What is another way to do the integer one you mentioned? $\frac{8}{2}$ can be thought of as how many sets of two things do you need to get eight things. This works for your second part. $\frac{10}{\frac{2}{5}}$. How many sets of 2/5's of things do you need to get 10 things, you would need 25 which is the answer.

Say you had a bunch of circles, each of these would constitute one full thing. Divide each one up into 5 parts. Now shade in 2 of those five parts and count as you do it. When you have ten circles, it would take you 25 times of filling in slices.

[quote]What is another way to do the integer one you mentioned?[/quote]
"another way" is this second way which I mentioned:
[quote]2) we divide eight into two parts and we get 4 as a result.[/quote]
once again: [b]first way[/b] - how many divisors are there in dividend; eg. 10/2 - there are 5 twos in 10. [b]second way[/b] - we divide a dividend into as many parts as divisor's value is... for example 10/2 - we divide 10 into two parts - each part is 5

[quote]Say you had a bunch of circles, each of these would constitute one full thing. Divide each one up into 5 parts. Now shade in 2 of those five parts and count as you do it. When you have ten circles, it would take you 25 times of filling in slices.[/quote]
yes this is true, this is the "first way" of understanding but how would you do it using the "second way". On another forums some people said that it's impossible but I can't (don't want to) believe it :)

[quote="peter111"]Hi.
I've been posting this on some forums but haven't got a satisfactory answer yet.
How to fully understand division by fraction? I don't want an answer such as "multiply dividend and divisor by a power of 10" :) or "multiply by a reciprocal". I want to "mentally" understand this as I understand division by integers!
When we divide any number by integer it can be understood in two ways: for example divide 8:2, then: 1) how many twos are there in eight - we get 4 as a result. 2) we divide eight into two parts and we get 4 as a result.
Now if we divide any number by fraction/decimal we can't use the second way of understanding (or can we??? how??!) for example 10/2,5; 2,5 fits exactly 4 times in 10 so first way works here... but how the heck can I imagine dividing 10 into 2,5 parts - is this possible to use this "second way of understanding" when dividing by fraction??[/quote]
When dividing by a fraction, you can do 1 of 2 things:

1. Simply remember that $\frac{a}{\frac{b}{c}}=\frac{ac}{b}$.

...or, if you can't remember that...

2. Do the same thing, but step-by-step. Instead of trying to divide by the fraction, multiply by the reciprocal. Yes, I know you said you don't want to hear that, but give me a shot at explaining it:

$\frac{x}y$ is the same as $x\div y$. When you multiply by a fraction, like $3\cdot\frac{x}y$, you are basically doing $3(x\div y)$. You can use the distributive property to obtain $3x\div3y$. When you divide by a fraction, you do the same thing, but with a different operation. Let's say you have $3\div\frac{x}y$. As before, we can express this as $3\div(x\div y)$. Use the distributive property again to get $3\div x\cdot y$, or $3\cdot y\div x$. We can express this as $3\cdot\frac{y}x$. As you can see, we went from $3\div\frac{x}y$, to $3\cdot\frac{y}x$. Dividing by a fraction is the same as multiplying by its reciprocal. When you have something like $\frac{4}{\frac32}$, you can write it as $4\div\frac32$, which becomes $4\cdot\frac23$, or $\frac83$. I hope this clears up your understanding of dividing by fractions.

I still think you could do it fine the way you want to. Divide 10 circles into 2/5 parts. Cut the circle into five slices, then pull slices off two at a time. Then you will have divided 10 things into 2/5 parts. You will have 25 of these double slices. It is essentially the same as the first way, with a slightly different emphasis on the way you think about it.

Well, when you divide a number by 8, you multiply it by 1/8. When you divide a number by 7, you multiply it by 1/7.

Notice that 8=8/1, and when you divide a number by 8, you multiply it by 1/8.

Do you know what a reciprocal is? The reciprocal of a fraction x/y is y/x, no matter what x and y are. When you divide a number by a fraction, you multiply it by it's reciprocal.

Well learning the reciprocal way is much much easier and also you could use them in some trig.

however, if you don't want that as an explanation then you can use what i_like_pie wrote; i found that it was the easiest explanation so far

$\frac{a}{\frac{b}{c}}= \frac{ac}{b}$

[quote="peter111"]
PS. I'm a little older than middle school :) and I know how to divide by a fraction but I'm just curious if it's possible to [b]imagine[/b] dividing by a number between 0 and 1 just like you imagine for example dividing 9 by 6 - you divide 9 into 6 equal parts; each part is 1,5. but when you want to divide 9 by 0,5 then how many equal parts you get? :/[/quote]

Do you mean

$\frac{2}{0.5}$?

That means that it would be

$\frac{2}{\frac{1}{2}}$ or

$2\times 2 = 4$...

The general rule is that if you are dividing by $\frac{1}{2}$ you just multiply the numerator by 2 and that is your final answer.

[quote="peter111"]thank you all for your answers.
[quote]Divide 10 circles into 2/5 parts. Cut the circle into five slices, then pull slices off two at a time. Then you will have divided 10 things into 2/5 parts. You will have 25 of these double slices.[/quote]
okay, you count those 2/5's until you reach 25*2/5=10 so this is "the first way" - how many 2/5's "go into" ten - answer: twenty five 2/5's.
I'm starting to think that it's impossible to [b]imagine[/b] dividing something by a number between 0 and 1. you must use the "first way" i.e. "how many times does a fraction/decimal "go into" the dividend". Or you can multiply by the reciprocal but can't understand it in its "original state" :/

PS. I'm a little older than middle school :) and I know how to divide by a fraction but I'm just curious if it's possible to [b]imagine[/b] dividing by a number between 0 and 1 just like you imagine for example dividing 9 by 6 - you divide 9 into 6 equal parts; each part is 1,5. but when you want to divide 9 by 0,5 then how many equal parts you get? :/[/quote]

u might think it impossible to imagine but me, and im sure many other ppl on this forum can imagine it

[quote="peter111"]
PS. I'm a little older than middle school :) and I know how to divide by a fraction but I'm just curious if it's possible to [b]imagine[/b] dividing by a number between 0 and 1 just like you imagine for example dividing 9 by 6 - you divide 9 into 6 equal parts; each part is 1,5. but when you want to divide 9 by 0,5 then how many equal parts you get? :/[/quote]

9/6=1.5 if u want to turn it around, 9/1.5=6
9/.5 equals 18, just think that their are 9 people and each person has 2 eyes. half of the amount of each persons eyes are one eye. so if u take all of the nine ppl and divide it by the half of each amount of eyes, u get 18 eyes

Here's one way you can look at it:

When you divide, say, $10$ things into $2$ groups, you have the same number of things in each group as if you divide $20$ things into $4$ groups, or $30$ things into $6$ groups...

In general,

$\frac{10}{2}= \frac{20}{4}= \frac{30}{6}= ... \frac{10x}{2x}= 5$

For any integer $x$.

The easiest way, I think, to think of dividing by a fraction is to extend this rule to fractions $x$; for example

$\frac{10}{2}= \frac{10 \cdot \left( \frac{1}{5}\right) }{ 2 \cdot \left( \frac{1}{5}\right) }= \frac{2}{ \frac{2}{5}}$

Where $x = \frac{1}{5}$ instead of an integer.

This way of thinking helps a lot once you start dividing by some more complicated things than fractions...

[quote="7h3.D3m0n.117"]The general rule is that if you are dividing by $\frac{1}{2}$ you just multiply the numerator by 2 and that is your final answer.[/quote]
As I said before, because $\frac{a}{\frac{1}{2}}=\frac{2a}1$, or just $2a$.

Notice that a/b = 10*1/b (1)

Now, you want to know what a/(x/b) is. Let x/b = c.

From (1), we know that a/c = a*1/c, so a/(x/b) = a*1/(x/b) = a * b/x, which is the recipricoal.

So the rule is multiply by the reciprocal.


Michael wants to mail a package using exact postage. If he chooses to use only 23-cent stamps, he needs 9 more stamps than if he uses only 32-cent stamps. How many dollars will it cost Michael to mail the package? Express your answer to the nearest cent.

We have $ \frac {100d}{23} \equal{} 9 \plus{} \frac {100d}{32}\implies\frac {900d}{736} \equal{} 9\implies d \equal{} \boxed{\$7.36}$.

Edit: Fixed an error.

Ebob you made an error.  It's supposed to be 207=9x, not 207+9x.   :-D


function
calculus
derivative
real analysis
real analysis unsolved

I know $\bigtriangledown^{2}f(r)=\frac{2}{r}f'+f''(r)$ for $R^{3}$,
what about Laplacian for$R^{n}$? Is there a general formula involving n?
and how to derive it?

The Laplacian of a radial function is radial itself ($\Delta$ commutes with orthogonal transformations). Thus, it's enough to consider the point $(r,0,0,\dots,0)$. The $\frac{\partial^{2}}{\partial x_{1}^{2}}$ derivative is $f''(r)$. The other $(n-1)$ derivatives are all equal to $\frac{f'(r)}{r}$ by the same calculation as in $2$ dimensions. Hence the answer: $\frac{n-1}{r}f'(r)+f''(r)$.


trigonometry
geometry
angle bisector
trig identities
Law of Sines

Proof in triangle ABC:

$sin{A/2} = \sqrt{ \frac{(s-b)(s-c)}{bc}}$

i understand the proof the solution manual gives, but can someone find what i did wrong in my method:

In triangle ABC, draw the angle bisector of A to BC. D is the intersection.

By the law of sines in ABD, $sin{A/2}=(s-c)/b * sin{ADB}$
By the law of sines in ACD, $sin{A/2}=(s-b)/c * sin{ADC}$

so i multipled those two, square rooted and got this instead for sin{A/2}:
$\sqrt{ \frac{(s-b)(s-c)}{bc} }*sin(ADB)$ (sin{ADB}=sin{ADC})

Edit:  Okay, I figured it out.  $BD \neq s - c$ and $DC \neq s - b$, that's the mistake.  The point where the angle bisectors intersect each side is not the same thing as the point where the incircle is tangent to each side!   :)


I am not sure if you can use the same. Give a perfect solution if you can.

umm anup, i think you are refering to a different sum. the sum i am talking about is problem 127 in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=181612&start=180][b]this[/b][/url] page.

[quote="hash_include"]hmm.. i'm getting the answer in terms of $ n$ where $ n$ is number of times B tosses the coin. is that allowed?[/quote]
Dei, they toss the coins only once da. But all at the same time.

i know. but the NUMBER of coins A and B toss are not given right?
i took that to be $ n\plus{}1$ and $ n$.
is that allowed i'm asking..

[quote="hash_include"]i know. but the NUMBER of coins A and B toss are not given right?
i took that to be $ n \plus{} 1$ and $ n$.
is that allowed i'm asking..[/quote]
i think you are rite da!!! i think we have to consider A have n number of coins and then considering the N+1 th coin toss of B and bring out a recurrsion and solve!! :D

we only state hypothesis we dont solve them!!!

SOLVE!!!! :rotfl:

Answer is 1/2.

i'm assuming strict inequality.

i'm not able to put the answer in a nice mathematical form and i would probably be failed in an exam if i posted this answer.
If A has n coins and B (n-1), then
No. of posiible combinations (of "heads" and "tails") for A is n+1
while that for B is n.

So, toltally there are n(n+1) situations that can arise.
write the various combiinations like $ (H_i,T_j)$ means i heads and j tails.
    A----------------------------->B
$ H_n,T_0$
$ H_{n\minus{}1},T_1$----------------->$ H_{n\minus{}1},T_0$
$ H_{n\minus{}2},T_2$----------------->$ H_{n\minus{}2},T_1$
$ H_{n\minus{}3},T_3$------------------>$ H_{n\minus{}3},T_2$
........
$ H_0,T_n$------------------------>$ H_0,T_{n\minus{}1}$

the first entry of A wen combined with all other entries of B always satisfies the condition. This gives us $ n$ combinations.
the second entry of A when combined with all except the first entry of B satisfies the condition. this gives $ n\minus{}1$ combinations.
And so on,......
the last entery wen combined with any entry of B never satsifies. this gives us $ 0$ combinations

Probablity that the condition is satisfied is $ \frac {\sum n}{n(n\plus{}1)}\equal{}\frac 12$

Right you are, no post on sundays. Yeah answer is right and your method is pretty acceptable. :D

Rohit, please don't put Jalra to other intelligent fellows who are capable of working and solving nice problems.

Its pro proof is a double summation
Let x be the no of heads a tosses and r be the no. of heads b tosses.
Then you just have to evaluate $ \sum_{x \equal{} 1}^{n} \sum_{r \equal{} 1}^{x \minus{} 1} f(x)$ .  :D

The problem i have is to find f(x)  :(

yeah, it's 1/2 only. stupid error, sorry about that.
anyway,[hide="a more mathematical solution:"]
let A have n+1 and B have n coins
take the case of the first n coins being tossed by A & B
i denote $ x$ as number of heads A gets and $ y$ as number of heads B gets in thhe n tosses.
evidently, $ P(x<y) + P(x=y) + P(x>y) = 1$.... (1)

$ P(x=y) = ^nC_0 (\frac1 {2^n}).^nC_0 (\frac1 {2^n}) + ^nC_1 (\frac1 {2^n}).^nC_1 (\frac1 {2^n}) + .... ^nC_n (\frac1 {2^n}).^nC_n (\frac1 {2^n})$
$ = \frac 1 {4^n} ^{2n}C_n$ (formula :P)

also, it can be understood that $ P(x<y) = P(x>y)$ (since the number of coins they toss are same)
so, substituting in (1)
$ => P(x>y) = \frac {1-\frac {^{2n}C_n ^} {4^n} }{2}$

the required probability is nothing but
$ (\frac 1 2)P(x=y) + P(x>y)$
$ = \frac 1 2$

[the mistake i had made earlier was that i 4got to multiply by half in the last step  :P ]

[/hide]

can someone help me find f(x) ?  :maybe:   :D


Prove this ineq for a,b,c≥0
$ \frac{a^2}{b^2\minus{}bc\plus{}c^2}\plus{}\frac{b^2}{c^2\minus{}ca\plus{}a^2}\plus{}\frac{c^2}{a^2\minus{}ab\plus{}b^2}$≥2

[quote="SUPERMAN2"]Prove this ineq for a,b,c≥0
$ \frac {a^2}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {b^2}{c^2 \minus{} ca \plus{} a^2} \plus{} \frac {c^2}{a^2 \minus{} ab \plus{} b^2}$≥2[/quote]
Suppose $ a \geq\ b\geq\ c \geq\ 0$
WE HAVE: $ b^2\minus{}bc\plus{}c^2 \le\ b^2$
                $ c^2\minus{}ca\plus{}a^2 \le\ a^2$
THERE FORE :LHS $ \geq\ \frac{a^2}{b^2}\plus{}\frac{b^2}{a^2}\geq\ 2$

That is a nice and simple solution,thanks.


geometry
integration
probability
expected value
calculus
calculus computations

I was working throught the section on Green's theorem in my calc book and got to this question.

Let $R$ be a plane region with area $A$ whose boundary is a piecewise smooth simple closed curve $C$. Use Green's Theorem to prove that the centroid $(\bar{x},\bar{y})$ of $R$ is given by

$\bar{x}=\frac{1}{2A}\oint_{C}x^{2}dy$ and $\bar{y}=\frac{1}{2A}\oint_{C}y^{2}dx$.

can someone plase explain how you would go about proving this.  Thanks.

We know that $\bar x=\frac{1}{A}\iint_{R}x\,dA$. By Green's theorem, $\int_{C}x^{2}dy=\iint_{R}2x\,dA$, if $C$ is oriented counterclockwise. That's it.

Thanks. But how do you show that

$\bar{x}=\frac{1}{A}\iint_{R}xdA$?

That's the definition of the centroid- it's the expected value of the position of a randomly chosen (uniform in area) point of the object.

thanks for that explanation [b]jmerry[/b]


How may I prove that:

2sin :theta: cos :theta:  = sin2 :theta:

[quote="pestich"]There is no way and no need to prove it.

    It is one of the  axioms of trigonometry.

     Maj.  Pestich[/quote]

Pestich, please stop spamming into topics with beginners' questions.

  darij

$sin(2\theta) = sin(\theta + \theta)$
$= sin(\theta)cos(\theta) + sin(\theta)cos(\theta)$
$= 2sin(\theta)cos(\theta)$

I dont know how to do it without using another identity though.

well, Sin (a+b) can be derived from the definition of sin so snowstorm explnation follows.

[quote="pestich"]There is no way and no need to prove it.

    It is one of the  axioms of trigonometry.

     Maj.  Pestich[/quote]It's an axiom of trigonometry?? That's actually pretty funny..

When the $\sin\theta$ function is defined (geometrically or with an infinite series or whatever else), it would be silly to just invent an identity concerning it with no proof or anything..

Well, we start by defining sin  :theta: .  The rest is trivial. :D

OK, seriously, there is a geometric proof involving triangles for sum of angles.  For the double angle formula specifically, there is a different but interesting proof that uses the following figure:

Right triangle ABC has right angle ABC.  Let D be the midpoint of AC.  Extend BD from D to E such that AE = DE.  Notice that  :ang: ADE = 2* :ang: CAB.  That should help.

There are also lots of other proofs, which you should have little trouble finding by going to this site called http://www.google.com :)

[quote="juggle7"]A non-calculus based proof can be found here:

http://www-istp.gsfc.nasa.gov/stargaze/Strig5.htm

But, this is what I find to be a more beautiful proof:

$\cos(a+b)+i\sin(a+b)=e^{i(a+b)}
=e^{ia}\times e^{ib}
=(\cos(a)+i\sin(a))(\cos(b)+i\sin(b))
=[\cos(a)\cos(b)-\sin(a)\sin(b)]+i[\sin(a)\cos(b)+\sin(b)\cos(a)]$

Since real parts and imaginary parts must be equal, $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$, and $\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$.

Euler's equation is very useful.  If you want a proof of that, I will have to prove Taylor series also, which requires calculus.[/quote]Nice proofs of the double angle formulas. The second one using Euler's equation is really nice. You just start with $\cos(a+b)+i\sin(a+b)$ and your forumlas just seem to come out of nowhere..

[quote="blahblahblah"][quote="juggle7"]If you want a proof of that, I will have to prove Taylor series also, which requires calculus.[/quote]

Nah, it's a lot easier if we define $e^x$ via its taylor series.[/quote]

Okay, but what about the series for sin(x) and cos(x)?  I meant proving Euler's equation without the aid of other facts that babie_juliet might not be able to prove yet.


What were the rules of the old AHSME's? (how they were scored, time limit, calculator allowed or not, etc.)
and how does their difficulty compare with current AMC12's? :?:

90 minutes
30 problems
 ~5 pts for each correct answer
 ~2 pts for each unanswered problem
 ~0 pts for each incorrect answer
No calculator

Difficulty is rougly about the same as the AMC 12 with variability from year to year, of course.

Are these still good for AMC 12 practice?

I have borrowed the Contest Problem Books I-III. These date from 1950 to about 1970-ish.


[hide]
x=2+root17
or 2-root17
i just let 4 = the first surd and 2root2 equal the second.. ><
i dont know how to do this..
ive tried expanding by squaring both sides.. not going 2 well..
any hints?
[/hide]

[hide="Solution"]\begin{eqnarray*}\sqrt{(x-1)(x-3)} - \sqrt{(x+1)(x-5)} &=& \sqrt{x^2-4x+3} - \sqrt{x^2-4x-5}\\ &=& \sqrt{(x^2-4x-1)+4} - \sqrt{(x^2-4x-1)-4}\end{eqnarray*}

If we substitute $x^2-4x-1=t$, the equation becomes

$\sqrt{t+4} - \sqrt{t-4}=4-2\sqrt{2}$

and we have

\begin{eqnarray*}\sqrt{t+4} - \sqrt{t-4}=4-2\sqrt{2} &\Rightarrow& \sqrt{t+4} = \sqrt{t-4}+4-2\sqrt{2}\\ &\Rightarrow& t+4=t-4+2(4-2\sqrt{2})\sqrt{t-4}+(4-2\sqrt{2})^2\\ &\Rightarrow& 16\sqrt{2}-16=2(4-2\sqrt{2})\sqrt{t-4}\\ &\Rightarrow& \sqrt{t-4}={{4\sqrt{2}-4}\over{2-\sqrt{2}}}\\ &\Rightarrow& \sqrt{t-4}={{(4\sqrt{2}-4)(2+\sqrt{2})}\over{2^2-\sqrt{2}^2}}\\ &\Rightarrow& \sqrt{t-4}=2\sqrt{2}\\ &\Rightarrow& t-4=8\\ &\Rightarrow& t=12\end{eqnarray*}

Now we have $x^2-4x-1=12$, which gives the solutions: $x_{1,2}=2\pm\sqrt{17}$.[/hide]


FTW MATHCOUNTS blogs LaTeX geometry probability AMC

there is no need to sign up
the rules are in the link in my sig, but the time is now like thanksgiving break (more details)
please post here if you are interested and will likely take the test.
this way, i can see if there will be enough people for a countdown round or such
(remember, your post is not a signup-- just a reference for me)

I will do it. Question: what day is it going to be on? We have 3 days.

uhh that will be a tricky one...
hopefully like the saturday if i have no math competition
the reason i am doing this is because i might need to write my own cd problems
maybe i should make my sprint round harder in order to spread out the scores.. or maybe i should allow partial credit on the targets :D

does that mean we have to show work for full credit on targets? :huh:

Will there be a team round?

If so, I'd like aleph0 to be on my team since we have the same avatar.  :D FTW!

You're talking.

How many people have signed up?

My score on the sprint $ x$ where $ \frac{x}{30}$ is the arithmetic mean of all the answers divided by the geometric mean of all the answers.

hmm
i have only 2 submissions
i will probably extend the deadline to around 3:00 p. m. sunday for all testsor something like that
you may resubmit your submissions if you feel it is necessary

I'm not in this, but clarification for #12: distinct?

Can I have an extended time?  I was in St. Louis yesterday from 4 Am - 12 AM...so I didn't have an access to an computer, and when I cam back, I was sleepy...  Right now I have to go out for dinner, so I was wondering if today night is okay?

[quote="math154"][quote="tinytim"]My score on the sprint $ x$ where $ \frac {x}{30}$ is the arithmetic mean of all the answers divided by the geometric mean of all the answers.[/quote]

Too bad that just gave all of the answers away. :P[/quote]

not necessarily... from that you can infer that all the answers are the SAME but not necessarily ____ .  :rotfl:

[quote="007math"]I was in St. Louis yesterday from 4 Am - 12 AM[/quote]

me too! :D

[quote="tinytim"][quote="math154"][quote="tinytim"]My score on the sprint $ x$ where $ \frac {x}{30}$ is the arithmetic mean of all the answers divided by the geometric mean of all the answers.[/quote]

Too bad that just gave all of the answers away. :P[/quote]

not necessarily... from that you can infer that all the answers are the SAME but not necessarily ____ .  :rotfl:[/quote]

[quote="math154"]Don't worry--once you do the first problem, you'll get the hang of it. :wink:[/quote]

Which is exactly why I said that. :P

[quote="tinytim"]

[quote="007math"]I was in St. Louis yesterday from 4 Am - 12 AM[/quote]

me too! :D[/quote]

Me too!!! Do I know you? :huh:

[quote="math154"][/quote][quote="tinytim"][quote="math154"][quote="tinytim"]My score on the sprint $ x$ where $ \frac {x}{30}$ is the arithmetic mean of all the answers divided by the geometric mean of all the answers.[/quote]

Too bad that just gave all of the answers away. :P[/quote]

not necessarily... from that you can infer that all the answers are the SAME but not necessarily ____ .  :rotfl:[/quote]

[quote="math154"]Don't worry--once you do the first problem, you'll get the hang of it. :wink:[/quote]

Which is exactly why I said that. :P

[quote="tinytim"]

[quote="007math"]I was in St. Louis yesterday from 4 Am - 12 AM[/quote]

me too! :D[/quote][quote="math154"]

Me too!!! Do I know you? :huh:[/quote]

Hmm...where were you guys?  I wen ton the tram to the top of the Arch at 10:25 AM, so we were at the arch form 10-11 and at the courthouse in front of it from like 1:30-12, and then at the science museum from like 1-3 or 4, and at the Hindu Temple and in the car to go back to Kansas.

[quote="math154"]Might you look at my location? :wink:  :P[/quote]

Hmm...so you're at St. Louis all the time.....

ok here is the full test
by the way, 1 team has already submitted answers
[hide]
1) What is the arithmetic mean of 666 and 2008?
2) Steven was twice as old as Meow 1 year ago. In 9 years, Meow's age will be 75% of Steven's age.
let S be Steven's age in years, and M be Meow's age in years. find $ S^3 \plus{} M$
3) evaulate the following where $ a \equal{} 3 \text{ and } b \equal{} 4$:
$ (2a \plus{} b)^3 \plus{} b^b \plus{} a^b$
4) Solve for x: $ \frac {3}{7}x \plus{} 53 \equal{} \frac {x \minus{} 23}{2} \minus{} 31$
5) CA Math needs to tile his 7' by 12' wall with 3" square tiles. CA Math goes to the store and luckily finds 7 3" square tiles, but he has to buy the rest. How many tiles did CA Math buy?
6) Square ABCD has side length 47, and square EFGC has side length 31. E is on BC. Find the area of ABEFG. Assume the squares do not share any interior points.
7) Square ABCD is inscribed in right triangle EFG with hypotenuse FG such that C and D are on FG and C is closer to G than F. If FD=7 and CG=191, what is the area of square ABCD?
8) Budi713 is writing all natural numbers from 1 to 2008. He crosses off all multiples of 3; then he crosses off all sixth powers. How many numbers remain?
9) Let p be the common fraction equal to $ \frac {5}{13} \minus{} \frac {1}{11} \minus{} \frac {1}{8}$. Find the sum of the numerator and the denominator of p.
10) Right triangle ABC with hypotenuse BC has legs of length 19 and 24. Right triangle BCD with hypotenuse BD has DC length 20. What is the area of square BDGH?
11) What is the area of the poygon with vertices (1,3); (17,42); (23,77); (13,37); and (3,216)?
12) What is the product of the biggest and 4th smallest prime factors of 5735730?
13) Stupidityismygam is good at programming. He programs his printer to print out the perfect squares of every natural number from 1 to 296. But, his printer jams on the last digit, and it does not print the last digit of $ 296^2$. Assuming it prints every other perfect square correctly, and it prints every digit in $ 296^2$ excuding the last, how many digits does the printer print?
14) Ra5249 is in big trouble! He has a book overdue for 18 days, another overdue for 12 days, 2 books overdue for 9 days, one book overdue for 1 week, and one last book overdue for 3 days. The local library has the following policy for returning overdue books:
There is a 1 cent fee for returning a book 1 day late.
For every day after the first, the fee doubles.
How much money does ra5249 owe the library? Assume he will return the books today, and he will return the books to his local library.
15) A tytia tree doubles in height every day. A particular tytia tree is 485.14 centimeters tall at noon on Wednesday. How much does it grow, in inches during 3 days (Noon Wedneday to Noon Saturday)?
16) Let p be the common fraction representing the probability that a randomly picked whole number less than 10000 is greater than 1999, has units digit divisible by 4, and whose middle 2 digits multiply to less than 6. Find the sum of the numerator and the denominator of p.
17) Tinytim is remodeling his living room. He wants to buy a 42 inch plasma TV, 2 leather sofas, and 10 new fluorescent lightbulbs for his light fixtures. Using the following chart, how many dollars will it cost him to remodel his living room? Assume there is no tax.
$ \begin{tabular}{|c|c|} \hline Item & Cost \\
\hline Fluorescent lightbulbs & \$ 9.99 \\
\hline Energy efficient fan & \$ 59.99 \\
\hline 42" Plasma TV & \$ 899.90 \\
\hline 48" Plasma TV & \$ 1149.90 \\
\hline 60" Plasma TV & \$ 1799.90 \\
\hline Leather Sofa & \$ 168.60 \\
\hline\end{tabular}$
18) James4l has the following algebra problem, and he needs to solve for the common fraction x: $ \frac {11}{5}x \plus{} \frac {2}{9} \equal{} \frac {4}{13}$
He thinks he has the right answer, but he is not sure, so he needs you to check it. You do not want to give him the answer, so you tell him the sum of the numerator and the denominator of x. What number did you tell him?
19) Yongyi can run quickly. He decides to try to run 831 miles in one day. To the nearest integer, how many kilometers is this?
20) There are 14 students in my Physics class. We all took a test, and we each scored higher than 88. If each student recieved a different (integer) score, what is the minimum possible sum of our test grades? Assume you may have grades higher than 100.
21) In Last City High School, every student studies at least one of the following: art, music, and USAMO proofing. If 713 students study art, 781 students study music, 666 students study USAMO proofing, 456 students study both art and music, 234 students study both art and USAMO proofing, 345 students study both music and USAMO proofing, and 212 students study all 3 subjects, then how many students attend Last Colony High School?
22) Find the 4 digit number abbc in the addition problem, where each number represents a distinct, nonzero digit. (ignore the dots)
. bab
. caa
[u]+bab[/u]
abbc
23) How many 1's are in the binary representation of the base-10 number $ \frac {\frac {3}{4} \cdot 2^{2008} \minus{} 5}{7}$?
24) Find the sum of the smallest 4-digit prime number, the smallest 3-digit prime number, and the $ 49^{th}$ prime number.
25) Just for fun, gauss1181 is renumbering the pages in her (yes, her) new book, but she does not renumber the first page since it is the cover. The book has 482 pages. Assuming she renumbers every page up to page 482 (except the first one) with the corresponding page number, how many digits does she write?
26) If x and y are integers, and $ \minus{} 3x^2 \plus{} 5x \minus{} 2y^2 \equal{} 10291$ find x+y
27) Let $ A,B,C, \text{and } D$ be whole numbers. You are given $ A \equal{} B^{ \minus{} 1} \plus{} C^{ \minus{} 1} \plus{} D^{ \minus{} 1}$ and $ C^2 \minus{} B \minus{} A \equal{} D$ Find $ \left(D \plus{} \frac {C}{B \plus{} A}\right)\left(CB^D \minus{} A\right)$.
28) Express the first four nonzero digits in the decimal expansion of the fraction $ \frac {23}{172}_{10}$ as a four-digit number.
29) A right triangle has legs of length 10696 and 1528. What is the length of a side of the largest square that can be inscribed in this right triangle?
30) The inhabitants of planet T0RA use base 84, and the inhabitants of planet JIR0U use base 40, but the inhabitants of planet T0RAJIR0U use base 38. Assume these 3 planets all use english as their language. People from planet T0RA tell people from planet T0RAJIR0U that the have 713 uber-nerds. people from planet JIR0U tell people from planet T0RAJIR0U that they have 666 uber-nerds. In planet T0RAJIR0U's numbering system, how many uber-nerds do planets T0RA and JIR0U have together?
1) Estimate the sum of the reciprocals of the first 90000 natural numbers to the nearest billionth.
2) A set of points $ a_n$ on the coordinate plane is defined as follows: $ a_n \equal{} (n,n^2)$
A primary triangle is the triangle $ a_0a_1a_2$ and a primary quadrilateral is the quadrilateral $ a_0a_1a_2a_3$
Find the area of a primary tetramegakaipentamyriakaihexachiliakaiheptahectakaioctacontakaienneagon.
3) Policecap writes out the 999-digit number 123456789123456789...123456789 in word form (like 42 is forty-two). How many letters does he write?
4) Define a 1-triangle to be a '1-dimensional triangle' or a line segment with length 1. Similarly, define a 2-triangle to be a 2-dimensional triangle, or an equilateral triangle with sides with length 1. Of course, a 3-triangle would be an equilateral tetrahedrom with edges of length 1 and, if you will, a 4-triangle would be a '4-dimensional tetrahedron, or hypertetrahedron' with sides of length 1. Find the 24-D volume of a 24-triangle.
5) The following iteration is executed on a 4 by 4 rubiks cube (note: the cube is oriented s. t. exactly 1 face is facing you):
turn the right 3 layers up 90 degrees
turn the right 2 layers up 90 degrees
turn the right-most layer up 90 degrees
re-orient the cube such that the original front face is now on the left side, and the original top face is now at the front.
How many times does the iteration have to be executed in order for the cube to be in its original position (EXACTLY the same as it was before the iterations were executed)?
6) Define p(n) to be the nth prime number. Find the smallest natural number, x, such that for all n, 0<n<21, $ x \equiv n \pmod{p(n)}$
7) A 3-D tesselation is made out of rhombicuboctahedra, cuboctahedra, and cubes. What percent of t he tesselation is occupied by rhombicuboctahedra?
8) Eggylv999 is playing the game "Beast". In the game "Beast", eggylv999 repeatedly flips a standard coin. He wins if he flips a number of consecutive heads greater that the total number of tails he has already flipped. For example (h=flipping heads, t=flipping tails): h makes eggylv999 win, thh makes eggylvwin, but ththh doesn't. Assuming eggylv999 flips the coin forever UNLESS he wins, what is the probability, rounded to the nearest billionth, that eggylv wins?
1) Evaluate: $ 1 \plus{} 2 \cdot 3$
2) Estimate $ \frac {1}{\pi}$ to the nearest hundredth.
3) Find the sum of the first 47 natural numbers.
4) Find the volume of a shpere that can inscribed in a cube with space diagonal of length $ 4\sqrt6$.
5) Given that a vertex from each of 3 regular polygons can meet at one point such that every ray with that point as en endpoint passes through either one of the regular polygons or an edge of 2 of the polygons, what is the largest number of sides on any one of these polygons?
6) Find the volume of a unit (regular) rhombicuboctahedron.
7) Find the volume of a tetrahedron with sides of length 2, 2, 2, 2, 2, and 3.
8) Find the hypervolume of a tesseract with hyperspace diagonal of length 42.
9) Find the shortest distance from the origin to [a point on] the graph of the equation $ y \equal{} x^2 \minus{} 1$.
10) All the natural numbers from 1 to one ducenunvigintillion are written down on a sheet of paper.
what is the sum of all the digits written down? Write your answer out in word form (ex: twenty-three, forty-two)
sorry for jumbling up the tests
[/hide]

Can teams be graded solely on the basis of the team round, as my team only has 3 people? When is this due? When do we get results?

[quote="stevenmeow"]hmm judging by how many (few) people submitted answers, i will give you until
[b]
Sunday, December 7 3:00 p.m. GMT-6:00
[/b][/quote]

'Nuff said. :)


Give x,y,z >0 . find Min:
 Q=(xyz+1)(1/x+1/y+1/z)+x/y+y/z+z/x-x-y-z
please hepl me

Q=(xyz+1)[(xy+yz+xz)/(xyz)]+[(x+y+z)/(xyz)]-(x+y+z)
setting a=x+y+z
b=xyz
c=xy+yz+xz
you get:
(b+1)(c/b)+a/b-a
=[c(b+1)+a-ab)/b]
=[c(b+1)+a(1-b)/b]
=[(c-a)(b+1)/b]

no idea what to do with that...

Tecnogram, that $\frac{x+y+z}{xyz}$ term looks wrong.

[quote="Math_Viet_Nam"]Give $x,y,z >0$ . find Min:
$Q = (xyz+1)( \frac{1}{x} +\frac{1}{y} + \frac{1}{z} )+ \frac{x}{y} + \frac{y}{z} + \frac{z}{x} - x- y -z$
please hepl me[/quote]

TeX'd up.


n is an integer
d(n) = number of digits in n
s(n) = sum of digits in n

For how many positive n < 1000000 does d(n) = s(n)?

[hide]If there's $m$ digits, then we have to find number of solutions of $a_1+\dots+a_m=m$ where $a_i$ are non-negative integers and $a_1\neq 0$.

By "dots and bars" argument, we find that the number of all possible solutions is ${{2m-1}\choose m}$ and the number of solutions with the leading zero is ${{2m-2}\choose m}$. Hence, the desired number of solutions is

\begin{eqnarray*}{{2m-1}\choose m}-{{2m-2}\choose m} &=& \frac{(2m-1)!}{m!(m-1)!}-\frac{(2m-2)!}{m!(m-2)!}\\ &=&\frac{(2m-2)!}{m!(m-2)!}\left(\frac{2m-1}{m-1}-1\right)\\ &=&\frac{(2m-2)!}{m!(m-2)!}\cdot\frac{m}{m-1}\\ &=&\frac{(2m-2)!}{(m-1)!^2}\\ &=&{{2m-2}\choose{m-1}}\end{eqnarray*}

Therefore, $N=\sum_{m=1}^{6}{{2m-2}\choose{m-1}}=1+2+6+20+70+252=351$[/hide]

351.
[hide]I'll use generating functions....

We are looking for solutions to $t_1+t_2+...+t_{k+1}=n$, where $t_1 \in \{1,2,...,9 \}$, and $t_i \in \{0,1,...,9 \}$ for $i=2,...,k+1$ Let $g(x)$ be the ordinary generating function for this situation. Then,

$g(x)= (x+x^2+...+x^9)(1+x+...x^9)^k$

$=\displaystyle \left \{ \frac{x-x^{10}}{1-x} \right \} \displaystyle \left \{ \frac{1-x^{10}}{1-x} \right \} ^k= x(1-x)^{-k}(1-x^{10})^k-x^{10}(1-x)^{-k}(1-x^{10})^k$

$=x \sum_{i \ge 0} \sum_ {j \ge 0} (-1)^j \binom{k+i-1}{i} \binom{k}{j}x^{10j+i}- x^{10}\sum_{s \ge 0} \sum_{t \ge 0} (-1)^t \binom{k+s-1}{s} \binom{k}{t} x^{10t+s}$, now we want the $nth$ coefficient of this expression:

$=[x^n] \displaystyle \left \{ x \sum_{i \ge 0} \sum_ {j \ge 0} (-1)^j \binom{k+i-1}{i} \binom{k}{j}x^{10j+i}- x^{10} \sum_{s \ge 0} \sum_{t \ge 0} (-1)^t \binom{k+s-1}{s} \binom{k}{t} x^{10t+s} \right \}$

$[x^{n-1}] \sum_{i \ge 0} \sum_ {j \ge 0} (-1)^j \binom{k+i-1}{i} \binom{k}{j}x^{10j+i}- [x^{n-10}] \sum_{s \ge 0} \sum_{t \ge 0} (-1)^t \binom{k+s-1}{s} \binom{k}{t} x^{10t+s}$

 $let, n-1=10j+i \Rightarrow n-1-10j=i$, and $n-10=10t+s \Rightarrow n-10-10t=s$, substituting gives:

$= \sum_ {j \ge 0} (-1)^j \binom{k+n-1-10j-1}{n-1-10j} \binom{k}{j}-\sum_{t \ge 0} (-1)^t \binom{k+n-10-10t-1}{n-10-10t} \binom{k}{t}$, but $k=n$, so

$= \sum_ {j \ge 0} (-1)^j \binom{2n-10j-2}{n-1-10j} \binom{n}{j}-\sum_{t \ge 0} (-1)^t \binom{2n-10t-11}{n-10-10t} \binom{n}{t}$ but we need $n-1-10j \ge 0$, and $n-10-10t \ge 0$

$\therefore$ we can ignore the second summation completely since,$1 \le n \le6$, and furthermore,$n \le6 \Rightarrow j=0$, giving:

$\binom{2n-2}{n-1}$ for $1\le n \le6$, so to get our desired answer we sum over these $n$

$\sum_{n=1}^{6} \binom{2n-2}{n-1} =351$.[/hide]


calculus
integration
conics
hyperbola
analytic geometry
calculus computations

Suppose the curve $ C: y \equal{} ax^3 \plus{} 4x\ (a\neq 0)$ has a common tangent line at the point $ P$ with the hyperbola $ xy \equal{} 1$ in the first quadrant.

(1) Find the value of $ a$ and the coordinate of the point $ P$.

(2) Find the volume formed by the revolution of the solid of the figure bounded by the line segment $ OP$ and the curve $ C$ about the line $ OP$.

[color=green][Edited.][/color]

1
$ P\equal{}(x_0,y_0)$ lies on both tangent lines $ y\equal{}(3ax_0^2\plus{}4)x\minus{}2ax_0^3\equal{}\frac{\minus{}1}{x_0^2}x\plus{}\frac{2}{x_0}$

so $ x_0^4\equal{}\frac{1}{\minus{}a}$ and so $ a\equal{}\minus{}4, x_0\equal{}\frac{1}{\sqrt 2}$ and $ y_0\equal{}\sqrt 2$

$ a\equal{}\minus{}4, P\equal{}(\frac{1}{\sqrt 2}, \sqrt 2)$

2
 what is axis l?

The answer is correct.

For 2, that means $ l$ is the axis of the rotation.

Oh, sorry, $ l$ is the line segment $ OP$. :blush:


An inclined plane of angle $\ \theta$ with a small block of mass $\ m$ resting on it, is given a horizontal acceleration $\ a$ with respect the ground.  Find $\ a$ such that the block does not accelerate with respect to the inclined plane.

Well, since the vertical forces have to balance, $N \cos \theta = mg$.

Then, the horizontal acceleration, $N \sin \theta = ma$.

Therefore, $ma = mg \tan \theta$, and therefore $a = g \tan \theta$.


Everybody write your names meaning please ...

Example : Vladimir - is a russian name coming from Vladey mirom which means take the world under the control. It is Jahongir in tajik language.

Well my name is Silouanos and it comes from the russian name Silvanos which means <<the human who lives in the forest>>. :)

[quote="Davron"]Everybody write your names meaning please ...

$Example$ : $Vladimir$ - is a russian name coming from $Vladey$ $mirom$ which means take the world under the control. It is $Jahongir$ in tajik language.[/quote]

Well, [b]Tarlan[/b] is [b]a kind of eagle[/b] in azeri language.

Please do not use text in $\LaTeX$ tags. Use the [b]bold[/b] BBCode if you want to make something strike out.

My name is Masoud and it is an arabic word meaning "celebration."  You usually don't see me celebrating with people. :P

Ravshan, means shunny like a sun shine :-)

Bekjan:
 [b]"Bek"[/b] - means [b]"hard"[/b](like a stone);
 [b]"jan"[/b] - means [b]"soul"[/b];
 So [b]"bek"+"jan"[/b] = [b]"hard"+"soul"[/b] = [b]"hardsoul" [/b]    :)
And just about the surname:
Djumabaev:
Came from the words [b]"Djuma"[/b]  $=$  [b]"Juma"[/b]  means Friday in many languages.
That [b]"D"[/b] in the beginning of [b]"Djuma"[/b]  and remaining [b]"baev"[/b] comes from USSR times.

Patryk, derived from the Latin name Patricius, which means "nobleman".

Well, why don't everybody write the meaning his/her name?! It is really nice to see them... THanKS! All of the names are great! :)

дл

OK , my name is Mihail , and it means 'Who is like God?' :roll:


Calculate sum
$ \sum_{n\equal{}1}^{\infty}{\text{sech} (\pi n)}$

Mathworld says to use [url=http://mathworld.wolfram.com/RamanujanCosCoshIdentity.html]the Ramanujan Cos Cosh Identity[/url]


inequalities
function
algebra proposed
algebra

Solve in real numbers the equation $ 9^x\plus{}4^x\plus{}1\equal{}6^x\plus{}3^x\plus{}2^x.$

Hello

let : $ X\equal{}2^x$ and $ Y\equal{}3^x$

and $ (E): 9^x\plus{}4^x\plus{}1\equal{}6^x\plus{}3^x\plus{}2^x$

$ (E) \Leftrightarrow X^2\plus{}Y^2 \plus{}1 \equal{} XY\plus{}X\plus{}Y$

on the other hand : $ X^2\plus{}Y^2 \plus{}1\geq 2XY\plus{}1$

Hence , $ (X\minus{}1)(1\minus{}Y) \geq 0$

therefore : $ X\minus{}1 \leq 0$ and $ Y\minus{}1 \geq 0$

Or  $ X\minus{}1 \geq 0$and $ Y\minus{}1 \leq 0$

From That  inequalities :  $ x\equal{}0$ By introducing expononciel function   ( $ 2^x\equal{} e^{xln(2)}$ )

*Excuse me for me poor english

Another way is to write it as

$ \frac{(3^x\minus{}2^x)^2 \plus{} (3^x\minus{}1)^2 \plus{} (2^x\minus{}1)^2}{2} \equal{} 0,$

after which $ x \equal{} 0$ is obviously the only solution.


geometry
circumcircle
incenter
geometric transformation
homothety
ratio
reflection

Is given a triangle ABC .The vertices B,C and the circumcircle are fixed. Prove that the locus of the orthocenter H, the incenter I and the centroid G are circumferences

In the attachment move only the point A on the circumference
[geogebra]a8ead54adbb448b2fec73655cd4ed047346475b2[/geogebra]

The orthocenter lies on the circle with center the symmetric P of the circumcenter O of ABC and radius OB. The centroid lies on the circle centered at a point J  on OP such that 2OJ=JP and the radius is OB/3. This can be easily proved using complex numbers. For the incenter we have that $ <BIC$=$ 90\plus{}<A/2$, so as $ <A$ lies on constant arc BC on a constant circle, $ <BIC$ also lies on a costant arc BC on a constant circle.

There is an easier way to show that the locus of $ G$ is a circumference without using complex numbers. Just see that this circumference is the homothetic circumference of the circumcircle of $ ABC$ in the homothecy with center in the midpoint of $ BC$ and ratio $ \plus{}1/3$.

$ \bullet$ [b][color=darkblue]The locus of $ H$[/color][/b]
Let $ (O)$ be the circumcircle of $ \triangle ABC$.  Let $ \triangle$ be the line through $ O$, such that $ \triangle \parallel{}BC$. Let us difine $ \mathcal {R}_{d}$ and $ \mathcal {T}_{\overrightarrow {u}}$ be the reflection through $ d$ and the translation through $ \overrightarrow {u}$ . We have $ \mathcal {R}_{\triangle}$ $ \circ$ $ \mathcal {R}_{BC}:$ $ A\mapsto A'\mapsto H$. But $ \mathcal {R}_{BC}$ $ \circ$ $ \mathcal {R}_{\triangle}$ $ \equal{}$ $ \mathcal {T}_{\overrightarrow {u}}$, where $ \overrightarrow {u} \equal{} 2\overrightarrow {OM} \equal{} const$, and $ M$ is the midpoint of $ BC$. But $ A\in (O)$, therefore $ H\in (O')$, where $ (O')$ is the image of $ (O)$ through $ \mathcal {T}_{\overrightarrow {u}}$, but $ (O')$ is fixed. Therefore $ (O')$ is the locus of $ H$.

$ \bullet$ [b][color=darkblue]The locus of $ G$[/color][/b]
Consider the homothety through center $ M$, ratio $ k \equal{} \frac {MG}{MA} \equal{} \frac {1}{3}$. We have $ \mathcal {H}^{k}_{M}: A\mapsto G$. But $ A\in (O)$ hence $ G\in (O_a)$ where $ (O_a)$ is the image of $ (O)$ through $ \mathcal {H}^{k}_{M}$. But $ (O_a)$ is fixed. Hence the locus of $ G$ is $ (O_a)$.

$ \bullet$ [b][color=darkblue]The locus of $ I$[/color][/b]
We have $ \angle BIC$ $ \equal{}$ $ 180^{\circ} \minus{} \frac {\angle B}{2} \minus{} \frac {\angle C}{2}$ $ \equal{}$ $ 90^{\circ} \minus{} \frac {\angle A}{2}$ $ \equal{} const$  since the locus of $ A$ is $ (O)$ hence $ \angle BAC \equal{} const$. Hence the locus of $ I$ is the arc $ BC$ that contains $ 90^{\circ} \minus{} \frac {\angle A}{2}$ which is fixed.


LaTeX
linear algebra
matrix
linear algebra unsolved

I haven't worked with matrices much, so I don't know if this is something well known, but would you know some elegant solution to this problem: Show that if $ A$ and $ B$  are $ n \times n$ matrices, and if elements of each column in $ A$ are the same as the ones in $ B$, but are reversely ordered, then:

$ \det(A) = (-1)^n \det(B)$

I do now know how to do matrices in $ \text{\LaTeX}$, so the general idea is that if the elements in a column were alphabetical letters, and sorted in $ A$ in a $ a \rightarrow z$ manner, then in $ B$ they would be ordered in the $ z \rightarrow a$ manner. Hope I have made my self clear enough.  :oops:

Blah... I thought it was something basic... But, I have another question. Suppose I have a $ n \times n$ matrix $ A$ that is very similar to transposed Vandermonde matrix, but it doesn't start with power $ 0$ but $ 1$. And suppose I have do calculate $ \det(A)$. How would I go about proving that:

$ \det(A) \equal{} ( \minus{} 1)^n a_1 \cdot a_2 \cdot ... \cdot a_n \det(V)$

where $ V$ is the "similar" Vandermonde matrix I talked about. Vandermonde matrix is described here:

[url]http://mathworld.wolfram.com/VandermondeMatrix.html[/url]

Thanks!

[b][EDIT] I got it.. Thanks any way! ;)[/b]


Of the 160 students in the seventh grade at Franklin Middle School, one-half of them belong to Drama Club. Of those, only ten participate in just Drama Club, while the rest also participate in Chorus. Only 30 of the seventh graders do not participate in either Chorus or Drama Club. How many seventh graders participate in only Chorus?

There's $ 160 \times \frac{1}{2}$, or 80 students in drama club. There are $ 160 \minus{} 30$, or 130 students in drama club, chorus, or both, and 80 of them are in drama club, so there are 50 students only in chorus, right?


[size=150]Suppose for a positive integer 'n' both '2n+1' and '3n+1' are perfect squares. Show that 'n' is divisible by
40.[/size]
Sorry No latex yet.  :|

[hide] let $ 2n\plus{}1\equal{}x^2$ and $ 3n\plus{}1\equal{}z^2$. then $ 3x^2\minus{}2z^2\equal{}1$. considering the various cases modulo 5, we must have $ x\equiv z \equiv 1 \pmod{5}$. plugging back into either of the original equations we see that $ n$ must be divisible by 5. now we look at $ 3x^2\minus{}2z^2\equal{}1$ modulo 8. every square is 0, 1, or 4 modulo 8. so $ 3x^2$ is 0, 3, or 4 modulo 8 and $ 2x^2\plus{}1$ is 1 or 3 modulo 8, and it follows that $ 3x^2\equal{}2z^2\plus{}1$ can only happen if $ x\equiv z \equiv 1\pmod{8}$. plugging back into $ 3n\plus{}1\equal{}z^2$, we get that $ n$ is divisible by 8. thus $ n$ is divisible by 5 and by 8 and hence by 40. [/hide]

[hide="Solution"]
Since all odd squares are of the form $ 8k \plus{} 1$ therefore we must have $ n\equiv 0 \mod 4$. Therefore $ 3n \plus{} 1 \equiv 1 \mod 8 \implies n \equiv 0 \mod 8$ Since $ \gcd (3,8) \equal{} 1$.
Observe that for any perfect square $ a^2$; $ a^2 \equiv 0; \pm 1 \mod 5$.
Case I: $ \boxed{2n \plus{} 1 \equiv 0 \mod 5 }$ Then $ 3n \plus{} 1 \equiv 2 \mod 5$ which is not possible.
Case II : $ \boxed{2n \plus{} 1\equiv \minus{} 1 \mod 5}$ Then $ 3n \plus{} 1 \equiv 3 \mod 5$, again a contradiction.
Therefore we must have $ 2n \plus{} 1 \equiv 1 \mod 5 \implies n \equiv 0 \mod 5$
We already have $ n\equiv 0\mod 8$; and multiplying we get the result. (because $ \gcd (8,5) \equal{} 1$) [/hide]
[edit] Beaten :) [/edit]


[quote="xsquaredster"]For the 1st card, there are 4 aces; for the second, 3; the third, 2; and the fourth, 1. There are 48 cards left in the deck at this point, so any of the 48 will do. There are 5 positions in which the non-ace could be.

As for the total number of hands that are possible, we have $52 \cdot 51 \cdot 50 \cdot 49 \cdot 48$ total hands.

Therefore, the probability is $\frac{5(4 \cdot 3 \cdot 2 \cdot 1 \cdot 48)}{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48}= \frac{1}{13 \cdot 17 \cdot 5 \cdot 49}= \frac{1}{54145}$.

Note that in my solution, I had permutations of hands count as distinguishable hands. This was to simplify the calculations. If one were to solve it not counting permutations as distinguishable, they would find $\binom{52}{5}= \frac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}= 13 \cdot 17 \cdot 5 \cdot 49 \cdot 48$ total hands, 48 of which include 4 aces. Therefore, the probability is $\frac{48}{13 \cdot 17 \cdot 5 \cdot 49 \cdot 48}= \frac{1}{54145}$.[/quote]
I dont think that's right.
Isn't thought $52C5=\frac{52!}{5!*47!}$?

[quote="dodobird"]I dont think that's right.
Isn't thought $52C5=\frac{52!}{5!*47!}$?[/quote]
That's what he wrote. :maybe:

[quote="burntonapike"]In a normal 52 card deck, whats the probability of drawing 4 Aces in a 5 card hand?[/quote]

[hide="solution"]
$ \frac{48}{{\binom{52}{5}}}=\frac{1}{54145}$[/hide]


algebra
polynomial
geometry
geometric transformation
function
superior algebra
superior algebra unsolved

[quote="Leva1980"]We consider only monic polynomials.

Suppose we have fixed some natural $ n>1$.
Prove that almost every polynomial of degree $ n $ does not commute with any
polynomial of degrees between $ 2 $ and $ n-1 $.[/quote]

What do you mean by commute? The commutativity in a ring $R[x]$ often is a consecuence of the commutativity of the ring $R$.

Just in case the problem is translation, [url=http://mathworld.wolfram.com/Commutative.html]here[/url] you can find a definition of [url=http://mathworld.wolfram.com/Commutative.html]Commutative[/url].

Best,

[quote="grobber"]Isn't it pretty clear that he means commuting wrt composition of functions? I mean $P(Q(x))=Q(P(x)),\ \forall x$.[/quote]

Well, it didn't seem that self evident for me at the beginning, but it has a lot of sence now.

Nevertheless, the statement is not true in general if we are talking about polynomials over finite rings.

Now, for polynomials over $\mathbb C$ (or $\mathbb Q$, $\mathbb R$), then it seems to be true.  Probably the best strategy (I guess) is to compare the coefficients of such polynomials. For example, if the two polynomials commute, then, suppose $P(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$ and $Q(x)=x^n+b_{m-1}x^{m-1}+\cdots+b_0$, then $P(b_0)=Q(a_0)$,  it may help, but also, note that the coefficient of degree $mn-1$ in $P(Q(x))$ is $ma_{n-1}b_{m-1}$, and in $Q(P(x))$ is $na_{n-1}b_{m-1}$. As by hypothesis $m\neq n$, then either $a_{n-1}$ or $b_{m-1}$ are zero. I suppose (but I am a little lazy right now to continue) that doing some kind of argument you can reach $P(x)=x^m$ and $Q(x)=x^m$.

You can probably try to play with the roots of both polynomials too, but I don't know if that will give much of a resoult.

Best,


[quote="dogcat"]Prove that $ \left(1\plus{}\frac 1n\right)^n\left(1\plus{}\frac 1{4n}\right)$ is strictly increasing for $ n>1$. Or in other words the recurrence equation is strictly increasing.[/quote]

ups, ugly mistake

hello, and how did you get the term
$\frac{4n+5}{4n+1}$ from
$\frac{a_{n+1}}{a_n}=\frac{\left(1+\frac{1}{n+1}\right)^{n+1}\left(1+\frac{1}{4(n+1)}\right)}{\left(1+\frac{1}{n}\right)^n\left(1+\frac{1}{4n}\right)}$?
Sonnhard.


Are aids like rules, protractors, and compasses allowed on the AIME? I found they were for the AMC10 and 12, but couldn't find any information for the AIME.

Yes, but ONLY those. No calcs/slide rules (lol)

It's on the cover of the AIME.


Ok, let's solve this 'old' one :
Using the convention $C_n^p = 0$ when $p >n$ or $p < 0$, and that $C_{n+1}^p = C_n^p + C_n^{p-1}$, we have :
$\dsp b_{n+1} = \sum_{k=0}^{n+1} (-1)^k [C_{n+k}^{2k} + C_{n+k}^{2k-1} ] = \sum_{k=0}^{n} (-1)^k C_{n+k}^{2k} + \sum_{k=0}^{n+1} (-1)^{k+1} C_{n+k+1}^{2k+1} = b_n - a_{n+1}$
and
$\dsp a_{n+1} = \sum_{k=0}^{n} (-1)^k [C_{n+k}^{2k+1} + C_{n+k}^{2k} ] = \sum_{k=0}^{n-1} (-1)^k C_{n+k}^{2k+1} + \sum_{k=0}^{n} (-1)^k C_{n+k}^{2k}  = a_n + b_n$

From these two relations, we easily deduce that $b_{n+1} = -a_n$ and that both $(b_n)$ and $(a_n)$ satifies the linear relation of order 2 :
$U_{n+2} = U_{n+1}- U_n$. (*)

Direct checking leads to $a_1 = 1,a_2 = 1$ so that it may be convenient to define $a_0 = 0$.
Solving (*), we deduce that $a_n = \frac 2 {\sqrt 3} sin( n \frac {\pi} 3)$ and that $b_n = - a_{n-1} = - \frac 2 {\sqrt 3} sin( (n-1) \frac {\pi} 3)$.

A direct consequence is that both sequences are 6-periodic.

Pierre.

You are right,Pierre.Thank you ! :)

kunny


geometry
circumcircle
trigonometry
search
geometry unsolved

The incircle of a triangle $ABC$ touches the side $AB$ and $AC$ at respectively at $X$ and $Y$. Let $K$ be the midpoint of the arc $\widehat{AB}$ on the circumcircle of $ABC$. Assume that $XY$ bisects the segment $AK$. What are the possible measures of angle $BAC$?

[quote="HTA"]sorry , i can't understand how "XY bisects the segment .. " , maybe I misunderstand somehow ?[/quote]
How I understood it: Line which goes through X and Y bisects the segment AK. I guess there is a triangle in which this can be achieved, but I couldn't construct such an example...

(The text of the problem is practically the same as the text on imo.org.yu...)

Where can we find all problems from this year?

Here you go, [b]Ahiles[/b]...

[url]http://www.imomath.com/othercomp/b348f/RusMO08.pdf[/url]

yup, imo.org.yu is where i got the question. i wonder if there's anyone knows the question from another source.

Here's another source: [url]http://olympiads.mccme.ru/vmo/34/2.pdf[/url]

But the text is same!

[quote]Вписанная окружность касается сторон AB и AC треуголь-
ника ABC в точках X и Y соответственно. Точка K  сере-
дина дуги AB описанной окружности треугольника ABC.
Оказалось, что прямая XY делит отрезок AK пополам. Че-
му может быть равен угол BAC?[/quote]

XY bisects a segment means that the line XY pass through the half point of the segment.

Sorry, there aren't any typos. :blush: I haven't got an elegant solution but with some calculations i solved the question.
[hide="Answer"]120[/hide]

Call D the point where XY meets AK. Then we must have:

$ 2 \overline{AD} = \overline{AK} \ \Longleftrightarrow \ 2 \frac{r \cot {\frac{A}{2}} \cos{\frac{A}{2}}}{\sin{\frac{B}{2}}} = R \sin {\frac{C}{2}} \ \Longleftrightarrow$
$ \Longleftrightarrow \ \frac{r}{R} \cos^2{\frac{A}{2}}= \sin{\frac{A}{2}} \sin{\frac{B}{2}} \sin{\frac{C}{2}} \ \Longleftrightarrow$
$ \Longleftrightarrow \ \cos^2{\frac{A}{2} = \frac{1}{4} \ \Longrightarrow \frac{A}{2} = \{ \frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3} \} \ \Longrightarrow \ A = \frac{2 \pi}{3}}$

[size=134][color=darkblue][b][u]Nice problem ![/u][/b][/color][/size]

[quote="anonymous1173"] [color=darkred]The incircle of a triangle $ ABC$ touches the side $ [AB]$ , $ [AC]$ at the points $ X$ , $ Y$ respectively. Let $ K$ be the midpoint of the arc $ AB$

in the circumcircle of $ ABC$ . Assume that $ XY$ bisects the segment $ AK$ . Ascertain the possible measures of angle $ \widehat {BAC}$ .[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote the midpoint $ T$ of the segment $ [AK]$ , the intersections $ \begin{array}{c}
D\in AB\cap CK\\\
S\in XY\cap CK\end{array}$ . If $ T\in XY$ , then prove easily that $ a>b$ , $ a>c$ .

Show easily that $ XD\equal{}\frac{(a\minus{}b)(p\minus{}c)}{a\plus{}b}$ . Apply the [b]Menelaus' theorem[/b] to the transversals :

$ \left\{\begin{array}{cccccc}
\overline {SYT}\mathrm {\ in\ }\triangle\ ACK : & \frac {SK}{SC}\cdot\frac {YC}{YA}\cdot\frac {TA}{TK}\equal{}1 & \implies & \frac {SK}{p\minus{}a}\equal{}\frac {SC}{p\minus{}c}\equal{}\frac {CK}{a\minus{}c} & \implies & SC\equal{}\frac {p\minus{}c}{a\minus{}c}\cdot CK\\\\
\overline {SYX}\mathrm {\ in\ }\triangle\ ACD : & \frac {SD}{SC}\cdot\frac {YC}{YA}\cdot\frac {XA}{XD}\equal{}1 & \implies & \frac {SD}{a\minus{}b}\equal{}\frac {SC}{a\plus{}b}\equal{}\frac {CD}{2b} & \implies & SC\equal{}\frac {a\plus{}b}{2b}\cdot CD\end{array}\right\|\implies$

$ \frac {p\minus{}c}{a\minus{}c}\cdot CK\equal{}\frac {a\plus{}b}{2b}\cdot CD\implies\frac {p\minus{}c}{a\minus{}c}\cdot CK\cdot CD\equal{}\frac {a\plus{}b}{2b}\cdot CD^2$ . From the well-known relations $ \boxed {\begin{array}{c}
CK\cdot CD\equal{}ab\\\\
CD\equal{}\frac {2}{a\plus{}b}\cdot\sqrt {abp(p\minus{}c)}\end{array}}$

obtain $ \frac {p\minus{}c}{a\minus{}c}\cdot ab\equal{}\frac {a\plus{}b}{2b}\cdot \frac {4abp(p\minus{}c)}{(a\plus{}b)^2}$ $ \implies$ $ b(a\plus{}b)\equal{}(a\minus{}c)(a\plus{}b\plus{}c)$ $ \implies$ $ a^2\equal{}b^2\plus{}c^2\plus{}bc$ $ \implies$ $ m(\widehat {BAC})\equal{}120^{\circ}$ .[/color]

Wow, guys, I know the trigonometric solutions are quite short, but this is only an angle-chaser.

I'll just give an outline of the solution (it's all angle chasing).

 [geogebra]d985fbfae4112099967f34b883f01a8be04a4de7[/geogebra]

First, let $ L$ be the midpoint of arc $ AC$. Then $ KL$ is parallel to $ XY$. Moreover, $ AX \equal{} XE$, $ E$ being the intersection of $ KL$ and $ AB$.

Second, $ KEIB$ is cyclic, so $ \angle IEA \equal{} 180^\circ \minus{} \angle A$.

Finally, since $ IX$ is both altitude and median of triangle $ AIE$, this triangle is isosceles, so $ \angle EAI \equal{} \angle IEA \iff \frac{\angle A}2 \equal{} 180^\circ \minus{} \angle A\iff \angle A \equal{} 120^\circ$.

PS: just an aside: for some reason, the Geogebra command 'incircle' doesn't work in my computer (and a forum search showed this to be the case for some other posters); I redid the picture by constructing the angle bisectors.

Too much work for nothing!!!
Just observe that $ \triangle AXS \sim  \triangle CIB$ ($ S\equal{}XY \cap AK$). Hence $ AS\equal{}\frac{AX.BC}{CI}$, but $ AS\equal{}1/2.KI \Longrightarrow 1/2KI.CI\equal{}\frac{AX.BC}{CI}.$ Also we know that $ KI.CI\equal{}2Rr \Longrightarrow Rr\equal{}a(p\minus{}a) \Leftrightarrow 1\equal{}4cos^2(\frac{A}{2})$ $ \Longrightarrow A\equal{}120^\circ$

Another synthetic approach.

Let $L$ be the midpoint of $\widehat{AC}$ and let $I$ be incentre of $\Delta ABC$.
By Fact 5, we have $IK = AK$ and $IL = AL$.
So, $KL$ is perpendicular bisector of $AI$. Also $XY$ is perpendicular to $AI$.
Thus $KL \parallel  XY$.
Let $AI \cap XY = T$.
It is not difficult to see that $3AT = IT$. So, $\boxed{\angle{BAC} = 120^{\circ}}$   $\blacksquare$


Euler
geometry
geometric transformation
rotation
vector
linear algebra
matrix

Hi,

When using Euler angles for rotations, you can either rotates the axes or rotate the vector - at least, that is what I have understood from various sources. When rotating the axes, the second and third rotation (around the y-axis and the x-axis) are replaced by "new" axes.
Is this necessary when rotating the vectors and leaving the axes fixed?

Links:

[url=http://upload.wikimedia.org/math/a/7/a/a7a46ed13c54cde539b70065e987cf0f.png]Matrix for rotating the axes[/url]
[url=http://upload.wikimedia.org/math/5/2/5/525d9c773a2ed0d6a41c48d9a3ea4291.png]Matrix for rotating the vectors[/url]
[url=http://www-structmed.cimr.cam.ac.uk/Course/MolRep/euler.gif]Illustration
with "new" axes [/url]

Thanks,
RicOh

When you rotate a vector and keep the axes fixed, there are no new axes to speak of. You may find it easier to see what is going on if you think of that scary $3\times 3$ matrix as the product
\[\begin{pmatrix}\cos\psi &-\sin\psi & 0 \\ \sin\psi & \cos\psi & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix}\cos\theta & 0 & \sin\theta \\ 0 & 1 & 0 \\-\sin\theta & 0 & \cos\theta \end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\ 0 & \cos\phi &-\sin\phi \\ 0 & \sin\phi & \cos\phi \end{pmatrix}\]

Thank you!
I did find that matrix by multiplying those three others (thanks to Derive), but it took me a while to find their product on the internet; the other one, using axis rotation, seems to be more popular!

If axes do stay fixed, is there still a possibility to get gimbal lock?

RicOh


Anyway, i think this inequality has been posted before. It's from RMO'05 proposed by Bogdan Enescu.Anyway, let $s_{k}=x_{1}+x_{2}+...+x_{k}$.It is obvoius that $s_1 <s_2 <...< s_n$ and $x_k=s_k - s_{k-1}$.Now for any $k\geq 2$, by Cauchy we have: $(\sum\frac{1}{1+s_k})^2\leq(\sum\frac{1}{x_k})\cdot(\sum\frac{x_k}{(1+s_{k})^2})$ and from here it is easy to see that $\sum\frac{x_{k}}{(1+s_{k})^2} <1$ so the conclusion follows. ;)


inequalities
calculus
integration
logarithms
induction
real analysis
real analysis unsolved

If $n$ is a positive integer,prove that $\frac{ln2}{n}<\int_{1}^{\infty}\frac{1}{1+x^{n+1}}dx<\frac{ln2}{n}+\frac{1}{4n^{2}}$.

I'm sorry.I made a mistake in my original post.Now it has been modified.

Proof:
notice:
$\int_{1}^{+\infty}\frac{1}{1+x^{n+1}}dx=\int_{0}^{1}\frac{x^{n-1}}{1+x^{n+1}}dx$
also we have:
$\frac{x^{n-1}}{1+x^{n}}<\frac{x^{n-1}}{1+x^{n+1}}<\frac{x^{n-1}}{4n}+\frac{x^{n-1}}{1+x^{n}}$($0 \le x \le 1$)  (1)
The left part is trivial.
For the right part,It is equivalent to:
$\frac{x^{n}(1-x)}{(1+x^{n})(1+x^{n+1}}<\frac{1}{4n}$
but we have:
$nx^{n}(1-x)<(1-\frac{1}{n+1})^{n+1}<\frac{1}{e}$
So:
$\frac{x^{n}(1-x)}{(1+x^{n})(1+x^{n+1}}<\frac{1}{2ne}<\frac{1}{4n}$
We complete the proof for (1)
Integral (1) form 0 to 1,you will get the desired inequality. :)

Sorry,I made a mistake here.
 :blush:
And Here is an ugly proof.Then it is equivalent to:
$(1+x^{n})(1+x^{n+1})>4nx^{n}(1-x)$
$\iff x^{2n+1}+(4n+1)x^{n+1}+1>4nx^{n}$
but we have
$n(\frac{4n+1}{n})x^{n+1}+1 \ge (n+1)(\frac{4n+1}{n})^{\frac{n}{n+1}}x^{n}$
So we should only prove:
$(n+1)(\frac{4n+1}{n})^{\frac{n}{n+1}}>4n$
$\iff (n+1)^{n+1}(4n+1)^{n}>n^{n}(4n)^{n+1}$
Can it proved by induction?I think so :)

Your last formula is equivalent to $(1+\frac{1}{n})^{n}(1+\frac{1}{4n})^{n}>\frac{4n}{n+1}$,but the limit of the left(when $n\to \infty$) is $e^{\frac{5}{4}}$,which is smaller than the limit of the right.This implies that for $n$ large enough your formula doesn't hold.Do you agree?


function
combinatorial geometry
combinatorics proposed
combinatorics

A [i]convex polytope[/i] in $ \mathbb R^{d}$ is the convex hull of finitely many points in $ \mathbb R^{d}$. A [i]face[/i] of a convex polytope is a proper subset on which some linear function $ f:\mathbb R^{d}\to\mathbb R$ attains its maximum. Finally, a [i]vertex[/i] of a convex polytope is a face consisting of a single point.

Now for the actual problem:

Let $ \mathcal K$ be a convex polytope in $ \mathbb R^{d}$, and $ P$ an arbitrary point in $ \mathcal K$. Let $ P_{1},\ \ldots,\ P_{t}$ be vertices of $ \mathcal K$, not all contained in a single face of $ \mathcal K$. Prove that one of the $ P_{i},\ i \equal{}\overline{1,t}$ can be removed so that the convex hull of the remaining vertices of $ \mathcal K$ still contains $ P$.

Something is obviously wrong: just take $ P_i$ to be the 3 vertices of a non-degenerate triangle on the plane and $ P$ to be any its internal point. :?

Darn.. I knew I had left something out :). The polytope is not a simplex. Terribly sorry; I'd edit the original message, but I can't (more than 24 hours have passed since I posted it).

Assume that some vertex $ Q$ of $ K$ is not neither of $ P_i$. Then the ray $ QP$ intersects the boundary of $ K$ in a point $ X$, which belongs to some facet of $ K$. There exist $ i$ such that $ P_i$ does not belong in this facet. Then remove $ P_i$ and it's ok.

So, consider the case when all the vertices of $ K$ are $ P_1,\,P_2,\dots$ Then the statement follows from Caratheodory theorem, but we may just prove it as in the first case. Consider arbitrary vertice $ Q$ of $ K$ and make the same thing. We may remove any vertice except $ Q$ and vertices of corresponding facets. So, all vertices except $ Q$ lie in the same facet. SInce $ Q$ is arbitrary, it implies that $ K$ is simplex.


Determine all values of k \in R for which the following inequality holds for every x,y,z \geq 0:
2(x^3+y^3+z^3)+3(3k+1)xyz \geq (1+k)(x+y+z)(xy+yz+zx)

In the case k = 1 this is just Schur's inequality ... so k = 1 is certainly a solution.

Yep my intuition says the same thing ... how about a proof ? :D

I think for -1<=k<=0 it's pretty easy to prove. We write the ineq as 2(x^3+y^3+z^3)+6kxyz >= (k+1) \sum x^2y where the sum is symmetric, not cyclic. We put 6kxyz on the right and I guess we can apply AM-GM a few times (x^3+y^3+z^3 is the greatest symmetric sum of deg 3 and 3 terms involving x, y, z). I'm not sure, but I think it works. I think it's also easy for k<-1.

k<=1!!! why??

well if we put in this ineq x=y=1 and then z=0 we get that k<=1.  All you have to do is just use Shur's ineq then. sum ( x^n(x-y)(x-z)>=0)

tschuesss!!! :D :D


function
induction
algebra proposed
algebra

Determine all the functions $f:N->N$ such that:
$f(n) = n - f(f(n-1)) $ , for all $n \geq 1$.

I asume $\mathbb N$ contains $0$, right? We get two possible beginnings for our sequence: (a): $f(1)=1,f(f(0))=0$ and (b): $f(1)=0,f(f(0))=1$. The first one yields (it can be shown by induction) the [url=http://mathworld.wolfram.com/HofstadterG-Sequence.html]Hofstadter G-sequence[/url], $a_n=\lfloor(n+1)\phi\rfloor-n-1,\ \phi=\frac{1+\sqrt 5}2$.

Let's look at the second possibility. We quickly get a contradiction: $f(2)=2-f(f(1))=2-f(0)\Rightarrow f(0)\in\{0,1,2\}$. If $f(0)=0\Rightarrow f(f(0))=0\ne 1$. If $f(0)=1\Rightarrow f(1)=1\ne 0$, and if $f(0)=2\Rightarrow f(2)=1\ne 2-f(0)=0$.

The answer must then be $a_n=\lfloor(n+1)\frac{1+\sqrt 5}2\rfloor-n-1$.


Find all polynomials f for which there exists a polynomial h such that f(2t) = h(f(t))

I determined that h must be linear with zero constant coefficient, but this implies that the only such polynomial f is the zero polynomial. Any ideas?

Nevermind. The set is all polynomials c(k)*t^k + d. This can be understood by first determining the degree of h and then explicitly writing f(2t)=h(f(t)) and determining the conditions for coefficients necessary for equality.


I have a APchem test coming up. The last one was kind of weighted in the fact that about 70% of the test is basic questions and the other 30% is challenge free response problems (that are pretty impossible). I want to improve my grade from the last test and I was looking around online since I know alot of high schools and colleges post the unit tests online. I couldn't find any though and I'm not sure how similar ap chem is to an actual college course.

This is what we have covered so far:
i. Electron configuration
ii. Bond energy and calculating delta H of a reaction using bond energies or heats of formation
iii. Lewis structures, resonance, and formal charges
iv. Hybridization (sp, sp2, sp3, dsp3, d2sp3)
v. very basic molecular electron model (bond order, etc.)

I know I understand the basic concepts but the challenge problems throw me for a curve and the questions in my textbook dont help much as they involve units we haven't done yet. I just need to practice some of the types of application problems that i might see.

[b]Problem 1.[/b] The atomic number of Chromium is 24. Write the electronic configuration of this atom in the ground state.

[b]Problem 2.[/b] Explain in your own words the concepts of sigma and pi bonds, and hybridization of orbitals.

[b]Problem 3.[/b] Indicate the type of hybridization of each carbon in 1,2-butadiene.

[b]Problem 4.[/b] The molecule of perchloric acid, $ \ce{HClO4}$, have:

(a) 13 non-bonding pairs of electrons, one pi bond, and 4 sigma bonds.

(b) 9 non-bonding pairs of electrons and 6 sigma bonds.

(c) 8 non-bonding pairs of electrons, 2 pi bonds, and 7 sigma bonds.

(d) 2 non-bonding pairs of electrons, 3 pi bonds, and 4 sigma bonds.

(e) 11 non-bonding pairs of electrons and 5 sigma bonds.

Select the correct option.

[b]Problem 5.[/b] Methyl isocyanate is an important intermediate material in chemical industry.

(a) Draw the Lewis structure for this molecule, including all non-bonding pairs of electrons.

(b) Indicate the hybridization scheme for each carbon and for the nitrogen and oxygen atoms.

(c) How many sigma and pi bonds are there in this molecule?

[b]Problem 6.[/b] Experimental evidence shows that the molecule of nitrous oxide, $ \ce{N2O}$, is linear and polar.

(a) On that basis, which of the arrangements $ \ce{NNO}$ or $ \ce{NON}$ is the correct one for this molecule?

(b) Based on your answer to (a), draw three possible ressonance structures for $ \ce{N2O}$.

[b]Problem 7.[/b] Classify as true or false each of the following sentences:

(a) Since an hybrid orbital sp3 results from the combination of four orbitals, it can have, at most, four electrons.

(b) In boric acid, $ \ce{B(HO)3}$, the central atom have sp hybridization.

(c) In hydronium ion, $ \ce{H3O\plus{}}$, the oxygen atom is sp2 hybridized.

(d) In a covalent double bond, both bonds can be sigma.

(e) In water, the hybridization scheme of oxygen is sp3.

(f) In the methane, the carbon atom uses sp3 hybrid orbitals to make bonds with the hydrogens.

(g) A double bond is longer than a single bond.

(h) Every carbon atom covalently bonded to two other atoms must have sp2 hybridization.

[b]Problem 8.[/b] Write the Lewis structure and predict the geometry for each of the following molecules. When possible, write ressonance structures: methane, $ \ce{NH3}$, water, carbon dioxide, $ \ce{N2}$, $ \ce{H2O2}$, carbon monoxide, $ \ce{Cl2O}$, borane, hydrazine, $ \ce{SOCl2}$, ethane, ethylene, acetylene, $ \ce{NH4\plus{}}$, cyanidric acid, nitrous acid, nitric acid, sulphuric acid, sulphur trioxide, phosphoric acid, $ \ce{N3\minus{}}$.

I also recently took a test on bonding and structure.

General tips:

1.) In order to predict the geometry of molecular structures, simply find out how many bonding pairs the central atom has (remember, double and trip bonds count as ONE pair) and use this to determine the hybridization bond and from the number of lone pairs, the actual geometry. You can easily memorize this from this year's Princeton Review book. I believe their section on the geometry is very straightforward. I neer thought I could memorize the geometries but their organization was very simple and efficient.

2.) Know the exceptions. In Carcul's Problem 1, chromium is an exception in the electronic configuration! Can you list the other element that has the exception? Also, my test on bonding included periodic trends. I got tripped up on some of the exceptions in the trends for ionization energy.

3.) All you have to know about enthalpy ($ \Delta H$) is that it's $ \text{reactants} \minus{} \text{products}$. The bond energies will be given to you. Hint: in many cases, the questions will ask you to predict products and/or reactants. You don't necessarily have to be good at this, but know your solubility rules but the bond energies they give you should help you determine if your product is correct or not.


function
calculus
derivative
calculus computations

Let x and y be functions of time t that are related by the equation y^2 = 2 +xy. At time t=5, the value of y is 3 and dy/dt =6. Find the value of dx/dt at time =5.

Note: The derivative of y^2 = 2 + xy = y / (2y - x)


Let $ f$ be a function with domain $ X$ and range $ Y$. Let $ A,B\subseteq Y$ and $ C,D\subseteq X$. Which of the following is not true?

[b](a)[/b] $ f^{\minus{}1}(A\cup B)\equal{} f^{\minus{}1}(A)\cup f^{\minus{}1}(B)$
[b](b)[/b]$ f^{\minus{}1}(A \cap B)\equal{} f ^{\minus{}1}(A)\cap f^{\minus{}1}(B)$
[b](c)[/b]$ f(C \cup D)\equal{} f(C)\cup f(D)$
[b](d)[/b]$ f(C \cap D)\equal{}f(C)\cap f(D)$

Please give explanation with your answer.Thanks.

The first three are all true; the fourth is $ f(C\cap D)\subseteq f(C)\cap f(D)$.  However, equality holds if $ f$ is injective.

To establish any of the equality ones, first assume that $ x\in f(C\cup D)$ (for example), and then show that $ x\in f(C)\cup f(D)$.  This establishes that $ f(C\cup D)\subseteq f(C)\cup f(D)$.  Then assume that $ x\in f(C)\cup f(D)$ and show that $ x\in f(C\cup D)$, which shows the other direction.  If $ x\in f(C)\cup f(D)$, then $ x$ is in one of the two (or both).  Let's say $ C$.  Then obviously, $ x\in f(E)$, where $ E\supseteq C$, so $ x\in f(C\cup D)$.  So that's one direction.  If $ x\in f(C\cup D)$, then $ \exists z\in C\cup D$ such that $ f(z)\equal{}x$.  Since $ z$ is in one (or both) of $ C$ and $ D$, $ x\in f(C)$ or $ f(D)$, hence $ x\in f(C)\cup f(D)$.


calculus
derivative
logarithms
search
calculus computations

Is the following statement true?

[b]A convergent sequence of reals can converge to a point that is not necessarily a limit point.[/b]

The definition can be found here:

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

I am very sure the statement is true, but someone told me it is not correct.

The issue here is sequences that are eventually constant, and it's a question of language- is the set or the sequence at the core of our limit point definition?

To be specific, consider the constant sequence $a_n=1$. It converges to 1. Is 1 a limit point?

[quote="jmerry"] is the set or the sequence at the core of our limit point definition?
[/quote]
What is the difference?

[quote="jmerry"]
To be specific, consider the constant sequence $a_n=1$. It converges to 1. Is 1 a limit point?[/quote]

I am guessing what you were saying above to answer this question:
a) No, 1 is not a limit point with respect to the [b]sequence[/b].
b) Yes if there exists infinite of points in the neighborhood of 1 for every $\epsilon > 0$ that belong to the [b]set[/b] (of which its closure contains the sequence). Otherwise, the answer is no.

Are they correct?

Limit point
A number $x$ such that for all$\varepsilon>0$ , there exists a member of the set $y$ $\bf different$ from $x$ such that$|x-y|<\varepsilon$.
So $1$ is not limit point...

Actually, the sequence definition in my mind is that a limit point is a limit of some subsequence. In that version, a convergent sequence always converges to its only limit point.

This limit may fail to be a limit point of the set of elements in the sequence, but only if this set is finite.

[quote="jmerry"]... a limit point is a limit of some subsequence.[/quote]
One way to keep the vocabulary distinct and unambiguous: use the name "cluster point" for this concept, the limit of a convergent subsequence of a given sequence. That would then leave "limit point" reserved for talking about sets.

[b]Definition.[/b] The point $l\in \overline {R}=[-\infty.\infty ]$ is a [u]limit point (accumulation point)[/u] w.r.t. the set $\oslash\ne A\subset R$ iff there is a sequence $x_n,\ n\in N^*$ so that $\boxed {l\not \in X}\equiv \{x_n\ |\ n\in N^*\}\subset A$ and $x_n\rightarrow l.$

If $A=\{x_n\ |\ n\in N^*\}$, then $l\in \overline R$ is a limit point w.r.t. $A$ iff there is a sequence [color=green](isn't necessarily be a subsequence !)[/color] $y_n,\ n\in N^*$ so that $\boxed {l\not\in Y}\equiv \{y_n |\ n\in N^*\}\subset A$ and $y_n\rightarrow l.$

In my opinion, the notion $\blacktriangleright$ the limit of the sequence $x_n,\ n\in N^*\blacktriangleleft$ and the notion $\blacktriangleright$ a limit point w.r.t. the set $X\equiv \{x_n |\ n\in N^*\}\blacktriangleleft$ are independently between them, i.e. in generally, one of them don't implicate the other.

[b]Examples.[/b]
$\blacksquare\ 1^{\circ}.$ A stationary sequence $x_n,\ n\in N^*$ (i.e. there are $a\in R$ and $n_1\in N^*$ so that for any $n>n_1,\ x_n=a$) is convergent ($x_n\rightarrow a$), but the point $a$ don't a limit point w.r.t. the set $X\equiv \{x_n\ |\ n\in N^*\}$. Exactly, if  $X'$ is the derivative of the set $X($ i.e. the set of the all limit points w.r.t. the set $X)$, then $X=\oslash.$
$\blacksquare\ 2^{\circ}.$ Let the set $X\equiv \{x_n\ |\ n\in N^*\}$, where ${x_n=\left\{\frac n2\right\}}\ln n+\frac{n}{n+1}$. Then the sequence $x_n,\ n\in N^*$ don't has limit, but $X'=\{1,\infty\}.$

I find the term [b]limit[/b] very confusing. Also a search in net, one lecture note uses the term [i]limit point[/i] and [i]cluster point[/i] interchangebly and in another, [i]limit point[/i] and [i]accumulation point[/i] interchangebly. So the three terms are equivalent.

In a book, it says:
[quote]A sequence of points $\{x_k\}$ in $\mathbb{R}^n$ is said to converge to a limit $x$(written $x_k \to x$) if the distance $d(x_k, x)$ between $x_k$ and $x$ tends to zero as $k$ goes to infinity, i.e., if for all $\epsilon > 0$, there exists an integer $k(\epsilon)$ such that for all $k > k(\epsilon)$, we have $d(x_k, x) < \epsilon$. A sequence $\{x_k\}$ which converges to a limit $x$ is called a [i]convergent sequence[/i].[/quote]

If I understand the statement correctly, it means if a sequence converges, it always converge to a [b]limit[/b] ( I am not sure if we can say if there is a limit for the sequence, it must converge). Also it does not state $x_k$ cannot take the value of $x$. So $x_k = x$ for all $k = 1, 2, \ldots$ converges to a [b]limit[/b] which is $x$. In this context, the example given by jmerry, the value 1 is indeed the [b]limit[/b] of the sequence. I presume the definition of [b]limit[/b] is not the [b]limit point[/b] as defined here: http://mathworld.wolfram.com/LimitPoint.html


function
integration
calculus
geometric series
real analysis
real analysis unsolved

Let $ f$ be a $ 2\pi$ periodic function and piecewise smooth.  Let $ m$ be a postive integer and let $ f_{m}(x)\equal{}f(mx)$.
Show that $ \hat{f}_m(n)\equal{}\hat{f}(\frac{n}{m})$ if $ m$ divides $ n$ and $ 0$ otherwise.

$ \hat{f}(n)$ is the nth fourier coefficient.

In my notes for the Fourier Analysis class I teach, I give a theorem for discrete Fourier transforms that is closely related to this. For this case:

$ \widehat{f_m}(n)=\frac1{2\pi}\int_0^{2\pi}f(mx)e^{-inx}\,dx.$

Make the substitution $ u=mx:$

$ \widehat{f_m}(n)=\frac1{2\pi m}\int_0^{2\pi m} f(u)e^{-i\frac{n}{m}u}\,du$

Chop this integral into pieces:

$ \widehat{f_m}(n)=\frac1{m}\sum_{k=0}^{m-1} \frac1{2\pi}\int_{2\pi k}^{2\pi(k+1)}f(u)e^{-i\frac{n}{m}u}\,du$

On each piece, make the substitution $ u=2\pi k+t,$ and use the fact that $ f$ is periodic of period $ 2\pi.$

$ \widehat{f_m}(n)=\frac1{m}\sum_{k=0}^{m-1}\frac1{2\pi} \int_0^{2\pi} f(t)e^{2\pi i\frac{kn}{m}}e^{-i\frac{n}{m}t}\,dt$

$ =\left(\frac1{m}\sum_{k=0}^{m-1}e^{-2\pi i\frac{kn}{m}}\right)\cdot\frac1{2\pi}\int_0^{2\pi}f(t)e^{-i \frac{n}{m}t}\,dt.$

What remains is to show that

$ \frac1{m}\sum_{k=0}^{m-1}e^{-2\pi i\frac{kn}{m}}= \begin{cases}1&\text{if }m\mid n\\0&\text{otherwise}\end{cases}.$

Then, if $ m\mid n,$

$ \widehat{f_m}(n)=\widehat{f}\left(\frac{n}{m}\right) \left(\frac1{m}\sum_{k=0}^{m-1}e^{-2\pi i\frac{kn}{m}}\right)= \widehat{f}\left(\frac{n}{m}\right).$

Aha. If you know more about Fourier series, you can just use the uniqueness theorem and the fact that $ e^{in(mx)}\equal{}e^{i(nm)x}$ for all $ x$ but Kent's solution is certainly the simplest one :).

As a side remark, note that Kent is using \widehat: $ \widehat f$ rather than  \hat $ \hat f$ for the Fourier transform. It looks prettier, doesn't it? :)

e^pi stated the hypotheses as "$ f$ is a $ 2\pi$ periodic function and piecewise smooth." My proof made no use of "piecewise smooth", and fedja's alternative argument also makes no use of that. So regard that condition as a convenience or a habit, but we can weaken the hypothesis to $ f\in L^1(\mathbb{T}).$ That is, that $ f$ is absolutely integrable over a period. That's all I needed.

[quote="Kent Merryfield"]

What remains is to show that

$ \frac1{m}\sum_{k = 0}^{m - 1}e^{ - 2\pi i\frac {kn}{m}} = \begin{cases}1 & \text{if }m\mid n \\
0 & \text{otherwise}\end{cases}.$

 [/quote]

how do you show this?
at first glance it seemed to be a simple geometric series, but that doesnt seem to work! :maybe:

It [b]is[/b] a simple (finite) geometric series.

$ \sum_{k\equal{}0}^{m\minus{}1}r^k\equal{}\begin{cases}\frac{1\minus{}r^m}{1\minus{}r}&\text{if }r\ne1\\m&\text{if }r\equal{}1\end{cases}.$

If $ m\mid n,$ then $ e^{2\pi i\frac{kn}{m}} \equal{}1$ and

$ \frac1{m}\sum_{k \equal{} 0}^{m \minus{} 1}e^{ \minus{} 2\pi i\frac {kn}{m}} \equal{}\frac1{m}\sum_{k\equal{}0}^{m\minus{}1}1\equal{}1.$

If $ m\,\not|\; n,$ then

$ \frac1{m}\sum_{k \equal{} 0}^{m \minus{} 1}e^{ \minus{} 2\pi i\frac {kn}{m}}\equal{}\frac1{m}\cdot\frac{1\minus{}e^{\minus{}2\pi i\frac{mn}{m}}}{1\minus{}e^{\minus{}2\pi i\frac{n}{m}}} \equal{} 0.$


Does anyone who has taken the AMC in previous years know when we will get our results?

Err...aren't you allowed a solutions manual where you can find out your results? I think once your teacher mails the tests, they are supposed to give you a solutions manual, and then you can find out on your own.

Ok, I will probably get that tomorrow. I'm just asking when will we get the "official" results.

Contests will be coming to us by mail starting Wednesday, February 11.  A majority will have arrived by Monday, February 16, and after organizing the packages, scanning will begin that week.  Scanning will continue through that week and the next week, and then the data processing to determine averages, 5% levels, 1% levels and so on will begin.  As soon as the data processing is completed, we will email results to the Teachers/Contest Managers who provided us with an email address.  That will be your earliest official notification.  A paper copy of the results will be postal-mailed soon after.  Based on previous year's  efforts, you can expect official results by email to teachers about the end of February.

That is, provided our sometimes cranky scanners work well, we don't have weather-related outages that slow the mail or even disrupt power, and a thousand other things that might slow us down.

Please bear with us, although the AMC staff is dedicated and experienced, we are a relatively small operation.  We will process somewhere around 200,000 score sheets from about 3,500 schools located world-wide to process at the same time we are shipping the B contests and doing many other things behind the scenes that students are not aware of.   We aim to provide you results as soon as we humanly can.

Steve Dunbar
AMC Director

our math teachers tells us preliminary scores the next day, i heard by the end of the month you'll get ur official scores

[quote="AMCDirector"]Please bear with us, although the AMC staff is dedicated and experienced, we are a relatively small operation.  We will process somewhere around 200,000 score sheets from about 3,500 schools located world-wide to process at the same time we are shipping the B contests and doing many other things behind the scenes that students are not aware of.[/quote]

After two years of participating in AMC contests I can say that I have been very pleasantly surprised by the speed and quality of information from AMC headquarters about contest results. But it's true: the weather on the Great Plains at this time of year is NASTY, so there may be unavoidable delays in processing. The last two times I've had young people in my homeschooling group test, I was the administrator, so I got results as soon as humanly possible. On 10B day, my son will test as part of another homeschooling group, of which I am NOT administrator, so I suppose the time lag may come from communicating with HER after she gets results--although she seems to be very efficient, perhaps after the example of AMC headquarters.

Keep up the good work!

I hope the young people who've just taken the test, or who will take it on the 25th, can bear the excitement while they wait.

you get it within 30 days after the test...


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a group and $ p$ the smallest prime dividing the order of $ G$. Prove that if $ H$ is a subgroup of $ G$ s.t. $ [G: H]\equal{}p$ , $ H$ is normal.

[hide="Hint"]
The normalizer of $ H$ in $ G$ contains $ H$, so what must it be? How many conjugates does $ H$ have? Call this number $ m$, and let $ S$ be the set of $ m$ conjugates. Let $ G$ act by conjugation on $ S$ (this is a standard action you should always look at when solving this kind of problems). This defines, as for any action, a homomorphism $ \phi : G \to S_m$. Consider the kernel $ K$ of this homomorphism. Show that $ K \leq H$ as well as $ K \neq H$, so $ [G: K] \equal{} [G: H][H: K]$ is a non-trivial factorization. Remembering that $ [G: K]$ divides $ m$, arrive at a contradiction.
[/hide]

If that's insufficient, I'll provide the details  :)

Thanks. Let's see if I can solve it based on this hint... Please correct me if I am wrong.

First assume that $ H$ is not normal. Consider the action of $ G$ on its subsets by conjugation. Since $ |Orb(H)| \dot |Norm(H)| \equal{} |G|$ and $ H \subset Norm(H)$. It follows that $ H\equal{}Norm(H)$ and the number of conjugates $ m \equal{} p$.

Now, the kernel $ K$ must be the intersection of all stabilizers of elements of $ S$, that is $ \cap Stab(s), \forall s \in S$. Since $ H \in S$ and $ Stab(H) \equal{} H$, follows $ K \subset H$. But $ \forall x , K \subset Stab(xHx^{\minus{}1}) \equal{}xHx^{\minus{}1}$. Since the intersection of all conjugates of $ H$ is not equal to $ H$ (otherwise $ H$ will be normal), $ K \neq H$.

I still don't understand why $ [G: H]$ divides $ m$.

I think I figured it out.

From the First Isomorphism Theorem, we have that $ [G: K] \equal{} |Im(G)|$ which divides $ p!$ because it is a subgroup of $ S_p$. On the other hand $ |G|\equal{}|K| \cdot [G: K]$, hence $ [G: K]$ divides $ (p!, |G|)$. But since $ p$ is the least prime dividing $ |G|$, we have $ (p!, |G|)\equal{}p$. Hence $ [G: K]$ divides $ p$, thus it is at most $ p$.

But, $ [G: K] \equal{} [G: H][H: K]$, where $ [G: H]\equal{}p, [H: K]>1 \Leftrightarrow [G: K]>p$. Contradition. Hence $ H$ is normal.

Now I see how the fact that $ p$ is the least prime dividing $ |G|$, can be used.


algebra
polynomial
search
algebra unsolved

Let P1 and P2 be two polynomials that take the same set of values on the rationals (i.e., P1(Q) and P2(Q) are the same set). Prove that there are constants a and b such that P1(x)=P2(ax+b) for every x.

He-he...
There was a long discussion on this problem on the forum. Try to search it.
I have a feeling that I was involved in this discussion too.

I can't find such a discussion on the forum, can you give me some searching hints,please ?

Take it http://www.mathlinks.ro/Forum/viewtopic.php?t=1890  :D  :D  :D


Given that the total surface area of a cube is 216 square inches, how many cubic inches are in the cube's volume?

Each face of the cube is $ \frac{216}{6}\equal{}36$ square inches. So the cube's edge length is $ \sqrt{36}\equal{}6$. Therefore the volume of the cube is $ 6^3\equal{}\boxed{216}$ cubic inches.


trigonometry
inequalities
function
inequalities proposed

I have simplified a number of inequalities to this one, but I am unable to prove it.  (I am not even completely sure that it is true).  Any help is appreciated.

Prove

$ \prod(1-\cos A)\ge\prod\cos A$

for $ A,B,C$ angles of a triangle.

$ \Pi\left(\frac {1}{\cos\alpha} \minus{} 1\right)\ge 1$ can be true only for acute triangles, since then $ 0\le \cos\alpha\le 1$. :wink:

Yes, I meant for $ 0<A,B,C<\frac{\pi}{2}$.  Sorry for not clarifying.

OK. No problem :arrow:

Let us get the function $ f(x) \equal{} \frac {1}{\cos x} \minus{} 1$.
$ f'(x) \equal{} \frac {\sin x}{\cos^2 x}$; $ f''(x) \equal{} \frac {\cos^3 x \plus{} 2\sin^2 x\cos x}{\cos^4x} > 0$ $ \Rightarrow$ the function is convex.

$ (\alpha,\beta,\gamma)\succ (\pi/3,\pi/3,\pi/3)$ $ \Rightarrow \Pi\left(\frac {1}{\cos\alpha} \minus{} 1\right)\ge \left(\frac {1}{\cos\frac {\pi}{3}} \minus{} 1\right)^3 \equal{} 1$ by Karamata :)

Wouldn't Karamata imply $ \sum f(\alpha)\ge 3$?


Let n > 1 be a positive integer and X be a set containing n elements. A1;A2; ... ;A101 are
subsets of X such that the union of any 50 of them has more than 50/51n elements.
Prove that among the given subsets it is possible to choose three, such that every two of them
have a non-empty intersection.

Bomb

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=5488

Pierre.


Prove the following inequality:

Let $ x_{i}>0$, $ S\equal{}x_{1}\plus{}x_{2}\plus{}x_{3}\plus{}......\plus{}x_{n}$. Show that
\[ \frac{S}{S\minus{}x_{1}}\plus{}\frac{S}{S\minus{}x_{2}}\plus{}\frac{S}{S\minus{}x_{3}}\plus{}.......\plus{}\frac{S}{S\minus{}x_{n}}\geq{\frac{n^2}{n\minus{}1}}\]

[quote="Sunkern_sunflora"]Prove the following inequality:

Let $ x_{i} > 0$, $ S \equal{} x_{1} \plus{} x_{2} \plus{} x_{3} \plus{} ...... \plus{} x_{n}$. Show that
\[ \frac {S}{S \minus{} x_{1}} \plus{} \frac {S}{S \minus{} x_{2}} \plus{} \frac {S}{S \minus{} x_{3}} \plus{} ....... \plus{} \frac {S}{S \minus{} x_{n}}\geq{\frac {n^2}{n \minus{} 1}}
\]
[/quote]
$ \Leftrightarrow T\equal{}\frac {S}{S \minus{} x_{1}} \plus{} \frac {S}{S \minus{} x_{2}} \plus{} \frac {S}{S \minus{} x_{3}} \plus{} ....... \plus{} \frac {S}{S \minus{} x_{n}}\equal{}\frac {S^2}{S^2 \minus{}S x_{1}} \plus{} \frac {S^2}{S^2 \minus{} Sx_{2}} \plus{} \frac {S^2}{S^2 \minus{} Sx_{3}} \plus{} ....... \plus{} \frac {S^2}{S^2 \minus{} Sx_{n}}$
apply Cauchy-Schwarz inequality we have
$ T\ge \frac{(nS)^2}{nS^2\minus{}S(x_1\plus{}x_2\plus{}..\plus{}x_n)}\equal{}\frac{(nS)^2}{nS^2\minus{}S^2}\equal{}\frac {n^2}{n \minus{} 1}$

Nice proof, bunhiacovski! Can you prove it without $ C.S.$, using simple $ A.M.\geq{G.M.}$? :wink:

It follows directly from AH-HM which is connected with C.S.
Applying AM-GM we obtain $ \sum_{i\equal{}1}^{n}\frac{S}{S\minus{}x_{i}}\ge n\prod_{i\equal{}1}^{n}\sqrt[n]{\frac{S}{S\minus{}x_{i}}}$. Now we have to prove that $ n(n\minus{}1)S\ge n^{2}\prod_{i\equal{}1}^{n}\sqrt[n]{S\minus{}x_{i}}$. Again by AM-GM $ nS\minus{}S\ge n\prod_{i\equal{}1}^{n}\sqrt[n]{S\minus{}x_{i}}$ we obtain $ n(n\minus{}1)S\equal{}n(n\minus{}1)S$. Done!

I think it can be proved by Jensen ineq
There $ \frac{S}{S\minus{}x}$ is a convex

[quote="Sunkern_sunflora"]Prove the following inequality:

Let $ x_{i} > 0$, $ S \equal{} x_{1} \plus{} x_{2} \plus{} x_{3} \plus{} ...... \plus{} x_{n}$. Show that
\[ \frac {S}{S \minus{} x_{1}} \plus{} \frac {S}{S \minus{} x_{2}} \plus{} \frac {S}{S \minus{} x_{3}} \plus{} ....... \plus{} \frac {S}{S \minus{} x_{n}}\geq{\frac {n^2}{n \minus{} 1}}
\]
[/quote]
We have:
By AM-GM
$ \frac {S}{S \minus{} x_{1}} \plus{}  \frac {n^2}{(n\minus{}1)^2}\frac {S \minus{} x_{1}}{S} \ge\ \frac {2n}{n\minus{}1}$
$ \rightarrow LHS \plus{} \frac {n^2}{(n\minus{}1)^2}\frac {nS \minus{} S}{S} \ge\ \frac {2n^2}{n\minus{}1}$
$ \rightarrow LHS \ge\ \frac {n^2}{n \minus{} 1}$
By C-S:
$ LHS \equal{} S (\frac {1}{S \minus{} x_{1}} \plus{} \frac {1}{S \minus{} x_{2}} \plus{} \frac {1}{S \minus{} x_{3}} \plus{} ....... \plus{} \frac {1}{S \minus{} x_{n}})$
$ LHS\ge\ S \frac {n^2}{nS \minus{} S}\equal{}\frac {n^2}{n \minus{} 1}$


$ a,b,c,d$ are real positive numbers
$ Proof$
$ \sqrt{a^2\plus{}b^2} \plus{} \sqrt{c^2\plus{}d^2} \leq \sqrt{a^2\plus{}c^2} \plus{} \sqrt{b^2\plus{}d^2} \plus{} \frac{2|ad\minus{}bc|}{\sqrt{a^2\plus{}c^2} \plus{} \sqrt{b^2\plus{}d^2} }$

:maybe: How about this one with the same codition?
\[ \sqrt {a^2 \plus{} b^2} \plus{} \sqrt {c^2 \plus{} d^2}\leq \sqrt {(a \plus{} c)^2 \plus{} (b \plus{} d)^2} \plus{} \frac {|ad \minus{} bc|}{\sqrt {(a \plus{} c)^2 \plus{} (b \plus{} d)^2}}\]

[quote="colorfuldreams"]$ a,b,c,d$ are real positive numbers
$ Proof$
$ \sqrt {a^2 \plus{} b^2} \plus{} \sqrt {c^2 \plus{} d^2} \leq \sqrt {a^2 \plus{} c^2} \plus{} \sqrt {b^2 \plus{} d^2} \plus{} \frac {2|ad \minus{} bc|}{\sqrt {a^2 \plus{} c^2} \plus{} \sqrt {b^2 \plus{} d^2} }$[/quote]

Try $ (a,b,c,d)\equal{}(4,1,8,2)$

$ 3\sqrt{17} > 5\sqrt{5}$ :wink:

[quote="crazyfehmy"][quote="colorfuldreams"]$ a,b,c,d$ are real positive numbers
$ Proof$
$ \sqrt {a^2 \plus{} b^2} \plus{} \sqrt {c^2 \plus{} d^2} \leq \sqrt {a^2 \plus{} c^2} \plus{} \sqrt {b^2 \plus{} d^2} \plus{} \frac {2|ad \minus{} bc|}{\sqrt {a^2 \plus{} c^2} \plus{} \sqrt {b^2 \plus{} d^2} }$[/quote]

Try $ (a,b,c,d) \equal{} (4,1,8,2)$

$ 3\sqrt {17} > 5\sqrt {5}$ :wink:[/quote]
In fact,I think colorfuldreams means this one
$ \sqrt {a^2 \plus{} b^2} \plus{} \sqrt {c^2 \plus{} d^2} \leq \sqrt {(a\plus{}c)^2\plus{}(b\plus{}d)^2} \plus{} \frac {2|ad \minus{} bc|}{\sqrt {a^2 \plus{} c^2} \plus{} \sqrt {b^2 \plus{} d^2} }$
And this a test of $ Belarus$ in 2002.