trigonometry
limit
calculus
calculus computations

the sum

$ \sum^\infty_{k\equal{}1}\sec^2{k}$ where $ k$ is in radian

converges or diverges?

Can someone enlighten me which test should I use here ?

The Divergence Test: if $ a_k\not\to 0$, then $ \sum a_k$ is divergent

$ \sec x$ is the secant of $ x$, defined as $ \frac1{\cos x}$.

So which means that as long as $ \lim_{n\rightarrow\infty} a_n\neq 0$ , not nessecary that we have to find the exact value for the limit $ L$ , then we can say that the sum of the sequence is divergent? Am I correct ?

I'm just confuse what if its limit does not exist such as the question I give above.Thanks

There is a difference between what I wrote ($ a_k\not\to 0$) and what you wrote ($ \lim_{n\to\infty} a_n\ne 0$). The first statement means that the sequence does not converge to zero. The second means that it converges to something that is not zero.

Since $ \sec^2 k\ge 1$ for all $ k$, the [b]sequence [/b]does not converge to zero, and therefore the corresponding [b]series [/b]is divergent.

which mean if we are given another sequence $ a_k\equal{}\cos k$ , then since it does not converge to $ 0$ so we can say that the series diverges?

Yes, that's right. Consider the series $ S \equal{} \sum a_n$ and let $ A_n \equal{} a_1 \plus{} a_2 \plus{} ... \plus{} a_n$. We have $ a_n \equal{} A_n \minus{} A_{n \minus{} 1}$. But if S is a convergent series then $ S \equal{} \lim A_n$ exists and its finite and we have $ \lim A_n \equal{} \lim A_{n \minus{} 1}$, which implies $ \lim a_n \equal{} 0$. So, if S is a convergent series then it must verify $ \lim a_n \equal{} 0$. On the other hand this implies that, if $ \lim a_n$ exists but its different from zero, or if it does not exist, then the series S diverges.

In your last example, since $ \lim \cos k$ does not exist, then $ S \equal{} \sum \cos k$ diverges.


Support MATHCOUNTS trigonometry algorithm algebra polynomial

I have been teaching math for about 4 years. Students in my algebra class have informed me that the textbooks do not have enough examples and lack detail. I am currently working on an example book just for ALGEBRA. The idea is to have a book where different type of problems are solved. I do plan on marketing this book for teachers, students as well home-schoolers. This book will provide students with detailed examples at different degrees of difficulty.

I realize there are some books out there such as Schaum's outlines, but they leave out some important examples and are sometimes not detailed enough.

It should not be mistaken for a textbook, but rather an example book to be used along with the required textbook.

I would like to hear from some of you and what you think. Any suggestions are appreciated.

Thank you,

[i]LazyEightBooks[/i]

I guess I never noticed that.  The algebra book I used had 4 examples per section, but I never really looked at them anyway...

I think it's the teacher's job to provide more examples and teach the lesson, that's what they're for.

But, more examples are always helpful.

I think it would be more useful for teachers and homeschoolers than for students.  As a student, I wouldn't want to carry another book around.  But, teachers could use the examples in the book for their lessons, maybe when they can't think of more types of problems.  Homeschoolers sometimes don't have a teacher to teach them, so the examples would be really helpful.

I support this endeavor, I hope you succeed!

If you do indeed plan to use examples, please make them somewhat mentally stimulating and different from the same old boring examples and word problems we see.. Be creative :)

Yes, as Anaxerzia said, don't be one of those people who make one problem and then just change the numbers to make 100 problems.

Also, try not to make TOO many problems use ridiculously tedious computations.. maybe at most, like 5% for each exercise...

I disagree a little bit with popular opinion on this one. Inculcation is basically how the "rest of us" learn mathematics, and of course you need a variety of mentally stimulating and challenging problems for the gifted ones. A mix of basic problems for students who have not completely grasped the concepts yet and then more involved problems for students who understand the material and are looking for a challenge seems like the best way to do it.

If you are making a book for algebra, I agree with Lazarus, in which you should have simple problems, and then some really hard ones. Maybe to have something to guide you, you could look at some old mathcounts tests. Just make sure to not make the same problems.

You might also want to check out the Problem Solver Series by Research & Education Assn., specifically

The Algebra and Trigonometry Problem Solver: A Complete Solution Guide to Any Textbook
    A volume in the series Problem Solver Series
    James R. Ogden
    Research & Education Assn, City of publication unknown: 1998
    Paperback. 915 pages.    ISBN 0-87891-508-7

Almost similar in format to the Schaum's outline series, it may nevertheless give you some ideas.  Especially inspiring is the introduction (virtually identical with all of their series but adapted to suit the subject at hand) which more or less echo your sentiments.  Good luck.

You want to get them to formulate their own examples. Make some of the questions open-ended (preferably after the students have mastered the material) -- i.e., "create-a-problem-with-these-conditions"-type-of-problems, or maybe a "whats-a-general-algorithm-for-solving-these-kinds-of-problems"-type-of-problem, or a "try-generalizing-this-result-and-write-it-in-formal-mathematical-language"-type-of-problem. In my opinion this is the most important part of the learning process.

Another important thing to make students think about is to what extent an idea they have learned can apply to things other than what it was introduced for (not sure how much this comes up in Algebra I, but for example polynomial factorization over R still works in prime mods, etc.).


For each of 8 colors, I have one shirt and one tie of that color.  How many shirt-and-tie outfits can I make if I refuse to wear a shirt and a tie of the same color?

There are $ 8$ choices for the shirt color and $ 7$ for the tie.  Thus, we have $ 8 \times 7\equal{}\boxed{56}$.

Actually, you multiply 8 and 8 to get 64. Then you subract 8, which represents the 8 outfit that are the same color. so the answer is 56. :wink:

I did 8x8 and subtracted the 8 times you have the same two colors. But there might be a flaw in my slution.

If you have eight eigths, and subtract one eight, then you have seven eights (or 7 x 8 = 56), which is correct.


calculus
integration
trigonometry
analytic geometry
linear algebra
matrix
algebra

Find $ \iint_D {\cos \frac{{x \minus{} y}}{{x \plus{} y}}\,da}$ where $ D$ is the triangle with vertices $ (0,0),$ $ (0,1),$ $ (1,0).$

After substitutions $ (u,v) \equal{} \left( {\frac{{x \minus{} y}}{{\sqrt 2 }},\frac{{x \plus{} y}}{{\sqrt 2 }}} \right)$ we have a new triangle with coordinates $ (0,0),\,\left( { \minus{} \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right),\,\left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right).$

And $ (x,y) \equal{} \left( {\left\{ {\frac{u}{{\sqrt 2 }} \plus{} \frac{v}{{\sqrt 2 }}} \right\},\left\{ {\frac{v}{{\sqrt 2 }} \minus{} \frac{u}{{\sqrt 2 }}} \right\}} \right),$ so the Jacobian matrix is

$ \frac{{\partial (x,y)}}{{\partial (u,v)}} \equal{} \left| {\begin{array}{lr}{\cfrac{1}{{\sqrt 2 }}} & { \minus{} \cfrac{1}{{\sqrt 2 }}}  \\{\cfrac{1}{{\sqrt 2 }}} & {\cfrac{1}{{\sqrt 2 }}}  \\\end{array} } \right| \equal{} \frac{1}{2} \plus{} \frac{1}{2} \equal{} 1.$

I've sketched the region, but I dunno if $ u$ goes from $ 0$ to $ \frac1{\sqrt2}$ or $ u$ goes from $ \minus{}v$ to $ v.$

$ u$ goes from $ \minus{} v$ to $ v,$ and $ v$ goes from $ 0$ to $ \frac1{\sqrt2}$. Use the order of integration $ dudv$.

Ahh, okay, I get it.

Selecting that integration order is by convenience?

Yes. Otherwise you would need to break-up the domain of integration into two parts. (Why?)

Draw the picture, in $ u$ and $ v$ coordinates. It's a triangle- where are the vertices?


The finite sequence $a_n$ (actually, from what I understand, we could simply call it a set) has values in the set $\{0,1,2\}$. A move consists of deleting two different numbers and placing another number instead, which differs from both of them (you may, for example, replace $1,2$ by a $0$). After a sequence of such moves we're left with a single number. Show that there's no sequence of moves which will leave another single number.

i think set is not used because in a set, every value can be only once, otherwise it's sometimes called a "multi-set", anyway, calling it a sequence makes it clear.
let $n_0,n_1,n_2$ be the number of 0's, 1's, 2's in $a_n$ respectively. then $3^{n_0}5^{n_1}7^{n_2}\; mod\; 8$ is an invariant.

Peter

Right... I'm sorry. I thought about set because the word "sequence" suggests an order, and this is not the case. Obviously, you're right. Anyway, it's only a question of terminology.

This is a very nice solution. Here's what I did: obviously, we can work with anything else instead of $0,1,2$, since these are only symbols. That's why I worked with $(1,0),(0,1),(1,1)$ as elements of $\mathbb Z_2^2$. The sum $\sum a_n$ is the invariant.


I've read some past threads that were similar to this subject, but all of them addressed what constituted a "well-known" formula, but this is a little different.  I've done some research online and I've found a statement that someone thought up and decided to prove that relates to a problem I'm working on.  Is it within the rules to use this person's statement and proof as a lemma in one of my solutions?  Should I reproduce it and use a link as a citation?  Thanks.

Basically, yes: if you find something online that doesn't involve directly asking another person, then you can cite it and use it.  The "internet" is not all that reliable a source, though, so you'll need to convince the graders that the result is indeed true.  (For example, if the result appeared in a peer-reviewed electronic journal, then it's pretty safe, but if the result appeared on somebody's personal web site, then a reader would be very skeptical unless you provided some details of the proof.)

You'll need to use your judgement on this.


calculus
integration
trigonometry
calculus computations

Evaluate $ \int_0^{\frac{\pi}{12}} \frac{dx}{(\sin x\plus{}\cos x)^4}$

[quote="kunny"]Evaluate $ \int_0^{\frac {\pi}{12}} \frac {dx}{(\sin x \plus{} \cos x)^4}$[/quote]

First $ \sin x \plus{} \cos x \equal{} \sqrt{2} \cos \left( \frac{\pi}{4} \minus{} x \right)$.

That means, $ \int_0^{\pi/12} \frac{dx}{(\sin x \plus{} \cos x)^4} \equal{} \int_0^{\pi/12} \frac{dx}{4\cos^4 \left( \frac{\pi}{4} \minus{} x \right)} \equal{} \int_{\pi/4}^{\pi/6} \frac{1}{4} \sec^4 t dt$

I can do the last integral but I am too lazy.  :P

[quote="kunny"]Evaluate $ \int_0^{\frac {\pi}{12}} \frac {dx}{(\sin x \plus{} \cos x)^4}$[/quote]

another way...  :|

$ \textbf I \;\equal{}\; \int_0^{\frac {\pi}{12}} \frac { \sec^4 x \;dx } { (1\plus{}\tan x)^4}$

$ \;\equal{}\; \int_0^{\frac {\pi}{12}} \frac { \sec^2 x\; dx } { (1\plus{}\tan x)^4} \;\plus{}\; \int_0^{\frac {\pi}{12}} \frac { \tan^2 x \cdot  \sec^2 x \;dx } { (1\plus{}\tan x)^4}$

$ \left[ \; \equiv  \int_1^{1\plus{}\tan \frac {\pi}{12}  } \;\frac { 1} { u^4} \; du \;\plus{}\; \int_1^{1\plus{}\tan\frac {\pi}{12}  }\; \frac {(u\minus{}1)^2 } { u^4} \; du \qquad ( \text{where} \quad u \equal{} 1\plus{}\tan x) \right]$

$ \;\equal{}\; \minus{}\dfrac{1}{3  (1\plus{}\tan x)^3 } \bigg| _0^{\frac {\pi}{12}}  \;\plus{}\; \left( \; \minus{}\dfrac{1}{  (1\plus{}\tan x)} \plus{} \dfrac{1}{ (1\plus{}\tan x)^2 }  \minus{}\dfrac{1}{3  (1\plus{}\tan x)^3 } \;  \right) \bigg| _0^{\frac {\pi}{12}}$

i vaguely recall from 6th grade trig....
$ \tan \frac {\pi}{12} \;\equal{}\; \dfrac { \sqrt 6 \minus{} \sqrt 2 } { \sqrt 6 \plus{} \sqrt 2 }$

$ \therefore \quad \textbf I \;\equal{}\;  \dfrac{2}{3} \minus{} \dfrac{ \sqrt 6 \plus{} \sqrt 2 }{2\sqrt 6} \plus{} \dfrac {( \sqrt 6 \plus{} \sqrt 2 )^2 } { 24 } \minus{} \dfrac {( \sqrt 6 \plus{} \sqrt 2 )^3 } { 72\sqrt 6}  \;\equal{}\; \boxed {  \dfrac{1}{3}\left( 1\minus{}\dfrac{5}{6\sqrt 3}\right) }$

Let $ t \equal{} \frac {1}{(\sin x \plus{} \cos x)^2}$


email geometry Support USAMTS AMC AIME function

To those who participated this round:

This is a thread where you can discuss the problems and any questions related to it. I have provided all of you with solutions so feel free to speak any question/comment. Because this is a discussion thread, not just a solution thread, it would be awesome if you guys can talk about just contest in general and overall feeling in comparison to actual AMC. Also, if there is something that you felt was inadequate, address that as well as long as the tone is moderated.  :)

And with the time period... I checked my thread again (announcement) and I DID say the time period was EST... It seems like that I never said it on emails so I guess it's still partly my fault but in future, all rounds' time periods will be on Eastern Time period. Thanks.

So... Talk!  :D

This contest was very fun. I feel it was slightly harder than the actual AMC, but not very much so. I noticed that there seemed to be fewer hard geometry problems than a typical AMC. For me, the problems spiked in difficulty around numbers 18-20. Problems before then did not seem very difficult while later problems seemed a lot harder. I think the problem difficulty chart supports this observation.

Pretty much agree with above, rationalize denominators, 21 was guessable, first half was easy, later questions were harder. Though I haven't really done math in a while so I can't say if it was harder or easier than a regular AMC.

[quote="xpmath"]I just have to say that I didn't really like #8. On basically every competition I've seen, all the answer choices are rationalized. Also, I was too lazy to actually rationalize the answer choices, so I spent about 20 minutes trying to find where I went wrong. Then I rationalized B. Yay.[/quote]

My apologies. Yes, that is true in most cases. I thought it was a very easy question (a free point) but I see the problem. I'll remember that for future.

[quote="Brut3Forc3"]
21 was highly guessable: even starting some casework, you'd find 1/7 chance of making a loop, and a 6/7 chance of connecting 2 ropes. Then, 105 is the only denominator divisible by 7.[/quote]

It's a recent USAMTS problem. I'm sure most people recognized the question anyway so oh well.. But I'll try to make the question trickier.

Couple of my inputs:

#12: I wanted to make this question an easy version of "substitution" algebraic problem. But turns out, it was way too easy problem that you could just substitute the first choice and it's right!  :rotfl: I didn't think to make the correct answer as say, C or D or E...

#20: This was the hardest problem because it was one of those problems that weren't really AMC-like. More likely, it should've been at AIME-level problem.

I do agree with lack of geometry. I'll remember to put some horrible geometry problems for next year then.  :)

I actually didn't find 8 too easy(didn't think about adding them)... It was one of the problems that took longer than I should have. I multiplied the top by sqrt(2) and bottom by sqrt(3) and solved for x and y and then did (x+y)^2.  I think rationalizing would have helped though.

That's exactly what I did. After getting that and seeing it wasn't an answer choice, I decided to go through everything again, but I got the same answer. As it has happened before, I thought I had made the same mistake twice or misread the question. After a while, I decided to plug in the numbers for x and y and found that they worked. It didn't take me too long, then, to rationalize the numbers. It wasn't a hard problem, but it basically helped kill my time, and so I phailed.

Besides that, I thought the problems were good. It seemed somewhat harder than a real AMC 12, but that's good practice. Thanks for making it :)

Ugh.  I didn't have time to take the test, but I looked at the problems.  Seemed a little bit harder than a real AMC 12, but like xpmath said, it's good practice.  Sorry for not taking it :(

I guess that is definitely true for questions like #8. It makes more sense now. I always try to ask this question, "Will I be able to do the same thing as solution right away?" It usually prevents problems from being too intricated but it didn't work well on that question.. I think I came up with that question like in couple minutes.

I'm glad that the problems were liked by many people. I have actually started on next year's rounds as well. I'll definitely remember to throw more geometry problems in and make less calculating ones.  :D

It seems likely that there will probably never be the third round. I don't know, I think that there are enough practice materials for USAMO.. It's usually AIME/AMC that are short in number of practice materials. At least that's what I think.

[quote="Altheman"]Did you post the problems somewhere? I cannot find them... Can I look at them?[/quote]

The problems are available only to those who competed in the contest, or if you requested them otherwise. I keep all the record of whereabouts in my email.

well, I did get better than what I got on the real AMC 12B last year :D

Anyway, personally, I didn't understand questions #11 and #21... so i couldn't attempt them..

I did attempt to find the pattern for #11, but I failed and just guessed :D

and, I think #18 was a pretty straightforward question applying Menelaus' Theorem :idea:

[quote="Silverfalcon"][quote="xpmath"]I got 13.5 points LESS than my AMC 12B score from last year. On #11, I think I accidentally counted wrong...darn. On #4, I somehow didn't read "real." In other words, I didn't make the AIME cutoff :o[/quote]

Don't feel bad for #4. A very large number of people got #4 wrong.. It is careful to read question carefully![/quote]

That makes me feel special :P

Haha... Well, good.  :lol:

Just an important announcement: I have finally found a way to use FTP in my website. So I can put on the documents in the internet now! No more you pleading me for documents ( :D although it was nice to check emails and get millions of "please" and "thank you").

The URL is:

http://www.hiddendiscoveries.com/

I'll update them often so keep checking back to see important announcements. The preliminary round problems, solutions, results, along with how to guide, are all available in the website under "Math Competition."

Merry Christmas~

perhaps I am missing something obvious, but on number 22, aren't there 2 possible points B? - the two points of tangency? and thus 2 possible lines, and thus 2 possible answers? one for each of the 2 tangent lines....I was extremely confused about this when I took this test...and still can't see what I am missing

Correct.

There are 2 answers to this problem because of two tangents. But note that only one of them is given so that is the correct answer. This answer, if I remember, was easier in calculation so that's why I chose this.

[quote="Brut3Forc3"][quote="azn_rocker0322"]Yes, there are two different points for B.
But, what the question requires us to do is to find, among the answer choices, the best fit answer.
So, of course you will get two answers, but you have to choose the one that's among the answer choices.

Remember that no contest will ever put TWO best answer choices among a multiple choice question.[/quote]
Actually, there was that one counting AMC question where it was unclear which of the two answers it was... (in the end, both were accepted).

When is Round 2 starting?[/quote]

Probably in January. I might end up making the Round 2 as two-day periods instead of one-week because I don't have much break in January (my college has a small course called Intercession in January and end around 23rd and spring semester starts in 28).

I originally wanted to make it after the real AMC but I don't think I can... Grading and making stat require lots of time.

Just out of curiousity, on number 15, was there a way to determine the actual function? Or are there infinitely many functions that satisfy the given info? I am horrible with functional equations.

On number 20, is there any reason why $ a_1,a_2,a_3,a_4, b_1,b_2,b_3,b_3,b_4$ needed to be integers? Do they need to be integers to satisfy the given info? Or was the fact that they were integers just there to throw us off?

I'm not sure about #15. I know that when I made this problem, I had the function before. It's certainly possible that there are more than one function which satisfy the properties. But I'm not good with functional equations as that extent to find that.

#20: I don't think it was a trick or anything. I probably (probably because I don't remember clearly) made the question like that because it may helped to be guessable. I honestly do not know. I don't  have the art of problem solving book with me (away at college now) so yeah.

I don't know if I checked to see if there were indeed 8 integers that satisfy the equation. Again, this was my first year writing the problems so certainly, it could've been flawed (as I did on one of the questions and was able to thankfully fix it!)


limit
real analysis
real analysis unsolved

$\sum \frac{1}{a_n}$ converges.$a_i \geq 0$ can we prove $\lim_{n\to \infty}\frac {n^2}{A_n}=0$?
where $A_n=a_1+a_2+\cdots+a_n$


i want to know who has the best technic in playing violin...

for example in flamenco guitar,paco has the best technic,hasnt he? ;)

I don't think you can really say one person has better technique than any other.  They all have different styles, and are good at different things.  The variance from one concert to another for a given person is probably enough to make it impossible to determine a best anyways.

[quote="ashegh"]
for example in flamenco guitar,paco has the best technic,hasnt he? ;)[/quote]

Juan Martin is better :D (Paco pena is good also, paco delucia is good too)

[quote="314159265358979323846"]Krishanu's right, all the great concert artists have vastly different styles and strengths...

... but in terms of sheer technique, Heifetz's is impeccable (that and every recording of his is like, a good 25% faster than the others out there)...[/quote]

Heh true...Heifetz also modifies stuff at will to his liking.
Insane bow control/speed...but then again, same for left hand...

As for people alive today...here's an example.
My brother has many pieces played by different artists.  There is a lot of overlap.  However, for each piece, he chooses the person who plays it best, based on their styles of playing.  For example, the recording of the Sibelius is played by someone other than for the recording of...say...the Tchaikovsky...

From what I know (I may be entirely wrong), Heifetz is most famous for his supreme vibrato.

I have seen the spelling "technic" (and perhaps a few other spellings) for "technique" in books about music. The origin of the word is Greek; it may have entered other European languages with more than one spelling in the Roman alphabet.

Dudes, Heifetz is like a god. No one else is comparable.

[quote="Aryth"]Heh, you spelled Technique incorrectly but that is besides the point.
I'm a violinist myself. Who is this "paco" person?[/quote]

Paco = Paco de Lucia

quote on Heifetz:"That boy's only happy when he can play fast." -Kathleen Parlow (had the same teacher as Heifetz)
Best technique? Does the person still have to be living?
if yes: Perlman
if no: Pagannini, Isaac Stern, Heifetz

The best violin dynasty was the Auer dynasty. Just google Auer.

[color=brown]
[size=200]

When you want to compare which one is better than other, you must set some criteria for the judgement.  Otherwise you will argue with each other.
Heifetz is not acclaimed to be the best because of his fast technic, but his interpretation is one of the  factor which contribute him  to his success.

[/color] [/size]

Isn't Joshua Bell good? And I forget the other guy ...But there is the person Ian Anderson's touring with,too.

(leftmost is best)

Alive: Itzhak Perlman, Hillary Hahn, Joshua Bell, Anne-Sophie Mutter, Maxim Venegrov, Ray Chen

Dead: David Oistrakh, Jascha Heifetz, Niccolo Paganini,Henryk Wieniawski, Joseph Joachim , Henryk Szeryng

Immortal: Ling Ling

hia, im glad to see im not the only person who plays violn... o wait... nvm  :(


function
geometry
geometric transformation
reflection
integration
real analysis
real analysis unsolved

[b]without [/b]using the [i]reflection [/i]formula for the gamma function, show :D

$ \Gamma\left(\frac{1}{5}\right)\cdot\Gamma\left(\frac{4}{5}\right)\;\; \equal{}\;\;\boxed{\pi\cdot\sqrt{\;2\plus{}\frac{2}{\sqrt 5}\;}}$

\begin{eqnarray*}\Gamma(z)\Gamma(1-z) & = &\Gamma(z)\int_{0}^{\infty}t^{-z}e^{-t}\; dt\\ & = &\int_{0}^{\infty}\int_{0}^{\infty}e^{-t}u^{z-1}e^{-tu}\; du\; dt\\ & = &\int_{0}^{\infty}\int_{0}^{\infty}u^{z-1}e^{-(u+1)t}\; dt\; du\\ & = &\int_{0}^{\infty}\frac{u^{z-1}}{1+u}\; du\\ & = &\frac{1}{z}\int_{0}^{\infty}\frac{1}{1+v^{1/z}}\; dv.\quad\quad (\text{where }v = u^{z})\end{eqnarray*}

Now substitute $ z =\frac{1}{5}$, then

$ \Gamma\left(\frac{1}{5}\right)\Gamma\left(\frac{4}{5}\right) = 5\int_{0}^{\infty}\frac{1}{1+v^{5}}\; dv.\quad\quad\cdots (1)$

Evaluating the right side of (1) is not so hard, if we use some techniques of complex analysis... or brute force would also work!


a cyclist starts from center 'O' of a circular track of radius 1 Km, reaches the edge P of the track and cycles along the circumference and stops at a point Q. if the total time taken is 10 minutes, what is the average velocity of the cyclist? (Angle POQ is 60 degrees).

a) 3 Km/hr  b) 6 Km/hr c) 9 Km/hr d) 12 Km/hr

Ans:

[hide]12km/hr[/hide]

Solution:

[hide]The cyclist first moves from the centre of the circle to the perimeter along a radius. And the radius of the circle is given as R = 1 km.
He then travels aound 60° of a circle and the distance of this arc is given by s = Rθ, where θ is in radians. 60° = pi/3 rads.
Therefore the 2nd part of his journey is R*pi/3.
Total distance travelled is D = R + R*pi/3
D = R(1 + pi/3)
Since pi is just over 3, then we can let pi/3 = 1 as an approximation. So,
D = 2R
D = 2 km
======

time is t = 10 mins = 1/6 hr

Therefore average velocity is v = 2km / (1/6)hr = 12 km/hr
v = 12 km/hr
=========[/hide]


Define the hyperstar $\text{HS}(n,k)$ as the graph whose vertex set is the set of all binary strings with $n$ digits and exactly $k$ $1$s. Two vertices are adjacent iff the first digit of one of the strings can be switched with a different digit in the string to form the other vertex's string (for example, in $\text{HS}(3,3)$, the vertices $111000$ and $011001$ are adjacent, since the first $1$ can be switched with the last $0$ in the first string to form the second.)

Conjecture: Is $\text{HS}(2k,k)$ Hamiltonian?

(This is a variation on the hypercube problem, which can be answered affirmatively, with a proof by induction. This conjecture seems to be harder.)

If the proof is in fact not difficult, a hint would be appreciated.

Tetrahedr0n, if you go onto Eddie Cheng's web page you'll see he still hasn't solved that :D

let $g$ be the graph with vertices $v_1,...,v_p$.when it has a cycle with lenght of $p$?


geometry
3D geometry
sphere
calculus
integration
analytic geometry
calculus computations

How do you find the surface area of the part of the sphere $ x^2\plus{}y^2\plus{}z^2\equal{}100$ that lies aboce the cone $ z\equal{}\sqrt{x^2\plus{}y^2}$

Step 1: describe the curve along which the cone and the sphere intersect. Once you have that, there are both Calculus II (area of a surface of revolution) and Calculus III (double integral for surface area, probably set up in some form of polar coordinates) solutions.


geometry
parallelogram
analytic geometry
graphing lines
slope
combinatorics proposed
combinatorics

[quote="hxy09"]Find all integers $ n$ for which a regular hexagon can be divided into $ n$ parallelograms of [b]equal area[/b].[/quote]

[geogebra]236531eba41779dee28be2c83af290b99c6d2c91[/geogebra]

By drawing lines parallel to the sides of the hexagon, we get this figure.

$ n\equal{}3k$ for any positive integers $ k$ are the solutions.

Can you prove that the hexagon can't be divided into $ n$ pieces when $ n$ is not divisible by $ 3$?I think this is the main difficulty of this problem. :)

the fact that $ 3|n$ can be seen by the following.
#1: All the parallellograms are of three types when you look at the slope of the sides.
#2: The total area of each type are the same.

#1 is easy to verify and
#2 can be seen by observing a single cross-section of a line parallel to one of the sides of the orginal hexagon.


Find all quadruples $(x, y, z,w)$ of positive integers such that
$x^{2}+y^{2}+z^{2}+w^{2}= 3(x+y+z+w).$

Consider mod 3. First, assume that not all four are congruent to zero.
Thus, exactly one is congruent to zero, say $x$. Then, define $3x'=x$ and we have:
\[9x'^{2}+\sum y^{2}=9x'+3\sum x \implies 3\sum y \geq \sum y^{2}\]
after using the trivial inequality for positive integers, $x^{2}\geq x$.
By Am-Gm, we have that $\sum y^{2}\geq \frac{\left(\sum y \right)^{2}}{3}$. After chaining the ineq's, we have that:
\[3\sum y \geq \sum y^{2}\geq \frac{\left(\sum y \right)^{2}}{3}\implies 9 \geq \sum y \]
Let $y,z,w$ be ordered $y\leq w\leq z$ so that $y\leq 3$, and since $y$ isn't $0$ mod 3 we have $y\leq 2$ so that $y=2$ or $y=1$.
Also, assume $x$ is $9$ or greater; then, $81+\sum y^{2}= 27+3\sum x \iff 3\sum x \geq 54 \iff \sum x \geq 18$ which is a clear contradiction.
So $x$ is also $3$ or $6$.
f $x$ is $6$, then notice that since each of $y^{2}$ is greater than or equal to 1 we have that $y+z+w\geq 7$ so that $y+z+w=7,8,9$ for the largest to hold, we must have equality in $y,z,w$ referring to the inequalities but this is impossible as each isn't divisible by three, so $\sum y =7,8$. First, try $y=2$, so $z+w=5,6$ thus $w=2$ thus $z=4$ (eliminate $3$) but $(6,2,2,4)=(x,y,w,z)$ which isn't a solution. Now try $y=1$, which yields solutions.

It remains the case $x=3$, which yields the other solutions.

Obviously, the last class of solutions is when $x=y=z=w=0$ mod 3 but then take $x'$, respectively, defined as before, and we find the solution $(3,3,3,3)$.

The advantage of the other solution is that right off the bat, each of $x,y,z,w$ is less than 4 by square positivity (and that $2x-3\not = 0$) so we can write out a proof as follows:

[hide="t0r proof explanation"]
\[x^{2}+y^{2}+z^{2}+w^{2}= 3(x+y+z+w) \iff 4\sum x^{2}= 12 \sum x \iff \sum (4x^{2}-12x+9) = 36 \iff \sum (2x-3)^{2}= 36 \]
Since $2x-3=0\iff 2x=3$ has no positive integer solutions and $2x-3$ is an integer, we have that $|2x-3|\geq 1\implies (2x-3)^{2}\geq 1$ for any of the variables in the problem, thus
\[(2x-3)^{2}+3\leq \sum (2x-3)^{2}= 36 \implies (2x-3)^{2}<36 \implies 2x-3<6\implies 2x<9 \implies x\leq 4 \]
for each of $x,y,z,w$. Clearly, we can't have two or more variables equal to $4$ since this would yield a sum over $50$. Assume some $x=4$; then $(2y-3)^{2}+(2w-3)^{2}+(2z-3)^{2}=11$; clearly the only way to express $11$ in this way is when the tuple $(x,y,w,z)=\text{up to permutation}\to(4,3,1,1),(4,3,1,2,),(4,3,2,2)$ just by considering ways to make three squares 11 (just case work).
Now it is clear that if all of $x,y,z,w$ are smaller than $4$, they must all be three to add to $36$ which yields $(3,3,3,3)$

which are all the solutions
[/hide]

Looks like your proof is not really one line at all  :wink: were you expecting those solutions?

[quote="me@home"]Looks like your proof is not really one line at all  :wink: were you expecting those solutions?[/quote]

Yes.  I didn't have enough time to write them out, but I thought they were obvious.   :wink:

[quote="PenguinIntegral"]How did T0r notice that factorization?[/quote]

The two-variable case is clearly the equation of a circle.  It's not hard to generalize from here.

he just completed the square...this is a common technique for these problems


function
limit
calculus
calculus computations

If f : R^n -> R differentiable convex function then I know
g(a) = (f(x+ay) - f(x)) / a
is nondecreasing hence sup{a>0} g(a) = lim{a->+oo} g(a)

Please prove lim{a->+oo} g(a) = lim{a->+oo} gradf(x+ay)'y
where gradf(x) is the gradient of f at x.

Thanks a lot!

This is a one-dimensional problem: $\lim_{x\to+\infty}\frac{f(x)-f(0)}{x}=\lim_{x\to+\infty}f'(x)$ for any convex function $f\colon\mathbb R\to\mathbb R$. We may assume $f(0)=0$. Let $L$ be the limit of $f'$, finite or not. Integrating $f'$ over the interval $[\sqrt{x},x]$, you will find $\lim_{x\to\infty}\frac{f(x)-f(\sqrt{x})}{x-\sqrt{x}}=L$, hence $\lim_{x\to\infty}\frac{f(x)-f(\sqrt{x})}{x}=L$. If $L$ is infinite, then we are done because $f(x)-f(\sqrt{x})\le f(x)$. If $L$ is finite, then $0\le f(\sqrt{x})\le L\sqrt{x}$, hence $\lim_{x\to\infty}\frac{f(\sqrt{x})}{x}=0$.


analytic geometry
calculus
integration
The Geometry of Numbers

Find coordinates of a set of eight non-collinear planar points so that each has an integral distance from others.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=150576 .
Take 8 arbitrary points from that circle and stretch it by an integral factor such that all rational distances get integers.


inequalities proposed
inequalities
High School Olympiads

Let a,b,c be positive real numbers.Prove that
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\le\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}+\frac{c^2}{a^2+b^2}$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=17207 and http://www.mathlinks.ro/Forum/viewtopic.php?t=34563 and the generalizations at http://www.mathlinks.ro/Forum/viewtopic.php?t=17230 , http://www.mathlinks.ro/Forum/viewtopic.php?t=23416 , http://www.mathlinks.ro/Forum/viewtopic.php?t=23752 .

  darij


in triangle $ ABC$ let $ B_1$ be on $ AC$ ,$ E$ on $ AB,G$ on $ BC$,and let $ EG$ parallel to $ AC$ .furthermore $ EG$ be tangent to the inscribed circle of the triangle $ ABB_1$ and intersect $ BB_1$ at $ F$ .let $ r,r_1,r_2$ be the inradii of the triangle $ ABC,ABB_1,BFG$ respectively.prove that
$ r\equal{}r_1\plus{}r_2$


floor function
logarithms
algebra
polynomial
vector
absolute value
number theory proposed

once I asked a similar question about primes, now it's time for Fibonacci numbers :)

so, my conjecture is:

for every $ n\in\mathbb{N}$ thete exists $ k\in\mathbb{N}$ such that the interval $ [n10^k,(n\plus{}1)10^k)$ contains at least one Fibonacci number.

is it true or not ?

$ F_{2m}=\left\lfloor\frac{\phi^{2m}}{\sqrt{5}}\right\rfloor$

So

$ n10^k\le\frac{\phi^{2m}}{\sqrt{5}}\le (n+1)10^k$
$ k+\log(n)\le m\log(\phi^2)-\log(\sqrt{5})\le k+\log(n+1)$
$ k+\log(n)+\log(\sqrt{5})\le m\log(\phi^2)\le k+\log(n+1)+\log(\sqrt{5})$

We need to find $ n$ such that the fractional part $ \{m\log(\phi^2)\}$ is between ${ \{\log(n)-\log(\sqrt{5})}$ and $ \{\log(n+1)-\log(\sqrt{5})\}$.

But since $ \log(\phi^2)$ is irrational $ \{m\log(\phi^2)\}$ will be in any nonzero interval in $ [0,1)$ for some integer $ m$, so we're done.

thank you

May I have one more question?
[quote="tjhance"]But since $ \log(\phi^2)$ is irrational[/quote]
this requiers $ \phi^j\notin\mathbb{N}$ for all $ j\in\mathbb{N}$.
I'm convinced it holds, but how to prove it ?

Well, suppose $ \phi$ is a root of the polynomial $ x^j \minus{} a \equal{} 0$, with $ a\in\mathbb{N}$. This a rational polynomial, so if $ \phi$ is a root, then its conjugate $ \overline{\phi} \equal{} \frac {1 \minus{} \sqrt {5}}{2}$ must also be a root. But $ 0 < |\overline{\phi}| < 1$ so $ \overline{\phi}^j$ cannot be an integer.

Alternatively, we could say that $ \phi$ and $ \overline{\phi}$ cannot both be solutions because they have different absolute values (but they should both have absolute value equal to $ \sqrt [j]{a}$).

Alternately, $ x^2 \minus{} x \minus{} 1$ is not a cyclotomic polynomial.  This is also a consequence of the identity $ \phi^j \equal{} F_j \phi \plus{} F_{j \minus{} 1}$.  (In other words, the rationality of $ \phi^n$ immediately implies the rationality of $ \phi$, and in other other words, $ \mathbb{Q}[\phi] : \mathbb{Q} \equal{} 2$.)

[quote="t0rajir0u"]$ \mathbb{Q}[\phi] : \mathbb{Q} \equal{} 2$.)[/quote]

Question:

What does that mean?

tjhance: this means that the mentioned field extension has degree 2, meaning that $ \mathbb Q[\phi]$ is a 2-dimensional $ \mathbb Q$ vector space.

[quote="t0rajir0u"]Alternately, $ x^2 \minus{} x \minus{} 1$ is not a cyclotomic polynomial.[/quote]
I don't see how this immediatly gives the result.

By the way, you also could prove it by trivial "calculation". Let $ x \plus{} y\sqrt 5 \equal{} \phi^n$ and show that $ y > 0$.

I don't understand the explanation with cyclotomic poly too

but others are clear, thanks very much :)

btw, here
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213412
is another one with Fibonacci's and intervals :wink:
maybe someone will have an idea :idea:


Checking my notes .... I found that:
----------
Let $a,b$ real numbers.
Prove that:
\[ \frac{a^2}{a^2 + b + a^3b} + \frac{b^2}{a + b^2 + ab^3} + \frac{a^2b^2}{a^3 + a^2b^2 + b^3} \leq 1. \]
-----------

Carlos Bravo :)
Lima-Peru

let $\frac{b}{a^2}=x$, $\frac{a}{b^2}=y$, $ab=z.$
then xyz=1 and should demonstrable, that
${\frac{1}{x+y+1}}+{\frac{1}{x+z+1}}+{\frac{1}{y+z+1}}\leq1,$

because $\frac{1}{x+y+1}\leq\frac{\sqrt[3]{z}}{\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}}$ is equivalent to $\frac{1}{x+y+\sqrt[3]{xyz}}\leq\frac{1}{\sqrt[3]{xy}(\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z})}$


Let $ S$ be the set of the $ 2005$ smallest multiples of $ 4$, and let $ T$ be the set of the $ 2005$ smallest positive multiples of $ 6$. How many elements are common to $ S$ and $ T$?

$ \textbf{(A)}\ 166\qquad
\textbf{(B)}\ 333\qquad
\textbf{(C)}\ 500\qquad
\textbf{(D)}\ 668\qquad
\textbf{(E)}\ 1001$

The elements that are common to both $ S$ and $ T$ must be multiples of $ 12$. The smallest and largest elements of $ S$ are $ 4 \cdot 1$ and $ 4 \cdot 2005$, respectively. The smallest and largest elements of $ T$ are $ 6 \cdot 1$ and $ 6 \cdot 2005$, respectively. The number of elements that are common to $ S$ and $ T$ is
\[ \lfloor \frac{4 \cdot 2005 \minus{} 6 \cdot 1}{12} \rfloor \plus{} 1 \equal{} \lfloor \frac{8020\minus{}6}{12} \rfloor \plus{}1 \equal{} 668 \rightarrow \boxed{\textbf{D}}.\]

^I did same thing as you did, but instead I looked at 4*2005 which is 8020, and I subtracted units of 4, until the digits added up to a multiple of 3. How did you do the short cut you used?

You can think of $S$ and $T$ as 'intervals' and find their intersection:

$S=\{4\cdot1,4\cdot2,\dots,4\cdot2005\}$ ---> $S=[4\cdot1,4\cdot2005]$
$T=\{6\cdot1,6\cdot2,\dots,6\cdot2005\}$ ---> $T=[6\cdot1,6\cdot2005]$

$S\cap T=[6\cdot1,4\cdot2005]=\{12\cdot1,\dots,12\cdot\left\lfloor\frac{4\cdot2005}{12}\right\rfloor\}=\{12\cdot1,\dots,12\cdot668\}$

The number of multiples of $m$ in a set $X=[a,b]$ is given by $\left\lfloor\frac{b-a}{m}\right\rfloor$. His added 1 worked out by 'luck' because if we were to find the number of multiples of 12 between 6 and 8022, this would have shown his formula was wrong. You only had to calculate $\left\lfloor\frac{4\cdot2005}{12}\right\rfloor=668$ to get the answer.

EDIT: See my next post for correction (this stays here for the record).

[quote="10000th User"]You can think of $S$ and $T$ as 'intervals' and find their intersection:

$S=\{4\cdot1,4\cdot2,\dots,4\cdot2005\}$ ---> $S=[4\cdot1,4\cdot2005]$
$T=\{6\cdot1,6\cdot2,\dots,6\cdot2005\}$ ---> $T=[6\cdot1,6\cdot2005]$

$S\cap T=[6\cdot1,4\cdot2005]=\{12\cdot1,\dots,12\cdot\left\lfloor\frac{4\cdot2005}{12}\right\rfloor\}=\{12\cdot1,\dots,12\cdot668\}$

The number of multiples of $m$ in a set $X=[a,b]$ is given by $\left\lfloor\frac{b-a}{m}\right\rfloor$. His added 1 worked out by 'luck' because if we were to find the number of multiples of 12 between 6 and 8022, this would have shown his formula was wrong. You only had to calculate $\left\lfloor\frac{4\cdot2005}{12}\right\rfloor=668$ to get the answer.[/quote]

I Don't think he added 1 by luck, your method works fine for say 1-N, however let's say if you have assorted numbers of 45-9876 (randomly numbers), then you'd need to subtract do the division and then add one again,

The proper way to count the number $N$ of multiples of $m$ between two random numbers $a<b$ is that

$N=\frac{(\text{largest multiple of }m\le b)-(\text{smallest multiple of }m\ge a)}{m}+1$

So his solution should be $\frac{4\cdot2004-12}{12}+1=668$ (4x2004 is the largest multiple of 12 in that set).

Sorry to be off topic, but Fermat's last theorem proves your signature impossible, so what's the wit and humor behind it?

It was featured on one of the [i]The Simpsons[/i] episode. It looks like it's true when you put in a standard calculator because the difference is so small.

EDIT PS: Moreover, FLT is an overkill to disprove that. Look at the parity on both sides of the equation. QED


How many ways are there to put 5 balls in 3 boxes if the balls are not distinguishable but the boxes are?

Sometimes, they simply want to introduce a vast amount of methods, and don't always put the most elegant solutions. You don't need to discuss every solution like this. :wink:

this problem can be thought of as a balls and urns problem.
all the balls before the first urn are those in the first box, all those after the second are those in the third box and all the stuff in between is in the second box. So the the answer is 7Choose2=21

Which is what jbatterson suggested, but as math154 said, they want to introduce a vast number of solutions.

I feel that this is the natural way of solving these types of painful distinguishability problems. Solution below:

[hide="My Solution"]

Since the balls are indistinguishable, what matters is the number of balls in each of the three distinguishable boxes. Then, it's a matter of casework:

Case 1: First box has $0$ balls:

There are $6$ ways to choose the number of balls in the second box (any number from 0-5), and then the number for the third box is decided. So, there are $6$ ways.

Case 2: First box has $1$ ball:

There are $5$ ways to choose the number of balls in the second box (any number from 0-4), and then the number for the third box is decided.

From there, it's pretty obvious that the possibilities for each case after that, up until Case 6 (the final case), is just the possibilities of the previous case minus $1$, making the final answer $6+5+4+3+2+1=\boxed{21}$ ways.

Please stop reviving threads, as you did with [url=https://artofproblemsolving.com/community/c63h287526p11341069]this[/url].

[hide=Easy Solution]
By stars and bars, this is simply $\binom{5+3-1}{3-1}=\binom{7}{2}=\boxed{21}$
[/hide]

[quote=Natsuki]noooooooooooo i silleid on this problem :mad:[/quote]

Well no need to bump and announce that you sillied...

[quote=Natsuki]noooooooooooo i silleid on this problem :mad:[/quote]

That wasn't the most productive bump ever. As a rule of thumb, don't bump threads which are twenty days old.

[quote=Natsuki]noooooooooooo i silleid on this problem :mad:[/quote]

Sad, but don't announce it to everyone like that! :(

[quote=ilovepizza2020][quote=HIA2020]Why was this posted b4 GameBots account was made?[/quote]

wdym? Gamebot is a bot.[/quote]

This was posted on jan 9, 2009, but this was posted in 2008

[quote=dragonborn56][quote=Jwenslawski][quote=Yiyj1][hide=solution] Since the balls are indistinguishable, we must only count the number of balls in the different boxes.

There are $3$ ways to arrange the balls as $(5,0,0)$ (specifically, box 1 can have 5, box 2 can have 5, box 3 can have 5).

There are $3! = 6$ to arrange $(4,1,0)$ and $3! = 6$ ways to arrange $(3,2,0)$; in each case, we must choose one of the 3 boxes to have the largest number of balls, and also one of the remaining two boxes to be left empty.

However, there are only $3$ ways to arrange $(3,1,1)$, and $3$ ways to arrange $(2,2,1)$; in each case, we must choose one box to have the `different' number of balls (3 in the $(3,1,1)$ case and 1 in the $(2,2,1)$ case).

This gives a total of $3 + 6 + 6 + 3 + 3 = \boxed{21}$ arrangements.[/hide][/quote]

NICE POSTCOUNT![/quote]

What?[/quote]

69... Nevermind

[quote=jbatterson]The solution should include the "balls and sticks/dividers" method.[/quote]

Did you write all of the Beast books?  I ADORE them.  Thanks so much.

[quote=cai40]stars and bars 7c2=21[/quote]

Thank you for the informative solution.

[quote=jbatterson]The solution should include the "balls and sticks/dividers" method.[/quote]

I did the question that way, but the answer is wrong.
First, there were 3 boxes, which means 2 sticks. There were 5 balls, and there were 6 gaps (since some boxes can have 0 balls in them, the gaps to the left of the first ball and to the right of the last ball is counted).
So the question was: "how to put 2 sticks in 6 gaps?" My answer is 6·5 divided by 2, which was 15.


[quote="conan_naruto236"]Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
$ abc \plus{} a^2b \plus{} b^2c \plus{} c^2a \leq 4$[/quote]
Because
\[ a^2 b \plus{} b^2 c \plus{} c^2 a \le \frac{4}{{27}}\left( {a \plus{} b \plus{} c} \right)^3  \minus{} abc
\]
It's very old my friend.

[quote="conan_naruto236"]Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
$ abc \plus{} a^2b \plus{} b^2c \plus{} c^2a \leq 4$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100960


induction
function
vector
algebra
functional equation

It was an exercise in my analysis homework last week.  Just let me copy my solution from the tex file.  By the way, continuity is absolutely required if we want our solutions to be of the form $f(x)=kx$.

...

This is Cauchy's functional equation.  By induction, it follows that:

$\displaystyle\ f\left(\sum^n_{i=1} x_1\right)=\sum^n_{i=1} f(x_i)$.

Thus, for any $n\in\mathbb{N}$, there exists $x\in\mathbb{R}$ such that $f(nx)=nf(x)$.  Thus, for rational $x$, say, $x=\frac {m}{n}$, we have $f(\frac {m}{n})=mf(\frac {1}{n})$.  However, $f(\frac {1}{n}\cdot n)=nf(\frac {1}{n})=f(1)$, so that $f(\frac {1}{n})=\frac {1}{n}f(1)$.  Therefore, we have $f(\frac {m}{n})=\frac {m}{n}f(1)$.  Thus, for rational $x$, we have $f(x)=f(1)x$, as desired.  However, if $c$ is irrational, we can take a sequence of rationals approaching $c$, which, by continuity of $f$, implies $f(c)=f(1)f(c)$, for $c$ irrational.  Thus, letting $k=f(1)$, we have that $f(x)=kx$ is the solution of the functional equation over $\mathbb{R}$.

[quote="Eugevir"]Does anyone know a simple proof that the only solutions to the functional equation f(a+b)=f(a)+f(b) are of the form f(x)=kx? It can fairly easily by shown true for the rationals using induction, but I recall seeing a very simple proof for all reals... any ideas?[/quote]

You can't. There exist solutions to f(a+b) = f(a)+f(b) where f(x) isn't kx for all x. So I assume thats where you are getting stuck :) If you are given any other condition, for example continuity, as blahblahblah showed, or something like the function being monotonic, etc, then you can prove it. But not otherwise.

yeah, you can "construct" a function that satisfies it that is not f(x) = kx.

I think this argument works, but it assumes the axiom of choice....

Let B be a basis for the vector space with R as the vectors and Q as the scalars.

For each element b in B, assign it an element f(b) in R.

Now, consider an element in R, say  x =a1b1 + a2b2 + ... + a_kb_k.

Then we define f(x) = a1 f(b1) + a2 f(b2) + ... + a_k f(b_k).

Checking the conditions, we can see this function works.

Yes, good point, there must be some condition I forgot. It might have been that the function is monotonic.

OK, so assuming the function is monotonic, how can you prove it must be f(x)=kx? I am fairly sure this is true; a continuity argument seems to work. But what would be a simple, more direct proof?

Well its just the same. You've proven it for the rationals, and for any irrational value you just take a rational on either side, and by the monotonicity it must lie somewhere between those values.


analytic geometry
geometry
circumcircle
conics
ellipse
geometry proposed

Dear Mathlinkers,
let ABC be a triangle, Q, Q' the first and second Brocard's points of ABC, K the Lemoine's point of ABC and K# the isotomcomplément of K wrt ABC.
Prove that K# is the midpoint of QQ'.

Sincerely
Jean-Louis

We'll use barycentric coordinates WRT $\triangle ABC.$ Let $\delta_{AB}$ be the circle passing through $A,B$ tangent to $BC.$ Its equation satisfies $A(1:0:0)$ and $B(0:1:0)$ $\Longrightarrow$ $\delta_{AB} \equiv a^2yz+b^2zx+c^2xy-rz(x+y+z)=0.$ Since $\delta_{AB}$ is tangent to $x=0$ at $B(0:1:0),$ then $z(a^2y-ry-rz)=0$ has double root equal to zero $\Longrightarrow$ $r=a^2.$

$\Longrightarrow \ \delta_{AB} \equiv a^2yz+b^2zx+c^2xy-a^2z(x+y+z)=0$

Similarly, we get the equation of the circle $\delta_{BC}$ passing through $B,C$ tangent to $CA$ and the circle $\delta_{CA}$ passing through $C,A$ tangent to $AB.$

$\delta_{BC} \equiv a^2yz+b^2zx+c^2xy-b^2x(x+y+z)=0$

$\delta_{CA} \equiv a^2yz+b^2zx+c^2xy-c^2y(x+y+z)=0$

Thus, circles $\delta_{AB},\delta_{BC}, \delta_{CA}$ concur at the 1st Brocard point $\Omega_1 \left (\frac{_1}{^{b^2}}:\frac{_1}{^{c^2}}:\frac{_1}{^{a^2}} \right).$ Analogously,  the another triad of circles $\delta_{BA},$ $\delta_{CB},$ $\delta_{AC}$ concur at the 2nd Brocard point $\Omega_2 \left (\frac{_1}{^{c^2}}:\frac{_1}{^{a^2}}:\frac{_1}{^{b^2}} \right).$

Midpoint of $\overline{\Omega_1\Omega_2}$ is then $\left (\frac{_1}{^{b^2}}+\frac{_1}{^{c^2}}:\frac{_1}{^{c^2}}+\frac{_1}{^{a^2}}:\frac{_1}{^{a^2}}+\frac{_1}{^{b^2}} \right),$ which indeed coincide with the coordinates of the isotomcomplement of the symmedian point $K(a^2:b^2:c^2).$

Let $K'$ be the  isotomic conjugate of $K$ with relation to $\triangle ABC,$ $\triangle A_1B_1C_1$ the First brocard triangle of $\triangle ABC.$ $M,$ $N,$ $L$ respectively the midpoints of $BC,$ $AC,$ $AB,$ $M',$ $N',$ $L'$ respectively the midpoints of $B_1C_1,A_1C_1,A_1B_1.$ Then $AA_1$ pass through $K'$ and easy to prove the midpoint of $QQ',$ $M$ $M'$ are collinear, it means that $MM',$ $NN',$ $LL'$ are concurrent at the midpoint of $QQ'$ and use congruent we can get $M'N=M'L.$ $\angle NM'L=\angle BA_1C,$ so $MM' \parallel AA_1$ i.e. $MM' \parallel AK',$ so the midpoint of $QQ'$ is the isotomcomplément of $K.$

PS. Let $O$ and $H$ respectively be the circumcenter and orthocenter of $\triangle ABC,$ then $OK \parallel HK'.$  I let Fang-jh posted my proof is here: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=182853[/url]

Notice that the isotomcomplement of $ P $ is the center of the conic which is tangent to $ BC, $ $ CA, $ $ AB $ at  $ AP $ $ \cap $ $ BC, $ $ BP $ $ \cap $ $ CA, $ $ CP $ $ \cap $ $ AB $ . Since the Brocard ellipse is an conic tangent to $ BC, $ $ CA, $ $ AB $ at the vertices of the cevian triangle of the Symmedian point, so the isotomcomplement of the Symmedian point is the center of the Brocard ellipse which is the midpoint of the first Brocard point and the second Brocard point .


ratio
geometric series
calculus
calculus computations

If the coefficients $ \left\{a_n\right\}$ of a power series $ \sum_{n\equal{}0}^\infty a_nx^n$ form a bounded sequence show that the radius of convergence is at least 1.

Suppose $ |a_n|\le M$ and $ |x|<1.$ Compare $ \sum_n|a_nx^n|$ to the convergent geometric series $ M\sum_n|x|^n.$ That's all you need.

All arguments about the radius of convergence of a power series ultimately come back to comparisons with geometric series.

(And if you want to quote the root test or ratio test here, you're also comparing to a geometric series.)


x,y,z>0 and x+y+z=1
prove that:
sqrt[x+((y-z)^2)/12)]+sqrt[y+((z-x)^2)/12)]+sqrt[z+((x-y)^2)/12]<=sqrt(3)
 with sqrt is a square root

I remember it! I was trying to solve it for 2 hours with null result.


Two sides of a triangle measure eight centimeters and three centimeters. If the length of the third side is j centimeters, how many possible integer values are there for j?

how would you do this problem?

[hide]Apply the triangle inequality. The sum of two sides of a triangle must be greater than the third side. Thus, the possible third side lengths are 6, 7, 8, 9, 10, and there are 5 such values.[/hide]

[hide]For a triangle to be a valid triangle, each set of two sides must be longer than the third side so that the sides can connect.

This is called the triangle inequality theorem, which states that if the sides of the triangle are $a$, $b$, and $c$, then $a+b>c$, $b+c>a$, and $a+c>b$.

This should be enough.[/hide]


geometry
perimeter
rectangle
ratio
trapezoid
trigonometry
trig identities

http://mcis.jsu.edu/mathcontest/geo02.pdf

Well. I did this test today, and I ghettoed a few of the ones I didn't know how to do on sight. Anyways, here are the ones I either missed, or ghettoed right and didn't know how to do.

Question Numbers:
15
24
43
44
42
40
45
48
49
50

Thanks!

EDIT: oops i forgot to post the link, woah woops

Either post the problems or move this to the Alabama forum. If you post the problems more people can help  :wink:

uhh they have pictures that are hard to draw without geometer's sketchpad, which I don't have.

so how in the world are we supposed to know how to do them if we dont have the problems???

do you have a link?

http://mcis.jsu.edu/mathcontest/geo02.pdf

oh snap. I forgot to post it oops!!!  :(

[hide]#24   Let w=width & l=length
(w+1)(l+1)=7wl
wl+w+l+1=7wl
w+l+1=6wl
2w+2l+2=12wl

(w+1)(l+2)=10wl
wl+2w+l+2=10wl
2w+l+2=9wl

(2w+2l+2)-(2w+l+2)=12wl-9wl
l=3wl
1=3w
w=1/3

From here, you should be able to calculate the length, which is 4/3. Therefore, the perimeter is 2w+2l or 10/3 or choice E. There might be a shorter way to do this, but I'm not so sure.[/hide]

[hide]15
Arc $GH$ is equal to arc $AH$.  This means that arc $AD$ is 130 degrees.  $\angle DEA$ is 80 degrees but i dont know how to get $\angle PAE$.

43
Let the left vertex be $A$, the top vertex be $B$ and the right angle be $C$.  Let the center of the larger circle be $O$ and the center of the smaller one be $P$.  Drop a line from $O$ to $AC$ at a right angle.  Drop a line from point $P$ to the line just drawn at a right angle and let them intersect at $D$.  $OD=9$ and $OP=15$ so $PO=12$.  Draw another perpendicular from $P$ to $AC$ and let the intersection be $E$.  $\triangle APE ~ \triangle POD$.  $AE=4$ so $AC=4+12+12=28$.

Drop another perpendicular from point $O$ to $BC$ and intersect at point $F$.  Let $FB=x$.

$28^{2}+(12+x)^{2}=(16+x)^{2}$ so $x=84$ so $BC=96$

$\frac{96\cdot 28}{2}=1334$

so $D$[/hide]

[quote="cincodemayo5590"][hide="44"]
WLOG $EF$=2. Label $DE=EC=x$. Drop perpendiculars from $B$ and $D$ to $CD$ and $AB$ respectively. We see two triangles form. Label $EC-FB=FA-DE=a$, so that the triangles have with area $\frac{1}{2}\cdot 2 \cdot a = a$. Now, the area of the rectangle determined by $D$, $E$, and $F$ is $2x$, and the smaller rectangle has area $2(x-m)$.

We want the ratio $\frac{AM}{MB}=\frac{x+a}{x-a}=R$. Time for some algebra:
\[p= \frac{2x+a}{2(x-a)+a}= \frac{2x+a}{2x-a}\implies a = 2x \left( \frac{p-1}{p+1}\right) \]

\[R= \frac{x+a}{x-a}=\frac{x+2x \left( \frac{p-1}{p+1}\right)}{x-2x \left( \frac{p-1}{p+1}\right)}= \frac{1+\frac{2p-2}{p+1}}{ 1-\frac{2p-2}{p+1}}= \boxed{\frac{3p-1}{3-p}\implies A}\]
[/hide] I have to go now, I'll get to the others later.  :)[/quote]

uhh...the trapezoid area yields (with h/2 canceling) DE+AF=p(FB+EC)...then CD=AB...it follows...

number 50
   since the angle on incidince equals the angle of refection each of the bounces will be simalr.  the fist takes use 3 ft in the second 6 and the third 9 ft.  the next bonuce goes only 1 ft and is simalr so the segment on BC is 1/3 of 3 so it is 5/3.  When it refects back the rest of the segment is 10/3, so the length of part on AB is 2.  Then the next bounce is 5 the next is on AB and is 8 back.  So the that segment is to meaning that AF is 10/3.

[hide="49"]Notice that the length of the belt equals the perimeter of the large hexagon plus the circumference of one of the circles.

So we have:

$d=6$

$a=3$

$s\cdot\frac{\sqrt{3}}{2}=3$

$P=6s = \frac{36}{\sqrt{3}}=12\sqrt{3}$

$C=\pi\cdot d =6\pi$

So is it $12\sqrt{3}+6\pi$?[/hide]

[hide="48"]Draw lines from the center to two adjacent verteces of the octagon.  the angle between them is 45.  by law of cosines, we get

$x^{2}=1+1-2\cos 45$

$x^{2}=2-\sqrt{2}$

$x=\sqrt{2-\sqrt{2}}$

This is the length of the hypotenuse of one of the right triangles.  divide by root 2 to get the side.

$\frac{\sqrt{2-\sqrt{2}}}{\sqrt{2}}=\sqrt{1-\frac{\sqrt{2}}{2}}$

there are 8 triangles so 4 squares.  square this and multiply by 4 to get the result:

$1-\frac{\sqrt{2}}{2}$

$4-2\sqrt{2}$

D[/hide]

[quote="rnwang2"]For some reason, I'm getting $\frac{4\sqrt{10}}{5}$ for #42.[/quote]
[hide] Since AB is a diameter, we have that Angle ACB is a right angle.  Additionally, since arc AC=arc BC, we have that AC=BC, so Triangle ABC is a right, isoceles triangle.

Thus, $AB^{2}=AC^{2}+BC^{2}\Rightarrow 36=2AC^{2}\Rightarrow AC=BC=3\sqrt{2}$.  Now, we have that Triangle ADE is similar to Triangle CDB, thus, $\frac{AD}{AE}=\frac{CD}{BC}\Rightarrow AE=\frac{2(3\sqrt{2})}{CD}=\frac{6\sqrt{2}}{CD}$.

Drop a perpindicular from C to AB; this will intersect AB at the center of the circle, call it O.  We have that OA=3 and AD=2.  Thus, DO=1.  Now, we also know that $CO=\frac{AC(BC)}{AB}=3$.  Thus, $CD=\sqrt{DO^{2}+CO^{2}}=\sqrt{10}$.

Now, we can say that $AE=\frac{6\sqrt{2}}{\sqrt{10}}=\frac{6}{\sqrt{5}}=\frac{6\sqrt{5}}{5}\Rightarrow \boxed{A}$[/hide]

[quote="rnwang2"]So what's wrong with this:

Let the center be O, and draw OC. Then COB = COD = 90. Since AB=6, the radius is 3. OC=3, and DO=1. Then DC=sqrt(10).

And from Power of a Point, ED=8/sqrt(10).[/quote]
It asked for AE, not ED.

Anyways, here is a solution for 45.
[hide="45"] Call the quaderilateral ABCD where A is the center of the small circle, B is the point of tangency between the two small circles, C is the center of the big circle, and D is the point of tangency between the two big circles.  Call the point where Circle A and Circle C are tangent "O."  We have that AO+CO=AC and AO=1 and CO=4.  Thus, AC=5.  We also have that AD is tangent to Circle C.  Thus, ADC is a right triangle where CD=4 and AC=5.  Thus, $AD=\sqrt{AC^{2}-CD^{2}}=3$.  So, $[ADC]=\frac{AD(CD)}{2}=6$.  Now, we can also say that BC is tangent to Circle A.  So, ABC is a right triangle where AB=1 and AC=5.  Thus, $BC=\sqrt{5^{2}-1^{2}}=2\sqrt{6}$.  Now, we can say that $[ABC]=\frac{BC(AB)}{2}=\sqrt{6}$.  Thus, ${[ABCD]=[ABC]+[ADC]=6+\sqrt{6}}\Rightarrow \boxed{A}$.  (By the way, if you don't know it already, then $[ABC]$ means the Area of Triangle ABC.)  [/hide]

[quote="The QuattoMaster 6000"][quote="rnwang2"]So what's wrong with this:

Let the center be O, and draw OC. Then COB = COD = 90. Since AB=6, the radius is 3. OC=3, and DO=1. Then DC=sqrt(10).

And from Power of a Point, ED=8/sqrt(10).[/quote]
It asked for AE, not ED. [/quote]
Oh, that would explain it.

[hide="15"]
Arc FG=30 (since FH and GA are parallel)
GD=30
DA=130
So wait, isn't $P=\frac{FH-DA}{2}$?
Someone point out what's wrong.[/hide]
[hide="24"]
$(x+1)(y+1)=7xy$
$x+y+1=6xy$
$x(6y-1)=y+1$
$x=\frac{y+1}{6y-1}$

$(x+2)(y+1)=10xy$
$2y+x+2=9xy$
$x(9y-1)=2y+2$

$\frac{2y+2}{9y-1}=\frac{y+1}{6y-1}$
$\frac{2}{9y-1}=\frac{1}{6y-1}$
$9y-1=12y-2$
$y=\frac{1}{3}$
$x=\frac{4}{3}$
E[/hide]

[hide="43"]
Connect the centers.
Note that the ratio of the segments tends from 24, to 6, and so on forming an infinite geometric series.
Name point D where the incircle is tangent to BC.
Using our infinte summation, AD=$24+6+...=32$.
Clearly BD=12.
So AB=$4\sqrt{55}$.
The rest follows from congruent tangents, too lazy to work, it out now, but it seems straight forward.[/hide]

I'll try the rest later, I'm going to be busy soon.

[hide="15"] Connect FA.  We have that arc FG=arc DG=arc AH=30.  Thus, arc AD=360-140-30-30-30=130.  Thus, we have that angle DFA=(arc AD)/2=65.  Now, connect DA.  We have that angle DAF=(arc DF)/2=30.  We also have that angle DAP=angle DFA=65, so angle PAF=30+65=95.  Thus, $\angle P=180-\angle PAF-\angle DFA=180-65-95=180-160=20\Rightarrow \boxed{D}$ [/hide]


linear algebra
matrix
search
abstract algebra
quadratics
geometry
3D geometry

1. s={matrices 2x2 , with natural values,det=1}. are there A,B,C in s sucvh that a)a^2+B^2=c^2 b)A^4+B^4=c^4?

2.matrix A nxn. prove that A has inverse if |aij|>sum(k=1 to n, k=/= i)|aij| for every i=1,2,...,n.

3.let A,B are two nxn matrices, symetrical with positive charactersitic roots. prove trAB>0.

4.A,B are from Mn(reals) and AB=BA. prove det(A^2+B^2)>=0.

5.A,B are 1999x1999 matrices, At means the transponsed matrix of A. AB=0. prove det[(A+At)(B+Bt)]=0.

6. A is real, invertible, symetrical nxn matrix(n>=2) with elemnet>0.what is the min number of not zero elements of A^(-1).

7.A,B nxn A+B=E, r(A)+r(B)=n. prove A^2=A, B^2=B.

8.find all possible values of rank of nxn matrix for which the elements on the diagonals are even and the others-odd.

9.A,B-nxn. A^2006=E,B^2007=E, AB=BA. prove A+B+E is invertable.

Many of those seem classical and should be already answered on that forum.

For instance 2) is a consequence of the Gershgorin circle theorem, mentioned there for instance http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=376047793&t=32062

3) must be on this forum as well

4) what about $\det(A^{2}+B^{2})= |\det(A+iB)|^{2}$?

you wrote that det(a^2+b^2)=|det(a+ib)|^2,so der(a+ib)=det(a-ib).but why is that true?

He didn't say that. As you can see:$\det( A^{2}+B^{2})=\det ((A+iB)(A-iB))=\det(A+iB)\cdot\det(A-iB)$.
If $\det(A+iB)=z\in C$ then $\det(A-iB)=\overline{z}$(you can write down the definition of the determinant if you want to check this).So $\det (A^{2}+B^{2})=z\cdot\overline{z}=|z|^{2}$ which is clearly greater than $0$.

#7: Suppose $x\in\ker A.$ Then $x=Ix=(A+B)x=Bx,$ so $Bx=x$ for $x\in\ker A.$ Similarly, $Ax=x$ for $x\in\ker B.$ These conditions also imply that if $x\in\ker A\cap\ker B,$ then $x=(A+B)x=0.$ Hence the two kernels have a trivial intersection. Then $\dim(\ker A\oplus\ker b)=\dim\ker A+\dim\ker B=n.$ That means that $\mathbb{F}^{n}=\ker A\oplus\ker b.$ (Letting $\mathbb{F}$ be whatever field we're working over - this theorem works in all fields, so it doesn't matter.)

Let $x$ be an arbitrary element of $\mathbb{F}^{n}.$ We can uniquely write $x=u+v$ for $u\in\ker A$ and $v\in\ker B.$ Then $Ax=Au+Av=0+v=v,$ and $A^{2}x=Av=v.$ Thus, $A^{2}=Ax$ for all $x.$ Similarly, $B^{2}x=Bx$ for all $x.$

[quote]8.find all possible values of rank of nxn matrix for which the elements on the diagonals are even and the others-odd.[/quote]
Consider this as a matrix over $\mathbb{Z}/2\mathbb{Z}.$ This is the matrix with zeros on the diagonal and ones elsewhere, which we can write as $J-I,$ where $J$ is the matrix of all ones. Considered as a matrix over $\mathbb{Q},$ $J-I$ has eigenvalues $n-1,-1,-1,\dots,-1$ and hence determinant $(-1)^{n-1}(n-1).$ If $n$ is even this is nonzero mod 2. Hence:

If $n$ is even, the rank of this matrix is $n.$

Now, if $n$ is odd, the rank in $\mathbb{Z}/2\mathbb{Z}$ is constrained by the fact that $n-1$ is a simple zero of the characteristic function. That makes the rank $n-1.$ If we go back to $\mathbb{Q}$ we could easily make the matrix nonsingular - the determinant could be a nonzero even number. Hence:

If $n$ is odd, the rank of this matrix is either $n-1$ or $n.$

3: (Proof provided by a colleague of mine who knows his matrix theory.)

We are given that $A$ and $B$ are real, symmetric, positive definite matrices and we are asked to prove that $AB$ has positive eigenvalues.

First: We may assume WLOG that $A=D$ is diagonal.

How to see that: There exists orthogonal $Q$ such that $Q^{T}AQ=D.$ Let $Q^{T}BQ=C.$ Then $A=QDQ^{T}$ and $B=QCQ^{T},$ so that $AB=QDQ^{T}QCQ^{T}=QDCQ^{T}.$ Since $AB$ is similar to $DC,$ they have the same eigenvalues, and $C$ is symmetric and positive definite.

So from now on, the problem is to show that $DB$ has positive eigenvalues given that $D$ is positive and diagonal and $B$ is symmetric and positive definite.

It makes sense to talk about $D^{1/2}$ and $D^{-1/2}.$ $DB$ similar to, hence has the same eigenvalues, as $D^{-1/2}DBD^{1/2}=D^{1/2}BD^{1/2}.$ Now consider the quadratic form generated by that matrix. For any $x\ne0,$

$x^{T}D^{1/2}BD^{1/2}x=(D^{1/2}x)^{T}B(D^{1/2}x)>0$ since $B$ is positive definite and since $D^{1/2}x\ne 0$ since $D^{1/2}$ is invertible. Hence, $D^{1/2}BD$ is positive definite, and it follows that $BD,$ which has the same eigenvalues, has positive eigenvalues.

Comment on #5: the fact that 1999 is an odd number is important. Suppose we have that $A$ and $B$ are $n\times n$ matrices when $n$ is even. Then the following block-structured matrices with square blocks provide a counterexample:

$A=B=\begin{bmatrix}0&I\\0&0\end{bmatrix}.$

But that counterexample can't be used if $n$ is odd.

[quote="Kent Merryfield"][quote]8.find all possible values of rank of nxn matrix for which the elements on the diagonals are even and the others-odd.[/quote][/quote]
Consider this as a matrix over $\mathbb{Z}/2\mathbb{Z}.$ This is the matrix with zeros on the diagonal and ones elsewhere

this is not true-on both diagonals there are even numbers

[quote="martin1985"]9.A,B-nxn. A^2006=E,B^2007=E, AB=BA. prove A+B+E is invertable.[/quote]
I assume this is over some subfield of $\mathbb{C}$.  Since $A$ and $B$ have finite order (not divisible by the characteristic of the field), they are diagonalizable.  Since they commute, they are simultaneously diagonalizable.  So we can assume $A+B+E$ is diagonal, with each diagonal entry of the form $\lambda+\zeta+1$, where $\lambda^{2006}=\zeta^{2007}=1$.  If $\zeta=-1-\lambda$, then $\lambda$ and $\lambda+1$ would both have norm 1, but this is only possible if $\lambda$ is a primitive cube root of unity, which it isn't.

[quote="martin1985"]1. s={matrices 2x2 , with natural values,det=1}. are there A,B,C in s sucvh that a)a^2+B^2=c^2 b)A^4+B^4=c^4?[/quote]
If $A=\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$, with $\det A = ad-bc=1$, then

$tr(A^{2})=a^{2}+2bc+d^{2}=a^{2}+d^{2}+2(ad-1)=(a+d)^{2}-2=(tr A)^{2}-2$,

and so $tr(A^{4})=(tr(A)^{2}-2)^{2}-2$, which is either 2 or 7 mod 8.  So $A^{4}+B^{4}=C^{4}$ is impossible.  I assume $A^{2}+B^{2}=C^{2}$ is possible, or you wouldn't have asked part (b).

Edit: For example, $A=\left(\begin{array}{cc}0&1\\-1&-1\end{array}\right)$, $B=\left(\begin{array}{cc}1&1\\-1&0\end{array}\right)$, $C=\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)$.

[quote="Eigenray"][quote="martin1985"]9.A,B-nxn. A^2006=E,B^2007=E, AB=BA. prove A+B+E is invertable.[/quote]
I assume this is over some subfield of $\mathbb{C}$.  Since $A$ and $B$ have finite order (not divisible by the characteristic of the field), they are diagonalizable.  .[/quote]

 A^2006=E=>A has finite order(because E is identity elemnent in the group of these matrices).why then A is diagonalizable?

[quote="Eigenray"][quote="martin1985"] A^2006=E=>A has finite order(because E is identity elemnent in the group of these matrices).why then A is diagonalizable?[/quote]
In general, a matrix is diagonalizable (over some field) iff its minimal polynomial splits into distinct linear factors (over that field).  And if $A^{n}=E$, then the minimal polynomial of $A$ divides $x^{n}-1$, which has distinct roots over $\mathbb{C}$.[/quote]

1)is it true that minimal polynomial of A is a polynomial, which has value at $A$ the zero matrix and is of min possible degree?
2)how do u know that the min.polynomial of $A$ divides any other polynomial with value the zero matrix at $A$.
3)how do u prove that if min.pol. of $A$ divides a polynomial with distinct real roots then A is diagonalizable?

[quote="martin1985"]1)is it true that minimal polynomial of A is a polynomial, which has value at $A$ the zero matrix and is of min possible degree?
2)how do u know that the min.polynomial of $A$ divides any other polynomial with value the zero matrix at $A$.[/quote]
The set of polynomials $f$ with $f(A)=0$ forms an ideal of the ring $F[x]$.  Since $F[x]$ is a principal ideal domain, this ideal is generated by a single polynomial $m$, which means that $f(A)=0$ iff $m|f$.  In particular it has the smallest degree of any annihilating polynomial of $A$.  This $m$ is the minimal polynomial of $A$, and is unique if we take $m$ to be monic.
[quote]3)how do u prove that if min.pol. of $A$ divides a polynomial with distinct real roots then A is diagonalizable?[/quote]
If the minimal polynomial splits into linear factors, then $A$ has a Jordan normal form, and the minimal polynomial of $A$ is the lcm of the minimal polynomials of each Jordan block.  Since the minimal polynomial of a Jordan block of size $k$ has minimial polynomial $(x-\lambda)^{k}$, it follows that the minimal polynomial of $A$ has distinct roots iff each Jordan block has size 1, i.e., iff $A$ is diagonalizable.

These facts should be in any linear algebra textbook.


inequalities
number theory proposed
number theory

$a,b,c,d$ are positive integers different  from $0$ and stisfying :
$a<b \leq c <d$    and    $ad=bc$   and    $\sqrt{d}- \sqrt{a} <1$
Prove that $a$ is a perfect square.

The problem is realted to an old problem from another olympiad wich states that there doesnt exist 4 numbers $a<b\leq c<d$ in  the set $N^2+1,N^2+2..., (N+1)^2$ with $ad=bc$. Now i think this problem is from Romania 1998....

Anyway, I have found long and boring solution!

I know this problem long enough, so Romania 1998 is not an original source  :?

Maybe I didn’t understand the problem.
But from condition we can suppose that there exists $x,y,z,k$ such that
$a = xy, b = xz, c = yk , d = zk$ and $y < z, x < k$.

Hence $1 \geq \sqrt {d}- \sqrt {a}= \sqrt {zk}-\sqrt {xy}\geq\sqrt {(x+1)(y+1)}- \sqrt {xy}$

But this is true iff $x = y$ or $a$- perfect square.

Myth! do you have a nice solution? or should I post mine?

Nice solution Prowler!, mine also sues same initial step, but then i solve it in more complicated way... Just to clarify something for other people, by Cauchy, the last inequality in Prowler's proof would be in the other way!, so we must have equality in this inequality and therefore $x=y$.

Indeed, it is based on four factors theorem.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=ad%3Dbc&t=5492

Pierre.


Καλή χρονιά σε όλους
Σε εγγράψιμο τετράπλευρο κάθε διαγώνιος ορίζει με τις πλευρές του δύο τρίγωνα με ακτίνες των εγγεγραμένων κύκλων τους τις r1,r2,r3,r4. Tα έγγεντρα των τριγώνων με κοινή πλευρά την κάθε διαγώνιο έχουν αποστάσεις d1 και d2 . Αν d είναι η μικότερη ή ίση απ τις d1,d2 ν.δ.ο. d^2>=4[(r1.r2.r3.r4)^(1/2)]
Για τον Sil

[color=blue]ΑΣΚΗΣΗ (κ.Χρήστος Πατήλας )
Σε εγγράψιμο τετράπλευρο κάθε διαγώνιος ορίζει με τις πλευρές του δύο τρίγωνα με ακτίνες των εγγεγραμένων κύκλων τους τις $ r_1,r_2,r_3,r_4$ . Tα έγγεντρα των τριγώνων με κοινή πλευρά την κάθε διαγώνιο έχουν αποστάσεις $ d_1$ και $ d_2$ .
Αν $ d$ είναι η μικότερη ή ίση απ τις $ d_1,d_2$  ν.δ.ο.  $ d^2\geq 4\sqrt{r_1r_2r_3r_4}$ [/color]

ΛΥΣΗ
Έστω $ I_1 ,\ I_2,\ I_3,\ I_4$ τα έγγεντρα των τριγώνων $ ADC,\ BAC,\ BAD,\ BCD$ .

Συνθετικά (ή και με τη χρήση μιγαδικών) μπορούμε να δείξουμε ότι το $ I_1I_3I_2I_4$ είναι ορθογώνιο . Επομένως θα έχουμε
$ d_1\equal{}d_2\equal{}d$ . Επιπλέον από την τριγωνική ανισότητα έχουμε $ d\geq r_1\plus{}r_2$ και όμοια $ d\geq r_3\plus{}r_4$

άρα από AM-GM  $ d^2\geq (r_1\plus{}r_2)(r_3\plus{}r_4)\geq 4\sqrt{r_1r_2r_3r_4}$ , το ζητούμενο

***
Ευχαριστώ πολύ για την αφιέρωση αυτής της όμορφης και κομψής άσκησης  :)
Αν δεν καταλάβει κάποιος ότι $ d_1\equal{}d_2\equal{}d$ η άσκηση είναι αρκετά πιο tricky . Σε εκείνη την περίπτωση θα έπρεπε να
πάρουμε από τριγωνική ότι (έστω d_1=min)  $ d_1\geq r_1\plus{}r_2$ από τριγωνική και μετά να δει κάποιος από το θεώρημα Ιαπώνων ότι $ r_1\plus{}r_2\equal{}r_3\plus{}r_4$  και να χρησιμοποιηθεί η ταυτότητα $ d_1^2\geq (r_1\plus{}r_2)^2\equal{}(r_1\plus{}r_2)(r_3\plus{}r_4)$ κτλ

και επειδή γνωρίζω και τον χρόνο που χρειάστηκες να την λύσεις ,μπράβο Sil

Σιλουανέ σε ποιά βιβλία πας και βρίσκεις αυτά τα θεωρήματά? :)  (Εκει που λες θεωρημα Ιαπωνων)

Σε κάθε κλασσικό βιβλίο γεωμετρίας υπάρχει ...  :wink:


When you add up the first 3 perfect numbers, what do you get?

530, the first perfect numbers are 6, 28, and 496

they are all factors kinda.

Numbers for which all of the divisors of the number added up are equal to the number. For $28, 1+2+4+7+14=28$

[hide]The first three perfect numbers are 6, 28, and 496. Then, when you add them up, you get 530.[/hide]

yep thats what i said earlier. I heard about perfect numbers a long time ago duno where since I Wasnt on math team back then.


function
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area of a triangle
calculus computations

Let S be the set of all continous function on [0,1], with norm:
||f(x)||= integral f(x) from 0 to 1. Show that S is not complete.

Have A Great Day!

The norm should be $ \|f\|\equal{}\|f\|_1\equal{}\int_0^1|f(x)|\,dx.$ Your statement should have included the absolute value.

Consider this sequence of functions:

$ f_n(x)\equal{}\begin{cases}0&0\le x\le \frac12\\ n\left(x\minus{}\frac12\right)&\frac12<x<\frac12\plus{}\frac1n\\ 1&\frac12\plus{}\frac1n\le x\le 1\end{cases}.$

(This only makes sense for $ n\ge 2.$)

Or to make it more pictorial: this is the piecewise linear continuous function with vertices at $ (0,0),\left(\frac12,0\right),\left(\frac12\plus{}\frac1n,1\right),$ and $ (1,1).$

The area between the curves is the area of a triangle and is easy to compute. In particular, $ \|f_n\minus{}f_m\|\equal{}\frac12\left|\frac1n\minus{}\frac1m\right|.$ Since this can be made arbitrarily small by making $ n$ and $ m$ sufficiently large, we have that $ \{f_n\}$ is Cauchy sequence in this normed vector space.

I'll let you think about why there is no $ f\in C[0,1]$ such that $ \|f_n\minus{}f\|\to 0.$


Can someone help me solve this one??? I've been trying forever...

A Magician asks you to draw 5 random card from a standard card deck.  You picked the 5 cards and give them to the magician's assistant.  The assistant puts a card back into the deck and lets you see which card she had put in, and then she puts the remaining 4 card into a pile on the table.  During these process the magician was blind folded and did not see anything.  The magician then pick up the pile, reads it, and find the card that the assistant had put in the deck.  (The deck is shuffled by you after the assistant puts the card back in).  Assume that neither the magician nor the assistant has supernatural power, what is the trick?

Harder version: you choose which card to put back in the deck.

This is a fun card trick.

[hide="my solution"]The card that you put back in the deck must be the same suit as one of the others and it must be 6 or less more than the other card of the same suit (where you can wrap around).
Put the card of the same suit on top. Then you will use the other 3 to form a number from 1-6 which will be added to the first card to get the card in the deck. Each of the three remaining cards will be numbered from lowest to highest. If 2 cards have the same number then the suit will determine which is lower. Here is the order of suits from lowest to highest: Clubs, diamonds, hearts spades. Each combination creates a number from 1-6: 123=1 132=2 213=3 231=4 312=5 321=6.[/hide]

yeah, that works for the first version

Anyone came up with a solution for the second one??? (where the audience/the person who watch the trick pick the card)


inequalities
number theory
prime numbers

Prove that for every natural number $n \geq 4$, there exists at least one natural number $m$, such that
\[n! < m < (n+1)! \]
and $n^{3}| m$.

Not always true, consider $n=2 \to n!=2,\; (n+1)!=6,\;n^{3}=8$.
Also consider $n=1$, where $n!=n^{3}=1$.
Also consider $n=3$, where $(n+1)!=24<27=n^{3}$

I think those are the only exceptions.

Apart from that, for composite numbers excluding 4, $n|(n-1)! \implies n^{2}|n! \implies n^{3}|n(n!)$, so for composite numbers, $m=n(n!)$ works. Checking, the case of $4$ works.

I'm working on the case for prime numbers excluding $2$ and $3$ :lol:

I'm very sorry, I just realised that I forgot to mention the constraint on $n$. In the problem, $n \geq 4$.

A non-constructive proof is much simpler.  Consider the set $A=\{n!+1,n!+2,\dots,(n+1)!-1\}$.  If the size of $A$ is at least $n^{3}$, then since all the integers in $A$ are consecutive, some multiple of $n^{3}$ must belong to $A$.  This means that

$n\cdot n!-1\ge n^{3}$

for all integers $n\ge 4$.  This is obvious, but if necessary, this inequality can be rigorously proved by a few cases of induction.


function
geometry
geometric transformation
reflection
trigonometry

I was just wondering what happens to a function, say f(x), when you take the absolute value of f(x). How does the graph change?

For example Lets say $f(x)=x$ It would be a diagonal crossing the origin. But with $f(x)=|x|$ when the values of x are negative it would be reflect over the x axis making a "v".

[quote="AstroPhys"]$f(x)$ can not take on negative values.[/quote]

to clarify,
any negative values output by f(x) will be reflected over the x axis and be positive

so take $f(x) = \sin x$

$|f(x)|$ would be a series of humps as the bottom half of each wave would get reflected up


[quote="lund"]Does anyone have an idea about the distribution of scores on the BC exam.  Do you have to get a perfect score for a 5? or can you afford to miss a couple? Any additional information on distribution of scores would be very helpful.[/quote]

lol...perfect score :D  I forgot what you have to get but it's fairly 'easy' (relative to other AP's) to get a 5...especially if you're good at math and motivated to study on your own.  But, yeah, you definitely don't have to get a perfect to get a 5...you probably have to get about 80% but I don't really know exactly.

even with a 70% you're in good standing for a five, i believe.

i remember on a practice one last year i omitted half the essay section bc i didnt feel like it, and still obtained a five. so, assuming that you have a strong aptitude for math by being on this forum, you shouldnt have to worry about much.  :)

Which book does your school use for AP Calculus?
I want to start studying in the summer.

We used [i]Calculus[/i] by Finney Demana Waits and Kennedy.

I wouldn't say it is the most difficult and helpful text but it gets the points accross as long as you work a little extra.

[quote="rain"]Is it harder to get a 5 on the AB exam than the BC? (that's what i heard)[/quote]

It COULDN'T be harder, because of how the tests are constructed and validated. It might be less common, statistically, just because of the differing populations of students who take the two different tests.

Well, BC covers more material.  But, in my experience, I found that the BC tests were easier once I had learned that extra material.  So once I learned all the curriculum, I found the BC tests easier than the AB tests.  I think this is because the problems exclusively from the BC curriculum are more straightforward (if you know the material).

You certainly don't need a high raw score to get a 5.  I know of people who scored in the 60%-70% range and still earned 5's.  [The same is true for AP Chemistry].

[url=http://www.collegeboard.com/student/testing/ap/calculus_bc/dist.html?calcbc]the college board BC distribution site[/url] shows that ~ 43% of the people taking the BC calc exam got a 5. you can browse through the site an see about other exams too :)

Yeah, "AB" and "BC" are fitting names because the BC is like 1.5*AB, so it's actually not necessary to take two years of calc for the purposes of AP Calc BC. You can miss a few questions, I think, but if you've taken 2 years of a reasonable calc course, I doubt there's anything you won't ace.

[quote="cubekid"]how many people here have taken ab and then bc? i know a lot of schools do that, but mine doesn't let you (we take either ab or bc, not both)[/quote]

At my school, most people do that. I skipped AB because I should've taken it freshman year, but only one or two do every year. But the math department at my school changed so now Advanced Calculus = Calc BC + a little multivariable. Pretty much if you can do math (and you try... a couple of practice tests mabye) you can get a 5 on the BC exam.

I was lazy and decided not to take AP Calc freshman year.  So at my school, we normally do AB one year then BC the next.  I was in AB last year (10th grade), but then decided that BC looked easy, so I just studied the additional material for about one weekend, signed up for the BC test instead of the AB, and got a 5 on both parts.  Yes, it really is that easy.  (and I got above 100 out of 108 on all practice test beforehand)  So you should be fine.

[quote="lund"] Do you have to get a perfect score for a 5? or can you afford to miss a couple? [/quote]

On the practice tests that we scored last year, out of 108 (the multiple choice and the free response, weighted sum) a 5 tended to be around 63-75 by year.  :)

[quote="rain"]are they picky on how u show ur work for the free response? im going to own the multiple choice but im scared if i skip a step or something for the frq[/quote]

What they do is assign point values to each important step (there are perhaps one or two per section of a Free Response question), and if that step is there they'll give you the point, and if the step isn't there they won't.  They won't be picky about skipping arithmetic steps, but the important results (citing the Mean Value Theorem or whatever) have to be there.

By the way, if you get a good enough score on the Multiple Choice you need less than half of the total Free Response points to guarantee a 5.  :)

what does good enough mean..like 40? i think i can do that including my stupid mistakes
what about 35/30? just curious

I've got Thomas and Finney's Calculus 9th Edition... I'm not really clear abt the syllabus ... can somone please tell precisely which chapters from this book need to be done for Calculus BC,,,,, Thanks!!!

I love the Thomas Finney book im pretty sure the bc material starts at methods of integration.

I wanted the same info as above.... but I couldnt find it on the link you gave --- can you kindly be more specific??! Thanks!

Well....look at the syllabus/course information in BC.  Compare that to the chapters in your book.  And there ya go!  If that's not enough, get the Princeton Review for AP Calculus and that book covers all the sections in AB and BC.  You can use it to get an idea of what chapters you should work in your book.

If I had any more specific info, I would give it to you.  But on the other hand...AP is supposed to be more like college...you need to learn to figure stuff on your own.

I got a 5 on last year's BC exam (and the AB subsection).  Many of the questions are not difficult at all, but it could be very difficult to get a 5, depending on the test.  We covered series in one week last year and did not cover polar.  Just my luck, I had one series and one polar problem out of the 6 free-response.  I think we also got jipped out of about 15 days of class time for standardized testing and assemblies and stuff too, so with those 15 days, we probably would have finished the material more comfortably.

You're looking at around 70% on the test to get a 5.  Time is a pretty big factor if you know how to do the stuff, don't spend too long perfecting one question.  "Down and dirty" then going back would probably work better since there is a point value assigned to each portion of each free response.  With the exception of those two free response questions though, I didn't think the rest of the test was too bad.  I think I got 5s on all of my practice exams as well.

If you study the material you've forgotten before the test, you shouldn't have too much trouble for as motivated of a student as you appear to be.  Calc really builds on itself, so if you learn it well the first time, it should stick.

Textbook:  Finney Demana Waits and Kennedy as well.  There's a big version (full text) and smaller version (first 15 chapters or so?).  We used the smaller version for BC, it covers all of the necessary material.

I suggest going through past AP free response sections.  I did that and immediately saw a trend that usually they have 1 question on sequences/series and another on polar/parametric/vector (one or more of the those 3 topics).  Thus I made sure I had that stuff down.  Come test day I basically didn't find any surprises on the test.

If you've already taken some calc, I don't see why not just do BC.

[quote="1/(ln x)"][quote="Silverfalcon"]How much is AP Calc BC emphasized? I plan to take AP Calc AB next year (I'm in calculus now but this year, we only covers half of the materials --> my school isn't really advanced). For my senior year, I won't be preparing for anymore calculus courses because I'll be taking class at local college. So, which one is BETTER? AP Calc AB or BC?[/quote]

I would say taking AP Calc BC is better. My math teacher said that 60% of the AP Calculus BC test is AB stuff. You'll do fine in BC, you probably already know most of the AB stuff.[/quote]

I have this really easy book called "Idiot's Guide to Calculus."  :D I'm just going to go through this book and cover basic stuffs (since my calculus class does not cover series and several stuffs at all) and study by my own or something. It sounds like it's not too bad to do 5 on this test.

However, they do release certain years' exams. I know my Calc BC teacher had the 1998 and 2003 tests, and it was legal. I believe he got it off AP Central.

[quote="paladin8"][quote="satyam"]Where can I find multiple choice questions for practice exams of AP Calculus? I checked on this website: [url]http://apcentral.collegeboard.com/members/article/1,3046,152-171-0-8031,00.html[/url], but they only have free-response questions there.[/quote]

College Board does not publish multiple choice questions; in fact, people are NEVER supposed to discuss them, and this would not be meaningful if they disclosed them. In the course manual, there are some practice multiple-choice problems. Review books also have some.[/quote]

why is that? i mean its a lil strange they are so liberal about the fqr but restraining on the mc

[quote="rain"][quote="paladin8"]why is that? i mean its a lil strange they are so liberal about the fqr but restraining on the mc[/quote][/quote]

Because it's difficult to come up with lots and lots of new multiple choice questions every year.  For the free response, they only need a handful of questions for each administration of the test.


calculus
derivative
function
limit
linear algebra
matrix
complex numbers

If f=u+iv where u and v are real valued functions.

Does it always hold that f'(x)=u'(x)+i v'(x)?

I am confused because for a real function of a complex variable, say g, g(a) is always 0 or it doesn't exist.  (Since $ \lim{\frac {g(a \plus{} h) \minus{} g(a)}{h}}$ should be equal to $ \lim{\frac {g(a \plus{} ih) \minus{} g(a)}{ih}}$ , h is real)
So f'(a)=u'(a)+iv'(a)=0 if f'(a) exists.

But from the definition of f'(x), I know it does not need to be the case. $ \lim{\frac {f(a \plus{} h) \minus{} f(a)}{h}}$ can very well be equal to $ \lim{\frac {f(a \plus{} ih) \minus{} f(a)}{ih}}$ without being zero.

Careful; you need to specify whether $ f : \mathbb{R} \to \mathbb{C}$ or $ f : \mathbb{C} \to \mathbb{C}$.  In the former case, that's perfectly valid.  In the latter case, neither $ u$ nor $ v$ satisfy the Cauchy-Riemann equations unless they're both zero.

The real part of a complex analytic function is smooth as a function from $ \mathbb{R}^2$ to $ \mathbb{R}$; the right notion of derivative is the gradient. It's not usually complex analytic.

Now, here's an interesting perspective:
Write the derivative of a function $ g(x,y)\equal{}(g_1(x,y),g_2(x,y))$ from $ \mathbb{R}^2$ to itself in teh usual way as $ g'\equal{}\begin{bmatrix}\frac{\partial g_1}{\partial x}&\frac{\partial g_1}{\partial y}\\ \frac{\partial g_2}{\partial x}&\frac{\partial g_2}{\partial y}\end{bmatrix}$.
Let $ u$ and $ v$ be differentiable functions from $ \mathbb{R}^2$ to the subspace $ \mathbb{R}\cdot (1,0)$. Let $ \mathbf{i}\equal{}\begin{bmatrix}0&\minus{}1\\1&0\end{bmatrix}$.
Then the derivative of $ u\plus{}\mathbf{i}\cdot v$ is $ u'\plus{}\mathbf{i}\cdot v'$. $ u\plus{}\mathbf{i}\cdot v$ can be interpreted as a function from $ \mathbb{C}$ to itself in the obvious way (the point $ (a,b)$ is $ a\plus{}bi$), and it has a complex derivative exactly when that real derivative $ u'\plus{}\mathbf{i}\cdot v'$ is of the form $ a\plus{}b\mathbf{i}\equal{}\begin{bmatrix}a&\minus{}b\\b&a\end{bmatrix}$

Thank you guys!!

Yes, f is a mapping from C to C.

So does it mean that if u'(a) and v'(a) both exist, then f'(a) must be zero?


floor function
inequalities
combinatorics unsolved
combinatorics

[quote="PaulS"]$ (k!)!$  is divisible by $ (k!)^{(k \minus{} 1)!}$.[/quote]

A number-theoretic proof:

The greatest power of $ p$ that divides $ (k!)!$ is given by

$ \sum_{n \equal{} 1} \left\lfloor \frac {k!}{p^n} \right\rfloor$

whereas the greatest power of $ p$ that divides $ (k!)^{(k \minus{} 1)!}$ is given by

$ (k \minus{} 1)! \sum_{n \equal{} 1} \left\lfloor \frac {k}{p^n} \right\rfloor$.

The inequality $ \left\lfloor \frac {k!}{p^n} \right\rfloor \le (k \minus{} 1)! \left\lfloor \frac {k}{p^n} \right\rfloor$ is obvious.  :)

This question has been asked numerous times  :)  Can the mods find it? i'll try to myself  :)
[hide="Hint"]
Think abt arranging the things in rows and columns
[/hide]

EDIT: Why isnt the hide thing working? What shd i do for that?  :)

Using the easily proved result, that the product of n consecutive numbers is divisible by n!

(n!)! consists of n! consecutive numbers, that is n (n-1)! consecutive numbers. That is the product of  (n-1)! sets of n consecutive numbers, each of which is divisible by n!

Hence (n!)! is divisible by $ (n!)^{(n\minus{}1)!}$

Equivalent to hsbhatt's solution, we have that the multinomial coefficient $ \binom{n!}{n;n;n;\ldots;n}$, where in the denominator we have $ (n \minus{} 1)!$ $ n$s, is an integer.

madness, don't use quotation marks in hide tags.

[hide="like this"]
[hide="not like this"]

And be careful of other punctuation, too.

[/hide][/hide]


Let positive real numbers $ a,b,c$ with $ abc\equal{}1$. Prove that
$ \frac{a\plus{}b\plus{}1}{a\plus{}b^2\plus{}c^3}\plus{}\frac{b\plus{}c\plus{}1}{b\plus{}c^2\plus{}a^3}\plus{}\frac{c\plus{}a\plus{}1}{c\plus{}a^2\plus{}b^3}\le \frac{(a\plus{}1)(b\plus{}1)(c\plus{}1)\plus{}1}{a\plus{}b\plus{}c}$

[quote="jerry_smith"]Let positive real numbers $ a,b,c$ with $ abc \equal{} 1$. Prove that
$ \frac {a \plus{} b \plus{} 1}{a \plus{} b^2 \plus{} c^3} \plus{} \frac {b \plus{} c \plus{} 1}{b \plus{} c^2 \plus{} a^3} \plus{} \frac {c \plus{} a \plus{} 1}{c \plus{} a^2 \plus{} b^3}\le \frac {(a \plus{} 1)(b \plus{} 1)(c \plus{} 1) \plus{} 1}{a \plus{} b \plus{} c}$[/quote]
Nice inequality :lol:
We have: $ (a\plus{}b^2\plus{}c^3)(a\plus{}1\plus{}ab) \ge (a\plus{}b\plus{}c)^2$
So $ LHS \le \frac{\sum \ (a\plus{}b\plus{}1)(a\plus{}1\plus{}ab) }{(a\plus{}b\plus{}c)^2}$.
On the other hand :$ \sum \ (a\plus{}b\plus{}1)(a\plus{}1\plus{}ab)\equal{} (a\plus{}b\plus{}c)(a\plus{}b\plus{}c\plus{}3\plus{}ab\plus{}bc\plus{}ca) \equal{}(a\plus{}b\plus{}c)((a\plus{}1)(b\plus{}1)(c\plus{}1)\plus{}1)$.
Then,we have Q.E.D :)


group theory
abstract algebra
symmetry
superior algebra
superior algebra unsolved

In a group ([i]G[/i], *), its center is the subset:

[i]Z[/i]([i]G[/i]) = { [i]x[/i] from [i]G[/i] | [i]xy[/i] = [i]yx[/i], for all [i]y[/i] from [i]G[/i] }

Prove that [i]Z[/i]([i]G[/i]) is a subgroup of [i]G[/i].

Because $ e\in Z(G)$ we have $ Z(G)\not \equal{} \emptyset$. Now, if $ x,y\in Z(G)$ then easy see that $ xy^ {\minus{} 1}\in Z(G)$, too.

From my understanding of the problem, y is an element of G but it does not have to be an element of Z(G).  Z(G) is the set of x, where
xy = yx for all y in G.

The way I thought it worked would be for example: Z(D4) = {e, r^2} because in D4, order does not matter when combining with r^2, but it does matter with other elements of D4.

D4 --> Dihedral group with 4 lines of symmetry (ie, square)

If I am interpreting this wrong, could someone please elaborate?

N.T.Tuan was not referring to the $ x$ and $ y$ in the set notation. The subgroup test was being used. Let $ a,b\in Z(G)$. Since $ Z(G)$ is nonempty, the subgroup test says that if $ ab^{\minus{}1}\in Z(G)$ then $ Z(G)$ is a subgroup of $ G$.


Find all functions  $f$ with  $|f(x) + f(y)| = |x + y |$ for all $x , y \in IR$

 [u]babis[/u]

Putting $y=-x$ we get

$|f(x)+f(-x)|=0\implies f(-x)=-f(x)$. Hence the function is odd and $f(0)=0.$

Now

$|f(x)+f(0)|=|x+0|\implies |f(x)|=|x|$. This means that for any given $x$, $f(x)$ is either $x$ or $-x$.

Assume that there exist non-zero $a,b$ such that $f(a)=a$ and $f(b)=-b$. Then $|a-b|=|a+b|$, which gives either $a=0$ or $b=0$. Hence, such $a,b$ don't exist, which means that for [i]all[/i] real numbers $f(x)=x\lor f(x)=-x$, and these are all solutions.

(Note to Stergiu: You can get the letter $\mathbb{R}$ by the code \mathbb{R})

Exactly my solution !
 But after having sent the problem , I found in a book one more solution , a little  better(!). i' ll  be waiting  for the answer(new solution) by other MLinkers  too.

 Thanks for        $\mathbb{R}$  , $\mathbb{Z}$ , $\mathbb{N}$ , $\mathbb{Q}$  :)

  [u]Babis[/u]

A simplification to my solution: putting $y=x$ we obtain $|f(x)|=|x|$ directly.

Let $y=-z$ and get $\frac{|f(x)-f(z)|}{|x-z|}=1$ for all $x,z\in\mathbb{R}$ and $x\ne z$. But if we take limit, we get $\lim_{z\rightarrow x}\frac{|f(x)-f(z)|}{|x-z|}=|f'(x)|=1$. So $f(x)=\pm x+c$. But $|f(0)+f(0)|=0+0\Rightarrow f(0)=0$, so $c=0$.
I think I assumed $f$ is differentiable :blush:

I'm afraid , NO! We have not this condition ,  but even then  , last part is wrong . We ought  have that $f '$ is continous , so it has constant sign. But we have not this condition , too. Keep in mind that from
$|f'(x)| = 1$ we can not deduce that $f'(x) = 1$ for all x or $f'(x) = -1$ for all x. This is a main problem in finding functions !This can be done only if we know the continouity of $f '$. If any question , please write.

[u] To Fahrenajt[/u]

 After the new relation that you wrote, square the initial condition to get $f(x)f(y) = xy$. Here put $x=y=1$ , so $f(1) = 1$  or $f(1) = -1$  ect.

 [u]Babis[/u]

[/quote]

[quote="stergiu"][u] To Fahrenajt[/u]

 After the new relation that you wrote, square the initial condition to get $f(x)f(y) = xy$. Here put $x=y=1$ , so $f(1) = 1$  or $f(1) = -1$  ect.

 [u]Babis[/u]
[/quote]

Neat! :)

[quote="stergiu"]I'm afraid , NO! We have not this condition ,  but even then  , last part is wrong . We ought  have that $f '$ is continous , so it has constant sign. But we have not this condition , too. Keep in mind that from
$|f'(x)| = 1$ we can not deduce that $f'(x) = 1$ for all x or $f'(x) = -1$ for all x. This is a main problem in finding functions !This can be done only if we know the continouity of $f '$. If any question , please write.

 [u]Babis[/u]

[/quote][/quote]I understand my mistake, thank you  :)


Hello  :)
I want a web site where can I find some national olympiad WITH SOLUTIONS  in pdf format or any other extension that allows to print.

Click the number of the problem. There's a link.  :)


geometry
rectangle
combinatorics unsolved
combinatorics

Every cell of a100x100 board is colored in one of 4 colors so that there are exactly 25 cells of each  color in every colum and in every row .Pro ve that one can  choose 2 rows and 2 colums so mthat 4 cells where they ihtesect are colored in 4 different colors.

well, lets see if this works:

Let $C$ be the number of pairs of different color and  in a rectangle. we have that every rectangle contributes at most $5$ to $C$ and by the other hand, every  pair of diferent color in the same row or column contributes at least $99$ to $C$, and every pair of different color not in the same row or column contributes at least 1 to  $C$.  so we must have:

I dont have the time right now, but there are actually many rectangles that only contribute $4$ to $C$, I think doing the calculations correctly we may reach a contradiction

I am really sorry for not posting a complete solution but I am away from home and i have no time!
 :(


What is the probability that flipping a fair coin three times produces fewer than two heads? Express your answer as a common fraction.

Two or three heads is 2 or more heads.
Zero or one heads is fewer than two heads.

These two cases are equally likely, and we want the probability of the latter. Therefore, my symmetry, the answer is $ \frac {1}{2}$

You only want one head to get fewer than 2 heads. So, the chance to get one head is [u][b]1/2[/b][/u]

[hide=help]
If you flip a head on the first flip I am needed to have 2 more heads
$\frac{1}{2}*\frac{1}{2}*\frac{1}{2}=\frac{1}{8}$
If i don't on the first flip the next one can be either a head or tail then the next one has to be the opposite
$\frac{1}{2}*\frac{1}{2}=\frac{1}{4}$
so the probability is
$\frac{1}{4}+\frac{1}{8}=\frac{3}{8}$
What????????
[/hide]
[hide=oh never mind]
when I said it was $\frac{3}{8}$ I said that probability of flipping one head
I forgot that 0 heads would work as well
the probability of that is $\frac{1}{8}$
$\frac{1}{8}+\frac{3}{8}=\frac{4}{8}=\frac{1}{2}$[/hide]

Why does everyone necr-
I'm done with this forum for now -_-

[quote=AC3131][quote=ilovepizza2020]Yay this was my 2000th problem done in Probability - Think About It![/quote]

congrats![/quote]

Thanks!

[hide=sweet post count]
Note that the probability of getting $0$ or $1$ heads is the same as getting $0$ or $1$ tails, so we have $\boxed{\dfrac{1}{2}}.$
[/hide]

[hide] This is the probability of flipping no heads (three tails) or one head (two tails). This is the same as the probability of flipping two heads (one tail) or three heads (no tails), since we can just interchange heads and tails. Since all outcomes are covered, both probabilities add up to 1, so the desired probability is $\boxed{1/2}$.[/hide]

[hide=Solution]
We can complement count. Getting two heads and one tail would have the probability of $\left(\frac{1}{2}\right)^3 \cdot \dbinom{3}{2} = \frac{3}{8}$. The probability of three heads is $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$. Adding these up, we get $\frac{1}{2}$. Since that is the part that we do not want, we subtract this from 1 to get $1-\frac{1}{2} = \boxed{\frac{1}{2}}$
[/hide]

[hide]If you get less than 2 heads when you flip 3 coins, that means you get either 1 or none. For none, the probability is 1/2^3 = 1/8. For 1, the probability is 3 * 1/8 = 3/8 because it's there are 3 ways (HTT, THT, TTH) which each have a 1/2^3 probability. 1/8 + 3/8 = 4/8 = [b]1/2[/b][/hide]