than one cut at the same division, that corresponds to zero terms in the sequence.

As the sequence may start or terminate with 0, cuts are also allowed at the two endpoints. The task is to find the number of ways to select n-1 out of the 9 division marks if repetitions are allowed.

Such selections are known as (n-1)-tuples from 9 elements with repetition, and there are  the combinations of 9+n-2 elements, class n-1 =  the combinations of n+7 elements, class n-1 (sorry, but I must write in the word form and it’s so difficult to write clear message with this form) :( . Thus there are the combinations of n+7 elements class n-1,  n-digit numbers in which the number of digits is 9n-8.

So final answer is $C_{n+7}^{n-1}$. Right?

I don't have latex so it's too difficult for me to write correct answers.
Sorry for the language I use now!!!!!!!
You must write up n+7 and down n-1

It is the same what I wrote, just another notation.

Ok if you are sure it's the same notation you are right


superior algebra
superior algebra unsolved

Let $ V_1, V_2 \subset \mathbb{A}^n(k)$ be Affine variaties.
Prove that $ I(V_1)\plus{}I(V2)\ne I(V_1 \cap V_2)$.

I know that "$ I(V_1)\plus{}I(V2)\subset I(V_1 \cap V_2)$".
But I can't find an example where equality does not hold.

$ V_1\equal{}${$ (x,y)|y\equal{}x^2\plus{}1$} $ \subset \mathbb{A}^2(\mathbb{R})$
$ V_2\equal{}${$ (x,0)|x \in \mathbb{R}$} $ \subset \mathbb{A}^2(\mathbb{R})$
$ I(V_1)\plus{}I(V_2)\equal{}<x^2\minus{}y\plus{}1>\plus{}<y>\equal{}<x^2\minus{}y\plus{}1,y> \ne \mathbb{R}[x,y]\equal{}I(V_1 \cap V_2)$


Expressed in terms of $ n$, what is the sum of all the terms in the arithmetic sequence: \[ (n\minus{}7),(n\minus{}2),\ldots,(n\plus{}128)\]

hello, we have $ \sum_{i\equal{}0}^{27}n\minus{}7\plus{}5i\equal{}1694\plus{}28n$.
Sonnhard.

The review seems to say the answer is $88n+18524$.

sorry for reviving but what is the answer then ?

To find how many groups there are, we do $5(n-1) - 7 = 128 \rightarrow n = 28. $ Therefore, we know we have a $28n.$ As for the sum $-7 - 2 + 3 + \cdots + 128$, we can say this sum is equal to $(-7 + 128)(28)/2 = 1694.$ Thus, the answer should be $1694 + 28n$, so the answer is wrong in FTW.


graph theory
combinatorics unsolved
combinatorics

Let a simple graph have valence at most 2005. Prove that there is a way to colour each vertex one of two coulours so that the difference of the number of colours of the arcs emitted by any vertex is at most 1 in absolute value.

Bomb

What does "valence" mean? The term is unknown to me..  :blush: As for the problem, I believe there must be something wrong. What if the graph is the complete graph with $5$ (or any small enough odd number) vertices, for example?

valence is the same as degree.
anyway i think that the edges are colored and not the vertices...
some clarification is needed.

Thanks, but it would still be false. Just take a $5$-cycle (or any odd cycle). I agree: we need some clarification :).

if the graph is simple [b]bipartite[/b] and the edges are colored then i guess the question makes sense...
and in that case it is also true(i.e the edges of a bipartite graph admits an 'equitable' color distribution)

This problem seems to be, essentially, problem B3 from [url=http://www.kalva.demon.co.uk/imo/imo86.html]IMO 1986[/url]. The vertices in one class are the rows which contain points in our finite set of lattice points, while the vertices in the other class are the columns which contain points from this set. The edges are the lattice points themselves, a certain point connecting the vertices corresponding to the row and column on which it lies.

I remembered this problem today, while doing some random brainstorming :).

We may as well assume that our graph is connected. Assume we have at least as many vertices of odd degree in class $A$ as we have in class $B$, and connect each odd vertex in $B$ with an artificial edge to an odd vertex in class $A$. There might be a few odd vertices left in $A$, so, for each two of them (we must have an even number), add a vertex in $B$ connected to the two of them (and to no other vertex of our graph). The result is an eulerian bipartite graph.

We can now consider an eulerian cycle through this graph and color its edges alternatively black and white (this is possible because the cycle is even, so the eulerian cycle is an even hamiltonian cycle in the line graph of our graph, which is thus $2$-colorable). After removing the artificial edges and vertices from our graph, we're left with the desired edge coloring (unless I'm doing some stupid mistake, of course :?).


probability
Ross Mathematics Program
calculus
MIT
college
email

1. Modern Control Engineering Solution Manual (4th Edition) by Katsuhiko Ogata

2. Probability Random Variables and Stochastic Processes 4th  Edition, Solution Manual, , Athanasios, Papoulis

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4. Managerial Accounting 12th Edition (Solution Manual) Garrison, Noreen, Brewer

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6. Fundamentals of Corporate Finance 8th edition Ross, Westerfield Solutions Manual

7. Cost Accounting 13Edition Horngren Solution Manual

8. Advanced Financial Accounting 5th Edition (Power Point) Including Solution Manual Baker/Lembke/King

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10. Physics, Student Solutions Manual, 7th Edition: John D.Cutnell

11. Auditing and Assurance Services: An Integrated Approach 12th Edition, (Test Bank/Answers) by Alvin Arens, Mark Beasley, Mark S. Beasley, Randal J. Elder

12. South-Western Federal Taxation: Comprehensive 2009 (Test Bank)  by Willis, Hoffman

13. South-Western Federal Taxation: Comprehensive 2009 (Solution Manual)  by Willis, Hoffman

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16. Advanced Accounting 9th Edition by Hoyle, Schaefer, Doupnik  (Testbank)

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27. Engineering Mechanics-Dynamics 5th Edition Instructor’s Manual Vol 2 J. L. Meriam and L. G. Kraige

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Hello,

Do you maybe have the answers to:

Intermediate Micro Economics  - A modern Approach - Seventh Edition by Hal R. Varian?

I really need these for my study. If anyone else who read this and has them please send me a pm or post a link here or maybe you can email me at jeroml89@gmail.com .

I would really appreciate this!


A drawer contains eight red, eight yellow, eight green and eight black socks.  What is the probability of getting at least one pair of matching socks when five socks are randomly pulled from the drawer?

[hide][color=indigo]The probability is one because there are only 4 types so with 5 socks you have to have at least one matching set.[/color][/hide]

[hide]Yeah. It's a basic application of the Pigeonhole Principle[/hide]

[hide]
Only 4 types so you take one of each and another sock of one color so you get a pair. The probabilityi is 1.
[/hide]

extension, but much easier

there are 24 red socks and 24 blue socks...

what is the smallest number of socks(100% confident) you would have to pick in order to get at least one same colored socks...????

hopefully, it should be fairly easy

you can try with actual socks, if you want... :D

[hide]on the 3th sock, you are sure to have a pair[/hide]

About the question above about the 1 probability, I was doing the problem and got confused on it too. it is logical to me why it's one now


if i get somewhere in a problem, but don't solve it, should I write it up and submit it?

yes, definitely, im pretty sure graders dont just take a look at your answer, but your thought process as well, so even if you dont get the answer, showing your work /how far you got means something.

but what if its semi-trivial, like writing an expression into an easier to use format?

so my parents said i can apply for this camp and im sooooooooooo   happy !!! yay

ok anyways, i would just put any work even if it isnt much because it tells the examiners that you have at least thought about it.

its not going to hurt, id write it..

yayyyy :D [for iYOA] lol