geometry
incenter
circumcircle
trigonometry
symmetry
trig identities

In triangle ABC, angle A=30, points S and I are ABC circumcentre and incentre respectively. Points D and E are on sides BA and CA respectively such that BD=CE=BC. Prove that SI is perpendicular and equal to DE.

so if there is anyone who can help me and post the solution?

[quote="cashaz"]Let $ ABC$ be a triangle with the circumcentre $ O$ and the incentre $ I$ . Suppose that exist $ D\in (AB)$

and $ E\in (AC)$ so that $ BD=CE=BC$ . Prove that $ \|\ \begin{array}{cc}
1. & OI\perp DE\\\
2. & A=30^{\circ}\stackrel{?!}{\Longrightarrow} DE=OI\end{array}$ .[/quote]

1) $ DE \perp OI$ is easy to prove, and already old; I think the easiest way is to prove $ DI^2\minus{}EI^2\equal{}DO^2\minus{}EO^2\equal{}a\cdot (c\minus{}b)$, by drawing the perpendiculars from $ I$ and $ O$ onto $ AB$ and $ AC$.
2) It's easy as well to see that $ \angle DIE \equal{} 45^\circ$ and $ \angle DOE \equal{} 135^\circ$, hence $ O$ is the orthocenter of $ \triangle DIE$, done.

Best regards,
sunken rock

There's a more general result coming from this configuration.

[b]Proposition.[/b] Let $ \triangle ABC$ be a scalene triangle and $ M,N$ are two points on $AC, AB,$ such that $ BN \equal{} CM \equal{} BC.$ Then $ MN$ is perpendicular to $OI,$ where $ O , I$ are the circumcenter and incenter of  $ \triangle ABC$ and $\frac {_{MN}}{^{BC}} \equal{} \frac {_{OI}}{^{R}}.$

[b]Proof.[/b] In the isosceles triangles $ \triangle OAC$ and $ \triangle OAB,$ we have

$ R^2 \minus{} OM^2 \equal{} AM \cdot MC , \ R^2 \minus{} ON^2 \equal{} AN \cdot BN$

$\Longrightarrow ON^2 \minus{} OM^2 \equal{} AM \cdot MC \minus{} AN \cdot BN$

$ \Longrightarrow ON^2 \minus{} OM^2\equal{} (CA \minus{} BC)BC \minus{} (AB \minus{} BC)BC \equal{} BC(CA \minus{} AB)$

Since $\triangle MIC$ and $\triangle MIB$ are isosceles, we have $ IM \equal{} IB,$ $IN \equal{} IC.$ Then

$ IN^2 \minus{} IM^2 \equal{} IC^2 \minus{} IB^2 \equal{} BC(CA \minus{} AB)$

$ \Longrightarrow ON^2 \minus{} OM^2 \equal{} IN^2 \minus{} IM^2 \Longrightarrow IO \perp MN$

On the other hand, $ \angle NBM \equal{} \angle OCI \equal{} 90^{\circ} \minus{} \frac {_1}{^2}\angle C \minus{} \angle A.$ If  $ L \equiv{} CI \cap BM,$ then it follows that  $ \angle NMB$ and $\angle OIC$ are supplementary, which implies that $ \triangle IOC$ and  $ \triangle MNB$ are pseudo-similar, thus

$ \frac {MN}{OI} \equal{} \frac {BN}{OC} \Longrightarrow \frac {MN}{BC} \equal{} \frac {OI}{R}.$

Note that  $ \angle A \equal{} 30^{\circ} \Longleftrightarrow  BC \equal{} R \Longleftrightarrow DE \equal{} OI $ (the proposed problem).

[quote]sunken rock Re: Lithuania TS 2009
Nice solution

--------------------------------------------------------------------------------

1)  is easy to prove, and already old; I think the easiest way is to prove , by drawing the perpendiculars from  and  onto  and .
2) It's easy as well to see that  and , hence  is the orthocenter of , done.

Best regards,
sunken rock [/quote]

can you explain me how you get DIE=45 and all yout second part of the solution.

Thnx luis, very interesting solution.

its interesting and easy to prove that $ OI\equal{}R_ADE$

In $ \triangle ABC$, $ \angle A = 30^\circ$, the points $ O$ and $ I$ are the circumcentre and incentre of $ \triangle ABC$ respectively. Points $ D$ and $ E$ are on the sides $ BA$ and $ CA$ respectively such that $ BD = CE = BC$. Prove that $ OI$ is perpendicular and equal to $ DE$.
(I'm used to $ O$ instead of $ S$ :P)

I have a solution using trigonometry:
Let $ \angle DBC = \beta, \angle ECB = \gamma$, then $ \beta + \gamma = 150^{\circ}$. We easily find $ \angle OBI = 60^\circ - \frac {\beta}{2} = \frac {\gamma}{2} - 15^\circ = \angle DCE$. Let $ X$ be the point on $ DC$, such that $ BI = CX$. Then $ \triangle CXE$ is congruent to $ \triangle BIO$. Let $ \angle ECD = u$. Then
$ IO^2 = EX^2 = CX^2 + EC^2 - 2 CX \cdot EC \cdot \cos u$, $ DE^2 = DC^2 + EC^2 - 2DC \cdot EC \cdot \cos u$.
So $ IO^2 = DE^2 \iff$
$ BI^2 + EC^2 - 2BI \cdot EC \cdot \cos u = DC^2 + EC^2 - 2DC \cdot EC \cdot \cos u \iff$
$ (BI - DC)(BI + DC - 2EC \cdot \cos u) = 0$
I will prove $ BI + DC = 2EC \cdot \cos u$, which implies $ IO = DE$.
Let $ BD = BC = CE = BO = OC = t$.
From the law of sines relation we get: $ BI = \frac {\sin \frac {\gamma}{2}}{\sin 75^\circ} t$, since $ \angle BCI = \frac {\gamma}{2}$, and $ \angle BIC = 105^\circ$.
Using the law of sines once again: $ CD = \frac {\sin \beta}{\sin 90^\circ - \frac {\beta}{2}} t = 2\cos \left ( \frac {\beta}{2} \right )t$.
Letting $ x = \frac {\beta}{2}$, and using $ \beta + \gamma = 150^\circ$, we see that $ BI + DC = 2EC \cdot \cos u \iff$
$ \frac {\sin 75^\circ - x}{\sin 75^\circ} + 2 \cos x = 2 \cos 60^\circ - x$.
But using the relations $ \sin a + b = \sin a \cos b + \sin b \cos a$, $ \cos a + b = \cos a \cos b - \sin a \sin b$, and $ \cot 75^\circ = 2 - \sqrt {3}$ it is easy to see that both sides equal $ \sin x + \sqrt {3}\cos x$.
Hence $ DE = IO$. But then we also have $ DE = EX$. Let $ v = \angle EDC = angle EDX = \angle DXE = 180^\circ - \angle EXC = 180^\circ - \angle BID$.
$ EC$ can be obtained by rotating and moving $ BC$ with an angle of $ 180^\circ - \ECB = 180^\circ - \gamma$ in positive direction. $ DE$ can be obtained by rotating, moving, and scaling $ CE$ with an angle of $ 180^\circ - \angle DEC = \angle EDC + \angle DCE$, so $ DE$ can be obtained by rotationg, moving, and scaling $ BC$ with an angle of $ 90^\circ + \left ( v + \frac {\beta}{2} \right )$
Similarily $ OI$ can be obtained by rotating, moving and scaling $ BC$ with an angle of $ v + \frac {\beta}{2}$. It follows that $ ED$ can be obtained by rotating, moving, and scaling $ OI$ with and angle of $ 90^\circ$, and hence $ ED \perp OI$. And we are done :)

To expand little bit my compressed solution:
Due to symmetry the following angle equalities hold:
$ \angle BID \equal{} \angle BIC \equal{} \angle CIE \equal{} 90^\circ \plus{} \frac{1}{2} \cdot m(\angle A)\equal{} 105^\circ$ $ \implies \angle DIE \equal{} 360^\circ \minus{} 3 \cdot 105^\circ \equal{} 45^\circ$.
$ \angle BOC \equal{} 60^\circ$, as $ \triangle BOC$ is equilateral; $ \angle OBD \plus{} \angle OCE \equal{} 180^\circ \minus{} \angle A \minus{} \angle OBC \minus{} \angle OCB \equal{} 30^\circ$. The triangles $ OBD$ and $ OCE$ are isosceles, hence $ \angle BOD \equal{} 90^\circ \minus{} \frac{1}{2} \cdot \angle OBD$, $ \angle COE \equal{} 90^\circ \minus{} \frac{1}{2} \cdot \angle OCD$, so $ \angle BOD \plus{} \angle COE \equal{} 180^\circ \minus{} \frac {\angle OBD \plus{} \angle OCE}{2} \equal{} 165^\circ$, consequently $ \angle DOE \equal{} 360^\circ \minus{} (\angle BOD \plus{} \angle COE \plus{} \angle BOC ) \equal{} 135^\circ$. Under the circumstances $ O$ is the orthocenter of $ \triangle DIE$ ($ O$ is placed on the altitude from $ I$ and $ \angle DIE \plus{} \angle DOE \equal{} 180^\circ$, hence it's true, as claimed).

I hope to be all clear now!

Best regards,
sunken rock

[quote]I hope to be all clear now!
[/quote] yes its all clear now, thanks a lot

$ \triangle ABC$, $ D$ is on $ AC$ and $ E$ is on $ BC$ such that $ AD=AB=BE$.$ O$,$ I$ are circumcenter and incenter of $ \triangle ABC$ respectively.prove that
1)$ OI \bot DE$
2)the radius of circumcircle of $ \triangle CDE$ is equal to $ OI$.

solution 1)
$ M$ is the midpoint of arc$ BC$ , $ K$ is the midpoint of $ BC$ , $ AM$ intersect $ BD$ at $ N$ , $ R$ is the radius of circumcircle of $ \triangle ABC$.
$ MI$=$ 2R$ $ \sin\frac{A}{2}$.
$ BD$=$ 2BN$=$ 2AB\sin\frac{A}{2}$.
Now it is easy to compute that $ \frac{BD}{MI}$ = $ \frac{BE}{OM}$  (1)
$ \triangle EDB$ and $ \triangle OIM$ : $ OM \bot BE$ , $ IM \bot DB$  (2)
(1),(2) $ \Longrightarrow$ $ \triangle IMO$ $ \sim$ $ \triangle DBE$
Now $ \triangle IMO$ and $ \triangle DBE$ are similar $ \Longrightarrow$ $ IM \bot BE$ , $ BE \bot OM$ $ \Longrightarrow$ $ OI \bot DE$
solution 2)
$ \triangle IMO$ $ \sim$ $ \triangle DBE$ $ \longrightarrow$ $ \frac{DE}{OI}$ = $ \frac{BE}{OM}$ = $ \frac{AB}{R}$
$ \frac{AB}{R}$ = $ 2\sinh$ $ \Longrightarrow$ $ \frac{DE}{OI}$=$ 2\sinh$. radius of circumcircle $ CDE$=$ R^'$.
$ \frac{DE}{R^'}$=$ 2\sinh$ , hence $ \frac{DE}{OI}$ = $ \frac{DE}{R^'}$ $ \Longrightarrow$ $ OI=R^'$

Dear Mathlinkers,
an article concerning some results about “two equal segments on two sides of a triangle” has been put on my website
http://perso.orange.fr/jl.ayme    vol. 6 Deux segments égaux sur deux côtés d'un triangle  p. 23
You can use Google translator
Sincerely
Jean-Louis


limit
function
Functional Analysis
absolute value
real analysis
real analysis unsolved

$g_n$ are bounded variations on $[a,b]$, n=1,2,...
$\{g_n(a)\}$ is a Cauchy sequence and we have
$\lim_{n\to\infty\atop m\to\infty} \bigvee_a^b (g_n - g_m) = 0$
Prove that there exists a bounded variation $g$ on $[a,b]$ such that
$\lim_{n\to\infty} \bigvee_a^b (g_n-g) = 0.$

In other words, the set of functions of bounded variation on $[a,b]$ is a Banach space, where the norm is the total variation plus the the absolute value at one point (such as $a$.) The result you are asking for here is the completeness.


[quote="conan6032"]Let $ p\geq3$ is a prime number and $ a_{1}, a_{2},..., a_{p \minus{} 2}$ are positive integers such that : $ a_{k}$ and $ a_{k}^{k} \minus{} 1$ don't divide by $ p$, for all $ k \equal{} 1,2,...,p \minus{} 2$. Prove that :
$ \forall m\in N\bigcap [1,p \minus{} 1], \exists S\subset N\bigcap [1,p \minus{} 2], \prod_{i\in S} a_i\equiv m (mod p)$[/quote]
Nobody can solve my problem??? :( . Pleases, help me  :(


Let $n$ be odd integer. What is the condition(s) of $n^{4}$ is sum of $n$ odd perfect squares?

Find all positive odd integers $n$ for which there exist odd integers $x_{1},x_{2}, . . . ,x_{n}$ such that $x_{1}^{2}+x_{2}^{2}+. . .+x_{n}^{2}=n^{4}$

Are there anybody who solve this question? :maybe:


Solve the ecuation x!(x!+y!)(x!+y!+z!)=(x!+y!+z!)!

[quote="Singular"]with equality when x, y, z = 1.[/quote]

You're being lazy.  The equality conditions are when $y! = z! = 1$ and $(x!)! = x!$, so there are twelve solutions, every combination of $y, z \in \{0, 1\}$ and $x \in \{0, 1, 2\}$


Welcome to the 4th rating game...... Rate the Location!!!! Rate the Location 1-10....

EDIT: Oh, sorry. Forgot to say that the location is under the post amount, which is under the joined date, which is under the word saying Online or Offline, which is under the Username of the person who posted.

1

how can you be at your house all the time?

9 creative

Why do I have this feeling that this will turn into the Hurt n' Heal of last year?

[quote="evilhamster"]1.

Advertising your site?[/quote]

its my aops blog, but the url wont fit

2 cause everyone says its Antarctica

and this will turn into the hurt and heal of some time ago...

1

How can you be in the middle of nowhere all the time?

7 for being poetic(?), but sadly too poetic for the stupid me to understand. (Are you talking about the ocean?)

$ 5$, how can you be lost in a sushi if you are bigger than a sushi.

2.7+3.4-O.32328 x 5- 5+6 ignoring ooo (order of operations)

3
BEE mor e CrEaTiVe\
IF you Know What I MEEAN


ceiling function
modular arithmetic
logarithms
number theory proposed
number theory

There have been two threads proving separately the existence of infinitely many $n$ with $n,n+1,n+2$ all squarefree (Proof: the density of squarefree numbers is $6/\pi^{2}> 1/2$), and infinitely many $n$ with $n^{2}+1$ squarefree (Proof: for arbitrarily large $N$, the proportion of numbers in $n$ in $[1,N]$ with $n^{2}+1$ not squarefree is less than $2\sum_{p \leq N}\lceil N/p^{2}\rceil /N \sim 2\sum_{p \leq N}1/p^{2}< 1$, where the sums are over the primes $1 \mod{4}$). Unify both results as follows: there are infinitely many $n$ for which $n,n+1,n+2$ and $n^{2}+1$ are simultaneously squarefree.

We will show there are infinitely many positive integers $k$ such that $n = 36k+1$ satisfies the desired properties.  By this choice, $n$, $n+1$, $n+2$, and $n^{2}+1$ are not divisible by 4 or 9.  So we can ignore the prime factors 2 and 3.

Let $p$ be a prime at least 5.  There is exactly one $k \pmod{p^{2}}$ such that $n$ is divisible by $p^{2}$, exactly one $k \pmod{p^{2}}$ such that $n+1$ is divisible by $p^{2}$, and exactly one $k \pmod{p^{2}}$ such that $n+2$ is divisible by $p^{2}$.  It is a little more work to see that there are at most two $k \pmod{p^{2}}$ such that $n^{2}+1$ is divisible by $p^{2}$.  All together, there are at most five $k \pmod{p^{2}}$ such that $n$, $n+1$, $n+2$, or $n^{2}+1$ is divisible by $p^{2}$.

Let $N$ be a large positive integer.  By the paragraph above, there at most $5 \left\lceil \frac{N}{p^{2}}\right\rceil$ values of $k$ between $1$ and $N$ such that $n$, $n+1$, $n+2$, or $n^{2}+1$ is divisible by $p^{2}$.  If $p > 36N+1$, then $p$ is too large for $n$, $n+1$, $n+2$, or $n^{2}+1$ to be divisible by $p^{2}$.  All together, the number of $k$ between $1$ and $N$ such that $n$, $n+1$, $n+2$, or $n^{2}+1$ is not squarefree is at most
\[\sum_{p=5}^{36N+1}5 \left\lceil \frac{N}{p^{2}}\right\rceil \, . \]

We can bound that sum as follows:
\begin{align*}\sum_{p=5}^{36N+1}5 \left\lceil \frac{N}{p^{2}}\right\rceil &\le \sum_{p=5}^{36N+1}5 \frac{N}{p^{2}}+5 \pi(36N+1) \\ &= 5N \sum_{p=5}^{36N+1}\frac{1}{p^{2}}+O(\frac{N}{\log N}) \\ &\le 5N \sum_{p \ge 5}\frac{1}{p^{2}}+o(N) \, . \end{align*}
So we will be done if we can show that
\[\sum_{p \ge 5}\frac{1}{p^{2}}< \frac{1}{5}\, . \]
Here's one way to bound that final sum:
\begin{align*}\sum_{p \ge 5}\frac{1}{p^{2}}&\le \sum_{m=2}^\infty \frac{1}{(2m+1)^{2}}\\ &\le \sum_{m=2}^\infty \frac{1}{(2m-1)(2m+1)}\\ &= \frac{1}{2}\sum_{m=2}^\infty \left(\frac{1}{2m-1}-\frac{1}{2m+1}\right) \\ &= \frac{1}{2}\cdot \frac{1}{3}\\ &= \frac{1}{6}\, . \end{align*}
So we have indeed bounded the sum by the desired amount.  That means at least one-sixth (asymptotically) of $k$ between $1$ and $N$ are such that $n$, $n+1$, $n+2$, and $n^{2}+1$ are squarefree.  Letting $N$ approach infinity, we get infinitely many such $k$.


I was bored, and decided to visit the AMC site again. I decided to look through the USAMO rosters for recent years to see the number of 12th graders who qualified for the USAMO (and will not be able to take it again next year) in those years:

2003 - 84/250 students (33.6 %)
2002 - (grades not posted)
2001 - 69/268 students (25.7 %)
2000 - 64/239 students (26.8 %)

It is clear that an exceptionally high number of students took the USAMO this past year, and so perhaps, because of that, more younger students will be able to compete for the USAMO spots this upcoming year.

Now, I know that this post is somewhat sick, but if this prediction of a .001 better chance for the reader to make USAMO next year makes that reader feel just .001 times better, then this post has served its use.  8-)

lol. i was just about to post that, when i saw yours. loooooooooooooooooooooooooooooooooooooool. awesome possum!

Not quite. It would be "1.001 times as good", but I didn't say that. I said ".001 times better". There is a big difference.

Your statement is quite ambiguous though.

Well, I think the term "better" implies an improvement, or addition.

[quote]i agree.[/quote]

Uh...with whom?

[quote]Umm...I'm not really comforted, and quite confused. But oka, whatever you say...[/quote]

Im saying that if an exceptionally large number of seniors took USAMO last year, an exceptionally large number of spaces for new kids would be available this year (though of course, it is certainly not certain that every other kid who took the USAMO last year will make it back this year, but hey, its just a bit of rough logic).

Okay.  I could argue that the differences between 2000 and 2001 were too small to really mean anything, but I'll leave it at that.  I think of it this way: if you qualify, you'll qualify.   :wink:

[quote]Okay. I could argue that the differences between 2000 and 2001 were too small to really mean anything, but I'll leave it at that. [/quote]

Not the difference between 2000/2001 and 2003...well..actually...sorta...15 ppl isn't that big of a deal...

[quote="Fierytycoon"][quote]i agree.[/quote]

Uh...with whom?[/quote]

since my post came immediately after yours, who else could i possibly have been agreeing with???

[quote]since my post came immediately after yours, who else could i possibly have been agreeing with???[/quote]

 :D


function
combinatorics unsolved
combinatorics

Given set $S: =\{1, 2, ..., n\}$, find the number of partitions of $S$ such that each subset in partition contains only $1$ or $2$ elements.

[quote="pbornsztein"]So you are looking for the number of involutions on $\{1,\cdots ,n\}$.
This is a classical problem.

Pierre.[/quote]

Can you explain a little bit, or give a link?

Let $Q(n)$ be the number of involutions on $S_{n}= \{1, \cdots , n \}$.

For any such involutions, either $f(n)=n$ so that there $Q(n-1)$ involutions in this case. Or, $f(n)=k$ for some $k\in S_{n-1}$. There are $n-1$ possible choices for $k$ and, for each of them, since $f(k) = n$, there are $Q(n-2)$ ways to define $f$ on $S_{n-1}-\{k\}$.
Therefore $Q(n) = Q(n-1)+(n-1)Q(n-2).$

Pierre.

[quote="pbornsztein"]$Q(n) = Q(n-1)+(n-1)Q(n-2).$
Pierre.[/quote]

Thank you. I got this recursion too but I thought there is maybe a closed formula or something like that. Looks like there isn't.

A number of different formulas for the sequence and properties of the associated generating function, as well as different characterizations of the sequence, are available at the OEIS: http://www.research.att.com/~njas/sequences/A000085


Determine all pairs $(h,s)$ of positive integers with the following property:

If one draws $h$ horizontal lines and another $s$ lines which satisfy
(i) they are not horizontal,
(ii) no two of them are parallel,
(iii) no three of the $h + s$ lines are concurrent,

then the number of regions formed by these $h + s$ lines is 1992.


AMC AIME geometry algebra polynomial trigonometry logarithms

I'm going to assume everyone knows how a marathon works. Iff successful, I will compile this into a .pdf for practice.

[b]Problem Criteria:[/b]
 AIME to easy USAMO.
Only Algebra and Equations(Single var, Double var, Diophantine, anything)
Good analyctic geometry, but only if it shows a cool algebraic principle.
[b]No Polynomials.[/b]

[size=150][b][i]Problem 1[/i][/b][/size]
[size=75]AIME1 2000 3[/size]
Integers m, n are relatively prime positive integers. The coefficients of $x^2$ and $x^3$ in the expansion of $(mx + n)^{2000}$ are equal. Find m + n.

[hide]

Equating coefficients, we get , which simplifies down to , or , so the relatively prime pair  is , and our answer is .[/hide]

Next problem:

Let  and  be any real number. How many ordered pairs  satisfy the equation ?

[quote="EFuzzy"]Um, technically (mx+n)^2000 is a polynomial.[/quote]
But we aren't using anything beyond really basic polynomial stuff.

Why no polynomials? That's one of the coolest parts of algebra!

Let $a$ and $b$ be any real number. How many ordered pairs $(a,b)$ satisfy the equation $(a+bi)^{2002}=a-bi$

[hide="solution"]
$z^{2002}=\overline{z}$, conjugate both sides, $\overline{z}^{2002}=z$
multiply, $z^{2003}=\overline{z}^{2003}$  $r^{2003}e^{2003i(\theta+2\pi k)}=r^{2003}e^{-2003i(\theta+2\pi k)}$

$r^{2003}\left( e^{2003i(\theta+2\pi k)}-e^{-2003i(\theta+2\pi k)}\right)=0$
$r^{2003}*2\sin(2003\theta)=0$

$r=0$ or $\theta=\frac{k\pi}{2003}$

then it is pretty obvious because of cyclicing angles, $k=0,1,2,...4005$ giving us $\boxed{4007}$ solutions[/hide]

[hide]Let $z=a+bi$, then $z^{2002} = \overline{z}$. Multiply both sides by $z$ to get $z^{2003} = z\overline z = |z|^2$. Clearly this is satisfied if $z$ is a 2003rd root of unity, and of course $z=0$ also works, which gives us $\boxed {2004}$ total solutions.[/hide]

[hide]after checking the angles back in the equation, i see that i had extra solutions, but i don't see why...the answers are $z=0, e^{\frac{2i\pi k}{2003}}$ for $k=0,1,...2002$ but the way that i did it with $\sin{2003\theta}=0$ should work too, but i guess that it does not, somehow i added solutions when i did what i did, so i am pretty sure you are right...it is 2004, also i am pretty sure that every step i took was rigirous, but there is some mistake

but i like your approach too, another thing to add is

$|z^{2002}|=|z|^{2002}=|\overline{z}|=|z|$, we solve that $|z|=0,1$ (it cannot be complex)

$z^{2003}=r^2=1$ and that obviously has $2003$ solutions, adding for $z=0$, we get $\boxed{2004}$[/hide]

[quote="Altheman"][quote="chess64"]Okay, here's one.

[size=150]Problem 3[/size]

Factor completely over reals: $x^{12} - 98x^8y^4 + 1393x^4y^8 - 1296y^{12}$.[/quote]

[hide="solution"]use the substitution: $z=\frac{x^4}{y^4}$ to help factor, then use difference of squares to finish it:
$y^{12}(z^3-98z^2+1393z-1296)$
$=y^{12}(z-1)(z-16)(z-81)$
$=(x^4-y^4)(x^4-16y^4)(x^4-81y^4)$
$=(x^2+y^2)(x-y)(x+y)(x^2+4y^2)(x-2y)(x+2y)(x^2+9y^2)(x-3y)(x+3y)$
(and since it is over reals, we do not factor the terms like $x^2+y^2$[/hide]

and is number theory allowed, because i made up a good algebra/number theory problem[/quote]
Sure.

$\{N\}$ is the set of the natural numbers and $N^*=N-\{0\}$.

Hence, in Romania zero is considered a natural number - good to know :)

Sorry, I didn't know the level of this "PreOlympiad" !. Soon, some other time I will offer a more easy problems.[hide="A generalization of the fifth problem."]
Solve the equation $a+{b\over {a+{b\over {a+_{{.}_{{.}_{{.}_{+{b\over{a+{b\over x }}}}}}}}}}}=x$, where $b\ne 0$ and the bar is repeated $n$ times.[/hide] [hide="Answer"] For any $n\in N^*$ the solutions are the roots of the equation $x^2=ax+b$, i.e. $x=a+\frac bx.$[/hide] [hide="Another problems."] $\blacksquare\ V.N.9^{\circ}$. Solve the equation $2^x+2^{\sqrt{1-x^2}}=3.$
$\blacksquare\ V.N.10^{\circ}$. If $y-x\le 1,\ y+x\le 1,\ x^2-1\le y$, then ascertain $\min y$ and $\max y.$
$\blacksquare\ V.N.11^{\circ}$. The system of the equations $x_1-x_2=x_3-x_4=a$, $x_1+x_2+x_3+x_4=1$ has at least a positive solution, i.e. $x_k>0,\ k\in \overline {1,4}$ if and only if $|a|<\frac 12.$
$\blacksquare\ V.N.12^{\circ}$. Let $0<a_1<a_2<\ldots <a_{n-1}<a_n$ and the set $S_n$ of the all permutations $\sigma : \overline {1,n}\rightarrow \overline {1,n}$. Ascertain the permutation $\sigma$ for which the sum $S_{\sigma}$ is maximum (minimum), where:
$12.1.\ S_{\sigma}=\sum\limits_{k=1}^n \frac{a_k-a_{\sigma (k)}}{a_k+a_{\sigma (k)}}.$
$12.2.\ S_{\sigma}=\sum\limits_{k=1}^n \ \ln\left(a^{\alpha}_k +a^{\beta}_{\sigma (k)}\right)$, where $\alpha \ne 0$ and $\beta \ne 0.$
$\blacksquare\ V.N.13^{\circ}.\ \left(\forall\right)\sigma \in S_n,\ \sum\limits_{k=1}^n\frac{\sigma (k)}{k^2}\ge \sum\limits_{k=1}^n \frac 1k.$[/hide]

[quote="Virgil Nicula"]
$\blacksquare\ 4^{\circ}.\ n\in 2N+1,\ \left(\forall\right)\{i,j\}\subset\overline {1,n},\ i\ne j,\ x_i\in Z,\ x_i\ne x_j\Longrightarrow$$\sum\limits_{k+1}^n x^2_k\ge \frac{n(n^2-1)}{12}.$
[/quote]

[hide="Proof"]Denote $k={{n-1}\over 2}.$ It is obvious that $S=\sum_{j=1}^{n}{x^2_j}$ is minimal for numbers $0, \pm 1,\pm 2,\dots,\pm k$, since in such a way we obtain the lowest possible squares with all the numbers different. Hence $S_{min}=2\sum_{j=1}^{k}{j^2}={{k(k+1)(2k+1)}\over 3}={1\over 3}{{n-1}\over 2}{{n+1}\over 2}n={{n(n^2-1)}\over 12}.$ QED[/hide]

Farenhajt, see in the anterior reply I added another three problems.

[quote="Virgil Nicula"]
$\blacksquare\ 7^{\circ}.\ \left(\forall\right) n\in N^*,\ \left(\exists\right) x_n\in (0,1)$, so that $n\left(x^{n-1}_n-x^n_n\right) >\frac 1e.$[/quote]

[hide="Proof"]Define $f(x)=n(x^{n-1}-x^n)=nx^{n-1}(1-x).$ For $x_0=1-{1\over n}$ we have $f(x_0)=(1-{1\over n})^{n-1}>e^{-1}$, since it's well known that the sequence $a_n=(1-{1\over n})^{n-1}$ is strictly decreasing and converging to $e^{-1}.$[/hide]

[quote="Virgil Nicula"]
$\blacksquare\ 11^{\circ}$.The system of the equations $x_1-x_2=x_3-x_4=a$, $x_1+x_2+x_3+x_4=1$ has at least a positive solution, i.e. $x_k>0,\ k\in \overline {1,4}$ if and only if $|a|<\frac 12.$
[/quote]

[hide="Proof"]It's very easy to find that $(x_1,x_2,x_3,x_4)=({1\over 2}-t, {1\over 2}-a-t,a+t,t), t\in\mathbb{R}$. (Put $x_1=x_2+a, x_3=x_4+a$, plug that into the third equation and put $x_4=t.$). Hence $x_4>0\Leftrightarrow t>0$ and $x_1>0\Leftrightarrow t<{1\over 2}.$ Now $x_3>0\Leftrightarrow a>-t>-{1\over 2}$ and $x_2>0\Leftrightarrow a<{1\over 2}-t<{1\over 2}.$ Therefore, $-{1\over 2}<a<{1\over 2}\Leftrightarrow |a|<{1\over 2}.$[/hide]

[quote="Virgil Nicula"]
$\blacksquare\ 10^{\circ}$. If $y-x\le 1,\ y+x\le 1,\ x^2-1\le y$, then ascertain $\min y$ and $\max y.$
[/quote]

[hide="Solution"]$y\le 1+x\land y\le 1-x\Leftrightarrow y\le 1-|x|.$ Hence $|x|^2-1\le y\le 1-|x|.$ Therefore the following must be valid: $|x|^2-1\le 1-|x|$. It's easily found that this is equivalent to $|x|\le 1$. Now it's obvious that $0-1\le |x|^2-1\le y\le 1-|x|\le 1-0 \implies -1\le y\le 1.$ The boundary values can be taken if $x=0.$[/hide]

[i]Dear Farenharjt[/i], I ask you if this last problems are in accordance with the level of this section "PreOlympiad" and if you wish another same problems. Please, answer me.

Well, some seem to be quite easy, and some seem difficult. Let's see what the others think :)

[quote="Virgil Nicula"]
$\blacksquare\ 8^{\circ}$. Study the nature of the sequence $a_1=0,\ a_{n+1}=a_n\sqrt 2+\sqrt{2+a^2_n},\ n\in N^*.$[/quote]

[hide="Detailed solution"]I'm not sure if the explicit formula for $a_n$ is required, since otherwise it's obvious that the sequence is strictly increasing and unbounded.

(Note: For the easier calculation, we'll study the sequence with the indices starting from zero, i.e. $a_0=0.$)

We have

$a_{n+1}=a_n\sqrt{2}+\sqrt{2+a_n^2}\qquad(*)$

Hence

\begin{eqnarray*}a_{n+2}&=&a_{n+1}\sqrt{2}+\sqrt{2+a_{n+1}^2}\\&=& 2a_n+\sqrt{4+2a_n^2}+\sqrt{2+2a_n^2+2a_n\sqrt{4+2a_n^2}+2+a_n^2}\\&=& 2a_n+\sqrt{4+2a_n^2}+\sqrt{4+2a_n^2+2a_n\sqrt{4+2a_n^2}+a_n^2}\\&=& 2a_n+\sqrt{4+2a_n^2}+a_n+\sqrt{4+2a_n^2}\\&=& 3a_n+2\sqrt{4+2a_n^2}\\&=& 3a_n+2\sqrt{2}\sqrt{2+a_n^2}\qquad(**)\end{eqnarray*}

Expressing the radical from $(*)$ and substituting into $(**)$ we get

\begin{eqnarray*}a_{n+2}=3a_n+2\sqrt{2}(a_{n+1}-a_n\sqrt{2})=2\sqrt{2}a_{n+1}-a_n\implies a_{n+2}-2\sqrt{2}a_{n+1}+a_n=0\end{eqnarray*}

The characteristic equation is $t^2-2\sqrt{2}t+1=0$ and the solutions are $t_{1,2}=\sqrt{2}\pm 1$

Now we can write

$a_n=A(\sqrt{2}+1)^n+B(\sqrt{2}-1)^n$, where $a_0=0, a_1=\sqrt{2}.$

Substituting $n=0,1$ and solving the resulting system for $A,B$ we get $A={1\over\sqrt{2}}, B=-{1\over\sqrt{2}}$

Hence $a_n={{(\sqrt{2}+1)^n-(\sqrt{2}-1)^n}\over\sqrt{2}}$

Now it's easy to see that $a_{2k}$ are integer numbers, since in the binomial expansions the terms NOT containing $\sqrt{2}$ will cancel out, and the remaining $\sqrt{2}$ will cancel out with the denominator. Conversely, in $a_{2k+1}$ the integer parts of the binomial expansions will add to one another, and since they are mutually equal, after dividing with $\sqrt{2}$ we'll have a number of the form $m\sqrt{2}, m\in\mathbb{N}.$

What else there's to say about $a_n$?[/hide]

I wish this was still in the intermediate forum, these problems are too hard for me... :( oh well

If after some time, any exercises wasn't solved, you can ask me to offer my its solutions. With the greatest pleasure !
[hide="Here are another exercises."]
$\blacksquare\ V.N.14^{\circ}.\ 1<x<y,\ 0<z\Longrightarrow \log_x y>\log_{x+z} (y+z).$
$\blacksquare\ V.N.15^{\circ}$. Ascertain the general term for the sequence $x_1\ge 0,\ x_{n+1}=|x_n-n|,\ n\in N^*.$[hide="Answer for V.N.15."]$x_n= \left \{ \begin {array}{cc} x_1-\frac 12\cdot n(n-1),\ &\ if\ \ n\in \overline {1,p}\\ (-1)^{n-p}\left(x_1-\frac{p^2}{2}\right)+\frac n2 -\left\{\frac{n-p}{2}\right\},\ &\ if\ \ n\ge p+1\end{array}\right.$,
where $p=\left[\frac 12\left(1+\sqrt{1+8x_1}\right)\right]$, i.e. $\frac{p(p-1)}{2}\le x_1<\frac{p(p+1)}{2}$ and $\{x\}=x-[x]$.
[u]Remark.[/u] $\lim_{n\to \infty} \frac{x_n}{n}=\frac 12.$[/hide] $\blacksquare\ V.N.16^{\circ}.\ a_k\in (0,1),\ k\in \overline {1,n}\Longrightarrow \sum_{k=1}^n \log_{a_k} \frac{na_1\cdot a_2\cdot \ldots \cdot a_n}{a_1+a_2+\ldots +a_n}\ge n(n-1).$
$\blacksquare\ V.N.17^{\circ}.\ x^3_k=x_k,\ k\in \overline {1,n}\Longrightarrow$ $\sum_{1\le i<j\le n}x_i x_j\ge -\left[\frac n2\right]$, where $z\equiv [x]\in Z,\ z\le x<z+1.$
$\blacksquare\ V.N.18^{\circ}.\ a,b,c,d>0\Longrightarrow 1<\frac{a}{a+b+d}+\frac{b}{a+b+c}+$ $\frac{c}{b+c+d}+\frac{d}{a+c+d}<2.$
$\blacksquare\ V.N.19^{\circ}.\ \left(\exists\right)x\in R,\ x^4+ax^3+bx^2+ax+1=0\Longrightarrow a^2+b^2\ge \frac 45.$
$\blacksquare\ V.N.20^{\circ}.\ n\in N^*\Longrightarrow \frac{2n}{3n+1}<\sum_{k=1}^n \frac{1}{n+k} <\frac{-1+\sqrt{1+8n}}{4n}.$
\[ \boxed {\ \mathrm {I\ shall\ continue\ in\ the\ following\ my\ reply}\ }  \][/hide]

[quote="Virgil Nicula"]
$\blacksquare\ V.N.13^{\circ}.\ \left(\forall\right)\sigma \in S_n,\ \sum\limits_{k=1}^n\frac{\sigma (k)}{k^2}\ge \sum\limits_{k=1}^n \frac 1k.$[/quote]

[hide="This easy?"]Isn't this obvious from Rearrangement Inequality? The initial permutation is $\sigma_0(k)=k$, hence the result.[/hide]

Denote $x\ .a.s.\ y\Longleftrightarrow xy>0\ \vee\ x=y=0$. You must to have in view the effect of a [u]transposition[/u] $\tau =(i,j)\in S_n$, i.e. $i\ne j$, $\tau (i)=j$, $\tau (j)=i$, $\tau (k)=k$, $k\not\in \{i,j\}.$

Denote $\boxed {\ S(\sigma )=\sum_{k=1}^n \frac{\sigma (k)}{k^2}\ }\ .$ Prove easily that

$S(\sigma \circ\tau )-S(\sigma )=\frac{\sigma (i)}{j^2}+\frac{\sigma (j)}{i^2}-\frac{\sigma (i)}{i^2}-\frac{\sigma (j)}{j^2}\ .a.s.\ [\sigma (i)-\sigma (j)](i-j)$.

Therefore, if the permutation $\sigma$ has not the inversion $(i,j)$, i.e. $[\sigma (i)-\sigma (j)](i-j)>0$, then $S(\sigma\circ\tau)>S(\sigma)$. So that, the sum $S(\sigma)$ is minimum if and only if the permutation $\sigma$ has not one inversion, i.e. $\sigma =\iota$, where $\iota (k)=k,\ k\in \overline {1,n}$ (the [u]identical permutation[/u]).

We came to the conclusion that for any $\sigma \in S_n$, $\boxed {\ S(\sigma)\ge S(\iota )=\sum_{k=1}^n \frac 1k\ }\ .$

[b]Remark.[/b] Similarly prove and the exercises $12.1$ and $12.2.$


I'm not sure if this applies to other sciences as well, but I've noticed that the reference section of most math papers is written like this:

[1] author last name, first name ....

[2] ...

[3] ...

that is, with square brackets. What is the accepted convention for this? Also, what is the order of everything after the author's name? Can anyone suggest to me a good reference, preferably on the Internet, that describes how to do this? I'm in the midst of writing a paper for Siemens, which is due very, very soon, and I wanted to know how to properly do the reference section. Thanks!

So, the square brackets don't matter at all, but if you think they're clear and helpful, use them.  Generally, I believe the order (when citing a paper published in a journal) is:

Last Name, First Name.  "Title of Article," [i]Name of Journal[/i], Date, Pages.

The punctuation doesn't matter much at all -- in fact, the order doesn't matter much, either, although that order that I just wrote seems to be sensible.  Just try and be consistent in your citing, making sure that you include the same amount of information for each source and that you give it in the same order and so on.  No one is going to throw you out of competition because you didn't format your bibliography by MLA rules.


this is question #18, on the 2004 AMC 10 B.

In a right triangle ACE, we have AC=12, CE=16, and EA=20. .......

The solution was to drop an altitude from point F to line DE, so that a 3,4,5 triangle forms.  How do you know weather the altitude is 3 or 4?

Thank you :lol:

Where is F and where is D?  I'm not sure by what you said.

Um do you mean from a point F on length AE to a point D on CE? Because the lengths of triangle ACE are in ratio 3:4:5

O! I see it.  yea I meant CE and point F on AE. Thanks.

Just put the whole question up there and then ask any additional questions.  :mad:


Are we allowed to started talking about this? Unless I am mistaken, the competition is over for everyone, correct?

I will only say that I improved my score from last year by 32!! :yup:

Well, here are my Chapter and State scores;

Chapter Sprint - 19
Chapter Target - 14
Chapter Total - 33
State Sprint - 13
State Target - 8
State Total - 21

I know, I did really bad.

I think that, even when the competition is running, that you are allowed to say your score and what you ranked.

[quote="hurdler"]Um, I don't believe this has anything to do with mathcounts, or does NATS have some sort of local competition?[/quote]

How does it not have anything to do with MC, because those are my scores. You posted something about your scores.  :huuh:

Wickedestjr, please pay attention to which forum you are posting in.

Oh, I thought that you guys were talking about MC, well, what is the fall start-up?

Hmmm, I found this years test to be especially easy...I finished with 8 minutes left, but alas I decided to make millions of stupid mistakes.

I think I got like an 89 =\

can you stop being annoying and research it yourself...last time i checked there is such a thing as google

Its just this competition that you take in your school every fall if your school does it.

Wow good job stupidityismygame. I only got a 72.

Hm, I believe the results come out this coming Monday, 10/13/2008.

Cant wait to see the results.

OH EM GEE

I got 58, and that placed 2nd at my local high school...im in 7th grade.

first place was a freshman, at 59.

Our school STINKS.  i never will finish taht test :(

STUPIDITY HOW DID YOU FINISH WITH 8 MINS LEFT!?!?!?!?  dude thats scary...13.2 sec per problem...

[quote="AIME15"]
STUPIDITY HOW DID YOU FINISH WITH 8 MINS LEFT!?!?!?!?  dude thats scary...13.2 sec per problem...[/quote]

I cheated

hehehe...

BTW wicked, if someone improved their mc score by 32, they must be genius this year or really dumb last year.

*AIME15 wants the answer key to next year's test!!!!!

On the scientific notation question, I accidentally put $ 10^2$ instead of $ 10^{\minus{}2}$ :(

ANd i have no chance on the last 10 q's, i dont know calc/precalc/alg 2

Ultimate phailure - 78.

I arrived 6 minutes late :(

78 IS NOT PHAIL.  Did you see my score?

Do you guys think it's a good idea to look at all the questions first?  So I don't stay on some stupid geometry problem when there's like Whats the range of the set etc. problems after that?

anything above 70 is top 5 in the nation, no joke. Google "2007 Fall Startup results" and look at the top scorers from each grade level...an 89 is nearly impossible.
I got a [b]64[/b]  and I'm in 12th grade...lol but I only got 35 last year

No, I dont think 70 is top 5.
In our school we had at least 6 ppl go >70

Yay the results are up. I guess this year just had an easier test. The highest score was a 94. That's pretty amazing. The sixth grader who won the middle school division is also apparently pretty amazing.

I wish my school had more people interested in math. The seniors' high score was a 49.........lower than all the other grades' highest at my school

I'm #4 for middle school!  Woot!

WHat place did you guys get in your grade?

argh dude I forgot this contest existed... I liked doing it too :(

GJ fan

3rd place for middle school.  u can see the results [url=http://www.natassessment.com/2008_FSE/2008_FSE_Results/2008_FSE_Results_FrameDef.htm]here[/url]

Fan how come your school is so beastly?  And you told me you lived in california....lier

22nd in 12th grade...do they give out awards for this place?

[quote="AIME15"]Uhh i don't think they give out awards.[/quote]

wrong

I think they give out plaques for top 15, I don't know about anything else. Sorry

oh really?? top 15 total or per grade??

awesome...but it doesnt say that on the site

Congrats to all…do they give out the same test to all grades?  That would make things nice and easy to compare…
AAST is BEAST.

I've gotten top 15 a number of times and never gotten a plaque of any sort. I have, however, received certificates for top 25 scores and "best in my school" scores.

Got 12th in Juniors this time around; not bad, not great.

First Middle school  :)

83, not bad.

I guessed 4 for the last like 20 problems. Must have gotten some right.

First as a freshmen.

85, not bad.

BOGTRO is like 3 years younger than me. How is this fair? -_-

2nd middle school, which is good considering i only answered about 70 and put 4 for about 30 questions.

[quote="hurdler"] The sixth grader who won the middle school division is also apparently pretty amazing.
[/quote]

Why thank you  :lol:

-_____- wow I feel stupid. I got a 57, but I skipped one and forgot to skip a space on the answer sheet until like 15 problems later >.< I would've been top 10 ish for freshmen, but alas, my teacher didn't grade the tests until a month after the results came out.

I'm sorry this is something of a revival, but is there any way I could get one of the 100 problem tests?

Well, I'm pretty sure I can't rope my school into signing up(ditto for the AMC8) but I really would like to see how I would've done


Given $ a, b, c \geq\ 0.$ satisfy $ ab\plus{}bc\plus{}ca\equal{}1.$
Prove that: $ (a \plus{} bc)(b \plus{} ca)(c \plus{} ab) \leq\ 1$

Very sorry because my problem was wrong. Honey_S checked it  :oops:
But the weaker is true:
$ \frac{1}{a\plus{}bc} \plus{} \frac{1}{b\plus{}ca} \plus{} \frac{1}{c\plus{}ab} \geq\ 3$
(with the same conditon)  :)

[quote="nguoivn"]Given $ a, b, c \geq\ 0.$ satisfy $ ab \plus{} bc \plus{} ca \equal{} 1.$
Prove that: $ \frac {1}{a \plus{} bc} \plus{} \frac {1}{b \plus{} ca} \plus{} \frac {1}{c \plus{} ab} \geq\ 3$[/quote]

$ \frac {1}{a \plus{} bc} \plus{} \frac {1}{b \plus{} ca} \plus{} \frac {1}{c \plus{} ab} \geq\ 3$

$ \Longleftrightarrow (a \plus{} bc)(b \plus{} ca) \plus{} (b \plus{} ca)(c \plus{} ab) \plus{} (c \plus{} ab)(a \plus{} bc)\geq 3(a \plus{} bc)(b \plus{} ca)(c \plus{} ab)$

$ \Longleftrightarrow (a \plus{} b \plus{} c)(abc \plus{} 1) \minus{} 3abc \plus{} 1\geq 3\{abc(a \plus{} b \plus{} c \minus{} 1)^2 \plus{} (abc \minus{} 1)^2\}$.

[quote="nguoivn"]Given $ a, b, c \geq\ 0.$ satisfy $ ab \plus{} bc \plus{} ca \equal{} 1.$
Prove that: $ \frac {1}{a \plus{} bc} \plus{} \frac {1}{b \plus{} ca} \plus{} \frac {1}{c \plus{} ab} \geq\ 3$[/quote]
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, it's obvious that our inequality is equivalent to $f(w^3)\leq0$, where $f$ is a convex function.
Hence, by $uvw$ we need to check two cases:
1. $b=a$. Hence, $a\neq0$ and the condition gives $c=\frac{1-a^2}{2a}$.
Hence, $f(w^3)\leq0\Leftrightarrow\frac{(1-a)(1-3a+3a^2+9a^3-6a^4)}{(1+2a-a^2)(2a^3-a^2+1)}\geq0$, which is true for $a\in(0,1]$;
2. $c=0$. Hence, $ab=1$ and we need to prove that $\frac{1}{a}+\frac{1}{b}\geq2$, which is obvious.


geometry
real analysis
AMC
AIME
number theory
AMC 10
combinatorics theorems

i am really good at (algebra,geometry) , a little at number theory but really unable at combinatorics, can anyone give me a reason, an advice??

How much of a combinatorics background do you have? Combinatorics may seem pretty low-tech but there are several important results that you should understand (Vandermonde convolution, hockeystick identities, summation identities, and their derivations). My school has a set of decently complete combinatorics lectures that go from the basics up to simple olympiad problems, if you want to take a look at them. Other than that, though, it's really a thought process, and the best way to get better is to gain intuition. The best way to do that is by doing problems, which these lectures also have. Past AIME combo problems are also nice. Don't try to start at the Olympiad level if that's what you're trying to do, take it slow.

Basic:
http://activities.tjhsst.edu/vmt/arml-lectures/2006/3b-basiccombo.pdf
Intermediate (both lectures are very similar; in fact, one basically copied off the other):
http://activities.tjhsst.edu/vmt/arml-lectures/5-cnp/5-cnp.pdf
http://activities.tjhsst.edu/vmt/arml-lectures/2006/3a-advancedcombo.pdf
Advanced:
http://activities.tjhsst.edu/vmt/arml-lectures/2006/11a-advancedcombolecture.pdf
Advanced Solutions (though the answer to 10 is wrong):
http://activities.tjhsst.edu/vmt/arml-lectures/2006/11a-advancedcombosolutions.pdf

Personally I think mastery of the topics covered in these lectures should be sufficient to solve any pre-olympiad combinatorics problem. Try to solve all the problems without looking at the solutions if you can, it is the best way to learn.

Please,does someone have these lectures ? the links do not work.

yes, can someone repost the links? they don't work and i also desperately need combinatorics help (as in, starting from amc 12 or maybe even amc 10 level)

i ill be waiting for these lectures too, i hope someone can repost it


probability
number theory
relatively prime
recursion

A solitaire game is played as follows.  Six distinct pairs of matched tiles are placed in a bag.  The player randomly draws tiles one at a time from the bag and retains them, except that matching tiles are put aside as soon as they appear in the player's hand.  The game ends if the player ever holds three tiles, no two of which match; otherwise the drawing continues until the bag is empty.  The probability that the bag will be emptied is $p/q,$ where $p$ and $q$ are relatively prime positive integers.  Find $p+q.$

[quote="dengmi"][hide]Let's count the probability the game ends and subtract from 1.
Obviously, if 4 pairs have been made, the game is a sure win since the remaining bit cannot be rearranged to end the game.
Let's set up 4 cases.  0 pairs, 1 pairs, 2 pairs, and 3 pairs.
Case 1: The first pick has no restrictions so has a probablity of 1.  Since we don't want pairs, the second pick cannot match the first so we have a probability of 10/11.  The third pick cannot match the first or second so we have 8/10.  Therefore the probabilty of ending the game with 0 pairs is 8/11.
Case 2: Again, the first pick is unrestricted.  The second pick can either pick a pair or a nonpair.  We split into 2 subcases.  The subcase where we make the pair has a probability of 1/11 for the second move.  The third move has no restrictions because all the remaining tiles in the bag will not make a pair.  The fourth move can't make another pair and so has a probability of 8/9.  The chances that the fifth move does not make another pair and ends the game for that subcase is 6/8.  The other subcase is when the second move makes a nonpair.  The probabilty of that happening is 10/11.  Then, since getting a third nonpair ends the game by repeating Case 1, a pair must be made.  The chances of this happening is 2/10.  The fourth move must create another nonpair as does the fifth so the chances are 8/9 and 6/8 respectively.  When these are multiplied we get $(1/11*8/9*6/8)+(10/11*2/10*8/9*6/8)=3*\displaystyle\frac{6}{11*9}$.
Let's generalize what we figured out from case one and two.  When there's 0 tiles in our hand, the probability is not restricted on the turn we draw to obtain a tile.  Also, pairing up as soon as possible can be done in only one way.  Pairing up by aiming to have two tiles in hand until the maximum allowed pairs is reached would always result in $2^n$ amount of possible ways where n is the number of maximum pairs.  This is because every time a pair must be formed, there's two ways to form it.  By forcing the tiles to pair early replaces ones with twos in the multiplication.  The addition of all of these possiblities is $2^{n+1}-1$.  That is the number multiplied by the fraction formed by the forced case.  The forced case for Case 3 is $\displaystyle\frac{4}{11*9*7}$ and the one for Case 4 is $\displaystyle\frac{2}{11*9*7*5}$.  The numbers that correspond to that from the power of 2 multiplier is 7 and 15.  We have the $1-(8/11+2/11+4/99+\displaystyle\frac{2}{11*3*7})$.  This is $1-(94/99+\displaystyle\frac{6}{99*7})$.  This is 1-(664/693) or 29/693.  These numbers are relatively prime so p+q=722.[/hide][/quote]

[hide="hmm"]Starting with the third case, I'm getting different probabilities. The probability of getting three different cards in your hand on the third and fifth draws are $\displaystyle \frac{8}{11}$ and $\displaystyle \frac{2}{11}$, respectively, as you showed. This means that the probability of having one card in your hand on the fifth draw is $\displaystyle \frac{1}{11}$ since they account for all cases and so their sum must be equal to one. Starting from one card after the fifth draw, the probability of getting a hand of three different cards by getting two different cards in the next two draws is then equal to $\displaystyle \frac{1}{11} \times \frac{6}{7} \times \frac{4}{6} = \frac{4}{77}$. Drawing a tree diagram, and using this method of counting, I get $\displaystyle \frac{6}{385}$ as the probability of getting a hand of three on the ninth draw.

So the total probability of getting a hand of three cards is equal to:

$\frac{8}{11} + \frac{2}{11} + \frac{4}{77} + \frac{6}{385} = \displaystyle \frac{376}{385}$.

The probability of emptying the bag is then equal to $1 - \displaystyle \frac{376}{385} = \frac{9}{385}$.

Did I do something wrong :?
[/hide]

Yes, you're right.
The multiplying factor is $3^{\frac{n-3}{2}}$ where n is the odd turn number because $\binom{\frac{n-3}{2}}{0}2^0+\binom{\frac{n-3}{2}}{1}2^1+...+\binom{\frac{n-3}{2}}{\frac{n-3}{2}}2^{\frac{n-3}{2}}=3^{\frac{n-3}{2}}$.  The first term represents the number of ways that a multiplier of 1 is added.  The second term represents the number of ways that a multiplier of 2 is added.  The third term represents the number of ways that a multiplier of 8 is added, and so on.

[hide="recursion"]

Let $ P_k$ be the probability of emptying the bag when it has $ k$ pairs in it. Let's consider the possible draws for the first three cards:

Case 1. We draw a pair on the first two cards. The second card is the same as the first with probability $ \frac {1}{2k \minus{} 1}$, then we have $ k \minus{} 1$ pairs left. So this contributes probability $ \frac {P_{k \minus{} 1}}{2k \minus{} 1}$.

Case 2. We draw a pair on the first and third cards. The second card is different from the first with probability $ \frac {2k \minus{} 2}{2k \minus{} 1}$ and the third is the same as the first with probability $ \frac {1}{2k \minus{} 2}$. We are left with $ k \minus{} 1$ pairs but one card already drawn. However, having drawn one card doesn't affect the game, so this also contributes probability $ \frac {P_{k \minus{} 1}}{2k \minus{} 1}$.

Case 3. We draw a pair on the second and third cards. This is pretty much the same as case 2, so we get $ \frac {P_{k \minus{} 1}}{2k \minus{} 1}$.

Therefore, we obtain the recursion $ P_k \equal{} \frac {3}{2k \minus{} 1}P_{2k \minus{} 1}$. Iterating this for $ k \equal{} 6,5,4,3,2$ (obviously $ P_1 \equal{} 1$), we get
\[ \frac {3^5}{11*9*7*5*3} \equal{} \frac {9}{385}.
\]
[/hide]

The recursive solution is nice  :) a minor note: the recursion should read:

[hide]$ P_k \equal{} \frac {3}{2k \minus{} 1}P_{k \minus{} 1}\; , \qquad \text{not} \qquad P_k \equal{} \frac {3}{2k \minus{} 1}P_{2k \minus{} 1}$[/hide]

Recursion without casework:

Let $ f(n)$ denote the number of ways one can empty a bag with $ n$ pairs. To construct the recursion, note that the first 3 tiles must contain one matching pair and some other tile. For $ f(n\plus{}1)$ there are $ n\plus{}1$ ways to pick the matching pair, and then $ 3! \equal{} 6$ ways to permute the first three tiles. Then what is remaining is $ f(n)$.

$ f(n\plus{}1) \equal{} 6(n\plus{}1)f(n)$.
$ f(1) \equal{} 2$, $ f(6) \equal{} 6(6)f(5)\equal{}6^5(6!)(2)$.

$ \frac{f(n)}{12!} \equal{} \frac{6^4}{11 * 10 * 9 * 8 * 7} \equal{} \frac{9}{385}$.

[quote=HamstPan38825]Sorry for the bump, but I don't quite understand the recursion solutions.

In particular, why are we just left with $f(n)$? What happens to the one tile that we draw out that still needs to be matched?[/quote]

It's fine. I just read the solution for recursion. I don't quite get it either. Perhaps stick with the first solution. It could just be the expression simplified.

EDIT: I figured it out. This is actually a quite important clarification.

Notice that the probability that we win given that $n$ pairs are left is the same as the probability that we win given that $n$ pairs are left and we hold one arbitrary ball.

This is because we can always draw an arbitrary ball from the $n$-pair condition with probability 1. (And the ball we originally drew was arbitrary). Therefore, the numerical probabilities are equal, and we can replace one with the other without changing the answer.

Counterintuitively, this means that we can put that arbitrary ball "back in the box" without affecting anything.

Here is a an attempt at a solution on my part that is not producing the correct answer. If anyone could fix it into the correct solution, that would be well-appreciated.

The probability the player wins the game is\[1 - \text{probability of player losing the game},\]so let's compute that instead. Let's now count the cases of losing and add them up:

 Probability of getting a sequence xyz from the start: $6(5)(4){{2^3}\over{12(11)(10)}} = {8\over{11}}$
 Probability of getting either sequence aaxyz or xaayz from the start: $2(6)(5)(4)(3){{2^4}\over{12(11)(10)(9)(8)}} = {4\over{33}}$
 Probability of getting either sequence aabbxyz, aaxbbyz, or xaabbyz from the start: $3(6)(5)(4)(3)(2){{2^5}\over{12(11)(10)(9)(8)(7)(6)}} = {4\over{231}}$
 Probability of either getting either sequence aabbccxyz, aabbxccyz, aaxbbccyz, or xaabbccyz from the start: $4(6)(5)(4)(3)(2)(1){{2^6}\over{12(11)(10)(9)(8)(7)(6)(5)(4)}} = {8\over{3465}}$

Therefore, I calculate the probability of the player winning the game to be\[1 - \left({8\over{11}} + {4\over{33}} + {4\over{231}} + {8\over{3465}}\right) = {{457}\over{3465}}\]However, the answer should be ${9\over{385}}$. So then where did I go wrong?

For instance, I'm missing ababxyz, and in general I'm assuming that two matching tiles are drawn consecutively, but that doesn't have to be the case. This means that I am undercounting with the ${4\over{231}}$ and ${8\over{3465}}$ above. What should they be instead?


Αγαπητέ   RΑΚΑR  ,
ξεφυλίζοντας λίγο το ελληνικό τμήμα , είδα την άσκηση που έβαλες. Κανονικά δεν πρέπει να ασχολούμαι με ανισότητες , γιατι ... μου αρέσουν πολύ και μπορεί να φάω κανένα βράδυ ολόκληρο. Δυστυχώς όμως έχω ανοιγμένες πολλές δουλειές(συγγραφικές) και πρέπει να ασχοληθώ και με αυτές.Τέλος πάντων , δίνω μια υπόδειξη παρακάτω - κρυμένη-για τους μικρότερους φίλους του forum.

Φιλικά - Μπάμπης

[hide]Βάζω αγγλικά γράμματα γιατί διευκολύνουν στο Latex.

Αν αντικαταστασήσουμε στους παρονομαστές το $1$ με $3(ab+bc+ca)$, τότε βλέπουμε ότι είναι θετικοί.

Αν πολλαπλασιάσουμε τα κλάσματα με $a , b, c$ αντίστοιχα και εφαρμόσουμε την ανισότητα Andreescu (BCS) , τότε αρκεί τελικά να αποδείξουμε ότι

    $(a+b+c)^{3}\geq a^{3}+b^{3}+c^{3}-3abc+a+b+bc = (a+b+c)(a^{2}+b^{2}+c^{2}+\frac{2}{3})$

  δηλαδή

   $(a+b+c)^{2}\geq a^{2}+b^{2}+c^{2}+\frac{2}{3}$

η οποία είναι , αν γίνουν οι πράξεις , ισότητα κλπ.Πότε ισχύει η ισότητα;[/hide]

Καλή :)

(σόρρυ για τα $\sum_{cyc}$ αλλά δεν αντέχω άλλο typing!)

CBS (επειδή οι παρονομαστές δεν είναι στη μόδα :lol: ) :

$\left( \sum_{cyc}a(a^{2}-bc+1) \right) \left( \sum_{cyc}\frac{a}{a^{2}-bc+1}\right) \geq \left( \sum_{cyc}a \right)^{2}$

δηλ:

$\sum_{cyc}\frac{a}{a^{2}-bc+1}\geq \frac{\left( \sum_{cyc}a \right)^{2}}{\sum_{cyc}a(a^{2}-bc+1)}$

Αλλάααα: $a^{2}-bc+1 = a^{2}-\frac{1}{3}+ab+ac+1 = a(a+b+c)+\frac{2}{3}= a(a+b+c)+2(ab+bc+ca)$ που σημαίνει ότι:

$\sum_{cyc}a(a^{2}-bc+1) = \sum_{cyc}a^{2}(a+b+c)+2a(ab+bc+ca) = (a+b+c)(a^{2}+b^{2}+c^{2})+2(a+b+c)(ab+bc+ca) = \left( \sum_{cyc}a \right)^{3}$

Άρα το ζητούμενο $\geq \frac{ \left( \sum_{cyc}a \right)^{2}}{ \left( \sum_{cyc}a \right)^{3}}= \frac{1}{a+b+c}$

Edit: φαίνεται ότι άργησα 2 λεπτά :)


trigonometry logarithms limit algorithm function integration calculus

$x_{0}=a$
$x_{n+1}=\sin^{2}x_{n}$
Find equivalent of $x_{n}$ when $n\to+oo$.

[quote="fedja"]There is nothing sacred about $\sin$ here. Generally, you can prove that a recursively defined sequence $x_{0}=a>0$, $x_{n+1}=e^{\delta_{n+1}}x_{n}^{2}$ with $\delta_{n}\to 0$ is asymptotically equivalent to $e^{c\cdot 2^{n}}$ with some constant $c\in \mathbb R$ determined by $a$ and $\delta_{n}$. The proof is fairly straightforward. Let $y_{n}=\log x_{n}$. Then $y_{n+1}=2y_{n}+\delta_{n}$. Let $z_{n}=\sum_{k>n}2^{n-k}\delta_{k}$. Then $y_{n+1}-z_{n+1}=2(y_{n}-z_{n})$, whence $y_{n}=2^{n}(y_{0}-z_{0})+z_{n}$. Since $|z_{n}|\le\max_{k>n}|\delta_{k}|\to 0$, we have $y_{n}=c\cdot 2^{n}+o(1)$. Of course, if the $\delta_{n}$ are not given explicitly but are determined by the recursion, there is no nice formula (except in some very special cases) to determine the value of $c$ exactly but you still can compute it with arbitrary precision by computing several first $\delta_{n}$. I understand that having an unknown constant factor in front of a huge number in the exponent in the answer is kind of disappointing, but this seems the best one can do: this $c$ seems to really depend on $a$ in a fairly unpleasant way.

So, what is sequ[b]a[/b]nce#2? :)[/quote]
In my case: What is answer?( and I don't understand: in my case $x_{n+1}=sin^{2}x_{n}$ and in your generraly case $x_{n}=e^{\delta_{n}+1}x^{2}_{n}$How do you get in particular my problem? I mean: for wich $\delta_{n}$ and $x_{n}$? I think there must be some other method by wich we will be can calculate this constant. Note: in my case $x_{n}$ not $\to 0$

[quote="Extremal"]
In my case: What is answer?
[/quote]
$e^{c\cdot 2^{n}}$ with some $c=c(a)<0$ for $a\ne \pi k$ and $0$ for $a=\pi k$.
[quote]
Note: in my case $x_{n}$ not $\to 0$
[/quote]
What do you mean? $|x_{1}|\le 1$ (unless you are willing to take $a$ complex, but that would be a completely different story). Hence $|x_{2}|\le|\sin^{2}1|<1$. From there on, $|x_{n+1}|\le |x_{n}|^{2}$, so $x_{n}$ does tend to  $0$ and very fast so.
[quote] in my case $x_{n+1}=sin^{2}x_{n}$ and in your generraly case $x_{n+1}=e^{\delta_{n+1}}x^{2}_{n}$
How do you get in particular my problem? I mean: for wich $\delta_{n}$ and $x_{n}$?
[/quote]
By taking the same $x_{n}$ and putting $\delta_{n+1}=\log(\sin^{2}x_{n}/x_{n}^{2})$, of course. Since $x_{n}\to 0$ (I hope you believe it now) and $\lim_{y\to 0}\sin y/y=1$, we get $\delta_{n}\to 0$ and that is the only thing that matters in my argument.
[quote]
I think there must be some other method by wich we will be can calculate this constant.
[/quote]
Here I choose to humbly disagree with you :).

Anyway, please, accept my apologies for not making all the details clear in the first place :blush:

Thanks, I will be think how can calculate this constant

Two things:
1) Please, do not just quote the previous post in full as it is; it never makes sense: you are answering to what is written there by default (I removed such a quote from your last post).
2) As to computation of $c$, I wish you good luck but you should understand clearly first what you mean by "compute". If you mean by that "to come with an algorithm that evaluates the constant with any given precision with an efficient estimate for the error", you already can do it by computing several first $\delta_{n}$. On the other hand, if you want an "elementary formula" expressing $c(a)$ in terms of $a$, keep in mind that most functions arising as sums of series have no elementary formulae associated with them. So, proving that $c(a)$ is not an elementary function may turn out to be a more reasonable thing to try ;). Of course, you may also try to represent $c$ as a different series/integral/limit but I do not see how that would be an improvement upon what we have already.

[quote="fedja"]
What do you mean? $|x_{1}|\le 1$ (unless you are willing to take $a$ complex, but that would be a completely different story). Hence $|x_{2}|\le|\sin^{2}1|<1$. From there on, $|x_{n+1}|\le |x_{n}|^{2}$, so $x_{n}$ does tend to  $0$ and very fast so.[/quote]
I mean $x_{n}$is not like continiuns function.
I have question: why we can't do next:
 we know $x_{n}\to 0$ thus we can use taylor for $\sin$ (([quote="fedja"] There is nothing sacred about $\sin$ here .[/quote]   :) ))  Really: for big $n$ we have $\sin{x_{n}}=x_{n}-\frac{x^{3}_{n}}{6}+O(x_{n}^{5}$ therefore $\sin^{2}{x_{n}}=x^{2}_{n}-\frac{x^{4}_{n}}{3}+O(x^{6}_{n})$ If we said $x_{n+1}-x_{n}=x'_{n}$ then we have $x'_{n}$~$x^{2}_{n}-x_{n}-\frac{x^{4}_{n}}{3}$And in final: $\int_{0}^{n}\frac{x'_{n}dn}{x^{2}_{n}-x_{n}-\frac{x^{4}_{n}}{3}}$~$n$ or $\int_{x_{0}}^{x_{n}}\frac{dy}{y^{2}-y-\frac{y^{4}}{3}}$~$n$  and after find this integral and find answer simpl.

We can also calculate equivalent of the next sequance $x_{n+1}=\sin{x_{n}}$
By this method easly find equivalent of $x_{n}$ which is $x_{n}$~$\sqrt{\frac{3}{n}}$ :wink:

(1) [quote="fedja"] $x_{n+1}=e^{\delta_{n+1}}x_{n}^{2}$ with $\delta_{n}\to 0$[/quote]
maybe $x_{n+1}=e^{\delta_{n}}x_{n}^{2}$ ?
because if you have this form $x_{n+1}=e^{\delta_{n+1}}x_{n}^{2}$ then  you can't  take this [quote="fedja"] $y_{n+1}=2y_{n}+\delta_{n}$.[/quote]

And whats about this case  $x_{0}=a>0$, $x_{n+1}=e^{\delta_{n}}x_{n}^{m}$ where $m\in{N}$

In generraly $x_{0}=a>0$, $x_{n+1}=e^{\delta_{n}}x_{n}^{m}$ with $\delta_{n}\to 0$ is asymptotically equivalent to $e^{c\cdot m^{n}}$ with some constant $c\in \mathbb R$ determined by $a$ and $\delta_{n}$. The proof is fairly straightforward. Let $y_{n}=\log x_{n}$. Then $y_{n+1}=my_{n}+\delta_{n}$. Let $z_{n}=\sum_{k>n}m^{n-k}\delta_{k}$. Then $y_{n+1}-z_{n+1}=m(y_{n}-z_{n})$, whence $y_{n}=m^{n}(y_{0}-z_{0})+z_{n}$. Since $|z_{n}|\le\max_{k>n}\frac{|\delta_{k}|}{m-1}\to 0$, we have $y_{n}=c\cdot m^{n}+o(1)$
 :wink:
Let's forget about my stupid question about constant$c$, I only don't  read good in english your text. I mean I only I have found in your text "$c$ constant" and ( i don't read full text) therefore ny first argument was to calculate this constant. Sorry ./ :oops:

[quote="Extremal"]
I have question: why we can't do next:
 $x_{n+1}-x_{n}=x'_{n}$
[/quote]
The problem is here. We can do this trick if $x(n)$ and $x'(n)$ are slowly changing functions. Now, if we try to do it for fast changing sequences, we'll be in trouble. Just consider the recurrence $x_{0}=\frac{1}{2}$, $x_{n+1}=x_{n}^{2}$ and see where your computation will lead you.
(if I understood you right, you will get $\int_{x_{0}}^{x_{n}}\frac 1{x^{2}-x}\,dx=n$, which gives you just an exponential decay instead of the correct double exponential one)
[quote]
We can also calculate equivalent of the next sequance $x_{n+1}=\sin{x_{n}}$
By this method easly find equivalent of $x_{n}$ which is $x_{n}$~$\sqrt{\frac{3}{n}}$ :wink:
[/quote]
Yes, but this would be a slowly changing sequence.
[quote]
because if you have this form $x_{n+1}=e^{\delta_{n+1}}x_{n}^{2}$ then  you can't  take this $y_{n+1}=2y_{n}+\delta_{n}$.
[/quote]
You are perfectly right, it should be $2y_{n}+\delta_{n+1}$.
[quote]
And whats about this case  $x_{0}=a>0$, $x_{n+1}=e^{\delta_{n}}x_{n}^{m}$ where $m\in{N}$
[/quote]
You've done it correctly. :)

Thanks, I was thinking so, because when I said [quote="Extremal"]
I mean $x_{n}$is not like continiuns function.
[/quote]
it is obviusly that we can't use this method "integration". But How do you determine : slowly changing functions and fast changed function? Can you said in generaly for $f(x)$ is slowly changed function if and only if when ?? or it we understand by intuition?I think it Depends from function and whats exactly result do you want to get.

In general, if $x_{n+1}=x_{n}-f(x_{n})$ with continuous $f(x)>0$ for $x>0$, you can use this trick when $f$ stays essentially constant (in the multiplicative sense) throughout the interval $[x_{n+1},x_{n}]$. If it is so, you can even replace $f(x)$ by its equivalent as $x\to 0$ and still be in good shape. This is true for $f(x)=x-\sin x$ but is false for $f(x)=x-x^{2}$, which makes all the difference.


I'm new to physics competitions and only recently found out about the IPhO, however when i visited the website it didn't mention anything about regional qualifying rounds or anything. Am i supposed to take it that you just show up and compete>?
The reason i ask is because i looked at the problems online and found them very challenging, this could be because i haven't taken AP physics, my school doesn't offer it, i am taking currently physics honors which is pretty easy though we do problems every now and then, rather the teacher does problems, that are as challenging as the IPhO.
I'm interested in physics and want to get involved in competing however if the problems are that hard as a first round i may have to wait. Could someone please enlighten me in respect to qualifying rounds, please?

If you're in the U.S., then unfortunately, it's too late to participate this year because the first three qualifying rounds for the U.S. Physics Olympiad are over.  These links have some more information:

http://www.aapt.org/olympiad2008/
http://www.aapt.org/contests/olympiad.cfm

You'll have to pester your physics teacher into arranging for you to take the tests, if you want to take them next year.


probability
vector
calculus
integration
invariant
conditional probability
probability and stats

I have two problems of probability, the first is easy but i'm searching for an elegant solution, the second is a litle bit difficult...
1) X and Y two aleatory real variables , independants and having same law ( iid ) ( not necessery having density or explicit distribution) why  P( X<=Y) =1/2 ?
2) Find IE( B_t2=b| B_t1=a , B_t3=c ),  (B_t) is a standard brownien motion and t1<t2<t3

1: is it ? try with $ X,Y$ iid such that $ P[X \equal{} 0] \equal{} P[X \equal{} 1] \equal{} \frac12$ ..
2: it is of course equal to $ 0$. You may want to compute the conditional probability $ P[B_{t_2} \leq b | B_{t_1} \equal{} a, B_{t_3} \equal{} c]$

So i was mistaking for the first one !! But in which case it's 1/2 ? just when they have a distribution ?
For the second : I meant IE( B_t2| B_t1=a , B_t3=c ) ! ( so if it was IE( B_t2=b| B_t1=a , B_t3=c ) then its 0 because B_t have distibution and that expectation is just a prbabilty of a point ? Right ? )

$ P(X\equal{}Y)\equal{}0$ exactly when $ P(X\equal{}a)\equal{}0$ for all real numbers $ a.$ (That is perhaps not immediately obvious, but I can see how to prove it.)

I think yassinus is saying "have a distribution" when he really means "have a density." Every real random variable has a distribution in the sense of having a cumulative distribution function. For a random variable, "having a density" is the same as being [i]absolutely continuous.[/i] The property we need here - no specific values of the variable having positive probability - should be called being [i]continuous.[/i] These names correspond to the corresponding properties of the cumulative distribution function.

Absolute continuity implies continuity, but the converse is not true. For an example, consider a random variable whose cumulative distribution is the Cantor "devil's staircase" function.

[quote="JBL"]We have $ 1 \equal{} \Pr(X < Y) \plus{} \Pr(X \equal{} Y) \plus{} \Pr(X > Y)$, and since they're iid the first and third summands are equal.  Thus ....[/quote]
thus $ P[X \leq Y]$ can be pretty much whatever one wants ..

Thank you all for your explanations, yes Kent, I meant having density ...is the cumulatif probability that should be continuous or the variable its self  :huh:   !! ?so if X and Y are iid so P(X<Y)=P(Y<X) ( whish equal to (1-P(X=Y) )/2 ) , but i still can't see how to prove this equality rigorously !
I see that no one reply for the seconde , the solution I made is a little bit  complicated  ( for me of course  :P )  ! i'm looking for an elegant one if it exist !

for the second one, it is just a conditional expectation for a normal random vector since $ X\equal{}(B_{t_1}, B_{t_2},B_{t_3})$ is a $ 3$-dimensional gaussian vector with covariance $ C(i,j)\equal{}\min(t_i,t_j)$.

You could also notice that
\[ E[B_{t_2} | B_{t_1}\equal{}a, B_{t_3}\equal{}c  ] \equal{} a \plus{} E[B_{t_2\minus{}t_1} | B_{t_3\minus{}t_1}\equal{}c\minus{}a  ] \]
so in fact you can study the $ 2$-dimensional gaussian vector $ (B_{t_2\minus{}t_1}, B_{t_3\minus{}t_1})$.
Anyway, as you have guessed, the answer is $ a \frac{t_3\minus{}t_2}{t_3\minus{}t_1} \plus{} c\frac{t_2\minus{}t_1}{t_3\minus{}t_1}$.

think you aleck but could you explain more the firste transformantion, when you change B_t3-t2=c-a ! you integrate the information B_t2 in B_(t3-t2)=c-a like that ! and how you use the covariance to calculate E(X|Y=y) for a given gausien vector (X,Y)  ? ( I can do it in that case in two wayes, using integral or using that (sB_1/s) is also brownien , you seem to have a general method whith covariance !  :idea:

1: the problem is really like if you were starting your brownian motion from $ a$, and then try to compute the conditional expectation $ E[\hat{B}_{t_2\minus{}t_1} | \hat{B}_{t_3\minus{}t_1}]$. The answer should only depends on $ t_3\minus{}t_1$ and $ t_2\minus{}t_1$
2: for a centered Gaussian vector $ (X,Y)$, remember that $ X,Y$ are independent iif $ Cov(X,Y)\equal{}0$. So try to find $ \alpha$ such that $ X\minus{}\alpha Y$ is independent of $ Y$. Then
\[ E[X|Y]\equal{}E[(X\minus{}\alpha Y)\plus{}\alpha Y|Y] \equal{} E[X\minus{}\alpha Y] \plus{} \alpha Y \equal{} \alpha Y \]
is that OK ?
Then $ E[X|Y\equal{}y]\equal{}\alpha y$

Yes , thank you, that was the missing point for me, the Cov therem for gaussien vector ! but for the first step , you seems that you changes B_t, by B_t+t1-B_t, which is also a brownien motien , but when you ignore the second information that B_t1=a is it because you are looking for the expectation of B_t2-B_t1 whis is indépendant of [B_t1=a] or its because we change reference ( starting our brownien from a ! )  in the second case I dont see the justification that subalgebra [ B_t3-B_t1=c-a] is the same as [ B_t3=c , B_t1=a ] !
I'm sorry for all that question , I really want to understand that rigorously ! thank you alekk for your precious explanations

oh, I understand your question. I would probably put it this way:
\[ E[B_{t_2} | B_{t_1}\equal{}a, B_{t_3}\equal{}c ] \equal{} E[B_{t_2\minus{}t_1} | B_{0}\equal{}a, B_{t_3\minus{}t_1}\equal{}c ] \]
because $ B_t$ is a Markov process, the situation is invariant under time translation. An then one can shift everything down by $ a$:
\[ E[B_{t_2\minus{}t_1} | B_{0}\equal{}a, B_{t_3\minus{}t_1}\equal{}c ] \equal{} a \plus{} E[B_{t_2\minus{}t_1} | B_{0}\equal{}0, B_{t_3\minus{}t_1}\equal{}c\minus{}a ] \]
\[ E[B_{t_2\minus{}t_1} | B_{0}\equal{}a, B_{t_3\minus{}t_1}\equal{}c ] \equal{} a \plus{} E[B_{t_2\minus{}t_1} | B_{t_3\minus{}t_1}\equal{}c\minus{}a ]\]
with the convention that a Brownian motion always starts from $ 0$. This is what i meant in my first post. Does it make sense ?

thanks, now things are clear, when you use that its markovien it justify the first step,, for the seconde its a little bit not obvious for me ( but i can do it in an other way so i will not ask for that again  :lol: )  thank you alot , you are a great man alekk , ( I guess u are from ENS ULM   :) )


p is prime  and p= 3( mod 4)

show that if m^2+n^2 is divisible by p then m,n is divisible by p

Assume that $m, n$ are not divisible by $p$.
Since $m^{2}\equiv-n^{2}(mod p)$
so $(\frac{-1}{p})=1$(Legandre symbol).
Contradiction!


What does USAMT year 16 mean? Does it mean that it is only for 16 years old or something???

The 16th year they are doing it.
This year is the 17th year USAMTS is running.


calculus
derivative
analytic geometry
graphing lines
slope
conics
parabola

Hello everyone,

I got the right answer for the following exercise but I do not understand how it is so. I would appreciate an explanation. Thank you!

---

1. Find the derivative of $ g(x) \equal{} 5x^2 \plus{} 7x \minus{} 6$.

I used the difference quotient and I ended up with $ g'(x) \equal{} 10x \plus{} 7$, which is the right answer according to the answer key.

Then I graphed both of these equations on my graphing calculator and I was confused. $ g'(x) \equal{} 10x \plus{} 7$ wasn't tangent to $ g(x)$. Instead, $ g'(x)$ just intersected $ g(x)$ at $ x \equal{} \minus{}1.34$ and plowed on through. It didn't appear to be tangent to $ g(x)$.

Could someone please explain this?

I'm not sure about this, but I think point of tangency would be at $ g(x)\equal{}g'(x)$

$ f'(x)$ denotes the [i]slope[/i] of a tangent line going through a point of tangency $ x$ on $ y \equal{} f(x)$, the derivative is not the actual tangent line.

Thanks for your replies.

@Xaero: How would I find the actual tangent line then?

There are infinitely many tangent lines to a curve, so it would make sense to make the line tangent to a point on the curve, say $ (x_{0},f(x_{0}))$. Then the line would be $ y \equal{} f'(x_{0})(x \minus{} x_{0}) \plus{} f(x_{0})$ in point slope form

Thank you very much for your reply.

So just to clarify, to find a tangent to $ (a, b)$ on $ f(x)$, I would have to find the derivative of $ f(x)$ first, then use that slope to find the tangent line to that $ (a, b)$.

Take a generic point on the curve: $ (a,5a^2\plus{}7a\minus{}6).$

The slope of the tangent line is $ 10a\plus{}7.$

That makes the tangent line

$ y\equal{}(10a\plus{}7)(x\minus{}a)\plus{}5a^2\plus{}7a\minus{}6\equal{}(10a\plus{}7)x\minus{}10a^2\minus{}7a\plus{}5a^2\plus{}7a\minus{}6$

$ y\equal{}(10a\plus{}7)x\minus{}5a^2\minus{}6.$

So for instance, if we let $ a\equal{}1,$ we get that $ y\equal{}17x\minus{}11$ is tangent to the curve at $ (1,6).$ Or if we let $ a\equal{}0,$ then $ y\equal{}7x\minus{}6$ is tangent to the curve at $ (0,\minus{}6).$

-------

Here's a challenge: find all lines tangent to this curve that pass through the point $ (\minus{}1,\minus{}15).$ (Note that that is not a point on the curve.)

Here's a different challenge: let $ f(x) \equal{} x^2 \plus{} bx \plus{} c$.  The parabola $ y \equal{} f(x)$ is in general not tangent to the line $ y \equal{} f'(x)$.  However, it may happen by accident that these two curves are tangent.  Under what conditions (on $ b, c$) does this actually happen?  (We could also consider $ f(x) \equal{} ax^2 \plus{} bx \plus{} c$, $ a \neq 0$, but all it does is make the algebra a little messier -- the approach is exactly the same.)

@Kent Merryfield

How would you do that challenge?

I am new to calculus and totally baffled.
Thank you.

I asked a question that was identical to that challenge except for the numbers on a midterm over the summer. It was a very hard question.

There are two things we know about the line: it passes through a particular point, and it's tangent to the curve at some other point. How would you express this information?

[hide="Challenge"]The equation of the tangent line at $ (x,y)\equal{}(r, s)$ is $ y\equal{} f'(r) (x\minus{}r) \plus{} s$

Here, we know that $ s\equal{}5r^2\plus{}7r\minus{}6$

Here, we know that $ (\minus{}1, \minus{}15)$ satisfies this equation. Thus, $ \minus{}15 \equal{} f'(r) (\minus{}1\minus{}r)\plus{}5r^2\plus{}7r\minus{}6$

$ f'(x) \equal{} 10x\plus{}7 \implies f'(r) \equal{} 10r\plus{}7$. thus, $ \minus{}15 \equal{} (10r\plus{}7)(\minus{}1\minus{}r)\plus{}5r^2\plus{}7r\minus{}6 \equal{} \minus{}5r^2\minus{}10r\minus{}13$

Thus, $ 5r^2\plus{}10r\minus{}2\equal{}0 \implies r \equal{} \frac{\minus{}5 \pm \sqrt{35}}{5}$

Now, you can use these two $ r$ values to find their appropriate $ s$ and $ f'(r)$ values, and use the original formula to find the equation of their respective tangent lines.[/hide]


If $ x\plus{}y\plus{}z\equal{}0$ and $ x^{4}\plus{}y^{4}\plus{}z^{4}\equal{}50$, then find $ xy\plus{}xz\plus{}yz$.

[hide]Let $ t\equal{}xy\plus{}yz\plus{}zx$.  Then

$ x^2\plus{}y^2\plus{}z^2\equal{}(x\plus{}y\plus{}z)^2\minus{}2t\equal{}\minus{}2t$

$ \Rightarrow x^4\plus{}y^4\plus{}z^4\equal{}(x^2\plus{}y^2\plus{}z^2)^2\minus{}2(x^2y^2\plus{}y^2z^2\plus{}z^2x^2)\equal{}4t^2\minus{}2((xy\plus{}yz\plus{}zx)^2\minus{}2xyz(x\plus{}y\plus{}z))\equal{}2t^2$

$ \Rightarrow t\equal{}\pm5$  [/hide]

[hide]We have $ 2(x^4 \plus{} y^4 \plus{} z^4) \equal{} (x^2 \plus{} y^2 \plus{} z^2)^2$ for all $ (x,y,z)$ if we have $ x \plus{} y \plus{} z \equal{} 0$. Using this equality we can easily prove that $ (x^2 \plus{} y^2 \plus{} z^2)^2 \equal{} 100$ and because $ x^2 \plus{} y^2 \plus{} z^2 > 0$ we have $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 10$. Because of the condition $ x \plus{} y \plus{} z \equal{} 0$ we imply that $ (x \plus{} y \plus{} z)^2 \equal{} 0$. Thus we obtain $ \frac{x^2 \plus{} y^2 \plus{} z^2}{ \minus{} 2} \equal{} xy \plus{} yz \plus{} xz$ and finally we find that $ xy \plus{} yz \plus{} xz \equal{} \minus{} 5$[/hide]

[quote="SAPOSTO"]Mine solution is almost the same. But the authors have written in their solution: Since $ ab \plus{} ac \plus{} bc \equal{} \frac {a^{2} \plus{} b^{2} \plus{} c^{2}}{ \minus{} 2} < 0$, then $ ab \plus{} ac \plus{} bc \equal{} \minus{} 5$. That's a bad mistake. However good work [b]cosinator[/b]![/quote]

well...  :blush:  that's only if you allow real numbers. :)


geometry
3D geometry
sphere
tetrahedron
ratio

23. Four balls of radius 1 are mutually tangent, three resting on a plane and the fourth resting on the others. A tetrahedron, each of whose edges has length q, is circumscribed around the balls. Find q.

i dont know if it helps but since its a tetrahedron you can draw a two-dimensional diagram of the base with the three circles..if i understand the question correctly..

connect the centers to form a smaller, similar tetrahedron, now the ratio of their sizes is the cube of the distance from the center, to a face, this is easy to calculate


In my textbook the following is stated:

[quote]If $ A\mathbf{x}\equal{}\mathbf{0}$ where $ A$ is a square matrix, then Cramer's rule implies that

$ \mathbf{x}\det{A}\equal{}\mathbf{0}$

Hence $ \mathbf{x}$ is a nonvero column vector if and only if $ \det{A}\equal{} 0$ which is a test for nontrivial solutions for systems in which the number of variables is the same as the number of equations.[/quote]

I don't understand why Cramer's rule implies this. Surely in the derivation of Cramer's rule we premultiply by $ A^{\minus{}1}$ thus assuming that $ \det{A}\not \equal{} 0$? I understand that if $ \det{A}\not \equal{} 0$ then $ \mathbf{x}\equal{}\mathbf{0}$ and there are no nontrivial solutions. This is because if $ \det{A}\not \equal{} 0$  we can apply Cramer's rule to obtain expressions for the components of $ \mathbf{x}$ each of which contain a determinant with a column of zeros (making $ \mathbf{x}\equal{}\mathbf{0}$). But why should the inverse be true: if $ \det{A}\equal{} 0$ then there are nontrivial solutions. Furthermore, why should there be an infinite number of these nontrivial solutions?

How about a different argument, which doesn't mention Cramer's rule:

If $ A$ is invertible and $ Ax\equal{}0$, then $ x\equal{}A^{\minus{}1}Ax\equal{}A^{\minus{}1}\cdot 0\equal{}0$, and the equation has no nontrivial solutions.

On the other hand, suppose that $ Ax\equal{}0$ has no nontrivial solutions. Choose $ n$ linearly independent vectors $ x_{1},x_{2},\dots,x_{n}$. If $ \sum c_{i}(Ax_{i})\equal{}0$, then $ A(\sum c_{i}x_{i})\equal{}0$, so $ \sum c_{i}x_{i}\equal{}0$ and $ c_{1}\equal{}c_{2}\equal{}\cdots\equal{}c_{n}\equal{}0$. This means that the $ Ax_{i}$ are linearly independent; since there are $ n$ of them, they form a basis of the space of column vectors. If $ B$ is the linear transformation mapping each $ Ax_{i}$ to $ x_{i}$ (possible since the $ Ax_{i}$ are a basis), $ B$ is an inverse for $ A$ and $ A$ is invertible.

Now, that other question, about how many solutions we get to $ AX\equal{}0$:
It's easy to show that the solutions form a vector space. A nontrivial vector space has at least as many elements as the field; if you're working over something familiar like $ \mathbb{R}$ or $ \mathbb{C}$, that's infinite. There are finite fields out there, and this particular conclusion doesn't work there.


Did anyone bother to give them? I wanna check my iiO result but I've lost my roll number.

Nevermind, I guessed it using a friends'

How did you fare in these Olympiads?
 :)

Got 121 rank in the International Olympiad in Informatics. I think there was a problem in checking my paper. It shows my marks around 75, though I only remember getting around 3 wrong. There will be a second round as well :) I dunno if I should bother to give considering how stupid(read fake) the paper was :P

Which of the olympiads did you write?
Math,Science and Informatics :?:

I don't remember really.I'm sure I did Informatics at least :P


geometry
geometric transformation
parallelogram
geometry proposed

On a plane when three discs with the radii $1$ respectively moves  in satisfying the following conditions $(a),\ (b),$ find the maximum value of the area $S$ of the union of sets for these three discs.

$(a)$ The center of all of the three discs lie on the circle centered at a fixed point $P$ with radius $1.$
$(b)$ The common point which all of these three disks have is only the point $P.$

Let (X), (Y), (Z) be the 3 unit disks. From (b), two of the unit disks are tangent at P, say (Y), (Z). Let the circles (X), (Y) intersecting at P meet again at Q and let the circles (X), (Z) intersecting at P meet again at R. Let (Y), (Z) meet their center line YZ again at A, B, respectively. Since the right angle triangle $\triangle PBR$ is a translation of the right angle triangle $\triangle APQ$ by the distance AP = PB = 2r in the direction of YZ, QR = 2r and YZRQ is a parallelogram with fixed sides YZ = QR = 2r, YQ = ZR = r. Sum of the areas of the intersection $(X) \cap (Y)$ and the intersection ${(X) \cap (Z)}$ is equal to

$S = S_{(X) \cap (Y)} + S_{(X) \cap (Z)} = S_{(X)} - S_{YZRQ}$

The area of the union $(X) \cup (Y) \cup (Z)$ is maximum, when this area sum is minimum, i.e., when YZQR is a rectangle. Then $S_{\text{min}} = \pi r^2 - 2r^2$ and the maximum area of $(X) \cup (Y) \cup (Z)$ under the condition (b) is $3 \pi r^2 - S_{\text{min}} = 2r^2 (\pi + 1).$