quadratics conics parabola AMC USA(J)MO USAMO probability

I think that NM should have its own marathon, because I sorta think its a good idea. Here its not as crowed, so when you trying to post a solution, the problem isn't already solved (yes, it's happened to me before)

We are looking for solutions, too.

Problem:

In solving a quadratic equation, Adam misread the coefficient of the first degree term and obtained 3 and -4 as solutions, while Zora misread the constant term, and obtained -1 and 5 as solutions. Find the correct solutions.

I'll solve the problem in just a moment (or maybe tomorrow).

Here are some proposed rules of marathon-

[b]New Mexico Marathon[/b]
1. Follow all general rules for AoPS marathons.
2. Post a solution, not just an answer, as well as another problem.
3. Keep problems at a reasonably simple level (no harder than late AMC10's, no easier than mid chapter MC)
4. I know there are like almost nobobdy who uses the NM forum, so problems may go for a long while unanswered. Don't make problems too difficult.[/b]

I may be wrong...... but here is what I have......

In the parabola with 3 and -4 as solutions, the "b" term is incorrect, but the "a" and "c" terms are correct. Thus, we know the "steepness/narrowness" of the correct parabola, as well as the y-intercept (aka the up-down position of the parabola). So, we then know the correct parabola has solutions 7 different from each other. This is because we know they have the same steepness and up-down position, making the difference of the solutions constant.

We now turn to the other clue. In this one with -1 and 5 as solutions, we know the steepness is still the same, and this time, the left-right position is correct, but the up-down is not. Fortunately, all we need now is the left-right position which is defined by the "b" term, which is said to be correct in this clue. Between -1 and 5 is 2, the axis of symetry.

So our final correct solutions are 7 apart and equally distant from 2. Adding and subtracting 3.5 from 2, we get $ \minus{}1.5$ and $ 5.5$ as solutions.

NP- kind of easy compared to the last one, i guess

Three students from the Land of Enchantment qualified for USAMO in 2009 ( actually true). Sylvester has a 47% chance of making IMO, while Glatis and Alfreda have a 91% and 97% chance, respectively, of not making IMO. All of them have a 57% chance of making MOP. What is the probability atleast two students from NM make the IMO team? Expess your answer has a percent to the nearest hundredth.

I'm sorry, but you're not right pinkmuskrat. click below to see the answer.

[hide]Adam solved \[ (x-3)(x+4)=x^2+x-12\]. Zora solved \[ (x+1)(x-5)=x^2-4x-5\]. Taking the correct components from each equation, the correct equation is \[ x^2-4x-12=(x+2)(x-6)\]. So, our roots are $ -2$ and $ 6$[/hide]

I'll answer you problem some other time...

Oh, whoops. I see that I was way majorly overthinking that problem. I guess you just use the "factor rule" with the solutions. I feel very stupid now....

First, Why did you put the information about MOP? That is irrelevant...

Second, I agree to those rules, though I think we should be able to post AMC 12 problems.

We can split it into 4 cases.

Case 1: All three make IMO. This is \[ \frac{47}{100} \times \frac{9}{100} \times \frac{3}
{100}\equal{}0.001269\]

Case 2: Sylvester and Glatis make IMO. This is \[ \frac{47}{100}\times\frac{9}{100}\equal{}0.0423\]

Case 3: Sylvester and Alfreda make IMO. This is \[ \frac{47}{100}\times\frac{3}{100}\equal{}0.0141\]

Case 4: Glatis and Alfreda make IMO. This is \[ \frac{3}{100}\times\frac{9}{100}\equal{}0.0027\]

Adding these, \[ 0.0423\plus{}0.0141\plus{}0.0027\plus{}0.001269\equal{}0.060369\equal{}6.0369\%\approx 6.04\%\]

NP: Express 1988 as a sum of two ar more consecutive positive integers in as many different.

In order to yield an even number like 1988, we need an odd number of consecutive numbers.
Factorize for 2*2*7*71
We want the most numbers possible, but still let it be odd.
The largest is 497. We make that the number of terms, and 1988/497=4 for the average of the consecutive terms. But of course, we cannot have so many consecutive numbers that are positive integers averaging 4. So we move on.
Try 71 terms. 1988/71=28. That still doesn't work.
Try 7 terms. 1988/7=284 as the average of the terms. That works.
We get: 281, 282, 283, 284, 285, 286, 287 They sum to 1988.

EDIT_ I did that wrong. Oh well. I know 8 works.... Lets make these probloems easier, shall we? Then we can make the NM Forum have lots of posts.
Okay, no ans, but when dividing by an even number of consec numbers, it needs to have a ".5" in it. For odds, quotient must be integer.

EDIT2- So I think that is the largest odd that will work, 7. But for evens, to get rid of the ".5", you can multiply so it is 1988*2=3976= 2*2*2*7*71.
Do the same thing maybe
3976 not work
1988 not work
994 not work
569 not work
no 284
no 142
yay 56?
3976/2=1988
1988/56= 35.5
56 terms from 8-63 i think is the best you can do.mayeb trial and error was better
Too hard..........

NP:Find the volume of half an icosohedron with edge length 2.

Note that for each face of the icosohedron, there is a tetrahedron under it.  Since there are 20 faces, there are 20 pyramids. However, you only want half, so 10 pyramids.  Therefore:

base=$ sqrt3$
height=$ (2sqrt6)/3$
tetrahedron=$ (2sqrt2)/3$
all ten of  them=[color=red]$ (20sqrt2)/3$[/color]

NP:$ Z$ is a complex number so that
$ z \equal{} 9 \plus{} i$

so that $ b$ is a postive integer and $ i^2 \equal{} \minus{} 1$.  If $ b^2$ and $ b^3$ have the same imaginary part, then what is $ b$?

EDIT:$ z\equal{}9\plus{}bi$

[quote="pinkmuskrat"]Um, wow there... Where's the "b" coming from?[/quote]

Sorry, I changed it.

if b is a positive integer, there is no imaginary part. If your saying z^2 and z^3, then that is solvable.

Solution: The expansion of $ z^2$ is $ 81 + 18bi + - b^2$, whose imaginary coefficient is $ 18b$. Similarly, the imaginary coefficient of $ z^3$ is $ 243b - b^3$. Thus,
\begin{align*} 18b = 243b - b^3 \\
\Rightarrow 18 = 243 - b^2 \\
\Rightarrow b^2 = 225 \\
\Rightarrow b = \pm 15 \end{align*}
But, when we divided by b, we got rid of a root, namely, b=0. This doesn't matter, though, as z is complex.
Also, b must be positive, so $ \boxed{15}$ is our answer. YAY!

NP: Let $ r, t,$ and $ s$ be the roots of the cubic equation $ x^3-2x+3=0$. Find $ r^2+s^2+t^2$.
[Hint:[hide]Use Vieta's Formulas[/hide]]

Since both pinkmuskrat and prezcoin had looked at my prob, and neither has posted a solution, I will create more hints. Each of the links below will be a hint. The hints tell a lot about the problem. Try it without the hints, please.

[hide]You don't need to find the individual roots[/hide]
[hide]Expand $ (x\minus{}r)(x\minus{}s)(x\minus{}t)$. What are the coefficients in terms of the roots?[/hide]
[hide]What product can we find that has $ r^2\plus{}s^2\plus{}t^2$ that has values we know, from the coefficients?[/hide]

If you are truly stuck, read this. I will advise you not to, though.
[hide]How would we find $ a^2\plus{}b^2\plus{}c^2$ if we knew $ a\plus{}b\plus{}c$, $ ab\plus{}bc\plus{}ac$, and $ abc$?[/hide]

After you have finished my problem, find $ r^3\plus{}s^3\plus{}t^3$. it should be eas after you have solved the first.

That doesn't mean I dont know how... besides i hate posting new probs here. i cant just make up a hard prob. have to look in a book or online.

Yes, VIETA TO THE RESCUE!!!

To find $ r^2\plus{}s^2\plus{}t^2$, you must know $ r\plus{}s\plus{}t$, $ rs\plus{}st\plus{}tr$, and $ rst$.  The $ r\plus{}s\plus{}t$ is $ 0$ and the $ rst$ is -3 from vietas.  When you multiply out $ (x\minus{}r)(x\minus{}s)(x\minus{}t)$, the coeficient of $ x$ is $ (rs\plus{}st\plus{}tr)$, or 2.

To find the final answer, use

$ (r\plus{}s\plus{}t)^2\minus{}2(rs\plus{}st\plus{}tr)\equal{}0\minus{}2(2)\equal{}\minus{}4$

Hard problems... hmm... how about a prob where you just think (a little or a lot depends on the person)
NP- You have 10 stacks of coins, each stack consisting of ten half-dollars. One entire stack is counterfeit, but you do not know which one. But you do know the weight of a genuine half-dollar, and you also are told each counterfeit coin weighs 1 gram more that it should. You may use a scale, if you want to. What is the number of the fewest weighings/how would you do it, to determine which is counterfeit.

Which one is it? :agent: :agent: :agent: :agent: :agent: :agent: :agent: :agent:

erm, this is a well-known Martin Gardner problem.

[hide]Well, you take 1 coin from the first pile, 2 from the second, 3 from the third,..., and ten from the tenth. Now, you weigh all of those coins. If every coin was genuine, then the weight would be $ \frac{10 \times 11}{2}\equal{}55$ times the weight of one genuine coin. But, because one of the piles is counterfeit, the weight is $ n$ grams more than 55 times the weight of the genuine coin, where $ n$ is the number of the pile that is counterfeit.[/hide]

NP: Prove all of the Millenium problems. jk...

Real NP: For how many natural numbers $ n$ is $ 12^{12}$ the lcm of the numbers $ 6^6, 8^8,$ and $ k$?

Prime factorize these puppies.
$ 6^6 \equal{} 2^6*3^6$
$ 8^8 \equal{} 2^{24}$
The LCM is
$ 12^{12} \equal{} 2^{24}*3^{12}$
By definition when we take the first two numbers mentioned and k, the third, we take the most of each prime's exponent available. We are only working here with 2's and 3's. So, based on our given LCM, we can have anywhere from 0-24 powers of two, since the amount needed is already met and we need to keep at that number or be less than it. That yields 25 possibilities. For our 3's, we have not met the requirement of 12 powers of 3, so we must include in k 12 powers of 3, or 1 choice.
Our answer is $ 25*1 \equal{} 25$.

NP. Simplify $ i^i$, where i is the square root of -1.

Yay... Remember, I like $ i$!

Solution: From Euler's Formula, $ i=e^{\frac{i\pi}{2}}$ by setting $ x=\frac{\pi}{2}$. (To learn more about Euler's formula, go here [url]http://mathworld.wolfram.com/EulerFormula.html[/url].) Now, plugging this in, we have
\begin{eqnarray*} i^i= e^{\frac{i\pi}{2}^i}\\= e^{\frac{i^2 \pi}{2}}\\= e^{-\frac{\pi}{2}}}\\ \approx  0.20788 \end{eqnarray*}
Thus ends the debacle of pinky's problem.

NP: In the year 2001, the US hosted the IMO (actually true). Let $ I, M,$ and $ O$ be distinct positive integers such that $ I \times M \times O = 2001$. What is the largest possible value of the sum $ I+M+O$?

Prime factor 2001:
$ 3*23*29$

If the numbers $ I$, $ M$, and $ O$ are distinct (which they are), then you want one of the three to be $ 1$.  Then, find the largest product consisting of 2 of the 3 numbers, which is $ 667$.  The answers is:

$ 1\plus{}3\plus{}667\equal{}671$

NP: A classic knights and knaves prob....

Three people are standing before you.  One is a knight (always tells the truth), another a knave (always lies to you), and an imposter (randomizes between truth and lies).  You must ask three questions directed to any of them to find the identities of the three people.  You don't know who they are, but they themselves know who is who.  What are the questions ($ 2$ scenarios; use both)?

 :P  :|  :maybe:

Here's a few hints:

[hide]Find out who's [b]not[/b] the randomizer first (then don't talk to him).[/hide]

[hide]List the 6 possibilities and use process of elimination for each answer you get.[/hide]

That question is stupid. Focus more on math.

I've memed. Ask person 1 if 2 is more likely to tell truth that C. You now know someone is not the randomizer. Ask him if he's the randomizer to see if he is true or false. Then you know his identity, so ask him about person A being randomizer or not.

NP- Write $ \sqrt{17}$ as a continued, infinite fraction.

To find the continued fraction, set up an equation:

$ \sqrt {17} \equal{} 4 \plus{} 1/(x \plus{} 1/(x \plus{} 1/....))$

After a little algebra, you get:

$ x \equal{} (32 \minus{} 8\sqrt {17})/(4 \minus{} \sqrt {17})$

use difference of the squares to get:

$ x \equal{} (128 \minus{} 136)/(16 \minus{} 17)$
$ x \equal{} \minus{} 8/ \minus{} 1$
$ x \equal{} 8$

Therefore, the fraction is:

$ 4 \plus{} 1/(8 \plus{} 1/(8 \plus{} 1/(...)))$

NP: Find the ratio of the surface area of an octohedron to the volume of the octohedron.

REAL EDITED NP: Find ratio of volume of a dodecahedron to volume of octohedron with the same side lengths

That probably belongs in the easier marathon...

hmmmm... Remember that problem that I suggested using Vieta's? well, Newton's sums makes it a lot easier... But Vieta's was doable.


[hide] We have $ x^n\equal{}1$ which rearranges to $ x^n\minus{}1\equal{}0$. We can now use Vieta's Formulas to get the product as $ \frac{(\minus{}1)^n*\minus{}1}{1}\equal{}(\minus{}1)^{n\plus{}1}$. Hence, if n is odd, it will be one, and if n is even, it will be -1.[/hide]

Er... the product of the nth roots of 1 is always equal to 1 regardless of the parity of n, no?  :maybe:

Am I missing something here...? Consider the square root of 1. It is 1 or -1, so the product is -1...


A space shuttle travels from Earth towards the moon at a speed of 30,000 kilometers per hour for 12 hours.  The total distance to the moon is 400,000 kilometers.  At the current velocity, how many additional minutes are needed to reach the moon?

[hide]Not sure what this is saying. The hours needed to be travelled to complete the trip is 160.

If it's asking for the remainder, the answer is 0.[/hide]

[hide]It has currently went 360000 km, so 40000 km are left.
40000/30000 = 1.3333 hours, or 80 minutes.[/hide]


geometry
geometric transformation
rotation
probability
ratio

1. Six smilies: Happy Face, Sad Face, Mr. Green, Razz, Oopsies!, and Mr. Mad are sitting at a round table where rotations are indistinguishable. How many ways can the smilies sit aruond the table if Mr. Green and Razz want to sit next to each other as a sign of their peace truce?

2. Darryl Shen and Bobby Wu are playing a game in which a player flips a fair coin, one side with  :D  and another with  :P  . A player loses if he flips a  :P  . If a  :D  is flipped, the game continues. Darryl goes first. Whats the probability that he loses?

3. :rotfl:   and Guassy are playing a game of tic-tac-toe. Guassy goes first and plays X, and  :rotfl:  goes second and plays O. If they played randomly, with no strategy, how many ways are there for Guassy to win?

1>  Let Mr.Green and Razz be considered one.  Then we have 5! ways to order the smilies.   But then we have to divide by 5 because the table can be rotated.  $ \frac {5!}{5} \equal{} 4! \equal{} 24$

2> Darryl has a 1/2 chance of losing on his first turn, a 1/8 chance of losing on his second turn, a 1/32 chance of losing on his third turn,... But Bobby also has a 1/2 chance of losing on his first turn, a 1/8 chance of losing on his second turn, a 1/32 chance of losing on his third turn,... Darryl's probability of losing added with Bobby's probability of losing should equal 1.

$ \frac {1}{2} \plus{} \frac {1}{2} \plus{} \frac {1}{8} \plus{} \frac {1}{8} \plus{} \frac {1}{32} \plus{} \frac {1}{32}... \equal{} 1$

$ 2(\frac {1}{2}) \plus{} 2(\frac {1}{8}) \plus{} 2(\frac {1}{32})... \equal{} 1$

$ \frac {1}{2} \plus{} \frac {1}{8} \plus{} \frac {1}{32}... \equal{} \boxed{\frac {1}{2}}$

It also just makes sense if you think about it.

3>  Um is it 8?  Solution:  There are only 8 ways to win in tic-tac-toe

[quote="suma_milli"]
$ \frac {1}{2} \plus{} \frac {1}{8} \plus{} \frac {1}{32}... \equal{} \boxed{\frac {1}{2}}$
It also just makes sense if you think about it.
[/quote]

really?

this is an infinite series with common ratio 1/4 so (1/2)/(1-1/4)so (1/2)/(3/4)=2/3.

Oh that's true..  But that means they both have a 2/3 chance of winning?  This problem is confuzzling.  :P


calculus
integration
function
logarithms
trigonometry
calculus computations

try to find integral of this function ln(sinx) in the interv al [0,pi/2]
 :P

A little correction:

Let be $I=\int_{0}^{\frac{\pi}{2}}{\ln sinx}dx$ then puting $\frac{\pi}{2}-x=t$ we got $I=\int_{0}^{\frac{\pi}{2}}{\ln cost}dt$ so we have $2I=\int_{0}^{\frac{\pi}{2}}{\ln(sinx cosx)}dx=\int_{0}^{\frac{\pi}{2}}{\ln \frac{sin2x}{2}}dx=\int_{0}^{\frac{\pi}{2}}{\ln{sin2x}}dx-\frac{\pi \ln2}{2}$
Puting $2x=t$ we got $2I=\frac12\int_{0}^{\pi}\ln sintdt-\frac{\pi \ln2}{2}$ . Note $J=\int_{\frac{\pi}{2}}^{\pi}\ln sintdt$ but puting $t-\frac{\pi}{2}=x$ we got the integral $J=\int_{0}^{\frac{\pi}{2}}{\ln sin\left(x+\frac{\pi}{2}\right)dx=I}$ so $\int_{0}^{\pi}\ln sintdt=2I$ therefore $I=-\frac{\pi \ln2}{2}$

Use \ln, \sin and \cos to make that look nicer.

$I=\int_{0}^{\frac{\pi}{2}}{\ln\sin x}dx$

Posted many times already:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=120089
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=116550

[quote="cckek"]that's right is'n it?[/quote]
Yes it's right .


I would like to sign up for a mathcouts class for the fall. I got an impression that there is a pretest which can tell if I need to take that class.  Where could I find the pretest probllem set for each class?  Thanks.

There is no pre-test for the Problem Series classes.

[quote="Course Recommendations"]We also have two Problem Series classes for MATHCOUNTS and AMC 8 preparation. If you are just getting started with these contests and find many of the problems very challenging, then we recommend the MATHCOUNTS/AMC 8 Basics course. If you have some experience with MATHCOUNTS and are preparing for the State or National MATHCOUNTS competition, you are probably ready for our Advanced MATHCOUNTS/AMC 8 course. [/quote]

Pick the one that is best suited to your abilities.  Alternatively, you could take the Basics class that starts on September 14, and the Advanced class that starts on November 11.


Find all $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that

$ f(x^2\plus{}f(y))\equal{}y\plus{}xf(x)$

for all reals $ x,y$

From $ x \equal{} 0$ we get $ f(f(y)) \equal{} y$. $ f$ is clearly bijective.

Let $ x \equal{} y \equal{} f(0)$, then $ f( [f(0)]^2 ) \equal{} f(0)$. Then $ [f(0)]^2 \equal{} 0$ by $ f$ injective, and $ f(0) \equal{} 0$.

Replace $ x$ with $ f(x)$ in the given equation and let $ y \equal{} 0$. We get $ f([f(x)]^2) \equal{} xf(x) \equal{} f(x^2)$. So by $ f$ injective $ [f(x)]^2 \equal{} x^2$. Therefore for all $ x$, $ f(x) \equal{} x$ or $ f(x) \equal{} \minus{}x$.

Suppose $ f(a) \equal{} a, f(b) \equal{} \minus{}b$, and $ a,b$ are both nonzero. Then $ x \equal{} a,y \equal{} b$ gives $ f(a^2 \minus{} b) \equal{} a^2 \plus{} b$. Because $ f(x)$ is plus or minus the input, $ a^2 \minus{} b \equal{} a^2 \plus{} b$ and $ b \minus{} a^2 \equal{} a^2 \plus{} b$. Either case implies one of $ a,b$ being $ 0$. Contradiction.

$ f(x) \equal{} x$ and $ f(x) \equal{} \minus{}x$ are the solutions.


$x\in\mathbb{R},n\in\mathbb{N},n\geq 3$ and $\{x\}=\{x^2\}=\{x^n\}$. Prove that $x\in\mathbb{Z}$.

Already proposed at least two times on ML.

Pierre.


The unit cells of an $ n\times n$ array are colored white or black in such a way that any $ 2\times 2$ square contains either exactly one white cell or exactly one black cell. Find all possible values of $ n$ for which such an array does not contain identical columns.

[quote="baysa"]$ n\leq4$. One can show 1st and 3rd column either coincide or opposite (that is different in each position), same result holds for 3-rd and 5th column. Thus it can not be $ n\geq5$.[/quote]
But in this figure with $ n\equal{}5$ and W-white cell, B-black cell:
[img]http://i227.photobucket.com/albums/dd72/TPFC/NewImage-5.png[/img]


Given x,y,z  positive real numbers satisfying xyz=32 find the minimum value of

   x^2 + 4xy + 4y^2 +2z^2

Source: United Kingdom 2000 mathematical olympiads


It's much easier to get yourself up to run and subject yourself to physical pain in general when you know a few other people are going to as well. Last year I managed to get up exactly once and this year I hope to do better.

Has anyone planned a time and place for morning runs this year? If not, I'll be outside the dorms to stretch (which are all next to each other) at... 7:00 A.M?  (according to this year's AMSP brochure, breakfast is 7:30-8:30). If you choose to run it'd obviously be at your own pace and you can run for as long/short as you want.

And if anyone else is wants to, I am willing to subject other people to the torture called the side plank (along with various other stuff).

.......

I heard some mentors make you run.

I wont :P

I think it's better after 7:00 p.m. thru 10 p.m. when you can do bunch of stuff like forums, swimming, playing tennis, frisbee, etc.

I would stretch if there was no "pain" and anyways we have to walk (5-10 min depending where you stay) to cafeteria.

You may want to say who you are, what you look like though......

I'm pretty sure to be running 6 days a week in the morning. Last year, I got up at more like 6:30. I don't know how much I'm going to run. I've been doing like 30 to 35 miles a week but I see how I feel in the Texas weather. One of the other assistants might be running too. But really, you can just run on your own. I might do more like four miles a day because I want to play frisbee after dinner too.

Lol.  I'll definately be up by then (I'm in EST, so I'll gain an hour, and I normally wake up at 6:20 anyways).  I'm not overweight, but I'm definately not in great shape either, so I guess I could use the exercise.  My goal is to be able to go (jog?) a full mile nonstop.

1, What's the side plank?

and 2, Does anyone know if there are rules against PMing each other names for something like this?  My name isn't too hard to guess, but...

Uhh... a side plank is like you lay on your forearm and you hold your body straight for however long...

I'm not a big fan of core stuff. I stopped once XC season ended.

And I'm pretty sure that you can just go outside at like 6:45 in the morning and see people. I don't know how names will help...

I doubt that you will be around your particular assistant all the time. I felt like I was always with some other students all the time (except when I ran...by the last week most people slept in). For the most part, mentors and assistants will be around. There are a few areas that you will pretty much stay around. Those are the dorms and the student union and the classrooms. But people did go off and play frisbee or soccer or whatever.

[quote="kl2836"]arent we required to be with our assistants/mentors wherever we go? especially in the morning?[/quote]

I don't think it's even possible to be with your mentor in the morning, as they wake up 5 minutes before they have to teach.

And I'll probably get up 5 minutes before I have to learn... :D

I haven't run in a month, since running to school.  Running to school taught me that I run at 3 MPH, since I can run 1/2 to school in 7 or 8 minutes and still barely make it to class without the teacher getting too upset.   :)

Meh, I want to run, but I'm slightly bowlegged and a little flat-footed.   :(  My doctor said so at my checkup last week.

[quote="alexhhmun"]...I run at 3 MPH...[/quote]

3mph = 1 mile / 20 minutes ???!!!!!????!!!!!!

I can run a 400m in 1:29  :D  but it takes me forever to run a mile, like 10 mins. kind of forever.

I can run 400m in 2:14.  I'm so much better than all of you!    :P

can't run; inherited heart weakness. :(

But I have a very light case of Restless Leg Syndrome, so I would be OK with walking. Or cycling.

When I did sprinting track as a freshman I think I got 58s for 400m...I am pretty sure that I cannot do that now lol

[quote="unimpossible"]I definitely will if my doctor says its okay for me to run now.

havn't run in over a week and now im out of shape again... :|[/quote]

Whoa were you injured or something...

Now why are we comparing running times...? I don't run, but I can do a 400m (time from 6th grade) in around 1:12... Not sure if it's gotten worse or better. And I guess my mile was on pace for a 6:00 but then my gym teacher was like you can do it... so I lost enthusiasm and started walking. =D. I'm lazy.

I have low endurance, and bad memory.  My 400 meter time is actually around 1:40.  I can't run or sprint the whole thing...

Haha...interesting times...

8th Grade Times...

400M: 1:07
Mile: 5:52
Endurance: Um...3.3 miles in 24 minutes...

9th grade doesn't really count, since we didn't practice these on a regular basis..

I don't really enjoy running, but it's not too bad if there's no pressure

hehe....i beat u all

eighth grade times--this past school year..:3

400: 57 secs..
Mile: 5 min 32 secs...

but i haven't exercised ina couple of weeks....soo..i'm definitely out of shape..

Cya all tomorrow.  I guess there's no point to talk more, is there?

Best Times in High School:

400: 59.7 (I'm not a 400 runner, and I have never done a fresh 400)
800: 2:11.8
1600 (Mile): 4:59.5

I didn't start this thread so everyone could start comparing their times... but w/e.


A body with P weight is hanging in the not dilatable AB wire with inconsiderable mass, which crosses a pulley with a static axis O. The weight of the block is F and its mass is steadily diffused in the disk with a r ray. The B end of the wire is tied with a spring with a C coefficient of elasticity.
Determine the waverings of the body if when starting the body is stationary ( his gravity is equilibrated by the spring's deformation ) and we give it an initial velocity of $V_{0}$ in the vertical direction. The frictions are negligible.
Illustration:
[url=http://img231.imageshack.us/my.php?image=rrotullayt6.jpg][img]http://img231.imageshack.us/img231/668/rrotullayt6.th.jpg[/img][/url]

I mean the oscillations of the body. I need the lagrange's equation of these oscillator movements.

Here is the solution:
Indicate :
$T = \frac{{m_{1}\mathop{x^{2}}^{.}}}{2}+\frac{{I\omega^{2}}}{2}$
$\omega = \frac{{\mathop x^{.}}}{r}$
We know that the moment of inertia of a homogeneous disc is:
$I = \frac{{m_{2}r^{2}}}{2}$
After substitutions we give:
$T = \frac{{(2m_{1}+m_{2})\mathop{x^{2}}^{.}}}{4}$
The potential energy of the spring is:
$V = \frac{{cx^{2}}}{2}$
$L = T-V$
$\frac{{\partial L}}{{\partial \mathop x^{.}}}= \frac{{(2m_{1}+m_{2})\mathop x^{.}}}{2}$
$\frac{d}{{dt}}\frac{{\partial L}}{{\partial \mathop x^{.}}}= \frac{{(2m_{1}+m_{2})\mathop x^{..}}}{2}$
$\frac{{\partial L}}{{\partial x}}=-cx$
$\mathop x^{..}+\frac{{2c}}{{2m_{1}+m_{2}}}x = 0$
$\omega^{2}= \frac{{2c}}{{2m_{1}+m_{2}}}$
$\mathop x^{..}+\omega^{2}x = 0$


Let $n$ be a positive integer that is not a perfect cube. Define real numbers $a$, $b$, $c$ by \[a=\sqrt[3]{n}, \; b=\frac{1}{a-\lfloor a\rfloor}, \; c=\frac{1}{b-\lfloor b\rfloor}.\] Prove that there are infinitely many such integers $n$ with the property that there exist integers $r$, $s$, $t$, not all zero, such that $ra+sb+tc=0$.


calculus
function
integration
calculus computations

$ let \ f(x) \ be \ a \ continous \ function  \ that  \ takes \ positive \ values \ for \\ x\geq 0 \ and \ satisfy \\\\\
\int_{0}^{x}{f(t).dt}=x\sqrt{f(x)}\\\\\
with \ f(1)=\frac{1}{2} \\ \\
then \ find \
f(1+\sqrt{2}) $

One approach:

Let $ y\equal{}F(x)\equal{}\int_{0}^{x}f(t)dt$. Then from the given hypothesis you get a differential equation $ x^2\frac{dy}{dx}\equal{}y^2$ with the initial condition $ y\equal{}\frac{1}{\sqrt{2}}$ when $ x\equal{}1$. This can be easily solved by variable separable method and the rest are easy.
I think the answer is $ \frac{1}{4}$.

There may be some other approach.

ur answer is there in one option
i think u r right  :)


There are three verticies of a triangle that a bug can walk on. The bug starts on the first vertex, and walks to an adjacent (any of the other two) vertex. The probability of going onto each adjacent vertex is 50%. What is the probability, that after 10 moves, the bug will return to its original vertex?

Problem was covered here:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9900[/url]

Oh, thanks. What a strange way to do it....makes perfect sense though.

There is actually another recursion you can create for this problem than what chinaboy did in his post.

(sorry for bringing this topic up btw but i realized that the 2003 AIME II is not on the contest directory  :o and I think this is easier than the solutions given in that topic (apart from JBL's which is pure genius :lol:) and kalva's solution using mods)

After each move the bug makes, it has a $\frac{1}{2}$ chance of going back to the vertex it started at but must also NOT be at the vertex it started at prior to making this move.

Let $P_{n}$ be the probability the bug will be at it's starting point after $n$ moves. We have that:

$P_{0}=1$ since the bug is already at it's starting vertex.

$P_{1}=0$ because the bug has to move off it's starting vertex.

$P_{2}=\frac{1}{2}$ because for the second move the bug can either go back to it's starting vertex or another vertex with equal probability.

From what I said earlier, we can generalize the probability through the recursion:

$P_{n+1}=\frac{1}{2}(1-P_{n})$

From this, we get that $P_{10}=\frac{171}{512}$.

[quote="breez"]How did you arrive at your recursion function?[/quote]

For the bug to be at its starting vertex after $n$ moves, it must be at a different vertex after $n-1$ moves, which occurs with probability $1-P_{n-1}$.  Then there is a $\frac{1}{2}$ probability that it will move from the different vertex to the starting vertex.

I used the same recursion.  :D

[hide="Hint"]Try to find a way of using recursion in this problem.[/hide]


In a bag, there are 8 balls, labeled from 1 to 8. After stirring these 8 balls sufficiently, two balls are drawn from the bag at a time. Let denote $ S$ the sum of the two balls.

(1) Find the probability such that $ S$ doesn't exceed in eight.

(2) Let $ f(n)$ be the probability such that $ S$ takes value $ n$. Find the value of $ n$ for which $ f(n)$ is maximal.

[hide="Part 1"]There are 56 ways to draw 2 balls from the bag (order matters). There are $ 8$ choices for the first ball and $ 7$ choices for the second, giving $ 8 \cdot 7\equal{}56$ in total.

Of these, there are $ 32$ that give a sum greater than 8. Therefore the probability that their sum does not exceed 8 is

$ 1\minus{} \frac{32}{56}\equal{} \frac{24}{56}\equal{} \frac{3}{7}$[/hide]

[hide="Part 2 (cheap solution)"]Drawing a sample space diagram reveals that 9 is the most likely sum with a probability of $ \frac{1}{7}$.[/hide]


On the sides $[AB],[BC],[CD]$ and $[DA]$ of a convex quadrilateral $ABCD$ squares are constructed outward this quadrilateral. $M_1, M_2, M_3, M_4$ are the midpoints of those squares. Show that $|M_1M_3| = |M_2M_4|$ and $M_1M_3 \bot M_2M_4.$

[b]Remark[/b]: In 1970 this was rather an unknown fact but it is not an original question. The theorem is due to Von Aubel.

[color=red][Moderator edit: This problem is really famous; see http://www.mathlinks.ro/Forum/viewtopic.php?p=162884 for a discussion.][/color]

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)


Find $ f(x): (0,\plus{}\infty) \to \mathbb R$ such that \[ f(x)\cdot f(y) \plus{} f(\frac{2008}{x})\cdot f(\frac{2008}{y})\equal{}2f(x\cdot y)\] and $ f(2008)\equal{}1$ for $ \forall x \in (0,\plus{}\infty)$.

This problem was proposed in the 2006 Spanish Mathematical Olympiad, except that the question stated that a $ \lambda\in\mathbb{R}$ existed such that
\[ f(x)f(y) \plus{} f\left(\frac {\lambda}{x}\right)f\left(\frac {\lambda}{y}\right) \equal{} 2f(xy),
\]
and $ f(\lambda) \equal{} 1$.

Same solution as #2, post for storage.

$P(1,1)$ gives us $f(1)=1$.
$P(1,y)$ gives us $f(y)=f(\frac{2008}{y})$.
So we got $2f(x)f(y)=2f(xy)$, which is just $f(x)f(y)=f(xy)$.
$P(x,x)$ gives us $f$ bounded, so $f(x)=1$ can be checked to be the only function that works.


calculus
derivative
function
calculus computations

If $ y\equal{}(\arcsin{x})^2$

Find $ y^{(n)}(0)$, here $ y^{(2)}$ means $ y''$,etc. :oops:

Write $ \arcsin x \equal{} \frac {e^x \minus{} e^{ \minus{} x}}{2}$.

Then you get $ \left(\arcsin x \right)^2 \equal{} \frac {1}{4} \left(e^{2x} \plus{} e^{ \minus{} 2x} \minus{} 2\right)$

Hence, $ f^{n}(x) \equal{} \frac {1}{4}\left(2^n e^{2x} \plus{} ( \minus{} 2)^n e^{ \minus{} 2x} \right)$

If n is odd, then you get $ f^{n}(0) \equal{} 0$ and if n is even you get $ f^{n}(0) \equal{} 2^{n \minus{} 1}$.

Ignore the above comment. A good exercise would be to compute the first few derivatives and evaluate them at 0.

Edited after Timestopper_STG

How come $ \arcsin x \equal{} \frac{e^{x}\minus{}e^{\minus{}x}}{2}$?

I think $ \sinh x \equal{} \frac{e^{x}\minus{}e^{\minus{}x}}{2}$, isn't it?

If $ y \equal{} (\arcsin x)^{2}$

Then $ y$ satisfies the differential equation (if I worked it out correctly):

$ y''(1 \minus{} x^{2}) \minus{} xy' \equal{} 2$.

Using the power series representation of $ y$ will give us the required coefficients.

Maple says:
$ \left( \arcsin \left( x \right)  \right) ^{2}\equal{}\sum _{k\equal{}0}^{\infty }{
\frac { \left( k! \right) ^{2}{4}^{k}{x}^{2\,k\plus{}2}}{ \left( 2\,k\plus{}1
 \right) !\, \left( k\plus{}1 \right) }}$

The differentiation equation gives this.

The equation I gave before seems right and it gives us the recurrence relation

$ a_{2k \plus{} 2} \equal{} \frac {(2k)^{2} a_{2k}}{(2k \plus{} 1)(2k \plus{} 2)}$  and $ a_{2} \equal{} 1$ which should be easy to solve.

Also, since function is even, $ a_{2i \minus{} 1} \equal{} 0$

We don't need Maple  :P


geometry
rectangle
circumcircle
algebra
polynomial
system of equations
national olympiad

3 days, 3 problems each, 4.5 hours per day. Ran from June 12 to June 14 in 2009.

[b]Day 1[/b]

1. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=289581

Let $ m$ and $ n$ be positive integers.  Mr. Fat has a set $ S$ containing every rectangular tile with integer side lengths and area of a power of $ 2$.  Mr. Fat also has a rectangle $ R$ with dimensions $ 2^m \times 2^n$ and a $ 1 \times 1$ square removed from one of the corners.  Mr. Far wants to choose $ m \plus{} n$ rectangles from $ S$, with respective areas $ 2^0, 2^1, \ldots, 2^{m \plus{} n \minus{} 1}$, and then tile $ R$ with the chosen rectangles.  Prove that this can be done in at most $ (m \plus{} n)!$ ways.

[i]Palmer Mebane.[/i]

2. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=289582

Let $ ABC$ be an acute triangle.  Point $ D$ lies on side $ BC$.  Let $ O_B, O_C$ be the circumcenters of triangles $ ABD$ and $ ACD$, respectively.  Suppose that the points $ B, C, O_B, O_C$ lies on a circle centered at $ X$.  Let $ H$ be the orthocenter of triangle $ ABC$.  Prove that $ \angle{DAX} \equal{} \angle{DAH}$.

[i]Zuming Feng.[/i]

3. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=289583

For each positive integer $ n$, let $ c(n)$ be the largest real number such that
\[ c(n) \le \left| \frac {f(a) \minus{} f(b)}{a \minus{} b}\right|\]
for all triples $ (f, a, b)$ such that

--$ f$ is a polynomial of degree $ n$ taking integers to integers, and
--$ a, b$ are integers with $ f(a) \neq f(b)$.

Find $ c(n)$.

[i]Shaunak Kishore.[/i]

[b]Day 2[/b]

4. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=289584

Let $ ABP, BCQ, CAR$ be three non-overlapping triangles erected outside of acute triangle $ ABC$. Let $ M$ be the midpoint of segment $ AP$. Given that $ \angle PAB \equal{} \angle CQB \equal{} 45^\circ$, $ \angle ABP \equal{} \angle QBC \equal{} 75^\circ$, $ \angle RAC \equal{} 105^\circ$, and $ RQ^2 \equal{} 6CM^2$, compute $ AC^2/AR^2$.

[i]Zuming Feng.[/i]

5. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=289585

Find all pairs of positive integers $ (m,n)$ such that $ mn \minus{} 1$ divides $ (n^2 \minus{} n \plus{} 1)^2$.

[i]Aaron Pixton.[/i]

6. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=289586

Let $ N > M > 1$ be fixed integers. There are $ N$ people playing in a chess tournament; each pair of players plays each other once, with no draws. It turns out that for each sequence of $ M \plus{} 1$ distinct players $ P_0, P_1, \ldots P_M$ such that $ P_{i \minus{} 1}$ beat $ P_i$ for each $ i \equal{} 1, \ldots, M$, player $ P_0$ also beat $ P_M$. Prove that the players can be numbered $ 1,2, \ldots, N$ in such a way that, whenever $ a \geq b \plus{} M \minus{} 1$, player $ a$ beat player $ b$.

[i]Gabriel Carroll.[/i]

[b]Day 3[/b]

7. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=289587

Find all triples $ (x,y,z)$ of real numbers that satisfy the system of equations
\[ \begin{cases}x^3 \equal{} 3x \minus{} 12y \plus{} 50, \\
y^3 \equal{} 12y \plus{} 3z \minus{} 2, \\
z^3 \equal{} 27z \plus{} 27x. \end{cases}\]
[i]Razvan Gelca.[/i]

8. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=289589

Fix a prime number $ p > 5$. Let $ a,b,c$ be integers no two of which have their difference divisible by $ p$. Let $ i,j,k$ be nonnegative integers such that $ i \plus{} j \plus{} k$ is divisible by $ p \minus{} 1$. Suppose that for all integers $ x$, the quantity
\[ (x \minus{} a)(x \minus{} b)(x \minus{} c)[(x \minus{} a)^i(x \minus{} b)^j(x \minus{} c)^k \minus{} 1]\]
is divisible by $ p$. Prove that each of $ i,j,k$ must be divisible by $ p \minus{} 1$.

[i]Kiran Kedlaya and Peter Shor.[/i]

9. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=289590

Prove that for positive real numbers $ x,y,z$,
\[ x^3(y^2 \plus{} z^2)^2 \plus{} y^3(z^2 \plus{} x^2)^2 \plus{} z^3(x^2 \plus{} y^2)^2 \geq xyz[xy(x \plus{} y)^2 \plus{} yz(y \plus{} z)^2 \plus{} zx(z \plus{} x)^2].\]
[i]Zarathustra (Zeb) Brady.[/i]

thank you very much.
the problem in day 1,2 really interesting

Isn't problem 3 the same as problem 14.3 in http://www.unl.edu/amc/a-activities/a6-mosp/a6-1-mosparchives/2007-ma/mosptests.pdf?

And this is the 2007 black tests...

Problem 5 is due to Aaron Pixton.  Problem 6 is due to Gabriel Carroll.  (I don't know the source of problem 2.)

Thanks for the missing sources.

pythag: The problem you cited is only the easier half of 3. Still a bit of a goof, but didn't seem to have great effect.

Also those weren't actual black tests, just their homework I think.


There are 14 red, 10 yellow, 5 green, and 3 purple candies in a bag. What is the probability that if a candy chosen, it is neither purple or red?

[hide]there are 32 total candies and there are 17 that are purple or red so there are 15 that are not purple or red so the probablity is
15/32
that seems wrong.[/hide]

[quote="pascal_1623"][hide]
There are 32 candies in the bag, and there are 24 candies that aren't red or purple.
$\frac{24}{32}=\boxed{\frac{1}{3}}$[/hide][/quote]
umm, 10+5+3 is not 24

[hide]picking red or purple is 17/32, so picking anything else is 15/32 :idea: thats our answer[/hide]


[quote="sharkman"]I have two problems below that my textbook does not explain how to solve.  Can someone explain, in easy steps, how to answer the following two questions?  I will provide the multiple-choice given in the math book.

(1) If a + c < b + d and b + d < 2a, which inequality MUST be true?

(a) a < b

(b) a > b

(c) a < c

(d) a > c

(2) If ab < 0 and c > 0, which statement MUST be true?

(a) abc > 0

(b) c - ab > 0

(c) (ab/c) > 0

(d) (a/b) > (c/b)

I need to know what is the logic behind the questions above.
I also would like to know what is the best way to answer such questions.[/quote]

[hide="(1)"] $a+c < b+d < 2a$. Then $a+c<2a$. Subtract $a$ from both sides, we get $c<a$, which is $\boxed{(d)}$.[/hide]

[hide="(2)"] Let's just consider all the choices given.

(a) $abc>0$.... we are given that $ab<0$ and $c>0$. When two numbers of opposite signs are multiplied, it is always negative, so this is false.

(b) $c-ab>0$.... We know that $ab<0$ and $c>0$. Then $-ab>0$ and $c>0$. Adding these shows this is true, so this is true, which means it's the ANSWER.

(c) $\frac{ab}{c}>0$... we are given $ab$ negative and $c$ positive. Dividing negative number by positive number gives negative roots, so it is false.

(d) $\frac{a}{b}>\frac{c}{b}$. There are two cases: when $b$ is positive and when $b$ is negative.
If $b$ is positive, then this implies $a>c$. But $a$ should be negative from $ab<0$ and $c$ is positive, so this is false in this case.
Because it says "MUST", we know this is not a good choice.

So the answer is... (b).[/hide]

Why must I subtract a from both sides in question 1?

eacause a+c<2a isn't an option in the muliple choice, we have to get one of those options.


I was looking for some problems on other country and saw the link to the Peru Olympiad but when I click on it, it led to the Yahoo! Page saying that no site is exist anymore.

?De donde es official site de Peru?

[p.s. Where is the Peru's official website?.. I know my spanish is incorrect.]

I don't know if this is the official one, but ..

[url]http://www.pucp.edu.pe/~matemat/olimpiadas[/url]

Would be good practice for your spanish, at least ;)

BOOM
19 YEAR BUMP!
IN YOUR FACE
LAVENDER HAZE!


Suppose you wanted to graph $y=\sin (2x+6)$. This is the way that my math teacher teaches it:

Factor out the 2 so it becomes $y=\sin 2(x+3)$.
Now start out with a table of values for $y=\sin x$.
Multiply the x-coordinates by $\frac{1}{2}$ ("compressing").
Add $-3$ to the x-coordinates ("shifting").
Then graph the equation.

What I don't get is why you should first "compress" the graph and then "shift" it. Please explain :help:

Because if you write it out mathematically, shifting before compressing will produce half the shift desired. :wink:

Your teacher is wrong. The order of the operations is always from inside outwards, hence FIRST you shift, THEN you squeeze/stretch. As for those two, it's important to know what do you perform them upon. If you do it on the AXIS, you just write $2x+6$ instead of every $x$ on it. If you do it on the GRAPH, then you should move it by $-3$ (that is $3$ to the left) and then SQUEEZE it $2$ times (which is equivalent to stretching the axis $2$ times)


let a,b,c are possitive numbers such as
    1/a+1/b+1/c=4
  provethat:
  1/(2a+b+c)+1/(a+2b+c)+1/(a+b+2c) \geq (1)[/code]

[quote="trannguyenquyND_VN"]let a,b,c are possitive numbers such as
    $ 1/a\plus{}1/b\plus{}1/c\equal{}4$
  provethat:
  $ 1/(2a\plus{}b\plus{}c)\plus{}1/(a\plus{}2b\plus{}c)\plus{}1/(a\plus{}b\plus{}2c) \geq (1)$[/quote]
Maybe $ 1/(2a\plus{}b\plus{}c)\plus{}1/(a\plus{}2b\plus{}c)\plus{}1/(a\plus{}b\plus{}2c) \leq 1$ $ ?$ :wink:

[quote="trannguyenquyND_VN"]let a,b,c are possitive numbers such as
    1/a+1/b+1/c=4
  provethat:
 $ 1/(2a\plus{}b\plus{}c)\plus{}1/(a\plus{}2b\plus{}c)\plus{}1/(a\plus{}b\plus{}2c) \geq (1)$[/quote]
It's not true for $ a\equal{}\frac{1}{2}$,$ b\equal{}1$,$ c\equal{}1$

$ \sum_{cyc}\frac {1}{2a \plus{} b \plus{} c}\leq1\Leftrightarrow\sum_{cyc}\left(\frac {1}{4a} \minus{} \frac {1}{2a \plus{} b \plus{} c}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {b \plus{} c \minus{} 2a}{a(2a \plus{} b \plus{} c)}\geq0\Leftrightarrow\sum_{cyc}\frac {c \minus{} a \minus{} (a \minus{} b)}{a(2a \plus{} b \plus{} c)}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \minus{} b)\left(\frac {1}{b(2b \plus{} a \plus{} c)} \minus{} \frac {1}{a(2a \plus{} b \plus{} c)}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {(a \minus{} b)^2(2a \plus{} 2b \plus{} c)}{ab(2a \plus{} b \plus{} c)(2b \plus{} a \plus{} c)}\geq0.$
The following stronger inequality is true too.

Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc \equal{} 3.$ Prove that:
\[ \frac {1}{2a \plus{} b \plus{} c} \plus{} \frac {1}{2b \plus{} a \plus{} c} \plus{} \frac {1}{2c \plus{} a \plus{} b}\leq\frac {3}{4}
\]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=247696


trigonometry
search
integration
calculus
calculus computations

Solve $ u_{tt} \equal{} a^2u_{xx}$
$ u(t,0) \equal{} \sin(\omega t),~u_x(t,1) \equal{} 0$
$ u(0,x) \equal{} 0,~u_t(0,x) \equal{} \omega \cos(\pi x)$.

[hide="My approach"]
Boundary values aren't that nice, so I try to change, them, assume $ u \equal{} v \plus{} w$, and define $ w(t,x): \equal{} \sin(\omega t)\cos(\pi x)$, and that seems a good way to go, we can notice that then $ v(t,0) \equal{} u(t,0) \minus{} w(t,0) \equal{} 0$, and so on $ v_x (t,1) \equal{} v(0,x) \equal{} v_t (0,x) \equal{} 0$. We notice that $ w_{tt}\neq a^2w_{xx}$, so $ w$ is not the solution, so we continue to find $ v$. We get $ v_{tt} \equal{} a^2v_{xx} \plus{} \sin(\omega t)\cos(\pi x)(\omega^2 \minus{} (\pi a)^2)$, we assume that $ v(t,x) \equal{} X(x)T(x)$, (plugging it into the equation, and using boundary values we get ) $ X_n (x) \equal{} \sin(\frac {2n \plus{} 1}{2}\pi x)$, that  we search the solution in the form $ v(t,x) \equal{} \sum_{n}X_n (x)T_n (t)$, $ T_n$ is unknown for the moment. We obtain $ \sum_{n}X_n(x)[T_n ''(t) \plus{} (a\frac {2n \plus{} 1}{2}\pi)^2 T_n (t)] \equal{} \sin(wt)\cos(\pi x)(\omega^2 \minus{} (\pi a)^2)$.
I don't know how to continue pass this point, the next thing to come in mind is to expand $ \cos(\pi x)$ in Fourier series ( by $ \sin (\frac {2n \plus{} 1}{2}\pi x)$), $ \cos(\pi x) \equal{} \sum_n a_n \sin (\frac {2n \plus{} 1}{2}\pi x)$, but then I get ( if my calculation isn't wrong ) that $ a_n \equal{} 0,~\forall n$, even if I continue with that,  $ T_n ''(t) \plus{} (a\frac {2n \plus{} 1}{2}\pi)^2T_n (t) \equal{} 0,~T_n(0) \equal{} T_n '(0) \equal{} 0$, and this gives me that $ T_n(t) \equal{} 0,~\forall n$, but that doesn't seem to make any sense. I hope someone can help with this problem.
[/hide]

[quote="Mr.Doe"]
I don't know how to continue pass this point, the next thing to come in mind is to expand $ \cos(\pi x)$ in Fourier series ( by $ \sin (\frac {2n \plus{} 1}{2}\pi x)$), $ \cos(\pi x) \equal{} \sum_n a_n \sin (\frac {2n \plus{} 1}{2}\pi x)$, but then I get ( if my calculation isn't wrong ) that $ a_n \equal{} 0,~\forall n$, [/quote]

No.
$ a_n \equal{} 2\int\limits_0^1 \cos(\pi x)\sin(\frac {\pi x (2n \plus{} 1)}{2})dx\ne 0$.


Let $ a,b,c,d$ are non-negative real numbers. Prove that
\[ a^{4}+b^{4}+c^{4}+d^{4}+2abcd\ge \sum_{cyc}a^{2}b^{2}+\frac{2}{3}\sum_{cyc}ab(a-b)^{2}\]
Hoping that $ \frac{2}{3}$ is the best constant :D

P/s: I edited it!

[quote="The soul of rock"]Don't sure that Tukevici or Tukervici[/quote]
He Turkevic. Not Turkevici, please! :wink:


search
inequalities
inequalities unsolved

If $a,b,c$ are real numbers such that $a^2+b^2+c^2=1$ prove that

$\frac{a^2}{1+2bc}+\frac{b^2}{1+2ac}+\frac{c^2}{1+2ab}\geq\frac{3}{5}$

If it was posted before please give a link. thanks

Maybe it was posted, anyway I could quicker solve it than search it in the forum.

Since $a^2+b^2+c^2=1$, we have

$\frac{a^2}{1+2bc}=\frac{a^2}{\left(a^2+b^2+c^2\right)+2bc}=\frac{a^2}{a^2+\left(b+c\right)^2}\geq\frac{a^2}{a^2+2\left(b^2+c^2\right)}$,

where in the last step, we used the inequality $\left(b+c\right)^2\leq 2\left(b^2+c^2\right)$ which is trivial from AM-QM or Cauchy-Schwarz. Now, if we denote $x=a^2$, $y=b^2$, $z=c^2$, then this becomes

$\frac{a^2}{1+2bc}\geq\frac{x}{x+2\left(y+z\right)}$.

Similarly,

$\frac{b^2}{1+2ca}\geq\frac{y}{y+2\left(z+x\right)}$;
$\frac{c^2}{1+2ab}\geq\frac{z}{z+2\left(x+y\right)}$.

Thus,

$\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab}\geq\frac{x}{x+2\left(y+z\right)}+\frac{y}{y+2\left(z+x\right)}+\frac{z}{z+2\left(x+y\right)}$
$=\frac{x^2}{x^2+2xy+2zx}+\frac{y^2}{y^2+2yz+2xy}+\frac{z^2}{z^2+2zx+2yz}$.

By [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15125]the Cauchy-Schwarz inequality in the Engel form[/url], we have

$\frac{x^2}{x^2+2xy+2zx}+\frac{y^2}{y^2+2yz+2xy}+\frac{z^2}{z^2+2zx+2yz}\geq\frac{\left(x+y+z\right)^2}{\left(x^2+2xy+2zx\right)+\left(y^2+2yz+2xy\right)+\left(z^2+2zx+2yz\right)}$
$=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+2\left(yz+zx+xy\right)}$.

Since $yz+zx+xy\leq\frac13\left(x+y+z\right)^2$ (this follows from $\frac13\left(x+y+z\right)^2-\left(yz+zx+xy\right)=\frac16\left(\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2\right)\geq 0$), we have

$\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+2\left(yz+zx+xy\right)}\geq\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+2\cdot\frac13\left(x+y+z\right)^2}=\frac35$.

Combining all the above, we get

$\frac{a^2}{1+2bc}+\frac{b^2}{1+2ca}+\frac{c^2}{1+2ab}\geq\frac35$,

and it's done.

  Darij

We have \[ (\sum \frac{a^2}{1 + 2bc})(\sum a^2(1 + 2bc)) \geq (\sum a^2)^2 = 1. \] Also, \[ \sum a^2(1 + 2bc)) = 1 + 2abc(a + b + c) \leq 1 + \frac{2}{3}(a^2 + b^2 + c^2)^2 = \frac{5}{3}. \] Combining these inequalities, we get the result.

Another solution :
 $\frac{a^2}{1+2bc}=\frac{a^2}{a^2+b^2+c^2+2bc}\geq \frac{a^2}{a^2+2b^2+2c^2}$
 Put $x=a^2+2b^2+2c^2$ . So $a^2=\frac{2y-2z-3x}{5}$
 And what we want to prove is :
 $\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{x}{z}+\frac{z}{x} \geq 6$ : which is obviously true

  Arne's solution using Cauchy in Engel form is the nicest !


calculus
integration
trigonometry
real analysis
real analysis unsolved

Can you prove the following formula?

$ \int_0^{\frac{\pi}{2}} \arccos\left(\frac{1}{1\plus{}2\cos x}\right) dx\equal{}\frac{\pi^2}{6}$

http://www.dma.ens.fr/culturemath/maths/pdf/analyse/coxeterInt.pdf

this is Coxeter integral, read the above article

I can't read that. :(

May be an english translation. (Not the whole paper of course.)

It will be useful to study french  :wink:

[quote=Moubinool]http://www.dma.ens.fr/culturemath/maths/pdf/analyse/coxeterInt.pdf
[/quote]
The link you offered for downloading is invalid, so valid links are given below.
https://sos440.tistory.com/attachment/cfile5.uf@1469651549AB9920378BD0.pdf

Quadrature
Magazine de mathématiques pures et épicées
La mathématique ouvre plus d'une fenêtre sur plus d'un monde
Les intégrales de Coxeter
par L.G. Vidiani

[quote=methylbenzene]valid links are given below.
[url=https://sos440.tistory.com/attachment/cfile5.uf@1469651549AB9920378BD0.pdf] see here[/url] [/quote]


geometry
3D geometry
sphere
vector
function
quadratics
calculus

What would be the minimal distance between any point on the sphere $ (x\minus{}2)^2 \plus{} (y\minus{}1)^2 \plus{} (z\minus{}3)^2 \equal{}1$ and any point on the sphere $ (x\plus{}3)^2 \plus{} (y\minus{}2)^2 \plus{} (z\minus{}4)^2 \equal{} 4$

Two [b]spheres[/b]...
Start by drawing a line through their centers.

Hopefully if that doesn't help too much, this will.

Consider "zooming" out of the $ \mathbb{R}^n$ co-ordinate system so far that these spheres, no matter how large look like dots. Essential just points in 3-Space. What would the distance between them be? How would you find that? "Zoom" in and you have the idea with drawing the line through their centers.

The shortet distance between 2 spheres is the distance between the centers of the 2 spheres minus the radius of each sphere.  Therefore, the distance between the centers of the first (call A) and center of second sphere (call B) is the distance of the vector going from A to B, Which is <-5,1,1>.  The distance of this vector is sqrt of 27.  sqrt 27 - 1(radius of first square) - 2(radius of second square) give you the shortest distance between the 2 mentioned sphere.

Just for the heck of it, I tried to see what would happen if I ignored the geometric solution and worked this as a Lagrange multiplier problem. That is:

Minimize $ (x\minus{}a)^2\plus{}(y\minus{}b)^2\plus{}(z\minus{}c)^2$ subject to $ (x\minus{}2)^2\plus{}(y\minus{}1)^2\plus{}(z\minus{}3)^2\equal{}1$ and $ (a\plus{}3)^2\plus{}(b\plus{}2)^2\plus{}(c\minus{}4)^2\equal{}4.$

That's six variables and two constraints. Since the functions are all quadratic, the partial derivatives are all linear.

Conclusion 1: This does work, if you have sufficient patience.

Conclusion 2: I'm not going to reproduce here the work in question.


please vote for taj mahal it is presently in 10th position in the list of new 7 wonders of world by UNESCO,
Each vote counts....

post your vote here [url=http://www.new7wonders.com/index.php]the new seven wonders of world[/url]-The [b]official UNESCO  site[/b]

send in as much friend requests as you can.The [b]voting finishes on july 7th..[/b].. :).

[quote="Padmavathi"]Somehow, I feel that once you've seen some of the AMAZING temples down South(I did only recently!!),the richness in architecture, the sculptures, the very atmosphere...and you compare them to the  acid-rainified Taj Mahal, it really does NOT deserve to even BE in the list of 7 Wonders... :)
That way they should include a lot more of Indian(North AND South) temples in the list...
But, since it's the only Indian monument there, I voted for it and I think we all should do our best and vote as much as we can to help it make the cut... :(
GO TAJ!!! :lol:[/quote]
yes i have been to most south indian temples ....
but i sincerely beleive though they are architeturally the best but in no way is taj mahal less ...just because we destroyed by acid rains doesn't mean it lost it's charm....it still is one of the best in the world.

P.S.-i am no way involved with  the tourism ministry of india :D
everyone can have their own opinion...just that we shouldn't let taj mahal slip out i posted this here :)

Which is exactly why I voted for it!!
Don't worry..I got about 57 people in school to vote for it and the process is still ongoing...but it would be cool if our Ministry of Tourism couls start a 7 Wonders Of India thing...cuz as the slogan goes,
$Incredible !ndia$
But I heard it's in 10th spot...do you think the Taj will make it?? :(

i am optimistic so i think it should make  :D but still it will be a tough competition pyramids,opera house,stonehenge,Acropolis,great wall,Easter islands,timbuktoo..... :rotfl:

More campaigning need to be done.And yes, some temples in the South are glowing examples of architectural virtuosity.
Sathej

I got fifty peple to vote!! :lol:
But I agree with Padmavathi, there should have been some better temples and stuff and more Indian contenders...
Oh well, maybe next time.  :D
Yay Taj!! But I got some wonky mail about it's being a Shiv Temple called Tejo Mahalay...and the mail gave a link to a BBC site!! Weird but would make a good sensational story...but approx. how many years old is it?

great work padmavathi(sorry about delaying this :( ) and ofcourse MissSmartyPants
let's hope we have some good news .... :lol:

thank you all who have voted for the Taj(including me  :D ). We made it  :clap2:  :thumbup:

We made it!
Thaks to everyone who voted.

 :jump:

We made it!!!
Heard some rich guy in Delhi spent Rs.10,000 getting people to vote for it...finally, it's in... :)

Rs. 10k is not that much to pay for something to become a wonder of the world

yes, Taj Mahal made it. If it didn't it would be a travesty.

Taj Mahal Made it...
Now our Tourism Ministry can go back to being it's callous old self and feel proud that it's Done It All without promoting other monuments and temples of great beauty... :)
Now India can go back to be called Land Of The Taj Mahal and be visited only for people to see it without their even BOTHERING to notice the rest of the country...  :)
Now whether people are littering the rest of the country's monuments or scribbling their names in places of cultural value, WHO CARES? After all, the Taj HAS MADE IT...What else do they want? :)
Sometimes, I just wish Taj hadn't won... :)

I agree with Maya...but it would have been a real SHAME if a country as big and culturally predominant as India did not have a single one of the seven wonders...now we have one...six more to go!!  :P  :lol:


let $A$ be matrix $m$x$n$ have range be 1
prove that exist two matrix $B_{mx1}$ and$C_{1xn}$ such that :
   $A=BC$

Let $A = ( a_{1}a_{2}\ldots a_{n})$. Because $rg A = 1$, there exist an $i$ such that for each $j \in \{ 1, 2, \ldots , n \}$ there is a  $\lambda_{j}\in \mathbb{R}$ so that $a_{j}= \lambda_{j}\cdot a_{i}$. Then $B = a_{i}$ and $C = (\lambda_{1}\lambda_{2}\ldots \lambda_{n})$


[quote="stephen38"][color=darkred]Prove that the equation $ f(x)\equiv x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+\frac{3}{4}= 0$ has no real zeros.[/color][/quote]
[color=darkblue][b]Proof (level **).[/b] $ f(-1)\ne 0$ and $ 4(x+1)f(x) =$ $ 4(x+1)(x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1-\frac{1}{4}) =$
$ 4(x^{7}+1)-(x+1) =$ $ 4x^{7}-x+3\equiv g(x)$. Observe that $ f(x) = 0$ $ \Longleftrightarrow$ $ g(x) = 0$ for any $ x\ne-1$.

$ \{\begin{array}{cccc}1\blacktriangleright & x <-1 &\Longrightarrow & 4x^{7}-x+3 < 4x-x+3 = 3(x+1) < 0\ .\\ \ 2\blacktriangleright &-1 < x < 0 &\Longrightarrow & 4x^{7}-x+3 > 4x^{3}-x+3 = (x+1)(4x^{2}-4x+3) > 0\ .\\ \ 3\blacktriangleright & 0 < x\le 3 &\Longrightarrow & 4x^{7}-x+3 = 4x^{7}+(3-x) > 0\ .\\ \ 4\blacktriangleright & 3 < x &\Longrightarrow & 4x^{7}-x+3 = x(4x^{6}-1)+3 > 0\ .\end{array}$[/color]

[quote="stephen38"]Prove that :
$ \left(x^{6}\minus{}x^{5}\plus{}x^{4}\minus{}x^{3}\plus{}x^{2}\minus{}x\plus{}\frac{3}{4}\right)$
 has no real zeros.[/quote]
If $ x\geq1$ or $ x\leq0$ it's obviously true.
Let $ 0 < x < 1.$ Hence, $ x^{6}\minus{}x^{5}\plus{}x^{4}\minus{}x^{3}\plus{}x^{2}\minus{}x\plus{}\frac{3}{4}\equal{}$
$ \equal{}\frac{3}{4}\minus{}x(1\minus{}x)\minus{}x^{3}(1\minus{}x)\minus{}x^{5}(1\minus{}x) \equal{}\frac{3}{4}\minus{}x(1\minus{}x)(1\plus{}x^{2}\plus{}x^{4}) > 0$
because $ \frac{1}{4}\geq x(1\minus{}x) > 0$ and $ 3 > 1\plus{}x^{2}\plus{}x^{4}> 0.$
$ x^{6}\minus{}x^{5}\plus{}x^{4}\minus{}x^{3}\plus{}x^{2}\minus{}x\plus{}\frac{2}{5}> 0$ true too.
I have found a nice elementary proof for this my inequality. :wink:


number theory
prime numbers
algebra open
algebra

Find all bijections $ f: \mathbb{N}\rightarrow \mathbb{N}$ such that if $n|f(m)$, then $ f(n)|m$.

[quote="pco"][quote="Number1"]Find all bijection $ f: \mathbb{N}\rightarrow \mathbb{N}$ such that:

if $ n|f(m)$   then   $ f(n)|m$.[/quote]

$ f(m)|f(m)$ and so $ f(f(m))|f(m)$ and so $ f(f(x))\le x$ $ \forall x$. So, since $ f(f(x))$ is a bijection : $ f(f(x)) \equal{} x$ $ \forall x$

If $ m|n$, then $ m|f(f(n))$, then $ f(m)|f(n)$ and, since $ f(f(x)) \equal{} x$, we get $ a|b$ $ \iff$ $ f(a)|f(b)$

It's then immediate to find a general solution :
Let $ A$ and $ B$ a split of the set of all prime numbers in two equinumerous subsets and $ h(x)$ a bijection from $ A\to B$.

$ f(x)$ is defined as :

$ f(1) \equal{} 1$
$ \forall$ prime $ p\in A$ $ f(p) \equal{} h(p)$
$ \forall$ prime $ p\in B$ $ f(p) \equal{} h^{[ \minus{} 1]}(p)$
$ \forall$ composite number $ > 1$ : $ f(\prod p_i^{n_i}) \equal{} \prod f(p_i)^{n_i}$[/quote]

Sorry, there is a mistake here. I missed some solutions :

The problem is equivalent to identify all bijections $ h(x)$ from the set $ \mathbb P$ of prime numbers into itself such that $ h(h(x)) \equal{} x$ and then we have  $ f(\prod p_i^{n_i}) \equal{} \prod h(p_i)^{n_i}$

And, to identify all $ h(x)$, we must split $ \mathbb P$ in [u][b]three [/b][/u]subsets $ A,B,C$ with $ A$ and $ B$ equinumerous and $ b(x)$ a bijection from $ A\to B$. Then :
$ \forall p\in A$ : $ h(p) \equal{} b(p)$
$ \forall p\in B$ : $ h(p) \equal{} b^{[ \minus{} 1]}(p)$
$ \forall p\in C$ : $ h(p) \equal{} p$  (I forgot this)


calculus
integration
trigonometry
parameterization
complex analysis

show the following... :D

$ \textbf{(i)}\;\int_{0}^{2\pi}\;e^{\cos \theta}\cdot\sin\; (\sin \theta)\;\;\textbf d\theta\;=\;\boxed{2\pi}$

$ \textbf{(ii)}\;\int_{0}^{2\pi}\;e^{\cos \theta}\cdot\cos\; (\textbf k\theta-\sin \theta)\;\;\textbf d\theta\;=\;\boxed{\frac{2\pi}{\textbf k!}}$

$ \textbf{(iii)}\;\int_{0}^{2\pi}\;e^{\sin \textbf k\theta}\cdot\sin\; (\theta-\cos \textbf k\theta)\;\;\textbf d\theta\;=\;\boxed{0}$

$ \textbf{(iv)}\;\int_{0}^{2\pi}\;\frac{e^{\cos \theta}\cdot\sin\; (\sin \theta)}{5-4\sin \theta}\;\;\textbf d\theta\;=\;\boxed{\frac{2\pi}{3}\cdot\sin 1}$

$ \textbf{(i)}\int_{0}^{2\pi}\;e^{cos\theta}\cdot \sin\;(\sin\theta)\;\;\textbf d\theta \;=\; Im \int_{0}^{2\pi}\;e^{e^{i\theta}}\textbf d\theta = 2\pi$

(II)

$ \oint_{|z|=1}\frac{e^{z}}{z^{k+1}}dz = \frac{2\pi i}{n!}$ by Cauchy's Integral Formula.

Using the parametrization $ \gamma(\theta) = e^{i \theta}$ from $ 0$ to $ 2\pi$ gives the desired result.

(iii) is similar to (ii):

The integral equals:  $ \Im \int_{0}^{2 \pi}e^{sin(k \theta)+i\theta-i\cos(k\theta)}\hspace{1 mm}d\theta = \Im \int_{0}^{2 \pi}e^{i\theta}e^{-i e^{ik\theta}}\hspace{1 mm}d\theta$.

Now let $ z=e^{i \theta}$:

$ \Im \oint e^{-i z^{k}}(-i) \hspace{1 mm}dz =0$  $ \hspace{3 mm}k \in \mathbb{N}$

For (iv), the correct answer is:

$ \int_{0}^{2 \pi}\frac{e^{\cos \theta}\cdot \sin(\sin \theta)}{5-4\sin \theta}\hspace{1 mm}d\theta = \frac{2 \pi}{3}\cdot \sin(\frac{1}{2})$  :wink:


It's not very surprising that $ (\minus{}1)^2 \equal{} 1^2$ yet $ \minus{}1 \ne 1$.

1) Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=312288[/url]

2) Do not make posts with such ridiculous titles.


With the following conditions,
\[ w\plus{}x\plus{}y\plus{}z\equal{}9\] and \[ w^{2}\plus{}x^{2}\plus{}y^{2}\plus{}z^{2}\equal{}21\],
for [b]real[/b] numbers $ w,x,y,z$.
Determine the maximum value of $ wxyz$.
 :arrow:

hello, with the method of lagrange Multipliers i have found that $ wxyz\le24$ for all real $ w,x,y,z$ with
$ w\plus{}x\plus{}y\plus{}z\equal{}9$ and $ w^2\plus{}x^2\plus{}y^2\plus{}z^2\equal{}21$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, with the method of lagrange Multipliers i have found that $ wxyz\le24$ for all real $ w,x,y,z$ with
$ w \plus{} x \plus{} y \plus{} z \equal{} 9$ and $ w^2 \plus{} x^2 \plus{} y^2 \plus{} z^2 \equal{} 21$.
Sonnhard.[/quote]
Can you post your solution my friend cause i'm trying to learn Langrange multipliers but i think i miss something :blush:

Thank you!
But I prefer the classical way more. Could anyone show me how to do that?  I would be pleased.
 :)


advanced fields
advanced fields unsolved

Trying to think of an order embedding of Z^3 into Q (the rationals)

And an order embedding from w^3 into Q  where w is the set of natural numbers.

what does "embedding" mean here ? just an injective map ?

An order-preserving injective map.  Presumably, you're taking the standard product orders on ${\bf Z}^{3}$ and ${\bf N}^{3}$.  I can describe a process one might use (i.e., show it is possible): just take planes of the form $x+y+z = n$ and imbed their lattice points into the interval $[n, n+1)$ in any way whatsoever.  This doesn't give an obvious nice expression for the embedding, however.

Given any countable ordered set S, there is an order embedding f of S into Q : Fix $q_{1},q_{2},q_{3},\cdots$ an enumeration of S. Inductively choose a value $f(q_{n})\in Q$ so that $f: \{q_{1},q_{2},\cdots,q_{n}\}\to Q$ is an order embedding for each $n$. Then $f: S\to Q$ is an order embedding.


search
number theory proposed
number theory

Let $x$ and $y$ be positive integers such that$3x^2 + x = 4y^2 + y$. Prove that
$x - y$is a square.

http://www.mathlinks.ro/Forum/viewtopic.php?t=32858
http://www.mathlinks.ro/Forum/viewtopic.php?t=38983
http://www.mathlinks.ro/Forum/viewtopic.php?t=38350
http://www.mathlinks.ro/Forum/viewtopic.php?t=30699

  darij

1. I'm not there to search for problems that have been posted $101$ times.
2.: e.g.: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=square+x-y&t=38983


algebra proposed
algebra
BritishMathematicalOlympiad

( BMO 2004)

Set  a, b, c are reals with sum zero.
Show that a^3 + b^3 + c^3 > 0 iff a^5 + b^5 + c^5 > 0

I already solved this on on ML, but I do not find it aymore.
The trick is that for $a,b,c$ such that $a+b+c=0$, we have :
$\frac {x^2+y^2 + y^2} 2 \cdot \frac {x^3 + y^3 + z^3} 3 = \frac {x^5 + y^5 + z^5}5$ from which the result follows easily.

Piere.


For $n$ a positive integer, denote by $P(n)$ the product of all positive integers divisors of $n$. Find the smallest $n$ for which

\[ P(P(P(n))) > 10^{12} \]

Using known formula :  $P(n) = n^{\tau (n)/2}$ with $\tau (n)$ being the number of positive divisors of $n$

If $p$ is prime $P^{(k)}(p) = p$ ... no comment :huh:

If $n = p^a$ with $p$ prime $P(n) = p^{a(a+1)/2}$ so $P^{(3)}(4)$ is not big enough.

$n = 6$ is the minimum. $P^{(3)}(6) = 2^{450} \cdot 3^{450}$ !


$ m_a, m_b, m_c$는 삼각형의 세 중선의 길이이고, $ r_a, r_b, r_c$는 삼각형의 세 방접원의 길이이다.

$ \frac{r_a r_b }{m_a m_b }\plus{}\frac{r_b r_c }{m_b m_c }\plus{}\frac{r_c r_a }{m_c m_a }\ge3$임을 증명하시오.

[hide]
먼저, 첫 번째 항에 대해서 생각해 봅시다.
일단, 삼각형의 각 변의 길이를 a=y+z, b=z+x, c=x+y로 치환합시다.
r_a=2[ABC]/(b+c-a)=[ABC]/x는 잘 아실 것이고,
그러면 r_ar_b=[ABC]^2/(xy)=(xyz(x+y+z))/(xy)=z(x+y+z)를 얻습니다.
이제, 사각형 M_A B A M_B에 톨레미의 부등식을 적용합시다.
그러면 m_am_b는 (a/2)(b/2)+c(c/2)=ab/4+c^2/2 보다 작거나 같습니다.
이것을 첫 번째 항에 대입하면
(r_ar_b)/(m_am_b)>=z(x+y+z)/(ab/4+c^2/2)=4z(x+y+z)/(2x^2+2y^2+z^2+5xy+zx+yz)이므로 이것의 cyclic sum이 3 이상임을 보이면 됩니다.
즉, (x+y+z) 곱하기 (cyclic sum of z/(2x^2+2y^2+z^2+5xy+zx+yz))>=3/4임을 보이면 됩니다.
코시-슈바르츠에 의해
(x+y+z) 곱하기 {(cyclic sum of z(2x^2+2y^2+z^2+5xy+zx+yz))(cyclic sum of z/(2x^2+2y^2+z^2+5xy+zx+yz))}>=(x+y+z)^3이므로,
(x+y+z) 곱하기 (cyclic sum of z/(2x^2+2y^2+z^2+5xy+zx+yz))>=(x+y+z)^3/(cyclic sum of z(2x^2+2y^2+z^2+5xy+zx+yz))입니다. 한편,
cyclic sum of z(2x^2+2y^2+z^2+5xy+zx+yz)=입니다.
즉, 이제
4(x+y+z)^3>=3(3(xy^2+yz^2+zx^2)+3(x^2y+y^2z+z^2x)+(x^3+y^3+z^3)+15xyz)임을 보이면 됩니다.
한편,
(x+y+z)^3=(x^3+y^3+z^3)+3(x^2y+y^2z+z^2x)+3(xy^2+yz^2+zx^2)+6xyz이고,
4(x+y+z)^3-3(3(xy^2+yz^2+zx^2)+3(x^2y+y^2z+z^2x)+(x^3+y^3+z^3)+15xyz)
=(x^3+y^3+z^3)+3(xy^2+yz^2+zx^2)+3(x^2y+y^2z+z^2x)-21xyz인데,
산술-기하에 의해x^3+y^3+z^3와 xy^2+yz^2+zx^2와 x^2y+y^2z+z^2x은 모두3xyz보다 크거나 같습니다.
이를 대입하면
(x^3+y^3+z^3)+3(xy^2+yz^2+zx^2)+3(x^2y+y^2z+z^2x)-21xyz>=0을 얻을 수 있습니다.
따라서 문제의 명제가 증명되었고, 등호조건은 마지막의 산술 기하에 의해 x=y=z, 즉 a=b=c로, 정삼각형일 때 성립합니다.
참고로 정삼각형에서는 아까 사용한 톨레미의 부등식이 등호로 사용되므로 성립한다는 것을 알 수 있습니다.
[/hide]


calculus
integration
calculus computations

Using integration by parts, find an expression for
$\int{\frac{e^{x}}{x+1}}\,dx$

Hence find
$\int{e^{x}\left(\frac{1}{x+1}+ln|{x+1}|\right)}\,dx$

The first integral can not be calculated by elementary methods. The second is done by Shyong.

  babis

Ignoring the intergration by parts request then

$\int{e^{x}(\frac1{x+1}+\ln|{x+1}|)\,dx}= e^{x}\ln|{x+1}|$

Does that mean $\int{e^{x}\ln(x+1)}\,dx = 0$?

It's because of what happens at $x=-1$ I think.

edit: Sorry, I realized what I said is a little unclear. I meant you can't assume that other integral is 0 because of what happens at $x=-1$.

The point of this question was just to get to the final result, and by expression I simply meant the inclusion of an integral on the RHS which could enable the finding of the more complex-looking integral requested at the end.  Naturally, submissions here are correct  :lol:


Given cards numbered from 1 to  $ n\geq 3$ in a box. When you mix up these cards and randomly draw a card, let  $ x_1$ be the number written on the card. Then you put the card back in the box and draw another card, ,let  $ x_2$ be the written on this card. Doing the similar procedure 2 more  times, let  $ x_3,\ x_4$be the numbers drawn the third and fourth time respectively.

(1) For $ n \equal{} 12$, find the probability such that $ x_1 < x_2$.

(2) For $ n \equal{} 12$, find the probability such that $ x_1 < x_2\leq x_3$.

(3) Let $ \frac {f(n)}{n^4}$ be the probability such that $ x_1 < x_2 < x_3$ and $ x_3 > x_4$. Find $ f(n)$ in terms of $ n$.

@brightzhu Thank you for helping me to translate the context.

[hide="1"]For $ n$ cards, there are $ n^2$ ways to pick 2 cards, and for $ n$ ways, we will have $ x_1 \equal{} x_2$. Otherwise, half the cases will have $ x_1 < x_2$, and the other half will have $ x_2 > x_1$. There are $ \frac {1}{2}(n^2 \minus{} n)$ total good ways, and $ n^2$ total ways, and for $ n \equal{} 12$, we get $ \frac {66}{144} \equal{} \frac {11}{24}$.
(Alternatively, there are $ \binom{n}{2}$ ways to pick 2 distinct numbers, and each of these pairs corresponds to one way to pick $ x_1 < x_2$.)[/hide]
[hide="2"]If $ x_2 \equal{} x_3$, then we can proceed as in part (1), finding $ \binom{n}{2}$ good ways to pick the cards. If $ x_2 < x_3$, then we similarly will have $ \binom{n}{3}$ other good ways to pick the cards. There are $ n^3$ ways to pick total, so the probability is $ \frac {\binom{n}{2} \plus{} \binom{n}{3}}{n^3}$, and for $ n \equal{} 12$, this is $ \frac {66 \plus{} 220}{1728} \equal{} \frac {143}{864}$.[/hide]

[hide]
1) the probability that $ x_1\equal{}x_2$ is $ \frac{1}{n}$ and the probability that $ x_1<x_2$ is the same as the probability that $ x_2<x_2$
$ \Rightarrow 1\equal{} \frac{1}{12} \plus{} 2p \Longrightarrow p \equal{} \frac{11}{24}$

2) $ P(x_1\neq x_2 \neq x_3) \equal{} \frac{n\minus{}1}{n}\cdot\frac{n\minus{}2}{n}$
And again we have equal probability for every permutation $ x_1,x_2,x_3$
$ \Rightarrow 3!p \equal{} \frac{11\cdot 10}{12^2} \Longrightarrow p \equal{} \frac{55}{432}$

3) From questions 1 and 2  we have $ P(x_3>x^4)\equal{}\frac{1}{2}\minus{}\frac{1}{2n}$ and  $ P(x_1< x_2 < x_3) \equal{}\frac{(n\minus{}1)(n\minus{}2)}{6n^2}$
$ \Rightarrow P(x_3>x^4 \plus{} x_1< x_2 < x_3) \equal{} \frac{n\minus{}1}{2n}\cdot \frac{(n\minus{}1)(n\minus{}2)}{6n^2}$

$ \Rightarrow f(n) \equal{} \frac{1}{12}n(n\minus{}1)^2(n\minus{}2)$[/hide]


Find all real and nonreal numbers $a$ and $b$ such that $a^b=b^{a}+1$

[hide]For every $a$, there is at least one $b$ (namely zero) that satisifies the equation.  Those trivial solutions aside, many solutions (of which there are an infinite number) will involve transcendental numbers, and so approximations will be the best we can do.[/hide]You might want to revise the question a little.  For example, the answer is more straightforward for integral $a$ and $b$.

P.S.  Sorry, Meghrabi! :)

Can you hide your solutions in the future ?

Thanks

[hide]There are an infinite number of trivial solutions again ($b=0$), and again transcendental numbers are going to be involved.  If we were just dealing with real numbers, there would only be a finite number of solutions for any given $a$ or $b$, but since we aren't, that won't always be the case (try $b=1$).[/hide]


analytic geometry
graphing lines
slope
geometry
geometric transformation
ratio
complex numbers

Let $w_1, w_2, \dots, w_n$ be complex numbers.  A line $L$ in the complex plane is called a mean line for the points $w_1, w_2, \dots, w_n$ if $L$ contains points (complex numbers) $z_1, z_2, \dots, z_n$ such that
\[ \sum_{k = 1}^n (z_k - w_k) = 0. \]
For the numbers $w_1 = 32 + 170i$, $w_2 = -7 + 64i$, $w_3 = -9 +200i$, $w_4 = 1 + 27i$, and $w_5 = -14 + 43i$, there is a unique mean line with $y$-intercept 3.  Find the slope of this mean line.

[hide]Let $ z{}_k{}_x$ be the x coordinate (real part) of $ z_k$, $ z{}_k{}_y$ be the y coordinate (imaginary part) of $ z_k$.

$ \sum_{k=1}^5z_k-\sum_{k=1}^5w_k=0$, and since we are given all $ w_$, we find that $ \sum_{k=1}^5z_k=3+489i$. So $ \sum_{k=1}^5z{}_k{}_x=3$, $ \sum_{k=1}^5z{}_k{}_y=504$.

Given that the y-intercept is 3, and letting the slope be m, for all k, $ z{}_k{}_y-3=mz{}_k{}_x$. Taking the sum of this expression $ \sum_{k=1}^5z{}_k{}_y-15=m\sum_{k=1}^5z{}_k{}_x \implies 504-15=3m \implies m=163$.[/hide]

[quote=Phelpedo][hide] Relabel the plane so that the line passes through the origin. This is a translation by $ (0, \minus{} 3)$, so we simply decrease the imaginary parts of the points by $ 3$. Then $ \Re(z_1 \plus{} z_2 \plus{} z_3 \plus{} z_4 \plus{} z_5) \equal{} \Re(w_1 \plus{} w_2 \plus{} w_3 \plus{} w_4 \plus{} w_5) \equal{} 3$ and $ \Im(z_1 \plus{} z_2 \plus{} z_3 \plus{} z_4 \plus{} z_5) \equal{} \Im(w_1 \minus{} 3 \plus{} w_2 \minus{} 3 \plus{} w_3 \minus{} 3 \plus{} w_4 \minus{} 3 \plus{} w_5 \minus{} 3) \equal{} 489$. But the ratio of the imaginary parts to the real parts of each $ z_k$ is constant and equal to the slope, so this is simply $ \frac {489}{3} \equal{} 169$. [/hide][/quote]

$ \frac {489}{3}=163$, not $169$.

[quote=muti66]Can someone please tell me how the y-intercept can be 3 in the complex plane? That doesn't make sense...[/quote]

[quote="muti66"]Isn't the "y-axis" the complex axis?[/quote]

Right. So if you had the number $4 + 3i$, it would have an x-coordinate of $4$ and a y-coordinate of $3$.

oh, I always thought of it as coordinate of 3i. :p

Sorry for reviving, but I was playing around with this problem and it seemed to me that the mean line always passes through the average of all the $w$ points. How would I prove or disprove this?

Well, if we take the original definition and divide everything by $n,$ we know that the average of the $z$ minus the average of the $w$ points is $0.$ But the average of a bunch of points on a line is also on the line, so the average of the $z$ is on the line. Thus, the average of the $w$ is on the line.

This means $z_1+z_2+z_3+z_4+z_5=3+504i$. Write as points on the real plane, for simplicity. From the $y$ intercept condition, the points are $(a,pa+3), (b,pb+3), (c,pc+3), (d,pd+3), (e,pe+3)$. Then $p(a+b+c+d+e)+3 \cdot 5=504$ and $a+b+c+d+e=3$, so the answer is $p=\frac{504-3 \cdot 5}{3} = \boxed{163}$.


geometry
3D geometry
sphere
advanced fields
advanced fields unsolved

[quote="Mr.Doe"]Let $ X,Y$ Hilbert's space $ A\in B(X,Y)$. Prove that then it's equivalent:

(a) $ A\in K(X,Y)$
(b) $ A^*\in K(Y,X)$
(c) $ A^*A\in K(X)$.

$ A^*$ denotes adjoint operator.[/quote]

1. $ A$ is compact if $ \overline{A(\overline{B_X})}\subset Y$ is compact, where $ B_X$ is the unit sphere in $ X$.

2. $ A$ is compact if for every $ x_n\subset X$ bounded sequence $ Ax_n\subset Y$ has a Cauchy subsequence.

3. $ \overline{A^*(\overline{B_Y})}\subset \overline{B_X}\implies A(\overline{A^*(\overline{B_Y})})\subset A(\overline{B_X})$, but $ A$ compact $ \overline{AA^*(\overline{B_Y})}$ compact, so $ AA^*$ compact.

$ a)\implies b)$
Let $ y_n\subset Y$ a bounded sequence.
$ AA^*$ compact so $ (AA^*y_n)$ has a convergent subsequence $ AA^*y_{n_k}$

$ \parallel{}A^*y_{n_j}\minus{}A^*y_{n_l}\parallel{}^2\equal{}\parallel{}A^*(y_{n_j}\minus{}y_{n_l})\parallel{}^2\le\parallel{}A(A^*(y_{n_j}\minus{}y_{n_l}))\parallel{}\cdot\parallel{}y_{n_j}\minus{}y_{n_l}\parallel{}$
but $ AA^*y_{n_k}$ Cauchy, and $ y_{n_k}$ bounded sequences, so  $ A^*y_{n_k}$ is a Cauchy sequence (2.) $ \implies$ $ A^*$ compact.

$ b)\implies c)$ The same demonstration like 3.


$ f : \mathbb{R} \to \mathbb{R} \\ f(f(x)) \equal{} x^2 \minus{} 2$

[quote="mathema*"]
PS: f(x)#x
[/quote]
If you mean that $ f(x) \equal{} x$, then it's obviously wrong - $ x \equal{} x^2 \minus{} 2$ doesn't hold for all real $ x$.
However, this problem is from an unfinished competition - [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=265021[/url] and we shouldn't discuss it here  :)

[quote="BG Yoda"]However, this problem is from an unfinished competition - [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=265021[/url] and we shouldn't discuss it here  :)[/quote]Sorry, I didn't know. That problem was posted without solution on one of polish math boards, nobody solved it yet there, so I was curious about solution.

But,It's an old problem in AMM,problem E984

so,can we discuss it here??

Sure. This also appeared in Tournament of Town 1996. It is already very famous.

we can prove that $ f(\minus{}x)\equal{}f(x)$ pair function

Let $ f_0(y)\equiv y, f_{k\plus{}1}(y)\equal{}f_k(y)^2\minus{}2\equiv f_1^{(k\plus{}1)}(y)\equiv f_k(f_1(y)\equiv f_k(y^2\minus{}2)$.
Obviosly $ f_k(y),k\equal{}0,1,...$ are solutions.

It's an old problem and I have seen it's solution long ago. As fas as I remember we should take a look at it's fixed points and the solution is not hard. ;)


Find all sets of intergers that solves $ 3^m\minus{}7^n\equal{}2$.

$ m\equal{}2$ and $ n\equal{}1$ :  only solutions... i think.....


Alguien sabe donde puedo encontrar examenes anteriores de la Olimpiada Peruana de Quimica o de otros paises :?:  :huh:

yo encontre algunos examenes en la pagina de la catolica ,

http://www.pucp.edu.pe/

http://www.pucp.edu.pe/quimica/sqp-olimpiadas.htm

bueno eso hasta el momento y creoq va a llevarse la primera fase este 9 de septiembre ( mi primera participacion en esta olimpiada  :D )


A 12'' x 18'' picture is surrounded by a 3'' border. What is the ratio of the number of square inches in the area of the picture itself to the number of square inches in the area of the picture and border together? Express your answer as a fraction.


What is the ratio of the numerical value of the area, in square units, of an equilateral triangle of side length 4 units to the numerical value of its perimeter, in units? Express your answer as a common fraction in simplest radical form.

It's well-known that the area of an equilateral triangle with side length $ a$ has area $ a^{2}\frac{\sqrt{3}}{4}$.  Then $ \frac{A}{P}\equal{}\frac{a^{2}\sqrt{3}}{4\cdot 3a}\equal{}\frac{a}{4\sqrt{3}}$.  Plugging in $ a\equal{}4$, we get that the ratio is $ \frac{\sqrt{3}}{3}$.


Prove that in a triangle $ ABC$, $ \cos^{2}A\plus{} \cos^{2}B \plus{} \cos^{2}C \geq \frac{3}{4}$.

substitute :
$ cos(A) \equal{} \frac {b^2 \plus{} c^2 \minus{} a^2}{2bc}$ and similarly..
the inequality is equivalent to :
$ \sum_{cyclic}\frac {(b^2 \plus{} c^2 \minus{} a^2)^2}{b^2c^2} \ge 3$
By C-S,
$ \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2} \ge 3$
which is equivalent to -
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} 2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \ge 3(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)$
or ,
$ a^4 \plus{} b^4 \plus{} c^4 \ge (a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)$
which is trivial.

Hi geniusbliss. I don't see how you get from your first inequality to your second one. (or where C-S comes in). Can you explain? Thanks.

Here my solution:
$ f(A,B,C)\equal{}\cos ^2A\plus{} \cos ^2 B \plus{}\cos^2 C$
$ A\plus{}B\plus{}C\equal{}180^{\circ}$ Hence
$ 4(f(A,B,C)\minus{}\frac{3}{4})\equal{}4(\frac{1\plus{}\cos 2A }{2}\plus{}\frac{1\plus{}\cos 2B }{2}\plus{}\frac {1\plus{}\cos 2C }{2}\minus{}\frac {3}{4})\equal{}2\cos2A \plus{}2\cos2B \plus{}2\cos2C\plus{}3\equal{}4\cos(A\plus{}B)\cos(A\minus{}B)\plus{}4\cos^2 (A\plus{}B)\plus{}1\equal{}(2\cos(A\plus{}B)\plus{}\cos(A\minus{}B))^2 \plus{}1\minus{}\cos^2(A\minus{}B)\ge 0$
$ f(A,B,C)\ge \frac{1}{4}$
With equality $ \cos(A\minus{}B)\equal{}1,\cos(A\plus{}B)\equal{}\minus{}\frac{1}{2}$
That is  $ A\equal{}B\equal{}C\equal{}60^{\circ}$

__________________________________________
Azerbaijan Land of the Fire  :rotfl:

let $ b^2\plus{}c^2\minus{}a^2 \equal{} x$ and so on..
u have -
$ \sum \frac{x^2}{b^2c^2} \ge \frac{(x\plus{}y\plus{}z)^2}{a^2b^2\plus{}b^2c^2\plus{}c^2a^2}$ by Cauchy-Schwarz in engel form or titu's inequality or watever..

substitute in the beginning the values for $ cos(A)$ and u see that 4 in the denominator of the RHS and the LHS gets cancelled.

[quote="AndrewTom"]Prove that in a triangle $ ABC$, $ \cos^{2}A \plus{} \cos^{2}B \plus{} \cos^{2}C \geq \frac {3}{4}$.[/quote]

Since thjis identity is true , solution of which you can find here http://www.mathlinks.ro/viewtopic.php?t=321070

 \[ \cos^2 \alpha\plus{}\cos^2\beta\plus{}\cos^2\gamma\plus{}2\cdot \cos \alpha \cdot \cos \beta \cdot \cos \gamma\equal{}1\]

the inequality becomes
\[ 1\minus{}2\cdot \cos \alpha \cdot \cos \beta \cdot \cos \gamma \ge \frac{3}{4}\]
which is equivalent with
\[ \cos \alpha \cdot \cos \beta \cdot \cos \gamma \leq \frac{1}{8}\]
which is true , therefore we're done. :)


LaTeX
symmetry
number theory
relatively prime

Determine all integers $x$ and $y$ that satisfy

$(x+2)^{4}-x^{4}= y^{3}$.

Let $y=2u$.
$(x+2)^{4}-x^{4}=8(x+1)(x^{2}+2x+2)=y^{3}=8u^{3}\iff (x+1)(x^{2}+2x+2)=u^{3}$

$\boxed{\text{Case}\ 1 \ \ (x+1=1)}$
$\left \{ \begin{array}{lll}x+1=1 \\ x^{2}+2x+2=u^{3}\end{array}\right.$
$\iff \left \{ \begin{array}{lll}x=0 \\ x^{2}+2x+2=u^{3}\end{array}\right.$
$\iff 0^{2}+2\cdot 0+2=2=u^{3}$
There is not a solution satisfying a meaning of a question.

$\boxed{\text{Case}\ 2 \ \ (x+1=u)}$
$\left \{ \begin{array}{lll}x+1=u \\ x^{2}+2x+2=u^{2}\end{array}\right.$
$\iff \left \{ \begin{array}{lll}x=u-1 \\ x^{2}+2x+2=u^{2}\end{array}\right.$
$\iff (u-1)^{2}+2(u-1)+2=u^{2}+1=u^{2}$
$\iff 1=0$
There is not a solution satisfying a meaning of a question.

$\boxed{\text{Case}\ 3 \ \ (x+1=u^{2})}$
$\left \{ \begin{array}{lll}x+1=u^{2}\\ x^{2}+2x+2=u \end{array}\right.$
$\iff \left \{ \begin{array}{lll}x=u^{2}-1 \\ x^{2}+2x+2=u \end{array}\right.$
$\iff (u^{2}-1)^{2}+2(u^{2}-1)+2=u^{4}+1=u^{2}$
$\iff u^{2}(u+1)(u-1)=-1$
There is not a solution satisfying a meaning of a question.

$\boxed{\text{Case}\ 4 \ \ (x+1=u^{3})}$
$\left \{ \begin{array}{lll}x+1=u^{3}\\ x^{2}+2x+2=1 \end{array}\right.$
$\iff \left \{ \begin{array}{lll}x+1=u^{3}\\ x=-1 \end{array}\right.$
$\iff (x, u)=(-1, 0)$

Therefore, $(x, u)=(-1, 0) \iff \left (x, \frac{y}{2}\right )=(-1, 0)$

Thus, $\boxed{\boxed{(x, y)=(-1, 0)}}$

[b]Kouichi Nakagawa[/b], your solution and your $\LaTeX$ are very nice!  :lol:
[hide="Another solution"]
Let $y=2z,$ thus we have: $(x+2)^{4}-x^{4}=y^{3}\iff x^{3}+3x^{2}+4x+2=z^{3}$

$1.\blacktriangleright\quad x\ge 0\implies (x+1)^{3}< z^{3}<(x+2)^{3}$
There is not a solution satisfying a meaning of a question.

$2.\blacktriangleright\quad x\le-2\implies x^{3}<z^{3}<(x+1)^{3}$
There is not a solution satisfying a meaning of a question.

$3.\blacktriangleright\quad x=-1\implies y=0$

Hence $\boxed{\boxed{(x,\,y)=(-1,\,0)}}$
[/hide]

kouichi -

Not very clear how one  can reduced the problem to these three cases. (Since   $u$  need not be  prime, it is not correct   to assume that those are the  only three cases.  Think of   $a b = 4^{3}= 2 \cdot 32$ and $a$ need  not  be $1$ or  $4^{2}$ or $4^{3}$

Your  "case 2 "     makes little sense in other ways, since the two factors on LHS are coprime to each other  while $u$ and $u^{2}$ are [b]not [/b] coprime to each other. (Ignoring trivial cases when a  factor is zero and other is 1).....

[hide="FWIW"]FWIW
Consider symmetry: let $z=x+1$ , the expression now becomes:  $8z(z^{2}+1)=y^{3}$
and it is easy to see,  the only solution is when z=0)[/hide]

[quote="Gyan"]
FWIW
Consider symmetry: let $z=x+1$ , the expression now becomes:  $8z(z^{2}+1)=y^{3}$
and it is easy to see,  the only solution is when z=0)[/quote]

Thought I might just elaborate on that.

Notice that $z(z^{2}+1)= (\frac{y}{2})^{3}$ now as $z$ and $z^{2}+1$ are relatively prime let: $z=a^{3}$ and $z^{2}+1 = b^{3}$. Now we have:

$a^{6}+1 = b^{3}$. Evidently $a=0$ and so $z=0$ and $z^{2}+1 = 1$. which gives us the only solutions $(x,y) = (-1,0)$.


Do you know who won the science bowl in MD? Was it Blair?

[quote="mdk"]What's your point?[/quote]

his point is...

WHY ARE YOU POSTING IN RANDOM OTHER FORUMS?

 :rotfl:

because I want to know who won from those 2 states. Is there anything wrong with that?

Blair wins...like they always do...

http://insidemc.montgomerycollege.edu/showStory.php?id=8040

btw do u know who was on the takoma team by any chance? im just wondering?

Sam Zbarsky, Charlie Pasternak, Daniel Zmir, Avikar Periwal, Myles Wagner


Let $a,b,c>0$ and$a+b+c=3$.
Find max and min of :$P=\sum \frac{a}{ab+1}=\frac{a}{ab+1}+\frac{b}{bc+1}+\frac{c}{ca+1}$.

$P\geq\frac{3}{2}.$
This is vasc's inequality. To proof use another vasc's inequality: $(x^2+y^2+z^2)^2\geq3(x^3y+y^3z+z^3x).$  ;)

Vasc indeed: http://www.mathlinks.ro/Forum/viewtopic.php?t=21689

  darij

Thank you.
Solution is very nice.
$x+y+z=3$ Prove that:
$\sum \frac{x}{xy+1}\ge \frac{3}{2}$.
 :) Solution of Vasc and Manlio:
$\frac{x}{xy+1}=x-\frac{x^2y}{xy+1}\to \sum \frac{x}{xy+1}=\sum (x-\frac{x^2y}{xy+1})=3-\sum \frac{x^2y}{xy+1}$.
We prove that $\sum \frac{x^2y}{xy+1}\le \frac{3}{2}$
Indeed, we have $xy+1\ge 2\sqrt{xy}$(Cauchy).
So $\frac{x^2y}{xy+1}\le \frac{x^2y}{2\sqrt{xy}}=\frac{1}{2}(\sqrt{x})^3\sqrt{y} \to$ we need prove that:$\LARGE P=\sum (\sqrt{x})^3\sqrt{y}\le 3$.
It's obvious by Vasc'ineq:
$(x^2+y^2+z^2)^2\ge 3(x^3y+y^3z+z^3x)$
because $P\le \frac{(x+y+z)^2}{3}=3$.


When Tracey's defective clock shows that 1 hour has gone by, a real clock shows 55 min gone by.  Tracey's clock and a real clock now both show 12:00.  Ignoring a.m/p.m, how many real hours go by before the two clocks show the same time again?

[hide] Every 55 real minutes, the clock adds 5 extra minutes.  So, on an average, every 11 minutes, the clock adds an extra minute.  We want the clocks to show the same time, so 12 hours worth of extra minutes must have gone by.  Therefore, 720 minutes must have gone by.  Therefore, 11*720 real minutes would have gone by for the clock to add an extra 12 hours.  This is 7920 minutes, which is 132 hours.  [/hide]

i remember this problem from MMM
our team got it wrong, thanks to me, Ubemaya, and Quatto.
now i know how to do it

In case Quatto's explanation is confusing (as they often are):

The ratio of the amount of time Tracey's clock covers to the amount of real time that goes by is 12 to 11. So $\frac{12x}{11}$ is the amount of time that Tracey's clock shows compared to x, the amount of real time. So in 11 real hours, Tracey's clock goes 12 hours, a difference of 1 hour. Multiply this by 12 to get 132, and the clocks will have a difference of 132 hours.

How many of you got 144 in Tracey-hours but forgot to make it real hours?

Our team did...

What school do you go to? What'd you get at M^3?


modular arithmetic
number theory
least common multiple

x = 3u+2 = 5v+3 = 7w+2 (x, u, v, w are natural numbers)
what is the smallest x?

need help!!

Meh I see it as modular arithmetic

\begin{eqnarray*} x & \equiv & 2\pmod{3},3\pmod{5},2\pmod{7} \\
x - 2 & \equiv & 0\pmod{3},1\pmod{5},0\pmod{7} \\
x - 2 & \equiv & 0\pmod{21},1\pmod{5} \\
x - 2 & = & 21 \\
x & = & 23 \end{eqnarray*}

[quote="MrRubix"]Wow.
[hide]The easiest way to do this would be to just list the first couple numbers
U=1 through 7: 5, 8, 11, 14, 17, 20, 23.
V=1 through 4: 8, 13, 18, 23.
W= 1 through 3: 9, 16, 23.
We see that 23 is in all of these, so [u]23 is the smallest x[/u], while u=7, v=4, and w=3 [/hide][/quote]

thanks!

[quote="hunter34"]Meh I see it as modular arithmetic[/quote]
Same here. And the title also reminds me of Chinese remainder theorem (you can do it, but it's more tedious than hunter34's way)

that was swift hunter34, it would have taken me a while

Another reason for y'all to learn modular arithmetic now.
 :)

haha modular arithmatic is fun and easy if you get the hang of it,
 but any way...
you could use the substitution method if u dont kno mods

i know it will take quite a lot of time and effort, but it is worth it  :roll:

er... sry  but im not very good at maths but i was wondering....... If we get the LCM of 7 and 3 we would get 21. Wouldnt that be correct? because  u=7 and w=3 . 5v+3 = 23 , 5v= 20 and  v= 4. So its still a natural number

lol thats y I dun understand wat are u typing because Malaysias maths form 1 dosent teach that type of LCM method......and moreover Im jz a normal form 3 student so i dun understand wat were u typing.

[quote="Nightmare124"]wth? that question is messed up[/quote]

no it isn't.  btw, don't post these one line posts that don't help much.


prove that 2 $ cot A \equal{} cot \frac {A}{2}\minus{} tan\frac {A}{2}$

[hide="Solution"]

By the double-angle formula for $ \tan$,

$ \tan A = \frac{2\tan \frac{A}{2}}{1-\tan^{2}\frac{A}{2}}$

$ \frac{2}{\tan A} = \frac{1-\tan^{2}\frac{A}{2}}{\tan\frac{A}{2}}$

${ \frac{2}{\tan A} = \frac{1}{\tan\frac{A}{2}} - \tan \frac{A}{2}}$

$ 2\cot A= \cot \frac{A}{2} - \tan \frac{A}{2}$

[/hide]


I am a student from Canada and I am learning Beginner Spanish this year. We have to do a research on a Spanish-speaking country. Can someone please check the grammar and spelling for me? Thank you so much!

Aquí está el mapa de Colombia.
Colombia está cerca de Panamá, de Venezuela, del Brasil, de Ecuador, y de Perú.
Colombia está cerca del Océano Pacífico y del mar del Caribe también.

El capital de Colombia es Bogotá.

Las ciudades principales son Medellín, Cali, Barranquilla, y Bucaramanga.

Aquí está la bandera de Colombia.
Los colores de la bandera son amarillo, azul, y rojo.
El color amarillo representa el oro de Colombia.
El color azules representa el Océano Atlántico que separa Colombia de España.
El color rojo representa independencia.

Aquí hay una foto de Cristóbal Colón.
Cristóbal Colón es el pico más alto de Colombia.

Fútbol es muy popular en Colombia.

La gente le gusta café.
Ellos beben veinte tazas de café diarias,

La gente utiliza Pesos colombianos. Dos mil y dosciento y setenta y nueve Pesos de colombianos es un dólar canadiense.

La población es 43,593,035.

Noventa por ciento de colombians son católicos.

Ellos comen empanadas y tamales.

Shakira es una cantante colombiana famosa.

El presidente es Álvaro Uribe Vélez.

Colombia produce muchos de café.

Los productos de la agricultura son café, plátanos, maíz, y arroz.

El himno nacional de Colombia es "¡Oh! ¡Gloria inmarcesible!"

La flor nacional es orquídea.

La lengua oficial es español.

El competir con la bicicleta es muy popular en Colombia.

[quote="chilly_snow"]

Fútbol es muy popular en Colombia.

La gente le gusta café.

Noventa por ciento de colombians son católicos.

Ellos comen empanadas y tamales.

Colombia produce muchos de café.

El competir con la bicicleta es muy popular en Colombia.[/quote]

El fútbol es muy popular...

A la gente le gusta el café

El noventa por ciento...

Los colombianos comen empanadas y tamales

En Colomba se produce mucho café

Las competiciones en bicicletas son muy populares en Colombia

[b]La [/b]capital de Colombia es Bogotá.

El color [b]azul[/b] representa el Océano Atlántico que separa Colombia de España.

La gente utiliza Pesos colombianos. [b]Dos mil doscientos setenta y nueve Pesos  colombianos[/b] es un dólar canadiense.

$Tipe$

[quote="chilly_snow"]thank you so much~![/quote]

You're welcome!

De nada!

[quote="chilly_snow"]Two more sentences:

Organizan las exhibiciones todos los meses.
Muestra los trabajo por famosos artistas colombianos también.[/quote]

Organizan exibiciones todos los meses

Muestran trabajos de famosos artistas colombianos también.

[quote="José"][quote="chilly_snow"]Two more sentences:

Organizan las exhibiciones todos los meses.
Muestra los trabajo por famosos artistas colombianos también.[/quote]

Organizan [b]exibiciones [/b] todos los meses

Muestran trabajos de famosos artistas colombianos también.[/quote]

jeje .. Jose ... aunque [b]exibiciones [/b] es correcto,
[b]exhibiciones [/b] es también correcto ..

revisar. ..

Carlos Bravo ;)
Lima - PERU