[quote="Mather"]1. Cho a-b+c=0.CMR: asinx+9bsin3x-25csin5x=0. CMR co it nhat 4 nghiem phan biet thuoc (0;2pi)
2. CMR: a(25sin5x-sinx)+b(49sin7x-9sin3x)=0 co it nhat 7 nghiem phan biet thuoc (0;2pi)

Ai nguoi Viet Nam dich. ho^ nhe'[/quote]

[color=darkblue]He is a new member, He can't post by English.

This is his problem:

$ 1$. Let $ a \minus{} b \plus{} c \equal{} 0$. Prove that this equation: $ asinx \plus{} 9bsin3x \minus{} 25csin5x \equal{} 0$, has at least 4 roots distributive $ \in (0; 2\pi)$.

$ 2$. Prove that this equation: $ a(25sin5x \minus{} sinx) \plus{} b(49sin7x \minus{} 9sin3x) \equal{} 0$, has at least 7 roots distributive $ \in (0; 2\pi)$.[/color]


You need to hold your breath to get out of a room. Assume that a room is in
the shape of a cube that was 5 inches on each side. You are located 1 inch
from the top and 2 inches from the side of one wall. The opening is on
the opposite wall 3 inches from the top and one inch from the side. Also, assume
you can hold your breath for five more seconds and can crawl 2 inches per
second. Are you able to escape? What is the shortest pathway out?

The problem is kind of obscure.
[hide]
Shortest path $3$.
Yes you can escape.
But that does not seem valid.
[/hide]

"You are located 1 inch from the top and 2 inches from the side of one wall". So if you put this in form
([i]x, y, z[/i]), you gave the x coordinates and the y coordinated but you never gave the z coordinates.

Also, you can "crawl" 2 inches per second. Don't you mean fly 2 inches per second? Because that looks like you have to travel in air.


An equilateral triangle has side length 6. What is the area of the region containing all points that are outside the triangle and not more than 3 units from a point of the triangle?

3 120 degree circle sectors with radii 3 and 3 3x6 rectangles:
$ (3)\dfrac{120}{360}3^2\pi \plus{} 3(3)(6) \equal{} 9\pi \plus{} 54$

2008 AMC 10 Problem 17

Surrounding the circle, there will be three 3*6 rectangles and at each corner there will be a 1/3 of a circle with radius 3. Therefore, area=   3(3x6) + 3(1/3pi3²) = 54 + 9pi


This song makes me laugh everytime i listen to it.

http://viknluda.com/foe/video/

hehehehe
 :rotfl:
"Oh my god look she's a white!"
So racist yet so funny.

[quote="Klebian"]So racist yet so funny.[/quote]

So we're locking this discussion.


Find all primes p such that each nonzero element of Z (little p) is its own multiplicative inverse.  You must prove that the primes you find work and are the only ones that work.

Is there something wrong with me, or is it trivial? For all $x \in \{1, 2, ..., p-1\}$ we must have $x^2 \equiv 1 \mod{p}$ but that's equivalent to $(x-1)(x+1) \equiv 0 \mod{p}$ so, all residues $\mod{p}$ are $1$ or $-1$ which implies that $p=2$ and $p=3$ are only primes which work.

for $p=2,3$ it obviously works

now $\mathbb{Z}_p$ is a field and $a=a^{-1}$ implies $a^2=1$ which can not have more than two solution, implying $p-1 \leq 2$ and we are done

Well, it doesn't get harder for nonprimes (looking at the invertible elements then):
When any number $k$ divides it, by reducing $\mod k$ shows that also $k$ itself works. Thus neither a prime $>3$ nor $9$ nor $16$ can divide a working number (the first by the above post, the other two by simple checking).
Thus the only numbers left are $2,3,4,6,8,12,24$, and all those work.


algebra
polynomial
function
functional equation
algebra proposed

Find all solutions $ f$: $ R$ → $ R$ of the functional equation
$ f(f(x) \plus{} f(y) \plus{} kxy) \equal{} xf(y) \plus{} yf(x)$, x, y ∈ R.
Consider the case when f is (a) a polynomial, (b) a continuous function, (c) an arbitrary
function.

Dear ghostchungho,

You are not allowed to post this functional equation.
It is a part of the list of problems suggested by the Jury of the International Tournament of Young Mathematicians (visit http://www.itym.org/). These problems are not to be widespread on any forum and it is prohibited to discuss or publish any possible solutions to them before the 6th of July 2009.

All rights reserved.

Sincerely,
The ITYM Organizing Committee

I'm so sorry!
So it is a probleme of a competition!!! :lol:

Apologies accepted.

Indeed, it is a research-type problem from our competition.


MATHCOUNTS
geometry
3D geometry
probability

Find a two-digit number such that the square of the sum of its digits is equal to the number.
Find a three-digit number such that the cube of the sum of its digits is equal to the number.
Generalize. (Find a x-digit number such that the number^x of the sum of its digits is equal to the number.

...*yawn* This'll be the last problem of the day, good night.
Announcer: And the record holds at 15! 15 folks! :lol:

[hide]1) 81. You know the number has to be a square number due to the fact that the sum of the digits is squared.

2) 512. Same theory as above. The number has to be a perfect cube.

3) No clue. Haven't learned how to generalize yet. Would someone please explain it to me?[/hide]

You're correct, maybe even for the third one. I was so sleepy when I wrote this problem that I didn't quite check if there existed a generalization :)

meh. basically just apply what you did but with variables!

you know how in this problem, it says (2-digit) and (3-digit)? Well, that's pretty specific, choosing one particular integer from the many many integers out there, so we generalize to an n-digit integer. I don't think a nice compact formula exists for this problem, nor am I sure if one such number always exists, but that would be a possible generalization of this problem.

[quote="MathFiend"]But how [b]do[/b] you generalize? You ask for generalizatoions on a lot of the other problems you post, but I've never been taught how to do them.[/quote]
Notice that I usually put two problems when I say "Generalize"
Just find the number that changed and replace them with x or y or whatever and solve the problem in term of x and y and such.
Example: I flip a coin 2 times. What is the probability of getting all heads? Answer: 1/2:^2:=1/4
I flip a coin 5 times, What is the probability of getting all heads? Answer: 1/25=1/32
Generalize. :arrow: I flip a coin x times, What is the probability of getting all heads? Answer: 1/2x

I'll go and change all the problem that just says "Generalize" into a question with variables just in case.

Could someone post a solution to this? It's a very interesting problem. I doubt it's possible, though.

Actually, it ONLY works for one more: 7^4's digits add up to 7.
After that, it never works. When you get to 5, only 7^5, 8^5, and 9^5 have 5 digits, so it's easy to check all those.

Curiously, for n>21, NO integer has n digits when you take it to the nth power...


How do you convert a base n number to base n^2?

Like how would you quikly convert (for example) 110110 in base 3 to a base 9 number?

Well, let $ N$ be the number in base $ b$, and let the digits be $ d_n,d_{n\minus{}1},\cdots,d_1,d_0$, so \[ N\equal{}d_0\plus{}b\cdot d_1\plus{}b^2\cdot d_2\plus{}b^3\cdot d_3\plus{}\cdots \plus{} b^n d_n\]
To convert to base $ b^2$, observe that \[ N\equal{}(d_0\plus{}b\cdot d_1)\plus{}b^2(d_2\plus{}b\cdot d_3)\cdots\]

So each digit in the base $ b^2$ number is the combination of two digits in the base $ b$ number, starting at the right side.

In your example, we group 11 01 10 and since it's originally in base 3 the digits are $ (3\cdot 1\plus{}1),(3\cdot 0\plus{}1),(3\cdot 1\plus{}0)$, so in base 9 the number is 413.


[hide]$R_n=\frac{1}{2}(a^n+b^n)=\frac{1}{2}((a+b)(a^{n-1}+b^{n-1})-ab(a^{n-2}+b^{n-2}))=6R_{n-1}-R_{n-2}$.

$R_0=1,\ R_1=3$. Then by recursion, look at the units digit of $R_n$: $R_2\equiv 7,\ R_3\equiv 9,\ R_4\equiv 7,\ R_5\equiv 3,\ R_6\equiv 1,\ R_7\equiv 3\pmod{10}$.

Since the units digits of $R_6$ and $R_7$ are the same as $R_0$ and $R_1$ and the recursion only looks at the last two numbers in the sequence, the units digit cycles every 6.

$12345\equiv 3\pmod{6}\Rightarrow$ units digit is 9 $\Rightarrow \boxed{E}$.[/hide]

[hide]
Let $w=3 + 2\sqrt2 i=r \text{ cis } \theta$ and $v = 3 - 2\sqrt2 i= r \text{ cis} (- \theta)$.  So we want $\frac{\Re(s) + \Im(s)}{2}$ where $s = w^n + v^n$.  By deMoivre, $s = 2r^n \cos n\theta$, so the thing we want just turns into $r^n \cos n\theta$.  We can find $\cos \theta$ by constructing a right triangle with sides $3$ and $2\sqrt2$ and hypotenuse $\sqrt{17}$, getting $\cos \theta = \frac{3}{\sqrt{17}}$ and $\sin \theta = \frac{2 \sqrt2}{\sqrt{17}}$.  Using our definition of $w$ from above, we see $3 + 2\sqrt2i = r \left( \frac{3}{\sqrt{17}} + \frac{2 \sqrt2}{\sqrt{17}}i \right)$, so $r=\sqrt{17}$.  Now we want the units digit of $17^{\frac{12345}{2}} \cdot \cos(12345 \theta)$... and I can't find the latter part.  I edited since I forgot about the 12345 coefficient before, which simplified very nicely and came out to one of the choices (B), but is probably not right...
[/hide]


induction
modular arithmetic
limit
ratio
logarithms
geometry
trigonometry

It seems that Slizzel already took the idea of Intermeidate Problem Solving Marathon so I'm trying to do something even "harder."

This is proof marathon so there will be a proof question and you have to give your answer with full proof. I think it would be really good idea to see a lot of good proof. I'm not sure how hard they have to be as long as someone can answer it. But let's not try to go too hard.

When posting problem, please post problem and hint in case it's too hard. :D And put source as well.

[size=150]Proof 1[/size]

FMO 2001

Prove that for every natural number $n$ greater than 1 satisfies:

$(n-1)^2$ is a divisor of $n^{n-1} - 1$.

[hide="hint"]
I think induction will be nice in this one.[/hide]

I like this proof, although it's probably not one that most people would think of.

We have

${n^{n-1}-1=(n-1)(n^{n-2}+n^{n-3}+\ldots+1})$

So we need to show that the last factor is divisible by $n-1$.  Towards that end, subtracting $2(n-1)$ from the last factor doesn't change anything.  After we do this, the last factor becomes

$n^{n-2}+n^{n-3}+\ldots +n^2-(n-3)$.  This can be rewritten as

$(n^{n-2}-1)+(n^{n-3}-1)+(n^{n-4}-1)+\ldots +(n^2-1)$

and it's obvious that all of those terms are divisible by $n-1$

So we're done!

wouldnt this be more appropriate for Pre-Olympiad??

[quote="Silverfalcon"]I got lost after the word "So" and where you subtracted $2(n-1)$.
[/quote]
blahblahblah's proof is perfectly valid.

To help you to understand his proof, I am trying to explain in a different way:

$n^k\equiv 1\pmod{n-1}$ since $n\equiv 1\pmod{n-1}$.

So ${n^{n-2}+n^{n-3}+\cdots+1}\equiv 1+1+\cdots+1\equiv n-1 \equiv 0 \pmod{n-1}$

Yes, the key is to break up $(n-3)$ into $(n-3)$ ones.  We can do this because the numbers of the form $n^j$ are indexed by $j=2$, $j=3$, $j=n-4$, $j=n-3$, $j=n-2$, etc.  As I said, it's not obvious, and its certainly not the way I would have done it back in high school when I was starting on proofs, but it's both direct and cute, in my opinion.

[quote="Silverfalcon"]I got lost after the word "So" and where you subtracted $2(n-1)$.

I was thinking of induction.. Except I stuck on $k+1$ because:

$(k+1-1)^2 = k^2$ and $k^{k+1-1} - 1 = k^k - 1$...

This was the final round of FMO... It should be hard.. :D[/quote]

[hide]I think you forgot that you are trying to prove $k^2$ is a divisor of $(k+1)^k-1$.

After this, you can do as blahblahblah did and make $(k+1)^k-1=(k+1-1)((k+1)^{k-1}+(k+1)^{k-2}+...+1)$.

Again, since the $k$ on top will cancel we are proving that $((k+1)^{k-1}+(k+1)^{k-2}+...+1)$ is divisble by $k$.  This obvious since every term will have a $k$ in it except for the $1$ (use binomial expansion).  But then since the first term  will make a $1^{k-1}$ and the second will make $1^{k-2}$ all the way to $1^0$, this adds up to $k$. So thus, it's divisible.[/hide]

Here's a classical one that everyone should like. :)

Uhh...I guess Proof 2:

Prove that there exists an infinite number of primes.

Suppose there are a finite amount of primes
$p_1, p_2, p_3, ... , p_k$
Then $p_1*p_2*p_3*....*p_k+1$ is clearly not divisible by any of the primes.
Thus $p_1*p_2*p_3*....*p_k+1$ is a new prime.
Proved by contradiction

[b]Proof 3:[/b]
Prove that :sqrt: 2 is irrational

Assume $\sqrt{2}$ is rational, so that $\sqrt{2}=\frac{a}{b}$ for relatively prime integers $a,b$.
Then $2=\frac{a^2}{b^2}$, and $a^2=2b^2$. But then $2|a$, so $a=2a_0$ for some integer $a_0$ and $4a_0^2|b^2$, so $2|b$, and $a,b$ are not relatively prime. Hence, $\sqrt{2}$ is not rational.

Proof 4:
Prove that the $n$th triangular number is composite for $n>2$.

[quote="K81o7"]The nth triangular number is equal to (1+2+...+n), which is equal to n(n+1)/2.  This is obviously a whole number, and it has two factors: either has n/2 and n+1, or n and (n+1)/2, depending on which is even.
Prove: Lim([n+1]/n)^nr=Lim([n+r]/n)^n as n goes to infinity.[/quote]
first, make it clear what exactly does "r" stands for, also, assuming "r" is a constant, then you need a parranthesis around "nr"--that's what i think... ;)

Okay..

Prove that Lim([n+1]/n)^[nr]=Lim([n+r]/n)^n as n goes to infinity given that r is a natural number

there we go :lol:
the $\lim$ on both sides is equal to $1$. because on the left side you will have something like this: $\frac{n^{nr}+....}{n^{nr}}$ where the $\lim$ is one (the ratio of the highest degree of coefficents)
same thing for the right side, the $\lim$ would be the same$=1$

OK i got it. writing $b=\frac{1}{a}$, and subing this in, multiply across by $a^3$ and we get something like this:

$a^6-2a^4+a^3-a+1 \geq 0$ By inspection and dividing out factors, we find three obvious roots: $-1$, and a double root $1$, taking these out we get:

$ \leftrightarrow (a^3+a^2+1)(a+1)(a-1)^2 \geq 0$
Which is obvious

that's it i did it the same way. there still alot more ways, i know of 4 elementary solutions.
nevertheless, next proof plz:

peeta

[quote="seamusoboyle"]Firstly note that one of $a$ or $b$ is $ \geq 1$
We can transform this inequality into:

$(a+b)(a-b)^2+1-a \geq 0$ if it is b that is greater than or equal to 1, this holds because a would then be less than or equal to 1.

Now the other way round: if b is less than equal to 1, set the $1$ in that expression to $ab$ and we get:

$(a+b)(a-b)^2+a(b-1) \geq 0$ which holds in this case

[size=150]Problem the next:[/size]

The series

$f(n)=\lfloor n + \sqrt{2n} + \frac{1}{2} \rfloor$ misses exactly the triangular numbers[/quote]

This one?

ok i got it now, finally.

the only problem is that i am using a lemma, which is kind of obvious, when u look at it, but it's just not fun to proof.

Lemma 1:
$g(n):=$ is the function that misses all triangle numbers
so i let $n=\frac{k^2+k}{2}-m$ with $0\le m <k;\;m,k\in\mathbb{N}$
then every $n\in\mathbb{N}^+$ can be "made" by $m$ and $k$, since $\frac{k^2+k}{2}-k=\frac{(k-1)^2+(k-1)}{2}$.
and then $g(x)=\frac{(k+1)(k+2)}{2}-m-1=\frac{1}{2}\cdot\left(k^2+3k-2m\right)$
(somebody else might proof this lemma, but i will use it now if u dont mind)

Lemma 2:
this is another obvious lemma, nearly always used with the floor-function
$\lfloor x\rfloor \le x<\lfloor x\rfloor+1$

Proof:
i only want to proof that $f(n)=g(n)$.
first we transform $f(n)$ a lil bit.
$f(n)=\lfloor n+\sqrt{2n}+\frac{1}{2} \rfloor=\lfloor \left(\sqrt{n}+\frac{1}{\sqrt{2}}\right)^2 \rfloor = \lfloor \frac{1}{2}\cdot\left( \sqrt{2n}+1\right)^2 \rfloor$

now we use lemma 2 to proof it.
1.
$\frac{1}{2}\cdot\left( \sqrt{2n}+1\right)^2\ge \frac{1}{2}\cdot\left(k^2+3k-2m\right)\\
k^2+k-2m+1+2\sqrt{k^2k-2m} \ge k^2+3k-2m\\
2\sqrt{k^2k-2m} \ge 2k-1\\
4k^2+4k-8m \ge 4k^2-4k+1\\
8k \ge 8m+1\\
8k>8m\\
k>m$
which is stated before

2.
$\frac{1}{2}\cdot\left( \sqrt{2n}+1\right)^2< \frac{1}{2}\cdot\left(k^2+3k-2m\right) +1\\
k^2+k-2m+1+2\sqrt{k^2k-2m}<k^2+3k-2m+2\\
2\sqrt{k^2k-2m}<2k+1\\
4k^2+4k-8m < 4k^2+4k+1\\
-1<0<8m$
this is true as well.

therefore $f(n)=g(n) \blacksquare$

if u allow the next to be proof is a geometrical problem (i chose it not too hard):

let in $\triangle ABC$ $AD$ ($D\in BC$) be a median, $E \in AC$ a point and $F$ the "junction"(where they cross) of $BE$ and $AD$.
proof that when $\frac{BF}{FE}=\frac{BC}{AB}+1$, then $BE$ is a bisector.

[hide](if too difficult tell me and i give some hints)[/hide]
peeta

When you say bisector, do you mean perpendicular bisector or angle bisector?

i mean the angle bisector of $\measuredangle CBA$.

o well it seems to be too difficult for intermediate as i didnt think
so here is a
[hide="hint"]draw a parallel line to AC through D and consider a fraction often used in proofs with angle bisectors[/hide]

peeta

Im crap at goemetry, could someone post a solution? Also, while im at it, whats the best book for learning oltmpiad level geomatry???

[quote="peeta"]
Let in $\triangle ABC$, $AD$ be a median.  Let $E$ be an arbitrary point on $AC$ and $F$ be the intersection of $BE$ and $AD$.
Prove that when $\frac{BF}{FE}=\frac{BC}{AB}+1$, then $BE$ is an angle bisector.
[/quote]

If WLOG $[ABC] = 1$, note that $[EDC] = \frac{EC}{2AC}$

$\frac{BF}{FE} = \frac{[ABD]}{[ADE]} = \frac{\frac{1}{2}}{\frac{1}{2} - \frac{EC}{2AC}} = \frac{1}{\frac{AC-EC}{AC}} = \frac{AC}{AE}$

Since we have $\frac{BF}{FE} = \frac{BC}{AB} + 1$, we must have

$\frac{AC}{AE} = \frac{BC}{AB}+1$
$\frac{EC}{AE} = \frac{BC}{AB}$

which implies that $BE$ is an angle bisector by the angle bisector theorem.

Next proof:

Prove that if squares on a chessboard are numbered 1-64 row by row from left to right, and the squares are then colored red and black such that every row and column contains an equal number of red and black squares, then the sum of the numbers on all the red squares equals the sum of the numbers on all the black squares. (long sentence lol)

I have what i think is a nice proof for this. Start with regular colourings, ie red/black on every second square. Note that what we are trying to prove holds true, (easily seen if you split it up into 2 by 2 squares, each has the same sum.) Now the tricky bit. Take an arbitary colouring say we have a square coloured red, in column and row $x,y$, and another red in $a,b$ note we can move the first one across to the column that the second point is in, while moving the second point across to the column that the first one was in. So the points moved to $x,b$ and $a,y$If our original colouring displayed the property that we seek to prove, so will this one, as we have add to one point what we have taken away from the other. Also, we will still have 4 points in each row and column. Now we need to prove we can make every single board from the regular one that we know works. so start with the first column. Clearly just by swapping with the second column only we can get whatever we want. Now look at the second column. We can make what we want here from the third and forth column, due to it being scrambled from making the first column. And ive just realised my final piece of logic is flawed :rotfl:

I will tell you that there is a good, easily explainable proof.  BTW proofs should generally not be a solid single paragraph *wink wink nudge nudge*.

I kinda had an idea of what i was doing and was filling it in as i went along, but i think i will hunt for this proof you speak of...

Ok i have it! Label the top of the grid with the numbers 1,2,3,4,5,6,7,8, and down the side of the grid as 0,8,16,24,32,40,48,56. Note that the number in any square is the sum of the numer at the top of the column plus the number at the side. Since ther is four reds in each row and column, the total sum of the reds is $4(1+2+3+4+5+6+7+8)+4(0+8+16+24+32+40+48+56)=4(36+224)=1040$ But the sum of all the squares is $ \frac{(64)(65)}{2}=2080$ Therefore the reds add up to exactly half the total $\rightarrow$ the sum of the reds=sum of the blacks

Yes! Go me!

I different version of the same problem was proved in the MATHCOUNTS forum, though I think it was slightly easier.

[size=150]Roll on problem the next!:[/size]

Prove that there are infinatly many powers of 2 in the sequence:

$\lfloor n \sqrt 2 \rfloor$


[quote="K81o7"]Well, I reduced the expression to $\sum ab(a-b)$, but I can't do much from there...[/quote]

This is equal to $(a-b)(b-c)(c-a)$, and the rest is posted here: http://www.mathlinks.ro/Forum/viewtopic.php?t=107821

Well, in order to reduce it, I just expanded it out, canceled some terms, substituted $1-c$ for $a+b$, and did some more cancellation...


geometry
incenter
trigonometry
angle bisector

If $ A\hat{B}C \equal{} 120^\circ,$ $ A\widehat{B}M\equiv C\widehat{B}M$ and $ K\widehat{C}B\equiv K\widehat{C}D,$ find $ A\widehat{P}M$.

[hide="Answer"]$ A\widehat{P}M\equal{}30^\circ$[/hide]

[hide="solution"]Call I the incenter of ABC, and $ E: CI \cap AB$, $ F: AI \cap CA$. So K is the harmonic cojugate of E w.r.t. BA and so MK pass trught F, then $ F \equiv P$. But $ \frac {BM}{MB} = \frac {\sin \gamma}{\ain 60} = \frac {AB}{AC}$. Call $ K : PM \cap CI$, then K is the incenter of CBM and $ \angle CKM = 120^\circ$. Then $ \angle APM = 60 - 30 = 30^\circ$[/hide]

$AB$ is the external angle bisector of $\angle CBM$, so $K$ is the $M-$ excenter of $\triangle BCM\iff MK$ is angle bisector of $\angle CMB$, but $BC$ is external angle bisector of $\angle ABM$, so $P$ is $A-$ excenter of $\triangle ABM$, so the required angle is $30^\circ$, indeed ($\angle BMC=60^\circ+\angle A\iff\angle KMC=30^\circ+\frac{\angle A}{2}$, but from $\triangle AMP$, $\angle KMC=\angle APM+\frac{\angle A}{2}$).

Best regards,
sunken rock


search
geometry
circumcircle
geometric transformation
reflection
inequalities
probability

The line $PQ$ have to cut the sides $AB,AC$ or $BC,BA$ or $CA,CB$ ..

Suppose for ex. that the line $PQ$ cut the sides $AB,AC$  in $M,N$.

Manifestly $|PQ| =< |MN|$ and $angle(AMN)+angle(ANM) = 120$
Suppose for ex. :
$\displaystyle angle(AMN) \leq angle(ANM) ==> angle(ANM) \geq 60 =angle(MAN) $

So $|AM| >= |MN|$ ... and so on ...

 :cool:  I like  the innocent questions  :cool:

All right. Thanks, Diogene, but it was the straightforward solution which is the ugly solution I was talking about.

Is there not an other one? I just can't believe it!!!

:cool:  Really ? What is it so ugly ?!!   :cool:

It is just a rather long solution. If you consider the same problem for an 2n-gon: you have a regular 2n-gon and two points inside and you try to find the largest possible distance between them.

Then you just  draw the circumcircle and the larges possible distance must be 2r(with r as radius).

I awaited a similiar prove for a (2n+1)-gon or at least for a triangle. Or perhaps an unawaited proof using projections, reflections or something like that.

The other disadvantage of our solution is that it can't be generalized for a (2n+1)-gon :( .

Misha

perhaps I dont understand the question: can't you draw a circle of radius 1 centered on p, it will contain the triangle...

Right, but you can't just assert that the circle contains the triangle; you have to prove it.

umm :P isn't this a bit very trivial? :P

I mean, the triangle ABC has a distance 1 between A and C, and it only decreases when moving the points away from the sides...

Plus you don't need to prove anything about the circle with radius 1, not on any good math contest. That's just too trivial :P

I agree that the problem is easy, but that doesn't eliminate the need to prove assertions, especially when that is what the original proposer is asking for.

Misha, there is absolutely no wonder that the problem you posted is not trivial. Concerning such problems from convexity geometry, I advise you to take a look at [url=http://groups.google.com/groups?hl=de&lr=&ie=UTF-8&threadm=2mc5ifFl2lufU1%40uni-berlin.de&rnum=3&prev=/groups%3Fq%3Dpolygons%2Bhave%2Bears%26hl%3Dde%26lr%3D%26ie%3DUTF-8%26selm%3D2mc5ifFl2lufU1%2540uni-berlin.de%26rnum%3D3]this de.sci.math thread[/url]. One of the problems discussed there was to prove that every polygon (not necessarily convex) has a diagonal which completely lies inside the polygon. Trivial? By no means; it could rather be a pretty good olympiad problem if one wouldn't have to expect so many students bombarding the jury with justified question on how the "interior" of a non-convex polygon is defined.

Your problem has a much simpler outset, an equilateral triangle, but it is clear that it is actually a special case of the following problem: Given an arbitrary polygon $A_1A_2...A_n$ (not necessarily convex, but not self-intersecting), and two points P and Q inside it. (Here, "inside" can also mean "on the boundary".) Prove that the distance PQ is not greater than the greatest of the pairwise distances $A_iA_j$ between two vertices of the polygon.

In order to prove it, we note that since the points P and Q lie inside the polygon $A_1A_2...A_n$, the line PQ intersects the boundary of the polygon at two points, which we may call U and V. (If the polygon is concave, the line PQ may intersect its boundary at more than two points; in this case, take the leftmost one and the rightmost one.) Clearly, PQ $\leq$ UV. So it remains to show that the distance UV is not greater than the greatest of the distances $A_iA_j$. In fact, there is a pretty easy lemma saying that if C is a point on a segment AB, and O is another point in the plane, then the distance OC is less or equal to the greater of the two distances OA and OB. Now, the points U and V both lie on the boundary of the polygon $A_1A_2...A_n$, i. e. each of these two points lies on one side of this polygon. Call $A_uA_{u+1}$ the side on which the point U lies, and call $A_vA_{v+1}$ the side on which the point V lies. Then, the lemma above, we see that the distance UV is less or equal to the greater of the distances $UA_v$ and $UA_{v+1}$. We assume, for instance, that the greater of the distances $UA_v$ and $UA_{v+1}$ is $UA_{v+1}$. Then, UV $\leq$ $UA_{v+1}$. Now, applying the lemma again (this time we use that the point U lies on the segment $A_uA_{u+1}$), we see that the distance $UA_{v+1}$ is less or equal to the greater of the distances $A_uA_{v+1}$ and $A_{u+1}A_{v+1}$. This time, let's assume that the greater of the distances $A_uA_{v+1}$ and $A_{u+1}A_{v+1}$ is $A_uA_{v+1}$. Then, $UA_{v+1} \leq A_uA_{v+1}$, and we get the chain of inequalities

$PQ \leq UV \leq UA_{v+1} \leq A_uA_{v+1}$.

And $A_uA_{v+1}$ is, of course, less or equal to the greatest of the pairwise distances $A_iA_j$. This completes the proof.

I hope I didn't make stupid errors above; I am really unexperienced in this topic of geometry...

  Darij

Here's another solution to darij's more general problem, using vectors.  Because $P$ is in the convex hull of the $A_i$, we have $\textstyle P = \sum_i p_i A_i$ for some probability distribution $p$.  Similarly $\textstyle Q = \sum_j q_j A_j$ for some probability distribution $q$.  Thus
\begin{eqnarray*}
|P - Q|
  & = & \bigl| \sum_i p_i A_i - \sum_j q_j A_j \bigr| \\
  & = & \bigl| \sum_{i, j} p_i q_j (A_i - A_j) \bigr| \\
  & \leq & \sum_{i, j} p_i q_j | A_i - A_j | \\
  & \leq & \max_{i, j} | A_i - A_j |.
\end{eqnarray*}

Thanks a lot for the solutions the more general problem: I heard of it before.

The story behind my first question here is: I considered how to explain the solution of my problem to a beginner and then I understood that I've never thought about it  properly (I just mentioned this fact as obvious in my solutions to some more complicated olympiad problems). Then I found the solution which was posted at the beginning of the topic and I couldn't believe that there is no beatiful argumentation as there often are in geo-combinatorial problems. But it seems that it was hopeless.

But it is still an interesting problem and I'll look up the link Darij posted.

Misha

[quote="santosh_blr"][quote="Misha123"]Hello!
Here is a problem which seems to be trivial. The straightforward solution I have seen is very ugly so I search for another(nice!!!) solutions.

You have an equilateral triangle of side length 1 and two points P, Q inside. Show that the distance PQ is less or equal to 1.

Sorry if there is a one-sentence solution. I just could not find it.

Misha[/quote]

Actually, we can go for a less rigorous geomerical proof .

Let's consider the following cases.

1> The points P, Q and any one of the three vertices are collinear.
Let's say A,P,Q in triangle ABC are collinear. Let APQ meet BC at D

Then  PQ < AD <AB = 1

2> P, Q are non-collinear with all the three vertices.

Produce PQ both ways . It will intersect exactly two sides . With no loss of generality, let's say PQ meets AB at D and BC at E. Then constructing an equilateral triangle with BD as one of the sides,we observe that PQ <DE <DB <AB <1

so PQ is always < 1, as long as the two pts P, Q are strictly inside the triangle[/quote]

OOps..the last inequality is PQ <DE <DB <AB = 1

not PQ <DE <DB <AB <1


How many ways are there to tile an 8x8 square using 2x2 "blocks" and 1x3 rectangles? The 1x3 rectangles may not be rotated.

[hide="Hints"]$ \blacktriangleright$ Show that the number of 1x3 rectangles is a multiple of 4

$ \blacktriangleright$ Show that the 1x3 rectangles must be aligned vertically

$ \blacktriangleright$ Show that 1x3 rectangles can be collected into 1x6 or 2x3 rectangles[/hide]

Those hints are a good way to get a 0 on this question :P.  See http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1142625#1142625 for jmerry's solution.


Ring Theory
superior algebra
superior algebra unsolved

Define a Principle Ideal Ring as a ring (not necessarily with identity) such that all ideals are principle. Then prove a PIR has an identity.

[quote="kouboy"]do you assume that the ring is commutative or not?[/quote]
Not.
=bmsiv-

[quote]
such that all ideals are principle.[/quote]

All proper ideals?

[quote="owk"]$2\mathbb Z$ is a counter-example. Are you sure you have [url=http://planetmath.org/encyclopedia/PrincipalIdeal.html]this[/url] definition of "principal ideal" in mind? owk[/quote]
I had a different definition of Principal Ideal:
if I is a principle ideal of R, then there is an element a of I such that for all x in I, x = ab for some element b of R.

The quantifiers are messing with me, but it seems like $2\mathbb{Z}$ should still be a counter-example...I'm probably just miss-interpreting bmsiv's definition though.

[quote="mlok"]$6\mathbb Z$ is a non-principal ideal in $2\mathbb Z$.[/quote]
Correct. 6 cannot be expressed as an element of I multiplied by an element of R, as it is irreducible.


It is well-known that  $ \frac{a_{n}}{a_{n\minus{}1}} \to \phi$ where $ a_{n}$ and $ a_{n\minus{}1}$ are consecutive numbers from the Fibonacci sequence. Are there any other sequences excpet those defined with the conditions:
1) $ a_{0}\equal{}0$
2) $ a_{1}\equal{}m$
3) $ a_{n}\equal{}a_{n\minus{}1}\plus{}a_{n\minus{}2}$
for which is true $ \frac{a_{n}}{a_{n\minus{}1}} \to \phi$

It is true for any $ m$.  Notice that $ a_n\equal{}a_{n\minus{}1}\plus{}a_{n\minus{}2}\Rightarrow b_n\equal{}1\plus{}\frac{1}{b_n\minus{}1}$ where $ b_n\equal{}\frac{a_n}{a_{n\minus{}1}}$.  Thus, we have

$ \lim_{n\to\infty}b_n\equal{}1\plus{}\frac{1}{1\plus{}\frac{1}{1\plus{}\cdots}}\equal{}\phi$

Yes, I know it is true for any m. But is there any other sequence for which this is true?

[quote="SAPOSTO"]It is well-known that  $ \frac {a_{n}}{a_{n \minus{} 1}} \to \phi$ where $ a_{n}$ and $ a_{n \minus{} 1}$ are consecutive numbers from the Fibonacci sequence. Are there any other sequences excpet those defined with the conditions:
1) $ a_{0} \equal{} 0$
2) $ a_{1} \equal{} m$
3) $ a_{n} \equal{} a_{n \minus{} 1} \plus{} a_{n \minus{} 2}$
for which is true $ \frac {a_{n}}{a_{n \minus{} 1}} \to \phi$[/quote]
yes,for example:$ a_n\equal{}(\phi\plus{}\frac{1}{n})^n$

[quote="SAPOSTO"]It is well-known that  $ \frac {a_{n}}{a_{n \minus{} 1}} \to \phi$ where $ a_{n}$ and $ a_{n \minus{} 1}$ are consecutive numbers from the Fibonacci sequence. Are there any other sequences excpet those defined with the conditions:
1) $ a_{0} \equal{} 0$
2) $ a_{1} \equal{} m$
3) $ a_{n} \equal{} a_{n \minus{} 1} \plus{} a_{n \minus{} 2}$
for which is true $ \frac {a_{n}}{a_{n \minus{} 1}} \to \phi$[/quote]
It mean $ a_n \equal{} mF_n$, therefore $ \frac {a_n}{a_{n \minus{} 1}} \equal{} \frac {F_n}{F_{n \minus{} 1}}.$


How do i achieve the following :
[code]
Regular
                         Centered
Regular
[/code]

?

You can center $\LaTeX$ with double dollar signs; I'm not sure about regular text.

The closest i can get so far is:
[code]
Line 1 Regular\\
\centering
Line 2 Centered

\noindent Line 3 Regular
[/code]

which produces:
[code]
Line 1 Regular
                        Line 2 Centered
                          Line 3 Regular
[/code]

which means i dont know how to "turn off" inline centering, even though I had started "Line 3" on a new paragraph to try to get latex to recognize the scope of centering ended on line 2....

a summary of what i've learned so far and my appeal for help : [url]http://tinyurl.com/2qa7zd[/url]

(read the second message, ive included a short latex document that demonstrates what i have so far)


algebra
polynomial
inequalities unsolved
inequalities

An advance :
Let $(a,b,c,d) = \( \frac{3}{13},\frac{6}{13},\frac{9}{13},\frac{2}{13}\)$

Then, $a^{2}+b^{2}+c^{2}+d^{2}= \frac{10}{13}$ is located between $\frac{1}{2}$ and $1$, and also satisfying $Q = 0$. Since $Q \geq 0$, then the minimum value of $Q$ is $0$.

[quote="chien than"]Let  $a,b,c,d$ be real numbers satisfying $\frac{1}{2}\leq a^{2}+b^{2}+c^{2}+d^{2}\leq 1$.
Find min and max:
$Q=(a-2b+c)^{2}+(b-2c+d)^{2}+(b-2a)^{2}+(c-2d)^{2}$ .[/quote]
I think, $\min Q=\frac{7-3\sqrt5}{4},$
which holds when $a=d=\frac{1}{\sqrt{10+2\sqrt5}},$ $b=c=\frac{1+\sqrt5}{2\sqrt{10+2\sqrt5}}$
and $\max Q=\frac{15+5\sqrt5}{2},$
which holds when $a=-d=\frac{1}{\sqrt{5+\sqrt5}},$ $b=-c=-\frac{1+\sqrt5}{2\sqrt{5+\sqrt5}}.$

Sorry, my solution was totally wrong..  :oops:
Can you tell me your solution, Mr. Arqady ?

RDeepMath91, use Lagrange multipliers method. Detail?

I'm sorry, but I ain't too familiar with Lagrange Multipliers method..
So, I think yes..  :wink:

Can you explain more clearly about the  polynomial $k^{4}-22k^{3}+131k^{2}-190k+25 = 0$ ?
About how to find it and what's exactly we need to find from the system of equation..
I'm sorry, I don't know too much about how to find the determinant with four variables or more..

When you prove the system $\begin{cases}2a-3b=0\\5a+8b=0\end{cases}$ you can do the following operations:
$\begin{cases}2a-3b=0/\cdot(-5)\\5a+8b=0/\cdot2\end{cases}\Leftrightarrow\begin{cases}-10a+15b=0\\10a+16b=0\end{cases}\Rightarrow31b=0.$
Similar: $31a=0.$ The number $31$ is determinant for this system.
Easy to see that determinant for system:$\begin{cases}a_{11}x+a_{12}y=0\\a_{21}x+a_{22}y=0\end{cases}$ is $\Delta=a_{11}a_{22}-a_{21}a_{12}.$
If we wish that for the system will be infinite pair of numbers $(x,y)$, which fulfill her, then $\Delta=0.$
By similar operations you will obtain:
$\begin{cases}(5-k)a-4b+c=0\\-4a+(6-k)b-4c+d=0\\a-4b+(6-k)c-4d=0\\ b-4c+(5-k)d=0\end{cases}\Leftrightarrow\begin{cases}\Delta\cdot a=0\\\Delta\cdot b=0\\\Delta\cdot c=0\\ \Delta\cdot d=0\end{cases},$ where $\Delta=k^{4}-22k^{3}+131k^{2}-190k+25 .$
There are other ways how to find a determinant. I only tried to explain, what is this.


calculus
integration
calculus computations

Let be given $ f: [0,1] \times [0,1] \to \mathbb R$ defined by

\begin{equation}
f(x,y)=
\begin{cases}
0 & {\text{ if $ x \notin \mathbb Q$}},\\
0 & {\text{ if $ x \in \mathbb Q$, $ y \notin \mathbb Q$}},\\
\frac{1}{q} & {\text{ if $ x\in \mathbb Q$, $ \mathbb Q \ni y \equal{} \frac {p}{q}$}}.
\end{cases}
\end{equation}

[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/6/5/b/65bdcf85a9d3c7eb2ed9746f56e10a7659358096.gif[/img]

Prove that $ f$ is integrable and calculate $ \int\limits_{\left[ {0,1} \right] \times \left[ {0,1} \right]} {f dV} \equal{} 0$.

I will post my solution at weekend.

what does $ Q \ni y \equal{} \frac {p}{q}$ mean?

[quote="0714446459923"]what does $ \mathbb Q \ni y \equal{} \frac {p}{q}$ mean?[/quote]

oh, this means that $ y \in \mathbb Q$ and $ y\equal{}p/q$ where $ p,q \in \mathbb N$, $ (p,q)\equal{}1$.


The sum of the first $ N$ positive odd numbers is 121. What is the value of $ N$?

The sum of the first $ N$ positive odd numbers, where $ N$ is any counting number, is $ N^2$. It is thus conclusive that $ N\equal{}\sqrt{121}\equal{}\boxed{11}$.


function
limit
real analysis
real analysis unsolved

Let {f_n} be a sequence of continuous functions which converges uniformly to a function f on a set E. Prove that
\[\mathop{\lim }_{n \to \infty }f_{n}(x_{n}) = f(x) \]
for every sequence of points x_n in E such that x_n -> x, and x in E. Is the converse of this true?

n.t.tuan also posted the same problem. its just few problems earlier than this. please check that out.

Check out this topic:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135394

What does the converse of that statement look like and how do we show that it is true?


Show that there exists an integer $n$ with the following properties:

(i) The binary expansion on $n$ has precisely 2004 $0$s and 2004 $1$s.
(ii) 2004 divides $n$.

Just note that the number of $1$s in $2004$ binary representation is $7$ and for $3*2004$ it's $9$.
Now write $2004=7a+9b$ for some $a,b\in \mathbb N$.
Now simply put $a$ copies of $2004$ binary representation in a row and put next to them $b$ copies of $3*2004$ binary representation.
Also note that number of $0$s in $2004$ binary rep. is less than number of $1$s, and this holds for $3*2004$ too.
Now just put enough $0$s right to the number you constructed already.

does this solution work?
Looking at the permutations of the number in binary, there are exactly $\binom{4007}{2003}$ numbers that satisfy the first requirement. We can assume that these numbers are spread out equally mod 2004, and are certain that at least one is divisible by 2005.


Is graph paper allowed in Mathcounts Competitions?

unless they suddenly change the rules for the 09 competitions, you can't use graph paper in mathcounts

You can always just take scrap paper and draw lines on it, though.

You're allowed as much scratch paper, so make use of it. There's nothing wrong about writing stuff down on your paper--make boxes for each of the problems, etc.


induction
inequalities
number theory proposed
number theory

$ x^3 \plus{} y^3 \equal{} (x \plus{} y)(x^2 \minus{} xy \plus{} y^2) \implies 2^n \plus{} n|8^n \plus{} n^3 \implies 2^n \plus{} n | 8^n \plus{} n^3 \minus{} (8^n \plus{} n) \equal{} n^3 \minus{} n$
And it implies that $ n^3 \minus{} n \equal{} 0$ or $ |n^3 \minus{} n| \ge 2^n$.
The first case gives $ n \equal{} 1$, which is a solution.
The second case: for $ n \ge 10$, $ 2^n > n^3 > n^3 \minus{} n$, by induction: $ 1024 > 1000$, and $ 2^n > n^3$ implies that $ 2^{n \plus{} 1} > 2n^3 \ge (n \plus{} 1)^3$ because $ 3n^2 \plus{} 3n \plus{} 1 \le 3n^2 \plus{} 3n^2 \plus{} 3n^2 \equal{} 9n^2 < n^3$ for $ n \ge 10$.
So we only need to consider $ 2 \le n \le 9$. The valid solutions are $ n \equal{} 2,4,6$. When we include the 1st case, the set of solutions is $ n \equal{} 1,2,4,6$.

[quote="kunny"]Find all positive integers $ n$ such that $ 8^n \plus{} n$ is devisible by $ 2^n \plus{} n$.[/quote]
We have $ 8^n \plus{} n \equiv n^3 \minus{} n \bmod 2^n \plus{} n$ so $ 2^n \plus{} n \mid n^3 \minus{} n$. From here it is easy to use inequalities to obtain all solutions.

[hide=Solution]Note that $$8^n + n\equiv (2^n)^3 + n\equiv (-n)^3 + n \equiv n - n^3 \pmod{2^n+n}.$$ Hence, we wish to find $n$ such that $2^n+n|n^3-n$. Note that for $n\ge 10$, we have that $2^n > n^3$. Thus, we only need to try values below $10$. Doing so, the only solutions that work are $\boxed{n=1,2,4,6}$.[/hide]

[quote=Japan MO Finals 2009 P1]Find all positive integers $ n$ such that $ 8^n \plus{} n$ is divisible by $ 2^n \plus{} n$.[/quote]
[b][color=#000]Solution:[/color][/b] Note that, $x+y \mid x^3+y^3$ Hence,
$$\frac{8^n+n}{2^n+n}=\frac{2^{3n}+n^3+n-n^3}{2^n+n} \implies \frac{n^3-n}{2^n+n} \in \mathbb{Z} \implies n^3 \geq 2^n+~2n$$
Which is [url=https://math.stackexchange.com/questions/438260/proving-by-induction-2n-n3-for-any-natural-number-n-9]true for 1 < n < 10[/url]. Also, $n^3-n$ $ =$ $0$ when $n=1$. Quickly checking the permissible values for $n$, we get $n=1,2,4,6$ $\qquad \blacksquare$


hello everyone.

Looking at a engine tuning software for a specific car.

In the software the engine computer is looking to do all mathematical calculations based on what injectors are in there currently. well to run larger injectors we have to do some mathematics, problem is that the injectors in there now flow 682cc/min at a fuel pressure of static 58psi. the company uses N-heptane which has a specific gravity of 685 kg/m^3 for their injector ratings.

Now the aftermarket larger injectors from RC engineering are 750cc/min at static 43.5psi. problem is that RC engineering uses a denser fluid which has a SG of 720 kg/m^3.

I need to convert the 750cc/min @ 43.5psi injector to flow using n-heptane which will be a lower flow in the end. I have turned the fuel pressure down to 43.5psi.

[b]So the test fluids:[/b]

N-heptane has a SG of 685 kg/m^3
Gasoline has a SG of 720 kg/m^3

[b]Injectors:[/b]

Factory injectors - 682cc/min @ static pressure of 58 psi. (test fluid, N-heptane)
After market injectors - 750cc/min @ static 43.5 psi. (test fluid, Gasoline)

Now here is a formula to get lb/hr using n-heptane:
682cc/min * 60min/hr * (1m/100cm)^3 * 685kg/m^3 *2.2lb/kg = 61.7lb/hr

Now here is a formula to get lb/hr using gasoline:
682cc/min * 60min/hr * (1m/100cm)^3 * 720kg/m^3 *2.2lb/kg = 64.8lb/hr

1 lb/hr gasoline = 10.5 cc/min (@720 kg/m3)
(1 (pound / hr)) / (720 (kg / (m^3)))

1 lb/hr N-heptane = 11.03 cc/min (@685 kg/m3)
 (1 (pound / hr)) / (685 (kg / (m^3)))

Now am I to assume:
750cc/min * 60min/hr * (1m/100cm)^3 * 685kg/m^3 *2.2lb/kg = 67.8lb/hr

to convert using n-heptane for lb/hr to cc/min:

67.8 lb/hr * 11.03 cc/min = 747.9 cc/min (doesn't make sense to me)

now most online calculators to convert lb/hr to cc/min will use a factor of 10.5. this translates to the aftermarket injector company's test fluid having a specific gravity of 720 kg/m^3 as seen above...

anyone know where im going wrong here.

my guess is that this is the wrong forum ..


Thomas J. Mildorf in his paper over the olympiad inequalities

http://www.artofproblemsolving.com/Resources/Papers/MildorfInequalities.pdf

write to prove $ 9a^2b^2c^2 \leq (a^2b \plus{} b^2c \plus{} c^2a)(ab^2 \plus{} bc^2 \plus{} ca^2)$ :

Rearrange the terms of each factor and apply Cauchy,

$ (a^2b \plus{} b^2c \plus{} c^2a)(ab^2 \plus{} bc^2 \plus{} ca^2) \geq (\sqrt [3]{a^2b^2c^2} \plus{} \sqrt [3]{a^2b^2c^2} \plus{} \sqrt [3]{a^2b^2c^2})^2 \equal{}$

$ 9a^2b^2c^2$.

I don't understand how rearrange the terms and apply Cauchy to obtain

this. Thanks in advance for some hint.

Instead of $ \sqrt [3]{a^2b^2c^2}$ I think is $ \sqrt {a^2b^2c^2}$?

[quote="Bob75"]

Instead of $ \sqrt [3]{a^2b^2c^2}$ I think is $ \sqrt {a^2b^2c^2}$?[/quote]

yes.

$ (a^2b\plus{}b^2c\plus{}c^2a)(ab^2\plus{}bc^2\plus{}ca^2)\geq 3\cdot 3\sqrt[3]{a^6b^6c^6}\equal{}9a^2b^2c^2$.
From AM-GM. I think the document had a small error in it.

From AM-GM is obvious, but my question is about using rearrangement and Cauchy? Thanks in advance.

We have to prove that $ (a^2b \plus{} b^2c \plus{} c^2a)(ab^2 \plus{} bc^2 \plus{} ca^2)\geq 9a^2b^2c^2$
$ (a^2b \plus{} b^2c \plus{} c^2a)(ab^2 \plus{} bc^2 \plus{} ca^2) \equal{} (a^2b \plus{} b^2c \plus{} c^2a)(bc^2 \plus{} ca^2 \plus{} ab^2)$
$ \geq (abc \plus{} abc \plus{} abc)^2\equal{}9a^2b^2c^2$
 :)

Thanks you very much for your answer, and Cauchy is useless because only with the rearrangement we solve it easy!


integration
function
Functional Analysis
real analysis
real analysis unsolved

I have been completely lost as I studied orthogonal projections. How to do this one:

Let $L^{2}([-3,3])$ be a Hilbert space with norm given by $\parallel f\parallel _{2}: =\left (\int_{-3}^{3}|f(t)|^{2}\,dt\right )^{1/2}$. Let $M$ be a subspace
\[M: =\{f\in L^{2}([-3,3])|f(t)=0 \text{ for almost every }t\in [0,3]\}\]
and $N$ 1-dimensional subspace spanned by the constant function 1.
Find orthogonal projections from $L^{2}([-3,3])$ to $M$ and to $N$.

Orthogonal projection to $M$ zeros the function on $(0,3]$ and leaves it alone on the other half; it's pretty easy to see that this gets you the minimum distance, and leaves a difference orthogonal to $M$. Orthogonal projection to $N$ maps a function $f$ to its average value. That requires a bit of calculation, but it's not hard to show it works.

[quote="jmerry"]Orthogonal projection to $N$ maps a function $f$ to its average value. That requires a bit of calculation, but it's not hard to show it works.[/quote]
Can someone elaborate this more. I'm relative new in functional analysis and I don't see how to prove this.

Let $u$ be an element of a Hilbert space $H$ - or inner product space, since completeness will play no role. We want to find the orthogonal projection from $H$ to the one dimensional subspace spanned by $u.$

I'll do this algebraically: Given $f\in H,$ we want to find $x$ that minimizes

$\|f-xu\|^{2}=\langle f-xu,f-xu\rangle=\|f\|^{2}-\langle xu,f\rangle-\langle f,xu\rangle-\langle xu,xu\rangle$

$=\|u\|^{2}x\overline{x}-x\overline{\langle f,u\rangle}-\overline{x}\langle f,u\rangle+\|f\|^{2}$

$=\|u\|^{2}\left(x\overline{x}-\frac{\overline{\langle f,u\rangle}}{\|u\|^{2}}x-\frac{\langle f,u\rangle}{\|u\|^{2}}\overline{x}\right)+\|f\|^{2}$

$=\|u\|^{2}\left\|x-\frac{\langle f,u\rangle}{\|u\|^{2}}\right\|^{2}-\frac{|\langle f,u\rangle|^{2}}{\|u\|^{2}}+\|f\|^{2}$

This will be minimized by setting $x=\frac{\langle f,u\rangle}{\|u\|^{2}}.$ The projection then sends $f$ to $\frac{\langle f,u\rangle}{\|u\|^{2}}\, u.$

For your particular case, you have $H=L^{2}(-3,3)$ and $u=1.$

Then $\|u\|^{2}=\int_{-3}^{3}1=6$ and $\langle f,u\rangle=\int_{-3}^{3}f.$ The orthogonal projection sends $f$ to the constant function whose value is $\frac16\int_{-3}^{3}f.$

(Which is exactly what jmerry said.)

The argument would be very similar for the orthogonal projection onto some finite dimensional space, provided that we have already gone to the trouble of finding an orthogonal basis for that finite dimensional space.


A sequence is defined by $ a_1\equal{}a_2\equal{}a_3\equal{}1$ and $ a_{n\plus{}3}\equal{}\minus{}a_n\minus{}a_{n\plus{}1}$ for all $ n \in \mathbb{N}$. Prove that this sequence is not bounded , i.e. that for every $ M \in \mathbb{R}$ there is an $ n$ for which $ |a_n|>M.$

$ a_n\equal{}Ax_0^n\plus{}Bx_1^n\plus{}Cx_2^n$, were $ x_0,x_1,x_2$ are roots of $ x^3\plus{}x\plus{}1\equal{}0$ and $ x_0<\minus{}1$ real root. Therefore complex roots $ |x_1|\equal{}|x_2|<1$.
If $ A\equal{}0$ $ a_n\to 0$, when $ n\to \infty$ give contradition with $ a_n$ is integer. Therefore for big n $ |a_n|to \infty$ as $ |Ax_0^n|$.


Let $ABC$, a triangle such that $P$ is an interior point, and $PA+PB+PC\sqrt2$ is minimum. Find $\angle APB$.

Let $Q$ the exterior point to the triangle $ABC$ such that $AQB$ is isosceles right angle at $Q$. We have $AQ\sqrt{2}=BQ\sqrt{2}=AB$.
Because Ptolomeo's inequality at cuadrilateral $BPAQ$,
$PA+PB+PC\sqrt{2}\geq (PQ+PC)\sqrt{2}\geq CQ\sqrt{2}$
Then, $C$, $P$ and $Q$ are collinear; and $BPAQ$ is cyclic.
In particular, $<APB=90^{0}$


i really need to know...

i always get 0....

im talkin about the problems...

how come people solve the problem in 5 secs and get it right--

i get nothin.......

is there a way to improve?

Thanky for replying (:

Well during a class always have a calc, paper and pen/pencil near by. Also reading the chapter the class is over beforehand always helps.

(He means FTW)
Practice.
Practice.
Learn.

you could do good by donating to the red cross...

sarcasm FTW.

FTW.  How ironic.

Uh, you should practice, and watch Alcumus videos, and maybe buy a book or 2.

Please move this, mods.


In a triangle $ABC$, the angle bisector of $A$ and the cevians $BD$ and $CE$ concur at a point $P$ inside the triangle. Show that the quadrilateral $ADPE$ has an incircle if and only if $AB=AC$.

[hide]

The center of an inscribed circle is the intersection of angle bisectors. Consider a circle and an arbitrary point outside of the circle. Draw tangents from the point to the circle. Then select an arbitrary point on the circle and draw the tangent. So far the whole thing is symmetric. The tangent crosses the bisector (the first line). From the intersection, draw the other tangent. Due to symmetry the resulting quadrilateral has pairs of adjacent equal sides. More symmetry, and triangle ABC is isosceles.

[/hide]

suppose that $ (AP)$ meet $ BC$ at $ Q$.
$ \bullet :$ if $ ADPE$ has an incircle . we have $ \widehat{APE}\equal{}\widehat{APD}$ so $ \widehat{APB}\equal{}\widehat{APE}\plus{}\widehat{EPB}\equal{}\widehat{APD}\plus{}\widehat{DPC}\equal{}\widehat{APC}$ and  $ \widehat{PAB}\equal{}\widehat{PAC}\equal{}\frac{\widehat{A}}{2}$ so $ \widehat{PBE}\equal{}\widehat{PCD}$ so $ \triangle PEB \sim \triangle PDC$
$ \Longrightarrow \frac{CD}{BE}\equal{}\frac{PD}{PE} (\star) .$

in other hand $ \frac{BP}{PD}\equal{}\frac{AB}{AD}$ and $ \frac{PE}{PC}\equal{}\frac{AE}{AC}$ but since $ \frac{BP}{PC}\equal{}\frac{BQ}{QC}\equal{}\frac{AB}{AC}$,  we have $ \frac{PE}{PD}\equal{}\frac{AE}{AD} (\star \star) .$

and by  ceva theorem's $ 1\equal{}\frac{BQ}{QC}\frac{DC}{AD}\frac{AE}{EB}\equal{}\frac{AB}{AC}\frac{DC}{AD}\frac{AE}{EB}$ and with $ (\star )$ and $ (\star \star)$ we have $ AB\equal{}AC$.

$ \bullet:$if$ AB\equal{}AC$ then $ 1\equal{}\frac{BQ}{QC}\frac{DC}{AD}\frac{AE}{EB}\equal{}\frac{AB}{AC}\frac{DC}{AD}\frac{AE}{EB}$
 $ \Longrightarrow \frac{DC}{AD}\equal{}\frac{BE}{EA}$

$ \Longrightarrow  \frac{AC\minus{}AD}{AD}\equal{}\frac{AB\minus{}EB}{EA}$
$ \Longrightarrow  AD\equal{}AE$ and $ AP\equal{}AP$ and $ \widehat{PAE}\equal{}\widehat{PAD}$ so $ PE \equal{}PD$ and in this case $ AD\plus{}EP\equal{}AE\plus{}PD$ so $ ADPE$ has an incircle .


[hide="Hint"]
For all $ x,y,a,b$ such that:

$ \frac{x}{y}=\frac{a}{b}$,

The following is true:

$ \frac{x}{y}=\frac{a}{b}=\frac{x+a}{y+b}$.[/hide]

[hide]Using the hint, we find $ \frac{x}{y}=\frac{(y)+(x+y)+(x)}{x+y}=\frac{2(x+y)}{x+y}=2$[/hide]

i remember this problem from some AHSME in AoPS vol 1!!! :lol:

[hide] as the hint given. u just add up all the numerators of the given proportions and u add up all the denominators of the given proportions..u get the answer is 2

if $ a/b=c/d$, then $ a/b=(a+c)/(b+d)=(a-c)/(b-d)=(a+kc)/(b+kd)$ just some properties of proportions that are very useful!!![/hide]

[quote="youdonotknowme"]If for three distinct positive numbers $ x$,$ y$,and $ z$,

$ \frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}$

then find the numerical value of $ \frac{x}{y}$[/quote]
are we assuming all these fractions are in lowest terms. either way im still stuck


function
greatest common divisor
number theory unsolved
number theory

[quote="Megus"]Let $n\geq 2$ be natural number
Define:

$f(x_1,...,x_n) = \frac{(x_1,S_1)+...+(x_n,S_n)}{x_1+...+x_n}$

where $S_i = (\sum_{k=1} ^n a_k) -a_i$.

What is the maximal value attained by $f$ as it varies over all $n$-tuples in $\mathbb{Z}_{+}^n$.

P.S. Looks nice but I haven't given it a much thought. Hope it won't turn out to be trivial  :)[/quote]

I didn't understand ...
Maybe $x_{i}$ are different ?
(In another case we can take $x_{i}=1$ then f=1.  But f<=1, so max(f)=1)

Oh yeah, thanks Ghavr for pointing out - I forgot this one  :oops:

[quote="Megus"]Let $n\geq 2$ be natural number
Define:

$f(x_1,...,x_n) = \frac{(x_1,S_1)+...+(x_n,S_n)}{x_1+...+x_n}$

where $S_i = (\sum_{k=1} ^n a_k) -a_i$.

What is the maximal value attained by $f$ as it varies over all $n$-tuples in $\mathbb{Z}_{+}^n$ with $x_i$ pairwise distinct.

P.S. Looks nice but I haven't given it a much thought. Hope it won't turn out to be trivial  :)[/quote]

The answer: for n=2 max(f)=$\frac{2}{3}$
                   for n>2 max(f)=1

Let $T(x_{1},..x_{n})= (x_1,S_1)+...+(x_n,S_n)$
Then $T \leq x_1+...+x_n \Rightarrow f\leq 1$
Also we can consider $(x_{1},..x_{n})=1$  (in another case we can devide each number
by the greatest common divisor and we'll get this condition)

Let n=2 then we have: $T=(x_1,x_2)+(x_2,x_1)=2 \Rightarrow f= \frac{2}{x_1+x_2} \leq \frac{2}{3} (\mbox {equality for } x_1=1 ,\ x_2=2)$

If n>2 then let's construct such sequence: $x_1=1 ,\ x_2=2 ;\ \mbox{ for } i \geq 3 : \ x_i=3*2^{i-3} \Rightarrow$
for $i \geq 3 : \ (x_i,S_i)=(x_i\mbox{ } , \mbox{ }x_i(2+2^2+...+2^{n-i})+3(2^{i-4}+...+1)+3)=$
$(x_i,3(2^{i-4}+...+1+1))=(x_i,3*2^{i-3}) =x_i$

for $i=2 : \ (x_2,S_2)=(2,2*(...)+3+1)=2$
for $i=1 : \ (x_1,S_1)=(1,S_1)=1 \Rightarrow$
$T=x_1+...+x_n \Rightarrow f=1$
V :)


Let P be an internal point of triangle ABC. The line through P parallel to AB meets BC at L, the line through P parallel to BC
meets CA at M, and the line through P parallel to CA meets AB at N. Prove that
$ \frac{BL*CM*AN}{LC*MA*NB} \le \frac{1}{8}$.
and locate the position of P in triangle ABC when equality holds.

[quote="Machu Picchu"]Let P be an internal point of triangle ABC. The line through P parallel to AB meets BC at L, the line through P parallel to BC
meets CA at M, and the line through P parallel to CA meets AB at N. Prove that
$ \frac {BL*CM*AN}{LC*MA*NB} \le \frac {1}{8}$.
and locate the position of P in triangle ABC when equality holds.[/quote]

I think this problem is not difficult
The line through P parallel to CA meets AB at N and meets BC at E
Let: $ x \equal{} BL, y \equal{} LE, z \equal{} EC$
$ \frac {BL.CM.AN}{LC.MA.NB} \le \frac {1}{8}$
 $ \Leftrightarrow \frac{xyz}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}  \le \frac {1}{8}$   :P


vector
algebra
polynomial
linear algebra
matrix
induction
invariant

Let V be a real vector space and T an operator on V. Prove that for any real a,b so that $ a^2<4b$, $ dimnull(T^2\plus{}aT\plus{}b)^k$ is even for any positive integer k.

Let $ P$ be a real polynomial with no real roots. We prove $ dim(ker(P(T)))$ is even.
Considering the Jordan's theory, it is sufficient to show this result in the 2 following cases:
1) $ T\equal{}\lambda{I_n}\plus{}N$ where $ N$ is the nilpotent matrix of order $ n$ ($ N^{n\minus{}1}\not\equal{}0$) and $ \lambda$ is a real number:
$ P(T)\equal{}P(\lambda)I_n\plus{}P'(\lambda)N\plus{}\cdots$. $ P(\lambda)\not\equal{}0$ then $ P(T)$ is invertible.
2) $ T\equal{}diag(\lambda{I_n}\plus{}N,\bar{\lambda}I_n\plus{}\bar{N})$ where $ \lambda$ is a non real number. $ P(T)\equal{}diag(P(\lambda)I_n\plus{}P'(\lambda)N\plus{}\cdots,P(\bar{\lambda})I_n\plus{}P'(\bar{\lambda})\bar{N}\plus{}\cdots)$.
If $ P(\lambda)\not\equal{}0$ then $ P(\bar{\lambda})\not\equal{}0$ and $ P(T)$ is invertible.
If $ P(\lambda)\equal{}0$ then $ P(\bar{\lambda})\equal{}0$ and we have the same result for the derivatives.
Thus or $ P(T)\equal{}0$ and $ dim(ker(P(T)))\equal{}2n$ is even or $ P(T)\equal{}diag(P^{(k)} (\lambda)N^k/k!\plus{}\cdots,P^{(k)} (\bar{\lambda})\bar{N}^k/k!\plus{}\cdots)$ where $ P^{(k)} (\lambda)\not\equal{}0,P^{(k)} (\bar{\lambda})\not\equal{}0$. Thus $ dim(ker(P(T)))\equal{}2k$ is even.  QED.

What is Jorden Theory?

Also, this is an exercise appear on a very basic linear algebra book, so is there any more elementary solution?

see the Jordan normal form in:
http://en.wikipedia.org/wiki/Jordan_normal_form
Here I use also this fact:
Let $ A$ be a real matrix; if in the Jordan normal form of $ A$ we find a non real Jordan block then its conjugate is also in the Jordan form with the same order of multiplicity.

Another method:
$ P(t)\equal{}u.\Pi(t\minus{}\lambda_k)^{a_k}$ and $ ker(P(T))$ is the direct sum of the $ ker((T\minus{}\lambda{I})^{a_k})$. It remains to see that $ ker((T\minus{}\lambda{I})^{a_k})$ and $ ker((T\minus{}\bar{\lambda}{I})^{a_k})$ are of same degree.

I rewrite the 2 last lines of my last post

$ ker((T\minus{}\lambda_k{I})^{a_k})$. It remains to see that $ ker((T\minus{}\lambda_k{I})^{a_k})$ and $ ker((T\minus{}\bar{\lambda_k}{I})^{a_k})$ are of same dimension.

A bit cleaner:

It suffices to show that, for $ c$ a nonzero real number, that $ \ker (T^2 \plus{} c^2)$ has even dimension (by induction and completing the square). Letting $ \phi  \equal{} T^2 \plus{} c^2$, we do this by showing that $ \ker \phi$ is actually a $ \mathbb{C}$ vector space. Endow $ \ker \phi$ with a $ \mathbb{C}$ action by the relation $ i v \equal{} (T/c) v$. Then since $ T^2 v \equal{} \minus{}c^2 v$, it follows that $ i^2 v \equal{} \minus{} v$; therefore we obtain a faithful $ \mathbb{C}$-action, so $ \ker \phi$ is even dimensional as a real vector space.

Nukular is right.
I assumed these facts:
1) Let $ v,\bar{v}$ be a complex basis of the complex plane $ \Pi$; then $ Re(v),Im(v)$ is also a basis of $ \Pi$. But $ Re(v),Im(v)$ is a real basis of a real plane $ P$ which is included in $ \Pi$.
2) If $ T$ is a $ \mathbb{R}$-linear application:$ \mathbb{R}^n\rightarrow\mathbb{R}^n$ then we can associate a $ \mathbb{C}$-linear application:$ \mathbb{C}^n\rightarrow\mathbb{C}^n$ as follows:
$ T(v)\equal{}T(Re(v))\plus{}iT(Im(v))$.
Moreover $ dim(ker(T))$ (as a $ \mathbb{R}$-linear application)=$ dim(ker(T))$ (as a $ \mathbb{C}$-linear application)

Your second induced map can also be written, for real vector spaces $ V$, as

$ T \otimes_{\mathbb{R}} id : V \otimes_{\mathbb{R}} \mathbb{C} \to V \otimes_{\mathbb{R}} \mathbb{C}$.

Secondary question: Say we are not working over $ \mathbb{R}$, but rather $ \mathbb{Q}$. If we have an irreducible polynomial $ P(T)$ of degree $ n$, does it follow that $ \dim \ker(P(T))$ is divisible by $ n$?

Hi Nukular,
Let $ T\in\mathcal{M}_k(\mathbb{Q})$ and $ P$ be an irreducible polynomial over $ \mathbb{Q}$ of degree $ n$.
Let $ E\equal{}ker(P(T))$; we assume $ E\not\equal{}\{0\}$; $ E$ is a $ T$-invariant $ \mathbb{Q}$-subspace. Let $ U\equal{}T_{|E}: E\rightarrow{E}$; then $ P(U)\equal{}0$. If $ \lambda$ is an $ U$-eigenvalue then $ P(\lambda)\equal{}0$ and $ P$ is the $ \lambda$-minimal polynomial. Thus the $ U$-characteristic polynomial (over $ \mathbb{Q}$) is in the form $ P^l$ where $ n.l\equal{}dim(E)$ and $ n$ is a divisor of $ dim(E)$.


algebra
polynomial
superior algebra
superior algebra unsolved

Hello,

I have yet another question about something in my algebra book.
The context is : projective plane curves.

Let $k$ be an algebraically closed field.  Let $f$ be a polynomial in $k[x,y,z]$, homogeneous, and of degree $d$.
We know we can write $f$ as $f_{0}z^{d}+\cdots f_{d}z^{0}$, where the $f_{i}$ are homogeneous polynomials in $x,y$ of degree $i$.

Let $R$ be the ring of fractions of homogeneous polynomials of the same degree, such that the denominator does not become zero for $x=0,y=0,z=1$
Let $m_{P}$ be the ideal in this ring generated by $\frac{x}{z}, \frac{y}{z}$

Prove that the smallest $i$ such that $\frac{f}{z^{d}}\in m_{P}^{i}$ is the smallest $i$ such that $f_{i}\neq 0$.

Seems trivial, I mean : if $f_{0},\cdots, f_{r-1}$ are all zero, then $\frac{f}{z^{d}}$ is in $m_{P}^{r}$, but what about the converse...

I feel silly for asking something like this, but I've been playing with these fractions for quite some time now and I still haven't found a satisfactory answer...
All hints welcome... :huh:

Hmm, I'm still in doubt about this. Is this trivial or very hard? :maybe:


[i][b]WIN ETERNAL AoPS FAME AND GLORY![/b][/i]

This is my rather difficult national competition that I wrote. Because it's a little harder than mathcounts(but covers the same topics) I'm not sure if I should post it here or not. So if the mods want to give me a little nudge, that's ok.

Anyways, you're on your own honor for this, and the same rules apply for the sprint and target as they do for any other competition. PM your answers to me using the answer sheet for sprint, [b]WHICH DOESN'T SAY THE ANSWERS, BUT HELPS YOU EXPRESS THEM IN THE WAY I WANT! [/b]but I never made an answer sheet for target, and have fun. Those of you who send me your answers will recieve the team round as a little bonus. [b]And your results will come to you as soon as possible after you take it.[/b] Requested solutions will also be PMed to you.
As well, for those of you who don't post here, you can email me at nma4@comcast.net

Eratta:
Sprint #7: Please eliminate the words "Can You".(Thanks Nathan :roll: )

Sprint # 10, assume what matters is who goes against who in the first round, who wins, who gets second, and who gets third.

Sprint#22: Someone who has taken it has pointed out that there is no answer, so credit for everyone!

Sprint #25:Assume that there are an infinite number of cereal boxes.

Target #3: A set of dominoes has every distinict combination of two integers from 0-6 once.

Target#6:A doolar is a dollar!

Target#8: There can only be one next closest city, if there isn't then no lines should be drawn.

It looks like not a lot of people want to do math after Nats, but if anyone else participates I'll join in.

awesome, it's good practice for next year as well if you're eligible, because it'll probably be closer to the level of difficulty than most old nats are

Also, if you only do one or two problems, you can still PM or email me the answers, or ask for solutions

So far, nobody has sent me their answers, so if you have them, send them, and you're results and solutions will be near instant!!! Also, it's really hard, so don't be embarassed if you only have a few answers. Sergei only got a three on the target, and we never officially did the sprint.

Do it whenever, it would just be good to have a sense of how hard it was.

I'll do it, but I haven't looked at any problems yet...

I have now made the tests (not the answer key) into .pdf format.  They also now have lines to write your answers on.  I hope that this will be a help to some people.

Thank you. [u][b]Just so everyone knows, the Answer page, doesn't have the answers to the questions, just it tells you how I want the answer expressed.[/b][/u]

So this is a competition (you'll post the results)?

Just because you edited your first post......

Okay, I'll be  [size=200][color=darkred][b]quiet![/b][/color][/size]:rotfl:

Are there any geometry questions on the test? :P JK I'll take it :D (but probably tomorrow or something, cause I'm kinda tired now  ;) )

Nathan, we know we share our hatred of geometry, so there's one question that's geometry, and it's fairly easy, and there's one question that one solution is geometry, so it's not bad at all.

Just some Errata for those who don't look at the first post:

Sprint #7: Please eliminate the words Can You.(Thanks Nathan :roll: )

Sprint # 10, assume what matters is who goes against who in the first round, who wins, who gets second, and who gets third.

Sprint#22: Someone who has taken it has pointed out that there is no answer, so credit for everyone!

Sprint #25:Assume that there are an infinite number of cereal boxes.

Target#6:A doolar is a dollar!

I just took this test and it is really good. So if you don't have anything better to do for 1 hour and 4 minutes, take this test!!!

:D

(It's really hard too   ;) )

[quote="mathgeek2006"]I'm exausted from all that "math" :sleeping:  :stretcher:  :starwars: .....I'm gonna retire for the night.[/quote]

That starwars emoticon is sick.

Also, those of you who send me your answers get the team round as a bonus. I might add bonus points to your score accordingly. Take half an hour for the team round if you do it by yourself, you won't finish it but still try!


Here's how it works: one person gives a problem, another solves it and gives another problem.  But all the problems must be numbered, and each problem must be associated with that number.  Some of the annual competitions would do something like this with the year.  For example, NYSML had this as an individual problem this year: "Find the sum of all the positive factors of 2006."  See what I mean?  Problem [i]n[/i] is centered around the number [i]n[/i].  Well, let's do it!

I'll start with problem 1, which will be centered around the number 1.

[b]PROBLEM 1[/b]
If $G=1+\sqrt{1+\sqrt{1+...}}$ and $R=1+\frac1{1+\frac1{1+...}}$, what is $GR$?

[hide="Hint"]$G=1+\sqrt{1+\sqrt{1+...}}=1+\sqrt{1+G}$
$R=1+\frac1{1+\frac1{1+...}}=1+\frac1R$[/hide]

[hide="Bigger Hint"]Solve for $G-1$, and then use that to find $G$.[/hide]

[hide="Gives It Away"]$x^2-x-1=0$[/hide]

[hide] We see that (g-1)^2=g, so g^2-3g+1=0, so g=$\frac{3+\sqrt{5}}{2}$.  $\frac{1}{r-1}=r$, so $r=\frac{1+\sqrt{5}}{2}$. Thus $gr=2+\sqrt{5}$[/hide]

[b]Problem 2[/b]

I draw 2 cards from a set of 2222 cards numbered 1 through 2222 and multiply the numbers on the cards.  What is the probability that the product obtained is divisible by 22 or 222 or both?


[quote="Γιώργος"]If $ x_{1},x_{2},...,x_{n}\in [a,b]$ with $ 0 < a < b$ and $ n\geq 2$ prove that $ (\sum^{n}_{i \equal{} 1}x_{i})\cdot (\sum^{n}_{i \equal{} 1}\frac {1}{x_{i}})\leq \frac {(a \plus{} b)^{2}}{4ab}n^{2}$ :D[/quote]
it's particular case of holder's inverse inequality look here :
http://www.mathlinks.ro/viewtopic.php?t=135954
just take  ($ p \equal{} q \equal{} 2$ and $ a_{i} \equal{} b_{i} \equal{} x_{i}^{2}$)
proof is elementary, and constant $ C$ you can find at google.ru

or proof is same as here http://www.mathlinks.ro/viewtopic.php?mode=attach&id=7600


[quote="4everwise"]Find the minimum and maximum values of $P=2x+3y$ if the constraints are $x+y\le 8$, $y-x\le 2$, $x\ge 0$, and $y\ge 0$.[/quote]
is there an easier way to solve this without graphing it and start coloring half-planes?

$y + x \le 8$
$y - x \le 2$
$2y \le 10 \Rightarrow 0 \le y \le 5$

Maximum value of P = 21 at the point (3,5).  Minimum value = 0 unless $x \neq y$.

I'm not sure, I use graphs to solve the answer.


logarithms
calculus
calculus computations

Don't get how can I solve it...  :huh:
Every positive number can be expressed as a power of 10, e.g. $ 325.8 \approx 10^{2.513}$

Express the number $ 0.0533$ in this form.  :?:

[quote="@petko"]$ 0.0533 \equal{} 10^x$ so $ \log_{10} 0.0533 \approx \minus{} 1.273 \equal{} x$
$ 0.0533 \approx 10^{ \minus{} 1.273}$[/quote]

Ye, that was so easy. I could have thought of that myself. Thanks a lot.


[quote="dyd"]$a,b,c,d \in \mathbb{R}$.
Prove that
$(a^2+b^2)\sqrt{c^2+d^2}\geq 2abc+d(a^2-b^2)$, and determine when the equality occur.[/quote]
$(a^2+b^2)\sqrt{c^2+d^2}=|(a+bi)^2(c+di)|=\sqrt{((a^2-b^2)c-2abd)^2+(2abc+d(a^2-b^2))^2}\geq$
$\geq2abc+d(a^2-b^2).$
Equality when $(a^2-b^2)c=2abd$ and $2abc+d(a^2-b^2)\geq0.$
I think this is very easy.

hmm, your solution is not fairly easy for me.. :(

I do understand the transforming of the expressions separated by $\geq$, but can you explain the using of $i$ and how to transform one expression to next expression (which separated by $=$)?

Or maybe there's anyone who can post the solution without complex transformation?

Sorry for my foolness..

wew
what a compressed solution!
thank you very much :)


probability
geometry
trapezoid
3D geometry
pyramid

Three numbers are chosen at random between 0 and 1. Wat's the probability that the difference between the greatest and the least is less than 1/3?

[hide="My (Wrong) Solution"]

We have continuous variables, so resort to geo.
It doesn't matter what th emiddle number is, so we simply graph the possibilities of the greatest and the least.

The least is the x-axis, the y is the greatest.
At any value of x, y can be up to x+1/3.
Thus we have a square with a trapezoid filled in.
The area of the trapezoid is $ 1 \minus{} \frac{\frac{2}{3}^2}{2} \minus{} (1)(1)(\frac{1}{2}) \equal{} \frac{5}{18}$

Thus the probability is $ \frac{5}{18}$[/hide]

Apparently, tho, the answer is.. [hide="..."] $ \frac{7}{27}$[/hide]  :maybe:

You can't ignore the middle number like that. There is a very small probability that $ 0.49$ and $ 0.51$ are the smallest and largest, but a large one that they are $ 0.01$ and $ 0.99$, because of that middle number.

[hide="Solution"]
Let $ z\ge x,y\ge0,z \minus{} \frac13$ be the largest variable. If $ 0\le z\le\frac13$, as long as $ x$ and $ y$ are less than or equal to $ z$, it is fine, so this happens with probability $ z^2$. Look at this as it ranges from $ 0$ to $ \frac13$; a pyramid with height $ \frac13$ and square base with sides $ \frac13$ is formed, so the probability that $ z\le\frac13$ is the largest number and the conditions in the question are satisfied is $ \frac13\cdot\frac13\cdot\left(\frac13\right)^2 \equal{} \frac1{81}$. If $ \frac13\le z\le1$, there will always be a range of $ \frac13$ that is necessary and sufficient for $ x$ and $ y$ to be in. Since the possible range for $ z$ for this to happen spans $ 1 \minus{} \frac13 \equal{} \frac23$, the probability that $ \frac13\le z\le1$ is the largest number and the conditions in the question are satisfied is $ \frac23\cdot\left(\frac13\right)^2 \equal{} \frac2{27}$. Combining these cases, the probability that $ z$ is the largest number and the conditions in the problem are satisfied is $ \frac1{81} \plus{} \frac2{27} \equal{} \frac7{81}$, but since any of $ x,y,z$ can be the largest number, the total probability is $ 3\cdot\frac7{81} \equal{} \boxed{\frac7{27}}$.
[/hide]


[i]With $ p$ is an odd prime , $ n$ is an positive integer

   Assume that the equation $ x^2 \plus{} y^2 \plus{} 1 \equiv 0 \ \ (mod \ \ p^n)$ has root

   Prove that the  equation $ x^2 \plus{} y^2 \plus{} 1 \equiv 0 \ \ (mod \ \ p^{n\plus{}1})$ has root[/i]


[size=150]
1. Suppose [b]x, y[/b] ∈ Rⁿ are nonparallel vectors.

(a) Prove that if s[b]x[/b] + t[b]y[/b] = [b]0[/b], then s = t = 0.
(b) Prove that if a[b]x[/b] + b[b]y[/b] = c[b]x[/b] + d[b]y[/b], then a = c and b = d.

2.

(a) Let [b]u, v[/b] ∈ R².  Describe the vectors [b]x[/b] = s[b]u[/b] + t[b]v[/b], where s + t = 1.
(b) Let [b]u, v, w [/b] ∈ R³.  Describe the vectors [b]x[/b] = r[b]u[/b] + s[b]v[/b] + t[b]w[/b], where r + s + t = 1.[/size]

$ 1.$ [url]http://planetmath.org/encyclopedia/LinearIndependence.html[/url]
$ 2.$ [url]http://planetmath.org/encyclopedia/AffineCombination.html[/url]


Hi, this is my first post. Below is a question i don't get on a test.

Box A contains 5 blue balls and 2 red balls. Box B contains 3 blue balls and 4 red balls.
a) a ball is randomly selected from Box A and placed in Box B. A ball is then randomly selected from Box B. What is the probability that the ball selected from Bag B is blue?

b) if a blue ball is selected from Box B, what is the probability that a red ball was transferred from Box A to Box B?

[hide="b"]
We have 6 ways of red then blue (2x3) and 20 ways of blueblue(5x4), so 6/26=3/13
[/hide]

Mewto's solution is better! (mine is unclear)

thanks. the teacher actually taught us the method RunpengFAILS used.


[quote="maokid7"]If you receive 12 phone calls in one week what is the probability that you received atleast 1 call each day.
[hide]
i was thinking find the number of solutions to $a_{1}+a_{2}+\dots+a_{7}=12$ where $a_{i}>0$ and are integers and let this be $Y$
and then let $X$ equal the number of solutions to $b_{1}+b_{2}+\dots+b_{7}=12$ where all $b_{i}\geq{0}$ and are integers.
so the probability would just be $\frac{Y}{X}$ but the book says that this is incorrect. what did i do wrong?
[/hide][/quote]

Here's what I did... the probability seems a little low... well actually very low

[hide]Count the number of ways to not receive a phone call every day. Exclude one day so you have 6 days and 12 calls. The number of ways is the number of non-negative solutions to $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=12$. Then you include the seventh day but you put a call in each day so that you know that every day you got a phone call. So you have 5 calls left and $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}+x_{7}=5$. The total number of ways to distribute the calls is $\binom{11}{6}+\binom{17}{5}$. The number of ways to not get a call one day is $\binom{17}{5}$. So then the answer is $1-\frac{\binom{17}{5}}{\binom{11}{6}+\binom{17}{5}}=\boxed{\frac{33}{475}}$?[/hide]

[quote="ch1n353ch3s54a1l"][quote="maokid7"]If you receive 12 phone calls in one week what is the probability that you received atleast 1 call each day.
[hide]
i was thinking find the number of solutions to $a_{1}+a_{2}+\dots+a_{7}=12$ where $a_{i}>0$ and are integers and let this be $Y$
and then let $X$ equal the number of solutions to $b_{1}+b_{2}+\dots+b_{7}=12$ where all $b_{i}\geq{0}$ and are integers.
so the probability would just be $\frac{Y}{X}$ but the book says that this is incorrect. what did i do wrong?
[/hide][/quote]

Here's what I did... the probability seems a little low... well actually very low

[hide]Count the number of ways to not receive a phone call every day. Exclude one day so you have 6 days and 12 calls. The number of ways is the number of non-negative solutions to $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=12$. Then you include the seventh day but you put a call in each day so that you know that every day you got a phone call. So you have 5 calls left and $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=5$. The total number of ways to distribute the calls is $\binom{10}{5}+\binom{17}{5}$. The number of ways to not get a call one day is $\binom{17}{5}$. So then the answer is $1-\frac{\binom{17}{5}}{\binom{10}{5}+\binom{17}{5}}=\boxed{\frac{9}{230}}$?[/hide][/quote]

I don't think that's quite right. For one thing, you're only distributing the last $5$ calls among $6$ days, which I don't get. Shouldn't it just be all $7$ days?

Here's how I did it (maybe I'm wrong?):

# of ways to get at least $1$ call each day: Assign $1$ call to each day. Then distribute $5$ among the rest of the $7$ days. This is $\binom{11}{5}$.

# of total ways to get $12$ calls in a week: Distribute $12$ among $7$ days. This is $\binom{18}{6}$.

Thus the probability is $\frac{\binom{11}{5}}{\binom{18}{6}}= \frac{11}{442}$.

Agh I screwed the second one up.....edited :ninja:

[quote="maokid7"]why do u do 11 choose 5 and 18 choose 5 and not 18 choose 6?[/quote]

Oops. You're right; I went through a little too quickly.


LaTeX
real analysis
real analysis unsolved

:) but I can't slove it. ;)
$f\in C^3[a,b]$
prove there exist a $k$ such that:
$f(b)=f(a)+\frac{(b-a)}{2} (f'(a)+f'(b) )-\frac{1}{12}(b-a)^3f'''(k)$
help me.thank you

Let $F(x)=f(x)-f(a)-\frac{1}{2}(x-a)(f'(a)+f'(x))$ and $G(x)=(x-a)^3$.
Try to apply Cauchy's mean-value theorem for two times..

[quote="zhaobin"]$f∈C^3[a,b]$[/quote]Use \in for that symbol in $\text{\LaTeX}$: $f\in C^3[a,b]$

thank you,liyi and jmerry.I used with TAYLOR formal to Expansive but I failed :)  ;)

[quote="the game"]I found this problem in one of my college exam & I am pretty sure that it can be done using finite Taylor series expansion.[/quote]
can you show the Taylor series expansion.
thank you in advance :)

[quote="Myth"]Zhaobin!

Please, read http://www.mathlinks.ro/viewtopic.php?p=140098 and name problems properly next time!

Myth[/quote]
thank you for reminding me.Myth!
but which name can be proper.
do you think "$a median problem$" will be better :)

Asked by Leitingok in 2013

[url= https://math.stackexchange.com/questions/295246/application-for-mean-value-theorem] solution by saz  [/url]
https://math.stackexchange.com/questions/295246/application-for-mean-value-theorem


[quote="Svejk"]Let $(x_{n})_{n \geq 1}$ be a squece of real numbers such that $\forall n\in \mathbb N$ ,$n\geq 2$ we have:
(*) \[x_{1}-C_{n}^{1}\cdot x_{2}+C_{n}^{2}\cdot x3-...+(-1)^{n}\cdot C_{n}^{n}\cdot x_{n+1}=0 \]
Prove that:
\[x_{1}^{k}-C_{n}^{1}\cdot x_{2}^{k}-...+(-1)^{n}\cdot C_{n}^{n}\cdot x_{n+1}^{k}=0 , \forall k\in N , n\geq k+1 \]
[/quote]
Must be (*) for all $n\ge 1$.


Product
number theory
IMO
imo 1976
maximization

Determine the greatest number, who is the product of some positive integers, and the sum of these numbers is $1976.$

Did I understand this question well:
 You have to find largest number which can be represented as $xy$ s.t. $x+y=1976$?
Because if that is a question then it is trivial:
 $x+y\geq 2\sqrt{xy}$ so $(x+y)^2\geq 4xy$ so that number is $\frac{1976^2}4=976144$
m@re

You can get a number which is MUCH bigger. For example, you could get $2^{988}$ :)

The biggest number you can find is $2.3^{658}$.

Proof: Assume there was any factor 5 or more. we can split 5=2+3 which restults in a factor 6>5. Same for numbers larger than 5. So there are none.

Thus we have the optimal number is of the form $2^m.3^n$. And since $3^2>2^3$, we need as much as possible into 3's, thus we end up with $2.3^{658}$.

Well, that's why I asked did I understand this question :D

Maybe we can find more: If $x_1,x_2,...,x_n$ are reals with $\sum x_i = 1976$, what is the maximum value $\prod x_i$ can obtain? Or maybe easier to start with, what is the value of $n$?

Simple computations gives that when $n$ is given, it's optimal to take $x_1=x_2=...=x_n$.
So we want $(\frac{a}{n})^n$ ($a=1976$ here) to be as big as possible.
Deriving $f(x)=(\frac{a}{x})^x=e^{x \ln \frac{a}{x}}$ gives $f'(x)=e^{x \ln \frac{a}{x}} (ln(\frac{a}{x})-1)$, so it attains it's only maximum at $x=\frac{a}{e}$. Now $n$ has to be the next greater or smaller integer (it's monotonous on both sides).

I just base cashed using AM-GM;

$ \left(\frac{1976}{n}\right)^n \ge a_1 a_2 ... a_n$

Then, as the product was an integer n must be a factor of 1976 and the prime factors of 1976 are;

2,2,2,13,19 then consider all combinations of these.

However, I don't know if I am missing something - it just seems a bit too simple

Exactly the same except that i explained a bit more on why we chose (2 and 3 and 3)not (4 and 4) or (5 and 3) or(1 and 4 and 3)

how do you know you must do an additional 2? how do you know that we should not separate the 2's and distribute them onto 3's?

what we know that is that $3^4>4^3$, but this does not mean $3\times3\times2>4\times4$ (without calculation)

[quote=Peter]Proof: Assume there was any factor 5 or more. we can split 5=2+3 which restults in a factor 6>5. Same for numbers larger than 5. So there are none.[/quote]

For this we first need to prove this claim. Anyone here please prove this.

[quote=starchan]If @above's solution is correct, then we have a nice way of giving explicit computations for $\lfloor e^{\frac{n}{e}}\rfloor$ for all positive integers.[/quote]
I have found error in my solution. Whatever. you can show the way of computing the value of $\lfloor e^{\frac{n}{e}}\rfloor$ what you have found.

[quote=Abdullahil_Kafi]The critical part is it says about positive integers. If positive reals, then the answer is $e^{\frac{1976}{e}}$. As it says for positive integers then have to work a little.
We get from AM-GM that $\left(\frac{1976}{n}\right)^n\geq t$ where $t$ is the product of $n$ integers. To maximize the value of $f(x)=\left(\frac{1976}{x}\right)^x$ we have to find the first derivative of this then have to make it zero.

Applying chain rule we have $f'(x)=e^{xln\frac{1976}{x}}\cdot f'(xln\frac{1976}{x})=e^{xln\frac{1976}{x}}\cdot (ln\frac{1976}{x}-1)$.
Hence we get $ln\frac{1976}{x}-1=0\Longrightarrow x=\frac{1976}{e}$

We have to do some case work. We have find the integer solutions of the plot of $f(x)$ and the highest $y$ value is finally the answer. The value $e^{\frac{1976}{e}}$ is useful for the reason that the nearest values of it will be actually the answer (I think). The rest is little boring therefore I didn't find.[/quote]

Umm you cannot have a nonintegral number of reals. $x$ must remain integer no matter what. You cannot have $x = \frac{1976}{e}$ because you cannot have those many positive reals. (this number should be an integer no matter what)

Suppose some positive integer $k\ge 5$ is there. Then replace $k$ with $2$ and $k-2$. We have $2(k-2)=2k-4>k$, so it is more optimal. If some $k=4$ is there, then replace it with two $2$'s. Thus, optimally, all these positive integers are either $2$ or $3$. Now we can replace three twos with two threes, because $3^2>2^3$. So the greatest is $\boxed{2\cdot 3^{658}}$.

Let $x_1, x_2, \dots, x_n$ be positives summing to $1976$ and $M$ be the max product of $x_1x_2 \dots x_n$. By observation we find that $M$ is of the form $2^r3^s$ implying that $2r+3s=1976$. It's obvious that to get max we must have as many 3's more than 2's possible (since $3^2> 2^3$). So, $2r \equiv 1976 ( \text{mod } 3) \implies (k,l)= (1,658) \implies M=2 \cdot 3^{658}.$               $\square$


algebra
polynomial
linear algebra
linear algebra unsolved

$E_n$ is the vectorial space of homogenous polynomial $P(X,Y) \in R[X,Y]$
of degree $n$.
$A_n$ the subset of $E_n$ of those polynomial that are multiple of $(X^2+Y^2)$
$H_n$ the subset of $E_n$ of polynomial with nul laplacian.

Show that $E_n$ is the direct product of $A_n$ and $H_n$

Alekk Does the problem is true with three variables

$E_n$ is the vectorial space of homogenous polynomial $P(X,Y,Z) \in R[X,Y,Z]$
of degree $n$.
$A_n$ the subset of $E_n$ of those polynomial that are multiple of $(X^2+Y^2+Z^2)$
$H_n$ the subset of $E_n$ of polynomial with nul laplacian.

then  $E_n$ is the direct product of $A_n$ and $H_n$


integration
real analysis
real analysis unsolved

Show that:

$ L(1/2, \chi) \equal{} \sum_{n\equal{}1}^A \frac{\chi(n)}{\sqrt{n}} \plus{} O(A^{\minus{}1/2})$

I will assume that $ \chi$ is a nontrivial Dirichlet character, so that $ s_n = \sum_{k}^{n} \chi(k)$ is bounded. That is, there is $ M > 0$ such that $ |s_n| \leq M$. Then for $ s > 0$, summation by parts shows that

\begin{eqnarray*}
\left| \sum_{k=m}^{n} \frac{\chi(k)}{k^s} \right|
& = & \left| \sum_{k=m}^{n} \frac{s_k - s_{k-1}}{k^s} \right| \\
& = & \left| \frac{s_n}{n^s} - \frac{s_{m-1}}{m^s} + \sum_{k=m}^{n-1} s_k \left( \frac{1}{k^s} - \frac{1}{(k+1)^s} \right) \right| \\
& \leq & \frac{|s_n|}{n^s} + \frac{|s_{m-1}|}{m^s} + \sum_{k=m}^{n-1} |s_k| \left| \frac{1}{k^s} - \frac{1}{(k+1)^s} \right| \\
& \leq & \frac{|s_n|}{n^s} + \frac{M}{m^s} + M \sum_{k=m}^{n-1} \frac{s}{k^{s+1}} \\
& \leq & \frac{|s_n|}{n^s} + \frac{M}{m^s} + M \int_{m-1}^{\infty} \frac{s}{x^{s+1}} \, dx \\
& = & \frac{|s_n|}{n^s} + \frac{M}{m^s} + \frac{M}{(m-1)^s}
\end{eqnarray*}

Thus taking $ \limsup_{n\to\infty}$ to both side, we have $ \sum_{k=m}^{\infty} \frac{\chi(k)}{k^s} = O\left( \frac{1}{m^s} \right)$.


MATHCOUNTS
probability
percent
ratio
conditional probability

(this is not a mathcounts problem, but a good probability problem from a math/ed class I had in college)

An auto accident occured outside the window on an elderly lady's apartment.  The car in question sped away but not before the woman looked out her window and saw the car.  When the police questioned her she told the police the car was definitely a Taxi.  Since in her city all cabs are either Green Cabs or Blue Cabs, the officer asked her which company the cab came from and she said the cab was a Blue Cab.

When the case eventually came to a jury trial against the Blue Cab company the lawyers in an effort to question the witness had her eyesight tested by having her identify the color of several cabs going past her window in a similar fashion.  She correctly identified the cab color 80% of the time.

In this city 85% of all cabs are Green and the other 15% are Blue.  It is your job as a member of the jury to determine the probability of the woman having correctly identified the cab in the accident.

I've seen the answer to this one in a book I read. Jump in here; try it out. Just remember to think carefully.

I would post, but if I did, frost (my coach) would kill me.
Anyways, hint:
[hide="hint-dont read unless REALLY stymed!!!"]
It says probability in percents right? But who says its x over 100?[/hide]

Well, maybe the old lady was color blind to green or something :D

not 72%, your best off as a common fraction

Try making a probability tree and thinking of it in terms of favorable/possible

Keep trying

I think this is a classic example of [b]conditional probability[/b].

Not 80% this is a problem that is not as easy as it looks, yet in the end not all that hard either :)

It is conditional probability.  The problem doesn't just ask for the probability that she is right, but the probability that she says "Blue" and is right.

[quote="frost13"]It is conditional probability.  The problem doesn't just ask for the probability that she is right, but the probability that she says "Blue" and is right.[/quote]

To nitpick, you're not really looking for a simple "and" here, what you're looking for, in the traditional language of conditional probability problems is the probability that the cab was blue, [i]given that [/i]she said "blue".

I hope frost won't mind if I add this hint:
[hide]So, you need to consider, what are all the cases in which she might say blue, and in what proportion of those is the cab actually blue.[/hide]

if she identifies the right color 80% of the time and 15% of all cabs are blue so would it no be: 4/5*3/20 = 3/25 = 12%? probably wrong...

Shake, you've got half the answer :)  Closest one yet.

Mathmom was right too!!  What you have is the probability she says blue and is right.

If you define probability as favorable / possible, or success / attempts, then Shake has the favorable or success part.  Keep going.

since we have the probability that she correctly identifies the cab and says its blue (which is correct)
now i guess we have to find the probability that she identifies the cab and says its green: .85 * .8 = 68%
so it seems that she correctly identifies the car as blue 12% of the time, correctly identifies the car as green 68% if the time, and is completely wrong 20% of the time...

getting close still shake, but if being RIGHT about BLUE is favorable, the possible wouldn't involve identifying GREEN.

she has a 12% chance of identifying the car correctly provide that she says blue.
but what about the probability that she identifies the car incorrectly but still says blue: 20% * 15% = 3%
so the ratio of these probabilities is:  4:1
is this any good?

shake, now you have the right idea but wrong multiplication.  If she incorrectly says blue, what color is the cab?  You put 15%, but those are the cabs that are blue.  If she says it wrong and says blue, that means the cab was green, not blue, so not 15%.  Almost there, but use a fraction for final answer.

keep going

85% of the cabs are green... so the probability that she incorrectly says that the cab was blue is: 1/5*17/20 = 17%
so our ratio is:
(3/20*4/5) / (1/5*17/20) = 12/17
so the probability that she correctly identified the car is 12/29....
so as a member of the jury... she does not seem that reliable

Good Job Shake, your persistence paid off!!!  That is correct

[quote="shake9991"]thanks...
to rcv: how about soroban's solution now that a middleschooler has gotten it.[/quote]
Soroban's solution has been restored.

[hide="Some comments on the solutions"]I think this was an excellent example of a problem, generally introduced at a high-school or college-level course, that [b]can[/b] be solved by middle schoolers, using middle school techniques.  Thanks to Coach Frost for providing the question and the necessary guidance to get to a solution.

Soroban's solution is very good.  However, it uses a theorem that I would not introduce to middle school students.[/hide]

[quote="Soroban"]It IS a conditional probability problem!

[hide]
Let $G$ = the cab is Green, $B$ = the cab is Blue.

We are given the following probabilities:
$Pr(G) = 0.85,\;Pr(B) = 0.15$
$Pr(\text{Right}) = 0.80,\;Pr(\text{Wrong}) = 0.20$

There are four cases:
[color=yellow]. . . [/color]$(1)\;Pr(G \land \text{said Green}) = (0.85)(0.80) = 0.68$
[color=yellow]. . . [/color]$(2)\;Pr(G \land \text{said Blue})\;\; = (0.85)(0.20) = 0.17$
[color=yellow]. . . [/color]$(3)\;Pr(B \land \text{said Blue})\;\; = (0.15)(0.80) = 0.12$
[color=yellow]. . . [/color]$(4)\;Pr(B \land \text{said Green}) = (0.15)(0.20) = 0.03$

In cases (2) and (3), she said the cab was Blue.
[color=yellow]. . . [/color]Hence, $P(\text{said Blue}) = 0.17 + 0.12 = 0.29$

The problem asks "What is the probability that the cab $is$ Blue, given that she said Blue?"
[color=yellow]. . . [/color]That is: $Pr(B\;|\;\text{said Blue})$

Bayes' Theorem says: $Pr(B\;|\;\text{said Blue}) \;= \;\frac{Pr(B \land \text{said Blue})}{Pr(\text{said Blue})} $

[color=yellow]. . . [/color]The numerator is Case (3): 0.12
[color=yellow]. . . [/color]The denominator was determined above: 0.29

Therefore: $Pr(B\;|\;\text{said Blue}) \;= \;\frac{0.12}{0.29} = 0.413793 \;\approx \;41.4\%$

[size=75]Solution restored by moderator rcv on Sun Feb 13, 2005 15:52 EST[/size][/hide][/quote]
that's elegant... Bayes' theorem probably could be helpful for National Mathcounts only

I have a similar problem at school that I will post soon and see if you can make the connection between the two.  Thanks to rcv for moderating the discussion and good job shake for sticking with it

No way i coulda solved these problems without help


Prove that if $ A\in M_{n\times n}$ has $ n$ distinct eigenvalues, then $ A$ is diagonalizable.

Choose $ n$ eigen-vectors $ v_1, v_2, ..., v_n$. Clearly they are linearly independent. And the matrix $ (v_1 v_2 ... v_n)$ diagonalises A.

Start by noting that if a matrix is diagonalizable that means that it is similar to a diagonal matrix. The complete theorem is:

A square matrix $ A \in M_{n \times n}$ is similar to a diagonal matrix if and only if A has n linearly independent eigenvectors.

Notice that it is not necessary for A to have n disctinct eigenvalues. To prove the above theorem, go through the following steps:

1) Show that two similar matrices share the same eigenvalues.

2) Show that every $ n \times n$ diagonal matrix has $ n$ linearly independent eigenvectors.

3) Show that if a matrix $ A \in M_{n \times n}$ is similar to a diagonal matrix, then it has $ n$ linearly independent eigenvectors.

4) Show that if a matrix$ A \in M_{n \times n}$ has $ n$ linearly independent eigenvectors, then A is similar to a diagonal matrix. This will complete the proof.


Find integers $ x,y,z$ such that $ xy\plus{}yz\plus{}zx\equal{}2\plus{}xyz$

Assume WLOG that x>=y>=z.  If z=1, then xy+x+y=2+xy, so x+y=2, x=y=1.
So (1,1,1) is a solution.  If z=2, then xy+2x+2y=2+2xy, so xy+2=2x+2y, (x-2)(y-2)=2.  Thus x=4, y=3.  So (4, 3, 2) is a solution.
If z>=3, then xyz+2>3xy>xy+xz+yz.

Therefore, the only solutions are (1,1,1), (4,3,2) and permutations of those.

[quote="abcak"]sorry, but what is WLOG?  Also, what do you mean by "permutations of these"?

Also, the problem just says "intergers", not "positive intergers[/quote]

WLOG = Without loss of generality

by permutations of these he means that $ (2,3,4),(4,2,3),(2,4,3)\cdots$ are also answers because they all have the same numbers but in different order.

Solution $ \text{\LaTeX}$'xed:

[quote="bpgbcg"]Assume WLOG that $ x\ge y\ge z$.  If $ z=1$, then $ xy+x+y=2+xy$, so $ x+y=2$, $ x=y=1$.
So $ (1,1,1)$ is a solution.  If $ z=2$, then $ xy+2x+2y=2+2xy$, so $ xy+2=2x+2y$, $ (x-2)(y-2)=2$.  Thus $ x=4, y=3$.  So $ (4, 3, 2)$ is a solution.
If $ z\ge3$, then $ xyz+2>3xy>xy+xz+yz$.

Therefore, the only solutions are $ (1,1,1), (4,3,2)$ and permutations of those.[/quote]

Actually, I think there are $ sideways8$ solutions.
i.e., z=1, x+y=1 and permutations as bpgbcg pointed out.

Also, i don't get the contradiction when z=3+

If $ z \geq 3$, then $ xyz \plus{} 2 > 3xy$. Furthermore, $ xy \plus{} xz \plus{} yz < 3xy$ (since $ x \leq y \leq z$), so $ xy \plus{} xz \plus{} yz \equal{} xyz \plus{} 2 > 3xy > xy \plus{} xz \plus{} yz$, a contradiction.

[quote="abcak"]$ sideways8$[/quote]

\infty = $ \infty$.

We get all sorts of solutions if we don't assume the numbers are positive. Some examples:

$ x\equal{}0 \implies yz\equal{}2$
$ x\equal{}1 \implies y\plus{}z \equal{} 2$

I don't see any other families immediately. Maybe it will help to rewrite the equation as:

$ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z} \equal{} 1\plus{}\frac{2}{xyz}$

By shifting the variables by $ 1$ (i.e. $ x\equal{}a\plus{}1$), then the condition is $ abc\equal{}a\plus{}b\plus{}c$.

So $ a(bc\minus{}1)\equal{}b\plus{}c$. i) $ bc\equal{}1$, $ b\plus{}c\equal{}0$ no solution.

ii) Else, $ |b|\plus{}|c|\ge |b\plus{}c|\ge |bc\minus{}1|\ge |bc|\minus{}1$ where we use the triangle inequality, the fact that $ |a|\ge 1$, and the triangle inequality again. Then we have

$ 2\ge (|b|\minus{}1)(|c|\minus{}1)$

So anyways, $ b\equal{}0$ gives $ a\equal{}\minus{}c$ so $ (t,0,\minus{}t)$ is a solution.

The rest is easy casework (and there are a finite number of other solutions).


function
geometry
calculus
derivative
real analysis
real analysis unsolved

given $ a,b,d>0$ consider function $ f(x)\equal{}\sqrt{a^2\plus{}x^2}\plus{}\sqrt{b^2\plus{}(d\minus{}x)^2}$ on $ [0,d]$

$ f$ continuous and $ [0,d]$ compact so $ f$ has minimum.

find it, but don't use the well known fact $ f'(x)\equal{}0$ at extreme-point

[hide="Hint."]Mirror. [hide="More explicitly."]Where does the line segment from $ (0,a)$ to $ (d,\minus{}b)$ go?[/hide][/hide]

[quote="primenumber"]given $ a,b,d > 0$ consider function $ f(x) \equal{} \sqrt {a^2 \plus{} x^2} \plus{} \sqrt {b^2 \plus{} (d \minus{} x)^2}$ on $ [0,d]$

$ f$ continuous and $ [0,d]$ compact so $ f$ has minimum.

find it, but don't use the well known fact $ f'(x) \equal{} 0$ at extreme-point[/quote]
Simple geometry :)
I would like to see a calculus explanation too though..

Let $ |AO| \equal{} a$,  $ |OB| \equal{} b$, $ |OD| \equal{} d$, $ |OX| \equal{} x$

So the $ |AX| \equal{} \sqrt {a^2 \plus{} x^2}$ and $ |DX| \equal{} \sqrt {b^2 \plus{} (d \minus{} x)^2}$ So $ |AX| \plus{} |XD| \ge |AD| \equal{} \sqrt {(a \plus{} b)^2 \plus{} d^2}$, so the minimum is $ \sqrt {(a \plus{} b)^2 \plus{} d^2}$

Edit: Dayum... How do I insert my geogebra picture..? "Insert picture" doesn't work... :S

[quote="Mathias_DK"]I would like to see a calculus explanation too though..[/quote]
I just ground through it. Take the derivative of that $ f,$ set $ f'(x)\equal{}0.$ Put one square root on each side of the equation. Square. Clear fractions. Cancel terms so that an apparent quartic equation turns out to be quadratic. Solve. Put results back into expression for $ x.$

Lots of algebraic manipulation, with not a lot of insight. If you really want to see it, do it yourself. Since you already know what the answer is, it should be possible to de-bug what you have.


number theory
superior algebra
superior algebra unsolved

hi

I have briefly started looking at 7-adic residues and have found that the only non-trivial ones of 29 are -12,12. Why is it that 29 has so few? Is it because it is the smallest prime of the form 7k+1? Any help or pointing to relevant theorems would be greatly appreciated.

Thank you very much

what do you mean by "$ 7$-adic residues of $ 29$"?

if there exists an x s.t. x^7 is congruent to a (mod 29) then a is a 7-adic residue of 29.

I think it can be explained elementarily by Fermat's Little Theorem.

Let $ x^7 \equiv a (mod 29)$,

Since by Fermat's Little Theorem, $ x^{28} \equiv 1 (mod 29)$ if $ x$ isn't a multiple of 29,
i.e. $ a^4 \minus{} 1 \equiv 0 (mod 29)$, so it has at most 4 roots, which are 1, - 1, 12, -12 if the calculation of amitie is correct.

I wouldn't call this a $ 7$-adic residue. It's an $ 7$-th power or something like that.

"$ 7$-th power residue" is the common nomenclature. Of course it's quite unrelated to anything $ 7$-adic, which is what you get by investigating solutions modulo $ 7$, $ 7^2$, $ 7^3$ etc. Here you are interested in solutions mod $ 29$.

ah ok. i was just using the definition from here: http://mathworld.wolfram.com/ReciprocityTheorem.html

Just an advice: MathWorld is very often a very unreliable source (for some topics, more than half of the entries have errors).


Find the max value of

$ \sqrt[3]{4x\minus{}3\plus{}\sqrt{16\minus{}24x\plus{}9x^{2}\minus{}x^{3}}} \plus{} \sqrt[3]{4x\minus{}3\minus{}\sqrt{16\minus{}24x\plus{}9x^{2}\minus{}x^{3}}}$

when x € (-1,1).

$ f(x)_{max,x\in [\minus{}1;1]}\equal{}2$ when $ x\equal{}1$  :wink:

[quote="zweig"]Find the max value of

$ \sqrt [3]{4x \minus{} 3 \plus{} \sqrt {16 \minus{} 24x \plus{} 9x^{2} \minus{} x^{3}}} \plus{} \sqrt [3]{4x \minus{} 3 \minus{} \sqrt {16 \minus{} 24x \plus{} 9x^{2} \minus{} x^{3}}}$

when x € (-1,1).[/quote]

$ \sqrt {16 \minus{} 24x \plus{} 9x^{2} \minus{} x^{3}}\equal{}\sqrt{(1\minus{}x)(4\minus{}x)^{2}}\equal{}[\minus{}1<x<1 ]\equal{}(4\minus{}x)\sqrt{1\minus{}x}$

$ \sqrt [3]{4x \minus{} 3 \plus{} \sqrt {16 \minus{} 24x \plus{} 9x^{2} \minus{} x^{3}}}\equal{}\sqrt[3]{4x\minus{}3\plus{}(4\minus{}x)\sqrt{1\minus{}x}}\equal{}\sqrt{1\minus{}x}\plus{}1$ --->

$ \sqrt [3]{4x \minus{} 3 \plus{} \sqrt {16 \minus{} 24x \plus{} 9x^{2} \minus{} x^{3}}} \plus{} \sqrt [3]{4x \minus{} 3 \minus{} \sqrt {16 \minus{} 24x \plus{} 9x^{2} \minus{} x^{3}}}\equal{}\sqrt{1\minus{}x}\plus{}1\plus{}1\minus{}\sqrt{1\minus{}x}\equal{}2$

If I understood it correctly, you proved it is [b]always[/b] 2, right?
[quote="zaya_yc"]
$ \sqrt [3]{4x \minus{} 3 \plus{} \sqrt {16 \minus{} 24x \plus{} 9x^{2} \minus{} x^{3}}} \equal{} \sqrt [3]{4x \minus{} 3 \plus{} (4 \minus{} x)\sqrt {1 \minus{} x}} \equal{} \sqrt {1 \minus{} x} \plus{} 1$ [/quote]
This step was a  nice observation  :lol:

We set $ A \equal{} 4x \minus{} 3$ and $ B \equal{} 16 \minus{} 24x \plus{} 9x^{2} \minus{} x^{3}$

Observing that $ \sqrt [3]{A \pm \sqrt {B}} \equal{} \sqrt [3]{(A \pm \sqrt {B})\cdot1\cdot1}$ then we obtain from AM-GM :

$ \sqrt [3]{A \plus{} \sqrt {B}} \plus{} \sqrt [3]{A \minus{} \sqrt {B}} \leq \frac {(A \plus{} \sqrt {B} \plus{} 2) \plus{} (A \minus{} \sqrt {B} \plus{} 2)}{3} \equal{} \frac {2A \plus{} 4}{3} \equal{} \frac {8x \minus{} 2}{3}$

But, $ \frac { \minus{} 10}{3}\leq\frac {8x \minus{} 2}{3}\leq 2$ when x € [-1;1], indeed by transitivity we obtain :

$ \sqrt [3]{A \plus{} \sqrt {B}} \plus{} \sqrt [3]{A \minus{} \sqrt {B}} \leq 2$,

which is the max, reached when $ x \equal{} 1$.


algebra
polynomial
number theory solved
number theory

Consider a polynomial P(x) with real coefficients. Prove that if there exist infinitely many pairs of rationals (a,b) s.t. P(a)=b then all of the coefficients of P are rational.

Probably I am idiot, but this sounds too easy. Just consider a suficiently large linear system ( that is, the number of equations is the number of coefficients) and express the coefficients in terms of Vandermonde determinants and other determinants will all entries rational. Probably I misuderstood the question.

Or we can use Lagrange formula.... :lol: 0

sorry i'm the idiot here  :blush:.  i was too concerned trying to find a counter-example to see the obvious solution.