MATHCOUNTS conics parabola algebra polynomial

What is your mathcounts index school+chapter+state=94(my score).

indianamath i guess you pwned me, BTW who was the 4th 7th grader who made it to aime. it was you, me , lynodon and someone else.

i'm guessing youkou homa (sp?) but i'm not sure if he made it.

I know I'm not supposed to be in an Indiana forum
but oh well

School: $44$
Chapter: $39$
State: $30$

so

$44+39+30=\fbox{113}$

It was Youkow Homma with a 135 on the 10B

My index:

School: 44
Chapter: 44
State: I can't reveal... greater than 38 shouldn't be that surprising though.

Total index: 126+

43 (school) + 39 (chapter) + 32 (state) = 114 (i think)

41 + 34 + 31 = 106

school, chapter, and state, respectively.

[quote="indianamath"]112 *sniff* i'm sad.[/quote]

you got 112?? i soo did not beat you! i couldn't have

i got 2 pm's from rcv(a mod) about spamming....

haha, but u weren't the only one spamming. at least i didn't spam that much (did I?). it's about time someone took control of this place.... :lol:

i'm pretty sure he's checking every comment i make. i got my second pm because i said i got 112 index and didn't make it.

oh yeah lol, that index was about chapter+state. i added my school too.lol

School: 44
Chapter: 44
State: -_- lol
Index: like over 88 <(^^)> lol...

[quote="mathcrazed"]
Index: like over 88 <(^^)> lol...[/quote]

that says a lot! lol. wow, major hint to what u got at states!! :lol:

[quote="mathcrazed"]yep it could be... possibly in the range from 0-46. Wow, we narrowed it down to only 47 numbers guys (and girls XD)![/quote]

more like [i]girl[/i]. lol. i am the only one right? i feel so special...(not ed. i'm one step ahead of all u ppl! :lol: )

Well no we've narrowed it down to 45. You couldn't have gotten a 45, because the paper said I got a "near perfect" so I couldn't have gotten a 46.

XD crazy logic.

Plus mwp probably knows her own score and you got a higher score

[quote="Fanatic"].
Plus mwp probably knows her own score and you got a higher score[/quote]

is it just me or did that kinda not make sense. obviously mathcrazed got a higher score than me...

[quote="mathcrazed"] XD aasd asd ASZX[/quote]

you ppl are crazy...

In my post earlier, my point was that mwp could further narrow down lyndon's score by using her score as the minimum for his.

but i think i already know the top 4, as i mentioned earlier...who knows, i could be wrong.

[quote="cognos599"]not necessarily you could have lost a tiebreaker from 13th to 4th.[/quote]

 :rotfl:  :rotfl:  :rotfl:

sorry yuqing, if that was the case i have deeply insulted you by saying you got the same score as my sister.

lol i'm not laughing at that. If 13th tied with 4th, it would REALLYREALLYREALLY suck to be 13th. lol lose all the tiebreakers lol....lol....lmao...

hey, i can see all this, just in case you didn't know! jk, but that would suck. terribly. i think i did lose to some ppl due to tiebreakers, though...

was that stevens? it's good that i came first in tiebreaker at least...

i have a question (math-related). it may sound stupid but bear with me...

just say you have a parabola $-3{x^{2}}+15x+30 = y$ . if you set $y = 0$.

then you can get $-3{x^{2}}+15x+30 = 0$.

if you factor out the $-3$ and divide throughout, you get ${x^{2}}-5x-10 = 0$.

now since they are both equal to zero, you can set them equal to each other:

$-3{x^{2}}+15x+30 ={x^{2}}-5x-10$

but by using common sense, you know the two sides aren't equivalent. i'm just confused how that works. could someone explain it to me, thanks

well, when you set them equal to 0, you found that the zeros of the two polynomials were equal, but the graphs actually equal-you could divide by -3 because you are finding the zeros.  To begin with,their "a" values are opposite-one is posite the other negative-so they arent even going in the same direction.  And when you graph parabolas like that, the equation isnt like ax^2+bx+c=0, it is just ax^2+bx+c, so in that case, you cant just cancel out the -3.


Not sure if this is the appropriate place to post this question. Does anyone know a good approximation to the trajectory of Halley's Comet?


[quote="blueshark"]I want to know, in simple language, why any number raised to the zero is 1.  I also want to know why any number raised to the one power is the base itself.

Sample:

100^1 = 100...WHY?

Sample:

100^0 = 1...WHY?

Please, do not explain using math textbook definitions.
I can read a math book for myself.  I am seeking a simple definition.  Okay?[/quote]First, why does $a^1=a$ for any real number $a$? This comes from the definition of $\text{positive integer exponents}$. $a^n$ means the product of $n$ factors of $a$. So..

$a^3=a\cdot{a}\cdot{a}\\\\a^2=a\cdot{a}\\\\a^1=a$

What else could $a^1$ be?

Next, why does $a^0=1$ for any nonzero real number $a$ (it's true for negative $a$ right?)? It is defined this way for consistency:

We know that $a^ma^n=a^{m+n}$, since \[a^ma^n=\underset{m}{\underbrace{a\cdot{a}\cdots{a}}}\underset{n}{\underbrace{a\cdot{a}\cdots{a}}}=\underset{m+n}{\underbrace{a\cdot{a}\cdots{a}}}=a^{m+n}.\]This clearly works for positive integers $m$ and $n$, but we can extend it to $0$ (the product of no $a\text{'s}$.) When, say, $m=0$ then we have $a^0a^n=a^{0+n}=a^n$, so when we divide both sides by $a^n$ we get $a^0=1$.

hey my method is....

$a^0$=
$a^{n-n}$
$a^n/a^n
let $a^n$ be x

hence x/x
=1

hence proved

I thank you both your explanation and examples.


What factoring trick is this called?  It looks like simmon's favorite factoring trick but its not.

$AMC+AM+MC+CA=(A+1)(M+1)(C+1)-(A+M+C)-1$

Thanks :P

$(x+\alpha)(x+\beta)(x+\gamma)=x^{3}+(\alpha+\beta+\gamma)x^{2}+(\alpha \beta+\beta \gamma+\gamma \alpha)x+\alpha \beta \gamma .$
Plugging $x=1,\alpha =A,\ \beta =M,\ \gamma =C.$ to this equation gives your equation.

I wouldn't even call it a factoring trick.  This is just a rewrite of

$(A+1)(M+1)(C+1) = AMC+AM+MC+CA+A+M+C+1$

Which is just a standard expansion.  (If you really wanted to, you could call this the "three-variable" case of Simon's.)

check out the video on aops wiki at the bottom of this page if u wish to know more  :lol:

http://www.artofproblemsolving.com/Wiki/index.php/Simon%27s_Favorite_Factoring_Trick

[quote="kunny"]Thank you!
I have just seen the video.

This trick is well known. :lol:

Alternative approach

Find all pairs of positive integers $(m,\ n)$ such that $mn+4m-2n=99.$

$mn+4m-2n=99\Longleftrightarrow m(n+4)=2n+99$

$\Longleftrightarrow m=\frac{2n+99}{n+4}$

$\Longleftrightarrow m=2+\frac{91}{n+4}.$

Since $n+4$ is positive integers such that $n+4\geq 5$ and divisor of $91,$ yielding $n+4=7,\ 13.\therefore (m,\ n)=(9,\ 9),\ (15,\ 3).$

By the way, How about this problem?

[color=blue]Find all pairs of positive integers[/color] $(m,\ n)$ [color=blue]such that [/color]$6mn+21m+10n-305=0.$ :D[/quote]
Well, your way can easily turn into Simon's Favorite Factoring Trick.  Just subtract two from both sides of $m=2+\frac{91}{n+4}$ and multiply everything by n+4, and that is the form of Simon's Favorite Factoring Trick.  Now, for the problem you gave:
[hide] $6mn+21m+10n-305=0\Rightarrow 6mn+21m+10n=305$.  We also have that $(3m+5)(2n+7)=6mn+10n+21m+35$.  Thus, we know that $(3m+5)(2n+7)-35=305\Rightarrow (3m+5)(2n+7)=340=2^{2}*5*17$.

We know that the factors of 340 are 1, 2, 4, 5, 10, 17, 20, 34, 68, 85, 170, and 340.  We find that the factors that will make n integral for 2n+7=x (if x is a factor) are all of the odd factors, which are 1, 5, 17, and 85.  However, 1 and 5 won't work because they will make n negative.

We can now say that 2n+7=17, so n=5.  Now, this means that 3m+5=340/17=20.  Thus, m=5.  So, one pair is (5,5).  Now, we can also say that 2n+7=85, thus n=34.  Now, we know that 3m+5=340/85=4.  This makes m negative, so that is not possible.

Finally, we can conclude that $\boxed{(5,5)}$ is the only solution.  [/hide]


Ok so this game goes like this.
I start with a country, town, state, or province.
You have to name a country, town, state, or province that starts with the letter that my country ends in.
Example:
USA
Andorra
Albania
Afghanistan
Nepal
Lebanon

And no repeats

Cyprus

(This is the 2012th topic in G&FF, the world is supposed to end in 2012, symbolic eh?)

Armenia
time for something other than A's.

Sydney

A side note, Antarctica is a continent not a country.

albania
antarctica
alberquerqe
el alamein
nubia
alaska
ancorage

-_-;;

only one country, and you can't repeat..

PLEASE read other posts before posting in an inconsiderate/improper manner....

Albania is STILL up for grabs...

Antarctica
Australia
United States of America
Alaska
Ann Arbor
Raleigh
Haiti
Italy
Yugoslavia
Austria
Antilles
Saskatchewan
Netherlands
Sri Lanka
Abkhazia
Argentina
Atlanta
Alabama
Acapulco
Orlando
Oman
Nairobi
Iceland
Denmark
Kenya
Albania
Afghanistan

Sorry, I just couldn't wait for someone else to post.

Antarctica
Australia
United States of America
Alaska
Ann Arbor
Raleigh
Haiti
Italy
Yugoslavia
Austria
Antilles
Saskatchewan
Netherlands
Sri Lanka
Abkhazia
Argentina
Atlanta
Alabama
Acapulco
Orlando
Oman
Nairobi
Iceland
Denmark
Kenya
Albania
Afghanistan
[b]Norway[/b]

Antarctica
Australia
United States of America
Alaska
Ann Arbor
Raleigh
Haiti
Italy
Yugoslavia
Austria
Antilles
Saskatchewan
Netherlands
Sri Lanka
Abkhazia
Argentina
Atlanta
Alabama
Acapulco
Orlando
Oman
Nairobi
Iceland
Denmark
Kenya
Albania
Afghanistan
Norway
[b]Yemen[/b]

Antarctica
Australia
United States of America
Alaska
Ann Arbor
Raleigh
Haiti
Italy
Yugoslavia
Austria
Antilles
Saskatchewan
Netherlands
Sri Lanka
Abkhazia
Argentina
Atlanta
Alabama
Acapulco
Orlando
Oman
Nairobi
Iceland
Denmark
Kenya
Albania
Afghanistan
Norway
Yemen
[b]Nebraska[/b]

Antarctica
Australia
United States of America
Alaska
Ann Arbor
Raleigh
Haiti
Italy
Yugoslavia
Austria
Antilles
Saskatchewan
Netherlands
Sri Lanka
Abkhazia
Argentina
Atlanta
Alabama
Acapulco
Orlando
Oman
Nairobi
Iceland
Denmark
Kenya
Albania
Afghanistan
Norway
Yemen
Nebraska
[b]Azerbaijan[/b]

[hide="We need lives..."]Antarctica
Australia
United States of America
Alaska
Ann Arbor
Raleigh
Haiti
Italy
Yugoslavia
Austria
Antilles
Saskatchewan
Netherlands
Sri Lanka
Abkhazia
Argentina
Atlanta
Alabama
Acapulco
Orlando
Oman
Nairobi
Iceland
Denmark
Kenya
Albania
Afghanistan
Norway
Yemen
Nebraska
Azerbaijan[/hide]

[b]Nicaragua[/b]

Antarctica
Australia
United States of America
Alaska
Ann Arbor
Raleigh
Haiti
Italy
Yugoslavia
Austria
Antilles
Saskatchewan
Netherlands
Sri Lanka
Abkhazia
Argentina
Atlanta
Alabama
Acapulco
Orlando
Oman
Nairobi
Iceland
Denmark
Kenya
Albania
Afghanistan
Norway
Yemen
Nebraska
Azerbaijan
Nicaragua
[b]Aruba[/b]

Antarctica
Australia
United States of America
Alaska
Ann Arbor
Raleigh
Haiti
Italy
Yugoslavia
Austria
Antilles
Saskatchewan
Netherlands
Sri Lanka
Abkhazia
Argentina
Atlanta
Alabama
Acapulco
Orlando
Oman
Nairobi
Iceland
Denmark
Kenya
Albania
Afghanistan
Norway
Yemen
Nebraska
Azerbaijan
Nicaragua
Aruba
[b]Amsterdam[/b]

Antarctica
Australia
United States of America
Alaska
Ann Arbor
Raleigh
Haiti
Italy
Yugoslavia
Austria
Antilles
Saskatchewan
Netherlands
Sri Lanka
Abkhazia
Argentina
Atlanta
Alabama
Acapulco
Orlando
Oman
Nairobi
Iceland
Denmark
Kenya
Albania
Afghanistan
Norway
Yemen
Nebraska
Azerbaijan
Nicaragua
Aruba
Amsterdam
[b]Mongolia[/b]

[quote="ASPEED"]Antarctica
Australia
United States of America
Alaska
Ann Arbor
Raleigh
Haiti
Italy
Yugoslavia
Austria
Antilles
Saskatchewan
Netherlands
Sri Lanka
Abkhazia
Argentina
Atlanta
Alabama
Acapulco
Orlando
Oman
Nairobi
Iceland
Denmark
Kenya
Albania
Afghanistan
Norway
Yemen
Nebraska
Azerbaijan
Nicaragua
Aruba
Amsterdam
Mongolia
[b]Albuquerque  [/b][/quote]
Eraguay

[hide]Antarctica
Australia
United States of America
Alaska
Ann Arbor
Raleigh
Haiti
Italy
Yugoslavia
Austria
Antilles
Saskatchewan
Netherlands
Sri Lanka
Abkhazia
Argentina
Atlanta
Alabama
Acapulco
Orlando
Oman
Nairobi
Iceland
Denmark
Kenya
Albania
Afghanistan
Norway
Yemen
Nebraska
Azerbaijan
Nicaragua
Aruba
Amsterdam
Mongolia
Albuquerque
Eraguay
[/hide]

I now invoke martial law; hide tags everybody.

And you meant Uruguay  :roll:

Since Eraguay isn't an actual place, I'm going back to Albuquerque.  (Actually, I've never been to Albuquerque... :rotfl: )

[hide]Antarctica
Australia
United States of America
Alaska
Ann Arbor
Raleigh
Haiti
Italy
Yugoslavia
Austria
Antilles
Saskatchewan
Netherlands
Sri Lanka
Abkhazia
Argentina
Atlanta
Alabama
Acapulco
Orlando
Oman
Nairobi
Iceland
Denmark
Kenya
Albania
Afghanistan
Norway
Yemen
Nebraska
Azerbaijan
Nicaragua
Aruba
Amsterdam
Mongolia
Albuquerque
[b]El Salvador[/b][/hide]


[quote="sharkman"]John now has two investments that produce a 150 dollars income each month. If
1,000 dollars more is invested at 9 percent than at 10 percent per year, how much was invested
at each percent? There were 12 months in that year.

I just need to know how to change this from English to Algebra?[/quote]

Edit: Wait that is inconsistent, sorry. I was in a hurry

But I think the problem has extra information that gives no solution to this problem...

I am not so clear on this.  Does this cover everything that the problem
is asking?  That is why I am stuck.  Please, explain a little further as
my bulb is not to bright.

Thank you so much for your time.

The solution given makes sense.  :maybe:

[quote="sharkman"]John now has two investments that produce a 150 dollars income each month.[/quote]

Is this [i]on average[/i] over the course of one year?

[quote="sharkman"]There were 12 months in that year.[/quote]

Because this seems to suggest that that is the correct interpretation, but the problem is quite unclear.

:huh: I just looked at kohjhsd's answer and it looked correct, thought once I did the problem it does not seem to work, the system of equations did not yeild consistent results.
Assuming that it is an average of 150 a month.

I'm sleepy but here's a shot:
[hide]Let $x$ be the amount invested at 10 percent.  We have:
$\frac{(x+1000)(1.09)+x(1.1)}{12}= 150 \implies (x+1000)(1.09)+x(1.1) = 1800$
$\implies 2.19x = 710 \implies x \approx 324.2$. Then we have approximately $\$324.20$ invested at $10\%$ and $\$1,324.20$ invested at $9\%$.[/hide]

Oh, so 150 dollars was sum of the income of each investments :blush:


calculus
integration
trigonometry
calculus computations

Find the integral.
$ \int{\frac{1}{1\plus{}x^4}dx}$

[quote="Patterns_34"]Just so it's known, there was another topic on evaluating this indefinite integral [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=200964]here[/url].[/quote]
Thank you.
But how I can use completing square to solve it?

[quote="sos440"]Some application of elementary complex shows that if we put $ \theta_k = \tfrac{\pi(2k + 1)}{n}$, then

\begin{eqnarray*} \frac {1}{1 + x^n} & = & - \frac {1}{n} \sum_{k = 1}^{n} \frac {x \cos \theta_k - 1}{x^2 - 2x \cos \theta_k + 1} \\
& = & - \frac {1}{n} \sum_{k = 1}^{n} \left[ \frac {\cos \theta_k}{2} \cdot \frac {2 x - 2 \cos \theta_k}{x^2 - 2x \cos \theta_k + 1} - \sin \theta_k \cdot \frac {\sin \theta_k}{(x - \cos \theta_k)^2 + \sin^2 \theta_k} \right], \end{eqnarray*}

Note that each summand is now integrable.

[quote="Patterns_34"]Just so it's known, there was another topic on evaluating this indefinite integral [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=200964]here[/url].[/quote]

Anyway, the method hash_include used is very impressive...![/quote]
Is there a way you can do it which only need to use basic integration formulas to solve this specific question, instead to n's power?


Hi,
I am new to this site and to this competition. It is also very late, so I hope you will forgive me if I violate some sort of protocol in the way I posted this.
I had a question about the USAMTS.

One of my solutions (completely ugly, but the best I could do) has a LOT of checking solutions. For that, can I simply say "checked by TI-89" and include as an appendix the precise formula I entered into the calculator, and get credit for the problem? That is, can I just say, "we test the following values into the following equation looking for this, and, if we use a calculator in the following manner, we get these results" and get credit for it? OTherwise, I'm probably going to have 4-6 pages to an otherwise 1.5 page problem.
I was not sure if the section in the FAQ regarding computers was totally applicable.

That should be fine - as long as you are clear how you did the checking and that you covered all possibilities, it is acceptable under current rules.

Thank you. One more question: How important are Lemma's? I'm not sure how to use them, and really not sure if I should be starting to learn how to use them now.

Thanks
sran.

[quote="sran"]I had a question about the USAMTS.

One of my solutions (completely ugly, but the best I could do) has a LOT of checking solutions. For that, can I simply say "checked by TI-89" and include as an appendix the precise formula I entered into the calculator, and get credit for the problem? That is, can I just say, "we test the following values into the following equation looking for this, and, if we use a calculator in the following manner, we get these results" and get credit for it? Otherwise, I'm probably going to have 4-6 pages to an otherwise 1.5 page problem.
I was not sure if the section in the FAQ regarding computers was totally applicable.[/quote]

If you wrote four pages of checking all cases, we graders would just scan the first few entries to check whether your test was an effective one. We would not read all four pages. While there is nothing bad about writing out all cases, except for paying extra postage to mail your submission, you may instead just show a few cases as examples of your test and then say that you continued in a like manner for the rest of the cases.

Alas, if your answer is as ugly as you say, you won't get a commendation. Often, once a mathematician answers a problem by difficult and tedious methods, the answer itself suggests a more elegant solution. Perhaps if you ponder your answer, you could find an elegant solution and earn a commendation.

[quote="sran"]Thank you. One more question: How important are Lemma's? I'm not sure how to use them, and really not sure if I should be starting to learn how to use them now.[/quote]

Lemmas are useful if your argument has a big side proof that is not part of the main flow of the proof. It makes the explanation clearer to set it apart as a lemma. For example, in problem 1/1/15 of last year's USAMTS, I could say, "Lemma: The function f is constant over all real numbers in the half-open interval [i/1000,(i+1)/1000) for all integers i.", and then prove that lemma before returning to the main proof. You do not have to call it a lemma. I myself often write such a side proof in one paragraph, beginning with the sentence, "I claim that truncating the value of x to three decimal places does not change the value of f(x).", and writing the proof as the rest of the paragraph.

Richard Rusczyk & Mathew Crawford provide a much more detailed description of solution-writing techniques with better examples at:
http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php

Erin Schram
USAMTS grader


calculus
integration
function
calculus computations

Find the exact length of the polar curve described by:
r = 5 e^(-t)
on the interval (5/4)pi to 9pi

I got:
∫ √(r² + (dr/dθ)²) dθ
∫ √(25e^(2θ) + 25e^(2θ)) dθ
∫ √(2*25e^2θ) dθ
∫ √2 * 5e^θ dθ
√2 * 5e^θ {5π/4 to 9π}
5√2 * (1.903*10¹² - 50.75)
about equal to:
 1.345 * 10¹³

which is wrong.

Any help?? Thank you!!!!

Kent Merryfield:
That's what I had at first, but this gives a negative number (about -0.13932) and I figured negative number here didn't make sense.

The curve spirals in, not out; the arclength should be short, not long.

Note that if $ r\equal{}5e^{\minus{}\theta},$ then $ \frac{dr}{d\theta}\equal{}\minus{}5e^{\minus{}\theta}$ and $ \left(\frac{dr}{d\theta}\right)\equal{}25e^{\minus{}2\theta}.$

$ r^2\plus{}\left(\frac{dr}{d\theta}\right)\equal{}50e^{\minus{}2\theta}.$

$ \sqrt{r^2\plus{}\left(\frac{dr}{d\theta}\right)}\equal{}5\sqrt{2}\,e^{\minus{}\theta}.$

You seem to have committed an exponent sign error in squaring.

The definite integral of a positive function must be positive. If you're getting something negative, then you're doing it wrong.

$ \int_{\frac{5\pi}4}^{9\pi}5\sqrt{2}\,e^{\minus{}\theta}\,d\theta\equal{} \minus{}5\sqrt{2}e^{\minus{}\theta}\bigg{|}_{\frac{5\pi}4}^{9\pi}\equal{} 5\sqrt{2}\left(e^{\minus{}5\pi/4}\minus{}e^{\minus{}9\pi}\right)\approx 0.139320351.$

Perhaps you missed the minus sign that comes from the chain rule?

Note also that the contribution of $ e^{\minus{}9\pi}$ is negligible. To 10-place accuracy this is the same as the arclength of the curve for $ \theta\in\left[\frac{5\pi}4,\infty\right).$


geometry
3D geometry
sphere
topology
Functional Analysis
real analysis
advanced fields

Let $X$ be a real Banach space, $X^{*}$ its dual. By the Hahn-Banach theorem for any element $x$ of the unit sphere $S_{X}$ there exists $x^{*}\in S_{X^{*}}$ such that $\langle x^{*},x\rangle=1$. In general we cannot choose $x^{*}$ so that it continuously depends on $x$ - consider $\ell_{\infty}$ or even its two-dimensional version $\ell_{\infty}^{2}$. Let's give ourselves a little more room in choosing $x^{*}$.

Prove: for any $\delta\in (0,1)$ there exists a norm-to-norm continuous map $\phi\colon S_{X}\to S_{X^{*}}$ such that $\langle \phi(x),x\rangle\ge \delta$ for all $x\in S_{X}$.

[hide="Hint"]The sets $U_{x^{*}}: =\{x\in S_{X}\colon \langle x^{*},x\rangle>\delta\}$, where $x^{*}$ ranges over $S_{X^{*}}$, form an open covering of the metric space $S_{X}$.[/hide]

Alas, this is not too much of a hint for me :(. If there were a locally finite continuous partition of unity associated with this covering, it would finish the story in no time but I do not see how to get such a partition in general. What is clear is that it would suffice to take $X=\ell^{\infty}(S)$ where $S$ is some set. Then the covering mlok mentioned can be reduced to the covering by the sets $U^{\pm}_{s}=\{x\,: \,\pm x_{s}>\delta\}$ but I do not even see how to make a desired partition of unity associated with this simpler covering even in the case when $S$ is countable. Seems like I'm missing something simple here  :maybe:  :blush: . Anyway, this was a very nice question to ask. Even if the solution for the "continuous" version turns out to be easy, one can then ask about the existence of a "uniformly continuous" or even "Lipschitz" mapping with these properties.

Isn't it true that for any open cover of a paracompact space there is a locally finite  continuous partition of unity subordinated to it?  :maybe: So says Proposition B.2 in the book I'm currently reading ("Geometric nonlinear functional analysis" by Benyamini and Lindenstrauss).

I've been working on the uniformly continuous case for a while. It's easy to prove that \[X\text{ is superreflexive}\implies \exists \phi\in UC(S_{X})\implies X\text{ has type }>1\] It's not easy to decide which converse fails (I might have a counterexample for the first one :maybe: ) Similarly,
\[X \text{ is 2-smoothable}\implies \exists \phi\in Lip(S_{X})\implies X\text{ has type }=2\] and this time I have no idea about either converse.

Yes, you're right. Stone's theorem... Seems I should refresh my memories of general topology  :oops:  :P.


geometry
function
integration
real analysis
calculus
calculus computations

I have just learned about calculating the area under the curve bounded by x-axis using Riemann Sum. And I'm stuck with the following problem:
Using Riemann Sum, calculating the area bounded by the curve and x-axis with x from 1 to 2: $ y\equal{}1/x^2$.
[b]Approach[/b]:
I used the Riemann Sum and need to calculate the limit of $ 1/n. sigma(1/(1\plus{}i/n)^2)$ with i from 1 to n. Probably it has something to do with the Riemann Zeta function, but I don't know how to calculate this limit.Wait! I used Squezee theorem and get the limit is 1? It should be 1/2 though! :(

$ U_n\equal{}\frac{1}{n}\sum_{k\equal{}0}^{n\minus{}1} f(1\plus{}\frac{k}{n}) \equal{}\frac{1}{n}\sum_{k\equal{}0}^{n\minus{}1}\frac{n^2}{(n\plus{}k)^2} \equal{}n\sum_{k\equal{}0}^{n\minus{}1} \frac{1}{(n\plus{}k)^2}\equal{} n\sum_{k\equal{}n}^{2n\minus{}1}k^{\minus{}2}$
$ \equal{}n[ \zeta(2,n)\minus{}\zeta(2,2n)] \equal{}\frac{n}{\Gamma (2)}\left[ \int_0^\infty \frac{t dt}{e^{nt}(1\minus{}e^{\minus{}t})}\minus{}\int_0^\infty \frac{t dt}{e^{2nt}(1\minus{}e^{\minus{}t})}\right]$

$ \equal{}n\int_0^\infty \frac{t}{1\minus{}e^{\minus{}t}}(e^{\minus{}nt}\minus{}e^{\minus{}2nt})dt$ .. which looks fun

but then again.. if we were trying to find a better way to integrate $ \frac{1}{x^2}$ we really shouldnt be looking at harder integrals.. should we?


[quote="nam_tran1237"]bạn nào có sách về dồn biến, pót lên cho minh với.
tôi cảm ơn nhiều  :roll:[/quote]
Bạn lên DDTH tìm bài của anh kimluan, có 1 file về dồn biến rất hay. Dồn biến có nhiều sách lắm, bạn có thể tham khảo STBDT của anh PKH hoặc SLVKP của thầy Thuận. Đây là pp dồn biến toàn miền và pp dồn biến mạnh :


For $ abc\equal{}1$, $ a, b, c>0$ prove the inequality:

$ \frac{a^{2}\plus{}b^{2}}{c^{2}\plus{}a\plus{}b}\plus{}\frac{b^{2}\plus{}c^{2}}{a^{2}\plus{}b\plus{}c}\plus{}\frac{c^{2}\plus{}a^{2}}{b^{2}\plus{}c\plus{}a} \geq 2$

Change $ a \equal{} x^{3}$, $ b \equal{} y^{3}$ and $ c \equal{} z^{3}$. Obviously, $ x,y,z > 0$ and $ xyz \equal{} 1$(I will use a,b,c instead of x,y,z and this a,b,c won't be the same as in the given condition). So we have to prove
    $ \sum \frac {a^{6} \plus{} b^{6}}{c^{6} \plus{} a^{4}bc \plus{} ab^{4}c} \geq 2$. If we use Cauchy-Shwarz inequality on fractions, we get that the LHS is more or equal to $ \frac {4(a^{6} \plus{} b^{6} \plus{} c^{6})^{2}}{2(\sum a^{6}b^{6}) \plus{} 2(\sum a^{10}bc) \plus{} 2(a^{7}b^{4}c \plus{} a^{4}b^{7}c \plus{} ab^{4}c^{7} \plus{} ab^{7}c^{4} \plus{} a^{7}bc^{4} \plus{} a^{4}bc^{7)}}$. So we need to prove that $ 2(\sum a^{12}) \plus{} 2(\sum a^{6}b^{6})\geq 2(\sum a^{10}bc) \plus{} (a^{7}b^{4}c \plus{} a^{4}b^{7}c \plus{} ab^{4}c^{7} \plus{} ab^{7}c^{4} \plus{} a^{7}bc^{4} \plus{} a^{4}bc^{7})$.
By AM_GM inequality, we get: $ a^{12} \plus{} 4a^{6}b^{6} \plus{} a^{6}c^{6} \geq 6a^{7}b^{4}c$. $ 4a^{12} \plus{} a^{6}b^{6} \plus{} a^{6}c^{6} \geq 6a^{10}bc$. Writing the symmetric inequalities and adding, we get the desired result.   Lasha Lakirbaia

I have used more powerful method (then Cauchy-Shwarz or AM_GM).

the solution could be finished in a slightly easier way, just as lasha did, to make it more convinient, we will consider $ a^3,b^3,c^3$ instead of  $ a,b,c$.

in each component of the sum, let's use the following inequality:

$ \frac{a^6\plus{}b^6}{c(c^5\plus{}a^4b\plus{}ab^4)}\geq\frac{\frac{a^6\plus{}b^6}{c}}{a^5\plus{}b^5\plus{}c^5}$

The rest is a consequence of the Muirherd inequality.


[quote="randomdragoon"]8 dice are rolled. What is the probability that all of the numbers 1-6 appear at least once?[/quote]
[hide]
$1-\frac{5}{6}^{8}$[/hide]?

(sorry, my 'g' key is broken, i have to copy and paste to make it work and i am too lazy for that)

So, i decided to think of each of the 6 numbers on the dice as slots and the 8 dice as balls. Whatever slot the ball drops in is the number the dice gets. This should wokr because each of the numbers appears with equal probability.

So it's the number of ways to have at least 1 ball in each slot over the number of ways to arrange the 8 balls randomly.

[hide]there are 28 ways to pick the individual dice for the rolls (123456xx, 12345x6x.. etc) and 6! ways to uniquely arrange the numbers 1 through 6 and a total number of 8^6 possabilities

therefore the answer is [b]6!(28)/8^6
8^2*3*5*3*7/8^6
315/8^4
315/4096[/hide][/b]

[quote="jli"]both of the solutions are incorrect.  bmps is just missing the parens  i think.  algosme, howd u get that???[/quote]
Ya, It was a typo. Kinda obvious that it's a typo, my current answer is over one :D
I meant
[hide]$1-(\frac{5}{6})^{8}$[/hide]
Why does everyone call me bmps?

It is bumpy.

I don't think that is right the answer, because, you are saying 1- same number each time.

That doesn't inclued stuff like 11111112

isn't it $\binom{7}{2}$ since 4 separators enerate only 5 slots?

[quote="randomdragoon"][hide="my solution"]
There are $\binom{6}{2}=15$ different combinations of dice rolls that satisfy the conditions (1,2,3,4,5,6 and then any two other rolls). There are $\binom{13}{5}=1287$ total combinations possible*. The probability is thus $\frac{15}{1287}= \frac{5}{429}$.

*This fact is far from obvious... but the argument is similar to the balls-and-walls argument found at the end of http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133029[/hide][/quote]

Would you explain how you got: $\binom{6}{2}=15$?
I think we need one more "wall"....

[quote="usaha"]i keep on getting 7/11664...[/quote]

great......

um......

were not gonna know what your mistake was if you dont tell us how you did it

[quote="usaha"]i keep on getting 7/11664...[/quote]

Your denominator seems too large. Did you mistakenly assume that the dice are distinguishable? Otherwise I don't know how you can get a number so high.

In this problem, what's important is how many dice are getting what number, not which dice gets which number.

So what exactly is the answer?

now i get 7/429 because its just (7C5)/(13C5)

Oh right, my bad, it should be $\binom{7}{2}$ in the numerator. :ewpu:

usaha, correct  :)


function
logarithms
calculus
calculus computations

Can someone give me some hint on how to solve for t?

$t = \omega^{t}$

I applied $ln$ to both sides but couldn't proceed.

It can't be solved completely in terms of the standard list of functions; the best we can do is something like $\frac{\ln t}{t}=\ln \omega$ or $t^{\frac1t}=\omega$. It isn't too hard to understand in this form.


1. The longest side of an isosceles right triangle has length $ \sqrt [3]{2}$.  Another side has length 2^{a}.  Find the value $ a$ expressed as a common fraction.
[hide]ANSWER: $ \frac { \minus{} 1}{6}$[/hide]

2. Twelve square tiles, on inch on a side, are arranged in a rectangle, without overlapping.  What is the number of inches in the sum of all possible distinct perimeters?
[hide]ANSWER: 56[/hide]

3. Which of these shapes will not tessellate the plane: an obtuse triangle, a trapezoid, a regular pentagon, or a regular hexagon?
[hide]ANSWER: Regular Pentagon[/hide]

Problem 3 is ambiguous. If I gave you an arbitrary trapezoid or an arbitary obtuse triangle, it would not necessarily tesselate the plane.

[hide="I think I have Solution for 1"]You can perceive that its an 45:45:90 triangle.
Therefore formula can work.$ \ (1: 1: \sqrt {2})$
Since $ \sqrt {2}$ can be written as $ \ 2^{\frac 12}$ and so can $ \sqrt [3]{2}$
Thus $ \ 2^{\frac12} * x \equal{} 2^{\frac 13}$
Then you simplify to $ \ x \equal{} \frac {2^{\frac 13 \minus{} \frac 12}}2$ which give you
$ \boxed {x \equal{} {\frac { \minus{} 1}6}}$[/hide]

[hide="I think I have Solution for 2"]You can decipher 12 tiles into 12 square units.
So, (1*12),(2*6),(4*3) are the pairs you can come up with.
Thus 2(13)+2(8)+2(7)= 26+16+14= [u][b]56 [/b][/u][/hide]

On question 3, what does tessellate mean??

Hope this helps.. :lol:


function
induction
trigonometry
algebra unsolved
algebra

Find all the continuous functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that:

$ f(x\plus{}y)\plus{}f(x\minus{}y)\equal{}2f(x)f(y)$, for all $ x,y \in \mathbb{R}$

[quote="moldovan"]Find all the continuous functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that:

$ f(x \plus{} y) \plus{} f(x \minus{} y) \equal{} 2f(x)f(y)$, for all $ x,y \in \mathbb{R}$[/quote]
Let $ P(x,y)$ the assertion $ f(x \plus{} y) \plus{} f(x \minus{} y) \equal{} 2f(x)f(y)$

$ P(0,0)$ $ \implies$ $ 2f(0) \equal{} f(0)^2$ and so either $ f(0) \equal{} 0$, either $ f(0) \equal{} 1$

$ f(0) \equal{} 0$ and $ P(x,0)$ $ \implies$ $ f(x) \equal{} 0$ $ \forall x$ and this is a first solution.

So we'll now consider $ f(0) \equal{} 1$.
$ P(0,x)$ $ \implies$ $ f(x) \plus{} f( \minus{} x) \equal{} 2f(x)$ and so $ f( \minus{} x) \equal{} f(x)$ and $ f(x)$ is even.
$ P(x,x)$ $ \implies$ $ f(2x) \equal{} 2f(x)^2 \minus{} 1$

Case 1 : $ \exists b$ such that $ f(b) < 1$.
======
Suppose now that $ \exists b$ such that $ f(b) < 1$. Since $ f( \minus{} x) \equal{} f(x)$, wlog consider $ b > 0$.
Obviously, the sequence $ u_0 \equal{} b < 1$, $ u_{n \plus{} 1} \equal{} 2u_n^2 \minus{} 1$ contains negative numbers and so (since $ f(0) \equal{} 1$ and $ f(x)$ is continuous), there exists $ c > 0$ such that $ f(c) \equal{} 0$. So the set $ Z \equal{} \{x > 0$ such that $ f(x) \equal{} 0\}$ has an infimum "$ a$" and $ f(a) \equal{} 0$ (since $ f(x)$ is continuous)

So we have $ f(0) \equal{} 1$, $ f(a) \equal{} 0$ and $ f(x) > 0$ $ \forall x\in[0,a)$
Then $ P(x \plus{} 2a,x)$ $ \implies$ $ f(x \plus{} 2a) \equal{} \minus{} f(x)$ and so $ f(x \plus{} 4a) \equal{} f(x)$ and so $ f(x)$ is a periodic function.

It's easy to show with induction over $ n$ (using $ P(\frac {a}{2^n},\frac {a}{2^n})$ and the fact that $ f(\frac {a}{2^n})\geq 0$ $ \forall n\geq 0$) that $ f(\frac {a}{2^n}) \equal{} \cos(\frac {\pi}{2^{n \plus{} 1}})$ $ \forall n\geq 0$

It's then easy to show with induction over $ p$ (using $ P(p\frac {a}{2^n},\frac {a}{2^n})$) that $ f(p\frac {a}{2^n}) \equal{} \cos(p\frac {\pi}{2^{n \plus{} 1}})$ $ \forall n\geq 0$ and $ \forall p\in[0,2^n]$

And, since the set $ \{p\frac {a}{2^n}$, $ \forall n\geq 0$ and $ \forall p\in[0,2^n]\}$ is dense in $ [0,a]$ and since $ f(x)$ is continuous :

$ f(x) \equal{} \cos(\frac {\pi x}{2a})$ $ \forall x\in[0,a]$
Now, since $ f(x \plus{} 2a) \equal{} \minus{} f(x)$, since $ f(x \plus{} 4a) \equal{} f(x)$ and since $ f( \minus{} x) \equal{} f(x)$, we have :

$ f(x) \equal{} \cos(\frac {\pi x}{2a})$ $ \forall x\in\mathbb R$

and it's easy to check that this function matches the original equation.

Case 2 : $ f(x)\geq 1$ $ \forall x\mathbb R$
======
$ f(1)\geq 1$ and so $ \exists u \geq 0$ such that $ f(1) \equal{} \cosh(u)$

It's easy to show with induction over $ n$ (using $ P(\frac {1}{2^n},\frac {1}{2^n})$, and the fact that $ f(\frac {1}{2^n})\geq 1$ $ \forall n\geq 0$) that $ f(\frac {1}{2^n}) \equal{} \cosh(\frac {u}{2^n})$ $ \forall n\geq 0$

It's easy to show with induction over $ n$ (using $ P(2^n,2^n)$, and the fact that $ f(2^n)\geq 1$ $ \forall n\geq 0$) that $ f(2^n) \equal{} \cosh(u2^n)$ $ \forall n\geq 0$

It's then easy to show with induction over $ p$ (using $ P(p2^k,2^k)$) that $ f(p2^k) \equal{} \cosh(pu2^k)$ $ \forall k\in\mathbb Z$ $ \forall p\in\mathbb N$

And, since the set $ \{p2^k$, $ \forall k\in\mathbb Z$ $ \forall p\in\mathbb N\}$ is dense in $ \mathbb R^ \plus{}$ and since $ f(x)$ is continuous :

$ f(x) \equal{} \cosh(ux)$ $ \forall x\in\mathbb R \plus{}$

And, since $ f(0) \equal{} 1$ and $ f( \minus{} x) \equal{} f(x)$, $ f(x) \equal{} \cosh(ux)$ $ \forall x\in\mathbb R$
and it's easy to check that this function matches the original equation.

Synthesis :
=========
All solutions are :
$ f(x) \equal{} 0$
$ f(x) \equal{} \cos(\alpha x)$
$ f(x) \equal{} \cosh(\alpha x)$

(and the solution $ f(x) \equal{} 1$ is obtained thru $ \alpha \equal{} 0$)


inequalities
vector
calculus
calculus computations

Consider the space ${\mathbf{R}}^{n}$(for $n\ge1$) with the three metrics:

1) $d_{1}((x_{1},.....,x_{n}),(y_{1},.....,y_{n}))=\sum_{i=1}^{n}{|x_{i}-y_{i}|}$

2)$d_{2}((x_{1},.....,x_{n}),(y_{1},.....,y_{n}))=\sqrt{\sum_{i=1}^{n}{(x_{i}-y_{i})^{2}}}$

3)$d_{\infty}((x_{1},.....,x_{n}),(y_{1},.....,y_{n}))= max_{i=1,2,...n}{\{|x_{i}-y_{i}|\}}$

a) Show that there is a constant $c>1$ such that for any two points $x,y$ $\in \mathbf{R}$ the following inequalities are satisfied: $\frac{1}{c}d_{1}(x,y)<d_{2}(x,y)<cd_{1}(x,y)$ and $\frac{1}{c}d_{1}(x,y)<d_{\infty}(x,y)<cd_{1}(x,y)$

b)Conclude that a sequence $(a_{k})_{k \in \mathbf{N}}\in{\mathbf{R}}^{n}$ converges with respect to any of the above metrics if and only if it converges with respect to any other.

I think part a) is easy .(I have solved it.)
Can you help me with this question.And please if you solve this problem and decide to post the solution explain your solution in a good way. :) (because I have started to solve this kind of questions from yesterday) :lol:

And can anyone give me link to similar questions about metrcis, or free books, so that I can train.?

Just a remark: if you want to avoid all the computation, you can just quote the fact that in a finite-dimensional real or complex vector space, all the norms are equivalents (and here, your distances all stem from norms).

From a pedogocial point of view, I would much rather do part (a) concretely, for these particular norms, before I ever mention the general theorem about norms on finite dimensional vector spaces.

(And you will see that the constant $c$ depends on the dimension $n,$ hinting that norms that look like these might not be equivalent on an infinite dimensional vector space.)

spider_boy: Part (b) will follow from part (a). Write out the full defnintion of the convergence of a sequence, with the metrics in there, and see how you could introduce one of the inequalities in (a) in an appropriate place to further the argument.


A section is cut out of a circular piece of paper having radius four inches, as shown. Points A and B are then glued together to form a right circular cone. What is the circumference of the base of the resulting cone? Express your answer in terms of $ \pi$.

[asy]import graph;
draw(Circle((0,0),42.4),linewidth(1));
draw((0,0)--(30,30),linewidth(2));
draw((0,0)--(30,-30),linewidth(2));
label("$A$",(30,30),E);
label("$B$",(30,-30),E);
label("4''",(15,15),NW);
draw((3,3)--(6,0));
draw((6,0)--(3,-3));[/asy]

Wow, there is actually a pic in this problem.

I am not sure about this problem, but i would find the circumference of the whole circle, which would be 8pi.  Then, I think you would have to find 3/4ths of that, so my guesstimation would be [b]6pi[/b].

Yes, ernine is correct. Basically, find major arc AB, which would be (360-90)/360 * 8pi, which is 6 pi.


integration
vector
real analysis
real analysis unsolved

Let $ \gamma : [a,b]\to \mathbb R^2$ be a $ C^1$ curve with nonzero derivative. Prove that
there exists a direction in the plane such that the direction of the tangent line to the curve coincides with that direction only in finite number of points.

So you assume that the curve is simple? Otherwise one can take $ \gamma(x)\equal{}\int_a^x g(t)\,dt$ where $ g$ is a continuous map from $ [a,b]$ onto $ \{(x,y)\in\mathbb R^2\colon 1\le x^2\plus{}y^2\le 4\}$.

I have put no assumptions other than what I stated on the problem. Can you explain about your choice of $ g$ and how do you know that it works?

$ g$ is a [url=http://en.wikipedia.org/wiki/Space-filling_curve]space-filling curve[/url]. By definition $ \gamma'(x)\equal{}g(x)$ for all $ x\in [a,b]$. For any unit vector $ \vec v$ there are uncountably many points $ x$ such that $ g(x)$ is a multiple of $ \vec v$, because $ g$ covers the segment $ \{\lambda \vec v\colon 1\le\lambda\le 2\}$.


Solve the equation \[ 2\sin\left(3x\plus{}\frac{\pi}{4}\right)\equal{}\sqrt{1\plus{}8\sin 2x\cos^2 2x}\]


limit
trigonometry
calculus
calculus computations

i want a solution without power series please. i'm not supposed to use this in my homework

And we are not supposed to do your homework :P. Anyway, the power series technique is the standard one to use here. Of course, you can do L'Hopital 5 times (it'll be a great mess but it will formally satisfy your demand for "no power series"). Otherwise I don't know: maybe there is some clever trick, but I do not see one.

yea, i thought there was some clever possible trick ... but i guess there isn't. power series are just fine then  :P thanks ..

Maybe this idea would work:

$\lim_{x\rightarrow 0}\frac{(\tan x-x)}{x^{3}}\lim_{x\rightarrow 0}\frac{\tan^{2}x+x\tan x+x^{2}}{x^{2}}+\lim \frac{x^{3}-\tan x^{3}}{x^{5}}=\frac{1}{3}\cdot 3+0=1$,

you calculate the first limit by using l'Hopital rule and the last limit is $0$ in view of the same rule.

[quote="didilica"]Maybe this idea would work:

$\lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}}\lim_{x\rightarrow 0}\frac{\tan^{2}x+x\tan x+x^{2}}{x^{2}}+\lim \frac{x^{3}-\tan x^{3}}{x^{5}}=\frac{1}{3}\cdot 3+0=1$,

you calculate the first limit by using l'Hopital rule and the last limit is $0$ in view of the same rule.[/quote]

hey... :o nice trick !


Hello! I've just been released after doing time... in the *insert unmentionable* army.
So here's my first contribution to this forum. (Finally!)

I came up with this question for one of our training sessions a few months back. No one solved it. i can only think of two reasons why this was been so. 1. 'Valiowk' was too busy explaining her math project to someone. AND/OR 2. I screwed up, and the question is wrong. I looked over it but it still looks correct. But maybe I'm blind or something.
So here it is:

A math contest was held in the planet Zorgland. Zorgland has 3 species – The Arv, Benz, and Cren. In this particular contest, 14 Arvs, 14 Benzs and 14 Crens took part. Each solved at most 4 questions, and every distinct triplet of Arv, Benz, and Cren solved at least 1 question in common. Show that there is a question which 3 Arvs, 3 Benzs, and 3 Crens solved.

Um... Any comments?
Even a 'Hey! Your so-called question is a cheap rip-off of IMO 2001/3!' would be nice.

I just knew someone would say that.
IMO 2001/3 v2.0, now in 3-D!


inequalities
quadratics
inequalities unsolved

:D Good day.!

This is the problem. I believe this has been used already as an Olympiad Problem. I just don't know whether this one was used as a Nat'l or an Int'l Problem... Here goes...

Help please.

For which m does the inequality mx^2 - 4mx + m^2 + 2m - 3 > 0 hold for all x?

Find the discriminant of the quadratic . What happens when it is negative ?

Also notice that its $mx^{2}$ and not $x^{2}$ .

$mx^{2}-4mx+m^{2}+2m-3=0 \Rightarrow m(x-2)^{2}+(m^{2}-2m-3) > 0$
i) $m \geq 0, m^{2}-2m-3 > 0 \Rightarrow m > 3$
ii) $m < 0$ , let $k =-m$, we have $k(x-2)^{2}< k^{2}+2k-3 \Rightarrow 0 < (x-2)^{2}< \frac{k^{2}+2k-3}{k}=-\frac{m^{2}-2m-3}{m}$ for all possible value of $x$, then
$\frac{m^{2}-2m-3}{m}\geq 0 \Rightarrow m <-1$, but $(x-2)^{2}$ can take any positive real value, so that for any $m <-1$, we can take $x$ sufficiently big to reverse the ineq..

Hence, the only possible value of $m$ is $m > 3$

I'm sorry if I wrote some mistakes  :maybe:

Thanks for the answer.  :D I really hope it's correct. Thanks guys.


[quote="Sinistra"]X    -3     -2      1      3

Y     1       2      5      7

Y = ????

Continue the equation[/quote]
[hide]
You add 4 to $x$ to get $y$ so $y=x+4$[/hide]

[hide]
i think if you add 4 to x you get y[/hide]

I don't get it. would someone explain it to me? how could you tell?

[hide]kBabe-Just compare the x's to the y's and a pattern is evident.

$x+4=y$[/hide]

whoa... I'm totally guessing, but are they x and y numbers suppose to line up? if they are then are you suppose to find the equation in slope-intercept form? I'm going to do the problem base on what i think it means.

[hide]to find the slope it's $\frac{difference      of       y}{difference      of       x}$ , if you random pick two numbers you can get $\frac{1}{1} = 1$ so the slope is $ 1$. Then to find the y-intercept, you should notice that the difference of x and y is always $4$... so if x is $0$ then y is $4$. so the equation of $y = x +4$
[/hide]
and i know I didn't really need to type all that... but I was kinda bored

I think the x and y he posted are supposed to be in a table.

[u] x    | y[/u]
-3 |1
-2 |2
 1    |5
 3    |7

Its a pretty bad table but i think this should help...

[hide]
-3+4=-1, -2+4=2...
So the answer is $\framebox{y=x+4}$
[/hide]


Which countries' olympiads/TSTs are of about IMO difficulty (harder is preferred, although a tad easier isn't too bad)?

USA, China, Russia, Japan, Romania TST's are definitely IMO level. a rule of thumb would be, if the country does consistently well at the IMO, then its TST's are comparable in difficulty to the IMO.

Russian Olympiad.

Here are a few problems solved by 0 or 1 Russian contestants. Some of the Russian contestants got 0 points in the following problems, but got perfect scores in IMO.

2002 grade 11 pr 4 --- 0
2002 grade 11 pr 7
2002 grade 11 pr 8
2005 grade 11 pr 8
2008 grade 10 pr 4 --- 0
2008 grade 11 pr 4 --- 0
2008 grade 11 pr 8

* 2005 grade 11 pr 8 was solved by only 1 contestant, Vasiliy Astakhov, who was the author of IMO 2007 pr 3.

I believe that China Olympiad is also very difficult, who knows maybe it's more difficult than IMO


The excircles of a triangle $ABC$ touch the sides $BC,CA, AB$ at $A',B',C'$ respectively. The circumcircles of triangles $AB'C',A'BC',A'B'C$ meet the circumcircle of triangle $ABC$ at $A_1,B_1,C_1$ respectively for the second time. Prove that the triangle $A_1B_1C_1$ is similar to triangle whose vertices are the points of contact of the incircle $ABC$ with  its sides.

We now that $BC'=CB'=P-a$
 $angleAC'A_1=angleAB'A_1$ =>$angleA_1C'B=angleA_1B'C$ and
 $angleABA_1=angleACA_1=angleB'CA_1$  => triangle $BC'A_1$
 is similiar to triangle  $B'CA_1$ but $BC'=CB'$
  => $1=BC'/CB'=A_1B/A_1C$ => $A_1B=A_1C$ similiar $B_1C=B_1A$  and $C_1A=C_1B$
                                           THE END :D
I  think do you finishing this solution :)

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=35313 .

  darij


geometry
3D geometry
analytic geometry
algebra
system of equations

These are from Vestavia 01

1. How many values of x satisfy the equation: log(x^2)=cos(x)

2. Find the value of x in the following system of equations:
3x+2y+z=3
3y+2q+w=2
3z+2w+x=1
3w+2x+y=0

3. Two circles have an external tangent with length 36 cm. The shortest distance between teh circles is 14 cm. Find the radius, in cm, of the larger circle if its radius is four times that of the smaller circle.

4. If A is the fourth perfect number, B is the least integer number greater than one which is both a cube and a square, and C ist he twelfth triangular number, find the value of sqr[(A/B)-C]

5. The graph of 2x^2-21xy+29x-62y+10y^2=0 is a(n)....

6. Find the volume of a hexahedron with vertices (1,0,5) (4,0,2) (0,0,0) (-1,-3,-1) (2,3,3)

7. Find the sum of the abscissas of all (x,y) coordinates which satisfy
2x^2+xy-y^2+3y-2=0 and 3x^2+3xy+y^2-6x-5y+4=0

8. If f(n) = n^8-2n^7+4n^6-8n^5+16n^4-32n^3+64n^2-128n+1 then what is f(1/128)?

9. Find the distance between teh point (5,1,3) and the line defined by the parametric equations
x=3, y=7, and z=1+t

10. Two ships leave port at the same time. One travels 20 mph at a bearing of N32degreesW. The second ship travels 30mph at a bearing of S28W. Find the distance in miles between the ships after 2 hours.

Sorry I didn't use LATEX.

These are all individual questions I am having trouble with.

Please post solution as soon as you know how to do one.


function
calculus
derivative
algebra
polynomial
Vieta
system of equations

$ f{(x)} \equal{} {x}^3 \minus{} 3{x}^2 \plus{} 2{x}$

The equation $ f{(x)} \equal{} k$ has exactly one positive solution and exactly one negative solution. Find $ k$.

Help needed  :D

[hide="hint"]
there must be a multiplicity since there is not a solution of 0[/hide]

dont know if that wil help though...
but there is a nice calc solution

like

$ ({x}^3 \minus{} 3{x}^2 \plus{} 2{x}) \equal{} (ax \minus{} k1)(bx \plus{} k2)$ ?

(I don't know how to do subscripts for the k1 k2 in latex).

It probably isn't as the coefficients don't work out.

it involves derivatives (might be too hard for intermediate forum)

but in regard to the multiplicity hint (multiplicity means a number is a double root of a polynomial-- it is a factor of the polynomial more than once) , i meant for you to think about $ (x \plus{} k_1)^2(x \plus{} k_2) \equal{} x^3 \minus{} 3x^2 \plus{} 2x \minus{} k$

then expand it and match coefficients and get equations and solve and find k
:D

[quote="JasonSullivan"]$ f{(x)} = {x}^3 - 3{x}^2 + 2{x}$

The equation $ f{(x)} = k$ has exactly one positive solution and exactly one negative solution. Find $ k$.

Help needed  :D[/quote]

First, $ k$ is not $ 0$ (we can check that because the roots of $ x(x - 1)(x - 2) = 0$ are $ \{0, 1, 2\}$).

Given that, we have an equation:
\[ x^3 - 3x^2 + 2x - k = 0
\]
We factor it with the three roots:
\[ (x - r_1)(x - r_2)(x - r_3) = 0
\]
Next, we know that since there are at least two real roots, all the $ r_i$ have to be real because if there were a complex root, the sum $ r_1 + r_2 + r_3$ would be complex, which violates Vieta's formulas because the coefficient of the $ x^2$ term is real.

So the last root either must be one of the original roots or 0. Plugging in $ x = 0$ gives $ x^3 - 3x^2 + 2x - k = 0 \implies - k = 0 \implies k = 0$.

Contradiction? Contradiction! So the last root cannot be 0. It follows that one of the roots must have a multiplicity of 2.

So we have two roots, and the equation turns into $ (x - r_1)^2(x - r_2) = 0$.
\[ (x - r_1)(x - r_1)(x - r_2) = x^3 + ( - r_2 - 2r_1)x^2 + (r_1^2 + 2r_1r_2)x - r_1^2r_2 = 0
\]
Guess what? We can now solve for $ r_1$ and $ r_2$ using this system of equations:
\begin{align*} - r_2 - 2r_1 & = - 3 \\
r_1^2 + 2r_1r_2 & = 2\end{align*}

And once we have those values (there will be two solutions), we choose the pair that has one negative and one positive and let $ k = r_1^2r_2$.

[color=darkred]Edit: I said let $ k = -r_1^2r_2$... actually it was supposed to be the positive of that, since we already had a $ - k$ in the original equation.[/color]

Thanks to you both. I know we are not supposed to post calculus here but I can do it and would appreciate the second take on the problem. Thanks.  :D

[hide="deadly calculus bomb"]
the derivative of $ x^3\minus{}3x^2\plus{}2x\minus{}k$ is $ 3x^2\minus{}6x\plus{}2$ for all k and we want to find where the derivative is zero, meaning where there MAY ba an extrema
so $ 3x^2\minus{}6x\plus{}2\equal{}0 \implies x\equal{}1\pm\frac{\sqrt3}{3}$
when $ x\equal{}1\minus{}\frac{\sqrt3}{3}$ the function is at its maximum, and if we substitute $ \minus{}f(x)$ for k, we well end up with 2 solutions, but they will both be positive
so we want the other extrema
so $ x\equal{}1\plus{}\frac{\sqrt3}{3},f(x)\equal{}2\plus{}\frac{10\sqrt3}{9} \implies k\equal{}\minus{}2\minus{}\frac{10\sqrt3}{9}$
[/hide]

why sub -f(x) for k when k=f(x)?

otherwise seems simpler than the other way  :D

[quote="stevenmeow"][hide="deadly calculus bomb"]
the derivative of $ x^3 - 3x^2 + 2x - k$ is $ 3x^2 - 6x + 2$ for all k and we want to find where the derivative is zero, meaning where there MAY ba an extrema
so $ 3x^2 - 6x + 2 = 0 \implies x = 1\pm\frac {\sqrt3}{3}$
when $ x = 1 - \frac {\sqrt3}{3}$ the function is at its maximum, and if we substitute $ - f(x)$ for k, we well end up with 2 solutions, but they will both be positive
so we want the other extrema
so $ x = 1 + \frac {\sqrt3}{3},f(x) = 2 + \frac {10\sqrt3}{9} \implies k = - 2 - \frac {10\sqrt3}{9}$
[/hide][/quote]

Umm... that doesn't work. Solving $ x^3 - 3x^2 + 2x = -2 - \frac {10\sqrt 3}9$ yields only one real solution, $ -.787576765619...$, and two complex solutions.

The answer can be retrieved from my post above. If we solve this system of equations: \begin{align*}-r_2 - 2r_1 &= -3 \\
r_1^2 + 2r_1r_2 &= 2\end{align*} We can substitute $ r_2 = 3 - 2r_1$ to get this equation: \[ r_1^2 + 2r_1(3 - r_1) = 2\] And solving for $ r_1$: \begin{align*}r_1^2 + 2r_1(3 - 2r_1) &= 2 \\
r_1^2 + 6r_1 - 4r_1^2 &= 2 \\
-3r_1^2 + 6r_1 - 2 &= 0 \\
3r_1^2 - 6r_1 + 2 &= 0\end{align*}
So $ r_1 \&= \frac {6 \pm \sqrt {36 - 4 \cdot 3 \cdot 2}}6 = \frac {3 \pm \sqrt 3}3$. Backsubstituting, this means that $ r_2 = 3 - 2r_1 = 3 - 2\left(\frac {3 \pm \sqrt 3}3\right) = 3 - 2 \mp \frac {2\sqrt 3}3 = \frac {3 \mp 2\sqrt 3}3$.

So our two possible pairs of solutions $ (r_1, r_2)$ are: \[ (r_1, r_2) = \left(\frac {3 + \sqrt 3}3, \frac {3 - 2\sqrt 3}3\right) \approx (1.58, -0.15) \\
(r_1, r_2) =  \left(\frac {3 - \sqrt 3}3, \frac {3 + 2\sqrt 3}3\right) \approx (0.42, 2.15)\] So it's the first combination we want, so letting $ k = r_1^2r_2$ yields $ k = -\frac {2\sqrt 3}9$.

We can check by plugging that value of $ k$ back into the original equation. I leave this up to you.


Okay, about the Königsberg bridge problem, I know that it is impossible to cross all seven bridges without "repeating" going over a bridge. I want to know why, I've searched on Google, but I don't really get it. I found two ways that you can find out that's it's impossible. Net Theory, and graphing. Could you explain one or both of them? If you want to see the problem I drew it (And if you're too lazy to look though that again to find the problem, you have to cross all bridges without going on one twice, you must only use the bridges to cross the water).

Theorem(I don't know if this theorem exists or not, but I know for a fact that this is true): You can completely draw a graph without taking your pencil off the paper if and only if the graph has 0 or 2 vertices with odd degrees.

In the Königsberg Bridge, 4 vertices have degrees of 3, 3, 3, and 5.

Since there are 4 vertices with odd degrees, it is impossible to cross all bridges without repeating any of them.

Application of this theorem was in somewhere between #34 and #36 on HMMT Guts Round...
Could've gotten it, but had no time... :(

[quote="math154"]Yes, that's true. :) If you have no odd vertices, you can start anywhere, and if you have two odd vertices, you have to start at one of them and end at another (why?).[/quote]

Blah, forgot to mention that. :|

We were trying to see which figures were traceable using a rule... apparently I wrote it wrong...  :blush: Then we're going to apply the rule to the Königsberg bridge tomorrow

Caveat: For no-odd vertex graphs, you can start anywhere, but you must end at the exact same point as where you started.

Out of curiousity, does this theorem have a name?

If so, what is it?

I'm not sure, but these are called [url=http://en.wikipedia.org/wiki/Eulerian_path]Eulerian paths[/url].

It's one of Euler's things... I think it's called Euler's tour, or Eulerian Circuit.

How do you know the DEGREES, did you use a protractor or something (3 is way too small for those though), and aren't there more than 4 vertices. ANd how did you come up with the theorems???

The degrees means the number of rays that touch that vertex right?
And there are exactly four vertexes. Look at the picture.  :)
And about the theorems, I don't know... >.> Anyways, the two theorems that I mentioned should be the same thing...

this problem was the very first graph problem ever solved

Well Euler found out why you couldn't go over the bridges which lead to the graphing, which he was the first to find out so this was/is the first problem that was solved by graphing....

Uh because I learned it?  *is getting off topic*


Let a,b,c>0 Prove that
$ \sum\frac{(4a\plus{}b\minus{}c)^2}{2a^2\plus{}(b\plus{}c)^2} \geq 8$

Do you have a proof for that? then please post it at http://www.mathlinks.ro/viewtopic.php?t=66493 .

  darij


The perimeter of an equilateral triangle exceeds the perimeter of a square by $1989$ cm. The length of each side of the triangle exceeds the length of each side of the square by $d$ cm. The square has perimeter greater than 0. How many positive integers are NOT possible values for $d$?

[hide]
The side length of the square is $s$.  The perimeter of the triangle is $3(s+d)$, and the perimeter of the square is $4s$.  We set up equations to solve for $d$ and $s$.

$3(s+d)-4s=1989$
$3s+3d-4s=1989$
$3d-s=1989$

Since the perimeter of the square is greater than zero, $3d$ must be greater than $1989$.

$3d-0>1989$
$3d>1989$
$d>663$

Otherwise $s$ would need to be a negative number, which is impossible.  So $d$ must be larger than $663$, which means that $663-1+1$ = $663$ values do not work.[/hide]

ubemaya, i remember seeing this question before, it's one of the mid 20's on the 1989 ahsme i believe.


A positive integer is equal to the sum of the squares of its four smallest divisors. What is the largest prime that divides this integer. A solution, not just answer please.

Thanks.

Anyone? The answer is

[hide="answer"]$13$[/hide]

but I don't have a solution.

A really cheap solution would be to assume the number is small and brute force  :D

That's what I would have done during the competition (because this is a free response answer and an integer between 1 to 999). Believe it or not, this is from the Junior Division so brute force it probably the way. But I still want a solution.

Junior section of what? 1-999 sounds like the AIME.

Nevertheless, I think I've gotten the solution. In my solution, I would be determining this number which is the sum of the squares of its four smallest positive divisors, hence finding its largest prime divisor.

First, we have 1 as the smallest divisor of the number (I'd call it x).

x cannot be odd, or all of its 4 smallest divisors would be odd, and the sum of their squares even.

x must thus be even, and its second smallest divisor must be 2.

Now we check if x is a multiple of 4. If x is a multiple of 4, the 4 smallest divisors must be 1, 2, 4, p where p is any odd prime. It cannot be 1, 2, 4, 8 as the sum of the squares of these 4 numbers is odd. Now we have 21 + p^2 = 4pk, where k is the product of the rest of the factors of x. p must thus be 3 or 7 (since p has to be a factor of 21, and must be one of the 4 smallest divisors). By checking these two cases, we realise that x cannot be a multiple of 4 since 21 + p^2 would not be divisible by 4 in either case.

So now we have 1, 2 and some q as the 3 smallest divisors of x (where q is an odd prime). If the 4th smallest divisor is also odd, we would have an odd sum of the squares of the 4 smallest divisors. Hence, the 4th smallest divisor has to be 2q. Thus, we have 5 + 5q^2 = 2qy, where y is the product of the rest of the factors of x. Since 5 clearly divides x now, p has to be 3 or 5. Testing 3, we realise that it would not work. Therefore, q = 5 and x = 130.

The largest prime that divides 130 is 13, and therefore we are done.


function
algebra
polynomial
limit
calculus
real analysis
real analysis solved

I think it's 1, but other might claim that it's 0. The reason I think it's 1:
The function [tex]f(z)=z^0[/tex] is continuous (in fact smooth) everywhere in [tex]\mathbb{C}\setminus\{0\}[/tex], and for any sequence [tex]c_n\rightarrow0, f(c_n)\rightarrow1[/tex].

I am not very sure about the behavior of [tex]f(z)=0^z[/tex] near [tex]z=0[/tex]...seems to me that there might be some sort of branch cut on the negative real axis?

Anybody?

$0^0$ is undefined.

You can also look at $F(z)=z^z$ and you will "conclude" that $0^0=1$ ;)

$0^0$ is undefined.

In the power series or polynomial expression $\sum_na_nx^n$ we understand that $x^0=1$ for all $x,$ including $x=0.$ This is merely a notational convention and is not meant to imply anything about $0^0.$

In $\lim_{x\to a}f(x)^{g(x)},$ where $\lim f(x)=0^+$ and $\lim g(x)=0$, the overall limit could be any positive number or nonexistent; we could construct examples to do anything we want. The calculus shorthand phrase for this is "$0^0$ is indeterminate."

I, and many authors, define $0^0$ to be 1.  The key reason is the one Kent mentioned -- it makes dealing with polynomials and power series easier.  I don't see why it's good to say $x^0$ is 1 when $x = 0$, but bad to say $0^0 = 1$ -- can't they both be viewed as notational conveniences?  To me, the calculus argument merely shows $x^y$ is discontinuous at (0, 0) no matter what we do.  But is continuity the final word?  I figure it's better to be continuous in one variable than neither.

Anyway, I don't want to get into an argument about it.  I agree it's a matter of definitional convenience.  We had a long discussion about it in a topic on a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11999]Limit Problem[/url].


A magic triangle is a triangle with distinct numbers on each side, where the sum of the numbers on each side are the same.  For example, [code]  5
 4 2
1 6 3[/code] is a magic triangle, since each side adds up to 10, and no numbers are repeated.

Proove that it is impossible to have a magic triangle with two numbers on each side.

Are you asking to prove that a triangle which looks like:
[code]
      a
    b  c
[/code]
Cannot exist?
[hide]
I'm kinda new to the whole proof business, but I'll give it a shot:

Say that there is such a triangle that exists:
[code]
      a
    b  c
[/code]
a+c would have to equal a+b, b+a would have to equal b+c, and c+a would have to equal c+b.  Every number is distinct, so it is not possible:
If a+c = a+b, by algebra:
a+c = a+b
a-a+c=b
c=b
Therefore all of the numbers cannot be distinct and it is impossible.
[/hide]

Well,

[hide]

a + b = a + c = b + c

a + b = a + c

b = c

Except the numbers must be distinct, so...[/hide]


limit
logarithms
probability
induction
calculus
calculus computations

(b) Is $n$ approaches 0? I think $n$ should be infinity, if so, the answer is $\frac{4}{e}$.

Yes, of course it should be as $n$ goes to $\infty.$

And this is just an application of this classical lemma :
if $u$ is a sequence of positive reals, and if the sequence $\left(\frac{u_{n+1}}{u_{n}}\right)$ converges towards some real $l,$ then the same goes for $(\sqrt[n]{u_{n}}).$

By the way, [b]kunny[/b], for (b), this yields that the limit is actually $4.$

But since you wanted to see $e,$ let's add one more exercise :
compute the limit of $\frac n{\sqrt[n]{n!}}.$

First problem : obviusly. ( by Taylor)
Seccond problem : Stirnling forever   :wink:

[quote="jmerry"]Stirling's formula? Overkill.

Look at the probability distribution of flipping $2n$ fair coins and counting the number of heads. The probability $p$ of exactly $n$ heads is $2^{-2n}\binom{2n}{n}$. Clearly, $p\le 1$. Also, $n$ heads are the most likely result, so $p\ge\frac1{2n+1}$.
Taking $n$th roots, $(2n+1)^{-1/n}\le p^{1/n}=\frac14\binom{2n}{n}^{1/n}\le 1$ and $\binom{2n}{n}^{1/n}\to 4$.[/quote]
Yes you are right. I agree with you. but
peoples who more work with stirling formula often, for them this problem is obviusly:
By stirling we only must consider main expression:
$(\frac{(\frac{2n}{e})^{2n}}{(\frac{n}{e})^{2n}})^{\frac{1}{n}}=4$

for part (b), you can also prove by induction that
\[\frac{4^{n}}{2\sqrt{n}}\leq \frac{(2n)!}{(n!)^{2}}\leq \frac{4^{n}}{\sqrt{3n+1}}\]
and then use $\lim_{n\rightarrow \infty}\sqrt[n]{n}=1$. This can be done with logarithms+L'Hopital, or difference of powers+AM-GM in the form
\[\sqrt[n]{n}-1=\frac{n-1}{\sum_{k=0}^{n-1}\sqrt[n]{n^{k}}}\leq \frac{n-1}{1+(n-1)\sqrt{n}}\]

For part a):
Let $h = \frac{2n}{1-n}$ and note that:
$\lim_{n \rightarrow 0}(\frac{1+n}{1-n})^{\frac{1}{n}}$
$= \lim_{n \rightarrow 0}(1+\frac{2n}{1-n})^{\frac{1}{n}}$
$= \lim_{h \rightarrow 0}(1+h)^{\frac{2+h}{h}}$
$= \lim_{h \rightarrow 0}(1+h)^{\frac{2}{h}}\times \lim_{h \rightarrow 0}(1+h)$
$= [\lim_{h \rightarrow 0}(1+h)^{\frac{1}{h}}]^{2}= e^{2}$

[quote="Mathmanman"]But since you wanted to see $e$ let's add one more exercise: compute the limit of $\frac{n}{\sqrt[n]{n!}}$[/quote]

$L = \lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}= \lim_{n\to\infty}\sqrt[n]{\frac{n^{n}}{n!}}= \lim_{n\to\infty}\frac{(n+1)^{n+1}n!}{(n+1)n!n^{n}}$
$= \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}= e$.


search
inequalities proposed
inequalities

Prove that $ (x^{2}\plus{}2)(y^{2}\plus{}2)(z^{2}\plus{}2)\geq 9(xy\plus{}yz\plus{}zx)$ for every $ x,y,z\in R^{\plus{}}$.

posted so many times...
http://www.mathlinks.ro/viewtopic.php?search_id=1615461828&t=82832


integration
calculus
derivative
analytic geometry
function
calculus computations

The lines not through the origin satisfy $ ax\plus{}by\equal{}1$ implying $ rf(\theta)\equal{}1$ for some function $ f$ such that $ f^{\prime \prime}(\theta)\equal{}\minus{}f(\theta)$. Now we have
\[ f(\theta)\equal{}\frac{1}{r},f^\prime (\theta)\equal{}\minus{}\frac{r^\prime}{r^2},\]
\[ f^{\prime \prime}(\theta)\equal{}\frac{2(r^\prime)^2}{r^3}\minus{}\frac{r^{\prime \prime}}{r^2}\]
Setting $ f^{\prime \prime}(\theta)$ equal to $ \minus{}f(\theta)$ gives
\[ r^{\prime \prime}r\equal{}2(r^\prime)^2\plus{}r^2\]

[quote="stealth"][b]1. The problem statement, all variables and given/known data[/b]

Problem from Arnold's "Mathematical Methods of Classical Mechanics" on page 59.

Find the differential equation for the family of all straight lines in the plane in polar coordinates.

[b]2. Relevant equations[/b]

$ \Phi \equal{} \displaystyle\int^{t_2}_{t_1} \sqrt {{\dot{r}}^2 \plus{} r^2{}\dot{\phi}^2}\,dt$

$ \frac {d}{dt}\frac {\partial{L}}{\partial{\dot{q_i}}} \minus{} \frac {\partial{L}}{\partial{q_i}} \equal{} 0$

[b]3. The attempt at a solution[/b]

L is the integrand.

We have two equations:

$ \frac {d}{dt}\frac {\dot{r}}{\sqrt {{\dot{r}}^2 \plus{} r^2{}\dot{\phi}^2}} \equal{} \frac {r\dot{\phi}^2}{\sqrt {{\dot{r}}^2 \plus{} r^2{}\dot{\phi}^2}}$

$ \frac {d}{dt}\frac {r^2\dot{\phi}}{\sqrt {{\dot{r}}^2 \plus{} r^2{}\dot{\phi}^2}} \equal{} 0$

The second gives: $ \frac {r^2\dot{\phi}}{\sqrt {{\dot{r}}^2 \plus{} r^2{}\dot{\phi}^2}} \equal{} c$

If c=0 we have the derivative of phi zero, that is phi would be constant and we are (essentially) done. If phi is constant, we have a bundle of lines passing through the origin. But what about r? Do we get problems in the origin? Polar coordinates are not defined there, are they?
If c is not zero, then the first equation can be rewritten as:
$ \frac {d}{dt}\frac {\dot{r}}{r^2\dot{\phi}} \equal{} \frac {\dot{\phi}}{r}$

I cannot get an idea of how to solve it for r-dot AND phi-dot. Tried to differentiate the left side and see what happens, but somehow nothing attractive comes out.
So how to proceed? Is it the right way? Or is there anything I can't see at a glance that helps?[/quote]

You've missed the point of the problem.

You do indeed have two differential equations, but the point is not to solve them necessarily but rather to expand them out and see what you get.

You have,
(1)   $ \frac{d}{dt}\frac{\partial L}{\partial \dot{r}} \minus{}\frac{\partial L}{\partial r} \equal{} 0$

(2)   $ \frac{d}{dt}\frac{\partial L}{\partial \dot{\phi}} \minus{}\frac{\partial L}{\partial \phi} \equal{} 0$

Actually do the indicated differentiations and leave the resulting expression equal to zero on their right hand sides. What you wind up with are homogeneous equations of motion.


Is there a special formula for problems like these?

1. There are 24 diagonals in a polygon. How many sides does this polygon have?

2. How many diagonals are in a polygon with 24 sides?

I just want to know, because whenever I encounter problems like these, it takes me a while to figure them out, while on AOPS FTW, I noticed, people solve these in a very short time (not memmed).

HELP PLEASE!!  :D

To find the number of diagonals in an x sided figure the formula is:
$ \frac{x(x\minus{}3)}{2}$

This is because in any figure for each point there can be 3 points to which a diagonal cannot be drawn. (the point itself and the adjacent points). And since you are counting each diagonal 2 times you must divide by 2

That's what I thought, too, but then for his question: How many sides are in a polygon with 24 diagonals, you don't get a whole number answer.

If a convex polygon has $ n$ sides, then it has $ \frac{n(n\minus{}3)}{2}$ diagonals.  (Like what SuerNerd said).  Another formula is also $ \binom{n}{2}\minus{}n$.  This is because you can chose $ 2$ points to draw a line through, but $ n$ of them will be sides, not diagonals.   For me, each take about the same time, so it's just your preference...
[hide="1)"]
There are no convex polygons with $ 24$ diagonals...
[/hide]
[hide="2)"]
We just enter $ 24$ for $ n$ in our equations and solve...
$ \frac{24\cdot21}{2}\equal{}12\cdot21\equal{}\boxed{252}$
OR
$ \binom{24}{2}\minus{}24\equal{}\frac{24\cdot23}{2}\minus{}24\equal{}276\minus{}24\equal{}\boxed{252}$
[/hide]

Ok so you have 24 diagonals. There is a fourmula for this. It is x(x-3)/2
so just plug the number 24 where the x goes. so now you have 24(21)/2
It has 252 sides. :)
Reventon

In that formula x stands for the number of sides not the number of diagonals... :)
So you would have to make an equation like this

$ \frac{x(x\minus{}3)}{2} \equal{} 24$

$ x(x\minus{}3) \equal{} 48$

We can see that no number qualifies...

Yeah, there are no polygons with 24 diagonals... :ninja:

actually, they could not me whole number sides.

I've never seen a polygon with a noninteger number of sides. Perhaps you can draw one for me, say a polygon with 4.5 sides?  :P

well ya good point izzy. :mad:   :(   :D

[asy]draw((0,0)--(1,0)--(1,1)--(0,1)--(1,0)--(0,-1)--(0,0));[/asy]

Draw all diagonals and tell me how many there are. Then plug it back into your fancy formulas.

I believe it has been established that the formula applies for only convex polygons.  I also believe that figure is neither convex nor a polygon.

I actually made a post about this on my wiki, and on my blog:

http://www.artofproblemsolving.com/Wiki/index.php/User:Dojo#Diagonal_Forumla

http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=273655


If you don't know how to do it, wouldn't that make it hard for [i]you[/i]?  Thus why do you call it 'easy'?  Also, the problem doesn't have to do with trig.

[hide="Solution"]
\begin{eqnarray}
F(x)+F\left(\frac 1{1-x}\right) &=& x\\
F\left(\frac 1{1-x}\right)+F\left(\frac{x-1}x\right) &=& \frac 1{1-x}\\
F\left(\frac{x-1}x\right)+F(x) &=& \frac{x-1}x
\end{eqnarray}

\[(1)-(2)+(3)=2F(x)=x+\frac 1{1-x}-\frac{x-1}x\Rightarrow F(x)=\frac{x^3-x+1}{2x(x-1)}. \ \ \boxed{C}\] [/hide]

BTW this problem (without options of course) was probably one of the easiest german pre-TST problems

This is a Pre-Olympiad Problem???? :o

This should be in Getting Started!!!! :rotfl:  :rotfl:  :rotfl:


geometry
geometric transformation
reflection
circumcircle
homothety
Gauss
perpendicular bisector

For a triangle ABC, let I be the incenter. Let the reflection (line) of BC in AI be L.
When midpoint of BC is M and tangent line of nine point circle at M is K,
prove that L is parallel to K.

I guess it is well known, isn't it ? afraid of its not being nice. Thank you for reading.

I'm not sure whether I saw this before, but anyway, here's how to do this:

Both lines are parallel to the tangent line to the circumcircle at the antipode of $ A$. Of course the tangent to the NPC is, because it is mapped to it by a homothecy with center $ H$ and factor $ 2$. The other line is too, because it is also perpendicular to $ AO$, since $ BC$ is perpendicular to $ AH$, the reflected line has to be perpendicular to $ AO$ because $ O$ and $ H$ are isogonal conjugates.

Note that $ O$ and $ H$ are the circumcenter and orthocenter of $ ABC$.

Here is my solution
Call P is midpoint of PO. By Gauss line theory, we have M, N, J is conlinear. Similar TST USA 2000, we get EF is perpendicular to MN, in the other hand, JE = JF. Hence, JMN is perpendicular bisector of MN……..


I can't for the life of me figure out how to do this:

0.6x-10=1.4x-14

the answer is 5 but I don't know how to get there. Please show your work.

Also -3(2x-1)<-4[2+3(x+2)]

I just can't remember the order for brackets and para theses.

Please show work. or at least show me the second half.

 :idea:

For the first one:

$ .6x\minus{}10\equal{}1.4x\minus{}14\Rightarrow 4\equal{}.8x\Rightarrow x\equal{}5$

All I did was subtract $ .6x$ from each side and add 14 to each side.

hello, we want to solve the equation
$ 0.6x\minus{}10\equal{}1.4x\minus{}14 |\minus{}0.6x$
$ \minus{}10\equal{}0.8x\minus{}14|\plus{}14$
$ 4\equal{}0.8x|: 0.8$
$ x\equal{}5$
Sonnhard.

I still don't get it.
How can 0.8x divided by 4 equal 5?

hello, given the following inequation
$ \minus{}3(2x\minus{}1)<4\left(2\plus{}3(x\plus{}2)\right)$
We have
$ \minus{}6x\plus{}3<8\plus{}12(x\plus{}2)$
$ \minus{}6x\plus{}3<8\plus{}12x\plus{}24 |\plus{}6x$
$ 3<8\plus{}24\plus{}18x|\minus{}32$
$ \minus{}29<18x|: 18$
$ \minus{}\frac{29}{18}<x$.
Sonnhard.

hello, no it must be $ 4: 0.8\equal{}5$.
Sonnhard.

It isn't .8 divided by 4, it's 4 divided by .8 that equals 5.

And for your second one:

$ \minus{}3(2x\minus{}1)<\minus{}4[2\plus{}3(x\plus{}2)]$

$ \minus{}6x\plus{}3<\minus{}4(2\plus{}3x\plus{}6)$

$ \minus{}6x\plus{}3<\minus{}8\minus{}12x\minus{}24$

$ \minus{}6x\plus{}3<\minus{}12x\minus{}32$

$ 6x<\minus{}35$

$ x<\minus{}\frac{35}{6}$

OOOOh,

Now that makes sense fishy.
Thanks a bunch.
Thanks to everyone for helping me.

PS.
Fishy could you do all of the first one too? The way you translate the problem makes sense to me.

Okay.

$ .6x\minus{}10\equal{}1.4x\minus{}14$

$ 4\equal{}.8x$ (I just added 14 and subtracted .6x from both sides)

$ 5\equal{}x$ (Divide both sides by .8)

So that's what I meant when I said that the division is by .8, not by 4.

I get the bracket distribution now but still having problems with decimals.

y+0.2=0.6(y+3)

The answer is 4. Please show work. THANKS

Okay, so first we distribute on the right side.

$ y \plus{} 0.2 \equal{} (0.6)(y) \plus{} (0.6)(3)$

$ y \plus{} 0.2 \equal{} 0.6y \plus{} 1.8$

Now bring the $ y$ terms on the left side and all the constants on the right side, like this.

$ y \minus{} 0.6y \plus{} 0.2 \equal{} 0.6y \minus{} 0.6y \plus{} 1.8$

$ 0.4y \plus{} 0.2 \minus{} 0.2 \equal{} 1.8 \minus{} 0.2$

$ 0.4y \equal{} 1.6$

Divide both sides by $ 0.4$ to get

$ y \equal{} \boxed{4}$

I was totally forgetting that the y on one side had a 1 attached to it! Duh!
Ah, Thank you so much :)

If that really seems to be a problem, don't be afraid to write 1 as a coefficient, even if other people don't. It's still correct. That might help you remember that there is still a 1 "attached" even if you don't normally see it.