Let $x,y,z$ be positive real numbers satisfying the conditions: $x^2+y^2+z^2+2xyz=1$. Prove:

$a)\ xyz \le \frac18$.
$b)\ xy+yz+zx \le \frac34$.
$c)\ x^2+y^2+z^2 \ge \frac34$.
$d)\ xy+yz+zx \le 2xyz+\frac12$.

[b]Solution to a)[/b]
By AM-GM inequality:
$\frac{x^2+y^2+z^2}{3}\geq (x^2 y^2 z^2)^{1/3}$
Substitute the given condition into this inequality:
$\frac{1-2xyz}{3}\geq (x^2 y^2 z^2)^{1/3}$
Denote $u=xyz$ for convenience:
$\frac{1-2u}{3}\geq u^{2/3}$ $\implies (1-2u)^3\geq 27u^2$.
Expand LHS and re-arrange terms:
$8u^3+15u^2+6u-1\leq 0 \implies (u+1)^2(8u-1)\leq 0 \implies u \leq \frac 1 8$.
That is, $xyz\leq \frac 1 8$.

[b]and hence solution to c)[/b]
$x^2+y^2+z^2 = 1 - 2xyz \geq 1- 2 \cdot \frac 1 8 = \frac 3 4$

[hide="b"]

By AM-GM, we have $\displaystyle \frac{x^2+y^2}{2}\geq  xy$, $\displaystyle \frac{x^2+z^2}{2}\geq  xz$, $\displaystyle \frac{y^2+z^2}{2}\geq  yz$

Adding them together yields:
$x^2+y^2+z^2\geq xy + yz + xz$

From $c$, we have $x^2+y^2+z^2\geq \displaystyle \frac{3}{4}$

So it follows that $xy + yz + xz\leq \displaystyle \frac{3}{4}$

[/hide]

Notice also that $x$, $y$ and $z$ are the cosines of the angles of a triangle. This is sometimes useful.

[quote="JBL"]Adunakhor, your argument doesn't work.  You have $a > b$ and $a > c$ and you are telling me that from that I can deduce a relationship between $b$ and $c$.[/quote]

Oh whoops  :blush: Thanks for pointing that out.


Homer gives mathematicians Patty and Selma each a different integer, not known to
the other or to you. Homer tells them, within each other’s hearing, that the number given to
Patty is the product $ab$ of the positive integers $a$ and $b$, and that the number given to Selma is the sum $a + b$ of the same numbers $a$ and $b$, where $b > a > 1.$ He doesn’t, however, tell Patty or Selma the numbers $a$ and $b.$ The following (honest) conversation then takes place:
Patty: “I can’t tell what numbers $a$ and $b$ are.”
Selma: “I knew before that you couldn’t tell.”
Patty: “In that case, I now know what $a$ and $b$ are.”
Selma: “Now I also know what $a$ and $b$ are.”
Supposing that Homer tells you (but neither Patty nor Selma) that neither $a$ nor $b$ is greater than 20, find $a$ and $b$, and prove your answer can result in the conversation above.

[hide="Answer"]We can reword Patty and Selma's conversation to a mathematical statement: We have a pair of numbers $(a,b)$ such that $ab=x$ has more than one possible pair $(a,b)$, and $a+b=y$ has every possible pair $(a,b)$ such that the $x$ for each of these pairs has more than one possible pair $(a,b)$. The only numbers $y$ that satisfy this condition are those such that when $a=2$, $b$ is not prime, so we have, based on the condition that $20\ge b>a>1$, that $y={11,17,23,27,29,35,37}$. If we write out a list of the possible $x$-values from each of these $y$-values, the only value that appears only once is $52$, so we have $(a,b)=\boxed{(4,13)}$.[/hide]


1. Factorise x3 +4x2 +x − 6 using the factor theorem.
Hence solve the equation
x3 +4x2 +x −6=0

hello, we have $ x^3\plus{}4x^2\plus{}x\minus{}6\equal{}(x\minus{}1)(x\plus{}2)(x\plus{}3)$
Sonnhard.

So the solutions are $ x_1\equal{}1,x_2\equal{}\minus{}2$ and $ x_3\equal{}\minus{}3$

A little more explanation of the factorization:

[hide]
From the rational roots theorem, possible candidates for roots are $ \pm 1, 2, 3, 6$. We immediately try $ 1$ because it is the simplest to plug in.

Plugging in $ x\equal{}1$ gives the sum of the coefficients, which is $ 1\plus{}4\plus{}1\minus{}6\equal{}6\minus{}6\equal{}0$, very conveniently. So $ x\equal{}1$ is a root and $ (x\minus{}1)$ is a factor. Now we divide by $ x\minus{}1$ using synthetic division to get:

$ \frac{x^3\plus{}4x^2\plus{}x\minus{}6}{x\minus{}1} \equal{} x^2\plus{}5x\plus{}6$.

$ x^2\plus{}5x\plus{}6$ clearly factors into $ (x\plus{}2)(x\plus{}3)$.

So our original polynomial is $ (x\minus{}1)(x\plus{}2)(x\plus{}3)$, and the roots are $ x\equal{}1,\minus{}2,\minus{}3$.
[/hide]


conics
ellipse
geometry
parallelogram
trapezoid
geometry proposed

Prove that if two chords of an ellipse bisect each other, then they're diameters.

May I use affine transformations?

Well, if yes then it's trivial. Some affine transformations takes the ellipse to a circle, and then we are left with a circle and two chords in this circle which bisect each other. Call these chords AC and BD. Then, since the segments AC and BD bisect each other, the quadrilateral ABCD is a parallelogram. Now, a parallelogram can be considered as a trapezoid, and actually in two ways: on one hand, the sides AB and CD are the bases, and the sides BC and DA are the legs, and on the other hand, the sides BC and DA are the bases, and the sides AB and CD are the legs. Now, the quadrilateral ABCD is not just a parallelogram, it's a [i]cyclic[/i] parallelogram. Hence it can be considered as a [i]cyclic[/i] trapezoid in two ways. But a cyclic trapezoid is always isosceles; thus, if we consider the cyclic trapezoid ABCD with the bases AB and CD and the legs BC and DA, we get < DAB = < ABC and < BCD = < CDA, and if we consider the cyclic trapezoid ABCD with the bases BC and DA and the legs AB and CD, we obtain < ABC = < BCD and < CDA = < DAB. Together this yields < DAB = < ABC = < BCD = < CDA; thus, the quadrilateral ABCD has equal angles and is a rectangle. Hence, its diagonals AC and BD are diameters of its circumcircle. In other words, the chords AC and BD are diameters in their circle. Qed..

  Darij

Of course you're allowed to do that (that's what I did :)). I don't like statements sounding like this: "prove.. without using..".

I know a quick way and will show you later

[quote="darij grinberg"]May I use affine transformations?[/quote]
I'm sorry but if you explain the "affine transformations" to me, I'll be very thankful.

[quote="darij grinberg"]May I use affine transformations?

Well, if yes then it's trivial. Some affine transformations takes the ellipse to a circle, and then we are left with a circle and two chords in this circle which bisect each other. Call these chords AC and BD. Then, since the segments AC and BD bisect each other, the quadrilateral ABCD is a parallelogram. Now, a parallelogram can be considered as a trapezoid, and actually in two ways: on one hand, the sides AB and CD are the bases, and the sides BC and DA are the legs, and on the other hand, the sides BC and DA are the bases, and the sides AB and CD are the legs. Now, the quadrilateral ABCD is not just a parallelogram, it's a [i]cyclic[/i] parallelogram. Hence it can be considered as a [i]cyclic[/i] trapezoid in two ways. But a cyclic trapezoid is always isosceles; thus, if we consider the cyclic trapezoid ABCD with the bases AB and CD and the legs BC and DA, we get < DAB = < ABC and < BCD = < CDA, and if we consider the cyclic trapezoid ABCD with the bases BC and DA and the legs AB and CD, we obtain < ABC = < BCD and < CDA = < DAB. Together this yields < DAB = < ABC = < BCD = < CDA; thus, the quadrilateral ABCD has equal angles and is a rectangle. Hence, its diagonals AC and BD are diameters of its circumcircle. In other words, the chords AC and BD are diameters in their circle.[/quote]

I can't understand why we need such long proof. ;) :)
Suppose AB and CD intersect in p.X, then AX=XB and CX=XD, moreover AX*XB=CX*XD; it yelds AX=CX, hence AB=CD and ABCD is a rectangle. Since angle ACB=90 we conclude AB is a diametr. The same for CD.


Putnam
AMC
AIME
MATHCOUNTS
calculus
linear algebra

It seems to me that most math geniuses I see at school and everywhere else (by genius, I mean by people who do extraordinarily well on contests) started seriously doing math on a regular basis since junior high. I used to think that my math was pretty decent at the elementary and highschool [i]curricular[/i] level and therefore never actually pushed myself. Here I am in first year uni and I'm surrounded by amazing students in my advanced math classes, some of whom I believe have even been in the national team before (for Canada). And the discrepancy in math brainpower is immense; some questions I need to spend hours to do, while the really crazy ones only take maybe half an hour. I've also realized that my performance on contests is simply pathetic compared to theirs.
Given these circumstances, is there any hope left for me to make any significant achievement in math? (eg. top 10-percentile in putnam - i really don't know how hard that is)

I know of many successful mathematicians who never participated in any contests and of some contests winners who later either quit mathematics or turned out to be quite mediocre mathematicians. What really matters for success in mathematics is whether you love it and are ready to think of it most of your time for the rest of your life. Also, you'll have to challenge yourself a bit above your "comfortable level" all the time not allowing yourself to relax, so if you see that your "curriculum performance" is "decent", go ahead and ask for more (read advanced books, solve extracurricular problems, do some research projects, whatever). The secret to a real success (in addition to having some minimal level of talent) is exactly in "pushing yourself" beyond your limits all the time. If you do it, in a couple of years you will see hardly any difference between you and the "geniuses around" and by the graduation time you may be well ahead of some of them. Of course, it is true that the earlier you start "doing math on a regular basis", the better, but it is not really late to start doing it in college. I wouldn't worry much about the fact that you need several hours to solve a problem somebody else can solve in 30 minutes as long as you can solve it by yourself. Speed here comes with experience.

As to "doing well on contests", this is by no means the main thing in mathematics (after all, mathematics is a science, not a sport). The contest prizes are just useful as a stimul for the everyday work throughout the rest of the time and it is the experience and skills you acquire during your preparation for Putnam or whatever, not the performance on the contest itself, that really matter. If you ask me, you lose nothing if you go to all the training sessions and then do not participate in the contest at all, though, of course, it is always fun to test your "brain muscles" when you feel that they are in good shape.

So, instead of getting intimidated by the "superiority" of the people around, just try to do slightly better than your best and keep in mind that despite those who do exactly the same but had a head start will always be a few steps ahead, there will be no point on the road they will achieve that you'll not be able to get to just a short time later.

[quote="plucesiar"]The reason I referred to ppl who do extraordinarily well on contests is because I always get this vibe that contests are viewed more highly than the math marks you get in school. K81o7, can you tell me more about when you started seriously training?[/quote]

I don't worry so much about my math grades in school.  To me, math copmetitions are more for fun than anything else: they've got interesting problems with elegant solutions, and they provide friendly competition.  I haven't done much serious training, but I joined AoPS in April 2005 (towards the end of my freshman year of high school), and since then, competitions and AoPS have served to keep me supplied with interesting things to think about, provoke me to think of math questions I've never thought of before, and direct me to other such people.  Although people who do well on contests have their names up and are admired for their achievements, it doesn't really mean as much as it seems it does.  Do what you love.

[quote="K81o7"][quote="plucesiar"](by genius, I mean by people who do extraordinarily well on contests)[/quote]

Is that what a genius really is?  If that's what you want to be, I guess you'd just have to practice (as long as you have the talent and passion).  For example, I didn't seriously start any contest math until around my freshman year.  But really, not being good at contest problems is no setback.[/quote]

can't anyone be a genius? All you have to do is work hard and even people that get C's can make AIME (this weird junior guy at my school did that)

Yes it is too late because you have valued achievments and rewards more than improving yourself.

Well just because they do well in competitions doesn't mean they are geniuses. They could just understand math at a practical level (i'm talking about MATHCOUNTs type questions) whereas some "math geniuses" take calculus and linear algebra at an early age. It really depends on what your definition of a "math genius" is. It is NEVER too late to start studying on a daily basis, you might be behind a couple years but if you work hard you should be able to catch up with them  :wink:  :)


AMC
AIME
geometry
trigonometry
probability
number theory

Now that the AIME is over (for most people), I'm wondering what I should do to prepare for next year's becuase I totally failed this year's. Any suggestions?

Hi despair,

Well, considering I got a 2 my freshman year and a 9 (and a possible 13) this year...I'd say my practice was pretty succesful. Definately brush up on some number theory and obscure geometry and trig, because that will certainly help (such as for the "star" problem on the AIME1 this year). Also do some counting and probability because that appears a lot. Learn some recursive tricks, working backwards, etc...

Also, taking the course at AoPS isn't a bad idea at all. If you are doing poorly, I'm sure it would help a lot, and even if you aren't, it really couldn't hurt.


geometry
circumcircle
perpendicular bisector
geometry unsolved

Let $ A$ be one of the two points of intersection of two circles with centers $ X, Y$ respectively.The tangents at $ A$ to the two circles meet the circles again at $ B, C$. Let a point $ P$ be located so that $ PXAY$ is a parallelogram. Show that $ P$ is also the circumcenter of triangle $ ABC$.

Excelent problem.
Please check my solution because i am not sure about it.
Consider the circumcircle of $ ABC$ with circumcenter $ O$.Let $ Q$ the second intersection point of $ (X,Y)$.In order to prove that $ P$ is the center of the circle $ ABC$ we have to prove that: $ AB,AC$ is perpendicular to $ PX$ and $ PY$ respectively.(because AB,AC,AQ are the radical axes of $ (X,O),(Y,O),(X,Y)$,so we will have that $ O \equiv P$.
$ AQ$ is perpendicular at $ XY$,so we only need to prove that $ \angle BAQ(\equal{}x)\equal{}\angle YXP(\equal{}y)$.
But $ y\equal{}\angle XYA$ (because of the parallelogram).So it suffices to prove that $ \angle XYA\equal{}\angle BAQ$ which is true because these angles have their lines perpendicular.
So we prove that $ \angle BAQ\equal{}\angle YXP$
Analogously we can prove that $ \angle CAQ\equal{}\angle PYX$. :)

$ PY//AX, AC\perp AX$ then $ PY\perp AC \Rightarrow P$ lies on perpendicular bisector of $ [AC]$, similar for $ AB$. We are done!

[b]Happy the Independence Day of Vietnam![/b]  :gathering:

Let $AB \cap XP=L,AC \cap PY=M$,$\angle BAY=90$ so $\angle L=90$,hence $XP$ is the perpendicular bisector of $AB$,similarly $YP$ is also perpendicular bisector of $AC$.hence done.

[hide=400th post!]Mark $AB \cap XP=Q,AC \cap PY=R$, $AR \cap XY=Z$ , let $\angle{XAQ}=\alpha$ and $\angle{AXY}=\beta$\\
notice since $\angle{XAR}=90^{\circ}=\angle{QAY}$ we get $\angle{XAQ}=\angle{RAY}=\alpha$\\
also in $\triangle{XAZ}$\\
$\angle{XZA}=90^{\circ}-\beta$\\
so in $\triangle{ZAY}$\\
we get $\angle{AYZ}=90^{\circ}-(\alpha+\beta)$\\
as $AXPY$ is parallelogram we have $\angle{PXY}=90^{\circ}-(\alpha+\beta)$\\
so in $\triangle{XAQ}$\\
$\angle{XQA}=90^{\circ}$\\
hence $\angle{BQP}=90^{\circ}$\\
in similar fashion one can prove $\angle{CRP}=90^{\circ}$ , hence we get $P$ as circum-center of  $\triangle{ABC}$ $\blacksquare$[/hide]


Start voting, I guess... i just want to see how evry1's self esteem is shaping up

Oh yeah, and someone else start a follow - up poll on sunday.... i will be too busy/lazy to do it......

I'm not predicting myself since it leads to jinxes. So I wrote doing nothing.

I predict that... pie tastes good :P

why bother?  we might as well see how we do at the real competition

Oh yeah, and don't ask why the poll got messed up... I really don't know myself.......... :(

do nothing! (with the yay)

cuz i didn't make states!!!!!

what do u need for a trophy? .... I'm clueless, this is my first time

Different states have different trophy rules, I think, but usually getting in the top 4 teams or the top 6 individuals before or after countdown will get you a trophy.

I should get first before and after countdown, but I'll probably get third or fourth and then lose the countdown.

[quote="mathcrazed"]do nothing. I'm sucking this year[/quote]

OMG, mathcrazed i keep telling you: you will come either 1st, 2nd, or MAYBE on if you have a very bad day and everyone else gets really lucky, you might might be 3rd but i highly doubt it. i on the other hand have a $ 1.243134 \times 10^{\minus{}10123901248}$ probability of making it ill be like 30th or something.

At this point, I see that the most popular answer is make nationals..... hmmmmmmmmmmmmmmmmmmmmmmmmmmmmm................. oh yeah, who wants to start a follow - up poll on Sun, March 2nd?

[quote="xpmath"][quote="#H34N1"][quote="xpmath"]I will most likely make Countdown unless I screw up...don't know about nats though.[/quote]

You know you are going to win first.[/quote]
I bet you 5 dollars I won't.
You have no clue how many stupid mistakes I make on every test I take.[/quote]

I'll take a 50 dollar bet.

[quote="xpmath"]Uh, I don't have 50 dollars. How about we just do this for fun?[/quote]

I'll take a 1:50 odd. If you win, you pay me a buck. If I win, I pay you 50. :P

Uh, okay.
I don't really care, since I suck at hard problems, so if this is a hard year, I'll fail. If it's easy, I'll make stupid mistakes and still fail. Yay.

[quote="MathAndKnowledge"]...
cognos or mathcrazed, what do you expect the cutoff to be for top 10 state in Indiana?

I expect to get a 38 in state (yes, cognos, I know I did bad at chapter).  Ugh...  I can't believe my score on chapter.  But I forget that score for now and focus on the score that I must achieve this Saturday.

By the way I put that I'd get almost get into countdown.  Of course, I don't know what the countdown cutoff is for Indiana... If top 16, then I think I'd almost get there if not get there, and if top 8, then I have a miniscule chance.[/quote]

i do remember last year the cutoff for nats was like 38 or 37, in IN or something right mathcrazed? the cutoff for top 10 last year was like 33 or 34, because as you know my sister got a 32 and she was 13th. i dunno what the cutoff for cd was.

[quote="D3m0n Shad0w"]hmm, if? so you haven't taken chapters yet?[/quote]

Sorry im so late, um i have had chapters but we got 2nd place and were wating to see if we got a wild card spot!  LOL its taking forever~

[quote="ra5249"]top 4 team...
at least top 200 at state in indivdual i hope[/quote]
oh darn. i cant make top 200.
i expect to get 36-38 in state ( i average 42 on practice state mcs without the pressure of it being the real thing)

I pumped up for it. It's tomorrow and I just got a 43 on a practice XD

[quote="ra5249"]
at least top 200 at state in indivdual i hope[/quote]

OH COME ON RA5249 YOU KNOW YOU'RE GOING TO GET 2ND OR 1ST
i, on the other hand, will be bad.

dude im totally placing 1st(if you add a few zeroes to the end of it) i hope top 6!!! i think probably make stage!

[quote="abacadaea"]im going to go from 10th place to 5th place, and lose the 4th place spot and not make nats....[/quote]
Ouch...nasty prediction.

I did that last year...sorta. I went from not on the team :( to 5th place State, which became 6th after cd. Still, improvement is nice. I just want more than a little (i.e. make the Texas team and hopefully get first).

EDIT: By not on the team I mean not on the school team  :(  :( . and I of course want to majorly improve [i]this year[/i].

eww..i wouldnt want to be in that position

lol abacadaea... it's wo hui zhuo not zhou.. zhou dui lol? walk right xD...

good luck on getting a perfect on the AIME lol =D


if $ a,b,c,d$ $ \in R^{\plus{}}$ $ a,b,c,d<1$ and $ a\plus{}b\plus{}c\plus{}d \equal{} 2$ prove that
\[ \frac{2}{9}(\frac{1}{\sqrt{(1\minus{}a)(1\minus{}b)}}\plus{}\frac{1}{\sqrt{(1\minus{}a)(1\minus{}c)}}\plus{}\frac{1}{\sqrt{(1\minus{}a)(1\minus{}d)}}\plus{}\frac{1}{\sqrt{(1\minus{}b)(1\minus{}c)}}\plus{}\frac{1}{\sqrt{(1\minus{}b)(1\minus{}d)}}\plus{}\frac{1}{\sqrt{(1\minus{}c)(1\minus{}d)}})\]
\[ \geq \frac{1}{1\plus{}a}\plus{}\frac{1}{1\plus{}b}\plus{}\frac{1}{1\plus{}c}\plus{}\frac{1}{1\plus{}d}\]

we have
$ \frac {1}{\sqrt {(1 \minus{} x)(1 \minus{} y)}} \ge \frac {2}{2 \minus{} x \minus{} y}$
 so
$ \frac {2}{9}(\frac {1}{\sqrt {(1 \minus{} a)(1 \minus{} b)}} \plus{} \frac {1}{\sqrt {(1 \minus{} a)(1 \minus{} c)}} \plus{} \frac {1}{\sqrt {(1 \minus{} a)(1 \minus{} d)}} \plus{} \frac {1}{\sqrt {(1 \minus{} b)(1 \minus{} c)}} \plus{} \frac {1}{\sqrt {(1 \minus{} b)(1 \minus{} d)}} \plus{} \frac {1}{\sqrt {(1 \minus{} c)(1 \minus{} d)}}) \ge$
$ \frac {2}{9}( \frac {2}{2 \minus{} a \minus{} b} \plus{} \frac {2}{2 \minus{} a \minus{} c} \plus{} \frac {2}{2 \minus{} a \minus{} d} \plus{} \frac {2}{2 \minus{} b \minus{} c} \plus{} \frac {2}{2 \minus{} b \minus{} d} \plus{} \frac {2}{2 \minus{} c \minus{} d})$
$ \equal{} \frac {2}{9}( \frac {2}{a \plus{} b} \plus{} \frac {2}{a \plus{} c} \plus{} \frac {2}{a \plus{} d} \plus{} \frac {2}{b \plus{} c} \plus{} \frac {2}{b \plus{} d} \plus{} \frac {2}{c \plus{} d}).$
and we have $ (x \plus{} y \plus{} z)(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}) \ge 9$  so $ \frac {2}{a \plus{} b} \plus{} \frac {2}{a \plus{} c} \plus{} \frac {2}{a \plus{} d} \ge \frac {18}{3a \plus{} b \plus{} c \plus{} d} \equal{} \frac {18}{2} (\frac {1}{a \plus{} 1})$
and $ \frac {2}{a \plus{} b} \plus{} \frac {2}{b \plus{} c} \plus{} \frac {2}{b \plus{} d} \ge \frac {18}{3b \plus{} a \plus{} c \plus{} d} \equal{} \frac {18}{2} (\frac {1}{b \plus{} 1})$
$ \frac {2}{c \plus{} b} \plus{} \frac {2}{a \plus{} c} \plus{} \frac {2}{c \plus{} d} \ge \frac {18}{3c \plus{} b \plus{} a \plus{} d} \equal{} \frac {18}{2} (\frac {1}{c \plus{} 1})$
 and   $ \frac {2}{a \plus{} d} \plus{} \frac {2}{d \plus{} c} \plus{} \frac {2}{b \plus{} d} \ge \frac {18}{3b \plus{} b \plus{} c \plus{} a} \equal{} \frac {18}{2} (\frac {1}{d \plus{} 1})$


vector
abstract algebra
combinatorial geometry
real analysis
real analysis unsolved

Let $f: E\to R$ be a linear application,where $E$ is a real vectorial normed subspace and let $H=Ker(f)$.Show that $f$ is continuous iff $E$ \ $H$ is not connexe par arcs.

I'll assume the functionals $f$ we're concerned with are not identically null. I'll also use the well-known characterization of continuous linear functionals on a normed vector space: a linear form $f: E\to\mathbb R$ is continuous iff its kernel is not dense.

If $f$ is continuous, then $E_+=f^{-1}((0,\infty))$ is open in $E$, and so is $E_-=f^{-1}((-\infty,0))$. $E\setminus H$ is the union of the two non-void, open and disjoint sets $E_+,E_-$, so it's not even connected, let alone path connected.

We now see what happens if $f$ is not continuous. Take any $x\in E_+$ and $y_0=y\in E_-$. If $(z_n)_{n\ge 1}\subset H$ is a sequence of points converging towards $x$ (we use the characterization mentioned in the first paragraph), then $(y_n)_{n\ge 1}=(2z_n-x)_{n\ge 1}\subset E_-$ is a sequence of points in $E_-$ converging to $x$. For every $n\in\mathbb N,\ n\ge 2$ map the interval $\left[0,1-\frac 1n\right]$ onto the segment $[y_{n-2},y_{n-1}]$ (the convex hull of the points $y_{n-2},y_{n-1}$ in $E$) continuously (this is clearly possible), and take $f(1)=x$. $f$ is a path in $E\setminus H$ from $y$ to $x$. The other cases, when $y,x$ both lie in $E_+$ or in $E_-$ is trivial, since the segment between them lies in the same set and it's a path between them. We have thus shown that $E\setminus H$ is path-connected.


function
limit
real analysis
real analysis unsolved

I have a few conceptual questions on limits that i need help with..

1. A student in your class says, "The limit of the sum of two functions is the sum of the limits of the functions." When is the statement not correct?

I'm not sure. I thought it was always correct because doesn't one of the theorems actually say that the limit of the sum of two functions is the sum of the limits of the functions. Is there an exception i'm missing?

2. Express the limit of (x-1)^3 as x approaches 3 in two different ways.

hmm.. i'm not sure what they want right here

On #1: there's another implicit assumption there. What are the hypotheses of that theorem?

[hide="Answer"]$\lim f+g=\lim f+\lim g$ if both $\lim f$ and $\lim g$ exist.[/hide]


logarithms
calculus
algebra
polynomial
irrational number

If an irrational number is raised to an irrational power, is the
resultant number always irrational?

Prove/disprove with examples.

this is quite difficult in some cases...look at wolfram on this topic...

[quote="Kurt Gödel"]You could also do something like this: if $\sqrt{2}^{\sqrt{2}}$ rational, we found a counterexample, otherwise consider $\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}}= 2$, which is a counterexample if $\sqrt{2}^{\sqrt{2}}$ isn't one.[/quote]

Edit: Dumb mistake.

i think $\sqrt{2}^{\log_{2}9}$ is the simpliest ;)

[quote="lingomaniac88"]$\pi i$ isn't irrational.  It's imaginary.

I think that $e^{\ln 2}=2$ is one of the simplest answers.[/quote]

Probably a dumb question, but how do you know $\ln 2$ is irrational?

[quote]  how do you know $\ln 2$ is irrational?[/quote]

Don't know of an easy proof but it can be proved (with the help of calculus) that  for any non-zero rational $p$ , $e^{p}$ is irrational.

For example look at Theorem 2, corollary 4  in the link:
[url=http://numbers.computation.free.fr/Constants/Miscellaneous/irrationality.html]link here[/url]

So with some what circular logic  (:) )
====>  since, $e^{\ln 2}$ is rational, $\implies \ \ \ln (2)$  is irrational :).

No such problem with the one in Kurt Gödel's, in fact there is some beauty here.

Another easy example:

[hide]$a=\sqrt{10}$  (easy to prove it is irrational)   (how?)
$b = \log_{10}(4)$  (again easy to prove it is irrational)  (how?)
$a^{b}$ is rational (value ?)[/hide]

http://numbers.computation.free.fr/Constants/Miscellaneous/irrationality.html
that's very interesting

I heared that even more : $e^{p}$ is transcendental for rational $p\neq 0$
have you seen the proof somewhere ?
i looked for it many times but i couldnt find

An algebraic number raised to a rational power is also algebraic. Therefore, it suffices to prove that [url=http://planetmath.org/encyclopedia/EIsTranscendental2.html]$e$ is transcendental[/url].

[quote="lingomaniac88"]$\pi i$ isn't irrational.  It's imaginary.[/quote]
It's imaginery _and_ irrational.

To $e$, there's the theorem of Lindemann-Weierstrass:
If $a_{1},a_{2},...,a_{n}$ are different algebraic numbers and $b_{1},b_{2}, ...,b_{n}$ are algebraic and not all $=0$, then
\[b_{1}e^{a_{1}}+b_{2}e^{a_{2}}+...+b_{n}e^{a_{n}}\neq 0. \]
Or equivalently: if $c_{1},c_{2},...,c_{n}$ are algebraic and linearly independent (meaning that there are no $r_{1}, r_{2}, ..., r_{n}\in \mathbb{Q}$, not all zero, with $r_{1}c_{1}+r_{2}c_{2}+...+r_{n}c_{n}=0$),
then $e^{c_{1}}, e^{c_{2}}, ...,e^{c_{n}}$ are algebraically independent (meaning that there is no nonzero rational polynomial $f(x_{1},x_{2},...,x_{n})$ such that $f(e^{c_{1}}, e^{c_{2}}, ...,e^{c_{n}})=0$).

This proves the transcendence of $\pi,e^{r},\pi,\mathrm{ln}(r),\sin(r)$ for $r \neq 0$ rational (or even algebraic) and a lot more.

[quote="ZetaX"][quote="lingomaniac88"]$\pi i$ isn't irrational.  It's imaginary.[/quote][/quote]
Oh. Okay.


function
limit
calculus
integration
real analysis
real analysis unsolved

Let $f: [a,b]\rightarrow \mathbb{R}$ be a function such that $\forall x_0 \in \mathbb{R} \text{ , } \exists \lim_{x \to x_0}f(x)<\infty$. Prove that f is Riemann integrable on [a,b].

Hint[hide="click text"]show that the function $g(x)=\lim_{y \to x}f(y)$ is continuous.[/hide]

Is $f: x \rightarrow \frac{1}{x}$,$f(0)=0$ Riemann-integrable on $[0,1]$ ?

In this usage, that limit has to be finite, and the function is defined on a closed interval. Your example does not satisfy the condition at zero, julien.

A (properly) Riemann integrable function has to be bounded- this limit condition should imply that.

Fix an $x_0$ arbitrarly. $\forall \epsilon>0, \exists \delta$ such that $|g(x_0)-f(x)|<\frac \epsilon 2, \forall |x-x_0|<\delta$     (1).
Assume that $\lim_{x\to x_0}g(x)\neq g(x_0)$. There will exist a sequence $(x_n)$ and $\alpha >0$ such that $x_n \to x_0$ and $|g(x_n)-g(x_0)|>\alpha$  (2)  $\Leftrightarrow|\lim_{x \to x_n}f(x)-g(x_0)|>\alpha$ . Now choose $x_n$ such that $|x_n-x_0| < \frac \delta 2$ . Since $\lim_{x \to x_n}f(x)=g(x_n)$ we have that for $|x-x_n|<\delta_1<\frac \delta 2$ , $|g(x_n)-f(x)|<\frac \epsilon 2$   (3) From (1) and (3)  we have that $|g(x_0)-g(x_n)|<|g(x_0)-f(x)|+|f(x)-g(x_n)|<\epsilon, \forall |x-x_n|<\delta1$. But from (2) we have that $|g(x_0)-g(x_n)|>\alpha$.Choosing $\epsilon$ sufficiently small we have a contradiction. So, g is continuous.
The next step is to prove that $h=f-g$ in continuous almost everywhere. The boundness of f is obvious.

jmerry: yeah I know  :P  ... I was actually trying to get a reaction from the author because I couldn't prove the statement  :D . Unfortunately, the part he has proved is trivial, and the interesting point is to show the equality a.e

you have to assume $f$ is bounded and that would imply in particular that the limits are finite, and not the other way around.

If  $\lim_{ x \to x_0}f(x)< \infty$ for all $x\in[a,b]$  I will prove that f is bounded.

Assume that f is not bounded. One can find a sequence $(x_n)\subset [a,b]$ such that $f(x_n) \to \infty$. Since the sequence $(x_n)$ is bounded t has a convergent subsequence. Denote it by $y_n \to y\in [a,b]$ and $f(y_n) \ to \infty$. contradiction. Why I'm wrong? Why I need to give the boundness of the function instead of the boundness of the limit in every point?

you are correct, you just proved that one implies the other, but it was not obvious to me.

Xirti, what book are you talking about? Is it my book of my father's? :)

The book is \emph{"Mircea Ganga-Teoreme si probleme de matematica-editura Tehnica 1991,colectia SERIE}. A romanian book (for those who didn't know it). By the way.. which is your book? I only have a red book "Tudorel lupu-Probleme de analiza matematica, Calcul integral". (another romanian book). Can you recomend any other book?

Yes, that's my father's book. Another book ( a very recent one :D ) is this one: "Probleme de Calcul Integral" by
Radu Miculescu, Corina Ciobotaru, Alexandru Rosoiu, Narcisa Banda and Cezar Lupu from GIL publishing house. :)  ;)

another way of seeing it (it might be there already and i could be overlooking it): the function has only jump discontinuities and hence countably many points of discontinuity; being bounded on top of it it must be r. integrable.

I do not know any elementary proof of the fact that any function which has only jump dicontinuities has a numerable set of points of dicontinuity.

I finished the proof of my problem.

Take $A_n=\{x|h(x)>\frac 1 n\}$. Assume that $A_n$ is not finite. Since $A_n$ is bounded it might have an "accumulation"(i'm not sure about this word :blush: ) point $a_n$. Let $(x_m)$ be a sequence such that $x_m\to a$ and $h(x_n)>\frac 1 n$, so $\lim_{x\to a}n(x) \ge \frac 1 n$. From the definition of h, we have that $lim_{x\to a}h(x)=0$. Contradiction.
Since every $A_n$ is finite, we have that $\cup_1^{\infty}A_n$ is numerable. Since $h=0,\forall x \in [a,b]\setminus \cup_1^{\infty}A_n$ we have that $h=0$ amost every where, so h is riemann integrable. And the problem is finished. :clap2:


[color=blue]Llegué, en un enunciado, a la siguiente ecuación funcional:
$ f: \mathbb{R}\to \mathbb{R}$
$ f\left({f\left( x \right)}\right) = f\left( x \right)$

Hallé, por el momento, estas soluciones:
$ f\left( x \right) = x$
$ f\left( x \right) = \pm \left| x \right|$
$ f\left( x \right) \equiv k$

Necesito alguna forma de demostrar que no existen más soluciones para la misma (intenté bastante tiempo, sin éxito).
Agradezco por adelantado.[/color]

Las funciones que buscas son llamadas [b]idempotentes[/b], hay infinitas funciones idempotentes, incluso se puede demostrar que hay infinitas familias de funciones idempotentes, cada familia con cierta propiedad. Ahora, las que tú has encontrado corresponderían (quizás) a todas las funciones idempotentes [b]continuas[/b].
Acá te muestro cómo generar infinitas funciones idempotentes (que no son continuas):
Sea $ \lambda$ un  real no nulo, definimos $ f$ por $ f(0)=0$ ,$ f(x)=\lambda$ cuando $ x \neq 0$, se comprueba fácilmente que esa función es idempotente, además hay tantas de ese tipo como $ \lambda$ reales no nulos hay, es decir, una cantidad no numerable de funciones.
Por otro lado, si esa función te parece "rara", puedes comprobar que la función [b]máximo entero[/b] también es idempotente.
Espero haberte sido de ayuda.

$ Tipe$

[color=blue]Muchas gracias, $ Tipe$.
Noto que hay una enorme cantidad de funciones idempotentes.
Entonces, sería importante demostrar en mi enunciado que la función es continua, o bien inyectiva o sobreyectiva.
Envío el documento en el cual he puesto mi razonamiento (Espera… Parece que no se puede. ¿Debo crear un nuevo tópico para adjuntar un archivo?).
Ruego que, cuando envíe el enunciado, no digan directamente la solución, sino que me den consejos, así como lo hizo $ Tipe$.

Muchas gracias.[/color]


MATHCOUNTS ARML USAMTS AMC AIME USA(J)MO USAMO

Hello, here put legitimate goals on contests like the AMC Contests, MATHCOUNTS, ARML, USAMTS, Purple Comet, etc.

To set a nice example, I'll post mine

AMC 8: 25
AMC 10: 150
AIME: 12
USAMO: 8
USAMTS: 80
MATHCOUNTS State written: 2nd
masters (if there is one): 1st
cd: 16th
Team:............ The world may never know
MATHCOUNTS Nats written: cd
Team: top 30
ARML: 8
UNM PNM: First overall

Notice how some of these are grossly high... yeah, that's because if you set your goals high, even if you make only a fraction of it, it will still probably be more than if you reached your low goals. And, some of these (believe it or not) actually have some backbone. like for MC State, these things MAY actually occur (although, i hope that i get better than 16th in cd  :maybe: )

k bleh here are teh lowly things

AMC10 139.5
AIME 6
Mandelbrot Individual 22
Mandelbrot Team 36
ARML 5
UNM PNM 1st division, 6th overall
Purple Comet 17

@pinky: how is that a reasonable AMC10 score? you can easily beat that...

... hmmmmmm... thinking again i may not be able to make nats cd

Also, I'm wondering if anyone wanted an oly marathon... because the mc level and up to AIME level kinda died, so we should make one that challenges us!! (don't worry, they will only go up to maybe USAMO #2... and it could even be subject specific...)

but anyways, here is a problem that is pretty easy :)

Find $ s(s(s(4444^{4444})))$.

(Note: $ s(n)$ is n's digit sum)

[hide="Hint"]Bound from above.[/hide]


geometry
3D geometry
induction
blogs
AMC
AIME
MATHCOUNTS

Is there a way to prove the sum of squares formula without induction?

[quote="math92"]what's induction? :blush:[/quote]

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42305

Don't worry though.

As far as I know, induction is necessary, because without it, or without something else that was proved using induction, one cannot prove that the formula or rule applies for every number.

The "proof without induction" probably requires identities that themselves need induction to be proved.

I proved it the long way [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=2166]here[/url]

While you're there, you can put some smilies in the shoutbox  :D

[quote="shobber"]Use this formula: $m^3-(m-1)^3=3m^2+3m-1$ to prove the square sum.
Similarly use $m^4-(m-1)^4$ to prove the cube sum.[/quote]

Can you show us the proof?  :)

[quote="anirudh"]It is impossible to do it without induction since induction is using math.  :D[/quote]

you sure?  :?

236factorial, your proof is interesting, and a lot of admirable hard work was obviously put into it, but the point on which the whole proof hinges is, as you said, the fact that
[quote="236factorial"]The fourth differences are the same. that means the polynomial form is $an^4+bn^3+cn^2+dn+e$[/quote]
which can only be proven using induction, or something else which requires induction to prove.  Furthermore, only induction can prove that all the fourth differences will be the same.

Nat mc's proof is also valid (and closer to the one I would use, although far from the same), but he requires induction to prove the statements that the representation of $x^2$ as the sum of the first $x$ odd integers, and to prove this : [quote="nat mc"]Then continue that until there are $2$ $2x-3$'s and $1$ $2x-1$.[/quote]

The proof suggested by shobber is the one I used to find the formula, but it requires induction to verify.

What we often see as common sense, e.g. what I have quoted of nat mc above, often requires induction to prove.

[quote="mathnerd314"]I thought I've seen a proof without induction before... but like Fermat, I don't have it.[/quote]
[quote="Boy Soprano"]As far as I know, induction is necessary[/quote]
Um, just look at my proof. It does not use induction. ?There are many easy ways of proving this. Telescoping, arts and crafts, gometry.
[quote="JesusFreak"]Does the picture proof on the left of the screen use induction? I don't entirely understand the induction thing; I might know what it is, but not have learned it by name.[/quote]
No, induction is where you prove that it works for one value, then you show that if it works for x, it also works for x+1

Here it is using induction:
plug in $1$,  $\frac{(1)(2)(3)}6=1$ True. Now The formula is
$\frac{x(x+1)(2x+1)}6$. Add $(x+1)^2$ and you get $\frac{(x+1)(x+2)(2x+3)}6=\frac{(x+1)((x+1)+1)(2(x+1)+1)}6$

[quote="Boy Soprano"]See my post above.  Note that deriving a formula (i.e., showing how you got it) and proving it (i.e., showing that it is true beyond doubt) are not necessarily the same.

"Telescoping, arts and crafts" are not adequate mathematical proof.  Nothing in the "real world" is.  The "real world" can provide applications that appear to confirm or contradict mathematical principles, but it cannot provide proof of any variety.  One could use a geometric derivation of the formula, but it would require induction for the proof.  Alternatively, one could use a geometric proof, but induction or a principle requiring induction to be proven would have to be included.  As I have demonstrated, the same goes for algebraic proofs.  The simple fact of the matter is that we cannot [i]prove[/i] that anything continues indefinitely without induction.

As a matter of fact, the side-bar picture does require induction to show that the sequence will continue indefinitely, just as the picture of the Pythagorean theorem requires a little geometry to show that it applies for all cases.

Many people find induction proofs to be unsatisfying, because it does not show how we got the answer.  That is why we show how we derive the formula also, or prove the formula using simpler consequences of the axiom of induction.[/quote]

You are thinking too deep. I think 236factorial was asking a simple question: Can we prove that formula without using [i]classroom[/i] mathematical induction? I remember this equation but I forget where I saw this: $\text{Natural numbers}=\text{Induction}$, but I don't think this is what 236factorial is looking for...

[quote="Boy Soprano"]236factorial, your proof is interesting, and a lot of admirable hard work was obviously put into it, [/quote]

Sorry to break your compliment, but I did it in the bathroom  :rotfl:

What's that equation natural numbers = induction?  :?

Does classroom mathematical induction mean "doing as many cases as you can until you get tired" or real mathematical induction?

Induction only works for the natural numbers. Proof:

Suppose it didn't. Then we could construct this false proof:
1 is an integer (definition of 1)
If k is an integer, then k+1 is an integer (integers are closed under addition)
Therefore all numbers are integers. QED

:rotfl:

236factorial's proof was exactly the one I was trying to remember, that I saw at Mathpath.

[quote="MCrawford"]In the AIME Problem Series we work this problem and there is in fact a nicer solution that results from the fact that the original equations can be expressed in a single common form.  Try rewriting all the equations at once using a parametric variable.  Do you see what options that gives us?[/quote]

Is that a little too advanced (on my blog)? AIME isn't Mathcounts level, so....  :huh:

$\sum\limits_{k = 1}^n {k^2 }  = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6},\forall n \in \mathbb{N}^* $

[color=red][b][u]Solution:[/u][/b][/color]

$\left( {k + 1} \right)^3  = k^3  + 3k^2  + 3k + 1,\forall k \in \mathbb{N} \Rightarrow $

$ \Rightarrow \left. \begin{array}{l}
 2^3  = 1^3  + 3 \cdot 1^2  + 3 \cdot 1 + 1 \\
 3^3  = 2^3  + 3 \cdot 2^2  + 3 \cdot 2 + 1 \\
 4^3  = 3^3  + 3 \cdot 3^2  + 3 \cdot 3 + 1 \\
  \cdots  \cdots  \cdots  \cdots \cdots  \cdots  \cdots  \cdots  \cdots  \cdots  \\
 \left( {n + 1} \right)^3  = n^3  + 3 \cdot n^2  + 3 \cdot n + 1 \\
 \end{array} \right\}\mathop  \Rightarrow \limits_{\mathop {\left( {\sum {} } \right)}\limits^ \uparrow  } $

$ \Rightarrow \left( {n + 1} \right)^3  = 3 \cdot \sum\limits_{k = 1}^n {k^2 }  + 3 \cdot \sum\limits_{k = 1}^n k  + \left( {n + 1} \right) \Rightarrow $

$ \Rightarrow \sum\limits_{k = 1}^n {k^2 }  = \frac{{\left( {n + 1} \right)n\left( {2n + 1} \right)}}{6}$

That looks very nice  :D , but I guess I'm not familiar with that $\sum$ symbol (I didn't get the last step).

[quote="236factorial"]Actually, I knew what it meant, but I don't get the last two lines of the solution.[/quote]
How could you not know?
Let $X=\sum_{k=1}^n k^2$ and $Y=\sum_{k=1}^n k$.

$(n+1)^3=1^3+3X+3Y+n$. Solve for $X$. You must know formula for $Y$ OK? :)

you don't have to use induction, what Shobber said is absolutly true....and it proves the formula very easily.

[quote="JesusFreak197"]That's absolutely true, actually, but your conclusion is incorrect, because you have the initial condition that $k$ is an integer.[/quote]

That's my point. My point was that if you drop the premise that k must be an integer initially, you can prove that all real numbers are integers. Obviously absurd.

you can prove it using combinatorics too i think

it was in the AOPs book, goes something like this

$\displaystyle\sum_{n=1}^{\infty} n^2 = \sum_{n=1}^{\infty} i_2 \binom{n}{2} + \sum_{n=1}^{\infty} i_1 \binom{n}{1}$

$n^2 = i_2\displaystyle\frac{n(n-1)}{2} + i_1(n)$

When $n=1$

$1 = i_1$

When $n=2$

$4 = i_2 + 2i_1 = i_2 + 2$

$2 = i_2$

Plugging this back in, you have:

$\displaystyle\sum_{n=1}^{\infty} n^2 = \sum_{n=1}^{\infty} 2\binom{n}{2} + \sum_{n=1}^{\infty} \binom{n}{1}$

$LHS = 2\binom{n+1}{3} + \binom{n+1}{2}$

$LHS = 2\displaystyle\frac{(n+1)(n)(n-1)}{6} + \frac{(n+1)(n)}{2}$

$LHS = (n+1)(n)\Big(\frac{2(n-1) + 3}{6}\Big)$

$LHS = \displaystyle\frac{(n+1)(n)(2n+1)}{6}$

now what i want to kno is why this method works cuz i nvr figured that out

so can ne1 help me out here?

i kno this method works for finding the sum of cubes, and etc

Excuse me, you don't take sums to infinity, gotztahbeazn.

[quote="gotztahbeazn"]oh oops.... is it supposed to be like

$\displaystyle\sum_{i=1}^{n} i^2$?

yea i was doing the proof thing off the top of my head[/quote]

Yes it is. or $1^2+2^2+3^2+4^2+...+n^2$


calculus
integration
LaTeX
function
calculus computations

Hey, I came across this formula, sum(i^(k+1))=(k+1)*int(sum(i^k))+an, as an integration method to find the general formula for the summation of a series of numbers raised to the kth power using integration. I apologize for writing this in ASCII and not having installed LATEX, being a new member, but anyways, can anyone show me exactly how this formula was derived? I have some ideas but not any satisfactory proofs. :( Note that the "an" term is NOT in the integrand. The variable "a" is similar to the constant C after one integrates. For example, sum(i^2)=2int(n^2/2+n/2)=(1/3)n^3+(1/2)n^2+an

Plug n=1, so the sum of the first square is 1.

1=(1/3)+(1/2)+a.......... a=1/6

sum(i^2)=(1/3)n^3+(1/2)n^2+(1/6)n=n(n+1)(2n+1)/6

You don't need $\LaTeX$ installed to use it here; just put your code inside dollar signs, and the forum software will interpret it.

I don't really understand where that formula is coming from, but I have a completely different formula:
$\sum_{j=1}^nj^k=\frac{n^{k+1}}{k+1}-n^k\sum_{m=0}^{k-1} \frac{\zeta(-m)n^{-m}}{\binom km}$.

This does imply your version; since $\frac{\binom km}{\binom {k+1}m}=\frac{k+1-m}{k+1}$, your integration and multiplication matches this. On the other hand, the derivation of my formula is a complicated algebraic trick, with the $\zeta$ function a later tweak.

haha... I don't really understand your formula either. I didn't recognize that greek symbol at first, found out it was the zeta function but the math world article on it just kinda blew me away at first glance because I didn't know half of what the other terms they were incorporating into the article meant. But thanks anyways.


show that :a,b,c:a,b,c>0 .
$\to$:
 $\sum\frac{b2}{c}\geq\ a+b+c+\frac{4(a-b)^{2}}{a+b+c}$
help me!

[quote="The soul of rock"]Just use $\frac{1}{a}+\frac{1}{b}\ge \frac{4}{a+b}$ and it's in BMO 2005?[/quote]

Yeah!
You can see here:
http://www.mathlinks.ro/Forum/resources.php?c=1&cid=19&year=2005&p=225004

[quote="The soul of rock"]Just use $\frac{1}{a}+\frac{1}{b}\ge \frac{4}{a+b}$ and it's in BMO 2005?[/quote]
thank you very much !
[hide]Fitim
I find it .[/hide]


It's $\geq 3$, not $\leq 3$. See

http://www.mathlinks.ro/Forum/viewtopic.php?t=53585
and http://www.mathlinks.ro/Forum/viewtopic.php?t=14567 (rather indirectly - see post #7 and the subsequent ones of this topic).

  Darij


How do I find all solutions in the interval 0 deg<x<360 deg in the equation 4cosx-2secx=0?

I just figured it out.  It's + or - at the end because of the sqrt.  Stupid me.

Oh score, that means I got it right? Self esteem + 5. :D

-BLt


[b]Source:[/b]Maritime Mathematics Competition 2006 Question #5

Let a "square pair" be defined as a set of integers which have a sum that is a square of an integer and a product which is a square of an integer.  For example, $(5,20)$ is a square pair because $5 +20=5^2$ and $5*20=10^2$.

Prove that no square pair exists which contains the number $3$.

Assume that a squarepair contains the number $\ 3$.  Let the other number be $\ n$.  So the pair is $\ (3,n)$.  So now we know $\ 3 + n = a^2$ and $\ 3a = b^2$.  Now we know that $\ n=4k + 1$ or $\ n = 4k + 2$ because squares must be congruent to $\ 0$ or $\ 1$ modulo $\ 4$.  Now applying that to the second condition of the problem we get $\ 12k + 3 = b^2$ or $\ 12k + 6 = b^2$ which is a contradiction, because once again squares are only congruent to $\ 0$ or $\ 1$ modulo $\ 4$.


On parallelogram ABCD, the line that bisects <ABC is extended until it meets line AD at point P. If DP=5, BP=6, and CP=6, what is AB?

[hide="I think"]The parallelogram is just a rectangle and AB would then be $\sqrt{11}$.[/hide]

That's not it either.

[hide="Hint"]Similar triangles[/hide]

[hide="haha, duh"]
Since BP bisects $\angle{ABC}$, we have

$\angle{ABP}=\angle{PBC}=\angle{PCB}=\angle{APB}$ because $\triangle{PBC}$ is isosceles and $\angle{PBC}$ and $\angle{APB}$ are alternate interior angles.  Therefore, $\triangle{APB}$ is similar to $\triangle{PCB}$.  So

$\overline{AB}=\overline{AP}$

$\overline{AP} + 5 = \overline{BC}$

$\overline{AB}=\frac{36}{BC}$

Set $\overline{BC}=x$

$\frac{36}{x}+5=x$

$x^2 - 5x - 36 = 0 \rightarrow x=9$

$AB = \frac{36}{x} = 4$[/hide]

Correct answer... and a much nicer way than mine: (Nasty angle-chasing). :)

dakryu's solution was for when P is inside parrellogram, I think mine's for when P's on an extension on AD :D

Oops... the problem stated P is in between A and D. Sorry.


One of the example problems in Aops asks,
"How many zeroes are there at the end of 103!?"

I don't quite understand the explanation they give.  Can someone post their solution and explain it a little?

well this was always a nice problem when i was in high school when they wanted the number of zeros in the end of the factorial of a number...
anyway notice that  a $0$ in the end of a number is produced by $2\times5$, for example $2050=2\times5^{2}\times41$, the number of $2$'s (lets name it $A$) is $1$ and the number of $5$'s (lets name it $B$) is $2$, and the number of zeros in the end of $2050$ is the number of multiplications $2\times5$ which is obviously:  $\left|A-B\right|=1$

so for a number $n$ we define:

so $A=\text{the biggest nonnegative integer }k \text{ such that}n \text{is divisible by }2^{k}$
$B=\text{the biggest nonnegative integer }l \text{ such that}n \text{is divisible by }5^{l}$

well we know that the biggest nonnegetive integer $m$, such that $n!$ is divisible by $p^{m}$ where $p$ is a prime number is:

$\sum^{\infty}_{i=1}\left[\frac{n}{p^{i}}\right]=\left[\frac{n}{p}\right]+\left[\frac{n}{p^{2}}\right]+\cdots$

so  for $n=103$,
$A=\left[\frac{103}{2}\right]+\left[\frac{103}{2^{2}}\right]+\left[\frac{103}{2^{3}}\right]+\left[\frac{103}{2^{4}}\right]+\left[\frac{103}{2^{5}}\right]+\left[\frac{103}{2^{6}}\right]+0+0+\cdots=50+12+6+3+1=72$

$B=\left[\frac{103}{5}\right]+\left[\frac{103}{5^{2}}\right]+0+0+\cdots=20+4=24$

as a consequence the number of zeros in the end of $103!$ is $\left|24-72\right|=48$

 :)

[quote="sinajackson"]well this was always a nice problem when i was in high school when they wanted the number of zeros in the end of the factorial of a number...
anyway notice that  a $0$ in the end of a number is produced by $2\times5$, for example $2050=2\times5^{2}\times41$, the number of $2$'s (lets name it $A$) is $1$ and the number of $5$'s (lets name it $B$) is $2$, and the number of zeros in the end of $2050$ is the number of multiplications $2\times5$ which is obviously:  $\left|A-B\right|=1$

so for a number $n$ we define:

so $A=\text{the biggest nonnegative integer }k \text{ such that}n \text{is divisible by }2^{k}$
$B=\text{the biggest nonnegative integer }l \text{ such that}n \text{is divisible by }5^{l}$

well we know that the biggest nonnegetive integer $m$, such that $n!$ is divisible by $p^{m}$ where $p$ is a prime number is:

$\sum^{\infty}_{i=1}\left[\frac{n}{p^{i}}\right]=\left[\frac{n}{p}\right]+\left[\frac{n}{p^{2}}\right]+\cdots$

so  for $n=103$,
$A=\left[\frac{103}{2}\right]+\left[\frac{103}{2^{2}}\right]+\left[\frac{103}{2^{3}}\right]+\left[\frac{103}{2^{4}}\right]+\left[\frac{103}{2^{5}}\right]+\left[\frac{103}{2^{6}}\right]+0+0+\cdots=50+12+6+3+1=72$

$B=\left[\frac{103}{5}\right]+\left[\frac{103}{5^{2}}\right]+0+0+\cdots=20+4=24$

[b]as a consequence the number of zeros in the end of $103!$ is $\left|24-72\right|=48$[/b]

 :)[/quote]

that last part isnt right..you dont need that last part.  the answer is just 24 (the number of 5s that divide it)

As said by da ban man, in problems like these, the 2's become irrelavent.
since for a 0, you need 10, or 2*5, you need a 2 for every 5. there are obviously more 2's than 5's, so you just need how many 5's.

[quote="sinajackson"]well this was always a nice problem when i was in high school when they wanted the number of zeros in the end of the factorial of a number...
anyway notice that  a $0$ in the end of a number is produced by $2\times5$, for example $2050=2\times5^{2}\times41$, the number of $2$'s (lets name it $A$) is $1$ and the number of $5$'s (lets name it $B$) is $2$, and the number of zeros in the end of $2050$ is the number of multiplications $2\times5$ which is obviously:  $\left|A-B\right|=1$

 :)[/quote]

Oh my God what a big mistake sorry sorry Guys, i was realized it exactly 30 minutes after5 posting this that what aq big mistake i've made:
the answer is $\text{min}\left\{A,B\right\}$

and

[quote="sinajackson"]as a consequence the number of zeros in the end of $103!$ is $\left|24-72\right|=48$[/quote]

as a consequence the number of zeros in the end of $103!$ is $\text{min}\left\{24,72\right\}=24$

also thanks [b]da ban man[/b]

and as [b]Walk Around The River[/b] said its obvious that the number of $2$'s is alwas more, and the answer is $B$.
thanks [b]Walk Around The River[/b] :)

[i]The following are based on a math team assignment I created about 2 years ago, and they happen to go along with the theme of this thread.[/i]

[b]Extensions:[/b]

i. Find a number whose factorial ends in 103 trailing zeros

ii. For which values of $n$ does there exist another integer $k$ such that $k!$ has $n$ trailing zeros?

iii. Consider the sequence of numbers that do not satisfy the above. Is there a formula for the nth such number?
[u]
Note:[/u] Tell me if you think this should go into intermediate for some reason


geometry
3D geometry
sphere
inequalities unsolved
inequalities

Let $ x, y, z >0$ and $ x^2\plus{}y^2\plus{}z^2\equal{}1$. Find minimum value of $ A\equal{}x\plus{}y\plus{}z$.

Judging from the condition for which in Euclidean space, the sphere $ x^2\plus{}y^2\plus{}z^2\equal{}1\ (x>0,\ y>0,\ z>0)$ has the intersection point with the plane $ x\plus{}y\plus{}z\minus{}A\equal{}0$ is $ 0<A\leq\sqrt{3}$, thus there don't exist the minimum value of $ A$.


Given $ a,b,c \ge 0$ and  $ 3(ab \plus{} bc \plus{} ca) \plus{} 2abc \equal{} 27$.Prove that:
$ a \plus{} b \plus{} c \ge \frac {9}{2}$
$ ab \plus{} bc \plus{} ca \ge \frac {27}{4}$
$ \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \le \frac {9}{2}$
$ 3(a \plus{} b \plus{} c) \ge 2(ab \plus{} bc \plus{} ca)$

[quote="babyloverain"]Given $ a,b,c \ge 0$ and  $ 3(ab \plus{} bc \plus{} ca) \plus{} 2abc \equal{} 27$.Prove that:
$ 3(a \plus{} b \plus{} c) \ge 2(ab \plus{} bc \plus{} ca)$[/quote]
If $ a \equal{} 0$ or $ b \equal{} 0$ or $ c \equal{} 0$ your inequality is true.
Hence, we can assume $ a\neq0,$ $ b\neq0$ and $ c\neq0.$
Let $ 3(a \plus{} b \plus{} c) < 2(ab \plus{} bc \plus{} ca),$  $ a \equal{} kx,$ $ b \equal{} ky$ and $ c \equal{} kz,$
 where $ k > 0$ and $ 3(x \plus{} y \plus{} z) \equal{} 2(xy \plus{} xz \plus{} yz).$
Hence, $ 3k(x \plus{} y \plus{} z) < 2k^2(xy \plus{} xz \plus{} yz),$ which gives $ k > 1$ and
$ 27 \equal{} 3(ab \plus{} bc \plus{} ca) \plus{} 2abc \equal{}$
$ \equal{} 3k^2(xy \plus{} xz \plus{} yz) \plus{} 2k^3xyz > 3(xy \plus{} xz \plus{} yz) \plus{} 2xyz,$ which is contradiction since,
$ 27\leq3(xy \plus{} xz \plus{} yz) \plus{} 2xyz\Leftrightarrow$
$ (xy \plus{} xz \plus{} yz)^2(x \plus{} y \plus{} z)^2 \plus{} xyz(x \plus{} y \plus{} z)^3\geq4(xy \plus{} xz \plus{} yz)^3\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^4y^2 \plus{} x^4z^2 \minus{} 2x^3y^3 \plus{} 3x^4yz \minus{} x^3y^2z \minus{} x^3z^2y \minus{} x^2y^2z^2)\geq0,$ which is obvious.

Nice solution,arqady,and $ \sqrt{ab}\plus{} \sqrt{bc}\plus{} \sqrt{ca} \le \frac{9}{2}$ also nice and easy.Try it.Sorry for my bad english.
But $ 3(a\plus{}b\plus{}c) \ge 2(ab\plus{}bc\plus{}ca)$,we can use schur.


topology
advanced fields
advanced fields unsolved

Let $ \omega_1$ be the first uncountable ordinal. Show: the space $ \omega_1\plus{}1$ is compact.

I see why it is Lindelof. Use the standard trick that the open set about $ \omega_1$ must "reach down" to some countable ordinal. All this says is that there is some countable subcover. It seems as if you could construct a cover that is something like $ \bigcup_{\alpha\in\omega_1} [0, \alpha]$ along with something about $ \omega_1$ that doesn't have a finite subcover.

Maybe the finite intersection property is a better technique here? The closed sets are kind of mysterious in this space in my mind, though.

[quote="HilbertThm90"](...) It seems as if you could construct a cover that is something like $ \bigcup_{\alpha\in\omega_1} [0, \alpha]$ along with something about $ \omega_1$ that doesn't have a finite subcover.
(...)[/quote]

That's what I thought at first too. But, for any limit ordinal $ \alpha$, the set $ [\alpha, \omega_1] \equal{} \{ \beta: \alpha \leq \beta \leq \omega_1 \}$ is not open. If you can see why, you're a step closer towards the solution.


"Participants of maths competition are doing 6 assignments, each marked one of marks : 6,4,2,0. We know that for each pair of participants (X,Y) we can choose two assignments - in both of them X got different mark than Y. For how many participants is this situation possible ? I mean the biggest number of participants "

Could you help me with this assignment ? It's very important for me and I suppose it's quite an interesting problem...

I will be grateful,
Paul

Moderator - problem already posted and locked.

[url]http://www.artofproblemsolving.com/Forum/topic-53644.html[/url]

[b]Problem 4 from current Polish MO due October 10![/b]

Pls ask Megus to have a look at this forum sometimes.


function
number theory
Diophantine equation
combinatorics unsolved
combinatorics

Hello to everybody!

On our team contest one fellow solved the problem from ISL-2007 using generation functions, but I can't completely get it. Could you explain to me some moments?

P.S. From the moment I've posted it here I've realized some points and made some corrections. So there's not much left to understand.

The problem.

Find all positive integers  for which the numbers in the set  can be colored red and blue, with the following condition being satisfied: the set  contains exactly  ordered triples  such that

(i)  are of the same color and

(ii)  is divisible by n.

***

I won't post complete solution, as the key point was to prove the fact that if there are  red numbers and  blue ones, than the number of monochromatic triples  with  is [hide]  [/hide]

Proof.[hide]

Let  be red numbers and  - blue ones. Consider the function

Then the sum of coefficients at  is exactly the number of triples, in which sum of elemens is divisible by , and it is equal to . (Now I see: simply )

Then we note that

Simply, the expression  is the same then

Then (it is difficult to understand what I've written here, as there was not much time to write) he says that for each

there exists exactly one  from  such that  is divisible by .

It means that  is equal to the sum of coefficients in

Then, by some magic way it follows that

I still can't understand how he got it, though Diophantine Equation and  look very similar

[/hide]

***

I will be very grateful if someone explains me the end of this solution

Sum of coefficients = plug in 1 for $ x$.  ;)

By the way, only five posts down on this forum is a discussion of this problem ([url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=214709[/url]) -- you could have posted this there.

Thanks, I have already understood while correcting my post  :)


inequalities
LaTeX
inequalities proposed

let $ x_1,x_2,...,x_n$ be non-negative reel numbers, find $ k_{max}$ such as:
$ k(x_1x_2\plus{}x_2x_3\plus{}...\plus{}x_nx_1) \leq (x_1\plus{}x_2\plus{}...\plus{}x_n)^2$

enndb0x: $ k$ should be independent of the $ x_i$'s. Also, check your last inequality.

I suspect $ k \equal{} 3$.

@bambaman:

If you set $ x_1\equal{}1, x_2\equal{} \lambda, x_3\equal{} \dots \equal{} x_n\equal{}0$, then $ k \le 2\plus{} \lambda \plus{} \frac{1}{\lambda}$ or $ k \le 4$.

Besides, I'm interested too how to center LaTeX  :roll:

I don't understand why you set $ x_{1} \equal{} 1,x_{2}\equal{}\lambda , x_{3}\equal{}x_{4}\equal{}\cdots \equal{}x_{n}\equal{}0$  :blush:

@enndb0x: Bambaman did that, but he edited his post while i was writing my post^^

by cauchy shwarz and am-gm inequality we see that k max is equal to n.

Would you please write down your solution  :wink:

@FelixD
If $ x_{1} \equal{} 2$ and $ x_{2} \equal{} 5$ then $ (x_{1} \plus{} x_{2})^{2} \equal{} 49\geq k\cdot2\cdot 5 \Rightarrow 4.9\geq k$

But my example shows, that $ k \le 4$. It doesn't matter if there are other examples where $ k$ is greater than $ 4$.  :P

I'll post my solution now.  :)
I'll show that $ k_{max} \equal{} 4$  iff $ n > 3$. If $ k > 4$, then $ x_1 \equal{} x_2 \equal{} 1, x_3 \equal{} \dots \equal{} x_n \equal{} 0$ gives $ 4 \ge k$, impossible for $ k > 4$. Thus it remains to show that $ k \equal{} 4$ works. Therefore, there are two cases to look at: $ n$ is even and $ n$ is odd. For $ n \equal{} 2m$, the estimations $ 4\left( \sum_{cyc} x_1x_2\right) \le 4\left(\sum_{i \equal{} 1}^{m \minus{} 1} x_{2i \plus{} 1} \right) \left( \sum_{i \equal{} 1}^{m} x_{2i} \right)$ and $ \left(\sum_{i \equal{} 1}^{m \minus{} 1} x_{2i \plus{} 1} \right) \left( \sum_{i \equal{} 1}^{m} x_{2i} \right) \le \left( \frac {\sum_{i \equal{} 1}^{2m} x_i}{2}\right)^2$ show that $ k \equal{} 4$ fulfils the inequality. If $ n \equal{} 2m \plus{} 1$, nearly the same estimation works. Assume that $ x_{l}x_{l \plus{} 1} \equal{} \min \{x_1x_2, \dots, x_{2m \plus{} 1}x_1\}$.
If $ l \equal{} 2j$, then $ \sum_{cyc} x_1x_2 \le (x_1 \plus{} \dots \plus{} x_{2j \minus{} 1} \plus{} x_{2j \plus{} 2} \plus{} \dots \plus{} x_{2m} )( x_2\plus{} \dots \plus{} x_{2j} \plus{} x_{2j \plus{} 1} \plus{} \dots \plus{} x_{2m \plus{} 1})$.
If $ l \equal{} 2j \plus{} 1$, then $ \sum_{cyc} x_1x_2 \le (x_2 \plus{} \dots \plus{} x_{2j} \plus{} x_{2j \plus{} 3} \plus{} x_{2m \plus{} 1} ) ( x_1 \plus{} \dots \plus{} x_{2j \plus{} 1} \plus{} x_{2j \plus{} 2} \plus{} \dots \plus{} x_{2m})$. Then go on with the same estimation as before to get the result.

[quote="enndb0x"]so $ k$ is not indepedent of $ x_{i}$'s ...  :blush:[/quote]
But isn't $ k$ ought to be a constant which is independent of $ x_i$'s?

@enndb0x:
It's our aim to find the maximal value for $ k$ such that for every choice of the $ x_i's$ the inequality is fulfilled. As I have shown above, this value is $ 4$.

[quote="enndb0x"]But we are wining different values of $ k$ for differnt values of $ x_{i}$'s.So $ k$ is dependent of $ x_{i}$'s :maybe:[/quote]

[quote="peine"]let $ x_1,x_2,...,x_n$ be non-negative reel numbers, find $ k_{max}$ such as:
$ k(x_1x_2 \plus{} x_2x_3 \plus{} ... \plus{} x_nx_1) \leq (x_1 \plus{} x_2 \plus{} ... \plus{} x_n)^2$[/quote]
It is meant of course:
find the maximal $ k$ such that \[ k(x_1x_2 \plus{} x_2x_3 \plus{} ... \plus{} x_nx_1) \leq (x_1 \plus{} x_2 \plus{} ... \plus{} x_n)^2\] for all non-negative $ x_1,x_2,...,x_n$.  :)

yes $ k_{max}$ must be independent of $ x_i$'s

[quote="enndb0x"]But we are wining different values of $ k$ for differnt values of $ x_{i}$'s.So $ k$ is dependent of $ x_{i}$'s :maybe:[/quote]
I hope you understand now that $ k$ [b][i]IS a CONSTANT[/b][/i] term , and as we know the difference between a constant and a variable; hence $ k$ is [i]not[/i] dependent of $ x_i$. :oops:

I think I understood now .Thanks to all of you !  :blush:

$ \Longrightarrow 4(x_1x_2 \plus{} x_2x_3 \plus{} ... \plus{} x_nx_1) \leq (x_1 \plus{} x_2 \plus{} ... \plus{} x_n)^2$


inequalities
calculus
inequalities proposed

well, (1) has been discussed even before (the general form of it) and all we need to check is that $3,0314331...=2^{1.6}>3$
as for the 3rd one I guess hungkhtn should have a proof as he posted it [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=103052]here[/url] before asking wether "is it hard" ?
I would like to see a solution of (3) [not only calculus, of course] as I've tried to solve it myself but couldn't

You are right about the last inequality. I didn't find either a nice solution to it.
But I found a nice solution for the first two inequalities. Of course, the general case of the first inequality is known; it is also proven in Algebraic Inequality by RCF way (chapter 3).

Well, for (c), Hungkhtn posted it and the conjecture in the link

http://www.mathlinks.ro/Forum/viewtopic.php?t=103052

but it hasn't any solution.  :wink:

There is an elementary solution to

${\sqrt{\frac{a}{pb+c}}+\sqrt{\frac{b}{pc+a}}+\sqrt{\frac{c}{pa+b}}\geq \frac 3{\sqrt{p+1}}}$

for $p \geq 19$. Consequently, we have

$\sqrt{\frac{a}{24b+c}}+\sqrt{\frac{b}{24c+a}}+\sqrt{\frac{c}{24a+b}}\geq \frac 3{5}$. :lol:

Vasc can you share the solution with us  :)

Using this method we can prove that for $p\geq \frac{9^{n}}{4}$:

$\sqrt[n]{\frac{a}{pb+c}}+\sqrt[n]{\frac{b}{pc+a}}+\sqrt[n]{\frac{c}{pa+b}}\geq \frac 3{\sqrt[n]{p+1}}$.

Actually, the best $p$ for this method is given by
$p+\frac 1{p}\geq \frac{9^{n}}{4}-2$. :)
This result cover the case $n=1$, too.

[quote="Vasc"]If $a,b,c$ are positive numbers, then
(3)* $\sqrt \frac a{8b+c}+\sqrt \frac b{8c+a}+\sqrt \frac c{8a+b}\geq 1$ (open).[/quote]

I was curious about this one and finally I solved it a few minutes ago.  :P

So we are waiting for you to post the solution :)

[quote="Vasc"]For (1) and (2) we may use Holder's Inequality.
For the last with $p \geq 19$, we can use the inequality
$\sqrt x+\sqrt y+\sqrt z \geq \sqrt{x+y+z+6xyz}$  for $xyz \leq 1$.[/quote]
But we can prove the last for any $p \geq 4$ if we use the inequality
$\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\geq \frac{xy+yz+zx+6xyz}{3}$  when $xy+yz+zx+6xyz \leq 9$.

[quote="Vasc"]If $a,b,c$ are positive numbers, then
(3)* $\sqrt \frac a{8b+c}+\sqrt \frac b{8c+a}+\sqrt \frac c{8a+b}\geq 1$ (open).[/quote]
Today it's easy!  :lol:


These are fairly easy problem even it's Team Round problems..  :?

1. Recall that $n! = n \cdot (n-1) \cdot (n-2) \cdot \cdots 2 \cdot 1$. Then find the last digit of the number $k$ if $k = 1!+2!+3!+ \cdots 10!$.

2. Find $(x,y)$ values such that $(x-y)^3 = 64$ and $x^2 y^2 = 144$.

well if the number is greater than or equal to 5, then the units is 0, so the only ones we have to look at ar 1,2,3,4:1,2,6,24  So $3$

[quote="nat mc"]:D[/quote]

" :D " what? What's so funny?  :? (How come the confused icon got changed  :mad: )

:lol: is a bad "lol" icon.  It should be something that actually looks like its laughing.


linear algebra
matrix
inequalities
linear algebra unsolved

$ A \in M_2(R)$ Prove that $ det(A^2\plus{}A\plus{}I_2) \geq\ \frac{3}{4} (1\minus{}detA)^2$
I have the following solution but i'm not sure of it.Please tell if i'm wrong[hide]

Let $ x$ $ y$ be the eigenvalues of A.Then
$ x^2\plus{}x\plus{}1$, $ y^2\plus{}y\plus{}1$ are the eigenvalues of $ A^2\plus{}A\plus{}I_2$We have $ Det (A)\equal{}xy, Det(A^2\plus{}A\plus{}I_2)\equal{}(x^2\plus{}x\plus{}1)(y^2\plus{}y\plus{}1)$
now the inequality is   $ (x^2\plus{}x\plus{}1)(y^2\plus{}y\plus{}1) \geq\ \frac{3}{4}(1\minus{}xy)^2$

 $ 1 \plus{} x \plus{} x^2 \plus{} y \plus{} x y \plus{} x^2 y \plus{} y^2 \plus{} x y^2 \plus{} x^2 y^2 \geq\ \frac{3}{4} (1\minus{}2xy\plus{}x^2y^2)$

$ \frac{x^2y^2}{4} \plus{}x^2\plus{}y^2\plus{}x^2y\plus{}xy^2\plus{}\frac{5}{2}xy \plus{}x\plus{}y\plus{}\frac{1}{4} \geq\ 0$

$ (\frac{1\plus{}xy}{2} \plus{}x\plus{}y)^2 \geq\ 0$[/hide]

isn't anyone to tell me if the solution is good?

If you haven't done something wrong on the calculation your solution is good. How much points you got on it?

Did you write excactly one you type here on your test paper?

From my point of view it should not be 0 for this kind of solution.
Of course $ A$ can have complex eigenvalues, but then $ xy$ and $ x\plus{}y$ would be real and all of your calculation remain valid (i assume that your calculation is true).

You could have worked with $ a,b\in \Bbb{R}$ as the eigenvalues first and then with$ z$,$ \overline{z}$ and $ Re(z)$ with $ z\in \Bbb{C}$, but other than that your proof is correct and somewhat more natural than the standard proof. Have you asked for a review by the committee?

i had my paper reviewed , and i got 7 points(max).:D


geometry
rectangle
function
calculus
integration
real analysis
real analysis unsolved

Let $\sum x_n$ be a divergente serie of real strictly positive terms.Show that $\sum \frac{x_n}{S_n^{\alpha}}$ converges if and only if $\alpha>1$,where $S_n=\sum_{k=0}^n x_k$.

One implication ($\alpha\le 1$ implies the divergence of the series) has been shown [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=186647#p186647]here[/url], in the second post. Actually, it's proven for $\alpha=1$ there, but the general result clearly follows. The other implication is mentioned in that same topic by Levon, but nobody replied :).

Anyway, write the series as $\sum\frac{S_n-S_{n-1}}{S_n^{\alpha}}$, and notice that this is the area of a disjoint union of rectangles (with width $S_n-S_{n-1}$ and height $\frac 1{S_n^{\alpha}}$) under the graph of the function $t\mapsto t^{-\alpha}$, and the integral of this function from $1$, say, to $\infty$ is finite.

Apply Mean-value Theorem to $\frac{1}{x^{\alpha-1}}$ in the interval $[S_{n-1},S_n]$, we know there exists $\xi\in (S_{n-1},S_n)$, such that $\frac{1}{S_{n-1}^{\alpha-1}} - \frac{1}{S_n^{\alpha-1}} = \frac{\alpha-1}{\xi^\alpha}$, hence
$\frac{a_n}{S_n^\alpha} = \frac{S_n - S_{n-1}}{S_n^\alpha} < \frac{1}{\xi^\alpha}(S_n - S_{n-1}) = \frac{1}{\alpha-1}\left(\frac{1}{S_{n-1}^{\alpha-1}} - \frac{1}{S_n^{\alpha-1}}\right)$.
The conclusion follows immediately.


Hi,

I have a suggestion on blog contributors.

All the time, I see, on my blog and others, requests to be a contrib. These are just pointless and clutter up the blog, so why not make a little link that says "Request to be contrib" or something like that.

The blog owner would then get a little pm saying "_____ has requested to be a contrib. Click here to confirm/ignore" or something like that.

Dojo


conics
ellipse
ratio
function
geometry proposed
geometry

Well, this does not look like an olympiad problem but in France some like this kind of problems so...

In the plane, a triangle $ABC$ is given. For each line $D$ not parallel to one of the sides of the triangle, we define $G_D$ to be the isobarycenter of the common points to $D$ with the lines $AB,BC,CA$.

The aim of the problem is to determine to set $F$ of the points $G_D$ when $D$ varies.

1) Prove that when $D$ moves while remaining parallel to a line $ \delta$, the locus of the points $G_D$ is a line $\Delta_{\delta}$.

2) In this question only, we assume that $ABC$ is equilateral.
Prove that when $\delta$ varies, the lines $\Delta_{\delta}$ are all tangents to a same circle, and determine the set $F$ in that case.

3) Back to the general case.
Prove that we may find an equilateral triangle $A'B'C'$ of the space such that its orthogonal projection onto the plane is $ABC$, and determine the set $F$.

Pierre.

This was definitely posted before by Moubinool, but he didn't post all the intermediate steps, so when I first saw it it appeared a lot harder. The locus seems to be the entire plane except for the interior points of the Steiner inellipse (the inscribed ellipse having the centroid of $ABC$ as its center). 1) is easy to prove. We just use the following lemma (easy to prove): if we have two lines $\ell_1,\ell_2$ intersecting at $O$ and a variable line of constant direction intersects them at  $X_1,X_2$ respectively, then the locus of the point $X$ which forms a constant ratio with $X_1,X_2$ is a line passing through $O$.

This will show us that if $X,Y,Z$ are the intersections of $D$ with $BC,CA,AB$ respectively, then $G_D$ (in the hypothesis of 1)) moves on a line passing through the intersection point of $BC$ and the line connecting $A$ to the midpoint of $YZ$ (analogous properties are true for the other vertices, of course).

It's not hard to prove what we want without making use of 2). Let $X',Y',Z'$ be the midpoints of $YZ,ZX,XY$ respectively. After we show that $BY',CZ'$ concur in $T$, on the line passing through the midpoints of $AB,AC$ (I didn't actually show this, but how hard can it be? :D) we find that the mapping $BT\leftrightarrow CT$ is a projective map between the pencils in $B,C$, so the function mapping $BY'\cap CA$ to $CZ'\cap AB$ is also a projective map between the lines $AC,AB$, so it's well-known that the line connecting the two points envelops a conic. After looking at some particular cases we realize that the conic is the Steiner inellipse.

2) is now merely a particular case.

> This was definitely posted before by Moubinool, but he didn't post all the intermediate steps,

I agree. That is definitely our Moubi  :D  :D

Pierre.


algebra
polynomial
complex analysis
calculus
calculus computations

Show that the polynomial $ p(x)\equal{}\frac{x^7}{7}\minus{}\frac{x^6}{6}\plus{}\frac{x^5}{5}\minus{}\frac{x^4}{4}\plus{}\frac{x^3}{3}\minus{}\frac{x^2}{2}\plus{}x\minus{}1$ has exactly one real root.

[hide]Rolle`s theorem.[/hide]

Suppose there are two real roots, $ a,b$.  Then $ f(a)\equal{}f(b)\equal{}0$.  Then by Rolle's theorem, there is some $ c\in[a,b]$ such that $ f'(c)\equal{}0$.  We see that $ f'(x)\equal{}x^{6}\minus{}x^{5}\plus{}x^{4}\minus{}x^{3}\plus{}x^{2}\minus{}x\plus{}1$.  But this has no real zeroes, contradicting Rolle's theorem.  Hence $ a\equal{}b$ and there is only one real zero.  Alternatively, if you know some complex analysis, you can use Rouche's theorem.

Thank you, JRav.

I solved the task the same way. But the most interesting part to me was how to prove that $ f'(x)\equal{}x^{6}\minus{}x^{5}\plus{}x^{4}\minus{}x^{3}\plus{}x^{2}\minus{}x\plus{}1$ has no real zeroes.

$ f'(x)\equal{}x(x\minus{}1)(x^{4}\plus{}x^{2}\plus{}1)\plus{}1$. If $ x\geq1$ or $ x\leq 0$, then $ f'(x)\geq 1$. Therefore we have to examine just $ x\in (0,1)$. Let us denote $ x(x\minus{}1)\equal{}r(x)$ and $ x^{4}\plus{}x^{2}\plus{}1\equal{}s(x)$. $ r(x) \in [\minus{}\frac{1}{4},0)$ and $ s(x) \in [1,3)$ on $ x\in (0,1)$. Hence $ |r(x)s(x)|<\frac{3}{4}$. So $ f'(x)>\frac{1}{4}$ on $ \mathbb{R}$.

How did you see that $ f'(x)$ has no real roots?

Multiply $ f'(x)$ with $ x\plus{}1$. You get $ x^7\plus{}1$, which has only one real root, and it's the one you multiplied with.

[quote="JRav"]Alternatively, if you know some complex analysis, you can use Rouche's theorem.[/quote]
Sometimes I wonder, JRav, whether you know what you are talking about or just make random shots. Can you tell us what exactly you meant by this comment :?.

Suppose that $ z_{0}$ is a root of the equation.  Then $ \bar{z_{0}}$ is also a root of the equation.  Now if $ z_{0}$ is in the right half of the complex plane, then so is $ \bar{z_{0}}$.  If we can use Rouche's theorem to prove that there is only one root in this region, then we conclude that $ z_{0}\equal{}\bar{z_{0}}$, hence it is real.

:what?: OK, let's, indeed, ask this question: how many roots does $ p$ have in the right half-plane $ \text{Re\,}z>0$?


induction
limit
real analysis
real analysis unsolved

a) Let $ n_{1},n_{2},\dots$ be a sequence of natural number such that $ n_{i}\geq2$ and $ \epsilon_{1},\epsilon_{2},\dots$ be a sequence such that $ \epsilon_{i}\in\{1,2\}$. Prove that the sequence: \[ \sqrt[n_{1}]{\epsilon_{1}\plus{}\sqrt[n_{2}]{\epsilon_{2}\plus{}\dots\plus{}\sqrt[n_{k}]{\epsilon_{k}}}}\]is convergent and its limit is in $ (1,2]$. Define $ \sqrt[n_{1}]{\epsilon_{1}\plus{}\sqrt[n_{2}]{\epsilon_{2}\plus{}\dots}}$ to be this limit.
b) Prove that for each $ x\in(1,2]$ there exist sequences $ n_{1},n_{2},\dots\in\mathbb N$ and $ n_{i}\geq2$ and $ \epsilon_{1},\epsilon_{2},\dots$, such that $ n_{i}\geq2$ and $ \epsilon_{i}\in\{1,2\}$, and $ x\equal{}\sqrt[n_{1}]{\epsilon_{1}\plus{}\sqrt[n_{2}]{\epsilon_{2}\plus{}\dots}}$

[hide="an idea for a)"]
[b]a)[/b]
Let $ a_{k}\equal{}\sqrt [n_{1}]{\epsilon_{1}\plus{}\sqrt [n_{2}]{\epsilon_{2}\plus{}\dots\plus{}\sqrt [n_{k}]{\epsilon_{k}}}}$
and $ b_{k\plus{}1}\equal{}\sqrt{2\plus{}b_{k}}$ ,$ b_{1}\equal{}\sqrt{2}$
It's obvious that $ a_{k}$ is increasing.$ (1)$
By induction it's easy to prove that :$ 1 < a_{k}\leq b_{k}$ $ \forall k\geq 1$ $ (2)$

From $ (1)$ and $ (2)$ and $ \lim_{k\rightarrow\infty}{b_{k}}\equal{} 2$ $ \Rightarrow$ $ a_{k}$ is convergent and $ \lim_{k\rightarrow\infty}a_{k}\in (1,2]$[/hide]

[quote="Omid Hatami"]a) Let $ n_{1},n_{2},\dots$ be a sequence of natural number such that $ n_{i}\geq2$ and $ \epsilon_{1},\epsilon_{2},\dots$ be a sequence such that $ \epsilon_{i}\in\{1,2\}$. Prove that the sequence:
\[ \sqrt[n_{1}]{\epsilon_{1}\plus{}\sqrt[n_{2}]{\epsilon_{2}\plus{}\dots\plus{}\sqrt[n_{k}]{\epsilon_{k}}}}\]
is convergent and its limit is in $ (1,2]$[/quote]

i *think* the result for a) will follow once you apply [i]Herschfeld's
Convergence Theorem[/i], which states --- for some $ p_{k}\in\mathbb R,\;\;\;0<p_{k}<1$
and $ \epsilon_{k}>0$ ($ k\in\mathbb Z,\;\;\;k>1$),
[color=white].............[/color]$ \lim_{k\to\infty}\left(\epsilon_{1}\plus{}\left(\epsilon_{2}\plus{}\left(\epsilon_{3}\plus{}\left(\epsilon_{4}\plus{}\dots\plus{}\left(\epsilon_{k}\right)^{p_{k}}\right)^{p_{4}}\right)^{p_{3}}\right)^{p_{2}}\right)^{p_{1}}$
converges [b]iff[/b] $ (\epsilon_{i})^{(p_{i})^{i}}$ is [i]bounded[/i] for some $ i\in\mathbb Z,\;\;\;i>1$.

in your example, $ \epsilon_{i}\in\{1,2\}$ is bounded, hence the result should follow immediately.

in other words, first try and prove the sequence in question is bounded.
and then if you can show it monotonically increases or decreases,
then you can [i]prove [/i]that the sequence is convergent.  :|

Thanks misan!
Could you tell me from where(which book) can I find about Herschfeld's Convergence Theorem?


Hi

Here are some easy and some hard questions, depending on the mathematician. However, they are very hard for me and i just wanted some solutions so i can try similar problems and try to develop the thinking style/method that you are using.

[b]1. [/b]When a positive integer $N$ is written in base $9$, it is a two-digit number. When $6N$ is written in base $7$, it is a three-digit number obtained from the two-digit number by writing digit $4$ to its right. Find the decimal representations of all such numbers $N$.

[b]2. Prove that the inequality[/b]
$(a+7b)(4a+b)$ $\geq$ $3a^2 + 33ab +2b^2$
holds for all real numbers $a$ and $b$

[b]3. [/b]

Two circles [i]C1[/i] and [i]C2 [/i]meet at points [i]A[/i] and [i]B[/i]. [i]P[/i] is a point on the circumference of [i]C1[/i] such that [i]PA[/i] is tangent to [i]C2[/i] at [i]A[/i]. [i]Q[/i] is a point on the circumference of [i]C2[/i] outside [i]C1[/i]. [i]QA[/i] is produced to meet [i]C1[/i] again at [i]T[/i] so that $<angle> PBT = 75degrees$. [b]Find the size of $<angle>ABQ$[/b]

[b]4. [/b]

For what values of $a$ there are exactly two real numbers that satisfy the equation
$x^3 + (a-1)x^2 + (9-a)x-9=0$
Find all posible answers
I think they are $0 or 10$ are they right, im happy to post the solution (rarther long) if someone has doubts.

Anyways thanks for your time
Keshin

I have solved problem two before in the Intermediate Forum.

Expand it:
\[4a^2+ab+28ab+7b^2 \geq 3a^2+33ab+2b^2\]
\[\Longleftrightarrow a^2-4ab+5b^2 \geq 0\]
\[\Longleftrightarrow (a-2b)^2+b^2 \geq 0\]
Which is obviously true. And we have equality if $a=b=0$.

Number one:

We have $n = 9a + b$, and $6n = 49a + 7b + 4$.

Linear combinations gives: $b = 5a + 4$

Both $a$ and $b$ are digits, so $a = 0$ or $a = 1$. This gives $b = 4$ and $b=9$ respectively. But it's two digits in base 9, so only 99 works.

[hide="Geometry Problem"]
$\triangle{PBT} \sim \triangle{ABQ}$. So I think $\angle{ABQ}=75^{o}$.[/hide]

$A, B, P, T$ lie on the same circle so $\angle{QAB}=\angle{TPB}$.
Also, because $PA$ is the tangent line, so $\angle{Q}=\angle{PAB}=\angle{BTP}$.
Therefore, $\triangle{TPB} \sim \triangle{QAB}$. Hence: $\angle{TBP}=\angle{QBA}$.

[quote="behemoth571"]For 4:

Using synthetic division, we find that
$ x^3 + (a - 1)x^2 + (9 - a)x - 9 = (x - 1)(x^2 + ax + 9) $

Since the cubic has only 2 roots (i.e. has one double root), $ x = 1 $ must be a root of both the monic and the quadratic. Plugging in,
$ 0 = (1)^2 + a + 9 $
$ a = -10 $

This can be confirmed: using $ a = -10 $, $ x = 1, x = 9 $[/quote]

If x=1, then wouldn't 'a' be any real number?

[quote="nr1337"][quote="behemoth571"]For 4:

Using synthetic division, we find that
$ x^3 + (a - 1)x^2 + (9 - a)x - 9 = (x - 1)(x^2 + ax + 9) $

Since the cubic has only 2 roots (i.e. has one double root), $ x = 1 $ must be a root of both the monic and the quadratic. Plugging in,
$ 0 = (1)^2 + a + 9 $
$ a = -10 $

This can be confirmed: using $ a = -10 $, $ x = 1, x = 9 $[/quote]

Also, a=-6 works, because then we would have x=1, x=3.[/quote]

Also, a=6 works, because then we would have x=1,x=-3.

Also, a=-10 works, because then we would have x=1, x=-9.


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

[quote="smith09"]Construct an injective resolution of a cyclic group (finite or infinite).

THks in advance.[/quote]

You mean, over the ring of integers, construct an injective resolution of a cyclic module?

It splits into two cases:

An injective resolution of Z is 0→Z→Q→Q/Z→0.

An injective resolution of Z/nZ is 0→Z/nZ→Q/Z→Q/Z→0 using the isomorphisms Z/nZ ≅ (1/n)Z/Z and Q/Z ≅ Q/(1/n)Z.

The first resolution is minimal, but the second is a bit sloppy.  Let K be the subring of Q containing Z and all 1/p where p does not divide n.  Then Z/nZ ≅ (1/n)K/K and Q/K ≅ Q/(1/n)K, and 0→Z/nZ→Q/K→Q/K→0 is a minimal injective resolution.

This same idea produces minimal injective resolutions for finitely generated modules over Dedekind domains.


Is there anyway to work latex into a chat or a chat that allows latex such as the math jams?

there is not a current latex-support for any chat environment, other than AoPS class.

maybe something will be developed in the future, but it is most unlikely, due to the security risks that it may present (the current software is java enabled, and security risks are minimal, because the latex engine resides on our server).


AMC
AIME
USA(J)MO
USAMO
MATHCOUNTS
algorithm
AMC 10

compared to other years, how hard do you think this year's aime was?
i am a 9th grader and got a 5 on the aime
do you think i will be able to make it to the usamo?

See the topic with the title, "AIME I Discussion" or something like that. This year was easy compared to others, and a 5 probably won't be enough. The floor is predicted at about 7-8.

im in tenth grade and i got a 7, im really happy, but im praying i get to USAMO

I'm in 9th grade, I got an 8 (from checking the test booklet answers I wrote with the website answers)

I relaly hope that the following two don't both happen:
-I marked a problem wrong in the answer form (or darkened something too much/too little)
-the floor is 8 or above.
because if both of them happen, that means i got a 7 and i won't qualify.

I heard there was a 144/7. So all of you guys who got 7s might just be in luck.

[quote="randomdragoon"]I heard there was a 141/7. So all of you guys who got 7s might just be in luck.[/quote]

that was The Zuton Force, I think.

I'm in 10th, and I got a 9, barring bubbling mistakes.
Thought it was a lot easier that the 2007 I, but that might just be me improving a lot in a year in AIME.
My AMC scores, however, have kinda suffered from neglect.  :P

Does anyone think that the secondary cutoff could be any higher than 197 for a floor of 7?

I'm a 9th grader, and I got an 8. I skipped a lot of easy problems because they looked hard.

I thought it was a lot easier than last year, considering I think I didn't improve much in math overall, but I doubled my score. Especially since the first 5 problems were way to easy for an AIME. Besides those, the problems were a little harder than normal. But overall, is was an easy AIME.

6th, 9/144. Probably good enough to pass.
This year was easier than normal, especially the first few and #15(which I got wrong because of a stupid mistake...), probably cutoff is going to be 7-8.

112.5/10 = 212.5 = probably good enough to pass... except that I didn't qualify for the AIME anyways! :mad: (The AIME I took was unofficial).

There's so many genius sixth graders around here...

pythag, do you have any like links to that? i've heard about some person like that, and i want to know more. And did that person make usamo?

EDIT: Oh wait, it's Alex Song, and he qualified for USAMO... :!: that's insane!

Is there any chance they would just stick the floor to an 8 to avoid the whole secondary-index thing?

That would not be the reason. But personally, I think the index will be 8 anyway.

Well the 144/7's might change things... (index above 214 is unlikely imo)

Yeah, but when the floor is 7, how high could the secondary index possibly be?  How many freshmen and sophomores got 7's?

[quote="TheHeinrich"][quote="pythag011"][quote="Zmastr"]The fact that middle schoolers even qualified for the AIME, let alone demolish it, still astounds me.[/quote]

Last year, a Canadian 4th grader destroyed the AIME.[/quote]

i heard from  that this year(grade 5), he got a 12. :!:  :!:[/quote]

lol...whos qualified for IMO the most times?

I got an 8 on the AIME (IF I did not make bubbling errors, and IF I got none of my guesses right), and my index would be 210.5. Last year, the secondary index requirement was 181.5 (?) for the AMC 10 people, so this year, considering that the floor is about 8, the index should hover between 200-203 (if there is such a requirement), depending on amc performances.

If the floor is 7, definitely expect to see an index cutoff, maybe around 195 (probably a high estimate).

yall bruthaz with 7z don tell me i di'nt try for ya  :)

I got a 132/13 this year (I'm in 11th).

I found this AIME to be easier than last year's.  I don't want to compare this one with the 2006 AIME, because I underwent major changes in terms of my math skills over the past two years.

um okay, i am quoting mathcrazed. it was EZ PZ lemon CHEEZ idk if i spelled the last word right. but seriously speaking i am in no position to say it was ez since i got a bad score of [size=34]6[/size]. so ya but it was surely easier than 2007 AIMEs I and II.

If the floor is an 8, I don't think they will have a secondary index cutoff.

[quote="236factorial"][quote="the future"]I think if the floor is 7, the index will be higher than the index if the floor is 8

So (7,196) or (8,192)

because the 8 will already get rid of alot of the people that got 7s so theres no need for a higher index to get rid of more people..

whereas a 7 floor will bring alot more people so they have to enforce a tougher index[/quote]

Index = 192 and AIME = 8 => AMC 10 = 112 => not qualifying for the AIME

There's clearly a problem there. That's why I think that there probably will not be an index cutoff if the AIME floor were 8.[/quote]

Nope because some people qualified through the 12- thats a flaw with the index-(infact I saw someone here with a 97.5 on the 12 and 10 on the AIME as a 10th grader-this would suck if the cutoff is 7 on AIME with a 198 index-and someone else got a 108 on the 12 and was worried aobut the index because they only got a 8 on the AIME)...and plus i know someone last year that got a 7 on the AIME as a 10th grader but didnt make USAMO since they took the 12 (although the cutoff was 6 and 181.5)-they got like 110ish/7

So basically if its 8, its really low index (obviously higher than 177.5 for the 97.5 AMC 12, but lower than if the floor is 7) or there will be no index-because the 8 might alreayd do the trick
---------------------
Anyways, if lets say onlly 470 make it if the cutoff is 8 but  thats not enough people (since its 500)...I think they should leave it at 470 because if they make it 7, then there will be like 550.  But now since there are too many, they will have to use an index and with this index someone with a 144/7 is considered to more favored tomake USAMO than a person with a 108/9 (108 on the 12), which is unfair so I hope they dont do that.

The index can be high. I think the first year I made USAMO the index was something like 230? I might be mistaken but I know it was >220.

the past years' qualification scores can be seen on wikipedia.  also, does anyone know if the bubbling in of your amc score matters?

Well that was also before the USAMO expansion...

I think that the USAMO should use the secondary index to cutoff people with similar AIME scores (kinda like a tiebreaker in MATHCOUNTS, between people with 120/7 and 126/7), but it shouldn't be possible for people with AIME scores greater than the floor to not make it.  As the future was saying, it's not fair for a 10th grader with a 108/9 to not make it, when a similar person with a 132/7 made it because they took the 10A and not the 12A.  This also affects me, because if the floor is 7, I might not make it as a sophomore with a 117/8, because the secondary index might too restrictive to let 197's in.

You could always give your suggestions to the AMC office, and if you're lucky, they'll take your suggestion.

[quote="AMCDirector"]Write your suggestion as an algorithm, and we will consider it.

Steve Dunbar
MAA Director, American Mathematics Competitions[/quote]

Not entirely sure what you want me to do  :blush:

[quote="MellowMelon"]He means that you should specify your proposal so exactly that it could be done mechanically. Note that there used to be a pinned thread full of such proposals; all were considered and none were chosen over the current system, so odds are against you.

If I understand, this is sort of the algorithm you are saying. (I don't know if the qualifying numbers are right, but this is easy to adjust)

(by 10th grader I mean anyone 10th grade and below)
- First, do the standard method of determining the standard index for everyone and getting about 350 people in this way.
- Now we want about 160 other 10th graders (regardless of which AMC was taken). Take the maximum AIME score $ M$ so that at least 160 10th graders that have not yet qualified got at least that score.
- Any 10th grader with AIME greater than $ M$ qualifies. Suppose there are $ x$ of them ($ x < 160$).
- Consider the AMC scores of the top $ 160 \minus{} x$ 10th graders with AIME score $ M$. The minimum AMC score of them is $ N$, the secondary cutoff. No distinction between 10 and 12.
- Any 10th grader with AIME score exactly $ M$ and AMC score greater than or equal to $ N$ qualifies.

This is still biased towards taking the AMC10 but does favor high AIME scores like you want.[/quote]

Yep, that's exactly what I was aiming for.  The secondary index is only determined after all people in grades less than or equal to 10 with a score above the floor make it.  All those with an AIME score equal to the floor are then subjected to a secondary index, depending on what lets ~500 people in.

[quote="the future"]Nope because some people qualified through the 12- thats a flaw with the index-(infact I saw someone here with a 97.5 on the 12 and 10 on the AIME as a 10th grader-this would suck if the cutoff is 7 on AIME with a 198 index-and someone else got a 108 on the 12 and was worried aobut the index because they only got a 8 on the AIME)...and plus i know someone last year that got a 7 on the AIME as a 10th grader but didnt make USAMO since they took the 12 (although the cutoff was 6 and 181.5)-they got like 110ish/7

.[/quote]

I believe that after deciding the 300-330 people with the highest index that took the AMC 12, the floor score is used to determine people who took the AMC 10 only.

For another algorithm, would we be allowed to suggest an extra bubbling portion in which we "bid" on how many points we would risk on one problem, but also a guessing penalty?  So that, if we are sure on a problem, we say we would wager 10 points on that, while if we were shaky on a question we wager more like 8 points or so.  That way, we would be able to not worry so much about guessing on the rest of the problems.

Note - I am not sure whether this would produce the requested results.

[quote="moogra"][quote="MellowMelon"]:huh:

These contests are supposed to be about how good you are at math. It's bad enough that meta-testing methods exist now; your proposal would make that problem tremendously worse.[/quote]

Meta-testing methods have always and will always exist. In every test, there are cheap strategies.[/quote]

Yeah, but he means that you shouldn't add more ways to change the basic nature of the test, because it's all about the math anyways  :).  You'll never get rid of basic strategies, but adding point scales for different questions just makes it way too complicated.

[quote="MellowMelon"]This is still biased towards taking the AMC10 but does favor high AIME scores like you want.[/quote]

I really wish that it wasn't biased towards the amc10, especially since in in 11th grade.

I think it would work better if the cutoff was based more on the AIME, which is closer to the USAMO.

Write an algorithm that meets your criteria.  We will test it.
I can tell you that last year we tested alternative algorithms suggested on this forum, and the different effects amounted to about a 1% difference in selection.

Steve Dunbar
MAA Director, American Mathematics Competitions


integration
inequalities unsolved
inequalities


let a,b,c is non-negative such that $ a^2\plus{}b^2\plus{}c^2\equal{}3$ prove that:
     $ \frac{a^3b}{1\plus{}ab}\plus{}\frac{b^3c}{1\plus{}bc}\plus{}\frac{c^3a}{1\plus{}ca} \leq \frac{3}{2}$

[quote="tuantam1lan"]let a,b,c is non-negative such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$ prove that:
     $ \frac {a^3b}{1 \plus{} ab} \plus{} \frac {b^3c}{1 \plus{} bc} \plus{} \frac {c^3a}{1 \plus{} ca} \leq \frac {3}{2}$[/quote]
Nice inequality :)
It 's equivalent to:
$ \frac{a^2}{1\plus{}ab}\plus{}\frac{b^2}{1\plus{}bc}\plus{}\frac{c^2}{1\plus{}ca} \ge \frac{3}{2}$
By Cauchy-Schwarz inequality we have:
$ LHS \ge \frac{(a^2\plus{}b^2\plus{}c^2)^2}{a^2\plus{}b^2\plus{}c^2\plus{}a^3b\plus{}b^3c\plus{}c^3a} \ge \frac{3}{2} \equal{} RHS$

[quote="quykhtn-qa1"][quote="tuantam1lan"]let a,b,c is non-negative such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$ prove that:
     $ \frac {a^3b}{1 \plus{} ab} \plus{} \frac {b^3c}{1 \plus{} bc} \plus{} \frac {c^3a}{1 \plus{} ca} \leq \frac {3}{2}$[/quote]
Nice inequality :)
It 's equivalent to:
$ \frac {a^2}{1 \plus{} ab} \plus{} \frac {b^2}{1 \plus{} bc} \plus{} \frac {c^2}{1 \plus{} ca} \ge \frac {3}{2}$
By Cauchy-Schwarz inequality we have:
$ LHS \ge \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{a^2 \plus{} b^2 \plus{} c^2 \plus{} a^3b \plus{} b^3c \plus{} c^3a} \ge \frac {3}{2} \equal{} RHS$[/quote]

such a nice solution!


Paul shared his baseball cards with three friends. He gave half his cards to one friend, one-third of the cards that were left to his second friend, and the remaining 12 cards to a third friend. How many cards did Paul have at the beginning?

Let $ x$ be our original amount.  We have $ \frac {1}{2}x\plus{}\frac {1}{2} \times \frac {1}{3}x\plus{}12\equal{}x$  This simplifies into $ \frac {2}{3}x\plus{}12\equal{}x$.  We subtract $ \frac {2}{3}x$ from both sides, and get $ 12\equal{}\frac {1}{3}x$.  Multiplying each side by $ 3$, we get $ x\equal{}\boxed {36}$


algebra
polynomial
inequalities open
inequalities

This one is a special case of the one I posted in http://www.mathlinks.ro/Forum/viewtopic.php?p=580614#580614 middle of first page (Tue Dec 06, 2005) :lol: :
[quote="spanferkel"]
[b] For partitions $\alpha,\beta\vdash n$ and $0\leq t\leq1$ we have $[\alpha]+[\beta]\geq [(1-t)\alpha\oplus t\beta]+[t\alpha\oplus(1-t)\beta]$. (*)[/b][/quote]


$25)$ Find the surface area of the sphere with center $(5,3,4)$ tangent to the plane $2x+2y-z=-6$.

so well you can find the distance between the point and that line..which i believe can be obtained by this formula

${D=\frac{|Ax+By+Cy+D|}{\sqrt{A^{2}+B^{2}+C^{2}}}=\frac{|2(5)+2(3)-1(4)+6|}{\sqrt{2^{2}+2^{2}+(-1)^{2}}}=\frac{18}{\sqrt{9}}}=6$

Surface Area of a Sphere is:
$4\pi{r^{2}=(4\pi){6^{2}}=\boxed{144\pi}}$