Hello,

I am interested in getting better at problem solving, but I am in a strange spot.  I am a senior math major about to graduate from my university.  I have taken all the basic undergraduate math courses (except for Abstract Algebra which I take in the spring).  However, my problem solving skills are not where I would like them.  I'm a little too old for competitions now (and my school doesn't offer the Putnam -- which I'd score a 0 on anyway), but if I become a teacher, I want to make sure that I give students the opportunity to excel in math -- something I never had.  So in order to teach, I have to increase my skills to a much higher level.  I am going to graduate school, but I only see that as resulting in my skills in a very specific high level area becoming more advanced.

[b]In a nutshell: what books should I get that will help me advance my problem solving skills?  It feels like everything is too trivial(basic algebra/geometry/number theory), or too advanced (Putnam level -- I rarely understand the solutions much less the problems) for me.  If you want a more detailed overview of my background, you can read below.[/b]

I started out clearly gifted, but I did not think anything of it, nor did my teachers.  In elementary school, when we were doing the "Addition/subtraction/etc... timed tests" I was completing 200-300 in the time everyone else did 100 (100 per sheet.  I remember doing it in about a minute, and when the teacher saw me w/ my head on the desk for the next 2 minutes, rather than giving me anything interesting, she just gave me more sheets).

In middle school, I did the standard Algebra I in 7th grade, but Geometry was not offered, so I took the same Algebra course again.  In 6th grade, me and my friend got #1 and #2 in the regional Math competition, but there was no after school program to help develop our skills, and we just learned the basic curriculum.  We would go again every year and perform worse and worse as the schools w/ better programs left us behind.

In high school, I was in the IB program and did the standard Higher Level math courses.  I learned the material, but did not develop any advanced problem solving skills besides being able to accomplish the basic Calc/Trig/Geometry expected of us.  By this time, I had lost the view that I was 'talented in math.'  Once a year we would, the 'top students ' in the IB Math class would go to some competition, but there was never any preparation; we did mediocre at best.  I saw the people at the competitions and simply thought they were much more gifted than I was.  I didn't realize that there were so many more techniques, theorems, and topics that other people were learning in high school -- techniques that you don't learn in college because you're too focused on learning differential equations, analysis, and so on.

I had never heard of the Putnam until a few months ago (I'm a 5th year senior math major this year).  No one in my classes and none of my profs have mentioned it.  I took a look at it, and I was amazed at the difficulty of the problems.  In almost all the textbooks I have used, even the most difficult were trivial for me -- something that lead me to the belief I was adequate or even above average.  However, I feel like my skills are several levels below ones needed for that test, and I would like to know how I should catch up.

P.S. No clue what forum to post this in, so move if needed.

You could be any of a hundred people I know - there is nothing in the slightest bit unusual about your experiences. You didn't say what level you would be teaching at, but I'm not sure it matters. Your interest in the problem-solving side does set you apart, and I commend you. Learning the difference between an "exercise" and a "problem" is worthwhile at any stage. (That particular distinction is explained in Paul Zeitz's [i]Art and Craft of Problem Solving[/i].)

My personal advice is to back up to the high school level. Read some of the [i]Art of Problem Solving[/i] books. Work through old AMC-12's and AIME's. Go hang out in the high school "Intermediate" and "Pre-Olympiad" forums on this site. There are vast riches at that level. Eventually you can come back to the college level and read Zeitz and Larson and Engel and look at Putnam and IMC problems - but don't start there.


Let $a_1, a_2, a_3, ..., a_n$ represent an arbitrary arrangement of the numbers $1, 2, 3, ..., n$.  Prove that, if $n$ is odd, the product:
\[ (a_1-1)(a_2-2)(a_3-3)...(a_n-n) \]
is an even number.

(From Eotvos competition 1906)

[hide]The product is odd only if all of the factors are odd. A factor $(a_i-i)$ is odd only if $a_i$ and $i$ are of opposite parities. Assume that the product can be odd if $n$ is odd. We need to pair every integer from $1$ to $n$ with another integer of the opposite parity from $1$ to $n$, and this means there are the same number of odd numbers as even numbers among the integers from $1$ to $n$. But $n$ is odd, so there is one more odd integer than even integer. Contradiction. Therefore the product is not odd, so it is even.[/hide]


[quote="shobber"][quote="djimenez"][quote="nano"]Hi, I need some advice on choosing a rigorous and challenging text on introductory algebra and geometry. I've done some browsing and found these names Lial/Miller, Foerster, Dolciani, Hall and Knight, C.V. Durell, and Weeks and Adkins. If anyone has opinions on these texts, preferences, level of difficulty, or knows of other texts worth looking at, please let me know.

Thank you[/quote]

I have given the same recomendation in the past. Earl Swokowski, Algebra and Trigonometry with Analytic Geometry, and it has to be an older version (prior 1996), the new ones are WAY too easy, but the old ones were great!

Best,[/quote]
can i find them online??[/quote]

Probably you can buy them online, but probably they are not going to be easy to find in the web in electronic format. You may try to find them on P2P networks, or may be in some IRC channels, but that is not exactly legal, at least not in the US!


Regarding college essays, how important is it that the events talked about in the essay are fairly recent?  I'm thinking about using an essay that talks about a specific event (involving myself) that occurred in kindergarten, and how it relates and is relevant to me now.  I clearly link the event to the present, but I am worried that admissions officers will frown upon the fact that the event occurred so long ago.  What do you think?

The date of the event is totally irrelevant.  All that matters is how well you say what you want to say about it.


function
vector
advanced fields
advanced fields unsolved

Let $X$ be a topological vector space and $f\colon X\to\mathbb R$ be a convex function. Prove that if for a certain $x_0\in X$ there is its convex open neighborhood in which $f$ is upper bounded, then this property is fulfiled for any $x\in X$.

[b]P.S.[/b] If you do not want $X$ to be topological vector space, assume it to be normed vector space for example...

Let $U_{0}$ be a convex open neighborhood of $x_{0}$ in which $f$ is bounded from above by $M$, say. Next, consider an arbitrary point $x$ in $X$, and let $y=2x-x_{0}$. The set $U=\left\{\frac{y+u_{0}}2\ |\ u_{0}\in U_{0}\right\}$ is easily seen to be a convex open neighborhood of $x$. Moreover, for every $u=\frac{y+u_{0}}2\in U$ we have $f(u)\le\frac{f(y)+f(u_{0})}2\le\frac{f(y)+M}2$, so $f|_{U}$ is bounded from above by $\frac{f(y)+M}2$.


Are you guys preparing for AIEEE? If yes, then how are you preparing.

I am not motivated at all do any serious study. Can anyone motivate me. :p

i m givin narayana test series thats all

in no mood to study new things

i think u guys shud study modern phy(semiconductors)

it wasn't there in JEE syllabus as far as i remember :oops:

[quote="Conjurer"]I am not motivated at all do any serious study[/quote]

No one is... :(

^ Are you sure out of the supposedly 10 lakh , noone is?

I mean amongst my friend circle. But I can bet good money that not all of the 100k are very enthusiastic about the exam.

certainly

ppl like me are not even going to give the exam (though i filled the form) :P

What will the marking scheme be this year? I suppose $ \plus{}3\minus{}0.75$? The AIEEE folks are reducing the max. marks each year. From $ 360$ to $ 315$ and now this time it'd be $ 300$. A few years later... :rotfl:

its a kind of exponential decay
so it will take infinite time to come to zero

no problems as far as $ \infty$ is not defined!!!!

me too,i was finally pushed by my father to give  the exam :o

yes
that was the first thing i tht on seeing the paper...

@Rituraj: Good. You must've enjoyed the exam. We were given papers 10min late but I managed to accelerate. They copied the JEE pattern. 'd say, if they were so keen on copying the pattern, also take a look at the number of pages for rough work they supply! You should see the "Space for Rough Work" at the bottom of each page! All you can draw there is the X-Axis  and each page was divided in two! I suppose the board is so financially starved that they had to save pages by adopting such tricks.

well the most trblsome thing 4 me is the thumb impression :mad:

'coz i m a lefty

Wait? You mean the AIEEE guys expect the students writing the exam to be angoothachaap?  :o

well actually most xams expect us to be angoothachaps :P

in fact we had 2 do it twice in JEE--the "left hand" thumb impression :rotfl:

[quote]
But why only left hand?
[/quote]

bcoz left and hand are together in indian government  :rotfl:

Plus they are all [i]angoothachaaps[/i]...

[quote="Sagar_Vaze"]bcoz left and hand [b]are[/b] together in indian government  :rotfl:[/quote]

were  :rotfl:  :rotfl:

gud 1

what is Hand...another party...idk :oops:

parasite???how can u kno much abt them??
i thought i was a bio student.... :maybe: .

 :rotfl:

hey PM me ur no. then
i may hav doubts :rotfl:

[hide]joking[/hide]

Don't waste money on study.

[hide]Instead, call the Axe Effect number...[/hide]

hey
i dont need that

i already have :rotfl:

rituraj's sig abt iit is applicable to axe too  :rotfl:

No I get calls from that number you see  :maybe:

Really? :huh:

[hide]I think their call centre girls (you see, placing that word "centre" there is crucial or the meaning changes entirely) quit so they forward all calls to you...[/hide]

I thought you were very innocent.
I do not want to incur the sin of corrupting your mind  :lol:

[quote="nuclear_alchemist"][quote="Sagar_Vaze"]I get calls inquiring abt some girls from some son of a b[hide]io teacher :rotfl: [/hide][/quote]

Looks AIEEE is really sharpening our minds!  :rotfl:[/quote]

gud for u.
a competitor was knocked off from the race for the coveted title of the king of fools  :P

sharpened my mind too!! :D
it just excessively lubricated the contact surfaces
and the friction increased :roll:

seems now i m a much bigger fool :rotfl:


[quote="gr8sk8r"]FOIL the following
$(45x-y^4)(x^7+5y)$[/quote]

$(45x-y^4)(x^7+5y) = \underbrace{45x\cdot x^7}_{\text{first}}+\overbrace{45x\cdot 5y}^{\text{outer}}-\underbrace{y^4\cdot x^7}_{\text{inner}}-\overbrace{y^4\cdot 5y}^{\text{last}}$

$ = 45x^8 +225xy -x^7y^4-5y^5$

hey gr8sk8r
i no where u got tat icon 4m! i like it!
 :D

[quote="mathprincess09"]foilin is easy! oooo! cool they got new smileys! dont no wats been going on this website for like months!  :)  :(  :?  :D  :lol:  ;)  :blush:  :P  :o  :mad:  :oops:  :roll:  :!:  :?:  :idea:  :arrow:  :|  :rotfl:  :huh:  :ninja:[/quote]
were have you been?  we've had these new icons for like a month

lol! i wasnt here! i hvent been on aops for a longgggggggggggggggg time!  :rotfl:


A Martian day, called a sol, is 24 hours, 39 minutes long in Earth time. If it is noon at a particular location on Earth and "noon" at the same moment at a particular location on Mars, after how many sols will it first be noon at these same locations at the same moment in time again?

Can you please explain where you got those numbers from?

[quote="pgpatel"]Sorry about that :blush:. I had the wrong answer too didn't I. Well there is editing for a reason.[/quote]

 :) It makes more sense now!  :)

[quote="bpms"][hide]480. Find the lcm of 24.65 and 24 (493 and 480) to get 236640. Divide that by 493 because that is the number of days to get 480 the number of sols.[/hide][/quote]
Was I right?


for what value of x we have $x^e+e^x= 2$  ?

i got the borrringgg answer,
e=x=1
i couldnt find any method for it, i know this is stupid and probably there are more solutions, can someone do it with some proper method??

Transcendental functions. Impossible to solve analytically.


calculus
integration
geometry
calculus computations

Prove that:
$ \int_0^1 {\int_0^1 {\max \left\{ {x^m ,y^n } \right\}dxdy} } \equal{} \frac{{m\plus{}n}}  {{mn \plus{} m \plus{} m}} \equal{} \frac{{\frac{1}{m} \plus{} \frac{1}{n}}}{{1 \plus{}\frac{1}{m} \plus{} \frac{1}{n}}}$

[quote="wingless"][hide="Solution"]Consider the integrated area. If we draw the curve $ x^m \equal{} y^n$ (or $ y \equal{} x^{\frac {m}{n}}$), points under this curve will return $ \max \left\{ {x^m ,y^n } \right \} \equal{} x^m$ and the points over this curve will return $ \max \left\{ {x^m ,y^n } \right \} \equal{} y^n$. We will divide the integration area into two regions and then integrate.

[img]http://img65.imageshack.us/img65/7996/grphka1.png[/img]

So, we can say that $ \int_0^1 {\int_0^1 {\max \left\{ {x^m ,y^n } \right\} \,dxdy} } \equal{} \int \int_{R_1} x^m \, dA \plus{} \int \int_{R_2} y^n \,dA$

There are lots of ways to solve the RHS. I'll do it this way:

$ \int_0^1 \int_0^{x^{\frac {m}{n}}} x^m \, dydx \plus{} \int_0^1 \int_{0}^{y^{\frac {n}{m}}} y^n \,dxdy$

$ \int_0^1 x^{\frac {m}{n} \plus{} m} \,dx \plus{} \int_0^1 y^{\frac {n}{m} \plus{} n} \,dy$

$ \frac {1}{\frac {m}{n} \plus{} m \plus{} 1} \plus{} \frac {1}{\frac {n}{m} \plus{} n \plus{} 1} \equal{} \boxed{\frac {m \plus{} n}{mn \plus{} m \plus{} n}}$[/hide][/quote]
This's  good solution.
 And general  pro
$ \idotsint_0^1 {\max \left\{ {x_1^{\alpha _1 } ,x_n^{\alpha _n },...,x_n^{\alpha _n } } \right\}dx_1 dx_2 ...dx_n } \equal{} \frac{{\sum\limits_{k \equal{} 1}^n {\frac{1}{{\alpha _i }}} }}{{1 \plus{} \sum\limits_{k \equal{} 1}^n {\frac{1}{{\alpha _i }}} }}$


hello, expanding your equation and moving all terms to the left we get
$32x^5-256x^2+24x^4-384x-x^8-144=0$.
This equation can be factorized as
$(x^2+2x+6)^2(x^2-2x-2)^2=0$
solving this we get
$x_1=1+\sqrt{3}=x_2$
$x_3=1-\sqrt{3}=x_4$
Sonnhard.


Can someone explain the method behind each of these problems? I just started learning random variables and am quite unfamiliar with them. Help is appreciated:

1. On a multiple choice test, there are 10 questions with five possible answers each. Dans knows the correct answers to five of these questions and randomly guesses the answers for the remaining questions. What is the probability that Dan gives exactly eight correct answers on this test?

2. Jane cought the flu on campus. She goes home for the weekend. Each of her five younger siblings has a 40% chance of catching the flu from Jane over the weekend. What is the probability that Jane will infect at most one of her younger siblings?

3. A fair coin is tossed eight times. Find the probability that heads comes up more often than tails.

For number 1:
[hide]
Since 5 of the questions are guaranteed, you are looking for the probability of getting 3 questions right out 5. This situation satisfies the binomial setting because all of the questions are independent, and they all have the same probability of success: 0.2. This means that the situation can be described by B(5,0.2). You can either use a calculator and enter binompdf(5,.2,3) or you can use the formula:
\[
P(X=k)={n \choose k} p^k (1-p)^{n-k}
\]

Either way, the result is .0512[/hide]

For number 2:
[hide]
\begin{eqnarray*}
P(X=0)+P(X=1)&=&{5 \choose 0} .4^0(1-.4)^5+{5 \choose 1} .4^1 (1-.4)^4\\
&=&.07776+.2592\\
&=&.33696
\end{eqnarray*}
[/hide]

Number 3:
[hide]
Since there are only 2 possibilities for outcomes, this simply means that you obtain 5 or more heads.
We have a binomial setting with p=.5 and n=8.
${P(x\geq5)=\sum_{i=5}^{n}P(x=i)}$
given the binomial probability formula
$P(X=k)={n \choose k} p^k (1-p)^{n-k}$
We see that this sum equals $(.5)^8 *\sum_{i=5}^{8} {8\choose i},$
which is one half.
(quicker version - the distribution of any binomial random variable is symmetric when p=1/2, therefore 5-8 should give exactly 1/2 of the distribution)[/hide]

Thanks for the explanations guys. Is there any informative webpage I can read to learn more about binomial distribution? I think i recently lost my stat. books  :(

Both of these websites describe the binomial distribution

http://www.herkimershideaway.org/apstatistics/ymmsum99/ymm881.htm

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

miraculouspostmaster, your method to number 3 is correct but the result is not 1/2.

There's no need to go too deeply into the theory of binomial distributions in order to solve most elementary probability problems. I've found that most of them can be solved by one (or some combination) of the following four methods.

1. Solving by counting. This is probably the first way that you learned probability. It usually entails counting the number of "good outcomes" and dividing by the "total number of possible outcomes."

2. Use symmetry. This works for example on your problem number 3. The probability of getting more heads than tails equals the probability of getting more tails than heads. So if we call this probability P, 2P + {the probability of getting an equal number of heads and tails} = 1.

3. Using recursion. As an example, try to solve the following problem: A particle moves among the vertices of an equilateral triangle, and on each move from a vertex has an equal probability of moving to either of the other vertices. Find the probability that a particle will return to its starting position after 7 moves. (This problem can also be solved by a clever counting reasoning).

4. Using geometry. This works for problems where there are an (uncountably) infinite number of possibilities. For example, solve the following problem by graphing the solutions and looking at the relative areas. Suppose X and Y are two numbers chosen independently and randomly from the interval [-1, 1]. What is the probability that |X + Y| > 1.

#3 without using stats' stuff (which I should know better than I currently do  :( ) :

[hide]
There are 3 possibilities:

(a) Heads comes up more times than tails.

(b) Heads comes up less times than tails.

(c) They come up an equal amount of times (i.e. 4 times for each).

By symmetry, (a) and (b) are equivalent in terms of the probability of them happening.  We evaluate (c) as follows:

There are $2^8=256$ total outcomes possible.  The number of outcomes with 4 heads and 4 tails is $\frac{8!}{4!\cdot4!}=70$.

So since (c) has 70 total outcomes, (a) and (b) have a combined $256-70=186$.  So the probability of (a) is $\frac{93}{256}.$[/hide]

Ok, using the binomial method, here's what I got for the three problems,c an anyone verify?

1. (5C3)(1/5)^3*(4/5)^2

2. 1053/3125 = 0.33696

3. 93/256 = (1/2)^8*(8C5+8C6+8C7+8C8)


Brand New to this Forum: Can anyone PLEASE help me obtain a copy of the MathCounts Competition broadcast on ESPN?
It can be from any year and can be video or dvd or cd -- any media! Please let me know if you have a copy you could lend or sell to me (for a reasonable price, of course), or if you know where I could go to obtain. I contacted ESPN and they said they never sell any footage, but maybe someone taped and kept a copy at home? I would be Very grateful if you could help. Thanks. Please leave contact info or a way I can reply back to you. Also, if I didn't post this right, please let me know.
Oh, I don't mind if I don't receive copy till after regional competition. I just would like to view what goes on - this is my first year participating.

Life1

[size=75][color=darkred]Please try to avoid using all caps. It's the equivalent of "SHOUTING!" Also, avoid using colored and large fonts. Thanks.
-nebula42[/color][/size]

people have asked in the past and i don't think they're available

jorian


calculus
integration
algebra
polynomial
trigonometry
calculus computations

Find the polynomials $ f(x),\ g(x)$ such that:

$ \frac{1}{\pi}\int_0^{\frac{1}{2}} \frac{tf'(t)\minus{}6g(t)}{\sqrt{1\minus{}t^2}}\ dt\equal{}f(x)\minus{}g(x)\plus{}x$

$ \frac{6}{\pi}\int_0^{\frac{1}{2}} \frac{8f(t)\minus{}5g'(t)}{\sqrt{1\minus{}t^2}}\ dt\equal{}2f(x)\minus{}3g(x)\minus{}x^2\plus{}2x$.

well assuming this isnt a typo, since the left terms have no x, lets say

$ c_1\equal{}f(x)\minus{}g(x)\plus{}x$ and $ c_2\equal{}2f(x)\minus{}3g(x)\minus{}x^2\plus{}2x$

$ \to  0\equal{}2f(x)\minus{}3[f(x)\plus{}x\minus{}c_1]\minus{}x^2\plus{}2x\minus{}c_2\equal{}\minus{}x^2\minus{}x\minus{}f(x)\plus{}3c_1\minus{}c_2$

$ \to f(x)\equal{}x^2\plus{}x\plus{}c_2\minus{}3c_1 \to g(x)\equal{} x^2\plus{}2x\plus{}c_2\minus{}4c_1$

now we just have to calculate

$ \frac{1}{\pi}\int_0^{\frac{\pi}{6}} A\sin^2 v\plus{}B\sin v \plus{}6C dv\equal{}C\plus{}\frac{2\minus{}\sqrt 3}{2\pi}B\plus{}A(\frac{1}{12}\minus{}\frac{\sqrt 3}{8\pi})$
[hide="first eq"]
$ c_1\equal{}\frac{1}{\pi}\int_0^{\frac{1}{2}} \frac{tf'(t)\minus{}6g(t)}{\sqrt{1\minus{}t^2}}\ dt$

$ \equal{}\frac{1}{\pi}\int_0^{\frac{1}{2}} \frac{6(4c_1\minus{}c_2)\minus{}4t^2\minus{}11t}{\sqrt{1\minus{}t^2}}\ dt$

$ (A,B,C)\equal{}(\minus{}4,\minus{}11,4c_1\minus{}c_2)\to c_2\equal{} 3c_1\plus{}\frac{12\sqrt 3\minus{}22}{2 \pi}\minus{}\frac{1}{3}$

[/hide]

[hide="second eq"]

$ c_2\equal{}\frac{6}{\pi}\int_0^{\frac{1}{2}} \frac{8t^2\minus{}2t\plus{}10\plus{}8c_2\minus{}24c_1}{\sqrt{1\minus{}t^2}}\ dt$

$ \equal{}\frac{6}{\pi}\int_0^{\frac{\pi}{6}} \frac{8\sin^2 v\minus{}2\sin v\plus{}10\plus{}8c_2\minus{}24c_1}dv$

$ (A,B,C)\equal{}(48,\minus{}12,10\plus{}8c_2\minus{}24c_1) \to 24c_1\equal{}7c_2\plus{}14\minus{}\frac{12}{\pi}$

[/hide]

so $ 24c_1\equal{}7c_2\plus{}14\minus{}\frac{12}{\pi}$ and $ c_2\equal{} 3c_1\plus{}\frac{12\sqrt 3\minus{}22}{2 \pi}\minus{}\frac{1}{3}$

$ \to c_1\equal{}\frac{84\sqrt 3\minus{}178}{6 \pi}\plus{}\frac{44}{9}$ and other guys just as nasty...

but then again maybe i just read this problem wrong?


probability
function
logarithms
graph theory
expected value
probability and stats

Let $ p(n)$ be a probability function on a random graph $ (G_n,p(n))$ such that $ p(n)\to 0$ for $ n\to \infty$. Prove that almost all $ G\in (G_n,p(n))$ have an isolated vertex.

Can you be more explicit about how your random graph $ G$ is constructed? From what you wrote, I cannot even understand how many vertices it has, not to say anything about the edges. :?

Sorry for not being specific. :) It is Erdos-Renyi model where all edges occur independently with the same probability. $ (G_n,p(n))$ stands for such a random graph with $ n$ vertices and probability $ p\equal{}p(n)$.

In the book I am reading (Diestel's Graph Theory) it is written that for $ p(n)\equal{}(1\minus{}\epsilon)\frac{\ln{n}}{n}$ ($ \epsilon \in (0,1)$) $ (G_n,p(n))$ will almost surely contain an isolated vertex. Can this be extended for all $ p(n)$ such that $ p(n)\to 0$ for $ n\to \infty$? In my proof I didn't use this particular function for $ p(n)$ just the property with limits so I don't know am I right.

Ah, I see. No, just tending to $ 0$ is not enough. The probability that a given vertex is isolated is  $ (1\minus{}p(n))^{n\minus{}1}$, so the expected number of isolated vertices is $ n(1\minus{}p(n))^{n\minus{}1}$. If $ p(n)\equal{}(1\plus{}\varepsilon)\frac{\ln n}{n}$ or above, this tends to $ 0$, so the result in the book is essentially sharp.

OK, could you find a flow in my reasoning.

Let $ X$ be the number of isolated vertices and let $ X_i$ be an indicator if the vertex $ i$ is isolated. Hence $ X \equal{} \sum{X_i}$. As you said $ E[X] \equal{} n(1 \minus{} p)^{n \minus{} 1}$.
We have that

$ P[X \equal{} 0]\leq P[|X \minus{} E[X]|\geq E[X]]\leq \frac {E[X^2] \minus{} E[X]^2}{E[X]^2} \equal{} k$.

If the latter expression tends to $ 0$ for $ n\to \infty$ then almost all graphs have at least one isolated vertex. For $ X^2$ we have

$ E[X^2] \equal{} E[(\sum X_i)^2] \equal{} \sum_{i\leq j}E[X_iX_j] \equal{} E[X] \plus{} 2\sum_{i < j}E[X_iX_j]$.

However $ X_iX_j \equal{} 1$ if both $ i$ and $ j$ are isolated and $ 0$ otherwise. Hence $ E[X_iX_j] \equal{} (1 \minus{} p)^{2n \minus{} 3}$. Therefore

$ E[X^2] \equal{} n(1 \minus{} p)^{n \minus{} 1} \plus{} n(n \minus{} 1)(1 \minus{} p)^{2n \minus{} 3}$

and

$ k \equal{} \frac {n(1 \minus{} p)^{n \minus{} 1} \plus{} n(n \minus{} 1)(1 \minus{} p)^{2n \minus{} 3} \minus{} n^2(1 \minus{} p)^{2n \minus{} 2}}{n^2(1 \minus{} p)^{2n\minus{}2}}$.

If $ p\to 0$ for $ n\to \infty$ than the numerator tends to $ 0$ while the denominator tends to $ \infty$. Therefore $ k\to 0$ and $ X\geq 1$ almost surely.

Where's the mistake?

You just forgot the power $ n \minus{} 1$ on $ (1 \minus{} p)$ in the denominator in the last displayed formula.

I see you corrected it, but now what makes you think that the denominator tends to $ \infty$. The first factor grows but the second one may decay much faster...


A point P is inside triangle $ A_1A_2A_3$ . $ PA_i$ (i=1--->3) intersect the opposite sides at $ B_i$ . $ C_i$ are  the midpoints of $ A_iB_i$ and $ D_i$ are the midpoints of $ BB_i$ . Prove that S($ C_1C_2C_3$) = S(  $ D_1D_2D_3$)

[quote="romano"]$ D_i $ are the central points of $ BB_i $.[/quote]

What's $B$?

In case you wanted to write $P$ instead of $B$, here's a solution (in fact, it's just a trivial application of the cross product of vectors).

We have to show that $\overline{D_1D_2}\times\overline{D_1D_3}=\overline{C_1C_2}\times\overline{C_1C_3}$. This is equivalent to $\overline{B_1B_2}\times\overline{B_1B_3}=(\overline{A_1A_2}+\overline{B_1B_2})\times(\overline{A_1A_3}+\overline{B_1B_3})$.

This, in turn, is equivalent to $\overline{A_1A_2}\times\overline{A_1A_3}+\overline{A_1A_2}\times\overline{B_1B_3}+\overline{B_1B_2}\times\overline{A_1A_3}=\overline 0$. This, however, is just a restatement of $S(A_1A_2A_3)-S(A_1A_2B_1)-S(A_1B_1A_3)=0$, which is obvious.

You meant $ S(ABC) = \overline{AB}  . \overline{AC} $ ?  I don't understand you , Grobber


Solve in complex numbers the equation:

                   $ (x\plus{}3)(x\plus{}4)(x\plus{}5)^{2}(x\plus{}6)(x\plus{}7)\equal{}\frac{5}{2}7!$

 :)

[hide="hint"]
Subsitution $ x^{2}\plus{}10x \equal{} t$, then:

$ (t\plus{}21)(t\plus{}24)(t\plus{}25) \equal{} 21\cdot 24\cdot 25$

$ t(t^{2}\plus{}60t\plus{}1629)\equal{}0$[/hide]


8 spherical balls of radius 1 are placed in a cylinder in two layers. with each layer containing 4 balls. Each ball is tangent to 2 balls in the same layer, 2 balls in another layer, one base, and the lateral surface of the cylinder. What is the height of the cylinder?

[hide="Analytic Approach"]
[asy]pair O,A,B,C,D,P,Q,X;
O=(0,0); A=(1,1); B=(-1,1); C=(-1,-1); D=(1,-1); P=(0,1); Q=(1,0); X=(-sqrt(2),0);
draw(Circle(O,1+sqrt(2)),linewidth(1));
draw(Circle(A,1));
draw(Circle(B,1));
draw(Circle(C,1));
draw(Circle(D,1));
draw(Circle(X,1));
draw(Circle(O,sqrt(2)),dashed);
draw(O--P--A--Q--O--A);
label("1",(0.5,0),S);
label("1",(1,0.5),E);
label("$\sqrt{2}$",(A+O)/2,NW);
draw(B--C--X--B);
label("$B$",B,N);
label("$C$",C,S);
label("$X$",X,W);
dot(O);[/asy]
Here is a top-down view of the bottom layer and one ball in the top layer.  The centers of the spheres lie on the "hollow cylinder" $ x^2+y^2=2$.  Let $ B=(-1,1,1)$, $ C=(-1,-1,1)$, and $ X=(-\sqrt{2},0,z)$ for some positive value of $ z$.  This way, the bottom of the cylinder lies on the plane $ z=0$.  Note that $ \triangle BCX$ is equilateral, and has side length of 2.  By the distance formula:
\begin{eqnarray*}
\sqrt{(\sqrt{2}-1)^2+1^2+(z-1)^2} &=& 2 \\
(\sqrt{2}-1)^2+1^2+(z-1)^2 &=& 4 \\
3-2\sqrt{2}+1+(z-1)^2 &=& 4 \\
(z-1)^2 &=& 2\sqrt{2} \\
z-1 &=& \sqrt[4]{8} \\
z &=& 1+\sqrt[4]{8}
\end{eqnarray*}
Thus the centers of the spheres in the top layer lie on the plane $ z=1+\sqrt[4]{8}$.  We must add one, since the radius of the sphere is 1.  Thus the top of the cylinder lies on the plane $ z=2+\sqrt[4]{8}$, hence its height is $ \boxed{2+\sqrt[4]{8}}$.
[/hide]


[quote="Interval"]Refer to the problem below to show that the arithmetic mean of a and b is EQUIDISTANT from a and b.

If a < b, show that a < (a + b)/2 < b. The number (a + b)/2 is called the ARITHMETIC MEAN of a and b.[/quote]
[hide]Since $a<b$, you can say $\frac{2a}{2}<\frac{a+b}{2}$ You can use the same reasoning for b. ($\frac{a+b}{2}<\frac{2b}{2}$)[/hide]

[quote="Interval"]Where did you get the fractions (2a/2) and (2b/2)?[/quote]
$\frac{2a}{2}=a$
$\frac{2b}{2}=b$


[quote="zaizai-hoang"]I think above problem very interesting  :) My opinion is use Mixing Variable method . But Maybe it is long and ugly (  :blush:
Is you use MV or another method?
$f(a,b,c)\ge f(a, \sqrt{bc},\sqrt{bc})\to ?!$  :D[/quote]
Nice idea even though I haven't tried it. I have a much nicer solution  :D !


If three times the larger of two numbers is four times the smaller and the difference between the numbers is 8, the the larger of two numbers is

$\text{(A)}\ 16 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 44 \qquad \text{(E)}\ 52$

Here is each of the answers and why they are wrong or right.

[hide="Why A Is Wrong Or Right"]
[b]A-16:[/b] $16\times3=48$ However $16-8=8$ and $8\times4=32$ [b]So it is not A.[/b][/hide]
[hide="Why B Is Wrong or Right"]
[b]B-24:[/b] $24\times3=72$ However $24-8=16$ and $16\times4=48$ [b]So it is not B.[/b][/hide]
[hide="Why C Is Wrong or Right"][b]C-32:[/b] $32\times3=96$ However $32-8=24$ and $24\times4=96$ [b]So it is IS $\boxed{C}$.[/b][/hide]
[hide="Why D Is Wrong or Right"][b]D-44:[/b] $44\times3=132$ However $44-8=36$ and $36\times4=144$[b]So it is not D.[/b][/hide]
[hide="Why E Is Wrong or Right"][b]E-52:[/b] $52\times3=156$ However $52-8=44$ and $44\times4=176$[b]So it is not E.[/b][/hide]
[hide="So The Answer is....."] $\boxed{C}$ [/hide]

[hide]
C, because A is small, so it just won't work.
I just went up the list and got B and C.[/hide]

Algebraic Solution:
[hide ] Set $x$ to the larger number.
We have $3x = 4(x-8)$
 => $3x = 4x -32$
 => $x = 32$ [/hide]


For how many nonnegative integers n does x^3-(n-1)x^2+(n-n^2)x-n^3 have all integer roots

[u]Ans[/u] 1 integer; n=0

[u]Solution:[/u]
$ x^{3} \minus{} (n \minus{} 1)x^{2} \plus{} (n \minus{} n^{2})x \minus{} n^{3}$
=$ (x \minus{} n)(x^{2} \plus{} nx \plus{} n^{2}) \minus{} (n \minus{} 1)x(x \minus{} n)$
=$ (x \minus{} n)(x^{2} \minus{} x \plus{} n^{2})$
So the roots of $ x^{3} \minus{} (n \minus{} 1)x^{2} \plus{} (n \minus{} n^{2})x \minus{} n^{3}$ are $ x \equal{} n, \frac { \minus{} 1\pm\sqrt {1 \minus{} 4n^{2}}}{2}$
In order to make $ \frac { \minus{} 1\pm\sqrt {1 \minus{} 4n^{2}}}{2}$ an integer, $ \sqrt{1 \minus{} 4n^{2}}$ must be an integer
$ 1 \minus{} 4n^{2}\ge\ 0$
$ (1 \minus{} 2n)(1 \plus{} 2n)\ge\ 0$
$ \minus{}\frac {1}{2}\le\ n\le\frac {1}{2}$
From the question that says n is a nonnegative integer, so n=0
Checks when n=0:
$ x^{3} \minus{} (n \minus{} 1)x^{2} \plus{} (n \minus{} n^{2})x \minus{} n^{3}$
$ \equal{} x^{3} \minus{} x^{2} \equal{} x^{2}(x \minus{} 1)$
which works.


Let $ ABC$ be a triangle and $ M$ is a point in the triangle. Denote $ A_1,B_1,C_1$ be the projection of $ M$ to $ BC,CA,AB$. Show that:

$ \frac{S_{A_1B_1C_1}}{S_{ABC}} \equal{} \frac {1}{4}(1 \minus{} \frac {OM^2}{R^2})$

where $ O$ and $ R$ are center and radius of the circumcircle of triangle ABC.

This problem is familiar, but I haven't solved it :(. If you have a solution, please explain it carefully. Thanks a lot.

This is Euler's theorem- a well-known result.
For an example, you can find the solution at here:
http://reflections.awesomemath.org/2007_2/pedal(2).pdf


trigonometry
function
calculus
calculus computations

What is the range of $ z$ when

\[ z\equal{}f(\theta, \phi)\equal{}\cos \theta \cos \phi \plus{} \sin \theta \cos \phi \plus{} \sin \theta \sin \phi
\]


geometry
trigonometry
angle bisector
geometry unsolved

Consider triangle ABC. D and E lie on BC and AC, respectively, such that BD=AE. Segments AD and BE intersects at point P. Bisector of angle ACB intersect segments AD and BE at Q and R, respectively. Proof that:

$ \frac{PQ}{AD}= \frac{PR}{BE}$

It isn't hard. My solution is not that "inspirational", there must be better solutions.

We want to show that \[ \frac{AP}{AD}+\frac{QD}{AD}= \frac{BR}{BE}+\frac{PE}{BE}.\] Notice that \[ \frac{QD}{AD}= \frac{QD}{AQ+QD}= \frac{CD}{AC+CD}\] by the angle bisector theorem and similarly \[ \frac{BR}{BE}= \frac{BR}{BR+RE}= \frac{BC}{BC+CE}.\] Also \[ \frac{AP}{AD}= \frac{\left[\triangle APC\right]}{\left[\triangle ADC\right]}= \frac{\left[\triangle APB\right]}{\left[\triangle ADB\right]}= \frac{\left[\triangle APC\right]+\left[\triangle APB\right]}{\left[\triangle ADC\right]+\left[\triangle ADB\right]}= \frac{\left[\triangle APC\right]+\left[\triangle APB\right]}{\left[\triangle ABC\right]}\] and similarly \[ \frac{PE}{BE}= \frac{\left[\triangle CEP\right]}{\left[\triangle CEB\right]}= \frac{\left[\triangle AEP\right]}{\left[\triangle AEB\right]}= \frac{\left[\triangle CEP\right]+\left[\triangle AEP\right]}{\left[\triangle CEB\right]+\left[\triangle AEB\right]}= \frac{\left[\triangle APC\right]}{\left[\triangle ABC\right]}.\] Hence, we must show that \[ \frac{\left[\triangle APB\right]}{\left[\triangle ABC\right]}= \frac{BC}{BC+CE}-\frac{CD}{AC+CD}.\] There are several ways to do this. Let me show you a very easy but boring one.

Name $ AC = b$, $ BC = a$, $ AE = BD = x$. Notice that \[ \frac{BC}{BC+CE}-\frac{CD}{AC+CD}= \frac{BC}{a+b-x}-\frac{CD}{b+a-x}= \frac{BC-CD}{a+b-x}= \frac{x}{a+b-x}.\] Hence we must show that \[ \frac{a}{x}+\frac{b}{x}-1 = \frac{\left[\triangle ABC\right]}{\left[\triangle APB\right]}.\] This is equivalent to \[ \frac{CD}{BD}+\frac{CE}{AE}= \frac{\left[\triangle APC\right]}{\left[\triangle APB\right]}+\frac{\left[\triangle BPC\right]}{\left[\triangle APB\right]}.\] But this is trivial since obviously \[ \frac{CD}{BD}= \frac{\left[\triangle APC\right]}{\left[\triangle APB\right]},\ \frac{CE}{AE}= \frac{\left[\triangle BPC\right]}{\left[\triangle APB\right]}.\] Hence we're done.

I found a better solution.

Let $ \ell$ be any line which is perpendicular to the angle bisector of $ \angle C$. For any point $ X$, let $ \mathcal{X}$ be the projection of $ X$ onto $ \ell$.

It is easy to see that $ \mathcal{AE}= \mathcal{BD}$ (because of the angle bisector!), so $ \mathcal{AD}= \mathcal{BE}$, and  $ \mathcal{Q}= \mathcal{R}$. Hence
\[ \frac{PQ}{AD}= \frac{\mathcal{PQ}}{\mathcal{AD}}= \frac{\mathcal{PR}}{\mathcal{BE}}= \frac{PR}{BE}\]
and we are done.

Yes, I thought that's hard problem...

My solution:

Let $ DE'\parallel BE$, $ BDE'E$ is parallelogram. Let $ \angle ACR=\angle BCR=\alpha$ and $ \angle EBC=\beta$ It's easy to show that $ \angle PRQ = \alpha+\beta$ But $ \angle E'DC=\angle EBC=\beta$. It's easy to show that $ \angle E'EC=2\alpha$ so $ \EE'A=\angle EAE' = \alpha$ and we see that $ \angle AE'D=\alpha+\beta=\angle PRQ$. Similary we show that $ \angle PQR=\angle E'AD$. Then $ \triangle PQR$ is similar to $ \triangle ADE'$. So:

$ \frac{PQ}{PR}=\frac{AD}{DE'}=\frac{AD}{BE}$.

One could also use menelaus in triangles $ QAC$ and $ PBD$ with transversals $ R-P-E$ and $ R-Q-C$ respectively+angle bisector thm twice. It`s just some light computation.

[quote="Przemo"][color=darkred] Consider $ \triangle ABC$ and the points $ D\in (BC)$, $ E\in (AC)$ such that $ BD=AE$. Denote the intersection $ P\in AD\cap BE$.
The bisector of the angle $ \widehat{ACB}$ intersect the segments $ (AD)$, $ (BE)$ at the points $ Q$, $ R$ respectively. Proof that $ \frac{PQ}{AD}= \frac{PR}{BE}\ .$[/color][/quote]
[color=darkblue][b]Proof (only using areas).[/b] Define the area $ [XYZ]$ of the triangle $ XYZ$ and the distance $ \delta_{XY}(Z)$ of the point $ Z$ to the line $ XY$.

For example, $ 2\cdot [RPQ]=[PR]\cdot\delta_{PR}(Q)=[PQ]\cdot\delta_{PQ}(R)$. Suppose w.l.o.g. that $ BC<AC$, i.e. the line $ \overline{CRQ}$ separates the points $ P$, $ B$.

Therefore , $ \{\begin{array}{ccccc}BD=AE\ ,\ \delta_{BC}(R)=\delta_{AC}(R) & \implies & [BRD]=[ARE] & \implies & [BRD]+[RECD]=[ARE]+[RECD]\\\ BD=AE\ ,\ \delta_{BC}(Q)=\delta_{AC}(Q) & \implies & [BQD]=[AQE] & \implies & [BQD]+[AQB]=[AQE]+[AQB]\end{array}\|$ $ \implies$

$ \begin{array}{c}[BEC]=[ADC]+[ARD]\\\ [ABD]=[ABE]+[BQE]\end{array}\|$ $ \implies$ $ [BEC]+[ABE]+[BQE]=[ABD]+[ADC]+[ARD]$ $ \implies$ $ [ABC]+[BQE]=[ABC]+[ARD]$ $ \implies$

$ [BQE]=[ARD]$ $ \implies$ $ BE\cdot\delta_{PR}(Q)=AD\cdot \delta_{PQ}(R)$ $ \implies$ $ \frac{BE}{AD}=\frac{\delta_{PQ}(R)}{\delta_{PR}(Q)}=\frac{PR}{PQ}$ $ \implies$ $ \frac{PQ}{AD}=\frac{PR}{BE}$.[/color]

[quote][color=darkred][size=117][b]A similar problem.[/b][/size] In an acute triangle $ ABC$ define the points $ \{\begin{array}{c}F\in (AB)\ ,\ CF\perp AB\\\\ \{U,V\}\subset (AB)\ ,\ F\in (UV)\ ,\ AU=BV\end{array}$

and the points $ \{\begin{array}{c}D\in (BC)\ ,\ DV\perp AB\\\\ E\in (AC)\ ,\ EU\perp AB\end{array}$. Denote the intersections $ \{\begin{array}{c}P\in AD\cap BE\\\\ Q\in CF\cap AD\\\\ R\in CF\cap BE\end{array}$. Prove that $ \frac{PQ}{AD}=\frac{PR}{AE}$.[/color][/quote]

[color=darkblue][b]Proof.[/b] Denote $ \{\begin{array}{c}m(\widehat{BCF})=\alpha\\\\ m(\widehat{ACF})=\beta\end{array}$. Observe that $ \{\begin{array}{c}AU=BV\implies BD\cdot\sin\alpha =AE\cdot\sin\beta\\\\ \delta_{BD}(R)=RC\cdot\sin\alpha\\\\ \delta_{AE}(R)=RC\cdot\sin\beta\end{array}\|$ $ \implies$ $ \frac{BD}{AE}=\frac{\sin\beta }{\sin\alpha }=\frac{\delta_{AE}(R)}{\delta_{BD}(R)}$ $ \implies$

$ BD\cdot\delta_{BD}(R)=AE\cdot\delta_{AE}(R)$ $ \implies$ $ \boxed{\ [BRD]=[ARE]\ }$. Prove similarly $ \boxed{\ [BQD]=[AQE]\ }$. Therefore, can continue as above.

$ \{\begin{array}{ccc}[BRD]=[ARE] & \implies & [BRD]+[RECD]=[ARE]+[RECD]\\\ [BQD]=[AQE] & \implies & [BQD]+[AQB]=[AQE]+[AQB]\end{array}\|$ $ \implies$ $ \{\begin{array}{c}[BEC]=[ADC]+[ARD]\\\ [ABD]=[ABE]+[BQE]\end{array}\|$ $ \implies$

$ [BEC]+[ABE]+[BQE]=[ABD]+[ADC]+[ARD]$ $ \implies$ $ [ABC]+[BQE]=[ABC]+[ARD]$ $ \implies$

$ [BQE]=[ARD]$ $ \implies$ $ BE\cdot\delta_{PR}(Q)=AD\cdot \delta_{PQ}(R)$ $ \implies$ $ \frac{BE}{AD}=\frac{\delta_{PQ}(R)}{\delta_{PR}(Q)}=\frac{PR}{PQ}$ $ \implies$ $ \frac{PQ}{AD}=\frac{PR}{BE}$.[/color]

[size=134][color=purple][b]P.S.[/b] I dedicate this proof to the my bosom friend, the teacher [b]Bogdan Enescu.[/b][/color][/size]


Prove that there exist infinitely many natural numbers $ n$ such that the numerator of $ 1 \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} \frac {1}{4} \plus{} ... \plus{} \frac {1}{n}$ in the lowest terms is not a power of a prime number.

Let $ f(n)=1+1/2+...+1/n $. Let $ X_p = \{ n : p|f(n) \}$. For big $n$, $n$ belongs to exactly one of these sets. Now let $a,x \in X_p$ with $a$ big and $x \le p-1$. Suppose $v_p(a)=1$. Then

$1<f(a)=\frac{p^{v_p(a!)+1}}{a!} < 1$

contradiction. Therefore $p^2 | f(a)$ and so

$p^2 | f(a)+p(\frac{1}{pa+1}+...+\frac{1}{pa+x})$

therefore $p | \frac{1}{p}+...+\frac{1}{pa} + \frac{1}{pa+1}+...+\frac{1}{pa+x}$

and since $p | f(m)-f(n)$ if $m=n+p-1$ (this is verified easily), then $p | f(pa+x)$. Therefore $pa+x \in X_p$. Using this and Wolstenholme we obtain $p^k-1 \in X_p$ and $p^k-p \in X_p$ for all $k$. Therefore, putting $k$ big enough we obtain $p^B \ f(p^k-1)-f(p^k-p)$ for $B$ arbitrarily big. If $B<k$ this means  $p^B | f(p-1)$ for big $B$, contradiction

If change a power of a prime number to a power of integer ,if the conclusion is still right?


Extending B perpendicular to AD on a specific point Q and C perpendicualr to AD at R, we draw BQ and CR to get 2 right triangles, the missing lengths beign 6 and 15 respectively. now using trapezoid formula

$ A\equal{}\frac{1}{2}(b_1\plus{}b_2)h$

we plug in values to get

$ 164\equal{}\frac{1}{2}(b_1\plus{}b_2)(8)$
$ 164\equal{}4(b_1\plus{}b_2)$
$ b_1\plus{}b_2\equal{}41$
$ b_2\equal{}b_1\plus{}6\plus{}15\equal{}b_1\plus{}21$
$ b_1\plus{}b_1\plus{}21\equal{}41$
$ BC\equal{}b_1\equal{}10$


floor function
number theory proposed
number theory

The positive integers $a_1,\ldots , a_n$ are such that each is less than $1000$ and $lcm(a_i,a_j)>1000$ for all $i\neq j$
Show that
\[ \displaystyle \sum_{i=1}^n \frac1{a_i} < 2 \]

[quote="DucHuynh"]Note that

\[ \lfloor \frac{1000}{a_1} \rfloor + \lfloor \frac{1000}{a_2} \rfloor + \lfloor \frac{1000}{a_3} \rfloor + \cdots + \lfloor \frac{1000}{a_n} \rfloor < 1000  \][/quote]
Why? :roll:

[quote="jin"]$\lfloor \frac{1000}{a_i} \rfloor$ means the number of the multiples of $a_i$ in$\{1,2,\ldots,1000\}$
And $lcm(a_i,a_j)>1000$,so count all multiples of some $a_i$ less than orequal to 1000,they are all distinct.
So we get it.[/quote]
Oh,thanks $jin$,I didn't know that it means number of the multiples of $a_i$ in$\{1,2,\ldots,1000\}$(I thought it is interger part :blush: ),now I see the solution.

It means integer part, but they are the same.


I find it annoying when I want to edit a template and can't.

Also, as far as I know, there hasn't been ANY vandalism on the AoPSWiki yet. Zero. Zip.

Here's a list of templates I'd like unprotected:

[[Template:Stub]]
[[Template:Disambig]]
[[Template:Wikify]]

There has, in fact, been at least one instance of vandalism.  Don't overstate your case ;)

All right, that particular part may have been overstated.  :oops: However, I haven't encountered any, and I regularly browse the recent changes section.


analytic geometry
advanced fields
advanced fields unsolved

Prove that a coordinate subspace is an irreducible variety when the field $ k$ is infinite.

And could you please give me a counterexample when the field $ k$ is finite.

Thanks in advance.

What is your definition of a [i]coordinate subspace[/i]? Of what space are they subspaces? :)


percent
geometry
perimeter
ratio
LaTeX
percent increase

The diagram consists of 4 congruent squares made with twelve "2" toothpicks. By moving exactly three toothpicks, three congruent squares can be formed so that each toothpick forms a side of a square. What is the percent increase in the perimeter of the figure? ( Assume that the toothpicks intersect at their endpoints.)

1. Show the diagram
2. What is a "2" toothpick?

When he says 2, I'll assume that's a side length. So for the first side length, it would be $ \frac{3}{2}$ because 24/16=3/2. And the second side length would be $ 2$.
Area 1: $ \frac{9}{4}
Area 2:$4$ Ratio:$\frac{4}{\frac{9}{4}}=\frac{16}{9}=1.77777...

[quote="jonathanchou711"]When he says 2, I'll assume that's a side length. So for the first side length, it would be $ \frac {3}{2}$ because $ \frac {24}{16} \equal{} \frac {3}{2}$. And the second side length would be $ 2$.
Area 1: $ \frac {9}{4}$
Area 2:$ 4$
Ratio:$ \frac {4}{\frac {9}{4}} \equal{} \frac {16}{9} \equal{} 1.77777...$[/quote]

latex fail


Let be $ ABC$ a triangle and $ M\in AC$ , $ N\in AB$ such that $ MA \equal{} m \cdot MC$ , $ NA \equal{} n \cdot NB$ and $ m \plus{} n \equal{} 2$. If $ \left\{ P \right\} \equal{} BM \cap CN$, show that $ area\left( {AMPN} \right) \geqslant \frac{{mn}}{3}area\left( {ABC} \right)$

[i]Virgil Nicula[/i]

Nice problem :)
I only used Euclidean geometry and AM-GM inequality for solving this problem.
In my solution I converted every area into $ m,n,(APC)$
Here $ (P)$ means the area of polygon $ P$
[color=red]Here goes my solution.[/color]

$ \frac{(APM)}{(PMC)}=m
\Leftrightarrow \frac{(APM)}{(PMC)}=\frac{1}{m}
\Leftrightarrow \frac{(PMC)}{(APM)}=\frac{1}{m}
\Leftrightarrow \frac{(APC)}{(APM)}=\frac{m+1}{m}
\Leftrightarrow (APM)=\frac{m}{m+1} (APC)$----(1)
Similarly, $ (APN)=\frac{n}{n+1} (APB)$
let the ray $ AP$ meets BC at $ O$

Using Ceva's theorem we get that, $ \frac{BO}{OC}=\frac{m}{n}$
Which implies, $ (APB)=\frac{m}{n}(APC)$
So, $ (APN)=\frac{m}{n+1} (APC)$---(2)

(1)+(2) implies,
$ (AMPN)=m(APC) {\left( \frac{1}{m+1} + \frac{1}{n+1} \right)}=m(APC)\frac{m+n+2}{(m+1)(n+1)}=m(APC)\frac{4}{mn+3}$---(3)

Now, $ \frac{(ABO)}{(AOC)}=\frac{m}{n}
\Leftrightarrow \frac{(ABC)}{(AOC)}=\frac{m+n}{n}=\frac{2}{n}
\Leftrightarrow (ABC)=\frac{2}{n}(AOC)$---(4)

Now we use Mennelaus's Theorem on $ \triangle AOC$ and transversal$ MB$ and get $ \frac{CM}{MA} \cdot \frac{AP}{PO}\cdot \frac{OB}{BC} =2$
Which implies, $ \frac{AP}{PO}=2$

So, $ \frac{(OPC)}{(APC)}=\frac{1}{2}\Leftrightarrow ((AOC))=\frac{3}{2}(APC)$
Putting this value in (4) we get,
$ (ABC)=\frac{3}{n}(APC)$

So the given inequality is just,
$ m(APC)\frac{4}{mn+3} \geq \frac{mn}{3}\cdot \frac{3}{n} (APC)$
$ \Longleftrightarrow \frac{4}{mn+3} \geq 1$ $ \Leftrightarrow 4\geq mn+3
\Leftrightarrow mn \leq 1$ ---(5)

using AM-GM inequality we get that,
$ 2=m+n \geq 2\sqrt{mn}$
$ \Leftrightarrow mn \leq 1$
so the ineq (5) is certainly true.

$ QED$


[quote="PenguinIntegral"]What would be the best way to do #21 of last years state countdown?

The algebraic expression $(3x-2y)^2-x(x+4y)$ can be written in the form $Ax^2+Bxy+Cy^2$. What is the sum of A+B+C?[/quote]

[hide]The best way to do this is just to simplify the expressions

$(3x-2y)^2-x(x+4y)$=$9x^2-12xy+4y^2-x^2-4xy$=$8x^2-16xy+4y^2$

So $A=8$, $B=-16$, and $C=4$

Therefore, A+B+C=-4[/hide]

I think you individually find the coefficients.

[hide]For $x^2$, there is $3x^2$ and $-x^2$.  :arrow: 8

For xy, there is -12 and -4.  :arrow: -16

For $y^2$, there is only 4.

8-16+4=-4[/hide]

i don't think this is right.

[hide]9x[size=59]2[/size][size=100] - 4y[/size][size=59]2[/size][size=100]-x[/size][size=59]2[/size][size=100]-4xy.  If you simplify it, you get 8x[/size][size=59]2[/size][size=100]-4xy-4y[/size][size=59]2[/size][size=100].  8 + (-4) + (-4) gives me 0.[/size][/hide]
are use supposed to use binomial theorem?

how did you get that pianoforte?

i got it the regular way

binomial theorem is FOIL (first outer inner last), isn't it, or is that something else?

no.
Binomial Theorem is this.

$(x+y)^n=\sum_{i=0}^{n}\binom ni x^i \binom{n-i}{n}y^{n-i}$

[quote="Ravi B"][hide="Slick_Solution"]Substitute $x = 1$ and $y = 1$ into the original expression.[/hide][/quote]
Can you prove that the answer after you plug in those variables is A+B+C?

$Ax^2+Bxy+Cy^2$ If you plug in $x=1$ and $y=1$, you get $A+B+C$. Since they give you $(3x-2y)^2-x(x+4y)=Ax^2+Bxy+Cy^2$, if you plug in $x=1,y=1$, you see that the LHS=$A+B+C$

[quote="Ravi B"][hide="Slick_Solution"]Substitute $x = 1$ and $y = 1$ into the original expression.[/hide][/quote]

That is a slick solution
Clever

O.o

That was clever. Did you win your countdown round?

Haha, I'm over 40 years old.  MathCounts didn't exist when I was a kid.  Now I coach my daughter's team in New York.


geometry perimeter trigonometry factorial function logarithms algebra

The perimeter of an equilateral triangle is 36. What is the length of a side of this triangle?

[hide]12[/hide]
doesn't deserve explaning

[hide]36/3 = 12... i hope isn't a MOEMS question >_<[/hide]

Let's say we have an equilateral triangle. If you draw a perpendicular line from each corner to its opposite side, and if you take their sin, cos, tan, etc, you will get the same results for the length of the lines (since all sides of the triangle are equal, and that the sum of the angles are 180 degrees, each angle is 60 degrees. Hence, all sides are equal.). This means that you have the right to divide 36 by 3 and get 12 for the length of each side. If you would like me to mathematically explain this in depth, I will be glad to do so. Remember that somethings are very easy to apply, but hard (for some people, depending on their education level) to explain in terms of mathematics in order to leave the people with no questions.

-Masoud Zargar

yeah, in equilateral triangles medians intersect at a point 2/3 the length of the median away from the corner... but most geometry & trigonometry probably doesn't belong here

[quote="math92"]it's supposed to easy...For grades 4-8[/quote]

I think I learned that in second grade...

[quote="nat mc"][quote="236factorial"][quote="4everwise"]3 sides equal, divide by 3!  :roll:[/quote]

I thought you meant divide by 3 factorial or 6.
I think I like factorials too much...[/quote]
Is there something special about your name, 236! or is it just a random number[/quote]

Going a little off topic, the original reason I joined this forum was because I wanted to know about the number of zeros ending n!. This was the thing that stumped me on one of the state countdown rounds (2004). The question asked for the number of zeros ending 236!. I had to think of a name quick, and I was thinking about that question just then, so the name applied.

[quote="236factorial"][/quote][quote="nat mc"][quote="236factorial"][quote="4everwise"]3 sides equal, divide by 3!  :roll:[/quote]

I thought you meant divide by 3 factorial or 6.
I think I like factorials too much...[/quote]
Is there something special about your name, 236! or is it just a random number[/quote][quote="236factorial"]

Going a little off topic, the original reason I joined this forum was because I wanted to know about the number of zeros ending n!. This was the thing that stumped me on one of the state countdown rounds (2004). The question asked for the number of zeros ending 236!. I had to think of a name quick, and I was thinking about that question just then, so the name applied.[/quote]
Heres the formula for number of zeros at the end of $n!$

Denote $\tau(n)=\text{\# of zeros at end of n!}$

Denote $[x]$ as greatest integer function

Let $a=[\frac{\log{x}}{\log5}]$

$\tau(n)=\sum_{i=1}^a{[\frac{n}{5^i}]}$

[quote="nat mc"][/quote][quote="236factorial"][/quote][quote="nat mc"][quote="236factorial"][quote="4everwise"]3 sides equal, divide by 3!  :roll:[/quote]

I thought you meant divide by 3 factorial or 6.
I think I like factorials too much...[/quote]
Is there something special about your name, 236! or is it just a random number[/quote][quote="236factorial"]

Going a little off topic, the original reason I joined this forum was because I wanted to know about the number of zeros ending n!. This was the thing that stumped me on one of the state countdown rounds (2004). The question asked for the number of zeros ending 236!. I had to think of a name quick, and I was thinking about that question just then, so the name applied.[/quote][quote="nat mc"]
Heres the formula for number of zeros at the end of $n!$

Denote $\tau(n)=\text{\# of zeros at end of n!}$

Denote $[x]$ as greatest integer function

Let $a=[\frac{\log{x}}{\log5}]$

$\tau(n)=\sum_{i=1}^a{[\frac{n}{5^i}]}$[/quote]

The formula seems to be very logical.

Seems more complicated than logical to me.

[quote="boxedexe"][/quote][quote="nat mc"][/quote][quote="236factorial"][/quote][quote="nat mc"][quote="236factorial"][quote="4everwise"]3 sides equal, divide by 3!  :roll:[/quote]

I thought you meant divide by 3 factorial or 6.
I think I like factorials too much...[/quote]
Is there something special about your name, 236! or is it just a random number[/quote][quote="236factorial"]

Going a little off topic, the original reason I joined this forum was because I wanted to know about the number of zeros ending n!. This was the thing that stumped me on one of the state countdown rounds (2004). The question asked for the number of zeros ending 236!. I had to think of a name quick, and I was thinking about that question just then, so the name applied.[/quote][quote="nat mc"]
Heres the formula for number of zeros at the end of $n!$

Denote $\tau(n)=\text{\# of zeros at end of n!}$

Denote $[x]$ as greatest integer function

Let $a=[\frac{\log{x}}{\log5}]$

$\tau(n)=\sum_{i=1}^a{[\frac{n}{5^i}]}$[/quote][quote="boxedexe"]

The formula seems to be very logical.[/quote]
What do you mean by logical, it makes a lot of sense?


function
algebra
domain
complex analysis
complex analysis unsolved

Let f be a non-constant entire function and (fof)(z)=f(z) for all z. prove that f must be identity mapping.

[fof]'(z) = f'(z) => f'(f(z))f'(z) = f'(z) => f'(f(z)) = 1 (f isn't constant)

Then f'(z) = 1 in an unbounded domain, so f(z) = z + c and c = 0 from the functional equation.


The third number in a geometric sequence is 1.5 and the common ratio is $ \frac{1}{2}$. What is the sum of the first five numbers in the sequence?

$ a_3 \equal{} ar^{3 \minus{} 1} \equal{} ar^2 \equal{} a\left(\frac {1}{2}\right)^2 \equal{} 1.5 \iff a \equal{} 6$.
Thus, $ \sum_{k \equal{} 1}^5 a_k \equal{} \sum_{k \equal{} 1}^5 6\left(\frac {1}{2}\right)^{k \minus{} 1} \equal{} 6 \cdot \frac {1 \minus{} \left(\frac {1}{2}\right)^5}{1 \minus{} \frac {1}{2}} \equal{} 6 \cdot \frac {31/32}{1/2} \equal{} \boxed{\frac {93}{8}}$

this is what I got, but the answer was in decimal. Why is it in decimal if it didn't say to be in decimal form?


The math-related portion of this puzzle is located in Getting Started, because it had to do more with math.

How many actual English words can you form by permuting the letters in ARTOFPROBLEMSOLVING? By permuting I mean that if a letter appears x times, you can use it in each word up to x times.

Still counting...

[hide]
A
ABLE
ABS
AGE
AGES
ALE
ALES
ALL
ALMS
ALOE
AM
AMP
AMPING
AMPS
AN
ANT
ANTS
APE
APER
APERS
APES
APING
APRON
APRONS
ARE
ARM
ARMING
ARMOR
ARMS
ARSE
ARSON
ART
ARTS
AS
AT
ATE
ATOLL
ATOM
ATOMS
BABE
BABES
BAG
BAGEL
BAGELS
BAGS
BAIL
BAILS
BAIT
BAITS
BALE
BALES
BALL
BALLET
BALLETS
BALLOT
BALLOTS
BALLPOINT
BALLPOINTS
BALLS
BALM
BALMS
BAN
BANE
BANES
BANG
BANGER
BANGERS
BANGLE
BANGLES
BANGS
BANISTER
BANS
BANTER
BANTERS
BAPTISE
BAR
BARE
BAREFOOT
BARES
BARF
BARFING
BARFS
BARGE
BARGES
BARING
BARN
BARNS
BARON
BARONS
BARREL
BARRELING
BARRELS
BARREN
BARS
BARTER
BARTERING
BARTERS
BASE
BASER
BASIL
BASIN
BASTE
BASTER
BASTILLE
BASTION
BAT
BATE
BATES
BATING
BATON
BATONS
BATS
BE
BEAM
BEAMING
BEAMS
BEAN
BEANS
BEAR
BEARING
BEARS
BEAT
BEATING
BEATS
BEG
BEGAN
BEGAT
BEGIN
BEGOT
BEGS
BELL
BELLS
BELT
BELTING
BELTS
BENT
BENTS
BERG
BERGS
BESON
BEST
BESTING
BET
BETA
BETS
BIER
BIERS
BIG
BILL
[/hide][/hide]

[quote="jmadsen"]Still counting...

A
...
BILL
[/quote]

lol, afterwards you can start counting the russian words that can be formed with those letters...


ratio
logarithms
calculus
integration
factorial
function
complex numbers

If you want to approximate n!..
You could use this formula :)
n!  :approx: e^(-n):sqrt:(2:pi:) n^(n+0.5)

The higher n gets, the closer to 1 the ratio between this and n!.. interesting, huh?

I've read the proof for this, very nice indeed.

erm. is that stirling's approximation? and if so, i've heard that you get that by taking the first two terms or someat of the stirling series. could someone explain?

hmm...TripleM or Spoon, mind getting a proof for that?

The proof of Stirling's formula is quite hard, but it's not hard to get a somewhat weaker form.

Take the logarithm of n!. That's the sum of the logarithms of k as k goes from 2 to n. Now we integrate. We bound the integral by saying log((n-1)!) < integral from 1 to n log(t)dt < log(n!), so n!>e*(n/e)^n. Similarly, n!=n(n-1)!<en*(n/e)^n. Thus e*(n/e)^n<n!<en*(n/e)^n.

A more accurate Stirling's formula is n!=(n/e)^n*sqrt(2:pi:n)(1+1/12n+1/288n^2-139/5140n^3+O(1/n^4)).

isnt this approximation based on the taylor expansion of e?

[quote="gauss202"]No, it is not based on the Taylor Series. The proof of Stirling's Approximation is not that "hard", but the starting point is an identity that most high school students are probably unfamiliar with. It involves using the integral definition for the factorial (or actually the Gamma Function, which is a generalized factorial function). It is:

n! = integral (0 to inf) [e^-x*x^n dx].

A student familiar with calculus should be able to prove this identity. And then after that, the proof requires basic calculus, though the steps are not obvious. Someone who's completed a course in BC Calculus could probably follow it.

A very brief proof could be found here: http://mathworld.wolfram.com/StirlingsApproximation.html

And a full proof of the Stirling Series for n! can be found in a famous and classic book called, [u]A Course in Modern Analysis[/u] by Whittaker and Watson, under the chapter on the Gamma Function.[/quote]
I can prove that by integration by parts for positive integer n. Wait...does that mean the factorial function is defined for non-integral n?

[quote="MysticTerminator"]
Wait...does that mean the factorial function is defined for non-integral n?[/quote]

Yes. The extension to non-integers (actually to complex numbers) is called the Gamma Function.

The gamma function has many nice characteristics in complex analysis. It's one of the coolest functions besides zeta.

[color=cyan]It's really nasty to compute for most values without complex integration, unfortunately.  Also, you need to use fancy things like analytic continuation to get negative number factorials.

Hey, on a random note:
I have seen a "sort-of" extension of the binomial coefficients to include things like C(r, n) where r is a real and n is a positive integer, given by
C(r, n) = r(r-1)(r-2). . .(r-n+1)/n!

But, if we have an extension of the ! function to reals, we could also define it as

G(n)/(G(r)*G(n-r)), where G means gamma.  Are the two equivalent?  If not, does one have more meaning than the other?[/color]

What I love most about the gamma function is its extention beyond the n! definition....

it has so many different properties, truely amazing... check out

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

The Beta function is pretty cool too :)

In college, i want to join BetaGammaZeta :):)


I'm typing some research in Microsoft Word right now, and occasionally I need to add in a few math expressions, so what I've been trying to do is go to the AoPS TeXer, produce the expression I need, and copy and paste it to my Microsoft Word document. However, when I copy and paste the image to my document, for some reason I get the negative of the image (the background is black, and the math is white). I think I found some way to fix this in the past, but I don't remember how I did it now.  :oops:
Can anyone tell me some way to fix this? Thanks.

EDIT: I found a way to solve the problem!  :)  Just paste the image into Paint, and copying works fine.

EDIT 2: No, it doesn't look anywhere near as good as my sig (before I found how to fix it). It looks like this:
[hide="Picture"][img]http://img220.imageshack.us/img220/7782/untitledgf4.png[/img][/hide]

After I fixed it, it looked more like this:
[img]http://img220.imageshack.us/img220/5240/4ade4fd0b237ead7dfdc93axp3.gif[/img]

There is an EXCELLENT soft (But sheareware!) who answer exactly your question, all you need is to install LaTeX system (MikteX) and the [url=http://www.nameless.ca/]ribbit[/url] software.
You can Try it for 30 days.

Hi,
TexPoint can do the job. It's a TeX add-on for Powerpoint that I use to generate equation images, but you should be able to copy the equation image from Powerpoint and paste it into Word.
SSN.

i didn't download the LaTeX stuff so what i do is post it on the test forum....take the print screen image and then just change the background color to white...and then paste on Word  :D

Saving the image rather than copying and pasting should solve the problem of colour reversal.

Or if you are working only in paint and copy and pasting, doesn't paint have an invert colors option? SO you invert once then the bug inverts it back :P

Try a [url=http://office.microsoft.com/en-us/default.aspx]trial of Microsoft Office 2007[/url], which supports math equations.  Go under Insert->symbols->equation (shortcut Alt N E I).  What's interesting about it is that it more or less supports $\LaTeX$ commands, so you can just type \sqrt and it will give $\sqrt{\ }$.

What's wrong with using the [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_TeXer.php]TeXer[/url]?

-jorian

you could just copy paste.

That was exactly what nebula42 did do and was why he asked the question. Copying and pasting sometimes inverts the image but saving and inserting may be fine. However, nebula42 did manage to solve the problem anyway.


[size=200]Arbelos 1982-1983[/size]
[size=75]
(the arbelos doesn't include solutions so i've included my results in hide tags, they may/may not be right)
[/size]
1. When the Easter vacation arrived, Hyde and Zeke decided to go to Florida. Hyde left in a transport of joy, but Zeke took the trip more calmly, on a higher plane, so to speak.

If the plane travled nine times as fast as the transport, and Zeke got 36 hours more than Hyde at the beach, how long did each take traveling?

[hide="my results"]
4.5 hrs and 40.5hrs
[/hide]

2. The area of a plane section of a circular cylinder is equal to $16\pi$. The area of a plane section of the same cylinder, which is perpendicular to the plane of the First section, is equal to $9\pi$. What is the radius of the cylinder?

3. If a yardstick is marked off into $\frac{1}{4}$ inches and a meter stick marked off into centimeteres are placed with their initial points together, which two marks on the two sticks will most nearly coincide ?(our almanac tells us that one meter is exactly $39.37$ inches long)
[hide="myresults"]
19.65
39.30
[/hide]

4. The e quation $x^2 - 97x + A = 0$ has roots equal to the fourth powers of the roots of $x^2 - x + B = 0$. What is the value of A?
[hide="my results"]
$a = 2$
$b = 3$
$B = 6$
$A = 6^4 = 1296$
[/hide]
5. A 5-digit number is a multiple of 41. If the highst order digit be removed and placed to the right of the units digit, the new number is a perfect cube. What was the original number?

[hide="my result"]
the number has to be 1 or 6 $\pmod{7}$
Otherwise I don't know
[/hide]

[hide="number 4 solution"]
Let the roots of the second equation be $a$ and $b$. We find that $a+b=1$.
The roots of the first equation are $a^4$ and $b^4$. They sum up to 97.

$a^4+b^4=97$
$a+b=1$

Trial and error gives us $a=3$, $b=-2$.

Thus the solutions of the first equation must be 81 and 16. $A$ is their product, or 1296. [/hide]

For number 5, since you have a restriction, you can just guess and check. There aren't too many numbers whose cubes are 5 digit numbers, while having a remainder of 1 or 6 when divided by 7.

[hide="Number 1 solution"]
Let $s$ be the speed of the transport and $t$ be Hyde's travel time. This would make the speed of the plane $9s$ and Zeke's travel time $t-36$.
$st=9s(t-36)$
$st=9st-324s$
$-8st=-324s$
$t=40.5$

Hyde's travel time was $40.5$ hours and Zeke's was $4.5$ hours.
[/hide]


64%of the homes for sale have a garage, 21% have swimming pools, and 17% have both features.
a.)If a home for sale has a garage, what's the probability that it has a pool too?


According to my wife, who is mexican, burritos and nachos are NOT Mexican foods (despite what you may have heard from Mexican restaurants in America). She is also very doubtful whether what you are thinking of as "tacos, enchiladas, and chalupas" are actally anything close to what they actually are in Mexico.

By the way, I am in Mexico right now, and I can confirm that most of the foods that are called Mexican foods in America are not anything like what people actually eat here. For example, what are called "chalupas" at Taco Bell are so far from the real thing that most Mexican's would not even recognize them. (The foods here are soooo good by the way, my favorites are tacos de carnitas, pozole, and sopes).

Edit: Here is a picture of chalupas.

Turkey in [i]mole poblano[/i]? Or anything else involving [i]mole poblano[/i]? (Chocolate and chile - plus many other spices - does make for quite an interesting combination.)

Or how about homemade tamales - make it a New Year's tradition!

I am very sorry for the inaccurate poll choices.  :rotfl:

I don't like Mexican food, chinese food ftw!

I like real mexican food and the americanized stuff - but that may be due to my addiction to capsaicin.

No Mexican food for me. I absolutely hate it, but its one of the only things people around here like...

A poll on Mexican food gets a bunch of replies, responses, and views, while a poll on fruits gets near to nothing.  :|

Was it all because of ftw advertising?

You learn well, my friend.  :D  :D  :rotfl:


function
geometry
combinatorics unsolved
combinatorics

The game of $pebbles$ is played on an infinite board of lattice points $(i,j)$. Initially there is a $pebble$ at $(0,0)$. A move consists of removing a $pebble$ from point $(i,j)$and placing a $pebble$ at each of the points $(i+1,j)$ and $(i,j+1)$ provided both are vacant. Show taht at any stage of the game there is a $pebble$ at some lattice point $(a,b)$ with $0 \leq a+b \leq 3$

This looks like a problem posted here a while ago, with three pebbles initially at $(0,0)$, $(0,1)$ and $(1,0)$. I think Pierre gave a nice solution to it, based on a simple weighting function, and I guess the same will work here: assign a weight $2^{-i-j}$ to the point $(i,j)$. Then clearly each move preserves the total weight of points with pebbles on them.  Now the initial weight is $1$, and the weight of the entire board except those points $(a,b)$ with $0\le a+b\le3$ is $\sum_{i=0}^\infty\sum_{j=0}^\infty2^{-i-j}-1-2.\frac{1}{2}-3.\frac{1}{4}-4.\frac{1}{8}=\frac{3}{4}<1$, so if that area is empty the weight of the stones must be less than one, which is impossible.


function
real analysis
real analysis unsolved

There are a lot of missing symbols in your problem so it's hard to completely understand what you want.

Try to fix your syntax.

[quote="Nischal"]Let  be an equicontinuous family of functions in C(K), where K is a compact subset of . Suppose that for each ,
 is finite. Prove that  is bounded.

[Hint]: Use equicontinuity to bound  by  on a ball about x. Suppose that  tends to +∞. Extract a convergent subsequence of .[/quote]You have to spend at least a minimum amount of effort when you copy and paste your homework problems.

Sorry about that. :blush: . ok here is the actual question:
Let  F be an equicontinuous family of functions in C(K), where K is a compact subset of R^n . Suppose that for each , x belongs to K,
sup{f(x), f belong to F} = M
 is finite. Prove that F is bounded.

[Hint]: Use equicontinuity to bound  F by  M+1 on a ball about x. Suppose that  mod(f_n(x_n))tends to +∞. Extract a convergent subsequence of .{x_n}


In a plane consider the circle with center, the point $ O$, radius $ r$. The angle formed by 2 tangent lines drawn the point $ A$ that is outside is $ \frac{\pi}{6}$. Find the value of $ \frac{r}{OA}$.

This is too easy. If P is a point of tangency, then $ \angle PAO \equal{} 15$, so $ \frac {r}{OA} \equal{} sin 15 \equal{} \sqrt {\frac {1 \minus{} cos{30}}{2}} \equal{} \frac {\sqrt {6} \minus{} \sqrt {2}}{4}$

That's correct.  :) I should have posted in high school basic forum.


[quote="vietnam2007"]let $a,b,c$ be non-negative real numbers such that $abc=1$

prove that :

$\frac{1}{\sqrt{ b+\frac{1}{a}+\frac{1}{2}}}+\frac{1}{\sqrt{ c+\frac{1}{b}+\frac{1}{2}}}+\frac{1}{\sqrt{ a+\frac{1}{c}+\frac{1}{2}}}\geq \sqrt 2$[/quote]
$\sum_{cyc}\frac{1}{\sqrt{ b+\frac{1}{a}+\frac{1}{2}}}=\sum_{cyc}\sqrt{\frac{2a}{2ab+a+2}}=$
$=\sum_{cyc}\sqrt{\frac{2}{2b+1+2bc}}\geq\sum_{cyc}\frac{\sqrt2}{1+b+bc}=\sqrt2.$ :)

[hide]
After substituting $a=\frac{x}{z},b=\frac{y}{x},c=\frac{z}{y}$, it becomes
\[\sum_\text{cyc}\sqrt{\frac{x}{x+2y+2z}}\geq 1\]
To deal with this, observe that $\sqrt{\frac{x}{x+2y+2z}}=\frac{x}{\sqrt{x(x+2y+2z)}}\geq \frac{x}{x+y+z}$ by AM-GM on $x$ and $x+2y+2z$.

Now just sum cyclically to get desired inequality.
[/hide]

[quote="scorpius119"]
After substituting $a=\frac{x}{z},b=\frac{y}{x},c=\frac{z}{y}$, it becomes
\[\sum_\text{cyc}\sqrt{\frac{x}{x+2y+2z}}\geq 1 \]
To deal with this, observe that [color=red]$\sqrt{\frac{x}{x+2y+2z}}=\frac{x}{\sqrt{x(x+2y+2z)}}\geq \frac{x}{x+y+z}$[/color] by AM-GM on $x$ and $x+2y+2z$.

Now just sum cyclically to get desired inequality.
[/quote]
I think you've made some mistakes.
First in the third line, $\sqrt{\frac{x}{x+2y+2z}}=\frac{\sqrt{x}}{\sqrt{x(x+2y+2z)}}$
And then, your inequality holds when $x+y=y+z=z+x=0 \Leftrightarrow x=y=z=0$ therefore $a,b,c$ is meaningless.

[quote="arqady"][quote="tunganh"]
First in the third line, $\sqrt{\frac{x}{x+2y+2z}}=\frac{\sqrt{x}}{\sqrt{x(x+2y+2z)}}$
[/quote]
$\sqrt{\frac{x}{x+2y+2z}}=\frac{x}{\sqrt{x(x+2y+2z)}}.$ :wink:[/quote]
Oh, thanks for correcting :)
But what about the condition when the inequality holds? I don't think it's true :D

[quote="tunganh"]
And then, your inequality holds when $x+y=y+z=z+x=0 \Leftrightarrow x=y=z=0$ therefore $a,b,c$ is meaningless.[/quote]
Actually, equality holds when any two of $x,y,z$ are 0 (since equality in the $\frac{x}{\sqrt{x(x+2y+2z)}}$ term holds when $x=0$ or $y=z=0$). This is meaningful: if we let $a=\epsilon ,b=1, c=\frac{1}{\epsilon}$ then we approach equality as $\epsilon \rightarrow 0$. But then again, the inequality could have been horribly weak in the first place (like if they had used $\frac{7}{5}$ instead of $\sqrt{2}$ for the right hand side).

[quote="FOURRIER"][quote="scorpius119"]
After substituting $a=\frac{x}{z},b=\frac{y}{x},c=\frac{z}{y}$, it becomes
\[\sum_\text{cyc}\sqrt{\frac{x}{x+2y+2z}}\geq 1 \]
To deal with this, observe that $\sqrt{\frac{x}{x+2y+2z}}=\frac{x}{\sqrt{x(x+2y+2z)}}\geq \frac{x}{x+y+z}$ by AM-GM on $x$ and $x+2y+2z$.

Now just sum cyclically to get desired inequality.
[/quote]
So whenever you have abc=1 you put a=x/y , b=y/z , c=z/x?[/quote]

its a useful trick, comes handy very often. :D

Let $a,b$ be positive real numbers such that $ab=1.$ Prove that
$$\frac{1}{\sqrt{a+\frac{1}{2}}}+\frac{1}{\sqrt{b+\frac{1}{2}}}>\sqrt{2}$$
[b]Solution:[/b]
$$\frac{1}{\sqrt{a+\frac{1}{2}}}+\frac{1}{\sqrt{b+\frac{1}{2}}}=\sqrt{2}\left(\frac{1}{\sqrt{2a+1}}+\frac{1}{\sqrt{2b+1}}\right)>\sqrt{2}\left(\frac{1}{a+1}+\frac{1}{b+1}\right)=\sqrt{2}$$Let $a,b,c$ be positive real numbers such that $abc=1.$ Prove that
$$\frac{1}{\sqrt{a+\frac{1}{b}+\frac{1}{2}}}+\frac{1}{\sqrt{b+\frac{1}{c}+\frac{1}{2}}}+\frac{1}{\sqrt{c+\frac{1}{a}+\frac{1}{2}}}>\sqrt{2}$$
[b]Solution:[/b]
$$ \sum_{cyc} \frac {1}{\sqrt {b + \frac {1}{a} + \frac {1}{2}}}=\sqrt 2\sum_{cyc}\frac{1}{\sqrt{2\left(b+\frac{1}{a}\right)+1}} >\sqrt {2}\sum_{cyc}\frac{1}{b+\frac{1}{a}+1}=\sqrt 2$$
Let $a,b,c,d$ be positive real numbers such that $abcd=1.$ Prove or disprove
$$\frac{1}{\sqrt{a+\frac{1}{b}+\frac{1}{c}+\frac{1}{2}}}+\frac{1}{\sqrt{b+\frac{1}{c}+\frac{1}{d}+\frac{1}{2}}}+\frac{1}{\sqrt{c+\frac{1}{d}+\frac{1}{a}+\frac{1}{2}}}+\frac{1}{\sqrt{d+\frac{1}{a}+\frac{1}{b}+\frac{1}{2}}}>\sqrt{2}$$
or
$$\frac{1}{\sqrt{a+\frac{1}{b}+\frac{1}{2}}}+\frac{1}{\sqrt{b+\frac{1}{c}+\frac{1}{2}}}+\frac{1}{\sqrt{c+\frac{1}{d}+\frac{1}{2}}}+\frac{1}{\sqrt{d+\frac{1}{a}+\frac{1}{2}}}>\sqrt{2}$$
[url=https://artofproblemsolving.com/community/c6h623834p3734308]here[/url]

Let $a,b$ be positive real numbers such that $ab=1.$ Prove that
$$\frac{1}{\sqrt{2a+\frac{1}{2}}}+\frac{1}{\sqrt{2b+\frac{1}{2}}}\geq 2\sqrt{\frac{2}{5}}$$
$$\frac{1}{\sqrt{4a+\frac{1}{2}}}+\frac{1}{\sqrt{4b+\frac{1}{2}}}\geq \frac{2\sqrt{2}}{3}$$
[url= https://artofproblemsolving.com/community/c6h402967p2245109]h[/url] [url=  https://artofproblemsolving.com/community/c6h1643258p10363042]h[/url]


solve the equation $\sqrt{a(2^x - 2) + 1} = 1 - 2^x$ for every value of the parameter a

[quote="Fermat -Euler"][hide=" idea"]if $2^x\leq1$ then
$a(2^x - 2) + 1 =( 1 - 2^x)^2$
$2^{2x}-2^x(a+2)+2a=0$
put $X=2^x$ then
$X^2-X(a+2)+2a=0$
and the equation it's easy now[/hide][/quote]

$X=\frac{(a+2)\pm\sqrt{(a+2)^2-8a}}{2}$
$\Rightarrow X=\frac{(a+2)\pm(a-2)}{2}$
either X=a or X=2.
Resppectively, x=\log_{2}{a} or x=1.
But  a should be a positive real number.


I need help to solve this:
The percent of concentration of a certain drug in the bloodstream x hr after the drug is administered is given by K(x) =(4x)/(x^2 + 25).
At what time is the concentracion a maximum?
Thank You

$ K(x) \equal{} \frac {4x}{x^2 \plus{}25}$
differentiating wrt $ x$ we get
$ K'(x) \equal{} \frac{4(x^2 \plus{} 25) \minus{} 2x*4x}{(x^2 \plus{}25)^2}$

for extrema, $ K'(x) \equal{} 0$
solving we get $ x \equal{} \plus{}5/\minus{}5$
obviously $ \minus{}5$ isnt possible, so at $ x\equal{}5$ hrs we have a extrema.

now, differentiating $ K'(x)$ wrt x again and substituting $ x\equal{}5$,
we get $ K''(5) <0$
so $ x\equal{}5$ is maxima.
hence, $ x\equal{}5$ hrs is the time at which concentration is maximum
:)


probability
search
probability and stats

Is there a probability distribution on $ {\bf Z}_{>0}$ such that the probability of selecting an integer divisible by $ p$ is $ \frac{1}{p}$ for all primes (or more strongly, all positive integers) $ p$?

I believe the answer to the first question is yes.  Let $ p_i$ be the $ i$th prime.  For convenience, set $ p_0 \equal{} 1$.  Let $ q_i$ be the product of the first $ i$ primes.  (In particular, $ q_0$ is 1.)  Assign the number $ q_i$ a probability of $ \frac{1}{p_i} \minus{} \frac{1}{p_{i\plus{}1}}$.  Every other number is assigned a probability of zero.  I believe this probability distribution satisfies your property.

I haven't thought about the second question.

I believe the answer to the second question is no.  We can't even make the condition work for all squarefree numbers.  Here is a sketch of the argument.

Suppose the condition did hold for all squarefree numbers.  We can then easily show that the events of being divisible by 2, 3, 5, etc., are mutually independent.  In particular, the probability of being nondivisible by 17, 19, 23, 29, 31, etc., is 0.  (Here I am using the fact that the sum of inverse primes is infinite.)  It follows that the probability of being 1 through 16 is 0.  By changing the 17 to an arbitrarily large prime, we get that the probability of each integer is 0, which is a contradiction.

Here's a remaining question.  Can we make the condition hold for all primes and products of two distinct primes?

As a side note, the "stronger" problem has been discussed [url=http://www.mathlinks.ro/viewtopic.php?search_id=421043866&t=152341]here[/url].


induction
polynomial
ratio
Gauss
number theory
prime numbers
modular arithmetic

Do there exist infinitely many primes in the form of (n!)2+1, where n is a positive integer?

As far as I know, this is an old open problem to determine if there are infinitely many primes of the form $n^2 + 1$, thus your question is still open too...

Pierre.

I thought this would be straight-forward? (or am I being too simplistic?)

[b]We start with this:[/b]
If we put $k$ as the product of all known consecutive primes, then $k + 1$ cannot have any of those lesser primes as a factor. This means $k + 1$ is either prime or, a product of primes larger than the largest prime used as a factor of $k$. No matter how large the prime we use to make $k$ larger, this is always true (so there are an infinite number of primes of the form $k + 1$)

(Note that while $k + 1$ (being very large) will 'more probably' be a composite number, there is nothing in the statement $k + 1$ that specifically forbids $k + 1$ from being prime when all possible values of $k$ are considered)

[b]By induction, we then go on to this:[/b]
Similarly, if we put $n = k$ = the product of all known consecutive primes, then $n^2 + 1$ cannot have any of those lesser primes as a factor either. This means $n^2 + 1$ is either prime or, a product of primes larger than the largest prime used as a factor of $n$. No matter how large the prime we use to make $n$ larger, this is always true (so there are an infinite number of primes of the form $n^2 + 1$)

(Note that while $n^2 + 1$ (being very much larger than $k + 1$) will even more probably be a composite number, there is nothing in the statement $n^2 + 1$ that specifically forbids $n^2 + 1$ from being prime when all possible values of $n$ are considered. {As a comparison, the same cannot be said for $n^2 - 1 = (n - 1)(n + 1)$ which is always composite for $n > 2$}).

[b]Using the preceding logic, we can now proceed to this much more generalized statement:[/b]
If there is nothing in the way that an infinite set of ordered numbers is expressed that specifically forbids that set of numbers from being prime (such as given in $n^2 - 1$ above) then there are an infinite number of primes in that set of numbers.

e.g._1
$2^m + 1$ (where m = 1, 2, 3, ..., infinity) is an infinite set of numbers that cannot have 2 as a factor, therefore $2^m + 1$ can only either be a prime or a product of any primes greater than 2.

e.g._2
$2^m.3^n + 1$ (where m, n = 1, 2, 3, ..., infinity)  is an infinite set of numbers that cannot have either 2 or 3 as a factor, therefore $2^m.3^n + 1$ can only either be a prime or a product of any primes greater than 3.

e.g._3
$2^m.3^n.5^p + 1$ (where m, n, p = 1, 2, 3, ..., infinity)  is an infinite set of numbers that cannot have either 2, 3, or 5 as a factor, therefore $2^m.3^n.5^p + 1$ can only either be a prime or a product of any primes greater than 5.

(Methinks the general case applies to the ordered infinite set of numbers given as Eisensteins conjecture as well, thus disproving it also :) )

John

[quote="ZetaX"]:huh: I agree with you till the point where you say "so there are an infinite number of primes of the form $n^2 + 1$", I see no way to get that from the previous statements. All you gave were heuristic reasons...
Note: there is as good as no chance for e simple prove of that statement...[/quote]If you agree up to that point, then you must agree with what I said immediately prior to that  [quote]$n^2 + 1$ is either prime or, a product of primes larger than the largest prime used as a factor of $n$. No matter how large the prime we use to make  larger, this is always true[/quote] i.e. [b]$n^2 + 1$ can only be either prime or composite[/b], if it's prime then it's quite clearly a prime of the form $P = n^2 + 1$ (like - ?QED?). It would be ludicrous to even suggest that $n^2 + 1$ must always be composite when the infinite number of possibilities for the value of $n$ are considered.

(N.B. What I gave before that (for $k + 1$) was essentially a restatement/formalization of the Euclidian proof of the infinitude of primes in the complete ([i]infinite[/i]) set of positive integers, and I hold that Euclids proof has equal validitity for the ([i]also infinite[/i]) subset of positive integers given by $n^2 + 1$ - Can I take it that you're holding that the subset $n^2 + 1$ may have something unique in the statement that precludes it from being prime past some unspecified point on the number line? What/where is this point? And what is the reason for holding that this is {or even [b]may[/b] be} true. Or do you perhaps hold this simply because you haven't found one such prime yet? - Keep looking). :)

[quote="blahblahblah"]You really should do a better job of defining the terms you use.  You 'define' $n$ to be the product of all known consecutive primes.  What do you mean by consecutive primes?  Do you mean twin primes?  Whatever is wrong with your 'proof' will doubtless become more transparent if you present it in more clarity.

As a litmus check, why don't you use your method to prove that there is an infinitude of primes of the form $4k+1$ and $4k+3$?[/quote]

I'm sorry, I had thought that the term 'consecutive primes' (although it's not [u]exactly[/u] the same meaning, I could've equally well have used 'successive') would be quite self-explanatory, 2 is the first prime, 3 is the next, 5 is the next, 7 the next, these are all the consecutive primes up to 7. If we go on from there, 11 is the next consecutive prime, 13, 17, 19, 23 and so on. (? twin primes weren't even mentioned ?)

Sorry, it's not 'my' method, the 'method' itself really belongs to Euclid...

$4k + 1$? assuming you're referring to $k$ in the sense that I used it above, one example is this - assume we only know all the primes up to 17, then $4.2.3.5.7.11.13.17 + 1 = 2^3.3.5.7.11.13.17 + 1$. Obviously 2, 3, 5, 7, 11, 13, and 17. cannot be a factor of $2^3.3.5.7.11.13.17 + 1$ hence$2^3.3.5.7.11.13.17 + 1$ is either a prime or the product of two or more primes greater than 17. We can put as many further consecutive primes (or even integer exponents of them) as we like in $k$ to make the sum $4k + 1$ larger, but the result always holds true, $4k + 1$ can only either be a prime or the product of two primes larger than the largest prime used as a product of $k$. (BTW - Did you even look at the examples I gave above before asking this?)

$4k + 3$?
$4k + 3$ is NEVER prime for $k = 2.3.5.7.11.13.17.19.23. .... etc$ as 3 can always be factored out to give $3.(2.5.7.11.13.17.19.23. .... etc + 1)$. This is what I meant by 'unless specifically forbidden from being prime' ([i]or something similar[/i]), however if 3 (and all integer exponents of 3) is omitted as a factor of $k$, the result holds true. For more on this I suggest you also read my own post http://www.artofproblemsolving.com/Forum/viewtopic.php?t=61433 (for which I'm still waiting for assistance).

When he says 'primes of the form $4k+3$ and $4k+1$', he means $p \equiv 1 \bmod 4$ or $p \equiv 3 \bmod 4$.

Your proof is invalid.  You've only shown that there probably exist some prime values, but you haven't actually shown that the number of prime values is infinite.  Moreover, the frequency of primes decreases faster than the frequency of numbers of the form $n^2 + 1$, so it is a given that as $n \to \infty$, the density of prime values must decrease.  The problem is to show that, even though the density decreases, there are still an infinite number of prime values (or there aren't).

[quote="t0rajir0u"]... the frequency of primes decreases faster than the frequency of numbers of the form $n^2 + 1$, so it is a given that as $n \to \infty$, the density of prime values must decrease.  The problem is to show that, even though the density decreases, there are still an infinite number of prime values (or there aren't).[/quote]Forgive me - but I recollect that as being the exact same argument that mathematicians of the time were using for the set of integers that can be expressed as n + 1 (where n = 1, 2, 3, ...., infinity) when Euclid resolved the question in the manner shown above. The actual relative density, or ratio of the primes to composites was a separate problem... As the density was later shown by Gauss and others to be logarithmic in nature, of course the number of primes never goes to zero. (This also applies if the density in other subsets were to be log of a log in nature or even log to some exponent in nature - whatever - it never goes to zero).

IMHO you should be looking at the problem from this perspective: In the infinite set of numbers expressed as n + 1 (n = 1, 2, 3, ...., infinity) there are an infinite number of primes.

If we look at any given ordered subset of those numbers, this subset may very well consist of only an exceedingly small fraction of the original set of numbers, but any fraction of infinity is still infinite in number. If the subset also contains a proportion of the primes taken from the original set, well - again - a fraction of infinity is still infinite in number.

You say it yourself:
[quote="johnske"][b][u]If[/u][/b] the subset also contains a proportion of the primes taken from the original set, [...] [/quote]
Who says that that this happens¿
You seem to claim the following:
If we cannot (on no way) prove that there are not infinetely many primes in the considered set, then there are infinetely many primes in it.
1. This sounds just funny, since it says nothing we can really check.
2. I'm not sure if this statement is really true, I think not.

[quote="t0rajir0u"]...Your proof is invalid.  You've only shown that there probably exist some prime values, but you haven't actually shown that the number of prime values is infinite.  Moreover, the frequency of primes decreases faster than the frequency of numbers of the form $n^2 + 1$, so it is a given that as $n \to \infty$, the density of prime values must decrease.  The problem is to show that, even though the density decreases, there are still an infinite number of prime values (or there aren't).[/quote]
(Have some time to respond to this with a little more rigour now)

"[b]It's a given[/b]" is it? Methinks you are making the common mistake of comparing apples to oranges here. I’ll explain as best I can…

If we look at how many primes of the form $n + 1$ (where n = 1, 2, 3, …, $\infty$) there are in the range between $N_1$ and $N_2$, and look at the number of primes of the form $n^2 + 1$ there are in the same numerical range, naturally there will be very many more primes of the form $n + 1$ than there are of the form $n^2 + 1$ (where n = 1, 2, 3, …, $\infty$).

For example ([i]on a very simple analysis[/i]) there are 19 primes of the form $n + 1$ between 1 and 70 and only 4 primes of the form $n^2 + 1$ between 1 and 70, there would thus seem to be a total of 23 primes in this region.

A more thorough analysis will show that the four primes also given by $n^2 + 1$ are included in the 19 primes given by $n + 1$, so they’re being counted twice ([i]see footnote[/i]). If we disallow those four primes from the number of primes given by $n + 1$, this then leaves 15 primes that are of that form and that form only.

Either way ([i]whether 19 or 15 primes[/i]) we could readily jump to the simple conclusion that the number of primes of the form $n^2 + 1$ that occur in a finite range is very much less than the number of primes of the form $n + 1$ in that same ([i]finite[/i]) range. The ratio of these being very approximately $($number of primes of form $n + 1): ($number of primes of form $n^2 + 1)^2$

Now, we already know the density on primes of the form $n + 1$ decreases logarithmically ([i]according to the approximation $NumPrimes(N) = N/lnN$[/i]) and can thus never go to zero, we could then perhaps wonder whether the density of the primes described by $n^2 + 1$ decreases at a different rate and becomes so very much less that it eventually reaches zero at some point on the number line.

Unfortunately what we’re doing here so far is really comparing “apples against oranges” and we’ll thus get quite erroneous results... The primes of the form $n + 1$ belong to the set of numbers that are described by $n + 1$ (where n = 1, 2, 3, …, $\infty$), whereas the primes that are of the form $n^2 + 1$ belong to the set of numbers described by $n^2 + 1$ (where n = 1, 2, 3, …, $\infty$) ) and this second set is a [u]completely different[/u] set of numbers that’s a subset of the first.

When we are to look at a finite set of numbers that are eventually to be used to later consider an infinite set of numbers, by rights we should be looking at how many primes of the form $n + 1$ there are when n = 1, 2, 3, …, 70 and how many primes of the form $n^2 + 1$ there are when n = 1, 2, 3, …, 70.

For this case there are approximately the [b]SAME[/b] number of primes of each of the two forms, 16 primes that are exclusive to the set $n +1$ and 15 primes that can be described by the subset $n^2 + 1$. ([i]For a total of 31 prime numbers[/i])

When we consider numbers between other values for n we get very similar results. For example, if we consider the number of primes between $n = 1.10 = 10$ and $n = 70.10 = 700$ (where n = 1.10, 2.10, 3.10, 4.10, …, 70.10) we find there are 30 primes of the form $n + 1$ and 30 primes of the form $n^2 + 1$. ([i]For a total of 60 prime numbers[/i])

So, does the density of primes of the form $n^2 + 1$ decrease at a greater rate than the primes given by $n + 1$? If you take those two examples alone, if anything the opposite appears true, the density appears to be decreasing at a slightly lesser rate!

However when enough examples are considered, the densities are found to decrease at the same rate, and, as the primes in the $n^2 + 1$ set of numbers also decrease in a logarithmic fashion ([i]i.e according to the approximation $NumPrimes(N^2) = N^2/lnN^2$[/i]) in an identical manner to the density given for the set of numbers given by $n + 1$, the number of such primes can never go to zero either, there are thus an infinite number of primes of the form $n^2 + 1$

[b][u]Footnote[/u][/b]

[b]Proposition I:[/b] Prime numbers that are of the form $n^2 + 1$ always belong to the set of numbers expressed as $n^2 + 1$. These particular primes also always belong to the set of prime numbers that can be expressed in the form $n + 1$. ([i]as the complete subset ($n^2 + 1$) is always a subset of the main set $n + 1$[/i]).

[b]Proposition ii:[/b] The larger body of prime numbers that are exclusive to the main set $n + 1$ and cannot be found in the subset ($n^2 + 1$) are unique to the main set $n + 1$, whereas ([i]from the definition given in Proposition i[/i]) prime numbers that can be found in both the subset and the main set properly belong to the subset $n^2 + 1$.

[b]Note:[/b] If we wish to count how many primes of the form $n + 1$ there are, and compare this against the number of primes of the form $n^2 + 1$, then, when considering the number of primes of the form $n + 1$, we really need to look at the primes that can only be found main set ($n + 1$) and not those that are obtained by delving into the subset ($n^2 + 1$) as these are part of the subset $n^2 + 1$.

PS: I am not familiar with modulo arithmetic, can see no point in using it for this discussion, and thus am, and have not been, using it :)

[quote="johnske"]However when enough examples are considered, the densities are found to decrease at the same rate, and, as the primes in the $n^2 + 1$ set of numbers also decrease in a logarithmic fashion ([i]i.e according to the approximation $NumPrimes(N^2) = N^2/lnN^2$[/i]) in an identical manner to the density given for the set of numbers given by $n + 1$, the number of such primes can never go to zero either, there are thus an infinite number of primes of the form $n^2 + 1$[/quote]

Nothing you have posted has given any indication why that result (which is in fact stronger than the unsolved problem as to whether there are infinitely many primes of the form $n^2+1$) may or may not be true.

As I said a couple of posts back, use your method to show that there are infinitely many primes of the form $4k+1$ and $4k+3$, and then we might take you more seriously, that is, there are infinitely many primes which leave remainder $1$ when divided by $4$, and infinitely many which leave remainder $3$ when divided by $4$.

Frankly, since nothing you've posted gives the indication that you know even rudimentary number theory, e.g, modular arithmetic, this thread is little more than good-spirited crankery (which, if you've ever browsed sci.math, is at least better than bad-spirited crankery).


[hide]Since 89 is the smallest number in this set, then that must be the case when the average of the smallest 3 are divided by the largest number.  But if that's true, then we have a contradiction because at least one of the three "smaller" numbers exceed the largest number.[/hide]

let $S$ be the sum, then

$\frac{S-a}{3a}=89$, etc

$\frac{S}{a}=3*89+1$

$\frac{S}{3*89+1}=a$

sum them...but then the S's cancel out... and that does not work since each is $<\frac{1}{4}$, so summing them $<1$

$\frac{1}{3*89+1}+\frac{1}{3*95+1}+...\ne 1$

(contradiction)

[quote="José"]I think you're missunderstanding the problem....[/quote]

The normal thing to do at this time would be to [i]explain what the problem means[/i], rather than just saying that they misunderstand.  Especially because, as far as I can tell, the problem is nonsense, for the reasons that they have mentioned.

[quote="ch1n353ch3s54a1l"][quote="José"]We have four positive integrer numbers. If we choose three of them, we get their average, and we divide this average by the fourth number, we get $89;95;101;117$. Find the four numbers.[/quote]

$\frac{a+b+c}{3d}=89$, $\frac{a+c+d}{3b}=95$, $\frac{a+b+d}{3c}=101$, $\frac{b+c+d}{3a}=117$ and then do you solve those...is that what you mean?[/quote]

Exactly!!!!! See JBL that the problem is not nonsense......

[quote="t0rajir0u"]The thing about this problem is that the values given are invariant under multiplication of the four original numbers by a constant, which means we can assume WLOG that $S = 1$.  This leads to the contradiction shown by Altheman.[/quote]

Well, but the problem is not nonsense... it HAS solution

[hide]
Calling the four numbers $a, b, c, d$, we have the equations:

$\frac{a+b+c}{3} + d = 89 \rightarrow a+b+c+3d = 267$
$\frac{b+c+d}{3} + a = 95 \rightarrow b+c+d+3a = 285$
$\frac{c+d+a}{3} + b = 101 \rightarrow c+d+a+3b = 303$
$\frac{d+a+b}{3} + c = 117 \rightarrow d+a+b+3c = 351$

Adding all four equations together:

$6(a+b+c+d) = 1206$
$a+b+c+d = 201$

Plugging this value into each of the four equations, $d = 33, a=42, b=51, c=75$
[/hide]

[quote="Severius"]There's an error in the original statement. It should say "we add the fourth number to this average", instead of "we divide this average by the fourth number".

Source: http://www.oma.org.ar/enunciados/oma20reg.htm[/quote]
That made the problem A LOT easier. :D

[quote="Severius"]There's an error in the original statement. It should say "we add the fourth number to this average", instead of "we divide this average by the fourth number".

Source: http://www.oma.org.ar/enunciados/oma20reg.htm[/quote]

Thanks


For no work shown, I guess that is okay. While your request for hiding posts course is reasonable, but I think that the proposed plan of action, deleting full, yet unhidden posts, oversteps your authority.

you could be like from now hide your posts because some people are new here

I agree with Altheman...what would t0rajir0u  say when he accidently forgot the hide tags in a very looong proof explanation?  :D   :P

I'm thinking that playing seek and destroy with the entire HSB forum is way too much effort. A better idea might be to make a stickied thread saying "Please hide all your solutions." or something along the lines of that.

[quote="13375P34K43V312"][quote="jli"]I'm deleting all posts that either dont have work shown or are not hidden[/quote]

Why don't you just edit the posts and make the solutions hidden, and if the person is a new member drop them a line via pm about hiding posts?[/quote]

because i have edited so many in the last week

srry...maybe i was overreacting a bit  :D but u guys get te point.  just hide ur posts

[quote="jli"]wondering where your post is[/quote]

NO, why would I?


I don't know why you're multiplying by (-2i)/(2i) in the first place.
To keep the value of your expression the same, multiply by 1, or (2i)/(2i).

Wait, after further perusal of your post, that must have been a typo. I think the calculator is wrong, then.

Last edit, I promise. :)
When you punch it in your calculator, put parentheses around 2i; otherwise, the calculator interprets it as ((8+20i)/2)*i.

I meant to put (-2i/-2i) but I guess 2i/2i works too. I just always thought to get rid of he denominator you had to multiply by the denominators conjugate so that's why I multiplied by -2i/-2i

I was going to mention that you need parentheses around 2i, but it looks like Bictor beat me to it.

Google gives [url=http://www.google.com/search?hl=en&q=%288%2B20i%29%2F2i+&btnG=Google+Search]This..[/url]	(8 + 20i) / (2 * i) = 10 - 4 i

Of course if you mean$((8+20i)/2) i  $  (That is what most  calculators would interpret, if you do not use '(' around  (2i)  you will[size=150] get   $4i -10 $  ... So make sure that '('  ')'  are in the right place(s) .[/size]

Depending of where is i  you get the change in sign ... that is what Ravi and otheres  are telling you...

[quote=seminolecc45]What I did was:
((8+20i)/2i) X (-2i)/(-2i)=
[b](-16i-40i^2 )/-4i^2 = [/b]
(-16i+40)/4=
10-4i
[/quote]

well... how can you multiply by different number on numerator and denominator and expect the result to be the same number as the original question?

well if i were u, i would simplify the fraction first, making it
$\frac{4 + 10i}{i}$
then multiply by the conjugate
$= \frac{4i - 10}{-1}$
$ = -4i + 10$

so THE CALCULATOR IS WRONG :D

[quote="seminolecc45"]Simplify:
(8+20i)/2i

What I did was:
((8+20i)/2i) X (-2i)/(2i)=
(-16i-40i 2 )/-4i 2 =
(-16i+40)/4=
10-4i

Yet, in my calculator te solution is -10+4i so I must have mde a rpoblem in the second step and it should be -4 after (2i)(-2i).
Can someone tell me what I am doing wrong?[/quote]\[\frac{8+20i}{2i}=\frac{4}{i}+10=10-4i,\] since $1=-i\cdot{i}\Longrightarrow\frac{1}{i}=-i.$ I'm not sure it that method has been posted yet, or if it's even right.

[quote="curl"]
reddevil,

$\displaystyle -4i + 10 = 10 - 4i$

[/quote]

that's why i said the CALCULATOR WAS WRONG... because after i finish typing up the problem, i realized that the seminolecc's answer WAS RIGHT.

The calculator wasn't wrong.  It's just that seminolecc45 typed incorrectly into his calculator.  Even his question here is stated incorrectly as (8+20i)/2i when he really means (8 + 20i)/(2i).


Let P is an arbitrary point in a quadrilateral $ ABCD$ such that  $ \widehat{PAB},  \widehat{PBA},  \widehat{PBC},  \widehat{PCB},  \widehat{PCD},  \widehat{PDC},  \widehat{PAD},  \widehat{PDA}$
are all acute angles. Let $ M,N,K,L$ are feets of the altitude from $ P$ on $ AB, BC, CD, DA$.
Find the smallest value of $ \frac {AB}{PM} \plus{} \frac {BC}{PN} \plus{} \frac {CD}{PK} \plus{} \frac {DA}{PL}$

All of your posts belong in other forums, specifically various places in the Olympiad section. This one is olympiad geometry, for example.


Let $ a$, $ b$, $ c$ be nonzero reals which are not all equal. Assume that
$ a\plus{}\frac{1}{b}\equal{}b\plus{}\frac{1}{c}\equal{}c\plus{}\frac{1}{a}$.
Show that $ \left|abc\right|\equal{}1$.

your equalities hold for any real number k=a=b=c...

The whole looks strange to me: Since the equation is invariant under cyclic permutation, I can assume a to be the greatest of the numbers a, b, c; then, $a + \frac 1b \geq c + \frac 1a$ with equality only for a = b = c. Now, this contradicts the initial assumption that "a, b, c are not the same". Hence, we have no solution, haven't we?

  Darij

Please be careful, Darij, your argument doesn't work. a,b,c are not necesarily positive!

Thanks, one has to distinguish some cases indeed. BTW, I don't see any solution yet.

  dg

I have a nice solution:
  We have $ a-b=\frac{b-c}{bc} $  and two other relations. Then multiply them.


Okay, I don't know how to type that star thing, but let's let ♠ denote that star thing.

Okay, so they want (2♠1)♠(3♠1). Using the table which you can see in the game, you get 2♠1 is 1 and 3♠1 is 2. So they want 1♠2. Using the table again you get 1.

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For those of you in a rush reading this part down here to find the answer... let me slow you down.

Here's the answer.

The answer is x-1073741823. The remainder when 2^7812530 is divided by 10^10 is equal to x.

If you can't solve this, then GO BACK UP AND READ THE SOLUTION I TOOK TIME TO POST!

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~R@d!0Act!vÉ