Why is my TI-84 plus silver edition so slow with graphing? All the memory is cleared, yet it graphs much slower than a normal TI-84.

There is no "normal" T-84.  It is the TI-84+, which is what I have.  I don't have much experience with the silver but I find that it tends to graph slowly only for exceedingly complex functions.  Does it graph slow for something like $ y\equal{}2x$?  I don't know what would cause it go slow.

change the graph speed. The difference between a 84+ and 84+ silver is only that silver has more memory. The difference between 83+ silver and 84+ silver I don't know.

[quote="JRav"]There is no "normal" T-84.  It is the TI-84+.[/quote]

Then what's the point on naming it "plus?" Is it just a cheap scam to get people to think they are buying a "better" product?  Or is this common practice and the "normal" version of a TI-89 is a TI-89 plus? :huh:

TI-83$ \Rightarrow$TI-83+$ \Rightarrow$TI-84+$ \Rightarrow$TI-84+ Silver

If you're in parametric mode, it might be the Tstep. Use a higher value.

[color=green]Moderator says: there is [i]never[/i] a good reason to block-quote the post immediately preceding yours.[/color]
I overclocked mine.

[quote="7h3.D3m0n.117"]The Silver Edition has a way of lagging in all aspects of calculations. I find that deleting [i]most[/i] of your applications (the ones that are useless to you) will speed up the overall process.[/quote]
Yeah, true. But I deleted all of my apps, and it's still really slow.

[quote="#H34N1"]I overclocked mine.[/quote]
It seems you can overclock all the TIs except the TI-84, how did you do it?

[color=green]Moderator says: there is [i]never[/i] a good reason to block-quote the post immediately preceding yours.[/color]
Unmounting some random chip. My friend unlocked mine for me. I'll ask him when I go back to school tomorrow.

EDIT: Nvm he overclocked my other calculator.

It runs fast enough for me. I like it more than java on my computer (for programs). I can go up to higher numbers, and it's less work. I have 1 app left though (Mirage OS lol).

I noticed my TI-89 Titanium executes programs much slower than my TI-83 plus does. :(

[quote="serialk11r"]I accidentally deleted all my applications. I also accidentally messed up all my programs too when they started going crazy (this factoring program was written incredibly inefficiently, testing all numbers up to the number instead of all primes up to the floor of the square root), which were only useful for random mathcounts stuff.
My calculator is fast :)[/quote]
How would you do that? To tes tall numbers to the square root use a for() loop
like for (a,1,sqrt(a),[1])
This means it takes variable A, starts at 1, goes to $ \sqrt{a}$ and increments 1. If it's increment is 1, the last part is optional, if you want to increase it by any number besides 1, you'll have to put another num there.
Of course you can also use an if statement


Three cards are dealt at random from a standard deck of 52 cards.  What is the probability that the first card is a Jack, the second card is a Queen, and the third card is a King?

Simple probability.

$ \frac{4}{52} \times \frac{4}{51} \times \frac{4}{50}\equal{}\boxed{\frac{8}{16575}}$.

Since it never specified whether the cards were drawn after one another or drawn at the same time, I think this problem should be modified.

It says FIRST and SECOND. That gives order and that is why it should be $52 \times 51 \times 50$

This problem, at least in my opinion, is a little too brute-forcey.


trigonometry
inequalities
inequalities proposed

very easy

again the famous cosine problem

let $x=\tan^{-1}a,y=\tan^{-1}b,z=\tan^{-1}c$
by hypothesis:
$x,y,z>0,x+y+z=\pi$ we wish to show:
$\cos{x}+\cos{y}+\cos{z}\leq \frac{3}{2}$ which is well known and true by Jensen.

so we are done. :)

Ok,here is a first step to the solution:The first relation is rewritten as: \[\frac1{ab}+\frac1{bc}+\frac1{ac}=1\] .Because of this,we may take $\frac1{ab}=\frac{x}{s}$ etc..with $s=x+y+z$.And after solving the sistem we easily get $c=\sqrt{\frac{xs}{yz}}$ so \[c^{2}+1=\frac{x(x+y+z)+yz}{yz}=\frac{(x+y)(x+z)}{yz}\] therefore our initial inequality becomes \[\sum \sqrt{\frac{yz}{(x+y)(x+z)}}\leq\frac32\] .Can someone finish it from here?

[quote="andyciup"] \[\sum \sqrt{\frac{yz}{(x+y)(x+z)}}\leq\frac32\] .Can someone finish it from here?[/quote]

Cauchy.

$\sum \sqrt{\frac{yz}{(x+y)(x+z)}}\leq \sqrt{\sum yz \sum \frac{1}{(x+y)(x+z)}}$
$=\sqrt{\frac{2\sum xy \sum x}{(x+y)(y+z)(z+x)}}$
$=\sqrt{2 \frac{\sum x^{2}y+3xyz}{\sum x^{2}y+2xyz}}$

$\sqrt{2 \frac{\sum x^{2}y+3xyz}{\sum x^{2}y+2xyz}}\leq \frac{3}{2}$
$\iff 8(\sum x^{2}y+3xyz) \leq 9(\sum x^{2}y+2xyz)$
$\iff 6xyz \leq \sum x^{2}y$ which is true by AM-GM, and this completes the proof.

[quote="andyciup"] \[\sum \sqrt{\frac{yz}{(x+y)(x+z)}}\leq\frac32\] .Can someone finish it from here?[/quote]

Also AM-GM .
$2\sqrt{\frac{yz}{(x+y)(x+z)}}\leq\frac{y}{x+y}+\frac{z}{x+z}$

 :wink:


Let $ ABC$ be an equilateral and $ C$ its incircle. If $ D$ and $ E$ are points on $ AB$ and $ AC$, respectively, such that $ DE$ is tangent to $ C$.
Show that $ \frac{AD}{DB} \plus{} \frac{AE}{EC} \equal{} 1$

Just use algebraic manipulation: $ AD^2 \plus{} AE^2 \equal{} (a\minus{}AD)^2 \plus{} (a\minus{}AE)^2$ where $ a$ is half the side of the triangle. Expand and we shall see that the equation to be proven is equivalence to this.

Denote M, N and P the contact points of AB, AC and DE with the incircle, then AM = a, MD = DP = x, EN = EP = y.
The relation to prove is equivalent to a^2 - a(x + y) - 3xy = 0.
Calculating DE from the triangle ADE with <DAE = 60 we get the same equality.

[b] Remarks:
[i] 1) Let D' and E' be the symmetrical of D and E w.r.t. M and N respectively; then D'E' passes through O, the circumcenter of ABC
    2) Old and known problem[/i][/b]

Best regards,
sunken rock


AMC
AIME
probability
geometry
geometric transformation
reflection

Say you are running out of time ... and you haven't even looked at #15 yet. What number are you going to guess?

Me ... lucky numbers are 025, 252, 729, and 899. :D

042, maybe the testwriters have a sense of humor.

eh I won't guess (don't guess on AIME dude) but if I did, probably 013

I will bubble in all the numbers :wink:.

wooo your probability of getting at least one guess correct if you guess on each problem is about 1.5% ONO

interestingly enough, there were two people who guessed correct answers on my mock test...

Maybe your test is just that guessable.  :P (Not really.)
I'll guess...hrmm...006.

[quote="Altheman"][quote="gimmemath"]why would 042 be funny??

 :rotfl: ?[/quote]

perfect score on usamo/imo

DUH...lol[/quote]

42 is also the meaning of life, the universe, and everything, if you didn't know.

Hm, last year I thought 025 would be a good guess, afterwards I actually read the question and it said $100<x<200$. But I'm going with 025...

I like 15, so I would guess 015 unless the answer is obviously big, in which case I would guess 515

Low numbers seem to show up more often than higher numbers; i.e. $1<x<100$, so I think I'll guess mostly low, but also, $462$ seemed to show up a lot on old AIME's, so if the answer looks like it will be high, I'll guess $462$.

As of 3 years ago, the most frequent answer was 025.  See
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=125555#125555[/url].

[quote="MysticTerminator"]hmm how about 013 I think we are not paying enough attention to 013[/quote]

Sheesh fine 13  :lol:

I say 006... I turned 16 on 6/6/06 no joke.  :|

[quote="usaha"]666...

Whats the lowest number that has ever appeared?  005?  I mean, 99.999% chance that 000 is probalby not going to appear..[/quote]

The lowest number that's ever appeared is the number $000$ which appeared on the AIME 2000 B question 9, check it out.  :lol:

actually read the question first...if it's like what is the largest prime that blah blah, then don't write like 024.

the ten numbers I would guess (one for each hundreds place)

077 181 236 394 405 519 634 792 831 947
unless of course the problem has to do with angles

also, what's the highest number an AIME answer has ever been? I don't remember seeing anything higher that I think 943 as problem 3 in '01 (I can't remember exactly - it was the fishing question)

[quote="sapphyre571"]also, what's the highest number an AIME answer has ever been? I don't remember seeing anything higher that I think 943 as problem 3 in '01 (I can't remember exactly - it was the fishing question)[/quote]

994 was the answer to question 10 in AIME 1989 I think.

The highest number an AIME answer was 997 and that was the answer to number 7 on the 1984 AIME.

Actually 997 was the answer to number 6 on the 2000 AIME as well.

what's the point of this thread anyways?

seriously, it's also kinda spammy....

"i'm gonna guess 002 cuz i have no brothers and no sisters and i had two pancakes this morning"

Just wondering (this is a retarded question) but do you have to write the extra zeros. Lets say the answer was 4 do you have to put 004? Or is 4 fine?

There's an answer form that you have to bubble in with a total of 3 digits per answer.  If the answer is 4, then you would have to write 004

oh I get it, it is my first time as you have probably guessed till now

The AIME is designed so it can be graded by machine- that's the reason for the weird answer format. A contest with on-site grading can use short-answer questions and human graders more efficiently, but the thousands of AIME papers make the machine-readable *** answer format the practical choice.

I don't think I ever bothered guessing on an AIME. I did once manage to get a question right by making two offsetting mistakes, but that's different.

024 for me... I have no earthly clue why....   :maybe:

i would guess large primes/prime-looking numbers if large

otherwise small, pretty numbers like 6, 12, 20 (no this is not n(n+1)),  25 (might as well), 48, 76 (>.<), 88, 102, and of course 144 and factors thereof

and of course 137 :D (because I like 137?)


prove $ lim_{x \minus{} > 6} 3 \equal{} 3$ via an epsilon-delta proof

-- my attempt --
$ Lim_{x \minus{} > 6} 3 \equal{} 3$
$ |3 \minus{} 3| < \varepsilon$
$ |0| < \varepsilon$
but i need to find
$ |x \minus{} 6| < k*{ \varepsilon}$, and [b]i can't see how i'd do that.[/b]
Then, let $ \delta \minus{} k \varepsilon$
Since all steps are reversible, whenever $ 0 < |x \minus{} 6| < \delta$, then $ |3 \minus{} 3| < \varepsilon$ and $ 3$ is the limit.

youre overthinking it

all epsilon delta proofs start with (even if they dont say it)

let $ \epsilon >0$

then we want to find a $ \delta>0$ s.t.

$ |x\minus{}6|<\delta \to |3 \minus{} 3| < \epsilon$

but of course such a delta exists, since 3 is constant no matter what we choose for x

so let $ \delta \equal{} 1$

then we know when $ |x\minus{}6|<\delta\equal{}1$ we have $ |3\minus{}3|\equal{}0 < \epsilon$

therefore $ \lim_{x\to 6} 3 \equal{} 3$


function
inequalities
limit
Functional Analysis
triangle inequality
topology
real analysis

Show that if X is a separable Banach space there exists an isometric inclusion, T, between X and the space l^(infinity), the space of real valued functions with the sup norm, which is also a Banach space. Hahn Banach could be used, but I am not sure how to get it. Much thanks!

The right trick here is a direct construction. Let $ x_1,x_2,\cdots$ be a dense sequence, and define $ f(x)\equal{}(\|x\minus{}x_1\|,\|x\minus{}x_2\|,\cdots)$. By the triangle inequality, $ \|f(x)\minus{}f(y)\|\equal{}\sup |\,\|x\minus{}x_n\|\minus{}\|y\minus{}x_n\| \,|\le \sup \|x\minus{}y\|$. Since we can choose a subsequence $ x_{n(k)}$ converging to $ y$, we also get that $ \sup |\,\|x\minus{}x_n\|\minus{}\|y\minus{}x_n\| \,|\ge \lim_k |\,\|x\minus{}x_{n(k)}\|\minus{}\|y\minus{}x_{n(k)}\| \,|\equal{}\|x\minus{}y\|$ and $ f$ is an isometry.

This actually proves a stronger result; we didn't use any norm properties, so in fact any separable metric space has an isometric injection into $ \ell^\infty$.


I found this problem and now I'm starring at it with lot's of what if's. I'm working on this problem but feel free to beat me at it It's not like I'm going back to school to take Algebra II again.

A * [(B + C)(D - E) - F(G*H) ] / J = 10

Knowing that each variable is a unique, single-digit, nonzero number, and that C - B = 1, and H - G = 3, what is the number ABCDEFGHJ, where each letter is a digit? For example, if A = 4, B = 2, and C = 7, ABC would equal 427.

Harder than it looks...

[hide="partial solution"]Since C-B=1, then B is one of 1,2,3,4,5,6,7,8 and C is one of 2,3,4,5,6,7,8,9 AND B+C is one of 3,5,7,9,11,13,15,17

Since H-G=3, H is one of 9,8,7,6,5,4 and G is one of 6,5,4,3,2,1 AND HG is one of 54,40,28,18,10,4

We can make a little distribution of the possible values of B, C, G, and H:

A|
B|   1,2,3,4,5,6,7,8
C|   2,3,4,5,6,7,8,9
D|
E|
F|
G|   6,5,4,3,2,1
H|   9,8,7,6,5,4
J|

We also know that $A\left(\left(B+C\right)\left(D-E\right)-FGH\right)=10J$[/hide]

And I thought I knew Intermediate Algebra! :P

This was supposebly "simple" algebra  :|


algebra
polynomial
inequalities
algebra proposed

A polynomial $ P(x)$ has degree at most $ 2k$, where $ k= 0, 1, ...$. Given that for an integer $ i$, the inequality $ -k\leq i\leq k$ implies $ |P(i)|\leq 1$. Prove that for all real numbers $ x$, with $ -k\leq x\leq k$, the following inequality holds: $ |P(x)|< (2k+1){2k\choose k}$


[quote="Yosh..."]Let $a_{1},a_{2},\cdots a_{N}$ be distinct positive integers that do not contain a $9$ in their decimal representation.

Prove that $\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\leq \frac{1}{30}$[/quote]

This is obviously wrong : take $N=1$ and $a_{1}=1$

I think that the problem is to show $\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}< 30$

Let $A=\{$ posive integers without any $9$ in their decimal representaion $\}$
Let $A_{k}=\{x\in A, 10^{k-1}\leq x<10^{k}\}$

Let $S_{k}=\sum_{x\in A_{k}}\frac{1}{x}$

$S_{k+1}=\sum_{x\in A_{k}}(\frac{1}{10x}+$ $\frac{1}{10x+1}+\frac{1}{10x+2}+\frac{1}{10x+3}+$ $\frac{1}{10x+4}+\frac{1}{10x+5}+\frac{1}{10x+6}$ $+\frac{1}{10x+7}+\frac{1}{10x+8})<9\sum_{x\in A_{k}}\frac{1}{10x}=\frac{9}{10}S_{k}$
So $S_{k+1}<S_{1}(\frac{9}{10})^{k}$
This shows that $\sum_{k\geq 1}S_{k}$ is a convergent sum and that ${\sum_{x\in A}\frac{1}{x}=\sum_{k\geq 1}S_{k}<S_{1}(1+\sum_{k\geq 1}}(\frac{9}{10})^{k})$

So $\sum_{x\in A}\frac{1}{x}<10S_{1}=10(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})=\frac{761}{28}<30$

And this shows that $\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}< 30$ for any $\{a_{i}\}\subset A$

(in fact we have $\sum_{x\in A}\frac{1}{x}\approx 22.920676619$)


AMC AIME geometry trigonometry probability function search

Exactly where  can I read about content on the AIME? [hide="Topics I am wondering about"] Are any of the following on the first half of the AIME?
-analytic geometry (on the level of the aops v2 chapter on it)
-complex numbers (not just knowing what they are and what i is)
-graph theory
-fermat's little theorem, wilson's theorem, other intermediate number theory topics above basic modular arithmetic
-trigonometry applications in geometry problems
-topics covered in the aops intermediate counting and probability book (1 to 1 correspondences, recursion, generating functions)
-cyclic quadrilaterals, other intermediate geometry topics
[/hide]

For starters, you could look at [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45]past AIME problems[/url] or the AMC website.

Out of the things you listed, graph theory is extremely unlikely to show up.  The rest are probably going to be useful.

Fermat's little theorem is a must-know for AIME problems in any part of the contest, but Wilson's Theorem tends to be needed only at the olympiad level.

The rest of your list is pretty good, though I would add polar complex numbers ($ re^{i\theta}$) and, to expand on analytic geometry, three dimensional geometry.

Either way, very few AIME problems that aren't the first two or three can be destroyed by a single mathematical concept, so don't just come in knowing lots of facts.

sorry, I hope I don't offend you with my lack of mathematical knowledge, but what is fermat's little theorem, and what was fermat's last theorem?

Haha. I like that google hyperlink =).  AIME problems obviously never include fermat's last theorem... I'm not sure about fermat's little theorem ~_-. Haven't seen one yet... At least I don't think so.  Perhaps rotation of axes? Usually not though.  Polar coordinates are nice, and using $ cis$ is also useful.  hmm. Most of the first 8-10 AIME problems are relatively simple and can be brute-forced, though it might take a while.  But if you're aiming for 10+, you'll have to think of some nice solutions.

[quote="t0rajir0u"]Such questions are easily answered by [url=http://www.google.com/search?hl=en&q=fermat%27s+little+theorem&btnG=Google+Search]Goo[/url][url=http://www.google.com/search?hl=en&q=fermat%27s+last+theorem&btnG=Search]gle[/url].[/quote]

no, i know what fermat's little theorem is i just wanted to see what kinds of AIME  problems require the use of fermat's little theorem.

I was responding to the poster above me.

I don't actually know if it'll be directly useful on the AIME, but a knowledge of modular arithmetic in general is certainly useful.  AIME I 2004 #1, for example, is much easier if you know that $ 10^3 \equiv 1 \bmod 37$.

[quote="cognos599"]really, maybe i haven't worked enough AIME probs, but i can't recall any problems that require the use of fermat's little theorem, could someone post a past AIME problem that requires or even uses this theorem?[/quote]

Quite ironically, the first time I saw Fermat's Little Theorem was when I read the solution to [[1989 AIME Problems/Problem 9]].

It (FLT) otherwise almost never shows up though, but I agree with t0rajir0u that modular arithmetic is helpful.

With topics like counting, I don't think that topics like generating functions appear often? But there are usually a number of combinatorics questions.

oh thanks i actually didn't use fermat's little theorem when i didi this problem, but my way wasn't very good.

you wont need to know about cis unless you're aiming for >10.

usually the problems involving complex numbers are actualy very simple, requiring more geometry and intuition than algebra.

In answer to the orginal question in this post, see our Frequently Asked Questions, AIME #1:
http://www.unl.edu/amc/f-miscellaneous/faq.shtml#aime

Steve Dunbar
MAA Director, American Mathematics Competitions


Here is the answer to the first one:

$2\cdot18-20+8=24$

Are you going to post several more after they have been answered?

I will kind of make this into a game so the scores are:

b-flat: 1

[b]Round 2.[/b]
Here are the next numbers. [b]Look at the top post for instructions.[/b]

[b]3, 5, 57, & 10 [/b](they don't have to be used in order)

P.S. If you have looked at the instructions look again because I added some stuff.

ok the score is:

[b]b-flat: 1
dogseatcheese: 1[/b]

[b]Next numbers[/b]. Look back for instructions
[b]4. 4. 5. & 8[/b]

Remember, you cannot use a number twice, unless it is listed twice, so you have to use 4 twice.

Next numbers????? i was the frst 1 2 answer =P(i don't like dah smiley)

You could make it the first person to reply gets 1 point, and if the second person posts a completely different (but still correct) solution, they get 1 point also.  Like Eru posted a completely different but correct solution so he/she should also get a point.  When are you posting the next problems?

SRY. I''ve had 2 many school projects these days.s Well, I don't believe Eru would be right because I'm pretty sure in the game, PEMDAS doesn't matter. However [b]I will make a new rule that if you solve it with PEMDAS you get half of a point[/b]. If there are multiple soulutions without PEMDAS both will get a point

[b]b-flat: 1
dogseatcheese: 1
Naonao: 1
Eru: .5[/b]

[b]Next number 5, 13, 20, 3[/b]

[b]b-flat: 1
dogseatcheese: 1
Naonao: 1
robinhe: 1
Eru: .5 [/b]

Come on...some 1 who has already
answered don't leave-No one has even reached 2
[b]
Next #'s
3, 3, 9, 7 [/b]

Here I am, with 2.  It is hard to find multiple solutions, and if someone gets there before you, you can't get any point.  It doesn't use PEMDAS.

$\frac93\cdot7+3=24$

There's no way to make 24 using those. I'm pretty sure.
23 I also got, 24 is next to impossible.
By the way, check out my game!
Link's in my sig!

[quote="ch1n353ch3s54a1l"][quote="R49917S"][b]Hey. Due to the new "Kleiban's 24", which has took all the attention away from mine, this whole 24 forum can be deleted by a moderator, because many people aren't replying anyway. And if a mod doesn't delete it, I'm not replying to anything or posting, so do whatever you want in thid forum.[/b]

[b]:bomb:    :mad:  :furious: THANKS KLEIBAN :furious:   :mad: :bomb: [/b]

yes, I show anger through emoticons.[/quote]

? You don't need someone to delete this. I would have responded to this 24, except i couldnt' find a solution to your numbers  :lol:[/quote]

I couldn't find an answer either so I just gave up on it.

[b]Okay. Have 6 other people say that they would answer and I will continue. Otherwise, make your own or something.[/b]

I started a new topic because this is getting way to confusing.
Some of your nubmers are off, and your rules are hard to understand.

[quote="R49917S"][b]Hi. This is based on a math game called "24." What you have to do is with four numbers, perform 3 different operations (only multiplication, division, addition, and subtraction) with all numbers, to equal 24. All parts must be whole. You can only use each number once, unless it is listed twice. The numbers don't have to be used in the order listed. Do not hide it since it is a game with points. Just type it, don't put in the "hide=Answer" stuff. Do not post once somebody has, it doesn't matter and you will not recieve points. If no one posts the answer in 3 days, I will post a new one. I mean exactly 3 days. The time is at the top of posts. If it's 3 days and 1 minute, it won't count. So if you solve one that is more than 3 days old, good for you but it doesn't count. :D  :D First person to 5 points earns the right to post 1 problem, whenever they want, as long as an answer has been posted and I have not made a new one yet (Write in the post that you earned it). And for people who are not new: PEMDAS no longer counts. If your are don't worry. At no point can the  number be a decimal of fraction

[/b][b]ex.[/b] the numbers are 2,4, 44 and 6. Notice how you can get 24 by doing $4\times6$. However, thats only using two numbers. The correct answer would be $44\div2+6-4=24$

[b]The first set of numbers are...[/b]
[b]8, 20, 2, and 18  (remember, the numbers do not have to be used in this order)[/b][/quote]

$24$

tz. You apparently have not read the previous page that says I have forever abandoned this post (well at least making problems). Type anything you want about anything. (For example-Duck are neat.

[quote="R49917S"] Type anything you want about anything. (For example-Duck are neat.[/quote]
wouldn't that be just spamming?

Penguins walk wobilly. Elephants have huge ears.

Please stop! If R49917S really doesn't want to continue, then leave this topic alone. Thank you.  :)


Let a,b,x,y,z be positive real numbers.
Show that

  x/(ay+bz) + y/(az+bx) + z/(ax+by)  \geq 3/(a+b)

Just a remark: This is a generalization of Nesbitt's inequality, for a=b we get his inequality.

it is a very easy application of a derived version of Cauchy-Schwartz inequality (sometimes called Cauchy-Buniakowski):

\sum ai / bi \sum aibi \geq ( \sum ai)2 , for all positive reals ai, bi.


probability
geometry
number theory
\/closed

Introduction to Algebra is the new Algebra 1 + Algebra 2 courses, so the next Algebra course would be Algebra 3 (the old Intermediate Algebra).

[quote="westiepaw"]What would be the next course to take at AoPS after the Into to Algebra course that began in March 2009 (assuming the student did fine in the course)?  I am confused by the new names of the AoPS Algebra courses.  Thank you, from the mom of a current Intro to Alg. student.[/quote]

I would recommend going through Intro to Number Theory and Intro to Counting & Probability (in either order) after Intro Algebra.  Then, Intro to Geometry, and then Algebra 3 (formerly Intermediate Algebra).  After that, the rest of the Intermediate classes in basically any order.


geometry
3D geometry
number theory
least common multiple
MATHCOUNTS
calculus
trigonometry

1.Find all those 5-digit positive whole numbers that are square and cube numbers also.

  2. How many 4-digit numbers are there in which the sum of the digits is 5?

  3 .Find a 6-digit number that gets 6 times greater when you move its last 3 digits to the front of the number without changing the order of these three digits.

   4.How many 3-digit numbers are there in which the digits are such that if you create all the different 3-digit numbers from them, and you add these numbers, the sum is divisible by 111?

can someone else post some problems -_-

[hide]
1) If a number is to be a perfect square and cube, it must be a perfect sixth power (6=LCM of (2,3)). The only numbers which are five digit sixth powers are 5^6=15625 and 6^6=46656. Thus, there are two numbers.

2) After a little bit of counting, you'll figure out that there are six partitions of 5 which can be expressed as 4 or fewer numbers (if fewer, then add enough zeroes to the end of the partition until it has four digits). These partitions are:

5+0+0+0
4+1+0+0
3+1+1+0
2+1+1+1
2+2+1+0
3+2+0+0

If you know about combinatorics, you should know that each of four numbers, if placed in the thousands place of a four digit number, adds six more different numbers to the total # of possibilities (provided all the numbers are distinct, and none are zero). But here, this is not the case. For the first partition, we have (4!-(3x6))/3!=1 combo.

For the second, we have (4!-(2x6))/2!=6 combos.
For the third, we have (4!-(1x6))/2!=9.
4th=4!/3!=4.
5th=(4!-(1x6))/2!=9.
6th=(4!-(2x6))/2!=6.

This is a total of 1+6+9+4+9+6=35 numbers.

3) Umm, I'm still working on it...

4) As long as the three digits are not zero, all possible arrangements added up will be divisible by 111, since their sum will be 222(a+b+c), where a,b, and c are the digits. Otherwise, you'll run into complications.

The total possible of such numbers is 9x9x9=729. [/hide]

1.[hide]These are just powers are 6... 5^6=15625 and 6^6=46656 are the only powers of 6 that are 5 digits.[/hide]

2.[hide]The ways to make the sum of the digits 5 are as follows: 5+0+0+0, 4+1+0+0 (can be repeated 6 times, 4100, 1400,4010, 4001, etc...), 3+2+0+0 (also 6 times), 3+1+1+0 (9 times), 2+2+1+0, (9 times) 2+1+1+1 (4 times). 9+9+6+6+4+1=35.[/hide]

3.[hide]600000a+60000b+6000c+600d+60e+6f=100000d+10000e+1000f+100a+10b+c
599900a+59990b+5999c=99400d+9940e+994f

Clearly, the unit's digits of 5999c and 994f must be the same. We want 994f as big as possible and 599900 as small as possible so we make f=7, and c=2. That puts 9 in the tens digits for the left side and 8 for the right side...
Edit: This is not a good method, read natmc's method.[/hide]

4.[hide]You have 3 digits, a, b, and c. You can have the numbers 100a+10b+c, 100a+10c+b... etc... and if you sum all of those you get 222a+222b+222c. Note that this will only work if none of the digits are 0. That leaves us with 9^3=729 numbers in which it works.[/hide]

[quote="eryaman"]ummm...what's with the title of this thread?[/quote]

Yeah, someone should change the title, because it will confuse people, and stop them from coming to see the problems, since I think we can assume that most people in Mathcounts aren't learning calculus or trig yet.

The title of this thread has been changed by moderator rcv.

G-UNIT:  The moderator requests that you use appropriate, meaningful titles.

[hide="1."]Must be powers of 6:
5^6=15625
6^6=46656
two.[/hide]
[hide="2."]
Possible combinations for the digits:
1,1,1,2
1,2,2,0
2,3,0,0
3,1,1,0
4,1,0,0
5,0,0,0

1st one: 4 combinations
2nd one: 9 combinations
3rd one: 6 combinations
4th one: 9 combinations
5th one: 6 combinations
6th one: 1 combination
9+9+6+6+4+1=[b]35[/b][/hide]

sorry about that, i didnt really think my choice of title mattered, i simply wrote down the first title i could think off... :)


Find all real solutions of the system:

$ x^2\plus{}2yz\equal{}x,$
$ y^2\plus{}2zx\equal{}y,$
$ z^2\plus{}2xy\equal{}z.$

[hide]Case I: one of $ x,y,z$ is $ 0$:

WLOG, let $ x\equal{}0$.

Then, $ yz\equal{}0$, $ y^2\equal{}y$, $ z^2\equal{}z$.

Thus, $ y,z \in \{ 0,1 \}$ and either $ y\equal{}0$ or $ z\equal{}0$.

Therefore, the only solutions in this case are: $ (0,0,0)$, $ (1,0,0)$, $ (0,1,0)$, $ (0,0,1)$.

Case II: $ x,y,z \neq 0$:

Then, multiplying the three equations by $ x$, $ y$, $ z$ respectively yields:
$ x^3\minus{}x^2 \equal{} y^3\minus{}y^2 \equal{} z^3\minus{}z^2 \equal{} \minus{}2xyz \equal{} K$ for some non-zero $ K \in \mathbb{R}$.

Thus, $ x,y,z$ are solutions of $ f(t) \equal{} t^3\minus{}t^2\minus{}K \equal{} 0$.

If $ x,y,z$ are distinct, by Vieta's $ xyz \equal{} K \equal{} \minus{}2xyz$ which is impossible for $ x,y,z \neq 0$.

WLOG, let $ y \equal{} z$.

Then:
$ y^2 \plus{} 2xy \equal{} y \Rightarrow y\equal{}1\minus{}2x$
$ x^2 \plus{} 2y^2 \equal{} x \Rightarrow x^2 \plus{} 2(1\minus{}2x)^2 \minus{} x \equal{} (3x\minus{}1)(3x\minus{}2) \equal{} 0$

Thus, $ x \equal{} \frac{1}{3} \Rightarrow y \equal{} z \equal{} \frac{1}{3}$ OR $ x \equal{} \frac{2}{3} \Rightarrow y \equal{} z \equal{} \minus{}\frac{1}{3}$

Therefore, the solutions in this case are: $ \left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$, $ \left(\frac{2}{3},\minus{}\frac{1}{3},\minus{}\frac{1}{3}\right)$, $ \left(\minus{}\frac{1}{3},\frac{2}{3},\minus{}\frac{1}{3}\right)$, $ \left(\minus{}\frac{1}{3},\minus{}\frac{1}{3},\frac{2}{3}\right)$. [/hide]


I can't get go2pdf to work on my computer, so I won't be able to include images in my solutions file.  Is it OK to submit images as a separate file?

Try CutePDF:

http://www.cutepdf.com/Products/CutePDF/writer.asp

Another option is to print out you solutions and hand-draw them.

You can also just make JPG files and include those the same way the instructions describe including PDF files.

We need to have a single document we can print and hand to graders.


Which of the following statements are true?

I. The set {-1,0,1} is closed under addition
II. The set {-1,0,1} is closed under multiplication
III. The set of rational numbers is closed under addition
IV. The set of rational numbers is closed under multiplication

The first two questions can be answered making a table to see if all the possible sums/products of elements of the set belong to the set. If so, we have closure.
[hide="I."] 1+1=2 by my definition, and 2 does not belong to {-1,0,1}.
The set {-1,0,1} is not closed under addition.
Of course, we could define an operation that set would be closed under, but I would not use the name addition.[/hide]
[hide="II."]All nine possible products are in {-1,0,1}. The set {-1,0,1} is closed under multiplication [/hide]
For the other two problems, it is a question of deciding if every time two rational number are added/multiplied the result is necessarily also rational.
[hide="III."] The set of rational numbers is closed under addition[/hide]
[hide="IV."] The set of rational numbers is closed under multiplication[/hide]


algebra
polynomial
Rational Root Theorem

Sometimes an approximation is good enough, say through Newton's Method.

[quote="tornado.adv4"]I know you can use the Rational Root Theorem to look for rational roots of a polynomial, but how would you find the roots if there are no rational roots, if all the roots are irrational or imaginary?

Thanks.[/quote]

That's a good question, although usually on AMC there are rational roots.

I had a similar question :blush:

As JRav said, sometimes approximation is good enough.

Maybe Intermediate Algebra can help, or ask at class if you're taking Algebra 3

[quote="tornado.adv4"]I know you can use the Rational Root Theorem to look for rational roots of a polynomial, but how would you find the roots if there are no rational roots, if all the roots are irrational or imaginary?

Thanks.[/quote]Depends. If you are looking for the roots in a 'nice' form, you should know by now polynomials degree of 5 or higher are not 'solvable' by radicals due to Galois Theory. There are other conditions that allow it to be 'solvable' but that's beyond the scope of this subforum.

JRav suggests a good way of finding an approximation.

EDIT: Descartes Rule of Signs is helpful if you want a rough idea about the nature of the roots. Other interesting topics are like Horner's method, Sturm sequences, etc.


What are some of the differences between American and foreign Universities? Would some of you students ever consider attending a foreign University?

What do you all think of the proposed changes to the British University system?

I'm afraid you'll  have to provide me a little background (or some good links) about the British system.

Canadian schools are cheap, and they have some good ones (McGill and U. Toronto pop to mind).

Under the currect system all British citizens (and I think even EU citizens) pay the same tuition for any British Univeristy. I think it's around $2000 US a year. Tony Blair wants to make it so that British Universities can charge more, but the students don't have to pay until after they graduate and their annual income becomes at least equal to the national average.

To us Americans this seems like a dream situation! But some of the British are fighting the proposal vehemently.

Which of them sounds like a dream situation - the old or the new format?  (Though I'd wager there are many who'd answer 'both'.)

And what are the grounds on which the complaining British object?

I think the main objection from some British is that they see this as a first step in the Americanization of the system. They fear that the tuitions will skyrocket, and that the elite Universities it will become exclusively for the rich.

Blair's argument is that in order to reach Britain's stated goal of having 50% of it's population go to University, they have to find a way to bring in more revenue. He also claims that the system as it is can't compete with the American system and that Britain is losing many of it's best professors to American Universities that can pay them better.

Heh when it comes to cost no one...er, country can beat Japan: 5K for all Japanese government-built universities (I apologize for that, I mistook "You can go to any college if you don't be choosy" for "You can go to any college no matter what" and applied it to all people...although in Japan most people are middle-classes and not a lot of people are really rich so I'm guessing most could afford it)

Tokyo University is still very cheap and it's very good too, around five grand (the best school in Japan). There are two other private great school in Japan, Waseda (around 20K) and Keiou (around 30K) depending on what you study. (All rates are annual and only includes the fees for the studying, does not include stuff like the living cost or transportation cost)

In general Japanese schools have this huge entrance exams which everybody studies for like maniacs (a.k.a. the Examination War) and then they slack off for the rest of the school until the end in which they start studying for The War again and this starts happening as soon as preschool :shocked:

This is the reason contributing to "you can at least go to somewhere for college" is the decreasing number of Japanese children. We have so much seniors and too little children such that we have 2:1 ratio for seniors versus children. Pretty soon we might perish...(I'm just having a shot in the dark here, don't take it seriously folks :wink:)

In fact the total number of total seats available in Japan exceeds the number of children in Japan thus this is possible.

What is the nature and set up of the entrance exams in Japan? Are they like the SATs, or like AP Exams, or something totally different?

Well there is this "Central Exam" (even though it's pronounced "center exam" I believe this is grammatically correct) but most college/university have their own tests and that is perhaps the only factor which will decide whether you get into that college/university or not. There is no scholarships, (maybe) retakes, very little looking over the other stuff (unless you have something big like a criminal record or an IMO winner :wink:). Basically if you fail the war you'll have to come back next year, even if you were just sick the day you took the test.
On the other hand there's no ridiculous requirement like "swim 100m" but at the same time there are no requirement...kinds extreme huh?

I wonder how much Russian university cost...

I mean what subjects do the exams usually cover - are they "aptitude" tests like the SAT or are they subject tests like the AP exams? And what is the level of difficulty of the tests.

Most colleges in other countries actually usually have an extremely simple process, compared to the U.S. In many countries, you just take a test(similar to the SAT/ACT), and submit your grades. Those may be the only considerations!

Gauss, the test usually depends on the country: each have their own test, and it usually covers all the material learned in high school only.


inequalities
trigonometry
geometry proposed
geometry

$1.\blacktriangleright\ \triangle ABC\Longrightarrow$ $\left(a^{2}+b^{2}+c^{2}\right)^{2}+\left(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\right)\le 12R^{2}\left(a^{2}+b^{2}+c^{2}\right)\ .$

$2.\blacktriangleright\ \sqrt 3\le\boxed{\frac{4R+r}{p}\le\sqrt{3+4\cdot \frac{R-2r}{r}}}\ .$
[hide="Hint."]Apply the first inequality in the (acute) triangle $A'B'C\ (\sqrt a,\sqrt b,\sqrt c)\ ,$ where $\cos A'=\frac{p-a}{\sqrt{bc}}$ a.s.o. and $R'=\sqrt{\frac{pR}{4R+r}}$ [/hide]

Indeed, the first inequality is a trivial consequence of the identity
\[OH^{2}=9R^{2}-a^{2}-b^{2}-c^{2}. \]


Problem2

i)Let n positive integer (n>= 0) find the sign of n^6 +5n^5.sin(n) +1

ii)For wich positive integer n (n>= 0)

(n^2 +5n.cos(n) +1)/(n^6+5n^5.sin(n)+1) >= 10^(-4)

holds ?

Without calcultrice

I suppose "angle n" (in sin(n)) is in degrees ?


trigonometry
inequalities
geometry
circumcircle
trig identities
Law of Sines
inequalities unsolved

Prove that for every acute triangle $ABC$ we have

$\sin A \sin 3A + \sin B \sin 3B + \sin C \sin 3C \leq 0$

It's really not difficult and It's a lemma of $\frac{a^4+b^4+c^4}{a^2+b^2+c^2} \ge 3R^2$
Any one post a solution,My solution use deriveate and I think It has other way.

[color=blue][b]Problem.[/b] Let ABC be an acute-angled triangle. Prove the inequality

$\sin A\sin\left(3A\right)+\sin B\sin\left(3B\right)+\sin C\sin\left(3C\right)\leq 0$.[/color]

[i]Solution.[/i] By a well-known formula from trigonometry,

$\sin\left(3A\right)=3\sin A-4\sin^3 A$.

Thus, $\sin A\sin\left(3A\right)=\sin A\left(3\sin A-4\sin^3 A\right)=3\sin^2 A-4\sin^4 A$. Similarly, $\sin B\sin\left(3B\right)=3\sin^2 B-4\sin^4 B$ and $\sin C\sin\left(3C\right)=3\sin^2 C-4\sin^4 C$. Thus, the inequality in question,

$\sin A\sin\left(3A\right)+\sin B\sin\left(3B\right)+\sin C\sin\left(3C\right)\leq 0$,

is equivalent to

$\left(3\sin^2 A-4\sin^4 A\right)+\left(3\sin^2 B-4\sin^4 B\right)+\left(3\sin^2 C-4\sin^4 C\right)\leq 0$.

If R is the circumradius of triangle ABC, then the extended law of sines yields $\sin A=\frac{a}{2R}$, $\sin B=\frac{b}{2R}$ and $\sin C=\frac{c}{2R}$, so that this inequality rewrites as

$\left(3\left(\frac{a}{2R}\right)^2-4\left(\frac{a}{2R}\right)^4\right)+\left(3\left(\frac{b}{2R}\right)^2-4\left(\frac{b}{2R}\right)^4\right)+\left(3\left(\frac{c}{2R}\right)^2-4\left(\frac{c}{2R}\right)^4\right)$
$\leq 0$.

After multiplication with $4R^4$ and simplification, this becomes

$\left(3R^2a^2-a^4\right)+\left(3R^2b^2-b^4\right)+\left(3R^2c^2-c^4\right)\leq 0$.

This is equivalent to

$3R^2\left(a^2+b^2+c^2\right)-\left(a^4+b^4+c^4\right)\leq 0$,

i. e. to

$3R^2\left(a^2+b^2+c^2\right)\leq a^4+b^4+c^4$.

Now, let's apply the Conway substitution (cf. http://www.mathlinks.ro/Forum/viewtopic.php?t=6523 post #4): Since the triangle ABC is acute-angled, we have $b^2+c^2>a^2$, $c^2+a^2>b^2$ and $a^2+b^2>c^2$; thus, the numbers $x=\frac12\left(b^2+c^2-a^2\right)$, $y=\frac12\left(c^2+a^2-b^2\right)$ and $z=\frac12\left(a^2+b^2-c^2\right)$ are all positive. Of course, we have $y+z=\frac12\left(c^2+a^2-b^2\right)+\frac12\left(a^2+b^2-c^2\right)=a^2$ and similarly $z+x=b^2$ and $x+y=c^2$, and thus $a^2+b^2+c^2=\left(y+z\right)+\left(z+x\right)+\left(x+y\right)=2\left(x+y+z\right)$.

Also, from $y+z=a^2$, $z+x=b^2$ and $x+y=c^2$, it follows that $\left(y+z\right)^2=\left(a^2\right)^2=a^4$ and similarly $\left(z+x\right)^2=b^4$ and $\left(x+y\right)^2=c^4$.

Finally, since the circumradius R of triangle ABC can be found using the formula $R=\frac{abc}{4F}$, where F is the area of triangle ABC, we have

$R^2=\left(\frac{abc}{4F}\right)^2=\frac{a^2b^2c^2}{16F^2}=\frac{\left(y+z\right)\left(z+x\right)\left(x+y\right)}{16F^2}$.

By the Heron formula for the area F of triangle ABC, we have

$16F^2=\left(a+b+c\right)\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)$;

on the other hand, using the equations $x=\frac12\left(b^2+c^2-a^2\right)$, $y=\frac12\left(c^2+a^2-b^2\right)$ and $z=\frac12\left(a^2+b^2-c^2\right)$ and performing some calculations (there is a synthetic proof, too, but I'm too lazy to write it down), we can find

$4\left(yz+zx+xy\right)=\left(a+b+c\right)\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)$.

Thus, $16F^2=4\left(yz+zx+xy\right)$. Hence,

$R^2=\frac{\left(y+z\right)\left(z+x\right)\left(x+y\right)}{16F^2}=\frac{\left(y+z\right)\left(z+x\right)\left(x+y\right)}{4\left(yz+zx+xy\right)}$.

Combining all the above results, we get

$a^2+b^2+c^2=2\left(x+y+z\right)$
$a^4=\left(y+z\right)^2$; $b^4=\left(z+x\right)^2$; $c^4=\left(x+y\right)^2$;
$R^2=\frac{\left(y+z\right)\left(z+x\right)\left(x+y\right)}{4\left(yz+zx+xy\right)}$.

With these formulas, the inequality that we must prove,

$3R^2\left(a^2+b^2+c^2\right)\leq a^4+b^4+c^4$,

becomes

$3\frac{\left(y+z\right)\left(z+x\right)\left(x+y\right)}{4\left(yz+zx+xy\right)}\cdot 2\left(x+y+z\right)\leq \left(y+z\right)^2+\left(z+x\right)^2+\left(x+y\right)^2$.

This simplifies to

$3\frac{\left(y+z\right)\left(z+x\right)\left(x+y\right)\left(x+y+z\right)}{2\left(yz+zx+xy\right)}\leq \left(y+z\right)^2+\left(z+x\right)^2+\left(x+y\right)^2$.

Upon multiplication with 2 (yz + zx + xy), this becomes

$3\left(y+z\right)\left(z+x\right)\left(x+y\right)\left(x+y+z\right)\leq 2\left(yz+zx+xy\right)\left(\left(y+z\right)^2+\left(z+x\right)^2+\left(x+y\right)^2\right)$.

But this inequality follows from

$2\left(yz+zx+xy\right)\left(\left(y+z\right)^2+\left(z+x\right)^2+\left(x+y\right)^2\right)-3\left(y+z\right)\left(z+x\right)\left(x+y\right)\left(x+y+z\right)$
$=yz\left(y-z\right)^2+zx\left(z-x\right)^2+xy\left(x-y\right)^2\geq 0$.

So the proof is complete.

[[b]EDIT:[/b] Now I see that this inequality has already been proved in http://www.mathlinks.ro/Forum/viewtopic.php?t=22866 , posts #10 and #14.]

  Darij

Since
$R^2=\frac{a^2b^2c^2}{16S^2}=\frac {a^2b^2c^2}{2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4}$,
using the substitution $a^2=\frac{y+z}{2}$ etc., the inequality
$\displaystyle \frac{a^4+b^4+c^4}{a^2+b^2+c^2} \ge 3R^2$
becomes as follows
$xy(x-y)^2+yz(y-z)^2+zx(z-x)^2 \ge 0.$


quadratics
modular arithmetic
probability
algebra
linear equation
number theory

Problem 2.18 is:

"Is 9409 prime?"

There is a hint, # 52, that gives the expression:

(100-x)^2 = 10000-200x+x^2

Now, this is great and I'm still working on the problem, but what I find confusing is how can one tell from the problem that algebra would be useful from the problem where it's not.
For example, most of the problems in this Challenge section are done with no algebra or arithmetic at all, just words. So, out of nowhere, comes this problem where this quadratic is given as a hint.
Problem 2.14 is similar, since the hint involves a linear equation, but there's nothing I can see from the problem that would lead me to believe algebra would be useful.
I realize I should just let go and trust that understanding will dawn at some point, but if anyone has any insight on this, I would be most appreciative.

Thanks,
Ed Beaugard

hello, solve the equation $ 10000\minus{}200x\plus{}x^2\equal{}9409$ for $ x$.
Sonnhard.

Algebra is useful because there are many factorization techniques that can be applied if you know that the number is of a special form, and factorizations are usually useful in number theory.

For example, [url=http://www.mathlinks.ro/viewtopic.php?t=150527]PEN E 4[/url] asks to prove that $ 1280000401$ is composite. By recognizing the special form $ 1280000401 \equal{} 20^7 \plus{} 20^2 \plus{} 1$ you may use the factorization $ a^7 \plus{} a^2 \plus{} 1 \equal{} (a^2 \plus{} a \plus{} 1)(a^5 \minus{} a^4 \plus{} a^2 \minus{} a \plus{} 1)$.

It's all a matter of familiarity with the technique.

Not only is 9409 not prime, it is a perfect square. $ 97^2\equal{}9409$.

I really appreciate the replies and patience with my Newbie questions.
I found ffao interesting, since he(I assume) relates factorization in algebra to factorization in number theory, which is exactly what I was wondering about, that is, when would one turn to algebra to solve these kinds of problems.

Thanks again.

[quote="AIME15"]...which is EXACTLY what the hint implied?

In general, when there's a question that says "Is ??? prime?", about 70% of the time (unless it's in a school textbook) it will be composite.  :)[/quote]

Lol , I like your way of thinking  :lol:

So, the key to recognizing when to apply algebra, such as special factorizations, to number theory problems is simply familiarity with the technique? Or is there a certain set of factorizations that we should learn to test on all prime/composite problems?

Also, note 9409 ends in a 9.
And the hint says 100^2-200x+x^2
So x must be 3 mod 10, otherwise you wouldn't have a 9 as the units digit.

If 9409 fits 10000 - 200x + x^2 you'll know that it is composite because then it is also a perfect square.  If it doesn't then the hint was pretty much useless.. :P

Generally, in my opinion, it would be helpful in the book if there were indications about when to use algebra and when not to on particular problems.

Thanks,
Ed Beaugard

But you aren't given these hints in real problems...

When and when not to use algebra to solve real problems are something we should know, I agree, but I'm talking about the textbook itself(which I like a great deal), where we're supposed to learn this. It's just something that could be thought about that might make a superb book even better.

Very sincerely,
Ed Beaugard

I think you have a good point. For instance, Introduction to Counting and Probability emphasizes that many times algebra and counting can be used together, but doesn't discuss too much about how and when. Perhaps it would be useful if a section or two were dedicated to explaining what clues to look for when approaching a counting problem that may require algebra.

Maybe someone could start a topic or forum about this, if there's enough interest.


in soal ro javab bedin.man ke khodam tv mibinam.shayad ham ba doostam beram biroon.gahi cinema raftam ham ba doostan hal mide :roll:

In the filming of the movie, you can watch the TV show. Maybe I have done a lot of  films like this?
rough translation

@aasta.sharma the guy is iranian not indian

Ya ik I plugged into google translate... XD
Also u spelled my name wrong :mad:
Lol

a dead post (and forum), that's what. i can't believe i'm the only poster on here for 5 years.


The character ⋯  is not properly displayed on IE 6 for some reason ...

I have IE6. What is that character (if I see it, I'll then know what is it)?

This was a note for myself ... no need to reply ;)


You need to have a complete, rigorous solution to get a 5 on a USAMTS problem, so a 5 on USAMTS would probably get a 7 on USAMO.

On the other hand, one could receive as much as 3/5 on a USAMTS problem without coming close to solving it. On the USAMO, this would probably result in 1-2 points if even that much.

[quote="le00327146"]I've know that the USAMO is probably much harder than the USAMTS in terms of how strictly it's graded, but does anyone know how much stricter? In other words, would a 5 on a USAMTS problem always get a 7 on a USAMO problem? Obviously a USAMTS problem wouldn't appear on the USAMO, but just compare the scoring.[/quote]

From my own experience (which is not much) I would say no, since USAMTS graders allow a certain degree of handwaving which the USAMO graders wouldn't. I do believe I received a couple of fives which probably would've reached a ~1/7 if the problem was on the USAMO.

Here's my weak attempt to match up the two grading scales (what constitutes a hole/minor is probably different for the two contests):

0 - No work, or completely trivial work
*0/5 = nothing presented that is credible.
*1/5 = evidence of understanding the problem
*2/5 = some good ideas but not much more;

1-2 - Progress on the problem, but not completely solved
*3/5 = larger holes in a promising solution;

3-4 - All steps are present, but may lack clarity. (These scores are very rare.)

5-6 - Complete solution with minor errors
*4/5 = minor shortcomings in the solution;

7 - Perfect solution
*5/5 = clear, concise, correct solution;

[quote="azjps"]
3-4 - All steps are present, but may lack clarity. (These scores are very rare.)
[/quote]

So does this mean you have a correct, valid solution, but it's messy or hard to understand?  Because that sounds like an awfully harsh point deduction for having the correct steps.

Basically, you [b]will not[/b] get a 3-4 on a one part problem. Even 2 and 5 are pretty rare, but much less so. On a two part problem, depending on the relative difficulties of the parts, it'll either be 3/4 or 2/5. On 3+ part problems, the middle points are much easier to obtain

I did USAMO last year, and I thought I had adequately solved two of them, but I ended up getting a one one both  :| . But I'm doing fine in USAMTS.

These are interesting answers. Having been through MOSP, I think I can say that I am familiar with the USAMO grading process, which favors very concise, to-the-point solutions. Many USAMTS graders, however, seem to like lots of elaboration. There have been instances where I have gotten 4 on USAMTS problems that I am quite certain I would have gotten 7 for on the USAMO. Of course, the same problems would not appear on the USAMO, but still... Basically, I think USAMO graders are much more likely to understand and like unusual arguments. On the other hand, they can be very hard on incomplete proofs, which doesn't seem to bother USAMTS graders as much (not sure about this part, though).

As I see it, USAMTS is aimed at those students who are just being introduced to proofs.  There are more problems that you can do by utter brute-force, the use of a computer is even allowed, and partial credit is given out gratuitously.  The only thing that really distinguishes a nice, elegant solution is the commending of solutions.

[quote="MellowMelon"]Did you do USAMTS last year? I'd be surprised if you haven't learned anything about writing solutions in the 5 months between the last USAMO and the beginning of this year's USAMTS.[/quote]

Well, no I didn't do USAMTS last year, but I don't feel like I learned much... :|

[quote="jb05"]These are interesting answers. Having been through MOSP, I think I can say that I am familiar with the USAMO grading process, which favors very concise, to-the-point solutions.[/quote]

So I guess this means the shorter the better, and points will be taken off for brute-force?

I wish you luck on brute-forcing a USAMO question (just kidding). Shorter indicates being more concise, but without losing any rigorosity, of course.

Last year the only (literally only) proof writing experience I had was this, and I somehow got a 12 on the USAMO ... still don't know how that happened ...

I didn't have any proof-writing experience for USAMO last year  :D but then, I only got a 4...

[quote="azjps"]but without losing any rigorosity, of course.
[/quote]

For future reference, there's no such word as "rigorosity."  There is "rigorism" or more simply just "rigor."

To go slightly off-topic... is "rigorize" a word? Probably not, but I use it a lot.

[url=http://dictionary.reference.com/browse/rigorize]Woah .. its not?[/url] [url=http://en.wiktionary.org/wiki/rigorize]Or maybe?[/url]

 :lol: I was too lazy to check if "rigorisity" was a word (it sounded like one), but I use 'rigorize' all the time. Apparently AoPS shows up in the top google results for 'rigorize'

Edit: this post is off-topic ..

According to Merriam-Webster's Collegiate Dictionary, rigorize is not a word.  Of course, I bet if you looked at one of those huge, 80 dollar dictionaries, you might find it.  Also, it doesn't really matter if it's a word if people can understand it.


function
inequalities
complex analysis
complex analysis unsolved

Suppose that f is analytic on complex plane except for a finite number of singularity and z*f(z) goes zero as z goes to infinity.
Prove that there exits M>0 such that |z^2*f(z)| < M for all z such that |z|>= M

Hint: $ f$ minus the sum of $ f$'s singular parts at these singularities is an entire function.  Consider $ z$ times this entire function, using the fact that each singular part is  is $ O(1/|z|)$ at $ \infty$.

Thanks for your help. I have tried the problem according to your suggestion and I did not get it,  could you please help me more

Now that I think of it, the problem is badly posed. A stronger result is easier to prove: Suppose $ f$ is analytic in $ \{|z|>R\}$ for some $ R > 0$. Show that if $ zf(z)\rightarrow  0$ at $ \infty$, then $ f(z) \equal{} O(1/|z|^2)$ as $ z\rightarrow \infty$. Hint: Consider the Laurent expansion of $ f$ in $ \{|z|>R\}$.


quadratics
number theory proposed
number theory

Show that the following equations have an infinite number of simultaneous solutions.

 $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 2d^4$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2d^2$ with $ \gcd(a,b,c,d) \equal{} 1$ and $ a,b,c,d \in \mathbb{N}$

There is a sign error in your second equation. It should be

$ a^4 \minus{} 2a^2(b^2 \plus{} c^2) \plus{} (b^2 \minus{} c^2)^2 \equal{} 0$

so your solution is incomplete

[quote="phymax"]thanks for the correction..
\[ a^{2} \equal{} \frac {2(b^{2} \plus{} c^{2})\pm\sqrt {2^{2}(b^{2} \plus{} c^{2})^{2} \minus{} 4(b^{2} \minus{} c^{2})^{2}}}{2}\]

\[ \equal{} b^{2} \plus{} c^{2}\pm \sqrt {4b^2c^2} \equal{} b^2 \plus{} c^2\pm 2bc \equal{} (b\pm c)^2\]
[/quote]

But this is not finished !

If $ a^2\equal{}(b\pm c)^2$, you get $ \frac{a^4\plus{}b^4\plus{}c^4}2\equal{}\left(\frac{a^2\plus{}b^2\plus{}c^2}2\right)^2$ but you are not sure that $ \frac{a^2\plus{}b^2\plus{}c^2}2$ is a perfect square ($ d^2$)

for example : $ b\equal{}2$, $ c\equal{}1$ and $ a\equal{}b\minus{}c\equal{}1$ imply $ \frac{a^2\plus{}b^2\plus{}c^2}2\equal{}3$ and $ \frac{a^4\plus{}b^4\plus{}c^4}2\equal{}9$ but there is no integer $ d$ such that $ a^2\plus{}b^2\plus{}c^2\equal{}2d^2$

IMHO

So close, but there are solutions.

E.G $ (a,b,c,d) \equal{} (3,5,8,7)$

You just have to show there are infinitely many.

So far you have shown that
$ a \pm b \pm c \equal{} 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2d^2 \implies a^4 \plus{} b^4 \plus{} c^4 \equal{} 2d^4$

[quote="PhilG"]Show that the following equations have an infinite number of simultaneous solutions.
 $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 2d^4$,$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2d^2$ with $ \text{gcd}(a,b,c,d) \equal{} 1$ and $ (a,b,c,d) \in \mathbb{N}^4$[/quote]
$ \displaystyle \left(\sum_{cyc}{a^2}\right)^2 \equal{} 2\left(\sum_{cyc}{a^4}\right)$ so $ \displaystyle \prod_{cyc}{(a \plus{} b \minus{} c)} \equal{} 0$. So it is enough to set $ c: \equal{} a \plus{} b$ and show that the equation $ a^2 \plus{} ab \plus{} b^2 \equal{} d^2$ iff $ \left(\frac {2a \plus{} b}{2d}\right)^2 \plus{} 3\left(\frac {b}{2d}\right)^2 \equal{} 1$ has infinitely many coprime solution in $ \mathbb{Z}^3$. But it is straghforward and well known that the elliptic curve $ X^2 \plus{} 3Y^2 \equal{} 1$ has infinitely many points in $ \mathbb{Q}^2$ (if more details are necessary look [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=150614]here[/url] for all class solutions).

Thank you for completing the problem  :)

The result can also be seen as an application of the multiplicative property of quadratic forms. I.e.

$ (x^2 \plus{} 3y^2)(z^2 \plus{} 3t^2) \equal{} (xz \minus{} 3yt)^2 \plus{} 3(xt \plus{} yz)^2$

We require a solution to $ u^2 \plus{} 3v^2 \equal{} 4d^2$.
To get one just take any odd integers $ x,y$ and take $ d$ from $ x^2 \plus{} 3y^2 \equal{} 2d$
Using the multiplicative identity we find $ u \equal{} x^2 \minus{} 3y^2, v \equal{} 2xy \implies  u^2 \plus{} 3v^2 \equal{} 4d^2$

In fact we can repeat the multiplication of solutions to get higher powers for the quadratic form which can be used to show that there are an infinite number of solutions for $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 2d^{2n}$ for any $ n \in \mathbb{N}$


GREECE ARE IN THE FINAL BEATING GERMANY :D  :D  :D
FRIDAY THE CUP AGAINST SPAIN

Tha to paroume xalaraa !  :lol:

Mas duskolepsan ligo sth fash twn omilwn, alla h ellhnikh omada den exei deiksei pros to paron ola ths ta opla   :wink:

:mad:  :mad:    spain-greece 1-0 instead of 1-5 gongrats to them but htey were awful  :wink:
next time we will win the future is us


I'm sorry if this was asked/answered before.

LaTeX => pdf.
In the folder, the pdf file is created (I can open it by double-clicking it), but
when I press F5 after pressing ctrl+F7, it says "Cannot excuse the command" and does not show the pdf file.

Thank you.

You have given no information about what program you are referring to, but on the assumption you are using Windows and TeXnicCenter, then see http://www.artofproblemsolving.com/Forum/topic-42360.html

Thank you very much.

However, now after I press F5,
adobe reader opens up, but it doesn't seem like it's opening the document.

Is this still wrong path?
C:\Program Files\Adobe\Acrobat 7.0\Reader\AcroRd32.exe

Once again, thank you very much.

The path looks ok, but check that all the other commands are set as at http://en.wikipedia.org/wiki/TeXnicCenter


Let $ k$ = a real number greater than $ 1$.

Given $ y \equal{} k^x \minus{} (k^2)^x$, what is the maximum value of the function?

hello, setting $ k^x\equal{}t$ so we get
$ f\left(\frac{\ln(t)}{\ln(k)}\right)\equal{}\minus{}\left(t\minus{}\frac{1}{2}\right)^2\plus{}\frac{1}{4}\le\frac{1}{4}$
the maximum will be obtained for $ x\equal{}\minus{}\frac{\ln(2)}{\ln(k)}$.
Sonnhard.


Τρια σημεια$ A,B,C$ μεταβαλονται σε τεταρτοκυκλιο κεντρου $ O(0,0)$ και ακτινας $ 2$.
Αν $ G$  ειναι το βαρυκενρο τoυ τριγωνου $ ABC$ (ακομη και εκφυλισμενου) να βρεθει το $ (OG)_{min}$
(Θα προτιμουσα λυση με μιγαδικους)

$ 9|z_G|^2\equal{}|z_1\plus{}z_2\plus{}z_3|^2\equal{}12\plus{}8(a\plus{}b\plus{}c)$ όπου $ a\equal{}z_1/z_2$ και κυκλικά
ομως $ 0<a\plus{}b\plus{}c,abc\equal{}1$ απο ΑΜ-ΓΜ $ a\plus{}b\plus{}c\ge 3$
αρα $ |z_G|_{min}\equal{}2$ πχ οταν a=b=c=1

[quote="r_boris"]$ 9|z_G|^2 \equal{} |z_1 \plus{} z_2 \plus{} z_3|^2 \equal{} 12 \plus{} 8(a \plus{} b \plus{} c)$ όπου $ a \equal{} z_1/z_2$ και κυκλικά
ομως $ 0 < a \plus{} b \plus{} c,abc \equal{} 1$ απο ΑΜ-ΓΜ $ a \plus{} b \plus{} c\ge 3$
αρα $ |z_G|_{min} \equal{} 2$ πχ οταν a=b=c=1[/quote]

Η ΑΜ-ΓΜ ισχύει για πραγματικούς, δεν ισχύει εδώ :wink:
Εξάλλου για οποιοδήποτε μη εκφυλισμένο τρίγωνο το $ OG<2$ αφού το G είναι στο εσωτερικό του τριγώνου!

Δημητρη εχει αριθμητικο ο Ροδόλφος το 8 βγαινει 4 και επισης οπως ανεφερες δεν ισχυει η ΑΜ-ΓΜ για μιγαδικους π.χ.$ z_1\equal{}1\plus{}i,z_2\equal{}1\minus{}i$ αι$ z_1/z_2\plus{}z_2/z_1>\equal{}0$ kai oxi toy 2
Ροδολφε το βιβλιο σου το ξεκοκαλιζω ακομη  Ομολογω οτι αρκετα απ τα θεματα που διαπραγματευεσαι ειναι εντυπωσιακα .......στο τελος του θα τα πουμε Συγχαρητηρια..

Ροδολφε δεν το εγραψα καλα θελει $ 8Re(z_1/z_2)\plus{}...$

Καλησπερα

Ειναι μια πολυ ωραια ασκηση, βαζω μια λυση αλλα δεν ειμαι 100% σιγουρος οποτε διορθωστε με αν εχω λαθος

Αν $ z_1, z_2, z_3, z_G$ οι εικονες των κορυφων του τριγωνου και του βαρυκεντρου στο μιγαδικο επιπεδο, τοτε:
$ |z_G| \equal{} \frac {1}{3}|z_1 \plus{} z_2 \plus{} z_3| \equal{} \frac {\sqrt {2}}{3}\sqrt {6 \plus{} \sum{ab} \plus{} \sum{\sqrt {4 \minus{} a^2}\sqrt {4 \minus{} b^2}}}$
οπου $ a \equal{} Re(z_1), b \equal{} Re(z_2), c \equal{} Re(z_3)$
Ομως ευκολα η συναρτηση $ f(a,b,c) \equal{} \sum{ab} \plus{} \sum{\sqrt {4 \minus{} a^2}\sqrt {4 \minus{} b^2}}$ ειναι κοιλη ως προς τις 3 μεταβλητες και επομενος παιρνει ελαχιστο στα ακρα της, οποτε κοιταζοντας στα ακρα ευκολα βγαζουμε οτι η ελαχιστη τιμη της ειναι ο αριθμος $ 4$
Αρα το ζητουμενο ελαχιστο ισουτε με $ {|z_G|}_{min} \equal{} \frac {\sqrt {2}}{3}\sqrt {6 \plus{} 4} \equal{} \frac {2\sqrt {5}}{3}$
με ισοτητα οταν τα 2 απο τα 3 σημεια ταυτιζονται με την τομη του τεταρτοκυκλιου με τον $ x'x$ και το 3ο ειναι η τομη του τεταρτοκυκλιου με τον αξονα $ y'y$ η αντιστροφα (εχουμε υποθεσει χβτγ οτι το τεταρτοκυκλιο οριζεται απο τους δυο θετικους ημιαξονες).

Διορθωστε με αν εχω κανει λαθος γιατι δεν ειμαι σιγουρος για οσα γραφω.

Συγνώμη παρασύρθηκα με τους μιγαδικούς

Λοιπον Nick δωσαμε παρομοιες λυσεις μονο που εγω βρηκα το ελαχιστο του $ 8Re(z_1/z_2)\plus{}...$ $ 8$ (δες τον τροπο του Ροδολφου) με την ισοτητα στο εκφυλισμενο τριγωνο......οπως προανεφερες
Η ασκηση κατασκευαστηκε στην προσπαθεια να βγει ασκηση για την Γ' κατευθυνση .Μονο που βγηκε για την Δ' κατευθυνση............
Αν καποιος εχει λυση με λυκειακες αυστηρα γνωσεις με ενδιαφερει.....

[quote="r_boris"]Συγνώμη παρασύρθηκα με τους μιγαδικούς[/quote]

ΠΡΟΚΕΙΜΕΝΟΥ ΝΑ ΕΞΙΛΕΩΘΩ για την χαζομάρα που βιάστηκα να πω δίνω στο συνημμένο μια ΚΑΘΑΡΑ ΓΕΩΜΕΤΡΙΚΗ ΛΥΣΗ


Let $ a,b,c$ be the length of the sides of an triangle. Prove that $ \sum_{cyc}\frac{a^{2}+b^{2}+c^{2}-4ab-4bc+6ca}{ab+bc-b^{2}}\leq 3$ and $ \sum_{cyc}\sqrt{\frac{a-b+c}{9a+b-c}}\leq 1$

We have that $ \sum_{cyc}\frac{a^{2}+b^{2}+c^{2}-4ab-4bc+6ac}{ab+bc-b^{2}}=-12+\sum_{cyc}\frac{(a+c)^{2}+4ac-3b^{2}}{ab+bc-b^{2}}$. Now we use $ 4ac\leq (a+c)^{2}$ so we are left to prove $ \sum_{cyc}\frac{2(a+c)^{2}-3b^{2}}{ab+bc-b^{2}}\leq 15$.This is equivalent after we use $ (a+c)^{2}-b^{2}=(a+b+c)(a+c-b)$ to $  2\left(\sum_{cyc}a \right)\left(\sum_{cyc}\frac{1}{a}\right) \leq 15+\sum_{cyc}\frac{a}{b+c-a}$.Now if $ a,b,c$ are sidelengths of a triangle we have $ a=y+z$,$ b=z+x$,$ c=x+y$ with $ x,y,z\geq 0$.The inequality becomes
$ 4\left(\sum_{cyc}x\right)\left(\sum_{cyc}\frac{1}{x+y}\right)\leq 15+\sum_{cyc}\frac{x+y}{2z}$.We substract $ 18$ from each side and after grouping we reach $ \sum_{cyc}\frac{(x-y)^{2}}{(x+z)(y+z)}\leq \sum_{cyc}\frac{(x-y)^{2}}{4xy}$.The last is easy to prove by SOS.

[quote="Vorelin Tancu"]Let $ a,b,c$ be the length of the sides of an triangle. Prove that $ \sum_{cyc}\frac{a^{2}+b^{2}+c^{2}-4ab-4bc+6ca}{ab+bc-b^{2}}\leq 3$[/quote]
Let $ a=y+z,$ $ b=x+z$ and $ c=x+y.$
Hence, $ \sum_{cyc}\frac{a^{2}+b^{2}+c^{2}-4ab-4bc+6ca}{ab+bc-b^{2}}\leq 3\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{y^{2}+z^{2}-4x^{2}+xy+xz}{(y+z)x}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{y^{2}+z^{2}-2x^{2}}{(y+z)x}+\sum_{cyc}\frac{y+z-2x}{y+z}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac{(z-x)(x+z)}{(y+z)x}-\frac{(x-y)(x+y)}{(y+z)x}\right)+\sum_{cyc}\left(\frac{z-x}{y+z}-\frac{x-y}{y+z}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac{(x-y)(x+y)}{(x+z)y}-\frac{(x-y)(x+y)}{(y+z)x}\right)+\sum_{cyc}\left(\frac{x-y}{x+z}-\frac{x-y}{y+z}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x-y)^{2}(xz+yz-xy)(x+y)z\geq0.$
Let $ x\geq y\geq z.$ Hence, $ S_{x}\geq0$ and $ S_{y}\geq0,$
where $ S_{x}=x(y+z)(xy+xz-yz)$ and $ S_{y}=y(x+z)(xy+yz-xz).$
Id est, it remains to prove that $ S_{y}+S_{z}\geq0.$
But $ S_{y}+S_{z}\geq0\Leftrightarrow x^{2}(y-z)^{2}+2y^{2}z^{2}+y^{2}xz+z^{2}xy\geq0.$

And the other inequality? :D

$ \sum_{cyc}\frac{1}{3a^{2}+\left( b-c\right)^{2}}\geq \frac{9}{\left( a+b+c\right)^{2}}$

[quote="Vorelin Tancu"]Let $ a,b,c$ be the length of the sides of an triangle. Prove that $ \sum_{cyc}\sqrt{\frac{a-b+c}{9a+b-c}}\leq 1$[/quote]
Let $ a=y+z,$ $ b=x+z$ and $ c=x+y.$
Then $ \sum_{cyc}\sqrt{\frac{a-b+c}{9a+b-c}}\leq 1\Leftrightarrow\sum_{cyc}\sqrt{\frac{x}{4x+5y}}\leq1,$ which known:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150022

[quote="Vorelin Tancu"]And the other inequality?
$ \sum_{cyc}\frac{1}{3a^{2}+\left( b-c\right)^{2}}\geq \frac{9}{\left( a+b+c\right)^{2}}$[/quote]
The same way:
it's equivalent to following inequality:
$ \sum_{cyc}\frac{1}{x^{2}+xy+y^{2}}\geq\frac{9}{(x+y+z)^{2}},$ which known:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6364


How many squares of any size appear in a 3x5 grid of unit squares?

How would you do these types of problems?

Is there any quick method?

Methodically count the number of each possible size of square:

[hide="like this"]
Number of $1\times1$ squares: 15
Number of $2\times2$ squares: 8
Number of $3\times3$ squares: 3

Total:26[/hide]

You are forgetting squares of noninteger lengths...

If we can make squares that have sides that are not parallel to the gridlines, we will have more squares.  If not, we are finished.  Usually the problem will specify, but if it does not, we assume sides do NOT have to be parallel to the gridlines.

i just counted them all (its pretty fast)
but you can consider (for positive sides only)
dimension 1 * dimension 2 + dimension1 * 1dimension 2-1 + dimension1-2 + dimension 2 -2. until one dimension = zero. then add them all up
so...
3x5=15 squares with side lengths 1
2x4=8 squares with side lengths 2
1x3=3 squares with side lengths 3

and so on...
if there are square root lengths, then thats 8 for root 2 and 4 for root 5.

[quote]How many squares of any size [b]appear[/b] in a 3x5 grid of unit squares?
[/quote]

I would argue that unless the sides of the squares were part of the grid, then the squares do not actually 'appear'.

But if we do count other squares, then there are an extra 12 making the total $\boxed{38}$


algorithm
geometry
perimeter
analytic geometry
conics
ellipse
search

Find a point P on equilateral triangle ABC such that (AP+BP+CP)/(AB+BC) is minimized, and prove your answer (preferebly, dont use the first fermat point algorithm)

(you mean on the perimeter of the triangle right?) AB and BC are constant, WLOG on AB, then AP+PB is constant, then minimize CP, pretty obvious the shortest distance is when C is P is the midpoint of AB, if you move it away from there, then by pythagorean theorem, it will be greater

if it is inside the triangle...not sure...a messy cooridinate solution could work...

P is inside the triangle
the solution (my solution at least) doesnt use coordinates but i suppose it could be done that way

[hide="hint"]use shearing (i think thats what the transformation is called[/hide]

[hide="Original thought"]Let $\triangle ABC$ be the equilateral triangle, and let $D, E, F$ be the midpoints of $AB, BC, CA$ respectively. If $P$ is the minimal point for $\triangle ABC$, then it also must be the minimal point for $\triangle DEF$, since they are similar and all the corresponding quotients for the two are the same. Therefore, $P$ must be inside $\triangle DEF$. Now take the midpoints of $DE, EF, FD$ and apply the same reasoning. Continue to infinity and you'll conclude that $P$ must be the centroid of $\triangle ABC$, because that's the only point that belongs to all the triangles, no matter how small they are.[/hide]

[hide="why that doesnt work"] Its the right answer for equilaterals but the reasoning is incorrect. How do I know it has to be inside all of the triangles? it just has to be in the same relative position for the triangles.[/hide]

[hide]it seems that it is the orthocenter, but i don't know how to prove it...i tried it for a right triangle as well, and i don't think shearing is the right word...that is for sheep  :lol:

another thing that might work would somehow have ellipses with the verticies as foci to show that it is a minimum sum of distances
[/hide]

one quck thing; its very difficult for the general triangle (search up first fermat point if you want to know). Use the fact that each pair of sides in the equilateral triangle are the same

[quote="Farenhajt"][quote="Scrambled"]it just has to be in the same relative position for the triangles.[/quote]

Right, and since it's given that $P$ should be inside the initial triangle, it must be inside all of them in order to be in the same relative position for all.[/quote]
uh, I dont exactly understand you
first of all, if K isnt in the triangle formed by the midpoints of the original triangle, that just means that its not the point for the centroid triangle; it doesnt imply anything about the original triangle


Prove that if $n$ is a positive integer number then:
a) $1.3.5...(2n-1)<n^{n}$
b) $n \left(\sqrt[n]{n+1}-1\right)<1+\frac{1}{2}+...+\frac{1}{n}<n\left(1-\frac{1}{\sqrt[n]{n+1}}+\frac{1}{n+1}\right)$

[hide="An one-line proof to a) without induction"]By $\mathrm{AM-GM}$:
$1\cdot 3\cdot\ldots\cdot (2n-1)< \left(\frac{1+3+\ldots+(2n-1)}{n}\right)^{n}= \left(\frac{n^{2}}{n}\right)^{n}= n^{n}.$[/hide]

[hide="simplest proof for a"]The LHS is equal to n^2. The only time that it is equal to the RHS is when n=2 or 1. Every other time it is less than the RHS.[/hide]

Try one and two for n in the first equation.

[quote="nutz_for2.718281828"]The LHS is equal to n^2. [/quote]
Correct me if I'm wrong, but I think that usually $1\cdot 3\cdot\ldots\cdot (2n-1)\neq n^{2}$.  :maybe:

Oh wait, I mixed up product and sum. :oops:

The solution of Rzeszut is right. Whatever sign "=" don't occur beacause all of numbers in sequence ins't same.
Who can solve part b of problem?

Part a can solved as following:
with $n=2$ easy to see thep problem is true.
Assume problem is true with $n=k\leq 1$. We have:
$1.3...(2k-1)(2k+1)<k^{k}.(2k+1)$
Remainder is that we have to prove $k^{k}(2k+1)<(k+1)^{k+1}$ but that is deduced directly by Newton binomial

I have a proof by using Caychy's inequality.
We have:
$\frac{1+\frac{1}{2}+\frac{2}{3}+...+\frac{n-1}{n}}{n}>\sqrt[n]{\frac{1}{n}}$
$\Rightarrow 1+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{n}\right)> \frac{n}{\sqrt[n]{n}}$
$\Rightarrow n-\frac{n}{\sqrt[n]{n}}>\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$, or:
$1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}<n\left(1-\frac{1}{\sqrt[n]{n}}+1\right)$
Clearly $n\left(1-\frac{1}{\sqrt[n]{n}}+1\right)<n\left(1-\frac{1}{\sqrt[n]{n+1}}+\frac{n}{n+1}\right)$
The right hand have proved.
Otherwhile we have:
$\frac{2+\frac{3}{2}+\frac{4}{3}+...+\frac{n+1}{n}}{n}>\sqrt[n]{n+1}$
$\Rightarrow 2+\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+...+\left(1+\frac{1}{n}\right)>\sqrt[n]{n+1}$
$1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}>n\left(\sqrt[n]{n+1}-1\right)$
The probelm has been almost proved almost

There are some problem is the same type:
[b]Problem 1[/b]:
$\frac{1}{1^{1}}+\frac{1}{2^{2}}+...+\frac{1}{n^{2}}>\sqrt[n+1]{\frac{n+1}{2}}$


Let $ x,y,z$ be non-negative real numbers such that $ x\plus{}y\plus{}z\equal{}3$. Prove that $ \sqrt[10]{x^{3}}\plus{}\sqrt[10]{y^{3}}\plus{}\sqrt[10]{z^{3}}\geq xy\plus{}yz\plus{}zx$.


The side of a square is increased by $ 50\%$. By what percent is the area of the square increased?

try to giver urself an example so let's say that the side is 4. when the side is 4, the area is 16. then u increase 4 by 50% making it 6. that area is 36, so the percent increase from 16 to 36 is 225%.

First thing here, the answer is incorrect. Secondly, when you formally prove something, you have to prove that it works for any side length $ x$, not just by giving a specific example.

Well, you can't just assume a definite side length of the square. In fact, this rule works out for any side length $ x$ of the square. Note that when you increase the side length by $ 50$ percent, you are multiplying the side length by 1.5, hence multiplying the original area by $ (1.5)^2 \equal{} 2.25$. So if you let $ x$ be the side length of the square, $ x^2$ the area, then $ 2.25x^2 \minus{} x^2 \equal{} 1.25x^2$, which means that the area of the square is increased by $ \boxed{125}$ percent.


geometry
3D geometry
analytic geometry
Euler
IMO Shortlist
combinatorics

Unit cubes are made into beads by drilling a hole through them along a diagonal. The beads are put on a string in such a way that they can move freely in space under the restriction that the vertices of two neighboring cubes are touching. Let $ A$ be the beginning vertex and $ B$ be the end vertex. Let there be $ p \times q \times r$ cubes on the string $ (p, q, r \geq 1).$

[i](a)[/i] Determine for which values of $ p, q,$ and $ r$ it is possible to build a block with dimensions $ p, q,$ and $ r.$ Give reasons for your answers.
[i](b)[/i] The same question as (a) with the extra condition that $ A \equal{} B.$

I am having trouble understanding the solution from IMO Compendium. Aren't we trying to look for hamiltonian cycles rather than eulerian ones?
Thanks

Solution by IMO Compendium:
Let us set a coordinate system denoting the vertices of the block. The
vertices of the unit cubes of the block can be described as {(x, y, z) | 0 ≤
x ≤ p, 0 ≤ y ≤ q, 0 ≤ z ≤ r}, and we restrict our attention to only
these points. Suppose the point A is fixed at (a, b, c). Then for every other
necklace point (x, y, z) numbers x − a, y − b, and z − c must be of equal
parity. Conversely, every point (x, y, z) such that x − a, y − b, and z − c are
of the same parity has to be a necklace point. Consider the graph G whose
vertices are all such points and edges are all diagonals of the unit cubes
through these points. In part (a) we are looking for an open or closed
Euler path, while in part (b) we are looking for a closed Euler path.
Necklace points in the interior of the (p, q, r) box have degree 8, points on
the surface have degree 4, points on the edge have degree 2, and points
on the corner have degree 1. A closed Euler path can be formed if and
only if all vertices are of an even degree, while an open Euler path can be
formed if and only if exactly two vertices have an odd degree. Hence the
problem in part (a) amounts to being able to choose a point A such that
0 or 2 corner vertices are necklace vertices, whereas in part (b) no corner
points can be necklace vertices. We distinguish two cases.
 (i) At least two of p, q, r, say p, q, are even. We can choose a = 1, b = c =
     0. In this case none of the corners is a necklace point. Hence a closed
     Euler path exists.
(ii) At most one of p, q, r is even. However one chooses A, exactly two
     necklace points are at the corners. Hence, an open Euler path exists,
     but it is impossible to form a closed path.
Hence, in part (a), a box can be made of all (p, q, r) and in part (b) only
those (p, q, r) where at least two of the numbers are even.

Solution from [i]Twitch Solves ISL[/i]: For (a), the answer is always yes; whereas for (b), the answer is only if at least two of $p$, $q$, $r$ are even.

Let the box $\mathcal B$ be located in space spanning $(0,0,0)$ to $(p,q,r)$. Then the string can move from $(x,y,z)$ to $(x \pm 1, y \pm 1, z \pm 1)$. So, the string is going to visit only points of two parity classes in ${\mathbb F}_2 \times {\mathbb F}_2 \times {\mathbb F}_2$, say $(\varepsilon_x \bmod 2, \varepsilon_y \bmod 2, \varepsilon_z \bmod 2)$ and $(1+\varepsilon_x \bmod 2, 1+\varepsilon_y \bmod 2, 1+\varepsilon_z \bmod 2)$. However, every unit cube has a string through it, so we find that the string visits every point of this form in $\mathcal B$. Let us denote the set of points by $V$ (which depends on the choice of $(\varepsilon_x, \varepsilon_y, \varepsilon_z)$).
Hence the question is asking when we can have a Eulerian path or circuit on the resulting graph $G$ (whose vertex set is $V$ and whose edges are the string).
However, the degrees of vertices of $G$ are always even [i]except[/i] for the corners of $\mathcal B$, i.e.\ the eight points \[ 	\Big\{ (0,0,0), \; (p,0,0), \; (0,q,0), \; (0,0,r), \; 	(p,q,0), \; (p,0,r), \; (0,q,r), \; (p,q,r) \Big\}. \] [list] 	[*]We can always ensure there are at most 	two corners in $V$ 	by choosing $\varepsilon_x$, $\varepsilon_y$, $\varepsilon_z$ randomly; 	the expected number of corners is $\frac14 \cdot 8 = 2$. 	[*]We can ensure there are zero corners 	if at least two of $(p,q,r)$ are even; 	otherwise we cannot. [/list]
Hence by the classical theorem on Eulerian paths and circuits, (a) is possible always, (b) is possible when at least two of $(p,q,r)$ are even.


I finally got a letter from the awesome people who make Olymon. Aparently, i made [b]many[/b] mistakes. Anyways, it had the questions for september, and all the solution up till september for 2006 (I think). I know we can't discuss the questions, but could i get a hint for the following question, thanks. :)

454. Let ABC be a non-isosceles triangle with circumcentre O, incentre I and orthocentre H. Prove that the angle OIH exceeds 90 degrees.

[quote="rem"]I am not sure if hints can be given out, please ask Mr. Barbeau if this is allowed. Just in case, this has to be done. :!:[/quote]
oh, sorry wrong question. And wrong wording :blush: . Could you explain the following question to me, thanks.

Let a and b be positive integers and let $u=a+b$ and $v=lcm(a,b)$. Prove that
$gcd(u,v)=gcd(a,b)$

This is from a problem set that is not due until October 31, so it should not be discussed.  The topic is locked.

Problem is Equivalent to Saying that $I$ lies inside the circle with diameter $HO$ .

Stronger result is true that is $I$ lies inside the circle with diameter $HG$.This is the best possible bound ie $I$ can be anywhere inside this Circle.

Just got the corrected solutions back the day before yesterday.

How did you guys do?


calculus
integration
function
calculus computations

The functions are positive so you just switch the integral and the summation sign? The sum even converges uniformly...


Don't know, but is it:

The first digit could be $9,8,7,6,5$, second $8,7,6,5,4$, third $7,6,5,4,3$, fourth $6,5,4,3,2$, fifth $5,4,3,2,1$, and sixth $4,3,2,1,0$, so would it just be $5^6$?

EDIT: Misunderstood question.

Well, since each two digit should have a number and a number smaller, the first out of the two digits can be a number between $1-9$ and then the second has to be a number between $0-8$ that goes for the other 4 digits too, so would it be $\boxed{9^6}$?

The hard thing is that you have to be careful about the numbers you change. Right now all I know is that the smallest is 543210 and the largest 987654.

I think my method makes sense though because for every place value, the digit is one less than the one to the left and the one to the left can not be $0$, so it should be $9^3$.

$9^6$ doesn't seem right because $9^6/(9 \cdot 10^5) = .59049$, and the answer is definitely $<.5$

[quote="236factorial"]The hard thing is that you have to be careful about the numbers you change. Right now all I know is that the smallest is 943210 and the largest 987654.[/quote]
??? Isn't the smallest 543210?
Hmm I got one answer as [hide]$\binom{9}{5}=126$ :?:[/hide]

[quote="10000th User"][quote="236factorial"]The hard thing is that you have to be careful about the numbers you change. Right now all I know is that the smallest is 943210 and the largest 987654.[/quote]
??? Isn't the smallest 543210?
[/quote]

Duhhhh  :blush:

[quote="jli"]how many 6 digit numbers are there such that out of each 2 digit, the one to the right is smaller?[/quote]

Once you pick the digits, they must be arranged decreasingly, so $\binom{10}{6}$ is right.

Say it is an 8 digit number...

Call a skip when it doesnt include a number

For example, if the skip is on 7, there wont be a 7 in the number

CASE 1: if the first digit is 8

There is only one way to arrang it.

CASE 2: if the first digit is 9

There has to be exactly one skip.

The number of ways to place the skip is ${8\choose 1}=8$

So the total is 9...

is this true?

If you're still talking about digits in descending order, that's right.

[quote="frt"]I got a similar answer to 10000th User's.

There are $\binom{10}{6}=210$ ways to pick 6 distinct numbers (if the order does not matter.)
For each possible set of 6 numbers, there's only one way to line up 6 numbers to satisfy the given condition.

Example: If the set was $\{1,2,3,5,7,9\}$, then there's only one possibility:$975321$

So I think the answer is $\boxed{210}$...[/quote]

oh...i didnt see your post

yes. this is correst


In a triangle, two altitudes are $ 1.2$ and $ 1.4$ times longer than the third, respectively. What's the ratio between the shortest altitude and the inradius?

[hide="Answer"]In a triangle, the product of an altitude and the side it touches is constant.  Thus, if the shortest altitude has length $ a$ and the side it touches has length $ x$, then the other sides have length $ x/1.2=5x/6$ and $ x/1.4=5x/7$.

Now, equate two different area formulas.  Note that $ r$ is the inradius and $ p$ is the perimeter:
\begin{eqnarray*}
A=\frac{xa}{2} &=& \frac{rp}{2}\\
\displaystyle xa &=& r(x+\frac{5x}{6} +\frac{5x}{7})\\
a &=& r(\frac{107}{42})
\end{eqnarray*}

So $ a: r = 107: 42$[/hide]

[hide="General formula"]We can derive the formula connecting the inradius and the altitudes. If the area of the triangle is $ P$ and its semiperimeter $ s$, then

$ 2s\equal{}a\plus{}b\plus{}c$

$ 2\cdot{P\over r}\equal{}{2P\over h_a}\plus{}{2P\over h_b}\plus{}{2P\over h_c}$

$ \boxed{{1\over r}\equal{}{1\over h_a}\plus{}{1\over h_b}\plus{}{1\over h_c}}$

Now $ {1\over r}\equal{}{1\over h}\plus{}{1\over 1.2h}\plus{}{1\over 1.4h}\iff {h\over r}\equal{}1\plus{}{5\over 6}\plus{}{5\over 7}\equal{}{107\over 42}$[/hide]


I will be attending a Scholar School in Canberra early next year for the Chemistry Olympiad (I hope to get selected  :D ). They will be using:

- Silberberg, Martin; Chemistry:  The Molecular Nature of Matter and Change, 3rd edition, ISBN0-07-239681-4
- Morrison and Boyd; Organic Chemistry, 6th edition, ISBN 0-13-643669-2

with us during the Scholar School but I was wondering if there are other books that I may use as well to strengthen and add to my current knowledge so I might a better chance of getting into the IChO Team for 2008 (or 2009).

I find "Chemistry: Molecules, Matter, and Change" from Peter Atkins a very good book. Take also a look at some other Organic Chemistry books, like Solomons or Carey. If spectroscopy is also part of the program, then I would strongly advise "Introduction to Spectroscopy" from Pavia, Lampman and Kriz (Brooks /Cole) - it is the book I use in my classes. Also, after studying the general chemistry books I would advise you to take a look at an Inorganic Chemistry book (like the one from Atkins).
You can also use this forum to improve your skills.

In any way, I wish you good look!


geometry
3D geometry
sphere
topology
real analysis
real analysis unsolved

H is infinite-dim. Why unit sphere S is weakly dense in unit ball B?
( I was given a hint that in fact, every x in B is weak sequential limit of points in S) Can someone explain? Thanks.

First, any infinite orthonormal sequence tends weakly to zero- can you show this?

Now, given $x$, let $V$ be the orthogonal complement of $x$. The intersection of $x+V$ with $S$ is a sphere of radius $\sqrt{1-\|x\|^{2}}$ in $V$ centered at $x$; find an infinite orthonormal set in $V$, scale appropriately, and add $x$.

[quote="alekk"]any open set is unbounded, so every open set that contains a point of the unit ball cuts the unit sphere.[/quote]

I am not sure alekk's method works: what if this open set is not "connected", i.e.: part of it is in unit ball, part of it outside unit ball. Then it can be unbounded and do not intersect unit sphere.

well, I should have said that each open set contains a line ($\{x+\lambda y : \lambda \in \mathbb{R}\}$)


algebra
polynomial
function
induction
superior algebra
superior algebra unsolved

Let $ A$ be a finite commutative ring with the property that the function $ f: A\minus{}>A,f(x)\equal{}x^2$ is injective.Prove that $ \forall n \in N^{*}$ and $ \forall a \in A$ the polynomial $ P(X)\equal{}X^{2n}\plus{}a$ is reducible in $ A$.

Since $ A$ is finite, $ f$ must also be surjective, so every element of $ A$ is a square.  By induction, we can write $ \minus{} a \equal{} b^{2n}$ for some $ b\in A$.  Then $ P(x) \equal{} X^{2n} \minus{} b^{2n} \equal{} (X^n \minus{} b^n)(X^n \plus{} b^n)$.

The induction doesn't work like that. For example take $ A\equal{}\mathbb F_4$ (field with $ 4$ elements), $ a \neq 0,1$ and $ n\equal{}3$. Then for any $ b$ it is $ b^6 \equal{}0,1$.

[quote="ZetaX"]The induction doesn't work like that. For example take $ A \equal{} \mathbb F_4$ (field with $ 4$ elements), $ a \neq 0,1$ and $ n \equal{} 3$. Then for any $ b$ it is $ b^6 \equal{} 0,1$.[/quote]

You're right, I got carried away.  We just need $ \minus{}a\equal{}b^2$, then $ X^{2n}\minus{}b^2\equal{}(X^n\minus{}b)(X^n\plus{}b)$.


[quote="bpms"][quote="Ignite168"]Find all values of x such that: $2log_{9}(15-x)-log_{3}(x-4)=3log_{27}(x+5)$.[/quote]
[hide]
$\log_{9}(15-x)^{2}-\log_{3}(x-4)=\log_{27}(x+5)^{3}$
$\log_{3}\frac{(15-x)^{2}}{x-4}=\log_{3}(x+5)$
$x^{2}+x-20=x^{2}-30x+225$
$31x=245$
$x=\frac{245}{31}$[/hide][/quote]
You messed up on your second line.  $log_{9}((15-x)^{2})-log_{3}(x-4)$ is not equal to $log_{3}(\frac{(15-x)^{2}}{x-4})$ because the two logarithms have different bases.  That must be changed first.

[quote="bpms"]
Oopsies.
[hide]
$\log_{3}\frac{|15-x|}{x-4}=\log_{3}x+5$
$|15-x|=x^{2}+x-20$
Assume the LHS is positive.
$15-x=x^{2}+x-20$
x=-7, 5
If the LHS is negative
$x-15=x^{2}+x-20$
$x^{2}=5$
$x=\sqrt{5}$, 5.[/hide][/quote] And so 3 roots are extraneous roots and the answer is $x=5$.


Who will win the NL West?  I think Padres.  I love the Dodgers, but don't think they have much of a chance :( I will only post for NL East, AL East, and AL West (and the wild cards)  because the Centrals both have dominant teams.

isn't one baseball topic enough? there are about six now. that's a bit much. no wonder why everyone now how a 1000 posts.  :D
anywho, go d-backs.

furious, i think you are right.  I will not make new posts for the AL teams :) maybe next week when this topic has all the votes it will get


Do they just choose 20 finalists from the 900 people (that qualified from local)  that take the national exam?
In addition, how are the 3 sections scored? Thanks in advance!

Scott said this in the other USNCO topic on this forum:

[quote]For everyone interested in how they grade the national exam, lets [i][sic][/i] run through the basics.

1) If you're not making above a 55/60 on the Multiple Choice, you may have some problems qualifying
2) If you're answering none of the higher-level (d through g) questions in the Free Response, you may have some problems qualifying
3) If you're absolutely terrible at labs, you'll have NO PROBLEM getting in. [/quote]

To answer the first part of your question, yes.  I don't know how exactly the three sections are scored, but I will say (in agreement with Scott) that the lab receives very little weight.  And by very little weight, I mean [b]very little weight[/b].  I [i]completely[/i] blew the lab, but still somehow qualified for camp.  The logic is that most high school chem students have very little lab expertise, I suppose.  As far as the relative weights of MC/FR, I don't know.  Try to do well on both, I guess.

To practice, take the practice tests they offer.  Many of the questions (especially multiple choice) repeat themselves by necessity; there are certain skills they have to test, and there are only so many ways to ask for the $E$ of a cell...

So that means that there is no semi-finalist thing for the Chemistry Olympiad like the USAMO is for the math olympiad, right? Anyways, thanks for the advice. I am currently studying with Mortimer's book, and I hope to be ready by test time :D.

Think of it this way: The local qualifier is the AMC, and the USNCO national exam is the USAMO.

And of course, both olympiads have the finals (the team selection test).

Hm.. the practice national exams seem tough, but do-able. Are the cutoffs for camp usually very high? I took a practice on the 2005 National Exam, and I got a 52/60 on the multipe choice, and explained the Free-response questions fairly well (messed up on part of the Organic chem problems) . Would that give me a good chance to be invited to the camp?

Wow, I read in the introduction topic that you made the USNCO camp in the same year you took AP Chem, inteluser. That's amazing!


What do you mean by the five distinct roots of 1?  If you mean the fifth roots of unity, then that would require knowledge of $ \sin36^\circ$, etc in terms of radicals.

[quote="boo"]1. Express the five distinct roots of 1 in terms of radicals.
2. Prove that the three perpendiculars dropped to the three sides of an equilateral triangle from any point inside the triangle have a constant sum.
3. For all $ n > 1$, show that $ (n \minus{} 1)^2$ is a factor of $ n^{n \minus{} 1} \minus{} 1$[/quote]
[hide="2"] Okay, so call the equilateral triangle to be ABC, the point on the interior to be P, and the perpindiculars to AB, BC, and AC to be PX, PY, and PZ respectively.  Also, say that AB=BC=AC=x.
\[ [ABC] \equal{} [PAB] \plus{} [PBC] \plus{} [PAC]
\]

\[ [ABC] \equal{} \frac {PX(AB)}{2} \plus{} \frac {(PY)(BC)}{2} \plus{} \frac {(PZ)(AC)}{2}
\]

\[ [ABC] \equal{} \frac {PX(x) \plus{} PY(x) \plus{} (PZ)(x)}{2}
\]

\[ [ABC] \equal{} (\frac {x}{2})(PX \plus{} PY \plus{} PZ)
\]

\[ PX \plus{} PY \plus{} PZ \equal{} \frac {2[ABC]}{x}
\]
Since x and [ABC] are constant, we have that PX+PY+PZ is constant.
[/hide]
[hide="3"] Start by dividing $ n^{n \minus{} 1} \minus{} 1$ by $ n \minus{} 1$.  It is quite easy to see that this quotient is $ n^{n \minus{} 2} \plus{} n^{n \minus{} 3}... \plus{} 1$.  We have to show that $ n \minus{} 1$ is a factor of this polynomial.  Look at the polynomial $ \bmod (n \minus{} 1)$.  So, we obtain that $ n^{n \minus{} 2} \plus{} n^{n \minus{} 3}... \plus{} 1\equiv 1 \plus{} 1 \plus{} 1...1\equiv n \minus{} 1\equiv 0\bmod (n \minus{} 1)$.  Thus, n-1 is a factor of the polynomial and it is proven.
[/hide]


What is,

$\sin \frac{\pi}{7}\cdot \sin \frac{2\pi }{7}\cdot ... \cdot \sin \frac{6 \pi }{7}=?$

$\sin \frac{\pi}{7}\cdot \sin \frac{2\pi }{7}\cdot ... \cdot \sin \frac{6 \pi }{7}=\frac{7}{2^{6}}.$
Posted before.

I tried searching it before for a long time, and still cant find it. Can you please post a link.


as apart of their preschool science project , mrs. joy pribble's class is holding snail races . the first snail, zippy , is known to move at a rate of 5 inches per hour .the second snail,lightning, is known to move at  4.5 inches per hour . if the snails race along a straight path, and if zippy finishes the race 0.25 hour before lightning , determine the time it takes lightning to run the race

a) 3 hours b) 2.5 hours c) 2.75 hours  d) 3.5 hours e) none of these

Let the time lighting takes equal $ t$. Then $ t\equal{}\frac9{10}t\plus{}.25$ \[ t\equal{}2.5\]The answer is $ B$.


[hide]
$\underbrace{11\ldots 11_9}_{n \text { ones}} = 1+9+9^2+\cdots+9^{n-1}$

$=\frac{9^n-1}{9-1}$

$=\frac{9^n-1}{8}$

We wish to show that $\frac{9^n-1}{8}$ can be written in the form $\frac{1}{2}(k)(k+1)$ for some positive integer $k$.

$\Longleftrightarrow 9^n-1=4k(k+1)$

$9^n-1=4k^2+4k$

$\iff 4k^2+4k-(9^n-1)=0$

$\Rightarrow k=\frac{-4\pm\sqrt{16+4(4)(9^n-1)}}{8}$

$=-\frac{1}{2}\pm \frac{1}{2}\sqrt{9^n}$

$=\frac{1}{2}(-1\pm 3^n)$

Since we want $k>0$, we will take $k=\frac{1}{2}(-1+3^n)$.

Because $3^n\equiv 1\pmod {2}$, $k$ must be an integer, as claimed. $\Box$
[/hide]

[quote="MCrawford"]Prove that every positive integer that can be written using only 1 as each digit in base 9 is a triangular number:
\begin{eqnarray*} 1_9 &=& 1 \\ 11_9 &=& 1 + 2 + 3 + 4 \\ 111_9 &=& 1 + 2 + \cdots + 13 \\ 1111_9 &=& 1 + 2 + \cdots + 40 \\ & \vdots & \end{eqnarray*}[/quote][hide]Theorem: \[ \underbrace{11\ldots11_9}_{n\text{ ones}}=1+2+\ldots+\sum_{i=0}^{n-1}3^i. \]The proof will be by induction upon the number of digits. Consider $n=1$ as the base case. If $n=1$, then, as demonstrated, \[ 1_9=1=\sum_{i=0}^{1-1}3^i. \]
Now, for the inductive hypothesis, suppose the theorem is true for some integer $k$. That is, \[ \underbrace{11\ldots11_9}_{k\text{ ones}}=1+2+\ldots+\sum_{i=0}^{k-1}3^i. \] Then \[ \underbrace{11\ldots11_9}_{(k+1)\text{ ones}}=\underbrace{11\ldots11_9}_{k\text{ ones}}+9^k. \] Notice that \begin{eqnarray*}9^k&=&\frac{(3^k)(3^k+1)}{2}+\frac{(3^k)(3^k-1)}{2}\\ &=&\sum_{j=1}^{3^k}j+(3^k)\left(\sum_{i=0}^{k-1}3^i\right)\\ &=&\sum_{j=1}^{3^k}\left(\sum_{i=0}^{k-1}3^i+j\right)\end{eqnarray*} Thus, \[ \underbrace{11\ldots11_9}_{(k+1)\text{ ones}}=\underbrace{11\ldots11_9}_{k\text{ ones}}+\sum_{j=1}^{3^k}\left(\sum_{i=0}^{k-1}3^i+j\right). \] But from the inductive hypothesis, \[ \underbrace{11\ldots11_9}_{k\text{ ones}}=1+2+\ldots+\sum_{i=0}^{k-1}3^i. \] Therefore, \[ \underbrace{11\ldots11_9}_{(k+1)\text{ ones}}=1+2+\ldots+\sum_{i=0}^{k-1}3^i+\sum_{j=1}^{3^k}\left(\sum_{i=0}^{k-1}3^i+j\right). \] Note that the final double sum is equivalent to \[ \left(\sum_{i=0}^{k-1}3^i\right)+1+\left(\sum_{i=0}^{k-1}3^i\right)+2+\ldots+\left(\sum_{i=0}^{k-1}3^i\right)+3^k. \] Indeed, \[ \underbrace{11\ldots11_9}_{(k+1)\text{ ones}}=1+2+\ldots+\sum_{i=0}^{k}3^i, \] and so the proof is complete.[/hide]

[hide]I just noticed something, any number that starts with a 1 and has the rest of its digits as zeros is a triangular number.  In base 9, for every 10 numbers in base ten, there is +1 in base 9, therefore all number that consist of 1 only end in the same number of 1's as the quotient of the original number divided by ten. So, all numbers that consist only of 1 in base 9 will be triangular. I'm sorry that the wording is confusing.[/hide]

[quote="maniacman384"]
I just noticed something, any number that starts with a 1 and has the rest of its digits as zeros is a triangular number.[/quote]

Last time I checked, 1000 was a triangle number, not in base 10 nor base 9...


function
floor function
algebra proposed
algebra

Suppose $f : \mathbb{N} \longrightarrow \mathbb{N}$ is a function that satisfies $f(1) = 1$ and
$f(n + 1) =\{\begin{array}{cc} f(n)+2&\mbox{if}\ n=f(f(n)-n+1),\\f(n)+1& \mbox{Otherwise}\end {array}$

$(a)$ Prove that $f(f(n)-n+1)$ is either $n$ or $n+1$.
$(b)$ Determine$f$.

Let $\phi=\frac{1+\sqrt{5}}{2}$, so that we have the identities $\phi+1=\phi^2$ and $\phi-1=\frac{1}{\phi}$.  Then $f(n)=\lfloor\phi n\rfloor$.

We prove this by strong induction.  The base case is given.  So suppose $f(n)=\lfloor\phi n\rfloor$ for all $n\leq k$, so that $f(f(k)-k+1)=f(\lfloor\phi k\rfloor -k+1)=\lfloor\phi(\lfloor\phi k\rfloor-k+1)\rfloor$.  It is easy to prove that this value is either $k$ or $k+1$.

To establish the inductive hypothesis, if $f(f(k)-k+1)=k$ (so that $f(k+1)=\lfloor\phi k\rfloor+2$) then
$k<\phi(\lfloor\phi k\rfloor-k+1)<k+1$
$k+\phi k-\phi<\phi\lfloor\phi k\rfloor<k+1+\phi k-\phi$
$\phi^2k -\phi<\phi\lfloor\phi k\rfloor<\phi^2 k-\frac{1}{\phi}$
Dividing by $\phi$ and adding 2,
$\phi k+1<\lfloor\phi k\rfloor+2<\phi k+2-\frac{1}{\phi^2}$
$\phi(k+1)-(\phi-1)<f(k+1)<\phi (k+1)-(\phi-1)+(1-\frac{1}{\phi^2})$
$\phi(k+1)-\frac{1}{\phi}<f(k+1)<\phi (k+1)$
$f(k+1)=\lfloor\phi(k+1)\rfloor$

Similarly if $f(f(k)-k+1)=k+1$, so that $f(k+1)=\lfloor\phi k\rfloor+1$, then we can deduce that $\phi(k+1)-1<f(k+1)<\phi (k+1)-\frac{1}{\phi^2}$, so again $f(k+1)=\lfloor\phi(k+1)\rfloor$.


Draw a square, side 15 units, labeled ABCD. Then label the midpoints of the sides as EFGH, so that E is between AB, F is between BC, G is between CD, and H is between AD. Draw in segments AF, BG, CH, and DE. What is the area of the square formed in the middle?

Be sure to hide your solutions!

Sure! No one else seems to have a clue.  As stated above, hide your solution.

$ 45$ $ units^2$
[hide="Solution"]I used simple manipulation to solve this
Label each piece with number and move them around
I got 5 congruent squares
Area of whole = 225 square units
225/5=45[/hide]

[quote="Eragon"]Really? :o  :wallbash_red:  :stretcher:
No alternate solutions?
this is a really basic approach btw
i was lazy
it shocks me that after this long, no one has bothered to post anything[/quote]

Well this isn't actually the most active state and I don't even know why we have a forum. I bet Lousiana and Hawaii could do better. And the solution has already been posted. If you want an alternate, then you post it.

[hide="Alternate solutoin im too lazy to type out fully"]
Similar triangle bash ftw.
[/hide]

Or

[hide="another one..."]
COORDINATES
[/hide]


calculus
integration
real analysis
real analysis unsolved

I wanted to finnish posting integrals in mathlinks.ro because I had no interesting new integrals (except in my book) but before 5-6 minutes i found these in Bulgarian math forum. There are lots of interesting topics in math - math10.com is the address of the site.

1. $ \int \frac{5x^{2}\plus{}4x\plus{}12}{\sqrt{(2x^{2}\plus{}7x\plus{}3)^{3}}} \,dx$
2. $ \int \frac{1\plus{}x^{2}}{x^{4}\plus{}3x^{3}\plus{}3x^{2}\minus{}3x\plus{}1} \,dx$
3. $ \int \frac{x^{3}\plus{}x}{3x^{6}\minus{}6x^{4}\plus{}30x^{3} \plus{} 6x^{2}\minus{}3} \,dx$

I believe you will like them.

There is a hint for the first one.
[hide]Split the integral to several integrals.[/hide]

For complete solutions - see:
http://www.math10.com/forumbg/viewtopic.php?p=16935#16935