I can't seem to solve this problem algebraically: Find in terms of x 2^x=x^2 or in general y^x=x^y.

Thanks in advance. This problem is driving me crazy while I'mtryign to solve a different problem.

I'm sorry for the confusion, but for the first problem I meant find in terms of x: x^2=2^x and I also don't understand the jump from your second step to your third step how 2^x turns into xlog2,

[quote="Iversonfan2005"]For the first one, $x \cdot 2^x=x^2$ is impossible unless $x=0$

$x \cdot 2^x=x^2$
$2^x=x$
$x\cdot \log{2}=x$
$\log{2}=1$ which is not true[/quote]

You forgot to take the $\log$ of the RHS.

EDIT: AS a result, it woudl be $x=\frac{\log x}{\log 2}$.


Find all positive integers $ a,b$ if $ a^{2}\plus{}ab\equal{}b^{2}\plus{}1$

Let $ a^{2}\plus{}ab\minus{}(b^{2}\plus{}1) \equal{} 0$
$ \Delta \equal{} b^{2}\plus{}4(b^{2}\plus{}1) \equal{} 5b^{2}\plus{}1$
So there exist an positive integer m such that
$ m^{2}\equal{} 5b^{2}\plus{}4$
$ m^{2}\minus{}5b^{2}\equal{} 4$
Then use Pell equation.


algebra
polynomial
induction
combinatorics proposed
combinatorics

Let $(a)_0: =1$ and  $(a)_k: =a(a+1)...(a+k-1)$ for $k=1,2,...$. [b] To prove that  for $n\in {\mathbb N}$  the numbers
\[ N_n(x,y) : =\sum\limits_{k=0}^{n}(-1)^{n-k}{n\choose k}(y+k)_k(x+y+2k)_{n-k} \; \; ,\; \; x\in {\mathbb Z}\; ,\; y\in {\mathbb C} \]
are integers. [/b]

Of course $N_n(x,y)$ is a polynomial in $x,y$ with integer coefficients.
It suffices to prove that all coefficents of non-zero powers of $y$ vanish, so that the polynomial would be in $x$ and then obviously integer.
To prove this we must prove $P(x,y)=const$ for fixed $x$ and variable $y$.
It suffices to prove this just for positive integer values of $y$ because it's a polynomial. Let's prove it for positive $x$ (analogously for negative).
Thus we perform by induction on $y+n$.
Basis is quite obvious.
Now induction step.
We write $N_{n}(x,y)=n! \sum_{k \in \mathbb{Z}} C_{y+k}^k C_{x+y+2k}^{n-k}$
Hovewer $C_{y+k}^k-C_{y+k-1}^{k-1}=C_{y-1+k}^k$.
Now summing over $k \in Z$ we get $N_n(x,y)+N_{n-1}(x+2,y)=N_n(x+1,y-1)$ so
$N_n(x,y)=N_n(x+1,y-1)-N_{n-1}(x+2,y)$ which is a polynomial only in $x$ by induction hypothesis.
QED

(Possibly I have made some errors  :blush: but I believe the main idea of the proof is correct)

[quote="diophantus"]What does your $(y+k)_k$ notation mean. :blush:[/quote]

Please make the substitution   $a\leadsto y+k $ in the definition of $(a)_k $ .

[quote="iura"]Of course $N(x,y)$ is a polynomial in $x,y$ with integer coefficients.
It suffices to prove that all coefficents of non-zero powers of $y$ vanish, so that the polynomial would be in $x$ and then obviously integer.
To prove this we must prove $P(x,y)=const$ for fixed $x$ and variable $y$.
It suffices to prove this just for positive integer values of $y$ because it's a polynomial. Let's prove it for positive $x$ (analogously for negative).
This we perform by induction on $y+n$.
Basis is quite obvious.
Now induction step.
We write $N_{x,y}=n! \sum_{k=0}^n C_{y+k}^k C_{x+y+2k}^{n-k}$
Hovewer $C_{y+k}^k=C_{y-1+(k+1)}^{k+1}-C_{y-1+k}^{k}$.
And now the induction step gets trivial.[/quote]
Please finish the proof ! Regards,Alex

[quote="flip2004"]Let $ (a)_0:=1 $ and  $ (a)_k:=a(a+1)...(a+k-1) $ for $ k=1,2,... $.  To prove that  for $n=0,1,... $  the numbers
\[ N_n(x,y) :=\sum\limits_{k=0}^{n}(-1)^{n-k}{n\choose k}(y+k)_k(x+y+2k)_{n-k}
\; \; ,\; \; x\in {\mathbb Z}\; ,\; y\in {\mathbb C}\]
are integers.[/quote]

I think it is a Taylor-like expansion of $f(x+y - y)$ but with respect to the difference operator $f(t+1)-f(t)$ in place of $\frac{d}{dx}$ (i.e. the operator that sends $(x)_k \to k(x)_{k-1})$.


real analysis
inequalities
real analysis unsolved

Prove that
$\limsup(a_{n}+b_{n}) \leq \limsup(a_{n})+\limsup(b_{n})$
and give an example where the inequality is strict.

The example

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=111011&highlight=limsup[/url]


calculus
integration
trigonometry
calculus computations

$ f(a) \equal{} \minus{} \int_0^a \sqrt [251]{(e^{ \minus{} x}\cos x)'}\ dx$, how  would I make proceed after that?

You don't have to actually evaluate the integral


trigonometry
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analytic geometry

Hint. [hide]Suppose csc<sup>-1</sup> (3/2) = alpha. Then csc (alpha) = 3/2. sin (alpha) = 1 / (csc (alpha)) = 1 / (3 / 2) = 2 / 3.[/hide]

[hide]realize the problem can be rewritten:

[tex]\sin(\arcsin{\frac{2}{3}}+\arctan{-\frac{1}{2}})[/tex]

which can be expanded to:

[tex]\sin(\arcsin{\frac{2}{3})\cos(\arctan{-\frac{1}{2}}) + \sin(\arctan{-\frac{1}{2}})\cos(\arcsin{\frac{2}{3}) [/tex]

Then it's just a matter of solving with triangles, don't forget the domain of acrtan, which makes the 1 negative and not the 2...

this simplifies to:

[tex]\displaystyle{\frac{2}{3}\frac{2}{\sqrt{5}}-\frac{1}{\sqrt{5}}\frac{\sqrt{5}}{3}}[/tex]

[tex]\displaystyle{\frac{4}{3\sqrt{5}}-\frac{1}{3}[/tex]

put it in whatever form you'd like.[/hide]

No, you don't need to know the angles. [tex]\csc^{-1}\frac{3}{2}=\sin^{-1}\frac{2}{3}[/tex]. [tex]\sin(\sin^{-1}\frac{2}{3})=\frac{2}{3}.[/tex]

This is all I got, then I became confused by the cos(tan:

HINT1:[spoiler] Use sum and difference identity to get = sin(arccsc(3/2))*cos(arctan(-1/2))+cos(arccsc(3/2))*sin(arctan(-1/2)).

HINT2:[spoiler]sin(arccsc(x))=x, right? b/c they are inverses.  So you get:

3/2 * cos(arctan(-1/2)) + 3/2 * sin(arctan(-1/2)

Note that this is not an identity arctan(x)=arcsine(x)/arccos(x), so you cant use it.

JS:

acrcsc x = arcsin (1/x)

so sin(arccsc x) = 1/x

As for cos(arctan(-1/2))...

Let [tex]\theta = \arctan{-\frac{1}{2}}, \text{then} \tan{\theta} = -\frac{1}{2} [/tex]

Contruct a right triangle with legs -1 and 2... for the -1, just make the leg in the negative direction a coordinate plane... Solve for the hypotenuse, which is [tex]\sqrt{5}[/tex] , now... use that triangle and solve for [tex]\cos{\theta}}[/tex].

Thanks!  That's embarassing since I just learned trig formally this year.


Give 10 red balls, 11 green balls, and 7 black balls, what is the probability that when you draw 6 balls you get 3 red balls, 2 green balls, and 1 black ball?

I can't figure this out because I always get confused about the order of the draw.

hmm

what i would do is

(10/28 * 9/27 *8/26) chance to get 3 red balls
(11/25*10/24)  chance to get 2 green balls
(7/23) chance to get a black ball

then just multiply everything together and time 6!/(2!3!1!) since the order of the balls doesnt matter

SIMPLY USE COMBINATION S::
u have 28 balls
P(3 red , 2 green ,1 black)={C(10,3)*C(11,2)*C(7,1)}/C(28,6)

[quote="SMCount"]Give 10 red balls, 11 green balls, and 7 black balls, what is the probability that when you draw 6 balls you get 3 red balls, 2 green balls, and 1 black ball?[/quote]
Ok the prob. the first ball is red is $ 10/28$ the probability the second is red is $ 1/3$ the probability the 3rd is red is $ 8/26$ the probability the fourth is green is $ 11/25$ the probability the 5th is green is $ 10/24$ the probability the 6th is black is $ 7/23$. There are $ \frac {6!}{3!2!} \equal{} 60$ ways to rearrange the balls so we multiply by 60 because we have to accuount for the number of ways this can happen. Therefore our answer is $ 10/28 * 1/3* 8/26 * 11/25 * 10/24 * 7/23 * 60 \equal{} \boxed{110/897}$

28C6 is the wrong denominator because clearly that assumes every ball is distinct which is not true. If your doing it with counting the denominator is C (12,2) becuase there are more than six of each ball.

heh?!

[quote]28C6 is the wrong denominator because clearly that assumes every ball is distinct which is not true. If your doing it with counting the denominator is C (12,2) becuase there are more than six of each ball.
[/quote]

28c6 is also right

we all got the same answer  110/897 ...
  :)


find  x and y  in $ R$ :

xy-x-y=1
2x^2-xy-y=0

Note that the second equation gives

$ 2x^{2}\equal{} (1\plus{}x)y$. If $ x \equal{}\minus{}1$, so solution so $ \frac{2x^{2}}{x\plus{}1}\equal{} y$.

But the second equation gives an irreducible cubic...?

Sketching the graphs $ y\equal{}1\plus{}\frac{2}{x\minus{}1},\ y\equal{}\frac{x^{2}}{x\plus{}1}$ gives $ x\equal{}\alpha\ (\frac{5}{2}<\alpha <3)$, where $ x\equal{}3$ is the x coordinate of the intersection of $ y\equal{}1\plus{}\frac{2}{x\minus{}1}$ and the line $ y\equal{}x\minus{}1$, the asymptote of $ y\equal{}\frac{x^{2}}{x\plus{}1}$.


C
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L
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how did you do in the states GBML???

Erm... real names scare me so I wont say any. His real name appears when clicking on his posts.


[quote="mdk"]let x and y be strictly positive real numbers such that x^2+y^3 ge x^3+y^4. Show that x^3+y^3 le 2.[/quote]
It's France 2006-2007,  dossier 4, exercice 4. :wink:
$x^{5}+y^{5}\leq2$ with same conditions true too, but it's not easy.

[quote="mdk"]Let $x$ and $y$ be strictly positive real numbers such that $x^{2}+y^{3}\ge x^{3}+y^{4}.$ Show that $x^{3}+y^{3}\leq 2.$[/quote]
[quote="arqady"]Let $x$ and $y$ be  positive real numbers such that $x^{2}+y^{3}\ge x^{3}+y^{4}.$ Show that $x^{5}+y^{5}\leq 2.$[/quote]
[quote="mdk"]it's not hard either.[/quote]
The proof for the first  inequality:
By AM-GM holds: $2x^{3}+1\geq3x^{2}$ and $3y^{4}+1\geq4y^{3}.$
Hence $x^{3}+y^{3}\leq2+3(x^{3}+y^{4}-x^{2}-y^{3})\leq2.$
What your proof for my inequality, mdk? Thank you.


Are there infinitely many numbers n such that $ 2^n-1$ divides $ (n+1)! $?

The answer is no. Schinzel has proved that for $n>12$ the largest prime factor of $2^n-1$ is greater than or equal to 2n+1.

Can you post a solution, Harazi ? It is interesting !

I don't have the article where he proved this result, unfortunately. I just read about this result yesterday. Hope that someone can provide the article.


Find  $ p(ab) , p(bc) $ and $ p(ca) \; . $

$p(ab)=abc\frac{ab-bc-ac+c^2}{a+b+c}$
$p(bc)=abc\frac{-ab+bc-ac+a^2}{a+b+c}$
$p(ac)=abc\frac{-ab-bc+ac+b^2}{a+b+c}$

Yes indeed it is really nice.

Really nice! From $P(ab)>0$, $P(ac)<0$, $P(bc)>0$ and $ab<ac<bc$ it follows that the roots of the polynomial are in $]ab,bc[$ which implies the statement.

Misha


trigonometry
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logarithms
inequalities proposed

Let $ x,y,z > 0$,with $ \sqrt {xy} \plus{} \sqrt {yz} \plus{} \sqrt {zx} \equal{} 1$.Prove that:

$ 8xyz\ge (1 \minus{} x)(1 \minus{} y)(1 \minus{} z)$

Let $ x \equal{} \tan^2\frac {A}{2}$ , $ y \equal{} \tan^2\frac {B}{2}$  , $ z \equal{} \tan^2\frac {C}{2}$    , $ A,B,C\in(0,\frac {\pi}{2})$

 , Inequality is equivalent with $ \prod {\frac {(1 \minus{} \cos A)}{\cos A}}\ge 1$ (*)

 But function $ f(x) \equal{} \ln (\frac {1 \minus{} \cos x}{\cos x})$ is convex on $ (0,\frac{\pi}{2})$ , and (*) is true from Jensen.

where $ A \plus{} B \plus{} C \equal{} \pi$

First of all, we introduce $ a,b,c$ as follows:

$ \sqrt{xy}\equal{}\frac {c} {a\plus{}b\plus{}c}, \sqrt{yz}\equal{}\frac {a} {a\plus{}b\plus{}c}, \sqrt{zx}\equal{}\frac {b} {a\plus{}b\plus{}c}$

$ \Rightarrow x\equal{}\frac {bc} {a(a\plus{}b\plus{}c)}, y\equal{}\frac {ac} {b(a\plus{}b\plus{}c)}, z\equal{}\frac {ab} {c(a\plus{}b\plus{}c)}$

After some reductions, the inequality is equivalent with:

$ \frac {a^2b^2\plus{}b^2c^2\plus{}c^2a^2\minus{}abc(a\plus{}b\plus{}c)} {abc} \ge \frac {(a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c)\minus{}9abc} {(a\plus{}b\plus{}c)^2}$

Now, W.L.O.G. we suppose that $ abc\equal{}1$ because the degree from the left hand is equal with the degree from the right hand.

We reduce again and finally we obtain the inequality:

$ (a^4b^2\plus{}1)\plus{}(a^2b^4\plus{}1)\plus{}(a^4c^2\plus{}1)\plus{}(a^2c^4\plus{}1)\plus{}(b^4c^2\plus{}1)\plus{}(b^2c^4\plus{}1)\plus{}$

$ \plus{}(a^3b^3\plus{}a^3c^3)\plus{}(b^3a^3\plus{}b^3c^3)\plus{}(c^3a^3\plus{}c^3b^3) \ge$

$ \ge 2a^2b\plus{}2ab^2\plus{}2a^2c\plus{}2ac^2\plus{}2b^2c\plus{}2bc^2\plus{}2a^3\plus{}2b^3\plus{}2c^3$

which is obviously true.


Call a sum of two squares interesting if at least one of them is not divisible by 5.  Prove that for any natural number $ n$, there is an interesting sum that adds to $ 10^n$.

Start with the pair $ (x,y)\equal{}(1,3)$ and generate the next pair by taking the previous pair $ (a,b)$, and transforming it into $ (3a\plus{}b,3b\minus{}a)$.

Now, we can prove by induction that the first is always congruent to 1 mod 5, and the second to 3 mod 5, so we are done (we don't have to make sure they are positive, since we can form the sequence, and then afterwards take the absolute value of every term in every pair).

what's up with the definition of "interesting"? if one isn't divisible by 5 and they sum to a multiple of 10, then obviously the other isn't either. actually, to be honest, dropping the "interesting" condition would have no effect on the problem whatsoever.


Evaluate:
$ \int \frac{x}{4x\sin x \plus{} (4 \minus{} x^2)\cos x} dx$

Mhmhm, are you sure the numerator is x and not x^2 :D , cause took me lot time and efford, also to many ppls :(
IF it  is so, then there is a solution

divide top and bottom of the integrand by "something" and then make a clever trig substitution!  :wink:


I assume you mean geometrically?

For all even: Draw a row of squares on the top and left sides, something like this:

[asy]draw((0,0)--(12,0)--(12,12)--(0,12)--(0,0));
draw((4,0)--(4,12));
draw((8,8)--(8,12));
draw((0,8)--(12,8));
draw((0,4)--(4,4));[/asy]

For all odd: Divide the above square like that, into n-3 squares (so that diagram would be drawn for n=9). Then, divide the big square into 4 parts.

[quote="hasan4444"]Show that for all $ n \ge 6$ a square may be divided into $ n$ smaller squares.[/quote]

Make that be "...divided into $ n$ smaller nonoverlapping squares" in your problem.

This is essentially 1999 MATHCOUNTS Team #9.

This problem was recently on [url=http://www.math.umd.edu/highschool/mathcomp/PartIISolns2008.pdf]a competition[/url] I participated in. Look at #1 for the solution.

Here is my solution:
If we have a solution for n squares we can just cut one square into four to get n+3 squares, so we have to show the problem for 6,7, and 8. See the picture. For the one with 8, the side of each square is divided into fifths. [geogebra]8a1b9d17afc9b3427bee4851889513cb2ad4b441[/geogebra]

Here is a solution presented by my coach:

We will use sized induction:
$ n=6$
\begin{eqnarray*}
x^2&=&\underbrace{\left(\frac{x}{3}\right)^2+\left(\frac{x}{3}\right)^2+\cdots+\left(\frac{x}{3}\right)^2}_{9\ times}\\
&=&4\left(\frac{x}{3}\right)^2+\underbrace{\left(\frac{x}{2}\right)^2+\cdots+\left(\frac{x}{2}\right)^2}_{5\ times} \\
&=& \left(\frac{2x}{3}\right)^2+\underbrace{\left(\frac{x}{2}\right)^2+\cdots+\left(\frac{x}{2}\right)^2}_{5\ times}
\end{eqnarray*}

$ n=7$
\begin{eqnarray*}
x^2&=&\left(\frac{x}{2}\right)^2+\left(\frac{x}{2}\right)^2+\left(\frac{x}{2}\right)^2+\left(\frac{x}{2}\right)^2\\
&=& \left(\frac{x}{2}\right)^2+\left(\frac{x}{2}\right)^2+\left(\frac{x}{2}\right)^2+\underbrace{\left(\frac{x}{4}\right)^2+\cdots+\left(\frac{x}{4}\right)^2}_{4\ times}\\
\end{eqnarray*}

$ n=8$
\begin{eqnarray*}
x^2&=&\underbrace{\left(\frac{x}{4}\right)^2+\cdots+\left(\frac{x}{4}\right)^2}_{16\ times}\\
&=& \left(\frac{3x}{4}\right)^2+\underbrace{\left(\frac{x}{4}\right)^2+\cdots+\left(\frac{x}{4}\right)^2}_{7\ times}\\
\end{eqnarray*}

Assume $ P_k$ true where $ k \ge 6$
\[ x^2=a^2_1+a^2_2+\cdots+a^2_{k-3}+a^2_{k-2}+a^2_{k-1}+a^2_k\]
Then we divide $ a^2_k$ to $ 4$ squares to get a number composed of $ k+3$ squares so $ P_{k+3}$ is true.


topology
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[b]My friend and I are still stuck on:[/b]

For each $ n \in \omega$, let $ X_n$ be the set $ \{0, 1\}$, and let $ \tau_n$ be the discrete topology on $ X_n$. For each of the following subsets of $ \prod_{n \in \omega} X_n$, say whether it is open or closed (or neither or both) in the product topology.

(a) $ \{f \in \prod_{n \in \omega} X_n | f(10) \equal{} 0 \}$
(b) $ \{f \in \prod_{n \in \omega} X_n | \text{ }\exists n \in \omega \text{ }f(n) \equal{} 0 \}$
(c) $ \{f \in \prod_{n \in \omega} X_n | \text{ }\forall n \in \omega \text{ }f(n) \equal{} 0 \Rightarrow f(n \plus{} 1) \equal{} 1 \}$
(d) $ \{f \in \prod_{n \in \omega} X_n | \text{ }|\{ n \in \omega | f(n) \equal{} 0 \}| \equal{} 5 \}$
(e) $ \{f \in \prod_{n \in \omega} X_n | \text{ }|\{ n \in \omega | f(n) \equal{} 0 \}|\leq5 \}$

Recall that $ \omega \equal{} \mathbb{N} \cup \{ 0 \}$

[b]Background:[/b]

If it helps you can think of $ \prod_{n\in \omega} X_n$ as $ \prod_{n \equal{} 0}^{\infty} X_n$. We define $ \prod_{n \equal{} 0}^{\infty}X_n$ to be the set of all functions $ f: \mathbb{N} \to \{ 0 , 1\}$ that satisfies $ f(n) \in \{ 0 , 1\}$.

There's a nice graphical representation of the product topology on $ Y^X$ (i.e. the product of the space $ Y$ $ |X|$ times). Namely, if we draw $ X$ as an "$ x \minus{}$axis" and $ Y$ as a "$ y \minus{}$axis", then elements in $ X^Y$ are "graphs of functions" in the $ X \minus{} Y$ "plane". An open nbhd of an element $ f$ is the set of all functions $ g$ whose graphs are close to the graph of $ f$ at finitely points. We get different nbhds by varying the closeness to $ f$ and/or the set of finite points.

In our case the product space is $ 2^\omega \equal{} 2^\mathbb{N}$, whose "plane" looks like two copies of the naturals $ \mathbb{N}$. In other words, if you were to imagine this as a 'subset' of $ \mathbb{R}^2$, it's just the set $ \{(n,i) : n \in \mathbb{N}, i \in \{0,1\}\}$.

[b]Progress:[/b]

So remember that open sets in the infinite product topology is really just having all but finitely many the whole space and the rest are open. Since the individual factors are discrete, you only need to check that all but finitely many are the whole space.

e.g. in (a) the 10th coordinate has a specific value, but all other coordinates can be whatever, so this is certain open.

Your post is here: http://www.mathlinks.ro/viewtopic.php?p=1340299#1340299 .
Don't repost something!


If i'm readfing the question right,it means x^(3^n),not 3*n

[quote="lotrgreengrapes7926"][quote="José"][quote="Sir.Aaron"]What does the $n$ in $1x^{3^n}$ mean?[/quote]

You maybe want to put $x^{3^n}$. This is: 1º calculate $3^n$, and this result is $k$. Then, do $x^k$.... am I wrong?[/quote]

That's correct. $x^{3^n}=x^{(3^n)}$ :D[/quote]

Yes, I think so although I don't understand that $1$ there...


The complement of an angle is $ 5^{\circ}$ more than four times the angle. What is the number of degrees in the measure of the angle?

Let the angle be $ x$.

\[ 90 \minus{} x \minus{} 5 \equal{} 4x \]

\[ x \equal{} \boxed{17} \]


combinatorics
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[quote="Xixas"]Find the maximal cardinality $|S|$  of the subset $S \subset A=\{1, 2, 3, \dots, 9\}$ given that no two sums $a+b | a, b \in S, a \neq b$ are equal.[/quote]
$S=\{1,2,3,5,8\}$ verify
If $|S|\ge 6$ there are 15 sums. but $3\le a+b\le 17$ (15 numbers)
Therefore the 15 sums are 3,4,...,17
therefore $1,2\in S$ and $8,9\in S$
1+9=2+8(contradiction)
Hence the maximal cardinality is 5

The answer is $5$. One possible construction is $S = \{1, 2, 3, 6, 9\}$, which clearly works.

Now, since $$1 + 9 = 2 + 8 = 3 + 7 = 4 + 6 = 10$$ we know our upper bound is (also) $9 - 4 = 5$, as at most one element from each of the pairs $$\{(1, 9), (2, 8), (3, 7), (4, 6)\}$$ can belong to $S$. $\blacksquare$


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1/a + 1/b + 1/c =13/20

When a,b,c, are all distinct, positive integers, what is the smallesr possible value of a+b+c?

[hide]We want to break 13 into 3 distinct numbers that are all divisible by 20 (so that simplifying each fraction results in 1/aninteger). Trial and error shows that 1, 2, 10 is the only triplet that works. So 13/20=(1+2+10)/20=1/20+1/10+1/2 and 20+10+2=32.[/hide]

what about [hide]1/15 + 1/2 + 1/12? your answer wasn't the only possible one (though it was the smallest)[/hide]

Oops. Forgot that you could 'un-simplify' 13/20 and get more answers that way.
Hmm. How would you do this rigorously then?

Here's an even better one:

[hide]$\frac13 + \frac14 + \frac{1}{15} = \frac{13}{20}.$[/hide]

Let's see if this is the best, using a rigorous proof method:

[hide]WLOG let a < b < c (since a, b, and c are distinct).  So $\frac{1}{a} > \frac{1}{b} > \frac{1}{c}$.  Thus $\frac{1}{a} \ge \frac13 \cdot \frac{13}{20} = \frac{13}{60}.$  (Why is this true?  Well, if it weren't true, then each of $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ would be less than a third of $\frac{13}{20}$, so their sum would be less than $\frac{13}{20}$, a contradiction.)  So what values can $a$ be?  Well, $a$ could be 2, 3, or 4; if $a \ge 5$, $\frac{1}{a} \le \frac{1}{5} < \frac{13}{60}$.  Now we split into three cases:

a = 2.  Then we have
\[\frac{1}{b} + \frac{1}{c} = \frac{13}{20} - \frac12 = \frac{3}{20}.\]
Multiplying this equation by 60bc, we have
\[60c + 60b = 9bc\]
\[9bc - 60b - 60c + 400 = 400\]
\[(3b - 20)(3c-20) = 400\]
We know that $3b-20$ and $3c-20$ are integers, so they must be pairs of factors of 400, and each must be > -20 (which rules out any pairs of negative factors, since at least one of them will be $\ge -20$.)  Thus

(3b-20, 3c-20) = (1,400), (2,200), (4,100), (5,80), (8,50), (10,40), (16,25), (20,20).
(3b,3c) = (21,420), (22,220), (24,120), (25,100), (28,70), (30,60), (36,45), (40, 40).
(b,c) = (7,140), (8,40), (10,20), (12,15)

(See your solutions among these?)  These yield a+b+c values of 149, 50, 32, and 29.

As an exercise to the reader, try doing this for a = 3 and a = 4.  Simon's favorite factoring trick (which was used to factor 9bc - 60b - 60c + 400) is a useful skill to know.

Move on to the next hide after you've tried it.[/hide]

[hide]For the a = 3, you should have done something like this: rewrite as 19bc = 60b + 60c.  How do we factor this?  Well, if we multiply by 19, we get 361bc - 60*19b - 60*19c = 0.  Now we can factor as (19b - 60)(19c - 60) = 60^2.

But this is really ugly, as 60^2 has...how many factors?  So instead, let's note that since $\frac{1}{b} + \frac{1}{c} = \frac{19}{60}$, $\frac{1}{b} > \frac{19}{120} > \frac{1}{7}$.  So we need only try b = 4 through 7.  Only four numbers - what a deal!  Doing so yields only (b,c) = (4,15) as a solution.  So the only possible solution when a = 3 is (a,b,c) = (3,4,15), which has a + b + c = 22.

Note that we could have used this method for a = 2 as well.  The other way just illustrates another powerful technique.

Finally, for a = 4, we have $\frac{1}{b} + \frac{1}{c} = \frac{2}{5}$, so $\frac{1}{b} > \frac{1}{5}$, or b < 5.  But this is impossible, since b > a = 4.  So there are no solutions for this case.

Conclusion: The best solution is (a,b,c) = (3,4,15), which has a + b + c = 22.  There are a few other solutions (and this method finds all of them!).[/hide]

[Edit]: Gack, LaTeX problems!  I keep forgetting to put dollar signs.

[quote="solafidefarms"]I get the smallest is [hide]1/10+1/4+1/5, for 19 :lol: Can't beat that![/hide]

Billy[/quote]

Except that equals 11/20...

[quote="solafidefarms"]I get the smallest is [hide]1/10+1/4+1/5, for 19 :lol: Can't beat that![/hide]

Billy[/quote]

1/4 + 1/5 + 1/10 = 11/20, not 13/20.  :lol:

Nice. That doesn't look like something that could be easily done in just three minutes though...or at least it isn't something I could do in just three minutes...

Mathfanatic, that's a great solution :)

To the author, are u sure this is a 1997 AMC question?  I have the 1997 AMC 12, and it's not there.

Perhaps they had split it into A and B already.  Although my memory tells me they hadn't yet.

By the way, since the answer is multiple choice, I would have tested all the 5 choices.  Cheap, but effective. :D

Yes, but probability, look what they asked for - the sum of the three variables. So if the answers were A. 20 B. 22 C. 26 D. 28 E. 42  how would you go about guess and check?

For numbers as small as those, it's not hard to figure out what a, b, and c must be.  And you start with the smallest of course.

AHA!!!!! <snip: a bunch of exclamation marks> *panting heavily*

LOOK AT ITEM 7A ON THIS GUIDE:  http://math.scranton.edu/monks/courses/ProblemSolving/MathCountsPlaybook.pdf

Um, can one of the mods delete all the exclamation marks?


MATHCOUNTS
analytic geometry
graphing lines
slope
combinatorics unsolved
combinatorics

I'll copy the problem first:
[b]Problem[/b]
Given a 3 × 7 rectangular array of dots, how many triangles can be formed whose vertices are dots in the array?

●  ●  ●  ●  ●  ●  ●
●  ●  ●  ●  ●  ●  ●
●  ●  ●  ●  ●  ●  ●

This should be in combinatorics or in pre-olympiad....
[b]Solution:[/b]
We can choose any set o f three points, there are $21 \choose 3$ such sets. Now we must substract sests of points which lie in one column (there are 7 such sets) or one row ($7 \choose 3$ for each row). So the result is
 ${21 \choose 3}-3{7 \choose 3}-7=1218$
Edit: we must substract another 18 triplets as written below, so the result should be 1200.

to generalize:
given an $m$ by $n$, we have:
${mn\choose3}- {m\choose3}- {n\choose3}$

[quote]This should be in combinatorics or in pre-olympiad....
Solution:
We can choose any set o f three points, there are ${21\choose 3}$ such sets. Now we must substract sests of points which lie in one column (there are 7 such sets) or one row (${7\choose3}$ for each row). So the result is ${21 \choose 3}-3{7 \choose 3}-7=1218$[/quote]

If I'm not mistaken, you must also subtract the triples of points that are collinear and [u]not[/u] vertical or horizontial.  So there are lines with slopes $-1,1,\frac{1}{2},\frac{-1}{2},\frac{1}{3},\frac{-1}{3}$, and so the number of triangles should be

$\displaystyle{21\choose 3}-3{7\choose 3}-7-2\cdot 5-2\cdot 3 -2\cdot 1=1200$.


Consider the three-digit number $a$. But also $3a$ and $5a$. A number $a$ is called mysterious if the representations of these three number in the decimal expansion contain all numbers from 1 till 9 and the number 0 does not occur.

I think
[hide]$100105a$ must contain nine digits one through nine.[/hide]

$a$ is a three-digit number !

[hide=""Answer""]

try to show that $a = 129$.

[/hide]


Suppose, for instance, that scissors scores one point against paper, paper scores two against rock, and rock scores three against scissors. In this situation, would you automatically form a rock and hope to score three, or would you expect your opponent to form a rock, which you could beat by forming paper?

thats tough.
If it were to people that were noligible with the rules, they would excpect the other person to do paper, to counter rock. Therefore they would do scissors, to beat paper. So i would do rock, to beat scissors.

It's an on-going cycle...

You want to do rock to get the most points if you win, but your opponent realizes this and does paper. However, you realize this and do scissors. He realizes this and does rock, and so forth...

Based on that, you could say it's best to do rock because you can't predict what your opponent will do, and you might as well go for the most points. However, this turns into a continuous cycle as well, and it really doesn't matter what you choose after all.

This is a pretty standard situation for the application of game theory (of the variety studied by Nash).  There's clearly no dominant pure strategy, but you might think about strategies of the form, "Choose randomly from among R, P, S, with probability $ r$ of choosing R and probability $ p$ of choosing P (with $ r \plus{} p \leq 1$)."  (These are called "mixed strategies.")


Putnam topology algebra polynomial videos calculus

At least around here. Save for one professor I've talked to (who ranked 11th back when he took it), pretty much everyone I've talked to thinks pretty lowly of the problems, hearing things like:

"They're contrived in a way that they're not at all natural like research problems, so they're pretty useless"
or "They're not really that hard, it just takes a lot of time an energy to them which I don't want to expend", etc.

What's with all the hate? The problems are (ultra) challenging, pretty, and worth solving...anyone else experience this?

I had only one professor that was really into problem solving and encouraged people to participate in the putnam, though I never heard comments such as the ones you received from other professors.

But I think they're partially right, the problems are pretty contrived and don't really resemble research problems at all. I wouldn't go as far as to say it's useless, doing these kinds of problems builds up an interesting skill set that can be useful in many circumstances.

Whenever I read a particularly clever proof, I always wonder where the author developed the skills to see such an elegant solution, the putnam is probably one such place.

Maybe some professors have been jaded by years of research to the point that they no longer strive for elegance, just results, who knows.

Maybe those who did not do well on it like to think of reasons it is unimportant.  And opposite for those who did well.  Even decades later.

I think this is a better question :  "Is it the case that most professors think poorly of the Putnam, and if this is the case, then why?"

I think there is reasonably good evidence that the professors who don't think poorly it are fairly widespread---otherwise, how could the Putnam possibly get administered to thousands of students at hundreds of schools across the country every year?

My personal experience is this :  At my university (a good state school), I have already been encouraged to take the exam, and there is a seminar to prepare for it---and it's not just one professor who runs it every year.  I have encountered American mathematicians who are a little lukewarm on the idea of math contests in general; however, I suspect that at least some of them would change their views if they knew the contests a little better.  My mother is a mathematician, and her opinion of the olympiads went up greatly after I started working at them.  In fact, there have been times when my own opinion of the contests sagged, but she insisted to me that they were a wonderful thing and reminded me of the big role that they have played in my own mathematical development. :)

I think it is hard to like math without having some appreciation (and desire) for elegance.  To be sure, Putnam problems can be quite beautiful sometimes.  But I submit that many modern mathematical theories achieve elegance on a scale that is simply not possible in the small space of a six-hour exam.  For instance, I am still staggering slightly from my first encounter with the Zariski topology on commutative rings a few months ago.  Some things are just too grand to fit (easily) into such a compact format.

[quote="BlueVelvet"][quote="Arachnotron"]"They're contrived in a way that they're not at all natural like research problems, so they're pretty useless"
or "They're not really that hard, it just takes a lot of time an energy to them which I don't want to expend", etc.

What's with all the hate?[/quote]What hate? Saying that Putnam questions are not very relevant to research math is not the same as hating it  :huh:
It's just an opinion about [i]one [/i] aspect. It doesn't mean it can be fun or challenging.[/quote]

And they are not right that Putnam questions are not very relevant to research math.

Any math problem can become research topic if you get enough off it.

One arguement of contrived is that you have an idea of what the solution is (no very advanced math or theorem and solvable) Thinking back on this year when I tried to use inverses of roots and expressing newtons sums in terms of coefficients of a polynomial for problem B1 then realizing that I couldn't prove the roots were real numbers I probably should have realized I was on the wrong path.

It reminded me of that video documentary of IMO 2006 when Zeb described his experience with problem 4 of how he initially tried to set 2^x=z and solve a Pell equation and recalling Zuming's words "you will never need Pells equation on the IMO, instead you should do something simple like factor"

As well as when an instructor described a student suddenly solving a problem after seeing that another student that was not that good at math had solved the problem.

Of course this mentality may not apply to all questions on the Putnam but it is something I'm trying to keep in mind.

[quote]
It reminded me of that video documentary of IMO 2006 when Zeb described his experience with problem 4 of how he initially tried to set 2^x=z and solve a Pell equation and recalling Zuming's words "you will never need Pells equation on the IMO, instead you should do something simple like factor"
[/quote]
I wish you knew how many nice and deep results were obtained on the way that starts with "That just cannot be as complicated as this guy presents it".
[quote]
As well as when an instructor described a student suddenly solving a problem after seeing that another student that was not that good at math had solved the problem.
[/quote]
Happens all the time in the "research math." too.

I remember seeing that an A6 from not too long ago was motivated by a technique that some recent paper of Terence Tao employed, so it's far from accurate to say that Putnam problems are inconsequential for research. I also have to agree that the Putnam gets more than its fair share of bashing from mathematicians, who generally demote "problem-solving" mathematics in favor of "theory-building."

Perhaps this reaction is exaggerated in order to discourage students from focusing too much on competitions. When there are contests, students tend to get absorbed in them and read too much into their results. Being good at the Putnam is a good thing. Eschewing all other math in order to prepare for the Putnam is probably not a good thing. This forum has a nice article on "The Calculus Trap", which is certainly valid...but there is a "competition trap" too.

Ultimately, solving problems is a big deal in mathematics, but most mathematicians would probably say that the true goal is to understand.

I remember last semester my Calculus I Professor gave us three of the problems for extra credit the Monday after the exam but then said I couldn't do them since I'd already looked up the answers after I took it on Saturday and that would be cheating. I later got him to give me the extra credit, however, since I had in fact taken the exam.


What is the principle of Inclusion-Exclusion


prove

tanA + tanB + tanC = tanAtanBtanC (A+B+C=180)

[hide]Put $C=\pi-(A+B)$. Then

\begin{eqnarray*}\tan А+\tan B+\tan C &=& {{\sin A}\over{\cos A}}+{{\sin B}\over{\cos B}}+{{\sin C}\over{\cos C}}\\ &=& {{\sin A}\over{\cos A}}+{{\sin B}\over{\cos B}}+{{\sin (\pi-(A+B))}\over{\cos (\pi-(A+B))}}\\ &=& {{\sin A}\over{\cos A}}+{{\sin B}\over{\cos B}}-{{\sin (A+B)}\over{\cos (A+B)}}\\ &=& {{\sin A\cos B+\cos A\sin B}\over{\cos A\cos B}}-{{\sin (A+B)}\over{\cos (A+B)}}\\ &=& {{\sin(A+B)}\over{\cos A\cos B}}-{{\sin (A+B)}\over{\cos (A+B)}}\\ &=& {{\sin(A+B)\cos(A+B)-\sin(A+B)\cos A\cos B}\over{\cos A\cos B\cos(A+B)}}\\ &=& {{\sin(A+B)(\cos(A+B)-\cos A\cos B)}\over{\cos A\cos B\cos(A+B)}}\\ &=& {{\sin(A+B)(\cos A\cos B-\sin A\sin B-\cos A\cos B)}\over{\cos A\cos B\cos(A+B)}}\\ &=& {{-\sin(A+B)\sin A\sin B}\over{\cos A\cos B\cos(A+B)}}\\ &=& {{\sin(A+B)\sin A\sin B}\over{\cos A\cos B(-\cos(A+B))}}\\ &=& {{\sin(\pi-(A+B))\sin A\sin B}\over{\cos A\cos B\cos(\pi-(A+B))}}\\ &=& {{\sin C\sin A\sin B}\over{\cos A\cos B\cos C}}\\ &=& \tan A\tan B\tan C\end{eqnarray*}[/hide]

$k\in Z,\ x+y+z=k\pi\Longrightarrow \tan x+\tan y+\tan z=\tan x\tan y\tan z.$

[u]Example:[/u] $\tan x\tan 2x\tan 3x=-\tan x\tan 2x\tan (-3x)=$
$-[\tan x+\tan 2x+\tan (-3x)]=-\tan x-\tan 2x +\tan 3x.$

For who is in the last class: $\int \tan x\tan 2x\tan 3x\ dx=$
$-\int \tan x\ dx-\int \tan 2x \ dx +\int \tan 3x \ dx=\ldots$ a.s.o.

[quote="shyong"][hide="shorter proof"]

Since $C=\pi-A-B$ we have

$\tan C=-\tan (A+B)$

$\tan C=-\frac{\tan A+\tan B}{1-\tan A\tan B}$

$\tan C-\tan A\tan B\tan C=-(\tan A+\tan B)$

which is obvious  :)
[/hide][/quote]

[hide="even shorter"]
Numerator  in the formula of $Tan(A+B+C)$ is $TanA+TanB+TanC-TanATanBTanC$.
As $A+B+C=\pi$, we get $Tan(A+B+C)=0$.  Hence the answer
[/hide]


geometry
3D geometry
sphere
rectangle
Pythagorean Theorem
calculus
calculus computations

find the dimensions of the right circular cylinder that can be inscribed in a sphere with radius of R

Many different cylinders can be inscribed. What quantity are you maximizing (or minimizing)?

Are you so clueless as to believe that answered the question? Read, and think.

you are maximizing the volume of the cylinder in the sphere

The cylinder is going to touch the inside of the sphere such that two "caps" are formed.

Consider a rectangular cross-section of the cylinder such that the rectangle is as wide as possible.  Then $ R$ will be half the length of each diaganol of this cross-section.  So, $ 2R$ is the length of each diagonal.  Call the height of the cylinder $ x$.  By the Pythagorean Theorem, the diameter of each of the circular bases is $ \sqrt{4R^2 \minus{} x^2}$, which means the radius is $ \frac{\sqrt{4R^2 \minus{} x^2}}{2}$.  The volume of the cylinder is given by $ V(x) \equal{} \pi \left(\frac{\sqrt{4R^2 \minus{} x^2}}{2}\right)^2 x \equal{} \pi (R^2 x \minus{} \frac{1}{4}x^3)$.

So, maximize $ V(x)$!


calculus
integration
function
geometry
real analysis
calculus computations

if $ f(x)\geq 0$ and is Riemman integratable on $ [a,b]$  $ \forall x\in[a,b]$ . Prove that $ \int^{b}_{a}f dx\geq 0$ .

Can someone give me an idea how to start the proof ? Thanks

I don't know what "Riemman integrable" means but if $ f(x)\geq 0$  in the interval $ [a, b]$ then the graph of the function lies on, or above the $ x$ axis in the interval. So the area under the curve is either $ 0$ or positive in the interval. Hence:

$ \int_{a}^{b}f(x)\,dx\geq 0$ as this integral gives the area under the curve, assuming continuity in the interval.

I am probably completely wrong, as the definition of "Riemman integrable" is probably different to the normal definitions of integrable.

[quote="sludgethrower"]I don't know what "Riemman integrable" means but if $ f(x)\geq 0$  in the interval $ [a, b]$ then the graph of the function lies on, or above the $ x$ axis in the interval. So the area under the curve is either $ 0$ or positive in the interval. Hence:

$ \int_{a}^{b}f(x)\,dx\geq 0$ as this integral gives the area under the curve, assuming continuity in the interval.[/quote]

thanks for replying .
Actually im seeking for a proof using 'partition' and condition for Riemann Integral .

[quote="jmerry"]Every Riemann sum (or upper/lower Darboux sum) is nonnegative. A limit of nonnegative quantities in nonnegative if it exists.[/quote]

I look up my book but it does not mention that "Every Riemann sum is nonnegative" ? Why can't we have a negative sum ?

The Riemann sum (or lower sum) is a sum of terms where each term is a value of the function somewhere (or an infinum of some of the values of the function), multiplied by the length of a subinterval. The length of the subinterval is positive. We were given that $ f(x)\ge0$ on the interval in question. Hence any value of the function must be nonnegative and any infinum of those values must be nonnegative. Hence, each term in the sum is nonnegative, and the sum of nonnegative terms is nonnegative. The final step is then what jmerry wrote.

Corollary: if $ f(x)\ge g(x)$ for all $ x\in[a,b]$ and $ f$ and $ g$ are both integrable, then

$ \int_{a}^{b}f(x)\,dx\ge\int_{a}^{b}g(x)\,dx.$

Proof: subtract, use the linearity of the integral, and quote the result above.

----

Here's a result of a different character, with a different proof. Suppose that $ f$ is continuous on $ [a,b],$ $ f(x)\ge 0$ on $ [a,b],$ and $ f$ is not identically zero on $ [a,b].$ Then

$ \int_{a}^{b}f(x)\,dx>0.$

[quote="Kent Merryfield"]The Riemann sum (or lower sum) is a sum of terms where each term is a value of the function somewhere (or an infinum of some of the values of the function), multiplied by the length of a subinterval. The length of the subinterval is positive. We were given that $ f(x)\ge0$ on the interval in question. Hence any value of the function must be nonnegative and any infinum of those values must be nonnegative. Hence, each term in the sum is nonnegative, and the sum of nonnegative terms is nonnegative. The final step is then what jmerry wrote.

Corollary: if $ f(x)\ge g(x)$ for all $ x\in[a,b]$ and $ f$ and $ g$ are both integrable, then

$ \int_{a}^{b}f(x)\,dx\ge\int_{a}^{b}g(x)\,dx.$

Proof: subtract, use the linearity of the integral, and quote the result above.

----

Here's a result of a different character, with a different proof. Suppose that $ f$ is continuous on $ [a,b],$ $ f(x)\ge 0$ on $ [a,b],$ and $ f$ is not identically zero on $ [a,b].$ Then

$ \int_{a}^{b}f(x)\,dx > 0.$[/quote]

Thanks Prof.
Now I understand .

As for your question , can we say that there exist an interval $ [c,d]\subset [a,b]$ such that $ f(x)>0$ in $ [c,d]$ (since $ f$ is not identically zero in $ [a,b]$) and the lower sum will be positive which implies the integral in the interval $ [a,b]$ to be positive?

Yes, although the existence of such an interval is nontrivial (you have to chase through the definition of continuity fairly carefully to get it, even though it's intuitively quite clear).

Unrelatedly, there is never a good reason to quote the immediately-preceding post in its entirety: all it does is to clutter up the thread.  Use quote when you're replying to some smaller piece of a larger post or when you're replying to some post which is not the most recent on the thread.


Show that if $ x$ and $ y$ are positive itnegers such that $ x^2\plus{}y^2\minus{}x$ is divisible by $ 2xy$, then $ x$ is a perfect square.

Thanks... I think there might be an algebraic solution as well as number theory but I can't find it

It's a number theory problem.  If by an "algebraic" solution you mean one by algebraic manipulation alone, that's highly doubtful.  And it's enough that $ x^2 \plus{} y^2 \minus{} x$ is divisible by $ xy$.

[hide="Solution"] Let $ x \equal{} ka, y \equal{} kb$ where $ \gcd(a, b) \equal{} 1$.  Then the condition is that $ ka^2 \plus{} kb^2 \minus{} a$ is divisible by $ kab$.

In particular, $ k | a$.  Let $ a \equal{} ka'$; then the condition is that $ k^2 a' \plus{} b^2 \minus{} a'$ is divisible by $ ka'b$.  Now $ a' | b^2$, but we stipulated $ \gcd(a, b) \equal{} 1$, hence $ \gcd(a', b) \equal{} 1$, so we conclude that $ a' \equal{} 1$ and $ x \equal{} k^2$ as desired. [/hide]


inequalities
geometry
Pythagorean Theorem
AMC

This doesn't apply to the well-known theorems such as the Pythagorean Theorem, but what if you just state the "British Flag Theorem" or something more obsecure and it completely trivilizes the problem? Is there an listing of all the theorems, corollaries, formulas etc. that you can use without proving?

[quote="MysticTerminator"]You are never allowed to cite a result that completely trivializes the problem.[/quote]

So you would have to prove that theorem?

[quote="MysticTerminator"]You are never allowed to cite a result that completely trivializes the problem.[/quote]

But is there some type of cutoff? what if it partially trivilizes the problem (in the eyes of the grader), but you wouldn't know since you assumed that everyone would state this result.

[quote="Phelpedo"]Would Schur's be allowed on this problem from IMO 2000?

Prove that $(a-1+\frac{1}{b})(b-1+\frac{1}{c})(c-1+\frac{1}{a}) \leq 1$,

since $(x+y-z)(y+z-x)(x+z-y) \leq xyz$ is pretty much straight from Schur?[/quote]

I think it would be allowed here because you made that substitution first. On the other hand, Schur is easy to prove, and you can also use triangle substitution + AM-GM, so if I were you I would do one of those two barring ridiculous circumstances.

Trivialize would generally be like

"Because of such and such a Theorem, the desired result immediately follows."

You are probably given a bit more leeway with well-known inequalities (Power Mean, Jensen, Schur, Cauchy). If you use Chebyshev or Nesbitt in a way that makes the problem a one or two-step problem, it may be good to offer a short proof if you can.

Additionally, the level of triviality you can reduce something to probably depends on how well-known the theorem is. For example, I don't think anyone will get mad at you for submitting a solution that is trivialized by something like the inscribed angle theorem (it probably means the solution is super-elegant or something).

Common sense is probably a good cutoff indication. If you feel like you didn't apply any ingenuity to the problem, but just cited theorems that you knew, you might want to prove those theorems first (and you should always be able to prove the theorems you know anyways).

If you find yourself at an olympiad, you really aren't sure whether you're allowed to cite something, and you can't seem to prove it easily, just right up the cited solution and move on to a fresh new problem, where you know there's ground to explore and points to be earned (unlike the problem that you cited something on, where you may have already garnered up some points and don't need to spend more time, but you're not completely sure). Then, once you feel there isn't much more to be gained from other problems (maybe because you already solved them all  :D ), try to originalize your proof.

In general, though, it's always a good idea to understand how to derive the formulas and theorems you know (assuming it draws upon an area of math that you're familiar with). In fact, it often solidifies a fact to actually derive it on your own. However, there are always going to be some theorems or formulas that you may not know, so you'll inevitably have to use common sense sometime or another. (In particular, there are quite a lot of neat geometry formulas that, although problem-solving and creativity are by default more useful than memorization, can be quite useful memorized.)

If you're unsure, think of it this way: If a problem says "Prove Schur's inequality," are you really going to say "Schur, QED"?
(Though it wouldn't be interesting to see the graders' faces if the person wrote an ingenious philosophical analysis of the reflexive property of logic  :P )


floor function
number theory proposed
number theory

Let n be the positive integer.Prove that if the number of factor 2 in $ n!$ is $ n\minus{}1$, then n is a power of 2.

$ Let 2^x \le n <2^{x\plus{}1}$, obviously $ \lfloor \frac{x}{2^x} \rfloor\equal{}1$
Then $ (n\minus{}1)\plus{}1\equal{}(\lfloor \frac{n}{2} \rfloor\plus{}\lfloor \frac{n}{2^2} \rfloor\plus{}...\plus{} \lfloor \frac{n}{2^x} \rfloor) \plus{} (\lfloor \frac{n}{2^x} \rfloor)(\dfrac{1}{2}\plus{}\dfrac{1}{2^2}\plus{}...) \le \dfrac{\dfrac{n}{2}}{1\minus{}\dfrac{1}{2}}\equal{}n$
equality holds only when $ n$ is a power of $ 2$, done.

[quote="stephencheng"]$ Let 2^x \le n < 2^{x \plus{} 1}$, obviously $ \lfloor \frac {x}{2^x} \rfloor \equal{} 1$
Then $ (n \minus{} 1) \plus{} 1 \equal{} (\lfloor \frac {n}{2} \rfloor \plus{} \lfloor \frac {n}{2^2} \rfloor \plus{} ... \plus{} \lfloor \frac {n}{2^x} \rfloor) \plus{} (\lfloor \frac {n}{2^x} \rfloor)(\dfrac{1}{2} \plus{} \dfrac{1}{2^2} \plus{} ...) \le \dfrac{\dfrac{n}{2}}{1 \minus{} \dfrac{1}{2}} \equal{} n$
equality holds only when $ n$ is a power of $ 2$, done.[/quote]

Correct a typing mistake. It should be $ \lfloor \frac{n}{2^x} \rfloor\equal{}1$ in the first row.


Find all functions $f: \mathbb{R}\rightarrow \mathbb{R}$ so that for every $x,y \in \mathbb{R}$: \[f(xf(x)+f(y))=(f(x))^{2}+y\]

Identity, f(x)=x, works. Maybe I'll look for more later. :)

[hide="I think I get this one"]
By fixing $x$ and varying $y$ we get that that $f$ is injective and surjective.

Hence there exists an $x$ such that $f(x)=0$. Substituting this in we get

$f(0+f(y))=0^{2}+y$
$f(f(y))=y$.

Then setting $x=0$ and $y=0$:

$f(f(0))=(f(0))^{2}=0$ and so $f(0)=0$

We substitue in $x: =f(x)$

$f(f(x)f(f(x))+f(y))=(f(f(x))^{2}+y$
$f(xf(x)+f(y))=x^{2}+y$ (using $f(f(x))=x$)

But $f(xf(x)+f(y))=(f(x))^{2}+y$, so
$(f(x))^{2}+y=x^{2}+y$
$(f(x))^{2}=x^{2}$
$f(x)=\pm x$

Now it only remains to show that $f(x)=x$ and $f(x)=-x$ are the only solutions (i.e. we can't have one for some values of $x$ and the other for the other)

Suppose that $f(x)=x$, but $f(y)=-y$.
Then
$f(x^{2}-y)=x^{2}+y$

But regardless of the sign of $f(x^{2}-y)$ we get that one of $x,y$ is zero so we conclude that
$f(x)=x$ and $f(x)=-x$ are the only solutions
[/hide]

[quote="tim1234133"]
We substitue in $x: =f(x)$

$f(f(x)f(f(x)+f(y))=(f(f(x))^{2}+y$[/quote]

EDIT: I have pointed out some mistakes in his solution, now it is OK.

Thanks conceive.

I have edited my solution accordingly.

[quote="tim1234133"]By fixing $x$ and varying $y$ we get that that $f$ is injective and surjective.[/quote] why is it injective? I think you need to do more than fixing x and varying y, or am I mistaken?

Here's my reasoning- I could quite well be wrong since I'm rather  inexperienced at functional equations.

Suppose $f$ is not injective. Then there exist $y_{1}$ and $y_{2}$, not equal, such that $f(y_{1})=f(y_{2})$. We fix $x$.

By substituting $y: =y_{1}$ and $y: =y_{2}$, we get

$f(xf(x)+f(y_{1}))=(f(x))^{2}+y_{1}$
$f(xf(x)+f(y_{2}))=(f(x))^{2}+y_{2}$                     [b] (1) [/b]

But since $f(y_{1})=f(y_{2})$,

$f(xf(x)+f(y_{1}))=f(xf(x)+f(y_{2}))$

Therefore by [b] (1) [/b] $(f(x))^{2}+y_{1}=(f(x))^{2}+y_{2}$

But since $x$ is constant, we get $y_{1}=y_{2}$, and so the function is injective.

Hmm... yes, you're right :) don't know why I didn't see that.


function
induction
quadratics
algebra
algebra proposed

I don't know why you say this problem is not hard, but everybody has its own opinion. So, let's give a monstruous solution:
   First, it is trivial to prove that f(0)=0, f(1)=1. Suppose e managed to show that f(2)=2. Put y=2 and you wil find that $ f(x \minus{} 1) \equal{} \frac {f(x) \plus{} f(x \minus{} 2)}{f(x) \minus{} f(x \minus{} 2)}$ and an easy induction shows that f(n)=n for any natural n. Now we use the fact that
 $ \frac {f(x) \plus{} f(y)}{f(x) \minus{} f(y)} \equal{} f(\frac {x \plus{} y}{x \minus{} y}) \equal{} f(\frac {1 \plus{} \frac {y}{x}}{1 \minus{} \frac {y}{x}}) \equal{} \frac {1 \plus{} f(\frac {y}{x})}{1 \minus{} f(\frac {y}{x})}$. Then, it's immediately to show that $ f(x) \equal{} x$ for any rational x and that $ f(x)\geq 0$ for any nonnegative x. So, for any $ x\geq y\geq 0$ we have $ f(x) \minus{} f(y)\geq 0$ since this one has the same sign as $ f(\frac {x \plus{} y}{x \minus{} y})$. Thus, f is increasing on the positive and also multiplicative. It follows that f(x)=x for any positive x. It is also trivial to show that f is odd, so f(x)=x for any x.
   Now, the main thing: proving that f(2)=2. Put $ f(2) \equal{} a$. Then $ f(4) \equal{} a^2$ and also $ f(3) \equal{} f(\frac {4 \plus{} 2}{4 \minus{} 2}) \equal{} \frac {a \plus{} 1}{a \minus{} 1}$. Using that $ f(x \plus{} 1) \equal{} \frac {f(x) \plus{} f(x \plus{} 2}{f(x \plus{} 2) \minus{} f(x)}$ we find that $ f(5) \equal{} \frac {a^2 \plus{} 1}{(a \minus{} 1)^2}$. But $ f(5) \equal{} f(\frac {3 \plus{} 2}{3 \minus{} 2}) \equal{} ... \equal{} \frac {a^2 \plus{} 1}{2a \plus{} 1 \minus{} a^2}$. Thus, $ (a \minus{} 1)^2 \equal{} 2a \plus{} 1 \minus{} a^2$ from where a=0 or a=2. If a=0, then f(3)=-1 so $ f^2(\sqrt(3)) \equal{} \minus{} 1$ false.

I didn't mean that the problem was easy, I just meant that it wasn't hard to the extent that any of those parts is not obtainable.

The (union of) the Singapore team got almost every part except showing that the function was increasing on positive (which, as we found out, is not difficult, it's just that we don't have enough experience with it).

Whatever the case is there are quite a number of things to be shown.  The official solution I have with all the details is 2 full pages long.

Small note regarding the ending: It's easy to observe that the function is injective (if f(x) = f(y) but x <> y, then the LHS is defined but the RHS is undefined).  Then since we have shown that f(0) = 0, a <> 0.

[quote="Valiowk"]Find all functions $ f: \mathbb{R} \to \mathbb{R}$ satisfying
\[ f\left(\frac {x \plus{} y}{x \minus{} y}\right) \equal{} \frac {f\left(x\right) \plus{} f\left(y\right)}{f\left(x\right) \minus{} f\left(y\right)}
\]
for all $ x \neq y$.[/quote]

Harazi, how do you show that f(x) >= 0 for x >= 0 ??

Here is all I have (not a complete solution, just some ideas which could be of use for finding a simpler solution):

We have given the equation

$ \text{\textbf{(1)\ \ \ \ \ \ \ \ \ \ }}f\left( \frac{x\plus{}y}{x\minus{}y}\right) \equal{}\frac{f\left( x\right) \plus{}f\left( y\right) }{f\left( x\right) \minus{}f\left( y\right) }$.

At first, our function f(x) must be injective; else, we would have x and y with f(x) = f(y) while $ x\neq y$, so that the right member of [b](1)[/b] would get undefined while the left member wouldn't.

Putting y = 0 in [b](1)[/b], we get

$ f\left( 1\right) \equal{}\frac{f\left( x\right) \plus{}f\left( 0\right) }{f\left(
x\right) \minus{}f\left( 0\right) }$,

what becomes (1 - f(1))f(x) = -f(0)(1 + f(1)). Now, if f(1) were not 1, we could divide by 1 - f(1), and f(x) would be a constant. This is not possible since f(x) is injective; hence, f(1) = 1. Therefore, -f(0)(1 + f(1)) = (1 - f(1))f(x) = 0. But since $ 1 \plus{} f\left( 1\right) \equal{} 2\neq 0$, we must have f(0) = 0. As a consequence, $ f\left( x\right) \neq 0$ for every $ x\neq 0$ (since f(x) is injective).

Putting y = -x in [b](1)[/b], we get

$ f\left( 0\right) \equal{}\frac{f\left( x\right) \plus{}f\left( \minus{}x\right) }{f\left(
x\right) \minus{}f\left( \minus{}x\right) }$.

Using f(0) = 0, this immediately becomes f(-x) = -f(x). Consequently, f(-1) = -f(1) = -1.

Now let $ a\neq 0$ be a real number; then,

$ \frac{f\left( x\right) \plus{}f\left( y\right) }{f\left( x\right) \minus{}f\left(
y\right) }\equal{}f\left( \frac{x\plus{}y}{x\minus{}y}\right) \equal{}f\left( \frac{ax\plus{}ay}{ax\minus{}ay} \right) \equal{}\frac{f\left( ax\right) \plus{}f\left( ay\right) }{f\left( ax\right)
\minus{}f\left( ay\right) }$,

what simplifies to f(x)f(ay) = f(y)f(ax). With y = 1, this becomes f(x)f(a) = f(1)f(ax) = f(ax). (This trivially holds for x = 1 and for a = 0, even if these cases were not allowed in the proof.) Hence, we have f(x)f(a) = f(ax). In other words, our function f(x) is multiplicative.

As a consequence,

$ f\left( x\right) f\left( \frac{1}{x}\right)  \equal{}f\left( 1\right) \equal{}1$

and

$ \text{\textbf{(2)\ \ \ \ \ \ \ \ \ \ }}f\left( \frac{1}{x}\right)  \equal{}\frac{1}{f\left( x\right) }$.

Now, substitute y = 1 in [b](1)[/b]. Then, with f(1) = 1 we obtain

$ \text{\textbf{(3)}\ \ \ \ \ \ \ \ \ \ }f\left( \frac{x\plus{}1}{x\minus{}1}\right) \equal{}\frac{ f\left( x\right) \plus{}1}{f\left( x\right) \minus{}1}$.

The quadratic equation

$ \text{\textbf{(4)\ \ \ \ \ \ \ \ \ \ }}\frac{x\plus{}1}{x\minus{}1}\equal{}x$

has two real solutions: $ x \equal{} 1 \plus{} \sqrt {2}$ and $ x \equal{} 1 \minus{} \sqrt {2}$. But [b](3)[/b] shows that if x is a solution of the quadratic equation [b](4)[/b], then f(x) must be a solution of [b](4)[/b], too. Hence, we have 4 possible cases:

[b]Case 1.[/b] $ f\left( 1 \plus{} \sqrt {2}\right) \equal{} 1 \plus{} \sqrt {2}$ and $ f\left( 1 \minus{} \sqrt {2}\right) \equal{} 1 \minus{} \sqrt {2}$.

[b]Case 2.[/b] $ f\left( 1 \plus{} \sqrt {2}\right) \equal{} 1 \minus{} \sqrt {2}$ and $ f\left( 1 \minus{} \sqrt {2}\right) \equal{} 1 \plus{} \sqrt {2}$.

[b]Case 3.[/b] $ f\left( 1 \plus{} \sqrt {2}\right) \equal{} 1 \plus{} \sqrt {2}$ and $ f\left( 1 \minus{} \sqrt {2}\right) \equal{} 1 \plus{} \sqrt {2}$.

[b]Case 4.[/b] $ f\left( 1 \plus{} \sqrt {2}\right) \equal{} 1 \minus{} \sqrt {2}$ and $ f\left( 1 \minus{} \sqrt {2}\right) \equal{} 1 \minus{} \sqrt {2}$.

The latter two cases are impossible because of the injectivity; hence, one of the Cases 1 and 2 holds. Wherever Case 1 holds or Case 2 is unimportant; in any case, we have

$ f\left( 1 \plus{} \sqrt {2}\right) \plus{} f\left( 1 \minus{} \sqrt {2}\right) \equal{} 2$

and

$ \left| f\left( 1 \plus{} \sqrt {2}\right) \minus{} f\left( 1 \minus{} \sqrt {2}\right) \right| \equal{} 2\sqrt {2}$.

Hence, applying [b](1)[/b] to $ x \equal{} 1 \plus{} \sqrt {2}$ and $ y \equal{} 1 \minus{} \sqrt {2}$, we find

$ \left| f\left( \frac{2}{2\sqrt{2}}\right) \right| \equal{}\frac{2}{2\sqrt{2}}$,

i. e.

$ \left| f\left( \frac{1}{\sqrt{2}}\right) \right| \equal{}\frac{1}{\sqrt{2}}$,

so that, using [b](2)[/b], we have

$ \left| f\left( \sqrt {2}\right) \right| \equal{} \sqrt {2}$.

Therefore, by the multiplicativity,

$ f\left( 2\right) \equal{}f\left( \sqrt{2}\right) \cdot f\left( \sqrt{2}\right)
\equal{}\left( f\left( \sqrt{2}\right) \right) ^{2}\equal{}\left| f\left( \sqrt{2}\right)
\right| ^{2}\equal{}2$.

Then, we may proceed as you did and show that f(x) = x for any rational x. But how do you get the positivity?

  Darij

Well, I think that we can derive that from the fact that $ f(xy)=f(x)f(y) $ for any x and y. Just put x=y.

[quote="harazi"]Well, I think that we can derive that from the fact that $ f(xy) \equal{} f(x)f(y)$ for any x and y. Just put x=y.[/quote]

LOL... I must have been really blind not to find this myself!

  Darij

It's a problem of the france correspondance 2002-2003 :lol:

See: http://www.animath.fr/tutorat/dossier_02036.html

[quote=Valiowk]Find all functions $ f: \mathbb{R} \to \mathbb{R}$ satisfying
\[ f\left(\frac {x \plus{} y}{x \minus{} y}\right) \equal{} \frac {f\left(x\right) \plus{} f\left(y\right)}{f\left(x\right) \minus{} f\left(y\right)}
\]
for all $ x \neq y$.[/quote]

Pretty neat.

[b][color=#00f]Solution.[/color][/b] As usual, let \(P(x,y)\) assert the above equation. Then,  we prove some claims.

[b][color=#f00]Claim 1.[/color][/b] \(f\) is injective.
[i]Proof.[/i] This holds because if \(f(x)=f(y)\) and \(x\neq y\), then the left hand side is defined while the right hand side is not, a contradiction.

[b][color=#f00]Claim 2.[/color][/b] \(f\) is an odd function.
[i]Proof.[/i] \(P(x,-x)\) gives us that \[f(0)=\frac{f(x)+f(-x)}{f(x)-f(-x)}\] so \[(f(x)+f(-x))(f(y)-f(-y))=(f(x)-f(-x))(f(y)+f(-y))\] or \[f(x)f(y)-f(x)f(-y)+f(-x)f(y)-f(-x)f(-y)=f(x)f(y)+f(x)f(-y)-f(-x)f(y)-f(-x)f(-y)\] for all \(x,y\) implying that \(\frac{f(x)}{f(-x)}\) is constant, say \(c\). Plugging this back into \(P(x,-y)\) gives us that \(c=-1\) and we are done.

[b][color=#f00]Claim 3.[/color][/b] \(f\) is multiplicative.
[i]Proof.[/i] Firstly, \(P(x,0)\) yields that \(f(1)=1\). Next, \(P\left(\frac{x+y}{x-y},1\right)\) gives us that \[f\left(\frac{\frac{x+y}{x-y}+1}{\frac{x+y}{x-y}-1}\right)=\frac{f\left(\frac{x+y}{x-y}\right)+1}{f\left(\frac{x+y}{x-y}\right)-1}\] implying that \(f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}\) and so we are done.

[b][color=#f00]Claim 4.[/color][/b] \(f\) is strictly increasing over \(\mathbb{R}^+\).
[i]Proof.[/i] We need to show that \(a<b\) implies \(f(a)<f(b)\) for all \(x,y\in\mathbb{R}^+\). \(P(a,-b)\) with \(a<b\) and \(a,b>0\) yields that \[0>f\left(\frac{a-b}{a+b}\right)=\frac{f(a)-f(b)}{f(a)+f(b)}\] and since \(f\) is multiplicative, \(x\geq0\) implies \(f(x)\geq0\) for all positive reals \(x\)(since \(f(x^2)=f(x)^2\)), we see that \(f(a)<f(b)\) and this completes the proof of this claim.

Coming back to our problem, since \(f\) is multiplicative, let \(f(x)=g(\ln x)\). Then, since \(x\) is a positive real, \(g(x+y)=g(x)g(y)\) for all \(x,y\in\mathbb{R}^+\). Let \(h(x)=\ln g(x)\). Then, we see that \(h\) is an additive function. Also, since \(f\) is strictly increasing over the positive reals, so is \(g\) and so is \(h\). Therefore, \(h\) is monotone and additive, implying that \(h(x)=kx\) where c is a constant. This means that \(g(x)=e^{kx}\) and so \(f(x)=e^{k\ln x}\) implying that \(f(x)=x^k\) for all \(x\). Plugging in the original equation yields that \(k=1\) or, \(f(x)=x\) for all \(x\in\mathbb{R}^+\). Since \(f\) is odd, we can conclude that \(f(x)=x\) for all \(x\in\mathbb{R}\) which indeed fits.

[hide=Remark]
My solution may seem similar to harazi's solution, but the key thing is I did not want to include any kind of non-trivial specific casework, like for P(2,1) or something like that. I hope this solution is correct.
[/hide]

We claim that the only such function is $f \equiv x$, plugging it in functional equation, we can see that it works. The second condition of our problem says that $f$ is injective. Let $P(x, y)$ be the assertion, we now proceed with a bunch of claims.
[b][color=#f00]Claim:[/color][/b] $f(0)=0.$
[b][color=#00f]Proof:[/color][/b] First of all notice that constant functions don't satisfy our functional equation. Now, $$P(x, 0) \rightarrow f(1) = \frac{f(x)+f(0)}{f(x)-f(0)}$$
$$\iff f(x)=f(0)\frac{f(1)+1}{f(1)-1}$$ so $f(0)$ must be zero since otherwise the function would be constant.  $\blacksquare$

[b][color=#f00]Claim:[/color][/b] $f(1)=1.$
[b][color=#00f]Proof:[/color][/b] $P(x, 0) \rightarrow f(1)=\frac{f(x)+f(0)}{f(x)-f(0)}=1$ $\blacksquare$

[b][color=#f00]Claim:[/color][/b] $f(x)+f(-x)=0$ i.e. $f$ is odd.
[b][color=#00f]Proof:[/color][/b] $P(x, -x) \rightarrow f(x)+f(-x)=0.$ $\blacksquare$

[b][color=#f00]Claim:[/color][/b] For any $x \in \mathbb{R}-\{0\}$, $f(x)f(\frac{1}{x})=1$.
[b][color=#00f]Proof:[/color][/b] $$P(x, -y) \rightarrow f\left(\frac{x-y}{x+y}\right)=\frac{f(x)+f(-y)}{f(x)-f(-y)}=\frac{f(x)-f(y)}{f(x)+f(y)}$$ Combining this with $P(x, y)$, we get our desired result. $\blacksquare$

[b][color=#f00]Claim:[/color][/b] $f(xy)=f(x)f(y)$ i.e. $f$ is multiplicative.
[b][color=#00f]Proof:[/color][/b] $$P(xy, 1) \rightarrow f\left(\frac{xy+1}{xy-1}\right) = \frac{f(x)+f(y)}{f(x)-f(y)}$$
$$\iff f(xy)=\frac{f\left(\frac{xy+1}{xy-1}\right)+1}{f\left(\frac{xy+1}{xy-1}\right)-1}$$
$$P(x, \frac{1}{y}) \rightarrow \frac{f\left(\frac{xy+1}{xy-1}\right)+1}{f\left(\frac{xy+1}{xy-1}\right)-1}=f(x)f(\frac{1}{y})=f(x)f(y)$$ So indeed we have $f(xy)=f(x)f(y)$. $\blacksquare$

[b][color=#f00]Claim:[/color][/b] $f$ is strictly increasing over $\mathbb{R}^{+}$
[b][color=#00f]Proof:[/color][/b] First of all, notice that $f(x^2)=f(x)^2 \geq 0 \forall x \in \mathbb{R}$ which means $f >0 \forall x \in \mathbb{R}^{+}$. Now suppose we have $a$ and $b$ such that $a>b>0$, $$f\left(\frac{x+y}{x-y}\right)=\frac{f(x)+f(y)}{f(x)-f(y)} >0$$ which means $f(x)>f(y)$. $\blacksquare$

Now let $g(x)=\ln(f(a^x))$, notice that $g(x+y)=g(x)+g(y)$. With all the conditions we have, $g(x)=cx$ and this means $f(x)=x^c$ and plugging it in our original equation, we get $c=1$ and $f \equiv x$. $\blacksquare$


[quote="StephenM"]I need help figuring out how this problem equals csc thetea

t=theta

(sec t + csc t)
------------------    =  csc t
     1+ tan t[/quote]

[hide]$\frac{\sec\theta+\csc\theta}{1+\tan\theta}=\csc\theta$

$\frac{\frac{1}{\cos\theta}+\frac{1}{\sin\theta}}{1+\frac{\sin\theta}{\cos\theta}}=\frac{1}{\sin\theta}$

$\frac{\frac{\sin\theta+\cos\theta}{\cos\theta\sin\theta}}{\frac{\cos\theta+\sin\theta}{\cos\theta}}=\frac{1}{\sin\theta}$

$\frac{\sin\theta+\cos\theta}{\cos\theta\sin\theta}*\frac{\cos\theta}{\cos\theta+\sin\theta}=\frac{1}{\sin\theta}$

$\frac{\cos\theta}{\cos\theta\sin\theta}=\frac{1}{\sin\theta}$

$\frac{1}{\sin\theta}=\frac{1}{\sin\theta}$

$\sin\theta=\sin\theta$[/hide]

[quote="StephenM"]any body?[/quote]

You are not allowed to do that, please be patient.

Also, try to learn LaTeX if you are planning to post again in the future.


inequalities
induction
function
inequalities proposed

[quote="soruz"]If   $ m,a_{1},a_{2},...a_{n}$ are positive numbers,  then

                     $ a_{1}^{m}\plus{}a_{2}^{m}\plus{}...\plus{}a_{n}^{m}\le(a_{1}a_{2}...a_{n})^{m}\plus{}n\minus{}1,\forall n\in N^{*},m > 0$.[/quote]

What if $ n\equal{}2, m\equal{}1, a_{1}\equal{}\frac{1}{2}, a_{2}\equal{}2$. What is the need for $ m$? The inequality is true if all $ a$ are less than 1 or all $ a$ are greater than 1.

This inequality is nonsense. The $ n\minus{}1$ can easily be scaled 'away'.

What if $ a_{1}\equal{}10^{\minus{}}{10}$ and $ a_{2}\equal{}12$, $ n\equal{}1$, etc. It is obviously false.

Sorry but I found'it in a book!!!It's a mistake in "If   $ m,a_{1},a_{2},...a_{n}$ are positive numbers"!Let's try to repair and to prove!I guess
$ m,a_{1},a_{2},...a_{n}\in (1,\infty)$
or
$ m,a_{1},a_{2},...a_{n}\in (0,1)$

 it should have been!

$ 0 < a_{i}^{m}< 1\ (i \equal{} 1,2,\ \cdots n)$ should be needed. :lol:
 I have almost found the simillar problem from Japanese university entrance exam as follows.

[color=blue]Ochanomizu University entrance exam/Domestic Science or Home Economics 1978[/color]

Let $ a_{k}\ (k \equal{} 1,\ 2,\ \cdots n)$ be real numbers with $ 0 < a_{k}< 1$. Prove that $ a_{1}a_{2}\cdots a_{n}> a_{1}\plus{}a_{2}\plus{}\cdots\plus{}a_{n}\plus{}1\minus{}n\ (n\geq 2)$.

Let $ S\equal{}f(a_{1})\equal{}a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n}\minus{}a_{1}a_{2}...a_{n}$,we keep $ a_{2},a_{3},...,a_{n}$ fixed.
Then $ f$ is a convex function and its max value attained when $ a_{k}\equal{}\left\{0,1\right\}$,for all k.
If $ a_{k}\equal{}1$,for all k,then $ S\equal{}n\minus{}1$.
If one of $ a_{k}\equal{}0$,then its max value attained when the others is 1,ie $ S\equal{}n\minus{}1$. :)

convex?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=48570&search_id=1817543315]See here[/url]

Dear kunny,is my solution true?? :maybe:
I can't figure which part is wrong.But i agree that this is a hard problem for an entrance exam.

It  call look at the end point.
$ f(x)\equal{}ax\plus{}b$

I think your answer is correct, but I wondered what did you mean by "convex"?

Maybe it seemed that I made a misread.

Geez,i was so foolish.:wallbash_red:
It can be proved by induction.
For n=2,$ (1\minus{}a_{1})(1\minus{}a_{2})\geq 0$
Suppose the statement is true for $ n$ numbers,for $ (n\plus{}1)$ numbers,we have
$ S \equal{} a_{n\plus{}1}(1\minus{}a_{1}a_{2}...a_{n})\plus{}(a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n})\leq 1\minus{}a_{1}a_{2}...a_{n}\plus{}a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n}\leq n$,
by induction hypotesis.

Kunny : I mean a convex function. :wink:

[quote="stephen38"]Geez,i was so foolish.:wallbash_red:
It can be proved by induction.
For n=2,$ (1\minus{}a_{1})(1\minus{}a_{2})\geq 0$
Suppose the statement is true for $ n$ numbers,for $ (n\plus{}1)$ numbers,we have
$ S \equal{} a_{n\plus{}1}(1\minus{}a_{1}a_{2}...a_{n})\plus{}(a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n})\leq 1\minus{}a_{1}a_{2}...a_{n}\plus{}a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n}\leq n$,
by induction hypotesis.[/quote]

That's right! :)  Induction leads to easy solution.

[quote="stephen38"]Kunny : I mean a convex function. :wink:[/quote]

I also thought you meant a convex function, but it seems strange to me. :wink:


This is my $400th$ post!!!! :D  :first:  :cool:     :w00tb:

$Problem 113$

Hope this is mathcount level....

For any three positive numbers a,b,c,the product $(a+b)(b+c)(c+a)$ will be greater than:

a)2abc
b)3abc
c)5abc
4)8abc

Sorry if this is not mathcount level...if it isn't i'll give the easiest question i can. :D

[hide="Answer"]$(a+b)(b+c)(a+c)=a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc$
Using AM-GM, we have $\frac{a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc}{8}\ge \sqrt[8]{a^8b^8c^8}=abc$
Therefore, $(a+b)(b+c)(a+c)=a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\ge 8abc \Rightarrow \boxed{4}$

Technically, it will be greater than [b]or equal to[/b] any of the four solutions you gave, though...[/hide]

[hide="ans"]
just use $(x+y)\geq2\sqrt{xy}$
Simple.
So we get $8abc$
Answer=$\boxed4$
[/hide]

What method did you use shady? same one? :lol:

The question is ridiculous since ALL ANSWERS ARE CORRECT.

 :huh:

No... In fact, we have \[ (a + b)(b + c)(c + a) \geq 8abc \geq 5abc \geq 3abc \geq 2abc \] so $(a + b)(b + c)(c + a)$ is greater than each of the four possibilities!

Do you see?

I think it should be for [b]all[/b] real numbers $a$, $b$, $c$, rather than positive numbers. In that case, you can only really be sure that it's $\ge 8abc$.

Oh... that's odd. So does AM-GM only work for positive reals?

Yes. It only works for positive reals.

There was some discussion about the negative reals, however. I'll give you direct link.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=344173#p344173

This post is in NOML (National Online Math League) forum.  :)

Yup all options are right,so only arne got it :D

Actually, I did mention that because it is $\ge 8abc$, it will be $\ge$ all four of the solutions...

Wait a minute, he didn't get it! The problem said "greater than," not "greater than or equal to," so it's only the first three that work. :lol:

actually the first thre dont hold either for non-negative reals:
take $a=b=c=0$

btw am-gm works definitly for all non negative reals not only positv ones ...  :D

$a+b\geq2\sqrt{ab}$
$a+c\geq2\sqrt{ac}$
$b+c\geq2\sqrt{bc}$

$(a+b)(a+c)(b+c)\geq8\sqrt{a^2b^2c^2}$
$(a+b)(a+c)(b+c)\geq8abc$

What's the point of your post? You're just repeating what someone wrote before.

Hold on...Is it valid to use AM-GM in this problem?Bcos it isn't given that a,b,c are in A.P or G.P :huh:  :huh:

[quote="Arne"]AM GM works for ANY positive real numbers :huh:[/quote]

ohh...I'm new to the concept...sorry :blush:

What does mean A.P and G.P?

Anyway I can use AM-GM because it is mentioned that $a, b, c$ are positive numbers

I guess he means Arithemetic Progression and Geometric Progression.


Let $n$ be a positive integer and  $N={1,2,..,n}$.
Denote by $a_{n}$  the number of vectors $(A_{0},A_{1},...,A_{n})$ such that $A_{i}$ is a subset of $N$ and $|A_{i}|=i$  for all
$i=1,2,...,n$; and  such that the collection ${A_{0},A_{1},...,A_{n}}$  is  closed with respect to intersection (i.e. for each
$i$ and $j$ the intersection of $A_{i}$ and $A_{j}$ has the form $A_{s}$ for some $s$).
Question: find a closed  formula for $a_{n}$.

The $A$ are in bijection with the set of orderings of $N$; look at $A_{j}\setminus A_{j-1}$. There are $n!$.

I'm sorry but i don't understand.
Do you have an idea for an closed formula?

No, I don't :(; I just questioned the validity of jmerry's argument (and moved the topic to the combinatorics subforum where it belongs :P).

You're right. It's more complicated than I thought.

So, $A_{0}= \emptyset$, $A_{1}$ is arbitrary and $A_{2}$ is arbitrary.  But after that there are cases and it's tricky.    As the first few values, I get 1, 1, 2, 9, 60, 590, which unfortunately doesn't exist in the OEIS.

Also, the next term is 7860.

I'm curious: if we change the problem so that our collection is closed under both intersection and union, does that give something more tractable?  The first few values in this case seem to be 1, 1, 2, 9, 48, 350.

The value for $ n \equal{} 7$ in the original question is 136185.  In my proposed variation (closed under both $ \cap$ and $ \cup$), the sixth and seventh values are 2940 and 29925.

Semi-relatedly: the [url=http://www.research.att.com/~njas/sequences/]OEIS[/url] does contain [url=http://www.research.att.com/~njas/sequences/A102897]A102897[/url], the number of subsets of $ \mathcal{P}([n])$ closed under intersection.  The given question corresponds to the subcollection of this set in which we require exactly one element of each size.  This suggests a number of other related questions, and I suspect that most of them are hard.  (The associated sequence asking for number of subsets of $ \mathcal{P}([n])$ closed under both intersection and union does not appear in the OEIS.  It begins 2, 4, 13, 74, 732.)

Also, one other possibility: how many orbits are there under the action of $ S_n$?  (The first few values when we're closed under both union and intersection are 1, 1, 1, 2, 3, 6, 10, 21 ([url=http://www.research.att.com/~njas/sequences/A008928]A008928?[/url]) while if we're closed only under intersection I get 1, 1, 1, 2, 4, 10, 29, 95 (no sequence).  This does, of course, assume all my programs do what I want them to.)

Interesting. Thanks for still posting!  :)


Two circles are located on the coordinate plane such that the distance between their centers is $ \sqrt{2}$ units. If both circles have a radius of $ 1$ unit, what is the area of the region enclosed by their intersection?

[quote="Illos"]Two circles are located on the coordinate plane such that the distance between their centers is $ \sqrt {2}$ units. If both circles have a radius of $ 1$ unit, what is the area of the region enclosed by their intersection?[/quote]
[hide]
Let the centers of the circles be $ A\equal{}(0,0)$ and $ B\equal{}(\sqrt{2},0)$.  Their midpoint is $ M\equal{}(\frac{\sqrt{2}}{2},0)$.  Extend a perpendicular at $ M$ from $ AB$ to meet the circles at $ P$ and $ P'$.

Note that $ AP\equal{}1$ and $ AM\equal{}\frac{\sqrt{2}}{2}$ so $ \angle PAM\equal{}45^\circ$.  Then $ \angle PAP'\equal{}90^\circ$.  The area of half of the intersection is:

$ \pi 1^2\minus{}\frac{1}{2}$ so the total area is $ \boxed{2\pi\minus{}1}$[/hide]

Sorry if I'm mistaken (since I'm new   :blush: ), but
[hide]I don't understand how it can be 2pi-1

The way I solved it-

Center of circle A-(0,0), circle B (1,1). This is valid because 2*1^2=2, thus the distance between the centers is sqrt(2)

Now, the two circles intersect at (0,1) and (1,0), thus making a square of 1u side lengths (with (1,1) and the origin). (see pic)

The area of both circles is pi, (1^2=1), and the area of the circle in the 1 by 1 square is pi/4 and the "outside" of the circle is 1-(pi/4). Now, the area of the intersection is a quarter of one of the circles (let's say A) (pi/4) minus the part of the quarter that is the "outside" of circle B (1-pi/4).

Thus, the area of the intersection is
pi/4-(1-pi/4)
pi/4-1+pi/4
(pi/2)-1

[img]http://img528.imageshack.us/img528/7955/53121680ky2.th.png[/img]
[/hide]

[quote="jli"][quote="Illos"]Two circles are located on the coordinate plane such that the distance between their centers is $ \sqrt {2}$ units. If both circles have a radius of $ 1$ unit, what is the area of the region enclosed by their intersection?[/quote]
[hide]
Let the centers of the circles be $ A \equal{} (0,0)$ and $ B \equal{} (\sqrt {2},0)$.  Their midpoint is $ M \equal{} (\frac {\sqrt {2}}{2},0)$.  Extend a perpendicular at $ M$ from $ AB$ to meet the circles at $ P$ and $ P'$.

Note that $ AP \equal{} 1$ and $ AM \equal{} \frac {\sqrt {2}}{2}$ so $ \angle PAM \equal{} 45^\circ$.  Then $ \angle PAP' \equal{} 90^\circ$.  The area of half of the intersection is:

$ \pi 1^2 \minus{} \frac {1}{2}$ so the total area is $ \boxed{2\pi \minus{} 1}$[/hide][/quote]

You messed up on the last step...

The area of half the intersection should be $ \frac{\pi}{4}\minus{}\frac{1}{2}$, so the answer is:

$ \frac{\pi}{2}\minus{}1$, as ceastybet said...

how do u know that the part of the triangles drawn at the intersection that is cut off by the intersection is equal to one?

Each triangle actually has area of 1/2, since they're isosceles right triangles with leg length 1.


We have $ 12$ balls that $ 11$ of them are completely equivalent, but one of them is different in wight from the others. We have a scale with $ 2$ palms ( Funny English, isnt it?! :D  ) and we can do the wighting action just $ 3$ times! How can we find the different ball?

Do we know if the ball is heavier/lighter than the others?[/hide]

I've heard this riddle before, and i don't know how to solve it :P you don't know whether its heavier or lighter

if the question stated if it were lighter or heavier the answer would be.

if the question asked you to find the heavier ball;

what you do is you put 6 on each scale

which ever side rises take those six

and put 3 of those 6 on each side

whichever side rises take those 3

and put a random 2 on each side.

if it is equal than the third ball is heavier than the rest

if one side is heavier then the ball on that side is heavier than the rest
and vice versa

now if the ball is supposedly lighter than the others

what you do is you put 6 on each scale

which ever side sinks take those six

and put 3 of the six on each side of the scale

whichever side sinks take those 3

and put a random 2 of those 3 on each side.

if it is equal than the third ball is lighter than the rest

if one side is lighter then the ball on that side is heavier than the rest
and vice versa

hope that helped :D

[b]Generalization 1:[/b] Prove that the most economical approach with $ n \minus{} 1$ light balls and $ 1$ heavy ball (or vice versa; provided it is known) uses $ \left \lceil \frac {\log n}{\log 3} \right \rceil$ moves.

[b]Generalization 2:[/b] Prove that the most economical approach with $ n \minus{} 1$ identical balls and $ 1$ different ball uses $ \left \lceil \frac {\log n}{\log 3} \right \rceil \plus{} 1$ moves.

[b]Follow-up:[/b] You have 2008 balls, 2001 of which have one particular weight and 7 of which all have a different, lighter weight. What is the minimum number of weighings needed to separate the balls into the two weight classes?


Two congruent right-angled isosceles triangles (with baselength 1) slide on a line as on the picture. What is the maximal area of overlap?

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=287[/img]

We note first that there are in fact two cases:

Case 1:  The area formed is a triangle

Then let the indicated length be $x$.  We are varying this length.  The base of the area overlapping can be determined as $1-x$ (the base length is 1 unit).  Due to the congruency between the two triangles, it can be easily determined that the height of the area overlapping is $\dfrac {1-x}2$.  The area of the triangle is then $A=\dfrac 12 bh= \dfrac14 (1-x)^2$ Then diffrentiating gets $x=1$, but verification by the second diffrential shows this is in fact the min (and this should be obvious anyway)

Case 2:  The area formed is a pentagon.  Let the indicated length be $x$.  Now, the pentagon can be split up into two trapeziums, such that the Area to be computed is now $A= \dfrac 12 h (a+b) *2=h(a+b)$.  Since the two triangles are congruent, it does not take much to get $h= \dfrac{1-x}2$.   By similarity, one can also show that $a=x$ and $b=\dfrac {1+x}2$.  Putting this together

$A= \dfrac14 (1-x)(3x+1)$ and diffrentiating yields the solution x=1/3.  Substitution then shows that $A=\dfrac13$ and the second diffrential shows that this in fact the max.

Two congruent right-angled isosceles triangles (with baselength $1$) slide on a line as on the picture. What is the maximal area of overlap?
[img]https://cdn.artofproblemsolving.com/attachments/a/8/807bb5b760caaa600f0bac95358963a902b1e7.png[/img]


Does there exist the function $ f(x)$, satisfying all of the following conditions:
(a) $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ for all real $ x$ and $ y$.
(b) $ f(1) \equal{} \sqrt{2}, f(\sqrt{2}) \equal{} 1$;
(c) $ f(x)$ is bounded on the interval $ [1;\sqrt{2}]$.

[quote="tdl"]Does there exist the function $ f(x)$, satisfying all of the following conditions:
(a) $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ for all real $ x$ and $ y$.
(b) $ f(1) \equal{} \sqrt {2}, f(\sqrt {2}) \equal{} 1$;
(c) $ f(x)$ is bounded on the interval $ [1;\sqrt {2}]$.[/quote]

No.
$ (a)$ is Cauchy's equation. It's well known that non-continuous solutions of this equation are unbound on any non-empty interval. So $ (c)$ implies we are looking for continuous solutions.

It is well known that continuous solutions of Cauchy's equation are $ f(x)\equal{}ax$

Then $ (b)$ implies $ f(x)\equal{}\sqrt{2}x$ in order to have $ f(1)\equal{}\sqrt{2}$ but then $ f(\sqrt{2})\equal{}2\neq 1$

So no solution.


Can anyone explain partial fraction decomposition to me?   I need a step-by-step explanation. :?:

It's simpler than it looks at first glance.  There are only two main concepts embodied in it:

[b]Concept 1:[/b]  When we add up a series of fractions, we use a common denominator.  For example, $\frac{A}{x+3}+\frac{B}{x-3}= \frac{A(x-3)+B(x+3)}{(x-3)(x+3)}$.

[b]Concept 2:[/b]  Two polynomials are equal if and only if each of their coefficients are equal.

Notice that in our first example we can rewrite the numerator of our end product as $(A+B)x+(3B-3A)$.  For different values of $A, B$ we will get different numerators.

[b]Observation:[/b]  We can produce any possible linear numerator of the form $Cx+D$ this way.

[b]Idea:[/b]  Given a numerator $Cx+D$, can we find our original $A, B$?  For example,

$\frac{2x+7}{(x+3)(x-3)}= \frac{A}{x+3}+\frac{B}{x-3}$

Can we do that?  Well, we already know that

$\frac{A}{x+3}+\frac{B}{x-3}= \frac{(A+B)x+(3B-3A)}{(x+3)(x-3)}$

So it makes sense to equate the numerators.  This is equivalent to multiplying by $(x+3)(x-3)$, and yields

$2x+7 = (A+B)x+(3B-3A)$

So much for concept 1.  Concept 2 tells us that this is equivalent to the system

$2 = A+B$
$7 = 3B-3A$

Which is just a system of linear equations that is quite easy to solve.  We find that

$B = \frac{13}{6}, A =-\frac{1}{6}$

So that we can conclude

$\frac{2x+7}{(x+3)(x-3)}= \frac{-\frac{1}{6}}{(x+3)}+\frac{ \frac{13}{6}}{x-3}$

This procedure works with more than two fractions, and also with nonlinear factors.  Keep in mind that we want a common denominator.  For example,

$\frac{A}{x-3}+\frac{B}{(x-3)^{2}}+\frac{C}{x^{2}+x+1}= \frac{ A(x-3)(x^{2}+x+1)+B(x^{2}+x+1)+C(x-3)^{2}}{(x-3)^{2}(x^{2}+x+1)}$

We're still using a common denominator, it's just that this time we don't just naively multiply all of the denominators together.

Again, we can rewrite the numerator as a polynomial (this time it's cubic) and use that to deduce that any cubic polynomial is a valid numerator.  We can then run the process in reverse, as before.  For example, try to solve

$\frac{x^{3}+3x^{2}-2x-7}{(x-3)^{2}(x^{2}+x+1)}= \frac{A}{x-3}+\frac{B}{(x-3)^{2}}+\frac{C}{(x^{2}+x+1)}$

(This is quite a complicated example, so you might want to play around with simpler ones instead.)

Thanks but my teacher explained it using $\frac{A}{x-p}+\frac{B}{x-q}$ and got an algorithm for a quick way to PFD it. So anyone got any other ideas because I lost the algorithm? :oops:

Hmm.  Usually people prefer concrete numbers.  Well, if you understood that explanation you should be able to work out problems like this reasonably fast on paper anyway.


Suppose that $\mu$ is a measure on a $\sigma$-ring $\mathcal{S}$ and that $E$ is a set in $\mathcal{S}$ of $\sigma$-finite measure. If $\mathcal{D}$ is a disjoint class of sets in $\mathcal{S}$, then $\mu(E \cap D) \neq 0$ for at most countably many sets $D \in \mathcal{D}$.

I first showed that $\mu(E_{i}\cap D) \neq 0$ for at most countably many sets $D \in \mathcal{D}$, where $E_{i}$ is one of the sets (with finite measure) that  covers $E$. Then, I observed
\[X_{n}= \{ D \in \mathcal{D}\ | \ \mu(E_{i}\cap D) \geq \frac1{n}\}\]
for all $n \in \mathbb{N}$. Every such $X_{n}$ can only contain finitely many disjoint $D \in \mathcal{D}$. If this were not the case, this would contradict the assumption that $\mu(E_{i})<\infty$. Now it is easy to conclude that $\bigcup_{n \in \mathbb{N}}X_{n}$ is countable. Because $E$ has $\sigma$-finite measure, we can cover $E$ with countably many $E_{i}$. Hence, there are at most countably many $D \in \mathcal{D}$ such that $\mu(E \cap D) \neq 0$.

Corrections or confirmations are very welcome :)


A function machine carries ot two operations on a number. The input number is first multiplied by itself; the answer is then divided by two to give the output number.

(i) What is the output if the input is $ 11$?

(ii) What is the input if the output is $ 98$?

[hide="(i)"] Our function is equivalent to $ \frac {x^2}{2}$. Plugging in $ x \equal{} 11$, we get the result of
                                                                             $ \boxed{60.5}$[/hide]

[hide="(ii)"] Reverse engineer the process (or take the inverse, if you will).
$ 98 \equal{} \frac {x^2}{2}$
$ 196 \equal{} x^2$
$ x \equal{} \boxed{\pm14}$.
[/hide]

[quote="gaussintraining"]Note also that this is not a function (since two different $ x$'s give the same output).[/quote]
It is a function, a sort of squashed parabola. It wouldn't be a function if the same $ x$ gave two different $ y$s.


I've been going through this one for ages but can't seem to figure it out... Can anyone help me?

simplify expression:

[1/(x^2+9x+20)]+[1/(x^2+11x+30)]

so far I have:

1/(x+4)(x+5) . 1/(x+5)(x+6)

Any help appreciated.
thanks

thanks for that!! much appreciated   :)

I'll work back from your solution.

[quote="tim1234133"]$\frac{1}{(x+4)(x+5)}+\frac{1}{(x+5)(x+6)}$[/quote]

Isn't is easier just continue
\begin{eqnarray*}&&\frac{1}{(x+4)(x+5)}+\frac{1}{(x+5)(x+6)}\\ &=&\frac{1}{x+5}\left (\frac{(x+6)+(x+4)}{(x+4)(x+6)}\right )\\ &=&\frac{1}{x+5}\left (\frac{2x+10}{(x+4)(x+6)}\right )\\ &=&\frac{1}{x+5}\left (\frac{2(x+5)}{(x+4)(x+6)}\right )\\ &=&\frac{2}{(x+4)(x+6)}\end{eqnarray*}

And furthermore, why on earth you hide your post? If you have a solution, please show it. Hiding is okay if there is some lenghty, boring manipulation in your solution, but your solution is not a lenghty one.

[quote="puuhikki]
And furthermore, why on earth you hide your post? If you have a solution, please show it. Hiding is okay if there is some lenghty, boring manipulation in your solution, but your solution is not a lenghty one.[/quote]

Usually it is good etiquette to hide a solution, so that others can work on it without seeing the solution.

[quote="not_trig"]Usually it is good etiquette to hide a solution, so that others can work on it without seeing the solution.[/quote]
If you are on the top of the page, you won't see the solutions.

I sometimes hide my solutions, not always. Not only does it prevent people from accidentally seeing the solution (though, as you say, you can't see it from the top) and also it keeps things neater (though again, as you say, it wasn't a long solution).

Anyway, it only takes one click to reveal the solution, and it's obviously there if you want to see it. As a general rule, I hide a solution if it is the first reply to the topic.

Other AoPS members may want to try to solve a challenging problem now and then but if you don't hide your answers you spoil their fun :mad: . LOL

[quote="7h3.D3m0n.117"]Other AoPS members may want to try to solve a challenging problem now and then but if you don't hide your answers you spoil their fun[/quote]
How do I spoil if I don't hide my solution? In my screen resolution, if I am on the top of the page, I won't see anyone else's solution.

yeah but also please note that not all the people around the world have the exact same screen resolution as you.

[quote="puuhikki"][quote="7h3.D3m0n.117"]Other AoPS members may want to try to solve a challenging problem now and then but if you don't hide your answers you spoil their fun[/quote]
How do I spoil if I don't hide my solution? In my screen resolution, if I am are on the top of the page, I won't see anyone else's solution.[/quote]

As a moderator, I would like to say that hiding solutions is greatly encouraged (for many reasons).

For many larger resolutions, you can see several posts on a single screen without scrolling down.

[quote="b-flat"]As a moderator, I would like to say that hiding solutions is greatly encouraged (for many reasons).[/quote]
As a mathematician, I would like to say that hiding solutions makes reading and learning mathematics slower.

By about, what, 3 seconds? How long does it take to click on the tag?

Besides, you learn mathematics better by [i]solving[/i] problems, not just reading solutions.

eh i'd hate to be a hypocrit but if a problem is beyond your comprehension then looking at the solution might help  :)

Posters are strongly encouraged to hide their solutions. If you view a thread and you accidentally see parts of the solution to the problem, there is a chance that the parts you saw would give away the solution to the problem, which is not cool. On the other hand, if you want to see the solution, just go ahead and click. It's not too difficult. ;)


[quote="Deng Qunyi"]show that  $ (1 + x)^a > {\frac {3}{\sqrt [3]{2}}a^{\frac {2}{3}}{(a - 1)^{\frac {1}{3}}x}}$,where $ 0 < x < 1 ,a > 1$.[/quote]

This inequlity should be called analytic inequality


Find all integer numbers $x, y, z$, with $1\le x< y< z\le 18$, such that $x^{2^{n}}+y^{2^{n}}+z^{2^{n}}\equiv 0(mod19), \forall n\in \mathbb{N},n\neq 0$.

Can you prove that those are solutions and they are the only one?

Well, the sequence (4, 16, 9, 5, 6, 17, 4, ...) repeats every time that the previous term is squared mod 19.

Also there are 6 terms in this sequence.

So picking every other term, starting from any, gives 2 different sets of 3 numbers, each both of which sum to $0 \, (\bmod \, 19)$.

Thus, $(4, 6, 9)$ is a solution because by squaring every term, you get $(16, 17, 5)$, which is also a solution (although the $x$, $y$, and $z$ are not order in the right way here, but the sum is still the same).

Squaring this gives $(9, 4, 6)$, simply a rearrangement of the original, which then still has the same sum.

That's probably kind of a bad explanation/proof, but I hope you get the idea.

Also, other than those, the only values that repeat are the pair $7$ and $11$, which alternate, and $1$, which is constant.

So in fact, $(x, y, z) = (1, 7, 11)$ is also a solution that I previously had missed.

When these terms are squared, $(1, 11, 7)$ is the result.

I think you could also use the generators of these values as solutions, since the case of $n = 0$ is not needed.


algebra
difference of squares
special factorizations
number theory unsolved
number theory

[quote="am2525"]prove that there exists a positive integer m such that for each positive integer n,
n>m-1.the following condition is true:
betwean the numbers:n,n+1,n+2,...,n+9 there is a number which has at least three different prime divisors. and find the minimal m.[/quote]
What is the relation between "10 numbers from n to n+9" and m?  :?:

i rewrite the question;
prove that there exists an integer like $ m$ such that if $ n$ is a positive integer and $ n > m$ then between the ten consecutive numbers : $ n,n \plus{} 1,n \plus{} 2,...,n \plus{} 9$ there exists a number which is divisible by at least three distinct prime.
also try to find the minimum of $ m$.

I dont have a full solution but since noone else has posted any progress, here is one approach that might work.

At least one of the numbers $ n,..,n \plus{} 9$ must be a multiple of $ 6$ (call it $ a$), another must be a multiple of $ 10$ (call it $ b$)
If neither has three primes factors then we have $ a \equal{} 2^x3^y, x,y > 1$ and $ b \equal{} 2^z5^t, z,t > 1$

So $ 2^x3^y \equal{} 2^z5^t \plus{} c$ where $ \minus{} 9 \le c \le 9$

The larger of the two powers of $ 2$ must be a factor of $ c$, so it can be cancelled out. this leaves one of two equations:

$ 3^y \equal{} 2^z5^t \plus{} c$ or  $ 2^x3^y \equal{} 5^t \plus{} c$ where $ \minus{} 5 \le c \le 5$

$ c$ can not have factors of $ 2,3$ or $ 5$, which leaves $ c \equal{} \pm 1$, unless $ x,z \equal{} 0$ then $ c \equal{} \pm1, \pm2, \pm4$.

So to solve the problem it is sufficient to show that there are only a finite number of solutions to all of
 $ 3^y \equal{} 2^z5^t \pm 1$,  $ 2^x3^y \equal{} 5^t \pm 1$ and $ 3^y \equal{} 5^t \pm1, \pm2, \pm4$

You have to show that for all sufficiently large $ n$, at least one integer in the range from $ n$ to $ n \plus{} 9$ inclusive has at least three different prime factors,
and determine what is the largest $ n$ for which they all have less than three prime factors.

Some of these exponential equations can be solved mod n.

trying mod 24 we have

$ 5^t \equiv 5,1 \bmod 24$
$ 2.5^t \equiv 10,2 \bmod 24$
$ 4.5^t \equiv 20,4 \bmod 24$
$ 2^z.5^t \equiv 16,8 \bmod 24, z > 2$

$ 3^y \equiv 3,9 \bmod 24, t > 0$
$ 2.3^y \equiv 6,18 \bmod 24,t > 0$
$ 4.3^y \equiv 12 \bmod 24, t > 0$
$ 2^x.3^y \equiv 0 \bmod 24, t > 0, x > 3$

This gives a number of cases that work mod 24 that we need to consider as follows

For $ 2^x3^y \equal{} 5^t \minus{} 1$ we find that $ t$ is even, then $ 5^t \minus{} 1$ factorises into two factors differing by 2 and this case can be eliminated

Other cases that can be eliminated by factorization of difference of squares
$ 3^y \equal{} 5^t \plus{} 4$ ($ y$ even $ t$ odd)
$ 3^y \equal{} 4.5^t \minus{} 1$, ($ t$ is even)

cases that can be solved mod n for some n

$ 2.3^y \equal{} 5^t \plus{} 1$,($ y,t$ both odd), (solutions mod $ 63$ give $ y \equal{} 0,1$)
$ 3^y \equal{} 2.5^t \plus{} 1$, ($ t$ even, $ y$ odd) (no solutions mod 20)
$ 3^y \equal{} 5^t \minus{} 2$ ($ t$,$ y$ odd) (solutions mod $ 315$ give $ y \equal{} 1$)

the case $ 3^y \equal{} 2^z5^t \plus{} 1$ ($ y$ is even) factorises and implies $ 3^y \equal{} 2.5^t \plus{} 1$ or $ 3^y \equal{} 2.5^t \minus{} 1$

two other cases still to be resolved

$ 3^y \equal{} 2.5^t \minus{} 1$ , ($ t$ odd, $ y$ even)
$ 3^y \equal{} 5^t \plus{} 2$ ($ y$ odd $ t$ even)

You can solve these kinds of exponential diophantine equations generically using the methods of Stormer and Lehmer.
See  http://en.wikipedia.org/wiki/St%C3%B8rmer%27s_theorem. However it would be nice to know if there are more elementary methods that can resolve the remaining two cases.


modular arithmetic
induction
number theory unsolved
number theory

consider a sequence $ x_{1}=1,x_{2}=12,....x_{1}2=123456789101112,...$ .Prove that there exist infinitely
many terms $ x_{i}$ of this sequence such that $ x_{i}$ is divisible by $ 2005$

[quote="Nbach"]consider a sequence $ x_{1} \equal{} 1,x_{2} \equal{} 12,....x_{12} \equal{} 123456789101112,...$ .Prove that there exist infinitely
many terms $ x_{i}$ of this sequence such that $ x_{i}$ is divisible by $ 2005$[/quote]

Let $ k\equal{}200p\plus{}152$ for any integer $ p\geq 0$
Consider then $ b\equal{}10^{k\minus{}1}\plus{}701$.
$ b$ has exactly $ k$ digits.
Let then $ a\equal{}10^k$. We have $ a\equal{}5\pmod{2005}$ and $ b\equal{}501\pmod{2005}$ Then :

$ x_{b\plus{}i}\equal{}5x_{b\plus{}i\minus{}1}\plus{}501\plus{}i\pmod{2005}$ $ \forall$ $ 0\leq i <9\cdot 10^{k\minus{}1}\minus{}701$
(this last constraint to be sure that length (in digits) of $ b\plus{}i$ is also $ k$).

It's rather easy to show (use induction) that $ 16x_{b\plus{}i}\equal{}16\cdot 5^{i\plus{}1}x_{b\minus{}1}\minus{}4(i\plus{}1)\pmod{2005}$
If we can find one value of $ i$ for which $ 4\cdot 5^{i\plus{}1}x_{b\minus{}1}\equal{}i\plus{}1\pmod{2005}$, then we have solved the problem because we would have $ 16x_{b\plus{}i}\equal{}0\pmod{2005}$ and so $ 2005$ divides $ x_{b\plus{}i}$

Let then $ u\equal{}x_{b\minus{}1}\pmod{401}$
If $ u\equal{}0$, then $ x_{b\minus{}1}$ is divisible by $ 401$ and also by $ 5$ (it ends with $ \ldots 00700$), so by $ 2005$.
If $ u>0$, then let $ i\equal{}158800u\minus{}1$

We have $ 5^{i\plus{}1}\equal{}5^{158800u}\equal{}(5^{400})^{397u}\equal{}1\pmod{401}$
So $ 4\cdot 5^{i\plus{}1}x_{b\minus{}1}\equal{}4u\pmod{401}$ and $ i\plus{}1\equal{}158800u\equal{}4u\pmod{401}$ and so
$ 4\cdot 5^{i\plus{}1}x_{b\minus{}1}\equal{}i\plus{}1\pmod{401}$ and so
$ 4\cdot 5^{i\plus{}1}x_{b\minus{}1}\equal{}i\plus{}1\pmod{2005}$ since $ 5$ divides both sides.
and so $ 2005$ divides $ x_{b\plus{}i}$

So, as a conclusion :
Let $ k\equal{}200p\plus{}152$ for any integer $ p\geq 0$
Let $ b\equal{}10^{k\minus{}1}\plus{}701$
Let $ u\equal{}x_{b\minus{}1}\pmod{401}$

Then $ 2005$ divides $ x_{b\plus{}158800u\minus{}1}$

And since we can choose as many $ p$ as we want, there exist infinitely many terms $ x_{i}$ such that $ x_{i}$ is divisible by $ 2005$.
Q.E.D.


In the quadrilateral $ ABCD$, $ AD$ is parallel to $ BC$. $ K$ is a point on $ AB$. Draw the line through $ A$ parallel to $ KC$ and the line through $ B$ parallel to $ KD$. Prove that these two lines intersect at some point on $ CD$.

[hide] Let the line through $ B$ parallel to $ KD$ intersect $ CD$ at $ X$ and let the line through $ A$ parallel to $ KC$ intersect $ CD$ at $ Y$.

If $ ABCD$ is a parallelogram then the claim is obvious:  $ BXDK$ is a parallelogram, thus $ BK\equal{}XD$.  $ KCYA$ is parallelogram, thus $ AK\equal{}CY$.  Thus, $ XD\plus{}CY\equal{}BK\plus{}AK\equal{}AB\equal{}CD$ and so $ X\equal{}Y$.

If $ ABCD$ is not a parallelogram, then let lines $ AB,CD$ intersect at point $ E$.  Without loss of generality, suppose $ E$ is closer to $ BC$ than to $ AD$ (if not, switch vertices $ A,B$ and $ C,D$).  Since $ AY\parallel{}KC$, we have $ \frac{EB\plus{}AB}{EB\plus{}BK}\equal{}\frac{EC\plus{}CY}{EC}$.  So, we have $ CY\equal{}\frac{EC(EB\plus{}AB)}{EB\plus{}BK}\minus{}EC$.

Also, since $ KD\parallel{}BX$, we have $ \frac{EB\plus{}BK}{EB}\equal{}\frac{EC\plus{}CD}{EC\plus{}CX}$.  Thus, $ CX\equal{}\frac{EB(EC\plus{}CD)}{EB\plus{}BK}\minus{}EC$.

Now, since $ BC\parallel{}AD$, we have $ \frac{EB}{EB\plus{}AB}\equal{}\frac{EC}{EC\plus{}CD}$.  Thus, $ EB(EC\plus{}CD)\equal{}EC(EB\plus{}AB)$ and so $ CY\equal{}CX$.  Thus, $ X\equal{}Y$. [/hide]