quadratics
complex analysis
number theory
superior algebra
superior algebra unsolved

Hi

is there an elementary proof that -2 is a square mod p
when p is a prime 3 mod 8

I mean with elementary, that stuff like little fermat is allowed, but not
something like quadratic reciprocity

thx

At least, also the things you named are elementary (not using deeper algebra or complex analysis etc.) and why didn't you post it to the Number Theory section when you want such a solution¿
But the technique used in http://www.mathlinks.ro/Forum/viewtopic.php?t=23681&highlight= should work.

Thanks

sorry for the misplacement
I am also having trouble deciding whether something should go in college playground or advanced

however, I am familiar with that criterion, it is used in leveque to actually prove reciprocity

But for p=1 mod 8 there is a trick : make p=8*v+1

then (x^(4*v)+1)*(x^(4*v)-1)=0 holds for all nonzero x

so there are x for which x^(4*v)+1=0 holds

or with x^(v)=y

(y+1/y)^2=2

is something like this possible here?


algebra
polynomial
induction
number theory proposed
number theory

Let $a_1,...,a_k$ algebraic integers and $S_n=a_1^n+...+a_k^n$. Prove that $n$ divides $\sum_{d|n}{ S_d f(\frac{n}{d})}$ where $f$ is the Moebius function. This generalizes many classical theorems, so it is of interest, I think.

so your statement means that $a_1 a_2...a_k$ are the roots of a monic integer polynomial? (all of the same polynomial?????)

Harazi if i am not wrong $p^t|a(a^{\phi(p^t}-1)$ also holds for the ring of algebraic integers.....(btw i cant remember the proof of this fact) using this property the problem becomes very easy!

Mayn I see how it is so easy, Pascual? :?

Honestly, I don't have a clue what you actually mean, Pascual  :huh:

i will ppost it this afternoon, i promess...but maybe i am a completely fool and it is just nonsense :(

Well, after you post a complete solution probably we'll see.  ;)

well I dont now if this is true but i am almost sure that $p^t|a^{p^t}-a^{p^{t-1}}$ int he ring of algebriac integers... So from here what can we get? just wathc separately the sum as $\sum_{d|n}a^df(n/d)$ now iff $p^t|n$ and $p^{t-1}|d$ we have that $a^d$ and $a^{dp}$ are both with coefficient $0$ or with different sign. but from the lemma $p^t|a^{dp}-a^{d}$ so n divides our sum in the ring of algebraic integers. Finally if $a_1, a_2...,a_k$ are associates, aour sum is a integer and therefore it is an integer multiple of $p$ since non integer rationals are not algebraic integers. (i dont know if i am just saying true facts or all of them are just an abuse of properties for integers) :blush:

I'm afraid my knowledge of Number Theory is not deep enough to judge this proof :blush:

mine neither, thats why i am not going to try and convince you it is true...

wella ctually now i am starting to believe my result is a fake and that the only useful thing are ideas as in this problem! [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=5452[/url]

I am sorry, I am dumb, but I really don't understand your solution. :(

ok sorry:

1. Lets assume what i say is in fact true, Namely for an algebraic integer $a$, there exist an algebraic integer $b$ such that $a^{p^t}-a^{p^{t-1}}=p^tb$ (this porperty holds if it holds for $t=1$ by induction on $t$).

If this property is true then we can keep going, otherwise what follows only works for integers.... ;)

2. we have: $\sum_{d|n}S_df(n/d)=\sum_{i=1}^{k}\sum_{d|n}a_i^df(n/d)$ so we will prove $n|\sum_{d|n}a_i^df(n/d)$ (in the field of algebraic integers, i mean the quotient is an algebraic integer).

3. The sum we want to analize has many cero terms, so we only deal with noncero terms, if $n=p_1^{a_1}p_2^{a_2}....p_l^{a_l}$ then $f(n/d)$ is different from $0$ only iff $n/d$ is square free... so $d$ must be divisible by $p_i^{a_i-1}$ for every prime divisor of $n$, call those $d$ "special divisors"

4. Now we prove $p_1^{a_1}$ divides our sum. Group the special divisors in the following way: take a special divisor $d$ such that $p_1^{a_1}$ doesnt divide it and group it with $dp$ it is clear that $f(n/d)=-f(n/dp)$. But also all termas are paired and by the property in (1) We have that all those pairs are multiples of $p^t$ since they are of the form $a^{dp}-a^d=a'^{p^t}-a'^{p^{t-1}}$. So their sum is also divisible by $p^t$ (in the RING of algebraic integers).

5. Finally we also prove $p_i^{a_i}$ divides the sum, and therefore $n$ divides it in the ring of algebraic integers...

6. Now if all of the algebraic integers given are  associates (Same minimal polynomial) then our sum is an integer, and therefore $n$ would divide it in the integers, since none of the quotients is rational! (rationals are not algebraic integers!)

Final Note: well if our algebraic integers are asociates, then we can also consider their minimal  polynomial $P(x)=x^m-\sum x^tb_t$ with $b_t$ integers. and we get $S_{m+h}=\sum S_{t+h}b_t$ so the sequence $S_n$ satisfies a linear recursion. So it is clear that for a prime power  not dividing the product of the roots, this sequence is purely periodic mod $p^t$, so we need somehow that $S_{t+n}-S_n$ to be a multiple of $p^t$ for a fixed $t$ and all $n$. We can also reduce $P(x)$ modulo $p$ and work with it or something like that....my guess is that $t=\phi(p^t)$ wich would also give a similar solution grouping $S_{dp}-S_d$...

Yes it is right!!!!! So we get that $a^p-a=pb$. Now suppose $a^{p^t}-a^{p^{t-1}}=p^tb$ then we also have:
$a^{p^t}=a^{p^{t-1}}+p^tb$ and raising both sides to the p-th power we get:

$a^{p^{t+1}}=a^{p^{t}}+Kp^{t+1}$ because $p|\binom{p}{k}$. So property $1$ follows!

Edit: a post by Harazi dissapeared where he sayed that $(1+a-1)^p-(1+a-1)$ is a multiple of $p$ because all multinomial coefficients are, so from here it follows that $p|a^p-a$, Nice!

No, I realized immeidately the mistake. It remains the number $\frac{(1+a-1)^p-(1+a-1)}{p}$. How can we guarantee that it is algebraic integer? Meanwhile, I think your proof is wrong, since the number $\frac{(1+\sqrt{2})^3-(1+\sqrt{2})}{3}$ is not algebraic integer. Or have I completely lost my mind?

It is not difficult to see that we only need to consider $n$ being a prime power. Suppose for example that we have $n = p^2 q^3$, then we need to show $n | S_{p^2q^3} - S_{p^2q^2} - S_{pq^3} + S_{pq^2}$. But $S_{p^2q^3} - S_{p^2q^2}$ and $S_{pq^3} - S_{pq^2}$ are both divisible by $q^3$ by considering the prime power cases $n' = q^3, a_i' = a_i^{p^2}$ and $a_i'' = a_i^{p}$. Thus $q^3 | \sum$ and similarly $p^2 | \sum$ once we prove the prime power cases.

Now let $n = p^m$ be a prime power, the problem becomes $p^m | \sum_{i = 1}^{k} (a_i^{p^m} - a_i^{p^{m-1}})$. We use induction on $k$ to show something stronger, that for any $k$-variable symmetric integer polynomial $F$, it holds that $p^m | F(a_1^{p^m}, \dots, a_k^{p^m}) - F(a_1^{p^{m-1}}, \dots, a_k^{p^{m-1}}).$ The original problem corresponds to $F(x_1, x_2, \dots, x_k) = x_1 + x_2 + \dots + x_k$. Now recall a well-known lemma:
[b]Lemma.[/b] Suppose $F$ is a symmetric integer polynomial, then $F(x_1^p, x_2^p, \dots, x_k^p) = F(x_1, x_2, \dots, x_k)^p + p \cdot G(x_1, x_2, \dots, x_k)$ with $G$ also a symmetric integer polynomial.

The proof of this lemma is simple: just expand  $F(x_1, x_2, \dots, x_k)^p$ and notice that all but few multinomial coefficients are divisible by $p$.

Going back to our claim, the base case $m = 1$ is immediately settled by the above lemma because $F(a_1^p, \dots, a_k^p) - F(a_1, \dots, a_k)^p = p \cdot G(a_1, \dots, a_k)$. Since $G(a_1, \dots, a_k)$, $F(a_1, \dots, a_k)$ and $F(a_1^p, \dots, a_k^p)$ are integers (by the fundamental theorem of symmetric polynomials), we deduce by Fermat's little theorem that $F(a_1^p, \dots, a_k^p) \equiv F(a_1,\dots, a_k)^p \equiv F(a_1, \dots, a_k) ~(mod~p)$. Assume we have proved the result for $m$, then by lemma we can write
$$F(a_1^{p^{m+1}}, \dots, a_k^{p^{m+1}}) - F(a_1^{p^{m}}, \dots, a_k^{p^{m}}) = \left[F(a_1^{p^m}, \dots, a_k^{p^m})^p - F(a_1^{p^{m-1}}, \dots, a_k^{p^{m-1}})^p\right] + p \cdot \left[ G(a_1^{p^m}, \dots, a_k^{p^m}) - G(a_1^{p^{m-1}}, \dots, a_k^{p^{m-1}})\right].$$
By induction hypothesis, $F(a_1^{p^m}, \dots, a_k^{p^m}) \equiv F(a_1^{p^{m-1}}, \dots, a_k^{p^{m-1}}) ~(mod~p^m)$. Thus we routinely have $F(a_1^{p^m}, \dots, a_k^{p^m})^p \equiv F(a_1^{p^{m-1}}, \dots, a_k^{p^{m-1}})^p ~(mod~p^{m+1})$. Also by induction hypothesis we have $p^{m+1} | p \cdot \left[ G(a_1^{p^m}, \dots, a_k^{p^m}) - G(a_1^{p^{m-1}}, \dots, a_k^{p^{m-1}})\right]$. The induction step is thus completed.

[hide=Maybe]It might not hurt to consider Dirichlet Convolution with the mobius function and a function giving the $n$th newton sum of a polynomial. [/hide]


conics
ellipse
analytic geometry
number theory solved
number theory

here is what you can find useful!! :D

http://www-gap.dcs.st-and.ac.uk/~history/HistTopics/Pell.html

cheers!

[quote="J"]The problem is: Find all pairs of positive rational numbers such that x^2 + 3y^2 = 1.

What I did was, I put x = a/b and y = c/d, got b = d = 2k and
a^2 + 3c^2 = 4k^2 where gcd(a,c) = gcd(k,c) = gcd(a,k) = 1

Maybe I'm in the wrong way or maybe I'm just missing something after a hard schoolweek, but I have no idea how to go furter, maybe some hints ? (I'm not expecting a full solutions, It would be great if someone could give me just a helping hint)[/quote]
=============
We can use technics of "cuts" to solve this problem. First, we reform the problem to: Find all rational points on the ellipse (E): x^2 + 3y^2=1 (1). Obviously, A(1,0) is one solution. Now, we take arbitrary point B (0, r) in Oy and find the second intersection point of (AB) with. By Viet theorem, coordinates of this point will be rational and we get the solution of (1). Inversely, if we have rational point C on (E) then AC meets Oy at rational poin. Answer: x = (3r^2-1)/(3r^2+1)
                     y = 2r/(3r^2+1)
where r is arbitrary rational number.

Have joy.

Namdung


inequalities
geometry
circumcircle
geometric inequality
inequalities proposed

Hello guys I'm stuck in this geometric inequality , I posted this problem in the Highschool section but seems like nobody can prove it geometrically , so I hope that someone here give another type of solution , here it is

[i]Show that in any acute-angled triangle there are two sides whose sum is always greater than the sum of the diameters of the incircle and circumcircle . [/i]

This means that we need to prove that one of these statements is an assertion:

$ a \plus{} c > 2(R \plus{} r) \ , \ b \plus{} c > 2(R \plus{} r)$ or  $ a \plus{} b > 2(R \plus{} r)$

hello, i think, this means
$ a\plus{}c>2(R\plus{}r)$ or $ b\plus{}c>2(R\plus{}r)$ or $ a\plus{}b>2(R\plus{}r)$.
Sonnhard.

I am really stuck in this problem :(   could someone give a hint or something ?


inequalities
geometry
circumcircle
incenter
inequalities proposed

Let a,b,c are positive numbers. Prove that
$ 4a^2b^2c^2\geq (a^3\plus{}b^3\plus{}c^3\plus{}abc)(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)$ :D

we can assume that $ a,b,c$ are the side lengths of a triangle (otherwise RHS would be negative and LHS would be positive so the inequality becomes trivial),so let $ a,b,c$ be the sidelength of triangle $ \triangle ABC$,and let $ p,r,R$ be the semiperimeter,inradus and circumradius of this triangle respectively,now we know that:

$ abc\equal{}4Rrp,a^3\plus{}b^3\plus{}c^3\equal{}2p(p^2\minus{}3r^2\minus{}6Rr),(p\minus{}a)(p\minus{}b)(p\minus{}c)\equal{}pr^2$

so after simplifying,our inequality will be equivalent to:

$ 4R^2\plus{}4Rr\plus{}3r^2\geq p^2$

which is equivalent to:

$ \overline{HI}^2\geq 0$

which is obviously true...

(note that $ H,I$ are the orthocenter and incenter of $ \triangle ABC$ respectively)

[quote="BaBaK Ghalebi"]we can assume that $ a,b,c$ are the side lengths of a triangle (otherwise RHS would be negative and LHS would be positive so the inequality becomes trivial),so let $ a,b,c$ be the sidelength of triangle $ \triangle ABC$,and let $ p,r,R$ be the semiperimeter,inradus and circumradius of this triangle respectively,now we know that:

$ abc \equal{} 4Rrp,a^3 \plus{} b^3 \plus{} c^3 \equal{} 2p(p^2 \minus{} 3r^2 \minus{} 6Rr),(p \minus{} a)(p \minus{} b)(p \minus{} c) \equal{} pr^2$

so after simplifying,our inequality will be equivalent to:

$ 4R^2 \plus{} 4Rr \plus{} 3r^2\geq p^2$

which is equivalent to:

$ \overline{HI}^2\geq 0$

which is obviously true...

(note that $ H,I$ are the orthocenter and incenter of $ \triangle ABC$ respectively)[/quote]

$ (a\plus{}b\minus{}c)(a\plus{}c\minus{}b)(b\plus{}c\minus{}a)>0$  <-> $ a\plus{}b\minus{}c>0,a\plus{}c\minus{}b<0,b\plus{}c\minus{}a<0$? --> a,b,c are triangles sides just mustn't be

@zaya_yc:
note that $ a,b,c$ are positive,now if we had $ a \plus{} c \minus{} b < 0$ and $ b \plus{} c \minus{} a < 0$ then by adding these two inequalities we get that $ c < 0$ which is a contradiction...

so we must have $ a \plus{} b \minus{} c > 0,b \plus{} c \minus{} a > 0,c \plus{} a \minus{} b > 0$ i.e. $ a,b,c$ are the side-lengths of a triangle


Show that for all $p \in \mathbb{P}$, there exists $n \in \mathbb{N}$ such that : $6n^2 + 5n + 1 \equiv 0 \pmod p$.

:)

[hide="One-line solution"]
$6n^2 + 5n + 1 = (3n + 1)(2n + 1)$ - the latter clearly cycles through all odd primes and $3n+1$ is even at least once.  :)
[/hide]


Let the operation $ f$ of $ k$ variables defined on the set $ \{ 1,2,\ldots,n \}$ be called $ \textit{friendly}$ toward the binary relation $ \rho$ defined on the same set if \[ f(a_1,a_2,\ldots,a_k) \;\rho\  \;f(b_1,b_2,\ldots,b_k)\] implies $ a_i \; \rho \ b_i$ for at least one $ i,1\leq i \leq k$. Show that if the operation $ f$ is friendly toward the relations "equal to" and "less than," then it is friendly toward all binary relations.

[i]B. Csakany[/i]

The case $n = 1$ is trivial.

We first look at the case $n \geq 3$. I will show by induction on $k$ that, if $f$ is friendly toward "equal to" and "less than", then there exists $i \in \{1, \dots, k\}$ such that $f(a_1, \dots, a_k) = a_i$ for any $a_1, \dots, a_k$.

For $k = 1$, we have a function $f$ from $\{ 1, \dots, n\}$ to itself, which satisfies $f(a) = f(b) \implies a = b$ (i.e. $f$ is injective) and $f(a) < f(b) \implies a < b$ (i.e. $a \geq b \implies f(a) \geq f(b)$, that is, $f$ is non-decreasing). It follows that $f(a) = a$ for all $a \in \{1, \dots, n\}$.

Now suppose that it's true for $k - 1$ and let's prove the result for $k$.
We define $g:(a_1, \dots, a_{k - 1})\mapsto f(a_1, \dots, a_{k - 1}, a_{k - 1})$. It is clear from the definitions that $g$ is friendly toward "equal to" and "less than". Thus by induction hypothesis, there exists $i \in \{1, \dots, k - 1\}$ such that $g(a_1, \dots, a_{k - 1}) = a_i$.

If $i \neq k - 1$, then I claim that $f(a_1, \dots, a_k) = a_i$ for any $a_1, \dots, a_k$.
In fact, for any $a_1, \dots, a_k$ and any $x \neq a_i$, we may choose $b_1, \dots, b_{k - 1}$ such that
- $b_i = x$;
- $b_j \neq a_j$ for any $1 \leq j < k - 1$;
- $b_{k - 1} \neq a_{k - 1}$ or $a_k$ (notice that $n \geq 3$).
It follows that $f(a_1, \dots, a_k) \neq f(b_1, \dots, b_{k - 1}, b_{k - 1}) = b_i = x$. Thus the only possibility is $f(a_1, \dots, a_k) = a_i$.

If $i = k - 1$, then the same argument as above shows that, for any $a_1, \dots, a_k$, we have $f(a_1, \dots, a_k) = a_{k - 1}$ or $a_k$ (for any $x\neq a_{k - 1}$ or $a_k$, choose $b_1, \dots, b_{k - 2}$ such that $b_j \neq a_j$ and consider $f(b_1, \dots, b_{k - 2}, x, x)$).
Now consider the two numbers $f(1, \dots, 1, 1, 2)$ and $f(2, \dots, 2, 2, 1)$. They must be different and both are equal to $1$ or $2$. Therefore one of them is equal to $1$.
Without loss of generality, we assume in the following that $f(1, \dots, 1, 1, 2) = 1$.

For any $a_1, \dots, a_{k - 2}$, we choose $b_1, \dots, b_{k - 2}$ such that $a_j \neq b_j$ and $b_j \neq 1$ (which is possible because $n \geq 3$).
It follows that $f(b_1, \dots, b_{k - 2}, 2, 1) = 2$ (because it is not equal to $f(1, \dots, 1, 1, 2) = 1$)  and hence $f(a_1, \dots, a_{k - 2}, 1, 2) = 1$ (because it is not equal to $f(b_1, \dots, b_{k - 2}, 2, 1) = 2$).

For any $a_1, \dots, a_{k - 1}$ such that $a_{k - 1} \neq 1$, we choose $b_1, \dots, b_{k - 2}$ such that $a_j \neq b_j$.
It follows that $f(a_1, \dots, a_{k - 1}, 1) = a_{k - 1}$, because it is not equal to $f(b_1, \dots, b_{k - 2}, 1, 2) = 1$.
This is also true for $a_{k - 1} = 1$, hence we have $f(a_1, \dots, a_{k - 1}, 1) = a_{k - 1}$ for any $a_1, \dots, a_{k - 1}$.

Finally, for any $a_1, \dots, a_k$ such that $a_k \neq 1$ or $a_{k - 1}$, we choose $b_1, \dots, b_{k - 2}$ such that $a_j \neq b_j$.
It follows that $f(a_1, \dots, a_k) = a_{k - 1}$, because it is not equal to $f(b_1, \dots, b_{k - 2}, a_k, 1) = a_k$.
This is also true for $a_k = 1$ or $a_{k - 1}$, hence we have shown that $f(a_1, \dots, a_k) = a_{k - 1}$ for any $a_1, \dots, a_k$.

-----
It remains to treat the case $n = 2$.

We view a relation $\rho$ as a subset of $\{1, 2\}^2$.
Note that, if $f$ is friendly toward two relations $\rho_1, \rho_2$, then $f$ is also friendly toward the union $\rho_1\cup \rho_2$.
Since every subset of $\{1, 2\}^2$ is a union of singletons, it suffices to prove that $f$ is friendly toward each singleton relation.

There are only four singleton relations:
- $\rho_1 = \{(1, 1)\}$, i.e. $a\textrm{ } \rho_1\textrm{ } b \iff a = b = 1$;
- $\rho_2 = \{(2, 2)\}$, i.e. $a\textrm{ } \rho_2\textrm{ } b \iff a = b = 2$;
- $\rho_3 = \{(1, 2)\}$, i.e. $a\textrm{ } \rho_3\textrm{ } b \iff a < b$;
- $\rho_4 = \{(2, 1)\}$, i.e. $a \textrm{ }\rho_4 \textrm{ }b \iff a > b$.
By assumption, $f$ is friendly toward $\rho_3$, and this implies that $f$ is also friendly toward $\rho_4$.

It remains to show that $f$ is friendly toward $\rho_1, \rho_2$. We prove it for $\rho_1$, the other proof being identical.

Since $f$ is friendly toward "equal to", for any $a_1, \dots, a_k \in \{1, 2\}$, we have $f(3 - a_1, \dots, 3 - a_k) = 3 - f(a_1, \dots, a_k)$.
Thus if $f(a_1, \dots, a_k)\textrm{ } \rho_1\textrm{ } f(b_1, \dots, b_k)$, then $$f(a_1, \dots, a_k) = 1 < 2 = f(3 - b_1, \dots, 3 - b_k)$$ which implies that there exists $i$ such that $a_i < 3 - b_i$.
This is only possible when $a_i = 1$ and $3 - b_i = 2$, or in other words, $a_i\textrm{ } \rho_1\textrm{ } b_i$.


[quote="cincodemayo5590"]We have
\[\sum_{k=0}^\infty \frac{1}{(3k+1)(3k+2)(3k+3)}= \sum_{k=0}^\infty \left( \frac{1}{6k+2}+\frac{1}{6k+6}-\frac{1}{3k+2}\right) \equiv \sum_{k=0}^\infty \left( \frac{1}{6k+6}-\frac{1}{6k+5}\right) \]
Now what?[/quote]
No, this is not correct.  Your result says that the original sum is negative. :wink:

Also, I think I have already posted a solution to this problem a while ago.

[quote="boxedexe"]
No, this is not correct.  Your result says that the original sum is negative. :wink:[/quote]
Oh, alright. I didn't notice that.  :)

But still, all reciprocals of the numbers that are $\equiv 2 \bmod 6$ will eventually be canceled out by the subtracted terms. I guess it matters "when" this happens, i.e. at what summation index.

So, is there an easy way to compute the value of this summation?

You can do it with calculus...it gets kind of messy.

[hide="calculus..."]
Let $f(x)=\sum_{k=0}^\infty \frac{x^{3k+3}}{(3k+1)(3k+2)(3k+3)}$. Observe that
\[f^{(3)}(x)=\sum_{k=0}^\infty x^{3k}=\frac{1}{1-x^{3}}\]
This means our sum is equal to
\[\int_{0}^{1}\int_{0}^{z}\int_{0}^{y}\frac{1}{1-x^{3}}\ dx \ dy \ dz\]
To evaluate this triple integral, use partial fractions (as $(1-x)(1+x+x^{2})=1-x^{3}$) to express $\int_{0}^{y}\frac{1}{1-x^{3}}\ dx$ as a linear combination of $\ln (1-y)$, $\ln (1+y+y^{2})$, and $\arctan \frac{2y+1}{\sqrt{3}}$. And that's just the first of three integrals! The others require several integrations by parts, and that's if $\ln (1+y+y^{2})$ even has nice antiderivatives. If it does, expect answers containing $\pi$ and $\ln 3$. Good luck... :rotfl:
[/hide]

[quote="scorpius119"]The others require several integrations by parts, and that's if $\ln (1+y+y^{2})$ even has nice antiderivatives. [/quote]

[hide="Fortunately..."] $\ln (y^{2}+y+1) = \ln (y-\omega)+\ln (y-\omega^{2})$ [/hide]


for a, b integers such that a+b=1, show that (a+1/a)^2+(b+1/b)^2 is > or equal to 25/2.

Maybe this problem is limited to $a>0,\ b>0.$

From $a>0,b>0,a+b=1$, using A.M.-G.M. inequality $2\sqrt{ab}\leq a+b=1\Longleftrightarrow \frac{1}{ab}\geq 4$

Thus since $y=x^{2}$ is concavedown, using Jensen's inequality

${\Longleftrightarrow \frac{\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}}{2}\geq \left(\frac{a+\frac{1}{a}+b+\frac{1}{b}}{2}\right)^{2}\Longleftrightarrow \left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}}\geq \frac{1}{2}\left(a+b+\frac{a+b}{ab}\right)^{2}$

$=\frac{1}{2}\left(1+\frac{1}{ab}\right)^{2}\geq \frac{1}{2}(1+4)^{2}=\frac{25}{2}$  the equality holds when $a=b\ and\ a+b=1\Longleftrightarrow a=b=\frac{1}{2}.$

[quote]for a, b [b]integers[/b] such that a+b=1[/quote]

But then again, your interpretation seems to be better, and at least it gives a correct problem.  :)

[color=darkblue]If $p=ab\ne 0$, then the relation $(*)\ \left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}\ge \frac{25}{2}$ $\Longleftrightarrow$ $a^{2}+b^{2}+\frac{1}{a^{2}}+\frac{1}{b^{2}}\ge \frac{17}{2}$ $\Longleftrightarrow$ $(a^{2}+b^{2})\left(1+\frac{1}{a^{2}b^{2}}\right)\ge \frac{17}{2}$ $\Longleftrightarrow$ $2[1-2(ab)]\left[(ab)^{2}+1\right ]\ge 17(ab)^{2}$ $\Longleftrightarrow$ $2(1-2p)(p^{2}+1)\ge 17p^{2}$ $\Longleftrightarrow$ $4p^{3}+15p^{2}+4p-2\le 0$ $\Longleftrightarrow$ $(4p-1)(p^{2}+4p+2)\le 0$ $\Longleftrightarrow$ $p=ab\in \left(-\infty\ ,\--2-\sqrt 2\right]\cup\left[-2+\sqrt 2\ ,\ 0\right)\cup\left(0\ ,\ \frac{1}{4}\right]$, i.e. $0\ne p\le\frac{1}{4}$ and $|p+2|\ge \sqrt 2\ .$
If $a>0\ ,\ b>0$ and $a+b=1$, then $p>0$ and $4p-1=4ab-(a+b)^{2}=-(a-b)^{2}\le 0$, i.e. $p\in \left(0,\ \frac{1}{4}\right]$ and the inequality $(*)$ is truly.[/color]

This was on the recent issue of Mathematical Horizons...

Well, specifically, it was prove $\sqrt{\sin^{2}{\theta}+\frac{1}{\sin^{2}{\theta}}}+\sqrt{\cos^{2}{\theta}+\frac{1}{\cos^{2}{\theta}}}\ge \sqrt{10}$.

Math Horizons is a magazine.  Sorry, its actual name is Math Horizons, not mathematical.


Please note that the Music & Art forum is intended for the musicians and artists among us to discuss techniques and their work. If you want to talk about your favorite band, please do it in Fun & Games.


geometry
rectangle
analytic geometry
ceiling function
function
number theory
prime factorization

Once again, for those who are curious (me), interested (me), or just wondering if they did it right (me)  :D
[i][b]
10.  Asaf is investigating rectangles drawn in the plane such that every vertex has integer coordinates. (The sides of the rectangle need not be parallel to the coordinate axes.) He considers two rectangles to be the same if one can be obtained by rotating and/or translating the other.

   1. How many distinct rectangles can Asaf draw with area 100?
   2. How many distinct rectangles can Asaf draw with a given integer area A? (See Note 3.)[/b][/i]

The formula I used is the well-known theory to find the number of factors for any given number $ N$, where the prime factors of $ N$ are $ N \equal{} A^a*B^b*C^c...$ with A, B, C... being unique prime factors and a, b, c... being the powers that they're raised to, there are $ (a \plus{} 1)*(b \plus{} 1)*(c \plus{} 1)...$ factors of $ N$.

After finding the number of factors, I divided it by two (to make pairs for the sides of the rectangles), and if there was an odd number of factors (if the number is a square), I bumped it up to the nearest integer.
Thus I was able to get $ 13$ rectangles for (1)  :D

Having seen the questions for the first time right now, - Edit - hmm I misread that you had $ N^2$ .. but don't you have to show that the side lengths are achievable, eg sum of two squares, at least for (2)?

Not returning  :(

Without loss of generality, let one vertex be at the origin. Let the other points be $ (a,b)$, and $ (c,d)$ where $ a,b,c,d$ are integers and $ a,b\ge 0$ (we can do this by rotating either $ 0$, $ 90$, $ 180$, or $ 270$ degrees. Clearly these conditions force the last vertex to be on a lattice point as well.

We have that $ |ad\minus{}bc|\equal{}A$ and $ ac\plus{}bd\equal{}0$.

$ ac\equal{}\minus{}bd\implies \frac{a}{d}\equal{}\minus{}\frac{b}{c}\equal{}\frac{m}{n}$ where $ (m,n)\equal{}1$. Then $ a\equal{}km$, $ d\equal{}kn$, $ b\equal{}lm$, $ c\equal{}\minus{}ln$ where $ k,l$ are integers.

(Also cover when $ b,c\equal{}0$, etc)

Then $ |ad\minus{}bc|\equal{}mn|k^2\plus{}l^2|\equal{}A$

Then I don't know about representation in terms of sum of squares. Part a is easy to do from here.

I didn't know about representation as sum of squares either. But the description said there was a fact from advanced number theory, so I figured that must be it, so I looked up "sum of squares" on http://www.mathworld.com, and found the "sum of squares function."

I don't want to paste my solution here unless prompted, because it's rather long, and I'll have to retype the bajillions of times I used Equation Editor. Instead, I'll post an image of the final formula, in all its ugly, difficult-or-impossible-to-simplify glory:

[img]http://img261.imageshack.us/img261/2330/mathcampquiz0810bansho6.jpg[/img]

Edit: Oh, I got 10 for part a, and I'm very certain I'm right, because I enumerated all the possibilities, along with the representation of the square of the smaller cofactor as a sum of squares (the larger one represented as a sum of squares is just a multiplication of the smaller one):

[img]http://img152.imageshack.us/img152/2928/mathcampquiz0810aanspd3.jpg[/img]

You did not have these cases:

$ 2\sqrt2\times25\sqrt2$, $ \left(2\sqrt2\right)^2\equal{}2^2\plus{}2^2$, $ \left(25\sqrt2\right)^2\equal{}25^2\plus{}25^2$
$ 4\sqrt5\times5\sqrt5$, $ \left(4\sqrt5\right)^2\equal{}4^2\plus{}8^2$, $ \left(5\sqrt5\right)^2\equal{}5^2\plus{}10^2$
$ 5\sqrt5\times4\sqrt5$, $ \left(5\sqrt5\right)^2\equal{}5^2\plus{}10^2$, $ \left(4\sqrt5\right)^2\equal{}4^2\plus{}8^2$

Oops, I forgot to mention that you have to apply the formula to $ 100^2$ (10,000).
I don't know why, but I'm guessing it has to do with your sum of the squares theory.

From mnmath's presentation of the sum of the squares', I now understand the idea behind it. It looks like the cases for each individual rectangle are the different are the different ways you can split the prime factors of 10,000.

Example: Above 3 cases. For the first, the squares are (2*2) + (2*2) and (25*25) + (25*25). Which makes up the 2^4*5^4 prime factorization of 10,000.

Hmm, it seems I just stumbled on the final formula to find the number of rectangles by chance, while having no idea of the theory behind it  :wink: Oh well. Go 13!

http://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares

The proof of this theorem that uses Gaussian integers is one of my favorite proofs learned at HCSSiM.

[quote="mnmath"]
$ 4\sqrt5\times5\sqrt5$, $ \left(4\sqrt5\right)^2 \equal{} 4^2 \plus{} 8^2$, $ \left(5\sqrt5\right)^2 \equal{} 5^2 \plus{} 10^2$
$ 5\sqrt5\times4\sqrt5$, $ \left(5\sqrt5\right)^2 \equal{} 5^2 \plus{} 10^2$, $ \left(4\sqrt5\right)^2 \equal{} 4^2 \plus{} 8^2$[/quote]

Those are the same case...

Hmm, I didn't notice that!

The last one is $ \sqrt{40}*\sqrt{250}$, or $ 2\sqrt{10}*5\sqrt{10}$.

Glad that's done with :P

My answer is that the number of squares is

$ \sum_{x\in \mathbf{B}_A}{\left\lceil \frac{\tau\left(\frac{A}{x}\right)}{2}\right\rceil}$

where $ \tau (n)$ is the divisor function, and

$ \mathbf{B}_A\equal{}\{x: x\equal{}1 \mbox{ or } x \mbox{ is squarefree and is a }\\\mbox{product of primes 2 and/or of the form } 4k\plus{}1,\mbox{ and } x|A, k,x\in\mathbf{N}\}$.

Plugging in $ A\equal{}100$, we get 13 for part a).


Let $z_1,z_2$ be two complex numbers satisifying: $z_1+z_2=2i$, $|z_1||z_2|=3$. Let $u=\frac{z_1}{z_2}$, $x=\arg u$. Find $|u|$ when $\cos x$ get maximum. Besides, what is the maximum value of $\cos x$?

does the x play any role in the problem?

EDIT: nevermind, i just read to the first part of the question

[quote="ifai"]No one want to try?[/quote]
...


Medians $BD$ and $CE$ of triangle $ABC$ are perpendicular, $BD=8$, and $CE=12$. The area of triangle $ABC$ is

[b](A)[/b] $24$ [b](B)[/b] $32$ [b](C)[/b] $48$ [b](D)[/b] $64$ [b](E)[/b] $96$

ATTACHMENT

[hide]since the diags are perpendicular, area of BCDE=1/2*8*12, and by similar triangles, BCDE is 3/4 of the whole triangle, so our answer is 4/3*1/2*8*12=64 or D[/hide]


geometry
calculus
trigonometry
inequalities
rectangle
circumcircle
triangle inequality

Solve, without Calculus, for the maximum possible area of a quadrilateral $ABCD$ in which $AB = BC = 3$ and $CD = 7$.

Okay, it's clear that we must have

$AC\perp CD$.  (That hint was free  :P )

[hide]Drop the perpendicular $P$ from $B$ and let $\angle PBC=t$

$K=\triangle_{ABC}+\triangle_{ACD}$
$K=2\cdot\frac{1}{2}(3\cos t)(3\sin t)+\frac {1}{2}\cdot 2 \cdot (3\sin t)\cdot 7$

So we need to maximize $9\cos t\sin t+21\sin t$

After some work, we get that $t=16\sqrt 2$, and yes, I may have used calculus to find that[/hide]

Pure geometry + AM-GM :)
Let $x=AD$. We know that area of quadrilateral with sidea $a$, $b$, $c$ and $d$ is maximal iff quadrilateral is inscribed (it is always possible to find inscribed quadrilateral with sides $a$, $b$, $c$ and $d$, if they satisfy triangle inequality, of course :) ).
Using Heron formula for inscribed quadrilateral, we obtain that
\[S^2=(p-a)(p-b)(p-c)(p-d)=\frac{1}{16}(x+7)^2(x-1)(13-x)=\]
\[=2\cdot\frac{x+7}{4}\cdot\frac{x+7}{4}\cdot\frac{x-1}{2}\cdot(13-x)\leq 2\cdot \left(\frac{\frac{x+7}{4}+\frac{x+7}{4}+\frac{x-1}{2}+(13-x)}{4}\right)^4=2\cdot 4^4\]
and equality holds iff $x=9$.
Therefore $S_{\max}=16\sqrt{2}$.

To complete my solution without calculus, can anyone prove that:

$9\sin t\cos t+21\sin t\leq 16\sqrt {2}$?

Equality occurs when $\cos t=\frac {1}{3}$ and $\sin t=\frac {2\sqrt {2}}{3}$

I was first trying to work pure geometrically. I simplify it a lot, but then, when it comes to find maximal area of one rectangle the same formula as blahblahblah wrote appeared to me, but I also can't find it's maximum without calculus.

Still no-one has come up with a completely geometric solution! ;) Nevertheless, I have learned a new way to do  this - with AM-GM.  Albeit well-known, I'm curious: how does one show that a polygon with given side lengths has maximal area iff it has a circumcircle?

[quote="ondrob"]Can you prove it pure geometrically from this point?[/quote]

Absolutely.

[quote="ondrob"]Let ABCD be any quadrilateral with AB=3,BC=3,CD=7.Now we don't decrease it's area, when we make from D-> D' such that CD'=7 and angle ACD is pi/2. Now we have quadrilatera ABCD'. We don't decrease it's area, when we make A->A' such that A'B=3 and angle A'BD pi/2.

From this we know that we have to just look at such ABCD, where angles ABD=ACD=pi/2 ( From this we know that this ABCD has circumcircle)

You can also see that angles CDB=BDA.

Can you prove it pure geometrically from this point?[/quote]

It is necessary for the quadrilateral for have right angles ABD and ACD. As there is just one such quadrilateral, this is sufficient. To get its area, take C' on its circumcircle such that BC' = 7 and C'D = 3. Observe that [ABCD] = [ABC'D]. AD = 2r is a diameter, and by Pythagoras and Ptolemy, we find that 4r 2 - 9 = 14r + 9 which is easily solved for the solution r = 9/2, and by a number of methods [ABCD] = 16 :sqrt: 2.

For Mildorf: If you want to prove that a given quadrilateral has maximal area if it is cyclic, with given sides,a,b,c,d, consider the area as

$\frac{1}{2}ab (Sin{\alpha})+\frac{1}{2}cd (Sin{\beta})$, and prove this is max if $\alpha+\beta=\pi$.

Then do the general case!

To prove it has max area if it is cyclic,

$Area=F(x)=\frac{ab}{2}(\sin{x})+\frac{cd}{2}(\sin{y})$

Where x and y are the angles between a,b & c,d, respectively.
Also, going off the diagonal $ac$ and the cosine rule, we get that

$a^2+b^2-2ab(\cos{x})=c^2+d^2-2cd(\cos{y})$

Diff $F(x)$ with respect to x.

$F'(x)=\frac{ab}{2}(\sin{x})+\frac{cd}{2}(\sin{y})(y')$

Also Diff the other condition:

$ab(\sin{x})=cd(\sin{y})(y')$

Re-writing this, we get $y'=\frac{ab(\sin{x})}{cd(\sin{y})}$

Subbing this into the first equation and simplifing, we get:

$F'(x)=\frac{ab}{2} (\frac{ \sin{(x+y)}} {\sin{y}})$

This is zero for a max, so ${\sin({x+y})=0}$

or

$x+y=2{\pi}$

[quote="ln(dx/dy)"]If the quadrilaterial is cyclic, then we can use $\cal{A}$$=\sqrt{(s-a)(s-b)(s-c)(s-d)}$.[/quote]

Very good point... we can used the generalized form, called Brahmagupta's formula, to prove what Mildorf wanted.

\[A=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\ \cos^2\left(\frac{B+D}{2}\right)}\]

The area is maximized when $\textstyle abcd\ \cos^2\left(\frac{B+D}{2}\right)$ is minimized.  Since this quantity is always nonnegative, we want it to be zero, which happens iff $B+D=180^\circ$, that is, when $ABCD$ is cyclic.


I got this when trying to post a message:
[code]
SQL requests not achieved

DEBUG MODE

SQL Error: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ' 19209) GROUP BY ug.user_id' at line 7

SQL Request: SELECT MAX(a.auth_value) as allowed, ug.user_id FROM phpbb_auths a, phpbb_user_group ug WHERE a.obj_type = 'f' AND a.obj_id = 132 AND a.auth_name = 'auth_watch_forum' AND a.group_id = ug.group_id AND ug.user_id IN (, 19209) GROUP BY ug.user_id

Line : 1189
File : functions_post.php[/code]

 :maybe:

EDIT: It happens in the MC forum...I know that...is it a bug or something?

I sometimes get that, particularly when I'm trying to load something long, complicated, and with lots of $\LaTeX$ to render. (Often it happens at the preview stage.) Then I click the "back" button on my browser, hit "preview" or "submit" again, and it goes through on the second try. I don't think the second try has failed me yet.

Something like that just happened to me when I tried to revive a topic that was over a year old. (I had my reasons - long story, and it involves a hidden forum.) Again, the second try worked.

Other than to speculate that it involves some kind of buffer overload or internal time-out, I don't think it particularly matters what the reason is as long as the second try works.

No. NO. It goes through when trying to post in a topic posted in like 1 minute ago. When I try teh secnod try, I get "Someone has already posted a reply" and it appears to be me. Just that that message is quite annoying.


geometry
analytic geometry
ratio
quadratics
inequalities
calculus
calculus computations

I know this has problem backround to it, but I dont have anywhere else to go.

Question Details:
I have two circles centered at the origin, one with radius A and the other with radius b.

Looking at the hemiwasher (area between) the circles form above the x axis, find the values of A and B that place the center of mass within the hemiwasher itself, not in the open middle space.

What i think i has solved so far: not necesseraly accurate:
I solved the Y value of the center of mass in terms of A and B to be:

Y=(4(A^3-B^3))/(3pi(A^2-B^2))

how can I use this to find values of a and b that put the y coordinate of the center of mass between a and b?

B is the smaller radii; the density is constant, so it is irrelvant.

Note that only the ratio of the two radii matters- it's probably a good idea to normalize and make one of them $ 1$.
This will lead to a quadratic equation.

But i am not given a ratio, or are you saying i assign one? like:

A:B becomes BX:B (x is how many times bigger a is than b)

And Im not seeing a quadratic, are you saying one will appear out of the equation I gave or from somewhere else?

The expression you gave is not in lowest terms; there's an obvious common factor of $ A \minus{} B$. (Note- please be consistent with capitalization of variables)

If we set $ A \equal{} rB$, we're trying to solve $ 1\le \frac {4}{3\pi}\frac {r^2 \plus{} r \plus{} 1}{r \plus{} 1}\le r$ with $ r\ge 1$. Clearing the denominator, $ r \plus{} 1\le \frac {4}{3\pi}(r^2 \plus{} r \plus{} 1)\le r^2 \plus{} r$.

Ok, this makes sense. Sorry, I didn't see the difference of cubics in the numerator. I'll try to use lowercase for variables.

But I'm not seeing how I can use this the find values (or range of values) for a and b specifically. Now I have a weird inequality that needs to be true for the ratio between a and b, but how can I convert that back to a and b?

Sorry for not grasping at the hints you're providing, I appreciate the help though, very greatly.

Ok, solving all of my equalities and making sure everything that has to be true stays true, and using the quad formular, I get

r> or = to 2.02

Thus, A>= 2.02 B

Thanks!


algebra
polynomial
inequalities
trigonometry
function
absolute value
algebra solved

Let $P(X)$ be a polynomial of degree $n$ satisfying
$|P(X)|\leq 1$ whenever $|X|\leq 1$.

Prove that $|P'(X)|\leq n^2$ whenever $|X|\leq 1$.
Does the equality hold only for Chebyshev polynomials?

It was posted by treegoner and solved by Maubinool,the genaral form of Markov inequality also wrote by flip2004 and Maubinool.Here is the link:
http://www.mathlinks.ro/Forum/topic-17044.html

Here is a magnificient proof that I found in a material from ENS:
   It suffices to prove the identity $ xf'(x)=\frac{n}{2}\cdot f(x)+\frac{1}{n}\cdot\sum_{k=1}^{n}{f(xz_k)\cdot\frac{2z_k}{(z_k-1)^2}}$ and the triangular inequality. Then, a simple identity based on Chebyshev's polynomials will solve the problem. I find it great!

I'm sorry I didn't get it. Could you please explain it in details, Harazi? I would be very grateful

Sorry, I was a little bit too expeditive with the proof, but here it goes:
   Consider $z_i$ the roots of the polynomial $ z^n+1$. If we write the interpolation formula for the polynomial $\frac{ f(xz)-f(x)}{z-1}$ in the points $z_i$, we will find an identity in z and x. Now, make z very close in the formula you obtain and you will find the beautiful identity I posted. Next, assume that $⎢f(x)⎥$ is smaller than 1 in the unit circle (or disk) and apply the triangular inequality in the identity. You will find a majoration for the absolute value of $ f'(x)$. You show then (exercise using the Chebyshev's polynomials) that in fact the majoration you find implies that the absolute value of $f'(x)$ is at most n. This is not hard.

A detailed explanation for harazi's post. Specially for Iura and for all other mathlinkers, who don't identify himself with Harazi :)

O'Hara P. J., Another proof of Bernstein's theorem, AMM 80 (1973), 673-674

[b]Theorem.[/b]
Let $\deg f=n$. Then $||f'||\leq n||f||$, where $||x||=\max_{|z|=1}||x(z)||$.

[b]Lemma.[/b]
Let $\deg f\leq n$ and $z_1,z_2,...,z_n$ are roots of $z^n+1=0$. Then
\[tf'(t)=\frac{n}{2}f(t)+\frac{1}{n}\sum_{k=1}^{n}f(tz_k)\frac{2z_k}{(z_k-1)^2}.\]

[b]Proof of lemma.[/b]
Let $g_t(z)=\frac{f(tz)-f(t)}{z-1}$. We see that $g_t(z)$ is a polynomial of degree at most $n-1$ and $g_t(1)=tf'(t)$. Using a Lagrange's interpolation formulae with points $z_1,z_2,...,z_n$ shows that
\[g_t(z)=\sum_{k=1}^n g_t(z_k)\frac{z^n+1}{(z-z_k)nz_k^{n-1}}=\frac{1}{n}\sum_{k=1}^ng_t(z_k)\frac{z_n^1}{z_k-z}z_k\]
(since $z_k^{n-1}=-1/z_k$).

Putting $z=1$ we have
\[tf'(t)=\frac{1}{n}\sum_{k=1}^ng_t(z_k)\frac{2z_k}{(z_k-1)}=\frac{1}{n}\sum_{k=1}^n\frac{f(tz_k)-f(t)}{(z_k-1)^2}=\] \[=\frac{1}{n}\sum_{k=1}^n f(tz_k)\frac{2z_k}{(z_k-1)^2}-\frac{f(t)}{n}\sum_{k=1}^n \frac{2z_k}{(z_k-1)^2}.\]

In order to calculate sum $\sum_{k=1}^n \frac{2z_k}{(z_k-1)^2}$, put $f(t)=t^n$. Then $f(tz_k)=-t^n$, so
\[nt^n=-\frac{2t^n}{n}\sum_{k=1}^n\frac{2z_k}{(z_k-1)^2},\]
i.e. [b](1)[/b]
\[\sum_{k=1}^n \frac{2z_k}{(z_k-1)^2}=-\frac{n^2}{2}.\]

[b]Lemma is proved.[/b]

Suppose $|t|=1$. We have
\[|f'(t)|\leq \left(\frac{n}{2}+\frac{1}{n}\sum_{k=1}^n\left|\frac{2z_k}{(z_k-1)^2}\right|\right)||f||.\]
Let's show that $2z_k/(z_k-1)^2$ is a real negative number. Indeed, $z_k=e^{is}\neq 1$, thus
\[\frac{2z_k}{(z_k-1)^2}=\frac{2eI{is}}{(e^{is}-1)^2}=\frac{2}{e^{is}-2+e^{-is}}=\frac{1}{\cos s-1}<0.\]
It follows that from [b](1)[/b] we obtain that
\[\frac{1}{n}\sum_{k=1}^n\left|\frac{2z_k}{(z_k-1)^2}\right|=-\frac{1}{n}\sum_{k=1}^n \frac{2z_k}{(z_k-1)^2}=\frac{n}{2}.\]

[b]Theorem is proved.[/b]

[b]Remark 1.[/b]
For analytic functions the following inequality holds
\[\max_{|z|\leq 1}|f(z)|=\max_{|z|= 1}|f(z)|.\]

[b]Remark 2.[/b]
Harazi, obviously, didn't solved Iura's problem. ;)

[b]Remark 3.[/b]
The mentioned link doesn't lead to a proof, just to hints ;)

Myth, these are just your opinions. Do you really think it is difficult to obtain a proof from these hints? Anyway,...


Fie un tetraedru in  care AB=1 , AC=4 si BD=7.Stiind ca exista un punct egal departat de cele sase muchii ale tetraedrului sa se calculeze lungimea segmentului CD.

 :) ASTEPT REZOLVARI :)

Daca punctul P este egal departat de laturile triunghiului ABC, el se afla pe perpendiculara pe plan dusa prin centrul cercului inscris acestuia. Analog pt a se demonstra ca P se afla pe perpendiculara dusa pe planul ABD prin centrul cercului inscris triunghiului ABD. Vom nota cu V pr lui P pe AD, cu R pr lui P pe CD, cu S pr lui P pe AC, cu T pr lui P pe AB si cu Q pr lui P pe BD. Se demonstreaza usor ca:
1. DR=DV=DQ=a (DO lat com ,OR si OV raze din T3P si ORD=OVD=90) unde o este centru ci al lui ABC
2. AS=AV=AT=b
3. BT=BQ=c
4. CS=CR=d
Din cele 4 relatii rezulta ca
1. AC=b+d=4
2. AB=b+c=1
3. BD=c+a=7
Si noi vrem sa calculam CD, adica d+a
care este BD-AB+AC=c+a-b-c+b+d=a+d
adica CD=7-1+4=10
Sperca nu am gresit prea tare :P


Thank you,Arne.This is what I want to find.
My solution is a bit more complex than yours.I used Heron's formula.So I think it abnormal.But in short time I can't make a normal solution which my classmate can accept.Maybe my brain itself is abnormal.

原是用三角形外切圆来做的。

这个题后来作为过高考题
前两天还有同学来问我的，是在高考复习材料里的……

[quote="dingdongdog"]已知x,y,z为正实数且xyz(x+y+z)=1,求（x+y）（y+z）最小值。
我有一个很abnormal的解法。[/quote]

$ (x\plus{}y)(y\plus{}z)\equal{}y(x\plus{}y\plus{}z)\plus{}zx \ge 2\sqrt{y(x\plus{}y\plus{}z)zx}\equal{}2$.

[quote="Fourier"]原是用三角形外切圆来做的。[/quote]

是内切圆吧?

呵呵, 那你的意思就是内切圆代换法咯, 这样吗?:

令$ a\equal{}x\plus{}y,b\equal{}y\plus{}z,c\equal{}z\plus{}x$, 则$ a,b,c$构成三角形, 并且由海伦公式易知其面积$ S\equal{}1$, 于是就有

$ (x\plus{}y)(y\plus{}z)\equal{}ab\equal{}\frac{2S}{\sin{C}}\equal{}\frac{2}{\sin{C}} \ge 2,$

当$ C\equal{}\frac{\pi}{2}$时取等号.


(a) P1 and P2 write down a number on a piece of paper, and only know the numbers they write down. They give their pieces of paper to a person P3. P3 writes on a piece of paper the sum of the numbers, and any other random number. Now P3 asks P1 if he knows P2's number; and we assume both P1 and P2 have complete, impeccable reasoning and neither lie; and P1 replies says 'yes' or 'no'; then P3 asks P2 if he knows P1's number who says "yes' or 'no'; and the process keeps going. Prove in finitely many steps, both will end up saying 'yes'.
(b) A more general version: exactly the same but instead of writing the sum and 1 random number, P3 writes down the sum and k other random numbers. Find all k for which the problem works (i think it's all values of k, but i have no idea).
(c) Do parts (a) and (b) but with the product of the two numbers instead of the sum (i think this is considerably easier).

do P1,P2 have to make guesses abou that number ?

i think the statement of the question is quite clear.
p1 and p2 try to work out each other person's number, and they're both super-intelligent. prove at some point each one manages to work out the other person's number.


[quote="math227"]consider a 1xn board.  suppose you color every square with two colors, white and black.  Let $ c_n$ be the number of colorings in which no two squares that are colored white are adjacent.  find a recurrence relation for $ c_n$ and then derive an explicite formula for $ c_n$.[/quote]

[hide]
Let $ a_{n}$ be the number of valid relations ending with a white one.
Let $ b_{n}$ be the number of valid relations ending with a back one.
$ c_{n}\equal{}a_{n}\plus{}b_{n}$
$ b_{n}\equal{}a_{n\minus{}1}\plus{}b_{n\minus{}1} \Rightarrow b_{n}\equal{}c_{n\minus{}1}$
$ a_{n}\equal{}b_{n\minus{}1}$ since the number of valid relations ending with white that is N blocks long is exactly equal to the number of varid ones, since obviously n-1 needs to be valid, and if n-1 ends with white it's no good.
$ c_{n}\equal{}b_{n\minus{}1}\plus{}b_{n}$
$ c_{n}\equal{}c_{n\minus{}1}\plus{}c_{n\minus{}2}$
Our first values are
1-2
2-3
The characteristic porynomiar is
$ x^{2}\equal{}x\plus{}1$
This is the fibonacci sequence, with the nth number being $ F_{n\plus{}2}$
You can use Binet's formula if you want to find the explicit value, I don't really remember it to be honest, and latexing it takes a while anyways.
Or, if you'd never seen the fibonacci before, you could solve the characteristic polynomial, and then that would give you the values needed to find the formula.[/hide]


ratio
arithmetic sequence
geometric sequence

[quote="nemomath"]Find 3 Arithmetic means between -25 and 5 _______, ______, ________
Find 2 Geometric means between 6 and 20.25. _____,_______

I could use some help. OK, I think what this means is find the nth term for the numbers in between But how do I get d? Also, what is the difference between Arithmetic and Geometric?[/quote]

Arithmetic mean of $a_{1},a_{2},\ldots,a_{n}$ is
\[\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\]
while the geometric mean is
\[\sqrt[n]{a_{1}a_{2}\cdots a_{n}}. \]
[hide="Generalization"]Let's say you are given numbers $a_{0}$ and $a_{n+1}$ and told to find the $n$ arithmetic means, $a_{1}$ through $a_{n}$ between them (in your context). Then we need the common difference $d$... And since $a_{1}=a_{0}+d, a_{2}=a_{0}+2d$, etc., it is true that $a_{n+1}=a_{0}+(n+1)d$. Therefore $d=\frac{a_{n+1}-a_{0}}{n+1}$. Then you can just add them... giving
\[a_{k}= a_{0}+kd = \boxed{a_{0}+\frac{(a_{n+1}-a_{0})k}{n+1}}. \]
If we were to find the $n$ geometric means between them, then we need common ratio $r$, and since $a_{1}=a_{0}\cdot r, a_{2}=a_{0}\cdot r^{2}$, etc., it is true that $a_{n+1}= a_{0}\cdot r^{n+1}$. Therefore $r=\sqrt[n+1]{a_{n+1}/a_{0}}$. Then
\[a_{k}= a_{0}\cdot r^{k}= \boxed{a_{0}\cdot (a_{n+1}/a_{0})^{\frac{k}{n+1}}}. \]
Worthless in plugging in numbers, but it is more interesting to do things in general.
[/hide]

[quote="nemomath"]Find 3 Arithmetic means between -25 and 5 _______, ______, ________
Find 2 Geometric means between 6 and 20.25. _____,_______

I could use some help. OK, I think what this means is find the nth term for the numbers in between But how do I get d? Also, what is the difference between Arithmetic and Geometric?[/quote]
$3$ Arithmetic means between $-25$ and $5$: $-25$, ___, ___, ___, $5$

[hide]So the formula for any term of an arithmetic sequence is $a_{n}=a+(n-1)d$ where $a$ is the first term, $n$ is the $n^{\text{th}}$ term, and $d$ is the distance between each term.   I'm sure you already know this though.

Note that $a_{5}=-25+4d$ and we already know that $a_{5}=5$ so set up and solve:

$a_{5}=-25+4d$
$5=-25+4d$
$4d=30$
$d=\frac{15}{2}=7.5$

Since $d=7.5$, our series goes like:
$-25,-25+7.5,-25+2 \cdot 7.5,-25+3 \cdot 7.5, 5$
$\equiv \boxed{-25,-17.5,-10,-2.5, 5}$[/hide]

$2$ Geometric Means between $6$ and $20.25$: $6$, ___, ___, $20.25$

[hide]Note that the the formula for geometric progression is $a_{n}=a \cdot r^{n-1}$ where $a$ ($a=6$) is the first term and $r$ is the common ratio between each term.

$a_{4}=a \cdot r^{3}$ and we know $a_{4}=20.25$

$a_{4}=6 \cdot r^{3}$
$20.25=6 \cdot r^{3}$
$r^{3}=\frac{27}{8}$
$r=3/2$

So the sequence is: $\boxed{6, 9, 13.5, 20.25}$.[/hide]


geometry
trigonometry
inequalities
rectangle
triangle inequality

To tile a region $R$ means to find a set $S$ of non-overlapping regions such that their union is $R.$ Suppose $R$ is a square with side length $1.$

$1.$ Is it possible to tile $R$ with finitely many circles?
$2.$ Is it possible to tile $R$ with finitely many $30-60-90$ triangles such that their longest side has a rational length?

[hide="Part 1"]
[b]
Claim: It is not possible to tile $R$ with finitely many circles.[/b]
[b]Proof:[/b]
Assume that we can tile $R$ in such a way.

Let the radius of the smallest circle, $\omega_{0},$ be $r.$ Consider circles $\omega_{1}, \omega_{2},$ that are externally tangent to $\omega_{0}$ and each other. In other words, $\omega_{0}, \omega_{1}, \omega_{2}$ are all externally tangent to each other. However, there is a region between these circles that hasn't been covered. In order to cover this area, we must use a circle of smaller radius than $\omega_{0}.$ Contradiction!

Now, assume that we have only two circles, $\omega_{0}$ and $\omega_{1}.$ Let these circles be externally tangent at point $X.$ Now, choose a point outside these circles and place it arbitrarily close to $X.$ This point must be covered, so, use a circle to do so. But there will be distance between this circle and $\omega_{0}$ and $\omega_{1}$ if the circles are not all externally tangent to each other, so let this circle be externally tangent to $\omega_{0}$ and $\omega{1}.$ From here, we proceed by following the earlier paragraph to reach a contradiction.

Thus, we cannot tile $R$ with finitely many circles.

$QED$

[/hide]
[hide="Part 2"]
[b]Claim: It is not possible to tile $R$ in such a way.[/b]
[b]Proof:[/b]
We will proceed by contradiction. Assume that $R$ can be tiled in such a way.
Let there be $n$ triangles whose hypoteneuse lengths are $s_{1}, s_{2}, \cdots, s_{n}$ for rational $s_{i}.$ For each $s_{i},$ we know that the area of the corresponding triangle is $\frac{s_{i}\sin 30 \cdot s_{i}\cos 30}{2}= \frac{s_{i}^{2}\sqrt{3}}{8}.$ If we add up the area of these triangles and equate the area to that of the unit square, we get:

$\frac{\sqrt{3}}{8}(s_{1}^{2}+s_{2}^{2}+\cdots+s_{n}^{2}) = 1$

This implies that at least one $s_{i}$ is irrational. Contradiction! Thus, $R$ cannot be tiled with finitely many $30-60-90$ triangles with rational hypoteneuse length.

$QED$

[/hide]

Your part 2 is good. Here's the solution I used for part one when we did this at school.

[hide]
<Insert a rigorous argument here>, so there must be a circle tangent to the square at every point. Then, since a circle can only be tangent to a square at at most one point, there are infinitely many circles in any such tiling.
[/hide]

[hide="1 with perhaps too much rigor"]
It's not possible. Suppose it is. Then there is a circle with least radius, let $m$ be this radius. Now consider the upper left corner (well, one of the corners) of the square, call it $P$. For any positive $\epsilon$, all points a distance of $\epsilon$ away from $P$ and inside the square better be covered. Therefore, by the Triangle Inequality, there exists the center of a circle $C$ with radius $r\geq m$ and distance at most $r+\epsilon$ from $P$.

Hmm... looks like this circle goes outside the square! Indeed, if we drop a perpendicular from $C$ to the left side of the square and let the foot be $H$, $CH$ will be at least $r$, but also $PC\leq r+\epsilon$, so that
\[PH=\sqrt{PC^{2}-CH^{2}}\leq \sqrt{(r+\epsilon )^{2}-r^{2}}\]
Similarly, if we drop a perpendicular to the top of the square and call it $K$, then $PK\leq \sqrt{(r+\epsilon )^{2}-r^{2}}$.

Wait! Quadrilateral $PHCK$ is actually a rectangle! This means $PK=CH$, so $\sqrt{(r+\epsilon)^{2}-r^{2}}\geq r$. Um, yea, we can just choose $\epsilon=\frac{m}{9^{9}}$ and get a really bad contradiction...
[/hide]

[quote="Altheman"]these are very nice problems!!!

@karth, how do you know that the circles don't make a ring or something to that effect...like w_1 touches w_2, w_2 touches w_3, but you have not shown that w_3 touches w_1...like what if there was a ring of circles or something[/quote]

Hmm... how did I forget to include that part in my solution? I had it somewhere here...  :(

EDIT: Aha! Found it  :D I'll edit my earlier solution.

[quote="Hamster1800"]
[hide]
<Insert a rigorous argument here>, so there must be a circle tangent to the square at every point. Then, since a circle can only be tangent to a square at at most one point, there are infinitely many circles in any such tiling.
[/hide][/quote]

<Rigorous argument>=Any point on the square must be contained on or within a circle.  The circle cannot be outside the square, so no part of it can be on the other side of the line.   Only a tangent circle would include a point on a line but be on one side of it.

Oh, it is possible for a circle to "be tangent to a square at [more than one] one point."  What is important is that the circle can only be tangent a finite number of points.


calculus
rotation
geometry
3D geometry
sphere
integration

The results you get from this problem are fairly interesting.

The circle with equation x :^2: + y:^2: = 25 is rotated around the x-axis to form a sphere.

a) Write the differential of surface area, dS, in terms of x.

b) Numerically find the area of the zone between
  i) x=0 and x=1
 ii) x=1 and x=2
iii) x=2 and x=3
 iv) x=3 and x=4
  v) x=4 and x=5

c) Prove that the surface area of a sphere of radius r is given by S=4 :pi:r:^2:.

d) Derive the formula for the volume of a sphere V= 4:pi:r:^3: /3 using spherical shells and the fact that the surface area is equal to 4:pi:r:^2:.

I'm not necessarily looking for a solution to be posted. But for those of you who know how to use calculus to do this, if you haven't done it before, have a look at it.

sigh long time since i've done this. hope this is right.

a. [hide]dS=:sqrt:(1+(dy/dx):^2:)dx, (dy/dx):^2:=x:^2:/(25-x:^2:), dS=5/:sqrt:(25-x:^2:)dx.[/hide]

b. [hide]A=2:pi:int(y*dS)=2:pi:int(:sqrt:(25-x:^2:)*5/:sqrt:(25-x:^2:)dx)=10:pi: (x from a to b).  for all, it is 10:pi:[/hide]

c. [hide]we use the same method as before except 5 is replaces by r. be 2r:pi: (x from -r to r) = 4:pi:r:^2:.[/hide]

d. [hide]letting S(r) be the surface area of a sphere with radius r, and using concentric spheres, we see that integral (S(r))=V(r), where V(r) is the volume, because the volume is pretty much all those little areas added up. So, V(r)=4/3:pi:r:^3:+c. since V(0)=0, c=0.[/hide]

Your answer for a) is dL, not dS, but you used the correct formula in part b), so it's okay.


Eight cubes form the figure shown. If the side length of each cube is 3 cm, how many square centimeters are in the surface area of the figure?

[asy]unitsize(1cm);
for(int i=0; i<4; ++i) {
draw((i,0)--(i,3)--(0.3+i,3.3),linewidth(0.7));
draw((0,i)--(3,i)--(3.3,i+0.3),linewidth(0.7));
}
draw((0.3,3.3)--(3.3,3.3)--(3.3,0.3),linewidth(0.7));
draw((1,1)--(1.3,1.3),linewidth(0.7));
draw((1.3,2)--(1.3,1.3)--(2,1.3),linewidth(0.7));[/asy]

This is (12+16+4)(9)=288, unless I miscounted.

If the eight cubes were separate, their total surface area would be $ 8\cdot 6\cdot 3^2\equal{}432$.  Wherever two faces are glued, $ 2(3^2)\equal{}18$ is taken from the surface area.  There are $ 8$ places where this occurs, so the surface area is $ 432\minus{}8\cdot 18\equal{}\boxed{288}$.

First, I counted the S.A. of the figure, disregarding the fact that each side length of a cube is 3 cm. I used the S.A. formula 2(lw+wh+hl). Notice that this is a 3 by 3 by 1 figure. Then, I subtracted 2 for the two squares missing (front & back) and added 4 because there are four faces in place. Last, I multiplied the expression by 9 because S.A. is 2-dimensional, so 3^2=9. Therefore, (2(9+3+3)-2+4)9=288, Q.E.D.


I'm not sure where $ \frac{1}{4}$ is, so I'm not sure what kind of pattern you're trying to form.  Could you clarify?


Let $ABCD$ be a trapezoid (without parallel sides). Let $a, \ b, \ c, \ d$ trapezoid sides and $d_{1}, \ d_{2}$ its diagonals respectively.

Is it true that $A_{ABCD}= \frac{1}{2}\sqrt{4(d_{1}\cdot d_{2})^{2}-(a^{2}-b^{2}+c^{2}-d^{2})^{2}}$??

That formula works for any quadrilateral.   :)

Cool.

I didn't know that  :o

Does anyone have a proof of this one?


[hide]
the sequence is
N, WxC, WxH, WxG, F, WaG, WaH. WaC, N, WxC, WxH, WxG, F, WaG, WaH. WaC,
so if it is WxC then 12 more phases makes it WaG :) [/hide]

[hide]colts18 made it easy wiht the list of phases he provided. Just trace it with your finger, to obtain WaG[/hide]

[quote="236factorial"][quote="math92"][quote="236factorial"]colts18 made it easy wiht the list of phases he provided.[/quote]
colts18 is a girl, by the way
 :D[/quote]

Whoops, there are too many boys on this site  :D  :P[/quote]
i agree that there are too many boys on this site.
I think that 96.141592653% of the people on AoPS are boys.
[size=75]that number is just a guess[/size]

[quote="colts18"]The phases of the moon alternate in the following sequence: new (N), waxing crescent (WxC), waxing half (WxH), waxing gibbous (WxG), full (F), waning gibbous (WaG), waning half (WaH), waning crescent (WaC), and back to new (N). This cycle can also be written: N, WxC, WxH, WxG, F, WaG, WaH. WaC, N, …. If the current phase is WxC, what phase will it be in if the phase sequence continues for 12 more phases??[/quote]

The answer is WaG by just counting.

[quote="math92"]
I think that 96.141592653% of the people on AoPS are boys.
[size=75]that number is just a guess[/size][/quote]
 :rotfl: yeah, just a guess... (a f**king precise guess  :P )


Let $ a,b,c,d,k$ be positive real numbers. Prove that

$ \left.\frac{a\plus{}b\plus{}c\plus{}d}{2}\cdot\left[\frac{a}{(b\plus{}c)^k}\plus{}\frac{b}{(c\plus{}d)^k}\plus{}\frac{c}{(d\plus{}a)^k}\plus{}\frac{d}{(a\plus{}b)^k}\right]\geq \frac{a}{(b\plus{}c)^{k\minus{}1}}\plus{}\frac{b}{(c\plus{}d)^{k\minus{}1}}\plus{}\frac{c}{(d\plus{}a)^{k\minus{}1}}\plus{}\frac{d}{(a\plus{}b)^{k\minus{}1}}\right.$

[quote="marin.bancos"]Let $ a,b,c,d,k$ be positive real numbers. Prove that

$ \left.\frac {a \plus{} b \plus{} c \plus{} d}{2}\cdot\left[\frac {a}{(b \plus{} c)^k} \plus{} \frac {b}{(c \plus{} d)^k} \plus{} \frac {c}{(d \plus{} a)^k} \plus{} \frac {d}{(a \plus{} b)^k}\right]\geq \frac {a}{(b \plus{} c)^{k \minus{} 1}} \plus{} \frac {b}{(c \plus{} d)^{k \minus{} 1}} \plus{} \frac {c}{(d \plus{} a)^{k \minus{} 1}} \plus{} \frac {d}{(a \plus{} b)^{k \minus{} 1}}\right.$[/quote]

This follows immediately from your lemma [url=http://www.mathlinks.ro/viewtopic.php?p=1671199]here[/url] with $ k\equal{}4$
[quote="marin.bancos"][b]MY LEMMA:[/b]

If: $ a_i\geq0,b_i > 0,i \equal{} 1,2,\dots,k$ and $ k,n\in N,k\geq1$, then:
$ \frac {{a_1}^{n \plus{} 1}}{b_1^n} \plus{} \frac {a_2^{n \plus{} 1}}{b_2^n} \plus{} \cdots \plus{} \frac {{a_k}^{n \plus{} 1}}{b_k^n}\geq\frac {a_1 \plus{} a_2 \plus{} \cdots \plus{} a_k}{b_1 \plus{} b_2 \plus{} \cdots \plus{} b_k}\cdot\left(\frac {{a_1}^n}{b_1^{n \minus{} 1}} \plus{} \frac {{a_2}^n}{b_2^{n \minus{} 1}} \plus{} \cdots \plus{} \frac {{a_k}^n}{b_k^{n \minus{} 1}}\right)$
[/quote]

(putting cyclically $ a_1\equal{}\frac {a}{b \plus{} c}$ etc., $ b_1\equal{}a$ etc. and $ n\equal{}k\minus{}1$),
followed by Shapiro for $ 4$ variables :lol:

Taking into account that my lemma was improved for $ k>0$, we get the following form of it:

For $ a_i\geq0,b_i>0,i\equal{}1,2,\dots,n,\ n\in N,n\geq1$ and $ k>0,$ then:
$ \left.\frac{{a_1}^{k\plus{}1}}{b_1^k}\plus{}\frac{a_2^{k\plus{}1}}{b_2^k}\plus{}\cdots\plus{}\frac{{a_n}^{k\plus{}1}}{b_n^k}\geq\frac{a_1\plus{}a_2\plus{}\cdots\plus{}a_n}{b_1\plus{}b_2\plus{}\cdots\plus{}b_n}\cdot\left(\frac{{a_1}^k}{b_1^{k\minus{}1}}\plus{}\frac{{a_2}^k}{b_2^{k\minus{}1}}\plus{}\cdots\plus{}\frac{{a_n}^k}{b_n^{k\minus{}1}}\right)\right.$

The inequality could be obtained for $ n\equal{}4$, $ a_1\equal{}a, a_2\equal{}b, a_3\equal{}c, a_4\equal{}d$ and $ b_1\equal{}a(b\plus{}c), b_2\equal{}b(c\plus{}d), b_3\equal{}c(d\plus{}a), b_4\equal{}d(a\plus{}b)$.
In fact, this is a new generalization for Shapiro's inequality ($ n\equal{}4$).


$ 3^{2457}\plus{}1\equal{}x$
without working out the answer , say how many numbers , and which number, how many times used in the answer $ (x)$.

[quote="binomial_4eva"]$ 3^{2457} \plus{} 1 \equal{} x$
without working out the answer , say how many numbers , and which number, how many times used in the answer $ (x)$.[/quote]

Simply put, WHAT!?

Are you looking for a table along the lines of

1 : 1123
2 : 12
3 : 123
4 : ....

Or what?

this is what i mean, i will give an example here

when $ 100$ is divided by $ 5$ the answer is $ 20$
there is one $ 2$ and one $ 0$  in $ 20$ .

how many of each numbers from the set $ {1,2,3,4...9}$ are used in the answer for this , $ 3^{2457}\plus{}1\equal{}?$ i hope its clear now.

Good luck with that, I have calculated it and its fairly extreme (there doesn't seem to be an obvious way to do it otherwise)

You want the number of digits in $ 3^{2457}\plus{}1$?

[hide="solution"]Well, since the units digit of $ 3^{2457}$ is not 9, we just need to find the number of digits in $ 3^{2457}$. We take the logarithm base 10:

$ \log_{10} 3^{2457}\equal{}2457\log_{10} 3$

Now you run for your calculator.[/hide]

[quote="1=2"]You want the number of digits in $ 3^{2457} \plus{} 1$?
[/quote]

He also wants this

[hide]
$ {132, 109, 123, 117, 111, 108, 113, 111, 126, 123}$
[/hide]

But I fail to see an elementary way of obtaining that

He's asking how many of each digit there are.

[quote="1=2"]
@SimonM: Are those the numbers of the digits?[/quote]

Indeed they are, yay mathematica!

[hide]DigitCount[3^{2457} + 1][/hide]

@1=2 ya i mean how do you find how many for example $ 2$ is in that long number. without using maple.

You use Mathematica instead.

Seriously, I really doubt there is a good way, known or unknown.


Let regular  polygon has $1999$ vertexs .$41$ of them are painted blue and remain are painted yellow.Find number of isoceles triangle which has tthree vertexs same colour.

Do you mean that this number is independant of the positions of the blue vertices?  :huh:

Pierre.

I think that there were many person in VN solved it


integration
real analysis
real analysis unsolved

If $f(x) \geq 0$ and $\int_0^{\infty} f(x) dx=0$, does this imply, $\int_0^{\infty} f(x) x dx=0$ ? or there is a counter example?

Thanks.

If $f\geq 0$ and $\int_{\mathbb{R}} f(x)dx=0$, then $f(x)=0$ a.e on $\mathbb{R}$.  It follows that $xf(x)=0$ a.e on $\mathbb{R}$, which in turn implies that
\[ \int_{\mathbb{R}} xf(x)dx=0  \]


induction
function
calculus
derivative
inequalities
inequalities proposed

The following general problem holds

For all positive real numbers $a_{1},a_{2},...,a_{n}$ with product $n$ then
\[\sum \frac{1}{ka_{i}+1}\ge \min\left\{1,\frac{n}{k+1}\right\}\]
\[\sum \frac{1}{ka_{i}+1}\le \max\left\{n-1,\frac{n}{k+1}\right\}\]

I solved it by "induction", "UMV" or LCF-RCF theorem. I think induction is most elementary way.

An idea that works is studying the function $f : \mathbb R \to \mathbb R$, $f(x) = \frac1{k e^{x}+1}$.
If you look at the second derivative, then you'll see that it's concave on one interval and convex on the rest.
Of course, some nasty computations remain, but they can be handled.

Mildorf solved a Romanian TST problem with this idea. I think you know which one I'm talking about.

[quote="edriv"][b]Problem 6[/b]
a) For each $n \ge 2$, find the maximum constant $c_{n}$ such that
$\frac 1{a_{1}+1}+\frac 1{a_{2}+1}+\ldots+\frac 1{a_{n}+1}\ge c_{n}$
for all positive reals $a_{1},a_{2},\ldots,a_{n}$ such that $a_{1}a_{2}\cdots a_{n}= 1$.[/quote]

For $a,\ b>0$, we have $\frac{a}{a+1}+\frac{b}{b+1}\geq \frac{a+b}{a+b+1}$, thus $\　a_{1}+a_{2}+\cdots+a_{n}\geq n\Longrightarrow \frac{a_{1}}{a_{1}+1}+\frac{a_{2}}{a_{2}+1}+\cdots+\frac{a_{n}}{a_{n}+1}\geq \frac{n}{n+1}$ holds inductively.

Rewriting this inequality gives

\[n-\sum_{k=1}^{n}\frac{1}{a_{k}+1}\geq 1-\frac{1}{n+1}\Longleftrightarrow \sum_{k=1}^{n}\frac{1}{a_{k}+1}\leq \frac{n^{2}}{n+1}\].

[quote="silouan"]Edriv could you post your solution ?Thank you[/quote]
Same idea http://www.mathlinks.ro/Forum/viewtopic.php?t=143678

[hide="a"]
For all $n\geq 2$, we have $c_{n}=1$. To see this, choose $a_{1}=\epsilon^{n-1}$ and $a_{k}=\frac{1}{\epsilon}$ for $k\neq 1$ and send $\epsilon \rightarrow 0$ to have the left hand side approach 1. Therefore, there cannot be a better constant than 1.

To actually prove the lower bound, WLOG let $a_{1}\leq a_{2}\leq \cdots \leq a_{n}$. Then $a_{1}a_{2}\leq 1$, so that
\[\sum_{k=1}^{n}\frac{1}{a_{k}+1}\geq \frac{1}{a_{1}+1}+\frac{1}{a_{2}+1}\]
\[=\frac{1}{a_{1}+1}+\frac{a_{1}}{a_{2}+a_{1}a_{2}}\geq \frac{1}{a_{1}+1}+\frac{a_{1}}{a_{1}+1}=1\]
as desired.
[/hide]
[hide="b"]
First, $d_{2}=\frac{2}{3}$, attainable when $a_{1}=a_{2}=1$. To prove the lower bound, let $a=a_{1}$, then $a_{2}=\frac{1}{a}$. The two-variable inequality is equivalent to
\[\frac{1}{2a+1}+\frac{a}{a+2}\geq \frac{2}{3}\iff 3(a+2)+3a(2a+1)\geq 2(2a+1)(a+2)\]
\[\iff (a-1)^{2}\geq 0\]
For $n\geq 3$, we have $d_{n}=1$. This is very similar to part (a): we can let $a_{1}=\epsilon^{n-1}$ and $a_{k}=\frac{1}{\epsilon}$ for $k\neq 1$ and send $\epsilon \rightarrow 0$ to show that there is no better constant than 1.

To prove the lower bound, we can WLOG let $a_{1}\leq a_{2}\leq \cdots \leq a_{n}$ and then forget about all terms beyond the third. We also have $a_{1}a_{2}a_{3}\leq 1$. Now let
\[x=\sqrt[9]{\frac{a_{2}a_{3}}{a_{1}^{2}}},\ y=\sqrt[9]{\frac{a_{1}a_{3}}{a_{2}^{2}}},\ z=\sqrt[9]{\frac{a_{1}a_{2}}{a_{3}^{2}}}\]
Then $a_{1}\leq \frac{1}{x^{3}},\ a_{2}\leq \frac{1}{y^{3}},\ a_{3}\leq \frac{1}{z^{3}}$, and $xyz=1$. Then
\[\sum_{k=1}^{n}\frac{1}{a_{k}+1}\geq \sum_{k=1}^{3}\frac{1}{a_{k}+1}\geq \frac{x^{3}}{x^{3}+2}+\frac{y^{3}}{y^{3}+2}+\frac{z^{3}}{z^{3}+2}\]
\[=\frac{x^{2}}{x^{2}+2yz}+\frac{y^{2}}{y^{2}+2xz}+\frac{z^{2}}{z^{2}+2xy}\]
\[\geq \frac{x^{2}}{x^{2}+y^{2}+z^{2}}+\frac{y^{2}}{x^{2}+y^{2}+z^{2}}+\frac{z^{2}}{x^{2}+y^{2}+z^{2}}=1\]
as desired.
[/hide]

[quote="kunny"]For $ a,\ b > 0$, we have $ \frac {a}{a + 1} + \frac {b}{b + 1}\geq \frac {a + b}{a + b + 1}$
[/quote]
How to prove this inequality? :?:  :maybe:

[quote="kunny"]$ \　a_{1} + a_{2} + \cdots + a_{n}\geq n\Longrightarrow \frac {a_{1}}{a_{1} + 1} + \frac {a_{2}}{a_{2} + 1} + \cdots + \frac {a_{n}}{a_{n} + 1}\geq \frac {n}{n + 1}$ holds inductively.

Rewriting this inequality gives
\[ n - \sum_{k = 1}^{n}\frac {1}{a_{k} + 1}\geq 1 - \frac {1}{n + 1}\Longleftrightarrow \sum_{k = 1}^{n}\frac {1}{a_{k} + 1}\leq \frac {n^{2}}{n + 1}
\]
.[/quote]
So...what's the correct value for $ c_n$? I think that, it is $ \frac{n}{n+1}$ But what about the proof of scorpioa119?

I am really sorry for bumping this old thread after years...But I really want to know the answer. Please help me. :(
Thanks in advance.

[quote=edriv]a) For each $n \ge 2$, find the maximum constant $c_{n}$ such that
$\frac 1{a_{1}+1}+\frac 1{a_{2}+1}+\ldots+\frac 1{a_{n}+1}\ge c_{n}$
for all positive reals $a_{1},a_{2},\ldots,a_{n}$ such that $a_{1}a_{2}\cdots a_{n}= 1$.

b) For each $n \ge 2$, find the maximum constant $d_{n}$ such that
$\frac 1{2a_{1}+1}+\frac 1{2a_{2}+1}+\ldots+\frac 1{2a_{n}+1}\ge d_{n}$
for all positive reals $a_{1},a_{2},\ldots,a_{n}$ such that $a_{1}a_{2}\cdots a_{n}= 1$.[/quote]
[b][u][color=#f00]Solutions :[/color][/u][/b]
a) I wanna prove that $c_n=1$, without loss of generality, assume that $a_1 \ge a_2 \ge ... \ge a_n$.

If $a_1 \ge a_2 \ge ... \ge a_{n-1} \ge 1 \ge a_n$ we have $\frac{1}{x+1}+ \frac{1}{x+1} \ge \frac{2}{\sqrt{xy}+1} \Leftrightarrow \frac{(\sqrt{x}-\sqrt{y})^2(\sqrt{xy}-1)}{(x+1)(y+1)} \ge 0$ which is obvious for $x,y \ge 1$.

Then by using a simple induction we can prove $\sum_{k=1}^{n-1} \frac 1{a_{k}+1} \ge \frac{n-1}{\sqrt[n-1]{a_1a_2....a_{n-1}}+1}$ therefore $\sum_{k=1}^{n} \frac 1{a_{k}+1} \ge \frac{n-1}{\sqrt[n-1]{a_1a_2....a_{n-1}}+1}+\frac 1{a_{n}+1}$

Let $b_i=\sqrt[n-1]{a_i}$ then $b_1b_2...b_n=1$, so, it remains to prove that $\frac{n-1}{b_1b_2....b_{n-1}+1}+\frac 1{b_{n}^{n-1}+1}>1$

Or, $\frac{(n-1)b_n}{1+b_n}+\frac 1{b_{n}^{n-1}+1}>1$, or  $\frac{a_n\left((n-2)b_n^{n-1}+n-1-b_n^{n-2}\right)}{\left(1+b_n\right)\left(b_{n}^{n-1}+1\right)} \ge 0$ which is obvious for $n \ge 2$ and $0<b_n \le 1$.
Equality occurs if and only if $n=2$

If $n>2$ then $\exists k \le n-1$, such that $a_1 \ge a_2 \ge ... \ge a_k \ge 1 \ge ... \ge a_n$

Therefore $\sum_{k=1}^{n} \frac 1{a_{k}+1} > \frac 1{a_{n-1}+1}+\frac 1{a_{n}+1} \ge  \frac 1{2}+\frac 1{2}=1$

b) We can do it with the same method using the cases :
For $n=2$ we get $\frac 1{2a_{1}+1}+\frac 1{2a_{2}+1}=\frac 23$,

[color=#00f][b]1) [/b][/color]If $a_1 \ge a_2 \ge ... \ge a_{n-1} \ge 1 \ge a_n$ for $n \ge 3$ we get $\sum_{k=1}^{n} \frac 1{2a_{k}+1} \ge \frac{n-1}{2\sqrt[n-1]{a_1a_2....a_{n-1}}+1}+\frac 1{2a_{n}+1} \ge 1$

[color=#00f][b]2) [/b][/color] If $ a_1 \ge a_2 \ge ... \ge a_{n-2} \ge 1 \ge a_{n-1} \ge a_n$ for $n \ge 3$ we get $\sum_{k=1}^{n} \frac 1{2a_{k}+1} \ge \frac{n-1}{2\sqrt[n-1]{a_1a_2....a_{n-2}}+1}+\frac 1{2a_{n-1}+1}+\frac 1{2a_{n}+1} \ge 1$

The last inequality is equivalente to $\frac{1}{2x+1}+\frac{1}{2y+1}+\frac{1}{2z+1} \ge 1$ for $xyz=1$ which is easy!!
[color=#00f][b]3) [/b][/color] If $a_1 \ge a_2 \ge ... \ge a_k \ge 1 \ge ... \ge a_n$ for $n \ge 4$ and $k \le n-2$

We get $\sum_{k=1}^{n} \frac 1{2a_{k}+1} > \frac 1{2a_{n-2}+1}+\frac 1{2a_{n-1}+1}+\frac 1{2a_{n}+1} \ge \frac13+ \frac13+ \frac13=1$.
Then $d_n= \frac23$ and equality holds if and only if $n=2$..


The value of $ \log_5 \frac {(125)(625)}{25}$ is equal to:

$\textbf{(A)}\ 725 \qquad
\textbf{(B)}\ 6 \qquad
\textbf{(C)}\ 3125 \qquad
\textbf{(D)}\ 5 \qquad
\textbf{(E)}\ \text{None of these}$

Notice that 125, 625, and 25 are all powers of 5! So, we can rewrite it to solve the problem. Answer: D.


geometry
ratio
trigonometry
geometric transformation
rotation
geometry proposed

You're given an acute triangle ABC. AD is height. There is a line x that passes through point D (different from BC). On the line x points E and F are taken such that the angles AEB and AFC are 90. L is midpoint of the segment EF and M is midpoint of BC. Find <ALM=?

A, B, D and E concyclic, same for A,D,C,F, this makes angle BAE = angle CAF = angle FDC and there4 there is a spiral similarity between the triangles EAF and BAC, the angle between the medians AL and AM being the angle between BC and the straight line x, while their ratio AL/AM = AE/AB, consequently the triangles AEB and ALM are similar and AL and ML are perpendicular, as AE and BE are.

Best regards,
sunken rock

Hi! sunken rock how did you get it

"the angle between the medians AL and AM being the angle between BC and the straight line x"

because: if we get it the problem is solved, so that is we need
Can you explain

Because of the similarity between triangles $ EAF$ and $ BAC$ ($ AE \equal{} AB \cos{\theta}$, $ AF \equal{} AC \cos{\theta}$ and there is a rotation $ \mathcal{R}$ by $ \theta$ which brings $ AB$ onto $ AE$ and $ AC$ onto $ AF$, where $ \theta \equal{} \angle{BAE} \equal{} \angle{CDF}$), rotation $ \mathcal{R}$ also brings median $ AM$ onto median $ AL$ and therefore $ AL / AM \equal{} AE / AB$.

Because of $ \angle{BAE} \equal{} \angle{MAL} \equal{} \theta$, triangles $ ABE$ and $ AML$ are also similar and finally $ \angle{ALM} \equal{} \angle{AEB} \equal{} 90^{\circ}$.

You can use the following remarkable geometrical locus :

Let $ \beta$ , $ \gamma$ be two fixed circles so that $ \beta\cap\gamma\equal{}\{A,D\}$ . Denote the fixed points $ B\in \beta$ , $ C\in \gamma$ for which  $ D\in BC$ and $ BC\perp AD$ . Let $ M\in \beta$ , $ N\in \gamma$ be two mobile points so that $ D\in MN$ . Then the geometrical locus of the midpoint $ L$ of the segment $ [MN]$ is the circle with diameter $ [AF]$ , where $ F$ is the midpoint of $ [BC]$ .


Let x,y,z be positive real numbers, $x+y+z\geq xyz$. Find the minimum value of $\displaystyle\frac{x^2+y^2+z^2}{xyz}$.

[quote="From Neptune to Mars"]I think it isn't difficult problem.
We have:     $ x \plus{} y \plus{} z \ge xyz.$
:arrow:$ (xyz)^2 \le xyz(x \plus{} y \plus{} z).$
We also have
:$ xyz(x \plus{} y \plus{} z)\le[(xy \plus{} yz \plus{} zx)^2]/3\le\frac {(x^2 \plus{} y^2 \plus{} z^2)^2}{3}$
:arrow:$ (xyz)^2\le\frac {[(x^2 \plus{} y^2 \plus{} z^2)^2]}{3}.$
:arrow:$ xyz < \equal{} (x^2 \plus{} y^2 \plus{} z^2)/ \sqrt 3$
Thus:
$ (x^2 \plus{} y^2 \plus{} z^2)/(xyz)\ge \sqrt 3.$
:arrow: $ min[(x^2 \plus{} y^2 \plus{} z^2)/(xyz)] \equal{} \sqrt 3.$
"=" happens when:x=y=z= \sqrt 3.
That is "dpcm".[/quote]


quadratics AMC AIME geometry FTW probability calculus

Kinda like Jeff Foxworthy's you know your a redneck if....
Reply with additions.
1.You know your a Math  geek if you wanted an N-spire for christmas.
2.You Know your a Math Geek if you prove your math teacher wrong on a daily basis.
3.You Know your a Math Geek if your favorite website is AoPS.
4. You Know your a Math Geek if one of your hobby is pwning people at mental math.

[quote="monkeygirl13"]3.You know you're a Math Geek if your home page is AoPS.[/quote]

Fixed. :P

1.You know your a Math geek if you wanted an N-spire for christmas.
2.You Know your a Math Geek if you prove your math teacher wrong on a daily basis.
3.You Know your a Math Geek if your homepage is AoPS.
4. You Know your a Math Geek if one of your hobby is pwning people at mental math.
5. You debate with your stuffed animals on how to REALLY extract a quaratic equation.

6. You know your a math geek if your new years resolution is to read to 3 AoPS books.
7. You know your a math geek if you read an AoPS book in class, while teh teacher is helping the others factor quads.

5. Change quadratic to quartic please.
6. Just no.

You know you're not the antithesis of a math geek if you spell "quadratic" correctly.

And will you guys actually post attributes of math geeks instead of attributes of normal people?

EDIT: Wait, nvrm. Didn't see the stuffed animals part.

you play with your stuffed animals???
 :P

You know if you are a math geek if...
1. Your homepage is Wolfram|Alpha
2. You have 50 dollars worth of AMC and AIME tests
3. You have about 10 geometry books, 5 of which are Chinese

Those are all true for me.

Stuffed Animals are beast.

11) You did all your addition in kindergarten without using counting blocks like the other kids.
12) You laugh when you see someone using their fingers to count how many hours are from 9 am to 5 pm.
13) You ask your mom if she can make pumpkin 3.14159... on the  sqrt400 + 2^2 day of November, 2005 + 1
14) You're awesome.

12. You know your a math geek if your math teacher searches your backpack for a calc cuz you finished 100 problems in 10 minutes or less.

13. you know ur a math geek if you did all the questions in the lesson because you forgot to write down the homework and you're too lazy to call ur friends.

[size=134]ARE YOU KIDDING ME, STUFFED ANIMALS [i]ARE[/i] AWESOME![/size]

[quote="piegraph"]13. you know ur a math geek if you did all the questions in the lesson just for the fun of it.
[size=134]ARE YOU KIDDING ME, STUFFED ANIMALS [i]ARE[/i] [color=red]NOT[/color] AWESOME![/size][/quote]

fixed.

I like stuffed animals :maybe:

- ...when you ask your engineering teacher for $ 12\pi$ square inches of wood for your project.

Yeah, that happened to me. It took me a while to realize.

Stuffed Animals FTW!
15. You know your a math geek if you named your stuffed cat pi.

16.  If you have memorized 134digits of pi, and organized a pi group with 5 followers.
Thats what happened to me.

[quote="eagle2"]mathlete. to tell you the truth you are sickkk. That girl is UGLY too.
for my it would
be nice + smart + not tempermental = good. so basically piegraphs things. sick minded creature.[/quote]
You are just saying that to piss me off...and yes she is nice. So that DOES work out.

ok the truth she is not the best i've seen. She is pretty ugly for a cheerleader. even my sister is better then her.

Oh goodness.
This is getting off topic.
You know you're a math geek if you had to prove to your math teacher that 'x cubed' means 'x to the third power', not 'x to the fourth power', and proved it to her by showing her why, by showing how it got it's name by the volume of a cube, and in this process, you were in mental facepalm mode.

You know your a math geek if you sign up for all of the AoPS classes at once.

If this gets really off topic again, can we please have it locked?

"......" you do math at lunch.

You know you are a Math Geek if....

1. People don't understand you.
2. You raise your hand in your math class in order to correct the teacher for about 10 times a day.
3. People do not get your joke.
4. You get about 10 percent higher in your calculus or similarly advanced class (for a middle school student) compared to somewhat nerdy student in your middle school gets in Honors Algebra class.
5. Your life is AoPS.
6. You would rather buy a full licence for Mathematica instead of WOW CD.

[quote="lightning"]I don't know if you need to, but did you get her permission to put the photo on the forum? Just in case she sues or something.....but since you said she was nice, it should be fine....

Sorry for going even more off topic.....[/quote]
She has already posted the photo publicly on the internet. And it is not like she is going to come here. She may be a math girl, but I doubt she has ever heard of aops, considering she is not even on my school's math team (she thought cheerleading was a bigger priority)

Can someone please lock this? It is getting extremely off-topic.

Ironically, this is a "math geek" thread and we're talking about girls.

I do advise that pictures are not posted on the forum without explicit permission. Also, posts like

[quote="eagle2"]mathlete. to tell you the truth you are sickkk. That girl is UGLY too.[/quote]

are completely unnecessary to make.  Insulting others' opinions is untasteful and spammy.

You know you're a math geek if when you saw "10!," you thought "What were the last two non-zero digits?," because AMC 2010 #24 is still in the back of you mind.

BTW, the answer is.. Math geek if you found this out or knew it already...[hide]88[/hide]

If you know random math facts and show them off at school.
If you get bullied every time after math.

You know you are a math geek if you are on Lehigh Valley Fire

NOTE THAT I JUST SAID "IF", NOT "IF AND ONLY IF"

You are a math geek if you corrected your teacher by pointing out 1/2>1/2 is a false statement in third grade.


geometry
analytic geometry
absolute value

i need help. what is
1) the hockey stick identity
2)the shoelace theorem. i saw it explained once but i didnt really get it

also, a complementary problem would be nice

thanks!

um i cant help you with the first one but the way i have understood the shoelace theorum is that you take the verticies of a polygon (not sure if it works with triangles) and list them. and multiply diagonally. e.g.  If a polygon has verticies of $ (a,b) (c,d) (e,f) (g,h) (i,j) (k,l)$ and $ (m,n)$ what is the area of this polygon?

We list them like this:

a,b
c,d
e,f
g,h
i,j
k,l
m,n

We then multiply diagonally from top right to bottom left, then subtract the sums of multiplying bottom right to top left yeilding --- (($ a\times{d}$) + ($ c\times{f}$) + ($ e\times{f}$) + ($ g\times{j}$) + ($ i\times{l}$) + ($ k\times{n}$)) - (($ m\times{l}$) + ($ k\times{j}$) + ($ i\times{h}$) + ($ g\times{f}$) + ($ e\times{d}$) + ($ c\times{b}$))

try looking up the hockey stick identity on the internet figure out what it is how to apply it and its origins in pascals triangle.

find the area of a quadrilateral with coordinates (0,0), (0,4), (4,4), (4,0),

well we know this is a square with side length 4 so the area is 16. so if we do that using shoelace we get

0         0
0         4
4         4
4         0
(remember you must put the points in order either clockwise or counterclockswise otherwise it won't work)

it is (0*4 + 0*0 + 4*0) - (0*0 + 4*4 + 4*4) = -32 now you take the absolute value of that and you get 32, now you divide that be 2 and VOILA 16.

I think you forgot something, cognos, although it didn't really matter since the first point was (0,0)

You have to...... write the first coordinate again.......

but you can't go like

0  0
4  4
4  0
0  4

try it. ITs messed up

OOPS THE FIRST COORDINATE I TOTALLY FORGOT. SORRY ABACADAE

[quote="mathcrazed"]PASCALS TRIANGLE. Draw a hockey stick. and add up the ones on the longer side. O.O its amazing[/quote]

how do you draw a hockey stick?

[quote]
HOCKEY STICK IDENTITY- It should discussed thoroughly in the Intro to Counting/Probability book (in the later chapters). You also might want to try looking it up in the AopsWiki.....[/quote]
[url=http://www.artofproblemsolving.com/Wiki/index.php/Combinatorial_identity]wiki doesnt really help...[/url]

yes, i would like to know the hockey stick identity as well. first of all i get a headache from looking at the wiki page and second i dont even understand what it does "summation" whats that???

Please any and all help for me and abacadaea would be greatly apreciated!!!

Ahem...Google does better

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=515429#515429

i under stand it now :lol: .

how do i used it??..... :huh:

Say they asked you what $ \binom{1}{n}\plus{}\binom{2}{n}\plus{}\binom{3}{n}\plus{}\binom{4}{n}\plus{}\binom{5}{n}\plus{}\binom{6}{n}$ equals.
You would use the hockey stick identity then.

for some reason it seems to me that from what i have heard the hockey stick identity is fairly rare in problems.  Could some one give me a problem that involves the use of the hockey stick identity but is not like the one posted before me.

[quote="ZhangPeijin"]Say they asked you what $ \binom{1}{n} \plus{} \binom{2}{n} \plus{} \binom{3}{n} \plus{} \binom{4}{n} \plus{} \binom{5}{n} \plus{} \binom{6}{n}$ equals.
You would use the hockey stick identity then.[/quote]

erm... zolojetto.....right there

and btw, could someone post a solution to that. sorry, im still confused[size=59]only a bit[/size]

Peijin, that's not Hockey stick unless $ n \equal{} 0$ or $ n\equal{}1$, which makes the sum easy to compute anyways


Which of the following correctly represents $ \text{CO}_{2}$, and why?  Choices $ (2)$ and $ (3)$ both look reasonable to me.  Thanks in advance!

the correct answer would be (3).
(2) is wrong because oxygen have only 6 electrons.
Every element must have at least eight electrons surround it ( with the exception if boron and beryllium)

[quote="Carcul"][quote="I_thirst_for_knowledge"](with the exception if boron and beryllium)[/quote]

And hydrogen. :)[/quote]

Thanks a lot! - I got it now.  :)

Note also that an element can have more than eight electrons "around it", provided it doesn't belong to the second row of the Periodic Table.


For each positive integer $ n\leq 49$ we define the numbers $ a_{n}\equal{}3n\plus{}\sqrt{n^{2}\minus{}1}$ and $ b_{n}\equal{}2(\sqrt{n^{2}\plus{}n}\plus{}\sqrt{n^{2}\minus{}n})$.Prove that there exist two integers A,B such that
$ \sqrt{a_{1}\minus{}b_{1}}\plus{}\sqrt{a_{2}\minus{}b_{2}}\plus{}...\plus{}\sqrt{a_{49}\minus{}b_{49}}\equal{}A\plus{}B\sqrt{2}$


1) Provide a synthesis of 1-Phenyl-3-bromopropene.

2) Starting from the compound prepared in part 1), outline reasonable synthesis of 2-Phenylcyclopropanecarboxilic acid.

1. benzene to toluene by Friedel and crafts
    etards to convert into benzaldehyde
    convert into cinnamic acid using perkins
    P/HI to convert COOH to CH3 ( i still dunno whether this rn can be used on paper )
    NBS to insert Br

Alternate

    benzene to propyl benzene using $ \ce{C2H5COCl and AlCl3}$ and then$ \ce{N2H4}$/$ \ce{HO\minus{}}$
    brominate using Br2 only to get 1 substituted propyl benzene
    alc. KOH
    NBS

First ,convert the benzene to benzaldehyde by the Gatterman-Koch reaction(treat it with CO in the presence of AlCl3).Then perform aldol condensation with Ethanal.Then reduce the aldehydic group by using LiAlH4.Then Use PBr5 for bromination to get the required product.

Is it correct :wink: ??????????????????

problems i think
1. gatterman koch rn requires HCl also
2. wouldnt you be reducing the double bond also in the process
And why do you want to use PBr5 ?
P.S. Chill on the question marks

No Only H2/Ni catalysis will reduce all the double bonds in the compound. Use LiAlH4 quantitatively i.e. 1 mole to get my reaction to happen

why shouldnt LAH reduce the double bond?

No i dont think what you say is true.
If you could accomplish things by mole concept or stoichiometry then why cant you monobrominate aniline?

That is because of the extensive electron density at the ortho and para positions due to the electron pair of the NH2 group involving in resonance.....Look at the acetanilide......SInce the lonepair on nitrogen is involved in resonance with the COCH3 group...We get only the paraproduct.........
Do you understand :wink: heyy!!!!!!!!!

Instead of just arguing over a never ending matter i accept to agree to Carcul's view on the topic
P.S. Carcul can you please recommend some books for the chem olympiad?

OK Let Carcul post the correct answer..................
p.s. Carcul--please recommend some chem olympiad books to me also

[quote="Madness"]1. benzene to toluene by Friedel and crafts
etards to convert into benzaldehyde
convert into cinnamic acid using perkins
P/HI to convert COOH to CH3 ( i still dunno whether this rn can be used on paper )
NBS to insert Br[/quote]

The problem is in P/HI.

[quote="Madness"]Alternate

benzene to propyl benzene using and then/
brominate using Br2 only to get 1 substituted propyl benzene
alc. KOH
NBS[/quote]

Very nice.

[quote="Valeriummaximum"]First ,convert the benzene to benzaldehyde by the Gatterman-Koch reaction(treat it with CO in the presence of AlCl3).Then perform aldol condensation with Ethanal.Then reduce the aldehydic group by using LiAlH4.Then Use PBr5 for bromination to get the required product.[/quote]

Very nice also. Just a remark: as Madness pointed out, the Gatterman formylation requires the presence of HCl (and also CuCl).

[quote="Madness"]wouldnt you be reducing the double bond also in the process[/quote]

No: for alpha-beta unsaturated aldehydes and ketones LiAlH4 reduces only the carbonyl.

[quote="Madness"]Carcul can you please recommend some books for the chem olympiad?[/quote]

The only books that I can recommend are the ones I use: "Chemistry: Molecules, Matter, and Change" from Atkins; for Organic Chemistry I have a lot of books, perhaps Solomons and Carey are my favourite for an introductory course.

[b]How about Part 2?[/b]

treat it with carbene, it adds to double bond.
aq KOH to substitute the bromo group for a OH group.
oxidise to acid by adding potassium dichromate in presence of sulphuric acid.

Right?  :D

[quote="valeriummaximum"]When you have a functional aldehyde group having greater nucleophilic charcter(since it is attached to a highly electro negative compound),LAH prefers to attack the aldehyde rather than the double bond.

Is my explanation satisfactory :wink: ????????????????[/quote]
actually the reducing nature of $ \ce{LiAlH4}$ is due to the $ \ce{H}^{\minus{}}$ it gives and that attacks a positive centre so there doesn't seem to be a reason of it attacking the already electron sufficient double bond
if u r so worried what about Meerwin-Verley-Pondorf reduction

[quote="Zimrock"]treat it with carbene, it adds to double bond.
aq KOH to substitute the bromo group for a OH group.
oxidise to acid by adding potassium dichromate in presence of sulphuric acid.[/quote]

Unfortunately, carbenes might also react with benzene rings. Also, is a cyclopropane ring stable under acidic/oxidative conditions?

But Chemrock told me that Cinnamaldehyde was a special case where Micheal addn takes place ( because of the nucleophilicity of the double bond ) and hence the double bond gets reduced  :maybe:

Michael addition occurs with weak nucleophiles, like cyanide, enolates, amines, enamines, and Gilman reagents; with strong nucleophiles (like hydride ion, organo lithiums and Grignard reagents) carbonyl addition is the main path.

So isnt Cinnamaldehyde a special case then?  :maybe:

What I said in my post above is the general case. Although, I was not aware of exceptions: so, I consulted one of my books and it turns out that cinnamyl systems, like cinnamaldehyde and cynnamic acid, are indeed special cases - with LiAlH4 [i]both[/i] the C-C double bond and the carbonyl are reduced.


For any three positive reals a, b, c, we have

$\frac{2a^3+abc}{b^2+c^2}+\frac{2b^3+abc}{c^2+a^2}+\frac{2c^3+abc}{a^2+b^2} \ge \frac{3}{2}(a+b+c)$
.

Another vornicu-schur type inequality.  :D
\[
\begin{gathered}
  \sum {\frac{{2a^3  + abc}}
{{b^2  + c^2 }}}  \hfill \\
   \geqslant \sum {\frac{{a^3  + a^2 b + a^2 c}}
{{b^2  + c^2 }}}  \hfill \\
   = \left( {a + b + c} \right)\sum {\frac{{a^2 }}
{{b^2  + c^2 }}}  \hfill \\
   \geqslant \frac{3}
{2}(a + b + c) \hfill \\
\end{gathered}
\]
by vornicu-schur and nesbitt

[quote="Andrei"][quote="siuhochung"]

 $ \sum {\frac{{2a^3  + abc}}
{{b^2  + c^2 }}}
   \geq \sum {\frac{{a^3  + a^2 b + a^2 c}}
{{b^2  + c^2 }}}$[/quote]

Could you please explain this step ?[/quote]
It is the vornicu-schur inequality as
\[
\begin{gathered}
  \sum {\frac{{a^3  + abc - a^2 b - a^2 c}}
{{b^2  + c^2 }}}  \hfill \\
   = \sum {\frac{{a(a - b)(a - c)}}
{{b^2  + c^2 }}}  \hfill \\
   \geqslant 0 \hfill \\
\end{gathered}
\]
You can read about vornicu-schur here.
http://www.mathlinks.ro/Forum/viewtopic.php?p=156212#p156212

The following more general inequality holds:

If $a_i$ are positive real numbers  then

$\frac{(n-1)a_1^n+g}{b_1}+\frac{(n-1)a_2^n+g }{b_2}+...+\frac{(n-1)a_n^n+g}{b_n} \ge \frac{n}{n-1}(a_1+a_2+...+a_n)$

where  $ g=a_1a_2...a_n$   and    $b_i=\sum_{k<>i}{a_k^{n-1}}$  for all $i$.
.


geometry
circumcircle
geometric transformation
rotation
trigonometry
angle bisector
geometry proposed

The ray $ [OA$ , the ray $ [OB$,  and a positive number $ k$ are given. Let $ P$ be a point on $ [OA$ and $ Q$ be a point on $ [OB$ such that $ OP\plus{}OQ \equal{} k$. Prove that the circumcircles of $ \triangle OPQ$ pass through a fixed point.

Let $ M,N$ be two points on the rays $OA,OB$ such that $ OM \equal{} ON \equal{} \frac {_1}{^2}k.$ Then

$ \frac {_1}{^2}k \minus{} MP \plus{} \frac {_1}{^2}k \plus{} NQ \equal{} k \ \Longrightarrow MP \equal{} NQ.$

Thus, circles $ \odot (OMN)$ and $ \odot (OPQ)$ meet at the center $ K$ of the rotation that takes the oriented segments ${MP}$ and $ NQ$ into each other. But $ K$ is the antipode of $ O$ WRT the circle $ \odot (OMN)$ because of $ KM \equal{} KN.$ Then $K$ is  fixed.

Let $ OM'\equal{}ON'\equal{}k/2$ and $ OM'' \plus{} ON'' \equal{} k$. Let $ P'$ be the intersection of $ (OM'N')$ and $ (OM''N'')$. $ \angle M''P'M' \equal{} N''P'N'$ and $ \angle OM''P' \plus{} \angle ON''P' \equal{} 180^\circ$. Also we know $ M'M''\equal{}NN''$. Use sinus theorem, then $ M'P' \equal{} P'N'$. $ OM'N'$ is fixed, so its angle bisector and its circumcircle, so their intersection point is fixed. So all $ (OMN)$s pass through a point on angle bisector of $ \angle MON$.

We will use ptolemy. Let $ P$ be the intersection point on angle bisector of $ \angle MON$ and $ (OMN)$. Use ptolemy. $ (OM \plus{} ON).MP \equal{} MN.OP$. Use sine law at $ \triangle MPN$, then $ \frac {MN}{sin(180^\circ \minus{} \alpha)} \equal{} \frac {MP}{sin(\alpha/2)}$. Combine two equations. $ OP \equal{} \frac {k}{2cos(\alpha/2)}$ which is fixed. The line $ OP$ is fixed, $ |OP|$ is fixed, so $ P$ is fixed. All $ (OMN)$s pass through the fixed point $ P$.


Are low carbs diet healthy? Why dont people just go out and excercise and eat healthy and well balanced diets?  and by healthy i dont mean a big mac w/ a DIET coke

Not good at all, just look at the data.

P.S. Did you know you can now use TeX? Thanks Mr. Rusczyk!

Low carb diets aren't healthy at all.  When you eat carbs, they are broken down into glucose monomers, which provide the main source of energy in aerobic respiration.  When you don't eat carbs, the body must break down fats and stuff.  This is the main idea behind the diet.

Problem is, the body also breaks down proteins, which are critical to virtually every process in the body.  Not only are proteins important in muscle structure and movement, but all enzymes are proteins, and as any biology student knows, enzymes facilitate almost all reactions.

I strongly recommend working out (particularly running or walking) and maintaining a healthy, but carb-containing diet.  You may not lose weight, but that's because you're gaining muscle mass at the same time as you're losing fat.

the dude that invented low-carb diets died fat. end of story.
i think my point is made.

[quote="white_horse_king88"]Fat is ESSENTIAL -  to all those people who have no clue.
It is important in the cell membrane, it is energy storage, it is lots of vital things your body would die without.
FAT is like ALCOHOL.
Bad in excess - good in moderation.[/quote]

No. I don't agree. I will claim that any alcohol is bad. No one ever listens to me though.

[quote="ComplexZeta"]I will claim that any alcohol is bad. No one ever listens to me though.[/quote]

I will listen to you long enough for you to lay out your evidence for your claim.

Well, first of all, studies tend to show that grape juice (which is nonalcoholic) helps the heart as much as wine does, so the alcohol gives no health benefit. But what concerns me more is the bias in published articles. People regularly see articles in, for example, [i]Scientific American[/i] claiming that alcohol in moderation is a health benefit. Well, ok, so maybe a few studies show that something in wine (most likely grapes, but we'll ignore that from now on in this post) helps stop heart attacks. But I worry about which articles get published. If someone does a study that shows that alcohol is harmful, which scientific magazine would publish the result? Everyone knows that alcohol is bad for people, so what's the points in publishing a result that everyone knows? But if a study (that possibly isn't performed with ``scientific rigor,'' whatever that means) shows that alcohol is [i]good[/i] for people, that sounds like a really amazing result because it contradicts the intuitions of all. Therefore, the result will be published all over the place, which makes it seem that loads of studies have verified the results because readers can read the results in so many different sources.

my grandfather drank one beer a day for a very good portion of his older years. he lived to be very very old. drinking a beer a day was a backwoods prescription. and like many backwoods prescriptions and wive's tales, people are finding out that there is some truth in it.

But I'll claim that he lived for a long time despite drinking the beer, not because of it. And there's nothing at all you can do to demonstrate that I'm wrong.

I'm pretty sure that the epidemiological evidence for a reduced mortality associated with MODERATE consumption of alcohol has already been shown to be independent of the source of the alcohol (that is, whether it is grape wine, rice wine, beer, distilled spirits, whatever). There are a lot of differing ethnic groups and religious groups with varied habits in this regard to provide natural experiments for testing hypotheses, and some very reasonable proposed mechanisms for why low-dose occasional ethanonl consumption is good for health. For all that, I often forget to take a dose of the wine that I began keeping around the house after I turned forty. (I also drink beer from time to time, but I don't have any in the house today.) I eat lots of fresh grapes, raisins, and grape juice too, so I ought to live to be 100  :wink: My maternal grandmother made it to age 99.

[quote="SirErnest"]someone told me that drinking grape juice makes you jump higher. is this true?[/quote]
Of course not.

How much higher? It might be possible to jump a few microns higher after drinking grape juice, but I doubt you'd be able to jump an extra meter.

[quote="SirErnest"]they said it had to do with the glucose in the grapes[/quote]
You mean fructose right? :)


conics
parabola
ellipse
inequalities unsolved
inequalities

Hello everyone

How do I identify the curves such as this one?

4x^2 + 9y^2 - 16x +54y +61= 0

is a parabola formula like that? circle? elipse? I AM LOST! How do I find center or vertex, and radius if applicable...

PLEASE HELP! it's urgent!

$ 4x^2 \plus{} 9y^2 \minus{} 16x \plus{} 54y \plus{} 61 \equal{} 0\Longleftrightarrow \frac {(x \minus{} 2)^2}{9} \plus{} \frac {(y \plus{} 3)^2}{4} \equal{} 1$ which is an ellipse.

Completing square for $ x$ and $ y$ at the same time.

Ok bear with me please. I am doing it this way...

4x^2 - 16 x + 9y^2 =54y = -61

4(x^2 - 16x) + 9(y^2 + 54y) = -61

completing square

4(x^2 - 16x +64) = 9( y^2+ 54y +729) = -61 + 4(64) + 9(729)

when i divide each side by 6756, it doesn't come out right!!!!!!!!!!!!!

In the second line, you made mistakes. :(

ohhhhhhhhhh silly me!

4(x^2 - 4x) + 9(y^2 + 6y) = -61

Now it works perfect!

Thanks a lot!  :lol: