Show that for each $r \geq 1$, there are infinitely many primes $p$ with
$p \equiv 1 \mod 2^{r}$

of course, without Dirichlet's theorem

[quote="Albatross Frankinfueter"]Try divisors of $(2n)^{2^{r}}+1$, they all work.[/quote]

sorry.. I don't know why  :blush:

It is easy. Let $p|(a^{2^{r}}+1)$  then $(a,p)=1$ and minimal period T, suth that $a^{T}=1(mod \ p)$ divisible $2^{r+1}$. Therefore $2^{r+1}|(p-1)\Longrightarrow p=1(mod \ 2^{r+1}).$
It give infinetely many primes this form, because if $n=p_{1},...,p_{k}$ distinct primes this form, prime divisors $n^{2^{r}}-1$ distinct ($n^{2^{r}}-1,n)=1$).


Mr. Jones has eight children of different ages.  On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times.  "Look, daddy!" she exclaims.  "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age."  Which of the following is not the age of one of Mr. Jones's children?

$ \textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

The $LCM$ of $1,2,3,4,5,6,7,8,9$ = $2520$.

From here, we see that we're going to have a multiple of $10$ IF $5$ is one of the ages. Since the last two digits are his age, we can NOT have $xx00$, where $x=1,2,3,4,5,6,7,8,$ or $9$. Therefore, IF 5 is one of the ages, it must be in the form $x0x0$. However, for any $x$, we see that it can factor into $(2)(5)(101)(x)$. We see that we need a multiple of $7,2^3,$ and $3^2$ for $x$, or else that number can not be divisible by all the ages. Since $x$ can not be any $2$ of those $3$ (we have to remove a number because he only has 8 children; therefore, you can take out one of the multiples), we know that 5 is not a possible age, which is $\boxed{B}$.

I completely forgot about the part that said "The last two digits just happens to be my age" when doing this...still came out with the answer though.

[hide]
So the oldest child is $9$, and he has seven other children of different ages. One way or another, this is going to have to be a even number. It also has to be divisible by $9$. Looking at the numbers, we see $5$ requires the last digit to be either $5$ or $0$, and it can't be $5$, as we've already eestablished. So what are the possibilities? Any combination of the digits $9$ and $0$, or $9090$, $9900$, $0909$, or $0990$.

Again, since the oldest child is $9$ and there are seven other children, then out of the ages $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, only one of them is not the age of a child. So if we're assuming $5$ works, then only one out of $1$, $2$, $3$, $4$, $6$, $7$, $8$ doesn't fit. If we try the four cases above, we see neither $7$ nor $8$ work for any of them, thus $5$ cannot possibly be the age of a child.
[/hide]

I must say, Mr. Jones has an awesome oldest child.

What if Mr. Jones is 0 years old?  :D  :rotfl:

[quote=shiningsunnyday][hide="Sorry for the revive, but here's my solution, in fact, an alternative proof that 5 cannot work"]
Using the fact the oldest child is aged $9$, I immediately think of the divisibility rule of $9$, since that really limits the possibilities.

The four-digit number we're dealing with has $2 A$'s and $2 B$'s (where $A$ and $B$ are distinct digits). So $2(A+B)$ must be divisible by $9$, meaning either both $A$ and $B$ are $9$ or $A+B = 9$. The former case cannot be true as they are distinct. So the only possible pairs of $A+B$ are $(1, 8) (2, 7) (3, 6) (4, 5)$.

Note that the four digit number has to be divisible by either/both of $4$ and $8$ (which is $2*4$, so clearly the last two digits must be divisible by $4$ (otherwise the entire number wouldn't be divisible by $4$. Looking at our possibilities the last two digits could only be $88$, $72$, $36$. Since $45$ violates this, $5$ cannot be the age of a child, so it's the answer.[/quote]

There is one case you missed: $(0,9)$. $9900, 9009, 9090$ could work. However, the divisibility by $4$ condition removes all of these except $9900$, which is invalid since the age of the father can't be $00$. So the last two digits can be $88,72,36,44$.

Now, to finish, clearly $5$ divides none of them so $5$ is the answer. But let us continue. $8$ is a factor, so last three digits can be $272,336,544$. This fixes the first digit as well. Note that $7272 = 72 \times 101$ so not divisible by $7$, and $6336 = 7 \times 900 + 36$ so again wrong. This gives us the numberplate to be $5544$ which is indeed a multiple of $7$.


[quote="Ligouras"]Prove that

$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{4(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)} \plus{} \frac {4abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\geq\frac {5}{2}$[/quote]
$ a\plus{}b \ge 2\sqrt{ab}$, etc. gives $ (a\plus{}b)(b\plus{}c)(c\plus{}a) \ge 8abc$
$ (a\plus{}b\minus{}c)(b\plus{}c\minus{}a) \le b^2$, etc. gives $ 4(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b) \le 4abc$
Hence $ 3\frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{16(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)} \ge \frac{3}{2}$ and we need only to prove:
$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{8(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)} \plus{} \frac {8abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\geq 2$
By AM-GM:
$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{8(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)} \plus{} \frac {8abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \ge 2\sqrt{\frac{abc}{(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)}} \ge 2$ and we are done.

[quote="Mathias_DK"][quote="Ligouras"]Prove that

$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{4(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)} \plus{} \frac {4abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\geq\frac {5}{2}$[/quote]
$ a \plus{} b \ge 2\sqrt {ab}$, etc. gives $ (a \plus{} b)(b \plus{} c)(c \plus{} a) \ge 8abc$
$ (a \plus{} b \minus{} c)(b \plus{} c \minus{} a) \le b^2$, etc. gives $ 4(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b) \le 4abc$
Hence $ 3\frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{16(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)} \ge \frac {3}{2}$ and we need only to prove:
$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{8(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)} \plus{} \frac {8abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\geq 2$
By AM-GM:
$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{8(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)} \plus{} \frac {8abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \ge 2\sqrt {\frac {abc}{(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)}} \ge 2$ and we are done.[/quote]

nice......  :wink:   :wink:


When is your state competition?

Oklahoma's is to be held on the 6th, which is next saturday.

Northern California: March 13th

not going though, so i go to different contest, where basically the Mathcount rejects are, so i gotta win that.

Indiana: March 6 [hide]it's too soon...6 more days...too soon...[/hide]

[quote="ml2"]washington april 25th[/quote]

are you serious?  april?!

[url=http://mnspe.org/mathcounts/]Minnesota[/url], March 20.

Can people go to Mathcounts state competitions as spectators, without having a connection to a team or an individual who is in the competition? Minnesota's state competition is a short drive away, and I'd love to see the countdown round.

[quote="ml2"]washington april 25th[/quote]

ml2, where are you from?!
Washington, is like, march 21st, in seattle

It was a joke, i was just kidding, i mean i think u would realize that state can't be [i][b]after[/b][/i] nats... its march 20, seattle, WA, USa,

I know it's perfectly all right in New York.  When I was in fifth grade I went to watch the countdown round, and it was quite exciting and inspiring.  The next year our whole team was going to go watch (despite coming nowhere near qualifying for states), but the weather was nasty and so I believe only I and another 6th grader went.  Next two years I was actually in the competition, but I came back to watch as an alumnus in 9th grade.  I've missed the competiton the last two years because of conflicts with high school math stuff, but I'm coming back this year as a coach!  Yay!  (New York is March 13.)

[quote="tokenadult"]Can people go to Mathcounts state competitions as spectators, without having a connection to a team or an individual who is in the competition? Minnesota's state competition is a short drive away, and I'd love to see the countdown round.[/quote]

GO Danny Zhu! Mathcounts Champion from NEw York last year! I know him.
Golly Gee

I know him, too -- he's a pretty cool little guy :)

Wasn't he only in 7th grade last year?

All i know is that he is the only kid who beat me in this one test...oh welll

lol, Danny is a cute kid...

Danny taking over the entire King sized bed
http://www.tripleintegral.org/pics/DSC00200.JPG

while

1) Mark slept on the chair
http://www.tripleintegral.org/pics/DSC00202.JPG

2) Jack slept in front of the door
http://www.tripleintegral.org/pics/DSC00195.JPG

lol, HMMT, some good times.. darn it Joel, where were you?!

Have fun at State Mathcounts tomorrow.  Hopefully you and I both make nats and have a jolly goode time.

georgia--march 20

Sheesh on the same day as one of the AIME seminars.


function
ceiling function
number theory unsolved
number theory

Let $ r,s$ be positive integers. Let $ t$ be the minimum of ${ 2^k(\lceil{\frac{r}{2^k}}\rceil+\lceil{\frac{s}{2^k}}\rceil-1})$ over all nonnegative integers $ k$.

Prove that $ \binom{t}{j}$ is even for all $ j$ such that $ t-s<j<r$, and prove also that $ t$ is the smallest integer with such property.

[url=http://en.wikipedia.org/wiki/Lucas%27_theorem]Lucas' theorem[/url] will help.


function
induction
algebra solved
algebra

Let f(x) be the function f(x)= Sum(i=1,...n){b(i)*sin(a(i)*x+c(i))}

You derivate 4m times and then : f<4m>(x) = Sum(i=1,...n){b(i)*a(i)4m*sin(a(i)*x+c(i))}

f(x)=0 ===> f<4m>(x)=0 ===> b(n)*sin(a(n)*x+c(n)) = - Sum(i=1,...n-1){b(i)*(a(i)/a(n))4m*sin(a(i)*x+c(i))}  for all m positive integer and all x real

but |a(i)/a(n)| < 1 for all i < n so lim(m-->+\infty ){(a(i)/a(n))4m} = 0  for all i < n, and finally b(n)=0

Now it's trivial to achieve the proof by induction ..


limit
real analysis
real analysis theorems

Please help me. How do  you find a suitable $ N \in Z$  for this question in term of $ \epsilon$. I just need  this example.
Prove by definition that the sequence $ \{\frac{n}{n\plus{}1}\}$ coverges to 1 as $ n$ approaches infinity.

For any $ \epsilon$ pick $ N > \frac{1}{\epsilon}$.

[quote="t0rajir0u"]For any $ \epsilon$ pick $ N > \frac {1}{\epsilon}$.[/quote]

but how do you do that? As I can remember, there is some derivation. For example, by definition, $ n > N$.  So this implies that $ n\plus{}1 > N \plus{} 1$ which implies that $ \frac{1}{n\plus{}1} < \frac{1}{N\plus{}1}$. Now im lost.

How am i going to show that $ \frac{n}{n\plus{}1} < \frac{N}{N\plus{}1}$. And after getting it, I remember that I have to equate it to $ \epsilon$ to get the value of $ N$ in terms of $ \epsilon$. And then how do I proceed knowing that if $ \frac{N}{N\plus{}1} \equal{} \epsilon$ which implies that $ N \equal{} \frac{\epsilon}{1 \minus{} \epsilon}$. But isn't this a fraction? Im really lost. I don't know if i remember it correctly. But that was one technique we were taught.

[quote="guile"]Please help me. How do  you find a suitable $ N \in Z$  for this question in term of $ \epsilon$. I just need  this example.
Prove by definition that the sequence $ \{\frac {n}{n \plus{} 1}\}$ coverges to 1 as $ n$ approaches infinity.[/quote]

We are to prove that $ \boxed{\forall {\epsilon > 0},\ \exists {N > 0}\ s.t.\ n\geq N\Longrightarrow |a_n \minus{} 1| < \epsilon}$.

$ |a_n \minus{} 1| \equal{} \left|\frac {n}{n \plus{} 1} \minus{} 1\right| < \epsilon \Longleftrightarrow \epsilon > \frac {1}{n \plus{} 1}\ \therefore n > \frac {1}{\epsilon} \minus{} 1$.

Thus no matter how $ \epsilon (0 < \epsilon < 1)$ is getting smaller, taking $ N$ s.t. $ N > \frac {1}{\epsilon}\minus{}1$,

we have $ |a_n \minus{} 1| < \epsilon$ for $ n\geq N$. $ \therefore \lim_{n\to\infty} a_n \equal{} 1$.


$ a;b;c>0$
$ abc\equal{}1$
Prove that
$ \frac{ab}{a^2c^2\plus{}b^2c^2}\plus{}\frac{ac}{a^2b^2\plus{}b^2c^2}\plus{}\frac{cb}{a^2c^2\plus{}b^2a^2}\ge \frac{3}{2}$

[quote="tranvanluan"]$ a;b;c > 0$
$ abc \equal{} 1$
Prove that
$ \frac {ab}{a^2c^2 \plus{} b^2c^2} \plus{} \frac {ac}{a^2b^2 \plus{} b^2c^2} \plus{} \frac {cb}{a^2c^2 \plus{} b^2a^2}\ge \frac {3}{2}$[/quote]
Try $ a\equal{}4$ and $ b\equal{}c\equal{}\frac{1}{2}.$ :wink:


floor function
vector
real analysis
real analysis solved

is there countable set $Y$ and uncountable family of its subsets $F$ such that for any two distinct $A,B\in F$ their interesection is finite?

Yes, trivially: consider the uncountable family
\[
F_a := \bigcup_{n \geq 1} \{n^2+\lfloor \alpha n \rfloor\}, a \in (0,1)
\]
of subsets of the positive integers.

--Vesselin

I had a nastier approach, but I'll post it anyway (it's not even complete :)):

Consider all such families of subsets of $\mathbb N$ (it's easier to regard them as characteristic vectors, I think). These families form an inductive set (each totally ordered set of families has the union of the families in the set as an upper bound). This means that the set of such families has a maximal element, and I think we can show that this maximal element cannot be countable by a diagonalization (a bit more elaborate, but not much). It's based on the fact that the intersection of the complements of every finite collection of sets in our maximal collection is infinite.

consider $Y=N$ .To any real number you associate a subset of $Y$.
For example you associate (3,31,314,31415,314159,..) with Pi.

Nice example, alekk! What about examples for the following related

Problem. Does there exist an uncountable collection $F \subset 2^{\mathbb{N}}$ such that for every $A,B \subset F$ we have that either $A \subset B$ or $B \subset A$?

Answer: Yes. Here's one example:
\[
F_{\alpha} := \bigcup_{n \geq 1} \{n^2,n^2+1,\ldots,n^2+\lfloor \alpha n \rfloor\}, \alpha \in (0,1).
\]
How about other examples?  :)

For the original problem, I was going to go with

$F_a=\{(x,y)\in\mathbb{Z}\times\mathbb{Z}\,:\,|y-ax|\le 1\}$ where $a\in\mathbb{R}.$

[quote="grobber"]I think it's been discussed before: $S_{\alpha}=$the set of all rational numbers $>\alpha\in\mathbb R$. They're not subsets of the naturals but we can construct the desired collection of sets of naturals through a bijection from $\mathbb Q$ to $\mathbb N$.[/quote]

 Yes, it is my proposed problem sometime ago. :D
 For the original problem, I think of the following example:

  For each $\alpha \in \mathbb{R}$ we let $F_{\alpha}$ be any sequence of rational numbers which converges to $\alpha$
 [Edit : sorry, mine is similar to alekk, I didn't see :blush: ]


Daca $a, b, c$ sunt numere reale, astfel incat $a+b+c=1,$ atunci \[10|a^{3}+b^{3}+c^{3}-1|\leq9|a^{5}+b^{5}+c^{5}-1|.\] Generalizare.

E cam usurica inegalitatea asta...
Se tine cont de indentitatile \[(a+b+c)^{3}-a^{3}-b^{3}-c^{3}=3(a+b)(b+c)(c+a)\]  \[\textrm{si}\]  \[(a+b+c)^{5}-a^{5}-b^{5}-c^{5}=5(a+b)(b+c)(c+a)(a^{2}+b^{2}+c^{2}+ab+bc+ca)\] Si se ajunge intr-un final ca inegalitatea nostra e echivalenta cu \[a^{2}+b^{2}+c^{2}+ab+bc+ca\geq\frac{2}{3}\] care conform ipotezei $\ a+b+c=1$ devine \[a^{2}+b^{2}+c^{2}\geq\ ab+bc+ca\] qed

O generalizare:

"Daca $a+b+c=m$, atunci:

$10m^{2}|a^{3}+b^{3}+c^{3}-m^{3}|\le9|a^{5}+b^{5}+c^{5}-m^{5}|$"

Ai nevoie de $m^{2}$ in primul membru pt. omogenizarea relatiei.


[hide]
the prime numbers less than ten are:
2, 3, 5, 7

Their sum is 17.

Is there a formula to solve this?[/hide]

[quote="gunes"]?? :( ??[/quote]
do you not understand the question or my answer?

[quote="gunes"]no no i see everything,but you asked for formula and i said that[/quote]

In this case, just add everything up, because the numbers are small. We see that 2+3+5+7=17. No formula needed.

But is there a formula?
What if we have to find the sum of the prime numbers under 100?

[quote="joejia"]What is the sum of all the prime numbers less than 10?[/quote]

[hide="Prime Numbers"]

[hide="Prime Numbers"]

[hide="Prime Numbers"]

The prime numbers under 10 are 2, 3, 5, and 7. The sum of them are 17.

[/hide]

[/hide]

[/hide]


Amy's grandmother gave her 3 identical chocolate chip cookies and 4 identical sugar cookies.  In how many different orders can Amy eat the cookies such that either she eats a chocolate chip cookie first, she eats a chocolate chip cookie last, or both?

There are $ \frac{7!}{3!4!}\equal{}35$ ways to eat.

HOWEVER

S?????S doesn't work

there are $ \frac{5!}{3!2!}\equal{}10$ ways to eat like that, so $ 35\minus{}10\equal{}25$

Er, that needs a lot of clarification :/

As mewto suggests, the way to count the total number of orderings is to count in how many different ways the 3 chocolate chip cookies are placed (the rest are automatically filled in, no decision to be made).  That would be 7C3=35 as stated.

In the next step, instead of counting the number of "successful" patterns, it's easier to count the number of unsuccessful patterns and subtract that from the total, 35.  The only way to have an unsuccessful pattern is to begin and end with sugar cookies: S?????S.

To count this, we only need to count how many ways ????? can be arranged with two sugars and the rest chocolate.  That's where 5C2=10 comes in, counting the number of ways sugars can be placed in those empty slots.

Out of 35 ways, we can't use those 10, so $ \boxed{25}$


Do you mean for the height to be variable, or what?

The surface area would be $2\pi r^{2}+2\pi rh$, if the height is $h$.

[quote="Interval"]A steel drum in the shape of a right circular cylinder is required to have a volume of 100 cubic feet.

(a) Express the amount A of material required to make the drum as a function of the radius r to the cylinder.

(b) How much material is required if the drum is of radius 3 feet?[/quote]

[hide]the volume is $\pi r^{2}\cdot h=100\Rightarrow h=\frac{100}{\pi r^{2}}$

$A=2\pi r^{2}+2\pi\cdot r\cdot h=2\pi r^{2}+2\pi\cdot r\frac{100}{\pi r^{2}}=2\pi r^{2}+\frac{200}{r}$
____________

just plug in 3 for $r$.[/hide]

b-flat, my questions are typed as they are given in the math study book.

jili,

Thanks for helping me.


In how many ways can 1992 be expressed as the sum of one or more consecutive integers?

[hide="Answer"]$1992=2^3\cdot 3\cdot 83$
We divide $1992$ by each of the positive factors of $1992$, and use those numbers as the middle of a consecutive series of positive integers. However, we cannot use even sums since $1992$ is an even integer. By the first restriction, we eliminate all factors of $1992$ with an even factor in them, so we are left with $1$, $3$, $83$, and $249$, so we have $\boxed{4}$ answers.[/hide]

[b]EDIT:[/b] Fixed.

[quote="dangvh"][quote="chess64"][hide]It's the same as the number of odd factors it has, minus 1.

$1992=2^3 \cdot 3 \cdot 83$. The odd factors are $1, 3, 83, \text { and } 3 \cdot 83 = 249$. There are 4 of those, so it can be written as the sum of consecutive integers in 3 ways.[/hide][/quote]

I would say $4$ because when you subtract one from the total number of odd divisors, you're finding the number of ways 1992 can be expressed as [i]two[/i] or more consecutive integers. So, I think that the answer is $\boxed{4}$[/quote]

Oh, I didn't read the question carefully :blush:

[hide] I am sure there are $4$ ways. There is $1992$, the sum of the numbers from $663$ to $665$, $18$ to $65$, and $117$ to $132$. You find these by dividing $1992$ by $1$, $3$, $83$, and $249$. This gives you the middle number of the string of numbers you need to add up to get to $1992$ with $1$, $3$, $83$, or $249$ terms in the string of numbers. When you divide by $83$ and $249$ however, you'll see that you'll have some negative numbers to add up, all of those negative numbers are cancelled out by there positive counterparts. So after all this, you see that the number of ways to express a number in one of more consecutive integers is just the number of odd divisors that number has. In this case, $1992$ has $\boxed{4}$[/hide]

[quote="JesusFreak197"][hide="Answer"]$1992=2^3\cdot 3\cdot 83$
We divide $1992$ by each of the positive factors of $1992$, and use those numbers as the middle of a consecutive series of positive integers. However, we cannot use even sums unless they include $0$, since $1992$ is an even integer. All of these must be positive, and since we will be using as many numbers as the number that $1992$ was divided by (e.g., $\frac{1992}{3}=664$, so $663+664+665=1992$), then we must have $\frac{1992}{x}=n$, and $n>\frac{x+1}{2}$. By the first restriction, we eliminate all factors of $1992$ with an even factor in them, so we are left with $1$, $3$, $83$, and the second restriction eliminates $83$, so we have only $\boxed{2}$ answers.[/hide][/quote]

Umm...

$-17+-16+-15+\cdots+24+\cdots+63+64+65=1992$, and that has $83$ integers in it...

I think it's because you're saying that they must be positive, but nowhere in the problem did it say "positive" ;)

It's definitely [hide]4[/hide].

Chess64, you see that all those negative numbers will be cancelled out right? like, the sum from $-17$ to $17$ will cancel to zero. So you just start from $18$ and sum to $65$, that way, you express them as positive integers  :D

Hmm, I was doing that hurriedly. I see where I went wrong now. ;)

I'm pretty sure the answer is $\boxed{8}$ (if 1992 just by itself is another answer since it says the sum of one or more consecutive integers -- I'm not too sure if just a single number alone satisfies that, but if not, the answer should be 7)

[hide]Assuming the first of these consecutive numbers is $a_1$ when we have $n$ numbers, we have $na_1 + \frac{n(n-1)}{2} = 1992 \Rightarrow n(2a_1 + n - 1) = 3984 = 2^4\cdot3\cdot83$

If $n$ is even, $2a_1 + n - 1$ is odd. Therefore, $n \in \{16, 48, 1328, 3984\}$. Solving for these for $a_1$ gives us $a_1 \in \{117, 18, -662, -1991\}$

If $n$ is odd, $n \in \{1, 3, 83, 249\}$. Solving for $a_1$ for each case of $n$, we find that $a_1 \in \{1992, 663, -17, -116\}$ respectively.

Therefore, we have $\boxed{8}$ different possible ways to do this:

$117 + 118 + ... + 131 + 132 = 1992$
$-116 + -115 + ... + 131 + 132 = 1992$

$18 + 19 + ... + 64 + 65 = 1992$
$-17 + -16 + ... + 64 + 65 = 1992$

$663 + 664 + 665 = 1992$
$-662 + -661 + ... + 663 + 664 + 664 = 1992$

$1992 = 1992$
$-1991 + -1990 + ... + 1991 + 1992 = 1992$

Hopefully, I haven't missed any :D[/hide]

Ohhh... lol, we've all been so stupid. :P

As I'm not sure how other people are getting their answers, I will post mine:
$\frac{b(b+1)}{2}-\frac{a(a+1)}{2} = 1992$
${(b-a)(b+a+1) = 2^4\cdot3\cdot83}$
For all parities of $a$ and $b$, one of either $(b-a)$ or $(b+a+1)$ is odd.
That means that you have $4$ possibilities for the odd term, $(1,3,83,3\cdot83)$.
As the odd term is one of those two terms, the answer is $8$.


I would like to collect competition problems that use double counting technique. Here are some that I know:
IMO'87 #1
IMO'89 #3
IMO'98 #2
IMO'05 #6
APMO'06 #5

Anyone else? You can list the question too.

(IMC 2002) Two hundred students participated in a mathematical contest. They had six
problems to solve. It is known that each problem was correctly solved by at least 120 participants. Prove that there must be two participants such that every problem was solved by at least one of these two students.

(IMO 1998) In a competition, there are a contestants and b judges, where $b \ge 3$ is an odd
integer. Each judge rates each contestant as either “pass” or “fail”. Suppose k is a number such that, for any two judges, their ratings coincide for at most k contestants. Prove that $\dfrac{k}{a} \ge \dfrac{b − 1}{2b}$

(Iran 1999) Suppose that C1, . . . , Cn (n ≥ 2) are circles of radius one in the plane such that
no two of them are tangent, and the subset of the plane formed by the union of these circles is
connected. Let S be the set of points that belong to at least two circles. Show that |S| ≥ n

 [USA TST 2005]
Let $n$ be an integer greater than 1. For a positive integer $m$, let $S_m =
{1, 2, …, mn}$. Suppose that there exists a $2n$-element set $T$ such
that
(a) each element of $T$ is an $m$-element subset of $S_m$
(b) each pair of elements of $T$ shares at most one common
element; and
(c) each element of $S_m$ is contained in exactly two elements of $T$.
Determine the maximum possible value of $m$ in terms of $n$.

[Iran 2010]
There are n points in the plane such that no three of them are collinear. Prove that the number of triangles whose vertices are chosen from these n points and whose area is 1 is not greater than $\dfrac{2(n^2 - n)}{3}$.

 [IMO 1987]
Let n and k be positive integers and let S be a set of n points in the
plane such that:
(i) No three points of S are collinear
(ii) For every point P in S, there are at least k points in S
equidistant from P.
Prove that $k < \dfrac{1}{2} + sqrt{2n}$

 [IMO Shortlist 2004, C1]
There are 10001 students at a university. Some students join
together to form several clubs (a student may belong to different
clubs). Some clubs join together to form several societies (a club
may belong to different societies). There are a total of k societies.
Suppose that the following conditions hold:
(i) Each pair of students is in exactly one club.
(ii) For each student and each society, the student is in exactly one club of the society.
(iii) Each club has an odd number of students. In addition, a club with 2m + 1 students (m is a positive integer) is in exactly m societies.
Find all possible values of k

 [APMO 1989]
Show that a graph with n vertices and k edges has at least $\dfrac{k(4k - n^2)}{3n}$ triangles

 [Indian TST 2001]
Let G be a graph with E edges, n vertices and no 4-cycles. Show
that $E \le \dfrac{n}{4}(1 + sqrt(4n - 3))$

[USAMO 1995]
Suppose that in a certain society, each pair of persons can be classified as either amicable or hostile. We shall say that each member of an amicable pair is a friend of the other, and each
member of a hostile pair is a foe of the other. Suppose that the society has n people and q amicable pairs, and that for every set of three persons, at least one pair is hostile. Prove that there is at least one member of the society whose foes include $q(1- \dfrac{4q}{n^2}) or fewer amicable pairs.

[IMO Shortlist 2010 C5]
$n \ge 4$ players participated in a tennis tournament. Any two players have played exactly one game, and there was no tie game. We call a company of four players bad if one player was defeated by the other three players, and these three players formed a cyclic triple (a set (A, B, C) such that A beat B, B beat C and C beat A). Suppose that there is no bad company in this tournament. Let $w_i$ and $l_i$ be respectively the number of wins and losses of the ith player. Prove that $\dsum_{i=1}^n (w_i - l_i)^3 = 0$


function
real analysis
real analysis theorems

Well, I need to find a generating function for these... where do I even begin?

$a) \text{ }1+3x+6x^{2}+9x^{3}+11x^{4}+11x^{5}+9x^{6}+6x^{7}+3x^{8}+x^{9}$
$b) \text{ }1+2x+4x^{2}+6x^{3}+8x^{4}+9x^{5}+10x^{6}+10x^{7}+9x^{8}+8x^{9}+6x^{10}+4x^{11}+2x^{12}+x^{13}$
$c) \text{ }1+x^{3}+x^{6}+2x^{9}+x^{12}+x^{15}+2x^{18}+...$
$d) \text{ }1+1x+2x^{2}+3x^{3}+5x^{5}+8x^{8}+13x^{13}+21x^{21}+...$
Notice the last one is the Fibonnaci numbers in power series form...
$\sum_{n\ge 0}F_{n}x^{n}$

So how do I find and solve for the generating function?
I guess I need the summation notation and what the summation notion is equal to...
$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$

Someone posted this link in another thread - it will probably be of use to you:

[url]http://www.math.upenn.edu/~wilf/gfology2.pdf[/url]


function
Support
integration
real analysis
real analysis unsolved

Let $ U, V$ be open sets with $ V \subset\subset U$. Show there exists a smooth function $ \zeta$ such that $ \zeta\equiv 1$ on $ V$, $ \zeta \equal{} 0$ near $ \partial U$.

I have an idea, take $ V \subset\subset W \subset\subset U$ and mollify the characteristic function of $ W$(i.e $ \chi_{W}$), but I need for detailed description.

Thanks!

Instead of a single set $ W$ in between, how about two sets? $ V\subset V_1\subset U_1\subset U.$

What I have in mind involves taking the distance from $ V$ to $ \del U$ and cutting that in thirds - but that would need a positive distance.

So start by assuming everything is taking place inside some bounded set. Split the distance in thirds; mollify (convolve with an appropriate smooth bump function of appropriately small support) the characteristic function of, say, $ V_1.$

Then you have to patch some things together with a partition of unity to deal with unbounded cases.

Thanks Kent Meeryfield!
What about this:

We choose $ \varepsilon > 0$ so small that $ |x \minus{} y|\geq 4\varepsilon$ when $ x\in V$ and $ y \in U^c$. Let $ \chi$ be the characteristic function of the set $ V_{2\varepsilon} \equal{} \{y: |x \minus{} y|\leq 2\varepsilon\}$. Then we can find a non-negative function $ \eta\in C^{\infty}_{c}(B)$, where $ B$ is the unit ball, such that $ \int\eta dx \equal{} 1$. Then $ \eta_{\varepsilon}(x) \equal{} \frac {1}{\varepsilon^n}\eta(\frac {x}{\varepsilon})$ has support in the ball $ B_{\varepsilon} \equal{} \{x: |x| < \varepsilon\}$ and $ \int\eta_{\varepsilon}dx \equal{} 1$, so $ \phi \equal{} \chi\ast\eta_{\varepsilon}\in C^{\infty}_{c} (V_{3\varepsilon})$ and $ \varphi \equal{} 1 \minus{} \phi \equal{} (1 \minus{} \chi)\ast\eta_{\varepsilon}$ vanishes in $ V_\varepsilon$. Also $ \phi \equal{} 0$ near $ \partial U$.  :)


inequalities function logarithms induction

Hi

Newbie here. Just looking for a proof of :

(xyz)^(1/3) < (x+y+z)/3< ((x^2+y^2+z^2)/3)^(1/3)

[quote="delta"][quote="Rzeszut"]The $\mathrm{AM-GM}$ inequality, i.e. $\sqrt[3]{xyz}\leqslant \frac{x+y+z}{3}$ is just Jensen inequality for convex function $f(t)=\mathrm{e}^{t}$:
Of course these proofs can be generalised to $n$ variables.
[/quote]
And how do u prove Jensen for 3 variables or $n$ variables? :maybe:[/quote]
I'll restate Jensen inequality:

[color=blue][i]Let $f$ be a convex function defined at interval $[a,b]$, $x_{1},\ldots,x_{n}\in [a,b]$ and $p_{1},\ldots,p_{n}$ be nonnegative reals whose sum is $1$. Then:
$f(p_{1}x_{1}+\ldots+p_{n}x_{n})\leqslant p_{1}f(x_{1})+\ldots+p_{n}f(x_{n})$.
If $f$ is concave, the sign of inequality should be reversed. [/i][/color]

We will prove it only for a convex function, the proof for a concave function is analogous.
Induction. The first step ($n=2$) is just the definition of a convex function. Let's assume that it works for $n$ variable, i.e. for any $p_{1},\ldots,p_{n}\geqslant 0$ whose sum is $1$ we have
$f(p_{1}x_{1}+\ldots+p_{n}x_{n})\leqslant p_{1}f(x_{1})+\ldots+p_{n}f(x_{n})$.
Let's consider nonnegative numbers $p_{1},\ldots,p_{n+1}$ whose sum is $1$. For $p_{n+1}$ the ineqaulity is obvious, so we can assume $0<p_{n+1}<1$. Then:
$f\left(\sum_{1\leqslant i\leqslant n+1}p_{i}x_{i}\right)= f\left((1-p_{n+1})\cdot\sum_{1\leqslant i\leqslant n}\frac{p_{i}}{1-p_{n+1}}\cdot x_{i}+p_{n+1}x_{n+1}\right)\leqslant$
$(1-p_{n+1})\cdot f\left(\sum_{1\leqslant i\leqslant n}\frac{p_{i}}{1-p_{n+1}}\cdot x_{i}\right)+p_{n+1}f(x_{n+1})\leqslant$
$(1-p_{n+1})\cdot \sum_{1\leqslant i\leqslant n}\frac{p_{i}}{1-p_{n+1}}\cdot f\left(x_{i}\right)+p_{n+1}f(x_{n+1})= \sum_{1\leqslant i\leqslant n}p_{i}f(x_{i})$
which ends the proof.

This is just a special case of Power Mean, which itself can be easily proven by an alternate form of Jensen's inequality:

Let $f(x), g(x)$ be two continuous, monotonic functions.  Define the function-mean of some reals $a_{1}, ... a_{n}$ as

$M_{f}(a_{i}) = f^{-1}\left( \frac{ \sum_{i=1}^{n}f(a_{i}) }{n}\right)$

[b]Jensen's inequality (Function-mean form)[/b]:  If $f(g^{-1}(x))$ is convex, then

$M_{f}(a_{i}) \ge M_{g}(a_{i})$

And if $f(g^{-1}(x))$ is concave, then

$M_{f}(a_{i}) \le M_{g}(a_{i})$

Letting $f(x) = x^{p}, g(x) = x^{q}$ produces Power Mean.  Letting $f(x) = \ln x$ produces the Geometric Mean.

Also, For three variables: (assuming $a,b,c$ are non-negative))
$a^{3}+b^{3}+c^{3}-3 abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$
$= (a+b+c)(1/2)((a-b)^{2}+(b-c)^{2}+(c-a)^{2})\ge 0$
so you have$(a^{3}+b^{3}+c^{3})/3 \ge abc$

Edited later: I see that it was already posted before...

Another easy way to show AM-GM for 3 (or for any number, for that matter) [hide]is (assuming you can prove it for 2 variables)
$\frac{a+b}2 \ge \sqrt{ab}$
$\frac{c+d}2 \ge \sqrt{cd}$

so $\frac{a+b+c+d}4 \ge \frac{\sqrt{ab}+\sqrt{cd}}2 \ge \sqrt[4]{abcd}$

IOW, you have proved it for 4 variables... and all you have to do now is:

[hide]Just substitute $d = (a+b+c)/3$ in above and you are done.
You can apply the same trick:  by proving this for a larger number (of the form $2^{n}$) and then  setting  unwanted ones to be equal to arithmetic mean of  the wanted ones..)
[/hide] [/hide]

[quote="Gyan"]Another easy way to show AM-GM for 3 (or for any number, for that matter) [hide]is (assuming you can prove it for 2 variables)
$\frac{a+b}2 \ge \sqrt{ab}$
$\frac{c+d}2 \ge \sqrt{cd}$

so $\frac{a+b+c+d}4 \ge \frac{\sqrt{ab}+\sqrt{cd}}2 \ge \sqrt[4]{abcd}$

IOW, you have proved it for 4 variables... and all you have to do now is:

[hide]Just substitute $d = (a+b+c)/3$ in above and you are done.
You can apply the same trick:  by proving this for a larger number (of the form $2^{n}$) and then  setting  unwanted ones to be equal to arithmetic mean of  the wanted ones..)
[/hide] [/hide][/quote]
It is the same as in [url]http://planetmath.org/?op=getobj&from=objects&id=2970[/url].

There is also a simple, but multistep proof (no Jensen, no convexity, just elementary methods and notions):

First prove it for two variables:

$\frac{x+y}{2}\geqslant\sqrt{xy}\iff x-2\sqrt{xy}+y\geqslant 0\iff (\sqrt{x}-\sqrt{y})^{2}\geqslant 0$, which is obviously true.

$\frac{x+y}{2}\leqslant\sqrt{{x^{2}+y^{2}\over 2}}\iff{x^{2}+2xy+y^{2}\over 4}\leqslant{x^{2}+y^{2}\over 2}\iff 0\leqslant(x-y)^{2}$, which is obviously true.

Now use two-variable case to show four-variable case:

${x+y+z+t\over 4}=\frac{{x+y\over 2}+{z+t\over 2}}{2}\geqslant\sqrt{{x+y\over 2}\cdot{z+t\over 2}}\geqslant\sqrt{\sqrt{xy}\sqrt{zt}}=\sqrt[4]{xyzt}$

${x+y+z+t\over 4}=\frac{{x+y\over 2}+{z+t\over 2}}{2}\leqslant\sqrt{\frac{\left(\frac{x+y}{2}\right)^{2}+\left(\frac{z+t}{2}\right)^{2}}{2}}\leqslant\sqrt{\frac{{x^{2}+y^{2}\over 2}+{z^{2}+t^{2}\over 2}}{2}}=\sqrt{{x^{2}+y^{2}+z^{2}+t^{2}\over 4}}$

Now make special use of four-variable case by putting $t={x+y+z\over 3}$ and simplifying it. But I'll leave that as an excercise for you.

[quote="me@home"]The first proof I learned of AM-GM (in fact powermean) was through mixing variables, which is in fact a very elementary concept.[/quote]
Non-weighted Jensen can be killed by mixing variables and usage of lemma posted at [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=643429#643429[/url].

The way Farehajt did it is the way [b]Cauchy[/b] proved AM-GM.
Step 1 : Prove for $n=2^{m}$ by induction (case n=2 is evident) and
Step 2 : Prove that if AM-GM is valid for some n its valid for n-1

Isn't AM-QM a special case of Cauchy-Schwarz as well?

[quote="PenguinIntegral"][quote="me@home"]The first proof I learned of AM-GM (in fact powermean) was through mixing variables, which is in fact a very elementary concept.[/quote]
Could you post that? I am unsure what mixing variables is.[/quote]

I think it is something like showing

$f(a,b,c) \ge f\left(\frac{a+b}{2}, \frac{a+b}{2}, c\right)$,

so you know the minimum occurs as $a$ and $b$ get closer and closer to each other (i.e. they are equal). If you apply a similar argument to $b$ and $c$, for example, you can then show $a = b = c$ at the minimum and just evaluate it. That's mixing variables as far as I know; it can be more general than the case I have given.

The links in [b]hungkhtn[/b]'s signature can also be useful:
[quote="hungkhtn"]The entirely mixing variables method
[url]http://reflections.awesomemath.org/2006_5/2006_5_entirelymixing.pdf[/url]
Mixing variables and SMV theorem
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=113248[/url][/quote]

[quote="Rzeszut"]The $\mathrm{AM-GM}$ inequality, i.e. $\sqrt[3]{xyz}\leqslant \frac{x+y+z}{3}$ is just Jensen inequality for convex function $f(t)=\mathrm{e}^{t}$:
$\mathrm{e}^{\frac{a+b+c}{3}}\leqslant \frac{\mathrm{e}^{a}+\mathrm{e}^{b}+\mathrm{e}^{c}}{3}$,
we just substitute $x=\mathrm{e}^{a},\; y=\mathrm{e}^{b},\; z=\mathrm{e}^{c}$ and we are done.
The $\mathrm{AM-QM}$ inequality, i.e. $\frac{x+y+z}{3}\leqslant \sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}}$ is just transformation of Jensen for convex function $g(t)=t^{2}$:
$\left(\frac{x+y+z}{3}\right)^{2}\leqslant \frac{x^{2}+y^{2}+z^{2}}{3}$.
It becomes $\mathrm{AM-QM}$ after taking the square root of both sides.
Of course these proofs can be generalised to $n$ variables.

[b]edit[/b]: We can also it prove using techniques working for the particular case of $3$ variables:

$\mathrm{AM-GM}$: we substitute $a=\sqrt[3]{x},\; b=\sqrt[3]{y},\; c=\sqrt[3]{z}$. Then we have to prove $a^{3}+b^{3}+c^{3}\geqslant 3abc$. It is true, because $a^{3}+b^{3}+c^{3}-3abc= \frac12\cdot (a+b+c)\left((a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right)\geqslant 0$.

$\mathrm{AM-QM}$: we transform the inequality equivalently:
$\sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}}\geqslant \frac{x+y+z}{3}$
$\frac{x^{2}+y^{2}+z^{2}}{3}\geqslant \left(\frac{x+y+z}{3}\right)^{2}$
$3(x^{2}+y^{2}+z^{2})\geqslant (x+y+z)^{2}$
$3(x^{2}+y^{2}+z^{2})\geqslant x^{2}+y^{2}+z^{2}+2(xy+yz+zx)$
$x^{2}+y^{2}+z^{2}\geqslant xy+yz+zx$.
This inequality is true, because $\mathrm{LHS-RHS}= \frac12\cdot\left((a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right)\geqslant 0$.[/quote]

Hi, just realised that at AM_QM inequality i posted in the opening question had to the power of a third- you have demonstrated to the power of 1/2? Was i wrong in my original post?

[quote="rodgab"]Was i wrong in my original post?[/quote]
Yes, you were wrong and there sould be $\frac12$. There's also a mistake in the subject - you mistoke $\mathrm{AM}$ with $\mathrm{GM}$.


limit
induction
function
algebra
binomial theorem
calculus
calculus computations

evaluate this limit:

$ \lim_{n\to\infty}\sum_{k\equal{}0}^{n}\frac{a^{k}}{k!}$

with $ a\in{\mathbb{R}}$.

thanx.

hello, prove by induction that
$ \sum_{k\equal{}0}^{n}\frac{a^k}{n!}\equal{}\frac{a^{n\plus{}1}\minus{}1}{(a\minus{}1)n!}$
Sonnhard.

This sum is, essentially, the [i]definition[/i] of $ e^a$.  At least if you think of the function $ e^x$ as being defined by its Taylor series.

Alternatively, if the definition you're working with is $ e^a \equal{} \lim_{n\to \infty}\left(1 \plus{} \frac {1}{n}\right)^{na}$, then you could apply the binomial theorem and show that the $ k$th coefficient approaches $ \frac {a^k}{k!}$.

[quote="Dr Sonnhard Graubner"]
$ \sum_{k \equal{} 0}^{n}\frac {a^k}{n!} \equal{} \frac {a^{n \plus{} 1} \minus{} 1}{(a \minus{} 1)n!}$
[/quote]

This isn't the same as the expression given in the original post.

[quote="reason"]evaluate this limit:

$ \lim_{n\to\infty}\sum_{k \equal{} 0}^{n}\frac {a^{k}}{k!}$

with $ a\in{\mathbb{R}}$.

thanx.[/quote]
we know that $ e^x \equal{} \sum_{n\equal{}0}^{\infty}{\frac {x^n}{n!}},x$ are arbitary reals, so

$ \lim_{n\to\infty}\sum_{k \equal{} 0}^{n}\frac {a^{k}}{k!} \equal{} e^a$


geometry
circumcircle
geometric transformation
homothety
conics
projective geometry
geometry proposed

Let k be an arbitrary real number. Let O be the circumcenter of a triangle ABC, and A', B', C' the midpoints of the sides BC, CA, AB. Further, let X, Y, Z be the feet of the altitudes of triangle A'B'C', and let X'Y'Z' be the image of the triangle XYZ under the homothety with center O and factor k. Show that the lines AX', BY', CZ' are concurrent.

[b]Notes.[/b]

[b]1.[/b] To see why this is equivalent to Bernard Gibert's remark in [url=http://groups.yahoo.com/group/Hyacinthos/message/10114?expand=1]Hyacinthos message #10114[/url] that "If you define A# as vec(A*A#) = k vec(A*A'), k $\neq$ 1, you will find a cubic [...] pass[ing] through [...] X3 [...]" is left as an exercise to the more advanced reader :-)

[b]2.[/b] In the problem above, I ask to show that the lines AX', BY', CZ' are concurrent. As a special case of this for k = 1, one can obtain the concurrence of the lines AX, BY, CZ. But this special case is quite easy; the general case is the real problem (although I have a nice synthetic proof  ;) ).

  Darij

Actually, there's a pretty standard way to solve these (it also works for Kiepert's theorem and Kariya's theorem). If we modify $k$ and look at the point of intersection of $BY',CZ'$, we obtain a conic because the map $BY'\leftrightarrow CZ'$ is projective. Then we look at certain particular positions of that intersection point: it passes through $A,B,C,O,H$ (this is not obvious, but we can work each on out pretty easily.. I think :D). We have thus obtained the locus of the intersection pt of $BY',CZ'$: it's the Jerabek hyperbola. We obviously obtain the same thing when we consider the other two pairs of cevians. Now assume $AX'$ cuts $BY',ZC'$ in two different points. Then these two points and $A$ lie on a non-degenerate conic, which is absurd.

Indeed I could imagine that there is such a solution. But there is also a different solution. It uses no conics and no projective geometry; it uses a nice lemma about two triangles in a special relationship... well, I think you can find it (I was delighted when I found this solution, and I hope you will be too). Note that this lemma generalizes this particular concurrence and Kariya, too (but not Kiepert!).

  Darij

Since $ \triangle X'Y'Z' $ and $ \triangle ABC $ are bilogic (with common orthology center $ O $ ) ,
so $ \triangle X'Y'Z' $ and $ \triangle ABC $ are perspective . i.e. $ AX', BY', CZ' $ are concurrent :)


AMC
AIME
AMC 10
AMC 12
USA(J)MO
USAMO
search

Hey guys,
I'm relatively new to problem solving and only got good at math last year by getting a tutor.  I'm a sophomore in high school currently, and want to qualify for at least AIME through AMC 10.  However, I've realized that over the years, there have been gaps in my math that need to be fixed (slight misconceptions that could be deadly).  Hence, I'm going through the Introduction series right now and solving contests from earlier ages to improve my problem solving abilities.  Specifically, I do MC and AMC 8s.  For example, on the AMC 8s I can pretty much nail the entire exam (or get stuck for a bit on the last question sometimes).  I'm working my way up to the AMC 10, but I realize it will take time and a lot of work.  Additionally, I don't think rushing is the way to go and neither is setting unachievable goals.  That's why I want to work slowly and steadily so that I don't burn myself out.  Honestly, does my plan sound reasonable?  Should I just solve older exams from easier competitions to first boost my problem solving abilities and then move on to doing the more advanced Intro books and then the Intermediate ones?

Generally, the older the competitions are, the more arithmetic and computation it has. The newer contests have more problem solving based problems. For this reason, they are steadily getting slightly harder. Also, going through easier competitions is fine until you get most questions right (like 23+ on AMC8). After that, you won't learn much (other than just plain practice with the format and silly mistakes), so you should go ahead to the AMC10/12.

I think going through the Intro Series is good, but you definitely won't make it through before the 2010 AMC10/12. I would suggest doing your weakest spot first, and then read the rest for next year (2011 AMC 12, since you will be a Junior then). The Intro series should be good up through AMC10 and part of the AMC12. The harder parts of the AMC12 come from subjects in the Intermediate series, which is good until about USAMO. Also try the classics, where Volume 1 should cover most of the Intro series and Volume 2 the Intermediate.

Good luck!

EDIT: Heh, yeah (to AIME15USAMO). Fixed.

Yeah, I agree with PowerOfPi, the most important things you should do at the time are probably the intro series and a lot of practice AMC 10's and AMC 12's rather than MC and AMC 8's.

Also, I think PowerOfPi meant that volume 2 covers the intermediate series, he accidentally said this...
[quote="PowerOfPi"]Volume 1 should cover most of the Intro series and Volume 2 the Intro. [/quote]

The best way to learn problem solving is to solve problems!  Work as many problems as can that are challenging, but not impossible for you (without burning yourself out).  Also work on an individual problem for at least 20-30 minutes before giving up.  The longer you struggle with a problem, the more you'll learn (even if you have to look at the solution).  The intro books are great - just make sure to do the problems instead of only reading the text.  No matter what, avoid the trap of memorizing formulas to try to cram for a test, and don't give up!

thanks so much, guys.  I am trying to push myself to complete the Intro series by mid-March.  I can do 1-2 books a month.  The reason I don't want to rush is because it's not good for the long-term.  Chances are I will need to review this material again if I don't work slowly to make sure all misconceptions are cleared.  I will buy the volume 1 book asap.  I will work through AMC 10s soon, covering one maybe every 2 days or so (I think this won't be a burnout).  I do also agree that for certain problems, it's good to struggle for a bit before looking at the solution.  Honestly, thanks so much for the insightful responses in less than one hour.  Btw, I just plan on getting maybe 110+ or so this year on the AMC 10.  I will have much more time to practice over the summer next year for the 12.  Thanks once again.

Wow, 1-2 books a month? It will probably take longer than that. There are lots of problems, and you have to solve [u]all[/u] or almost all of the problems to get something out of it. I would recommend to slow down a bit, maybe 1 book per month at max.

I also prepared for the AMC 10 in some months. What I would recommend is this:

Do an AMC 10 and an AMC 12 and see where you are. I would recommend doing this under timed conditions and from different years. I would also try to think about what you need to qualify for AIME. You need about 20 questions from the AMC 10 and only about 15 -16 from the AMC 12. You might find that you are better at accomplishing one task in comparison with the other.

For reading material, I would recommend what has already been recommended. I found First Steps for Math Olympians helpful since it has 10 problems from easy to hard difficulty, all from old AMC competitions. The explanations for the theorems are much more indepth in the AoPS books though, while the First Steps book includes most, if not all of them that you need. After learning the basics, you should just do problems at about your level to improve the most.

Most of all, just have fun. I hope you gain an interest in math and an interest in how nice these problems are.

[quote="lvslugger36"]...and want to qualify for at least AIME through AMC 10.[/quote]

If your sole goal is to qualify for the AIME, take the AMC 12.

Why? It's easier to qualify through the AMC 10.

You need a 120+ on the AMC 10 to qualify and a 100+ on the AMC 12 to qualify (excluding the cases where it changes based on percent).

True, but in my opinion, it's easier to get a 120 on the AMC 10 than it is to get a 100 on the AMC 12. This was even more true when I had first started looking at AMC's (coming out of the MC-mode)

I don't think Intro to NT will be good enough for USAMO (I think it only goes up to AMC10/12). On the other hand, I don't know of a good intermediate/olympiad NT book either.

Okay. This is a point you've reached where you've got to start doing AIME problems. Even if they are harder, once you look the AMC 12's later on you'll think they're much easier. Do the intro books while doing AIME problems (great for hiding under a desk in class :) ).

I'm in a similar situation, except I'm in grade 8.  I'm going through the Introduction series, going through Volume 1, going through Intermediate, and then Volume 2.  Optimally, I think this is a good course of events, but since you don't have much time, I would suggest going through Volume 1, and if you have trouble, go to those topics in the intro series.

Just my two cents.

[quote="PowerOfPi"]On the other hand, I don't know of a good intermediate/olympiad NT book either.[/quote]

Number Theory Structures, Examples, and Problems by Andreescu and Andrica.

To qualify for the AIME from the AMC 12 all you need to do is answer the 14 questions you are sure you will get and leave the rest balnk. Automatic 100.5

[quote="lvslugger36"]Thanks guys.  For me, I think it looks easier (as of today) to get a 120 on AMC 10 than 100 on AMC 12.  Also, this may be a stupid question but is Intro to Number Theory enough for USAMO type problems, or will I need additional NT books (not problem books, just ones that teach)?[/quote]

Uh dude, if you're studying to make AIME, I don't think you should worry about studying for USAMO... In fact, study to make USAMO before that...

Also, I recommend just AoPS 1. You don't NEED the intro series to make AIME. Truthfully, there is a much bigger difference in Intermediate vs. AoPS II than Intro Series vs. AoPS I. So I only recommend the supplmental texts for after you make AIME and start studying for AMC 12 and studying the AIME.
And as already mentioned, do loads of problems. Mock AMCs, etc. I recommend doing slightly above average AMC 10 mock tests, than in January, start doing the old AMC 10s (save more recent ones the longest). The fact that it's slightly harder in the beginning will make it feel easier when you take the actual ones. For me, Mildorf's set of above average AIMEs really boosted up my score, combined with rigorously doing AoPS II, a solid 50 points on the AIME.

I'm thinking of making a sort of AMC "11", so I recommend you try it out. Also look in AoPS wiki, type mock amc and search for mock AMC 10s, which some people have written. These often have surveys in the beginning that result in "Harder than average" for the most part. Though, I don't think there are that many AMC10s, most are AMC 12).


A firm of photocopying bears a fixed monthly cost from $ \$$ 1200.00 per month and variable costs from $ \$$ 0.06 per sheet reproducing. Express the function of the total cost in terms of profit "x" of copied pages per month . Express, too, the profit function. If consumers pay $ \$$ 0.12 per sheet, the minimum number of leaves that the firm has to play not to make a loss?

Formulas: CT(x ) = CF + CV( x )
And profit: L( x) = R ( x) - CT( x )

cost function f(x)= .06x+ 1200

profit function=p(x)=.12x-.06x+1200


trigonometry
symmetry
geometry
function
area of a triangle
Heron\u0027s formula
cyclic quadrilateral

Prove the Heron's formula

$ S\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$ where $ s\equal{}\frac{a\plus{}b\plus{}c}{2}$

hello, here you will find a nice proof
http://jwilson.coe.uga.edu/EMT668/EMAT6680.2000/Umberger/MATH7200/HeronFormulaProject/GeometricProof/geoproof.html
Sonnhard.

[quote="bundler"]Prove the Heron's formula

$ S \equal{} \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$ where $ s \equal{} \frac {a \plus{} b \plus{} c}{2}$[/quote]
Since
\[ \sin \frac {\alpha}{2} \equal{} \sqrt {\frac {(s \minus{} b)(s \minus{} c)}{bc}}\]

\[ \cos \frac {\alpha}{2} \equal{} \sqrt {\frac {s(s \minus{} a)}{bc}}\]
observe that
\[ \sin \alpha \equal{} 2 \sin \frac {\alpha}{2} \cdot \cos \frac {\alpha}{2} \equal{} 2 \cdot \frac {\sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}}{bc}\]
It's easy to see that:
\[ S \equal{} \frac {bc \sin \alpha}{2} \implies \frac {bc \cdot 2 \frac {\sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}}{bc}}{2} \cdot \implies \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}\]
Hence we're done.

[quote="TZF"]You're missing a factor of $ \sqrt {s}$ in your final result. Take a look at the equation in the original post. I think it's a quick-fix, though.[/quote]

Yes I made a typo  :oops:  , writing $ \sqrt{s\minus{}a}$ instead of $ \sqrt {s(s\minus{}a)}$. Now it's edited.

Thank you TZF.

[quote="Obel1x"][quote="bundler"]Prove the Heron's formula

$ S \equal{} \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$ where $ s \equal{} \frac {a \plus{} b \plus{} c}{2}$[/quote]
Since
\[ \sin \frac {\alpha}{2} \equal{} \sqrt {\frac {(s \minus{} b)(s \minus{} c)}{bc}}\]

\[ \cos \frac {\alpha}{2} \equal{} \sqrt {\frac {s(s \minus{} a)}{bc}}\]
[/quote]

Could you explain me a bit , thanks.

[quote="bundler"]Could you explain me a bit , thanks.[/quote]

see here http://www.mathlinks.ro/viewtopic.php?t=322695

Alternately you can prove Brahmagupta's formula and just use the case where one side of the quadrilateral is 0. :rotfl:

(1) Does anyone knows how to prove Brahmagupta's formula WITHOUT trigonometry? I do...  :)

If ABCD is a cyclic quadrilateral then

$ [ABCD] \equal{} \sqrt {(p \minus{} a)(p \minus{} b)(p \minus{} c)(p \minus{} d)}$

(2) You can prove Heron's formula using Pythagoras...

The proof in the link above is supposed to be the original one, ie, the one Heron found himself (Journey Thru Genius, William Dunham).

I once read a nice simple proof using symmetry proporties of the area function f(a,b,c) = [ABC].

I think I've seen the proof you're talking about, it's not really a real proof but more of an intuitive argument. It just says that the area must be $ 0$ when $ a\plus{}b\equal{}c$ or symetrically the others, so it is likely that the formula contains factors of $ (a\plus{}b\minus{}c)$, $ (\minus{}a\plus{}b\plus{}c)$, and $ (a\minus{}b\plus{}c)$, and that since area in two dimensions is usually a product of two factors, it is likely the square root of four factors. Again, this isn't a proof, but it gives an easy way to understand the formula.


i have 273 friends!?

this is wrong

my number of friens is  only imaginary!!!!

According to this I have 7 friends.  That's only 7 more friends than I actually have... :)

EDIT: just kidding.  lol.

i like this, cuz for me, wen i plug in my values
($x=10$, $y=3$, $c=130$ as the constant), i get $144$ friends!

however, it is still $4856$ less than my real number of friends.

Hehe! Long names rock!

DiscreetFourierTransform:
$x = 24$; $y = 9$; $c = 130?$

$3|(27)^{2}+9^{2}-130|$
$3|680|$
$2040$

Wow!
I'm super popular all of a sudden?
That's more than five times larger than all the people in my highschool.

[quote="Treething"]Say the number of letters in your AoPS username is $x$ and the number of vowels in your AoPS username is $y$ (ignoring all spaces and numbers).

Scientific studies have shown that the amount of friends one has is determined by the formula $3|(x+3)^{2}+y^{2}-c|$. If this means Treething has 69 friends, how many friends does Interval have?[/quote]

c=176

$3|6^{2}+1^{2}-176|=417$

YES!!!

lmao treething :rotfl:


This is a video that I came across recently about 9/11,and i dont know if its been posted before here. Please watch the WHOLE movie and post your comments. I dont really know what to think at this moment, I dont want to believe its true, but its pretty convincing stuff.

http://www.policestateplanning.com/loose_change_ii.htm

just watched it. Very interesting. I perticularly enjoyed all about who benifited. I was supprised with how much money was attached to it all.

I reccomend it to everyone.


function
analytic geometry
complex analysis
complex numbers
complex analysis unsolved

Consider the punctured coordinate plane $ D \equal{} \{(x,y) \in \mathbb R^2 | (x,y) \ne (0,0)\}$, and let $ h : D \to \mathbb R$ be a harmonic function such that
\[ h(x,y) > \minus{} 1 \quad {\rm for \; all }\quad (x,y) \in D.\]
Is it true that $ h$ must be a constant function on $ D$?

Any ideas? Thank you very much  :)

$ h(e^z)$ is harmonic in the entire plane and bounded from below. If you make two holes (or three if you count infinity) there are non-constant positive harmonic functions though, because there is a positive Green's function.

Well I identify $ (x,y)$ with $ x\plus{}iy$. Harmonicity is preserved when composing with analytic functions.

[quote="Kalle"]Well I identify $ (x,y)$ with $ x \plus{} iy$. Harmonicity is preserved when composing with analytic functions.[/quote]

I am wondering since $ h$ is given and you are proving that $ h(e^z)$ is a non-constant harmonic function instead of $ h$. Am I wrong?

The function $ z \mapsto h(e^z)$ is harmonic everywhere on $ \mathbb{C}$ (since $ e^z$ misses the value 0) and is bounded from below. It follows that $ h(e^z)$ is constant (if you do not know this, consider the entire function which has real part $ h(e^z)$). Hence $ h$ itself is constant.

[quote="transseries"][quote="Kalle"]The function $ z \mapsto h(e^z)$ is harmonic everywhere on $ \mathbb{C}$ [/quote]

??? h is defined only on D[/quote]

What Kalle did mean is the following: we consider the function $ g : \mathbb C \to \mathbb C$ defined as $ g : z \mapsto h(e^z)$. Then $ g$ is bounded from below by $ \minus{}1$ and harmonic. In other words, $ g$ must be a constant and so is $ h$. Is it true?  :lol:

[quote="transseries"][quote="bookworm_vn"]consider the function $ g : \mathbb C \to \mathbb C$ defined as $ g : z \mapsto h(e^z)$. [/quote]

How to define $ g(z)$ when $ e^z \not\in D$ so that $ h(e^z)$ not defined ?[/quote]

Please keep in mind that $ D \equal{} \mathbb C \minus{} \{ (0,0) \}$.

[quote="bookworm_vn"]
Please keep in mind that $ D \equal{} \mathbb C \minus{} \{ (0,0) \}$.[/quote]
Stupid me, I was thinking $ D$ is disk.


You need java....

Otherwise, I think it should be compatible...

I've accessed the classes using linux and it works fine...

[quote="Hathelin Darkshadow"]Hello,
I was wondering if the classes are mac-compatible...[/quote]

Yes, they are.  Many of our students use Macs.


algorithm
continued fraction
number theory
Euclidean algorithm

Reduce the fraction $\frac{2121212121210}{1121212121211}$ to its simplest form.

One staightforward (and often the fastest) way is simply to  use continued fraction or Euclidean algorithm ..

May spoil the  fun part but it is very fast ...

[hide]For example this is  simply
$1+\cfrac{1}{1+\cfrac{1}{8+\cfrac{1}{4}}}= \frac{70}{37}$
(Literally took about 30 seconds without a calculator or spenidng any time in thinking  or looking for any pattern(s) -
[/hide]


Find all $ m,n\in Z^\plus{}$ such that:$ (1\plus{}m^{3^n}\plus{}m^{2.3^n})\vdots n$

He means : Find all positive integers $ m,n$ such that $ n$ divides $ (1 \plus{} m^{3^n} \plus{} m^{2.3^n})$

Obviosly $ n$ is odd. Let $ p|n$ and $ p > 3$, then $ (p,m) \equal{} 1, p|1 \plus{} m^{3^n} \plus{} m^{2*3^n} \equal{} \frac {m^{3^{n \plus{} 1}} \minus{} 1}{m^{3^n} \minus{} 1}.$
Therefore $ 3^{n \plus{} 1}|p \minus{} 1$ contradition with $ p|n$. Only case for n is $ n \equal{} 3^k$.
It is easy to chek, that for any $ k\ge 1, m\equal{}1\mod 3$ $ n \equal{} 3^k$ work.
$ n \equal{} 1 (k \equal{} 0)$ work for any m.

[quote="Rust"]Obviosly $ n$ is odd. Let $ p|n$ and $ p > 3$, then $ (p,m) \equal{} 1, p|1 \plus{} m^{3^n} \plus{} m^{2*3^n} \equal{} \frac {m^{3^{n \plus{} 1}} \minus{} 1}{m^{3^n} \minus{} 1}.$
Therefore $ 3^{n \plus{} 1}|p \minus{} 1$ contradition with $ p|n$. Only case for n is $ n \equal{} 3^k$.
It is easy to chek, that for any $ k\ge 1, m \equal{} 1\mod 3$ $ n \equal{} 3^k$ work.
$ n \equal{} 1 (k \equal{} 0)$ work for any m.[/quote]
OK.Good sloution  :)


Two boxes are each filled with colored socks. Box A has 2 red socks and 4 blue socks. Box B has 4 red socks and 4 blue socks. A box is chosen at random, and a sock is drawn from the box. Given that the sock chosen is blue, what is the probability that it came from box A?

(p.s.  sorry about the multiple questions)

Okay.. so, there are four cases to draw socks: draw blue from first, draw red from first, draw blue from second, or draw red from second. Since we are given that the sock is blue, that eliminates the second and fourth cases. We are left with two cases: the blue sock came from Box A, or it came from Box B. We can now calculate this as we would most probability problems. P(successful)/P(total=P(A)/(P(A)+P(B)), where P(A) is the probability of drawing from Box A and P(B) is the probability of drawing from Box B. Now we only have to find the probability of these two cases happening. Let's start with drawing a blue sock from Box A. There is a 1/2 chance that we select A, and a 4/6 chance we draw a blue sock. Therefore, the probability of drawing a blue from A is equal to 1/2*4/6=1/3. Let's do this for B too. There is a 1/2 chance we pick B, and a 4/8 chance we draw a blue. This is equal to 1/2*4/8=1/4. We now see that P(A)/[P(A)+P(B)] equals (1/3)/[(1/3+1/4)]. This is equivalent to 4/7, which is the answer.


We know that:

$ \frac{x_{1} \plus{} x_{2} \plus{} \dots \plus{}x_{n}}{n} \geq (x_{1} x_{2} \dots x_{n})^{\frac{1}{n}}$

then $ (\frac{1}{3})(\sum \frac{\sqrt{a}}{b\plus{}ac}) \geq (\prod{\frac{\sqrt{a}}{b\plus{}ac}})^{\frac{1}{3}}$

With equality when $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$

et voila!

[quote="can_hang2007"][quote="shfdfzhjj"]I think it is a nice problem. :)[/quote]
Yes, It is a nice problem. To prove it, my solution use the following lemma
\[ ab \plus{} bc \plus{} ca \ge \frac {243}{64}(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2
\]
:)[/quote]

How can you find the lemma? It's really wonderfui! :lol:
My proof is too long. :(


I need some inequaliti is prove by nesbit's inequaliti
 Who can help me ? Thank

Nesbitt inequality
 If x,y,z >0
\[\sum_{cyc}\frac{x}{y+z}\geq \frac{3}{2}\]
We put $a=x+y$ ,$b=y+z$ ,$c=x+z$ a,b,c >0 and we the inequality is equivalent to:
\[\sum_{cyc}\frac{a+b-c}{2c}\geq \frac{3}{2}\]
$\leftrightarrow$
\[\sum_{cyc}(\frac{a}{b}+\frac{b}{a})\geq 6 \]
which is true because $a+\frac{1}{a}\geq 2$.EDITED

[quote="Svejk"]We put $x=a+b$ ,$y=b+c$ ,$z=c+a$ a,b,c >0 [/quote]
I think you mean $a=x+y, b=y+z, c=z+x$.

Yes you're right Rzeszut.Sometimes I'm such a robot :blush: .Thank you !

Here's another proof of Nesbitt using Cauchy:
$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^{2}}{ab+ac}$
But by Cauchy $\rightarrow$ $\sum_{cyc}\frac{a^{2}}{ab+ac}\geq \frac{(a+b+c)^{2}}{2(ab+bc+ac)}\geq \frac{3}{2}$

See [i]Topics in Inequalities- Theorems and Techniques[/i] at [url=http://ultrametric.googlepages.com/]Hojoo Lee's website[/url]. There are at least ten proofs in the booklet.

[quote="hien"]the following inequality is stronger than Nesbit. It's dedicated by Vasc:
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{13}{6}-\frac{2(ab+bc+ca)}{3(a^{2}+b^{2}+c^{2})}$[/quote]
I think if $a=b=c=1$,this inequality not true

Chien Than: $\frac32=\frac{13}6-\frac23$ since $\frac32+\frac23=\frac{2^{2}+3^{2}}{6}=\frac{13}{6}$....
mathboy:
\[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{13}{6}-\frac{2(ab+bc+ca)}{3(a^{2}+b^{2}+c^{2})}\geq \frac32 \]
where the last: \[\iff \frac{13}{6}\geq \frac23\frac{\sum ab}{\sum a^{2}}+\frac32 \iff \sum a^{2}\geq \sum ab \]
which is obvious... I'm not sure how you guts got confused..

Hien: Can you prove your's ineq?
Thank you ; it very nice :D

I think it has been posted before somewhere but i don't know exactly where is it. Oke i rewrite the solution by using S.O.S method here.

Rewrite the inequality as:
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-\frac{3}{2}\ge\frac{2}{3}\left(1-\frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}\right)$
Since:
$\sum\left(\frac{a}{b+c}-\frac{1}{2}\right)=\sum\frac{(a-b)+(a-c)}{2(b+c)}=\sum\frac{a-b}{2(b+c)}+\sum\frac{b-a}{2(c+a)}$
$=\sum\frac{a-b}{2}\left(\frac{1}{b+c}-\frac{1}{c+a}\right)=\sum\frac{(a-b)^{2}}{2(b+c)(c+a)}$
and $\frac{2}{3}\left(1-\frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}\right)=\sum\frac{(a-b)^{2}}{3(a^{2}+b^{2}+c^{2}}$
the inequality becomes
$\sum (a-b)^{2}\left(\frac{1}{2(b+c)(c+a)}-\frac{1}{3(a^{2}+b^{2}+c^{2})}\right)\ge 0$
It is true because
$3(a^{2}+b^{2}+c^{2})-2(b+c)(c+a)=(a+b-c)^{2}+2(a-b)^{2}\ge 0$
Equaliy holds if and only if $a=b=c$

Isnt it useful?

[url]http://web.mit.edu/~tmildorf/www/Inequalities.pdf[/url]

[quote="Svejk"]Nesbitt inequality
 If x,y,z >0
\[ \sum_{cyc}\frac {x}{y \plus{} z}\geq \frac {3}{2}
\]
We put $ a \equal{} x \plus{} y$ ,$ b \equal{} y \plus{} z$ ,$ c \equal{} x \plus{} z$ a,b,c >0 and we the inequality is equivalent to:
\[ \sum_{cyc}\frac {a \plus{} b \minus{} c}{2c}\geq \frac {3}{2}
\]
$ \leftrightarrow$
\[ \sum_{cyc}(\frac {a}{b} \plus{} \frac {b}{a})\geq 6
\]
which is true because $ a \plus{} \frac {1}{a}\geq 2$.EDITED[/quote]
This solution is wrong.

[quote="stephen38"]I think it is true.Could you please explain where is the wrong part??
I think this is true too
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{b \plus{} a}\ge \frac {21}{16} \plus{} \frac {27(a^{3} \plus{} b^{3} \plus{} c^{3})}{16(a \plus{} b \plus{} c)^{3}}$. :wink:[/quote]
Suppose that $ a\plus{}b\plus{}c\equal{}3$
and we have the inequality:
$ \frac{a}{3\minus{}a}\ge \frac{a^3\plus{}9a\minus{}2}{16}\leftrightarrow \frac{(a\minus{}1)^2(a^2\minus{}a\plus{}6)}{3\minus{}a}\ge 0$

Done!

another proof for nessbit inequality :
$ S\equal{}\frac{a}{b\plus{}c}\plus{}\frac{b}{a\plus{}c}\plus{}\frac{c}{b\plus{}a}$

$ S\equal{}\frac{a\plus{}b\plus{}c}{b\plus{}c}\plus{}\frac{a\plus{}b\plus{}c}{a\plus{}c}\plus{}\frac{a\plus{}b\plus{}c}{b\plus{}a}\minus{}3$

$ S\equal{}(a\plus{}b\plus{}c)(\frac{1}{b\plus{}c}\plus{}\frac{1}{a\plus{}c}\plus{}\frac{1}{b\plus{}a})\minus{}3$

$ S\equal{}(\frac{(a\plus{}b)\plus{}(a\plus{}c)\plus{}(b\plus{}c)}{2}(\frac{1}{b\plus{}c}\plus{}\frac{1}{a\plus{}c}\plus{}\frac{1}{b\plus{}a})\minus{}3$

Now by caushy schwartz :

$ ((a\plus{}b)\plus{}(a\plus{}c)\plus{}(b\plus{}c))(\frac{1}{b\plus{}c}\plus{}\frac{1}{a\plus{}c}\plus{}\frac{1}{b\plus{}a})\,\succeq\,9$

So :   $ S\,\succeq\,\frac{9}{2}\minus{}3\equal{}\frac{3}{2}$

Cordialement

[quote="Svejk"]Here's another proof of Nesbitt using Cauchy:
$ \sum_{cyc}\frac {a}{b \plus{} c} \equal{} \sum_{cyc}\frac {a^{2}}{ab \plus{} ac}$
But by Cauchy $ \rightarrow$ $ \sum_{cyc}\frac {a^{2}}{ab \plus{} ac}\geq \frac {(a \plus{} b \plus{} c)^{2}}{2(ab \plus{} bc \plus{} ac)}\geq \frac {3}{2}$[/quote]

Can someone please explain this? I'm not following the last line at all.

EDIT: Nevermind, I'm just a complete newb at inequalities >.>

I forgot that Cauchy can be stated both as $ (a_1^2 \plus{} a_2^2)*(b_1^2 \plus{} b_2^2) \geq (a_1*b_1 \plus{} a_2*b_2)^2$ and the equivalent, but more relevant $ (b_1 \plus{} b_2)*(\frac {a_1^2}{b_1} \plus{} \frac {a_2^2}{b_2}) \geq (a_1 \plus{} a_2)^2$   :blush:

One step proof of nesbitts inequality can be by rearrangement theorem as done by engel :)

[quote="chien than"][quote="hien"]the following inequality is stronger than Nesbit. It's dedicated by Vasc:
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\geq \frac {13}{6} \minus{} \frac {2(ab \plus{} bc \plus{} ca)}{3(a^{2} \plus{} b^{2} \plus{} c^{2})}$[/quote]
I think if $ a \equal{} b \equal{} c \equal{} 1$,this inequality not true[/quote]
why?? if a=b=c=1, I am sure the inequality is true

The Nesbit's ineqaulity can be easily proved by the Andreescu-Eneescu lemma. See the topic under the same name to see the lemma! :lol:


induction
number theory proposed
number theory

Prove that if $ a_{1}, \dots, a_{k}, b$ are positive reals, then the equation $ \sum{\frac{a_{i}}{x_{i}}}\equal{}b$ with variables $ x_{i}$ has only finitely many natural solutions.

It can be prove by induction on n.
For $ n=1$ then it is true.
Suppose it is true for $ n$
We will prove for $ n+1$
$ x_1\geq x_2\geq x_3\...\geq x_{n+1}$
Then $ (\sum_{i=1}^n a_1).\frac{1}{x_{n+1}}\geq b$
So $ x_{n+1}$ take only finite value.
Problem claim.

if you're not gonna post a thorough explanation, don't post anything. it makes it hard to check for mistakes. people don't like to sift through other people's messes. seriously now.


Term by Term integration of a convergent Fourier series results in a Fourier series that converges more rapidly than the original Fourier series.

Can anyone please give me a link to the above theorem on Fourier Series?

Regards,
Kolahal.

This should have been posted in the Calculus Computations and Tutorials Forum.


When an open metal pipe is cut into two pieces, the lowest resonance frequency for the air column in one piece is 256 Hz and that for the other is 440 Hz. a) What resonant frequency would have been produced by the original length of pipe? b) How long was the original pipe?

Let $ v$ be the speed of sound and let $ L_1$ and $ L_2$ be the lengths of the two parts (I am ignoring end correction).
a) $ \frac{v}{2L_1}\equal{}\nu_1$ and $ \frac{v}{2L_2}\equal{}\nu_2$
Therefore, $ \nu\equal{}\frac{v}{2(L_1\plus{}L_2)}\equal{}\frac{\nu_1\nu_2}{\nu_1\plus{}\nu_2}\equal{}162\text{Hz}$
b) $ L_1\plus{}L_2\equal{}\frac{v}{2\nu}$ As length is explicitly dependant on speed of sound, I think that should be given.


inequalities
geometry
circumcircle
inequalities proposed

Prove that

$ 18rR\leq a^2\plus{}b^2\plus{}c^2\leq9R^2$

[quote="Ligouras"]Prove that

$ 18rR\leq a^2 \plus{} b^2 \plus{} c^2\leq9R^2$[/quote]

The following inequality is equivalent to

$ 18rR\leq 2(s^2 \minus{} 4Rr \minus{} r^2)\leq9R^2$

$ 2(s^2 \minus{} 4Rr \minus{} r^2)\leq9R^2\Longleftrightarrow 9R^2 \plus{} 8Rr \plus{} 2r^2\geq 2s^2$

from O.Kooi  we only prove

$ 9R^2 \plus{} 8Rr \plus{} 2r^2\geq 2\frac {R(4R \plus{} r)^2}{2(2R \minus{} r)}$

$ \Longleftrightarrow \frac {(R \minus{} 2r)(2R \plus{} r)(R \plus{} r)}{2R \minus{} r}\geq 0$

$ 18rR\leq 2(s^2 \minus{} 4Rr \minus{} r^2)\Longleftrightarrow s^2\minus{}13Rr\minus{}r^2\geq 0$

from Gerretsen we only prove  $ 16Rr\minus{}5r^2\minus{}13Rr\minus{}r^2\geq 0\Longleftrightarrow R\geq 2r$

obvious true

Although the above proof is correct, it is like using a canon to kill a mosquito.
The left inequality follows from the obvious (using AM-GM)
\[ (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2) \geq 9abc,\]
upon substituting $ a \plus{} b \plus{} c \equal{} 2s,abc \equal{} 4sRr$.

The right inequality follows from
\[ OG^2 \equal{} R^2 \minus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{9},\]
where $ G$ is the barycenter and  $ O$ the circumcenter of triangle $ ABC$.

$ a^2 \plus{} b^2 \plus{} c^2\geq18Rr$
$ < \equal{} a^2 \plus{} b^2 \plus{} c^2\geq2(h_a \plus{} h_b \plus{} h_c)R$
$ < \equal{} > s^2 \minus{} 12Rr \minus{} 3r^2\geq0$
$ < \equal{} > (s^2 \minus{} 16Rr \plus{} 5r^2) \plus{} 4r( \minus{} 2r \plus{} R)\geq0$,
clear.
BQ


Divide both equations by $xy$ to obtain \[ \left(x+2+\frac1x\right) \left(y+2+\frac1y \right)=27\] and
\[\left(x+\frac1x \right)\left(y+\frac1y \right)=10.\]
Set $a=x+\frac1x,b=y+\frac1y$, etc.


A prime $p\geq 3$ is given.We define a seqence $(a_{n})$:
$a_{n}=n$ for $n=1,2,\ldots ,p-1$ ,
$a_{n}=a_{n-1}+a_{n-p}$ for $n\geq p$.
Determine $a_{p^{3}}modp$.

Do you mean $a_{0}$ ?
Actually, we need that because $a_{p}= a_{p-1}+a_{0}$  :maybe:

:blush:  sorry,I made a mistake.It schould be $a_{n}=n$ for $n=0,1,\ldots,n$.

There is a typo again..
It must be for $n = 0,1,2,\cdots,p-1$  :lol:

Posted before :lol: Polan test, you can find in Resources


While randomly testing the strength of a new material in a pair of skis, seven pairs of skis out of 2,000 were found defective. What is the probability that if you buy a pair of these skis, they will not be defective? Express your answer as a decimal to the nearest 10,000th.

...except we need the probability the skis are [b]not[/b] defective, so $ 1\minus{}.0035\equal{}\boxed{0.9965}$


geometry
circumcircle
incenter
inradius
perimeter
geometry unsolved

Let $O$ be the circumcenter, $I$ the incenter and $H$ the orthocenter of a triangle $ABC$. Denote by $R$ the circumradius and by $r$ the inradius of triangle $ABC$. Prove that the triangle $OHI$ is isosceles if and only if $\frac{a^{3}+b^{3}+c^{3}}{3abc}=\frac{R}{2r}$.

[color=darkblue][b]STEP 1.[/b] $\boxed{\ \frac{a^{3}+b^{3}+c^{3}}{3abc}=\frac{R}{2r}\Longleftrightarrow p=\sqrt 3\cdot (R+r)\ }\Longleftrightarrow 60^{\circ}\in \{A,B,C\}\ .$

[b]STEP 2.[/b] $\boxed{\ IH^{2}=4R^{2}+4Rr+3r^{2}-p^{2}\ }\ \ ;\ \ \boxed{\ IO^{2}=R^{2}-2Rr\ }\ .$

[b]STEP 3.[/b] $\boxed{\ IH=IO\Longleftrightarrow p=\sqrt 3\cdot (R+r)\ }\ .$[/color]

Virgil Nicula thanks for solution. But what is p? Is perimeter? Can you expand your solution?

Actually, $\angle{A}=60^{o}$ directly implies that $HI=OI$: http://www.mathlinks.ro/Forum/viewtopic.php?p=580375#580375


[quote="zvi"]For n a positive integer, let P(n) denote the product of the decimal digits of n, and let S(n) denote their sum.
Find the number of solutions of the equation P(P(x))+P(S(x))+S(P(x))+S(S(x))=1984.

Thanx  :)[/quote]

If you consider that $ P(0)$ and $ S(0)$ are defined and are $ 0$, there are infinitely many solutions :

Let $ a\equal{}\sum_{i\equal{}1}^{1984}10^i$ and let $ b\equal{}\sum_{i\equal{}1}^{a}10^i$ and let $ k\geq 0$

$ P(P(10^kb))\equal{}P(0)\equal{}0$
$ S(P(10^kb))\equal{}S(0)\equal{}0$
$ P(S(10^kb))\equal{}P(a)\equal{}0$
$ S(S(10kb))\equal{}S(a)\equal{}1984$

So $ 10^kb$ is solution of $ P(P(x))\plus{}P(S(x))\plus{}S(P(x))\plus{}S(S(x))\equal{}1984$ $ \forall k$


trigonometry
calculus
integration
function
calculus computations

In general, what is

$1+\frac{x^4}{4!}+\frac{x^8}{8!}+\frac{x^{12}}{12!}+\cdots$ ?

The answer can be written in several ways, most compactly as

[hide]$\frac12(\cos x + \cosh x)$ so that the particular answer here is

$\frac12(\cos 1+\cosh 1)=\frac12\cos 1+\frac14\left(e+\frac1e\right).$[/hide]

[quote="Kent Merryfield"]In general, what is

$1+\frac{x^4}{4!}+\frac{x^8}{8!}+\frac{x^{12}}{12!}+\cdots$ ?
[/quote]

Generally one may ask for:

$1+\frac{x^k}{k!}+\frac{x^{2k}}{(2k)!}+\frac{x^{3k}}{(3k)!}+\cdots$
which is the solution of $f^{(k)}(x)=f(x)$ and $f(0)=1$, $f^{(i)}(0)=0$, $1 \leq i \leq k-1$.

I still don't get if there is any nice formula for:

$1+\frac{x}{1*1!}+\frac{x^2}{2*2!}+\frac{x^3}{3*3!}+\cdots$
or:
$1+\frac{x}{(1!)^2}+\frac{x^2}{(2!)^2}+\frac{x^3}{(3!)^2}+\cdots$
Does anyone know?

[quote="heartwork"]I still don't get if there is any nice formula for:

$1+\frac{x}{1*1!}+\frac{x^2}{2*2!}+\frac{x^3}{3*3!}+\cdots$
or:
$1+\frac{x}{(1!)^2}+\frac{x^2}{(2!)^2}+\frac{x^3}{(3!)^2}+\cdots$
Does anyone know?[/quote]

Unfortunately, there does not seem to be any. Usually in these cases we refer to several useful handbooks by Gradshteyn and Rhyzhik or Prudnikov, Brychkov and Marichev. The first series has an integral representation:

\[ 1+\frac{x}{1*1!}+\frac{x^2}{2*2!}+\frac{x^3}{3*3!}+\cdots = 1 + \int_0^x \frac{e^t - 1}{t} dt \]

The second one is actually a modified Bessel function of the 2nd kind (rather long-winded term). The MBF2 of order $\nu$ and argument $x$ is defined by
\[K_v (x) = \sum_{k=0}^\infty \frac{(x/2)^{2k+\nu}}{k! \Gamma (\nu + k + 1)}\]
where $\Gamma(x)$ is the gamma function and if $x$ is an integer then $\Gamma(x) = (x-1)!$.
By choosing $\nu = 0$ and taking $x^2/4 = u$, we see that the series is $K_0 (2\sqrt{u})$. The Bessel functions are often used in physics and also in distribution theory. Try checking http://www.mathworld.com  for more details since these functions fall under a category known as "special functions", which may not be familiar to students until they are in the 2nd year (probably) of undergraduate studies.


scoring above 100 on AMC12

or be the top 5% ??????

BOTH qualify right??

Getting a 100 is easier.

You can solve 12 questions on an AMC 12 and get a 100

EDIT: Although they're basically equivalent.

It's almost always easier to score 100 points.  The top 5% rule is used in extremely rare occasions to lower the cutoff score.  This happened in 2000, when the cutoff was something like 84 due to the AMC12 being insanely difficult.  I'm not sure if the 5% rule was in place prior to that year.

In any case, it would take a very difficult AMC for the top 5% score to be lower than 100.

Yeah, I think it has been 120 every year except for one.  (Someone correct me if I'm wrong.)

[quote="qweretyq"][quote="hockeyadi23"]Is the same true for the AMC 10s? Is 120 easier or the top 1%. I didn't know anyone who made the AIMEs lower than a 120 but then again I could be wrong...[/quote]

Actually two years ago, AMC 10 A was unusually hard, and qualifying score was less than 120. I know a few people (including myself) who got under 120 and made it.
Qualifying was 110 I believe[/quote]

That explains why 2004 seemed pretty hard compared to '05 and '03.

[quote="hockeyadi23"][quote="h_s_potter2002"]Actually, most of the time, the qualifying score is less than 120.  But l wouldn't count on it though.[/quote]

So was it 120 last year? 10A was an easy test, but I could be wrong.[/quote]

Yea, it was 120 last year.


[quote="Altheman"]Here is another problem that I liked from my old aime that was interesting:

9. Zeus, Athena, and Posideon arrive at Mount Olympus at a random time between 12:00 pm and 12:00 am, and stay for 3 hours. All three hours does not need to fall within 12 pm to 12 am. If any of the 2 gods see each other during 12 pm to 12 am, it will be a good day. The probability of it being a good day is $\frac{m}{n}$ where $m$ and $n$ are coprime positive integers, find $m+n$.[/quote]
I've seen two variable versions of this, where it is graphed on a square plane. I assume you project the times onto a cube. Would you care to enlighten us with your solution?

[quote="Altheman"]well, first a few things to get started, later i will post a solution if nobody can get it

[hide="..."]first make the arrival times x,y,z and make a set of inequalities using the arrival times

then it is not very easy to compute at least 2 meetings, so it is easier to count the complement, when none of them see each other and subtract that from 1[/hide][/quote]
Could you post it now? Or someone else?


People need to stop posting contest math in Classroom Math.  That's not what this forum is for.

actually this is not a contest math question, it came from the textbook [b]Holt Algebra I, chapter 14, pg.735, College Entrance Exam Practice, number 31[/b]

and you are not to judge what this forum is/isnt for, that is for the moderatrrs.

also if you do not want to answer, dont. i would care less...

[quote="DaReaper"][b]College Entrance Exam Practice[/b][/quote]

In a Middle School forum...

well.. it is in a middle school (8th grade) book.. so i suppose so...

[quote="ckck"][quote="DaReaper"]Find the sum of the $3$ roots of the equation

$x^{3}-13x+12$

*please show all work*[/quote]

How would you do this problem?[/quote]

The only way I would do it is by graphing  :lol:

isnt it just...

[hide="am i being dumb?"]$0$...since $\frac{-b}{a}=\frac{0}{1}=0$?[/hide]

[quote="MathTeX"][quote="DaReaper"]Find the sum of the $3$ roots of the equation

$x^{3}-13x+12$[/quote]

Just see: $x^{3}-13x+12=(x-1)(x^{2}+x-12)$

Then it's done...[/quote]

Ok can you tell me how you got there?

[quote="DaReaper"]actually the answer is $0$....[/quote]
Theres more than one solution. I tried 1 and it worked.

[quote="DaReaper"]actually the answer is $0$....[/quote]
Yes, because $(x-1)(x^{2}+x-12)$ can be further factored into $(x-1)(x-3)(x+4)$. The solutions are $x={1,\,3,\,-4}$. The sum is $0$.

but does the formula stated above work?  \[\frac{-b}{a}\]

and does it always work?

Oh it said sum of the roots. I didn't see that.

So do you think the -b/a would always work for the sum of the roots?

kind of old but...does the \[\frac{-b}{a}\] only work when the problem is in the quadratic form $ax^{2}+bx+c$  ???

no. There's something in the [b][size=150]high school basics[/size][/b] about that. I can give the link once i find it.
EDIT:here it is[quote="The QuattoMaster 6000"]Say the roots of $ax^{3}+bx^{2}+cx+d$ are m, n, and p, so $x^{3}+\frac{bx^{2}}{a}+\frac{cx}{a}+\frac{d}{a}$ also has roots m, n, and p.  This means that $(x-m)(x-n)(x-p)=x^{3}-(m+n+p)x^{2}+(mn+np+mp)x-mnp=x^{3}+\frac{bx^{2}}{a}+\frac{cx}{a}+\frac{d}{a}$.  Corresponding the $x^{2}$s, we get that $\frac{bx^{2}}{a}=-x^{2}(m+n+p)$.  This means that m+n+p=-b/a.[/quote]

but it doesnt work... because then it would be

\[\frac{-b}{a}=\frac{-(-13)}{1}=\frac{13}{1}=13\]

13 isn't b. It's c. 0 is b. Imagine that ther is a "+0x^2" in there

what do you mean Imagine that ther is a "+0x^2" in there

and how is b 0?

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh...that explains it alllll!!!!! :lol:  :D


It is obvious that both sums (the one in $a_k$ and the one in $b_k$) are both greater than 1, or both smaller than 1.
Assume that those sums are less than 1
$\sqrt{(1-a_1^2-a_2^2-...-a_n^2)(1-b_1^2-b_2^2-...-b_n^2)}\le \frac{(1-a_1^2-a_2^2-...-a_n^2)+(1-b_1^2-b_2^2-...-b_n^2)}{2}=$
$\frac{2-(a_1^2+b_1^2)-...-(a_n^2+b_n^2)}{2}\le \frac{2-2a_1b_1-2a_2b_2-...-2a_nb_n}{2}=(1-a_1b_1-a_2b_2-...-a_nb_n)$

Squaring both sides we contradict the given inequality.


Where is a proof for this statement:

The area of a regular n-gon of side length s is $ \frac{ns^2*\tan{(90\minus{}\frac{180}{n})}}{4}$.

Also, what is the best way to learn theorems and formulas?
In the past, I've tried several ways to learn formulas
- understanding the proof for it
- trying to "see" why it is true intuitively
- practicing it with examples until I'm convinced it works

um. first of all, don't focus just on learning theorems and formulas: learn approaches and methods of problem solving! if u just learn a formula for a specific type of problem, than u would have to learn a formula for every type of problem every conceived!!!

btw, there is a simpler formula for the area of a regular n-gon
it goes as follows:
$ \dfrac {a\cdot p}{2}$ where $ p$ is the perimeter of the polygon and $ a$ is its apothem.

[hide]
Divide the polygon into $ n$ triangles, each with one vertex at the center of the polygon, plus two adjacent vertices of the polygon.  This is an isosceles triangle, as the radii are equal.  The other side has length $ s$.  The base angles of the triangle are each half of an angle of the polygon, so they have measure

$ \frac{1}{2}\frac{180(n\minus{}2)}{n}\equal{}\frac{180n\minus{}360}{2n}\equal{}90\minus{}\frac{180}{n}$

The altitude to the base of the triangle is equal to half the base times the tangent of the base angle: $ \frac{s}{2}\tan{(90\minus{}\frac{180}{2})}$.  Thus the area of the triangle is $ \frac{s^2\tan{(90\minus{}\frac{180}{2})}}{4}$.   There are $ n$ triangles, so the total area is $ \frac{ns^2\tan{(90\minus{}\frac{180}{2})}}{4}$.
[/hide]

About the theorems and formulas, I think you have already found the best way to learn them.


induction
combinatorial geometry
combinatorics proposed
combinatorics

Let S be a set of n points in the plane. Suppose that every two points in S are at distance at least 1 apart.
Prove that there are at most 3n pairs of points in S such that each pair of points is at distance exactly one apart.

i thought i posted this like 4months ago but just couldnt find it in my posts...

We draw the graph $G$ having the points as vertices and in which two vertices are joined iff the distance between them is $1$. We want to show that $G$ has at most $3n$ edges.

Suppose we show that there is a vertex $x$ with at most $3$ neighbours. Then, by induction, $G-x$ has at most $3(n-1)$ edges, and since $x$ adds at most $3$ edges, we are done.

All we need to do now is show that there is a vertex with $\le 3$ neighbours. Consider the convex hull $\mathcal P$ of the $n$ points. It's easy to see that every vertex $A$ of the polygon $\mathcal P$ has $\le 3$ neighbours (otherwise, the angle at $A$ would be $\ge\pi$, which is absurd, since $\mathcal P$ is convex).

Consider the graph $G$ whose vertices are the $n$ points two joined by an edge if and only if their mutual distance is $1$. Note that this distince is the minimal positive one in the set of the $n$ points.
We will prove that $G$ is planar, from which the desired result will follow immediately because it is a classical result (just use Euler's formula) that a planar graph with $n$ vertices has at most $3n-6$ edges (and this gives a slighty better bound than desired).

Assume that $G$ is not planar. That is the open segments $]AB[$ and $]CD[$ are edges and have a common point, say $O$. Note that since $AB = CD = 1$ they only have $O$ in common. Thus :
$OA+OC > AC$ and $OB+OD > BD$. Summing, it leads to $2 = OA+OB+OC+OD > AC+BD$ so that $AC$ or $BD$ is less than $1$, which contradicts that $1$ is the minimal positive distance, and we are done.

Pierre.

Nice! :) I saw that it was planar, but didn't make the connection. You can also prove the bound $3n-6=3(n-2)$ in the exact same way as I did above, by induction. The only thing changing is the first induction step.


For those of those into it, prove that a made rubiks cube with one corner twisted isnt solvable.

This should be really easy.
Let white and yellow faces be a 'light coloured' face.
Each corner has one light coloured face. If the corner's light coloured face is matched with a centre cube that also has a light colour, we assign a value, 0, to the corner. If it is turned clockwise, assign 1. If it is turned counterclockwise, assign -1. The sum of the 8 assigned values are 0 when we have the solved cube. Whatever turn you make, the sum is invariant in mod 3. QED


I am sure many of you participate in the brain bowl. If not, here is a short description: Brain Bowl is a tournament with teams that go against each other by solving question. The questions can be anything to test your knowledge. I will start:

[b]What is the name of the hat worn by bishops?[/b]

Whenever a problem is solved, the next user must post a new one.
THANKS!

Miter?

I'm not sure I spelled that right.

NP:
[b]What jeweler to Nicholas II, the last Tsar, is famous for his highly decorated Easter eggs?[/b]

Faberge right?

I watched a documentary about that, lol.

Just to keep up the mathyness of this forum,
Find the third derivative, with respect to x, of sin (2x).


inequalities
trigonometry
geometry
circumcircle
inequalities solved

hey
i need some help with this somwhat innocuous inequality(i cam somehow not able to solve it,though it seems easy) :D
sin A +sin B +sin C > 2 for A,B,C being acute angles
plz help
cheers!

just a question: are A, B, C the angles of a triangle?

oh i am  sorry for not mentioning that they are angles of a triangle
cya

This is a very nice solution, although I have to admit it's 'inspired' from another proble. The triangle ABC is acute, so O(the circumcenter) is inside the triangle. We denote by M, N, P the midpoints of the sides BC, CA, AB. Then O must be in one of the quadrilaterals BCNP, CAPM, ABMN. Let's assume it's in the interior of BCNP. Then 2R = BO+OC < BP+PN+CN = (AB+BC+AC)/2, so 4R<AB+BC+CA, and by dividing with 2R we get 2 < sinA+sinB+sinC.

The problem from which I 'borrowed' ( :D ) the idea is this: M is a point inside triangle ABC. Then MA+MB+MC+min(MA, MB, MC) < AB+BC+CA. It was proposed at a selection examination in Romania, and you can see that your problem is a particular case of this one.

Very nice solution,Grobber. This is inequality is the same with the following:

                                          $ a+b+c>4R$

Just wanted to note that the inequality in question was also shown by me in http://www.mathlinks.ro/Forum/viewtopic.php?t=5379 post #3.

  darij

I think here is a proof for it
   We know a well-known identity $\sum sin^2{A} = 2 + 2cosAcosBcosC$. Since $ABC$ is acute, we have$\sum sin^2{A} > 2$ (1). Also we have $2ab+2bc+2ca > a^2+b^2+c^2$, and this implies that $2sinAsinB + 2sinBsinC + 2sinCsinA > 2$ (2). By adding (1) and (2), the inequality we need to show follows  :)


Let $ ABCD$ be a convex quadrilateral. Prove that

$ (AB\plus{}CD)^2\plus{}(BC\plus{}DA)^2\geq (AC\plus{}BD)^2$.

I think that it's pretty easy with Ptolemy's theorem and inequality.


modular arithmetic
number theory proposed
number theory

Prove that there can't exist integers $ x$ and $ y$ such that $ 15x^{2} \minus{} 7y^{2}\equal{} 9$

[quote="ridgers"]Prove that there can't exist integers $ x$ and $ y$ such that $ 15x^{2} \minus{} 7y^{2} \equal{} 9$[/quote]

$ 3|RHS$ $ \implies$ $ 3|y$ $ \implies$ $ 9|7y^2\plus{}9$  $ \implies$ $ 9|15x^2$ $ \implies$ $ 3|x$.

So, let $ x\equal{}3u$ and $ y\equal{}3v$ and the equation becomes $ 15u^2\minus{}7v^2\equal{}1$

Then $ 15u^2\equal{}0\pmod 3$ and $ \minus{}7v^2\equal{}0$ or $ 2\pmod 3$ so $ LHS\equal{}0$ or $ 2\pmod 3$ while $ RHS\equal{}1\pmod 3$, so no solution.

[quote="pco"][quote="ridgers"]Prove that there can't exist integers $ x$ and $ y$ such that $ 15x^{2} \minus{} 7y^{2} \equal{} 9$[/quote]

$ 3|RHS$ $ \implies$ $ 3|y$ $ \implies$ $ 9|7y^2 \plus{} 9$  $ \implies$ $ 9|15x^2$ $ \implies$ $ 3|x$.

So, let $ x \equal{} 3u$ and $ y \equal{} 3v$ and the equation becomes $ 15u^2 \minus{} 7v^2 \equal{} 1$

Then $ 15u^2 \equal{} 0\pmod 3$ and $ \minus{} 7v^2 \equal{} 0$ or $ 2\pmod 3$ so $ LHS \equal{} 0$ or $ 2\pmod 3$ while $ RHS \equal{} 1\pmod 3$, so no solution.[/quote]

I solved almost like this but in the last part, here, $ 15u^2 \minus{} 7v^2 \equal{} 1$ I solved in another way.
If now we consider $ modulo 5$ then we get that $ 7v^{2} \equal{} 4\pmod 5$ so $ v^{2} \equal{} 2\pmod 5$ which is impossible.

[quote="ridgers"] If now we consider $ modulo 5$ then we get that $ 7v^{2} \equal{} 4\pmod 5$ so $ v^{2} \equal{} 2\pmod 5$ which is impossible.[/quote]

Quite OK too  :)


Let a sequences: $ a_1\equal{}0,|a_i|\equal{}|a_{i\minus{}1}\plus{}1|\forall i\equal{}1,...,n$. Prove that: $ a_1\plus{}a_2\plus{}...\plus{}a_n\geq \minus{}\frac{n}{2}$

Rewrite as $ a_{k\plus{}1}^2\equal{}(a_k\plus{}1)^2\equal{}a_k^2\plus{}2a_k\plus{}1$, so that
\[ a_1^2\plus{}2(a_1\plus{}a_2\plus{}\cdots \plus{}a_{n\minus{}1})\plus{}(n\minus{}1)\equal{}a_n^2\]
and using the fact that $ a_1\equal{}0$,
\[ a_1\plus{}a_2\plus{}\cdots \plus{}a_n\equal{}\frac{a_n^2\plus{}2a_n\minus{}(n\minus{}1)}{2}\]
so the problem reduces to $ (a_n\plus{}1)^2\geq 0$.


limit
trigonometry
search
real analysis
real analysis solved

V.I.ARNOL'D, a russian mathematician, wrote:
" We should not understimate the ingenuity of Newton's predecessors. Thus Huygens and Barrow could find the value of the limit:

\[
\mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\tan x} \right) - \tan \left( {\sin x} \right)}}{{\arcsin (\arctan x) - arctan(\arcsin x)}}
\]

instantly from geometrical considerations. There are a few contemporary mathematicians who could evaluate this limit within an hour. What do you think?

Where in the forum  can I find the solution Moubinool?

It seems Kent gave a solution,   sowhere in solved problem analysis.

He used Taylor expansion near zero

N the numerator $N(x)=-\frac{x^7}{30}+o(x^7)$

D the denominator $D(x)=-\frac{x^7}{30}+o(x^7)$

[quote]It seems Kent gave a solution, sowhere in solved problem analysis.[/quote]
I've been using the search feature on the site, trying to find it. jmerry had his own interesting take on it as well. But I don't think it's there. I'm starting to suspect it was in the week of the crash and is one of those topics which no longer exists.

V.I.ARNOL'D so calculated the limit:
Let $ \ f = sin(tanx); \ g = tan(sinx); \ g^{-1}= arcsin(arctanx);$
$\ f^{-1}=arctan(arcsinx)$; then
\[
\frac {f-g} {g^{-1}-f^{-1}}
\]
and via the Hospital rule we get:
\[{
\frac {f'-g'} {\displaystyle \frac {1} {g'}-\frac{1} {f'}}}= g' f' \to 1
\]


Is there a way to "do" AOPS vol. 1 or 2 at school, like work problems in a book in spare time, or something, without bringing the book to school?

Share ideas =D I seem to have spare time in a lot of instances where there is nothing else to do, and all work is done in several periods.

Any random ideas? =)

Lunch, Recess, on the bus, after homework is done, spring break, summer break, weekends.

Just stare at a blank piece of paper all day.
That's what I do.
:|

You could just write down or photocopy the problems you want to do that day... that's what I do anyway.

Get books smaller in size like CPIG or ACOPS (not really that much smaller but it actually avoids detection better than AoPS Vol.2, as I found out by having Vol.2 confiscated for 30 minutes by my math teacher) :D
Another solution is find one or two beastly problems, and it'll occupy you for a while.

I do it in math class because I'm allowed to :lol:

If your school has a free photocopier, make copies! They're flat and look like the worksheets that teachers tend to assign.


1.Show that if any two consecutive primes (greater than 2) are added, the sum can be expressed as a product of at least three prime numbers greater than 1; for example: 23+29=52, and 52=2x2x13.

2. Show that, except for 2 and 3, every prime number can be written in the form 6n+1 or 6n-1, where n is a natural number.

I think number two, has something to do with twin primes..........

2.
Analyzing all possibilities 6n,6n+1,6n+2,6n+3,6n+4,6n+5
we deduce that numbers 6n,6n+2,6n+3,6n+4 are not primes for n $\geq4$
because they are either multiples of 2 or of 3.
So every prime number $\geq4$  can be written in the form 6n+1 or 6n+5 (that is 6n-1)
where n is a natural number.

the problems already told you the answers

[hide="1"]suppose that $p$ and $q$ are consecutive primes such that $p>2$ and $q>2$. then they are both odd primes. $p+q$ is an even number and so it is divisible by 2. now note that $p<\frac{p+q}2<q$ and since $p,q$ are consecutive primes, $\frac{p+q}2$ is not prime and therefore it is composite that is a product of at least 2 primes. thus, $p+q$ is a product of at least 3 primes. easy[/hide]

[hide="2"]every positive integer is of one of the form $6n-2,6n-1,6n,6n+1,6n+2,6n+3$. from these forms we have
$6n-2=2(3n-1)$, $6n=2\cdot 3\cdot n$, $6n+2=2(3n+1)$, $6n+3=3(2n+1)$. therefore, every prime $>4$ is either $6n-1$ or $6n+1$. again easy[/hide]

G-UNIT: nah, it is almost related to twin primes (like the sum of twin primes greater than 3 is always divisible by 12) but no, these problems are solvable without knowing twin primes ;)


To show:
$\sum_{cyc}(b^{2}(a-b)^{2}+c^{2}(a-c)^{2}+2abc(b+c-a)) \ge 0$

I think we can use SOS but it seems weird: already almost that form but not obvious how to rearrange it into the right form. I wonder if there is some other more obvious way that doesn't require so much clever algebra? (but that I can't find because I am teh noob, that is)

em, unless i misunderstand your notation, your ineq seems to be
$(a^{2}+b^{2})(a-b)^{2}+(b^{2}+c^{2})(b-c)^{2}+(c^{2}+a^{2})(c-a)^{2}+2abc(a+b+c) \ge 0$?

Oh I guess it does! Wow I miss a big simplification by trying to work with just one term of the cyclic sum and forgetting there is such repetition...

ok but we want it true for all $a, b, c$ so how do we show the first three bits are bigger than the last one when some may be $<0$? Somehow it is still not obvious to me (you must take into account that I am not experienced with inequalities!) ;) Still my only solution is to expand everything and introduce some extra terms and re-factor it into SOS form...

Actually, it is easy. Obviously if something is 0, then it is true. So we can assume $a, b >0$, $c =-1$. Then expanding out, we get
$2a^{4}+2a^{3}+a^{2}+2b^{4}+2b^{3}+b^{2}+2a^{2}b^{2}+2ab+2 \ge 2a^{3}b+2a^{2}b+2ab^{2}+2ab^{3}$

Then noting by AM-GM
$a^{4}+b^{4}\ge a^{3}b+ab^{3}$
$a^{3}+b^{3}\ge a^{2}b+ab^{2}$

it is over.  :|

My other way was to smash it into $\sum_{cyc}(a+b-c)^{2}(a-b)^{2}$ ...


inequalities
limit
calculus
calculus computations

$ \sum_{n \equal{} 1}^{ \plus{} \infty}\frac {1}{(\sqrt {n \plus{} 1} \plus{} \sqrt n) \plus{} (\sqrt {n \plus{} 2} \plus{} \sqrt n)(\sqrt {n \plus{} 2} \plus{} \sqrt {n \plus{} 1})}$

$ \sum_{n \equal{} 1}^{\plus{}\infty}\frac{1}{(\sqrt{n\plus{}1}\plus{}\sqrt n)\plus{}(\sqrt{n\plus{}2}\plus{}\sqrt n)(\sqrt{n\plus{}2}\plus{}\sqrt{n\plus{}1})}  \geq
\sum_{n\equal{}1}^{\infty} \frac{1}{6 (n\plus{}2)} \equal{} \infty$

$ \sqrt{n\plus{}1}<\sqrt{n\plus{}n}\equal{}\sqrt{2n}<2\sqrt{n}$
So we get
$ \sqrt{n\plus{}1}<2\sqrt{n}$
This way we can show
01) $ \sqrt{n}<2\sqrt{n}$
02) $ \sqrt{n\plus{}1}<2\sqrt{n}$
03) $ \sqrt{n\plus{}2}<2\sqrt{n}$

So we get

$ (\sqrt{n\plus{}1}\plus{}\sqrt n)\plus{}(\sqrt{n\plus{}2}\plus{}\sqrt n)(\sqrt{n\plus{}2}\plus{}\sqrt{n\plus{}1})$
$ <(2\sqrt{n}\plus{}2\sqrt{n})\plus{}(2\sqrt{n}\plus{}2\sqrt{n})(2\sqrt{n}\plus{}2\sqrt{n})$
$ \equal{}4\sqrt{n}\plus{}(4\sqrt{n})(4\sqrt{n})\equal{}4\sqrt{n}\plus{}16n<20n$

So $ \frac{1}{(\sqrt{n\plus{}1}\plus{}\sqrt n)\plus{}(\sqrt{n\plus{}2}\plus{}\sqrt n)(\sqrt{n\plus{}2}\plus{}\sqrt{n\plus{}1})}> \frac{1}{20n}$

$ \sum_{n \equal{} 1}^{\plus{}\infty}\frac{1}{20n}\equal{}\plus{}\infty$

So $ \sum_{n \equal{} 1}^{\plus{}\infty}\frac{1}{(\sqrt{n\plus{}1}\plus{}\sqrt n)\plus{}(\sqrt{n\plus{}2}\plus{}\sqrt n)(\sqrt{n\plus{}2}\plus{}\sqrt{n\plus{}1})}\equal{}\plus{}\infty$

$ \frac{1}{(\sqrt{n\plus{}1}\plus{}\sqrt{n})\plus{}(\sqrt{n\plus{}2}\plus{}\sqrt{n})(\sqrt{n\plus{}2}\plus{}\sqrt{n\plus{}1})}$

$ \equal{}\frac{1}{\sqrt{n\plus{}1}\plus{}\sqrt{n}\plus{}n\plus{}2\plus{}\sqrt{\left( n\plus{}2\right) \left( n\plus{}1\right) }\plus{}\sqrt{n\left( n\plus{}2\right) }\plus{}\allowbreak \sqrt{n\left( n\plus{}1\right) }}$

$ >\frac{1}{\sqrt{(n\plus{}1)^{2}}\plus{}\sqrt{n^{2}}\plus{}n\plus{}2\plus{}\sqrt{\left( n\plus{}2\right) \left( n\plus{}2\right) }\plus{}\sqrt{(n\plus{}2)\left( n\plus{}2\right) }\plus{}\allowbreak \sqrt{(n\plus{}1)\left( n\plus{}1\right) }}$

$ \equal{}\frac{1}{6n\plus{}8}>\frac{1}{14n}$

Hence series diverge.

You don't really need to be that careful, and you don't really need inequalities that are valid for all $ n;$ large enough $ n$ would suffice. The terms $ a_n$ "look like" $ \frac{C}n$ in the sense that $ \lim_{n\to\infty}na_n$ exists and is some nonzero constant. We can see that just by look at it: the first term of the denominator looks like $ 2\sqrt{n},$ while the second term of the denominator looks like $ (2\sqrt{n})(2\sqrt{n})=4n$ for large $ n.$ The second term then dominates the first term, and in the long run we have that $  a_n$ looks like $ \frac1{4n}.$

From there, the claim that the series converges is an instance of the limit comparison test, comparing to the harmonic series.