There is a triangle ABC such that <B=90. We take a point D of  extention of line AC (from C) and AB=CD with <CBD=30. Find AC/CD?

[quote="Ulanbek_Kyzylorda KTL"]There is a $ \triangle ABC$ such that $ \angle B\equal{}90$. We take a point $ D$ of  extention of line $ AC$ (from $ C$) and $ AB\equal{}CD$ with $ \angle CBD\equal{}30^o$. Find $ \frac{AC}{CD}$?[/quote]

Let's determine $ AC\equal{}x$, $ AB\equal{}y$, $ \frac{AC}{CD}\equal{}\frac{AC}{AB}\equal{}\frac{x}{y}\equal{}a$, $ \angle ACB\equal{}\alpha$
We have that:
$ BD\equal{}\frac{y\cdot sin(\alpha)}{sin30^o}\equal{}\frac{2y^2}{x}$

$ (x\plus{}y)^2\equal{}AD^2\equal{}AB^2\plus{}BD^2\minus{}2AB\cdot BD \cdot cos120^o\equal{}y^2\plus{}\frac{4y^4}{x^2}\plus{}\frac{2y^3}{x}$
or
$ \frac{x^2}{y^2}\plus{}2\frac{x}{y}\equal{}4\frac{y^2}{x^2}\plus{}2\frac{y}{x}$
$ \leftrightarrow$
$ a^2\plus{}2a\equal{}\frac{4}{a^2}\plus{}\frac{2}{a}$
$ \leftrightarrow$
$ (a^3\minus{}2)(a\plus{}2)\equal{}0$
therefore
$ a\equal{}\sqrt[3]{2}$

The perpendicular through D to AD intersects AB at E; from the power of A to the circle (CDEB) we get $ BE \cdot c \equal{} a^2 \plus{} b \cdot c$  ( 1 ), a, b and c being the sides of $ \triangle ABC$. The triangle CDE is rightangled at D and has $ \angle CED \equal{} \angle CBD \equal{} 30^o$, hence $ DE \equal{} c \cdot \sqrt 3$ ( 2 ) , that is, from $ \triangle ADE$ rightangled at D we get $ AE^2 \equal{} b^2 \plus{} 2 \cdot b \cdot c \plus{} 4 \cdot c^2$( 3 ). But from (1) we can get $ AE \equal{} c \plus{} \frac {a^2 \plus{} b \cdot c} {c}$ ( 4 ). From (3) and (4), with $ a^2 \equal{} b^2 \minus{} c^2$ we get an equation similar to the one above, and from the same, the value for the required ratio.

Best regards,
sunken rock


Show that for each point (a, b, c) on the surface $x^{2}+y^{2}+z^{2}= 1$, there is a neighborhood of (a, b, c) in which the surface may be represented as a smoothly parameterized 2-surface. Hence, there is a tangent plane to this surface at every point.

You can always parametrize by two of $x,y,z$ locally.


rotation
geometry
trapezoid
analytic geometry
geometry proposed

Let $ABCD$ a trapezoid. Prove that $AC^{2}+BD^{2}=AD^{2}+BC^{2}+2.AB.CD$

We can rotate the trapezoid etc etc ... to get more simply coordinates of this vertices ... So

${A=\{a,0\}}$
${B=\{b,0\}}$
${C=\{c,e\}}$
${D=\{d,e\}}$

So, the length

${\text{AB}=\sqrt{(a-b)^{2}}}$
${\text{AC}=\sqrt{(a-c)^{2}+e^{2}}}$
${\text{AD}=\sqrt{(a-d)^{2}+e^{2}}}$
${\text{BC}=\sqrt{(b-c)^{2}+e^{2}}}$
${\text{BD}=\sqrt{(b-d)^{2}+e^{2}}}$
${\text{CD}=\sqrt{(c-d)^{2}}}$

Lets Proof

${\text{AC}^{2}+\text{BD}^{2}=\text{AD}^{2}+\text{BC}^{2}+2 \text{AB}\text{CD}}$

Substituing

${(a-c)^{2}+(b-d)^{2}+2 e^{2}=(b-c)^{2}+(a-d)^{2}+2 e^{2}+2 \sqrt{(a-b)^{2}(c-d)^{2}}}$
${b (d-c)+a (c-d)+\sqrt{(a-b)^{2}(c-d)^{2}}=0}$

If We Assume   $(a<b\land d<c)\lor (a>b\land d>c)$  The equation become true.

Just assuming this, because the localization of the vertices

A basic approach is by calculations based on Pythagorian theorem.

Another solution is as follows:

We denote as $E,$ $F,$ the orthogonal projections of $A,$ $B$ respectively, on the sideline $CD.$ If $M,$ is the midpoint of the segment $CD,$ from the triangles $\bigtriangleup ACD,$ $\bigtriangleup BCD,$ we have that $($ second theorem of medians $):$

$(AC)^{2}-(AD)^{2}= 2(CD)\cdot (ME)$

$(BD)^{2}-(BC)^{2}= 2(CD)\cdot (MF)$

$(AC)^{2}-(AD)^{2}+(BD)^{2}-(BC)^{2}= 2(CD)\cdot (ME+MF)$

$(AC)^{2}+(BD)^{2}= (AD)^{2}+(BC)^{2}+2(CD)\cdot (AB)$ and the proof is completed.

Kostas Vittas.


How many of the triangles whose vertices are chosen from the vertices of a cube are equilateral?
(A) 4
(B) 8
(C) 12
(D) 24

My first AOPS solution... hope I didn't screw up :)

[hide]There are three different lengths connecting vertices of a cube: the sides of the cube with length s, the diagonals on the faces of the cube with length s, and the 'space diagonals' connecting two opposite vertices with length s. Neither the sides nor the space diagonals of a cube will form a triangle (sides are always connected in fours, and space diagonals intersect), so we are left with only the face diagonals.

There are 12 face diagonals -- 2 on each face for 6 faces. Each diagonal will form two equilateral triangles with the vertices in the opposite corners of the parallel face. That makes 12*2 = 24 triangles; however, they are counted for each of the three diagonals, so the number of distinct triangles is 1/3 of that. B) 8[/hide]


Let $P,Q$ be some unitary real polynomials and $n\geq 1$ such that $P^{[n]}=Q^{[n]}$ (where this denotes the composition of $P$ with itself $n$ times). Then $P=Q$.

I remember Marius Cavachi proposed a similar problem, with complex polynomials but $n=2$ for a recent TST. I wonder if the one here also holds for complex polynomials.

For real polynomials it's pretty easy, I think, to prove that if, say, $P(x)-Q(x)\to\infty$ as $x\to\infty$, then $P^{[n]}(x)-Q^{[n]}(x)\to\infty$ as $x\to\infty$, and since $P\ne Q$ implies $x\to\infty\Rightarrow|P(x)-Q(x)|\to\infty$, that's that.

I hope I didn't say something stupid. :?


linear algebra
matrix
linear algebra unsolved

Let $A,B$ in $M_n(R)$
Prove that :
1/ $r(A+B)\leq r(A)+r(B)$
2/ $r(A)+r(B)\leq r(AB)+n$

The rank of a square matrix is the dimension of the space spanned by its rows, and is equal to the dimension of the space spanned by its columns.  The space spanned by the rows of $A+B$ is contained in the space spanned by the rows of $A$ and the rows of $B$, so the first assertion follows on taking dimensions.

Think of $A$ and $B$ as linear operators, then
\[ rank(AB) = \dim (Im(AB)) = \dim(Im(A)) - \dim(\ker(B)\cap Im(A)) \geq \dim(Im(A)) - \dim(\ker(B)) = rank(A) - (n - rank(B)) = rank(A) + rank(B) - n. \]


inequalities
trigonometry
inequalities proposed

Suppose $ n \in \mathbb{N},x_0=0,x_i>0$ for $ i=1,2,...,n,$ and $ \displaystyle\sum_{i=1}^{n}x_i=1.$ Prove that:

${ 1\le \displaystyle\sum_{i=1}^{n}\frac {x_i} {\sqrt{1+x_0+x_1+...+x_{i-1}} \cdot \sqrt{x_i+...+x_n}}}<\frac {\pi} {2}$

[quote="Mathias_DK"][quote="moldovan"]Suppose $ n \in \mathbb{N},x_0 = 0,x_i > 0$ for $ i = 1,2,...,n,$ and $ \displaystyle\sum_{i = 1}^{n}x_i = 1.$ Prove that:

${ 1\le \displaystyle\sum_{i = 1}^{n}\frac {x_i} {\sqrt {1 + x_0 + x_1 + ... + x_{i - 1}} \cdot \sqrt {x_i + ... + x_n}}} < \frac {\pi} {2}$[/quote]

${ \displaystyle\sum_{i = 1}^{n}\frac {x_i} {\sqrt {1 + x_0 + x_1 + ... + x_{i - 1}} \cdot \sqrt {x_i + ... + x_n}}} \ge \displaystyle\sum_{i = 1}^{n}\frac {2x_i} {1 + x_0 + x_1 + ... + x_n} = \sum_{i = 1}^{n} x_i = 1$.

Honey_S: $ x_n = 1,x_1 = x_2 = ... = x_{n - 1} = 0$ gives $ 1$ and not $ \frac {1}{\sqrt {2}}$..... (Btw. $ x_1,..,x_n > 0$ too)[/quote]

The $ x_1,..,x_n > 0$, it doesn't matter my friend. And sorry because I remember that the problem give $ x_0 = 1$

[quote="Mathias_DK"][quote="Honey_S"]The $ x_1,..,x_n > 0$, it doesn't matter my friend. [/quote]
Yes it does! That means that in the case $ x_0 \equal{} 1$ there is no minimum - just a lower bound! And that is actually a difference.[/quote]

In my opinion, I don't think so.  :wink:  :|


Basically a thread to list good websites with games or links to specific games. 1

1. [url]kongregate.com[/url]

Dang, this one is awesome!

http://notdoppler.com/magnetizr.php

Major ownage game:

http://www.kongregate.com/games/Xdragonx10/medieval-rampage-2

[quote="evilhamster"]So as soon as I see this thread, I think "oh, I'm going to post Kongregate". Dang.

[url]http://www.armorgames.com[/url][/quote]
me too

Yeah, same here.

Maybe Kongregate deserves its own thread.

[url]http://www.gamefudge.com/Age-of-War[/url]

pretty weird but addicting (kind of)

I also thought "I'll post Kongregate"

Miniclip: miniclip.com

FFR: flashflashrevolution.com

I knew a guy who was incredibly beast at FFR once...it was cool

[url]http://www.kongregate.com/games/polymerrabbit/frantic[/url]

[url]http://www.kongregate.com/games/ArmorGames/sonny-2[/url]

Not my absolute favorites, but pretty close.

my favorite game, maplestory.

cgyao15, i expect you to agree with me right?

i do :P wats ur highest lvl? mine's 134 closing in on 135

scania not my main right now, working on guy called pwnlight DW lvl 28 and what are u sin? dit? mage? cleric?

hmm, i used to play that in 5th grade.
it took me a year to get to level 16.
so i would advise you not to waste your time.

[quote="AIME15"]You're probably not that efficient :D

I got to 40-ish on 3 different accounts in a year, on about 2 hours a week.[/quote]
that's probably why. i played for like 15 minutes a week while doing homework.

@cgyao15:

I'm in Broa.

I wish that you were in Broa so you could donate to me (I'm broke)

I have a lvl 31 sin, 43 mage, and lvl 53 thunder breaker

congratz my frined haves 3m there but i dunno..

I know an awesome one.
[url]http://www.bored.com[/url]

Another good one:

[url]http://www.gamesmate.com/free-action-games/free-game-playing-with-fire2/free-game-playing-with-fire2.asp[/url]

awesome game.

Similar to bomb man


Suppose $(a_{i})_{i \ge 1}$ is a sequence of positive integers satisfying $\gcd(a_{i}, a_{j}) = \gcd(i, j)$ for $i \neq j$. Show that $a_{i}= i$ for each $i$.

[hide]
Assume to the contrary, so let's say $a_{m}\neq m$. Note that $m \mid a_{m}$ as $gcd(a_{m}, a_{2m}) = m$. So, $a_{m}=km$ for some $k$. Note that $mk \mid a_{mk}$ (for the same reason as above). $gcd(m,mk)=m$.
But $gcd(a_{m},a_{mk}) = mk$. Contradiction.
[/hide]


function
quadratics
algebra
polynomial
algebra proposed

Find all the functions f:R->R such that fog=gof for every g:R->R, g a degree two function.

What's a degree two function? Do you mean a compositely periodic function of order 2? :?

No, a degree two function is a function of the form ax^2+bx+c

you can call that a quadratic function or rather a second degree polynomial function.

plain degree 2 usually refers to the derivation degree :D.

We have f(x 2 ) = (f(x)) 2  and f(x 2  - x) = (f(x)) 2  - f(x).
Then for x = 0, we deduce that (f(0)) 2  = f(0) and f(0) 2  = 2f(0), so that f(0) = 0.

Let a be a real number. We have f(x 2  + a) = (f(x)) 2  + a. For x = 0, it leads to f(a) = a.
Then f = id, which is indeed a solution.

Pierre.


Solve the following system for real $a , b, c, d, e$: \[ \left\{ \begin{array}{ccc} 3a & = & ( b + c+ d)^3 \\ 3b & = & ( c + d +e ) ^3 \\ 3c & = & ( d + e +a )^3 \\ 3d & = & ( e + a +b )^3 \\ 3e &=& ( a + b +c)^3. \end{array}\right.  \]

Without loss of generality, we may assume that $a$ is the greatest among the five given numbers. Then $d + e + a \geq c + d + e$, and hence $c = (d + e + a)^3/3 \geq (c + d + e)^3/3 = b$. Also, $e = (a + b + c)^3/3 \geq (b + c + d)^3/3 = a$, and since $a \geq e$, we get $a = e$. Hence also $e \geq b$, and $b = (c + d + e)^3/3 \geq (b + c + d)^3/3 = a$ - giving $a = b$. Also, since $c \geq b$, we get $a = c$, and finally we also get $a = d$. So, we need to solve $3a = (3a)^3$ which gives the following solutions: $(1/3,\dots,1/3)$, $(0,\dots,0)$ and $(-1/3,\dots,-1/3)$.

[quote="tranthanhnam"]let $g(x)=3x$ and $f(x)=(k-x)^3$
 because , $g'(x)=3$ and$f'(x)>0$ .
  therefore,$a=b=c=d$
, $\iff a=b=c=d=0$or $a=b=c=d=1/3$[/quote]
you are wrong!Or i false?

$k$ is...?

Just look at the given system - there are FIVE variables, and you dealt with only four of them. And what is your $k$?

He may let k=a+b+c+d,but he really leave out e

That is a frivolous question. Have you read post #2 ?


Let $P$ and $Q$ be points on the sides $BC$ and $CD$ of a convex quadrilateral $ABCD$, respectively, such that $\angle{BAP}=\angle{ DAQ}$. Prove that the triangles $ABP$ and $ADQ$ have equal area if and only if the line joining their orthocenters is perpendicular to $AC.$

Let H1 and H2 be the orthocenters of AQD and ABP respectively. Let a=|DÂQ|=|BÂP|.
[ADQ]=[ABP] <=> 0,5.|AD|.|AQ|.sin a = 0,5.|AB|.|AP|.sin a <=> |AD|.|AQ|.cos a = |AB|.|AP|.cos a <=> AD.AQ - AB.AP = 0  (this is some vectors)

AD.AQ - AB.AP = (AB + BH2 + H2H1 + H1D) (AC + CQ) - AB (AB + BH2 + H2P)

= AB.AC + AB.CQ + BH2.AC + BH2.CQ + H2H1.AC + H2H1.CQ + H1D.AC + H1D.CQ - AB.AB - AB.BH2 - AB.H2P

= AB.BC + AH1.CQ + BH2.BC + H2H1.AC + H1D.AQ - AB.H2P = BC.AH2 + H2H1.AC = H2H1.AC

So [ADQ]=[ABP] <=> H1H2 is perpendicular to AC. :D

Is any body have a alternate solution ???

Let $H_1,H_2$ be the orthocentres of $\triangle ABP,\triangle ADQ$ respectively. Let $X,Y$ be the feet from $A$ to $CB,CD$ respectively. Let $Z_1,Z_2$ be the feet of the perpendiculars from $H_1,H_2$ to $AC$.

Now we observe:
$$AZ_1=AH_1 \cos{\angle CAH_1}=BP \cdot \frac{AX}{AC} \cdot \cot{\angle BAP}=\frac{2[ABP]}{AC \cdot \tan{\angle BAP}}$$
So using $\angle{BAP}=\angle{ DAQ}$ we get:
$$\frac{AZ_1}{AZ_2}=\frac{[ABP]}{[ADQ]}$$
Hence as $H_1H_2 \bot AC \Leftrightarrow AZ_1=AZ_2$ we see $H_1H_2 \bot AC \Leftrightarrow [ABP]=[ADQ]$ as desired.

Let $H_1,H_2$ be the orthocentres of $\triangle ABP,\triangle ADQ$ respectively. Let $X,Y$ be the feet from $B$ to $AP$ and $D$ to $AQ$. Call $Z$ the intersection of $AC$ and $H_1H_2$. If we prove that $BX$ and $DY$ intersects on $AC$, we are done. This result follows from radical axis of circle with diameter $AH_1$ and $AH_2$.

Now note that $\triangle ABP = \triangle ADQ$ means that $AB \cdot AP = AD \cdot AQ$, since $\angle ABP = \angle ADQ$. This also gives us the spiral similarity of $\triangle ABQ$ and $\triangle APY$. Note that $BH_1$ and $CH_2$ are all lines drawn from the original points of the triangle to its corresponding sides. I do not know how to continue, due to my extensive lack of knowledge in geometry. Anyone care to help?


cyc[root(x^2 + y^2)root(y^2+z^2)] <= (x+y+z)^2

Is this true?

${\sum_{ cyc}{[\sqrt{(x^2 + y^2)}\sqrt{(y^2+z^2)]}}} \leq (x+y+z)^2$
is the expression in Latex language.

This is wrong: Take $y=z=1$ and $x \to \infty$.


function
trigonometry
algebra unsolved
algebra

How many times changes the sign of the function \[ f(x)\equal{}\cos x\cos\frac{x}{2}\cos\frac{x}{3}\cdots\cos\frac{x}{2009}\] at the interval $ \left[0, \frac{2009\pi}{2}\right]$?

[quote="April"]How many times changes the sign of the function
\[ f(x) \equal{} \cos x\cos\frac {x}{2}\cos\frac {x}{3}\cdots\cos\frac {x}{2009}
\]
at the interval $ \left[0, \frac {2009\pi}{2}\right]$?[/quote]

Very nice problem !

[hide="My solution proposal"]
We are looking for the number of roots of $ f(x) \equal{} 0$ in $ (0,\frac {2009\pi}{2})$ whose order is odd. (I take the open interval since there is no sign change at the bounds, even if this is a root).

Roots are $ x \equal{} kp\frac {\pi}{2}$ for any $ k\in[1,2009]$ and odd $ p$ such that $ 0 < kp < 2009$

So roots are $ r_n \equal{} n\frac {\pi}{2}$ with $ n\in[1,2008]$ and n having some odd divisors.

Order of the root $ r_n$ is the number of odd divisors of $ n$.

So, if $ n \equal{} 2^m\prod_{odd\cdot prime\cdot p_i}p_i^{k_i}$, the order of $ r_n$ is $ \prod (k_i \plus{} 1)$
So the only roots whose order is odd are roots whose all $ k_i$ are even and so roots which may be written $ 2^ma^2$ (with odd a).

So the required number $ X$ is the number of integers in $ [1,2008]$ of the form $ n^2$ or $ 2n^2$

So $ X \equal{} [\sqrt {2008}] \plus{} [\sqrt {1004}] \equal{} 75$
[/hide]


vector
linear algebra
linear algebra theorems

Ok I think I am missing something the size of an iceberg on this problem.

But here it is:

Let V and W be vector spaces over a field F

Let T be a linear transformation with  T: V -> W

Show:    dim( image(T) ) + dim( ker(T) ) = dim( V )

The definitions I am using are:

image(T) = col(T) = column space of T = { w in W | there exists v in V such that T(v)=w }

ker(T) = Null(T) = { v in V | T(v)=0 }

dim(V) = dimension(V) = | basis(V) | = number of vectors in the basis of V

I have been reviewing linear algebra (2 yrs since my last class) and this problem has stopped me dead in my tracks, I am stuck on something here but I can't put my finger on it. Too much abstract since i had linear i think...

thanks
matthew

This is not a problem, but a fundamental theorem in linear algebra. Any course of linear algebra should have this theorem proved.

the rank-nullity theorem - got it now
 thanks

from this $dim(ker(V))+dim(im(V))=dim(V) $ can someone deduce Grassmannn's theorem? :?

About the theorem in question, this link may be useful: http://planetmath.org/encyclopedia/ProofOfRankNullityTheorem.html .

[quote="Cezar Lupu"]from this $dim(ker(V))+dim(im(V))=dim(V)$ can someone deduce Grassmannn's theorem? :?[/quote]

What's that, Cezar? (Graussmann theorem)


[color=darkred]Let $ABC$ be a triangle and let $\{D,E,F\}\subset (BC)$ be three points so that $D\in (BF)$, $E\in (FC)$.

For a mobile point $M\in AF$ denote $\{\begin{array}{c}U\in BM\cap AD\ \ ,\ \ V\in CM\cap AE\\\ X\in FU\cap AB\ \ ,\ \ Y\in FV\cap AC\end{array}$. Prove that

the lines $UV$, $XY$, $BC$ are concurrently in a fixed point $S$ and $\{\begin{array}{c}\frac{SB}{SC}=\frac{BD}{CE}\cdot\frac{FE}{FD}\\\\ \frac{SD}{SE}=\frac{BD}{CE}\cdot\frac{FC}{FB}\end{array}$.[/color]

Consider a central proiection which put line parallel to $BC$ and passing through $A$ to infty.
Let $S'=U'V' \cap A'B'$ then $\frac{S'B'}{S'C'}=\frac{U'B'}{U'M'}*\frac{V'M'}{V'C'}=\frac{D'B'}{D'F'}*\frac{E'F'}{E'C'}$.
Also $\frac{X'B'}{Y'C'}=\frac{X'B'}{M'F'}*\frac{M'F'}{Y'C'}=\frac{D'B'}{D'F'}*\frac{E'F'}{E'C'}=\frac{S'B'}{S'C'}$ So $R'$,$U'$,$V'$ are colinear and lines
$U'V'$,$C'B'$ and $X'Y'$ are concurent.So are the lines $UV$,$CB$ and $XY$.
Also our proiection saves simple ratio of segments paralell to $BC$.That why holds $\frac{SB}{SC}=\frac{DB}{DF}*\frac{EF}{EC}$


[quote="RDeepMath91"]Let $x,y,z$ be nonnegative real numbers with $x^{2}+y^{2}+z^{2}= 1$.
Prove that $1 \leq \frac{z}{1+xy}+\frac{x}{1+yz}+\frac{y}{1+zx}\leq \sqrt{2}$.[/quote]
I have the prove for$\sum_{cyc}\frac{x}{1+yz}\geq1$
We have that the ineq is cyclic, so I can suppose that $z=0$. Then, we need to show $x+y\geq1$, where $x^{2}+y^{2}=1$.
$x+y\geq 1$ or $(x+y)^{2}\geq 1^{2}$ or $2xy\geq 0$ which is true, with equality if $x=0, y=1$ or $x=1, y=0$. So we have equality if two of our numbers are $0$ and the third is $1$.
Now, if$x,y,z\ne 0\Rightarrow  \sum_{cyc}\frac{x}{1+yz}=\sum_{cyc}\frac{x^{4}}{x^{3}+x^{3}yz}\geq\frac{(x^{2}+y^{2}+z^{2})^{2}}{x^{3}+y^{3}+z^{3}+3xyz(x^{2}+y^{2}+z^{2})}$ so we need to prove that $1\geq x^{3}+y^{3}+z^{3}+3xyz=(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)+3xyz+xyz=(x+y+z)(1-xy-yz-zx)+4xyz$ or $1+\sum_{cyc}x^{2}y+\sum_{cyc}x^{2}z\geq xyz+x+y+z$
But $\sum_{cyc}x^{2}y\geq 3xyz>xyz$ and $x(1-x)+y(1-y)+z(1-z)>0$ equivalent to $1>x+y+z$, so $1+\sum_{cyc}x^{2}y+\sum_{cyc}x^{2}z>xyz+x+y+z$ and we are done with the left ineq.Am I wrong somwhere?

[quote="Zamfirmihai"][quote="RDeepMath91"]Let $x,y,z$ be nonnegative real numbers with $x^{2}+y^{2}+z^{2}= 1$.
Prove that $1 \leq \frac{z}{1+xy}+\frac{x}{1+yz}+\frac{y}{1+zx}\leq \sqrt{2}$.[/quote]
I have the prove for$\sum_{cyc}\frac{x}{1+yz}\geq1$
[color=red][b]We have that the ineq is cyclic, so I can suppose that $z=0$.[/b][/color] Then, we need to show $x+y\geq1$, where $x^{2}+y^{2}=1$.
$x+y\geq 1$ or $(x+y)^{2}\geq 1^{2}$ or $2xy\geq 0$ which is true, with equality if $x=0, y=1$ or $x=1, y=0$. So we have equality if two of our numbers are $0$ and the third is $1$.
Now, if$x,y,z\ne 0\Rightarrow \sum_{cyc}\frac{x}{1+yz}=\sum_{cyc}\frac{x^{4}}{x^{3}+x^{3}yz}\geq\frac{(x^{2}+y^{2}+z^{2})^{2}}{x^{3}+y^{3}+z^{3}+3xyz(x^{2}+y^{2}+z^{2})}$ so we need to prove that $1\geq x^{3}+y^{3}+z^{3}+3xyz=(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)+3xyz+xyz=(x+y+z)(1-xy-yz-zx)+4xyz$ or $1+\sum_{cyc}x^{2}y+\sum_{cyc}x^{2}z\geq xyz+x+y+z$
But $\sum_{cyc}x^{2}y\geq 3xyz>xyz$ and $x(1-x)+y(1-y)+z(1-z)>0$ equivalent to $1>x+y+z$, so $1+\sum_{cyc}x^{2}y+\sum_{cyc}x^{2}z>xyz+x+y+z$ and we are done with the left ineq.Am I wrong somwhere?[/quote]
Yes, in the red part, unless you are willing to supply a general theorem that applies to this case (I think a vietnamese user is trying to develop one for symmetrics). Also, this very inequality has been posted before but on an exclusive forum, so i'll post a hint here.
[hide="hint"]Isolated fudging[/hide]

[quote="Zamfirmihai"]$x(1-x)+y(1-y)+z(1-z)>0$ equivalent to $1>x+y+z$[/quote]

I think it should be $x(1-x)+y(1-y)+z(1-z)>0 \leftrightarrow x+y+z>x^{2}+y^{2}+z^{2}=1$

What is isolated fudging?
Thx..

In the red part I think it`s ok, because I took a case. But $1>x+y+z$ so I was wrong. I was really tired :D

[quote="RDeepMath91"]Let $x,y,z$ be nonnegative real numbers with $x^{2}+y^{2}+z^{2}= 1$.
Prove that $1 \leq \frac{z}{1+xy}+\frac{x}{1+yz}+\frac{y}{1+zx}\leq \sqrt{2}$.[/quote]
huhm... There are a solution is simpler than  Zamfirmihai's proof :wink:
By UCT we can find the following inequality easily...
\[\frac{x}{1+yz}\le \frac{\sqrt{2}x}{x+y+z}\iff (y+z-x)^{2}+2y^{2}z^{2}\]
Equaly occurs : $(x,y,z)=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)$
And
\[1 \leq \frac{z}{1+xy}+\frac{x}{1+yz}+\frac{y}{1+zx}\]
It was solved by the following inequality: $x+xyz\le 1$
We are done.

[quote="zaizai-hoang"]

By UCT we can find the following inequality easily...
\[\frac{x}{1+yz}\le \frac{\sqrt{2}x}{x+y+z}\iff (y+z-x)^{2}+2y^{2}z^{2}\]
[/quote]
General, we need to feed $p$ such that
\[\frac x{1+yz}\leq\frac{\sqrt2 x^{p}}{x^{p}+y^{p}+z^{p}}.\]
How did you know $p=1$?


Let the  polynomial  $p$ be defined by

$ p(x) = 2x^{3}-3x^{2}+4x-5$

Then
$p(1)=$ ,
$p(2)=$ , and
$p(-1)=$ .

Suggestions?

Edit: Sorry first one was a mis-paste.  hehe.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=145200

 :mad:

This seems a little easy, eh?

$p(x) = 2x^{3}-3x^{2}+4x-5$

[hide]I'm not going to solve any of the things you wanted outright but, for example, say you were asked to find $p(0)$.  All you would do is plug $0$ in where you see $x$

$p(x) = 2x^{3}-3x^{2}+4x-5$
$p(0) = 2(0)^{3}-3(0)^{2}+4(0)-5$
$\boxed{p(0) =-5}$[/hide]


[quote="Interval"]Two congruent 10-by-10 squares overlap so that a vertex of one is the center of the other.  What is the largest possible overlapping area?[/quote]
[hide]
It can be shown that no matter how the overlap is shaped it is always $25$

$25$[/hide]

[quote="lotrgreengrapes7926"][quote="mathgeniuse^ln(x)"][hide]
It can be shown that no matter how the overlap is shaped it is always $25$

$25$[/hide][/quote]
How would you prove that?[/quote]

this is a cute problem, basically, let the two squares intersect at X and Y, the center/vertex point be O, take a perpendicular from O to the sides containing X and Y of the fixed square, to A, and B, it is easy to show that AOX is congruent to BOY

Yep, that is the way.

And you can't just rotate it to show that it will fit, you have to prove those triangles are congruent.


vector
induction
linear algebra
linear algebra unsolved

Let V be a finite dimensional vector space over an infinite field F.Prove that if
S_1,...,S_k are subspaces of V of equal dimension, then there is a subspace T
for which V = direct sum ( S_i + T ) for all i = 1,...,k.

Proof by induction on the co-dimension.

If $ \dim S_i\equal{}\dim V$ we can set $ T\equal{}\{\bar{0}\}$. So assume $ \dim S_i<\dim V$ and let $ \bar{v}\in V\setminus\bigcup_{i\equal{}1}^kS_i$ (here we need that $ V$ is defined over an infinite field). Then we have $ <\bar{v}>\cap S_i\equal{}\{\bar{0}\}$ and now apply the induction hypothesis to the subspace $ R_i\equal{}S_i\oplus <\bar{v}>$, i.e. $ \exists T'\leq V$ such that $ \forall i\equal{}1,\ldots,k: V\equal{}R_i\oplus T'$ from which we conclude that $ T\equal{}T'\oplus<\bar{v}>$ will do the job


function
algebra
polynomial
vector
calculus
derivative
real analysis

Decide if the following asserts are true:
$ K \equal{} \mathbb{R}$, $ K \equal{} \mathbb{Q}$
If $ f : K \times K \rightarrow K$ is a function such that for each $ x_{0} \in K$, $ f(x_{0},y) : K \rightarrow K$ is a polynomial in $ K[y]$, and for every $ y_{0} \in K$, $ f(x,y_{0}) : K \rightarrow K$ is a polynomial in $ K[x]$, then $ f$ is a polynomial in $ K[x,y]$.

If $ K\equal{}\mathbb R$, the answer is true:

Let $ V_{\mathbb R}$ be the vector of functions $ f: \mathbb R^2\to \mathbb R$ to be considered. Since every $ f\in V$ has partial derivatives (w.r.t. $ x$ and $ y$, resp.) of every order, we get $ V_{\mathbb R}\subset C^\infty(\mathbb R^2)$.

1 - If $ g\in V_{\mathbb R}$ vanishes on a set $ I\times J\subset\mathbb R^2$ with infinite subsets $ I,J\subset \mathbb R$, then $ g$ vanishes identically.

2 - The union of the closed sets \[ F_n: \equal{}\{(x,y)\in\mathbb R^2\mid \partial_x^mf(x,y)\equal{}\partial_y^mf(x,y)\equal{}0 \,\,\forall m\ge n\}\quad (n\in\mathbb N)\] covers $ \mathbb R^2$, hence there are a natural number $ n_0>0$ and open non-void intervals $ I,J\subset\mathbb R$ with  $ I\times J\subset F_{n_0}$, by Baire. Then apply the preliminary remark (1) to $ g\equal{}\partial_x^{n_0}\partial_y^{n_0}f$ to see that $ g\equiv 0$. This shows that $ f\in\mathbb R[x,y]$.

Now the question: Does the approach above generalize to $ K\equal{}\mathbb Q$? ;)

I have a completely different solution, at first a had the same idea than you but i remember that I had a problem with this and changed my mind.... For the second case i have a counterexample, I'll post both things later...


calculus
integration
algebra
polynomial
function
inequalities
complex numbers

A nice problem from me:  :)

Let $a_1,a_2,\ldots,a_k; c_1,\ldots, c_k \in \mathbb{Z}$ be integers with the property that, for every positive integer $n$, the number
\[
b_1a_1^n + b_2a_2^n + \cdots + b_ka_k^n
\]
is a perfect square. Prove that there exist integers $c_1,\ldots,c_k,d_1,\ldots,d_k$ such that
\[
b_1a_1^n + b_2a_2^n + \cdots + b_ka_k^n = \big( d_1c_1^n + d_2c_2^n + \cdots + d_kc_k^n
 \big)^2
\]
for all $n$.

--Veselin

Vess, HAVE YOU SOLVED THIS ONE? I'm struggling for more than a year to solve it and I did not manage. Please, show me the solution.

Harazi, this special case of Pisot's square root conjecture is actually much easier than you suspect (here the $a_i$ are integers, not arbitrary complex numbers for which the power sums are integral; what we gain from this apparently weak assumption is that there is a unique $a_i$ with maximal absolute value, i.e. we are in the case known as the [i]dominant root case[/i].) I will present a proof below. But first, for the others, I'd like to give a brief introduction to this beautiful problem.

Many times on this forum the following problem has appeared: if $f \in \mathbb{Z}[x]$ is an integer polynomial which is a perfect square on certain infinite sets with some properties, then $f$ is the square of another integer polynomial. In this spirit, Pisot conjectured the following:

[b]Pisot's $d$th Root Conjecture. [/b] If a power sum
\[
a(n) = \sum_{1 \leq i \leq k} P_i(n)\alpha_i^n, \quad P_i \in \mathbb{C}[x], \alpha_i \in \mathbb{C}
\]
is the $d$th power of a rational number for $n=0,1,2,\ldots$, then $a(n)$ is the $d$th power of a function of the same form.

Equivalent versions of Pisot's conjecture are the following:

[b]Alternative Version 1. [/b] If $a(n) \in \mathbb{Q}$ are such that the function $\sum_{n \geq 1} a(n)x^n$ is rational, and if $a(n)$ is a perfect $d$th power for each $n$, then one can choose $d$th roots $b(n)$ of $a(n)$ such that $\sum_{n \geq 1}b(n)x^n$ is also a rational function.

[b]Alternative Version 2. [/b] Obviously, if $\{b(n)\}$ is a linear recursive sequence of rationals, then $\{b(n)^d\}$ is also a linear recursive sequence. The conjecture is that the converse also holds: if $\{a(n)\}$ is a linear recursive sequence of rationals and $a(n)$ is a perfect $d$th power for all $n$, then we may choose a $d$th root $b(n)$ of $a(n)$ such that $\{b(n)\}$ is also a linear recursive sequence.

There are analogous statements in fields more general than $\mathbb{Q}$, but I won't plunge into details here.

The Pisot conjecture turns out to be surprisingly intractable. For many years it has remained unsolved, except in some trivial cases (such as the one that I'll prove below), and only recently, in 2000, it was solved completely by Umberto Zannier ([i]Annals of Mathematics, 151 (2000), 375--383[/i]); the proof, which uses Galois theory and Chebotarev's density theorem, seems quite complicated  :lol: .

While the general case is absolutely monstrous, there is a particularly simple special case which is relatively easy to resolve:

[b]THE DOMINANT ROOT CASE.[/b]

[b]Theorem. [/b] [i]Assume that there is a unique $\alpha_i$ for which $|\alpha_i|$ is maximal. If $a(n)$ is a $d$th root for all sufficiently large $n$, then it is the $d$th power of another exponential polynomial (i.e. a function of the same form). [/i]

The work of Perelli and Zannier allows us to assume without loss of generality that the $P_i$ are constant polynomials (which is the case in our particular problem). But, in fact, we will only need that $P_{i_0} \equiv \text{const}$ for the unique $i_0$ for which $|\alpha_{i_0}|$ is maximal. So, I will here present a [b]complete[/b] (self-contained, i.e. not using any advanced results) elementary proof of the following special case of the above theorem:

[b]Claim. [/b] Let $c(n)$ be a generalized power sum with poles of absolute value less than $1$ (that is, $c(n) = \sum Q_i(n)\beta_i^n$ for some polynomials $Q_i$, where $|\beta_i|<1$ for all $i$). If, for all sufficiently large positive integers $n$, $a(n):= 1 + c(n)$ is the $d$th power of a rational number, then $a(n)$ is identically the $d$th power of another generalized power sum.

But let us first see why this implies our initial problem; after all, we may not assume that $\max |a_i|$ in our problem is attained for a unique $i$. Sure, as the $a_i$ are (different) reals, there is a unique $i_+$ with $a_{i_+}$ maximal and a unique $i_-$ with $a_{i_-}$ minimal, but it may happen that $a_{i_-} = - a_{i_+}$. However, we may escape this problem by looking at the even values of $n$ only; that is, replace the constants $a_i$ by $a_i^2$; then, after 'merging' the resulting equal values into a single constant, it follows from the above claim that (if $a(n)$ is the power series in the initial problem) we have $a(2n)=b(n)^2$ for all $n$, where $b(n)$ is some other generalized power sum; analoguously, $a(2n+1)=c(n)^2$ for all $n$, and it is easily seen after comparing coefficients that $a(n)$ is identically the square of another power series.

Now --- the promised

[b]Proof of the Claim. (!!!!) [/b]  ;)  ;)  ;)  We have $a(n) = 1 +c(n)$, where $c(n)$ is a power sum with poles of absolute value less than $1$. Hence, by the expansion of $(1+x)^{1/d}$, we may write, for sufficiently large $n$,
\[
b(n):= a(n)^{1/d} = (1+c(n))^{1/d} = 1 + \binom{1/d}{1}c(n)  + \binom{1/d}{2} c(n)^2 + \cdots + \binom{1/d}{\ell} c(n)^{\ell} + R_{\ell}(n) = C_{\ell}(n) + R_{\ell}(n),
\]
where the number of terms in the power sums $C_{\ell}(n)$ increases linearly in $\ell$, and the remainder satisfies $R_{\ell}(n) = O(c(n)^{\ell + 1})$.

Now, of course, the power sums $C_{\ell}(n)$ satisfy some integral linear recurrence relation whose size is bounded by a linear function of $\ell$. That is, if $E$ denotes the 'shifting' operator $E:f(n) \to f(n+1)$, there exist polynomials $P_{\ell} \in \mathbb{Z}[x]$ of degree bounded by $c\ell$ for some constant $c$ (and thus totally independent of $n$) such that the operator $P_{\ell}(E)$ 'annihilates' $C_{\ell}(n)$. By the above equation, we have
\[ (*)     \quad
P_{\ell}(E)(b(n)) = P_{\ell}(E)(R_{\ell}(n)).
\]
We now claim that if (*) holds for all $\ell,n$ and if $b(n)$ (and hence $R_{\ell}(n)$) is rational all (sufficiently large) $n$, then $P_{\ell}(E)$ [i]annihilates[/i] $b(n)$ for all sufficiently large $\ell,n$; this will obviously imply the claim, because then the $b(n)$ satisfy a linear recurrence relation.

Indeed, it is easily seen that the right hand side of (*) is $O(\delta^{\ell n})$ for some constant $\delta \in (0,1)$. On the other hand, since $P_{\ell} \in \mathbb{Z}[x]$ and $b(n) \in \mathbb{Q}$ for all $n>n_0$, it also is a rational number. Thus, for every $n > n_0$, either this rational number is $0$, or its denominator exceeds $\lambda^{\ell n}$ for some constant $\lambda > 1$. We now show that this is a contradiction with the left hand side, unless $P_{\ell}(E)(b(n)) = 0$ for all sufficiently large $\ell, n$.

Because the $a(n)$ satisfy a linear recurrence relation with integer coefficients, it is clear that they all belong to a finite extension of $\mathbb{Z}$, and hence there is a non-zero integer $m$ such that $m^na(n) \in \mathbb{Z}$ for each $n$. As $a(n)^d = b(n)$ for $n>n_0$, it is clear that $m^nb(n) \in \mathbb{Z}$. As $P_{\ell} \in \mathbb{Z}[x]$ is an integer polynomial of degree $< c\ell$, one easily sees that $P_{\ell}(E)(b(n))$ is a rational number with denominator dividing $m^{n+\deg{P_{\ell}}} < m^{n+c\ell}$. By the previous paragraph, either the rational number $P_{\ell}(E)(b(n)) = P_{\ell}(E)(R_{\ell}(n))$ is $0$, or else we have $m^{n+c\ell} > \lambda^{n\ell}$. Because $c,m,\lambda$ are fixed, it is clear that the latter inequality cannot hold if both $n$ and $\ell$ are sufficiently large. We thus conclude that $P_{\ell}(E)(b(n)) = 0$ for all sufficiently large $n$, as claimed. [b]Q.E.D.[/b]

PHEW!

--Vesselin

P. S.  I hope that I did not write this in vain  :lol: ; please read this proof!  :D I hope that an interesting discussion will spring from this thread.  :P

[quote="vess"]actually, it has been proven that a non-square polynomial can take squares for only [i]finitely many[/i] integers)[/quote]

 :?
What do you mean? What about $P(x)=x$ or $P(x) = x^3$...?

Anyway, nice post Vess.

Pierre.

Yes, sorry,  I think that it is proven in an 'essential case,' that is, with the exception of some self-evident cases (for example, I think it's shown if $P$ if irreducible of degree $>1$).

So now, of course, the assertion labeled 'Theorem' does not hold as stated, for the reasons you said. But I think that it holds in the non-trivial cases. Anyway, I haven't seen any papers in which this result is proven, I've just seen the following sentense from Zannier's paper in the [i]Annals of Mathematics[/i]: 'We remark that [...] contains proofs of an analogous statement assuming the much weaker fact that $a(n)$ is a $d$th perfect power for infinitely many $n$. However, that method requires (at least) the existence of a dominant $\beta_i$.' So --- I now realize this --- perhaps there are other non-trivial conditions to be imposed (other than the root domination) ...

Thanks for the remark --- I'll edit the post accordingly. By the way, I now realize that the proof I wrote cannot be modified as I thought, i.e. it essentially depends on the assumption that $a(n)$ is a $d$th root for [b]all[/b] large $n$ (and not just infinitely many).

--Vesselin


[quote="bob123"][quote="G-UNIT"]Pick a 3-digit number, and divide it by the sum of its digits. In case of which number will this quotient be the greatest, and the smallest?[/quote]

[hide]For greatest, any multiple of 100...
For smallest, 199[/hide][/quote]
i agree

[quote="bob123"][quote="4everwise"]You're agreeing with your own answer.  ;)[/quote]

I think that's how he gets so many posts... all his posts that I have seen say "I agree" or some variation of it, except for the problems that he posts[/quote]

Hey, that's a good idea!

I agree  :lol:

[quote="G-UNIT"]No, the first time "math92" was agreeing to bob123's solution.  The second time he was agreeing with "4everwise", that Bobs's solution is logical and deserves a "10/10".  But nontheless,

I agree, LOL[/quote]

I agree  :) 370 posts

[quote="bob123"][quote="4everwise"]You're agreeing with your own answer.  ;)[/quote]

I think that's how he gets so many posts... all his posts that I have seen say "I agree" or some variation of it, except for the problems that he posts[/quote]

I agree with myself :D

i just thought of something wonderful to say that you won't expect.....

[hide]i agree!! [/hide]

this is getting really fun  :rotfl:   :rotfl:


[b][color=darkblue]Prove that: $ x^n.\sqrt {1 \minus{} x} < \frac {1}{\sqrt {2ne}}$, $ \forall x\in (0;1)$ and $ n\in Z^\plus{}$.[/color][/b]

[quote="micf"][quote="quykhtn-qa1"]
By AM-GM[/quote]

Could you show me that step?[/quote]
Ok,my friend :)
We have:$ LHS^2\equal{} \frac{1}{2n} x^{2n} (2n\minus{}2nx) \le \frac{1}{2n} (\frac{2nx\plus{}2n\minus{}2nx}{2n\plus{}1})^{2n\plus{}1} \equal{}\frac{1}{2n}(\frac{2n}{2n\plus{}1})^{2n\plus{}1}$
I think now, you understood :)

[quote="quykhtn-qa1"][quote="thanhnam2902"][b][color=darkblue]Prove that: $ x^n.\sqrt {1 \minus{} x} < \frac {1}{\sqrt {2ne}}$, $ \forall x\in (0;1)$ and $ n\in Z^ \plus{}$.[/color][/b][/quote]
Nice inequality :)
ByAM-GM,we have:$ LHS^2 \le \frac {1}{2n}(\frac {2n}{2n \plus{} 1})^{2n \plus{} 1}$
We will must prove that:
$ < \equal{} > \frac {1}{2n}(\frac {2n}{2n \plus{} 1})^{2n \plus{} 1} < \frac {1}{2ne}$
$ < \equal{} > e < (\frac {2n \plus{} 1}{2n})^{2n \plus{} 1}$
$ < \equal{} > \frac {1}{2n \plus{} 1} < ln(2n \plus{} 1) \minus{} ln 2n$
Put $ f(x) \equal{} ln x$ .By Lagrange 'theorum we have$ a\in (2n,2n \plus{} 1)$ such that:
$ f(2n \plus{} 1) \minus{} f(2n) \equal{} f '(a)$
$ ln(2n \plus{} 1) \minus{} ln 2n \equal{} \frac {1}{a}$
$ \equal{} > \frac {1}{2n \plus{} 1} < \frac {1}{a} < \frac {1}{2n}$
$ \equal{} > \frac {1}{2n \plus{} 1} < ln(2n \plus{} 1) \minus{} ln 2n$ (Q.E.D) :)[/quote]

[b]Very good.....[/b]


I am stuck on this problem

0=18000-4.5x+6(x^2+950^2)^.5

If anyone could help me I would apreciative it.  Thanks

Bring the squareroot to one side. Square both sides and then smash and bash with quadratic formula.  :wink:


Find the of factors of $ 18900$ having exactly $ 4$ factors [hide="themselves."] including $ 1$ and itself.[/hide]

$ 18900 \equal{} 2^2 \cdot 3^3 \cdot 5^2 \cdot 7$.  A number has four factors when it's the product of two distinct primes or when its a prime cubed.  So it should be 4 choose 2, and one more for 27, giving a total of 7.

edit: probability?

Correct.
Thank You.

-$ Rajiv$ $ Patki.$


See all the problems from 5-th Kyiv math festival [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=506789#p506789]here[/url]

Let $x,y>0$ and $xy\ge1.$ Prove that $x^3+y^3+4xy\ge x^2+y^2+x+y+2.$

Let $x,y>0$ and $xy\ge1.$ Prove that $2(x^3+y^3+xy+x+y)\ge5(x^2+y^2).$

For the first we have : $x^3+y^3=(x+y)(x^2-xy+y^2)\geq 2x^2-2xy+2y^2$
So it suffices to prove that $x^2+y^2+2xy\geq x+y+2$ .
But we have  $2xy\geq 2$  and  $x^2+y^2\geq\frac{(x+y)^2}{2}\geq x+y$

[quote="rogue"]

Let $x,y>0$ and $xy\ge1.$ Prove that $x^3+y^3+4xy\ge x^2+y^2+x+y+2.$

[/quote]

Inspire from Arqady's solution. For the first one:

$x^3+y^3+4xy-x^2-y^2-x-y-2=(x+y-2)(x^2-xy+y^2+x+y+1) \geq 0$

[quote="rogue"]
Let $ x,y > 0$ and $ xy\ge1.$ Prove that $ x^3 \plus{} y^3 \plus{} 4xy\ge x^2 \plus{} y^2 \plus{} x \plus{} y \plus{} 2.$[/quote]

Let $ x \plus{} y \equal{} s,\ xy \equal{} t$, we have $ x > 0,\ y > 0\Longleftrightarrow s > 0,\ t > 0,\ s^2\geq 4t$

Since $ t\geq 1$, the condition becomes $ s > 0,\ t\geq 1,\ s^2\geq 4\Longleftrightarrow s\geq 2,\ t\geq 1.$

$ \therefore x^3 \plus{} y^3 \plus{} 4xy \minus{} (x^2 \plus{} y^2 \plus{} x \plus{} y \plus{} 2)$
$ \equal{} s^3 \minus{} 3st \plus{} 4t \minus{} (s^2 \minus{} 2t \plus{} s \plus{} 2)$
$ \equal{} 3t(2 \minus{} s) \plus{} s^3 \minus{} s^2 \minus{} s \minus{} 2$
$ \geq 3(2 \minus{} s) \plus{} s^3 \minus{} s^2 \minus{} s \minus{} 2\ \because t\geq 1$
$ \equal{} s^3 \minus{} s^2 \minus{} 4s \plus{} 4$
$ \equal{} (s \minus{} 1)(s^2 \minus{} 4)\geq 0.$

Analogously,

[quote="rogue"]

Let $ x,y > 0$ and $ xy\ge1.$ Prove that $ 2(x^3 \plus{} y^3 \plus{} xy \plus{} x \plus{} y)\ge5(x^2 \plus{} y^2).$[/quote]

$ 2(x^3 \plus{} y^3 \plus{} xy \plus{} x \plus{} y) \minus{}5(x^2 \plus{} y^2)$
$ \equal{} 6t(2 \minus{} s) \plus{} 2s^3 \minus{} 5s^2 \plus{} 2s$
$ \geq 2s^3 \minus{} 5s^2 \minus{} 4s \plus{} 12$
$ \equal{} (s \minus{} 2)^2(2s \plus{} 3)\geq 0$ :lol:

not so dantern.it suffices to let
$x+y=p,xy=q$
and use derivatives.


algebra
binomial theorem
complex numbers

Try to find a formula for factoring $x^{n}\pm y^{n}$.  For example, $x^{2}-y^{2}=(x+y)(x-y)$ (everybody knows that) and $x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2})$ (everybody knows that too).  But how do you factor the general thing?
HINT:For $x^{n}+y^{n}$ one factor is $x+y$.

[quote="mgao"]HINT:For $x^{n}+y^{n}$ one factor is $x+y$.[/quote]
So is $x+y$ a factor of $x^{2}+y^{2}$?

[quote="mgao"]all even n where $n>2$.[/quote]

Plainly false.  For example, $(2+3) \not | (2^{4}+3^{4})$

Seems too easy for Intermediate.

a^n - b^n is divisible by a-b and a^(2n+1) + b^(2n+1) is divisible by a+b.

ok factoring $x^{n}-y^{n}$ is easy
the binomial theorem says that it's $x^{n}-y^{n}=(x-y)(x^{n-1}+n^{n-2}y+...+xy^{n-2}+y^{n-1})$, I believe.

For factoring $x^{n}+y^{n}$, it's almost the same things except there's a factor of $i$ thrown in a few places. But I dont know jack about complex numbers so I wont try

[quote="sapphyre571"]For factoring $x^{n}+y^{n}$, it's almost the same things except there's a factor of $i$ thrown in a few places. But I dont know jack about complex numbers so I wont try[/quote]

$x^{n}+y^{n}$ is in general not factorable with integer coefficients.  If you wish to factor over complex coefficients,

$x^{n}-y^{n}= \prod_{k=0}^{n-1}\left( x-e^{ \frac{ 2k \pi i}{n}}y \right)$


quadratics
number theory unsolved
number theory

Find all non-negative integers $ m$ and $ n$ that satisfy the equation:
\[ 107^{56}(m^2\minus{}1)\minus{}2m\plus{}5\equal{}3\binom{113^{114}}{n}\]
(If $ n$ and $ r$ are non-negative integers satisfying $ r\le n$, then $ \binom{n}{r}\equal{}\frac{n}{r!(n\minus{}r)!}$ and $ \binom{n}{r}\equal{}0$ if $ r>n$.)

Something odd with this formulation. The version I remember was
$$
107^{56}(m^2-1)+2m+3 = \binom{113^{114}}{n}.
$$
I will assume this is the correct one, and will provide a proof. Let $n\in\{1,2,\dots,113^{114}-1\}$. Then, $113\mid  \binom{113^{114}}{n}$. Hence,
$113\mid 107^{56}(m^2-1)-2m+5$. Now, $107^{56}\equiv 6^{56}\pmod{113}$. Note that by Fermat's theorem, $6^{56}\in\{-1,1\}\pmod{112}$, with being one iff $6$ is a quadratic residue modulo  $113$, and $-1$ otherwise. I now show $6$ is not a quadratic  residue modulo $113$. Observe that, $113\equiv 1\pmod{8}$, and $113\equiv -1\pmod{6}$. Thus, $2$ and $-1$ are both quadratic residues, whereas $-3$ is not. Thus, $6$ is not, either. Therefore, $6^{56}\equiv -1\pmod{113}$. With this, we conclude that $-m^2+1+2m+3\equiv 0\pmod{113}\iff (m-1)^2\equiv 5\pmod{113}$. Now using quadratic reciprocity, it follows that $5$ is not a quadratic residue modulo $113$. Hence, this is impossible.

The remaining possibilities are $n\in\{0,113^{114}\}$. Both cases are easily studied to show to yield no solutions.


modular arithmetic
number theory
relatively prime

What is the remainder when 10! is divided by 13?

I solved this problem by actually doing the division and got the correct answer. I was wondering if there is any quicker way to do this problem?  :lol:

\[ 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \pmod {13}\]

Notice that $ 2 \cdot 7 \equal{} 14 \equiv 1 \pmod {13}$, so the product is equivalent to \[ 10 \cdot 9 \cdot 8 \cdot 6 \cdot 5 \cdot 4 \cdot 3\] $ 3 \cdot 5 \equal{} 15$ can be replaced with 2: \[ 10 \cdot 9 \cdot 8 \cdot 6 \cdot 4 \cdot 2\] $ 2 \cdot 8 \equal{} 16$ can be replaced with 3: \[ 10 \cdot 9 \cdot 6 \cdot 4 \cdot 3\] $ 4 \cdot 10 \equal{} 40$ can be replaced with 1: \[ 9 \cdot 6 \cdot 3\] $ 9 \cdot 6 \equal{} 54$ can be replaced with 2: \[ 3 \cdot 2 \equiv \boxed 6 \pmod {13}\]

[hide] Look up Wilson's Theorem. It states that for a prime p, $ (p\minus{}1)!\equiv\minus{}1\pmod{p}$. As 13 is prime, we have $ 12!\equiv12\pmod{13}$. As 12 and 13 are relatively prime, we can divide by 12 to get that $ 11!\equiv1\pmod{13}$. We can now add multiples of 13 until the RHS is divisible by 11. We find that 1+65=66 is divisible by 11. Hence, $ 10!\equiv6\pmod{11}$.[/hide]

Or, you could discover 10! and then calculate.
But of course, with large numbers, using mod saves a lot of time.


geometry
trigonometry
LaTeX
geometric transformation
reflection

[geogebra]0f6e2bdbcfeeb09cde116fe8039481acee1c6233[/geogebra]
Given AB//DC, AB=DB and AC=BC. Find angle ABD.

[hide]If $ a: \equal{}AC\equal{}BC$, then $ AB\equal{}BD\equal{}a\sqrt{2}$.

If $ x: \equal{}CD$, then from $ \triangle BCD$: $ x^2\plus{}a^2\minus{}2ax\cos 135^\circ\equal{}2a^2\iff x^2\plus{}ax\sqrt{2}\minus{}a^2\equal{}0$, which yields $ x\equal{}a{\sqrt{6}\minus{}\sqrt{2}\over 2}$

If $ b: \equal{}AD$, then from $ \triangle ACD$: $ b^2\equal{}x^2\plus{}a^2\minus{}2ax\cos 45^\circ\equal{}a^2(4\minus{}2\sqrt{3})\iff b\equal{}a(\sqrt{3}\minus{}1)$

If $ \phi: \equal{}\angle ABD$, then from $ \triangle ABD$: $ \cos\phi\equal{}{2(a\sqrt{2})^2\minus{}b^2\over 2(a\sqrt{2})^2}\equal{}{\sqrt{3}\over 2}\iff\phi\equal{}30^\circ$[/hide]

[hide="Alternatively"]
Since triangles $ ACB$ and $ ADB$ have the same base and height, they have the same area.  So we have:
$ [ADB] \equal{} \frac{1}{2} (\sqrt{2}a)^2 \sin\phi \equal{} \frac{1}{2} a^2$
$ \sin\phi \equal{} \frac{1}{2} \Rightarrow \phi \equal{} 30^\circ$
[/hide]

[quote="kflam"]
Given AB//DC, AB=DB and AC=BC. Find angle ABD.[/quote]

Use $ \text{\LaTeX}$, parallel command is just \parallel:  $ \parallel$

Reflect C and D over AB to C' and D'. Then ACBC' is a square, so $ DD'\equal{}CC'\equal{}AB\equal{}BD\equal{}BD'$, so BDD' is equilateral, hence $ \angle ABD\equal{}30$.


Find the range of the function, $f(x) = \frac{2x}{\sqrt{x^{2}+x+1}}$

f(x) is undefined when x^2+x+1=0 or x^2+x+1<0.

The latter is impossible, so we look at the first one.

Is there really anyway x^2+x+1=0?
No, unless you include complex numbers.

Hence, (-inf,inf) is the domain.

Now, for the range:

We rewrite f(x) as (4x^2/(x^2+x+1))^(1/2)

Divide x^2 from the top and bottom.

(4/(1+1/x+1/x^2))^(1/2)

As x goes to infinity in positive direction, f(x) goes to 2; in the negative, -2.
Because (4/(1+1/x+1/x^2))^(1/2) is clearly greatest/less at infinities, range is:

(-2,2)

Actually, graphing it, we see that $\lim_{x\to-\infty}f(x)=-2^{-}$, so $-2$ is not the minimum.

So what you're saying is the range of f is $-2.3 \leq y \leq 2.0$?

[quote="lotrgreengrapes7926"][quote="prime_lover"]The limit of f(x) as x approaches $-\infty$ is -2...the function has horizontal asymptotes at $y = \pm 2$[/quote]Yes, but that value is not the minimum. So this statement:[quote="neelnal"]Because (4/(1+1/x+1/x^2))^(1/2) is clearly greatest/less at infinities[/quote] is not true.
I'm not sure how to prove it but... [hide] the minimum is at $x=-2$ giving $f(x)=-\frac{4}{\sqrt3}$ [/hide][/quote]

oh yea true.

anytime for a negative number when

abs(1/x) > abs(1/x^2), there is a value lower than -2.

WAAAAAAA

I wanna use calculus.

EDIT: since i can't post twice within 1 hour, the range is wat prime_lover said in the last post. -2.3 inclusive to 2 exclusive

[quote="jli"][quote="prime_lover"]So what is the range?[/quote]

its the possible y values.

to neehal: using calculus would be really hard because of the fraction and the square root.[/quote]

Not really. I would do it here, but then I would get banned lol.

You use the quotient rule or the chain rule.


function
induction
calculus
derivative
algebra unsolved
algebra

Find all the functions $ f: N^* \mapsto N^*$ that satisfy $ f(2x \plus{} 3y) \equal{} 2f(x) \plus{} 3f(y) \plus{} 4$ for any naturals $ x,y\geq 1$.

No solutions? There should be a functions expert to solve this...

Consider $ g(x)\equal{}f(x)\plus{}1$

$ g(2x\plus{}3y)\equal{}2g(x)\plus{}3g(y)$

x=y gives  $ g(5x)\equal{}5g(x)$

Now, let $ f(1)\equal{}a$

$ g(25)\equal{}25a$ but we also have $ g(25)\equal{}f(2*2\plus{}3*7)\equal{}2g(2)\plus{}3g(7)\equal{}2g(2)\plus{}3(2g(2)\plus{}3g(1))\equal{}8g(2)\plus{}9a$

Hence, $ 25a\equal{}8g(2)\plus{}9a$ so $ g(2)\equal{}2a$

$ g(25)\equal{}g(2*8\plus{}3*3)\equal{}2g(8)\plus{}3g(3)\equal{}2(2g(1)\plus{}3g(2))\plus{}3g(3)\equal{}16a\plus{}3g(3)$

Hence, $ 25a\equal{}16a\plus{}3g(3)$ so $ g(3)\equal{}3a$

$ g(15)\equal{}5g(3)\equal{}15a$ but we also have $ g(15)\equal{}g(2*6\plus{}3*1)\equal{}2g(6)\plus{}3g(1)\equal{}2g(6)\plus{}3a$

Hence, $ f(6)\equal{}6a$.

Now use the chicken mcnugget theorem...

From:
$ g(1)\equal{}1a$
$ g(2)\equal{}2a$
$ g(3)\equal{}3a$
and use induction to get:
$ g(k)\equal{}ka$

then $ f(x)\equal{}kf(1)\minus{}1$

Let $ f(1)\equal{}a, f(2)\equal{}b, f(3)\equal{}c, f(4)\equal{}d$ and $ f(6)\equal{}e$.

$ \bullet$ $ f(24)\equal{}f(2 \cdot 6\plus{}3 \cdot 4)\equal{}2f(6)\plus{}3f(4)\plus{}4\equal{}2e\plus{}3d\plus{}4$

$ f(24)\equal{}f(2 \cdot 3\plus{}3 \cdot 6)\equal{}2f(3)\plus{}3f(6)\plus{}4\equal{}2c\plus{}3e\plus{}4$

$ \Rightarrow e\plus{}2c\equal{}3d$

$ \bullet$ $ f(30)\equal{}f(2 \cdot 6\plus{}3 \cdot 6)\equal{}2f(6)\plus{}3f(6)\plus{}4\equal{}5e\plus{}4$

$ f(30)\equal{}f(2 \cdot 3\plus{}3 \cdot 8)\equal{}2f(3)\plus{}3f(6)\plus{}4\equal{}5e\plus{}4\equal{}$

$ \equal{}2f(3)\plus{}3(2f(1)\plus{}3f(2)\plus{}4)\plus{}4\equal{}2c\plus{}6a\plus{}9b\plus{}16$

$ \Rightarrow 5e\equal{}12\plus{}6a\plus{}9b\plus{}2c$

$ \bullet$ $ f(15)\equal{}f(2 \cdot 3 \plus{} 3\cdot 3)\equal{}2f(3)\plus{}3f(3)\plus{}4\equal{}5c\plus{}4$

$ f(15)\equal{}f(2 \cdot 6\plus{}3 \cdot 1)\equal{}2f(6)\plus{}3f(1)\plus{}4\equal{}2e\plus{}3a\plus{}4$

$ \Rightarrow 5c\equal{}2e\plus{}3a$

$ \bullet$ $ f(21)\equal{}f(2 \cdot 6\plus{}3 \cdot 3)\equal{}2f(6)\plus{}3f(3)\plus{}4\equal{}2e\plus{}3c\plus{}4$

$ f(21)\equal{}f(2 \cdot 3\plus{}3 \cdot 5)\equal{}2f(3)\plus{}3f(5)\plus{}4\equal{}$

$ \equal{}2f(3)\plus{}3(5f(1)\plus{}4)\plus{}4\equal{}2c\plus{}15a\plus{}16$

$ \Rightarrow 2e\plus{}c\equal{}15a\plus{}12$

$ \bullet$ $ 5c\equal{}2e\plus{}3a$

$ 2e\equal{}15a\plus{}12\minus{}c$

$ \Rightarrow 5c\equal{}15a\plus{}12\minus{}c\plus{}3a \Rightarrow 6c\equal{}18a\plus{}12 \Rightarrow c\equal{}3a\plus{}2$

$ \Rightarrow 2e\equal{}15a\plus{}12\minus{}3a\minus{}2\equal{}12a\plus{}10$

$ \Rightarrow e\equal{}6a\plus{}5$

$ \bullet$ $ 5e\equal{}12\plus{}6a\plus{}9b\plus{}2c$

$ \Rightarrow 30a\plus{}25\equal{}12\plus{}6a\plus{}9b\plus{}6a\plus{}4$

$ \Rightarrow 30a\plus{}25\equal{}16\plus{}12a\plus{}9b \Rightarrow 18a\plus{}9\equal{}9b$

$ \Rightarrow b\equal{}2a\plus{}1$

$ \bullet$ $ e\plus{}2c\equal{}3d$

$ \Rightarrow 6a\plus{}5\plus{}6a\plus{}4\equal{}3d \Rightarrow d\equal{}4a\plus{}3$

We observe that $ f(x)\equal{}(a\plus{}1)x\minus{}1$ satisfy the required conditions. We will use induction for the rest of the demonstration. Suppose that $ \forall n<N, f(n)\equal{}n(a\plus{}1)\minus{}1$. $ \exists x,y \in \mathbb{N}^\ast$ such that $ N\equal{}2x\plus{}3y$ when $ N \not\equal{} 1,2,3,4,6$.

$ f(N)\equal{}f(2x\plus{}3y)\equal{}2f(x)\plus{}3f(y)\plus{}4\equal{}2((a\plus{}1)x\minus{}1)\plus{}3((a\plus{}1)y\minus{}1)\plus{}4\equal{}$

$ \equal{}(a\plus{}1)(2x\plus{}3y)\minus{}1\equal{}(a\plus{}1)N\minus{}1$

Finally, $ f(x)\equal{}(a\plus{}1)x\minus{}1$ where $ a \in \mathbb{N}^{\ast}$ is arbitrary.

I forgot to prove $ f(4)\equal{}4a$ which can be done by considering f(15).

Induction is easier than any theorems at this point, so nvm. the chicken theorem (also called the postage stamp theorem).

We have $ 2(x \minus{} 3) \plus{} 3(y \plus{} 2) \equal{} 2x \plus{} 3y$, so:

$ 2f(x) \plus{} 3f(y) \plus{} 4 \equal{} 2f(x \minus{} 3) \plus{} 3f(y \plus{} 2) \plus{} 4$

Hence $ k \equal{} 2(f(x) \minus{} f(x \minus{} 3)) \equal{} 3(f(y \plus{} 2) \minus{} f(y))$.

Therefore $ f(x \plus{} 3) \minus{} f(x) \equal{} \frac {k}{2}$ and $ f(x \plus{} 2) \minus{} f(x) \equal{} \frac {k}{3}$.

From the above, we get $ f(2 \plus{} 3y) \equal{} f(2) \plus{} \frac {yk}{2}$.

But $ f(2 \plus{} 3y) \equal{} 2f(1) \plus{} 3f(y) \plus{} 4$.

Combining these, we get: $ f(y) \equal{} \frac {f(2) \minus{} 2f(1) \plus{} 4}{3} \plus{} (\frac {3k}{2})y$. So $ f(x) \equal{} c_{1} \plus{} c_{2}x$.

Sub this in to the given equation, and we get $ c_{1} \equal{} \minus{} 1$. So $ f(x) \equal{} cx \minus{} 1$, $ c > 1$.

You cannot differentiate because it is a natural function. Even if it was real valued, condition of continuity is not given, so you cannot differentiate.

What about the more general problem, given $ p, q, c \in \mathbb{N}^{*}$ find all functions $ f: \mathbb{N}^{*} \mapsto \mathbb{N}^{*}$ such that $ f(px\plus{}qy) \equal{} pf(x)\plus{}qf(y)\plus{}c$ for all $ x,y \in \mathbb{N}^{*}$? (the method from my post works, but is there a more elegant way?)

About rajat's approach, couldn't we perhaps argue that we can differentiate as there is some differentiable function $ g: \mathbb{R} \mapsto \mathbb{R}$ such that $ g(n)\equal{}f(n), \forall n$ (I also can't justify this offhand, but I think it's true)? Either way, the proof (s)he posted is incorrect where it states:

$ 2kx\plus{}3ky\plus{}c\equal{}2kx\plus{}2ky\plus{}4$

Which should be:

$ 2kx\plus{}3ky\plus{}c \equal{} 2(kx\plus{}c) \plus{} 3(ky\plus{}c) \plus{} 4$

(But this small problem aside, from here we get $ c\equal{}\minus{}1$ as wanted).

Differentiation is never valid for functions on discrete sets like $ \mathbb{N}$. The differentiable function $ g$ can do almost [i]anything[/i] in between integer values, so the derivative gives no information.

[quote="Fermat's Little Turtle"]What is the chicken mcnugget theorem, rofler?[/quote]
If you have two numbers, i.e. 7 and 11, the Chicken McNugget Theorem will show you the HIGHEST IMPOSSIBLE NUMBER you can make with the formula
$ ab\minus{}(a\plus{}b)$
hope this helps


In a small town of 100 men, 85 are married, 70 have a telephone, 75 own a car, and 80 own their own home. On this basis, what is the smallest possible number of men who are married, have their own telephone, own their own car, and own their own home?

EDIT: Oops, double post. Can a mod delete this?

DO NOT REPLY TO THIS

too bad and it is dah 42

        P.S. its my favorite number :arrow: 42

                 :starwars:  :spam:


This was on Canada 1969 (http://www.kalva.demon.co.uk/canada/can69.html):

If x is a real number not less than 1, which is larger: sqrt(x+1) - sqrtx or sqrtx - sqrt(x-1)?

However, I found out about the square root inequality at the http://mathworld.wolfram.com/SquareRootInequality.html . So, could one potentially just say "due to the square root inequality sqrtx - sqrt(x-1) > sqrt(x+1) - sqrtx , so QED"?  :lol:

What sort of understanding would it show on your part to say that? I wouldn't expect to get points for that answer. On an olympiad, you actually have to do math to get the points. A better way to get points would be to compare sqrt(x+1)+sqrt(x-1) and 2sqrt(x) and then just square both sides. Or you can use the concavity of sqrt(x).

Yeah, but its like asking you on an olympiad to prove the AM-GM inequality. How weird is that.

Fierytycoon

Sometime in the early 90s there was a problem at the IMO that was a simple result of a recent theorem.  Lenny Ng (and some other students from other countries) wrote a one-line proof that was scored 7/7.  There was nothing else the committee could do.

However, in general I would expect that graders would not be happy to see such an answer.  The idea of writing a proof is based on the need to justify some assertion(s).  At this level of problem solving, I wouldn't suggest writing "the square-root" inequality.  I would expect that they are asking you to prove it yourself.

[color=cyan]I don't think the name "Square-root inequality" is in common enough usage to make it a usable theorem.  Anyhow, as Simon showed, you can prove it without much difficulty.  It's important to note that older contests are, well, older, and they have a much different level of expected knowledge that more recent ones.  Generally, though, they don't want answers that state a theorem whose result is the answer to the question.  Although, if you state and prove such a theorem, they would be very happy.  Err on the side of writing more, always.[/color]

[quote="MCrawford"]Sometime in the early 90s there was a problem at the IMO that was a simple result of a recent theorem.  Lenny Ng (and some other students from other countries) wrote a one-line proof that was scored 7/7.  There was nothing else the committee could do.
[/quote]

Do you remember what the problem/theorem was?

Sorry, I didn't even know at the time.  I heard Titu Andreescu talking about it and I wasn't really involved in the conversation.

I'm pretty sure that at the international mathematical olympiad (not sure about others) if you quote a theorem and possibly where you found it (if its not that well known), then you can use it.

[quote="Brian"]
Do you remember what the problem/theorem was?[/quote]

Lenny Ng was on the '91, '92, and '93 IMO Teams. So it would have been one of trhe questions from those years.

You can find them here -- http://www.kalva.demon.co.uk/imo.html
Or here -- http://www.unl.edu/amc/a-activities/a7-problems/problemarchive.html


trigonometry
inequalities
triangle inequality
geometry
angle bisector
trig identities
Law of Cosines

Solve the equation $ \sqrt{2\minus{}2\cos x}\plus{}\sqrt{5\minus{}4\cos x}\equal{}\sqrt{5\minus{}4\cos 2x}.$

hello, $ x\equal{}0$ is a solution.
Sonnhard.

hello, after squaring both sides of the equation you will get:
$ \sqrt {2(5 \minus{} 9\cos(x) \plus{} 4\cos^2(x))} \equal{} 6\cos(x) \minus{} 6 \plus{} 8\cos^2(x)$
with $ \cos(2x) \equal{} 1 \minus{} 2cos^2(x)$.
Squaring one more time and setting $ t \equal{} \cos(x)$ gives:
$ 32t^4 \plus{} 48t^3 \minus{} 34t^2 \minus{} 27t \plus{} 13 \equal{} 0$.
Sonnhard.

Exactly
$ \sqrt {2(5 \minus{} 9\cos(x) \plus{} 4\cos^2(x))} \equal{} 3\cos(x) \plus{}1\minus{} 4\cos^2(x)$
with $ \cos(2x) \equal{} 1 \minus{} 2cos^2(x)$.
Squaring one more time and setting $ t \equal{} \cos(x)$ gives:
$ (t\minus{}1)(t\plus{}1)(4t\minus{}3)^2 \equal{} 0$.

hello, yes, that's right, it must be
$ \cos(2x)\equal{}2\cos^2(x)\minus{}1$ and so you get
$ (t\minus{}1)(t\plus{}1)(4t\minus{}3)^2\equal{}0$.
Sonnhard.

$ t \equal{} \cos{x} \equal{} \minus{} 1$ yields $ 2 \plus{} 3 \equal{} 1$ in the original equation, and is therefore a degenerate solution.

[hide="Geometric Solution"]Let $ A,B,C,D$ be points in a plane such that $ AB \equal{} AC \equal{} 1$, $ AD \equal{} 2$, and $ AC$ bisects $ \angle BAD$.
Let $ x \equal{} \angle BAC \equal{} \angle CAD$.

Using the Law of Cosines:
$ BC \equal{} \sqrt {2 \minus{} 2\cos{x}}$
$ CD \equal{} \sqrt {5 \minus{} 4\cos{x}}$
$ BD \equal{} \sqrt {5 \minus{} 4\cos{2x}}$

By the triangle inequality, $ BC \plus{} CD \ge BD$ with equality iff $ B,C,D$ are collinear and $ BD$ is the longest side.
It is easy to see that $ B,C,D$ are collinear for $ x \equal{} 0^\circ, 180^\circ$, however, $ x \equal{} 180^\circ$ yields $ CD$ as the longest side, and is not a solution.
The only other possibility for $ B,C,D$ to be collinear is if $ AC$ is a cevian in $ \Delta ABD$.
Using the angle bisector formula: $ BD \equal{} \frac {3}{2}\sqrt {2}$. So $ BC \equal{} \frac {\sqrt {2}}{2}$, and $ CD \equal{} \sqrt {2}$.
From the law of cosines, $ \cos{x} \equal{} \frac {3}{4}$.

Thus, the solutions to the original equation are $ x \equal{} 0, \pm \arccos{\frac {3}{4}}$ and any equivalent angles. [/hide]

Squaring both sides and let $ t\equal{}\cos x$ and squaring again gives $ (t\minus{}1)(t\plus{}1)(4t\minus{}3)^2\equal{}0 \left(\minus{}\frac{1}{4}\leq t\leq 1\right)\Longleftrightarrow t\equal{}1,\ \frac{3}{4}$. The answer is $ x\equal{}2m\pi\ ,\ \cos ^ {\minus{}1} \frac{3}{4}\plus{}2n\pi$ for $ m,\ n\in\mathbb{Z}$.


What is the angle of elevation of the sun when a 35-ft mast casts a 20-ft shadow?

Please show all work in detail.

Thank you.

[quote="Jeff219"][hide]
Use Trig to find the angle of elevation.
$\tan^{-1}(\frac{7}{4})\approx 60.255$[/hide][/quote]

Excellent.  Thank you, Jeff  :)


71 sticks (all but one are unmarked) are placed inside a barrel. 71 people are to take sticks out randomly one after the other. If a person takes out a marked stick, that person wins and the game is over. However, if the person takes out an unmarked stick, that person marks it and tosses it back in the barrel. Which person out of the 71 has the greatest chance of taking out a marked stick?

EDIT: Whoops, clicked submit before finishing the message... or solving the problem.

[hide=" "]The nth person has a $\frac{70\cdot 69\cdot 68 \cdot...\cdot(71-n)\cdot n}{(71^{n})}$ chance of winning.[/hide]

[hide="some ideas"]in order for the kth person to win, all the people before him had to not win, and he has to get one of the k marked sticks.  so we get this recursive definition, with $P_{k}$ as the probability that the kth person wins:
$P_{k}=(1-P_{k-1})\left(\frac{k}{71}\right)$
$P_{1}=\frac{1}{71}$

now i can't seem to find an explicit definition, but maybe we could figure out when the sequence is incresing and decreasing.  though i'm not sure how.[/hide]

This is kind of like Russion Roulet  :P .
Where the farthest away you are from going first, the greater the possiblilty that someone before you takes it but if they dont, then the greater the possibility you take it.

[hide="maybe...? o_O"]
The chance of you winning if your the $nth$ person to go:

$(\frac{71!}{71^{n}(71-n)!})(\frac{n}{71})$

Now you just need to solve the maximum of this equation.

After plugging in numbers, the maximum probability occurs when your the $8th$ guy.[/hide]

[hide] I thiks it's need to be the first person, I got $\frac{70!}{(71-n)!\cdot71^{n-1}\cdot n}$[/hide]

[hide="my idea"]
The first person has a $\frac{1}{71}$ chance of winning.

For the second person to win, the first person must lose (70/71 chance), and he must draw one of the two marked sticks (the first person marked another one when he lost).
So he has a $\frac{70}{71}\cdot\frac{2}{71}=\frac{140}{5041}$ chance of winning.

For the third person to win, the first two people must lose ($\frac{70}{71}\cdot\frac{69}{71}$), and he must draw one of the three marked sticks.
He has a $\frac{70}{71}\cdot\frac{69}{71}\cdot\frac{3}{71}$ chance of winning.

It seems like the nth person to win has a
$\frac{70\cdot69\cdot68....\cdot(72-n)}{71^{n-1}}\cdot\frac{n}{71}=\frac{70!n}{(71-n)!71^{n}}$ chance of winning.

Substituting a few numbers gives that person #8 has the greatest chance of winning (about 0.0748).
[/hide]

236factorial and Chocoboom got the same equation, but different answers. And I don't think your equation works, srulikbd, try plugging in 1.

EDIT: I agree with 236factorial's answer.

You all seemed to get a little caught trying to figure out which player has the best chance -- you could calculate the probability for each of them, but no one had a "pretty" method for figuring out when it was highest.  This would be bad if the number 71 were replaced with something really big, like 10,000, and the calculations would have fried your calculator.  It would be even worse if 71 were just replaced by a general $m$.

But I noticed that formula has lots of exponents and factorials -- try calculating the ratio of consecutive terms.  When the ratio is bigger than 1, the probability is going up, and when it's less than 1, the probability is going down.  So in fact, it shouldn't be that hard at all in the general case -- why don't you give it a try?  Then you can check if it works in the particular case of 71.


Could someone explain why a progression with difference a/b, where a and b are coprime integers, gives the same number of integers as a progression with difference 1/b?

you consider two aritmetic sequences (or arithmetic pregressions).
In one sequence, consecutive terms (you may call them elements or some other name) differ by $ 1/b$. One term and the one coming $ k$ terms later will differ by $ k(1/b)\equal{}k/b$. Only when $ k$ is a multiple of $ b$ will that difference be an integer. Hence, integers come at $ b$ intervals. Every $ b$th tern after the first integer is an integer.
In the other sequence, consecutive terms differ by $ a/b$. One term and the one coming $ k$ terms later will differ by $ k(a/b)\equal{}ka/b$. Only when $ ka$ is a multiple of $ b$ will that difference be an integer. However, because $ a$ and $ b$ are relatively prime (or coprime, or gcd(a,b)=1), $ ka$ is a multiple of $ b$ only when $ k$ is a multiple of $ b$. Hence, integers come at $ b$ intervals in this case too.


Show that any representation of 1 as the sum of distinct reciprocals of numbers drawn from the arithmetic progression $ {2, 5, 8, 11,\ldots}$ such as given in the following example must have at least 8 terms: \[ 1 \equal{}\frac{1}{2}\plus{}\frac{1}{5}\plus{}\frac{1}{8}\plus{}\frac{1}{11}\plus{}\frac{1}{20}\plus{}\frac{1}{41}\plus{}\frac{1}{110}\plus{}\frac{1}{1640}.\]

Hi!
Let n be the total number of fractions in the set. And $ x_{i}$ be the numbers 2,5,8...
there are a few methods, the easiest is to note that all the $ x_{i} \equiv \minus{} 1 (mod 3)$!

so we get
\[ ( \minus{} 1)^{n}\equiv (n)( \minus{} 1)^{n \minus{} 1} (mod 3)
\]

\[ n\equiv \minus{} 1 (mod 3)
\]
then brute out cases 2 and 5. so done! hahaha

You seem to have found all the old posts. Lol..

But how do you get $ (\minus{}1)^n \equal{} n(\minus{}1)^{n\minus{}1} (\text {mod } 3)$ out of nowhere?

Multiply by the product of the denominators. Then on the RHS there are $ n$ products of numbers $ \equiv \minus{}1\pmod 3$. Du you see it now? :)

[quote="powerchess"]
then brute out cases 2 and 5. so done! hahaha[/quote]

Excuse me, but can you explain to me how can you prove that? (I mean how you can exclude these cases...)

The largest sum of two terms possible is $ \dfrac 12 \plus{} \dfrac 15 \equal{} \dfrac 7{10}<1$, so 2 won't work. Similarly, $ \dfrac 12 \plus{} \dfrac 15 \plus{} \dfrac 18 \plus{} \dfrac 1{11} \plus{} \dfrac 1{14} \equal{} \dfrac {3041}{3080} < 1$.


4. area
I heard that it can be scalar or vector depending on context. Could anyone put more light?

Hello
U r absolutely right.
Area can considered to be a vector quantity whose magnitude is that of the area of an surface &whose direction is that of a normal to the surface .
Thank u.

Hello
Sorry couldn't get ur question.Plz explain it again.

Which question? I mean that if the eqn. of the plane is $ z\equal{}0$, then what is the direction of the vector? positive or negative direction of the $ z$ axis?[hide=Index][/hide]

hello
well as per conventions it is taken as +ve z axis only.
But look I m telling u as per my phy. knowledge.
Thank u.

[quote="Akashnil"]Thanks. What do you think the answer should be on a test with no context?
Edit: Normal in which direction? below or above?[/quote]

First question: It depends on the fact how much physical knowledge the tested students have: if they could have encountered area as a vector (i.e. Maxwell's and Gauss' Law, some pressure-related laws etc.) then the question would be ambiguous (with a slight advantage for the "scalar" answer). Otherwise, it would clearly be scalar.

Second question: Right-hand rule. Of course, there's a need for orientation of the surface boundary in that case. By convention, if the area is a part of closed surface, the vector is directed to the exterior. BTW, if the surface isn't flat or made very small (dS), then area vector makes no sense. That's another argument for stating that generally speaking, area is a scalar.


abstract algebra
vector
real analysis
real analysis theorems

f is a linear functional on a normed vector space.
Prove that if the kernel of f, {x: f(x)=0}, is closed, then f is bounded.

Thanks

you can find your answer at:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87856[/url]


A farmer has a rectangular field that measures 100 feet by 150 feet.
He plans to increase the area of the field by 20%. He will do this by
increasing the length and width by the same amount, x. Which equation
represents the area of the new field?

(1) (100 + 2x)(150 + x) = 18,000
(2) 2(100 + x) + 2(150 + x) = 15,000
(3) (100 + x)(150 + x) = 18,000
(4) (100 + x)(150 + x) = 15,000

[hide]The new area is $18000$, and the new sides are $100+x$ and $150+x$. The equation is $(100+x)(150+x)=18000$.[/hide]

That is exactly what I was looking for.

Thanks!


function
calculus
derivative
limit
real analysis
real analysis unsolved

Does anyone know of a function that is differentiable over a dense subset of the reals and not differentiable anywhere else?

Motivation: Consider the function f(x)={ sin(x)+1 if x is rational
                                                     -sin(x)-1  if x is irrational}
This function has countable many differentiable points, namely all the N*Pi/2 where the Ns are integers.

Plenty of them. For an "explicit" example, take something like $ \min_{q\in\mathbb Q} D(q)^{10}(x\minus{}q)^2$ where $ D(q)$ is the denominator of $ q$. The result will be differentiable at every rational point with derivative $ 0$ but it'll be discontinuous at every point where it is not $ 0$ (and, since most real numbers $ x$  satisfy $ |x\minus{}\tfrac pq|\ge cq^{\minus{}5}$ with positive $ c\equal{}c(x)>0$, you'll have a lot of such bad points).

Sorry fedja, but could you elaborate more on your example. I don't understand the part about the min of all q over the rationals  \min_{q\in\mathbb Q} D(q)^{10}(x-q)^2. Thank you.

Call it an infimum- that matters in some cases. That's a countable collection of nonnegative functions; at any point, they're bounded below by zero and have a greatest lower bound.
Call $ D(q)^{10}(x\minus{}q)^2$ $ f_q$, and call $ \inf_{q\in \mathbb{Q}}f_q$ $ f$. At any rational point $ r$, we have $ f(r)\equal{}f_r(r)\equal{}0$, and $ f_r'(r)\equal{}0$. By squeezing, $ 0\le \lim_{x\to r}\frac{f(x)\minus{}0}{x\minus{}r}\le \lim_{x\to r}\frac{f_r(x)\minus{}0}{x\minus{}r}\equal{}0$, and $ f'(r)\equal{}0$.
On the other hand, since $ f$ is zero on a dense set, it must be discontinuous at every point at which it's not zero. How many of those points are there? It's a set of full measure; while the rationals aren't all of $ f^{\minus{}1}(0)$, the full inverse image of zero has measure zero. By the full measure, the points at which $ f$ is positive are also a dense set.


number theory
greatest common divisor
relatively prime

your gcd restriction really does nothing, since the sheer fact that $ x\plus{}y\equal{}n$ implies that their gcd divides $ n$.

The restriction that $ x$ and $ y$ are nonnegative integers is more useful.  Start with $ x\equal{}0$, and we see $ y\equal{}n$.  Since $ y\equal{}n\minus{}x$, $ y$ is nonnegative for $ x\equal{}0,1,2,3 \ldots n$.  So we have a total of $ n\plus{}1$ ordered pairs.

[quote="facis"]your gcd restriction really does nothing, since the sheer fact that $ x \plus{} y \equal{} n$ implies that their gcd divides $ n$.

The restriction that $ x$ and $ y$ are nonnegative integers is more useful.  Start with $ x \equal{} 0$, and we see $ y \equal{} n$.  Since $ y \equal{} n \minus{} x$, $ y$ is nonnegative for $ x \equal{} 0,1,2,3 \ldots n$.  So we have a total of $ n \plus{} 1$ ordered pairs.[/quote]The problem is easy,indeed. But, you have made a mistake. Your solutions don't satisfy the condition $ \gcd{(x,y)} \equal{} p$.
For example, if $ n \equal{} 100$ and $ p \equal{} 5$ then, you are saying that, $ \gcd{(1,99)} \equal{} 5$....
So,try again. :)

[quote="Moonmathpi496"]Find the number of ordered pairs $ (x,y)$ satisfying,
$ x \plus{} y \equal{} n$
with $ \gcd{(x,y)} \equal{} p$ such that, $ p|n$

[Here $ x,y$ are non-negative integers][/quote]

[hide]Let $ n\equal{}n'p,x\equal{}x'p,y\equal{}y'p$. Then $ (x',y')\equal{}1$ and we have $ x'\plus{}y'\equal{}n'$. Call a pair $ (x,y)$ of non-negative integers good if $ (x,y)\equal{}1$ and $ x\plus{}y\equal{}n'$. Note that $ (x',y')\equal{}(x',x'\plus{}y')\equal{}(x',n')\equal{}1$. Hence the number of good pairs equals the number of integers $ x'$ such that $ (x',n')\equal{}1$ and $ x'\le n'$. This number is obviously $ \phi(n')$. Hence our answer is $ \phi(n/p)$.[/hide]

Oh $ p$ is a specific given number?  I thought we were just defining $ p$ to be the gcd of $ x$ and $ y$...

Anyway I guess I agree with Nayel now.

And for an interesting extension:  Prove that $ \sum_{d|n} \phi(d) \equal{} n$

Yes facis you're right. The number of ordered pairs $ (x,y)$ of non-negative integers satisfying $ x \plus{} y \equal{} n$ is obviously $ n \plus{} 1$. On the other hand the number of such pairs with the additional constraint $ (x,y) \equal{} d$ is $ \phi(n/d)$. Now as $ d$ ranges through the divisors of $ n$ we get all possible solutions. But note that when $ d \equal{} n$ we get two solutions, namely $ (n,0)$ and $ (0,n)$; however in this case $ \phi(n/d) \equal{} \phi(1) \equal{} 1$. This contradiction is because of $ 1$ being the only positive integer which is relatively prime to itself. Hence the total number of solutions equals
\[ n \plus{} 1 \equal{} 1 \plus{} \sum_{d\mid n}\phi(n/d) \equal{} 1 \plus{} \sum_{d\mid n}\phi(d)
\]
yielding $ \sum_{d\mid n}\phi(d) \equal{} n$, as desired.


greatest common divisor
number theory theorems
number theory

Prove that $(p^{k-1}, p^{k+1}-1)=1$ for all primes $p$ and all natural numbers $k$. Here $(m,n)$ denotes the greatest common divisor of $m$ and $n$.

[quote="ZetaX"]Are you sure about how it is stated¿ That's a bit very trivial since $p \nmid p^{s}-1$ for all $s$...[/quote]

Thanks a lot. You may remove my posting.

Hi.
p^{k-1} does not have any divisor other than p, so if we can prove that p^{k+1}-1 is not divisible by p, we will be the same as we prove the corollary. So,
                                           p^{k+1} is obviously divisible by p, and suppose p^{k+1}-1 is divisible by p, too.
                                           So then 1 is divisible by p, which is impossible.
Suggestion is proven.
                                  Tural Mahmudov (madshiet)

[code]p^{k+1}[/code] is an integer times of [code]p^{k-1}[/code] so [code]p^{k+1}-1[/code] cannot be, also.


Find all  numbers $ x,y \in Z$ such that
$ a$) $ y^2 \equal{} x^3 \plus{} 1$.
$ b$) $ y^2 \equal{} x^3 \minus{} 1$.

see http://www.mathlinks.ro/viewtopic.php?t=303498
and http://www.math.leidenuniv.nl/~jdaems/scriptie/Catalan.pdf section 3.3


Hello every1,

Plz solve this

There is a 5*5 array. All its 25 cells are set to 0. WE need to change all of them to1 in minimum moves under the following constraints:-

   A move is changing a cell from 0 to 1 (or from 1 to 0). A move can affect 4 neighbors (if they exist) of a cell namely the one above,below,to left and to right of the cell where we make the change. The effect is the change of values contained in these neighboring cells.

So, plz find the minimum moves neede to change all cells to 1.

Also if possible generalize this result for a n*n board.

Thanx and have a nice day.....

does the move change all neighbours or could it change exactly 0, 1, 2, 3 or 4 neighbours?

Naphthalin

All neighbours are changed.

Here's a program.


For $a,b,c>0$ and $a+b+c=1$, prove or disprove:

$\frac{1}{a+bc+3abc}+\frac{1}{b+ca+3abc}+\frac{1}{c+ab+3abc} \leq \frac{2}{ab+bc+ca+abc}$

Dehomogenizing, multiplying by the RHS denominator and rearranging, this becomes $\sum_{cyc} \frac{a(a+b)(a+c)+abc}{(a+b+c)(a+b)(a+c)+3abc} \geq 1$. Multiplying out and cancelling (lots of cancelling) this works out as

$6a^2b^2c^2(a(a+b)(a+c)+b(a+b)(b+c)+c(a+c)(b+c))$
$+2abc(a+b+c)^3(a+b)(a+c)(b+c)$
$\geq 3abc(a+b+c)(a+b)(a+c)(b+c)(2ab+2ac+2bc)+9a^2b^2c^2(a+b)(a+c)(b+c)$.

This follows from
$a(a+b)(a+c)+b(a+b)(b+c)+c(a+c)(b+c)$
$=a^3+b^3+c^3+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+3abc$
$\geq \frac{3}{2}(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)+3abc$
$= \frac{3}{2}(a+b)(b+c)(c+a)$
and also $(a+b+c)^2 \geq 3(ab+ac+bc)$.

Dear Fiachra, your solution is very nice and to check it I spent a lot of time.

How did you guess the right factorization of LHS and RHS ?

By intuition, by trial or by computer.

Thank you very much.

I largely left things factorized when multiplying the fractions out, just to keep the algebra manageable. I was lucky that they happened to be useful in the end.

Thanks for the compliment  :)


vector
function
calculus
integration
probability and stats

I have the following question:

$ S$ is a vector space, say, the space of semimartingales. $ \phi, \psi$ are measurable functions such that the following integral exists, e.g. $ \phi(x) \equal{} \psi(x) \equal{} x^{1/3}$. $ L$ is the Levy Process.
$ x_{ \plus{} } : \equal{} \max\{x,0\}$, $ x_{ \minus{} } : \equal{} \minus{} \min\{x,0\}$
We know now
$ a \equal{} \int_{ \minus{} \infty}^{t} \phi((t \minus{} s)_{ \plus{} }) \minus{} \psi(( \minus{} s)_{ \plus{} }) d L(s) \notin S$, i.e. $ a$ is not a semimartingale.
and let
$ b \equal{} \int_{ \minus{} \infty}^{t} \phi((t \minus{} s)_{ \minus{} }) \minus{} \psi(( \minus{} s)_{ \minus{} }) d L(s)$.

can we see if $ b \in S$ or not?
Thanks for any help!

Hi mathmean

Your integral looks like a generalisation of fractionnal brownian motion (which is not a semi martingale except for hurst index equal to 1/2)
I don't know to actually prove that "a" is not a semi-martingale but I would be interested in having a look at your proof of this fact (which might led you to the answer for "b").

Regards

Thanks, TheBridgeTwo. It seems that one can not direct use the property of vector space to show if $ b \in S$ or not.
You are right. it is a generalisation of fractionnal brownian motion.
$ a$ is actually the linear fractional stable motion:
$ a \equal{} \int_{ \minus{} \infty}^{t} (t \minus{} s)_{ \plus{} }^{H\minus{}1/\alpha} \minus{} ( \minus{} s)_{ \plus{} }^{H\minus{}1/\alpha} d L_{\alpha}(s),  t \geq 0,$
where $ L_{\alpha}(t)$ is a $ \alpha$-stable Levy process.
Then $ a$ is a fractionnal brownian motion for $ \alpha \equal{} 2$.

I've read a paper, there one has showed that  $ a$ is a semimartingale if and only if $ H \equal{} 1/\alpha$. (See the attachment of another post of mine ("Riemann-Liouville fractional integral"),  on page 18).

What i want is to show that the "well-balanced" fractional stable motion:
$ c \equal{} \int_{ \minus{} \infty}^{t} |t \minus{} s|^{H\minus{}1/\alpha} \minus{} |s|^{H\minus{}1/\alpha}d L_{\alpha}(s)$
  $ \equal{}a\plus{} \underbrace{\int_{ \minus{} \infty}^{t} (t \minus{} s)_{ \minus{} }^{H\minus{}1/\alpha} \minus{} ( \minus{} s)_{ \minus{} }^{H\minus{}1/\alpha}) d L_{\alpha}(s)}_{\equal{} b}, t \geq 0,$
is a semimartingale if and only if $ H \equal{} 1/\alpha$.
is that makeable? thanks for any hint!


Evaluate: 3*3+3(3+3)-3^3

Please post and explain

If you're asking about the FTW question, I'm pretty sure the answer on there is wrong (it's in the 200s i think).


Find the radius of convergence of the power series $\sum\frac{z^{p_{k}}}{{p_{k}}!}$ where $p_{k}$ denotes the $k$th prime.

If you fill in the missing terms, you get the exponential; $e^{|z|}$ is larger, and this series converges for all $z$.


limit
trigonometry
calculus
integration
absolute value
complex analysis
least common multiple

can we prove $\lim_{n \to \infty}(\cos n)^\frac{1}{n}=1$?

Yes, we can (I hope, you meant $|\cos n|^{1/n}$ rather than $(\cos n)^{1/n}$ :) ). To this end, we'll have to use Hata's theorem saying that $|\pi-\frac pq|\geqslant \frac 1{q^9}$ for all sufficiently large denominators $q$. It follows that $|\cos n|\geqslant \frac C{n^8}$ from where the statement is easy to derive.
Of course, the exact power in Hata's theorem doesn't matter here and we can use the earlier result by Mahler (with power $43$, if I remember it right) or by Beukers (I don't remember the exact power one can extract from his proof of irrationality of $\zeta(2)=\frac {\pi^2}6$ but it is better than Mahler's and worse than Hata's; personally, I like Beuker's approach most for the clarity of the proof).

You need an absolute value in the statement there- we don't want to be taking roots of negative numbers.
It's hard. I'll quote an advanced theorem I have no idea how to prove (which is also referenced in this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22093]thread[/url] in Solved problems).

For any integer $p$, and sufficiently large $q$, $\left|\pi-\frac pq\right|>\frac1{q^9}$ (Hata's theorem).
Choose $a$ between 0 and 1 arbitrarily.
In order to have $|\cos n|<a^n$, we must have $\left|n-q\frac\pi2\right|<2a^n$ for some integer $q$. Equivalently, $\left|\pi-\frac{2n}q\right|<\frac{4a^n}{q}.$ Obviously, $\frac n2<q<n$ for large $n$. For sufficiently large $n$, $4a^n<q^{-8}$, so Hata's theorem applies and there are only finitely many such pairs $n,q$. Thus $\sqrt[n]{|\cos n|}>a$ for large enough $n$, and we are done.

[As I was composing this, fedja posted]

Fedja, you said Beukers approach is nice and clear. Is it elementary, I mean if I started to read the proof, would I understand anything? I tried once to read a proof of Maurice Mignotte, but it used complex analysis and I don't know yet this field.

Yes, harazi, it is elementary. Nothing more then integration by parts.

[quote="harazi"] Is it elementary, I mean if I started to read the proof, would I understand anything? I tried once to read a proof of Maurice Mignotte, but it used complex analysis and I don't know yet this field.[/quote]
The only catch is that one needs the prime number theorem (more precisely, one needs to know that the least common multiple of $1,2,\dots,n$ does not exceed $3^n$ for large $n$). And the easiest proofs of it use either complex or Fourier analysis. On the other hand, if the prime number theorem is assumed known, the rest of the proof is so short and simple that I'm even thinking of posting it in "Formulas and Theorems" section (I tried to find an online version of Beukers's paper but failed, otherwise I would just give a link right here). Meanwhile, whoever is interested in these matters may want to read his survey paper [url=http://www.math.leidenuniv.nl/~naw/serie5/deel01/dec2000/pdf/beukers.pdf]A rational approach to $\pi$[/url], which doesn't contain the formal proof but rather discusses various ideas that may be used in constructing such a proof (surprisingly, Burkers's own approach does not appear in the survey).

thanks to all above.you mean is make a rational estimate of $\pi$,but hate's theorem is not elementary,i will read Beukers's approach. :)
thanks again


HOW DID DUKE GO... AWESOME :winner_second:  PLACE FOR GEORGIA A AND GEORGIA B SERVED SC LIKE A TENNIS BALL :evilgrin:

lol nice. congratulations to everyone that went

[quote="cleansmyheart"]karthik, you amuse me. one day, i'll pwn you at ping pong. just wait and see.[/quote]

Who are you?  :huh:

karthik, i think you meant served like a ping pong ball, i don't think you know what tennis is yet. :P

and amy, i don't think too many ghpers can beat karthik, he is pretty darn good, exception: that scheisse guy

[quote="cleansmyheart"]Amy. From GHP/ARML/Facebook. Surely you could've guessed that.[/quote]

oh my god! s

dude

i never knew you used aops

haha you only have 6 posts lol

but look how long i've been a member..

i usually only read, lots of good math on here. but i thought i might as well participate since i know you guys.

who said you know us, but anyway, what's your name? I'm Shing-shing, nice to meet you!

Karth, amy is not a guy. :P

it hurts my eyes to see the Lord's name used in such a manner

funcia


Sets A and B are equidecomposable if you can partition set A into a finite amount of subsets and rearrange them by rigid motions only to form set B.
Are a unit circle and a unit circle with one point missing equidecomposable?

how exactly is a "rigid motion" defined?

[hide="Hint"] Is it possible to partition $ B$ into a finite number of subsets, all of which are closed? [/hide]


What is the measure of an angle, in degrees, if its supplement is six times its complement?

Simply set up an equation, with $ x$ being the angle, $ 90\minus{}x$ being the complement, and $ 180\minus{}x$ being the supplement.

$ 6(90\minus{}x)\equal{}180\minus{}x$
$ 540\minus{}6x\equal{}180\minus{}x$
$ 5x\equal{}360$
$ x\equal{}\fbox{72}$


trigonometry
SFFT
geometry
rectangle
geometric transformation
dilation
calculus

The expression can be factored into $(x+1)(y+1)$. Then, obviously, to maximize the product, we must have $x=y=\frac{\sqrt{2}}{2}$, hence the expression is equal to $\frac{1}{2}+\sqrt{2}+1$, or $\frac{3}{2}+\sqrt{2}$.

[quote="frt"]Real numbers $x$ and $y$ satisfy $x^{2}+y^{2}=1$. What is the largest possible value of $xy+x+y+1$?[/quote]
[hide]
Let $x=\sin \theta$ and $y=\cos \theta$
$.5\sin2\theta+(\sqrt{1+\sin2\theta})+1$
Notice that if we maximize the first term, then we maximize the second term as well.
Since $\sin2\theta \leq 1$ we see that our maximum value is
$1.5+\sqrt{2}$
Wow, this is the first problem I've ever seen with xy+x+y+1 that doesn't involve SFFT[/hide]
@ davidyko how is it obvious that x=y maximizes?

Hrm...well, that's one good way to actually prove my "obviously" statement. I thought about it like maximizing the area of a rectangle...but in retrospect, I suppose it doesn't make too much sense. Ah well, maybe it was intuition.

EDIT: I think it was intuitively obvious from a sketch of a unit circle...the rectangle (or in this case, a square) with sides $\sin{\theta}$ and $\cos{\theta}$ had the largest area when $\theta=\frac{\pi}{4}$...since the diagonal of the rectangle remains constant, I felt that the area was obviously maximized when the rectangle was a square. I don't really know how to explain it.

[quote="junggi"][hide="i suck at ineqs so im probably wrong =( or being dumb and doing this the long way. oh well"]
We want to maximize $xy+x+y+1=(x+1)(y+1).$
By AM-GM
$\frac{(x+1)^{2}+(y+1)^{2}}{2}=x+y+\frac{3}{2}\ge (x+1)(y+1)$
By Cauchy,
$(x^{2}+y^{2})(1+1)\ge (x+y)^{2}$
$\sqrt2\ge (x+y)\ge-\sqrt2$
So
$\sqrt2+\frac{3}{2}\ge x+y+\frac{3}{2}\ge (x+1)(y+1)$
equality is achieved when $x=y=\frac{\sqrt2}{2}$ which maximizes the expression to $\sqrt2+\frac{3}{2}$

@diophantient:I guess my soln involves SFFT? =P(unless I'm being stupid)
[/hide]

er what's up with this "obviously" proof[/quote]
I meant that my solution didn't involve it

Very nice solution by the way :)

Challenge:
Given that $x^{2}+y^{2}=1$ then maximize $xy+2x+y+1$

[quote="t0rajir0u"][hide="Guys"] $\frac{x^{2}+y^{2}}{2}+2 \sqrt{ \frac{x^{2}+y^{2}}{2}}+1 \ge xy+x+y+1$

By AM-GM and QM-AM.  Of course, this argument wouldn't work at all unless the equality case were $x = y$, but it is, so woohoo, the upper bound is achievable.  [/hide]

diophantient, do you have a solution to your problem?  I doubt it is easy without calculus.  Unless you guess the equality case correctly.[/quote]
No I don't. I thought that there would be an elegant given the variety of solutions presented, but maybe not.
My intuition is telling me $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$ giving $\frac{5\sqrt{3}}{4}+\frac{3}{2}$
However, I have no idea at all why I'm getting this idea, if it's right I'll be shocked.

@davidyko: QM-AM-GM-HM is an extention of AM-GM

Check it out on aops wiki, I'm sure it exists

[quote="t0rajir0u"][hide="Guys"] $\frac{x^{2}+y^{2}}{2}+2 \sqrt{ \frac{x^{2}+y^{2}}{2}}+1 \ge xy+x+y+1$

By AM-GM and QM-AM.  Of course, this argument wouldn't work at all unless the equality case were $x = y$, but it is, so woohoo, the upper bound is achievable.  [/hide]

diophantient, do you have a solution to your problem?  I doubt it is easy without calculus.  Unless you guess the equality case correctly.[/quote]
I believe you can substitute $x=\sin\theta\wedge y=\cos\theta$ and then use the Cauchy-Schwarz inequality.

I always thought it was RMS-AM-GM-HM. Are they the same thing?

(Root Mean Square)$\equiv$ (Quadratic Mean)

Alright, cool. I just got confused for a moment. Didn't know what QM stood for.
I've just never seen it written that way.

[quote="Bictor717"][quote="diophantient"]
Challenge:
Given that $x^{2}+y^{2}=1$ then maximize $xy+2x+y+1$[/quote]
If you think of this geometrically, $x^{2}+y^{2}=1$ is a circle, while $xy+2x+y+1=(x+1)(y+2)-1=k$ is a family of equilateral hyperbolae centered at (-1,-2), where each member is a dilation of another from (-1,-2). Wouldn't this boil down to finding the point on the circle furthest from (-1,-2)? Certainly there are non-calculus methods of producing a solution, although they might border on calculus ideas.[/quote]

no calculus needed there...consider a locus of distance constant from (-1,-2)...this is a circle...we make this as large as possible...it becomes tangent to the unit circle... then the point of tangency and the two centers are collinear! we easily compute this from here...

[quote="1=2"][quote="diophantient"]Challenge:
Given that $x^{2}+y^{2}=1$ then maximize $xy+2x+y+1$[/quote]

[hide="solution"]$xy+2x+y+2=(x+1)(y+2)$

To maximize xy+2x+y+1, we need to maximize xy+2x+y+2.

x+1=y+2
x=y+1

$2y^{2}+2y+1=1$

$2y^{2}+2y=0$

$y=0,-1$

(x,y)=(1,0),(-1,0),(0,-1)

The first gives 3, the second gives -1, the third gives 0.

3[/hide]

I don't know if this is right....[/quote]
That's definitely not right.
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=796082#796082
$\frac{5\sqrt{3}}{4}+\frac{3}{2}>3$ ;)

I don't think this one can be so easily guessed.


[hide="Solution"]
Start from the bottom.

\begin{align*}\frac{1}{1+\frac{1}{1+\frac{2}{3}}}&=\frac{1}{1+\frac{1}{\frac{5}{3}}}\\&=\frac{1}{1+\frac{3}{5}}\\&=\frac{1}{\frac{8}{5}}\\&=\boxed{\frac{5}{8}}.\end{align*} Sorry, not much to explain here.[/hide]


vector
function
calculus
derivative
parameterization
trigonometry
calculus computations

Suppose the vector $ \displaystyle \vec {T} \equal{} a_1 {i} \plus{} a_2 {j}\plus{}a_3 \dot{k}$ is tangent to the curve, $ \text{C}.$
The curve $ \text{C}$ is the intersection of 2 surfaces.

$ \displaystyle \vec {T} \cdot \nabla {u} \equal{} 0$

$ \vdots$

$ \frac{dx}{a_1} \equal{} \frac{dy}{a_2} \equal{} \frac{dz}{a_3}$ (1)

Is it correct to say that the 3 equations (1) represent the gradient of the curve C?

How do you define "the gradient of a curve"?

Functions have gradients; curves have tangent vectors.

[quote="mlok"]How do you define "the gradient of a curve"? [quote]

See sketch

That's an impressive illustration, but I fail to see how the yellow vector is obtained from the rest of the data.

(Made the sketch to be sure we could be sure we were talking the same thing.)

I think I see...

Solutions to Eq. (1) above give curves from which the NORMAL (yellow) may be obtained...?

I have no idea what your last comment meant or had to do with the rest of the conversation -- curves don't have gradients, as mlok said.  (They do have normals.)  Since a curve is the intersection of many different pairs of surfaces, whatever vector you end up with from your process (which is unclear: in his last post, mlok was asking how you actually find that yellow vector given the other data) is not a function of the curve itself (as the terminology "the gradient to a curve" would suggest) but rather a function of two intersecting surfaces.  If you're just looking for a way to pick a particular normal to a curve, take the tangent vector, normalize it to unit length and take its derivative (or in other words, take a constant-speed parametrization of the curve and look at its second derivative).

For example, suppose $ \displaystyle \vec {T} \equal{} 0 {i} \minus{}4 z {j}\plus{} 4 y \dot{k}$ is tangent to C, represented by $ u \equal{} f(y,z)$.

The normal to C is $ \nabla u \longrightarrow \nabla u \equal{}0 {i}\plus{} u_y {j} \plus{} u_z {k}$

If $ \displaystyle \vec {T}$ is $ \bot\; \nabla u$ then

$ \displaystyle \vec {T} \cdot \nabla {u} \equal{} 0$

$ \displaystyle \vec {T} \cdot \nabla {u} \equal{} \minus{}4z u_y \plus{} 4y u_z \equal{}0$

Solutions to:

$ \frac{dx}{a_1} \equal{} \frac{dy}{a_2} \equal{} \frac{dz}{a_3}$ (1)

Give $ u_1 \equal{} y^2 \plus{}z^2$ and $ u_2\equal{}$ Constant.

$ \nabla u_1 \equal{} 0 {i} \plus{} 2y{j} \plus{} 2z {k}$

$ \displaystyle \vec {T} \cdot \nabla {u} \equal{} 0$

Thus my last question:
"Solutions to Eq. (1) above give curves from which the NORMAL (yellow) may be obtained...?"

The particular problem requires finding the curve C $ \rightarrow u$.

I appreciate your comments... Thanks

Probably, it will help if you back up a bit and post the problem [b]exactly[/b] as it appears in your textbook or wherever else you took it from. Then we will, at least, be able to figure out what was asked :roll:.

[quote="JBL"]If you're just looking for a way to pick a particular normal to a curve, take the tangent vector, normalize it to unit length and take its derivative (or in other words, take a constant-speed parametrization of the curve and look at its second derivative).[/quote]

Would you mind defining "constant-speed" parametrization?  I'm assuming you mean $ t$?  Suppose the parameter is $ \theta$, does second derivative give a normal as well?

A constant-speed parametrization is one in which the first derivative of this parametrization (tangent vector) is of constant length.  Given such a parametrization, the second derivative is always normal to the tangent.  (Essentially, if it had a nonzero component parallel to the tangent then the tangent would be forced to vary in length.)  The particular letter that you choose as your parameter has no bearing whatsoever on this result.  An arclength parametrization is a constant-speed parametrization in which the first derivative is always of length 1.  You don't need to know such a parametrization [i]a priori[/i] in order to get this normal, however: as I wrote in the post you quoted, for any parametrization with nonzero derivative it suffices to normalize the derivative (tangent) and then take the derivative of that.

E.g., consider the curve given by $ T \equal{} (t, t^2, \frac{2}{3}t^3)$.  The tangent to this curve is $ T' \equal{} (1, 2t, 2t^2)$, not of constant length.  So, to we take the constant-length tangent $ \frac{T'}{\| T' \|} \equal{} (\frac{1}{2t^2 \plus{} 1}, \frac{2t}{2t^2 \plus{} 1}, \frac{2t^2}{2t^2 \plus{} 1}).$  To get a normal to the curve, just take another derivative.

I think I understand...

$ \theta$ as used in a helix of form $ R(\theta) \equal{} a \cos \theta \textbf{i} \plus{} a \sin \theta \textbf{j} \plus{} b \theta \textbf{k}$ would represent a constant-speed parametrization.

$ | R'(\theta) | \equal{} \sqrt{a^2 \plus{} b^2}$  because this is constant, $ \theta$ is a constant speed parametrization.  Correct?


If $ a$ and $ b$ and $ c$ are integer  and $ ab^{2} \plus{} 1 \equal{} 0$ and $ abc$ is different by $ 0$
solve $ \frac {x^{3}}{a} \plus{} \frac {x^{2}}{b} \plus{} \frac {x}{c} \equal{} bc$  :)

Er... if $ a, b, c$ are integer and $ ab^2 \plus{} 1 \equal{} 0$ then $ a \equal{} \minus{} 1, b \equal{} \pm 1$.  Are you sure this is what you intended?


calculus
derivative
inequalities
function
analytic geometry
graphing lines
slope

Let $a$ be a positive real, and let $p \geq q$ be positive reals.

Prove that $\frac{a^p - 1}{p} \geq \frac{a^q - 1}{q}$.

You should realize that the participants in this selection test didn't know calculus, so I encourage you to look for a pretty solution without using calculus :)

I can see how to do it using calculus, but I don't see a non-calculus proof.  Can you give a hint?

First prove that result for positive integers p and q, then for rationals, and then for all reals.

I see, makes sense.  But to pass from rationals to reals, you need the concepts of limits and continuity, which is calculus, no?

True, but the South African students are used to the concept of a limit (in an "informal" way), so they are able to define a sequence of rationals approaching an irrational number, and so on. You should not ask them to calculate a derivative though :)

[quote="Arne"]Let $a$ be a positive real, and let $p \geq q$ be positive reals.

Prove that $\frac{a^p - 1}{p} \geq \frac{a^q - 1}{q}$.[/quote]

Why not just solve the problem? Since the reals p and q are positive and satisfy $p\geq q$, there exists a real $u\geq 1$ such that p = qu. Also, denote $b=a^q$. Then, $\frac{a^p - 1}{p}=\frac{a^{qu}-1}{qu}=\frac{\left(a^q\right)^u-1}{qu}=\frac{b^u-1}{qu}$; on the other hand, $\frac{a^q-1}{q}=\frac{b-1}{q}$. Thus, the inequality in question, $\frac{a^p - 1}{p} \geq \frac{a^q - 1}{q}$, becomes $\frac{b^u-1}{qu}\geq\frac{b-1}{q}$; upon multiplication with qu, this becomes $b^u-1\geq u\left(b-1\right)$. This is equivalent to $b^u\geq 1+u\left(b-1\right)$. But this is just Bernoulli's inequality, noting that $u\geq 1$.

  Darij

Again, I'm new to this inequality stuff, so I don't know if we can do this:

Since $a^x$ is a convex function, then for $0\le 0<q\le p$, we have $\frac{(a^q)-(a^0)}{q-0}\le\frac{(a^p)-(a^0)}{p-0}$.
Or $\frac{a^p-1}{p}\ge\frac{a^q-1}{q}$.

Sure, that's fine.  This is how that inequality is derived.

A function is convex iff $f(\lambda x+(1-\lambda)y)\leq \lambda f(x)+(1-\lambda)f(y)$ for every $\lambda\in (0,1)$

Let $a<b<c$ be inside the interval on which $f$ is convex

Write $b=\frac {c-b}{c-a}a+\frac {b-a}{c-a}c$ and apply the definition of convexity.  This gives

$(c-a)f(b)\leq (c-b)f(a)+(b-a)f(c)$

and rewriting gives us

$(c-a)(f(b)-f(a))\leq (b-a)(f(c)-f(a))$ and another similar inequality.

[quote="Palytoxin"][quote="towersfreak2006"]\[\begin{equation*}
\frac{f(c)-f(a)}{c-a}\le\frac{f(d)-f(b)}{d-b}
\end{equation*}\][/quote]

slope of the secant line in that interval :!:  :idea:[/quote]
that can be a geometric interpretation.

[quote="towersfreak2006"][quote="blahblahblah"]
$(c-a)(f(b)-f(a))\leq (b-a)(f(c)-f(a))$ [/quote]

Thanks! :)
I saw something of that form while reading Kedlaya's notes. I think it was:
\[\begin{equation*}
\frac{f(c)-f(a)}{c-a}\le\frac{f(d)-f(b)}{d-b}
\end{equation*}\]
for $a\le b<c\le d$ in an interval $[a,d]$ in which $f$ is convex. Is this inequality well-known and does it have a name?[/quote]

Yes, it is well-known, and no, it doesn't have a name.  As wonderful as Kedlaya's notes are, he doesn't try to prove everything in them, so I was just giving you a proof.


June 2-3.

This sentence is only here because my message was too small otherwise :P

[quote="darkquantum"]geehoon!! aww too bad you are at MIT now, u can't come to ARML

do we get ice cream if NYC A beats/ties TJ A?[/quote]

I might be there -- I'll think about it :-)

oh man if we provide moral support to NYC A in their winning effort can we get ice cream too

[quote="JBL"]Haha, you can go buy yourself icecream.

(By the way, nice job at HMMT.)[/quote]

Oh right, wait, Mystic, u are NJ right?

A kid from my school visited Boston before I did, and when she got back the most first thing she saw fit to tell me about MIT was that they had [i]really good ice cream![/i]  :rotfl: Is it true? (p.s. she never did get to the academics... :roll: )

well, i never had the ice cream.
but if you visit Harvard Square, there's this bubble tea place. The bubble tea is amazing! :)

[quote]i bet TJ A, B, AND C will beat NYC [/quote]
Just like you dominated ARML Power?

The Harvard Square bubble tea is not noticeable more wonderful than bubble tea I've had elsewhere.

[quote="JBL"]
The Harvard Square bubble tea is not noticeable more wonderful than bubble tea I've had elsewhere.[/quote]
that's cause NY bubble tea is the best :)
TJ didn't seem to do too well at HMMT this year! good luck to NYC :)

(and to everyone participating in ARML this year!)

[quote="keeyounglee"]lol
i bet TJ A, B, AND C will beat NYC

i dunno why, but tj might take 3 teams of 15 this year because our coach found a bigger bus. o.O[/quote]

haha some friendly rivalry, very cool.  Though TJ hasn't been a force @ ARML for a while now.. should NYC even have TJ on its radar anymore? ;)