geometry
conics
hyperbola
geometric transformation
reflection
circumcircle
incenter

Upon a request, I am posting theorems related to Poncelet points and antigonal conjugates in Euclidean plane geometry. This is going to be a bunch of theorems, but nothing stops you from seeing it as a bunch of exercises and writing the proofs under my post!

[b]1. Poncelet points[/b]

[b]Theorem 1.[/b] Let A, B, C, D be four points in the plane. The nine-point circles of triangles ABC, BCD, CDA, DAB concur at one point.

This point is called the [b]Poncelet point[/b] of the four points A, B, C, D. Note that there are two cases when the Poncelet point of four points is not defined: In fact, if the quadrilateral ABCD is orthocentric (this means that $DA\perp BC$, $DB\perp CA$ and $DC\perp AB$; equivalently, each of the four points A, B, C, D is the orthocenter of the triangle formed by the other three), then the nine-point circles of triangles ABC, BCD, CDA, DAB coincide, and hence the Poncelet point of the four points A, B, C, D cannot be uniquely defined. Also, if two of the points A, B, C, D coincide, then the Poncelet point of the four points A, B, C, D is not defined.

In the following, a [i]non-orthocentric quadrilateral[/i] will mean a quadrilateral that is not orthocentric and has four pairwisely distinct vertices.

[b]Theorem 2.[/b] For every non-orthocentric quadrilateral ABCD, there is exactly one rectangular hyperbola passing through the points A, B, C, D, and the center of this hyperbola is the Poncelet point of the points A, B, C, D.

Note that Theorem 2 almost won't be needed anywhere below, by the way - all the results that can be formulated without conics can be proven without them as well, except for Theorem 5 maybe.

[b]Theorem 3.[/b] Let ABCD be a non-orthocentric quadrilateral, and let H be the orthocenter of triangle ABC. Then, the Poncelet point of the points A, B, C, D is simultaneously the Poncelet point of the points H, B, C, D, as well as the Poncelet point of the points H, C, D, A, as well as the Poncelet point of the points H, D, A, B, and lies on the nine-point circles of triangles HAD, HBD, HCD.

The [b]pedal circle[/b] of a point P with respect to a triangle ABC is defined as the circle passing through the orthogonal projections of the point P on the lines BC, CA, AB.

[b]Theorem 4.[/b] The Poncelet point of four points A, B, C, D lies on the pedal circle of the point A with respect to triangle BCD, on the pedal circle of the point B with respect to triangle CDA, on the pedal circle of the point C with respect to triangle DAB, and on the pedal circle of the point D with respect to triangle ABC.

A harder property to which I don't have an elementary proof yet is:

[b]Theorem 5.[/b] The Poncelet point of four points A, B, C, D lies on the circle through the three points $AB\cap CD$, $BC\cap DA$ and $AC\cap BD$.

[b]2. Antigonal conjugates[/b]

Let ABC be a triangle, and D a point which differs from its vertices A, B, C and from its orthocenter. Then, the reflection of the point D in the Poncelet point of the four points A, B, C, D is called the [b]antigonal conjugate[/b] of the point D with respect to triangle ABC. That this "antigonal conjugate" transformation is really a conjugation can be seen from the following fact:

[b]Theorem 6.[/b] In the plane of a triangle ABC, select a point D that doesn't lie on any of the circumcircles of the triangles BHC, CHA, AHB, ABC, where H is the orthocenter of triangle ABC. Let D' be the antigonal conjugate of the point D with respect to triangle ABC. Then, the antigonal conjugate of the point D' with respect to triangle ABC is the point D again. [If the point D would lie on one of the circumcircles of the triangles BHC, CHA, AHB, ABC, then its antigonal conjugate D', if it exists, would be one of the points A, B, C, H, and thus the antigonal conjugate of this point D' would be not defined.]

Note that it can be hard to prove Theorem 6 just by means of the above definition of antigonal conjugation, but it will easily follow from the next property:

[b]Theorem 7.[/b] If D' is the antigonal conjugate of a point D with respect to a triangle ABC, then < BD'C = - < BDC, < CD'A = - < CDA and < AD'B = - < ADB.

Hereby, we use directed angles modulo 180°. Again, we can show:

[b]Theorem 8.[/b] The antigonal conjugate D' of a point D with respect to a triangle ABC is simultaneously the antigonal conjugate of the point D with respect to any of the triangles BHC, CHA, AHB, where H is the orthocenter of triangle ABC.

As a consequence of Theorems 7 and 8, we can get:

[b]Theorem 9.[/b] If H is the orthocenter of triangle ABC, and D' is the antigonal conjugate of the point D with respect to triangle ABC, then < AD'H = - < ADH, < BD'H = - < BDH, < CD'H = - < CDH.

From Theorem 6, we can conclude:

[b]Theorem 10.[/b] If D' is the antigonal conjugate of a point D with respect to a triangle ABC, then the Poncelet point of the four points A, B, C, D is simultaneously the Poncelet point of the points A, B, C, D', and thus it also lies on the nine-point circles of triangles BCD', CD'A and D'AB.

Another crucial property of antigonal conjugates is:

[b]Theorem 11.[/b] Let E be the isogonal conjugate of a point D with respect to a triangle ABC, let E' be the inverse of the point E in the circumcircle of triangle ABC. Then, the isogonal conjugate of the point E' with respect to triangle ABC is the antigonal conjugate D' of the point D with respect to triangle ABC.

In other words, if two points are inverse to each other with respect to the circumcircle of triangle ABC, then their isogonal conjugates with respect to triangle ABC are two points which are antigonal conjugates of each other with respect to triangle ABC.

[b]Theorem 12.[/b] If A', B', C' are the reflections of a point D in the sidelines BC, CA, AB of a triangle ABC, then the circumcircles of triangles A'BC, B'CA, C'AB and A'B'C' pass through the antigonal conjugate D' of the point D with respect to triangle ABC.

[b]Theorem 13.[/b] If A", B", C", D" are the centers of the circumcircles of triangles A'BC, B'CA, C'AB, A'B'C', respectively, then the point D' is the inverse of the point D" in the circumcircle of triangle A"B"C".

[b]3. An application: The Antigray point[/b]

As an application of the above theory, we can prove some results about the "Antigray point" of a triangle.

Let's start with the basic facts that don't need any theory:

[b]Theorem 14.[/b] Let $I_{a}$, $I_{b}$, $I_{c}$ be the excenters of a triangle ABC, and let $A_{a}^{\prime}$, $B_{b}^{\prime}$, $C_{c}^{\prime}$ be the reflections of these excenters $I_{a}$, $I_{b}$, $I_{c}$ in the lines BC, CA, AB. Then, the circumcircles of triangles $A_{a}^{\prime}BC$, $B_{b}^{\prime}CA$, $C_{c}^{\prime}AB$ and the lines $AA_{a}^{\prime}$, $BB_{b}^{\prime}$, $CC_{c}^{\prime}$ all have one common point K.

This point K will be called the [b]Antigray point[/b] of triangle ABC.

Note that this point K is X(80) in [url=http://faculty.evansville.edu/ck6/encyclopedia/]Clark Kimberling's ETC[/url], and I am calling it Antigray point because of the analogy to the definition of the Gray point X(79).

Now, we denote by I the incenter of triangle ABC, and apply the above theory to D = I. We get:

[b]Theorem 15.[/b] If I is the incenter of a triangle ABC, then the Poncelet point P of the points A, B, C, I is the Feuerbach point of the triangle ABC, that is, the point of tangency of the nine-point circle and the incircle of triangle ABC. This Feuerbach point P lies on the nine-point circles of triangles IBC, CIA, ABI, HAI, HBI and HCI, on the pedal circle of the point A with respect to triangle IBC, on the pedal circle of the point B with respect to triangle CIA, on the pedal circle of the point C with respect to triangle ABI, and on the circle through the points where the angle bisectors of the angles CAB, ABC, BCA meet the sides BC, CA, AB.

This was quite a plenty of properties of the Feuerbach point, but now we can consider the antigonal conjugate of the point I and we obtain:

[b]Theorem 16.[/b] The Antigray point K of triangle ABC is the antigonal conjugate of the incenter I with respect to triangle ABC. This Antigray point K is the reflection of the incenter I in the Feuerbach point P of triangle ABC. The isogonal conjugate of the Antigray point K with respect to triangle ABC is the inverse of the incenter I in the circumcircle of triangle ABC.

  Darij

Just after I looked the post I took a piece of paper and started looking for some anothers...  :D

What about

$\bigstar$ Given $\triangle ABC$, $PP'$ is a diameter of the rectangular hyperbola circunscribed $\triangle ABC$.

$\bigstar$ Given $\triangle ABC$ and $P$and $P'$ two antigonal conjugates wrt $\triangle ABC$. Then the midpoint of $\overline{PP'}$ lies on the nine-point circle of $\triangle ABC$.

$\bigstar$ Given two points with the same tripolar coordinates, then its isogonal conjugates are antigonal conjugates.

$\bigstar$ Corollary. $F^{+}$ and $F^{-}$ are antigonal points (because the isodinamyc points have the same tripolar coordinates) and the midpoint of $F^{+}F^{-}$ lies in the nine-point circle of $\triangle ABC$.

José Carlos Chávez Sandoval

[quote="hucht"]$\bigstar$ Given $\triangle ABC$ and $P$and $P'$ two antigonal conjugates wrt $\triangle ABC$. Then the midpoint of $\overline{PP'}$ lies on the nine-point circle of $\triangle ABC$.[/quote]

This follows from the definition of antigonal conjugates and Poncelet points.

[quote="hucht"]$\bigstar$ Given two points with the same tripolar coordinates, then its isogonal conjugates are antigonal conjugates.[/quote]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=46344]If two distinct points have the same homogeneous tripolar coordinates, then they are inverse to each other with respect to the circumcircle of the triangle.[/url] So we get Theorem 11.

  Darij

I'll reformulate the theorem

[b]Theorem 11.[/b] Given two points $P_{1}$ and $P_{2}$ which are inverses wrt the circumcircle of $\triangle ABC$. Their isogonal conjugates $P_{1}^{*}$, $P_{2}^{*}$ of $P_{1}$ and $P_{2}$ respectively are antigonal conjugates wrt $\triangle ABC$.

[b]Proof. [/b]Let $O$ be the circumcenter of $Ítriangle ABC$, let $P_{1}$, $P_{2}$ be two inverse points wrt the circumcircle of $\triangle ABC$ and $X$, $Y$ two points on this circumcircle wich are collinear with $P_{1}$ and $P_{2}$. Since $P_{1}$ and $P_{2}$ are inverses wrt the circumcircle, $m\angle AP_{2}O=m\angle P_{1}AO$ and $m\angle P_{2}AO=m\angle AP_{1}O$. Similary we get $m\angle CP_{2}O=m\angle P_{1}CO$ and $m\angle P_{2}CO=m\angle CP_{1}O$. Adding $m\angle AP_{1}C+m\angle AP_{2}C=m\angle AP_{1}O+m\angle OP_{1}C+m\angle AP_{2}O+m\angle OP_{2}C=m\angle P_{2}AO+m\angle OCP_{2}+m\angle AP_{2}O+m\angle OP_{2}C=m\angle AOY+m\angle YOC=2m\angle ABC=2m\angle AXC$.

From the definition of isogonal conjugates we get $m\angle CAP^{*}_{2}=m\angle P_{2}AB$ and $m\angle ACP_{2}^{*}=P_{2}CB$ and adding $m\angle CAP_{2}^{*}+m\angle ACP_{2}^{*}=m\angle P_{2}AB+m\angle P_{2}CB=m\angle ABC-m\angle AP_{2}C=m\angle AXC-AP_{2}C$. Similary $m\angle P_{1}^{*}CA+m\angle CAP_{1}^{*}=m\angle AP_{1}C-m\angle AXC$. Thus $m\angle AP_{2}C+m\angle AP_{1}C=2m\angle ABC=2m\angle AXC$ and hence $m\angle CAP_{2}^{*}+m\angle ACP_{2}^{*}=m\angle P_{1}^{*}CA+m\angle CAP_{1}^{*}$ thus $m\angle AP_{2}^{*}=m\angle CP_{1}^{*}A$. Similary fot the othe vertices we get the result. $\blacksquare$


Here is the IMO shortlist problems 1994, I got it from internet.

Here are the links to some of these problems discussions:

The sequence $x_0, x_1, x_2, ...$ is defined by $x_0 = 1994$, $x_{n+1} = \frac{x_n^2}{x_n + 1}$. Show that $[x_n ] = 1994 - n$ for $0 \leq n \leq 998$.
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=31736[/url]

There are $n+1$ cells in a row labeled from $0$ to $n$ and $n+1$ cards labeled from $0$ to $n$. The cards are arbitrarily placed in the cells, one per cell. The objective is to get card $i$ into cell $i$ for each $i$. The allowed move is to find the smallest $h$ such that cell $h$ has a card with a label $k > h$, pick up that card, slide the cards in cells $h+1$, $h+2$, ... , $k$ one cell to the left and to place card k in cell k. Show that at most $2^n - 1$ moves are required to get every card into the correct cell and that there is a unique starting position which requires $2^n - 1$ moves.
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=31733[/url]

Define the sequence $a_1, a_2, a_3, ...$ as follows. $a_1$ and $a_2$ are coprime positive integers and $a_{n+2} = a_{n+1}a_n + 1$. Show that for every $m > 1$ there is an $n > m$ such that $a_m^m$ divides $a_n^n$. Is it true that $a_1$ must divide $a_n^n$ for some $n > 1$?
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=31735[/url]

This are the problems from 1994 I have posted until now...

Thanks to Orl for helping me keeping the armony of the site... :lol:


[quote="arqady"]I think, the following inequality is more interesting:
For non-negative $ a,$ $ b$ and $ c$ prove that:
\[ (a^3 \plus{} b^3 \plus{} c^3)^2\geq(a^4 \plus{} b^4 \plus{} c^4)(ab \plus{} ac \plus{} bc)
\]
[/quote]
This inequality is equilvalent to :
$ \sum (a\minus{}b)^{4}(a\plus{}b)^{2} \plus{} \sum (a^{4}b^{2}\plus{}a^{4}c^{2}\minus{}2a^{4}bc) \geq 0$
which is true

[quote="khashi70"][quote="arqady"]
For non-negative $ a,$ $ b$ and $ c$ prove that:
\[ (a^3 \plus{} b^3 \plus{} c^3)^2\geq(a^4 \plus{} b^4 \plus{} c^4)(ab \plus{} ac \plus{} bc)
\]
[/quote]
This inequality is equilvalent to :
$ \sum (a \minus{} b)^{4}(a \plus{} b)^{2} \plus{} \sum (a^{4}b^{2} \plus{} a^{4}c^{2} \minus{} 2a^{4}bc) \geq 0$
which is true[/quote]
Wow! Very nice proof, khashi70! :lol:
My proof see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106975


(1) Let $a,b,c$ be distinct complex numbers such that $\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=k.$
Find the value of $k.$

(2) Let $a,b,c$ be distinct complex numbers such that $\frac{a}{1-b}=\frac{b}{1-c}=\frac{c}{1-a}=l.$
Find the value of $l.$

1)
The product of the three expressions is 1, so we have
$1=k^3$. Thus $k=-\frac 1 2 \pm \frac{\sqrt 3}{2}i$

We have that $a = \ell-\ell b = \ell-\ell^{2}+\ell^{2}c = \ell-\ell^{2}+\ell^{3}-\ell^{3}a$, so $a \left( 1+\ell^{3}\right) = \ell-\ell^{2}+\ell^{3}$.

If $1+\ell^{3}\neq 0$, then $a = b = c$, contradiction.
Else we that $\ell^{3}=-1$, so $\ell = \exp \left( \pm \frac{\pi i}{3}\right)$ ($\ell =-1$ doesn't "work").


function
real analysis
real analysis unsolved

If $ f(x)$ satisfy $ xf(y)\plus{}yf(x)\equal{}(x\plus{}y)f(x)f(y)$, $ x,y \in \mathbb{R}$

Find $ f(x)$ :D

[quote="Xingyuan"]If $ f(x)$ satisfy $ xf(y) \plus{} yf(x) \equal{} (x \plus{} y)f(x)f(y)$, $ x,y \in \mathbb{R}$

Find $ f(x)$ :D[/quote]

Put $ x \equal{} 0$ we get $ yf(0)[1 \minus{} f(y)] \equal{} 0 \forall y \in R$. So either $ f(0) \equal{} 0$ or $ f(y) \equal{} 1 \forall y \neq 0$ (in this case the solution is $ \begin{cases} f(0) \equal{} c \\
f(x) \equal{} 1 \forall x \neq 0 \end{cases}$, where $ c$ is any real constant.)

If $ f(0) \equal{} 0$, then

[b]Case 1:[/b] there exist $ x_0 \neq 0$ so that $ f(x_0) \equal{} 0$, then put $ x \equal{} x_0$ we have $ x_0f(y) \equal{} 0 \forall y \in R \Rightarrow f(x) \equal{} 0 \forall x \in R$

[b]Case 2:[/b] $ f(x) \neq 0 \forall x \neq 0$ , then $ \frac {x}{f(x)} \plus{} \frac {y}{f(y)} \equal{} x \plus{} y \forall x,y \neq 0 \Rightarrow f(x) \equal{} 1 \forall x \neq 0$

So $ \begin{cases} f(0) \equal{} c \\
f(x) \equal{} 1 \forall x \neq 0 \end{cases}$, where $ c$ is any real constant , or
$ f(x) \equal{} 0 \forall x \in R$ , are the only solutions.

an other solution:
for $ -x=y$ we have: $ xf(x)=f(-x)x$,then $ \forall x\in{\mathbb{R^{*}}}$ :

$ f(x)=f(-x)$ $ \Rightarrow$ $ f$ is even.

from the f.e we have: $ xf(y)(1-f(x))=yf(x)(f(y)-1)$,we replace $ y$ with $ -y$ we get:

$ xf(y)(1-f(x))=yf(x)(f(y)-1)=-yf(x)(f(y)-1)$ ($ y\neq 0$) $ \Rightarrow$ $ f(x)=1$ and $ f(0)=d$ $ (d\in{\mathbb{R}})$ or $ f(x)=0 (\forall x\in{\mathbb{R^{*}}})$and $ f(0)=d'$ $ (d'\in{\mathbb{R}})$.


with a counting argument show:

\[ \binom{n\plus{}m\minus{}1}{n}\equal{}\sum_{i\equal{}1}^{m}\binom{m}{i}\binom{n\minus{}1}{i\minus{}1}\]

\[ \binom{n \plus{} m \minus{} 1}{n} \equal{} \sum_{i \equal{} 1}^{m}\binom{m}{i}\binom{n \minus{} 1}{n \minus{} i}
\]The LHS denotes the number of ways we can pick $ n$ objects out of a group of $ (n\minus{}1) \plus{} m$ total objects. The RHS denotes the same quantity, except we count the number of ways to pull $ i$ objects from a $ m$-group, and pull the remaining $ n\minus{}i$ objects from the $ n\minus{}1$-group, counting through all $ i$ (note also we cannot pull $ 0$ objects from the $ m$-group).

Sorry,
Only prove the argument of summation from i=1 up to i=n.
Cannot notmake sure whelther the orginal statement  is correct or not.
My argument is written in the attachment.

$ \text {Consider }$ $ (1\plus{}x)^{n\plus{}m\minus{}1} \equal{} (1\plus{}x)^m(1\plus{}x)^{n\minus{}1 },$ $ \text {where m,n are positive integers ,} n{\ge }m$

$ (1\plus{}x)^{n\plus{}m\minus{}1} \equal{} (1\plus{}x)^m(1\plus{}x)^{n\minus{}1 }$

$ (1\plus{}x)^{n\plus{}m\minus{}1} \equal{} \sum_{j\equal{}0}^{n}\binom{n\plus{}m\minus{}1}{j} x^j\equal{}1\plus{}\sum_{j\equal{}1}^{n\plus{}m\minus{}1}\binom{n\plus{}m\minus{}1}{j} x^j$..............(1)

$ \text {By using binomial expansion}
(1\plus{}x)^{m} (1\plus{}x)^{n\minus{}1} \equal{} \sum_{r\equal{}0}^{m}\binom{m}{r} x^r\cdot\sum_{j\equal{}0}^{n\minus{}1}\binom{n\minus{}1}{i} x^i$..............(2)

$ \text {Consider the term of  } x^n$

$ \text {In(1), } \text {the coeff. of }   x^n  \equal{}\binom{n\plus{}m\minus{}1}{n}$

$ \text {In(2),}  \text {the coeff. of }   x^n \equal{}\binom{m}{1} \binom{n\minus{}1}{n\minus{}1}\plus{}\binom{m}{2}\binom{n\minus{}1}{n\minus{}2}\plus{}\binom{m}{3}\binom{n\minus{}1}{n\minus{}3}\plus{}\cdots\plus{}\binom{m}{m}\binom{n\minus{}1}{n\minus{}m}$

$ \equal{}\sum_{i\equal{}1}^{m}\binom{m}{i}\binom{m\minus{}1}{n\minus{}i}$

$ \text {Thus,}  \binom{n\plus{}m\minus{}1}{n}\equal{}\sum_{i\equal{}1}^{m}\binom{m}{i}\binom{m\minus{}1}{n\minus{}i}$


I'll do the second one.

I don't feel like doing the first one. :P

Ok. This problem is called composite function unless I'm mistaken.

So, we have:

$f(g(x)) = f(3x-2) = 4(3x-2)^2-9(3x-2)+2$

Hence:

The result is $36x^2-75x+36$.

[quote="dk2"][hide]
Hmm... I wonder if the first one would be 8x-9+h...
If only h equaled zero...
Or even better, if h approached zero... :P
[/hide][/quote]

close...your just missing a 4 infront of the $h$

[quote="cindy"]f(x)=4x 2 -9x+2
g(x)= 3x-2

find:      [u]f(x+h)-f(x)[/u] divided by h

and:      f[g(x)][/quote]Cindy,

Two questions: are you taking an introductory calculus class right now? and, what is it about these problems that you're having trouble with? These questions both seem to deal with function notation, so maybe that's the issue.


Find the ordered pairs (x,y) such that $x^{2}-2xy+y^{2}=25$ and $x^{2}+xy-y^{2}=-1$

[hide="answer"]
taking the first equation $x^{2}-2xy+y^{2}= 25$, it is clear that

$(x-y)^{2}= 5^{2}$
$x = 5+y$

subbing into the other equation we get

$y^{2}+15y+26 = 0$
$(y+13)(y+2)$
thus $y =-2,-13$ and $x =-8, 3$

making our ordered pairs $(3,-2)$ and $(-8,-13)$
[/hide]

x - y = -5  is another case from first equation

  x = y - 5

  Substituting into second equation and simplifying gives

   y^2 - 15y + 26 = 0

 Which factors as

  (y - 2)(y - 13) = 0

  From this we get two other ordered pairs of solutions

   (-3,2)  and (8,13)


Find all functions $ f: \mathbb{Z}\rightarrow\mathbb{Z}$ such that

(i) $ f(f(m \minus{} n)) \equal{} f(m^2) \plus{} f(n) \minus{} 2nf(m)$ for all $ m,n\in\mathbb{Z}$

(ii) $ f(1) > 0$

Is there any?

For $ n\equal{}0$, it gives $ f(f(m))\equal{}f(m^2)\plus{}f(0)$ for all $ m$.
Thus, the initial equation leads to
$ f((m\minus{}n)^2)\plus{}f(0)\equal{}f(m^2)\plus{}f(n)\minus{}2nf(m)$ for all $ m,n$. (1)

For $ m\equal{}n$, we get $ 2f(0) \equal{} f(m^2)\plus{}f(m)\minus{}2mf(m)$ for all $ m$. (2)
Since the LHS of (1) is symmetric with respect to $ m$ and $ n$, then so is the RHS, which gives that
$ f(m^2)\plus{}f(n)\minus{}2nf(m)\equal{}f(n^2)\plus{}f(m)\minus{}2mf(n)$ for all $ m,n$. (3)

For $ n\equal{}0$ and (3), we have $ f(m^2)\equal{}f(m)\minus{}2mf(0)$ for all $ m$.
Thus, (2) can be rewritten as $ 2f(0) \equal{} (2\minus{}2m)f(m)\minus{}2mf(0)$ for all $ m$
which is $ f(m) \equal{} \minus{}f(0) \plus{} \frac {4f(0)}{2\minus{}2m}$ for all $ m \neq 1$.
Since $ f(m)$ is supposed to be an integer, it forces $ 4f(0)$ to be a multiple of $ 2\minus{}2m$ for all $ m\neq 1$. Thus $ f(0)\equal{}0$.
Then $ f(m)\equal{}0$ for all $ m\neq 1$.

Back to (3) for $ n\equal{}1$ and $ m\equal{}2$, we obtain $ 4f(1) \equal{} 0$, a contradiction.

Pierre.


Calculate this

$ T \equal{} \tan{20} \minus{} \tan{40} \plus{} \tan{80}$
 (in degree)
[hide="and ..."]and of course I know it is $ 3\sqrt3$[/hide]

[hide]After a lot of algebra with identities and what not, we get the equation:
$ x^3\minus{}3\sqrt3x\minus{}3x\plus{}\sqrt3\equal{}0$, with roots at tan240/3, tan60/3, and tan420/3.
The desired expression is tan20-tan40+tan80. We see that tan240/3=tan80, tan60/3=tan20, and tan420/3=tan140. Because tan(180-x)=-tanx, tan140=tan40, and so tan20-tan40+tan80 is equal to the sum of the roots of the polynomial, which is $ 3\sqrt3$, taking the second coefficient.[/hide]

Wow, how did you do that?  :) Can you be a little more specific?

Btw, I think we have a simplier answer for this. (since it is given in a very normal quiz  :D )

[hide="This is what I did in the quiz ..."]

We know that $ 3\sqrt{3} \equal{} 3\tan60$
So we are to prove $ P \equal{} \tan20 \minus{} \tan40 \plus{} \tan80 \minus{} 3\tan60 \equal{} 0$

$ P \equal{} (\tan20 \minus{} \tan60) \minus{} (\tan40 \plus{} \tan60) \plus{} (\tan80 \minus{} \tan60)$
$ \equal{} \frac { \minus{} \sin40}{\cos20\cos60} \minus{} \frac {\sin100}{\cos40\cos60} \plus{} \frac {\sin20}{\cos80\cos60}$

Here we need to prove $ \minus{} \frac {\sin40}{\cos20} \minus{} \frac {\sin100}{\cos40} \plus{} \frac {\sin20}{\cos80} \equal{} 0$
Or $ \minus{} \sin20 \minus{} \sin40 \plus{} \cos10 \equal{} 0$
and that is obviously true  :)
[/hide]

Triple Angle formula? (had to derive, which took a bit of time because of all the expanding)


Να αποδείξετε ότι η ακολουθία $\alpha_{\nu+1}=1-e^{-\alpha_{\nu}}$ με $\alpha_{1}=1$ με $\nu\in N$ συγκλίνει..

Ναι και συγκλινει στο $0$. :)

i akolu8ia apodikniete oti siglini eukola..ine fragmeni sto $(0,1)$+gnisios monotoni afu $a_{n+1}=1-e^{-a_{n}}\leq a_{n}$  apo tin $e^{x}\geq x+1$ pali... :P
opote pernontas oria stin arxiki vriskume ontos oti siglini sto $0$.. :)
endiaferon 8a itan an i antistixi sira siglini..... :huh:

 :D


Let $ABCD$ and $ABEF$ be two squares situated in two perpendicular planes and let $O$ be the intersection of the lines $AE$ and $BF$. If $AB=4$ compute:

a) the distance from $B$ to the line of intersection between the planes $(DOC)$ and $(DAF)$;

b) the distance between the lines $AC$ and $BF$.

[hide]u must draw the cube ABCDTSFE
a) $(DOC)\in(DMNC)$  where M,N re the middlepoints of AF,BE=>the line of intersection between (DOC) and (DAF) is DM...and the distance from B to DM is the high in the isoscel triangle BDM
b)was easier.. $ AC\in(CAS)$, and   $BF\in(BFT)$, $(CAS)\|(BFT)$  CSAD and BFTE are regular pyramids$\Rightarrow \ TQ\bot(BFT)$ and $DR\bot(CSA)$ (then u must prove that D,R,Q,E $\in (DE)$ AND THAT DR=RQ=QE) =>the distance between the lines AC and BF is  QR=$\frac{4\sqrt{3}}{3}$[/hide]


Anyone on the board from southern Africa is invited to contact me if you are looking for math contests and programs to get involved with.  Earlier this year i helped organize some programs in that region and have some contacts who could help get you involved.  In addition, i would appreciate hearing of any programs anyone knows about in South Africa and adjacent countries..

I know nothing myself but I know Arne is moving to South Africa in a couple of days, I'm sure he'll be very interested! (if he doesn't know everything yet :))


Hey, I was playing cards today, and for some sad reason, I made up a math problem to do while playing. Anyways, I made some headway on it, but I don't have it solved yet.

The Ace of Spades is at the very bottom of the deck(of 52 cards). The deck is shuffled in a way so that you first cut the deck in half, then alternately pile cards from the bottom of the two halves, starting with the bottom card of the top half. In other words, the way everybody shuffles. The cards alternate perfectly so that any two consecutive cards that are now in the pile are from two different halves. If you continue to shuffle like this, will the Ace of Spades ever reach the top, and if so,  how many times do you need to shuffle before the Ace of Spades becomes the top card?

I like this problem. Fun to work with...

Edit:I solved this, I think.[hide]There are 13 cards below it at the 15th shuffle and 13 cards below it after the 24th shuffle. It'll cycle.[/hide]
Well then, for which card positions(0 cards from the bottom, 1 card from the bottom, etc.) will the card become the top card after some number of shuffles?

no why?

wait, is it the one with relatively prime numbers below it?

[hide]Let the position of a card be $n$.

We can clearly see that the new position of a card after a shuffle can be split into two cases: if $1\leq n\leq 26$ and $27\leq n\leq 52$, where card 1 is the top card and card 52 is the bottom.

Trying out we see:
Card in position $26\Rightarrow 52$
Card in position $25\Rightarrow 50$
Card in position $24\Rightarrow 48$

Card in position $52\Rightarrow 51$
Card in position $51\Rightarrow 49$
Card in position $50\Rightarrow 47$

So we can form equations:
If $1\leq n\leq 26$, then after a shuffle the new position will be $2n$.
If $27\leq n\leq 52$, then after a shuffle the new position will be $2n-53$.

Applying this rule multiply times (we won't have to do it more than $52$ times since the cycle must repeat before then)

52,51,49,45,37,21,42,31,9,18,36,19,38,23,46,39,25,50,47,41,29,5,10,20,40,27,[b]1[/b].

So yes it does reach $1$, after $26$ shuffles.[/hide]

huh, I must have messed up somewhere, but I got different formulas than you did.


Gauss
logarithms
real analysis
real analysis unsolved

Discuss the convergence or divergence of the below series

$ \sum_{n\equal{}1}^{\infty}(\frac{(2n\minus{}1)!!}{(2n)!!})^{k}$
 :roll:

[quote="Xingyuan"]Discuss the convergence or divergence of the below series

$ \sum_{n \equal{} 1}^{\infty}(\frac {(2n \minus{} 1)!!}{(2n)!!})^{k}$
 :roll:[/quote]

$ \frac {(2n \minus{} 1)!!}{(2n)!!}\equal{} \prod_{m \equal{} 1}^{n}(1 \minus{} \frac {1}{2m})$. Apply $ \ln$, use the fact that $ \ln(1 \minus{} x)\approx \minus{} x$ for small $ x$, and then the fact that $ \sum_{m \equal{} 1}^{n}\frac {1}{m}\approx \ln n$ to see this expression is on the order of $ \frac {1}{\sqrt {n}}$. So as anik stated, the series converges for $ k > 2$, otherwise it diverges.


calculus
integration
trigonometry
real analysis
real analysis unsolved

Find \[ \int \sin x \sqrt{\sin 2x} dx.  \]

Is more interestingly $\int^{\frac{\pi}{2}}_0 \sin x\sqrt{\sin 2x}\ dx\ !.$

Here are two more intrresting integrals to compute: \[ I_{1}=\int_{0}^{\frac{\pi}{2}} \sin 3x \sqrt{\sin 2x} dx,  \]
and \[ I_{2}=\int_{0}^{\frac{\pi}{2}} \cos 3x \sqrt{\sin 2x} dx.  \]

Changing $x=\frac \pi 4 -t$ we get that:

\[ \int_0^{\frac{\pi}{2}} \sin x \sqrt{\sin 2x}dx=-\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin(\frac{\pi}{4}-t)\sqrt{\cos 2t}dt= \]
\[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin t \cos \left(\frac{\pi}{4}\right)\sqrt{\cos 2x}-\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos t \sin \left(\frac{\pi}{4}\right)\sqrt{\cos 2x}= \]
\[ -\frac{\sqrt 2}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos t \sqrt{1-2sin^2 t}dt = \]
\[ -\frac{\sqrt{2}}{2}\int_{-1}^1 \sqrt{1-2t^2}dt \]

Which is cauchy integral and can be calculated easily. I maybe mistake some signs but i willl fix them later.


[quote="HTA"][hide]
no offence but i dont like this kind of inequality , it looks ugly and it tests one only about calculations and nothing else .
anyway , the key point is that the equality occurs when (x,y,z) = (1/6,1/3,1/2) [/hide][/quote]
Yes,but I don't need calculations :wink:


there are 1000 locks.there are 1000 persons numbered1,2,....1000.each one closes & opens(like a person opens if it is closed& locks it if it is open)  the locks on the basis of the multiples of his number.find the number of locks opened?

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=321505[/url]


[quote="anyone"]Source is [url=http://www.geocities.com/grissom_math/2003_tests/Comprehensive_Test.doc]here[/url].
7.	A ball is dropped from a height of 4096 ft.  Each time it bounces, it rebounds to 1/16 of its previous height.  Find the distance the ball has traveled in feet if it is caught at the highest point after the third bounce.
	A. 4640	B. 4641	C. 4642	D. 4643	E. 4644
I keep getting 4674....

  9.	Given parallelogram ABCD, with AB = 10, AD = 12, and BD =$2\sqrt{13}$  , find AC.
	A.  10	B.  $2\sqrt{13}$	C.  $12\sqrt{3}$ 	D. $2\sqrt{109}$  	E. cannot be determined[/quote]

[hide="#7"]7. $4096=2^{12}$
The distance traveled is equal to
$2^{12}+\frac{1}{2^{4}}\cdot 2 \cdot 2^{12}+\frac{1}{2^{8}}\cdot 2 \cdot 2^{12}+\frac{1}{2^{12}}\cdot 2^{12}$
$=2^{12}+2^{9}+2^{5}+1=4096+512+32+1=\boxed{4641}$.

Thus $\boxed{B}$.

(How did you get 4674?)[/hide]

[hide="#9"]9. In a parallelogram, two adjacent angles are supplementary.

$\cos(\angle BAD)=\frac{100+144-52}{2 \cdot 10 \cdot 12}$
$\cos(\angle ADC)=\frac{100+144-AC^{2}}{2 \cdot 10 \cdot 12}$

Therefore $100+144-AC^{2}=-(100+144-52)=-192$
$AC^{2}=436$
$AC=\boxed{2 \sqrt{109}}$.

Thus $\boxed{D}$.[/hide]


limit
real analysis
real analysis solved

For problem (1):

I only have the convergence of a_n up until now. It's obvious that a_n is in (0,1) for all n, because a_1 3 - a_1 2 is in (-1,0) and we use induction. a_(n+1)-a_n=(a_n-1)(a_n 2 -1) >=0, so the sequence a_n is increasing and bounded, so it's convergent. Actually, it's convergent to 1, because we have a_n->l, so l 3  - l 2 -l+1=0, so (l-1)(l 2 -1)=0, so l=1.

The 1st one was given by my teacher for a homework or smth like that, here is what I have done:

a_(n+1)-1=(a_n-1)(a_n)^2 and giving values from n=1 to n we get that
a_(n+1)-1=(a1-1)*a1^2*..*(a_n)^2 from where we get that the product:
a1^2*..*(a_n)^2=(a_(n+1)-1)/(a1-1) and we get that lim(a1*..*an)=lim sqrt((a_(n+1)-1)/(a1-1))=0 so the limit is 0

[quote="Lagrangia"]we get that the product:
a1^2*..*(a_n)^2=(a_(n+1)-1)/(a1-1) and we get that lim(a1*..*an)=lim sqrt((a_(n+1)-1)/(a1-1))=0 so the limit is 0[/quote]

Well let's continuous

Give an asmptotic expansion of a1*..*an when n->+oo, I mean hwo fast
does it tends to 0 ?

for the 2nd part observe that it's easy to prove that a_n>=0  and also:
a1-a0=((a0-1)(a0-sqrt(2)(a0+sqrt(2))/(sqrt(a0^3+3)+a0+1)

and now take 5 cases:
a0>sqrt(2); a0=sqrt(2); a0=1; a0 is in (1,sqrt(2)) and a0 is in [0,1) and see what happens! :D :D
then you can see that the only values for which the sequence converges are [0,sqrt(2)].

hope it's okay! :D

cheers!

Moubinool, what do you mean by "how fast does it tend to 0" ?

I explain with an example take x_0>0 and (1) x(n+1)=x(n)+1/x(n)
it is easy to see x(n) -->+oo
But how fast, this mean find an asymptotic expression of x(n)
take the square of x(n+1)^2=x(n)^2+2+1/x(n)^2=>
(3) x(n+1)^2 - x(n)^2 --> 2
with Cesaro theorem if u(n)->k then (u(1)+...u(n))/n->k
(3)=> (x(n)^2 - x(1)^1)/n -> 2 => x(n) is equivalent to  \sqrt (2n)
we knwo that x(n) tend to +oo, like \sqrt (2n)

Oh, so we want a closed expression for a sequence b_n s.t.
( \pi {k=1->n}a_n)/b_n -> 1 as n->00.

with the condition b(n)--> 0 because (a1)a(2)...a(n)-->0


P is a point inside a given triangle ABC; D, E, F are the feet of the perpendiculars from P to the lines BC, CA, AB, respectively. Find all P for which BC/PD+CA/PE+AB/PF is minimal

I think Cauchy?s inequality works just fine here:

(BC/PD+CA/PE+AB/PF)*(BC*PD+CA*PE+AB*PF) >= (BC+CA+AB)^2. But it?s obvious that BC*PD+CA*PE+AB*PF = 2S where S is the area of triangle ABC. We have BC/PD+CA/PE+AB/PF >= 4p^2/S where p is the semiperimeter. We have equality iff the triplets (BC/PD, CA/PE, AB/PF) and (BC*PD, CA*PE, AB*PF) are proportional (it?s well known that this is the equality case for Cauchy-Buniakovsky-Schwarz inequality). This translates to PD=PE=PF, so P=I (the incenter).


Show that a natural number $ p>1$ is a prime if and only if there exists a natural number $ n<p$ such that $ (n\minus{}1)!(p\minus{}n)!\equiv(\minus{}1)^{n}(modp)$

Evidently $ (p\minus{}n)! \equal{} (p\minus{}n)(p\minus{}n\minus{}1)...(2)(1)\equiv (\minus{}1)^{p\minus{}n}n(n\plus{}1)...(p\minus{}1)$, hence

$ (n\minus{}1)! (p\minus{}n)!\equiv (\minus{}1)^{p\minus{}n}(p\minus{}1)!$

For $ p \equal{} 2$ we choose $ n \equal{} 1$.  The rest follows by Wilson's theorem.


calculus
integration
logarithms
calculus computations

That diverges. The value is $ \plus{}\infty.$

Note that as $ x\to\infty,$ the denominator goes to $ 1,$ so the integrand behaves like $ \ln x.$ That integral diverges.

Perhaps you meant to write something else?


Prove:
\[ \sqrt{ab(a+b)}+\sqrt{ab(a+b)}+\sqrt{ca(c+a)} \ge \sqrt{10abc+(a+b)(b+c)(c+a)}  \]
For $a,b,c \ge 0$

[quote="Fermat -Euler"][quote="zhaobin"]Prove:
\[ \sqrt{ab(a+b)}+\sqrt{ab(a+b)}+\sqrt{ab(a+b)} \ge \sqrt{10abc+(a+b)(b+c)(c+a)}  \]
For $a,b,c \ge 0$[/quote]
???
it's
$\sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ac(a+c)} \ge \sqrt{10abc+(a+b)(b+c)(c+a)}$   :?:[/quote]
yes,sorry :blush:

[quote="zhaobin"][quote="Fermat -Euler"]
it's
$\sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ac(a+c)} \ge \sqrt{10abc+(a+b)(b+c)(c+a)}$   :?:[/quote]
yes[/quote]
$\sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ac(a+c)} \ge \sqrt{10abc+(a+b)(b+c)(c+a)}\Leftrightarrow$
$\Leftrightarrow(\sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ac(a+c)})^2 \ge (\sqrt{10abc+(a+b)(b+c)(c+a)})^2\Leftrightarrow$
$\Leftrightarrow\sum\sqrt{a(a+b)(a+c)}\geq6\sqrt{abc}.$ This is obviously true by AM-GM. ;)


Could someone give me a quick solution to [tex]\sin{10}\sin{50}\sin{70}?[/tex]

Thank You!

$\sin(90-x)=\cos(x)$
$\cos(x)\cdot\cos(2x)\cdot\cos(4x)=1/8$
In this case $x=20\textdegree$. $\blacksquare$


Two circles with radii 1 and 2 touch each other and a line as in the figure. In the region between the circles and the line, there is a circle with radius $ r$ which touches the two circles and the line. What is $ r$?

[img]http://i250.photobucket.com/albums/gg265/geometry101/GeometryImage2.jpg[/img]

A. 1/3
B. $ \frac {1}{\sqrt {5}}$
C. $ \sqrt {3} \minus{} \sqrt {2}$
D. $ 6 \minus{} 4 \sqrt {2}$
E. None of these

[hide]Pythagorean theorem gets you to the answer, D.[/hide]

[hide="Answer"]Apply the pythagorean theorem and connect the lines. This results in choice $ \boxed{D}$.[/hide]

It's classic but can you guys put a solution to this problem? It'll help me a lot when I later add the solutions to them.  :)

Given two touching circles with radii $ R$ and $ r$ their common tangent has length $ 2\sqrt{Rr}$. So we obtain $ 2\sqrt{Rr} \equal{}2\sqrt{Rx} \plus{}2\sqrt{rx}$ and a nice formula:

$ \frac{1}{\sqrt x} \equal{}\frac{1}{\sqrt R} \plus{}\frac{1}{\sqrt r}.$

[quote="delta"]Given two touching circles with radii $ R$ and $ r$ their common tangent has length $ 2\sqrt {Rr}$. So we obtain $ 2\sqrt {Rr} \equal{} 2\sqrt {Rx} \plus{} 2\sqrt {rx}$ and a nice formula:

$ \frac {1}{\sqrt x} \equal{} \frac {1}{\sqrt R} \plus{} \frac {1}{\sqrt r}.$[/quote]

hmm whats x?

[quote="unimpossible"]odd, the forumla doesn't work for this problem.

am i missing something?[/quote]

It works.

$ \frac{1}{\sqrt{r}}\equal{}\frac{2\plus{}\sqrt{2}}{2}\\
\sqrt{r}\equal{}\frac{2}{2\plus{}\sqrt{2}}\\
r\equal{}\frac{4}{6\plus{}4\sqrt{2}}\equal{}\frac{2}{3\plus{}2\sqrt{2}}\equal{}\boxed{6\minus{}4\sqrt{2}}$


[quote="236factorial"]If I drink 60 gallons of Vodka (some type of drink) a year, how many fluid ounces do I drink every hour, to the nearest hundreth?

tip: don't try this at home :D[/quote]
[hide]
there are 24x365 hours = 8760 hours a year.

60/8760 = 0.01 to the nearest hundreth

[/hide]

Well, this topic is certainly appropriate.  :D

[quote="math92"]your answer is in gallons and he wanted the answer in fluid ounces. ;)
Also, do you really drink that much 236!  I think you should change Vodka to Coke or something. :)[/quote]

I hate coke, and you can solve the problem while you're wondering why I drink so much distilled spirits.  :D


function
Vieta
quadratics
algebra
polynomial
Rational Root Theorem
quadratic formula

Hello everybody  :)
could someone interpret me in how many ways we can solve equations which are in this form :

         $ ax^{4} \plus{} bx^{3} \plus{} cx^{2} \plus{} dx \plus{} e \equal{} 0$ ???

There's a quartic formula out there somewhere, but trust me, you don't want to use it.

Use the [url=http://en.wikipedia.org/wiki/Rational_root_theorem]Rational root theorem[/url] and factoring methods.

[hide="Hint"]You use [b]Ferrari for equation: $ ax^4+bx^3+cx^2+dx+e = 0 $[/b][/hide]

First convert it to a depressed quartic:

[url]http://en.wikipedia.org/wiki/Quartic_function#Converting_to_a_depressed_quartic[/url]

[quote="grn_trtle"]There's a quartic formula out there somewhere, but trust me, you don't want to use it. [/quote]

Here it is
http://planetmath.org/encyclopedia/QuarticFormula.html

Most of the ways to solve it perfectly are pretty ugly, but your best bet would be Rational Root Theorem, factoring (perfect square, .....etc.), and using Vieta's formulas for special cases. For example if $e=0$, then you would know for sure that one factor is $x$.

The quartic formula is overkill given that this is the High School Pre-Olympiad forum. I second grn_trtle's suggestion of the Rational Root Theorem, and splitting the middle term to factor into two quadratics often works when the rational root theorem fails.

Also, in contest math scenarios, there is usually other information given which simplifies the problem, or there is no need to find the roots because [url=http://www.artofproblemsolving.com/Wiki/index.php/Vieta's_Formulas]Vieta's Formulas[/url] or [url=http://www.artofproblemsolving.com/Wiki/index.php/Newton%27s_Sums]Newton's Sums[/url] can be used instead.

IF A=K.Y ; E=M.N; C=K.N + Y.M + I.J; D= K.J + Y.I ; D= M.J +N.I

WE HAVE THAT
(KX^2 + I.X + M)(YX^2 + J.X + N)=AX^4 + BX^3 +...+E = 0

[hide="clearer version of above post"]
If:$(kx^2 + ix + m)(yx^2 + jx + n)=ax^4 + bx^3 +cx^2+dx+e = 0$
then
$ a=ky \\
 e=mn \\
 c=kn + ym + ij \\
 b= kj + yi \\
 d= mj +ni$

[/hide]
my interpretation of the problem:
[hide="is here"]define $x^4+\frac{b}{a}x^3 + \frac{c}{a}x^2 + \frac{d}{a}x +\frac{e}{a} $ as $P(x)$
define $f,g,h,j$ as follows:
$(x^2+fx+g)(x^2+hx+j)=P(x)$
Thus,
$fh+g+j=\frac{c}{a} \\
gj=\frac{e}{a} \\
fj+hg=\frac{d}{a} \\
h+f=\frac{b}{a} $
define $n$ as $\frac{c+e-d-b}{a}$and $m$as $ \frac{b+c+d+e}{a}$ then,
$fh+gj+g+j-h-f-fj-hg=n \\
(f+g)(h+j)-(f+g)-(h+j)=m \\
m+n=2fh+2gj-2f-2h \\
\frac{c+e}{a}=(fh+g+j)-g-j+gj-f-h \\
\frac{c+e}{a}=\frac{c+e}{a} -g-j-f-h \\
h+f=g+j \\
\frac{b}{a}=\frac{c}{a}-fh \\
fh=\frac{c-b}{a} \\
h=\frac{b}{a}-f \\
\boxed{f\frac{b}{a}-f^2=\frac{c-b}{a}}$
use the quadratic formula for $f$ and $h$ give a different one of the two values, if there are two,  that come out to each of them. if there is one, then give it to both. plug in a bunch to find the rest. use the quadratic formula in the end, then you are done. [/hide]


induction
number theory unsolved
number theory

Consider the sequence, whose first two terms are the numbers 1 and 2 respectively, and each next term is the least natural number, which has not yet appeared in the sequence and which is not coprime with the foregoing term of this sequence.

Prove that each natural number is a term of this sequence.

Let have a look here: http://mathlinks.ro/Forum/viewtopic.php/?t=2076
 Maybe it's easy for you but I can't see any ray of light :(

Let the sequence be $a_1, a_2, ...$. Suppose the result is false. Let $k$ be the smallest number that never appears in the sequence. It is divisible by at least one prime, say $p$. All the numbers smaller than $k$ must appear after a finite number of $a_i$, in other words, there must be a positive integer $i$ such that $1, 2, ..., k-1$ are all among $a_1, a_2, ..., a_i$. Now consider the $a_j$ with $j>i$. If any of them is a multiple of $p$, the next $a_j$ will be $k$. But we've assumed that $k$ never appears, so there's no multiple of $p, a_j$ with $j>i$. Then there are at most finitely many multiples of $p$ in the sequence.

This implies that there's some $C$ such that all multiples of $p$ in the sequence are smaller than $C$. So there's some $c$ such that none of $cp, (c+1)p, (c+2)p, ...$ appear in the sequence. Either $c$ or $c+1$ is even, so there's an even number which doesn't appear.

Let $2n$ be an even number that never appears. After a finite number of $a$, all the numbers among $1, 2, ..., 2n-1$ that will ever appear, will have appeared. So let $i$ be such that all the numbers among $1, 2, ..., 2n-1$ that will ever appear are among $a_1, ..., a_i$. Now if there's some even number $a_j$ with $j>i$, the next number that appears will either be $2n$ or some smaller number. Since we've assumed that all the numbers that would appear after a finite number of $a$ have appeared, we must have $2n=a_{j+1}$, contradiction. So $a_j$ is odd, that is, there are at most finitely many even numbers. Then there's some $C$ such that all even numbers appearing are smaller than $C$.

Suppose we're working with the $a_j$ with $j$ big enough, such that no new even numbers will appear; so that all the following $a_j$ are odd. Let $q$ be the smallest prime divisor of $a_j$. Either $a_{j+1}$ is smaller than $a_j$, or it is $a_j+q$. But $a_j+q$ is even, so $a_{j+1}<a_j$. By induction $a_j>a_{j+1}>a_{j+2}>...$ Contradiction, because there are finitely many numbers smaller than $a_j$.

The meat of the proof lies in the last paragraph, as I see it... once you know that an infinite number of even numbers are hit, you know that every even number is hit (because there are finitely many numbers smaller than it, so once all those smallers numbers that will go are gone and you hit some even $a_n$, it will be $a_{n+1}$).  Once you know that every even number is hit, you know that every number is hit (similar reasons.)


function
inequalities
triangle inequality
algebra unsolved
algebra

Find all functions $f: \mathbb{R}\to \mathbb{R}$ such that: \[\forall r\in\mathbb{Q}, ~~|f(x)-f(r)|\leq 2(x-r)^{2}\]

Let $g(x)=f(x)-f(0)$. Then the same inequality holds for $g(x)$, and we have $|g(kx)-g((k-1)x)|\le2x^{2}$. Summing from $k=1$ up to $n$ and using the triangle inequality, we get $|g(nx)|\le2nx^{2}.$ Letting $nx=y$, we have \[|g(y)|\le\frac{2y^{2}}{n}\] for all positive integers $n$, and so we must have $g(y)=0$ for all $y$ (by fixing $y$ and letting $n$ become arbitrarily large).

So, we have $f(x)=f(0)$ for all $x$, and hence the only functions for which this inequality holds are constant functions.


calculus
derivative
quadratics
calculus computations

For what values of $ x$ is the graph of $ y\equal{}e^{\minus{}x^2}$ concave down?

[hide="what I did"]My first derivative of $ y$ is $ \minus{}2x \cdot e^{\minus{}x^2}$.
My second derivative of $ y$ is $ \minus{}4e^{\minus{}x^2} \minus{} 2x$.

I know that the graph is concave down when the second derivative is less than zero, but I don't know how to simplify my expression...[/hide]

look like
[img]http://img149.imageshack.us/img149/498/32739563qy9.jpg[/img]

 :rotfl:

how do you solve for this algebraically?

[quote="jeez123"]My first derivative of $ y$ is $ - 2x \cdot e^{ - x^2}$.
My second derivative of $ y$ is $ \boxed{- 4e^{ - x^2} - 2x}$.[/quote]

This is incorrect, like that $ -2e^{-x^2}+4x^2e^{-x^2}$.

$ - 2e^{ - x^2} + 4x^2e^{ - x^2 < 0}$
$ (e^{ - x^2})( - 2 + 4x^2) < 0$
And we can set each quantity less than zero... but I don't know how to do it.

[quote="jeez123"]$ (e^{ \minus{} x^2})( \minus{} 2 \plus{} 4x^2) < 0$[/quote]

Clearly, $ e^{\minus{}x^2}$ will be positive for all real $ x$. So... just solve $ \minus{} 2 \plus{} 4x^2 < 0$ (you can do this by sketching the graph of the lefthand side since it's just a quadratic and/or solving $ \minus{} 2 \plus{} 4x^2 \equal{} 0$ for $ x$ and then looking to the "sides" of each zero and seeing over what intervals the expression is negative).


How do you prove that $\sum_{k=0}^{n}\binom{n}{k}^{2}=\binom{2n}{n}$ without knowing the sum equals $\binom{2n}{n}$?

(Edited)

[quote="ch1n353ch3s54a1l"]How do you prove that $\sum_{k=0}^{n}\binom{n}{k}=\binom{2n}{n}$ without knowing the sum equals $\binom{2n}{n}$?[/quote]

I think you mean $\sum_{k=0}^{n}\binom{n}{k}^{2}=\binom{2n}{n}$.

[hide="Proof"]This is equal to $\sum_{k=0}^{n}\binom{n}{k}\binom{n}{n-k}$.  Suppose we want to form a $n$-person committee from a $2n$-person senate of $n$ men and $n$ women.  There are $\binom{n}{k}\binom{n}{n-k}$ committess with $k$ men.  As we evaluate this expression from $k=0$ to $k=n$, we cover all possible committees, hence the sum is equal to $\binom{2n}{n}$.[/hide]

[hide="Polynomial method"]We also can consider the coefficient by $x^{n}$ in the polynomial $(x+1)^{2n}$. It of course equals $\binom{2n}{n}$. But $(x+1)^{2n}= (x+1)^{n}\cdot (x+1)^{n}= \left(\sum\limits_{k=0}^{n}\binom{n}{k}x^{k}\right)\cdot \left(\sum\limits_{k=0}^{n}\binom{n}{k}x^{k}\right)$. After bunching, $x^{n}$ can appear only in constituents of form $\binom{n}{k}x^{k}\cdot \binom{n}{n-k}x^{n-k}$, so the coefficient by $x^{n}$ is sum of coefficients in such constituents, i.e. $\sum_{k=0}^{n}\binom{n}{k}\binom{n}{n-k}= \sum_{k=0}^{n}\binom{n}{k}^{2}$. The coefficient by $x^{n}$ in $(x+1)^{2n}$ is uniquely defined, so $\sum_{k=0}^{n}\binom{n}{k}^{2}= \binom{2n}{n}$.[/hide]

Here's a way of looking at it that uses pascal's triangle.
$\binom{n}{k}$ is the number of ways to get from the top of the triangle to the entry $\binom{n}{k}$. It is also the number of ways to get from $\binom{n}{k}$ to $\binom{2n}{n}$ (easy to see if you actually draw it).
So the number of ways to get from $\binom{0}{0}$ to $\binom{2n}{n}$, going through $\binom{n}{k}$ on the way, is $\binom{n}{k}^{2}$. Then \[\sum_{k=0}^{n}\binom{n}{k}^{2}\] is the number of ways to get from $\binom{0}{0}$ to $\binom{2n}{n}$ going through any number on the $n$th row, which are $all$ the ways to get to $\binom{2n}{n}$.


algebra
function
domain
inequalities
complex numbers

Let us write each rational number in the domain (0,1] as a fraction a/b with gcd(a,b)=1. Now cover each a/b with the interval

[tex]
\displaystyle \left[\frac{a}{b}-\frac{1}{4b^2},\frac{a}{b}+\frac{1}{4b^2}\right]
[/tex]

Show that [tex]\frac{\sqrt{2}}{2}[/tex] cannot be covered.

I have a solution, but it seems very contrived to me. Can any mortal here come up with a nice one?

[quote="Ravi B"]Step 1: Can you rewrite the condition that :rt2: / 2 is not in that interval as an inequality?[/quote]

[tex]\left|\frac{\sqrt{2}}{2} - \frac{a}{b}\right| \geq \frac{1}{4b^2}[/tex] or [tex]\left|\frac{\sqrt{2}}{2} - \frac{1}{4b^2}\right| \geq \frac{a}{b}[/tex].

Okay, your first inequality is fine (with a > instead of :ge: ).  Your second inequality is not right.

So, our goal is to prove your first inequality.  It has a square root on the left side.

Step 2: Do you know any techniques to get rid of a square root?

[quote="Ravi B"]Okay, your first inequality is fine (with a > instead of :ge: ).  Your second inequality is not right.[/quote]

Ah, right. Sorry, my bad.

[quote]So, our goal is to prove your first inequality.  It has a square root on the left side.

Step 2: Do you know any techniques to get rid of a square root?[/quote]

Square it. :) I don't think that that is the answer you want, hehe. Of course, we *could* multiply the LHS by [tex]\frac{\sqrt{2}}{2} + \frac{a}{b}[/tex] , but I couldn't think of that myself. Maybe I just need more algebra experience or something. Did you manage to come up with that expression to multiply by yourself?

Yes, but then I've seen problems like these before.  Whenever you see an annoying square root, think about multiplying by the conjugate.  It's the same thing you do when trying to rationalize the denominator in say 1 / ( :rt2: - 1 ).

Every time you see a+ib in complex numbers, you think a-ib something. Every time you see [tex]a+b\sqrt c[/tex], you think [tex]a-b\sqrt c[/tex]. For exactly the same reason - believe me, 99% of problems involving some of that form will involves its 'conjugate'.

Ok, I will try to be more aware of the uses of conjugates. Thanks for the help.


let $ n $ be an interger $ n \models 0 $
let $ a_1, ... , a_n  $ and $ b_1, ... , b_n $ be postive reals such that :
                  $ a_1 + ... + a_n = b_1 + .... b_n = 1 $
what is the minimum of the expression $ \frac {a_1^2}{a_1 + b_1} + ... + \frac {a_n^2}{a_n + b_n}$

i found an other easy solution, and i wish it's true this time  :(

[hide]by cauchy we have
$ (\frac {a_1^2}{a_1 + b_1} + ... + \frac {a_n^2}{a_n + b_n}) ( (a_1 +b_1) + ... + (a_n + b_n) \geq ( a_1 + ... +a_n)^2$
then
 $ \frac {a_1^2}{a_1 + b_1} + ... + \frac {a_n^2}{a_n + b_n} \geq  \frac 12 $
the equality occurs for $ a_1 = ... =a_n = b_1 = ... = b_n = \frac 1n $
so the minimum is $ \frac 12$[/hide]

sorry, i checked that the question and my solution were proposed before :oops:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=19559\[/url]


inequalities
inequalities proposed
algebra

$ a,b,c\ge 0$ prove
$ (\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})(\frac{1}{1\plus{}a}\plus{}\frac{1}{1\plus{}b}\plus{}\frac{1}{1\plus{}c})\ge\frac{9}{1\plus{}abc}$
P.S. I do it by pqr

[quote="dima ukraine"]$ a,b,c\ge 0$ prove
$ (\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})(\frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b} \plus{} \frac {1}{1 \plus{} c})\ge\frac {9}{1 \plus{} abc}$
P.S. I do it by pqr[/quote]
Nice, dima ukraine. :) In the book: [b][i]Old and New Inequalities 2[/i][/b], Vo Quoc Ba Can - Cosmin Pohoata, I presented my proof with only AM-GM for 3 variables to solve this inequality. :)

[quote="dima ukraine"]$ a,b,c\ge 0$ prove
$ (\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})(\frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b} \plus{} \frac {1}{1 \plus{} c})\ge\frac {9}{1 \plus{} abc}$
P.S. I do it by pqr[/quote]
It's also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=145655

[quote="matex (L)(L)(L)"]Is this book Old and New Inequalities 2, Vo Quoc Ba Can - Cosmin Pohoata available? I'd like to buy it or at least see sample pages.[/quote]
Yes, it is now available at GIL publishing house. :)

Could someone gove link concerning the book "Old and New Inequalities 2" because I'm very interested in this book and I'd like to buy it.

And what are other book connected with inequalities?(apart from "Old and New Inequalites"

Let $a,b,c,d>0$.[url=https://artofproblemsolving.com/community/c6h149117p1928912]Prove that:[/url]
$$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+d}+\frac{1}{d+a}\right) \geq \frac{16}{1+abcd}$$

Let $ a,b,c> 0$ and $ a+b+c=3.$ Prove that
$$\left(\frac1a+\frac1b+\frac1c\right)\left(\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\right)\ge\frac1a+\frac1b+\frac1c\ge\frac{9}{1+2abc}$$


let $ a$ , $ b$ , $ c$ , $ d$ $ \geq 0$  prove that :

$ \frac{ab}{(c\plus{}d)^2}\plus{}\frac{bc}{(d\plus{}a)^2}\plus{}\frac{cd}{(a\plus{}b)^2}\plus{}\frac{da}{(b\plus{}c)^2} \geq 1$

$ \frac{ab}{c\plus{}d}\plus{}\frac{bc}{d\plus{}a}\plus{}\frac{cd}{a\plus{}b}\plus{}\frac{da}{b\plus{}c} \geq \frac{a\plus{}b\plus{}c}{2}$

[quote="y-a-s-s-i-n-e"]let $ a$ , $ b$ , $ c$ , $ d$ $ \geq 0$  prove that :
$ \frac {ab}{c \plus{} d} \plus{} \frac {bc}{d \plus{} a} \plus{} \frac {cd}{a \plus{} b} \plus{} \frac {da}{b \plus{} c} \geq \frac {a \plus{} b \plus{} c}{2}$[/quote]
I think it is wrong,try $ d\equal{}b\equal{}0, a>0,c>0$

[quote="y-a-s-s-i-n-e"]sorry !!

$ a$ ,$ b$, $ c$ ,$ d$ $ > 0$[/quote]
It is still wrong!try $ b\equal{}d\rightarrow0^{\plus{}} , a>0 , c>0$!

[quote="y-a-s-s-i-n-e"]let $ a$ , $ b$ , $ c$ , $ d$ $ \geq 0$  prove that :
$ \frac {ab}{(c \plus{} d)^2} \plus{} \frac {bc}{(d \plus{} a)^2} \plus{} \frac {cd}{(a \plus{} b)^2} \plus{} \frac {da}{(b \plus{} c)^2} \geq 1$
[/quote]
and also this one!try $ a\equal{}c\rightarrow0^\plus{} , d>0 , b>0$


function
LaTeX
logarithms
geometry
geometric transformation
real analysis
real analysis unsolved

Are there functions $f: (0,\infty)\mapsto \mathbb{R}$ which are continuous in infinitely many points and such that $ f(x)=0$ if and only if $ f(2x)$ is non-zero?

[Moderator edit: inserted a space after the ":". Watch out for this - the system interprets ": - something" as an emoticon, and that causes much trouble inside $LaTeX.$ Inserting spaces fixes the problem.]

[quote="harazi"]Are there functions $f: (0 , \infty )\to\mathbb R$ which are continuous in infinitely many points and such that $f(x)=0$ iff $f(2x)$ is non-zero?[/quote]

The characteristic function of the set $\bigcup_{n\in\mathbb Z} [2^{2n},2^{2n+1})$ works, doesn't it? :?

P.S. That $\LaTeX$ mistake appears to have originated from putting $:$ right before $($. I don't know why, but I put a space between the two, and it worked. :?

Maybe that should be "all but finitely many points"?

For convenience, take logarithms in the argument, so $f(x)$ is zero iff $f(x+1)$ is nonzero. If $f$ is continuous on an interval $I=[a,b]$, then $\{x: f(x)=0\}$ is closed in $I$, and $\{x: f(x)\neq0\}$ is closed in $[a-1,b-1]$ and also closed in $[a+1,b+1]$ by the translation condition. Taking unions and intersections, if $b-a\ge2$, both $\{x: f(x)=0\}$ and $\{x: f(x)\neq0\}$ are closed in $I$ and of course nonempty. Since an interval in $\mathbb{R}$ is connected, every interval of length at least 2 must contain a point of discontinuity, and the function has infinitely many points of discontinuity.


Let $f: \mathbb{R}\rightarrow\mathbb{R}$ and $g: \mathbb{R}\rightarrow\mathbb{R}$ such that both $f$ and $g$ are surjective funcions, and $h: \mathbb{R}\rightarrow\mathbb{R}$, $h(x) = f(x) + g(x)$. Show that $h$ is a surjection.

$\left. \begin{array}{l}
 f:\mathbb{R} \to \mathbb{R},f\left( x \right) = x,\forall x \in \mathbb{R} \\
 g:\mathbb{R} \to \mathbb{R},g\left( x \right) =  - x,\forall x \in \mathbb{R} \\
 \end{array} \right\} \Rightarrow $

$ \Rightarrow h:\mathbb{R} \to \mathbb{R},h\left( x \right) = f\left( x \right) + g\left( x \right) = 0,\forall x \in \mathbb{R}$

Sorry

Why $f(x) = x$ ?

$f$ is said to be a surjection (or surjective map) if, for any $b \in \mathbb{R}$, there exists an $a \in \mathbb{R}$  for which $b = f(a)$.
http://mathworld.wolfram.com/Surjection.html

I understand the statement of the problem like that:
$f$ and $g$ are surjection implies that $f+g$ is a surjection

And I show you the counter-example with $f=Id$ and $g=-Id$

Where I'm wrong ?
Further more, $f(x)=x$ is a surjection since for all $b$, there exists $a$ ($a=b$ !!!) such that $f(a)=b$ since $b=b$ !!

[quote="Rodolphe2005"]I understand the statement of the problem like that:
$f$ and $g$ are surjection implies that $f+g$ is a surjection

And I show you the counter-example with $f=Id$ and $g=-Id$

Where I'm wrong ?
Further more, $f(x)=x$ is a surjection since for all $b$, there exist $a$ ($a=b$ !!!) such that $f(a)=b$ since $b=b$ !![/quote]

You are correct, the my proposition is completely wrong.  :blush: Sorry...
The set of real numbers is closed for sum, then I thought the proposition above was true.  :?

My intention was to create an "auxiliar theorem" to solve the problem below...

[b](IME - 1986)[/b] Let a bijective function $f: \mathbb{R} \to \mathbb{R}$ and a function $h$ such that:

$\left\{ \begin{array}{l} h: \mathbb{R}^2 \to \mathbb{R}^2 \\ h(x, y) = \left(x^3, x - f(y)\right) \end{array} \right.$

Show that h is bijective.

Maybe it works for $f\times g$ with the property that $f, g$ are continuous

$f$ is bijective so $f^{-1}$ exists
then for all $(\alpha,\beta)\in\mathbb{R}^2$, $h(\sqrt[3]{\alpha},f^{-1}(\sqrt[3]{\alpha}-\beta))=(\alpha,\beta)$
injectivity is easy


algebra
function
domain
analytic geometry
advanced fields
advanced fields unsolved

Let $ \Omega$ be bounded open domain in $ R^n$ and $ u\in W^{1,1}(\Omega)$. The level sets coresspond to $ u$ are defined as $ U_t: \equal{}\{x\in\Omega: u(x)>t\}$. Prove that for almost every $ t\in R$ we have \[ \partial U_t\equal{}\{x\in\Omega: u(x)\equal{}t\}\]

Pointwise operations on Sobolev functions can be tricky. For example, let $ \Omega$ be the unit ball and $ u(x)\equal{}x_1$ (the 1st coordinate): this is a very nice function and everything works as you said. But if I redefine $ u(x)\equal{}100$ on the line $ x_2\equal{}x_3\equal{}\dots\equal{}x_n\equal{}0$, then for $ \minus{}1<t<1$ we have $ \partial U_t\equal{}\{x\in \Omega\colon x_1\equal{}t\}\ne \{x\in \Omega\colon u(x)\equal{}t\}$.

A function $ u\in W^{1,p}$ has a $ p$-quasicontinuous representative (i.e., continuous except for a set of small $ p$-capacity). But this does not seem to help at all in dimensions $ n\ge 3$, since $ p$-capacity roughly corresponds to dimension $ n\minus{}p$. With $ p\equal{}1$ you can hope to control the behavior of $ u$ on $ (n\minus{}1)$-dimensional subsets, not on lines.

Since the statement appear in a text book of Enrico Giusti, I believe that it is true in some sense which I do not know now. I think that the level set of a function in a Sobolev space can be defined in some sense, something may similar to the trace on the boundary.


inequalities
algebra
IMO Shortlist
calculus

WLOG assume that  $p\geq q\geq r\geq s$ .
Easily $pq+rs+pr+qs+ps+qr\leq 30$  and from the first $pq+rs\geq 10$ . Now put $p+q=t$

Then $t^{2}+(t-9)^{2}\geq 41$  so $(t-4)(t-5)\geq 0$  and since
$t\geq r+s$ we find that  $25\leq 21-r^{2}-s^{2}+2pq\leq 21+2(pq-rs)$ and QED

use that $2(x^{2}+y^{2})=(x+y)^{2}+(x-y)^{2}$. then,

$84=4(p^{2}+q^{2}+r^{2}+s^{2})=(p+q+r+s)^{2}+(p+q-r-s)^{2}+(p+r-q-s)^{2}+(p+s-q-r)^{2}$

this implies that $(p+q-r-s)^{2}+(p+r-q-s)^{2}+(p+s-q-r)^{2}=3$, then

$\max\{|p+q-r-s|,|p+r-q-s|,|p+s-q-r|\}\geq 1$.

suppose wlog that $|p+q-r-s|\geq 1$, then this implies that

$2(p+q)-9=p+q-r-s\geq 1$ or $2(r+s)-9=r+s-p-q\geq 1$

the first case implies that $p+q\geq 5$, while the second that $r+s\geq 5$. suppose wlog that $p+q\geq 5$

then, $21+2(pq-rs)=(p+q)^{2}+(r-s)^{2}\geq 25$, from where we conclude that $pq-rs\geq 2$.

[quote="who"]Four real numbers $ p$, $ q$, $ r$, $ s$ satisfy $ p+q+r+s = 9$ and $ p^{2}+q^{2}+r^{2}+s^{2}= 21$. Prove that there exists a permutation $ \left(a,b,c,d\right)$ of $ \left(p,q,r,s\right)$ such that $ ab-cd \geq 2$.[/quote]
Let $a \ge b \ge c \ge d$.
$ab-cd \ge \frac{1}{2}(c+d)(a+b-c-d) \iff$
$2ab+c^2+d^2 \ge ac+ad+bc+bd$. Since $c^2+d^2 \ge 2cd$ it suffices to prove:
$2ab+2cd \ge ac+ad+bc+bd \iff$
$(a-d)(b-c) + (a-c)(b-d) \ge 0$.
And:
$3(a+b-c-d)^2 \ge 3 = 4(a^2+b^2+c^2+d^2)-(a+b+c+d)^2 \iff$
$2ab+2cd \ge ac+ad+bc+bd$ is true, so $a+b-c-d \ge 1$.
Let $a+b-c-d = t$. Then $\frac{1}{2}(c+d)(a+b-c-d) = \frac{1}{4}(2c+2d)(a+b-c-d) = \frac{1}{4}(9-t)t$, which is increasing on $(-\infty;4.5]$ so $\frac{1}{4}(9-t)t \ge \frac{1}{4}(9-1)\cdot1 = 2$, so $ab-cd \ge 2$.
(There is equality only for $(a,b,c,d) = (3,2,2,2)$)

Here's another solution. It's not nearly as elegant, but in my opinion it is much more straightforward:

WLOG, let $p \geq q \geq r \geq s$. I claim that $pq - rs \geq 2$. I also claim that the result true not only when $p^2 + q^2 + r^2 + s^2 = 21$, but also when $p^2 + q^2 + r^2 + s^2 \geq 21$, provided that $s > \frac{1}{4}$.

But we may always suppose that $s > \frac{1}{4}$! If $s \leq \frac{1}{4}$, then $p+q+r = 9-s \geq \frac{71}{4}$. It follows that $p^2 + q^2 + r^2 \geq \left(\frac{p+q+r}{3}\right)^2 \geq \left(\frac{71}{12}\right)^2 > 5^2 = 25 > 21$, which is a contradiction, so we may indeed always suppose that $s \geq \frac{1}{4}$.

Replacing $p$ and $q$ with $p_0 = p+q - r$ and $q_0 = r$ preserves $p_0 \geq q_0 \geq r \geq s$, $p_0 + q_0 + r + s \geq 2$, and $p_0^2 + q_0^2 + r^2 + s^2 \geq 21$. In addition, $p_0q_0 - rs = pr + qr - r^2 - rs \leq pq - rs$ (since $(p-r)(q-r) \geq 0$.) Hence, we may suppose without loss of generality that $q=r$.

We reformulate our problem as follows: when $p \geq q \geq s \geq \frac{1}{4}$, $p+2q+s = 9$, and $p^2 + 2q^2 + s^2 \geq 21$, then $q(p-s) \geq 2$.

Let $k = \frac{q-s}{2}$. Let $p' = p + k$, $s' = s + k$, and $q' = q - k$. We see that $p' + 2q' + s' = 9$, $q'(p-s') = (q-k)(p-s) \leq q(p-s)$, and that
\begin{align*}
p'^2 + 2q'^2 + s'^2
&= p^2 + 2q^2 + s^2 + 4k^2 + 2k(p+s-2q) \\
&= p^2 + 2q^2 + s^2 + (p-s)^2 + (p-s)(p+s-2q) \\
&= p^2 + 2q^2 + s^2 + 2(p-q)(q-s) \\
&\geq p^2 + 2q^2 + s^2 \geq 21. \end{align*}
Observing that $s' = q'$ and that $s' \geq s$, we see that it is sufficient to prove this inequality when $q=s$.

We reformulate our problem again: when $p \geq q \geq \frac{1}{4}$, $p+3q = 9$, and $p^2 + 3q^2 \geq 21$, then $pq - q^2 \geq 2$. $9 = p + 3q \leq 4q$, so $q \leq \frac{9}{4}$. $p^2 + 3q^2 = (9-3q)^2 + 3q^2 \geq 21$, so $12q^2 - 54q + 81 \geq 21$, so $2q^2 - 9q + 10 \geq 0$, so $(q-2)(2q-5) \geq 0$, so $q \leq 2$ or $q \geq \frac{5}{2}$. But $q \leq \frac{9}{4}$, so we must have that $q \leq 2$.

$pq - q^2 = (9-3q)q - q^2 = 9q - 4q^2$. Since $9q - 4q^2$ is concave in $q$, it attains its minimum when $q$ is either maximized or minimized. $\frac{1}{4} < q \leq 2$; but at both $q = \frac{1}{4}$ and $q = 2$, $9q - 4q^2 = 2$, so we see that $pq - q^2$ must always be greater than or equal to 2, which completes our proof.

Assume WLOG that $ p \ge q \ge r \ge s $.  Note that $ pq + pr + ps + qr + qs + rs = \frac{(p + q + r + s)^2 - (p^2 + q^2 + r^2 + s^2)}{2} = 30 $.  By the rearrangement inequality, we have that $ pq + rs = \max(pq + rs, pr + qs, ps + qr) $ and so $ pq + rs \ge 10 $.  Letting $ x = p + q $ we have that $ x^2 + (x - 9)^2 = 21 + 2(pq + rs) \ge 41 $ which becomes $ (x - 4)(x - 5) \ge 0 \Longrightarrow x \ge 5 $.  This means that $ 2\sqrt{rs} \le r + s \le 4 \Longrightarrow rs \le 4 $.  But since $ pq + rs \ge 10 $ this means that $ pq \ge 6 $ and so we are done.

Now I want to discuss motivation.  After some playing around we see the only equality case is $ (p, q, r, s) = (3, 2, 2, 2) $.  Now, looking at this, we want the extreme values of the sum of the two biggest or the two smallest to be greater than $ 5 $ and less than $ 4 $ respectively, so letting $ x = p + q $ we want to get exactly the equation $ (x - 4)(x - 5) \ge 0 $.  Now we want an $ x - 9 $ in there somewhere, and rearranging it turns out we want $ x^2 + (x - 9)^2 \ge 41. $  But this is the same as $ pq + rs \ge 10 $, and when looking at a sum like this, the rearrangement inequality should come to mind immediately.

Solved with [b]eisirrational[/b]. We believe that this solution, although ``ugly" (compared to, say, post 2), is straightforward. (Of course, here, the solution is presented backwards.)

Without loss of generality assume $p \leq q \leq r \leq s$. The central claim is $p+q \leq 4$. Set $(x,y) = (p+q,p^2+q^2)$. Then
\begin{align*}
q &= \frac{1}{2}((p+q) + |p-q|) = \frac{1}{2}\left(x + \sqrt{2y-x^2}\right), \\
r &= \frac{1}{2}((r+s) - |r-s|) = \frac{1}{2}\left((9-x) - \sqrt{2(21-y) - (9-x)^2}\right).
\end{align*}
Then we have $q \leq r$, which yields
$$x + \sqrt{2y-x^2} \leq (9-x) - \sqrt{2(21-y) - (9-x)^2} \implies \sqrt{(2y-x^2)(2(21-y) - (9-x)^2)} \leq 3(x-4)(x-5),$$
so $3(x-4)(x-5) \geq 0$, id est $x \leq 4$ (as $p+q \leq \tfrac{1}{2}(p+q+r+s) = 4.5$).

Then by Cauchy-Schwarz,
$$(p-4.5)^2 + (q-4.5)^2 \geq \frac{(p+q-9)^2}{2} \geq 12.5.$$
But this expands to $p^2 + q^2 - 9p - 9q \geq -28$, which yields
\begin{align*}
2 &\leq p^2 + q^2 - 9p - 9q + 30 \\
&= \frac{1}{2}((9-p-q)^2 - (21 - p^2 - q^2)) - pq \\
&= \frac{1}{2}((r+s)^2 - (r^2+s^2)) - pq \\
&= rs - pq,
\end{align*}
as desired. Equality holds when $p+q = 4$, id est when $(p,q,r,s) = (2,2,2,3)$.

WLOG $p\ge q\ge r\ge s$. If $s$ is negative, then $p+q+r\ge 9\implies p+q\ge 6$. So, $$(p+q)^2+(r-s)^2>36>25\implies 2(pq-rs)>4$$ Hence, we can assume that $s\ge 0$. Now, the permutation which maximizes $ab-cd$ is $(p,q,r,s)$, so assume the contrary, namely that $pq-rs<2$. Now, consider replacing $(r,s)\to\left(\frac{r+s}{2},\frac{r+s}{2}\right)$, and $(p,q)$ with suitable numbers such that both conditions are still preserved. Obviously, we still have $p\ge q\ge r\ge s\ge 0$, and this decreases $pq-rs$, hence we only need to prove the case where $r=s$.

In this case, $p+q=9-2s$, $p^2+q^2=21-2s^2\implies 2pq=6s^2-36s+60\implies (p-q)^2=-8s^2+36s-39$. Solving, $q=\frac{1}{2}\left(9-2s-\sqrt{-8s^2+36s-39}\right)$. On the other hand, $pq<2+s^2\implies 3s^2-18s+30<2+s^2\implies s^2-9s+14<0\implies s>2$. As $s$ is the smallest, it lies in the interval $2<s<2.25$. As $s$ increases in this range, both $2s$ and $-8s^2+36s-39$ increase, so $q$ decreases. Hence, $q<\frac{1}{2}\left(9-4-\sqrt{1}\right)=2$, and we get $q<s$, which is a contradiction, as desired.

Let $a, b, c,$ and $d$ be real numbers such that $a \geq b \geq c \geq d$ and $a+b+c+d = 13,a^2+b^2+c^2+d^2=43.$ [url=https://artofproblemsolving.com/community/c6h2496042p21040074]Show that [/url]$$ab-cd \geq 3 .$$
Equality holds for $(4,3,3,3).$
[b]2021 Philippine[/b]

[quote=TheUltimate123]Let \(p\ge q\ge r\ge s\). Observe that \[(pq+rs)+(pr+qs)+(ps+qr)=\frac{(p+q+r+s)^2-\left(p^2+q^2+r^2+s^2\right)}2=30,\] but by Rearrangement, \(pq+rs\ge pr+qs\ge ps+qr\), so \(pq+rs\ge10\).

Note the following: \begin{align*}     (p+q)+(r+s)&=9\\     (p+q)^2+(r+s)^2&\ge41. \end{align*} Since \(p+q\ge r+s\), we must have \(p+q\ge5\).

Finally, \[25\le(p+q)^2+(r-s)^2=21+2(pq-rs),\] so \(pq-rs\ge2\), as needed. Equality holds at \((p,q,r,s)=(3,2,2,2)\)[/quote]

"$(p+q)+(r+s)=9   (p+q)^2+(r+s)^2 \ge41.$   Since $ (p+q\ge r+s )$, we must have $(p+q\ge5).$"

How do you get $p+q \geq $ ?

[quote=Wizard_32][quote=who]Four real numbers $ p$, $ q$, $ r$, $ s$ satisfy $ p+q+r+s = 9$ and $ p^{2}+q^{2}+r^{2}+s^{2}= 21$. Prove that there exists a permutation $ \left(a,b,c,d\right)$ of $ \left(p,q,r,s\right)$ such that $ ab-cd \geq 2$.[/quote]
Assume that $\max\{pq,pr,ps,rq,rs,qs\}=pq,$ and assume on the contrary that $pq<rs+2.$ Since $pq+pr+ps+rq+rs+qs=60,$ hence $pq \geqslant 10.$ In particular, this means $2+rs>10 \implies rs>8>0.$ Then since $p^2,q^2,r^2,s^2$ are positive reals, hence by AM-GM
\begin{align*}
    8^2 < (rs)^2 \leqslant pqrs \leqslant \left( \frac{p^2+q^2+r^2+s^2}{4}\right)^2 \implies 8^2 \cdot 4^2 < 21^2
\end{align*}
which is clearly false. $\square$[/quote]

$pq+pr+ps+rq+rs+qs=60$ ; this is false, actually we have $(p+q+r+s)^2 = p^2 + q^2 + r^2 + s^2 + 2(pq + pr+ps+rq+rs+qs ) = 81 = 21 + 2(pq + pr+ps+rq+rs+qs )$, and hence $pq+pr+ps+rq+rs+qs=30$. Notice this implies $5 \leq pq$, which means your solution is wrong.

Here's a different solution with a weird substitution.
WLOG suppose that $p \geq q \geq r \geq s$. I will show that $pq-rs \geq 2$. Since $p+q+r+s=9$, we can substitute
\begin{align*}
p&=\frac{9}{4}+k+x\\
q&=\frac{9}{4}+k-x\\
r&=\frac{9}{4}-k+y\\
s&=\frac{9}{4}-k-y,
\end{align*}
where $k,x,y\geq 0$. Since $q \geq r$, we require $2k \geq x+y$. Using the substitution, the condition $p^2+q^2+r^2+s^2=21$ becomes
$$\frac{81}{4}+\frac{9}{2}(k+x+k-x-k+y-k-y)+(k+x)^2+(k-x)^2+(k+y)^2+(k-y)^2=\frac{81}{4}+4k^2+2x^2+2y^2=21\implies 2k^2+x^2+y^2=\frac{3}{8}.$$
Finally, the inequality $pq-rs\geq 2$ is equivalent to
$$9k+y^2-x^2\geq 2$$
Suppose now that we fix $k$. Then it is clear that
$$9k+y^2-x^2\geq 9k-4k^2,$$
and since the inequality $9k-4k^2\geq 2$ holds for $\tfrac{1}{4} \leq k \leq 2$, $pq-rs\geq 2$ holds in that range of $k$ as well. If $k>2$, then
$$2k^2+x^2+y^2\geq 2k^2>8>\frac{3}{8},$$
so we cannot have $p^2+q^2+r^2+s^2=21$ in that case. Thus we can discard the case of $k>2$. If $k<\tfrac{1}{4}$, then we have
$$\frac{3}{8}=2k^2+x^2+y^2<\frac{1}{8}+x^2+y^2 \implies x^2+y^2>\frac{1}{4} \implies (x+y)^2>\frac{1}{4} \implies x+y>\frac{1}{2}>2k,$$
which contradicts $2k\geq x+y$. Thus we can also discard $k<\tfrac{1}{4}$, leaving only $k \in [\tfrac{1}{4},2]$ which we already proved $pq-rs\geq 2$ for. Thus we're done. $\blacksquare$

WLOG $p\ge q\ge r\ge s$. We claim that $pq-rs \ge 2$.
If $s\le 0$, then, by Cauchy-Schwarz, $21 = p^2+q^2+r^2+s^2 \ge \frac{1}{3}(p+q+r)^2+s^2 \ge \dfrac{1}{3}(9)^2+s^2 \Rightarrow s^2\le -6$ which is clearly absurd. Hence, $p,q,r,s \in \mathbb{R}^+$.
We have
\begin{align*}
	&(9-q-r-s)^2+q^2+r^2+s^2=21 \\
	\Leftrightarrow \quad &2q^2-2(9-r-s)q+(9-r-s)^2+r^2+s^2-21=0 \\
	 \Leftrightarrow \quad &q=\dfrac{2(9-r-s)\pm \sqrt{42-(2r^2+2s^2+(9-r-s)^2)}}{2}
\end{align*}
If $q=\dfrac{(9-r-s)+ \sqrt{42-(2r^2+2s^2+(9-r-s)^2)}}{2}$, then
$q\le p = 9-q-r-s = q-\sqrt{42-(2r^2+2s^2+(9-r-s)^2)} \le q$ so $42-(2r^2+2s^2+(9-r-s)^2) = 0$ which will be included in the other case.\\
Therefore, we can assume that $q=\dfrac{(9-r-s)- \sqrt{42-(2r^2+2s^2+(9-r-s)^2)}}{2}$. We will show that $r+s\le 4$. Suppose, ftsoc, that $r+s > 4$. We have
\begin{align*}
	&\dfrac{(9-r-s)- \sqrt{42-(2r^2+2s^2+(9-r-s)^2)}}{2} \ge r \\
	\Leftrightarrow \quad &3r^2+(2s-18)r+(s^2-9s+30)\ge 0 \\
	\Leftrightarrow \quad &3\left(r+\dfrac{s-9}{3}\right)^2+\dfrac{2s^2-9s+9}{3}\ge 0
\end{align*}

[b]Claim:[/b] [i]$\dfrac{3}{2}\le s \le \dfrac{9}{4}$[/i]

[b]Proof:[/b] If $s< \dfrac{3}{2}$, then $p+q+r+s \ge 3r+s > 3(4-s)+s > 12-2(\dfrac{3}{2}) = 9$, a contradiction. So $s \ge \dfrac{3}{2}$. Also, since $p+q+r+s=9$, $s\le \dfrac{9}{4}.$ $\blacksquare$

By the claim, $2s^2-9s+9= (2s-3)(s-3)\le 0$. We can now consider 2 possible cases.

[b]Case 1: [/b] $r\ge\dfrac{9-s}{3}+\sqrt{\dfrac{-2s^2+9s-9}{9}}$\\
$9=p+q+r+s \ge 3r+s \ge (9-s)+3\sqrt{\dfrac{-2s^2+9s-9}{9}} +s =9+3\sqrt{\dfrac{-2s^2+9s-9}{9}}$ so $2s^2-9s+9=0 \Leftrightarrow s=3, \dfrac{3}{2}$ and $p=q=r$. By claim, $s= \dfrac{3}{2}$ giving $p=q=r=\dfrac{5}{2}.$ Thus, $r+s = 4$, a contradiction.

[b]Case 2: [/b]  $r\le\dfrac{9-s}{3}-\sqrt{\dfrac{-2s^2+9s-9}{9}}$\\
$s\le \dfrac{9-s}{3}-\sqrt{\dfrac{-2s^2+9s-9}{9}} \Leftrightarrow 2s^2-9s+10 \Leftrightarrow (2s-5)(s-2)\ge 0$. By claim, $s\le 2$.
However, $4-s < r \le\dfrac{9-s}{3}-\sqrt{\dfrac{-2s^2+9s-9}{9}} \Rightarrow 2s^2-7s+6>0 \Leftrightarrow (s-2)(2s-3)>0$, which is a oontradiction since $\dfrac{3}{2}\le s\le2$.

We therefore have that $r+s \le 4$ and
\begin{align*}
	pq-rs &= (9-q-r-s)q-rs\\
	&= \dfrac{81-((9-q-r-s)^2+q^2+r^2+s^2)}{2}+(r^2+s^2)-9(r+s) \\
	&= 30 +(r^2+s^2)-9(r+s) \\
	&= 22 + (r^2+4) +(s^2+4) -9(r+s) \\
	&\ge 22 -5(r+s) \\
	&\ge 2
\end{align*}
as desired.

Here's a different solution (which is a bit longer, though every step is easily motivated):
[rule]
WLOG $p \le q \le r \le s$. We will show $pq - rs \ge 2$. Observe,
$$\sum_{\text{sym}} pq = \frac{p^2 - 21}{2} = 30 \implies \sum_{\text{sym}} (p-q)^2 = 3 \cdot 21 - 2 \cdot 30 = 3 \qquad \qquad (1)$$
Let $p-q = x, p - r = y, p -s = z$. Observe $0 \le x \le y \le z$. Then $p = \frac{p+x+y+z}{4}$, thus
\begin{align*}
T &= pq - rs = p(p-x) - (p-y)(p-z) = p^2 - px - p^2 + p(y+z) - yz = p(y+z-x) - yz \\
&= \frac{(9 + x + y +z)(y+z-x)}{4} - yz = \frac{\left( y +z + \frac{9}{2} \right) - \left(x + \frac{9}{2} \right) }{4} - yz \\
&= \frac{ y^2 + z^2 + 9y + 9z + 2yz - x^2 - 9x }{4} - \frac{yz}{2} = \frac{9(y+z-x) + y^2 + z^2 - x^2}{4} - \frac{yz}{2} \\
&= \frac{9(y+z-x) + (z-y)^2 - x^2}{4} \ge \frac{9(y+z-x) - x^2}{4}
\end{align*}
So it suffices to show $S = 9(y+z-x) - x^2 \ge 8$.

[b][color=#f00]Claim: [/color][/b] $x \le 1$ and $3z^2 + y - x \ge 3$.
[b][color=#00f]Proof:[/color][/b] Using $(1)$ we have $3 \ge x^2 + y^2 + z^2 = 3x^2$, giving $x \le 1$. Also,
\begin{align*}
3 - 3z^2 &= x^2 + y^2 + z^2 + (x-y)^2 + (y-z)^2 + (z-x)^2 - 3z^2 = 3(x^2 + y^2) - 2z(x+y) - 2xy  \\
&= x(3x - 2z - 2y) + y(3y - 2z) \le x(3x - 4y) + y^2 = (y-x)(y-3x) \le y-x
\end{align*}
This proves our claim. $\square$

Now if $z \ge 1$, then $S \ge 9z - x^2 \ge 9 - 1 = 8$. Otherwise, if $z \le 1$ then
\begin{align*}
S \ge 9(3 + z - 3z^2) - 1 = 26 + 9z - 27z^2 = 8 + 9(2 + z - 3z^2) = 8 + 9(1-z)(2+z) \ge 8
\end{align*}
This completes the proof. $\blacksquare$

Nothing new here.

WLOG assume that $p\ge q\ge r\ge s$.

Note that
$$pq+rs+pr+qs+ps+qr=\frac{(p+q+r+s)^2-(p^2+q^2+r^2+s^2)}{2}=30.$$
Since $p\ge q\ge r\ge s$ we have $pq+rs=\max(pq+rs,pr+qs,ps+qr)$. Hence $pq+rs\ge 10$. This proves that
$$(p+q)^2+(r+s)^2=(p^2+q^2+r^2+s^2)+2(pq+rs)\ge 21+2\cdot 10=41.$$
Now let $t\doteqdot p+q$. Then $t^2+(9-t)^2\ge 41$, or equivalently $(t-4)(t-5)\ge 0$. As $t\ge 4.5$ we obtain $t\ge 5$. Therefore
$$(p+q)^2\ge 25=p^2+q^2+r^2+s^2+4\ge p^2+q^2+2rs+4.$$
This rearranges to $pq-rs\ge 2$.

Note that we have $pq+pr+ps+qr+qs+rs=30.$ Let $p\ge q\ge r\ge s$ and consider $pq-rs.$ Note that $pq+rs\ge pr+qs\ge ps+qr$ so $pq+rs\ge 10.$ Let $a=p+q$ and $b=r+s$ then $(p+q)^2\ge 4pq$ gives:
\[a+b=9\]
\[a^2+b^2\ge 40\]
Thus, $-(a-b)^2=(a^2+b^2+2ab)-2(a^2+b^2)\le -1$ so $a-b\ge 1.$ This implies $a\ge 5.$ Therefore, we have $(p+q)^2+(r-s)^2\ge 25.$ On the other hand $(p+q)^2+(r-s)^2=p^2+q^2+r^2+s^2+2(pq-rs)=21+2(pq-rs)$ so $pq-rs\ge 4$ as desired.

ISL Marabot Solve

WLOG, $p\geq q\geq r\geq s$. We will prove, $pq-rs\geq 2$.

FTSOC, let $pq-rs < 2$.
 Now $\sum_{sym} pq = \frac{(p+q+r+s)^2-(p^2+q^2+r^2+s^2)}{2}=30$, So, by rearrangement inequality, $pq+rs\geq 10$. If $rs < 4$ then $pq-rs= pq+rs - 2rs > 10-8=2$, But as we assumed, this is not true. So, $rs\geq 4 \implies \frac{r^2+s^2}{2}\geq rs \geq 4\implies r^2+s^2\geq 8$ (By, AM-GM). Now, $p^2+q^2-r^2-s^2=\sum p^2 - 2(r^2+s^2)\leq 21-2\times 16=5$.

Now let $p+q=x, r+s=9-x$ So, $\sum p^2 + 2(pq+rs)\geq 21+2\times 10=41 \implies (p+q)^2+(r+s)^2=(x^2)+(9-x)^2\geq 41$ $\implies (x-5)(x-4)\geq 0 \implies x\geq 5$ or $x\leq 4$, Since $p\geq q\geq r\geq s$ and $\sum p = 9$, so, $x=p+q\geq 5$ and $9-x=r+s\leq 4$

So, $(p+q)^2-(r+s)^2= p^2+q^2-r^2-s^2+2(pq-rs)\geq 5^2-4^2=9 \implies 2(pq-rs)\geq 9-(p^2+q^2-r^2-s^2)\geq 9-5=4$ $\implies pq-rs\geq2$. But we guessed, $pq-rs<2$. So, by contradiction, $pq-rs\geq 2$

[hide=Solution] Let $p \geq q \geq r \geq s$ and want to prove that $pq-rs \geq 2$. Motivated by the fact the equality holds when $r=s$, we notice that $21=p^2+q^2+r^2+s^2=(p+q)^2-2pq+r^2+s^2 \geq (p+q)^2-2pq+2rs$, so $2(pq-rs)\geq (p+q)^2-21$, i.e. we need $p+q \geq 5$.

Now we should make use of the assumed arrangement of the variables. We have $(pq+rs)+(pr+qs)+(ps+qr)=30$, so by Rearrangement inequality, $pq+rs$ is the largest of the three summands, so $\frac{(p+q)^2+(r+s)^2}{4}\geq pq+rs\geq 10$. Using $r+s=9-p-q$, it rearranges to $(p+q-4)(p+q-5) \geq 0$, so $p+q \geq 5$ as it is greater than $4.5$.

[quote=who]Four real numbers $ p$, $ q$, $ r$, $ s$ satisfy $ p+q+r+s = 9$ and $ p^{2}+q^{2}+r^{2}+s^{2}= 21$. Prove that there exists a permutation $ \left(a,b,c,d\right)$ of $ \left(p,q,r,s\right)$ such that $ ab-cd \geq 2$.[/quote]

This problem reminds me of several similar ones relating $a_1+...+a_n=k$ with $a_1^2+...+a_n^2=l^2$ and trying to maximize an element lol (for example in particular that AMSP test 3 Level 2 question or smth or that USAMO 1978 or smth like that
WLOG $p\ge q\ge r\ge s.$ We see that $$pq+rs+pr+ps+qr+qs=30\implies pq+rs\ge 10$$ by rearrangement ineq. Then note that we want to find maximum of rs so that we can subtract it. To do this, we want to maximize the sum r+s, so we are motivated to set t=r+s, otherwise it would be hard to show a maximum sum that also satisfies the second equality. Now, $$t^2+(9-t)^2=p^2+q^2+2pq+r^2+s^2+2rs=41\implies (t-4)(t-5)\ge 0\iff 4\ge t=r+s,$$ where the last step is because $r+s\le 9/2<5$ so it couldn't be bounded below by 5. Indeed, we see that maximum of rs=4 when r=s=2, and the inequality still holds since $$pq+4\ge 10\iff pq\ge 6\implies pq-rs\ge 2. \blacksquare$$


linear algebra
matrix
calculus
derivative
algebra
polynomial
function

If $ F \in C^p (R^{m \plus{} n} ;R^n )$ and the last n columns of the Jacobi matrix of F at a point x form an invertible matrix then why does this condition hold in a neighborhood of x? My book states this without proof.

det at that point not 0 (with those n columns), so there is a neighborhood with det not 0 too

Because you said the partial derivatives were continuous, and the determinant, being a polynomial function of its entries (those partial derivatives), is continuous. A continuous function that is nonzero at a point is nonzero on a neighborhood of that point.

Thanks Kent. This is after exploring determinants as tensors over modules in a very algebraic setting. It's easy to forget these things :)


Dear everyone.

Please think of the following.
Let two circles s and t meet at O and P, and let each diameter OS and OT of the two circles cut the other circle at A and B. Then chord OP passes through the center of circle OAB.

I think this is nice. :)
Thank you for reading.

$ \angle APO\equal{} \angle ATO$
$ \angle OPB\equal{}\angle BSO$
$ \angle APS\equal{}\angle SOT\equal{}\angle BPT$
$ T,P,S$ are collinear.
$ \implies \angle APO\equal{}\angle OPB \implies \angle BSO\equal{}\angle ATO \implies \angle OSP\equal{}\angle OBA$
$ \angle OSP\equal{}90^\circ\minus{} \angle POS$.
so we r done!

Obviously, T, P and S are collinear and $ OP \perp TS$
From $ AT \perp OS$ and $ BS \perp OT$ $ \implies  ABTS$ cyclic and, with the previous observation, the circumcenter of $ \triangle OAB$ lies on OP.

Best regards,
sunken rock


When I press F5 it says "Cannot execute the command" when the PDF would open before and I just added a little bit more of text. There weren't any errors or anything in the text i had added, so...what's wrong?

This is way too vague a description of your problem for anybody to be able to give you any advice.


I discovered it many years ago. I was very proud I invent such inequality :P (Don't blame me - I was very young).

Consider a sum $1\cdot n+2\cdot (n-1)+...+n\cdot 1=S$. You can easily check $S=\frac{n(n+1)(n+2)}{6}$. So applying AM-GM for numbers $1\cdot n,2\cdot (n-1),...,n\cdot 1$ we obtain:
\[\left(\frac{(n+1)(n+2)}{6}\right)^n\geq n!^2.\]

look beautiful, but too hard for me

 :(

congratulation!!!!!!!!!!!!!!!!!!!!!!!1 :lol:


algebra
polynomial
inequalities
complex numbers
algebra proposed

Let $ a,\ b$ be real numbers such that $ a\geq 1,\ b\geq 1$. Prove the following equations don't have common root.

$ ax^{2}\plus{}(a\plus{}1)x\plus{}(a\plus{}2)\equal{}0$
$ bx^{3}\plus{}(b\plus{}1)x^{2}\plus{}(b\plus{}2)x\plus{}(b\plus{}3)\equal{}0$

Assume that a common root $ r$ exists, we will reach a contradiction.

If common root $ r$ is complex, since both equations have real coefficients, then the complex conjugate of $ r$, is also a common root.  Then the second degree polynomial divides the 3rd degree polynomial.  The 1st degree coefficient in the remainder is found to be $ \frac{\left(a\minus{}b\right)\left(a\minus{}1\right)}{a^{2}}$, or it must be either $ a\equal{}b$ (absurd, since substracting the first given equation multiplied by $ r$ from the second yields $ b\equal{}\minus{}3$), or $ a\equal{}1$, resulting in $ r^{2}\plus{}2r\plus{}3\equal{}0$.  Substracting $ br\minus{}b\plus{}1$ times this last equation from the second given equation yields $ b\equal{}0$, absurd.  The common root cannot be complex.

Writing the first equation as $ r^{2}\minus{}1\equal{}\left(a\plus{}1\right)\left(r^{2}\plus{}r\plus{}1\right)$, and the second equation as $ r^{3}\minus{}1\equal{}\left(r\plus{}1\right)\left(\left(b\plus{}1\right)r^{2}\plus{}b\plus{}2\right)$, we find that $ \left(a\plus{}1\right)\left(r\plus{}1\right)\left(\left(b\plus{}1\right)r^{2}\plus{}b\plus{}2\right)\equal{}\left(a\plus{}1\right)\left(r^{3}\minus{}1\right)\equal{}\left(r\plus{}1\right)\left(r\minus{}1\right)^{2}$.  If $ r\equal{}\minus{}1$, then from the first equation $ a\equal{}\minus{}1$, absurd.  So $ r^{2}\minus{}2r\plus{}1\equal{}\left(a\plus{}1\right)\left(b\plus{}1\right)\left(r^{2}\plus{}1\right)\plus{}a\plus{}1$, or $ \left(ab\plus{}a\plus{}b\right)\left(r^{2}\plus{}1\right)\plus{}2r\plus{}a\plus{}1\equal{}0$.  Obviously $ r$ cannot be a positive real number.  If it is a negative real number, AMGM inequality yields $ r^{2}\plus{}1\geq\minus{}2r$, or since $ ab\plus{}a\plus{}b\geq3$, we find $ 2r^{2}\plus{}a\plus{}3\leq0$, absurd.  Then $ r$ cannot be real.  So no common roots exist, qed.


A certain amount of work can be done by seven identical machines in 30 hours.  How long will it take 10 of these same machines to do the same amount of work?

10 machines can completely the job $ \frac{10}7$ as fast, which means they do the job in $ \frac7{10}$ of the time.

$ 30*\frac7{10}\equal{}\boxed{21}$


Prove that if the equation , $x^{4}+ax^{3}+bx+c=0$
has all its roots real,then $ab\leq0$.

The sign of $ab$ does not dictate the number of real roots the function will have.

For example, consider $f(x)=x^{2}-\frac{1}{10}x^{3}-20x^{2}+1$ where, $ab=2>0$.

This function has roots at $x\approx-4.4167,\;x\approx-0.224,\;x\approx 0.2238,\;\text{and}\;x\approx 4.4569$, and I strongly believe all four of these are real. :wink:


[quote="mr. math"]how would you do this problem?

Circle O is inscribed in a right triangle ACB with hypotenuse segment AB. If AC= 3, BC= 4, and segment OB intersects circle O at point P, then find the length of segment BP.

please include an explanation.  :arrow:   :idea:  :alien:[/quote]

[hide]The inradius of the circle is 1. The radius parallel to AC is drawn and intersects BC. This length is 1 and the point where the radius crosses BC to B is length 3. So OB=$\sqrt{10}$ and PB=$\boxed{\sqrt{10}-1}$...is that correct?[/hide]

[quote="easyas3.14159..."]I got
[hide](rt73/2)-2. This is right triangle with hypotenuse 5 and area 6,and you know the semiperimeter is 6. As semiperimeter*inradius=triangle's area, the inradius is one and the diameter is 2. To find BP, subtract two from rt73/2.[/hide][/quote]

Why $\sqrt{73}/2$?

Apparently the median is not the same thing as the angle bisector.
Me and my careless errors. :blush:


algebra polynomial geometry ratio quadratics

$ 1)$ Solve the following equation for $ x$: $ x/50b\plus{}24a^2/x\equal{}a/b$

$ 2)$ The eight term of a geometic progression is $ 2187/64$ and the second term is 3. If the 5th terim is in the form $ \plus{}$ or $ \minus{}r/s$, find $ r\plus{}s$.

For the first one: Multiply by 50bx and get x^2-50ax+1200a^2b=0=>x=25a+-5sqrt[25a^2-48a^2b]

For the second one:u8=u1*a^7 and u2=u1*a where a is the common ratio=>u8/u2=a^6=>729/64=a^6=>a=+-3/2
if a=3/2 then u1=2 whereas if a=-3/2 u1=-2
u5=u1*a^4=+-3^4/2^3, so r=3^4=81 and s=2^3=8 so r+s=81+8=89

[quote="salonikios"]For the first one: Multiply by 50bx and get $ x^2\minus{}50ax\plus{}1200a^2b\equal{}0\rightarrow x\equal{}25a\plus{}/\minus{}5\sqrt{25a^2\minus{}48a^2b}$

For the second one:$ u_8\equal{}u_1\cdot a^7$ and $ u_2\equal{}u_1\cdot a$ where a is the common ratio $ \rightarrow \frac{u_8}{u_2}\equal{}a^6\rightarrow \frac{729}{64}\equal{}a^6\rightarrow a\equal{}\plus{}/\minus{}3/2$
if $ a\equal{}\frac{3}{2}$ then $ u_1\equal{}2$ whereas if $ a\equal{}\frac{\minus{}3}{2}$, $ u_1\equal{}\minus{}2$

$ u_5\equal{}u_1\cdot a^4\equal{}\plus{}/\minus{}\frac{3^4}{2^3}$, so $ r\equal{}3^4\equal{}81$ and $ s\equal{}2^3\equal{}8$ so $ r\plus{}s\equal{}81\plus{}8\equal{}894$[/quote]

Latexed for you, and it seems correct.
[hide="hint for 1"]
Quadratic form should be whispering in your earrrrrr

[/hide]


algebra
polynomial
number theory proposed
number theory

Let $A$ be a set with at least $3$ integers, and let $M$ be the maximum element in $A$ and $m$ the minimum element in $A$. it is known that there exist a polynomial $P$ such that: $m<P(a)<M$ for all $a$ in $A$. And also $p(m)<p(a)$ for all $a$ in $A-(m,M)$. Prove that $n<6$ and there exist integers $b$ and $c$ such that $p(x)+x^2+bx+c$ is cero in $A$.

Two questions: what is $A-(m,M)$ and what is $n$?

[quote="Pascual2005"]$n=|A|$, and $A-(m,M)$ is all the elements in $A$ different from $M$ and $m$[/quote]
I.e. $A-(m,M)=A\setminus\{m,M\}$ ;)

thanks! now solve it Myth! I know you will enjoy it!

I guess $P(x)\in \mathbb{Z}[x]$ (otherwise consider $P(x)=x/2$, $m=-100,M=100$).

But if $P(x)\in \mathbb{Z}[x]$ then I got complete absurd.

We have $|P(M)-P(m)|<M-m$ and on the other hand $M-m\mid P(M)-P(m)$. Thus $P(M)=P(m)$.
Therefore we can write $P(x)=P(m)+(x-m)(M-x)Q(x)$, where $Q(x)\in \mathbb{Z}[x]$ and $Q(x)>0$ on $A\setminus\{m,M\}$.
Consider $a\in A\setminus\{m,M\}$. Then $M-1\geq P(a)=P(m)+(a-m)(M-a)Q(a)\geq (m+1)+1(M-m-1)\cdot 1=M$ -- contradiction... Thus $A\setminus\{m,M\}=\varnothing$.

Maybe there are some typos in signs in your statement? Or my solution is wrong...

Sorry Myth! $P$ has integer coeficients as you said!


If $m$ and $n$ are integers, show that

\[ \frac{(2m)!(2n)!}{m!n!(m+n)!} \] is also an integer.

2m!2n!/(m!n!(m+n)!)
wLOG m>=n.
when 2m! and m! cancel out, we are left with 2m(2m-1)(2m-2)...(m+1) and when 2n! and n! cancel out, we're left with 2n(2n-1)(2n-2)...(n+1)
since m>=n, then 2m>=m+n>=2n
Thus we are left with 2m(2m-1)(2m-2)...(2m-n+1)*2n(2n-1)(2n-2)...(n+1)/n(n-2)(n-3)
since the second part is 2nCn, the whole expression equals 2mP(2m-n)*2nCn which is obviously an integer

I do not believe that final expression is right...let $m=3$ and $n=2$. the original expression is 12, but yours is 2160.

ummm does the solution branch from the fact that the given expression equals:

$\frac{\binom{2m}{m} \cdot \binom{2n}{n}}{\binom{m+n}{m}}$

? im not sure wat to do from there tho.... some.... clever combinatorial argument or i ono

Yea, I'm trying to fit them into binomial coefficients, but I keep ending up with two binomials on top and one on the bottom.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=21614]IMO 1972 #3 ;)[/url]

Ooh, didn't know it was an IMO problem

Nice solution though, only mildly thought about considering an arbitrary prime, but I guess binomial fitting doesn't work here


[quote="Snailboto"]#1 of States
What is the number of distinct  ways of arranging the letters in the word average?

Basically I'm asking is how you switch letters around with doubles.
I know how to do it with distinct but I'm having trouble with doubles. Thanks for your help.[/quote]
[hide]Well, there are 7 letters in average. And there are 2 a's and 2 e's. So it looks like $\frac{7!}{2!2!}=7*6*5*2*3=1260$.[/hide]

so if 2 numbers repeated thrice, you would divide by 36 (3!3!)?

-jorian

yes, and if 3 letters were repeated twice like aabbcc

it would be $\frac{6!}{2!2!2!}$

[hide="Answer"]
The word is $AVERAGE$. There are $7$ letters. There are two $A$'s and two $E$'s. You may think that it is just $7$ choices for the 1st letter, 6 for the second, and so on for a total of $7!=5040$ possibilities, but it is wrong.

Lets label the A's as $A_{1}$ and $A_{2}$.

Then, we could have arrangements of $A_{1}VERA_{2}GE$ and $A_{2}VERA_{1}GE$, but they count as two in our counting, but they are the same word.

There are $2$ ways to arrange the $A$'s. Likewise, there are $2$ ways to arrange the $E$'s.

So, the answer would be $\frac{7!}{2\times 2}=1260$.

Now, you try it with more letters.

You may figure that in larger words like $AAAB$, there are these ways to arrange the $A's$:

$A_{1}A_{2}A_{3}$
$A_{1}A_{3}A_{2}$
$A_{2}A_{1}A_{3}$
$A_{2}A_{3}A_{1}$
$A_{3}A_{1}A_{2}$
$A_{3}A_{2}A_{1}$

or a total of $3!$ ways to arrange. So, you must use factorials, not just simple multiplication.

So, for the followup, the answer would be $\frac{4!}{3!}=4$ ways.

For the original problem, the answer would be $\frac{7!}{2!2!}=\boxed{1260}$.[/hide]


Let $ a,b,p\in R$, and $ p\ge 1$, show that:
$ |ta\plus{}(1\minus{}t)b|^p \le t|a|^p \plus{}(1\minus{}t)|b|^p,0\le t \le1$

Let be $ f(x) \equal{} |x|,g(x) \equal{} x^p ;$
$ |ta \plus{} (1 \minus{} t)b|^p  \le t|a|^p  \plus{} (1 \minus{} t)|b|^p  \Leftrightarrow (g \circ f)(ta \plus{} (1 \minus{} t)b) \le tg(f(a)) \plus{} (1 \minus{} t)g(f(b));$
We can easily demonstrate that $ {\rm (}g \circ f)''(x) > 0,\forall x \in R$ and the function is convex.


I'm currently writing a mock AMC, and I'm sure many people will be interested because winter break is coming soon so it's time to start studying for AMC 12. Anyone who's interested just post here, and also note that I will need someone to check over my problems. I'm trying to make the difficulty level a bit on the high side for an AMC 12, so keep that in mind. I'm hoping to finish the problems sometime mid-next week.

I can create an HTML form like how i did for my mock. It exports it to an XLS file and then after the deadline, you can view it.

Better yet, I can regulate teh amount of submissions from 1 ip address.

Anirudh, talk to me, that would be great.

Okay, I talked to beta and he agreed to check it over. Right now, it looks like the problems should be ready by around Wednesday.

I like mocks. This will be my first time taking an AMC 10 test, so AMC 12s would be cool for most people.  :D

[quote="calc rulz"]Okay, I finished the test and sent it over to beta for checking, he is still checking right now.[/quote]

Does that mean it'll be out today?

Dunno, depends on how long it takes beta to correct it and me to edit it. I'm also waiting for Anirudh to reply about making an HTML form for it.

I'm in...

I need to keep track of all these mocks.

When is it due?


geometry
rectangle
rotation
combinatorics unsolved
combinatorics

Is it possible to remove one square from a $ 5X5$ board so that the remaining $ 24$ squares can be covered by eight $ 3X1$ rectangles? If yes, find all such squares.

Color it like this:

R-G-B-R-G
G-B-R-G-B
B-R-G-B-R
R-G-B-R-G
G-B-R-G-B

Each rectangle covers and R,G,B, and there are currently 8,9,8 of them respectively, so we need to remove a G.

Now, color the board the same way, but rotated 90 degrees. Then we must similarly remove a G in this new coloring. But the only G common to both colorings is the middle square, and indeed, removing the middle square works:

1-2-3-3-3
1-2-4-4-4
1-2-X-6-5
7-7-7-6-5
8-8-8-6-5


function
integration
calculus
limit
real analysis
real analysis unsolved

let f a function strictly increasing and continuos on [a,b]
 show that int(f,a..b)+int(f-1,f(a)..f(b))=bf(b)-af(a)

Maybe you  mean
\[ \int\limits_{a}^{b}f(t)dt+\int\limits_{f(a)}^{f(b)}f^{-1}(t)dt = bf(b)-af(a)? \]

Then, it must be obvious from the geometric interpretation :)

may be it's obvious , but for witch interpretition do you talk about , and even if it's geometricly obvious we must show it algebricly  :)

Yes , it's clear , i see what mistakes i have made in my graphic solution ( i made the symetric of f by y=x to get f-1 but i ....)
what about an algebric proof , i try with sequence but without result  :(

[quote="xirti"]You can use the change of variable $x=f(t)$ in the second integral. Then integrate by parts. Good luck ;)[/quote]

If we do so, then (i think) we must assume that there exists $f'$. Because this condition is need for integrate by parts, and for theorem about change of variable :!:

We cannot use such thereom as above for Riemann integral, but if we will write integral Riemann-Stieltjes :)
$(RS) \int\limits_{f(a)}^{f(b)} f^{-1}(t)dt = |f(x)=t| = \int\limits_{a}^{b} x df(x) =$
by parts $= \int\limits_{a}^{b} g(x) df(x) = g(b)f(b) - g(a)f(a) - \int\limits_{a}^{b} f(x) dg(x)=$
$=bf(b) - af(a) - \int\limits_{a}^{b} f(x) dx.$
And we obtain the result :)

What is "Riemann-Stieltjes" integral?

Who is function g which apears after the integration by parts?

This is a very old result in calculus. In fact, I'm pretty sure that it's due to Leibniz. (Well, not the Stieltjes version.)

to xirti:
$g(x)=x.$
and in a few words
\[ (RS) \int\limits_{a}^{b} f(x)\ dg(x)= \lim_{\begin{array}{cc} n\to \infty && x_{k+1}-x_k\to 0 \end{array}} \sum_{k=0}^{n} f(\xi_k)(g(x_{k+1})-g(x_k)) \]

I will simulate the integration by parts.

Denote $f^{-1}=g$
$\int_{f(a)}^{f(b)}g(x)dx=\lim_{n \to \infty \\ \max(y_{i+1}-y_i)\to 0} \sum_{i=0}^ng(\tau_i)(y_{i+1}-y_i)$ which becomes after substitutions : $\tau_i=f(\xi_i);y_i=f(x_i)$

\[ \sum_{i=0}^n \xi_i(f(x_{i+1})-f(x_i))= \]
\[ [\xi_0 f(x_1)-\xi_0 f(x_0)+\xi_1f(x_2) -\xi_2 f(x_1)+...+\xi_{n-1}f(x_{n})-\xi_n f(x_{n-1})]= \]
\[ -[ \xi_0f(x_0)+f(x_1)(\xi_1-\xi_0)+f(x_2)(\xi_2-\xi1)+....+f(x_n)(\xi_n-\xi_{n-1})+ f(x_n)\xi_n]= \]
\[ -\sum_{i=0}^n f(x_i)(\xi_{i+1}-\xi_i)+f(b)\xi_n-f(a)a \text{ (1)} \]

From our substitutions we get that:

-since $f^{-1}$ increasing $\tau_i \in [y_i,y_{i+1}]\Rightarrow \xi_i\in [x_i,x_{i+1}]$
-since $f^{-1}$ is continuous on a compact it is uniformly continuous on it and then since $\max(y_{i+1}-y_i)\to 0 \Rightarrow \max (x_{i+1}-x_i) \to 0$
-$x_0=a,x_n=b$

From these two facts, tending n to infinity in (1) we get that $\int_a^bg(x)=-\int_a^bf(x)+xf(x) -af(a)$. q.e.d.