geometry
geometric transformation
homothety
congruent triangles
geometry unsolved

Hi ! ABC is a triangle . Where we should choose [img]http://www.texify.com/img/%5CLARGE%5C%21E%2CF.gif[/img] on [img]http://www.texify.com/img/%5CLARGE%5C%21AB%20%2C%20AC.gif[/img] so that [img]http://www.texify.com/img/%5CLARGE%5C%21BE%3DEF%3DFC.gif[/img] ?

mahanmath I know that you are new here on AoPs , I am new too , but I know that there is no need to reply for help when you just have posted the topic today. Nobody is under obligation of replying immediatly

But never mind ,  :) I think that this is a well-known construction , but I cannot find any references

[quote="livetolove212"]Use [b]Thales[/b]'s theorem.[/quote]

Livetolove212 could you explain in detail how  Thales theorem helps , because EF is not parallel to BC in general unless ABC is isosceles with legs AB,AC
Thanks   :)

Livetolove212 i couldn`t solve it with Thales but i think Homothety can help us .

I heard the solution 1 week ago.. but I cannot remember all of the sol.....  :|

Let P as an intersection of BC's perpendicular at midpoint and ABC's circumscribed circle.

Let M as an intersection of AB and a trace of point X s.t PC:CB = PX:XB (circle of Apollonios)

Let N as an intersection of AC and a trace of point Y s.t PB:BC = PY:YC (circle of Apollonios, too)

Then PB=PC, ∠PBM = ∠PCN, PN:NC = PB:BC = PC:CB = PM:MB.

Therefore PMB and PNC is congruence. -> MB = NC

And, PMN and PBC is similar. From this, BM=MN also can be made. (<< I cannot do this now..)

My idea: Construct the angle bisectors of $ \angle BEF$ and $ \angle EFC$. These intersect at $ X$. Then $ AX$ bisects $ \angle BAC$ and you get three congruent triangles and $ \angle BXC\equal{}\frac{3}{2}(180\minus{}A)$. So you can pretty easily construct $ X$ from these two properties and then finding $ E$ and $ F$ is easy.


geometry
3D geometry
sphere
pyramid
algebra
polynomial
function

The number of spheres needed to create a triangular pyramid is a polunomial function given by $ N(x)\equal{}\frac{1}{6}x^3\plus{}\frac{1}{2}x^2\plus{}\frac{1}{3}x$, where $ x$ is the number of layers and $ N(x)$ is the number of spheres. How many spheres are in the eighth layer?

Eight layer = # of spheres needed to make $ 8$ layer pyramid - #of shperes needed to make $ 7$ layer pyramid
Plugging into equation gives $ \frac168^3 + \frac128^2 + \frac138 - \left(\frac167^3 + \frac127^2 + \frac137\right) =$(calc)$ = 36$
Also, you can use logic and see that each layer is the traingle number.

[color=blue]Fixed $ \text{\LaTeX}$. :) - Izzy[/color]


Let $ x\equal{}a\plus{}\frac1{b},y\equal{}b\plus{}\frac1{c},z\equal{}c\plus{}\frac1{a}$.Show that:
$ xy\plus{}yz\plus{}zx\ge 2(x\plus{}y\plus{}z)$

[quote="duongptnkvn"]Let $ x \equal{} a \plus{} \frac1{b},y \equal{} b \plus{} \frac1{c},z \equal{} c \plus{} \frac1{a}$.Show that:
$ xy \plus{} yz \plus{} zx\ge 2(x \plus{} y \plus{} z)$[/quote]
It's obviously wrong. Try $ a\equal{}b\equal{}5$ and $ c\equal{}\minus{}0.001.$ :wink:


When are the definite articles [b]el[/b] and [b]la[/b] not used in spanish?

How do you say "stop sign" in spanish? My book says that it's "senal de stop", but I think that's a typo.

In most cases it is there.

Some cases that it isn't:
When stating someone's job.
Mi padre es bombero.

On SOME countries, but it is disputable.

Yo soy de Mexico.

I don't realy know Spanish so don't kill me if I am wrong :nhl:

Why don't they use the article in this sentence?

Usamos ingredientes puros solamente.

(We only use pure ingredients. )

In addition, many books say that the definite article is used before a noun in a [i]general sense[/i]... what exactly does that mean?

I think its because you don't say "we use the pure ingredients"

but you have to ask a spanish dude to be sure

Sometimes, in spanish we don't use articles, because the article is obvious when you read the verbs.

e.g.: Cocinamos comida (we cook food). The verb "cocinamos" is always for nosotros (us).

In spanish, verbs can be infinitive or "conjugados". In english, verbs are only in infinitive.

Let me explain you.

Let's put the verb love (amar). In english you say: I love. You love. He love. We love. The person or article change, but the verb is always the same (the spelling does not change). In spanish, we say: Yo amo, tu amas, el ama, nosotros amamos. The verb (spelling) change you see?? (amo, amas, ama, amamos). So, let's say again: We love our grandma. Now, let's say "amamos a nuestra abuela". We don't have to put "nosotros amamos a nuestra abuela" becuase the ver "amamos" is only for the person "us". In conclusion, in spanish, a verb can change it spelling to change the person or article we're talking about. In english, the verb is the same, but you have to include the article.

[quote="236factorial"]No.. I mean the definite article el or la as in

el libro ([i]the[/i] book) or

la almohada (the pillow)

And I think the stuff about the verb conjugations is very cool  :D[/quote]

Are you talking why do we say "el libro" en vez de "la libro", for example??

[quote="236factorial"]I'm talking about something like:

¿Por qué no usa usted "el" en esta oración: Usamos ingredientes puros solamente. ?
¿Por qué no es : Usamos [i]los[/i] ingredientes puros solamente. ?

[size=75]I hope that made sense...[/size]
(Does "usa" mean "you use"?)[/quote]

"usa" means "he/she uses" .

Yo can use "el" in that sentence, but, when you're talking, the meaning of the sentence is obvious, so it's not necessary to put it.

[quote="José"]
"usa" means "he/she uses" .

Yo can use "el" in that sentence, but, when you're talking, the meaning of the sentence is obvious, so it's not necessary to put it.[/quote]

How do you say "you use"?

Can you give an example of a sentence where the meaning is not obvious, and "el", "la", "los", or "las" is necessary?

[quote="José"]
Yo can use "el" in that sentence[/quote]

Sorry, it was: you can use "los" in that sentence

[quote="236factorial"][quote="José"]
"usa" means "he/she uses" .

Yo can use "el" in that sentence, but, when you're talking, the meaning of the sentence is obvious, so it's not necessary to put it.[/quote]

How do you say "you use"?

Can you give an example of a sentence where the meaning is not obvious, and "el" is necessary?[/quote]

..you use=vos usás...

do you use pen? (vos) usás lapicera???.

About the other thing: ¿EL sombrero está en su lugar correcto? (is the hat in it's correct place?) [you can't say:¿sombrero está en su lugar correcto?   ...  Sorry 236factorial if I'm not to good to explain...

you have to put articles when they're before a noun, but it isn't necessary to put them when they're before a verb.

[quote="236factorial"]Do you mean after a verb? Después de un verbo? Or otherwise, the sentence that I mentioned wouldn't follow your rule (ingredientes is a noun and I didn't [i]have[/i] to put an article before it).

Sorry if I'm making this too hard  :D[/quote]

Becuase ingredientes is plural I think... if not, you would have put ingrediente (singular)

Usamos (los) ingredientes puros solamente... Putting or not LOS doesn't change the meaning, because by only looking at ingredientes you see that it's plural... Sorry if I made a very huge mess in your head.

[quote="José"]Becuase ingredientes is plural I think... if not, you would have put ingrediente (singular)

Usamos (los) ingredientes puros solamente... Putting or not LOS doesn't change the meaning, because by only looking at ingredientes you see that it's plural... Sorry if I made a very huge mess in your head.[/quote]

Los espárragos son saludables.

The "los" is for some reason necessary...

[quote="236factorial"]
Los espárragos son saludables.

The "los" is for some reason necessary...[/quote]

Do you know something about "sujeto y predicado"?? (subject and object). The "los" is necessary because you need it (obligation) for studying and dividing that sentence in sujeto y predicado. Moreover, when el, la, los, las are starting a sentence, they're almost ever necessary.

Los espárragos son saludables.
Espárragos son saludables. (The 1º sound best).

Me gusta comer espárragos.
Me gusta comer los espárragos (The 2º sound best-don't you think that los here is unnecessary??).

The problem is that one get used to it when you speak every day, and it's dificult to study without talking and talking fluently every day.

1°
Can mean to things: 1 degree (un grado) or First (Primero)
By the way, You don't say "Usamos los ingredientes puros solamente" for the same reason why you don't say "We use the pure ingredients only"
The definite article is used to determinate the noun. For example:
"Voy a comprar libros" ("I'm going to buy books")
"Voy a comprar los libros" (I'm going to bue the books")
When you say the first one, the books that you are going to buy are not so clear, I mean, If you are in the middle of a conversation and you say that, someone will not be sure about which books are your going to buy. But if you use the second one (and in the conversation you have already mentioned the books) it is clear that the books that you are going to buy are those wich you have already said.

To make more clear the diference between "Usamos ingredientes puros solamente" and "Usamos los ingredientes puros solamente" you will have to put them into a contex, for example "Al usar ingredientes para nuestro pastel, usamos los ingredientes puros solamente" (At using ingredients for our cake, we use the pure ingredents only) And: "En nuestro pastel, usamos ingredientes puros solamente" (In our cake, we use pure ingredients only)

[quote="236factorial"]Thanks for replying. By the way, what does 1º mean?[/quote]

1º=First.


calculus
integration
limit
real analysis
real analysis unsolved

Proove that:
\[ \lim_{n\to\infty }\,\int_{0}^{\infty }\left(\sum_{i\equal{}1}^{\infty }\frac{i\plus{}x}{\left(i^{n}\plus{}x^{n}\right)^{n}}\right)\, dx\equal{}\frac{3}{2}\]
Or the generalization:
\[ \lim_{n\to\infty }\,\int_{0}^{\infty }\left(\sum_{i\equal{}1}^{\infty }\frac{\alpha i^{a}\plus{}\beta x^{b}}{\left(i^{n}\plus{}x^{n}\right)^{n\plus{}\kappa }}\right)\, dx\equal{}\alpha\plus{}\frac{\beta }{b\plus{}1}\]
$ b > 0$ and the others constants dont have restrition

Here are some calculations:

$ I \equal{}\sum_{i \equal{} 1}^{\infty}\int_{0}^{\infty}\frac{i\plus{}x}{(i^{n}\plus{}x^{n})^{n}}dx$

Using the substitutions $ x^{n}\equal{} y$ and $ y \equal{} i^{n}t$ we get that

$ I \equal{}\sum_{i \equal{} 1}^{\infty}\frac{1}{n}\frac{1}{i^{n^{2}\minus{}2}}\int_{0}^{\infty}\left(\frac{t^{1/n\minus{}1}}{(1\plus{}t)^{n}}\plus{}\frac{t^{2/n\minus{}1}}{(1\plus{}t)^{n}}\right)dt$

$ \equal{}\frac{1}{n}\left(\beta(1/n,n\minus{}1/n)\plus{}\beta(2/n,n\minus{}2/n)\right)\zeta(n^{2}\minus{}2)$

$ \equal{}\left(\frac{\Gamma(1/n)}{n}\frac{\Gamma(n\minus{}1/n)}{\Gamma(n)}\plus{}\frac{\Gamma(2/n)}{n}\frac{\Gamma(n\minus{}2/n)}{\Gamma(n)}\right)\zeta(n^{2}\minus{}2)$

Now use Stirling's formula combined with $ n\minus{}\Gamma(1/n)\rightarrow\gamma$, $ \zeta(\infty) \equal{} 1$ and the Legendre's formula

$ \Gamma(1/n)\Gamma(1/n\plus{}1/2) \equal{}\frac{\sqrt{\pi}}{2^{2/n\minus{}1}}\Gamma(2/n)$,

and you get that the first term converges to 1 and the second one to 1/2.

As for the generalization, it follows by similar computations.


The wiki automatically converts to PNG for display; get the thumbnail from MediaWiki; it'll be PNG.


college
search
superior algebra
superior algebra theorems

I'm find a free version of this book, but i can't fint out. Please help me, Thanks a lot


how many shortest paths are there in space from (0,0,0) to (6,6,6)?  A path consists in moves of increasex,y or z coordinate by exactly 1.

You want to find out how many paths of length exactly 18 there are given that each step can only be of length 1 in the positive $ x$, $ y$, or $ z$ direction. This is a counting problem. For any configuration of 18 steps, 6 of them will be $ x$'s, 6 of them will be $ y$'s, and 6 of them will be $ z$'s. Thus the problem is equivalent to finding the number of ways to arrange:

XXXXXXYYYYYYZZZZZZ,

which is just $ \frac{18!}{6!6!6!}\equal{}\boxed{17153136}$.

To add to that, there's pretty much a general formula for counting the number of paths when moving a units among x-axis, b units among y-axis, and c units among z-axis, which is just $ \frac{(a\plus{}b\plus{}c)!}{a!b!c!}$, which also works together with the similar formula for the paths on a grid formula for 2-dimensional walks (with x horizontal, y vertical) $ \frac{(x\plus{}y)!}{x!y!}$.

Similar formulas also work for higher dimensions.


calculus
integration
logarithms
trigonometry
calculus computations

Answer the following questions.

(1) Evaluate $\int_e^{e^e} \frac{\ln (\ln x)}{x\ln x} dx.$

(2) Let $\alpha,\beta$ be real numbers.Find the values of $\alpha,\beta$ for which the following equality holds for any real numbers $p,q.$

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(p\cos x+q\sin x)(x^2+\alpha x+\beta) dx=0.  \]

For 1)
$\int_{e}^{e^e} \frac{\ln (\ln (x) )}{x \ln (x)} dx = \int_{e}^{e^e} (\ln (\ln (x) ) )' (\ln (\ln (x) ) ) dx = [\frac{1}{2} (\ln (\ln (x) ) )]_{e}^{e^e}=\frac{1}{2}$

For 2)
let's make an integration by parts :
$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (p \sin x -q \cos x)'(x^2+\alpha x + \beta )dx$
$=p(\frac{\pi^2}{2} +2\beta ) + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (p \cos x +q \sin x)'(2x+\alpha )dx$
$=p(\frac{\pi^2}{2} +2\beta ) + 2q \alpha -2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (p \cos x -q \sinh x)dx$
$=p(\frac{\pi^2}{2} -4 +2 \beta ) + q (2 \alpha) = 0$
since this is verified for all $p,q \in \mathbb_R$ then :
$\beta = 2- \frac{\pi^2}{4}$ and $\alpha = 0$


is it true that if f is monotonous and surjective, then f is continuous?

Can you post the proof? please. thank you

It's an immediate proof. Suppose $ f$ is growing. Fix a point $ a \in \mathbb{R}$. Since $ f$ is growing it has a limit $ A \leq f(a)$ as $ x \to a^{ \minus{} }$*. But we cannot have $ A < f(a)$ because then the range of $ f$ would not contain any number between $ A$ and $ f(a)$. This means $ \lim_{x \to a^{ \minus{} }} f(x) \equal{} f(a)$. Analogously,  $ \lim_{x \to a^{ \plus{} }} f(x) \equal{} f(a)$

*I looked at your previous posts, and maybe you're not so familiar with analysis. So let me explain this part. It is an important property of the real numbers that every non-empty bounded from above set $ M \subset \mathbb{R}$ has a least upper bound $ B \equal{} \sup M$. This is equivalent to the so called completeness of $ \mathbb{R}$. To prove it you need to go down to the construction of the real numbers. Anyway, in this case we let $ A \equal{} \sup \{f(x) : x < a \}$. It's easy to verify that $ \lim_{x \to a^{\minus{}}} f(x) \equal{} A$

oh, it was kind of basic property of f. thank you for your help. :)


The set {1, 2, 3, ..., 2005} is partitioned into two sets: the sets of numbers having an odd digit sum and the set of numbers having an even digit sum. Let A be the sum of the numbers in the first set and let B be the sum of the numbers in the second set. Determine A - B.

Out of 2 consecutive numbers, one must have a even sum, and one must have an odd sum.

ie, if n has an odd sum, n+1 has an even sum
if n has an even sum,  n+1 has an odd sum

So, its' (1-2)+(3-4)+...+(2003-2004)+2005=-1002+2005=1003

Oh hm, I found a little mistake that I didn't consider, stuff like, 1999 and 2000, hehe

[hide]In the set, there are 9 numbers with only 1 digit. From those numbers, there are $5$ numbers with an odd digit sum.

There are 90 numbers of 2 digits in the set.
To obtain an odd sum, the number must exist out of an odd number and an even number. This is possible in two ways: OE or EO (E=even, O=odd)
There are 5 odd digits, and 5 even digits, so realising that the even digit in the second possibility can not be 0, the number of odd sums is $5.5+4.5=45$

There are 900 numbers of 3 digits in the set. The odd sums can be obtained like this:
EEO, EOE, OEE, OOO. Thus the ammount of odd sums is: $4.5^{2}.2+5^{3}.2=450$

Then we get the numbers with four digits, and starting with 1. There are thousand such numbers, and to obtain an odd sum, we must have 1OOE,1OEO,1EOO or 1EEE.
This gives us $4.5^{3}=500$

There are only 6 numbers with 4 digits and greater or equal to 2000 in the set. Of them, 3 have an odd sum.

So in the entire set, there are $1003$ numbers with an odd digit sum. That means that there are $1002$ numbers with an odd sum.

Therefor, A-B$=1003-1002=1$

Shorter an better way:
We have the following possibilities to obtain an odd sum in the set: 1OOE,1OEO,1EOO,0OOO,EEO,EOE,OEE,EEE.
Then there are the 3 odd sums that start with a 2
So that means $5^{3}.3+5.5^{3}+3=1003$

Correct?[/hide]

On a first glimp: for 1..1999

When is SoD(n) = SoD(n+1) [mod 2] ?
every jump 9-->10, except when n ended in an even number of 9's.

So that means 200 - 18 = 182. Since this is 91 of each parity, same for the 18 (=9 of each parity), here we have a 0.

Above, we have 2005-2004,...,2001-2000, which makes me believe the answer is 3? Why do you all say 1003?

I woke up at 4 am this night, realising how easily and nice this problem could be solved:

For 1..1999, we have that SoD(k) and SoD(2000-k) always have the same parity, and since all parities appear equally much, we have 0 for 1..1999

And from there on the other 6 terms give a trivial 3 indeed :)

[quote="a_vakilian"]I think $0$ because consider $\overline{xa}$ which $x \in \{0,1,...,200\}$ and ${a \in\{0,1,...,9}$ .if x is even so $A-B$ which $A,b \in \overline{xa}$ is $5$ and it is $-5$ if it is odd and we know sum of the digit of $x$ is even in half cases. so $A-B=0$.
I think I explain it badly :blush:[/quote]well, in your example you go from 0 to 200, so you have more even. ;) so you have part of a +5. That part is {0,...,5} and not {6,...,9}, which is 3/5 of 5 = 3. You again confirm my claim. I am very sure now it is 3.

Yes Peter I want to say there is little wrong withme because I consider $\{1,2,..,2009\}$.So It is $3$.


Let $a,b,c$ are positive number such that:
$a+b+c=\frac1a+\frac1b+\frac1c$
we have $a^{2}+b^{2}+c^{2}\geq \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}$
 :)

After homogenizing, it's enough to prove that \[\sum \frac{a^{2}}{bc}\geq \sum \frac{ac}{b^{2}}\] But this is just Muirhead; $[2,-1,-1]\succ[1,1,-2]$.

I have proved the general inequality but certainly, I doubt its valid. May you justify it?
Let $a,b,c$ are positive number such that $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Prove that
$a^{r}+b^{r}+c^{r}\geq \frac{1}{a^{r}}+\frac{1}{b^{r}}+\frac{1}{c^{r}}$ for all of positive real $r > 1$.

It is equivalent to
$\frac{\sqrt[r]{a^{r}+b^{r}+c^{r}}}{a+b+c}\ge \frac{\sqrt[r]{(ab)^{r}+(bc)^{r}+(ac)^{r}}}{ab+bc+ac}$

Let $(a, b, c) = v$ and $(b, c, a) = w$. It is
$\frac{\|v\|_{r}}{\|v\|_{1}}\ge \frac{\langle v, w\rangle_{r}}{\langle v, w \rangle_{1}}$

In other words, Cauchy-Schwartz introduces more error when the powers are higher. I conjecture that if $1 \le q \le r$, we can replace 1 by $q$ in the denominators and it's still true. I suspect this should be obvious to me, but some dude hit me really hard in the head today at frisbee, and I have a terrible headache now and can't think.


In triangle ABC, AX=XY=YB=BC, and the measure  of angle ABC=120 degrees. What is the measure of angle BAC?

Hide your answers!

Billy

[hide]We can make angle YBC = x, and angle XBY = 120 - x. Since XY = YB, then it is an isoscoles triangle which means that the base angles are equal. BXY must then also be 120 - x. Angle AXY = 180 - (120 - x) = 60 + x. Angle A = [180 - (60 + x)]/2 = (120 - x)/2. Angle C = (180 - x)/2. Angle A + B + C = 180 = (120 - x)/2 + (180 - x)/2 + 120. Solving this makes x = 90. Plugging it into the equation for angle A, we get 15 degrees.[/hide]


A permutation $ \{x_1, \ldots, x_{2n}\}$ of the set $ \{1,2, \ldots, 2n\}$ where $ \mbox{n}$ is a positive integer, is said to have property $ T$ if $ |x_i \minus{} x_{i \plus{} 1}| \equal{} n$ for at least one $ i \in \{1,2, \ldots, 2n \minus{} 1\}$. Show that, for each $ \mbox{n}$, there are more permutations with property T than without.


If n=1(mod 2),n>1 prove it is impossible that  m^(n-1)=-1 (mod n)

Just note that if the order of $ m$ modulo $ n$ is $ d$ then $ d$ divides $ 2(n\minus{}1)$ and $ d$ not divides $ n\minus{}1$ (because $ n\minus{}1$ is even. Then $ 2(n\minus{}1)$ divides $ \phi{n}$, which is impossible because $ 2(n\minus{}1)>n>\phi{n}$ fo $ n>1$

[quote="gilcu3"]Just note that if the order of $ m$ modulo $ n$ is $ d$ then $ d$ divides $ 2(n \minus{} 1)$ and $ d$ not divides $ n \minus{} 1$ (because $ n \minus{} 1$ is even. Then $ 2(n \minus{} 1)$ divides $ \phi{n}$, which is impossible because $ 2(n \minus{} 1) > n > \phi{n}$ fo $ n > 1$[/quote]
but why $ 2(n \minus{} 1)$ divides $ \phi{n}$ ? i don't understand :blush:  :blush:

It is the good solution.
Let $ 2^c$ the maximum power of $ 2$ that divides $ n\minus{}1$. Note that like $ d$ divides $ 2(n \minus{} 1)$ and $ d$ not divides $ n \minus{} 1$ then $ d\equal{}2^{c\plus{}1}q$ for some $ q$ odd.
Then for each prime $ p$ (not $ 2$ because $ n$ is odd) that divides $ n$ the order $ s$ of $ m$ modulo $ p$ is of the form $ s\equal{}2^{c\plus{}1}r$ for some $ r$ odd. And then like $ s|p\minus{}1$. $ p\equal{}2^{c\plus{}1}v$ for some $ v$ odd.
Then $ n$ is of the form $ 2^{c\plus{}1}j\plus{}1$ and this contradict the choice of $ c$

[quote="gilcu3"]It is the good solution.
Then for each prime $ p$ (not $ 2$ because $ n$ is odd) that divides $ n$ the order $ s$ of $ m$ modulo $ p$ is of the form $ s \equal{} 2^{c \plus{} 1}r$ for some $ r$ odd.
[/quote]

I didn't understand that either. Why does it hold for every prime divisor of n?

For each prime $ p$ that divides $ n$
$ m^{n\minus{}1} \equal{} \minus{}1 (mod p)$
Then if $ s$ is the order, $ s$ divides $ 2(n\minus{}1)$ and $ s$ don't divides $ n\minus{}1$.


Here's a go at reducing it to something solvable:
multiplying through by e and taking the natural logarithm of both sides, and letting b = (k + 1)a, k > 0, and simplifying, we get
1 + ln((k+2)a/2) >? ln(a) + (k + 1)ln(k)/k
1 + ln(a) + ln((k + 2)/2) >? ln(a) + ln(k) + ln(k)/k
1 + ln((k + 2)/2) >? ln(k) + ln(k)/k

We might perhaps reduce this to two cases: where 1 - ln(2) > ln(k)/k (which is for large k, and since we know ln(k + 2) > ln(k), we will be done) and alternatively the alternate case (in which case k is small and so ln(k + 2) - ln(k) is sufficiently large to overcome the differences in the other terms.)  I'm not sure about this, particularly whether or not it will work.

Here is the inequality in a better form:

:sqrt:((1/3)(a^2+ab+b^2)) > (a+b)/2 > (1/e)(b^b/a^a)^(1/(b-a)) > (b-a)/ln(b/a)


Can you guys help me with this problem?  Thanks in advance. :lol:

 A rectangular fish tank has a base 2 feet wide and 3 feet long. When the tank is partially filled with water, a solid cube with an edge length of 1 foot is placed in the tank. If no overflow of water from the tank is assumed, by how many inches will the level of the water in the tank rise when the cube becomes completely submerged?

A. 1/6

B. 1/2

C. 2

D. 3

E. 4

Well when the cube goes in, assuming that it's a paper cube it wouldn't be able to sink below the surface so the answer is 0.  :lol:
Nah just joking, this is how you do it.
The cube has a volume of 1 cubic foot.
The fish tank has a base of 6 square feet.
That means that in order to get 1 cubic feet of water in the tank, you need an rectangular prism with a base of 6 and a height of $ \frac{1}{6}$
Therefore, the answer is $ \frac{1}{6}$.

However, the problem asked for inches, so it would be 1/6(12)=2 inches.

This should have been in the AMC forum. It came from the AMC 10 and 12.

@Z-coli and Zhangpeijin: thnx for helping me solve it  :)

@ gauss: really? it was a homework problem for me.


[quote="Wumbate"]A deck of cards is shuffled.  Then you draw from the top until you draw an ace.  What is the average amount of cards you draw until the ace comes up?[/quote]
[hide]..I suppose it's not 4 is it..because expected value is the same for all numbers, so...52/13=4[/hide]

I think the average cards drawn before an ace is between the 6th and 7th card (52/4/2=6.5)

[hide]I think the answer is $\frac{53}{5}$[/hide]

number of cards left, percentage that it's an ace, number of aces out of 100 attempts
52 7.6% 7
51 7.8% 13
50 8.0% 19
49 8.1% 25
48 8.3% 30
47 8.5% 35
46 8.6% 40

Edit: This is probably wrong, I forgot to take into account that the more aces you find the lower the chances of aces in the future become.  The more I look at this problem the more I agree with the 10th card.  If you only have 13 cards (all of one suit) then you will find the ace on average around 6th or 7th draw.  If you increase the base set to 52 cards (four of each suit) then the chances probably decrease but not very much.

If "average amount" means expected value, then SplashD is correct.  More generally,

[hide]if you shuffle a deck of $m$ cards, $n$ of which are aces, the first ace is on average the $\frac{m+1}{n+1}$th card from the top of the deck.

Here's one way of showing that:

How many ways can an ace be first?  Well, there are $m-1 \choose n-1$ ways to choose the places for the other aces.  What if the first ace is second?  There are $m-2 \choose n-1$ ways to arrange the other aces.  What if the first ace is $m-n+1$st (that is, $n$th from the end)?  Well, there are only $n-1 \choose n-1$ ways to arrange the other $n-1$ aces (they have to be the last $n-1$ cards).

So, the expected value is just $E = \frac{1}{{\binom m n}}\sum_{i = 1}^{m-n+1}i \cdot{m-i \choose n-1}$.  How to evaluate this?  Well, one way is peel this sum apart, the same way one can peel a sum like $r+2r^{2}+3r^{3}+4r^{4}+\ldots$ into many geometric sums.  Equivalently, we can re-write the sum as $\sum_{i = 1}^{m-n+1}i \cdot{m-i \choose n-1}= \sum_{i = 1}^{m-n+1}(m+1) \cdot{m-i \choose n-1}-\sum_{i = 1}^{m-n+1}(m+1-i) \cdot{m-i \choose n-1}= (m+1)\sum_{i = 1}^{m-n+1}{m-i \choose n-1}-n\sum_{i = 1}^{m-n+1}{m-i+1\choose n}$.

From here, you apply the hockey-stick identity and everything works out nicely and you get the answer that I gave.

[/hide]

I got [hide]$\frac{53}{5}$[/hide] with many combinations and summations like JBL mentioned. I concur with SplashD   :wink:

A slight generalization:  if there are $m$ cards, $n$ of which are aces ($1 \leq n \leq m$), what is the expected value of the position of the $k$th ace ($1 \leq k \leq n$) in the deck?

That formula gives the answer in the case $k = 1$, which I gave in my first post.  The question I just asked is more general.

oh sorry. i didn't see that.

i found that vendermonte's identity comes into the mix but i'm not sure of how to evaluate the sum since you have to worry about multiplying respective probabilities by the corresponding number of cards pulled.

[hide]There are $n$ aces and $m-n$ non-aces.
Each of the $m-n$ non-aces can be inserted either above the top ace, below the bottom ace, or between any two consecutive aces.
So, there are $n+1$ space to insert a non-ace card.
So, the average number of non-ace cards in each space is $\frac{m-n}{n+1}$.
The $k$th ace has $k$ spaces above it.
So, on average, there are $\frac{k(m-n)}{n+1}$ non-ace cards above the $k$th ace.
There are always $k$ aces that are above the $k$th ace or are the $k$th ace.
So, the expected value of the position of the $k$th ace is $\frac{k(m-n)}{n+1}+k=\boxed{\frac{k(m+1)}{n+1}}$[/hide]
1st post in Intermediate Topics


Okay. We need to get our act together for DMM. (which happens to be on Halloween this year... coincidence?) Let's not get our butts kicked by TJ again and let's make sure Hamster has a miserable time if he goes.

Well, not really, but you know what I mean.

to North Carolina, looks like. can't let those Virginians or Georgians or South Carolinians win.

btw good luck making Hamster feel "miserable"

Besides Enloe, what NC schools are sending teams?

Edit: what NC schools are sending teams that have a legit shot at being top ten?

[quote="Hamster1800"]Woohoo, I'm glad I left an impression on North Carolina. But you know guys, maybe you should just accept defeat. Knowing when to gg is a good skill to have. So is making carriers. GLHF

Just kidding. TJ's team isn't as good this year as it was last year.[/quote]

bm is also a good skill to have

also [img]http://www.ghetto-overlord.com/Blog/Pictures/ICCUP/idracup.JPG[/img] YEAH

but dude we'll see

[quote="Hamster1800"]Woohoo, I'm glad I left an impression on North Carolina. But you know guys, maybe you should just accept defeat. Knowing when to gg is a good skill to have. So is making carriers. GLHF

Just kidding. TJ's team isn't as good this year as it was last year.[/quote]

we pwned you peoples xP i wonder if this is a beginning of a new era, hmmm

Darn... Our team lost to AAST B by 0.5 points... It was our first time at Duke, though.
Individual finals was so fun-although slightly intimidating.

I spent most of the time in power explaining the orbit-stabilizer theorem and Burnside's Lemma to my teammates.

[quote="33286"]How do you do number 10?[/quote]

hmm... $ \left(a,b,c\right)\implies\left(\frac{a\plus{}b\plus{}c}{3},b,c\right)\implies\left(\frac{a\plus{}b\plus{}c}{3},\frac{a\plus{}4b\plus{}4c}{9},c\right)\implies\left(\frac{a\plus{}b\plus{}c}{3},\frac{a\plus{}4b\plus{}4c}{9},\frac{4a\plus{}7b\plus{}16c}{27}\right)\equal{}\left(a_1,b_1,c_1\right)$. It's easy to show that $ a\plus{}2b\plus{}3c\equal{}a_1\plus{}2b_1\plus{}3c_1$, and so we have found an invariant.

[quote="bpgbcg"]Darn... Our team lost to AAST B by 0.5 points[/quote]

that was just too close. -phew-

when they said that we got 5th by a sliver, mu A thought that they got edged out XD

[quote="Hamster1800"]Woohoo, I'm glad I left an impression on North Carolina. But you know guys, maybe you should just accept defeat. Knowing when to gg is a good skill to have. So is making carriers. GLHF

Just kidding. TJ's team isn't as good this year as it was last year.[/quote]

Oops i didn't accept defeat.

P.S. Your cousin goes to my school.

Dude, i just want to make it clear that the answer for #10 is $ \frac{40}{6}$, not $ \frac{20}{3}.$

Also, I failed mandelbrot, so i should be in good shape for PuMaC :)

So does anyone have the full results?  My team failed.  But we did beat some teams  :P

I guessed 48/7 ((1a+2b+4c)/7) for 10. I guess I wasn't far off...

I'd like to see full results as well.

[quote="PI-Dimension"][quote="not_trig"][quote="Hamster1800"]Darn I lost to the Chinese Taylor. That was pretty sad. Also the first two tiebreakers weren't that bad....although I forgot the (2,q,q) cases at first.[/quote]

darn I lost to myself

that was even sadder

and the first tiebreak was pretty easy except I kinda went an extra step
and the second I was pretty dumb or something
and the third I was screwing around and got something like x+y+z=8 1/x+1/y+1/z = -3/4 but then I was like oh noes I probably want another 3var guy so I am not dead or something

also darn we lost to TJ B? what? darn [though someone on TJ B got #10 hmm][/quote]

WHAT FAIL #3 was just newton sums[/quote]

yes i was aware but i don't know that crap


I heard you guys talking about it earlier, how can I get involved?

Wait until February. I'll probably send you something or get Lucian to. Basically, there's going to be a test at a school in your area. If you do well (even with only MathCounts-level knowledge you'd be fine) you make the team. The actual meet is in late May/early June; see [url]http://www.arml.com[/url].


limit
logarithms
integration
real analysis
real analysis unsolved


logarithms
limit
number theory unsolved
number theory

for each number $ N$, suppose that Bob chooses a random number $ n$ in the interval $ \frac{1}{2}\leq n\leq \frac{3}{2}N$. If he repeats this process many times, prove that approximately $ \frac{1}{\ln{N}}$ of his numbers will be prime.

In other words, given

$ P(N)\equal{}($prob that an integer $ n$ in the interval $ \frac{1}{2}\leq n\leq \frac{3}{2}N$ is a prime number)

then prove that

$ \lim_{N\to\infty}\frac{P(N)}{1/\ln{N}}\equal{}1$

More generally, given

$ P(c_1,c_2,N)\equal{}($Prob that an integer $ n$ in the interval $ c_1N\leq N\leq c_2N$ is a prime number)

Give the correct term for $ X$ for the following equation:

$ \lim_{N\to\infty}\frac{P(c_1,c_2,N)}{X}\equal{}1.$

Am I wrong, or is it the same as
http://www.mathlinks.ro/Forum/viewtopic.php?t=170399

Pierre.

so does anyone know what should go in $ X$?

Same answer : $ \frac 1 {ln(N)}$, and same way as in the given link.

Pierre.


I got the same answer as SplashD, but used a different method [hide]there is a 5/11 probability that you will pick a ten first, which guarantees success.  if you pick a one, two, or five first, which has a 6/11 chance , you will only win if you then pick a ten, which has a probability of 5/10=1/2, so if you pick anything but a ten first, you have a 6/11*1/2=3/11 chance of succeeding.  So, the answer is 5/11+3/11=[color=red]8/11[/color][/hide]

[quote="bpms"][quote="SplashD"][quote="usaha"]In a envelope, there are 2 one dollar bills, 2 two dollar bills, 2 five dollar bills, and 5 ten dollar bills.  What is the probability that if i draw any 2 bills, without replacement, that i get a sum of more than 10 dollars.[/quote]

[hide]You can count and get 40 sums that are more than 10. There are a total of $\binom{11}{2}=55$ ways to choose 2 bills. So the answer is $\frac{40}{55}=\boxed{\frac{8}{11}}$[/hide][/quote]
[hide]You are aware that you have to pick a 10 aren't you?[/hide][/quote]

Umm... yeah. Why? Is my answer wrong or something?

[quote="SplashD"][/quote][quote="bpms"][quote="SplashD"][quote="usaha"]In a envelope, there are 2 one dollar bills, 2 two dollar bills, 2 five dollar bills, and 5 ten dollar bills.  What is the probability that if i draw any 2 bills, without replacement, that i get a sum of more than 10 dollars.[/quote]

[hide]You can count and get 40 sums that are more than 10. There are a total of $\binom{11}{2}=55$ ways to choose 2 bills. So the answer is $\frac{40}{55}=\boxed{\frac{8}{11}}$[/hide][/quote]
[hide]You are aware that you have to pick a 10 aren't you?[/hide][/quote][quote="SplashD"]

Umm... yeah. Why? Is my answer wrong or something?[/quote]
No your answer is right. (I think)


Does someone know when it will be released??? I know it's from mit to late march, but does someone know the specific day?

Uh, it says mid to late March on the MIT admissions website.  So that's the best you're going to get.  It depends on how quickly they can get everything done...

Didn't they get it out pi day last year  (3/14) ? I figure it's probably around the same time this year

it's usually around pi day, if not exactly pi day. last year was a couple days late due to some weird shipping error they had with the physical letters themselves.

Good luck to everyone who applied (myself included)!


number theory
least common multiple
induction
inequalities
number theory proposed

[quote="nhat"]Consider $k$ positive integers $a_{1},a_{2},...,a_{k}$ satisfying $1 \leq a_{1}<a_{2}<...<a_{k}\leq n$ and $[a_{i},a_{j}] \leq n$ for all $1 \leq i \leq j \leq k$.Prove that:$k \leq 2[\sqrt{n}]$[/quote]
What is mean $[a_{i},a_{j}]$?

I think it means the LCM of $a_{i}$ and $a_{j}$.

By descending induction we prove that $a_{i}\leq \frac n{k+1-i}$ :

This is clear for $i=k$. Assume it holds for $i>1$. Then :
$a_{i}-a_{i-1}\geq \gcd (a_{i}, a_{i-1}) = \frac{a_{i}a_{i-1}}{[a_{i}, a_{i-1}]}\geq \frac{a_{i}a_{i-1}}n$.
Thus $a_{i-1}\leq \frac n{1+\frac n{a_{i}}}\leq \frac n{k+1-(i-1)}$, as desired (the last inequality follows from the induction hypothesis).

From the inequality, it easily follows that if $i \leq k-[ \sqrt n ]$ then $a_{i}\leq [\sqrt n ]$. Hence, since the given sequence is increasing, we have $k-[ \sqrt n ] \leq [ \sqrt n ]$, which leads to the result.

Pierre.


$\frac{\sqrt{2n(n+1)}}{4n+1}$ or rather $\frac{\sqrt{2n(2n+1)}}{4n+1}$ ?

Because $\frac{\sqrt{2n(n+1)}}{4n+1} < 1/2$ which makes the problem obvious.

What's the 'easy way' to see that it is <1/2?

I will use induction:
Firstly we check if it's true for $n=1$
$\frac{\sqrt{6}}{5}<\frac{1}{2}$
It's true.
Now I'll show that if it's true for $n$, then it's true for $n+1$
Assumption :
$\frac{\sqrt{6}}{5}+\frac{\sqrt{20}}{9}+\frac{\sqrt{42}}{13}+..............+\frac{\sqrt{2n(n+1)}}{4n+1}<\frac{n}{2}$
Thesis:
$\frac{\sqrt{6}}{5}+\frac{\sqrt{20}}{9}+\frac{\sqrt{42}}{13}+..............+\frac{\sqrt{2n(n+1)}}{4n+1}+\frac{\sqrt{2(n+1)(n+2)}}{4n+5}<\frac{n+1}{2}$

Now I add $\frac{\sqrt{2(n+1)(n+2)}}{4n+5}$ to the both sides of our thesis, and I get this:

$\frac{\sqrt{6}}{5}+\frac{\sqrt{20}}{9}+\frac{\sqrt{42}}{13}+..............+\frac{\sqrt{2n(n+1)}}{4n+1}+\frac{\sqrt{2(n+1)(n+2)}}{4n+5}<\frac{n}{2}+\frac{\sqrt{2(n+1)(n+2)}}{4n+5}$

Now I'm going to prove, that:After I do so I'll have the thesis

$\frac{n}{2}+\frac{\sqrt{2(n+1)(n+2)}}{4n+5}<\frac{n+1}{2}$
After I do so I'll have the thesis

$2\sqrt{2(n+1)(n+2)}<4n+5$
$(n+1)(n+2)<(4n+5)^2$
$8n^2+16n+9>0$, which is true for all $n$, because $\Delta<0$


I've gotten a fun!

Your inequality is equivalent to
\[5\sum a^6b^2+10\sum a^5b^3+10\sum a^4b^4\geq 6\sum a^6bc+2\sum a^5b^2c+4\sum a^4b^2c^2+8\sum a^4b^3c +10\sum a^3b^3c^2.\]
Applying several times AM-GM we obtain result:
1) $\sum a^6b^2 \geq 2\sum a^6bc$;
2) $\sum a^5b^3 \geq \sum a^5b^2c$;
3) $\sum a^4b^4\geq \sum a^4b^3c$;
4) $\sum a^4b^4\geq 2\sum a^4b^2c^2$;
5) $\sum a^6b^2 \geq \sum a^3b^3c^2$;
6) $\sum a^6b^2 \geq  \sum a^3b^3c^2$.


limit
advanced fields
advanced fields unsolved

Let A be a c*-algebra. I tried to show that given a positive a in A, the closure of the set aAa is the smallest hereditary C*-subalgebra containing a, I was able to show that it contains a^2, but not a... and why is it hereditary?

it's not easy at all :-). the proof can be found in
B. Blackadar, Operator Algebras: Theory of C*-Algebras and von Neumann Algebras, Proposition II.3.4.2 (ii)

Thanks a lot!
I actually got the book and understood the proofe, but now I don't get another thing -
in the same proposition - part (iii) - why these the element h =$ \sum_{n \equal{} 1}^{\infty}{2^{ \minus{} n} x_{n}^{*} x_n}$
make $ B \equal{} \overline{hAh}$?

(I assume it has something to do with the fact that: $ b \in \overline{x^{*}Ax}$ if and only if
$ lim_{\alpha \rightarrow 0} {(xx^{*})^{\alpha}b(xx^{*})^{\alpha}} \equal{} b$, is that correct?)

Take $ \{u_{n}\}$ to be any countable approximate unit in B (there is such an approximate unit as B is separable)
and take $ a=\sum 2^{-n}u_{n}$, so $ a \in B^{+}$ and as $ \overline{aAa}$ is the smallest hereditary subalgebra containning a, $ \overline{aAa} \subseteq B$.
As $ 2^{-n}u_{n} \leq a$, and $ \overline{aAa}$ is hereditary, $ u_{n} \in \overline{aAa}$,
and that is why if $ b \in B$ we get ${ b=\lim{u_{n}bu_{n}}}$ and ${{ u_{n}bu_{n}}} \in \overline{aAa}$, so $ b \in \overline{aAa}$, so fimally $ B = \overline{aAa}$.


Seven students in a class compare their marks in 12 subjects studied and observe that no two of the students have identical marks in all 12 subjects. Prove that we can choose 6 subjects such that any two of the students have different marks in at least one of these subjects.

Let $\{X_i\}_{i=1}^{12}$ be the subjects, and $(a,b,c,d,e,f,g)$ be the students.

If (in example) $(a,b) \in X_1$, then one of $(a,j), (b,j)$ is in $X_1$, for j = c,d,e,f,g.

Our strategy is as follows.  Pick a subject containing (a,b) [there is atleast one].  From our lemma above, we guarantee 5 other unique pairs in that subject.  Now pick a subject containing a pair we havent got to yet.  We guarantee 4 other unique pairs, because only one can repeat (it is easy to show **)  Continuing in this fashion, we guarantee 6+5+4+3+2+1 =  21 unique pairs, and we are done.

** a short proof with handwaving: if in example we used (a,b), then we have also used either (x,a) or (x,b).  So when we use (i, j), we use either (y,i) or (y,j).  So if (x,a) = (y,i) in example, we have a,b,i,j,x fixed, so that when y varies it can only match up with (x,a) at most once.

Clarification -- what do the ordered pairs represent?  And why would "only one of $(a,j),(b,j)$" be in $X_1$ if $(a,b) \in X_1$?


integration
function
calculus
real analysis
real analysis unsolved

let $f: [0,1]\to\mathbb{R}$ be integrable. let $f(x)=0$ everywhere except the cantor set. then  show that $\int_{0}^{1}f(x)dx=0$. Also  my friend said that if $f=0$ everywhere except $\mathbb{Q}\cap [0,1]$ then this is not true. i think it might be something to do with the fact that the characteristic function of $\mathbb{Q}$ not being integrable, but i'm not able to prove both these statements. can someone help? perhaps a hint or something. thanks in advance

yep, riemann integral. i dont know lebesgue  :)

See [b]Theorem 11[/b] [url=http://www.math.cuhk.edu.hk/~mat2060/mat2060b/Notes/notes2.pdf]here[/url].

EDIT: jmerry's approach is indeed better.

ya i know this theorem but how is it going to help me with the problem i posted in post no 1? sorry if i didnt get the entire meaning you intended to convey

so let me state more clearly, i'd like to have hints for the problem in post 1, the $\int_{0}^{1}f(x)dx=0$ thing

For this sort of problem, you're probably best off working with upper and lower sums, which depend only on the partition. The function is integrable if and only if the upper sums can be made arbitrarily close to each other.

For the Cantor set, try a partition with $3^{n}$ parts of equal length (assign the endpoints to the interval with members of the Cantor set). The lower sums are all zero. The upper sums aren't, but the function is zero on most of the pieces. Calculate the upper sums and show they go to zero.

For the characteristic function of $\mathbb{Q}$, the upper sums are all 1 and the lower sums are all zero for any partition.


inequalities
trigonometry
calculus
calculus computations

Prove that $ 2\plus{}2x\sin\frac{1}{x}>\cos\frac{1}{x}\ \forall x\in\mathbb{R}\setminus\{0\}$.  Find the minimum value of $ M\in\mathbb{R}$ for which $ M\plus{}2x\sin\frac{1}{x}>\cos\frac{1}{x}\ \forall x\in\mathbb{R}\setminus\{0\}$.

Can this be shown analytically? The behavior around 0 is highly irregular.

For $ |x| < \frac {1}{2}$ we have $ 2 \plus{} 2x\sin \frac {1}{x} > 1$ and $ \cos \frac {1}{x}\leq 1$.  For $ |x|\geq \frac {1}{2} > \frac {1}{\pi}$ we have $ x\sin \frac {1}{x} > 0$ so the inequality clearly holds.

To find $ M$, let $ x_1 > x_2 > ...$ be the positive points at which $ f(x) \equal{} \cos \frac {1}{x} \minus{} 2x\sin \frac {1}{x}$ obtains local maxima and show that $ f(x_1) > f(x_2) > ...$.  Then $ M \equal{} f(x_1)\approx 1.05498$, where $ x_1\approx 0.16834$ is the second greatest solution of $ f'(x) \equal{} 0\Leftrightarrow \left(1 \plus{} 2x^2\right)\sin \frac {1}{x} \equal{} 2x\cos \frac {1}{x}$. (though strictly speaking, no $ M$ satisfies the condition as stated; the $ >$ should be $ \geq$).


[quote="porte"]Let $n$ be a positive integer,prove (1) to (3)

(1) $1111^{n}-1109^{n}$ can be divided by $2$

(2) $11^{n}-8^{n}-3^{n}$ can be divided by $24$

(3) $2450^{n}-1370^{n}+1150^{n}-250^{n}$ can be divided by $1980$

This is an Japan Medical University entrance exam in [b]1980[/b] . not so difficult :P

[hide="Hint"]
$a,b$ ; positive integer$(a>b)$,then $a^{n}-b^{n}$can be divided by $a-b$ :lol:
[/hide][/quote]

[hide="3"]
$180|(2450^n-1370^n)+(1150^n-250^n)$ and $11|2450^n-250^n)-(1370^n-1150^n)$

[/hide]

thank you for reply , Both are best answer !shadysaysurspammed and shyong :P

[quote="ArcticMonkey"][quote="porte"](1) $1111^{n}-1109^{n}$ can be divided by $2$[/quote]
It's obvious that the difference of two odd numbers is an even number. :P[/quote]

Yeah that's right but I gave that identity beacuse it is helpful for the other questions.

[quote="shadysaysurspammed"]
Now $11^{n-1}-3^{n-1}$ is divisible by $8$

[/quote]

Can anyone prove that $11^n-3^n$ is divisible by $8$?

[quote="smellyfart"]
Can anyone prove that $11^n-3^n$ is divisible by $8$?[/quote]

$11^n-3^n=(11-3)(11^{n-1}+\cdots+3^{n-1})$ :lol:

[quote="porte"]Let $n$ be a positive integer,prove (1) to (3)

(1) $1111^{n}-1109^{n}$ can be divided by $2$

(2) $11^{n}-8^{n}-3^{n}$ can be divided by $24$

(3) $2450^{n}-1370^{n}+1150^{n}-250^{n}$ can be divided by $1980$

This is a Japan Medical University entrance exam in [b]1980[/b] . not so difficult :P

[hide="Hint"]
$a,b$ ; positive integer$(a>b)$,then $a^{n}-b^{n}$can be divided by $a-b$ :lol:
[/hide][/quote]

What does this have to do with medicine?  :huh:

Many Japanese universities offer their own entrance exams in addition to the standarized test. Students usually need to take Japanese, math, English and one more subject (either social studies or science, depending on the major). Although math (and English, Japanese etc.) has nothing to do with medicine, the students still need to take the exams for all these different subjects.


search calculus LaTeX limit trigonometry geometry derivative

How to show $ lim_{x \longrightarrow 0} \frac {sinx}{x} \equal{} 1$ without L'Hospital rule?

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1450063994&t=242582
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1450063994&t=211925

(I found these by searching for the word "circular" in the Calculus & Analysis forum.)

Some [[LaTeX]] suggestions: the code you should have used is \lim_{x \to 0} \frac{\sin x}{x} = 1.

Kent Merryfield's posts [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=229706]here[/url] has some very good general comments, and there's more related discussion of L'Hopital and how you might want to use it or why you might want to avoid it in certain circumstances [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=256605]here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=190611]here[/url].

Actually we can use the seris expansion of  [b]sin x[/b]
[img]http://upload.wikimedia.org/math/2/9/c/29cb648e96291f52707c7225630c1a17.png[/img]
and then we can divide by x.as x tends to 0 we see that substituting o in the series gives us the ans as 1

Unless that series is the definition you're using, that's as circular as anything else.

There is a proof of the limit using the Squeeze theorem. Shouldn't be too hard to find ;)

the one using the squeez theorem is quite easy . its an elegant proof using geometry .

it given in all ncert textbooks followed in india here is the link for the proof..

[url]http://ncert.nic.in/textbooks/testing/Index.htm[/url]  jus go here and in the top select the class as X1 and subject mathematics,a pdf opens and the go to chapter limits and derivatives


calculus
integration
geometry
3D geometry
parameterization
analytic geometry
calculus computations

Crunch time finals next week...oy. I got this and not sure on second step.
4) Evaluate the surface integral $\int\int_{\sigma}(y^2+z^2)dS$ where $\sigma$ is the portion of the cone $x=\sqrt{3(y^2+z^2)}$ for $0\leq x\leq3$.
My book says that a special form of the surface integral can be simplified if you use x=x y=y and z=g(x,y).
I used this with: $y=y$ $z=z$ $x=\sqrt{3(y^2+z^2)}$
So I could use the formula: $\int\int_{D}f(y,z,g(y,z))\sqrt{\frac{\partial x}{\partial y}^2+\frac{\partial x}{\partial z}^2+1}\,dA\Rightarrow$
$\frac{\partial x}{\partial y}=\frac{\sqrt{3}y}{\sqrt{y^2+z^2}}$ and $\frac{\partial x}{\partial z}= \frac{\sqrt{3}z}{\sqrt{y^2+z^2}}$
Plug in the variables and you get: $2\int\int_{D}y^2+z^2\,dA$
My problem is I am not sure how to find the area D.

The set $D$ is the projection of your cone onto the $yz$ plane, and your $dA$ is $dy\,dz.$ You've made the minor adjustments needed to turn a standard formula for a surface given by $z=h(x,y)$ into formulas that apply to a surface given by $x=h(y,z).$ Since you're using $y$ and $z$ as your parameters, you need to find out what values $(y,z)$ can take on.

$x=\sqrt{3(y^2+z^2)}\le3$ so $\sqrt{y^2+z^2}\le\frac3{\sqrt{3}}=\sqrt{3}$ or $y^2+z^2\le3.$ Your set $D$ is a circular disk of radius $\sqrt{3}.$

Your integral is $2\int\int_{y^2+z^2\le3}y^2+z^2\,dy\,dz.$

Of course, now you're going to use polar coordinates. OK, it's not quite standard looking polar coordinates, since this is $y$ and $z,$ but given what you did at the beginning of the problem, you should be able to see what I mean.

Not quite sure how I could keep this as x=x y=x and z=g(x,y)....

Maybe x=z y=y z=g(x,y) ???

or is it legitimate to just continue by making it:
$2\int_{0}^{2\pi}\int_{0}^{3}r\cdot r\,dr\,d\theta$

Just don't worry about the names of the coordinates - work by analogy.

And it is, as you say, "legtimate to just continue by making it ..." only please get it right:

$2\int_0^{2\pi}\int_0^{\sqrt{3}}r^2\cdot r\,dr\,d\theta$

doh!  :blush: wish I could say that was a typo.............just typing without thinking...oy.


function
real analysis
real analysis unsolved

Prove that if A is a Jordan region and f and g are integrable functions on A, then fg is also integrable on A.

Riemann integrable, of course. Since $fg=\frac14((f+g)^{2}-(f-g)^{2})$, it suffices to prove that $f^{2}$ is integrable whenever $f$ is. Being Riemann integrable, $f$ is bounded: $|f|\le M$, say. Therefore, $|f(x)^{2}-f(x')^{2}|\le 2M|f(x)-f(x')|$ for any $x,x'\in A$. This implies, after some routine work, that Riemann sums for $f^{2}$ have a limit.


man ke hichi nafahmidam.
va khodaavand latex aafarid... :chief:

hads mizanam ke masalat ghashange vali hichish maloom nis

kheili tablooye ke az ye jaei copy kardi. ye jaei mesle kava ,ye keshvari ,ye sali ,soale 3 jabr(fekrkonam shortlist)

fekr konam soorate doroste soal ine

$n>2\ ,\ x_1,x_2,\dots,x_n$ are real numbers such that$2\le x_i\le 3$.show that:

$\frac{x_1^2+x_2^2-x_3^2}{x_1+x_2-x_3}+\frac{x_2^2+x_3^2-x_4^2}{x_2+x_3-x_4}+\dots+\frac{x_{n-1}^2+x_n^2-x_1^2}{x_{n-1}+x_n-x_1}+\frac{x_n^2+x_1^2-x_2^2}{x_n+x_1-x_2}\le 2(x_1+x_2 + \dots+x_n)-2n$

hahahahah   belakhare fahmidam soalo az koja cop zadi   :trampoline: :clap:

inham ye link:
http://www.kalva.demon.co.uk/short/sh95.html

khob bashe ghanool mikonam
manam rahehalesho cop mizanam

[quote="amir2"]akhe adam ingadr tablo cop mizane[/quote]
baba hala mage chiye... khaaste ye maseleye ghashang matrah kone... :lol:

[quote="amir2"]The ith term on the lhs is (xi2 + xi+12 - xi+22)/(xi + xi+1 - xi+2) (with cyclic subscripts). We can write this as (xi + xi+1 + xi+2) - 2xixi+1/(xi + xi+1 - xi+2). Now (xi - 2)(xi+1 - 2) ≥ 0, so -2xixi+1 ≤ -4(xi + xi+1 - 2) = -4( (xi + xi+1 - xi+2) + (xi+2 - 2) ). Hence the ith term ≤ (xi + xi+1 + xi+2) - 4(1 + (xi+2 - 2)/(xi + xi+1 - xi+2) ). But the constraints on xi imply that (xi + xi+1 - xi+2) and (xi+2 - 2) are positive and (xi + xi+1 - xi+2) ≤ 4. So the ith term <= (xi + xi+1 + xi+2) - 4(1 + (xi+2 - 2)/4) = xi + xi+1 - 2. Hence lhs ≤ 2s - 2n.[/quote]
let me guess....kalva? :D

ramtin az babate del garmit mamnoon vali man natoonestam hal konam va be hamin dalil matrah karmam albate soal male hamon jaee ke amir goft bood

lot fan ye rahe hale ghashang bedid


summer program
Mathcamp
geometry
MATHCOUNTS
AMC
USA(J)MO
USAMO

Is mathcamp more about contest math or about learning difficult math?

If all you want to do is do contest math all day, go to AwesomeMath.

Mathcamp will give you a deep understanding in topics you've never heard of. The classes are in areas you might not see again until your graduate years (though, because of the 1-week timeframe, they're often very specific).

Of course, we certainly do do contest math at Mathcamp, but it's not the main focus (though you can make it your main focus if you really want). Every week there are problem-solving classes (usually two -- one "intro class," which last year was taught by a different visitor every week and ranged in difficulty from Mathcounts to USAMO with them), and an advanced problem class). Every Wednesday there's Team Problem Solving, where teams of about 8 people (the same team every week -- scores are cumulative) are given 2 hours to collaborate on a USAMO-style test. Finally, every Saturday there are Relays, where teams work on two questions at a time to compete for a bag of candy. Relays can get quite wacky -- questions usually just ask you to find some number like an ARML individual round, but other times they might ask your team to stand with 3 feet total on the ground, or play a game blindfolded. The last Relay included a competition to make up the best rap on a mathematical topic for bonus points (Laura Zehender has a video of this on Facebook). One Relay was done in the swimming pool area, where a team member had to swim across to turn in an answer.

The video mentioned above is here: http://www.facebook.com/video/video.php?v=73908337943 :)

I really would say that in general, the answer to your question is "about learning difficult math." (Or I might replace it with "abstract" or "advanced.")

then i think I will definately attend because i don't like contest math very much and am not very good at it  :)
will people look down on me if im not good at it? (right now i cud prb get 7 on the AIME)
As for the video no offense but i could freestyle ALOT better than that, i'd post link to some vids but real names scare me  :P  im PROMYS this summer but maybe ill do mathcamp next summer if I get in

Last year I got a 4 on the AIME, and I did not feel "looked down on" at all.

I believe some people last year did not make AIME. I know at least one of this years' campers did not make AIME. And no, we don't really care. It's not something that comes up much.

Personally, if someone can't get at least a 12 or 13 on the AIME, I don't find it worth trying to have an intelligent conversation with them.  Math ability is a good indicator of intelligence; and besides, the best mathematicians are usually the ones who do contests in high school.

12 or 13 on the AIME? Why aim so low!? I personally refuse to talk to anyone in real life who isn't either a member of my immediate family or a perfect scorer on the IMO.

[hide]To be honest, when I first saw benzi's post the only reason I took it for sarcasm was because it was kinda outlandish; hence, I made it even worse to emphasize that we are, in fact, being sarcastic. And then I made this hidden message because I realize that sarcasm can be hard to detect on the internet sometimes <_< (and actually in real life, apparently. Only my group of friends can recognize when I'm being sarcastic; random people I talk to think I'm being serious usually)[/hide]

I sure hope so :P .

(And I'm one of those people who's pretty oblivious to sarcasm. Once I actually thought my friend was deaf because of that...heh :roll: . but I was able to catch on with Zeph's post, at least.)

Lol.  Thank you for your help, Zeph.
[quote="solafidefarms"]I too am curious to know if Benzi's post was sarcastic.[/quote]
It was.

[quote="perfect628"]Hmm is associating AIME score with math ability also sarcastic? :P[/quote]
Yes.  So there were a few points of sarcasm.  First of all, I cannot get a 12 or 13 on the AIME.  Second, math ability is often no indicator of intelligence, and AIME score is often no indicator of math ability.  Third, mathematicians are often not the ones who do contests in high school.

[quote="LadyKn1ght"](And I'm one of those people who's pretty oblivious to sarcasm. [...])[/quote]
Lol.  I'm going to need to work on that.

Yeah, after I posted that message I got to thinking that maybe I'd trolled too hard.  But hey, it's all in good fun; and besides, a lot of the time I'm the guy who pushes the limit on that anyway. :lol: :spam: :rotfl:

[quote="benzi455"]  Third, mathematicians are often not the ones who do contests in high school. [quote]
Is that really true? *cough Terence Tao
while I kno wjust b/c one mathematician did well on contests that is not the rule i am interested to know if most top mathematicians did math contests in HS

I think he's saying that while there is a correlation between contest skill and higher-level math success, it's not as prominent as some people make it out to be. I'd say it also depends on the contest; for instance, the ARML Power Round is probably a better indicator than Mathcounts Countdown.


If a little car with initial velocity $v_{0}= 2 \sqrt{gR}$ rides through a loop with radius $R$, at which height does the car leave the track. You can neglect friction.

Hint: Using the Newton's second law and the law of conservation of energy to find the answer.

[hide]
Newton's second law:

$\large N+mg\cos\theta=\frac{mv^{2}}{R}$

$\Longrightarrow N=\frac{mv^{2}}{R}-mg\cos\theta$

When the car leaves the track

$\large N=0 \Longrightarrow \frac{mv^{2}}{R}=g\cos\theta_{0}$

$\large \cos\theta_{0}=\frac{h-R}{R}$ where $h$ is the height we need.

The law of conservation of energy:

$\large \frac{mv_{0}^{2}}{2}=mgh+\frac{mv^{2}}{2}$

With $\large v_{0}=2\sqrt{gR}$ we have $\boxed{h=\frac{5R}{3}}$

[/hide]

Thanks alot.

That was more or less what I found. A pitty that I forgot the cosine on my exam :)

[quote="phucnv87"]

Newton's second law:

$\large N+mg\cos\theta=\frac{mv^{2}}{R}$

$\Longrightarrow N=\frac{mv^{2}}{R}-mg\cos\theta$

[/quote]

This is not quite correct.

The correct expression is:

$ma_{R}-m(\frac{v}{R})^{2}R+mg(1-\cos\theta) = 0$

$\Rightarrow a_{R}= \frac{v^{2}}{R}-g(1-\cos\theta)$

The car leaves the track if $a_{R}< 0$

$\Rightarrow \frac{v^{2}}{R}-g(1-\cos\theta) < 0$

$\Rightarrow \frac{v^{2}}{R}< g(1-\cos\theta)$

$\Rightarrow \frac{v^{2}}{gR}-1 <-\cos\theta$

$\Rightarrow 1-\frac{v^{2}}{gR}> \cos\theta$

From the conservation of energy:

$\frac12 mv_{0}^{2}= \frac12 mv^{2}+mgh$

$v^{2}= v_{0}^{2}-2gh$

$\Rightarrow 1-\frac{v_{0}^{2}-2gh}{gR}> \cos\theta$

$\Rightarrow 1-\frac{4gR-2gh}{gR}> \cos\theta$

$\Rightarrow \frac{-3R+2h}{R}> \cos\theta$

$\Rightarrow \frac{-3R+2h}{R}> \cos\theta \ge-1$

$\Rightarrow-3R+2h >-R$

$\Rightarrow 2h > 2R$

$\therefore$ If $h > R$ then the car leaves the track.


Evaluate: $ \frac{2}{7} \cdot \frac{5}{9} \plus{} \frac{5}{7} \cdot \frac{5}{9}$. Express your answer as a common fraction.

Instead of multiplying it out, we see that we can factor out $ \frac{5}{9}$.  This gives us $ \frac{5}{9}(\frac{2}{7}\plus{}\frac{5}{7})\equal{}\frac{5}{9}(1)\equal{}\boxed{\frac{5}{9}}$

hello, we get $ \frac{10}{63}\plus{}\frac{25}{63}\equal{}\frac{35}{63}\equal{}\frac{5}{9}$
Sonnhard.

[quote="jxl28"]Instead of multiplying it out, we see that we can factor out $ \frac {5}{9}$.  This gives us $ \frac {5}{9}(\frac {2}{7} + \frac {5}{7}) = \frac {5}{9}(1) = \boxed{\frac {5}{9}}$[/quote]

really random $ \text{\LaTeX}$ hint. Use [code]$\left(\right)$[/code] to make the parentheses bigger and actually cover the fractions.


Some Mock AIME questions are sufficiently poorly worded that I would like to re-word them.  I obviously am not going to go around doing this for "real" contests, but is it okay to do this for the Mock AIME?

Yeah, go ahead. I don't see why anyone would have problems with that.  :lol:

I guess it would be ok if the original wording is unavailable.

[quote="Boy Soprano II"]On a similar note, would it be acceptable to enter paraphrased USAMO questions from before 1989 (since I don't think we have the original wording from those years), with a disclaimer that the question is paraphrased?[/quote]Unavailable?  :huh: I don't think you should post the problems if you don't have the original version.

I have 1972-1986. If I have time I'll try to post some, although I'm going to be very busy this year.


A plane has a speed of "p" km/hr. when flying in still air. The plane flies from New York City to Washington D.C., and then back to New York again, with and against a wind of speed "w" km/hr. find the average speed of the plane.

To solve this, we must use [url=http://www.artofproblemsolving.com/Wiki/index.php/Harmonic_mean]harmonic mean[/url].
\[ \dfrac{2}{\frac {1}{p \plus{} w} \plus{} \frac {1}{p \minus{} w}} \equal{} \frac {p^2 \minus{} w^2}{p}
\]
Thus, our answer is $ \frac {p^2 \minus{} w^2}{p}$ km/hr.

You should hide your solution so as to give others the opportunity to solve it. It's $ \frac {p^2 \minus{} w^2}{p}$, not \frac {p^2 - w^2}{2p}. The 2's in the numerator and denominator cancel out.

You should've used the "*" symbol if you already knew the answer. Then, you are supposed to hide your solution as it is of no use to anyone else. I assumed you didn't know the answer which is why I didn't hide my solution.

Wait,is it against regulations to put problems you made on the Middle School section. In that case, I should stop.

I don't think it's against the regulations... :ninja:


How to type the dollar signs in the forums (not in LaTeX)?

It would be nice if you would read the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=51330]announcements[/url] in this forum and then ask questions (if there are any left :P).


Let $n$ be a positive integer. An [i]a-stump[/i] of $n$ is the integer gained by removing one or more digits from the right hand side of the base $a$ representation of $n$. Let $T_{a}(n)$ be the sum of all the [i]a-stumps[/i] of $n$. If $S_{a}(n)$ is the sum of the digits in the base $a$ representation of $n$, prove that $n=S_{a}(n)+(a-1)T_{a}(n)$

Let the digits on $n$ in base $a$ be $n_{k}, n_{k-1},\cdots,n_{2},n_{1},n_{0}$. Then $n=\sum_{i=0}^{k}a^{i}n_{i}$.
We say that the jth a-stump of $n$ is the one which removes the last $j$ digits. Then the jth a-stump is $\sum_{i=j}^{k}a^{i-j}n_{i}$. So we have
\[T_{a}(n)=\sum_{j=1}^{k}\sum_{i=j}^{k}a^{i-j}n_{i}=\sum_{i=1}^{k}n_{i}\sum_{j=1}^{i}a^{i-j}=\sum_{i=1}^{k}n_{i}\left(\frac{a^{i}-1}{a-1}\right).\]
So finally,
\[(a-1)T_{a}(n)=\sum_{i=1}^{k}\left(a^{i}-1\right)n_{i}=\sum_{i=0}^{k}a^{i}n_{i}-\sum_{i=0}^{k}n_{i}=n-S_{a}(n).\]

exactly. it can be solved with just plain manipulations. so isn't the prob too easy? i mean, for APMO...and in the exam it was to be proven only for base 10.


Say you have a huge speech due on Monday (wow reminds me of something that happened last year). You just finished writing it on Sunday and now you have to memorize it. How would you go about doing this?

I really don't know what I would do. Since palm cards don't help me much. I guess I will just try and memorize as much as possible (after all I am the one who wrote it), get an early night sleep and hope for the best. :P

If I wasn't winging it (which is what I normally do) I would almost certainly read it line by line and repeat it until I was confident on that section. Then I'd move onto the next section (although still saying the first section out loud again)

However, if you are pushed for time. Learn the flow of the argument and talk naturally following the flow.

Huh. Well...that's a problem.
Just go with the flow, remember what your main points and statistics/facts and go for it.
That's what I usually do, if it's not complete improv. I manage to fool the teachers...:D

Another thing I tend to do is use humour so I remember it better. :D

Hmm, in reading and study skills (yes, it's a class) we discussed ways to memorize things. Apparently, the best way is to start with what it begins with, create a weird/interesting (so you remember it) picture in your head of the next thing, do it again for the next, and keep doing that for all the major parts.

Also, if you just say it over and over, it gets recorded in your first level memory, if someone keeps saying it to you, it gets recorded in your second level memory, if you write it down in addition to the others, it gets recorded in your third level memory, and I forgot what to do to keep it in your fourth level memory :roll:


The problem is equivalent to determining $n$ such that
\[ 5^2 \cdot 7 | 6^n - 1  \]
Observe that
\[ 6^n - 1 \equiv ( - 1 )^n - 1 ( \mbox{mod } 7 )  \]
So $n$ has to be even. Moreover
\[ 6^n \equiv 1 ( \mbox{mod } 25 ) \Longleftrightarrow 5| n  \]
Therefore $n$ must be a multiple of $10$.


function
algebra
domain
complex analysis
topology
complex analysis unsolved

Hi all,

Below are two problems in which one is asked to prove the constancy of a given function, here we go:

(1) Suppose the function f is analytic on and inside a simple closed curve C, f(z) is never zero for all z on and inside C, and |f(z)| is constant on C. Prove that f is a constant function.

(2) Suppose that f(z) is an entire function such that |f(z)| = 1 for all |z| = 1, and f(z) is never 0 inside the open unit disc centred at the origin. Prove that f is a constant function.

I was thinking if it was possible to come up with functions (or composite of functions) to show that in fact f is bounded on the whole complex plane and then conclude by Liouville. But I haven't managed to solve these problems so far. It's not hard to see that the two problems are in fact very similar.  Another problem I encountered and solved is as follows, though I'm not sure if the solution is correct:

(3) Suppose that g is an entire function and Im g(z) + Re g(z) > 1 for all z in the complex plane. Prove that g is a constant function.

My solution runs as follows: Im g + Re g >  1 implies that g maps the complex plane to the half plane above the line y = -x + 1, which we shall denote by U. As U is a simply connected domain, we can find an analytic isomorphism from U to the open unit disc centred at the origin, say h.

Thus h(g(C)) is contained in h(U), the open unit disc. Observe that h(g(z)) (viewed here as a composite function) is entire and bounded. We conclude by using Liouville's Theorem that g is in fact constant throughout the complex plane.

Can anyone help me with these 3 problems?

(2) is a special case of (1). Let's consider (1). If $ |f(z)| \equal{} K$ on $ C$, then $ |f(z)|\le K$ inside $ C$, by the maximum principle. But $ 1/f$ is analytic too, so $ |1/f(z)|\le 1/K$ inside $ C$. Thus $ |f(z)| \equal{} K$ everywhere inside $ C$. However, the image of an open set under a nonconstant analytic function is open. (Either that, or use the form of maximum principle with an equality case.)

Hi,

  I got the first part too, but I do not quite get your last part of your argument. Could you please clarify this a little?


Let G be a finite group and let I ={g in G: g^2 = e} \ {e} be its subset of involutions. Show that G is abelian if card(I) => (3/4)card(G).

A solution is given at http://mathforum.org/library/drmath/view/66879.html when you take T to be the identity automorphism.

This is a theorem of G.A.Miller from around 1906.

G. A. Miller. "Note on the Possible Number of Operators of Order 2 in a Group of Order 2^m"
The Annals of Mathematics, Second Series, Vol. 7, No. 2 (Jan., 1906), pp. 55-60
http://www.ams.org/mathscinet-getitem?mr=1502317
http://dx.doi.org/10.2307/1967420

Miller's proof is:  Such a G is generated by its involutions.  Suppose by way of contradiction that H is not abelian.  Then G must have a non-central involution, t.  Consider H=C_G(t). If g an element of order 2 in G-H, then gt cannot have order 2, since <a,b:aa=bb=(ab)^2=1> is abelian.  Hence for every element of order 2 in G-H, there is an element of order higher than 2 in G-H.  Since G-H contains at least |G|/2 elements, at least 1/4 of the elements of |G| are of order larger than 2, but this is a contradiction.

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=27046]this thread[/url].


give  0<x,y,z < \pi /2    and  x+y+z= \pi /4 ,tg x=1/a, tg y=1/b, tg z=1/c
   a,b,c is integer ,0<a,b,c  to find a,b,c

1 = tg(x+y+z) = (tg x + tg y + tg z - tg x tg y tg z)/(1 - tg x tg y - tg x tg z - tg y tg z) = (1/a + 1/b + 1/c - 1/abc)/(1 - 1/ab - 1/ac - 1/bc) = (bc+ac+ab-1)/(abc - c - b - a)

bc+ac+ab-1 = abc-c-b-a

abc-bc-ac-ab-a-b-c+1 = 0

The only positive integer solutions of this (found by probing, i found no other way to this diophantic equation) are (2,5,8), (2,4,13), (3,3,7) and their permutations.


function
inequalities
inequalities unsolved

let $ a;b;c > 0$ and $ a \plus{} b \plus{} c \equal{} 3$; prove that:
$ (3a \plus{} \sqrt {9a^2 \plus{} 16})(3b \plus{} \sqrt {9b^2 \plus{} 16})(3c \plus{} \sqrt {9c^2 \plus{} 16})\leq 512 \equal{} 8^3$

let $ 3a\equal{}x,3b\equal{}y$ and $ 3c\equal{}z$.
consider the function $ f$ defined for every positive real number $ t$ by $ f(x) \equal{}ln(t \plus{} \sqrt {t^2 \plus{} 16})$ we have $ f''(x)\equal{}\minus{}1/2(x^2\plus{}16)^{\minus{}\frac{3}{2}}2x<0$ so $ f$ is concave and by jensen's inequality $ f(x)\plus{}f(y)\plus{}f(z) \le 3f(3)\equal{}3ln(8).$

[quote="tchebytchev"]let $ 3a \equal{} x,3b \equal{} y$ and $ 3c \equal{} z$.
consider the function $ f$ defined for every positive real number $ t$ by $ f(x) \equal{} ln(t \plus{} \sqrt {t^2 \plus{} 16})$ we have $ f''(x) \equal{} \minus{} 1/2(x^2 \plus{} 16)^{ \minus{} \frac {3}{2}}2x < 0$ so $ f$ is concave and by jensen's inequality $ f(x) \plus{} f(y) \plus{} f(z) \le 3f(3) \equal{} 3ln(8).$[/quote]

I thinks we should have a solution by Cauchy or Bunhiacopxki; ...

[quote="nguoivn"]I think if the stronger is trues, it'll be easier
$ \sum\ (3a \plus{} \sqrt {9a^2 \plus{} 16})(3b \plus{} \sqrt {9b^2 \plus{} 16}) \leq 675$
Is it trues ?  :wink:[/quote]
I think 192 not 675 because of a=b=c=1
can you write your solution?
thank you

Sorry, I posted wrong. 192, not 675.
I don't have solution for it. I try to expand and uue Am-Gm but I don't obtain any result.
However, there are many ways to use Am-Gm which I haven't to try. So, I think somebody 'll find a solution for it  :)


geometry
3D geometry
tetrahedron
geometric transformation
dilation

Let $ \mathcal Q$ be a unit cube. We say that a tetrahedron is [b]good[/b] if all its edges are equal and all of its vertices lie on the boundary of $ \mathcal Q$. Find all possible volumes of good tetrahedra.

This has nothing to do with combinatorics at all.  Also, it's very easy: any such tetrahedron has side length at most $ \sqrt{2}$, by comparison of circumradii.  There is such a tetrahedron, $ T$: take 4 vertices of the cube such that no two are adjacent.  Then we can get tetrahedra of any smaller volume by a dilation of $ T$ centered at any of its vertices.


USAMTS
Ross Mathematics Program
induction
quadratics
AMC
AIME
ratio

What were your round 2 scores?

Mine were 455[b]55[/b].  I lost a point for #1 because I didn't explicitly say why $ n\leq 7$ doesn't work.

Got 25 again, but no commended this time.

553-5

Got a "good execution" for number 1 but no commended.

Is the following a valid lemma for 3?

"Two 4-row remarkable triangular arrays with the same top row are identical."

5 5 5 5 [b]5[/b]

My solution to number one was a pure program. lol

5 5 5 5 4

had a nice grader for #5, I should try to get commends...

total: 24+24=48

I could get gold if I try hard enough...

5/5/5/5/3

my friend got a 4 for #5, and it turns out we made the same mistake...=(

25+23=48. Still in contention for gold =D

woohoo

5 5 [b]5 5[/b] 5

didn't take off for typos on #5 :)

[quote="lingomaniac88"]I got 21 on the first round, so I have 45 now.  I'm hoping that I can pull off two 25s in the last two rounds in order to make gold.[/quote]

The cutoff for gold is 96.

[quote="worthawholebean"]"Is the following a valid lemma for 3?

Two 4-row remarkable triangular arrays with the same top row are identical."[/quote]

Well yeah, isn't this trivial?

[b]5[/b]5[b]55[/b]4

Wow... 6 commends .. (guess who I am!)  :lol:

[b]5[/b] 5 [b]5[/b] 5 [b]5[/b] !

Wow...I didn't expect so many (to me) commends, especially for no.1...it's sheerly bruteforce...

[quote="CatalystOfNostalgia"][quote="worthawholebean"]"Is the following a valid lemma for 3?

Two 4-row remarkable triangular arrays with the same top row are identical."[/quote]

Well yeah, isn't this trivial?[/quote]
Grader said it was invalid.

[b]5[/b] 5 5 [b]5[/b] [b]5[/b]

Now I have to look at the third round...

5 5 5 [b]5 5[/b]

Amazingly, my four page solutions both got commended, which I thought was kind of strange.

[quote="buzzer11"]I don't know, I could be wrong as well. I don't know what ordered/unordered counting is, either.[/quote]

Considering the talk about complex sums that are hard to calculate, it looks as if your proof would qualify as ordered counting.

[quote="mountain.dew"]Does anyone know the difference between unordered counting and ordered counting?  :huh:[/quote]
azjps's solution, and case 2 of buzzer11's solution are ordered counting. The nature of ordered counting is that the exact position of each point matters. Each point is indexed with a number that describes its position, and the number of triangles is expressed as a nested summation that sums over the indicies.

Case 1 of buzzer11's solution is unordered counting. The nature of unordered counting is to look at all the points involved in forming a triangle and see how many other triangles use the same set of points in different roles. Then the problem of counting triangles reduces to the problem of counting sets of points on the arcs.

[quote="buzzer11"]Here's what I did: Diagram is attached.[/quote]
I can stop talking about about the rubric and how a typical grader would grade buzzer11's solution. It looks familiar, so I think I graded that one myself. (Note: I always initial my comments "EJS", so people can tell what I graded.) That problem was so close to earning four points that I started to expand the triple summation at the end so that I cound spot the computational error and take only one point off. But due to the quadratic terms in the summation, the expansion was getting quite messy and I realized that duplicating the calculations would take at least 15 minutes. I could not afford 15 minutes. So I stuck to the rubric and took two points off.

Erin Schram
USAMTS grader

Yes, you did grade my solution ("EJS" was written in the comment box).
I think I understand unordered vs. ordered counting now.
And oh well about the three points. :) I'll just try and do well enough this round to make the AIME cutoff.

You could just take AMC to make AIME. I think 100 in AMC 12 or 120 on AMC 10 is a lot easier than 68 on USAMTS.

Actually that depends.  Considering the time allotment for USAMTS, everyone has enough time to do serious research if necessary, and some people just work better if there's a lot of time.  Also, it's good too have a safety net just in case, say, you get the flu on both AMC test dates. (Highly unlikely, just a hypothetical example)  The cutoffs of both competitions are pretty equivalent.  Someone who can get a 100 on the AMC 12 should be able to get 68 on USAMTS.

I tend to think that scoring a 100 on the AMC 12 is much easier than scoring a 68 on the USAMTS. After all, about 10000 people a year make it through the AMC 12, versus a couple hundred (at most) for the USAMTS. Naturally that statistic is skewed due to the difference in the total number of total participants, but still.. USAMTS requires effort that many students wouldn't be willing to give during their free time but the AMCs only take up an hour and half of time.

Yes, I agree that the AMC 12 is easier. However, considering that I screwed up badly on last year's AMC 12 and missed the cutoff by a few points, I want to have a back up way into the AIME.

[quote="azjps"]After all, about 10000 people a year make it through the AMC 12, versus a couple hundred (at most) for the USAMTS. Naturally that statistic is skewed due to the difference in the total number of total participants[/quote]

Consider the ratios of students making it from both competitions.  Assuming a low estimate of 100 people from USAMTS as you stated, and considering about 500 USAMTS participants, we've got around 20% of people making it to AIME.  However, from the AMC 12, it's only supposed to be the top 5% or so, so from this statistic, a much higher percentage of people make it from USAMTS.  I still think both are comparable when considering both difficulty and time allotment.

There's a really big difference though between the average person who takes the AMC 12 and the average person who does the USAMTS.

Hmm, looks like we're not talking about Round 2 anymore...

You can't really compare percent of qualifiers from the two tests. The people that do AMC are from all around the nation and probably aren't as bright as those who do USAMTS.

And while we're at it...What happens if you take the AMC 10 and AMC 12?

[quote="vishalarul"][quote="le00327146"]And while we're at it...What happens if you take the AMC 10 and AMC 12?[/quote]

It's possible to qualify through both ways. But you can only take one of each (like 10A and 12B). Obviously, you need to earn a higher score on the 10 compared to the 12 in order to qualify for AIME.[/quote]

They take your better score right?

They'll take whichever score makes you qualify more easily into the upper rounds, since you can't exactly find a correlation between an AMC 10 and AMC 12 score.

5,3,3,3,2

i'm so horrible at this contest, but it's my first year doing this -.-;

Don't worry, you're not horrible. It's just that all the people who post on AOPS are the ones who get all the high scores. There's a bunch of people with lower scores who don't bother with (or don't know about) the forum.


$a,b,c>0,\ abc=1\Longrightarrow 27\le \overline {\underline {\left|\ 3\sum a\cdot \sum \frac 1a\le 21+\sum \left(a^3+\frac{1}{a^3}\right)\ \right| }}.$


Other than old AIME problems and mock AIME's, what are some good practice problems around the level of AIME problems 7-12? any suggestions for books or other problems?

thanks. what are you guys doing to prep for AIME?

AoPS v2 if you haven't done it already is great practice for AIME.

The Intermediate/Pre-Olympiad forums have plenty of problems all the time to do, too.

The AIME Problem Series (and Seminar in the spring I think) give good practice, but you might encounter problems you have already solved (not sure about this).

... and I'm sure there are plenty of other things to practice with that I can't think of right now.

[quote="probability1.01"]From what I've seen, ARML problems are good too.  But trust me, just doing the old AIME problems especially with recent years having I and II, you will have lots of practice, especially if you work on most of them as if you were taking the actual test.[/quote]

Agree. Emphasis on "as if you were taking the actual test".

What about Thomas and other people's awesome Mock AIME?!!

Also, USAMTS types are very good because they usually require thinking. Some problems may not be hard as later AIME problems but again, if they were to be solved in 3 hours, who knows? (assuming the calculation works out nice way in integers)

There are many contests out there in different nations as well. You can even use them to prepare for AIME. Well, at least that's what I would do except I'm mainly concentrating on scoring high on AMC-12 now.  :lol:

[quote="Silverfalcon"]Also, USAMTS types are very good because they usually require thinking. Some problems may not be hard as later AIME problems but again, if they were to be solved in 3 hours, who knows? (assuming the calculation works out nice way in integers)[/quote]

I disagree. USAMTS problems are usually much harder than AIME problems. If you just want to do the actual problem, not the proof, then maybe you could use some USAMTS. But in general, I don't think USAMTS are very good for AIME practice. Just stick to AIME. There are so many of them that you can do one every week and still not run out by the time you take the real AIME. (If you start now.)


algebra
polynomial
function
continued fraction
algebra proposed

The question is to find the value of this continued fraction:

$1+\frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{5+...}}}}$

It goes on forever in this arithmetic sequence.

I need someone's help to find an exact value. I suspect it's either the solution of an irreducible polynomial of degree 3 or greater, or transcendental.

I was able to get the following approximation:
$\frac{7489051}{5225670}$

Wait, according to the site:
[url]http://mathworld.wolfram.com/ContinuedFractionConstant.html[/url]
this is equal to:
$\frac{I_0(2)}{I_1(2)}$.

Although I don't understand modified Bessel functions (can someone explain?), and mathematica refuses to simplify this result, I suppose this may be as exact an answer as I'll get. Too bad.

OK, looks like that is equal to:

$\frac{\sum_{k=0}^\infty \frac{1}{(k!)^2}}{\sum_{k=0}^\infty \frac{1}{(k)!(k+1)!}}$

Of course, this is not evaluable in exact form. I'll just leave it at that... Ugh.


Let $ P$ be a convex polygon with sides of integer lenght and odd perimeter. Prove that the area of $ P$ is greater than or equal to $ \frac{\sqrt{3}}{4}$.

The $ P$ of minimal area is the equilateral triangle of side 1, perimeter 3 and area $ {{\sqrt 3} \over 4}$.
É isso ?


floor function
combinatorics proposed
combinatorics

There are several scientists collaborating in Niichavo. During an $ 8$-hour working day, the scientists went to cafeteria, possibly several times.It is known that for every two scientist, the total time in which exactly one of them was in cafeteria is at least $ x$ hours ($ x>4$).
What is the largest possible number of scientist that could work in Niichavo that day,in terms of $ x$?

I guess the answer is $ \left\lfloor\frac{16}x\right\rfloor$.

Assume that $ \frac{16}3<x\le8$. We will prove that there are at most 2 scientists. Assume that there are 3 scientists. Draw a Venn diagram as below.

Note that $ a\plus{}f\plus{}b\plus{}e>\frac{16}3,a\plus{}d\plus{}c\plus{}e>\frac{16}3,b\plus{}d\plus{}c\plus{}f>\frac{16}3$. Sum up the three equations to get $ 2(a\plus{}b\plus{}c\plus{}d\plus{}e\plus{}f)>16$, or $ a\plus{}b\plus{}c\plus{}d\plus{}e\plus{}f>8$. On the other hand, we must have $ a\plus{}b\plus{}c\plus{}d\plus{}e\plus{}f\plus{}g\le8$, contradiction. Therefore, there are only 2 scientists at most. But $ \left\lfloor\frac{16}x\right\rfloor\equal{}2$ for $ \frac{16}3<x\le8$. So our answer is true for $ \frac{16}3<x\le8$.

I'm too lazy to check the case $ 4<x\le\frac{16}3$, but I'm sure it can be solved analogously.

I think the answer is wrong because for example for $x=8$ we can have 6
man1{0-4}
man2{4-8}
man3{8-12}
man4{12-16}
man5{0-12}
man6{6-18}

What does man6{6-18} mean?  He is there from the 6th hour to the 18th hour?  But the day is only 8 hours long!

the scientists have 18 hours free time I divide this 18 hours in to {0-18}
and for example {0-4} means that man 1 from hours 0 until being in cafeteria

Please re-read the question and my remark: the day is only 8 hours long, not 18!

Most likely sahadian assumed that they go to the cafe when they are not working (aka in the 24-8=16 hours of the day), although I don't know how sahadian has a 26-hour day.

I will post my solution since jgnr's seems to be pretty far from complete.

Instead of finding the maximum number of scientists for a given time, I will find the maximum time for a given number of scientists. Suppose there are $n$ scientists. First we break down the $8$ hour work day into intervals based on when scientists join and leave (so in each interval the set of scientists stays constant). Then we add up the times for each possible set of $n$ scientists. Let $t_{i,j}$ where $j>i$ represent the time for which exactly one of the scientists $i$ or $j$ is in the cafeteria. Let $T=\sum t_{i,j}$. Now for each interval we assign a weight based on its length of time. For every interval with $k$ scientists and weight $w$, this interval contributes $wn(n-k)$ to the sum $T$. For example, suppose there is the set $\{1,2\}$ consisting of $2$ elements and taking time $w$. Then this contributes $w$ to each of $t_{1,j}$ where $j\ne 2$ and $t_{i,2}$ where $i\ne 1$ so it is $w(2)(n-2)$ overall. Since we are trying to maximize the time for the given number of scientists, we are really trying to maximize $\min(t_{i,j})$. Therefore, the way to maximize is to first maximize the sum $T$, then make all of the $t_{i,j}$ equal so the minimum is maximized too. However to maximize $T$, since the total time has to be $8$ hours and an interval with $k$ scientists contributes $wk(n-k)$, the best we can do is maximize $k(n-k)$ by setting $k= \left\lfloor \frac {n}{2}\right\rfloor$ and using no other intervals, so essentially being greedy. Then the sum will be $8\left\lfloor \frac {n}{2}\right \rfloor(n-\left\lfloor \frac {n}{2} \right\rfloor)$. So $\min(t_{i,j})=\frac {16\left\lfloor \frac {n}{2}\right \rfloor(n-\left\lfloor \frac {n}{2} \right\rfloor)}{(n)(n-1)}$ which is the maximum number of hours for a given number of scientists. The construction is simply to divide the $8$ hours into $n\choose {\left\lfloor \frac {n}{2} \right\rfloor}$ equal intervals and do all possible sets of $\left\lfloor \frac {n}{2} \right\rfloor$, this guarantees that the $t_{i,j}$ will all be equal.

Now the original problem asks for $n$ in terms of $x$. For  $\frac{8k+8}{2k+1}<x\le \frac{8k}{2k-1}$ and $k\ge 1$ The maximum number of scientists is $2k$, which is the same as the first interval given by jgnr. Note in addition that as $k$ increases $x$ approaches $4$ which is why $x>4$ condition is given.


calculus
modular arithmetic
number theory unsolved
number theory

[quote="epsilon07"]1^1+2^2+3^3+4^4+....+n^n = ? mod ( n )

find the solution for n=1000....

Is there any general case for this problem ?[/quote]

I mean n=1000

Can anybody solve this? I need the solution...

[quote="epsilon07"]1^1+2^2+3^3+4^4+....+n^n = ? mod ( n )

find the solution for n=1000....

Is there any general case for this problem ?[/quote]

I dont think there is a general method.

Direct calculus gives the result : $ 700$

You can improve the calculus speed writing $ (100n\plus{}a)^{100n\plus{}a}\equal{}a^aa^{100n}(100n\plus{}1)\pmod{1000}$

[quote="pco"][quote="epsilon07"]1^1+2^2+3^3+4^4+....+n^n = ? mod ( n )

find the solution for n=1000....

Is there any general case for this problem ?[/quote]

I dont think there is a general method.

Direct calculus gives the result : $ 700$

You can improve the calculus speed writing $ (100n \plus{} a)^{100n \plus{} a} \equal{} a^aa^{100n}(100n \plus{} 1)\pmod{1000}$[/quote]

ı got it, thanks ... is there any other solution?


calculus
integration
geometry
function
derivative
analytic geometry
real analysis

Let $f$ be a continuous function on the plane with real values such that the integral of $f$ over each square of area 1 in the plane is 0. What can we say about $f$?

Let $g(x,y)=\int_x^{x+1}\int_y^{y+1}f(u,v)\, dv\, du$.
By the Fundamental Theorem, $g_{yx}(x,y)=f(x+1,y+1)-f(x,y+1)-f(x+1,y)+f(x,y)$. Since we know that $g=0$, its derivatives are zero.

Let $p(x,y)$ be any continuous function on $[0,1]^2$ with integral 0 and $\int_0^1 p(0,y)-p(1,y)\, dy=0=\int_0^1 p(x,0)-p(x,1)\, dx.$
Let $h(x,y)$ be any continuous function on $[0,1]^2$ with $\int_0^1 h(x,y)\, dy=0$ for all $x$, and $h(0,a)=h(1,a)=p(1,a)-p(0,a), h(b,0)=h(b,1)$ for all $a,b.$
Let $v(x,y)$ be any continuous function on $[0,1]^2$ with $\int_0^1 v(x,y)\, dx=0$ for all $y$, and $v(0,a)=v(1,a), v(b,0)=v(b,1)=p(b,1)-p(b,0)$ for all $a,b.$

Now, for $x,y\in [0,1)$, let $f(x+m,y+n)=p(x,y)+mh(x,y)+nv(x,y)$. This $f$ satisfies the conditions, and any $f$ that satisfies the conditions can be uniquely expressed in this form.

This still allows for some rather strange examples.

Hmmm,... My understanding was that the sides of square are not necessarily parallel to the coordinate axes, which makes the problem completely different...

I considered an analogous problem some years ago (with disks instead of squares), but I didn't find a final answer. My progress with the disk was the following: the assumption means that the convolution of $f$ with the characteristic function of a disk is zero, and so its Fourier transform is zero. But the Fourier transform of a convolution is just a product and so the Fourier transform of $f$ must be zero (you need to prove the the Fourier transform of the characteristic function is non-zero). The problem is that some summability assumptions on $f$ are needed in order to compute the transform. I would also appreciate a more elementary approach.

There is a beautiful (but not really elementary) theorem that says that any shift-invariant closed subspace $E$ of $C(\mathbb R)$ is just the closure of all exponential polynomials contained in $E$. Unfortunately, I do not remember whether the result holds in $\mathbb R^2$.  :blush:  :(


Prove that the sides $a,b,c$ of a triangle satisfythe inequality

$5(ab^2+bc^2+ca^2)\ge 3(a^2b+b^2c+c^2a)+6abc$

Suppose $a\ge b,c$. Rewrite the inequality as\[3(a-b)(a-c)(c-b)+2(\sum ab(b-c))\ge0\]
If $c\ge b$, then we are done.
Assume $a\ge b\ge c$. Let a=c+x+y,b=c+x,$x,y\ge0$. Since a,b,c are sidelengths of a triangle, $c\ge y$.
Now substitue this into the inequality, we obtain \[(c+x)(c+x+y)x+cy(x+y)\ge3xy(x+y)\]This is true since $(c+y)(c+x+y)x+cy(x+y)\ge(x+y)(y+x+y)x+y^2(x+y)=(x+y)(x^2+y^2+2xy)\ge4xy(x+y)$.

P.S. Probably there are some computational errors in the proof.


[color=darkblue][size=150]Let equation: $ (x\plus{}1).lnx\minus{}x.ln(x\plus{}1)\equal{}0$.

Prove that this equation have only one root?[/size][/color]

I can prove it ^^ .
$ x \ge 0$
$ f(x) \equal{} (x \plus{} 1)lnx \minus{} xln(x \plus{} 1)$
$ f'(x) \equal{} ln(\frac {x}{x \plus{} 1} ) \plus{} \frac {x \plus{} 1}{x} \minus{} \frac {x}{x \plus{} 1}$
Setting :$ t \equal{} \frac {x}{x \plus{} 1} < 1$
$ g(x) \equal{} lnt \plus{} \frac {1}{t} \minus{} t$
$ g'(x) \equal{} \frac {1}{t} \minus{} \frac {1}{t^2} \minus{} 1 \le 0 \forall t \in R$
=>$ g(x) \ge g(1) \equal{} 0$
=>$ f'(x) \ge 0$
=>...
But we have :$ f(...) \equal{} ... > 0$
Done ;)
[hide="@thay_Quy"] Đơn điệu viết sao hả thầy ?[/hide]

Thank [b]Mitdac123[/b]. You are very good stupid boy.

[hide="For only mitdac123"] Đơn điệu tiếng Anh là: Monotone [/hide]

[quote="thanhnam2902"]Thank [b]Mitdac123[/b]. You are very good [b]stupid[/b] boy.

[hide="For only mitdac123"] Đơn điệu tiếng Anh là: Monotone [/hide][/quote]Is this the kind of language allowed in AoPS? Very degrading...

EDIT: Nevermind, he called himself that way too  :what?:


function
logarithms
calculus
calculus computations

Is there a curve that bounds the nth prime function from the left? For instance, $ P_n \geq 2n\plus{}1$, would be the right bounds.

Are you looking for something like $ 2^n>p_n$?

A consequence of the [url=http://en.wikipedia.org/wiki/Prime_number_theorem]prime number theorem[/url] is that $ p_n\leq \frac {3}{2}n\log_2 n$ for $ n\geq 2$.  Albanian Eagle's assertion also follows from the more elementary [url=http://en.wikipedia.org/wiki/Proof_of_Bertrand%27s_postulate]Bertrand's postulate[/url].