Triangle
geometry
construction
excircles
IMO Shortlist
IMO Longlist

Construct a triangle given the radii of the excircles.

We construct a segment $x$, then a segment $y$ s.t. $xr_a=yr_b$, then a segment $z$ s.t. $xr_a=zr_c$. We then construct the triangle with sides $a'=y+z,b'=z+x,c'=x+y$, construct its $a'$-exradius $r_a'$, and rescale it to sides $a,b,c$ s.t. $\frac a{a'}=\frac {r_a}{r_a'}$.

[hide=Solution]Let the $A$-exradii be $r_A,$ and define $r_B,r_C$ in the same manner. Also, let $r$ be the length of the inradius of $ABC.$ Thus,
\begin{align*}
\frac{s}{s-a}\cdot r&=r_A,\tag{1}\\
\frac{s}{s-b}\cdot r&=r_B,\text{ and }\tag{2}\\
\frac{s}{s-c}\cdot r&=r_C.\tag{3}
\end{align*}
Dividing equations (1) and (2) and equations (1) and (3), we find that
\begin{align*}
\frac{s-b}{s-a}&=\frac{r_B}{r_A}\text{ and }\tag{4}\\
\frac{s-c}{s-a}&=\frac{r_C}{r_A}.\tag{5}
\end{align*}
Adding equations (4) and (5), we find that
\begin{align*}
\frac{a}{s-a}=\frac{r_B+r_C}{r_A}&\Leftrightarrow\\
\frac{s}{s-a}=\frac{r_B+r_C+r_A}{r_A}&\Leftrightarrow\\
s-a=s\cdot \frac{r_A}{r_A+r_B+r_C}&\Leftrightarrow\\
a=s\cdot\frac{r_B+r_C}{r_A+r_B+r_C}&.
\end{align*}
Therefore, we have know know that $a:b:c=r_B+r_C:r_A+r_C:r_A+r_B.$ Now, we can construct the triangle using standard techniques. (Fix $BC,$ draw circles of radius $AB,AC$ centered at $B,C,$ respectively. Their intersection (it doesn't matter which) is point $A$. Thus, we have all three vertices, which means that we have constructed the triangle).[/hide]


AMC
USA(J)MO
USAMO
inequalities
geometry
induction
pigeonhole principle

I know Kalva has a load of proof problems, but is there any site that actually "teaches" important things to know in order to solve proofs (like major inequalities, theorems, and concepts)?

Thanks

Well, AoPS offers classes (which in my experience are very very good), but if you need resources for this year's USAMO, it's a little late for that. (And they/we/wey have Olympiad problems too [click "Resources"], which are sometimes a little easier to understand than Kalva's.)

Arthur Engel's [i]Problem-Solving Strategies[/i] and, of course, the Art of Problem Solving books aren't websites, but are fairly good resources nonetheless. Again, though, unless Barnes and Noble or your local library carries it, it might be a little late to use for the USAMO.

For geometry resources, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81150]my thread[/url], and there are several threads which list good inequalities and other concepts to know. I'll try to reiterate what I can, though:

*Trivial Inequality (x^2>0)
*RMS-AM-GM-HM (or General Mean if you really think it'll help)
*Cauchy-Schwarz (or Holder if you really think it'll help)
*Triangle (or Minkowski if you really think it'll help)
*Rearrangement (or [T]Cheb[y/i][c/s]he[v/ff] ... actually, [i]and[/i] Tchebiysheffv. It'll help.)
*Jensen

You should also be fairly comfortable with WLOGging (which you can do to a symmetric inequality, plus a lot of other stuff), induction, Pigeonhole Principle, basic Diophantine equations and methods for solving them, and...that's about all I can think of at the moment. If I left something major off, someone else is sure to post it.

Hey, I'll just use this space to advertise [url=http://artofproblemsolving.com/Forum/weblog.php?w=307]my blog[/url], which contains lecture handouts from my lectures on geometry, a bit of combo, and proof methods. Of course, we're spending this week on inequalities and next week on random other stuff we haven't gone over, and that will be on the site soon. Enjoy!  :lol:

[quote="randomdragoon"]Spend your math class reading the Art of Problem Solving book, volume 2.

Heck, spend your English, history, art, and world language classes doing that too.[/quote]

heh... i might skip school just to staty inside and do USAMO problems. dang, that sounds lame.

[quote="amitavadhar"]
heh... i might skip school just to staty inside and do USAMO problems. dang, that sounds lame.[/quote]

Well, that's my spring break....

And my week after spring break, except I won't actually be skipping, just not paying attention.

I want to restate a suggestion someone posted before - go to your local library.
They might not have any book that is useful for Olympiad Level Problem Solving, but if you give
them a specific title or two or three, they will inter-loan it for you (if another local library has it) or they will purchase it for you.  Yes, that's right.  Public library = free books :)
Put those tax payer dollars to good use.  Make them buy you math books!

I compiled a list of useful books in the attached file.  Go to Amazon.com if you want to read reviews.
My High School library has Winning Solutions by Edward Lozansky, Cecil Rousseau and it is exactly what you are looking for.

[quote="Thorn"]I want to restate a suggestion someone posted before - go to your local library.
They might not have any book that is useful for Olympiad Level Problem Solving, but if you give
them a specific title or two or three, they will inter-loan it for you (if another local library has it) or they will purchase it for you.  Yes, that's right.  Public library = free books :)
Put those tax payer dollars to good use.  Make them buy you math books!
[/quote]

And if your public library isn't cooperative or is hard to reach or whatever, you can always talk to your school librarian. Chances are it'll be easier to reach, and some school media specialists might even have kids or relatives who've qualified for USAMO in the past or present!   ;)

[quote="Fermatprime"]And if your public library isn't cooperative or is hard to reach or whatever, you can always talk to your school librarian. Chances are it'll be easier to reach, and some school media specialists might even have kids or relatives who've qualified for USAMO in the past or present!   ;)[/quote] ...or they'll remember you from last year and do anything you want! :)


geometry
3D geometry
ratio
prism
percent
percent increase
area of a triangle

[quote="vishalarul"]I'll post a bunch of problems for you to solve:

1)The sum of four integers is $439$. The three smaller are consecutive integers and the two larger are consecutive even integers. Find the product of the smallest and the largest of the four integers.

2)The difference between two numbers is $7$, and the sum of the squares of these two numbers is $569$. What is the value of the product of the two numbers?

3)What is the sum of the elements in the following sequence
$-1,3,-5,7,-9,11, ...,-321$?

4)Square $SQUA$ with midpoints $M$ and $N$ of sides $SQ$ and $SA$ respectively, has an area of $400$ square units. What is the number of square units rounded to the nearest integer, in the area of the largest circle that can be inscribed in the pentagon $MNAUQ$? Take Pi=$3.14$.

5)The outside faces of a large cube are painted, The large cube is then divided into unit cubes. There are exactly $216$ unit cubes that have no paint. How many unit cubes were used to create the large cube?

When all the problems are done, someone please pm me or something. I'll post more problems. I'm too busy with my other posts...[/quote]
[hide="1"]The numbers are x, x+1,x+2,x+4. $4x=432$
$x=108$
$108\cdot112=12096$[/hide]

[quote="vishalarul"]I'll post a bunch of problems for you to solve:

1)The sum of four integers is $439$. The three smaller are consecutive integers and the two larger are consecutive even integers. Find the product of the smallest and the largest of the four integers.

2)The difference between two numbers is $7$, and the sum of the squares of these two numbers is $569$. What is the value of the product of the two numbers?

3)What is the sum of the elements in the following sequence
$-1,3,-5,7,-9,11, ...,-321$?

4)Square $SQUA$ with midpoints $M$ and $N$ of sides $SQ$ and $SA$ respectively, has an area of $400$ square units. What is the number of square units rounded to the nearest integer, in the area of the largest circle that can be inscribed in the pentagon $MNAUQ$? Take Pi=$3.14$.

5)The outside faces of a large cube are painted, The large cube is then divided into unit cubes. There are exactly $216$ unit cubes that have no paint. How many unit cubes were used to create the large cube?

When all the problems are done, someone please pm me or something. I'll post more problems. I'm too busy with my other posts...[/quote]
[hide="5"]The ones that are painted are the ones on the faces.
There are 216 that aren't on faces. If you cut of the faces you get a 6by6by6 cube. Add two to the faces to get 8. $8^{3}=\boxed{512}$.[/hide]

[hide]5 is 512 unit cubes[/hide]

[size=75][color=darkred]Please hide your answers when posting in reply to a problem. Thanks.
-nebula42[/color][/size]

[quote="vishalarul"]4)Square $SQUA$ with midpoints $M$ and $N$ of sides $SQ$ and $SA$ respectively, has an area of $400$ square units. What is the number of square units rounded to the nearest integer, in the area of the largest circle that can be inscribed in the pentagon $MNAUQ$? Take Pi=$3.14$. [/quote]

I wouldn't use the word "inscribed," but rather "contained."

[hide="4"]
Draw the diagram, the largest circle must have tangent lines $MN$, $AU$, and $QU$, let's name the center O. Connect O to points of tangency on $MN$, $QU$, and $AU$, and connect $SU$, the diagonal of the square.

since the square has area of $400$ s.u., the side length is $20$ units. $AU=20\sqrt2$

assume circle O has radius r, you can see:
$5\sqrt2+r+r\sqrt2=20\sqrt2$
$r=30-15\sqrt2$

Area of circle O=$Pi*r^{2}=3.14*(30-15\sqrt2)^{2}=242.43{unit}^{2}$

Therefore, the area of circle is about $\boxed{242}$ ${unit}^{2}$[/hide]

[quote="alex__ib"][hide="2, my favorite"]
we have x and y
$x-y=7\Rightarrow x=y+7\\ y^{2}+(y+7)^{2}=569\\ 2y^{2}+49+14y=569\\ 2y^{2}+14y=520\\ y^{2}+7y=260\\ y\times\ (y+7)=260\\ We\ know%Error. "that" is a bad command.
\ 260=2^{2}\times5\times\13=260\\ 2^{2}\times\5=20=13+7\\ \Rightarrow y=13x=20$[/hide]

[hide="and 3"]
we group like:
$(-1+3)+(-5+7)......+(-317+319)-321=2\times80-321=160-321=-161$[/hide]....

~~~pm me if there is any mistake, please!~~~[/quote]

ya there is some mistake at #2.. it asked the value of the product of the two numbers, not those two numbers. By the way, there is one more possible answer: $-13$ and $-20$.

First set two numbers as x and y, and x>y
$x-y=7$ or $(x-y)^{2}=49$ or $x^{2}-2xy+y^{2}=49$
$x^{2}+y^{2}=569$

you want to find $xy$...

so take the second equation minus the first.. you get:
$2xy=520$ or $xy=\boxed{260}$

[quote="dodobird"][quote="vishalarul"]I'll post a bunch of problems for you to solve:

1)The sum of four integers is $439$. The three smaller are consecutive integers and the two larger are consecutive even integers. Find the product of the smallest and the largest of the four integers.

2)The difference between two numbers is $7$, and the sum of the squares of these two numbers is $569$. What is the value of the product of the two numbers?

3)What is the sum of the elements in the following sequence
$-1,3,-5,7,-9,11, ...,-321$?

4)Square $SQUA$ with midpoints $M$ and $N$ of sides $SQ$ and $SA$ respectively, has an area of $400$ square units. What is the number of square units rounded to the nearest integer, in the area of the largest circle that can be inscribed in the pentagon $MNAUQ$? Take Pi=$3.14$.

5)The outside faces of a large cube are painted, The large cube is then divided into unit cubes. There are exactly $216$ unit cubes that have no paint. How many unit cubes were used to create the large cube?

When all the problems are done, someone please pm me or something. I'll post more problems. I'm too busy with my other posts...[/quote]

[hide]On #3 you can just add the sum of every 2 numbers so the sum would be $-2$. So then just multiply that by $\frac{321-3}{2}$ so i think the answer is $-318$[/hide][/quote]

dodobird, you made a mistake on $\frac{321-3}{2}$..

$\frac{321-3}{2}+1$ is how many numbers from $3$ to $-321$..
you have to divide that by $2$ to get how many groups. which is $80$ groups. Each group have sum $-2$, and there is an extra $-1$, so the answer is $-2*80-1=\boxed{-161}$

[quote="vishalarul"]I'll post a bunch of problems for you to solve:

1)The sum of four integers is $439$. The three smaller are consecutive integers and the two larger are consecutive even integers. Find the product of the smallest and the largest of the four integers.

2)The difference between two numbers is $7$, and the sum of the squares of these two numbers is $569$. What is the value of the product of the two numbers?

3)What is the sum of the elements in the following sequence
$-1,3,-5,7,-9,11, ...,-321$?

4)Square $SQUA$ with midpoints $M$ and $N$ of sides $SQ$ and $SA$ respectively, has an area of $400$ square units. What is the number of square units rounded to the nearest integer, in the area of the largest circle that can be inscribed in the pentagon $MNAUQ$? Take Pi=$3.14$.

5)The outside faces of a large cube are painted, The large cube is then divided into unit cubes. There are exactly $216$ unit cubes that have no paint. How many unit cubes were used to create the large cube?

When all the problems are done, someone please pm me or something. I'll post more problems. I'm too busy with my other posts...[/quote]

[hide="1"]
12096[/hide]

[quote="vishalarul"]Here are more problems:
1. The ratio of the height of a triangle $ABC$ to the height of a triangle $MNP$ is $\frac{7}{12}$. If the two triangles are similar, what is the ratio of the area of the triangle $ABC$ to the area of the triangle $MNP$? Give your answer as a common fraction.

2. The ratio of the hypotenuse of a right triangle to one of the other sides is $37: 12$. What is the smallest possible area of the triangle if the length of each side of this triangle is an integer?

3. The length of the sides of a rectangular prism measuring $4$  $\textrm{by}$  $6$  $\textrm{by}$  $8$  $\textrm{cm}$ are each increased in length by  $30\%$. What is the percent increase in the volume of the prism? Express your answer as a decimal to the nearest tenth.

4. The three sides of a triangle measure $14$, $10$  $\textrm{and}$  $12$ cm . Find the numbes of square centimeters in the area of this triangle. Express your answer the nearest integer.

5. How many four-digit integers greater than $2000$ for which the thousands digit equals the sum of the other three digits?[/quote]
[hide="1"]If the heights are in a ratio of $\frac{7}{12}$ then so is the rest of a triangle. If the ratio of the sides of two triangles are $\frac{x}{y}$ then the ratio of the areas are $\frac{x^{2}}{y^{2}}$. $(\frac{7}{12})^{2}=\frac{49}{144}$[/hide]
[hide="2"]It just so happens that 12/35/37 is a pythagorean triple. That makes the area $210$.[/hide]
[hide="3"]In essence, the problem doesn't require 4, 6, and 8.
If the sides are x, y, and z then a 30% increase in each is 1.3x, 1.3y, and 1.3z. The volume is 2.197xyz. So the answer is 219.7% regardless of the sides.[/hide]
[hide="4"]Heron's formula states that the area of triangle ABC is $\sqrt{(s-AB)(s-BC)(s-AC)(s)}$ where s is the semiperimeter (half the perimeter). So the area is $\sqrt{18(4)(8)(6)}=24\sqrt{6}=$about 59 (I think. I didn't use a calc)[/hide]
[hide="5, the hardest by far :D"]Let's ignore the first digit and make the second third and fourth x y and z. So we need the number of combos for
x+y+z=3/4/5/6/7/8/ and 9 each.
Then we need the sum of them
Lets break it down.
$x+y+z=1$ 3 ways
$x+y+z=2$ 6 ways
$x+y+z=3$ 10 ways
So these are all triangular numbers.
So for $x+y+z=a$ there are $\binom{a+2}{2}$ ordered pairs (x,y,z).
So the answer is
$\binom{4}{2}+\binom{5}{2}+...+\binom{11}{2}$
Which works out to be 216 [/hide]


Did they seriously change it from Top 25 to Top 8?  They are gonna be so many less people this time. :|

[quote="Ihatepie"]If I had to guess, I would say the majority of AOPSers are going to state.[/quote]

Yeah, true.  But they narrowed it down a lot from last year.  In fact, if I had placed the same as last year, I wouldn't have made it into state.

What did you get last year? Anyways, good luck there. Be sure to have fun.

[quote="Ignite168"]Wasn't it always like top 3 teams + next 8 highest individuals or something?[/quote]It's different for each chapter.

I believe our team is coming too.
The problem is, I heard the CT Econ Challenge is going to be on the same day or something, so we would be missing 4 people if I can't arrange any rides.

[quote="ernie"][quote="Ignite168"]Wasn't it always like top 3 teams + next 8 highest individuals or something?[/quote]

For my chapter, yeah.

Also, vinle, which school do you participate/coach in?[/quote]

Well, let's just say I'm not one of those genius peoples at Hopkins, or Choate, or Glastonbury ;). I barely made AIME this year (and by barely, I don't mean AoPS barely, I mean normal people barely). Heck, I still wonder why I'm president.

I'm not going to state (9th grade) but I am going on http://www.powertolearn.com/challenge2008/index.php

well, people who are going, see you there!

(NOTE: yes, this means I am going)

Not many middle schoolers in CT. :|

What is your goal?  My goal is to make State CD Round, and possibly even Nats. :)

I got 18th at states!!!!! Yay I am really happy!

Didn't get to go to Nats, but oh well, I wasn't expecting to do nearly this well.

[quote="Ihatepie"][quote="ernie"]Well, the only AoPS-er who made it to nats was scai1 (yeah scai!), and it turned out I got 5th in a tiebreaker situation.  Beat scai1 in CD championship though. :)[/quote] You won cd?
Congratz![/quote]

The only reason I won cd was because I basically froze my competitiors by hitting the buzzer like 2 seconds after the problem came.  Haha, my correct pct. was only 50%.

Yeah, I noticed that....

A lot of people up there worried about speed more than correctness......

[quote="spiderboy95"]A lot of people up there worried about speed more than correctness......[/quote]

Such as I.  Glad it was a winning strategy! :D

Good point.

Wasn't there a conversion one that was quite easy? But then there was that one about the sequence that was pretty hard.

[quote="spiderboy95"]Wasn't there a conversion one that was quite easy? But then there was that one about the sequence that was pretty hard.[/quote]

I don't think you can discuss these yet, please edit your post.

Also, when I get back home, I'll post a complete list of CT participants (although I'll cut out some names :)).

Shoot, can't edit post anymore....

If any moderator sees this, can you clarify if my vague descriptions are a rule breach? If they are, please delete my above post w/ the descriptions.


probability
geometry
ratio
probability and stats

[quote="KBriggs"]One in 10 people carry a certain gene. What is the probability that 4 or more people will have to be tested before 2 with the gene are found?

I can solve the problem for the case where we are looking for only one person:

P(X>=4)=1-P(X=3)-P(X=2)-P(X=1)
=1-0.9*0.9*0.1-0.9*0.1-0.1 = 0.729

But I am not sure how to generalize the method.

Can someone help?[/quote]
geometric probability is widely seen,such as lengh,area,volume,you  only need  follow definition of geometric probability.Take area as an example,probability can be calculated like this,it's a ratio between the erea in which events happen and the total erea.

The probability that 2 out of the first 3 have it is $ {3 \choose 2}\cdot \frac{1}{10^2}\cdot \frac{9}{10}\equal{}\frac{27}{10^3}$.  The probability that 3 out of the first 3 have it is $ \frac{1}{10^3}$, so the probability that at least 2 out of the first 3 have it is $ \frac{28}{10^3}$.  The event that at least 4 people are required is the complement of the event that at least 2 out of the first 3 have it, so its probability is $ 1\minus{}\frac{28}{10^3}$.


Is the second-to-last expression supposed to be 1/9 2

[hide] I got C. I just set it up and took a minute or two and crossed out all of the factors[/hide]

Well, that way works but I think theres a better way.

[hide]We see that $\ 1-\frac{1}{n^2}=\frac{(n-1)(n+1)}{n^2}$  So taking the product of this expression from all integers from 2 through 10, we see we can multiply each individual part by itself.  So  the $\ n-1$ term multiplies out to 9!.  The $\ n+1$ term multiplies out to 10!/2.  The $\ n^2$ term multiplies out to be 10!*10!.  So simplifying that expression we get 11/20, choice C.[/hide]

[hide]Using $a^{2}-b^{2}=(a+b)(a-b)$ we can rewrite the expression as:

$\left(1+\frac{1}{2}\right)\left(1-\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1-\frac{1}{3}\right)...\left(1+\frac{1}{10}\right)\left(1-\frac{1}{10}\right)=$

$\frac{3}{2}\times\frac{1}{2}\times\frac{4}{3}\times\frac{2}{3}...\times\frac{11}{10}\times\frac{9}{10}=$

$\frac{1}{2}\left(\frac{3}{2}\times\frac{2}{3}\right)\left(\frac{4}{3}\times\frac{3}{4}\right)...\left(\frac{10}{9}\times\frac{9}{10}\right)\times\frac{11}{10}$
We can see that all terms will cancel except $\frac{1}{2}$ and $\frac{11}{10}$, so the answer is $\frac{1}{2}\times\frac{11}{10}=\frac{11}{20}$ or [b]C[/b].[/hide]

I did a problem like this and I remembered that in a problem like this, the answer is always a little more than 1/2, so it has to be C  :D Maybe, I'm just too lazy to work it out.

[hide]

Notice that $1 - \frac{1}{n^2} = \frac{n^2-1}{n}$. Now the product can be rewritten as: $\frac{3}{4} \times \frac{8}{9} \times \cdots \times \frac{80}{81} \times \frac{99}{100}$.

\begin{eqnarray*}
\frac{3}{4} \times \frac{8}{9} \times \cdots \times \frac{80}{81} \times \frac{99}{100} &=& (\frac{\sqrt{3 \times 8 \times \cdots \times 80 \times 99}}{2 \times 3 \times \cdots \times 9 \times 10})^2\\
&=& (\frac{\sqrt{3 \times 2 \times 4 \times 3 \times 5 \times 2^3 \times 3 \times 7 \times 5 \times 2^4 \times 3 \times 3^2 \times 7 \times 2^4 \times 5 \times 3^2 \times 11}}{10!})^2\\
&=& (\frac{81 \times 64 \times 70 \times \sqrt{55}}{10!})^2\\
&=& \frac{3^8 \times 2^{14} \times 5^3 \times 7^2 \times 11}{(10!)^2}\\
&=& \frac{3^8 \times 2^{14} \times 5^3 \times 7^2 \times 11}{(5 \times 2 \times 3^2 \times 2^3 \times 7 \times 2 \times 3 \times 5 \times 2^2 \times 3 \times 2)^2}\\
&=& \frac{3^8 \times 2^{14} \times 5^3 \times 7^2 \times 11}{(5^2 \times 2^8 \times 3^4 \times 7)^2}\\
&=& \frac{3^8 \times 2^{14} \times 5^3 \times 7^2 \times 11}{5^4 \times 2^{16} \times 3^8 \times 7^2}\\
&=& \frac{11}{5 \times 2^2}\\
&=& \frac{11}{20}
\end{eqnarray*}
Finally...  :( That was a very inefficient solution.
[/hide]

Here is bit harder problem.

[url=http://artofproblemsolving.com/Forum/viewtopic.php?p=194622#194622]Number 7 on this page[/url]

This is the link to COP Shortlist-1. :) Like it says, number 7 is very similar.


Hi, I'm a freshman in high school right now, and I was wondering what kind of things I should do to get into U Penn for finance. Any help or pointers would be great. Thanks!

:D

Getting some sort of internship would probably be useful.  Demonstrating that you have been interested in and involved with finance for several years should help you tremendously.

You're a freshman - you should wait longer before making decisions, in my opinion.

If you're interested in money consider doing one or more of the following:
1. Get a job. Making money supports your claim that you are interested in learning about money. It will also give you some initial funding for the next suggestions, as well as strengthen your application in general.
2. Start a club/organization that involves funding. Play in integral role in managing money for starting up and sustaining the club. Apart from fund raising, consider getting funding through grants and by other means (preferably not by asking members to pay large dues). Maintain an organized budget
3. Invest. This does not necessarily imply stocks.
4. Start your own little "business" -- some means of generating money. Appear professional and organized.
5. Get an internship somewhere that involves finance/accounting/budgeting/handling money.

Of course, you have to do this with sincerity, and not just because it will help your application. Not only will your motive be transparent, you will likely fail at what you're trying to do. So make sure you don't focus too much on something like this from so early on unless you are genuinely interested in finance and how money works on a theoretical/conceptual level (as a opposed to a practical/superficial desire for wealth).


find all the ternes (x, y, z) of positive integers solutions that:

1 +  4^x   + 4^ y = z^2

[quote="diegu"]find all the ternes (x, y, z) of positive integers solutions that:

$ 1 \plus{} 4^x \plus{} 4^ y \equal{} z^2$ [/quote]

WLOG, $ x \le y$,
then the equation $ \Leftrightarrow (\frac {z \plus{} 1}{2})(\frac {z \minus{} 1}{2}) \equal{} 4^{x \minus{} 1}(4^{y \minus{} x} \plus{} 1)$, so one of $ \frac {z \plus{} 1}{2},\frac {z \minus{} 1}{2}$ is divisible by $ 4^{x \minus{} 1}$ (i.e. $ \ge 4^{x \minus{} 1})$, and another one $ \le 4^{y \minus{} x} \plus{} 1$

Now if $ y \minus{} x \equal{} x \minus{} 1$ , we have $ (x,y,z) \equal{} (x,2x \minus{} 1,2^{2x \minus{} 1} \plus{} 1)$ as solutions.

if $ y \minus{} x < x \minus{} 1$, then $ \frac {z \minus{} 1}{2} \plus{} 1 \le (4^{y \minus{} x} \plus{} 1) \plus{} 1 < 4^{x \minus{} 1} \le \frac {z \plus{} 1}{2}$, contradiction.

if $ y \minus{} x \ge x$, i.e.$ y \ge 2x$, then

$ 4^y \plus{} 4^x \plus{} 1 \equal{} (2^y)^2 \plus{} 2^{2x} \plus{} 1 \Rightarrow (2^y)^2 < (2^y)^2 \plus{} 2^{2x} \plus{} 1 < (2^y)^2 \plus{} 2(2^y) \plus{} 1 \equal{} (2^y \plus{} 1)^2$,
suggests that $ 4^y \plus{} 4^x \plus{} 1$ cannot be a square, contradiction.

So the only solutions are $ (x,y,z)\equal{}(k,2k\minus{}1,2^{2k\minus{}1}\plus{}1),(2k\minus{}1,k,2^{2k\minus{}1}\plus{}1)$, where $ k$ is any positive integer.

I don't understand why you write if y-x= x-1  and y-x<x-1 and y-x>x


I'm stuck on this problem for a week
Please help. My algorithm is too slow. basically i take the smallest number that has not been multiplied by the prime numbers and multiply them. insert them back to the linked list, then do the same thing again

and my program fails in the 4th testcase when it needs to find the 1000th number

please help. Thanks!

Well this problem is a bit tricky!

Surely the first smallest number is the smallest given prime.

Now keep a stack which will increase and keep that smallest answer first. and keep different pointer of that stack for each prime.

then take the smallest possible next one and update the stack.

for example:
2 3 5 7

in this case:
stack: 2
pointers: 1 1 1 1
Next number: min(2*2,3*2,5*2,7*2)=4
We have taken it from 2 so update its pointer.
stack: 2 4
pointers: 2 1 1 1
Next number: min(2*4,3*2,5*2,7*2)=6

Thus go on..


ABCD is a parallelogram and P is a point inside triangle BAD. If the area of triangle PAB is 2 and the area of triangle PCB is 5, what is the area of triangle PBD?

oo nice problem

[hide]

letting $ [ \cdots ]$ denote area, we see that $ [PBD] = \frac{1}{2}[ABCD]-[PAB]-[PAD]$

Then, drawing the height of the parallelogram, $ h$, through point $ P$, we label the height from $ P$ to $ BC$ as $ h_{1}$ and $ P$ to $ AD$ as $ h_{2}$ and hence $ h = h_{1}+h_{2}$

Now, we know:

$ \frac{1}{2}\cdot h_{1}\cdot BC = 5 \implies h_{1}\cdot BC = 10$

Then:

$ h \cdot BC = (h_{1}+h_{2}) \cdot BC = 10+h_{2}\cdot BC = 10+2[PAD] = [ABCD]$

Thus:

$ [PAD] = \frac{1}{2}[ABCD]-5$

Substituting this back into our original equation gives $ [PBD] = \fbox{3}$

[/hide]


This is one of those cases where I hate see induction invoked. The other (as simple if not simpler proof sheds more light on the reason). Apply AM/GM to all factors to get a lesser value of $ 2^n\sqrt{x_1x_2\cdots x_n} \equal{} 2^n$.

Sorry, the questions were sent to me as induction questions and I was in a great rush and didn't read them carefully. I wanted the person who sent them to get answers but I had to go out.

OK, let's see. For $ n\equal{}1$ we have $ x_1 \equal{} 1$, therefore $ 1\plus{}x_1 \equal{} 2$. Assume true up to $ n$. For $ n\plus{}1$, group $ x_n$ and $ x_{n\plus{}1}$ into one, so we can apply the induction hypothesis:

$ \prod_{k\equal{}1}^{n\plus{}1} (1\plus{}x_k) \equal{} \left (\prod_{k\equal{}1}^{n\minus{}1} (1\plus{}x_k) \right )(1 \plus{} x_nx_{n\plus{}1})\cdot \frac {(1\plus{}x_n)(1\plus{}x_{n\plus{}1})} {1 \plus{} x_nx_{n\plus{}1}} \geq 2^n \frac {(1\plus{}x_n)(1\plus{}x_{n\plus{}1})} {1 \plus{} x_nx_{n\plus{}1}}$.

Unfortunately,  $ \frac {(1\plus{}x_n)(1\plus{}x_{n\plus{}1})} {1 \plus{} x_nx_{n\plus{}1}} \geq 2$ is not guaranteed. So we need an idea. Well, since $ x_1x_2\cdots x_{n\plus{}1} \equal{} 1$, there must be some $ x_i \leq 1$ and some $ x_j \geq 1$. Redo all of the above with $ x_i,x_j$ instead of $ x_n,x_{n\plus{}1}$. We reach that last desirable inequality, equivalent to $ (1\minus{}x_i)(1\minus{}x_j) \leq 0$, which now is clearly true!

But why the hassle ...


symmetry function induction vector geometry parallelogram inequalities

a) find the necessary and sufficient geometric conditions on a subset of $\mathbb{R}^2 $ that it be unit ball for some norm.
b)Given necessary and sufficient geometric conditions that a subset be unit ball for a norm  arising from an inner product.
c)Generalize to $\mathbb{R}^m$.

your problem has certainly interested me. The choice of norm is not imp. & thats the most notable feature of this problem..
I will giv it a try....

Hello,

[quote="Myth"]I invented a) for my students :). We need two conditions: central symmetry + convexity[/quote]

You're right Myth, at least for $\mathbb{R}^n$ it is totally correct. Just a hint, think about the boundary in two dimension, and try to define the norm there. After that, the generalization to several dimensions it quite easy.

Best,

Yes, for (a) we can take the origin to be the center of the set $\mathcal S$, $O$, we can define the norm to be $1$ on the boundary, and $|a|$ for a point $Y$ situated on the ray $(OX$, with $X$ on the boundary of $\mathcal S$, s.t. $\overrightarrow{OY}=a\cdot\overrightarrow{OX}$. It’s not hard to check that if $\mathcal S$ is, as mentioned by Myth, centrally symmetric wrt $O$ and convex, the function we have defined is indeed a norm.

For higher dimensions we can use induction on the number of dimensions by looking at sections of smaller dimension through the set.

The interesting part is (b) :). I believe the answer is that the set must be an ellipsoid.

Take an ellipsoid $\mathcal E$ centered at $O$ which contains the set and start shrinking it until it’s no longer posible. There must be two points $A,B$ common to the boundaries of $\mathcal S,\mathcal E$ s.t. $O,A,B$ are not collinear (otherwise we could shrink $\mathcal E$ some more).

For a vector $x$, define $\|x\|_{\mathcal S}$ to be the norm defined in the first part corresponding to $\mathcal S$, and $\|x\|_{\mathcal E}$ the norm defined in the same way, but corresponding to $\mathcal E$.

For all vectors $v$ we have $\|v\|_{\mathcal S}\ge\|v\|_{\mathcal E}$, with equality iff the ray starting at $O$ and passing through $v$ (we can also regard vectors as points in an obvious way) cuts $\mathcal E$ and $\mathcal S$ in the same point.

Let $x=\overrightarrow{OA},y=\overrightarrow{OB}$. We now use the parallelogram identity (a necessary and sufficient condition for a norm to arise from an inner product) $\|u+v\|^2+\|u-v\|^2=2(\|u\|^2+\|v\|^2)$ for $u=tx,v=(1-t)y$ for all $t\in[0,1]$, both for $\|\cdot\|_{\mathcal S}$ and $\|\cdot\|_{\mathcal E}$, the inequality stated above, and the fact that for all $\alpha\ge 0$ we have $\|\alpha x\|_{\mathcal S}=\|\alpha x\|_{\mathcal E}$ (and the same for $y$) to conclude that the ray passing through $O$ and $tx+(1-t)y$ cuts both $\mathcal S$ and $\mathcal E$ in the same point. We do the same for the pairs $(-x,y),\ (x,-y),\ (-x,-y)$ instead of $(x,y)$ to reach the conclusion that $\mathcal S$ and $\mathcal E$ must coincide, so $\mathcal S$ must be an ellipsoid.

Conversely, it’s clear that every ellipsoid can be regarded as the unit ball for a norm arising from an inner product.

This is easy to extend to higher dimensions, I think, and the answer is the same: the sets we’re looking for are the ellipsoids.


For $a,b,c>0$ prove that

$\root 3\of{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}} \geq \frac{9}{(a+b+c)^2+2}$

By Cauchy-Swarts,we have:
Left side>=[9/(ab+bc+ca)]^(1/3).
But: ab+bc+ca<=[(a+b+c)^2]/3.
=>Left side>=[27/(a+b+c)^2]^(1/3).
put:x=[(a+b+c)^2]^(1/3)
=>x>0.
left side>=3/x.And:Right side=9/(x^3+2)
We shall prove:
                             3/x>=9/(x^3+2);
<===> x^3+2>=3x.
This can be proved by Cauchy iqe.
===>dpcm.
the "=" happens when x=1 and a=b=c.
=>a=b=c=1/3.


[hide]$x^{2}+x+1 = \frac{1}{x}$
$x^{3}+x^{2}+x-1=0$
$\frac{x^{4}-1}{x-1}=2$
$x^{4}-2x+1=0$

Let the root be $\alpha$. So, that means, the intersection point of the two graphs is $(\alpha,\frac{1}{\alpha})$. The distance from this point to the origin is:

$\sqrt{\alpha^{2}+\frac{1}{\alpha^{2}}}=$
$\sqrt{\frac{\alpha^{4}+1}{\alpha^{2}}}$

Since $\alpha^{2}-2\alpha+1=0$,

$\sqrt{\frac{\alpha^{4}+1}{\alpha^{2}}}= \sqrt{2\alpha}{\alpha^{2}}= \sqrt{2}{\alpha}$

So if we can prove that $\alpha > 0.5$, we have our proof to be true.

Taking the derivative (sorry) of $x^{4}-2x+1$ with respect to $x$, we have $3x^{3}-2$. We set this to $0$ to find the local maxima/minima. But since the second derivative is always greater than or equal to zero, this graph has no relative maxima. Thus, we have the minima to be at $x = \sqrt[3]{\frac{1}{2}}$. Substituting back into $x^{4}-2x+1$, we get the value to be some negative number. Now, we substitute $x=0.5$ into the equation to get some very small positive number. Thus, we know that there must be some value between $[0.5, \sqrt[3]{1/2}$ that there is a zero. Thus, $\alpha$ is more than 0.5. Q.E.D.[/hide]

[quote="delta"]Let $M$ be the intersection point of the grafics of $f(x)=x^{2}+x+1$ and $g(x)=\frac1x$. Prove that the distance from $M$ to the origin $O(0,0)$ is less than 2[/quote]
Here is a non-calculus based solution though I guess using calculus would be a smarter choice because the problem appeared in a university entrance exam.

Let $M\equiv (h,\frac{1}{h}).$ Then we have $1+h+h^{2}= \frac{1}{h}.$

[b]Claim:[/b] $\frac{10}{19}< h < \frac{3}{5}.$

$1^{\circ}:$ If $h \ge \frac{3}{5},$ then $\frac{5}{3}\ge \frac{1}{h}= 1+h+h^{2}\ge 1+\frac{6}{10}+\frac{36}{100}= 1.96,$ a contradiction.

$2^{\circ}:$ If $h \le \frac{10}{19},$ then $1.9 \le \frac{1}{h}= 1+h+h^{2}\le 1+\frac{10}{19}+\frac{100}{361},$ a contradiction (this can be verified but I am not doing it here.)

Thus, our claim is true.

$\therefore{|OM|}^{2}= h^{2}+\frac{1}{h^{2}}< \frac{36}{100}+\frac{361}{100}= \frac{397}{100}< 4.$
$\Rightarrow |OM| < 2.$

Yeah, I guess so, but I think that a non-calculus based solution might be plausibly better due to the fact that it illustrates creativity  :wink:

[quote="delta"]Here is a nice solution found by a coleague of mine [b]Wl. Popov[/b][hide]Observe that $f(x)+g(x)=x^{2}+f(x)g(x)$ and let $M(u,v)$ be an intersection point. Then $2v=u^{2}+v^{2}$ or $u^{2}+(v-1)^{2}=1$ which is a circle with center $C(0,1)$ and radius 1. Obviously all the points from this circle except $A(0,2)$ are at a distance les that 2 from the origin. :wink: [/hide][/quote]
This is a very creative solution and I must congratulate your colleague for coming up with it  :)  There is only one small clarification I wish to seek, but before that I want to ask, how did he come up with this solution at all?

I mean on the surface it is quite non-intuitive. For example, the solution I presented above is quite straight forward.  The distance of $M$ from $O$ is $x^{2}+\frac{1}{x^{2}}$ and from the relation $1+x+x^{2}= \frac{1}{x}$, it clear that I need to and can find some strict upper and lower bounds on $x$, and these bounds can be found after some rough work. But, what ideas were running in your colleague's head when he came up with that solution?

Lastly, the clarification I want to seek is notwithstanding the excellent solution, don't we still have to prove that $M(u,v)$ cannot be the point $A(0,2)$. Else, our solution would remain incomplete.

OKay, I now see why $M(u,v)$ cannot be the point $A(0,2).$ Thanks for your solution.

Here's a relatively simple and very straightforward one.

Let the intersection point be $(t,{1\over t})$. Since $(\forall x\in\mathbb{R})x^{2}+x+1>0$, the intersection point must line in the first quadrant, hence $t>0$

The distance from the origin is given by $d=\sqrt{t^{2}+{1\over t^{2}}}$. From $t^{2}+t+1={1\over t}$ we get $t^{3}+t^{2}+t=1\implies t^{3}+t^{2}+t+1=2\implies t^{4}-1=2(t-1)\implies t^{4}=2t-1$

(Note: $(1,1)$ obviously is not the intersection point, hence $t\neq 1$ and multiplication by $t-1$ is valid.)

Now $t^{2}+{1\over t^{2}}-4=\frac{t^{4}-4t^{2}+1}{t^{2}}=\frac{2t-1-4t^{2}+1}{t^{2}}=-\frac{2(2t-1)}{t}$

Assume $t<{1\over 2}$. Then $f(t)=t^{2}+t+1<{1\over 4}+{1\over 2}+1={7\over 4}<2$ and $g(t)>2$, hence there can be no intersection. Consequently, $t>{1\over 2}$ which gives $-\frac{2(2t-1)}{t}<0\implies t^{2}+{1\over t^{2}}<4\implies d<2$

A solution similar to Farenhajt's

$f(x)>0, \ \forall x\in \mathbb{R}$, but $g(x)>0$ only in $(0,+\infty)$. So if $f(x)=g(x)$ then $x>0$

Also, in $(0,+\infty), \ f$ is strictly increasing and $g$ is strictly decreasing. So $f'-g'>0 \Rightarrow (f-g)$ is strictly increasing.
So if there is an intersection point then it is unique

$x^{2}+x+1=\frac{1}{x}\Rightarrow$

$\boxed{x^{3}+x^{2}+x-1=0}\ \ (1) \Rightarrow$
$\cdots \Rightarrow$
$\boxed{x^{4}-2x+1=0}\ (2)$

The function $h(x) = x^{3}+x^{2}+x-1$ is continuous with $h\left(\frac{1}{2}\right)<0<h(1)$. So there is a $x_{o}\in \left(\frac{1}{2},1\right)$ such that $h(x_{o})=0$

$x_{o}$ satisfies $(1)$, so it also satisfies $(2)$, that is $x_{o}^{4}-2x_{o}+1=0$

$x_{o}^{2}+y_{o}^{2}<4 \Leftrightarrow x_{o}^{2}+\frac{1}{x_{o}^{2}}<4 \Leftrightarrow x_{o}^{4}-4x_{o}^{2}+1<0$

But

$x_{o}^{4}-4x_{o}^{2}+1 =$

$(x_{o}^{4}-2x_{o}+1)+(2x_{o}-4x_{o}^{2}) =$

$0+2x_{o}(1-2x_{o}) <0$ because $x_{o}\in \left(\frac{1}{2},1\right)$


err... i have never really posted in this section.. so i hope this problem is appropriate for this part of the forum:

Let A = 444...44 (2n digits) and B = 888...88 (n digits).  Show that A + 2B + 4 is a perfect square.


vector
abstract algebra
geometry
rectangle
inequalities unsolved
inequalities

I think this is the right section for this problem

[quote="jokeccil"]If $a,b,c,d \in R$ satisfy $(a^{2}+b^{2}-1)(c^{2}+d^{2}-1)>(ac+bd-1)^{2}$

 Show that $a^{2}+b^{2}>1$ and $c^{2}+d^{2}>1$.
[/quote]

If you make a little computation you find

$ (ad-bc)^2 \geq (a-c)^2 +(b-d)^2 $ so if you consider the plane vectors

$ \vec{A}= (a,b) $ and $\vec{B} =(c,d) $ it can be reformulated as follow

If $2S > T$ then module of $ \vec{A}$ and $\vec{B}$ is greater than 1.

($S$ is area of  triangle $OAB$ and $T$ is area of rectangle of sides $AB$ and unit lenght)


function ARML email HMMT probability geometry

When we created the first local community it was for the local math circle that Richard, Dave and I run here in San Diego.  We kept the group hidden because there was no need for the whole site to see and it might make some of the students shy to post.

As we added a few other communities we followed this lead and kept them hidden.

Now, what do any of you think about unhiding the local communities.  Will it help them grow faster and flourish more or will it take away from the community feel and make students shy to post who shy away from the main forums for that reason?

Would unhiding the communities make any difference?  Will many people really take the time to look through all the different communities other than their own?

Well, I personally don't think there is too much private about math. I would say they all should be read-only until you join, so that people not as advanced would still not be discouraged and have a chance.

Actually my idea is not to make them hidden, but rather un-accesible if you are not part of the group. That is you can actually see the community exists, however you can't read or post anything, untill you join it. This is because many people don't really know about this, and if the forums are hidden, you can't really find out about them :).

[quote="Valentin Vornicu"]Actually my idea is not to make them hidden, but rather un-accesible if you are not part of the group.[/quote]

Even though I am a fairly active member, I didn't know local communities existed until about mid-December, when there was a post about them.  I now understand these have existed for quite some time.  So Valentin's idea seems pretty reasonable.  I would suggest that if a non-member attempts to access such a forum, you might allow access only to a special announcement, associated with the forum, giving the purpose of the forum, the qualifications for joining the group (if any), and the person to PM to be added to the group.

As a programmer, I understand it may be a major change to the way the board works to allow certain posts (such as the special announcement) to be viewable by the world, while maintaining a cloak over the rest of the forum.

As far as allowing local communities to be viewable by everyone, I have mixed feelings.  Here are some thoughts off the top of my head...  As a new local moderator, I appreciate being able to see how some existing groups have been handled, and the kinds of discussions that take place on local groups.

On the other hand, I think shy members, and members whose parents have safety concerns, will be more likely to post, if they know their post is visible to a only few dozen users.

Another advantage of having the posts cloaked to non-members:  Some of the same problem sets have been seeded by MCrawford into multiple local groups.  Each group can tackle the problems sets independently and at their own pace.  [Contrast this with the open forums, where many (most) students ignore even the best challenge problems once a solution is posted.]

One more *possible* disadvantage to uncloaked messages:  The public schools in my community have strict rules when students are accessing the Internet from school.  One of the rules prohibits posting messages on "bulletin boards".  [I presume other districts have similar rules.]  Some school districts might be inclined to grant an exemption to local communities that had limited viewership.

Yes, I think we can make an annoucement in the us-local category (where state forums will be re-located) that can be viewed. If not, we'll find other solutions.

However there is no need to PM anyone. If you want to register into an open group, the moderator of the group must first approve your entry. If the moderator already has reasons to approve he will do so. If not, of course, he can ask the user for more informations.

I'd rather keep the status quo, and keep my local forum hidden. If there were some way to publicize an announcement, I would slightly rewrite the announcement that is now at the top of the local forum and go along with it, but even that isn't something I'm asking for.

I still view my primary function here as the coach and organizer of the Southern California ARML team. The team has existed, with me as coach, since the fall of 2001. I only made contact with Richard last year, and he only set the local forum up for me last summer, after 2004 ARML. I saw it as another means (besides the email group I already have) of communicating with the team.

Yes, since then I've let myself be sucked into involvment with AoPS/ML outside of the local forum, but it's still why I'm here.

Even within the local forum: a significant number of the students who attend ARML practices have never registered with AoPS, so I must still maintain my other forms of communication. Furthermore, of the 50 or 60 who have registered, only about 5 post with any regularity, even on the private forum, and most have never posted at all.

One thing that everyone understands: if they post on the local forum, I (as coach) will read it and very likely respond to it. Even that probably inhibits some people.

One extra privacy issue: as coach, I do know the true identities behind nearly all of the screen names within the closed group. I see that as another reason to keep it closed.

It seems that there are some language confusions here. What does hidden mean? It means that nobody knows about it, except when people are contacted by the moderator.

What I propose: one regular user (maybe not guests, will see about that), sees the forums actually exist (ie appears on the /index.php page). However he cannot even see the names of the topics inside it.

You can still maintain a hidden usergroup, if you don't want other people to find out which users are populating those forums.

Or, for special cases (like yours) in which coaches want hidden forums, we might be able to provide that for them. But this is different from what we are trying to do with the local / national communities.

We all agree, in some form or another, that many students are too shy to even consider posting on the main forums (some think the others are too smart in respect to them, others are just afraid of other's reactions, others are just lazzy, etc.). But maybe creating local communities, would help overpass this "shiness" kids are experiencing.

But for that they must know that these forums exist! I didn't even knew that they exist for a long time, and I am an admin! :)

Chiming in with my own two cents here.

Im the moderator of the Florida group, and the way I see it, I think that it would be a good idea to make the groups visible so that people would know that they exist. I am also (to a degree) in the same boat as Ken, except the reverse situation. The florida group was first set up as a way to reach the Florida Community and bring them to here rather than having so much focus on Mu Alpha Theta. Lately, though, I have organized a team to send to HMMT, and that is going to be the focus of the group for the next couple of months. So even though that is somewhat exclusive (being that there are only 8 going), I still welcome new members to the group. Making the group visible would help that.

O, and people that arent in the group do find out (how I do not know). Im not referring to people who are in my state, but actually to people who are out of it. (MysticTerminator asked me this morning if he could join.)

[quote="jelyman"]
O, and people that arent in the group do find out (how I do not know). Im not referring to people who are in my state, but actually to people who are out of it. (MysticTerminator asked me this morning if he could join.)[/quote]

Mr. Crawford has a list of all the communities and their moderators [url=http://artofproblemsolving.com/Forum/viewtopic.php?t=20872]here[/url].   So that's how some people find out but the problem is that there are a lot of people who don't frequent the Other Problem Solving Topics forum and thus would only find out about this if someone told them about it.  Letting people see that there is a community, but only allowing those in the Usergroup to see the contents and post, would be, in my opinion, very helpful for these Local Communities.  Right now the Carolinas community is very inactive.  Hopefully once school starts back up and when the math competitions start back up more people will be informed about AoPS and these communities.

[quote="joml88"]Mr. Crawford has a list of all the communities and their moderators [url=http://artofproblemsolving.com/Forum/viewtopic.php?t=20872]here[/url].   So that's how some people find out but the problem is that there are a lot of people who don't frequent the Other Problem Solving Topics forum and thus would only find out about this if someone told them about it. [/quote]That's exactly the point the new coming National and Local Communities category. Everyone will have the chance to find out that their community exists. Using a simple probability check and along with the pigeon-hole principle, we can definitely assume that most of the people do not know that these exist :)

In a way, I think it might be nicer if the word of the local forums was spread through word of mouth within the local community.  But then I live in a pretty small state (Vermont). ;-)

However, I do see the point that it would be nice if established AoPS community members could easily see that there is a forum for their local area, and be able to easily get information about how to join it.

So, I'm happy to go either way, either keeping the forum completely hidden or making the existence of the forum visible but the posts invisible until a person joins.

[quote="Valentin Vornicu"] Everyone will have the chance to find out that their community exists. Using a simple probability check and along with the pigeon-hole principle, we can definitely assume that most of the people do not know that these exist :)[/quote]

I completely agree. I just discovered these local forums. I definitely think they must be visible. Otherwise they will not get any new members other than the respective moderators recruits.  :)

Well it's done. They are visible :) But just that ;)

[quote="captcha000"]is it possible for someone not in a group, someone that hasn't been added by the moderater, to post in that group?[/quote]What do you mean post in that group?! You cannot post in a group. You can only post in topics (which are located in forums).

sorry it was kind of worded badly. nevermind, i get it now.

[quote="mathmom"][quote="Valentin Vornicu"]What I propose: one regular user (maybe not guests, will see about that), sees the forums actually exist (ie appears on the /index.php page). However he cannot even see the names of the topics inside it.

You can still maintain a hidden usergroup, if you don't want other people to find out which users are populating those forums.[/quote]

Valentin,

I noticed that now that the existence of the local groups is visible, it is also possible to see who all the members of the group are, in the usergroups screen.

From what you wrote above, it sounds like there is a way to hide the list of users, without hiding the group.  Is this so?  If so, perhaps that should be done by default until the local group moderators decide, possibly in consultation with their members, if they want the membership of their group to become public.

(Also, is it possible for me to add a description to my Vermont group?  If that is something an admin needs to do, could someone please set it to: "A community for math and problem solving enthusiasts from Vermont".  Thanks!)[/quote]It's not possible at the time (nor I think it will be soon enough) to have the group visible but not it's members. It's either completely hidden (for good reasons - ie like arml team training as Kent has requested), or open (with the restraints above). :)

Only registred users have access to the member lists and they have access to all the members anyway, so hidding them but not the group would just mean  that people are hidding that they're from Vermont ... end to which I see no point :)

The main point of these forums is to make people (read kids) communicate more among themselves mathematical or / and non-mathematical issues. If they can't do that on the main site, they might be proned to do it in a more "local" (thus supposedly friendly) enviroment. Having to discover hidden stuff, etc. seems way complicated (it's almost impossible to figure out quickly how to join one of these forums and it's disappointing - kids don't have a lot of patience now-a-days :D).

[quote="Valentin Vornicu"]Only registred users have access to the member lists and they have access to all the members anyway, so hidding them but not the group would just mean  that people are hidding that they're from Vermont ... end to which I see no point :) [/quote]

While it is not a big deal, it is an additional level of privacy that some of the parents to whom I plan to recommend these forums may prefer for their children.

I have children as young as 9 or 10 to whom I plan to recommend these forums, and this will be a first foray into electronic bulletin boards for most of these families.  The more their children's privacy/identity/location can be protected, the more likely these parents are to allow their children to participate.

Again, while I think AoPS is one of the last places an internet predator would come looking for children to prey upon,  a wacko in Vermont would likely first turn to the Vermont forum to get a short list of kids who might be in the local area.

I don't want paranoia to interfere with the best interests of the board, but I do think it would be better if the list of subscribers to the local forums could be kept private.  People can post their location in their profile if they so choose.  If they choose not to, they should not have to avoid their local forum to keep that private.

As Mathew said a little paranoia never hurts, but a lot of might have opposite effects. Even if someone is activating in the Vermont usergroup that doesn't make him from Vermont. And even if ... so what?! I imagine Vermont has more than a couple of streets in a village right?! :)

I mean I state that I'm from Bucharest, Romania - I've got all sorts of personal info on the site, email in footer of each page, however it's really hard (if not impossible) to find out where exactly do I live in Bucharest. And if I'd be a maniaq this would be the last place on earth I'd end up looking for ... [whatever they are looking for].

In plus even hidden, you still have to let people in, that's the whole point of having a forum on the net. If you only have around neighboars which you can always invite in your backyard, then there's no point to it.  And if you let people in, they can easily mislead you as being from Vermont and you have the same situation.

I thus see absolutely no logical reason to hide a state forum. Exception is the case of a particular ARML team training, which *needs* to be hidden to not reveal to the other teams the training strategies :D

Don't think of me as someone trying to drive you crazy, think of me as giving some insight into the parents who end up not letting their kids sign up for forums such as this.

You are right that anyone, with some scrutiny but really no guarantees, could join the forum and then find out at least who is active there.  So you are right, that hiding the membership does not do much good (though it does do some -- as someone "suspicious" might not be admitted inot the forum).

To be clear, I'm not trying to demand anything here, I'm just exploring the issues, and exposing the group to some of the kind of thinking I hear from many parents just getting into internet-related issues with their pre-teens.

btw, I think the state of privacy in North America and Romania is quite different.  With a full name and state or province, it is pretty easy to get an actual address for a North American resident who has a listed phone number or other such information "on record".  (Especially if the name is not too common.)

What I am trying to say that hidding the usergroup members does not help with *maniaks* if they would ever take a step on our site. I can prove that to you simply by infiltrating without you ever knowing as a normal user in the Vermont group. It's not that hard to do, and once in you still get all the discussions and the members, so in the end we arrive at the same thing (but stop 99% of the *non maniaks* from joining, or at least making it a lot harder for them).

So it's either having a hidden forum that no-one knows about and you only invite kids / people you personally know and can verify (which kinna goes around what online communities are all about), or go with it as we propose we do (and make sure kids know not to divulge too personal information on any forums, not only the vermonts).

I'm not thinking of you as someone trying to get me crazy :), nor do we wish to use the Admin privilegies to create / hide / unhide forums without consulting relevant users / people on the site before doing anything like that. That is why we are still discussing this :) ... everything we do is for the comfort and advantage of the users of these forums and personally, besides the fact that it's not a build-in feature (requires addional coding), I personally don't think there are consistent reasons for hidding members in the usergroup (without hidding absolutely everything else too).

[quote="Valentin Vornicu"]What I am trying to say that hidding the usergroup members does not help with *maniaks* if they would ever take a step on our site. I can prove that to you simply by infiltrating without you ever knowing as a normal user in the Vermont group. It's not that hard to do, and once in you still get all the discussions and the members[/quote]

Yes, I've conceded that hiding the list of users does not gain much since one could easily join the group to see who is there.

The best approach is certainly to advise kids to be careful about what personal details they post, and to discuss this with their parents.

Thanks for all your hard work on this site!

[quote="mathmom"]Yes, I've conceded that hiding the list of users does not gain much since one could easily join the group to see who is there.
[/quote]
Finally :P
[quote="mathmom"]The best approach is certainly to advise kids to be careful about what personal details they post, and to discuss this with their parents.[/quote]We also strongly advise all users, especially minor ones (i.e. young kids) not to post ANY personal details. This should be a general rule, in both internet and real life (the 21th century "don't talk to strangers" :D).
[quote="mathmom"]Thanks for all your hard work on this site![/quote]You're welcome. We appreciate the kind words.


[quote="byrdgyrl"]What is the last digit in 9^352476598?[/quote]

The last digit in 9^1 = 9
9^2 = 1
9^3 = 9
9^4 = 1
in general, when the power is even the last digit is 1,
so 1.


Well, why don't you post your proofs for the first two so we can see how they might be expanded on.

well the first two are rather trivial.
Sqrt(a)+sqrt(b)=c
subtract sqrt a from both sides and square
b=c^2-2c*sqrt(a)+a
b-c^2-a=-2c*sqrt(a)
c^2+a-b=2c*sqrt(a)
dividing both sides by 2c, we see that the RHS is irrational while the LHS is rational
sqrt(a)+sqrt(b)+sqrt(c)=d
subtract sqrt a from both sides and squar
b+2sqrt(bc)+c=d^2-2d*sqrt(a)+a.
moving all square roots to one side and rationals to the other and dividing by two, we get:
sqrt(bc)+sqrt(ad 2 )=(d^2+a-b)/2, and we have already verified the sum of two roots cannot be an integer (and sum of two roots cannot be a rational directly follows from that because we can multiply by the rationals denominator)

[quote="3X.lich"][quote="Scrambled"]...sqrt(bc)+sqrt(ad 2 )=(d^2+a-b)/2...[/quote]
That's the only part that you got wrong, it should be $\sqrt{bc}+\sqrt{ad^2}=\frac12(d^2+a-b-c)$. The whole argument is still sound and good.

Anyone got a proof that $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}=e$ is impossible?[/quote]
apparently not

We'll let the pre-Olympiad people take a shot at it.

[quote="Ravi B"]Anyone have a proof or counterexample?  I thought about it for 30 minutes, but didn't get anywhere.  I could show that if the sum of 4 square roots of integers is rational, then the sum is in fact an integer.  But that doesn't seem to help.[/quote]

You have?  In that case, you could just use infinite descent, and the remainder of the proof becomes fairly trivial.

Hmm, well take a look at this link:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11610[/url].

There, zscool and I showed that the sum being rational implied it was an integer.  But I don't see how to answer Scrambled's original question, with infinite descent or any other technique.  Can you finish it off?  Thanks.

Eh, no.  I was just about to edit my post... :blush:

I just realized that it really isn't trivial.  However, it might still be possible to prove it with infinite descent-suppose that we have the smallest rational number that works, and prove that this implies a smaller rational number.

dts, just so you know, it can be tricky applying infinite descent to the rationals.  This is because the rationals are dense - we can create infinite decreasing sequences of positive rationals (obvious - take (1/2)^n) but we cannot do the same with the naturals because of Well-Ordering.

[quote="Scrambled"]help here...
sqrt(a)+sqrt(b)+sqrt(c)+sqrt(d) cannot equal e when all the numbers are integers and a isnt a perfect square. Is this statement true?
the proof for sqrt(a)+sqrt(b)+sqrt(c)=d is impossible when a isnt a perfect square and all are integers is pretty trivial.  As with sqrt(a)+sqrt(b)=c.  But unfortunately, ive been unable to prove the first statement, so im beginning to suspect it is untrue, seing the triviality of the other cases...
is it?[/quote]

If you want to prove the problem by induction or the method of infinite descent, you need to rephrase the problem into a more general form.

First, a definition: an integer is [i]square-free[/i] if it has no factors of the form $n^2$, where $n$ is an integer greater than one.

Prove that if
$r\sqrt{a}+s\sqrt{b}+t\sqrt{c}+u\sqrt{d}+v\sqrt{e}=0$
where r, s, t, u, and v are rational numbers and a, b, c, d, and e are distinct positive square-free integers, then r=s=t=u=v=0.

We need square-free integers to avoid solutions that dodge the real question, such as $1\sqrt{8}-2\sqrt{2}=0$.

To convert Scrambled's problem to this form, if any of his a, b, c, or d is not square-free, we factor out the maximal square and move it outside the radical; and we set the last radical in my problem to $\sqrt{1}$, so that the last term is rational.

For induction or  infinite descent, we want any number of terms, so we index the variables:

Prove that if
$\sum_{i=1}^nr_i\sqrt{a_i}=0$
where the $r_i$ are rational numbers and the $a_i$ are distinct square-free integers, then $r_i=0$ for all $i=1,2,\ldots,n$.

See if you can solve it from there before I give any hints. This result is well-known, but the proof is not well-known, so it is worth pursuing.

Erin Schram


Write each expression in the standard form a + bi.

(1) (3 - 4i) - (-3 - 4i)

(2) 3i(-3 + 4i)

[hide]
1. Just combine like terms. (3 - -3) +(-4i - -4i). The answer is [6].

2.Using the distributive property, you get -9i + 12(i^2). by definition i^2 = -1, so the answer is [-12 - 9i].[/hide]

You said i^2 = -1.

We also know that sqrt{-1} = i, right?

Some important things to know about $i$:
$i=\sqrt{-1}$
$i^{2}=-1$
$i^{4}=1$
And just for fun... $\sqrt[4]{i^{4}}=1$ or $-1$


If you are given four numbers (that are part of a bigger number - but that fact is not relevant to this question) and the restriction that not all can be zeros at the same time, then many possible four-digit numbers are there?

It's obvious from the fact that there is only one number that has a four zeros in it, and that's 0000, so by a counting principle, it's 10*10*10*10 - 1 = 9999.

What happens it you use the counting principle early on, so because three of the digits can be 0, you get 10*10*10*9  = 9000, which is clearly not 9999. A further contradiction can be reached by knowing that because the zero can be anywhere, you multiply 9000 by 4, getting 36000. But that's not the point.

Any help with either "contradiction" would be greatly appreciated.

Well since its a 4 digit number, there can be zeroes anywere excepts in the first position, the thousands position of the number.  So actually there would be 9000 possible combinations.

And for your second contradiction, simply changing the position of the zero wont double, triple, or quadruple the number of combinations since the 9000 that you calculated include every possible combination with a zero in each place.

The fact that it is part of the bigger number DOES matter because that allows the left most digit to be 0.

Look at all permutations of $ 4$ digits (I won't call them numbers because you cant start a number with $ 0$. Permutation= Different arrangements). By multiplying by $ 9$ early on, you didn't eliminate just one number, but $ 1000$ of them. The multiplication represents the number of different digits that can occupy that position ($ 10\times10\times10\times10 \minus{} 9\times10\times10\times10\equal{}1\times10\times10\times10$, which equals $ 1000$). Using the counting principle first eliminates values such as $ 0499$, $ 0322$, $ 0077$, $ 0987$, etc; any value that starts with a $ 0$. However, you're trying to only eliminate the $ 0000$, one term.

So, using the principle first is not the same as using it last. The reason? Order of operations.  :wink:
(It might be worth noting that $ 0$ is one term, so $ 0000$ has $ 1\times1\times1\times1 \equal{} 1\: \text{possible values}$, the counting principle again, although you might not see it.)

sorry, i'm not that quick to understand...
But why would it be leaving out values starting with 0s? The 10*10*10*9 represents the ten digits, which includes 0. But if you say that I am excluding any values that have a 0 at the end, I am not (I think), because you would just use the commutative field axiom to move it to somewhere else...

Right???

The numbers starting with 0 are 3 digit numbers. IE $ 0123\equal{}123$

There are indeed 9000 4-digit sequences of the form $ XABC$ in which $ X$ cannot be zero, but the other digits have no restriction. There are also some sequences that start with zero, but are not $ 0000$. You'd have to count them and add the result to your $ 9000$. May be you could work on that. It would be the same problem , but reduced to 3-digit sequences. You can get to the same answer down that longer road, but tread carefully.
Your problem is that, when you try to use the road less traveled, it's hard not to count some sequences twice. For example, when you consider the other 4 positions for your $ X$ (non-zero digit), you think that would mean multiplying by 4, but $ XABC$ can be the same number as $ AXBC$,  $ ABXC$, and $ ABCX$. In fact they could all be $ 1111$, or $ 5555$.

It's like this:

There are 10 different digits that can be in the leftmost position (thanks, happyme, for pointing that out)

For each of those, we can place any of 10 digits in the 2nd spot = 10*10 = 100 combinations so far

For each of those, we can place any of 10 digits in the 3rd spot = 10*10*10 = 1000 combinations so far

For each of those. we can place any of 10 digits in the 4th spot = 10*10*10*10 = 10000 combinations so far

But there is 1*1*1*1 = 1 way to get 0000 so we subtract 1 leaving 9999.

The important thing that you're missing is that this method has counted ALL possible combinations of numbers INCLUDING the different orders. So there is no moving a number around to get a new number. We've just counted all the combinations.

[quote="redsky"]But why would it be leaving out values starting with 0s? The 10*10*10*9 represents the ten digits, which includes 0. But if you say that I am excluding any values that have a 0 at the end, I am not (I think), because you would just use the commutative field axiom to move it to somewhere else... [/quote]

I think this is where you're confused.  Multiplying by 9 implies that there are [b]only nine[/b] possible digits that can go in that position. Which means [b]you're leaving out all of the numbers that have the 10th digit in that position.[/b]

but the nine digits would be 1,2,3,4,5,6,7,8,9; right redsky?

[quote="redsky"]I'm sorry, I still don't get it...
Can't you just commute the non-zero digit around???[/quote]
There are $ 1000$ 4-digit sequences of the form $ 1ABC$ (and $ 1000$ of $ 2ABCD$, and $ 1000$ of $ 3ABCD$, etc). There's a total of $ 9000$ that have a non-zero digit on the left. Among those sequences there are many of the $ 9000$ sequences that have a nonzero digit in the second place, and you do not want to count them more than once. You could say the same for the third and the fourth places. Similarly, shuffling the digits around, will lead you to count the same sequence multiple times. Be careful of what you add to that first $ 9000$ you counted.
Here's a way to count every sequence just once.[hide]There are $ 900$ 4-digit sequences that start with zero with a non-zero digit in the second place, including many that have a nonzero digit in the third and/or the fourth place.
There are another $ 90$ 4-digit sequences that start with $ 00$ with a non-zero digit in the third place, including $ 9$ that have a nonzero digit in the fourth place.
Add the $ 9000$ plus the $ 900$, plus the $ 90$ plus the $ 9$ and you arrived at the result through a less traveled, difficult (and inefficient) way. However, trying different ways to a solution and confirming that there were no holes in your reasoning is fun.[/hide]

so use the inclusion-exclusion principle...got it...
Thanks everyone!!!

Would anyone listen to me or happyme?

The leftmost digit CAN be a 0, because in the original problem, it said that the 4 digits were part of a bigger number. (It incorrectly said that this is not relevant, unless the 4 digits happen to be the 4 leftmost digits of the number.)

sorry for the confusion, everyone   :blush:
I thought it was not relevant, but I guess I was wrong...
The four digits are at the right of the big number
Thanks for correcting me  :roll:


function
inequalities
quadratics
algebra
algebra solved

find all function $f:(0,\infty)\rightarrow (0,\infty)$ st:$f(x+\sqrt{f(x)})\leq f(x)+\sqrt{f(x)}$

Are you sure this is the statement? I find it too difficult...

at last one problem I found that it is difficult for you :D .according to my material the statement is right.

NO, all problems that you post I find very difficult.  :D  :D

The original statement in Pi in the Sky (Sept 2003 issue, page 32, Problem 1)  is different :

Find all functions $f: (0,\infty)\rightarrow (0,\infty)$ such that

$f(x+\sqrt{x}) \leq x \leq f(x)+\sqrt{f(x)}$

for all $x \in (0,\infty)$

which is probably easier to solve !

Ha ha, this is trivial. But what samn posted is incredibly difficult and probably unsolvable.

Well, this lasy problem is indeed easy, so maybe sam-n was referring to something else.

We just write $x$ in terms of $t=x+\sqrt x$ for the first inequality and we get $f(t)\le \frac {1+2t-\sqrt{1+2t}}2,\ \forall t>0\ (*)$. We now solve the $2$'nd degree inequality $f(x)+\sqrt{f(x)}\ge x$ to find $f(x)\ge \frac {1+2x-\sqrt{1+2x}}2,\ \forall x>0\ (**)$. Now $(*),(**)\Rightarrow f(x)=\frac {1+2x-\sqrt{1+2x}}2,\ \forall x>0$.

oops!! I'm not guilty :blush: some part of this journal is translated in our monthly journal and it doesnt mentioned $\leq x\leq$ so it seems difficult to me( whne it seems difficult to Harazi what about me :D ) btw I promise that I kill the translator :D

Another way to see the problem :
Let $g(x) = x + \sqrt {x}$.
It is clear that $g$ is an increasing bijection from $(0;+ \infty)$ onto itself. The given conditions may be rewritten as $f(g(x)) \leq x \leq g(f(x))$.

Using the monotony of $g$, and the first inequality, it follows that $g(f(g(x))) \leq g(x)$. But from the second inequality for $g(x)$ instead of $x$, we have $g(x) \leq g(f(g(x)))$. Thus $g(x) = g(f(g(x)))$.
Since $g$ is bijective, it follows that $f(g(x)) = x = g(f(x))$. Thus $f$ is the inverse of $g$.
Since $g(x)=a$ is a quadratic equation in $x$, we easily deduce the form of $f(x)$ as Grobber's did, but all I wanted to say is that way generalize to other functional equations of the same form.
As an example, a recent Crux problem was to determine the function $f$ from $R$ onto $R$ such that $f(x^3 + x) \leq x \leq (f(x))^3 + f(x)$.
The above method works even we do not know how to solve a third degree equation (yes, yes, I know you all know solving it :D)

Pierre.


let $A$ be matrices have $range(A) =1$
   and $B$  be matrices such that : $trace(AB)=0$
   prove that :
  $ABA=0$
                 :blush:

Let,s think about matrix A as a linear operator.Then dim(Im(A))=1.
We can write matrix A as A=[q_1*a,q_2*a,...,q_n*a],where q_1,...,q_n are reals
and a=(a_1,a_2,...,a_n) vector-column.Then Im(A)=<a>.We are to prove that ABA=0.
It is sufficient to prove that AB(a)=0.Let's AB=C , C=[c_1,c_2,...,c_n].
Then it is easy to see , that vector-column c_i is a linear combination of vector-columns of matrix A.So, we obtain, that c_i have a form c_i=s_i*a.
Then AB(a)=C(a)=[s_1*a,s_2*a,...,s_n*a]*(a)=v=[v_1,v_2,...,v_n].
Because of tr(AB)=0 we have that s_1*a1+...+s_n*a_n=0.
v_i=s_1*a_i*a_1+s_2*a_i*a_2+...+s_n*a_i*a_n=a_i*tr(AB)=0.
So we have v=0.
 We obtain , that Im(ABA)=0.
 The problem is solved.

[quote="Slon"]Let,s think about matrix $A$as a linear operator.Then $dim(Im(A))=1$.
We can write matrix A as$A=[q_{1}*a,q_{2}*a,...,q_{n}*a]$,where q_1,...,q_n are reals
and$a=(a_{1},a_{2},...,a_{n})$ vector-column.Then $Im(A)=<a>$.We are to prove that$ABA=0.$
It is sufficient to prove that$AB(a)=0$.Let's$AB=C , C=[c_{1},c_{2},...,c_{n}]$.
Then it is easy to see , that vector-column$c_{i}$ is a linear combination of vector-columns of matrix $A$.So, we obtain, that$c_{i}$ have a form $c_{i}=s_{i}*a.$
Then $AB(a)=C(a)=[s_{1}*a,s_{2}*a,...,s_{n}*a]*(a)=v=[v_{1},v_{2},...,v_{n}].$
Because of $tr(AB)=0$ we have that$s_{1}*a1+...+s_{n}*a_{n}=0.$
$v_{i}=s_{1}*a_{i}*a_{1}+s_{2}*a_{i}*a_{2}+...+s_{n}*a_{i}*a_{n}=a_{i}*tr(AB)=0.$
So we have $v=0.$
 We obtain , that $Im(ABA)=0$.
 The problem is solved.[/quote]
 thanks , ur solution is good


floor function
combinatorics proposed
combinatorics

For $ S\equal{}\{ x_1,x_2,...,x_n \}$ a set of $ n$ real numbers, all at least $ 1$, we count the number of reals of the form: $ \displaystyle\sum_{i\equal{}1}^{n} \epsilon_ix_i, \epsilon_i \in \{0,1\}$ lying in an open interval $ I$ of length $ 1$. Find the maximum value of this count over all $ I$ and $ S$.

Cute: it's actually a famous problem in disguise.  For any $ S$ and any $ M \subseteq [n]$, define $ f(S, M) \equal{} \sum_{i \in M} x_i$.  So we want to know for fixed $ S$ what the largest number of these $ f(S, M)$ that can fall into a unit interval is.  Note that if $ M \subsetneq N$ then for any $ S$, $ f(S, N) \minus{} f(S, M) \geq 1$.  So if $ |f(S, X) \minus{} f(S, Y)| < 1$ then neither $ X$ nor $ Y$ may contain the other.  So, the size of a maximal (here meaning "of largest possible size") antichain of subsets of $ [n]$ is an upper bound on our answer.

It's well-known that one maximal antichain of subsets of $ [n]$ is formed by all subsets of size $ \lfloor n/2\rfloor$, and if we choose the $ x_i$ sufficiently close together then $ \{f(S, X) \mid X \subset [n], |X| \equal{} \lfloor n/2\rfloor\}$ can certainly be covered by an open interval of size 1 (or, in fact, arbitrarily small size).


A lot of mathematical types  in some posts are invisible.

What is the trouble?

Thank you very much for an answer.

It's not trouble - it's spoiler!!  :)  A lot of the time, questions are posted as challenge problems and a lot of people want to try it.  So, to give everyone a chance to try it without seeing the solution, the solution is put in what is called "spoiler," which makes the text blend with the background.  You can view the text by highlighting it.

For example, say someone asked a question like "for which integers is it true that |x-5| < 2?"  Since I don't want to spoil the problem for other people who may also want to solve it, I can put my answer in spoiler using the [ spoiler ] tag (no spaces between the brackets).  So I might type something like:

[spoiler]Since |x-5| < 2,
-2 < x-5 < 2
3 < x < 7
x = 4, 5, 6[/spoiler]

If you wanted to see how I did the problem, you simply have to highlight the blank space.

Thank you very much.

Now , that's is clear!!


symmetry
combinatorics unsolved
combinatorics

Two rooks on a chessboard are said to be attacking each other if they are placed in the same row or column of the board.
a. There are eight rooks on a chessboard, none of them attacks any other. Prove that there is an even number of rooks on black fields.
b. How many ways can eight mutually non-attacking rooks be placed on the $9.9$ chessboard so that all eight rooks are on squares of the same color

My solution to A:

Let's number the columns from 1 to 8 from left to right, and let's number lines from 1 to 8 from down to up. Then when we count the sum of number of the column and line of the unit square, it will be odd for white unit squares and even for black unit squares. Although, there is a rook in every column and in every line. So if we sum the numbers of columns and lines in which the rooks are standing, we'll get 72. It follows now, that there cannot be an odd number of rooks standing on white unit squares. So there is an even number of rooks standing on unit squares, and an even number of rooks standing on black squares.

[hide]b. Consider the columns 1,2,3,4,5,6,7,8,9 and the rows 1,2,3,4,5,6,7,8,9. By symmetry, we may assume that square is black if and only if the column and the row have the same parity. Placing eight mutually non-attacking rooks on $ 9\times 9$ is like pairing up the columns (1,2,3,4,5,6,7,8,9) with (1,2,3,4,5,6,7,8,9), but one pair is not used. If they are on the black squares, we pair up odds with odds, evens with evens. If they are on the white squares, we pair up odds with evens. The rest is just dull counting. :D [/hide]


Let $m,n >2$ be integers.Prove that there exist a sequance of different integers like $a_{0}=m,a_{1},a_{2},...,a_{k-1},a_{k}=n,a_{i}>2$ such that \[a_{i}+a_{i+1}|a_{i}a_{i+1}+1\] for $0 \leq i \leq k-1$ .

Umm... doesn't the sequence

m, 1, n

always work, since x+1 | x*1 + 1????

I'm assuming m != n, otherwise the "different" condition can never hold.

I think you're leaving out some detail of the problem... does this have to hold for all k?  Does each ai have to be > 2 or something?

[quote="Elyot"]
 Does each $a_{i}$ have to be bigger than two?[/quote]
Yes that was what i forgot to mention.sorry for that.
[quote="Elyot"]
Does this have to hold for all k?
[/quote]
No, just for some natural number,Of course bigger than one.
[quote="Elyot"]

I'm assuming $m != n$, otherwise the "different" condition.[/quote]
I dont quite understand this one.What do you mean ?

[hide="My Solution"]
Write $m \longrightarrow n$ and say $n$ is reachable by $m$  if there exist such a sequance ,satisfing the problem condition.
Clearly if $m \longrightarrow n$ then $n \longrightarrow m$ .

Now we show that if $m$ is odd then $m \longrightarrow m+2$
because \[m+m+2=2(m+1)|m(m+2)+1=(m+1)^{2}\] Which is true while $m$ is odd.
So for every two odd numbers like $a,b$ we have $a \rightarrow b$
Now if $m$ is even then we have $m \longrightarrow m^{2}-m-1$
because \[m+m^{2}-m+1=m^{2}-1|m(m^{2}-m-1)+1=(m-1)(m^{2}-1)\] So each even number is reachable by at least one odd number .

These arguments show that for natural numbers like $m,n$ We have $m \longrightarrow n$ .
Proof complete.[/hide]


calculus
integration
logarithms
trigonometry
function
geometry
calculus computations

hello, do you mean $ \int \frac{dx}{x^2\plus{}2x}?$
Sonnhard.

forget it...
i need help with this other integral:
integral(1/(x*(ln x)*sqrt((ln x)^2+2ln x+4))dx
thanks

hello, the answer to your first question is
$ \int\frac{dx}{x^2\plus{}2x}\equal{}\ln\sqrt{\left|\frac{x}{x\plus{}2}\right|}\plus{}C$.
Sonnhard.

Please use Latex. Is this your primitive, $ P \equal{} \int \frac{1}{x \ln x \sqrt{(\ln x)^2 \plus{} 2 \ln x \plus{} 4}}\,dx$?

yes this is the integral...
i'm sorry, but i dont know how to use latex....
anyway how do we solve this integral?

hello, use the substitution $ t\equal{}\ln(x)$.
Sonnhard.

[hide]Make $ \ln x = u$, $ x = e^u$, $ dx = e^u\, du$, and so

$ P = \int \frac {1}{e^u \cdot u \cdot \sqrt {(u + 1)^2 + 3}} \cdot e^u\,du = \int \frac {1}{u \cdot \sqrt {(u + 1)^2 + 3}}\,du$

$ = \int \frac {1}{(\sqrt {3} \tan t - 1) \cdot \sqrt {(\sqrt {3} \tan t)^2 + 3}} \cdot \sqrt {3} \sec^2t \,dt$

$ = \int \frac {\sec t}{\sqrt {3} \tan t - 1}\,dt = \int \frac {1}{\sqrt {3}\sin t - \cos t}\,dt$.

Now the substitution $ \tan \frac {t}{2} = y$ convert this into

$ P = \frac {2}{y^2 + 2 \sqrt {3}y - 1}\,dy = \int \frac{2}{(y + \sqrt{3})^2 - 4}\,dy$

$ = \int\frac{2}{(y + \sqrt{3} + 2)(y + \sqrt{3} - 2)}\,dy = \frac12 \int \left(\frac{1}{y + \sqrt{3} + 2} - \frac{1}{y + \sqrt{3} - 2}\right)\,dy$

$ = \frac12 \ln \left|\frac{y + \sqrt{3} + 2}{y + \sqrt{3} - 2} \right| + C$.

Now just substitute back.[/hide]

Carcul, first of all thank you very very much!!!
but when i substitute back i dont know what to do with the arctan function [because i dont controll this function well...]
what i get is:
[img]http://i36.tinypic.com/29xivba.jpg[/img]
my question how do i open that?
i tried to find what is tan(t/2) :
[img]http://i33.tinypic.com/20qlwuc.jpg[/img]
how do i proceed?
thanks

$ \tan \frac{t}{2} \equal{} \frac{\sqrt{1 \plus{} \tan^2t} \minus{} 1}{\tan t}$.

SO IS THIS CURRECT??
[img]http://i35.tinypic.com/2qdn2gw.jpg[/img]

hello,here is another approach. Setting $ t \equal{} \ln(x)$ we have $ dt \equal{} \frac {1}{x}dx$ and your integral is:
$ \int\frac {dt}{t\sqrt {t^2 \plus{} 2t \plus{} 4}}$. Now put $ t \equal{} \frac {1}{u}$ and we get:
$ dt \equal{} \minus{} \frac {1}{u^2}du$  and our integral will be
$ \minus{} \int\frac {\frac {1}{u^2}du}{\frac {1}{u}\sqrt {\frac {1}{u^2} \plus{} \frac {2}{u} \plus{} 4}} \equal{} \minus{} \int\frac {du}{\sqrt {4u^2 \plus{} 2u \plus{} 1}}$. Now we write
$ 4u^2 \plus{} 2u \plus{} 1 \equal{} 3\left(\left(\frac {2}{\sqrt {3}}(u \plus{} \frac {1}{2})\right)^2 \plus{} 1\right)$ and
our integral has the form
$ \minus{}\frac {1}{\sqrt {3}}\int \frac {du}{\sqrt {\left(\frac {2}{\sqrt {3}}(u \plus{} \frac {1}{4})\right)^2 \plus{} 1}}$.
With $ z \equal{} \frac {2}{\sqrt {3}}\left(u \plus{} \frac {1}{4}\right)$ we have $ du \equal{} \frac {\sqrt {3}}{2}dz$ and the integral is
$ \minus{}\frac {1}{2}\int\frac {dz}{\sqrt {z^2 \plus{} 1}} \equal{} \minus{} \frac {1}{2}arsinh(z) \plus{} C$.
Sonnhard.

hello, i have the same result as you.
Sonnhard.

i have a problem!
i want to find the area bewean the lines x=1/e x=1/e^2 y=0 and the function's graph but in x=1/e the integral's denominator  is zero!!!
why is that??


Two consecutive positive even numbers are each squared.  The
difference of the squares is 60.  What is the sum of the original
two numbers?

Let $ x$ be the smaller number, then we want $ 2x+2$.  We get
\begin{align*}
(x+2)^2-x^2=60 &\Rightarrow (x+2-x)(x+2+x)=60 \\
&\Rightarrow 2(2x+2)=60 \\
&\Rightarrow 2x+2=\boxed{30} \\
\end{align*}

The difference of the squares of 2 consecutive numbers A and B is A+B, so between 2 consecutive even integers being 60 means that A+2(A+1)+B=60, 3A+B=59, and A+2(B-1)+B=60, A+3B=61. Adding these two equations gives us 4A+4B=120. Divide by 4 gives you A+B=$ \boxed{30}$


I realy can't seem to get this question right!

At a price of $20 per 1000cups, 400 000 cups per week will be sold, but for every price rise of$1 per 1000 cups sales will drop by 2000 cups per week.

QUESTION -- Determine the number of cups to be manufactured per week, which will maximise the revenue for the company. Display your results with calculus with a full algebraic analysis

So I figured I needed a function which will determine the revenue from the sale of the cups per week .. as below

R = (20+X)*(400000-(2000*X))

Where R is revenue and X is dollars increase

And from there on I just have to do what exactly ...

Thanx
Telon

1) This question should be in the calculus tutorials forum, not the unsolved + proposed problems forum.

2) You now have a function $R$ in terms of a variable $x$.  Go read the question -- what does it want you to find?

[quote="Telon"]Its asking which number of cups need to be manufactured to maximise the revenue ... using algebra and what not[/quote]

Good.  So, you're trying to maximize $R$.  How do you think you should find where the maximum is?

So why don't you do that and carry on from there?

Oh, yea I think I got it now

Thanx a tonne  :blush:


I have been level 40. But now I am very tired of powerleveling.At first,I thought it is very

interseting and I am fresh in it.But later I found that it is a so hard and long work and it

took me so much time.I nearly want to give up my powerleveling.
But at that time I found a website some days ago.It says that it can give World of War

player help.It also provides powerleveling service,and promises safty and security to all

the accounts.And my character will be leveled up in a short time and the price is cheaper than any other i know .
I want my character to be level 60,so that I can enjoy the most wonderful WoW experience.

I'd like to choose this website to help me,but I don't know how is it.Please go and see the

website:[color=red]http://www.playerturbo.com[/color], and give me some advices. and u can contact the seller

via ICQ：108239029
AIM：gilssales
YAHOO：sellgilstoyou
MSN：gils_sales@hotmail.com   or   richplayer@hotmail.com
network  games： US WOW ; EURO WOW ;  FFXI ; DAoC ; EQ2 ; Guild Wars ; Eve online ; Matrix

and so on or Power Leveling .

Erm, yeah.

Please don't advertise.

MODS, this is an advertisement for some powerleveling thing, so it shouldn't be on AOPS so I would think it should be deleted.

Wow. You must have gotten lost of something. Let me inform you of your current whereabouts. This is The Art of Problem Solving Forum, a place where people discuss math mostly, and some other random stuff. I believe you are looking for a different location, because you seem to assume we have the background knowledge of what you are talking about as well as some advice to offer. It is to my knowledge that neither of these is true of the community as a whole and that perhaps you may have more success at another community, perhaps a gaming one. Have a nice trip. :P  :D

lol, maybe he interpreted "The Round Table" as meaning something to do with a game.

maybe the community title isn't clear enough. this is the art of problem solving. and apparently this guy has a problem.
and my advise, just play the game and level up normally. this "power-leveling" seems kind of shady.

lol i wonder how a post of this nature got here lol... he had a problem and wanted it solved i guess...  :P  but WoW is terribly addictive... lol

*blinks* uh.... huh. didnt get a word of that.


Let $ a,b,c,x,y,z$ be positive real numbers such that $ a\plus{}x\equal{}b\plus{}y\equal{}c\plus{}z$.

Prove that

$ \frac{(a\plus{}b\plus{}c)(x\plus{}y\plus{}z)}{3}\geq ax\plus{}by\plus{}cz$

[quote="Georg-A."]Let $ a,b,c,x,y,z$ be positive real numbers such that $ a \plus{} x \equal{} b \plus{} y \equal{} c \plus{} z$.

Prove that

$ \frac {(a \plus{} b \plus{} c)(x \plus{} y \plus{} z)}{3}\geq ax \plus{} by \plus{} cz$[/quote]
$ (a,b,c)$ is sorted oppositely of $ (x,y,z)$ and the conclusion follows by chebyshev :)


http://www.nbc11.com/news/10470178/detail.html

Uh-oh, I do hope that Mr. Kim is found alive. (If you don't know who James Kim is, he's a CNET editor.) I've really enjoyed his articles before.... :(

Yes, I heard about this.

It is unfortunate.

[quote="solafidefarms"][quote="moogra"][quote="solafidefarms"]:(  :noo: http://news.com.com/James+Kim+found+deceased/2100-1028_3-6141498.html?tag=cnetfd.ld1 They found him dead. He was such a nice fellow too...[/quote]

he WHAT? NOOOOO! my english calss was all talking about him and all he did... :([/quote]

Are you expressing regret that he is dead or disbelief that I thought he was a nice fellow?[/quote]
i think the emoticon expresses this person's sorrow :(

Hmmm, I thought this was only local news because he was from around here, but apparently not.  Strange, some couple was missing and was found like a week before.  Or something.  James Kim famous or something?

I heard about it. I was really hoping he was alive after the rest of his family though. Well, I guess not everything has a happy ending.  :(

[quote="Piggypi314"] James Kim famous or something?[/quote]

Being a CNET reporter person - their most active product reviewer - is not a low-profile job ;).

What is CNET? Isn't that the Satanic radio station in Kentucky?

[quote="bpms"]Is he Korean (James Kim)?[/quote]

That should probably say was....

[quote="asmallgoat"]hello treething i worship you[/quote]

I can not think of anybody worse.

dude, asmallgoat's avatar, sig (except for the bold) is the exact same...


Write IMO web-sites. What's a web-site of IMO2006?

Official website of IMO 2006:
http://imo2006.dmfa.si/
 :blush:

IMO 2007 website:
http://www.imo2007.edu.vn/

IMO 2008 website:
http://www.imo-2008.es/

IMO 2009 website:
http://www.imo2009.de/

Informations related to IMO: http://imo.math.ca/


What is MOEMS? im doing mathcounts but not sure what MOEMS are. where can i get infomation on it?

Generally, before posting, it's a good idea to read stickies and announcements. There's a reason why they're there, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=71119]right?[/url]

yay! now we have 400 topics in MOEMS! :D

[quote="cgyao15"]useless post= :spam: and :furious:[/quote]
technically I was giving news about MOEMS. Now we do have 400 topics :D

[quote="cgyao15"][quote]cgyao15 wrote:
useless post=  and

technically I was giving news about MOEMS. Now we do have 400 topics
[/quote]
Actually i was talking about mathlovers post after yours. :D[/quote]

oh okay, :D

[quote="qwertythecucumber"]By the way, this question has been asked before. It's a good idea to search before asking the question. :D[/quote]
lol, so true :D

AAAHHHH! we're being invaded by :D's!


function
limit
real analysis
real analysis theorems

Our teacher taudht us smth about the omega function and its properties which i understand easily, but then i find out a lot of questions...
the defenition was like that:

Any number $x\in [0;1]$ can be written as $x=0.x_1x_2\cdots$
denote  $y_n=0.x_1x_2\cdots x_n$ and $m_n$ the number of digits equal to the zero among $x_1....x_n$
then
$\omega (x)=\lim_{n\to\infty}{\frac{m_n}{n}}$ if this limit exists and $\omega (x)$ undefined if limit does not exist

he proved the following properties:

1. For any rational x this function is defined.
2. There are some irrational numbers such that at this points the function is not defined.
3. For every point y between the 0 and 1 there exist such number x(infinetly many) such that $\omega (x)=y$
4. $\forall a\in [0; 1] , t , \delta$ there exist $x: |x-t|<\delta \Rightarrow \omega(x)=a$

My teacher forget to tell us that in defenitio of omega function we assume that x is a number not in the form $0,x_1x_2...x_k(9)$
This function is interesting


This is a problem from Martin Gardner. I've altered it a little bit to focus on the part that I can't figure out.
Let's say we're playing roulette, and I offer you a bet. Yan can pick any triplet of black and red- say red/red/black or red/black/red. Then I'll pick a different one. At the starting point, we'll watch each spin of the roulette wheel until one of our triples appears as a run. If yours come first, you win. If mine comes first, I win. You always have the first choice.
Suppose I am a very intelligent man. When your choice is black/red/black, what is my chance of winning?

If each triple has the same chance of coming up, what other information could an omniscient player possible have in this problem, even knowing my triple? Wouldn't the game be fair, no matter what triple I pick?

You might think so, but there are dependencies between the triples.  Consider, for example, the game in which you choose BBBB and I choose RBBB -- your odds of winning on the first four rolls are equal to mine.  But your odds of winning using rolls two through five are actually 0: if you fail to win on the first four rolls, I always win, so actually I'm something like a 15-1 favorite in this game.

JBL is right.
But I can only see the very obvious ones like RRR and BBB.
What triplet should I choose if the other player chooses BBR or BRB?

I'm sure this has appeared before, so it might be worthwhile spending some time searching for previous discussions to see if someone (look especially for posts by Ravi B, I bet) has given a more thorough intro than what follows:

Choose two patterns, say BBR and BRB.  Draw a state diagram with six states: BB, BR, RB, RR, BBR and BRB.  Then draw arrows in the natural ways: an arrow labeled B from BR to BRB; an arrow labeled R from BR to RR; etc.  Then the four positions BB, BR, RB, and RR all have two arrows leaving them, while the positions BBR and BRB each have one arrow entering.  Define a function f by f(BBR) = 1 and f(BRB) = 0 and for any other state X with arrows from X to states Y and Z, set f(X) to be the average of f(Y) and f(Z).  This gives you some algebra problems to solve to compute f(BB), f(BR), f(RB) and f(RR).  The value f(X) is the probability that the player BBR wins once you're already in state X.  When you start, you're equally likely to end up in each of the two-letter states after two moves, so just take the average of the four values of f to get your answer.

Here are some past threads where similar ideas are mentioned.  (I found them via a search for "state diagram*".)
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=715838328&t=202582
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=715838328&t=198960
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=715838328&t=158355
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=715838328&t=133655

Oh, I think I know now.
I somehow managed to solve it with a simple diagram.
Thanks!


What is the least possible positive integer with exactly five distinct positive factors?

You have to know that the number of factors is taking the exponet of the prime factorization of the number and multiplying it and adding 1. So its 5-1=4
So its 2^4=16

Alternatively, note that only perfect squares have an odd amount of factors, so we can go through all the perfect squares. 1 obviously doesn't have five factors, nor does 4 or 9, but $ \boxed{16}$ does.

what i dont understand is that you need 5 [b]Distinct[/b] factors. wouldn't 2^4 be indistinct?


what's the correct answer for this question? It's asking what fraction of the grades are satisfactory. Should it be 4 out of 5? they aren't asking what fraction of the students. Am I right?

You just add up the satisfactory grades for the numerator which are.
$5$, $4$, $3$, and $3$ which equals $15$.

The denominator should be the total number of grades so you add up all of them.

$5 + 5 + 4 + 3 + 3$ is $20$

So the answer is $\frac {15}{20}$ which simplifies to $\frac {3} {4}$

The wording is a bit weird, but buzzer11's probably right (the fraction of grades [i]on the graph[/i] who got satisfactory grades, not the fraction of grades considered satisfactory).

ok, thanks. I thought they were asking the grades ratio (grade A, B ,C...) instead of the number of students ratio. LOL

Solution:

Let the number of satisfactory grades be $n$, and the total number of students $t$.  The problem is asking for the value of $$\frac{n}{t}$$  First we find $n=5+4+3+3 \rightarrow n=15$.  Then we find $t=5+4+3+3+5 \rightarrow t=20$.  Going back to our fraction we find that $$\frac{n}{t}=\frac{15}{20}$$
$$=\boxed{\frac{3}{4}}$$

5 bad grades, 20 grades in total so the answer is 1-5/20=3/4


Dear Mathlinkers.
      When does your winter vacation begin? How long does it lasts? I am interested in the informations from all over the world.

Thanks.

[quote="h_s_potter2002"]Ours began yesterday, Decmber 22, and we will go back to school on Janurary 3rd.  SO that's 12 days of vacation.[/quote]
Oh that's a really short vacation......but I guess you have a much longer summer vacation?

[quote="Dave Vather"]Mine began on november 29 and last until twenty-(something) of January[/quote]

   :huuh:  :o   :huuh:  :cursing:

  I'm assuming this ALMOST 2-MONTH WINTER VACATION is compensated by shorter summer/spring breaks...?

I think that we are very lucky. We Chinese don't have Christmas holiday, but have a 3-days break for the New Years day. Thus we will have a big party with our classmates and teachers at 12-31!  :D

[quote="dogseatcheese"]Mine began on Dec. 16 and starts on Jan. 9. I go to school in California.[/quote]

Same here!

My Christmas break began on December 7 and ends on January 3. :lol:

[quote="Dave Vather"]Mine began on november 29 and last until twenty-(something) of January[/quote]
What school do you go to?!?! :what?: Or did you mean December rather than January?

Our winter break is from 12/23 to 1/2. We resume school on 1/3, giving us only 11 days of relief.

I probably have it the shortest.  Last day was the 23rd, and we resume school on Monday, January 2nd.


function
integration
inequalities
real analysis
real analysis theorems

Suppose $f(x)$ is a convex function on $[a,b]$. Prove that $\forall x_1, x_2 \in [a,b], x1<x2$, it holds that

\[
f(\frac{x_1+x_2}{2}) \leq \frac{1}{x_2-x_1}\int_{x_1}^{x_2} f(t)dt \leq \frac{f(x_1)+f(x_2)}{2}.
\]

[quote="Alison"]Well, by weighted Jensen or majorization, $f(x_1) + f(x_2) \ge f(x_1 + a) + f(x_2 - a) \ge 2 f(\frac{x_1 + x_2}{2})$ for $0 \le a \le x_2 - x_1$.  Integrate over that range, divide through, and you're done.[/quote]

For a convex function, it is obvious that f(x1) + f(x2) >= 2f([x1+x2]/2)

It is also obvious that f(x1 + a) + f(x2 - a)  >= 2f([x1+x2]/2)

I can see that the majorization inequality implies f(x1) + f(x2) >= f(x1 + a) + f(x2 - a), however, I cannot figure out how Jensen's or weighted Jensen's implies this.

I came up with a proof on my own via considering:

WOLOG, a <= (x2-x1)/2

[f(x1) - f(x1 + a)]/[-a] = f'(b1), b1 E (x1, x1+a)
[f(x2 - a) - f(x2)]/[-a] = f'(b2), b2 E (x2-a, x2) via the MVT.

f'(b2) >= f'(b1), so the result follows.

Anyone want to tell me how weighted Jensen's implies the first part of the above inequality?

[quote="iura"]Weighted Jensen or Karamata:
If $x_1>x_2$ and $y_1>y_2$ and $x_1+x_2=y_1+y_2$ and $x_1>y_1$ then
$f(x_1)+f(x_2)>f(y_1)+f(y_2)$ because we can easily find $\alpha$ with
$y_1=\alpha x_1+(1-\alpha)x_2$ and $y_2=\alpha x_2+(1-\alpha)x_1$ now apply Jensen and sum[/quote]

Ah.  I knew that as majorization.