geometry
perimeter
3D geometry
sphere
trigonometry
geometric transformation

About 3 years ago, Japanese board of education changed some policies on mathematics, and one of the most controversial issues was regarding the estimation of $\pi$. Before it was changed, we used $\pi\approx 3.14$ in elementary school. (We use approximation because we don't learn the symbol $\pi$ until we enter junior high.) However, surprisingly, the board of education decided to change the approximation value from $\pi\approx 3.14$ to $\pi\approx 3$! The main reason of the change was because they belived that kids can "save time" by not doing "ridiculus" calculations like $3.14\times 16$.

Although it was 3 years ago, I still feel very upset about this decision. What do you think?

P.S. We rarely use any calculator in elementary/junior high/high school in Japan. We do all calculations by hand!

well, in egypt $\pi$ is taught as $\frac{22}{7}$ in elementary schools then you're supposed to just keep the product as a fraction, i guess multiplying by $22$ is not seen as bad!
officially setting the approximation of $\pi$ to $3$ is very weird though...

man, that should just be for people from grades 1-4 since most 5th graders and higher know how to use the pi key and have a calc

Well, we don't usually use any calculator in Japanese elementary schools (or even junior high/highschool)...

I doubt most people in primary school even know what an inscribed hexagon is! :D

[quote="frt"]I see your point. But sometimes it causes some problems. For example, consider a regular hexagon and its circumcircle. Obviously their perimeters are different. However, using $\pi=3$, they become [b]the same[/b]! :([/quote]

  Yes, just get the teachers to make a note that when they are using apporxamations like $3$ for $\pi$ they will always come out to answers that are smaller than their actual values.  I think $3.1$ is the bext approxamation, its close to the actual value of $\pi$ and its not very difficult to handle in calculations.

[quote="yif man12"]Wasn't this also the case in some state some years ago? Like Indiana or somewhere like that?[/quote]

The full story is in [url=http://db.uwaterloo.ca/~alopez-o/math-faq/node45.html#SECTION00610000000000000000]the sci.math FAQ[/url].

In my math class, they allow me to simply leave it as $\pi$.  The teacher otherwise says "If you want to put it into decimal form, give it to 3 decimal places after the decimal point.  But if you're going to do that, you can't use the $=$ sign.  You have to use $\approx$."
He prefers that we leave it in exact form, so our method follows clearly (of course, the method is more important than the answer).  So I'm happy...

Malaysia too employ the value of [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/f75c927137d43dcb48ae34570098a42d.gif[/img]

[quote="K81o7"]In my math class, they allow me to simply leave it as $\pi$.  So I'm happy...[/quote]

Yep, that's how it's always been for me, that's how I like it best - keeping the pi's and rt(2)'s and rt(3)'s just as they are, in order to get the most exact answer.

My great uncle, who helped found the school of areospace medicine at The Ohio State University told, when he was still alive, of an effort by some of the OSU staff with too much time on their hands, to get the state of Ohio to change the value of PI to 3.  The report talked about how it wouldn't change the volume in cans and the like too much and it would give the students in Ohio a leg up, as they could calculate easier.

The professors got concerned when the Ohio Senate passed the bill with no opposition or discussion.  They told the legislature that it was a joke and the bill died before it could get to the Ohio House and be enacted.

Some ideas are persistent, even if they are not good ideas.

[quote="G-UNIT"][quote="236factorial"][quote="G-UNIT"]I think that if kids are taught about the actual value of pi, approxamations like "$3$" should very well be acceptable.  I remember as I child doing long tedious calculations and scoring very poorly on tests because the ".14" made the calculations unbearable.[/quote]

Same here, but memorzing 3.14 times any one digit number can save time.[/quote]

  Would you actually considering doing that?[/quote]

I've done it so much that I already have  :D

In South African it's frustrating because each teacher just decides what he wants his kids to do... so you end up using a different value for $\pi$ every year.

READ MY SIGNATURE :lol:  It's from the book [i]Going Postal[/i] by Terry Pratchett. A professor from a South African university commented on an extract from that book and said:

"Actually, some cars are fitted with wheels like that. These wheels have five or so stout curved spokes, and the plane in which the wheel makes contact with the butt of the axle is not the same as that in which the spokes join the rim. One can idealize such a wheel as lying on the surface of a sphere, with the rim being a circle and the spokes arcs of great circles. If the hub of the wheel is at a pole, the rim is at latitude 30 degrees. The universe in question is thus one with elliptic geometry, in which pi is not constant but can take any value between 0 and Roundworld-pi."

[quote="fleeting_guest"][quote="stupidkid"][quote="yif man12"]Wasn't this also the case in some state some years ago? Like Indiana or somewhere like that?[/quote]
uhh...I live in Indiana and we use 3.14 for $\pi$ in elementary schools.[/quote]

The famous episode he is referring to is of the Indiana state legislature passing a bill declaring the value of Pi to be 3.   It was a favor from one of the state senators to somebody in his district who had been claiming a method for squaring the circle.[/quote]

Um.  You're joking, right?  Who would do something as silly as declaring $\pi$ to be 3 as a favor to someone?

[quote="Saul"]I really don't see the point, from a mathematical perspective, of having any numerical value for pi... pi exists, and that's really all we need to know.[/quote]

Well, we need to find some circumfrences, and that's when it helps (wheel circumfrences, etc.)

[quote="frt"][quote="Derek"]In India, people use 3.1415926535 as the estimation for pi.

Jk.[/quote]
Wow, that's a lot :o
So when you calculate $16\pi$, for example, you need to do $16\times 3.1415926535$?[/quote]

he said Jk....

[quote="EUCLA"]we didn't use to calculate it we just put a number followed by $\pi$[/quote]
That's what we do in China. But in some physics problems, we will still have to use $3.14$.

well, i think pi should be 3.14 because of the fact that it is rounded and not exact and never will be exact! so using 3 is not really that smart, though thats just me :?

We set $\pi=3$ is to reduce the work done by the students in an exam.

[quote="EUCLA"]we didn't use to calculate it we just put a number followed by $\pi$[/quote]
Uh yeah, like any mathematician worth his or her salt does it this way! (unless he/she is also a physicist :P)
$\pi \approx 3$? What on earth?

[quote="Mariag"]a couple of weeks ago, i was tutoring a bunch of middle school kids about circles and such, and they almost attacked me when i said that $\pi$ isn't exactly $3.14$. they seemed to really dislike the fact that there could exist irrational numbers....[/quote]So they were actually [i]taught[/i] that $\pi$ is [i]exactly[/i] $3.14$? That's a little extreme...

I don't know about you guys, but I learned that it was 3.141592654 when I was about 9. So... yeah. Turns out that that's a little bit off, but who cares?

However, for some reason I was also taught that the square root of two is 1.41414141414141 repeating... strange world.....

[quote="ccy"][quote="fleeting_guest"]It was a favor from one of the state senators to somebody in his district who had been claiming a method for squaring the circle.[/quote]Um.  You're joking, right?  Who would do something as silly as declaring $\pi$ to be 3 as a favor to someone?[/quote]

Every geometric construction that attempts to "square the circle" amounts to some approximation for $\pi$.   If the person in the senator's district had a construction that, given a circle of radius $R$, produces a square of side $R \sqrt{3}$, the correctness of this construction (which he believed, published, etc) would imply that the true value of $\pi$ was $3$ and had remained unknown for centuries.

In the book, [i]More Joy of Mathematics[/i](by Theoni Pappas), they said that, "In 1700 B.C., the Egyptians value for pi, 256/81=3.16050...", they also said that archimedes proved pi was between 3 10/71 and 3 1/7 or 3.140845... and 3.142857....

The bible said pi was exactly 3.

Chinese mathmaticians in 125 A.D. said pi was sqrt(10)=3.162...

William Shanks in 1873 carried pi out to 707 places, but messed up at the 528th place without a calculator.

[quote="frt"]Although it was 3 years ago, I still feel very upset about this decision. What do you think?[/quote]Actually, it was $\pi$ years ago.

[quote="kyyuanmathcount"]

Chinese mathmaticians in 125 A.D. said pi was sqrt(10)=3.162...

.[/quote]

With all due respect, may be  [b]some [/b] chinese  did say that, but  $many$ good mathematicians (In China or elsewhere in word) knew much more exact value even    MANY thousands years ago .. and certainly by 125 AD.

I have read some Indian mathematicans (I know enough Sanskrit  to read some of the work in the original) and  it is clear from their books on astronomy ...way back  many centuries ago, that they   knew this  fairly accurately.   (at least 5 significant figure) ( . They have to -  -  to do astornomical calculations. (Eg  sin, cos tables -    $\sin 1^o$ is very nearly $\pi/180$ ...etc .... and they had these tables. ...

Funny thing is - many historians would make a mistake in reporting who knew when ...  and the others will just quote them .. repeating the same mistake.

Case in point -  I saw it being reported many times that Brhamgupta's  (600 AD)'s  value  of pi was $\sqrt{10}$ and in one book I even saw a "picture  page from his manuscript"  "supporting" this.  What the page said (In rough translation was)  something like- " An interesting fact is - $\sqrt{10}$ is very near to $\pi$  so it may help in calculation to replace $\pi^2$ as $10$ .. however, if you really want a more accuracy you should take .. " and he gives the value of pi accurate up to 6-7 decimals!!  - Of course,  I don't know how, but  the author missed it compeletly.

Ptolemy (150 AD) certainly knew pi  more than 4 figures.. and many   astronomers and mathematicians (Indian, Chinese etc)  before too knew this.

Madhava of Kerla (1400 AD) had Pi calculated up to 12-13 digits --- to demonstrate his $\arctan$ series ...!   (He used what  is some times  known as  Gregory series few hundred years before Gregory).

A Chinese Mathmetician got $\pi \in (3.1415926, 3.1415927)$ before 500. And he also got $\frac{22}{7}$ and $\frac{355}{113}$.

[quote="catcurio"] So they were actually [i]taught[/i] that $\pi$ is [i]exactly[/i] $3.14$? That's a little extreme...[/quote]
i dont' remember exactly how i learned about $\pi$ but i seem to remember using $22/7$ as an approximation if the radius was a multiple of $7$ and $3.14$ otherwise... i think the book/teachers pretty much said, $\pi$ is either $22/7$ or $3.14$, depending on the value of the radius!

[quote="zeb"]However, for some reason I was also taught that the square root of two is 1.41414141414141 repeating... strange world.....[/quote]
funny, cause i decided to let $\sqrt2$ equal $1.\overline{4142}$ because it looked like that :D

[quote="stupidkid"]"In 1896, an Indiana physician promoted a legislative bill that made $\pi$ equal to 3.2 exactly. The Indiana House of Rep. approved the bill unanimously,  67 to 0. The Senate, however, deferred the debate aoubt the bill 'until a later date.'"
--A Passion for Mathematics

 :rotfl:
Wow, that must be some stupid physician and why not 3.1 instead of 3.2?[/quote]

Not only Indiana (which, as I heard it,  "defered" it by some accident ) but Arkansas actually passed the law.
Well US Congress  actaully passed a bill which effectivly outlawed  "Ohm's law"


Does anyone know the name or how to prove the formula found on this webpage?

http://www.seriesmathstudy.com/generalprod_sum.htm

Specifically this one: http://www.seriesmathstudy.com/generalprod_sum_files/image012.jpg

I wanted to reference it in a proof.  It is really cool but I couldn't find it anywhere else.

Thanks,

If you divide through by m!, it's just the hockey-stick identity. Check the AoPSWiki.

Alternatively, you can do a super-induction, where you prove $ \dfrac{m!}{0!} \plus{} \dfrac{(m\plus{}1)!}{1!} \equal{} \dfrac{(m\plus{}2)!}{(m\plus{}1)(n\minus{}1)!}$ by induction as a base case, then prove the given statement for $ n\plus{}1$ substituted in for $ n$, which would prove the whole given statement. :D


Find all triangles rectangles, with sides of whole measures in the area which has numerical value equal to the perimeter.


Square $ABCD$ has sides of length $1$. Points $E$ and $F$ are on $BC$ and $CD$, respectively, so that triangle $AEF$ is equilateral. A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $AE$. The length of a side of this smaller square is $\frac{a-\sqrt{b}}{c}$, where $a$, $b$, and $c$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b+c$.

[hide] Call $EB$ $x$. So,
$\sqrt{1^{2}+x^{2}}=(1-x)\sqrt{2}$
Simplify to $x^{2}-4x+1=0\iff x=2-\sqrt3$, since $2+\sqrt3>1$.
If we plot the square onto a Cartesian coordinate system with $A$ at the origin, $B(1,0)$, and $D(0,1)$, we get the equation for $AE$: $y=(2-\sqrt3)x$. If the side length of the small square is $s$, then we have $s=(2-\sqrt3)(1-s)\iff s=\frac{2-\sqrt3}{3-\sqrt3}=\frac{3-\sqrt3}{6}$.

$3+3+6=\boxed{12}$. [/hide]

where does this come from? [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/f101f1db9683f0b3c7051072a691ccff.gif[/img]

He is using the pythagorean theorem on right triangles (the RHS is a 45-45-90 triangle).


Prove that
$ \frac{a^2}{b^3\plus{}c^3\plus{}2} \plus{} \frac{b^2}{a^3\plus{}c^3\plus{}2} \plus{}\frac{c^2}{a^3\plus{}b^3\plus{}2} \ge \frac{2}{3}$
where $ a;b;c$ be positive real numbers satisfying $ abc\equal{}8$

Try $ a\equal{}b\equal{}3,c\equal{}\frac{8}{9}$  :wink:


If $ a(n)\equal{} \frac{1}{1}\plus{} \frac{1}{2}\plus{} \frac{1}{3}\plus{}\frac{1}{4}\plus{}.......\plus{} \frac{1}{y^{n\minus{}1}}$

Then find the most appropriate range for $ a(n)$ in terms of $ y$ and $ n$.


group theory
abstract algebra
algebra proposed
algebra

Let $G$ be a group with at least one proper subgroup. Prove that the followings propositions are equivalent:
a)$G$ is the reunion of any two proper subgroups
b)there is a $p$ prime s.t. $G$ is isomorphic with $\mathbb{Z}_{p^2}$

I think it should go in the Algebra section from College Playground..

Anyway, the first sentence is equivalent to "$G$ has exactly one proper subgroup", because it's well-known that $G$ cannot be the union of two proper subgroups. It's bookwork from here. First we show that $G$ is cyclic (it's pretty clear that an element outside the proper subgroup must generate $G$), then that it's finite, so it's isomorphic to some $\mathbb Z_n$, etc.


superior algebra
superior algebra unsolved

Show that with a compass, a straight-edge and a trisector (allowing you to construct the third of an angle), it is possible to construct a $7$-gon.

It is also possible to construct 9-gons, 13-gons, etc. The regular polygon with $n$ sides is constructible with these tools iff $\phi(n)$ is the product of a power of 2 and a power of 3.


Find all real polynomials that \[p(x+p(x))=p(x)+p(p(x))\]

[quote="harazi"]Observe that if $deg(p)\geq 2$ then by the mean value theorem $p(x+p(x))-p(p(x))$ grows like $x\cdot p'(c_{x})$ for  some $c_{x}$ between $p(x)$ and $x+p(x)$. This cannot hold, because $p(x)$ grows much smaller. Thus the degree is 1 at most and then it's easy.[/quote]
didn't understand, can anyone give a detailed answer?

An elementary solution: $ p(x)=\sum_{i=0}^{n}a_{i}x^{i}$,$ p(x+p(x))-p(p(x))=p(x)$, if $ n\ge 2$ on the left side we will have an element  $ na_{n}^{n}x^{n(n-1)+1}$, while the right side has $ grad=n$, so $ n\le 1$, we get $ p(x)=ax$, $ a\in \mathbb{R}$

It is enough to solve this problem by noting that Q(x)=P(x)-x has only one root 0 or Q(x)=0


algorithm
algebra
polynomial
number theory
modular arithmetic

[hide]Evaluate it in $\mod7$ and find that only $1,5\mod7$ work so the highest multiple of $7\le100$ is $98$. $98/7 = 14.$ $14*2+1 = \boxed{29}$.[/hide]

[quote="demji"][hide="solution"]

after messing around with mods a bit we find that solutions are found such that: x=1(mod7) and also x=6(mod7).  Thus every x one more and 1 less than a multiple of 7 are solutions giving $28+1=29$ such values for x that satisfy the initial conditions.

[/hide][/quote]

I think you meant 5(mod 7) instead of 6(mod 7).

:blush:  yeah i ment 5(mod7) which means one more than a multiple of 7 and 2 less than on still giving 29.  I just got messed up by the 6+6 thingy and forgot that x=5 for that $x^{3}+4x$.  I was very lucky that somehow i still got the right number.

[quote="vvr1590"][quote="ShoeFactory"]Evaluate it in $\mod7$ and find that only $1,5\mod7$ work so the highest multiple of $7\le100$ is $98$. $98/7 = 14.$ $14*2+1 = \boxed{29}$.[/quote]

How do you 'evaluate it in mod 7'?[/quote]

You plug in values in mod 7

the only possible values mod 7 are 0 through 6.  So by plugging in these numbers to the equation and solving and then taking it mod 7 we can find classes of values of x which work.  Check out modular arithmetic on aopswiki.

[quote="Altheman"][quote="tiredepartment"]How many integers $x$ are there such that $1 \le x \le100$ and $x^{3}+4x+2$ is divisible by 7?[/quote]

$x^{3}+4x+2\equiv x^{3}+4x-5\equiv (x-1)(x^{2}+x+5)$
$\equiv (x-1)(x^{2}+x-2)\equiv (x-1)^{2}(x+2)\mod 7$

therefore it is divisible by 7 for $x\equiv 1\mod 7$ or $x\equiv 5\mod 7$

98=7*14, so there are 2*14 values of x, then don't forget 99, so 29 values[/quote]

That's what I did except I subtracted $7x$ instead from the original equation. Is there a name for this technique?

[quote="ShoeFactory"]Modding out. No real name.

Like I would say "I modded out the 6x" or something. dunno.[/quote]

Modding :rotfl: That's a funny word. I tried that after I couldn't factor the original equation.

[quote="ShoeFactory"]You dont need to factor it since its just mod 7. But it would be smart to do so if it were mod 26 or something.[/quote]

yes, the factoring establishes an algorithm that you can follow for many problems where as testing values does not work well for this type of problem....i wonder if a generalization of this exists

namely a monic polynomial, $f(x)$, with integer coefficients of degree $n>0$ can be $0 \mod m$ for integers $0\le x<m$ for most $n$ values


Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

This was actually in the contest, wasn't it?

It's kind of hard to put into words. The red points consists of a series of $n$ columns of decreasing height (I mean decreasing starting from left to right, and I don't mean strictly decreasing). The blue points consist of the completion of the red columns up to the border of the triangle having vertices $(0,0),(0,n-1),(n-1,0)$. At the same time, the blue and red sets can also be regarded as rows instead of columns.

$A$ is simply the product of the lengths of all blue columns, while $B$ is the product of the lengths of all blue rows. If we show that the same numbers, with the same multiplicities, appear in the collections of a) blue column lengths and b) blue row lengths, then we are clearly done. Call this assertion $(*)$.

The blue set can be regarded as the union of some maximal right triangles with the legs parallel to the axes of the plane (maximal in the sense that they aren't part of any other similar right triangles). The triangles are obtained as follows: from a point on the diagonal $x+y=n-1$ go down as far as you can without taking any red points. Now go to the right until you meet the diagonal again. The maximal triangles of this set are the ones we're looking for.

We can now prove the assertion $(*)$ by induction on $n$.

If there are points on the diagonal $x+y=n-1$ which are red, then it's easy to see that the number of blue columns of length $0$ is equal to the number of blue rows of length $0$, and we can look at the picture as the union of some smaller pictures (for smaller $n$'s), by breaking the diagonal where it has red points. We use the induction hypothesis on these smaller pictures.

On the other hand, if no points on $x+y=n-1$ are red, then we can look at the points [b]strictly[/b] below the diagonal $x+y=n-1$, and use the induction hypothesis on these. After we append the diagonal $x+y=n-1$ all lengths (of blue columns and rows) increase by $1$, so we have shown what we wanted: the number of blue columns of length $t$ is the same as the number of blue rows of length $t$ for all $0\le t\le n$.

Watch out: I might have reversed "red" and "blue" in some places. :)

Yes, this problem was on the IMO.

I did a bijection: assign to all lattice blue points on the line $x + y = n - k$ the number $k$. Let $n_k$ the number of blue points labeled with number $k$. Let $M$ be the maximum $k$ such that $n_k > 0$. Then $A$, as well as $B$, is equal to the product of $n$ numbers: $n_k$, $n_{k-1}-n_k$, $\ldots$, $n_1-n_2$ (draw a diagram and do the number assignment to see why).

If $ n\equal{}4$, construct the set of points like this (or for any other $ n$).
$ (3,0) (2,1) (1,2) (0,3)$
         $ (2,0) (1,1) (0,2)$
                $ (1,0)  (0,1)$
                         $ (0,0)$

Let $ C(n)$ be the number of blue points in column $ n$, and $ D(n)$ be the number of blue points in diagonal $ n$ (starting from the diagonal that cuts through only the top left point). It is clear that the number of ways to choose an x coordinate is $ C(1)C(2)...C(n)$.

Call a coloring good if $ C(1),C(2),...C(n)$ is a permutation of $ D(1), D(2)...D(n)$. It is clear that a good coloring would satisfy the problem conditions. Obviously for $ n\equal{}1$ any coloring is good. Suppose for some $ n$ any coloring is good.

To construct the set $ x\plus{}y<n\plus{}1$, attach the row of points $ (n,0)...(0,n)$ above the set $ x\plus{}y<n$. For example,
$ (3,0) (2,1) (1,2) (0,3)$
$ (2,0) (1,1) (0,2)$
$ (1,0)  (0,1)$
$ (0,0)$

is constructed by adding points $ (3,0)...(0,3)$ above the set $ x\plus{}y<4$. Since any coloring of $ x\plus{}y<n$ is good, constructing $ x\plus{}y<n\plus{}1$ adds $ 1$ to each column and diagonal so it is also good. Now suppose we color a point red in the row $ (n,0)...(0,n)$. But coloring $ (a,n\minus{}a)$ is equivalent to coloring $ (a, n\minus{}a\minus{}1), (a\minus{}1,n\minus{}a\minus{}1)$ and $ (a,n\minus{}a)$ red. Note that the first two points lie within $ x\plus{}y<n$, which remains good. However, by coloring $ (a,n\minus{}a)$, $ C(a)$ and $ D(n\minus{}a)$ are $ 0$. Therefore, removing that point does not affect the goodness of $ x\plus{}y<n\plus{}1$. This completes the induction, and since all colorings are good, $ A\equal{}B$.

Wait... I think my solution is correct, but thinking I got it in 5 minutes makes me think otherwise.

You create a bijection. Suppose you have a configuration where the points $ B_{1}$, ..., $ B_{k}$ are blue and the points $ R_{1},...,R_{l}$ are red. Then you just switch the $ y$ coordinate by the $ x$ coordinate in each point. It's easy to see that all the points that were in the original line $ x\plus{}y\equal{}n$ are in the transformed line ($ x\plus{}y\equal{}y\plus{}x$) and the red points still meet the conditions $ x'\leq x$ and $ y'\leq y$, but now the blue points have distinct $ y$ coordinates instead of distinct $ x$ coordinates.

Its trivial proving that this transformation is equivalent to rotating the plane $ 90$ degrees counter-clockwise and then reflecting it over the $ y$-axis. Finally, since each of the planes with blue points having distinct $ x$ coordinates each correspond to exactly one plane with distinct blue $ y$ coordinates, and also each plane having blue points with distinct $ y$ coordinates corresponds to a plane having blue points with distinct $ x$ coordinates (for this case just perform the reverse transformation), then we can easily conclude that $ A\equal{}B$.

I agree with grobber... it IS hard to put into words, but I hope it's clear enough.

Call P a critical point if it is red and there are no red points above or to the right of it. Call Q, the set of points P, "x-good" if it leads to a configuration where all the blue points have distinct x coordinates. Let Q' be the set of points P with their x and y coordinate points switched. Therefore, Q' is "y-good", where "y-good" is analogously defined. Since this is a bijection, QED.

I thought of the bijection solution, but we can also use recursion by seeing what happens if add a row, right?

If we take the n+1 by n+1, take out the bottom row, and assume the result for n, we get that for n+1, if the bottom row has k blue squares, the number of ways to choose distinct vertical blue points gets multiplied by k and the number of ways to choose distinct horizontal blue points gets multiplied by (2/1)...(k/(k-1))=k.

Don't mean to spam or anything, but isn't this too easy even for IMO P1?
Just drawing a neat figure makes the bijection pretty obvious!

Confusedhexagon, it's pretty easy for an IMO P1, but then again people weren't as smart ten years ago.  :P  Also remember that in the early days of the IMO, there was not much combinatorics at all.

Just find a down-left-ward pointing corner of the blue and then the bijection does follow.

I found that solution, but also I found this other solution that I posted...

Will anyone tell me if it is correct?

I really want to know this lol.

Let $A(p,.)$ be red point with $p$ maximal, $B(.q)$ be the point with $q$ maximal among red point. ($A$ and $B$ can be equal).
Point $C(h,k)$ is blue iff $h>p$ or $k>q$. To choose $n$ blue points with distinct  $x-$ coordinates $\{(0,k_0),...,(n-1,k_{n-1})\}$ it necessarily that : \[n>k_0>q, n-1>k_1>q,...,n-p>k_p>q\] and  \[k_{p+1}<n-(p+1),...,k_{n-1}<n-(n-1)\] which gives a number (if it is not equal to zero i.e $p+q$ not large enough) of possible configurations \[\prod_{i=1}^{p+1}(n-i-q)\prod_{i=p+1}^{n-1}(n-i)=\frac{(n-(q+1))!(n-(p+1))!}{(n-p-q-2)!}\]
This is symmetric with respect to $p$ and $q$ so it is also equal to the number of configurations with distinct  $y-$ coordinates.

Ok....
Hard to put in words.........

Still I'll give it my best shot.

Let $a_i$ denote the number of blue points with coordinate $x_i$
and     $b_i$ denote the number of blue points with coordinate $y_i$

Now we have to prove $\prod_{i=0}^{n-1}a_i=\prod_{i=0}^{n-1}b_i$

Let $(x_i,y_i)$ denote the rightmost red point of the form $(k,y_i)$
Observe that $x_i \ge x_{i+1}$
Then quite easily, the number of blue points of the form $(x_i,a)$ is $(n+1-x_i-y_i)$ (since $x+y \le n$, the above lines)

Thus  $\prod_{i=0}^{n-1}a_i= \prod_{i=0}^{n-1}(n+1-x_i-y_i)$

The same result is true when we replace $x_i$ by $y_i$
That is,
 $\prod_{i=0}^{n-1}b_i= \prod_{i=0}^{n-1}(n+1-x_i-y_i)$

$\implies \prod_{i=0}^{n-1}a_i=\prod_{i=0}^{n-1}b_i$
$\implies A=B$

[hide=Solution (possibly flawed)]Call a line [i]integral[/i] if its equation is $x=k$ or $y=k$, where $k$ is a non-negative integer less than $n$. Consider one such configuration of red and blue points in the plane. Let $x_m$ be the maximum possible $x$-coordinate of a red point, and $y_m$ be the maximum possible $y$-coordinate of a red point, where $0 \leq x_m,y_m \leq n-1$. Then, from the given condition, we can draw a broken line $\kappa$, with all points of $\kappa$ lying on [i]integral[/i] lines, and which starts from $P(x_m,0)$ and ends at $Q(0,y_m)$, such that all red points lie either on or inside the region bounded by the coordinate axes and $\kappa$. Note that if $\kappa$ is symmetric about the line $y=x$, then $A$ and $B$ must also be equal (by symmetry). So, from now on, assume that $\kappa$ is not symmetric about the line $y=x$.

Choose $n$ blue points with different $x$-coordinates, and name them $A_1(0,a_1),A_2(1,a_2), \dots ,A_n(n-1,a_n)$. Then this selection is a part of $A$ (as given in question). Similarly, choose one such selection $B_1(b_1,0),B_2(b_2,1), \dots ,B_n(b_n,n-1)$, which is a part of $B$. Now, reflect the whole figure about the line $y=x$. Then we get another configuration which satisfies all the given conditions, with the only difference being that now $\kappa$ is present from $P'(y_m,0)$ to $Q'(0,x_m)$. In this case the previously chosen points become $A'_1(a_1,0),A'_2(a_2,1), \dots ,A'_n(a_n,n-1)$, i.e. they become blue points with distinct $y$-coordinates. Then these points form one way out of $B'$, as these points all lie outside $\kappa$. Similarly, we get the points $B'_1(0,b_1),B'_2(1,b_2), \dots ,B'_n(n-1,b_n)$, which are being counted in $A'$. Call this plane the [i]inverse[/i] of the configuration chosen initially. As $\kappa$ is not symmetric about the line $y=x$, so the [i]inverse[/i] of each configuration must be different from itself. Thus, [i]inversion[/i] of the plane is a bijection. But, as the [i]inverse[/i] of the plane is symmetric to the original configuration about the line $y=x$, the number ways of choosing blue points with distinct $x$-coordinates must be equal for both configurations, or equivalently $A'=A$. In a similar fashion, we get $B'=B$. Thus, Combining everything we have assimilated so far, we get $B'=A=A'=B$, thus giving $A=B$. Hence, done. $\blacksquare$

REMARK (for those who gave TOT 2018 Senior O-Level :D): This problem is quite easy, but expressing the whole idea in words is quite difficult tbh (as you can see from my solution :P). In the end, this idea is quite similar to the one that I used to solve this year's Tournament of Towns Senior O-Level P5, in which case we had to join two 50X100 boards to form a 100X100 board, and then reflect those kings present on white cells of the first 50X100 board to the black cells of the second 50X100 board.
[/hide]

EDIT (26 Jan 2020): This solution seems incorrect to me now that I read it. It might be possible that I am missing something which I had in mind while writing this solution, but in any case this is a really badly written solution. I'll try to correct this later if I get time :)

Let $a_x$ denote the number of blue points with a given $x$-coordinate, and define $b_y$ similarly.

We actually claim that

[b][color=red]Claim:[/color][/b]  The multisets $\mathcal A \overset{\text{def}}{=} \{ a_x \mid x \}$ and $\mathcal B \overset{\text{def}}{=} \{ b_y \mid y \}$ are equal.

[i]Proof.[/i]  By induction on the number of red points. If there are no red points, then $\mathcal A = \mathcal B = \{1, \dots, n\}$. Now suppose we color one point $P = (x,y)$ red. Before the coloring, we have $a_x = b_y = n-(x+y)$; afterwards $a_x = b_y = n-(x+y)-1$ and no other numbers change, as desired. Since every permissible set of red points can be built inductively this way (indeed it's a Young tableaux!) this implies the result. $\blacksquare$

Finally, \[ A = \prod_{x=0}^{n-1} a_x = \prod_{y=0}^{n-1} b_y = B. \]

[quote=orl]Let $n$ be a positive integer. Each point $(x,y)$ in the plane, where $x$ and $y$ are non-negative integers with $x+y<n$, is coloured red or blue, subject to the following condition: if a point $(x,y)$ is red, then so are all points $(x',y')$ with $x'\leq x$ and $y'\leq y$. Let $A$ be the number of ways to choose $n$ blue points with distinct $x$-coordinates, and let $B$ be the number of ways to choose $n$ blue points with distinct $y$-coordinates. Prove that $A=B$.[/quote]

Notice that the red points form a "staircase". For each $0 \le i, j <n$, let $x_i$ denote the number of blue points with $x$ coordinate $i$ and $y_j$ denote the number of blue points with $y$ coordinate $j$. We need to show that $\prod^{n-1}_{i=0} x_i = \prod^{n-1}_{j=0} y_j$, to conclude.

We induct on the number of red points. If there is none; we're done. Otherwise, any time a red point is added, it must be added to the North-East of an "L" vertex of the staircase, to increase the number of red points by $1$. Then, suppose $(a, b)$ was now marked red; except $x_a$ and $y_b$, none of the $x_i$'s or the $y_j$'s change. Also, $x_a \mapsto x_a-1$ and $y_b \mapsto y_b-1$ but $x_a=y_b$, if $(a, b)$ was added to the NE neighbor of an "L" vertex, anyway, so we're done.

We will prove that the sets of the counts of the number of blue points in the rows and columns coincide. We do this by induction on the number of red points. Add them repeatedly starting with the lower ones and going upwards in a staircase manner, i.e.  if we have just added $(x,y)$, then $x+y$ is not less than the sum of the coordinates of the previously added point (note that the red points form an staircase due to the problem statement). Suppose that we have added $(x,y)$; then the number of blue points is decreased by $1$ in row $x$ and column $y$ and remains the same in the other lines, so the two sets of counts remain the same due to the I.H. (note that the counts in row $x$ and column $y$ were equal to $n-x-y$ before the addition of the red point).

Let $r_i$ be the number of blue points in the row with points $(0, i), (1, i), ..., (n-1-i, i)$ and let $c_i$ be the number of blue points in the column with points $(i, 0), (i, 1), ..., (i, n-1-i)$. $A=c_0c_1\cdots c_{n-1}$ and $B=r_0r_1\cdots r_{n-1}$. We can create any coloring of red and blue points by first coloring all the points blue. Then, we start at the point $(i, 0)$ where the $i$th column is the leftmost column that doesn't have any red points, and we color the points $(i, 0), (i, 1), ..., (i, k_i)$ all red for some $k_i$ and repeat the process. This ensures that any time we add a point, all the points underneath it and to the left of it must be red, which creates a valid coloring. Now we'll show that each time we color a point red, $A$ must remain equal to $B$. Let's say we color the point $(x, y)$ red. Then $A$ is being multiplied by $\frac{c_x-1}{c_x}$ and $B$ is being multiplied by $\frac{r_y-1}{r_y}$. We get that $c_x = (n-x-1) - (y) + 1 = n-x-y$ and $r_y=(n-y-1) - (x) + 1 = n - x - y$, so $A$ and $B$ are being multiplied by the same amount each time. Therefore, $A=B$ when the coloring is complete.

Say that blue dot is primitive if all dots in the rectangle with the bottom-left dot and that blue dot are red except for the blue dot itself.\\

Note that for any primitive blue dot, the number of blue dots in its row and the number of blue dots in its column are equal.\\

We claim that the number of blue dots in the columns is a permutation of the number of blue dots in the rows. This is clearly true of the configuration where all dots are blue. Now, consider flipping a primitive blue dot into a red dot. This maintains the condition in the problem, but decreases the numbers of the row of the flipped dot and the column of the flipped dot, and since they were originally equal, they are still a permutation of each other. We can repeat this procedure to generate any valid configuration (since any allowed configuration can be achieved by starting from all blue and repeatedly flipping primitive blue dots).\\

Since permutations have the same product, we are done.

This problem is harder to understand than to solve.

If we colour all points $(x, y)$ satisfying $x+y \leq n - k$ as red, (with $1 \leq k \leq n - 1$), we colour the remaining points contained in the square (formed by the points $(0, 0), (n - 1, 0), (0, n - 1), (n - 1, n - 1)$), as blue.

Let $c_i$ denote the number of blue points in column $i$, and similarly let $d_i$ denote the number of blue points in row $i$. Then observe that $c_i = d_i$. Now as $A = \prod c_i$ and $B = \prod d_i$, we have indeed $A = B$, as desired. $\blacksquare$

[quote=MagicalToaster53]If we colour all points $(x, y)$ satisfying $x+y \leq n - k$ as red, (with $1 \leq k \leq n - 1$) [/quote]
And if not? This is not a complete solution...

What do you mean? As in we may have used blue instead of red?

No. I mean that the reds can lie in a more complicated configuration than the ones covered by your cases (they do not need to form a "complete triangle").

omg i actually solved a combo. marabot solve btw

We further claim that the number of blue points of each $x$ coordinate is the same as the number of blue coordinates of each $y$ coordinate up to permutation. We will prove this using strong induction on $n,$ with the base case $n = 1$ being fairly obvious.

Now assume that the result is true for $n = 1, 2, \dots, k.$ There are 2 cases:

[b]Case 1:[/b] There are no blue points on the line $x + y = k - 1.$ Then just shift one to the left and apply the inductive hypothesis; this does not affect the ordering of the blue points.

[b]Case 2:[/b] There is some blue point on the line $x + y = k - 1.$ Then this blue point separates the red and blue points into two halves, and apply the inductive hypothesis on each half.

Therefore, we have finished the induction. As a result, multiplying all of the number of blue points of each $x$ coordinate will yield the same result with the $y$ coordinates, so $A = B,$ as desired.

[quote=huashiliao2020]Let $a_k,b_k$ be the number of blue points with x (resp. y) coordinates k; we'll prove that $A=\prod a_k=\prod b_k=B$ by showing the sequence b is a permutation of the sequence a using strong induction, with base case n=1 trivial, so assume now the inductive hypotheses, and for convenience let $\iff$ denote permutation of each other. If all points on the line x+y=n are blue, we're done since $\forall1\le k\le n-1$, $a_k\iff b_k$ due to inductive hypothesis and just adding 1 for every point (since there's an extra blue point now on every horizontal/vertical line). However, if some point (m,n-m) is red, then $a_m=0=b_m$, meaning A=0=B. We conclude. :surf:

[b]Remark.[/b] Looking at the extremal line x+y=n is well motivated by induction.[/quote]

Just want to point out that the induction is not complete because your induction’s hypothesis is that the the sequence $a$ is a permutation of $b$. However, in the end of your solution, you only show that $A=B=0$. Fortunately, this is not very hard to fix.

Let $c_x$ and $d_y$ denote the number of blue points of column $x$ and row $y$, respectively. Let $(x, y)$ be red with $x + y = n - k$, with $k > 1$.

We prove by strong induction that $A = \prod c_x = \prod d_y = B$. The base case with $x + y = 2$ is trivial. Assume now that $A = B$ for all $x + y = 2, 3, \ldots, n - 2$. Adding in the (new) blue line $x + y = n - 1$ turns the product
\begin{align*}
c_0c_1 \cdots c_{n - 2} &= d_0d_1 \cdots d_{n - 2} \\
\implies (1)(c_0 + 1)(c_1 + 1) \cdots (c_{n - 2} + 1)(1) &= (1)(d_0 + 1)(d_1 + 1) \cdots (d_{n - 2} + 1)(1),
\end{align*}
so that $A' = B'$. $\blacksquare$
[hide = Remark]If we have a red point on line $x + y = n - 1$ instead of an all blue point line, we have by strong induction that $x + y = n$ as an all blue line also finishes this problem.[/hide]

[hide=first combo ISL (probably fakesolve)]
[img]https://cdn.artofproblemsolving.com/attachments/c/3/b07fa45cdab5dac822d292098b21f6e06b1457.png[/img]
[b]Claim[/b]: The # of blue for each y coordinate is a permutation of the # of blue for each x coordinate.

[b]Base Case[/b]: If $n=1$, there is only one point. If it is red, $A=B=0$. If it is blue, $A=B=1$.

[b]Induction[/b]: Take the tallest rectangle and draw an isosceles right triangle on top and on the right with the same length/width. If a rectangle contains the entire column, then $A=B=0$ If there are no rectangles, then $A=B=n!$. Let $x$ be the length of the tallest rectangle and $y$ be the height. Call the triangle with side length $x$ "X" and the one with side length $y$ "Y". Because all pervious values of $n$ work, triangles X and Y satisfy the claim, so the triangle made by combining triangle X, triangle Y, and the tallest rectangle also works because the triangles have no overlap in the x or y coordinate. So $n$ would also work because changing from the one with side length $x+y$ to the entire triangle increases the number of blue in each row and column increases by $n-x-y$, so they would still be permutations of each other, and $n$ also works.

Define $a_c$ as the number of blue points with $x$-coordinate $c$, and $b_c$ as the number of blue points with $y$-coordinate $c$. We claim the stronger statement that $\{a_i\}$ and $\{b_i\}$ are permutations of each other through induction on $n$, with trivial base case $n=2$.

Assuming that $n=k$ satisfies the condition, we claim that adding the $k+1$ points on the line $x+y=k$ to prove the case $n=k+1$, as
\[(a_0+1) \cdot \ldots  \cdot (a_k+1) \cdot 1 = (b_0+1) \cdot \ldots \cdot (b_k+1) \cdot 1.\]

Now we just need to consider the cases where the added line contains a red point $(m, k-m)$, which can be handled by noting a configuration where we can add the desired red point such that no other red points are forced. This just changes the pair $(a_m, b_{k-m}) = (1,1)$ to $(0,0)$, as desired. $\blacksquare$

We will do induction on $n$ with our base cases of $n = 1$ and $n = 2$ being true. We can do induction by adding in the set of points on the line $x + y = n$ from which our total set of points becomes $x + y < n + 1$.
If $x_k$ is the number of blue points on row $k$ and $x_y$ is the number of blue points on column $k$, the problem is equivalent to showing that the set $\{x_1, x_2, \dots, x_n\} = \{y_1, y_2, \dots, y_n\}$.
If after adding in the points on $x + y = n$ we don't add any extra red points on the new line, we have $\{x_1 + 1, x_2 + 1, \dots, x_n + 1, 1\} = \{y_1 + 1, y_2 + 1, \dots, y_n + 1, 1\}$. If we do add an extra red point on the line $x + y = n$ then we don't have to add any other new points since it is equivalent to just adding the one red point in a different configuration(where no new red points would be added). Clearly the sets $x_i$ and $y_i$ are equal since we are just changing one of the variables in each set from $1$ to $0$, and we are done by the induction hypothesis.

Let $r_i, c_i$ be the number of blue points in each row and column. It satisfies to prove multisets $\{r_i\} = \{c_i\}$ as $A = \prod r_i = \prod c_i = B.$ We prove this by induction. The base case of $n = 1$ is trivial.

	Assume for $n = k$ and consider $n = k+1.$ Observe that for $T' \subset T$ bounded by $x + y < k$ with row/column blue values $r'_i, c'_i$, by induction $\{r_i'\} = \{c_i'\}.$ Now for each point $x + y = k$, if $(x-1, y)$ or $(x, y-1)$ were blue, then $(x,y)$ must be blue. So the corresponding $r'_i, c'_i$ are incremented by $1$ to get $r_i, c_i$. Otherwise, if both (or only 1 for edges) are red, the new point can be red or blue. In the former, the corresponding $r_i = c_i = 0.$ In the latter, $r_i = c_i = 1.$ Either way, the inductive step is complete as the multisets  $\{r_i\} = \{c_i\}$ (since non-zero terms increment and zero terms have preserved equality).


Does anybody have a results pdf like the ones for '04 and '05? I'm mostly interested in the individuals and the Teams #11-25

here you go.

That was the year New Hampshire(my team) beat out Mass, Florida, Michigan, New York, and was just 2 spots behind Texas!!  Oh the memories :)

[quote="Klebian"]Haha, look at 37th place![/quote]

2nd place is freakier, he has the exact same name as my dad.

Hey look, a Jeffrey Chen from VA is 51st.

#11 :D.
There are [b]4[/b] chens :rotfl: and 2 Wus.

[quote="pgpatel"]There are [b]4[/b] chens :rotfl: and 2 Wus.[/quote]

So we can write something like "Wu's On First"... :rotfl:

That was the year Missouri got second :w00t:

Aw, no one's last name is 'easy' or 'pie' or something like that.

[quote="chess64"][quote="Klebian"]Haha, look at 37th place![/quote]

Hey, that's scorpius119 :P[/quote]
He's located in PA...  :ninja:


vector
integration
geometry
3D geometry
sphere
inequalities
function

Suppose that $ \xi_j$ ($ j\in\mathbb Z$) are standard complex Gaussians (density $ \frac 1\pi e^{\minus{}|z|^2}$ on the plane) and the covariances $ E\xi_i\bar\xi_j$ are bounded by $ \delta_{|i\minus{}j|}$ where $ \delta_k>0$ is a fast decreasing sequence (you can go all the way down to $ \delta_k\equal{}e^{\minus{}ck^2}$, if you want). Let $ b\equal{}E\log|\xi|$. I want to prove (or disprove) that for any $ t_j\ge 0$, we have
\[ E\prod_j |\xi_j|^{t_j}\le e^{b\sum_j t_j}\cdot e^{C\sum_j t_j^2}\]
with some $ C$ depending on the sequence $ \delta_k$. Any ideas?

this is nothing and may be completely useless, but what if one tries first the finite dimensional case, and $ 1$-dimensional Gaussian random variables instead of complex ones. This means that there exists unit vectors $ v_k$ such that $ |(v_i,v_j)| \leq \delta_{|i\minus{}j|}$ for all $ 1 \leq i,j \leq n$, say. One then has to study $ P(\omega) \equal{} \prod_k |(v_k,N(\omega))|^{t_k}$ where $ N$ is a standard $ n$-dimensional Gaussian random variables. This is also equal to $ \|N\|^{\sum t_k} \prod_k |(v_k,\frac{ N(\omega)}{\|N(\omega)\|})|^{t_k}$ so that one easily obtains that $ E[P]\equal{} f(N) \frac{\sigma(S^{n\minus{}1})}{(2\pi)^{n/2}} \int_0^{\infty} r^{\sum_k t_k \plus{} (N\minus{}1)} \exp(\minus{}\frac{r^2}{2}) \, dr$ where $ f(N)$ is the expectation of $ \prod_k |(v_k,X)|^{t_k}$ with $ X$ uniformly distributed on the unit sphere of $ \mathbb{R}^n$. One can see that this quantity very small since every time that $ X$ is near $ v_k$, then $ (X,v_j) \approx (v_k,v_j) \leq \delta_{|j\minus{}k|}$. Nevertheless, though this quantity looks more geometric, it does not seem much easier to find a good upper bound  :blush: One could also say that this expectation is something like $ \int_{O_n} \prod_k |(Uv_k,e_1)|^{t_k} d \mu(U)$ avec $ \mu$ uniform measure on $ O_n$.

Ah, of course, you can restate this question in a million ways. :roll: Unfortunately, I haven't been able to find a reformulation that makes it any simpler :(. Thanks for an attempt though :).

I can prove a more weak version of the inequality claimed. Namely, this inequality holds for every $ \tilde b>b$. Here is the sketch of the argument.
1) Consider r.v. $ \xi$ being a function of i.i.d. standard normal r.v.'s $ \{\epsilon_k\}$: $ \xi\equal{}f(\epsilon_1,\epsilon_2,\dots)$. Assume $ \nabla f: \equal{}\left(\sum_{k} (f'_k)^2\right)^{1/2}$ is bounded by a constant $ F$. Then
\[ Ee^{\xi}\leq \exp[E\xi\plus{}F^2/2].\quad \quad \quad(*)\]
This comes from the Gaussian concentrarion inequality; e.g. http://www.math.univ-toulouse.fr/~ledoux/Berlin.pdf, section 2.
2) Consider real-valued Gaussian r.v's $ \{\xi_n\}$ with covariances $ c_{nm}$ and r.v. $ \xi\equal{}g(\xi_1,\dots)$. By Karhunen's theorem one can write $ \xi_n\equal{}\sum_ka_{nk}\epsilon_k$ with i.i.d. standard normal $ \{\epsilon_k\}$. Then $ \xi\equal{}f(\epsilon_1,\epsilon_2,\dots)$, and
\[ f'_m(\{x_k\})\equal{}\sum_{n}a_{nm}g'_n(\{y_k\}), \, y_n: \equal{}\sum_ka_{nk}x_k.\]
Note that $ c_{nm}\equal{}\sum_{k}a_{nk}a_{mk}$. Then
\[ \sum_k(f'_k(\{\xi_j\}))^2\equal{}\sum_k\sum_{n, m}a_{n k}a_{m k}g'_{n}(\{\epsilon_j\})g'_{m}(\{\epsilon_j\})\equal{}\sum_{n,m}c_{n m}g'_{n}(\{\epsilon_j\})g'_{m}(\{\epsilon_j\})\]
Assume $ |c_{nm}|\leq \delta_{n\minus{}m}$ and  $ |g'_n|\leq T_n$. Then  $ \sum_k(f'_m(\{\xi_j\}))^2\leq F^2\equal{}\left(\sum_{r\in\mathbb{Z}}\delta_r\right)(\sum_nT_n^2)$. Consequently,  (*) holds. In particular, for every Lipschitz fumction $ \psi: \Re\to \Re$ one has
\[ E\exp\left[\sum_k t_k \psi(\xi_k)\right]\leq \exp[\sum_{k}t_k E\psi(\xi_k)\plus{}C\sum_{k}t_k^2]\quad\quad \quad(**)\]
with $ C\equal{}(1/2)\left(\sum_{r\in\mathbb{Z}}\delta_r\right)Lip(\psi)$, where $ Lip(\psi)$ is the Lipschitz constant for $ \psi$.

3) At last, let us adapt these arguments to the problem under discussion. I'd like to write $ \[ E\prod_{j}|\xi_{j}|^{t_{j}}\equal{}E\exp\left[\sum_j t_j \psi(\xi_j)\right]$
with $ \psi(z)\equal{}\log|z|$ and then use (**). Two troubles arise. First, $ \xi_j$ are complex-valued. This is not a trouble, since we can consider every $ \xi_j$ as a two-dimensional vector and us the same argument based on the Karhunen theorem. Second, $ \psi$ is not Lipschitz -- and this is serious. What can be done: one can dominate $ \psi$ by Lipschitz $ \tilde \psi$. Then
$ \[ E\prod_{j}|\xi_{j}|^{t_{j}}\leq \exp[\tilde b\sum_{j}t_j \plus{}\tilde C\sum_{j}t_j^2]$ with $ \tilde b\equal{}E\tilde \psi(\xi_k)$ and $ C\equal{}(1/2)\left(\sum_{r\in\mathbb{Z}}\delta_r\right)Lip(\tilde \psi)$.
Clearly, one can make $ \tilde b$ arbitrarily close to $ b$. Unfortunately, if $ \tilde b$ approaches $ b$,  then $ \tilde C$ tends to $ \infty$  :(

Thanks! :) I have proved the same statement, though by completely different means. This was enough for my immediate purposes but left me wondering if the clean inequality holds true as well. You are right: the main problem is that $ \log|z|$ has singularity at $ 0$, which kills many good approaches. :(


[quote="Bovy11"]Find all $ n\in \mathbb N$ for which $ n$ is equal to twice the number of its positive divisors.[/quote]
Write $ n \equal{} p_1^{a_1}...p_r^{a_r}$ Then the number of divisors is $ (a_1 \plus{} 1) \plus{} ... \plus{} (a_r \plus{} 1)$. So we have:
$ p_1^{a_1}...p_r^{a_r} \equal{} 2(a_1 \plus{} 1) \plus{} ... \plus{} (a_r \plus{} 1)$
$ r \equal{} 1$ gives $ n \equal{} 8$.
$ r \equal{} 2$ gives $ n \equal{} 12$
$ r \ge 3$ gives $ p_1^{a_1}...p_r^{a_r} > 2(a_1 \plus{} 1) \plus{} ... \plus{} (a_r \plus{} 1)$
So the solutions are $ 8$ and $ 12$.

[quote="Bovy11"]Mathias,your solution is rough a little i think...more detail please if possible.[/quote]
Ok, then :)

Write $ n \equal{} p_1^{a_1}...p_r^{a_r}$ Then the number of divisors is $ (a_1 \plus{} 1) \plus{} ... \plus{} (a_r \plus{} 1)$. So we have:
$ p_1^{a_1}...p_r^{a_r} \equal{} 2(a_1 \plus{} 1) \plus{} ... \plus{} (a_r \plus{} 1)$
$ r \equal{} 1$: As $ 2 \mid n$ we know that $ p \equal{} 2^k$ for some $ k$. This gives the equation $ 2^k \equal{} 2(1 \plus{} k)$. and it gives the solution $ n \equal{} 8$.

$ r \equal{} 2$: Again $ p \equal{} 2^k \cdot q^m$ where $ q$ is a prime. Assume $ q \equal{} 3$ this gives $ 2^k * 3^m \equal{} 2(1 \plus{} k)(1 \plus{} m)$ And it has the only solution $ n \equal{} 12$, because the $ LHS$ is growing much faster than $ RHS$. If $ q \ge 5$, we know that either $ k \ge 4$ or $ m \ge 4$, but then the $ LHS$ is bigger than the $ RHS$. (It's easy to see that if $ p_1^{a_1}...p_r^{a_r} > 2(a_1 \plus{} 1) \plus{} ... \plus{} (a_r \plus{} 1)$ and $ b_1 \ge a_1$, ..., $ b_r > a_r$ then $ p_1^{a_1}...p_r^{a_r} \ge p_1^{b_1}...p_r^{b_r} > 2(a_1 \plus{} 1) \plus{} ... \plus{} (a_r \plus{} 1)$)

$ r \ge 3$: gives $ p_1^{a_1}...p_r^{a_r} > 2(a_1 \plus{} 1) \plus{} ... \plus{} (a_r \plus{} 1)$ (With the same argument as before. We know that $ 2*3*5 > 2(1\plus{}1)(1\plus{}1)(1\plus{}1)$ and incrementing either $ p_i$ or $ a_i$ won't help us. Thus no solution.)
So the solutions are $ 8$ and $ 12$


inequalities
function
inequalities unsolved

Let $a,b,c$ are positive real such that $abc=1$. Pove that:
$\frac{1}{\sqrt{b+\frac{1}{a}+\frac{1}{2}}}+\frac{1}{\sqrt{c+\frac{1}{b}+\frac{1}{2}}}+\frac{1}{\sqrt{a+\frac{1}{c}+\frac{1}{2}}}\geq\sqrt{2}$

[quote="Megus"]it should be $3\sqrt{2}$ instead of only $\sqrt{2}$. We can substitute $a=\frac{x}{y}$, $b=\frac{y}{z}$,$c=\frac{z}{x}$ and $xy+yz+zx=1$ as inequality is homogenous. Function $f(x)=\frac{1}{\sqrt{1-x}}$ is convex as $f''(x) \geq 0 \iff 2(h'(x))^{2}\geq h''(x) h(x)$ where $h(x)=\sqrt{1-x}$, which is true. Hence

$LHS = f(\frac{xy}{2})+f(\frac{yz}{2})+f(\frac{zx}{2}) \geq 3f(\frac{1}{2})=3\sqrt{2}$[/quote]
But let$a=b=c=1$.
the LHS is obvious<3...

This was posted before, http://www.mathlinks.ro/Forum/viewtopic.php?t=50805

I somehow must have messed up something during substition  :blush:


Let $a,b,c$ be non-negative real numbers with sum $2$. Prove that
\[\sqrt{a+b-2ab}+\sqrt{b+c-2bc}+\sqrt{c+a-2ca}\ge 2.\]
It's inspired from hitek's one but nicer.

Hitek even had the following before

Prove or disprove that if $a,b,c\geq0$ such that $a+b+c=1$, then
\[\sqrt{a+b-ab}+\sqrt{b+c-bc}+\sqrt{c+a-ca}\leq\sqrt5+\Bigl(\frac{2\sqrt2+\sqrt3-2\sqrt5}4\Bigl)q, \]
where $q=(|a-b|+|b-c|+|c-a|).$

To hitek. The first problem is "prove or disprove", so you shouldn't say that "Hitek even had the following before".

My problem is different from you because there are two case of equalities and therefore 2 is the best constant.

We may write the inequality in the form
\[\sqrt{\frac{a+b}2-ab}+\sqrt{\frac{b+c}2-bc}+\sqrt{\frac{c+a}2-ca}\geq\sqrt2. \]
Squaring both sides of the inequality gives
\[a+b+c-(ab+bc+ca)+2\sqrt{\Bigl(\frac{a+b}2-ab\Bigl)\Bigl(\frac{b+c}2-bc\Bigl)}\\+2\sqrt{\Bigl(\frac{c+a}2-ca\Bigl)\Bigl(\frac{a+b}2-ab\Bigl)}+2 \sqrt{\Bigl(\frac{b+c}2-bc\Bigl)\Bigl(\frac{c+a}2-ca\Bigl)}\geq2, \]
which reads as follows after some simple manupilations
\[\sqrt{\Bigl(\frac{a+b}2-ab\Bigl)\Bigl(\frac{b+c}2-bc\Bigl)}+\sqrt{\Bigl(\frac{c+a}2-ca\Bigl)\Bigl(\frac{a+b}2-ab\Bigl)}\\+\sqrt{\Bigl(\frac{b+c}2-bc\Bigl)\Bigl(\frac{c+a}2-ca\Bigl)}\geq \frac12(ab+bc+ca). \]
Put
\[t=\sqrt{\Bigl(\frac{a+b}2-ab\Bigl)\Bigl(\frac{b+c}2-bc\Bigl)}. \]
Notice that
\[\frac{a+b}2-ab=\frac14(a-b)^{2}+\frac14c(a+b). \]
By Cauchy Schwarz inequality, we have
\[t\geq \frac12|(a-b)(b-c)|+\frac12\sqrt{ca(a+b)(b+c)}. \]
Now the rest is no problem I think because
\[|(a-b)(b-c)|+|(b-c)(c-a)|+|(c-a)(a-b)|+\sqrt{ca(a+b)(c+b)}\\+\sqrt{ab(a+c)(b+c)}+\sqrt{bc(b+a)(c+a)}\geq2( ab+bc+ca). \]
So the art of problem solving is to get a problem no problem. :P

[quote="GacKiem"]How to prove the rest? I think it's not "no problem".[/quote]
OK, no problem :P

By Cauchy Schwarz,
\[(a+c)(a+b)\geq (a+\sqrt{bc})^{2}.\]
Thus, it is sufficient to prove that
\[\sum_{\rm cyclic}|(a-b)(b-c)|+ab+bc+ca+\sqrt{abc}(\sqrt a+\sqrt b+\sqrt c)\geq 2(ab+bc+ca).\]
Note that
\[a^{2}+b^{2}+c^{2}-ab-bc-ca\leq \sum_{\rm cyclic}|(a-b)(b-c)|.\]
To finish the proof, we show that
\[a^{2}+b^{2}+c^{2}+\sqrt{abc}(\sqrt a+\sqrt b+\sqrt c)\geq 2(ab+bc+ca),\]
which is a little bit well known.

Still, there is another nice proof for the second step.  :P

Yes. You are right, Thuan. It's also my solution. For the rest, we can use Schur or let $a=b+y,b=c+x$.

Very nice proof, pvthuan. Congratulations!

[quote="hitek"]Prove that if $a,b,c\geq0$ and $a+b+c=1$, then
\[2\leq \sqrt{a+b-ab}+\sqrt{b+c-bc}+\sqrt{c+a-ca}\leq\sqrt5. \][/quote]
I don't think the right-hand side is true. For the left-hand side one can prove a stronger result:
\[\sum_{cycl}\sqrt{a+b-ab}\geq \sum_{cycl}\sqrt{a^{2}+b^{2}+3ab}\]

[quote="ductrung"][quote="hitek"]Prove that if $a,b,c\geq0$ and $a+b+c=1$, then
\[2\leq \sqrt{a+b-ab}+\sqrt{b+c-bc}+\sqrt{c+a-ca}. \]
[/quote]
 For the left-hand side one can prove a stronger result:
\[\sum_{cycl}\sqrt{a+b-ab}\geq \sum_{cycl}\sqrt{a^{2}+b^{2}+3ab}\]
[/quote]
It's standard:
$\sum_{cycl}\sqrt{a+b-ab}\geq \sum_{cycl}\sqrt{a^{2}+b^{2}+3ab}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{ac+bc-2ab}{\sqrt{a+b-ab}+\sqrt{a^{2}+b^{2}+3ab}}\geq0.$
But $\sum_{cyc}\frac{ac+bc-2ab}{\sqrt{a+b-ab}+\sqrt{a^{2}+b^{2}+3ab}}=$
$=\sum_{cyc}\left(\frac{b(c-a)}{\sqrt{a+b-ab}+\sqrt{a^{2}+b^{2}+3ab}}-\frac{a(b-c)}{\sqrt{a+b-ab}+\sqrt{a^{2}+b^{2}+3ab}}\right)=$
$=\sum_{cyc}\left(\frac{c(a-b)}{\sqrt{b+c-bc}+\sqrt{b^{2}+c^{2}+3bc}}-\frac{c(a-b)}{\sqrt{a+c-ac}+\sqrt{a^{2}+c^{2}+3ac}}\right)=$
$=\sum_{cyc}\frac{c(a-b)\left(\sqrt{a+c-ac}+\sqrt{a^{2}+c^{2}+3ac}-\sqrt{b+c-bc}-\sqrt{b^{2}+c^{2}+3bc}\right)}{\left(\sqrt{a+c-ac}+\sqrt{a^{2}+c^{2}+3ac}\right)\left(\sqrt{b+c-bc}+\sqrt{b^{2}+c^{2}+3bc}\right)}=$
$=\sum_{cyc}\frac{c(a-b)^{2}\left(\frac{a+b}{\sqrt{a+c-ac}+\sqrt{b+c-bc}}+\frac{a+b+3c}{ \sqrt{a^{2}+c^{2}+3ac}+\sqrt{b^{2}+c^{2}+3bc}}\right)}{\left(\sqrt{a+c-ac}+\sqrt{a^{2}+c^{2}+3ac}\right)\left(\sqrt{b+c-bc}+\sqrt{b^{2}+c^{2}+3bc}\right)}\geq0.$ :)

I know many people ask you this.
1) Can you look at that step one of yours and go to step two, quickly? How is it done?
2) Is there a "standard" method you use which solves these kind of problems easily, or is it just experience?

May I answer your question? Solution of Arqady is basing on SOS method that once posted in Mathlinks by Hungkhtn, my brother. It's quite standard.

Here is the article related to some results in this post.


The equation $ \displaystyle {\parallel{}|x|-2|-3|=a}$, where $ a$ is a real parameter, has 5 real roots for what value of $ a$?

[hide="Answer"]$ x \equal{} q$ is a root iff. $ x \equal{} \minus{} q$ is a root. Thus, if there are an odd number of roots, $ x \equal{} 0$ must be a root. Thus, we must have $ \parallel{}|0| \minus{} 2| \minus{} 3| \equal{} a \implies a \equal{} 1$.

Verify that $ a\equal{}1$ indeed produces 5 roots: $ 0,\pm 4, \pm 6$, and we are done.[/hide]

Edit -- yeah, that was silly xD

Your answer is correct and the equation has 5 roots when $ a \equal{} 1$?  :maybe:

[hide="Including..."] $ x \equal{} \pm 6$.  :) [/hide]


Amy works for 36 hours per week for 10 weeks during the summer, making $ \$3000$. If she works for 30 weeks during the school year at the same rate of pay and needs to make another $ \$3000$, how many hours per week must she work?

She works three times as much during the school year as in the summer, so if she wants to make the same amount of money, she should work $ \frac{1}{3}$ as much as she did during the summer. The answer is thus $ \frac{36}{3}\equal{}\boxed{12}$ hours per week.

[quote="Textangle"]She works three times as much during the school year as in the summer... she should work $ \frac {1}{3}$ as much as she did during the summer.[/quote]

Er, that really needs rewording.

What he's saying is that since the [b]number of weeks worked[/b] has increased by a factor of three, then the [b]number of hours worked[/b] can be 1/3 of what it originally was, in order to accumulate the same total time worked.

That is:
$ 10\cdot 36\equal{}30\cdot 12$ working hours.


[quote="GameBot"]It is possible to earn 0, 1, 3, 7 or 10 points with each shot in the game of \emph{Blippy}. How many positive scores less than 30 cannot be made in 3 shots?[/quote]
There are 15, 19, 22, 25, 26, 28, and 29, so it should be [b]7[/b].

WE can make 15 (771) but we can't make 24 so it is 7.

we can make 24 (10, 7 , 7) so its 6....Is there a formula for this????and/or can someone derive it?

I don't think there is any relatively simple method other than organized casework...

Is there a formula to these type of questions?

The answer is 6 because 15 is possible: 7,7,1.


$a,b,c$ are sides of a triangle. Prove that:

$(\tan{\frac{B}{2}}+2)(\tan{\frac{C}{2}}+2)=\frac{7a+5b+5c}{a+b+c}\Longleftrightarrow A=\frac{\pi}{2}$

$(2+\tan\frac{B}{2})(2+\tan{\frac{C}{2})=\frac{7a+5b+5c}{a+b+c}}$ $\Longleftrightarrow$ $2p\cdot(2+\tan\frac{B}{2})(2+\tan\frac{C}{2})=10p+2a$ $\Longleftrightarrow$

$8p+4p\cdot(\tan\frac{B}{2}+\tan\frac{C}{2})+2p\cdot\tan\frac{B}{2}\tan\frac{C}{2}=10p+2a$ $\Longleftrightarrow$ $\|\begin{array}{c}2p\cdot (\tan\frac{B}{2}+\tan\frac{C}{2})+p\cdot\tan\frac{B}{2}\tan\frac{C}{2}=p+a\\\\ \tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{B}{2}\tan\frac{C}{2}+\tan\frac{C}{2}\tan\frac{A}{2}=1\end{array}$ $\Longleftrightarrow$

$2p\cdot(\tan\frac{B}{2}+\tan\frac{C}{2})=p\cdot\tan\frac{A}{2}(\tan\frac{B}{2}+\tan\frac{C}{2})+a$ $\Longleftrightarrow$ $p\cdot(\tan\frac{B}{2}+\tan\frac{C}{2})(2-\tan\frac{A}{2})=a$ $\Longleftrightarrow$

$\cot\frac{A}{2}(2-\tan\frac{A}{2})=1$ $\Longleftrightarrow$ $\tan\frac{A}{2}=1$ $\Longleftrightarrow$ $A=\frac{\pi}{2}\ .$


AMC USA(J)MO USAMO Harvard college MIT search

Yeah, well the beginner's corner is defined like the place for the people that feel that do not belong to advanced section (yet  :P), so you fit it.

On the other hand, the thing is to solve as many problems as possible. Or at least try. Those you can / can't solve post them (maybe if they are known problems from known competition a search after problem source will reveal the they have been already posted somewhere) to get other's opinions / ideas.

It very much depends on what do you want to do with it? I mean do you want to reach USAMO? or do you want to perform better at Putnam? or are you even an American? :P

Haha, yeah, I'm an American, and uh, damn, I have to QUALIFY for the USAMO?  Gr, yeah, I'd do practically anything.   :P   Well, I never heard of the Putman?  Uh, what the hell is Putman?   ;)  All my hours of searching online for math competitions hasn't paid off!   :(

actually it's putnam, not putman :D. the largest, most important math competition for college students in USA.

the best thing you should do is to download these books:
[url=http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc96-97-01feb.pdf]book 1[/url] and [url=http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc97-98-01feb.pdf]book 2[/url] and start solving / reading.

mainly you have to start by solving problems, catching new ideas, etc. easy problems WON'T do the trick for you. it may be hard at the beginning, but you have the beginner's corner.

also you can use the list located [url=http://www.kalva.demon.co.uk/book1.html]HERE[/url] to grab some of the books there. anymore specific questions, just post them in the right forum, and people (even me :P) will answer you.

yes, what do you think I did when I was 10? (you are 10!?! :P:D) ...

[quote="SingTheSorrow"]Haha, I browsed some of those problems... and I guess I can SORT OF follow some of the logic that they're doing, but I can't SOLVE anything.  :P  My math homework was multiplying and dividing rational algebraic expressions - did you people do that when you were 10 or something?   ;)[/quote]

I usually do this stuff in my spare times on math classes (when I was 10 and in grade 4) :D

[quote="SingTheSorrow"]Hey everyone, I was just wondering if AYNONE has any insight on like actually understanding and completing the problems featured here and the USAMO and IMO...   :) Haha, I thought I was mad smart in math, but ever since I looked at some of the IMO problems, I'm, beginning to think I'm really stupid.  I look at the solutions for the problems here, and I can't even follow the LOGIC of it.   :? Granted, I am only a freshmen, but I'd still like to uh, be able to comprehend some of the stuff you guys are talking about!  I doubt I'm smart enough even to be in the Beginner's Corner on this site!   I'm in the most advanced math course offered at my school (Algebra II), and I wanted to double up, or even triple up, on math classes, but of course, my school doesn't allow that.  :maybe:  Haha, anyway, anyone got any information on "Learn-It-Yourself" math?  Oh, and I looked at Harvard-MIT Mathematic Competition, and I showed it to my teacher, but uh, yeah, she couldn't even figure it out!  Of course, she did got a problem right, but she gave me a lame excuse how she never needed to learn High School students this...  Haha, anyone got [b]any[/b] ideas?    ;)[/quote]

Hey SingTheSorrow (lets call you STS),

see here:

"My story is probably quite unique compared to the rest of you . . .

Throughout my life, I've never been encouraged in math at all. In elementary school (grades 1-5), I was pushed to the lowest math class because my parents hadn't forced me into doing math before I entered school (the blame is not on my parents though). Throughout elementary school, I never learned anything besides addition, subtraction, multiplication, and division.

When I got to junior high school, I got an "A" in the 6th grade math class. As much as I tried, the school system would not let me move into the accelerated "gifted" math program. During the next 3 years at junior high school, all I learned was addition, subtraction, multiplication, and division (fundamentals of arithmetic). It was such a shock knowing that my friend had already passed algebra, geometry, and number theory while I was still learning the basics (he was in the accelerated math program). During these years, I did quite poorly because I had no motivation to do good because they wouldn't allow me in the upper class no matter how good I performed.

Upon entering into high school, I entered Prealgebra/Algebra, 1 a very NON-simulating math class that taught the students how to do one-variable equations the whole year. Again, I felt like I was in the wrong class especially since the students in it had no care for math whatsoever. I begged my teacher to put me into Geometry Accelerated but to no avail I was placed in Geometry "regular." Again, I felt quite depressed that no matter how hard I tried to move up in the maths and the sciences, I was kept down. In my Sophomore year, I had a nervous breakdown . . . my grades slipped and my interest in school and life in itself waned. After doing things I won't mention and learning from the consequences of my depressed actions, I pulled out of my doldrums and in my junior year I took 3 sciences (I forced the school system to allow me) and got my first "A" in math since junior high school. After receiving all "A's" my junior year, I forced the school system to allow me into 3 maths, 2 sciences, and 2 languages . . . all at the accelerated advanced level, and that is where I am now. My senior year of high school.

Great pains have I suffered, great pains that I am suffering now, and great pains I will suffer in the future . . . but that's life . . . DEAL WITH IT!

REMEMBER . . . the best lesson in life is failure.

"Far better it is to dare mighty things, to win glorious triumphs, even though checkered by failure, than to take rank with those poor spirits who neither enjoy much nor suffer much, because they live in the gray twilight that knows not victory nor defeat." -Theodore Roosevelt, 26th president of the United States of America."
via http://www.mathlinks.ro/viewtopic.php?t=3665&start=0

Know that you know about me STS.  Let me relate more to you.  Yes, I'm very understanding of the politics behind getting an education in the states.  Its basically a business, even though your in school learning for the sake of "learning" your really "learning" because its part of the social system.  You have to come to realize that nobody really cares if you succeed nor do they care if you fail.  In fact, people want you to fail so they can push ahead of you.  Therefore, in order to get better at math, your going to go beyond what's expected of you.  You might think your smart but if you haven't be recognized yet in the papers then no one really cares if your the best student in ALgebra II.  So, in order to get better at math, hear my advice:  GET BOOKS, USE WEBSITES, GO TO THE LIBRARY, and DONT WAIT FOR SOMEONE TO TELL YOU WHAT YOU CAN AND CANNOT DO.

Here are some websites that can help you:

http://www.mathlinks.ro/viewtopic.php?t=3419

Choose one's suited for your ability.  Here are some especially good one's that I think can benefit you right now.

http://www.purplemath.com
http://home.alltel.net/okrebs/

Finally, don't let people discourage you.  Find out what you want by yourself while staying aware of the many different facets of life around you (hence, no narrow-mindedness).I'm no longer a senior in high school.  No.  I'm a graduate of high school.  I won a full scholarship to the University of Maryland due to my persistence to be something people said I couldn't.

Your not alone.  I'm still learning.  Every morning I look in the mirror when I wake up and I tell myself "your dumb so do something about it." and I do . . .

Thanks!  haha, I think it's much easier to complain about your school system, then actually go out there and make libraries and the internet, your school.   ;)  I mean, I do try to make the best of my school (well, at least in math) and I pop up with 100s on all the tests, and I come here and I can't help but thinking, "Um... is my school LYING to me?"  I looked at Valentin's problems (that he suggested) for a hour, and I just kept thinking, "Uh-huh, how am I suppose to UNDERSTAND THAT?!"  You have a good mentality, although a little mean.  Thanks for the websites, they DO look like they're on my level...

Is it just in the states where grades 1-6 all you learn is basically adding, subtracting, dividing, and multiplying?  Haha, what's really irritating to me is that the US feels that we shouldn't learn other languages until middle school!  Well, I'm done.  :P  Congrats on your[b] FULL [/b]scholarship!  :cool:

I too have complained on this website about the same problem.....and I felt and I kind of stil feel like you too.....but this year I learned an important lesson that no one cares about you if you don't....so my advice too is....please don't give up!...it was easier for me too to blame the system.....and I did...I think i stil do.....but if you continue there will come a time when you'll really be able to feel proud of yourself.....despite of the first impression difiicult problems created about you.

about the system here...I can tell you that if you don't live in an important city.....you're chances at the NMO are low.....many people admit this...because thre are't people who care and there isn't enough information....
luckly for us this website exists...and also...I dare to say people who care....hope you'll do better...bye!


linear algebra
matrix
superior algebra
superior algebra unsolved

$ \mathbb{A}^n(k)$ = Affine space of dimension $ n$ over $ k$
Let $ \pi : \mathbb{A}^n(k) \to \mathbb{A}^m(k)$ be an linear transformation given by the matrix formula $ \pi (x) \equal{} Ax\plus{}b$.
Show that $ \pi$ is a $ k$-algebra homomorphism.

I can't show the following:
(1)$ \pi(kx)\equal{}k\pi(x)$
(2)$ \pi(x\plus{}y)\equal{}\pi(x) \plus{} \pi(y)$
(3)$ \pi(xy)\equal{}\pi(x) \pi(y)$
What is a '$ k$-algebra homomorphism'?

It's not a homomorphism. A $ K$-algebra homomorphism between two $ K$-algebras $ A$ and $ B$ is a mapping $ \pi: A\to B$ satisfying
1. $ \pi (kx) \equal{} k\pi (x)$
2. $ \pi (x \plus{} y) \equal{} \pi (x) \plus{} \pi (y)$
3. $ \pi (xy) \equal{} \pi (x)\pi (y)$
for all $ k\in K,x,y\in A$.

In the first line of your post you are defined $ k$ to be a field. But in the condition (1) you are using $ k$ as a scalar. Be careful what is a field and what is a scalar!


I was watching this:

http://www.youtube.com/watch?v=9iaoypSrIT0

MIT OCW Introduction to Biology. Eric Lander talks about genes and the work of Mendel.

He comments that the terms "Dominant/Recessive" refer only to phenotypical traits and that strictly speaking alleles should not be dubbed in this way (even though all my Biology textbooks still use the terms dominant/recessive alleles).

Now I am confused by something. My textbooks seem to say that each allele (or gene) is responsible for a single trait while Lander seems to say that each allele can code for multiple traits. So which one is right?

So is he saying that say we have alleles $ R$ and $ r$ then traits are called "dominant" if $ Rr$ (heterozygous combination) results in that phenotype?

I haven't taken bio in two years but what I recall is that each allele codes for a different trait and can turn its expression on or off.  He is correct about "Dominant/Recessive"; they really only do apply to phenotypical traits, like when you're computing a Punnett Square.

Yes, there are genes that can control multiple phenotypic traits. This is known as pleotropy.For example, coat color in cats is due partly to the gene that produces melanin in melanocytes. These melanocytes are also responsible for transmitting chemical signals from the ear to the brain. So, defects in the melanin gene produce white cats with a tendency to deafness.

Also there are traits which are controlled by multiple genes. An example is skin colour.

^^It's pleiotropy, you forgot the i. :wink:

[quote="BanishedTraitor"]So is he saying that say we have alleles $ R$ and $ r$ then traits are called "dominant" if $ Rr$ (heterozygous combination) results in that phenotype?[/quote]

This is basically saying that a heterozygous individual will look phenotypically the same from a homozygous dominant. There can be no differentiation between the two. I think that's what he meant for the definition of dominant.

Ah yes after I looked up a bit more information I found what Lander was saying about sickle cell anaemia. Thanks everyone.


I'm trying to solve for log(i).
I so far have Bz*log(i) - Bh*log(i)
There is more to the problem but it has nothing to do with this part
As you can tell I'm really bad at Algebra.

The book has log(i) = (1/ (Bz-Bh))

well, if you mean that you have $ Bz\cdot\log{i}\minus{}Bh\cdot\log{i}\equal{}1$, then you can factor log(i):

$ \log{i}(Bz\minus{}Bh)\equal{}1$; divide by Bz-Bh

$ \log{i}\equal{}\dfrac{1}{Bz\minus{}Bh}$, which is it.

That's assuming that the expression equals 1, and that indeed nothing else in the problem has to do with this...


Solve the equation $ \frac{1}{x^{2}} \plus{} \frac{1}{(2\minus{}x)^{2}} \equal{} 1$.

Thanks modularmarc101. I let $ x\equal{}1\minus{}y$ to get the same thing.


I'm planning on getting a TI-84, since I will definately need a good graphing calculator for school next year. It would be great if you could provide me some info about the reliability of the calculator and how it compares to other calculators.

Please define "reliable". Of course it will give the right answer if you put in the right question, but I'm sure you mean reliable in some other sense.

As far as i know, the ti-84 plus and silver editions are just ti-83s with a lot more memory (this statement may be false, u might have to verify it). If this is true...I really do not like the way the ti-83s are set up. I prefer the ti-89 and ti-89 titanium as they have an OS built into them and IMO they are just easier to learn to use. Of course, if u get a ti-84, u can download ur own OS......

Note: I probably wouldn't get a ti-89 now because of the fact that the ti-84s and ti-89 titanium use USB lines to connect to computers.

See ticalc.org for some pretty detailed specs, in particular

[url]http://www.ticalc.org/basics/calculators/ti-84plus-se.html[/url]
and
[url]http://www.ticalc.org/basics/calculators/ti-89-ti.html[/url]

For a comparison of all the TI graphing calculators, go to [url]http://education.ti.com/downloads/pdf/us/graphing.pdf[/url].

You're not allowed to use TI89s on some standardized tests and competitions... but TI83+'s and 84+'s are OK.

The TI-83 Plus Silver Edition and TI-84 Plus Silver Edition (the best of both kinds) are exactly the same except in the amount of memory you have (with the 84 having more).  But I never came anywhere near the limit of my memory on my 83+SE, it should not be a problem.  The TI-84 Plus (not SE) is slower than the TI-83 Plus SE.  I did a test seeing which is faster between TI-84+SE and TI-83+SE, and surprisingly the 83 is just a tiny bit faster.  I have not tested, but I think the 84 graphs faster though.  The TI-84's also have a clock on it (which is neat), and time functions for use in BASIC programs (which are completely unnecessary).

In terms of ease to use, I find the 83 family to be the best.  They are very simple and very powerful, and writing programs for them is fun and easy (and good on any tests whenever you can use them.  Being able to tell it to iterate thru all numbers between x and y and find whatever is needed is a lot easier than using mods or whatever you'd need normally  :lol: ).

In terms of reliability, I do not know enough about the 89s to compare to, but the 83's are fairly reliable.  If you download tons of applications and ASM programs, you're bound to get your ram cleared every once in a while.  Most people stock their calc with an ASM program called ION, which allows them to play ASM games, which are far superior in terms of speed than BASIC games.  But even the most experienced ASM programmers can have a bug, which will crash your calculator, deleting everything you haven't archived.  This is why I refuse to have any ASM programs on my calculators, and I deleted all my applications, to make sure there's NO chance of my ram getting cleared.  However I still archive my stuff because there are jerks at school who will get onto your calc and clear your ram manually, as there is an option of doing that.

I refer to the 83's as all the 83's and all the 84's, as they are basically the same, varying only in speed and memory, and the clock on the 84's.

[quote="Priapus"]Actually, the 89 series can be used wherever the 83 and 84 series can be used because standardized tests such as the SATs say that only non-QWERTY type calculators are allowed. The 83 series runs faster than the 84 series, and the 89 runs faster than the 89 titanium.

About the calculator crash issue.....
either archive all ur stuff
or
back it up on ur computer[/quote]

Right...

Sorry. :blush: I guess it was the 92 that I was thinking of.

Don't you find it ironic that the older calculator runs faster than the newer one?

I have a 82 :mad: and my calculator doesn't seem to crash. Unless we're shading below a line and we break it too early.

Wow, I thought the limit on calculators was if they could factor.  I guess they changed it... oh well.  I think I'll still be using my TI-84 anyway.

Your statement on the speeds is incorrect.  TI83+SE = TI84+SE, TI83+ < TI83+SE, TI84+< TI84+SE.  So TI83+SE > TI84+, but that can't be generalized to 83s are faster than 84s.

If you just use the calculator for math, it should not crash.  But I am the only one in my school who does so, everyone else loads tons of games onto their calculators, and most of them can crash if you do something wrong, which messes up your calculator.  My friend's TI83+ was so bad, it kept deleting his OS... He could use it for stuff on the home screen, but if he tried to graph or use a program, it crashed and deleted his OS, and had to have it transferred back.

I have a TI 89 and I think the 89's are probably the best TI calculators around. (I even prefer my TI 89 above my TI 200 Voyager.) That's actually surprising because the TI 89 are a little bit "old" :)

I think you should first inform how much you're actually using it. We had to buy an expensive calculator in school, and worked with it for about 2 hours in the whole year. That was a waste of money.
 (luckily I won mine for free :))

Honestly, if we now have an exam to where we can take a calculator, I'll usually take my non-graphical one. Just because there's no use for all the fancy stuff on there, calculators rule for statistics or numerical computations, but any calculator doing that equally good will be equally good in my eyes.

I think on some math or science contests they don't let you use Ti 89/92 or higher because it can perform symbolic operations.  That also means that the ti-89 is alot better, but the price is not that much more.  I looked on amazon and it was $142 vs $128.

There aren't very many such contests (that I know of.) At least you can use it on the AMC contests, the SATs, AP exams, etc..

Yeah, that's true.  The ones I were thinking of were local.


Hi!

I think equation (3) in MathWorld link for "Isogonal Conjugate" (http://mathworld.wolfram.com/IsogonalConjugate.html) is wrong.

In the image attached, x = d(Q, QAC), y = d(Q, QBC), r = d(P, PAC) and s = d(P, PBC), but x/y is not equal to s/r.

Am I doing something wrong?

Thanks, Humberto.

You are right. It would be nice if you point out the mistake to Eric Weisstein.

What was probably meant by (3) is Theorem 14 in http://www.mathlinks.ro/Forum/viewtopic.php?t=18472 .

  Darij


You mean the sum is not in $ N$, I hope.

Anyways, the cleanest ways I've seen to do this:

1. Show that the sum is in an interval that doesn't contain an integer. This is usually done by converting denominators to powers of 2.

2. Assume Bertrand's postulate. Show that if $ n < p < 2n$, then the sum clearly can't be an integer. The conclusion follows immediately.


I've been self-studying (*cough* window-gazing) for the AP Electricity and Magnetism exam, and I don't understand any of it. Here are a few things I need explained to me:

"Gauss's Law for Magnetic Fields.. The fact that magnetic field lines are always close upon themselves tells us that the magnetic flux through any closed surface must be zero..." (From the better of the two books I've been using X_X) It just had me solve problems where I calculated flux! It seems to that if the closed surface were symmetric and the axis of the dipole passed perpendicularly through the center of the surface (unless the dipole were actually through the surface) this would be true, but there are plenty examples of there being net flux through another area. This is pretty much representative of my entire experience with this course--I read something that makes no sense and I don't know whether I'm misunderstanding or whether the book's just not properly defining its case.

"Below shows a bar magnet moving down, through a circular loop of wire. What will be the direction of the induced current in the wire?" (The North side is a little above the loop.) They said this would create "a change of more flux downward" but the magnetic fields are currently denser going downward (to form wide loops out back to the South pole) but as the magnet goes down, the net flux should be becoming more positive--net flux upward--since the field loops get close enough to the magnet that they're going up again within the circle of wire. Am I wrong?

How do you know whether Ampere's Law applies/is useful or not in a certain physical situation? For instance, my review book reports a formula (B=permeability x number of turns / length x I) for the magnetic field inside a solenoid, but no current's enclosed in a solenoid. On the other hand, we're supposed to have no compunctions about applying Ampere's Law to the inside of a toroidal solenoid. Why?

I've looked at various sources and textbooks online and used the two review books that I have, but I haven't found anything that explains the theory worth anything. So, finally, any good links?

First, we'll talk about Gauss's Law for Magnetic Fields.  It says that the net flux through a closed surface is 0.  This means that there is no such thing as a magnetic monopole (of course there COULD be, but no one has found one yet so this equation is perfectly valid).  The simplest example is a bar magnet.  The magnetic field lines extend out from the north pole and then close back in on the south pole so the net flux is zero.  If we put some surface, say a sheet of paper, above the magnet, then there IS a non-zero net flux because the surface is not closed.  That is why when we write Gauss's Law, we write, $ \oint_{S}\textbf{B}\cdot\ d\textbf{a}\equal{}0$, not $ \int\textbf{B}\cdot\ d\textbf{a}\equal{}0$.

Now for Ampere's Law.  Ampere's Law (the one not ammended by Maxwell) is applicable for time-independent magnetic fields and situations with a certain type of symmetry.  For example, we may use Ampere's Law for an infinite wire, an infinite solenoid, and even a toroidial solenoid.  Note that all three of these are very symmetric.  In the case of the infinite solenoid, we construct a rectangle around the edge.  The two edges perpendicular to the solenoid cancel each other, and the edge outside experiences no magnetic field so the only one we care about is the one inside the solenoid.

I hope this clears this up, at least a little.  Feel free to ask some more questions.

[quote="akkaren"]...How do you know whether Ampere's Law applies/is useful or not in a certain physical situation? For instance, my review book reports a formula (B=permeability x number of turns / length x I) for the magnetic field inside a solenoid, but no current's enclosed in a solenoid. On the other hand, we're supposed to have no compunctions about applying Ampere's Law to the inside of a toroidal solenoid. Why?
[/quote]

I not sure where your difficulty lies with the first part of your question, but for this part I believe you have misinterpreted the formula for $ \mathbf{B}$.

Ampere's law (without Maxwell's amendment) is:

\[ \nabla \times \mathbf{H} \equal{} \mathbf{J} \]

from which it follows by Stoke's theorem:

\[ \Rightarrow \oint_{C} \mathbf{H}\cdot\, d \vec{\ell} \equal{} \iint_{S} \nabla \times \mathbf{H}\cdot \mathbf{\hat{n}}\, da \equal{}\iint_S \mathbf{J}\cdot \mathbf{\hat{n}}\, da \equal{} I_{\text{en}} \]

For a solenoid: $ d\vec{\ell}\equal{} d\vec{z}$ and $ da \equal{} \sum_{i \equal{} 1}^{N}2\pi r\, dr$.

Therefore, $ \int_{0}^{L} \mathbf{H}\cdot\, d\vec{z} \equal{} I_{en} \equal{} N\, I$, where $ I$ is the current through one loop of the solenoid.

\[ \Rightarrow H_{z} \equal{} \frac{N}{L}\, I\]

Let $ n \equal{}\frac{N}{L}$.

We also know that $ \mathbf{B}\doteq \mu_0\mathbf{H}$.

Therefore,
\[ B_{z} \equal{} \mu_0 n I\]

I hope that helps.


Two complementary angles, A and B, have measures in the
ratio of 7 to 23, respectively. What is the ratio of the measure
of the complement of angle A to the measure of the
complement of angle B? Express your answer as a
common fraction.

We have $ \frac ab\equal{}\frac a{180\minus{}a}\equal{}\frac7{23}$.
We want to find $ \frac{180\minus{}a}{180\minus{}b}\equal{}\frac{180\minus{}a}a\equal{}\boxed{\frac{23}7}$.

If A and B are complements, then the complement of A=B and the complement of B=A. So we are looking for B/A which is the reciprocal of A/B=7/23. So our answers is 1/(7/23)=23/7.


Solve the following diophantine equation:
$ 3^x \minus{} 2\equal{}z^2$
$ x,z \in \mathbb{N}$

I suspect the only solutions are $ (x, z) \equal{} (1, 1)$ or $ (3, 5)$ but I can't see a proof yet.


algebra
polynomial
modular arithmetic
Ring Theory
superior algebra
superior algebra unsolved

Let $ f\in \mathbb{Z}[x]$ and $ f(0) \equal{} 1$. Prove that for every $ n$ there is a $ g(x) \in \mathbb{Z}[x]$ such that $ f(x) \cdot g(x) \equiv 1 \mod x^n$.

Well it suffices to show that $ a\equal{}f(x) \plus{} (x^n)$ is invertible in $ S\equal{}\mathbb{Z}[x]/(x^n)$.
The latter is a local ring since obviously the only maximal ideal is $ M\equal{}(x)/(x^n)$.
If$ a$ weren't invertible then it would be contained in a maximal ideal of $ S$ and since the unique maximal ideal of $ S$ is $ M$ we conclude $ a \in M$, which is a contradiction since $ f(0)\equal{}1$

$ f(x)\equal{}xh(x)\plus{}1\\
\Rightarrow 1\equal{}(f(x)\minus{}xh(x))^n\equal{}\sum_{k\equal{}0}^n\binom{n}{k}f(x)^k(\minus{}xh(x))^{n\minus{}k}\equiv f(x)g(x)\pmod{x^n}$
for $ g(x)\equal{}\sum_{k\equal{}1}^n\binom{n}{k}f(x)^{k\minus{}1}(\minus{}xh(x))^{n\minus{}k}$.


$\sin a \cdot \sin b \cdot \sin c = \left(\frac{e^{ai}-e^{-ai}}{2i}\right)\left(\frac{e^{bi}-e^{-bi}}{2i}\right)\left(\frac{e^{ci}-e^{-ci}}{2i}\right)$.

If you carry out the calculations, you will get that

$\sin a \cdot \sin b \cdot \sin c =-\frac{1}{4}\left[\sin(a+b+c)-\sin(a+b-c)-\sin(a-b+c)+\sin(a-b-c)\right]$.

Now make $a = \frac{\pi}{24}$, $b = \frac{5 \pi}{24}$, and $c = \frac{11 \pi}{24}$ and so

$\sin\left(\frac{\pi}{24}\right) \cdot \sin\left(\frac{5 \pi}{24}\right) \cdot \sin\left(\frac{11 \pi}{24}\right) =$

$=-\frac{1}{4}\left[\sin\left(\frac{17 \pi}{24}\right)+\sin\left(\frac{5 \pi}{24}\right)-\sin\left(\frac{7 \pi}{24}\right)-\sin\left(\frac{15 \pi}{24}\right)\right]$.

But:

$\sin\left(\frac{17 \pi}{24}\right) = \sin\left(\frac{24 \pi}{24}-\frac{7 \pi}{24}\right) = \sin(\pi) \cos\left(\frac{7 \pi}{24}\right)-\cos(\pi) \sin\left(\frac{7 \pi}{24}\right)$
$= \sin\left(\frac{7 \pi}{24}\right)$.

$\sin\left(\frac{5 \pi}{24}\right) = \sin\left(\frac{\pi}{4}-\frac{\pi}{24}\right) = \frac{\sqrt{2}}{2}\left[\cos\left(\frac{\pi}{24}\right)-\sin\left(\frac{\pi}{24}\right)\right]$.

$\sin\left(\frac{15 \pi}{24}\right) = \sin\left(\frac{2 \pi}{3}-\frac{\pi}{24}\right) = \frac{1}{2}\left[\sqrt{3}\cos\left(\frac{\pi}{24}\right)-\sin\left(\frac{\pi}{24}\right)\right]$.

So:

$\sin\left(\frac{\pi}{24}\right) \cdot \sin\left(\frac{5 \pi}{24}\right) \cdot \sin\left(\frac{11 \pi}{24}\right) =$
$= \frac{1}{2}\left[(\sqrt{2}-\sqrt{3}) \cos\left(\frac{\pi}{24}\right)-(\sqrt{2}-1)\sin\left(\frac{\pi}{24}\right)\right]$.

Finally, we have:

$\cos\left(\frac{\pi}{24}\right) = \frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{3}}}$

$\sin\left(\frac{\pi}{24}\right) = \frac{1}{2}\sqrt{2-\sqrt{2+\sqrt{3}}}$.


quadratics
number theory proposed
number theory

$n$ is a positive integer. Prove that there is a prime $p<cn^{\frac{3}{4}}$ ($c$ is an absolute constant) such that $p$ does not divide $k^{2}+n$ for all $k \in {\mathbb Z}$.

I think we shall prove that there is an $c$ independent from $n$ with that property.

I think it has the same meaning as just "constant"..

[b]dzeta[/b]: Could you show your solution, please? I bother with your beatiful problem for over a week and I've already used my ideas in vain (I tried to bound $n(x)$ (number of primes less than $x$ for which $-n$ is quadratic nonresidue) from above and below and get conclusion from this but I guess my estimates were too weak or it just simply doesn't lead to a solution  :oops: ).

I think you will be dissapointed with this, but here is goes. Take the largest nonnegative integer $a$ such that $N=4n-\left(4a^{2}+1\right)^{2} \geq 0$. By straightforward calculations $N<cn^{3/4}$ for some constant $c$. Since $N>0$ and $N \equiv 3(\bmod 4)$, there is a prime divisor $p \equiv 3(\bmod 4)$ of $N$. I claim $p$ has the desired property. Of course, $p \leq N < cn^{3/4}$. Suppose by way of contradiction that there is a $k \in {\mathbb Z}$ such that
\[ p|k^{2}+n \Rightarrow p|4k^{2}+4n = 4k^{2}+\left(4a^{2}+1\right)^{2}+N \Rightarrow p|4k^{2}+\left(4a^{2}+1\right)^{2}, \]
and since $p \equiv 3(\bmod 4)$, $p|2k$ and $p|4a^{2}+1$ which is impossible since $p \equiv 3(\bmod 4)$.

I'm not dissapointed but rather suprised by such a short solution. And I expected something more "analytic"  :)

Anyway could you show your calculations in evaluating $N<cn^{\frac{3}{4}}$ as well? I don't know how you're getting $n$ with exponent $\frac{3}{4}$...

Megus, $0\geq 4a^4+2a^2+(1-n)$, so $a\leq\sqrt{\frac{-1+\sqrt{4n-3}}{4}}$, so $4a^2+1\geq 4+\sqrt{4n-3}-4\sqrt{-1+\sqrt{4n-3}}$.

Sqaring this, we get terms of order $n$, $n^{\frac{3}{4}}$, $n^{\frac{1}{2}}$, and $n^{\frac{1}{4}}$.  The term of order $n$ cancels when subtracted from $4n$, so $N$ is of order $n^{\frac{3}{4}}$.


Ross Mathematics Program
email
summer program
MathPath
geometry
number theory

If anybody is still looking for a great summer opportunity, the Ross Program at Ohio State

University is still accepting applications.  I will be returning to the program, for my third year

as a counselor.  I highly recommend it as a way to develop mathematical talent and ability.

Please visit the webpage at:

[url=http://www.math.ohio-state.edu/ross/]www.math.ohio-state.edu/ross/[/url]

and feel free to PM me, or send me email at: zls2@cornell.edu

If I am accepted as a first year student, am I required to take the number theory course, if I already have significant experience with number theory already? Is it possible to advance, if I find that the material is too easy?

I'm sure you will find material that you do not know, although there will be an abstract algebra course, and a course for Junior Counselors that you are more than welcome to participate in.

- Zach

I can only go for 4 weeks, is that okay??

The number theory course goes for all 8 weeks with a problem set a weekday. The most one can reasonably miss of the beginning and still have a chance at catching up is one or two weeks. And the earliest anyone leaves is one or two weeks before the end. (Yes, Ross really is worth it for all 8 weeks.)

[quote="zscool"]xxreddevilzxx:

I'm not really sure if this is feasible, but the best person to ask would be Dr. Shapiro, who

runs the program, at: ross@math.ohio-state.edu[/quote]

okay I will do as you say. Thanks.

[quote="tokenadult"]What would you recommend as a way to prepare for the Ross Program? I'm asking on behalf of some MathPath participants this year who may want to do Ross Program when they are old enough to participate in that.[/quote]

There really is nothing that you can do to prepare for the Ross program.  As long as you go

with a curious attitude and some determination, you will have a great time and learn some

very interesting mathematics.

im going to ross!!!  :lol:  im 9th grade so it'll be my first year... is anyone else on this board going? oh, & we can take an extra course if we want to in abstract geometry?


[quote="mohd_ahsan"]
and also can ya show how the formula  A = P ( 1 + r/100)^n  is derived? I'm confused.[/quote]

You never really need to think about a formula if you just think about what is happening. You start off with P dollars, and the interest you are getting is r% a year, right? So how much do you have next year? You have r/100 * P interest, added to that P, which gives (1 + r/100) * P. Now how much do you have the year after that? Again, we just times by (1+r/100), so we have P*(1+r/100)^2. Each year we just have to multiply by (1+r/100).

yeah got that. But wat about the problem? what if for example it asks for a period of 30 years, it would take too long without a formula.

thats too little a hint.................

[quote="mohd_ahsan"]A special form of endowment insurance requires an initial investment of 500 dollars, and then further deposits of 200 dollars each year. If the interest is compounded annually at 6%, what is the total value of the investment after a period of 10 years?

and also can ya show how the formula  A = P ( 1 + r/100)^n  is derived? I'm confused.[/quote]

[b]Part 1. Compound Interest[/b]

How much investment of $500$ dollars will be in $10$ years?

$P_n = P(1+r)^n$

Where $P_n$ is an amount after $n$ periods (here after 10 years)
$P$ is a principal $P = 500$,
$r$ is interest rate per compounding period (in decimals) $r = 0.06$
$n$ is number of periods $n = 10$

Substituting and evaluating:

$P_{10} = 500(1+0.06)^{10}\approx 895.42$

[b]Part 2. Ordinary Annuities.[/b]

What will be the amount of annuity in 10 years?

$S_n = M\cdot \left(\frac{(1 + r)^n - 1}{r}\right)$

Where $S_n$ is the accumulated value of the payments at the time of the $n^{th}$ payment,
$M$ is the size of the periodic payment,
$n$ is the number of payments,
$r$ is the fraction interest rate per period.

Substituting and evaluating:

$S_{10} = 200\cdot \left(\frac{(1 + 0.06)^{10} - 1}{0.06}\right)\approx 2,636.16$

[b]Part 3. Total.[/b]

$895.42 + 2,636.16 = 3,531.58$

[b]Note:[/b]

If the initial principal investment is $P$, then interest earned in the fist period is $r\cdot P$. Adding this to $P$ to create a new principal $P_1$, we have
$P_1 = P + Pr = P(1 + r)$
$P_2 = P_1 + Pr = P_1(1 + r) = [P(1 + r)](1 + r) = P(1 + r)^2$
$\vdots$
$P_n = P(1 + r)^n$

[i][b]Good luck with your investment![/b][/i]

help is appreciated bro.  I was studying geometric sequences and one application of it , the book said was in calculation depreciation, inflation e.t.c. so it introduced the compound interest formula. Things went alright till this question suddenly appeared at the end of the exercise. and i couldn't do it.

The answer provided in the book is 3331.58 dollars and ur answer is 3531.58 which is exactly 200 more. so you must have counted it once more i guess.

I searced the net for "annuity" but didn't find something which explained it. can you suggest a site

I think that thats because they meant for you to add another $200 at the end of the time as you are adding 200 at the end of each year and this amount would be interest-less since it would have been added right at the end


function
trigonometry
inequalities
real analysis
real analysis unsolved

In which points function $ f(x,y) \equal{} (x \plus{} y)\sin\frac {1}{x}\sin\frac {1}{y}$ when $ xy\neq 0$ and $ f(x,y) \equal{} 0$ when $ xy \equal{} 0$, isn't continuous

[hide]
Note that : $ 0 \leq sin(\frac {1}{x}).sin(\frac {1}{y}) \leq 1$ Then $ 0 \leq x.sin(\frac {1}{x}).sin(\frac {1}{y}) \leq x$
Note that : $ 0 \leq sin(\frac {1}{x}).sin(\frac {1}{y}) \leq 1$ Then $ 0 \leq y.sin(\frac {1}{x}).sin(\frac {1}{y}) \leq y$
Combining the two inequalities to obtain that :
$ 0 \leq (x\plus{}y).sin(\frac {1}{x}).sin(\frac {1}{y}) \leq (x\plus{}y)$
Take the limit $ x$ and $ y$ attains $ 0$, the $ LHS$ is valued $ 0$ and the $ RHS$ is also valued $ 0$, then $ Lim((x\plus{}y).sin(\frac {1}{x}).sin(\frac {1}{y}))\equal{}0$ when $ x$ and $ y$ attains $ 0$.
Because $ f(0,0)\equal{}0$, $ f$ is continuous at $ (x,y)\equal{}(0,0)$
It is also obvious that the function is continuous at $ x$ that is not equal to $ 0$ or $ y$ that is not equal to $ 0$. Then, the function is continuous at every point in $ R^2$.
I am sorry if I am wrong..
[/hide]

What is the product of discontinuous and continuous 'non-zero' functions?