help me solve an equation :
   (sin2x)^2 + sin2x + sinx + 1 = 0  :(

[hide="Hint"]
$ \sin 2x \equal{} 2 \sin x \cos x$

$ (2 \sin x \cos x)^2 \plus{} 2 \sin x \cos x \plus{} \sin x \plus{} 1 \equal{} 0$

$ 4 \sin^2 x \cos^2 x \plus{} 2 \sin x \cos x \plus{} \sin x \plus{} 1 \equal{} 0$

I think you can do something with $ \cos^2 x$ with Pythagorean Identity, but that $ \cos x$ is going to be bit annoying. Try.  :) [/hide]

i have  dissected as you do . but i didn't have result  :(
it' s 15' exercise of my class

I just entered the original equation into the solving subsection of
the equation section of http://www.quickmath.com

It states "Warnings  An exact solution could not be found."

And if you go to "Plot Equations"

Equations: Type in:  y = (sin(2x))^2 + sin(2x) + sin(x) + 1 = 0

Tick Marks (check this)
Grid lines (check this)

Suggested:
x:  -20 to 20
y:  -1 to 4

Automatic origin (check this)
Height : weight = 8:8  . . . (can vary)
Text output size:  leave at 20 points

---    ---    ---    ---      ---    ---     ---     ---

The solutions look to be irregularly spaced apart on the x-axis.

Note the solution $ \frac{3\pi}{2}\plus{}2\pi\cdot i$, $ i\in \mathbb{Z}$. However, the other solution modulo $ 2\pi$ is transcendental. Indeed,  in general, a trigonometric equation does not have any nice solutions. (In this case, you may find an equation in $ \sin{x}$ with a root of $ \minus{}1$.)

hello, rewriting your equation in the form
$ 4\sin(x)^2\cos(x)^2+2\sin(x)\cos(x)+\sin(x)+1=0$ and converting $ \sin(x)$ and $ \cos(x)$ in $ \tan(\frac{x}{2})$ we get
$ {16\,{\frac { \left( \tan \left( 1/2\,x \right)  \right) ^{2} \left( 1-
 \left( \tan \left( 1/2\,x \right)  \right) ^{2} \right) ^{2}}{
 \left( 1+ \left( \tan \left( 1/2\,x \right)  \right) ^{2} \right) ^{4
}}}+4\,{\frac {\tan \left( 1/2\,x \right)  \left( 1- \left( \tan
 \left( 1/2\,x \right)  \right) ^{2} \right) }{ \left( 1+ \left( \tan
 \left( 1/2\,x \right)  \right) ^{2} \right) ^{2}}}+2\,{\frac {\tan
 \left( 1/2\,x \right) }{1+ \left( \tan \left( 1/2\,x \right)
 \right) ^{2}}}+1=0}$
substituting $ t=\tan(\frac{x}{2})$ and factoring we have
$ (t+1)(t^7-3t^6+23t^5-21t^4-5t^3+15t^2+5t+1)=0$
to solve.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, rewriting your equation in the form
$ 4\sin(x)^2\cos(x)^2 + 2\sin(x)\cos(x) + \sin(x) + 1 = 0$ and converting $ \sin(x)$ and $ \cos(x)$ in $ \tan(\frac {x}{2})$ we get
$ {16\,{\frac {( \tan ( x/2) ) ^{2} ( 1 - ( \tan ( x/2) ) ^{2} ) ^{2}}{( 1 + ( \tan ( x/2 ) ) ^{2} ) ^{4 }}} + 4\,{\frac {\tan ( x/2 ) ( 1 - ( \tan ( x/2 ) ) ^{2} ) }{( 1 + ( \tan ( x/2 ) ) ^{2} ) ^{2}}} + 2\,{\frac {\tan ( x/2 ) }{1 + ( \tan ( x/2 ) ) ^{2}}} + 1 = 0}$
substituting $ t = \tan(\frac {x}{2})$ and factoring we have
$ (t + 1)(t^7 - 3t^6 + 23t^5 - 21t^4 - 5t^3 + 15t^2 + 5t + 1) = 0$
to solve.
Sonnhard.[/quote]

The above of what I typed in this quote
box is more consistent with the top right
(same quote box) of the tangent of half of
x form than of what you typed out prior.

Else, the equivalent of $ \tan ( 1/2  x)$ . . . (1)
showing up on the screen doesn't work because
the variable needs to be immediately adjacent to
the right of the constant.  And, in this horizontal
style, that would require most likely parentheses
around the fraction of 1/2 or the use of a decimal
such as .5:

$ \tan(1/2)x$, or possibly $ \tan((1/2)x)$, for the fraction examples . . . (2)

or $ \tan(.5x)$ as a decimal example . . . (3)

I don't recommend the forms in (2) because of more characters,
and I don't want to use the form in (3) because of the
aesthetics of the decimal point.


abstract algebra
category theory
superior algebra
superior algebra theorems

Show that if $M$ is a projective module, then there exists a module $N$ such that $M\oplus N$ is a free module.

Also, show that any projective module is flat.

The first one is the definition of projective modules, or did you use another¿

I think he's referring to the following definition (which actually gives a hint as to where that "projective" comes from):

$_RM$ is projective (as a left module over the ring $R$) if for any $R$-module epimorphism $f: A\to B$ and any $R$-module homomorphism $g: M\to B$, we can find an $R$-module homomorphism $\varphi: M\to A$ such that $g=f\circ\varphi$.

This is equivalent to $M$ being a direct summand of a free $R$-module. To see the implication baz is interested in, notice that, since every module is the homomorph of a free module, we can find an epimorphism $f: F\to M$, where $F$ is free. Now apply the definition to $A=F$, $f$ as defined above, $B=M$, and $g$ being the identity on $M$. The homomorphism $\varphi$ that the definition gives us is an embedding of $M$ in $F$ as a direct summand, as can easily be seen. The other implication is also easy.

Let $P$ be a projective left $R$-module. We want to prove that $P$ is flat. This is equivalent to proving that if $0\to K\to M$ is a very short exact sequence of right $R$-modules, then $0\to K\otimes_RP\to M\otimes_RP$ is also exact.

It is known and I won't prove it to you no matter how nice you ask that for every left $R$-module $N$, $-\otimes_RN$ is a covariant right exact functor on the category of right $R$-modules. This is a fancy way of saying that if $0\to K\to M\to L\to 0$ is a short exact sequence of right $R$-modules, then
$K\otimes_RN\to M\otimes_RN\to L\otimes_RN\to 0$ is also exact.

[b]Lemma 1:[/b] If $K$, $M$ and $L$ are right $R$-modules and $0\to K\to M\to L\to 0$ splits, then the sequence $0\to K\otimes_RN\to M\otimes_RN\to L\otimes_RL\to 0$ is exact and splits.
Proof: Since $0\to K\to M\to L\to 0$ splits, you also have the same exact sequence with the arrows reversed. Apply the right exactness of $-\otimes_RN$ for these sequences.

[b]Corollary:[/b] Under the conditions necessary for this to make sense, $(A\oplus B)\otimes_R C\simeq (A\otimes_RC)\oplus(B\otimes_RC)$.

Back to our problem: If $P$ is a projective left $R$-module, then there exists $n\in\mathbb N$ and a left module $Q$ such that $P\oplus Q\simeq R^n$.
Let $0\to K\to M$ be an exact sequence of right $R$-modules. Then $0\to K\otimes_RR^n\to M\otimes_RR^n$ is exact, hence $0\to (K\otimes P)\oplus (K\otimes Q)\to (M\otimes P)\oplus(M\otimes Q)$ is exact. Conclude that $0\to K\otimes_RP\to M\otimes_RP$ is exact.

With the same proof it follows that a direct sum of modules is flat if and only if every summand is flat.

[quote="artin"]Show that if $M$ is a projective module, then there exists a free module  such $F$ that $F\oplus M$ is a free module.[/quote]

Extra assumptions other than $R$, the base ring, be any ring with $1$ and $M$ any projective module, finitely generated or not?

Take a possibly infinite free resolution $F^\cdot$ of $M$. Inductively, we get $F^{n}= K^{n}\oplus K^{n-1}$ with $K^{0}= M$.

Now, $M \oplus K^{1}\oplus K^{2}\oplus \ldots$ is free: $(M \oplus K^{1}) \oplus (K^{2}\oplus K^{3}) \oplus \ldots$ and $(K^{1}\oplus K^{2}) \oplus \ldots$ is also free.

In general, such a free module can't be finitely generated.

That is a wonderful proof sissa.

You may wish to add that $\oplus_{i\geq 1}K^{i}$ is also free because it is
$\oplus_{i\geq 1}F^{2i}$.


AMC
USA(J)MO
USAMO
pigeonhole principle
inequalities

I was thinking we could compile the solutions to some USAMO problems, and because I don't want to retype everthing, I'm just going to link to the .pdf and hope for some answers. :wink:

So take a look at the following problems and post a few solutions:
http://www.tjhsst.edu/~tmildorf/math/2004TJUSAMO.pdf

Problem 3:
$\angle{APY}+\angle{A}=180^{o}$, $\angle{ZPX}+\angle{B}=180^{o}$. Add them togather we get $\angle{APY}+\angle{ZPX}+180^{o}-\angle{C}=360^{o}$. Since $\angle{APY}+\angle{ZPX}=360^{o}-\angle{XPY}$, hence $\angle{C}+\angle{XPY}=180^{o}$. Therefore $C, X, P, Y$ are concyclic.

#1 - Pigeonhole on (1,2), (3,4), etc. n and n+1 are relatively prime.

#2 - basic factoring.

[hide="5"]It's pretty simple to find a counterexample for $n=4$.

Here's a nice proof from the WOOT forum that any $6\times 6$ board can be divided into two pieces with a straight cut:

[quote="ThAzN1"]There are 5 + 5 = 10 lines within the chessboard parallel to one of the sides. Assume for the sake of contradiction that all of these pass through a domino. Note that each domino has exactly one line passing through it. Consider any one of the 10 lines (let's call it L). Note that L must pass through an even number of dominoes (just count areas). So, each of the 10 lines cuts through at least 2 dominoes -- that gives us a total area of at least 40. Woah, and I think we're done.[/quote]

The answer is $n=3$.[/hide]

Problem 4:

[hide]
Very similar to the solution Peter VDD gave in 'The attack of inequalities' with rearrangement.

We know that $a^{5}+b^{5}\geq ab(a^{3}+b^{3})$ so \[ \sum_{cyc}{\frac{xy}{x^{5}+xy+y^{5}}\leq \sum_{cyc}{\frac{1}{x^{3}+y^{3}+1}}}\] And now the problem reduces to USAMO 1997[/hide]

It's funny how a problem from 1996 can reduce to a problem from 1997. :)

Indeed, I think we can guess where USAMO got the inspiration  :wink:


calculus
limit
topology
search
real analysis
real analysis solved

An outline of a path:

[hide]Assumption: every bounded increasing sequence converges.

Simple corollary: every bounded monotone sequence converges.

Nested interval theorem:

If $[a_0,b_0]\supset[a_1,b_1]\supset[a_2,b_2]\supset\cdots$ then $\bigcap_{n=0}^{\infty} [a_n,b_n]\ne\emptyset.$

Furthermore, if $\lim_{n\to\infty}(b_n-a_n)=0,$ then the intersection consists of a single point.

(The proof follows quite directly: $\{a_n\}$ and $\{b_n\}$ are bounded monotone sequences, after all.)

Next theorem: Suppose $A$ and $B$ are disjoint nonempty subsets of $\mathbb{R}$ such that $A\cup B=\mathbb{R}$ and for every $a\in A,\,b\in B,\, a<b.$ Then either $A$ has a greatest element or $B$ has a least element.

We start with $a_0\in A,\,b_0\in B.$ Let $x=\frac{a_0+b_0}2.$ If $x\in A,$ then let $a_1=x,\,b_1=b_0.$ Else if $x\in B,$ then let $a_1=a_0,\,b_1=x.$ Iterate this process. By the nested interval theorem, there exists a unique $y$ such that $a_n\le y\le b_n,\,\forall\,n.$ If $y\in A,$ then it is the greatest element of $A.$ If $y\in B,$ then it is the least element of $B.$

Final proof: Suppose $S$ is a nonempty set with an upper bound. Let $B$ be the set of all upper bounds of $S.$ Let $A=\mathbb{R}\setminus B.$ Verify that these sets satisfy the hypotheses of the previous theorem, and chase down the consequences of that.[/hide]

[quote="Arne"]Can you prove that any non-empty bounded set of reals has a supremum assuming the fact that any increasing bounded sequence converges?[/quote]

This question does not make much sense if you do not precise your definition of the real numbers. Taking a standard definition like the completion of $\mathbb{Q}$ for instance will give the result almost immediately (and you will never need your hypothesis that every increasing bounded sequence converges). Same thing if you say that $\mathbb{R}$ is the maximal archimedean totally ordered field, or isomorphic to any totally ordered Dirichlet-complete field.
To sum it up, the question makes sense only if you specify what you call a real number.

[quote="julien_santini"]
This question does not make much sense if you do not precise your definition of the real numbers. [/quote]
Well, what was, probably, meant is that the axioms of an ordered field together with the postulate that every bounded increasing sequence has a limit imply the completeness axiom as it is stated in most textbooks (existence of the supremum). Understood this way, the question makes perfect sense.

Exactly. And Kent Merryfield gave a correct solution :)

Ok, seen like that this does make sense.
(By the way there is a typo in KM's post: the inclusions in the nested interval theorem should be the other way). [color=red][Fixed now. - K.M.][/color]

Note that the nested interval theorem is a particularization of the fact that a metric space is complete iff the intersection of a decreasing sequence of nonempty closed sets (whose diameters tend to $0$) is reduced to one point. This also can be seen by saying that the intersection of compact subspaces of an Hausdorff topological space is compact (and nonempty here). However for the latter we need to know that closed bounded intervals of $R$ are compact, which is nontrivial.

[quote="Arne"]In most calculus courses, one proves that any increasing bounded sequence converges using the fact that any non-empty bounded set of reals has a supremum.

Can you prove that any non-empty bounded set of reals has a supremum assuming the fact that any increasing bounded sequence converges? :D[/quote]

The subtlety is that the "easy" direction is canonical whereas the "hard" direction requires some form of the axiom of (countable) choice; given a set of reals you have to somehow select a sequence converging to its supremum, and it is hard to believe that this can be done in a uniform way.   I don't remember clearly the difference between Countable Choice (CC) and the stronger axiom of countable Dependent Choice (DC) but I believe the following argument uses DC:

given a nonempty set of reals (it contains some number $a$) and a strict upper bound for this set ($b$), repeatedly bisect the interval $[a,b]$ to search for the supremum (i.e. select subinterval so that the smaller endpoint is never a strict upper bound and the larger endpoint always is one).

I doubt it can be done with pure CC, but maybe somebody here knows for certain.

I think this is also related to the fact that different completeness axioms are inequivalent in the generality of partially ordered sets or (maybe) linearly ordered sets, so the easy theorems are the ones that are true for posets and the harder ones are the statements that are false in the generality of posets (so one must use some specific properties of the real numbers).    Just a guess.


Let S be the sum of the base 10 logarithms of all the proper divisors of 1000000. What is the integer nearest to S?

$1000000=10^{6}=5^{6}2^{6}$
So 1000000 has 48 proper divisors.  Each of these divisors except 1 and 1000 have a pair with which they multiply to 1000000.  Using the property $\log a+\log b=\log ab$ we sum these pairs first so that we have $23\log1000000+\log1+\log1000$.  We know $\log1000000=6$ and $\log1=0$ and $\log1000=3$ so the sum is $23(6)+0+3=141$.


abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a finite group whose order is not divisible by $3$, suppose $(ab)^{3}= a^{3}b^{3}$   $\forall a,b \in G$.
Prove that $G$ must be Abelian.

Since the order of $G$ is not divisible by $3$, there are integers $u,v$ s.t. $u|G|+3v = 1$. Then for every $x \in G$

$x = x^{u |G|+3v}= (x^{v})^{3}$

In particular, the map $G \to G ,~ x \mapsto x^{3}$ is surjective. But then its also injective. Hence its an isomorphism. This suffices to show (independantly from the assumptions before), that $G$ is abelian:

For all $x,y \in G$, we have $(xy)^{3}= x^{3}y^{3}$, i.e. $(yx)^{2}= x^{2}y^{2}$. Thus,

$y^{3}x^{3}= (yx)^{3}= (yx)^{2}yx = x^{2}y^{2}yx = x^{2}y^{3}x$
$y^{3}x^{2}= x^{2}y^{3}$
$y^{3}= (x^{2}y x^{-2})^{3}$
$y = x^{2}y x^{-2}$
$y x^{2}= x^{2}y$
$(xy)^{2}= y^{2}x^{2}= x^{2}y^{2}$
$yx = xy$


[quote="E^(pi*i)=-1"]This is a little easy even for getting started . . .

[hide] Raise 4 to both sides to get $x=4^5=1024$.[/hide][/quote]This is OK because logarithms are usually not taught in middle schools and this is the lowest high school forum, so this is OK.


This game is so dumb, I thought it belonged here.

Who plays sticks?

I resumed after 2 idol months of not playing sticks.

How many different variations are there of this?

umm.. you mean like pick up sticks??
confuzzled, lol.

Are you talking about that one game that resembles the wii, except its much worse/cheaper?

No no and no.
Well, I can't really explain right here...

Do you mean the game where you play with your fingers and hit them on each other, and add the sum, and you lose 1 hand when it gets to 5?

Yes.
Of course its much cheaper than wii you dont need to buy it.

I thought that game was called chopsticks...

Yeah...its called chopsticks :P
When you said sticks...I thought you meant: [url=http://www.wired.com/reviews/product/golive2_stix_200]Stix[/url]

well...
um...
this will get a lot of spam...
and will probably get locked down as soon as dragon96 sees it...
i thinks there is an easy way to win chopsticks.


inequalities logarithms limit induction function calculus derivative

[quote="kimix"]$ x$ is a positive real number.
Prove that $ 2^{ \minus{} x} \plus{} 2^{ \minus{} \frac {1}{x}} \le 1$[/quote]
AM-GM: $ 2^{\minus{}x} \plus{} 2^{\minus{}\frac{1}{x}} \le 2 \sqrt{2^{\minus{}x\minus{}\frac{1}{x}}} \le 1 \iff$
$ 2^{\minus{}x\minus{}\frac{1}{x}} \le 2^{\minus{}2} \iff$
$ 2^{x\plus{}\frac{1}{x}} \ge 2^2 \iff$
$ x\plus{} \frac{1}{x} \ge 2$ which is just AM-GM again.

[quote="Marius Mainea"][quote="kimix"]$ x$ is a positive real number.
Prove that $ 2^{ \minus{} x} \plus{} 2^{ \minus{} \frac {1}{x}} \le 1$[/quote]

 Let $ f(x) \equal{} 2^{ \minus{} x} \plus{} 2^{ \minus{} \frac {1}{x}}$

 Then x=1 is maxim absolute.[/quote]

How do you prove it?

Easy :wink:

 Compute f' and write variation to f.

 $ f^{\prime}(x) \equal{} \frac {\ln 2\cdot2^{ \minus{} x}}{x^2}(2^{x \minus{} \frac {1}{x}} \minus{} x^2)$

 $ g(x) \equal{} (x \minus{} \frac {1}{x})\ln 2 \minus{} 2\ln x$

 $ g^{\prime}(x) \equal{} (1 \plus{} \frac {1}{x^2})\ln 2 \minus{} \frac {2}{x}$ ..............................

 Then f' is increasing from $ (0,x_1)\cup (x_2,\infty)$ wher $ x_1,x_2$ is the roots of ecuation $ \ln 2x^2 \minus{} 2x \plus{} \ln 2 \equal{} 0$.................................

[quote="Marius Mainea"]Easy :wink:

 Compute f' and write variation to f.

 $ f^{\prime}(x) \equal{} \frac {\ln 2\cdot2^{ \minus{} x}}{x^2}(2^{x \minus{} \frac {1}{x}} \minus{} x^2)$

 $ g(x) \equal{} (x \minus{} \frac {1}{x})\ln 2 \minus{} 2\ln x$

 $ g^{\prime}(x) \equal{} (1 \plus{} \frac {1}{x^2})\ln 2 \minus{} \frac {2}{x}$ ..............................[/quote]

Your calculation is correct so far, but how do you reach the conclusion?

For $ x\geq 1$, we have $ 2^{ \minus{} x}\leq \frac {1}{2}\cdot \frac {1}{x\ln 2 \plus{} 1 \minus{} \ln2}$ by the convexity of $ 2^{x}$, so

$ 2^{ \minus{} x} \plus{} 2^{ \minus{} \frac {1}{x}}\leq \frac {1}{2}\left(\frac {1}{x\ln 2 \plus{} 1 \minus{} \ln 2} \plus{} \frac {1}{(1 \minus{} \ln 2)x \plus{} \ln 2}\right)$

$ \leq \frac {1}{2}\left(\frac {1}{\ln 2 \plus{} 1 \minus{} \ln 2} \plus{} \frac {1}{1 \minus{} \ln 2 \plus{} \ln 2}\right) \equal{} 1$. The equality holds when $ x \equal{} 1$.

[quote="kunny"][quote="Marius Mainea"]

$ 2^{ \minus{} x} \plus{} 2^{ \minus{} \frac {1}{x}}\leq \frac {1}{2}\left(\frac {1}{x\ln 2 \plus{} 1 \minus{} \ln 2} \plus{} \frac {1}{(1 \minus{} \ln 2)x \plus{} \ln 2}\right)$

[/quote][/quote]

Your inequality is not true. take $ x\minus{}>\infty$

[quote="kimix"]you've used AM-GM in the wrong way !!![/quote]
Yes I did. Thanks for noticing. I'll give the problem a second try later. :)

No answer ?  :wink:
Actually there is a more general problem:
if $ x_i \ge 0$, $ i\equal{}1,...,n$ such that $ \sum_{i\equal{}1}^{n}{\frac{1}{1\plus{}x_i}}\le1$
Then $ \sum_{i\equal{}1}^{n}{2^{\minus{}x_i}} \le 1$

Let $ y_i \equal{} 2^{ \minus{} x_i}$, $ \sum \frac {1}{1 \minus{} \ln y_i}\le 1$. We want $ \sum y_i \le 1$. But $ x(1 \minus{} \ln x)\le 1\Leftrightarrow x\ln x \plus{} 1 \ge x$ is true, so we have $ 1\ge \sum \frac {1}{1 \minus{} \ln y_i} \ge \sum y_i$.

EDIT: Oops.  :oops: We just take log base 2, then; I believe it works then.

[quote="fwolth"]Let $ y_i \equal{} 2^{ \minus{} x_i}$, $ \sum \frac {1}{1 \minus{} \ln y_i}\le 1$. We want $ \sum y_i \le 1$. But $ x(1 \minus{} \ln x)\le 1\Leftrightarrow x\ln x \plus{} 1 \ge x$ is true, so we have $ 1\ge \sum \frac {1}{1 \minus{} \ln y_i} \ge \sum y_i$.[/quote]
$ \ln y_i \equal{} \minus{} \ln 2 \cdot x_i$ :)

[quote="kimix"]$ x$ is a positive real number.
Prove that $ 2^{ \minus{} x} \plus{} 2^{ \minus{} \frac {1}{x}} \le 1$[/quote]

$ x \ge 0$
$ 2^x \ge 1\plus{}x \Rightarrow \frac{1}{2^x}\plus{}\frac{1}{2^{\frac{1}{x}}} \le \frac{1}{1\plus{}x}\plus{}\frac{x}{1\plus{}x}$

[quote="Sergic Primazon"][quote="kimix"]$ x$ is a positive real number.
Prove that $ 2^{ \minus{} x} \plus{} 2^{ \minus{} \frac {1}{x}} \le 1$[/quote]

$ x \ge 0$
$ 2^x \ge 1 \plus{} x \Rightarrow \frac {1}{2^x} \plus{} \frac {1}{2^{\frac {1}{x}}} \le \frac {1}{1 \plus{} x} \plus{} \frac {x}{1 \plus{} x}$[/quote]

That's wrong. Try $ x\equal{}\frac{1}{2}$.

[quote="Marius Mainea"]Easy :wink:

 Compute f' and write variation to f.

 $ f^{\prime}(x) = \frac {\ln 2\cdot2^{ - x}}{x^2}(2^{x - \frac {1}{x}} - x^2)$

 $ g(x) = (x - \frac {1}{x})\ln 2 - 2\ln x$

 $ g^{\prime}(x) = (1 + \frac {1}{x^2})\ln 2 - \frac {2}{x}$ ..............................

 Then f' is increasing from $ (0,x_1)\cup (x_2,\infty)$ wher $ x_1,x_2$ is the roots of ecuation $ \ln 2x^2 - 2x + \ln 2 = 0$.................................[/quote]

 Because $ \lim_{x\yo0}f^{\prime}(x)=-\ln 2$ , $ f^{\prime}(1)=0$ , $ \lim_{x\to\infty}f^{\prime}(x)=0$ and $ x_1<1<x_2$ , result  that  f' have  two roots $ y_1\in (0,1)$ and $ y_2\in (1,\infty)$

 Since  $ \lim_{x\to 0}f(x)=1$ , $ f(1)=1$ and $ \lim_{x\to\infty}f(x)=1$ we have  that $ f(y_1)$ , $ f(y_2)$ is minimum local from f and f(1) is maximum global.

 Done.

I think i got a solution! Suppose the case $ n \equal{} 2$ proved. and the inequality true for $ n \minus{} 1$ Then if we suppose that

$ x_i\ge1$ for every $ i$ then $ 2^{x_i}\ge 1 \plus{} x_i$ and the result follows.

else there is at most one x_i less than 1. Suppose $ x_1 \le 1 \le x_2 \le... \le x_n$

Since $ \frac {1}{1 \plus{} x_1} \plus{} \frac {1}{1 \plus{} x_2} \le 1$ then $ \frac {1}{x_1} \le x_2$.

denote $ \alpha \equal{} \frac {x_1}{1 \plus{} x_1}$.

we have $ \sum_{i \equal{} 2}^{n}{\frac {1}{1 \plus{} \alpha x_i \plus{} \alpha \minus{} 1}} \le 1$ and $ \alpha x_i \plus{} \alpha \minus{} 1 \ge 0$ for $ i \equal{} 2,...,n$

Then by induction hypothesis $ \sum_{i \equal{} 2}^{n}{2^{ \minus{} (\alpha x_i \plus{} \alpha \minus{} 1)}} \le 1$

Or $ \sum_{i \equal{} 2}^{n}{2^{ \minus{} \alpha x_i}} \le 2^{\alpha \minus{} 1}$

then $ \sum_{i \equal{} 2}^{n}{2^{ \minus{} x_i}} \le 2^{ \minus{} \frac {1}{x_1}}$

So $ \sum_{i \equal{} 1}^{n}{2^{ \minus{} x_i}} \le 2^{ \minus{} \frac {1}{x_1}} \plus{} 2^{ \minus{} x_1} \le 1$

if there's is an $ \alpha$ that satisfies $ 2^x\geq x^{\alpha}\plus{}1$, then $ \alpha\equal{}2\ln 2$...

consider the function, $ f_r(x)\equal{}2^x\minus{}x^r\minus{}1$... we know that $ x\equal{}1$ is a "zero" of the function.. so, if we want that $ f_r(x)\geq 0$ for all $ x$, we should have also that $ f'(1)\equal{}0$, which implies that $ r\equal{}2\ln 2$...

i haven't proved this inequality yet...

i think i have proved it.. let $ f(x) \equal{} 2^x \minus{} x^{2\ln2} \minus{} 1$... then $ f'(x) \equal{} \ln 2(2^x \minus{} 2x^{2\ln2 \minus{} 1})$ and $ f''(x) \equal{} \ln 2(2^x\ln 2 \minus{} 2(\ln2 \minus{} 1)x^{2\ln2 \minus{} 2})$...

note that $ f''(x)$ is increasing and continuous on the positive reals... $ f''(x)\rightarrow \minus{} \infty$ when $ x\rightarrow 0$ and $ f''(1) > 0$, so there exists exactly one real $ t$, $ 0 < t < 1$, such that $ f''(t) \equal{} 0$...

in $ (t,\infty)$ $ f$ is convex, which implies that $ f'$ is increasing, therefore $ 1$ is the only value in $ (t,\infty)$ for which $ f'(x) \equal{} 0$... then, the minimum in $ (t,\infty)$ is $ f(1) \equal{} 0$... we get also that $ f(t) > 0$, since $ f(t)\neq f(1)$ because this would imply that the derivative would vanish for a real in $ (t,1)$, and this is impossible...

in $ (0,t)$ $ f$ is concave... we have that all the points lie above the line through $ (0,0)$ and $ (t,f(t))$ which implies that $ f(x) > 0$ in this case, and we're almost done..  :D

then, $ 2^{ \minus{} x} \plus{} 2^{\frac { \minus{} 1}{x}}\leq \dfrac{1}{x^{2\ln2} \plus{} 1} \plus{} \dfrac{1}{\frac {1}{x^{2\ln2}} \plus{} 1} \equal{} 1$, and now we're done  :D


floor function
ceiling function
function
calculus
integration

[quote="electron-proton"]Find the $ 2007^{th}$ term of the series
1,2,2,3,3,3,4,4,4,4.........[/quote]

$ \frac {62 \times 63}{2} < 2007 < \frac {63 \times 64}{2}$

So the 2007th term $ \equal{} 63$

This should be moved to High School Basics.

yes u r right but can u tell me how u found that bounds??

is it it related to the fact that the sum of n natural numbers
is $ \frac {n(n \plus{} 1)}{2}$?

It's because at first we are summing 1 one, then two 2s, and so on so forth.  So the number of numbers is given by
$ \frac{n(n\plus{}1)}{2}$

Then we are merely finding which number occurs for the 2007th term, and that is 63 by the means of the previous post.

you find it by noticing that the last time a number $ n$ shows up in the sequence it's the $ \dfrac{n(n \plus{} 1)}{2}$th term.

For example the last time the number $ 2$ shows up it's the $ 3$rd term or $ 1 \plus{} 2$. Another example would be the last time the number $ 3$ shows up, and that's the $ 6$th term which is also $ 1 \plus{} 2 \plus{} 3$.

In conclusion if you want to find what number is at the $ n$th term, you need to find two consecutive numbers $ k$ and $ k \plus{} 1$ such that $ \dfrac{k(k \plus{} 1)}{2} < n < \dfrac{(k \plus{} 1)(k \plus{} 2)}{2}$ then the number at the $ n$th term is $ k \plus{} 1$. If $ n$ is in the form $ \dfrac{k(k \plus{} 1)}{2}$ then the number in that term is $ k$.

Edit: Oops cipos2 already showed why.

Technically, given the first few terms of a sequence and trying to find the $ n^{th}$ term can get you infinitely many answers. You need to have the definition of the sequence.
See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=264274[/url]

How to derive a closed-form for the nth term of the sequence (w/ def. provided, of course)?

$ \left\lfloor\frac{1}{2} \plus{} \sqrt{2n}\right\rfloor$ or $ \left\lceil\frac{\sqrt{1\plus{}8n}\minus{}1}{2}\right\rceil$. How to derive it? Hints: use the $ \frac{n(n\plus{}1)}{2}$ and solve the quadratic. Add the floor/ceiling function to keep the answer in integers.

this years mmpc part II question number 2 was actually this exact question-finding a closed form. [url=http://www.svsu.edu/mmpc/mmpcep2c.pdf]here is the link to the solution[/url].

Nice link! How did you find it? I only got to [url=http://gauss.math.oakland.edu/mmpc/previousexams.html]this page[/url] but unfortunately it doesn't have 2009 problems.

By the way how to show that $ \left\lfloor\frac {1}{2} \plus{} \sqrt {2n}\right\rfloor \equal{} \left\lceil\frac {\sqrt {1 \plus{} 8n} \minus{} 1}{2}\right\rceil$?
I've got $ \left\lfloor\frac {1 \plus{} \sqrt {8n}}{2}\right\rfloor \equal{} \left\lceil\frac {\sqrt {1 \plus{} 8n} \minus{} 1}{2}\right\rceil$

and then use the fact that for non-integral $ x, \lfloor x \plus{} 1\rfloor \equal{} \lceil x\rceil,$ but how to account for $ \sqrt {8n}$ and $ \sqrt {8n \plus{} 1}$?

I am looking for the MMPC pdfs - part2  of years 2008-09, 2011 and 1968 (in order to complete [url=https://artofproblemsolving.com/community/c5h2760506p24143309]this[/url]) ,  does anyone have them?

Only with problem wordings (I do not care if it has solutions)


suppose a circle with diameter $AB$ .we have been given a chord $CD$ under $AB$.
(suppose $AB$ is horizental).  .find $M$ above $AB$ on circle so that the incircle of $MCD$ is tangent to $AB$

lets say the problem inother word:
[color=blue]let $AB$ be a daimeter of a circle and $CD$ is a chord not intersecting $AB$ find point $P$ such that incircle of tiangle $PCD$ be tangent to $AB$[/color]

we know that if $I$ be inceter of $\Delta PDC$ then $\measuredangle DIC=90^{\circ}+\frac{\angle CPD}{2}$ so let $AB\cap CD\equiv X$ draw bisector of $\angle CXA\ \ (\ell)$

and draw arc $CID$ such that $\measuredangle DIC=90^{\circ}+\frac{\angle CPD}{2}$
intersection point of that arc and $\ell$ is incenter of $\Delta DPC$

There is not always a solution for this problem.

If the chord $CD$ is too small and too far from $AB$ we can't find a point $M$ with these properties.

Also, what happens if $AB$ and $CD$ are parallel?

yeah there isn't always that point.

the problem has solution  when that arc and $\ell$ intersect eachother.

when $AB\parallel CD$ , $\ell$ will be a line with equal distance from $AB$and $CD$


Prove that $ \forall a,b,c,d>0$ we have:

$ \frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b} \geq 2$.

cheers! :D :D :cool:

Let S denote the the LHS. The

  S=(a(d+a)+c(b+c))/(b+c)(d+a) + (b(a+b)+d(c+d))/(c+d)(a+b)

Since (x+y) 2  \geq 4xy

then S \geq 4[ (a(d+a)+c(b+c))/(a+b+c+d) 2 +(b(a+b)+d(c+d))/(a+b+c+d) 2 ]=2((a+b+c+d) 2 +(a-c) 2 +(b-d) 2 )/(a+b+c+d) 2  \geq 2

Equality holds iff a=c and b=d

Very interesting solution. Another possible solution is the following:
   Use Cauchy-Schwarz to obtain that the expression is at least (sum a)^2/(ab+bc+cd+da+2ac+2bd). This one is at least 2, since it reduces to (a-c)^2+(b-d)^2>=0.


[hide="hint"]Let the amount of coffee sold for $\$2.75$ be $x$, and the total amount of coffee be $y$.  What equations can you write?[/hide]

[quote="Interval"]A coffee manufacturer wants to market a new blend of coffee that will cost 3.90 per pound by mixing two coffees that sell for 2.75 and 5.00 per pound, respectively.  What amounts of each coffee should be blended to obtain the desired mixture?

Hint: Assume that the total weight of the desired blend is 100 pounds.

This is a considered a mixture problem.  I have a hard time with mixture questions.

What is the easiest way to learn mixture applications?[/quote]
Usually you would have a total amount already set... but...

[hide]You should probably set up a table
................amount.....price per unit.....total price
coffee 1.......x...............2.75..............2.75x
coffee 2...100-x............5.00...........5.00(100-x)
mixture.....100.............3.90...............390
(sorry about the dots, I couldn't space it)
Then set up an equation
$2.75x+5(100-x)=390$
$275x+50000-500x=39000$
$11000=225x$
$x=48.\overline{8}$
Then 48.88... of the 2.75 coffee and then 51.11... of the 5.00 coffee[/hide]

Well, you can do it without a table. You can set up some equations that look like [hide="this:"]
Calling $x$ and $y$ the first and second types, respectively, you know that:
$x+y=100$
$2.75x+5.00y=3.90(x+y)$.
Thus, substituting,
$2.75x+5(100-x)=3.90(100)$
$2.75x-5x=-110$
$x=48.888...$
 [/hide]

[quote="Interval"]Quevvy,

Basically, there is not other way to solve mixture questions without making a table?

Also, what clue in the word problem indicated to subtract x from 100?

You wrote:

100 - x.

Where did 100 - x come from?[/quote]
A table just organizes it better.
100-x is saying that when the two types of coffee are added, there will be 100 pounds.

I did not know that equations can be set up to answer mixture questions.  I also did not know about the table arranegments?

Tell me, is there a way to answer mixture questions using circles?

I can't think of how circles could be used to represent this kind of mixture problem...

Here's how I'd do it:

Let $x$ be the number of pounds of $\$2.75$ per pound coffee in one pound of coffee.  So there are $1-x$ pounds of $\$5.00$ per pound coffee in one pound of coffee.

Now solve for $x$:
$2.75x+5(1-x)=3.9$


number theory
greatest common divisor
search
complex numbers
algebra proposed
algebra

$ x,y,z$natural$ (x^2 \plus{} 1)(y^2 \plus{} 1) \equal{} z^2 \plus{} 1find x,y,z$

There are infinitely many solutions, for example (which doesn't mean there are other solutions - I don't know if these are the only ones), $ (t, t\plus{}1, t(t\plus{}1)\plus{}1)$. Let's see if it's indeed true:
$ (t^2\plus{}1)((t\plus{}1)^2\plus{}1)\equal{}(t(t\plus{}1)\plus{}1)^2\plus{}1$ equals
$ t^2\plus{}t^2\plus{}2t\plus{}1\plus{}t^2(t\plus{}1)^2\equal{}t^2(t\plus{}1)^2\plus{}2t(t\plus{}1)\plus{}1$, which equals
$ 2t^2\plus{}2t\equal{}2t(t\plus{}1)$, which is, of course, true.

Well we can factor in Gaussian integers:
\[ (x^2\plus{}1)(y^2\plus{}1)\equal{}z^2\plus{}1\iff (x\plus{}i)(x\minus{}i)(y\plus{}i)(y\minus{}i)\equal{}(z\plus{}i)(z\minus{}i)\]
mornik's solution set corresponds to the case $ (x\plus{}i)(y\minus{}i)\equal{}(z\plus{}i)$
Matching coeff's yields $ x\minus{}y\equal{}1$ and $ z\equal{}xy\plus{}1$, as in mornik's solution

Notice there's also going to be the case $ (x\plus{}i)(y\plus{}i)\equal{}z\plus{}i$ which is slightly different.

Now to prove that each solution corresponds to a factorization like the one above, we have to use the prime factor-ability of gaussian integers. Basically, suppose a prime divides the right hand side of the initial equation. Then it divides exactly one of the two factors unless it divides the gcd, which is $ 2i$. But $ z$ is real so this can't happen, so it divides exactly one of the two factors. Then the prime must divide some of the factors on the left hand side, which leads to the same cases as above.

That's the idea, fill in the holes :D

[quote="Euclid's Turtle"]Well mornik, how did you arrive at your solution? As opposed to just testing.[/quote]

I saw that $ 2\cdot 5\equal{}10$. I also saw that $ 5\cdot 10\equal{}50$, which gave me the idea. There was nothing smart about it, really. As you would put it, just testing.

[quote="me@home"] Notice there's also going to be the case $ (x \plus{} i)(y \plus{} i) \equal{} z \plus{} i$ which is slightly different.[/quote]

Doesn't this lead to $ xy\minus{}1\equal{}z$, $ x\plus{}y\equal{}1$? The latter would be obviously wrong since they are both natural.

A Gaussian integer is a complex number whose real and imaginary part are both integers. (source: [url=http://en.wikipedia.org/wiki/Gaussian_integers]Wikipedia[/url])

You know, Google has a lot of friends. If you treat him nicely and search him every so often, he will become your friend too. Then, he will stop being so shy and he'll show you his magic. :D


So here I am, trying to fall asleep, and this question pops into my head. Sorry if the question is of so-so quality... ^^"

Suppose you have a regular n-gon, n is even. In how many ways can you put m dots (2m < n) on the vertices of this n-gon such that when you join any two dots, the line does [b]not[/b] go through the centroid?

Centroid of a polygon... is that medians all connected?

The centroid of a regular polygon is the point right in the middle.  (An even-sided polygon doesn't have a nice object called a median, since there is no "opposite side" as there is for odd-sided polygons.)

Ok..  I think its n(n-4)/2....  I can't prove it right now, but so far, its worked for a 4,6, and 8 gon.  Help anyone?

Your function should be in 2 variables -- both the number n of the n-gon and the number of points m placed down on it.  You may be correct for a special case you are considering, however.

I used mathematica to look at it recursively, and so I've got the answer.  I haven't gone ot actually prove it yet, so I won't say anything until I do.

Ahh.. I read the question wrong.. I thought he meant from each vertex of the polygon.

Let us write 2k instead of n to make the notation easier. As long as we don't have two opposite points among the m dots, we won't have a line passing through the centroid. So we consider each set of opposite points as a unit. We can pick any m of them out of k. That's (k choose m). But from each of the m dots, we could use either of the opposite vertices, so we must multiply by 2^m. Thus there should be (k choose m)*2^m ways to satisfy the problem's conditions.


Using the thermodynamic parameters of $S_{solution}, S_{solvent}, S_{solid}, and \Delta{H_{freezing}}$ and the Gibbs Free Energy equation explain why the freezing point of a solution is always lower than the freezing point of pure solvent.

This is easily Raoult's Law, but we need to prove this using thermodynamics here.

sorry for being like 3 years late on this.... :blush:

So we first begin by stating the obvious:

$ \Delta G = \Delta H - T\Delta S$

Specifically, if we talk about a system:

$ \Delta G_{sys} = \Delta H_{sys} - T_{sys}\Delta S_{sys}$

At a freezing point, the system reaches equilibrium; therefore, at the systems freezing point $ T_f$:

$ \Delta G_{sys}=0 \rightarrow T_f = \Delta H_{sys}/ \Delta S_{sys}$

Further, a system comprised of a solvent alone satisfies the following:

$ \Delta S_{solvent}=\Delta S_{sys}$

Whereas a solution requires an additional parameter:

$ \Delta S_{sys} = \Delta S_{solvent} + \Delta S_{solute}$

Substituting these expressions in, we see that for a solvent, the freezing point is:

$ T{}_f{}_{solvent} = \Delta H_{sys}/ \Delta S_{solvent}$
Whereas for a solution, the freezing point is defined as:

$ T{}_f{}_{solution}= \Delta H_{sys}/ (\Delta S_{solvent} +\Delta S_{solid})$

Hence we see that if the statement above is true, then:

$ T{}_f{}_{solvent} > T{}_f{}_{solution}$

Or:

$ \Delta H_{sys}/ (\Delta S_{solvent} +\Delta S_{solid})< \Delta H_{sys}/ \Delta S_{solvent}$

$ \Delta H$ cancels out and we are left with:

$ \Delta S_{solvent} +\Delta S_{solid}>  \Delta S_{solvent}$

Or

$ \Delta S_{solid} > 0$

Quoting the 3rd law of thermodynamics, this must always be true unless at absolute 0, in which case entropy would be 0. However, this is not possible since at absolute 0, the system will be well below its freezing point. Hence the statement always holds that the freezing point of a solution is always lower than the freezing point of pure solvent.

$ Q.E.D$  :D


calculus
integration
trigonometry
function
real analysis
real analysis unsolved

how can i prove that the continuous curve $ \gamma: [0,1]\rightarrow \mathbb{R}^2$ with $ \gamma(0)\equal{}(0,0), \gamma(t)\equal{}(t,tcos(1/t))$ for $ 0<t \leq 1$ is not rectifiable?

Set up the integral for the length and show that it diverges at $ 0$.

try increasing partitions

$ P_n \equal{} \cup_{k \equal{} 1}^n \{\frac {1}{k\pi}\} \cup \{1\}$

and show as $ n\to \infty$, we have $ V(\gamma, P_n) \to \infty$

editted because i cant read the difference between cosine and sine. thanks kent

I'll assume that kenn4000 meant to use $ \frac1{k\pi}$ in his partition. (It is the cosine, not the sine, and you do want the function values to alternate in sign.) And check the details on each end, not that they're particularly important.

You will encounter something comparable to the harmonic series, or to the equivalent integral, if you do it fedja's way.


[quote="altusi"]hi all,i thing our community is begenning to die!! :mad:
here is a first and easy inequality to began:

a,A,b,B,c,C are positive realls such that :a+A=b+B=c+C=k,prouve that :

a*A+b*B+c*C>=k²

(*means product). ;)[/quote]
it is clear is not it?
(a+A)^2>=4a*A
so a*A<=k^2/4
and a*A+b*B+c*C<=3k^2/4<=k^2
ps:happy eid fetr!!!

[quote="ebn_mahdi"][quote="altusi"]hi all,i thing our community is begenning to die!! :mad:
here is a first and easy inequality to began:

a,A,b,B,c,C are positive realls such that :a+A=b+B=c+C=k,prouve that :

(*means product). ;)[/quote]
it is clear is not it?
(a+A)^2>=4a*A
so a*A<=k^2/4
and a*A+b*B+c*C<=3k^2/4<=k^2
ps:happy eid fetr!!![/quote]

It seems like your proof shows that a*A+b*B+c*C[b]<=[/b]3k^2/4[b]<=[/b]k^2 but not that a*A+b*B+c*C[b]>=[/b]k², which is what we had to show.

But I think you are right. Here is another way one could have done it:
Observe that (a * A) <= (a^2 + A^2).
Then we have k^2 = (A + a)^2 = 2(a*A) + (a^2 + A^2) >= 2(a*A) + (a*A) = 3(a*A) = a*A+b*B+c*C.
Hence a*A+b*B+c*C <= k^2 NOT the other way.

Thanks,
Akech

Observe that $(a * A) \leq  (a^{2}+A^{2}).$
Then we have $k^{2}= (A+a)^{2}= 2(a*A)+(a^{2}+A^{2}) \geq  2(a * A)+(a * A) = 3(a * A) = (a * A)+(b * B)+(c * C).$
Hence $(a * A)+(b * B)+(c * C) \leq  k^{2}.$

My point was that the preceding proof is contrary to what was supposed to be proved unless the original author mistyped some inequalities.

Can you wrie the symbol for the word GOAT

Actually, just use the fact that the expression is linear in its variables and hence reaches its maximum at the endpoints!


Let $ a,b,c$ be the sides of a triangle and $ S$ 's the area.Prove that:
$ 2(ab\plus{}bc\plus{}ca)\minus{}a^2\minus{}b^2\minus{}c^2 \ge \frac{2a^2(b\plus{}c\minus{}a)}{b\plus{}c}\plus{}\frac{2b^2(c\plus{}a\minus{}b)}{c\plus{}a}\plus{}\frac{2c^2(a\plus{}b\minus{}c)}{a\plus{}b} \ge 4\sqrt{3}S$.

With the first inequality,we only need $ a,b,c$ be none-gative,no two of which are zero.
 :)

The first ineq  follows from the well-known: $ \sum\ \frac {a}{b \plus{} c} \plus{} \frac {6(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2} \geq\ \frac {7}{2}$
The second ineq is obviously trues by Am-GM and CS  :maybe:

The first inequality is equivalent to

$ \sum_{cyc}(b\plus{}c)(b^2\plus{}c^2\minus{}a^2)(b\minus{}c)^2 \ge 0$

Now, S.O.S does the work gently.

We are done!

[quote="great math"]The first inequality is equivalent to

$ \sum_{cyc}(b \plus{} c)(b^2 \plus{} c^2 \minus{} a^2)(b \minus{} c)^2 \ge 0$

Now, S.O.S does the work gently.

We are done![/quote]
Nice solution  .
This inequality 's a funy and relax :maybe:   :blush:

[quote="quykhtn-qa1"]Let $ a,b,c$ be the sides of a triangle and $ S$ 's the area.Prove that:
 $ \frac {2a^2(b \plus{} c \minus{} a)}{b \plus{} c} \plus{} \frac {2b^2(c \plus{} a \minus{} b)}{c \plus{} a} \plus{} \frac {2c^2(a \plus{} b \minus{} c)}{a \plus{} b} \ge 4\sqrt {3}S$.
 :)[/quote]
How did you prove this ineq, Quy? I only have a way to solve it :|
PS: [hide]Sao cai' cau. filomen ko hieu? ng` VN hay ng` nuoc ngoai` ma` post lung tung the ko biet. Post toan may bai` vo van?, lam` n~ topic hay cua? moi. ng` troi het' xuong duoi', kho' tim` qua' !~~![/hide]

[hide="nguoivn"]we can solve the second inequality by Cauchy-Schwartz, AM-GM and substitution.[/hide]

I think our proof is the same, greatmath. I also use Am-Gm and next, CS to solve it.
The last ineq is $ (ab\plus{}bc\plus{}ca)^2 \ge 3abc(a\plus{}b\plus{}c)$

[quote="gigaman"][quote="nguoivn"]I think our proof is the same, greatmath. I also use Am-Gm and next, CS to solve it.
The last ineq is $ (ab \plus{} bc \plus{} ca)^2 \ge 3abc(a \plus{} b \plus{} c)$[/quote]

Can you post your full prove for this solution
First I assume $ a \equal{} x \plus{} y;...$ so we have this ineq

$ \Leftrightarrow \sum \frac {a(b \plus{} c)^2}{2a \plus{} b \plus{} c}\ge \sqrt {3abc(a \plus{} b \plus{} c)}$

Use Am-GM, we need to prove $ LHS\ge ab \plus{} bc \plus{} ca$

If your solution like this so I think your solution was wrong

Sr for my bad English[/quote]
Of course, my proof isn't so simple like this, gigaman  :wink:

[quote="gigaman"]
Can you post your full prove for this solution
[/quote]

Here is my solution, Hung  :)
We need to prove $ \Leftrightarrow\sum\frac {x(y \plus{} z)^{2}}{2x \plus{} y \plus{} z}\ge\sqrt {3xyz(x \plus{} y \plus{} z)}$
Using the well-known ineq $ (a \plus{} b \plus{} c)^2 \ge 3(ab \plus{} bc \plus{} ca)$, we have
$ LHS^2 \ge 3\sum\ \frac {xy(x \plus{} z)^2(y \plus{} z)^2}{(2x \plus{} y \plus{} z)(2y \plus{} z \plus{} x)} \ge 3xyz(x \plus{} y \plus{} z)$
<=> $ \sum\ \frac {(x \plus{} y)^2(x \plus{} z)^2}{z(2x \plus{} y \plus{} z)(2y \plus{} z \plus{} x)} \ge x \plus{} y \plus{} z$
Now, using Cauchy Schwarts, we obtain:
$ LHS \ge \frac {(\sum\ x^2 \plus{} 3\sum\ xy)^2}{\sum\ z(2x \plus{} y \plus{} z)(2y \plus{} z \plus{} x)} \ge RHS$
<=> $ (xy \plus{} yz \plus{} zx)^2 \ge 3xyz(x \plus{} y \plus{} z)$ (obviously trues by Am-Gm). We have done!  :)


trigonometry
Pythagorean Theorem
geometry
geometry solved

Let ABC be a right angled triangle(<B=90 ).Let BH is perpendicular to AC side at H point.Construct BCDE square on BC side. [BH] and [AD] line segments are intersection at point P.
Show that

1/BP=1/BH+1/AC

Hi mathquark.
I have solved your problem,but i can't post the solution now.I'll post it soon.

I have computational solution using center of mass.

Let $E$ be point of intersection of $AD$ and $BC$. Triangles $DEC$ and $ABE$ are similar, it follows that $CE:EB=c:a$, where $c=AB$ and $a=CB$.
We have $2S_{ABC}=BH\cdot AC=AB\cdot CB$, so $BH=\frac{ac}{\sqrt{a^2+c^2}}$. Using Pythagorean theorem for $BHC$ and $BHA$ we obtain $CH:HA=a^2:c^2$.
Consider mass-points $(c^2)C$, $(a^2)A$ and $(ac)B$. Grouping $(c^2)C$ and $(a^2)A$ we see that center of mass of whole system lies on $BH$, on the other hand, grouping $(c^2)C$ and $(ac)B$ we see that center of mass lies on $AE$. It follows that $P$ is a center of mass of $(c^2)C$, $(a^2)A$ and $(ac)B$. Moreover, we immediately obtain $BP=\frac{a^2+c^2}{a^2+c^2+ac}BH$, i.e. $BP=\frac{ac\sqrt{a^2+c^2}}{a^2+c^2+ac}$.
Now it is routine to show $\frac{1}{BP}=\frac{1}{AC}+\frac{1}{BH}$.

Here is my solution.
We have to prove that $1-\frac{BP}{BH}=\frac{BP}{AC}$ or that
$\frac{PH}{BH}=\frac{BP}{AC}$. Using the triangles $BAH$ and $PHA$ we get that
$\frac{PH}{BH}=\frac{\tan{\alpha}}{\tan{(\alpha+\beta)} }$  where $\alpha=\angle{PAH}$ and $\beta=\angle{BAP}$.On the other hand we have that $\frac{BP}{AC}=\frac{\sin{\beta}\cdot\cos{(\alpha+\beta)}}{\cos{\alpha}}$ using  $AB $.
Equalizing this equalities we get that we have to prove $\frac{\sin\alpha}{\sin\beta}=\sin{(\alpha+\beta)}$.
$DC$ is parallel to$ AB$ so$ \angle{CDA}=\beta$.
Using  the sine rule in the triangle CDA we have $\frac{\sin\alpha}{\sin\beta}=\frac{CD}{AC}=\frac{BC}{AC}=\sin(\alpha+\beta)$.
The problem is solved.


Indeed easy.
But not that ugly. this problem was given in JBMO (2000 i think)

Anyway the way to go is not unusual: put $x+y=a$ , $xy=b$ and factorize!!

 :)

[quote="riddler"]I think I did it differently but another question:
What ages is the JBMO for?[/quote]

For kids under 15,5 years old!
I took part in this competition when I was in 9th form....... :lol:

I am willing to see your approach. :)  :)

[quote="riddler"][hide="Sol, dont see if you havent tried"]
$[x^3+y^3-10^3+30xy]+[(x+y)^3-1000]=0$
And factorise, :D
[/hide][/quote]

Very nice!!
 :D  :lol:

[quote="riddler"][quote="socrates"][quote="riddler"]Sorry I got the forum wrong! :blush:[/quote]

What do you mean??
 :?[/quote]
Some moderator moved this post to intermediate forum, dont ask me, I dont know why.[/quote]

Which section was it firstly in??

[quote="riddler"]GS forum[/quote]

I think intermediate section is the appropriate place for it!!
Although some would consider it pre olympiad........ ;)

I moved it. Can you really expect someone who's just [b]getting started[/b] with problem solving to be able to solve this? ;)

My 2 cents:  this problem is definitely not GS and at least Intermediate.  Please take note of this...the difficulty level of the forums is often confused.

Seeing that no understandable solution has been posted, I will post one (for the sake of the Contests Section):

[hide="Solution"]
Let $a=x+y$ and $b=xy$, then we have
\begin{align*} x^3 + y^3 + (x + y)^3 + 30xy &= 2000 \\
(x+y)^3-3xy(x+y)+(x+y)^3+30xy&=2000 \\
a^3-3ab+a^3+30b&=2000 \\
2a^3-3ab+30b-2000&=0 \\
(a-10)(2a^2+10a-3b+200)&=0
\end{align*}
We get the factorization using synthetic division and the rational root theorem.  Then, we have the only two possible solutions that are
\begin{align*} a&=10\\ 2a^2+10a-3b+200&=0\end{align*}
The first case is just $a=x+y=10$ and we're done.

The second case is just a tiny bit trickier.  We use the quadratic formula to get \[a\&=\frac{-10\pm\sqrt{100-8(200-3b)}}{2} \\
\&=-5\pm\sqrt{25-400+6b} \\
\&=-5\pm\sqrt{6b-375}
\]
Since $x$ and $y$ are positive, we have $a>0$ as well.  Thus, the only possibility is $a=-5+\sqrt{6b-375}$ because $-5-\sqrt{6b-375}<0$.  Thus, we have
\begin{align*}a&>0 \\
-5+\sqrt{6b-375}&>0 \\
\sqrt{6b-375}&>5 \\
6b-375&>25 \\
6b&>400 \\
b&>\frac{200}{3} \\
xy&>\frac{200}{3}
\end{align*}
However, we have $x^3 + y^3 + (x + y)^3 + 30xy \&= 2000$ but $30xy>30\left(\frac{200}{3}\right)=2000$ and every term is greater than $0$ therefore we have the quadratic providing faulty solutions.

Thus, the only solution is $a=x+y=10$ as desired.
$\blacksquare$
[/hide]

^^^ Perhaps a little clearer in the above:
Once you have
$x+y=a,xy=b$ AM>GM gives..
[hide]$\implies  a^2\ge 4b $   This means the second factor ($2a^2+10a-3b+200$) is greater than zero, and only
a=10 is possible solution.[/hide]

What I did was I showed 10 would work. Then I showed that $10+a$ and $10-a$ would not work for $a>0$

[quote=riddler]Let $x$ and $y$ be positive reals such that \[ x^3 + y^3 + (x + y)^3 + 30xy = 2000. \] Show that $x + y = 10$.[/quote]

We have $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ and $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2- ab - bc - ca)$.
From the example we have that $[x^3 + y^3 - 10^3 + 3*10xy] + [(x + y)^3 - 10^3] = 0$,and we apply the formulas and we get
 $(x + y - 10) * (x^2 + y^2 + (x+y)^2 + 10(x + y) + 200 - 10x - 10y + xy) = 0$ and because the second factor > 0
   we obtain $x + y - 10 = 0$ or $x + y = 10$.
[quote=ionbursuc][quote="riddler"]Let $x$ and $y$ be positive reals such that \[ x^3 + y^3 + (x + y)^3 + 30xy = 2000. \] Show that $x + y = 10$.[/quote]
Generalization
Let $x,y,k,r$ be positive reals numbers such that
	\[{{x}^{3}}+{{y}^{3}}+r{{(x+y)}^{3}}+3kxy=\left( r+1 \right){{k}^{3}}\]
 Show that $x+y=k$.[/quote]

Similar we do the generalisation:
 $[x^3 + y^3 - k^3 + 3kxy] + r[(x+y)^3 - k^3] = 0$ or
$(x + y - k)(x^2+y^2 + 2k^2 - kx - ky +xy + r(x + y)^2 + kr(x+y)) = 0$ and we get $x + y - k = 0$ or $x + y = k$.

We could obviously prove that [code]x+y<10[/code] and[quote]x+y<10[/quote] can't be happened.

$2(x^3+y^3-1000)+3xy(x+y-10)=(x+y-10)t$. t is a big expression so I didn't wrote it. But obviously it can't be 0. So x+y=10

Here is a highly non-standard solution to this really easy problem. (The factorization solution is cute but a bit routine)

We shall apply the Method of Lagrange Multipliers!!! (Scared eh...? Trust me it's gonna be magical)

Indeed, consider the constraint function as

\begin{align*}
g(x,y)=x^3+y^3+(x+y)^3+30xy-2000
\end{align*}

and the objective function

\begin{align*}
f(x,y)=(x+y-10)^2
\end{align*}

We choose to maximize $f$ subject to $g$ and then see that indeed $f \le 0$ but in reality $f \ge 0$ and thus, $x+y=10$

 Note that $f,g$ are continuous with continuous partial derivatives.

\begin{align*}
\nabla f=(2(x+y-10),2(x+y-10))
\end{align*}
and also,

\begin{align*}
 \nabla g =(6x^2+3y^2+30y+6xy,6y^2+3x^2+6xy+30x)
\end{align*}

Evidently, $\nabla g \not= 0$

We define these functions over $U=(0,9000)^2$ for certainly $x,y \in U$.
We shall prove that the result holds for any $x,y \ge 0$ satisfying $g$ instead of just positive reals. (This for the informed viewer being just an excuse to execute our notions of topology and compact sets over $\mathbb{R}^2$)

Define the [i]closure[/i] of the set $U$ as $\bar{U}=[0,9000]^2$.

Then consider set $S=\{x \in \bar{U} : g(x)=0\}$

Then, clearly, we have that $S$ is compact and so $f$ achieves a global maxima $x_0$ over $S$.

Now, every global maxima is a local one as well and so by the theorem of Lagrange Multipliers

Case 1. $x_0 \not \in \bar{U}\cap U$

In this case one of $x,y$ is zero which means that the other is automatically equal to $10$ by our original equation.

Case 2.  $x_0 \in \bar{U}\cap U$

Thus, we have $\nabla f = \lambda \nabla g$ and therefore we conclude that

$2(x+y-10)=\lambda (6x^2+3y^2+6xy+30y)=\lambda (6y^2+3x^2+6xy+30x)$ and so we get

$x=y$ or $x+y=10$.

While still being ignorant we proceed assuming $x=y$ which from $g$ implies $x=y=5$ and we are done again.

Thus, Lagrange Multipliers assure that $f \le 0$ and common sense assures that $f \ge 0$.

Thus, $f=0$. $\square$

In my own humble opinion this solution deserves a special prize with a lot of criticism for being overkill.

The only insight I intended to pass by writing all this non-sensical stuff was to apply LM in situations where it is unexpected. I would post a note on this soon when I am free on my blog. Do check it out! (Its private as of yet)

P.S.- Yay! 400 th post...

I think I hab a shorter solution by case work (seems ironic lol)
[hide=Quick 5-liner]
Note that $x^3+y^3=(x+y)[(x+y)^2-3xy]$
Case 1: $x+y>10$. Plugging this in, we get LHS>2000
Case 2: $0<x+y<10$. Plugging again, LHS <2000
Case 3: $x+y=10$ Plugging once more, we get LHS=2000
Hence, we must have $x+y=10$

[/hide]
Also nice idea there using the $a^3+b^3+c^3-3abc$ factorization (I think it might be getting a bit too slightly overused tho). I will go try it.

This thread started in 2005, then was revived in 2011, then in 2014, 2015, and now 2018.

No... I see December 31, 2015. Maybe we are in different time zones.

Easy problem. Let $x=a+b$ and $y=a-b$. Rewriting the equation and dividing by $2$ gives:
$5a^3+3ab^2+15a^2-15b^2=1000$. Therefore $(3a-15)b^2=(1000-15a^2-5a^3)$. However, $3a-15$ is negative in the range $[0,5)$, while $(1000-15a^2-5a^3)$ is positive. Similarly, $3a-15$ is positive for $a>5$, while $1000-15a^2-5a^3$ is negative. This implies that $b^2$ is negative, contradiction. Hence $a=5$, which means $x+y=(a+b)+(a-b)=10$.

Strange solution:
Denote $x+y=S$, and $xy=P$. So we get: $3S=\frac{2S^{3}-2000}{S-1}$=$2(S+S+1)-\frac{1998}{S-1}\in Z$, so we must have $S-1\in D_{1998}$ where $1998=2*3^{3}*37$ and testing these, we found $S=10$

Note that $x^3+y^3+(x+y)^3+30xy-2000=0$ factors as $(x+y-10)(2x^2+xy+2y^2+20x+20y+200)=0$

If $x,y$ are both positive then the second factor is positive, so that we are forced to conclude $x+y=10$.

AM GM Solution!!!!!!!

We can factor our given expression into $2(x+y)^3-3xy(x+y)+30xy=2000$. Let $x+y=a, xy=b$. Then we get that $2a^3-3ab+30b=2000$. Solving for b, we get that $b=\frac{2a^3-2000}{3a-30}$. If a is not equal to 10, this can be simplified as $b=\frac{2a^2+20a+200}{3}$. By AM-GM, we know that $\frac{a^2}{4} \ge b$. When we plug in what we found for b, we get that this simply cannot occur, as the inequality won't hold when a is positive. Now let us see if a=10 works. Then we get b=0, meaning that our possible ordered pairs of $(x, y)$ are $(10, 0), (0, 10)$, which indeed do work.

Here's my work.

Rearrangement gives $$x^3+y^3+(x+y)^3+30xy-2000=0.$$ Then, let $a=x+y, b=xy.$ We get $$x^3+y^3+a^3+30b-2000=0 \implies (x+y)^3-3x^2y-3xy^2+a^3+30b-2000=0 \implies a^3-3(by+bx)+a^3+30b-2000=0 \implies a^3-3ab+30b+a^3-2000=0.$$ Applying Difference of Cubes, $$2(a-10)(a^2+10a+100)+3a+30b=0.$$ Then, substituting back $a=x+y, b=xy,$ we have $$2(x+y-10)[(x+y)^2+10(x+y)+100]-3xy(x+y) \implies (x+y-10)(2x^2+2y^2+xy+20x+20y+200)=0. ~ ~ ~ \text{(By factoring)}$$ Clearly, the bracket on the right is non-zero, and the desired result follows. (Because $x+y-10=0$ implies that $x+y=10.$)

Feel free to give any constructive feedback below! I would really appreciate it!

[quote=KRIS17]After some manipulation we get:
$3xy(10 - (x+y)) = 2((x+y)^3 - 10^3)$

Case 1: $x+y > 10:$
$RHS > 0$, So $LHS$ has to be $> 0$, so $10 - (x+y) > 0$ or $x+y < 10 => contradiction!$

Case 2: $x+y < 10:$
$RHS < 0$, So $LHS$ has to be $< 0$, so $10 - (x+y) < 0$ or $x+y > 10 => contradiction!$

Thus $x + y = 10$[/quote]

You can factor the RHS as a difference of two cubes and then you can divide both sides by 10-xy which then leads to a contradiction so 10-xy has to be 0.

Why have I seen this before somewhere?

[hide=Solution]Write $x+y=a, xy=b$. Then we have $x^3+y^3=(x^2+y^2-xy)(x+y)=a(a^2-3b)$, and $a(a^2-3b)+a^3+30b=2000$. This simplifies to $a^3+a^3-3ab+30b=2000$, or $2a^3-3ab+30b=2000$. We can see that $a=10$ is a root, and factoring out gives $(a-10)(2a^2+10a-3b+200)=0$. Obviously the first case gives $x+y=10$. By the Quadratic Formula, the second case gives $a=\frac{-10 \pm \sqrt{100-8(200-3b)}}{4}=\frac{-10 \pm \sqrt{24b-1500}}{4}$, or $\frac{-5 \pm \sqrt{6b-375}}{2}$. This must be greater than $0$, so $a=\frac{-5+\sqrt{6b-375}}{2}$. Solving $-5+\sqrt{6b-375}>0$ gives $b>\frac{200}{3}$, or $xy>\frac{200}{3}$. However in itself $30xy>2000$ in that case and $x^3+y^3+(x+y)^3$ is positive, so this is not possible. Therefore our answer is $x+y=\boxed{10}$, and we're done.[/hide]

Denote $a=x+y,b=xy$. The equation is:
$$2a^3-3ab+30b=2000$$
$$\Leftrightarrow2(a^3-1000)-3b(a-10)=0$$
$$\Leftrightarrow(a-10)(2a^2+20a+200-3b)=0$$
Assume FTSOC that $a\ne10$, so $2a^2+20a+200=3b$. Then by AM-GM, $a\ge2\sqrt b$, so $2a^2+20a+200\ge8b+40\sqrt b+200>3b$, contradiction reached. $\square$


Is AOPS going to make a book on Calculus? And if they do, when will they?

I think not, no, because calculus typically does not appear on competitions, and definitely doesn't appear on the AMC, AIME, USAMO, or IMO.
Then again, they have a class for it, so you never know.

I hope they do so I can use it to fully grasp the fundamentals of calculus while being challenged. This would help me in the Physics Olympiad

I'm not sure this is the right forum
But, to answer the question, with the addition of a Calculus class, a calculus book is perhaps coming soon?  Similar thing happened with Trig/Complex numbers, class first, book is coming out this fall i believe?  not completely sure

[quote="math154"]I'm pretty sure David Patrick is working on Calculus, and, along with the Trig/Complex Numbers book, it should come out some time this year or next year?[/quote]

DPatrick is working on Calculus, which we expect to be out by the end of 2009.  I am working on Precalculus, which I hope to have finished by the end of the summer.


http://planetmath.org/encyclopedia/KummersTheorem.html

When they say "Finally, we introduce ...", there is a claim about "the power of $ p$ dividing $ \binom{n}{m}$ ".

Can anyone explain where it comes from?

See [url=http://www.cecm.sfu.ca/organics/papers/granville/paper/binomial/html/binomial.html]this online paper[/url].


Bridget bought a bag of apples at the grocery store. She gave half of the apples to Ann. Then she gave Cassie 3 apples, keeping 4 apples for herself. How many apples did Bridget buy?

$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 11 \qquad \textbf{(E)}\ 14$

[hide="Solution"]
Suppose Bridge bought $ x$ apples.  Then, after giving apples to Ann, she has $ \dfrac{x}{2}$ apples left.  After giving Cassie 3 apples, she has $ \dfrac{x}{2}\minus{}3$ apples.  Because she has 4 apples left, $ \dfrac{x}{2}\minus{}3\equal{}4 \implies x\equal{}14$, or $ \boxed{\textbf{(E)}}$.
[/hide]

You can also go backwards

She has four apples
then you +3 to take back Cassie's
then you x2 to take back Ann's

Answer: 14

Obviously he has the problem set. (Ask your teacher tomorrow for it) As do I....  :)

OMG I THINK I GOT THIS WRONG DUE TO A READING ERROR!!!!!!!!!!! :censored:  :wallbash_red:

[quote="CRICKET229"]hey pi where are you getting the problems?[/quote]
I got the booklet from my coach. You should be able to get it too. Also, if you marked your answers in your booklet, your teacher should have a solutions booklet where he/she can check your answers.

[quote=AwesomeToad]Well, you could say the USA is irrelevant, but USA "specifies" it is from the US.

Although AMC stands for American Math Competitions[/quote]

And it is mainly for 8th graders, although anyone younger can also write it.


analytic geometry
graphing lines
slope
calculus

Point P is on the y-axis. From P, two lines are drawn to be tangent to $x^2-6x+y^2+6y+14=0$ at Q and R. If Q is $(4,-3+\sqrt3)$, find the length QR.

My answer looks too ugly to be right, but here's what I did:

Complete the square to get

$(x-3)^2+(y+3)^2=4$

We know a point $P$ with coordinates $(0,p)$ is tangent to the upper half of the circle at $(4,-3+\sqrt{3})$

The equation for the upper half of the circle is $y=\sqrt{-x^2+6x-5}-3$

$\frac{d}{dx} \left( \sqrt{-x^2+6x-5}-3 \right)=\frac{3-x}{\sqrt{-x^2+6x-5}}$

At $x=4$, the slope of the circle is $-\frac{\sqrt{3}}{3}$, so we know that the slope of the line joining $P$ and $(4,-3+\sqrt{3})$ is $-\frac{\sqrt{3}}{3}$.

We now have $\frac{-3+\sqrt{3}-p}{4}=-\frac{\sqrt{3}}{3}\Rightarrow p=\frac{7\sqrt{3}}{3}-3$

Now we have to find the point on the lower half of the circle that is tangent to the other line from $P$.

The equation for the lower half of the circle is $y=-\sqrt{-x^2+6x-5}-3$

$\frac{d}{dx}\left(-\sqrt{-x^2+6x-5}-3\right)=\frac{x-3}{\sqrt{-x^2+6x-5}}$

From the point slope formula, we have $-\sqrt{-x^2+6x-5}-\frac{7\sqrt{3}}{3}=\frac{x(x-3)}{\sqrt{-x^2+6x-5}}\Rightarrow x=\frac{20}{19}$

Thus $Q=\left(4,-3+\sqrt{3}\right)$ and $R=\left(\frac{20}{19}, \frac{-5\sqrt{3}}{19}-3\right)$

Applying the distance formula, we get $QR=\frac{16\sqrt{19}}{19}$.

Sorry for using calculus, but I didn't see any other way.

Actually I think we can find the first tangent line, which, combined with the fact that P is on y-axis, gives us the co-ordinate of point P. Then we can find point R and then the distance QR. But it's quite complicated as well. ;)
So I'm wondering if there is a solution which gives us the length QR without having to find the point R.


Prove that if $x$ and $y$ are rational numbers, satisfying the equation  $x^{5}+y^{5}= 2x^{2}y^{2},$ then $1-xy$ is the square of a rational number.

[Note: If you know the solution, please don't write it. Just provide a hint or two to those who wish to attempt to solve it.]

[u][b]Hint:[/b][/u] Set $t = x/y$ and find a parametrization $x = f(t), \; y = g(t)$ of the curve given by the equation $x^{5}+y^{5}= 2x^{2}y^{2}.$

oo thanks for the hint  :lol:

[hide]

letting $\frac{x}{y}= t$ for some rational $t$, we divide both sides of the equation by $x^{2}y^{2}$ to get:

$\frac{x^{3}}{y^{2}}+\frac{y^{3}}{x^{2}}= 2 \implies xt^{2}+\frac{y}{t^{2}}= 2$

now, we know that $x = yt$ and $y = \frac{x}{t}$, and substituting those in we get:

$y = \frac{2t^{2}}{t^{5}+1}$

$x = \frac{2t^{3}}{t^{5}+1}$

hence:

$1-xy = 1-\frac{4t^{5}}{(t^{5}+1)^{2}}= \frac{(t^{5}-1)^{2}}{(t^{5}+1)^{2}}$

and we are done
[/hide]


Prove that if $n\in\mathbb{N}$ and $a\in\mathbb{R}$ and $n<a<n+1$ then $a\notin\mathbb{N}$.

I know this is obvious but how would you prove this without using the fact that $(n+1)-n\in\mathbb{N}$.

We have defined $\mathbb{N}$ as
1) every natural number is a real number
2a) $0\in\mathbb l{N}$
2b) $n+1\in\mathbb{N}$ for all numbers $n\in\mathbb{N}$
2c) $n-1\in\mathbb{N}$ for all $n\in\mathbb{N}$ such that $n\neq{0}$
3) Every nonempty subset of $\mathbb{N}$ has a least member.

I think that this is the definition of natural numbers-the successive number to $n$ is $n+1$ ..

I thought we define the naturals as following:

1. $\mathbb{N}\subset \mathbb{Z}$
2. $\forall x \in \mathbb{Z}$, exactly one of the following holds:
$x \in \mathbb{N},-x \in \mathbb{N}, x=0$ (Trichotomy)
3. Every nonempty subset of $\mathbb{N}$ has a least element (WOP)
4. $\mathbb{N}$ is nonempty (Nontriviality)
5. $\mathbb{N}$ is closed under addition and multiplication.

Following this set of definitions, what you want to prove will follow from the fact that there exists no integer between 0 and 1.  Proving that is simple: suppose there is a natural $x$ s.t. $0<x<1$.  Then it's a simple matter to show that $0<kx<k$ for each natural $k$.  Then, because the naturals are closed under multiplication, $x^{n}$ is a member of the naturals (by induction).  We then consider the subset of the naturals $\{1, x, x^{2}, x^{3}, ...\}$.  Hey, no least element!  Contradiction.  Then no such $x$ exists.

i feel like we need the definition of real numbers to prove this...

Suppose there exists $n, a$ natural numbers such that $n < a < n+1$.  Then consider the sequences $\{n, n-1, n-2, n-3, \ldots \}$ and $\{a, a-1, a-2, \ldots\}$ where "$\ldots$" means "continue forever, unless you hit 0."  Only one of the two sequences can include 0 (this needs some justification).  Then the other sequence is a nonempty set of natural numbers with no smallest element, contradicting point 3.

There are, of course, many ways to define the natural numbers.  Constructions that take the real numbers as given strike me as fundamentally backwards (the naturals seem, well, far more "natural," and indeed far more "real," than the reals), but such is life.

thanks JBL and everyone. I never thought about proofs of those forms. but i did know that i would have to use $WO$ somewhere along the line.


vector
arithmetic sequence
combinatorics proposed
combinatorics

The numbers 1,2,...,N are colored by black or white. One moviment consists in change a color of 3 number in arithimetic progression. Find all N such that ever is possible have all numbers with same color.

I don´t find this problem in ARO 2000  :(

Please, give me a solution.

clearly if it is possible for $N$, then it is possible for $N+1$. We prove it is not possible for 7, consider the numbers $2,3,5,6$, any arithmetic progression contains none or two of them, so the number of black numbers between this 4 numbers mantains its parity. So if initial there is an odd number, we cant get all numbers white. Now we must prove for 8 it is always possible. and believe me it is true, show it is enough to considere those cases where the only black numbers are $1$,$2$, or both of them. And just solve this 3 cases.

He-he! Pascual! It is the problem I was solving (I told you) :)
Since I have solved it, I should write it here.
Consider in $\mathbb{Z}_2$ vectors (1,1,1,0,0,0,...), (0,1,1,1,0,...), and so on, where three '1' are placed in arithmetic sequence (we consider all possible cases). Then the problem will be solved if we prove that these set of vectors contains linear independent subset of $n$ vectors. For $n=3,4,5,6,7$ the set contains less than $n$ independent vectors (easy direct check).
For $n\geq 8$ consider
(1,1,1,0,0,0,...,0)
(0,1,1,1,0,0,...,0)
(0,0,0,1,1,1,...,0)
(all possible vectors with three consequtive '1' - $n-2$ vectors)
(1,0,0,1,0,0,1,0,...)
(0,1,0,0,1,0,0,1,...)

This problem from ARO round IV.

Pascaul2005

please, give me a proof that n = 8 is true.

pleease, give me a proof that n = 8 is ok  :?

I don´t understand Myth´s Proof.
I prove that if is valid to n, is valid to n+1 and that n <= 7 is not valid.
Please, prove that n = 8 is valid.

ok, Myth already did this! but here goes my idea:
let the 8 numbers be on a line $1,2,3...,8$ now look for the biggest $i$, such that $i$ is black, and then change $i,$,$i-1$,$i-2$. do this again, and again, and again. so we get that by this procedure we have $3,4,5...,8$ white, and $1,2$ are arbitrarely colored.  now we have only to considere 4 cases:
(in the diagrams a $1$ means that term is black and a $0$ means it is white)
case 1:
$11000000$
case 2:
$10000000$
case 3:
$01000000$
case 4:
$00000000$
And please dont be laizy! finish them by yourselves!


geometry
analytic geometry
geometric transformation
reflection
circumcircle
ratio
Euler

Let $DEF$ be the orthocentric triangle of $ABC$ (so $D$ is the feet of the perpendicular from $A$ on $BC$, $E$ is the feet of the perpendicular from $B$ on $CA$ and $F$ is the feet of the perpendicular from $C$ on $AB$). Let $A'$ be the orthogonal projection of $A$ on $EF$, $B'$ the orthogonal projection of $B$ on $FD$ and $C'$ the orthogonal projection of $C$ on $DE$. Let $K$ be the midpoint of $DA'$, $L$ the midpoint of $EB'$ and $M$ the midpoint of $FC'$. Show that the lines $AK$, $BL$, $CM$ are concurrent.

I have a proof which has an ugly part and a nice part. I won't post the solution for the ugly part (I'll post it tomorrow if you really want me to):

Ugly part: I used barycentric coordinates to prove that the lines AK, BL, CM pass, in fact, through the midpts of EF, FD, DE respectively (I think it's ugly because it involves calculations and I don't like calculations). To get nicer barycentric coordinates I reformulated the problem, like this: DEF becomes our original triangle and ABC is seen as IdIeIf (the triangle formed by the centers of the exinscribed circles of DEF).

Nice part: Let X, Y, Z be the midpts of EF, FD, DE respectively. We're trying to prove that AX, BY, CZ are concurrent. But it's easy to prove that AX, BY, CZ are, in fact, the symmedians of the triangle ABC (EF is antiparallel to BC etc.), so they're concurrent. Notice that we've found our concurrency point: K, Lemoine's pt.

I tried to simplify grobber's solutin and  ended up in combining the 2 parts (ugly and nice) into one "medium-ugly" part. It's a long computational solution to prove that if M' is where AM find BC then AM' is a simedian
(BM'/CM')=(c/b)^2  ...

I finished the computations and used Steiner th only 3 times and Menelaos once :)
so i will surely won't write down the solution. But i think trying for yourself to finish the computations will be a good exercise (not...) or a very annoying one (probably...).
(i tried for 1/2 hour) :(  :?

...
"Let A' be the orthogonal projection of A on EF"
it must be because of my vocab... what exactly does "orthogonal projection" mean?  mirrors A in the line EF? in other words, the line EF is the perpendicular bisector of the segment AA' ?

anyhow, if that is not what Arne intended, this still makes a new and nice problem, seeing how the lines AK, BL, CM pass through the same point.

[color=red][[b]Moderator edit:[/b]
The author of this posting misunderstood the problem: he interpreted the points A', B', C' as the [b]reflections[/b] of the points A, B, C in the lines EF, FD, DE, rather than as the [b]orthogonal projections[/b] of the points A, B, C on the lines EF, FD, DE (= feet of the perpendiculars from A, B, C to EF, FD, DE) as Arne intended. However, as he noted, this misunderstanding generates a new problem: Show that the lines AK, BL, CM concur, where the points A', B', C' defined as the reflections of A, B, C in EF, FD, DE. Note that, with this definition of A', B', C', we have two more triples of concurring lines:

- The lines DA', EB', FC' concur.
- The parallels to the lines DA', EB', FC' through the points A, B, C also concur.
][/color]

The orthogonal projection of a point on a line is the foot of the perpendicular from that point to that line.

For example, D is the orthogonal projection of A on BC in the given situation.

Btw, What is Steiner's theorem ?  :?

ahh kool, i discovered something new then  :D  :D  :D can any1 prove that ;)

The problem of bugzpodder sounds the same as the initial problem just that instead of saying K is midpoint of DA'  we would say that DK/A'K=2 (this is an easy menelaos)

I ask something bigger so the new problem would be:
"
Let DEF be the orthocentric triangle of ABC. Let A' be the orthogonal projection of A on EF, B' the orthogonal projection of B on FD and C' the orthogonal projection of C on DE. Let K be on [DA'], L on [EB'] and M on [FC'] so that [b]DK / KA' = EL / LB' = FM / MC' = t  [/b]. Show that the lines AK, BL, CM are concurrent.
"

Steiner th is, as far as i know (but then again i always get something like "oh... i thought that was called somehow else...")
  In triangle ABC we have D and D' on BC so that <BAD=<CAD' (in romanian AD and AD' are called izogonals - i don't know if this is the translation) then
  (BD/DC)(BD'/D'C)=AB 2 /AC 2

The proof is easy using surfaces expresed in 2 ways. We can, of course calculate AD/AD' also. This is sometimes useful...

So i ask one of you people who have euclid to veify the above problem for different values of t. Thank you. I am very curious... :?  :?

[color=red][[b]Moderator edit:[/b] Yes, your general problem is correct for all values of t. There is a synthetic proof, which I will post later when I have got the time. - Darij][/color]

[quote]DK/A'K=2 [/quote]
you meant DK/A'K=1

In your problem (so where EF is bisector for AA') let AA' meet EF at A1 and AK meet A1D at K1
Then menelaos says

(DK/KA') (AA'/AA1) (A1K1/K1D) = 1 but

(DK/KA') = 1 and (AA'/AA1) = 2.

Ad literam you are right bugzpodder but you understand now what i want to say, right? Anyway i formulated the whole problem in my last post so there will be no missunderstandings.

I used my initial idea of looking at DEF as ABC and ABC as IaIbIc and then using barycentric coordinates to prove Bugz' last problem. The reason for this is that the barycentric coordinates of Ia, Ib, Ic are easy to express: Ia:[-a/2(p-a)]*A+[b/2(p-a)]*B+[c/2(p-a)]*C, where A, B, C are seen as the affixes (is this the word?) of A, B, C respectively. The rest is pure computation.

From another view point: $A$ is the excenter of $DEF$, $A'$ is the touch point of excircle $(A)$ with $EF$ wrt $DEF$, $K$ is midpoint of $DA'$, then $KA$ passes trough the midpoint  of $EF$ (see: http://www.mathlinks.ro/Forum/topic-15932.html ) and Lemoine's point of $ABC$, similarly for $BL$ and $CM$. Hence the concurrency point is the Lemoine's point of $ABC$.

Thanks for reminding me again that I promised to write up my solution for Kappa's general problem from post #7:

[quote="kappa"]I ask something bigger so the new problem would be:
"
Let DEF be the orthocentric triangle of ABC. Let A' be the orthogonal projection of A on EF, B' the orthogonal projection of B on FD and C' the orthogonal projection of C on DE. Let K be on [DA'], L on [EB'] and M on [FC'] so that [b]DK / KA' = EL / LB' = FM / MC' = t  [/b]. Show that the lines AK, BL, CM are concurrent.[/quote]

The first step of my solution is to find a reasonable characterization of the point of concurrence. This means extending the problem, but extending it in a way that it becomes easier to solve. This extension is as follows:

[color=blue][b]Extended problem.[/b] Let D, E, F be the feet of the altitudes of triangle ABC, and let A', B', C' be the orthogonal projections of the points A, B, C on the lines EF, FD, DE, respectively. Let t be an arbitrary real number, and let K, L, M be three points on the lines DA', EB', FC' such that $\frac{DK}{KA^{\prime}}=\frac{EL}{LB^{\prime}}=\frac{FM}{MC^{\prime}}=t$.

Now, let H be the orthocenter and O the circumcenter of triangle ABC, and let P be a point on the line HO such that $\frac{OP}{PH}=\frac{t}{2}$. Then, prove that the lines AK, BL, CM concur at the isogonal conjugate of the point P with respect to triangle ABC.[/color]

[i]Solution of the extended problem.[/i] We will use directed angles modulo 180°.

Let the perpendicular to the line CA at the point C meet the line AB at U, and let the perpendicular to the line AB at the point B meet the line CA at V. Then, since < UBV = 90° and < UCV = 90°, the points B and C lie on the circle with diameter UV. Thus, < VCB = < VUB. In other words, < ACB = - < AUV. Similarly, < ABC = - < AVU. Thus, the triangles ABC and AVU are indirectly similar; hence, there exists an indirect similitude mapping $\phi$ mapping the triangle ABC to the triangle AVU.

Now we are going to consider the three altitudes of triangle AVU. Two of these altitudes, namely the ones issuing from the vertices U and V, are simply the lines UC and VB (since $UC\perp AV$ and $VB\perp AU$). Thus, the point of intersection T of these lines UC and VB is the orthocenter of triangle AVU. The third altitude of triangle AVU, this is the altitude issuing from the vertex A, must also pass through the orthocenter T; let the foot of this altitude be denoted by X.

Note that thus, in the triangle AVU, the point T is the orthocenter, the points X, B, C are the feet of the altitudes, and the point D is the orthogonal projection of the point A on the line BC. Hence, the points T, X, B, C and D play the same role in the triangle AVU as the points H, D, E, F and A' in the triangle ABC. Thus, the indirect similitude mapping $\phi$, which maps the triangle ABC to the triangle AVU, must carry the points H, D, E, F and A' to the points T, X, B, C and D, respectively. Of course, this mapping $\phi$, mapping the triangle ABC to the triangle AVU, must also carry the points A, B and C to the points A, V and U, respectively. Thus, we get a lot of interesting proportions, in particular $\frac{HA}{AD}=\frac{TA}{AX}$.

The point T satisfies < ABT = 90° and < ACT = 90°; thus, the points B and C lie on the circle with diameter AT. Hence, the circumcircle of triangle ABC is the circle with diameter AT; hence, the center O of the circumcircle of triangle ABC is the midpoint of the segment AT. Thus, $TA=2\cdot OA$, and the relation $\frac{HA}{AD}=\frac{TA}{AX}$ becomes $\frac{HA}{AD}=\frac{2\cdot OA}{AX}=2\cdot\frac{OA}{AX}=2\cdot\frac{AO}{XA}$.

Since the similitude mapping $\phi$ transforms the points D and A' into the points X and D, the image of the point K, which lies on the line DA' and satisfies $\frac{DK}{KA^{\prime}}=t$, under this mapping $\phi$ must be a point K', which lies on the line XD and satisfies $\frac{XK^{\prime}}{K^{\prime}D}=t$. Thus, $\frac{K^{\prime}D}{XK^{\prime}}=\frac{1}{t}$. In other words, $\frac{DK^{\prime}}{K^{\prime}X}=\frac{1}{t}$.

Now, we will use the [b]Menelaos theorem for quadrilaterals[/b]:

[color=blue][b]Menelaos theorem for quadrilaterals.[/b] Let ABCD be a quadrilateral, and let X, Y, Z, W be four points on its sidelines AB, BC, CD, DA, respectively. Assume that three of the points X, Y, Z, W lie on one line. Then, the fourth one lies on the same line if and only if the equation $\frac{AX}{XB}\cdot\frac{BY}{YC}\cdot\frac{CZ}{ZD}\cdot\frac{DW}{WA}=1$ holds.[/color]

The proof of this theorem is left to the reader (it is readily done).

Returning to the problem, consider the quadrilateral DXOH with the points K', A, P, A on its sidelines DX, XO, OH, HD, respectively. We have

$\frac{DK^{\prime}}{K^{\prime}X}\cdot\frac{XA}{AO}\cdot\frac{OP}{PH}\cdot\frac{HA}{AD}=\frac{1}{t}\cdot\frac{XA}{XO}\cdot\frac{t}{2}\cdot\left(2\cdot\frac{AO}{XA}\right)=1$.

Hence, after the Menelaos theorem for quadrilaterals, since three of the points K', A, P, A, lie on one line - namely, the three points A, P and A (trivial ;) ) -, the fourth point K' must lie on the same line; hence, the points A, P and K' are collinear. Thus, < PAC = < K'AV. But since the indirect similitude mapping $\phi$ carries the points K, A, B to the points K', A, V, respectively, we have < K'AV = - < KAB (indirect similitude mappings preserve angles but change their sign), so that we arrive at < PAC = - < KAB.

This equation yields that the lines AP and AK are symmetric to each other with respect to the angle bisector of the angle CAB. Hence, the line AK is the reflection of the line AP in the angle bisector of the angle CAB. Similarly, the lines BL and CM are the reflections of the lines BP and CP in the angle bisectors of the angles ABC and BCA. Thus, the lines AK, BL and CM concur at the isogonal conjugate of the point P with respect to triangle ABC.

The extended problem is solved!

Note one particular case of the extended problem, namely the case when t = 1. In this case, $\frac{DK}{KA^{\prime}}=\frac{EL}{LB^{\prime}}=\frac{FM}{MC^{\prime}}=t$ becomes $\frac{DK}{KA^{\prime}}=\frac{EL}{LB^{\prime}}=\frac{FM}{MC^{\prime}}=1$, so that the points K, L, M are the midpoints of the segments DA', EB', FC', respectively. On the other hand, $\frac{OP}{PH}=\frac{t}{2}$ becomes $\frac{OP}{PH}=\frac12$, so that $\frac{PH}{OP}=2$; in other words, the point P divides the segment HO in the ratio 2 : 1. Thus, by the properties of the Euler line, the point P must be the centroid of triangle ABC. The isogonal conjugate of the centroid is the symmedian point of triangle ABC. Thus, the extended problem yields that the lines AK, BL, CM concur at the symmedian point of triangle ABC. This gives one more solution of the initial problem of this thread.

  Darij


number theory
least common multiple
greatest common divisor

The difference between the lowest common multiple and greatest common divisor of the numbers 5, 10, and 35 is:

a. 1745 b. 35 c. 65 d. 5 e. NG

[hide]The LCM of $ 5$, $ 10$, and $ 35$ is $ 2(5)(7) \equal{} 70$. The GCD is simply $ 5$. Hence the difference is $ 70 \minus{} 5 \equal{} 65$, or C. [/hide]

We see that the GCD is 5 and the LCM is 70, so LCM-GCD=70-5=65, or choice $ \boxed{\text{c}}$ (gaussintraining, you said d).


geometry
circumcircle
incenter
power of a point
radical axis
geometry unsolved

Let $ ABC$ be a triangle and let $ P$ be an arbitrary point on the side $ BC$. Denote by $ P^{\prime}$ the second intersection of $ AP$ with the circumcircle $ \mathcal{O}(O,R)$ of triangle $ ABC$ and let $ \mathcal{T}_{A}^{b}$, $ \mathcal{T}_{A}^{c}$ be the circles simultanously tangent to $ AP$, $ BP$, $ \mathcal{O}$ and $ AP$, $ CP$, $ \mathcal{O}$, respectively (These two circles are called the [i]Thebault circles[/i] of the cevian $ AP$). If $ A_{b}$, $ A_{c}$ are the tangency points of the side $ BC$ with $ \mathcal{T}_{A}^{b}$, $ \mathcal{T}_{A}^{c}$, respectively, prove that the second intersection of the circumcircle of triangle $ P^{\prime}A_{b}A_{c}$ with $ \mathcal{O}$ coincides with the tangency point of the $ A$-mixtilinear incircle with $ \mathcal{O}$.

I think that the circle $ \mathcal T_b^A$ is tangent to $ AP, BP$ and the circle $ \mathcal T_c^A$ to $ AP, CP,$ otherwise they would be the B-, C-mixtilinear incircles with no relation to an arbitrary cevian $ AP.$ :read:

Of course, you are right. I've edited now.

$ Q_b, Q_c$ are tangency points of $ \mathcal T_b^A, \mathcal T_c^A$ with $ AP.$ By Thebault theorem, $ A_bQ_b, A_cQ_c$ intersect at the incenter $ I.$ The $ \triangle PA_bQ_b$ is isosceles with $ PA_b \equal{} PQ_b.$ Let $ u$ be tangent to $ (I)$ at $ U$ from $ A_b$ other than $ BC.$ Then $ \angle UA_bI \equal{} \angle IA_bP \equal{} \angle Q_bA_bP \equal{} \angle PQ_bA_b$ $ \Longrightarrow$ $ u \equiv A_bU \parallel PQ_b \equiv PA.$ Similarly, tangent $ v \equiv A_cV$ to $ (I)$ at $ V$ from $ A_c$ is parallel to $ PA.$ The infinite intersection $ X^* \equiv A_bU \cap A_cV$ is on $ PA.$ By the problem [url]http://www.mathlinks.ro/viewtopic.php?t=209898[/url], the cirumcircle of the $ \triangle P'A_bA_c$ passes through the tangency point $ A^*$ of the A-mixtilinear incircle of the $ \triangle ABC$ with its circumcircle $ \mathcal O.$

Dear Cosmin and Mathlinkers,
a new way for a proof of this result can be found on on my website :
http://perso.orange.fr/jl.ayme   vol. 4   A new mixtilinear incircle adventure III, p. 28
Sincerely
Jean-Louis

1：Well known $ST$、$I_1II_2$、$BC$ are concurrent，$\angle EIF=90$ . Let $AA' \parallel BC$，then $A^*$ is the A-mixtilinear point，$A^*P'PD$ is a circle .

2：Let $PL \perp I_1I_2$，then easy to see $\angle ELF=90$，so $ILEF$ is a circle，so $QE \cdot QF=QI \cdot QL=QD \cdot QP=QS \cdot QT$ . Hence $SDPT$ is a circle .

3：Since 3 radical Axis must be concurrent，so $ST$、$DP$、$A^*P'$ are concurrent . Hence $QA^* \cdot QP'=QE \cdot QF$，so $A^*P'EF$ is a circle . Done .


[color=blue]Se refiere a que $x;y$ son dos lados cualesquiera del triángulo.[/color]

[quote="aev5peru"]HOLA A TODOS, ACA UNA DESIGUALDAD MUY INTERESANTE.
Sea $a,b,c$ lados de un triangulo, $R$ es el circunradio, $r$ es el inradio.
${x,y}$es un sub-conjunto de ${a,b,c}$.
demostrar que:
$R/r\geq x/y+y/x$.[/quote]

Buenos dias,
 esta es una inecuación muy interesante...

Aki les muestro un articulo sobre inecuaciones de este tipo, escrita por el Profesor:
[color=blue][b]Amengual Covas, Miguel
La desigualdad de Euler a partir de otras desigualdades. [/b][/color]

donde revisa inecuciones desde un punto de vista geométricos, aunq su tratamiento es algo algebraico....  :huh:

Bueno, revisenlo y denme sus comentarios...

PD: Más lecturas ,aki: http://www.oei.es/oim/saladelectura.htm

Saludos,

CARLOS Bravo.


logarithms
inequalities
quadratics
rearrangement inequality
number theory proposed
number theory

The first one is trivial since $ d_{i}\cdot d_{k-i+1}= n$.

If $ n\not= d^{2}$ then we can pair the devisors in the couples $ (d_{i},\frac{n}{d_{i}})$ for $ i=1,...,k$ with product in each pair $ n.$ Therefore $ d_{1}...d_{k}=n^{\frac{k}{2}}.$ If $ n=d^{2}$ we can do the above procedure for all the devisors but $ d$ and in this case $ d_{1}...d_{k}=n^{\frac{k-1}{2}}.n^{\frac{1}{2}}=n^{\frac{k}{2}}.$

Now we have to prove that \[ d_{1}\log d_{1}+...+d_{k}\log d_{k}=(d_{1}+...+d_{k})\log n-nrk\] for some $ 0\leq r\leq \frac{\log 3}{3}.$ The above equality is equivalent to $ (d_{1}+...+d_{k})\log n\geq d_{1}\log d_{1}+...+d_{k}\log d_{k}\geq (d_{1}+...+d_{k})\log n-nk\frac{\log 3}{3}.$ But since $ d_{i}\leq n$ we have that $ (d_{1}+...+d_{k})\log n\geq d_{1}\log d_{1}+...+d_{k}\log d_{k}.$ Thus it remains to prove that \[ S=d_{1}\log d_{1}+...+d_{k}\log d_{k}\geq (d_{1}+...+d_{k})\log n-nk\frac{\log 3}{3}.\] We have that $ d_{1}<d_{2}<...<d_{k}$ and $ \log d_{1}<\log d_{2}<...<\log d_{k}$ thus by the Rearrangement inequality we have that $ S\geq \sum_{i=1}^{k}d_{i}\log \frac{n}{d_{i}}=\sum_{i=1}^{k}d_{i}(\log n-\log d_{i}).$ Therefore we have to prove that \[ nk\frac{\log 3}{3}\geq \sum_{i=1}^{k}d_{i}\log d_{i}.\] On the other hand by Cauchy we have that $ (\sum_{i=1}^{k}d_{i})(\sum_{i=1}^{k}\log d_{i})\geq (\sum_{i=1}^{k}d_{i}\log d_{i})^{2}.$ Also by the first part of the problem $ \sum_{i=1}^{k}\log d_{i}=\frac{k}{2}\log n$ so it is enough to prove that \[ \frac{n^{2}k.2\log^{2}3}{9\log n}\geq \sum_{i=1}^{k}d_{i}.\] But since $ d_{i}\in [1,n]$ we have that $ d_{i}+\frac{n}{d_{i}}\leq n+1$ (you can prove this by solving the quadratic inequality) and therefore $ \sum_{i=1}^{k}d_{i}\leq \frac{k}{2}(n+1)$ (this is true even if $ n=d^{2}$!). So it remains to prove that $ \frac{n^{2}.4\log^{2}3}{9\log n (n+1)}\geq 1$ which have to be true (it is true for not so big $ n$ and it can be checked for the other values).

Hope it is correct.


ratio
function
limit
geometric series
complex numbers
calculus
calculus computations

find whether the series is convergent or divergent using d'alembert's ratio test

Sigma[((10+5i)^2)/(n!)],where n goes from 0 to infinity,i hope u can the series expression is understandable

if it cant be found out by above test,then find with any other

the answer given was convergent

You wrote $ \sum_{n\equal{}0}^{\infty}\frac{(10\plus{}5i)^2}{n!},$ which isn't very interesting.

Perhaps you actually meant $ \sum_{n\equal{}0}^{\infty}\frac{(10\plus{}5i)^n}{n!}.$

If it is the latter, then note that this is simply a special case of $ \sum_{n\equal{}0}^{\infty}\frac{z^n}{n!}.$

Yes, the ratio test shows that this converges for all $ z\in\mathbb{C}.$ Furthermore, we define the function it converges to as $ e^z.$

yes i wrote wrong now how  could u prove that ,is it a special case of  ration test?

The ratio test is a comparison test (comparison to a geometric series) and hence a test for absolute convergence. The absolute values of complex numbers do behave themselves as you would expect. In particular, $ | z^n|=|z|^n,$ so $ |(10+5i)^n|=(5\sqrt{5})^n.$

For general $ z,$ we test the convergence of $ \sum_n\frac{z^n}{n!}$ by taking

$ \lim_{n\to\infty}\frac{|z^{n+1}|}{(n+1)!}\cdot\frac{n!}{|z^n|}= \lim_{n\to\infty}\frac{|z|}{n+1}=0<1$ no matter what $ z$ is.

Hence the series converges (absolutely) for all $ z,$ including $ z=10+5i.$


geometry
rectangle
analytic geometry
pigeonhole principle

Let A be a region in the plane $\mathbb{R}^{2}$, with area strictly greater than one.
Show that A contains two distinct points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ such that $x_{2}$ − $x_{1}$ and $y_{2}$ − $y_{1}$ are both integers.

My Attempted Solution:

This problem seems pretty trivial,

A is the some area in $\mathbb{R}^{2}$ bigger than one.

Pick any two points so that the $x_{2}$ − $x_{1}$ = 1and $y_{2}$ − $y_{1}$ = 1

...

This just seems so trivial that it's hard to prove.

[quote="1337h4x"]Your solution is incorrect. Suppose $A$ is a rectangle with vertices $(0,0)$, $(3,0)$, $\left(0,\frac{1}{2}\right)$, and $\left(3,\frac{1}{2}\right)$.The area of $A$ is $\frac{3}{2}>1$ so it satisfies the intitial conditions. However, there are no two y coordinates which are 1 unit apart. Thus, your two points do not exist.[/quote]

Umm... $(3-0) \in \mathbb{Z}$ and $(0-0) \in \mathbb{Z}$...

[quote="Altheman"]once you see the solution, you are like wtf...but this is a really nice problem...you need to use... *gasp* pidgeonhole![/quote]

That is what first came into my mind...

The pigeon hole is really starting to bug me.

This question seems poorly defined to me. Area? This is a difficult issue to discuss. Region? Does this mean what it usually does in analysis - i.e. open and connected set?

I doubt any solution at a high school level is correct.

Kalle, I think you are overanalyzing the given problem... especially for its intended difficulty level.

[quote="problem_solver"]Let A be a region in the plane $\mathbb{R}^{2}$, with area strictly greater than one.
Show that A contains two distinct points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ such that $x_{2}$ − $x_{1}$ and $y_{2}$ − $y_{1}$ are both integers.
[/quote]

I'm not sure how to make use of the Pigion Hole Principle, but I came up a proof - Could somebody please veryify them for me.

Proof:

Take any area in $\mathbb{R}^{2}$ greater than one.

Divide this area up into sub areas which are squares with area 1.

Then choose the points on the verticies on any square in the given area and you have two distinct points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ such that $x_{2}$ − $x_{1}$ and $y_{2}$ − $y_{1}$ are both integers.

How does this sound ?  :weightlift:

Can you always divide the region into squares of area 1? Careful.

[quote="paladin8"][hide] What happens if we take all of the lattice squares and... *gasp* stack them all together! [/hide][/quote]

Take all the square lattices,

Stack them up vertically.

Then as long as their are two points an any vertex, then we are done?

[quote="Kalle"]There are very reasonable sets that do not contain any single square but have large area.[/quote]I do not understand what you are trying to point out. Perhaps consider this is not a topology/analysis problem.

So stack the squares? If two points overlap, we are done. If not, the 'stacked' area fills up the square but since the area is larger than 1, there must exist two points that must overlap.


analytic geometry
graphing lines
slope
geometry
Pythagorean Theorem
AMC

A cubical cake with edge length $ 2$ inches is iced on the sides and the top. It is cut vertically into three pieces as shown in this top view, where $ M$ is the midpoint of a top edge. The piece whose top is triangle $ B$ contains $ c$ cubic inches of cake and $ s$ square inches of icing. What is $ c\plus{}s$?
[asy]unitsize(1cm);
defaultpen(linewidth(.8pt)+fontsize(8pt));

draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle);
draw((1,1)--(-1,0));
pair P=foot((1,-1),(1,1),(-1,0));
draw((1,-1)--P);
draw(rightanglemark((-1,0),P,(1,-1),4));

label("$M$",(-1,0),W);
label("$C$",(-0.1,-0.3));
label("$A$",(-0.4,0.7));
label("$B$",(0.7,0.4));[/asy]$ \textbf{(A)}\ \frac{24}{5} \qquad
\textbf{(B)}\ \frac{32}{5} \qquad
\textbf{(C)}\ 8\plus{}\sqrt5 \qquad
\textbf{(D)}\ 5\plus{}\frac{16\sqrt5}{5} \qquad
\textbf{(E)}\ 10\plus{}5\sqrt5$

Ok here's the way I did it (I doubt it is optimal)

[hide]So make a coordinate plane for the top and have the bottom left corner (0,0) Top left (0,2) and so on. You want to find where the 3rd corner of B is on the coordinate plane. Lets call this point z. The line from M to the top right is from (0,1) to (2,2) so the equation is y=x/2+1. Another line comes from the BR (bottom right) and meets it at z. Since it is perpendicular, the slope is -2 and the equation is y=-2x+4. So z is (1.2,1.6)

Now use shoelace. Your three points are (2,2), (1.2,1.6), and (2,0).
2    2
1.2 1.6
2    0
2    2

You end up getting 7.2-5.6 or 1.6/2=.8 in area Since the height is 2, .8*2=1.6+.8=2.4 or 12/5. Icing is also on the side, so 2*2=4=20/5 20/5+12/5=32/5 or

[B]

[/hide]

I just thought of it when looking it over. Sadly I only did so at the end and couldn't complete it :(

Alternately note that triangles A and B are similar, and then proceed from there.
But I also like Ihatepie's solution.

Triangle B is exactly 1/5 the area of the entire square, since it is similar to triangle A.

Well, triangle $ A$ is $ \frac{1}{4}$ the area of the entire square, and triangle $ B$ is $ \frac{1}{4} \cdot \left(\frac{2}{\sqrt{5}}\right)^2$=$ \frac{1}{5}$ the area of the entire square.

[quote="daman71692"]Well written solution. Where did you get the idea to use a coordinate plane?[/quote]
thats what i did too but i crapped out and did not finish cuz i did the pythagorean theorem wrong  :rotfl:

[quote="daman71692"]Well written solution. Where did you get the idea to use a coordinate plane?[/quote]

Whenever I can't solve a problem that involves "perpendicular", you're given lengths, and circles aren't there, I like to use coordinates :)

[hide="not bash"]
Let Z be the upper right vertex. We have $ [A]\equal{}2*1/2\equal{}1$. We see that $ [A]/[B] \equal{} ((MZ)/2)^2 \equal{} 5/4$, so $ [B] \equal{} 4/5$. Since the area of the face to the side is 4, $ s \equal{} 4/5 \plus{} 4 \equal{} 24/5$. Since the height of the cake is 2, $ c \equal{} 4/5*2 \equal{} 8/5$. Thus, $ c \plus{} s \equal{} 32/5\Rightarrow \boxed{B}$
[/hide]

i always use coordinate plane systems but for this problem i think the similar triangles method is much quicker.

Usually I find that using coordinate planes take longer than finding similar triangles. But that's just me :) .


the least and greatest numbers in a list of 7 real numbers are  2 and   20 ,respectively. the median of the list is 6, and the number 3 occurs most often in the list which of the following could be the average (arithmetic mean ) of the numbers in the list

I. 7
II 8.5
III 10

(A)I only  (B) I and II only   (C) I and III only
(D) II  and  III only   (E)  I, II, and III

If the median is 6, that means 6 is the 4th number. 2 is the first, and 20 is the last. Hence, we have:

2 _ _ 6 _ _ 20

Obviously, if the 3 occurs most often, it can't occur just once, so it has to occur twice:

2 3 3 6 _ _ 20

So far, this is a sum of 34. The two unknown numbers have to be between 7 and 19, inclusive, so the smallest mean would be $ \frac{7\plus{}8\plus{}34}{7} \equal{} 7$ and the largest $ \frac{18\plus{}19\plus{}34}{7} \equal{} 10\frac{1}{7}$. Therefore, all three of I, II, and III are possible: E.


If $ \triangle ABC$ has points $ D, E$ and $ F$ on sides $ AB$, $ BC$, and $ AC$,respectively, and the cevians $ AE$, $ CD$, and $ BF$ are concurrent at point $ P$, then how do you find the ratio, using mass points, of $ \frac {AD}{DB}$ if you know the lengths of $ CP, BP, DP, FP$ and $ FC$?

I know how to use mass points to find the ratio of the segments the cevians are split up in given the ratios of the segments the sides are split up in, but I am getting confused on how to use the ratio of the cevians segments to find the ratio of the segments the sides of the triangle are split up in.

I would appreciate any help.
Thanks.

i think this is a little too hard for HSB...
CAUSE I CAN"T DO IT!!!


abstract algebra
group theory
superior algebra
superior algebra unsolved

I've problem when doing these questions under the topic of Sylow Theorems....

1) Classify the groups which have order 1,225.

2) Prove that a finite abelian group is isomorphic to the product of the finite number of abelian groups the order of each of which is a power of a prime.

3) If every Sylow subgroups of F is a normal subgroup, show that G is isomorphic to the product of its Sylow subgroups.

Thanks...

1) this question is easy when you use the two following
[url]http://www.math.gatech.edu/academic/graduate/GP/Comp/compFall04AlgSol.pdf[/url]
Google-ing is so nice sometimes....

2)that is a theorem
you should find that in books

3) first prove this lemma, if A and B are normal subgroups of G and their intersection is trivial, every element in A commutes with every element with B

now try to prove this theorem, if $N_{i}$ are all normal subgroups and $G=N_{1}\cdots N_{m}$ and for all i $N_{i}\cap (N_{1}\cdots N_{i-1} N_{i+1}\cdots N_{m})=\{e\}$ then G is isomorphic with their cartesian product

now try to understand why the sylow groups satisfy all the conditions in the previous theorem


calculus
integration
trigonometry
calculus computations

Calculate the following integrals.

(1) $ \int e^{2x \plus{} 3}\ dx$.

(2) $ \int \sin \left( \minus{} \frac {x}{2} \plus{} \frac {\pi}{3}\right)\ dx$.

(3) $ \int x^2e^{ \minus{} x}\ dx$.

(4) $ \int e^x\sin x\ dx$.

(5) $ \int \frac {dx}{x^2 \minus{} x \minus{} 2}$.

(6) $ \int \frac {dx}{x(x^2 \minus{} 1)}$.

(7) $ \int \frac {1}{x^2(x \minus{} 1)}\ dx$.

(8) $ \int \sqrt {1 \plus{} x^2}\ dx$ by setting $ \sqrt {1 \plus{} x^2} \plus{} x \equal{} t$.

(9) $ \int \frac {3x^2 \plus{} x \plus{} 4}{x^3 \plus{} x^2 \plus{} 2x \plus{} 2}\ dx$.

(10) $ \int \frac {\cos ^ 2 x \minus{} \sin ^ 2 x}{1 \plus{} 2\sin x\cos x}\ dx$.

The first word of mine is to say thank to Kunny, you got a favor for maths and contribute every effort in this forum, especially to help the new comers. Those contribution should be appreciated. By the way, can you post some full entrance exams to Universities in Japan? 'cause I am curious to explore about the level of difficulty there. Thanks a lot.

Thank you for your admiration. :lol:

For example, you can see the some problems [url=http://www.mathlinks.ro/resources.php?c=87]here[/url]. :)

hello
[hide="1"]
$ \frac{e^{2x\plus{}3}}{2}\plus{}C$
[/hide]
[hide="2"]
$ 2 cos( \frac{\pi}{3}\minus{}\frac{x}{2})\plus{}C$
[/hide]
[hide="3"]
$ \minus{}\frac{x^2\plus{}2x\plus{}2}{e^{x}}$

[/hide]
[hide="4"]
we have proved this many times
$ I \equal{} \int e^{ax}sin(bx) dx \equal{}\frac{ e^{ax}}{a^2\plus{}b^2}( a sin(bx)\minus{}bcos(bx))\plus{}C$
so answer is $ \frac{e^{x}}{2}( sin(x) \minus{} cos(x))\plus{}C$
[/hide]
[hide="5"]
"I think there is dx only one time"
$ \frac{1}{x^2\minus{}x\minus{}2}\equal{} \frac{ 1}{3(x\minus{}2)}\minus{}\frac{1}{3(x\plus{}1)}$
$ \frac{1}{3}log{\frac{x\minus{}2}{x\plus{}1}}\plus{}C$
[/hide]
[hide="6"]
it can be written as $ \frac{1}{2}[ \frac{1}{x\minus{}1}\plus{}{1\minus{}x}]\minus{}\frac{1}{x}$
answer=$ \frac{1}{2}log(x^2\minus{}1) \minus{}log (x) \plus{}C$

[/hide]
[hide="8"]
U have suggested one method .
I m doing by Integration by parts .
$ I \equal{}\int \sqrt{1\plus{}x^2} dx$
$ I \equal{} x \sqrt{1\plus{}x^2} \minus{} \int \frac{x^2}{\sqrt{1\plus{}x^2}} dx$
$ I \equal{} x \sqrt{1\plus{}x^2} \minus{} \int \frac{x^2\plus{}1\minus{}1}{\sqrt{1\plus{}x^2}}$
$ I \equal{} x \sqrt{1\plus{}x^2} \minus{} I \plus{} log ( x\plus{}\sqrt{1\plus{}x^2})$
$ I \equal{} \frac{1}{2} [ x\sqrt{x^2\plus{}1}\plus{}log ( x \plus{}\sqrt{1\plus{}x^2})\plus{}C$
[/hide]
[hide="7"]
$ \frac{1}{x^2(x\minus{}1)}\equal{}\frac{1}{x\minus{}1}\minus{}\frac{1}{x}\minus{}\frac{1}{x^2}$
$ I \equal{} log (\frac{x\minus{}1}{x})\plus{}\frac{1}{x}\plus{}C$[/hide]
[hide="9 "]
$ \frac{3x^{2}\plus{}x\plus{}4}{x^{3}\plus{}x^{2}\plus{}2x\plus{}2} \equal{} \frac{Ax\plus{}B}{x^2\plus{}2}\plus{}\frac{C}{x\plus{}1}$
$ A\equal{} 1 , B\equal{}0 , C \equal{}2$
$ I \equal{}\frac{log( x^2\plus{}2)}{2} \plus{}2log (x\plus{}1)\plus{}C$

[/hide]
[hide=" 10"]
$ \int\frac{\cos^{2}x\minus{}\sin^{2}x}{1\plus{}2\sin x\cos x}\ dx$
$ \int \frac{cos(x)\minus{} sin(x)}{cos(x)\plus{}sin(x)}$
$ \equal{} log ( cos(x) \plus{}sin(x)) \plus{}C$[/hide]
sorry if some where i have done mistake .
thank u