This is from a previous post of mine:
 [quote="cincodemayo5590"] [hide="Balls and Urns"]

[u]Balls and Urns[/u]

Say we draw a diagram depicting this situation, where each candy is a dot (indistinguishable from the others) and each person is an urn, which will be the space between two divider lines. The diagram looks like this:
\[| \cdot | \cdot | \cdot | \cdot | \cdot | \cdot | \cdot | \ \ \cdot \cdot \ \cdot \]
Those last three dots are the extra candies to be distributed. Now, we need to devise a system to count all of the possible places where we can add these three candies, while not counting anything we don't want, like one person not having any candies. The method I use is to make larger elements out of the balls and walls, so that no matter what I move, each person (urn) will have at least one candy. The elements are boxed:
\[\boxed{\boxed{ | \cdot }}\boxed{ | \cdot }\boxed{ | \cdot }\boxed{ | \cdot }\boxed{ | \cdot }\boxed{ | \cdot }\boxed{ | \cdot }\cdot \cdot \ \cdot \boxed{\boxed{|}}\]
Now, we cannot move the double-boxed elements on the left and right, since they determine the outermost urns. Note also that one of the two double-boxed elements needs one ball, so that the left-most or right-most urn always will have at least one ball, along with the others. The single-boxed elements and the extra balls can move around freely; the number of permutations of these is $\frac{ 9!}{6! 3!}= \dbinom{9}{6}$. Now, if do not place the restriction that all children must have candy, the diagram looks thus:
\[\boxed{ \boxed{|}}\cdot | \cdot | \cdot | \cdot | \cdot | \cdot | \cdot \cdot \cdot \cdot \boxed{ \boxed{ | }}\]
The number of permutations in this situation is $\frac{ 16!}{10! 6!}= \dbinom{16}{6}$, and since we want the probability, we divide... (etc.)
[/hide][/quote]

The non-negative integers case is simply treating all balls (number space, instead of candies as was the question this post was answering) and all walls as separate elements; the positive integers is the case where each urn must have at least one ball.

Note that the non-negative integers case is a bit more easy to deal with; and we can reduce your second equation to the first by making the substitution $z_{i}= y_{i}+1$.

[quote="123s"]Is there any way to explain why:
y_1, y_2, y_3, y_4 are non-negative integers
y_1+y_2+y_3+y_4=47
so there are 50C3 different ordered pairs of (y_1,y_2,y_3,y_4)

better than that we can say that this is equivalent to:
y_1, y_2, y_3, y_4 are positive integers
y_1+y_2+y_3+y_4=51
and then use balls and walls?[/quote]

For the first one, the formula for

$x_{1}+x_{2}+\cdots+x_{y}=z$

for non-negative integers is

$\binom{z-y+1}{y-1}$

I don't know how to prove it.

I forgot the second one...

Isn't the first one (non-negative) $\dbinom{z+y-1}{y-1}$, and the second one (positive) $\dbinom{z-1}{y-1}$?


Suppose I tell you that the number 577,934,211 is the product of two distinct primes; and also that phi(577,934,211) = 577,883,992. Find the two prime factors of 577,934,211.

In that case, try it with 577934213, which is the product of two primes. phi(577934213)=577322968.

Oops, I apologize for that. That was a pretty big typo. It should have been 577,932,211. With phi(577,932,211) = 577,883,992. Very good job on factoring that number though fairyofwind!!

Try either mine or Zeta's number.

Very good! This exact type of problem is closely related to the RSA Public Key encryption method.

They don't tell you what phi(n) is in that case though.

Yes, and this is exactly why. Because if you know phi(n) then you can factor n very easily.

Here's a further challenge. Now that you know the factorization of my number, can you write it as the difference of two squares?

[hide]If x and y are the prime factors, then the number = xy = (x/2 + y/2)2 - (x/2 - y/2)2[/hide]


How did scores run in the seattle and moses lake regions?
In Spokane region, our school got first with a high 50s low 60s score and the indies ran 26, 24, 23, 22, 21, 20 (calc division).

I believe the top two team scores in the 11/12 division were around 75 and 72, namely Garfield and Bellevue, respectively.

Well, I'm not surprised to see the scores run higher over there. That's how it usually goes.
Looks like little Pullman's got it's work cut out. We will be making some slight changes on our first team for masters (most likely for the better). I guess we'll see you guys in January.

I did not take note for 11th/12th (probably what you are looking for), but here is 9th / 10th top 7 teams (top 4 won trophy):

1. Lakeside School 47.75
2. Bellvue HS         44.50
3. Pullman HS       43.25
4. Interlake HS      39.00
5. Hanford HS         37.25
6. Prime Learning Center 36.50
7. Garfield              31.50

Here you go the top teams in 11th and 12th grade:

1. Belleve HS  66.75
2. Garfield      63.75
3. Pullman HS 50.75
4. Lewis & Clark HS 49.00
5. Redmond HS  42.50
6. Mt Rainier HS 36.50
7. Hanford HS    34.75
8. MOses Lake HS 34.25

Individual result in Calculus similar: Jeffrey Wang took first while Adam Hesterberg took 2nd.  Jonathan Hung Hung and Qiaochu Yuan from BHS took 3rd and 4th, reflecting an impressive talent pool at BHS.

The team scores were something like 66ish Bellvue, 63ish Garfield, 50.75 Pullman, 49.00 Lewis and Clark.
10th place individual was 28 points.

I just edited the 11th/12th posting with scores and added #8 spot.  The rest are below 30's.

[quote="minghu"]Individual result in Calculus similar: Jeffrey Wang took first while Adam Hesterberg took 2nd.  Jonathan Hung Hung and Qiaochu Yuan from BHS took 3rd and 4th, reflecting an impressive talent pool at BHS.[/quote]

Yay-yuh! :lol: If only you guys knew how close we were to not winning... on multiple occasions, the pressure, team, and college bowl rounds for example :wink:.

ahhh
i was wondering why i didnt see you or meir at the awards ceremony  :wink:  :(
that really sucks. the drive from pullman was perfect.

[quote="samath"]:( Our school wouldn't let us go because of the snow.
Even though the roads up at the pass were completely clear  :mad:[/quote]

Eh? Why did the 11/12 team get to go? Meir was there.

Sam was referring to RJH (9th/10th grade) team.
Redmond High School participated (11th/12th) was there (5th place).

o my
i missed him  :blush: now i feel kinda stupid
oh well
what school are you from, minghu

[quote="samath"]:( Our school wouldn't let us go because of the snow.
Even though the roads up at the pass were completely clear  :mad:[/quote]

same here... thats a shame but i guess we will see what happens at the northwest math championship

[quote="samath"]
The district withdrew their approval for the field trip.
So if we weren't allowed to go, he shouldn't have been allowed to go ...
There's probably going to be some trouble at the high school tomorrow when school starts again, since they went without permission.
That is, assuming somebody finds out ... :ninja:[/quote]

I don't remember the last time we had/cared about district approval at Bellevue...

yeah ...
We would get in a LOT of trouble if we tried it though  :(

Lake Washington School District is extremely risk adverse.  I can't tell you how many instances that I have been running into red tapes in getting after school activities going.  They tend to think everything in risk/litigation perspective.  For example, how many ways can you think of a student get injuried in school chess club, and in turn sue the school district for damage?  So it is not hard to imagine that when decision has to be made about your trip, it goes like this: well the road could get icy, which could lead to a car accident, a student might get hurt, and if they sue us, we are liable if we give them the permission.  Therefore let's proactively "mitigate" the risk by not allowing any opportunity for this to happen at all.

It would be interesting to find out if Redmond High team got themselves into trouble by attending MIC Master.

ha, just noticed this discussion.

"It would be interesting to find out if Redmond High team got themselves into trouble by attending MIC Master."

Assuming our high school knew we had a math team and knew that we attended competitions.. they might possibly be mad, but seeing as that is not the case, we didn't get in any trouble.

And frankly, i don't even recall telling anybody at the JH when i went to competitions.[/quote]

Don't you notice that all these people have quit AoPS like five years ago?

[quote=nihao4112]68 out of 98 1st place![/quote]

which division were u in

[quote=nihao4112]68 out of 98 1st place![/quote]
First of all that's not really something to brag about because in your grade I got 93. And secondly you really shouldn't brag about accomplishments in dead threads, go make a new thread.


For a particular peculiar pair of dice, the probabilities of rolling a 1,2,3,4,5, and 6 are in the ratio 1 : 2 : 3 :4 :5 : 6. What is the probability of rolling a total of 7 on the two dice?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=75349]This problem has been discussed in the AMC-Competition section of the forum[/url]

 :)

Weight each probability accordingly and match up the pairs.


what is the volume of a cube formed by center of the sides of a octahedron.
Edge of the octahedron = s

Assuming it's a regular octahedron, right?
[hide]In any triangle, the centroid is 2/3 of the way from the vertex any median starts from to where that median meets the opposite side. So if you draw a line through the centroid (note that equilateral triangles are the only time you can get away with just saying 'center' instead of centroid/incenter/circumcenter/orthocenter) parallel to an edge, you get a similar triangle with each length 2/3 of the original. So the line segment between the center and an edge parallel to another edge is s/3. The edge of the cube has a length equal to the hypotenuse of a 45-45-90 triangle with legs s/3, or $\frac{s\sqrt{2}}{3}$. Then you cube that and find $\frac{2s\sqrt{2}}{27}$ as the volume.[/hide]
And as usual, my explanation was horrible...


how to sketch

$ Re(z)\equal{}|z\minus{}1|$?

Answer

Let $ z \equal{} a\plus{}ib$ where $ a,\,b\in R$

$ \Rightarrow a \equal{} |(a\minus{}1)\plus{}ib|$

$ \Rightarrow a^{2}\equal{} (a\minus{}1)^{2}\plus{}b^{2}$

$ \Rightarrow a^{2}\equal{} a^{2}\minus{}2a\plus{}1\plus{}b^{2}$

$ \Rightarrow a \equal{}\frac{1}{2}(b^{2}\plus{}1)$

Hence the locus in the complex plane is a parabola, with axis of symmetry on the $ Re$-axis, with the left turning point at $ \left(\frac{1}{2},\,0\right)$

[quote="sludgethrower"]
$ \Rightarrow a \equal{} |(a\minus{}1)\plus{}ib|$

$ \Rightarrow a^{2}\equal{} (a\minus{}1)^{2}\plus{}b^{2}$

[/quote]

How did you get the 2nd step?

[quote="A M Y"]That's the distance formula being used on the complex plane.

$ |x\plus{}yi| \equal{}\sqrt{x^{2}\plus{}y^{2}}$ and $ Re(x\plus{}yi) \equal{} x$

so that results follows from those definitions, after squaring.[/quote]

so $ \sqrt{(x\minus{}1)^{2}\plus{}iy^{2}}$ then simplify??


floor function
LaTeX
inequalities proposed
inequalities

$a_1,a_2,...,a_n$ are non-negative real numbers such that $\sum_{i=1}^na_i=1,$and$0<k\leq n,k\in\mathbb N.$
Find $minmax\{\sum_{i=1}^ka_{j+i}\},$ where $j=0,1,2,...,n-k.$ :)

The answer is $\boxed{\frac{1}{\lfloor n/k \rfloor}}$ :lol:

Yes!
For instance, $\lfloor\frac{10}{3} \rfloor=4,\lfloor \frac{10}{5} \rfloor=2.$
Zhaobin,where you have found these markings? I mean to $\lfloor\frac{n}{m} \rfloor.$
 In a quick LATEX course they are not present. Thanks.

Here they are : http://www.mathlinks.ro/LaTeX/AoPS_L_GuideSym.php :)


linear algebra
matrix
geometry
3D geometry
sphere
vector
function

Prove that a square matrix with nonnegative elements has at least one real eigenvalue. This is really the nicest problem in linear algebra that I have ever seen.

if i remember correctly it was discussed in forum long time ago and i didnt see linear algebraic solution to it do you know such a soln.

Sorry, I did not know it was discussed. I have twwo solutions for this beautiful problem and one is algebraic (of course, It was not me who found them!). Anyway, it is a good problem, we can post it twice.  ;)

perhaps you know this magnificient solution:
Let $A$ be the matrix with positive coefficient.
If $A$ is non invertible then $0$ is an eigenvalue.
If $A$ is invertible:
Let $D$ be the compact subset of the unit sphere whose elements
are the vectors $x$ such that: $|x|=1$ and $x_i \geq 0$ for every $1 \leq i \leq n$
Define the function $f$ from $D$ towards $D$ by:
$f(x)=\frac{A(x)}{|A(x)|}$ ($f(x) \in D$ because the coefficients of $A$ are positive).
Then $f$ has a fixed point (Brouwer theorem !), conclusion follows.
of course if the examinator asks you to prove brouwer theorem, you are in trouble :)

Yes, this is a classic :). So, has anyone ever seen a linear algebra solution? I saw it as an exercise in Munkers' "Topology: a first course" it said "this is for the students who think it ewould be a trivial exercise in linear algebra" :). However, the problem was discussed there only for $n=3$.

Wow, I have never seen this solution, but it is really great. And you know what, alekk, damn on the mad examinator who asks a proof for Brower's theorem after giving in an oral examination such a monster problem!  :D . Ok, here is an elementary really nice solution, but very hard to find:
      Let first a matrix A such that all coefficients are strictly positive and let $ r(A)$ be the largest absolute value of any eigenvalue of A. We will prove that in fact $r(A)$ is an eigenvalue of A. We start with:
    Lemma
        We have $ r(A)\geq min_{1\leq i\leq n}{a_{i1}+...+a_{in}}$.
     Proof:
        Let $b=RHS$ and let the matrix $ b_{ij}=\frac{ba_{ij}}{a_{i1}+...+a_{in}}$. Then we have $A\geq B\geq 0$ and thus $ A^k\geq B^k$ for all k. Since $ ||A^k||^{\frac{1}{k}}$ tends to $r(A)$ for any norm $||$ (exercise) and since $ r(B)=b$ (exercise!) we conclude by taking k very large.
    From here we find
      Corollary:
      For any $x_i>0$ we have $ r(A)\geq min_{i=1,n}{\frac{1}{x_i}({a_{i1}x_1+...+a_{in}x_n})}$
    This follows by applying the lemma to the matrix $ S^{-1}AS$ where $S$ is the diagonal matrix with $x_i$ on the diagonal.
    Now, the main result. COnsider a non-zero vector $x$ such that $ A= px$ and $|p|=r(A)$. Then $ r(A)|x|=|p||x|=|px|=|Ax|\leq A|x|$ where of course $ |x|$ is the vector whose components are the absolute values of the components of x. Put $ y=A|x|-r(A)|x|$. From the lemmas, we have $ r(A)>0$. If $y$ is not 0, then let $ z=A|x|>0$. Then we have $ Az>r(A)z$ and this contradicts the corollary.
    If some elements of A are 0, we consider a sequence of matrices with positive elements that converge towards A.


function
algebra
polynomial
quadratics
calculus
calculus computations

Q1 Let the roots of f(x)=x be α and β where f(x) is a quadratic polynomial. It is true that α,β also satisfies f(f(x))=x. Let the other roots of the eqn. f(f(x))=x be γ and δ. prove that
 If α+β=γ+δ then none of α β γ δ are real.

someone explain this with a detailed solution please

[b]There seems to be sth wrong in your problem statement[/b]. Pls recognize the following:

Make the Ansatz $ f(x)\equal{}C(x\minus{}\alpha)(x\minus{}\beta)\plus{}x$ with $ C\not\equal{}0$. Then $ f(f(x))\minus{}x\equal{}[f(x)\minus{}x][C(f(x)\plus{}x\minus{}\alpha\minus{}\beta)\plus{}2]$ and we can conclude $ C(f(x)\plus{}x\minus{}\alpha\minus{}\beta)\plus{}2\equal{}C^2(x\minus{}\gamma)(x\minus{}\delta)$. Comparison of coefficients corresponding to $ x$ yields the identity: $ \minus{}C^2(\alpha\plus{}\beta)\plus{}2C\equal{}\minus{}C^2(\gamma \plus{}\delta)$. Now, $ \alpha\plus{}\beta\equal{}\gamma\plus{}\delta$ gives the contradiction $ 2C\equal{}0$, [b]i.e. no such $ \alpha,\beta,\gamma,\delta$ exist[/b].


When are participants who scored 40+ on the USAMTS supposed to choose their books?  I have heard that the time varies by what award you received, but when does this process start?

I presume it's after the last round is scored.

I think gold chooses first, then silver choosese from what's left, etc...

Here's an old thread that dealt with prize dates, prizes, etc.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=145413

[quote="123456789"]How many prizes can each tier of winners choose? Ex. Gold winner, Silver winner, etc.[/quote]

I heard you can pick two book/software prizes, and a T-shirt Size, along with a medal.  However I'm unsure how to receive it.  Is it through mail as in a package?

Also are prizes different every year, or is it the same prizes :P?

It's not a physical medal (or, it wasn't last year), it's a certificate saying which "medal" you won.


Here are a list of last names. THey are arranged in a certain order.  Can you give a logical explanation for the order of the names?
here are the names:
Deoul
Traub
Green
Margolies
Morris
Peltz
Zimrin
Soule
Resar
Rothen
Brooks
Kraus
Patel
Winner
Desai

Can you help me ASAP?!?!?!?

From where is this? Not a take-home exam?

its from my math teacher
it was just a fun thing
and its due today!!!


quadratics
algebra
quadratic formula
polynomial
sum of roots
product of roots

can somebody explain step by step how to solve this problem?

2x^2+6x-20=0

[hide="hint"]
You can factor out the 2 to get $ 2(x^{2}+3x-10)$
[/hide]

[hide="Solution"]
From my hint you get that:
$ 2(x^{2}+3x-10)$
Now you list the factors of ten when subtracted from each other give the number 3.
The factors of ten are $ \pm1,\pm2,\pm5,\pm10$ By testing this out you see that -5 and 2 work. So solving this will give you:
\begin{eqnarray*}2(x-2)(x-(-5))\\ 2(x-2)(x+5)\end{eqnarray*}
So the answer is $ \boxed{2(x-2)(x+5)}$
[/hide]

Thanks ShockDrake for catching my error.

Okay, if you want to know how to factor look [hide="http://en.wikipedia.org/wiki/Factorization"]here[/hide].

-Hope it helps!

[quote="hunter34"][hide="hint"]
You can factor out the 2 to get $ 2(x^{2}+3x-10)$
[/hide]

[hide="Solution"]
From my hint you get that:
$ 2(x^{2}+3x-10)$
Now you list the factors of ten when subtracted from each other give the number 3.
The factors of ten are $ \pm1,\pm2,\pm5,\pm10$ By testing this out you see that -5 and 3 work. So solving this will give you:
\begin{eqnarray*}2(x+3)(x-(-5))\\ 2(x-3)(x+5)\end{eqnarray*}
So the answer is $ \boxed{2(x-3)(x+5)}$
[/hide][/quote]

It should be:

[hide="Solution"]2(x - [b]2[/b])(x + 5) = 0

x = {[b]2[/b], -5}

Not 3.[/hide]

You could also use the quadratic equation :)

Exactly. Couldn't of said it better myself.

[quote="redsmartie"]There are 3 ways.
[hide][b]formula: $ \frac{\minus{}b\pm\sqrt{b^{2}\minus{}4ac}}{2a}\equal{}\frac{\minus{}6\pm\sqrt{6^{2}\minus{}4\times 2\times (\minus{}20)}}{2\times 2}$.[/b]

[b]Factorization: $ 2x^{2}\plus{}6x\minus{}20 \equal{} 0$
$ (x\plus{}5)(2x\minus{}4)$[/b]

[b]The sum of the 2 roots is $ \minus{}\frac{b}{a}$ and the product is $ \frac{c}{a}$
supposing $ m$, and $ n$ are the 2 roots,
$ m\plus{}n \equal{}\minus{}\frac{b}{a}$, and $ mn \equal{}\frac{c}{a}$.

Either way, the 2 roots are:
$ \boxed{x_{1}}\equal{}\minus{}5$
$ \boxed{x_{2}}\equal{} 2$[/b][/hide][/quote]

how do you get the third one?

[quote=abacadaea]how do you get the third one?[/quote]
[hide="sum of roots"]One way is that the sum is basically the quadratic formula added together with the minus and plus (you know, the one before the square root) added seperately.
$ \frac{\minus{}b\plus{}\sqrt{b^{2}\minus{}4ac}}{2a}\plus{}\frac{\minus{}b\minus{}\sqrt{b^{2}\minus{}4ac}}{2a}\equal{}\frac{\minus{}2b}{2a}\equal{}\frac{\minus{}b}{a}$.[/hide]

[hide="product of roots"]Just multiply the plus and minus seperately (as the $ (a\plus{}b)(a\minus{}b)$ way, and you simplify to $ \frac{b^{2}\minus{}b^{2}\plus{}4ac}{4a^{2}}\equal{}\frac{c}{a}$[/hide]


Find a number with $3$ digits, knowing that the sum of its digits is $9$, their product is $24$ and also the number read from right to left is $\frac{27}{38}$ of the original.

[hide]It's easy.If i don't have any mistake then the solution is:
Call that number is abc We have:
$\ abc=\frac{27}{38}\ cba$ so $\ cba\equiv0\mod 38$
ANd $\ a+b+c=9$ so $\ cba\equiv0\mod9$
So cba have the form:9.38.k(0<k<3)
Try with k=1 and 2 we have k=1.
So abc=243[/hide]

MAth is my life :D

PPonte you say "a" is congruent to "b" mod 38 when 38 | a - b, congruence is usefull, you use it in most of number theory problems, read about it

Let the 3-digit number be abc.
We have a+b+c=9 and abc=24.
Since 100c+10b+a=(27/38)(100a+10b+c),38 divides 100a+10b+c.
Thus,by listing multiples of 38 we easily see that 342 satisfies all the conditions.
Hence,the number is 342.

abc = 24. So, one of them is a multiple of 3. Of course putting 6 won't satisfy. So, one of them is 3. Now the sum of the other two is 6 and their product is 8. So the other two numbers are 4 & 2. Also right to left is a multiple of 27. So, right to left will be 243. Hence the original number is 342.


quadratics
modular arithmetic
calculus
integration
algebra
difference of squares
special factorizations

I learned from AoPS volume 2 on how to solve linear congruences.

But what happens if I get equations like this one?

$\sqrt{47+5l}$ is an integer? All I know is since squaring the LHS and RHS gives $47+5l = a^2$ (arbitary integer $a$) and since we want the square, we associate this with mod 4 so $47+5l$ must equal to 0 or 1.

But here's a part I don't understand. Since $l$ is not specified, $5l$ equals various numbers in mod 4. Like

$5*1 = 1 \mod 4$
$5*2 = 2 \mod 4$
$5*3 = 3 \mod 4$
$5*4 = 0 \mod 4$

So how can I solve this equation? If it is possible I mean. I'm pretty sure guess-and-check might work but I'm not sure how to say there's no solution if it doesn't work...

The reason to use $\mod 5$ is because we have the term $5l$

On second hand, the way I did it is stupid and unnecessary! Just take the initial equation $\mod{5}$ and we're done!

and for such equations which work like $46+5l=a^2$ u do the same thing and get $a^2\equiv 1\pmod 5 $ and there is an integral solution for every $a\equiv \pm1 \pmod 5$ so thats quite easy as well ;)

[quote="paladin8"]I'm not sure $\mod{4}$ is the way to go here, since $5$ cycles through every possibility. When I see a square, I try to use difference of squares to factor... here's what I did for this one:

[hide]
$47+5l = a^2 \Rightarrow 5l-2 = a^2-49 = (a+7)(a-7)$

Taking this $\mod{5}$, we have

$3 \equiv (a+2)(a-2) \mod{5}$
$2 \equiv a^2 \mod{5}$

But $2$ is not a quadratic residue $\mod{5}$ so we see that there are no solutions.[/hide][/quote]

I'm bad at number theory so how did you get LHS to be 3 in mod 5?

And I'm not so sure how to show that 2 is not a quadratic residue in mod 5. Thanks for all your help!  :)

[quote="Silverfalcon"]

I'm bad at number theory so how did you get LHS to be 3 in mod 5?

And I'm not so sure how to show that 2 is not a quadratic residue in mod 5. Thanks for all your help!  :)[/quote]

He means the equation $5l-2=(a+7)(a-7)$ - we see $5l-2 = -2 = 3(\mod 5)$

And well... $0^2 = 0(\mod 5), 1^2=1, 2^2=4, 3^2=4, 4^2 = 1$.  And $5^2 = 0^2 = 0, 6^2 = 1^2 = 1, 7^2 = 2^2 = 4$, and so on, in $\mod 5$, so you can see there's no $2$. (Also, you can try and play with other mods and see which values the squares can be and cannot be)

But as paladin mentioned, we can see $47+5l = 2(\mod 5)$ and since it's under a square root, the conclusion follows immediately.  :)


logarithms
geometry
limit
geometric transformation
rotation
analytic geometry
integration

$ \boxed{1}$ Let $ C\ : y \equal{} \ln x$. For $ n \equal{} 1,\ 2,\ , \cdots$, $ A_n$ is the area of part bounded by the line passing through 2 points $ (n,\ \ln\ n) ,\ (n \plus{} 1,\ \ln\ (n \plus{} 1))$ on $ C$ and the curve $ C$, $ B_n$ is the area of the part bounded by the tangent line at the point $ (n,\ \ln\ n)$, the curve $ C$ and the line $ x \equal{} n \plus{} 1$. Let $ g(x) \equal{} \ln (x \plus{} 1) \minus{} \ln x$.

(1) Express $ A_n,\ B_n$ in terms of $ n,\ g(n)$ respectively.
(2) Use $ A_n > 0,\ B_n > 0$ to find $ \lim_{n\to\infty} n\{1 \minus{} ng(n)\}$.

$ \boxed{2}$ Let $ C\ : xy \equal{} \minus{} 2$ and $ Q$ be the point obtained by rotation in such a way that the point $ P\left(t,\ \minus{} \frac {2}{t}\right)\ (t\neq 0)$ on $ C$ in counter clockwise by the angle $ \theta$ about the origin.

(1) Express the coordinate of $ Q$ in terms of $ \theta$ and $ t$.
(2) Let $ \theta$ fix, when $ P$ moves on $ C$, find the equation of the locus of $ Q$.
(3) Find the values of $ \theta$ such that  the curve found in (2) passes through the point $ (\sqrt {3} \plus{} 1,\ \sqrt {3} \minus{} 1)$ for $ 0 < \theta < 2\pi$.

[hide="1"]
$ A_n \equal{} \int_n^{n\plus{}1} \ln x \minus{} \frac{\ln n \plus{} \ln (n\plus{}1)}{2}\equal{}x(\ln x \minus{}1) |_n^{n\plus{}1} \minus{}\frac{\ln n \plus{} \ln (n\plus{}1)}{2}$

$ \equal{} (n\plus{}1) \ln (n\plus{}1) \minus{}1\minus{} n \ln n \minus{} \frac{\ln n \plus{} \ln (n\plus{}1)}{2}$

$ \equal{} (n\plus{}\frac{1}{2}) (\ln (n\plus{}1) \minus{}\ln n)\minus{}1\equal{}(n\plus{}\frac{1}{2})g(n)\minus{}1$

$ y(x) \equal{} \ln x \to y'(x) \equal{} \frac{1}{x}$

$ l(x)\minus{}y(n)\equal{}y'(n)(x\minus{}n) \to l(x) \equal{} \frac{x}{n} \plus{} \ln n \minus{} 1$

having problems understanding what region you mean for B_n
[/hide]

:winner_first: 50+50=100

:winner_second: 36+50=86

:winner_third: 50+18=68


advanced fields
advanced fields theorems

[quote="AngelaS"]Hi,
How can we prove that the eigenvalues of a regular sturm liouville problem are simple?[/quote]

I found the answer. Thanks! It was very easy!


function
inequalities
logarithms
real analysis
real analysis unsolved

this question is related to a real analysis problem,
but maybe those guys can crack this tough nut.

[quote="fedja"]It is not tough, just not very attractive. The inequality is, actually, very rough. Let $m=\max(k,j)$. Then $f(j,k)\le m+(m-1)(2m-1)=2m^{2}-2m+1<2m^{2}$. Thus, the left hand side is at most $1+\log 2+2\log m<2(1+\log m)$, which is not greater than the right hand side.[/quote]
 :clap2: nice proof.
I met this one while reading the proof of some space is logconvex, which is used to prove pointwise
convergence (almost everywhere) of Fourier series on the space $LlogL$ (space near $L1$).


Super Woman has a HUGE crush on Super Man (because SuperMan is really hot! :lol: ).  One day, she was walking past a grafiti wall at her school and she saw the initials SM & SW written on a wall with a heart around it.  She quickly looked at her school's yearbook and saw that there are 3 boys with the initials SM, 10 girls with SM as initials, 5 SW boys, and 8 girls with the initials SW (including Super Man and Super Woman).  What are the chances that her the initails were spraypainted on the wall by Superman about SuperMan and SuperWoman?

I have no idea what the level of this is...probably upper level sprint or medium level target...i don't know the answer

The message could have been written by any of the people with initials $SM$ or $SW$ except SuperWoman, because since she is wondering who wrote the message, she must not have written it. :lol:  So, the number of people who could have written the message is $3+10+5+8-1=25$.  Therefore, there is a $\frac{1}{25}$ probability that SuperMan wrote the message.  Then, assuming that SuperMan did write the message, he must have written the message to a girl with the initials $SW$.  Therefore, the number of people he could have written the message to is $8$.  So, there is a
\[\frac{1}{25} \times \frac{1}{8} = \frac{1}{200}\] probability that SuperMan wrote the message to SuperWoman.

[quote="sirorange"]Now, there are 13 SMs and 13 SWs, so therefore there are 169 combinations, so the probability is 1/169.[/quote]

I believe you overlooked the fact that a boy would most likely not write his initials in a heart with the initials of another boy, but I guess the problem does not point that out ;)

Oh yeah, but you are still wrong. 13 SM, and 13 SW. So anyone of the 13 SM could have written it to one of the 13 SW, or anyone of the 12 non-superwomen SW could have written it to any of the thirteen SM. So we have 1/(13*13+13*12)=1/325

In your new solution, you did recognize that the message could have been written by a person with the initials $SM$ or $SW$, and you recognized that SuperWoman could not have written the message.  However, the question of whose solution is correct depends on whether or not the possibility exists of a boy writing the message to another boy or a girl writing the message to another girl.

Could you please clarify which situation you intended to put in the question?

it could have been written about anyone...what are the chances that the SM is Super Man and the SW is Super Woman?

If you want it with a boy and a girl, it is 3*8+10*5  (1/74), else it is 13*13 (1/169)

No, because you know superwoman didn't write it. Unless she came down with a sudden case of amnesia. . . .

assume she didn't have amnesia...

you want it with a boy and girl....

So 1/74, because it is 3*8+5*10.  I don't get the part about Superman writing it.  How many people are allowed to write it?  Is it 1/(74*26)  because 26 people can write it or 1/(74*2) because only 1 of the 2 could write it?

anybody except for super woman can write it because she is the one who saw it

look at the attachment its soo cool

i dont know i didnt do it...

i guess thats right it makes sense

[sarcasm]ha haha hahahahahaha hahahahahah[/sarcasm]

This problem can be looked at from varying levels of depth.  The actual problem says "What are the chances that her the initails were spraypainted on the wall by Superman about SuperMan and SuperWoman?".  Let $p$ represent the population of the school.  Then the chance that Superman spraypainted it is $\frac1{p-1}$, assuming that Superwoman doesn't have amnesia and didn't spraypaint the message herself.

We do not no that the SM stood for superman in the grafiti even if supermad did write it.  If you do NOT assume that SM and SW are of opposite genders (they could be homosexual), then the chance that SM is Superman, it was written by superman, and SW is Superwoman is

$\frac1{p-1}*\frac1{13}*\frac1{13} = \frac1{169(p-1)}$.

However, if SM and SW ARE assumed to be of opposite genders (boy & girl) then the chance is

$\frac1{1-p}*\frac1{13}*\frac1{8}=\frac1{104(p-1)}$

 because the SW must be female.  This is the type of math problem where human insight and english class can come in handy... "did Romeo really love Juliet?" ... "did SuperMan really love SuperWoman?".  Probrably not, both mathematically and psychologically.  Again, problems that involve human emotion involve an amazing number of variables, and we cannot account for every possibility within reason.

assume that only a sm or sw can write it


topology
geometry
geometric transformation
homothety
advanced fields
advanced fields unsolved

Let $A \subset \mathbb{R}^2$ be the union of the circles with centre $(1/n,0)$ and radius $1/n$, with $(0,0)$ removed ($n \in \mathbb{N}$). Let $B \subset \mathbb{R}^2$ be the union of the circles with centre $(n,0)$ and radius $n$, with $(0,0)$ removed ($n \in \mathbb{N}$). Are these two spaces homeomorphic? They are the disjoint union of circles with the origin removed, and each such pair of $S^1 \setminus 0$ is homeomorphic.

Indeed, they are homeomorphic as one can check that both maps
$f: A\to B$, sending the circle with radius $\frac 1n$ to the circle with radius $n$ by a homothety with center (0,0), and
$g: B\to A$, sending the circle with radius $n$ to the circle with radius $\frac 1n$ by a homothety with center (0,0)
are continuous (and inverse to each other). For this you mainly have to check that each point has some neighborhood with only points of one circle inside it.

Note that the map $f$ was not continuous we did not delete the point $(0,0)$.

They are homeomorphic. To map $A$ to $B$, scale the $n$-th circle by $n^2$. That's continuous on $A$ because each circle is both open and closed.

If we included the origin in both spaces, they would not be homeomorphic.


function
algebra
binomial theorem
combinatorics unsolved
combinatorics

Prove the following 2 statements using the multinomial theorem (do NOT give combinatorial proofs and do NOT use the formula for binomial coefficient!):

$\binom{k}{m}\binom{n}{k} = \binom{n}{m}\binom{n-m}{n-k}$

and

$\sum_{k=m}^{n}\binom{k}{m}\binom{n}{k} = \binom{n}{m}2^{n-m}$

I'm new to combinatorics so I'm having trouble with these easier problems; any help is appreciated.  Thanks!

Hi,

First one is so easy using binomial formula  :?

For the second one

rewrite $\sum_{k=m}^{n}\binom{k}{m}\binom{n}{k}$ as $\sum_{k=0}^{n-m}\binom{k+m}{m}\binom{n}{k+m}$, apply first result, extract  $\binom{n}{m}$, then apply binomial theorem

Hey, thanks for the help.  But... I'm still having trouble with the first one! Can anyone elaborate a little more? I dont know how you can get so many coefficients in the binomial theorem.  thanks.

Just write $\binom{n}{m} = \frac { n! } { m! (n-m)! } $

I swear it's not difficult (it's called Newton's identity sometimes)

Hmm there's  supposed to be a way without using the formula for the binomial coefficient.  im so bad at this stuff -_-. how does binomial/multinomial theorem help here?  thanks

Yes, there is a way to do this without using the "formula". I'll just do the first part. You can try to do the second part in a similar way.

Consider a group of $n$ people. Suppose you need to choose a committee of $k$ people from this group, and suppose that you want to give $m$ of these $k$ people a "special role". We'll count the number of ways to do this in two ways.

Obviously, you can choose the committee in ${n \choose k}$ ways, and then you can choose the people with a special function in the committee in ${k \choose m}$ ways. So, there are ${n \choose k}{k \choose m}$ ways to do is.

But, you could also pick the $m$ people with a special function directly from the group of $n$ people (which can be done in ${n \choose m}$ ways) and then choose the rest of the committee - that is, $m - k$ people - (those without a special function) from the remaining $n - k$ people. Obviously, this means that you need to pick $n - k - (m - k) = n - m$ people which are NOT in the committee, so this can be done in ${{n - k} \choose {n - m}}$ ways. So, that gives you a total ${n \choose m} {{n - k} \choose {n - m}}$ ways to do this.

Now, obviously, we must have equality here: ${n \choose k}{k \choose m}  = {n \choose m} {{n - k} \choose {n - m}}$, and we're done.

Hi,
THanks for the solution but I already got this one :).  I'm looking for one that uses neither combinatorial argument nor formula for coefficient, but rather uses binomial theorem or multinomial theorem (i.e. expansion for (x+y+z)^n).  i know there is one, i just can't find it. thanks!


Some general advice for doing anything in life: Do it very carefully.

In your case, practice where there is a very low risk of anything being broken.

Try using eggs first as they are easier to juggle. Glass bottles work well too :) .

[quote="tongos"]does anybody here juggle four balls? if so, can you give me some tips on juggling four balls?[/quote]first learn to juggle with 2 balls into one hand. then you will learn step by step to juggle with 4 balls.
but it's easier (and looks a lot nicer) to juggle with 5 balls.

I remember that at IMO in 2002 we had in the opening ceremony a professional mathematician and juggler. He explained to us that he is close to classifying juggling tricks on a mathematics basis, such that people can recreate them just by looking at formulaes. He showed us a lot of tricks, and then a huge jugglin current appeared. I also learned to juggle at that IMO, but only with 3 balls.

I can juggle three balls. Four balls with 0-40 (rarely 100) cathces. On IMO 2004 there were at least 5 sets of juggling balls. Many tricks are at [url]http://www.jugglingdb.com/[/url].

[img]http://img.photobucket.com/albums/v259/drspyder/Juggling.gif[/img]

:) .

[quote="ondrob"]I can juggle three balls. Four balls with 0-40 (rarely 100) cathces. On IMO 2004 there were at least 5 sets of juggling balls.[/quote]I know that in IMO 2003, A. Kaseorg use to juggle a lot with 5 balls at a time :)

hehe, I know a juggling mathematician, he even gave us a lecture on the "Mathematics of Juggling" :D

as a result, i learned that to juggle four balls, you just juggle two balls on each hand and then make it look fancy :P

There are some other nice 4-ball tricks as well.  One that I am (half-heartedly) trying to learn is what you would get if you juggled 5 balls, one of which is invisible.  It breaks the usual juggling pattern, though, because the throws are right-left-left-right-right-left-left-right-right etc., although generally I've lost it before I get that far.  I think Valentin is right about 5 looking much nicer, but I'm not convinced about it's being easier.  Practice doing 2 in your weaker hand if you want to get 4 -- that's really the key.

[quote="JBL"]I think Valentin is right about 5 looking much nicer, but I'm not convinced about it's being easier.  Practice doing 2 in your weaker hand if you want to get 4 -- that's really the key.[/quote]Now it depends what you understand by "standard 4 ball juggling". If by that you mean 2 balls juggle in left, 2 in right, without interaction between the balls (ie each hand catches the same two balls at all times) it easier than 5 balls. If you mean interaction between the two hands, I think it's harder (it might be what you were describing as 5 balls with 1 invisible :D).

Yeah, it's true, some 4-ball tricks are much harder then 5-ball cascade :)

And what about other juggling instruments?
Have you ever tried diabolo?
I've got one and it's really fun to play with it...

Or have you ever been juggling with knifes?
I can juggle with 2 balls and 1 knife without being hurt, cause you have just to look at knife and when it goes wrong then run away from it  :D

It's also real fun to juggle with everything from oranges, pens, sticks to buckets or mobile phones (over bed :)). I tried it :), it's funny especially with mobile phones of others :D :D

[quote="ondrob"]Or have you ever been juggling with knifes?
[/quote]For the record: Kids, [b]don't try this at home[/b] :)

You guys should have warned me earlier!
You make it sound soooo easy
It's hard to touch type with one hand!

[quote="somedude"]... I can't juggle at all!  I've tried 3 balls, well oranges today infact, but I can't do it... any pointers?[/quote]Practice makes perfect ... try more, until you succeede. That's what I did anyway. I admit it took me a couple months or so :D You can make this faster by watching movies with jogglers. There are plenty to find with [url=http://www.google.com]Google[/url].

The weird thing with me is that I tried really hard to learn to juggle from when I was 10 until I was 12. Then, I stopped and didn't think about it for a year, and tried again last year and did it! So maybe you j ust need a breatk from working on it.

http://jongl.home.pages.de/

Very cool juggling simulator.

Anyone interested in math behind juggling should search for "siteswap."

In order to do that, you would most certainly need to use 0 hands ...


geometry
geometric transformation
homothety
incenter
geometry unsolved

[quote="monkeyman2"]Let $ K$ be the incircle of $ ABC$ and let $ A_1, B_1, C_1$ be tangent point to the sides $ BC, CA, AB$. Inscribe circles into $ AB_1C_1, BC_1A_1, CA_1B_1$. Prove that the comon tangent of the incircles $ AB_1C_1, BC_1A_1$ , that is not a site of the triangle $ ABC$ is paralell to the tangent to $ K$ in the midpoint of the arc $ A_1B_1$, closer to C.

Can somebody prove this using homothety?[/quote]
Do you mean the [b]external tangent[/b]??

anybody? I really need a short solution for this one.

I could not come up with a solution using homothety like you want.

This is my solution using angle chasing:

Let $ t$ denoted the tangents mentioned in the problem, $ I_a,I_b,I_c$ are the incenters of $ \triangle AB_1C_1,\triangle BC_1A_1, \triangle CA_1B_1$, i.e. they are the midpoints of arc $ B_1C_1,C_1A_1,A_1B_1$ of cirle $ (I)$. Besides, let $ X,Y$ denotes the points of tangency between $ (I_a),(I_b)$ with $ t$ and let $ A_1B_1,t$ intersects $ AB$ at $ D,E$. We have:

$ \widehat{C_1EX}\equal{}180^o \minus{} 2\widehat{EC_1L}\equal{}180^o \minus{} \widehat{A_1C_1B_1} \minus{} 2\widehat{A_1B_1C_1} \equal{} \widehat{B_1A_1C_1} \minus{} \widehat{A_1B_1C_1}$
$ \Rightarrow \widehat{C_1EX} \equal{} \frac{180^o \minus{} \widehat{BAC}}{2} \minus{} \frac{180^o \minus{} \widehat{ABC}}{2} \equal{} \frac{\widehat{ABC} \minus{} \widehat{BAC}}{2}$

On the other hand, $ \widehat{C_1DB} \equal{} \frac{180^o \minus{} \widehat{ACB}}{2} \minus{} \widehat{BAC} \equal{} \frac{\widehat{ABC} \minus{} \widehat{BAC}}{2} \equal{} \widehat{C_1EX}$, this implies that $ XY // A_1B_1$, and this is exactly what we have to prove. :)


Point $ O$ is the center of a circle circumscribed about isosceles $ \triangle ABC$. If $ AB\equal{}AC\equal{}7$, and $ BC\equal{}2$, find $ AO$.

[hide]Using Heron's, $ [ABC] \equal{} 4\sqrt{3}$

Using the formula $ [ABC] \equal{} \frac{abc}{4R}$:

$ 4\sqrt{3} \equal{} \frac{7*7*2}{4R}$

$ 4R \equal{} \frac{98}{4\sqrt{3}}$

$ R \equal{} \frac{98}{16\sqrt{3}} \equal{} \boxed{\frac{49\sqrt{3}}{24}}$[/hide]


hi guys
know the slots machine?
you can find them at all the casinos and aslo at some air ports..
you can also find them in [url=http://casino-on-net.com]online casino[/url] sites..
anyway, i know its all about statistics but where can i find the mathematicle answer for the winning odds in slots?

I don't know, but it's certainly not in the games and fun factory.

Try somewhere else...like not here...maybe round table

Spam is tastey.  The post looks like an advertisement.

Slot machines are programmed so that the expected value is something crappy.

i would think that the percentage depends on the machine or the company who makes the machine. i wouldn't think they are all the same.

[quote="Lil_mad_dog"]Gambling is a tax on the stupid... no offense[/quote]
more precisely, gambling is a tax on people who can't do math.  :lol:


I realize that it means "1 day, 1.5 days, 2 days, 3 days." But you never hear somebody write on a note, "You have 168 hours to give me a million dollars." to represent a week. Then they just say, "You have a week."

It's the same deal with "30 minutes, 60 minutes, 90 minutes" to represent "half an hour, an hour, an hour and a half". You never hear anyone say "You have 120 minutes to do this test", they just say "2 hours".

Where is the line drawn?

It depends on the person. There's a teacher here who always gives times in seconds. "You have 600 seconds left for this test."

[quote="ZetaX"]Whenever you fail to calculate the exact number in a very short amount of time  :wink:[/quote]

I'm sure that most people can't do 3 x 24 = 72 in three seconds or less. :wink:

I'm pretty sure everyone who uses that number for three days can.

In many cases, someone who gives an ultimatum wants to seem threatening and/or give a sense of urgency.

If Terrorist Tina were to say, "You have 36 hours to give me the money or I shoot your wife," she's counting in hours, a much smaller unit than the day, so it seems more urgent. Not to mention that the term "day" doesn't have the sense of exactness that "hour" does.

But it doesn't always work that way. If she said, "You have 168 hours to pay me back, or you never hear from your son again," it doesn't work, since the size of the number 168 prevents it from giving that sense of urgency.

Similarly, if she said, "If you don't return the files within 1/52 of a year, consider yourself dead!", it would have little effect, since the term year prevents it from having that sense of urgency.

So ultimately it boils down to picking numbers and units to convey a sense of urgency and seriousness (and usually threat), based on the kinds of numbers and units we encounter in day-to-day life.

[color=green]Moderator says: there is never a good reason to quote the entire post immediately preceding your own.[/color]

I see. Well, I think the vastness of the number 72 also stops a few people from taking it seriously. It seems kinda iffy at 72, but I guess 168 just pushes it too far.

[color=green]Moderator says: there is never a good reason to quote the entire post immediately preceding your own.[/color]
Well to be honest, I almost never hear people saying 72 hours. I hear 3 days.

I use big numbers only when I want to bother people, that is, if I get annoyed with someone, I will start using quantities such as 31,536 kiloseconds (yay for making up units) to represent a year.  :rotfl:

For me, 72 hours, 168 hours, etc. mean the same thing to me as the "real" unit used to measure that length of time, simply because I can almost instantaneously convert back and forth in my head. Others that I know cannot do this, and as such, have problems with using "big" numbers. It is much the same with people I know who cannot really gauge 10 centimeters, but can easily estimate 4 inches.

[quote="sdkudrgn88"][color=green]Moderator says: there is never a good reason to quote the entire post immediately preceding your own.[/color][/quote]

Ehrm, sorry. But yeah, it happens with the news too. (cf.[?] http://xkcd.com/558/)

I think it's because of how many syllables it has. 168 hours has at least 6 syllables, while 1 week has 2.

It probably also depends on what it's about. If you're taking AMC10, you would use minutes (because it's not too much over an hour) but not hours (b/c of syllable thing).


The sum of 5 consecutive even integers is $ 4$ less than the sum of the ﬁrst $ 8$ consecutive odd counting numbers. What is the smallest of the even integers?

$ \textbf{(A)}\ 6 \qquad
\textbf{(B)}\ 8 \qquad
\textbf{(C)}\ 10 \qquad
\textbf{(D)}\ 12 \qquad
\textbf{(E)}\ 14$

the sum of the first $ n$ odd numbers is always $ n^2$ (this can be proved by induction). thus, the sum of the first 8 odd numbers is $ 64$, so the sum of the even numbers is $ 60.$ The middle even number is clearly $ 12$, so the smallest is $ 12\minus{}4\equal{}\boxed{8}$, or choice $ (B)$

[hide="Solution"]The first even integer is $ a$. The sum of the 5 consecutive even integers is $ 5a\plus{}20$.
The sum of the first 8 odds is $ \sum_{k\equal{}0}^{7}2k\plus{}1\equal{}7\cdot8\plus{}8\equal{}64$.
We have $ 5a\plus{}20\equal{}64\minus{}4\implies a\equal{}\boxed{\textbf{(B)}\ 8}$.[/hide]

[hide="Proof that the sum of the first $ n$ odds is $ n^2$"]
$ \sum_{k\equal{}0}^{n\minus{}1}2k\plus{}1\equal{}2\cdot\frac{(n\minus{}1)n}{2}\plus{}n\equal{}n^2\minus{}n\plus{}n\equal{}n^2$.

[hide="Proof by induction"]
Let $ P(n)\equal{}\sum_{k\equal{}0}^{n\minus{}1}2k\plus{}1$.

$ \textbf{Base case.}$ $ P(0)\equal{}1\equal{}1^2$.

We claim $ P(m)\equal{}m^2$.

$ \textbf{Inductive step.}$
$ P(m\plus{}1)\equal{}\sum_{k\equal{}0}^{m}2k\plus{}1\equal{}(2m\plus{}1)\plus{}\sum_{k\equal{}0}^{m\minus{}1}2k\plus{}1\equal{}m^2\plus{}2m\plus{}1\equal{}(m\plus{}1)^2$.
[/hide]

$ \mathbb{QED}$.
[/hide]

@ grn turtle your proof for the sum of the first n positive odd numbers cannot be seen and just show that n odds = n^2.

[hide="Proof that the sum of first N odd numbers equal N^2"]
Every odd number can be written as $ 2k\minus{}1$
The sum of the first N numbers is
$ (2(1)\minus{}1)\plus{}(2(2)\minus{}1)\plus{}....(2(N)\minus{}1)$
$ 2(1\plus{}2\plus{}...N)\minus{}N$
$ 2(\frac{(N)(N\plus{}1)}{2})\minus{}N$
$ 2(\frac{(N)(N\plus{}1)\minus{}2N}{2})$
$ (N)(N\plus{}1)\minus{}2N$
$ N^2\plus{}2N\minus{}2N$
$ N^2$[/hide]

[quote="grn_trtle"][hide="Solution"]The first even integer is $ a$. The sum of the 5 consecutive even integers is $ 5a \plus{} 20$.
The sum of the first 8 odds is $ \sum_{k \equal{} 0}^{7}2k \plus{} 1 \equal{} 7\cdot8 \plus{} 8 \equal{} 64$.
We have $ 5a \plus{} 20 \equal{} 64 \minus{} 4\implies a \equal{} \boxed{\textbf{(B)}\ 8}$.[/hide]

[hide="Proof that the sum of the first n odds is n^2"]
$ \sum_{k \equal{} 0}^{n \minus{} 1}2k \plus{} 1 \equal{} 2\cdot\frac {(n \minus{} 1)n}{2} \plus{} n \equal{} n^2 \minus{} n \plus{} n \equal{} n^2$.[/hide]

[hide="Proof by induction"]
Let $ P(n) \equal{} \sum_{k \equal{} 0}^{n \minus{} 1}2k \plus{} 1$.

$ \textbf{Base case.}$ $ P(0) \equal{} 1 \equal{} 1^2$.

We claim $ P(m) \equal{} m^2$.

$ \textbf{Inductive step.}$
$ P(m \plus{} 1) \equal{} \sum_{k \equal{} 0}^{m}2k \plus{} 1 \equal{} (2m \plus{} 1) \plus{} \sum_{k \equal{} 0}^{m \minus{} 1}2k \plus{} 1 \equal{} m^2 \plus{} 2m \plus{} 1 \equal{} (m \plus{} 1)^2$.
[/hide]

$ \mathbb{QED}$.
[/quote]

Sorry, I guess that's what comes of not using the Preview button. Thanks, enndb0x.


I have a question about the USAMO SIII form from the USAMO 2007 Teacher's Manual.

My question is in the blank where it says "Signed" and below -- USAMO Participant, are we supposed to write our name or just sign it?

Because if I am Canadian, then if I just sign that blank and I do not fill out the Section B my name will not appear anywhere on the form...

Thank you very much for your help.

Please direct all informational and procedural queries about AMC forms, awards, certificates, procedures etc directly to the AMC office at amcinfo@maa.org, or call our toll-free number at 800-527-3690.  It makes much more sense to contact us directly with a question and have it answered by the responsible staff member than to post your question here for people who are not reponsible for executing the procedures to speculate about.

Now to answer your question:  Write legibly in the space labeled "signed".  Use the same name as appears on your AMC and AIME contests, and we will figure out who you are.  This same form has woked well,  unchanged, for as long as I have been here.  I am sure it will continue to work as well this year.

Steve Dunbar
AMC Director


During a single day at radio station WMZH, the probability that a particular song played is .38. How do you express the probability that this song will be played on exactly 5 days out of 7 days?

[quote="superwizard188"][hide]$ (0.38)^5*(0.62)^2$[/hide][/quote]

I'm pretty positive that you have to multiply that by 10.

http://en.wikipedia.org/wiki/Binomial_distribution

If you call the playing of the song (the event) as a success , then we need to have 5 successes in 7 attempts.
This can be given by $ 7C5 . (0.38)^5 . (0.62)^2$
 :P  :)


calculus
integration
trigonometry
calculus computations

$ \int \frac {x^4 \plus{} 1}{x^2 \sqrt {x^4 \minus{} 1}} \equal{} ...$

What I've done is substituting $ x^2 \equal{} \sec t$, but I can't solve it..
I've also tried $ x^2 \equal{} \tan t$, still it doesn't work..
TT

hello, i have got
$ \int\frac{x^4\plus{}1}{x^2\sqrt{x^4\plus{}1}}\,dx\equal{}\frac{\sqrt{x^4\minus{}1}}{x}\plus{}C$.
Sonnhard.

I've just discovered a clever method for computing this integral.
Divide the numerator and the denominator by $x^3$.
We'll get
\[\int \frac{x + x^{-3}}{\sqrt{x^2 - x^{-2}}} dx\]
This can be easily solved by a simple u-substitution method.
And I realized how easy this can be done, but the idea to divide the numerator and the denominator by $x^3$ is so "expensive".. =)

The answer to your problem, I see nothing new, but why say divided by $x^3$ is expensive?
It seems, you're a funny.
Carried out with integral irrational, the common method is to solve the base.


vector
linear algebra
matrix
geometry
parallelogram
real analysis
real analysis unsolved

hello;

I have 2 points, ptA and ptB.

ptA also has a direction vector, dirA;

I am able to determine if ptB is 'behind' ptA by checking the sign of [b]dirA DOT ( ptB - ptA ) [/b];

but I would also like to check if ptB is to the left or right of ptA relative to dirA;

dot product does not seem to give this type of information;

any thoughts or suggestions?

thanks,
Shannon

If you're thinking that way, it's the sign of the cross product you want. Note that in two dimensions, the cross product lives on one axis outside the plane, and can be read as a single number.

Jmerry, thanks for your response;

but a cross product returns a vector, which does not have a sign??

though it's components have signs: <-1,0,0> for instance;

In three dimensions, this question doesn't make sense at all.

Jmerry, yes I am working in 3d;

thanks,
Shannon

OK. There is no answer, because "left" and "right" don't make sense.

so is there a such thing as a cross product for 2d .. that returns a signed number?

jmerry addressed your latest question in his first post. Consider two vectors (a,b) and (c,d) as the 3d vectors (a,b,0) and (c,d,0) and take their cross product. Only its third component is nonzero, so you can use that.
(This is the same as taking the determinant of the matrix $ \begin{pmatrix}
a & c\\
b & d
\end{pmatrix}$.)

Bictor717, thanks for your response;

if I am not mistaken, the cross product of <a,b,0>x<d,e,0> is still going to be a vector? and thus will not have a sign associated with it;

I am not a mathematician so my question might be ridiculous;

MY GOAL: to determine to which side a ptB is relative to a dirA located at ptA; everything is projected onto a plane;
another way of saying it might be: is ptB cw or ccw from dirA;

thanks,
Shannon

[quote="dsdsdsdsd"]Bictor717, thanks for your response;
if I am not mistaken, the cross product of <a,b,0>x<d,e,0> is still going to be a vector? and thus will not have a sign associated with it;
[/quote]
For the third time, I think: yes, it's a vector, but you only care about the sign of its third component. In your notation, you only need the sign of $ ae\minus{}bd$.

randomgraph, thanks for your response;

you are the FIRST :)  person to say that I need the sign of [i]ac-bd[/i];

however did you mean [i]ad-be[/i] considering my notation is: <a,b,0>x<d,e,0> ??

thanks for your time,
Shannon

I may be the first who said it in so many words, but if you read jmerry's and Bictor717's responses carefully, you'll find it there ("Only its third component is nonzero, so you can use that.").

And no, I meant what I wrote: $ (a,b,0) \times (d,e,0) \equal{} (0,0, ae\minus{}bd)$.

randomgraph, thanks for your time;

Shannon

Most of the time, I think of this two-dimensional cross product as the area operator; the number we get is the signed area of a parallelogram with those two vectors as sides. Some of my thoughts on it are [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1044687#1044687]here[/url].


Establish each identity.

1) Show that tan(sin^-1(v)) = v/sqrt{1 -v^2}

2) Show that cos(tan^-1(v)) = 1(sqrt{1 + v^2})

[hide]
#1:
Let $\alpha =\sin^{-1} v$.

Consider a right triangle ABC with $\angle A=\alpha$ and $\angle C=90^\circ$.

Let $AB=1$. Then $BC=v$, and we find $AC=\sqrt{1-v^2}$ by using Pythagorean Theorem.

Therefore $\tan(\sin^{-1} v)=\tan\alpha=\frac{v}{\sqrt{1-v^2}}$

You can use the same method for #2 :)
[/hide]


MATHCOUNTS conics hyperbola Asymptote rotation articles geometry

[color=darkblue][size=75]<split from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39795]parent thread[/url] by moderator rcv>[/size][/color]

[quote="Original Question"][color=darkblue]This is a countdown problem from this year: I believe they said on TV that it isn't hard if you have calculus lol.

If $y=\frac{3x}{x-2}$ and x is allowed to be any real number other than 2, what is the only value that y cannot be?

I came up with a solution, but if you guys would tell me if there is a better one I would appreciate it.[/color][/quote]
...

[quote="Hamster1800"]Right I see. I did graphs of hyperbolas generated by rational functions like these in algebra I so that's how I recognized it. Thank you for all your insights.[/quote]
Uhhh, I don't think this would be a hyperbola.
By definition, the equation of a hyperbola is:
${\frac{(x-h)^2}{a^2}}-\frac{(y-k)^2}{b^2}=1$ or
${\frac{(y-k)^2}{a^2}}-\frac{(x-h)^2}{b^2}=1$
with center at $(h,k)$, asymptotes $(y-k)=\pm\frac{b}{a}(x-h)$
and foci at $(h\pm2a,k)$ or $(h,k\pm2a)$ and vertices at $(h\pma,k)$ or $(h,k\pma)$ and directrix at $x=h$ or $y=k$.  It is the set of all points equidistant from the foci and the directrix.

Try to find these for the equation you've given, I don't think it will work.

The standard form you have given is only for hyperbolas that are parallel to the axes.  But hyperbolas can also be rotated.  The original equation is a hyperbola, rotated 45 degrees.

I don't think a hyperbola is the set of points equidistant from a point (focus) and a line (directrix).  That is a parabola.  In a hyperbola, the distance to the focus is some constant times the distance to the directrix, but that constant (the eccentricity) is greater than 1.

I split and moved this topic, so you could feel comfortable discussing the finer points of hyperbolas, which are beyond the scope of MATHCOUNTS.

[quote="Hamster1800"]By definition, the equation of a hyperbola is:
${\frac{(x-h)^2}{a^2}}-\frac{(y-k)^2}{b^2}=1$ or
${\frac{(y-k)^2}{a^2}}-\frac{(x-h)^2}{b^2}=1$
with center at $(h,k)$, asymptotes $(y-k)=\pm\frac{b}{a}(x-h)$
and foci at $(h\pm2a,k)$ or $(h,k\pm2a)$ and vertices at $(h\pma,k)$ or $(h,k\pma)$ and directrix at $x=h$ or $y=k$.[/quote]
Ravi has corrected some misconceptions.  Please add that I don't believe those are the possible locations of the directrix.

[quote="nat mc"]Well in this case the asymptotes are x=2 and y=3.  this makes the center (2,3).  This makes the directrix y=-x+5 and one of the foci (-2,-3).  A random point (-4,2) is on the graph and is not equidistant from the foci and directrix[/quote]
I think the foci are at $(2+\sqrt{12},3+\sqrt{12})$ and $(2-\sqrt{12},3-\sqrt{12})$.  I think the directrix is $y=-x+5+\sqrt{12}$.

The definition of a hyperbola is that the difference in distances from any point on the hyperbola and the two fixed foci is a constant.  Rather than using the directrix and the eccentricity, try using the distance formula with a few points on the curve to convince yourself.

[quote="tarquin"]Hyperbola = set of points such that the difference of the distances from the center is always constant.[/quote]
No.  See above.

Ok, so in this case, the directrix is NOT the perpendicular bisector of the line created by the vertices, since it seems buck closer to the top curve

[quote="nat mc"]Ok, so in this case, the directrix is NOT the perpendicular bisector...[/quote]
For a hyperbola, there are two distinct directrices.  The directrix is [b]never[/b] the perpendicular bisector of the line segment connecting the foci.  See [url=http://mathworld.wolfram.com/ConicSectionDirectrix.html]this MathWorld article[/url] for more information.


I was just wondering if there were any other homeschoolers here. For a time, I was homeschooled, but now I go to a very small private school. One of the best things about homeschooling is that I was able to get ahead in math, and last year I took calculus.

I am! :)

Incidentally, how old are you!

Billy

[color=green]*high five* Sixteen. And you?[/color]


search
linear algebra
matrix
linear algebra unsolved

K commutative field

$W$ an $K$ subvector space of nilpotent matrices in $M_n(K)$

What is maximum of $dim_{K}W$

Hard problem. I suspect the answer is $\frac{n^2-n}{2}$ but I don't have good ideas to prove it. Some hints, Professor Moubi?  :blush:

Oh people, again with the hints? :) Can't he PM the hints to you? :) There are problems on the forum which we haven't touched for a year or more, but you want these solved in like two days? Give people a chance to think about them, please.

Ok, ok, grobber, you don't have to be so impatient and angry. Let me tell you what happens to the problems you are speaking about: they are lying there dead. And nobody seems to be interested in them. I just didn't what to add some nice problems posted in the last few days in the nice list I was speaking about.

[quote="grobber"]There are problems on the forum which we haven't touched for a year [/quote]

which problems ?

just search the forum and see how many have only the statement and no solution

Hmmm... I was right... Now, because the problem is far too nice and hard, I think it deserves to be brought again in discussion. Anyone, anything? Moubinool, do you have a solution?

This is a particular case ,but for $K=R$,as $W \cap S_n(R)=\emptyset$(since they are diagonalizable)we get the result.

What is a K subvector space? Just a Subspace?

Yes, it is a subspace. Now, this was only an attempt to bring this (again???) under attention. ;)  :P

harazi,
Do you actually have a proof that the max dimension is $\frac{n(n-1)}{2}$? Because I have the feeling that it's more :oops:

Well, there are more nilpotent matrices than can fit into any one subspace.

I can be kind of nose-to-the-ground literal-minded sometimes. What's the very smallest example I can think of? I'm doing this not to solve the problem but to gather ways of thinking about it. The smallest example I can think of is $M_{2}(\mathbb{Z}_{2}),$ which is 4-dimensional with a cardinality of 16. The 16 members of this set can be classified as follows:

The zero matrix
6 rank-1 idempotents
3 rank-1 nilpotents
6 invertibles

(The six invertibles are isomorphic as a group to $S_{3}.$)

The three rank-1 nilpotents are:

$\begin{bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\1&0\end{bmatrix}, \begin{bmatrix}1&1\\1&1\end{bmatrix}$

These three matrices are linearly independent, but the span of any two of them contains an invertible matrix. Hence the largest subspace of nilpotents is one-dimensional, containing the zero matrix and exactly one of the three matrices above.

I know, it doesn't prove anything. I'm just trying to understand what the problem is. It's obvious that you can have a subspace of dimension $\binom{n}{2}$ - just take all strictly upper triangular matrices. It's also obvious that there are more nilpotent matrices than just those, but in this tiny example, we couldn't make a larger subspace.

Side question: what is the relationship between the linear span of all nilpotent matrices and the set of matrices of trace zero? In this tiny example, they were the same.

the "side question" question has already been answered on the forum: they are the same.(make the matrix with zero trace a matrix with null diagonal trough a similarity)


Find all integer solutions (n,m) to -
$ n^4\plus{}2n^3\plus{}2n^2\plus{}2n\plus{}1 \equal{} m^2$

Hello,

not sure about this, but if we factor the left side of the equation, we obtain

$ (n \plus{} 1)^2(n^2 \plus{} 1) \equal{} m^2$

Now $ n^2 \plus{} 1$ needs to be perfect square, because $ (n \plus{} 1)^2$ and $ m^2$ are perfect squares.

From $ n^2 \plus{} 1 \equal{} x^2$ we get $ n \equal{} 0$ and $ x \equal{} \plus{} \minus{} 1$, from there $ m \equal{} \plus{} \minus{} 1$

And second solution would be for $ m\equal{}0$, then we have $ n\equal{}\minus{}1$.

That's a great way to solve it!

This problem is factoring practice.

[hide="How to Factor"][b]Way 1:[/b]
$ n^4 \plus{} 2 n^3 \plus{} 2 n^2 \plus{} 2 n \plus{} 1$
$ \equal{} (n^4 \plus{} 2 n^3 \plus{} 3 n^2 \plus{} 2 n \plus{} 1) \minus{} n^2$
$ \equal{} (n^2 \plus{} n \plus{} 1)^2 \minus{} n^2$
$ \equal{} (n^2 \plus{} 2n \plus{} 1)(n^2 \plus{} 1)$
$ \equal{} (n \plus{} 1)^2 (n^2 \plus{} 1)$

[b]Way 2:[/b]
$ n^4 \plus{} 2 n^3 \plus{} 2 n^2 \plus{} 2 n \plus{} 1$
$ \equal{} (n^4 \plus{} 2n^3 \plus{} n^2) \plus{} (n^2 \plus{} 2n \plus{} 1)$
$ \equal{} (n^2\plus{}1)(n^2 \plus{} 2n\plus{}1)$
$ \equal{} (n^2 \plus{} 1)(n\plus{}1)^2$
[/hide]

[quote="Tomekk"]Hello,

not sure about this, but if we factor the left side of the equation, we obtain

$ (n \plus{} 1)^2(n^2 \plus{} 1) \equal{} m^2$

Now $ n^2 \plus{} 1$ needs to be perfect square, because $ (n \plus{} 1)^2$ and $ m^2$ are perfect squares.

From $ n^2 \plus{} 1 \equal{} x^2$ we get $ n \equal{} 0$ and $ x \equal{} \plus{} \minus{} 1$, from there $ m \equal{} \plus{} \minus{} 1$

And second solution would be for $ m \equal{} 0$, then we have $ n \equal{} \minus{} 1$.[/quote]
thats exactly what i did .. but i wasnt confident.. now i am sure.. thnx also to TZF


function
integration
topology
real analysis
real analysis unsolved

$f \in L^{1}(\mathbb{R}^{n})$ an integrable function over $\mathbb{R}^{n}$. For $0 \leq s < n$, one can define:
$E_{s}= \{x \in \mathbb{R}^{n}: \limsup_{r \to 0}\frac{1}{r^{s}}\int_{B(x,r)}|f(x)| dx > 0\}$
Prove that the $s$-Hausdorff measure of $A_{s}$ is null.


ratio
geometry
analytic geometry
vector
geometry proposed

Let $ ABC$ be a triangle $ M,P\in AB$, $ N,Q\in AC$, $ \left\{X\right\} \equal{} MN\cap PQ$ and $ \left\{D\right\} \equal{} AX\cap BC$.
If $ \frac {MB}{MA} \equal{} m$, $ \frac {NC}{NA} \equal{} n$, $ \frac {PB}{PA} \equal{} p$, $ \frac {QC}{QA} \equal{} q$ and $ \frac {DB}{DC} \equal{} d$ then prove $ d \equal{} \frac {p \minus{} m}{n \minus{} q}$.

Nice problem. It provides a way to determine the ratio DB/DC without all that Ceva's stuff.

Unfortunately, I couldn't solve it. I've tried many approaches, including extra parallels and area comparison, but nothing worked. [b]Petry[/b], can you please post your solution?

Hello!

My first post is incomplete. I am sorry. :blush:

Let $ ABC$ be a triangle, $ M,P\in\left(AB\right)$, $ N,Q\in\left(AC\right)$, $ \left\{X\right\} \equal{} MN\cap PQ$, $ X\in Int\left(\Delta ABC\right)$ and $ \left\{D\right\} \equal{} AX\cap BC$.
If $ \frac {MB}{MA} \equal{} m$, $ \frac {NC}{NA} \equal{} n$, $ \frac {PB}{PA} \equal{} p$, $ \frac {QC}{QA} \equal{} q$ and $ \frac {DB}{DC} \equal{} d$ then prove $ d \equal{} \frac {p \minus{} m}{n \minus{} q}$.

Solution:

$ \frac {DB}{DC} \equal{} \frac {AB\cdot sin\left(\angle BAD\right)}{AC\cdot sin\left(\angle CAD\right)} \equal{} \frac {AB}{AC}\cdot\frac {\frac {XM\cdot sin\left(\angle AXM\right)}{AM}}{\frac {XN\cdot sin\left(\angle AXN\right)}{AN}} \equal{} \frac {AB}{AC}\cdot\frac {AN}{AM}\cdot\frac {XM}{XN} \equal{}$
$ \equal{} \frac {AB}{AC}\cdot\frac {AN}{AM}\cdot\frac {QA}{QN}\cdot\frac {PM}{PA}$
So $ \frac {DB}{DC} \equal{} \frac {AB}{AC}\cdot\frac {AN}{AM}\cdot\frac {QA}{QN}\cdot\frac {PM}{PA}$   (1)

$ \frac {MB}{MA} \equal{} m\Rightarrow\frac {AB}{AM} \equal{} m \plus{} 1$   (2)
$ \frac {NC}{NA} \equal{} n\Rightarrow\frac {AN}{AC} \equal{} \frac {1}{n \plus{} 1}$   (3)
(2),(3)$ \Rightarrow\frac {AB}{AC}\cdot\frac {AN}{AM} \equal{} \frac {m \plus{} 1}{n \plus{} 1}$    (4)

$ \frac {PM}{PA} \equal{} \frac {AM}{AP} \minus{} 1$   (5)
$ \frac {AM}{AB} \equal{} \frac {1}{m \plus{} 1}$ and $ \frac {AP}{AB} \equal{} \frac {1}{p \plus{} 1}$ $ \Rightarrow\frac {AM}{AP} \equal{} \frac {p \plus{} 1}{m \plus{} 1}$   (6)
(5), (6) $ \Rightarrow\frac {PM}{PA} \equal{} \frac {p \minus{} m}{m \plus{} 1}$   (7)

$ \frac {QA}{QN} \equal{} \frac {1}{1 \minus{} \frac {AN}{AQ}}$   (8)
$ \frac {AN}{AC} \equal{} \frac {1}{n \plus{} 1}$ and $ \frac {AQ}{AC} \equal{} \frac {1}{q \plus{} 1}$ $ \Rightarrow\frac {AN}{AQ} \equal{} \frac {q \plus{} 1}{n \plus{} 1}$   (9)
(8), (9) $ \Rightarrow\frac {QA}{QN} \equal{} \frac {n \plus{} 1}{n \minus{} q}$   (10)

(1), (4), (7), (10) $ \Rightarrow\frac {DB}{DC} \equal{} \frac {p \minus{} m}{n \minus{} q}$.

Best regards, Petrisor Neagoe :)

According to the given ratios, we express the barycentric coordinates of $ M,N,P,Q$ as follows

$ M \ (m: 1: 0) \ , \ N \ (n: 0: 1) \ , \ P \ (p: 1: 0) \ , \ Q \ (q: 0: 1)$

Lines $ MN \equiv x \minus{} ym \minus{} nz \equal{} 0$ and $ PQ \equiv x \minus{} yp \minus{} qz \equal{} 0$ meet at

$ X \ (mq \minus{} np: q \minus{} n: m \minus{} p)$

Hence, ray $ AX$ cuts $ BC$ at $ D \ (0: n \minus{} q: p \minus{} m)$ $ \Longrightarrow$ $ \frac {BD}{DC}\equal{} \frac {p \minus{} m}{n \minus{} q}$

[quote] [color=darkred]Let $ ABC$ be a triangle . Given are the points $ \{M,P\}\subset AB$ (sideline !) and  $ \{N,Q\}\subset AC$ (sideline !) .

Denote $ \left\|\begin{array}{c}
X\in MN\cap PQ\\\\
D\in AX\cap BC\end{array}\right\|$ and $ \frac {\overline {MB}}{\overline {MA}} = m\ ,\ \frac {\overline {NC}}{\overline {NA}} = n\ ,\ \frac {\overline {PB}}{\overline {PA}} = p\ ,\ \frac {\overline {QC}}{\overline {QA}} = q$ .

Then there are the relations  $ \frac {\overline {DB}}{\overline {DC}} = \frac {m-p}{n - q}$ and $ \frac {\overline{XD}}{\overline{XA}}=\frac {mq-np}{(m-n)-(p-q)}$ . [/color] [/quote]

[color=darkred][b][u]Lemma (well-known !)[/u].[/b] Consider in $ \triangle ABC$ the points $ D\in BC$ , $ E\in CA$ , $ F\in AB$

(sidelines !) and $ X\in EF\cap AD$ . Then $ \frac {\overline {FB}}{\overline {FA}}\cdot\overline {DC}+\frac {\overline {EC}}{\overline {EA}}\cdot\overline {BD}=\frac {\overline {XD}}{\overline {XA}}\cdot \overline {BC}\ \ (*)$ .[/color]

[hide="Proof."]
[color=darkblue] Let $ d$ be the line for which $ A\in d\parallel BC$ . Denote $ P\in EF\cap BC$ , $ R\in EF\cap d$ .

Therefore, ${ \frac {\overline {FB}}{\overline {FA}}\cdot\overline {DC}+\frac {\overline {EC}}{\overline EA}}\cdot\overline {BD}=$ $ \frac {\overline{BP}}{\overline {AR}}\cdot\overline {DC}+\frac {\overline{CP}}{\overline{AR}}\cdot \overline {BD}=$

$ \frac {\left(\overline {BD}+\overline{DP}\right)\cdot\overline {DC}+\left(\overline{CD}+\overline{DP}\right)\cdot\overline{BD}}{\overline{AR}}=$ $ \frac {\overline{DP}\cdot\left(\overline {DC}+\overline{BD}\right)}{\overline{AR}}=$ $ \frac {\overline{XD}}{\overline {XA}}\cdot \overline {BC}$ .

In conclusion, $ \boxed {\ \frac {\overline {FB}}{\overline {FA}}\cdot\overline {DC}+\frac {\overline {EC}}{\overline {EA}}\cdot\overline {BD}=\frac {\overline {XD}}{\overline {XA}}\cdot \overline {BC}\ }$ .[/color][/hide]

[color=darkblue][b][u]Remark[/u].[/b] Prove easily that the remarkable relation $ (*)$ is equivalently with $ \frac {\overline {AB}}{\overline {AF}}\cdot\overline {DC}+\frac {\overline {AC}}{\overline {AE}}\cdot\overline {BD}=\frac {\overline {AD}}{\overline {AX}}\cdot \overline {BC}$ .[/color]

[color=darkblue][b][u]Proof of the proposed problem[/u].[/b] Apply the above lemma ([u]I used always  the oriented segments[/u] !) :

$ \left\|\begin{array}{ccc}
\{M,X,N\} & \implies & \frac {MB}{MA}\cdot DC+\frac {NC}{NA}\cdot BD=\frac {XD}{XA}\cdot BC\\\\
\{P,X,Q\} & \implies & \frac {PB}{PA}\cdot DC+\frac {QC}{QA}\cdot BD=\frac {XD}{XA}\cdot BC\end{array}\right\|\implies$

$ \frac {XD}{XA}\cdot BC=m\cdot DC+n\cdot BD=p\cdot DC+q\cdot BD$ $ \implies$

$ (n-q)\cdot BD=(p-m)\cdot DC\implies \boxed {\ \frac {\overline{DB}}{\overline{DC}}=\frac {m-p}{n-q}\ }$ . Observe that $ \frac {XD}{XA}\cdot BC=m\cdot DC+n\cdot BD\implies$

$ \frac {XD}{XA}\cdot (BD+DC)=m\cdot DC+n\cdot BD\implies$ $ \frac {XD}{XA}\left(\frac {BD}{DC}+1\right)=m+n\cdot\frac {BD}{DC}\implies$

$ \frac {XD}{XA}\left(-\frac {m-p}{n-q}+1\right)=m-n\cdot\frac {m-p}{n-q}\implies$  $ \boxed {\ \frac {\overline{XD}}{\overline{XA}}=\frac {mq-np}{(m-n)-(p-q)}\ }$ . [/color]

Take A as origin of a system of vectors. All single capital letters below mean the position vector from A (hence W means the vector AW). Given the problem conditions,
M = B/(m + 1), P = B/(p + 1), Q= C/(q + 1), R = C/(r + 1)   ( 1 )
The equation of the line MN is given by the position vector of a moving point X from this line:
X = kM + (1 - k)N  ( 2 ) while the equation of PQ is X = lP + (1 - l)Q  ( 3 ).
Replace in (2) and (3) the relations (1) and, to get the position of the common point X of the lines MN and PQ, we have to equalize the coefficients of B and C: X[(n+1)(p+1) - (m+1)(q+1)] = (n-q)B + (p-m)C  ( 4 ).
The equation of BC is given by the position vector of the moving point D: (d + 1)D = B + dC  ( 5 ), while the equation of AX is given by a moving point D from AX: D = eX ( 6 ). The point common point of AX and BC will be found by equalizing the coefficients of B agardsnd C, from this we find d = (p - m)/(n - q), while $ \boxed {\ \frac {\overline{XD}}{\overline{XA}}\equal{}\frac {mq\minus{}np}{(m\minus{}n)\minus{}(p\minus{}q)}\ }$ .

Best regards,
sunken rock


Let p denote a prime number of the form 4k+3 and let x be any positive integer. Prove that the integer $ \displaystyle x^2  -xy+\frac {p+1}{4}y^2 $  does not have any divisor of the form $kp-1$.

I remember that grobber posted this problem not long time ago and I solved it. Check it in the solved problems forum.


limit
LaTeX
real analysis
real analysis solved

Compute the limit:

$\displaystyle \lim_{n \to \infty} \sum_{k=1}^{n} \arccos \frac{1-\sqrt{(k-1)k(k+1)(k+2)}}{k(k+1)}$.

have fun! :D

Use $\arccos(xy+ \sqrt {1-x^2}\sqrt{1-y^2})= \arccos(y) - \arccos(x)$ with $x=1/k$, $y=1/(k+1)$

you will get a telescoping sum

[b] Moderator edit: Edited for $\LaTeX$[/b]


algebra
polynomial
number theory
greatest common divisor
algebra solved
Polynomials

Fin all polynomial solutions of the functional equation
\[f(x)f(x+1) = f(x^2+x+1)\]

If $f$ is constant, then $f(x) = 0, 1$.

Otherwise, it has some complex zero $z$ of maximal modulus.  Then setting $x=z$ gives $f(z^2+z+1) = 0$ and setting $x=z-1$ yields $f(z^2-z-1)=0$.  Obviously, $z=0$ is a contradiction because then it gives another root $x=1$.  Further, if $z^2+1 <> 0$, ${z, -z, z^2+z+1, z^2-z+1}$ is a ||gram whence $|z^2+z+1|>|z|$ or $|z^2-z-1|>|z|$, either one of which contradicts our initial assumption.  Therefore, $f(x) = (x^2+1)^n g(x)$ with $\text{gcd}(x^2+1, g(x))=1$ where $g$ is some polynomial.  We easily see that $g$ also satisfies the original equation.  But since $\text{gcd}(x^2+1,g(x))=1$, $g(x)=1$.

It follows that $f(x)=0$ and $f(x)=(x^2+1)^n$ for all non-negative integers $n$.


let a,b,c >0:abc=1.prove that
sum\frac{a^3c}{(b+c)(c+a)}\ge\frac{3}{4}

Bad
See at http://www.mathlinks.ro/Forum/viewtopic.php?p=954176#954176
[quote="rem"]After clearing denominators we obtain:
$ 4\sum_{cyc}(a^4c \plus{} a^3cb)\ge 3(\sum_{cyc}(a^2b) \plus{} \sum_{cyc}(a^2c) \plus{} 2abc)$ multiply $ LS$ by $ 1 \equal{} a^{\frac {2}{3}}b^{\frac {2}{3}}c^{\frac {2}{3}}$
$ \Leftrightarrow 4\sum_{cyc}(a^4c) \plus{} 4\sum_{cyc}(a^3cb)\ge 3\sum_{cyc}(a^{\frac {8}{3}}b^{\frac {5}{3}}c^{\frac {2}{3}}) \plus{} 3\sum_{cyc}(a^{\frac {5}{3}}c^{\frac {8}{3}}b^{\frac {2}{3}}) \plus{} 6a^{\frac {5}{3}}b^{\frac {5}{3}}c^{\frac {5}{3}}$
$ \Leftrightarrow \sum_{cyc}(a^4c \plus{} a^3bc \plus{} b^4a \minus{} 3a^{\frac {8}{3}}b^{\frac {5}{3}}c^{\frac {2}{3}}) \plus{} \sum_{cyc}(a^4c \plus{} a^3bc \plus{} c^3ab \minus{} 3a^{\frac {8}{3}}b^{\frac {2}{3}}c^{\frac {5}{3}}) \plus{} (a^4c \plus{} b^4a \plus{} c^4b \minus{} 3a^{\frac {5}{3}}b^{\frac {5}{3}}c^{\frac {5}{3}}) \plus{} (a^3bc \plus{} b^3ca \plus{} c^3ab \minus{} a^{\frac {5}{3}}b^{\frac {5}{3}}c^{\frac {5}{3}})\ge 0$  (*)
Each of the expressions in the brackets is greater than or equal to $ 0$ by AM-GM and therefore inequality (*) is true, hence the original inequality is true.[/quote]


Let triangle ABC.
[b] Pro1[/b]
Prove that $ m_a\plus{}m_b\plus{}m_c\geq p\sqrt{3}$

[b] Pro2[/b]
Prove that $ 8m_am_bm_c\leq\sqrt{(2a^2\plus{}bc)(2b^2\plus{}ca)(2c^2\plus{}ab)}$
good luck@

[quote="Algadin"]Let triangle ABC.
[b] Pro1[/b]
Prove that $ m_a \plus{} m_b \plus{} m_c\geq p\sqrt {3}$
[/quote]
I think, it's not true. Try $ a\rightarrow b \plus{} c$ and $ b \equal{} c.$ :wink:
[b] Pro2[/b]
[quote="Algadin"]Prove that $ 8m_am_bm_c\leq\sqrt {(2a^2 \plus{} bc)(2b^2 \plus{} ca)(2c^2 \plus{} ab)}$
good luck@[/quote]
Yes, it's true, but my proof is very ugly: after squaring I used SOS.

OK. My ugly proof:

$ 8m_am_bm_c\leq\sqrt {(2a^2 \plus{} bc)(2b^2 \plus{} ca)(2c^2 \plus{} ab)}\Leftrightarrow$
$ \Leftrightarrow(2a^2 \plus{} 2b^2 \minus{} c^2)(2a^2 \plus{} 2c^2 \minus{} b^2)(2b^2 \plus{} 2c^2 \minus{} a^2)\leq(2a^2 \plus{} bc)(2b^2 \plus{} ac)(2c^2 \plus{} ab)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(4a^6 \minus{} 6a^4b^2 \minus{} 6a^4c^2 \plus{} 4a^3b^3 \plus{} 2a^4bc \plus{} 2a^2b^2c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(2(a^6 \minus{} a^4b^2 \minus{} a^2b^4 \plus{} c^6) \minus{} 4(a^4b^2 \minus{} 2a^3b^3 \plus{} a^2b^4) \minus{}$
$ \minus{} 4(a^3b^3 \minus{} a^2b^2c^2) \plus{} 2(a^4bc \minus{} a^2b^2c^2))\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \minus{} b)^2(2(a \plus{} b)^2(a^2 \plus{} b^2) \minus{} 4a^2b^2 \minus{} 2(ab \plus{} ac \plus{} bc)c^2 \plus{} (a \plus{} b \plus{} c)abc)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \minus{} b)^2(2a^4 \plus{} 4a^3b \plus{} 4ab^3 \plus{} 2b^4 \plus{} a^2bc \plus{} b^2ac \minus{} c^2ab \minus{} 2c^3a \minus{} 2c^3b)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \minus{} b)^2S_c\geq0.$
Let $ a\geq b\geq c.$ Then $ S_c\geq0,$  $ S_b\geq0$ and $ (a \minus{} c)^2\geq(b \minus{} c)^2.$
Hence, $ \sum_{cyc}(a \minus{} b)^2S_c\geq(a \minus{} c)^2S_b \plus{} (b \minus{} c)^2S_a\geq(b \minus{} c)^2(S_a \plus{} S_b) \equal{}$
$ \equal{} 2(b \minus{} c)^2\left((a \minus{} b)^2(a^2 \plus{} ab \plus{} b^2) \plus{} (a^3 \plus{} b^3)c \plus{} abc^2 \plus{} 2(a \plus{} b)c^3 \plus{} 2c^4\right)\geq0.$
See also the nice proof:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=151719
I  forgot about it! :D


calculus
function
real analysis
Harvard
college
vector
topology

Could one study single and multivariable calculus from a standard textbook like stewart, then go on to linear algebra, and then use Spivak's book as an intro to analysis text (i.e. followed by a book about Real Analysis say by Royden?). Wouldn't it make sense to read the book once you know how to do proofs properly?

For starters, I'm not a big fan of Royden, but it seems to me like the jump in sophistication between Spivak and Royden is probably far too big.  You need some exposure to things like metric spaces and a good treatment of sequences and series of functions before you jump to measure theory.

So what would you recommend a student do if he is taking the "plug and chug" calculus II course? Should he go on and study calc III from a standard text like Stewart, and then go to Spivak? Or should he go to Spivak before going on to multivariable calculus? What is the standard sequence of subjects/courses that a math major studies (examples from your colleges?)

The textbook sequence at my university goes something like

Calc I-III : Stewart
Intro to Analysis I : (generic introduction to analysis text)
Intro to Analysis II : Blue Rudin
Real Analysis : Folland (measure theory)

I don't think that this is the best possible way of doing things (either of the first two texts could certainly be omitted for strong students), but the jump from your generic intro analysis text (e.g, without metric spaces) to abstract measure theory is huge, and you need a good, tough text to bridge that gap (and I think Rudin does a spectacular job of precisely that).

when do you take linear algebra/ differential equations? And the generic intro to analysis text would be something like Spivak, right?

Linear Algebra, Calculus III, and the first Intro to Analysis course can all be taken around the same time.  There is a second linear algebra course that can be taken after you have an abstract algebra course under your belt.  Differential equations can be taken as soon as you have linear algebra and calculus II.

This was the text I used for Intro to Analysis I; it's alright for what it tries to do, but is super basic. [url=http://www.amazon.ca/Analysis-Introduction-Steven-R-Lay/dp/0131481010/sr=1-3/qid=1161761449/ref=sr_1_3/702-8660062-3334416?ie=UTF8&s=books]link[/url]

so the spivak, apostol, and courant calculus books arent really required then? the intro to analysis course basically substitutes these books then? In other words, going back to Spivak right now would be a waste of time when I could go and learn linear algebra/ intro to real analysis?

My understanding is that Spivak, despite the title, is (at least) on the same level as most introductory analysis texts.  I don't have a copy, however, so there are people who could speak more knowledgably on the topic than I.

You probably could start picking up linear algebra now, though.

I recommend Pugh's book [i]Real Mathematical Analysis[/i] for an undergraduate analysis text. He doesn't assume too much knowledge, he writes really well, and he has lots of interesting topics and problems.

And linear algebra is always a good idea.

I was somewhat surprised to read you all talking about Spivak for analysis, because I am homeschooled and using Spivak to learn Calculus I and II.  Is that a bad idea?

Spivak's Calculus is a great book, period. My son is in a calculus course based on the Stewart textbook, but he has been reading along in Spivak for supplementation for the past three semesters. It is never a bad idea to revisit the "same" topics from the point of view of more than one author, especially if the author you learn from is Spivak. There used to be a tendency to use Spivak's textbook for a first calculus course for first-year university students, back when few high school students took calculus and high school mathematics courses emphasized proof more than they do now. Today, because most strong math students have a watered-down calculus course in high school, Stewart has become the dominant college calculus textbook, and Spivak is used, if at all, mostly as a first real analysis textbook. But Spivak's textbook covers a lot of ground from a very higher math perspective, and is useful any time you haven't MASTERED everything in it.

I appreciate blahblahblah's comments on other textbooks and on a typical sequence of courses beyond calculus II. My son's second semester of calculus II will use the Bretscher linear algebra textbook, which was developed when Bretscher taught at Harvard, and which is a very understandable, interesting "reform" linear algebra textbook.

Another book that would fit into the years just after calculus is Hubbard and Hubbard's Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach. At my son's university, that book is used in an honors mathematics course for math majors that follows the calculus sequence he is now in. That is a very beautiful and very interesting book.

I agree little Rudin is still the king of introductory real analysis books. It's not easy, but it's a classic.

What do you think about Courant's Calculus books? I have heard that Spivak actually tried to copy his style (i.e. Spivak learned from Courant). Also, would you recommend doing every single problem? Or follow a sample syllabus (i.e. U of Chicago has a syllabus using Spivak with questions and readings)

One caution after reading tokenadult's post: there never really was a mythical golden age in which high school students arrived at college comfortable with proofs and Spivak was the dominant calculus book.

Spivak's [i]Calculus[/i], and its rival, Apostol, only sold in one very narrow market niche: proof-intensive "honors" calculus courses. I'll illustrate with my case: I was a freshman in 1971, at a highly selective, math/science/engineering intensive college (Rice), and I took that honors calculus course. Perhaps 20 people lasted the year in that course, out of a freshman class of 600, 400 of whom were in science or engineering. (In this particular case, the textbook for the course was Apostol in the first semester and no textbook at all in the second semester. But that's the same market as for Spivak.)

The "regular" calculus course, the one taken by the other nearly 400 freshmen, used Thomas for a textbook - that would be the "Alternate Edition" of Thomas, and that truly was the dominant calculus book in the U.S. at the time. The Rice faculty had a couple of theoretical add-on's (in supplemental notes) to Thomas, but, to be honest, the add-on's didn't have all that much impact on the students. The sentiment of "Can we get something like Thomas, only a little bit more rigorous" brought forth various other texts - one I remember was by Salas and Hille. And there have been lots of "mainstream" calculus books: Swokowski, Edwards & Penney, and so on. Stewart has the dominant market position now; it's a descendent of Thomas in some ways.

The honors sequence continued for a second year, with one semester being mostly a differential equations course and the other being a vector calculus course that used two textbooks: one by Fleming, and the other Spivak's [i]Calculus on Manifolds[/i]. (I'm not familiar with the Hubbard and Hubbard book that tokenadult mentions - sounds interesting.) I'll also say this: marvelous as that skinny little Spivak book is in hindsight, I did [i]not[/i] really understand or grasp it at the time I took the course. And then I took an abstract algebra course (Herstein's [i]Topics in Algebra[/i]), and then finally another analysis course: upper division undergraduate number, "Real Analysis" as a title, and Royden-level measure and integration course.

I survived, of course, and went on to graduate school, and became an analyst. But there are a lot of holes there. I didn't really learn the "techniques of integration" part of the calculus course until I became a tutor/grader for the regular calculus course over the next two years. To this day, I'm not sure where or when I got an understanding of uniform convergence, or of metric spaces - some of that would have been in the freshman course, but how much? And the only real exposure to linear algebra I got as an undergraduate was the back half of Herstein's book - and that's no place to start. I think I learned Gaussian elimination the first time I had to teach it, to business students.

Of course, I now teach a sophomore-level introductory linear algebra course fairly frequently, and a junior level "real analysis" course: basic epsilontics, continuity, the Riemann integral, uniform convergence, metric spaces, and so on. At this point the text I'm most likely to use is the little Springer book by Kenneth Ross. So it doesn't have every topic in it that I want to teach? I'm comfortable with adding my own notes when needed - better that than swamping students with more than they need.

The books I used to learn calculus/multivariable calculus/linear algebra/analysis were:

High School:
For Calculus I and II (introduction to Calculus): Thomas' [u]Calculus[/u], Barron's [u]AP Calculus[/u], and Shaum's Outline of [u]Advanced Calculus[/u] (this book is awesome. its great for anybody who's taken basic calc and wants to see some more advanced and challenging calculus problems.)

For Multivariable Calculus/Calculus III: Marsden and Tromba's [u]Vector Calculus[/u] (I really liked this book. I finally understood physics!!)

College:
Multivariable Calculus and Linear Algebra: Apostol, [u]Calculus, Vol II[/u] (I found this Apostol book to be pretty difficult, but not impossible. It helped a lot that I had done the Marsden and Tromba book before I had this course, otherwise I don't think I would have [i]really[/i] had an intuitive understanding of multivariable calculus. Definitely better than Spivak's [u]Calculus[/u] as an introduction to rigorous mathematics because it goes beyond just single variable calculus. I really like the linear algebra part of Vol. II also. Its not as "algebraic" in its approach as some other books, but it integrates well with the "Multivariable Calculus" topics.)

Introduction to Analysis: Rudin, [u]Principles of Mathematical Analysis[/u] (loved this book. it prepared me well for later courses in analysis) and Spivak, [u]Calculus on Manifolds[/u] (I understood very little of this book at the time. It wasn't until I took a later course in Differential Geometry that I really understood what it was talking about).

Real Analysis: Royden's [u]Real Analysis[/u] (I couldn't understand this book at all. I ended up dropping the class), Rudin's [u]Real and Complex Analysis[/u], and Zygmund's [u]Measure and Integration[/u] (I understood Real Analysis much better when I took the class using Rudin and Zygmund's books. It also helped that I took a class in basic topology before trying to take Real Analysis again. I have a feeling that none of these books are [i]ideal[/i] for learning Real Analysis for the first time, but I couldn't recommend a better book specifically).

As the last two posts mentioned Spivak's Calculus on Manifolds, I would like to ask about it...

I am taking a course using this book, we are working on Chapter 4, and I'm highly confused. What is differential forms actually? (sorry if it's a bit off topic) Spivak has defined many things, he has done many tedious and horrible calculations, and he said stoke's theorem falls out. I don't understand the logic behind. Does that mean differential form is simply something created to match the classical theory? Or what is it at all?

(I don't know differential geometry at all)

For Rudin's RCA, I don't know if it's because it's the first time I really touch these stuff, but I simply can't grasp the intuition behind at all.. (I have just finished the first two chapters)

Read the preface. This is simply the nature of the subject.

The difficulty of Stokes' theorem is entirely in the definitions. Stokes' theorem for a box in $\mathbb{R}^{n}$ is an easy calculation; for any other region, we need to work even to define an integral. Differential forms make more sense in the context of smooth manifolds (introduced Ch. 5), where we know things look like $\mathbb{R}^{n}$ locally, but there is no canonical choice of coordinates. In the manifold context, we know what it means for a function to be differentiable, but not what a derivative actually is; the usual definitions in $\mathbb{R}^{n}$ depend too much on specific coordinates. Trying to match the classical theory without coordinate dependence naturally leads to differential forms.
Once we make the appropriate definitions of orientation, integration, and everything else, it turns out everything we can integrate on is diffeomorphic to a union of finitely many boxes, and the integrals are diffeomorphism-invariant.

For what it's worth, Folland's Real Analysis text (read: measure theory) seems to have become the most prevalent text of its type.

I found it a very hard text to learn from (the proofs are hard to read, typos abound, and whatnot), but the organization is good, the exercises are hard, and I don't think you can choose the topics any better than he did (roughly speaking: measures, integration, differentiation, topology, functional analysis, L^p spaces, fourier analysis, distributions, pde, probability, hausdorff measure).

At Harvey Mudd, we use Spivak in our accelerated single-var calculus course. That is, it is definitely ok to be learning calculus from Spivak.

Two words: Tom Apostol.

Tom Apostol's "Calculus Volume 1" ought to be used as an intro to calculus, and when you are done with it, you are ready for real analysis. It's very rigorous but does not assume that the reader has already been through a calculus course.

This is the book we use here at Carnegie Mellon as part of the Honors math track during freshman year. Sophomore year we proceed on to a math 55-like class that covers the rest of undegrad math, i.e. real analysis, in single and multiple variables, and algebra. Highly recommend it even though our prof makes jokes about T. Apostol non-stop in lecture.

I'm a fan of using Marsden's [i]Elementary Classical Analysis[/i] as an intro-to-analysis text.  It takes a metric space approach and the writing is clear enough that one can self-study from it.

tokenadult, I've recently heard some negative things about Bretscher (haven't looked at it in detail myself) regarding the text not treating general dimension or general vector spaces well.  I believe that Andrilli & Hecker's book is comparable in level (it's what I use, and is extremely readable), and might be worth checking out from the library to see if it's a good complement/supplement to Bretscher.

Thanks for the recommendation of the Andrilli & Hecker linear algebra book. I have never seen it. Through the kindness of emeritus professor A. Wayne Roberts of Macalester College, I now have some linear algebra books he was clearing from his office. Bretscher is good, but I agree it actively avoids the more abstract topics. Better for function spaces is Olver and Shakiban. I happen to have the Lay textbook too, from a different source, but that is indeed elementary. And I really like how Hubbard and Hubbard tie together linear algebra and vector calculus in their textbook on those topics.

I need to learn a lot of linear algebra for my research on psychology, so I am always glad to hear about current linear algebra books, as well as about acknowledged classic titles.

Axler's [i]Linear Algebra Done Right[/i] is a nice little (cheap!) linear algebra text, which is more rigorous than the usual texts (enough so that Noam Elkies used it when he taught Math 55).  It's written by an analyst as well so it handles things that segue naturally into functional analysis (e.g. 'function spaces') pretty well.  I generally don't think function spaces are that important to linear algebra texts, since infinite-dimensional vector spaces are usually the domain of functional analysis, and require significantly more of an analysis background than can be reasonably assumed of an linear algebra student.

It does lack some things (dual spaces, rational canonical forms, etc) but these can be found in Hoffman and Kunze or Dummit and Foote for example.  One linear algebra text which I suspect is very, very good (and significantly more advanced than any of the previously mentioned texts) is Peter Lax's text, but I haven't had a chance to look at it yet.


trigonometry
ratio
rotation
geometry unsolved
geometry

A triangle $ABC$ is given in the plane with $AB = \sqrt{7},$ $BC = \sqrt{13}$ and $CA = \sqrt{19},$ circles are drawn with centers at $A,B$ and $C$ and radii $\frac{1}{3},$ $\frac{2}{3}$ and $1,$ respectively. Prove that there are points $A',B',C'$ on these three circles respectively such that triangle $ABC$ is congruent to triangle $A'B'C'.$

Let $ a = BC = \sqrt {13},\ b = CA = \sqrt {19},\ c = AB = \sqrt 7$ be the triangle sides opposite to the vertices A, B, C. Let F be the 1st Fermat point of the triangle $ \triangle ABC$ and x = AF, y = BF, z = CF. The angles $ \angle AFB = \angle BFC = \angle CFA = 120^o$ are all equal. Using the cosine theorem for the triangles $ \triangle AFB, \triangle BFC, \triangle CFA$ with the sides (x, y, c); (y, z, a); (z, x, b) and the angles $ \phi = \widehat{(x, y)} = \widehat{(y, z)} = \widehat{(z, x)} = 120^o$, $ \cos \phi = - \frac 1 2$,

$ \triangle AFB: \ \ x^2 + y^2 + xy = c^2 = 7$

$ \triangle BFC: \ \ y^2 + z^2 + yz = a^2 = 13$

$ \triangle CFA: \ \ z^2 + x^2 + zx = b^2 = 19$

Subtracting the 1st equation from the 2nd one, 2nd from 3rd and 3rd from 1st,

$ z^2 - x^2 + yz - xy = (z - x)(x + y + z) = a^2 - c^2 = 6$

$ x^2 - y^2 + zx - yz = (x - y)(x + y + z) = c^2 - b^2 = - 12$

$ y^2 - z^2 + xy - zx = (y - z)(x + y + z) = b^2 - a^2 = 6$

Dividing any pair of equations always yields nothing else but z = 2x - y. Substistuting z into the original set of equations,

$ x^2 + y^2 + xy = 7$

$ 4x^2 + y^2 - 2xy = 13$

$ 7x^2 + y^2 - 5xy = 19$

The 3rd equation is a linear combination of the 1st and 2nd, hence, we can disregard it: $ (3) = 2 \times (2) - (1)$. Substracting the 1st equation from the 2nd one, in order to eliminate at least one square, we get

$ 3x^2 - 3xy = 6,\ \ y = x - \frac 2 x$

Substituting y into the 1st equation,

$ x^2 + x^2 - 4 + \frac {4}{x^2} + x^2 - 2 = 7$

$ 3x^4 - 13x^2 + 4 = 0$

$ x^2 = \frac {13 \pm \sqrt {169 - 48}}{6} = \frac {13 \pm 11}{6} = 4$ [color=white].[/color] or [color=white].[/color] $ \frac 1 3$

Since we are not interested in negative roots, x = 2 or $ \frac {\sqrt 3}{3}$. Calculating y, z from the previous equations

$ y = x - \frac 2 x = 1$ [color=white].[/color] or [color=white].[/color] $ - \frac {5 \sqrt 3}{3}$

$ z = 2x - y = 3$ [color=white].[/color] or [color=white].[/color] $ \frac {7 \sqrt 3}{3}$

Since we are interested only in the solution with x > 0, y > 0, z > 0, the distances of the 1st Fermat point from the vertices A, B, C are

x = AF = 2, y = BF = 1, z = CF = 3 and x : y : z = 2 : 1 : 3

On the other hand, radii of the circles $ (A), (B), (C)$ are in the ratios $ r_A : r_B : r_C = \frac 1 3 : \frac 2 3 : 1 = 1 : 2 : 3$. I have solved the problem before and it came with exactly the same error - it should be $ r_A = \frac 2 3, r_B = \frac 1 3,\ r_C = 1$, which implies $ r_A : r_B : r_C = 2 : 1 : 3$. If this was the case, then

$ \frac {r_A}{x} = \frac {r_B}{y} = \frac {r_C}{z} = \frac 1 3$

i.e., the circles $ (A), (B), (C)$ would be seen from the 1st Fermat point F under the same angle $ 2 \theta = 4 \arcsin{\frac {r_A}{2x}} = 4 \arcsin{\frac 1 6}$. If the triangle $ \triangle ABC$ was then rotated by the angle $ \pm \theta = \pm 2 \arcsin{\frac 1 6}$ around the point F into a congruent riangle $ \triangle A'B'C'$, the vertices $ A', B', C'$ would simultaneously fall on the circles $ (A), (B), (C)$.

Alternately, the problem could be cured by keeping the radii of the circles $ (A), (B), (C)$ as $ r_A = \frac 1 3,\ r_B = \frac 2 3,\ r_c = 1$, but specifying the triangle side lengths as $ a = BC = \sqrt {19},\ b = CA = \sqrt {13},\ c = AB = \sqrt 7$. Then we would get x = AF = 1, y = BF = 2, z = CF = 3 and $ x : y : z = 1 : 2 : 3 = r_A : r_B : r_C$.

EDIT: See [url]http://www.mathlinks.ro/viewtopic.php?t=194847[/url] for the correct solution.