inequalities
logarithms
floor function
function
triangle inequality
inequalities open

I am wondering if anybody has seen this problem:
take $ n>1$ positive numbers $ x_1,\ldots, x_n$.
Then
\[ \sum_i\ln(x_i)\minus{}2\sum_{i<j}\ln\left({x_i\plus{}x_j\over2}\right)\plus{}3\sum_{i<j<k}\ln\left({x_i\plus{}x_j\plus{}x_k\over3}\right)\minus{}\ldots\le0\]

This is equivalent to

$ \frac{ \prod_i x_i \prod_{i<j<k} {(\frac{x_i\plus{}x_j\plus{}x_k}3)^3} ... } {\prod_{i<j} {(\frac{x_i\plus{}x_j}2)^2} ...} \leq 1$

or $ \prod_i x_i \prod_{i<j<k} {(\frac{x_i\plus{}x_j\plus{}x_k}3)^3} ... \leq {\prod_{i<j} {(\frac{x_i\plus{}x_j}2)^2} ...}$

I don't think that it's true: Couting the degree of both sides must give: ($ t \equal{} \lfloor \frac n2 \rfloor$)
$ \sum_{i\equal{}0}^{t} {(2i\plus{}1) \binom{n}{2i\plus{}1} } \equal{} \sum_{i\equal{}1}^{t} {2i \binom{n}{2i}}$
But this equality obviously does not hold.

The correct one might be:
\[ \sum\ln(x_i)\minus{}\sum \ln\left({x_i\plus{}x_j\over2}\right)\plus{}\sum\ln\left({x_i\plus{}x_j\plus{}x_k\over3}\right)\minus{}\ldots\le0\]

[quote="mathangel"]

I don't think that it's true: Couting the degree of both sides must give: ($ t \equal{} \lfloor \frac n2 \rfloor$)
$ \sum_{i \equal{} 0}^{t} {(2i \plus{} 1) \binom{n}{2i \plus{} 1} } \equal{} \sum_{i \equal{} 1}^{t} {2i \binom{n}{2i}}$
But this equality obviously does not hold.

[/quote]

But it does hold:
\[ \sum_k(\minus{}1)^{k\minus{}1}k\binom{n}{k}\equal{}\left({d\over dx}(1\plus{}x)^n\right)\Big|_{x\equal{}\minus{}1}\equal{}0\]
(here I am using that $ n>1$).

In fact, if you take exponents,  the "beatiful inequality" becomes
$ n\equal{}2$: the AM-GM inequality,
$ n\equal{}3$: A nice inequality
\[ \big(\frac {a \plus{} b}{2}\big)^2\big(\frac {b \plus{} c}{2}\big)^2\big(\frac {c \plus{} a}{2}\big)^2\geq abc\big(\frac {a \plus{} b \plus{} c}{3}\big)^3\]
mentioned many times on these Forums.

When $ n\equal{}2$ or $ 3$, one can take any concave function in the "beatiful inequality" instead of the logarithm. Indeed, concave functions can be approximated by linear combinations of linear functions and functions of the form $ \minus{}|x\minus{}\alpha|$, and for the latter "modulus" function, the "beautiful inequality" becomes (with $ y_i\equal{}x_i\minus{}\alpha$)
$ n\equal{}2$: $ |y_1\plus{}y_2|\le |y_1|\plus{}|y_2|$ (triangle inequality)
$ n\equal{}3$: $ |y_1\plus{}y_2|\plus{}|y_1\plus{}y_3|\plus{}|y_2\plus{}y_3|\le |y_1|\plus{}|y_2|\plus{}|y_3|\plus{}|y_1\plus{}y_2\plus{}y_3|$ (Hlawka inequality)
But the problem is that an obvious analogue of Hlawka for $ n\ge4$ (just keep all even sums on the left and all odd sums on the right) does not hold (this was an old conjecture of Freudenthal, easy to disprove). So there is no chance that the "beautiful inequality" holds for arbitrary concave functions. I am stuck with $ n\equal{}4$ case.


Suppose that $ x_n \ge 0$ for n $ \in$ N. Under the interpretation that 1/0 = $ \infty$ and 1/$ \infty$ = 0, prove that
$ \displaystyle\limsup_{x\to\infty}\frac{1}{x_n}=\dfrac{1}{\liminf_{x\to\infty}{\(x_n}}$

More problems can be found here http://anhngq.files.wordpress.com/2007/09/gt1-03.pdf.


[quote="xxxxx"]If f(x) is a linear function such that f(3)=8 and f(4)=10,
find f(14).[/quote]
[hide]It's easy to find the original funtion, $f(x)=2x+2$ plugging in $x=14$, you get $14(2)+2=28+2=30$[/hide]

[quote="jli"][hide]or you could notice that it increases 2 for every 1 unit so we must move 10 units right which gives us 2*10=20 so 20+10=30[/hide][/quote]
But how would you express that as a function?

[quote="Quevvy"][quote="jli"]ummm, thats the point of using my method, so you dont have to spend time looking for the function[/quote]
But if you have f(x), that denotes a function that can be expressed in terms of x.[/quote]

i know but that doesnt mean you need to find the function.  All you need to know is that its linear.

If the function is linear, then can take the form $f(x)=mx+n$, and agree with the data, we've got:

$\begin{equation*}\begin{aligned}3m+n&=8\\ 4m+n&=10 \end{aligned}\end{equation*}$

Solving that sistem, we find that $(m,n)=(2,2)$

$\therefore \ f(x)=2x+2=2(x+1)=2 \cdot 15=30 \ \blacksquare$

[quote="cincodemayo5590"]Rofl, nobody other than [b]jli[/b] is noticing that it isn't necessary to find the explicit formula for the function. His/her solution is a bit less work.

[i]Follow-up question:[/i]
If $G(x)$ is a quadratic in $x$ and has $G(1)=2, \ G(2)=3, \ \text{and }G(3)=7$, find $G(4)$.

[hide="The point of this:"]
Finite differences, not solving a system for the function's coefficients.  :wink: [/hide][/quote]
[hide="Differences"]3-2=1, 7-3=4, x-7=9 x=16[/hide]

That's not how you do finite differences.

[hide="Heres how..."]
Since the function is quadratic, the second differences will be constant.
$[G(4)-G(3)]-[G(3)-G(2)] = [G(3)-G(2)]-[G(2)-G(1)]$
$G(4)-11=3 \Longrightarrow G(4)=14$[/hide]

BTW, does anybody know of a proof that a for an n-degree function, the nth differences will be constant?

[hide](14,y) is on the same line as (3,8) and (4,10), whose equation is y=2x+2, so $y=14\times2+2=30$.[/hide]


I know this is a very broad question, and it has been probably posted in other places. What is the best way to prepare for AIME? I heard AoPS Volume 2 is the most important AoPS book for it. And obviously, old problems are important. Any input? Thanks

Well, also taking the class (which conveniently is filled up, darn) is a good way.

Mostly though, I think its just doing old contests. Especially for me, now that I've been cutting down my careless mistakes (Ive never gotten a #1 correct though) and getting some harder problems correct.

3, 2, 4, 5, 6. I'd say I'm making good progress  :)

AoPS Volume 2 is also helpful I think. There are some theorems and methods in there that I'm sure I would find useful in the AIME's.

Oh, and if you want, do some problems posted here on the forums. Some of them are really good, and especially read the alternate solutions.

any sites with such "methods and theorems"?

Just do lots of problems and you'll learn the "methods and theorems" on your own. If you don't know the theorem for it, you can ask on the forums or just look it up by yourself.

For AIME preparation, please read the pdf available in my website titled "Problem Solving Techniques." This includes which chapters are helpful.

http://hiddendiscoveries.com/mathcompetition.html


Hi guys,

Im sure you all have heard of Hurricane Katrina and would love to help. Just following this link:

<snipped url>

And enter your information. This is not a spam! It is a campaign sponsored by Oxygen. For each person who registers for the campaign, Oxygen will donate $1 to the victims.

Thanks guys.
Help the unfortunate!

Anh Nguyen.

If it's not spam, why don't you use the official address for it? :?
For the campain, please go to http://speakup.oxygen.com/campaigns/neworleans/

Hello,

All over the Internet there are links to help victims of Katrina. Please do!!!

-hwenterprise


superior algebra
superior algebra unsolved

Is Q(square root 2) isomorphic to Q(1 - square root 8)?

Thanks a lot in advance  :)

Yes. $ \mathbb{Q}(1\minus{}\sqrt{8}) \equal{} \mathbb{Q}(\sqrt{8}) \equal{} \mathbb{Q}(2\sqrt{2}) \equal{} \mathbb{Q}(\sqrt{2})$.


function
analytic geometry
geometry
geometric transformation
reflection
floor function

i know how to graph  the regular greatest integer notation f(x)=[x], but how would u graph something like f(x)=2[x], or f(x)=[x]-1, or f(x)=1/2[x], or f(x)=-2[x]-3

(could you tell me how to set up the table used and how to graph it?)

Well, basically, for any $ f(x)$...
$ f(x) \plus{} a$ shifts the graph $ a$ units upward.
$ f(x \plus{} a)$ shifts the graph $ a$ units to the left.
$ af(x)$ multiplies each y-coordinate of the graph by $ a$.
$ f(ax)$ multiplies each x-coordinate of the graph by $ \frac{1}{a}$.
$ \minus{}f(x)$ reflects the graph about the x-axis.
$ f(\minus{}x)$ reflects the graph about the y-axis.

Hopefully, you can use these facts for when $ f(x) \equal{} \lfloor x \rfloor$. If this isn't the answer you were looking for, say so and I or somebody else could explain further. :)

ohh, so ur saying take the generic graph, and move it accordingly to the the function looks like? which btw looks like this http://www.mathwords.com/f/f_assets/floor_graph.gif

if anyone was wondering

Yeah. That's what I was saying. I'm glad you understood.  :lol:


calculus
integration
logarithms
function
calculus computations

show :D

$ \int_0^\infty\;\frac{\ln\,x}{1\plus{}x^a}\;\;\textbf dx\;\;\equal{}\;\;\boxed{\minus{}\left(\frac{\pi}{a}\right)^2\;\cot\left(\frac{\pi}{a}\right)\text{cosec}\left(\frac{\pi}{a}\right)}$

where $ 1<a<\infty$

do [b]not[/b] use techniques from complex analysis.

by substituting $ \frac {1}{1 \plus{} x^{a}} \equal{} z$ it's can be expressed as beta function with first derivetive.


Water leaves the nozzle of hose so that all particles of water move in the direction in which the nozzle is pointing, with speed V.Initially the nozzle is on ground level pointing horizontally.At time t=0 it begins to turn in a vetrtical plane so that the angle  P  which the issuing jet makes with the horizontal is given by SinP =at,where a is a constant, the nozzle remaining at ground level.
Assuming that the ground is a horizontal plane and neglecting air resistance find the speed and direction of motion of the point where the stream of water hits the ground.

IS this question from the 'Guide to Mechanics' by some British authors???

Answer to Rushil
I have never seen the book you mention.The question was used in a school level examination in N.Z. about forty years ago.It probably came from UK in the first place.

Michael.


In geometry class they teach us that triangles inscribed in semicircles are right. However, this seems to be true if and only if the diameter of the circle is a side of the triangle.

What if the diameter of the circle isn't a side of the triangle? The rule seems to no longer apply. The textbooks should be more specific.

Right?

Here, I believe that inscribed triangle means "largest possible triangle," so the book would be correct. The important part is understanding it; the textbooks are never going to be good whatever they do.


Prove that there exists a rational number $\frac{c}{d}$ with $d<100$ such that

$\left\lfloor k\frac{c}{d}\right\rfloor = \left\lfloor k\frac{73}{100}\right\rfloor$.

for for $k=1,2,3...100$.

You're going to want to use \lfloor and \rfloor, or just plain old square brackets.


[quote="Spacecow"]Hi all

This is from the Taiwan 1995 Olympiad, also ACoPS as I mentioned

Consider the operation which transforms the 8 term sequence x_1, x_2, ..., x_8 to
|x_1 - x_2|, |x_2 - x_3|,..., |x_8 - x_7|, |x_1 - x_8|

Find all 8 term sequences of integers such that after finitely many applications of this operation, one is left with a sequence, all of whose terms are equal.

Now I have worked on this problem but I am stuck on the last step. I would postulate that all 8 term sequences are eventually all zero. First note that after the first application of the operation, all terms are positive so WLOG assume all terms are positive in the initial sequence. Then the initial max value of the 8 term sequence is in fact the max value for all 8 term sequences after any finite number of operations. Hence there are only finitely many sequences we can achieve from the operation. Thus we have either eventually all terms are constant ( in fact 0) and the operation does nothing, or we get a repeated value after a finite number of operations that is not a constant, and hence we get a repeating cycle.

Now I am stuck on how to show that the latter is not possible. This question is in the invariant section of the text, so I have been trying to look for a monovariant that must eventually decrease / increase but I have been unable to find a satisfactory solution.[/quote]

[hide="Solution"]Consider the sum of the squares of all the numbers. It is a monovariant if all the x's are positive. Also, it will be strictly decrease unless we have something with 0,a,0,b,0,c,0,d or a,0,b,0,c,0,d,0 which is easily taken care of.[/hide]

Lol, S.O.S. definately not a monovariant.

The actual idea is to consider the numbers mod 2. Now, $ |a\minus{}b|$ is congruent to $ a\plus{}b$

After 8 operations of a+b, you get each term is a multiple of 2. Then you can divide all the numbers by 2, since that wont affect whether it stops or not, and proceed. Since the distance between the highest and lowest term goes down every time we divide by 2, and either stays the same or goes down otherwise, we have either it stops or you have the distance between the highest and lowest number can keep being lowered and divided by 2 indefinately, while staying a positive integer, which cannot happen. Q.E.D.


Jessie has $5.10 worth of stamps. She has equal numbers
of 50-cent, 20-cent, 10-cent, and 5-cent stamps. She has
no other stamps. How many 50-cent stamps does she have?

[hide]Since the number of every kind of stamp is equal, we can set up the equation x(.50+.20+.10+.05)=5.10. Solving we find x=6. There are 6 50-cent stamps.[/hide]

[quote="vishalarul"][hide="It is easy..."]Adding the total values of stamps, we get $50+20+10+5=85$.
$\frac{510}{85}= 6$. She has $6$ of each stamp. The answer is $6$ $50$-cent stamps. [/hide][/quote]

MOEMS questions are supposed to be easy...
I think it is now the easiest forum...

Yes it is easy I find that you could do this with division but...[hide] guess 5..little too small 6..that was easy but who cares? I ..[/hide]actually you start to think that reading other's responses after solving might enlighten us for harder problems....sometimes it is better to go back to the root of a problem.

this is trivial  :wink:
[hide]$\frac{510}{85}=\boxed{6}$[/hide]


$ \lim_{n\to\infty}\frac{1}{\sqrt{\frac 12}\cdot\sqrt{\frac 12\plus{}\frac 12 \sqrt{\frac 12}}\cdot \sqrt{\frac 12\plus{}\frac 12\sqrt{\frac 12\plus{}\frac 12 \sqrt{\frac 12}}}\cdots n\  \text{terms}}$

The denominator approaches $ \frac{2}{\pi}$, so the whole expression approaches $ \frac{\pi}{2}$.  See [url=http://en.wikipedia.org/wiki/Vi%C3%A8te%27s_formula]Viete's formula[/url].


$ x,y$ real number . solve system:

$ x\plus{}\frac{3x\minus{}y}{x^{2}\plus{}y^{2}}\equal{}3$

$ y\minus{}\frac{x\plus{}3y}{x^{2}\plus{}y^{2}}\equal{}0$

$ x^2 \plus{} y^2 \equal{} \frac {3x \minus{} y}{3 \minus{} x} \equal{} \frac {x \plus{} 3y}{y} \equal{} \frac {x(3x \minus{} y) \plus{} y(x \plus{} 3y)}{x(3 \minus{} x) \plus{} y^2} \equal{} \frac {3(x^2 \plus{} y^2)}{3x \minus{} x^2 \plus{} y^2}$
$ \implies 3x \minus{} x^2 \plus{} y^2 \equal{} 3(*)$
$ \frac {3x \minus{} y}{3 \minus{} x} \equal{} \frac {x \plus{} 3y}{y}\implies 3xy \minus{} y^2 \equal{} 3x \plus{} 9y \minus{} x^2 \minus{} 3xy$
$ 6xy \minus{} 9y \equal{} 3x \minus{} x^2 \plus{} y^2 \equal{} 3($ From $ *)\implies 2xy \equal{} 3y \plus{} 1\implies y \equal{} \frac {1}{2x \minus{} 3}$
Putting in $ (*)\implies 4x^4 \minus{} 24x^3 \plus{} 45x^2 \minus{} 27x \plus{} 2 \equal{} 0\implies (4x^2 \minus{} 12x \plus{} 1)(x \minus{} 2)(x \minus{} 1) \equal{} 0$
The solutions: $ \boxed{(2,1),(1, \minus{} 1),(\frac {3 \minus{} 2 \sqrt {2}}{2}, \minus{} \frac {\sqrt {2}}{4}),(\frac {3 \plus{} 2 \sqrt {2}}{2},\frac {\sqrt {2}}{4})}$

P.S.I didn't guess any answer.


Joe wants to final all the four-letter words that begin and end with the same letter. How many combinations of letters must he examine?

We want to find all combinations with the first and fourth letters the same.

There are 26 choices for the first letter, 26 for the second letter, 26 for the third, and 1 choice for the fourth (given the first).

This is $ 26^3 \equal{} 17576$


Let $ p(x)$ be the cubic polynomial $ 7x^3\minus{}4x^2\plus{}K$. Suppose the three roots of $ p(x)$ form an arithmetic progression. Then the value of K is...

a) $ \frac{4}{21}$

b) $ \frac{16}{147}$

c) $ \frac{16}{441}$

d) $ \frac{128}{1323}$

[hide]Let the three roots be $ c\minus{}d, c, c\plus{}d$.

Then by Vieta's: $ 3c\equal{}\displaystyle\frac{4}{7}\Rightarrow c\equal{}\displaystyle\frac{4}{21}$.

Also by Vieta's: $ c(c\minus{}d)\plus{}c(c\plus{}d)\plus{}(c\minus{}d)(c\plus{}d)\equal{}0$
$ c(2c)\plus{}c^2\minus{}d^2\equal{}0$

$ 3c^2\minus{}d^2\equal{}0$

$ d^2\equal{}3c^2$

We know that $ \frac{\minus{}K}{7}\equal{}c(c^2\minus{}d^2)$ by Vieta's.

$ c^2\minus{}d^2\equal{}\minus{}2c^2$ so $ \frac{K}{7}\equal{}2c^3$

$ K\equal{}14c^3\equal{}\displaystyle14\cdot\frac{64}{343\cdot 27}\equal{}\frac{128}{49\cdot 27}\equal{}\frac{128}{1323}$

$ \boxed{D}$[/hide]


graph theory
combinatorics proposed
combinatorics

We have a bipartite graph $G$ (with parts $X$ and $Y$). We orient each edge arbitrarily. [i]Hessam[/i] chooses a vertex at each turn and reverse the orientation of all edges that $v$ is one of their endpoint. Prove that with these steps we can reach to a graph that for each vertex $v$ in part $X$, $\deg^{+}(v)\geq \deg^{-}(v)$ and for each vertex in part $Y$, $\deg^{+}v\leq \deg^{-}v$

Call an edge "bad" if it is oriented from $Y$ to $X$ now if we have a vertex in $X$ s.t $deg^{+}v>deg^{-}v$ (or alternately in $Y$ s.t $deg^{+}v<deg^{-}v$) Hessam can choose that vertex and thus have less "bad" edges in the graph but this procedure must come to an end, and that means we reach the state required in the problem.


Henry flips 10 coins and lays them on the desk. He then chooses one of the coins at random, and if it is tails, he flips it to heads. What is the expected number of heads showing up?

I did not understand a step in the book's [hide="solution"] Let $ x$ be the number of heads before Henry chooses one of the coins, and let $ y$ be the number of heads after Henry has possibly flipped a tail to a head. Then $ y\equal{}x$ if Henry chooses heads, and $ y\equal{}x\plus{}1$ if henry chooses tails.

The book then uses a function $ E(K)$ to define something... maybe some sort of expected value related to $ K$? I skipped some chapters so I'm not familiar with the notation.

I don't if this helps but said that $ E(y)\equal{}0.5*E(x)\plus{}0.5*E(X\plus{}1)\equal{}E(x)\plus{}0.5$
and $ E(x)\equal{}5$ so our answer is $ 11/2$

What does the function E represent and how do they come up with the equation. [/hide]

$ E(K)$ denotes the expected value of $ K$. Thus $ E(y)$ is what the question asks for, the expected number of heads after Henry possible flips over a coin.

The expected value of something is computed by multiplying each of the possibilites by their respective probabilities and then adding them up. There are two possibilities to consider for $ E(y)$: if he picked a heads-up coin (in which case he does not flip it over, and the number of heads remains $ x$), or if he picked a tails-up coin (in which case he flips it over, and the number of heads would become $ x\plus{}1$). These have an equal probability of $ 1/2$. Thus we add the probabilities of each case:
$ E(y)\equal{}0.5\cdot E(x)\plus{}0.5\cdot E(x\plus{}1)$
but since he flips ten coins, $ E(x)\equal{}5$,
$ E(y)\equal{}0.5\cdot5\plus{}0.5\cdot 6 \equal{} 11/2$

Ok.
Forget the $ E(k)$ thingy.
After Henry flips the 10 coins, he will expect to have 5 heads and 5 tails.
Then, he chooses one of the coins at random. It is tails with probability $ \frac{1}{2},$ so the probability that the number of heads goes up by 1 is $ \frac{1}{2},$ and so the expected increase is also $ \frac{1}{2}.$

So now he is expected to have $ 5\plus{}\frac{1}{2}\equal{}\frac{11}{2}$ heads.
Not fully rigorous, but easy to understand (in my opinion).


What is the slope of the line containing the midpoint of the segment with endpoints at (0, 0) and (2, 2) and the midpoint of the segment with endpoints at (5, 0) and (6, 2)? Express your answer in simplest form.

First find the two midpoints by averaging coordinates, then find the slope between them.

Midpoint of (0,0) and (2,2): (1,1)
Midpoint of (5,0) and (6,2): (5.5, 1)

These two points have the same y-coordinate and different x coordinates, so they lie on a horizontal line, which has a slope of $ \boxed{0}$.


I got question.. Um.. It's not for this class question but I hope you can answer this for me... I'm really proud of myself for state but I would be really happy if I made to nats.. It's this Friday... I'm not asking a lot.. What should I do to do my best on state competition.. I just want to know your opinion, instructor..

Silver - I split your question out of the thread it was in because it is its own topic.

As for how to get to Nats, it's too late to do anything notable now.  That's a long road, and you can't wait until the last minute.

Primarily the only thing you can really do anything about now is game-day (and day before) strategy.  Don't do too much math the day before the contest.  Maybe do a few simple things (even a card game might do) to warm up your brain right before the contest.

Try to get plenty of sleep the night before.

The difference between 3rd and 7th place is almost always careless errors.  Throughout the contest, you must focus on reading the question carefully, rereading the question after you get your answer, and making sure you answer the question that is asked in the form it is asked.

Above all, try to enjoy it.

yes, read very very carefully and try not to get down if you know you missed a target question or something.

Just remember that there is always hope in the countdown round (if it is official).  I knew I had done terrible during the written round, so I was going to go in with the strategery of "give 'em a goode show".

Chinaboy, are you hinting something?

Oh.... Man! I took the test today and read this good messages from instructor today..Oh, well.. I didn't make on state neither so it doesn't matter...Um.. One more question (not about Mathcounts), what does test AIME is? and how hard is it?


real analysis
Euler
number theory solved
number theory

number one
if y is even then x^2 = 7 (mod 8)
not possible
y^3+8=x^2+1

(y+2)(y^2-2y+4)=x^2+1
we know that x^2+1 cannot have any factors of the form 4k+3
so suppose y+2=4(k+1)+1
so y=4k+3
but y^2-2y+4 = 3 (mod 4)
therefore no integers satisfy the equation

The following fact:

A number of the form x2 + 1 cannot have prime factors of the form 4k + 3

seems to be quite useful. I saw it already a few times in the solutions of olympiad - style problems. The proof is as follows. Suppose the opposite is true, i.e. suppose there exists a prime p = -1 (mod 4) such that x2 = -1 (mod p). Then we have (x2)(p - 1)/2 = (-1)^(p - 1)/2 (mod p) or xp - 1 = -1 (mod p) (since (p - 1)/2 is odd) but this is a contradiction since xp - 1 = 1 (mod p) by Fermat's little theorem.

Well, here's how you use it for the second prb.:
(4x-1)y=z^2+x iff (4x-1)4y=(2z)^2+4x iff (4x-1)(4y-1)=(2z)^2+1. Because 4x-1 is a M4-1 (this means multiple of 4 -1) it must have a prime divisor of the form 4k-1 which divides (2z)^2+1, which is impossible. However, if x=-a and y=-b with a and b natural then it's obvious that the eqn has an infinity of solns. All we have to do is get infinitely many z such that 5/(2z)^2+1 and we take a=1, and then 4b+1=((2z)^2+1)/5 and we find b, then we put x=-a and y=-b and we found infinitely many solns in Z^3.


trigonometry
geometry
trig identities
Law of Sines

If x + 5 is a tower, find the height of the tower.

[img]http://img9.imageshack.us/img9/208/provacl.jpg[/img]

The horizontal line we can call y. So $ \tan{30}\equal{}\frac{5}{y}$ and $ y\equal{}\frac{5}{\tan{30}}\equal{}5\sqrt{3}$
$ \tan{60}\equal{}\frac{x}{5\sqrt{3}}$ so $ x\equal{}15$ and the height of the tower is $ 15\plus{}5\equal{}\boxed{20cm}$

Oh, I forgot to mention, this would be assuming that the horizontal line and vertical line are perpendicular to each other. It would otherwise be impossible, I think, without assuming so.

I'm not sure how you would use Law of Sines in this case. Also, based on the aforementioned assumption that it's a right triangle, trigonometry is much easier that the Law of Sines.

I tried this:

$ \frac{x}{\sin60}\equal{}\frac{5}{\sin30}$

$ \frac{x}{\frac{\sqrt{3}}{2}}\equal{}\frac{5}{\frac{1}{2}}$

$ x\equal{}5\sqrt{3}$

Using sines law, but I think I'm wrong.  :(

This is because the rule of signs relates lengths and sines of angles in one triangle, whereas you have taken each side of your equation from a different triangle.

[quote="luiseduardo"]I tried this:

$ \frac {x}{\sin60} \equal{} \frac {5}{\sin30}$

$ \frac {x}{\frac {\sqrt {3}}{2}} \equal{} \frac {5}{\frac {1}{2}}$

$ x \equal{} 5\sqrt {3}$

Using sines law, but I think I'm wrong.  :([/quote]
y my dear its wrong
here are asimple solution: and there are another simple solution with simillarity too
[url=http://www.servimg.com/image_preview.php?i=316&u=12899977][img]http://i89.servimg.com/u/f89/12/89/99/77/almonk10.gif[/img][/url]
mr mohamed
 :)

Using almonkez notation, $ AD\equal{}2CD$ (30-60-90 triangle), and do the same for $ \triangle ABD$ and get $ BD\equal{}2AD\equal{}4CD\equal{}20$.

Oh, men ... @#%@#%@%  :oops:
I'm wrong  :(
I need to know that's sines law is IN the same triangle.
Thanks for the help.


After I got some new lessons, I did a few problems, then the video lessons suddenly disappeared!!! Is this a glitch? Or a bug?

You likely lost "ready" status for some subjects by getting problems wrong.


algebra
polynomial
function
superior algebra
superior algebra unsolved

Hello,

I'm reading Abel's proof of the solution of polynomials by radiacals. Here the treatment is in $ n$ variables, and one shows that polynomials of degree 5 or higher are in general not solvable. I don't quite understand the proof as yet, but it seems wierd that one can have a polynomial in say $ 4$ variables $ x,y,z,w$, something like $ x^4 \plus{} 4y^4 \plus{} z^4 \plus{} 3w^3 \plus{} x^2 \plus{} y^3 \plus{} z^3 \plus{} w \plus{} 1$ and still a closed form solution is guaranteed to exist, by virtue of it being under degree 5? Is this corrent?

Thanks in advance,

Los

The solution is parametric. You can treat all variables as constant and then solve for $ x$ for example. You can do this because we can solve quartics like in http://en.wikipedia.org/wiki/Quartic_function


If $ q$ is an integer that can be expressed as the sum of two integer squares, show that both $ 2q$ and $ 5q$ can also be expressed as the sum of two integer squares.

Well, actually if $ q$ and $ r$ are integers that can be expressed as the sum of two perfect squares, the number $ qr$ can be expressed in that form as well.
This follows from the identity $ (a^2\plus{}b^2)(c^2\plus{}d^2) \equal{} (ac\plus{}bd)^2 \plus{} (ad\minus{}bc)^2$.

In your case, if $ q\equal{}a^2\plus{}b^2$, we have
$ 2q\equal{}(a\plus{}b)^2\plus{}(a\minus{}b)^2$, and
$ 5q\equal{}(a\plus{}2b)^2\plus{}(2a\minus{}b)^2$.


AMC AIME USA(J)MO USAMO USAMTS email Support

Richard Rusczyk said that he had fielded some questions about MOSP 2004 and asked me to post some information:

The selection criteria for MOSP are set (the CAMC approved a
recommendation from me at Phoenix on January 10.)  The
Teacher's Manual for the AIME and USAMO with all the directions, instructions and rules and it will go to print by the end of the week.  Teachers won't have it in hand until about the end of Feb, but no harm in announcing the rules now.

From the USAMO Teacher's Manual, page 4, Section VI:
"The top 12 scoring students on the USAMO will be invited to attend the Mathematical Olympiad Summer Program regardless of grade.  In addition to the top 12 USAMO winners, an additional 12-18 high-scoring USAMO participants not graduating form high school, who are current juniors and below, will also be invited to participate in the Mathematical Olympiad Summer Program.  Thanks to a grant from the Akamai Foundation, we will invite an additional (approximately) 30 ninth-grade USAMO participants for a
MOSP program aimed at preparing them for Olympiad style competitions.

The Mathematical Olympiad Summer Program will be held on the campus of the University of Nebraska-Lincoln June 13-July 3, 2004.  Students will be treated to classes on advanced mathematical topics, special guest speakers, mathematical contest training, and related social and mathematical activities."

Is it 9th grade and below or exclusivly 9th graders?

I would think it's 9th grade or below, since 7th and 8th graders have even more potential because they have more years ahead of them.  But you'll have to wait for Mr. Dunbar to reply before you have an official answer...

[quote="Fierytycoon"]So it has been decided that MOP will not be expanding to 60 people this year?[/quote]

"The top 12 scoring students on the USAMO will be invited to attend the Mathematical Olympiad Summer Program regardless of grade. In addition to the top 12 USAMO winners, an additional 12-18 high-scoring USAMO participants not graduating form high school, who are current juniors and below, will also be invited to participate in the Mathematical Olympiad Summer Program. Thanks to a grant from the Akamai Foundation, we will invite an additional (approximately) 30 ninth-grade USAMO participants for a
MOSP program aimed at preparing them for Olympiad style competitions."

Um, 12 plus 18 plus 30 equals 60 where I come from. If it ends up being a few less, that's still 60 to a reasonable degree of approximation.

blah, I misread, was in a hurry, sorry about that guys, and thanks for the catch tokenadult.

Does this mean that people who are currently in 9th grade, and when they qualify for USAMO, they automatically get into MOSP? I mean, the precentage of 9th graders out of all USAMO participants is pretty low....

You do not automatically get into MOP, usually there are more 9th graders alone taking USAMO than the number of students invited, 30.

Actually, last year there were 33 9th graders and 12 8th graders or below.  I would expect that since they're choosing from among 9th graders and below (although we'll have to wait for Mr. Dunbar to reply to be sure), there will be some 9th graders or below who won't make it, but those people will be few.

Bummer, too bad they're inviting just extra 9th graders (for the pool of extra 30 students) and not simply the top 60 non-seniors.

[quote="tetrahedr0n"]it is quite sad, isn't it? (sophomore speaking)[/quote]

Indeed (junior speaking). Now we need to get 3 questions completely right, perhaps even a couple points more.

On the other hand, I (freshman speaking) only need to make it to USAMO and I'll have about 75% chance of getting into MOSP..... I know about 7 people in my region who are capable of doing that..... However, they probably wouldn't make it in their first year in HS if MOSP didn't expand.... (I won't make it anyway because I'm not a US citizen :-x )

No, it's excellent news.  (Freshman speaking).

Fiery, you'll only have to do as well as you would have had to last year, which was probably 2 questions right.  At least you're not a senior.  *nod matt*

[quote="mathfanatic"]Fiery, you'll only have to do as well as you would have had to last year, which was probably 2 questions right.  At least you're not a senior.  *nod matt*[/quote]

Naw, more like 3+ questions. There were 27 people who scored 22 or above last year. 2 questions would have been what I needed had the MOP expansion be just for the top 60 USAMO scorers.

:(:(:(:(:(:(:(:(:(:( i wonder if i could pretend i'm a 10th grader. much fun.

In answer to the questions posted above, the intent is to select approximately 30 ninth-graders to invite to a section, or division, of the MOSP which is directed at ninth-grade students capable of reaching the USAMO, and helping them to develop their abilities further.

The last two years, there have been about 30 ninth-graders qualifying for USAMO each year.  One of the previous posts noted the precise figures from last year.  I anticipate the numbers will be similar again this year.  By inviting a group of this size which is also relatively uniform in age and mathematical background we will be better able to create a curriculum for that group.

Steve Dunbar
AMC Director

[quote="MysticTerminator"]ack. will 9th graders be able to qualify into the non-9th grader division of MOP?[/quote]

[quote="AMCDirector"]From the USAMO Teacher's Manual, page 4, Section VI:
"The top 12 scoring students on the USAMO will be invited to attend the Mathematical Olympiad Summer Program regardless of grade.  In addition to the top 12 USAMO winners, an additional 12-18 high-scoring USAMO participants not graduating form high school, who are current juniors and below, will also be invited to participate in the Mathematical Olympiad Summer Program. "[/quote]

Looks like yes to me.

what is the selection criteria for the 12-18 non winners/non seniors in the main section of mop? is it based on usamo score only?

No, other tests are factored in.  I believe it's by the classic MOP Index: AMC + 10 * AIME + 9 * USAMO.

[quote="JBL"]Always USAMO score only.[/quote]

[quote="churchilljrhigh"]then how will you explain for the people who don't take the AMC (USAMTS).[/quote]

USAMTS counts as 100 for your AMC score.

[quote="AMCDirector"]The last two years, there have been about 30 ninth-graders qualifying for USAMO each year.  One of the previous posts noted the precise figures from last year.  I anticipate the numbers will be similar again this year.  By inviting a group of this size which is also relatively uniform in age and mathematical background we will be better able to create a curriculum for that group.[/quote]

When you say that 30 ninth-graders will be invited to MOP, does that mean exactly 30, or a few more or less?  (Say 32 ninth-graders make USAMO - will all of them be invited to MOP, or only the top 30?  What if significantly more than 30, say 45, ninth-graders make USAMO?)

[quote="mathfanatic"]
When you say that 30 ninth-graders will be invited to MOP, does that mean exactly 30, or a few more or less?  (Say 32 ninth-graders make USAMO - will all of them be invited to MOP, or only the top 30?  What if significantly more than 30, say 45, ninth-graders make USAMO?)[/quote]

Same with the main MOP group. How do you decide how many people other than the top 12 will be invited (the 12-18 range you gave)?

[quote="AMCDirector"]In answer to the questions posted above, the intent is to select approximately 30 ninth-graders to invite to a section, or division, of the MOSP which is directed at ninth-grade students capable of reaching the USAMO, and helping them to develop their abilities further.

Steve Dunbar
AMC Director[/quote]

Does this mean 30 students entering the ninth grade or those who were in 9th grade when they tested for qualifying?  They would officially be 10th graders at the time of MOSP, right?

MOP is over the summer. Hence the name. You're not in any grade over the summer. You can spend all your time doing maths.

[quote="cobra"]
Does this mean 30 students entering the ninth grade or those who were in 9th grade when they tested for qualifying?  They would officially be 10th graders at the time of MOSP, right?[/quote]

I'm sure he means "in 9th grade" when they tested to qualify (hence they'd be rising 10th graders during the Summer Program). I would actually bet (though it wasn't stated explicitly) that he really means "9th grade or below" for the additional 30 spots.

Very strange. That doesn't make much sense to me, but maybe they have their reasons.

He explained his reasoning here:

[quote="AMCDirector"]By inviting a group of this size which is also relatively uniform in age and mathematical background we will be better able to create a curriculum for that group.[/quote]

For the past four years (with the exception of '02, of course) there has always been exactly 30 MOPpers.  Perhaps the 12-18 is to give some slack in case funding runs short... I don't know.

[quote="Fierytycoon"]
Same with the main MOP group. How do you decide how many people other than the top 12 will be invited (the 12-18 range you gave)?[/quote]

[quote="ml2"]he doesn't, i asked him and he stated it is exactly 30 9th graders in 2003-2004 school year.[/quote]

ml2, are you saying you sent an email to Mr. Dunbar? The quoted text from him in this thread says, "we will invite an additional (approximately) 30 ninth-grade USAMO participants for a MOSP program aimed at preparing them for Olympiad style competitions." There, he says "approximately" as to the number of students, and the unwritten implication regarding age sure seemed to be that young people WITH THE REQUISITE SCORES who happened to one grade younger (or even two grades younger?) than ninth grade might have a chance to come on board. The main point is to have a special program for a select group of YOUNG math contestants.

Occasionally, there is a superstar who scores the old-fashioned way into MOP when still very young, but there is evidently now an attempt to find a farm team for future IMO contests early, and give that group of 30 or so an early summer program that they can digest while of high school age. Maybe the United States team is picking up some talent development ideas from China.

Good luck to everyone trying to get on board MOP this year, through whichever of the three channels seems suitable.

Tokenadult, I think the "approximately" referred to the number of ninth-graders, not the grade of the participants.  As far as I understood, the "thirty young students" will consist of thirty ninth-graders [i]only[/i] - not eight-graders, not seventh-graders, etc.

In my opinion, the unwritten implication is not that younger (i.e. below 9th grade) students can qualify; rather, it is that some younger students will miss out on making MOP through this special program because they are younger.  The AMC committee decided that the benefits of having a relatively uniform group (in terms of age, maturity, mathematical exposure, etc.) outweight the drawback of excluding one or two students, as Mr. Dunbar explained when he said

[quote]By inviting a group of this size which is also relatively uniform in age and mathematical background we will be better able to create a curriculum for that group.
[/quote]

And while it may seem unfair, remember that these younger students will get their chance to qualify for this program when they are 9th graders.

[quote="ml2"]Umm, i PM'ed him and he  said that you must be a 9th-grader during 2003-2004, this is what the exact was atteached too, and it is aprox. 30 people. I quote this "3. Be a ninth-grader in the 2003-2004 academic year who qualified for, and took the USAMO. If there are more than 30 such students, select the 30 with the highest USAMO scores from this group. "[/quote]

Okay, so the "ninth grade" (at time of taking the test) is exact, and the "thirty students" is approximate. That's clear enough. And, yes, as said above in the thread, an eighth grader always has a second chance when he reaches ninth grade age, and always has the theoretical, lightning-strike chance of testing into one of the higher groups. That sounds like a good overall plan to build for the future.

Again, best wishes to anyone trying to get in through one channel or another.

yes, but what about older people? argh.... if i was a year older maybe i could have made mop 2002, and if i was a year younger maybe i'd be in this freshman group.... class of 06 is just bad.

I will try to address some of the questions that are being raised here.

Naturally, we will invite the twelve top-scoring students on the USAMO, regardless of grade.  I think that much is clear.

After those twelve, we will invite 12 to 18 of the next highest scoring students who are not graduating seniors.  We will invite hese students regardless of grade, based on their "USAMO index" as mentioned in one of the posts.  As an example, a student who was just done with eighth-grade (a "rising ninth-grader") was invited to MOSP 2003.  We also invited some students who were ninth-graders in 2002-2003 to MOSP 2003.   And so on.  We had a mix of grade levels last year.  This portion of MOSP 2004 remains the same as it has been for several years.  I trust that this much also is clear, it has not changed signifcantly in several years.

As Alison mentioned, in 2001 and 2003 there were 30 at MOSP.  In the early 90's the numbers at MOSP were closer to 24, but over the 90's the number slowly crept up to 30.  That was possible by stretching the MOSP budget and "borrowing" from other AMC activities.  It is not clear how much longer that budget stretching can continue, so I have to reserve some flexibility.  Naturally, I will continue to try to have many students as possible, but I also have meet a budget.

After inviting the 24 to 30 as we have done for several years, with the new special program support from the Akamai Foundation, we will invite approximately 30 USAMO participants who were ninth-graders in the 2003-2004 academic year (so-called "rising tenth-graders"). The reasons for focusing on ninth-graders were stated above: a desire to have a reasonably uniform group in terms of age, background and experience.  The reasons for "approximately 30" are again budgetary and additional reasons of available dorm space.  I need the flexibility to invite 28,29,30,31, or 32 ninth-graders depending on factors that I will only know in May, including last minute details of cost.

I hope this answers your questions.

Steve Dunbar
AMC Director


integration
calculus
calculus computations

Given $ f: [0,1] \to R$ continuous and two sequences $ (a_n),(b_n)$ so that $ \displaystyle\int^1_0 |f(x) \minus{} x.a_n \minus{} b_n|\,dx \equal{} 0$ and $ lim a_n\equal{}a, lim b_n\equal{}b$
Define $ f(x)$

Since the convergence under the integral sign is uniform (and the interval of integration is finite), we can pass to the limit: $ \int_0^1 |f(x)\minus{}ax\minus{}b|\,dx\equal{}0$. This and the continuity of $ f$ imply that $ f(x)\equal{}ax\plus{}b$.

Of course, the original poster left some things out- as stated, we must have $ a_n\equal{}a,b_n\equal{}b$ for all $ n$ as well.


function blogs geometry search probability inequalities sussyamongus

How many girlfriends have you had in the last 4 years? Don't count girls from elementary school. A girl is considered a girlfriend if you have dated her for at least 3 days.

Are you counting girlfriends in 3rd and 4th grade??  I am just finishing 7th grade, and I know that there are many people who will still be able to include the "girlfriends" they had in elementary school.  Could you please be more specific on what makes someone a "girlfriend"?  Thanks.

He says IN THE LAST FOUR YEARS. SO IF SINCE IF YOU ARE 7TH GRADE, THEN YOU INCLUDE 7TH, 6TH, 5TH, AND 4TH GRADE.

Well, a lot of people do not consider people you "go out with" in the elementary school to be true girlfriends/boyfriends.  This is what I mean; what does the poster of this topic (drunner2007) consider to be a girlfriend?  Going out for less than a week is not a girlfriend, in my opinion.

I'm going to edit it. Girlfriends since the beginning of 6th grade. Elementary school barely counts. Also, to count a girl as an actual girlfriend in this poll you have to have dated her for at least 3 days.

Thank you.  That is much clearer on what you mean.  I have had a whole whopping
[hide]0[/hide]  What about you Drunner2007?

Well, I have been writing letters to my friend for the past year and we talk to each other often through IM. We have been playing together for many years BUT I never thought of her as 'girlfriend'. I like her but not what some people think of 'love relationships'. Is she considered to be on my GF list? If not, my answer is 0.

0 for me too...there have been girls that, we kind of like each other but never anything "official"

how to unterstand if someone is a mathematician: ask him how many girlfriends he had and if he asks you how you define a girlfriend he is  :)
i suppose it is a random value of the function with the following ability: f'=f and there is no e in her type

What do you mean by 3 days? You wouldn't normally go out with a girlfriend every night, (or at least in my limited middle school experience) so I'm guessing you mean something more along the lines of 3 dates? As there's usually always more than a 3 day gap between dates.

[quote="rem"][quote="MysticTerminator"][hide]174[/hide][/quote]
Nice... Mystic Terminator, a girlfriend is a human :D

btw, for me, it's 0

I will calculate the average number of girlfriends that aopsers had over the past 4 years. So far, we have:
[b][size=134]0[/size][/b]
with:
4 ppl with 0 girlfriends.

I did not include mystic terminator's answer because it's unrealistic. :!:[/quote]

you discriminatory person you

[quote="kahsi"]Now, do we include complex girlfriends?[/quote]
Yes, can we include nonreal ones?

 :D

Dude this topic is dying.

DON'T LET IT DIE!

[quote="MysticTerminator"][quote="wobster109"][quote="Peter VDD"]Mystic, you're also 11... That's the AM of your [size=0]non-mathematical [/size]mental and official age. :mad:[/quote]

so true, MT.  you had it coming. :roll:[/quote]

wtf is wrong with you

can you also not do math

this disappoints me[/quote]

Yeah, reallly, the AM of 18 and 2 is 10, not 11. *Sigh*
I understand your disappointment, Mystic.

[quote="Klebian"][/quote][quote="MysticTerminator"][quote="wobster109"][quote="Peter VDD"]Mystic, you're also 11... That's the AM of your [size=0]non-mathematical [/size]mental and official age. :mad:[/quote]

so true, MT.  you had it coming. :roll:[/quote]

wtf is wrong with you

can you also not do math

this disappoints me[/quote][quote="Klebian"]

Yeah, reallly, the AM of 18 and 2 is 10, not 11. *Sigh*
I understand your disappointment, Mystic.[/quote]

dude you got both the ages wrong. like I could see your inability to comprehend my incredibly mature genius and thus underestimating by mental age but you have to be pretty retarded to miss my physical age, wtf

hmm, so MT is officially 17, i'd guess?  i think he turns 18 in november?  so if the AM of his two ages is 11, that does seem a bit high.  MT is mentally 3, i'd say.  so klebian is right. :P

[quote="Tbyrd1992"]all my friends dates have been equations[/quote]

don't be like me a long time ago and bring back old topics. I have a feeling you searched it up

The solution to $ x^{2}\equal{}0$ or the number of real solutions to $ x^{2}\plus{}1\equal{}0$, oh wait I mean $ 0$.

$ \left(\frac{\int_6^7 \arctan x^2\,dx\plus{}\log_\pi 2007}{\cot^3\frac{\ln 2}{\sqrt[e]{7i}}\minus{}\sin^{1337}6}\right)^0\minus{}1\equal{}\boxed{0}$

I came close a couple of times, but it didn't work out in the end.


Let ABC be an acute triangle.
Let U and T the intersection of the (tangent of A and the tangent of C) and (tangent of B and the tangent of C). Hereby, "tangent" means "tangent to the circumcircle of triangle ABC". Let P the intersection of AT and BC, and Q the midpoint of AP. Let R the intersection of BU and AC, and S the midpoint of BR. Prove that $\angle{BAS}$= $\angle{ABQ}$

If I am not mictaken, it is enough to prove that the intersection point $M,$ of the segment lines$AS,$ $BQ,$ lies the segment line $OY,$ where $Y,$ is the intersection point of the tangent lines of the circumcircle $(O)$ of $\bigtriangleup ABC,$ at vertices $A,$ $B.$

kostas Vittas.

Let $ABC$ be an acute triangle and $w$ your circuncircle.
Let $U$ the intersection of the tangents of $w$ on the points $A$ and $C$ and let $T$ the intersection of the tangents of $w$ on the points $B$ and $C$.
Let $P$ the intersection of $AT$ and $BC$. And let $R$ the intersection of $BU$ and $AC$. Let $Q$ the midpoint of the segment $AP$. And let $S$ the midpoint of the segment $BR$. Prove that $angle{ABQ}$ $=$ $angle{BAS}$
 :huh:

This is exactly this problem:
http://www.mathlinks.ro/Forum/viewtopic.php?t=19806

  darij


function
trigonometry
calculus
derivative
calculus computations

Find the real value function that is equal to

$\frac{x^{3}}{3!}+\frac{x^{9}}{9!}+\frac{x^{15}}{15!}+\dots$

[hide="my solution"]
i use the laplace transform (not sure if i was justified in doing so) and got

$\frac{e^{x}}{6}-\frac{e^{-x}}{6}+e^{\frac{x}{2}}\left(\sin\left(\frac{x\sqrt 3}{2}\right)-\cos\left(\frac{x\sqrt 3}{2}\right)\right)-e^{\frac{-x}{2}}\left(\sin\left(\frac{x\sqrt 3}{2}\right)-\cos\left(\frac{x\sqrt 3}{2}\right)\right)$

or something similar to that
[/hide]

The third derivative is $\sum_{k=0}^{\infty}\frac{x^{6k}}{(6k)!}$. To obtain this series from $\exp(x)$, use the standard trick to kill the powers not divisible by $6$:
\[\frac16\sum_{j=0}^{5}\exp(e^{2\pi ij/6}x) =\sum_{k=0}^{\infty}\frac{x^{6k}}{(6k)!}\]
Integrate three times to find
\[f(x)=\frac16\sum_{j=0}^{5}(-1)^{j}\exp(e^{2\pi ij/6}x)\]

Or there's the differential equations approach: this function $y(x)$ the solution to this sixth degree initial value problem:

$y^{[6]}-y=0;y(0)=y'(0)=y''(0)=0,y'''(0)=1,y^{[4]}(0)= y^{[5]}(0)=0.$

That certainly gives rise to the same ingredients as in the previous two posts. I'm not quite sure what maokid7 was using the Laplace transform on, but you could use that transform on this differential equation. I don't feel like checking the details at the moment.


Prove that if the polynomial $ P(x)$ with integer coeficients admits value $ 1$ for more than $ 3$ distinct integer values of $ x$, then $ P(x)$ cannot admit the value $ \minus{}1$ for integer values of $ x$ .

if $ a < b < c < d$ are four distinct values at which $ P(x)$ takes the value 1, and $ e$ is a value at which $ P(x)$ takes the value -1, then $ |e \minus{} a|, |e \minus{} b|, |e \minus{} c|, |e \minus{} d|$ must all be equal to 1 or 2. the only way this can happen is if $ a < b < e < c < d$ are 5 consecutive integers. we may assume wlog that these 5 integers are $ 0,1,2,3,4$. then $ P(x) \equal{} Q(x)x(x \minus{} 1)(x \minus{} 3)(x \minus{} 4) \plus{} 1$, where $ Q\in \mathbb{Z}[x]$. plugging in 2, we get $ \minus{} 1 \equal{} 4Q( 2) \plus{} 1$, or $ \minus{} 1 \equal{} 2Q(2)$, which is impossible.


Fie ABC un triunghi ascutitunghic si O centrul cercului circumscris.Fie o drepata care trece prin O si intersecteaza AB in M si AC in N.Fie mijloacele lui BN si CM , R respectiv S.Demonstrati ca $\angle ROS=\angle BAC$

da... intr-adevar problema e draguta ... A fost de mai multe ori postata (chiar recent) la "geometry unsolved problems". Rezolvarea pe care am inteles-o eu ( :oops: ) foloseste teorema lui pascal :wink:


USAMTS algebra polynomial inequalities induction function calculus

I'll put the three scores for the first day I have in few hours but I'll post the solution first.

[hide="Week 1"]
$\begin{tabular}{|c|c|c|c|c|c|c|c|c|} \hline
\multicolumn{9}{|c|}{USAMTS Practices Scoreboard Week 1}\\ \hline
& Day 1 & Day 2 & Day 3 & Day 4 & Day 5 & Day 6 & Day 7 & Total \\ \hline
zanttrang & - & - & 5 & 5 & 5 & 5 & 5 & 25\\ \hline
1234567890 & - & - & 5 & - & 5 & - & - & 10\\ \hline
LynnelleYe & 1 & - & 4 & - & 4 & - & - & 9\\ \hline
no one 2000 & - & - & - & 5 & - & - & - & 5\\ \hline
chess64 & - & -1 & 5 & - & 1 & - & & 5\\ \hline
Scrambled & - & - & - & - & - & - & 3 & 3\\ \hline
agolsme & 1 & 1 & - & - & - & - & - & 2\\ \hline
tetrahedr0n & 1 & - & - & - & - & - & - & 1\\ \hline
\end{tabular}$ [/hide]

$\begin{tabular}{|c|c|c|c|c|c|c|c|c|} \hline
\multicolumn{9}{|c|}{USAMTS Practices Scoreboard Week 2}\\ \hline
& Day 8 & Day 9 & Day 10 & Day 11 & Day 12 & Day 13 & Day 14 & Total \\ \hline
zanttrang & 5 & 1 & 5 & 2 & 5 & & & 18\\ \hline
JesusFreak197 & - & - & - & - & 5 & & &5\\ \hline
chess64 & 5 & 0 & - & - & - & & & 5\\ \hline
agolsme & - & 0 & - & - & - & & & 0\\ \hline
LynnelleYe & - & 0 & - & - & - & & & 0\\ \hline
\end{tabular}$

Scores..

[b]Day 1 Solution[/b]

All 3 people's answer were basically based on same idea.

[quote="tetrahedr0n"]
Solution:

Let the number of pirates be $x$.  Each pirate handles $11$ treasure chests, so the total number of "handles" is $11x$. Each treasure chest was handled by $7$ pirates, so there are $11x/7$ treasure chests. The number of chests must be an integer, and since $7$ does not divide $11$, $7$ must divide $x$. But we are given $60<x<69$, and in this interval there is only one multiple of $7$, $63$. So we have that $x=63$ and the number of chests is $(63)(11)/(7)=99$.[/quote]

If this was real USAMTS question, I think they probably got the all the credit because this is how the contest answer looked like.

I did write my answer but I think it's really poor.. I'll still put it here to see people's opinion though. Critics will be great as praise to me because I rather get critics when I'm doing a practice than getting them after I sent the answer for real thing.

[hide="My answer"]
[quote="I"]

Let $P$ be the number of the pirates.
Let $T$ be the number of treasure chests.

By given information, I know $60 \leq P \leq 69$. Because I do not know neither $P$ or $T$, I decided to start off with the most simple case: 7 pirates and 11 treasure chests. Each pirate will handle all 11 treasure chests as shown below:

\[P_1: T_1,T_2,T_3,T_4,T_5,...T_{11}\\
P_2: T_1,T_2,...T_{11}\\
\cdot\\
\cdot\\
\cdot\\
P_7: T_1,T_2,...T_{11}\]

Now, if I consider 7 pirates and 11 treasure chests as one unit, the two units will have 14 pirates and 22 treasure chests and will satisfy the condition on the problem because 14 pirates and 22 treasure chests can be broken down into one unit, which already works as shown above. Using this idea, we can have 9 units, or 63 pirates and 99 treasure chests. If they're broken down as units, this is what we'll have:

\[P_1^1: T_1^1,T_2^1,T_3^1,T_4^1,T_5^1,...T_{11}^1\\
P_2^1: T_1^1,T_2^1,...T_{11}^1\\
\cdot\\
\cdot\\
\cdot\\
P_7^1: T_1^1,T_2^1,...T_{11}^1\]

\[P_1^2: T_1^2,T_2^2,T_3^2,...T_{11}^2\\
\cdot\\
\cdot\\
\cdot\\
P_7^2:T_1^2,T_2^2,T_3^2,...T_{11}^2\\
\cdot\\
\cdot\\
\cdot\]

Using this, the last one in our unit would be:
\[P_7^9: T_1^9,T_2^9,T_3^9,...T_{11}^9\]

And so forth where for for $P_n^m$ and $T_n^m$, $m$ stands for unit number and $n$ is the number of pirate or treasure chest (like first pirate, second pirate, et cetera).Because there is only one value of $m$ that has $P$ satisfying $60 \leq P \leq 69$, this is only answer. Since our 9 units, or 63 pirates and 99 treasure chests, has $P$ values of 63, the number between the interval of the problem, this implies that the number of treasure chests is 99. [/quote][/hide]

Interesting way of looking at it. I see two problems--you didn't prove 99 was the only answer that works (it may not be possible to break the pirates and treasure chests down into units, since the groups could overlap), and it looks like you're trying to show your thought process as well as your actual solution. I've never done an olympiad before but I would think that the graders wouldn't like that--they'd only want your final solution, not the path you took to it, which could be extremely twisted.

I agree.

I knew my solution may have been interesting but not the kind graders look for because it's extra complicated.

I'll get better though. Hopefully. :)

[b]Day 2 Solution[/b]

Thanks to agolsme who was the only one who got this problem right on Day 2. His answer, in my opinion, would receive full credit in real USAMTS because it's simple and easy-to-follow and do not seem to miss any obvious points. Also, I like the fact that he also gave non-matice way answer because I'm not sure on how to use matrices for system of equation (I should learn it soon :blush: )

[quote="agolsme"]we can write this as:
\[
\left(
\begin{array}{cccccc}
1 & 1 & 1 & 1 & 1 & 2\\
2^4 & 2^3 & 2^2 & 2^1 & 1 & 4\\
3^4 & 3^3 & 3^2 & 3^1 & 1 & 6 \\
4^4 & 4^3 & 4^2 & 4^1 & 1 & 8\\
5^4 & 5^3 & 5^2 & 5^1 & 1 & 34
\end{array}
\right)\]

Then we can reduce it into reduced row-echelon form(which, after a lot of messing around, returns:
 \[
\left(
\begin{array}{cccccc}
1 & 0 & 0 & 0 & 0 & 1\\
0 & 1 & 0 & 0 & 0 & -10\\
0 & 0 & 1 & 0 & 0 & 35\\
0 & 0 & 0 & 1 & 0 & -48\\
0 & 0 & 0 & 0 & 1 & 24
\end{array}
\right)\]

giving $a -b + c -e +d = 1 + 10 + 35 + 48 + 24 = 118$

Alternatively, not using matrices the solution can be reached via subtracting each equation from the one below it and repeat.
It looks like this
\[
15a + 7b + 3c + d = 2\\
65a + 19b + 5c +d =2\\
175a + 37b + 7c +d =2\\
369a + 61b + 9c + d = 26
\]
repeat
\[
50a + 12b + 2c = 0\\
110a + 18b + 2c =0\\
194a + 24b + 2c = 24
\]
finally
\[
60a + 6b = 0\\
84a + 6b = 24\\
\]

so $a = 1$ and you substitute back to get same thing as before for $b,c,d,e$. Eitherway

$\fbox{118}$ is the answer[/quote]

Aghgghghgh! *!@#$%^! STUPID ADDING ERROR!  :mad:

 :D  :D  :D

From Day 3's problem I'll post today (which will be very soon.. I'm still looking for good problem), I'll be using 5/5 scale that USAMTS graders do.

See the post below for some details I forgot. :D

[quote="zanttrang"][quote="Silverfalcon"]From Day 3's problem I'll post today (which will be very soon.. I'm still looking for good problem), I'll be using 5/5 scale that USAMTS graders do.

So any computer solution won't be worth much point.[/quote]
That doesn't make sense - on a real USAMTS problem, computer solutions are given full credit if the procedure used by the computer is fully explained and does indeed give the correct solution.  The solutions may be boring, and will never get a commendation, but many real-world math problems can be and are solved by computers, and so to discourage this route by taking off points for a correct application of technology seems arbitrary and anachronistic to me.[/quote]

Oops. That came out wrong way. I remembered seeing this one problem said "computer solution can worth maximum one point."

Yes, I'll give my best shot on grading computer solution as well as the other ones. I'm not sure how people grade the computer solution though. :huh:

[quote="Silverfalcon"]
Oops. That came out wrong way. I remembered seeing this one problem said "computer solution can worth maximum one point."

Yes, I'll give my best shot on grading computer solution as well as the other ones. I'm not sure how people grade the computer solution though. :huh:[/quote]
Ah, okay, if the original problem said it, that's different - I'm fine with computer solutions not getting credit as long as it's stated in the problem, and not applied arbitrarily afterward.  For this particular problem, though, I don't see how a computer solution could get credit - saying that you used Mathematica or another CAS to solve the equation and it gave you only one solution doesn't show any mathematical talent.

[b]Day 3 Solution[/b]

Simple solution.

[quote="zanttrang"]Here's my solution to day 3 of your USAMTS practice:

Note that $(4x^2+6x+4)(4y^2-12y+25) = 28$ can be converted into

$\left(4\left(x+\frac{3}{4}\right)^2+\frac{7}{4}\right)\left(4\left(y-\frac{3}{2}\right)^2+16\right) = 28$

by completing the square.  Pulling out a $\frac{1}{4}$ from the first factor and a $4$ from the second yields

$\left(16\left(x+\frac{3}{4}\right)^2 + 7\right)\left(\left(y-\frac{3}{2}\right)^2 + 4\right) = 28$

For $(x,y) = \left(-\frac{3}{4}, \frac{3}{2}\right)$, we have
$7 \cdot 4 = 28$, which fulfills the given equation.  For all other real $(x,y)$, we have

$\left(16\left(x+\frac{3}{4}\right)^2 + 7\right)\left(\left(y-\frac{3}{2}\right)^2 + 4\right) > 28$

proving that
$(x,y) = \left(-\frac{3}{4}, \frac{3}{2}\right)$
is the only solution to the given equation.  QED[/quote]

Sidecomment from me:

Although it is fine, I think we could simply complete square in what we were given

$(4x^2+6x+4)(4y^2-12y+25) = 28$

This equals $\left(\left(2x+\frac{3}{2}\right)^2 + \frac{7}{4} \right)((2y-3)^2+16) = 28$ by completing square.

Note that $\frac{7}{4} \cdot 16 = 28$ so this implies that the equation holds true if values with $\left(2x+\frac{3}{2}\right)^2$ and $(2y-3)^2$ equal to 0.

Thus, setting them equal to zero, we find $(x,y) = \left(-\frac{3}{4}, \frac{3}{2}\right)$.

Alternate solution that some people used.

[quote="chess64"]$\text {We are given that }$
\[f(x) \cdot g(y) = 28\]
$\text {such that } f(x)=4x^2+6x+4 \text { and } g(y)=4y^2-12y+25. \text { Rearranging the equation, we have }$
\[4y^2-12y+25 - \frac{28}{f(x)}= 0\]
$\text {Because we have a quadratic equation with only one solution, we can set the discriminant equal to zero.}$
\begin{eqnarray*}
b^2-4ac&=&144-4(4)(25 - \frac{28}{f(x)}) = 0\\
144&=&400 - \frac{448}{f(x)}\\
256&=&\frac{448}{f(x)}\\
f(x)&=&\frac{7}{4}\\
\frac{7}{4}&=&4x^2+6x+4\\
\end{eqnarray*}
$\text {Solving for } x \text {, we see } x=-\frac{3}{4} \text {. Substituting into the original equation, we have }$
\begin{eqnarray*}
\frac{28}{g(y)}&=&\frac{7}{4}\\
g(y)&=&16\\
4y^2-12y&=&-9\\
y^2-3y&=&\frac{-9}{4}\\
y^2-3y+\frac{9}{4}&=&0\\
(y-\frac{3}{2})^2&=&0\\
y&=&\frac{3}{2}\\
\end{eqnarray*}
$\text {Thus the only solution is } (x,y)=(-\frac{3}{4},\frac{3}{2}). \ \ \ \blacksquare$[/quote]

[quote="Silverfalcon"][b]Day 2 Solution[/b]

Thanks to agolsme who was the only one who got this problem right on Day 2. His answer, in my opinion, would receive full credit in real USAMTS because it's simple and easy-to-follow and do not seem to miss any obvious points. Also, I like the fact that he also gave non-matice way answer because I'm not sure on how to use matrices for system of equation (I should learn it soon :blush: )
 [/quote]

Here is a non-matrices yet simple solution

Let $p(x) = ax^4 + bx^3 + cx^2 + dx + e$, Notice that $p(1)=1, p(2)=4, p(3)=6, p(4)=8, p(5)=34$.
Consider the polynomial $g(x)$, where $g(x)= p(x)-2x$. Since $g(x)$ is still a 4th degree polynomial with leading coefficient $a$, and roots $1, 2, 3, 4$.

Thus $g(x)=a(x-1)(x-2)(x-3)(x-4)$, and $p(x)=a(x-1)(x-2)(x-3)(x-4)+2x$. When $x=5$, we have
$p(5)=24a+10=34$, thus $a=1$. Expand $p(x)$ we have

$p(x)=x^4 - 10x^3 + 35x^2 - 48x + 24$, so $a-b+c-d+e =1+10+35+48+24=118$

[quote="Silverfalcon"]Hmm..

One question: Why did you have $g(x) = p(x) - 2x$? Why not $3x,4x$ or something other one?

I'm bad at polynomial manipulation so good explanation will be really grateful. Thanks for posting alternate solution by the way. :)[/quote]

I chose p(x)-2x because of the pattern, p(1)=2(1)=2, p(2)=2(2)=4, p(3)=2(3)=6, p(4)=2(4)=8,

[b]Day 4 Solution[/b]

[quote="zanttrang"]My day four solution:

Let the sum we are attempting to find be represented as

$S = \sum_{k=1}^{4012008} \frac{1}{(k+1)\sqrt{k} + k\sqrt{k+1}}$

We can manipulate the general term of the series into a simpler form that will allow us to quickly find the solution.

$\frac{1}{(k+1)\sqrt{k} + k\sqrt{k+1}} = \frac{(k+1)\sqrt{k} - k \sqrt{k+1}}{((k+1)\sqrt{k} + k\sqrt{k+1})((k+1)\sqrt{k} - k \sqrt{k+1})}$

$= \frac{(k+1)\sqrt{k} - k \sqrt{k+1}}{(k+1)^2k - k^2(k+1)}$

$ = \frac{(k+1)\sqrt{k} - k \sqrt{k+1}}{k(k+1)}$

$ = \frac{\sqrt{k}}{k} - \frac{\sqrt{k+1}}{k+1}$

We can now see that our sum can be represented by a telescoping series as

$S = \left(\frac{\sqrt{1}}{1} - \frac{\sqrt{2}}{2}\right) + \left(\frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{3}\right) + ... + \left(\frac{\sqrt{4012008}}{4012008} - \frac{\sqrt{4012009}}{4012009}\right)$

Cancelling terms gives us

$S = 1 - \frac{\sqrt{(2003)^2}}{(2003)^2}$

$S = 1 - \frac{1}{2003}$

$S = \frac{2002}{2003}$

Therefore, we have the rational number $\frac{m}{n} = \frac{2002}{2003}$ such that $S = \frac{m}{n}$
QED[/quote]

[b]Day 5 Solution[/b]

[quote="zanttrang"]Day 5 solution:

We can expand the given cosine expression using the double-angle and addition identities for sine and cosine.

$\frac{\cos 3x}{\cos x} = \frac{\cos 2x \cos x - \sin 2x \sin x}{\cos x}$

$= \frac{(2 \cos ^2 x - 1)\cos x - 2\sin ^2 x \cos x}{\cos x}$

$= \frac{(2 \cos ^2 x - 1)\cos x - 2(1 - \cos^2 x) \cos x}{\cos x}$

$= 2 \cos^2 x - 1 - 2(1 - \cos^2 x)$

$ = 4 \cos^2 x - 3$

which gives us the equation

$4 \cos^2 x - 3 = \frac{2}{5}$

Now let

$\frac{\sin 3x}{\sin x} = k$

We can manipulate the sine expression in the same way we manipulated the cosine expression.

$\frac{\sin 3x}{\sin x} = \frac{\sin 2x \cos x - \cos 2x \sin x}{\sin x}$

$ = \frac{2 \sin x \cos^2 x + (2 \cos^2 x - 1) \sin x}{\sin x}$

$ = 2 \cos^2 x + 2 \cos^2 x - 1$

$ = 4 \cos^2 x - 1$

which gives us the equation

$4 \cos^2 x - 1 = k$

We now have the two equations

$4 \cos^2 x - 1 = k$

$4 \cos^2 x - 3 = \frac{2}{5}$

Subtracting the second equation from the first gives

$2 = k - \frac{2}{5}$

Solving for $k$ gives

$k = \frac{12}{5}$

We therefore have our solution

$\frac{\sin 3x}{\sin x} = \frac{12}{5}$

$\tex{QED}$[/quote]

When are you going to update the scores?

Soon. I don't think you sent me the answer to Day 6 though.

I have some work to do so I posted Day 7's problem first. I'll post solution and edit scoreboard tonight. Till then, you're allowed to send Day 6 AND Day 7's solution.

Hope this cleared up! :idea:

[b]Day 6 Solution[/b]

Thanks to zanttrang again for posting correct solutions.

[quote="zanttrang"]Note: this problem can be solved in at least two ways: with calculus or without.  I won't present the calculus solution here, as it is dull and monotonous.  Instead, I'll give the rather clever non-calculus solution I came up with.

Day six solution:

First, re-arrange the given inequality.

$\sec^6x + \csc^6x + (\sec^6 x) (\csc^6 x) \geq 80$

$\sec^6x + \csc^6x(\sec^6x + 1) \geq 80$

$(\csc^6 x + 1)(\sec^6x + 1) - 1 \geq 80$

$(\csc^6 x + 1)(\sec^6x + 1) \geq 9^2$

Now, we apply Cauchy-Schwartz to the left side of the inequality to obtain

$(\csc^6 x + 1)(\sec^6x + 1) \geq ((\sec^3 x) (\csc^3 x) + 1)^2$

Our original inequality then reduces to proving that

$(\sec^3 x) (\csc^3 x) + 1 \geq 9$

(note that the positive case must be taken because $(\sec^3 x) (\csc^3 x) +1 > 0$ on the given interval).  Re-arranging this gives (remembering that the following manipulations are valid because $\sec x > 0$ and $\csc x > 0$ on the given interval)

$(\sec^3 x) (\csc^3 x)  \geq 2^3$

$(\sec x)(\csc x) \geq 2$

$1 \geq 2 (\sin x)(\cos x)$

$1 \geq \sin 2x$

The above inequality is trivially true (and has equality for $x = \frac{\pi}{4}$ on the given interval), allowing us to make the statement that

$(\csc^6 x + 1)(\sec^6x + 1) \geq ((\sec^3 x) (\csc^3 x) + 1)^2 \geq 81$

which proves our original inequality that

$\sec^6x + \csc^6x + (\sec^6 x) (\csc^6 x) \geq 80$

for $0 < x < \frac{\pi}{2}$, with equality when $x = \frac{\pi}{4}$

$QED$[/quote]

I did with simple AM-GM. I don't know how to use Cauchy so here's my idea.

[quote="Silverfalcon"]

AM-GM Inequality (Arithmetic Mean-Geometric Mean) states that for $a$ and $b$, $\frac{a+b}{2} \geq \sqrt{ab}$ and equality holds true when $a=b$.

Since we have an Pythagorean Identity $\sin^2 x + \cos^2 x = 1$, let's set $a = \sin^2 x$ and $b = \cos^2 x$.

$\frac{\sin^2 x + \cos^2 x}{2} \geq \sqrt{\sin^2 x \cos^2 x}\\
\frac{1}{2} \geq \sin x \cos x\\
2 \leq \csc x \sec x\\
2^6 \leq \csc^6 x \sec^6 x\\
64 \leq \csc^6 x \sec^6 x$

Now, we can apply AM-GM again on $\sec^6 x$ and $\csc^6 x$.

$\frac{\sec^6 x + \csc^6 x}{2} \geq \sqrt{\sec^6 x \csc^6 x}\\
\sec^6 x + \csc^6 x \geq 2 \sqrt{64}\\
\sec^6 x + \csc^6 x \geq 16$

Note that we substituted $\sqrt{\sec^6 x \csc^6 x}$ as $\sqrt{64}$ using our first inequality from AM-GM.

Adding the two inequalities, we get $\sec^6 x +\csc^6 x + \sec^6 x \csc^6 x \geq 80$. Hence, the original inequality is proven. QED [/quote]

I think you made a small typo in your work: you forgot to switch the direction of the inequality sign when manipulating your AM-GM statement.  Your final statement should read:

$\csc^6x \sec^6x \geq 64$

Other than that it looks good, and possibly even simpler than mine.

Yeah. I forgot to switch signs when I raise to $^{-1}$ power on $\frac{1}{2}$.

Thanks for the typo.

[b]Day 7 Solution[/b]

[quote="zanttrang"]My Day 7 solution:

We first assert that

$a_n > \sqrt{2n - 1}, \ n \geq 2$

This assertion can be proven by induction as follows.

For $n = 2$,

$a_2 = 2 > \sqrt{2 \cdot 2 - 1} = \sqrt{3}$

Now assume $a_n > \sqrt{2n - 1}$.  This must imply that

$a_{n+1} > \sqrt{2n + 1}$

$a_n + \frac{1}{a_n} > \sqrt{2n + 1}$

$a_n\left(a_n - \sqrt{2n + 1}\right) + 1 > 0$

$\left(a_n - \frac{\sqrt{2n+1}}{2}\right)^2 + \frac{3-2n}{4} > 0$

Our assumption allows us to state that

$a_n\left(a_n - \sqrt{2n + 1}\right) + 1 > \sqrt{2n - 1}\left(\sqrt{2n - 1} - \sqrt{2n + 1}\right) + 1$

assuming that $\sqrt{2n - 1}$ is larger than the value giving the minimum of the parabola.  We see that this is true (from the completed square given above) if

$\sqrt{2n - 1} > \frac{\sqrt{2n + 1}}{2}$

Notice that this is true on our given interval of $n \geq 2$, meaning that our inequality is valid.  Now note that

$\sqrt{2n - 1}\left(\sqrt{2n - 1} - \sqrt{2n + 1}\right) + 1 = 2n - \sqrt{4n^2 - 1} > 0$

on our given interval of $n \geq 2$, meaning that

$a_n\left(a_n - \sqrt{2n + 1}\right) + 1 > 2n - \sqrt{4n^2 - 1} > 0$

for $n \geq 2$.  So, we have proven that

$a_n > \sqrt{2n - 1}$

implies

$a_{n+1} > \sqrt{2n + 1}$

for $n \geq 2$, completing our proof by induction.  Now, using what we have proven, we can make the statement

$a_{10000} > \sqrt{2 \cdot 10000 - 1} \approx 141.418$

Therefore, $a_{10000} > 141$

$QED$[/quote]

EDIT: I can't find any better way to solve this.

[quote="Silverfalcon"]
EDIT: I can't find any better way to solve this.[/quote]

Where did this problem come from?  If it's from the USAMTS archives, then there should be a solution given for each problem.

Not this one.  :D

It came from one of the state math competitions where you have to write proofs.

 :)

P.S. They had answers but it was basically same as yours. I was wondering if there is any other way to do without using induction or using completely different ideas.

[quote="Silverfalcon"]I was wondering if there is any other way to do without using induction or using completely different ideas.[/quote]

Well, I originally got my function for an inductive proof from a calculus topic.  The recursive equation you gave is the "discrete analogue" of a differential equation.  You can see this because the equation can be re-arranged to give

$a_{n+1} - a_n = \frac{1}{a_n}$

So, the difference between terms of the sequence is equal to the reciprocal of the current term.  The function analogy for this is

$\frac{df}{dx} = \frac{1}{f(x)}$

The discrete analogues of differential equations often have similar properties to the function represented by the equation, so I decided to investigate this function further.  Solving this differential equation gives

$f(x) = \sqrt{2x + c}$

where $c$ is an arbitrary constant.  Using $f(1) = 1$ (from $a_1 = 1$), I found $c = -1$.  I then noted that for the next several $a_n$, $a_n > f(n)$.  I wondered if this held for all $a_n$, and the best way to prove something of this sort is to use induction.  This doesn't really provide a different method for solving the problem, but it does do a better job of motivating my inductive proof.

I only received two answers for yesterday's problem.  :? And I think it's one of the easiest problem I posted in here.

I'll give more time on it (since I want to get more replies) and since I have to find the next problem. :D

It will be posted in few hours.. I just have some leftover work to do.. :lol:

[b]Day 8 Solution[/b]

Most simple but easy way to do this problem.

[quote="chess64"]Let's see if we can find a pattern for the first few $n$ for $1776^{n} \mod {2000}$:

$\begin{array}[b]{c|c}
1776^{1} & 1776\\
1776^{2} & 176\\
1776^{3} & 576\\
1776^{4} & 976\\
1776^{5} & 1376\\
1776^{6} & 1776\\
1776^{7} & 176\\
1776^{8} & 576\\
\end{array}$

Obviously there is a pattern; the remainder mod 2000 repeats every 5. Because $1492!$ is divisible by $5$, $1776^{1492!} \equiv 1776^5 \equiv \boxed {1376} \mod {2000}. \ \ \ \ \blacksquare$[/quote]

Little bit different approach.. Not much different but pretty interesting.

[quote="zanttrang"]Note the following two congruences

$1376^2 \equiv 1376 \ (\text{mod} \ 2000)$

$1776^5 \equiv 1376 \ (\text{mod} \ 2000)$

These can be easily verified by calculator or computer.

From this we can see that

$1776^5 \cdot 1776^5 \equiv 1376 \cdot 1376 \equiv 1376 \ (\text{mod} \ 2000)$

$1776^{10} \cdot 1776^5 \equiv 1376 \cdot 1376 \equiv 1376 \ (\text{mod} \ 2000)$
.
.
.
$1776^{5n} \equiv 1376^2 \equiv 1376 \ (\text{mod} \ 2000)$

for positive integer $n$.  As $1492!$ is clearly divisible by 5, we have

$1776^{1492!} \equiv 1776^{5n} \equiv 1376 \ (\text{mod} \ 2000)$

$QED$[/quote]

If you weren't sure how to calculate $1776^5 \mod 2000$ or the values that appeared large on the calculator, remember the important properties of modular arithmetic:

$(a \mod b) \cdot (c \mod b) = ac \mod b$

For example:

$(2 \mod 5) \cdot (7 \mod 5) = (14 \mod 5) = 4 \mod 5$

But note that $7 \mod 5 = 2 \mod 5$ so $(2 \mod 5) \cdot (7 \mod 5) = (2 \mod 5) \cdot (2 \mod 5) = 4 \mod 5$

If you knew this, I think this problem was quite easy.  :)

[b]Day 9 Solution[/b]

I'm not completely sure about this one because this problem isn't in one of the archives section on USAMTS page. It was in Round 27 of IMTS.

Anyway.. This is what zanttrang got (because it agrees with my answer)

[quote="zanttrang"]$f(2) = f(1) + f(1) + 3(4 - 1) = 9$

$f(3) = f(2) + f(1) + 3(4 \cdot 2 - 1) = 30$

$f(4) = f(2) + f(2) + 3(4 \cdot 2^2 - 1) = 63$

$f(8) = f(4) + f(4) + 3(4^3 - 1) = 315$

$f(16) = f(8) + f(8) + 3(4 \cdot 8^2 - 1) = 1395$

$f(19) = f(16) + f(3) + 3(4 \cdot 16 \cdot 3 - 1) = \fbox{1998}$[/quote]

Below is my answer:

[quote="Silverfalcon"]
$f(2) = f(1+1) = f(1) + f(1) + 3(4 \cdot 1 \cdot 1 - 1) = 0+0+9 = 9$
$f(4) = f(2+2) = f(2) + f(2) + 3(4 \cdot 2 \cdot 2 - 1) = 9+9+45 = 63$
$f(8) = f(4+4) = f(4) + f(4) + 3(4 \cdot 4 \cdot 4 - 1) = 63+63+3(63) = 315$
$f(9) = f(1+8) = f(1) + f(8) + 3(4 \cdot 1 \cdot 8 - 1) = 0+315+93 = 408$
$f(18) = f(9+9) = f(9) + f(9) + 3(4 \cdot 9 \cdot 9 - 1) = 408+408+969 = 1785$
$f(19) = f(1+18) = f(1) + f(18) + 3(4 \cdot 1 \cdot 18 - 1) = 0+1785+3(71) = 1785+213 = 1998$. [/quote]

I'll post the other people's solution in next thread (the ones with different answer). I'm pretty sure 1998 is the correct answer but I'll post the other ones too. Hope that clears up on why some people have different answers..

oops. I multipled by the wrong number in f(16) it was $3(4*8*8-1)$

Probably I made some funny calculation error. Will edit this post if I find the specific mistake.
EDIT: note to self: f(4)=45+18=63, not 65. *bonks self on head* Can I still get 4 points?

I'm pretty sure 1998 is the correct answer.

As for the credit, I'll only worth this 1 point (maximum). The reason is because the problem only requires you to know how to use the function. ;)

If you actually had to do any work with the function (such as rearranging etc), then it will worth more but I think I would only give 1 point.

That's easier.

Oh, that explains it! I misread the function! It was $3(4mn-1)$, not $3(mn-1)$! :wallbash_red:

[b]Day 10 Solution[/b]

I'll post mine using simple AM-GM soon.

This is only solution I received.

[quote="zanttrang"]I couldn't really figure out how to solve this problem completely (I'm probably missing something really obvious), but I do have a partial solution:

note that

$(a + b +c + d)^2 = a^2 + b^2 + c^2 +d^2 +2(ab + ac + ad + bc + bd + cd) = 144$

$a^2 + b^2 + c^2 + d^2 + 2(abcd - 27) = 144$

$a^2 + b^2 + c^2 + d^2 + 2abcd = 198$

Applying Cauchy-Schwartz, we have the inequality

$(a^2 + b^2 + c^2 + d^2)(1 + 1 + 1 + 1) \geq (a + b + c + d)^2$

leading to

$a^2 + b^2 + c^2 + d^2 \geq 36$

with equality when $a = b = c = d = 3$

Now, applying AM-GM we have the inequality

$\frac{a + b + c + d}{4} \geq (abcd)^{\frac{1}{4}}$

$81 \geq abcd$

with equality when $a = b = c = d = 3$

Note that when we have equality, the equation

$a^2 + b^2 + c^2 + d^2 + 2abcd = 198$

is fulfilled, because

$36 + 2 \cdot 81 = 198$

Thus, one solution is $(a,b,c,d) = (3,3,3,3)$

[/quote]

[quote="Silverfalcon"][b]Day 10 Solution[/b]

I'll post mine using simple AM-GM soon.
[/quote]

Brilliant!  That "hint" did it for me.  I've solved it, as follows:

First, apply AM-GM to obtain

$\frac{a+b+c+d}{4} \geq (abcd)^{\frac{1}{4}}$

Substituting $a+b+c+d=12$ gives

$81 \geq abcd$

with equality iff $a = b = c = d = 3$.  Now apply AM-GM again to obtain

$\frac{ab + ac + ad + bc + bd + cd}{6} \geq (a^3b^3c^3d^3)^{\frac{1}{6}}$

which simplifies to

$ab + ac + ad + bc + bd + cd \geq 6\sqrt{abcd}$

Now substituting $ab + ac + ad + bc + bd + cd = abcd - 27$ gives

$abcd - 27 \geq 6\sqrt{abcd}$

This factors as

$(\sqrt{abcd} - 9)(\sqrt{abcd} + 3) \geq 0$

Note that this implies (for positive $a,b,c,d$)

$\sqrt{abcd} \geq 9$

$abcd \geq 81$

So, we now have the inequality

$81 \geq abcd \geq 81$

which is satisfied iff $a = b = c = d = 3$.  Therefore, our only solution is

$(a,b,c,d) = (3,3,3,3)$

$QED$

:D You were faster.

Yes, that was basically what the main idea of this problem was.

By the way, it was one of the problem in British Math Olympiad (Not USAMTS so don't worry if you found it pretty hard)

[b]Day 11 Solution[/b]

I posted this later even though I got the answer on the time. I was hoping for more replies..  :?

[quote="zanttrang"]Part 1:

There are two ways we can go about solving this problem.

Method 1: square, square, square,....

We have been given the equation

$2 = \sqrt{6 *_1 \sqrt{6 *_2 \sqrt{6 *_3 \cdots}}}$

Squaring and re-arranging gives

$-2  = *_1 \sqrt{6 *_2 \sqrt{6 *_3 \sqrt{6 *_4 \cdots}}}$

The square root will be positive, so we must have $*_1 = -$

Now square and re-arrange again:

$-2 = *_2 \sqrt{6 *_3 \sqrt{6 *_4 \sqrt{6 *_5 \cdots}}}$

Again, we have $*_2 = -$.  It is now obvious that all $*_n = -$, meaning we have

$2 = \sqrt{6 - \sqrt{6 - \sqrt{6 - \cdots}}}$

Method 2: test convergence points

First, assume that $*_n = -$.  If the nested radical converges, the convergence point $x$ has the property that

$x = \sqrt{6 - x}$

Solving this gives $x = 2$.  So, assuming that the nested radical converges, we have

$2 =  \sqrt{6 - \sqrt{6 - \sqrt{6 - \cdots}}}$

It is a well-known property that a nested radical with constant coefficients will converge, so our proof is complete.

Part 2:

Again, both methods can be used.

Method 1: square, square, square,...

We have been given the equation

$2 = \sqrt{7 *_1 \sqrt{7 *_2 \sqrt{7 *_3 \cdots}}}$

Squaring and re-arranging gives

$-3  = *_1 \sqrt{7 *_2 \sqrt{7 *_3 \sqrt{7 *_4 \cdots}}}$

The square root will be positive, so we must have $*_1 = -$

Now square and re-arrange again:

$2 = *_2 \sqrt{7 *_3 \sqrt{7 *_4 \sqrt{7 *_5 \cdots}}}$

Now we must have $*_2 = +$.  Now we note that this pattern will recur indefinitely, so we have $*_{2n} = +$ and $*_{2n - 1} = -$ for $n \geq 1$

This gives us

$2 = \sqrt{7 - \sqrt{7 + \sqrt{7 - \cdots}}}$

Method 2: convergence points

Assume that $*_{2n} = +$ and $*_{2n - 1} = -$ for $n \geq 1$.  If the nested radical converges, the convergence point $x$ has the property that

$x = \sqrt{7 - \sqrt{7+x}}$

Solving this gives $x = 2$.  So we have, assuming convergence (which is a well-known property, as noted above):

$2 = \sqrt{7 - \sqrt{7 + \sqrt{7 - \cdots}}}$

$QED$[/quote]

I got the same answer except I wasn't sure how to explain my answers.  :(

[b]Day 12 Solution[/b]

zanttrang and JesusFreak197 basically had same answer (or very close from what I saw)

[quote="JesusFreak197"]$7x^2-3y^2+4z^2=8 \Rightarrow y^2=\frac{7x^2+4z^2-8}{3}$
$16x^2-\frac{7}{3}(7x^2+4z^2-8)+9z^2=-3 \Rightarrow x^2+z^2=65$

$7x^2-3y^2+4z^2=8 \Rightarrow z^2=\frac{3y^2-7x^2+8}{4}$
$16x^2-7y^2+\frac{9}{4}(3y^2-7x^2+8)=-3 \Rightarrow x^2-y^2=-84$
$(x+y)(x-y)=-84$

Since $x,y,z$ are positive integers, the only possible multiples that could result in this value are $(42,-2)$ and $(14,-6)$. If it is $(42,-2)$, then $x=20$ and $y=22$, which is impossible because $x^2+z^2=65$, meaning that the correct multiples are $(14,-6)$, and we have $x=4$, $y=10$, and $z=7$, which all fit the previous equations, and $x^2+y^2+z^2=165$.[/quote]

There are actually many other ways to do this problem. For instance, I think I got the equation $y^2+z^2 = 149$ (I think) and from there, I got two $(y,z)$ which were $(10,7)$ and $(7,10)$. But when I plugged in $(7,10)$, I think you get negative $x$. But you can't have negative values for $x,y,z$ so this immediately makes $(10,7)$ as the answer.

I could've had little bit more trouble if there were other pairs but from what I see, all methods basically test if you can deal with the system of the equation or not.

I was hoping for more replies though.  :D

Yeah, I got $y^2+z^2=149$ as well, but that was from the second equation, and I realized that, once I had $x^2+z^2=65$ and $(x+y)(x-y)=-84$, I could just use those.


Hello all ,
It may be posted before ,but however ...

Prove that   $ \frac{(a_1 \plus{} a_2 \plus{} ...\plus{}a_n)!}{a_{1}! a_{2}! \cdots a_{n}!}$ is integer if $ a_i$'s are nonegative integers.

[url=http://en.wikipedia.org/wiki/Multinomial_theorem]Multinomial coefficients[/url] are integers.


geometry
3D geometry
sphere
advanced fields
advanced fields unsolved

Prove: $K\subseteq \mathbb{R}^{n}$ is a convex set if and only if  $\forall~\alpha,\beta \geq 0,~(\alpha+\beta)K=\alpha K+\beta K$.

Also,
Let $S,K\subseteq \mathbb{R}^{n}$ and $K$ is a convex set .Prove that set $A: = \{a \in \mathbb{R}^{n}| a+S \subseteq K \}$ is a convex set.

Regards

1) $K \subseteq\alpha K+\beta K$ is obvious. If $\alpha+\beta=1$, then $\alpha K+\beta K\subseteq K$ is just a restatement of convexity. General case reduces to this by dividing both sides by $\alpha+\beta$.

2) Suppose $\alpha+\beta=1$ and $\alpha,\beta\ge 0$. If $a,b\in A$, then
$(\alpha a+\beta b)+S\subseteq (\alpha a+\beta b)+\alpha S+\beta S=\alpha(a+S)+\beta(b+S)\subseteq \alpha K+\beta K=K$
hence $\alpha a+\beta b\in A$

Let $K\subseteq \mathbb{R}^{n}$ a convex set and $r>0$.Lets define $K_{r}: =\{x\in \mathbb{R}^{n}| B(x,r)\subseteq K\}$ where $B(x,r)$ is a sphere around x radius r.Prove that $K_{r}$ is a convex set.
(I know it easy,but please help,or at least give me a hint)

Setup: $a,b\in K_{r}$ and $\alpha,\beta\ge 0$, $\alpha+\beta=1$.

Must prove: $B(\alpha a+\beta b,r)\subseteq K$.

Idea: $B(\alpha a+\beta b,r)=\alpha B(a,r)+\beta B(b,r)$.

[quote="mlok"]
Idea: $B(\alpha a+\beta b,r)=\alpha B(a,r)+\beta B(b,r)$.[/quote]

This statement isnt true.

$\supseteq$: if $|x-a|\le r$ and $|y-b|\le r$, then $|(\alpha x+\beta y)-(\alpha a+\beta b)|\le \alpha |x-a|+\beta|y-b|\le r$.

$\subseteq$: if $|z-(\alpha a+\beta b)|\le r$, let $x=a+z-(\alpha a+\beta b)$ and $y=b+z-(\alpha a+\beta b)$. Then $x\in B(a,r)$, $y\in B(b,r)$, and $z=\alpha x+\beta y$.

See you in the court.

You are right! I thought if I choose $r: =\frac{1}{n},n\in \mathbb{N}$ ,that then $\exists m\in \mathbb{N},m>n \pitchfork B(\alpha a+\beta b ,\frac{1}{m}) \nsubseteq \alpha B(a,\frac{1}{m})+\beta B(b,\frac{1}{m})$


[quote="mdk"]Prove that the equation $ a^{2}+b^{2}=c^{2}+3$ has infinitely many integer solutions[/quote]

$ a=2x$
$ b=2x^{2}-2$
$ c=2x^{2}-1$

satisfies $ a^{2}+b^{2}=c^{2}+3$ for all positive integral $ x$.

If you are curious just how one guesses the above:
[hide]$ (c^{2}-b^{2})=a^{2}-3$
$ (c+b)(c-b) = (a^{2}-3)(1)$
so if you write $ (c+b)=a^{2}-3$ and $ c-b = 1$
Just add (and subtract) these two equations and and write  $ c$ and $ b$ in terms of $ a$  (and you are done ... you may  like to take $ a=2k$ so that you may  avoid fractions ...:) ($ 2c=4k^{2}-2$ and $ 2b=4k^{2}-4$ etc.. [/hide]


Team A and B play a series. The first team to win 3 games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game?

[hide="Probably wrong"]
First, you can see that the teams play a maximum of 5 games.

There is one possible way of B winning the first game and A winning the series:
BBAAA

If A won the first game we can have:
ABAA
ABBAA
ABABA

I think those are all the possible ways.

So is it $\displaystyle\frac{1}{4}$?
[/hide]

I think this problem was from 2005 AMC 10A.

Anyway, I'm horrible at probability, but I'd hazard a guess that you find all the possibile set-ups where team B wins the second game but A wins three, then find the number where B wins the first (which would be 1 way), making sure to factor in the probability for each possibility.

[hide]
Like ccy said, there are at most 5 games.

_B_ _ _

Like robinhe said, the only way B could win the first game is
BBAAA.

So lets count how many ways A could win first game
AB_ _ _
ABAAB (which is in our case identical to ABAAA)
ABABA
ABBAA

So, I also get $\frac14$[/hide]

Weird way to get the answer... prob. wrong

[hide]If team B won the first game, then the only way for A to win the series is by A winning the last three games -- which would be 1/2 * 1/2 * 1/2, so the probability of team B winning the first game is 1/8[/hide][/hide]

But the fact that A wins the series is a given.
So you don't need to calculate that.
In fact, [quote] Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent.[/quote]
Is basically uneeded information.

[quote="robinhe"]Ccy, the probability that the series goes ABAA is not the same as if it went ABABA.

[hide]This is probably wrong too, but the only way a series in which team B gets the first two games and team A wins the series is BBAAA.  Since BB is already given, would the probability be $\frac12 \times \frac12 \times \frac12 = \frac18$?[/hide][/quote]

at first i got $\frac{1}{4}$, but then i read this post. so then i got
[hide="probably wrong"]$\frac{3}{40}$[/hide]?

How'd you get $\frac3{40}$?
Your first ans. seemed right..

I also got 1/4...what's a little confusing is that the answer is actually 1/5...something to do with weighted probability.
Basically, the solution says that the probability of ABAA is twice the probability of, say, ABABA. The problem I have with this is that I'm sure I've solved problems like this without weighting probabilities...and still gotten it right :oops:  :mad:

Hey, cool, I was wondering if that was the case. :P

Basically, it's because ABAA has one fewer game, and therefore has twice the probability of occuring. There's a $\frac{1}{16}$ chance of that, and $\frac{1}{32}$ of the other three, so you have $\frac{\frac{1}{32}}{\frac{5}{32}}=\frac{1}{5}$ once you calculate the total probability.

Ok...so take a look at this problem: A box contains 5 chips, 3 red and  2 white. Chips are removed one at a time without replacement until all the chips of one color are drawn. What is the probability that the last chip drawn is white?

Why can't you just solve it like this:

ww= 1/4 chance
rww = 1/8 chance *2 = 1/4
rrww=1/16 *3 = 3/16.

By the above method the answer is 1/2, but it's not correct.

wait, you said 5 chips 3 red and white? should that be 6?

Anyway, you didnt count all possible ways.
3 Ws=$\frac 12 \times \frac 25 \times \frac 14 =\frac{1}{20}$
1 R and 3 Ws=$\frac 12 \times \frac 25 \times \frac 34 \times \frac 13 \times 3=\frac{3}{20}$
2 Rs and 3Ws=$\frac 12 \times \frac 25 \times \frac 34 \times \frac 23 \times \frac 12 \times \binom 42 = 6= \frac{3}{10}$
Adding gives $\frac{1}{20}+\frac{3}{20}+\frac{3}{10}=\frac 12$
But that's not the answer?
It doesnt make sense unless you meant 3 white and 2 red, because if not they have to have equal probability as there is no dinstinguishibility between them. (basically if i calculated the probability for white, it would have to be the same.)

He left out a number. I'm pretty sure that it's 3 red and 2 white.

Basically, look at the different possibilities:

WW (1/10)
RWW (1/10)
WRW (1/10)
RRWW (1/10)
RWRW (1/10)
WRRW (1/10)

So, the total chance should be $\frac{3}{5}$. You can't treat this problem the same as the other one, since this one involves removing chips without replacing them, and therefore they don't have equal probabilities of happening every time. Notice that the probability is the same for every possible combination. That's because of how it works with removal. If you had an even number of both red and white chips and you used replacement (although you would obviously have to modify the problem to do that), then the method would be the same as the other problem.

Yeah I meant 3 red and 2 white...the answer 3/5 makes sense logically. Ok I see what I did wrong- I accidentally assumed that both had an equal chance of being picked. I think my answer does work if I go back.

WW- 2/5*1/4 = 2/20=1/10
RWW- 2(3/5*2/4*1/3)= 2/10
RRWW- 3(3/5*2/4*2/3*1/2)= 6/20

which sums to 3/5.

Yeah, I read the post... your point? Did you mean to write something else?

[b]EDIT:[/b] Okay, I see.

Sorry clicked the wrong button...I've edited it now.


How do you do these ones, their completely confusing

 $Q(y)=4(xy)^3+256$
 $Q(y)=2(a-b)^2+5(a-b)+3$
 $Q(y)=x^2+10x-y^8+25$
 $Q(y)=a^2+b^2+4b-4$

Plz as soon as possible ^_^

Just to help you a little :

1- think about $x^3+y^3$
2- polynomial is null when $a-b = -1$
3- look for $(x+5)^2$
...


So it's apparently okay to discuss the problems now.  How do you guys think you did?

I got the answers all right, though my proof for number 3 was uglified and my proof for number 5 was an all-out java program.  The rest of my solutions were pretty solid.

Is it okay if we cited Catalan's conjecture (which was proven by some person whose name escapes me) as the reason that 8 and 9 are the only two nontrivial consecutive powers?

Ummm..See my thread with the poll that I posted before you did...(by 2 minutes) :wink:

My bad.

I decided to post this before anyone else posted anything, but it took some time for me to write it out.


(1) for any postive real $a,b,c$ such that $a+b+c=1$ ,prove that

$abc+\frac{1}{abc}\geq \frac{730}{27}$

(2) for positive real $a$ , prove that

$a+\frac{1}{a}+\frac{1}{a+\frac{1}{a}}\geq \frac{5}{2}$

 :)

the firs is false try for $a=b=c=\frac{1}{3}$ and the second exercice is very easy,u must majorite the function $f(x)=x+1/x$
$f'(x)>0$ on $[2,+\infty[$ so $f(x)>\frac52$ $\forall x \in [2,+\infty[$

[quote="marocleverness"]the firs is false try for $a=b=c=\frac{1}{3}$ [/quote]
It isn't false... the case you tell us is just the equality case :)

Also, for #2 you need to prove that it is true for all $x \in (0, \infty)$, not just 2 or greater.

Sorry i said mistakes.
in fact $\frac{1}{abc} \geq 27$ so we deduce as in second case that $abc+\frac{1}{abc} \geq 27+\frac{1}{27}=\frac{730}{27}$

How do you prove that $\frac{1}{abc} \ge 27$ when $a^2+b^2+c^2=1$?

no Hokkage it's known that $a+\frac1a \geq 2$

[quote="marocleverness"]SORRY
$abc \leq (\frac{(a+b+c)}{3})^3$ so $\frac{1}{abc}\geq 27$[/quote]

that only tells you that $abc\leq \frac{1}{27}$ and $\frac{1}{abc}\geq 27$  .... unfortunately , they cannot been add up since the sign is different ... :(

don't u uderstand functions.
just consider $f(x)=x+\frac1x$ so $f'(x)=1-\frac{1}{x^2}>0$on $[27,+\infty[$ .so $abc+\frac{1}{abc}\leq f(27) \leq \frac{730}{27}$

[quote="marocleverness"]SORRY always problems with LaTeX
don't u uderstand functions.
just consider $f(x)=x+\frac1x$ so $f'(x)=1-\frac{1}{x^2}>0$on $[27,+\infty[$ .so $abc+\frac{1}{abc}\geq f(27) \geq \frac{730}{27}$[/quote]

actually , I would rather it to be done in am-gm (due to the title up there :) ) instead of calculus  :D though your method has nothing wrong  :lol:

Eg: A hint,  (It is easier than one  woud guess)
[hide]if $X \ge 27$
then  $(X-27) \ge \frac{X-27} {27X}$ or ${X-27 \ge \frac 1 {27} - \frac 1 X}$
or $X+\frac 1 X \ge 27 + \frac 1 {27}$  and we are free to choose    X=(1/abc) )  etc..[/hide]

[quote="shyong"][quote="prototypex24"]The second question:

[hide]

By AMGM,
$\frac{a+\frac{1}{a}}{2} \geq \sqrt{(a)(\frac{1}{a})}$  $=1$
$\therefore a+\frac{1}{a} \geq 2$

So $a+\frac{1}{a}+\frac{1}{a+\frac{1}{a}}\geq 2+\frac{1}{2}=\frac{5}{2}$ [/hide][/quote]

in my opinion , if we consider the function $f(x)=x+\frac{1}{x}$ , we can see that the min is when $x=2$ . But if we want to apply am-gm on this case , I doubt it can be stated in such way since $\frac{1}{a+\frac{1}{a}}\leq 2$ while $a+\frac{1}{a}\geq 2$ and both sign is different so we cannot add together . :)[/quote]
Yes,  one needs something like (see my last post)
let $y=x+1/x$
$(x+1/x) \ge 2$  so $y \ge 2$
hence $(y-2) \ge \frac{(y-2)} {2y}$   (because (y-2)  is nonnegative  and (2y)  is greater than 1)
which gives $y+1/y \ge 2+1/2$


Let $ f$ be a linear transformation denoted by $ A \equal{} \left( \begin{array}{cc} a & b \\
c & d \end{array} \right)$ and $ g$ by $ B \equal{} \left( \begin{array}{cc} a & c \\
b & d \end{array} \right)$.

(1) Show that $ f(\overrightarrow{u})\cdot \overrightarrow{v} \equal{} \overrightarrow{u}\cdot g(\overrightarrow{v})$ for any vectots $ \overrightarrow{u},\ \overrightarrow{v}$.

(2) Suppose that the line $ l$ passing through the origin $ O$ is transformed to itself by $ f$.Take a point $ P\neq O$ on $ l$,  let $ Q\equal{}f(P)$  and $ R\equal{}g(P)$. Show that either (a) or (b) of the following contexts holds.

(a) $ Q \equal{} R$
(b) 3points $ Q,\ R,\ O$ become the vertices of a right triangle.

The statement of (2) is a little off- it should say that $ R\equal{}g(P)$, not $ Q\equal{}g(R)$.

#1 in inner product language: $ \langle Ax,y\rangle\equal{}\langle x,A^ty\rangle$; the abstract transpose with respect to the usual inner product is the usual transpose.

For #2, the right angle is at $ Q$; we want to show that a certain inner product is zero, using the fact that $ P$ is an eigenvector. The properties of the transpose established in #1 are essential.


We define addition in a different way than usual; an addition statement is true only if the letters in the addends is a rearrangement of the letters in the sum.  For example,
10 + 6 = 16?
TEN + SIX = TENSIX = SIXTEN, but to be 16 it would need another E.
Find a "true" addition
a + b = c + d.

fourteen + six = sixteen + four

fourteen + seven = seventeen + four

fourteen + nine = nineteen + four

sixteen + nine = nineteen + six

seventeen + nine = nineteen + seven

(I was able to post these without an accompanying green smiley face.)

Whoops, hehe  :blush:
$a, b, c, d < 14$.

[quote="tjhance"][hide="Solution"]
twelve+one=eleven+two
[/hide][/quote]That's it!


Hi folks, I would like to know the criterion for getting a  job at investment firms like D.E.Shaw,etc?

Firstly, I know that recruitment process at D.E.Shaw is very selective , often recruiting Putnam winners, IMO gold medalists, etc; I feel why do these guys recruit these (Putnam winners,IMO medalists) people, I know that they want extremely intelligent people, but I feel ordinary ( less focused) people can do equally well.

Can anyone justify (to me) why do investment banks(especially D.E.Shaw) opt for Putnam winners,etc ( leaving the fact they are good at math)?

And finally can anyone tell me what are the criterion for selection into D.E.Shaw?

[quote="D.E.Shaw Group Website"]The D. E. Shaw group offers opportunities for individuals of all backgrounds and levels of experience who display evidence of [b][color=blue]outstanding ability[/color][/b] and have a [b][color=blue]track record of exceptional accomplishment[/color][/b]. We hire gifted mathematicians and computer scientists as well as extraordinary generalists with liberal arts or other non-technical backgrounds. (Some of the firm's managing directors hold degrees in fields as disparate as English, psychology, and Russian studies.) Many of our employees have little knowledge of finance when they join the firm, but [b][color=blue]all share uncommon intelligence, analytical ability, and drive[/color][/b].[/quote]

It's so much that they're looking for math as it is that they're looking for brilliance of mind.

They need to make the best models, the best predictions, and the best decisions, and they need people who they think will be good for that. Obviously having a Putnam award isn't going to make someone a shoe-in to D.E. Shaw, and that certainly is not the first thing they look at. But oftentimes, people who have exhibited a high level of creativity and analytical ability through such competitions are the ones that are best suited for the job, for obvious reasons. However, there are myriad other ways of displaying uncommon intelligence, analytical ability, and drive. How do you think the Russian studies major got in? Certainly not through the Putnam.

You may want to explore the [url=http://www.deshaw.com/Recruiting.html]Job Opportunities section of deshaw.com[/url] to see what the various D.E.Shaw positions are like. Obviously they will hire those who they believe will be best suited for the relevant positions.