inequalities
geometry
circumcircle
cyclic quadrilateral
perpendicular bisector

On the circle $K$ there are given three distinct points $A,B,C.$ Construct (using only a straightedge and a compass) a fourth point $D$ on $K$ such that a circle can be inscribed in the quadrilateral thus obtained.

[hide="Solution"]Note that a circle can be inscribed within a quadrilateral $ABCD$ if and only if $AB+CD = AB+CD.$ Also, note that given a triangle $ABC$ and a point $D$ such that $AB+CD = AB+CD,$ the locus of $D$ is a line. So, given points $A,B,C$ on the quadrilateral, let $AB = b, BC = a.$ Now, draw a circle with centre $A$ and radius $2b+a$ and another circle with centre $C$ and radius $2a+b.$ Choose the intersection points of these circles that is farther away from $B$ and call this point $D_{1}.$ Then, draw a circle with centre $A$ with radius $3b/2+a/2$ and another circle with centre $C$ with radius $3a/2+b/2.$ Take the intersection points of these circles that is farther away from $B$ and call this point $D_{2}.$ Draw $D_{1}D_{2}$ and extend this segment infintitely. Take the intersection point between $D_{1}D_{2}$ and the circle circumscribing $ABC$ which makes $ABCD$ convex and call this point $X.$ $X$ is the desired point.[/hide]

Ah, yes, my mistake. I shall change it.  :)

Here's another solution:

[hide]
Rearrange the equation $AB+CD=AD+BC$ (1) into $BC-AB=CD-AD$ (2). Construct a circle with radius AB and center B and let its intersection with BC be E (WLOG assume that BC>=AB and CD>=AD, if the inequalities are reversed then a similar situation results). Construct a circle with radius CE which intersects CD at F (CD hasn't been constructed yet) We can now reduce equation (2) to $CE=CF$. Subtracting equation (1) from equation (2), we find that $AB+DF=AD+BE$. However, $AB=BE$, so $DF=AD$. Let angle $ABC=\theta$. Because ABCD is a cyclic quadrilateral, angle $CDA=180-\theta$. Since triangle FDA is isosceles, angle $AFC=180-\frac{\theta}{2}$. Construct the perpendicular bisector of $AC$ and let its intersection above the line with the circumcircle of triangle ABC be G. Now construct a circle with center G passing through A and C. Mark its intersection with circle C as H. Note that $H=F$ because angles AHC and AFC are congruent and they both lie on circle C. In order to complete the construction, draw a line through C and F. The intersection of line CF with the circle (besides C itself) must be D, so we are done.
[/hide]

[quote="knexpert"][quote="Karth"] $AB+CD = AC+BD$ [/quote]
This is incorrect. It should be $AB+CD=AD+BC$.

EDIT: You still haven't corrected it. (Nothing special about $AB+CD=AB+CD$  :D ).[/quote]

Wow, that was pretty stupid  :blush: I'll go back and correct it correctly now  :D

EDIT: Gah, it won't let me edit it... dang it. Oh well, you know what I mean, I guess...


inequalities
rearrangement inequality
inequalities unsolved

Let $a>0,b>0,c>0$. Prove the following inequality.

\[\frac{1}{a^2(b+c)}+\frac{1}{b^2(c+a)}+\frac{1}{c^2(a+b)}\geqq \frac{3}{2abc}\]

Rewrite the inequality like:
                        bc/[a(b+c)] + ac/[b(a+c)] + ab/[c(a+b)] >= 3/2

Now, using the rearrangement inequality, It’ll be true.

Sorry, I am not accustomed to use rearrangement inequality.
So I can't confirm your solution. :(

Let $a=\frac{1}{x},b=\frac{1}{y},c=\frac{1}{z}$. Then the ineq becomes \[\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\ge\frac{3}{2}\]which is exactly the Nesbitt's Inequality.


A somewhat interesting problem:

A long rectangular conducting loop of width $ L$, resistance $ R$, and mass $ m$, is hung in a horizontal, uniform magnetic field $ \vec{B}$ that is directed perpendicular to the plane of the rectangular loop and that exists only above a certain horizontal line parallel to the plane of the loop.  The loop is then dropped and the gravitational force of the Earth acts on it with downward acceleration $ \vec{g}$.  During its fall, it accelerates unitl it reaches a certain terminal speed $ v_t$.  Ignoring air drag, find an expression for $ v_t$ in terms of the given information.

the terminal velocity can only be obtained once a part of  it is above that fixed horizontal line
and another part below
then the current in it is $ \frac{Bvl}{R}$ hence force is  $ \frac{B^{2}l^{2}v}{R}$.So to acheive terminal speed
$ \frac{B^{2}l^{2}v}{R}\equal{}mg$
hence $ v\equal{}\frac{mgR}{B^{2}l^{2}}$


When the number $n$ is written in base b its representation is the two-digit number AB where A=b-2 and B=2.  What is the representation of n in base (b-1)?  (MA 1991)

this is what I've got so far:
[hide] $AB=(b-2)(b)+2=b^{2}-2b+2=b^{2}+2(b-1)$  to convert to base b-1  do I divide $b^{2}-2(b-1)/b-1$?  but it dosen't really get me anywhere[/hide]  and how do you make a divide(straight line, not slash) in latex?

thanks

To make a fraction in $\LaTeX$, use \frac{numerator}{denominator}.  $\frac{b^{2}-2(b-1)}{b-1}$

[hide="hint"]I would try using long division (or synthetic division) here.  Think about what the remainder would equal and what the quotient would be (in terms or the base $b-1$ number).[/hide]

Well, to simplify $\frac{b^{2}-2(b-1)}{b-1}$, just expand the parentheses in the numerator, then factor the quadratic.

[hide]think of how number bases are represented in base 10.
for example ABCDE in base n would be represented as $(A$x$n^{4})+(B$x$n^{3})+(C$x$n^{2})+(D$x$n^{1})+(E$x$n^{0})$.
so you know that if it's in base b-1, you'll have something like
$(A$x$(b-1))+B$ and then it'll be $AB_{b-1}$
lalala =] what if it was $(b-1)(b-1)+$ something[/hide]


trigonometry
function
geometry
angle bisector
trig identities
Law of Cosines

[quote="unimpossible"]i've been trying but i can't find a proof from the angle bisector theorem AD^2=AB*AC-BD*CD with AD the angle bisector.

i've been using law of cosines but i can't find a way to cancel out the cosines.

thanks for posting.[/quote]

[hide]Given $ \triangle{ABC}$ with angle bisector of $ \angle{B}$ being $ BD$, draw a line parallel to $ BD$ through $ A$. Now extend $ BC$ so that it intersects the parallel line through $ A$ to create a new $ \triangle{ACF}$ .  Now we know that $ \angle{ABD}\equal{}\angle{CBD}$ and by alternalte interior angles, and corresponding angles we know that $ \angle{AFC}\equal{}\angle{CBD}$ and $ \angle{BAF}\equal{}\angle{ABD}$ so $ \angle{BAF}\equal{}\angle{AFC}$. So its an isoscles triangle, so $ FB\equal{}AB$. Now since we know that $ AF\parallel BD$ $ \frac{FB}{BC}\equal{}\frac{AD}{CD}$ and by substitution $ \frac{AB}{BC}\equal{}\frac{AD}{CD}$[/hide][/hide]

We have cos(CDA)=-cos(BDA), pluging for those cos functions from law of cosines yields the result (using AB:AC=DB:DC).

[quote="unimpossible"][quote="unimpossible"]i've been trying but i can't find a proof from the angle bisector theorem AD^2=AB*AC-BD*CD with AD the angle bisector.

i've been using law of cosines but i can't find a way to cancel out the cosines.

thanks for posting.[/quote]

i need a solution to AD^2=AB*AC-BD*CD

thanks.[/quote]

why use laws of cosines, you can just use stewarts..

[hide="proof"]
let $ AD$ be the angel bisector .
$ \cos ADB \equal{}\minus{}\cos ADC$
this gives by the cosine rule
$ \frac{AD^{2}\plus{}BD^{2}\minus{}c^{2}}{2 AD .BD}\equal{}\minus{}\frac{AD^{2}\plus{}CD^{2}\minus{}b^{2}}{2 CD .AD}$
but $ \frac{BD}{CD}\equal{}\frac{c}{b}$
so $ AD^{2}\plus{}BD^{2}\minus{}c^{2}\equal{}\minus{}(AD^{2}\plus{}CD^{2}\minus{}b^{2})\frac{c}{b}$ *
also $ \frac{(b\plus{}c)BD}{CD}\equal{}\frac{c(b\plus{}c)}{b}\equal{}\frac{bc}{c^{2}}\plus{}c\equal{}\frac{b BD^{2}}{CD^{2}}\plus{}c$
this gives $ \frac{bBD^{2}\plus{}cCD^{2}}{b\plus{}c}\equal{}BD.CD$ **
simplifying *
gives
$ (b\plus{}c)AD^{2}\plus{}(b\plus{}c)\frac{(bBD^{2}\plus{}cCD^{2})}{b\plus{}c}\equal{}bc(b\plus{}c)$
using ** we get
$ AD^{2}\equal{}bc\minus{}BD.CD$[/hide]


[quote="Moubinool"]
a(n+1)= a(n) + 1 -n/(n+1)! [/quote]
Ok, the correct relation is a(n+1)=a(n)+1/(n+1)!
thanks Lagrangia


ohhh... So now I'm a zombie? But I feel normal  :D

AAAAAAA I don't know how I'm typing this since I can't even think.

[quote="noneoftheabove"]so what was the point of that site[/quote]

It's a massivly multiplayer game thing.

that site was weird
[url=http://kevan.org/brain.cgi?math92]my proof for fermats last theorem[/url]

Change your title to something more appealing, such as "my proof for fermat's last theorem."

I have eatten 10 brains :rotfl: I'm full! :D

Since we all like posting links to sites check out this graphiti site while your having your brains eaten...

[url=http://kevan.org/brain.cgi?Black_stripe*36]Graphiti[/url]

Yeah, I really don't know how to spell graphiti, graffiti.... It has funny jokes all the same.


Find all integer solutions of \[ x^{4}\plus{}y^{2}\equal{}z^{4}\]

Already discussed before (especially not new)...

Easy to check the equation not have solution (x,y,z) where all $ x,y,z$ different from 0
Case: x=0
$ y^{2}\equal{} z^{4}$
Case 2 y=0
$ x^{4}\equal{} y^{4}$

I think it appear in DIOPHANTE EQUATION,A book of Dr.ANDREESCU

Why is this very famous problem in an Olympiad Cotest? :huh:

We will prove that the only solutions are ones where $xyz = 0$ (it's easy to solve the problem in this case).

Suppose, for contradiction, that there existed a solution where $x, y, z > 0$. Take the one where $z$ is minimized.

Then we clearly have that $x, y, z$ are relatively prime. We consider two cases.

[b] Case 1. [/b] $z, x$ are both odd.

Then we know that $z^2 - x^2 = 2a^2, z^2 + x^2 = 2b^2$ for some $a, b \in \mathbb{N}.$ This means that $z^2 = a^2 + b^2$ and $x^2 = b^2 - a^2$, so that $a^4 + (zx)^2 = b^4.$ However, $b < z$ and so this contradicts the minimality of $z.$

[b] Case 2. [/b] $z, x$ are not both odd.

Then we must have that $z$ is odd and $x$ is even.

We have two subcases here. In the first case, there are integers $a, b \in \mathbb{N}$ so that $a^2 + b^2 = z^2$ and $a^2 - b^2 = x^2$. Then we have $a^4 - b^4 = (xz)^2$ and $a < z$, contradicting the minimality of $z$.

Else there are integers $a, b \in \mathbb{N}$ so that $a^2 + b^2 = z^2$ and $2ab = x^2.$ Note that $\gcd(a, b) = 1$.

The first equation yields that there are integers $t, u \in \mathbb{N}$ so that $\{a, b\} = \{t^2 - u^2, 2tu\}$. This means that $x^2 = 2 \cdot (t^2 - u^2) \cdot 2tu$, and so $tu (t^2 - u^2)$ is a square. We have that $t, u, t^2 - u^2$ are pairwise relatively prime and so they're all squares. Hence, $(\sqrt{t})^4 - (\sqrt{u})^4$ is a square. As $\sqrt{t} < z$, this contradicts the minimality of $z$.

We've derived a contradiction in all cases, and so our assumption that there is a solution where $xyz \neq 0$ was incorrect. So all solutions are those where $xyz = 0$ (easy to find).

$\square$


inequalities
symmetry
inequalities unsolved

By AM_GM we have

$ \frac {x_1^2}{x_2}\ge 2x_1 \minus{} x_2$

OK could we say $ \left(\minus{}1,2,0\right) \succ \left(1,0,0\right)$  :?:

[quote="hasan4444"]Prove that
\[ \frac {x_1^2}{x_2} \plus{} \frac {x_2^2}{x_3} \plus{} \ldots \plus{} \frac {x_n^2}{x_1} \geq x_1 \plus{} x_2 \plus{} \ldots \plus{} x_n\]
for all positive numbers $ x_1,x_2,\ldots,x_n$[/quote]

Nobody seemed to solve it as the title says. $ x^{2}_{1}\frac {1}{x_{2}} \plus{} x^{2}_{2}\frac {1}{x_{3}} \plus{} ... \plus{} x^{2}_{n}\frac {1}{x_{1}}\geq x^{2}_{1}\frac {1}{x_{1}} \plus{} x^{2}_{2}\frac {1}{x_{2}} \plus{} ... \plus{} x^{2}_{n}\frac {1}{x_{n}} \equal{} x_{1} \plus{} x_{2} \plus{} ... \plus{} x_{n}$

To Hasan->> How do you come to this conclusion? $ \left( \minus{} 1,2,0\right)\succ\left(1,0,0\right)$. It's false but...

Thanks "Γιώργος" I think my post was completely false I forget about that symmetry thing.


calculus
integration
trigonometry
calculus computations

Calculate

\[\int^{\frac{3\pi}{2}}_{\frac{\pi}{2}} \left|\left(\frac{2}{x^3}+\frac{1}{x}\right)\sin x\right|dx\]

$\left(-\frac1{x^2}\sin x-\frac 1x\cos x\right)'=\left(\frac 2{x^3}+\frac 1x\right)\sin x$. The rest is routine.


Let $ \left(\begin{array}{c} U\\ V\\ W\end{array}\right)=\left(\begin{array}{ccc}a&b&c\\
b&c&a\\
c&a&b\end{array}\right)\left(\begin{array}{c}x\\
y\\
z\end{array}\right) $ and denote $ E(\alpha,\beta,\gamma)=\alpha^3+\beta^3+\gamma^3-3\alpha\beta\gamma. $
Prove that $ E(U,V,W)=E(a,b,c)E(x,y,z)\; . $
 [Apply this equality in order to obtain some identities/inequalities in the  geometry of triangle.]

$U=ax+by+cz$
$V=bx+cy+az$
$W=cx+ay+bz$

$E(U,V,W)=a^3x^3+b^3y^3+c^3z^3+b^3x^3+c^3y^3+a^3z^3+c^3x^3+
a^3y^3+b^3z^3+9abcxyz-3abcx^3-3abcy^3-3abcz^3-3a^3xyz-3b^3xyz-3c^3xyz$

Expand $E(a,b,c)E(x,y,z)$ gives the same result

------------------------------------
Taking the determinant of

$\left(\begin{array}{ccc}U&V&W\\V&W&U\\W&U&V\end{array}\right)=
\left(\begin{array}{ccc}a&b&c\\b&c&a\\c&a&b\end{array}\right)\left(\begin{array}{ccc}x&y&z\\y&z&x\\z&x&y\end{array}\right)$

Gives also the result


hello
please help explaining it step by step
thanx

Solve the recurrence relation

an = an−1 + 20an−2

with a0 = −1 and a1 = 13.

Woah, how did you see that you were supposed to make that substitution?

[quote="cincodemayo5590"][hide="Or...."]
You can use the [u]characteristic polynomial[/u]. (Look it up; I don't have the time to write up an explanation right now.) :)
\[a_{n+2}-a_{n+1}-20 a_{n}= 0 \implies x^{2}-x-20=0 \implies x_{1}=5, \ x_{2}=-4 \]

\[\implies a_{n}= A \cdot 5^{n}+B \cdot (-4)^{n}, \ \{ A,B \}\subset \mathbb{R}, \ n \in \mathbb{Z^{*}}\]
Now, the initial terms give a linear system in $A$ and $B$, which can be solved to obtain the explicit formula for the recurrence, starting with $a_{0}$:
\[\boxed{ a_{n}= 5^{n}-2 \cdot (-4)^{n}}\]
[/hide][/quote]

Hmmm Interesting. I looked up Characteristic Polynomial and I read about it and understood the gist of it. I also understand Eigenvalue and Eigenvectors etc.

Can you please explain how this method helps to find the solution for this sequence? I sort of have an idea of the reasoning but not sure of it.

[quote="me@home"][quote="cincodemayo5590"][hide="Or...."]
You can use the [u]characteristic polynomial[/u]. (Look it up; I don't have the time to write up an explanation right now.) :)
\[a_{n+2}-a_{n+1}-20 a_{n}= 0 \implies x^{2}-x-20=0 \implies x_{1}=5, \ x_{2}=-4 \]

\[\implies a_{n}= A \cdot 5^{n}+B \cdot (-4)^{n}, \ \{ A,B \}\subset \mathbb{R}, \ n \in \mathbb{Z^{*}}\]
Now, the initial terms give a linear system in $A$ and $B$, which can be solved to obtain the explicit formula for the recurrence, starting with $a_{0}$:
\[\boxed{ a_{n}= 5^{n}-2 \cdot (-4)^{n}}\]
[/hide][/quote]

That's exactly what Godel used, except using it to facilitate his own steps in the proof in creating an elementary understandable proof, instead of just citing it as you have :wink:[/quote]

I understand that. I am just offering a quicker (with respect to work) solution for those who are interested, i.e. one that would give a result in less time, on a non-proof contest or the like. I actually prefer [b]Kurt Gödel[/b]'s solution to my own; but I know that the method I have used will give a result in under a minute or two on the AMC. :)

Find $(\alpha ,\ \beta)$ such that $a_{n}=a_{n-1}+20a_{n-2}\Longleftrightarrow a_{n}-\alpha a_{n-1}=\beta (a_{n-1}-\alpha a_{n-2}).$
Comparing the coefficients of $a_{n-1},\ a_{n-2}$ gives $\alpha+\beta =1$ and $\alpha \beta =-20,$ yielding $(\alpha ,\ \beta)=(-5,\ 4),\ (4,\-5).$

[quote="hyderman"]thnax for help.... however i could not understnad the characteristic polynomial. can some one explain this in steps ...please this type of questions will be in the test and i need some help thanx[/quote]

[url=http://www.cs.cmu.edu/~adamchik/21-127/lectures/recursion_4_print.pdf]This[/url] PDF, by Victor Adamchik, will hopefully help.

hello
please check my answer and let me know if there are any errors and correct it......

Solve the recurrence relation

an = an−1 + 20an−2

with a0 = −1 and a1 = 13.

Step – I : Find Characteristic equation and solve for its roots:

an = an−1 + 20an−2 ------equation (0)

First of all make corresponding characteristic equation of equation 0 and find roots of that equation

let
an= r2
an-1 =r
an-2 =r0

Substituting above values in equation (0)
r2 =r + 20

r2-r-20=0 is the characteristic equation.

r2-5r+4r-20=0
r(r-5)+4(r-5)=0
(r-5 ) (r+4)=0

r1 = 5 , r2 = -4 are the roots of this characteristic equation

Step– II : Find Constants’ value using Initial Condition

So solution of this polynomial would be

an = c1 r1n + c2 r2n ------------------equation (1)

Where c1 & c2 are some arbitrary constants .

Now we have to find values of c1 and c2 using Initial Condition.

As a0 = −1 and a1 = 13.

So put n=0 in equation (1)

a0= c1 r10 + c2 r20
a0= c1 + c2
-1 = c1 + c2 ---------equation (2)

Now put n=1 in equation (1)
a1 = c1 r11 + c2 r21

13= c1 (5)1 + c2 (-4)1 --------------equation (3)

Multiply equation (2) by 4

-4= 4c1 + 4c2 -------------------------------------------------equation (4)

Add equation (3) and (4) to get the values of c1 & c2.
-4= 4c1 + 4c2
13= 5c1 - 4c2

9 = 9c1
c1=1

put c1=1 in equation (2)

c2=-2

Step-III ..Substitute values of c1 , c2 , r1 and r2 to get the final solution

so substituting these values in equation
an = 1.5n + (-2) 4n

an = 5n – (2).4n is the final solution of this recurrence relation


Find all ordered triple of primes$ (p,q,r)$ such that

$ p|q^r \plus{} 1\ \ ,\ \ q|r^p \plus{} 1\ \ ,\ \ r|p^q \plus{} 1$

USA TST 2002  :P

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=357351#357351[/url]


There are a lot of pirates on their island. Every pirate has the same habits which are
1. They don't want to die.
2. They love money.
3. They like to see the others die.
Habit no.1 is more important than no.2, e.g. if a pirate has to choose between getting money but being killed and getting no money but being alive, they will choose the second one. Assume that all of them are so clever that they will choose the best way according to their habits. If there are n pirates, each of them has his own number in {1,2,3,...,n} which are all different for each one. To clarify, there are P1,P2,P3,...,Pn on the island if there are n pirates. The pirate with more number has more power.

Whenever they find a treasure(usually 100 coins), the one with the most power will decide how to divide 100 coins, e.g. n=3, no.3 gives 30, 30, 40 coins for number 1,2,3(himself). And everyone(including himself) will vote whether he agree with that idea or not. If there are A HALF OR MORE THAN agreement, that idea from the most powerful pirate will be accapt. But if there are LESS THAN A HALF agreement, the most powerful pirate will be killed, and the second most will be the one who decides how to divide that 100 coins. And so on until there are a half or more than agreement.

There are n pirate on the island. The question is...(Do it one by one)
[hide="Problem1"]n=2  How will pirate no.2 divide 100 coins? and How much will he get?[/hide][hide="Problem2"]n=3 How will pirate no.3 divide 100 coins? and How much will he get?[/hide][hide="Problem3"]n=4 How will pirate no.4 divide 100 coins? and How much will he get?[/hide][hide="Problem4"]n=5 How will pirate no.5 divide 100 coins? and How much will he get?[/hide][hide="Problem5"]n=100 How will pirate no.100 divide 100 coins? and How much will he get? You can try 6,7,8,.. first if you can't do 100.[/hide][hide="Problem6"]n=200 How will pirate no.200 divide 100 coins? and How much will he get?[/hide][hide="Problem7"]n=201 Will no.201 die?[/hide][hide="Problem8"]n=202 Will no.202 die?[/hide][hide="Problem9"]n=203 Will no.203 die?[/hide][hide="Problem10"]n=204 Will no.204 die?[/hide][hide="Problem11"]n=205 Will no.205 die?[/hide][hide="Problem12"]n=206 Will no.206 die?[/hide][hide="Problem13"]n=207 Will no.207 die?[/hide][hide="Problem14"]n=208 Will no.208 die?[/hide][hide="Problem15"]Who's the next one who can survive?[/hide]

[hide="n=2"]The pirate #2 suggests 100 coins for himself and no coins for #1. He votes his suggestion, so he has the 50 percent[/hide]
[hide="n=3"]The pirate #2 knows that, if #3 die, he can take 100 coins. So he is not going to vote the suggestion of #3.

#3 knows this, so he suggests 99 coins for himself and 1 coin for #1
#1 prefers to get 1 instead of 0 coins, so he will vote #3's suggestion
[/hide]
[hide="n=4"]#3 is not going to vote #4's suggestion
So #4 wants to get a vote from #1 or #2.
It's better for #4 to offer 1 coin to #2 and keep 99 coins for himself.

Indeed, if he give 1 coin to #1, then:
#1 knows that he can take (such or else) 1 coin from #3, so he won't vote #4's suggestion, because he wants to see #4 die
So, if #4 wants the vote of #1 he must give to him #2 coins. [/hide]
[hide="n=5"]#4 is not going to vote #5's suggestion.
#5 wants 2 votes (except of his vote). It's better for him to offer coins to the pirates who took less money in the previous step.
#1,#3 took nothing, so he offers 1 coin to each and keeps 98 for himself[/hide]
[hide="n>5"]If we continue this procedure we'll see that
the pirate with number $2k+1$ gives 1 coin to each the k odd numbered pirates $1,3,5, \cdots , 2k-1$. He keeps $100-k$ for himself
the pirate with number $2k+2$ gives 1 coin to each the k even numbered pirates $2,4,6, \cdots , 2k$. He keeps $100-k$ for himself

The above can be done if $k\leq 100 \Rightarrow $
$2k+1\leq 201$ and
$2k+2\leq 202$

Notice that the pirate #201 keeps no money for himself, but he stays alive.
The same for #202

Now, the pirate #203 wants the vote of the odd-numbered pirates $1,3, \cdots ,199,201$.
They are 101 persons, but he can offer to them only 100 coins.

So, I think that every pirate with number $n\geq 203$ will die[/hide]

[quote="pontios"][hide="n=2"]The pirate #2 suggests 100 coins for himself and no coins for #1. He votes his suggestion, so he has the 50 percent[/hide]
[hide="n=3"]The pirate #2 knows that, if #3 die, he can take 100 coins. So he is not going to vote the suggestion of #3.

#3 knows this, so he suggests 99 coins for himself and 1 coin for #1
#1 prefers to get 1 instead of 0 coins, so he will vote #3's suggestion
[/hide]
[hide="n=4"]#3 is not going to vote #4's suggestion
So #4 wants to get a vote from #1 or #2.
It's better for #4 to offer 1 coin to #2 and keep 99 coins for himself.

Indeed, if he give 1 coin to #1, then:
#1 knows that he can take (such or else) 1 coin from #3, so he won't vote #4's suggestion, because he wants to see #4 die
So, if #4 wants the vote of #1 he must give to him #2 coins. [/hide]
[hide="n=5"]#4 is not going to vote #5's suggestion.
#5 wants 2 votes (except of his vote). It's better for him to offer coins to the pirates who took less money in the previous step.
#1,#3 took nothing, so he offers 1 coin to each and keeps 98 for himself[/hide]
[hide="n>5"]If we continue this procedure we'll see that
the pirate with number $2k+1$ gives 1 coin to each the k odd numbered pirates $1,3,5, \cdots , 2k-1$. He keeps $100-k$ for himself
the pirate with number $2k+2$ gives 1 coin to each the k even numbered pirates $2,4,6, \cdots , 2k$. He keeps $100-k$ for himself

The above can be done if $k\leq 100 \Rightarrow$
$2k+1\leq 201$ and
$2k+2\leq 202$

Notice that the pirate #201 keeps no money for himself, but he stays alive.
The same for #202

Now, the pirate #203 wants the vote of the odd-numbered pirates $1,3, \cdots ,199,201$.
They are 101 persons, but he can offer to them only 100 coins.

So, I think that every pirate with number $n\geq 203$ will die[/hide][/quote][hide]All are correct except one. $204$ will not die like ZataX said.[/hide]

Could someone explain why? And what happens with the next pirates?

Note that number $203$ will vote for $204$ even if he gives him nothing, because if $204$ is killed, $203$ knows he will be killed, too.

Nice.

But will the pirates 2,4,6,... vote for 204?

They know that, if $204$ is killed $\Rightarrow  203$ is killed $\Rightarrow  202$ gives them 1 coin to each.

and they like very much to see 203,204 been killed!!

Yes!  :)
Who's the next one who survives?

[quote="ZetaX"]He will give the coins to number $1,3,5,7,...$.[/quote]

I see now  :)

The pirates $1,3,5,..$ know that, if don't vote for him, then $202$ will give the coins to the pirates $2,4,6,...$

It's too complicated after 200  :maybe:

I can't find who's the next  :(

I think I got it now!

204 will survive. He gives coins to 1,3,5,...,199.

205,206,207 will die (3 persons)
208 will survive
he gets 100 votes (from those who get the coins) + 1 (from himself) + 3 (from those who don't want to die)
But he gives coins to 2,4,...,200

209,...,215 will die (7 persons)
216 will survive
he gets 100 votes (from those who get the coins) + 1 (from himself) + 7 (from those who don't want to die)
He gives coins to 1,3,...,199

217,...,231 will die (15 persons)
232 will survive
he gets 100 votes (from those who get the coins) + 1 (from himself) + 15 (from those who don't want to die)
He gives coins to 2,4,...,200

and so on..


calculus
integration
algebra
partial fractions
calculus computations

$ \text{(i)}\quad \int\;\sqrt {\; \sqrt {\; x \;}  \plus{}x \;} \;\textbf dx$

$ \text{(ii)}\quad \int\;\sqrt {\; \sqrt {\; \sqrt {\; x \;}  \plus{}x \;}  \plus{}x \;} \;\textbf dx$

 :D

[quote="misan"]$ \text{(i)}\quad \int\;\sqrt {\; \sqrt {\; x \;} + x \;} \;\textbf dx$[/quote]

\begin{eqnarray*}
\int \sqrt{\sqrt{x}+x}dx
& = & \int \left( \frac{1}{16 \sqrt{x} \sqrt{\sqrt{x}+x}} + \frac{-1 + 16x + 16x^{3/2}}{16 \sqrt{x} \sqrt{\sqrt{x}+x}} \right) dx \\
& = & \frac{\sqrt{\sqrt{x}+x} \ \text{Arc}\sinh{x^{1/4}}}{4\sqrt{1+\sqrt{x}} \ x^{1/4}} - \frac{1}{4} \sqrt{\sqrt{x}+x} + \frac{1}{6} \sqrt{x} \sqrt{\sqrt{x}+x} + \frac{2}{3} x \sqrt{\sqrt{x}+x} \\
\end{eqnarray*}

[b]Kouichi Nakagawa[/b], how did you get this "partial fractions" :huh: :?:

The first one I would do that way:
$ \int \sqrt {\sqrt {x} \plus{} x} \, \textbf d x \equal{} \int x^{1/4} \left( 1 \plus{} x^{1/2} \right)^{1/2} \, \textbf d x$
and substitute $ t \equal{} \sqrt {\frac {1 \plus{} \sqrt {x}}{\sqrt {x}}} \, \ldots$


geometry circumcircle rectangle trapezoid rhombus trigonometry perimeter

I'm sorry if i can't explain this problem better in English.
Source: Indonesian province selection

A quadrilateral  $ABCD$ with $a$ as the largest side, and the circumradius of the triangle $ABD$ is 1. Determine the smallest value of $a$ and when it is occur!

The smallest value of $a$ is when $ABCD$ is a square and so all sides are equally long.
In this case $\Delta ABD$ is rectangle and $a = \sqrt{2}$

mmh... it should be wrong  :?

Well, shouldn't it be $2$, all the sides can't be equal and the largest side can't be equal to another side based on his wording, thus, the smallest it can be is a trapezoid like figure with the largest base being the diameter of the circle.

[quote="krustyteklown"]I'm thinking we can take a to be infinitely small.
Here's how I'd construct it:

Draw a circle with radius 1.

We pick 3 points on the circle such that the "middle" point is the same distance from both other points. This is triangle ABD. It's an isosceles. To complete the quadrilateral, we just complete the rhombus, with all sides "a".
Since these 3 points can be indefinitely close together on the circle, a is indefinitely small?[/quote]

Actually, I think you are right because as I said before that if you take a trapezoid for example, you can keep making the top base smaller, and the sides farther and farther, thus the bottom base (largest side) can keep getting smaller, so you can't calculate it's length.

if you draw a triangle in a circle, you can have an obtuse triangle where all the sides approach 0, the one side that gets bigger while the others get smaller can be the diagonal BD, say $\angle A = 180 - 10^{-15}$, them BD = is essentially 0, then the other sides would be tiny as well

the answer is $a=\sqrt{2}$ and it occurs when all sides are equal.
I use cosine law. Here is my effort:
$a^2=2-2cos(A)$
$b^2=2-2cos(B)$
$c^2=2-2cos(C)$
$d^2=2-2cos(D)$
to minimize $a$ , which are the largest side, is equivalen to minimize the Perimeter of quadrilateral or minimize $a+b+c+d$ ....
and so we have to minimize
$\sqrt{2-2cos(A)}+\sqrt{2-2cos(B)}+\sqrt{2-2cos(C)}+\sqrt{2-2cos(D)}$

Hm, maybe the question was worded wierd then, cuz I'm pretty sure my construction works, same with Altheman's


[url=http://imageshack.us][img]http://img474.imageshack.us/img474/6008/determinantub8.gif[/img][/url]
Shot at 2007-08-14

This is just straightforward calculation. You (probably) can't reduce it to a form where the answers come out nicely. You can tell because the expected results are not given as arithmetic combinations, which all determinants are.


There are two paticles with equal mass and charge .they are thrown with an horizontal speed $ v$ from heights $ h_{1}$ and $ h_{2}$ from the earth along the same vertical .The particle one reaches ground at a distance $ l$ from the vertical.Find the height $ H_{2}$ at which the second particle is when the first one reaches the ground.Neglect air drag and any charge induced on the ground.

The system is conservative :)

$ m g  h_1 \plus{} \frac{1}{2} m v^2 \plus{} K \frac{q^2}{h_2\minus{}h_1} \equal{}
  m g 0 \plus{} \frac{1}{2} m {V_1}^2 \plus{} K \frac{q^2}{H_2}$

$ m g  h_2 \plus{} \frac{1}{2} m v^2 \plus{} K \frac{q^2}{h_2\minus{}h_1} \equal{}
  m g H_2 \plus{} \frac{1}{2} m {V_2}^2 \plus{} K \frac{q^2}{H_2}$

Solving for $ H_2,v$

$ H_2 \equal{} \frac{(V_1\minus{}V_2)(V_1\plus{}V_2)}{2 g}\plus{}h_2\minus{}h_1$

I don't know any good way to determine the final velocity of each particle. To Solve the euler equation for this system will be a little bit difficult.

[hide="hint"]
consider the motion of the centre of mass [/hide]

This isn't a hint, is the solution.

Newton is really a good boy

The trajectory of CM is a parabola.

The equation of the movement is:

$ y_1\plus{}y_2 \equal{} \frac{1}{2} g t^2 \plus{} \frac{1}{2}( h_1\plus{}h_2)$
$ x \equal{} v t$

For the problem $ x \equal{} l$ and $ y_1 \equal{} 0$

And doone. I was trying to use Lagrange for this problem,   :rotfl:

everything right except that it will be $ \frac{\minus{}gt^{2}}{2}$

$ g < 0$ :)

I'm not the type of physician that write $ \minus{}g$  :D

For me is a retrocesse use only "positive variables" 'cause we have the negative ones.

So don't say it is wrong, I never assumed $ g>0$, in the contex is clear that $ g<0$

if  u aren't that type of [b]physicist[/b] (not [b]physician[/b] which means a doctor) then please use $ \vec{g}$ instead   :wink:

$ \mathbf{g}$ is a more fashioned way to express this :) (but take time, and is just a retrocesse

perhaps, you are wrong gallant pardersi, you said that my answer was wrong, when it was just a misunderstand of yourself.

just apologize and don’t make a fuss about that

[quote="EulerIntegral"]
perhaps, you are wrong gallant pardersi, you said that my answer was wrong, when it was just a misunderstand of yourself.[/quote]
i never said the word wrong. :huh:
[quote="pardesi"]everything right except that it will be $ \frac { \minus{} gt^{2}}{2}$[/quote]

[quote="EulerIntegral"]just apologize and don’t make a fuss about that[/quote]
apologize  for what????? pointing out a mistake.
Either u should have written in the vector form or since u think ur a ...physiciAN  :rotfl:  u should have explained the meaning of the "new" symbols u r using and please stop asking people to do what they are supposed not to  :ninja:

[quote="Immanuel Bonfils"]Eu acho que o erro e seu[/quote]
immanuel i beleive most people don't understand that please can u translate that

a thing that inst right, by induction (fermat) is wrong, QED

I used $ g<0$ and any kid can see that $ g<0$.

Is a OBVIOUSLY thing, the error was your to think wrongly that I used $ g>0$

Just like when i put the electric field $ E \equal{} \frac{q^2}{r^2}$ it's clear that $ E$ is normalized.

First - One can't just put g in a vector form on an energy expression (unless in a scalar product or something of this kind...); it'll mess out scalars with vectors.

Second - The "retrocession using only positive variables" doesn't applies to g, a physical constant; it is definitely positive.

Third - It shouldn't be $ \;\;{{y_1 \plus{} y_2}\over 2}\;\;$ ?

Note: I've used  Portuguese cause it seems EulerIntegral could understand better...

[quote="EulerIntegral"]a thing that inst right, by induction (fermat) is wrong, QED[/quote]
nice proof  :P

[quote="EulerIntegral"]I used $ g < 0$ and any kid can see that $ g < 0$.

Is a OBVIOUSLY thing, the error was your to think wrongly that I used $ g > 0$[/quote]
sorry i am not a kid nor  am i kidding . :huh:  i haven't seen a book where $ g < 0$ and probably everywhere in this world in physics $ g \equal{} 9.8 ms^{ \minus{} 2}$  and nothing else.just to have ur answer right right u cannot change meanings. :noo:

[quote="Immanuel Bonfils"]First - One can't just put g in a vector form on an energy expression (unless in a scalar product or something of this kind...); it'll mess out scalars with vectors.
[/quote]
that wasn't an energy conservation equation.it was the trajectory of the centre of mass.so no problem in putting vector sign through out
[quote="Immanuel Bonfils"]Note: I've used Portuguese cause it seems EulerIntegral could understand better...
[/quote]
ok

Let's Prove That $ g>0$ or $ g<$

$ \minus{}9.8 \equal{} (\minus{}1)(9.8)$

So the minus signal can be outside the constant or inside.

$ a\equal{}7$ and you anwser is $ \minus{}a$

this is the same that $ b\equal{}\minus{}7$ and you answer $ b$

Minus operator is a magical operator.

And for immanuel "O erro é meu é a ... Eu nunca assume que eu usei uma outra configuração


consider the function f(x)=x/3-1/2. if we were to graph this function, on what point would the x and y value be the same?

this is like the tenth time today that you are correct. :D

It helps for both of us to be online from 8am this morning.  :D

[quote="math92"]wow, don't you guys have a camp or something? I am gone from 7:45 to 5:00.
that's why 236! has a trillion posts i the last couple of days.[/quote]

I had 93 posts yesterday  :rotfl:


在这个forum上是否要用WORD公式编辑来写公式?在上面能够不用打开WORD就写的方法吗?比如内部提供公式编辑的图片?

:huh: 兄弟，快去学学LaTex编辑软件吧！好处多多。

可是我依然没有弄太明白怎么用!

用MathType打出来然后转换成代码就行了~

[quote="RichardGao"]用MathType打出来然后转换成代码就行了~[/quote]

转出来的全是“大”公式，即居中的。很多时候要将前后两个括改成美元符号

貌似···我直接上截图的···

[quote="DoctorRoyal"]貌似···我直接上截图的···[/quote]
上传截图可以，不过这个不方便引用。当然几何图形就另当别论了，这里是指公式。

貌似我得花上一段时间来研究下公式的语法···

这帖子居然还在首页上，距离我上次发帖都4年了啊

[quote="shobber"]有一年多没发帖了……

编辑公式很容易，只需要在公式代码两端加上 $ \$$ 就行了。比如在f(x)=ax^2+bx+c两边加上Dollar Sign, 就成了 $ f(x)\equal{}ax^2\plus{}bx\plus{}c$

剩下的就是多学习多练习啦！[/quote]

LS肯定是參加過什麽聯賽冬夏令營集訓隊這類

你说我么？我没有啊

[quote="colorfuldreams"]Ls肯定是參加過什麽聯賽冬夏令營集訓隊這類[/quote]

For learn about latex, see [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:About]LaTeX Guide[/url]. And, for upload the image use Attachment.

[quote="youarebad"]Thx, so the writer not online now ?? Sorry, i am use google translate to translate your word.[/quote]

becuse the post is at  Oct 12, 2007, 11:02 am, 4 years ago. I think if he was a high school student , and today he becomes to a university student

兴趣虽然还有，可是已经完全不会做题了……大学念了跟数学完全不搭的专业，以前学过的东西忘得一干二净了

[quote="colorfuldreams"][quote="shobber"]你说我么？我没有啊
[/quote]

難得對數學那麽有興趣。話說我見過不少大學生來玩高中競賽的。

ps , 這個論壇的 Olympiad Section 區的問題感覺跟競賽這類問題很不同地說。。。或者説比競賽難且計算量海量，競賽一般不會那麽考。感覺有集訓隊訓練題的味道。

ps . High School Pre-Olympiad (Ages 14+) 這個區的問題很貼近競賽區，不過有不少簡單的問題偶爾又會出現非常難的問題。有點像聯賽。

ps , 除了以上2個區，別的區我都沒去過。

ps . youarebad 是一個好人。(可惜樓主不知道已經還在不在了。)[/quote]

[quote="lssl"]:D  :D  :D
哈哈！shobber， 又见到你了，你好像已经很长时间没来Mathlinkers 了，你应该是这里最有名的Chinese mathlinker  :) 了！[/quote]

嘿，确实很久没来了！不过现在这里是不是彻底改成Art of Problem Solving了？Mathlinks彻底被吞并了？到底发生了什么呀


MATHCOUNTS
geometry
percent
ratio
3D geometry
parallelogram
LaTeX

I am thinking of starting the math chats back up.  If anyone would like to participate in one of these post here.  AIM seems to be the easiest way to do this since it seems that most ppl use it.  I am starting to put together some problems to go over.  The level of the problems I am getting are mathcounts to AMC 10.  The level is aimed at true beginners.  Other people can participate but don't expect this to go much past AMC 10 level.  Right now I am thinking about having it go for about an hour (it doesn't have to be exact but I am going to try to get enough problems for one hour).  AS always you can stay after and go over different problems etc.  As for the date and time I don't have an exact one yet but I am looking at later this week or on the weekend.  I wont make it during one of the Mock AMC's so don't worry about that.  I would like to hear some of your thoughts on time availability.

Sure, AIM screen name [u]Feiqi J1989[/u], anytime

I'd love to do one.  It would be especially helpful when I take the AMC 10 this year, and for my MATHCOUNTS team.  The only problem is that I don't have AIM, and my dad is crazy about me downloading stuff, but hopefully, I can get it.

My screenname will probably h_s_potter2002, it's usually like that for everything.

I can participate/help out! My IM is whitehorseking88.

Bringing them back sounds pretty good. My SN is caocoprez

I'd be interested aim sceen name- BHorseMath.

Oh yeah I forgot to give my sn. It is BaseballJoe2002.

I should be able to participate...should fun.  Screen name is lonelytraveler8.

Man, am I the only person on here who doesn't have AIM?  I feel so alone right now.

I don't think that you have to download anything to use AIM express  [url]http://www.aim.com/get_aim/express/aim_expr.adp?aolp=[/url]

For the time I'm thinking either friday, saturday, or sunday night.  If it is friday it would have to end before 7:30 because I have a class at that time.  I think I ll make a poll to see what is good.

joml88 is right, you don't need anything to use AIM express.
I'm not sure when I can do it this weekend because one of the days I'm going to be going to six flags and it's still undecided exactly what day that is.  But it'll be one of the three choices posted.

I'd really be happy to join a math chat!  s/n = KilliingInSleep

Note:  I voted for Friday night, but now that I think about it, Saturday night is better since there isn't MC Counting then.

I'd be glad to help, reaperofthegrave is my sn.

i'd like to join! i have 3 sn's.  first try suenified.  if that doesn't work, try pinkksunsets or xX caro0 Xx

I'd be willing to help/participate

AIM ID: hezgotzbeenz

so math is hard im a dumn person, and i really need help, what is the answer to 888z*59x-45y=78 ;)  ;)  ;)  ;)  :blush:  :maybe:

Depends on what x, y, and z are :P . Although I seriously doubt you did anything other than type random numbers o_O ...

I might b able to do it... depending on what time it is... I'll give u my sn and if I'm on at that sec, I can probably to it. My sn is: luver of rain
And also, if some of my friends wants to do it, can they come to ?

no answer to I hate genius2154 problem.  You can only set the answer up as an inequality and you should be able to set it up as a conjunction. (I think please correct me if I'm wrong) and yes it does depend on what x, y, and z are.

I'll do it for sure, but it's just that my dad kida blocked the pop-up thing for aim, and ummm.. i just can't seem to find a way to get the thing unblocked.
Soo.. umm if anyone would like to help mua out here, then pleaseee do!
Anyways my AIM sn is Kristy24seven, and my e-mail's me4everfun@yahoo.com.

If anyone wants to do a "freestyle" math chat this friday from about 5:30-6:30 EST (of course this will probably go later for those who want to stay) I ll be willing to do it.  What I mean by freestyle is that there won't really be much of a certain topic covered.  So if you want I can just get some old ARML, Mandelbrot, AHSME or AIME problems to discuss.  Of course, if anyone has any problems they would like to discuss that is fine also.

I apologize but I don't think I ll be able to do a mathchat tomorrow.

joml88, are you still too busy to be hosting math chats?

If you are, would someone start hosting math chats.

I really enjoyed them.

Silverfalcon is hosting one this Friday at 6 eastern - look at AMC 10 mathchat.

EST = GMT - 5:00

At least, I think so...

Me too.  I would also like to have a math chat.

I won't be here for a week. So, I'm sure Silverfalcon can manage things on his own for now. I'm skiing in Vermont.  :D  Happy Holidays!


Find all functions $f: \mathbb Z \to \mathbb Z$  such that \[f(x + y + f(y))=f(x) + 2y ,\] for all integers $x$ and $y$.

[quote="Vova-LFML"]Find all f (x + y + f(y))=f(x) + 2y , f : Z:Z, x,y : Z[/quote]

$ f(x\plus{}y\plus{}f(y))\equal{}f(x)\plus{}2y$ $ \implies$ $ Q(x,y,p)$ : $ f(x\plus{}p(y\plus{}f(y))\equal{}f(x)\plus{}2py$ $ \forall x, y, p\in\mathbb Z$

$ Q(0,x,1\plus{}f(1))$ : $ f((1\plus{}f(1))(x\plus{}f(x))\equal{}f(0)\plus{}2(1\plus{}f(1))x$
$ Q(0,1,x\plus{}f(x))$ : $ f((x\plus{}f(x))(1\plus{}f(1))\equal{}f(0)\plus{}2(x\plus{}f(x))$

And so $ 2(1\plus{}f(1))x\equal{}2(x\plus{}f(x))$

And so $ f(x)\equal{}f(1)x$

Putting back this value in the original equation, we get $ f(1)(f(1)\plus{}1)x\equal{}2x$ and so $ f(1)\equal{}1$ or $ f(1)\equal{}\minus{}2$

Hence the two solutions :
$ f(x)\equal{}x$
$ f(x)\equal{}\minus{}2x$


calculus
integration
geometry
rectangle
function
complex analysis

First of all, in what situations is it valid to evaluate an integral of the form given by Perron's formula by instead considering a rectangular contour which has its rightmost side along the line of integration? The line integral along the other three sides of the rectangle must evaluate to zero, right? So how does this work with the zeta function and the explicit formula for Chebyshev's psi function given by Riemann?

Secondly, if you have a complex function F(x) = g(x)/h(x), and you want to use the residue theorem on F(x) for a contour integral, I know you must consider the zeros of h(x); but, should you consider the singularities of h(x) at all? Technically, the function is not defined there, right? But would its residue there be 0 anyway, or what?

A (nonessential) singularity of $ h$ means that $ h$ maps the point to $ \infty$ with some well-defined multiplicity. So taking the reciprocal means you're going to have a zero of some multiplicity, which you don't need to worry about when considering residues, unless, of course, you have a singularity of $ g$ of a higher order at the same exact spot.

Unfortunately I don't know Perron's formula so I can't help with that. :(

I think I understand now; if the singularity is a removable one, the residue there is zero anyway, so it may be ignored.
Thanks, Mystic.

As for Perron's formula...

Well, the way I use it is this; if you have $ g(s) \equal{} \sum_{n\equal{}1}^{\infty} {{f(n)} \over {n^s}}$, then Perron's formula can be used to find an explicit formula for a finite sum of f(n) values for n not exceeding x. The only catch is that the answer you get is at a "halfway point" for integers x because it's trying to act like a step function.

For example...

Let $ A(x) \equal{} \sum_{n\equal{}1}^{x} f(n)$.

Let $ A^{\star}(x)$ be the function with "halfway points" at the integers to make it better behaved.

Then Perron's formula implies (because we chose s = 0) $ A^{\star}(x) \equal{} {1 \over {2\pi i}}\int_{c\minus{}i\infty}^{c\plus{}i\infty} g(z){{x^z} \over z}\, dz$.

You can choose the value of c you use, but you have to look at where the function is convergent.
http://en.wikipedia.org/wiki/Perron%27s_formula


Determine all rational u, v satisfying the following equation:

[tex]u^3-v^2+1 = 0[/tex]

I think it's already been proven that (2,3) is the only pair of consecutive square and cubic powers (in either direction)... :D

$a^3+1=b^2$
$(1+a)(1-a+a^2)=b^2$
Obviously $a,b>1$ (except the trivial solution $(a,b)=(0,1)$ so $a^2>a^2-a+1>(a-1)^2$ so $a^2-a+1$ can't be a perfect square. Results that $(a+1,a^2-a+1)=d>1$. From $d|a+1$ and $d|a^2-a+1$ results that $d|3$ which leads to $d=3$.
So $a\equiv 1(\mod 3)$ and $b\equiv 0(\mod 3)$.
Let $a=3x+1$ and $b=3y$.
 The equation is equivalent to
$x(3x^2-3x+1)=y^2$. Since $(x,3x^2-3x+1)$ results that both $x$ and $3x^2-3x+1$  are perfect squares.
Let $x=q^2\ge1$. If $q=1$ we get the solution $(a,b)=(2,3)$. If $q>1$ then $3q^4-3q^2+1=t^2>1$.
$3q^2(q^2-1)=t^2-1$ which leads to $3q^2|t^2-1$ (1).
$(3q^2-1)(q^2-1)=t^2-q^2$ which leads to $3q^2-1|3t^1-1$ (2).
From (1) and (2) we deduce that there exist $m,n>0$ s.t.
$\left\{\begin{array}{l}
t^2-1=3mq^2\\
3t^2-1=n(3q^2-1)\end{array}\right.\Leftrightarrow\left\{
\begin{array}{l}
3t^2-1=9mq^2+2\\
3t^2-1=3nq^2-n
\end{array}\right.\Rightarrow 9mq^2+2=3nq^2-n\Rightarrow n+2=3q^2(n+3m)$
 but $q>1$ and $n+3m>n+2$ (because $n,m>0$ so we can't have a solution for $q>1$.

The solutions remain $(a,b)\in\{(0,1),(2,3)\}$

Did I miss something or both answers only deal with integers instead of rational numbers? :?

Pierre.

[quote="pbornsztein"]Did I miss something or both answers only deal with integers instead of rational numbers? :? [/quote]
It is unimportant thing...
 :lol:  :lol:  :lol:  :lol:  :lol:

Did Ulysses realy meant rationals, or should it be replaced by integers?

I think, Ulysses meant rational numbers!

In fact, it doesn't change anything about the solutions, even if they omit $(-1,0)$.
But, it seems to be a bit harder :
Let $u = \frac p q, v = \frac a b$ where $\gcd(p,q)= \gcd (a,b) = 1$, and $b,q > 0$.
Clearing the denominators, it is easy from the equation to deduce that $b^2$ divides $q^3$ and that $q^3$ divides $b^2$, so that $q^3 = b^2 = r^6$ for some positive integer $r$, coprime with $a$ and $p$.
Thus, the equation becomes :
$a^2 + (-p)^3 = r^6$.

This last equation is quite well-known, but it would be too lengthy to solve it here, so that I give the appropriate link :
http://www.mathpages.com/home/kmath213.htm

Note that they have omitted the case $x = +/- k^2, y = 0, z = +/- k$ in their solutions.
Using that $r$ is coprime with $a$ and $b$ and is positive, it leads to the solutions $(-1,0),(0,1),(2,3)$.

Pierre.


In the math jam, Zuming said something about writing about a problem that you solved in the past, but I can't find anything about that in the application information on the website. Can someone give me some more information?

Thanks

Zuming was referring to the second prompt on the personal information form.

Complete application information is here: [url]http://www.awesomemath.org/application.shtml[/url].

The form in question is here: [url]http://www.awesomemath.org/application/Personal_Information_Form_electronic.doc[/url].

Hope this helps!

Zac Cox
Program Coordinator, AwesomeMath


Let $A=\{1,2,\ldots, 2006\}$. Find the maximal number of subsets of $A$ that can be chosen such that the intersection of any 2 such distinct subsets has 2004 elements.

it's ok that every subset has at least 2004 elements, and at most 2006 elemnts.

If we have subset with 2006 elemts, then the maximal number of subsets is 2.

If we don't have subset with 2006 elemnts:

1) If we don't have subset with 2004 elemnts maximal number is 2006. Subsets with 2005 elements.
2) we have subset with 2004 which doesn;t contain numbers $a$ and $b$. Now we can at most 2004 subsets with 2005 because it can't contain $a$ and $b$.

Hence 2006 is max


geometry
trapezoid
power of a point
radical axis
similar triangles
geometry proposed

Nice problem. :)

First, we'll prove a lemma:

Let $A, B$ be fixed points, and let $r_1, r_2$ be variable reals such that $r_1+r_2 \geq AB$. Let $C_1, C_2$ be circles with centers at $A, B$ and radii $r_1, r_2$ respectively. Let $P$ be one of the points of intersection of the circles, and let $X$ be a fixed angle. We prove that if the line through $P$ such that its angle with line $AB$ is $X$ intersects the circles at points $M, N$, then $MN$ is fixed.

To prove it, fix $r_2$, take $r_1'$ not equal to $r_1$ and take the circle with center $A$ and radius $r_1'$. Let $P', M', N'$ be defined analogously. Then it's easy to show that $MPP'M'$ is an isosceles trapezoid and the same for $PNN'P'$, and thus $MNN'M'$ is a parallelogram. That's it.

Now, back to the problem. Let the circle centers be $O_1, O_2$. The angle $PAQ$ is semi-inscribed in both circles, so it is equal to both $AMQ$ and $ANP$. Then the angles $AMQ, ANP$ are equal. Thus angles $PBA, QBA$ are equal, because they are supplementary of $AMQ, ANP$. Then substracting equal angles $MBQ, NBP$, $NBA=MBA$. So the bisector of angle $MBN$ is the radical axis of $C_1, C_2$.

Now, by applying the lemma, $PM, QN$ are equal to the corresponding segments we get when we take $r_1=O_1O_2-r_2$. But then since angles $NBA=MBA$, the segments are symmetric and thus equal.

It's an easy problem :D
It's easy to see that triangles $ APB$ and $ ABQ$ are similar
then : $ PA/QA=PB/AB=AB/BQ$
so $PA^2/QA^2=PB/QB$
But $PB.PM=PA^2$ and $QB.QN=QA^2$
So$PM=QN$ QED

[quote="marko avila"]Suppose that the circles $C_1$  and $C_2$ of different radii intersect in the points A and B . The tangents to these circles from A intersect $C_1$  and $C_2$  at the points P and Q, respectively. Extend PB  and QB  until they intersect  $C_2$  and $C_1$ in M  and N respectivly. prove that PM=QN.[/quote]

The problem is really straightforward:

Using directed angles modulo 180°, we have < AQB = < AMB, since the points A, B, Q and M lie on one circle (namely, on the circle $C_2$). In other words, < AQN = < AMP. Similarly, < ANQ = < APM. Thus, the triangles AQN and AMP are similar.

Since the line AP is the tangent to the circle $C_2$ at the point A, we have < (AP; AB) = < AQB by the tangent-chordal angle theorem. Similarly, < (AQ; AB) = < APB. Thus,

  < ABP = < (AB; BP) = < (AP; BP) - < (AP; AB) = < APB - < (AP; AB)
            = < (AQ; AB) - < AQB = < (AQ; AB) - < (AQ; BQ) = < (BQ; AB) = < NBA.

Since the points A, B, P and N lie on one circle (namely, on the circle $C_1$), we have < ABP = < ANP and < NBA = < NPA; thus, this becomes < ANP = < NPA. Thus, the triangle NAP is isosceles with AP = AN. Hence, the similar triangles AQN and AMP must be congruent. Thus, PM = QN. Problem solved.

  darij

[quote="marko avila"]i just have a question. i solved this problem but im not sure if its ok. what i proved is the following:

triangles APN and AQM  are similar
triangle APN  is isoceles
hence triangle AQM is also isoceles
angleNAQ=anglePAM
hence since having two corresponding sides equal and the angle between also equal we have
APM and  ANQ are congruent so were done. im correct right darij??? its just that in your proof you avoid proving the isoceles and similar triangles APN  and AQM SO IM HAVING SOME DOUBTS :blush:[/quote]

Hmm, how exactly do you prove that the triangles APN and AQM are similar, and, more importantly, that the triangle APN is isosceles? I mean, I don't see any really [i]obvious[/i] reason for that, but maybe there is one and I am just blind. However, if you prove this using an angle chase, then your solution does not differ really much from mine.

  Darij

well here is my proof :
construct lines AB, AM and AN . so we have anglePBN=angleQBM=angleQAM=anglePAN= :theta: .  also angleAQB=anglePAB=  :varepsilon: because AP is a tangent to C2. analogously angleBAM=angleANB= :beta:  because AQ  is tangent to C1. now by pure angle chase calculated with the angles already given we hace anglePBA= :beta: + :varepsilon: now angleANP=anglePBA= :beta: + :varepsilon:. angleAPN=angleAPB+angleBPN= :beta: + :varepsilon:   because they subtend the same arcs. also angleAQN= :beta: + :varepsilon:  analogously. so angle APN=angleAQM  and angleQAM=anglePAN   APN and AQM are similar but angleAPN=ANP= :beta: + :varepsilon: so APN  is isoceles and hence AQM  to now angleNAQ=angleMAP  so since we have AQ=AM and AP=AN  we have that triangles QAN and  MAP ARE CONGRUENT   so were done. by the way i think ypu proved that APN so i know im right im pretty sure AQM IS  TOO  :blush:


calculus
integration
logarithms
calculus computations

\[ \int\limits_{0}^{4}{\frac {x + 1}{(x^2 + 2x + 2)^{2}} dx
}\]

EDIT: Problem has been fixed.

Sorry, I copied the problem incorrectly. It has been fixed.


function
combinatorics unsolved
combinatorics

"In how many ways can a total of 1001 be made from numbers 1-100 without repeating any number"
remember-using all numbers is not necessary but no number should be repeated...

plz anyone reply...
this is very imp for me...

this same problem was posted a couple of days ago by someone else. where is it from?

i dont know...just encountered it while solving another question....
if it was solved earlier can u provide the link as i cant find it...

there's no nice way to calculate the number of partitions of a number... but this may help you a bit:

http://en.wikipedia.org/wiki/Partition_of_a_number

Hi arshad;

That problem requires generating functions and the use of a computer. The number of unrestricted partitions of 1001 is 25032297938763929621013218349796.

Since your problem is a restricted partition problem the number of partitions of 1001 will be way too large.

You start with the linear diophantine equation.

a1 + a2 + a3 + a4 + ... a100 = 1001

The ordinary generating functions for this are:

$ (1 \plus{} x ) (1 \plus{} x^2) (1 \plus{} x^3 ) \dots (1 \plus{} x^{100})$

or another one:

$ \frac {(1 \minus{} x^2)(1 \minus{} x^4)(1 \minus{} x^6)(1 \minus{} x^8)(1 \minus{} x^{10})\dots(1 \minus{} x^{200})}{(1 \minus{} x)(1 \minus{} x^2)(1 \minus{} x^3)(1 \minus{} x^4)(1 \minus{} x^5)\dots(1 \minus{} x^{100})}$

You must find the coefficient of x^1001 in either of those expansions and that is the answer.

Normally it can be done by binomial tricks or residues but finding the coefficient of x^1001 here turned out to be difficult so I used a computer. The answer was found to be.

646972680324577041212 ways without repeating any number.


[hide="A proof of the Butterfly theorem."][b]The Butterfly theorem.[/b] Given a chord $PQ$ of a circle $w=C(O,r)$ draw any other two chords $AB$ and $CD$ passing through its midpoint $M$. Call the points $X$, $Y$ where $AD$ and $BC$ respectively meet $PQ$. Then $M$ is also the midpoint of $XY$.

[u]Proof.[/u] Call the projections $U$, $V$ of the center $O$ on the chords $AD$, $BC$ respectively.
$\triangle ADM\sim\triangle CBM\Longrightarrow \widehat {AUM}\equiv \widehat {CVM}\ .$ The quadrilaterals $OMXU$, $OMYV$
are cyclically (are inscribed in the circles with the diameters $[OX]$, $[OY]$ respectively).
Therefore, $\widehat {AUM}\equiv \widehat {XOM},\ \widehat {CVM}\equiv \widehat {YOM}\Longrightarrow \widehat {XOM}\equiv \widehat {YOM}\Longrightarrow MX=MY\ .$[/hide]

Dear Mathlinkers,
for the last problem, you can see
Http://perso.orange.fr/jl.ayme  vol. 7 Butterfly p. 50-53

You can also see at the end of the article references and historical notes...
Sincerely
Jean-Louis

But I can not find the solution to this whole

[quote="XmL"][quote="AnhIsGod"]Using butterfly theorem, prove this problem:
Problem: Let triangle $ABC$, $BK\perp{AC}$, $CL\perp{AB}$, intersect in $H$. A line through $H$ cut $AB$, $AC$ in $P$, $Q$ respectively. Prove that: $HP=HQ$ iff $MP=MQ$.[/quote]

what's $M$?[/quote]
Sorry, M is midpoint BC

I saw this problem in Turkish olympiad P-5(think it TST).  :wink:

Dear Mathlinkers,
have a look tp
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=192515
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=260415
Sincerely
Jean-Louis

[quote="SayanRoychowdhuri"]It is clear that $ LBCK $ is cyclic, and also observe that ,$ M $ is the center of the circle through the points $ L,B,C,K $ , now we draw the circle with these point, applying butterfly theorem we get, $ H $ is the midpoint of $ PQ $ iff $ H $ is the mid point of $ XY $ , which means $ PH=HQ $ iff $ HM \perp XY $ , because $ M $ is the center of the circle,  so, finally  the desired condition will be satisfied iff the perpendicular condition situates , and if    $ HM \perp XY $ proves $ MP=MQ $ as $ \triangle PXM $ and $ \triangle YMQ $ will be congruent..[/quote]

What is $X,Y$? Intersections $PQ$ with $k(LBCK)$?


I also think that Christian Reiher is one of the favourits of this fantasy competition!!! :D

But we will never know who will win...cause I think that there will never be a thing like that( never say never :D).

cheers!

Probably there would be an acerbic fight between Reid Barton and Ciprian Manolescu :)

I think that such a competition would be hard to organize.
Cuz some problems are of such type (Fermats TH. or others) that may require years even for them, but other problems would be easy for all of them. I don't thnk that there wud be a problem, such that one of thm wud solve it, but the other wudnt. Maybe only if it were very specific on some topic. ( I mean R. Barton is the best at informatics, so maybe in combinatorics he would now something that the others dont know )
Hope u didn't get tired reading this.
:)
:)

Laurent Lafforgue got 42/42 two times in 1990 and 1991 I think. (his brother got the Fields Medal!).
There's also a norvegian who did 6 or 7 times the IMO !!! (1st place with Olivier Benoist at the Concours General 2003)

This norvegian is David Kunszenti - Kovacs I think.

I thought the "concours gnral" was only for French students ?

You know, Olivier (26 p) beats David K - K (23 p) at the IMO. :D

Yes, but David Kunszenti - Kovacs was in a french school, so he could do it.

yes, and he isn't even french, nor novegian... I think he was from eastern europe, but i can't remember from where.

Well, I don't think that the best is the contestant that got perfect score on the maximum number of IMOs.  For me the best is  :maybe: Grigorij Perelman. He participated only in one olympiad (IMO 1982) and got the perfect score there. Now he seems to have proved the Poincare Conjecture.

David Kunszenti-Kovacs is from Hungary.

The greatest mathematician for the moment and who did extremely well at IMOs is Laszlo Lovasz (from Hungary). He took part at IMO from 1963 to 1966 winning three gold medals and a silver medal.

Kantor

Now we are playing fantasy IMO now? My fantasy basketball team is enough to manage. . . =)

But to actually stay on the topic. . . it could be YOU! So work hard =D

To be serious, these things are hard to judge, for a variety of reasons:

1) IMOs are getting harder, but the rate at which this is happening is hard to determine, so how do you know how much to handicap them?
2) People know more nowadays going to the IMO (correlated to #1) thanks to information transmission, books, great sites like this one (go Valentin), peers, and the evolution of the art of olympiad problem solving. So people on average are much better - but it doesn't mean that someone from 10 years ago, given the same situations, might not be as good as others are now if they were born later.
3) depends on the problems, styles, etc. so much. Judges too.
4) in more competitive countries, even the best people get knocked down sometimes just because the pool is so high (i.e. China) - you might see people that would otherwise go to the IMO 5 times or something only be on one Chinese team since the selection process is so rigorous and all. . .

Dang that's a long first post.

Later,

-Y.Z.

P.S. Haha this site makes me feel warm and fuzzy sorta

I would wish that the winner will be Christian Reiher. But I don't belive that. And I think participantes from IMO 10 years ago would have a chanche too.  Because they had 10 years more to improve their skills in Problem Solving. I think the young 11year-old (at that time) Australian who participated in the IMO 1986-1988 (I don't know his name anymore) would be one of the favourites.

Alas. . . 10 more years to improve, yes. But olympiad problem solving skill is not something that necessarily goes up with skill in mathematics - so once someone is done with the IMO and such they (probably) won't do their training at nearly an intense level as they did before - unless their job is to train IMO kids or something =P if anything, I think this skill deteriorates in college/grad school when you are crunching away at more real yet less nice math problems.

As a side note, a big portion (a majority by now???) of kids near the top of IMO level seem to end up doing something BESIDES pure math - this phenomenon is also seen with the Putnam exam. Hmm. . . that probably calls for a different philosophical topic.

One last muse - this topic is iffy since it might just become a combat of nationalism ^_^

But I'll leave it at that.

-Yan

[quote="Igor"]Laurent Lafforgue got 42/42 two times in 1990 and 1991 I think. (his brother got the Fields Medal!).
[/quote]

I do believe this is Vincent who got the gold medal in 1990 and 1991.
Laurent attended the IMO contest in 1984 and 1985.

--
Fabrice

Manolescu X Barton...?

Actually, I will be happy to participate in the IMO once time in my life...

[quote="spiritrock"]Lê Hùng Việt Bảo ![/quote]

Hey, I suppose that Le Hung Viet Bao is brilliant, and possibly very brilliant among the Vietnamese IMO prize winners. Could you please explain why you choose Le Hung Viet Bao ?
Thanks

[quote="imam12"]Hi.   If all the outstanding contestants at IMO from past to present  would compete in a harder than IMO competition , who would be the ultimate winner? Give your comments.
                                                              THANKS[/quote]

Phan Thi Ha Duong

She is the most brilliant person I have ever met.
And the way she outstands is that she decides to come back to Vietnam to help Vietnamese young talents to win.

to me I belıeve the chinese are better in terms of skills and flair in mathematics....but unfourtunately,most of the contestants from chına only participate once and they go away wıth gold medal.....pls guys,take this into consideration

They say the questions are getting harder but that's because more people get good training which makes a difference.
When I went in 1977 and 1978 as part of the British team we had no training at all. So I must make a special mention for
two of my team mates, John Rickard who had three golds and Richard Borcherds, 1 gold + 1 silver who is a Fields medalist.
I am sure they would do very well against more recent winners, but you would have to hold the contest in another platonic
realm because John has sadly left this one.

I kinda of read an article about Artur Avila's excentric life (http://www.imo-official.org/participant_r.aspx?id=3801) and he seems a monster.

He was young when he got gold on his first attempt on IMO and he refused to represent Brazil again because it seems he already conquered what was to be conquered in a competition of this level. He refused to go again saying: "Problems that have a certain solution don't interest me. The uncertain is much more fun" (or something like that, the article was in portuguese and was too long so I read bits of it) and to follow the path of a mathematician. Apparently what shocked the most was the young boy strong resolution about his future.

What I find most interesting about him was that when he was 6th grade and he first participated an mathematic olimpyad he got bronze in the national competition (sounds like he is human after all), this turned out gold medal the following years needless to say...

Either Terence Tao or Reid Barton or Iurie Boreico.

[quote=RunyangWang]Terence Tao I guess? The youngest bronze, silver, and gold medalist in IMO history. Won Fields Prize in 2006.[/quote]

agree.


integration
trigonometry
analytic geometry
geometry
calculus
graphing lines
slope

Given that, for non-negative integers $ n$,

$ I_n \equal{} \int_0^\frac{\pi}{2} \sin ^{n} \theta\, d\theta$,

show that, for $ n\ge2$, $ nI_n \equal{} (n\minus{}1)I_{n\minus{}2}$.

The equation of a curve, in polar coordinates, is

$ r\equal{}\sin^{3}\theta,$ for $ 0\le\theta\le \pi$.

Find the equations of the tangents at the pole and sketch the curve.

Find the exact area of the region enclosed by the curve.

For the first one, use integration by parts.

That's the easy part:

$ I_{n}\equal{}\int_{0}^\frac{\pi}{2}\sin^{n}\theta\, d\theta$

$ \equal{} \int_{0}^\frac{\pi}{2}\sin^{n\minus{}1}\theta\sin\theta\,d\theta$

Integrating by parts, we have

$ (n\minus{}1)\int_{0}^\frac{\pi}{2}\sin^{n\minus{}2}\theta\cos^{2}\theta\,d\theta$

$ \equal{}(n\minus{}1)\int_{0}^\frac{\pi}{2}(\sin^{n\minus{}2}\theta \minus{} \sin^{n}\theta)\, d\theta$

$ \Rightarrow nI_{n} \equal{} (n\minus{}1)I_{n\minus{}2}$.

The area of the region bounded by the curve is given by $ A\equal{}\int_{\theta_{1}}^{\theta_{2}}\frac{1}{2}(r(\theta))^{2}\,d\theta$.

Thanks, JRav. I know that, but do I assume the limits $ \theta_1$ and $ \theta_2$ to be $ 0$ and $ \pi$ respectively or do I need to find the tangents at the pole first? Are these tangents $ \theta \equal{} 0$ and $ \theta \equal{} \pi$?

For the graph I have a single loop.

I have seen the following (are there other ways of finding tangents at the pole?):

If $ f(\alpha) \equal{} 0$ but $ f(\theta) > 0$ in an interval $ \alpha < \theta < ...$ or $ ... < \theta < \alpha$, then the line $ \theta \equal{} \alpha$ is a tangent to the graph of $ r \equal{} f(\theta)$ at the origin (pole).

If so, I have

$ \frac {1}{2}\int_{0}^{\pi} \sin ^{6}\theta\,d\theta$

$ \equal{} \int_{0}^\frac {\pi}{2} \sin^{6}\theta\, d\theta$

which by Wallis' formula is

$ \frac {5}{6}\times\frac {3}{4}\times\frac {1}{2}\times\frac {\pi}{2}$

$ \equal{} \frac {5\pi}{32}$.

Is this correct?

Can someone post a sketch of the curve and/or tell me how to post graphs?

I would also like to learn more about finding the equations of tangents at the pole (and to curves given in polar form in general).

I'm fairly sure my sketch is correct, but I'm still bothered about tangents at the pole. There must be a number of different ways of approaching this.  For instance, can we change to Cartesian coordinates and differentiate implicitly and then substitute $ x\equal{}0$ and $ y\equal{}0$?

I get (and I'm rushing again, so I've probably made some mistakes)

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

$ \Rightarrow\, y' \equal{} \frac{\minus{}4x(x^{2}\plus{}y^{2})}{4x^{2}y \plus{} 4y^{3} \minus{}3y^{2}}$

$ \Rightarrow$ the slope of the tangents at the origin and at $ (0, 1)$ are $ 0$ and so the tangents at the pole are $ \theta \equal{} 0$ and $ \theta \equal{} \pi$.

Is this correct?

How else could one approach this problem?

I'm hoping that different countries have different ways of approching the problem of tangents at the pole.


[quote="SUPERMAN2"]Supposing $ a,b,c > 0$,prove that:
$ (a \plus{} 1 \minus{} \frac {1}{b})(b \plus{} c \minus{} \frac {1}{c}) \plus{} (b \plus{} 1 \minus{} \frac {1}{c})(c \plus{} 1 \minus{} \frac {1}{a}) \plus{} (c \plus{} 1 \minus{} \frac {1}{a})(a \plus{} 1 \minus{} \frac {1}{b})$≥$ 3$[/quote]

That's wrong: try $ a\equal{}b\equal{}c\equal{}\frac{1}{2}$
I think you mean:
$ (a \plus{} 1 \minus{} \frac {1}{b})(b \plus{} 1 \minus{} \frac {1}{c}) \plus{} (b \plus{} 1 \minus{} \frac {1}{c})(c \plus{} 1 \minus{} \frac {1}{a}) \plus{} (c \plus{} 1 \minus{} \frac {1}{a})(a \plus{} 1 \minus{} \frac {1}{b})$≥$ 3$

but it still wrong, try the same...$ a\equal{}b\equal{}c\equal{}\frac{1}{2}$

[quote="SUPERMAN2"][quote="SUPERMAN2"]Supposing $ a,b,c > 0$,prove that:
$ (a \plus{} 1 \minus{} \frac {1}{b})(b \plus{} c \minus{} \frac {1}{c}) \plus{} (b \plus{} 1 \minus{} \frac {1}{c})(c \plus{} 1 \minus{} \frac {1}{a}) \plus{} (c \plus{} 1 \minus{} \frac {1}{a})(a \plus{} 1 \minus{} \frac {1}{b})$≥$ 3$[/quote]
I am really sorry.This problem is from "Secrets in ineq" written by Pham Kim Hung(I don't know how to solve it).He must make some mistakes in his book.[/quote]
This inequality 's true with condition $ abc\equal{}1$ :)

[quote="quykhtn-qa1"][quote="SUPERMAN2"][quote="SUPERMAN2"]Supposing $ a,b,c > 0$,prove that:
$ (a \plus{} 1 \minus{} \frac {1}{b})(b \plus{} c \minus{} \frac {1}{c}) \plus{} (b \plus{} 1 \minus{} \frac {1}{c})(c \plus{} 1 \minus{} \frac {1}{a}) \plus{} (c \plus{} 1 \minus{} \frac {1}{a})(a \plus{} 1 \minus{} \frac {1}{b})$≥$ 3$[/quote]
I am really sorry.This problem is from "Secrets in ineq" written by Pham Kim Hung(I don't know how to solve it).He must make some mistakes in his book.[/quote]
This inequality 's true with condition $ abc \equal{} 1$ :)[/quote]
If $ abc\equal{}1$, so, is my solution right?

$ \sum (a\plus{}1\minus{}\frac 1b)(b\plus{}c\minus{}\frac 1c) \ge 3$,

$ \sum abc(a\plus{}1\minus{}\frac 1b)(b\plus{}c\minus{}\frac 1c) \ge 3$,

$ \sum a(ab\plus{}b\minus{}1)(bc\plus{}c^2\minus{}1) \ge 3$,

$ \sum (a^2b^2c \plus{} a^2bc^2 \minus{} a^2b \plus{} ab^2c \plus{} abc^2 \minus{} ab \minus{}abc \minus{} ac^2 \plus{} a)\ge 3$,

$ \sum (a\plus{}b\plus{}c \plus{} ac \plus{} \frac ac \plus{} \frac cb) \ge \sum (3\sqrt[3]{abc} \plus{} ac \plus{} \frac ac \plus{} \frac cb) \ge 6$.

 :wink:

[quote="SUPERMAN2"]Supposing $ a,b,c > 0$,prove that:
$ (a \plus{} 1 \minus{} \frac {1}{b})(b \plus{} c \minus{} \frac {1}{c}) \plus{} (b \plus{} 1 \minus{} \frac {1}{c})(c \plus{} 1 \minus{} \frac {1}{a}) \plus{} (c \plus{} 1 \minus{} \frac {1}{a})(a \plus{} 1 \minus{} \frac {1}{b})$≥$ 3$[/quote]

Can you tell me the number of page of this problem in the book?


Medians $ BD$ and $ CE$ of triangle $ ABC$ are perpendicular, $ BD \equal{} 8$, and $ CE \equal{} 12$. The area of triangle $ ABC$ is
[asy]defaultpen(linewidth(.8pt));
dotfactor=4;

pair A = origin;
pair B = (1.25,1);
pair C = (2,0);
pair D = midpoint(A--C);
pair E = midpoint(A--B);
pair G = intersectionpoint(E--C,B--D);

dot(A);dot(B);dot(C);dot(D);dot(E);dot(G);
label("$A$",A,S);label("$B$",B,N);label("$C$",C,S);label("$D$",D,S);label("$E$",E,NW);label("$G$",G,NE);
draw(A--B--C--cycle);
draw(B--D);
draw(E--C);
draw(rightanglemark(C,G,D,3));[/asy]$ \textbf{(A)}\ 24\qquad \textbf{(B)}\ 32\qquad \textbf{(C)}\ 48\qquad \textbf{(D)}\ 64\qquad \textbf{(E)}\ 96$

[hide="Click for solution"]
We have $ CG\equal{}\frac{2}{3}CE\equal{}8$ and $ [BCD]\equal{}\frac{1}{2}(BD)(CG)\equal{}\frac{1}{2} \times 8 \times 8\equal{}32$.  Because $ BD$ is a median, $ [ABD]\equal{}[DBC]$, so the area of the triangle is $ 64$, or $ \boxed{\textbf{(D)}}$.
[/hide]


Here's a very delightful problem (from MOP) that I enjoyed.
Let $ABCD$ be a rectangle with an interior point $P$. If $\angle APD+\angle BPC=180^{\circ}$, what is the sum of $\angle DAP$ and $\angle BCP$?

[hide="Solution"]
Move $\triangle DPC$ on top of $AB$ and call the new triangle $AQB$.  $QP$, $AD$, and $BC$ are parallel. It is clear that $AQBP$ is cyclic due to the fact that $\angle AQB+\angle APB=180^{\circ}$.  $\angle DAP=\angle APQ$ (PLT-Z) and $\angle BCP=\angle PQB$. $AB$ and $PQ$ intersect at $O$ at $90^{\circ}$; hence, $\angle POB=90^{\circ}$. $\angle DAP+\angle BCP=\angle APQ+\angle PQB$ is equal to half the sum of the arcs of $AQ$ and $PB$. $\angle POB=90^{\circ}$ is also equal to half the sum of the arcs of $AQ$ and $PB$, thus $\angle DAP+\angle BCP=\angle APQ+\angle PQB=90^{\circ}$.
[/hide]
Masoud Zargar

[hide]
You can also note that since
$a + b = c + d = a + c = b + d = 180$, letting $a = APB, b = BPC, c = CPD, d = DPA$, we see $a = b = c = d = 90$, and the result follows.
[/hide]


Prove that any polynomial with real coefficients that takes only nonnegative values can be written as the sum of the squares of two polynomials.

[hide]Let the polynomial be $ Z(x)$.  Clearly we can assume that $ Z(x)$ and all of its factors are monic.  The given conditions imply that $ Z(x) \equal{} P(x)^2Q(x)$ where $ Q(x)$ has no real roots but all real coefficients.  Note that every real root of $ Z(x)$ must be a double root otherwise $ Z(x)$ would change signs.  Because $ Q(x)$ has real coefficients, every root must have a corresponding conjugate root.  Therefore $ Q(x) \equal{} \left(\prod^n_{i \equal{} 0}(x \minus{} \alpha_i)\right)\left(\prod^n_{i \equal{} 0}(x \minus{} \bar{\alpha}_i)\right)$.  Through induction we can prove that if $ \left(\prod^n_{i \equal{} 0}(x \minus{} \alpha_i)\right) \equal{} a(x) \plus{} ib(x)$, then $ \left(\prod^n_{i \equal{} 0}(x \minus{} \bar{\alpha}_i)\right) \equal{} a(x) \minus{} ib(x)$ where $ a(x)$ and $ b(x)$ are polynomials with real coefficients.  I would actually show this but I can't seem to format the latex nicely.  Do it yourself, it's not that hard.  Once we have this, we have $ Q(x) \equal{} a(x)^2 \plus{} b(x)^2$ so $ Z(x) \equal{} P(x)^2Q(x) \equal{} (P(x)a(x))^2 \plus{} (P(x)b(x))^2$.

EDIT:  I should at least acknowledge the case where $ Z(x)$ is a positive constant for which the statement is true.  Also, my first edit was pointless. [\hide]


S28

A [i]lucky number[/i] is a positive integer which has a value 19 times the sum of its digits. How many different [i]lucky numbers[/i] are there?

Not many. I'm not sure (maybe I made mistake in my calculations) but it's just 399, 266 and 133.

EDIT: So It's little bit more. My solution was same as DiscreetFourierTransform's
but I wrote the last two digits in wrong order and didn't release one thing  :blush: .

Two digit:
$ 19(a+b) = 10a+b$
$ 19a+19b = 10a+b$
$ 9a =-18b$

meaning either a or b is negative, which is impossible (digits themselves are never negative, only the number they represent.
No Two Digit perfect numbers.

Three Digit: (a not equal to zero
$ 19(a+b+c) = 100a+10b+c$
$ 81a-9b-18c = 0$
$ 9a-b-2c=0$
$ 9a = b+2c$

Cases:
$ b+2c \equiv 0 \pmod 9$

$ b+2c = 9$ (9, 0) ; (7, 1) ; (5, 2) ; (3, 3) ; (1, 4) --> Five Solutions (for a = 1)
----------------------
$ b+2c = 18$
B must be even for $ 2c$ to divide $ 18-B$.
(0, 9) ; (2, 8) ; (4, 7) ; (6, 6) ; (8, 5) --> Five Solutions. (for a = 2)
---------------------
$ b+2c$ = 27
B must be odd for $ 2c$ to divide $ 27-B$.
2c is maximally 18, so B must be minimally 9. However, this limits us to once case, because B is also maximally 9.
(9, 9) --> One Solution (for a = 3)
---------------------
We can't have any more solutions:
Because $ 2c$ is at most 18, and B is at most 9, the highest digital sum they can produce is 27. Therefore, there are

5 + 5 + 1 solutions = 11.

EDIT: Oops, forgot to consider four digits:
$ 19(a+b+c+d) = 1000a+100b+10c+d$
$ 981a+81b-9c-18d = 0.$
$ 981a+81b = 9c+18d$
$ 109a+9b = c+2d$
And it's clear that for maximal c and d, we have $ 109a+9b = 27$ which is clearly unreachable: Therefore there are no four digit lucky numbers.

shoot this is an ARML 2007 Team Question T2

Looks like the ARML is a good source of problems.. also there was a problem on the American MC on how many lucky numbers were 13 times the sum of their digits i think


MATHCOUNTS trigonometry AMC AIME calculus integration geometry

I know trig/log is a must, but anyone have a list of useful formulas/theorems for the olympiad? (especially ones involving circles)

Ugh, now that you brought MAML up again, I have found out that our school may not be doing it since we don't have classes next Tuesday (Its Parents' Weekend, we had Parent's Day yesterday).

[quote="cats..."]i dont think this test is about how many theorems you know.  I'll bet all the formulas you learned for mathcounts would suffice.

But while we're on this topic, how long is the test and how many points for each correct and unanswered question?[/quote]
Details [url=http://www.artofproblemsolving.com/wforum/viewtopic.php?t=1773]here[/url]

Um, sorry to contradict you but judging from the fact that most problems need some kind of formula whether it be n(n+1)/2 or power of point...I really need to know this stuff [b]fast[/b] you guys, the comp is in three days and I must make it, or else! (and I don't even want to think about the "or else"... :( )

You can't learn to use things quickly in three days.  It's a skill acquired with time, as a result of doing many different problems.  Anyhow, calm down.

Mmm, maybe you need to know about the "or else"...

Sorry to act jitterish, but the "or else" is scaring the heck out of me...

"or else": (military) camp

sorry, but i really agree with joel about not trying to cram new stuff in.  I don't think it'd be worth it.

I really don't think it's about how many theorems/formulas you know, anyways.  Like I said, I wouldnt be surprised if you learned all the necessary formulas in mathcounts.  You cited the sum of an arithmetic progression and the power of a point theorem... that is stuff (i think) that you probablly already learned in your preparation for mathcounts.  The theorems/formulas on this test don't get much more advanced than that.  It's more about how well you can think, so you'd be better off doing past tests.

*sigh* trig formulas and circle stuff (as in problem 19 for the last year's ([b]2002[/b]-2003) test) we don't learn for MATHCOUNTS...I'm doomed...

not to be mean, but if you don't do well its not because you didnt know enough math, but because you were too nervous.  Seriously, calm down... if you're too nervous, regardless if you know how to do every question on there, you won't be able to do too well.  Im sure you have all the tools necessary to do well.  And you will do well if you calm down.

For trig identities, you really only need to know the sum and difference formulas:  sin(x+y)=sinx*cosy+siny*cosx.  cos(x+y) = cosx*cosy-sinx*siny.  tan(x+y) = (tanx+tany)/(1-tanx*tany) .

And this might actually matter for 1 question.  Maybe 2...
But that really it for trig...

Then as far as that circle question goes, all you needed to know is that when 2 chords intersect in a circle, ... im not sure how to articulate this exactly, but the products of the segments up until their intersection points are equal to each other.  Going by the diagram given in the question, this means that AE*EB=CG*GD.  The rest of that question can be solved just by knowing that, and then by thinking your way to the solution.

Actually, power of a point comes from knowing other things that you should know.  It's just similar triangles.  You have to know that the inscribed angle of any arc of a circle is constant.  And that gives you similar triangles (when you draw two additional chords), and then you're done.  Besides which, I think that was fair game for Mathcounts.  And I agree that you may not have been expected to know trig yet, but it is unlikely to feature heavily at all.

I must say though, that not knowing the math needed to solve some problems is a major hit on one's confidence. Sometimes, when I try to solve some problems on national olympiads, I give up (too easily) just because I think I do not know the math needed to solve a problem, even though I may in fact know enough. On the other hand, when I do AMC12's, I am completely confident that I know all the math needed to solve the problems on the exams, and so I do well.

that happens to me, but for a slighly different reason. i quickly get discouraged and if i dont solve a problem in a matter of minutes, i give up and say its too hard for me. when i practice aimes or more recently olmypiads, i tend to have a larger score when i take it under the official test conditions than at home, which is funny. but the key thing is not too get discouraged, at home , or at the contest. tare, probably you already know enough math to solve a majority of the problems. if you cram too much, you might actually become less confident in your knowledge than you are now. and if you go in too nervous (like me before the amc 12 last year), you will make a myriad of stupid mistakes. go in confidently (but not overly so) and do your best on each problem, knowing that you can solve all of them if you keep trying.

Okay, thanks everyone, guess as long as i think I'm doomed I'm doomed and if I think not then I'm not :)

I dont know if you guys are allowed to keep your tests or not, but i dont think my teacher lets us.  Can you guys keep track of your answers somehow (write it on a piece of scrap) so that we can have some idea of our scores afterwards?

It wasn't too bad, although I made a very stupid mistake on one of them.  I answered 26, at least 1 of which was incorrect.

Hmm. It was alright. I got 22 right, 2 wrong, and omitted the rest. I don't know this officially of course, but I compared with all the other people. The last couple questions were, for me, pretty impossible (the hexagon with the squares thing? Distance? GRR!)
What was everyone else's score and experience with it?
Also, when do they announce the Level 2 qualifiers?

I got E - Can't determine anything about it, but I'm not sure if I got that right. A lot of people got different answers.

Hold on, everyone is done taking the test right? If they aren't, then I'll have to take my answer down.

[quote="cats..."]what was the one with ax^2+bx+c=0 and b^4-a^2*c^2=65[/quote] The key to this problem is that a, b, and c are integers.

We have b^4-a^2*c^2=65.
So (b^2-4ac)(b^2+4ac)=65.  So one of these factors is 13 and the other is 5 or 1 is -5 and the other is -13.  Suppose that both are negative.  Than b^2-4ac<0 and b^2+4ac<0, so adding yields 2b^2<0, a contradiction.  So b^2-4ac is either 5 or 13, and the solutions are therefore irrational.

oops. my mistake, you're right. i'm wrong. :(

EDIT: what integral a, b, and c values produce 65 and 1? I can't think of any.
EDIT 2: now i'm pretty sure there aren't any. if b^2-4ac=65 and b^2+4ac=1 (or the other way around), 2b^2=66 and b=sqrt 33.
So my logic was bad, but i think i'm right now.

Now it's officially over. I got 18 and guesstimated 1...should've guessed more :(
The hexagon thing I guessed since it must be smaller than the side of the square+side of the hexagon+diagonal of the square=x+x+x:rt2:=2x+x:rt2: (I don't remeber the side but the side is x), so two of the answers didn't make sense. Thus I guesstimated it was the second largest answer...am I right?

P.S. You do need to memorize all the formulas...the half-angle theorem was a must on this test and such.

Is it perhaps possible to get some of these questions posted, as opposed to commontary on them?  Are they online somewhere?

they're not online and i dont have them.  But i think this is the question, though i dont remember the answer choices.

We have a regular hexagon of side 4.  The midpoints of each side are connected to form a smaller hexagon, and this process is repeated.  Find the sum of all areas of these hexagons.

Tare, if you're referring to the trig one where you had to simplify, that was just 1 question.  Also, that circle question w/ the chords was essentially the same as the one in 2002.

Edit: and the question above that i posted was ax^2+bx+c=0 and b^4-a^2*c^2=65.  The question asked what can you determine about the roots: i dont remember all the choices, but some of them are a)they're imaginary b) they're irrational c)they're the same d)the roots can't be determined.

i remember some more now, so i'll post them.  z_1=1-ai, z_2=1+ai, z_3 = 2+ai.  Given that z_1*z_2*z_3 is a real number, find (z_1*z_2*z_3)^10.

How many ways can you make 2 dollars out of nickels, quarters, and dimes, given that the absolute difference between the number of quarters and nickels is 2.

log_a 21 = x, log_a 49 = y , find an expression for log_a 27

If you put $500 in your bank account today, and every 3 days after today, you take out $12, and every 17th day after today you take out $25, and every 51st day you put in $100, on what day will you have a negative amt. of money.

These sound like difficult questions.  How much time do you have to solve them?

90 minutes, 30 questions. See my post about the official rules somewhere else...you know isn't it annoying when you type up this really detailed thing and then you can't find/remember where you posted it? It's so much "real life"ish it's giving me the left-over creeps from Halloween :lol:


[hide="Hint"] Well, just set $N=800$ and solve. Do you need any hints beyond that? [/hide]

[quote="myc"]Can anyone give me a [u]hint[/u] on this problem.

The formula $N=8\times10^{8}\times x^{-3/2}$ gives, for a certain group, the number of individuals whose income exceeds $x$ dollars.  What is the smallest possible value of the lowest income of the wealthiest 800 individuals? (AHSME 1960)

   thanks

[size=75][color=red]b-flat:  Check your $\LaTeX$.  I fixed it for you.  Use \times for *, and when your exponents have more than one character, surround them all by the braces.[/color][/size][/quote]

I set n=800...

[quote="ch1n353ch3s54a1l"][quote="myc"]Can anyone give me a [u]hint[/u] on this problem.

The formula $N=8\times10^{8}\times x^{-3/2}$ gives, for a certain group, the number of individuals whose income exceeds $x$ dollars.  What is the smallest possible value of the lowest income of the wealthiest 800 individuals? (AHSME 1960)

   thanks

[size=75][color=red]b-flat:  Check your $\LaTeX$.  I fixed it for you.  Use \times for *, and when your exponents have more than one character, surround them all by the braces.[/color][/size][/quote]

I set n=800 to get [hide]$10000$[/hide][/quote]

He asked for a hint.  Please only give him a hint, not the answer.  Thanks.  :)

Sorry. That's why I hid the answer, in case he might eventually want it...

o yea it works out, subsituting 800 for N , how come I didn't think of that.  Thanks  :lol:

[quote="myc"]Can anyone give me a [u]hint[/u] on this problem.

The formula $N=8\times10^{8}\times x^{-3/2}$ gives, for a certain group, the number of individuals whose income exceeds $x$ dollars.  What is the smallest possible value of the lowest income of the wealthiest 800 individuals? (AHSME 1960)

   thanks

[size=75][color=red]b-flat:  Check your $\text{\LaTeX}$.  I fixed it for you.  Use \times for *, and when your exponents have more than one character, surround them all by the braces.[/color][/size][/quote]
[hide] \begin{eqnarray*}N & = & 8\times10^{8}\times x^{-3/2}\\ 800 & = & 8\times10^{8}\times x^{-3/2}\\ x^{3/2}& = & 10^{6}\\ \log_{10}{x}& = & \frac{6}{3/2}\\ \log_{10}{x}& = & 4 \\ x & = & 10000 \end{eqnarray*} [/hide]


limit
calculus
integration
real analysis
calculus computations

The question is as follows (pardon my formatting):

Evaluate:

lim n-> infinity    n(sum)i=1  [root(1+(8i/n)) * 8/n]

Perhaps this?

$\lim_{n\to\infty}\sum_{i=1}^n\sqrt{1+\frac{8i}n}\cdot\frac8n$

[hide="Two word hint."]Riemann sum.[/hide]

Yup, that's the question.... and I figured it was a Riemann Sum, but I wasn't sure how to "evaluate" it exactly... Do I simply express it as an integral? I'm just not sure where to go with it... thanks.

You're looking for this to be of the form $\sum f(x_i)\Delta x.$

In looking at this, you see the square root, which looks like a candidate for $f(x),$ and you see the extra factor of $\frac8n.$

So let's start by trying to pin down $\Delta x.$ Try the possibility $\Delta x=\frac8n.$

Then as $i$ runs from $1$ to $n,$ the net increase in $x$ will be $n\Delta x=8.$ We're looking for $x$ to increase by $8.$

Inside the square root, you see $\frac{8i}n.$ Try letting that be $x.$ As $i$ ranges from $1$, to $n,$ this would range from close to $0$ to $8.$ That looks like our interval.

Putting the pieces together we get that this is a Riemann sum for

$\int_0^8\sqrt{1+x}\,dx.$

This answer is by no means unique, and by no means the only reasonable answer. We could just as reasonably have written

$\int_1^9\sqrt{x}\,dx$ or $\int_0^1\sqrt{1+8x}\cdot8\,dx$

Of course, these various forms of the integral are interconnected by substitutions and represent the same number.

Beautiful...thanks so much for breaking it down so well!


Consideram un sir $\ s_{n}=1,9,7,7,...$ in care numarul urmator este suma celorlalte patru din fata luate modulo $\ 10$.
$\ (a)$Poate aparea secventa $\ 1977$ dupa un numar finit de miscari?
$\ (b)$Daca sirul incepe cu patru numere pare, demonstrati ca sirul e periodic si aflati perioada sa.
$\ (c)$Daca sirul incepe cu patru numere impare, demonstrati ca sirul e periodic si aflati perioada sa.

[i]OMUS[/i]

Fie $\left\{ s_{n}\right\}_{n \geq 1}$ un şir astfel încât $s_{n}\in \overline{0,9}, \forall n \geq 1$ şi $s_{n+4}\equiv s_{n+3}+s_{n+2}+s_{n+1}+s_{n}\left( \bmod \, 10 \right), \, \forall n \geq 1$.

Fie $a_{n}= \left( s_{n}, s_{n+1}, s_{n+2}, s_{n+3}\right)$. Exista un numar finit de posibilitati pentru valorile lui $a_{n}$, $10^{4}$. Dar cum sirul $\left\{ a_{n}\right\}_{n \geq 1}$ e infinit, exista $u<v$ astfel incat $a_{u}= a_{v}$. E destul de simplu de demonstrat ca $a_{u-1}= a_{v-1}$, $a_{u-2}= a_{v-2}$, $\ldots$, $a_{1}= a_{v-u+1}$ (folosim inducţie finită).

Deci $\left\{ a_{n}\right\}_{n \geq 1}$ e periodic cu una din perioade egala cu $v-u$.

Pentru restul problemei, o sa editez mesajul asta (daca reusesc s-o fac).

Mie chiar mi s-a parut grea problema asta, asa ca n-am stiut-o. :(


This is funny  :mrgreen:

http://www.collegiatechoice.com/myaccept.htm

[quote="mirebeka"]...This one is also funny, especially for those who haven't applied yet but are busy getting all these recruiting letters...
[url]http://www.netjeff.com/humor/item.cgi?file=CollegeLetter.txt[/url][/quote]

That is sooooo funny  :mrgreen:   :mrgreen:

[quote="applepuppy"][quote="mirebeka"]I liked that!  This one is also funny, especially for those who haven't applied yet but are busy getting all these recruiting letters...
[url]http://www.netjeff.com/humor/item.cgi?file=CollegeLetter.txt[/url][/quote]

that's hilarious... but it's not real, is it? (the acceptance letter i mean, not the reply...)[/quote]
I know for a fact that MIT sends out that letter to potential applicants; whether that reply was real or not does not matter much.

Humor continues after high school! How would you like to have these professors in college?

http://www.ratemyprofessors.com/Funniest.jsp


calculus
integration
real analysis
real analysis unsolved

Find:
1. $ \int y^{3}\sqrt {2y^{4} \minus{} 1}\,dx$
2. $ \int \frac {\sqrt [3]{x}}{(\sqrt [3]{x^{4}} \plus{} 9)^{2}}\, dx$
They are from sosmath.com.

[quote="borislav_mirchev"]
2. $ \int \frac {\sqrt [3]{x}}{(\sqrt [3]{x^{4}} + 9)^{2}}\, dx$
[/quote]
The first one is quite trivial.

\begin{eqnarray*}\int\frac{\sqrt[3]x}{\left(\sqrt[3]{x^4}+9\right)^2}\,dx&=&\frac34\int\frac{\cfrac43\sqrt[3]x}{\left(\sqrt[3]{x^4}+9\right)^2}\,dx\\&=&-\frac34\int\left(\left(\sqrt[3]{x^4}+9\right)^{-1}\right)'\,dx\\&=&-\frac3{4\left(\sqrt[3]{x^4}+9\right)}+k\end{eqnarray*}

I had in my book difficult irational integrals but it doesn't make sence to send integrals who 2-3 people only can solve. I agree that is trivial enough but it was my last discovery... last three days I spent lots of time finding interesting integrals on the net.


Let $ a,b,c$ be positive real numbers such that $ a^2\plus{}b^2\plus{}c^2\equal{}3$. Prove that
$ a^3b^2\plus{}b^3c^2\plus{}c^3a^2\le3$

I knew a proof by Lagrange, a proof by Cauchy-Schwarz. Now, I know one more another proof addition, very simple :)

[quote="dduclam"]Let $ a,b,c$ be positive real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Prove that
$ a^3b^2 \plus{} b^3c^2 \plus{} c^3a^2\le3$

I knew a proof by Lagrange, a proof by Cauchy-Schwarz. Now, I know one more another proof addition, very simple :)[/quote]
Yes, my friend. There exists a nice solution by using the fact
\[ c(ab\plus{}bc\plus{}ca)(b\minus{}c)(b\minus{}a) \le 0\]
where $ b$ is the number between $ a$ and $ c$.
:)

[quote="can_hang2007"][quote="dduclam"]Let $ a,b,c$ be positive real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Prove that
$ a^3b^2 \plus{} b^3c^2 \plus{} c^3a^2\le3$

I knew a proof by Lagrange, a proof by Cauchy-Schwarz. Now, I know one more another proof addition, very simple :)[/quote]
Yes, my friend. There exists a nice solution by using the fact
\[ c(ab \plus{} bc \plus{} ca)(b \minus{} c)(b \minus{} a) \le 0
\]
where $ b$ is the number between $ a$ and $ c$.
:)[/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?t=146030

I will check it later, because I'm too busy now. If it's true, then it's very nice ! :roll:


If $ a,b,c>0$ then $ \frac{\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}}{2\sqrt{abc}}\ge\frac{ab}{a^3\plus{}b^3}\plus{}\frac{bc}{b^3\plus{}c^3}\plus{}\frac{ca}{c^3\plus{}a^3}$

[quote="moldovan"]If $ a,b,c > 0$ then $ \frac {\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}}{2\sqrt {abc}}\ge\frac {ab}{a^3 \plus{} b^3} \plus{} \frac {bc}{b^3 \plus{} c^3} \plus{} \frac {ca}{c^3 \plus{} a^3}$[/quote]

obvious by AM-GM:  $ \sum\frac {bc}{b^3 \plus{} c^3} \le\sum\frac{bc}{2bc\sqrt {bc} }\equal{}\frac{\sqrt a}{2\sqrt {abc}}$  :)


geometry
3D geometry
combinatorics unsolved
combinatorics

[quote="moldovan"]Eight singers participate in an art festival where $ m$ songs are performed. Each song is performed by $ 4$ singers, and each pair of singers performs together in the same number of songs. Find the smallest $ m$ for which this is possible.[/quote]

[hide="here is my solution"]Each song brings together $ 6$ pairs of singers so the $ m$ songs together bring together $ 6m$ (not necessarily distinct) pairs of singers.  There are a total of $ \binom{8}{2} \equal{} 28$ distinct pairs of singers.  So given that each distinct pair collaborate an equal number of times it follows that $ 28|6m$ or $ 14|3m$ or $ 14|m$.  Hence in particular $ m\ge 14$ and if we can find a specific example for which $ m \equal{} 14$ we will be done.  To do this, imagine the $ 8$ singers to correspond to the $ 8$ vertices of a cube.  The $ 6$ faces of the cube give us $ 6$ sets of $ 4$ singers.  The $ 6$ face diagonals give us another $ 6$ sets of $ 4$ singers.  To get the remaining $ 2$ sets of $ 4$ singers, color the vertices alternatively black and white, and pick the $ 4$ black vertices as one set and the $ 4$ white vertices as the other set.  This gives us a total of $ 14$ sets of $ 4$ vertices among the $ 8$ vertices and it is fairly easy to show (I won't provide the details) that each pair of vertices is represented together in exactly $ 3$ such sets.

So the smallest possible $ m$ meeting the problem conditions is $ m \equal{} 14$.[/hide]