Let $ ABCD$ be a convex quadrilateral, $ P \equal{} AB\cap CD, Q \equal{} AD\cap BC$. Let $ O$ be an interior point of $ ABCD$ such that $ \angle BOP \equal{} \angle DOQ$. Prove that $ \angle AOB \plus{} \angle COD \equal{} 180^{\circ}$.

We will use Trigonometric Ceva four times to the triangle $ OBD$. Denote:
$ \angle BOD \equal{} \gamma, \angle OBD \equal{} \alpha, \angle ODB \equal{} \beta,$
$ \angle BOP  \equal{} \angle DOQ \equal{} x,$
$ \angle BOA \equal{} \gamma_1, \angle DOC \equal{} \gamma_2,$
$ \angle OBA \equal{} \alpha_1, \angle CBD \equal{} \alpha_2,$
$ \angle ODA \equal{} \beta_1, \angle ODB \equal{} \beta_2.$
(I haven't marked all angles in the picture)
The trigonometric Ceva applicated to the triangle $ OBD$ and the points $ P, Q, A, C$ respectively yields four equations:

\[ \frac {\sin{x}}{\sin{(\gamma \minus{} x)}} \cdot \frac {\sin{(\beta\plus{}\beta_2)}}{\sin{\beta_2}} \cdot \frac {\sin{(\alpha\plus{}\alpha_1)}}{\sin{\alpha_1}} \equal{} 1.\]

\[ \frac {\sin{x}}{\sin{(\gamma\minus{}x)}} \cdot \frac {\sin{(\alpha\plus{}\alpha_2)}}{\sin{\alpha_2}} \cdot \frac {\sin{(\beta\plus{}\beta_1)}}{\sin{\beta_1}} \equal{} 1.\]

\[ \frac {\sin{\gamma_1}}{\minus{}\sin{(\gamma\plus{}\gamma_1)}} \cdot \frac {\sin{\beta_1}}{\sin{(\beta\plus{}\beta_1)}} \cdot \frac {\sin{(\alpha\plus{}\alpha_1)}}{\sin{\alpha_1}} \equal{} 1.\]

\[ \frac {\sin{\gamma_2}}{\sin{(\gamma\minus{}\gamma_2)}} \cdot \frac {\sin{(\alpha\plus{}\alpha_2)}}{\sin{\alpha_2}} \cdot \frac {\sin{\beta_2}}{\sin{(\beta\plus{}\beta_2)}} \equal{} 1.\]

After multiplying out the first equation with the third and then dividing by second and fourth we obtain
$ \frac{\sin{\gamma_1}}{\minus{}\sin{(\gamma\plus{}\gamma_1)}} \equal{} \frac {\sin{\gamma_2}}{\sin{(\gamma\minus{}\gamma_2)}}.$
Easy transformations show that this is equivalent to $ \sin{\gamma} \sin{(\gamma_1\plus{}\gamma_2)} \equal{} 0$ and the result follows.

Looks like a good solution, but the official solution also uses a lot of sines, like this...  :(

So I was wondering if there's a synthetic solution.  :)

Actually, the Ceva theorem is a part of the synthetic geometry, but I agree that using it in such fashion doesn't reveal many geometric properties of this configuration. I think that it is somehow related to the Miquel point of quadrilateral but I'm not sure if this is the right trace to find more interesting solution.


Im not sure this applies everywhere in the world, but usually here

$\sqrt[3]{-8}=-2$ While $(-8)^{\frac{1}{3}}$ would mean all values. If you use the sqrt sign then it implies the principle value

yeah, but actually I believe in both ways it is still -2.

[quote="The Zuton Force"]Say I write $\sqrt[3]{-8}$ on my paper.

Does that imply the principle third root $2\text{cis}\frac{\pi}{3}$ or the real third root $-2$?

Thanks[/quote]

why is that the principle root? [cite somewhere that says the principle root is the root with the smallest positive argument, etc]

I always thought the principle cuberoot was the real one, but if I try to graph cube-root on Mathematica, it only shows up over non-negative x.

TI-89's also do the same thing. What's even more confusing is that if you set the calculator's Complex Mode to "Real", it returns -2 for the cube-root of -8, but if you set the Complex Mode to either "Rectangular" or "Polar", then it gives you the $2\text{cis}\frac{\pi}{3}$

So I'm really at a loss.

The answer, as far as I know, is that it depends on context.


Express $\sqrt{17+12\sqrt{2}}$ in $a+b\sqrt{c}$ form where a,b, and c are integers.

[hide="hint"]
$17=8+9$
$12\sqrt{2}= 2*3*2\sqrt{2}$
[/hide]
Wow I can't believe this was a nationals#30...seems too easy.

[quote="13375P34K43V312"]Ah, I remember this.

[hide]Square both sides to obtain $17+12\sqrt{2}=a^{2}+cb^{2}+2ab\sqrt{c}$. Comparing irrational and rational parts, we note that $c$ must be $2$, as it is the only radical in the LHS. Thus, seting the integer and radical parts equal and simplifying, we are left with

$a^{2}+2b^{2}=17$
$ab=6$

We have $a=\frac{6}{b}$ and substituting, we have
$\frac{36}{b^{2}}+2b^{2}=17$.
Now, let $b^{2}=n$.
$\frac{36}{n}+2n=17$
$2n^{2}-17n+36=0$
$(n-4)(2n-9)=0$
$n=4,\frac{9}{2}$
Since we want $b$ to be an integer, we take the root with an integer square root, $4$. Thus, $b=2$, and $a=3$.

This leads us to the final answer,
$\sqrt{17+12\sqrt{2}}=3+2\sqrt{2}$, which we can square and verify is true. [/hide][/quote]

Correct!! :lol:  :lol:

[quote="mathgeniuse^ln(x)"]If the asked for the fourth root of this equation...I would bet many people wouldn't consider rewriting the problem as the sqrt(sqrt(blah))...

But that is just me saying it...[/quote]

Yeah they should have asked for the 4th root, because $3+2\sqrt{2}$ is also a "perfect square."

That relly is the best, most clear explanation...

What is confusing?

By the way, I messed this up SO STUPIDLY. I switched the 2 and 3!

Hmmm...well, there you go...you can't always be perfect...sorry for the bad pun...

You have next year though?

Luckilly, yes. Also, HILARIOUS! and so original. I wish they'd asked for the sum of the coefficients or something, and I'd have gotten it...Unfortunately, this wasn't the only dumb mistake I made...

I made dumb mistakes to...thinking that 27)16*1=48...don't ask, and thinking that 20) 8/22=1/4...don't ask...and that 13)54+14=61...don't ask...

But I had fun while it lasted...

I did Pythagorean backwards on the state Target wit the rt. triangle...some of my stupidest nat. mistakes:the above; 1/5 an hour is NOT 20 minutes; isosceles isn't equilateral; why would they ask the sum of the y intercepts given equations in y=mx+b?; there aren't 5! ways to arrange 1x2 rects. in a 2x5 one; finally, more backwards Pythagorean. Others I can't remember.

Hmm...

I do horrible at school (37, 45, 43); (8th, 1st, 1st), written round only
I do horrible at chapter (36, 43, 44); (27th, 3rd, 4th), written, (27th, 4th, 1st), countdown
I do okay at states (-,40,43); (-,3rd,1st), written, (-,1st,2nd), countdown
I do so bad at nationals that bad is an understatement, (-,23,35); (-,69th,14th), I think...

Well, you guys are better...

[quote="jhredsox"]i don't really get leet's explanation, can someone else explain?

-jorian[/quote]

can someone please do this?

-jorian hoover (thanx)

[quote="perfect628"]Considering I only did MC last year(7th grade), i'd say you've done considerably better than me. I got, in order, 46, 43, 39(don't ask), and 26(PLEASE don't ask!), and 1,3,2,102(you know what I want to say).[/quote]

Dude I did worse

I got 41, 44, 41, 25
which turned out to be 3, 2, 1, 112

[quote="mathgeniuse^ln(x)"]Jorian, leet assumed that a+bsqrt(c) was the answer, so he squared both sides...and then it is self-explanatory from his work there...

I don't get it...how do we get the same scores at School and State...

There are so many Ss.[/quote]

HEY it's an alliteration
idk, I bombed school, I think Target #1 from last year was poorly worded, it didn't say you could glue stuff together...

But, you made it to chapters...then you made it to states...then you made it to nats...as a 7th Grader...

This is fairly easy:
[hide]$c^{2}=a^{2}-b$
In our case $a^{2}=289$ b=288 so c=1.
The result is $\sqrt{\frac{a+c}{2}}+\sqrt{\frac{a-c}{2}}=\sqrt{9}+\sqrt{8}=3+2\sqrt{2}$[/hide]

Hmmm...

Nice way to manipulate the problem...


Can somebody recommend some excellent, introductory books on Computer programming keeping the informatics olympiads in mind? I did do programming earlier but left it last year. Any books on programming software learning?
Also, I beleive Pascal and C++ are the basic programs used int he IOI. Can somebody recommend free downloads of the best programs for the above languages. Can somebody also recommend online docs etc. on fully learning these 2 langages. Any paper book recommendation will also do!

I suppose you work under Windows. In that case, I would recommend using FreePascal (you can get it from http://www.freepascal.org) and Dev-C++ (also goes by dev-cpp, see http://www.bloodshed.net/dev/devcpp.html). There's a nice page I've learned C++ from - http://www.cplusplus.com (documents section). Pascal comes with documentation of all it's libraries and base commands although I'm not sure if you'll be able to learn programming in Pascal from these.
In the way of learning algorithms, a very well known introductionary book is "Introduction to Algorithms". ([url=http://www.amazon.com/exec/obidos/tg/detail/-/0262032937/qid=1126949283/sr=8-1/ref=pd_bbs_1/102-2812339-4788933?v=glance&s=books&n=507846]This is the one I have.[/url]) It contains pretty much everything you need to know for IOI for example. I think it's actually used as a textbook at some colleges but it really is introductionary (at the beginning, at least), don't worry. If you really want to learn all the algorithms properly, I would suggest you do some practicing. The best way is to do online competitions and training - see http://ace.delos.com/usacogate , that's where I had started.

Do you have some online books on algorithms, computing etc...???

There is a lot of various texts, papers, lectures, and occasionally you can also find books. Try ocw.mit.edu, they have something there, maybe google will do a bit.

But as Bus has already said and what I really want to emphasize, it is not a good idea to learn (especially olympiad-style) programming by reading books. Nothing is better than practice. There are several good sites (I'd also recommend the USACO training pages to start) to learn. They have quite well-designed structure with increasing difficulty of problems and also some texts as you proceed to higher levels. Doing individual problems is the best way to learn - also because of the fact that there is MUCH more material online on specific topics than "general learning books". If you encounter a difficulty while attempting to solve a problem, do some research and finally solve it, that's much much more valuable than if you just read the theory in a book, however ecxellent book it may be...

And anytime you have trouble with some topic or a specific problem, you have the great opportunity to ASK someone - your friends, teachers, coaches, people running the sites, and of course, US here at mathlinks:)

Good luck!

thanks for your wondeful post! It really seems to be the correct thing! Anyways, I just need to learn C and basic alogirthms right now and then I'll get to paractice. Regarding the USACO site, is it only for Americans? I registered and just th einto was accessible to me. Have the other chapters not benn released yet or are they blocked for international people???

The other chapters open up when you solve the problems in your current chapter.


A tree grew from a height of 6 feet to 15 feet. By what percent did
its height increase?

The height increased by $ 9$, so our answer is $ \frac96\equal{}\frac32\equal{}\boxed{150\%}$


i know the basic one:

$\frac{\pi}{4} = 1 - \frac{1}{3}+\frac{1}{5}-\frac{1}{7} ...$

which i was able to get by using the power series for arctan(x).

but there's series like ramanujan's method:

$\frac{1}{\pi} = \frac{\sqrt{8}}{9801}\sum_{n\to\infty}\frac{(4n)!(1103 + 26390n)}{(n!)^4*396^{4n}}$

that blow my mind, i dont even know where to begin.

does anyone know what sort of tactics are taken in deriving pi?  is is all just manipulation of the basic series above? the power series for arctan was very lenghty, does anyone know of a better way to get that series?

The infinte sum of $1/x^2$ is $\pi^2/6$
The infinte sum of $1/x^4$ is $\pi^4/90$
The infinte sum of $1/x^6$ is $\pi^6/945$

etc.


Rezultati Srbije:

Nikola Sibalic - zlato (8.mesto - valjda najbolji rezultat u istoriji)
Aleksandar Vasiljkovic - srebro
Ognjen Ivkovic - bronza
Aleksandra Dimic - bronza
Ugljesa Stojanovic - bronza

Cestitke svima ne fenomenalnom uspehu!

rezultati ostalih eX Yu zemalja...Bosna me zanima :) ???

http://www.dfs.org.yu/takmicenja/predhodne%20godine/09/Olimp09Sr/Olimp09Sr.htm

BiH - bez medalja, 3 ima ovo zeleno...
Hrvatska - 2 bronze, 1 zeleno...

Slovenija ima (minimalno) jedno srebro, to znam.

Please, can you post all the results? Thank you.

[quote="hsiljak"]Yup, zeleno je pohvala. Pretpostavljam da znaš da je ja IPhO sistem dodjele:

    a) Gold Medals should be awarded to 8% of the contestants (rounded up the nearest integer)..
    b) Gold or Silver Medals should be awarded to 25% of the contestants (rounded up the nearest integer).
    c) Gold, Silver or Bronze Medals should be awarded to 50% of the contestants (rounded up the nearest integer).
    d) An Olympic Medal or Honourable Mention should be awarded to 67% of the contestants (rounded up the nearest integer).[/quote]
nisam znao, ali pretpostavio sam da je nešto slično matematici, što i jeste :D

[quote="Bugi"]The results will be posted as soon as the final ceremony ends on the official site. Until then, go to the site
http://www.dfs.org.yu/takmicenja/predhodne%20godine/09/Olimp09Sr/Olimp09Sr.htm

Click the text in blue (it says ''ОВДЕ'' (without apostrophes) ), you will get an Excel table. In it, click the tab ''TABELA SVI'', and there are the results under country codes.

Romania got 3 golds, and 2 silvers.[/quote]

Hi I want to know if your post its about the results of 2009 [b]IPho[/b] held in Mexico.
I having problems trying to translate the link, is it russian or hungarian? I want to know the results of Colombia
anyway thanks for your answer :)

I uploaded the excel file from the link. Click on ''tabela svi'' on the bottom, to get the sheet with results of all contestants. Colombia: 1 bronze, 1 honorable mention (if it's called like in IMO).


[quote="sharkman"]The students in a gym class all belong to either the school choir or the school baseball team or are not on either.  If 15 students are in the choir, 18 students are on the baseball team, 11 students are on BOTH choir and baseball and 14 students are NOT on either choir or baseball, how many students are in the class?

I got 36 students in all.

Book's answer is 28 students in all.

How do I get 28?

Also, how can I use venn diagrams to solve such questions?

Is there an easier method other than venn diagrams?[/quote]
dude, what if you take negatives than you have negatives in all three man.
That's not cool.
Be a playa and use PIE

[hide="You were right"]Let $a$ be the number of students that do neither choir nor basketball, $b$ the number of students that do basketball but not choir, $c$ the number of students that do choir but not basketball, and $d$ the number that do both. Then we are given:

$a = 14$
$d = 11$
$b+d = 18 \Rightarrow b = 7$
$c+d = 15 \Rightarrow c = 4$
Thus the total number of students is $a+b+c+d = 11+7+4+14 = 36$.

Your solution book is wrong. :huh: [/hide]

I get $36$ too.

What I did:
Number in choir only $=15-11=4$
Number on baseball only $=18-11=7$
Number on both $=11$
Number on neither $=14$

Sum $=\boxed{36}$

Maybe we're both missing something or the book is just wrong.  For a venn diagram just draw two intersecting circles.  Label one choir and one baseball.  Then the intersecting part will be both which you fill in with $11$.  Then fill in as explained above and outside the circle put $14$ for neither group.


[quote="N.T.TUAN"][quote="new_man"]$Z_x$.[/quote] What is this?  ;)[/quote]
I believe it's the field of integers over $x$ (i.e., this polynomial cannot be factored into two polynomials both in terms of $x$ such that every coefficient of $x$ is an integer).

Normally it should be either $\mathbb{Z}$ or $\mathbb{Z}[x]$.
But you shouldn't say 'irreducible over $\mathbb{Z}[x]$' since that would concern polynomials with coefficients from $\mathbb{Z}[x]$ and not polynomials in $x$.

And $\mathbb{Z}[x]$ is neither a field nor the integers over $x$: it's the ring of polynomials over $\mathbb{Z}$ in the variable $x$.

[quote="new_man"]Prove that polynomial $P(x)=(x^2+1)^{2^n}+1$ is irreducible over $Z_x$.[/quote]

As I know this was a question '' Prove that polynomial $P(x)=(x^2+x)^{2^n}+1$ is irreducible over $\mathbb{Z}[X]$''and plz note that :
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=49799]www.mathlinks.ro/Forum/viewtopic.php?t=49799[/url]

For $P(x)=(x^2+1)^{2^n}+1$, just multiply it out. The free coefficient is $2$, the leading coefficient is $1$, and all other terms in the binomial expansion of $x^2+1$ have coefficients of the form $\binom{2^n}{k}$ for some $0<k<2^n$. These numbers are known to be all even, so by Eisenstein's criterion for $p=2$, we're done.


What is the percent probability that a two-digit number selected at random will have its tens digit at least two more than its units digit?

If the units digit was one, the possible tens digits could be 3, 4, 5, 6, 7, 8, and 9, so 7 possibilities. If it was two, they would be 4, 5, 6, 7, 8, and 9, or 6 possibilities. And so on. Thus, the number of two-digit numbers that satisfy this condition is $ 7 \plus{} 6 \plus{} 5 \plus{} \dots \plus{} 2 \plus{} 1 \equal{} 28$ or $ \frac{28}{90} \equal{} 31\frac{1}{9}\%$.

[quote="isabella2296"]If the units digit was one, the possible tens digits could be 3, 4, 5, 6, 7, 8, and 9, so 7 possibilities. If it was two, they would be 4, 5, 6, 7, 8, and 9, or 6 possibilities. And so on. Thus, the number of two-digit numbers that satisfy this condition is $ 7 \plus{} 6 \plus{} 5 \plus{} \dots \plus{} 2 \plus{} 1 \equal{} 28$ or $ \frac {28}{90} \equal{} 31\frac {1}{9}\%$.[/quote]
The units digit could also be $ 0$, which would give us the possible tens digits $ 2,3,4,5,6,7,8,9$, which is $ 8$ choices.  So we have $ 8\plus{}7\plus{}6\plus{} \dots \plus{}2 \plus{}1\equal{}36$.  $ \frac{36}{90}\equal{}\frac{4}{10}\equal{}\boxed{40}\%$


induction
floor function
search
strong induction
graph theory

Prove Dirac's Theorem:

If $ G$ is a connected graph with $ n$ vertices, and each vertex of $ G$ has degree at least $ n/2$, then $ G$ has a Hamiltonian cycle.

[hide]We prove the assertion using strong induction on $ n$.  For $ n \equal{} 3$ the claim is obvious.  Assume the claim is true for all $ n\le m \minus{} 1$.  Suppose we have a graph satisfying the given conditions with $ m$ vertices.

First, we prove that there must exists some cycle in the graph.  Choose some vertex $ v$.  Let $ v$ be connected by an edge to $ a_1,a_2,...,a_{\frac {m}{2}}$.  Consider the vertices which $ a_1$ is connected to.  If it is connected to some other $ a_i$, then we have the cycle $ v,a_1,a_i$.  If not, then since there are only $ m \minus{} \frac {m}{2} \minus{} 1\le m \minus{} 1$ vertices, it is connected to all of them.  Considering the vertices one of these is connected to results in a cycle of $ 3$ or $ 4$.

Choose the cycle with the most vertices $ C_1: \equal{} (v_1,v_2,...,v_k)$.  If this includes all vertices, we are done.  If not, let $ u_1,u_2,...,u_{m \minus{} k}$ be the vertices not part of $ C_1$.  It is clear that each vertex $ u_i$ is connected by an edge to at most $ \lfloor\frac {k}{2}\rfloor$ vertices of $ C_1$ or else, it would be connected by an edge to two consecutive vertices in the cycle, and so including $ u_i$ in the cycle, would make a bigger cycle.  Thus, each vertex $ u_i$ is connected by an edge to at least $ v \minus{} \lfloor\frac {k}{2}\rfloor$.  But, $ v \minus{} \lfloor\frac {k}{2}\rfloor\ge\frac {v \minus{} k}{2}$ and so by the inductive hypothesis, we have that there exists a Hamiltonian Cycle $ C_2 \equal{} (u_1,u_2,...,u_{v \minus{} k})$.  Since $ C_1$ is the maximal cycle, we have $ k\ge v \minus{} k$.  Then, $ v \minus{} k\le\frac {v}{2}$ and so, since each vertex of $ C_2$ has degree $ \frac {v}{2}$, it is connected by an edge to at least one vertex of $ C_1$.  So, let $ v_1$ be connected to $ u_i$ and $ v_2$ be connected to $ u_j$ where without loss of generality, $ i\le j$.  Then the vertices $ v_1,u_i,u_{i \plus{} 1},...,u_{j \minus{} 1},u_j,v_2,v_3,...,v_k$ form a cycle larger than $ C_1$.  [/hide]


What are they?  I missed Round 1, and I found about USAMTS recently and had a couple days to solve Round 2.  Therefore, I don't see myself getting 0-25-25-25, so silver is out of the question, and gold already is.

I'm stuck with Bronze, Honorable Mention, or nothing.  What are the prizes for Bronze and HM?

[quote="AdithyaGanesh"][quote="BOGTRO"]In both cases, you will get a USAMTS T-shirt and 1 prize (Book, possibly Mathematica, etc.) that has not already been exhausted.[/quote]

What are the chances of getting Mathematica for Bronze and HM?  You said a prize that has not already been exhausted.  How often is Mathematica left until then?[/quote]

Mathematica is usually in high demand, but it has been known for Bronze winners to scrape a mathematica, although you'd have to stay up until about 3 AM to make it.

HM generally will not get mathematica.

I got bronze last year and easily got a copy of Mathematica. My friend who got HM also got Mathematica, and reportedly it was still available a few days after HMs got to choose prizes. There's more people this year, though.

I don't think that there was mathematica for HM, since I was HM and checked on the first day, but I wasn't looking for it, since I already have a few copies.

I got HM last year and got Mathematica, but I stayed up until midnight to get it, on the first possible day.

And then I gave it away  :P


If $ a,b,c\in(0,\infty), m\in N^{*}$ prove that
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\le (\frac{a}{b})^{m}\plus{}(\frac{b}{c})^{m}\plus{}(\frac{c}{a})^{m}$.

the given inequality is equivalent to
$ \sum_{sym}x\leq\sum_{sym}x^{m}$ where $ xyz\equal{}1$
$ \rightarrow\sum_{sym}x^{\alpha\plus{}1}(yz)^{\alpha}\leq\sum_{sym}x^{m\plus{}\beta}(yz)^{\beta}$
which is true for $ \alpha,\beta$ such that $ \alpha\minus{}\beta\equal{}\frac{m\minus{}1}{3}$ by Muirhead since
$ (\alpha\plus{}1,\alpha,\alpha)$ majorises $ (m\plus{}\beta,\beta,\beta)$ for all such $ \alpha$ and $ \beta$.

My solution only use AM-GM inequality:
$ (\frac{a}{b})^{m}\plus{}1\plus{}1\plus{}...\plus{}1\geq m\frac{a}{b}$ ($ m\minus{}1$ number $ 1$)
$ \Rightarrow (\frac{a}{b})^{m}\plus{}(m\minus{}1)\geq m\frac{a}{b}$
Then $ (\frac{a}{b})^{m}\plus{}(\frac{b}{c})^{m}\plus{}(\frac{c}{a})^{m}\geq m(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})\minus{}(3m\minus{}3)$
Use $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\geq 3$ we have solution!

[quote="nktp"]
Then $ (\frac{a}{b})^{m}\plus{}(\frac{b}{c})^{m}\plus{}(\frac{c}{a})^{m}\geq m(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})\minus{}(3m\minus{}3)$
Use $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\geq 3$ we have solution![/quote]
form $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\geq 3$ we get that:

$ (m\minus{}1)(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})\geq 3(m\minus{}1)\equal{}3m\minus{}3$

$ \Rightarrow (m\minus{}1)(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})\minus{}(3m\minus{}3)\geq 0$

$ \Rightarrow m(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})\minus{}(3m\minus{}3)\geq\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}$

therefore:

$ LHS\geq m(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})\minus{}(3m\minus{}3)\geq\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}$

as wanted... :maybe:


What are the next 2 symbols in this sequence?

q-W0e9

[hide] R8 because the numbers or - goes right to left at the top

and the letters goes left to right starting at q[/hide]

I decided they must alternate between lower case and capitals

This question is REALLY along the lines of

"I'm thinking of a pattern and I want you people to guess it."

But there is NO set algorithm/rule that would give any ONE answer, so the solution could pretty much BE any two characters (symbols).  It's WHAT the
poser wants that matters.

That said,

from the posts accumulated above, "QWER" on a typewriter keyboard COULD
supply the letters.  "890-" also are on a keyboard in that order, so reversing
them would give "-098."  With the letters alternate lower case beginning with
the first letter:  "qWeR."  Then move each of the "-098" symbols to the right
of each of one of the letters:  "q-W0e9R8." (The period is not included.)

This is just a possible explanation however.

"R8" is what YOU were looking for.

My prior post is logical and true regardless of how I feel about it.


floor function
combinatorics proposed
combinatorics

Two players $A$ and $B$ have one stone each on a $100 \times 100$ chessboard. They move their stones one after the other, and a move means moving one's stone to a neighbouring field (horizontally or vertically, not diagonally). At the beginning of the game, the stone of $A$ lies in the lower left corner, and the one of $B$ in the lower right corner. Player $A$ starts.
Prove: Player $A$ is, independently from that what $B$ does, able to reach, after finitely many steps, the field $B$'s stone is lying on at that moment.

How do the two stones interact? Is the rule that neither stone can enter an occupied square?

If that's it, the proof is very short and simple:
[hide="Solution"] Before player A moves the stones are at an even (taxicab) distance from each other. This means that the stones are not adjacent and that player A can make any move he wishes. After player A moves, the stones are at an odd distance, which may restrict player B's mobility.[/hide]

[hide]The problem is closely related to [color=red]opposition[/color] in chess. Initially, $ A$ and $ B$ occupy fields of different color; moreover, at each move they change color. In order for $ A$ to win, he has to be placed just next to $ B$. He achieves this by reducing whenever possible (or keping constant otherwise) the $ \parallel{} \cdot \parallel{}_{\infty}$ distance to $ B$, i.e. the $ \max\{|x_A \minus{} x_B|,|y_A \minus{} y_B|\}$. This pushes $ B$ into a corner, where $ A$ realizes a distance of $ 1$ by being just diagonally next to $ B$. Then $ B$ has to move orthogonally adjacent to $ A$ and loses.
Of course, some more formalized proof must be produced in order to ensure this happens.

Let initially have $ A(1,1)$ and $ B(1,2n)$.
Probably the best way to see it is to have $ A$'s strategy to minimize $ \delta(A,B) \equal{}$
$ \max\{|x_A \minus{} x_B|,|y_A \minus{} y_B|\} \plus{} (|x_A \minus{} x_B| \plus{} |y_A \minus{} y_B|) \plus{} \frac {1} {x_A\plus{}x_B\plus{}y_A\plus{}y_B}$.
Using this strategy, $ A$ always remains left and lower to $ B$, and so it reaches a moment when $ \delta(A,B) \equal{} 3 \plus{} \frac {1} {8n\minus{}2}$ (anytime $ \lfloor \delta(A,B) \rfloor \equal{} 2$ and it's $ A$'s turn to move, he wins by moving over $ B$). Then $ B$ moves to some $ \lfloor \delta(A,B) \rfloor \equal{} 2$, and so $ A$ wins.
[/hide]

I thing you are wrong. If the stones are on a different color, it's A to turn. So, if B stay one his first and second places, A could be not able to reach to lower left coner.
So, the problem is wrong. It could be correct if it's a 101x101 chessboard.

No, if originally $ A$ and $ B$ sit on the same color, $ A$ can never win! Read carefully: $ A$'s goal is not to reach the [color=red] initial[/color] position of $ B$, but the one $ B$ occupies precisely [color=green]at that moment[/color], when $ A$'s turn to move comes. The only way for $ A$ to win is to "push" $ B$ into a corner,
as he moves diagonally adjacent to it, and then at his next move $ B$ becomes orthogonally adjacent to $ A$, so $ A$ can move on it.

oh...ok
Thank you...I read initially instead of "of that moment"


[quote="Shinraehee"]Let $ a,b,c > 0$ . Prove that:
$ \frac {abc(a \plus{} b \plus{} c)}{{a}^{4} \plus{} {b}^{4} \plus{} {c}^{4}} \plus{} \frac {12({a}^{3} \plus{} {b}^{3} \plus{} {c}^{3})}{(a \plus{} b \plus{} c)({a}^{2} \plus{} {b}^{2}  {c}^{2})} \geq 5$[/quote]
$ \frac {abc(a \plus{} b \plus{} c)}{{a}^{4} \plus{} {b}^{4} \plus{} {c}^{4}} \plus{} \frac {12({a}^{3} \plus{} {b}^{3} \plus{} {c}^{3})}{(a \plus{} b \plus{} c)({a}^{2} \plus{} {b}^{2}\plus{}{c}^{2})} \geq 5\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(7a^7 \minus{} 10a^6b \minus{} 10a^5b^2 \plus{} 14a^4b^3 \minus{} 6a^4b^2c \plus{} a^5bc \plus{} 2a^3b^3c \plus{} 2a^3b^2c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(7a^7 \minus{} 10a^6b \minus{} 11a^5b^2 \plus{} 14a^4b^3 \plus{} 14a^3b^4 \minus{} 11a^2b^5 \minus{} 10ab^6 \plus{} 7b^7) \plus{}$
$ \plus{} \sum_{cyc}(a^5b^2 \plus{} a^5c^2 \plus{} 2a^5bc \minus{} 6a^4b^2c \minus{} 6a^4c^2b \plus{} 4a^3b^3c \plus{} 4a^3b^2c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \minus{} b)^2(a \plus{} b)(7a^4 \minus{} 3a^3b \minus{} 7a^2b^2 \minus{} 3ab^3 \plus{} 7b^4) \plus{}$
$ \plus{} \sum_{cyc}(a \minus{} b)^2c(c^4 \minus{} 2ab(a \plus{} b)c \plus{} 2ab(a^2 \plus{} b^2))\geq0,$ which true because
$ 7a^4 \minus{} 3a^3b \minus{} 7a^2b^2 \minus{} 3ab^3 \plus{} 7b^4\geq7a^4 \minus{} 3a^3b \minus{} 8a^2b^2 \minus{} 3ab^3 \plus{} 7b^4 \equal{}$
$ \equal{} (a \minus{} b)^2(7a^2 \plus{} 11ab \plus{} 7b^2)\geq0$ and
$ c^4 \plus{} 2ab(a^2 \plus{} b^2) \equal{} c^4 \plus{} 3\cdot\frac {2ab(a^2 \plus{} b^2)}{3}\geq4c\sqrt [4]{\left(\frac {2ab(a^2 \plus{} b^2)}{3}\right)^3}\geq2ab(a \plus{} b)c.$


1) How'd you like our SC rap?

2) Anyone have a solution for number 6 or 8?

hahaha our song was sick. keep working on ur rapping for next year and we'll own.

[hide="Problem 6"]Standard Vieta's (Shobhit  ;)  ).

We have $y=x^4-5x^2-x+7$ (I don't remember the constant, but it doesn't matter) and $y=x^2-3x$. Thus, $0=x^4-6x^2+2x+7$. We want the sum of the y coordinates, or the sum of $x^2-3x$. We know from the above that the sum of the roots is 0 and the pairwise sum is -6. We want the sum of the roots squared minus twice the pairwise sum (this is the sum of the squares) minus three times the sum of the roots. This yields [b]12[/b].[/hide]

[hide="Problem 8"]Brute force with partitioning of 5. The only way to get this is to not waste your time on #7 and to be incredibly focused and not make mistakes. My partitions are seperate groups of computers (the only case not covered being all computers connected):

11111     1 partition, 1 connection scheme
1112       10 partitions, 1 connection scheme
122         30 partitions, 1 connection scheme
113         10 partitions, 4 connection schemes
23           10 partitions, 4 connection schemes
14           5 partitions, n connection schemes

To find n we repeat for partitionings of 4:

1111       1 partition, 1 connection scheme
112         6 partitions, 1 connection scheme
22           6 partitions, 1 connection scheme
13           4 partitions, 4 connection schemes

This makes for 29 ways to leave 4 computers unconnected, so 35 ways to leave them all connected. Thus, there are 296 ways to leave 5 computers unconnected, so 728 ways to connect them. Then 728/1024=[b]91/128[/b] is the answer.[/hide]

[quote="jb05"]Standard Vieta's (Shobhit  ;)  ).
[/quote]

Ha :lol: , too bad I didn't notice that there wasn't an $x^3$ term, so I took the sum of the roots as $6$  :rotfl:

[quote="arkmastermind"]i got 38 ways for four to be connected, but maybe i'm just remembering wrong.[/quote]

yeah, you did get $38$  :?

H3y SC people...I m3t you guys at ARML. I th3nk I m3t you guys on Th3rsday


algebra
polynomial
geometry
3D geometry
Ring Theory
number theory
prime numbers

Let $p$ be prime. When there is no integer $a$ such that $a^3 \equiv 2 (mod p)$, prove that there are no integers $a,b,c$ such that
$a^3 +2b^3 +4c^3 - 6abc = p$.

We consider the field $\mathbb{F}_{p}$ (that one $\mod p$) and it's extension $\mathbb{F}_{p^{3}}$ given by $\mathbb{F}_{p}[\sqrt[3]2]$.
We want that $a^{3}+2b^{3}+4c^{3}-6abc = 0$ and it suffices to show that this only happens for $a=b=c=0$ in $\mathbb{F}_{p}$.
Let $\zeta \neq , \zeta^{3}=1$,
so let the roots of $x^{3}-2$ be $\sqrt[3]2, \zeta \sqrt[3]2 , \zeta^{2}\sqrt[3]2$ (clearly $\zeta \in \mathbb{F}_{p}$). Then we have
$0=a^{3}+2b^{3}+4c^{3}-6abc =$
$= (a+\sqrt[3]2 b+\sqrt[3]4 c)(a+\zeta \sqrt[3]2 b+\zeta^{2}\sqrt[3]4 c)(a+\zeta^{2}\sqrt[3]2 b+\zeta \sqrt[3]4 c)$,
so one factor is zero (we are in a field!).
WLOG let be $a+\sqrt[3]2 b+\sqrt[3]4 c=0$. Since $a,b,c \in \mathbb{F_{p}}$, when they would not be all [b][i]0[/i][/b] we would have a polynomial of second degree over $\mathbb{F}_{p}$ with root $\sqrt[3]2$, namely $a+bx+cx^{2}$. But the latter is impossible.
Thus $a=b=c=0$ and we are done.
And we have shown even more: only numbers can be of that form for that all prime divisors for that $2$ is not a perfect cube occure a mutltiple of three times in the factorisation;
The natural question occurs: is this criterion sufficient¿
Answer: yes (but I don't have an elementary proof or even have some dumb mistake...).

So I propose:

Let $P$ be the set of prime numbers such that $x^{3}\equiv 2 \mod p$ has no solution. Then $n$ is of type $n=a^{3}+2b^{3}+4c^{3}-6abc$ iff every prime factor from $P$ occurs $0,3,6,9,...$ times in $n$.
(note that this is very similar to the condition on $n$ being the sum of two squares and this is no random event)

Nice solution  :lol:
I also proved your proposed problem, but I used that $Z[\sqrt[3]2]$ is UFD, which is not an elementary fact.  :(

I used that it has class number $1$, but thats equivalent.


$ x \plus{} 2y \plus{} xy \equal{} xy \plus{} x \plus{} y \plus{} 1 \plus{}y \minus{} 1\equal{}(x \plus{} 1)(y \plus{} 1) \plus{} y\minus{}1 \equal{} 4$

i think there would be no greatest value of $ x^2 \plus{} y^2$, since maximizing this means maximizing the absolute values of $ x$ and $ y$. you could plug in any large value of $ y$ in the above equation and solve for $ x$, so there is no greatest value of $ P$.

[hide=" solution"]

$ x^2\plus{}y^2 \ge 2xy$, equality iff $ x\equal{}y$.
$ x\plus{}2x\plus{}x^2\equal{}4 \rightarrow x^2\plus{}3x\minus{}4\equal{}0 \rightarrow (x\plus{}4)(x\minus{}1) \rightarrow x\equal{} \minus{}4, 1$.
$ (1)^2\plus{}(1)^2\equal{}2$[/hide]

2xy is not a constant so I think azn was wrong.

[quote="Kyoshiro"][quote="L_Euler"][quote="Kyoshiro"]Let $ x \plus{} 2y \plus{} xy \equal{} 4.$
Find the least values of $ P \equal{} x^2 \plus{} y^2$[/quote]
We can use Graph method!!![/quote]
Can you post you solution?
I have tried. :|[/quote]

$ (1) \leftrightarrow x(y \plus{} 1) \equal{} 4 \minus{} 2y$

With $ y \ne  \minus{} 1$ we have $ x \equal{} \frac{{4 \minus{} 2y}}{{y \plus{} 1}}$

So $ P \equal{} y^2  \plus{} \left( {\frac{{2y \minus{} 4}}{{y \plus{} 1}}} \right)^2$

[hide="@@@Kyoshiro"]Đến đây thì khảo sát hàm số là ra thôi mà! :wink: [/hide]

So have you tried computing which value of $ y$ we want from there?

There's a reason [b]Kouichi Nakagawa[/b] expressed his answers approximately. The exact answer, simplified, is attached.

To the OP: what motivated this question?

[quote="L_Euler"][/quote][quote="Kyoshiro"][quote="L_Euler"][quote="Kyoshiro"]Let $ x + 2y + xy = 4.$
Find the least values of $ P = x^2 + y^2$[/quote]
We can use Graph method!!![/quote]
Can you post you solution?
I have tried. :|[/quote][quote="L_Euler"]

$ (1) \leftrightarrow x(y + 1) = 4 - 2y$

With $ y \ne - 1$ we have $ x = \frac {{4 - 2y}}{{y + 1}}$

So $ P = y^2 + \left( {\frac {{2y - 4}}{{y + 1}}} \right)^2$

[hide="@@@Kyoshiro"]Đến đây thì khảo sát hàm số là ra thôi mà! :wink: [/hide][/quote]

So

$ P = y^2 + \left( 2 - \frac {6}{y + 1} \right)^2 = y^2 + 4 - \frac {24}{y + 1} + \frac {36}{(y + 1)^2}$

$ \frac {dP}{dy} = 2(y + 1) - 2 + \frac {24}{(y + 1)^2} - \frac {72}{(y + 1)^3} = \frac {g(y + 1)}{(y + 1)^3}$

where $ g(t) = 2t^4 - 2t^3 + 24t - 72$

Solving
$ g(t) = 0$
is possible but  so hard, I think.


anybody notice that the canada problems in aops2 are never usually with the needle and thread?
Are canadian comps easier?
only other countries like bulgaria have the needle

I think everyone knows that the problems used in selecting the Bulgarian team for the IMO are usually much harder than the ones used to select the Canadian team for the IMO ...  :?

[quote="GoBraves"]I didnt know that..[/quote]

Not to worry. What Valentin means is that everyone who practices regularly with problems from an assortment of national olympiads from around the world will eventually get a feel for the general difficulty level of a particular country's problems in comparison with those of other countries.

I think that may have been a misuse of the word "everyone". I'm sure he was only referring a subset of AoPS-ML users, specifically those who are constantly in the Olympiad boards and a few select others.

But just fyi, the Canadian olympiad is considered one of the easier ones. The Bulgarian, along with, say, the Chinese and the Vietnamese, are considered much more difficult.

[quote="MithsApprentice"][quote="GoBraves"]I didnt know that..[/quote]
I think that may have been a misuse of the word "everyone". I'm sure he was only referring a subset of AoPS-ML users, specifically those who are constantly in the Olympiad boards and a few select others.[/quote]Yes by everyone I ment everyone which works with this kind of probelms (but not only)  :)


Please help me solve by elimination

A box contains $ \$$7.15 in nickels, dimes, and quarters. There are 42 coins in all, and the sum of the numbers of nickels and dimes is 2 less than the number of quarters. How many coins of each kind are there?

Thanks :!:

A box contains 7.15 in nickels, dimes, and quarters. There are 42 coins in all, and the sum of the numbers of nickels and dimes is 2 less than the number of quarters. How many coins of each kind are there?

[hide="Hint 1"]
Let $ N$ be the number of nickels and $ D$ be the number of dimes. Since there are $ 42$ coins in all, how many quarters are there?
[hide="Hint 2"]
How do you express the sum of the numbers of nickels and dimes?
How do you express 2 less than the number of quarters?
Set these equal, and you have one equation in two variables, $ N$ and $ D$.
[hide="Hint 3"]
In terms of money, what is the worth of $ N$ nickels? $ D$ dimes? $ Q$ quarters?
The sum of these must equal $ \$$$ 7.15$, or $ 715$ cents. This gives you a second equation, which will allow you to solve for the two variables.

$ Q = 42 - N - D$, so:

$ N + D = (42 - N - D) - 2$
$ 5N + 10D + 25(42 - N - D) = 715$

$ N = 7$
$ D = 13$
$ Q = 22$

[/hide]
[/hide]
[/hide]


trigonometry
inequalities
inequalities proposed

With $ a,b,c$ are sides of triangle. Prove that
1.$ \sqrt {\frac {a}{b + c - a}} + \sqrt {\frac {b}{c + a - b}} + \sqrt {\frac {c}{a + b - c}} \geq \sqrt {\frac {(a + b + c)^3}{3abc}}$
2.${ \frac {\sqrt {a}}{b + c - a} + \frac {\sqrt {b}}{c + a - b} + \frac {\sqrt {c}}{a + b - c}} \geq \frac {(a + b + c)^2}{3abc} \sqrt {\frac {a + b + c}{3}}$

(1) is obviously trues by CS and Schur.
(2) is obviously trues be (1) and $ \sqrt{3(a\plus{}b\plus{}c)} \ge (\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})$  :)
PS: [hide]Giau' cai' bdt dau` tien di chu' em, ai lai that. tha` the'  :) [/hide]

1.  $ \Leftrightarrow \sqrt 3 ( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} + \sin \frac{\gamma}{2}) \ge \sin \alpha +\sin \beta +\sin \gamma$ - which is true by Chebyshev :

 $ 2 (\sin\frac{\alpha}{2} \cos\frac{\alpha}{2}+ \sin\frac{\beta}{2} \cos\frac{\beta}{2}+\sin\frac{\gamma}{2} \cos\frac{\gamma}{2}) \le \frac{2}{3}(\sin\frac{\alpha}{2} +\sin\frac{\beta}{2} +\sin\frac{\gamma}{2} )$ $ (\cos\frac{\alpha}{2} +\cos\frac{\beta}{2} +\cos\frac{\gamma}{2} ) \le \sqrt 3 ( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} + \sin \frac{\gamma}{2})$

2. $ \Leftrightarrow \frac{3 \sqrt 3}{\sqrt{2}}( \frac{\sin^{3/2} (\frac{\alpha}{2})} {\cos^{1/2} (\frac{\alpha}{2})} +\frac{\sin^{3/2} (\frac{\beta}{2})} {\cos^{1/2} (\frac{\beta}{2})}  +\frac{\sin^{3/2} (\frac{\gamma}{2})} {\cos^{1/2} (\frac{\gamma}{2})} ) \ge ( \sin \alpha + \sin \beta + \sin \gamma)^{3/2}$

${{ LHS \ge \sqrt\frac{3}{2}} (\sin^{3/2} (\frac{\alpha}{2}) +  \sin^{3/2} (\frac{\beta}{2})} + \sin^{3/2} (\frac{\gamma}{2}) )$ $ ( \frac{1}{ \cos^{1/2} \frac{\alpha}{2}} + \frac{1}{ \cos^{1/2} \frac{\beta}{2}} + \frac{1}{ \cos^{1/2} \frac{\gamma}{2}} ) \ge$ $ 3^{3/4} (\sin \frac{\alpha}{2} +\sin \frac{\beta}{2} +\sin \frac{\gamma}{2} )^{3/2}\ge$ $ ( \sin \alpha + \sin \beta + \sin \gamma)^{3/2}$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a subgroup of $(\mathbb{R},+)$ such that $G_{+}=( x \in G | x>0)$ has $0$ as it's infinum.
Prove that $G$ is dense in $\mathbb{R}$

This is pretty obvious, isn't it? Take any interval $(a,b)\subset\mathbb R$, and some $x>0,x\in G$ with $x<b-a$. Then the set $\{nx\ |\ n\in\mathbb Z\}$ must intersect $(a,b)$.

now find all the subgroups of $(R,+)$ (describe all the facts you can obtain about this subgroups)
it seems to me it has already been discussed on the forum-sorry if it is so

The problem of finding the [i]closed[/i] subgroups of $(\mathbb R^n,+)$ has been discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=measurable&t=47161]here[/url].


Let $ p$ be a prime number between 40 and 60. What is the
probability that $ p \plus{} 12$ is also a prime number? Express your
answer as a common fraction.

the primes between 40 and 60 are 41, 43, 47, 53, and 59. Of these, 41+12, 47+12, and 59+12 are also prime, giving an answer of $ \boxed{\frac{3}{5}}$


analytic geometry
graphing lines
slope
number theory proposed
number theory

Let $m,n$ be integers. On the $xy$ plane, a point $(m,n)$ is called a lattice point.Taking each lattice point as a center,

draw a circle of radius $r$. All of line with slope $\frac{2}{5}$ intersects with some of these circles.

Find the minimum value of real numbers $r$ having this property.

Okay, these are the thoughts I've had, but some of the explanations may not be clear...

First of all it's easy to see that a line intersects a circle iff it intersects the diameter of slope $-\frac{5}{2}$. So we can forget the circles and just take these diameters.

(*): Consider the points $(0, 0)=A, (-2, 5)=B, (2, -5)=C, (29, 0)=D, (27, 5)=E, (31, -5)=F$. It's easy to see that $BC, EF$ have slope $-\frac{5}{2}$. Also, $CE$ has slope $\frac{2}{5}$. Now divide all coordinates by $29$, so that $AD=1$. Then $B=(-\frac{2}{29}, \frac{5}{29})$. And $AB^2=\frac{2^2+5^2}{29^2}=\frac{1}{29}$. So $AB=\frac{1}{\sqrt{29}}$.

Now if $r \geq \frac{1}{\sqrt{29}}$, take any line of the given slope, and a set of circles with centers at $(i, 0)$ for all integers $i$. Consider the first circle (looking from the left) such that the line is above its center. It's easy to see that the line will intersect the previous circle, by the above reasonings.

I don't know yet if $r<\frac{1}{\sqrt{29}}$ allows for a line that doesn't touch any circles, though...


videos
Alcumus
parameterization
probability
geometry
algebra
binomial theorem

Hi,

I'm a parent and volunteer math problem solving teacher and have opened a class for the kids I work with on Alcumus.  I am also working on the Alcumus problems myself to refresh and learn a few topics that I wasn't familiar with.  One topic that I am working on is the Binomial Theorem, which I never learned in school, but have mostly picked up around here. ;-)  So... I was watching those videos and they refer to Pascal's Identity.  I'm kind of vaguely familiar with that as well, but figured I'd like to see the video on it.  When I looked for my list of videos, however, the Pascal's Triangle videos are not available.  How can it think I am ready for the Binomial theorem without thinking I am ready for or passed Pascal's triangle?

I can access the videos through the teacher's panel, so that's not an issue for me, but this is more a question about whether the "readiness" parameters for those topics need to be tweaked so that students don't find themselves in the same position.

Also... as a teacher, it would be great to be able to "push" a video or set of videos to my students relating to the subject we were planning to cover in class that week.

Thanks for Alcumus!  It is truly an awesome resource!

Hello and thanks for trying out Alcumus.

We have just made a change to our report page which should now allow you to see the Pascal's Triangle topic and will allow you to view the related videos.

Thanks.  It still doesn't show up on my "Lessons" page but  I can access it through the Reports page.

Is there a way for a student and/or a teacher to view the "network" of topics mastered, passed and "ready for" in terms of a graph of pre-requisites?  Now that there are some "untouched" topics showing up on the reports, I am trying to figure out how they relate to the current topics being worked on and mastered, and why they haven't shown up at all yet.

Also, it would be good to be able to start a student "at the beginning" if they have no background in Counting and probability etc. without them having to do a 60-problem placement test, which would be frustrating for a beginner. Perhaps some way to cut off the placement test earlier and move on to lessons if it is clear that the student does not have mastery of the earliest topics.

Finally, a small point, but there doesn't seem to be any notification when the student has completed the required 60 problems and now has video lessons available to him/her.

Thanks again for a wonderful resource!

Some great ideas here; we'll look into them (and keep them coming, please!).

One thing that would be really nice would be a skip/save this problem button.  Sometimes you're just not in the mood for a particular type of question at a given time. :)  Of course depending on the set of problems one is ready for, the available problems might cycle back fairly quickly, but in any case, I think it would be nice to be able to "pass" on a problem temporarily without giving up sometimes.

Also it would be interesting for a teacher to be able to see the particular problems a student has attempted in a given topic, and what their answer(s) were.

Thanks for Alcumus!

By any chance do you know when you expect to have Algebra added?

[quote="mathmom"]One thing that would be really nice would be a skip/save this problem button.  Sometimes you're just not in the mood for a particular type of question at a given time. :)  Of course depending on the set of problems one is ready for, the available problems might cycle back fairly quickly, but in any case, I think it would be nice to be able to "pass" on a problem temporarily without giving up sometimes.[/quote]
Thanks for the suggestion. We'll add it to our list of suggestions.

[quote="mathmom"]Also it would be interesting for a teacher to be able to see the particular problems a student has attempted in a given topic, and what their answer(s) were.[/quote]
From your teacher page, click on Report next to the user's name. Then click on a particular topic in their report and you should be able to see their problems along with the answers they included. (Note: there was a bug in which if you were also the student's friend it would not display the problems. This has been fixed.)

[quote="mathmom"]By any chance do you know when you expect to have Algebra added?[/quote]
Sorry, we currently have no estimate for when Algebra will be added.

[quote]From your teacher page, click on Report next to the user's name. Then click on a particular topic in their report and you should be able to see their problems along with the answers they included. (Note: there was a bug in which if you were also the student's friend it would not display the problems. This has been fixed.) [/quote]

Thanks.  I thought I had seen this earlier.  I think it wasn't working this morning for some reason.

I also have one more possible bug:  I have one student whose report currently does not show any areas as passed, even though he has several badges.  He used to have areas showing as both passed and a couple as mastered.

As for a "pass" button -- are you basically looking for something along the lines of "I can't do this now, but don't show me the answer".

We want to know when a student can't do a problem, since this helps us determine where a student is mathematically.  So, we wouldn't give an option of "save this one for me until later, but don't mark me wrong just yet!"  The "give up but don't show me the answer" option tells us that the student doesn't know how to do it now, and the student will still possibly see that problem again in the future.

(I do like the idea of "give up but don't tell me" -- we'll see what we can do about implementing it.)

As for algebra, I have a question: how good/bad/indifferent is having just problems and no (or sparse) video lessons?  I think we can have a pretty large population of problems for Intro Algebra by the end of the summer, but the lessons are much more time-consuming.

[quote="rrusczyk"]As for a "pass" button -- are you basically looking for something along the lines of "I can't do this now, but don't show me the answer".  [/quote]

I was more thinking along the lines of "I know how but don't feel like it right now" or "I'm tired of that type of problem".  In that case, I would treat it as if it hadn't been presented at all from the point of view of the AI.  I'm thinking more along the lines of "shuffle that one back in for the moment".  I've been working through the program myself to fill a few gaps and keep my skills sharp. :)  But at the current time, if I get to a problem I don't feel like solving, I am just quitting for that session.  (I don't want to "give up" because I do know *how* to solve the problem even if I don't have the patience for it at that moment.)

[quote](I do like the idea of "give up but don't tell me" -- we'll see what we can do about implementing it.)[/quote]

As explained above, that's not quite what I was looking for, but would still be nice functionality.

[quote]
As for algebra, I have a question: how good/bad/indifferent is having just problems and no (or sparse) video lessons?  I think we can have a pretty large population of problems for Intro Algebra by the end of the summer, but the lessons are much more time-consuming.[/quote]

Better than nothing. :)  Much better, actually.

Of course in the ideal world, you'd have the video lessons too, but having the problems and the AI to select them appropriately, along with the pointers to where the related lesson is in the text would still be really nice.

And sparse video lessons would be a lot better than no video lessons, especially if you choose wisely and fill in first in the topics where you think students are most likely to get stuck if using only the text.

I have son who is ready to do an independent study of "Algebra I" and we have the AoPS text.  But just running through the text on his own is not that engaging for him, whereas he's been very enthusiastic about Alcumus.  So, Alcumus for Algebra would be awesome for him, even without the videos, especially since he's picked up a sort-of random selection of algebra skills already.  It would be nice for me to know that once he worked through the program, I could be fairly confident that he didn't have any gaps, whereas if we just pick and choose through the text, we might miss stuff.

Thanks again!

[quote="rrusczyk"]As for algebra, I have a question: how good/bad/indifferent is having just problems and no (or sparse) video lessons?  I think we can have a pretty large population of problems for Intro Algebra by the end of the summer, but the lessons are much more time-consuming.[/quote]

This would be [b]VERY GREAT[/b]!!!!!!!

Anyway, I think that a lot of people don't look at the lessons anyway.  (not me, of course :))

[quote="rrusczyk"]As for algebra, I have a question: how good/bad/indifferent is having just problems and no (or sparse) video lessons?  I think we can have a pretty large population of problems for Intro Algebra by the end of the summer, but the lessons are much more time-consuming.[/quote]
Having the algebra problems would be great, even if there weren't many (or any) videos. :coolspeak:


inequalities
function
inequalities proposed

Let $x,y,z$ be positive reals with sum $1.$ Prove that \[\sum_{\mathrm{cyclic}}\frac{x+1}{x^{2}+1}\leq\frac{18}{5}.\]

Q: for x,y,z > 0 with x + y + z = 1, show:
    (x+1)/(x^2 + 1) + (y+1)/(y^2 + 1) + (z+1)/(z^2 + 1) <= 7^(1/7).
A: I think the r.h.s is a bit small for it to be true. The least upper bound should be around 3.6 = 18/5 if not bigger. Now if you let x -> 0 ,
y -> 0 , and z -> 1, then the l.h.s -> 3 and 3 > 7^(1/7) so the ineq is no longer valid. I offer the following: replace the r.h.s 7^(1/7) by 18/5 and it should be fine. Here is how:

New Proof:
________

The function f(x) = (x+1)/(x^2 + 1) is indeed concave on (0 , 1) because
f''(x) = -2(1-x)(x^2 + 4x + 1)/((x^2 + 1)^2) < 0. So we can use Jensen's ineq here and get:
l.h.s <= 3f((x+y+z)/3) = 3f(1/3) = 18/5 = 3.6.
____________________________________________________________


inequalities
induction
real analysis
real analysis solved

Prove the inequality $e^{n-1}n!<n^{n+\frac{1}{2}}$

In order to prove it by induction, it suffices to show that
$(1+\frac{1}{n})^{n+\frac{1}{2}} > e$
We prove that $a_n(1+\frac{1}{n})^{n+\frac{1}{2}}$ decreases when $n$ increases, and its limit is $e$, hence every term is larger than $e$.

$\frac{a_{n+1}}{a_n} = \left(1+\frac{1}{n+1}\right)\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+\frac{1}{2}} = \left(1+\frac{1}{n+1}\right)\left(1-\frac{1}{(n+1)^2}\right)^{n+\frac{1}{2}}$

By Bernolli's inequality,
$\left(1-\frac{1}{(n+1)^2}\right)^{n+\frac{1}{2}} \leq 1-\left({n+\frac{1}{2}}\right)\frac{1}{(n+1)^2}$

hence
$\frac{a_{n+1}}{a_n} \leq \frac{2n^3+6n^2+5n+2}{2(n+1)^3}<1$


logarithms
number theory unsolved
number theory

Prove that number of distinct prime divisors of $n$ cannot exceed $\frac{\log n}{\log 2}$

Let [tex]n = p_1^{\alpha_1} \cdot \ldots \cdot p_k^{\alpha_k}[/tex] where [tex]p_j[/tex] are prime factors of [tex]n[/tex], then obviously [tex]p_j \geqslant 2[/tex], for [tex]j = 1, \ldots, k[/tex]. Thus [tex]2^k \leqslant p_1 \cdot \ldots \cdot p_k \leqslant p_1^{\alpha_1} \cdot \ldots \cdot p_k^{\alpha_k} = n[/tex], take logarithms on both side, to obtain [tex]k \leqslant \ln_2 n[/tex]. Recall that [tex]k = \omega ( n )[/tex] where [tex]\omega ( n )[/tex] is the number of distinct prime divisors of [tex]n[/tex], and we are done. :)


inequalities
geometry
3D geometry
conics
ellipse
geometric transformation
rotation

find reels a,b,c such that :

$ \Large\ a^2\plus{}b^2\plus{}c^2\plus{}3 \leq ab\plus{}3b\plus{}2c$

[hide="Hint"]
Rewrite as:

$ \left( a \minus{} \frac{b}{2} \right)^2 \plus{} \frac{3}{4} (b\minus{}2)^2 \plus{} (c\minus{}1)^2 \leq 1$

1. What is the geometric relationship between that region and the region defined by $ a^2\plus{} \frac{3}{4} (b\minus{}2)^2 \plus{} (c\minus{}1)^2 \leq 1$?
2. What is that second region?
3. Therefore, what is the first region?
[/hide]

May be a solution. I really don't know what is the expected answer.
[hide]
Suppose that $ a$, $ b$ and $ c$ are reals numbers that satisfy the inequality.
Rewrite it as
\[{ (a-\frac{b}{2}})^2+3(\dfrac{b}{2}-1)^2+(c-1)^2\le 1\]

Introducing $ d=\dfrac{b\sqrt{3}}{2}$ and shaking the inequality a little it becomes
\[{ (a-\frac{d}{\sqrt{3}}})^2+(c-1)^2\le 1-(d-\sqrt{3})^2\]

Since the left side is nonnegative we must have that $ |d-\sqrt{3}|\le 1$ wich is the same as $ 2-\dfrac{2}{\sqrt{3}}\le b\le\dfrac{2}{\sqrt{3}}+2$.

For a given value of $ d$ in this interval the set of points $ (a,c)$ that satisfy the inequality (when represented in a coordinated plane) form a circle (not a circunmference) with center $ (\dfrac{d}{\sqrt{3}}, 1)$ and radius $ \sqrt{1-(d-\sqrt{3})^2}$.

As $ d$ moves in its interval from the least value to the greatest value, the radius of the circle is increased from $ 0$ until it reaches a maximun at $ d=\sqrt{3}$, then it is decreased until the circle becomes a point abecoming a point again.

Let $ r$ represent the radius of the circle at each time. We have that $ r^2+\dfrac{3}{4}(b-2)^2=1$. Thus the points $ (r, b)$ form an ellipse.

From that I suspect that the set of points $ (a,b,c)$ that satisfy the inequality when represented in a cordinated 3D euclidean space form a symmetric figure that seems like a double cone, moreover, as the radius varies within a cuadratic relation ship (wich forms an ellipse), it seems elypticaly curved, like an egg.
[/hide]

[hide="Note"]I guess that the region I mentioned before is a the result of all rotations of certain ellipse about one of its axes but I'm not sure on how to verify that, and, if it is true, I don't know how to determine the exact ellipse.[/hide]

:)


geometry
calculus
derivative
3D geometry
sphere
real analysis
real analysis theorems

Is it a coincidence that the circumference of a circle is the derivative if it's area? And the surface area of a sphere the derivative of the volume? Or the surface area of a sphere the square of the circumference of the circle?

[quote="1=2"]Is it a coincidence that the circumference of a circle is the derivative if it's area? And the surface area of a sphere the derivative of the volume?[/quote]
No coincidence. Consider the difference between the area of a circle of radius $ r\plus{}h$ and of radius $ r.$ Visualize the picture. Same for the volume of the ball and area of the sphere.

In fact, we can say this: if the $ n$-dimensional measure of the ball of radius $ r$ in $ \mathbb{R}^n$ is $ V_n(r)$ and the $ n\minus{}1$-dimensional measure of the sphere of radius $ r$ in $ \mathbb{R}^n$ is $ A_n(r),$ then $ \frac{d}{dr}V_n(r)\equal{}A_n(r).$

[quote]Or the surface area of a sphere the square of the circumference of the circle?[/quote]
This isn't even true. You're overlooking a factor of $ \pi.$


analytic geometry
Support
superior algebra
superior algebra solved

Let x_1,x_2,x_3 be  three pozitive numbers <>0 which are given! and we have the series x_n which is given like so:

x_(n+2)=(x_(n+1)^2+5x_n^2+x_(n-1)^2)/(x_(n+1)+5x_n+x_(n-1)).

prove that the series converges!!

bye!

Denote by m_n= min{x_n, x_(n-1), x_(n-2)} and M_n=max{x_n, x_(n-1), x_(n-2)}.
It is a well known result that if a,b,c are reals and x,y,z positive reals then
min{a,b,c} \leq (xa+yb+zc)/(x+y+z) \leq max{a,b,c}.

I have a nice solution for this:
Take 3 objects of coordinates a \leq b \leq c on a line of positive weights x,y,z. Then m=(xa+yb+zc)/(x+y+z) is the mass center of the 3 objects.
Remember that m is the equilibrium point for the system of 3 objects, so if we support the line of the 3 objects || to the groud in m, it doesn't move. It is quite obvious that if we support the line in a point outside [a,c] it will move.
I know there is a mathematical proof for this, but this is a solution I like.

Back to the problem notice that m_n+1 \leq x_n+2 \leq M_n+1.
This proves m_n is an increasing sequence and M_n is decreasing, so x_n is bounded.
m_n and M_n are bounded and monotonous, so they converge to points m and M.

We have (xn+1 2 +5*xn 2 +xn-1 2 )/(xn+1+5*xn+xn-1) \geq (xn+1+5*xn+xn-1)/7. This is the well known CBS inequality.

It is easy to see that we have m_n=x_n for infinitely many n.
So m_(n+2) \geq (xn+1+5*xn+xn-1)/7 \geq (M_(n+1)+6*m_(n+1))/7. This is because M_n+1 is one of x_n+1, x_n and x_n-1 and m_n+1 is also one of them. The number I wrote the least from a possible list of (M_n+1+6m_n)/7 and (2m_n+1+5M_n)/7.

So we've got m_(n+2) \geq (M_(n+1)+6*m_(n+1))/7 for infinetely many n. We tend n to infinity and obtain m \geq (M+6m)/7 or better said m \geq M.
It is obvious that we have M_n \geq m_n for all n, so M \geq  m.
These 2 show M=m, so x_n converges.

[quote="amfulger"]Back to the problem notice that m_n+1 \leq x_n+2 \leq M_n+1.
This proves m_n is an increasing sequence and M_n is decreasing, so x_n is bounded.
m_n and M_n are bounded and monotonous, so they converge to points m and M.
[/quote]
how can u prove the monotonity of $m_n$ and $M_n$?


number theory
relatively prime
2002 AIME I 8

Find the smallest integer $k$ for which the conditions
$(1)$ $a_1, a_2, a_3, \ldots$ is a nondecreasing sequence of positive integers
$(2)$ $a_n=a_{n-1}+a_{n-2}$ for all $n>2$
$(3)$ $a_9=k$
are satisfied by more than one sequence.

[hide="Answer"]$a_9=a_8+a_7=a_5+2a_6+a_7=...=13a_1+21a_2$
Since $13$ and $21$ are relatively prime, all terms of the sequence are positive, and the sequence never decreases (although the first two terms can be equal), the first time that we can have two distinct sequences with the same $a_9$ is when we can subtract $21$ from the value of $a_1$ (and add $13$ to the value of $a_2$, but that is never a problem). So, we first have this for $a_1=22$, $a_2=22$, and so $k=\boxed{748}$.[/hide]

sorry to bring up an old topic, but I'm not really understading how the 1st time we have a value of k that is satisfied by more than one sequence is when $ a_1$ and $ a_2$ are both 22?

You want to have at least two ordered pairs $ (a_1, a_2)$ that have equal values of $ 13 a_1 \plus{} 21 a_2$.

We note that $ (a_1\plus{}21,a_2\minus{}13)$ yields the same value of that linear combination as $ (a_1, a_2)$

So now, let $ a_1 \equal{} 1$ in the first pair. Then, the second pair is $ (1\plus{}21,a_2\minus{}13) \equal{} (22,a_2 \minus{} 13)$.

To minimize $ k$, we want to minimize $ a_2\minus{}13$. Since $ a_2 \minus{} 13 \geq 22$, as the sequence is non-decreasing, we set $ a_2\minus{}13 \equal{} 22$.

Thus, $ (a_1\plus{}21, a_2\minus{}13) \equal{} (22,22)$. This, as well as $ (a_1, a_2) \equal{} (1, 35)$, produce the same $ k$ value, which is the minimum value of $ k$ that corresponds to multiple ordered pairs of the first two terms.

Isn't this exactly the same as 2013 AMC12B #14, but just with a different index?

uh no this is 2002 AIME I #8

I think this is the thread that the contest collection was supposed to link to. Maybe 2013 AMC 12B #14 was similar to this ig MAA reuses problems a lot.

p2: We have $a_9=13a_1+21a_2$, so $a_9$ is in the form $13a+21b=13c+21d$, or $13(a-c) = 21(d-b)$ so $a-c$ is a multiple of $21$, $d-b$ is a multiple of $13$. The minimum would be at $a - c = 21, d - b = 13$, but $a \leq b$ and $c\leq d$, so $a = 22, b = 22, d = 35, c = 1$, giving $\boxed{748}$.


What is the smallest prime number greater than 120?


system of equations
algebra
polynomial
IMO Shortlist
IMO Longlist

Find all real solutions of the system of equations:
\[\sum^n_{k=1} x^i_k = a^i\] for $i = 1,2, \ldots, n.$

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]  :)

The case $ n\equal{}1$ yields trivially $ x_1\equal{}a$ (only equation in the system).

Assume $ n\geq4$.  Then, $ a^4\equal{}(x_1^2\plus{}...\plus{}x_n^2)^2\equal{}x_1^4\plus{}...\plus{}x_n^2$, yielding $ \sum x_i^2x_j^2\equal{}0$ for $ 1\leq i<j\leq n$, or at most one of the $ x_i$ is nonzero, and it must equal $ a$.

If $ n\equal{}3$, we find $ a^2\equal{}x_1^2\plus{}x_2^2\plus{}x_3^2\equal{}(x_1\plus{}x_2\plus{}x_3)^2$ yielding $ x_1x_2\plus{}x_2x_3\plus{}x_3x_1\equal{}0$, or using that $ x_1^3\plus{}x_2^3\plus{}x_3^3\minus{}x_1x_2x_3\equal{}(x_1\plus{}x_2\plus{}x_3)(x_1^2\plus{}x_2^2\plus{}x_3^2\minus{}x_1x_2\minus{}x_2x_3\minus{}x_3x_1)$, we conclude that $ a^3\minus{}x_1x_2x_3\equal{}a(a^2\minus{}0)$, or $ x_1x_2x_3\equal{}0$, and wlog at least $ x_3\equal{}0$.  But then $ x_1x_2\equal{}0$, and again at most one of the $ x_i$ is nonzero, and it must equal $ a$.

Finally, if $ n\equal{}2$, then $ a^2\equal{}x_1^2\plus{}x_2^2\equal{}(x_1\plus{}x_2)^2$, and $ x_1x_2\equal{}0$, and again at most one of the $ x_i$ is nonzero, and it must equal $ a$.

$S_k=x_1^k+x_2^k+...x_n^k$ and let $\alpha_k, k=0,1,...,n$ denotes the $k$th elementary symmetric polynomial in $x_1,x_2,...x_n$ Using Newton's formula, $k\alpha_k=S_1\alpha_{k-1}-S_2\alpha_{k-2}+......+(-1)^kS_{k-1}\alpha_1+(-1)^{k-1}S_k$ This equation has a root for $\alpha_1=a,\alpha_k=0$ so by vieta $x_1,x_2,...x_n$ are the roots of the polynomial $x^n-ax^{n-1}$. So the roots are permutation of $a, 0,0,...,0$


geometry
analytic geometry
trapezoid
graphing lines
slope
number theory
relatively prime

The coordinates of the vertices of isosceles trapezoid $ABCD$ are all integers, with $A=(20,100)$ and $D=(21,107).$  The trapezoid has no horizontal or vertical sides, and $\overline{AB}$ and $\overline{CD}$ are the only parallel sides.  The sum o f the absolute values of all possible slopes for $\overline{AB}$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m+n.$

Ok, so this problem involves number theory along with a sense of geometry. First of all, consider that the two legs must be equal. Let
          B= (a,b)               C=(c,d)

Then (a-c)^2 + (b-d)^2 = 50

There are only a set number of possibilities in integers mainly (let (a-c)=x and (b-d)=y
              x= (1,-1)        y= (7,-7)
              x= (7,-7)        y= (1,-1)
              x= (5,-5)        y=(5,-5)

Also, from the slopes we know that (b-100)/(a-20)=m=(d-107)/(c-21).

If you follow through and solve for b and d in terms of a,m, and c we have

   b= (a-20)m+100             and         d= (c-21)m +107

Substituting this in the equation (a-c)^2 + (b-d)^2 = 50 we have

      (a-c)^2 + (7-m(-1+c-a))^2 =50

Using the values for (a-c) or x above you can solve for all values of m. However the second leg is parallel when (x,y) =(1,7) or (-1,-7) so disregard these.

You are left with 12-2 =10 cases of which 2 are horizantal or vertical and the sum of the slopes comes out to be (119/12). So answer is 131

[hide="Solution"]

Since $ \overline {AB} $ and $ \overline {CD} $ are parallel, let the gradient of these two lines be $ m $. Let the line containing $ \overline {AB} $ be $ y_{1} $ and the line containing $ \overline {CD} $ be $ y_{2} $. By doing some calculation, we have

\[y_{1}=mx+100-20m \]

\[y_{2}=mx+107-21m \]

Let the $ x $ coordinate of $ B $ and $ C $ be $ x_{1} $ and $ x_{2} $ respectively.

\[ B= (x_{1},mx_{1}+100-20m) \]

\[ C= (x_{2},mx_{2}+107-21m) \]

We shall let $ M_{1} $ and $ M_{2} $ be the midpoint of $ \overline {AB} $ and $ \overline {CD} $ respectively. Since the quadrilateral is an isosceles trapezoid, $ \overline {M_{1}M_{2}} $ must be perpendicular to $ \overline {AB} $ and $ \overline {CD} $. Hence the gradient of $ \overline{M_{1}M_{2}}=-\frac{1}{m} $

\[M_{1}=(\frac{x_{1}+20}{2}, \frac{mx_{1}+200-20m}{2}) \]

\[M_{2}=(\frac{x_{2}+21}{2}, \frac{mx_{2}+214-21m}{2}) \]

After rearranging and simplifying,

\[\frac{m(x_{1}-x_{2})+m-14}{(x_{1}-x_{2})-1}=-\frac{1}{m} \]

Let $ x_{1} -x_{2} $ be $ A $. [b]Since the coordinates of $ B $ and $ C $ are integers, $ m $ must be rationale.[/b]

\[\frac{mA+m-14}{A-1}=-\frac{1}{m} \]

\[(A+1)m^{2}-14m+(A-1)=0\]

\[m=\frac{7 \pm \sqrt{50-A^{2}}}{A+1} \]

Since $ A $ is an integer, easy calculation will lead to only $ 6 $ possible values of $ A $, which are $ \pm 1 $, $\pm 3 $ and $ \pm 5 $. $ \pm 1 $ is not feasible as the question states that the trapezoid has no horizontal or vertical sides.

Now, substitute the respective values of $ A $ and we will get the sum of the absolute values of all possible slopes for $ \overline{AB} =\frac{119}{12} $.
[/hide]


We know that there exist infinite prime in the form $ k.2^n \plus{} 1$
Here is problem :
Consider a prime in the form $ k.2^n \plus{} 1$ where $ k\equiv 1 (\mod 2)$
Know that $ k|2^{2^m } \plus{} 1$
Prove that $ p|k^{2^{n \minus{} 1}} \minus{} 1$

what's $ p$ and $ m$
if $ m$ is arbaitrary in IN and $ p\equal{}k2^n\plus{}1$
we take $ n\equal{}2,k\equal{}3,m\equal{}0$
we have$ k\equal{}3|2^{2^{0}\plus{}1}\plus{}1\equal{}2^{2^{m}\plus{}1}\plus{}1$
but $ p\equal{}13\not|8\equal{}3^{2^{n\minus{}1}}\minus{}1$

Oh sorry I have some problem when type .
$ 2^{2^m}\plus{}1$ is the Fermat number . :wink:


How many five letter distinguishable code words can be created from the letters in TRIGONOMETRY such that the first letter is an R, the third letter is a vowel, and the last letter is a Y?
(A) 200
(B) 160
(C) 144
(D) 80
(E) 72
[hide="Answer"]B[/hide]

Is there a fast way to do this?

[quote="hydro"]why is B the right answer? because we want to find how many distinct words in the form R_ _ _ Y we can compose, so we can choose among 10 different other letters to put in the word. The third is a vowel, and we can choose it in 4 different way. Chosen the vowel, in the other two empty spaces have to put a letter among 9 and then 8 different letters. Then we have to divide $9*8*4$ by the product of the permutations of the letters that appears more than one time, that are O and T. The product of their permutations is 4, so the result I found is 72... where is my solution wrong?[/quote]

You can't simply divide by $9 \times 8 \times 4$ by $4$ because the letters O and T do not necessarily appear twice, or at all in that set of $3$ letters.

A solution that doesn't actually work:  :lol:

After taking away the R and Y, there is 1 E, 1 I, 1 G, 1 M, 1 N, 2 O's, 1 R, and 2 T's remaining.

Case 1:  The third letter is an E or I.  If the second and fourth letters are the same, there are $2$ possibilities.  If they are different, there are $7\cdot 6=42$ possibilities.

Case 2:  The third letter is an O.  If the second and fourth letters are the same, there is $1$ possibility.  If they are different, there are $8\cdot 7=56$ possibilities.

Hmm... this gives me $145$, which obviously isn't right.  What did I do wrong?


Triangle $ ABC$ has sides $ 8$, $ 15$ and $ 17$. A point $ P$ is inside the triangle. Find the minimum value of $ PA^2\plus{}PB^2\plus{}PC^2$.

Please post your solutions in detail  :)

Thanks.

Minimun value occurs when $ P$ is the centroid of $ \triangle ABC,$ due to Leibniz theorem.

$ PA^2 \plus{} PB^2 \plus{} PC^2 \equal{} \frac {_1}{^3}(a^2 \plus{} b^2 \plus{} c^2) \plus{} 3 \cdot PG^2$

$ PA^2 \plus{} PB^2 \plus{} PC^2$ is minimum $\Longleftrightarrow \ PG \equal{} 0 \Longleftrightarrow \ P \equiv G$

$\Longrightarrow \ PA^2 \plus{} PB^2 \plus{} PC^2 \equal{} \frac {578}{3}$


I think we should get more competent mods for G&FF now that Klebian was away. I posted a link to a freakin NEWS STORY from ABC and then it was removed with this message and the thread locked.

"MithsApprentice wrote:
1) No commercial advertising. No exceptions.

Please don't spam, Ignite."

Umm, since when is posting a link to a news story advertising? And how the heck is it commercial?

This was the link if any of you are interested:
http://abc.net.au.newsitems.200706.999584001215588457.851154.html.macksfoto.com/s1942069.htm

which I've found out now is false and the site is a satire site but that still doesn't make it advertising or commercial. This is why 6th graders shouldn't mod. I mean they may be in Algebra II or whatever but obviously they can't understand the meanings of the words "advertising" or "commercial."

[quote="Ignite168"][quote="junggi"][quote="Ignite168"]now that Klebian was away
[/quote]
shouldnt you be away also[/quote]

treething brought a laptop[/quote]
thats cool

and didnt that demon dude lock your thread not klebian?

[quote="7h3.D3m0n.117"][quote="Ignite168"]This was the link if any of you are interested:
http://abc.net.au.newsitems.200706.999584001215588457.851154.html.macksfoto.com/s1942069.htm

which I've found out now is false and the site is a satire site but that still doesn't make it advertising or commercial.[/quote]

If you found out that the site was false, then you should have no problem accepting the fact that your thread was locked.

Okay, if you just found out then I would not call it advertising but still, double check your information  :wink: .[/quote]
dude jokes$\ne$advertisements
if you're not sure what advertisements mean, here you go
[url]http://en.wikipedia.org/wiki/Advertisement[/url]

[quote="7h3.D3m0n.117"]Before this, the last time someone posted a joke that some AoPSer died no one found it funny hmm?[/quote]
your point being...?
not all jokes have to be funny
HOWEVER, most jokes are not advertisements, and therefore not spam and should not be subject to lockation.
also, there's a difference in joking abot fellow aopsers and joking about a useless woman

[quote="junggi"]HOWEVER, most jokes are not advertisements, and therefore not spam and should not be subject to lockation. [/quote]

His "joke" included an advertisement, so don't go off topic.

We are not debating what the definition of a joke is, it is why there was an ad in one of Ignite's posts. Im all for trashing Paris Hilton, but the presence of such resistance tells me that you are okay with spamming G&FF with ads.

I just opened it and it shows nothing but a few ads (one is a World's best mom smtg smtg ad, and one is a set of google ads)

I don't see any articles

[quote="7h3.D3m0n.117"][quote="junggi"]HOWEVER, most jokes are not advertisements, and therefore not spam and should not be subject to lockation. [/quote]

His "joke" included an advertisement, so don't go off topic.
[/quote]
in short, we are forbidden to post ANY KINDS OF LINKS that includes advertisements?
and, i don't recall seeing advertisements in his links?
all i remember are abc news-like page with a (fake) article about paris hilton

the link is broken right now

[quote]We are not debating what the definition of a joke is, it is why there was an ad in one of Ignite's posts.[/quote]
wtf?
by no advertisements, im sure miths meant posts of the nature

[quote]Come to this website RIGHT NOW and buy our stuff![/quote]

not

[quote]hey look cool this link has funny stuff in there (but it happens to have ads on the sides, oh well)[/quote]

and to repeat:
i don't recall seeing advertisements in his links?
all i remember are abc news-like page with a (fake) article about paris hilton

[quote="junggi"]and to repeat:
i don't recall seeing advertisements in his links?
all i remember are abc news-like page with a (fake) article about paris hilton[/quote]

[img]http://img2.freeimagehosting.net/uploads/d6b4a60077.jpg[/img]

When I hit the link, something like this came up. I assumed that the article didn't exist and that Ignite just put an ad there. Sorry. I did not realize that there was a previous "article." If you want to trash Hilton please just make another topic and don't joke about people's deaths.


geometry
ratio
ARML
AMC
USA(J)MO
USAMO
analytic geometry

No idea, maybe because there was a thread where people were using it to solve a problem here. But maybe not, because in that thread was a link explaining it:

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

[quote="jelyman"]could someone please explain it in simpler terms?[/quote]

Mass point geometry is based on the idea of imagining that there are weights on the vertices of a figure, and finding the point at which the figure would balance. That's the simple idea. The hard idea, which takes practice, is the mathematics which will show where that point is for any particular figure, without having to build a model and try balancing it.

You can imagine where the balance point of an equilateral triangle would be, right? Now think about what would happen if you stretched the triangle into varying obtuse or acute triangles. Where is the balance point on those?

Does anyone mind if I move this to "Other Problem Solving Topics"?  I think that's where it belongs.

Mass point geometry is closely related to vectors.  It's a useful technique, but I don't know off-hand a nice way of describing it.

Go ahead and move this thread, by the way.

I'll see if I can dig up a couple of good mass-point example questions.  If other people could do this, too (find simple questions that involve mass points), that would be nice.  I think Ceva can be proved mass-pointily,* right?  I'll see if I can find that.

*note: this is a made up word.

There are [url=http://mathcircle.berkeley.edu/mpgeo.pdf]BMC problems[/url] on mass points.

In triangle ABC, points D and E are on AB and AC, and angle bisector AT intersects DE at F.  If AD=1, DB=3, AE=2, and EC=4, find AF/AT .

Does this just require using mass-points or is there some more information we need to find first (like the actual length of BT or TC and not just the ratio of BT:TC?)  The answer is [hide] 5/18 [/hide]

Oh I remember this problem! It's from ARML a long time ago, right? I remember not solving it at an ARML practice.

I don't think any problem will ever require mass points.

Spoiler: [hide]One way to solve this problem is to use Stewart's theorem, the angle bisector theorem, and the law of cosines. I think it should work out nicely because ABC is isosceles. Keep in mind that you can let BC be anything you want because this problem comes from ARML and not from USAMO, but you should also be able to justify it reasonably easily just by dealing with angle BAT, for example.[/hide]

Sorry, i worded my question poorly.  I meant if I wanted to use mass points, could I just jump into it, or do I need to know the actual length of BT or TC first, not just the ratio of BT:TC ?
You're right, this is from 1992 ARML individual.

The beauty of mass points is that you absolutely do not need to know exact lengths.  In fact, they are quite irrelevant.  The reason for this is that if you have a balanced setup, any similar (lengths in proportion, weights in proportion) setup will also be balanced.

Alright, then i must be doing something wrong then.  Could someone point out where my mistake is?

Going by the diagram i posted...

Place a mass at vertex A of 3, and a mass at vertex B of 1.  Then 3*2=4*mass at C, so mass at C = 3/2.  Then we place a mass at T that is the sum of the masses at B and C, or 3/2 + 1 = 5/2.  This means that AF is 5x/2, and FT is 3x.  (5x/2)/(5x/2 + 3x) = AF/AT = 5/11.

Well, I personally use a version of mass points that is slightly different.  I suggest that the fault almost certainly lies in using F as the balance point of AT.  You never used that F is on DE, which seems troubling to me.  I suspect that the flaw lies there.

How is your method different?  Could you show me how you would solve it using yours?

Sure.  Your method uses only one coordinate.  My method uses 2.  (Well, actually 3, but their sum is always 1, so we can just get rid of the third entirely.)  If we have a point X in between (on the same line as) Y and Z, and YX = t*YZ for some value of t, then X = tZ + (1 - t)Y.  This is really much more vectory than the alternative, but I believe that it works in essentially the same fashion.

So, let A = (0, 0), B = (1, 0) and C = (0, 1).  Then D = (1/4, 0) and E = (0, 1/3).  Using the angle bisector theorem to find the ratio BT:TC, we get that T = (3/5, 2/5).  Using the angle bisector theorem in ADE, we get that F = (1/6, 1/9) = (3/18, 2/18).  Thus AF:AT = 5/18.  (Since they're on a straight line, we can just subtract and divide and that's what we get, and subtracting A is easy since it's (0, 0).)

So, looking further at your solution, you need to set up in advance somehow that F is a balance point.  I note that your solution actually makes it so that F isn't even a balance point of DE.  I would suggest that if you redo your weights, maybe starting with weights on A and D (instead of A and B), that it would be more likely to turn out correctly that way.  Use A and D to get B, use D and the ratio DF : FE to get E, use A and E to get C, and then make sure that T is the balance point of BC, and I think (hope) things will turn out okay.

JBL... is there a name for the type of coordinates you're using? are they barycentric? trilinear? (ive heard these words but I have no idea what any of them are)

I call them barycentric, and the fact that I keep them always adding to 1 may give them a new name, either Normalized Barycentric or Areal.  All of these things are on [url=http://mathworld.wolfram.com]Mathworld[/url], if you do a search for them.  Looking at it briefly, I think I'm using what they call areal coordinates.  Anyhow, see if you can decipher the Mathworld entries on those topics.

I don't think i really understand what's going on.  I'm stuck here:
[quote]Using the angle bisector theorem in ADE, we get that F = (1/6, 1/9) [/quote]

When we use the angle bisector theorem on :tri: ADE, EA=1/3 and AD=1/4.  Then 1/(3*EF)=1/(4*FD), which means EF=4*FD/3.  But going by the diagram I attached with the problem, 1/FD=2/EF, or EF=2*FD.  And, i'm stuck on how to get F=(1/6, 1/9) ?

Looking again, maybe I'm using Exact Trilinear coordinates.  I'm not really sure.  They do work, though :)

Okay, here we go.  In triangle ADE, we have (lengths here) AD = 1, AE = 2, right?  Thus DF : FE = 1 : 2.  Thus DF = DE/3, FE = 2DE/3.  So F = 1/3*E + 2/3*D = 1/3*(0, 1/3) + 2/3*(1/4, 0) = (0, 1/9) + (1/6, 0) = (1/6, 1/9).  Remember, these coordinates don't reflect actual lengths -- they reflect ratios.  So the fact that E has a coordinate (0, 1/3) and D has a coordinate (1/4, 0) tells you nothing about the actual lengths from any point to any other.  What it does tell you is that E is 1/3 of the way from A to C and D is 1/4 of the way from A to B.  The lengths we have to use here are the lengths that are given to us, although actually all that matters is what their ratio is.

These are very different from Cartesian coordinates, even though I'm using the same notation as one uses for cartesians.

ComplexZeta, thanks for pointing out the link to the Berkeley Math Circle notes on mass-point geometry: [url]http://mathcircle.berkeley.edu/mpgeo.pdf[/url].  I enjoyed them, and highly recommend them to others.

Joel, sorry it's taken so long to respond.

Thanks for your help, I think I understand now - that was very helpful.

find the weights you put on each vertices of triangle so that the centroid coincides with
a) circumcenter
b) orthocenter
c) incenter
d) Nagel point
e) Lemone Point
f) Nine-pt center
g) Fermat's 1st pt
h) Fermat's 2nd pt


Let $ A_{1},\ A_{2},\ \ldots,\ A_{n}$ be $ n$ points inside (or on the boundary) of a given unit circle (with unit radius). Prove that
\[ \prod_{1\leq i < j \leq n}{A_{i}A_{j}} \leq n^{\frac {n}{2}}.
\]
[hide]Viewing my proof, I would rather say that this problem belongs to the Algebra forum, but I really wish to see some more combinatorial approaches.[/hide]


For any positive integer $ n$ the sum $ 1\plus{}\frac{1}{2}\plus{}\cdots \plus{}\frac{1}{n}$ is written in the form $ \frac{P(n)}{Q(n)}$, where  $ P(n)$ and $ Q(n)$ are relatively prime.

a) Prove that $ P(67)$ is not divisible by 3;

b) Find all possible $ n$, for which $ P(n)$ is divisible by 3.

(sorry if this one already has been posted before)

http://www.mathlinks.ro/Forum/viewtopic.php?p=19008#19008

Have a look there

Daniel


analytic geometry
geometry
graphing lines
slope
calculus
calculus computations

1)Find the coordinates  of the point where the tangent and normal to teh curve y=x+1/x,$x \neq0$ cross the x nd y axes and hence determine the area enclosed by the x-axis,the y-axis,the tangent and the normal.

2)Show that every normal to teh curve $y=\sqrt{a^{2}-x^{2}}$ will always pass through the point (0,0).

[quote="you are not Him"]2)Show that every normal to teh curve $y=\sqrt{a^{2}-x^{2}}$ will always pass through the point (0,0).[/quote]

This curve is the upper semicircle of radius $a$ centered at the origin, the normal lines to the curve point radially outward from the origin (this is why, when you spin around on a merry-go-round, you tend to fly-off radially outward from the center of it lest you hold on). Enough physics though, here the math: differentiating $y=\sqrt{a^{2}-x^{2}}$ gives the slope of the tangent line to the curve, namely $y^\prime=\frac{-x}{\sqrt{a^{2}-x^{2}}}$. The slopes of perpendicular lines, say $m_{1}$ and $m_{2}$, multiply to give $m_{1}m_{2}=-1$ so that $m_{2}=-\frac{1}{m_{1}}$ and hence we know that the slope of the normal line to the curve is

\[m_\perp =-\frac{1}{y^\prime}=\frac{\sqrt{a^{2}-x^{2}}}{x}.\]

To write the equation of the normal line we use point-slope form noting that the point must be on the given curve, we may choose the point to be $\left( x_{0}, \sqrt{a^{2}-x_{0}^{2}}\right) ,$ upon which choice we must use the slope for the normal at the same $x$ value, which is $m_\perp = \frac{\sqrt{a^{2}-x_{0}^{2}}}{x_{0}}$ whence the equation of the normal line (in point-slope form) is

\[y-\sqrt{a^{2}-x_{0}^{2}}= \frac{\sqrt{a^{2}-x_{0}^{2}}}{x_{0}}(x-x_{0})\]

which simplifies to

\[y= \frac{\sqrt{a^{2}-x_{0}^{2}}}{x_{0}}x\]

so it is clearly seen that (0,0) is a point on every normal line.

2) Another method.
$y^{2}+x^{2}=a^{2}.$ Hence this is circle, statement's trivial
 :D

Benorin, in regards to your solution, we can conclude that every normal line goes through the origin because the family of normal curves are in the fomr y = mx + b whre b = 0, correct?

[quote="benorin"]when you spin around on a merry-go-round, you tend to fly-off radially outward from the center of it lest you hold on[/quote]
Have you done this experiment yourself? :) I think you would fly off along a tangent, not along a normal.

Yes, you would most definitely fly off along the tangent. The myth that you would fly outward radially is based on the nonexistent (for the most part) [url=http://en.wikipedia.org/wiki/Centrifugal_force]centrifugal force[/url].

ROLEX4life: yes, your statement is correct, except that there is a vertical normal line, which cannot be written in the slope form. But it also passes through $(0,0)$.

Going off on a tangent...
[img]http://imgs.xkcd.com/comics/centrifugal_force.png[/img]

Problem #1 appears ill-posed, since in general both normal and tangent lines intersect both coordinate axes, and the collection of lines encloses more than one region.


Let  ABC  a  triangle  and  D  midlepoint  of  AB , E  middlepoint  of  BC . M  is  a  point  on  AC  such  that  ME>CE. Proove  that  AD>MD.
Thanks

What if M is really far away on the ray (CA? Maybe you mean M on [AC]?  If that's the case, here it goes:

Let B' be the foot of the altitude from B. ME>CE translates to M being outside the segment [CB1] (I'll let you figure out why). If M is on [AC] but outside [CB1] then it must be inside [AB1], so it should be closer to the perpendicular from D to AC than A, so AD>MD.

I assumed A and C <90, butit works for all the other cases as well if we take M to be on [AC]. If M can be newhere on AC we can have lots of counterexamples.


[quote="croatian_dan"]New York has a population of 20,000,000 in a 50,000 sqaure mile area.
Illinois has a population of 12,000,000 in a 80,000 square mile area.

How many people from New York would have to move to have the same population density as Illinois.[/quote]

$\text{NY: }20,000,000 \text{ people in }50,000 \text{ sq. miles}$
$\text{IL: }12,000,000 \text{ people in }80,000 \text{ sq. miles}$

Problem where they move from NY to IL:
[hide]$\text{NY: }\frac{20,000,000}{50,000}= 400 \frac{\text{people}}{\text{mile}^{2}}$
$\text{IL: }\frac{12,000,000}{80,000}= 150 \frac{\text{people}}{\text{mile}^{2}}$

$\frac{20,000,000-x}{50,000}=\frac{12,000,000+x}{80,000}$
$400-\frac{x}{50,000}=150+\frac{x}{80,000}$
$250=\frac{x}{50,000}+\frac{x}{80,000}$
$50,000x+80,000x=1,000,000,000,000$
$130,000x=1,000,000,000,000$
$13x=100000000$
$x=\frac{100000000}{13}\approx 7,692,307.692$

We cannot have $.692^{th}$ of a person, thus we use $\boxed{7,692,307 \text{ people}}$ must move from NY to IL giving us proportions in both states to be $\approx 246.1538 \frac{\text{people}}{\text{mile}^{2}}$.[/hide]

Just out of NY and NOT into IL:
[hide]$\frac{20,000,000-x}{50,000}=\frac{12,000,000}{80,000}$
$400-\frac{x}{50,000}=150$
$250=\frac{x}{50,000}$
$x=12,500,000$

Thus $\boxed{12,500,000 \text{ people}}$ must move out of NY and NOT into IL in order for NY to have the same population density as IL which is $150 \frac{\text{people}}{\text{mile}^{2}}$[/hide]

I did it the first way then realized what you were asking.  So there are two different ways to think about it.


geometry
geometric transformation
rotation
trigonometry
geometry proposed

Let $ ABCD$ be a convex quadrlateral such that $ AB\equal{}BC\equal{}CD\equal{}a$, $ \angle BAD\equal{}75^0$, $ \angle ADC\equal{}45^0$. Find the length of $ AD$.
p/s: Can anyone solve this using rotation to kill this problem? If it is possible, would you please give me a proof in detailed? :lol:

[color=darkblue]From [url=http://www.mathlinks.ro/viewtopic.php?p=827187#827187]BMO 2007[/url] we conclude that $ EA \equal{} ED$ and $ \angle AED \equal{} 120^\circ$.  Then $ \angle ABD \equal{} \angle BAD \equal{} 75^\circ$, so $ AD \equal{} BD \equal{} \sqrt {2a^2 \minus{} 2a^2 \cos 150 } \equal{} a \sqrt {2 \plus{} \sqrt {3}}$[/color]


geometry
3D geometry
sphere
inequalities
geometry unsolved

http://www.mathlinks.ro/Forum/viewtopic.php?p=945797#945797
$ A,B,C,D,E$ are 5 points on a sphere with radius $ R$
such that $ \widehat{BAC}\equal{}\widehat{CAD}\equal{}\widehat{DAE}\equal{}\widehat{EAB}\equal{}\frac{2}{3}\widehat{BAD}\equal{}\frac{2}{3}\widehat{CAE}$
Prove the inequality $ AB\plus{}AC\plus{}AD\plus{}AE \leq 4\sqrt{2}R$

http://www.mathlinks.ro/Forum/viewtopic.php?t=170558
$ M,N,P$ are 3 points lying respectively on the edges $ AB,BC,CA$ of a triangle ABC such that
$ AM\plus{}BN\plus{}CP\equal{}MB\plus{}NC\plus{}PA$. Prove that $ S(MNP) \leq \frac{1}{4}S(ABC)$


[quote="EFuzzy"]In the fifth edition solutions manual, it uses the formula (a+b-c)/2 for the inradius of a right triangle, on an 8-13-17 triangle. But 8-13-17 triangles aren't right. Is this a typo?[/quote]

Notice that r and s [b]don't[/b] stand for the radius and semiperimeter. I think it's right.

Yes, the book is right.  EFuzzy, see Example 17-3.

Oh OK, I somehow overlooked example 17-3.  :blush: Thanks!


[hide="hint"]true or false:  the train is out of the tunnel once its front is out of the tunnel.
now think about it  :D [/hide]

[quote="Ihatepie"][quote="helpmeplz"]This seems easy and I'm sure the answer is very simple but I dont get it. Please give me the answer guys.

A 5 KM train has to cross a 4 KM tunnel by traveling at 1 KM/HR.

How much time will be passed from the train's arrival until its completely crossed over?

Its reallly bugging me guys!![/quote]

[hide]9 hrs[/hide]

btw that is a really slow train[/quote]

How did you find 9?

[hide="different hint"]If the front of the train starts at the beginning of the tunnel, how many km from the beginning of the tunnel is the back of the train then? How many miles does the back need to travel to go through the entire tunnel?[/hide]


Let $ ABC$ be a triangle and let $ M\in (AB)$ , $ N\in (AC)$ so that  $ O\in CM\cap BN$ and $ OB \equal{} OC$ , $ AM \equal{} AN$ . Prove that $ \triangle\ ABC$ is isosceles.

No, the point $ O$ is the intersection of the lines $ BN$ and $ CM$ .

[color=brown]Actually this problem was given at [b]Russian MO 1993 Grade 9.[/b][/color]

[color=darkblue]Suppose $ \angle{B} > \angle{C}$. Let $ K\in AC$, such that $ BK \equal{} CK$. Let $ T \in MN \cap BK, L \in BK \cap CM$. Then $ \angle{KLN} \equal{} \angle{KNL}$.
But $ \angle{KNT} < \angle{KNL} \equal{} \angle{KLN} < \angle{KTN} \equal{} \angle{BTM} < \angle{AMN}$. But $ \angle{KNT} \equal{} \angle{AMN}$, contradiction, so $ \angle{B} \equal{} \angle{C}$ and $ \triangle{ABC}$ is isosceles. [/color]

See and [url=http://www.mateforum.ro/viewtopic.php?t=1947&highlight=][color=red][b]here[/b][/color][/url] (in Romanian language).


Let $ a_n \equal{} \sum_{k\equal{}1}^n k^k$ and $ b_n \equal{} n^n$. As lim (b_n) = infinity then we can write:

$ L \equal{} \lim_{n \to \infty} \frac{a_n}{b_n} \equal{} \lim_{n \to \infty} \frac{a_{n\plus{}1} \minus{} a_n}{b_{n\plus{}1} \minus{} b_n} \equal{} \lim_{n \to \infty} \frac{(n\plus{}1)^{n\plus{}1}}{(n\plus{}1)^{n\plus{}1} \minus{} n^n}$

$ \equal{} \lim_{n \to \infty} \frac{1}{1 \minus{} \frac{n^n}{(n\plus{}1)^{n\plus{}1}}}$.

Can you proceed from here?


Let $R$ and $r$ be the circumradious and the inradious of a triangle $ABC$ respectively.

If $a,b,c$ are the sides of the triangle prove that

$\frac{R}{r}\geq\frac{b}{c}+\frac{c}{b}$ .

 :)

$\displaystyle \frac{R}{r}\geq\frac{b}{c}+\frac{c}{b}\geq2$

This is equivalent to $\frac{abc}{4(p-a)(p-b)(p-c)}\geq\frac{b}{c}+\frac{c}{b}$

Put $p-a=x$,  $p-b=y$,  $p-c=z$  (or....$a=y+z$,  $b=x+z$,  $c=x+y$)

then we have to prove that $\frac{(x+y)(y+z)(z+x)}{4xyz}\geq\frac{x+z}{x+y}+\frac{x+y}{x+z}$

and use the fact that $(x+y)^2\geq 4xy$..........

is this ok???   :)  :P

Yeah it´s ok  :D ... it´s really nice your solution, i tried it all the day but i really suck in inequalities and besides i start getting it bigger with trig...

can u help me to finish mine...

i have to proof that  a*s >= sen^2 B + sen^2 C

quite cool yours :idea:  :lol:

This is V.Bandila's inequality. Now, it's too late, but tommorow I'll post another solution. ;)

[quote="socrates"]Let $R$ and $r$ be the circumradious and the inradious of a triangle $ABC$ respectively.

If $a,b,c$ are the sides of the triangle prove that

$\frac{R}{r}\geq\frac{b}{c}+\frac{c}{b}$ .[/quote]

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=15808 and http://www.mathlinks.ro/Forum/viewtopic.php?t=15807 .

  darij


[quote="zsydtc"]Well,can someone offer me some ebooks of polynomials,specially the book published by Springer,thanks very much.[/quote]

Most of the top books by Springer, you may have to buy. Its illegal to download or ask for such e-books.

However, you should like this: http://www.springerlink.com/mathematics-and-statistics

Oh,I see.,but how about some olympiad training materials on it?

Check out Reid Barton's handouts they're not bad. IMO Compendium has a very nice writeup on polynomials as well.

ashwath.rabind ranath,
can you give some links of the handouts?thanks


At the golf club, a player scored a hole in one on the 18-hole course. i asked six members to tell me which number hole it was. these were the replies:

A said: "It was an odd number"
B said: "It had two digits"
C said: "It was a prime number"
D said: "It was divisible by 5"
E said: "It was a single-digit number"
F said: "It was between 5 and 15"

One of them lied. which hole was it ?

im going with hole #5, to which member B lied.

I am led to hole #5 because if it were to be a prime number, it is only divisible by one adn itself, and if it is divisible by 5 then the answer must be 5.

members B and E contradict each other, one saying its a single digit number and the other a double digit number. If it were a double digit number and divisible by 5 then it would not be a prime.,

[quote="Glen Mann"]im going with hole #5, to which member B lied.[/quote]

yeah because you can't have a two digit number thats divisible by 5 and be prime  :lol:


If x,y,z are pozitive numbers.
Solve:
xy^3 + yz^3  + zx^3 >= xyz(x+y+z)


let $ X$ be an orientable space, $ Y$ is a non-orientable space, my question : why $ Y$ cannot be a covering space of $ X$ ?

thanks so much !

The orientation form on $ X$ pulls back to an orientation form on $ Y$. Oops.

jmerry,

thanks for your reply, I also thought about that, but I am not convinced myself, could you  give  more details?

First, pullbacks make sense in general. If the map is a local diffeomorphism, they send (locally) nonzero forms to (locally) nonzero forms. What is an orientation form on an $ n$-manifold? It's simply an $ n$-form which is never zero; no other requirements.


The topic explains itself.  I just want to get a feel for the music that AoPS-ers listen to.

I'm currently addicted to 'Till I Collapse by Eminem.

Engraved Within by Serenity... which is an amazing song!

Same as poincare
ode to joy
[hide=" the song ..."]Freude schoner gotter funken tochter aus elysium, vir be treten foyer trunken him le scheiden hii ligtum, dinite zauber binden vieder was de mode strengetteilt, alle menschen ver den bruder vordin sanfter flugelveilt, dinite zauber binden veider was de mode strengetteilt, alle menschen ver den bruder vor dein sanfter flu gelveilt.[/hide]

[quote="darkdieuguerre"][url=http://www.youtube.com/watch?v=3HM9tyNkUAg]Brooklyn Rage[/url] by LittleKuriboh.[/quote]
YU GI OH ABRIDGED  :!:

21 guns by Green Day

Clair de Lune from Suite Bergamasque by Debussy  :D

used to be addicted to Fireflies by Owl City

Bring me to life - Evanescene
Right now - Fort minor :D

Tatakau Monotachi (Those Who Fight) by Nobuo Uematsu.

I wish I could play this on the piano.  :cursing:

"Barney is a dinosaur who comes from imagination..."

elmos world by elmo. . . . which is strange. . . .

whatcha say- jason derulo
replay- iyaz
tik tok- kesha
down- jay sean
do you remember- jay sean
dead and gone- timabaland

Fireflies by Owl City

[size=84][color=red]Mod-notes: Rickrolling is completely inappropriate. Don't even think about it. [/color][/size]

Revised

Savior-Rise Against
Behind Closed Doors-Rise Against
Prayer of a Refugee-Rise Against

Note: haha rickrolling

[quote="monkeygirl13"]Fireflies by Owl City
[/quote]

Same.

K.
I just got a new song stuck in my head!
Maithreem bhajatha- Listen to the one sung by M.S Subbalakshmi

On my new and improved playlist;

Fireflies - Owl City
Rocketships & Radios - Nickasaur!
Empire State of Mind - Jay Z
Here in your arms - Hellogoodbye
Baby, it's fact - Hellogoodbye
Tik Tok - Ke$ha
Replay - Iyaz


A basic cellular phone plan costs $ 20 per month for 60 calling minutes. Additional time costs$0.40 per minute. The
formula C = 20 + 0.40(x − 60) gives the monthly cost for this plan, C, for x calling minutes, where x > 60. How many
calling minutes are possible for a monthly cost of at least $ 28 and at most$40?

A. between 80 & 110 minutes
B. between 75 & 105 minutes
C. between 60 & 90 minutes
D. between 30 & 60 minutes
E. none of these

Let $ C\equal{}28$. Solving this for $ x$, we have that $ x\equal{}80$. Now, Let $ C\equal{}40$. Solving this for $ x$, we have $ x\equal{}110$. So the answer is A.


integration
function
real analysis
real analysis unsolved

Let $f\in L^1(\mathbb{R}^n)$ and $f\neq 0$. We define $Hf(x): =\sup_{r>0} \frac{1}{m(B(r,x)})\int_{B(r,x)} |f(y)|\,dy$.  Prove that there exist $C,R$ such that for every $|x|>R$ we have
\[ Hf(x)\geq C|x|^{-n}  \]

My old nemesis, the Hardy-Littlewood maximal function!

Here's your hint -- you know $f \in L^1$ and $f \ne 0$. What does that tell you about the behavior of $f$ for large $R$? Then, think about what you might be able to do by selecting an appropriately sized ball.