inequalities
geometry
inequalities unsolved

IN triangle prove that
(ma*mb)/(ra*mc)+(mb*mc)/(rc*ma)+(mc*ma)/(rb*mb)>=4/3
ma,mb,mc medium of triangle ABC
ra,rb,rc :radius of the circles inscribed in A,B,C and tangent to BC,CA,AB

I think you have many typos in your post, [b]blue allrise[/b]. First of all I suppose that you wanted to write symmetric sum. I think here is the right inequality:
$\sum_{cyclic}{\frac{m_{a}\cdot{m_{b}}}{r_{c}\cdot{m_{c}}}}\geq{3}$.
I have a very nice proof for this one. We have to prove that:
$\frac{1}{2}\sum_{cyclic}{(\frac{m_{a}\cdot{m_{b}}}{r_{c}\cdot{m_{c}}}+\frac{m_{b}\cdot{m_{c}}}{r_{a}\cdot{m_{a}}})}\geq{3}.$
Now: ${(\frac{m_{a}\cdot{m_{b}}}{r_{c}\cdot{m_{c}}}+\frac{m_{b}\cdot{m_{c}}}{r_{a}\cdot{m_{a}}})=\frac{m_{b}}{S}(\frac{m_{a}(p-a)}{m_{c}}+\frac{m_{c}(p-c)}{m_{a}}})$
${\frac{m_{b}}{S}(\frac{m_{a}(p-a)}{m_{c}}+\frac{m_{c}(p-c)}{m_{a}}})\geq{\frac{2m_{b}}{S}\cdot{\sqrt{(p-a)\cdot{(p-c)}}}}=\frac{2m_{b}}{\sqrt{p(p-b)}}=\frac{\sqrt{2a^{2}+2c^{2}-b^{2}}}{\sqrt{p(p-b)}}\geq{\frac{\sqrt{(a+c-b)(a+c+b)}}{\sqrt{p(p-b)}}}=2.$
We can do the others in similar way and then add them up to get the desired result.
Hereby $m_{a}$ is the median from vertice $A$, $r_{a}$ is the exradius, $S$ is the area of the triangle, $a,b,c$ are the sides of the triangle. :)

Well  i also think  that  blue  allrise's  inequality  needs  3 as  a  RHS
Here is  my  solution :
Use  [tex] m_a\geq\sqrt{s(s-a)} [/tex]  to  find  [tex]m_am_bm_c\geq s^2r =r_ar_br_c[/tex] . The rest  is  AM-GM  to  the  initial inequality .

[tex]  s [/tex]  is  semiperimeter , and the  other  symbols  are  the  well  know  ones.


Between $ 1$ and $ 8000$ inclusive, find the number of integers which are divisible by neither $ 14$ or $ 21$ but divisible by either $ 4$ or $ 6$.

can't we some how use the fact that every 7th of 4's multiple is divisible by 14 and every 21st one divisible by 21 and every 7th multiple of 6 is divisible by 14 and 21.then to exclude double counting eve

Yes, but it's awfully hard to keep track of all the details. My answer is [hide]2287[/hide] but I'm not sure if that's correct.

yep the answer is 2287.there r 2667 nums that r multiples of either 4 or 6.there are 285 multiples of 4 that are also divisible by 14 and 95 that r divisible by 21 & 95 that are div by 42(l.c.m. of 14&21).thus 285 multiples of 4 that r divisible by either 14 or 21.doing the same we find 190 nums for 6.there r 95 multiples of 12(l.c.m of 4&6) that r also divisible by 14,95 for 21 & 95 for 42.
so the answer is 2667-285-95+95-190-190+190+95=2287.


Hi,

This problem is from the book AoPS Vol. 1. It states:

Two dogs, each traveling $ 10ft/sec$, run toward each each other from $ 500$ feet apart. As they run, a flea flies from the nose of one dog to the nose of the other at $ 25ft/sec$. The flea flies between the dogs in this manner until it is crushed when dogs collide. How far did the flea fly?

I first tried:

Let $ x \equal{}$ the number of seconds until the dogs collide

Therefore, we have $ 10x \plus{} 10x \equal{} 500 \Rightarrow 20x\equal{}500 \Rightarrow x\equal{}25$

So there are $ 25$ seconds until the dogs collide. In that $ 25$ seconds, the flea flies: $ 25(25) \equal{} 625ft$

So the flea flies $ 625ft$ until it is crushed. And according to the Solution manual $ 625ft$ is the right answer. But how does that make sense? Cant the flea's distance not exceed $ 500ft$ because the collision happens in that range?

Your help will be appreciated!

Regards,
limac

[quote="limac"]Hi,

This problem is from the book AoPS Vol. 1. It states:

Two dogs, each traveling $ 10ft/sec$, run toward each each other from $ 500$ feet apart. As they run, a flea flies from the nose of one dog to the nose of the other at $ 25ft/sec$. The flea flies between the dogs in this manner until it is crushed when dogs collide. How far did the flea fly?

I first tried:

Let $ x \equal{}$ the number of seconds until the dogs collide

Therefore, we have $ 10x \plus{} 10x \equal{} 500 \Rightarrow 20x \equal{} 500 \Rightarrow x \equal{} 25$

So there are $ 25$ seconds until the dogs collide. In that $ 25$ seconds, the flea flies: $ 25(25) \equal{} 625ft$

So the flea flies $ 625ft$ until it is crushed. And according to the Solution manual $ 625ft$ is the right answer. But how does that make sense? Cant the flea's distance not exceed $ 500ft$ because the collision happens in that range?

Your help will be appreciated!

Regards,
limac[/quote]

Not nessacarily.
x----------------------x
If I were to move back and forth between these two x's, I could move 500,1000,1500, etc.

[quote="limac"]Thank you for the reply, BOGTRO! :)

Ah, so we are considering that the fly will move back and forth to gain a distance a distance of 625? But the problem says that the fly wants to go from the nose of the first dog to the nose of the other dog. So did that mean that the fly will be traveling back and forth across the two dogs' noses?

Regads,
limac[/quote]

Yes. If you consider infinite geometric series this should beocme more clear to you.

Gotcha! Thank you very much for your time BOGTRO  :D


How many positive integers less than 1000 are multiples of 7 and have a units digit of 8?

The explanation was very confusing and I don't understand how to do it. Can you please help?

Thanks!

Hint:
The first number that matches those terms is 28, and it repeats in cycles of 10.
i.e first is 28, then 98, and there are 10 multiples of 7 between 28 and 98, so next number would be 168.
 :D

Well, first you find the number of multiples of seven less than a thousand.
1000/7=142.(some random digits)  so the largest multiple of 7 under a thousand is 142*7=994, and smallest is 7, so there are 994/7=142 multiples.

now we have to find the ones that end in eight, and if you multiply 7x1 7x2... a bit, you'll find that the units digit is only affected by the other numbers units digit. So for it to have a units digit of 8, it needs 7 times a number that ends in 4. (4, 14, 24, etc.) because 4x7=2[8].  you can count these out: 4, 14, 24, 34, 44, 54,...134.
so there are fourteen.

I'm not sure if I explained well enough, sorry if thats the cause.

Please post your questions in the questions selection of the Forum, please.


quadratics
algebra
difference of squares
special factorizations

Do You Have Math Anxiety?
A Self Test

Rate your answers from 1 to 5; add them up and check your score below.  (1) = Disagree, (5) = Agree.

*I cringe when I have to go to math class. 1 2 3 4 5
*I am uneasy about going to the board in a math class. 1 2 3 4 5
*I am afraid to ask questions in math class. 1 2 3 4 5
*I am always worried about being called on in math class. 1 2 3 4 5
*I understand math now, but I worry that it's going to get really difficult soon.1 2 3 4 5
*I tend to zone out in math class. 1 2 3 4 5
*I fear math tests more than any other kind. 1 2 3 4 5
*I don't know how to study for math tests. 1 2 3 4 5
*It's clear to me in math class, but when I go home it's like I was never there. 1 2 3 4 5
*I'm afraid I won't be able to keep up with the rest of the class. 1 2 3 4 5

CHECK YOUR SCORE:

40-50 Sure thing, you have math anxiety.
30-39 No doubt! You're still fearful about math.
20-29 On the fence!.
10-19 Wow!

What is math anxiety?

Math anxiety is an emotional reaction to mathematics based on a past unpleasant experience which harms future learning. A good experience learning mathematics can overcome these past feelings and success and future achievement in math can be attained.

Do you agree or disagree with this statement?

Why?  Why not?

10 :D

have to do my homework now, so I won't expand on this

I got a 24, but I think i answered "5" for different reasons than intended; for example on the question "I don't know how to study for math tests." I said 5 because I can never find any way that studying would do anything since everything all makes sense and seems natural without memorization needed or studying required. It seems that 5's are suppose to be negative, yet that is clearly a positive thing... similar reasons for my other scores above 1.

[quote="tjhance"]I said 5 because I can never find any way that studying would do anything since everything all makes sense and seems natural without memorization needed or studying required.[/quote]

Haha I was thinking the exact same thing; none of the math classes at my school pose any challenge...

18....

1
1
1
1
1
5...Uhhh i know it already
1
5...I don't need to study
1
1

[quote="sharkman"]Do You Have Math Anxiety?
A Self Test

Rate your answers from 1 to 5; add them up and check your score below.  (1) = Disagree, (5) = Agree.

*I cringe when I have to go to math class. 1 2 3 4 5

hmm sometimes 2

*I am uneasy about going to the board in a math class. 1 2 3 4 5

 hmm I don't go to the board

*I am afraid to ask questions in math class. 1 2 3 4 5

 uhh kinda 3

*I am always worried about being called on in math class. 1 2 3 4 5

wtf you don't get called on

*I understand math now, but I worry that it's going to get really difficult soon.1 2 3 4 5

uh I only somewhat understand math now and it is definitely going to get very difficult soon 5

*I tend to zone out in math class. 1 2 3 4 5

yeah sometimes 2

*I fear math tests more than any other kind. 1 2 3 4 5

no I fear them about the same as any other kind 3

*I don't know how to study for math tests. 1 2 3 4 5

no, I can study for them. I'm not very good at it, but I can 2

*It's clear to me in math class, but when I go home it's like I was never there. 1 2 3 4 5

no this is stupid 1

*I'm afraid I won't be able to keep up with the rest of the class. 1 2 3 4 5

haha no 1

CHECK YOUR SCORE:

40-50 Sure thing, you have math anxiety.
30-39 No doubt! You're still fearful about math.
20-29 On the fence!.
10-19 Wow!

What is math anxiety?

Math anxiety is an emotional reaction to mathematics based on a past unpleasant experience which harms future learning. A good experience learning mathematics can overcome these past feelings and success and future achievement in math can be attained.

Do you agree or disagree with this statement?

Why?  Why not?[/quote]

so uhh like 21 or so maybe

1
3
3
3
2
2
1
1
1
1

I answered 3's on #2-4 because I usually never really concentrate on the questions in class much

5 I cringe not from the actual class but from the teacher
1
1
1
1
1
1
1 but I never study.
1
1

=14

I don't have math anxiety.

1
1
1
1
1
5
1
1
1
1

hmm I should stop reading algebra books in math class...

[quote="max_tm"]All those who are scoring in the math anxiety range are getting it for reasons other than the test intended (if you get a 5 for cringing when you go to math class, it's not supposed to be because the math class will bore you so much that you don't want to go  :P). I had math anxiety when I was a young'n, I can remember shaking with nervousness if the teacher asked me a question, or walking into exams (...yeah, exams in the 5th grade) on the brink of tears. It's not so much that I was bad at math, but I was in constant fear of making a mistake, or not being able to solve a problem the first time I saw it, or having kids in the class understand before me. None of you guys have math anxiety, you wouldn't be here if you did.[/quote]
Well for example, I always feel uneasy when going to the board because sometimes the method in the textbook isn't the best way to solve the problem, and I have to force myself to show the generic solution.
Like in solving $(x+25)(x+27)=8$ the textbook solution is to expand, combine, and plug into the quadratic, I can't bear to go through all those ugly steps when the simple substitution x=n-26 trivializes the problem.

[quote="pianoforte"]Don't you just love those problems that are like, solve for x in $(x-7)(x+3)=0$ by expanding and using the quadratic formula.[/quote]
 :rotfl:

in school, I'd have said 14 (I zone out in any class whatsoever, me not zoning out in a class is a rare thing, one of the rarest in the history of classes I should say)

in the past year (especially in the last sem), everything's changed. :( It's all maths now (next year we have a physic course though), and I'm doing miserably. I carried the idea that 'nobody studies for [i]maths[/i] tests, pooh, it's [i]maths[/i] after all' into college and have been shocked. I must review my policies for the coming year. :(

That is a large part of the problem with high schools.  College courses are so much more rigorous than high-school courses that many students, even those in "college prep" curricula, are shocked by the sudden increase in difficulty between high school and college.

[hide="Breakdown"]I cringe when I have to go to math class.
2  We keep it entertaining

I am uneasy about going to the board in a math class.
1   Meh

I am afraid to ask questions in math class.
2  I usually don't get answers anyway... unless it doesn't have anything to do with math.

I am always worried about being called on in math class.
2  It's distracting

I understand math now, but I worry that it's going to get really difficult soon.
1  I hope it does though.

I tend to zone out in math class.
4  Every now and then

I fear math tests more than any other kind.
5  It means I can't do homework that class

I don't know how to study for math tests.
5  I don't know how I would

It's clear to me in math class, but when I go home it's like I was never there.
5  Both parts of the sentence are true

I'm afraid I won't be able to keep up with the rest of the class.
1  No...[/hide]

[hide="Total"]28[/hide]

But yeah, I agree in part with bpms...
I remember in 8th grade, we had the problem $x^{2}=256$
So I did $x=\pm \sqrt{256}=\pm 16$ and the teacher yelled at me.

Apparently he wanted:
$x^{2}=256$ $\implies x^{2}-256=0$ $\implies (x+16)(x-16)=0$ $\implies x+16 = 0 \text{ or }x-16=0$ $\implies x=-16 \text{ or }x=16$

How was I supposed to know we were on the difference of squares chapter? Is it even a reasonable expectation for students to know what chapter their class is on? Not in that class.

It just shows how mathematics is taught as a set of procedures to be followed, rather than a set of tools to be used and applied in efficient and creative ways. I don't think that promotes clean thinking.

$x^{2}=256 \implies x^{2}-256=0 \implies (x+16)(x-16)=0 \implies x+16 = 0 \text{ or }x-16=0 \implies x=-16 \text{ or }x=16$
 :o How did you do that?!?  :o
Sorry, just spazzing, don't mind me...

I got an 18, don't know how to study, can't pay attention in class, too boring, play (my own version!!!) of the arithmetic game on my calculator (or read math books.)

[hide="Breakdown"][b]I cringe when I have to go to math class. [/b]
1. I love math class.

[b]I am uneasy about going to the board in a math class.[/b]
1. I don't mind it.

[b]I am afraid to ask questions in math class.[/b]
2. Afraid that I'll ask a stupid question.

[b]I am always worried about being called on in math class.[/b]
3. I don't know the reason...

[b]I understand math now, but I worry that it's going to get really difficult soon.[/b]
1. It's never going to get difficult.

[b]I tend to zone out in math class.[/b]
1. No.

[b]I fear math tests more than any other kind.[/b]
1. I don't fear math tests at all.

[b]I don't know how to study for math tests.[/b]
1. I know how to study, but I never do...  :D

[b]It's clear to me in math class, but when I go home it's like I was never there.[/b]
1. It's always clear to me.

[b]I'm afraid I won't be able to keep up with the rest of the class.[/b]
1. I go ahead of the class.[/hide]

[hide="Total"]13[/hide]

Here's my breakdown:

I cringe when going to math class.
(1) I enjoy it.  My math teachers have been very nice.  I can't say anything about next year, but I'm sure it'll still hold.

I am uneasy about going to the board in math class.
(1) I feel comfortable with going to the board.  I've done so many times.

I am afraid to ask questions in math class.
(1) I don't usually ask questions, but if there's something I'm confused about, I'm not afraid to ask.

I'm always worried about being called on in math class.
(1) I don't mind being called on.

I understand math now, but I worry that it's going to get really difficult soon.
(2) I feel a little nervous about the level of difficulty of math in years to come, but I'm sure that I'll be able to catch on very quickly.

I tend to zone out in math class.
(5) I know most of the stuff.

I fear math tests more than any other kind.
(1) Nope.  They're easy to me.

I don't know how to study for math tests.
(3) I've never found the need to study for a math test, so I can't say I absolutely know how.  But I have a pretty good guess.

It's clear to me in math class, but when I go home it's like I was never there.
(1) It's always clear to me.

I'm afraid I won't be able to keep up with the rest of the class.

My score: 17
(1) I feel that I'm ahead of most people in my class.


integration
calculus
derivative
calculus computations

Given the voltage drops for a resistor and capacitor are $ E_R \equal{} iR$ and $ E_c \equal{} \frac {1}{c}\int{i.dt}$ respectively, show that an RC-circuit with resistance R, capacitance C, voltage E and current i has equation $ R.\frac {di}{dt} \plus{} \frac {1}{c}i \equal{} \frac {dE}{dt}$

how to proof this formula?

The sum across the voltage is 0 ("The Loop Law" I believe its called, that's not important really).

Thus $ E\minus{}E_R\minus{}E_C\equal{}0$ or $ E\minus{}iR\minus{}\frac{1}{c}\int i dt\equal{}0$.

Take the derivative with respect to time to get $ \frac{dE}{dt}\minus{}\frac{di}{dt}R\minus{}\frac{i}{c}\equal{}0$. Now rearrange terms.


[quote="myc"]True or False?

a) 15 =35 (mod 20)

TRUE. 35/20 has a remainder of 15. Thus 15 is congruent to 35 (mod 20).  ( this is from [url=http://www.cs.usask.ca/resources/tutorials/csconcepts/1999_7/tutorial/trail/tp03TFsoln.html]this website[/url])

Is 15=35 (mod 20)  the same as 35=15(mod 20?

Thanks[/quote]

(15-35)/20 is an integrer, so it's true I think...

What do mean by "are they the same thing"? They're both true, if you want to know...

how is 15 equal to 35 (mod 20)?    doesn't  15=5(mod 20)?   how did they get 35.   35= 15(mod 20), because 35 divided by 20 =1 R 15.  or am I wrong?

Thanks

For any integer $n$, we have $15+20n\equiv 15\pmod{20}$.
It is because $20n$ is divisible by $20$ no matter what $n$ is.

This is true even if $n$ is a negative integer; for example, if  $n=-1$, then $15+20n=-5\equiv 15\pmod{20}$.

Properly, you shouldn't use the equals sign (you should use 'congruency,' or $\equiv$), and you shouldn't think of it in terms of remainders (which is what it looks like you're thinking).

$x \equiv y \bmod z$ is true if and only if $x - y$ is divisible by $z$ (or, actually, if you still want to think about it in terms of remainders, if $x$ and $y$ [i]both give the same remainder upon division by $z$[/i]).  For example, $15 \equiv 35 \bmod 20$ because $15 - 35 = -20$, which is divisible by $20$.

But you can also say, for example, that $15 \equiv 35 \equiv 55 \equiv 75 \equiv -5 \equiv -25 \bmod 20$ - each of those statements is true because each of those numbers differ from each other by some multiple of $20$.

An equivalent way of thinking about it:  $x \equiv y \bmod z$ if $x$ and $y$ end in the same digit in base $z$ (applies only if they all have the same sign).  For example, $3 \equiv 13 \equiv 23 \equiv 33 \equiv 43 \bmod 10$ because they differ from each other by $10$.

Also, when you say $15 \equiv 5 \bmod 20$, this is incorrect because $15 - 5 = 10$.

Yeah, I'm sure you meant $15 \equiv -5 \pmod 20$


X, Y and Z are pairwise disjoint sets of people.  The average ages of people in the sets X, Y, Z, X $ \cup$ Y, X $ \cup$ Z and Y $ \cup$ Z, respectively, are as follows: 37, 23, 41, 29, 39.5, 33

Find the average age of the people in set X $ \cup$ Y $ \cup$ Z

[hide="Answer (for reference)"]$ A\equal{}34$[/hide]


analytic geometry
Mandelbrot Competition
probability
expected value

ok.... here's the answer to your question to regional level for round three:
The question writers probably found 88, and rinted it; but afterwards they changed the answer to 6... both answers would have counted (they should have).... the test never stated if a single digit integer is a palindrome

i think both answers to the palindrome question are accepted.

as for the lightbulb problem, the answer should indeed be 13/9.  You interpreted the problem wrong (Or at least the wording is too vague).  Our team actually thought the same way you did, at first, and also got an answer of 1.  But here's why it's incorrect: looking at part iii (i think it was), they phrased the problem something like: "prove that [some result] is true for all k >= 2"  And the thing is, with YOUR assumption that ALL lights go off, this result (i forgot what it was)would NOT be true, making that entire question faulty.
Since that is obviously not the intent of the question writers, we assumed the other possible interpretation: that once a lightbulb has been touched twice, that particular light goes off and that's it.  It turns out that with this interpretation part iii became solvable.

but yea, i agree, the wording of the setup was pretty vague.  Our team argued about this very thing for a long time, before coming to our [ultimately correct] interpretation

[quote="Mandelbrot Contest"]A row of n lights is situated above the stage of the school auditorium, each controlled by a separate switch on the lightning panel offstage. It is late in the afternoon after math team practice, and all the lights are currently off. Using a calculator, an idle team member decides to choose a number from 1 to n at random, then flip the corresponding switch on the panel. He then repeats this process; if he happens to choose the same switch as before then all the lights will again be off, otherwise there will be two lights on. He continues picking numbers at random and flipping switches until an irate custodian puts an end to the activity after a total of k turns. On average, how many lights are on at this point?[/quote]The question seems pretty clear to me.  The first sentence says that each light is controlled by a separate switch.  The fourth sentence is referring only to the second turn -- otherwise why would there have to be zero or two lights on?  If the switches are separate, then how can one switch turn many lights off?  These are just ordinary switches.

I agree with Ravi -- that wording isn't ambiguous at all, you just decided it was more complicated than it really was.  I doubt you'll get any credit for it.  Sorry.

Our team had no problem with the interpretation as far as I know. Make sure to read problem statements carefully.


A window frame consisting of 6 equal rectangles is shown

[url=http://img162.imageshack.us/my.php?image=40617920db0.png][img]http://img162.imageshack.us/img162/1614/40617920db0.th.png[/img][/url]

a) Let the entire frame have height $ h$m and width $ y$m. If 12m of frame is available, show that y=$ \frac{1}{4}(12-3h)$

[hide]So basically, you get the equation $ 3h+4y=12$. Next, you plug in $ 3h+4y=12$ into the equation $ y=\frac{1}{4}(12-3h)$ and you get $ \frac{1}{4}(4y)$, which obviously equals $ y$.  [/hide]

[hide]I'm guessing the "frame" means all the lines in the picture..
Since there are three vertical lines and four horizontal lines, $ 3h+4y=12$.
$ 4y=12-3h$
$ y=\frac{1}{4}(12-3h)$
Is there a b)?[/hide]


In the pulley system attached to this post, shown are two masses of 10kg each. The pulley on the left is fixed to the ceiling while the pulley on the right is attached to the right mass of 10kg. What will be the accelaration of each mass? (assume the pulleys and rope are frictionless and the rope has negligible mass)

[hide="Answer"]If I am in no mistake, $ a_{left}\equal{}\frac{2g}{3}$ and $ a_{right}\equal{}\frac{g}{3}$.[/hide]

I'm not sure if your answers are correct, but you did you arrive with those answers?

Let T be the tension along the rope. The left mass is labeled 1 and that on the right labeled 2. Then I used

$ P\minus{}T \equal{} ma_{1}$, $ P\minus{}2T \equal{} ma_{2}$, and $ a_{1}\equal{} 2a_{2}$.

What is $ P$? By the way, my teacher says the answer is $ 6g$

P is the weight of each mass. The answer is 6g? Then I have done something quite wrong. Can you provide your reasoning? By the way, how can the acceleration be greater than g if each body is restricted in its motion due to another body?

Well, the bodies have a mass of 10kg each, so it's possible for the acceleration to be 6g? Because gravity exerts a force of 9.81 Newtons per kilogram.

I will ask my teacher for the working

remember gravity is not the only force acting on the block....


I think it would be an extremely convenient feature if when a post was made in one of someone's "Favorites" threads then the button would light up (similar to the private message icon).

Thanks,
-diophantient

It used to be that the favorites were arranged in the order of the time of the last post (like it is in all fora) and just a quick look was enough to see if someone posted in one of your favorite threads but now thay are arranged according to when you added them. There is some logic in this arrangement, of course, but I would prefer to have it the old way or, even better, to be able to sort the favorites in the same way as the usual forum threads :).

I will try to add both options (although the one about lighting up the button is not that easy to implement, while the second is not that complicated) at some point in the future.


Consider parallelogram $ABCD$. Point $E$ is chosen such that $BE=BC$ and $B$ lies on a segment $AE$. Line passing through $C$ and perpendicular to $BD$ intersects line through $E$ perpendicular to $AB$ in $F$. Prove that $\angle BAF =\angle DAF$.

This is very simolar problem. I hope it will help you.

            http://www.mathlinks.ro/Forum/viewtopic.php?highlight=parallelogram+AND+perpendicular&t=56498

[color=red][Moderator edit: This is actually the same problem, stated in a different way.][/color]

 [u]Babis[/u]


vector
group theory
abstract algebra
superior algebra
superior algebra solved

Hi, I really need help for this question

Let G be the group $\mathbb{Z}^n_2$, the set of binary vectors ($a_1$,$a_2$,...,$a_n$) in which $a_i$=0 or 1. The binary operation is vector addition modulo 2, in which 1+1=0.

(i) Let H be the subset of G consisting of all binary vectors with an even number of 1's. Show that H is a subgroup of G. Determine the order of G in terms of n.

(ii) Construct the group table for H for n=4.

Thanks before

For (i)
Consider $v_1,v_2\in H$. $v_1$ has $2a$ 1's and $v_2$ has $2b$ number of 1's. Let $u$ be the number of 1's which are on the same position in $v_1$ and $v_2$.
Then $v_1+v_2$ will have $1$ on a position $i$ iff $v_1$ and $v_2$ have different values on position $i$ and 0 iff $v_1$ and $v_2$ have the same value on position $i$.
So, the number of positions where $v_1$ has $1$ and $v_2$ has $0$ is $2a-u$ and the number of positions where $v_1$ has $0$ and $v_2$ has $1$ is $2b-u$, hence the number of positions where $v_1+v_2$ has $1$ is $2a+2b-2u$ which is even.
So, if $v_1,v_2\in H\Rightarrow v_1+v_2\in H$, so $H$ is a stable part(i'm not sure how to call it in english) of $\mathbb{Z}_2^n\Rightarrow$ H is a subgroup of $\mathbb{Z}_2^n$
The order of $G$ is $2^n$.


Let $P$ be a convex polygon with no two parallel sides. For each side of $P$ we consider the angle under which this side is seen from the farthest vertex in $P$. Prove that the sum of all these angles (for each side of $P$) is $180^\circ$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=40281 .

  darij


A gram of fat contains 9 calories, and a gram of carbohydrates and protein contains 4 calories. 10 pieces of McDonald's Chicken selects contains 66 grams of fat, 92 grams of carbohydrates, and 77 grams of protein. What percentage of the total amount of calories come from protein? Express your answer to the nearest tenth.

[hide="answer"]There are a total of $9\times66+4\times92+4\times77$ calories.  Of those, $4\times77$ are from protein.  That makes a total of 1270 calories, with 308 from protein.  Therefore, $\frac{1270}{308}\approx0.2425$, which is $24.3\%$.[/hide]

That is some sieriously unhealthy mcdonalds food.
The chicken selects are really filling.


Find all functions $ f : \mathbb{R} \to \mathbb{R}$ such that \[ f(x\plus{}y)f(f(x)\minus{}y)\equal{}xf(x)\minus{}yf(y)\] for all $ \ x,y \in \mathbb{R}.$

[hide="First solution  :D "]
$ f(x)\equal{}x$
$ f(x)\equal{}0$
[/hide]

$ f(0) \equal{} c$, $ f(x) \equal{} 0$ for all real number $ x$ except 0 could be solution too :P

I'll rewrite my solution in another way:
Solution:
$ x \equal{} 0$ in the main equation gives us:
$ f(y)f(f(0) \minus{} y) \equal{} \minus{} yf(y)$,
so it is either $ f(y) \equal{} 0$,or $ f(c \minus{} y) \equal{} \minus{} y$,where $ c \equal{} f(0)$.
Hence $ f(c \minus{} y) \equal{} \minus{} y$,for all $ y\in\mathbb{R}$,such that $ f(y)\neq 0$.
$ y\rightarrow c \minus{} y$,so $ f(y) \equal{} y \minus{} c$,for all $ y$,such that $ f(c \minus{} y)\neq 0$.
If $ f(x) \equal{} x \minus{} c$,for some $ x\neq c$ then we make substitution
$ y \equal{} 0$:
$ (x \minus{} c)f(x \minus{} c) \equal{} x(x \minus{} c)$.
If $ f(x \minus{} c) \equal{} x$ and $ x\neq 0$,then $ c \equal{} 0$.
If $ x \equal{} 0$,then $ c \equal{} 0$.
So $ f(x) \equal{} x$,if $ f( \minus{} x)\neq 0$.
Suppose that there exist $ x_0,y_0\in\mathbb{R}$,such that $ f(x_0) \equal{} 0,f(y_0) \equal{} y_0,y_0\neq 0$.
If we put $ x \equal{} x_0,y \equal{} y_0$:
$ f(x_0 \plus{} y_0)f( \minus{} y_0) \equal{} \minus{} y_0^2$
It is obvious that $ f( \minus{} y_0),f(x_0 \plus{} y_0)\neq 0$,thus $ f(x_0 \plus{} y_0) \equal{} x_0 \plus{} y_0$,hence $ x_0 \equal{} 0$.
So if $ f(x) \equal{} x$,for some $ x\in\mathbb{R}$,then it is true for all $ x\in\mathbb{R}$.
If $ f(y) \equal{} 0$,then it is true for all $ y\in\mathbb{R}$,except $ y \equal{} 0$.
So the answer is $ f(y) \equal{} y$ and $ f(y) \equal{} 0,f(0) \equal{} c,y\neq 0$.

[quote="ringos"][quote="Erken"]
now since $ f$ is injective and $ f(2x)\neq 0,$,then
$ f(f(x) \minus{} x) \equal{} 0 \equal{} f(0)$.
[/quote]

Why is $ f$ injective?

In fact, there are strange solutions:

$ f(x) \equal{} 0 (x \neq 0), f(0) \equal{} c$[/quote]
It is injective,because $ f(f(x)) \equal{} x$.
Oh yes,i forgot to consider $ f(0)$,when $ f(x)\equiv 0$,for all $ x\in\mathbb{R},x\neq 0$.So $ f(0)\equal{}c,f(x)\equal{}0,\forall x\in\mathbb{R},x\neq 0$ is a solution too.Thank you.

[quote="Erken"][quote="ringos"][quote="Erken"]
now since $ f$ is injective and $ f(2x)\neq 0,$,then
$ f(f(x) \minus{} x) \equal{} 0 \equal{} f(0)$.
[/quote]

Why is $ f$ injective?

In fact, there are strange solutions:

$ f(x) \equal{} 0 (x \neq 0), f(0) \equal{} c$[/quote]
It is injective,because $ f(f(x)) \equal{} x$.
Oh yes,i forgot to consider $ f(0)$,when $ f(x)\equiv 0$,for all $ x\in\mathbb{R},x\neq 0$.So $ f(0) \equal{} c,f(x) \equal{} 0,\forall x\in\mathbb{R},x\neq 0$ is a solution too.Thank you.[/quote]

$ f(f(x)) \equal{} x$ for some $ x$, not for all $ x$.

[quote="Erken"][quote="kunny"][quote="Erken"]Very strange,that an easy problem like this was at Japan MOF.[/quote]

I agree with you.[/quote]
Where is the other problems?
Thank you.[/quote]

Here are.

[url=http://www.mathlinks.ro/viewtopic.php?t=188109]Problem 1[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=188111]Problem 3[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=188110]Problem 5[/url]

[quote="primoz2"]This problem isnt so easy as it looks like, because just like ringos wrote f(f(x))=x just for some x.  I will post my solution as soon as i have time. I just hope i didnt make a mistake.[/quote]
I've already changed my solution,i didn't use $ f(f(x))\equal{}x$,i replaced the old proof,by new,see above.

[quote="Erken"]I've posted another solution,i hope it is clear now.See above.[/quote]

[color=green]Where is the first one, Erken?[/color]

I replaced it by a new proof few minutes ago.

[quote="primoz2"]This problem isnt so easy as it looks like, because just like ringos wrote f(f(x))=x just for some x.  I will post my solution as soon as i have time. I just hope i didnt make a mistake.[/quote]

At first sight, I thought that this problem is not difficult, but when I began to solve, it seemed to be unable to solve so easily.

[quote="ringos"][quote="Erken"]
If $ x \equal{} 0$,then $ c \equal{} 0$.
[/quote]

Could you explain this part?

I couldn't solve this problem in the competition.[/quote]
We have that $ f(x)\equal{}x\minus{}c$,so if $ x\equal{}0$,then $ f(0)\equal{}\minus{}c\equal{}\minus{}f(0)$,hence $ c\equal{}0$.So in any case $ c\equal{}0$.

[quote="Erken"]We have that $ f(x) \equal{} x \minus{} c$,so if $ x \equal{} 0$,then $ f(0) \equal{} \minus{} c \equal{} \minus{} f(0)$,hence $ c \equal{} 0$.So in any case $ c \equal{} 0$.[/quote]

Thank you, I understood. Nice solution.

[quote="quangpbc"][color=green]Uhm, and what about it ? Did you have a mistake with this solution? Show it again please, friend.[/color][/quote]
No,i think this solution is correct,friend. :)
P.S:I've got a nice number of posts:$ 666$. :lol:

[quote="Erken"]
If $ f(x) \equal{} x \minus{} c$,for some $ x\neq c$ then we make substitution
$ y \equal{} 0$:
$ (x \minus{} c)f(x \minus{} c) \equal{} x(x \minus{} c)$.
[/quote]

Where do you substitute $ y\equal{}0$?

[quote="Jumbler"][quote="Erken"]
If $ f(x) \equal{} x \minus{} c$,for some $ x\neq c$ then we make substitution
$ y \equal{} 0$:
$ (x \minus{} c)f(x \minus{} c) \equal{} x(x \minus{} c)$.
[/quote]

Where do you substitute $ y \equal{} 0$?[/quote]
In the main equation

Thank for your post. Thank you very much.

[hide="Answer"]$ f(x)\equal{}x$ and $ f(x)\equal{}\left\{
\begin{array}{ll}
c\ (x\equal{}0)\ c: arbiterary\ real\ numbers  &\quad  \\
0\ (x\neq 0) &\quad
\end{array}
\right.$[/hide]

[quote="kunny"][hide="Answer"]$ f(x) \equal{} x$ and $ f(x) \equal{} \left\{ \begin{array}{ll} c\ (x \equal{} 0)\ c: arbiterary\ real\ numbers & \quad \\
0\ (x\neq 0) & \quad \end{array} \right.$[/hide][/quote]
I've already solved this problem,see the previous list :wink:

[quote="silouan"]See here (at the last post ) for a nice straightforward solution

http://www.mathlinks.ro/viewtopic.php?t=194064[/quote]

 Thank you Silouan!
 Good luck by the IMO 2008 !!!

 [b]NOTE[/b]

 The solution at the last post(in the greek forum) is in english , so everybody can read it.

 Babis

[quote="kunny"]Find all functions $ f : \mathbb{R} \mapsto \mathbb{R}$ such that $ f(x \plus{} y)f(f(x) \minus{} y) \equal{} xf(x) \minus{} yf(y)$ for all $ \ x,y \in \mathbb{R}.$[/quote]

+ $ f \equal{} 0$

+ $ f \neq 0$

$ y \equal{} 0\ \equal{} > \ f(x).f(f(x)) \equal{} x.f(x) \ \equal{} > \ f(f(x)) \equal{} x \ \ \ (1)$

$ y: \equal{} f(x)\ \equal{} > \ f(x \plus{} f(x)).f(0) \equal{} x.f(x) \minus{} f(x).f(f(x))\ \ \ (2)$

$ (1)\ \ vs\ \ (2) \ \ f(0) \equal{} 0$

$ x \equal{} 0\ \equal{} > \ f(y).f( \minus{} y) \equal{} \minus{} y.f(y) \ \equal{} > \ f( \minus{} y) \equal{} \minus{} y\ \equal{} > \ f(x) \equal{} x$

Two solutions :  $ f(x) \equal{} 0 \ \ ; \ \ f(x) \equal{} x$

[quote="hophinhan"]
Two solutions :  $ f(x) \equal{} 0 \ \ ; \ \ f(x) \equal{} x$[/quote]

You should read the entire thread.
There exist other solutions :

$ f(0)\equal{}a\neq 0$ and $ f(x)\equal{}0$ $ \forall x\neq 0$

[quote="stergiu"][quote="Erken"]I'll rewrite my solution in another way:
Solution:
$ x \equal{} 0$ in the main equation gives us:
$ f(y)f(f(0) \minus{} y) \equal{} \minus{} yf(y)$,
so it is either $ f(y) \equal{} 0$,or $ f(c \minus{} y) \equal{} \minus{} y$,where $ c \equal{} f(0)$................
......................................
[/quote]

 I'm afraid that this conclution is not valid. Generally, if $ f(x)g(x) \equal{} 0 , \forall x \in \mathbb R$ , we can not take $ f(x) \equal{} 0 , \forall x \in \mathbb R$ or $ g(x) \equal{} 0 , \forall x \in \mathbb R$ .Do you have something more in mind ?

 Babis[/quote]

Please read it through more carefully, I didn't make such a conlusion.

we have 2 case first is f(x)=0 second is not.
for second;
let y=x so we have $ f(2x).f(f(x)\minus{}x)\equal{}0$ and plugging y=0 we have $ f(f(x))\equal{}x$
so taking f of two sides in first equation we have     $ f(x)\equal{}f(0)\plus{}x$ writing it to first function we get $ f(x)\equal{}x$ or from first case $ f(x)\equal{}0$

I didn't know that this problem is from Japan.

Thank you for posting the problem that I wanted.

Oh!!I must mention that there are too many useless posts of this problem.

Let us denote the given statement by $P(x,y)$. Then,
$P(x,f(x))\Rightarrow f(x+f(x))f(0)=xf(x)-f(x)f(f(x))$.
$P(x,0)\Rightarrow f(x)f(f(x))=xf(x)$.
So, $f(x+f(x))f(0)=0 \forall x\in \mathbb{R}$.
So we have three cases here.

[b]Case 1:[/b] When $f(x+f(x))=0 \forall x\in \mathbb{R}$.
$P(x+f(x),-x)\Rightarrow f(f(x))f(x)=xf(-x)$. But $f(f(x))f(x)=xf(x)$. So, $f(x)=f(-x)\forall x\in \mathbb{R}$.
$P(x,0)\Rightarrow f(0)f(x+f(x))=0=2xf(x)$. This yields us the solution:
$f(x)=0\forall x\neq 0$ and $f(0)=c\in \mathbb{R}$. In particular, if $c=0$, then it is the constant solution to the given equation.

[b]Case 2:[/b] When $f(x+f(x))\neq 0\forall x\neq 0$.
Then $f(0)=0$. The only constant solution belongs to the family of solution in case $1$. So let $f$ be non-constant.
$P(x,-x)\Rightarrow f(x)=-f(-x)\forall x\in \mathbb{R}$.
$P(0,y)\Rightarrow f(y)(f(y)-y)=0 \forall y\in \mathbb{R}.....(*)$
Now since $f$ is non-constant, there exists $d\neq 0$ such that $f(d)\neq 0$. Then from equation$(*)$, $f(d)=d$.
If there is a $k\neq 0$ such that $f(k)=0$, then,
$P(k,d-k)\Rightarrow df(k-d)=(d-k)f(k-d)\Rightarrow kf(k-d)=0\Rightarrow f(k-d)=0$. So,
$P(k-d,d)\Rightarrow 0=-df(d)\Rightarrow d^{2}=0\Rightarrow d=0$ which is a contradiction.
So, $r=0$ if and only if $f(r)=0$. Thus from $(*)$ and from $f(0)=0$, we have that $f(y)=y\forall y\in \mathbb{R}$.

[b]Case 3:[/b]  When $\exists a,b$ such that $f(a+f(a))=0$ and $f(b+f(b))\neq 0$, where both $a,b\neq 0$.
According to our previous results, $f(b+f(b))f(0)=0$. This means that $f(0)=0$. So,
$P(0,y)\Rightarrow f(y)[f(y)-y]=0$. Set $y=b+f(b)$, then we have $f(b+f(b))=b+f(b)$.
Let $d=b+f(b)$ and $k=a+f(a)$. Now unless $k=0$, we will have $d=0$, i.e. $f(b+f(b))=b+f(b)=0$ which contradicts our assumption. So, $k=f(a)+a=0$. Thus,
$P(a+f(a),-a)\Rightarrow f(a)=f(-a)$. Then $f(-a)=f(a)=-a$. So,
$P(-a,a)\Rightarrow 0=-af(-a)-af(a)=2a^{2}\Rightarrow a=0$ which is also a contradiction. So no such $a,b$ exist.
Thus either $f(x+f(x))=0\forall x\in \mathbb{R}$, or $f(x+f(x))$ is non-zero for all non-zero $x$.

So finally the solutions to the given equations are:
$f(x)=0\forall x\neq 0$ and $f(0)=c\in \mathbb{R}$
or,
$f(y)=y\forall y\in \mathbb{R}$.

It is easy to check that these are indeed the solutions.

y=0 => f(f(x))=x  (1)
y=f(x) => f(0)=0 (2)
y=-x => f(x)= -f(-x) (3)
If you change x and y =>
f(x+y)f(f(y)-x)=yf(y)_xf(x)
And with compare to the general equation and (3)
=> f(x-f(y))=f(f(x)-y)
And if you  put y=0
=> f(x-f(0))=f(f(x)) => (2) f(x)=f(f(x)) => (1) f(x)=x

[quote=Erken]I'll rewrite my solution in another way:
Solution:
$ x \equal{} 0$ in the main equation gives us:
$ f(y)f(f(0) \minus{} y) \equal{} \minus{} yf(y)$,
so it is either $ f(y) \equal{} 0$,or $ f(c \minus{} y) \equal{} \minus{} y$,where $ c \equal{} f(0)$.
Hence $ f(c \minus{} y) \equal{} \minus{} y$,for all $ y\in\mathbb{R}$,such that $ f(y)\neq 0$.
$ y\rightarrow c \minus{} y$,so $ f(y) \equal{} y \minus{} c$,for all $ y$,such that $ f(c \minus{} y)\neq 0$.
If $ f(x) \equal{} x \minus{} c$,for some $ x\neq c$ then we make substitution
$ y \equal{} 0$:
$ (x \minus{} c)f(x \minus{} c) \equal{} x(x \minus{} c)$.
If $ f(x \minus{} c) \equal{} x$ and $ x\neq 0$,then $ c \equal{} 0$.

sorry but I dont understand why here you get c=0
explain it please  :oops:

[quote=silouan]See here (at the last post ) for a nice straightforward solution

http://www.mathlinks.ro/viewtopic.php?t=194064[/quote]
in order not to miss the solution mentioned by [b]silouan [/b] as link went dead, here is the correct [url=https://artofproblemsolving.com/community/c267h194064p1066876]one[/url]
and the solution given by [b]savvas_michael[/b] [hide=follows]
[quote=savvas_michael]Set $ y \equal{} 0$ in the given equation and get $ f(x) \cdot f(f(x)) \equal{} x \cdot f(x) \ (*)$.

Therefore there exists at least one $ c \in R$ so that $ f(c) \equal{} 0$.

Now plug in $ y \equal{} f(x)$ to get according to (*) that $ f(0) \cdot f(f(x) \plus{} x) \equal{} 0$

and then $ y \equal{} \minus{} x$ to get $ f(0)f(f(x) \plus{} x) \equal{} x(f(x) \plus{} f( \minus{} x))$ and thus

$ f( \minus{} x) \equal{} \minus{} f(x)$, for every $ x \neq 0$.

Plugging in now $ x \equal{} c$ and $ y \neq 0$ we get $ f(y) \cdot (f(y \plus{} c) \minus{} y) \equal{} 0$.

As a result for every $ x \neq 0$ we have either $ f(x) \equal{} 0$ or $ f(x \plus{} c) \equal{} x$.

Let those $ x \neq 0$ satisfying the first equation form set $ A$ and the remaining $ x \neq 0$ form set $ B$.

We distinguish cases:

1. $ A$ is empty. Then we must have $ c \equal{} 0$ and therefore $ f(x) \equal{} x,\ \forall x \in R$.

2. $ A$ is non-empty. If $ B$ is empty then $ f$ is identically zero everywhere except for its value at $ 0$ which can be arbitrary.

Suppose now that $ B$ is non-empty. We will reach a contradiction.

Pick $ a \in A$ and $ b \in B$ and plug in $ x \equal{} a \plus{} b$ in (*) to get

$ f(b) \equal{} b\plus{}a$.  Therefore it is obvious that $ A$ has a single element $ a$. Then $ f(x)\equal{}x\plus{}a, \forall x \neq 0,a$

and hence as $ f$ is odd in $ R*$ we obtain $ a\equal{}0$ which is a contradiction.[/quote]

[/hide]

[quote=CanVQ][quote="kunny"]Find all functions $ f : \mathbb{R} \mapsto \mathbb{R}$ such that $ f(x\plus{}y)f(f(x)\minus{}y)\equal{}xf(x)\minus{}yf(y)\quad (1)$ for all $ \ x,y \in \mathbb{R}.$[/quote]
Let $a=f(0).$ Replacing $y=0,$ we get \[f(x)\cdot f\big(f(x)\big)=x\cdot f(x),\quad \forall x \in \mathbb R. \quad (2)\] Now, replacing $y=f(x)$ and using $(2),$ we get \[a\cdot f\big(x+f(x)\big)=0.\quad (3)\] There are two cases to consider:

[b]Case 1: [/b]$\mathbf{a=0}.$ Replacing $x=0$ in $(1),$ we get \[f(y)\cdot f({-y})=-y\cdot f(y),\quad \forall y \in \mathbb R. \quad (4)\] Replacing $y$ by $-y$ in $(4),$ we get \[-y\cdot f(y)=f(y)\cdot f({-y})=y\cdot f({-y}),\] so by combining with the fact $f(0)=0,$ we get \[f({-y})=-f(y),\quad \forall y \in \mathbb R. \quad (5)\] Using this result in $(4),$ we obtain \[f(y)=0\vee f(y)=y,\quad \forall y \in \mathbb R. \quad (6)\] Assume that, there exist $u,\,v \ne 0$ such that $f(u)=u$ and $f(v)=0.$ Replacing $x=u,\,y=v$ in $(1),$ we get \[f(u+v)\cdot f(u-v)=u^2.\] Since $u^2 \ne 0,$ we must have $f(u+v) \ne 0$ and $f(u-v)\ne 0.$ Therefore, $f(u+v)=u+v,\, f(u-v)=u-v.$ It follows that \[u^2=(u+v)(u-v)=u^2-v^2,\] and hence $v=0,$ a contradiction. So, we must have $f(x)=0,\, \forall x \in \mathbb R$ or $f(x)=x,\, \forall x \in \mathbb R.$ It is easy to check that these functions satisfy our condition.

[b]Case 2: $\mathbf{ a\ne 0}.$[/b] From $(3),$ we get \[f\big( x+f(x)\big) =0,\quad \forall x \in \mathbb R. \quad (7)\] Now, by replacing $x$ by $x+f(x),$ $y$ by $x$ in $(1),$ we get \[f\big(2x+f(x)\big)\cdot f({-x})=-x\cdot f(x),\quad \forall x \in \mathbb R. \quad (8)\] Replacing $y$ by $x+ f(x),$ we also have \[f\big( 2x+f(x)\big)\cdot f({-x})=x\cdot f(x),\quad \forall x \in \mathbb R. \quad (9)\] Combining $(8)$ and $(9),$ we get \[f(x)=0,\quad \forall x \ne 0.\] Therefore, \[f(x)=\begin{cases} 0 &\text{if } x \ne 0 \\ a &\text{if } x =0 \end{cases}\] It is easy to check that this function satisfies the given condition.[/quote]

CanVQ, Perfect !


logarithms trigonometry function calculus derivative

[quote="mathpk"]$0 \leq x \leq \pi$ find $x$ such that,
\[ \log \sin x = \sin \log x  \][/quote]
I'm sure that the proof uses function, but I cannot find the function's derivative now, just because I have no paper and ball point pen!

[quote="mathpk"]$0 \leq x \leq \pi$ find $x$ such that,
$\log(\sin(x))=\sin(\log(x))$[/quote]
If you graph these functions, you'll see that this is never true.  Is that what you want us to prove?


Let $ a$, $ b$, and $ c$ be positive real numbers such that $ ab + bc + ca = 3$. Prove the inequality
\[ 3 + \sum_{\mathrm{\cyc}} (a - b)^2 \ge \frac {a + b^2c^2}{b + c} + \frac {b + c^2a^2}{c + a} + \frac {c + a^2b^2}{a + b} \ge 3.
\]

[quote="Raja Oktovin"]Let $ a$, $ b$, and $ c$ be positive real numbers such that $ ab + bc + ca = 3$. Prove the inequality
\[ 3 + \sum_{\mathrm{\cyc}} (a - b)^2 \ge \frac {a + b^2c^2}{b + c} + \frac {b + c^2a^2}{c + a} + \frac {c + a^2b^2}{a + b} \ge 3.\]
[/quote]

I kinda remember that the left side could be changed into the following stronger inequality:
\[ 3\plus{}\frac12\sum_{cyc}(a\minus{}b)^2\geq\sum_{cyc}\frac{a\plus{}b^2c^2}{b\plus{}c}\]
But then I can't prove it. Anybody?

[quote="wangsacl"]I kinda remember that the left side could be changed into the following stronger inequality:
\[ 3 \plus{} \frac12\sum_{cyc}(a \minus{} b)^2\geq\sum_{cyc}\frac {a \plus{} b^2c^2}{b \plus{} c}\]
But then I can't prove it. Anybody?[/quote]
This ineq is equivalent to :$ {a^2} \plus{} {b^2} \plus{} {c^2} \ge \sum {\frac {a}{{b \plus{} c}} \plus{} } \sum {\frac {{{b^2}{c^2}}}{{b \plus{} c}}}$
Easy to see that : $ a \plus{} b \plus{} c \ge 3$
Notice :
$ \frac {a}{{b \plus{} c}} \equal{} \frac {{a(ab \plus{} bc \plus{} ca)}}{{3(b \plus{} c)}} \equal{} \frac {{{a^2}}}{3} \plus{} \frac {{abc}}{{3(b \plus{} c)}} \le \frac {{{a^2}}}{3} \plus{} \frac {{a(b \plus{} c)}}{{12}}$
And $ \frac {{{b^2}{c^2}}}{{b \plus{} c}} \le \frac {{bc(b \plus{} c)}}{4}$
And we need to prove that:
$ \frac {{2({a^2} \plus{} {b^2} \plus{} {c^2})}}{3} \ge \frac {1}{2} \plus{} \frac {{ab(a \plus{} b) \plus{} bc(b \plus{} c) \plus{} ca(c \plus{} a)}}{4}$
 $ \Leftrightarrow \frac {{2({a^2} \plus{} {b^2} \plus{} {c^2})}}{3} \plus{} \frac {{3abc}}{4} \ge \frac {1}{2} \plus{} \frac {{(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}}{4} \equal{} \frac {1}{2} \plus{} \frac {{3(a \plus{} b \plus{} c)}}{4}$
By Cchur ineq, we have : $ \frac {1}{4}({a^2} \plus{} {b^2} \plus{} {c^2} \plus{} 3abc) \ge \frac {1}{4}({a^2} \plus{} {b^2} \plus{} {c^2} \plus{} \frac {{9abc}}{{a \plus{} b \plus{} c}}) \ge \frac {3}{2}$
The proof'll complete if we can show that
$ \frac {{5({a^2} \plus{} {b^2} \plus{} {c^2})}}{{12}} \plus{} 1 \ge \frac {{3(a \plus{} b \plus{} c)}}{4}$
Easy to see that $ LHS \ge \frac {{5{{(a \plus{} b \plus{} c)}^2}}}{{36}}$
And from the ineq $ \frac {{5{{(a \plus{} b \plus{} c)}^2}}}{{36}} \plus{} 1 \ge \frac {{3(a \plus{} b \plus{} c)}}{4} \Leftrightarrow (a \plus{} b \plus{} c \minus{} 3)(a \plus{} b \plus{} c \minus{} \frac {{12}}{5}) \ge 0$  ( which is true )
Show that the ineq is true. So we are done !


Find all positive integer solutions to $a^{2}=b^{3}+1$.

AFAIK it's definitely NOT easier then the previous one. The previous one is an excercise for these people who are familiar with $Z[i]$ but for this one probably all solutions are based on the elliptic curves (if I remember correctly one can also make it elementary but the idea will be the same and it won't be easy).

you're right. my solution wasn't correct. :blush:

[quote="RDeepMath91"]
i) $q = 3k-1 \Rightarrow p^{2}(p^{2}-1) = k(3k-2) \Rightarrow (2p^{2}-1)^{2}= s(3s+2)$ with $k \in \mathbb{N}, s = 2k-1$
Since $(s,3s+1) = 1$, [/quote]

you wrote $3s+1$ instead of $3s+2$ in the last line above. so your solution has some gaps to be filled in.

Yup, thx for correcting..
But the answer is $1$ since $s = 2k-1$ and $(s,2) = 1$, so $(s,3s+2) = 1$  :roll:

[quote="RDeepMath91"]I think we use the similar way Andrei Negut used in proving the previous problem... One of the solution is $(3,2)$[/quote]
Sorry, I can't understand this step,Could you explain more ,RDeepMath91?

I mean if you read the link given by someone (I forgot) in the previous topic : 'diophantine equation'. There are 2 solutions, but Andrei Negut gave a solution with a lemma concerning Pell's equation.. So, I think we can use a similar way he used to find the Pell's form since in my solution, it hasn't been in the correct form of Pell's equation.  :)


two triangles are similar one with hypotenuse x and leg y+1, the other one has a hypotenuse x+1 and leg y. is this possible? if not, then why not?

My solution, not sure if its correct:

[hide]

Not possible.

x / y+1 =x+1 / y

(x+1)(y+1) = (xy)

xy + y +x + 1 = xy

x+y = -1

If their sum is -1 then one of them has to be negative, and negative lengths don't exist in euclidian geometry.[/hide][/hide]

That looks ok so far, but I think you've only looked at one of two possible cases.

Good point, but I think the problem is a little ambiguous.

However, it seems like  :sqrt: (y+1) :^2: - x :^2: and  :sqrt: y :^2:  - (x+1) :^2:  are really hard expressions to work w/.  Maybe it works for a few cases?[/hide][/quote]

Well you have to consider the case where y and y+1 are not corresponding sides.  Then, work out the missing two legs by similar triangle proportions, and set up ratios using the side lengths u found, checking to see if it all makes sense.  I did find that this case works.


$\text{Let}\; a,b,c,d\; \text{ be positive real numbers, prove that}$
\[ \frac1{a^2+b^2+ab+cd}+\frac1{b^2+c^2+bc+da}+\frac1{c^2+d^2+cd+ab}+\frac1{d^2+a^2+da+bc}\leq\frac1{\sqrt{abcd}}.  \]

please edit it.It is wrong when $a=b=c=d=1$ :)


How are the stats calculated?

For ex: (Karma) Dojo 2060 under daily top 10. My total well... 2295.

Whats going on?

The stats represents karma earned, not total karma. You earn karma by rating problems, videos, and submitting comments to videos. That being said, looking at the karma right now (11 pm pacific, 2 am eastern), it shows you with the most karma earned today. "Today" has only been 2 hours according to the server and it shows you with over 2000 karma earned? That seems like an awful lot and I'm wondering if we got a bug in the system.

Probably because now BOGTRO has 4,330 karma and it's only 7:30

Yeah, and, on the high scores, he has 4330 total...

On some days, Jongao has like 11K karma.   His percentage would be MUCH lower if he really answered that many problems and rated, and only got however many points he has.

[quote="levans"] That being said, looking at the karma right now (11 pm pacific, 2 am eastern), it shows you with the most karma earned today. "Today" has only been 2 hours according to the server and it shows you with over 2000 karma earned? That seems like an awful lot and I'm wondering if we got a bug in the system.[/quote]

Yeah when the stats refreshed at the new day, it started me off with 2000 some karma.

If I may, I would suggest just make the rankings for daily the amount earned that day, monthly be amount earned per month and overall be actual karma.

Yes, and Dojo  now has 4755 karma for today, 6 minutes into the day :P

And he has 2 points.

Yeah I've realized that my Monthly and Daily Karma earned are the always the same and I'm always on the leaderboard as like 4495 Karma earned before I do any quesiton (or rate anything)


Hi ,
I have a problem about system of equation:

Below are partially reduced augmented matrices for systems of equations. In each case, give the number of solutions to the system and when consistent express the soution as linear combination of n*1 vectors:

[b][1 1 2 1 | 6    ]   This is a matrice . I just dont know how to present it
[0 1 6 2 | 17]
[0 0 2 4 | 30][/b]

 "[b]|[/b]" in the matrix, marks the right side of the system.

I need help to find the number of solution.
Can we say that because the last line is not  all  0 0 0 0 0, the system is inconsistent ( no solutions)??

Bertrand

Absolutely not; you can only say that if all of the entries on the coefficient side of such a row are zero.
In this case, there are infinitely many solutions (a translated 1-dimensional solution space).


Five coins look the same, but one is a counterfeit coin with a different weight than each of the four genuine coins. Using a balance scale, what is the least number of weighings needed to ensure that, in every case, the counterfeit coin is found and is shown to be heavier or lighter?

[quote="vallon22"]take four of the coins, and put two on each side
if they balance, then the one you didn't weigh is the counterfeit

however, if one side is heavier, take the lighter side and divide it into two parts again
the lighter side is the counterfeit

therefore, the least number of weighings is 2[/quote]

However, we do not know if it is heavier or lighter. And, the second weighing might have both sides equal, so you would have to weigh again. By the third weighing, we would know which pile of 2 coins have different weights, so you randomly take 1, and weigh with one of the other ones.

I can't find a solution under 4...

????????

like, the first weighing, just randomly choose 4 coins, and put two on each side

if one side is heavier than the other, take the lighter side and split it
the lighter side after that weighing will have the lighter coin?

how does that not work?

We do not know if the counterfeit coin is heavier or lighter than real coins.

OH
i see

then we take the lighter side and measure
if it's equal, we take the heavier side and measure

then we would have to compare it to another coin and see....
hmmm...

i think the answer's 4... but i'm not sure

I believe that I asked this question a few days ago and recall that the correct answer was 3.

Thanks...yea...I thought 2 as well, but I forgot that I did not know whether the mystery coin was heavier or light than the others

[quote="5849206328x"]I think I got 3...

Let A,B,C,D,E be the coins

first weighing:
A B | C D
If the two are equal, then E is the counterfeit.  If the two are not equal, assume that A B is higher than C D

second weighing:
A C | E D
If A C is higher than E D, then either A is lighter than a regular coin or D is heavier than a regular coin.  This case would then take 3 weighings (third weighing could be A | E).  If A matches E, then D is heavier, and if A is lighter than E, then we're done.
If A C is the same as E D, then B is obviously the counterfeit.  The first weighing tells you that it's lighter.
If A C is lower than E D, then C must be the counterfeit coin (think it through a bit).  It also follows that C is heavier.

sorry if that was unclear...

did anyone find a way to get it in two?[/quote]

No. That is correct. I have done this problem before, and there is a formula for any amount of coins, including 999999999999999999999999999999999999. :)

You have piqued our curiosity. What is the formula?

In general, $ n$ weighings are sufficient to find a bad coin among $ \frac{3^n\minus{}1}2$ coins (this reminds us of dividing into $ 3$ groups, which always is the minimum for numbers that can be obtained from this formula). Note that this formula only finds the MAXIMUM number of coins that these weighings can find, and many values can't be obtained.

Thus, if we let $ \frac{3^n\minus{}1}2$, then we need to solve for $ n$ in terms of $ x$.

$ 2x\plus{}1\equal{}3^n\implies n\equal{}\log_3{2x\plus{}1}$.

But this is not always an integer, so the formula is $ \boxed{n\equal{}\lceil\log_3{2x\plus{}1}\rceil}$.

In this particular problem, we have $ n\equal{}\lceil\log_3{11}\rceil\equal{}\boxed3$.


I need help on this problem. My book doesn't show the solution except its answer. :(

Sorry but it is just impossible to view the image.

From Next Time Please Make a Neatly Labelled Figure.

I guess I May be right . However ,What Answer Does The Book Show???


geometry
number theory
national olympiad

[i]1st Balkan Student Mathematical Competition was organized by Young talented mathematicians "Marin Getaldic", a group of high-school students and university students (which act as mentors to high-school students and try to improve their knowledge of mathematics and their result in various mathematical competitions) based in Zagreb. It was held in November 2008, with help from various people from 5 countries (Bosnia and Herzegovina, Croatia, Macedonia, Montenegro and Serbia), who acted as local organizers. 103 students from 11 schools participated altogether.[/i]

[b]Problems[/b]
Grade 9
1. (Algebra) http://www.mathlinks.ro/viewtopic.php?p=1339430#1339430
2. (Combinatorics) http://www.mathlinks.ro/viewtopic.php?p=1339433#1339433
3. (Geometry) http://www.mathlinks.ro/viewtopic.php?p=1339436#1339436
4. (Number Theory) http://www.mathlinks.ro/viewtopic.php?p=1339440#1339440

Grade 10
1. (Algebra) http://www.mathlinks.ro/viewtopic.php?p=1339444#1339444
2. (Combinatorics) http://www.mathlinks.ro/viewtopic.php?p=1339456#1339456
3. (Geometry) http://www.mathlinks.ro/viewtopic.php?p=1339459#1339459
4. (Number Theory) http://www.mathlinks.ro/viewtopic.php?p=1339446#1339446

Grades 11 and 12
1. (Algebra) http://www.mathlinks.ro/viewtopic.php?p=1339448#1339448
2. (Combinatorics) http://www.mathlinks.ro/viewtopic.php?p=1339451#1339451
3. (Geometry) http://www.mathlinks.ro/viewtopic.php?p=1339462#1339462
4. (Number Theory) http://www.mathlinks.ro/viewtopic.php?p=1339453#1339453

All the problems, official solutions, results and statistics can be seen (English version is included, of course) in files contained in the following zip file: http://www.2shared.com/file/4401926/280ec27f/BSMC.html (click on "Save file to your PC: click here" near the end of the page)

[i]Writing time was 240 minutes and each problem was worth 10 points. We are especially proud of the fact that 9 of 12 problems were original.
If there are any questions or suggestions about the competition or if there are any problems with the zip file above, please, PM me and we'll see how I can help you.[/i]

Moderators, how about putting this competition into "Resources" page? (there are 9 original problems out of 12 :) )

I asked it in FY forum, and the answer was I think: your school must participate and then everybody who is interested can participate in their own school.

Yes, Dimitris, Bugi is correct (mostly  :lol: ), as I've answered in the (Former) Yugoslavia community.

The original idea was to make a fun, interesting and simple competition. As the competition was completely unofficial (i.e. lacking any funding whatsoever  :D ) and the organizers (both locally and globally) were usually high-school teachers and university students (like myself), it was obvious that it could not be held at a single place. Therefore, we have decided to ask all the people we knew who hold preparations for competitions in their schools if they'd like to organize the competition at their school.

There were no formal prerequisites for competing (although it was obvious the problems were tough and ment for more advanced competitors), so anybody who wanted to compete, could.

In the end, 9 schools held the competition. However, 2 more students (from two other schools) came, on their own initiative, to nearby schools which were holding the competition and competed there. Thus, students from 11 schools participated in the contest.

I hope this explains it. I know, it's kind of confusing, but, in the beginning (and partly until the end), BSMC was not ment to be a really serious, formal competition - it was more of a way to motivate students and allow them to see what areas of mathematics they're good at, and what areas they need to work on more.

If you've got more questions, please, do that (you can also PM me). I'm here to answer (and I also like to talk about this competition way too much :D ).

Sorry, it seems the link in the opening post has timed out. This is the new link: http://www.2shared.com/file/5357102/45bfdefc/BSMC.html .


[quote="Gyan"]This is an old problem we actually put a skit about it when I was in elementary school (that was many decades ago)

It was lunch time and there were only three young men in the car of the train.  When the luch boxes were  opened, Ashok had $3$  rotis (packed by his mom), Bob had $5$ roties, but no one packed a lunch  for Chou. But in any case, all three divided their  roties   equally among  all three and had  a  nice luch.

Next stop, Chou  got down from the train,  thanked  A & B for  sharing their lunch with him  and handed  Rs $8$ to them.

Bob said something like " Well $8$  Ruppes for $8$  rotis  so I will keep $5$ and you keep $3$"

Ashok on the ohter hand wanted to  equally divide the money  and wanted $4$ Rupees.
(In the skit we put, here brief intermission  at this stage after  a big argument  erupted  among the two young men - and audience was asked by the narrator  "how the money should be divided - assuming it is to be done in a mathematically correct way")

So how the money should be divided?

(Funny - After so many decades I saw the same problem  -without a skit and the nice story above - in an old mathcounts test)[/quote]

[hide="answer"]
Well, if they shared it equally, all of them got 8/3 rotis. So Ashok gave 1/3 of a roti and Bob gave 7/3 of a roti. So 1 rupee goes to Ashok and 7 rupees goes to Bob.
[/hide]

Its a kind of bread, very popular in India,  (and other places):
[url=http://www.saveurs.sympatico.ca/ency_6/triperie/foie-roti.htm]Check it out , for example...[/url]

Or [url=http://www.lasan.co.uk/photos/press/Roti%20sag.jpg]For a close up view ...[/url]

With that filling it looks really yummy :) .

Except... how do you split that into thirds :P ?


Points $ A$, $ B$, $ C$, $ D$, $ E$, $ F$, and $ G$ are all consecutive vertices on an inscribed 7-gon. $ \angle A \equal{} 120^\circ$, $ \angle B \equal{} 160^\circ$, and $ \angle E \equal{} 154^\circ$. Find the measure of minor arc $ FG$.

Is a measure something other than distance? because I think we need more info

[geogebra]b2d1e7ebeac1b9d7350d92ae3f83a655b7a6820e[/geogebra]

Those sides of the septagon can just be slid round the circle, therefore the minor arc $ FG$ can take a range of values. Also, depending on how big you scale the circle can the minor arc take values $ [0,\infty)$?

I think i must be missing something  :?:

The problem was asking for [i]angle[/i] measure. (I assume this because they usually ask for the length when referring to the length of the arc and measure when referring to the angle measure.) You also have to consider that the interior angles of septagon always sum up to one value. Maybe you should also take a look at the exterior angles too.

By the way, the problem is right, it's exactly as it appeared on the paper. (But of course, there might be chance that there was a mistake in writing the problem itself.)

Well, it doesn't make a difference. We can slide $ \widehat{DEF}$, giving different answers.


I think this has something to with the Theory of Symmetric Polynomials.  :huh:

Yea basically, just brute force this problem.

[quote="neelnal"][quote="Imperial Effect"]In addition, the use of the Theory of Symmetric Polynomials and an understanding of elementary symmetric polynomials are also used when solving this problem.[/quote]

Symettric polynomials? Um....

Please provide a solution to back up your claim.[/quote]
An example has already been provided. Look at the way besttate manipulates the equations.
“In mathematics, a symmetric polynomial is a polynomial P(X1,X2,...,Xn) in n variables, such that if some of the variables are interchanged, the polynomial stays the same.”

Here are some links to help understand symmetric polynomials:
http://en.wikipedia.org/wiki/Symmetric_polynomials
http://planetmath.org/encyclopedia/ElementarySymmetricPolynomial.html
http://everything2.com/index.pl?node=symmetric%20polynomial

I’m sure there are a lot more just search on [url]http://google.com[/url] and you will find more.

P.S Do not Google "[b]Symettric[/b] polynomials" instead try "[b]symmetric [/b]polynomials"

Ok, I just thought you had a less tedious way :)

Yea. The way you wrote, I thought there was a less brute-force solution waiting to be revealed.

It would be pretty cool if this could be done in an easier way.

I was gonna do that, but then i thought it'd take too long to figure out the coefficients of the polynomial.

EDIT: Come to think of it, you can work out the coefficients pretty easily. Then work out S4 from that.

[quote="Franz Joseph"][quote="t0rajir0u"]http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums[/quote]

That's an awesome way of doing it! Thanks [b]t0rajir0u[/b].[/quote]

Definetely. Thanks t0rajir0u.


search
combinatorics proposed
combinatorics

Van der Waerden number $ V(2,4) \equal{} 35$...

Pierre.

[quote="Friendly1"][quote="Friendly1"]In a circle,let $ 35$ points$ A_1,A_2....A_n$ satisfying $ A_1A_2 \equal{} A_2A_3 \equal{} ...A_{34}A_{35} \equal{} A_{35}A_1$ (it mean here is a regual $ 35$-gon,and thay were colored by $ 2$ colors:red or green.Prove that there are $ 5$ point (in them )were colored by a similar color,and they make two isosceles triangles(such that two one have a common point)
It is hard.who can help me[/quote]
Ha ha ha ha  :rotfl:  :rotfl:
I think it is too hard for all Mathlinks member ,I am sure no one can solve it :lol:  :lol:[/quote]
don't spam all names!
@pbornsztein:Can you post your full solution?

Label the vertices $ 1,2,\cdots,35$ in clockwise sense.
Any two colouring of the vertices induces a two colouring of the integers from $ 1$ to $ 35$. But, since Van der Vaerden's number $ V(2,4)\equal{}35$ it means that for any two colouring of the integers $ 1,2,\cdots,35$ there exists a monochromatic sequence of length $ 4$, say $ a,b,c,d$ in that order.
Therefore the vertices corresponding to these labels give two isoscele triangles, namely $ abc$ and $ bcd$, as desired.

Pierre.

[quote="Math pro"]

don't spam all names!
@pbornsztein:Can you post your full solution?[/quote]
all names,who is he?,and who are you,mathpro  :rotfl:  :rotfl:
Any way I am quite  :mad: ,with your reply,a post without any information and I think it is exactly spam post(Although I dont know all names  I guess he is not a spamer as you)
And I think you should think more before post any post because our site is Mathlinks where is not the "land" for a spamer(as you again)
 @Add :rotfl:  :rotfl: :I have just checked your profile and found "I'm not a spamer!" in your signature
Morever,I dont see no post of you http://www.mathlinks.ro/search.php?search_author=Math+pro
arent spams post,
But I am very thank to pbornsztein,the first who care my hard problem,thank you,
But can you show me what is Van der Vaerden's number ,and who have a conceivable  proof?,I really want to know all your help


conics
ellipse
geometry
geometric transformation
homothety
circumcircle
inradius

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14596[/img]

 :(

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14607[/img]

 :(

the point P* that you draw in you figure is the perspector, not the center, for obtain the center call Q the isotomic conjugate of P* and take an homothety with radio $ \minus{}\frac{1}{2}$ with center the centroid and Q go in P that is the center of the ellipse.

[quote="¬[ƒ(Gabriel)³²¹º]¼"]call $ P: AT_1 \cap BT_2$ and $ T_3: CP \cap AB$. Then the conic must pass through $ T_3$ (Carnot + Ceva theorem). Also it's know that the center of an inscribed conic with perspector P it's the complement of the isotomic conjugate of P and so we get other 3 point.[/quote]

Hello, what carnot's theorem are you referencing? I am only familiar with the carnot's theorem that relates the lengths of the projections of the circumcenter to the inradius and circumradius. Surely there is something else that you are refering to. Could you please explain?

[quote="Altheman"]Hello, what carnot's theorem are you referencing? I am only familiar with the carnot's theorem that relates the lengths of the projections of the circumcenter to the inradius and circumradius. Surely there is something else that you are refering to. Could you please explain?[/quote]
[b]Carnot's Theorem[/b]

[i]if a conic intersect the sides BC at X, X', CA at Y, Y' and AB at Z, Z' then[/i]

$ \frac {BX}{XC}\cdot\frac {BX'}{X'C}\cdot\frac {CY}{YA}\cdot\frac {CY'}{Y'A}\cdot\frac {AZ}{ZB}\cdot\frac {AZ'}{Z'B} \equal{} 1$

[i]and if the relation is true then the 6 points are on a conic.[/i]

[b]"for obtain the center call Q the isotomic conjugate of P* and take an homothety with radio - 1/2  with center the centroid and Q go in P that is the center of the ellipse."[/b]

you mean that center of ellipse is midpoint of GP* ? but it's not true, so [u][b]where's the center of the ellipse[/b][/u]?

 :(

...see at this figure:

[img]http://img100.imageshack.us/img100/8445/geometria36eg6.png[/img]


Noone posted a solution for this in the intermediate section so I'm reposting it here:

Show that if the root of the algebraic equation [tex]f(x)=0[/tex] is of the form [tex]a+bi[/tex] and [tex]a,b\in\mathbb{R}[/tex]
 [tex]f(x)\in\mathbb{R}[x][/tex], then the number [tex]a-bi[/tex] is another root of the equation.

???

We can write $0=\overline{f(a+bi)}=f(a-bi)$. That's all.

Let be $z$ any root of the equation $\sum_{k=0}^n a_k x^k = f(z)=0$. For the conjugate of $z \in \mathbb{C}$ I will write $z'$.
Then $0=f(z)'=(\sum_{k=0}^n a_k z^k)'=\sum_{k=0}^n (a_k z^k)'=\sum_{k=0}^n a_k' (z^k)'=\sum_{k=0}^n a_k z'^k=f(z')$.

Maybe a proof which does not use the properties of the conjugate is what Xixas wants:
Let $P(x)=(x-a-bi)(x-a+bi)=x^2-2ax+a^2+b^2\in \mathbb {R}[X]$.
The long division algorithm gives the existence of $c,d\in \mathbb {R}$ and $Q\in \mathbb {R}[X]$ such that $f(x)=P(x)Q(x)+cx+d$. Plugging $x=a+bi$ yields $c(a+bi)+d=0$, hence $ac+d=bc=0$.
We have $f(a-bi)=c(a-bi)+d=(ac+d)-bci$ hence $f(a-bi)=0$, as well.


Solve:

$ x(16 \minus{} y) \equal{} 78$
$ y(16 \minus{} x) \equal{} 14$

FIND $ (x,y)$

[hide]The first one can be rewritten as $ 16x \minus{} xy \equal{} 78$ whereas the second one as $ 16y \minus{} xy \equal{} 14$
subtracting the second from the first we get that $ 16x \minus{} 16y \equal{} 64 \equal{} > x \minus{} y \equal{} 4 \equal{} > y \equal{} x \minus{} 4$
So substituting y in the first one we get $ x(16 \minus{} x \plus{} 4) \equal{} 78 \equal{} > 20x \minus{} x^2 \equal{} 78 \equal{} > x^2 \minus{} 20x \plus{} 78 \equal{} 0$ which has two solutions: $ x \equal{} 10 \minus{} \sqrt22 and x \equal{} 10 \plus{} \sqrt22$
for the first one we get that $ y \equal{} 6 \minus{} \sqrt22$ and for the second $ y \equal{} 6 \plus{} \sqrt22$
so $ (x,y) \equal{} (10 \minus{} \sqrt22,6 \minus{} \sqrt22) or (x,y) \equal{} (10 \plus{} \sqrt22,6 \plus{} \sqrt22)$[/hide]


Suppose a runner moving at $ 0.75c$ carries a horizontal pole 15m long toward a barn that is 10m long. The barn has front and rear doors. An observer on the ground can instantly and simultaneously open and close the two doors by remote control. When the runner and pole are inside the barn, the ground observer closes and then opens both doors so that the runner and the pole are momentarily captured inside the barn and then proceed to exit the barn from the back door. Do both the runner and the ground observer agree that the runner makes it safely through the barn?

This is a very famous problem and the answer is that in the pole frame, the observer seems to open the doors in the wrong order.

A variation: The back door cannot be opened. The observer ('farmer') closes the front door when the runner ('his son') is inside. What will each observe when the runner comes to rest? There are two ways to go about this.
(Of course, technically, this cannot be done in special relativity but give a qualitative answer.)

The solution resolves it by saying that by relativity of simultaneity, the doors do not open and close and open at the same time from the pole carrier's point of view. The pole carrier gets through safely but he and the observer see different things happening =P

That variation is tricky.. will the length contraction stop and the pole expand or something? Lol..

You are on the right track.
The farmer will indeed see the length contraction stop and the rod expands till it is held  compressed (under tension) by the door.
What will the son see?

He will see the door get farther away? Because in his frame, the length contraction of the barn stops. Counterintuitive!

Yes. He will also see the rod in his hand contract due to it's own inertia, so that it finally fits in the barn.
Note the use of [u]psuedo force[/u] in a non onertial frame (that of the son).

What's the pseudo force here? By the way, in the twin paradox, they argue that the travelling twin is in a non-inertial frame and there's the pseudo forces acting on him (he speeds up and slows down while turning his spacecraft around). But could it not also be that the twin on earth is accelerating while he is not? It's like the inertial and non-inertial frames could be swapped. Correct me if i'm wrong  :P

When the son reaches the end, he accelerates backwards; hence in his frame, there is a psuedo force acting forwards (on the pole for instance).

The travelling twin experiences a [i]real[/i] force, that of the rocket engines. This makes his frame non inertial.


Consider two circles in a plane $ S_1$ and $ S_2$ that intersect at the points $ A$ and $ B$. Let $ Q$ be a point on $ S_1$ and outside $ S_2$, and let $ D$- the intersection of $ QA$ with $ S_2$ and let $ C$- the intersection of $ QB$ with $ S_2$. Suppose that the tangents to $ S_1$ at $ A$ and $ B$ intersect at $ P$. Prove that $ QP$ halves $ CD$.

Denote L is the midpoint of AB
KAQB is a harmonic quadrilateral   => ∠LQB=∠KQA =>⊿LQB∽⊿MQD
M is the midpoint of CD

Dear Mathlinkers,
1. according to Reim's theorem, the tangent Tq at Q is parallel to CD
2. according to Boutin's theorem, P is the midpoint of the segment C'D' determined by the parallel to Tq passing through P with endpoints on QB and QA
3. Consider the trapez C'D'DC....
for more see
http://perso.orange.fr/jl.ayme   vol. 1 A propos du théorème de Boutin
Sincerely
Jean-Louis

It’s a well known fact that PQ is simmedian in triangle ABQ, hence median in $ \triangle CDQ$.

Best regards,
sunken rock

I solved it the same way sunken rock did, but I would like to post the details on why that fact is true.
Draw the parallel through P to CD, let it meet DQ at R and CQ at S. Let K be any point on PB, extended past B. Then $ \angle PBS\equal{} \angle QBK\equal{} \angle QAB$. Also, since ABCD is cyclic, $ \angle PSB\equal{}\angle DCB\equal{}\angle QAB$. So $ \triangle PBS$ is isoceles. Similarly $ \triangle PAR$ is isoceles. So $ PS\equal{}PB\equal{}PA\equal{}PR$. So $ QP$ bisects $ RS$, but $ CD \parallel RS$, so $ QP$ bisects $ CD$.


vector
real analysis
real analysis theorems

In a normed vector space $ (V,\left\|\cdot\right\|)$, show that $ B_{r}(x)\equal{}x\plus{}B_{r}(0)\equal{}\{x\plus{}y\mid \left\|y\right\|<r\}$ and that $ B_{r}(0)\equal{}rB_{1}(0)\equal{}\{rx\mid \left\|x\right\|<r\}$, where $ B_{r}(x)\equal{}\{y\mid d(x,y)<r\}$ is the open ball centered at $ x$ with radius $ r$.

Of course it's intuitively obvious, but I'm having some trouble showing this rigorously.

First part. You want to prove
$ \{z\colon |z\minus{}x|<r\} \equal{} \{x\plus{}y\colon |y|<r\}$
Inclusion from left to right: let $ y\equal{}z\minus{}x$.
Inclusion from right to left: $ |(x\plus{}y)\minus{}x|\equal{}|y|<r$.


superior algebra
superior algebra unsolved

Prove:if G is a group for which G/Z(G) is cyclic, where Z(G) denotes the center of G, then G is abelian.

Let $ H \equal{} Z(G)$. Now as $ G/H$ is cyclic, there is a coset $ aH$ which generates the group $ G/H$. Let $ x,y\in G$. Then $ xH,yH\in G/H$ thus $ xH \equal{} a^nH$ and $ yH \equal{} a^mH$. This means that $ x \equal{} a^n z_1$ and $ y \equal{} a^mz_2$ where $ z_1,z_2\in H$. But then $ xy \equal{} a^n z_1 a^mz_2 \equal{} a^{n \plus{} m}z_1z_2$ and $ yx \equal{} a^m z_2 a^n z_1 \equal{} a^{n \plus{} m}z_1z_2$ because $ z_1,z_2$ commute with everything. So $ G$ is abelian.


\[ \text{Let}\; x,y,z\; \text{be positive real numbers such that}\; x+y+z=1,\;\text{prove that}  \]
\[ \frac1{\sqrt{ x+yz}}+\frac1{\sqrt{ y+zx}}+\frac1{\sqrt{ z+xy}}\geq\frac92.  \]
This one is not too hard, but it has a very nice solution.

Thuan

\[ \frac1{\sqrt{ x+yz}}+\frac1{\sqrt{ y+zx}}+\frac1{\sqrt{ z+xy}}\geq\frac{9}{\sqrt{3\sum x+3\sum yz}}\ge\frac{9}{2}  \]

Since $x+y+z=1$, we have $x+yz=x(x+y+z)+yz=y(x+z)+x(x+z)=(x+z)(y+x)$. Applying AM-GM inequality to get $\sqrt{(x+z)(y+x)}\leq\tfrac12(2x+y+z)$. Two other similar inequalities and AM-HM inequality solve the desired inequality.


function
inequalities
real analysis
real analysis unsolved

Does  there   exist     such   a   function  $ f(x)$   on  $ \mathbb{R}$,   $ f(x)$  is  not  continuous  at  least  one  point,   and   every   point  $ x \in \mathbb{R}$  is  an   Extreme Value  point   of   $ f(x)$    ,  that  is ,  for every  $ x$,  there  is  a open  interval  $ (a, b)$,  $ x \in (a, b)$,  $ f(x) \geq f(y)$  or  $ f(x) \leq f(y), \forall y \in (a, b)$?

The characteristic function of the rationals. Ho hum.

Thank you  very much!  how  about  the  following:

Does  there   exist     such   a   function  $ f(x)$   on  $ \mathbb{R}$,   $ f(x)$  is  not  continuous  at  least  one  point,   and   every   point  $ x \in \mathbb{R}$  there  is  a open  interval  $ (a, b)$,  $ x \in (a, b)$,  $ f(x) > f(y)$  or  $ f(x) < f(y), \forall y \in (a, b), y \neq x$?

First, a side remark. When I see ???????? at the end of a question (as well as other completely unnecessary excessive punctuation, capitalization, usage of large fat fonts in bright colors, etc.), I am strongly tempted to dismiss the entire post. As far as I know, I'm not the only one here with this attitude. ;)

Now, closer to the subject. When $ y\equal{}x$, your strict inequalities obviously cannot hold, so you must have meant something else.


find m.n is integer such that with p be a prime number and 2^p+3^p=m^n :D  :?

I think this one has already been discussed. The idea is that $ 2^p+3^p$ is divisible by 5, but not by 25 for odd prime p.


Hi,

How to extract data from a .tex file to SQL Server and then display it on ASP page with all the printable/mathematical symbols, what is available on that .tex file ?????? Extraction is not the big deal but how to display the symbols on ASP pages ???

Thanks,
Alok.

A couple of possibilities come to mind:
1. Convert the TeX file to HTML first using tools like [url=http://hutchinson.belmont.ma.us/tth/]TtH[/url] or [url=http://www.cis.ohio-state.edu/~gurari/TeX4ht/mn.html]Tex4ht[/url]. If this is to be done dynamically you will have to use TtH; Tex4ht is slower but uses images to give better results.

2. Assuming you are happy with delving into PHP and ASP code, you could convert [url=http://www.mayer.dial.pipex.com/tex.htm#latexrender]LatexRender[/url], which powers the on-the-fly [tex]\LaTeX[/tex] conversion on this board, from PHP to ASP. [url=http://www.php-asp-convertor.com/index.php]The PHP to ASP converter[/url] may help. You will need [tex]\LaTeX[/tex] to be installed on the server.

I would be very interested if you do manage to convert LatexRender to ASP  :)


Find $ f : R \to R$ satisfying
$ xf(y)\minus{}yf(x)\equal{}f(\frac{y}{x})$ with all $ x;y \in R;x \neq 0$

Is this also cheating?
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=259079


im just wondering if anyone has some national tests from mathcounts that are free to download. if so, can some please pm them to me.

note  that they need to be ones that can be downloaded on to acrobatic reader or word.

p.s. sorry about the first post it was a mistake.

You know those weeks when Mathcounts posts sample nationals problems on the Problem of the Week? I compiled all of those and saved them in a file. Anyone?

[quote="mr. math"]but do you have any regular ones? :ninja:   :roll:[/quote]

What are "regular ones?"

mathnerd314, can you pm me those problems :ninja:
 :rotfl:

They've been posted in my blog. Please, no more PM-ing--I'm too lazy to copy and paste a hundred million times.

Why the  :ninja: ?


How do i change my username. Suppose i tell an admin to change it for me, will the online classes, currently MATHCOUNTS, be terminated? Currently, i don't want to change it though.

I thought you could, that's what happened to another user in the support forum.

*I* (or any other admin) can change a username, but *you* cannot.

Since it's a pain for us to change them, we strongly discourage it (i.e. you had better have a good reason for changing your username).


[quote="meinstein"]Να λυθεί στους ακεραίους :

$ x^3 \minus{} y^3 \equal{} xy \plus{} 61$[/quote]

[hide]Pollaplasiazeis epi 27 kai meta euler.[/hide]

hahahhahaha :lol:

pou to thimithikes????ego entaksei molis tin eida, katalava amesos oti einai gnosti mou alla den thymomoun to post klp klp...pantos exei plaka ee kai thn prohgoumenh fora me ton idio tropo eixa apanthsei...

[quote="Nick Rapanos"]pou to thimithikes????[/quote]

Ξέρεις ότι έχω αρκετά καλή μνήμη . :D
Όντως είχες απαντήσει το ίδιο  :P


geometry
perimeter
rectangle
angle bisector
geometry unsolved

Let $ ABCDE$ be a pentagon whose perimeter is $ 4$, $ AB\equal{}DE\equal{}1$ and $ \widehat{EAB}\equal{}\widehat{BCD}\equal{}\widehat{DEA}\equal{}90^o$. Prove that the angle bisector of $ \widehat{BCD}$ bisects the area of the pentagon.

Extend CB & CD to intersect AE @ P & Q resp. Let CR be angle bisector of C, R on AE.
Let CD=[i]x[/i], CB=[i]y[/i] & AE=[i]z[/i].
Then it is sufficient to prove that ar(CRP)-ar(CRQ)=ar(ABP)-ar(DEQ).
This reduces to proving [i]x[/i] + [i]y[/i] = [i]xy[/i] + [i]z[/i], which can be proved using [i]x+y+z[/i]=2 & [i]x^2+y^2=z^2[/i].

[quote="Ajinkya Dahale"]Then it is sufficient to prove that ar(CRP)-ar(CRQ)=ar(ABP)-ar(DEQ).[/quote]

Well, to say this you need to prove that the bisector of $ C$ intersects the segment $ AE$, right?

[quote="Ajinkya Dahale"]This reduces to proving x + y = xy + z[/quote]

Can you give more detais? In my picture, it is not true that $ x^2 \plus{} y^2 \equal{} z^2$... But my picture is not very good.

Feliz, first, your pictures are always not good  :P

I will explain x^2+y^2=z^2, it's easy.

Note that BA//DE and BA=DE=1, by <BAE=90º, we get that BAED is a rectangle.Then, BD=AE and in the rectangle triangle CBD, follows the relation.

Bye :D


geometry
topology
real analysis
real analysis solved

[quote="grobber"]Assume that you have three mutually tangent circles (externally tangent) in $A,B,C$. Then, at each step, you take three of the circles which are mutually externally tangent, and inscribe a (closed) disk in the region formed between them. If you repeat this for each region formed, can you eventually cover the entire $ABC$ region? By this I mean: will every point of $ABC$ be covered after a finite number of steps (different numbers of steps for different points, of course)?

I hope it's clear enough.[/quote]

Yes, it is clear enough. The answer is: No in a finite number of steps, but the area that is not covered after $n$ steps tends to zero as $n$ tends to infinity.

Best,

[quote="grobber"]Hmm... Maybe I didn't make myself understood, or maybe I'm misunderstanding you :).

We say a point is covered if it's covered after a finite number of steps, but the finite number of steps may vary from point to point. According to this definition, are all the points of the region $ABC$ covered? There may be points (a set of measure $0$) which always stay uncovered.[/quote]

Well, that changes the definition a little. For sure you cannot cover that region completely in a finite number of step. Actually, after $n$ steps, the non-covered area is always positive. How to prove it, well, a little scketch. Prove that if you have more than three circles that are mutually tangent (that is, every circle in the collection is tangent to every other circle in there) then, the region in between the circles has an interior that is disconnected. After that, prove that you cannot do it in a single step and use induction.

Now, it is a little trickly but not too hard to prove that you can cover every single point in the interior of the region you defined originally can be covered in a finite number of steps, if you choose wisely those steps (and the boundary has mesure 0). Finally, if you proved this, prove that the process, done in the right order, gives you a decreasing area not covered, that goes to zero.

A "wise" order could be just, you start with one region uncovered, and after the first step you have three, choose the greatest of those and use it in the second step (in case of a tie, choose any), in the third step you have five regions, choose the greatest and do the same, and so forth. Drawback: every point is covered by the closed set of the circles you construct, that means, there are an uncontable (but mesure zero) set of points that are not covered by the open circles!

Something interesting could be: are such circles contructible using only compass and ruler? Probably yes.

Best,

Actually, I believe the answer to be "no" :). I say "I believe" because I'm never too sure of my proofs. There are points which are never covered, no matter how many steps you take. That set might have measure zero, but that has nothing to do with it. The order in which you put disks down has nothing to do with the problem either, I think :?.

[quote="grobber"]The order in which you put disks down has nothing to do with the problem either, I think :?.[/quote]

Yes it does. Suppose you always pick the subregion with lower area!!!, think about it!

Best regards,

[quote="grobber"]Ok, so you have $ABC$ at first. After you inbscribe a disk in it, you have three regions. If you choose to inscribe a disk in the one with the lowest area, and then disregard the other two, then you'll obviously have some uncovered points in those regions. So what does this show?[/quote]

That the order matters!! In that case the set of points that you don't cover has positive mesure.

Now, you are making my doubt. I am going to think about it, because I still think that the closure of the points that are covered is the full region if you always pick the greatest region on each step.

Best,

The order matters in the sense that you can get an uncovered region of positive measure, but that is not my point. I'm saying that no matter what you do, there always are uncovered points, may the uncovered region have positive or null measure. There's a difference, of course, but when I said it didn't matter I meant to say that it didn't affect the answer to the problem, which I still believe to be "no". This is the important issue here. When I asked what your example showed, I meant to say that it did not disprove the fact that the answer is "no".

The closure of the covered region can be made to be the entire $ABC$, I'm not arguing with that. However, that still doesn't show that there are no uncovered points, does it? :) It's easy to disregard this problem by rushing into it :).

Given three circles $ P,Q,R $, let us inscribe into the region between them
the circles $ P',Q',R' $, as drawn in the picture "circles"
attached.
 Now starting with three original circles $ P_{1},Q_{1},R_{1} $,
applying this procedure, we get a sequence of triples of circles $
P_{n},Q_{n},R_{n} , n=1,2,\ldots $. Denote by $ D_{n} $ the (open)
region between $ P_{n},Q_{n},R_{n} $. So $ D_{1} \supset D_{2}
\supset \ldots $, and I claim that the intersection of them is
nonempty. Indeed, the region $ D_{n} $ always contains the closure
of $ D_{n+1} $ for each $ n $, so intersection of all $ D_{n} ,
n=1,2,\ldots $ is equal to the intersection of their closures,
which is clearly nonempty. On the other hand, it is quite clear
that the point of intersection of them cannot be covered by the
discs after any number of steps.

[quote="Leva1980"]the region $ D_{n} $ always contains the closure
of $ D_{n+1} $ [/quote]

I really am not sure about this step. Are you sure? :? The way I see it, $D_n$ and $D_{n+1}$ share a piece of their boundary, and this piece of common boundary belongs to none of them. This is unless I'm misunderstanding your definition of $D_n$.

ok Leva your proof is fine that their intersection is non-empty(Cantor's intersection theorem).
If we are to do the intersection then we are left with only 1 pt & a singleton set is trivially closed.So their's no problem.Will grobber please tell me explicitly what he's trying to express.

Ok, I got what Leva was trying to say :). I had a more complicated solution. Sorry about that nonsense I wrote :).

O never mind that.sometimes we do certain tough problems & after that we tend to think in the similar complicated fashion.Its difficult to tone down to the easy ones.
Thats why I am doing so well in the college exams(from the back of course) ;)


inequalities
logarithms
inequalities unsolved

Prove that if $a,b,c\geq2$, than $\log_{a+b}c^{2}+\log_{b+c}a^{2}+\log_{a+c}b^{2}\geq3$

Should go in inequalities... Accidentaly posted it here, someone could move it...

a(b/2-1)+b(a/2-1)>=0 ,so ab>=a+b ,then lna+lnb>=ln(a+b) ===>
lnc/(lna+lnb)<=lnc/ln(a+b)
so its easy now by adding this 3 inequality and using Nesbit inequality


modular arithmetic
number theory proposed
number theory

Does there exist a square-free positive integer $ n$ such that

$ n\plus{}1,n\plus{}2,...,n\plus{}2007$ are not square-free, and $ n\plus{}2008$ is not a power of some positive integer?

Call $ p_1,..,p_{2007}$ is 2007 distant prime number .
Consider the system of congruence equation :
$ x\equiv \minus{}i (\mod p_i^2)$
From chinese remainder theorem this system have a root . Call smallest natural root is a .
$ M\equal{}\prod _{i\equal{}1}^{2007}p_i^2$
Then $ n\equal{}a\plus{}mM$ is satisfy for all m .
Easy  prove that exist m for $ a\plus{}mM\plus{}2008$ is not a power of an interger .

[quote="nguyentrang"][quote="TTsphn"]
Then $ n \equal{} a \plus{} mM$ is satisfy for all m .
[/quote]

How do you know $ n$ is square-free?[/quote]
From your problem it is not square free. Because $ p_i^2|x\plus{}i$

I said $ n$ is square-free, but $ n\plus{}1,n\plus{}2,...,n\plus{}2007$ is not  :)

[Edited to fix some mistakes.]

I will rely on the following lemma.

[b]Lemma:[/b] Let $ m$ be a positive integer and $ a$ be an integer such that $ \gcd(a, m)$ is square-free.  Then there is a square-free positive integer that is $ a \pmod{m}$.

The proof of the lemma is by adapting the usual proof of the density of square-free integers.

Now, let $ p_1$, $ p_2$, ..., $ p_{2007}$ be distinct odd primes such that $ p_i^2$ is not a divisor of $ i$ (for all $ i$ from 1 to 2007).  Consider the equations
\begin{align*} n & \equiv - i \pmod{p_i^2} \\
n & \equiv 2\pmod{4}. \end{align*}
Let $ m = 4 p_1^2 p_2^2 \dots p_{2007}^2$.  By the Chinese Remainder Theorem, there is an integer $ a$ such that the solution to the above system of congruences is $ n \equiv a \pmod{m}$.  Then $ n + i$ is a multiple of $ p_i^2$, and so is not square-free.  Also $ n + 2008$ is $ 2 \pmod{4}$, and so is not a power of an integer.

Finally, we can check that $ \gcd(a, m)$ is square-free, because $ p_i^2$ is not a divisor of $ i$.  So the Lemma says we can choose $ n \equiv a \pmod{m}$ such that $ n$ is square-free.

[quote="Ravi B"]
[b]Lemma:[/b] Let $ m$ be a positive integer and $ a$ be an integer such that $ \gcd(a, m)$ is square-free.  Then there is a square-free positive integer that is $ a \pmod{m}$.

The proof of the lemma is by adapting the usual proof of the density of square-free integers.

[/quote]

I think it's a very nice lemma. Please show its proof.

Sorry, I don't really have time now.  One high-powered way is to use Dirichlet's theorem on primes in arithmetic progressions.  That theorem gives the lemma almost immediately.

In fact, we can modify our construction to make $ \gcd(a, m) \equal{} 1$, not merely square-free.  Back to the original problem, that means we can make $ n$ prime, not merely square-free.

To modify the construction, we may have to change $ n \equiv 2 \pmod{4}$ to $ n \plus{} 2008 \equiv 3 \pmod{9}$.

[quote="Ravi B"]
In fact, we can modify our construction to make $ \gcd(a, m) \equal{} 1$, not merely square-free.  Back to the original problem, that means we can make $ n$ prime, not merely square-free.

[/quote]

What do you mean? How to modify it?

First, change $ n \equiv 2 \pmod{4}$ to $ n \plus{} 2008 \equiv 3 \pmod{9}$.  Let $ p_i$ be distinct primes different from 3 such that $ p_i$ (not just $ p_i^2$) is not a divisor of $ i$.  Finally, change $ m$ to $ 9 p_1^2 p_2^2 \dots p_{2007}^2$.  With these modifications, we can check that $ \gcd(a, m) \equal{} 1$.