[quote="Iversonfan2005"][quote="vidyamanohar"][quote="Iversonfan2005"]Suppose triangle $ABC$ is a triangle with area $24$ and there is a point $P$ inside triangle $ABC$ which has a distance of $2$ from each of the sides of the triangle. What is the perimeter of $ABC$?[/quote]

A point that is at equal distance from sides of a triangle is called incentre of the triangle.  The distance is called the inradius  denoted by r.  If we denote the perimeter by 2s,

the relation between Area, inradius and semi-perimeter s is given by

Area = inradius  x  semiperimeter

Using this formula, you can calculate s as 12[/quote]

Yes and the perimeter is $24$. Simple problem, however, difficult if you don't know the $rs$ formula, or if you couldn't see that the distance was the inradius.[/quote]

It is not difficult if you dont know the $rs$ formula because you can do it without that.
See the figure below. You can find the area of the small triangles and then get the $area=a+b+c$ where $a,b,c$ are the sides of the triangle.
So $area=a+b+c$ so $perimeter=24$ YOU CAN DO IT WITHOUT THE FORMULA
 :rotfl:  :rotfl:  :rotfl:  :rotfl:

Hmm, I understand what you are getting at but the way you said it is flawed.  For one, you said $area=a+b+c$... :?: Since when can you add the lengths of sides and end up with an area!  Instead you should have said $area=\frac 12 ra +\frac 12 rb+\frac 12 rc$.

And just to note, you aren't really doing it without the formula.  I mean, the way you did it was the same exact way that formula is derived.  But doing it your way is much more instructive however.  It's much more important to know the math behind formulas than the formulas themself.  So aside from those little mistakes, very good job! :)


probability
expected value
algebra unsolved
algebra

Happy Meals from McDonalds offer 4 different toys that are equally probable. Every meal includes 1 toy. What is the expected number of meals I would need to buy in order to get all 4 toys?

Let the random variable $X$ be the number of happy meals bought that finally resulted in having all $4$.  Then for any integer $n\ge 4$, we have that

$P(X=n)=\frac{{4\choose 3}[3^{n-1}-{3\choose 2}*2^{n-1}+3]}{4^{n}}=\frac{3^{n-1}-3\cdot 2^{n-1}+3}{4^{n-1}}$.

Hence we have that

$E[X]=\sum_{n=4}^{\infty}n\cdot P(X=n)=\sum_{n=4}^{\infty}[n \left (\frac{3}{4}\right )^{n-1}-(3n)\left (\frac{1}{2}\right )^{n-1}+(3n)\left (\frac{1}{4}\right )^{n-1}]$

$=\frac{25}{3}$.

is it possible to generalise this problem to n toys?

would the expected number of trips be
n * (1 + 1/2 + 1/3 + 1/4 + ... + 1/n) ?[/code]

[quote="Phipps_Phipps"][quote]is there a limit to how many toys one can get of a certain type?[/quote]
Nope. I figured it out though.
[hide]In general, the expected value is the reciprocal of the probability. After the first purchased Happy Meal, you're pleased with any of the four toys since the probability of getting a desired toy then is 4/4. The expected value is also 4/4, or 1. After the Happy Meal, you need 1 of 3 prizes. The probability of getting a desired prize is 3/4, and the expected value is 4/3. The total expected number of trips, then, is the sum: 4/4 + 4/3 + 4/2 + 4/1. This works out to be about 8.33, or 8.[/hide][/quote]

i am not sure i follow...all it seems is that you identified a pattern...


advanced fields
advanced fields unsolved

The following holds in a somewhat more general setting, but this formulation is good enough:

Suppose $ X$ is a set and denote by $ \mathcal{P}(X)$ its power set, partially ordered by set-theoretic inclusion. Prove that if $ f: \mathcal{P}(X) \rightarrow \mathcal{P}(X)$ is order preserving, then it has a fixed point.

$ X$ itself could be a fixed point. In this case we're done. If it is no fixed point, then $ X\supseteq f(X) \implies X\supseteq f(X)\supseteq f^2(X) \supseteq f^3(X)\supseteq\ldots$ Now $ A: \equal{}\bigcap_{n\in\mathbb{N}} f^n(X)$ is a fixed point.

[quote="Gockel"]Now $ A: \equal{} \bigcap_{n\in\mathbb{N}} f^n(X)$ is a fixed point.[/quote]
Why? I have a somewhat more involved solution, I've tried claiming that this set is a fixed point but somewhat couldn't see it; can you elaborate?

This particular set may fail to be a fixed point. Let $ X\equal{}\mathbb N$. Consider the mapping that removes the least number in the set among those that have the smallest power of $ 2$ in their prime decomposition if the set is non-empty and does nothing if it is empty. It is easy to show that this mapping preserves the order but the intersection in question is the set of all even positive integers, which is not fixed. Still, the idea is OK, only we need the number of steps well beyond countable. I guess this is what Xixas means by a "more involved solution".

[quote="fedja"]Still, the idea is OK, only we need the number of steps well beyond countable. I guess this is what Xixas means by a "more involved solution".[/quote]

My solution is different. I assume the contrary and analyze the non-empty ($ f(\varnothing)$ is in there) set $ \mathcal{F} : \equal{} \{ A \in f(\mathcal{P}(X)) \mid \exists B \in f^{\minus{}1}(A) \text{ s.t. } B \varsubsetneq A\}$. Then $ T : \equal{} \bigcup_{A \in \mathcal{F}} A$ is a fixed point. Indeed, $ f(T) \supset A$ for any $ A\in\mathcal{F}$ by monotonicity, so that $ f(T) \supset T$ and thus $ f(T) \in \mathcal{F}$. But then $ f(T) \subset T$ and the conclusion is immediate.

You probably had another idea in mind, can you elaborate?

My original thought was the following:

Let us call by a chain the subset $ \Gamma\subset P(X)$ such that $ X\in \Gamma$, $ A\in\Gamma\implies f(A)\in\Gamma$ for all $ A$ and $ \Gamma$ is closed under intersections. Since the intersection of every family of chains is a chain, there is the minimal chain $ \Gamma_0$. Now, for every $ A\in\Gamma_0$, the set $ \{B\in\Gamma_0: f(B)\subset B\}$ is, again, a chain. Thus, it coincides with $ \Gamma_0$. Now just take the intersection $ I$ of all elements in $ \Gamma_0$. Since it is an intersection of the family of sets with the property $ f(A)\subset A$, we have $ f(I)\subset I$. On the other hand, $ f(I)\in \Gamma_0$, so it cannot be smaller than $ I$.

As it turns out, only the closedness of the property $ f(A)\subset A$ under intersections is really relevant in the above argument (shamelessly copied from the proof of total ordering, of course). So, it can be simplified to
\[ I\equal{}\bigcap_{A: f(A)\subset A}A\,\]
which is very similar to your approach.


Ok, this game is very simple.

Open iTunes / iPod. Enable/go to shuffle. Post the song that starts playing.

(you can do this with any music player that has something comparable to the above).

Brahms: Sonata for Cello & Piano #2 in F, Op. 99

Clarinet Concerto in A, K. 622:I.Allegro
Jack Brymer, London Symphony Orchestra and Sir Colin Davis

Handel: Messiah - Pastoral Symphony
Oratorio Society Of New York

The Fair Day
Boston Pops Orchestra(The Celtic Album)

The Twelve Days Of Christmas
New York Choral Artists

Chit-Chat (Quick Polka), op. 214
Zubin Mehta: Wiener Phiharmoniker

*skips 6 classical songs*

Empowered (Club Remix)
David & Steve Gordon

Bhaja Hure Ma
Jaya Lakshmi

Day Tripper (The Beatles)

Black Tambourine (Beck)

Symphony No. 40 in G Minor, K 550: II. Andante
Chicago Symphony Orchestra
Wolfgang A. Mozart

Winter: Largo
Antonio Viivaldi

Concerto in G Minor for 2 Cellos , strings and Basso Continuo, RV531, II. Largo
Yo-Yo Ma
Antonio Vivaldi

Bach: Cello Suite 2 in d, Gigue

The Flashbulb: Lawn Wake IX

Strauss II - Unter Donner und Blitz, Polka schnell, Op. 324
Chopin - Nocturne No. 2 in E flat major, Op. 9, No. 2: No. 2 in E flat
Chopin - Minute Waltz.

Kv247 Divertimento in F major "Lodronische NachtMusiken" -3. Menuetto
Thomas Füri/Camerata Bern
W. A. Mozart

ready to fall ---rise against

whats left of the flag---flogging mary

**** her gently---tenacious d

mulan theme song...?!?!

[quote="jli"]mulan theme song...?!?![/quote]

hehehe...

Shashkin - Omar Faruk Tekbilek

The Sharpest Lives ----My Chemical Romance

Layla (Deric and the Dominos)

Hey Jude (Beatles)

Sgt. Pepper's Lonely Hearts Club Band (The Beatles)

Pinball Wizard (The Who)

alrite now its

Here I Stand-Madina Lake

Everything's Magic-Angels and Airwaves

and finally

Stand By Me-Shinee

Super Freak - Rick James
 :rotfl:
All Star - Smash Mouth
Because of You - Kelly Clarkson

All oldies...

:rotfl:
Poor Izzy

Baby It's fact - Hellogoodbye
My life would suck without you - Kelly Clarkson
and er...
Goodbye - Kristina Debarge

Bleeding Love- Leona Lewis

George Gershwin-Piano Concerto in F-London Symphony Orchestra

Boom Boom Pow- Black Eyed Peas

Johannes Brahms- Academic Festival Overture, Op. 80- Nuremberg Symphony

Mozart -Piano Sonata No. 16 In C

Beethoven Minuet In G

Shake it- Metro Station.

You belong with me - Taylor Swift
Good Girls go Bad - Cobra Starship
On the BrightSide - Nevershoutnever!

Yeah....  :P

Now to list songs and artists no one has ever heard of.

Lord of the Universe --- Pia
Baila Verena --- Potosch Potschka
Hidden Forces All Around --- David & Steve Gordon
Inspiral --- Bobby Cochran
Touba - Daru Salaam --- Youssou N'Dour

Don't Look Back - Boston
Just That Girl - Drew Steely
Hey There Delilah - Plain White T's
Let It Rock - Kevin Rudolf

wow izzy, KEVIN RUDOLF?!?!?!?!?

heh, i didnt get too much better

amazing- kanye west

haha me too!

dance dance- fall out boy
amazing- kanye west (wow.)
down with the sickness- disturbed :D

Photogragh-Def Leppard
Don't Stop Believin'-Journey
Light My Fire-Doors
Day Tripper-The Beatles

Pour some sugar on me - Def Leppard
Hello Goodbye - Beatles
I don't care - Fall Out Boy

I suggest you the concerto in G Minor for 2 Cellos , strings and Basso Continuo, RV531, II.

By deaf people shouting in a microphone, yes I mean metal.


Let n be an integer. Show that, if $ 2+2\sqrt{1+12n^{2}}$ is an integer, then it is a perfect square.

[quote="riddler"]i dont get why k_2 is 3a^2 why cant it be 3a why a^2?[/quote]
[hide="answer"]If we have $ kl=3n^{2}$, where $ k,l,n\in\mathbb{Z}_{+}$ and $ gcd(k,l)=1$, then must be: $ k=3a^{2}\wedge l=b^{2}$ or $ k=a^{2}\wedge l=3b^{2}$ for some $ a,b\in\mathbb{Z}$.[/hide]


Let $ P$ be a point on side $ AB$ of triangle $ ABC$. Consider all pairs of points $ X\in BC, Y\in AC$ such that $ \angle PXB\equal{}\angle PYA$. Prove that the midpoints of all such segments $ XY$ lie on a single line.

Let $ Q,$ $ R$ be, the orthogonal projections of $ P,$ on the sidelines $ BC,$ $ AC$ respectively and we denote as $ D,$ $ E,$ the intersection points of $ BC,$ $ AC,$ from the lines through $ P$ and parallel to $ AC,$ $ BC,$ respectively.

It is easy to show that the segments $ QR,$ $ DE,$ have the property as the problem states $ ($ $ \angle PQB \equal{} \angle PRA \equal{} 90^{o}$ and $ \angle PDB \equal{} \angle PEA \equal{} \angle C$ $ ,(1)$ $ )$

Because of $ \angle PXB \equal{} \angle PYA,$ we conclude that the right triangles $ \bigtriangleup QPX,$ $ \bigtriangleup RPY,$ are similar and so from $ (1),$ we conclude that $ \frac {QD}{DX} \equal{} \frac {RE}{EY}$ $ ,(2)$

Hence, from $ (2)$ and based on the [b][size=100]Divisionic theorem[/size][/b](*) already has been posted in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=97493]Collinearity of orthocenters[/url] (post #7), we conclude that the midpoint $ M$ of the segment $ XY,$ lies on the constant line connecting the midpoints of the segments $ QR,$ $ DE$ and the proof is completed.

Kostas Vittas.

(*) Temporarily, I have such that named this (well known) powerful theorem and I am looking forward in your advices, for another better name, or if it has already a well known name, but unknown to me.


calculus
derivative
function
real analysis
real analysis unsolved

Prove that if $t > 0$ is sufficiently large, the equation
$xe^{x}= \frac{1}{t}$
has exactly one positive real solution $x$ which is given by
$x = \sum_{k = 1}^\infty \frac{(-1)^{k-1}k^{k-1}}{t^{k}k!}$

 :o  :D  :lol:

Easy with the [url=http://en.wikipedia.org/wiki/Lagrange_inversion_theorem]Lagrange inversion theorem[/url].

That's how the author of the book I read solves it. But I was wondering if there's any other way to approach this kind of problem. Maybe some iterative method would work but it seems like it would be really tedious... (I guess those often are)

You could start writing out formulas for derivatives of an inverse function, but that would be a boring calculus exercise.

Derivatives of inverse functions are not always boring, of course: see the [url=http://en.wikipedia.org/wiki/Jacobian_conjecture]Jacobian conjecture[/url].


How many different two-letter monograms are possible if
letters may be repeated?

There are 26 letters, so our answer is $ 26 \cdot 26 \equal{} \boxed{676}$.


limit
complex analysis
complex analysis unsolved

Suppose $ \mathop {\lim a_n }\limits_{n \to \infty }  \equal{} a$ and $ \mathop {\lim b_n }\limits_{n \to \infty }  \equal{} b$. Prove that:
$ \mathop {\lim }\limits_{n \to \infty } \frac{{a_1 b_n  \plus{} a_2 b_{n \minus{} 1}  \plus{} ... \plus{} a_n b_1 }}{n} \equal{} ab$

It's easier to prove in the case $ b\equal{}0$ and then using that result to prove the general form. Hint: 1) $ a_n,b_n$ are bounded sequences. 2) Cesaro-Stoltz


Here is another one:

P and  Q are the mid-points of sides BC and CD of a parallelogram ABCD and the lines AP and AQ intersect the diagonal BD at the points K and L resp.Prove that BK=KL=LD= BD/3.

Thank you.You have already helped me a lot.

[hide]Let $ R$ be the mid point of $ AB$ and $ S$ the mid point of $ AD$
Connect $ AC$, whose intersection with $ BD$ is let be $ X$

Now $ CR$ and $ PA$ intersect at $ K$ because $ AR$, $ BX$ and $ AP$ are medians of triangle $ ABC$
So $ BK: KX\equal{}2: 1$. Similarly $ DL: LX\equal{}2: 1$.
Hence $ BK: KL: LD\equal{}1: 1: 1$, as desired. [/hide]

nice, Chrichris.Is there some other solution?

Thank you.

Oh no! it is not just nice, but excellant. I just realized that KX=LX. Otherwise, the sum would not be valid.Of course, KX=LX.

Thanks again. :icecream:  :omighty:


How do we join Fairfax math circle?

A google search returns some competitions that they have competed in, but I cannot find their website, and am unsure where to look for brochures.
http://www.google.com/search?q=fairfax+math+circle&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a

Try David Levonian, dlevonian at cox.net

Thanks.

What is his relationship to fairfax math circle?

If you go to Tj than there might be a few issues. I don't know why but there is something.

unimpossible, you don't know the Fairfax Math Circle :gasp: ?

The Fairfax Math Circle has some negative relationship with TJ(Where u go).


linear algebra
matrix
linear algebra unsolved

Let $ A \equal{} \|a_{ij}\|$ be matrix with order $ p \plus{} q$ such that  $ a_{ij} \equal{} 0$ for $ 1\leq i,j\leq p$ and $ p \plus{} 1\leq i,j\leq p \plus{} q$. Show that if $ \lambda$ is eigenvalue for $ A$ then so is $ \minus{} \lambda$.

$ A\minus{}\lambda{I_{p\plus{}q}}\equal{}\begin{pmatrix}\minus{}\lambda{I_p}&V\\U&\minus{}\lambda{I_q}\end{pmatrix}$. Let $ c$ be the $ A$-charpoly.
If $ p\equal{}q$ then $ c(\lambda)\equal{}det(\lambda^2{I_p}\minus{}UV)$ and we are done.
If $ p\not\equal{}q$ then change the signum of the $ p$ first rows and after change the signum of the $ q$ last columns of $ A\minus{}\lambda{I}$. We obtain $ c(\minus{}\lambda)\equal{}(\minus{}1)^{p\plus{}q}c(\lambda)$ and we are done.


Support articles percent geometry MATHCOUNTS search trigonometry

Purchase advanced workbooks for your child.  In school, he may complain that he is bored, and you may be tempted to stop your child from learning at home.  This is what happened to me.  Don't do this.  Not only does this not work, but in the future when your son begins to compete in math competitions, he will be at a disadvantage.  Perhaps your child's teachers can help him individualy.  Older students are even better at offering support.  But don't push your son too hard, he will only resent it.  Instead let him have fun doing math, like I do. Good luck!  Even if your not that good at math, you should still be able to help him for a number of years.

And by the time you cant help him...he can use this site! :lol:

I higly recommend the Johns Hopkins programs. My parents used their distance education program for math while I was homeschooled, and I ended up doing algebra in 5th grade  :D. Yeah. If you don't use the distance ed, use something extracurricular. Good luck...

-Isaac

About skipping grades.. I think it's OK if your child is born before the end of the year.  My second one is an October kid.  And he is doing fine.   Though my older is a May child, he was forced to skip a grade because while he was given advanced curriculum in language arts and Math in the second grade, the third grade teachers refused to do the same.  He is doing fine but I am not very happy with him being so much younger than the rest.

If your school allows to advance your child's curriculum, make sure that it will be supported all the way through.  The school agreed to give my younger one 5th grade math in the 4th grade.  I got a committment from the principal and the 5th grade teachers that they will support advanced curriculum when he moves to the 5th.  We did not have the experience to do this with the first.

I am from KS too.

More on skipping grades. . .

I feel that people (teachers, school psychologists, etc) tend to overestimate the impact of age difference on social interactions.  It's not as if being 12 automatically means you are best suited to interact with other 12 year olds.

I was skipped a grade in primary school and then took above grade level course work in middle school.  Luckilly, I attended a school that regularly practiced this sort of acceleration.  The teaching staff was prepared for problems arising from age differences.  The school also boasted a number of "8th grade" math classes composed entirelly of 6th and 7th graders.

The ability of a child to adapt to skipping one or multiple grades is dependent upon his/her disposition and the disposition of teachers/administrators.  Teachers can be downright hostile to parents (and their children) if they believe a parent is challenging some cherished educational dogma.

I think that children should be placed in the classes that are most appropriate to challenge them, but also where they can learn and feel successful. Social interactions can be hard regardless of grade level and age differences. Anything within two years of their peers' tends not to be too big of a difference, especially if they stay in classes with them all along. It's more the stigma of being "advanced" that causes isolation in most cases. It really depends on the childs individual personality though, and on how common this sort of thing is at the school.

It goes without saying though, to just be supportive and encouraging of whatever your childs interests may be. The best way to encourage a child is lead them to interesting books on whatever they may be interested in. As they get older, a mentor of some type can be really helpful also -- maybe a friend or relative with a forte in that subject that they can talk to and learn from from time to time.

I have heard good things about the John's Hopkins Program but I don't know anything about it first hand. I can't see any harm in looking into it though. I find it useful on many levels not to refer to children too often as gifted or talented though. It reinforces the idea that things should always come easily to them - which is not the case. You want to reinforce that things come to those who have a passion for knowledge and who apply themselves. That no matter how much you know, there is always more out there. There is always another level of knowledge and understanding. I think that is an important psychological impression to make on them early.

Peggyv asked,

[quote]I'm not sure where I should post this, so I'll start here. I have a six-year-old kindergartener who is testing at the 99.9% in math and language skills. I can help him with the language skills, but I'm clueless when it comes to math. We're considering enrolling him the Johns Hopkins University's Center for Talented Youth internet program. Do any of you have experience with this program? Do you have any other tips on how to deal with a math wiz -- from a parent's or kid's perspective? How do you feel about skipping grades?[/quote]

Here is a FAQ file for parents who think their children are advanced in math. I'm a Chinese major myself, so I can take care of the verbal side of my son's education, but I have been researching math education for the past few years because of my oldest son's strong interest in math.

Here are suggestions for parents whose children who appear to be bright at a young age in math.

Suggestion 1: Run, don't walk, to get and read a copy of Liping Ma's book [url=http://www.amazon.com/exec/obidos/ASIN/0805829091/learninfreed]Knowing and Teaching Elementary Mathematics[/url]. You can request it by interlibrary loan if it is not in your local library. Ma's book makes apparent what kind of foundation is necessary at the beginning for a child to go "as high as he can go" in math. Try solving the teaching problems in that book for yourself, and you'll see what I mean.

Suggestion 2: LOOK UP some of the research Professor Tony Gardiner in Britain is doing comparing acceleration to enrichment as a strategy for preparing bright children for advanced study of mathematics. See, for example,

http://www.m-a.org.uk/association/presidents_page/able_students/

(Gardiner's definitive article on the subject is not on the Web, alas, and no longer in print.) Gardiner has shown what he means in a series of books called Maths Challenge published in Britain

http://www.singaporemath.com/supplement_others.htm#Gardiner%20Order

which I have bought and which are full of problems to develop a STRONG understanding of math. The Maths Challenge books are designed by Gardiner for seventh-graders and older students who are in the top 10 percent of the British population--younger students with good preparation could start using them at a younger age.

Suggestion 3: Surf over to Professor Hung-hsi Wu's Web site

http://math.berkeley.edu/~wu/

and make sure to download and read the draft chapters "Whole Numbers (Draft)" and "Fractions (Draft)" to get more than 100 pages each on those "easy" subjects from a thoughtful mathematician with a deep interest in math education. Then read his "How to Prepare Students for Algebra" for more insights.

Suggestion 4: Get and read the book [url=http://www.amazon.com/exec/obidos/ASIN/0486284247/learninfreed]Concepts of Modern Mathematics[/url] by Ian Stewart. If your child is an advanced reader that book might be readable by your child solo. This will show you what your child will be thinking about if he or she takes university-level math courses.

Suggestion 5: Get and read [url=http://www.amazon.com/exec/obidos/ASIN/0821813986/learninfreed]How to Teach Mathematics[/url] (2nd edition, 1999) by Steven G. Krantz, which is a book pertaining mostly to university-level math study, but with some interesting comments by Krantz on the Saxon math program and on other topics. The book includes essays by other professors of mathematics. Think about what kind of primary and secondary mathematics education (an issue Krantz hardly addresses in his book) would be fit preparation for university study of mathematics.

Suggestion 6: Having done the above, ponder what materials you are using for primary instruction in mathematics. My top recommendation for a first mathematics program is Miquon Math,

http://www.sonlight.com/miquon.html

http://www.keypress.com/catalog/products/supplementals/Prod_Miquon.html

a program designed for use over three years that covers almost all of elementary school mathematics from a higher math perspective. For people who have already gone through early elementary math, my number-one recommendation is the Singapore Primary Mathematics series

http://www.sonlight.com/singapore.html

http://www.singaporemath.com/primary_math_US_ED.htm#primary%20mathematics%20Order

which is described by many mathematicians as the "best mathematics textbook series available in English," an accurate description. The Singapore Primary Mathematics series is followed by other series from Singapore that take a learner up to all the mathematics needed for A level examinations in the British university entrance system.

Sometimes these two programs, Miquon and Singapore, contain problems that are confusing to American parents who had more conventional math instruction. My friendly suggestion is to take the confusing parts of those books as learning opportunities. Mathematicians linger and ask "why?" and an alternative representation of a mathematical operation (and you can count on the Primary Mathematics series to have ACCURATE representations of mathematical operations in visual, verbal, and other forms) is an opportunity to THINK about why the content was presented that way. Sure, not every child "gets" the content first from the same kind of presentation, but knowing why all the presentations relate to the same idea is part of understanding mathematics thoroughly.

Many of the illustrations in the Singapore books show examples of manipulatives that could be used in the classroom or at home as a first introduction to a topic. In my house, we USUALLY just went straight to the book, figuring our real life and earlier use of Miquon Math had already provided the "concrete" examples that fit into the Singapore "concrete --> pictorial --> abstract" model of instruction. But the concrete examples are latent in the coursebook (for example, baking cookies, inviting guests to parties, etc.) and may be helpful for many learners.

The Education Program for Gifted Youth (EPGY)

http://epgy.stanford.edu

mathematics program is probably wholly unnecessary at the very earliest age level, but it is a great way to move ahead for young people who like that kind of computer-based instruction, especially beginning at about the fourth- or fifth-grade level. EPGY can take learners all the way up to university-level math at their own pace. The ALEKS online program

http://www.aleks.com/

is a useful supplement to any of the other recommended programs, being almost as good as and a lot less expensive than EPGY, with a fairly complete K-12 mathematics sequence but not continuing to advanced undergraduate mathematics.

Many parents have tried out many other kinds of math programs to help their precocious math learners move ahead with a good foundation. Math programs that I personally do NOT recommend, based on the desirability of a) truly challenging word problems, b) multiple representations of mathematical ideas, and c) clear, CORRECT explanations of mathematical concepts include 1) Saxon Math, 2) Math-U-See, 3) any old, traditional program (e.g., A Beka), 4) exclusive use of "gifted learner" worksheet books or other books that consist mostly of exercise sets, or 5) any "reform" math program used in United States public schools, although the best of these are better than some of the other programs I don't recommend. These negative recommendations are not intended to offend any parent who has used these programs in a good-faith belief that they are useful math programs, but are mentioned to suggest trying out the truly superior programs if you haven't done so already.

The very best mathematics textbook series in the world, as best I can ascertain, is the Hua Loo-keng School Mathematics Textbook series published in China. I may have to turn that series into English to give American students an opportunity to learn from the very best. The Hua Loo-keng series makes the curriculum expectations of the EPGY series look like slow learner expectations. The Hua Loo-keng School series doesn't just go faster but also deeper.

Suggestion 7: Get involved in "competition culture" for a reality check on how your child is doing in math. There is a great variety of mathematics competition programs these days, unlike the days when I went to school, with many programs of differing characteristics. See what local math competitions there are in your area, and what the requirements are for forming a team or joining as an individual. The American Mathematics Competition programs

http://www.unl.edu/amc/

are readily available worldwide and start at the prealgebra level and go up to qualifying tests for the International Mathematics Olympiad. The better math competitions (MATHCOUNTS

http://www.mathcounts.org/

is also in this category) are an excellent reality check on how much math your child knows as second nature.

Suggestion 8: The talent search tests

http://www.ditd.org/Cybersource/record.aspx?sid=12649&scat=902&stype=110

http://www.hoagiesgifted.org/talent_search.htm

are another way to get a reality check on a child's math level. The talent search tests are typically a standardized achievement test normed for one age group and given to a younger age group. Most of the regional talent search centers will give you DETAILED information about where your child stands in test performance ranking compared to other children who show up to take the test. The tests vary in format, and thus check whether your curriculum is developing a well-rounded approach to solving (simple) mathematical problems.

Suggestion 9: The Art of Problem Solving Web (AoPS) site

http://www.artofproblemsolving.com/

includes a treasure trove of resources for young math learners, and has long been the online home of this FAQ. The online forum is a great way for young people with interest in math to meet one another and improve their skills. There are many [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Articles.php]helpful articles[/url] on AoPS as well.

Hope this helps! Best wishes to your child.

[End of FAQ]

I'm not sure how young your son is, but you could play some simple number games with him occasionally.  He'll learn basic math without really realizing it... just randomly ask him things like "if I had two apples, and then you gave me three more apples, how many apples would I have?"  If he's tolerant enough for it, you could also try flash cards.  But I'm sure you know addition through multiplication enough that you can teach him that stuff without much difficulty as fast as he is ready for it.  If you don't know basic algebra, you could probably learn it fairly easily by reading a book or two, and then you could teach him that too.  Oh, and throw in stuff like how to find the area of simple shapes in some creative way too.  If you can succeed in teaching him all that stuff at a fairly early age, he'll have a great start and you/he should be able to figure out where to go from there without much difficulty.  I learned algebra "officially" in seventh grade in our advanced program, but I had done a lot of it earlier because I was ready for it.  We did a little bit of right angle trig in sixth grade, and my parents had told me very basic things about variables.  Programming also helped with the concept of variables, because they are used all the time.  TI-Basic for the calculator and QBasic for the computer were not beyond the capabilities of my friends and I to understand in fourth and fifth grade, and programming is great for thinking logically.  If he shows an interest in computers, you may want to consider introducing him to programming as well.  I would not suggest C, C++, or Java until he was older since they are far more advanced and difficult to use, and would not suggest any other programming language unless he shows a true interest.  Feel free to contact me if you want more information about getting a kid started in programming, because it isn't exactly something a lot of people know how to start.  Personally, I asked my dad how programs were made and he handed me a book, telling me that he hadn't read it because he didn't understand it.  I read the book and the help files, and had a go at programming ;)  Anyway, I hope I've been of help.


I'm using a program in java that uses wilson's theorem to determine primeness. Unfortunately, it tells me that 20 is prime, because doubles don't give enough digits of accuracy.
What variable can I use to get more accurate than a double?

[quote="tjhance"]well, you could try modding at each step. what i mean is, instead of calculating 19!, and then taking it mod 20, you could do:

[code]int a = 1;
for(int i = 2; i <= 19; i++) {
    a = a * i;
    a = a % 20;
}
[/code]

so you wouldn't have accuracy issues[/quote]
Thanks, that should work :)
It should also give me more efficiency.

For the record, there are Java classes that allow you to do arbitrary-precision arithmetic (namely BigInteger and BigDecimal, I believe).  I used the BigDecimal class for a project of mine, and it worked out fairly well.  That being said, for your specific problem it's of course best to mod at every step.


I know this is probably easy, but I am having trouble thinking about the following...

Is it always true that $ cl(A) \equal{} A\cup \partial A$, where $ cl(A)$ denotes the closure of $ A$ and $ \partial A$ denotes the boundary of $ A$?

I know that the closure of a set is that set together with all points of accumulation, and the assertion seems to follow from this, but I am not sure if I am thinking properly. If this is incorrect, is it true in $ \mathbb{R}^{n}$? Or is there some condition that can be imposed on $ A$ that makes it true, say if $ A$ is open?

Sorry for posting such an easy question. All replies are appreciated.

It's true in any topological space, by definition
$ \partial A \equal{} cl(A) \minus{} int(A)$
and $ int(A)  \subset A$

Expanded answer.

If $ p$ is a point if some abstract set $ X$ and $ A$ is a subset of $ X$, then there are two possibilities: either $ p$ is in $ A$ or it's not. Nothing else to be said.

Situation changes when $ X$ is a topological space. Now we have three possibilities:
(1) "$ p$ is safely within $ A$": there exists a neighborhood of $ p$ that is contained in $ A$. Formally, $ p$ is an interior point of $ A$.
(2) "$ p$ is safely outside of $ A$": there exists a neighborhood of $ p$ that does not intersect $ A$. Formally, $ p$ is an exterior point of $ A$.
(3) neither (1) nor (2) holds, which means that any neighborhood of $ p$ meets both $ A$ and its complement. Formally, $ p$ is on the boundary of $ A$, $ p\in \partial A$.

If $ p\in A$, then either (1) or (3) happens - $ p$ is either in the interior or on the boundary.

Once the above is understood, it should become clear that in both cases (1) and (3) $ p$ is a point of accumulation of $ A$, but in case (2) it's not. So, the closure of $ A$ is the union of interior and the boundary. We can also say that the closure of $ A$ is the union of $ A$ and the boundary, because all points of $ A$ are included in the union either way.


[hide][img]http://img513.imageshack.us/img513/4955/tilingyv2.png[/img]

The above tiling is a complete 5x9 rectangle. It can be placed in a corner of the table, and the rest of the table can be filled up with hundreds of rows of "flat" pieces.

Of course there are much more interesting patterns, but this is math, not art. :)[/hide]


A circle is tangent to rays $ OA$ and $ OB$ at $ A$ and $ B,$ respectively. Let $ K$ be a point on minor arc $ AB$ of this circle. Let $ L$ be a point on $ OB$ such that $ OA\parallel KL.$ Let $ M$ be the intersection $ ($distinct from $ K)$ of $ AK$ and the circumcircle $ \omega$ of $ \triangle KLB.$ Prove that $ OM$ is tangent to $ \omega$

Proceed by angle chasing; denote the angles: $ \angle KAO=x$, $ \angle BAK=y$.

$ \angle MKL=x$ since $ AO//KL$.
$ \angle KBM=x$ since $ MLKB$ is cyclic.

Hence $ MOAB$ is cyclic since $ \angle OBM=\angle OAM$.

$ \angle MOB=y$ since $ MOAB$ is cyclic.
$ \angle KBO=y$ by tangent-secant angle theorem.
$ \angle LMK=y$ since $ LMKB$ is cyclic.

$ \angle KBA=x$ by tangent-secant angle theorem.
$ \angle BKM=x+y$ by the exterior angle theorem.
$ \angle BLM=x+y$ since $ MLKB$ is cyclic.
$ \angle LOM+x=x+y\implies \angle LOM=x$ by the exterior angle theorem.

Since $ \angle MLO=\angle MBL$, $ OM$ is a tangent to the circumcircle of $ \triangle LKB$, as desired.

Thanks for the nice problem.

Comment: $ \triangle KBA\sim\triangle LKM\sim\triangle MBO\sim\triangle LMO$ by AA.

[b]Thank you for your interest, Altheman![/b]  :)
Here is my solution.

[b]Proof.[/b]  We have: $ \angle OAM = \angle LKM = \angle OBM$ $ \implies OABM$ is cyclic $ \implies KB \parallel OM \quad(1)$
$ \implies \angle BKM = \angle KMO = \angle ABO$
On the other hand, $ \angle BMK = \angle AOB$
Thus $ \triangle BMK \sim \triangle AOB \implies MB = MK \implies PM \perp KB\quad (2)$

From $ (1)$ and $ (2),$ we have: $ PM \perp OM$
Hence $ OM$ is tangent to $ \omega$


MATHCOUNTS inequalities quadratics algebra

[hide]Multiply both sides by x to get x :^2: +1 :le: -2x.  Then, add 2x on both sides to get x :^2: +2x+1 :le: 0, a quadratic equation we like.  It can be easily factored to become (x+1)(x+1) :le: 0  x+1 :le: 0  x :le: -1

Ergo, x :le: -1[/hide]

_____________________

   -Alex

Tisk tisk tisk, it's not that easy...think about what will happen if x was negative.

Also, (x + 1)(x + 1) :le: 0 does not imply that (x + 1) :le: 0.

[hide](x+1)^2 is not non-positive, unless we use complex numbers of x is -1.[hide][/hide][/hide]

Yes.  But, anyhow, he shouldn't ever have gotten to that equation in the first place.

[hide]if we assume that x is positive, we solve for x and get x :le: -1/3, so x is negative

if we assume that x is negative, we solve for x and get x :ge: -1/3

since x is negative,  -1/3 :le: x <0 (can't be :le: 0 because 0 can't be in the denominator)[/hide]

Posted by MrMcGeek

I don't think your 1/3 is quite right there.. if you show your working we might see why..

edited with the right solution finally, i hope [hide](x+1)(x+1) is less than or equal to 0 (if x is positive this equations works), which is impossible. If x is negetive, then (x+1)(x+1) more than or equal to 0, which means that x is x<0 because anything squared is bigger than zero [/hide]

by the way, i think people who get 1/3 multiplied first side by x and got x+1 instead of x^2+1 (they probably thought it is (x+1)/x))

I see my problem:

I thought (x+1)/x :le: -2

it's x+ 1/x :ie: -2

so now when i solve for x is negative, i get x<-1
when i solve for positive i get x>1

Posted by MrMcGeek

Mmm, not quite, that's what I got at first too. Listen to Joel: [quote="Joel"]Also, (x + 1)(x + 1):le:0 does not imply that (x + 1):le:0.[/quote] and you shall be unstopapble :lol:

As for your answer ml2 double check it for x<0.

yeah, it's been corrrected. We could also note that x must be negetive since a positive plus it's inverse are also positive. And then just say x<0 without any algebra, since any number + it's inverse is bigger than or equal 2 or smaller than equal -2


ratio
trigonometry
algebra proposed
algebra

hi,
if there exists an $ x \in \mathbb{R}$ s.t.
\[ \frac{\sin{x}}{a} \equal{} \frac{\sin{2x}}{b} \equal{} \frac{\sin{3x}}{c} \equal{} \frac{\sin{4x}}{d},\]
where $ a,b,c,d \in \mathbb{R}$ prove
\[ 2a^3 (2b^2 \minus{} d^2) \equal{} b^4(3a \minus{} c).\]

$ \frac{\sin{x}}{a} \equal{} \frac{\sin{2x}}{b} \equal{} \frac{\sin{3x}}{c} \equal{} \frac{\sin{4x}}{d}\equal{}k$
$ 2a^3 (2b^2 \minus{} d^2)k^5 \equal{} b^4(3a \minus{} c)k^5$  or  $ k\equal{}0$
$ \Leftrightarrow 2 sin^3 x (2sin^2 2x\minus{}sin^2 4x)\equal{}sin^4 2x(3sinx\minus{}sin 3x)$
$ \Leftrightarrow 2 sin^3 x(2\minus{}2cos^2 2x)\equal{}sin^2 2x(3sinx\minus{}sin 3x)$  or  $ sin 2x\equal{}0$
$ \Leftrightarrow 4sin^3 x\equal{}3sinx\minus{}sin3x$ , which is obviously true.


Compute S+ T where
\[
\begin{array*}
S= \sin^4 2^\circ+\sin^4 7^\circ+\cdots +\sin^4 87^\circ \\
 T= \sin^4 3^\circ+\sin^4 8^\circ+\cdots +\sin^4 88^\circ
\end{array*}
\]


Hi

The following problem:
[img]http://img40.imageshack.us/img40/9915/problvy.jpg[/img]

I am new to the Putnam contest and I was trying to solve this sample problem I got. I spent a lot of time making a table of values and trying to see the pattern.

Any way, how would you rate the difficulty of this Putnam problem relative to others? It seems that the solution to this is pretty elementary (doesn't involve any more advanced mathematics; at least I am solving it using just basic algebra).

Alternatively, can this one actually be much more easily solved if I know some more advanced maths beyond algebra?

Thank you!

No comment on the difficulty- but this is Example 2.2.5 in Zeitz' ACoPS, and that's a very good exposition of how to solve it. The elementary approach, started by just playing around and looking for patterns, is best.

Below is my solution

[img]https://nrich.maths.org/discus/messages/27/146862.jpg[/img]

Is my solution considered satisfactory under the Putnam contest setting?

Please help

thanks[/img]

Any one please help with regard to my previous post????  :(


$ a,b,c \geq 0$ prove
$ (\sum_{cyc}{(a^2b \plus{} ab^2)} \minus{} 6abc)^3\geq 27abc(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2$

[quote="hedeng123"][quote="ye109"]$ a,b,c \geq 0$ prove
$ (\sum_{cyc}{(a^2b \plus{} ab^2)} \minus{} 6abc)^3\geq(abc(\sum_{cyc}(a \minus{} b)^2)$[/quote]
[/quote]
i have edit it


Hey ppl
any of u doing ur 10th boards next yr? CBSE and ICSE only..pls reply..this is  extremely important..there may be a COLOSSAL change in portions...i am so freaked out...plz let me know if u r then i can tell u guys the whole thing.
Thanx,
Padma

well I am in cbse and am doing my 10th board exam this academic year... (2006-2007).... What are the changes by the way ??

WELL..FOR 1 thing..CBSE..i am not quite sure but i heard that you are going to have
a) 10 marks of questions from current affairs in any one paper
 b) 40 marks objective
c) the marks for the hindi essay are going to be increased to 30.
d) an additional 10 marks of internal assessments in all subjects...
is it true? i had to confirm..this had come in Deccan Herald and I was not sure. I have friends in really good schools whose teachers have confirmed this so I thought I'd pass it around...and also get confirmation.
I was thinking what I would do if i was forced to join CBSE next year (my 10th) and  then this came as a major shock...
Anyway...waddya think? is it true?
(Don't scream if it's not!!)  :D

I really dont know about those things..... one good part is, we write an autonomous maths and science paper, so atleast those two things stay the same.... :D

yes, the autonomous paper scheme is extremely helpful, I don't think that will change. :D
The other things..my friend's teacher said they may be true..she will tell me on Wednesday...anyway..
I am more scared what will happen to ICSE in all this! ICSE has also got a biiiig change coming...
What was the highest percentage in CBSE last year, by the way?

[quote="ashwath.rabindranath"]I'm going to be writing my 10th CBSE board in March.[/quote]
good luck! :D
soo...are any of these things about the change of syllabus and pattern of the question papers true, or has NPS bangalore taken us all for a big ride?
what did your school tell u about the changes of the paper?
I am writing my boards ICSE (but i may switch over to CBSE if it seems easy and unchanged...dunno) in March 2008 and I think i'll be [b]DEAD 10 times[/b]with [b]tension[/b] by then!  :)

Ok when I said I'll die 10 times..I was mistaken..MAKE IT A 100!!
[b]I just got it straight from the class teachers of STD X in both ICSE and CBSE schools...there IS gonna be a change!![/b]
Either the devil or the deep sea....dunno which one to take..is CBSE easy?
(I have been in ICSE for 9 yrs so...)

Hello,
   I have been reading this thread for some time.I have a brother who is writing his Class X boards this year in CBSE.However,I don't understand why a change in 'pattern' or even syllabus for that matter should cause so much concern(as to make people die a 100 times).After all,they are not going to ask a Class X student to study a university level course.Even then,it is the same for everyone,isn't it?
Sathej

Sathej,
Hey!
Well, you could say that was a BIT of an exaggeration  :blush:  :lol: but then it is quite bad if ur forced to change ur syllabus in Std.X and people spring it on you that you have a BIG load more of work...in MY case...and this will happen only if ICSE changes... I have atleast 4 completely NEW textbooks to lean..IN EACH SUBJECT..I dunno if u wud classify that as a 100 deaths...that's why I wanna clarify abt the changes... but otherwise also..u have a lot more to study...and a LOT MORE to forget..when I was told this I didn't believe it..but when I saw the seniors in the exam hall...that feeling...that I will be there next year..and only HALF as prepared.....anyway!!
U finished 10th right? so it's ok 4 u..no parents and teachers pressurizing u to hell!! Can you believe that one of my teachers is actually organizing BETS among the teachers..on who will do well??
Anyway...I guess life just goes on..no sweat...i have stopped freaking out...will start again in March! hehehe

Hello,
  Yes,there is a lot more to face.Class X Boards would appear so miniscule and irrelevant in the long run.And pressure?Well,it all depends on how you deal with it.Take heart in the fact that,as I said,it is the same for so many people.Best of luck,anyway.
Sathej

Thanx!
It's just sooo scary when people from all quarters tell u that ur WHOLE LIFE depends on it..coz which college/job/ u will get will be based on ur marks. Thanx anyway, I am not so freaked out now, after learning that I have to stick in ICSE...CBSE wud have been so bad!
So finally whether there IS a change or not, I don't care!!  :D  :lol:  :lol:  :D
The fact that we are starting std X portions only in march also helps de-stress!
So now i am freeeeeeeeeeeeeeeeee!!!!

this may sound absurd but is it true that in CBSE u don't have logs and trig until u get to 11th????  :huh:  :o

Hello,
    CBSE has a bit of trigonometry in X.But, logarithms are not introduced.Anyway,logarithms are not that different from exponents,are they?

hey
thanx
that's weird...even though logs and exponents r similar, the log tables part is so much fun! and i didn't think that CBSE had it only later.. u c...when i had gone 2 delhi 4 a seminar, this math teacher who was in charge kept saying "ICSE learns so much more...we(CBSE) don't have logs and trig till 11th...oh my...oh my...I wish there was equality in Indian education...it would promote peace and harmony.." :rotfl:  :huh:  and when she said this i replied "weird!!...can that be true?" so i HAD 2 satis5 my curiosity...thanx! :D

:D cool...i am not freaked out any more since the teacher who keeps bugging me and giving me pressure and basically ruining my life ...has resigned!!! :clap:
(Bio teachers are evil...) and life is so much more fun now...
Oh-and shadysaysurspammed-i like ur display pic!

yeah the main thing is not to get ternsed n ur xams go better than every1s...enjoy


calculus
integration
inequalities
topology
real analysis
real analysis unsolved

If $ Y$ is a metric space and $ f: [a,b]\times Y\rightarrow \mathbb{R}$ is continuous,show that
$ F(y)\equal{}\int_a^bf(x,y)dx$

if $ y_0$ is fixed and $ \epsilon > 0$, use the compacity of $ [a,b]$ to find $ \eta$ small enough such that for all $ |y \minus{} y_0| < \eta$ the inequality $ |f(x,y) \minus{} f(x,y_0)| < \epsilon$ is true for all $ x \in [a;b]$.


Check if there is a permutation $(x_{1},x_{2},...,x_{p-1})$ of a set $\{1,2,3,...,p-1\}$ that all expressions $x_{1}, x_{1}x_{2},...,x_{1}x_{2}x_{3}...x_{p-1}$  give different residues when divided by $p$ , where $p$ is a prime

I took it from another thread, but now it's after deadline  :D

Take:
$x_{1}\equiv 1 \mod{p}$
$x_{k}\equiv \frac{k}{k-1}\mod{p}$
Then: $x_{1}x_{2}...x_{t}\equiv t \mod{p}$ for every $1 \leq t \leq p-1$
Which obviously works.

I've had the same solution - kinda easy problem. Another way was to use the fact that multiplicative group of integers mod p is isomorphic to additive group mod p-1 and find some trivial example of desired sequence in that.


[hide=":ninja:"]The time is 9:00. How many minutes will pass until the minute hand and hour hand make a 28 degree angle for the first time?[/hide]

[hide]At $9: 00$, the number of degrees between the minute hand and the hour hand is $90$, but since the minute hand is faster and will eventually overtake the hour hand we must look at it as $270$ degrees.

Every minute, the minute hand moves $6$ degrees, but also every minute, the hour hand moves $30/60=\frac{1}{2}$ degrees so you subtract $\frac{1}{2}$ from $6$ and get $\frac{11}{2}$.  Now we subtract $270-28=242$ as we need the minute hand to catch up that much ground.

$242*\frac{2}{11}=22*2=44$

so $\displaystyle{\boxed{44}}$[/hide]

[hide]
28 degrees is approximately 30 degrees. 30 degrees is about one hour or 5 minutes (between two numbers on the clock).
45 seems to be pretty close:
If it has been 45 minutes, the minute hand is on the 9. 45 minutes is $\frac{3}{4}$ of an hour. $\frac{3}{4}\cdot60=22.5$
Try 44:
44 is $\frac{44}{60}=\frac{11}{15}$ of an hour. $11\cdot2=22$ because 15 times 2 is 30 which is the number of degrees in an hour.
$360-(44\cdot6)=96$, so below the nine, there is 6 degrees.
$22+6=28$
So, the answer is 44 minutes.[/hide]


trigonometry
calculus
calculus computations

Find the condition that $P=ax^2+2bxy+cy^2$ has the same order of $\rho^2$ in a small neighbor of the origin, where $\rho=\sqrt{x^2+y^2}$.

That asks us to prove that $\rho^2/P(x)$ and $P(x)/\rho^2$ are bounded.
Let $x=\rho\cos \theta$ and $y=\rho\sin\theta$,
we have $P(x)/\rho^2=a\cos^2\theta+2b\sin\theta\cos\theta+c\sin^2\theta$, denote it by $f(\theta)$. The condition requires that $f(\theta)\neq 0$ for all $\theta\in [0,2\pi]$.

Write $f(\theta)$ as $f(\theta)=\frac{a+c}{2}+\frac{c-a}{2}\cos 2\theta+b\sin 2\theta$, hence it requires that \[\left|\frac{a+c}{2}\right| > \sqrt{\left(\frac{c-a}{2}\right)^2+b^2} \]
and we finally obtain that $ac>b^2$.

[quote="liyi"]
Write $f(\theta)$ as $f(\theta)=\frac{a+c}{2}+\frac{c-a}{2}\cos 2\theta+b\sin 2\theta$, [/quote]

You are wrong, liyi. But the answer is O.K.

oops... It is $\frac{a+c}{2}+\frac{a-c}{2}\cos 2\theta+b\sin 2\theta$


function
induction
algebra proposed
algebra

Find all functions $ f: \mathbb{R}\rightarrow \mathbb{R}$ such that for all reals $x$ and $y$,
\[f(x+y)f(x-y)=f(x^2)-f(y^2).\]

[quote="popolux"]Find all functions $ f: \mathbb{R}\rightarrow \mathbb{R}$ such that $ \forall x,y\in\mathbb{R},f(x \plus{} y)f(x \minus{} y) \equal{} f(x^2) \minus{} f(y^2)$
(no continuity assumption on f)[/quote]

Yes, it's a nice problem.

Let $ P(x,y)$ be the assertion $ f(x\plus{}y)f(x\minus{}y)\equal{}f(x^2)\minus{}f(y^2)$

$ P(0,0)$ $ \implies$ $ f(0)\equal{}0$
$ P(x,0)$ $ \implies$ $ f(x^2)\equal{}f(x)^2$
If $ f(a)\equal{}0$, then $ f(\minus{}a)^2\equal{}f((\minus{}a)^2)\equal{}f(a^2)\equal{}f(a)^2\equal{}0$ and $ f(\minus{}a)\equal{}\minus{}f(a)$
If $ f(a)\neq 0$, then $ P(0,a)$ $ \implies$ $ f(a)f(\minus{}a)\equal{}\minus{}f(a^2)\equal{} \minus{}f(a)^2$ and so $ f(\minus{}a)\equal{}\minus{}f(a)$
So $ f(\minus{}x)\equal{}\minus{}f(x)$ $ \forall x$

Then, from $ f(x^2)\equal{}f(x)^2$, we get $ f(x)\equal{}f(\sqrt x)^2$ $ \forall x\geq 0$ and so $ f(x)\geq 0$ $ \forall x\geq 0$
Then, from $ f(\minus{}x)\equal{}\minus{}f(x)$, we have $ f(x)\leq 0$ $ \forall x\leq 0$

Then, for $ x\geq y\geq 0$, we have $ x\plus{}y\geq x\minus{}y\geq 0$, so $ f(x\plus{}y)\geq 0$ and $ f(x\minus{}y)\geq 0$, so $ f(x)^2\geq f(y)^2$ and so $ f(x)\geq f(y)$ since these are two positive numbers.
Then, for $ 0\geq x\geq y$, we have $ \minus{}y\geq \minus{}x\geq 0$ and so $ f(\minus{}y)\geq f(\minus{}x)$, so $ \minus{}f(y)\geq \minus{}f(x)$ and so $ f(x)\geq f(y)$
Then, for $ x\geq 0\geq y$, we obviously have $ f(x)\geq 0 \geq f(y)$

So, $ f(x)$ is  a non decreasing function.

Then, $ P(1,0)$ $ \implies$ $ f(1)^2\equal{}f(1)$ and so either $ f(1)\equal{}0$, either $ f(1)\equal{}1$

1) $ f(1)\equal{}0$
Since $ f(x)$ is non decreasing, we get $ f(x)\equal{}0$ $ \forall x\in[0,1]$
Then $ P(x\plus{}1,x)$ $ \implies$ $ f(2x\plus{}1)f(1)\equal{}f(x\plus{}1)^2\minus{}f(x)^2\equal{}0$ and so $ f(x\plus{}1)^2\equal{}f(x)^2$ $ \forall x$
An immediate induction implies $ f(x)\equal{}0$ $ \forall x$ and it's easy to check that this value fit the equation.

2) $ f(1)\equal{}1$
Let $ f(2)\equal{}a$ (we have $ a\equal{}f(2)\geq f(1)\equal{}1$
$ P(2,1)$ $ \implies$ $ f(3)\equal{}a^2\minus{}1$
$ P(3,1)$ $ \implies$ $ af(4)\equal{}(a^2\minus{}1)^2\minus{}1\equal{}a^2(a^2\minus{}2)$ and so $ f(4)\equal{}a(a^2\minus{}2)$ (since $ a\geq 1$)
But we also have $ f(4)\equal{}f(2)^2\equal{}a^2$ and so $ a^2\equal{}a(a^2\minus{}2)$ and so $ a\equal{}\minus{}1$ or $ a\equal{}2$ so $ f(2)\equal{}2$

Then, an easy induction, using $ P(n,1)$ shows immediately that $ f(n)\equal{}n$ $ \forall n\in \mathbb{Z}$

We'll now show that $ f(x)\equal{}x$ $ \forall x$.
First, we must establish that $ |f(x)\minus{}x|<1$.
Suppose we have, for some $ x$ : $ |f(x)\minus{}x|>1$. Then we have, for some $ n\in\mathbb{Z}$, either $ f(x)<n<x$, either $ f(x)>n>x$, but :
$ f(x)<n<x$ $ \implies$ $ x>n$ and $ f(n)>f(x)$, which is impossible, since $ f(x)$ is non decreasing
$ f(x)>n>x$ $ \implies$ $ x<n$ and $ f(n)<f(x)$, which is impossible, since $ f(x)$ is non decreasing
Q.E.D.

Suppose now that $ \exists u$ such that $ f(u)\neq u$. We'll show first that it exists then some $ w>2$ such that $ f(w)\neq w$
Let $ v\equal{}|u|$. Obviously, we have $ f(v)\neq v$ and $ v>0$

$ P(v\plus{}2,v)$ $ \implies$ $ 2f(2v\plus{}2)\equal{}f(v\plus{}2)^2\minus{}f(v)^2$
If $ f(2v\plus{}2)\equal{}2v\plus{}2$ and $ f(v\plus{}2)\equal{}v\plus{}2$, then we have $ f(v)^2\equal{}(v\plus{}2)^2\minus{}2(2v\plus{}2)\equal{}v^2$ and $ f(v)\equal{}v$, which is wrong (choice of v).
So let $ w\equal{}v\plus{}2$ if $ f(v\plus{}2)\neq v\plus{}2$ and $ w\equal{}2v\plus{}2$ if $ f(v\plus{}2)\equal{}v\plus{}2$ (and so $ f(2v\plus{}2)\neq 2v\plus{}2$

So we have now $ w>2$ such that $ f(w)\neq w$ Let $ b\equal{}f(w)\minus{}w$. Since $ |f(x)\minus{}x|<1$ $ \forall x$, we have $ |b|<1$
We have $ f(w^2)\minus{}w^2\equal{}b(2w\plus{}b)$ and, since $ w>2$ and $ |b|<1$, we have $ 2w\plus{}b > 2$ and $ |f(w^2)\minus{}w^2|>2|b|$
repeating this process, we'll obtain some t such that $ |f(t)\minus{}t|>1$, which is impossible.

So, no $ u$ such that $ f(u)\neq u$ exists and $ f(x)\equal{}x$ (which indeed match the original assertion).

So the only two solutions are :
$ f(x)\equal{}0$
$ f(x)\equal{}x$

Nice proof!(sorry for my late answer)
Mine was more complicated.However,i think i can simplify the end.Indeed,you proved that f is monotone and that f(n)=n.We can easily deduce that $ f(x)/x\rightarrow 1$ when $ x\rightarrow\infty$.So,if we fix x>1,then $ \left(f(x)/x\right)^{2^n}\equal{}f(x^{2^n})/x^{2^n}\rightarrow 1$ when $ n\rightarrow\infty$,so f(x)=x.

[quote="popolux"]Nice proof!(sorry for my late answer)
Mine was more complicated.However,i think i can simplify the end.Indeed,you proved that f is monotone and that f(n)=n.We can easily deduce that $ f(x)/x\rightarrow 1$ when $ x\rightarrow\infty$.So,if we fix x>1,then $ \left(f(x)/x\right)^{2^n} \equal{} f(x^{2^n})/x^{2^n}\rightarrow 1$ when $ n\rightarrow\infty$,so f(x)=x.[/quote]

Oh yes ! quite nice improvement!

Thanks :)


analytic geometry
graphing lines
slope
MATHCOUNTS
search
trigonometry
rotation

[hide="PROBLEM 3"]We have a Diophantine-Type Equation.  Because we know that there are thirteen pencils and three erasers, converting everything to cents, $ 100 \minus{} 13x \equal{} 3y$, when $ x$ represents the cost in cents of pencils and $ y$ represents the cost in cents of erasers.

After some logic and trial & error work, we can see that:

Pencils can be 4 cents, and erasers can be 16 cents

or

[b]Pencils can be 7 cents and erasers can be 3 cents.[/b]

We must used the bolded option because the problem states that
$ \text{Cost of Pencil} > \text{Cost of Eraser}$

Therefore, the cost of one pencil and one eraser is $ 7 \plus{} 3 \equal{} \boxed{10 \ \text{Cents}}$. [/hide]

[hide="or"]
We have $ 13p\plus{}3r\equal{}100$. Taking mod 3 of both sides, we get
$ p\equiv 1$ (mod 3), so $ p\equal{}1,4,7,...$  Anything larger than 7 won't work, and for each of 1,4,7, we get $ r$ can be equal to 29, 16, and 3.  Only $ (7,3)$ satisfies the last condition, so the answer is 7+3, or $ \boxed{10}$.
[/hide]

#1 can be solved with slope

[hide="Another Solution of Mine"]
Call the cost of a pencil $ y$ and the cost of a pencil and eraser $ s$.
$ 10y\plus{}3s\equal{}100$
Therefore, since 10y has to be a multiple of 10, s has to be a multiple of 10 in order for the sum of them to be 100. Now, knowing this, s could either be 10, 20, or 30.
Now, using this information, we can find that either:
Pencils cost:
7 cents
4 cents
1 cent
Now, if pencils costed 7 cents,
Erasers would cost:
3 cents
16 cents
29 cents
However, the only way for pencils to be worth more than erasers would be for pencils to be worth 7 cents and erasers 3 cents.
Therefore, the total of one pencil and one eraser would be [b]10 cents[/b].
[/hide]

Seeing as you had gotten the prices, what was the point of using $ s$ for the sum?

[quote="ZhangPeijin"]1) A line passes through point A, (1,1) and point B, (100,1000). How many points with integer coordinates are located on this line?

[/quote]

Look at that wording, that implies the answer is inifity.

I think you mean between A and B inclusive? (or not inclusive?  I cant tell because the wordinig is ambigious)

[quote="alanchou"]segment, maybe?

and $ A$ and $ B$ are the endpoints, I would assume

Mathcounts has never had a problem where the answer was $ \infty$[/quote]

I belive that was an AMC question..I think I remember it..(it doesnt really matter all AMC questions can be mathcounts questions though)

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1263958893&t=60022

heres the link, infact the answer is 8.

Its only stricly between the points.

For #2, draw a circle with radius of 10, and find the integer coordinates. My guess is [u]143[/u]. Probably wrong.

Umm, no that wouldn't work.
@Alan, Yesh.

Number 2 is 121, just look for a pattern

start with 0,1,2,3 moves...

the pattern is easy to find.

[quote="alanchou"]except for the ones with trig :wink:

@Peijin - did you put in #1 cause you know I hate fractions?[/quote]
Does ZhangPeijin not like you?

For #1, pretending that it's a segment, the site that the future put explained it this way.

Since the slope is $ \frac{111}{11}$, you increase the first ordered pair by 11. That gives you the total of 8. (12 through 100)

[quote="DonkeyKong"][quote="alanchou"]except for the ones with trig :wink:

@Peijin - did you put in #1 cause you know I hate fractions?[/quote]
Does ZhangPeijin not like you?

For #1, pretending that it's a segment, the site that the future put explained it this way.

Since the slope is $ \frac {111}{11}$, you increase the first ordered pair by 11. That gives you the total of 8. (12 through 100)[/quote]

actually 12 through 89 is 8

but the way it is worded^^, the answer is 1 though 100, which is 10.

@future - More importantly, you have to prove the pattern holds.

And drawing a good diagram actually kinda trivializes the problem.

[quote="DonkeyKong"]Parden me for double posting, but on second thought, does #2 make a square when drawn?

EDIT: It happens that 3 posts beat me.[/quote]

yea, good observation

a slanted square and yea if you shift that back to origin (rotate it) there will be (n+1)^2

[quote="alanchou"]You mean you don't even know?  Which would imply you memorized the answer...

How many points are there that are10 up from the x-axis?  how about 9? 8?  continue down to 1, then multiply by 2 and then consider the points on the x-axis

@future - have any explaination for why that formula's true?  It seems to simple to be so...[/quote]
?
To me?
Anyways, you solve this because the object could end up in a place 10 points left, right, up, or down of the origin.
So then you basically have 21 points you could end up at on the y axis, 19 to the left and right of it, 15 to 2 units to the left and right of the y axis, etc.
$ 21 \plus{} 2(19) \plus{} 2(17) \plus{} .... \plus{} 2(1)$
Which I'm too lay to compute.

[quote="the future"][quote="alanchou"]You mean you don't even know?  Which would imply you memorized the answer...

How many points are there that are10 up from the x-axis?  how about 9? 8?  continue down to 1, then multiply by 2 and then consider the points on the x-axis

@future - have any explaination for why that formula's true?  It seems to simple to be so...[/quote]

Umm no but its easy with observation..

anyways i know that an even move will result in a point (a,b) such that a+b is even.

odd, (a,b), a+b=odd..

Thats easy to prove, then just use that square and find all lattice points inside of it, then subtract the lattice points which have odd sum for its coordinates..[/quote]
Oh, I get it. I get what you are talking about. Since it's 10 by 10 square, there will be one more line of points, so $ (n\plus{}1)^2$ holds true. If you don't understand start out with 1 by 1 square with 4 dots. That's how I got it.

[quote="DonkeyKong"][quote="the future"][quote="alanchou"]You mean you don't even know?  Which would imply you memorized the answer...

How many points are there that are10 up from the x-axis?  how about 9? 8?  continue down to 1, then multiply by 2 and then consider the points on the x-axis

@future - have any explaination for why that formula's true?  It seems to simple to be so...[/quote]

Umm no but its easy with observation..

anyways i know that an even move will result in a point (a,b) such that a+b is even.

odd, (a,b), a+b=odd..

Thats easy to prove, then just use that square and find all lattice points inside of it, then subtract the lattice points which have odd sum for its coordinates..[/quote]
Oh, I get it. I get what you are talking about. Since it's 10 by 10 square, there will be one more line of points, so $ (n \plus{} 1)^2$ holds true. If you don't understand start out with 1 by 1 square with 4 dots. That's how I got it.[/quote]
Dude, the square has a diagonal of 21...
10 going up,
10 going down,
1 on the origin.

@ZhangPeijin: I was only talking about the formula, not the answer. I got what you said.

[quote="Ihatepie"][quote="ZhangPeijin"]1) A line segment has points A, (1,1) and B, (100,1000) as endpoints. How many points with integer coordinates are located on this line?

----------------------------------------------------------------

2) An object on the plane moves from one point with integer coordinates to another. At each step, the object may move one unit up, down, right, or left. If the object starts at the origin and takes 10 steps, how many different points could be its final point?

----------------------------------------------------------------

3) 13 pencils and 3 erasers cost a total of 1USD. Assuming that both items costs a whole number of cents and a pencil is worth more than an eraser, what is the cost of one pencil and one eraser?[/quote]

[hide]

1.
So there is a 99 difference in between the pt.A x coord. and the pt.B x.coord
There is a 999 difference between the pt. A y coord. and the pt. B y coord.

So you prime factorize each one.

99=11*3^2
999=37*3^3

You can see the GCF of the numbers is 9.

So there are 9 points with integer lengths. But don't forget The first point or 1,1.

So that makes 10 points.

2.
I have no idea.

3.
So 13 pencils and 3 erasers equals 1USD or 1.00

To make it easier, I will change it to 100 cents.

So what you have to do is have a certain cost per pencil so the remainder of the dollar is divisible by 3.

Since a pencil has to cost more than an eraser, you start as high as possible.

7 cents each per pencil or 91 for all pencils so 9 cents for erasers.

Luckily, I got it right on the first try
So pencils are 7 cents and erasers are 3 cents.

It asks what the cost is for a pencil and an eraser is so

7+3 equals 10 cents

[/hide]

@Zhang Wouldn't the square have a diagonal of 20 then?

If there are 21 points then the length is 20.[/quote]
Oh yeah...
Dangit!


Γεια σας και απο μενα,
γνωριζει κανεις σας προβληματα (ιδιως σε ανισοτητες) που μοιαζουν να λυνονται αποκλειστικα με χρηση αλγεβρας αλλα να λυνονται και γεωμετρικα; Ειχα πετυχει πριν λιγο καιρο σε ενα παμπαλαιο τευχος του Ευκλειδη  μια ανισοτητα ,σε ΙΜΟ νομιζω ειχε μπει, που δινοταν και η λυση ενος ρωσου διαγωνιζομενου ο ποιος εκανε γεωμ. κατασκευη και την απεδειξε με ανισοτητα Πτολεμαιου! :ninja: . Τρομερα ανορθοδοξος ο τροπος βεβαια αλλα απιστευτος...
ολες οι προτασεις ευπροσδεκτες.
       Ιακωβος

Στέλνω ένα μέρος της συλλογής που έχω σε μιγαδικούς για το θέμα που ζητάς Υπάρχει και το αντίστροφο, δηλαδή  Εκφώνηση γεωμετρική,λύση αλγεβρική καθώς και θέματα που αφορούν νιοστές ρίζες της μονάδας και πολύγωνα

Ωραίες οι ασκησεις κ. Μπόρη! Σας ευχαριστώ πολύ. Απλώς ψάχνω εκφωνήσεις που να είναι ''ολυμπιακού" πνεύματος ιδιώς όσον αφορά ανισοτικές σχέσεις που μοιάζουν να λύνονται με κλασικές ανισότητες... Θα ανεβάσω με την πρώτη ευκαιρία εκείνο το συγκεκριμένο πρόβλημα. Ιάκωβος

Μια πολύ κλειστή ερώτηση σε μια πολύ ανοιχτή απάντηση! Ξεκίνα απο το βιβλίο Recent Advances in Geometric Ineqs, D. S Mitrinovic, Kluwer, 1989. Δεν γνωρίζω αν υπάρχει η όχι σε ηλεκτρονική μορφή, πάντως η έντυπη θα σε ενθουσιάσει.


you could....picking up trash....

i have to think of some community serivce things also... :)  to  make my application look good..


function
algebra
domain
LaTeX
absolute value

Given f(x) = (x^2 + 2)/(x + 4), answer a through d.

(a) What is the domain of f?

(b) Is the point (1, 3/5) on the graph of f?...Why?  Why not?

(c) If x = 0, what is f(x)?  What point is on the graph of f?

(d) If f(x) = 1/2, what is x?  What point is on the graph of f?

Thanks!

[hide]

(a)

Domain is $x\in\mathbb{R}$ such that $x\neq-4$

(b)

$f(1)=\frac{1+2}{1+4}=\frac{3}{5}$

Therefore the Point $\left(1,\frac{3}{5}\right)$ is on the graph $f(x)$

(c)

$f(0)=\frac{2}{4}=\frac{1}{2}$

Therefore the Point is $\left(0,\frac{1}{2}\right)$

(d)

$\frac{1}{2}=\frac{x^{2}+2}{x+4}$
$\Rightarrow x+4=2x^{2}+4$
$\Rightarrow 2x^{2}-x=0$
$\Rightarrow x(2x-1)=0$
$x=0,\frac{1}{2}$

Therefore the points are: $\left(0,\frac{1}{2}\right)$ and $\left(\frac{1}{2},\frac{1}{2}\right)$
[/hide]

 :maybe:  how do u write the "such that" sign in latex is it just the absolute value: "|"

I follow you through and through.

I can now work out similar questions in my study booklet.


How many rectangles of any size are on an 8 x 8 checker board?

If you mean rectangles whose sides are on the lines:

There are many solutions; here is the one I'm most familiar with:

[hide="Solution"]A rectangle is defined by two vertical and two horizontal lines. There are $ \binom{9}{2}$ ways to choose two vertical lines, and the same number of ways to choose two horizontal lines.

$ \binom{9}{2}^2 \equal{} \boxed{1296}$[/hide]

That's what I do.
So basically, in an $ m\cdot{n}$ board (chessboard in this case), since there are $ m\plus{}1$ vertical lines, and $ n\plus{}1$ horizontal lines, the total number of rectangles is $ \binom{m\plus{}1}{2}\cdot{\binom{n\plus{}1}{2}}$.


Prove that the area of a triangle is $rs$ where r is the incenter and s is the semiperimeter.

http://artpad.art.com/?j5cd7a18epxo
$\frac{ABr}{2}+\frac{BCr}{2}+\frac{ACr}{2}=$area.
Factoring gives $r(\frac{AB+BC+AC}{2})=rs$.
By the way the formula is rs not $\frac{rs}{2}$

[quote="daermon"]its $rs$ not $\frac{1}{2}rs$.  you can easily prove it by drawing radii perpendicular to the sides and using base$\times$height[/quote]That is what I did if you click on the link ;)

Proofs should never be in middle school forums, even if they're easy.

I know this one is easy because I can do it... I suck at proofwriting...


prove that for any positive real $x$, $\lfloor{nx}\rfloor\ge \sum_{i=1}^{n}\frac{\lfloor{kx}\rfloor}{k}$
the solution at kalva is long a bit, I think there is a shorter one.

Hermite's identity:
\[[nx]=[x]+\left [ x+\frac{1}{n}\right ]+...+\left [ x+\frac{n-1}{n}\right ] \]
And we have $k\left [x+\frac{k-1}{n}\right ]\geq k[x]\geq [kx],\forall k \in \{1,2,..,n\}$

Your last inequality $ k[x]\ge[kx]$ is wrong. Take, for example, $ k\equal{}4,x\equal{}2.3$.

Anyway, this result is a special case of a problem from [url=http://www.mathlinks.ro/viewtopic.php?t=79787]APMO 1999[/url], by setting $ a_k\equal{}\minus{}\lfloor kx\rfloor$.


Can someone please explain the change in sntropy of universe,system surrounding in revresible irreversible processes..Can absolute value of Entropy be setermined

In short: $ \Delta S_{univ} \geq 0$ for any process, where the equality holds for reversible processes; for a given change of state, $ \Delta S$ is the same wether the process is reversible or irreversible (because entropy is a state function).

Carcul can u give me a example where equality doesn't hold

Consider expanding a piston against a vacuum.  Then we get no change in the entropy of the surroundings but an increase in entropy of the system (more positions -> more entropy).

Regarding absolute entropy, yes it can be determined.  This comes from the Third Law of Thermodynamics: the (absolute) entropy of an elemental crystal at 0K is 0.  Then take the $ \Delta S$ from such a state to the desired state, and one can calculate absolute entropy.

[quote="Gregx007"]Then we get no change in the entropy of the surroundings[/quote]

Only if the system is surrounded by adiabatic walls (Joule experiment).

[quote="Pardesi"]why is there increae in entropy of the system[/quote]

Consider, as a simple example, the Joule experiment for an ideal gas. As the gas expands to a vacuum, w = 0. As the system is surrounded by adiabatic walls, q = 0. As the gas is ideal, $ \Delta U \equal{} 0$ and so $ \Delta T \equal{} 0$. Therefore, we have

$ dU \equal{} TdS \minus{} PdV \equal{}> dS \equal{} \frac{dU}{T} \plus{} \frac{PdV}{T} \equal{} nR \frac{dV}{V}$,

and integrating we get $ \Delta S \equal{} nR \ln\frac{V_2}{V_1}$.

As the volume increases, entropy also increases.


conics ellipse geometry calculus integration

Whats the circumference of an ellipse? Its not stated in AOPS2, so I was wondering what it was....
Just in case you wanted to know, it came up on an algebra2 vestavia test.

Better yet, [[ellipse]].

It has example problems.

[quote="ckck"]Whats the circumference of an ellipse? Its not stated in AOPS2, so I was wondering what it was....
[/quote]

Although Aops2 didn't explicitly give the circumference, it did show a method of finding the area of an ellipse through projections, which I'm quite sure can be modified slightly to find the circumference.

Unfortunately no, there is no closed form for the circumference of a ellipse (in terms of elementary functions). If you try to find it using calculus you end up with an elliptic integral (of the second kind I beleive), and alas elliptic integrals generally cannot be computed using elementary functions.


Hi all,

I've been looking at the Physics Olympiad website and can't find any information on how to register for the tests or when they occur.  Does anyone know?

Thanks!

Hehe, there have been a billion people asking about this.  Information about when the tests will take place should become available later.  Until then, see:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=202627
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=196710


algebra
system of equations
algebra unsolved

slove the following system of equations:
$ x,y \in \mathbb{C}$,
$ x^4\plus{}y^2\minus{}xy^3\minus{}\frac{9}{8} x\equal{}0$
$ y^4\plus{}x^2\minus{}yx^3\minus{}\frac{9}{8} y\equal{}0$

Please help me, thanks.

First of all, see that if one of the variables is equal to $ 0$, then the other is equal to $ 0$ too and $ (0,0)$ is a solution.
So we can assume that $ xy\neq0$.
Multiplying first equation by $ x^2$ and second by $ y^2$ we have:

$ x^6\plus{}x^2y^2\minus{}x^3y^3\minus{}\frac{9}{8}x^3\equal{}0$,
$ y^6\plus{}x^2y^2\minus{}x^3y^3\minus{}\frac{9}{8}y^3\equal{}0$.

So $ 0\equal{}(x^6\minus{}y^6)\minus{}\frac{9}{8}(x^3\minus{}y^3)\equal{}(x^3\minus{}y^3)(x^3\plus{}y^3\minus{}\frac{9}{8})$.

We have two possibilities:
$ (1)$, $ x^3\equal{}y^3$;
These equations turn into $ y^2\equal{}\frac{9}{8}x$ and $ x^2\equal{}\frac{9}{8}y$ so $ x^4\equal{}\frac{81}{64}y^2\equal{}\frac{729}{512}x$ so $ x^3\equal{}y^3\equal{}\frac{729}{512}.$
So $ x\equal{}\frac{9}{8}\cdot \left(\frac{1\plus{}\sqrt{\minus{}3}}{2}\right)^k, y\equal{}\frac{9}{8}\cdot \left(\frac{1\minus{}\sqrt{\minus{}3}}{2}\right)^k$ where $ k\in \{0,1,2\}$ - next three solutions.

$ (2), x^3\plus{}y^3\equal{}\frac{9}{8}$;
Equations turn into $ y^2\equal{}2xy^3$ and $ x^2\equal{}2x^3y$ so $ xy\equal{}\frac{1}{2}, x^3y^3\equal{}\frac{1}{8}$ so we have next six solutions:
$ x\equal{}\left(\frac{3}{4}\plus{}\frac{(\minus{}1)^n}{4}\right)\left(\frac{1\plus{}\sqrt{\minus{}3}}{2}\right)^k, y\equal{}\left(\frac{3}{4}\minus{}\frac{(\minus{}1)^n}{4}\right)\left(\frac{1\minus{}\sqrt{\minus{}3}}{2}\right)^k$ where $ n\in\{0,1\}, k\in\{0,1,2\}$.

[quote="mazur89"]First of all, see that if one of the variables is equal to $ 0$, then the other is equal to $ 0$ too and $ (0,0)$ is a solution.
So we can assume that $ xy\neq0$.
Multiplying first equation by $ x^2$ and second by $ y^2$ we have:

$ x^6 \plus{} x^2y^2 \minus{} x^3y^3 \minus{} \frac {9}{8}x^3 \equal{} 0$,
$ y^6 \plus{} x^2y^2 \minus{} x^3y^3 \minus{} \frac {9}{8}y^3 \equal{} 0$.

So $ 0 \equal{} (x^6 \minus{} y^6) \minus{} \frac {9}{8}(x^3 \minus{} y^3) \equal{} (x^3 \minus{} y^3)(x^3 \plus{} y^3 \minus{} \frac {9}{8})$.

We have two possibilities:
$ (1)$, $ x^3 \equal{} y^3$;
These equations turn into $ y^2 \equal{} \frac {9}{8}x$ and $ x^2 \equal{} \frac {9}{8}y$ so $ x^4 \equal{} \frac {81}{64}y^2 \equal{} \frac {729}{512}x$ so $ x^3 \equal{} y^3 \equal{} \frac {729}{512}.$
So $ x \equal{} \frac {9}{8}\cdot \left(\frac {1 \plus{} \sqrt { \minus{} 3}}{2}\right)^k, y \equal{} \frac {9}{8}\cdot \left(\frac {1 \minus{} \sqrt { \minus{} 3}}{2}\right)^k$ where $ k\in \{0,1,2\}$ - next three solutions.

$ (2), x^3 \plus{} y^3 \equal{} \frac {9}{8}$;
Equations turn into $ y^2 \equal{} 2xy^3$ and $ x^2 \equal{} 2x^3y$ so $ xy \equal{} \frac {1}{2}, x^3y^3 \equal{} \frac {1}{8}$ so we have next six solutions:
$ x \equal{} \left(\frac {3}{4} \plus{} \frac {( \minus{} 1)^n}{4}\right)\left(\frac {1 \plus{} \sqrt { \minus{} 3}}{2}\right)^k, y \equal{} \left(\frac {3}{4} \minus{} \frac {( \minus{} 1)^n}{4}\right)\left(\frac {1 \minus{} \sqrt { \minus{} 3}}{2}\right)^k$ where $ n\in\{0,1\}, k\in\{0,1,2\}$.[/quote]

Yes, thank you. The answers are correct. But why do you want "Multiplying first equation by $ x^2$ and second by $ y^2$", I want to know. Can you tell me?


quadratics
Pythagorean Theorem
geometry
algebra

[b][size=150]Problem has been solved.
No further assistance is required.[/size][/b]
I'm looking at the pythagorean theorem, and something catches me:

$a^2-6a-27=0$
$(a-9)(a+3)=0$
somehow it turns into

[b]a-9=0 (==) a=9
or
a+3=0 (==) a=-3[/b]

Which it then goes with 9.

How did it get to the parts in bold?

[quote="ShoeFactory"]What do you mean?

The bold parts is finding the zeros of the function by some theorem that I don't remember the name of.

Then it throws away -3 because its extraneous.[/quote]

Yeah, side lengths can't be negative.

How does this:
$(a-9)(a+3)=0$

turn into this:
[b]a-9=0 (==) a=9
or
a+3=0 (==) a=-3[/b]

Were the words you were looking for quadratic equation?

[quote="ShoeFactory"]What do you mean?

The bold parts is finding the zeros by some theorem that I don't remember the name of. It is because a product of two things = 0 so you can set each of them = to 0.

Then it throws away -3 because its extraneous.

If you're not asking about that then disregard my post.[/quote]

It's the Zero-Product Property: If $ab=0$, then $a=0$ or $b=0$.

Set each product part thingy to 0; thats how you find every solution.

Does this question really belong in the high school forum?

To SplashD:
http://www.mathlinks.ro/Forum/viewtopic.php?p=551704#551704

That's my response since the webmaster is too stupid, yes stupid--not ignorant, to figure these things out.

[b]BTW[/b]
I think i see what's going on:

(a-9)(a+3)=0

The two grouped numbers are split apart.
I guess I didn't notice this until the fifth time.
But... Why are they split apart?   :|
I don't understand that..

[quote="Cyberman"][quote=SplashD] Whine whine whine..[/quote]
[/quote]

That's mis-quoting(which is illegal).

[quote="Cyberman"][quote="lotrgreengrapes7926"]
It's the Zero-Product Property: If $ab=0$, then $a=0$ or $b=0$.[/quote]

Hmm.. this must be what I'm looking for. I've never heard of this property, even though I've already taken algebra. Interesting.

I question why these things are still split apart, though. What proof exists?[/quote]

This property is included in the basic algebra curriculum.

anything times 0 is 0!!!

Yes, but I don't see where the consideration to split them into their own equations came from. However, the basic algebra concept that anything times 0 is 0 is a concept I understand very well. When factored materials were split into individual equations, my reactions were stumped; for I knew not what direction to take next. However, it seems that the zero property came to be an oddity at this point of idea:

8*9=0
8=0
9=0

Of course, such things are contradictions, of which I notice. To have seen factored materials in a different tune was not usual.

I assume the zero property had a different accordance in this particular problem because of an unknown variable and its ability to equal something other than zero.

[quote="Cyberman"]Yes, but I don't see where the consideration to split them into their own equations came from. However, the basic algebra concept that anything times anything is 0 is a concept I understand very well. When factored materials were split into individual equations, my reactions were stumped; for I knew not what direction to take next. However, it seems that the zero property came to be an oddity at this point of idea:

8*9=0
8=0
9=0

Of course, such things are contradictions, of which I notice. To have seen factored materials in a different tune was not usual.

I assume the zero property had a different accordance in this particular problem because of an unknown variable and its ability to equal something other than zero.[/quote]
You only use the zero-product property for variables, not for given constants.

Maybe it was covered and I didn't notice it.
Either way, thanks all for the help.

[quote="Cyberman"]To SplashD:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=551704#551704

That's my response since the webmaster is too stupid, yes stupid--not ignorant, to figure these things out.
[/quote]

These forums are geared toward problem solving, not homework help.

Most of us take algebra in middle school, so it should be there. If you take a look at the content in the high school forums, none of it is as simple as this.

If you say you use SOSmath or whatever why don't you post there? I'm sure V.V. wouldn't appreciate you posting insults here.

If you have the equation $(x-1)(x-2)(x-3)(x-4)=0$, then any one of the equations
$x-1=0$, $x-2=0$, $x-3=0$, $x-4=0$
will solve the system. Try plugging in $x=1$, $2$, $3$, and $4$ to see how it works. However, this property is for a situation when you are trying to solve for variables; something like $5\times8\times4=0$ isn't what the property is about (and doesn't even make sense either).

And finally,
[quote]To SplashD:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=551704#551704

That's my response since the webmaster is too stupid, yes stupid--not ignorant, to figure these things out.[/quote]
With such a large forum here, the admins are usually very busy and may not have time to help every person with every problem he/she has right away. If he didn't understand what you meant the first time you asked, try explaining again [i]politely[/i] to get your message through. There's no reason to call anyone "stupid."

[quote="Ignite168"]I never learned the Zero Product Property explicitly either but wouldn't it be obvious? Two non-zero numbers cannot multiply to zero so one of them or both must be 0.[/quote]

That's what we've been saying all along.

His question has been answered, so there is no need for unnessessary spam, so this topic is now locked.


function
algorithm
number theory
relatively prime
number theory unsolved

I will give an example ;
φ(16) =8  since 1, 3, 5, 7, 9, 11,13, 15 are relatively prime to 16.
I wonder that what is the reason of
1+15=3+13=5+11=7+9=16

I'm not sure I fully understood your question...
anyways, if $ (x,a)\equal{}1$, then $ (x\minus{}a,x)\equal{}1$ (it follows from Euclid's Algorithm), and vice versa (when x>a). This is why we can partition the set of numbers less than natural x and relatively prime to x into pairs of numbers whose sum is x (except for x=2).


geometry
integration
calculus
calculus computations

In class, my teacher was going over finding the area under the curve of a graph as well as finding the volume when the graph is revolved around a line. So, we were given the following area to compute:

$ \int_{2}^{\infty}\frac{1}{x-1}dx=\infty $

Similarly, if you revolve the graph around the $ x $-axis, with the same bounds you arrive at the following volume:

$ \pi \int_{2}^{\infty}\left(\frac{1}{x-1}\right)^{2}dx=\pi $

So, my question is how can something have an infinite surface area, but when revolved around a line, have a definite volume?

At the time this was discovered, it was considered a paradox. The apparent paradox has been described informally by noting that it seems it would take an infinite amount of paint to coat the interior surface, but it also seems that it would be possible to simply fill the interior volume with a finite amount of paint and so coat the interior surface. The resolution of the paradox is that the implication, that an infinite surface area requires an infinite amount of paint, presupposes that a layer of paint is of constant thickness; this is not true in theory in the interior of the horn, and in practice much of the length of the horn is inaccessible to paint, especially where the diameter of the horn is less than that of a paint molecule. - If the paint is considered without thickness, it would further take infinitely long time for the paint to run all the way down to the "end" of the horn.

Another way this "paradox" is often proposed is by the statement that one could "fill the horn with paint, but not have enough to paint the outside".
copied from
http://en.wikipedia.org/wiki/Gabriel's_horn


Prove that $\frac{tan \theta}{1-cot\theta}+\frac{cot \theta}{1-tan\theta}= 1+tan\theta+cot\theta$

Hello :)

[hide]

$\frac{\frac{sin x}{cos x}}{1-\frac{sinx}{cosx}^{-1}}\, = \, \frac{sin x}{cosx-\frac{cosx \cdot sinx}{cosx}^{-1}}\, = \, \frac{sin x}{cosx-(1+\frac{sinx}{cosx})^{-1}}$

$\frac{sin x}{cosx-(1+\frac{sinx}{cosx})^{-1}}+\frac{1}{\frac{sinx}{cosx}-\frac{sin^{2}x}{cos^{2}x}}\, = \, 1+\frac{tan^{2}x+1}{tanx}$

Note that: ${\frac{1}{\frac{sinx}{cosx}-\frac{sin^{2}x}{cos^{2}x}}\, = \, (tan x-tan^{2}x})^{-1}$

$\frac{\frac{1}{tanx}}{1-tanx}\, = \, 1+tan x+\frac{1}{tanx}-\frac{tanx}{1-\frac{1}{tanx}}$

$\frac{1}{tanx}= 1-\not{tanx}+\not{tanx}-tan^{2}x+\frac{1-tanx}{tanx}-\frac{tanx-tan^{2}x}{1-\frac{1}{tanx}}$

$\not\frac{1}{tanx}= \not 1-tan^{2}x+\not cotx \not-1-\frac{tanx-tan^{2}x}{1-\frac{1}{tanx}}$

$tan^{2}x(1-\frac{1}{tanx}) \, = \,-tanx+tan^{2}x$

$tan^{2}x-tanx \, = \,-tan x+tan^{2}x$

[/hide]

By the way, how can I cut a whole thing, like $\frac{5}{x}+1$ ?

[hide] \[\begin{aligned}\frac{\tan x}{1-\cot x}+\frac{\cot x}{1-\tan x}& = \frac{\tan x\cdot \sin x}{\sin x-\cos x}+\frac{\cot x \cdot \cos x}{\cos x-\sin x}= \frac{\tan x\cdot \sin x-\cot x\cdot\cos x}{\sin x-\cos x}\\ & = \frac{\sin^{3}x-\cos^{3}x}{\sin x\cdot\cos x\cdot\left(\sin x-\cos x\right)}= \frac{\sin^{2}x+\sin x\cdot\cos x+\cos^{2}x}{\sin x\cdot \cos x}\\ & = 1+\frac{\sin^{2}x+\cos^{2}x}{\sin x\cdot\cos x}= 1+\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}= 1+\tan x+\cot x\end{aligned}\] [/hide]

[hide="Sorry yet another solution but I promise it is different..."] \[\frac{\tan x}{1-\cot x}+\frac{\cot x}{1-\tan x}{? \choose =}1+\tan x+\cot x\] $\begin{aligned}\equiv (\tan x)\cdot (1-\tan x)+(\cot x)(1-\cot x){? \choose =}\ (2-\tan x-\cot x)\cdot (1+\tan x+\cot x) \\ \tan x-\tan^{2}x+\cot x-\cot^{2}x{? \choose =}\ (1+1-(\tan x+\cot x))\cdot (1+\tan x+\cot x)= (1+\tan x+\cot x)+1-(\tan x+\cot x)^{2}\\ \tan x-\tan^{2}x+\cot x-\cot^{2}x = 1+1+\tan x+\cot x-(\tan^{2}x+\cot^{2}x+2)_{OREGON}\end{aligned}$
Yay done[/hide]

[quote="Tiago Guerreiro"]Hello :)

By the way, how can I cut a whole thing, like $\frac{5}{x}+1$ ?[/quote]

How exactly do you want to "cut" the whole thing ?? By my interpretation, the simple way out would be to take an L.C.M and then cut.... so you should rather cut $\frac{5+x}{x}$

[quote="Tiago Guerreiro"]Hey

I'm seeing I was not clear :P

I want to cut it with latex, because if I type: \not{\frac{5}{x}+1} will appear this: $\not{\frac{5}{x}+1}$[/quote]
I don't think that's necessary. We can understand your equations and steps without the canceling out written in.


The height of a trapezoid is 10 meters and the length of the longer base is 12 meters. Given that its area is 105 square meters, what is the number of meters in the length of the shorter base?

Let $ x$ be the length of the shorter base.

$ (\frac{1}{2})(10)(12\plus{}x)\equal{}105$
$ 5(12\plus{}x)\equal{}105$
$ 12\plus{}x\equal{}21$
$ x\equal{}\boxed{9}$ meters

9 Meters would be the correct answer. :)


I wanted to see the 2008 but mathcounts.org took it off their website.  Where is it?

Do you have a link?  I'm not finding it..........

[quote="cgyao15"]Look somewhere down there in the posts, that's where i saw it.[/quote]

[Sarcasm]Nice Description[\Sarcasm]

[url=http://lmgtfy.com/?q=2008+Mathcounts+Highlight+Video]Let Me Google That For You[/url]

Also, try using archive.org or something..idk? :maybe:

Ah, and that was quite helpful as well, I'm sure.  :P

They took it off, so you can't view it anymore.

@King1 pls don't use the sarcasm brackets; the are very offensive.

Yeah, I did find a link to but it was "out of date", just like practice MC Tests on Mathacle.com(those people just decided to "take it down" :()

The first link in the google search was this topic...

They take it down before the next year's come up, just like the tests, countdowns, and other stuff from just that year.

Just save it if you want to keep it.

But I never got to see it in the first place!  Is there any way I can see it?


AMC
AIME
real analysis
ARML
graph theory

After I fail the AIME tomorrow and can finally bring myself to learn new math, rather than just AIME math, I'd like to learn some graph theory.  What's the book for this?  An undergraduate level introduction textbook would be great.

[quote="ComplexZeta"]The book by Diestel is nice, although it's a graduate-level text. If you go to Diestel's website, you can download a copy, but I think printing is disabled.[/quote]

I had a feeling you would know this :)

This is the text used at my university.  I've also seen it recommended at least once on the MathLinks forums.

http://www.amazon.com/exec/obidos/tg/detail/-/0130144002/104-0413591-1871908?v=glance

I learned the little graph theory I know out of Grimaldi's discrete math text and another text whose name I can't remember off hand.

[quote="Magnara"]The amazon peoples don't like that book.  They very rarely steer me wrong on these things.[/quote]

I just read the reviews and the errors would probably scare me off, now that I look at it.  However, I think you need to be careful about what the 'amazon peoples' say - in some cases, a book like Rudin which is nearly universally acknowledged as 'THE' introductory analysis book can have its rating dragged down by people who weren't very interested in mathematics and hence had their grades dragged down by Rudin.

A Course in Combinatorics by van Lint and Wilson is a nice read. I guess you can amazon it if you want. cya at arml practice.


For positive integers $t,a,b,$a $(t,a,b)$-[i]game[/i] is a two player game defined by the following rules. Initially, the number $t$ is written on a blackboard. At his first move, the 1st player replaces $t$ with either $t-a$ or $t-b$. Then, the 2nd player subtracts either $a$ or $b$ from this number, and writes the result on the blackboard, erasing the old number. After this, the first player once again erases either $a$ or $b$ from the number written on the blackboard, and so on. The player who first reaches a negative number loses the game. Prove that there exist infinitely many values of $t$ for which the first player has a winning strategy for all pairs $(a,b)$ with $a+b=2005$.

I just found that if for $x$ Player 1 wins, then he wins for $x+2005k$.

To prove it suppose Player 1 wins for $x$ between $1$ and $2005$, and call $P_1$ his sequence of moves. Now if they start from $2005k+x$ let Player 1 start with the firs step of $P_1$ and then if Player 2 substracts $a$, he substracts $b$ or backwards until they reach $x-$first step of $P_1$, from here, the game is analogus to the one started with $x$, becuase it is player 2 turn and player 1 can do $P_1$ from step two on!

Finally $x=2004$ works, because player 1 plays $max[a,b]$ and clearly he wins!

[quote="Pascual2005"]I just found that if for $x$ Player 1 wins, then he wins for $x+2005k$.

To prove it suppose Player 1 wins for $x$ between $1$ and $2005$, and call $P_1$ his sequence of moves. Now if they start from $2005k+x$ let Player 1 start with the firs step of $P_1$ and then if Player 2 substracts $a$, he substracts $b$ or backwards until they reach $x-$first step of $P_1$, from here, the game is analogus to the one started with $x$, becuase it is player 2 turn and player 1 can do $P_1$ from step two on!
[/quote]

all seems fine except for a small point which is not clear from the post: if for $x+(a+b)k$ it is a player 1 win then if he plays $x+(a+b)k\rightarrow x+(a+b)(k-1)+b$,say(i.e.he removes $a$) then inductively it is not clear that it reduces to $x$ since the second player could again remove $a$ instead of $b$ which will inductively reduce the problem.
but that is not a very major problem because we can induct on $k$ and all it suffices to show is that for $k=1$ the problem is reducible,i.e., if $f(x)$ denotes the winner of the game starting with $x$($f$ takes values $1$ or $2$) then $f(t+a+b)=f(t)$.if $f(t)=2$ then Pascual2005 's argument works fine so that case is easy.Now $f(t)=1$ iff either $f(t-a)=2$ or $f(t-b)=2$. suppose then that $f(t)=1$ but that both $f(t+a)=f(t+b)=1$(so that $f(t+a+b)=2$) then clearly we have $f(t+a-b)=f(t+b-a)=2$. but that means $f(t-b)=f(t-a)=1$ and that contradicts $f(t)=1$ and that finishes the proof.

If $A$ wins with $t$, then he wins with $t+a+b$ as follows: he makes the first move that he would make in order to win with $t$, say, he removes $a$. If $B$ removes $a$ as well, then $A$ removes $b$, and they are in the position $t-a$, with $B$ to move, i.e. just as if we had a game with $t$, and $A$ has just removed $a$, which is a winning strategy. Isn't it as simple as this? :?

well,i guess you are right...
and i think that's what Pascual was saying as well.i just didn't understand it properly..

yes thats what i was saying, easy problem right?

Is the problem really that easy? Or is the problem trying to say that there are infinite values of t such that all (t, a, b) games with a+b=2005 are [b]simultaneously[/b] good for player 1?

too late, but another solution:

for a sake of contradiction assume for some finite values of $t$ player $1$ has the winning strategy. so for some infinte values of $t$ player $2$ has the winning strategy.let $t=t_{0}$ be one of the values that player $2$ has the winning strategy. so player $1$ has the winning strategy for both $t_{0} +a,t_{0}+b$ as he can subtracts $a,b$ from the initial number and then continue as player $2$ continue  for the initial value $t_{0}$. by our assumption, we get that player $1$ also has winning strategy for some infinite values of initial numbers $t$. a contradiction.

You are required to prove that infinitely many such $t$ exist, where each $t$ wins for all $(a,b)$. You have only proven that for all $(a,b)$, infinitely many such $t$ exists; you haven't proven that each such $t$ works for all $(a,b)$.


Bob rows at a rate of 4 kilometers per hour in a stream without a current.  He rows under one bridge and then rows under a second bridge 10 kilometers upstream. As he rows under the second bridge, his hat falls off. After rowing 5 more kilometers, he realizes his hat is gone so he turns around catching the hat exactly as it passes under the first bridge. How fast is the current (in kilometers per hour)?

solution and reasoning please!

[hide="Answer?"] 0km/h, because you told us in the first sentence. [/hide]

uhh, no..........

it has a current when he goes back....

Let the speed of the current be $ x$ km/h.
He went 5 km farther from 2nd bridge, which took $ \frac{5}{4\minus{}x}$ hours for him to do that.

He realized that he dropped his hat, so he went back total of 5+10=15 km, which took $ \frac{15}{4\plus{}x}$ hours.

And, the hat just floated 10 miles by the speed of the current, which is x kmph.

Thus, $ \frac{5}{4\minus{}x}\plus{}\frac{15}{4\plus{}x}\equal{}\frac{10}{x}$

Solving for $ x$, we have $ x\equal{}2$

Answer: 2

There was a similar problem in ACoPS, but I can't remember the simpler solution...

actually that is incorrect.

the answer is 4

but i have no clue how....
ANY SUGGESTIONS?

Clarification: distance between the 1st bridge and the 2nd one is 10 km right?

i would assume so considering it says "rows under one bridge and then rows under a second bridge 10 kilometers upstream"

so yes...

I did the problem the same way and got the same answer as FantasyLover.

Also, if the current was 4 km/hr, Bob's upstream speed would be 0 km/hr, meaning he couldn't travel upstream.

actually, my bad. i think the answer is 2. sorry fantasylover :blush:

now, how does the denominator in the times change from plus to minus?
also, where does the denominator come from?

Lol.

At first, Bob is going upstream(through 2nd bridge), so you have to subtract Bob's speed from the speed of the current.

However, when Bob noticed that his hat was gone, he turned around, going downstream. So, now you have to add speed of the current to Bob's speed.

oh

and why is that equal to 10 over x. why is it over the current x?

$ \text{Distance}\equal{}\text{Speed}\times \text{Time}$

It took $ \frac{10}{x}$ hours for the hat to float 10 km.


At Pizza Perfect, Ron and Harold make pizza crusts. When they work separately Ron finishes the job of making 100 crusts 1.2 hours before Harold finishes the same job. When they work together they finish making 100 crusts in 1.8 hours. How many hours, to the nearest tenth of an hour, does it take Ron working alone to make 100 crusts?

$ \frac{1}{R}\minus{}\frac{1}{H}\equal{}1.2$

$ \frac{1}{R}\plus{}\frac{1}{H}\equal{}1.8$

$ \frac{2}{R}\equal{}3\\\\R\equal{}\frac{2}{3}\text{ hours}\approx\frac{7}{10}\text{hours}$

Are you letting the rate be $ R$ and the time be $ \frac{1}{R}$?

If so, then $ R\equal{}\frac{2}{3}$ implies that the answer should  be $ \frac{3}{2}\equal{}1.5$

However, this solution seems correct, but does not seem to match mine.............

I let $ h$ be the number of hours it takes harold to complete the job, and $ r$ be the number of hours is takes ronald to complete the job.

I set it up as $ h\minus{}r\equal{}\frac{6}{5}$ and $ \frac{h\plus{}r}{hr}\equal{}\frac{9}{5}$ and got something different --can someone tell me what I am doing wrong?

I randomly guessed [hide="answer"]3.1[/hide]  and that was the correct answer. Can someone tell me how to get this?

Let the [u]times[/u] be $ h,r$. From the first statement, $ h\minus{}r\equal{}1.2\equal{}\frac65\implies h\equal{}r\plus{}\frac65$. From the second statement, $ \frac1{\frac1h\plus{}\frac1r}\equal{}1.8\equal{}\frac95\implies\frac1h\plus{}\frac1r\equal{}\frac{h\plus{}r}{hr}\equal{}\frac59$. Substituting $ h\equal{}r\plus{}\frac65$ and simplifying results in $ \frac{2r\plus{}\frac65}{r^2\plus{}\frac65r}\equal{}\frac{10r\plus{}6}{5r^2\plus{}6r}\equal{}\frac59\implies90r\plus{}54\equal{}25r^2\plus{}30r\implies25r^2\minus{}60r\minus{}54\equal{}0\implies r\equal{}\frac{60\pm\sqrt{9000}}{50}\equal{}\frac{60\pm30\sqrt{10}}{50}\equal{}1.2\pm0.6\sqrt{10},\ r>1.2$
$ \implies r\equal{}1.2\plus{}0.6\sqrt{10}\approx1.2\plus{}0.6\cdot3.162\approx0.6(2\plus{}3.162)\approx0.6(5.162)$
$ \approx3.097\implies\boxed{3.1}$.


Prove that 4 is divisible $ C_{2n}^n\equal{}\frac{2n!}{n!n!}$ iff $ n$ isn't a power of 2

[hide="Hint"]$ C_{2n}^n\equal{}2C_{2n\minus{}1}^n$, use Lucas's theorem now.[/hide]

Exactly $ ord_2(C_{2n}^n)\equal{}ord_2(n\plus{}1)\plus{}ord_2(Ca_n)$ and $ ord_2(Ca_n)\equal{}m\minus{}1$, were m is number of nontrivial 2-adic digits n+1.
Therefore $ ord_2(C_{2n}^n)\ge 1$ and $ ord_2(C_{2n}^n)\equal{}1$ if and only if $ n\equal{}2^k$.


[quote="Quevvy"]$d_{n}=3n-2$ (if n is even)
$d_{n}=\frac{2}{15}n+3$ (if n is odd)
What is $d_{75}+d_{78}$?[/quote]

[hide]
$d_{75}=13$

$d_{78}=232$

edit: oops forgot it was addition so

$232+13=245$

[/hide]

is that right?

Extention: Using the same function "d", what is $d_{(d_{64})}$?

i believe Ubemaya is correct.
i get the same answer