Albatross was very angry that Frankinfueter killed his pets, so he decided to go to the local supermarket and buy poison to poison Frankinfueter. When he goes there, he sees rat poison, which costs three dollars and has a 40% chance of poisoning people. He also sees cat poison, which costs six dollars fifty cents and has an 80% chance of poisoning people. Finally, he sees a poison making kit that has a 20% chance of killing people but only costs one dollar and twenty-five cents. Which poison should he buy to be the most cost-effective?

Assuming we're using simple ratios and not some other complex form of math that doesn't belong in getting started:

[hide="My Solution"]40/3=13.3...
80/6.5=12.3...
20/1.25=16
The cat poison is most cost-effective.[/hide]

What I did was set up ratios to find out which one would kill in the least amount of money.

[hide]
So we have:
rat poison:
40%x = 3
x = 7.5

cat poison:
80%x = 6.5
x = 8.125

make it yourself kit:
20%x =1.25
x = 6.25

So the kit is the most cost effective!

On a side note, is there even such thing as cat poison?  :lol:

[/hide]

Yes, I patented cat poison  .

It's just simple ratios.

RC-7th:

[hide]

RC-7th, your method is valid, except you're looking for the highest number in your case I think  . If human poison poisoned 100% of the time and cost 3 dollars, it would be rated as 33.3 with your method. Obviously, the higher the number, the better it is if you do it your way.

[/hide]

[quote="RC-7th"][quote="Treething"]Yes, I patented cat poison :P .[/quote]

Have you patented other items which don't work well? I'd like to send them to some people.[/quote]

Yes, I've patented K'Nex crossbows that fire up to 20 feet and use blunt bolts :P .

[quote="bradpreston"]
I don't think that this is as easy as it seems, consider the following simple example:
At the casino I can play on two slot machines,
on machine A I have a 40% chance of winning the jackpot, each attempt costs 1 dollar.
on machine B I have an 80% chance of winning the same jackpot, each attempt costs 2 dollars.

I can only win the jackpot once, but I have to decide how many times i'm going to play each machine at the start and I have to pay that amount upfront. Are the two choices equivalent ?

Suppose I spend 2 dollars on machine A and 2 dollars on machine B. I play one dollar on machine A, if I win then I stop, If I lose then I play again. What chance do I have of winning?

Then I play machine B, I only have one attempt with an 80% chance of winning.

Which machine represents the best value for money?

(I've tried to make this as equivalent to the previous question as possible)
[/quote]

Well no, in this problem, he only buys one container.

[quote="Treething"]
Well no, in this problem, he only buys one container.[/quote]

Yes, but do you think a 40% chance of achieving a certain outcome is worth half of an 80% chance of acheiving the same outcome?

The probablity of winning on the first machine is $4/10+6/10*4/10=16/25$. The probablity of winning on the second machine is $8/10=20/25$. So the second machine is worth more.

,but in LA casino, there will be alot of fake ilegal machines that never makes the jackpot...

and even if they did have jackpot, the actual probability that u can get to have a jackpot is 50%because it's like yes or no stuff.. even it's 99% machine,there is still probability that u might lose.. you win or lose.. which makes everything 50% for these kind of problems

What?

OK, let's put it this way. In your terms, if you have 5,000,000 red coins and 1 black coin, you have a 50-50 shot of getting the black one, as there are two possibilites. Explain to me how that's true.

simple... I remove 4,999,999 red coins before I actually draw.
well: you theorectically could look at all probability as 50-50 chance... but it would be very wrong... take this problem for instance:
(i' ve borrowed the characters)
Albatross convinced Frankinfueter to jump off of a bridge. The probability that Frankinfueter jumps is 10% , the probability that he  severely injures himself(not neccessarily from jumping) is 60%. Assuming that if Frankinfueter does not jump he pushes Albatross off,and Albatross gets severly injured with a 75% chance(assume that this probability is prob. that he falls and gets injured). What is the probability that SOMEONE gets severely injured? definitely not 50-50
Hint:
[hide]first see prob that frankinfueter falls and gets injured then do it for albatross<- dependent of to things:1) Frankinfueter NOT jumping/pushing Albatross, 2)Albatross' prob that he gets hurt (severely injured)
[/hide]

i answered the original question by removing the chance of complete failure from 1. oh and by complete failure i mean assuming that Albatross spent the same amout of money for diffrent number of doses of each poison (say 6.50 for 1 dose of cat poison, 6.50 for 2 1/6 doses of rat poison and 6.50 for 5.2 doses of selfmade) and used all doses on Frankinfueter.

so....[hide]the chance of complete failure is the chance of one failure to the power of the number of doses
the chance of success is the chance of complete failure subtracted from 1

cat poison-- 80%
rat poison-- 1-0.6^(13/6)=67%
selfmade-- 1-0.8^5.2=69%

cat poison is the most cost efficient[/hide]

as to the casino problem...
[hide]machine A-- 1-.6^2=64%
machine B-- 80%

so machine B is better
...but if you dont stop after winning on machine A, you can win twice, 0.4 ^2 chance
[/hide]

well, I think of it ina different way..

 :blush: but when I first thought about it, I thought it was like healthy, or death, problem.. .

[quote="shinwoo"],but in LA casino, there will be alot of fake ilegal machines that never makes the jackpot...[/quote]

Calculating the probablity that a machine is illegal is too hard for getting started though...

@shake9991: Do people jumping get severly injured?

ha ha ha

it's a special laugh to RC-7th

your funny


algebra
polynomial
system of equations
algebra unsolved

Solve in real numbers the following system of equations:

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

$x^2y+xy^2+x^n+y^n=21$

Let $u=x+y$ and $v=xy$
The first equation becomes $u^3-3v=7$.
Therefore, $v=\frac{u^2-7}{3}$.
The second equation is a polynomial equation in $u$ and $v$ because we can express $x^n+y^n$ as a polynomial of $u$ and $v$.
For example,
n=1, $x+y = u$
n=2, $x^2+y^2=u^2-2v$
n=3, $x^3+y^3=(x+y)(x^2+y^2-xy)=u(u^2-3v)$
...........
and so forth.
**************************************************
As an example, for n = 1,
$u^2-3v=7$ and $uv+u=21$
Eliminate v:  $u^3-4u-63=0$
Solving the equation: $u=4.313418976$ and $v=3.868527754$
Then recover $x$ and $y$ from $T^2-uT+v=0$:
$x=3.0415078$ and $y=1.2719112$, or
$x=1.2719112$ and $y=3.0415078$.
**************************************************
The idea is simple.
But there is no general formula for x and y if n is not specified.
Besides, I plotted the graph for different values of n and I noticed some interesting patterns:

For n = 1, 3, 5, 7, .....,there are two solutions.
For n = 2, there are two solutions.
For n = 4, there are [b]four [/b]solutions.
For n = 6, 8, 10, 12, .... , there are [b]no solutions[/b].
**************************************************
I wonder if your friend really has a neat solution for the system of equations of this thread.   :)


Suppose that f(z) is analytic on $ |z|<1$ and that $ |f(z)| \leq 1$. What estimate can be made about $ |f'(0)|$?


Ignite168 is in the nicest airport he's ever been in. There is a free place to sleep, a place to use free internet, a room filled with about 100 arrcade machines to play video games, a place to shower, a place to get a massage, as well as many as other things. Ignite's plane got in at 3:47AM, he slept for 4 hours and doesn't want to sleep anymore. If his flight departs at 9:50AM to Shanghai and he wants to spend the rest of the time doing every activity at the airport but sleep (including "other things") equally, to the nearest minute, how many minutes willl he have for each activity?

Assume it takes no time to go between activities or get on the airplane.

Korean is +13 from EDT and it's Monday 7:49AM here. We had a 6 hour layover to go on a flight that will only take 1 hour -_-

Edit: Ok I have no idea what day it is. The copmuter says it's the 12th which I think would be a Wednesday but someone I'm talking to on MSN in California says it's the 10th there so it should be the 11th here, a Tuesday. Ahh well who cares, I have to go.

Are you considering the 100 arcade games different activities?

So are there 4 total activites or 103?

[quote="b-flat"]Are you considering the 100 arcade games different activities?[/quote]

Yes, he surely does

Each arcade game is not a different activity, and there are more than 4 activities.

There are five activities.

[hide]9:50 - 7:47 = 2:03 = 123 minutes divided by 5 activities = ~25 min. per activity [/hide]

[hide="expected answer"]
doing the exact same thing that fanatic did, I get $24\frac35$ minutes
[/hide]
[hide="realistic answer"]
if you consider ignite waiting in front of the ladies' room as his sixth activity, I get $20\frac12$ minutes - this has to be the answer because it's much nicer than $24\frac35$.
[/hide]

Every time I fly to China my plane stops in Japan...

[hide]
The plane got in at 3:47.
Ignite168 slept for 4 hours until 7:47 ([b]hey that's a plane!!!!!!!!![/b] :o :o :o)
So there are 2 hours and 3 minutes=123 minutes until 9:50.
123/5=24.6
[/hide]


This isn't math club-related, but I'm curious anyway.  Class schedules are up now on LionLinks...what do you all have?  Something bumped my sport (probably math, since there's only one section) so now I don't have one.  But when I do, it will be GH.  Can you say no waiting in line for dinner?  And I got Ms. Waterman again for Physics.

A english, T/B architecture, and then spaaaaaz, e physics with cosgrove, f math feng, G free!!!, and Q/H chinese.

Interesting schedule.  I have A english as well, then BU physics, CV spanish, D history, E health (=free), F math (says Feng, but probably Wolfson?), and GH... whatever sport I end up in.

Howard, do you know if everyone from fall T3X is still in winter T3X?

I know Catherine's still going to be in T3X, I don't know about anyone else Perhaps i'll ask dan later.

Hmm.....yea...Dan here...
HQ ceramics....E English.... BU French....G Health...FY t3x....N Band....CDV  Spaz.....and AT Bio....

><


What is the least three-digit whole number, the product of whose
digits is 6?

The factors of $ 6$ are $ \{1,2,3,6\}$. We obviously want the hundreds digit to be $ 1$ to minimize the number. Now we want the minimize the tens digit, by making it $ 1$. Now the only digit left for the units digit is $ 6$, and the desired number is $ \boxed{116}$.


conics
ellipse
geometry
geometric transformation
reflection
AMC
AIME

[i]"An elliptical snooker table is constructed without pockets. A ball is placed at one focus and struck by the player's cue in any direction. Show that the ball passes through the other focus after a single rebound.

(Hint: Let $ F_1$ and $ F_2$ be the two foci and $ P$ be any point on the ellipse. Let $ l$ be the tangent line to the ellipse at $ P$. Show that as a point $ Q$ moves along $ l$, $ |QF_1| \plus{} |QF_2|$ takes its minimum value when $ Q \equal{} P$. Let $ F'_1$ be the reflection of $ F_1$ in $ l$. Show that $ F_2, P, F'_1$ lie on a straight line.)"[/i]

So I've drawn out a diagramatic representation of the information given in the hint but I don't know where to go from here. I was thinking perhaps something to do with differentiating to find the minimum value but I have the tangent to consider as well as the ellipse. I'm not sure. Any help? Thanks!

The collinearity part is actually a hint to the first part. See [[1985 AIME Problems/Problem 11]] for a related problem/solution.


function
combinatorics unsolved
combinatorics

Let $ P_n$ be the set of subsets of $ \{1,2,\dots ,n \}$. Let $ c(n,m)$ be the number of functions $ f: P_n \to \{1,2,\dots ,m\}$ such that $ f(A\cap B)\equal{}\min \{f(A),f(B)\}$. Prove that
\[ c(n,m) \equal{} \sum_{j\equal{}1}^m j^n.
\]

http://www.mathlinks.ro/Forum/viewtopic.php?t=126241

  darij


inequalities
algebra
polynomial
inequalities proposed

Let $a, b, c>0$ such that $ab+bc+ca=3.$ Prove that the following inequality holds:

             \[ \frac{a^{3}+2abc}{(b+c)^{2}}+\frac{b^{3}+2abc}{(c+a)^{2}}+\frac{c^{3}+2abc}{(a+b)^{2}}\geq\frac{9}{4}.  \]

let $S$ the left side.
suppose $a \ge b \ge c$.
by chebschev we have

$3S \ge (a^3+b^3+c^3+6abc)(\frac{1}{(b+c)^2} + \frac{1}{(a+c)^2} + \frac{1}{(b+a)^2})$.

by the iran 96 inequality, we have $(\frac{1}{(b+c)^2} + \frac{1}{(a+c)^2} + \frac{1}{(b+a)^2}) \ge 3/4$.

by the polish inequality (posted this week) we have $a^3+b^3+c^3+6abc \ge 9$.

so, the problem is solved.  :)

With the same conditions, the generalized inequality

$ \frac {a^3 \plus{} kabc}{(b \plus{} c)^{2}} \plus{} \frac {b^3 \plus{} kabc}{(c \plus{} a)^{2}} \plus{} \frac {c^3 \plus{} kabc}{(a \plus{} b)^{2}}\geq\frac {3}{4}(1 \plus{} k)$

holds if and only if $ \minus{} 284.03\cdots \equal{} k_0\leq k\leq \frac {8}{\sqrt {3}} \minus{} 1 \equal{} 3.6188\cdots,$

where $ k_0$ is a root of the following irreducible polynomial

$ 768k^{10} \plus{} 222396k^9 \plus{} 1321340k^8 \plus{} 31180224k^7 \minus{} 166652564k^6 \plus{} 310036455k^5$

$ \minus{} 308089883k^4 \plus{} 205256046k^3 \minus{} 109744594k^2 \plus{} 44910243k \minus{} 10137703;$

with equality if $ k \equal{} k_0,a \equal{} 2.9720\cdots$ is a root of the following irreducible polynomial

$ 172s^{10} \plus{} 1635s^9 \plus{} 825s^8 \plus{} 7920s^7 \minus{} 31608s^6 \minus{} 28782s^5 \minus{} 238626s^4 \minus{} 237168s^3$

$ \minus{} 588708s^2 \minus{} 371061s \minus{} 465831,$

$ b \equal{} c \equal{} 0.46788\cdots$ is a root of the following irreducible polynomial

$ 43t^{10} \minus{} 753t^9 \minus{} 1530t^8 \minus{} 3384t^7 \minus{} 3762t^6 \minus{} 4266t^5 \minus{} 3348t^4 \minus{} 2592t^3 \minus{} 81t^2$

$ \plus{} 243t \plus{} 486.$

When $ \minus{}1\leq k\leq 3$, see :  http://www.mathlinks.ro/Forum/viewtopic.php?t=63872

I'm expecting nice solutions for $ k \equal{} \minus{} 284:$

If $ x,y,z$ be nonnegative numbers such that $ yz \plus{} zx \plus{} xy \equal{} 3,$ then

$ \frac {x^3 \minus{} 284xyz}{(y \plus{} z)^{2}} \plus{} \frac {y^3 \minus{} 284xyz}{(z \plus{} x)^{2}} \plus{} \frac {y^3 \minus{} 284xyz}{(x \plus{} y)^{2}}\geq \frac {849} { \minus{} 4},$

with equality if and only if $ x \equal{} y \equal{} z \equal{} 1.$

My proof for $k=3$.
Let $x+y+z=3u$, $xy+xz+tz=3v^2$, where $v>0$ and $xyz=w^3$.
Hence, $\sum_{cyc}\frac{a^3+3abc}{(b+c)^{2}}\geq3\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(x^3+3xyz)(x+y)^2(x+z)^2\geq3v(x+y)^2(x+z)^2(y+z)^2\Leftrightarrow$
$\Leftrightarrow f(w^3)\geq0$, where $f(w^3)=$
$=729u^7-1215u^5v^4+405u^3v^4+270u^4w^3-144u^2v^2w^3+19uw^6-v(9uv^2-w^3)^2$.
We see that $f'(w^3)=270u^4-144u^2v^2+38uw^3-2v(w^3-9uv^2)\geq0$.
Thus, by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=278791]uvw[/url] it remains to prove that $\sum_{cyc}\frac{a^3+3abc}{(b+c)^{2}}\geq\sqrt{3(xy+xz+yz)}$
in two cases:
1. $y=1$, $z=0$;
2. $y=z=1$,
which gives that our inequality is true.


calculus
integration
algebra
polynomial
function
domain
Ring Theory

I have solved this problem and I want another solution.
Prove that all maximal ideals are prime

[i]Edited by Myth[/i]

yes it is wellknown assume that $f.g\in I$ but $f$ and $g$ is not in $I$ then $<f>+I$ contains $I$ so it should be whole ring so $1\in <f>+I$ then $1=hf +d$ such that $d$ is polynomial in $I$ so $g=hfg+df$ we know that $hfg\in I $and$df\in I$ so g is in $I$ contradiction.

Proving those theorems is a triviality:

1) $I$ is prime iff $A/I$ is an integral domain: $I$ is prime means that $ab\in I\Rightarrow a\in I$ or $b\in I\ (*)$. On the other hand, $A/I$ integral domain means $\hat a\hat b=\hat 0\Rightarrow \hat a=\hat 0$ or $\hat b=\hat 0$, which is clearly equivalent to $(*)$.

2) $I$ is maximal iff $A/I$ is a field: $A/I$ is a field iff any $\hat a$ is invertible, i.e. iff there is some $\hat b$ s.t. $\hat a \hat b=\hat b\hat a=\hat 1$ iff for any $a\not\in I$ we have $1\in (a)+I$ iff there is some $b$ s.t. $1=ab+t$ for some $t\in I$, which is the necessary and sufficient condition for $I$ to be maximal.

[quote="Omid Hatami"]Well I didn't know the theorem you said and I solved it is some longer and I believe it's beautiful.
Supose there exist $f,g$ that $fg \in I$ but $f,g$ are not in $I$.
Suppose $I=<r_1,...,r_k>$ (Hilbert basis theorem)Because $I$ is maximal so there exist $a_0,...,a_k$ that $q=a_0p+a_1r_1+...+a_kr_k$
Then $q^2 \in I$ now $q \in \sqrt{I}$ and because $I$ is maximal $\sqrt{I}=I$ so $q \in I$Contradiction[/quote]

Huh? I suppose there should be more to the problem text for your solution to make sense, such as the ring to be Noetherian.
Also, there should be more to your solution. For example, who is p in $q=a_0p+\ldots$?

[quote="Omid Hatami"]Well I didn't know the theorem you said and I solved it is some longer and I believe it's beautiful.
Supose there exist $f,g$ that $fg \in I$ but $f,g$ are not in $I$.
Suppose $I=<r_1,...,r_k>$ (Hilbert basis theorem)Because $I$ is maximal so there exist $a_0,...,a_k$ that $f=a_0g+a_1r_1+...+a_kr_k$
Then $f^2 \in I$ now $f \in \sqrt{I}$ and because $I$ is maximal $\sqrt{I}=I$ so $f \in I$Contradiction[/quote]

 I don't know why you do this!! :?  What textbook in algebra did you read??

I don't know why you don't understand this ??? :?

In your third line you say that $I$ is finitely generated. Not all maximal ideals, in all rings, are finitely generated. :?

Oh,I proved my theorem for $k[x_1,...,x_n]$

[quote="grobber"]Isn't it a bit too well-known? First of all, an ideal $I$ in a ring $A$ is prime iff $A/I$ is an integral domain. On the other hand, it's maximal iff $A/I$ is a field. It's clear that the latter implies the former, so..[/quote]
Eh...I think we need $A$ to be commutative for your two "iff's.


[quote="M.A"]let a,b,c>=1 and a+b+c=9 then prove that

 ab+bc+ca <= ( (a)^1/2 +(b)^1/2 +(c)^1/2 )^2 :)[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=105823

Let a,b,c$ \geq 1$ and a+b+c=9 then prove that

$ ab\plus{}bc\plus{}ca <\equal{} ( a^{1/2} \plus{}b^{1/2 }\plus{}c^{1/2})^2$


vector
linear algebra
linear algebra unsolved

beg me a pardon if I don't express myself well...

V is an inner multipication real space of dimension n.
1<=k<=n.

given two groups of vectors {v1,v2,....,vk},{w1,w2,...,wk} which are linearly independant and fulfill : for every 1<=j<=k,1<=i<=k . <vi,vj>=<wi,wj>.

prove that there exists an orthogonal linear transformation T:V->V such that  for every vj (1<=j<=k):  T(vj)=wj

thanks!

If $T(v_j)=w_j,$ then it follows by direct calculation that if $x,y\in\text{span} \{v_1,v_2,\dots,v_k\}$ then $\langle T(x),T(y)\rangle=\langle x,y\rangle.$

The only remaining issue is that $\text{span} \{v_1,v_2,\dots,v_k\}$ may not be all of $V.$

Extend $\{v_1,v_2,\dots,v_k\}$ to a basis $\{v_1,v_2,\dots,v_k,u_{k+1},\dots,u_n\}$ of $V.$

Use the Gram-Schmidt process to produce an orthonormal basis $\{e_1,e_2,\dots,e_n\}$ such that $\text{span} \{e_1,e_2,\dots,e_k\}=\text{span} \{v_1,v_2,\dots,v_k\}$

For $1\le j\le k,$ define $f_j=T(e_j).$ This is well-defined, since $T$ is defined on $\text{span} \{v_1,v_2,\dots,v_k\}.$ And since we already have that $T$ is an orthogonal mapping on that set, that means that $\{f_1,f_2,\dots,f_k\}$ is an orthonormal set. Extend it to an orthonormal basis of $V,$ written as $\{f_1,f_2,\dots,f_n\}.$ We can do this, again, by the Gram-Schmidt process. Now define $T(e_j)=f_j$ for $k<j\le n.$ This defines $T$ on a basis of $V,$ and since it maps an orthonormal basis to an orthonormal basis, it is an orthogonal mapping.

Note how much the proof depended on the Gram-Schmidt orthogonalization construction, including the detail that the span of the first $k$ vectors of the constructed orthonormal set is the same as the span of the first $k$ vectors of the original independent set.


I got a 5!!!!!!!!! :)

How'd everybody else do who took it?

5 :)
Though I was pretty confident of it anyway (would be quite mad if I did not get it!)


1)Find the sum of all the numbers that can be formed with the digits $2,3,4,5$ taken all at a time.

2)Serial numbers for an item produced in a factory are to be made using two letters followed by four digits(0 to 9).If the letters are to be taken from six letters of english alphabet without repetition and the digits are also not repeated in a serial number,how many serial numbers are possible?

What do you mean by "taken all at a time" for Problem 1? Here is my interpretation:
[hide="Problem 1"]
You look at one digit at a time. For example, look at 2. It can be in any the four positions of a number, like:
_ _ _ 2,  _ _ 2 _,  _ 2 _ _, and 2 _ _ _  which can be represented as:
 $2\cdot 10^{0},2\cdot 10^{1},2\cdot 10^{2},\ \text{and}\ 2\cdot 10^{3}$
The number of times each of the $2\cdot 10^{n}$ repeats is $3!$, because you permute the remaining 3 numbers into the three remaining blanks. Therefore, the sum of all of the 2-digits in all of the numbers is:
$3! \cdot (2\cdot 10^{0}+2\cdot 10^{1}+2\cdot 10^{2}+2\cdot 10^{3}) = 2 \cdot 3! \cdot 1111 = 2 \cdot 6666$

This same approach can be taken for each number, so the sum of all of the numbers would be:
$(2+3+4+5)(6666)=\boxed{93324}$
[/hide]

[hide="Problem 2"]I don't know if I understand your question properly, but $\binom{6}{2}$ is the number of letter combinations possible and $\binom{10}{4}$ is the number of digit combinations possible. The letters can be arralged $2!$ ways for each combination, and the digits can be arranged in $4!$ ways for each combination. The total number of digits possible is $\binom{6}{2}\binom{10}{4}2!4!=\boxed{151200}$.
[/hide]

I see what the above poster did for your first problem; maybe the original poster could clarify the question?  This was my interpretation:

[hide="Solution"]The number of permutations is $4!=24$.  Each of these will have the four digits that you listed, giving a total sum of $13\cdot 24=312$.[/hide]

[hide="what an idiot"]Number 2, after editing, is

6*5*10*9*8*7=151200 [/hide]

No 236!,that's wrong.Both the solutions of Vendiddy are right(if i interpreted the 1st question right).

:roll:  :roll:  it would help if I had read the problem as "four digits", not "two digits".

[hide="Number 1"]Since each digit is used the same number of times and since the sum of any special pair is 7777, there are $7777(12)=\boxed{93334}$[/hide]


To find the tangent to the circle x^2 + y ^2 +4x -6y - 12 = 0 at the point (2,6), write the equation as (x-2)^2 + (y-6)^2 + 8x +6y -52 = 0. The tangent is 8x + 6y -52 = 0. Why does this work?

[hide="Iffy"]Implicit differentiation on the circle gives $ 2x\plus{}2yy'\plus{}4\minus{}6y'\equal{}0$ so $ y'\equal{}\frac{2x\plus{}4}{6\minus{}2y}$.

Thus at the point $ (h,k)$, we have the equation of the line being $ y\minus{}k\equal{}\frac{2h\plus{}4}{6\minus{}2k}*(x\minus{}h)$

OR

$ y\equal{}\frac{2h\plus{}4}{6\minus{}2k}x\plus{}\frac{2h^2\plus{}4h\minus{}2k^2\plus{}6k}{6\minus{}2k}$

Also, at the point $ (h,k)$, we have $ (x\minus{}h)^2\plus{}(y\minus{}k)^2\plus{}(4\plus{}2h)x\plus{}(2k\minus{}6)y\plus{}h^2\plus{}k^2\minus{}12\equal{}0$.

Then we see that removing the $ (x\minus{}h)^2\plus{}(y\minus{}k)^2$ terms gives us: $ y\equal{}\frac{2h\plus{}4}{6\minus{}2k}x\plus{}\frac{h^2\plus{}k^2\minus{}12}{6\minus{}2k}$

It suffices to show that $ h^2\plus{}k^2\minus{}12\equal{}2h^2\plus{}4h\minus{}2k^2\plus{}6k$
$ h^2\minus{}3k^2\plus{}4h\plus{}6k\plus{}12\equal{}0$

In other words, $ (h\plus{}2)^2\minus{}3*(k\minus{}1)^2\plus{}11\equal{}0$
Or $ \frac{3(k\minus{}1)^2}{11}\minus{}\frac{(h\plus{}2)^2}{11}\equal{}1$, which is the equation of a hyperbola...

Since we assumed $ (h,k)$ was originally on the circle, it must also be on the hyperbola.

So there are exactly 4 points that this works for. One of them being $ (2,6)$ (at least I think there are 4 points...)[/hide]

For a circle centred at (a, b), radius r and point of tangency (c,d),

rewriting the circle (x-a)^2 + (y-b)^2 = r^2 as

(x-c)^2 + (y-d)^2 +2(c-a)x + 2(d-b)y + a^2 - c^2 + b^2 - d^2 - r^2 = 0,

implicitly differentiating the first equation and finally substituting r^2 = (a-c)^2 + (b-d)^2,

the tangent is indeed 2(c-a)x + 2(d-b)y + a^2 - c^2 + b^2 - d^2 - r^2 = 0.


group theory
abstract algebra
complex numbers
superior algebra
superior algebra solved

Give an example of a infinite group in wich all the nontrivial subgroups are infinite, and of an infinite group in which all the nontrivial subgroups are finite.
Give an example of a noncomutative group wich has a comutative subgroup.

For the first, take Z (the aditive integers group).
For the third take GL(n,R) the multiplicative group of matrices with real entries and order n.
For the second I need more time. My intstincts tell me no.

Okay let's see.
Supose we have such an infinite group G.
Take a in G an element different from 1 (the neutral element of G) and <a> the subgroup of G generated by a.

Lemma1: all elements of G have finite orders. The proof is obvious.
Lemma2: G is not finitely generated. The proof uses lemma1

I construct an increasing sequence of subgroups G_i of G.
G_1 is generated by an element x of G-<a>.
G_i+1 is the group generated by G_i and an element from G-(<a> \cup G_i). This construction goes on forever.

Notice that for all i, G_i is different from G, because it has i generators and by lemma2, G is not finitely generated.
Also notice that G_i is strictly included in G_i+1.

Take H= \cup G_i for all positive integers i. Prove that H is a subgroup of G. Because a is not in G_i, for all i, a is not in H, so H is not G.
It is quite obvious that H is infinite (it had infinitely many generators)
The conclusion is now obvious.

As xxxxtt noticed, my previous proof is obsolete.

Take U_k the group of complex numbers with p^k elements and take U= \cup U_k. p is a fixed prime.
Then every proper subgroup of U is finite.

The mistake in my proof is that the construction proposed there might not be possible forever. What I mean is that it is not always possible to get en element x such that <x> and G-<a> are disjoint. The example above works for this.

Sorry for my mistake, and thanks xxxxtt for spotting the error.


The Mexican Olympiad has no round table here in MathForums. How can I build a Mexican Olympiad Round Table??? The MMO has just been and no one knows here in MathLinks... how can I build it?

Try the [url=http://www.mathlinks.ro/index.php?f=255]Spanish Communities[/url]


geometry
circumcircle
Euler
incenter
geometry solved

Let $O_T$ be the circumcenter of tangential triangle of $\triangle ABC$ (i.e. the triangle formed by the lines tangent to the circumcircle of $\triangle ABC$ at its vertices). If $O, H$ are the circumcenter and the orthocenter of $\triangle ABC$ prove that $O_T, O, H$ are collinear.

For the solution of your problem, I will use a lemma:

[b]Lemma 1.[/b] If the incircle of a triangle ABC touches the sides BC, CA, AB of triangle ABC at the points M, N, P, then the circumcenter of triangle ABC lies on the Euler line of triangle MNP.

[i]Proof of Lemma 1.[/i] In the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=6228 , it was proven the orthocenter of triangle MNP, the incenter of triangle ABC, and the circumcenter of triangle ABC are collinear. In other words, the circumcenter of triangle ABC lies on the line joining the orthocenter of triangle MNP with the incenter of triangle ABC. But since the incenter of triangle ABC is clearly the circumcenter of triangle MNP (in fact, the incircle of triangle ABC is the circumcircle of triangle MNP, since the points M, N, P lie on the incircle of triangle ABC), the line joining the orthocenter of triangle MNP with the incenter of triangle ABC is the line joining the orthocenter of triangle MNP with the circumcenter of triangle MNP, and thus it is the Euler line of triangle MNP. So we obtain that the circumcenter of triangle ABC lies on the Euler line of triangle MNP. And Lemma 1 is proven.

Now, I will solve your problem for the case of an acute-angled triangle ABC:

Let XYZ be the tangential triangle of the triangle ABC. Then, the lines YZ, ZX, XY are the tangents to the circumcircle of triangle ABC at the points A, B, C, respectively, and moreover, since the triangle ABC is acute-angled, these points A, B, C lie inside the segments YZ, ZX, XY. Hence, the circumcircle of triangle ABC is the incircle of triangle XYZ. The points where this incircle touches the sides YZ, ZX, XY of triangle XYZ are the points A, B, C. Hence, applying Lemma 1 to the triangle XYZ, we see that the circumcenter of the triangle XYZ lies on the Euler line of triangle ABC. In other words, the circumcenter of the tangential triangle of triangle ABC lies on the Euler line of triangle ABC.

This solves your problem in the case of an acute-angled triangle ABC. In the case of an obtuse-angled triangle ABC, we have to use not Lemma 1, but rather an analogue of Lemma 1 for an excircle instead of the incircle.

I am realizing that the above sounds like a formless boring sequence of trivialities, but I can't do anything about this as I'm half asleep and writing more or less automatically. Hope that, at least, my proof was comprehensible.

  Darij


Let $a,b,c$ be non-negative numbers. Prove that
$\sum \sqrt{\frac{2a(b+c)}{(2b+c)(b+2c)}} \geq 2$.

Yeah it may be hard if we dont know the lehma:
$\sum \sqrt{ \frac{a}{b+c}}\geq \frac{3}{ \sqrt{2}}$
Also for all positive x :
$\frac{(x+1)^{2}}{(x+2)(2x+1)}\geq \frac{4}{9}$

[quote="Kummer"]Yeah it may be hard if we dont know the lehma:
$\sum \sqrt{ \frac{a}{b+c}}\geq \frac{3}{ \sqrt{2}}$[/quote]
false.

[quote="Vasc"]Let $ a,b,c$ be non-negative numbers. Prove that
$ \sum \sqrt {\frac {2a(b \plus{} c)}{(2b \plus{} c)(b \plus{} 2c)}} \geq 2$.[/quote]
By Holder we obtain:
$ \left(\sum_{cyc} \sqrt {\frac {a(b \plus{} c)}{(2b \plus{} c)(b \plus{} 2c)}}\right)^2\sum_{cyc}\frac{a^2(2b \plus{} c)(b \plus{} 2c)}{b\plus{}c}\geq(a\plus{}b\plus{}c)^3.$
Thus, it remains to prove that $ (a\plus{}b\plus{}c)^3\geq2\sum_{cyc}\frac{a^2(2b \plus{} c)(b \plus{} 2c)}{b\plus{}c}.$
But $ (a\plus{}b\plus{}c)^3\geq2\sum_{cyc}\frac{a^2(2b \plus{} c)(b \plus{} 2c)}{b\plus{}c}\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(a^5b\minus{}a^4bc\minus{}a^3b^3\plus{}a^3b^2c)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(ca^4\minus{}ca^4b\minus{}cab^4\plus{}cb^5)\minus{}\sum_{cyc}(c^3a^3\minus{}c^3a^2b\minus{}c^2ab^2\plus{}c^3b^3)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a\minus{}b)^2c(a\plus{}b)(a^2\plus{}b^2\minus{}c^2)\geq0\Leftrightarrow\sum_{cyc}(a\minus{}b)^2S_c\geq0.$
Let $ a\geq b\geq c.$ Then $ S_c\geq0,$ $ S_b\geq0$ and $ (a\minus{}c)^2b^2\geq(b\minus{}c)^2a^2.$
Hence, $ b^2\sum_{cyc}(a\minus{}b)^2S_c\geq(a\minus{}c)^2b^2S_b\plus{}(b\minus{}c)^2b^2S_a\geq(b\minus{}c)^2(a^2S_b\plus{}b^2S_a)\equal{}$
$ \equal{}(b\minus{}c)^2(ab(a^2\minus{}b^2)^2\plus{}abc(a\plus{}b)(a\minus{}b)^2\plus{}abc^2(a^2\plus{}b^2)\plus{}abc^3(a\plus{}b))\geq0.$

[quote="Vasc"]Let $ x,y,z$ be nonnegative numbers such that $ yz \plus{} zx \plus{} xy > 0$. Prove that

$ \sqrt {\frac {x(y \plus{} z)}{(2y \plus{} z)(y \plus{} 2z)}} \plus{} \sqrt {\frac {y(z \plus{} x)}{(2z \plus{} x)(z \plus{} 2x)}} \plus{} \sqrt {\frac {z(x \plus{} y)}{(2x \plus{} y)(x \plus{} 2y)}} \geq \sqrt {2}.$[/quote][quote="arqady"]By Hölder we obtain:

$ \left[\sum{\sqrt {\frac {x(y \plus{} z)}{(2y \plus{} z)(y \plus{} 2z)}}}\right]^2\sum\frac {x^2(2y \plus{} z)(y \plus{} 2z)}{y \plus{} z}\geq(x \plus{} y \plus{} z)^3.$

Thus, it remains to prove that

$ (x \plus{} y \plus{} z)^3\geq2\sum\frac {x^2(2y \plus{} z)(y \plus{} 2z)}{y \plus{} z}.$[/quote]But $ (x \plus{} y \plus{} z)^3 \minus{} 2\sum\frac {x^2(2y \plus{} z)(y \plus{} 2z)}{y \plus{} z}\equal{} \sum{\frac {(y \minus{} z)^2(y \plus{} z \minus{} x)^2}{2(y \plus{} z)}}\geq0.$

[quote="Ji Chen"]$ (x \plus{} y \plus{} z)^3 \minus{} 2\sum\frac {x^2(2y \plus{} z)(y \plus{} 2z)}{y \plus{} z} \equal{} \sum{\frac {(y \minus{} z)^2(y \plus{} z \minus{} x)^2}{2(y \plus{} z)}}\geq0.$[/quote]

Using $ (\sum x)^3 \equal{} \sum x^3 \plus{}3(x\plus{}y)(y\plus{}z)(z\plus{}x)$  and

$ (2y\plus{}z)(y\plus{}2z)\equal{}2(y\plus{}z)^2\plus{}yz$

I wrote

$ (x \plus{} y \plus{} z)^3 \minus{} 2\sum\frac {x^2(2y \plus{} z)(y \plus{} 2z)}{y \plus{} z}\equal{}\frac {1}{2}\sum ((x\plus{}y\minus{}z)\minus{}\frac {4xyz}{(y \plus{} z)(x\plus{}z)})(x\minus{}y)^2\equal{}\sum{\frac {(y \minus{} z)^2(y \plus{} z \minus{} x)^2}{2(y \plus{} z)}}$

but I cannot prove last equality. :(   How do you prove it, Ji Chen?

Thank you very much. :)

[quote="Ji Chen"]$ (x \plus{} y \plus{} z)^3 \minus{} 2\sum\frac {x^2(2y \plus{} z)(y \plus{} 2z)}{y \plus{} z} \equal{} \sum{\frac {(y \minus{} z)^2(y \plus{} z \minus{} x)^2}{2(y \plus{} z)}}.$[/quote][quote="manlio"]I cannot prove last equality. :(   How do you prove it, Ji Chen?[/quote]$ 2\sum\frac {x^2(2y \plus{} z)(y \plus{} 2z)}{y \plus{} z}\plus{} \sum{\frac {(y \minus{} z)^2(y \plus{} z \minus{} x)^2}{2(y \plus{} z)}}$

$ \equal{}\sum{\left\{x^2\left[\frac {2(2y \plus{} z)(y \plus{} 2z)}{y \plus{} z}\plus{}\frac{(y \minus{} z)^2}{2(y \plus{} z)}\right]\minus{}x(y\minus{}z)^2\plus{}\frac{(y\plus{}z)(y\minus{}z)^2}{2}\right\}}$

$ \equal{}\sum{\left[\frac{9x^2(y\plus{}z)}{2}\minus{}x(y\minus{}z)^2\plus{}\frac{(y\plus{}z)(y\minus{}z)^2}{2}\right]}\equal{}(x \plus{} y \plus{} z)^3.$

[quote="Ji Chen"][quote="Ji Chen"]$ (x \plus{} y \plus{} z)^3 \minus{} 2\sum\frac {x^2(2y \plus{} z)(y \plus{} 2z)}{y \plus{} z} \equal{} \sum{\frac {(y \minus{} z)^2(y \plus{} z \minus{} x)^2}{2(y \plus{} z)}}.$[/quote][quote="manlio"]I cannot prove last equality. :(   How do you prove it, Ji Chen?[/quote]$ 2\sum\frac {x^2(2y \plus{} z)(y \plus{} 2z)}{y \plus{} z} \plus{} \sum{\frac {(y \minus{} z)^2(y \plus{} z \minus{} x)^2}{2(y \plus{} z)}}$

$ \equal{} \sum{\left\{x^2\left[\frac {2(2y \plus{} z)(y \plus{} 2z)}{y \plus{} z} \plus{} \frac {(y \minus{} z)^2}{2(y \plus{} z)}\right] \minus{} x(y \minus{} z)^2 \plus{} \frac {(y \plus{} z)(y \minus{} z)^2}{2}\right\}}$

$ \equal{} \sum{\left[\frac {9x^2(y \plus{} z)}{2} \minus{} x(y \minus{} z)^2 \plus{} \frac {(y \plus{} z)(y \minus{} z)^2}{2}\right]} \equal{} (x \plus{} y \plus{} z)^3.$[/quote]

Thank you very much, Ji Chen. :)


calculus
integration
function
calculus computations

Does it help that the two integrands are inverses of each other?

Then we get I+J = 1

Well done, [b][color=red]misan[/color][/b]...  :D Another way to state the answer is:

$ \frac {1}{2006}\mathbf{B}\left(\frac {1}{2006},\frac {2008}{2007}\right) \minus{} \frac {1}{2007}\mathbf{B}\left(\frac {1}{2007},\frac {2007}{2006}\right).$

Regards,

Hey....how can we say that the sum of the two integrals is equal to 1....?? :maybe:

Consider this...

(1-x^2006) < (1-x^2006)^1/2007  for all  0<x<1

so integral  from 0 to 1 of (1-x^2006)^1/2007 > 1-(1/2007)

and similarly integral from 0 to 1 of (1-x^2007)^1/2006 > 1-(1/2008)

and thus the sum of the two integrals is greater than 1...

but it was a nice observation...that the two functions are inverses of each other...:)

can some one find out the numerical difference of the two integrals..??

[quote="misan"]= $ \frac {1}{2006}\;\cdot \frac {\Gamma\left( \frac {1}{2006} \right)\; \cdot \; \Gamma\left( \frac {2008}{2007} \right)\; } {\Gamma\left( \frac {1}{2006} \plus{} \frac {2008}{2007} \right)\;} \; \minus{} \; \frac {1}{2007}\;\cdot \frac {\Gamma\left( \frac {1}{2007} \right)\; \cdot \; \Gamma\left( \frac {2007}{2006} \right)\; } {\Gamma\left( \frac {1}{2007} \plus{} \frac {2007}{2006} \right)\;}$[/quote]
This simplifies to $ 0$.

More generally,
\[ \int_0 ^ 1{(1 \minus{} x^a)^{\frac {1}{b}}dx} \equal{} \int_0^1{(1 \minus{} x^b)^{\frac {1}{a}}dx}
\]
if $ a,b \in \mathbb{R}^ \plus{}$.  The proof is straightforward, using hsbhatt's observation that the integrands are inverses.  Indeed, for any invertible function $ f$, integration by parts gives
\[ \int_0^1 {f(s)ds} \equal{} f(1) \minus{} \int_{f(0)}^{f(1)}{f^{ \minus{} 1}(s)ds}
\]
Letting $ f(x) \equal{} (1 \minus{} x^a)^{\frac {1}{b}}$ and noting $ f(1) \equal{} 0$, $ f(0) \equal{} 1$, yields the desired result.


Refer to attached diagram.

Quadrilateral ABCD is a square. Point E lies on AB such that the perimeter of triangle ADE is 30, and the perimeter of BCDE is 44. Find the length of DE.

Set up a system.  Let the side length of the square be $ s$, the hypotenuse of the right triangle $ h$, and the length of $ \overline{EB}$ be $ z$.

$ s^2 \plus{} (s\minus{}z)^2 \equal{} h^2$

$ s^2 \plus{} s^2 \minus{} 2sz \plus{} z^2 \equal{} h^2$

$ 2s^2 \minus{} 2sz \plus{} z^2 \equal{} h^2$

$ s \plus{} (s\minus{}z) \plus{} \sqrt{2s^2 \minus{} 2sz \plus{} z^2}\equal{} 30$

$ 2s \minus{} z \plus{} \sqrt{2s^2 \minus{} 2sz \plus{} z^2} \equal{} 30$

$ 2s \plus{} z \plus{} h \equal{} 44$

$ 2s \plus{} z \plus{} \sqrt{2s^2 \minus{} 2sz \plus{} z^2} \equal{} 44$

Solve this by subtracting the first equation from the second.

$ 2z \equal{} 14$

Thus, $ z \equal{} 7$.

By plugging $ z \equal{} 7$ into one of the two above equations, we get:

$ 2s \plus{} 7 \plus{} \sqrt{2s^2 \minus{} 14s \plus{} 49} \equal{} 44$

$ 2s \plus{} \sqrt{2s^2 \minus{} 14s \plus{} 49} \equal{} 37$

$ \sqrt{2s^2 \minus{} 14s \plus{} 49} \equal{} 37 \minus{} 2s$

$ 2s^2 \minus{} 14s \plus{} 49 \equal{} (37 \minus{} 2s)^2$

$ 2s^2 \minus{} 14s \plus{} 49 \equal{} 4s^2 \minus{} 148s \plus{} 1369$

$ \minus{}2s^2 \plus{} 136s \minus{} 1320 \equal{} 0$

Solving for this gives $ s \equal{} 12$.  Thus, the hypotenuse is $ \boxed{13}$.

Thanks Ernie. This problem was part of a 10-minute 3-question set; I was hoping to find a fast systematic solution other than guess-and-check but it appears there isn't one.

Oh, it was a speed problem?  Then for these, it's best to know some basic Pythagorean triplets.

hmm, nice solution ernie, allowed me to not post anothr solution  :P

Also, this 10 minute-3 proble set sounds like GHNML. hmm? Is it?


let $ p,n$ two positive integers,

compute:

$ \sum_{i \equal{} 0}^{n} \prod_{j \equal{} 1}^{p} (i \plus{} j)$

hello, i have $ \sum_{i \equal{} 0}^{n} \prod_{j \equal{} 1}^{p} (i \plus{} j)\equal{}\frac{(n\plus{}1)(n\plus{}1\plus{}p)!}{(p\plus{}1)(n\plus{}1)!}$.
Sonnhard.


Find the limit of the following sequence $(x_{n})_{n\geq 2}$,
\[x_{n}=\frac{1}{2\sqrt{n}}+\frac{1}{3\sqrt [3] n^{2}}+\ldots+\frac{1}{n\cdot\sqrt [n]{n^{n-1}}}. \]

Hi,
The limit is $0$.

$n^{1/2}<n^{\frac{k-1}{k}}$ for all $2\leq k\leq n$.
Therefore we get that
$0<x_{n}<\frac{\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}}{\sqrt{n}}\rightarrow 0$ since the limit of the right hand sequence is $0$ via Cesaro Stolz lemma.
Didi


[quote="hoangclub"]Let a,b,c be nonegative numbers satisfying a+b+c=1 find the max value of
                            P =a(b^2+c^2+b(c^2+a^2)+c(a^2+b^2)[/quote]
$ (a\plus{}b\plus{}c)^3\geq4\sum_{cyc}(a^2b\plus{}a^2c)\Leftrightarrow3abc\plus{}\sum_{cyc}(a^3\minus{}a^2b\minus{}a^2c\plus{}abc)\geq0,$
which true by Schur. The equality holds when $ a\equal{}b\equal{}\frac{1}{2}$ and $ c\equal{}0.$
Id est, $ \max P\equal{}\frac{1}{4}.$


algebra
polynomial
calculus
derivative
algebra proposed

a,b,c real numbers satisfying: a<b<c  a+b+c=6 ab + bc + ca = 9

Prove that 0<a<1<b<3<c<4

$ a, b, c$ are roots of the polynomial $ P(x) \equal{} x^3 \minus{} 6x^2 \plus{} 9x \minus{} abc$.  We have $ P'(x) \equal{} 3x^2 \minus{} 12x \plus{} 9$ which has roots $ x \equal{} 1, 3$, and it is well-known that the roots of the derivative of a polynomial are "interweaved" between its roots; that is,

$ a < 1 < b < 3 < c$.

It remains to prove the left and right bounds.  I'm still working on it.  (I have not yet been able to derive a contradiction in the case that $ a$ is negative.)

The sentence the roots of the derivative of a polynomial are "interweaved" between its roots is a consequence of Roll's theorem am I right?

Yes, but it can also be proven from the identity $ \frac{P'(x)}{P(x)} \equal{} \sum_{P(r) \equal{} 0} \frac{1}{x \minus{} r}$.

It seems you forgot that $ P(1)>0$ and $ P(3)<0$ must be satisfied in order to have real roots.

$ 9 \equal{} ab \plus{} bc \plus{} ca \equal{} a(b \plus{} c) \plus{} bc \equal{} a(6 \minus{} a) \plus{} bc < a(6 \minus{} a) \plus{}$ $ (b \plus{} c)^2 /4 \equal{} a(6 \minus{} a) \plus{} (6 \minus{} a)^2 / 4 \implies 0 < a < b < c < 4$  :wink:


Hiyah, just put this for disscusion of the U.S. Open at Bethpage Black. The second round has finished, and leader (last time I checked) is Mike Weir. Phil Mickelson has a chance, and the famed Tiger Woods is waaaay behind. The question is...... WHO? WILL? WIN?

EDIT: Hmmm.... Apparently my machine will not allow me to include Other.

I'm a little worried for Tiger. He might not make the cut this year  :( .


induction
limit
inequalities
articles
function
calculus
triangle inequality

Prove using induction that, for $ A, B \in \mathbb{N}$,

$ \sqrt {A} \plus{} \sqrt {B} > \sqrt {A \plus{} B}.$

EDIT: Even if you don't use induction, squaring both sides and then simplify is still not valid because you are only allowed to work on one side of it, but not both; you can start from something that is accepted true and then work on it to transform it into what we are asked to prove.  For this one, we can't do what you suggested.  Instead, we start from the fact that $ (2\sqrt {A}\sqrt {B}) \equal{} 2\sqrt {A}\sqrt {B} \plus{} A \plus{} B \minus{} A \minus{} B > 0$; or $ (\sqrt {A} \plus{} \sqrt {B})^2 > A \plus{} B$, and then take the positive square root of both sides to get to what we wish to prove.  However, prove it by induction.

Do you specifically need induction? Because you can just square both sides and QED.

For real $ a$, $ b$, $ a \plus{} b < (a\plus{}b) \plus{} 2\sqrt{ab} \equal{} (\sqrt{a}\plus{}\sqrt{b})^2$. Since $ a,b>0$, $ \sqrt{(a\plus{}b) \plus{} 2\sqrt{ab}} \equal{} \sqrt{a}\plus{}\sqrt{b}$ so $ \sqrt{a\plus{}b} < \sqrt{a}\plus{}\sqrt{b}$.

Here's a follow-up question:

If $ a,b$ are positive reals, determine the range of values that $ \frac{\sqrt{a}\plus{}\sqrt{b}}{\sqrt{a\plus{}b}}$ can take.

[hide="limits are in AoPS volume 2"]Let $ a \equal{} 1$. $ \lim_{b\Rightarrow \infty} \dfrac{\sqrt {a} \plus{} \sqrt {b}}{\sqrt {a \plus{} b}}$. Since $ a$ is a constant, we can get rid of that since $ a \plus{} b$ goes to $ \infty$. $ \lim_{b\Rightarrow \infty} \dfrac{\sqrt {b}}{\sqrt {a \plus{} b}}$. We use [[L'Hopital's Rule]] to get $ \lim_{b\Rightarrow \infty} \dfrac{\sqrt {a \plus{} b}}{\sqrt {b}}$. Since it's the reciprocal of the limit just before, the limit must go to 1.

And now we must find the maximum of $ \dfrac{\sqrt {a} \plus{} \sqrt {b}}{\sqrt {a \plus{} b}}$, since we know that the lower bound is 1
(but it never reaches 1)[/hide]

the \to command is much more standard for limits then the arrow you're using.  Also, the limit argument gives you nothing in terms of the questions being asked: in particular, it doesn't tell you that 1 is a lower limit of that expression.  (L'Hopital is also overkill.)

Another proof of the original result (also not inductive): this is just the triangle inequality for a right triangle with legs $ \sqrt{A}$ and $ \sqrt{B}$.

Write $ \frac {\sqrt {a} \plus{} \sqrt {b}}{\sqrt {a \plus{} b}} \equal{} \sqrt {1 \plus{} \frac {2\sqrt {ab}}{a \plus{} b}}$ and note that $ 0 < \frac {2\sqrt {ab}}{a \plus{} b} \le 1$.

[quote="blahblahblah"][quote="mathwizarddude"]EDIT: Even if you don't use induction, squaring both sides and then simplify is still not valid because you are only allowed to work on one side of it, but not both;[/quote]

of course it's valid, if $ x,y\geq 0$ and $ x^2\geq y^2$ then $ x\geq y$.[/quote]
Also, you can assume the opposite, work on both sides, then find the contradiction.

[quote="mathwizarddude"][quote="blahblahblah"][quote="mathwizarddude"]EDIT: Even if you don't use induction, squaring both sides and then simplify is still not valid because you are only allowed to work on one side of it, but not both;[/quote]

of course it's valid, if $ x,y\geq 0$ and $ x^2\geq y^2$ then $ x\geq y$.[/quote]

Read this
http://mathforum.org/library/drmath/view/51867.html[/quote]
@blahblahblah: mathwizarddude made that comment in reference to uldivad9's reply on the post #2, but I agree about your comment since they are positive numbers.

@mathwizarddude: I doubt blahblahblah needs to read that article, his expertise is beyond that level  :whistling:

[quote="blahblahblah"][quote="mathwizarddude"]
Read this
http://mathforum.org/library/drmath/view/51867.html[/quote]

if you made an effort to understand when it is and when it is not permissible to operate on both sides of some expression, you wouldn't need to give me silly links like that.  why do i care what 'doctor pete' has to say when i understand the math involved?[/quote]

he didn't get his Ph.D to be trash talked by kids one-third of his age!

"squaring both sides" is not the most accurate way of phrasing the logic used, but it gets the point across. By "squaring both sides," he means to say:

The function $ f(x) \equal{} x^2, \; x \in \mathbb{R}^ \plus{}$ is strictly increasing. Therefore, for all $ x,y \in \mathbb{R}^ \plus{}$, $ x > y \iff f(x) > f(y)$. Since they are equivalent conditions, it suffices to prove the latter in place of the former.

That being said, the Math Forum is a fine resource, and we all should respect those who volunteer their time towards a noble cause.

It's more a matter of style than anything else.  It's bad form to do anything that sounds like you're working from the desired statement to a known result and any actual proofs should be written in the correct direction (taking the positive square root of another expression).  I don't really understand why a proof by induction is desired since the result obviously holds for positive reals.

[hide="Roundabout Inductive Proof"]wlog let $ a \equal{} b \plus{} k \geq b$
[b]
Base case, k=0:[/b]

Note that $ \sqrt {b}$ is positive, and therefore $ 2 > \sqrt {2} \iff 2 \sqrt {b} > \sqrt {2} \sqrt {b} \iff \sqrt {a} \plus{} \sqrt {b} > \sqrt {a \plus{} b}$

[b]Inductive step:[/b]

Consider the case $ k \equal{} t$, and $ \sqrt {b} \plus{} \sqrt {b \plus{} t} > \sqrt {2b \plus{} t}$ is true.

[b]Lemma:[/b]$ x < y \implies \sqrt {x \plus{} 1} \minus{} \sqrt {x} > \sqrt {y \plus{} 1} \minus{} \sqrt {y}$

Using $ x \equal{} b \plus{} t$ and $ y \equal{} 2b \plus{} t$, this means that $ \sqrt {b \plus{} t \plus{} 1} \minus{} \sqrt {b \plus{} t} > \sqrt {2b \plus{} t \plus{} 1} \minus{} \sqrt {2b \plus{} t}$

Adding the above to the case of $ k \equal{} t$ yeilds the case of $ k \equal{} t \plus{} 1$, and the induction is complete.

[b]*Proof-ish:[/b]This is true of any twice-differentiable function that is concave but strictly increasing.

Because it is concave and strictly increasing, $ x < y \iff x \plus{} r < y \plus{} r \iff f'(x \plus{} r) > f'(y \plus{} r) \iff f'(x \plus{} r) \minus{} f'(y \plus{} r) > 0$. Integrate that as a function of $ r$ between $ 0$ and $ 1$, or something like that.[/hide]

[quote="Lazarus"]he didn't get his Ph.D to be trash talked by kids one-third of his age![/quote]

'doctor pete' may not be a doctor after all!

[quote="About Ask Dr. Math"]
In fall of 1994 the Geometry Forum (now the Math Forum) discovered a dormant project called "Ask Prof. Maths," where K-12 students could send in math questions and get personal answers. We decided to revive the program, using Swarthmore College math students as 'Math Doctors' - we called them the 'Swat Team' - students who loved to answer questions from other students.

By the fall of 1995, the rapidly increasing volume of questions required that new members of the staff be recruited from other colleges around the country. By the year 2000, there had been over 300 volunteer 'Doctors' from all corners of the globe. The service continues to grow in popularity and has received a number of Internet awards.
[/quote]

don't let this shake your conviction that the majority of people on the internet calling themselves 'doctors' have either Ph.Ds or medical degrees though!

e: by the way if you were posting ironically i tip my hat to you good sir but i'm pretty sure you weren't and if you were you did a bad job of it

e2: i'm not a 'kid' although i may be a third of 'doctor pete's' age if he's in his sixties

sarcasm is misconstrued/unappreciated here

wah wah i made an unfunny post and no one laughed :(

Enough with the sarcasm/trolling. If you guys want to continue this behavior, please take it on PM's, not here.

It is possible to prove The Zuton Force's lemma without calculus:

Prove that $ x<y\Rightarrow\sqrt{x\plus{}1}\minus{}\sqrt{x}>\sqrt{y\plus{}1}\minus{}\sqrt{y}$ [b]without[/b] calculus.

Since $ x < y$, $ \sqrt{x} < \sqrt{y}$ and $ \sqrt{x \plus{} 1} < \sqrt{y \plus{} 1}$. Hence, $ \sqrt{x} \plus{} \sqrt{x \plus{} 1} < \sqrt{y} \plus{} \sqrt{y \plus{} 1}$. Hence, $ \frac{1}{\sqrt{x} \plus{} \sqrt{x \plus{} 1}} > \frac{1}{\sqrt{y} \plus{} \sqrt{y \plus{} 1}}$. Rationalizing the denominators yields $ \sqrt{x \plus{} 1} \minus{} \sqrt{x} > \sqrt{y \plus{} 1} \minus{} \sqrt{y}$, as desired.

sqrt(x+1)<sqrt(y+1)
sqrtx<sqrty
we have to prove that,ceiling[sqrtx]-sqrt(x+1)<[sqrty]-sqrt(y+1).
which is easily done as x,y E N with [sqrtx]<[sqrty] and(1).

sorry 4 not using Latex.


geometry
geometric transformation
reflection
circumcircle
geometry proposed

Let $ ABC$ be a triangle and $ P,Q$ be isogonal conjugate points wrt triangle $ ABC$.Define by $ X$ the reflection of circumcenter of triangle $ QBC$ on side $ BC$ and defince $ Y,Z$ similarly. Let $ D,E,F$ be the projection of $ P$ on $ BC,CA,AB$. Prove that
a) triangle $ XYZ$ and $ DEF$ are similar
b) $ P$ and isogonal conjugate of $ P$ wrt triangle $ DEF$ is corresponding points of $ \triangle XYZ\sim \triangle DEF$ respectively.

△O12O13O23∽△EDF  and  △O12O13O23∽△XYZ (*)
so we solve （a）
later I will post solution to (*) I have in mind

We redefine the points here
D= O23    E= O12    F= O13
1.∠Z’ FD=∠Z’FC-∠DFC
∠Y’ED=∠BED -∠BEY’
But we have ∠DFC+∠BED=1/2∠BEQ+1/2∠CFQ=∠A
And ∠Z’FC=90-∠C     ∠BEY’=90-∠B
∠Z’ FD=∠Z’FC-∠DFC=90-∠C-∠A+∠BED=∠BED-90+∠B=∠BED-∠BEY’=∠BEY’ (1)

2.EY’/FZ’=EBsin∠B/FCsin∠C=EB/FC*AC/AB
 ∠AFC=2(180-∠AQC)
 AC=2FC sin∠AFC=2FC*sin(1/2∠AFC)=2FC*sin∠AQC=2FC*sin∠EFD
 AB=2EB*sin∠FED
 AC/AB= FC/EB* sin∠EFD/ sin∠FED= FC/EB * ED/FD
 So ED/FD= AC/AB * EB/FC
    =>  EY’/FZ’= ED/FD (2)
3. according to (1)(2)  =>   △EDY’∽△FDZ’ =>  △DY'Z'∽△DFE
4. .△DY'Z' and △XYZ are congruence
   =>  △DFE∽△XZY
Done!


Sa se rezolve ecuatia:

   $x\cdot 2^{x-1}+\frac{1}{\sqrt{x}}\cdot 2^{\frac{1}{\sqrt{x}}} =3.$ ;)

se inmulteste cu 2 , se aplica de 2 ori mediile si se obtine x=1

Ideea este buna, dar daca ai si scrie ar fi mai bine. ;)

[quote="perfect_radio"]avem $x 2^x + \frac{1}{\sqrt{x}} 2^{1/\sqrt{x}} + \frac{1}{\sqrt{x}} 2^{1/\sqrt{x}} \geq 3 \sqrt[3]{2^{x+1/\sqrt{x}+1/\sqrt{x}}}$

ramane de dem ca $x+\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x}} \geq 3$, care e "am-gm"[/quote]
Da, ideea este buna, dar se putea face si altcumva.

[quote="Cezar Lupu"]Mah, dar tu nici n-ai aterizat bine pe site-ul asta si deja si vad ca deja dai reply-uri la toate posturile posibile si imposibile. N-ar fi o idee sa postezi si o solutie daca tot zici ca se poate face si altcumva. ;)[/quote]
    Bine.Se observa ca membrul stang este un numar par, in timp ce 3 este un numar impar.Atunci unul din numerele x-1 sau sqrtx/x este nul.Cum sqrtx/x este diferit de zero, vom obtine ca x-1=0, deci x=1. ;) Asta ca sa fii tu multumit.

[quote="Valentin Vornicu"][quote="stancioiu sorin"][quote="spix"]Se pare ca spam-detectorul lui Cezar a intrat in functiune :D  :spam:[/quote]
  Cred ca mai degraba a intrat in functiune spam-detectorul meu. :spam:[/quote]Sorin, mai usor cu spam-ul. Cezar are dreptate. Daca ai ceva interesant de spus, spune-o, dar nu mai posta doar de dragu de a scrie ceva. Ai un blog pentru chestia asta.[/quote]
    Am inteles Valentin. :surrender:

[quote="perfect_radio"][quote="stancioiu sorin"]Bine.Se observa ca membrul stang este un numar par, in timp ce 3 este un numar impar.Atunci unul din numerele x-1 sau sqrtx/x este nul.Cum sqrtx/x este diferit de zero, vom obtine ca x-1=0, deci x=1. ;) Asta ca sa fii tu multumit.[/quote]

Cred ca (scuzati cacofonia) conceptul de paritate al numerelor reale inca n-a fost inventat.[/quote]
  Scuze...atunci numerele sunt divizibile cu 2.Sunt de-abia la inceputuri.Sunt in clasa a VII-a, mai am destul pana sa invat tainele matematicii.La varsta asta mai greseste orice om, asa cred...

[quote="Cezar Lupu"][quote="stancioiu sorin"]Bine.Se observa ca membrul stang este un numar par, in timp ce 3 este un numar impar.Atunci unul din numerele x-1 sau sqrtx/x este nul.Cum sqrtx/x este diferit de zero, vom obtine ca x-1=0, deci x=1. ;) Asta ca sa fii tu multumit.[/quote]

 Mai bine nu mai zic nimik de acum incolo ca daca este sa fac observatie tuturor noilor veniti, ajung eu primul la balamuc.  :wacko:

EDIT: Pai, daca tot zici ca esti in clasaa-7-a, mai bine stai si invatza mai intai.[/quote]
     De ce zici asta?Nu e buna solutia?

In general, se spune despre un numar intreg ca e divizibil cu 2, nu despre unul real.

Hint: Nu mai da quote la mesajele care sunt "ultimul post". Quote in quote in quote ... = inutil.


Solve the system of equations :
 $x+y+(z^2-8z+14)\sqrt{x+y-z} =1$ and $2x+5y+\sqrt{xy+z} = 3$

this question is in the mathscope, or no?


search
quadratics
number theory proposed
number theory

Let $F_n$ be the $n$-th Fibonacci number. Prove that if $p>5$ is a prime, then
\[ F_p\equiv \left(\frac{p}5\right) \mod p \]

Because of the Law of Quadratic Reciprocity: $(\frac{p}{5}) = (\frac{5}{p})$.

Suppose that $5$ is quadratic residue mod $p$.
Then, $\sqrt{5}$ is an element of $Z_p$, so:
$F_p=\frac{(\frac{1+\sqrt{5}}{2})^p - (\frac{1-\sqrt{5}}{2})^p}{\sqrt{5}} \equiv \frac{\frac{1+\sqrt{5}}{2}-\frac{1-\sqrt{5}}{2}}{\sqrt{5}} = \frac{\sqrt{5}}{\sqrt{5}}=1=(\frac{5}{p})$
as desired

Now, suppose that $5$ is not quadratic residue mod $p$.
We extend field $\mathbb{Z}_p$ by the new element $\sqrt{5}$. We see that:
$(1+\sqrt{5})^p \equiv 1 + (\sqrt{5})^p$
(other terms cancel $\mod{p}$)
Also:
$1+(\sqrt{5})^p=1+5^\frac{p-1}{2}*\sqrt{5}=1-\sqrt{5}$
because $5$ is not a quadratic residue.
Analogously: $(1-\sqrt{5})^p \equiv 1 + \sqrt{5}$
So:
$F_p=\frac{(\frac{1+\sqrt{5}}{2})^p - (\frac{1-\sqrt{5}}{2})^p}{\sqrt{5}} \equiv \frac{\frac{1-\sqrt{5}}{2} - \frac{1+\sqrt{5}}{2}}{\sqrt{5}}=\frac{-\sqrt{5}}{\sqrt{5}}=-1=(\frac{5}{p})$
as desired...

[quote="Yimin Ge"]Sorry if this has been posted before but I don't want to search it since searching for 'Fibonacci' is a bit depressing (too many results obviously)[/quote]

Try searching for Fibonacci and Legendre ;)

I don't know more about the number theory in $Z_n$,where can we find a nice book or article or post which introduce it?


[quote="nguoivn"]Given $ a, b, c > 0$ and $ ab + bc + ca = 1$.

Prove that: $ \frac {(1 + a^2)^2}{b^2 + c^2} + \frac {(1 + b^2)^2}{c^2 + a^2} + \frac {(1 + c^2)^2}{a^2 + b^2} \geq\ 8$
 :)[/quote]

Assume $ a^2+b^2+c^2=x\ge 1$

$ LHS\ge \frac{8}{9}\sum \frac{1+3a^2}{b^2+c^2}\ge \frac{8(3x+3)^2}{9\sum (1+3a^2)(b^2+c^2)}=\frac{8(x+1)^2}{3(x+\sum a^2b^2)}$

We will prove

$ \frac{(x+1)^2}{x+%Error. "suma" is a bad command.
^2b^2}\ge 3$

$ \Leftrightarrow x(x-1)\ge 3\sum a^2b^2-1$

$ \Leftrightarrow \sum a^2\left(\sum a^2- \sum ab\right)\ge 3\sum a^2b^2-\left(\sum ab\right)^2$

$ \Leftrightarrow (a^2+b^2-c^2)(a-b)^2+(b^2+c^2-a^2)(b-c)^2+(c^2+a^2-b^2)(c-a)^2\ge 0$

I think the last ineq true by SOS

[hide]Mấy cái tiêu chuẩn SOS em kô biết mấy nên mong anh thông cảm, đừng quát em, em dễ bị sốc lắm  :D   :rotfl: [/hide][/hide]

I checked this proof and I obtained $ x^2+1 \geq 6\sum a^2b^2$ or $ \sum a^4 +2abc\sum a \geq 3\sum a^2b^2$ which in SOS form is $ \sum (a-b)^2(a^2+b^2+2ab-2c^2) \geq 0$ and I don't think it can be solved by SOS method!!

[quote="gigaman"][quote="nguoivn"]Given $ a, b, c > 0$ and $ ab + bc + ca = 1$.

Prove that: $ \frac {(1 + a^2)^2}{b^2 + c^2} + \frac {(1 + b^2)^2}{c^2 + a^2} + \frac {(1 + c^2)^2}{a^2 + b^2} \geq\ 8$
 :)[/quote]

Assume $ a^2 + b^2 + c^2 = x\ge 1$

$ LHS\ge \frac {8}{9}\sum \frac {1 + 3a^2}{b^2 + c^2}\ge \frac {8(3x + 3)^2}{9\sum (1 + 3a^2)(b^2 + c^2)} = \frac {8(x + 1)^2}{3(x + \sum a^2b^2)}$

We will prove

$ \frac {(x + 1)^2}{x + %Error. "suma" is a bad command.
^2b^2}\ge 3$

$ \Leftrightarrow x(x - 1)\ge 3\sum a^2b^2 - 1$

$ \Leftrightarrow \sum a^2\left(\sum a^2 - \sum ab\right)\ge 3\sum a^2b^2 - \left(\sum ab\right)^2$

$ \Leftrightarrow (a^2 + b^2 - c^2)(a - b)^2 + (b^2 + c^2 - a^2)(b - c)^2 + (c^2 + a^2 - b^2)(c - a)^2\ge 0$

I think the last ineq true by SOS[/quote]

[quote="manlio"]I checked this proof and I obtained $ x^2 + 1 \geq 6\sum a^2b^2$ or $ \sum a^4 + 2abc\sum a \geq 3\sum a^2b^2$ which in SOS form is $ \sum (a - b)^2(a^2 + b^2 + 2ab - 2c^2) \geq 0$ and I don't think it can be solved by SOS method!!

[quote="gigaman"][quote="nguoivn"]Given $ a, b, c > 0$ and $ ab + bc + ca = 1$.

Prove that: $ \frac {(1 + a^2)^2}{b^2 + c^2} + \frac {(1 + b^2)^2}{c^2 + a^2} + \frac {(1 + c^2)^2}{a^2 + b^2} \geq\ 8$
 :)[/quote]

Assume $ a^2 + b^2 + c^2 = x\ge 1$

$ LHS\ge \frac {8}{9}\sum \frac {1 + 3a^2}{b^2 + c^2}\ge \frac {8(3x + 3)^2}{9\sum (1 + 3a^2)(b^2 + c^2)} = \frac {8(x + 1)^2}{3(x + \sum a^2b^2)}$

We will prove

$ \frac {(x + 1)^2}{x + %Error. "suma" is a bad command.
^2b^2}\ge 3$

$ \Leftrightarrow x(x - 1)\ge 3\sum a^2b^2 - 1$

$ \Leftrightarrow \sum a^2\left(\sum a^2 - \sum ab\right)\ge 3\sum a^2b^2 - \left(\sum ab\right)^2$

$ \Leftrightarrow (a^2 + b^2 - c^2)(a - b)^2 + (b^2 + c^2 - a^2)(b - c)^2 + (c^2 + a^2 - b^2)(c - a)^2\ge 0$

I think the last ineq true by SOS[/quote][/quote]
You are right,try c=0 and a=b.

I solved the ineq by cauchy shwarz  , but infortunately I cannot post the solution  for some private reasons :maybe:  if someone want to see it please pm me :roll:  :wink:

[quote="Evariste-Galois"]Show them your nice proof anas  :lol:  !!![/quote]

I cannot sorry evariste  :blush:  :blush:

[quote="leedt26"][quote="nguoivn"]It's look weak but not so weak. I created it by Am-Gm and Voni Schur  :)[/quote]
Can u give some hints how to use Voni Schur, nguoivn ?[/quote]
Using Am-Gm, we only need to prove: $ \sum\ \frac {a(b \plus{} c)(a^2 \plus{} bc)}{b^2 \plus{} c^2} \geq\ 2(ab \plus{} bc \plus{} ca)$
(obviously trues by Voni Schur)  :)

[quote="manlio"]How to solve

$ \sum\ \frac {a(b \plus{} c)(a^2 \plus{} bc)}{b^2 \plus{} c^2} \geq\ 2(ab \plus{} bc \plus{} ca)$

by Vornicu-Shur?

Thank you very much. :)[/quote]
I think it's solved by CS .
Because we have $ LHS \equal{} \sum {\frac{{a\left( {b \plus{} c} \right)\left( {{a^2} \plus{} bc} \right)}}{{\left( {{b^2} \plus{} {c^2}} \right)}}}  \equal{} \sum {{a^2}.\frac{{b\left( {{a^2} \plus{} {c^2}} \right) \plus{} c\left( {{a^2} \plus{} {b^2}} \right)}}{{a\left( {{b^2} \plus{} {c^2}} \right)}} \equal{} \sum {\frac{{{a^2}\left( {y \plus{} z} \right)}}{x}}  \ge 2\left( {ab \plus{} bc \plus{} ca} \right)}\equal{}RHS$
by Cauchy-Schwarz, where $ x \equal{} a\left( {{b^2} \plus{} {c^2}} \right);y \equal{} b\left( {{c^2} \plus{} {a^2}} \right);z \equal{} c\left( {{a^2} \plus{} {b^2}} \right)$

[quote="manlio"]How to solve

$ \sum\ \frac {a(b \plus{} c)(a^2 \plus{} bc)}{b^2 \plus{} c^2} \geq\ 2(ab \plus{} bc \plus{} ca)$

by Vornicu-Shur?

Thank you very much. :)[/quote]

Sorry, it's trues by Am-Gm, not Voni Schur. I didn't remember my solution  :oops:
We have: $ LHS \equal{} \sum\ ab(\frac {b^2 \plus{} c^2}{c^2 \plus{} a^2} \plus{} \frac {c^2 \plus{} a^2}{b^2 \plus{} c^2}) \geq\ 2(ab \plus{} bc \plus{} ca)$  :)

[quote="manlio"][quote="leedt26"][quote="manlio"]How to solve

$ \sum\ \frac {a(b \plus{} c)(a^2 \plus{} bc)}{b^2 \plus{} c^2} \geq\ 2(ab \plus{} bc \plus{} ca)$

by Vornicu-Shur?

Thank you very much. :)[/quote]
I think it's solved by CS .
Because we have $ LHS \equal{} \sum {\frac {{a\left( {b \plus{} c} \right)\left( {{a^2} \plus{} bc} \right)}}{{\left( {{b^2} \plus{} {c^2}} \right)}}} \equal{} \sum {{a^2}.\frac {{b\left( {{a^2} \plus{} {c^2}} \right) \plus{} c\left( {{a^2} \plus{} {b^2}} \right)}}{{a\left( {{b^2} \plus{} {c^2}} \right)}} \equal{} \sum {\frac {{{a^2}\left( {y \plus{} z} \right)}}{x}} \ge 2\left( {ab \plus{} bc \plus{} ca} \right)} \equal{} RHS$
by Cauchy-Schwarz, where $ x \equal{} a\left( {{b^2} \plus{} {c^2}} \right);y \equal{} b\left( {{c^2} \plus{} {a^2}} \right);z \equal{} c\left( {{a^2} \plus{} {b^2}} \right)$[/quote]

Dear leedt26,

can you please explain how to apply CS?
Thank you very much.[/quote]
Ok,my friend.
We have $ \sum {\frac{{{a^2}\left( {y \plus{} z} \right)}}{x} \equal{} \left( {x \plus{} y \plus{} z} \right)\left( {\sum {\frac{{{a^2}}}{x}} } \right)}\minus{} \left( {{a^2} \plus{} {b^2} \plus{} {c^2}} \right)$
then applies CS we retains that.

It is very nice :)

Thank you very much. :)

[quote="nguoivn"]Given $ a, b, c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 1$.

Prove that: $ \frac {(1 \plus{} a^2)^2}{b^2 \plus{} c^2} \plus{} \frac {(1 \plus{} b^2)^2}{c^2 \plus{} a^2} \plus{} \frac {(1 \plus{} c^2)^2}{a^2 \plus{} b^2} \geq\ 8$
 :)[/quote]

We can use cauchy and AM-GM to solve it.(by my brother)
$ LHS \ge \frac{[(a^3\plus{}b^3\plus{}c^3\plus{}3abc\plus{}ab(a\plus{}b)\plus{}bc(b\plus{}c)\plus{}ca(c\plus{}a)]^2}{2(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)} \ge \frac{2(a^3\plus{}b^3\plus{}c^3\plus{}3abc)[ab(a\plus{}b)\plus{}bc(b\plus{}c)\plus{}ca(c\plus{}a)]}{a^2b^2\plus{}b^2c^2\plus{}c^2a^2} \ge 8(ab\plus{}bc\plus{}ca)$
Finally,we need to prove that:
$ (a^3\plus{}b^3\plus{}c^3\plus{}3abc)[ab(a\plus{}b)\plus{}bc(b\plus{}c)\plus{}ca(c\plus{}a)]\ge 4(ab\plus{}bc\plus{}ca)(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)$
<=>$ \sum ab(a^4\plus{}b^4)\plus{} \sum a^2b^2(a^2\plus{}b^2) \ge 4(a^3b^3\plus{}b^3c^3\plus{}c^3a^3)$(obivious true by AM-GM)
Our proof are completed.Equality occur if and only a=b=c.
We can prove the general(in the same way)
$ \frac{(a^2\plus{}k)^2}{b^2\plus{}c^2}\plus{} \frac{(b^2\plus{}k)^2}{c^2\plus{}a^2}\plus{} \frac{(c^2\plus{}k)^2}{a^2\plus{}b^2} \ge \frac{(3k\plus{}1)^2}{2}$
true for $ 0 \le k \le 1$ with the same condition.


floor function
ratio
number theory unsolved
number theory

Let $ A$ be a non-empty subset of $ \mathbb N$ satisfying the following condition: If $ a \in A$, then $ 4a$ and $ \left[\sqrt {a}\right]$ are also in $ A$ (where $ \left[x\right]$ denotes the integer part of $ x$). Prove that $ A \equal{} \mathbb N$.

By applying $ \lfloor \sqrt{a} \rfloor$ repeatedly, we eventually get that $ 1 \in A$ if any element is in $ A$. Now we suppose there is an $ A$ such that the problem is false. Take the smallest element $ s > 1$ that is not in $ A$. Then no elements in between $ s^2$ and $ (s\plus{}1)^2 \minus{} 1$ inclusive are in $ A$. This in turn implies that no elements in between $ s^{2^k}$ and $ (s\plus{}1)^{2^k} \minus{} 1$ are in $ A$. The ratio of these two values is unbounded, so eventually we must have that $ (s\plus{}1)^{2^k} \minus{} 1 > 4s^{2^k}$. For such a $ k$, there must be an $ m$ such that $ 4^m$ is in the interval, and we know $ 4^m$ is in $ A$ since $ 1$ is in $ A$. Contradiction.

If this solution is correct, $ 4$ can be replaced with any positive integer.

@mods: Move this to temp, please (MY active problem).

let $ n\in\mathbb{N}$,we take for $ a\in\mathbb{N}: \ f(a) \equal{} [\sqrt {a}],g(a) \equal{} 4a$
we have $ {a\in[n^{2^k},(n \plus{} 1)^{2^{k}}[\Longrightarrow f(a)\in[n^{2^{k \minus{} 1}},(n \plus{} 1)^{2^{k \minus{} 1}}[\Rightarrow...f_k(a) \equal{} fof..k\ times...of(a)\in[n,n \plus{} 1[}$
then if $ a\in A$: $ n \equal{} f_k(a)\in A$
it's evident that for some $ k\in\mathbb{N},\exists h\in\mathbb{N}: \ n^{2^{k}}\le 4^h < (n \plus{} 1)^{2^{k}}$ $ \boxed{because\ \exists M,\forall k\ge M: \ (n \plus{} 1)^{2^{k}} \minus{} n^{2^{k}} > 4^{k \plus{} 1} \minus{} 4^{k}}$
we take $ b\in A$ we have $ f_m(b) \equal{} 1$ for some $ m\in\mathbb{N}$ so $ 4^h \equal{} g_h(1)\in A$
then $ n \equal{} f_k(4^{h})\in A$

The other thread containing this problem was locked because this is a problem from an ongoing contest, so this one should be too.  And $ A$ is non-empty.

But as long as this thread already contains the correct solution, I should mention that the first condition can, amusingly enough, be replaced with the condition "$ A$ contains the primes."


Let $ a_1,...,a_n,b_1,...,b_n$ be positive real numbers, such that $ a_1 \plus{} ... \plus{} a_n \equal{} b_1 \plus{} ... \plus{} b_n$.Prove that $ \dfrac{a_1^2}{a_1 \plus{} b_1} \plus{} ... \plus{} \dfrac{a_n^2}{a_n \plus{} b_n}\geq\dfrac{a_1 \plus{} ... \plus{} a_n}{2}$.


We have the numbers $a=\sqrt[3]{60}$ and $b=2+\sqrt[3]{7}$
Which number is bigger?

[[b]EDIT[/b]: this was nonsense. Sorry. ]

Silouan, I posted this in the intermediate forum a few days ago... :|

Jensen for $f(x) = x^{1/3}$ gives \[2 + \sqrt[3]{7} = f(8) + f(7) \leq 2 f\left(\frac{15}{2}\right) = \sqrt[3]{60}\] so $2 + \sqrt[3]{7} < \sqrt[3]{60}$. So, Kondr, you were wrong :)

How did you get
f(a)+f(b)>=2f((a+b)/2)
Is it through calculus.

It's using Jensen's ineq ( http://mathworld.wolfram.com/JensensInequality.html ).
For $f(x)=x^{1/3}$ we have $f''(x)=-\frac 2 9 x^{-\frac 5 3}<0$, so ineq
$\frac {f(a)+f(b)} 2 \leq f\left( \frac {a+b} 2 \right)$ holds.
A  non-calculus solution is on
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=324803#p324803

Arne - good solution.
Varun -Iit is sort of intutive. If you  think of the curve ($y=x^{(1/3)}$) , it is concave, so if you join two points (x=7, and x=8) the line (all points on the line except the end points)  is always "below" the curve, and the mid point on the line is below the the point on the curve (where x=7.5). (BTW you may have a typo when you said " f(a)+f(b)>=2f((a+b)/2) " the sign ">" should be "<")

Also,  FWIW  The "straight forward metod"  for many of us who grew up before the age of calculators (but after the age of binomial expansion)  , is  to think
$60^{(1/3)} = (64-4)^{1/3}$  and  $7^{1/3} = (8-1)^{1/3}$    and the difference between these two
 = $4 (1 - 1/48 -(1/9)(1/16)^2 \cdots) - 2 ( 1 - 1/24 -(1/9)(1/8)^2 \cdots )$ is greater than $2$ .     QED.


Prove positive integer $n$ is prime iff $\varphi (n)|n-1\ ,\ n+1|\sigma (n)$

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=lehmer&t=14533

Pierre.


Ratios always trip me up, could someone solve this problem and show me easy way to do ratios?

The sum of four numbers is 771. The ratio of the first to the second is 2:3. The ratio of the second to the third is 5:4. The ratio of the third to the fourth is 5:6. Find the second number.

Let the first number be a. Then work out the second number in terms of a, then the third one, then the fourth one. Then you just get an equation in terms of a.

If its the actual ratios you are having problems with, then if you have a ratio of x:y and you know the first number is a, then the second number must be.. (in spoiler) [hide]ay/x, because x:y is the same as 1: (y/x)[/hide]

After you have done that, you could possibly make your calculations a bit easier by letting the first number, say, be 50a, just to keep everything out of fractions.


Albatross has his friends Bobert, Caddlebot, Deberg, Eustine and Frankinfueter over at his house for New Years. The friends get restless so they decide to go see a movie. The row is 6 chairs long. Frankinfueter hates everyone there except Albatross, so the only person he will sit next to is Albatross. How many seating arrangement s are there so that Frankinfueter is happy.

[hide]

There are two cases: where F sits on the very left and where F sits on the very right.

Case 1: F sits on the very left. You have F A _ _ _ _. 4! can be used to describe the 4 blanks.
Case 2: F sits on the very right. _ _ _ _ A F. 4! can be used again.

So the answer is 2 x 4! = 48

[/hide]

Well, if Frankinfueter hates them enough to steal Albatross's poison kit...

[hide]

As all of the other people are eliminated, Albatross has six seats to choose from and Frankinfueter has five, so the answer is 30.

[/hide]

Assuming Albatross used all of the poison on Frankinfueter (or leeched it into the water supply)...

[hide]

F A _ _ _ _

24 possibilities

_ _ _ _ A F

24 possibilities

24 + 24 = 48, like aznphatso said.

Actually, Frankinfueter is never happy. If he has fun pushing his friend off buildings, he probably derives happiness from other murderous things, like poisoning Albatross's friends (see above).

[/hide]

Is Caddlebot a person?

[quote="Treething"]Okay then :) .

Caddlebot is cool :) . He bounces to 1/4 of his previous height when dropped.[/quote]
You made that fact up.

Treething wrote:Well, if Frankinfueter hates them enough to steal Albatross's poison kit...

[hide]
As all of the other people are eliminated, Albatross has six seats to choose from and Frankinfueter has five, so the answer is 30.
[/hide]

Except that Frakinfueter still has to sit next to Albatross even if there are 6 seats. I don't think I sit next to someone if there's four seats between us.

It says the only [b]person[/b] Frankinfueter will sit next to is Albatross. This means that he can sit next to an empty space, as an empty space isn't a person.


Will there be an Intermediate Counting and Probability book to go along with the class?

There is no text at this time. According to the Book List, no release of such a book is planned this year.

Eventually, but not this summer.  The book will likely be out sometime between Dec, 2006 and Mar, 2007.

I'm taking intermediate counting and probability this summer. There is enough problems from the intermediate book ( comming out later ) that I can still justifiy buying it as a problem book/reference? Or not worth the effort?

Not only will the book have additional exercises, but there will be topics in the book that are covered more thoroughly than they will be in the class.


email
summer program
Mathcamp
geometry
floor function

I got an email saying that one of my solutions (#7) was good (unexpected), and that I would be lucky (:wink:) enough to present it. I'm assuming other people have gotten similar?

Confirmation is necessary by July 3rd, so I thought I'd just check over here that I won't be working myself into a nervous wreck over this. Alright, "nervous wreck" is an exaggeration, it can't be *that* bad.  :lol:

So is anybody else planning on presenting?

Aha! So you're one of the other lucky winners.  I will present a solution, but I won't say which problem.  :P

I personally did not get such an email, but I am not entirely sure what you're worried about. If it was good, just look over your solution (you kept it, right :-) ), and essentially read it.

Out of curiosity, what did you get? At this point, I think we can openly discuss answers, since they are not accepting any more applications. I am pretty sure I got the right answer, but I did not simplify it from a form with a summation (I could, but the deadline was nearing, and I did not want to risk making the mistake. I got in, so I guess it was OK!) What did you get?

(Speaking as an Alum)

So at some point in week 2 (I believe it's week two, at least.  It's definitely not like... day 2 of camp or anything), most of the camp goes to an assembly at which all 10 qualifying quiz solutions are presented by new campers.  Thus, if you want to present, you won't be alone ;).  There will be 9 other new campers in your shoes.  I'd highly recommend you take them up on the offer, but that's just me...

Anyway, congrats on getting a "commended" solution, and I look forward to seeing you in a couple of weeks.

I will be presenting number 10 (despite the fact that the solution I came up with used a calculator).

[quote="hahafaha"]Out of curiosity, what did you get? At this point, I think we can openly discuss answers, since they are not accepting any more applications. I am pretty sure I got the right answer, but I did not simplify it from a form with a summation (I could, but the deadline was nearing, and I did not want to risk making the mistake. I got in, so I guess it was OK!) What did you get?[/quote]

I assume we're allowed to, as applications are over? Actually, can we compile a list of all the correct answers (perhaps not the proofs, since I assume we're going over them at mathcamp)? I want to know how many I actually got correct.

knexpert, I used the computer to do #10 ... not really a good sign if you're using the computer to solve a geometry problem. Oh well, it worked  :lol:

Here are my answers, which have been checked with others:

1. $ \lfloor \frac{3n}{2}\rfloor$
2. 1, 2, 3, 4, and 91
3. (a) yes (b) no
4. 59
5. (a) unsure, most work (b) prove it!
6. 2
7. (a) $ \frac{2}{3}$ (b) $ \frac{(n-k+3)!k!}{3!n!}$ (except when $ k \le 4$)
8. $ n^{2}-k^{2}$
9. Define all configurations inductively, either add a facedown card or flip over the rightmost card and add a face up card
10. a=12, b=15, c=13

[quote="calc rulz"]Here are my answers, which have been checked with others:

(snip)

6. 2

(snip)

[/quote]

Really? Which two configurations worked?

I'll make things easier, I got everything right except #4 I think.

So, for problem #6, I got 0. Here is a brief description of a counter-claim to the proof (I'm not going to bother looking at all cases, just one of the ones Tom presented):

When we multiply by a power of 2, the color changes if the power is odd and remains the same if it was odd (because every re-group essentially corresponds to a color change, and to a power). Tom suggested a situation in which 9 = green, which would mean that 4 and 5 are both blue (4+5=9; blue+blue=green). 1 is then also blue. Now through fairly simple arithmetic, we can deduce that 40 is blue. Also, 2048=1x2^11, and since 11 is odd, 2048 is green.

Now if blue+blue=green, then green-blue=blue. Thus, 2008=2048-40=green-blue=2008=blue.

Now if 2007=green, then 2009=2007+2=green+green=blue. However, 2009=2008+1=blue+blue=green, so there is a contradiction. Similarly, if 2007=blue, then 2008=2007+1=blue+blue=green. However, 2008 was previously determined to be blue, so there is still a contradiction.

I am too lazy to check it for the other one, as well as to write out the proof for all the other cases, but I just thought I'd share this. I'd be interested in feedback/discussion!

Here are my solutions. I got everything right except for #5a, and in #6 I made an unproved claim, which I think is not too hard to prove using the division algorithm.

So, the issue I see with 2007 being the first non-red (green, specifically) positive integer is this (note that I may be completely wrong):

It is easy to see that green+red=green, because 0 has to be red (since -0=0, and -green=blue and vice-versa, we'd have a contradiction otherwise). Since x+0=x, a green x+0 must equal green. We assumed that all positive integers before 2007 were red, and 2007 itself was green. However, we then have:

n=2007
n = green
n+1=green+red=green
n+2=(n+1)+2=green+red=green
n+3=(n+2)+1=green+red=green
...
2n+3=(2n+2)=green+red=green

However, 2n+3=(n+1)+(n+2)=green+green=blue, not green. Thus, there is a contradiction. In general, by this rule, I don't think 1 can be red successfully.

[quote="hahafaha"]
It is easy to see that green+red=green, because 0 has to be red (since -0=0, and -green=blue and vice-versa, we'd have a contradiction otherwise). Since x+0=x, a green x+0 must equal green. We assumed that all positive integers before 2007 were red, and 2007 itself was green. However, we then have:
[/quote]

Yes, green+red=green is true when the red number is 0, but it need not hold in general  :wink:

[quote="calc rulz"]Yes, green+red=green is true when the red number is 0, but it need not hold in general  :wink:[/quote]

Oh, that is a good point I forgot to account for. I stand corrected.

:o I got 8 answers correct! (idk about proofs though)

In #6, the two cases are where 9 and 2007 are the first green, non-red integers.

[quote="azjps"]In #6, the two cases are where 9 and 2007 are the first green, non-red integers.[/quote]

Yeah. I realized that my mistake stemmed from a flawed assumption.


abstract algebra
LaTeX
superior algebra
superior algebra unsolved

Let V be a countable dimensional vectorspace over a field F .
Let R denote End_F V .
Prove that V is a simple R-module. If e1 , e2 , . . . is a basis
of V , then we have a module homomorphism φ_j from R to V ,
sending f
∈ R to f (e_j ).
Find the ker(φ_j) .
Find Jac(R).   here I mean jacobson radical.
Prove that there is exactly one non-trivial twosided ideal,
namely
{f ∈ R| dimf (V ) < ∞} Prove that R

It sholud be

Let V be a countable dimensional vectorspace over a field F .
Let R denote End_F V .
Prove that V is a simple R-module. If $ e1 , e2 , . . .$ is a basis
of V , then we have a module homomorphism φ_j from R to V ,
sending f in  R to f (e_j ).
Find the ker(φ_j) .
Find Jac(R).   here I mean jacobson radical.
Prove that there is exactly one non-trivial twosided ideal,
namely
$ {f ∈ R| dimf (V ) < \infty}$ Prove that R is isomorphic to R + R here + means direct sum . thanks.

[quote="peteryellow"]Let $ V$ be a countable dimensional vectorspace over a field $ F$.  Let $ R$ denote $ End_F V$ .
Prove that $ V$ is a simple $ R$-module.

If $ e_1, e_2 , \cdots$ is a basis of $ V$ , then we have a module homomorphism $ \phi_j$ from $ R$ to $ V$, sending $ f \in R$ to $ f (e_j )$.  Find the $ ker\ \phi_j$.

Find $ Jac(R)$.   here I mean jacobson radical.

Prove that there is exactly one non-trivial twosided ideal, namely $ \{f \in R | dim\ f(V ) < \infty\}$. Prove that $ R$ is isomorphic to $ R\oplus R$.[/quote]
Frequently these threads are read hundreds of times, so you may want to learn a thing or two about readability.  Here's a start: dollar signs and line breaks go a long way.

You might as well use $ \ker$ and $ \dim$ while you're at it.  (To say nothing of the \rm command for things like End, Jac, etc.)

See [[LaTeX]] and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 for information about using the mathematical typesetting system LaTeX.


algebra
polynomial
linear algebra
linear algebra unsolved

1.Show that $A$ is nilpotent if and only if  $\forall k \in (1,..,n) ,Tr(A^k)=0$
2.Suppose that the set $(\lambda|A+\lambda B$ is nilpotent$)$ contains at least $n+1$ elements.Show that $A$ and $B$ are nilpotent.
(All the matrices belong to $M_n(C)$)

[hide]2) From result 1) all $P_k(x) = Tr ((A + x B)^k)$ are polynomials of degree at most $n$ with $n+1$ roots : they are all null, in particular $P_k(0) = 0$ and A is nilpotent.[/hide]


The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is $ 20$ cents. If she had one more quarter, the average value would be $ 21$ cents. How many dimes does she have in her purse?

$ \textbf{(A)}\ 0\qquad
\textbf{(B)}\ 1\qquad
\textbf{(C)}\ 2\qquad
\textbf{(D)}\ 3\qquad
\textbf{(E)}\ 4$

[hide]

let $w,x,y,z$ represent the number of [b]pennies, nickels, dimes, and quarters[/b] respectively

So you get two equations:

1. $w+5x+10y+25z=20(w+x+y+z)$
2. $w+5x+10y+25(z+1)=21(w+y+z+1)$
   $\Rightarrow 5x+10y+25z+4=21(w+x+y+z)$

Subtracting 1 from 2;

3. $w+x+y+z=4$, therefore there are $4$ coins in total

Substitute (3) into 1.

$\Rightarrow w+5x+10y+25z=80$

If there are $2$ quarters then you need $3$ more dimes to get a total of $80$, which would amount to $5$ coins we can only have four, so therefore:

$\Rightarrow w+5x+10y+25(3)=80\implies w+5x+10y=5\implies x=1 \ w=0 \ y=0$

Thus she has $\boxed{0\implies\text{A}}$ dimes in her purse.
[/hide]

[hide="slightly less calculations"]
when adding a quarter increases the average by 21,
subtract 21 from 25(the value of the quarter) to get 4
then multiply 4 by 20(the original average) to get 80.
that means she had 80 cents
then use smart's way to find the number of coins
and then get the answer of 0.....[/hide]

[quote="#H34N1"]The average value of all the pennies, nickels, dimes, and quarters in Paula's
purse is 20 cents. If she had one more quarter, the average value would be 21
cents. How many dimes does she have in her purse?
(A) 0 (B) 1 (C) 2 (D) 3 (E) 4[/quote]

[hide]Let there be $n$ coins in her purse so there are a total of $20n$ cents in her purse.

Now we can set up the equation $20n+25=21(n+1)$ so $n=4$.

This means that she orgionally had $80$ cens in her purse.  The only way to make $80$ cents with $4$ coins is with $3$ quarters and $1$ nickel.

Therefore, there are $\boxed{0}$ nickles.[/hide]

You mean 0 dimes not 0 nickels (slight typo..nice solution)


inequalities
rearrangement inequality
inequalities unsolved

More generally, for any 2n reals $a_1$, $a_2$, ..., $a_n$, $x_1$, $x_2$, ..., $x_n$, we have

$\sum_{i=1}^n a_ix_i+\sqrt{\sum_{i=1}^n a_i^2\cdot\sum_{i=1}^n x_i^2}\geq\frac{2}{n}\cdot\sum_{i=1}^n a_i\cdot\sum_{i=1}^n x_i$.

In fact, this inequality rewrites as

$\sqrt{\sum_{i=1}^n a_i^2\cdot\sum_{i=1}^n x_i^2}\geq\frac{2}{n}\cdot\sum_{i=1}^n a_i\cdot\sum_{i=1}^n x_i-\sum_{i=1}^n a_ix_i$;

denoting $X=\sum_{j=1}^n x_j$, we have

$\sum_{i=1}^n\left(\frac{2}{n}X-x_i\right)^2=\sum_{i=1}^n\left(\frac{4}{n^2}X^2-\frac{4}{n}Xx_i+x_i^2\right)$
$=n\cdot\frac{4}{n^2}X^2-\frac{4}{n}X\underbrace{\sum_{i=1}^n x_i}_{=X}+\sum_{i=1}^n x_i^2$
$=\frac{4}{n}X^2-\frac{4}{n}X^2+\sum_{i=1}^n x_i^2=\sum_{i=1}^n x_i^2$

and

$\frac{2}{n}\cdot\sum_{i=1}^n a_i\cdot\sum_{i=1}^n x_i-\sum_{i=1}^n a_ix_i=\sum_{i=1}^n a_i\cdot\frac{2}{n}\sum_{j=1}^n x_j-\sum_{i=1}^n a_ix_i$
$=\sum_{i=1}^n a_i\left(\frac{2}{n}\sum_{j=1}^n x_j-x_i\right)=\sum_{i=1}^n a_i\left(\frac{2}{n}X-x_i\right)$;

thus, our inequality is equivalent to

$\sqrt{\sum_{i=1}^n a_i^2\cdot\sum_{i=1}^n\left(\frac{2}{n}X-x_i\right)^2}\geq\sum_{i=1}^n a_i\left(\frac{2}{n}X-x_i\right)$;

but this obviously follows from the Cauchy-Schwarz inequality.

  Darij

[quote="spider_boy"]Let $a,b,c,x,y,z$   be real numbers.Prove that
$ax+by+cz+\sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)}\ge\frac{2}{3}(a+b+c)(x+y+z)$[/quote]

 The problem is nice i think that you can solve it yourself. Its really easy.
Expand the terms and use Cauchy Swarz then prove a stronger version of this then for the last step use the rearrangement Inequality hope that you will enjoy

Davron

Dear Darij Grineberg

Why do you prefer to solve problems by the hard ways.
I think mine is shorter.
1)C-S
2)rearrangment
3)The problem is killed

Davron

Dear Davron,

How exactly do you want to apply rearrangement? We have no information about how our numbers are sorted...

  Darij

Dear Darij

yes sure you are right i stupidly assumed that a is max and x max and i am wrong  ;)
But dont you have any other solutions
Try you it please i believe in you can find !

Davron

[quote="arqady"][quote="spider_boy"]Let $a,b,c,x,y,z$   be real numbers.Prove that
$ax+by+cz+\sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)}\ge\frac{2}{3}(a+b+c)(x+y+z)$[/quote]
I think this is vasc's inequality. ;)[/quote]
If you mean vasc has created this inequality ,then i ' m not sure.As I have stated
in the source it is an old inequality from famous [b]KVANT[/b] journal. :lol:

What about the following, even more general conjecture:

  If $a_i, b_i, c_i \ge0$ for $i=1,...,n$ (this will only work for non-negative variables), prove
$\sum a_ib_ic_i+\sqrt[3]{\sum a_i^3\cdot\sum b_i^3\cdot\sum c_i^3}\ge \frac2{n^2} \sum a_i\cdot\sum b_i\cdot\sum c_i$ :?:
and similarly for $k$ rows of variables, the fraction on the RHS becoming $\frac2{n^{k-1}}$.

In the [i]very [/i]particular case that the $a_i, b_i, c_i$ are equally sorted, this follows of course easily from Hölder and Chebyshev. But no idea how to prove it otherwise. :roll: Darij's nice method doesn't seem to work with more than two sums...

The case $n=2$ and $k=3$ of that, yielding
$abc+xyz+\sqrt[3]{(a^3+x^3)(b^3+y^3)(c^3+z^3)}\ge\frac12(a+x)(b+y)(c+z)$,
smells again like rearrangement but maybe not that easy.  :? Any ideas :?:


Suppose $ A, B, C, D, E, F$ are six distinct points in the plane, and suppose $ AB, DE, CF$ are concurrent and $ AD, BC, EF$ are concurrent. Prove that $ AF, BE, CD$ are also concurrent.

[hide]It is easy applying the [b][size=100]Pappos theorem[/size][/b].[/hide]
Kostas Vittas.


Let $ a_1,a_2,..,a_{2000} \in Z$, $ |a_i| \leq 1000 \ \forall i, \ \displaystyle\sum_{i\equal{}1}^{2000} a_i \equal{}1$
Prove $ \exists \ K \subset (1,2,..,2000), K \neq \phi$ so that $ \displaystyle\sum_{i \in K}^{} a_i \equal{}0$

Does $ K \subset (1,2,..,2000)$ means that $ K$ is a subset of consecutive terms of that??(like K=(5,6,7.8))

 :?:

 :)

this is from Canada 2000...
I know a solution using pigeon hole principle

to ElChapin : yes, $ K$ is a subset of $ (1,2,..,2000)$

I use a lemma to prove this problem
Lemma: If $ x_1,x_2,..,x_m,y_1,y_2,..,y_n \in N (N\equal{} (1,2,....))$
and $ \displaystyle\sum_{1}^{m}x_i\equal{}\displaystyle\sum_{1}^{n}y_i\equal{}V<mn$
Then $ \exists \ K \subseteq (1,2,..,m)$ and $ L \subseteq (1,2,..,n)$ that $ \displaystyle\sum_{i \in K}x_i\equal{}\displaystyle\sum_{i \in L}y_i$


Let $ a_i>p>0$, $ a_1\equal{}a_{n\plus{}1}$ and $ \sum_{i\equal{}1}^{n}{a_i}\equal{}k.$ Find the minimum of
\[ \sum_{i\equal{}1}^{n}{\frac{a_i}{a^2_{i\plus{}1}\minus{}p^2}}\]

Ok. Now I can post my solution, hope it will be interesting...
We know an inequality for $ a > 0$, $ b > 0$, $ x,y\in R^{ \plus{} }$ and $ x\ne y$:
\[ \frac {a^x}{b^y}\geq \frac {xa^{x \minus{} y} \minus{} yb^{x \minus{} y}}{x \minus{} y}
\]
It is equals only if $ a \equal{} b$.

Now:
\[ \frac {a_i}{a_{i \plus{} 1}^2 \minus{} p^2} \equal{} \frac {a_i}{2a_{i \plus{} 1}}\left(\frac {1}{a_{i \plus{} 1} \plus{} p} \plus{} \frac {1}{a_{i \plus{} 1} \minus{} p}\right)\rightarrow
\]

\[ \rightarrow \frac {a_i}{2a_{i \plus{} 1}}\left( \frac {1}{\left(\frac {k}{n} \plus{} p\right)^2}\cdot\frac {\left(\frac {k}{n} \plus{} p\right)^2}{a_{i \plus{} 1} \plus{} p} \plus{} \frac {1}{\left(\frac {k}{n} \minus{} p\right)^2}\cdot\frac {\left(\frac {k}{n} \minus{} p\right)^2}{a_{i \plus{} 1} \minus{} p} \right)\geq
\]

\[ \rightarrow \frac {a_i}{2a_{i \plus{} 1}}\left( \frac {1}{\left(\frac {k}{n} \plus{} p\right)^2}\cdot \left(2\left(\frac {k}{n} \plus{} p\right) \minus{} (a_{i \plus{} 1} \plus{} p)\right) \plus{} \frac {1}{\left(\frac {k}{n} \minus{} p\right)^2}\cdot\left( 2\left(\frac {k}{n} \minus{} p\right) \minus{} (a_{i \plus{} 1} \minus{} p)\right) \right)\rightarrow
\]

\[ \rightarrow \frac {a_i}{2a_{i \plus{} 1}}\left( \frac {\frac {2k}{n} \plus{} p \minus{} a_{i \plus{} 1}}{\left(\frac {k}{n} \plus{} p\right)^2} \plus{} \frac {\frac {2k}{n} \minus{} p \minus{} a_{i \plus{} 1}}{\left(\frac {k}{n} \minus{} p\right)^2} \right)\rightarrow
\]

\[ \rightarrow \frac {a_i}{2a_{i \plus{} 1}}\left( \frac {\frac {4k^3}{n^3}}{\left(\frac {k^2}{n^2} \minus{} p^2\right)^2} \minus{} \frac {\frac {2k^2}{n^2} \plus{} 2p^2}{\left(\frac {k^2}{n^2} \minus{} p^2\right)^2} \cdot a_{i \plus{} 1}\right)\rightarrow
\]

\[ \rightarrow \frac {\frac {2k^3}{n^3}}{\left(\frac {k^2}{n^2} \minus{} p^2\right)^2}\cdot \frac {a_i}{a_{i \plus{} 1}} \minus{} \frac {\frac {k^2}{n^2} \plus{} p^2}{\left(\frac {k^2}{n^2} \minus{} p^2\right)^2} \cdot a_{i}
\]
Now we can write:
\[ \sum_{i \equal{} 1}^{n}{\frac {a_i}{a_{i \plus{} 1}^2 \minus{} p^2}}\geq \frac {\frac {2k^3}{n^3}}{\left(\frac {k^2}{n^2} \minus{} p^2\right)^2}\cdot \sum_{i \equal{} 1}^{n}{\frac {a_i}{a_{i \plus{} 1}}} \minus{} \frac {\frac {k^2}{n^2} \plus{} p^2}{\left(\frac {k^2}{n^2} \minus{} p^2\right)^2} \cdot \sum_{i \equal{} 1}^{n}{a_{i}}
\]

\[ \sum_{i \equal{} 1}^{n}{\frac {a_i}{a_{i \plus{} 1}}}\geq n
\]

\[ \sum_{i \equal{} 1}^{n}{\frac {a_i}{a_{i \plus{} 1}^2 \minus{} p^2}}\geq \frac {\frac {2k^3}{n^3}}{\left(\frac {k^2}{n^2} \minus{} p^2\right)^2}\cdot n \minus{} \frac {\frac {k^2}{n^2} \plus{} p^2}{\left(\frac {k^2}{n^2} \minus{} p^2\right)^2} \cdot k \equal{} \frac {k}{\left(\frac {k}{n}\right)^2 \minus{} p^2}
\]
Equality holds only if $ a_1 \plus{} p \equal{} a_2 \plus{} p \equal{} ... \equal{} a_n \plus{} p \equal{} \frac {k}{n} \plus{} p$ and $ a_1 \minus{} p \equal{} a_2 \minus{} p \equal{} ... \equal{} a_n \minus{} p \equal{} \frac {k}{n} \minus{} p$, so if $ a_1 \equal{} a_2 \equal{} ... \equal{} a_n \equal{} \frac {k}{n}$

Answer:
\[ \boxed{\min\sum_{i \equal{} 1}^{n}{\frac {a_i}{a_{i \plus{} 1}^2 \minus{} p^2}}\equal{}\frac {k}{\left(\frac {k}{n}\right)^2 \minus{} p^2}}
\]


At the end of intro to geometry, it says it's coming out in 2009.

Unfortunately, I think it's going to be a bit later than that :(

why?
i mean any serious problem
 :maybe:  :maybe:

How long does it take you to write one book?

[quote="rrusczyk"]I've been [url=http://www.artofproblemsolving.com/Alcumus/Introduction.php]a little busy[/url], and I have to take care of writing the trig/complex numbers book first, which I'll start in a few weeks.[/quote]

Yes!  Trig/complex numbers!  Must buy!

Hm...does this mean that algebra is going into Alcumus soon?


function
real analysis
real analysis unsolved

Let $ f$ be a continuous, nonconstant, real function, and assume the existence of an $ F$ such that $ f(x\plus{}y)\equal{}F[f(x),f(y)]$ for all real $ x$ and $ y$. Prove that $ f$ is strictly monotone.

second problem, here: [url]http://www.mathematik.uni-muenchen.de/~yarotsky/jarnik/Jarnik_daniel_harrer.pdf[/url]

Since the link above isn't working (at least for me), I'll go ahead and post a solution.

Assume for the sake of contradiction that $f$ is not strictly monotone.
That means that $f$ has a local extrema, which is attained at some real number, call it $n_0$.

For arbitrarily small $\epsilon > 0$ we can find $x_1,$ $x_2 \in (n_0-\epsilon,n_o+\epsilon)$ such that $f(x_1)=f(x_2)$.
Now, see that $$f(x+x_2-x_1) = F[f(x-x_1), f(x_2)] = F[f(x-x_1), f(x_1)] = f(x)$$
for any $x \in \mathbb{R}$.

It follows that $f$ is a continuous function that has arbitrarily small periods, therefore $f$ is constant, contrary to the hypothesis.


The product, $ \log_a b \cdot \log_b a$ is equal to:
$ \textbf{(A)}\ 1 \qquad\textbf{(B)}\ a \qquad\textbf{(C)}\ b \qquad\textbf{(D)}\ ab \qquad\textbf{(E)}\ \text{none of these}$

[hide="Solution"]
Change of base gives $ \frac{\log a}{\log b}\cdot \frac{\log b}{\log a}\equal{}1$.
[/hide]


A sequence with first two terms equal 1 and 2 respectively is defined by the following rule:
each subsequent term is equal to the smallest positive integer which has not yet occurred in
the sequence and is not coprime with the previous term. Prove that all positive integers occur
in this sequence.

Consider a prime $ p$. We claim that if infinitely many multiples of $ p$ occur in the sequence then all multiples of $ p$ occur in the sequence. Otherwise, let $ kp$ be the smallest multiple of $ p$ which does not occure in the sequence. If $ p|a_n$, then $ a_{n\plus{}1}<pk$, which contradicts the fact that infinitely many multiples of $ p$ occur in the sequence.

Now assume that not all even numbers occur in the sequence. By our claim, only finitely many even numbers occur in the sequence. Hence the sequence only contains odd integers starting at some point. Let $ a_m$ be odd for all $ m>x$. Note that there exist infinitely many $ m$ such that $ a_{m\plus{}1}>a_m$, where $ m>x$. Let $ d\equal{}gcd(a_m,a_{m\plus{}1})$. But $ d<a_{m\plus{}1}\minus{}a_m$, or $ a_m\plus{}d<a_{m\plus{}1}$. So $ a_m\plus{}d$ is also a term in the sequence, but $ a_m\plus{}d$ is even, a contradiction. Therefore all even numbers occur in the sequence.

Finally, assume an odd integer $ q$ does not occur in the sequence. All numbers of the form $ 2kq$ must be in the sequence. Each of these numbers must be followed by a number less than $ q$, there are only finitely many of them, a contradiction. So all odd numbers occur in the sequence, which completes our proof.


A car starts at point P from rest, with zero initial acceleration, but with a constant rate of change of tangential acceleration of 2m/s^3. The car is constrained to travel on a circular track of unknown radius. When the car reaches point Q, the angle between the acceleration and velocity vectors is 30 degrees. The center of the track is point O. What is the angle (in radians) POQ?

The vector of the velocity of the car will be colinear with the vector of tangential acceleration. Thus, the angle between the total acceleration and its tangential component is $30^\circ$. The ratio of the radial (centripetal) and the tangential acceleration will then be $\frac{a_{\rm cp}}{a_{\rm t}} = \tan 30^\circ = \frac{1}{\sqrt{3}}$. Since $a_{\rm cp} = \omega^2 R$ and $a_{\rm t} = \dot{a} \tau$ we get that $\omega^2 R\sqrt{3} = \dot{a}\tau$, where $\omega$ is the current angular velocity, $R$ is the radius of the car's trajectory and $\tau$ is the time that passed since the car started moving.

Now, since $\omega = \int\alpha dt$ where $\alpha$ is the car's angular acceleration given by $\alpha = a/R = \dot{a}t/R$, we get that $\omega = \frac{\dot{a}t^2}{2R}$. Plugging this result for $t=\tau$ back in the above equation we get $\tau^3 = \frac{4R}{\dot{a}\sqrt{3}}$. Finally, \[ \angle POQ = \int_0^\tau\omega dt = \frac{\dot a}{6R}\tau^3 = \frac{2}{3\sqrt{3}} \approx 22^\circ.  \]

I don't guarantee that the solution's right because I wrote it down in a hurry, but I think it's ok...

I think you have the  $\sqrt{3}$ on the wrong side of the fraction: $\tau^3 = \frac{4 R \sqrt{3}}{\dot{a}}$. So $\angle POQ = \frac{2 \sqrt{3}}{3}$.

Hmmm, I don't think that it's wrong, since $\tan 30^\circ = 1/\sqrt{3}$...


inequalities
function
inequalities unsolved

The second one is an easy one. I believe you compose it yourself after a couple of minutes of thinking to the first one... at least that's how it happened to me.
Use Cauchy for $\sum\frac 1{ab+1}\sum(ab+1)\geq 9\Rightarrow \sum\frac 1{ab+1}(3+\frac {(a+b+c)^2}3)\geq 9\Rightarrow S(3+\frac13 S^2)\geq9$. And $(ii)$ is equivalent to this.
As for the first one, the equality doesn't hold for equal numbers. More than that I see now that the inequality doesn't hold at all for $a=b=c$.

I don't think the first one is correct.
Let a=b=c, the restriction turns to
3/(a^2+1)=3a
From the equation we get a^3+a-1=0. Since this is an increasing function, and f(0)<0, so a>0  *
The ineq is 9a^3+3a=9(1-a)+3a=9-6a>=9
Obviously , the ineq is not correct.

it's no use to say "I think".
Why not give us some ideas?