Show that there are infinitely many primes in form $6k+5$


linear algebra
matrix
linear algebra unsolved

Dear friends,

I have noticed that

If $A=A_{r}+jA_{i}$ is a $n\times n$ complex matrix with full rank, then
\[A_{r}+A_{i}{A_{r}}^{-1}A_{i}={B_{r}}\]
where $B={A}^{-1}=B_{r}+jB_{i}$

How to prove it mathematically?

regards

You mean $B_{r}^{-1}$. It is a straightforward calculation from $(A_{r}+iA_{i})(B_{r}+iB_{i})=1$. owk


The following may or may not be trivial; it hasn't been the subject of extensive effort from my part  :lol: :

Do there exist infinitely many pairs $(a,b), a \neq b$ of distinct positive integers such that the numbers in each of the pairs $\{a,b\}$ and $\{a+1,b+1\}$ have the same set of prime divisors?

I expect to spend more time on this and some related questions tomorrow; nevertheless, I post this here in advance, as I haven't posted anything for quite some time  :) .

--Vesselin

They exist. It is an old problem. Take $b+1=(a+1)^2$, where $a+2$ is a power of 2.

Ah, thank you, indeed it is simple (but the solution is cute  :) ); I didn't know it was a well-known problem. Ok, let's try an extension: for every $k$, do there exist $a \neq b$ such that the numbers in each of the pairs $\{a,b\}, \{a+1,b+1\}, \ldots, \{a+k,b+k\}$ have the same set of prime divisors?

Of course, the answer should be 'yes,' but certainly it doesn't look trivial. Or is it?  :?  Anyway, I think that's a nice question to ponder; I'll think about it tomorrow.


geometry
rectangle
combinatorics open
combinatorics

Is it possible to dissect a rectangle into 3 polygons (not rectangles) that are similar but not congruent?

If the rectangle is not a square, you may dissect it into three right-angled pairwise similar but non congruent triangles. Just draw the diagonal, then dissect one of the parts by the height from a corner to the diagonal.

Pierre.

Yes, but if the rectangle is a square? you have to prove there that it's impossible...

I gave an answer for the general but one exception case.... :D

Yes you skipped the most important and most difficult case :D

Do you know for sure it's impossible (in which case this would be a proposed problem)?

Hum...I never thought much about the square case, it DOES seem fairly difficult...

But my original question was to ask if it's possible to dissect a rectangle into 3 CONGRUENT polygons (not rectangles). Then I saw something like "If k congruent polygons tile up a rectangle (with k>1), then it is NOT YET KNOWN if k is necessarily even." So I was forced to modify my problem, but it seems still interesting...at least the square case... :lol:

The problem of odd dissections of a rectangle into congruents polygons is not open. Klarner has found a dissection of a rectangle into 11 congruents hexominos, and into 15 congruent trominos.

Pierre.


A,B,C,D are for concyclic points  M is the midpoint of BC, ME⊥BC,E=EM⊥DE
H1,H2,H3 are orthocenters of △ADE, △BDE,△CDE
Show that ∠H3H1E=∠H2H1E

I hope someone can solve the stubborn problem
 :)

[quote="mstoenescu"][quote="plane geometry"] ME⊥BC,E=EM∩DE
[/quote]
You mean that MED is a right angle, like you have done in your figure ?[/quote]
Yes,I am sorry for making the mistake
But I have edited the problem,hope you will enjoy it :)

thank you for editing but...I have no idea :blush:  to solve it ! i'm waiting for GREAT GEOMETERS here...


IN the sides BC,  CA  of  triangle ABC points D and E are taken such that

BD:DC =1:2 , and  CE:EA = 3:1. Let AD and BE intersect at O .Find these ratios :

1. AO:OD

2.BO:OE

3.[DOE]:[AOB]  where [DOE] means the area of triangle DOE

[hide="solution1"]

denote $[abc] = k$

$[ADC] = \frac{2k}{3}$

$[ADE] = \frac{2k}{3} * \frac{1}{4} = \frac{k}{6}$

$\displaystyle \text{ since ADE and AOC have the same area, }$

$[ADE] = [AOC] = \frac{k}{6}$

$[ODC] = [ADC] - [AOC] = \frac{2k}{3} - \frac{k}{6} = \frac{k}{2}$

$\displaystyle \text{ by looking at the picture, since AOC and DOC have the same height, we can conclude that } AO:OD = [ADC]:[DOC] = 1:3$[/hide]

[hide="solution2"]

denote $[ABC] = k$

$[ODC] = \frac{2}{3} * \frac{2}{3} * k = \frac{4k}{9}$

$[BOD] = \frac{1}{2} * [ODC] = \frac{2k}{9}$

$ \text { then BOC is equal to ODC plus BOD or } \frac{2k}{3}$

$[OEC] = [BEC] - [BOC] = \frac{3k}{4} - \frac{2k}{3} = \frac{k}{12}$

$ \text { since OEC and BOC have the same height, }$

$BO:OE = [BOC] : [OEC] = 8:1$[/hide]

[hide="solution 3"]
$[DOE]:[AOB]=\frac{AO\cdot BO}{DO\cdot EO}=\frac83$[/hide]

[quote]\displaystyle \displaystyle \text{ since ADE and AOC have the same area, }[quote]

Why is this so?
[/quote][/quote]

Wow I messed up that last post.  Anyway, Im just asking why the area of triangle ADE must be equal to the area of triangle AOC?


In a seminar today, I was discussing how jazz pianist Art Tatum, arguably one of the most innovative performers of all time in his style and approach, is often criticized for replacing musicality with his ostentatious virtuosity.  I was just wondering if anybody had any thoughts on the limits of "showing off" one's technical facility with respect to compromising musical expression.

Personally, I think nothing is lost in Tatum's style of playing.  Though it's extremely fast (up to 1000 notes a minute in certain recordings), and often hard to follow upon first listening, each note seems meticulously placed, with respect to its dynamic and articulation.

If you're not familiar with Tatum, I would highly suggest listening to some of his recordings.  Some of my favorites are "Get Happy" and "Aunt Hagar's Blues."

I like Art Tatum a lot. I personally happen to like even better Thelonius Monk, who had a very different approach to playing and composing jazz piano works.


how would one go about prooving mathematically that launching a projectile at an angle of 45 degrees produced the maximum horizontal distance?

It follows from the equation for the horizontal range $R$ of a projectile launched at an angle $\theta$ to the horizontal: $R = \frac{u^2sin2\theta}{g}$, where $u$ is the initial velocity and $g$ is the acceleration due to gravity.

Do you know how to derive this equation?

Thanks for giving me that equation cuz i really had no idea where to start but i think i got it. It took me awhile (obviously, i'm posting at 4 AM lol) because i had to look up the equation that said 2 sin x cos x = sin 2x. I did not know that seeing as though my trig class hasn't covered identities yet but at least i'll know it when we start learning it :D


[quote="yrudoy"]What is used as the tiebreaker for the 12 winners in the USAMO? What happens if, say, the 11th, 12th, 13th, and 14th top scores are all 23's?[/quote]

Maybe from how clearly the solutions are written (even though both are correct), whether solutions are elegant or even hand writing (I think this happened last year)

I thought they had the option to name more than 12 winners in the event of a tie (like Putnam does).  But if not, the teachers' manual specifically says that elegance of proofs may be used in breaking a tie.

The teachers' manual also says, incidentally, that a "carefully stated non-trivial extension" of a problem may be used as a tiebreaker.  However, you shouldn't do this.  The likelihood of it benefiting you as a tiebreaker is very small, whereas it's likely to take up time that you could be spending on other problems.  In addition, ANY incorrect math you write on your paper, even if it's part of an extension and not relevant to your main proof, can get you docked points.

They must have exactly 12 winners.  They will go to half-points if necessary to break ties.

I should add that if the AMC has to split hairs to break a tie at 12th place, they sometimes invite the 13th, 14th, etc. person to take the team selection test.  However, this is the exception, not the rule, and there are no exceptions to there being exactly 12 USAMO winners.

[quote="HiDN428"]i'm just curious(kinda new :oops: ) how the usamo is graded?[/quote]

7 points for each problem.  42 max.
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=270356[/url]
This thread might help a little.

Or read the Frequently Asked Questions post at the top of the form.

[quote="yrudoy"]What is used as the tiebreaker for the 12 winners in the USAMO? What happens if, say, the 11th, 12th, 13th, and 14th top scores are all 23's?[/quote]

Don't worry about it. My personal rule with math competitions is: do the problems, let the contest people worry about results, then look them up when they're available.

[quote="DPatrick"]I...they sometimes invite the 13th, 14th, etc. person to take the team selection test.  However, this is the exception, not the rule...[/quote]I believe the exception only happens if some of the winners choose not to take the TST (e.g. they are going to RSI), or are not eligible for the US Team (e.g. Canadians).

Actually they invited a 13th person to take it last year before knowing that two people wouldn't be taking it. I don't believe they have any cut and dried rules at all in these tiebreaking procedures.


function
integration
real analysis
real analysis unsolved

$ f: I \minus{} > R$ a derivable function with $ f''$ continuos .If exist $ a,b \in I$  $ a < b$ s.t $ f'(a) \equal{} f'(b) \equal{} 0$ then exist $ c \in (a,b)$ s.t
$ |f(a) \minus{} f(b)| \le \frac {(b \minus{} a)^2}{4} |f''(c)|$

Solution#1. $ f(b)\minus{}f(a)\equal{}\int_a^b\left(\frac{a\plus{}b}2\minus{}x\right)f''(x)\,dx$. If $ |f''(c)|<\frac{4|f(b)\minus{}f(a)|}{(b\minus{}a)^2}$, for $ c\in(a;b)$, then $ |f(b)\minus{}f(a)|<\frac{4|f(b)\minus{}f(a)|}{(b\minus{}a)^2}\int_a^b\left|\frac{a\plus{}b}2\minus{}x\right|\,dx\equal{}|f(b)\minus{}f(a)|$. That's impossible.

Solution#2. The existence of $ f''$ is enough (it does not have to be continuous or even integrable). Suppose that $ |f''(c)|<\frac{4|f(b)\minus{}f(a)|}{(b\minus{}a)^2}\equal{}: M$, for $ c\in(a;b)$. Then:
for $ a<x\leqslant\frac{a\plus{}b}2$, $ |f'(x)|\equal{}|f'(x)\minus{}f'(a)|\equal{}|f''(\xi)|(x\minus{}a)<M(x\minus{}a)$;
for $ b>x\geqslant\frac{a\plus{}b}2$, $ |f'(x)|\equal{}|f'(b)\minus{}f'(x)|\equal{}|f''(\zeta)|(b\minus{}x)<M(b\minus{}x)$.
Hence $ |f(b)\minus{}f(a)|<M\int_a^{\frac{a\plus{}b}2}(x\minus{}a)\,dx\plus{}M\int_{\frac{a\plus{}b}2}^b(b\minus{}x)\,dx\equal{}|f(b)\minus{}f(a)|.$


arithmetic sequence
combinatorics theorems
combinatorics

I dont know about VV theorem.Can you help me?

There are two versions of VdW's theorem. The finite one and the infinite one. These two versions are equivalent by compacity argument.

Infinite case :

Let $r,k$ be two positive integers.
If you color each positive integer with one color among $r$ possible colors, then you can find an monochromatic arithmetic progression of length $k$.

Finite case :

Let $r,k$ be two positive integers.
There is an integer $n_0 = n_0(k,r)$ such that for all $n \geq n_0$ and all $r$-colouring of the integers $1, \cdots, n $, there is an monocromatic arithmetic progression of length $k$.

The least integer $n_0$ which satisfies the above property is the Van der Waerden number $W(k,r).$ As usual in Ramsey theorey, only few non-trivial values of $W(k,r)$ ar known, and the current bounds are quite large.

Pierre.

and    proof???  .i cannot find in Wolfam website


calculus
integration
trigonometry
calculus computations

Show that
$ \displaystyle \int\limits_0^{\frac {\pi }{6}} {\arctan \left( {\frac {{\sqrt 2 \cos \left( {3x} \right)}}{{\left( {2\cos \left( {2x} \right) \plus{} 3} \right)\sqrt {\cos \left( {2x} \right)} }}} \right)dx} \equal{} \minus{}\int\limits_{\frac {\pi }{6}}^{\frac {\pi }{4}} {\arctan \left( {\frac {{\sqrt 2 \cos \left( {3x} \right)}}{{\left( {2\cos \left( {2x} \right) \plus{} 3} \right)\sqrt {\cos \left( {2x} \right)} }}} \right)dx}$

Heh   :)  , Andrew Tom this was the origin problem... to show that int from 0 to π/4 is equal to 0.
I saw with graph that the point π/6 is the root and that the integrals from 0 to π/6 and from π/6 to π/4 are oposites but i coundt prove it. I made a substitution x=2t-π/3 and thought of identity $ arctanx \equal{} 1/2*arctan(2x/(1 \minus{} x^2))$ though was  difficult for me to continue it


induction
inequalities
inequalities solved

If a,b,c >=0 and n is natural >=2 prove that:

(n-1)/2 * (a^n+b^n)+c^n>=nabc*((a+b)/2)^(n-3).

cheers! :D

Lemma: Let a,b>=0. Then (2^(n-1))(a^n+b^n)>=(a+b)^n (A generalization of C.B.S. inequality).
Solution: Induction on n. The case n=1 is obvious.
We have 2(a^(n+1)+b^(n+1))>=(a+b)(a^n+b^n) <=> (a-b)(a^n-b^n)>=0. The last part is obvious since the two terms have the same sign.  From this the lemma follows immediately.

Back to the problem! Multiplying the two terms by 2 we get ((n-2)a^n+b^n+c^n)+(a^n+(n-2)b^n+c^n) >= {by AM-GM inequality} n(a^(n-2))bc+na(b^(n-2))c=nabc(a^(n-3)+b^(n-3))>= {by the Lemma} 2nabc((a+b)/2)^(n-3).


Find all natural numbers m,n s.t :
   (m+n-5) 2 =9mn

It's pell

the solutions are:

(m,n)=((L_(2k))^2,(L_(2k+2))^2) and (m,n)=(5(F_(2k))^2,5(F_(2k+2))^2)

L_n= Lucas sequence;
F_n= Fibonnaci sequence.

cheers! :D

yes,That's right! :)  :)
But Markov or Pell is true! :D

And? That is all? No one interested how he get the answer? :)


A digital clock shows hours and minutes. How many times during a 12-hour period will the sum of the digits on this clock be 3?

There aren't that many possibilities of times where the digits add up to $ 3$, so we could just count:

$ 1: 02$,
$ 1: 11$,
$ 1: 20$,
$ 2: 01$,
$ 2: 10$, and
$ 3: 00$,

making $ \fbox{6}$ such times.


function
logarithms
number theory unsolved
number theory

It's well known that  if $\pi(\cdot)$ is the counting prime function
\[\frac{ \pi(n)\ln n}{n}\to 1, \quad {\rm when } \quad  n\to \infty\]
but is in general $ \frac{ \pi(n)\ln n}{n}\le  1 \ $  for every positive integer $n$?.  Answer: [b]{n=19}[/b]   :blush: .

What I know is that although the estimate of prime counting function $\frac n{\ln n}$ is an underestimate for every number that our computer can deal with, It is already proved that for some numbers that are large enough, $\frac n{\ln n}>\pi\left(n\right)$. In early 20th century, someone proved that you are able to find such an example if your number is greater than $10^{10^{10^{34}}}$, which is an extraordinary, and famous number.


MATHCOUNTS Duke college articles percent search Harvard

[quote="richpang"]:)
I'm also from LA (12th grade) and I tutored him (not frequently though), though he already has a much higher SAT score than me and did much better than me at nats Good luck and congrats to everyone else[/quote]

if you and neal don't mind giving out his SAT score, what was it? I failed horribly on the SAT math section this year.... only a 670. I was expecting a 770 at least.  :(

...That's absolutely incredible. I'm proud of my 1580 in 9th grade. :P I can't believe I got one wrong on math, though. >_< At least I got perfect on verbal. :lol:

My brother Seungjun didn't make nationals, but got 1560 (perfect math and 760 verbal) in 7th, age 12. I got 1520 (750 math, 770 verbal) in 7th grade as well.

This year on the old tests I got 800 math and 500 verbal=1300= :(

Lol, you guys suck at verbal. I got 670 + 670 = 1340 (in 7th grade), but I screwed up on the math real badly. On the practice tests, I got 720-750 on math. I should have gotten at least 1390. *sigh*

Last year in 8th grade 770 math, 610 verbal
darn I definitely could've done better.
My score is pathetic compared to all yours.

For previous years, I got 650 Mat / 600 Verbal in 7th grade, and 750 Math / 720 Verbal in 8th grade. I'm very close together in the two, surprisingly; I would have expected to do better in Math.

Lol. I did it in the seventh grade. I was positively awful then at both math and verbal. I hadn't done RDWPC or Mathcounts back then. I got a 720/730. I won't be doing it again.

I wouldn't call a seventh grader getting a 720/730 "awful!" Bujeezus!

I wonder how this "Congratulations, Neal Wu!" turned into "What did you get on the SAT?" :P  And you shouldn't be worried about your SAT score.  Those tests aren't accurate anyways, and you have like at least 3 more years before you take them for real.

Does anyone want to know Neal's score or would that be giving away too much personal information about him?  :?

[quote="richpang"]ummm, idk if he cares but ill post it. i apologize neal, i feel like bragging about you. 1550, but this was on the old one.[/quote]
1550

Lovely. I got a grand 780 as a junior. And I call myself a math nerd. For shame... :blush:
At least I increased 50 points in 4 years? Yay?
Being optimistic gets hard sometimes. ;)

You need to take the new SAT; starting in 2007 (I think), all colleges require the new SAT.

So what? Do you *really* think they'd turn me down? And I made sure that the College Board did *not* delete my score.

I hope to be in college by 07 anyway. ...

Really? How do you plan to do that? Early entrance program?

[quote="solafidefarms"]Do you *really* think they'd turn me down?[/quote]

I can't say what they would or wouldn't do, but they may well. If you did that well in 7th grade, then why not just take it once more to make sure you can get into whatever college you want?

And as for being in college in '07... what grade are you in right now? I wouldn't skip more than one year of high school... there are some very valuable things you can do in high school that you can't really do in college, and it's also nice to not be two years younger than everyone else.

I'm not sure if anyone said this, but one place said that the writing section was made so that the SAT scores wer more balanced for boys and girls. The girls had the advantage in the verbal section, but the boys had twice the advantage in the math section. With the new verbal section, things can change.

I'm taking it in January, my expected score is

700+520+480 = 1700...

OK, ya'll have convinced me. But I *don't* plan to study one word beforehand ;)

I don't really care about that... I never study for the SAT. :P

How did you guys get such high verbal scores? what special things did you study? And did you guys do formal SAT prep?
Ryan

I really have no idea. I'm very surprised that I scored as well as I did. Although, now that analogies are gone, it should be easier... I remember one in particular last year where I basically guessed because two options were almost equally likely.

As for the stories, don't read them. Look at the questions, then skim through to find out where in the story it talks about that. Gives you more time to think about the questions, and hopefully some time to review them, as well. On each verbal section, I just barely had time to go back through all the questions by doing that.

RIght now on the writing section, I can't write an essay in 25 minutes. I don't write that fast. They should make it 30 minutes.

I'm glad those analogies are gone. They were just a test for logic and vocabulary. Now only sentence completions are there to stump you for vocabulary.  :D

I actually liked analogies. They're better than the essay at least. I'm not really that good with "impromptu" essays.

I know that some universities at least don't even look at the writing section unless you did really badly on the other two.

BTW guys - i checked, collegeboard deleted my scores. Is there any way to get them back? I wanna send them to the high schools im considering and i have no proof now.

Yeah, it sucks. I forgot to ask them to save my scores, and now they're gone. :mad:


I guess we needed a thread...

For Wisconsin, we really haven't decided yet but we'll get a decision by the end of the week (deadline).

our pin was awful last year...but i think we (meaning the pin-makers) came up with a kinda ok pin. at least it was better than last year.

a awful pin would have been better than what we had.

Weren't you both in the same teams with the same pins last year?

[quote="solafidefarms"]Connecticut pins are worthless. Don't accept anything less than 4 of their pins for one of yours. :P ;) [2 years ago I traded 12 Ct pins for one of my Georgia pins :)...][/quote]
well yeah
maybe you can buy more pins for less money per pin.
and so for georgia

Ours were sooooooooooooo bad. It generally took 2-3 pins of ours to get one of another state that wasn't too rare. Then the other CT team members couldn't get any because that team shared the CT pins they received and they would no longer trade for them.

i thought dan u and jon had all the pins?

didn't you have hundreds though?

jorian

ok, i will reveal the real reason that i really wanted to go

i was gonna buy like a thousand of corny pins off of ebay (disney, etc.) and trade them

if i were competing, i'd personally rather want a corny one. it's kind of funny going around with a bunch of random pins :P

you can take my idea if u want. should be pretty cool.

jorian

[quote="perfect628"]We're lucky anyone would take ours at all. I had to take other pins, pin mine upside down in their hands, and RUN.[/quote]
so that explains why I have Multiple of those pins...

Wait, wait.

You're telling me that MATHCOUNTS has official pins (like the ones at $\text{Disneyland}$), that people actully COLLECT?

Thats absurd...  :ninja:

[quote="xxazurewrathxx"]I have all the pins from Nationals 2004. Anyone wanna buy the collection (+ the hat too)? I'll throw in ten of those really lame florida pins I got.  :wink:

I think back then, the Guam and CA pins were the best. I also received a bunch of random Geico pins and those of other businesses.

[quote="perfect628"]We're lucky anyone would take ours at all. I had to take other pins, pin mine upside down in their hands, and RUN.[/quote]

Err.. you pinned it on their hands?[/quote]

ill give you 5 bucks tomorrow for them.

[quote="diophantient"][quote="pianoforte"][quote="solafidefarms"]Connecticut pins are worthless. Don't accept anything less than 4 of their pins for one of yours. :P ;) [2 years ago I traded 12 Ct pins for one of my Georgia pins :)...][/quote]
It seemed that Georgias were worthless too last year, I got around 10 for 1 of my extra Nebraskas  :P .

Edit: I went and counted, it was 8 GAs. I know a teammate had about 15.[/quote]
What are you talking about?
Georgia was one of the most valuable ones, because the kids only had 20 or so each.

I got lucky and got a georgia pin early on for 1:1, but I remember on Friday some kids traded 7 of theirs for a Georgia.[/quote]

Really? Saturday night (after banquet): "I'll trade you these [handful] Gerogia's for any pin that I don't already have [Nebraska]"...My teammate traded 2 pins for 15 GA's.

Well yeah because at the end their coach finally gave them more. In the beginning I was "trading" with this one Georgia kid and he just walked off with my pin and I'm like hey ese! What you doin' jackin' my pimpin' pins? He's like no ese we ain't got none so then I'm like aww foshizzle. Kinda felt bad for the kid and let him have it. I had lots of extras anyway. Last year, everyone's reaction to Wisconsin's...

"Oh those are worthless"
"I got one of them last year"
"Your pins are boring"
"Ooooh...a Wisconsin...pssh*sarcasm*"

But this year our pins are going to be better thanks to Darwin.

they've had those little blue pins with the outline of Alaska in them, before at least

I have an Alaska pin from last year...
The only state without is department of defense, I think.

Alaska had both a blue alaska pin, and a walrus pin last year (i had both at some point).

[quote="Phelpedo"]I hate Rhode Island. Every year they lie and say they only have 10 pins each or something, so they want 2 for 1. Luckily, 2 CT were more useless than 1 RI.[/quote]
Rhode Island didn't do that last year, or maybe I'm just being forgetful...

If you want to get a lot of cool pins, challenge your teammates at nationals to poker.  I won every single pin from one of my teammates last year at five card draw.  Now I've got tons of extras.  Might pull it off this year, too.

[quote="kstan013"]we're doing the cheese pi one, although I liked the real math one.[/quote]

Hahaha!  Cheese Pi Power!  Cheese Pi Owns!  What was it, 2 to 1??

[quote="kstan013"]WHAT'S SO GOOD ABOUT CHEESE PI???[/quote]

It's Cheese Pi.

EDIT: Bucky Badger would have been cool but it's too expensive.  I think it's around $\$5.00$ for [i]one[/i] pin.  [b]ONE PIN[/b]

Yeah well real math cheese is cool because it's cheese real math :P Oh yeah Frost, btw, Wisconsinites like cheese and most people that I know of that come from out of state into Wisconsin say they have to have cheese curds every time they come over. There's a city that my grandma lives close to that claims to be the "cheese curd capital of the world". Have you ever heard of Colby cheese? Yeah. That originated in a town called Colby about 45 miles west from where I live. They don't call Wisconsin the dairy state for nothing, so we're exploiting that in our pins.

[quote="frost13"]I drove cross country a couple years ago, and the second I was in Wisconsin there were signs everywhere for cheese. Cheese this and cheese that.  It was crazy.

We stopped at this "cheese factory" (for lack of a better name, just a big house where they made different types of cheese) and got some cheese.  I got talked into trying some free Cheese Curds.  Never has there been anything more gross!  But the cheese was the real deal :D

Oh yeah, and Nice Pin![/quote]

Cheese curds are good if you're used to them.  Also the dairy here is very inexpensive...

Oh yeah i forgot.  You have to have deep fried cheese curds.  Nothing like it.

[quote="footballrocks41237"]Oh yeah i forgot.  You have to have deep fried cheese curds.  Nothing like it.[/quote]

uuhh.. wut's a cheese curd???

No really.. The only kind of "chip in a bag" we've had for a while is Lays chips. I haven't had cheese curds since '01(I think)

A deep fried cheese curd is this:
[img]http://img61.imageshack.us/img61/4351/friedcurdstw5.jpg[/img]

A cheese curd is this:
[img]http://www4.mailordercentral.com/carrvalleycheeseco/images/1948curd-b.jpg[/img]

i know this might make me sound stupid, but what r u guys talking about (pins?)does each state gets their own down at nats? how much is WVs worth.  not a lot, i bet

 :D 100th post   :D

Nobody's pin is worth more or less than another at the beginning... the less WV pins people have, the more they want then....

Everybody gets pins...

to have fun trading them while meeting new people.

But isn't it just a distraction from the actual competition?

No, unless you focus too much on the pins


In a class of 50 students, 28 take science, 21 take French, and 5 students take both classes. How many students take neither class?

Using PIE: $ 50\minus{}28\minus{}21\plus{}5\plus{}x\equal{}0 \implies x\equal{}\boxed{6}$.


Let P be an arbitrary point in a triangle ABC and let X, Y, Z be foots of perpendiculars from P on AB, BC and CA. Show that AP + BP + CP >= 2(PX + PY + PZ).


trigonometry calculus integration inequalities

For $n=1,\ 2,\ 3,\ \cdots,$ let $a_{n}=\tan (11n).$ Prove the following contexts. Here $\pi =3.14159265\cdots$ is a circular constant.

(1) $\frac{\pi}{711}<11-\frac{7\pi}{2}<\frac{\pi}{709}.$

(2) $a_{1}<0<a_{2}.$
(3) $a_{1},\ a_{3},\ a_{5},\ a_{7},\ \cdots ,\ a_{707},\ a_{709}$ is an increasing sequence.
(4) The infinite sequence $a_{1},\ a_{3},\ a_{5},\ a_{7},\ \cdots$ isn't an increasing sequence.

1. $\iff \frac{4979\pi}{1422}<11\iff \pi<\frac{1422\cdot 11}{4979}<\frac{2}{7}\cdot 11=\frac{22}{7}=3.\overline{142857}$
but we are given the first 9 decimal digits of pi so it is apparent (or use integral trick :P )
b. $\iff \frac{(7\cdot 709+2)\pi}{2\cdot 709}>11\iff \pi>\frac{22\cdot 709}{7\cdot 709+2}=\frac{15598}{4965}=3.141591...$ I can't find a better way for the second part...

[quote="kunny"]For $n=1,\ 2,\ 3,\ \cdots,$ let $a_{n}=\tan (11n)$.
(2) $a_{1}<0<a_{2}.$[/quote]

[hide="2."]We must show that:
$\tan(11)<0<\tan(22)$

[hide="Proof of the left inequality"]Notice that:

\[3\pi<11<\frac{7\pi}{2}\]

Since the period of $\tan(\theta)$ is $\pi$, we may apply tangents to obtain:

\[-\infty<\tan(11)<0\]

Therefore, $\tan(11)<0$.
[/hide][hide="Proof of the right inequality"]Notice that:

\[\frac{13\pi}{2}<22<7\pi\]

Since the period of $\tan(\theta)$ is $\pi$, we may apply tangents to obtain:

\[0<\tan(22)<+\infty\]

Therefore, $\tan(22)>0$[/hide][/hide]


ill post my thoughts later (im sleepy), but post yours

if you really care this can extend to all 3 of the movies

I knew this was going to come up... sooner or later... except it was later than i thought it would be...

I think the original was actually good...

Original was good. The others...eh. I watch them for the special effects. The philosophy I can read on my own, after all.

my friends who have seen it say #1>#2>#3, and that can't be good, considering #2 was pretty stupid compared to the brialliance of #1. i guess i'll see it anyways...

[quote="mathfanatic"]Still haven't seen the 2nd one (yes, so slap me), and parents probably will discourage me from going to see the 3rd one.[/quote]

Yay! At least im not the only one that hasn't seen 2...

and sequels are usually not as good as the originals...

i think the 2nd sucked. the 1st was the greatest and the 2nd was inbetween, but it was closer to the 1st's greatness.

1st one was good because it uncovered the matrix, it was interesting. they made a sequal with too much action, nothing new, so it sucked. i heard 3rd one was the same way...

[quote="Chinaboy"]

and sequels are usually not as good as the originals...[/quote]

That's not the case with LoTR

me either...the ROTK is gonna be AWESOME!!! especially if the follow the book...

what's ROTK? is it a new movie similar to the LoTR?

"I'm going to see it in New York, ha ha, ha ha, I'm going to see it in New York so ***** all you ****" (from South Park, edited by me about the destination and asterisized all the "bad" words)

Also Terminator, although the third suxorz...I mean if you haven't seen the third you'd be happy right about now but [b]nooooo[/b], Warner Brothers weren't satisfied with the gazillions they already had so they decided to make the ending bad instead of good, me really :-x
The second was actually pretty good, and please differ all you want with this one.

Terminator = original idea
Terminator 2 = twist on the original idea
Terminator 3 = [i]exactly[/i] the same as T2, except with a hot chick

I agree, Tare.  The first one was fairly good (although I liked Robocop better, but oh well), and the second one was good because it had some fairly new ideas (e.g. terminator fighting terminator, sarah having to trust the "bad guy") in it too.  But the third one had basically nothing new.  The bad terminator was still basically unstoppable, and could change shape, etc.  The only thing new was the concept of an underground bunker, and even that was terrible.  Hey, as long as it was going to suxorz, they might have thrown in a bunch of action scenes.  (I thought the wars between humans and robots were actually kinda cool.)

[quote="rippledance"]Terminator is okay. Personally I prefer Robocop. =)

Hm...Matrix => LoTR => Terminator...?[/quote]
Yup, Matrix => LoTR => Terminator => MysticTerminator :)

I vote one for Christina...er, that person...what's her name...T-ReX, let's just leave it at that :lol:

Yes, it looks like spam, it sounds like spam, heck it even tastes like spam, maybe IT IS SPAM :lol: </jk>

[quote="rippledance"]

Hm...Matrix => LoTR => Terminator...?[/quote]

She must be quite confused...the correct order is:

LotR > Matrix > Terminator

[quote="MithsApprentice"]She must be quite confused...the correct order is:

LotR > Matrix > Terminator[/quote]

yup yup yup... :D


induction
combinatorics unsolved
combinatorics

Let $\gamma = \{a_1,a_2,...,a_k\}$ be a subset of the first n positive integers which has been arranged in increasing order: $1 \leq  a_1 < a_2 < .... < a_k  \leq n$.

a.) How many such $\gamma$ are there that begin with an odd number and thereafter alternate in parity: $\gamma  = \{odd, even, odd, even, \ldots \}$ ? For convenience let the empty set be included in the count.

b.) How many $\gamma$ are there of size $k$ ?

For a
Denote by f1(n) the number of sequences for such an n that end up in an odd number.
Denote by f0(n).................................................................................. even number.
See that f1(2n)=f1(2n-1) and f0(2n)=f1(2n-1)+f0(2n-1).
And that f1(2n+1)=f0(2n)+f1(2n+1) and f0(2n+1)=f0(2n).
These result from dividing the sequences for an n as sequences that end up in n and sequences that don't end up in n.
We must find f0(n)+f1(n) for each n.
Take F_n be the Fibonacci sequence F_1=1, F_2=1
The sequences f0(n) and f1(n) go like this
f0 1 2 2 5 5 13
f1 1 1 3 3 8 8
I considered f0(1)=1 (the only seqence to be the empty set)
It can be proven by induction that f0(n)=F_(2[n/2]+1) and f1(n)=F_(2[(n+1)/2]).
Them the required number is F_(n+2), because {2[n/2]+1, 2[(n+1)/2]}={n,n+1}.
By the way [x] is trunc(x), the greatest integer  \leq x.

Can anyone please tell me why the answer is not $\binom nk$ for part [b]b)[/b]? What am I misunderstanding here?

I think that we have to find the number of $\gamma$ sequences from $a)$ that have exactly $k$ terms. If not, the answer is what grobber said.

oooohh. I see. Thanks for the suggestion. That should be stated clearly though :).

It is time to obtain appropriate link...  ;)


This is probably useful, so I'd put this in the forum as a relatively worthless gift to you all :lol:

let's say we want to type $P_{r_{i_{1}}}$

Mouse over the actual code to see the LaTeX input.

Now why the LaTeX would interpret it like that is beyond my understanding. I suppose it take a look at the innermost brackets first, but that just my guess.

I'm not sure what you're pointing out.  That's what I would expect to see  :maybe:

whoops... looks like this isn't useful to you guys after all. My mistakes :p


Find the number of non-negative integer solutions to $ x\plus{}2y\plus{}4z\equal{}50$

It's actually easier to do it without generating functions:

Note that x has to be even; therefore we have

$ x' \plus{} y \plus{} 2z \equal{} 25$

to which the number of solutions is

$ 26\plus{}24\plus{}...\plus{}2 \equal{} 182$

That's a nice solution.  I saw this problem somewhere else, and they started to use a generating functions method, but they never finished it.  I got the same answer, but I wasn't sure if it was right.  If nobody else posts, I'll give my solution.

[quote="iniquitus"]It's actually easier to do it without generating functions:

Note that x has to be even; therefore we have

$ x' \plus{} y \plus{} 2z \equal{} 25$

to which the number of solutions is

$ 26 \plus{} 24 \plus{} ... \plus{} 2 \equal{} 182$[/quote]

From where do you get the series 26 + 24 + ... 2 = 182?

if z is 0, there are [color=red]26[/color] solutions to $ x' \plus{} y \equal{} 25$.
If z is 1, there are [color=red]24[/color] solutions to $ x' \plus{} y \equal{} 23$.
If z is 2, there are [color=red]22[/color] solutions to $ x' \plus{} y \equal{} 21$.
...
If z is 12, there are [color=red]2[/color] solutions to $ x' \plus{} y \equal{} 1$.

[hide="complete generating functions solution"]
The $ x$ can give you any integer from 0 onwards, so it has a GF of $ 1\plus{}x\plus{}x^2\plus{}x^3...\equal{}\frac{1}{1\minus{}x}$.

The $ 2y$ can give you any integer multiple of 2 from 0 onwards, so it has a GF of $ 1\plus{}x^2\plus{}x^4\plus{}x^6...\equal{}\frac{1}{1\minus{}x^2}$.

The $ 4z$ can give you any integer multiple of 4 from 0 onwards, so it has a GF of $ 1\plus{}x^4\plus{}x^8\plus{}x^{12}...\equal{}\frac{1}{1\minus{}x^4}$.

We multiply these GFs:

$ (\frac{1}{1\minus{}x})(\frac{1}{1\minus{}x^2})(\frac{1}{1\minus{}x^4})$

Now we want to make the exponents of this GF easier to evaluate, so we manipulate it a little bit:

$ \frac{1}{(1\minus{}x)(1\minus{}x^2)(1\minus{}x^4)} \equal{}$

$ \frac{1\plus{}x}{(1\minus{}x^2)^2(1\minus{}x^4)}\equal{}$

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

We can use binomial theorem on the denominator, and we see we can multiply out the numerator to make things simpler.  We do so and get:

$ \frac{x^5\plus{}x^4\plus{}2x^3\plus{}2x^2\plus{}x\plus{}1}{(1\minus{}x^4)^3} \equal{} (x^5\plus{}x^4\plus{}2x^3\plus{}2x^2\plus{}x\plus{}1)(1\minus{}x^4)^{\minus{}3}$

We are looking for the coefficient of $ x^{50}$, and we realize that every term in $ (1\minus{}x^4)^{\minus{}3}$ will be of the form $ 1, x^4, x^8, x^{12}...$.  Thus the only way we can combine terms to get an $ x^{50}$ is to multiply the $ 2x^2$ and then some $ ax^{48}$.

So now all we have left to do is find the coefficent of $ x^{48}$ in $ (1\minus{}x^4)^{\minus{}3}$.  This is equivalent to the coefficent of $ x^{12}$ in $ (1\minus{}x)^{\minus{}3}$

By binomial theorem, this will be $ \dbinom{12\plus{}3\minus{}1}{12} \equal{} 91$.

So we know we have $ 91x^{48}$.  Multiply this with $ 2x^2$ and we get $ 182x^{50}$.  So the coefficient of $ x^{50}$ is $ \boxed{182}$.

[/hide]


Putnam
algebra
polynomial
floor function
inequalities
function
college contests

Find a nonzero polynomial $P(x,y)$ such that $P(\lfloor a\rfloor,\lfloor 2a\rfloor)=0$ for all real numbers $a.$

(Note: $\lfloor v\rfloor$ is the greatest integer less than or equal to $v.$)

In case anyone wants motivation for that, note that we have the inequality chain $2\lfloor a\rfloor \leq \lfloor 2a\rfloor\leq 2\lfloor a\rfloor +1$, and that one of the two inequalities is always an equality, since the middle term is an integer between two integers, inclusive.

We conclude that

$(\lfloor 2a\rfloor - 2\lfloor a\rfloor)(\lfloor 2a\rfloor - 2\lfloor a\rfloor-1)=0$

for any $a\in\mathbb{R}$.  Put $y=\lfloor 2a\rfloor$, $x=\lfloor a\rfloor$, so that

$P(x,y)=(y-2x)(y-2x-1)$

satisfies the conditions of the problem.

[quote="Kent Merryfield"]Undoubtedly the easiest problem on the test. $P(x,y)=(2x-y)(2x+1-y)$ will do.[/quote]

Just curious. Did you still have to prove that it satisfies for every value? Or is it sufficient to state the function?

[quote="Kent Merryfield"]You should put some kind of argument. Had I been writing it up, I would have considered the two cases $k\le a<k+\frac12$ for some integer $k$ and $k+\frac12\le a<k+1.$

Or you could write it up as blahblahblah did.[/quote]

That's what I did. It was one page, so it didn't take that long. (It did take me quite long to realize that if y-2x is always 0 or 1, then (y-2x)(y-2x-1) is always 0 :( ). I hope I didn't leave out some little condition.

Either $\lfloor2a\lfloor=2\lfloor a\rfloor$ or $\lfloor2a\lfloor=2\lfloor a\rfloor+1$, so $P(x,y)=(2x-y)(2x-y+1)$ works.


geometry
incenter
search
geometry proposed

Let $ ABC$ be a triangle with $ (I)$ its incircle. Denote $ B_0,C_0$ repsetively by the tangency points of $ (I)$ with $ AC,AB$. Let $ B_1,C_1$ be the projections from $ B,C$ onto $ AC,AB$ and $ B_2,C_2$ respetively are the foot of the inner bisectors from $ B,C$ onto $ AC,AB$. Denote $ I_1,I_2,I_3$ respectively by the projections from $ I$ onto $ B_0C_0$, $ B_1C_1$, $ B_2C_2$. Prove that $ I\in (I_1I_2I_3)$.

we have to show B0C0,B1C1,B2C2 are concurrent,like dear[b] jayme[/b] mentioned above
but I don't know too much about the A-point of Pelletier point

[quote="jayme"]Dear Mathlinkers,
the Pelletier's points was pointed by Yetti in one of my last problem...
For a first contact see
http://www.mathlinks.ro/Forum/viewtopic.php?t=289088
where another line passes through the point of concurs...
Sincerely
Jean-Louis[/quote]

Historical note:

A trivial search will show that Pelletier triangle with vertices in all 3 Pelletier points

was first mentioned by the P-name on this site by my dear friend Maj. Pestich

years ago in the post

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=285398675&t=26000,

M.T.

Another result:
Let $ A_1$ be the projection of $ A$ onto $ BC$. $ A_1C_1\cap A_0C_0 \equal{} \{S\}, A_1B_1\cap A_0B_0 \equal{} \{T\}. C_1B_1\cap C_0B_0 \equal{} \{P\}, SB_0\cap TC_0 \equal{} \{W\}$. Prove that $ W$ lies on $ (I)$ and $ W,P,Z_0$ are collinear.
Similar the second Schröder's point .
Is it well-known?

Dear Livetolove,
you are right and these results are only mentioned as generalization in my article
http://perso.orange.fr/jl.ayme  vol. 2  Les deux points de Schroeter.
Be carefull with the name because there exist another geometer named Schröder...
Sincerely
Jean-Louis

Dear Mark,
I haven't try the "search"  but now with your answer, I remember this perspectrix that is tangent to the incircle.
From where comes this nice result? Any ideas of a proof?
Sincerely
Jean-Louis

[quote="jayme"]Dear Mathlinkers,
note that the point of concurs of B0C0, B1C1 and B2C2 (named A-point of Pelletier) is also on this circle.
Sincerely
Jean-Louis[/quote]

My proof of your fact: http://www.mathlinks.ro/viewtopic.php?p=1571697#1571697


I am trying to show that if $ X \subset C([a,b],\mathbb{R})$ is convex, i.e. if for all $ x, y, \in X$ there is $ z \in X$ such that $ d(x,y)\equal{} d(x,z)\plus{}d(z,y)$, then the mapping $ F(t)\equal{} (1\minus{}t)x\plus{}ty$ on $ [0,1]$ has its image properly contained in $ X$.  The space has the uniform metric.  Can someone help?

It isn't true.
Let $ f_{q}(x) : \equal{} q$ for $ x \in [0,1]$ ($ q \in Q$)
And let  $ X \equal{} \{f_q: q \in Q \}$


[quote="imedeen"]Let $ a,b,c \ge 0$ such that $ a \plus{} b \plus{} c \equal{} 2$ ,Prove that
\[ \frac {a}{{{b^2} \plus{} {c^2} \plus{} 3}} \plus{} \frac {b}{{{a^2} \plus{} {c^2} \plus{} 3}} \plus{} \frac {c}{{{a^2} \plus{} {b^2} \plus{} 3}} \ge \frac {1}{2}\]
[/quote]

 :oops:

WLOG $ c \equal{} \max\{a,b,c\}$. Because :
\[ a(b^2 \plus{} c^2) \plus{} b(c^2 \plus{} a^2) \plus{} c(a^2 \plus{} b^2) \equal{} ab(a \plus{} b \minus{} 2c) \plus{} c(a \plus{} b)(a \plus{} b \plus{} c)\]
\[ \le 2c(a \plus{} b) \le \frac {(c \plus{} a \plus{} b)^2}{2} \equal{} 2\]
Then, use the Cauchy–Schwarz inequality :
\[ \frac {a}{{{b^2} \plus{} {c^2} \plus{} 3}} \plus{} \frac {b}{{{a^2} \plus{} {c^2} \plus{} 3}} \plus{} \frac {c}{{{a^2} \plus{} {b^2} \plus{} 3}}\]
\[ \equal{} \frac {a^2}{{a({b^2} \plus{} {c^2}) \plus{} 3a}} \plus{} \frac {b^2}{{b({a^2} \plus{} {c^2}) \plus{} 3b}} \plus{} \frac {c^2}{{c({a^2} \plus{} {b^2}) \plus{} 3c}}\]
\[ \ge \frac {(a \plus{} b \plus{} c)^2}{a(b^2 \plus{} c^2) \plus{} b(c^2 \plus{} a^2) \plus{} c(a^2 \plus{} b^2) \plus{} 3(a \plus{} b \plus{} c)}\]
\[ \equal{} \frac {4}{a(b^2 \plus{} c^2) \plus{} b(c^2 \plus{} a^2) \plus{} c(a^2 \plus{} b^2) \plus{} 6} \ge \frac {1}{2}\]
Equality holds if $ (a,b,c) \equal{} (0,1,1)$

@Thanx Mathias_DK

[quote="hophinhan"]
\[ a(b^2 \plus{} c^2) \plus{} b(c^2 \plus{} a^2) \plus{} c(a^2 \plus{} b^2) \minus{} 2 \equal{} ab(a \plus{} b \minus{} 2c) \plus{} c(a \plus{} b)(a \plus{} b \plus{} c)\]
[/quote]

Very nice proof, but there is a typo :P

Who have other solution use mixing variable?
[i]Hophinhan[/i] 's solution is very nice


search
combinatorics unsolved
combinatorics

Let $ A_1,A_2,...,A_r$ be a collection of distinct subsets of $ B \equal{} \{1,2,...,n\}$ and $ A_i \cap A_j$ is even for $ i,j \in B$($ i,j$ is not necessarily distinct).
Find the [b]max[/b] value of $ r$

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=815421098&t=13941

Pierre.


This might be little bit hard for the beginning MC people but once you know what my question is saying, it's really easy question.

MAO Alpha

I'll change the wording little bit.

If the ratio $\frac {a}{b} = 0.37545454545454...$ where $a$ and $b$ are relatively prime natural numbers (meaning only 1 divides $a$ and $b$), find $b-a$.

A) 687
B) 1513
C) 4019
D) 6183
E) None of those answers

[hide]but you forgot about the statement that $a$ and $b$ are relatively prime. actually, 375417/999900=413/1100 (cancel 909). therefore the answer is A) 687[/hide]


ARML
geometry
calculus
integration
trigonometry
rhombus
3D geometry

1.
In right triangle $ABC$, $\angle C = 90^\circ$, $c=3$, and $a+b=\sqrt{17}$.  Compute the area of the triangle.

2.
If $(a+b):(b+c):(c+a) = 6:7:8$, and $a+b+c=14$, compute the value of $c$.

3.
Let $z$ be a root of $z^5-1=0$, with $z \not= 1$.  Compute the value of

\[z^{15} + z^{16} + z^{17} + \cdots + z^{50}. \]

4.
A regular polygon of $2n$ sides is inscribed in a circle of radius 5.  The sum of the squares of the distances from any point on the circle to the vertices of the polygon is 600.  Compute the value of the integer $n$.

5.
The circles, each of radius 5, lie in perpendicular planes.  They intersect in two points, $X$ and $Y$, thus determining a common chord $\overline{XY}$.  If $XY = 8$, compute the distance between the centers of the circles.

6.
How many positive integers less than 54 are equal to the product of their proper divisors?

7.
Find all positive integers $n$, less than 17, for which $n! + (n+1)! + (n+2)!$ is an integral multiple of 49.

8.
In triangle $ABC$, $a \ge b \ge c$.  If $\displaystyle \frac{a^3+b^3+c^3}{\sin^3{A} + \sin^3{B} + \sin^3{C}} = 7$, compute the maximum possible value of $a$.

There are a lot of errors in some of your solutions.

1. $2ab=8\implies \dfrac{ab}{2}=2$ actually.

4. A complex number proof should suffice to show that for $2n$, the square of the distances is $100n$.

5. AXBY isn't actually a rhombus since it is in two planes, and the actual answer is $\sqrt{3^2+3^2}=3\sqrt 2$.

6. There are 14 such numbers: $2\times 3, 2\times 5, 2\times 7, 2\times 11, 2\times 13, 2\times 17, 2\times 19, 2\times 23, 3\times 5, 3\times 7, 3\times 11, $$3\times 13, 3\times 17, 5\times 7$, but it can also be the cube of a prime, also giving two more: $2^3$ and $3^3$. So the answer is $16$.

7. There are actually $5$ such numbers, but it asks you for the numbers themselves. In fact, the answers are actually $5, 12, 14, 15, 16$. It factors as $n!(n+2)^2$.

8. By the Law of Sines, it's clear that we must have $8R^3=7$, where $R$ is the circumradius, so $R=\dfrac{\sqrt[3] 7}{2}$. $a$ is maximized when it's the diameter, so our answer is $\sqrt[3] 7$.

EDIT: Darn I did all of these without paper but hopefully no errors.


Does anyone have any graphical displays of square residues modulo m?

My friend and I conjectured the other day that if you were to plot integers along the x-axis, and the sq. residues modulo m on the y-axis, then you would get a pattern of different portions of a parabola, becoming slimmer and slimmer, and then, at around m/2, the parabolas would be reversed (exhibit vertical symmetry).

Does anyone know what I am talking about?

The remainder when the square of the integer,k (say) is divided by m, is that what you mean?
Anyway, it seems a good topic to do a bit of investigation.

Sanjith

So basically you just want to graph $y=x^2 \mod r$ for some $r$.  First, you need to think about when and how this graph differs from the regular $y=x^2$ graph.  Every time $x^2$ hits the next multiple of $k$, the graph starts back at zero.  So basically, what you have is the same parabola except that it is discontinuous at $x=k\sqrt{r}$ for $k\in \mathbb{Z}$.

So to describe the graph, I would say start with the normal $y=x^2$.  Then mark each multiple of $\sqrt{r}$.  Now, cut vertically at each of these marks and place every piece of the graph so that the leftmost piece of each portion is at 0.

Yeah, that's what i'm talking about. I think it's really interesting.

I was bored and I made a graph of that in mathematica...

It doesn't totally match your description, but it's close.

To make it bigger you can click on the pic below

 :D

wow, its beautiful. i knew there would be symmetry about x=50, since you can show that $(50-k)^2 \equiv (50+k)^2$ . To see this, note that $(50-k)^2 \equiv 2500-100k+k^2 \equiv 2500+100k+k^2 \equiv (50+k)^2$ since we can add $200k$ (a multiple of 100, where we are working with mod 100) to the second expression.
I just realized that this works for 25 as well, as the graph shows (symmetry about $x=25$ ). Just note that $(25-k)^2 \equiv 625-50k+k^2 \equiv 625+50k+k^2 \equiv (25+k)^2$ . In fact, it works for $75$ , and, more generally, it works for any integer $z$ such that $100|(4z)$ .
This is not hard to show either. If we want $(z-k)^2 \equiv (z+k)^2$ for $k\leq{z}$ , then we need to be able to add a multiple of 100 to the middle term of $(z-k)^2$ to get a middle term of $2kz$ . This implies that $2z-(-2z)=4z$ must be a multiple of 100. In this case, the numbers 25, 50, and 75 satisfy this condition.
Suppose instead that we have modulo m. Then $z$ is a pt. of symmetry iff $m|(4z)$ . Now of course, this is not a rigorous proof; maybe someone else should make it rigorous.

I noticed one more thing. The condition $m|(4z)$ implies $4z=mn$ for some integer $n$ . Thus $z=\frac{mn}{4}\leq{m}$ for $n\leq{4}$ . The values $n=1,2,$ or $3$ give exaclty 3 values of $z$ not equal to 0 and m, that exhibit symmetry in the plot.

lol, my description holds if you extend the definition of mod to all real numbers :D (that's actually what I was thinking)


A boat heads in the direction N40 degrees W at a speed of 24 knots. The current runs at 6 knots in the direction N25 degrees E. Find the resultant speed and direction of the boat.

Thanks

[hide="hint"]Make a triangle out of the vectors and solve for the missing side.
[/hide]

I did a little sketch. I was looking through some notes and did the following:

take sqrt of (24 sqrd and 6sqrd) to get 24.74 knots.

arctan (24/6) = 75.96

90 - 75.96 = 14.04

the direction is N14.04W

is this correct???


inequalities
triangle inequality
inequalities unsolved

$ \left|a \right|\plus{}\left|b \right|\plus{}\left|c \right|\plus{}\left|a\plus{}b\plus{}c \right|\geq\left|a\plus{}b \right|\plus{}\left|b\plus{}c \right|\plus{}\left|a\plus{}c \right|$

This seems like it should be easy but I don't see how to do it.  I keep thinking the triangle inequality should help but I don't see how.

Take a look at inequalities marathon at pre-olympiad :wink: .
I have solve this inequality for $ a,b,c \in \mathbb{C}$ (problem 74)....


how do you find an the equation of a plane given three points?
Example: $ (2, \minus{}1, 3$, $ (\minus{}2, 1, \minus{}5)$ and $ 1, 0, 1$.

Find the normal vector of the plane, which will be orthogonal to any two vectors chosen between those three points, i.e. a multiple of the vectors' cross-product. Then, the vector between any point $ (x,y,z)$ on the plane and one of the points (choose one) will be orthogonal to the normal vector, so the cross product between those two vectors will be zero.

The most basic method is by a system of equations. Suppose our plane is of the form $ z \equal{} mx \plus{} ny \plus{} b$. We have three sets of points to plug in, and three parameters to set, which yields a unique solution provided the given points are not collinear.

There are techniques we can use to simplify computation along the way.

- If A, B, C, etc are on the plane, then any weighted average of them is also on the plane. For example, if $ (2, \minus{} 1,3)$ and $ ( \minus{} 2, 1, \minus{} 5)$ are on the plane, then their midpoint, $ (0,0, \minus{} 1)$, is also on the plane. Can you see how this can be helpful?

- We can use a change of variables. While this is not necessarily the fastest way of solving for the plane, it's a good thing to be familar with, and will help in higher math courses.

For example, in this case, let's define $ (a,b,c) \equal{} (x \plus{} 2y,5y \plus{} z,z \minus{} x \plus{} 1)$. Then we get the three new points $ U(0, \minus{} 2,2)$, $ V(0,0, \minus{} 2)$, and $ W(1,1,1)$. Averaging $ U$ and $ V$ gives $ T(0, \minus{} 1,0)$. Now, let the equation be $ c \equal{} pa \plus{} qb \plus{} r$. Plugging in $ V,T,W$ painlessly gives $ (p,q,r) \equal{} (5, \minus{} 2, \minus{} 2)$, thus, the plane is defined by $ c \equal{} 5a \minus{} 2b \minus{} 2$ $ \implies (z \minus{} x \plus{} 1) \equal{} 5(x \plus{} 2y) \minus{} 2(5y \plus{} z) \minus{} 2$ $ \implies \boxed{2x \minus{} z \equal{} 1}$
[i]
The one catch is that the new components have to be independent. That is, the determinant $ \left| \begin{array}{ccc} 1 & 2 & 0 \\
0 & 5 & 1 \\
\minus{} 1 & 0 & 1 \end{array} \right| \neq 0$, where each row gives the coefficients of $ x,y,z$, respectively in the definitions of $ a,b,c$.[/i]


For example:
http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44&year=2005
#17 of the A contest.
The diagrams apparently are not on the server anymore, maybe an admin could put them back on?

The problem is that people put up diagrams that are hosted on another site; when the other site gets taken down for whatever reason the diagrams disappear. Fixing this requires two people to cooperate:

1. Someone needs to update the diagram to Asymptote, preferably.

2. A mod needs to edit the diagram into the original post.

If you want to fix this, make an asymptote version of the diagram and put it in the thread. Then PM a mod and ask them to edit it into the first post.

Of course, if you're a mod you can do it by yourself.

The problem is finding the diagrams... :(

[quote="Temperal"]Moderators can't edit the resource center. Only Resource Managers can.[/quote]You don't need to. You just need to edit the post that it's being based on (click on the number). And yes, that's why I didn't want people to submit pictures on foreign sites, as they usually go down.


geometry MATHCOUNTS rotation AMC AIME USA(J)MO USAMO

Hi people,

I'm an officer for my school's math club this year and was wondering how other people advertise the AMC/AIME/USAMO.  Do you guys have flyers or posters, or use a trophy case showing AMC certificates from previous years?  It would be a big help to me if you could share some examples.

Thanks!

My school gives honors math students extra credit based on their AMC score.  Almost all the honors students take the AMC because of this.

In my school, the math teachers just ask their classes who wants to sign up. To interest kids in actually taking the tests, maybe you can hang up some posters that emphasize how you get to miss classes to take the tests   :P

Well, in our school, trophies from tournaments are displayed on shelves in our math team sponsors' classrooms, and they are always recruiting from their classes.  We don't have a well-defined set of people who are the "math team", but we invite everybody who wants to come to tournaments.  Also, they had many of their high honors classes take the AMCs; e.g., my algebra class last year took the AMC 10 (except for me and some others who took the 12) and counted their scores as a grade.  Many of them cleared 100 on the 10, so I think it was a positive experience for them.

For our school, we just send invitations to top 5-10 students in each math class from honors algebra II and up near the test date, and so being able to take the test means you're smart in a way.

We don't advertise, unfortunately. Half the guys that take the AMC make the USAMO.

We took the AMC during the day, while classes were going on. Getting out of 5 of your classes was enough incentive for most people.

Why do you want to advertise anyway? It costs the school more money for more people to take it.

I'm wondering, how can you advertise when you live in a small town that has little or no interest in math(little, because of me(If I wasn't here, there may be no interest in math))?
Also, I have no website of my very own, I have no AMC tee-shirt, and I'm out of ideas. Not to mention I'm YOUNG for this sort of thing!

P.S. I can't paint up one of my tee-shirts to advertise the AMC, and my parents won't approve of putting it into the newspapers.

You offer the opportunity to miss English class and people sign up like crazy.

[quote]
And, according to some (including me), "supergeniuses" are merely people who have invested a certain measure of thought and energy well in a particular field.
[/quote]

COUND NOT AGREE MORE :D

[quote="Boy Soprano II"][quote="Scrambled"][quote="k8reindeer"]if there's no interest, nothing you just described is going to create interest. The only people who do math competitions are math-competition-people. Any people you get to take it advertising that way will only be taking it because thier parents pressured them to do it.[/quote]
However, math contest people were at some point in time non-math contest people. I'm willling to bet some people who arent interested in math due to the inadequacy of the average curriculum to generate interesting maths are really supergeniouses waiting to happen.[/quote]
Yes, I think you might be surprised.  I remember that I showed up at math team in sixth grade almost on a whim.

And, according to some (including me), "supergeniuses" are merely people who have invested a certain measure of thought and energy well in a particular field.[/quote]
Well, I have been a math competition person since third grade when I saw my brother dominate in countdown...
I was amazed and determined to do the same...

But unfortunately, this may discourage people who can score in about the 80-90 range (which is not bad) from taking the test. I'd recommend something more like a few (maybe 5) bonus points for just taking the test, or maybe something like 3 points for just taking it, 4 points for scoring from 70-90, 5 points for scoring 90-110, etc.

[quote="Boy Soprano II"][quote="Phelpedo"]An idea: suggest to your math teacher that it should be counted as an extra test grade. Since the score is out of 150, it will encourage guys to take it (good chance to increase average A LOT), as well as encourage them to practice for it.[/quote]
Yes, this is what our school does (we even make the grade optional, I think).  And matt276eagles, I think that Phelpedo meant that even though the maximum score is 150, the score you receive counts as your score out of 100 (which is what we do).[/quote]

Yes, that's also what I thought Phelpedo meant.  But at my school, where pretty much all honors math students take the AMC, a great majority of people (excluding people in Calc BC) get less than 100 on the AMC.

Well if we make it, it gets put in our schoolnewspaper...

At my school, the AMC test is completely voluntary, but costs something like 2 dollars. The school advertises it a bit, particularly in honors math classes (I think in Geometry H the class even did a practice test). A reasonable number of kids take it, 17 kids took AMC 12 (all of whom qualified for AIME) and I forget how many took AMC 10 (none of whom qualified for AIME).

That extra credit sounds like a great idea. My idea for the values might be:

1: 80-89
2: 90-99
3: 100-109
4: 110-119
5: 120-129
6: 130-139
7: 140-149
10: 150

[quote="Boy Soprano II"][quote="matt276eagles"]Yes, that's also what I thought Phelpedo meant.  But at my school, where pretty much all honors math students take the AMC, a great majority of people (excluding people in Calc BC) get less than 100 on the AMC.[/quote]
Really? :o   Most people in my algebra class last year got at least As, and many got over 100—admittedly, that was the AMC 10, though.[/quote]

In my particular class last year, Honors Algebra 2, everyone in the class took the AMC 10 A, and only me and one other person qualified for the AIME.  I don't know exactly how many people scored more than one hundred points, but I would guess that it was about 5 or 6.

Since everyone seems to be offering possible extra credit systems, I'll mention that the teachers at my school divide each student's AMC score by a predetermined number, and then take the floor.  This predetermined number is chosen by the teacher, and is usually 10 or 15. This encourages people to take the AMC because it is possible to earn a decent amount of extra credit even if they can only answer five questions or so.

in my school different teacher have different systems but every teacher gives extra credit for any of the math contests, so my freshmen teacher (head of math department ) is for everpoint you get above national average, I get one extracredit point, so i end up slack off on test and hw and still have mroe than 100%
at my school people just go take it cuz we are like a nerd school( 70% asian with parents following us around to remind us to study >.<)
so every year each student pay 3 dollars to take the test, and we have about 200 people taking it
and about 50-60 makes it to AIME
and 7-10 to USAMO (with scores ranging from 20s to zero XP)

[quote="rainynightstarz"]we have about 200 people taking it
and about 50-60 makes it to AIME
and 7-10 to USAMO (with scores ranging from 20s to zero XP)[/quote] Those numbers are amazing. Probably comparable to Exeter or TJ, unless you go to one of those.  :wink:

My teacher has a wierd extra credit system. Every extra point counts as a $100\%$. In other words, if you get 95 out of 100 and 15 bonus on a test, your score is $\frac{95+15}{100+15}$, so bonus points are given liberally. :D

my school has about 1800 people so 200 isn't near how much math teacher want! he wants EVERYONE to take it!

I go Lynbrook , its in silicon valley. the people who get into USAMO  are mostly the poeple who were in mathcount states or nationals.

1/4 of my class take calc bc in their junior year, 7 people including me, took it as sophmore and 2 took as freshmen
we just have very pushy parents

I'd rather go to a school like that.  At my school, I'm in Calc AB as a junior right now, and I'm trying to get them to let me switch to Calc BC, but they don't want to let me, because I'm already a year ahead of the regular honors math schedule...

Last year, I had the highest AMC 12 score with 115, and the 12 people we had qualify for AIME was 3 times the most we'd ever had before...

interesting
my school even switches the schedule around JUST so people can take AMC, AIME without missing too many class time

and oh yeah, this summer, 2 people from my school went to MOP, that includes the only girl that was at MOP (there should be more girls)
they are super smart

Well, our school has a television channel of its own so it is posted there.

We drew quite a lot of people. 14 of us went into the AIME and 2 into the USAMO.

Wow... our school usually only has about 20 people who take the AMCs (but then all 20 of us qualify  :D ). Nobody has every made the USAMO from our school (our school is only 4 years old), and the highest AIME score has been a 5.

Hey, you're beating us.  Our school has been around for a long time (like 150 years if you include the different buildings it's been at), and we've still never had anyone make USAMO.  The highest index anyone from our school has ever had is a 174, and that was with a 4 on the AIME.


A couple of days after CCU, I got sick.  Talking to my friends and other people who went to CCU, I noticed [b]everyone[/b] who went on the trip is sick.  Symptoms are similar, so I thought it might have been something passed around on the bus ride.  However, Asif said he is sick too.  Do you think there was something in the food?  Is anyone else who went to CCU sick?

what did you eat? i ate the pizza and am feeling fine.

On second thought, it probably wasn't the food itself.  But I'm convinced SOMETHING went around at CCU.  The entire math team is out sick.  Ask Joe if he's sick.  Our symptoms were fevers, headaches, coughing, and stuffiness.   Everyone was on Tylenol for our Academic Bowl today. :?

It's probably just a flu going around in general.  A bunch of people I know (not on the math team) were pretty sick last week and this week.

Well, if it were the food, then it might be because of all the people who got their food before you.  Summerville got there pretty quickly so we got our food first before it would have been contaminated :)  (maybe we poisoned your food  :o ...jk, :D )  I ate just about everything but the pizza, lol

It might have also been the cold weather + wetness...but you would think that would get more of us from the lower state sick...hmph.

Anyway, I feel fine.  Hope you all feel better cause I know how much it stinks being sick : (

yeah i skipped school today bc i was sick. i doubt it was due to ccu though.

I think this is the first time I [b]haven't[/b] gotten sick at a math contest.

must have been you then, john. you must have spread it. probably it was still in your system and you thought it would be nice to take out the competition.

haha, im only just kidding! oh yeah, congratulations on the win at CCU btw.


Points on a grid are numbered as shown. Movements from point number A to point number B can only be made if $B>A$ and the two points are in the same row or column. Find the total number of ways to go from point 1 to point 30.

I would get a picture, but it's late and I'm dead. I'll make a picture tomorrow.
But, it's like finding the number of ways to get from 1 to 30 if 30 were on the other side. Do it the same way, except kind of like in reverse.

 :huh: I just realized that solution is a piece of crap.  :rotfl:

EDIT: Do you know what theta test this is? And if it is nationals or state?

[hide="my solution"]
For each point you write down how many ways are to get to this point from the first. And for each next point number of ways is sum of the ways to get to upper point (it is allowed to go down anytime) and to left or right point (depends on row). At the point 30 number of ways is $671$.
[/hide]


inequalities
trigonometry
inequalities proposed

Let $ z \in \mathbb{C}$ , prove that the following inequality is true : \[ (|z|^2 \plus{}1)^n \geq |z^{2n}\minus{}1| .\]
When the equality case holds ?

I suppose that $ n \geq 1$  :D  . $ z\equal{} |z| (\cos (t) \plus{}i\sin (t)) \Longrightarrow (|z|^2 \plus{}1)^n \geq \sqrt {|z|^{4n}\plus{}1 \plus{} 2|z|^n}$ $ \geq \sqrt {|z|^{4n}\plus{}1\minus{}2|z|^{2n}\cos(2nt)}\equal{}|z^{2n}\minus{}1|$

Manifestly the equality needs $ (|z|^2 \plus{}1)^n \equal{} \sqrt {|z|^{4n}\plus{}1 \plus{} 2|z|^n} \equal{} \sqrt {|z|^{4n}\plus{}1\minus{}2|z|^{2n}\cos (2nt)}$
Finally $ z\equal{}0$ or $ (n\equal{}1)\& (z\equal{} i A)\& (A\ real)$
 :cool:


Because LHS and RHS are positive.
We can prove like following.
$ \sqrt {d(d \plus{} c)} \plus{} \sqrt{d(d \plus{} c \plus{} b)} \plus{} \sqrt{d(d \plus{} c \plus{} b \plus{} a)} \plus{} \sqrt{(d \plus{} c)(d \plus{} c \plus{} b)} \plus{} \sqrt{(d \plus{} c)(d \plus{} c \plus{} b \plus{} a)} \plus{} \sqrt{(d \plus{} c \plus{} b)(d \plus{} c \plus{} b \plus{} a)}\ge d \plus{} d \plus{} d \plus{} (d \plus{} c) \plus{} (d \plus{} c) \plus{} (d \plus{} c \plus{} b) \equal{} 6d \plus{} 3c \plus{} b$

[quote="behdad.math.math"]If $ a,b,c,d\geq 0,$ prove that $ \sqrt{d} \plus{}\sqrt{d\plus{}c} \plus{}\sqrt{d\plus{}c\plus{}b} \plus{}\sqrt{d\plus{}b\plus{}c\plus{}a}\geq\sqrt{16d\plus{}9c\plus{}4b\plus{}a}.$[/quote]$ P(n): \sqrt{a_{n}}\plus{}\sqrt{a_{n}\plus{}a_{n\minus{}1}}\plus{}\cdots\plus{}\sqrt{a_{n}\plus{}a_{n\minus{}1}\plus{}\cdots\plus{}a_{1}}$

$ \geq \sqrt{n^2a_{n}\plus{}(n\minus{}1)^2a_{n\minus{}1}\plus{}\cdots\plus{}a_{1}},$

where $ a_{1},a_{2},\cdots,a_{n}\geq0.$

$ \Longleftrightarrow Q(n): (b_{n}\plus{}b_{n\minus{}1}\plus{}\cdots\plus{}b_{1})^2$

$ \geq n^2b_{n}^2\plus{}(n\minus{}1)^2(b_{n\minus{}1}^2\minus{}b_{n}^2)\plus{}\cdots\plus{}(b_{1}^2\minus{}b_{2}^2)$

$ \equal{}(2n\minus{}1)b_{n}^2\plus{}(2n\minus{}3)b_{n\minus{}1}^2\plus{}\cdots\plus{}b_{1}^2,$

where $ b_{1}\geq b_{2}\geq\cdots\geq b_{n}\geq0.$

[b]Proof[/b] $ Q(1): b_{1}^2\equal{}b_{1}^2.$

$ Q(n)\Longrightarrow Q(n\plus{}1): (b_{n\plus{}1}\plus{}b_{n}\plus{}\cdots\plus{}b_{1})^2$

$ \equal{}b_{n\plus{}1}^2\plus{}2b_{n\plus{}1}(b_{n}\plus{}\cdots\plus{}b_{1})\plus{}(b_{n}\plus{}\cdots\plus{}b_{1})^2$

$ \geq b_{n\plus{}1}^2\plus{}2nb_{n\plus{}1}\plus{}(2n\minus{}1)b_{n}^2\plus{}\cdots\plus{}b_{1}^2$

$ \equal{}(2n\plus{}1)b_{n\plus{}1}^2\plus{}(2n\minus{}1)b_{n}^2\plus{}\cdots\plus{}b_{1}^2.$


Very nice one  :)

By Holder we have  $ (\sum\frac {a^2}{b})^2(a^2b^2 \plus{} b^2c^2 \plus{} c^2d^2 \plus{} d^2a^2)\geq (a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)^3 \equal{} 64$

But we also have $ a^2b^2 \plus{} b^2c^2 \plus{} c^2d^2 \plus{} d^2a^2 \equal{} (a^2 \plus{} c^2)(b^2 \plus{} d^2)\leq\frac {(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)^2}{4} \equal{} 4$

so we are done . Which is your solution arqady ?

[quote="arqady"]Let $ a,$ $ b,$ $ c$ and $ d$ are positive numbers such that $ a^2 + b^2 + c^2 + d^2 = 4.$ Prove that
\[ \frac {a^2}{b} + \frac {b^2}{c} + \frac {c^2}{d} + \frac {d^2}{a}\geq4
\]
[/quote]

We have $ \sum\frac {a^2}{b}\ge\frac {(a^2 + b^2 + c^2 + d^2)^2}{\sum a^2b}$.

Hence it remains to show that $ a^2b + b^2c + c^2d + d^2a\le 4$.

From Cauchy-Schwarz again we have $ \sum a^2b\le\sqrt {\left(\sum a^2\right)\left(\sum a^2b^2\right)}$. Therefore we'll be done if we can show that $ \sum a^2b^2\le 4$, which I guess silouan already showed above.. :blush:

Very nice proofs, silouan and nayel!  :lol:
silouan, I have proved it by squaring and AM-GM. Your proof more better.
I think, AM-GM does not work for the following inequality:
Let $ a,$ $ b,$ $ c$ and $ d$ are positive numbers such that $ a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \equal{} 4.$ Prove that:
\[ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{d} \plus{} \frac {d^2}{a}\geq4
\]
By the way, the following inequality is not true:
Let $ a,$ $ b,$ $ c$ and $ d$ are positive numbers such that $ a^5 \plus{} b^5 \plus{} c^5 \plus{} d^5 \equal{} 4.$ Prove that:
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{d} \plus{} \frac {d^2}{a}\geq4$

[quote=SaMeH][quote="arqady"]

I think, AM-GM does not work for the following inequality:
Let $ a,$ $ b,$ $ c$ and $ d$ are positive numbers such that $ a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \equal{} 4.$ Prove that:
\[ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{d} \plus{} \frac {d^2}{a}\geq4
\] [/quote]

What's the solution for this one ??   :lol:  :lol:[/quote]

I have an idea:
By Vasc,
$$(LHS)^2 \geq 4( \sum_{cyclic} \frac{a^3}{\sqrt{bc}})$$
But what to do after that??

[quote=arqady]Let $ a,$ $ b,$ $ c$ and $ d$ are positive numbers such that $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}4.$ Prove that
\[ \frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{d}\plus{}\frac{d^2}{a}\geq4\][/quote]
Let $a,b,c,d$ be positive real numbers satisfying $a+b+c+d=4$. [url=https://artofproblemsolving.com/community/c6h1626529p10197761]Prove that [/url]
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\geq 4+(a-d)^2$$

Let $a,b,c,d$ be positive real numbers . Show that
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\ge \frac{4(a^2+b^2+c^2+d^2 )}{a+b+c+d}$$

[hide=Thanks.][quote=Soham50]Let a,b,c,d be positive real numbers . Show that a^2/b  +  b^2/c  + .....>=4(a^2+b^2+....)/a+b+c+d
Solution:           We know   a1^2/b1  +  ......+an^2/bn  >= (a1+a2+........+an)^2/b1+b2+.....+bn for all positive real numbers a1,a2,.....,an,b1,b2,.....,bn.
                           so a^2/b+b^2/c+....>=(a+b+c+d)^2/a+b+c+d>=4(a^2+b^2+...)/a+b+c+d
                                                                                (It follows from expanding)[/quote][/hide]

[quote=sqing][quote=arqady][hide=*]Very nice proofs, silouan and nayel!  :lol:
silouan, I have proved it by squaring and AM-GM. Your proof more better.
I think, AM-GM does not work for the following inequality:
Let $ a,$ $ b,$ $ c$ and $ d$ are positive numbers such that $ a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \equal{} 4.$ Prove that:
\[ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{d} \plus{} \frac {d^2}{a}\geq4
\][/hide]
By the way, the following inequality is not true:
Let $ a,$ $ b,$ $ c$ and $ d$ are positive numbers such that $ a^5 \plus{} b^5 \plus{} c^5 \plus{} d^5 \equal{} 4.$ Prove that:
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{d} \plus{} \frac {d^2}{a}\geq4$[/quote]
Let $ a,$ $ b,$ $ c$ and $ d$ are positive numbers such that $ a^5 \plus{} b^5 \plus{} c^5 \plus{} d^5 \equal{} 4.$ [url=http://tieba.baidu.com/p/4305472771]Prove that:[/url]
$$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{d} \plus{} \frac {d^2}{a}\geq3.987022.$$[/quote]
When  does the inequality hold?

[quote=jasperE3]It says in the baidu post, it's when the roots of a $70$th degree polynomial occur.[/quote]

That problem is very difficult. If the condition is $a^4 + b^4 + c^4 + d^4 = 4$ instead (to prove $\frac {a^2}{b} + \frac {b^2}{c} + \frac {c^2}{d} + \frac {d^2}{a} \ge 4$ ), there are several proofs available.


linear algebra
matrix
algorithm
number theory

Hi guys,I've learned these days about matrices and I was thinking : How can I use matrices to solve a problem.
As I saw until now ,there are just tables without any speacial meaning (I know I'm wrong  :) )
Can anyone of you find me some interesting problems to solve using matrices.

Matrices are extremely important.  One very simple yet very important application is the solving of a system of equations.  See [url=http://en.wikipedia.org/wiki/Gaussian_elimination]Gaussian elimination[/url].

There are a lot of problems with matrices in [u]College Playground[/u], here on the forum.

Another nice application is that a matrix can show you how a linear transformation acts on basis elements (or any vectors).

The easiest thing you may do it with matrices is solving system of $ 2$ equations:
http://www.math.csusb.edu/math110/src/matrices/sol-sys.html

And a very interesting thing in matrices is the cryptography. I made an excel spreadsheet for $ 2\cdot7$ messages and matrices.

Well its by creating a message with a special code so no body can know the message except you. Its a hard thing to explain you may see it on the net I will try to find you something soon.

Cryptography is making and breaking codes. A code is a series of changes to letters or symbols. For example, it can be as simple as reversing the order of the letters, such that

"HELLO, HOW ARE YOU"

becomes

"UOY ERA WOH, OLLEH".

It could also be a=1, b=2, ..., z=26, such that

"HELLO"

becomes

"8,5,12,12,14"

A caeser shift is a typical type of code. This creates A -> B, B-> C, etc.

Such that:

Normal:
ABCDEFGHIJKLMNOPQRSTUVWXYZ
Code:
BCDEFGHIJKLMNOPQRSTUVWXYZA

So

"HELLO, HOW ARE YOU TODAY"

becomes

"IFMMP, IPX BSF ZPV UPEBZ".

Obviously, to anyone not the sender or reciever, this looks like complete nonsense.

There are an infinite number of codes. Yet another one is to move one space to the right on a keyboard. For example,

HELLO, HOW ARE YOU

becomes

JR;;P, JPE STR UPI.

Make up your own codes. You and the person that you are sending it to can figure out the message, if both of you know the rule of the change. Make sure to make it complicated, or people might break it!

Thanks a lot to all of you for your time ,but I still don't understand how to use matrices in cryptography.I saw how to create codes,that none can break without knowing algorithm,but I'm still noob in using matrices. :oops:

[quote="JRav"][url=http://people.richland.edu/james/lecture/m116/matrices/applications.html]Here[/url] is one link (scroll down a little bit) and [url=http://aix1.uottawa.ca/~jkhoury/cryptography.htm]here[/url] is another.[/quote]
Yes the first site explains everything and its a real good way for communicating your friends sometimes  :read:

If you are responding to the person in front of you, you don't need to quote them.

Number theory is also very important to cryptography.  For example, read "The Big Picture" in chapter 23 of AoPS Volume 2.

OK can you say how NT is important to cryptography?
I don't have AoPS V2 book.

Lucky for you, I managed to come across an entire course dedicated to [url=http://www.math.umn.edu/~garrett/crypto/]Crypto and Number Theory[/url].


geometry parallelogram trapezoid function ratio

Let a right-angled parallelogram ABCD. Let K the midpoint of BC and L the midpoint of AD. The perpendicular line from B to AK intersects the AK at E and the CL at Z.
Prove that the AKZL is an isosceles trapezoid.
Prove that: $(ABKZ)=\frac{1}{2}(ABCD)$ (We symbolize with (......) the area)
If the ABCD is a square with AB=BC=CD=AD=a  find the area of isosceles trapezoid  AKZL as a function of a.

Let $F$ be the point on $AK$ such that $AL\parallel FZ$.
Then $AFZL$ is a parallelogram since $AL\parallel FZ$ and $AF\parallel LZ$.
So $AL=FZ(=BK)$.

Since $FZ=BK$ and $FZ\parallel BK$, $\triangle FEZ\cong\triangle KEB$.
Thus $FE=EK$.
In $\triangle FKZ$, the perpendicular line from $Z$ to $FK$ (which is $ZE$) bisects $FK$, so $FKZ$ is an isosceles triangle with $FZ=FK$, which implies $AKZL$ is an isosceles trapezoid with $AL=KZ$ ($\because AL=FZ=KZ$).

Note that $BFZK$ is a parallelogram.
Since $BF=FZ$ and $\angle AFB=\angle AFZ$, $\triangle ABF\cong\triangle AZF$
So $(ABF)=(AZF)=(AKZ)=(DLZ)$ (parenthesis means area.)
Also, $(BFZ)=(BKZ)=(CKZ)$

Therefore,
$(ABKZ)=(ABF)+(AFZ)+(BFZ)+(BKZ)$
$=(ALZ)+(DLZ)+(CKZ)+(BKZ)$
$=(ADZ)+(BCZ)$
$=\frac{1}{2}(ABCD)$

For the last problem:

To find the ratio $AE: EK$, look at $\triangle ABK$.
Since $\frac{AE}{EK}=\frac{AB^2}{BK^2}=4$, $AE: EK=4: 1$.
So $AF: FE: EK=3: 1: 1$ ($\because FE=EK$)

$(AKZL)=(AFZL)+(FKZ)$
$=\frac{3}{5}(AKCL)+\frac{1}{2}(FKCZ)$
$=\frac{4}{5}(AKCL)$
$=\frac{4}{5}\cdot\frac{1}{2}a^2$
$=\frac{2}{5}a^2$


can we say iff$\varphi(n)|n-1$ then $n$ is prime


Find the smallest natural number which can be represented in the form $ 19m\plus{}97n$, where $ m,n$ are positive integers, in at least two different ways.


Find the gratest real number $A$ such that for all positive real numbers $x, y, z$ is \[ \frac{x}{\sqrt{y^2 + z^2}} + \frac{y}{\sqrt{x^2 + z^2}} + \frac{z}{\sqrt{y^2 + x^2}} > A  \]

Yes,it has been posted many many times.

 :roll:


What is the smallest product one could get by multiplying two different numbers from the set $ \{\minus{}8,\minus{}6,2,5,7\}$?

[hide="Solution"]
Well when you multiply a negative number and a positive number you get a negative product.
Knowing that we take the smallest and biggest numbers and multiply them to get -56.[/hide]

Shouldn't it be "least"? Smallest makes me think least magnitude, though in the context of the problem you can guess easily enough.

Smallest only makes you think of smallest in magnitude. Consider straightening out your mathematical vocabulary. xD


Given triangle ABC with the Inscribed circle (I)  . AI intersect the circle (I) at A1 . A2 is the midpoint of BC .Similarly for B1 , B2 ; C1 , C2 . Prove that A1A2 ; B1B2 and C1C2 have 1 common point .

Dear Mathlinkers,
this problem has been presented on this site without success for a synthetic proof.
Eric Danneels Wrote a nice article on this subject in Forum Geometricorum with another approach: http://forumgeom.fau.edu/FG2006volume6/FG200629.pdf
Sincerely
Jean-Louis


integration
trigonometry
calculus
calculus computations

Method 1) Use $ \sec^3 x \equal{} \frac{\cos x}{\cos^4 x} \equal{} \frac{\cos x}{(1 \minus{} \sin^2 x)^2}$.

Method 2) Use $ \sec^3 x \equal{} \sqrt{1\plus{}\tan^2 x} \, \sec^2 x$ on $ [\minus{}\tfrac{\pi}{2}, \tfrac{\pi}{2}]$.

I don't see method 2 being useful. That integral after substitution is just as bad.
The usual method in calculus texts uses integration by parts to relate this to the integral of $ \sec$, which is known.

Method 3. Integrate by parts, by detaching $ \sec^2 x$ and using the facts that $ \int \sec^2x \,dx \equal{} \tan x$ and $ (\sec x)' \equal{} \sec x \tan x$.

[hide]$ P \equal{} \int \sec^3x\,dx \equal{} \int \sec^2 x \cdot \sec x \,dx \equal{} \tan x \sec x \minus{} \int \tan x \cdot \sec x \tan x\,dx$

$ \equal{} \tan x \sec x \minus{} \int \tan^2x \sec x\,dx \equal{} \tan x \sec x \minus{} \int (\sec^2 x \minus{} 1) \sec x\,dx$

$ \equal{} \tan x \sec x \minus{} \int \sec^3x\,dx \plus{} \int \sec x \,dx$

$ \equal{}> P \equal{} \tan x \sec x \minus{} P \plus{} \ln|\sec x \plus{} \tan x| \plus{} C$

$ \equal{}> P \equal{} \frac12 \tan x \sec x \plus{} \frac12 \ln|\sec x \plus{} \tan x| \plus{} C$.
[/hide]

The method Carcul showed is what I would teach in my ordinary calculus class. It's a particular case of a reduction formula.

Can you use the same ideas to express $ \int \sec^nx\,dx$ as the sum of an algebraic expression in $ \sec x$ and $ \tan x$ and a multiple of $ \int\sec^{n\minus{}2}x\,dx\ ?$

This particular integral tends to arise in a number of places, including the computation of the arclength of a parabola.


inequalities
calculus
geometry
inequalities proposed

Is this a true inequality?

  For $a,b,c>0$ we have $2(a+b+c)(ab+bc+ca)\leq9abc+2(a^3+b^3+c^3)$ :roll:

Replace the 2 by 3 for the coefficient of $\sum a^3$ then it's true by schur and muirhead. Otherwise take $a=b=c=1$

Well,that's it. I was scared because clearly I made a mistake in my calculations. ;)

[b]"The MATH, in the its multiple appearancies, is a monstruos tautology."[/b]

From the wellknown relation $9(4R+r)^2\cdot G\Gamma ^2=4[s^2 (2R-r)(2R+5r)-r(4R+r)^3]$,

where $G$ is the centroid and $\Gamma$ is the Gergogne's point results the strong inequality \[s^2\ge \frac {r(4R+r)^3}{(2R-r)(2R+5r)}\ \ (*).\] The your inequality, with the substitutions $a:=s-a,\ b:=s-b,\ c:=s-c$ becomes (after a easy calculus) \[s^2\ge 8Rr+11r^2\ \ (**)\]. I will prove that $(*)\Longrightarrow (**).$ Indeed, with the substitution $R=tr, t\ge 2$, we have:

${\frac{r(4R+r)^3}{(2R-r)(2R+5r)}\ge 8Rr+11r^2\Longleftrightarrow (4t+1)^3\ge (2t-1)(2R+5)(8t+11)}\Longleftrightarrow$

$8t^3-15t^2-9t+14\ge 0\Longleftrightarrow (t-2)(t+1)(8t-7)\ge 0$ what is true because $t\ge 2.$

Do you like, Caesar ? If you want to use this method, I can give you and the inequality (for right hand):

\[s^2\le \frac{R(4R+r)^2}{2(2R-r)}\] what results from the distance between orthocentre $H$ and the Gergogne's point $\Gamma$.

[i](See my book of geometry)[/i]

Yes,indeed,Levi.This is so cool.I like it beacause from these geometric configurations we can obtain many inequalities. But,I think the main idea is to lok for configurations to find strong inequalities.Mybe when you'll be back to Bucharest we will talk about it. ;)


induction
number theory solved
number theory

Can we use the induction method contrary...

for ex.: -if $n=1$ we have a contradiction
           - if $n=2$ we have a contradiction
           - suppose $n=k$ we have a contradiction
(*)now prove that we have a contradiction when $n=k+1$ , too...

I think its not true...   :(  ...

Davron

Please be more specific and explain how it is meant...

You can clearly prove the contraposition of a statement by induction sometimes.

As far as I see, proving that a statement $P$ is false for every $n\in\mathbb{N}$ is equivalent to proving that $\lnot P$ is true for all $n\in\mathbb{N}$, so i'd say there's no fault in your reasoning.

Most times we don't get a true contradiction but just a proof of the contraposition.
Let's say yes, but well, everything that does only use the logic we have is acceptable.


[quote="[url=http://www.artofproblemsolving.com/Wiki/index.php/Mock_AIME_4_2006-2007_Problems/Problem_12]Mock AIME 4 2007[/url]"]The number of partitions of 2007 that have an even number of even parts can be expressed as $a^{b}$ where $a$ and $b$ are positive integers and $a$ is prime. Find the sum of the digits of $a+b$.[/quote]

chess64 believes that the above question is reasonable and that the answer takes the given form.  I believe that it is effectively incalculable without substantial computer resources, and that even verifying that the answer takes the given form is similarly intractable.  Your thoughts?

[quote="boxedexe"]What does the problem mean by "the number of partitions of 2007?"  Could you please give me an example?[/quote]

The number of partitions of $n$ is the number of ways that $n$ can be written as the sum of positive integers. Also, order is irrelevant. So, the partitions of 5 are:
5
4+1
3+2
3+1+1
2+2+1
2+1+1+1
1+1+1+1+1

[quote="krustyteklown"]Have you consulted the writer of the question about his solution?[/quote]

chess64 is the writer I think. See here (question 12): [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=125025[/url]

My thoughts:

Well, the number of partitions of $2007$ is:
$5749318143678144230778676663789672984169195116$.

The number of partitions with an even number of parts is equal to the total minus the number of partitions with an uneven number of parts.

It's known that the number of partitions with an uneven number of parts is the same as the number of partitions with distinct parts. This should be less than half of the total partitions, so we can say that the number of partitions with an even number of parts is greater than half of that large number above. So we're talking about a number with $46$ digits.

Of course there still could be a nice way to work out $a$ and $b$ without working out $a^{b}$. chess64: what solution did you have in mind? (I assume the mock-AIME competition is over by now).

rjt: it's an even number of [i]even[/i] parts.

krusty: chess64 is the author.

[quote="JBL"]rjt: it's an even number of [i]even[/i] parts.
[/quote]

Oh! Well, nevermind then.

Hmm, I think I saw this problem somewhere, but it was last summer so I don't remember where, and I can't find it now. (Maybe the problem is just wrong...)


Let's take the quadrilateral $\ ABCD$ so that $\ AC\cdot\ BD=AB\cdot\ CD+AD\cdot\ BC$.If the sides of the $\ ABCD$ are $\ a,b,c,d$, prove the inequality $\ 2R^2(a+b)(b+c)(c+d)(d+a)\geq\ (ab+cd)(ac+bd)(ad+bc)$, where the notation $\ R$ is well-known.

Inequality is equivalent with $(a+b)(b+c)(c+d)(d+a)\geq \prod_{cyclic}(O-2d)$, ( where $O=a+b+c+d$) because $R=\sqrt{\frac{(ab+cd)(ac+bd)(bc+ad)}{2\prod_{cyclic}(O-2d)}}$. We take $x=a+b+c-d$ and similar for $y,z,w$. Now the inequality becomes $(x+y)(y+z)(z+w)(w+x) \geq 16xyzw$ which i true by AM-GM.

q.e.d.


An equilateral triangle has sides of length 1 cm. Five points lie in this triangle (on the sides or in the
interior).
(a) Show that two of the points lie no farther than .5 cm apart.
(b) Show that .5 (in part (a)) cannot be replaced by a smaller number even if there are 6 points.
(c) If there are eight points, can .5 be replaced by a smaller number? Prove your answer.

hint for part a: cut the triangle up into four smaller equilateral triangles. What can you say if two points are inside the same small triangle?