geometry
incenter
perimeter
angle bisector

Let $J, K, L$ be the feet of the altitudes of an acute triangle.
(a) Prove that the incentre of $\Delta JKL$ coincides with the orthocentre of $\Delta ABC$.
(b) Prove that $A, B, C$ are the ex-centres of $\Delta JKL$

The angle bisector of $\angle C$ in $\Delta ABC$ intersects the side $AB$ in $D$.
Prove that the segment $CD$ is less than the geometric mean of the sides $CA$ and $CB$.

Well known and related to above: Prove that $\Delta JKL$ is triangle with minimum perimeter (of any triangle formed whose one point is on $AB$, one on $BC$ and one on $CA$)

Yes.  :)
For the second we remind this fact. In $\Delta ABC$ if $CD$ is the angle bisector of $\angle C$ then,
$CA \cdot CB = CD^2 + AD \cdot BD$. In Italy it's called the angle bisector theorem.

[quote="Andreas"]
For the second we remind this fact. In $\Delta ABC$ if $CD$ is the angle bisector of $\angle C$ then,
$CA \cdot CB = CD^2 + AD \cdot BD$. In Italy it's called the angle bisector theorem.[/quote]
It looks like Stewart Theorem.


[quote="quytx"][u]Find the greatest value of the expression:[/u]
\[P=\frac1{2a+b+3}+\frac1{2b+c+3}+\frac1{2c+a+3}\]
where $a,\;b,\;c$ are real positive numbers satisfying the condition $abc=1$[/quote]
$P=\sum_{cyc}\frac{1}{2a+b+3}=\sum_{cyc}\frac{1}{2\left(\frac{a+1}{2}+\frac{a+b}{2}+1\right)}\leq\sum_{cyc}\frac{1}{2\left(\sqrt a+\sqrt{ab}+1\right)}=\frac{1}{2}.$


On all the contests with images, when you open the "Download PDF Version" link, the image does not show up, but an image tag does.  Sometimes this screws up the entire document's formatting.  Is there something wrong with the way the script generates the PDF?

Yes. It'll be fixed (probably) by some point, but it's not a top priority. If you want to help, find another copy of the diagram and convert it to Asymptote. Then send it to a mod to edit into the original post.


search
function
number theory proposed
number theory

If $a,b\in\mathbb{N},\ a,b>1$, and forall $n\in\mathbb{N}$ there is
$\frac{a^{n}-1}{b^{n}-1}\in\mathbb{Z}.$
Prove that $a=b^{k}\ (k\in\mathbb{N})$.
Another one is, if $a,b\in\mathbb{N},\ a,b>1$, and forall $n\in\mathbb{N}$ there is
$\frac{a^{n}+1}{b^{n}+1}\in\mathbb{Z}.$
Prove that $a=b^{k}\ (k\in\mathbb{N},\ 2\nmid k)$.
I'm sure they're right because they've been solved, but the solutions are way too long. Do you have any ideas about them? (Especially for the second one)

Opinion: use search function, try "masterpiece".

But what about the second problem above?
I couldn't find it in the forum

http://www.mathlinks.ro/Forum/viewtopic.php?t=4556
second pro : I think  you search : a^n+1. because i have ever read it.


In the planer hexagon $ABCDEF$, $AB$ is parallel to $DE$, $BC$ is parallel to $EF$, and $CD$ is parallel to $AF$. Prove that $\triangle ACE$ and $\triangle BDF$ have equal areas.

The hexagon ABCDEF with the opposite sides parallel can be only convex or reflex. Ignoring the reflex case.

Let P be the intersection of the alternate sides BC, AF. Since ABCDEF is convex, P lies on the opposite side of the line AB than the remaining hexagon vertices C, D, E, F. Let the line AB move, remaining parallel to opposite side ED, so that the hexagon ABCDEF remains convex. At one limit, the points $A_1, B_1, P$ coincide and $A_1F \equiv PF > CD$. At the other limit, either the points $A_2, F$ or the points $B_2, C$ coincide, whichever comes first. WLOG, assume that the points $A_2, F$ coincide first and $0 = A_2F < CD$. By the continuity principle, a line $A_0B_0$ exists, such that $A_0F = CD$. Since in addition, $A_0F \parallel CD$, $A_0CDF$ is a parallelogram and $A_0C = DF$. Similarly, $A_0B_0 = ED$ and $A_0B_0 \parallel ED$, so that $A_0B_0DE$ is a parallelogram and $B_0D = A_0E$. In exacly the same way, $B_0F = CE$. As a result, the triangles $\triangle A_0CE \cong DFB_0$ are congruent and therefore,  they have the same areas

$S(\triangle A_0CE) = S(\triangle DFB_0)$

At the 1st limit, the points $A_1, B_1, P$ coincide and the following triangle areas are equal:

$S(\triangle A_1CE) \equiv S(\triangle PCE) = S(\triangle PCF) = S(\triangle PDF) \equiv S(\triangle B_1DF)$

Since the triangle areas $S(\triangle ACE), S(\triangle BDF)$ are linear functions of the positions of the points $A \in PF, B \in PC$, i.e., of the position of the line $AB \parallel ED$ and since they are equal for 2 dfifferent positions $A_0B_0 \parallel ED, A_1B_1 \parallel ED$ of this line, they are equal for all positions of the line $AB \parallel ED$, $S(\triangle ACE) = S(\triangle BDF)$.


Normally, I would be studying math right now. Or working on the math training program with solafidearms.(I really hope Mr.Rusczyk and Mr.Crawford know about that. I need to impress them. They are like my idols.) Or maybe trying to find mathcounts national problems( I have no idea where to get them. Google turns up null.)

But instead, I have photoshopped an anti-meth poster. Enjoy.
I wanted the blocky text to look real, but I couldn't copy and paste any A's.
[img]http://tinypic.com/b4hh8h.jpg[/img] :(

after you try math, you'll go to jail and die. :rotfl:

[quote="Derek"]You forgot the edit the e on the bottom. :P[/quote]
But would editing the e create any extra funny?

Uh...moving to Graphics board?

Although to be honest I have absolutely no idea where this belongs.

I think the font they used is called stencil, mabye you can screenshot an 'A' from a random website :lol:

I got it an image with the text..

and also the font itself...doesn't upload...

Hope this helps.


For all of those who are excellent in chemistry, I have some questions:

1.) What are the 5 postulates of kinetic molecular theory?
2.) What is an ideal gas equation?
3.) What is Charles' law?

4.) The molecular gas constant (R) is $ .0821 \left(\frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}}\right)$. What is the value of the gas constant in $\frac{\text{mL} \cdot \text{Pa}}{\text{mol} \cdot \text{K}}$?

for the ideal gas equation, is it:
\[pV=nTR\]?

Yeah, I think so...

I found the answers to the rest of them except for the last question. Any ideas?

charles's law states that if the pressure in a reaction is constant, the ratio of the volume of the gas and the temperature will remain constant... i am prett sure, cause Boyle's law states that if temperature is constant and Gay-Lussak's law states that volume is constant

[quote="mcalderbank"][quote="sky9073"]R = 8.31 J/mol · K
Since 1 J = 1 N · m, and mL · Pa = 10^-6 N · m = 10^-6 J,
shouldn't R = 8.31 x 10^-6 mL · Pa / mol · K?[/quote]

I thought fanaticism was referring to that same molecular gas constant (.0821), so all you would have to do is unit conversions   :?[/quote]

The difference in your answer and mine are huge, though....
Wouldn't it make more sense if they were slightly off but still close (at least in the order of magnitude)? :?
Hmm..

It's because mcalderbank should have multipled by 1000 and multiplied by 101325 (instead of divide).  If you do this, you get 8.31 x 10^6.  Sky, I think you made a similar error (divided by10^6 instead of multiplying).


There are infinetely many sequences.
For example let $a_{p}$ all any sequence (p - prime), suth that $(a_{p},a_{q})=1$. We can define $a_{n}=\prod_{p|n}a_{p}$.

Yes, but there exists a beautiful characterization of all such sequences. However, this is a very difficult problem! I've seen it in some Iran Olympiad and the proof it had spanned over 3 pages. I found another one, much shorter, but I lost my paper...  :blush:

harazi, where did you find the proof consisting of 3 pages :?:

See also :
http://www.mathlinks.ro/Forum/viewtopic.php?t=54800
http://www.mathlinks.ro/Forum/viewtopic.php?t=63078
http://www.mathlinks.ro/Forum/viewtopic.php?t=74322
http://www.mathlinks.ro/Forum/viewtopic.php?t=76151

Problems concerning these sequences appeared in both the indian mo2001 and iranian mo2001. :D

[quote="dondigo"]Find all sequences $a_{n}\quad n=1,2,3,\cdots$ satisfying:

$gcd(a_{n},a_{m})=a_{gcd(n,m)}$ for all $n,m$[/quote]

Fibonaccie's sequence $F_{n}$ has this property too.


In an examination, a student  must answer 65% of  the question correctly.  But it was found that after answering 7 questions correctly out of the first 14, the student qualified .  What is minimum number of questions on the exam?

a)22
b)18
c)20
d) None of the above

My Work:

35% of total no. of questions = (14 - 7) = 7
or, total no. of questions = (7*100)/35=20.
Is it correct?

[hide="hint"]$\dfrac{7+k}{14+k} \ge 65 \%$, where $k$ is the number that he got right after the first fourteen problems.[/hide]

(btw, this is better :))

[hide]How can that be....
7 he answered correctly right?
so $\dfrac{7+x}{7+k}\geq65\%$
And this is impossible
Im confused :huh:[/hide]

Where did you get that inequality from, riddler? And what are $x$ and $k$? Look at my hint for the right one... :)

sorry if Im dumb but,
I think it cant be possible....

[quote="10000th User"][quote="ccy"]I don't think you can qualify for $65\%$ if you answered 7 out of 14 questions right..

If $x$ equals the number of questions after the 14th (positive) then
$\dfrac{7}{14+x}<\dfrac{7}{14}$
and $\dfrac{7}{14}=50\%$

How can $\dfrac{7}{14+x}$ be more than 50% when $x>1$?[/quote]
You are thinking the wrong thing...

[hide="Trying not to spoil it"]"a student must answer 65% of the question correctly" this means that 65% of the entire test must be correct.

"But it was found that after answering 7 questions correctly out of the [b]first 14[/b], the student qualified" please read the highlighted part and think about it.
[hide="More spoiling"]Is 14 the [b]total[/b] number of questions? Is it? [b]Is it?[/b]

[hide="Last spoil"]What do you know about the rest of the questions?  :P [/hide][/hide][/hide][/quote]

What I was saying was that If there were 14+x questions on the test, and the student got only 7 right, then the percentage right will decrease as x increases.  So the maximum the person could have gotten with 7 out of 14+x right was 50% when x=0.

[quote="YooSam"]ccy, I get what you mean. The question is quite confusing in a way saying that the student is qualified only after having 7 correct answers. It is like he doesn't have to do anything more after the first 14 questions, which results in the descreasing of $\dfrac{7}{x+14}$.

However, I guess here the student on the other hand does all the rest of the questions right. So that's why we have the inequality

$\dfrac{7+k}{14+k} \ge 65 \%$

[hide]$\boxed{(C)}$[/hide][/quote]
Thank you now I understand
 :D


geometry
circumcircle
parallelogram
geometric transformation
reflection
cyclic quadrilateral
geometry proposed

Let acute triangle ABC inscribed in circumcircle O. E, F are the points on AB, AC such that O is the midpoint of EF. Let A' be the second intersection of AO with (ABC). Prove that EF, BC, and the tangent of the circle (O) by the point A' intersect at 1 point.

Denote by $ H$ the middle point of $ BC$.

We have the quadrilateral $ AEA'F$ is a parallelogram, so $ \angle EA'A \equal{} \angle A'AC \equal{} \angle A'BC$. And we also have $ \angle EAA' \equal{} \angle BAA' \equal{} \angle A'CB$, it implies that $ \triangle A'BC\sim\triangle EA'A$, i.e. $ \triangle A'HC\sim\triangle EOA$ $ \Rightarrow \angle A'HC \equal{} \angle EOA \equal{} \angle A'OK$. It follows that the quadrilateral $ OHA'K$ is cyclic, so $ \angle OA'K \equal{} \angle OHK \equal{} 90^\circ$. Hence $ KA'$ is tangent to $ \mathcal C(O)$ at $ A'$. Our proof is completed.

[b]EDIT.[/b] Using this problem as a lemma, I have proposed a problem for Mathematical Reflections sometime ago. [quote]Let $ M,\,N$ two points interior to the circle $ \mathcal{C}(O)$ such that $ O$ is the midpoint of $ MN$. Let $ S$ an arbitrary point lies on $ \mathcal C(O)$, and $ E,\,F$ the second intersections of the lines $ SM,\,SN$ with $ \mathcal C(O)$, respectively. The tangents in $ E,\,F$ with respect to the circle $ \mathcal C(O)$ intersect each other at $ I$. Prove that the perpendicular bisector of the segment $ MN$ passes through the midpoint of $ SI$.
[i]
Proof.[/i] Denote $ K\in MN\cap EF$, $ J$ the second intersection of $ SO$ with $ \mathcal C(O)$ and $ P$ the intersection of the perpendicular bisector of segment $ MN$ with $ SI$.

Applying the above Problem, we have $ KJ$ is tangent to $ \mathcal C(O)$ and notice that $ K$ lies on $ EF$ the polar of $ I$ relative to circle $ \mathcal C(O)$, we have $ IJ$ is the polar of $ K$ relative to $ \mathcal C(O)$, i.e. $ IJ\perp OK$. It follows that $ IJ\parallel OP$. On the other hand, $ O$ is the midpoint of $ SJ$, so $ P$ is the midpoint of $ SI$, i.e. the perpendicular bisector of the segment $ MN$ passes through the middle point of $ SI$ as we wanted.[/quote]

CL//EF   =>  ∠GDC=∠ABC=∠AA’C   =>  G,D,A’,C are concyclic
=>  O,D,A’,K are concyclic  => KA’⊥AA’
Done!

Dear Mathlinkers,
my synthetic proof:
1. let B' the second intersection of BO with (ABC) and Ta' the tangent to (ABC) at A'.
2. According to the butterfly theorem, A', F and K are collinear
3. According to Pascal' theorem applied to BB'A' Ta' ACB,      O, F and K are collinear…
...
Sincerely
Jean-Louis

Hello!

Your problem can be generalized:
Let $ A'$ be a point on the circumcircle $ \left(O\right)$ of the triangle $ ABC$.
$ E\in AB$, $ P\in AA'$ and $ F\in AC$ such that $ E,P,F$ are collinear points and $ EP \equal{} PF$.
Prove that $ EF$, $ BC$ and the tangent to the circle $ \left(O\right)$ at the point $ A'$ are concurrent.

Solution:

Let $ K$ be the intersection point of the tangent to the circle $ \left(O\right)$ at the point $ A'$ with $ BC$.
Let's prove that $ K\in EF$.
$ \left\{X,Y\right\} \equal{} EF\cap \left(O\right)$ and $ \left\{B'\right\} \equal{} BP\cap\left(O\right)$

From the butterfly theorem $ \Rightarrow$ the points $ A',F,B'$ are collinear.
$ EA\cdot EB \equal{} EX\cdot EY \equal{} FY\cdot FX \equal{} FA'\cdot FB' \equal{} FA\cdot FC \Rightarrow\frac {FA}{EA} \equal{} \frac {EB}{FC}$   (1)

$ \frac {KB}{KC} \equal{} \frac {A'B\cdot sin\left(\angle KA'B\right)}{A'C\cdot sin\left(\angle KA'C\right)} \equal{} \frac {A'B}{A'C}\cdot\frac {sin\left(\angle BCA'\right)}{sin\left(\angle CBA'\right)} \equal{} \left(\frac {A'B}{A'C}\right)^{2}$   (2)

$ \frac {A'B}{A'C} \equal{} \frac {\frac {AB\cdot sin\left(\angle BAA'\right)}{sin\left(\angle BA'A\right)}}{\frac {AC\cdot sin\left(\angle CAA'\right)}{sin\left(\angle CA'A\right)}} \equal{} \frac {AB}{AC}\cdot\frac {sin\left(\angle BAA'\right)}{sin\left(\angle CAA'\right)}\cdot\frac {sin\left(\angle CBA\right)}{sin\left(\angle BCA\right)} \equal{}$
$ \equal{} \frac {sin\left(\angle BAA'\right)}{sin\left(\angle CAA'\right)} \equal{} \frac {\frac {EP}{EA}}{\frac {FP}{FA}} \equal{} \frac {FA}{EA}$   (3)

(1), (2), (3) $ \Rightarrow\frac {KB}{KC} \equal{} \left(\frac {FA}{EA}\right)^{2} \equal{} \frac {FA}{EA}\cdot\frac {EB}{FC}\Rightarrow\frac {KB}{KC}\cdot\frac {FC}{FA}\cdot\frac {EA}{EB} \equal{} 1\Rightarrow K\in EF$.

Best regards, Petrisor Neagoe :)

$ \triangle ABC$inscribed in circumcircle $ O$. $ P \in O$, tangent of $ O$ by $ P\cap BC\equal{}K$
$ AP\cap$circumcircle of $ \triangle OKP\equal{}P,Q$, $ KQ \cap AB\equal{}E, KQ\cap AC\equal{}F$.
Prove that $ QE\equal{}QF$.

Oh :lol:. Nice work!!! Actually, the problem that I posted was the lemma that I used to prove this problem: "Let acute triangle ABC inscribed in circumcircle O. E, F are the points on AB, AC such that O is the midpoint of EF. Let A' be the second intersection of AO with (ABC). P is the projection of A' onto EF. Prove that ∠BPA' = ∠CPA'."
[b]Proof[/b]
Now by that lemma, let K be the intersection of EF, BC, tangent of circumcircle (O) by A'. Obviously, A'P is the polar of K through (O). Hence, if we call S is the intersection of AP' with BC, we have (B,C;S,K)= -1 then P(B,C;S,K)= -1, PS is perpendicular with PK so PS is the bisects ∠BPA, wich leads to the result of the above problem  :blush: .
[/list]

[quote="Petry"]Hello!

Your problem can be generalized:
Let $ A'$ be a point on the circumcircle $ \left(O\right)$ of the triangle $ ABC$.
$ E\in AB$, $ P\in AA'$ and $ F\in AC$ such that $ E,P,F$ are collinear points and $ EP \equal{} PF$.
Prove that $ EF$, $ BC$ and the tangent to the circle $ \left(O\right)$ at the point $ A'$ are concurrent.
[/quote]
This is another express:
Given triangle $ ABC$ and the circumcircle $ (O)$. Let $ M$ be the midpoint of $ BC. AM\cap (O)\equal{}\{D\}$. The tangency of $ (O)$ at $ D$ cuts $ BC$ at $ E. EFG$ is a secant of $ (O)$. $ AF, AG$ cut $ BC$ at $ X,Y$. Prove that $ MX\equal{}MY$.
[b]Solution:[/b]
Let $ ES$ be the second tangency of $ (O)$. Since $ A(BCDS)\equal{}\minus{}1$ and $ MB\equal{}MC$ we get $ AS//BC$
Denote $ I$ the intersection of $ AD$ and $ EF$. A line which passes through $ I$ and parallel to $ BC$ cuts $ AF, AG$ at $ U,T$.
Because $ A(FGDS)\equal{}\minus{}1$ then $ IU\equal{}IT$. By [b]Thales[/b]'s theorem we have $ MX\equal{}MY$

We can reverse, instead of proving $ MX \equal{} MY$ we will show that if have $ MX \equal{} MY$ then $ BC, FG$ and the tangency of $ (O)$ at $ D$ are concurrent. Then we have [b]another proof[/b]:
Let $ E$ be the intersection of $ BC$ and $ FG$. We will show that $ DE$ is the tangency.
Draw a line which passes through $ G$ and parallel to $ BC$, it cuts $ AD, AF$ at $ N,L$.
Denote $ H$ the midpoint of $ FG$.
We have $ NL \equal{} NG$ so $ NH$ is midparallel line of triangle $ LGF$ then $ NH//AF$.
Since $ AFDG$ is cyclic quadrilateral and $ NH//AF$ we get $ NHDG$ is cyclic.
Thus $ \angle GHD \equal{} \angle DNG \equal{} \angle DME$
Therefore $ MHDE$ is cyclic.
But $ OMHE$ is cyclic then $ \angle ODE \equal{} 90^o$.
We are done!  :wink:


What is the remainder when 5^207 is divided by 7?

I wasn't quite sure how to solve this. Can anyone explain it to me?

Compute the remainders  of the powers of $ 5$ when divided by $ 7$.

$ 5^1, 5^2, 5^3, 5^4, 5^5, 5^6, \ldots$ yield remainders $ 5,4,6,2,3,1, \ldots$. (There exists a famous result, Fermat's little theorem, that ensures that $ 5^{7\minus{}1}$ yields remainder $ 1$, since $ 7$ is a prime and does not divide $ 5$).

This means $ 5^{204}$ yields remainder $ 1$ (can you see why?), so we are left with the remainder of $ 5^3$, which is $ 6$.


I don't know, Fibonacci? Or are you alluding to $ 42$? Or possibly both?

Yep. Fibonacci. And that guest who got an incorrect answer of 4 was me. I feel special.

A better Fibonacci guest would have been "Guest215589" since that is symmetrical on the keyboard and for other reasons.

I think a guest would be cool if it was 1234567 or something.


A(0, A(1,2n+2))

using the ackerman's function:

Ackermann function - Wikipedia, the free encyclopedia

I tried to follow the pattern, but eventually it will lead us nowhere, can this be solved?

[quote="t0rajir0u"]$ A(0, x)$ and $ A(1, x)$ are both fairly straightforward functions that are given in the Wikipedia article.  What are they?  What do you get when you compose them?[/quote]

well that is easy $ A(0,x)$ is just x +1 and $ A(1,x) is x\plus{}2$ right?

so solving the inner part of the equation first:

$ A(1,2n \plus{} 2) \equal{} 2n \plus{} 4$, this will give me
$ A(0, 2n \plus{} 4)$, so therefore this will result in $ 2n\plus{}4\plus{}1 \equal{} 2n\plus{}5$, am I right?


find all integer solutions to:

$x^2+y^2+z^2=x^2y^2$

[quote="shyong"][hide]
Looking at mod 4 so $x,y,z$ must be all even . Then making substitution for $x=2x',y=2y',z=2z'$ and plug back and take mod 4 again . And $x',y',z'$ is again even . So we keep on reducing them until $x=y=z=0$
[/hide][/quote]

Nice way of solving!

Incidentally, this also shows that the product of two numbers, each one less than a perfect square, can never be one more than a perfect square, except in the case $-1 \cdot -1 = 1$.

[quote="mathmanman"]Yes, it is Fermat's [url=http://en.wikipedia.org/wiki/Infinite_descent]infinite descent[/url].[/quote]

New for me! Good link!! Thanks.....


Let m,n be natural numbers such that
$ A\equal{}\frac{(m\plus{}3)^{n}\plus{}1}{3m}$
is an integer.Prove that A is odd.

[hide="A Start..."]To prove that $ A$ is odd, we can show that there is no pair $ (m,n)$ such that $ \frac{(m\plus{}3)^{n}\plus{}1}{6m}$ is an integer.[/hide]

[hide]

Suppose $ 2^{k}$ is the greatest power of $ 2$ dividing $ m$.  Now it suffices to show that $ 2^{k\plus{}1}$ does not divide the numerator.

The $ n \equal{} 1$ case is trivial.  So WLOG, $ n\ge 2$.  Moreover, the $ m$ odd case is trivial, as the numerator is odd.  So Wlog, $ 2|m$.  Now we take the numerator modulo $ 2^{k\plus{}1}$.
\[ (m\plus{}3)^{n}\plus{}1\equiv 3^{n\minus{}1}mn\plus{}3^{n}\plus{}1\equiv 2^{k}n\plus{}3^{n}\plus{}1\]
But notice that because $ 3^{n}$ is always $ 1$ or $ 3$ modulo 8, we have that the greatest power of $ 2$ dividing $ 3^{n}\plus{}1$ is $ 2^{2}$.  Okay, so clearly the $ k\ge 3$ case is out the window.

If $ k \equal{} 1$, then we want to show that $ 4$ does not divide $ 3^{n}\plus{}2n\plus{}1$.  For $ n$ odd, this expression is $ 3\plus{}2\plus{}1\equiv2\mod{4}$, and for $ n$ even, the expression is $ 1\plus{}0\plus{}1\equiv 2\mod{4}$.

If $ k \equal{} 2$, then we have a problem.  Will be back with a solution.

EDIT: Okay, here it is, as promised.  However, I used quadratic reciprocity...I will look for a cleaner way.
If $ 2^{2}$ is the greatest power of $ 2$ dividing $ m$, then let $ m\equal{}4x$, where $ x$ is odd.  Clearly $ x\equiv 2\mod{3}$.  It is also clear that $ n$ must be odd.  We have
$ x|3^{n}\plus{}1$
We will prove that every odd prime divisor of $ 3^{n}\plus{}1$ where $ n$ is odd must be $ 1\mod{3}$.
We have $ p|3^{n}\plus{}1\implies 3^{n}\equiv\minus{}1\mod{p}\implies\left(\frac{\minus{}3}{p}\right)\equal{}1$
$ \implies\left(\frac{p}{3}\right)\equal{}1\implies p\equiv 1\mod{3}$, as desired.
[/hide]

[quote="K81o7"][hide]

Suppose $ 2^{k}$ is the greatest power of $ 2$ dividing $ m$.  Now it suffices to show that $ 2^{k\plus{}1}$ does not divide the numerator.

The $ n \equal{} 1$ case is trivial.  So WLOG, $ n\ge 2$.  Moreover, the $ m$ odd case is trivial, as the numerator is odd.  So Wlog, $ 2|m$.  Now we take the numerator modulo $ 2^{k\plus{}1}$.
\[ (m\plus{}3)^{n}\plus{}1\equiv 3^{n\minus{}1}mn\plus{}3^{n}\plus{}1\equiv 2^{k}n\plus{}3^{n}\plus{}1\]
But notice that because $ 3^{n}$ is always $ 1$ or $ 3$ modulo 8, we have that the greatest power of $ 2$ dividing $ 3^{n}\plus{}1$ is $ 2^{2}$.  Okay, so clearly the $ k\ge 3$ case is out the window.

If $ k \equal{} 1$, then we want to show that $ 4$ does not divide $ 3^{n}\plus{}2n\plus{}1$.  For $ n$ odd, this expression is $ 3\plus{}2\plus{}1\equiv2\mod{4}$, and for $ n$ even, the expression is $ 1\plus{}0\plus{}1\equiv 2\mod{4}$.

If $ k \equal{} 2$, then we have a problem.  Will be back with a solution.

EDIT: Okay, here it is, as promised.  However, I used quadratic reciprocity...I will look for a cleaner way.
If $ 2^{2}$ is the greatest power of $ 2$ dividing $ m$, then let $ m \equal{} 4x$, where $ x$ is odd.  Clearly $ x\equiv 2\mod{3}$.  It is also clear that $ n$ must be odd.  We have
$ x|3^{n}\plus{}1$
We will prove that every odd prime divisor of $ 3^{n}\plus{}1$ where $ n$ is odd must be $ 1\mod{3}$.
We have $ p|3^{n}\plus{}1\implies 3^{n}\equiv\minus{}1\mod{p}\implies\left(\frac{\minus{}3}{p}\right) \equal{} 1$
$ \implies\left(\frac{p}{3}\right) \equal{} 1\implies p\equiv 1\mod{3}$, as desired.
[/hide][/quote]
Thank's a lot  :)


AMC AIME AoPSwiki USA(J)MO USAMO USAMTS ARML

The other topic has seemingly ground to a halt, at a tremendous length of 12+ pages.

However, it is clear that great points were made, from both sides of the issue. I'm mostly referring to the "keep same" and the "something has to change" viewpoints.

I'd like to have people who believe they made substantial points in that previous thread to repeat them here. I ask that people actually do NOT comment on other's ideas directly, but rather just state their own first. Also, you may refer to an "argument against" and refute that point. I am trying to create a more compact, serious, and readable version of the first thread, so that it would be easier for people like AoPS admins, and the AMC Director to read.

Current Summary List:
[quote="xxazurewrathxx"]
[u]
[b]USAMTS[/b][/u]
[i]For:[/i]
1. USAMTS is a proof-based competition, so more similar to the USAMO than the AIME is.
[i]Against:[/i]
1. USAMTS is not proctored. If it becomes a standard for USAMO qualification, one could easily cheat on it.
[u]
[b]AIME as Multiple-Choice[/b][/u]
[i]For:[/i]
1. It will be harder to make careless errors on.
[i]Against:[/i]
1. It will become [b]SIGNIFICANTLY EASIER[/b] to guess on.
2. It will make the test overall a lot easier.
3. It will make the AIME even more different from the USAMO.
[u]
[b]MOPpers take USAMO again[/b][/u]
[i]For:[/i]
1. MOPpers (blue and black) are probably already at the level of Olympiad and should not have to undergo the AMC/AIME process again.
2. The selection of the IMO team requires one to be good at proof-based questions, not multiple-choice or free-response.
[i]Against:[/i]
1. Everyone should have to prove himself/herself again each year in order for the system to be fair.
2. Allowing MOPpers back into the USAMO with a free pass may induce them into laziness.
3. The idea of an annual competition would suggest that each year the competitors start over again without any preferential treatment.[/quote]

[quote="solafidefarms"]How about on the AoPSWiki, so that each side can work together to make a compelling argument instead of having bits and pieces?[/quote]

That's a really good suggestion. However, I'm not too sure about how many people read AoPS Wiki, and how many people know how to use (edit) it. I for one, do neither

[quote="solafidefarms"]How about on the AoPSWiki, so that each side can work together to make a compelling argument instead of having bits and pieces?[/quote]
I don't read AoPS wiki either.

In a nutshell:
[b]Arguments for:[/b]
[list][*]AIME does not correlate well with the USAMO; many "bubble" people are capable of taking the USAMO
[*]Easier to make silly errors on AIME than USAMO
[*]Possible to have a bad day on AIME that ruins USAMO chances[/list]

[b]Arguments against:[/b]
[list][*]Not fair for people who only have resources to take one AIME
[*]Mistakes are part of competition
[*]Proposal will require extra resources and money[/list]

[b]Alternate proposals:[/b]
[list][*]Take average of two scores, instead of maximum (or the one score if only one test was taken)
[*]USAMTS?
[*]Previous blue/black moppers = auto-qualify
[*]Anyone can take USAMO unofficially, but may be recognized / accepted into MOP for exceptionally high scores
[*]Others...?[/list]

Tell me if I missed any major arguments.

Another alternate proposal: on future AIME tests, try to write questions that are more open to ingenious solutions, rather than brute-force computation.

dude I want to weigh in guys guys can I weigh in oh man I want to weigh in

alternate proposal of writing good AIME questions: dude this is an awesome idea. I think this is pretty much what they try to do though, just some times it doesn't work out? e.g. this year. this year was sort of crappy? usually though AIMEs are pretty good

alternate proposal of letting blue/black MOPpers qualify for USAMO: this one is sort of obvious. my dad actually thought this existed and was pretty surprised when he found out it didn't. seriously, there's no reason against this one.

alternate proposal of USAMTS: uhh, what about usamts? it's really easy, isn't it? it's probably fine for AIME qualification, as is the system now, but not for USAMO

alternate proposal of unofficial USAMO: so my first thought is dude how the heck do you take the USAMO unofficially someone has to grade the damn thing. but actually this might work. for example, I know on USAPhO that teachers grade all the initial entries, and then only the really high scores get regraded again by the national folks in charge of the contest. It depends on whether teachers or other graders on a local level are willing to get involved.

alternate proposal of two tests: man what the heck that is way too much work. I don't even take two tests for the AMC. THAT IS WEAK, MAN, WEAK.

hmm all the other arguments seem to deal with two tests, which I already said is PRETTY WEAK, MAN, WEAK

As for allowing someone to take the USAMO unofficially and still get into MOP, there are two reasons against it.

1. There will be no resources to grade that many USAMOs.
- Arnav mentioned getting local teachers to grade the USAMO like the USAPhO is graded. Well keep in mind that the free-response sections of the USAPhO exam is much much easier compared to the USAMO (although if you're really good, I suppose you wouldn't notice a difference). How many local teachers do you know who could grade USAMO proofs?

2. Allowing them to go to MOP... hmm..  then why do we even have AMC/AIME? If you can go straight to the USAMO and make MOP, this pretty much eliminates the point of preliminary exams.

As for allowing blue/black MOPpers back into the USAMO the next year...

It seems reasonable, and initially I didn't really feel strongly against it. However, it seems to me that this would encourage students to slack off or make their senioritis worse. If you did well one year, what will prevent you from doing well again? If you really want to make MOP, then why don't you study for the preliminary exams as well? Personally, I'd be more in favor of just letting black MOPpers in again, if at all. I think it'd be more fair to have everyone proves him/herself again each year before going to the USAMO or MOP. The blue/black MOPpers who didn't make the USAMO the next year, how many are there each year anyway? One or two?

dude what is this MOsP business I am finding it very difficult to reply to someone who prominently features such an abhorrent construct in his post

dude yeah if you could actually orchestrate such an effort on a local scale, AMC/AIME would be pointless. meaning don't have them. I'm not actually expecting this to happen next year or anything - it was hard enough for me to find someone to administer the USAPhO - I'm just sayin that it's actually a pretty good idea were it to become feasible at some point (maybe through AoPS? who knows AoPS is pretty beastly man they can do [i]anything[/i])

dude that is a junk argument against letting real MOPpers back into USAMO

JUNK

why the heck should I have to study for AMC

that is JUNK

I will emphasize this by saying "JUNK" several more times

JUNK JUNK JUNK JUNK JUNK

[quote="MysticTerminator"]dude what is this MOsP business I am finding it very difficult to reply to someone who prominently features such an abhorrent construct in his post
[/quote]

Rofl. MOP, my bad. Fixed.

Why should you have to study for the AMC? Obviously, you don't have to. Then why give MOPpers a free pass? They don't have to study for it anyway. In the off chance that they did have to study but chose not to, that's a lesson learned.

Anyway, I said I don't particularly stand against that proposal, at least not as much as allowing two AIMEs or letting USAMTS count for the index.

Wow. It doesn't matter who you are. It doesn't make you God... cough. So don't yell and respect other opinions.

I don't think the argument of having people qualify each year is bogus at all. Each year a brand new slate. Sounds reasonable. If you are really good, AIME ought to be a breeze for most. As was said... a couple of exceptions. But then again, there are exceptions for almost everything. Nothing stunning.

MOSP is awesome. Right. MOSP. Go tell it to the mountain. I love this MOSP business, man. What is MOP. Such abhorent construction that's not even the right acronym.

[quote]2. Allowing them to go to MOP... hmm.. then why do we even have AMC/AIME? If you can go straight to the USAMO and make MOP, this pretty much eliminates the point of preliminary exams. [/quote]

I thought the point of the AMC/AIME was to narrow down the pool so there aren't 40,000 papers to grade come USAMO time. If somehow there was a way to grade 40,000 papers there wouldn't be much point for the AMC/AIME. (Not that one exists, but who knows someone might invent a time machine next year)
[size=59]

(and by might I mean will not)[/size]

[quote="white_horse_king88"]I don't think the argument of having people qualify each year is bogus at all. Each year a brand new slate. Sounds reasonable. If you are really good, AIME ought to be a breeze for most. As was said... a couple of exceptions. But then again, there are exceptions for almost everything. Nothing stunning.[/quote]
so those few exceptions should not even get to take the usamo even though they are perfectly capable of making black mop?
and you say there aren't a lot of people who made blueblack mop and doesnt qualify for usamo. so what's the harm in letting those few people take the usamo?
i really dont see any reason why people are arguing against this in the first place

I've always thought a harder USAMTS (like something that actually models USAMO...?) would be a good idea as qualification for USAMO. I mean, they have a hard 9-hour test and an easier 30 day test that's not common sense dude wtf?
And if you want USAMTS for the beginners....how about make the last 2-3 problems harder.

I also think that multiple choice would help to make AIME model the USAMO more closely. Like you could still have integer answer, but give 5-10 choices so that if someone makes a stupid mistake, they're much less likely to choose an answer. Or, if the goal is to take the best USAMO people, why not just make AIME longer and make the questions harder?

However, I think selecting the best performers for the USAMO is not necessarily the best idea - if new people qualify for it, that should motivate them to do well on it.

[quote]I don't think the argument of having people qualify each year is bogus at all. Each year a brand new slate. Sounds reasonable. If you are really good, AIME ought to be a breeze for most. As was said... a couple of exceptions. But then again, there are exceptions for almost everything. Nothing stunning.[/quote]
I guess this just comes down to philosophy...I feel like the ability to do well on the AIME and the ability to do well on the USAMO are so obviously different that eliminating people we KNOW can do well on the USAMO just because they mess up on the AIME is a really bad idea. I know people with very high USAMO scores and low AIME scores are rare, but they do exist, and it seems to me that when they show up, they should be able to demonstrate again how well they can do on the USAMO, even if they're weak on the AIME.

I still think it would be a good idea to have some kind of USAMO qualification based on APMO, because it makes MOPpers work for their qualification, yet if you are truly good at olympiad problems, you'll be getting a second chance. However, apparently the AMC guys don't really like making the APMO anything official, so...

Yeah, you were joking. I'm am being serious, er... seriously joking about complaining about how you edited my statement. It doesn't matter, now. In fact, we have three threads that are hitting the same stumbling blocks...

[quote="white_horse_king88"]HAHAHA I SAY MOSP WHAT A SILLY ACRONYM MOSP MOSP MOSP[/quote]

yeah pretty much

MOP is awesome

white_horse_king88, let me tell you about my personal experience with AIME: without ever seeing AIME type of questions in my life before, but with extensive experience of Olympiad style proof style problems I was able to completely solve an past AIME test (from 98 or 99 don't remember) in about one hour and 20 minutes back when I was in my senior year in highschool.

What I'm trying to say is that if a person has any chances of making MOP or doing reasonable at an IMO Team Selection Test, passing AIME (I'm not taking about making a 15, i'm taking about passing the test!) should be absolutely no problem.

If you have problems in passing AIME, then I'm 99.9% sure you have no chance in making IMO (or top 12 for that matter), even MOSP.

[quote="Valentin Vornicu"]white_horse_king88, let me tell you about my personal experience with AIME: without ever seeing AIME type of questions in my life before, but with extensive experience of Olympiad style proof style problems I was able to completely solve an past AIME test (from 98 or 99 don't remember) in about one hour and 20 minutes back when I was in my senior year in highschool.

What I'm trying to say is that if a person has any chances of making MOP or doing reasonable at an IMO Team Selection Test, passing AIME (I'm not taking about making a 15, i'm taking about passing the test!) should be absolutely no problem.

If you have problems in passing AIME, then I'm 99.9% sure you have no chance in making IMO (or top 12 for that matter), even MOP.[/quote]

whoa I'm impressed I saw Valentin's name here and figured all these threads would be locked

that seems to happen with threads I post in

but hey I guess not

awesome

Valentin, I totally agree.
However, there is a debate here, since alot of others feel like the AIME is too computational or something.

[quote="Valentin Vornicu"]What I'm trying to say is that if a person has any chances of making MOP or doing reasonable at an IMO Team Selection Test, passing AIME (I'm not taking about making a 15, i'm taking about passing the test!) should be absolutely no problem.[/quote]

This is simply not true. An example:
In 1998, as a 9th grader, Oaz Nir got 122 on the AHSME and at most 5 on the AIME. This was below the qualifying standard for 9th graders (the number for 10th graders was a 190 index; 9th was somewhat lower), but he qualified anyway as the state representative for Missisippi.
On the USAMO, he earned Honorable Mention and went to MOP. In both 2000 and 2001, he was on the IMO team.

Now, his AIME scores did go up after he went to MOP- he had a 10 AIME in 1999.
There's a similar pattern in my own history- I made MOP as a minimal USAMO qualifier with HM in 1998, then doubled my AIME score to 14 the next year. MOP did far more for my AIME scores than for my USAMO scores.

Your perspective is much like mine after MOP- you knew all the tricks from your olympiad training, and that made the AIME seem much easier. That's why giving free USAMO qualification to past MOP attendees wouldn't make much difference. The real issue is finding those people in the first place; many who might have a shot never qualify at all.

In response to jmerry's last post.

I think it would be clearer if you voiced what you think of the difference between "studying/preparing for USAMO" and "MOP". It seems to me that the two should be quite similar in terms of impact on AIME capabilities.

For example, I think many people's caveat is that they studied for olympiad type problems too much, but that might have negatively impacted their AIME scoring, and hence USAMO invitation chances.

[quote="Zellex"]They should have a 15 minute reading period where we just spend 15 minutes reading the question to the AIME. Nothing can be written down, but the reading period is there to prevent people from misreading questions, and it allows people time to just quietly think about the problem, instead of attacking it straight away.

Most of us who attack the problems right away probably misread the question anyway or try to skim through it as fast as possible to not lose any time. I think that's where most of the mistakes come from (at least for me).

Hrm, I think this should be more of a strategy than a change to the AIME testing :P.[/quote]

The thing I hate most is when I can read a problem but not write stuff down. I proceed to try to do the computations in my head, with the result that I make more stupid mistakes from being unfocused.

[quote="Zellex"]They should have a 15 minute reading period where we just spend 15 minutes reading the question to the AIME. Nothing can be written down, but the reading period is there to prevent people from misreading questions, and it allows people time to just quietly think about the problem, instead of attacking it straight away.

Most of us who attack the problems right away probably misread the question anyway or try to skim through it as fast as possible to not lose any time. I think that's where most of the mistakes come from (at least for me).

Hrm, I think this should be more of a strategy than a change to the AIME testing :P.[/quote]

Hm, I might actually do that next year. Maybe not 15 minutes, but maybe 5 minutes. Or something like that. That way, I wouldn't have fallen prey to problems like #2. Gah.

[quote="krustyteklown"]... if you voiced what you think of the difference between "studying/preparing for USAMO" and "MOP".[/quote]
The difference? When I went through, I didn't know the first existed- it was certainly much less common ten years ago. I'm not sure I even knew the USAMO existed until I qualified for it in 1999. I didn't intensively prepare for any contests- everything I did was a contest at some level. The most useful USAMO practice was the mail-in proof contests- those helped my writing skills enormously.
It took months for the effect of MOP to fully set in, but it was huge when it did.

[quote="Zellex"]They should have a 15 minute reading period where we just spend 15 minutes reading the question to the AIME. Nothing can be written down, but the reading period is there to prevent people from misreading questions, and it allows people time to just quietly think about the problem, instead of attacking it straight away.

Most of us who attack the problems right away probably misread the question anyway or try to skim through it as fast as possible to not lose any time. I think that's where most of the mistakes come from (at least for me).

Hrm, I think this should be more of a strategy than a change to the AIME testing :P.[/quote]

Reading through questions, and related organization skills, are important skills to learn. If forced in this manner, those skills could not be tested.

$Volume$ $1$: $The$ $Art$ $of$ $Problem$ $Reading$
$Volume$ $2$: $The$ $Art$ $of$ $Problem$ $Checking$
I should learn Latex.........

I've been reading through these threads, and I find the opinions on both sides of the discussion interesting.  Also, the civility of this particular discussion thread is much appreciated.

I'm not sure, though, what the purpose is for this topic.  If you are intending to put together a proposal for change, I would urge you to look at the situation as much as possible from the viewpoint of those you will present with the proposal.  Such a proposal should be civil and acknowledge the great amount of time, effort, and dedication put into creating and managing these tests by the MAA.

The administrators and staff of the AMC programs deserve thanks not only for their time and effort, but also for the fact that they find ways, on a limited budget, to make these programs as affordable as possible for schools, and for students to find materials to study with.

As someone who has been the president of two non-profit corporations and chaired too many committees, I can assure you a proposal is more likely to be considered if its writers make clear they acknowledge the efforts of the people they wish to convince.

And please keep in mind that what is intended as neutral or mildly angry or frustrated can come across as much more intense than intended.  Once the post is made, it's hard to go back and say, "Well I didn't mean that the way it sounded." It's always better to go on the side of courtesy.

That said ... speaking only for myself, as someone who has been on many committees receiving proposals, I appreciate thoughtful presentations where the writers have taken into consideration both sides of the issues, done their homework, and come up with reasonable solutions to potential problems.  Mind you, committees won't always vote to accept the proposal; you can work for months on putting one together and still have it turned down.  But sometimes they do.

You not only have to convince them your idea is worth considering; you also have to convince them the idea is viable.

I will use the two-AIME's discussion to illustrate what I mean.  The following are the types of problems that have to be addressed to make the suggeestion viable.  These are only examples: I'm sure you can think of others.

1. If students were allowed to take both AIMEs, it would only be fair if

a) every school and/or student could be made aware of this policy in a timely manner,

b) the administrators of the AIME program could do it without undue stress on their time or resources, and

c) students who qualified for the AIME could participate even if their school didn't do the AIME II (e.g., by having local colleges administer the test, as with the AMC-10/12).

2. It is unlikely enough time currently exists in the schedule to get the AIME II out and have everything graded and decided by the deadlines if students were allowed to take both AIMEs.  The AIME II has a fee because the test has to be expressed mailed to get in to the teachers in time, and teachers must get the tests back to the office by March 30, which means they probably also have to express mail them.

This is a web site full of highly talented problem solvers. When you all put your heads together, I've no doubt you can do great work and come up with solutions, if they exist.

One suggestion, if you all decide to brainstorm.  Comment only on the pros and cons of ideas, never on the people making them.  Also -- avoid inflammatory words.  Denigrating the tests or those who criticize the test may make those who agree with the poster express their agreement, but it isn't likely to convince anyone who isn't certain, on top of which, it may turn off people.  Please avoid making inflammatory assessments of the personality, work ethic, or mathematical ability of posters you don't agree with.

As far as whether or not students should be allowed to take both AIMEs, in the past, I would have said no.   After reading the comments here and other topics in this forum, I have reconsidered.  Please remember this is only my opinion.  It in no way represents that of anyone else.

But my response is:  Yes, I think they should, if the process can be made viable.

It is my experience, after 30 years of teaching, that on any test difficult enough to separate out the top students, roughly one third of the students will under-perform their ability by anywhere from a small to a large amount.  It has nothing to do with ability; it happens to the best students.  If students have a chance to try twice, with their top score counted, about 95-97% will perform to their ability.

I doubt it is coincidence that this corresponds roughly to the first and second standard deviations of a Gaussian distribution.

This is why some teachers allow students to drop the lowest score in their homework, quizzes, even tests sometimes.  It's why the grade in most classes doesn't depend solely on one test.  For those of you who have taken SAT tests, you've probably noticed your test scores include a range of possible values centered on your actual score.  That is to take into account such variances.

It is also why the College Board says that if you retake the test, the likelihood your score will not improve is greater than the chance it will improve; if roughly one third under-perform, two thirds don't.  However, if you felt your performance on the test was lower than your ability, you do have the option of taking it again.

In real life, as a mathematician, you can do things over as much as you want.  Andrew Wiles spent years working out the Fermat proof.  Math competitions are a world unto their own, however, where time and speed are more important.  The purpose of these contests is multi-fold.  It is to encourage math students and to pick the best possible team for the IMO.  The need for timed tests is important; it is a necessary part of the process.  It is also important to maximize the chance of getting the best possible IMO team in a manner that is as fair as possible.

(Edits were to fix typos and stream)

Thanks, Mrs. Asaro. That makes tons of sense (that bit with the Gaussian distribution was particularly interesting). I think you are right: people who want change ought to make a proposal. There really isn't much else to hash out in these three-four threads.


Why are there infinitely many number with the same sum of digits in a given arithmetical sequence ?  :)

[hide]

If the A.P. has first term a and difference b, then 10^k * b + a has a fixed sum of digits for k > number of digits of a.

[/hide]

This can be generalized to any polynomial with integer coefficients. An arithmetical progression is just the linear case.
See :
http://www.mathlinks.ro/Forum/viewtopic.php?t=3572&highlight=polynomial+digits

Pierre.


trigonometry
function
logarithms
ratio
calculus
calculus computations

Evaluate

\[ \sum_{n=2}^\infty \frac{1}{2^n}\tan \frac{\pi}{2^n}. \]

first we note that \[ \prod_{n=2}^{N} \cos\left(\frac{x}{2^n}\right)=\frac{2^{N-2}\cos(x/2^2)\ldots \cos(x/2^N) \sin(x/2^N)}{2^{N-2}\sin(x/2^N)}= \]
\[ =\frac{\sin(x/2)}{2^{N-2}\sin(x/2^N)}\to \frac{\sin(x/2)}{x/4},\ N\to \infty \]
Now we define function $f(x)=-\sum_{n=2}^{\infty}\ln(\cos(x/2^n))$. Thus \[ f(x)=-\ln\prod_{n=2}^{\infty} \cos\left(\frac{x}{2^n}\right)=-\ln\left(\frac{\sin(x/2)}{x/4}\right)= \]
\[ =-\ln(\sin(x/2))+\ln(x)-\ln(4). \]
We claim that $\sum_{n=2}^{\infty}\frac{tan(x/2^n)}{2^n} \rightrightarrows$, when $x\in (0;\pi+\varepsilon)$. This statement follow from Weierstrass ratio test.
And it follows that  \[ f'(x)=\sum_{n=2}^{\infty}\frac{tan(x/2^n)}{2^n}=-\frac{ctg(x/2)}{2}+\frac{1}{x}. \]
If we take $x=\pi$ we obtain the answer $f'(\pi)=\frac{1}{\pi}$.

I'm happy norwegian universities don't have entrance exams. :ninja:


counting
distinguishability
number theory
least common multiple
Purple Comet

These problems are from the Purple Comet Math Meet 2007. I've already had the answers, but I need the full solutions.
Problem 1:
Last Sunday at noon the date on the calendar was 15 (April 15, 2007). What will
be the date on the calendar one million minutes after that time?
 *I had 9 for this answer, but it was wrong.

Problem 9:
Purple College keeps a careful count of its students as they progress each year
from the freshman class to the sophomore class to the junior class and, finally, to
the senior class. Each year at the college one third of the freshman class drops out
of school, 40 students in the sophomore class drop out of school, and one tenth of
the junior class drops out of school. Given that the college only admits new
freshman students, and that it wants to begin each school year with 3400 students
enrolled, how many students does it need to admit into the freshman class each
year?

Problem 11
The alphabet in its natural order ABCDEFGHIJKLMNOPQRSTUVWXYZ
is T0. We apply a permutation to T0 to get T1 which is
JQOWIPANTZRCVMYEGSHUFDKBLX. If we apply the same permutation
to T1, we get T2 which is ZGYKTEJMUXSODVLIAHNFPWRQCB. We
continually apply this permutation to each Tm to get Tm+1. Find the smallest
positive integer n so that Tn = T0.

Problem 12
If you alphabetize all of the distinguishable rearrangements of the letters in the
word PURPLE, find the number n such that the word PURPLE is the n-th item
in the list.

Problem 15
We have some identical paper squares which are black on one side of the sheet and
white on the other side. We can join nine squares together to make a 3 by 3 sheet
of squares by placing each of the nine squares either white side up or black side up.
Two of these 3 by 3 sheets are distinguishable if neither can be made to look like
the other by rotating the sheet or by turning it over. How many distinguishable 3
by 3 squares can we form?[/code][/hide]

1.  There are 1440 minutes in a day.  1 million divided by 1440 is equal to 694.44444.  So we simply take 694.  We divide that by 365 to get 1 r 329.  We make that 1 r 328 because 2008 is a leap year.  328 days after April 15, 2008 is March 9, 2009.  Correct me if I'm wrong.

Wow!  These problems are pretty difficult.  Is this the middle school purple comet or the high schooL?

[quote="niels.henrik.abel"]
Problem 1:
Last Sunday at noon the date on the calendar was 15 (April 15, 2007). What will
be the date on the calendar one million minutes after that time?
[b] *I had 9 for this answer, but it was wrong.[/b]
[/quote]
[quote="maybach"]

1. There are 1440 minutes in a day. 1 million divided by 1440 is equal to 694.44444. So we simply take 694. We divide that by 365 to get 1 r 329. We make that 1 r 328 because 2008 is a leap year. 328 days after April 15, 2008 is March [b]9[/b], 2009. Correct me if I'm wrong. [/quote]
You're wrong  :(
[hide="My Solution"]
As maybach pointed out, there are $ 1440$ minutes in a day, meaning that there are $ 694\frac {4}{9}$ days in one million minutes.  The $ \frac {4}{9}$ will be important.  In $ 366$ days (because 2008 is a leap year), it will be April 15, 2008 at noon.  We are left with $ 327\frac {4}{9}$ days.  In $ 365$ days, it will be April 15, 2009 at noon.  However, we have over counted by $ 37\frac {5}{9}$ days.  $ 15$ days back means it would have been March 31 at noon, leaving $ 22\frac {5}{9}$ days.  Thus, $ 22$ days before would be on March 9 at noon.  However, $ \frac {5}{9}$ days is more than $ 12$ hours, meaning it would have been March $ \boxed{8}$, 2009.[/hide]

I believe these were all Middle School probs from the 07 Purple Comet.

For #9, we can say that there are $ x$ freshmen at the start of each year. One third leave that year, giving us $ \frac{2x}{3}$ sophomores. If forty sophomores leave, we have $ \frac{2x}{3} \minus{} 40$ juniors. If 10% of juniors leave, we have $ \frac{9}{10}\Big(\frac{2x}{3} \minus{} 40\Big) \equal{} \frac{3x}{5} \minus{} 36$ seniors.

Adding all the numbers for each year, we have a total of
$ \frac{44x}{15} \equal{} 3476$ students. Solving this equation yields $ x \equal{} 1185$ freshmen.

For #11, we need to find the number of permutations that occur for each substring of letters to return to their original positions. We see that one substring includes the letters A, J, Z, X, B, Q, and G in that order of permutations. These 7 letters cycle to their original positions again every 7th permutation.

Another substring has the letters C, O, Y, and L in that order, so they return every 4th permutation.

A third substring has the letters D, W, K, R, S, H, N, M, and V in that order. These letters return every 9th permutation.

The final substring has the letters E, I, T, U, F, and P in that order. These letters return every 6th permutation.

Thus, we are looking for the smallest permutation on which they all return to $ T_0$. This answer is $ \text{lcm}(4, 6, 7, 9) \equal{} 4\cdot9\cdot7 \equal{} \boxed{252}$.

For #12, we are looking to see how many arrangements come before PURPLE alphabetically.
With first letter E, we have $ \frac{5!}{2!} \equal{} 60$ arrangements.
With first letter L, we also have 60 arrangements.
Within the list with first letter P, we have $ 5! \equal{} 120$ arrangements. Examining the letters, we see that PURPLE is the last possible word alphabetically that begins with P, and so it is number $ 60 \plus{} 60 \plus{} 120 \equal{} 240$ on the list. Thus $ \boxed{n \equal{} 240}$.

For #15, we can do some casework.
We can denote $ f(x)$ as the number of ways to have $ x$ white squares facing up.

For $ f(0)$: Obviously, there is exactly 1 way to arrange all white side up, which is the same paper as all black side up.

For $ f(1)$: With 1 white square up, it can be in the center, in a corner, or in the middle of an edge (non-corner), for a total of 3.

For $ f(2)$: We can have 1 in the center and 1 in the corner, 1 center-1 edge, 2 opposite corners, 2 opposite edges, 2 adjacent corners, 2 adjacent edges, 1 corner-1 adjacent edge (twice) or 1 corner-1 opposite edge (twice). Thus $ f(2) \equal{} 10$.

For $ f(3)$:We can have
1Center-2OppositeCorners,
1Center-2OppositeEdges,
1Center-2AdjacentCorners,
1Center-2AdjacentEdges,
1Center-1Corner-1AdjacentEdge (twice),
1Center-1Corner-1OppositeEdge (twice),
3Corners,
3Edges,
2OppositeCorners-1Edge (twice),
2AdjacentCorners-1Edge (four times),
2OppositeEdges-1Corner (twice),
2AdjacentEdges-1Corner (four times).
This list gives us a total of 22, so $ f(3) \equal{} 22$.

For $ f(4)$, we can have
1Center-3Corners
1Center-3Edges
1Center-2OppositeEdges-1Corner (twice)
1Center-2AdjacentEdges-1Corner (four times)
1Center-2OppositeCorners-1Edge (twice)
1Center-2AdjacentCorners-1Edge (four times)
2OppositeCorners-2OppositeEdges (twice)
2OppositeCorners-2AdjacentEdges (twice)
2OppositeEdges-2AdjacentCorners (twice)
2AdjacentEdges-2AdjacentCorners (four times)
3Corners-1Edge (four times)
3Edges-1Corner (four times)
4Edges
4Corners
This list shows our 34 options for $ f(4)$.

Because the problem states that an arrangement is the same if it can be flipped over to become the same, we know that $ f(x) \equal{} f(9 \minus{} x)$. Thus, we don't have to count $ f(5)$ through $ f(9)$, because they just show the flips of our previous cases.
$ f(0) \plus{} f(1) \plus{} f(2) \plus{} f(3) \plus{} f(4) \equal{} 1 \plus{} 3 \plus{} 10 \plus{} 22 \plus{} 34 \equal{} \boxed{70}$.

If anyone else has a quicker method for #15, please post it, as brute casework was my only solution.
BTW, I have verified all these answers with the Purple Comet website to ensure accuracy.


BTW: Regarind LaTex, how do i make the +/- symbol so that i can write

$x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}$

properly?

Use \pm (p stands for plus, and m for minus):
\[ x = {-b\pm\sqrt{\Delta}\over 2a} \]

$x=\frac{-b \pm\sqrt{b^2-4ac}}{2a}$

Ah ok, i see, thx for the help...

Now why does it show the fraction like it has been pressed together?

That's because it's long. If you want it to be larger, use \displaystyle

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

[quote="Silverfalcon"]That's because it's long. If you want it to be larger, use \displaystyle[/quote]On the forum you need not use \displaystyle (it is automatically added).

Well, the first time i made the fraction it didn't resize it, but it did the second time. Now i'm on my stationary PC and it shows the fraction in the correct size. But well, nvm with that.


number theory
greatest common divisor
limit
number theory proposed

I really HOPE I havent posted this before!!! I also searched a lot in my posts but couldnt find it! I just solved it today and its no doubt a cute one! :)

Let $ \eta_n(m)$ be the number of postivie integers smaller than or equal to $ m$ and coprime to $ n$.

$ \eta_n(m) \equal{} \left|\left\{ x\;|\;x\in\mathbb{N}^{\leq m},\;\;\text{gcd}(x,n) \equal{} 1\right\}\right|$

Prove that:

$ \lim_{m\rightarrow\infty}\frac {\eta_n(m)}{m} \equal{} \frac {\phi(n)}{n}$

[quote="sinajackson"]I really HOPE I havent posted this before!!! I also searched a lot in my posts but couldnt find it! I just solved it today and its no doubt a cute one! :)

Let $ \eta_n(m)$ be the number of postivie integers smaller than or equal to $ m$ and coprime to $ n$.

$ \eta_n(m) = \left|\left\{ x\;|\;x\in\mathbb{N}^{\leq m},\;\;\text{gcd}(x,n) = 1\right\}\right|$

Prove that:

$ \lim_{m\rightarrow\infty}\frac {\eta_n(m)}{m} = \frac {\phi(n)}{n}$[/quote]
Is it obvious ?
The first we have :
\[ \gcd(x,n) = \gcd(x + kn)
\]
Therefore two complete resides mod n have equal number of $ x$ such that $ x,n$ are relative prime.
Now consider $ m = kn + r$ with $ r\in \{0,...,n - 1\}$
Consider $ k + 1$ set with k set is a complete reside mod n and the remain is set of $ \{kn + 1,..,kn + r\}$ (it can have no element).
We have in each complete reside have exactly $ \phi(n)$ element $ x$such that  $ \gcd(x,n) = 1$
Therefore total x is relative prime with n in first k sets is $ k.\varphi (n)$
The remain have at least 0 and at most $ \phi(n - 1)$ element .
So we have :
$ \frac {k.\varphi(n)}{kn + r}\leq \frac {\eta_n(m)}{m}\leq \frac {k.\varphi(n) + \varphi(n - 1)}{m}$
Two sequence tend to $ \frac {\varphi(n)}{n}$ so the sequence $ \frac {\eta_m(n)}{m}$ is also .
Problem claim .

Nice! Yeah it kind of looks obvious at first look!
but this was my solution (which seems a bit similar to yours too):
obviously we can consider $ m>n$, then there exist positive integers $ q,r$ such that $ m\equal{}nq\plus{}r$ and $ 0\leq r<n$.
Then I proved that:
$ \eta_n(m)\equal{}q\varphi(n)\plus{}\eta_n(r)$
(its easy to do...)
and then we hvae:
$ \lim_{m\rightarrow\infty}\frac{\eta_n(m)}{m}\equal{}\lim_{q\rightarrow\infty}\frac{\varphi(n)q\plus{}\eta_n(r)}{nq\plus{}r}\equal{}\frac{\varphi(n)}{n}$
Done!


ΑΣΚΗΣΗ  1

Το  άθροισμα  κάποιων  θετικών  αριθών  είναι  10 και  το άθροισμα  των τετραγώνων  τους  είναι μεγαλύτερο απο  20. Να αποδείξετε  οτι το άθροισμα  των κύβων  τους  είναι μεγαλύτερο απο  40.

Λοιπον
νομίζω την έλυσα.

Έστω ν το πλήθος των αριθμων.
Έστω επίσης
$S_1$ το άθροισμα τους
$S_2$ το άθροισμα των τετραγώνων τους και
$S_3$ το άθροισμα των κύβων τους.

Απο BCS έχουμε $S_1 \cdot S_3\geq S_{2}^2>400$
Άρα $S_3>40$

οκ??
Να θεσω την επόμενη?? :)

Φυσικά και είναι οκ!!  Ρίξε  την επόμενη!

Επίσης να προτείνω  κάτι. Επειδή προς  το παρόν δέν  γίνεται να  μας  χωρίσουν  το  Ελληνικό  φορουμ  σε  υποφόρουμ,,,,αν κάποιος  θέλει να προτείνει  άσκηση  απο  άλλο  ''κλάδο''  να  βάλει  νεο  τοπικ  με  κεφαλαία  όπως  αυτό  με  τις ανισότητες.(Πχ ΓΕΩΜΕΤΡΙΑ , ΣΥΝΔΙΑΣΤΙΚΗ , ΑΛΓΕΒΡΑ....) . Εκει  θα  βάζουμε  ότι  ασκήσεις  θέλουμε απο  τον ''κλάδο'' αυτό, και ας  πιάσει  όσες  σελίδες  θέλει  το κάθε  τόπικ. Μέχρι να  μας  χωρίσουν  το φορουμ...οκ?

Aν α, β ,γ ,δ μη αρνητικοί αριθμοί  τέτοιοι ώστε
                                 α^5+β^5 μικρότερο η ίσο του 1
                                 γ^5+δ^5 μικρότερο η ίσο του 1
νδο
                      α^2.γ^3 + β^2.δ^3 μικρότερο η ίσο του 1

Πολυ ωραία η πρόταση του Νικόλα. Πρέπει να υπάρχει τάξη και ταξινόμηση στις ασκήσεις ώστε να τις βρίσκουμε εύκολα και να τις μελετάμε κατα τομείς!

Επίσης συνιστώ σε όλους να μάθουν το Λάτεχ.Είναι εύκολο και απλό. Οδηγος για τη χρήση του υπάρχει στο [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=5261[/url]   (δέιτε κυρίως το αρχείο pdf ).

Στην άσκηση τώρα...

Έχουμε $a^2c^3=\sqrt[5]{a^5a^5c^5c^5c^5}\leq\frac{2a^5+3c^5}{5}$ απο AM-GM
ομοίως έχουμε $b^2d^3=\sqrt[5]{b^5b^5d^5d^5d^5}\leq\frac{2b^5+3d^5}{5}$

Mε πρόσθεση κατα μέλη λαμβάνουμε $a^2c^3+b^2d^3\leq\frac{2(a^5+b^5)+3(c^5+d^5)}{5}\leq\frac{2+3}{5}=1$

Eπίσης μπορούμε να δούμε οτι $2+3=5$ ή $\frac{2}{5}+\frac{3}{5}=1$ και να εφαρμόσουμε την ανισότητα σταθμισμένου AM-GM :
$(a^{5})^{\frac{2}{5}} \cdot (c^5)^\frac{3}{5}\leq \frac{2}{5}a^5+\frac{3}{5}c^5$ κτλ... όπως προηγούμενα...

Οι δυο προσεγγίσεις είναι ασφαλώς ισοδύναμες.

ok?
 :)

Aκολουθεί η.....

ΑΣΚΗΣΗ 3

Αν $a,b,c\geq0$
δείξτε ότι $2(a^5+b^5)+c^5\geq\frac{5}{4} \cdot abc \cdot (a+b)^2$.

 :)

[quote]Γρήγορα άρχισε να ανεβαίνει ο πήχυς . Καλό αυτό .[/quote]

[quote="socrates"]

ΑΣΚΗΣΗ 3

Αν $a,b,c\geq0$
δείξτε ότι $2(a^5+b^5)+c^5\geq\frac{5}{4} \cdot abc \cdot (a+b)^2$.

 :)[/quote]

Εχουμε $a^5+b^5\ge 2\left( \frac{a+b}{2} \right)^5 \Longrightarrow$
 $2(a^5+b^5)+c^5\geq  a^5+b^5+c^5+2\left( \frac{a+b}{2} \right)^5\geq 5\sqrt[5]{  a^5b^5c^5\left( \frac{a+b}{2} \right)^5\left( \frac{a+b}{2}\right)^5}= \frac{5}{4} \cdot abc \cdot (a+b)^2$.

.

Επιτρέψτε μου να δόσω μια συνθετική λύση σε αυτήν την πολύ σφιχτή ανισότητα.
Θα επικαλεστώ την ορθότητα των παρακάτω ανισοτήτων. Τις αποδείξεις αυτών μπορεί να βρεί κανείς στο βιβλίο του Δ. Κοντογιαννη "Ισότητες και Ανισότητες στο τρίγωνο".
$ 16Rr -5r^2 \leq p^2 $ (1)
$ p^2 \leq 4R^2 + 4Rr + 3r^2 $ (2)
$ R \geq 2r $ (3) .
Βάση της (3) και με απλές πράξεις παίρνουμε ότι :
$ r^2 +16R^2 + 8Rr \geq 3\cdot (4R^2 + 4Rr + 3r^2)$
Τότε από την (2) έχουμε ότι :
$ r^2 +16R^2 + 8Rr \geq 3\cdot p^2 $ άρα
$ r^2 \cdot (r^2 +16R^2 + 8Rr - 3\cdot p^2) \geq 0$ (4)
Επίσης από την (1) έχουμε :
$ p^2 \cdot ( p^2 -16Rr +5r^2) \geq 0 $ (5)
Και με πρόσθεση των (4) και (5) κατα μέλη παίρνουμε ότι :
$2p^4 +2r^4+32R^2r^2-12p^2r^2 +16Rr^3-16p^2Rr\geq 16p^2 \cdot (Rr-r^2)$.
Η σχέση αύτη αποδεικνύει το ζητούμενο διότι με μερικές προφανείς αλλά πολύ κουραστικές για να τις γράψω με λατεχ πράξεις παίρνουμε ότι :
$2p^4 +2r^4+32R^2r^2-12p^2r^2 +16Rr^3-16p^2Rr = a^4 +b^4 +c^4$ .

Φυσικά η πορεία που ακολούθησα για να φτάσω στην λύση ήταν αναποδή απο αυτήν που παρουσίασα παραπάνω. Αξίζει να αναφέρω ότι η ανισοτητα δεν ισχύει εαν υποβιβάσουμε τις δυναμεις των όρων στο αριστερό μέλος με κάποια ανισότητα μέσης δύναμης. Θα ήθελα πολύ να δώ μια διαφορετική λύση.

Ας είναι το επόμενο πρόβλημα η αποδείξη των (1) και (2).

[quote="socrates"]ΑΣΚΗΣΗ 4

Να δείξετε ότι σε κάθε τρίγωνο ΑΒΓ , με ημιπερίμετρο $p$, ακτίνα περιγεγραμμένου κύκλου $R$, ακτίνα εγγεγραμμένου κύκλου $r$ και εμβαδό $S$ ισχύει

$a^4+b^4+c^4\geq16pS(R-r)$

 :)[/quote]

Γνωρίζουμε οτι $abc=4SR, pr=S$ και $16S^2=2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)$
Η ζητούμενη ανισοτητα ειναι ισοδυναμη με την
$2(a^2b^2+b^2c^2+c^2a^2)\ge 2abc(a+b+c)$ που ισχυει λογω της $x^2+y^2+z^2\ge xy+yz+zx$
.

[quote="nikos"]Aν a ,b ,c ,d θετικοί αριθμοί τέτοιοι ώστε α +b +c +d = 1 , νδο
                    $abc + bcd + cda + dab \leq \frac{1+176abcd}{27}$[/quote]

Βλέπε http://www.mathlinks.ro/Forum/viewtopic.php?highlight=%2Aabc%2Bbcd%2Bcda%2Bdab%2A&t=32472 Post #12
.

Έστω x ,y ,z θετικοί πραγματικοί , αν
                          Μ = $\frac{x}{y+2005z} + \frac{y}{z+2005x}+\frac{z}{x+2005y}$
να βρείτε την ελάχιστη τιμή του Μ .

Λοιπόν , αγαπητοί φίλοι

       ας βαδίσουμε στο πλάνο του Νικόλα. Αυτό τον καιρό είμαι λίγο μπλοκαρισμένος με κάτι γεωμετρικά θέματα στο Forum , αλλά θα έρθω σε λίγο και μαζί σας. Ωραία  η δουλειά  σας μέχρι τώρα.  Αν κάποια άσκηση είναι από το αγγλικό Forum , καλό είναι να βάζουμε τη λύση (τουλάχιστον την πιο κατανοητή) και στα Ελληνικά. Έτσι θα μπορούν όλοι να τη μελετήσουν  και κυρίως αυτοί που δεν έχουν ακόμα εξοικειωθεί με τους αγγλικούς όρους. Δε βλάπτουν όμως και οι παραπομπές στα αγγλικά. Προτείνω ,  τις πιο καλές ασκήσεις από το φόρουμ , σε όλα τα πεδία , να τις περάσουμε σιγά - σιγά και εδώ. Να γράφουμε όμως στα ελληνικά τουλάχιστον τις εκφωνήσεις.

 Μπάμπης

Galeε στην ουσία η λύση σου πετυχαίνει ακόμα καλύτερη τιμή στην πρωτη ανισότητα, γιαυτό είναι και τόσο περίπλοκη.
Θα δόσω παρακάτω την λύση που είχα κατα νου.
1)
$\sqrt{a^2 +(\frac{b-c}{2})^2} < a + \frac{b+c}{2}$ και $ \sqrt{b} + \sqrt{c} \leq \sqrt{2} \cdot \sqrt{b+c} $.
Άρα αρκεί να φράξω το :
$a + \frac{b+c}{2} + \sqrt{2} \cdot \sqrt{b+c} = \frac12 + \frac{a}{2} + \sqrt{2} \cdot \sqrt{1-a} = f(a)$.
Είναι πλέον εύκολο να πετύχουμε ανω γράγματα για αυτήν την παράσταση, είτε επειδή είναι τριώνυμο ως προσ το 1-α, ειτε επειδη είναι φθίνουσα ως συνάρτηση του α (π.χ $f'(a) < 0$) άρα $f(a) \leq f(0) = \frac12 + \sqrt{2}$.
2)
Είναι προφανές ότι $\sqrt{a^2 +(\frac{b-c}{2})^2} < a + \frac{b+c}{2}$ ομοίος και οι υπόλοιποι όροι, και απλα προσθέτουμε.

Επόμενη ανισότητα ?

eprepe sonei kai kala na parousiaseis ti lisi sti deuteri anisotita????  :(
tsampa to latex pu egrafa  :mad:

mia anisotita kai apo mena:
$x,y,z$ pragmatikoi, $x^2+y^2+z^2+2xyz=1$. Na deixtei oti $x^2+y^2+z^2\geq \frac{3}{4}$

kai alli mia pio diskoli:
$x,y\geq0$ : $x+y+\sqrt{2x^2+2xy+3y^2}=k$, $k\geq0$ (statheros). Na vrethei h megisti timi tu $x^2y$

Η λύση σου στην πρώτη ανισότητα ήταν περίπλοκή και μάλλον επιρρεάστηκες για την δεύτερη ανισότηατα.  ;) Η μέθοδος σου όμως είναι πολύ ενδιαφέρουσα.

Λοιπόν , η πρώτη ανισότητα που προτήνεις έχει μια πολύ εύκολη αλγεβρική λύση και μια πολύ ενδιαφέρουσα γεωμετρική λύση. Δεν θα γράψω την λύση μου. Την αφίνω για όσους θέλουν να ασχολειθούν.

Η δεύτερη όμως είναι λίγο περίεργη.
Εάν δεν έκανα κάπου λάθος στους υπολογισμούς μου, τότε $x^2y \leq \frac{(7-2\sqrt{10})(\sqrt{5} -1)}{4} \cdot k^3 $.

Για να καταλήξω εκεί εργάσθηκα ως εξής : έθεσα $a=x^2y$ (1) και επειδή απο την σχέση που δίνεις συνεπάγεται ότι $x^2+2y^2+2k(x+y)=k^2$. Αν κάνουμε μια φορα αντικατάστση του y απο την (1) και θεωρήσουμε το τριώνυμο ως προς a με μη-αρνητική διακρίνουσα παίρνουμε ότι $x\leq (\frac{\sqrt{5}}{\sqrt{2}} -1)\cdot k$. Ομοίως και για το χ. Και επείδη η ισότητα είναι δυνατή λαμβάνουμε το ζητούμενο μέγιστο.

[quote="nikos"]Γεια σου Στέργιο .

Αν x , y , z πραγματικοί αριθμοί νδο
$x^4(1 + y^4) + y^4(1 + z^4) + z^4(1 +x^4) \geq6x^2y^2z^2$ .[/quote]

Άμεση  εφαρμογή  της  ΑΜ-GM  για έξι όρους :  $LHS=x^4+x^4y^4+y^4+y^4z^4+z^4+z^4x^4\geq6\sqrt[6]{x^{12}y^{12}z^{12}}=6x^2y^2z^2$

[quote="Anto"] Και επείδη η ισότητα είναι δυνατή λαμβάνουμε το ζητούμενο μέγιστο.[/quote]
H ισότητα δεν είναι δυνατή παρόλο που οι πράξεις είναι σωστές. ;)
Μια δυνατή τιμή για το $ x $ εξασφαλίζει την ύπαρξη του α, το ίδιο και για το $ y $, αλλά δεν εξασφαλίζουν το ίδιο α.

Η αλήθεια είναι ότι δεν κάθησα να κάνω τις πράξεις...  :D  :D
Ηλπιζα ότι θα είναι σωστές :D  :P  :blush:  :D
Θα το ξαναδώ αύριο γιατί τώρα είναι αργά.

Δεν άντεξα να κοιμηθώ χωρίς να το σκεφτώ λιγο... :P
Το μέγιστο λαμβάνετε όταν x=2y.
και $x^2y \leq 4 k^3 (\dfrac{\sqrt{15}-3}{6})^3$. Ελπίζω να μην έκανα κάποιο λάθος στις πράξεις μου, γιατί αυτήν την φορά η μέθοδος μου δουλεύει σίγουρα.
Σκιαγραφώ την λύση μου.
Τα x,y κινούνται πάνω σε τμήμα έλλειψης. 'Αρα (μπορούμε να αποδείξουμε ότι) υπάρχει γωνία t του πρώτου τεταρτημορίου ώστε $x=k(\dfrac{\sqrt{5}}{\sqrt{2}} cos(t) -1) , y=\dfrac{k}{2} (\sqrt{5}sin(t) -1)$. Θεωρώ τώρα την συνάρτηση $f(t) =x^2y$  και με εφαρμογή του διαφορικού λογισμού παίρνω ότι το ζητούμενο μέγιστο λαμβάνεται όταν $x=2y$. Λεπτομέριες θα γράψω (αν χρειαστεί) αύριο που θα έχω καθαρότερο μυαλό.
Στέργιος

:coolspeak:
To periergo me auti tin askisi einai oti vgainei k me Cauchy!

:)
Πράγματι υπάρχει μια πολυ απλή λύση με στοιχειώδεις ανισότητες. Πολύ εύκολη να βρεις αν γνωρίζεις ποτε ισχύει η ισότητα. Αλλα αν όχι τότε είναι πολύ δύσκολο να πάει το μυαλό σου εκεί.  :roll:
Την γράφω παρακάτω :
Έστω ότι $a= x+y$.
Απο caychy έχω ότι $2a = x+x+2y \geq (x^2 \cdot 2y)^{\dfrac{1}{3}}$ άρα $x^2y\leq 4a^3$(1) .
Επειδή όμως $a^2 = (0.5x +0.5x +y)^2 \leq 3 \cdot (\dfrac{1}{2}x^2 +y^2) = \dfrac{3}{2}(k^2-2ka)$. Παίρνουμε την μέγιστη τιμή του α. Και λόγω της (1) το ζητούμενο μέγιστο.
Προφανώς η ισότητα διατηρείται σε όλες τις ενδιάμεσες ανισότητες.

oraios!
mia alli lisi pu eixa dei se vivlio kai einai adinato na tin skefteis einai
$x+y\geq3\sqrt[3]{\dfrac{x^2y}{4}}$
$2x^2+2xy+3y^2\geq15\sqrt[15]{(\dfrac{2x^2}{8})^8(\dfrac{2xy}{4})^4(y^2)^3}$ $=15(\sqrt[3]{\dfrac{x^2y}{4}})^2$

Για την  ιστορία να αναφέρω  οτι η άσκηση αυτή  είναι  το  crux problem , no 358.

Για να  μην  ξεχνιόμαστε....προτείνω  το εξής  πρόβλημα  (έχω  ήδη  λύση  του).

Αν  $0\leq x\leq2$ να  προσδιορίσετε την  μέγιστη τιμή της  παράστασης : $A=\sqrt{4x-x^3}+\sqrt{x+x^3}$

To μέγιστο είναι $4$ μήπως?

[quote="nickolas"]Σε  πρώτη  φάση  : $(2A)^2=(\sqrt{2}\sqrt{8x-2x^3}+2\sqrt{x+x^3})^2$. Μέτα....BCS.[/quote]

Ακριβώς και εγώ έτσι το έκανα.


Here are a couple of old state problems (2001) that I couldn't get and didn't find using the search function. I'd appreciate it if people could give me hints (not solutions, please) for them.

1. The $ n$th term of a sequence is $ a_n \equal{} (\minus{}1)^{n\plus{}1}(3n \plus{} 2)$. What is the value of $ a_1 \plus{} a_2 \plus{} \dots \plus{} a_{100}$?

For this one, I realized that the terms are alternating positive and negative, and $ a_{100}$ is negative. That's pretty much as far as I got...

2. Two numbers are called mirror numbers if the digits of the 2 numbers are in reverse order. For example, 421 and 124 are mirror numbers. What is the sum of the two mirror numbers whose product is 110,080?

If the digits are $ a, b, c$, then $ (100a \plus{} 10b \plus{} c)(100c \plus{} 10b \plus{} a) \equal{} 110,080$. Solving that didn't look fun, and the only other thing I can think of to do is guess-and-check.

[hide="Problem 1"]
What is the sum of $ a_1$ and $ a_2$?
What about $ a_3$ and $ a_4$?
[/hide]
[hide="Problem 2"]
Prime factor 110,080.
What must the numbers begin and end with?
[hide="Bigger Hint"]
One number is a multiple of 5,
but it can't be a multiple of 2 or it would end in 0.
[/hide]
[/hide]

[hide="1"]If n is odd, what is $ a_n \plus{} a_{n\plus{}1}$? extrapolate[/hide]

[hide="2"]one of the numbers must be

even # _ 5 and the other 5 _ even #.

even 405 * 504 is way too big so even # must be 2.

[/hide]

Thanks.  :)

For Problem 1:

$ a_1 \equal{} 5$ and $ a_2 \equal{} \minus{}8$, so their sum $ \minus{}3$. $ a_3 \equal{} 11$ and $ a_4 \equal{} \minus{} 14$, so their sum is $ \minus{}3$...gasp! There are 50 pairs of numbers in total, each with a sum of $ \minus{}3$, so the answer is $ 50 \times \minus{}3 \equal{} \minus{}150$?

For Problem 2:

$ 110,080$ prime factorizes as $ 2^9 \cdot 5 \cdot 43$. That means one number is a multiple of 5 that ends in 5 (since 0 would mean the other number would be a two-digit one), so the numbers look like 5 _ _ and _ _ 5. The first number has to be even, so it ends in 2, 4, 6, or 8. I guess-and-checked to get 512 and 215, so the sum is $ 512 \plus{} 215 \equal{} 727$, I think.

Can you please confirm if my answers are right or wrong?

[quote="isabella2296"]Thanks.  :)

For Problem 1:

$ a_1 \equal{} 5$ and $ a_2 \equal{} \minus{} 8$, so their sum $ \minus{} 3$. $ a_3 \equal{} 11$ and $ a_4 \equal{} \minus{} 14$, so their sum is $ \minus{} 3$...gasp! There are 50 pairs of numbers in total, each with a sum of $ \minus{} 3$, so the answer is $ 50 \times \minus{} 3 \equal{} \minus{} 150$?

For Problem 2:

$ 110,080$ prime factorizes as $ 2^9 \cdot 5 \cdot 43$. That means one number is a multiple of 5 that ends in 5 (since 0 would mean the other number would be a two-digit one), so the numbers look like 5 _ _ and _ _ 5. The first number has to be even, so it ends in 2, 4, 6, or 8. I guess-and-checked to get 512 and 215, so the sum is $ 512 \plus{} 215 \equal{} 727$, I think.

Can you please confirm if my answers are right or wrong?[/quote]

Those are both right. I know the first one for sure is right.

just a thought:

do we need to really check randomly to see that the difference is $ \minus{}3$? not really. we can see that the first number of any given pair will be positive, so let this be $ 3x\plus{}2$. the next pair will be $ \minus{}(3(x\plus{}1)\plus{}2)$. the sum of these is $ (3x\plus{}2)\minus{}(3x\plus{}5)$, which equals $ 2\minus{}5\equal{}\minus{}3$. just a way of verifying that our assumption that every pair sums to $ \minus{}3$.

by the way, izzy, both of your answers are correct. :)

For question #2:
We can prime factor $ 110,080$.
We find that it is $ 2^9*5*43$.
Since the number that is a multiple of $ 5$ is not even it must be $ 5*43\equal{}215$.
$ 215$ happens to be $ 2^9\equal{}512$ backwards.

[quote="akhil0422"]Question 1 was in FTW. Wasn't there a solution there for you to look at? (This topic was what made me get it correct :P ).

I'm just curious. Most all of the problems have solutions.[/quote]

If it was asked, I certainly didn't see it.


given n numbers x1,x2,....,xn each of them is equal to +1 or to -1.find n if we know that 1997<=n<= 2003 and x1x2+x2x3+....+xnx1=0

[quote="DuongChinh_K41SPHN"]given n numbers $ x_1,x_2,....,x_n$ each of them is equal to $ \plus{} 1$ or to $ \minus{} 1$.find $ n$ if we know that $ 1997 \le n\le 2003$ and $ x_1x_2 \plus{} x_2x_3 \plus{} .... \plus{} x_nx_1 \equal{} 0$ [/quote]
$ \exists(a_1,.a_2,...,a_m,e)\in{\mathbb{N}^*}^m\times \{ \minus{} 1,1\}: \ a_1 \plus{} a_2 \plus{} ... \plus{} a_m \equal{} n$ such that:
$ x_1 \equal{} x_2 \equal{} ... \equal{} x_{a_1} \equal{} e$
$ x_{a_{1} \plus{} 1} \equal{} x_{a_1 \plus{} 2} \equal{} ... \equal{} x_{a_1 \plus{} a_2} \equal{} \minus{} e$
$ x_{a_1 \plus{} a_2 \plus{} 1} \equal{} x_{a_1 \plus{} a_2 \plus{} 2} \equal{} .... \equal{} x_{a_1 \plus{} a_2 \plus{} a_3} \equal{} e$
...
...
$ x_{a_1 \plus{} a_2 \plus{} .. \plus{} a_{m \minus{} 1} \plus{} 1} \equal{} x_{a_1 \plus{} a_2 \plus{} .. \plus{} a_{m \minus{} 1} \plus{} 2} \equal{} .... \equal{} x_{a_1 \plus{} a_2 \plus{} .. \plus{} a_{m}} \equal{} ( \minus{} 1)^{m \plus{} 1}e$
then we have:
$ {0 \equal{} x_1x_2 \plus{} x_2x_3 \plus{} .... \plus{} x_nx_1 \equal{} \minus{} (m \minus{} 1) \plus{} (\sum_{k \equal{} 1}^{m}(a_k \minus{} 1)) \plus{} ( \minus{} 1)^{m \plus{} 1} \equal{} n \minus{} 2m \plus{} 1 \plus{} ( \minus{} 1)^m}$
then $ n \equal{} 2m \minus{} 1 \plus{} ( \minus{} 1)^m$
$ m \equal{} 2k$ gives $ n \equal{} 4k$
$ m \equal{} 2k \plus{} 1$ gives $ n \equal{} 4k \plus{} 2 \minus{} 2 \equal{} 4k$
so $ n \equal{} 2000$ because $ 1997 \le n\le 2003$


I need help with a probability problem. There are 30 random people in the room. What is the probability that two of them share a birthday? I need a step-by-step solution like this:
[hide="Step 1."]
blah blah blah.
[hide="Answer."]
blah blah blah.
[hide="Step 2."]
blah blah blah.

blah blah blah.

...

[/hide]
[/hide]
[/hide]
Thanks. :lol:

count the complement...what is the probability that nobody has the same birthday?

[quote="SM4RT"]umm is it [b]atleast two[/b] or only two (cuz thats not as simple)[/quote]
At least two. :rotfl:

This is really famous, in fact the smallest $n$ such that $n$ people share a birthday with probability greater than $.5$ is $23$, i.e. among $23$ random people in a room, it is likely that two share a birthday.

[hide]

Find the probability that no two people share a birthday.

It doesn't matter what the first person is, but then there is $\frac{364}{365}$ chance that he'll have a different birthday. The third person will have a chance of $\frac{363}{365}$ of not sharing a birthday. Keep going through all the people. Find the product of these probabilities, and then subtract it from one to get your answer.

[/hide]

[quote="1=2"]uuhh... clue me in: leap year???

[hide="solution"]Find the probability that no two people share a birthday:

$1*\frac{364}{365}*\frac{363}{365}*\cdots*\frac{336}{365}$

If it's a non-leap year.

So that's

$\frac{364*363*362*\cdots*337*336}{365^{29}}$

So the probability that at least two share a birthday is

$\frac{365^{29}-364*363*362*\cdots*337*336}{365^{29}}$
[/hide]

IF it's a non-leap year![/quote]
Can you express that like a percentage? Or something easier.

[quote="SM4RT"]So basically for $n$ people its just (for a non-leap year):
$1-\frac{365 P n}{365^{n}}\implies1-\frac{365!}{(365-n)!\cdot365^{n}}$

So in this case $n=30$;

$1-\frac{365!}{335!\cdot365^{30}}\approx{0.7063}$

Thus thats approximately $\boxed{71\%}$[/quote]
Is there any easy way to derive that percentage other than multiplying it out?

[quote="bigboy82892"]
Is there any easy way to derive that percentage other than multiplying it out?[/quote]

Most people would use [url=http://mathworld.wolfram.com/StirlingsApproximation.html]Stirling's approximation[/url].


A standard six-sided die is rolled, and $ P$ is the product of the five numbers that are visible. What is the largest number that is certain to divide $ P$?

$ \textbf{(A)}\ 6\qquad
\textbf{(B)}\ 12\qquad
\textbf{(C)}\ 24\qquad
\textbf{(D)}\ 144\qquad
\textbf{(E)}\ 720$

[hide="Solution"]
All possible values of $ P$ are $ \frac{6!}k\equal{}\frac{720}k$ for $ 1\le k\le6$. For every $ k$, the most we can "take away" is the least common multiple of $ \{1,2,3,4,5,6\}\equal{}60$. Thus, it must always be divisible by $ \frac{720}{60}\equal{}\boxed{\textbf{(B)}\ 12}$. Note that even though $ 60$ is not achievable, in order for $ \frac{720}k$ to be an integer for [i]all[/i] possible values of $ k$, we must account for the maximum possible prime powers for each of these.
[/hide]


Just to figure out how often others study math.

Please comment as well.

I think my time has gone from like 2-2.5 -> .5-1

now I suck :(

Well, it depends what you consider math. Does FTW count? If so, 4-6 hours. If not, 2-3 hours.

Does looking at random TST problems that seem interesting count?

(r15s11z55y89w21 is about to sound like a very stupid person)

What is TST?

team selection test methinks
dunno, it was this mean thing they used at arml to place me on silver team
30 minutes. i do it during math class at school :D
there's no 45 minute choice, so i chose 30minutes to account for weekends and holidays

TST = Team Selection Test. Used to select the USA's IMO Team.

God you middle schoolers (not you pythag) need to find out about what math is. It seems like you haven't figured out Mathcounts doesn't count as real math.

the title of this thread makes me angry
slightly


If n is an integer and $20<2^n<200$, what is the sum of all the possible values of n?

Know your powers of 2,

[hide]5+6+7=18[/hide]

Like 236factorial said, you shuold know your powers of 2 up to $2^{12}$.  It makes a lot of problems easier.

[hide]
The 3 possible choices for n are 5, 6, and 7.
5+6+7=18[/hide]

[hide]2^5 = 32
2^7 = 128
so the answer is 18[/hide]

Everyone else got the answer that I got... ^^^^

Even if you don't know the powers of 2 firsthand, this problem is pretty easy since it works with the smaller powers. However, it ~definitely~ is good to know them for other problems.


ARML
HCSSiM
summer program
Mathcamp
geometry
MATHCOUNTS

Hi, it's been kind of quite here. I have been trying to organize a math club locally in the Twin Cities. What activities related to math in Minnesota are you aware of?

Whoa, never knew a MN forum existed.

I guess the next thing coming up for a lot of people (including myself) is ARML. The team was just announced at [url]http://www.macalester.edu/mathleague/2005%20arml%20team.html[/url]. There was talk of a get-together for ARMLites before the first practice, but I'm not sure where thats going.

Thanks for posting the link, Lekan.  I made it!  So happy....

As for the original post, perhaps the reason no one's talking on the forum is that not much is going on?
I know that that applies to me. ;)

your welcome and congrats! yep. There's definitely a lull between the State Tournament and ARML every year. I'm so psyched for ARML.  :D

Danbert suggested reviving this forum,  so here I am posting. What would everyone like to talk about?

I guess I wasn't thinking when I posted that.
This is the part where we realize that there really is very little interesting in MN math between June 3rd and mid-November.  Oh well.

Hey, an idea!  Is anyone else going to some sort of math camp / thing this summer (MITY, HCSSiM, Mathcamp, etc.) ?
(I'm probably going to HCSSiM)

I'm just glad I go to a high school with Math Team, I know of area schools that have no idea what math team is. I did go to a grade school without mathcounts though. apparently that hasn't hurt me though because I was the first sophmore ever from my school to qualify for the AIME.


If AA' , BB',CC' concur then the six points of intersection of the pairs of lines (AB,A'B'), (BC,B'C'),(CA,C'A'),
(A'B,AB'),(B'C,BC') , (C'A,CA') are in groups of 3 on 4 lines.


If $ a,b,c\geq 0$ and $ a\plus{}b\plus{}c\equal{}3$ prove that:
 \[ \frac{1}{2\plus{}a^2b}\plus{}\frac{1}{2\plus{}b^2c}\plus{}\frac{1}{2\plus{}c^2a}\leq \frac{a^3\plus{}b^3\plus{}c^3}{2\plus{}abc}\]

I am pretty sure that $ \frac{1}{9}\cdot (2a^3\plus{}3b^3\plus{}4c^3)\ge \frac{2\plus{}abc}{2\plus{}a^2b}$. I do not have an easy proof for it.


I have a few word problems I am trying to figure out that I don't have a clue how to go about setting up.  I have gotten really good at solving the equations once I have that part down but actually figuring them out is really difficult for me for some reason.  Its not that I don't understand how to translate the English into algebreese, it’s that the problems are rather difficult to setup and don't give you a lot of info to go on.  (And they call this book Algebra: The Easy Way!)  I have the answers so I can solve them that way but I will never learn how to set these up by working backwards from them.  Can anyone offer some suggestions?

Exponents Section:

1. If the interest rate is 5 percent, how many years will it be until the amount of money in your bank account doubles?

Fractions/Rational Numbers Section:

2. The value of your stock declined by 12 percent on Monday, but then on Tuesday the value went back up to the original value.  What was the percentage increase on Tuesday?

3.  If you buy a new car with 20 percent better gas mileage than your old car, what will the percentage reduction in the amount you spend on gas?

4. If the average rainfall for the first 11 months of the year equals 3 inches per month, what must the rainfall be in December in order to make the average for the years equal to 3.5 inches per month?

5.  If your refrigerator currently uses 14 percent of the electricity in your home, by what percentage can you reduce your electricity budget if you buy a new refrigerator that uses 24 percent less electricity than the old one? Assume that non-refrigerator electricity use remains constant.

EDIT: blah, it was suggestions.

#1

$ x\times\left(\frac {105}{100}\right)^n \equal{} 2x$

[hide="answer"]
$ x\times\left(\frac {105}{100}\right)^n \equal{} 2x \implies \left(\frac {105}{100}\right)^n \equal{} 2 \implies n \equal{} \log_\frac {21}{20}{2} \equal{} 14.2067$

answer : 14.2067 years[/hide]

#2

$ x\times\frac {88}{100}\times y \equal{} x$

[hide="answer"]$ x\times\frac {88}{100}\times y \equal{} x \implies \frac {22}{25}\times y \equal{} 1 \implies y \equal{} \frac {25}{22}$

so increased $ \frac {3}{22}\times100 \equal{} 13.6$

answer : 13.6 %[/hide]

#3

$ x\times\frac {6}{5}\times y \equal{} x$

[hide="answer"]$ y\equal{}\frac{5}{6}$, so increased by $ \frac{1}{6}$, which is about 16.7 percent.

answer : 16.7 %[/hide]

#4

$ 11*3\plus{}x\equal{}12*3.5$

[hide="answer"]that implies $ 33\plus{}x\equal{}42 \implies x\equal{}9$

answer : 9[/hide]

#5

$ 14*\frac{76}{100}$

[hide="answer"]answer : 10.64%[/hide]

Geez how are you able to do that? Well here is the answers that the book has.  Not bad, although I still am not sure how to put these into formulas and these answers the book gives are confusing.  Anyone else think these things are pretty hard for a title of "The Easy Way".  I'll study what you did and maybe that will help.  Btw sorry for making you working all of these out.  I could solve them pretty well but trying to get the equation was my problem.  :blush:

[hide="1. Answer"]15 years, since $ 1.050^{15}$ = 2.08[/hide]
[hide="2. Answer"]Let $ V_{1}$ be the original value and $ V_{2}$ be the value after the decline on Monday.  Then $ V_{2}$ = $ V_{1}$ - 0.12$ V_{1}$ = 0.88$ V_{1}$.  Let g represent the percent gain from Monday to Tuesday.
Then $ V_{2}$(1 + $ \frac {g}{100}$) = $ V_{1}$.

\[ 0.88V_{1}(1 \plus{} \frac {g}{100}) \equal{} V_{1}
\]
\[ 1 \plus{} \frac {g}{100} \equal{} \frac {V_{1}}{0.88V_{1}} \equal{} 1.1364
\]

\[ g \equal{} 100 \cdot (1.1364 \minus{} 1) \equal{} 100 \cdot 0.1364 \equal{} 13.64\%
\]
[/hide]
[hide="3. Answer"]Let x be the amount you originally spent on gasoline per year, p be the price of gasoline, d be the distance you drive per year, and m represent the miles-per-gallon value for your old car.  Then x = $ \frac {pd}{m}$.  Let $ x_{2}$ represent the new amount you spend on gasoline: $ x_{2}$ = $ \frac {pd}{(1.2m)}$ (assuming p and d don't change).

\[ \frac {x_{2}}{x} \equal{} \frac {\frac {pd}{(1.2m)}}{\frac {pd}{m}} \equal{} \frac {1}{1.2} \equal{} 0.8333
\]
\[ percentage\ change \equal{} 100 \cdot \frac {x_{2} \minus{} x}{x} \equal{} 100 \cdot (0.8333 \minus{} 1) \equal{} \minus{} 16.67\%
\]
[/hide]
[hide="4. Answer"]Let x equal the December rainfall.

\[ \frac {11 \cdot 3 \plus{} x}{12} \equal{} 3.5
\]
33 + x = 12 $ \cdot$ 3.5 = 42

x = 42 - 33 = 9 inches[/hide]
[hide="5. Answer"]Let x be the current amount spent on refrigerator electricity.  Let B represent your total electricity budget; 0.14B = x.   B = $ \frac {x}{0.14}$ = 7.1429x.
Let N represent non-refrigerator electricity use: N = B - x = $ \frac {x}{0.14}$ - x = x($ \frac {1}{0.14}$ - 1) = 6.1429x.
Let $ B_{2}$ and $ x_{2}$ be the new values for total budget and refrigerator budget, respectively.

\[ B_{2} \equal{} N \plus{} x_{2} \equal{} 6.1429x \plus{} x(1 \minus{} 0.24) \equal{} 6.9029x
\]
\[ \frac {B_{2}}{B} \equal{} \frac {6.9029x}{7.1429x} \equal{} 0.9664
\]
\[ percent\ change \equal{} 100 \cdot (\frac {B_{2} \minus{} B}{B}) \equal{} 100 \cdot (0.9664 \minus{} 1) \equal{} \minus{} 3.36\%
\]
[/hide]

Whew.... after two hours of coding all that I think I can skip preparing for a Bachelor's in computer programming!  :D

whoa why do you even need those formula

once you understand the problem, you dont need to memorize those formulas...

Yeah I am not too sure about this book.  Its written in a way that is like a fantasy novel to make learning this stuff easier but there is a hugh lack of teaching you what you need in order to understand all these word problems they throw at you.  I can work the normal problems just fine and get the majority right (with a few errors on my part for not showing all my work).  Then they have tons of problems like these that are way over my head based on what the chapter taught.

Maybe I should learn this stuff somewhere else instead of this book.  Either way I have till the end of the year to relearn all this stuff because I am applying to go back to college for a BS in Computer Science and they are going to start me with Pre-Calculus and its been 10+ years since I have had Algebra and I didn't really pay attention then.

Any thoughts on the subject??  :huh:

I can see it is definitely not a good book.

Don't read books that claim you can learn maths "the easy way". They are normally terrible, asking you to memorise formulas and such.

Buy AoPS books. They are the best I can find. Or get some MAA contest problem books.

[b]Want more problems? Found a problem you like? Contribute/Solve by clicking on the link below![/b][/b]

Thanks for the suggestion on the book.  I just placed the order for the Intro to Algebra one.  Hopefully that one will be much better than the current one I have.


LaTeX
function
AMC
AIME
linear algebra
matrix
geometry

Let $F(z) = \frac{z + i}{z - i}$ for all complex numbers $z \neq i$, and let $z_n = F(z_{n-1})$ for all positive integers $n$.  Given that $z_0 = \frac{1}{137} + i$ and $z_{2002} = a + bi$, where $a$ and $b$ are real numbers, find $a + b$.

Ouch. I can't do so much arithmetic without making mistakes. Maybe it's easier to show F(F(F(F(z))))=intimidating-complex-fraction-that-simplifies-to-F(z).

(F(z))^2 looks nice but in (F((F(z))^2)^2), the fraction is a bit difficult to combine and factor. It ends up equal to F(z). Too complicated-looking, even LaTeX'd.

I commend your persistance and careful arithmetic, something that I desperately need.

well showing that it repeats itself was the only way I could come up with answers so I tried.

It isnt bad as you think it is.  I was just lucky in terms of that.

I think bictor is right that it's easier to simplify in the general case, f(f(f(z))), then it is in the particular case of z = i + 1/137 or whatever.  Then you only have to apply f to actual numbers once, so less chance for arithmetic mistakes.  Plus, it does tell you something cool about the general behavior of the function rather than specific behavior at a point.  (Not that that's useful for solving the problem so much as it is a cool fact about that particular function.)

Nice solution, Geehoon, to a rather tough AIME question.  And yes, Bictor and Joel are right that showing $F(F(F(z))) = z$ for an arbitrary z would make the calculations a little easier.

By the way, functions such as the one given are called linear fractional transformations (LFTs).  More generally, an LFT is a function of the form $\frac{az + b}{cz + d}$.  They have interesting properties.  For example, if f and g are LFTs, then their composition $f \circ g$ is an LFT.  Similarly, if f is an LFT, then $f^{-1}$ (if it exists) is an LFT.

More surprisingly, there is a correspondence between LFTs and 2-by-2 matrices.  Let $\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$ be a 2-by-2 matrix.  Let's call its LFT the function $\frac{az + b}{cz + d}$.  Suppose that f is the LFT of matrix M, and suppose that g is the LFT of matrix N.  Then $f \circ g$ is the LFT of the product matrix $MN$.  Similarly, if M is invertible, then $f^{-1}$ is the LFT of the inverse matrix $M^{-1}$.  Finally, notice that the LFT of the identity matrix (or any of its nonzero multiples) is the identity function.

In that way, we can translate many questions about LFTs to questions about 2-by-2 matrices.  For example, in our AIME question, our matrix is $\left( \begin{array}{cc} 1 & i \\ 1 & -i \end{array} \right)$.  The square of our matrix is $\left( \begin{array}{cc} 1 +i   & 1+i \\ 1-i & i-1 \end{array} \right)$.  The cube of our matrix is $\left( \begin{array}{cc} 2+2i & 0 \\ 0 & 2+2i \end{array} \right)$, which is a (nonzero) multiple of the identity matrix.  Thus $F \circ F \circ F$ is the identity function.

While this matrix stuff is probably overkill for this particular problem, it can be useful in both theory and practice.  Because we are all used to multiplying 2-by-2 matrices, the calculations come out quite simple.  When I first saw this AIME question, I just computed the square and cube of the matrix, and bingo!


factorial
calculus
integration
algebra
binomial theorem

is there any expansion formula for $(a^m+b^n)^o$ which works for any rational number $m$, $n$, and $o$?

Just use the binomial theorem:

${(x+y)^z=\sum_{k=0}^z {z \choose k} (x^{z-k})(y^k)}$

with $x=a^m$ and $y=b^n$.

Could you simplify that for me and write it out as a formula (I'm not particularly a fan of long expansions and simplification).

That's about as simple as it gets...

${z \choose k}=\frac{z!}{(z-k)!k!}$

[url=http://mathworld.wolfram.com/BinomialTheorem.html]Mathworld- Binomial Theorem[/url]
[url=http://en.wikipedia.org/wiki/Binomial_theorem]Wikipedia- Binomial Theorem[/url]

As cincodemayo5590 said, we need to use binomial expansion...

$(a^m+b^n)^o = (a^m)^o + \binom{o}{1} (a^m)^{o-1}(b^n) + \binom{o}{2} (a^m)^{o-2} (b^n)^2 + ... + \binom{o}{o-1} (a^m) (b^n)^{o-1} + (b^n)^o$

[quote="miyomiyo"]But given that the two numbers are $a^m$ and $b^n$, could you simplify it?[/quote]

Umm, its pretty simple... use what i showed you and exponent rules.

Note that he wanted rational powers, so you have to use the rational binomial theorem, namely:

$(x+y)^n = \sum_{i=0}^{\infty}\binom{n}{i}x^iy^{n-i}$

Where $\binom{n}{k}$ is no longer defined as $\frac{k!}{n!(k-n)!}$ but rather as $\frac{(n)(n-1)(n-2)(n-3)\cdots(n-k+1)}{k!}$ because $n$ may not be an integer and factorials with non-integers is a pain.

Or, in product form:

$\binom{n}{k} = \frac{\prod_{i=0}^{k-1}(n-i)}{\prod_{j=1}^{k}j}$

Note that the denominator could be represented as $k!$

$.333....=\frac{3}{9}$

Rational binomial theorem or substitution of integrals, I believe.

[quote="miyomiyo"]As an example, what is $(a^3+b^3)^.{}^3{}^3{}^3{}^3^.^.^.$?[/quote]

$(a^3+b^3)^{\frac{1}{3}}$

Plugging into our formula:

$(x+y)^n = \sum_{i=0}^{\infty}\binom{n}{i}x{}^iy{}^{n-i}$

$(a^3+a^3)^{\frac{1}{3}} = \sum_{i=0}^{\infty}\binom{\frac13}{i}a^{3i}b^{1-3i}$

So it is:

$\binom{\frac13}{0}b + \binom{\frac13}{1}\frac{a^3}{b^2} + \binom{\frac13}{2}\frac{a^6}{b^5}+\cdots$

Where $\binom{\frac13}{0} = 1$ and $\binom{\frac13}{1} = \frac{\frac13}{1} = \frac13$ and

$\binom{\frac13}{2} = \frac{\frac13\cdot(\frac13-1)}{2\cdot1} = \frac{-\frac{2}{9}}{2} = -\frac19$ and so on.

Note that the rational form works with integer powers too. All the terms with $i>n$ would be equal to zero, which is why it terminates.

So what is a good approximation of the answer?

Well, using the first 3 terms will give you about 1-2 decimal places. Using the first 5 terms is pretty accurate. Using the first 10 terms will probably give you very close to the same as a calculator extracting the roots by its method.

For example: With $a=2$ and $b=3$

$\binom{\frac13}{0}b + \binom{\frac13}{1}\frac{a^3}{b^2} + \binom{\frac13}{2}\frac{a^6}{b^5}+\cdots$

$\sqrt[3]{2^3+3^3}\approx\binom{\frac13}{0}3 + \binom{\frac13}{1}\frac{2^3}{3^2} + \binom{\frac13}{2}\frac{2^6}{3^5}$

$(1)(3) + (\frac13)(\frac89) - (\frac19)(\frac{64}{243})$

$3 + \frac8{27} - \frac{64}{2187}$

$3 + \frac8{3^3} - \frac{64}{3^7}$

$\frac{6561 + 8\cdot3^4 - 64}{2187}$

$\frac{6497 + 8\cdot81}{2187}$

$\frac{6497 + 648}{2187}$

$\frac{7145}{2187} \approx 3.27$

Now to check (with a calculator)

$3.27^3 \approx 34.96$

$2^3+3^3 = 35$

By continuing it with more terms, you get a more accurate answer.

I apologize for any (if any) computational mistakes.


what is the remainder when 100! is divided by 103.  explain your answer.  note: 103 is prime

Since $0\leq 51 \leq 102$, the remainder is $51$. What else¿ :?

thanks for your help i understand that problem now.  would you possibly be able to reply to the other posts that i submitted please i would appreciate it alot

[quote="mathdork"]so the remainder would be 51? or would the 51mod103 be the whole solution[/quote]

Both. Remember that if you have that $c$ is the remainder of the division of $a$ by $b$, then there exists some integer $n$, such that $a = bn + c$, therefore $a-c=bn$, and then $n|a-c$ that it is equivalent (just the last proposition) to $a\cong c \mod b$. Hence, if the remainder is 51, then, the number is congruent with 51 mod 103. The other way is not necessarly true, because, for example, $100! \cong 154 \mod 103$, but obviusly 154 can't be the remainder.