I made a small calculation error on #3 at the end when I was calculating the final result on both part A and B. Would I lose a point?

[quote="durt"]I did that too. I found the correct formula for part B, then ended up with 90 instead of 98 somehow for part A.  :wallbash:[/quote]
There's no way they'd dock you for that - if you got the right formula, it would be clear to the graders that you made a silly mistake in calculating part (a).

[quote="zanttrang"]
There's no way they'd dock you for that - if you got the right formula, it would be clear to the graders that you made a silly mistake in calculating part (a).[/quote]

I think there is a good possibility of losing a point for a mistake like that.  I remember last year (my first year) that on Problem #1 Round #1, I has a correct procedure for testing whether an infinitite arithmetic solution has a perfect square, but I somehow miscalclated for one of the sequences.  My answer was off by one, and I lost a point.

I imagine that it depends on the grader and the clarity of your solution.  If the grader is feeling nice and/or your solution is very clear, then you probably won't lose a point.  I've personally had an incorrect calculation and still gotten a 5, but I've also heard of many people who make calculation errors and lose one point.  Could go either way, but I'd tend to agree with zanttrang, it probably won't cost you a point if the formula is correct.


Problem: Show that $ [ABC]\equal{} 2R^2 sin A sin B sin C$

[hide="My solution"]we can write the RHS as $ 2R*2R sin A sin B sin C$ so that we have $ \frac{a^2}{sin^2 A} sin A sin B sin C$ we can write this as $ \frac{a}{sin A}*sin B*a*sin C$ then we get this to be $ ab*sin C$. this is very close to [ABC] because of the formula $ \frac{1}{2}ab sin C$ but not quite right. Where did I go wrong?[/hide][/hide]

You said that $ 2R^2\sin A\sin B\sin C\equal{}2R\cdot2R\sin A\sin B\sin C\equal{}4R^2\sin A\sin B\sin C$, which is clearly incorrect.

It should be $ 2R^2\sin A\sin B\sin C\equal{}R\cdot2R\sin A\sin B\sin C$, and then the rest of your solution is correct. :)

oooh, thank you very much math154, i see my mistake :D


The mean age of the members of a baseball team is 20. The mean age
   of the members of a softball team is 45. The mean age of the members
   of both teams combined is 30. Assuming the teams do not share any
   players, find the ratio of the number of baseball players to the number
   of softball players. Express your answer as a common fraction.

20x+45y=30(x+y)
4x+9y=6x+6y
2x=3y

therefore the answer is [b]3/2[/b]...

take 6 baseball, 4 softball for example, the sum of the ages is 120+180=300, the average is 30, so it works!


geometry
3D geometry
prism
perimeter
parallelogram

I think that formula would work for any prism, since it would be the area of several parallelograms, which have an analogous formula for area.

[quote="redsky"]Isn't the lateral area of any prism equal to the perimeter of a base times the height of the prism rather than just a right prism???
If it is just for a right prism, could you give a simple yet convincing argument that disproves the statement for an oblique prism?[/quote]
The height is considered the perpendicular distance from one base to the other.

In a prism that is not right, the height of the prism isn't necesarily the height for each lateral face (Slant height is not always equal to actual height). So, for at least 1 lateral face, the area is not calculated correctly.

You have to find the height on any face that isn't perpendicular to the base from its bottom to top and use that in the calculation, not the height of the prism.

So are you saying that the lateral area of any oblique prism is equal to the perimeter of a base times the height (the perpendicular distance from the plane containing a base to the other) of the prism?

No, he's saying the opposite. The height of a prism is the [b]perpendicular[/b] line going from a point on one base to the corresponding point on the other.

With an oblique prism, the slant height is not the same as the height you're thinking of.

Think of it this way...

You have a right triangle, with the legs going vertically and horizontally.
The vertical leg is the regular height, and the hypotenuse is the slant height. See how that would change the value, depending on which one you used?
The hypotenuse is longer.


For any polynomial F(x) with integer coefficients, does there necessarily exist an integer a so that F(a) has at least n distinct prime divisors for any integer n?

Unlike blahblahblah's "Don't FIB, kids" thread, I actually did like this problem. The pun came after. Has anyone worked on it? hint: [hide]prove that for a polynomial f(x) with integer coefficients,  if f(k) is divisible by a prime p then f(k+mp) is also divisible by p for any integer m.[/hide]

Mathematical puns are the first sine of madness...


Prove that there are no positive integers $x$ and $y$ such that $x^2+y+2$ and $y^2+4x$ are perfect squares

[hide]The next perfect square after $y^2$ is $(y+1)^2 = y^2+2y+1$.

We are given that $x$ and $y$ are positive integers, so $y+2 > 0$. Assuming (for the sake of reaching a contradiction) that $x^2+y+2$ was a perfect square, $y+2 \geq 2x+1$.

Similarly the next perfect square after $x^2$ is $(x+1)^2 = x^2+2x+1$. Thus, (as $4x>0$) again assuming that $y^2+4x$ is a perfect square $4x \geq 2y+1$.

We have inequations $y+2 \geq 2x+1$ and $4x \geq 2y+1$. Manipulating, $2y+2 \geq 4x \geq 2y+1$. $4x$ is obviously even, so $4x=2y+2$. But then $y^2+4x = y^2+2y+2 = (y+1)^2 + 1$ is not a perfect square for integer $y$. Contradiction. Initial assumption that both $x^2+y+2$ and $y^2+4x$ could be perfect squares for positive integer $x$ and $y$ is false.[/hide]

Really nice :!: , how did you came up with this idea?

[color=#00f][b]Case 1:[/b][/color] $x\geq y$
We get $$x^2< x^2+y+2\leq x^2+2x+1=(x+1)^2$$
So, there is no answer in this case.
[color=#00f][b]Case 2:[/b][/color] $x<y$
We get $$y^2< y^2+4x< y^2+4y< (y+2)^2$$
So $y^2+4x=y^2+2y+1\Longrightarrow 4x=2y+1$ but $2\nmid 1$. So, there is no answer in this case
So, we are done.

[hide=nice problem]first of all we notice a trivial observation that since $x,y \in \mathbb{Z^{+}}$ we  must have \\

$x^2<x^2+y+2$ and $y^2+4x>y^2$\\

[color=#f00][b]Claim:-[/b][/color]$x^2+y+2<(x+1)^2$ and $(y^2+4x)<(y+1)^2$\\

[color=#00f]pf:-[/color]FTSOC assume , $x^2+y+2>(x+1)^2 \iff y+2>2x+1 \iff y>2x+1$, similarly $y^2+4x>(y+1)^2\iff 4x>2y+1$\\

so we observe:\\

[color=#ff9a00]Case1:-[/color] $x>y$\\

in this case we get:\\

$x>y>2x-1\implies 0>x-1$ , which is clearly not possible for $x\in \mathbb{Z^{+}}$\\

[color=#ff9a00]Case2:-[/color]$y>x$\\

in this case we observe , since $y>x\implies y^2>x^2$\\

so we have $x^2+y+2<y^2+y+2<(y+1)^2$ which from our assumption is not correct which brings a  contradiction for our assumption\\

also , for $x=y$ we have $x^2+x+2$ and $x^2+4x$ to be perfect square which is clearly not possible  for any $x\in \mathbb{Z^{+}}$\\

hence our claim follows , and hence we have $x^2<x^2+y+2<(x+1)^2$ and $y^2<y^2+4x<(y+1)^2$\\

hence for $x,y \in \mathbb{Z^{+}}$ we don't have any (x,y) such that these two stuffs are perfect squares $\blacksquare$[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Is this form of Cayley's Theorem Wrong?
Let $ G$ be a group . Then $ G$ is isomorphic into the symmetric group $ S_G$ .

I think it should be $ G$ is isomorphic into a subgroup of  symmetric group $ S_G$

The term "isomorphic into" makes no sense. The correct wording would be something like "Every group $ G$ can be embedded into the symmetric group $ S_G$".
Your last sentence is correct after replacing "into" with "to".


How do I change the css for the top header part? How about for the entries? I just want the part that replaces body here:

[code]
body
{
background-image: clouds.gif
}
[/code]

You should try using the Firefox Web Developer Extension. Very helpful ;)

Whoa! Thanks a lot. :) Much easier to edit css now. :)


$18)$ The picture is here.

In Quadrilateral $ABCD$, diagonals $BD$ and $AC$ are perpendicular and $FG$ is perpendicular to $HG$. Given $m<EFH=67$, find $m<EGF.$

Do you have the actual picture to scan?

I think that might be better.

you know that you can draw straight lines in paint by just clicking the line button...

better picture made by geogebra
btw use \angle to make angles like this $m\angle$

[quote="mathgeniuse^ln(x)"]I don't think you can attach paint documents though?[/quote]

not you...the person who made the picture in paint...

You can attach a paint document if you save it as a GIF

I think you gotta edit your problem specifying where does come $\overline{FG}$ and $\overline{HG}$.

I don't think there's enough information to solve this problem. :oops:


search calculus integration algebra function domain

This problem is created by my Algebra 2 teacher.  It has already been solved except he wants the numbers to be smmmallll.   Meaning that cool graphing calculator should not used to used to factor some 4 or 5 digital scary freako-nomials.  Please help!  :(

7 :sqrt: (3x+10) - 4 :sqrt: (9x+19) =  :sqrt: (6x 2 -31x+14)

Thanks Bunches
PS.  I'm not sure where this should be, so excuse me if i put it on the wrong board. :blush:

Square both sides then isolate the radical on the left hand side and then square once more and simplify/  You'll get this nasty equation:
$
36x^4-3864x^3+9652x^2+41328x+12560=0
$
I plugged this into my Computer Algebra System (but there is a method to solve the general quartic, by reducing it to a cubic, but I'm lazy) and it shot out these answers
{5.,104.666666666666,-2.,-0.333333333333}

Testing these I found $104.\overline{6}$ to be extraneous, but the other three work.

what he said.
friend123: that's too big
friend123: we need a more sensable way

is that even possible?
and i think we got something like that and stoped there because it is too big to factor manually

I don't know any "general" approach to this question, but if you only want to search for rational roots, after you square it out one time, you see that (3x + 10)(9x + 19) must be the square of some rational number.  You can just try some numbers to make it integral squares, at that point.  It's not a fool-proof method by any means, but it will work in this case if you guess to set that expression equal to 4 or 144 (or 6400), for example.  Actually, in this case you can take the completely unwarranted step of trying to pick a value for x that makes both pieces of that expression a square, which definitely will not always work (because sometimes there jsut won't be numbers like that).

:? how do u go about doing that?  and is there a way to use substitution?  what else can i do to prevent the numbers from going too large?

If we were promised [b]rational[/b] roots, we have a chance.

themonster's equation can be reduced: $9x^4 - 966x^3 + 2413x^2 + 10332x + 3140 = 0$

Since 9 has factors {1, 3, 9} and 3140 has {1,2,4,5,10,20,157,314,628,785,1570,3140},
we have 'only' 72 possible rational roots to test.

Some can eliminated by examining the domain of $x.$

From the first radical: $x > -\frac{10}{3}$
From the second: $x > -\frac{19}{9}$
From the third: $x > \frac{14}{13}$ or $x < \frac{1}{2}$

Hence, the domain of $x$ is: $\left(-\frac{19}{9},\frac{1}{2}\right) \;\cup \;\left(\frac{14}{13},\infty\right)$

[i]Good luck![/i]

JBL: what do you mean "you can take the completely unwarranted step of trying to pick a value for x that makes both pieces of that expression a square"?  How do you do that?

Well, figure a value of x to make (3x + 10) and (9x + 19) both squares ... ones I see are 2 and -1/3, for example (just by trying small squares for each of them and seing if the other becomes a square).  Obviously, if both terms are squares, their product is a square.  However, this method is weak, because the opposite is certainly not true: there are plenty of products of numbers that are square with out the factors themselves being squares.

[quote="Soroban"]
From the third: $x > \frac{14}{13}$ or $x < \frac{1}{2}$

Hence, the domain of $x$ is: $\left(-\frac{19}{9},\frac{1}{2}\right) \;\cup \;\left(\frac{14}{13},\infty\right)$

[i]Good luck![/i][/quote]
Are you sure? Because $6(2)^2-31(2)+14=-24<0$ and $2\in\left(\frac{14}{13},\infty\right)$

I think you meant $\frac{14}3$ and [b]NOT[/b] $\frac{14}{13}$


Let $ x,y,z>0$ and $ xyz\equal{}1$ . Prove :

$ \sum_{cyc}\frac{1}{x^5\plus{}y^2\plus{}z^2}\le \frac{x\plus{}y\plus{}z}{xy\plus{}yz\plus{}zx}\le \frac{xy\plus{}yz\plus{}zx}{3}$

[quote="Pain rinnegan"]Let $ x,y,z > 0$ and $ xyz \equal{} 1$ . Prove :

$ \sum_{cyc}\frac {1}{x^5 \plus{} y^2 \plus{} z^2}\le \frac {x \plus{} y \plus{} z}{xy \plus{} yz \plus{} zx}\le \frac {xy \plus{} yz \plus{} zx}{3}$[/quote]

Generally,we have
$ \sum_{cyc}\frac {1}{x^{m} \plus{} y^{n} \plus{} z^{n}}\le\frac {x \plus{} y \plus{} z}{xy \plus{} yz \plus{} zx}\le\frac {xy \plus{} yz \plus{} zx}{3} ,$
it  is holds for
                             $ m \equal{} 0,    n \equal{} 2,3,...;$
 or
                             $ m \equal{} 1,2,3,...,      n \equal{} 1,2,3,....$


The proof is in the attachment

Please help me.

Open as many of them as you really need. If you get a feel of it halfway through, you may as well figure it out yourself.

[hide="1"]
Construct a line parallel to AB through C. Call the intersection of this line and segment DE point "G".
[/hide]
[hide="2"]
By similar triangles, BD/CD=BF/CG
[/hide]
[hide="3"]
Since F is the midpoint of AB, BF=AF, and
BF/CD=AF/CG
[/hide]
[hide="4"]
We can show that triangles EFA and EGC are similar because CG is parallel to AF and angle AEF is vertical to angle GEC. This similarity gives AF/CG=AE/EC
[/hide]
[hide="5"]
We have:
BD/CD=BF/CG
BF/CD=AF/CG
AF/CG=AE/EC

Transitively, $BD/CD=AE/EC$
[/hide]

[hide]
Let $[ABC]$ denote the area of $\triangle ABC$. Firstly, since $F$ is the midpoint, we know $[AFE] = [BFE]$ and $[AFD] = [BFD]$ and consequently $[AED] = [BED]$.

Then

$\frac{BD}{CD}= \frac{[BED]}{[CED]}= \frac{[AED]}{[CED]}= \frac{AE}{EC}$.[/hide]

EDIT: Looks like I was beat to it.

[quote="paladin8"][quote="Altheman"]or you could be like this follows directly from menelaus' theorem[/quote]

You could... but that doesn't constitute much of a proof, does it?[/quote]

huh what is wrong with:

we apply menelaus' theorem on transveral DEF, triangle ABC, this gives

$\frac{BD}{DC}*\frac{CE}{EA}*\frac{AF}{FB}=1$

using $AF=FB$, we get $\frac{BD}{CD}=\frac{AE}{EC}$ as desired

[quote="paladin8"]Isn't the point of proofs not to use stronger results to prove weaker ones? Clearly the result here is just a special case of menelaus, so using menelaus to show it is basically circular reasoning.[/quote]

...it is useful to know menelaus...this result is useless because of the midpoint condition...

but just to offer a proof...

--
let $h_{a}$, $h_{b}$, $h_{c}$ be the heights to $DF$ from $A$, $B$, $C$ , respectively, using similar triangles...
$1=\frac{AF}{BF}=\frac{h_{a}}{h_{b}}$ and $\frac{BD}{CD}=\frac{h_{b}}{h_{c}}$ and $\frac{CE}{EA}=\frac{h_{c}}{h_{a}}$
multiply these, to get the desired result
--

anyway...as long as i know the proof, there is no reason not to use a stronger result...of course, menelaus is assumed on any olympiad...i don't know of any other proof situation that would not use menelaus (maybe school...but you could just tell your teacher what you did)


Can anyone help me with a website or sth which talks about: how to instruct a car( for example a toy Car) which works with the energy of sun( using a sunny sheet).....

Thanks for that Dude :)
but i killed myself  :wallbash_red:  :wallbash: to find some more information about for example what is the best voltage for this solar car or some of it's calculations and formulas that we use to instruct it...
but i couldn't :(
can you help me more?
thanks

I'm sure you can use the particle-wave duality of light to define the energy the light transfers to the car based on intensity and the size of the solar panel.


If $ a,b,c$ positive, show that:
\[ \frac{a(a^2\plus{}bc)}{b^3\plus{}c^3} \plus{} \frac{b(b^2\plus{}ac)}{a^3\plus{}c^3} \plus{} \frac{c(c^2\plus{}ab)}{a^3\plus{}b^3} \leq \frac{a^4}{b^2c^2} \plus{} \frac{b^4}{a^2c^2} \plus{} \frac{c^4}{a^2b^2}\]

[quote="hendrata01"]If $ a,b,c$ positive, show that:
\[ \frac {a(a^2 \plus{} bc)}{b^3 \plus{} c^3} \plus{} \frac {b(b^2 \plus{} ac)}{a^3 \plus{} c^3} \plus{} \frac {c(c^2 \plus{} ab)}{a^3 \plus{} b^3} \leq \frac {a^4}{b^2c^2} \plus{} \frac {b^4}{a^2c^2} \plus{} \frac {c^4}{a^2b^2}
\]
[/quote]

my solution , I don't know it's right or wrong ^^!
$ abc\equal{}1,$ and
we have$ \frac{a\plus{}1}{b\plus{}c}\plus{}\frac{b\plus{}1}{a\plus{}c}\plus{}\frac{c\plus{}1}{a\plus{}b}\leq a^2\plus{}b^2\plus{}c^2$
$ a\equal{}\sqrt{a}$ for easy (my english is very bad )

$ <\equal{}> a^3\plus{}b^3\plus{}c^3\plus{}3(ab\plus{}bc\plus{}ca)\plus{}(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca) \leq (a^2\plus{}b^2\plus{}c^2)(ab^2\plus{}ba^2\plus{}ca^2\plus{}ac^2\plus{}bc^2\plus{}cb^2)\plus{}a^2\plus{}b^2\plus{}c^2$

Expand , we use $ a^4b\plus{}a^4c\plus{}a^2\geq 3a^3$
and $ \sum a^2b^2(a\plus{}b) \equal{} \sum b^2(a^3\plus{}c^3)\geq acb^2(a\plus{}c)\equal{}\sum b(a\plus{}c)$

we have here an homogenous inequality so we can suppose that abc=1

the inequality become like "fesha3t" said

$ \frac{a\plus{}1}{b\plus{}c}\plus{}\frac{b\plus{}1}{a\plus{}c}\plus{}\frac{c\plus{}1}{a\plus{}b}\leq a^{2}\plus{}b^{2}\plus{}c^{2}$

we have $ \frac{a\plus{}1}{b\plus{}c}\plus{}\frac{b\plus{}1}{a\plus{}c}\plus{}\frac{c\plus{}1}{a\plus{}b} \leq \frac{a\plus{}1}{2\sqrt{bc}}\plus{}\frac{b\plus{}1}{2\sqrt{ca}}\plus{}\frac{c\plus{}1}{\sqrt{ab}}$

so we must to prove that:
$ a\sqrt{a}\plus{}b\sqrt{b}\plus{}c\sqrt{c}\plus{}\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c} \leq 2\sum{a^2}$
$ \Leftrightarrow \sum{x^3}\plus{}\sum{x} \leq 2\sum{x^4}$
we have $ x^2\plus{}y^2\plus{}z^2 \geq \frac{1}{3}(x\plus{}y\plus{}z)(x\plus{}y\plus{}z) \geq 3\sqrt[3]{xyz}\frac{1}{3}(x\plus{}y\plus{}z)\equal{}(x\plus{}y\plus{}z)$
and we have $ \sum{x^3\plus{}x} \geq 2\sum{x^2}$
so $ x^3\plus{}y^3\plus{}z^3 \geq (x^2\plus{}y^2\plus{}z^2)\plus{}(x^2\plus{}y^2\plus{}z^2) \minus{}(x\plus{}y\plus{}z) \geq x^2\plus{}y^2\plus{}z^2$
and $ \sum{x^4\plus{}x^2} \geq 2\sum{x^3}$
so $ x^4\plus{}y^4\plus{}z^4 \geq (x^3\plus{}y^3\plus{}z^3) \plus{}(x^3\plus{}y^3\plus{}z^3)\minus{}(x^2\plus{}y^2\plus{}z^2) \geq x^3\plus{}y^3\plus{}z^3$

so $ x^4\plus{}y^4\plus{}z^4 \geq x^3\plus{}y^3\plus{}z^3$ is true  (1)
and we have $ \sum{x^4\plus{}y\plus{}z} \geq \sum{3\sqrt[3]{x^4yz}}\equal{}3\sum{x}$
then it 's clear that $ x^4\plus{}y^4\plus{}z^4 \geq x\plus{}y\plus{}z$ (2)

by (1) and (2) we 'll have $ 2(x^4\plus{}y^4\plus{}z^4)\geq x\plus{}y\plus{}z\plus{}x^3\plus{}y^3\plus{}z^3$
and we deduce the solution :lol:


We have:
\[\left \{\begin{array}{l}a+b>c,\,b+c>a,\,c+a>b\\ a+b+c=1 \end{array}\right \| \Longrightarrow 0<a,\,b,\,c<\frac{1}{2}\\ \Longrightarrow (1-2a)(1-2b)(1-2c)>0\\ \Longrightarrow 1>2(a+b+c)-4(ab+bc+ca)+8abc\\ \Longrightarrow\frac{1}{2}>(a+b+c)^{2}-2(ab+bc+ca)+4abc\quad\left(\text{because}\; a+b+c=1\right)\\ \Longrightarrow\boxed{\frac{1}{2}>a^{2}+b^{2}+c^{2}+4abc}\]
[b]Remark.[/b] [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=752820&sid=fd917c1b74690d9582fb6717ea2fd872#752820]Similary problem[/url]  :D

[quote="Otávio_Menezes"]If $a,b,c$ are sides of a triangle, with $a+b+c=1$, prove that

$a^{2}+b^{2}+c^{2}+4abc<1/2$[/quote]

similary , like in April's link, we can do $a^{2}+b^{2}+c^{2}=2p^{2}-2r^{2}-8Rr$, and $abc=4Rrp$, and from $p=\frac{1}{2}$, your inequality is equiv. with $r>0$.

[quote=April]We have:
\[\left \{\begin{array}{l}a+b>c,\,b+c>a,\,c+a>b\\ a+b+c=1 \end{array}\right \| \Longrightarrow 0<a,\,b,\,c<\frac{1}{2}\\\]
\[ \Longrightarrow (1-2a)(1-2b)(1-2c)>0\\ \Longrightarrow 1>2(a+b+c)-4(ab+bc+ca)+8abc\\ \]
\[\Longrightarrow\frac{1}{2}>(a+b+c)^{2}-2(ab+bc+ca)+4abc\quad\\ \Longrightarrow\boxed{\frac{1}{2}>a^{2}+b^{2}+c^{2}+4abc}\]
[b]Remark.[/b] [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=752820&sid=fd917c1b74690d9582fb6717ea2fd872#752820]Similary problem[/url]  :D[/quote][quote=Otávio_Menezes]If $a,b,c$ are sides of a triangle, with $a+b+c=1$, prove that

$$a^{2}+b^{2}+c^{2}+4abc<1/2$$[/quote]
$$\iff$$
$$ab+bc+ca-2abc>\frac{1}{4}$$
[url=https://artofproblemsolving.com/community/c6h444973p2505024]Uzbekistan 2005:[/url]
Let $ABC$ be a triangle and $a+b+c=2$. Prove that $$abc+1<ab+bc+ca<abc+ \frac{28}{27}$$
[url=https://artofproblemsolving.com/community/c6h591327p3503481]here[/url] [url=https://artofproblemsolving.com/community/c6h132992p752820]here[/url] [url=https://artofproblemsolving.com/community/c6h482348p2702151]here[/url]


An algebrical solution:

$ \frac{a}{b+c}+\frac{r}{R}\geq 1 \Longleftrightarrow \frac{a}{b+c}+\frac{4S^{2}}{pabc}\geq 1 \Longleftrightarrow \frac{a}{b+c}+\frac{4(p-a)(p-b)(p-c)}{abc}\geq 1$.

Now by plugging $ a=x+y$, $ b=y+z$, $ c=z+x$, $ x$, $ y$, $ z$ becoming positive reals, where the condition of $ a = \max\left\{a,b,c\right\}$, goes into $ z=\min\left\{x,y,z\right\}$. $ (1)$

Therefore, the inequality rewrites as $ \frac{x+y}{2z+x+y}+\frac{4xyz}{(x+y)(y+z)(z+x)}\geq 1$

Which is equivalent with $ 4xyz(2z+x+y)+(x+y)^{2}(y+z)(z+x) \geq (x+y)(y+z)(z+x)(2z+x+y)$

$ \iff 2xy(2z+x+y) \geq (x+y)(y+z)(z+x)$

$ \iff 2xyz+xy(x+y) \geq yz(y+z)+xz(x+z)$, which is true, because $ 2xyz \geq z^{2}(x+y)$, due to $ (1)$, and $ xy(x+y) \geq z(x^{2}+y^{2})$ as well.

[quote="Virgil Nicula"]$ \left\{\begin{array}{c}\triangle\ ABC\\\ a=\max\{a,b,c\}\end{array}\right\|\implies \boxed{\frac{a}{b+c}+\frac{r}{R}\ge 1}$ (standard notations).
[b]A nice April's remark.[/b] $ \frac{a}{b+c}+\frac{r}{R}\ge 1$ $ \Longleftrightarrow$ $ \cos B+\cos C\ge 1$ $ \Longleftrightarrow$ $ \frac{1}{b}+\frac{1}{c}\ge \frac{1}{a-b+c}+\frac{1}{a+b-c}$.[/quote]
[b]Proof.[/b] $ \left\{\begin{array}{c}\sum a^{2}(p-b)(p-c)=4p^{2}r(R-r)\\\\ \sum a(p-b)(p-c)=2pr(2R-r)\end{array}\right\|$ $ \implies$ $ \frac{\sum a(p-b)(p-c)\cdot a}{\sum a(p-b)(p-c)}=2p\cdot\frac{R-r}{2R-r}$ $ \implies$

$ 2p\cdot \frac{R-r}{2R-r}\le a$ $ \implies$ $ \frac{2R-r}{R-r}\ge 1+\frac{b+c}{a}$ $ \implies$ $ \frac{R}{R-r}\ge \frac{b+c}{a}$ $ \implies$ $ \frac{R-r}{R}\le\frac{a}{b+c}$ $ \implies$ $ 1\le\frac{a}{b+c}+\frac{r}{R}$.

[quote="Virgil Nicula"][color=darkred]$ \{\begin{array}{c}\triangle\ ABC\\\ a=\max\{a,b,c\}\end{array}\|\implies \boxed{\frac{a}{b+c}+\frac{r}{R}\ge 1}$ (standard notations).
[b][u]A nice April's remark.[/u][/b] $ \frac{a}{b+c}+\frac{r}{R}\ge 1$ $ \Longleftrightarrow$ $ \cos B+\cos C\ge 1$ $ \Longleftrightarrow$ $ \frac{1}{b}+\frac{1}{c}\ge \frac{1}{a-b+c}+\frac{1}{a+b-c}$. [b]Indeed[/b],
$ \frac{a}{b+c}+\frac{r}{R}\ge 1$ $ \Longleftrightarrow$ $ \frac{r}{R}\ge \frac{2(p-a)}{b+c}$ $ \Longleftrightarrow$ $ \frac{r}{p-a}\ge \frac{2R}{b+c}$ $ \Longleftrightarrow$ $ \tan\frac{A}{2}\ge \frac{1}{\sin B+\sin C}$ $ \Longleftrightarrow$
$ 2\cos\frac{A}{2}\cos\frac{B-C}{2}\cdot\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}}\ge 1$ $ \Longleftrightarrow$ $ 2\cos\frac{B+C}{2}\cos\frac{B-C}{2}\ge 1$ $ \Longleftrightarrow$ $ \cos B+\cos C\ge 1$.[/color][/quote]
[color=darkblue][b][u]Proof I.[/u][/b] $ \cos B+\cos C\ge 1$ $ \Longleftrightarrow$ $ \frac{a^{2}+c^{2}-b^{2}}{2ac}+\frac{a^{2}+b^{2}-c^{2}}{2ab}\ge 1$ $ \Longleftrightarrow$ $ b(a^{2}+c^{2}-b^{2})+c(a^{2}+b^{2}-c^{2})-2abc\ge 0$ $ \Longleftrightarrow$

$ (a-b)(a-c)(b+c)+b^{2}(a-b)+c^{2}(a-c)\ge 0$, what is truly because $ a\ge b$ and $ a\ge c$.

[b][u]Proof II.[/u][/b] $ \{\begin{array}{c}\sum a^{2}(p-b)(p-c)=4p^{2}r(R-r)\\\\ \sum a(p-b)(p-c)=2pr(2R-r)\end{array}\|$ $ \implies$ $ \frac{\sum a(p-b)(p-c)\cdot a}{\sum a(p-b)(p-c)}=2p\cdot\frac{R-r}{2R-r}$ $ \implies$

$ 2p\cdot \frac{R-r}{2R-r}\le a$ $ \implies$ $ \frac{2R-r}{R-r}\ge 1+\frac{b+c}{a}$ $ \implies$ $ \frac{R}{R-r}\ge \frac{b+c}{a}$ $ \implies$ $ \frac{R-r}{R}\le\frac{a}{b+c}$ $ \implies$ $ 1\le\frac{a}{b+c}+\frac{r}{R}$.

[b][u]Proof III[/u] (April's).[/b] $ \cos B+\cos C=$ $ \frac{a^{2}+c^{2}-b^{2}}{2ac}+\frac{a^{2}+b^{2}-c^{2}}{2ab}\ge$ $ \frac{a^{2}+c^{2}-b^{2}}{a^{2}+c^{2}}+\frac{a^{2}+b^{2}-c^{2}}{a^{2}+b^{2}}$ because $ \{\begin{array}{c}2ac\le a^{2}+c^{2}\\\ 2ab\le a^{2}+b^{2}\end{array}$.

Therefore, $ \cos B+\cos C\ge 2-(\frac{b^{2}}{a^{2}+c^{2}}+\frac{c^{2}}{a^{2}+b^{2}})\ge$ $ 2-(\frac{b^{2}}{b^{2}+c^{2}}+\frac{c^{2}}{c^{2}+b^{2}})=1$ because $ a\ge b$ and $ a\ge c$.

[b][u]Proof IV[/u] (Pohoatza's).[/b] $ \frac{a}{b+c}+\frac{r}{R}\ge 1$ $ \Longleftrightarrow$ $ \frac{a}{b+c}+\frac{4pr^{2}}{4Rpr}\ge 1$ $ \Longleftrightarrow$ $ \frac{a}{b+c}+\frac{4(p-a)(p-b)(p-c)}{abc}\ge 1\ \ (1)$.

Now I"ll use the classic substitutions $ \{\begin{array}{c}a=y+z\\\ b=z+x\\\ c=x+y\end{array}$, where $ \{x,y,z\}\subset (0,\infty )$, $ \{\begin{array}{c}x=p-a\\\ y=p-b\\\ c=p-c\end{array}$ and $ x=\min \{x,y,z\}$.

Therefore, the relation $ (1)$ becomes $ \frac{y+z}{2x+y+z}+\frac{4xyz}{(y+z)(z+x)(x+y)}\ge 1$ $ \Longleftrightarrow$ $ \frac{4xyz}{(y+z)(z+x)(x+y)}\ge \frac{2x}{2x+y+z}$ $ \Longleftrightarrow$

$ 2yz(2x+y+z)\ge (y+z)(z+x)(x+y)$ $ \Longleftrightarrow$ $ 2yz(x+y)+2yz(x+z)\ge y(x+y)(x+z)+z(x+y)(x+z)$ $ \Longleftrightarrow$

$ y(x+y)[2z-(x+z)]+z(x+z)[2y-(x+y)]\ge 0$ $ \Longleftrightarrow$ $ y(x+y)(z-x)+z(x+z)(y-x)\ge 0$, what is truly because $ z\ge x$ and $ y\ge x$.

[b][u]Question[/u] (for April, Pohoatza & Co).[/b] Is truly $ \cos B+\cos C\ge 1\Longrightarrow a=\max \{a,b,c\}$ ?!.[/color]

[size=167][color=darkred][b][u]Question[/u] (for April, Pohoatza & Co).[/b] Is truly $ \boxed{\cos B+\cos C\ge 1\Longrightarrow a=\max \{a,b,c\}}$ ?!.[/color][/size]

May I welcome :D
The answer is false see the figure.

[color=darkblue]Yes, Gemath, the answer is [b]NO[/b] !

Generally, for $ 0<c-b<a<\frac{bc+\sqrt{b^{4}-b^{2}c^{2}+c^{4}}}{b+c}<c$ we have $ \frac{a}{b+c}+\frac{r}{R}\ge 1$ and $ a<\max \{a,b,c\}$.  Example :  $ \{\begin{array}{c}a=192\\\ b=100\\\ c=200\end{array}$.[/color]


Let (I) be an incircle of triangle ABC. (I) touch BC,CA,AB at M,N and P. Let O be a point lie in triangle MNP.MO,NO,PO intersect (I) at $ A_1,B_1,C_1$.Prove that $ AA_1,BB_1,CC_1$ are concurrent.

Dear Mathlinkers,
AA1, BB1, CC1 are concurrent at the Steinbart's point...
See for example,
http://perso.orange.fr/jl.ayme   vol. 3  les points de Steinbart et de Rabinowitz...
Sincerely
Jean-Louis

Can you post your solution for this problem?
Pr.2:
$ MO,NO,PO$ cut $ NP,MP,MN$ at $ A_1,A_2,A_3$.Prove that $ AA_1,AA_2,AA_3$ are concurrent :D

Dear Mathlinkers,
for your new problem applies the"cevian nests theorem" that you can find a proof on
http://perso.orange.fr/jl.ayme   vol. 3
Sincerely
Jean-Louis


Given $ a, b, c \geq\ 0$ satisfy $ a\plus{}b\plus{}c\equal{}3.$
Prove that: $ \sqrt[]{a} \plus{} \sqrt[]{b} \plus{} \sqrt[]{c} \plus{} \frac{6}{ab\plus{}bc\plus{}ca} \geq\ 5$
 :)

[quote="nguoivn"]Given $ a, b, c \geq\ 0$ satisfy $ a \plus{} b \plus{} c \equal{} 3.$
Prove that: $ \sqrt []{a} \plus{} \sqrt []{b} \plus{} \sqrt []{c} \plus{} \frac {6}{ab \plus{} bc \plus{} ca} \geq\ 5$
 :)[/quote]
The stronger inequality still holds
\[ 1 \plus{}\frac{6}{ab\plus{}bc\plus{}ca} \ge 2\left( \frac{1}{a\plus{}1}\plus{}\frac{1}{b\plus{}1}\plus{}\frac{1}{c\plus{}1}\right).\]
;) :)

[quote="can_hang2007"]The stronger inequality still holds
\[ 1 \plus{} \frac {6}{ab \plus{} bc \plus{} ca} \ge 2\left( \frac {1}{a \plus{} 1} \plus{} \frac {1}{b \plus{} 1} \plus{} \frac {1}{c \plus{} 1}\right).
\]
;) :)[/quote]
I agree with you. Your inequality is true. But I don't see a nice proof for this one.


Let $a\in \mathbb{N}_{>1}$ show that $a^{k}+1$ is never prime for an odd exponent $k>1$.

it's easy to prove that $a^{k}+1$ is prime => $k$ is power of $2$
so, $a^{k}+1$ is never prime if $k=ab$ with $a$ odd $>1$ in particular for b=1 and $a$ odd


function
real analysis
real analysis unsolved

a) Give an example of a bijection from $\mathbb{R}$ to $\mathbb{R^{*}}.$
 b) Prove that such a function can't be continous on $\mathbb{R}.$

a continuous function maps a connex set into a connex set, isn't it .

Another one : find  an injective continous function from a circle to a line !
:cool:

[quote="ZetaX"][quote="zhaobin"][quote="tµtµ"]a)

I think this works :

$f(x) = x$ if $x \notin \mathbb{N}^^{\circ}$ = ($0, 1, ...$)

$f(n) = 1+n$ if $x \in \mathbb{N}^^{\circ}$

b)

$\mathbb{R}$ is connected (= connex from Alekk probably ?) but not $\mathbb{R}^*$[/quote]
the problem does not means $\left[0,+\infty \right)$ to $\mathbb{R}$ :P[/quote]

Where did he say that it's just from $[0, \infty[$ to $\mathbb{R}$¿[/quote]
I mean that the example tµtµ give doen't meet the problem.
the example he gives is only from $\left[0,+\infty \right)$ to $\mathbb{R}^*$
am I wrong?

[quote="zhaobin"]
I mean that the example tµtµ give doen't meet the problem.
the example he gives is only from $\left[0,+\infty \right)$ to $\mathbb{R}^*$[/quote]
I don't think so. "my" function maps (at least tries to) [b]every[/b] numbers to the same number except for $\geq 0$ integer which are shifted to the right.

can I ask a stupid question.
is $\mathbb{R}^*$ means $\mathbb{R}-\{ 0 \}$?


geometry
circumcircle
3D geometry
tetrahedron
prism
rhombus

Given four points $ A_1, A_2, A_3, A_4$ in the plane, no three collinear, such that
\[ A_1A_2 \cdot A_3 A_4 \equal{} A_1 A_3 \cdot A_2 A_4 \equal{} A_1 A_4 \cdot A_2 A_3,
\]
denote by $ O_i$ the circumcenter of $ \triangle A_j A_k A_l$ with $ \{i,j,k,l\} \equal{} \{1,2,3,4\}.$ Assuming $ \forall i A_i \neq O_i ,$ prove that the four lines $ A_iO_i$ are concurrent or parallel.

[i]Nikolai Ivanov Beluhov, Bulgaria[/i]

[hide="hint"]Inverse with center $ A_i$ we get $ A_jA_kA_l$ becomes a equalateral triangle. Now the calculation of angle becomes quite easy thus the application of Ceva is possible.[/hide]

The approach that comes to mind is that if you have a tetrahedron in space and inscribe it in a parallelpiped, then those conditions mean that it is actually inscribed in a rectangular prism. Then you would take the limit as height of this prism went to zero. Possibly it would be easier to prove in 3 space (but maybe it is false then...).

My approach is different than the rchlch's  :oops:

It is enough to prove that $ O_1A_1, O_2A_2, O_3A_3$ are concurrent, because of the simetry.
Now, the condition $ A_1A_2.A_3A_4 \equal{} A_1A_4.A_2A_3$ is equivalent to $ \frac {A_1A_2}{A_2A_3} \equal{} \frac {A_1A_4}{A_4A_3}$, from that, the angle bisectors of $ < A_1A_2A_3$, $ < A_1A_4A_3$ and the line $ A_1A_3$ are concurrent, let's say in $ T$.

Now let's consider $ S$ the center of the Apolonius circle $ \omega$ of triangle $ A_1A_2A_3$ respect to $ A_2$.Therefore $ B$ is in $ \omega$ iff $ \frac {A_1B}{BA_3} \equal{} \frac {A_1A_2}{A_2A_3}$, then it is obvious that $ A_2$ and $ A_4$ belong to $ \omega$.

The point $ S$ is the center of the Apolonius circle of the triangle $ A_1A_2A_3$ and $ A_1A_4A_3$, thus, $ S$ belongs to the bisectors of the segments $ A_2S$ and $ A_4S$, i.e. $ S$ is the circumcircle of the triangle $ A_2A_4T$, then $ S$ belongs to the bisector of $ A_2A_4$ as well. But because $ O_1$ and $ O_3$ are the circumcenters of $ A_2A_4A_3$ and $ A_2A_4A_1$, then $ O_1O_3$ is perpendicular to $ A_2A_4$. That means that the intersection of $ O_1O_3$ and $ A_1A_3$ is a center of the Apolonius circle of the triangle $ A_1A_2A_3$.

Therefore, the other two intersections between: $ O_1O_2,A_1A_2$ and $ O_2O_3,A_2A_3$ are also center of the Apolonius circles of the triangle $ A_1A_2A_3$. It is well known that these three centers are collinear, then by the Desargues Theorem, the result follows.

[quote="orl"]Given four points $ A_1, A_2, A_3, A_4$ in the plane, no three collinear, such that
\[ A_1A_2 \cdot A_3 A_4 \equal{} A_1 A_3 \cdot A_2 A_4 \equal{} A_1 A_4 \cdot A_2 A_3,
\]
denote by $ O_i$ the circumcenter of $ \triangle A_j A_k A_l$ with $ \{i,j,k,l\} \equal{} \{1,2,3,4\}.$ Assuming $ \forall i A_i \neq O_i ,$ prove that the four lines $ A_iO_i$ are concurrent or parallel.[/quote]
Or equivalently, prove that the isodynamic points lie on the [url=http://pagesperso-orange.fr/bernard.gibert/Exemples/k001.html]Neuberg cubic[/url]  :wink:.

I forgot that there're "no three collinear", so a found a configuration that satisfys the problem condicions :wink:, the bissector case...

I cannot but recommend to read the solution(s) in the RMC2009 (Romanian Mathematical Competitions 2009), using Apollonius circles followed by Desargues theorem.

Can it be something but a triangle with a point inside? Did someone proved that it can be a quadrilateral?

Yes, it could easily be some kite (even the rhombus) with the equal angles equal to $ \pi/6$.

I thought that I had concluded the problem... I think I've made a mistake...
Well, is the triangle with his incenter a valid case?

I have a slightly different approach. Denote by $\omega_{ij}$ the Apollonius circle of points $A_k$ and $A_l$ and ratio $\dfrac{A_iA_k}{A_iA_l}=\dfrac{A_jA_k}{A_jA_l}$, which passes through $A_i$ and $A_j$. It is somewhat well-known that $\omega_{ij}$, $\omega_{il}$ and $\omega_{ik}$ are coaxal and orthogonal to $\odot{A_jA_lA_k}$, so that the radical axis $r_i$ of these three circles passes through $O_i$. Hence $r_i$ is actually the line $A_iO_i$. So $A_iO_i$, $A_jO_j$ and $A_kO_k$ are respectively the radical axes of $\omega_{ij}$ and $\omega_{ik}$, $\omega_{ij}$ and $\omega_{jk}$, $\omega_{ik}$ and $\omega_{jk}$, and therefore concur. It is enough to prove that three of them concur due to symmetry.

Let $ A_1=A, A_2=B, A_3=C, A_4=P, O_1=O_A, O_2=O_B, O_3=O_C$, it suffices to prove $AO_A, BO_B, CO_C$ concur. Notice that $ \angle BPC = \angle BAC + 60$. This is done by taking APA'A'' a cyclic quadrilateral with A'' and A' foots of ange bisectors, and the analogous cyclic quads. And then finish with Trigonometric Ceva

My solution:

[b]Lemma[/b] (well-known) [b]:[/b]

Let $ P, Q $ be the isogonal conjugate of $ \triangle ABC $ .
Let $ O_P, O_Q $ be the circumcenter of $ \triangle BPC, \triangle BQC $, respectively .

Then $ AO_P, AO_Q $ are isogonal conjugate of $ \angle BAC $ .
____________________________________________________________
[b]Back to the main problem:[/b]

WLOG $ A_4 $ is in $ \triangle A_1A_2A_3 $ .

Let $ \triangle N_1N_2N_3 $ be the outer Napoleon triangle of $ \triangle A_1A_2A_3 $ .
Let $ F, N $ be the 1st Fermat point, 1st Napoleon point of $ \triangle A_1A_2A_3 $, respectively

Since $ A_4 $ is the 1st Isodynamic point of $ \triangle A_1A_2A_3 $ ,
so $ F $ is the isogonal conjugate of $ A_4 $ WRT $ \triangle A_1A_2A_3 $ .
From lemma $ \Longrightarrow A_1O_1, A_2O_2, A_3O_3 $ are concurrent at the isogonal conjugate of $ N $ WRT $ \triangle A_1A_2A_3 $ .

Similarly, we can prove $ A_2O_2, A_3O_3, A_4O_4 $ are concurrent $ \Longrightarrow A_1O_1, A_2O_2, A_3O_3, A_4O_4 $ are concurrent .

Q.E.D

Assume wlog $A_1$ lies inside triangle $A_2A_3A_4$, it is clear that it is the first isodynamic point of this triangle. Clearly, triangle $O_2O_3O_4$ is orthologic with triangle $A_2A_3A_4$ and the centers of orthology are obviously, $O_1$ and $A_1$. Thus, showing that $A_iO_i$ for $i=2,3,4$ concur automatically establishes that the concurrency point lies on $A_1O_1$ by the Sondat's Theorem.

For the concurrency, use the "Distance Ceva" as follows: Let $\angle A_i=\theta_i$ for $i=2,3,4$. Then, we want \begin{align*} \Pi_{i =2,3,4} \left(\frac{d(O_i,A_iA_{i+1})}{d(O_i,A_iA_{i+2})}\right)=1  \end{align*} where $A_5=A_2$ and $A_6=A_3$. This holds since the above mentioned product in fact equals the quantity \begin{align*} \Pi_{i=2,3,4} \left(\frac{\sin (\theta_{i+2}-60^{\circ})}{\sin (\theta_{i+1}+60^{\circ})} \right) \end{align*} which is obviously equal to $1$. The result follows.

Note that $\frac{A_2A_4}{A_3A_4}=\frac{A_1A_2}{A_1A_3}$, so $A_4$ lies on the $A_1$-Appolonius circle of $A_1A_2A_3$. Similarly, it lies on the other two Appolonius circles as well. It is well-known that the Appolonius circles of a triangle are coaxial and that its circumcenter lies on its common radical axis. Hence, the four lines we want to prove the concurrency of are the radical axes of the Appolonius circles of the four points taken three at a time. Call these radical axes $\ell_1,\ell_2,\ell_3,\ell_4$, where $\ell_i=A_iO_i$. Call the Apollonius circles $C_{12},C_{13},C_{14},C_{23},C_{24},C_{34}$, where $C_{ij}$ is the Appolonius circle of $A_i$ and $A_j$ that passes through the other two points. Let $\ell_1$ intersect $\ell_2$ at $P$. Then, $p(P,C_{23})=p(P,C_{34})=p(P,C_{24})$ and $p(P,C_{13})=p(P,C_{34})=p(P,C_{14})$, so $p(P,C_{24})=p(P,C_{34})=p(P,C_{14})$ and $p(P,C_{23})=p(P,C_{34})=p(P,C_{13})$, which respectively mean that $P$ lies on $\ell_3$ and $\ell_4$, as desired.

Note that $\frac{A_2A_4}{A_3A_4}=\frac{A_1A_2}{A_1A_3}$, so $A_4$ lies on the $A_1$-Apollonius circle of $A_1A_2A_3$. Similarly, it lies on the other two Apollonius circles as well.
Also since the Apollonius circles of a triangle are coaxial and are orthogonal to the circumcircle, so the radical axis passes through the circumcenter. By radical axis theorem on these Apollonius circles twice, we get that $O_1A_1, O_2A_2, O_3A_3$ are concurrent. Now by similarly repeating this argument or by Sondat's theorem on $O_1O_2O_3$ and $A_1A_2A_3$ we get our result.

EDIT: realised that this is almost the same as the previous solution.

It is enough to prove lines $A_1O_1$, $A_2O_2$, $A_3O_3$ are concurrent or parallel. Rename the points to get the following statement:
[quote]
Let $P$ be a point in the plane of $\triangle ABC$ (not lying on any sideline of $\triangle ABC$) such that
\[AP \cdot BC = BP \cdot CA = CP \cdot AB.\]
Let $O_A$ denote the circumcenter of $\triangle BPC$; define $O_B$ and $O_C$ similarly. Prove that lines $AO_A$, $BO_B$, $CO_C$ are concurrent.
[/quote]

Let $A'$ denote the inverse of $A$ in $(BPC)$; define $B'$ and $C'$ similarly.

[b]Claim.[/b] $BCB'C'$ is cyclic.
[i]Proof.[/i] Note $\tfrac{B'A}{B'P} = \tfrac{BA}{BP} = \tfrac{CA}{CP} = \tfrac{C'A}{C'P}$ and so the four points lie on an Appolonius circle of $\overline{AP}$. $\square$

Likewise $CAC'A'$ and $ABA'B'$ are cyclic. It is easy to see that $A$, $B$, $C$, $A'$, $B'$, $C'$ are not all concyclic (note $\measuredangle BA'C = \measuredangle BAC \pm 120^{\circ}$), so lines $AA'$, $BB'$, $CC'$ are concurrent or parallel; i.e. lines $AO_A$, $BO_B$, $CO_C$ are concurrent or parallel.

I am unable to find a possible configuration for the parallel case. Does anyone elaborate on it? :oops:

It suffices to prove that $A_2O_2$, $A_3O_3$ and $A_4O_4$ are concurrent. Let $B_i$ be the image of $A_i$ under the inversion with center $A_1$ and radius $1$ for $i\in\{2,3,4\}$. Since $B_2B_3=\frac{A_2A_3}{A_1A_2\cdot A_1A_3}$, $B_2B_3B_4$ is equilateral. Under our inversion, $O_2$ goes to the point symmetric to $A_1$ wrt $B_3B_4$, so $A_2O_2$ goes to the circle through $A_1$ and $B_2$ with center on $B_2B_3$. It suffices to prove that three centers of such circles are collinear. If $P_2$, $P_3$, $P_4$ are circumcenters of $A_1B_3B_4$, $A_1B_2B_4$ and $A_1B_2B_3$, this is equivalent to $P_2P_3P_4$ being perspective to $B_2B_3B_4$. However, $B_iP_i$ concur at the center of $B_2B_3B_4$, so we're done by Desargues's theorem.


calculus
derivative
function
parameterization
calculus computations

Hi people!!

I don´t know if this question is a idiot question, but... I'm newbie here and i didn't find the aswer in wikipediwe neither in google  :oops:

i have  that $\frac{dx}{dy}= \frac{1}{ \frac{dy}{dx}}$

the inverse derivative. but, If x is dependent of 2 variables? is possible to find the inverse derivative for a funtion with many variables?

Sorry if this is a easy question. and thanks

It's an issue of definitions.

Your statement, in more precise language:
Suppose $Y(x)$ is a differentiable function and $X(y)$ is a differentiable inverse: $X(Y(x))=x$ for all $x$ in some interval $I$, and $Y(X(y))=y$ for all $y$ in $I'=Y(I)$.
Then $X'(Y(x))=\frac1{Y'(x)}$; this is really just the chain rule. There is also the inverse function theorem, which states that a differentiable function with nonzero (invertible) derivative has a differentiable inverse on some interval.

The problem with trying to extend this to more variables is that we don't have straightforward inverse functions. If I have three functions $X(y,z),Y(x,z),Z(x,y)$, what's the relation among them? If we have something like $X(Y(x,z),z)=x$ and similar relations for the other choices, we've essentially duplicated the 1-variable picture, with an extra parameter. If that extra parameter is held constant, the two remaining functions are inverses of each other. That leads to partial derivatives, of course.

Try looking up the Implicit Function Theorem and the Inverse Function Theorem.


advanced fields
advanced fields unsolved

an automorphism on a group takes more than 3/4 of it's elements to their inverses.
prove that it takes all the elements to their inverse.


For given n in N, Find all polynomials p(x) s.t. :

p(x)^3 = p(x^3 ) + 3 x^n p(x)

well for odd values of n it's simple: there are none (as we'll see n=2m)

elseway say P(x) = ax^m+bx^(m-1)+...+cx+d

then a^3 . x^3m + ... = a. x^3m+ 3. x^n. (ax^m+...)
so n=2m follows, as this is necessary to get equalness for the highest grade

next thing we see is that 4.a = a^m, thus that a^(m-1)=4
same for d³=d thus d=1 or d=0
I also found that x^(n-2m) = a², and a lot more stuff

The options are endless here and I'm pretty sure I'll get there, but it's waaaay to long for being a good solution.

Do you have a better (read: shorter) one?

Well, I remember that there are several p(x)s such that meat condition whether n is odd or not.

Example : n=5, p(x)=x^5 + 1

LHS = x^15 + 3x^10 + 3x^5 + 1

RHS = x^15 + 1 + 3x^5(x^5 + 1)=LHS

If p is non zero polynomial
(*) p(x)^3 = p(x^3 ) + 3 x^n p(x) => p(x) divides p(x^3)
p(x)=x^k+1 satisfy p(x)|p(x^3) Is there other solution ?

p(x)=x^n+1 verify (*)

In fact there are many p(x) for natural number n

p(x) can be x^n +1, x^(n-1) +x, x^(n-2) +x^2, ...

And I believe there are no more solutions without them.. :)


According to my text book the first step to the opening of stoma is:

"A proton pump actively (using ATP) pumps $ \ce{H\plus{}}$ out of the guard cell"

The purpose is to decrease $ [\ce{H\plus{}}]$ and negative charges build up inside the cell.

How does this step happen? As in in there some sort of enzyme catalyzed process (like a control system or something) that determines when hydrogen ions are pumped out (after all guard cells don't think for themselves I don't think  :P )?

You can find some nice animations about this online; this is honestly not a hard question to answer on on your own.  See, for example, http://www.maxanim.com/biochemistry/Proton%20Pump/Proton%20Pump.htm or http://highered.mcgraw-hill.com/olc/dl/120068/bio05.swf .

Thanks for the animations  :)  but that's not what I am asking. I am not asking how the proton pump works. I am asking what triggers the proton pump to work. After all the stoma only opens under certain conditions, they are not opening all the time.

K+ efflux triggers H+ influx--that's levitt's theory i suppose

We just did this chapter in biology last week  :) .

ATP synthase pumps protons ($ H^\plus{}$ ions) out of the guard cells, while potassium and chloride ions go in. This creates an electrochemical gradient that facilitates the movement of water into the guard cells. This causes the guard cells to be turgid, and conform it's shape into an opening.


[quote="ye109"][quote="Ji Chen"]If $ x,y,z$ are nonnegative numbers such that $ x \plus{} y \plus{} z\leq 3$, then

$ 8(3 \minus{} x)^2(3 \minus{} y)^2(3 \minus{} z)^2\geq xyz(9 \minus{} x)(9 \minus{} y)(9 \minus{} z).$[/quote]
let $ x \plus{} y \plus{} z \equal{} p,xy \plus{} yz \plus{} zx \equal{} \frac {p^2 \minus{} q^2}{3},xyz \equal{} r$

it's equal to $ f(p,q,r) \equal{} 8(27 \minus{} 9p \plus{} p^2 \minus{} q^2 \minus{} r) \minus{} r(729 \minus{} 81p \plus{} 3p^2 \minus{} 3q^2 \minus{} r)$

[/quote]
i got a edit mistake of $ f(p,q,r)$ ,the right  answer is
$ f(p,q,r) \equal{} 8(27 \minus{} 9p \plus{} p^2 \minus{} q^2 \minus{} r)^2 \minus{} r(729 \minus{} 81p \plus{} 3p^2 \minus{} 3q^2 \minus{} r)$


logarithms
algebra
function
domain
limit
real analysis
real analysis unsolved

How many real roots does the following equation have:

$log_x{(2x+1)} = lg(x^2 +2 )$?

hello, we want to solve the equation $ \log_x(2x+1)=\lg(x^2+2)$ over the domain $ x \in \mathbb{R}$ and $ x>0$ and $ x\neq1$. We write the given equation in the form $ \frac{\ln(2x+1)}{\ln(x)}-\frac{\ln(x^2+2)}{\ln(10)}=0$ and define $ f(x)=\frac{\ln(2x+1)}{\ln(x)}-\frac{\ln(x^2+2)}{\ln(10)}$ It is clear that$ \lim_{x \to 1+0}f(x)=\+\infty$ and $ \lim_{x \to \infty}f(x)=-\infty$.
For $ 0<x<1$ is $ f(x)<0$ and the equation $ f(x)=0$has no solution in this case.
$ f^'(x)=\frac{2}{(2x+1)\ln(x)}-\frac{\ln(2x+1)}{\ln(x)^2x}-\frac{2x}{(x^2+2)\ln(10)}$ For $ x>1$ is $ -\frac{2x}{(x^2+2)\ln(10)}<0$ and
$ \frac{2}{(2x+1)\ln(x)}<\frac{\ln(2x+1)}{\ln(x)^2x} \Leftrightarrow$
$ 2x\ln(x)<(2x+1)\ln(2x+1) \Leftrightarrow 2x(\ln(x)-\ln(2x+1))<\ln(2x+1)$
$ \Leftrightarrow 2x\ln\left(\frac{x}{2x+1}\right)<\ln(2x+1)$. For $ x>1$ is
$ \ln(2x+1)>0$ and $ \ln\left(\frac{x}{2x+1}\right)<0 \Leftrightarrow \frac{x}{2x+1}<1 \Leftrightarrow$ $ 0<x+1$, thus the function $ f(x)$ is for $ x>1$
strictly monotonously decreasing and hase one intersection point with the $ x$-axis and our equation has exactly one solution, it is $ x \approx 5.2597318360822073148591048819021860$.
Sonnhard.


How do I let directory listing for a directory on a webserver that doesn't allow directory listing?

Like on http://mathprob.com/code/ (MathProb, my website), it gives me a forbidden message. I want it to list the contents that are there by using HTACCESS? How would I do that?

Also, when there is a 404, it redirects to my webhost's page. How do I make it go to error.php, which says a 404 error?

Thanks.
#H34N1

You'd be best asking on a specialized forum.

To allow directory listings in a folder, edit that folder's .htaccess file to include the line:

[code]
Options +Indexes
[/code]

For custom error documents, in your root .htaccess file, add:

[code]
ErrorDocument404 /path/to/error_page.php
[/code]


[img]http://i2.tinypic.com/7weq9fr.jpg[/img]
The plane area shown in the figure consists of an isosceles trapezoid (non-parallel sides equal) and a segment of a circle. If the non-parallel sides are tangent to the segment at points A and B, find the area of the composite figure.

Someone helped me with a good answer, so I thought I would respond in kind.
Refer to attached PDF.  I didn't look for a specific formula for the isos. trapizoid.
I broke down the complex shape into smaller shapes that have mensuration formula available.

I elected to use triangles and sectors.

To construct the sector:
1) Draw perpendiculars to the tangents of the circle at both A and B. Tangent implies 90 degree angles.  Also, they will intersect at the center.
2) Use "AIA" alternate interior angles to find the central angle of the sector  --> central angle = 60 degrees  = 30 + 30.
3) using the chord ----  3 = 2*R*sin 30 -->  R=3.0
4) Area of sector is theta * r^2 =  9.4247 in this problem. Theta = 60 degrees.

To construct the triangle
1)  The 2 radius and the 3" chord create an equilateral triangle -->  Area = 7.7942

Subtract the triangle area from the sector's total area to arrive at the area between the 3" chord and the circle segment.  -->  1.6305

At this point, you're home free...

The "middle ordinate" of the sector is M = R * (1- cos 30) --> 0.4019.  The middle ordinate is the vertical line between the 3" chord and the circle segment.

The distance between the 3" chord and the "longest" side is 5" minus 0.4019 -->  4.598

Hope you can finish from here.

My solution is somewhat sketchy so feel free to ask if there is doubt :)

[b]Strategy[/b]: Divide the whole figure into two areas, as suggested by your drawing and vector2's post.

[b]Area of segment[/b]: $ A_\text{segment}\equal{}A_\text{sector}\minus{}A_\triangle$

So here we start off by drawing perpendicular lines at $ A$ and $ B$ and these perpendicular lines intersect at the point $ O$. Since the nonparallel equal sides of the isosceles trapezoid are tangent to the circle containing the points of tangency $ A$ and $ B$, it follows that point $ O$ is the center of the circle with radii $ OA\equal{}OB$. Note the 60-degree angle shown. Taking complements and using right angles, $ \angle OAB\equal{}\angle OBA\equal{}60^\circ$. So $ \triangle AOB$ is equilateral.

$ A_\text{sector}\equal{}\frac16\pi r^2\equal{}\frac32\pi$

$ A_\triangle\equal{}\frac{r^2}4\sqrt{3}\equal{}\frac94\sqrt3$

$ A_\text{segment}\equal{}\frac32\pi\minus{}\frac94\sqrt3$

[b]Area of isosceles trapezoid[/b]: $ A_\text{trapezoid}\equal{}\frac12(b_\text{small}\plus{}b_\text{large})h$

So far we only have $ b_\text{small}\equal{}3$.

We find the height of the trapezoid first. The height of the whole figure is 5 and is greater than the radius of circle, which is 3, so the center $ O$ is inside the figure. So the distance from the center to base $ b_\text{large}$ is 2. Now the perpendicular distance from the center to the chord $ AB$ is the height of the equilateral triangle and that is equal to $ \frac32\sqrt3$. Thus, $ h\equal{}2\plus{}\frac32\sqrt3$.

Now, we look at $ b_\text{large}$. It's composed of two legs of two congruent 30-60-90 triangles and the rectangle formed with chord $ AB$ (bases of a trapezoid are parallel). Each leg of the triangle on $ b_\text{large}$ is equal to $ h\sqrt3$. Hence, $ b_\text{large}\equal{}b_\text{small}\plus{}2h\sqrt3\equal{}3\plus{}2\left(2\plus{}\frac32\sqrt3\right)\sqrt3\equal{}12\plus{}4\sqrt3$

$ A_\text{trapezoid}\equal{}\frac12(b_\text{small}\plus{}b_\text{large})h\equal{}24\plus{}\frac{61}4\sqrt3$

[b]Total area[/b]: $ A_\text{total}\equal{}A_\text{segment}\plus{}A_\text{trapezoid}\equal{}\frac32\pi\plus{}24\plus{}13\sqrt3$

We agree with the total area given above.  We confirmed the area using our cadd package.   Soon as we figure out LaTex we'll post better solutions.

thanks for the solutions... I can solve it now myself..  :D  :D  :D


induction
superior algebra
superior algebra solved

Do you guys know how to prove that : "In a boolean ring, any finitely generated ideal is principal"


algebra
polynomial
induction
binomial theorem

[b]1.[/b] Calculate $ \sum_{k \equal{} 0}^n 2^k \cdot \dbinom{n}{2n \minus{} k}$

[b]2.[/b] Prove that this equality exists,for any $ n \in N$:
\[ \dbinom{n}{0} x_0 \plus{} \dbinom{n}{1} x_1 \plus{} ... \plus{} \dbinom{n}{n}x_n \equal{} x_{2n}
\]
where $ (x_n)_{n\geq 0}$ is the Fibonacci row.

[hide="Hint for number 2"]Finite differences on the Fibonacci series[/hide]

For number 2, try this
[hide]
Write $ x_n \equal{} \frac {1}{\sqrt {5}}(\varphi^n \plus{} \phi^n)$ (this is just the explicit formula for the Fibbonacci linear recurrence) where $ \varphi$ and $ \phi$ are both the roots of $ x^2 \minus{} x \minus{} 1 \equal{} 0$. Now factor out $ \frac {1}{\sqrt {5}}$ and we can split the sequence into two polynomials which are just the binomial theorem, and so we have in total $ \frac {1}{\sqrt {5}}((\varphi \plus{} 1)^n \plus{} (\phi \plus{} 1)^n)$ on the left hand side. But since both $ \varphi$ and $ \phi$ are the roots of $ x^2 \minus{} x \minus{} 1 \equal{} 0$, we know $ \varphi \plus{} 1 \equal{} \varphi^2$ and $ \phi \plus{} 1 \equal{} \phi^2$. Substituting we have $ \frac {1}{\sqrt {5}}(\varphi^{2n} \plus{} \phi^{2n})$, which is of course equivalent to $ x_{2n}$, on the right hand side. [/hide]

I'm sure there is a nice induction argument there too :) (I haven't tried it yet...)

[quote="skyhog"]Is the first one just $ 2^n$?

Aren't the $ \dbinom{n}{something bigger than n}$ all equal to 0?[/quote]
At the first problem is $ \binom{2n\minus{}k}{n}$ and the sum isn't $ 2^n$ but $ 2^{2n}$

the second problem see here:
[hide]It is know that $ x_n \equal{} \displaystyle\frac {1}{\sqrt {5}}[(\displaystyle\frac {1 \plus{} \sqrt {5}}{2})^{n \plus{} 1} \minus{} (\displaystyle\frac {1 \minus{} \sqrt {5}}{2})^{n \plus{} 1}]$,for any $ n$ natural.
Then
\[ \sum_{i \equal{} 1}^n \binom{n}{i} x_i \equal{} \frac {1}{5} [ \sum_{i \equal{} 0}^n \binom{n}{i}(\displaystyle\frac {1 \plus{} \sqrt {5}}{2})^{i \plus{} 1} \minus{} \sum_{i \equal{} 0}^n \binom{n}{i}(\displaystyle\frac {1 \minus{} \sqrt {5}}{2})^{i \plus{} 1}] \equal{}
\]

\[ \equal{} \frac {1}{\sqrt {5}}[\frac {1 \plus{} \sqrt {5}}{2}\cdot (\frac {1 \plus{} \sqrt {5}}{2} \plus{} 1)^n \minus{} \frac {1 \minus{} \sqrt {5}}{2}(\frac {1 \minus{} \sqrt {5}}{2} \plus{} 1)^n] \equal{}
\]

\[ \equal{} \frac {1}{\sqrt {5}}[\frac {1 \plus{} \sqrt {5}}{2} \cdot (\frac {3 \plus{} \sqrt {5}}{2})^n \minus{} \frac {1 \minus{} \sqrt {5}}{2}(\frac {3 \minus{} \sqrt {5}}{2})^n] \equal{}
\]

\[ \equal{} \frac {1}{\sqrt {5}}[\frac {1 \plus{} \sqrt {5}}{2} \cdot (\frac {1 \plus{} \sqrt {5}}{2}^{2n} \minus{} \frac {1 \minus{} \sqrt {5}}{2} \cdot (\frac {1 \minus{} \sqrt {5}}{2})^{2n} \equal{}
\]

\[ \equal{} \frac {1}{\sqrt {5}}[(\frac {1 \plus{} \sqrt {5}}{2})^{2n \plus{} 1} \minus{} (\frac {1 \minus{} \sqrt {5}}{2})^{2n \plus{} 1}] \equal{} x_{2n}
\]
[/hide]

and for the first one see here:

[hide]$ \binom{2n\minus{}k}{n}$ is the  $ x^n$ coeficient in $ (1\plus{}x)^{2n\minus{}k}$,therefore the general term of the sum, $ 2^k\cdot \binom{2n\minus{}k}{n}$, is the  $ x^n$ coeficient in $ 2^k(1\plus{}x)^{2n\minus{}k}$
\[ 2^k(1\plus{}x)^{2n\minus{}k}\equal{}2^{2n}\displaystyle\frac{(1\plus{}x)^{2n\minus{}k}}{2^{2n\minus{}k}}\equal{}2^{2n}(\frac{1\plus{}x}{2})^{2n\minus{}k}\]
From where $ \sum_{k\equal{}0}^n 2^k \binom{2n\minus{}k}{n}$ is the $ x^n$ coeficient in \[ S_1\equal{}2^{2n} \sum_{k\equal{}0}^n (\frac{1\plus{}x}{2})^{2n\minus{}k}\equal{} 2^{2n} \sum_{k\equal{}n}^{2n} (\frac{1\plus{}x}{2})^k\]
Let $ S_2\equal{}2^{2n} \sum_{k\equal{}0}^{n\minus{}1} (\frac{1\plus{}x}{2})^k$. Observe that $ S_2$ doesn't contain $ x^n$ terms, therefore the sum from our problem is the $ x^n$ coeficient from \[ S\equal{}S_1\plus{}S_2\equal{}\displaystyle\frac{2^{2n\plus{}1}\minus{}(1\plus{}x)^{2n\plus{}1}}{x\minus{}1}\]
At the begining of the problem, x was arbitrary.We can make $ x<1$ and $ m\rightarrow \infty$.
It is well known the formula $ 1\plus{}x\plus{}x^2\plus{}...\plus{}x^m\equal{}\frac{x^{m\plus{}1}\minus{}1}{x\minus{}1}$
\[ \frac{1}{1\minus{}x}\equal{}1\plus{}x\plus{}x^2\plus{}...\plus{}x^{n\minus{}1}\plus{}x^n\plus{}x^{n\plus{}1}\plus{}...\]
In conclusion $ S\equal{}(2^{2n\plus{}1}\minus{}(1\plus{}x)^{2n\plus{}1})(1\plus{}x\plus{}x^2\plus{}...\plus{}x^n\plus{}...)$ from where our sum is $ 2^{2n\plus{}1}\minus{}T$,where \[ T\equal{}1\plus{}\binom{2n\plus{}1}{1}\plus{}\binom{2n\plus{}1}{2}\plus{}...\plus{}\binom{2n\plus{}1}{n}\]
But $ \binom{2n\plus{}1}{k}\equal{}\binom{2n\plus{}1}{2n\plus{}1\minus{}k}$ and $ T\equal{}2^{2n}$ and finaly our sum is $ 2^{2n\plus{}1}\minus{}2^{2n}\equal{}2^{2n}$[/hide]


AMC AIME percent USA(J)MO USAMO HMMT quadratics

Discuss the index and your AIME score here. I only put the poll for scores though.

assuming i missed the last one i got a 10
so my index is 250 :)

I think I got a 13. Of the two I missed, one (#11) I never figured out a way to calculate it, and on the other  (#12), I was about 90% there, but was $9^\circ$ off on one of the angles, so I missed it. (That is, assuming my second method is correct). I thoroughly checked the other 13, so I'm pretty sure I got them.

The two problems I spent the most time on I didn't get. My index is (assuming I was correct w/everything I thought I was) 263.5.

the answers are posted [url=http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2006-aa/06answers-aime1.html]here[/url]

btw, wow, nice score :D

Yeah, I didn't get a 13. I wonder how I missed those three problems... I really got a 10.

That's weird, I was sure on two of those.

I got an 8 :). Now I just have to hope it is good enough for the floor.
Answered 1, 2, 3, 4, 7, 9, 11, and 13 right. On 6 I forgot to read distinct. :(

Apparently, I missed #'s 8 (my 63, their 89), 10 (14 vs. 65), 11+12 (see above), and 14 (95 vs. 183). Needless to say, I'm disappointed, especially on 10 and 14. On 8, I think I just made an incorrect assumption that gave me a wrong result. Oh well.

For all of us taking the AIME II, around when should the test be posted? Sorry if this is quick I just got back on but I just want to know so that I can take it as a practice before tommorow night's math jam...

Thanks...

By the way I'm hearing mixed things: some saying it was hard, others saying it was easy... etc. How would you compare it to AIMEI 2005?

Ahhhh, a stupid mistake on 4 got me a 7.  HOpefully that'll be good enough for the floor.

[quote="amirhtlusa"]im 90% sure that the floor will NOT be higher than 8 :)
as a matter of fact pretty sure it will be a 8, (7 at lowest)[/quote]

I'll be convinced that it's higher than 7 only if this poll starts showing more high scores.  Remember, there are 50 percent more people qualifying this year.

I agree that the floor will be 7.  This was not a particularly easy AIME and the expansion makes an 8 floor unlikely.  Of course, I got a 7, so I'm a little biased. :P

I dont have my answers with me, but I think I got a 7-8.  Missed #9 by 1  :(
Silly rabbit.

I got a 9.  That'll put me at a 219 index... right around where the cutoff will be I think.  :|

EDIT: 1-6, 9, 10, 12

[quote="amirhtlusa"]

thats not your index  ;)  (index is based on amc12 not 10)[/quote]

I've seen a lot of people say that index is based only on amc12. What happens if you only take the amc10- you don't get an index? This is what is says on the website:
    *  Selection for the USAMO will be made according to the follwing rules:

         1. The goal is to select about 250 of the top scorers from the prior AIME and AMC 12A, AMC 12B, AMC 10A and AMC 10B contests to participate in the USAMO.
       [b]  2. Selection will be based on the USAMO index which is defined as 10 times the student’s AIME score plus the student’s score on the AMC 12 or the AMC 10.[/b]
         3. The first selection will be the approximately 160 highest USAMO indices of students taking the AMC 12A or AMC 12B contest.
         4. The lowest AIME score among those 160 first selected will determine a floor value. The second selection of USAMO participants will be from the highest USAMO indices among students who took the AMC 10A or AMC 10B and the AIME, and got an AIME score at least as high as the floor value.
         5. The student with the highest USAMO index from each state, territory, or U.S. possession not already represented in the selection of the first and second groups will be invited to take the USAMO.
         6. To adjust for variations in contest difficulty, the number of students selected from A & B contests will be proportional to the number of students who took the (A & B) Contests.
         7. The selection process is designed to favor students who take the more mathematically comprehensive AMC 12A and AMC 12B contests.

As for the poll, I do think that AoPS represents the higher end of the scores, partly because AoPSers receive great mathematical training and partly because it attracts the more devoted math students. Also, obviously the lower scoring members would be less inclined to vote than the higher-scoring members, and I think we can be safe to assume that there aren't really 7 15's (this happened last year in Mathcounts when 6 or 7 people claimed to have gotten 46's on the state contest).

[quote="Phelpedo"][quote="amirhtlusa"]

thats not your index  ;)  (index is based on amc12 not 10)[/quote]

I've seen a lot of people say that index is based only on amc12. What happens if you only take the amc10- you don't get an index? This is what is says on the website:
    *  Selection for the USAMO will be made according to the follwing rules:

         1. The goal is to select about 250 of the top scorers from the prior AIME and AMC 12A, AMC 12B, AMC 10A and AMC 10B contests to participate in the USAMO.
       [b]  2. Selection will be based on the USAMO index which is defined as 10 times the student’s AIME score plus the student’s score on the AMC 12 or the AMC 10.[/b]
         3. The first selection will be the approximately 160 highest USAMO indices of students taking the AMC 12A or AMC 12B contest.
         4. The lowest AIME score among those 160 first selected will determine a floor value. The second selection of USAMO participants will be from the highest USAMO indices among students who took the AMC 10A or AMC 10B and the AIME, and got an AIME score at least as high as the floor value.
[/quote]

Look at number 3. They select the people with the top indices from the [b]12a and 12b[/b]. If you took the 10 you can only qualify through the floor value.

If you take the AMC10, then to qualify for the USAMO:
I. You must have an AIME score higher than the floor.
II. Your USAMO Index must be in the top 120* that are in the 10th grade and below.

So your index does help you qualify, but only with a good AMC10 score.

Top 120 will be the top from each AMC test.

[quote="MithsApprentice"][quote="daine"]If you took the 10 you can only qualify through the floor value.[/quote]

Not anymore (probably). It still seems like 7 would include too many people, but I could be wrong.[/quote]

Why would 7 be too many people?

Last year, ~400 people got above the floor (thus 250/400=62.5% of those above floor qualifying), and right now the poll shows ~200 people above 7 for aime I.  And with the 50% expansion, 375/.65~575 expected people scoring above the floor.  Asuming that aops has at least 1/3 of the total qualifiers this year, 7 would be just about right.......
but im hoping that aops has even more than 1/3 of the best and drop the floor to 6...... :)  :(

[quote="daine"]

Look at number 3. They select the people with the top indices from the [b]12a and 12b[/b]. If you took the 10 you can only qualify through the floor value.[/quote]

My point was that the AMC10 does count toward your index, but it can't be used to set the floor. Some people have been talking about how any AMC10 qualifying score will be converted to an AMC12 score of 100.
[quote="pieterminate"]
Why would 7 be too many people?

Last year, ~400 people got above the floor (thus 250/400=62.5% of those above floor qualifying), and right now the poll shows ~200 people above 7 for aime I. And with the 50% expansion, 375/.65~575 expected people scoring above the floor. Asuming that aops has at least 1/3 of the total qualifiers this year, 7 would be just about right.......
but im hoping that aops has even more than 1/3 of the best and drop the floor to 6...... Smile Sad
[/quote[

Don't forget that the AIME II is yet to take place. Before I looked on AoPS, I talked to some friends who took the AIME I and they all thought it was easier than usual; many were worried that even a 9 on the AIME would not make it, but obviously the consensus on AoPS is a little bit different.

[quote="Phelpedo"][quote="daine"]

Look at number 3. They select the people with the top indices from the [b]12a and 12b[/b]. If you took the 10 you can only qualify through the floor value.[/quote]

My point was that the AMC10 does count toward your index, but it can't be used to set the floor. Some people have been talking about how any AMC10 qualifying score will be converted to an AMC12 score of 100.
[quote="pieterminate"]
Why would 7 be too many people?

Last year, ~400 people got above the floor (thus 250/400=62.5% of those above floor qualifying), and right now the poll shows ~200 people above 7 for aime I. And with the 50% expansion, 375/.65~575 expected people scoring above the floor. Asuming that aops has at least 1/3 of the total qualifiers this year, 7 would be just about right.......
but im hoping that aops has even more than 1/3 of the best and drop the floor to 6...... Smile Sad
[/quote]

Don't forget that the AIME II is yet to take place. Before I looked on AoPS, I talked to some friends who took the AIME I and they all thought it was easier than usual; many were worried that even a 9 on the AIME would not make it, but obviously the consensus on AoPS is a little bit different.[/quote]

True, but last year was still easier than this year's and we can calculate from that, assuming that the overall average of the everyone hasn't improved.  And (l believe) much less people take the II than the I even if TJ is taking the I this year.

[quote="MithsApprentice"][quote="pieterminate"]Why would 7 be too many people?

Last year, ~400 people got above the floor (thus 250/400=62.5% of those above floor qualifying), and right now the poll shows ~200 people above 7 for aime I.  And with the 50% expansion, 375/.65~575 expected people scoring above the floor.  Asuming that aops has at least 1/3 of the total qualifiers this year, 7 would be just about right.......
but im hoping that aops has even more than 1/3 of the best and drop the floor to 6...... :)  :([/quote]

Intuition. They may take all the people who scored 7, but that seems like it would easily exceed 375.[/quote]

Intuition? That only works for math problems  :) :lol:

[quote="calc rulz"]How does AMC 10 score help USAMO qualification whatsoever? To qualify by AMC 10 index, you must also get above the floor value. However, if you're someone who can take the AMC 10, that is, someone in 10th or below, you automatically qualify by getting the floor value, correct?[/quote]

That's what we're discussing. *pokes* Keep up with the conversation.

"3. The first selection will be the approximately 160 highest USAMO indices of students taking the AMC 12A or AMC 12B contest. "
"4....The second selection of USAMO participants will be from the highest USAMO indices among students who took the AMC 10A or AMC 10B and the AIME, and got an AIME score at least as high as the floor value...
------------------------------------------------------------------

We know that there are some tenth and lower grade students took AMC 12A/B contests and got high scores. [b]Will they also be included in the 160 highest indices of students, or the 160 are only selected from juniors and seniors[/b]? If they are included in this 160 group, then the total 10th and lower graders, included those 120 second selection group, may be more than juniors and seniors.

they are included in the initial selection as well

[quote="calc rulz"]How does AMC 10 score help USAMO qualification whatsoever? To qualify by AMC 10 index, you must also get above the floor value. However, if you're someone who can take the AMC 10, that is, someone in 10th or below, you automatically qualify by getting the floor value, correct?[/quote]

Suppose the floor is a 7 but not all 9th and 10th graders that took the AMC10 will make the USAMO since that will mean too many people qualify. Then the USAMO index will determine which people that got a 7 will make it. This means that the AMC 10 does help in tiebreaker situations. But not in others.

Gahh, this is looking bad for me.  I have a 7 AIME (pretty sure), but 112 AMC, which is probably lower than most other's AMC scores.  So I hope they give at least some extra consideration for those who took the AMC 12.


you know fermat's little theorum, (or the manipulated version)

a^p = a (mod p)

well I was wondering is there an equation where in the place of a we have a prime number

eg. s is prime
s^x = ? mod x
or
what is the remainder when 23^24 is divided by 24?

I've come across this problem before and hit a dead end. Probably something simply that I'm missing.

[quote="blahblahblah"]Euler's Theorem states that $a^{\varphi(n)}\equiv 1\bmod n$.

For the example you gave though, it's obvious that the answer is 1.[/quote]

sorry, could you explain that formula? I don't know what the squiggle means.

and also maybe I'm missing something but I don't see why it's 1

are you allowed to say:

23 = -1 (mod 24)
23^2 = 1 (mod 24)

by squaring both sides?
and then since 24 is even the remainder will be 1?
is the remainder -1 when it is 23^23?

[quote="blahblahblah"]Euler's Theorem states that $a^{\varphi(n)}\equiv 1\bmod n$.

For the example you gave though, it's obvious that the answer is 1.[/quote]

Here  $\varphi(n) $ means   count of natular numbers less than n, and prime to n.

When n is prime,  $\varphi(n)  = n-1 $
When n is not prime, $\varphi(n) < n-1 $

when n=24, it is  8 . ( These  8  numbers are prime to 24 {1,5,7,11,13,17,19,23}

And  for any x (which is prime to  24) ,  you have $x^8 \equiv 1 \bmod 24 $

Hope this helps.

PS in the case of  23, even 23^2  is 1 mod 24   so of course 23^8 would be 1 too.

(Strange as it may seem, for  the special case of $24$ ,  for any number $x$, prime to 24, $x^2 \equiv  1 \bmod  24 $  - But that is beside the point :)
For any  number ,  $\varphi(n) $ will always work  -

P.S. A better function  than $\phi $ is , for this purpose is  $ \lambda(n) $  (Called  Carmichael Function) (which is  some times smaller than  $\phi $
$\varphi (24)  = 8  $ but $\lambda (24) = 2 $

etc ...

and  $a^{\lambda(n)} \equiv 1 \bmod n $ where a is prime to n ..

[quote="JBL"][quote="mikewd"]sorry, could you explain that formula? I don't know what the squiggle means.[/quote]

The squiggle is the greek letter phi.  In this case, it's being used as the Euler phi (or [url=http://mathworld.wolfram.com/TotientFunction.html]totient[/url]) function.  Usually, I think one uses the other written phi for the function, but I don't know how to make it in Latex.

[/quote]

By the way, the squiggly phi is \varphi, the straighter phi is \phi (or \Phi for uppercase).  I guess it's what you're used to, my algebra text last semester used \varphi and I've just gotten used to it.  I personally use $\phi$ for the golden ratio and $\varphi$ for the totient function.

[quote="JBL"][quote="mikewd"]sorry, could you explain that formula? I don't know what the squiggle means.[/quote]

The squiggle is the greek letter phi.  In this case, it's being used as the Euler phi (or [url=http://mathworld.wolfram.com/TotientFunction.html]totient[/url]) function.  Usually, I think one uses the other written phi for the function, but I don't know how to make it in Latex.

[quote]are you allowed to say:

23 = -1 (mod 24)
23^2 = 1 (mod 24)

by squaring both sides?
and then since 24 is even the remainder will be 1?
is the remainder -1 when it is 23^23?[/quote]

Yes.  It's just multiplication: we know that for all $a, b, c, m$, $a \equiv b \mod m$ implies $ac \equiv bc \mod m$.  That powers preserve equivalence follows easily.  I don't have any idea what 24 being even has to do with it, though.[/quote]

if n is even then it would mean 23^n = 1 (mod 24), yeah?
if n is odd it would mean 23^n = -1 (mod 24) ?

Does anyone have a link to any problems that use euler's theorum? I could do with some practice with it!
anyway no problem if there aren't many, thanks for all your helps. :lol:

[quote]if n is even then it would mean 23^n = 1 (mod 24), yeah?
if n is odd it would mean 23^n = -1 (mod 24) ?
[/quote]

Yes,  $23$ and $-1$ are same modulo $24$    so you  get the same answer if you substitute $-1$ for $23$ in almost   all  such airthmatic expressions ..


trigonometry
inequalities proposed
inequalities

[hide]Since $ \sin x$ and $ \tan x$ are increasing when $ 0 < x < \pi/6$, we have

$ \sum_{k\equal{}1}^n(\sin^{2k \minus{} 1} x \plus{} \tan^{2k} x) < \sum_{k\equal{}1}^n\left(\sin^{2k \minus{} 1} \frac {\pi}6 \plus{} \tan^{2k}\frac {\pi}6\right).$[/hide]


Prove that the shortest distance from a point $(x_0, y_0)$ to the line $Ax+By+C=0$ is

$\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$

without using vectors.

I was thinking about finding the equation of the line through $(x_0, y_0)$ perpendicular to the given line, finding the intersection, and then finding the distance between the intersection and the given point. That would involve equations nasty beyond belief (not that bad, but close).

Your method will work, but like you said it will take a while and will be messy.  Why don't you just go with vectors?  It would be much easier.

Take the vector $\vec{v}=\left(x_{0}-x_{1},y_{0}-y_{1}\right)$.  The direction of the desired vector is $\vec{w}=\left(A,B\right)$.  So your distance is

\[ d=\frac{\left|\vec{v}\cdot\vec{w}\right|}{\left|\vec{w}\right|}=\frac{\left|A\left(x_{0}-x_{1}\right)+B\left(y_{0}-y_{1}\right)\right|}{\sqrt{A^{2}+B^{2}}}  \]

\[ d=\frac{\left|Ax_{0}+By_{0}-\left(Ax_{1}+By_{1}\right)\right|}{\sqrt{A^{2}+B^{2}}}=\boxed{\frac{\left|Ax_{0}+By_{0}+C\right|}{\sqrt{A^{2}+B^{2}}}}  \]

Using this approach, you can easily extend it to problems such as finding the shortest distance between a point and a plane.

you want to use the pythagorean theorem to prove that the minimum distance is the perpendicular distance

[hide][quote]I was thinking about finding the equation of the line through $(x_0,y_0)$  perpendicular to the given line, finding the intersection, and then finding the distance between the intersection and the given point. That would involve equations nasty beyond belief (not that bad, but close).[/quote]

i tried doing that and it did get quite messy.
first of all, the equation of the line is
$y=-\frac{A}{B}x-\frac{C}{B}$

so, a line perpendicuar to this that runs through the given coordinates would be

$y=\frac{B}{A}x+y_0-\frac{B}{A}x_0$

then u have

$\frac{B}{A}x+y_0-\frac{B}{A}x_0=-\frac{A}{B}x-\frac{C}{B}$

when isolating, i kept forgetting the subscript so i kept messing up. :huuh: anyways, i get

$x=\frac{B^2x_0-ABy-AC}{B^2+A^2}$

$y=\frac{B^2x_0-A^2x-AC-B^2x_0}{AB}$

i have a feeling that's wrong so i wont continue yet. [/hide]

Although this method is a bit too advanced for this forum, the use of Cauchy's Inequality yields a particularly elegant solution to this.


Factor the determinant!

$\left| \begin{array}{ccccc} 1&a_1&(a_1)^2&...&(a_{1})^n\\ 1&a_2&(a_2)^2&...&(a_{2})^n\\ ...&...&...&...&...\\ 1&a_{n+1}&(a_{n+1})^2&...&(a_{n+1})^n\\ \end{array} \right|$

[hide="hint 1"]try a simple case[/hide]

[hide="hint 2"]find some factors, and use symmetry to find the others[/hide]

[quote="ZetaX"]Just one more word or hint ;)
[hide]Vandermonde[/hide][/quote]

i thought i was original for thinking of this...oh well  :lol:


in triangle $ ABC, CC'$ and $ BB'$ are altitudes.$ B'C' and BC$intersect at D.
prove that $ DM(AB^2\minus{}AC^2)\equal{}BC^3/2$ where M is the midpoint of BC :ninja:

In this problem, I think shoud be an algebraical prove. And this is my solution
Give DB= x. We'll calculate x follow a,b,c.
First, B'C'/BC= HC'/HB= cos A so B'C'= a.cosA.
DC'/DB= DC/DB'= CC'/BB'= b/c. Then DC'= DB.b/c, so DB'= x.b/c +a.cosA,
so DC=(x.b/c+a.cosA).b/c(1)
But DC= x+a(2)
From (1) and (2) we have x=a.(c^2-b^2+a^2)/2(b^2-c^2).
then DM= x+a/2= a^3/2(b^2-c^2). (which we need prove) :)

[quote="ishfaq420haque"][color=darkred]In triangle $ ABC, CC'$ and $ BB'$ are altitudes.$ B'C' \cap BC\equal{}D$.$ M$ is the midpoint of $ BC$.Let, $ DM\equal{}k$
Prove that, $ k(c^2 \minus{} b^2) \equal{}\frac {a^3}{2}$
[/color][/quote]
[u][b]
Solution:[/b][/u]
Using Menelaus's theorem on $ \triangle ABC$ and transversal $ B'C'D$ we get,
$ \frac{k\minus{}a/2}{k\plus{}a/2}\equal{}\minus{}\frac{c \cdot \cos B}{b \cdot \cos C}$
Now using cos law we get,
$ \frac{k\minus{}a/2}{k\plus{}a/2}\equal{}\minus{}\frac{a^2\plus{}c^2\minus{}b^2}{a^2\plus{}b^2\minus{}c^2}$
which implies that,
$ \frac{k}{\minus{}a/2}\equal{}\minus{}\frac{a^2}{c^2\minus{}b^2}$


Prove that infinitely many triangular numbers are perfect squares.

We obviously have $n^2 + n = 2m^2$, whence $(2n+1)^2 - 8m^2 = 1$.  Put $(a,b) = (2n+1,2m)$.  Then we have $a^2 - 2b^2 = 1$.  This is a pell equation with infinitely many positive integer solutions.


I'm having trouble with a few problems from Hatcher's book on Van Kampen.

1. Let $ X \in \mathbb{R}^{m}$ be the union of convex open sets $ X_1, X_2, ..., X_n$ such that the triple intersections of $ {X_i, X_j,X_k}$ is always non-empty. Show that $ X$ is simply connected.

2. Let $ X$ in $ \mathbb{R}^{3}$ be the union of $ n$ lines through the origin. Compute $ \pi_{1}(\mathbb{R}^3 \minus{} X)$.

I don't quite understand the following questions, so it would be great if someone could explain them (but not the solutions):

3. Let $ X \in \mathbb{R}^{2}$ be a connected graph that is the union of a finite number of straight line segments. Show that $ \pi_{1}(X)$ is free with a basis consisting of loops formed by the boundaries of the bounded complementary regions of $ X$, joined to a basepoint by suitable chosen paths in $ X$.  -- What are bounded complementary regions?

4. Let $ X$ be the quotient of $ S^2$ obtained by identifying the north and south poles to a single point. Put a cell complex structure on $ X$ and use this to compute $ \pi_{1}(X)$.  - What does it mean, formally, to identify two points together, and how would I visualise this space concretely?

5. Compute the fundamental group of the space obtained from two tori $ S^1 \times S^1$ by identifying a circle $ S^1 \times {x_0}$ in one torus with the corresponding circle in the other torus. - What does this mean again, and how would I visualise this, if possible?

---
My attempts at #1,2:
[hide] #1: Since I need all the spaces to contain a basepoint to apply Van Kampen's Theorem, I tried choosing an arbitrary basepoint $ T$. Then for each space $ X_i$, I tried to form a space $ X'_i$, which is the union of $ X_i$ and the point $ T$ with, for each point $ x \in X_i$, a path connecting $ x$ to $ T$. While $ X$ is now the union of the spaces $ X'_i$ which I think have double and triple intersections path-connected and contain the basepoint, I'm not sure that $ \pi_1(X'_i) \equal{} 1$, and not sure how to modify this construction to ensure that $ \pi_1(X'_i)$. I'm also not completely sure how to use the fact that the triple intersections of $ X_i$ are path-connected, since the first part of Van Kampen which says the map from the free product of $ \pi_1(X_i)$ to $ \pi_1(X)$ is injective would suffice if we can modify the $ X_i$ so that they contain the basepoint, and still have $ \pi_1(X_i) \equal{} 1$.
#2: When $ n \equal{} 1$, I tried by proving that $ \mathbb{R}^3 \minus{} X$ deformation retracts onto a cylinder $ S^1 \times \mathbb{R}$ , so $ \pi_1(\mathbb{R} \minus{} X) \equal{} \pi_1 (S^1 \times \mathbb{R}) \equal{} \pi_1 (S^1) \times \pi_1 (\mathbb{R}) \equal{} \mathbb{Z}$. When n=2, I tried placing $ 2$ planes to separate $ \mathbb{R}^3$ as the union of $ 4$ parts, and each part containing only one line (e.g. the $ 2$ planes being $ x \equal{} 0$ and $ y \equal{} 0$, and the two lines being $ z \equal{} 0, x \equal{} y$ and $ z \equal{} 0, y \equal{} \minus{} x$), then applying Van Kampen to show that $ \pi_1(\mathbb{R} \minus{} X)$ is the free group on $ 4$ generators; and a very similar argument to show that with $ n$ lines, it is the free group on $ 2n$ generators. I think this is wrong and the answer should be the free group on $ n$ generators, but I can't find my mistake in amplying Van Kampen. [/hide]

What you're asking about is the [url=http://en.wikipedia.org/wiki/Quotient_space]quotient topology[/url]. Formally, if X is a space and $ \sim$ is an equivalence relation on $ X$, then there is a canonical topology on the set $ X / \sim$, which is the 'weakest' possible such that the natural projection $ \pi : X \to X / \sim$ is continuous. Specifically, a subset $ U \subset X / \sim$ is open iff $ \pi^{ \minus{} 1}(U) \subset X$ is open. Maps out of $ X / \sim$ can be shown to be precisely those maps out of $ X$ which are indistinguishable on $ \sim$-equivalent points.

You can imagine the quotient topology by ``gluing'' together points. For the antipodal poles of $ S^2$, imagine a beachball where you push together the top and bottom, leaving something like a donut with no hole.

Often, its easiest to work with the formal properties when you have to whip out algebraic topology.

I think the answer to #2 is the free group on $ 2n\minus{}1$ generators. (Hint: I thought about it as homotopy equivalent to $ S^2$ with $ 2n$ holes in it).

[quote="HilbertThm90"]I think the answer to #2 is the free group on $ 2n \minus{} 1$ generators. (Hint: I thought about it as homotopy equivalent to $ S^2$ with $ 2n$ holes in it).[/quote]

That it is, which is homeomorphic to $ \mathbb{R}^2$ with $ 2n\minus{}1$ pts removed.


Hey nice job guys you got 37th in the A division

Yes, you must have made Yasha proud... because, you know, Yasha B K is from Connecticut.

thanks

we'd like to dedicate it to our captain, Matt manedlberg

who got a 2 on the indiviudal


ok, here is an extra credit problem that my math class had, (i was like the only one to get it...)
simplify

$1+ \frac{1}{1+ \frac{1}{1+ \frac{1}{1+ \frac{1}{x}}}}$

of course, there is a clever way to do it, and a clumsy way to do it, but here is a problem that i thought of to go along with this that requires a little more thought (not necessarily that much)

let $f(x) = 1 +\frac{1}{x}$, and using the general notation for compound functions, such as $f^2(x) = f(f(x))$
find a simplified formula for $f^n(x)$ that contains only one fraction bar
(so like the problem that i had for my class would be $f^4(x)$

ummm... $\frac{F_{n-1}x+F_{n-2}}{F_{n}x+F_{n-1}}$ or something like that, proven by induction.  I can't think of any better way...

yes, that is close to what i got, not necessarily that hard to see the fibbonacci sequence, but this stuff is beyond my math class, then only thing is that the number of the fibbonacci sequence is wrong for you expression, and yes, this is an easy thing to prove by induction, if you get the sequence numbers correct,


Simply refer from a lemma of Hungkhtn post in

http://www.mathlinks.ro/Forum/viewtopic.php?t=110851

However, I can not prove above statement at that time. Ways $a=c+x,b=c+y$ works but too long.


floor function
number theory proposed
number theory

Find the largest positive integer $n>10$ such that the residue of $n$ when divided by each perfect square between $2$ and $\dfrac n2$ is an odd number.

I have a solution but I think it is no good.

Let $\lfloor\sqrt{n}\rfloor=a$.  Since $n>10$, we have $2<a<\frac{n}{2}$, so $n\equiv b\mod{a}$, and $b$ must be odd.  Also, $n\equiv1\mod{2}$ (since $n\equiv1\mbox{ or }3\mod{4}$), so $a$ must be even.

But now $n\equiv 2a+b-1\mod{(a-1)^2}$. So if $2a+b-1\le(a-1)^2$, then the condition is violated.  But we know $b\le2a-1$ since $n<(a+1)^2$ and $b$ is odd, so $2a+b-1\le4a-2$.  But $4a-2<(a-1)^2$ for $a\ge6$, so we must have $a=4$, $b\le7$.

But it is easy to check that $n=4^2+7=23$ satisfies the given condition, and so it is the maximal solution.

You are wrong because you don't understand that we have division.For example:
If we have the division $\frac{1251}{625}$. In this case the perfect square is $25^2$ and it is smaller from $\frac{n}{2}$ as the n in my example is $n=1251$. And the residue from this division is 1 which is odd. Am I wrong?

Oops - apparently I did misunderstand the problem - I thought you wanted numbers with odd remainders on division by the square of each number from $2$ to $\frac{n}{2}$, not just each square in that range.

Incidentally, $1251$ is still not a solution, since it leaves even remainders upon division by $7^2,9^2,13^2,15^2$ and $19^2$.

Yes there is no solution. But why Valentin Vornicu said in the forum which is posted the problem that the solution is $505$? Valentin could you give us a solution?

Well, [tex]505[/tex] [b]is[/b] a solution, just check it ! :)

Well, [tex]505[/tex] [b]is[/b] a solution, but $n=505$ is not the largest!!!!! :)

I will show that the largest such $n$ is 505.  It has already been checked that 505 is a solution.  Let $n$ be an integer bigger than 505.

First, if $n$ is even, then when $n$ is divided by 4, the remainder will be even, violating the property.  Thus we may assume that $n$ is odd.

I claim that there is an odd integer $k$ such that $3k^2 \le n < 4 k^2$.  For $k \ge 13$, we can check that $3(k + 2)^2 \le 4k^2 - 1$.  Thus the intervals $[3k^2, 4k^2 - 1]$ over all odd integers $k \ge 13$ cover all numbers from 507 and up.  Thus we have proved the claim.

Clearly, $k^2$ is between $2$ and $\frac{n}{2}$.  When $n$ is divided by $k^2$, the remainder is $n - 3 k^2$.  But $n$ and $k$ are both odd, so the remainder $n - 3k^2$ is even.  Thus $n$ does not satisfy the property.

But my counterexample $n=1251$? We know $1251>505$. What a strange problem :noo:

Diarmuid already pointed out that your "counterexample" 1251 is not a counterexample.  [i]Every[/i] remainder has to be odd, not just one remainder.

Edit: Oops, Ravi beat me to it...

I already told you Silouan, you're counterexample is not a counterexample.

[quote="Diarmuid"]Incidentally, $1251$ is still not a solution, since it leaves even remainders upon division by $7^2,9^2,13^2,15^2$ and $19^2$.[/quote]

Ok i understand this thank you very much for the solution Ravi


Hypotenuse $AB$ of right triangle $ABC$ is divided into three equal parts by lines dran from the vertex. If these lines are $7$ units long and $9$ units long, how long is the hypotenuse?


$x_1, x_2, \cdots, x_n$ are positive numbers such that $\sum_{i=1}^{n} x_i = 1$.

Prove: $\left( \sum_{i=1}^{n}\sqrt{x_i} \right) \left( \sum_{i=1}^{n}\frac{1}{\sqrt{1+x_i}} \right) \leq \frac{n^2}{\sqrt{n+1}}$


geometry
angle bisector
geometry proposed

Let $ ABC$ be a triangle with the incircle $ w$ . Denote $ \left\|\begin{array}{c} D\in BC\cap w \\
\ E\in CA\cap w \\
\ F\in AB\cap w\end{array}\right\|$

and $ \left\|\begin{array}{c} M\in (BD)\ \ ,\ \ \frac {MB}{MD} \equal{} \frac {a}{p \minus{} c} \\
 \\
N\in (DC)\ \ ,\ \ \frac {NC}{ND} \equal{} \frac {a}{p \minus{} b}\\\\
P\in EF\ \ ,\ \ DP\perp EF\end{array}\right\|$ . Prove that $ \widehat {MPB}\equiv\widehat {NPC}$ .

I am sorry if I am mistaken, but I think must be $ \frac {MB}{MD} \equal{} \frac {BD}{BC} \equal{} \frac {p \minus{} b}{a}$ $ ,(1)$ and $ \frac {NC}{ND} \equal{} \frac {CD}{BC} \equal{} \frac {p \minus{} c}{a}$ $ ,(2)$

From $ (1),\ (2)$ $ \Longrightarrow$ $ \frac {MB}{MD}\cdot \frac {ND}{NC} \equal{} \frac {BD}{CD}$ $ ,(3)$ $ \Longrightarrow$ $ \angle MPB \equal{} \angle NPC$ because the line segment $ PD,$ is the angle bisector of $ \angle BPC$ $ ($ from $ \frac {QB}{QC} \equal{} \frac {DB}{DC}$ and $ PD\perp PQ,$ where $ Q\equiv BC\cap EF$ $ ).$

Greetings from Hania of Kretan, Kostas Vittas.

PS. I will post here later more details.

[quote="vittasko"]I am sorry if I am mistaken, but I think must be $ \frac {MB}{MD} \equal{} \frac {BD}{BC} \equal{} \frac {p \minus{} c}{a}$ $ ,(1)$ and $ \frac {NC}{ND} \equal{} \frac {CD}{BC} \equal{} \frac {p \minus{} b}{a}$ $ ,(2)$

From $ (1),\ (2)$ $ \Longrightarrow$ $ \frac {MB}{MD}\cdot \frac {ND}{NC} \equal{} \frac {BD}{CD}$ $ ,(3)$ $ \Longrightarrow$ $ \angle MPB \equal{} \angle NPC$ because the line segment $ PD,$
is the angle bisector of $ \angle BPC$ $ ($ from $ \frac {QB}{QC} \equal{} \frac {DB}{DC}$ and $ PD\perp PQ,$ where $ Q\equiv BC\cap EF$ $ ).$ [/quote]
My hypotesis is correctly, sorry ! $ BD \equal{} p \minus{} b$ , $ CD \equal{} p \minus{} c$ ... Greetings from Saint Petersburg, Florida !

You are right dear [b][size=100]Virgil[/size][/b], the notations of your nice problem are correctly and thank you for your remark about my typo ( it has been corrected ).

The familiar conditions I mentioned before is true, as another case of the proposed problem ( also, another case of the problem was the conditions you posted at first ).

The relation of $ \frac {MB}{MD}\cdot \frac {ND}{NC} \equal{} \frac {BD}{CD}$ is again enough, to conclude that $ \angle MPB \equal{} \angle NPC$ as your problem states.

Have a nice day, Kostas Vittas.

[b][size=100][color=DarkBlue]LEMMA. - A triangle $ \bigtriangleup ABC$ is given and let $ M,\ N$ be, two points on its side-segment $ BC,$ such that $ \angle MAB \equal{} \angle NAC$ $ ($ $ M,$ between $ B,\ N$ $ ).$ Prove that $ \frac {MB}{MD}\cdot \frac {ND}{NC} \equal{} \frac {BD}{CD},$ where $ D$ is the trace on $ BC,$ of the internal angle bisector of $ \angle A.$[/color][/size][/b]

[b][size=100]PROOF OF THE LEMMA.[/size][/b] - From the triangles $ \bigtriangleup AMB,\ \bigtriangleup ANC$ and because of $ \angle MAB \equal{} \angle NAC,$ we have $ \frac {MB}{NC} \equal{} \frac {(AMB)}{(ANC)} \equal{} \frac {(AB)\cdot (AM)}{(AC)\cdot (AN)}$ $ ,(1)$

Similarly, from the triangles $ \bigtriangleup AMD,\ \bigtriangleup AND$ and because of $ \angle MAD \equal{} \angle NAD,$ we have $ \frac {MD}{ND} \equal{} \frac {(AMD)}{(AND)} \equal{} \frac {(AM)\cdot (AD)}{(AN)\cdot (AD)}$ $ ,(2)$

From $ (1),\ (2)$ $ \Longrightarrow$ $ \frac {MB}{MD}\cdot \frac {ND}{NC} \equal{} \frac {AB}{AC}$ $ ,(3)$

From $ (3)$ and because of  $ \frac {AB}{AC} \equal{} \frac {BD}{CD}$ $ \Longrightarrow$ $ \frac {MB}{MD}\cdot \frac {ND}{NC} \equal{} \frac {BD}{CD}$ and the proof of the [b][size=100]Lemma[/size][/b] is completed.

$ \bullet$ Return now in the proposed problem, we have $ \frac {MB}{MD} \equal{} \frac {a}{p \minus{} c} \equal{} \frac {BC}{CD}$ $ ,(4)$ and $ \frac {NC}{ND} \equal{} \frac {a}{p \minus{} b} \equal{} \frac {BC}{BD}$ $ ,(5)$

From $ (4),\ (5)$ $ \Longrightarrow$ $ \frac {MB}{MD}\cdot \frac {ND}{NC} \equal{} \frac {BD}{CD}$ $ ,(6)$

It is easy now to show that the points $ Q,\ B,\ D,\ C,$ where $ Q\equiv BC\cap EF,$ are in harmonic conjugation, from the complete quadrilateral $ AEG_{e}FBC,$ where $ G_{e}\equiv AD\cap BE\cap CF,$ as the [b][size=100]Gergone point[/size][/b] of $ \bigtriangleup ABC$ as well.

So, from the harmonic pencil $ P.QBDC$ and because of $ PD\perp PQ,$ we conclude that the line segment $ PD,$ bisects the angle $ \angle BPC.$

In the triangle $ \bigtriangleup PBC,$ we define the point $ N'$ between $ D,\ C,$ such that $ \angle N'PC \equal{} \angle MPB$ $ ,(7)$ and we will prove that $ N'\equiv N.$

From $ (7),$ based on the above [b][size=100]Lemma[/size][/b], we have that $ \frac {MB}{MD}\cdot \frac {N'D}{N'C} \equal{} \frac {BD}{CD}$ $ ,(8)$

From $ (6),\ (8)$ $ \Longrightarrow$ $ \frac {N'D}{N'C} \equal{} \frac {ND}{NC}$ $ \Longrightarrow$ $ N'\equiv N.$

Hence, we conclude that $ \angle MPB \equal{} \angle NPC$ and the proof of the proposed problem is completed.

Kostas Vittas.

See [url=http://www.mathlinks.ro/viewtopic.php?t=294577][color=red][b]here[/b][/color][/url] my lemma.


calculus
integration
logarithms
function
algorithm
derivative
Galois Theory

It doesn't have an elementary antiderivative.  Don't post an integral unless you have reason to believe that it exists.

Im sorry to have posted this but this was one of my home work problems.

Hello JRav, how do we find if a func has an elementery antiderivative or not (Also, please explain me how we define 'elementary' here)? I've another ques too. Is there an elementary antiderivative for logx? Thankyou

[quote="lajanugen"]antiderivative for $ \log x$?[/quote]

$ \int x^n \ln x^m \text{d}x \equal{} \frac {m}{n \plus{} 1}\left(x^{n \plus{} 1}\ln x \minus{} \frac {x^{n \plus{} 1}}{n \plus{} 1}\right) \plus{} c$

...that's my [url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=49258]signature[/url]! :wink:

Call $ f(x) \equal{} \log(x)$, $ g'(x) \equal{} 1$. $ g(x) \equal{} x$.

Then

$ \int \log(x) \,dx \equal{} \int f g' \, dx \equal{} f g \minus{} \int f' g \, dx \equal{} x \log x \minus{} \int 1 \, dx \equal{} \boxed{x \log x \minus{} x \plus{} C}$.

[quote="mamama"]The antiderivative of $ log x$ can be derived using integration by parts.[/quote]

Yes, the formula too is derived using integration by parts.

In the formula, $ m\equal{}1, n\equal{}0$, we get $ x\log x \minus{}x \plus{}c$.

Minor notational aside: in the last post, Tyl is using the symbol $ \log x$ to mean the natural (base $ e$) logarithm of $ x.$ While this is a very widespread and time-honored notation, it's not universal, and it's not what the buttons on my calculator say. Two posts above, Rajiv used the notation $ \ln x$ for the same function.

Now to address lajunugen: while there is a [url=http://en.wikipedia.org/wiki/Differential_Galois_theory]theory[/url], I would not use that language with a beginning calculus student. What is an elementary function? Start with a modest "approved" list of functions: constants, powers of $ x,$ exponentials, logarithms, trigonometric functions, and inverse trigonometric functions. An elementary function is any function that can be assembled from these pieces in a finite number of steps, where the steps involve addition, subtraction, multiplication, division, and composition.

Fact: every elementary function has an elementary derivative. How do we know this? We have a "machine," an algorithm we can apply in an automatic fashion. We know the elementary derivatives of each of our building block functions, and we know the rules of combination, such as the product rule, quotient rule, and chain rule, for being able to differentiate the appropriate combinations of functions.

But the reverse does not hold. There are many elementary functions that do not have elementary antiderivatives. Saying that is essentially saying that there's no "machine", no universal automatic algorithm for finding elementary antiderivatives. In fact I would dare to say that "most" (in some sense that I don't care to specify) elementary functions do not have elementary antiderivatives.

The practical import: if you can't see any good reason why a function should have an elementary antiderivative, then assume that it most likely doesn't. Let the [i]default[/i] setting for "Does it have an elementary antiderivative?" be "No, probably not." Only change the setting when you can see a reason to.

Alternatively, you've got the [url=http://en.wikipedia.org/wiki/Risch_algorithm]Risch algorithm[/url] but I wouldn't actually use this unless I had a lot of free time.

If you have at least one elementary function $ f$ without an elementary antiderivative (EA), you can create new elementary functions without EA just by taking any elementary function $ g$, and then defining $ h : \equal{} g \plus{} f$.
Though this might be considered as silly, it does show that there are at least as many elementary functions without EA as there are with EA.

As for the log integral formula, wouldn't it be simpler to just consider it as a function of m and n, rewrite it in terms of their derivatives? There's no need to go to the trouble of integration by parts.

Let $ \frac {1}{x} \equal{} t\Longrightarrow \minus{} \frac {1}{x^2}dx \equal{} dt$, we have $ \int e^{\frac {1}{x}}dx \equal{} \minus{} \int \frac {1}{t^2}e^tdt$.

Let denote $ I_n \equal{} \int \frac {1}{t^n}e^tdt$,

$ I_n \equal{} \frac {1}{t^n}e^t \plus{} n\int e^t\cdot \frac {1}{t^{n \plus{} 1}}dt$, yielding $ \boxed{I_n \equal{} \frac {e^t}{t^n} \plus{} n I_{n \plus{} 1}}$.

[quote="kunny"]$ \boxed{I_n \equal{} \frac {e^t}{t^n} \plus{} n I_{n \plus{} 1}}$[/quote]

Would that not be an infinite series?

Yes, it would be a recursive formula for the integral.  Better than nothing I suppose.

Yeah! I have heard of reduction formulae, this seems to be just the opposite ($ I_{n+1)}$)! :rotfl:

[quote="b555"]is  there  any name  of such formulae?[/quote]

Are you asking about reduction formulae?

[url=http://en.wikipedia.org/wiki/Integration_by_reduction_formulae]Here[/url] is a small example of a reduction formula.  It allows you to evaluate a more complicated integral (usually with high powers) in terms of the same (or really similar integrals) in lower powers.


[hide]We have r,R,w,W
R=w+W, so since all are prime w=2
now we have r,W+2,2,W
r+W+2=5z
2W+r=63
W+2=prime
r=prime
from this we can get W=5
r=53

so
r=53
R=7
w=2
W=5[/hide]

[hide][quote="G-UNIT"]There are 67 ( white and red) balls in a bowl.  There are small ones and big ones among them.  We know that:
a) the number of red balls is divisble by 5;
b) the number of big red balls is the same as the number of white balls;
c)  the number of small white balls is the smallest;
d)  the number of each kind is prime
  How many of each of the balls are there in the bowl?[/quote]
By finding 4 prime numbers that add up to each other, i tget the folllowing: 53 is small red, 2 is small white, 7 is big red, and 5 is big white[/hide]

[quote="Plagued Templar"][quote]R=w+W, so since all are prime w=2 [/quote]
Why do we assume w=2?[/quote]

  Because every prime number except 2, is odd.  Now lets suppose we were dealing with 2 odd-prime numbers the result would be an even number, which is not prime.  Thus, at least 1 (not both) must be an even prime number, 2.  2 is the smallest prime number, so in this case the smallest sum would correspond to the smallest prime, thus, the number of small white balls must be 2.


How can I set such that the two numbers can be "aligned"?

$ \int_{\minus{}1}^{1}\frac{1}{x}\,dx$

The integral sign is at an angle so limits aren't usually lined up* but if you insist then add \hspace before the upper limit as in $ \int_{\minus{}1}^{\hspace{0.2em}1}\frac{1}{x}\,dx$

* See for example the bottom of the second page of [url=http://media.pearsoncmg.com/aw/aw_thomas_calculus_11/mmt/ch05/tcu11_05_03.pdf]The Definite Integral[/url], a nicely printed book I took at random, though the -2 and 2 do look lined up on the last page :)


calculus
integration
logarithms
trigonometry
calculus computations