1.
Let $r_{1},r_{2}....r_{n}$ be real numbers each greater than 0. Prove that for any real number $x>0$,
 $(x+r_{1})(x+r_{2})......(x+r_{n})\geq(x+({r_{1}\cdot r_{2}\cdot.....r_{n}})^{\frac{1}{n}})^{n}$

2.
If $a,b,c$ are positive and $a+b+c=6$, show that
$(a+\frac{1}{b})^{2}+(b+\frac{1}{c})^{2}+(c+\frac{1}{a})^{2}\geq \frac{75}{4}$

JBL-yeah I fgot! thanks

The first one fails for $n = 1$ if $r_{1}< 1$.  And it fails for $n \geq 3$ if I make the $r_{i}$ very large, for instance.  You probably want it to be the $n$th root.

[quote="riddler"][quote="paladin8"][hide]The first one is just a special case of the generalized Holder inequality.

The second we can say

$\sum_{cyc}\left(a+\frac{1}{b}\right)^{2}\ge \frac{1}{3}\left(\sum \left(a+\frac{1}{a}\right)\right)^{2}$

by AM-GM and

$\sum \left(a+\frac{1}{a}\right) = 6+\sum \frac{1}{a}\ge \frac{15}{2}$

by AM-HM/Cauchy. Then just plug in and we have

$\sum_{cyc}\left(a+\frac{1}{b}\right)^{2}\ge \frac{1}{3}\left(\frac{15}{2}\right)^{2}= \frac{75}{4}$.[/hide]

I hope my inequalities still go the right way...[/quote]
Okay, thanks for posting but I dont understand how you got the 1st line of your solution. :maybe:[/quote]

From $x^{2}+y^{2}+z^{2}\ge \frac{(x+y+z)^{2}}{3}$, where $x = a+\frac{1}{b}$, $y = b+\frac{1}{c}$, $z = c+\frac{1}{a}$?

[quote="riddler"]
2.
If $a,b,c$ are positive and $a+b+c=6$, show that
$(a+\frac{1}{b})^{2}+(b+\frac{1}{c})^{2}+(c+\frac{1}{a})^{2}\geq \frac{75}{4}$

JBL-yeah I fgot! thanks[/quote]

i am just learning inequalities so bear with me...

[hide="solution"]By am-hm \[\frac{ \sum a}{3}\ge \frac{3}{ \sum \frac{1}{a}}\Longleftrightarrow \sum\frac{1}{a}\ge 1.5 \Longleftrightarrow\]  \[\sum a+\sum\frac{1}{a}\ge 1.5+6=7.5\] By cauchy, \[\left[ (a+\frac{1}{b})^{2}+(b+\frac{1}{c})^{2}+(c+\frac{1}{a})^{2}\right]\cdot (1^{2}+1^{2}+1^{2}) \ge \left( \sum a+\sum\frac{1}{a}\right)^{2}\ge (7.5)^{2}\]  \[\Longleftrightarrow (a+\frac{1}{b})^{2}+(b+\frac{1}{c})^{2}+(c+\frac{1}{a})^{2}\ge \frac{75}{4}\] [/hide]

Thanks for your solutions. Can anyone post the proof of the 1st one?

[hide="Inelegant proof of #1"]Expand the LHS. The general term is of the form

$\sum_{1\leqslant i_{1}<i_{2}<\dots<i_{k}\leqslant n}x^{n-k}r_{i_{1}}r_{i_{2}}\dots r_{i_{k}}$

There's ${n\choose k}$ terms in the sum and every $r_{j}$ appears exactly ${{n-1}\choose{k-1}}$ times. Hence by AM-GM we have

$x^{n-k}\sum_{1\leqslant i_{1}<i_{2}<\dots<i_{k}\leqslant n}r_{i_{1}}r_{i_{2}}\dots r_{i_{k}}\geqslant x^{n-k}{n\choose k}\left((r_{1}r_{2}\dots r_{n})^{{n-1}\choose{k-1}}\right)^{\frac{1}{{n\choose k}}}={n\choose k}x^{n-k}\sqrt[n]{r_{1}r_{2}\dots r_{n}}^{k}$

because $\frac{\binom{n-1}{k-1}}{\binom{n}{k}}=\frac{(n-1)!}{(k-1)!(n-k)!}\cdot\frac{k!(n-k)!}{n!}={k\over n}$

Summing up all such inequalities for $0\leqslant k\leqslant n$, on the RHS we get the expanded form of $(x+\sqrt[n]{r_{1}\dots r_{n}})^{n}$, and the proof is thus completed.[/hide]

The first one also follows by the same method as [hide]the conventional proof of AM-GM.[/hide]


probability function algebra polynomial domain

"We say that the sets A and B are in a 1-1 correspondence if there exists a function f, mapping leements of A to elem,ents of B, that satisfies both of teh following properties:
(a) f is one-to-one or 1-1: if f(x) = f(y), then x = y (this also goes by the fancier name injective)
(b) f is onto: for every b (symbol for in the set) B, there exists a (same symbol) A such that f(a) = b (this also goes by teh fancier name surjective)
The function f is also sometimes called a bijection."

Don't we also have to say that for an element in A cannot correspond to more than one element in B?

I mean, I understand that that fact might be included in the definition of a function, but when I looked at the solutions of problems (in aops) showing that functions were in a 1-1 correspondence, they did NOT prove that the relation was a function...

I thank you in advance...

BTW - all polynomials are functions, right? Can you prove it?????

Well, let's say two symbols $ x$ and $ y$ in set $ A$ map to the same element in set $ B$. We therefore have that $ f(x) \equal{} f(y)$, so $ x \equal{} y$ because the mapping is injective, so we don't need your claim?

How often do you write down a function for which it's not obvious that what you've written actually is a function?

(Answering for myself: occasionally.  Usually, this happens when the domain is a set of equivalence classes, and the function is defined in terms of some member of the class -- then one has to take the time to show that using a different member still results in the same function value.  This is probably not the situation you're facing.)

Unrelatedly: it's rather rude not to take the time to present your question carefully.  In this case, this means
1) Not making many annoying typos.
2) Learning a little bit of LaTeX.  In particular, typing \in is just not very difficult, and neither is putting dollar signs around your math symbols or equations.
3) Choosing an informative title: for example, you could have titled this post "when do we need to show a bijection is a function" or something, which would be much more useful.


calculus
integration
calculus computations

Hint: Expand the expression by $1-\sqrt{x+2}$ and split it into simpler integrals.


calculus
function
absolute value
real analysis
calculus computations

Problem:

Show by $\epsilon-\delta$ definition such that $lim_{x\to1}f(x) = 3$ if $f(x) = 2x+1$ if $x \neq 1$ and $2$ if $x = 1$

Preliminary Analysis:
for $0 <|x-1|< \delta$

$|f(x)-L| = |2x+1-3|\\ = |2x-2|\\ = 2|x-1|\\ = 2 \delta$

Question: Why did $x-1$ became equal to $\delta$?

Proof:
Let $\epsilon > 0$
Choose $\delta = \frac{\epsilon}{2}$
then for
for $0 <|x-1|< \delta$
$|f(x)-L| = |2x+1-3|$
$= |2x-2|$
$= 2|x-1|$
$< 2 \delta = 2(\frac{\epsilon}{2})= \epsilon$

Actually, x-1 is not equal to epsilon! This is just a preliminary analysis (as you state) so that you can get x - 1 after working with f(x) - L.
The reason you need x-1 is due to the condition you have :
the absolute value of x - 1 is less than epsilon.

[quote="maokid7"]could you also do

$|x_{1}-x_{2}|<\delta$ and
$|f(x_{1})-f(x_{2})|=|2x_{1}+1-2x_{2}-1|=2|x_{1}-x_{2}|<\epsilon$
so
just choose $x$ values that are withing $\frac{\epsilon}{2}$[/quote]

Do you know any good site with this kind of stuff?

I got that from ACoPS and I know it is in an apendix of my calculus book.

basically all it says is that for a [b]uniformly continuous function[/b] there is a $\delta$ that satisfies

$|x_{1}-x_{2}|<\delta$ and $|f(x_{1})-f(x_{2})|<\epsilon$

and then you just have to play around w/ $|f(x_{1})-f(x_{2})|$ to obtain a $|x_{1}-x_{2}|$.

and the value of $\delta$ depends only on $\epsilon$ and not on the values of $x_{1}$ or $x_{2}$.


Person A fills up a sugar cone with sprinkles and is about to devour it. However, he suddenly has to use the bathroom. Person B comes along and sees the cone sitting upright in a glass and is very tempted. He eats some, then some more, and by the time Person A exits the bathroom the top 3/4 of the cone has been devoured, and there are only 75 sprinkles left in it.

Person A loves his sprinkles, and of course will be very unhappy upon seeing the tiny bit left. He will yell at Person B in a loudness of one decibal for every 25 sprinkles Person B has eaten. How many decibals is the yelling that Person B receives?

wow, wat a random problem

[hide]$25\times25=\boxed{625}\text{earsplitting decibels}$[/hide]

thats the fun of making up problems. You can come up with anything, for example, the person C could come in and go shoot everyone and eat a third of the remaining sprinkles and give the rest of them to person D. How many sprinkles does person D get?

see!

intresting problem...  did you make it up....

[hide="person D eats..."]50 sprinkles[/hide]

I'm sorry, I just had to...  :D


calculus
integration
logarithms
calculus computations

How do I integrate exp(-2*sqrt(t))

It's been a while I don't remember how to do these. Thanks

Step 1: make the substitution $ x\equal{}2\sqrt{t},$ which you can write as $ t\equal{}\frac{x^2}4.$ Differentiate that to get $ dt\equal{}\frac{x}{2}\,dx.$

Now we have $ \int e^{\minus{}2\sqrt{t}}\,dt\equal{}\frac12\int xe^{\minus{}x}\,dx.$

Step 2 will be an integration by parts. Can you see how to proceed from here?

[color=darkblue]Denote $ I\equiv\int e^{ - 2\sqrt t}\ \mathrm {dt}\ .$ Apply the [b]substitution[/b] $ \{\begin{array}{c} x = \phi (t) = e^{ - 2\sqrt t} \\
 \\
(\ \ln x = - 2\sqrt t\ ) \\
 \\
\phi '(t) = \frac { - 1}{\sqrt t}\cdot e^{ - 2\sqrt t}\end{array}$

Thus, $ I = \int e^{ - 2\sqrt t}\cdot\frac { - \sqrt t}{e^{ - 2\sqrt t}}\cdot \underline {|\ \phi '(t)\ \mathrm {dt}}$ $ \Longrightarrow$ $ \boxed {I = \frac 12\cdot F(e^{ - 2\sqrt t}) + \mathcal C}\ ,$

where $ F(x) = \int \ln x\ \mathrm {dx}\ .$ Apply [b]the integration by parts[/b] :

$ \{\begin{array}{ccc} u(x) = \ln x & \Longrightarrow & u'(x) = \frac 1x \\
 \\
v'(x) = 1 & \Longrightarrow & v(x) = x\end{array}\|$ $ \implies$ $ \boxed {F(x) = x\ln x - x + \mathcal C}\ .$

In conclusion, $ \boxed {\boxed {I = - (\sqrt t + \frac 12)\cdot e^{ - 2\sqrt t} + \mathcal C}}\ .$[/color]

[quote="Kent Merryfield"]Step 1: make the substitution $ x \equal{} 2\sqrt {t},$ which you can write as $ t \equal{} \frac {x^2}4.$ Differentiate that to get $ dt \equal{} \frac {x}{2}\,dx.$

Now we have $ \int e^{ \minus{} 2\sqrt {t}}\,dt \equal{} \frac12\int xe^{ \minus{} x}\,dx.$

Step 2 will be an integration by parts. Can you see how to proceed from here?[/quote]

yes thank you

$ \int e^{\minus{}2 \sqrt{t}}\,dt \equal{} \minus{} \int \sqrt{t} \cdot \left(\minus{} \frac{1}{\sqrt{t}}\right) e^{\minus{}2 \sqrt{t}}\,dt \equal{} \minus{}\left(e^{\minus{}2 \sqrt{t}} \cdot \sqrt{t} \minus{} \int e^{\minus{}2 \sqrt{t}} \cdot \frac{1}{2 \sqrt{t}}\,dt\right)$

$ \equal{} \minus{} \sqrt{t} \cdot e^{\minus{}2 \sqrt{t}} \minus{} \frac{1}{2} \int \left(\minus{} \frac{1}{\sqrt{t}}\right) e^{\minus{}2 \sqrt{t}}\,dt \equal{} \minus{}e^{\minus{}2 \sqrt{t}} \left(\sqrt{t} \plus{} \frac{1}{2} \right) \plus{} C$.

I've got one more

Integrate:
exp((.02*t)/(1+.05t))


geometry
quadratics
inequalities
algebra
quadratic formula

This problem is really cool!

ABCD is a convex quadrilateral with AC and BD intersecting at M. If the areas of AMB and CMD are 4 and 9 respectively, what is the minimum area of ABCD?

[hide]A(MBC)/A(AMB)=A(DMC)/A(MAD) because of sharing altitude. (A(AMD) means area of triangle AMD)

sub 4 and 9 in

we have A(MBC)/4=9/A(MDA). Let A(MBC) be x,A(MDA) be y.

xy=36. y=36/x

Let x+y=k

x+36/x=k

x :^2: +36=kx

x :^2: -kx+36=0

Quadratic Formula

(k+ :pm:  :sqrt: (k :^2: -144))/2

 so k  :ge: 12. therefore 12 is the minimum value of x+y. 12+4+9=25= minimum value of the quad.[/hide]

good job!

Note: [hide]an easier way to minimize x+36/x is with AM-GM inequality: x+36/x >= 2 sqrt(x*36/x) = 12.[/hide]


superior algebra
superior algebra unsolved

(Galois closure of $ \mathbb{Q}(\sqrt[4]{2},\zeta_6)$ over $ \mathbb{Q}$)$ \equal{} \mathbb{Q}(\sqrt[12]{2},\zeta_{12})$?

Is it right?

Why the $ \sqrt[12]2$¿ It should be a $ \sqrt[4]2$.


Gauss
number theory proposed
number theory

Let $a,b$ be integers, and $p,q\ (q\neq0)$ be rational numbers.
Prove that if $p+qi$ is the root of $x^2+ax+b=0$,then show that both $p$ and $q$ are integers.

Since second root is $p-qi$, from Viet theorem (am I correctly wrote it?) we obtain $2p=-a$ and $p^2+q^2=b$. Suppose $a$ is odd, then $4b=a^2+(2q)^2$, so $2q$ is a rational and $(2q)^2$ is an integer. It follows that $2q$ is integer. But in this case $a^2+(2q)^2=1$ or $2$ modulo 4 -- contradiction. Therefore, $a$ is even, $p$ is an integer, and $q$ must e integer too.

Naturally you are right as usual.

kunny


Aici sunt subiectele de la etapa judeteana  http://scoala18.xhost.ro/. Au fost aspru criticate. Ce parere aveti ??? Va multumesc.


This is just a technicality to which i am unsure.

 If $ A \vee B \Rightarrow C$

Then is it correct to say

$ A \Rightarrow C$ or $ B \Rightarrow C$

If you're unsure if a logic formula is a tautology, you can always write its truth table:

$ \begin{matrix}A & B & C & A\lor B & (A\lor B)\implies C & (A\implies C) & (B\implies C) & (A\implies C)\lor (B\implies C) \\
T & T & T & T & T & T & T & T \\
T & T & F & T & F & F & F & F \\
T & F & T & T & T & T & T & T \\
T & F & F & T & \boxed{F} & F & T & \boxed{T} \\
F & T & T & T & T & T & T & T \\
F & T & F & T & \boxed{F} & T & F & \boxed{T} \\
F & F & T & F & T & T & T & T \\
F & F & F & F & T & T & T & T\end{matrix}$

As you can see, the boxed cases disprove the equivalence between the two formulas.

You can use elementary transformations of basic logic operations to get to the correct tautology. Having in mind that $ (p\implies q)\iff(\lnot p\lor q)$, and that $ \land$ and $ \lor$ are distributive to one another, we get

$ [(A\lor B)\implies C]\iff[\lnot(A\lor B)\lor C]\iff[(\lnot A\land\lnot B)\lor C]\iff [(\lnot A\lor C)\land(\lnot B\lor C)]\iff [(A\implies C)\boxed{\boxed{\land}}(B\implies C)]$

From Farenhajt's truth table, if $ A \lor B \implies C$, then $ A \implies C$ or $ B \implies C$. However, the converse is not true. In symbolic notation:

$ (A \lor B \implies C) \implies ((A \implies C) \lor (B \implies C))$

But:

$ \lnot ((A \lor B \implies C) \iff ((A \implies C) \lor (B \implies C)))$


calculus
derivative
articles
geometry
number theory

Hey folks,

I have a series of second-round quant interviews for a summer internship position coming up later this week and am wondering whether any of you who have gone through such a process have any advice on how to prepare and do well in the interviews. Any input would be greatly appreciated. Thanks!

i just had my second interview for jane street capital this thursday. i think i did pretty well on the first interview, and then i bombed the second by completely missing the first brainteaser question. :oops_sign:

at the end he said I would hear back next week.

how do you think you did?

Since it’s a not a full time role it will not be quite as tough, but I’d still recommend preparing the same way.  You might want to pick up “Heard On the Street”, “C++ Design Patterns and Derivatives Pricing”, and “Options, Futures, and Other Derivatives” (Hull).  It’s probably a bit late for the second two, but definitely read through the first.

I just got a job as a quantitative analyst, and it surprised me a bit, because I did pure maths, so am not as well qualified as a lot of people.

I think it helped that I read the annual report of the company, and google-searched them and read every news article about them that I could find, and then carefully prepared answers to the common questions like "why do want to work here?" and "what are your skills?" and so forth.  So I would recommend doing that.  I think it helps to show enthusiasm about the specific company you're applying to.

[quote="Dominic Connor"]As a headhunter, I am struck by the number of people who get the "why do you want do do this" question wrong.[/quote]

What is the correct/best answer?

The answer should be honest...

It should show that you've thought about this kind of work, and this particular job.

Money, is not a great motivation, though it does not hurt to say something like
"obviously the money is a factor, but I'm enthusiastic about..."

Send the message that your enthusiasm is for this job.
It is an opportunity for some naked bragging of the form "this lets me apply my abilities in XX"
That's more important than you might think.

Zeta4g2 was surprised to get past, and I can see the issues he faced. He had deep skills, but not the ones the interviewers might have asked about. Thus if you feel part of your light has been shaded, it's a chance to bring it in, as useful.
Thus everything your hear about the job has value, even if it's the standard BS about "being a good team player".

With all due respect to Zeta, he didn't know what they wanted, few entry level people do with precision. I recall one very well known quant, famous in his area who understands classical quant maths at a deep level. Sadly in his first job he found that he needed number theory, which is of course pure maths.

Some wrong answers we hear are quite tragically bad.
Shocked silence, expressing an interest in a totally different sort of work, and mumbling random things.

mb1551- i am doing an internship with a Quant Team. I got it by just sending them a resume, and with no financial skills. Its pretty good- at N A B, but what skills do I need to know to get an internship at a bigger bank? how did u guys find out about these? i just did b/c my family member worked at the particular bank.


geometry
3D geometry
Pythagorean Theorem

In the picture attached, there is a cube, having three vertices called A, B, and C. Find the measure of angle ABC that is formed by the dashed segments. Anyone got a solution?

[hide="Hint"]
A and C are just begging to be connected.[/hide]

Hmm, looking at the post below, it is a special triangle...but what kind of triangle is it?

Is it not an equilateral triangle?

[hide="Solution"]Once we connect A and C, what do we discover? A special triangle![/hide]

[quote="fanaticsm"]So am I right or no?[/quote]

[hide="Answer"]You're right. Everyone of the sides is a diagonal of a face of the cube, so unless it's not a perfect cube,you can be pretty confident in your answer.[/hide]

[hide]
connect A and C, which makes an equilateral triangle.  If you don't get this, assume that this is a unit cube.  Using pythagorean theorem or the formula for a 45-45-90 triangle, you wil find that each side is equal to sqrt(2).  So it is an equilateral triangle which has each angle equal to 60 degrees.[/hide]


HCSSiM
Columbia
AMC
AMC 8
AMC 10
AMC 10 B

I had the highest score on the AMC 10B in my state, but someone had a higher score on the 10 A. will i get a state winners plaque, or is it just who had the highest score on either test. The scores on the A were much higher than on the b so i think it would be a little unfair if I didnt.

No, you won't.  I know from being in the exact position that you are in.

[quote="Tenoreoz"]hm I had the highest score on the amc 10 B in my state last year and the director at the site I took at it told me she got a plaque for me. However, I haven't received it yet so I'm not sure  :huh:[/quote]In one year? lol

hrm... i think it's just your score that matters, and that you don't really need awards.
i mean.. what use are they??
maybe i just feel bad cuz im in CA :( , but still, awards don't really matter.

I had the highest 10B score last year, but someone else had a higher score on the 10A. I still got a plaque for the highest 10B scorer.

[quote="E^(pi*i)=-1"]Really?  In that case I earned 2 plaques that I never got . . .  :mad:[/quote]

now you know how I feel, except worse lol.

According to the AMC website, intramural awards are given for each test, but regional and national awards are for the score in a year.  Hence, if one is beaten by someone on the other test, one should not get a plaque.

I hope that seperate plaques are awarded, b/c otherwise i won't get one :(\

Is it possible that this varies by state?

Do ties get plaques? (ie. we have 3 141's on the 12A and 1 141 on the 12B in MA, do we all get plaques)

[quote="ProtestanT"][quote="xscapezaer"]You know, sometimes it really sucks to live in CA... :([/quote]
you can look at it in a more positive manner: If someone does achieve very well in CA, he'd probably feel better than if he did in some less populated state.[/quote]
hehe yeah same way I felt when I got first in Oregon with a 24 AMC 8.
EDIT: err I meant I didn't feel as accomplished as I would've for getting first in a more populated state..

and The Zuton Force - I got a plaque for the AMC 8 mentioned above, even though it was a three-way tie.

Cant we just ask the amc director? there seems 2 b alot of confusion

a) why the triple post
b) i have an AMC 8 top scorer plaque  :10:

[quote="Quickster94"]i'm happy, i did end up getting a plaque for top AMC-10 A scorer  :lol:[/quote]

i only got a pin :(

[quote="Quickster94"]i'm happy, i did end up getting a plaque for top AMC-10 A scorer  :lol:[/quote]

did you get a 120? i got a 121.5 on the 10A on i was like 10th in my state, we had 2 perfects.  :(

I got a perfect on the 10 but i haven't gotten anything yet...

[quote="Quickster94"]hmmm, yes in fact i did get a 120, how did u know?...[/quote]

you are from vermont and you said you got a 10A plaque so i looked at the stats.

EDIT: 800th POST!!!!!!!!!

It's wierd. People said that you get a plaque if get first in the state. But at our school the state champion got a trophy.

i won AMC 10A and 10B at my school and i only got 1 pin.  :(

Maybe they give different awards in different states or counties. :maybe:

i got my plaque today!!!
AMC 10B winner :lol: finally

Congratulations!! They r very nice  :lol:

Strange...  I'm the highest scorer in GA on the 10A, but I never even got so much as a "school winner's pin" for it...

oh, ur school probably has them somewhere...  I didn't get my plaque until june, idk when they got it though


algebra
polynomial
number theory unsolved
number theory

for positive integers n and k, show that\[(n^{4}-1)(n^{3}-n^{2}+n-1)^{k}+(n+1)n^{4k-1}\]
is divisible by $n^{5}+1$

if it's "legal" then it follows very easy when using roots of unity...

now i ask: is it legal to use them? i mean, since we're working with polynomials with integer coefficients, does their use affect in something?

A polynomial divides another iff the dividend contains all of the roots of the divisor; this is over their splitting field, regardless of the coefficients.


What is the greatest possible number of points of intersection for eight distinct lines in a plane?

[hide]1 line = 0 points
2 lines = 1 point
3 lines = 3 points
4 lines = 6 points
5 lines = 10 points
6 lines = 15 points
7 lines = 21 points
8 lines = 28 points

See a pattern? Because to maximize intersection points, the nth line must cross through all (n-1) other lines on the plane. So each 'term' increases by n-1. Therefore, A(n) [# intersection points where n is the # of distinct lines in the plane] = 1 + 2 + 3... (n-1) = (n)(n-1)/2. 8(7)/2 = [b]28.[/b]
[/hide]
lol I must be really bored.

I did this one in my math club. Fundamentally, you start off with pictures, slowly changing to patterns at a destined time. Just like, Superwizard188, you have to use patterns, and also pictures, which he probably did as well (just not included, obviously).

[hide]Whenever we choose a pair of lines, they should meet. So $\binom{8}{2}=28$.[/hide]


ratio
limit
inequalities
algebra
polynomial
arithmetic sequence
geometric sequence

Is it possible to partition N, the set of all positive integers, into two disjoint sets A and B such that:

i) There exists no infinite (non-constant) arithmetic sequence in A, and
ii) There exists no infinite (non-constant) geometric sequence in B?

 :D

Yes it is possible. And, if I'm not wrong the following partition excludes any infinite (non constant) arithmetical or geometrical progression in each set.

Consider the sequence $(U_n)$ defined by $U_1 = 1$ and $U_{n+1} = n(U_1 + \cdots + U_n).$

Now, color the positive integers alternatively red and blue by groups of size $U_1$ (red), $U_2$ (blue), $U_3$ (red), and so on...
Let $A$ be the set of red integers and $B$ be the set of blue integers.

Let $a,d>0$ be the first term and the ratio, respectively, of an arithmetical sequence. Since $\lim U_n = +\infty$, it is clear that there are arbitrary large strings of consecutive red (resp. blue) integers after $a$. Thus at least one term of the sequence must be red and at least one must be blue.
Therefore there is no monochrome arithmetical sequence.

Let $a,q$ be the first term and the ratio, respectively, of a geometrical sequence.  Now, let $n>q$ be even such that $aq^k $ belongs to the string of $U_n$ for some integer $k$. Wlog, we may assume that $k$ is maximal, that is $aq^{k+1}$ does not belong to this string.
But, since $aq^k$ belongs to the string, we clearly have $aq^k \leq U_1+U_2 + \cdots + U_n$. Thus $aq^{k+1} \leq q(U_1 + \cdots + U_n) <  U_{n+1}$ so that $aq^{k+1}$ must be red.
We have a similar reasoning for to prove that any geometrical sequence cannot be all red.
Therefore, there is no monochrome geometrical sequence.

Pierre.

Pierre, please correct me if I am wrong, but from your argument, it doesn't seem like A :cup: B = N, the entire set of positive integers because {U_n} = {1, 1, 4, 18, ... }.... :(

No, you don't color the numbers $U_n$. You color the positive integers by groups of consecutive integers, and the size of the $n^{th}$ group is $U_n$.

Pierre.

[quote]No, you don't color the numbers U_n. You color the positive integers by groups of consecutive integers, and the size of the nth group is U_n.

Pierre[/quote]

Oh! Oops, then it's an ingenius solution! Thanks a lot! :)

Actually, Pierre, if you are interested to know, the original problem is still open on CRUX:

Is it possible to partition N, the set of all positive integers, into two disjoint sets A, B such that:

1) No three numbers in A make an (non-constant) arithmetic sequence.
2) There is no infinite (non-constant) arithmetic sequence in B.

I don't know, did I change it to an easier one?  :lol:

I know solution for this one, it can't be open.

So, it is your turn to answer Harazi! :P

Pierre.

Let $ x_1, x_2,...$ be an enumeration of all the infinite arithmetic progressions of natural numbers. Take any number $a_1$ from $x_1$, then take $ a_2>2a_1$ from $ x_2$ then $ a_3>2a_2$ from $x_3$ and so on. Put all numbers $ a_k$ in a class A and the rest in B. Then B clearly does not have infinite arithmetic progressions. And since $ a_k+a_i>2a_{k-1}$, no three elements in A are in arithmetic progression.

Using the same idea, here is another partition that avoids in both classes infinite arithmetic or geometric progressions. Let $ a_i$ be all infinite arithmetic progressions and $g_i$ all infinite geometric progressions. Take $ x_1$ in $a_1$ and $ z_1>x_1$ in $g_1$. Take $ y_1>z_1$ in $ a_1$ and $ t_1>z_1$ in $ g_1$. Then take $ x_2>t_1$ in $ a_2$, $ z_2>x_2$ in $ g_2$, $ y_2>z_2$ in $ a_2$, $t_2>y_2$ in $ g_2$ and continue the process. Then, put all numbers $ x_i, z_i$ in B and all the other numbers in A. Any infinite arithmetic or geometric progression will have elements both from A and B.

Yes, I think this kind of tools can be used for quite any countable set of things which has to be avoided.

Pierre.

Amazing solutions, harazi! Many thanks to you and Pierre!  :lol:

btw, I didn't renew my CRUX subscription, so it was written on the September of 2002 issue that the problem was open...(bet it isn't now!)  :lol:

Actually, now that I think about this again, I am a bit confused :? ...why is there no arithmetic sequence in B? First time I thought it was because one term has been taken out of every infinite sequence, but that doesn't mean it can't be replaced because there are other sequences containing that term without it being removed. So suppose we took out a_1 from x_1, there are surely other sequences containing a_1. But we take out a_i (different from a_1) from each of them, so when putting them altogether in B, a_1 exists as well as every other term in x_1. So we get an infinite sequence, don't we? I can see I'm really confused now... :?  :lol:  :?

Any infinite arithmetical sequence is defined by its first term and its ratio, which both are positive integers. Thus, the set of infinite arithmetical sequence is countable, so that you can enumerate it with the positive integers, and that is what Harazi said.
Then, following this ordering, we choose (at least) one term in each of this sequences and put it in $A$, therefore none of these sequences is completely contained in $B$.
The choices made have to be done carefully to avoid that a three-term arithmetical sequence appears in $A$. That's why Harazi add the inequality condition in his choices. Note that he can do that because the initial arithmetical sequences are supposed to be infinite.

The key is that they form a countable set. That's why I said that this method can be extended to other such sets. For example, instead of arithmetical sequences, which are degree 1 case, you may forbid any sequence of the form $P(0),P(1), \cdots$ where $P$ ranges over the set non-constant polynomials with integer coefficients. You may also add the geometrical sequences (since an union of two countable sets is countable) and so on...

I hope it is clearer.

Pierre.

[quote]Any infinite arithmetical sequence is defined by its first term and its ratio, which both are positive integers. Thus, the set of infinite arithmetical sequence is countable, so that you can enumerate it with the positive integers, and that is what Harazi said.
Then, following this ordering, we choose (at least) one term in each of this sequences and put it in A, therefore none of these sequences is completely contained in B.[/quote]

I understand the enumeration part, and that we take term(s) out of each these arithmetic sequences. But as long as we take out finite number of terms (in fact, I think Harazi indicated that we take one term out of the sequence), the problem still remains. Consider the following:

{1, 3, 5, 7, 9, 11, 13, 15, ... } is an infinite arithmetic sequence. So we take term(s) out of it (say 3, 7) and put them in A, then the rest in B. But we STILL have the infinite arithmetic sequence (9, 11, 13, 15, ... ) in B, don't we? Recall that a sequence, though might not be "complete", can still be infinite.

Sorry for my extreme slowness.  :blush:

No, because the infinite remaining sequence is counted as another one in the enumeration : The first one is associated to $(1,2)$ and the remaining one to $(9,2)$ (first term and ratio). Thus, the remaining sequence will appear somewhere in the enumeration, so that at least one of its term will be put into $A$, and therefore the remaining sequence will not be entirely contained in $B$.

Pierre.

I got it. Thanks Pierre, for your vast patience.  :)


ratio
limit
Euler
geometric series
calculus
calculus computations

Prove that $ \frac{2}{1} \times \frac{3}{2} \times \frac{5}{4} \times \frac{7}{6} \times \cdots \times \frac{p}{p \minus{} 1} \times \cdots$ diverges.

Well we can tell that each term is greater than 1 and with infinite terms slightly greater then 1 multiplied by each other the result will be infinite and not 1. Everthing diverges unless the numbers form a geometric series that have fractional common ratio like 1+1/2+1/4... which we know equals 2. As p tends to infinite the value of the fraction is merely one because infinite/infinite +1 still equals 1 which we learn in limits.

[quote="varunrocks"]Well we can tell that each term is greater than 1 and with infinite terms slightly greater then 1 multiplied by each other the result will be infinite and not 1.[/quote]
Nope.  For example, $ \lim_{n \to \infty} \left( 1 \plus{} \frac {1}{n} \right)^n \equal{} e$.  If you don't count that as a legitimate product, $ \prod_{i \equal{} 1}^{\infty} \left( 1 \plus{} \frac {1}{r^i} \right)$ converges for all $ r > 1$.

One way to do this problem is to write the product as $ \prod_{i \equal{} 1}^{\infty} \left( \frac {1}{1 \minus{} \frac {1}{p_i}} \right)$, which "converges" to the harmonic series.  This is the Euler product.


parameterization
function
advanced fields
advanced fields unsolved

I have the parametetrization $t \mapsto (t^2 - 1, t(t^2-1))$ of the curve $y^2 = x^2(x+1) = x^3+x^2$. How does this help me to find an isomorphismn $K[t] \to K[x,y]/\langle{y^2 -x^2(x+1)}\rangle$?

They are not isomorphic - Indeed, $K[t]$ is integrally closed, the other ring not. Another possibility was to see that the local ring of $K[x,y]/(y^2-x^2(x+1))$ at $(0,0)$ is not regular, whereas all local rings of $K[t]$ are regular. But you DO get a morphism $K[x,y]/(y^2-x^2(x+1)) \to K[t]$ given by $x\to t^2-1, y\to t(t^2-1)$. The fact that this is not an isomorphism corresponds to the fact that the preimage of the point $(0,0)$ contains two points.

[quote="Peter Scholze"]They are not isomorphic - Indeed, $K[t]$ is integrally closed, the other ring not. [/quote]How can I show the latter?

BTW, I have $K = \mathbf{R}$.

The field you work on does not really matter, but let's take $K=\mathbb{R}$. $R=\mathbb{R}[x,y]/(y^2-x^2(x+1))$ is not integrally closed, since $\frac{y}{x}$ is an algebraic integer in the field of fractions of $R$: $(\frac{y}{x})^2 - (x+1) = 0$. But clearly $\frac{y}{x}$ is not contained in $R$.


Oh. My gosh.
 :rotfl:
Google won't search for Chuck Norris because it knows you don't find Chuck Norris, he finds you.

Ah... 'let me google that for you.'  I use that site all the time. :)


[size=150]You have received a $2,000 invoice from the Acme Swamp Water Company for raw materials to use in your beer manufacturing plant. The terms are 3/10, net 40. If yor fail to take advantage of this discount. You are paying what interest rate?

1.22.7%
2.36,735%
3.37.11%
4.55.67%
5.None of the above

Thanks[/size]

[quote="mathnerd314"]Where's the problem from?

I should know this, but what is meant by 3/10 and net 4?[/quote]

Thanks.

Look at this Problem.

If credit terms of "2/10, net 40" are offered, the approximate cost of not taking the discount and paying at the end of the credit period would be closest to which of the following? (Assume a 365-day year.)

1.18.6%

2.4.3%

3.24.8%

4.30.0%

[2/(100-2)]*[365/(40-10)]=.248

[quote="Palytoxin"]well can you tell us more about it? like can you explain what is terms 2/10 and net 40? what do they mean? and what subject is this? finance?  :?[/quote]

 Yes, finance.


No one said it was. We can only do the best we can with what we have.
Your comments about IMO can be applied to anything else in life. There is no way to make things perfectly equal for everyone because we can't measure quantitatively how much one person has that another doesn't.

I don't think we share the same deffenition for the word 'fair'.  :wink:

[quote="JBL"]Maybe I should begin lobbying again for the creation of a Self Pity forum.[/quote]

It would pretty much instantly become the most popular fourm. Math is hard, and peoples egos are fragile.    :P  :roll:

Life is ruled by the rules of nature. They are what they are, we can not judge them, though we are all subjects to it. This applies to absolutely all segments of our lives. Which are just a small part of ubiquitous nature :)

And of course, whatever you do, you can't fool nature  :D

[quote="setout"]It isn't.

Some people are born in rich families, some aren't. Some people are born in family in with their parents has Uni degree, some aren't.

That could be also some reason why whole IMO isn't fair... You have there equal conditions, that is right, but because the life before the IMO fair isn't, whole IMO isn't fair eighter.

Life isn't fair!!

What do u think about it?[/quote]
Was it ever assumed life [i]was[/i] fair? Are you delusional?

[quote="fredbel6"]Wait a minute, setout, isn't he the same user who started that topic about Peter Scholze's clicking of pens and who made at least three separate topics about acquiring romantic partners? :|[/quote]
yeah, he is.

Life was never fair, never is fair, and never will be fair. There is completely no doubt about that. :!:
The thing that Life is not fair is pretty much a fact accepted by nearly everyone in the world.

I hope you understand what I mean when I say "The greatest inequity is complete and total equality".

First of all we are differend; clever or not beautifull or not, etc

[quote="orl"]Maybe a bit unrelated. But there was a user with the signature something like "Life is unfair. Otherwise I could not get what I do not deserve.".  :D[/quote]

http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=11966

[quote="setout"]It isn't.

Some people are born in rich families, some aren't. Some people are born in family in with their parents has Uni degree, some aren't.

That could be also some reason why whole IMO isn't fair... You have there equal conditions, that is right, but because the life before the IMO fair isn't, whole IMO isn't fair eighter.

Life isn't fair!!

What do u think about it?[/quote]

Hello setout. How was IMO? And did you find the girl you were looking for? (I actually ask these because you seem upset)

IMO tests how well people can solve math problems.  well, some people will be better than others, and that's competition.  it's unfair because some people are naturally more effective with math problems than others and were born with more math aptitude.  but society does not celebrate mediocrity.  if life were completely fair, competition is obsolete.  read the book Anthem by Ayn Rand.  very intense book.

that's why you should sell all your things and become a hobo and eventually end up in an asylum. life isn't fair and it never will be...sometimes you have the "unfair" advantage and sometimes not.

edit: please don't think i meant the hobo part...

[quote="wobster109"] read the book Anthem by Ayn Rand.  very intense book.[/quote]

barf

the last thing we need is more randroids

[quote="Elemennop"]
barf

the last thing we need is more randroids[/quote]

at least it's better than watching so much tv that you hallucinate pokemon watching tv. :P

[quote="Bictor717"][quote="Cesario"][quote="wobster109"]
society does not celebrate mediocrity.  [/quote]

I completely disagree with you there. Unfortunately, in most cases, society DOES celebrate mediocrity :!:[/quote]

Perhaps you can offer some proof or example? I haven't heard of anyone being rewarded for being normal or for doing something ordinary.[/quote]

Examples are all around you, Bictor. I did not mean there were actually prizes given for mediocrity, but generally, average people are always better accepted in and by the society than geniuses are. Don't you agree?

[quote="wobster109"][quote="G-UNIT"]All men are created equal.  All men shall rightly share in the pursuit of life, liberty, and happiness.  All goods and services shall be subject to the prices assigned to them by the government and, thus, no man shall stand richer nor any man poorer in the wake of his neighbour.  In the last few sentences I have adressed the problem proposed by the originator of this thread and, at the same time, created a form of government free of imperfection.[/quote]

in short, the ideal communist government.  but is that really fair?  suppose you work much harder than your neighbor.  should you two really be at absolute economic equality?[/quote]

Yes! Really no...communism is actually wrong however appealing so when you become a power hungry dictator don't do it. Communism can't affect brains... doctor and bagger at grocery store..same?

[quote="Bictor717"][quote="Cesario"][quote="wobster109"]
society does not celebrate mediocrity.  [/quote]

I completely disagree with you there. Unfortunately, in most cases, society DOES celebrate mediocrity :!:[/quote]

Perhaps you can offer some proof or example? I haven't heard of anyone being rewarded for being normal or for doing something ordinary.[/quote]

What about awards for perfect attendance? They reward you for consistantly showing up every day.


function
trigonometry
probability and stats

suppose $ \{\xi_{n} \}_{n \geq 1}$  be sequence of independent random variables such that :
$ \mathbb{P} \{\xi_{n} \equal{} \minus{} n \} \equal{} \mathbb{P} \{\xi_{n} \equal{} n \} \equal{} \frac {1}{n}$
and
$ \mathbb{P} \{\xi_{n} \equal{} 0 \} \equal{} 1 \minus{} \frac {2}{n}$
Show that such sequence of independent random variables  don't satisfy the strong law of large numbers .
I know it's simple question for the peoples who   are familiar with Probabilities and Statistics.
This is what I have done.
At first let's check that it satisfay to the law of large numbers.
Since $ \mathbb{E}\xi_{n} \equal{} 0$ hence by P.Levy's theorem  it's enough to show that
$ f_{\eta_{n}}(t) \to f(t) \equal{} 1$
where $ f_{n}(t)$ and $ f(t)$ is correspondent characteristic function of $ \eta_{n} \equal{} \frac {1}{n}\sum_{1\leq j\leq n} \xi_{j}$ and $ 0$.
note that $ f_{\xi_{j}} (t) \equal{} 1 \minus{} \frac {2}{j} \plus{} \frac {2\cos jt}{j}$ thus we have
\[ f_{\eta_{n}}(t) \equal{} \prod_{j \equal{} 1}^{n}f_{\xi_{j}/n}(t) \equal{} \prod_{j \equal{} 1}^{n}\left( 1 \minus{} \frac {2}{j} \plus{} \frac {2\cos jt/n}{j}\right) \approx \exp \left( \sum_{j \equal{} 1}^{n} \minus{} \frac {2}{j} \plus{} \frac {2 \minus{} (tj)^{2}/n^{2}}{j} \right) \approx \exp \left( \minus{} \frac {t^{2}}{2} \right)
\]
thus it don't satisfy the (weak ) law  of large numbers and hence strong.

I'd like to see any other solutions . (simple solutions )

or you can use the (second) Borel-Cantelli lemma:
http://en.wikipedia.org/wiki/Borel-Cantelli_lemma

I am doubting that it will help. :wink:
Can you write a part of your variant of  this theorem proof ?

the events $ A_n \equal{} \{ \xi_n\equal{}n \}$ occur infinitely many times, almost surely, since $ \sum P(\xi_n\equal{}n)\equal{}\plus{}\infty$ and these events are independant (second Borel Cantelli lemma). Hence $ S_n \equal{} \frac{\xi_1 \plus{} \ldots \plus{} \xi_n}{n}$ is not a Cauchy sequence since $ S_{n\plus{}1}\minus{}S_n$ does not go to $ 0$.


Will someone give me problems including Venn Diagrams and then explain how to use them?
Thanks.

Look on alcumus. There is a whole section of problems and even videos explaining the concept.

I don't think I have unlocked that part yet...
I just need one or two example problems and solutions.

[hide="Easier Problem"]Suppose every student in the school takes French, Spanish, both, or neither. There are 33 French, 33 Spanish, and 12 neither. With 66 people in the school, how many students take French and Spanish?[/hide]
Try this.

By the way, most problems (like mine) won't include a direct instruction to use a venn diagram. As you get more advanced you won't even need one (PIE)

If set $ A$ has $ n$ elements, compute the total number of pairs of sets $ (X, Y)$ such that $ \emptyset \subseteq X \subseteq Y \subseteq A$.
[hide="Solution"]The Venn diagram is made of three regions: $ X$, $ Y \setminus X$ and $ A\setminus Y$. Any of the $ n$ elements may belong to any of these regions, so the answer is $ 3^n$.[/hide]

Most problems in Mathcounts will not use that notation.

Actually, most high school contests will not either.

Easy problem;

A school teaches the languages Spanish and French. There are 40 students that go to the school. Five students don't learn any languages. There are 6 people who decided to learn both languages, and the spanish class is seven students larger than the French class. How many people are taking French?

Hard problem;

At Mallard High School there are three intermural sports leagues: football,
basketball, and baseball. There are 427 students participating in these sports: 128
play on football teams, 291 play on basketball teams, and 318 play on baseball
teams. If exactly 36 students participate in all three of the sports, how many
students participate in exactly two of the sports?

I got that one from a competition. Enjoy! :)

[hide=" Easy problem"]

$ 40 \minus{} 5 \equal{} 35$ study at least one language $ x$ study Spanish but not French, so $ x\minus{}7$ study French but not Spanish.

$ x\plus{}6 \plus{}x\minus{}7 \equal{} 35$

$ x\equal{}18$

$ 17$ study French.

[/hide]

[hide=" Hard problem"]

$ IEP$ gives $ 427 \equal{} 128 \plus{} 291 \plus{}318 \minus{} x \plus{}36$

$ x\equal{}346$.

[/hide]

Hi bull98. You're right. I forgot to subtract the $ 36$'s.


search
function
number theory unsolved
number theory

Prove that the number$2^{2^{n}}+1$ has a prime factor greater than $2^{n+2}(n+1)$

This problem posted (and somebody solved) by harazi. Please search or PM to he  :wink:

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=factor+prime&t=98980
i don't think this.LINK? Do you know? :lol:

n.t.tuan: could you post the link? i don't have nice results with search function

the topic http://www.mathlinks.ro/Forum/viewtopic.php?highlight=factor+prime&t=98980 don't have solutions  :blush:

see http://www.mathlinks.ro/Forum/viewtopic.php?t=46125  :wink:


Find all positive integer solutions to $abc - 2 = a + b + c$.

Here's how I'd try approaching this problem.

[hide="Possible Method"]$abc=a+b+c+2$, so we're looking for a cubic polynomial with the sum of the roots 2 larger than the product.[/hide]

4everwise, i do not see how relating them to the roots of a polynomial would help. could you please explain a little more.

[hide="Solution"]
$abc = a + b + c + 2$

We will do a little casework before proving the nonexistence of large solutions.  First, assume WLOG that $a \ge b \ge c$.

Case:  $c = 1$.  Our equation becomes $ab = a + b + 3 \implies (a-1)(b-1) = 4$, which has solutions $\boxed{a = 5, b = 2, a = 3, b = 3}$.

Case:  $c = 2$.  We have $2ab = a + b + 4$ which has solution $\boxed{a = 2, b = 2}$.  Now, suppose we increment either $a$ or $b$.  The LHS increases by at least two while the RHS increases by only one.  Hence there cannot be any larger solutions.

Case:  $c > 2$.  Suppose we begin with $a = b = c = 3$ (we have already verified that no smaller solutions exist) and attempt to find a larger solution by incrementing any of $a, b, c$.  The LHS increases by at least $9$ (from its initial value of $27$) while the RHS increases by one (from its initial value of $11$).  Hence there cannot be any larger solutions.

QED.
[/hide]


The sides of $\triangle ABC$ are $AB=8$, $BC=12$, and $AC=9$. if $\overline r{GF}$, $\overline{IE}$, and $\overline{HD}$ are respectively perpendicular to the sides as shown, and $GH=4$, find the ratio of the area of $\triangle GHI$ to that of $\triangle ABC$.

Okay this is what Elemmnop said. And I get it now  :D

Since AB= 8 and GH= 4, and since the two triangles are similar, the area ratio from the small to large triangle is 4 squared to 8 squared. Therefore the ratio is 16:64 which simplifies to 1:4.

1:4 is then the answer.

How do you know the triangles are similar?

triangles in problems liket this are similar most of the times hehe  :P

Ok, I figured out why they are similar, but how do we know $GH$ is the smallest side?

[quote="Farenhajt"]By establishing which angle is equal to which, determine the pairs of the corresponding vertices.[/quote]
Oh, of course! I should have realized that. Thank you all for your help. :)

yeah, farenhajt is awesome at explaining  :D

I asked like several people about a similar problem from 1991 Vestavia Tournament it was like number 21. except the measures were different. But exact same picture,  exact same segments, just different measure for GH AB and BC


limit
real analysis
real analysis theorems

Let $ X,Y$ be metric spaces and $ f: X\rightarrow Y$. How to prove following equivalence:
 $ x\in cl(A) \Rightarrow f(x)\in cl(f(A)) \Leftrightarrow x\equal{}\lim x_n \Rightarrow f(x)\equal{}\lim f(x_n)$ ?

cl(A) is closure of A

The $ \Leftarrow$ direction - If $ x \in cl(A)$, then there is $ x_n \in A$ with $ x_n \to x$ so by assumption $ f(x_n) \to f(x)$, but $ f(x_n) \in A$ so $ f(x) \in cl(A)$

I dont think the other direction is true. For example let $ f: \mathbb{R}\to \mathbb{R}$ given by $ f(x) \equal{} x$ except $ f(0) \equal{} 1$ and $ f(1) \equal{} 0$. Then take $ A\equal{} (0,1)$ and notice that $ x \in cl(0,1)\equal{}[0,1] \Rightarrow f(x) \in cl(f(0,1)) \equal{} [0,1]$, but, for example $ x_n \equal{} 1/n \to 0$, but $ f(1/n)\to 0 \not \equal{} f(0) \equal{} 1$


I need some help with solving a trig problem using difference formulas.

cosxcosy(tanx+tany)

I know it equals sin(x+y) cause its in the back of my book but I need to see the steps to actually find that out.

\[\begin{aligned}\cos x\cos y\left(\tan x+\tan y\right) & = \cos x\cos y\left(\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}\right) \\ & = \cos y\sin x+\cos x\sin y = \sin(x+y).\end{aligned}\]

Yeah...I dont get how that works out at all. Is there any more work?


In R, show that all intervals are Borel sets.

I'm a bit confused here. If I understand correctly, a Borel set is the smallest sigma-algebra of set S that contains all the elements of set S. As a sigma-algebra, it must contain the empty set, it must contain the complement of any element of the set and it must contain the union of any elements of the set.

I just don't see how to prove this statement.

Thanks for any help


inequalities
topology
triangle inequality
real analysis
real analysis solved

This is very easy, but it's sort of interesting, I think:

Find the necessary and sufficient conditions which a real set must satisfy in order to be the set of distances between pairs of points in some metric space.

Necessary and sufficient condition: the set contains only positive numbers. (we ignore trivial segments and exclude zero as a distance)

For a set $A$ of positive numbers, let our metric space be $A\cup0$, with metric $d(x,y)=\max(x,y)$. Since every triangle is isosceles, the triangle inequality is satisfied. Obviously, the possible distances are exactly the elements of $A$.


If  3x+2y+5z=1 then find the minumum value of (1/x)+(1/y)+(1/z)

By Cauchy, we have:

$\displaystyle \left( 3x+2y+5z \right)\left(\frac {1}{x}+\frac {1}{y}+\frac {1}{z}\right)\geq (\sqrt 3 + \sqrt 2 +\sqrt 5)^2$

Since $\left( 3x+2y+5z \right)=1$, we see that $(\sqrt 3 + \sqrt 2 + \sqrt 5)^2$ is the minimum possible value (it shouldn't be hard to show that it's attainable.)

However, this is only valid for positive numbers.  In fact, if we allow negative numbers, it's not hard to show that we can make the quantity go to negative infinity.

mmr:

The Cauchy-Schwarz (Buniakowski) inequality states that, for any $x_i,y_i\in\mathbb{R}$

$\displaystyle\left(\sum^{n}_{i=0} x_i^2\right)\left(\sum^{n}_{i=0} y_i^2\right)\geq\left(\sum^{n}_{i=0} x_iy_i\right)^2$ with equality iff $\frac {a_i}{b_i}=\frac{a_j}{b_j}$ for all $i,j$ such that $0\leq\ i,j,\leq n$.

actually, I used Cauchy to finish off an ugly geometry problem (which I coordinatized) on Putnam 96 A2 the other day.


AMC
USA(J)MO
USAMO
geometry
circumcircle
analytic geometry

Let $A_1A_2A_3$ be a triangle and let $\omega_1$ be a circle in its plane passing through $A_1$ and $A_2.$ Suppose there exist circles $\omega_2, \omega_3, \dots, \omega_7$ such that for $k = 2, 3, \dots, 7,$ $\omega_k$ is externally tangent to $\omega_{k-1}$ and passes through $A_k$ and $A_{k+1},$  where $A_{n+3} = A_{n}$ for all $n \ge 1$. Prove that $\omega_7 = \omega_1.$

When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Furthermore, please do not "hide" any portion of the solution. Please use LaTeX for posting solutions. Thanks.

Clearly, $\omega_i$ is determined by its center $O_i$, since we know that it passes through $A_i,A_{i+1}$. We also know that for all $i,\ O_i,A_{i+1},O_{i+1}$ are collinear. We must prove that in these conditions, $O_7=O_1$.

Suppose we prove that for all $i$, the angles $\angle OA_iO_i$ and $\angle OA_iO_{i+2}$ are equal (where $O$ is the circumcenter of $A_iA_{i+1}A_{i+2}$). The result would then follow, since $O_i\mapsto O_{i+3}$ would be an involution on the perpendicular bisector $d_i$ of $A_iA_{i+1}$, mapping a point on this line to its harmonic conjugate wrt $O$ and the intersection of the tangents in $A_i,A_{i+2}$ to the circumcircle of $A_iA_{i+1}A_{i+2}$. Since $O_1\mapsto O_4\mapsto O_7$, we must have $O_7=O_1$, as desired.

All that’s left to prove is:

Let $O_1$ be any point on $d_1$, and put $O_2=O_1A_2\cap d_2,\ O_3=O_2A_3\cap d_3$. Then, $A_1O_1,\ A_1O_3$ make equal angles with $OA_1$.

This, however, is a simple angle chase :).

[quote="MithsApprentice"]Let $ A_1A_2A_3$ be a triangle and let $ \omega_1$ be a circle in its plane passing through $ A_1$ and $ A_2.$ Suppose there exist circles $ \omega_2, \omega_3, \dots, \omega_7$ such that for $ k \equal{} 2, 3, \dots, 7,$ $ \omega_k$ is externally tangent to $ \omega_{k \minus{} 1}$ and passes through $ A_k$ and $ A_{k \plus{} 1},$  where $ A_{n \plus{} 3} \equal{} A_{n}$ for all $ n \ge 1$. Prove that $ \omega_7 \equal{} \omega_1.$[/quote]
Sorry to revive an old topic, but I have a rather different solution than the one posted previously:
[hide="Solution"]
Invert with arbitrary radius about $ A_1$.  Let the image of point $ P$ map to $ P'$.  Now, notice that $ \omega_3$ and $ \omega_4$ are both tangent and pass through $ A_1$, so they are tangent at $ A_1$.  The same applies for $ \omega_6$ and $ \omega_7$.  Now, since $ \omega_3$ passes through $ A_3$, $ \omega_3'$ is a line through $ A_3'$ and similarly, $ \omega_4'$ passes through $ A_2'$.  Furthermore, $ \omega_3'\parallel \omega_4'$.  Now, $ \omega_5$ passes through both $ A_2$ and $ A_3$ and is tangent to $ \omega_4$, so $ \omega_5'$ passes through $ A_2'$ and $ A_3'$ and is tangent to $ \omega_4'$ (at $ A_2')$.  $ \omega_6$ is tangent to $ \omega_ 5$ and passes through $ A_3$ and $ A_1$, so $ \omega_6'$ is a line passing through $ A_3'$ and tangent to $ \omega_5'$.  $ \omega_7$ passes through $ A_1$ and $ A_2$ and was tangent to $ \omega_6$ at $ A_1$, so $ \omega_7'$ is a line parallel to $ \omega_6'$ passing through $ A_2'$.  $ \omega_2$ passes through $ A_2$ and $ A_3$ and is tangent to $ \omega_3$ at $ A_3$, so $ \omega_2'$ is a circle tangent to $ \omega_3'$ at $ A_3'$.  $ \omega_1$ is tangent to $ \omega_2$ at $ A_2$ and passes through $ A_1$, so $ \omega_1'$ is a line through $ A_2'$, tangent to $ \omega_2'$ and $ A_2'$.  Let $ \omega_4'$ and $ \omega_6'$ meet at $ P$ and let $ \omega_1'$ and $ \omega_3'$ meet at $ Q$.  Furthermore, let $ \omega_5'$ meet $ \omega_1'$ and $ \omega_3'$ at $ B$ and $ A$ respectively and let $ \omega_2'$ meet $ \omega_6'$ and $ \omega_4'$ at $ C$ and $ D$ respectively.  Since $ QA_3'$ and $ QA_2'$ are tangents to $ \omega_5'$, we have that $ \angle QA_3'A_2' \equal{} \angle QA_2'A_3'$.  Since $ \omega_4'\parallel \omega_3'$, $ \angle QA_3'A_2' \equal{} \angle PA_2'A_3'$.  For a similar reason as before,
\[ \angle PA_2'A_3' \equal{} \angle PA_3'A_2'\implies \angle PA_3'A_2' \equal{} \angle QA_2'A_3'\implies QA_2'\parallel PA_3'\implies \omega_1'\parallel \omega_6'\implies \omega_1' \equal{} \omega_7'\]
since $ \omega_7'\parallel \omega_6'$ and $ \omega_7'$ passes through $ A_2'$.  Finally, we get the result that $ \omega_1 \equal{} \omega_7$, as desired. [/hide]

[hide="Easy Coordinates"]
Let the centers of $ \omega_1, \omega_2, \ldots, \omega_7$ be $ O_1, O_2, \ldots, O_7$.  Since $ \omega_k$ and $ \omega_{k\plus{}1}$ are externally tangent at $ A_{k\plus{}1}$, we see that $ A_{k\plus{}1}$ is the midpoint of $ O_k$ and $ O_{k\plus{}1}$.  Let $ A_1\equal{}(0,a)$, $ A_2\equal{}(b,0)$, and $ A_3\equal{}(c,0)$, and let $ O_1\equal{}(x,y)$.  Then it is easy to calculate that

$ O_2\equal{}(2b\minus{}x,\minus{}y)$

$ O_3\equal{}(2c\minus{}2b\plus{}x,y)$

$ O_4\equal{}(2b\minus{}2c\minus{}x, 2a\minus{}y)$

$ O_5\equal{}(2c\plus{}x,y\minus{}2a)$

$ O_6\equal{}(\minus{}x,2a\minus{}y)$

$ O_7\equal{}(x,y)$

as desired.
[/hide]

[quote="E^(pi*i)=-1"]Since $ \omega_k$ and $ \omega_{k \plus{} 1}$ are externally tangent at $ A_{k \plus{} 1}$, we see that $ A_{k \plus{} 1}$ is the midpoint of $ O_k$ and $ O_{k \plus{} 1}$.[/quote]

Why?

  darij

Oops, you're right, it's wrong.   :oops:

It seems like there should be a constant ratio between radii of circles through $ A_1$ and $ A_2$ and those through $ A_2$ and $ A_3$, though . . . maybe there is a way to multiply the coordinates by these factors such that it works out?

Hmm I think you can do this with just trivial angle chasing: $ A=\angle{A_{3}A_{1}A_{2}},B=\angle{A_{1}A_{2}A_{3}},C=\angle{A_{2}A_{3}A_{1}},X=\angle{O_{1}A_{1}A_{2}}$. Then, measures of the angles $ \angle{O_{i}A_{i}A_{i+1}}$ are easy to get, we find that $ \angle{O_{7}A_{1}A_{2}}=X$, and we also need $ O_{1},O_{7}$ on the same side of $ A_{1}A_{2}$ (which follows from the fact that we need $ A_{i}$ between $ O_{i+1}$ and $ O_{i+2}$, blah.

We let the radius of the circle $\omega_i$ by $r_i$.As usual denote the side opposite to vertex $A_i$ by $a_i$ for $i=1,2,3$.
Let $O_1A_2A_1=\theta$,where $O_1$ is the centre of $\omega_1$.
By applications of sine rule we may easily deduce the following relations:

$\frac{r_1}{r_2}=-\frac{a_1}{a_3}\frac{\cos\theta}{\cos(A_2+\theta)}$

$\frac{r_2}{r_3}=-\frac{a_2}{a_1}\frac{\cos(A_2+\theta)}{\cos(A_2-A_3+\theta)}$

$\frac{r_3}{r_4}=-\frac{a_3}{a_2}\frac{\cos(A_2-A_3+\theta)}{\cos(A_1-A_2+A_3-\theta)}$

Multiplying these out we get

$\frac{r_1}{r_4}=-\frac{\cos\theta}{\cos(A_1-A_2+A_3-\theta)}$.

Similarly it is easy to observe that

$\frac{r_4}{r_7}=-\frac{\cos(A_1-A_2+A_3-\theta)}{\cos(A_1-A_2+A_3-(A_1-A_2+A_3-\theta))}=-\frac{\cos(A_1-A_2+A_3-\theta)}{\cos\theta}$.

Multiplying these equalities we obatain $r_7=r_1$.Also because $7 \equiv 1\pmod{3}$ we note that $\omega_7$ passes through $A_1$ and $A_2$.We thus obtain $\omega_7=\omega_1$.

[hide=Hint]You only care about the circle centers, which happen to be uniquely determined by the previous center ($O_2=\overline{O_1A}\cap \ell$, where $\ell$ is the perpendicular bisector of the next side).

The idea is to keep track of the subtended arc $\widehat{A_i A_{i+1}}$ of $\omega_i$ for each $i$. To this end, let $\beta = \measuredangle A_1 A_2 A_3$, $\gamma = \measuredangle A_2 A_3 A_1$ and $\alpha = \measuredangle A_1 A_2 A_3$.

[asy] size(8cm); defaultpen(fontsize(9pt)); pair A2 = (0,0); pair O1 = (-8,0); pair O2 = (5,0); pair A1 = abs(A2-O1)*dir(285)+O1; pair A3 = abs(A2-O2)*dir(245)+O2; pair O3 = extension(O2, A3, midpoint(A1--A3), midpoint(A1--A3)+dir(A1-A3)*dir(90));

filldraw(circle(O1,abs(A1-O1)), invisible, deepgreen); filldraw(circle(O2,abs(A2-O2)), invisible, deepgreen); filldraw(circle(O3,abs(A3-O3)), invisible, deepgreen); filldraw(A1--A2--A3--cycle, invisible, red);

pair J = (0,-3.2); draw(O1--O2--O3, deepcyan+dotted);

dot("$O_1$", O1, dir(90), deepcyan); dot("$O_2$", O2, dir(90), deepcyan); dot("$O_3$", O3, dir(-90), deepcyan); dot("$A_1$", A1, dir(A1-J), red); dot("$A_2$", A2, dir(A2-J), red); dot("$A_3$", A3, dir(A3-J), red);

label("$\alpha$", A1, 2.4*dir(J-A1), red); label("$\beta$",  A2, 2*dir(J-A2), red); label("$\gamma$", A3, 2*dir(J-A3), red); [/asy]

Initially, we set $\theta = \measuredangle O_1 A_2 A_1$. Then we compute \begin{align*} 	\measuredangle O_1 A_2 A_1 &= \theta \\ 	\measuredangle O_2 A_3 A_2 &= -\beta-\theta \\ 	\measuredangle O_3 A_1 A_3 &= \beta-\gamma+\theta \\ 	\measuredangle O_4 A_2 A_1 &= (\gamma-\beta-\alpha)-\theta \\ \end{align*} and repeating the same calculation another round gives \[ \measuredangle O_7 A_2 A_1 = k-(k-\theta) = \theta \] with $k = \gamma-\beta-\alpha$. This implies $O_7 = O_1$, so $\omega_7 = \omega_1$.

I think you can do this with moving points, but here is the angle chase approach.

Let $O_i$ be the center of $\omega_i$ for $i=1,2,\dots,7$.

Observe
\[\measuredangle O_7A_1A_2=-\measuredangle A_2A_1A_3-\measuredangle A_3A_1O_6=-\measuredangle A_2A_1A_3-\measuredangle O_6A_3A_1=-\measuredangle A_2A_1A_3+\measuredangle A_1A_3A_2+\measuredangle A_2A_3O_5=\]
\[-\measuredangle A_2A_1A_3+\measuredangle A_1A_3A_2+\measuredangle O_5A_2A_3=-\measuredangle A_2A_1A_3+\measuredangle A_1A_3A_2-\measuredangle A_3A_2A_1-\measuredangle A_1A_2O_4.\]
A similar angle chase finishes.

Let $O_i$ be the center of $\omega_i$. Also let $\theta_1=\measuredangle A_1A_2A_3,\theta_2=\measuredangle A_2A_3A_1,\theta_3=\measuredangle A_3A_1A_2$. Observe that $\omega_1$ is uniquely determined by $\measuredangle O_1A_2A_1$, and $\omega_7$ is uniquely determined by $\measuredangle O_7A_2A_1$, so if we can show those are equal we're done. Now, let $\measuredangle O_1A_2A_1=\theta$. By basic angle chasing (since $O_i,A_{i+1},O_{i+1}$ are collinear), we can find:
\begin{align*}
\measuredangle O_1A_2A_1&=\theta\\
\measuredangle O_2A_3A_2&=-\theta_1-\theta\\
\measuredangle O_3A_1A_2&=\theta_1-\theta_2+\theta\\
\measuredangle O_4A_2A_1&=-\theta_1+\theta_2-\theta_3-\theta\\
\measuredangle O_5A_3A_2&=\theta_1-\theta_2+\theta_3-\theta_1+\theta\\
\measuredangle O_6A_1A_2&=-\theta_1+\theta_2-\theta_3+\theta_1-\theta_2-\theta\\
\measuredangle O_7A_2A_1&=\theta_1-\theta_2+\theta_3-\theta_1+\theta_2-\theta_3+\theta=\theta.
\end{align*}
hence $\omega_1=\omega_7$ as desired. $\blacksquare$

Interesting problem. We will aply an inversion on $\Gamma$ with center $A_1$ and ratio $R$ (it's not important). Define $Inv _{\Gamma} (\omega _i)=\Omega _i$ and $Inv _{\Gamma} (A_i)=B_i$. Note that $\Omega _i$ is a line for $i=1, 3, 4, 6, 7$ because $\omega _i$ pass through $A_1$. Also note that $\Omega _i$ and $\Omega _{i+1}$ are parallels for $i=3, 6$ because $\Omega _i$ and $\Omega _{i+1}$ are tangents in $A_1$. We want that $\Omega _1=\Omega _7$, it's sufficient to prove that $\Omega _1 \parallel \Omega _6$. Now we will use directed angles. Let $\ell=B_2B_3$:
$$\measuredangle (\Omega _1,\ell)=\measuredangle(\ell,\Omega _3)=\measuredangle(\ell,\Omega _4)=\measuredangle (\Omega _6,\ell).$$

So we have that $\Omega _1$ and $\Omega _6$ are parallels as desired.

Not sure if this works.

Through some manipulations we find that it suffices to show $\angle O_1A_2O_4=\angle O_7A_1O_4$, but this is true because $$\angle O_1A_2A_4=\angle O_5A_2O_2=\angle O_5A_3O_2=\angle O_3A_3O_6=\angle O_3A_1O_6=\angle O_7A_1O_4$$

wait i think this works only when using directed angles (because $A_1O_4$ bisecting $\angle O_1A_1O_7$ is possible)

Let $O_i$ be the center of $\omega_i$. Define $\angle A_3A_1A_2=\alpha,\angle A_1A_2A_3=\beta,\angle A_2A_3A_1=\gamma.$ Define $\angle A_iA_{i+1}O_i=\angle A_{i+1}A_iO_i=\theta_i.$ It suffices to show $\theta_1=\theta_7.$ In fact, note that
\begin{align*}
\theta_2 &= 180^\circ-\beta-\theta_1 \\
\theta_3 &= 180^\circ-\gamma-\theta_2 \\
&= \beta-\gamma +\theta_1 \\
\theta_4 &= 180^\circ - \alpha -\theta_3 \\
&= 180^\circ - \alpha - \beta + \gamma - \theta_1 \\
&= 2\gamma - \theta_1
\end{align*}
Note that this process is the exact same for $\theta_4\to \theta_7$, and since function $f(k)=2\gamma -k$ is cyclic with order $2$, we are done.

Let $\theta=\angle A_1O_1A_2$, where $O_i$ is the center of $\omega_i$. Furthermore, let $\alpha, \beta, \gamma$ be the angles of the triangle. Then, check that we have the sequence $$\theta \to 2\beta - \theta \to \theta+2\gamma-2\beta \to 2\alpha-2\gamma+2\beta-\theta=360^\circ-4\beta-\theta.$$ Upon applying another sequence of three transformations, this returns the identity $\theta$, so the centers of $\omega_1$ and $\omega_7$ coincide, which suffices.

Define $O_{k}$ as the centers of each respective circle. Let $\alpha = \angle{O_{1}A_{2}A_{1}}$.
\begin{align*}
    \angle{A_{2}O_{4}A_{1}} &= 180-\angle{A_{1}}-\angle{O_{3}A_{1}A_{3}} \\
    &= 180-\angle{A_{1}}-(180-\angle{A_{3}-\angle{A_{2}A_{3}O_{2}}}) \\
    &= \angle{A_{3}}+\angle{A_{2}A_{3}O_{2}}-\angle{A_{1}} \\
    &= \angle{A_{3}}+180-\angle{A_{2}}-\alpha-\angle{A_{1}}
\end{align*}
We can apply this transformation twice to get that,
\[\angle{A_{2}A_{1}O_{7}} = \alpha\implies \omega_{1} = \omega_{7}\]

Consider if we extend the internal tangent of all of the pairs of circles $\omega_{k-1}$ and $\omega_k$ for $k = 2, 3, \dots, 8$. Name these $\ell_1, \ell_2, \dots \ell_7$. Since $\ell_x$ is just $\ell_{x-1}$ reflected over the perpendicular bisecter of some side, and because the internal tangent is not dependent on the circle (it is in fact what makes the next circle), we just have to prove that $\ell_1 = \ell_7$.

Call $\angle A_2A_1A_3$ $a$, $\angle A_3A_2A_1$ $b$, and $\angle A_1A_3A_2$ $c$
We know that the angle formed by $A_1$, $A_2$, and $\ell_1$ is less than $a$ (otherwise $\omega_2$ does not exist). Thus we call this $x$ and we know that the angle formed by $\ell_1$, $A_1$, and $A_3$ is $a-x$ we can see that this is equal to the angle formed by $A_1$, $A_3$, and $\ell_2$, so we know that the angle formed by $\ell_2$, $A_3$, $A_2$ is $b-a+x$. Continuing this pattern for subsequent angles we find that the next angles are $180-2b-x$, $a+2b+x - 180$, $180 - a - b - x$, and $x$. Thus we find that $\ell_7 = \ell_1$ and we are done. $\blacksquare$

For $1 \le i \le 7$, let $O_i$ be the center of $\omega_i$ and $\theta_i = \measuredangle O_iA_iA_{i + 1} = \measuredangle A_iA_{i + 1}O_i$. Then for $2 \le i \le 7$, the tangency condition tells us that $O_{i - 1}, A_i$, and $O_i$ are collinear, so
$$\theta_i = \measuredangle O_iA_iA_{i + 1} = \measuredangle O_{i - 1} A_iA_{i + 1} = \measuredangle O_{i - 1}A_iA_{i - 1} + \measuredangle A_{i - 1}A_iA_{i + 1} = -\theta_{i - 1} + \measuredangle A_{i - 1}A_iA_{i + 1}.$$
Therefore, applying this identity in succession, we have
$$\theta_7 = \theta_1 + \sum_{i = 2}^7 (-1)^{7 - i} \measuredangle A_{i - 1}A_iA_{i + 1} = \theta_1 + \sum_{i = 2}^4 ((-1)^{7 - i} + (-1)^{4 - i})(\measuredangle A_{i -1}A_iA_{i + 1}) = \theta_1.$$
Thus, $\measuredangle O_1A_1A_2 = \measuredangle O_7A_1A_2$ and $\measuredangle A_1A_2O_1 = \measuredangle A_1A_2O_7$. This is enough to imply that $O_1 = O_7$ since we would otherwise have $\angle O_1A_1A_2 + \angle O_7A_1A_2 = 180^{\circ}$ (using non-directed angles), which is impossible since $\angle O_1A_1A_2, \angle O_7A_1A_2 < 90^{\circ}$. So, we are done.


does anyone know the mathcounts practice site, that you choose which kind of problems you want (target, countdown, sprint, etc) and how many problems you want to do? and the site scrambles the numbers in the problems and you submit your answers...? sorry, I know this is kind of vague. I had this site last year but I haven't visited it in a year so I don't know where it went?

any help would be appreciated :D

:D  Heeh, Im pretty sure this is it: http://mathcounts.saab.org/mc.cgi

  Just type "mathcounts drills" on google.


1.  Vijay takes 5 hours to mow the lawn.  His sister Deepali takes 3 hours to mow the same lawn.  Working together, how long will it take them to mow the lawn?

2.  Wei drives from Chicago to New York, averaging 45 miles per hour.  He then drives back the same route, but averages 60 mph on the return trip.  What is his average speed?

3. Yan takes 90 seconds to walk up an up escalator.  Walking at the same pace down the up escalator, it takes Yan 150 seconds to reach the bottom.  Walking at the same pace, how long would it take Yan to climb the same escalator if the escalator was turned off?  How long would it take Yan to ride up the escalator (if she wasn't walking, but the escalator was on?)

4.  Alex and his two brothers, Rob and Paul, are going to paint the 4 walls of Alex's cubical house.  The decide that Alex and Rob will paint the first side, Rob and Paul will paint the second, Paul and Alex will paint the third, and all three together will paint the 4th.  If Alex paints 1 wall in 6 hours, Rob paints one wall in 2 hours, and Paul paints 1 wall in 3 hours, what total fraction of the four walls will each of the brothers paint?

5. Jooho is two thirds of the way from one end of a railway tunnel when he sees a train 100 feet away approaching the tunnel.  Calculating quickly, he realizes that no matter which way he runs, he will escape from the tunnel at exactly the same moment he would be plowed down by the train.  How long is the tunnel?  How many times faster does the train travel than Jooho?

6. David travels from A to B and then from B to C.  He realizes that if he makes the first segment of the trip at 4 mph and the second at 4 mph, the trip will take him 3 hours.  If he travels the first at 2 mph and the second at 6 mph, the trip will take him 3 hours.  If A, B, and C are in the same plane, what is the shortest possible distance between A and C?  The largest?

7. Runners A and B race along a hexagonal track, always running at constant speeds.  B begins to fall behind, and when he is exactly half done with the race realizes he can't possibly win.  So he cuts directly across the center of the racetrack, catching A just as she crosses the finish line.  If A runs at 18 kilometers per hour, how fast does B run?

8. Dongbo always runs 5 miles every morning at a constant speed of 5 miles per hour before leaving for work.  One day, he decided that he would start out at 7 miles per hour, and after every mile he ran, he would decrease his speed by 1 mile per hour.  (So he ran his first mile at 7 mph, the second at 6 mph, and the last at 3 mph.)  He was surprised to find that under this system, he ended up being late for work.  How much more time did his run take under this system?

9. Jim drove from Lawrence, KS, to Amherst, MA taking an out-of-the-way route in order to travel at 60 mph.  Returning home to Lawrence, he took a more direct route but was only able to travel 40 mph.  He discovered to his surprise that the two routes had taken the same amount of time.  What was his average speed for the round trip?

Nice choice of problems Joel :twisted:

By the way if the number's a fraction/decimal is the unit in singular form or plural? What if it's 0 or negative? imaginary?

Hmm ... if it's rational fractional, you say "one third of a mile per hour," or some such.  But if it's irrational and still between 0 and 1, you say "square root of two minus 1 mile[b]s[/b] per hour."  For the first case, you could also use decimals to say "point three-repeating mile[b]s[/b] per hour."  I hope none of the answers are zero, irrational, or imaginary, although I did just make the numbers up off of the top of my head, so it's theoretically possible.

Please put your solutions in spoiler!

Don't be lazy -- write up your solutions!

Well. It is generally the same solution for 1,2, and 4.

1)[hide]Vijay takes 5 h and Deepali takes 3h. Each hour, Vijay mows 1/5 of the lawn and Deepali mows 1.3 of the lawn. Together in one hour they mow 1/3+1/5 of the entire lawn. Since the lawn is one whole, their time together is 1/ ( 1/3 + 1/5) = 15/8.[/hide]

2)[hide] Set the distance from the cities to be x. Then it takes Wei x/45 hours for one way and x/60 on the way back. His entire time is x/45 + x/60. The distance traveled is 2x. 2x / (x/45+x/60) = 360/7[/hide]

3) [hide]x is the person's speed, and y is the escalator speed. Z is the escalator length. z/(x+y) = 90 and z/ (x-y) = 150 . Do some simultaneous equations to end up with what you want : z/x and z/y. They turn out to be 225/2 and 450.[/hide]

4)[hide]This one takes some time. You want the fractions of the 4 walls each person paints. Alex and Rob together take 1/(1/6 + 1/2) hours, Alex and Paul together take 1/(1/6+1/3) h, Paul and rob together take 1/(1/2+1/3)h, and all three brothers take 1/(1/6+1/3+1/2)h. You solve and for A and R 3/2 hours, for A and P 2 hours, for R and P 6/5 hours, and for A, P, and R 1 hour. Since rate*time = total distance or job done, you solve for each person.

Alex did. 3/2 * 1/6 + 2 * 1/6 + 1* 1/6 for the three walls he did altogether. Simplified, Rob did 3/4 + 3/5 + 1/2 of all the walls, and Paul did 2/3 + 2/5 + 1/3 of all the walls. Since there are 4 walls, the sum of A + R +P = 4, so you divide the portion of the 4 walls they did by 4 to find the fraction of the total they painted.[/hide]

Haha... I feel sorry for the guy in problem 5. Why did he waste time calculating? If he ran absentmindedly in one direction before calculating, he might have made it out.

5) Maybe he was a mathematician and calculated really quickly

[hide]

Ok so let y = length of tunnel, x = speed of Jooho, and kx = speed of train (k times faster than Jooho)

We can make two equations with these:

(y/3)/x = (100-y/3)/kx time it takes to get to closer end of tunnel

(2y/3)/x = (100+2y/3)/kx time it takes to get to further end

Do a bit of simplifying and cross-multiplying to get:

3kyx = 900x-3xy

and

6kyx = 900x+6xy

Subtract the first from the second to get 3kyx = 9xy wihch means k=3. Plug it back into the first equation and solve for y because the x cancels out. You get 75.

So the tunnel is 75 feet long and the train is 3 times faster.[/hide]

7)[hide] so assume the hexagon is regular to calculate it more easily. and call the side lengths 3. So at 18kph it will take 1 hour. now, half of it is obviously 9, and if you draw equilateral triangles you can see that cutting across will be 6. Therefore, it is 15 km in one hour. So the speed is 15kph.[/hide]

8)[hide] well for the first mile, it takes 1/7 of an hour, then 1/6, 1/5, 1/4, and 1/3. Add them up and you get 459/420, which is 13/140 hours later, approximately 5.5 minutes.[/hide]


inequalities
geometry
trigonometry
symmetry
geometry proposed

Given are $\triangle ABC$ area $S$ with any $M$ on plane prove that
$\cos\frac{A}{2}MA^{2}+\cos\frac{B}{2}MB^{2}+\cos\frac{C}{2}MC^{2}\ge 2S$
$\cos\frac{B}{2}MA^{2}+\cos\frac{C}{2}MB^{2}+\cos\frac{A}{2}MC^{2}\ge 2S$

[color=darkred][b]A classical and remarkable geometrical inequality.[/b]

For any real numbers $x\ ,\ y\ ,\ z$ for which $x+y+z> 0$ and for any point $M$ from the plane of the triangle $ABC$ there is the following inequality :

$\boxed{\ x\cdot MA^{2}+y\cdot MB^{2}+z\cdot MC^{2}\ge \frac{yza^{2}+zxb^{2}+xyc^{2}}{x+y+z}\ }$ with equality if and only if $M: =P\ (\ x\ ,\ y\ ,\ z\ )\ .$[/color]

Can you show show clearly Virgil, actually I tried but I can't solve by this way! :)

[quote="gemath"][color=darkred]Can you [b][u]show , show clearly[/u][/b] Virgil? Actually [u]I tried but I can't solve by this way ![/u][/color][/quote]
[color=darkblue][b][u]All done ![/u][/b] I found a proof using the well-known inequality $\boxed{\ x+y+z>0\Longrightarrow\sum xMA^{2}\ge\frac{\sum xya^{2}}{\sum x}\ \ (*)\ }$ and the strong inequality
$\boxed{\ \sum a\sin\frac{A}{2}\ge p\ \ (**)\ }$ ( [u]see the topic[/u] http://www.mathlinks.ro/Forum/viewtopic.php?t=80834 [u]where appears and a nice, easy and geometrical proof of it[/u] ).

[b]Proof.[/b] $\left\{\begin{array}{c}p\ge 3r\sqrt 3\Longrightarrow\boxed{\ p^{2}\ge 3S\sqrt 3}\\\\ \boxed{DOWN\equiv\sum\cos\frac{A}{2}\le \frac{3\sqrt 3}{2}}\mathrm{\ (\cap-Jensen\ for\ Cos)}\\\\ \boxed{UP\equiv\sum a^{2}\cos\frac{B}{2}\cos\frac{C}{2}=p\cdot\sum a\sin\frac{A}{2}\ge^{(**)}p^{2}}\end{array}\right\|$ $\Longrightarrow$ $\sum MA^{2}\cos\frac{A}{2}\ge^{(*)}\frac{UP}{DOWN}$ $\ge$ ${\frac{2p^{2}}{3\sqrt 3}}\ge 2S$ $\Longrightarrow$ $\boxed{\boxed{\ \sum MA^{2}\cos\frac{A}{2}\ge 2S\ }}\ .$

[b]Remark.[/b] $\boxed{\sum MA^{2}\cos\frac{A}{2}\ge \frac{2\sqrt 3}{9}\cdot p^{2}}\ge 2S\ .$ If you please, I "pester" you a bit : mine is [u]stronger[/u] than yours ![/color]

ok thank Virgil for your stronger ineq, it gives me an interesting idea! :)

Congratulations Virgil ! The beautiful solutions one after the other :lol:
(This was really surprising)
@gemath: Your inequalities are nice but how come you never post your solutions? For example this one has been solved now , why don't you post yours, too?  :P  :wink:

I agree with you dear Albanian Eagle.

In my opinion, when a different solution of an our proposed problem, is presented in the forum, then the solver directly (at first, by previous message) and later, by post message all the other mathlinkers, must know the solution we have in mind (if we have), as a minimum intication of honorific.

Kostas Vittas.

yes vittasko and Albanian Eagle I know I need to explain more than only say thank you  :D .
Really I found the second before it base on [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=114248]my inequality[/url], my solution of that like of msceok,
and by simple (but most important) inequality
$aMA^{2}+bMB^{2}+cMC^{2}\ge abc$ add them and use law of sin we have:
$(\sin A+\sin C)MA^{2}+(\sin B+\sin A)MB^{2}+(\sin C+\sin B)MC^{2}\ge\frac{2abc}{2R}=4S$
and note that
$\sin A+\sin C=2\sin\frac{A+C}{2}\cos\frac{A-C}{2}\le 2\cos\frac{B}{2}$
similar we will have the second mine.
I think that method can't solve the non-symmetry problem as the second!
when I found the second I think the first is true and post both of them and actually it is true ! :P


LaTeX
inequalities proposed
inequalities

Let $ a,b,c$ be three real numbers so that $ a,b,c>0$ and $ a^2\plus{}b^2\plus{}c^2 \le 12$.

Find $ \min s\equal{}\frac{{a^6 }}{{ab \plus{} 2\sqrt {1 \plus{} c^3 } }} \plus{} \frac{{b^6 }}{{bc \plus{} 2\sqrt {1 \plus{} a^3 } }} \plus{} \frac{{c^6 }}{{ca \plus{} 2\sqrt {1 \plus{} b^3 } }}$.
 :D  :D  :D


Find the smallest positive integer $n$ such that \[0< \sqrt[4]{n}-\lfloor \sqrt[4]{n}\rfloor < 0.00001.\]

$\sqrt[4]n$ has to be slightly bigger than an integer. Thus $n=a^4+1$ (replacing $1$ by a bigger number always results in a greater difference and bigger n, thus is contraproductive).

We want $\sqrt[4]{a^4+1} - a < \epsilon$ (here $\epsilon=0.00001$). This means $a^4+1 < (a+\epsilon)^4=a^4+\epsilon(4a^3+6a^2\epsilon+4a\epsilon^2+\epsilon^3)$.
Now this requires that $\frac 1{4\epsilon} < a^3$, giving (using my PC :D) that $a \geq 30$. A small calculation then shows that $a=30$ works (we don't need to plug it in the first, but rather in the third inequality).


limit
real analysis
real analysis unsolved

Let $ a_{k}$, $ k \equal{} 0, 1, 2, \dots$, be a sequence of non-negative real numbers such that

(a) $ a_{m \plus{} n} \le a_{m}\cdot a_{n}$ for all $ m$, $ n \ge 0$;
(b) there exists an $ m_0$ such that $ a_{m_{0}} < 1$.

Show that $ \displaystyle\sum_{k \equal{} 0}^{\infty}a_{k} < \infty$, i.e. the $ \displaystyle\lim_{N\rightarrow\infty}\displaystyle\sum_{k \equal{} 0}^{\infty}a_{k}$ exists and is finite.

$ a_{p \plus{} km_0} \leq a_pa_{km_0}\leq a_pa_{m_0}^k$ , so $ \displaystyle\sum_{n \equal{} 0}^{\infty}a_{n} \equal{} \sum_{p \equal{} 0}^{m_0 \minus{} 1}\sum_{k \equal{} 0}^{\infty}a_ {p \plus{} km_0} \leq  \sum_{p \equal{} 0}^{m_0 \minus{} 1}a_p( \displaystyle\sum_{k \equal{} 0}^{\infty}a_{m_0}^k) \equal{} \displaystyle (\sum_{p \equal{} 0}^{m_0 \minus{} 1}a_p)/(1 \minus{} a_{m_0})$  :cool:


[quote="#H34N1"]$S$  is the set of all points with coordinates  $(m,n)$  such that  $m$  and  $n$  are positive integers with  $m < 4$  and  $n < 4$ . Two points from  $S$  are chosen at random. What is the probability that the midpoint of the segment of the two points is also in the  $S$ ? Give your answer as a common fraction in lowest term.[/quote]
Would this be 1/1 because the midpoint will always be in S?

500th post!

[hide]That would be 9 points total. There are 36 ways to pick two out of 9. 8 pairs have their midpoints also in S so it's 8/36, 2/9.[/hide]

[hide] Your choices are 1,2 or 3 for both M and N as the person above said there are 36 different pairs then think of it like tic-tac toe, there 8 different lines. 8/36= 2/9[/hide]


My parents won't let me buy it so I am looking for an ebook but I can't find one. Anyone have it? Thanks.

Post full data (full author name, exact title, edition and publication year (if you know them), etc.). Also, specify what languages you can read. Then, maybe, someone will be able to help you :).

Apparently it is a very famous book, but here are the details.

Aptitude test problems in physics
SS Krotov
MIR Publication

That is all I know about it. I know English.


linear algebra
matrix
algebra
polynomial

Let M be a square matrix of order n.

Then show that there exists an integer k>0 such that for all n>k,
$M-\lambda\textbf{I}_{n}$ is invertible.

how do u add an $n \times n$ and $m \times m$ matrix? :?

Should that be $\lambda>k$? The question makes no sense as written.

Alright then. I had the same doubt. But here is the solution given.
Can someone clearly explain this?

Soln: Since |$M-\lambda \textbf I_{n}$| is a polynomial in $\lambda$, it has finitely many roots.
Thus as soon as k is an integer which is greater than the absolute values of the roots, we have n > k $\Longrightarrow$  |$M-\lambda \textbf I_{n}$| $\not=0$.

Therefore $M-\lambda \textbf I_{n}$ is invertible.

That solution would be correct if we replaced "$n>k$" with "$\lambda>k$".


[quote="jhredsox"]What would happen if not all of your team is at States for the Team Round (even if you have 2 team, 1 ind or something)???

Can you pair up with a person from another school???

What if your whole team is at states, but you are the only individual? What happens then?

-jorian[/quote]
If your team has three members they don't take the average, they divide by four. Otherwise, my team last year probably would have had three people.  :|

(unless you purposely throw off getting team to get indiv's so u plan who ur with)

-jorian

it'd be cool if u got together with the best people around the state and did individuals...and they counted it :P

-jorian

It's up to the states, I know CT did different things in '05 than '06

Oh, I ended up pretty much doing the whole team round by myself.  I made my usual dumb mistakes and got a 7.  Yeah, they divide by 4 no matter how many members you have.

[quote="Ubemaya"][quote="easyas3.14159..."]Individuals pair up with three other individuals to make a team...but they don't usually turn out well.[/quote]
How would you know?

I don't think that they count the scores for teams made up of individuals...[/quote]

You don't have to go to nationals to figure that 4 strangers probably wont work as good as 4 teammates.

unless last year

you had

andrew
neal
kevin
daesun

they would own (even though they prob weren't top 4 written...but still)

didn't daesun get 5th, neal 1st (im sure about neal)

-jorian

How'd you forget Daniel? He's like the biggest to forget.

I think he had the top 4 from countdown.

[quote="13375P34K43V312"][quote="perfect628"]I was ind. last year at states, and we got 7. they weren't that bad...[/quote]

I thought you said they were bad and you did the whole thing...oh well that's better than what we got as a team last year, bleh[/quote]

That was my school team. This year they're not TOO bad.

[quote="perfect628"][quote="Klebian"]I'm pretty sure they all missed number 10 on the team round.

1st: Neal
2nd: Daniel Li
3rd: Kevin
4th: Nathan Benjamin[/quote]

So did basically everyone else.[/quote]
We still don't know who got number 10 right...
But a team of those four could've had the time to spare to go through 10.

I thought a bunch of people established that the Virgin Islands guessed #10 right?

[quote="pianoforte"]I seriously doubt that...[/quote]
*cough* *cough*
[quote="Ignite168"]Virgin Islands got #10 right, this is how it went with 1 minute left
"Hey, man, I have no idea what the answer is"
"How about we put that answer to everything from that one book"
"Yeah, that's a great idea, what's that number man?"
"Uhh.. I forgot"
"Hmm.. I don't know either"
"Beats me"
"Wait isn't it like 52 or something?"
"Yeah that's it!"
"Let's write it"

And they got it right :D[/quote]


inequalities
induction
algebra
polynomial
quadratics
difference of squares
special factorizations

If [tex]a_0=1[/tex] and [tex] a_i < a_{i+1} \leq a_i + 1[/tex] for [tex] 0 \leq i \leq n[/tex] , show that

[tex]\displaystyle \left(\sum_{i=0}^n a_i \right)^2 \geq \sum_{i=0}^n a_i^3[/tex]

Ideas?

I think induction works.  Figure out what inequality you need to make the induction go through.  That inequality in turn might need its own inductive proof.

A solution by indirect induction:

Also, for convenience, I set a_0 = 0. This includes the original as a special case a_1 = 1.

[hide]

Let  S(m) = :Sigma: (a_k) : 0 :le: k :le: m, and C(n) = :Sigma: (a_k :^3:) : 0 :le: k :le: m

Then C(m+1) - C(m) = a_(m+1):^3: and

S(m+1):^2: - S(m):^2: = a_(m+1)*(a_(m+1) + 2*S(m)) = a_(m+1)*(-a_(m+1) + 2S(m+1))

We use induction on the hypothesis that a_(m+1):^3: :le: S(m+1):^2: - S(m):^2:.

This is obvious for m=0, as a:^3: :le: a:^2: for a :le: 1

We can write the hypothesis for m-1 as

a_m :^3: + a_m :^2: - 2*S(m)*a_m :le: 0, or equivalently a_m :^2: + a_m - 2S(m) :le: 0.

Let the roots of x:^2: + x - 2*S(m) = 0 be r and -1-r, with r :ge: 0. The inequality  a_m :^2: + a_m - 2*S(m) :le: 0 implies a_m :le: r.

By definition of a_m, a_(m+1) :le: r+1

Then the roots of  x:^2: - x - 2*S(m) = 0 are 1+r and -r, so

 a_(m+1):^2: - a_(m+1) - 2*S(m) :le: 0 and

a_(m+1):^3: - a_(m+1):^2: - 2S(m)*a_(m+1) :le: 0.

This gives a_(m+1):^3: :le: S(m+1):^2: - S(m):^2: as desired, so the inequality is true for all m.

Finally, S(n):^2: - C(n) =  :Sigma: (S(m+1):^2: - S(m):^2: - a_(m+1):^3:) : 0 :le: m :le: n-1, and we are done.

Equality is only possible if a_(m+1) = a_m + 1 for all m < n.[/hide]

There is a simple and elegant solution using a variation on the formula that alison gave here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12175
The formula is
[tex]n^3 = (a_1 + a_2 + ... + a_n)^2 - (a_1 + a_2 + ... + a_{n-1})^2[/tex]
For the conditions of this particular inequality, we can use the fact (easily proven by induction) that
[tex](a_1 + a_2 + ... + a_n)^2 - (a_1 + a_2 + ... + a_{n-1}) \ge a_n^3[/tex] (the only equality condition being when [tex]a_{i+1} = a_i + 1[/tex])
We can add the series of inequalities having [tex]a_i^3[/tex] as the right side, and the left side will telescope and become the desired expression.

Thanks for the help folks.

Yeah, I tried induction on it, but I didn't realize to perform the necessary 'induction within the induction'.

By the way, jmerry, when you wrote

[quote="jmerry"]
Then the roots of  x:^2: - x - 2*S(m) = 0 are 1+r and -r, so
 a_(m+1):^2: - a_(m+1) - 2*S(m) :le: 0 and
a_(m+1):^3: - a_(m+1):^2: - 2S(m)*a_(m+1) :le: 0.
This gives a_(m+1):^3: :le: S(m+1):^2: - S(m):^2: as desired, so the inequality is true for all m.
[/quote]

did you actually mean to write the following?

[quote]
Then the roots of  x:^2: + x - 2*S(m+1) = 0 are 1+r and -r, so
 a_(m+1):^2: + a_(m+1) - 2*S(m+1) :le: 0 and
a_(m+1):^3: + a_(m+1):^2: - 2S(m+1)*a_(m+1) :le: 0.
This gives a_(m+1):^3: :le: S(m+1):^2: - S(m):^2: as desired, so the inequality is true for all m.
[/quote]

Hmm, the solution still seems to be difficult to discover though. How long did it take you to solve the problem?

I meant what I wrote. Since I hadn't calculated anything about S(m+1), I wasn't going to use it in an inequality. Your second quote is actually wrong- the roots of that polynomial are between -r-1 and -r and between r+1 and r+2. It's not enough to prove anything.
I also don't believe schulmannerism about the ease- the induction he refers to is most of my proof.

I originally worked backwards to find that -a_m and a_(m+1) were between the two roots of x:^2: - x - 2*S(m) = 0; this could be written formally as the induction I needed. The key was that I could write the factor S(m) + S(m+1) as either 2*S(m) + a_(m+1) or 2*S(m+1) - a_(m+1)

The solution didn't take me long to find, but I'm not sure anyone else would get it quickly. The step that you questioned is one of my favorite tricks- "flipping the quadratic" by switching focus from one root to the other. I'd never applied it to an inequality before, but it was obvious to me.

[quote="jmerry"]
I also don't believe schulmannerism about the ease- the induction he refers to is most of my proof. [/quote]

Here's how the induction goes.
[tex](a_1 + a_2 + ... + a_n)^2 - (a_1 + a_2 + ... + a_{n-1})^2 \ge a_n^3[/tex]
Is manipulated (using difference of squares) to become
[tex]2(a_1 + a_2 + ... + a_{n-1})  \ge a_n^2 - a_n[/tex]
This is our induction hypothesis. Add [tex]2a_n[/tex] to both sides
[tex]2(a_1 + a_2 + ... + a_{n-1} + a_n)  \ge a_n^2 + a_n[/tex]
Now we just need to prove that [tex]a_n^2 + a_n \ge a_{n+1}^2 - a_n[/tex]
[tex]a_n^2 + a_n = a_n(a_n + 1) \ge (a_{n+1}-1)(a_{n+1}) \ge (a_{n+1}-1)(a_{n+1})[/tex] as desired.

BTW, making your posts nice with TeX is easy. Just putting the begin and end tags around a_1 + 2s^2 turns it into [tex]a_1 + 2S^2[/tex] Also, if you  quote someone, you can see their TeX.

Another way to prove that inner inequality is via telescoping sum:
[tex]\begin{eqnarray*}
a_n^2 &  =   & \sum_{i=0}^{n-1} (a_{i+1}^2 - a_i^2) \\
            & =   & \sum_{i=0}^{n-1} (a_{i+1} - a_i)(a_{i+1} + a_i) \\
            & \le & \sum_{i=0}^{n-1} (a_{i+1} + a_i) \\
            & =   & a_n + 2 \sum_{i=0}^{n-1} a_i \, .
\end{eqnarray*}
[/tex]


modular arithmetic
algorithm
number theory
Diophantine equation
Euclidean algorithm

I tried this problem but my solution was very "painful". Here is my solution:

Clearly $ 13^{\varphi(1000)} \equiv 1 \pmod{1000} \equal{}> 13^{400} \equiv 1 \pmod{1000}$  $ \equal{}> 13^{398}*169 \equiv 1 \pmod{1000} (1)$

What I did now was find a solution to the congruence $ 169x \equiv 1 \pmod{1000}$ to get rid of the 169 on (1). I had a hard time finding the solution because I dont know how to solve this type of congruence. Finally I got x= 929, so $ 13^{398} \equiv 929 \pmod{1000}$

Id like to see other solutions or how do you solve the congruence I mentioned.

The congruence you were trying to solve amounts to finding the multiplicative inverse of 169 (mod 1000).  This can be done using the Euclidean Algorithm.

By definition, $ 169x\equiv1 (mod1000)$ means that $ 169x\minus{}1000y\equal{}1$ for integers $ x$ and $ y$.  Now we have a diophantine equation and we can use the Euclidean Algorithm:

$ 1000\equal{}5\cdot169\plus{}155$
$ 169\equal{}1\cdot155\plus{}14$
$ 155\equal{}11\cdot14\plus{}1$
$ 14\equal{}14\cdot1\plus{}0$

Running this in reverse, we get

$ 1\equal{}155\minus{}11\cdot14$
$ 1\equal{}155\minus{}11(169\minus{}155)\equal{}12\cdot155\minus{}11\cdot169$
$ 1\equal{}12(1000\minus{}5\cdot169)\minus{}11\cdot169\equal{}12\cdot1000\minus{}71\cdot169$

This gives the solution $ x\equal{}\minus{}71$, $ y\equal{}12$, but we want a positive integer solution for $ x$.  To do this, add 1000 to $ x$ and subtract 169 from the (meaningless) value of $ y$ to get $ x\equal{}929$, the solution you found.  There might have been a more direct way to arrive at the answer, but that's the quickest I know off the top of my head.

[hide]When congruencies don't work, this is how u can ponder with it....
$ 13^{398}\equal{}(170\minus{}1)^{199}$
Now try....[/hide]


Factorize $ n^4\plus{}n^2\plus{}1$ then use Telescoping Method.

I've tried that, I'd like a solution if possible.

I have managed to factor it, i have transformed it to sum of terms twice, and the result still doesn't cancel out. This is what I ended up with (or something like that, I don't remember fully, the test was yesterday:

$ \frac{1^2}{1\cdot3}\plus{}\frac{2^2}{3\cdot7}\plus{}\frac{3^2}{7\cdot 13}\plus{}\cdots$

Again, I need a [i]solution[/i] , thanks in advance for whoever takes some time to post it.

[quote="triplebig"]What is the answer?[/quote]

$ \frac{2019045}{4038091}$

[quote="triplebig"]Could you express it in terms of the general sum formula?[/quote]

If you turn 2009 into k, the answer will be:

$ \frac{k(k\plus{}1)}{2k(k\plus{}1)\plus{}2}$


I'm sorry if this has already been discussed but there r a number of problems in my blog like:-

1)Edit/delete an entry
2)Contributors can't do anything

are these common problems or just with my blog?

Yeah, it would be really helpful to be able to edit/delete blog entries :P

You can edit entries, just it takes some work. :D

<snipped for now>


i have a major doubt...

let $ \omega_r$ be the resonant frequency

and $ \omega_1$ and $ \omega_2$ be the frequencies when power is half the max. power

now.....
[b]1.[/b] acc. to NCERT,the graph is symmetrical
so $ \omega_1 , \omega_2 \equal{} \omega_r \pm \delta \omega$

i.e. $ \omega_r \equal{} \frac{\omega_1 \plus{} \omega_2}{2}$

[b]2.[/b] graph is actually not symmetrical

an alternate way to find $ \omega_r$
at resonance  $ L\omega_1 \minus{} \frac{1}{C\omega_1} \equal{}\frac{1}{C\omega_2} \minus{}  L\omega_2$

solving,we get $ \omega_r \equal{} \sqrt{\omega_1 \omega_2}$

which one is true???
if 1st is wrong ,it means whole derivation of NCERT is wrong (the most possible case)

if both formulae are applicable ,solve the 2 equations
it will give $ \omega_1 \equal{}\omega_2 \equal{} \omega_r$
 :rotfl: not possible....

I dont think graph is symmetrical. The function 1/wC - wL doesnt look symmetric at all. f(wo+w') =/ f(wo - w')

then NCERT is wrong...

and so is the formula $ Q \equal{} \frac{\omega_r L}{R}$???

meanwhile,
the problem is 2 days,i had two tests
both hav the same questions
but diff formula had to be applied to get the answer :rotfl: now what??

fk the derivation. remember the result and its significance.

actually for a medical streamer ppl say prefer wat is given in NCERT
 :rotfl:


Prove that if $a,b,c\geq 0$ then the following inequality is true

$a^{3}+b^{3}+c^{3}+2a^{2}b+2b^{2}c+2c^{2}a+6abc\geq 5ab^{2}+5bc^{2}+5ca^{2}$

If (a,b,c)=(1,1,0), then  sum(a^3+2*a^2*b)+6*a*b*c-5*sum(a*b^2)=-1,the inequality does not hold! We have

         sum(a^3+6*a^2*b)>=sum(5*a*b^2)+6*a*b*c.

[quote="fjwxcsl"]sum(a^3+6*a^2*b)>=sum(5*a*b^2)+6*a*b*c.[/quote]

Could you post your solution thank you ,

fjwxcsl just wants to say that your inequality is wrong...take $a=1,b=1,c=0$


limit
logarithms
calculus
derivative
function
algebra
binomial theorem

Evaluate $ \lim_{x\to\infty} \frac{x\plus{}1}{x\minus{}1}$.

[hide]$ \lim_{x\to\infty}\frac {x \plus{} 1}{x \minus{} 1} \equal{} \lim_{x\to\infty} 1 \plus{} \frac {2}{x \minus{} 1} \equal{} 1 \plus{} 0 \equal{} 1$[/hide]

[hide]Since $ \lim_{x\to\infty}x\plus{}1 \equal{} \lim_{x\to\infty}x\plus{}1 \equal{} \infty$, we can use L'Hopital's Rule:
$ \lim_{x\to\infty}\frac{x\plus{}1}{x\minus{}1} \equal{}  \lim_{x\to\infty} \frac{1}{1} \equal{} 1$[/hide]

HEre is a way . whenever we have these forms like $ \lim_{x\to\infty}$  if possible always take the highest powers of the variable $ \rightarrow \infty$ outside . then it becomes easy .

For example in this case take $ x$ outside and write it as   $ \lim_{x\to\infty}\frac { x(1\plus{}1/ x ) }{x( 1 \minus{}1/ x )  }$

As $ x \rightarrow \infty$  therfore $ 1/x \rightarrow  0$.  hence the limit is $ 1$ .

Just don't take "highest powers" too literally; think about what's big and what's small, and what's bigger than what else.

$ \lim_{x\to\infty}\frac{x^2\plus{}x\plus{}1\plus{}2^x}{x^2\minus{}x\plus{}1\minus{}2^x}\equal{}?$

[quote="ThetaPi"]Evaluate $ \lim_{x\to\infty} \frac {x \plus{} 1}{x \minus{} 1}$.[/quote]

Oops, the question is to evaluate $ \lim_{x\to\infty} \left(\frac {x \plus{} 1}{x \minus{} 1}\right)^x.$

The interesting challenge then is to do that in your head without writing anything down.

[hide="Short answer."]$ e^2$[/hide]

I should have realized that the problem was too simple.
[hide]
$ \lim_{x\to\infty}\left(\frac {x \plus{} 1}{x \minus{} 1}\right)^{x} \equal{} \lim_{x\to\infty}\left(1 \plus{} \frac {2}{x \minus{} 1}\right)^{x}$

Let $ y \equal{} \frac {x \minus{} 1}{2}$, then $ x \equal{} 2y \plus{} 1$, and as $ x \to \infty$, $ y \to \infty$.

Therefore:
$ \lim_{x\to\infty}\left(\frac {x \plus{} 1}{x \minus{} 1}\right)^{x} \equal{} \lim_{x\to\infty}\left(1 \plus{} \frac {2}{x \minus{} 1}\right)^{x}$

$ \lim_{x\to\infty}\left(\frac {x \plus{} 1}{x \minus{} 1}\right)^{x} \equal{} \lim_{y\to\infty}\left(1 \plus{} \frac {1}{y}\right)^{2y \plus{} 1}$

$ \lim_{x\to\infty}\left(\frac {x \plus{} 1}{x \minus{} 1}\right)^{x} \equal{} \lim_{y\to\infty}\left(1 \plus{} \frac {1}{y}\right)\left[\left(1 \plus{} \frac {1}{y}\right)^y\right]^2$

$ \lim_{x\to\infty}\left(\frac {x \plus{} 1}{x \minus{} 1}\right)^{x} \equal{} \left[\lim_{y\to\infty}\left(1 \plus{} \frac {1}{y}\right)\right]\left[\lim_{y\to\infty}\left(1 \plus{} \frac {1}{y}\right)^y\right]^2$

$ \lim_{x\to\infty}\left(\frac {x \plus{} 1}{x \minus{} 1}\right)^{x} \equal{} 1 \cdot e^2 \equal{} e^2$[/hide]

However, the problem I face is that I [i]do[/i] know that the value is $ e^2$, but I have no idea how to tackle the exponent. Using the binomial theorem does not seem to work either.

The logarithm of your limit is the limit of $ x\ln\left(\frac{x\plus{}1}{x\minus{}1}\right)\equal{}x\ln\left(1\plus{}\frac2{x\minus{}1}\right).$

We can expand out the Taylor series of that logarithm:

$ \ln\left(1\plus{}\frac2{x\minus{}1}\right)\equal{}\frac2{x\minus{}1}\plus{}O(x^{\minus{}2})$

$ x\ln\left(1\plus{}\frac2{x\minus{}1}\right)\equal{}\frac{2x}{x\minus{}1}\plus{}O(x^{\minus{}1})\equal{}2\plus{}O(x^{\minus{}1})$

Let $ x\to\infty$ and that tends to $ 2;$ hence our answer.

I am sorry, but as you can see, I have only just started calculus at school; I am still in the midst of understanding product rule, chain rule, intermediate value theorem, continuity and differentiability etc.

A simple solution would be good. I do not understand why such a question has appeared in my homework.

All right, here's the elementary version of that:

As before, we take the logarithm $ x\ln\frac{x\plus{}1}{x\minus{}1}$. We don't like limits at $ \infty$ very much, so we substitute $ t\equal{}\frac1x$; it becomes $ \frac1t\cdot\ln\frac{\tfrac1t\plus{}1}{\tfrac1t\minus{}1}\equal{}\frac{\ln\frac{1\plus{}t}{1\minus{}t}}{t}$

OK, now it's $ \lim_{t\to 0}\frac{\ln\frac{1\plus{}t}{1\minus{}t}}{t}$. That's a derivative; if we define $ f(t)\equal{}\ln\frac{1\plus{}t}{1\minus{}t}$, it's $ \lim_{t\to 0}\frac{f(t)\minus{}f(0)}{t}\equal{}f'(0)$. If the logarithm is on the list of functions you know how to differentiate, you now have the tools.

I don't understand why people take the logarithm.

The limit

$ \lim_{x \to \infty} (1 \plus{} \frac {a}{x})^x$ is one of the fundamental ones and equals $ e^a$.

In fact, some texts use this to define the constant $ e$...

I agree with you. Neither L'Hopital nor Landau sign do need for the limit. :wink: