geometry
MATHCOUNTS
3D geometry
quadratics
circumcircle
modular arithmetic
trigonometry

So the question was(modified to my choice): An equilateral triangle sits on top of a square using one edge of the square as its base. A circle passes through all vertices that the two polygons do not share. If the area of the square is 64 square inches, how many inches long is the radius of the circle. I don't have a computer drawing abilities so a picture would be appreciated. Thanks in advanced.  :)

yep, i think that is probably it, thanks!

Since I don't want create a new topic, I will just post another problem here from my coach. Jan has 4 dogs. She ties each of them to corner of her shed which has a square base of 144 square feet. Each dog is tied to a different corner of the shed by using a rope that is $ 6\sqrt2$ feet long. The dogs have no way to get in the shed. What is the total area where the two dogs could play together? Express your answer in terms of pi. It seems to me you can't do this with regular Mathcounts methods....thanks.

[hide]I assume that you mean where two of the four dogs can play. I think you know what it looks like, but don't understand how to find the intersection areas. Here is a diagram...

(sorry for my bad asymptote...just showing one side and the circles there =P)

[asy]pair A=(0,0);
pair B=(12,0);
pair C=(6,0);
pair D=(0,-12);
pair E=(12,-12);
pair F=(6,6);
pair G=(8.485,0);
MP('A',A,SW);
MP('C',C,S);
MP('B',B,SE);
MP('F',F,N);
MP('G',G,NE);
D(A--B--E--D--cycle);
D(C--F--A);
D(circle(A,8.485));
D(circle(B,8.485));[/asy]

Hm don't wanna spend time with labels. So...
$ AB\equal{}12\Rightarrow AC\equal{}6$
$ r\equal{}AF\equal{}6\sqrt2$
$ \angle{ACF}\equal{}90^\circ\Rightarrow CF\equal{}6\Rightarrow \angle{FAC\equal{}45^\circ}$
Now we know that one of the small areas we are looking for is $ [\text{Sector} AGF]\minus{}[\Delta ABC]\equal{}9\pi\minus{}18$.
Since we have 8 of these small areas (2 are shown in the diagram, there are 4 sides though), our final answer is $ \boxed{72\pi\minus{}144}$. I bet I made a computational error somewhere...
[/hide]

Sounds good to me, except you said sector AGF-triangle ABC. Triangle ABC is a straight line! I think you meant triangle ACF.

I have a few more from a different coach.

1.Two different circles are each tangent to both the x-axis and the y-axis. One of the points of intersection of the circles is (2,7). What is the product of the lengths of their radii? A picture gives it away way to easily, so I wondering if there was another way using algebra.

2.What is the area of the circle that passes through (-2,2)(12,6)(2,-12)? Express your answer in terms of pi. I expanded (x-a)^2+(y-b)^2=r^2 for each set of points and subtracted two different pairs and solved for a and b. Is this the only way or is there a faster way?

3.Muriel bought some blodgets at $ 29 each, some wickets at$14 each, and some doggens at $ 9 each. If she bought 34 items in all and spent$556, how many solutions are there? I made two equations 29a+9b+14c=556 and a+b+c=34. Multiplied a+b+c=34 by 9 and subtracted getting 20a+5c=250-->4a+c=50. But then I had to check if any values didn't work. Any easier ways?

4.How many three-digit prime numbers are two more than a perfect cube?

1. The equation of the circles can be written as $ (x \minus{} a)^2 \plus{} (y \minus{} a)^2 \equal{} a^2$ where $ a$ is the radius and $ (a,a)$ is its center since it is tangent to both axes. Plugging the point (7,2) into that equation gives $ a^2 \minus{} 18a \plus{} 53$. Since we are only concerned with the product of the radii, the answer is just the product of the roots to the quadratic equation which is 53.

2...

We are only concerned with finding the circumradius, which easily leads to the area. The sides are of lengths $ 2\sqrt {106}$, $ 2\sqrt {53}$, and $ 2\sqrt {53}$. And yay we find that this is an isosceles right triangle! :jump: The circumradius is just half the hypotenuse or $ \sqrt {106}$...so $ 106\pi$.

edit: I didn't know it was a right triangle before I was going to use that formula (and was too lazy to change it)...:P the formula was useless there

Edit: 3

Continuing from where you left off, just express the equation in modulo 4 ($ c\equiv2{\pmod4}$). We also know that $ 2\leq c\leq34$...it shouldn't be too hard from here.

Try to avoid using something like $ R\equal{}\dfrac{4abc}{[ABC]}$ if you're [b]practicing[/b] MathCounts. The reason is that you want to look for elementary solutions (by MathCounts standards), and as far as I know, the only way to prove that formula is using the law of sines (or something very annoying that's equivalent to sines).

In particular, for #2, there is indeed a more elementary way to do the problem. What's special about the triangle those points make? Draw a good diagram and it may become more clear.

Thanks to everyone, very good solutions. How about number 4?

And I have some more
1. What is the smallest positive integer n such that n! ends in at least 2003 zeros?

2.Seven friends all put their car keys in a bowl. At random they reach in and take one set of keys. WHat is the probability that exactly 4 of the friends get the wrong keys?

3. Similar to last one. 5 greedy kids were eating pizza. After they had each taken one bite of their respective slices they were scared into dropping their slices on the floor. Being the vultures that they are, they each grabbed a piece off the floor and continued eating. In how many ways could they have all grabbed a piece other then the one they had first taken a bite from? I know that there is a formula for this, but I still don't get how to do this.

more to come....

[hide="1"]the formula for the number of 0's at the end of $ n!$ is $ \lfloor{\frac {n}{5}}\rfloor \plus{} \lfloor{\frac {n}{25}}\rfloor \plus{} \lfloor{\frac {n}{125}}\rfloor \plus{} \cdots$. Now be guess and check, we know it must be less than 10000. So lets try 8000!. We get 1998 0's. Now lets add 25 to that to see what we get. 8025! ends in 2004 zeros. That's the closest we'll get to 2003 so the answer is $ \boxed{8025}$.[/hide]

[hide="2"] First we choose the 4 people that get the wrong keys. $ {7\choose{4}} \equal{} 35$. Now we must figure out the number of ways 4 people can all get wrong keys. There are a total of $ 4! \equal{} 24$ ways to  distribute the keys without restrictions. Now we must count the number of ways that at least one of the 4 people get their correct key and subtract that from 24.

case1: 1 person gets correct key- $ {4\choose{1}}\times{2} \equal{} 8$(choose the person that gets the correct key and then you have 2 ways to give the remaining people wrong keys)

case2: 2 people get correct key- $ {4\choose{2}}\times{1} \equal{} 6$(if two of the people get the right keys, then there is only one way for the remaining two to get wrong keys)

case3: everyone gets correct keys- 1.

Obviously, there is no way for 3 people to get correct keys because the 4th person would also get the their key.

So the total number of ways for each of the 4 people to get wrong keys is $ 24 \minus{} 8 \minus{} 6 \minus{} 1 \equal{} 9$

The total number of ways for 4 out of 7 people to get wrong keys is $ 35\times{9} \equal{} 315$. And there are a total of $ 7!$ ways to distribute the keys so the answer is $ \frac {315}{7!} \equal{} \boxed{\frac {1}{16}}$ hmmm... pretty simple answer, wonder if there is easier method to do it...[/hide]

[hide="3"]There are $ 5!$ total ways. Now we must subtract the cases were at least one kid gets their original pizza.

case1: 1 kid gets correct pizza- $ {5\choose{1}}\times{9} \equal{} 45$(in the previous problem we found that there are 9 ways for 4 people to get the wrong keys)

case2: 2 kids get correct pizza- $ {5\choose{2}}\times{2} \equal{} 20$

case3: 3 kids get correct pizza-$ {5\choose{3}}\times{1} \equal{} 10$

case4: everyone gets correct pizza-$ 1$

$ 5! \minus{} 45 \minus{} 20 \minus{} 10 \minus{} 1 \equal{} \boxed{44}$[/hide]

That was pretty lucky guessing and checking on the first one.....
I did like your answer to number two though, which was basically the same as number 3.

[quote="tinytim"]hmmm... pretty simple answer, wonder if there is easier method to do it...
[/quote]

For 2 and 3, refer to [url=http://en.wikipedia.org/wiki/Derangement]this[/url]. So you might wanna memorize some of those to save about 15 seconds. :P But that's how I would do it. (was this what you were looking for shentang?)

@shentang, for your previous number 4, there's only 5 cubes within the range...so just brute force it...


[color=green]Go to khanacademy.com to learn all this math up to linear algebra[/color]

Personally, I think miniclip is kind of crappy. Miniclip has all these useless games. And FTW is not a game. I think it's an educational tool for MC!!!! And I personally think that playing video games is the real nerdy thing. Which everyone calls me a nerd for doing math at my school....  :maybe:

EDIT: personally, I like go.hrw.com!!! It's an awesome site that has a lot of textbooks and gives you homework help on almost every sub. Cool...

I kind of get addicted to math so I appreciate for the win more.

I also get called nerd.

For me, it is most definitely FTW all the way.

Why is this in the MOEMS forum?

Is this considered reviving? Cause this is pretty old.  :maybe:

I don't think so; he was asking whether or not it was supposed to be in the MOEMS forum. :|  :huh:  :(

I know what he was asking. But this is purdy old. You don't really need to ask that if the topic is dead.

okay new topic.

solve this

2x+y=12

2y+x=?

posting in a topic that has not been posted in for like a week :wink:


Some cards have one side white and the other side black. A card is placed on each square of an m x n board. All cards are white side up, except for the card in the top left corner. A move is to remove any black card and to turn over any cards on squares which share an edge with the square from which the card was taken. For which m, n can one remove all the cards from the board by a sequence of moves.

This problem is IMO Shortlist 1998 #28. I could post the solution here, but I don't want to claim it as my own.


function
greatest common divisor
number theory unsolved
number theory

Find all $ k \in Z$ so that there is a function $ f:N^* \longrightarrow Z$ satisfying:
	1) $ f(2004)=2005$
	2) $ f(mn)=f(m)+f(n)+kf((m,n))$ with (m,n) is the greatest common divisor of m and n.

I think this one is adapted from one using in Taiwan 1997.

Let $k$ be an integer such that a function $f$ satsfying the statements does exist.
Then, for $a=b=1$ we have $(k+1)f(1)=0$. Thus, $k=-1$ or $f(1)=0$.

If $f(1)=0$ : let $p=2004$. We have :
$f(p^2) = f(p)+f(p) + kf(p) = (k+2)f(p)$.
As above, we successively deduce that $f(p^3)=f(p)+f(p^2)+kf(p)=(2k+3)f(p)$, and $f(p^4)=f(p^3) + f(p) + kf(p) = (3k+4)f(p)$.
In another hand $f(p^4)=f(p^2)+f(p^2)+kf(p^2) = (k+2)^2f(p)$.
Since $f(p) \neq 0$, we deduce that $3k+4 = (k+2)^2$ which leads to $k=-1$ or $k=0$.

Conversely :
- If $k=-1$, we only have to consider $f(x) = 2005$ for all positive integers $x$.
- If $k=0$, the desired condition is then $f(ab) = f(a)+f(b)$ for all positive integers $a,b$. It is equivalent to :
 $f( \prod_{i=1}^k p_i^{a_i})= \sum_{i=1}^k  a_i \cdot p_i$ (1)
for all primes $p_1, \cdots, p_k$, and positive integers $a_1, \cdots , a_k$.

Now, $2004 = 2^2 \cdot 3 \cdot 167$ and $2005 = 5 \cdot 401$.
Then, let $f(1)=0=f(2)=f(p)$ for all prime $p$ except $3$ and $167$.
Let $f(3)= 5$ and $f(167)=401$
Then define $f(n)$ using its prime decomposition and (1) to define a function $f$ satisfying the statements of the problem.

Therefore the solutions are $k=0$ and $k=-1$.

Pierre.

Thanks, Pierre.
Your proof seems to be better than PNT's.

That's kind, but the end is wrong because $f(2004) \neq 2005$ since $f(ab) = f(a) + f(b)$ and not $f(a)f(b)$.

Fortunately, it can be fixed easily with $f(1) = f(p) = 0$ for all primes $p \neq 3$ and $f(3) = 2005$.

Pierre.


Hi,
I'm attending a school based on the ISC curriculum(Indian system) in Dubai and Im in my final year which ends in March. I most probably want to pursue maths(applied) in college. I scored an 85/100 in math last term and and I normally maintain a 90+ in computers. My marks as such havent been commendable( due to a lack of studying uptil now but im hoping my marks in the next next few sets of exams will overshadow my past record) I scored a 1900 in my SAT which is horrible I will be taking them again and I'm aiming for a 2200.
I had a few doubts which I wanted to clear:-
1. Should I take a gap year?( I really want to pursue maths in college but I haven't really done much to prove my capabilities)
2. What math programs can I take during a gap year ( The toughest possible as being in an Indian system the highest level of maths is taught)
3. What kind of a project or research can I do to prove the same?
4. What job opportunities are there in the feild of mathematics during a gap year ( I worked in a software company the previous summer where i wrote a couple of codes and learnt a few languages such as javascript, HTML, AJAX and found rather boring)
5. How can I improve my horrible chances of getting into a good college? (I know its a bit late but I would really appreciate any kind of help)

- Muskan

Have you taken Linear Algebra and/or single variable calculus? Perhaps you could find a nearby proffessor to do research with.


inequalities
induction
inequalities proposed

i saw my mistake.
1/X1+1/X2+...+1/Xn>=n^2/(X1+X2+...+Xn) and not the other way around.

It's good. Now, the solution... Is this problem so simple?

If we put a_1=..=a_n-1=x tends to 0 we get close to n-2.
Now to prove another part:
Firstly (a_1^2+a_2)(a_2^2+a1)>=(a_1^2+a_1)(a_2^2+a2).
So we replace to prove sum{1/(a_i+1)(a_(i+1)+1)}<=n-2.
For n=3 it's simple replacing a_i=x/y.
For n>3 we prove by induction:
If a_1a_2<1 and a_2a_3>1 we can replace a_2 and a_3 with a_2a_3 and prove the value decreases by at most 1 so we can use induction step.
This problem didn't look very simple to me


trigonometry
ratio
trig identities
Law of Cosines
Pythagorean Theorem
geometry

Can anyone give me a proof to the Law of Cosines without involving/using pythagorean's theorem?

I doubt very much. In France  this Law of Cosines is named "Al-Kashi Theorem"  or "Generalized Pythagore".
I think that [b]cosinus[/b] is less important here then the three squares of the sides . :huh:
Al-Kashi was a great arabian mathematicien who died in 1429.

i doubt that you could completely not use the pythagorean theorem, but i suppose for every part of the proof that you use it, just substitute pythagorus' proof, (the one with the squares and triangles) so you are technically not using it

Trigonometric ratios are derived from the pythagorean theorem, so it's pretty hard to avoid it.


algebra
polynomial
number theory unsolved
number theory

Show that

$ (\sqrt{3} \plus{} \sqrt{2})^{1/100} \plus{}(\sqrt{3} \minus{} \sqrt{2})^{1/100}$

is irrational.

The sum clearly between 2 and 3 and it's also an algebraic integer so it must be irrational.
Proof that it's an algebraic integer:
$ \sqrt3\plus{}\sqrt2$ and $ \sqrt3\minus{}\sqrt2$ are both algebraic integers, and they share a minimal polynomial (although that isn't important). If you take their minimal polynomial $ f(x)$ and plug in $ x^{100}$ we get another monic with integer coefficients which has roots the $ (\sqrt3\plus{}\sqrt2)^{1/100}$ and $ (\sqrt3\minus{}\sqrt2)^{1/100}$, so they are algebraic integers, so their sum is.


1. Let $ C_{[0,1]}^{1}$ is the collection of all continuous function $ x: [0,1]\rightarrow \mathbb{R}$ and have derivative $ x'$ on $ [0,1]$.
$ \forall x,y \in C_{[0,1]}^{1}$ set $ d(x,y)\equal{}sup\{|x(t)\minus{}y(t)|: t\in [0,1]\}\plus{}sup\{|x'(t)\minus{}y'(t)|: t\in [0,1]\}$. Show that $ (C_{[0,1]}^{1},d)$ is a complete metric space.

1. use the fact that $ \mathbb{R}$ is complete and if $ (x_{n})$ Cauchy  $ \in C^{1}[0,1]$  then $ \forall t \in [0,1] : (x_{n}(t))$ is Cauchy in $ \mathbb{R}$ so there is  $ x(t)$   st :  $ x_{n}(t)\to x(t)$.

2. Prove that  the above $ x$ is in $ C^{1}[0,1]$ by showing that $ x_{n}\to x$ uniformly


$ \sum_{m \equal{} 1,(m,n) \equal{} 1}^n 1$

How can you put the second condition (m,n)=1 under the first one?

also, how can you make the -1 and 0 line up in the following

$ \mu(p^\alpha)\equal{}
\begin{cases}
\minus{}1\,\,\,\text{if} \,\,\,\alpha\equal{}1 \\
0\,\,\,\text{if} \,\,\,\alpha\geq 2
\end{cases}$

[quote="aspiringchemlete"]How can you put the second condition (m,n)=1 under the first one?[/quote]Use \substack (which requires the amsmath package)
[code]\sum_{\substack{m = 1\\(m,n) = 1}}^n 1[/code]$ \sum_{\substack{m \equal{} 1\\(m,n) \equal{} 1}}^n 1$
[quote="aspiringchemlete"]also, how can you make the -1 and 0 line up[/quote]Use \hphantom{-}. This puts the space for a - sign before the 0 without actually printing the - sign.
In the cases environment, use & to tab the conditions so the 'if's line up, and use a space in \text{if } to line up the alphas.
[code]\mu(p^\alpha) =
\begin{cases}
 -1 &\text{if } \alpha = 1 \\
\hphantom{-}0 & \text{if } \alpha\geq 2
\end{cases}[/code]$ \mu(p^\alpha) \equal{} \begin{cases}
 \minus{}1 &\text{if } \alpha \equal{} 1 \\
\hphantom{\minus{}}0 & \text{if } \alpha\geq 2 \end{cases}$

Another possibility is \hfill:
$ \mu(p^\alpha) \equal{}
\begin{cases}
 \minus{}145432 &\text{if } \alpha \equal{} 1 \\
\hfill M10 & \text{if } \alpha\geq 2
\end{cases}$
The advantage is that you do not need to mimic the width yourself.


vector
analytic geometry
geometry
parallelogram
real analysis
real analysis unsolved

If a vector $ \vec C$ divides equally the angle between two vectors $ \vec A$ and $ \vec B$ ,how do I determine the vector $ \vec C$ by using   vectors $ \vec A$ and $ \vec B$ ?

$ \vec A \cdot \vec B \equal{} \left \| \vec A \right \|\left \| \vec B \right \| cos\theta$

$ cos\frac {\theta}{2} \equal{} \sqrt {\frac {1}{2}(1 \plus{} cos\theta)}$

Is it possible that the vector $ \vec C$  is as below ? How do I get this ?
$ \vec C \equal{} \frac {1}{2}(\frac {\vec A }{\left \| \vec A \right \|} \plus{} \frac {\vec B}{\left \| \vec B \right \|})$

Yes, $ \frac {1}{2}(\frac {\vec A }{\left \| \vec A \right \|} \plus{} \frac {\vec B}{\left \| \vec B \right \|})$ gives the direction of $ \vec C$ (the length, of course, is not unique). $ \frac {\vec A }{\left \| \vec A \right \|}$ is the unit vector in the direction of $ \vec A$ and $ \frac {\vec B}{\left \| \vec B \right \|}$ is the unit vector in the direction of $ \vec B$. The midpoint of two points in a plane is found by taking the average of their coordinates, so $ \frac {1}{2}(\frac {\vec A }{\left \| \vec A \right \|} \plus{} \frac {\vec B}{\left \| \vec B \right \|})$ gives you a vector that ends at the midpoint of $ \vec A$s unit vector and  $ \vec B$s unit vector, hence bisecting the angle between $ \vec A$ and $ \vec B$.

I don't understand that can the new vector which is got by the addition of the two vectors bisect the angle between the two vectors ?The addition of the two unit vectors can get a new vector by parallelogram law but a diagonal line can or cannot bisect the angle in the parallelogram . I'm still fairly confused .
Could someone help me figure out where is my problem ?


We have the test, but the website is down. Does anyone know the range of dates? I'm pretty sure all this week is fine...

[quote="pianoforte"]I heard last year's cutoff was like 14 so I think you'll do fine... (and how do you know your score already?)
[/quote]
Well pretty much every single question was pretty darn trivial, and my answers for the slightly harder ones match the answers of people who are good at physics so yay.

[quote="pianoforte"]I heard last year's cutoff was like 14 so I think you'll do fine... (and how do you know your score already?)

The website is back on line, next round "Administer the exam between February 23 and March 4. Exam papers must be received back at AAPT by March 5."[/quote]

14?! That can't be right.

Well, I think I am not going to round 2 for sure.  I can't do 5 of the questions and (most likely) made some random mistake here and there (as I tend to do).

[quote="serialk11r"]Just wondering, what is the approximate date of the next round? Also is 21.25 going to get me to the next round? I kinda made 3 dumb mistakes :([/quote]

For some reason, it bothers me when people who are sure enough that they will qualify ask questions so that they can brag about how smart they seem. If you got answers and compared with people at your school who are similarily involved in physics, I sincerely doubt that one of them would not have known about the cutoff.

[quote="serialk11r"]Actually I think 14 is possible because so few people take the first round.[/quote]

Thank you. I guess that's a little reassuring. (slightly). Now I should just have my fingers cross and hope for the best.

[quote="serialk11r"][quote="abejnood"]
For some reason, it bothers me when people who are sure enough that they will qualify ask questions so that they can brag about how smart they seem. If you got answers and compared with people at your school who are similarily involved in physics, I sincerely doubt that one of them would not have known about the cutoff.[/quote]
Well actually I compared with my friend who was getting solutions from a USAPHO camper (Andy Tsao). I think someone who made the summer camp is going to know what he is talking about. Our school is doing USAPHO for the first time, so no one knows about the cutoff or whatever.

Also, sorry if it seems to you that I am bragging about how smart I seem, but seeing that the overall level of people here is quite high, I think my post is definitely reasonable. In addition, if you were trying to make a reference to the other thread where you posted, I'm sorry if it makes you feel that way, but I don't see anything wrong with the way I am posting.[/quote]

I actually didn't realize that you were the one in the other thread. But now that you point that out, you definately did know about the cutoff range. Your post is analogous to saying, "Oh dear, I got a 2370 on my SAT, do you think it's enough?" I didn't mean to pick on you or anything, it's just that at my school, most people (myself included) get scores like 14 or 16 and do, in fact, worry about the cutoff range. Seeing someone, who obviously knows physics very well, act as though the test is "too easy" and worrying about whether a question here or there will doom them does, in fact, seem a bit arrogant to me, although it's true that many at AOPS act in this manner.

Does anyone know when and how we can get the results of the first round? thx

[quote="abejnood"]

I actually didn't realize that you were the one in the other thread. But now that you point that out, you definately did know about the cutoff range. Your post is analogous to saying, "Oh dear, I got a 2370 on my SAT, do you think it's enough?" I didn't mean to pick on you or anything, it's just that at my school, most people (myself included) get scores like 14 or 16 and do, in fact, worry about the cutoff range. Seeing someone, who obviously knows physics very well, act as though the test is "too easy" and worrying about whether a question here or there will doom them does, in fact, seem a bit arrogant to me, although it's true that many at AOPS act in this manner.[/quote]

Sorry I think there is a misunderstanding. I am actually pretty confused, because someone at school said you need like 23+, then I heard 20, then I heard 14, then I heard 16, so I am quite confused. In the case that you need about 20, I would be in pretty big trouble, as one more incorrect answer could be the difference.

A friend of mine claims that last year's cutoff was 13.  I know that the two people at my school who made it (me and someone else) had 14.25 and 15.5, each.  So the cutoff cannot have been more than 14.25.

The test did seem a bit easier this year, but I don't think the cutoff will be much higher.  Anyone 20+ should be safe.

"In order to maintain exam security, do not communicate any information about the questions (or their answers or solutions) on this contest until after February 20, 2009." - rules


A .145 kg baseball travelling 35.0 m/s strikes the catcher's mitt, which in bring the ball to rest, recoils backword .11 m. What was the average force applied by the ball on the glove?

[hide]I got about 810 newtons[/hide]

thanks. I did Vf^2 = Vi^2 + 2ad to find the accel and used F = .145 * a  to get the force... that's an interesting way to do it.

another problem:
Mass A is on a table and Mass B is hanging off the side. They are connected by a rope on a frictionless pulley.

Problem says "The coeffient of sliding friction between the block A and the table is .4. Mass A = 35 kg, Mass B = 25 kg. How far will the block B drop in the first 3 s after the system is released?"

[hide]Answer teacher gave was 11 m.
I'm getting 8.1 m using this.
Friction is = coefficient * normal force = .4 * 35 * 9.81 = 137 N. Tension = Mass of B * gravity  = 25 * 9.81 = 245 N.
When the system starts the forces (only 2 are friction and gravity on the mass B?) is 245-137 = 108 N

108 N = ma = 60*a, a = 1.8. d = vt + 1/2a(t^2)

d = 0t + .9 * 3^2 = 8.1 m [/hide]


For how many whole numbers a is from 1 to 10,000 is it that $ 2^a \minus{} a^2$ is a multiple of 5?

The answer I got is below, don't know if it's right..
[hide]$ a\equal{}{0,2,4,...10 000} \equal{} \boxed{5001}$ numbers.[/hide]

[hide="not complete"]modular residues $ \pmod{5}$ cycle $ 1,4,4,1,0$, powers of 2 residues are $ 2,4,3,1$; we need $ 4 \minus{} 4$ or $ 1 \minus{} 1$, hence numbers 2 or 3 (mod 5) and 2 (mod 4) or 1 or 4 (mod 5) and 0 (mod 4) work; we can count this easily but I am feeling rather tired and/or lazy right now.[/hide]

EDIT: wow obviously that gives 4 per 20 so 2000 per 10000. argh I cannot think at night.

also could you show how you got that? I am pretty sure it is wrong.


What is the sum of the mean, median and mode of the elements in
the following data set?
\[ \{42,56,48, 56,53\}\]

The mode=# that occurs most often=56.
The median=middle # when arranged in numerical order=53.
The mean=sum of data/# of pieces in data=255/5=51.
51+53+56=160.

Ill answer this in a series of steps

[hide="THIS IS MY EXPLANATION"]1.MEAN is the total #'s added and divided by the total #'s divided by-51

2.MEDIUM is putting all the #'s in order and find the middle # -53\

3.MODE- is the # that occurs most inside a set of #'s-56

4.adding all these together we get

 51+53+56=160

5.[hide="ANSWER"]160[/hide][/hide]

[hide="."][size=200][color=#004080]PLEASE RATE MY POST THANK YOU SO MUCH!![/color][/size][/hide]

[size=50][b][i][u]please be fair when you're rating my post and please PM me if there is a better way to explain this thank you so much!![/u][/i][/b][/size]

[quote="Kestzh"]Is there a way to do mean without adding them all together, because that slowed me so that I missed the time limit  :([/quote]
my way: find the mean of differences from 50, since 50 looks pretty close to the mean. Below is the mean.
$\frac{-8+6-2+6+3}{5}+50$

evaluating this (pretty simple) gives 51.


modular arithmetic
number theory unsolved
number theory

Dear friend,

The question may be is not clear. You mean want to know the parity of numbers a,b,c,d? In case note that b can be odd or even.
In fact, we can work out all such positive integers a,b,c,d satisfy the equation
\[(2^a+1)^b=(2^c-1)^d\]
Here is my ugly solution,
At first we will prove that $a$ is odd, $c,d$ are even. Indeed, it's obvious to see how $d$ is even. Cause $c>1$ then $2^c-1$ is $4t-1$, so it has an 4t+3 prime divisor $p$. From the fact that $p|2^a+1$ we implies that $a$ must be odd, ideed if not then $-1$ is QR modulo $p$. But $2^a+1$ is divisible by $3$ so is $2^c-1$. This happens when $c$ is even.
The second step, put $b=2^mb_1$ and $d=2^nd_1$ where $b_1,d_1$ are odd, then we must have $m<n$. So now we consider
\[(2^a+1)^b=(2^c-1)^d\]
where $c,d$ are even and $a,b$ are odd numbers. The RH is a perfect square so $2^a+1$ is a perfect square. Put $2^a+1=u^2$ then $u$ is odd and $(u-1)(u+1)=2^a$. Note that $gcd(u-1,u+1)=2$ so we obtain $u-1=2$ and $u+1=2^{a-1}$. Subtracting these equation we find $a=3$.Then
\[3^{2b}=(2^c-1)^d\]

so $2^c-1=3^t$ and $td=2b$. But $d$ is even and $b$ is odd, so $t$ must be odd. If $c>2$ then $2^c$ is divisible by $8$. Note that $3^t+1\equiv3+1\equiv4\pmod{8}$, this contracts! So we have $c=2$. Now we have $t=1$ and $d=2b$.
Ingeneral, we get
$(a,b,c,d)=\bigl(3,2^m(2x+1),2,2^{m+1}(2x+1)\bigl)$ where $x,m$ are nonegative integers. These are all we want to find!
Comment. The first step relates to the following simple but beutiful theorem deu to J.Browkin :
${\bf Theorem.}$ If $m,n$ are positive integers and $m$ is odd then $gcd(2^m-1,2^n+1)=1$


Given $ a, b, c > 0$ and $ a\plus{}b\plus{}c\equal{}3.$ Prove that
\[ 11\left(\sqrt[4]{a}\plus{}\sqrt[4]{b}\plus{}\sqrt[4]{c}\right)\plus{}3 \ge 12(ab\plus{}bc\plus{}ca)\]
 :)

Can_hang and me had a nice proof for it  :)
The similar results can be solved by same way (with the same condtion):
$ 5\left(\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}\right) \ge 4(ab \plus{} bc \plus{} ca) \plus{} 3$
And  $ \sqrt [3]{a} \plus{} \sqrt [3]{b} \plus{} \sqrt [3]{c} \ge ab \plus{} bc \plus{} ca$
They'll appear in our new book  :)


[quote="jacob8888888"]Find the cube, rounded to the nearest integer, of the sum of the square roots of the first four positive integer multiples of 3

Thanks[/quote]

are we supposed to solve this without a calculator? hehe.. :?  :oops:  :?

You should be able to if you just know the first two or three digits after the decimal for $\sqrt{2}$ and $\sqrt{3}$, but for the length of this, if it was actually in one of the competitions, I would expect it to be target or team, in which case you can use a calculator.

[hide]$\sqrt3 + \sqrt 6 +\sqrt 9 + \sqrt{12}$ is $\sqrt3 + \sqrt 3 * \sqrt 2 + 3 + 2 * \sqrt 3$

if we estimate square root of 3 to be about 1.73 and sqrt 2 to be about 1.41, we get:

$1.73 + 1.73*1.41 + 3 + 2*1.73$

which is: $8.19 + 1.73*1.41$ (about 2.44)

that's about $(10.63)^3$. which is (I'm estimating) 1200. And I'm absolutely sure that there's no way that's right.

This isn't a hard question, but you can't figure it out without calculator. So I'm sure that that's a target/team.

now to check what the answer is: 1206. Wow. I was close. But yeah, only you can do it without multiplying 10.63 by itself 3 times is using calculator.[/hide]


quadratics
algebra
polynomial
Vieta
quadratic formula

Show that the product of the roots of a quadratic equation is
c/a.

What is a more simple way of expressing this question?

[hide]If we have the equation $ax^{2}+bx+c$, then the roots are (by the quadratic formula) $\frac{-b+\sqrt{b^{2}-4ac}}{2a}$ and $\frac{-b-\sqrt{b^{2}-4ac}}{2a}$.

Multiplying, we see that it becomes $\frac{b^{2}-b^{2}+4ac}{4a^{2}}$Thus, we simplify to $\frac{4ac}{4a^{2}}= \frac{c}{a}$[/hide]

A more simple way of expressing this question could be to say
What is the product of the roots/zeros of any quadratic equation?

- NAWAHD WERDNA

Good reply.  I understood your steps easily.

[hide]Let the quadratic be $a(x-r)(x-s)=0$ where $r,s$ are the roots.  If we multiply it out, we get $a(x^{2}-rx-sx+rs)=ax^{2}-a(r+s)x+ars=0$.  In this case, $a=a, \ b=-a(r+s),$ and $c=ars$.

From here, it's clear that the product of the roots, $rs$, is $c/a$.[/hide]

[quote="I"][hide="some useful formulas that you might want to know"]We have the roots of a quadratic equation

$\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. We sum them:

$\frac{-b+\sqrt{b^{2}-4ac}}{2a}+\frac{-b-\sqrt{b^{2}-4ac}}{2a}=\frac{-2b}{2a}=\frac{-b}{a}$.

We take the product of the roots:

$\frac{-b+\sqrt{b^{2}-4ac}}{2a}*\frac{-b-\sqrt{b^{2}-4ac}}{2a}=\frac{b^{2}-b^{2}+4ac}{4a^{2}}=\frac{c}{a}$[/hide][/quote]

[quote="mathgeniuse^ln(x)"]Um, yeah, I just have one thing to say.

This can be like generalized to an nth degree polynomial too.[/quote]

Vieta's Theorem (or something along those lines). Quite useful.


MATHCOUNTS
AMC
AIME
summer program
MathPath
Mathcamp
calculus

For those of you that don't have school. where are you traveling this summer?
What kinds of academic things are you working on?
**************************************************
For me, I'm not traveling this summer, although I will be going to Japan this fall. (YAY!!!)

I think I'll work on some AOPS stuff, read a lot, and do some old MATHCOUNTS stuff just for the heck of it. I also might make a science display board and work on some social studies stuff for the competitions at the state fair in my state, as well as science, math and spelling. (first place gets a $100 savings bond, second gets  a $50 savings bond.)

Oh, yeah. And study my math books from the past 2 years. (our class gets to keep our books) Our teacher has already said that we'll have a test the end of the first or second week.

lol. I'll be lucky to get around to any of that. Wow, summer goes fast....

First I go to Vienna for two weeks, then I go to HCSSiM. Maybe afterwards I will work some more on AoPS Vol. 2, but my summer is very scheduled. I doubt I will be able to do half the things I plan to do.

well, i have IS, but i'm also practicing for amc's and aime (can't prepare too much :D ) and i have summer tennis camp/league/whatever. Then i'm going to taiwan and china for 2 weeks, so my summer will be fun. i have 3 whole months of sumer, which is 2 weeks more than usual so i'm happy.

Uhh lets see, I'm not traveling anywhere.

First I'm reading some math books (pre-cal, maybe a bit of a calc book too, and AOPS books), while volunteering at an Alzheimer's assisted living place.
Then I'm going to Tx. State Math camp, which is 6 weeks long.  On the weekends, which are mostly free during the 6 weeks I have to study for PSAT.
When I come back from math camp, I'm doing my summer reading and the assignments that go along with it (book is Catch-22) plus some math stuff, probably.

Right now, I'm studying pre-calc, so I can do do calc once summer starts. Then I'm going to mathpath  :)

My plans are grand, as always.

- Do AP summer assignments (art museum visit, 1000-word bio essay, etc)
- Start writing colllege essays.. and I think a major scholarship application's due in August, so work on that, too..
- Study for the October SAT if I didn't do well today.. (but hopefully not)
- Mathcamp! :D
- Work on AoPS books
- Do some artwork (sketches, pastels, watercolors, etc)
- Catch up on all the movies I missed (been waiting for that...lol)

I highly doubt I'll get everything done (if anything at all), but it's nice to have good plans. It feels very productive. :)

I have NEMC and then CTY... that's about it.

[quote="Sunny"]6 weeks of my summer will be spent doing cell bio research at the university of Calgary.  yay![/quote]
Whattt... I am so jealous :mad:  :mad:

- Read some AoPS books
- Read some calculus books
- Go volunteer
- Go to Hawaii

June 28 - July 10 - Korea
July 11 - late Aug - Working at CCNY (TA)

I would like to visit mathcamp during puzzle hunt weekend too.

AoPS Ind study (May 13 - Aug 14)
Scouts Camp (July 3 - July 10)
Stanford University Summer Institute(July 11-July 30)
MAT NAT (Aug 1-Aug 7)
Korea (Aug 8 - Aug 20)
Tennis tryout (Aug 22-23)

rest preparing for PSAT/SAT/USAMO/ etc.

busy summer... some volunteer work while in Korea, as well as more studying of SAT/PSAT/AoPS Ind study  ;)

-study AoPS v2 and ARML problem solving books to improve my math abilities
-read the pile of books that miraculouspostmaster let me borrow so i can hopefully improve my SAT reading score by 60 points and therefore get into clemson honors w/out having to get violent  :evilgrin:
-video games to relax from the intense math studying
-and if bordom gets unbearable, a summer job might be an option. i really don't want to get a job tho.

[quote="LynnelleYe"]~AIME Problem Series and Intermediate Algebra at AoPS
~Finish up EPGY Geometry and English courses
~Summer school--badminton, two hours a day
~Hopefully get to swim a lot
~Work through AoPS Vol. 2
~CTY Physics
~Go to Tahoe near the end of summer vacation
Goal for the summer: be able to double-digit AIMEs by the start of school.[/quote]

Yay. Badminton is cool. But tennis is cooler :D

-Tennis practice.
-Learn to drive!
-Research project of some sort.
-Mu Alpha Theta Nats.
-USAMO practice.
-USACO training pages.
-Some AP preparation.
-Study for SAT II's.

That's pretty much it, I guess.

Oops forgot to add goal for the summer:

by Fall or Winter, master:

800MATH SAT, 700VERB SAT, 700 WRITING SAT
Double-Digit AIME
135+ AMC12
Triple-Digit Volunteer Hours
20+ USAMO

funny how all my goals are set on a number scale  :D

i'm just gonna keep BREATHING... :D
that's my plan :)

I plan to sleep a lot, play soccer and frisbee, and ignore everything school-related for as long as possible.  I get enough of that during the school year. :)

tennis camp for 1 wk
mathcamp for 5 wks
kiawah island for 2 wks
school for 40 wks.  (ugh)

wow...

seen many similar events here guys :D

my summer would be perfect if you don't have to go the stupid tennis camp like some peeps here

list of thing

Tennis - so useless pathetic sport
Math Summer School - Algebra I - forced by mom, I hate moving~!!!!! why can't just a tutor come to my house?
Indenpendent Study with AoPS text and some other - yeahiness!!
Korea - Finally!! and meeting a new math tutor - convinced my mommy!

and the summer will be over

[quote="paladin8"][quote="shinwoo"]
Tennis - so useless pathetic sport
[/quote]

 :mad: Tennis is awesome. Don't be dissin' tennis. :P[/quote]

ditto. although tennis is not THE BEST sports, it is certainly awesome.

i'm mostly going to camp. and i might go to florida :lol:

Odd, tennis seems to be the sport of the thread, heh.

I'll be practicing tennis this summer (about 4 days a week) so I can make my school's team in fall.
Taking ~6-7 college courses
Taking some AOPS courses
Reading through this killer 900 page SAT/PSAT review.
Go to New Jersey (and visit Princeton!)
I want to periodically play my sax/piano so I don't lose some ground over the summer (doubt I'll get this done though, so lazy with my instruments)
Learn to drive (get my car in a few weeks!.. but no license 'til Nov.)

Well that's about it.. we'll see how it all goes.

For the past two years, my entire summer has been spent on tour with The Cadets.
This year, however, I decided to take a break from that and focus more on Piano, going to the Southeastern Piano Festival and practicing a whole lot.  Besides that, I'll probably just chill and enjoy the time off before college.

2 weeks of basketball camp
the rest at home

[quote="lightrhee"]My plans:
June 12 to June 18: Waterloo Seminar
June 25 to July 19: IMO2005
July 21 to late August: Europe trip

There goes my summer vacation.[/quote]

where in europe? :lol:

[quote="xxreddevilzxx"][quote="xfactor"]Given your name reddevil, I imagine your favorite sport is football, aka soccer.  The red devils are my favorites too.[/quote]

Do u know the red devils????? Are u korean??? (not many non-koreans, besides other Asians, know the red devils very well...)[/quote]

Let me make a distinction. xfactor has a British flag, so I can only surmise that s/he is referring to Manchester United.
You on the other hand are talking about the South Korean national team. They both have red devils as a nickname.

-just got back from ARML
-at USC (South Carolina...not Southern California) for an intership for 6 weeks
-keep up my running schedule while at USC
-go to a distance runners camp for a week immediately after USC

[quote="yif man12"]Anyone here have a summer job or plan to get one? I'm trying to find a semi-good paying job for the summer.[/quote]

ooh, me, i've got one! but then, i've had one since i was legally old enough (14). this is my second job... and it pays better... and its more fun because its almost exclusively outside... although its HOT. but yes, my summer plans:
m-w-f (and some weekends): work. and swimming
t-t: class. and watching foreign films with my friend.
weekends/nights: movie watching, book reading, chore doing, errand running. hanging out with significant other. college searching.

[quote="yif man12"]Anyone here have a summer job or plan to get one? I'm trying to find a semi-good paying job for the summer.[/quote]
I have a perfect semi-good paying job.
I'm going to work at the city college of new york as a teaching assistant.
I get paid for doing math in the class of great teachers like putnam fellow, Jan!

yay ccny!  im gonna be there (as a student)

[quote="Geehoon"]
I'm going to work at the city college of new york as a teaching assistant.
[/quote]

Me too, yay!
also, i hope to be able to spend some time in egypt (after finishing at CCNY and before orientation)

I'll be a junior counselor at PROMYS. (Effectively, all summer)

Tennis for me. THE best sport. Exept for all those other ones people say are better.  :nhl:  :football:  :juggle:  :surf:


i have no idea but this looks like a 24 game problem  :D

[quote="Interval"]How do you get these numbers to equal 24?

Here are the numbers: 3, 3, 7, 7.

Using the numbers
once, you can also put the numbers together like this: 37,33,77,73. and using any combanation of BEDMAS.

What is BEDMAS?

I did not say PEMDAS.  Okay?[/quote]
BEDMAS stands for
[b]B[/b]rackets (Parentheses)
[b]E[/b]xponentiations
[b]D[/b]ivision
[b]M[/b]ultiplication
[b]A[/b]ddition
[b]S[/b]ubtraction

[quote="davidlizeng"]i think a better way to remeber it is
[b]P[/b]lease (parentheses)
[b]E[/b]xcuse (exponents)
[b]M[/b]y
[b]D[/b]ear
[b]A[/b]unt
[b]S[/b]ally[/quote]
That's how I learned it but BEDMAS also works.

[quote="Interval"]How do you get these numbers to equal 24?

Here are the numbers: 3, 3, 7, 7.

Using the numbers
once, you can also put the numbers together like this: 37,33,77,73. and using any combanation of BEDMAS.

What is BEDMAS?

I did not say PEMDAS.  Okay?[/quote]
[hide=":D"]$7\left(3+\frac37\right)=24$[/hide]


Let $ x\equal{}A \mid B, x'\equal{} A' \mid B'$ be cuts in $ Q$. We defined

$ x\plus{}x'\equal{}(A\plus{}A') \mid$ rest of $ Q$

Show that although $ B\plus{}B'$ is disjoint from A+A', it may happen in degenerate cases that Q is not the union $ A\plus{}A'$ and $ B\plus{}B'$

if x and x' are both irrational, but x + x' = q is rational, then q will not be in A + A' or B + B'


geometry
analytic geometry
vector
LaTeX
USAMTS

i edited my pdf a little bit  (such as wording, simplification, etc) but here it is

I think my #1 should be commended but we'll see ;) :P

EDIT: I missed 2b so if anyone wants to tell me what how many pts i can expect out of it (if any) that'd be good

This is my first solution ever... let me know what you think.  I wonder what sort of partial credit my numbers 2b and 5 will earn..?

Dear TheAmazingOne:
Really nice solutions. ine were about twice longer and twice less elegant then yours. But then this is my first year, and I'm only in 9th grade.

Here are mine - they're all right, but none are particularly elegant.  I got lazy on 2b and didn't put diagrams in, so it's a bit of a chore to read through unless you already know basically how to proceed to solve it.  #5 I didn't feel like doing real geometry so I just pulled out my coordinate geometry and vectors and killed the problems.

Mine are probably very hard to follow, and #5 is BS, but here they are anyway.

[quote="TheAmazingOne"]Post your solutions and critique each other's approaches in this topic![/quote]

Your number 2 is amazing. So is the generalization made in #5.

I'm pretty happy with my solutions.  The algebra in my #3 is a bit icky, and I didn't explain myself well on #2b (as a result of only finding the solution with 1 hour until the deadline), but they are all correct, and #1, 2a, 4, 5 are nice, in my opinion.

However... AmazingOne's stuff is superb.  Generalizations, conciseness, and methods somewhat (or completely) different from mine on three of the problems.  Awesome work!

Here are my solutions.  #1-4 are right, and I didn't even get an answer for #5 but wrote what I had anyway for partial credit.  I think #1 is nice and is different from the solution from the math jam.

My solutions. I think 1,2a,3,4, and 5 are correct. Correct me if I am wrong. I thought 3 was a really nice problem especially part b.

I'll post mine later when I have access to my computer.  Be warned: there are NO diagrams :D

Edit: here they are!

omg no i just realized i sent my solutions to the wrong address :( :( :( :( :( i didn't realize i was [i]that[/i] sleepy (it was 2:29AM here on the east coast)

well i guess doesn't really matter because i didn't have enough time to write my solution for #2 and #3b, and i was rushed so i messed up the answer to #3a, so i only got #1 and #4. even my solution to #1 is really random :(

Expect commended for number 5. It was almost as pwnage as mine:

[quote]The length of the portion of $l$ inside ABCD is 3.[/quote]
No need to congratulate me. Just be sad that my solution was totally awesome.

Do you have to use latex to get commended? I know that you have to in order to get your solution posted at the site.

chess64, how did you tex that triangle?! that was amazing. I'm guessing it is just an embedded image, with letters placed around it based on absolute location? I am not very good with latex, teach me how to do it! (Or just post your .tex file)

I used [url=http://en.wikipedia.org/wiki/Metapost]MetaPost[/url]. It sounds hard to learn but it's not. :) If you don't want to install metapost or something, you can use the [url=http://www.tlhiv.org/cgi-bin/MetaPostPreviewer/index.cgi]metapost previewer[/url]. For example, try typing [code]draw (0,0)--(30,0)--(30,40)--(0,0)[/code]

Aargh I completely forgot about USAMTS after I got the letter in the mail.  I worked on it for like 30 minutes and only did #1.  My solution for it might not be commendable or anything, but it was certainly concise.  I don't have it on file so I'll just latex it to the best of my ability.

[hide]Okay, basically, I simply took the given and everything it implied systematically to find other digits.

I started out with this because 4 is the units digit of the original digit, and the result of the slide is 4 times the number.

$...XXXXXX4 * 4$  This is easier to write on paper in "Universal 3rd Grade Format""

$4*4 = 16$, so I wrote 6 underneath the equals sign and carried the one.  Since the number under the equals sign was 4 times the original number, and the result of a digit slide, the 6 (now in the units position) would have been in the 10's position.  So I wrote 6 next to the 4 in the original number.  I now had something like this:

$...XXXXX64 * 4 = ...XXXXX6$.  The one was still carried as it would have been, and more or less the multiplication continued until I had filled in 6 digits, and my paper looked like this:

$...102564 * 4 = ....410256$
So 102564 will fit the requirements and will be the smallest possible, because all I did was further the given.

Well, if I had remembered to finish the rest of the problems and entered Round 1, my proof would have been cleaner (and probably used Microsoft word to show the Universal 3rd Grade Multiplication format), and clearer, but the method worked and I apparently got the correct answer.[/hide]

Well, I think my #4 should be commended for an AMAZING generalization  :D

My 2b was brute forced :(

and I thought #5 was really easy, and it was [i]wrong[/i]...

Here's mine  :D

My #1, 3, and 5 were probably the best.

Hmm... well, here are mine, which are all amazingly long and amazingly abundant with diagrams :D

-- EDIT -- Darn, I thought they would at least be a Meg big!!  :blush:

by the way, Carlos, I originally had solutions that looked quite a bit like yours (including the 4x82 for number... 4? i think?) but then I decided that I would write extremely explanatory I-don't-think-you-can-say-I-guessed solutions for all of the questions, so they ended up amazingly... amazing-wordy


cho so n la tu nhien duong .va tap A={1,2,...,n} .tim so f(n) lon nhat sao khi chia tap A thanh f(n) khong rong ,khong giao nhau luon ton tai mot tap hop con co ba so x,y,z, thoa man x<y<z va x+y>z

ai good english thi gui vao mathlinks. ho cai


linear algebra
matrix
algebra
polynomial
vector
linear algebra unsolved

Suppose that $ \mathsf{A}$ is an $ n \times n$ square matrix and that $ \mathsf{A}^{\minus{}1}$ exists. Show that if $ \mathsf{A}$ has characteristic equation $ a_0 \plus{} a_1t \plus{} \cdots \plus{} a_nt^n \equal{} 0$, then $ \mathsf{A}^{\minus{}1}$ has characteristic equation
\[ (\minus{}1)^n \det (\mathsf{A}^{\minus{}1}) (a_n \plus{} a_{n\minus{}1}t \plus{} \cdots \plus{} a_0t^n) \equal{} 0.\]

$ \phi _A(\lambda) \equal{} \det(\lambda I \minus{} A) \equal{} \det(\lambda A A^{ \minus{} 1} \minus{} A) \equal{} \det(A) \det(\lambda A^{ \minus{} 1} \minus{} I)$

$ \equal{} ( \minus{} 1)^n \lambda^n \det(A) \det(\frac {1}{\lambda} I \minus{} A^{ \minus{} 1}) \equal{} ( \minus{} 1)^n \lambda^n \det(A) \phi_{A^{ \minus{} 1}} \left(\frac {1}{\lambda} \right)$

$ \equal{} > \phi_{A^{ \minus{} 1}} (\lambda) \equal{} \frac {1}{\det{A}}( \minus{} 1)^n \lambda^n \phi_A \left(\frac {1}{\lambda}\right)$.

Here, $ \phi_P$ denotes the characteristic polynomial of matrix $ P$, and $ A \in M_{n \times n} (\mathbb{K})$.

Oh, I forgot to ask: why don't we need to worry about zero eigenvalues?

If there are any zero eigenvalues, there is no inverse (and conversely).

I thought the condition for invertibility was that the determinant is nonzero - how do you prove that no eigenvalue is zero if the matrix is invertible?

One way: determinant is  product of eigenvalues (with multiplicity), though I suspect that is unsatisfying given the original post.

Another way: a zero eigenvalue means a nontrivial solution $ Av \equal{} 0$, which means non-invertible.

And so there is at least one vector in the nullspace! Thanks :)


Let a,b,c be non-negative numbers, such that $ a\equal{}max${$ a,b,c$}, $ b\plus{}c \ge a$ (or a,b,c are 3 sides of triangle)
Prove that
$ 2(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca) \ge a^3\plus{}b^3\plus{}c^3\plus{}15abc$

If we take $ a\equal{}z\plus{}y , b\equal{}x\plus{}z , c\equal{}x\plus{}y$;

Inequality equivalents to $ x^3\plus{}y^3\plus{}z^3\plus{}3xyz \geq x^2y\plus{}xy^2\plus{}x^2z\plus{}xz^2\plus{}y^2z\plus{}yz^2$ which is Schur.  :wink:

Let $ a \equal{} x \plus{} y,b \equal{} y \plus{} z,c \equal{} z \plus{} x$. Then $ x \equal{} \frac {a \minus{} b \plus{} c}2,y \equal{} \frac {a \plus{} b \minus{} c}2,z \equal{} \frac { \minus{} a \plus{} b \plus{} c}2$ are all non negative. We have
\[ 2(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \\
\equal{} 2(2x \plus{} 2y \plus{} 2z)((x \plus{} y)(x \plus{} z) \plus{} (y \plus{} z)(y \plus{} x) \plus{} (z \plus{} x)(z \plus{} y)) \\
\equal{} \sum4x(x^2 \plus{} y^2 \plus{} z^2 \plus{} 3xy \plus{} 3yz \plus{} 3zx) \\
\equal{} \sum4x^3 \plus{} 4xy^2 \plus{} 4xz^2 \plus{} 12x^2y \plus{} 12x^2z \plus{} 12xyz \\
\equal{} \sum 4x^3 \plus{} 16xy^2 \plus{} 16x^2y \plus{} 12xyz\]
and
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 15abc \\
\equal{} \sum(x \plus{} y)^3 \plus{} 5(x \plus{} y)(y \plus{} z)(z \plus{} x) \\
\equal{} \sum2x^3 \plus{} 3x^2y \plus{} 3xy^2 \plus{} 10xyz \plus{} 15x^2y \plus{} 15xy^2 \\
\equal{} \sum2x^3 \plus{} 18x^2y \plus{} 18xy^2 \plus{} 10xyz\]
Thus the inequality is equivalent to
\[ \sum 4x^3 \plus{} 16xy^2 \plus{} 16x^2y \plus{} 12xyz\ge\sum2x^3 \plus{} 18x^2y \plus{} 18xy^2 \plus{} 10xyz\]

\[ \Leftrightarrow \sum 2x^3 \plus{} 2xyz\ge\sum2x^2y \plus{} 2xy^2\]
which is true by Schur.

[quote="leviethai"]Let a,b,c be non-negative numbers, such that $ a \equal{} max${$ a,b,c$}, $ b \plus{} c \ge a$ (or a,b,c are 3 sides of triangle)
Prove that
$ 2(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \ge a^3 \plus{} b^3 \plus{} c^3 \plus{} 15abc$[/quote]

$ 2(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} (a^3 \plus{} b^3 \plus{} c^3 \plus{} 15abc)$

$ \equal{} (3a \minus{} b \minus{} c)(b \minus{} c)^2 \plus{} (b \plus{} c \minus{} a)(a \minus{} b)(a \minus{} c)$

by $ a \equal{} \max\{a,b,c\}$ and $ b \plus{} c \ge a$, ineq clearly true.


trigonometry
limit
calculus
derivative
function
LaTeX
calculus computations

I'm having trouble finding the limit of x sin(1/x) as x-> infinity.
It seems like 1/infinity is zero, and sin(0)= 0. So the limit itself has to be zero. But graphically, I see that it's 1. :?

Even when I differentiate, I still get zero. using the product rule: sin(1/x) + cos(1/x)*-x^-2*x

simplified: sin(1/x) - cos(1/x)*(1/x) which still equals 0.

What's wrong with my logic? And how would you find the limit correctly?

I know these seem easy, but I'm having a little trouble. Hopefully this is my last post on limits for a while. :lol:

$x\sin\left(\frac{1}{x}\right) = x\left(\frac{1}{x} + O(x^{-3})\right) = 1 + O(x^{-2})$

So the limit is $1$.

Edit: You can't differentiate the expression as it is. You are perhaps thinking of L'Hopitals rule. In order to do that, rewrite the expression into a $\left[\frac{0}{0}\right]$ indeterminate form (for instance):

$\frac{\sin(1/x)}{1/x}$

Here both numerator and denominator approaches $0$, so you can differentiate both, take the limit, and you'll get $1$ then too.

[quote="skibum14"][quote="kunny"]$\lim_{x\to\infty} x\sin \frac{1}{x}=\lim_{x\to\infty} \frac{\sin \frac{1}{x}}{\frac{1}{x}}=1.$[/quote] Wait, how in the world does that equal 1? As x goes to infinity, 1/x = 1/infinity = 0. Right? (I really need to know if this is right or not!) Assuming I'm right, this is sin(0)/0. How do you get 1 out of that?[/quote]

Just use $\boxed{\lim_{\theta \rightarrow0} \frac{\sin \theta}{\theta}=1}$

[quote="TheBB"]$x\sin\left(\frac{1}{x}\right) = x\left(\frac{1}{x} + O(x^{-3})\right) = 1 + O(x^{-2})$[/quote] Excuse my ignorance, but I've never seen the notation ${O}$. What does that mean?

[quote]Edit: You can't differentiate the expression as it is. You are perhaps thinking of L'Hopitals rule. [/quote] Why not? I asked my math teacher this also, but he said he didn't know. He said it would be on the calculator part of the exam and I should just check it graphically. :dry:

[quote]In order to do that, rewrite the expression into a $\left[\frac{0}{0}\right]$ indeterminate form (for instance):

$\frac{\sin(1/x)}{1/x}$

Here both numerator and denominator approaches $0$, so you can differentiate both, take the limit, and you'll get $1$ then too.[/quote] Ahh, I get it! I've never heard of indeterminate form before. That means the last post is complete bogus. Sorry kunny.

[quote="kunny"] Just use $\boxed{\lim_{\theta \rightarrow0} \frac{\sin \theta}{\theta}=1}$[/quote] Jeez! Your genius! I never even thought about it like that before. :coolspeak: Thanks!

No problem, skibum14. :) I dare to say that this problem doesn't need to use de l'Hopital's theorem and Landau sign in particular. ;)

you can't use l'hoptial's rule right off the bat becuase it doesnt fall into one of the conditions in which that rule can be applied. as $x$ tends to $\infty$ for $x\sin\left(\frac1x\right)$ you get the indeterminate form $\infty\cdot 0$. If you want to use l'hopital's you need to have $\frac00$ or $\displaystyle \frac{\infty}{\infty}$

[quote="skibum14"]Your right. But I still would like to know what stops me from simply differentiating or plugging in infinity to the original equations like I first did. :)[/quote]
Because it doesn't tell you anything. It's a meaningless non-method.

Let's take a simple example:

Let $f(x)=\frac1x+313.24$

It's clear that $\lim_{x\to\infty}f(x)=313.24.$ It's also clear that nothing about the derivative would tell you that - that constant is simply invisible to the derivative.

At least the derivative of $f(x)$ goes to zero. Does having the derivative go to zero at least mean that some finite limit exists?

Well, no. Here's another example:

Let $g(x)=\sqrt{x}.$

Then $g'(x)=\frac1{2\sqrt{x}}$ and $g'(x)\to0$ as $x\to\infty.$ But $g(x)\to\infty$ as $x\to\infty.$

How about the other way around? If the function tends to zero (or to some finite limit), does the derivative have to go to zero?

Well, no. Once more, an example:

Let $h(x)=\frac1x\sin(x^3).$ Then $h(x)\to0$ as $x\to\infty.$

But $h'(x)=3x\cos(x^3)-\frac1{x^2}\sin(x^3).$ And as $x\to\infty,\,h'(x)$ is unbounded and oscillates wildly.

[quote="Kent Merryfield"]
Because it doesn't tell you anything. It's a meaningless non-method.

Let's take a simple example:

Let $f(x)=\frac1x+313.24$

It's clear that $\lim_{x\to\infty}f(x)=313.24.$ It's also clear that nothing about the derivative would tell you that - that constant is simply invisible to the derivative.

At least the derivative of $f(x)$ goes to zero. Does having the derivative go to zero at least mean that some finite limit exists?

Well, no. Here's another example:

Let $g(x)=\sqrt{x}.$

Then $g'(x)=\frac1{2\sqrt{x}}$ and $g'(x)\to0$ as $x\to\infty.$ But $g(x)\to\infty$ as $x\to\infty.$

How about the other way around? If the function tends to zero (or to some finite limit), does the derivative have to go to zero?

Well, no. Once more, an example:

Let $h(x)=\frac1x\sin(x^3).$ Then $h(x)\to0$ as $x\to\infty.$

But $h'(x)=3x\cos(x^3)-\frac1{x^2}\sin(x^3).$ And as $x\to\infty,\,h'(x)$ is unbounded and oscillates wildly.[/quote] Okay, Mr. Merryfield, I understand why I can't simply differentiate now. But what I'm wondering about more (and what I'm finding the most difficulty understanding) is why I can't simply plug in infinity originally. To do so would give me zero, right? Because (here's a first attempt at LaTex) $\lim_{x\to\infty}f(x)=x\sin\left(\frac1x\right)$ gives you   $\frac1{\infty}\sin\left(\frac1{\infty}\right)$ which is $\infty \cdot 0$ which is $0$.

Besides the LaTex writing ;), what's wrong with that?

[color=blue]Moderator edit: I fixed your LaTex; click "edit" for this post to see the details. One key point: use curly brackets {} to enclose things for LaTeX purposes, such as the denominator of a fraction.[/color]

[quote]... which is $\infty \cdot 0$ which is $0.$ [/quote]
No, that's not what it is. It's not  (literally infinity) times (literally zero).

It's (something that tends to infinity) times (something that tends to zero).

Let's see: a very large number times something ought to be very large,

but,

a very small number times something ought to be very small.

Those are two conflicting tendencies, and we cannot resolve it at a glance. We could find examples (like $xe^{-x}$) that go to zero; we could find examples (like $x^2\left(\sqrt{x^2-1}-x\right)$) that go infinity, we could find examples (like the one the one that started this topic) that go to some finite nonzero limit, and we could find examples that do other things.

Two conflicting tendencies make a "intederminate case." There are a number of indeterminate cases:

$\frac00,\,\frac{\infty}{\infty},\,0\cdot\infty,\,1^{\infty},\,0^0,\,\infty^0.$

None of these are meant to be taken literally; they're all meant as limits. All the designation "indeterminate" is really good for is this: you can't just read off what the answer is. You must do some more work.

I think this thread is a good illustration why teaching limits in high school by intuitive reasoning from graphs, and then introducing L'Hopitals rule as some sort of magical way to compute limits should be banned. L'Hopitals rule is just a confusion factor if you don't know what a limit really is.


Find the coefficient of $x^{2}$ in the polynomial $(1-x)(1+2x)(1-3x)...(1+14x)(1-15x)$.

[hide]
Clearly you have to choose 2 x's out of 15. 105 ways to do that. Therefore each x term will be selected 105/15 = 7 times. Adding all  x coefficients in the factored form gives -8. -8 * 7 = -56.
[/hide]

[quote="JavaMan"]
Adding all  x coefficients in the factored form gives -8. -8 * 7 = -56.
[/quote]

I don't get what "factored form"?

Wow. I just brute force expanded this and the answer is
[hide]
The coefficient is -588.
[/hide]
I wonder if there is a better way to do this.

Is my solution correct?
Also, if it is, I'd like to see a more elegant solution.

Are you sure yours is right, I don't see what's wrong with my solution.

Brute forcing gave me an explicit polynomial. Want me to post it?
It's rather long, but you can check it in Mathematica if you want.

OK, i found my mistake. You should multiply each pair of coefficients, then add them.

You add these up:
-1 * -7
2 * -9
-3 * -6
4 * -10
-5 * -5
6 * -11
-7 * -4
8 * -12
-9 * -3
10 * -13
-11 * -2
12 * -14
-13 * -1
14 * 15
= -588

This works because its the same as -1 * (2 - 3 + 4 - 5 ... - 15) + 2(-3 + 4 - 5 ... -15) + ... + 14(15). Notice the numbers on the right hand side have a pattern, minus 2, plus 3, minus 4 etc


A square $ABCD$ and a circle $\Gamma$ with diameter $AB$ are given Let $P$ be an arbitrary point on the side $CD$, $M$ and $N$  the intersection of the lines $AP$ and $BP$ with $\Gamma$ respectively, and $Q$ is the intersection of the lines $DM$ and $CN$ (Assume that $(M,N) \cap (A,B) = \Phi$). prove that $Q \in \Gamma$ and that $AQ: QB = DP : PC$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=35359 , particularly the EDIT in post #7.

  darij

Thanks, darij!!! Does your site contain all such theorems???


superior algebra
superior algebra unsolved

Let $ R$ be a ring with identiry,given $ n>1$,and for every $ x\in R$,We have $ x^n\equal{}x$,prove the $ R$ is commutative.

[quote="4dspace"]Let $ R$ be a ring with identiry,given $ n > 1$,and for every $ x\in R$,We have $ x^n \equal{} x$,prove the $ R$ is commutative.[/quote]

This is Jacobson's theorem.  A textbook proof is (12.10) page 196 of Lam's First Course in Noncommutative Rings.

Jacobson, N. "Structure theory for algebraic algebras of bounded degree."
Ann. of Math. (2) 46, (1945). 695-707.
http://www.ams.org/mathscinet-getitem?mr=14083
http://dx.doi.org/10.2307/1969205

[quote="coquitao"]Take a look at the last post in this [url=http://www.mathlinks.ro/viewtopic.php?t=259395]this[/url] thread.[/quote]

That won't help because that thread is only about the simpler case n = 3. That proof does not generalize to arbitrary n. In fact it was a longstanding open problem to try to find a purely equational proof that works for general n (vs. the model-theoretic proof by Jacobson). By the completeness theorem for equational logic, such a proof must exist. Burris told me that John Lawrence discovered such a proof, but afaik it is not yet published. See their earlier joint paper [1] for some partial results.

--Bill Dubuque

[1] S. Burris and J. Lawrence, Term rewrite rules for finite fields.
International J. Algebra and Computation 1 (1991), 353-369.
http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PAPERS/fields3.pdf


What would the sin equation be if a tennis ball starts at 5 ft in the air, hits 0 ft, come back up 2.5 feet. Repeat process. Also, what is the amplitude, period, vertical shift, and the phase shift?  THANKSSS!!!!!!!!!!!!!!!!!!! :D  :lol:  :blush:

anyone!??! I really need some help on this...Thanks!

this isnt a sinosoidal graph...
i dont think

Yeah it is. It's because I start the ball at 5 feet in the air, drop it, it hits the floor (0feet) and then comes up to 2.5 feet  (repeat process)

my question has always been:

Who cares? (you should tell that to your teacher)

I mean, its not like Im going to be in a tennis match and start calculating the equation of a tennis ball...or how the graph of a home run looks like

Anyways, more seriously, you should read the chapter in your book covering that...reading is boring but you can learn a lot from it then asking people.

Well, i need this for a project....thanks for your reply. Helped tons....really.


Given x^2 - ax + 2 = 0, with solutions s and r, find (in terms of a):
a)  s^2 + r^2,  s^3 + r^3
b)  s^3 + r^3

Can we use Newton's sums for these? I'm just about to learn about them in AoPS II.

EDIT (So I CAN do these with Newton's Sums, it's pretty much the same thing.):
$s_{1}+(-a)=0,$ so $s_{1}=a.$
$s_{2}+(-a)(a)+4=0,$ so $s_{2}=a^{2}-4$.
$s_{3}+(-a)(a^{2}-4)+2a=0$, so $s_{3}=a^{3}-6a$.

I have the same solution as Samath. Maybe I should review this semi-hourly.


1) if a, b, c, d are positive integers for which ab=cd, show that a:^2: +b:^2:+c:^2:+d:^2: is composite.

2) Ma and Pa and brother and me
The sum of our ages is eighty-three
Six times Pa's is seven times Ma's
And Ma's is three times me.

Find the ages: no fractions please

3) Prove that the sum of the digits of every multiple of 2739726 up to the 72nd is 36.

4) Show that without using a calculator that 71/2+71/3+71/4<7 and 41/2+41/3+41/4>4

5) Prove that, for all natural numbers n, 22n+24n-10 is divisible by 18.

6) let n be an integers. show that the greatest common divisor of n^2+1 and (n+1)^2+1 is either 1 or 5.
ill post more later

have fun!

[quote="Syntax Error"]4) Show that without using a calculator that 71/2+71/3+71/4<7 and 41/2+41/3+41/4>7[/quote]

Is this perhaps not correct?

sorry bout that, i had just copied and pasted the equation with 7's in it because i didnt feel like writing all those superscripts again. ill edit it in a sec

2. Ma is 6; Pa is 7. They have adopted two children: you are 2 and your brother is 68. (nice happy family)

2. how bout Ma is 30, Pa is 35, you are 10, and your brother is 8?

yep, another solution would be where a son is 1 year younger than the mom, but thats not really possible.

5. [hide]22n + 24n - 10 is even.

If n :equiv: 0 mod 3, 22n + 24n - 10 :equiv: 1 + 0 - 1 :equiv: 0 mod 9

If n :equiv: 1 mod 3, 22n + 24n - 10 :equiv: 4 + 6 - 1 :equiv: 0 mod 9

If n :equiv: 2 mod 3, 22n + 24n - 10 :equiv: 7 + 3 - 1 :equiv: 0 mod 9

Therefore 22n + 24n - 10 is always divisible by 2*9 = 18.[/hide]

Syntax Error wrote:1) if a, b, c, d are positive integers for which ab=cd, show that a:^2: +b:^2:+c:^2:+d:^2: is composite.

[hide]

A try for this one.

ab=cd can be rewritten a/c=d/b=x/y with pcg(x,y) = 1

Then y <= c and y <= b. (*)

A = a:^2: +b:^2:+c:^2:+d:^2: = (c+b) * (x+y) / y

Since gcd(x,y)=1, gcd(x+y,y)=1 (easy) so since A is an integer y | c+b. But because of (*) y < c+b.

So A = (c+b)/y * (x+y) with (c+b)/y and (x+y) both positive integer > 1.[/hide]

Syntax Error wrote:6) let n be an integers. show that the greatest common divisor of n^2+1 and (n+1)^2+1 is either 1 or 5.

[hide]

Let p prime s.a. p |  N1=n^2+1 and N2=(n+1)^2+1.

Then p | N2-N1=2n+1

Then p | (2n+1) - 4*N1 = 4n-3

Then p | 2(2n+1) - 4n-3 = 5

So gcd(n^2+1,(n+1)^2+1)) = 1 or 5

Both are possible : (1+1,2+1)=1, (2+1,3+1)=5[/hide]

#1 is from a German olympiad. We came across this problem in Putnam practice and found the following beautiful solution (in spoiler):

[hide]Add zero creatively: a2+b2+c2+d2=a2+2ab+b2+c2-2cd+d2=a2-2ab+b2+c2+2cd+d2=(a+b)2+(c-d)2=(a-b)2+(c+d)2. Thus a2+b2+c2+d2 can be written as a sum of two squares in two ways. It's not hard to show that these two ways must be distinct if a2+b2+c2+d2 is to be prime. But then we have two different ways to write a2+b2+c2+d2 as the sum of two squares, but primes don't work like that, since primes congruent to 1 (mod 4) can be expressed as the sum of two squares in exactly one way, and primes congruent to 3 (mod 4) can be expressed as the sum of two squares in exactly zero ways. Thus a2+b2+c2+d2 is composite.[/hide]

[quote="Syntax Error"]3) Prove that the sum of the digits of every multiple of 2739726 up to the 72nd is 26. [/quote]

Isn't it rather 36  :wink: ??


inequalities
algebra
polynomial
inequalities unsolved

For positive reals $a,b,c$ find maximal $k$ such that:
$b\frac{(a-b)^2}{a^2}+c\frac{(b-c)^2}{b^2}+a\frac{(c-a)^2}{c^2}\geq
k\left(\frac{(a-b)^2}{a}+\frac{(b-c)^2}{b}+\frac{(c-a)^2}{c}\right)$

This ineq is equivalent to:
$(1+k)(a+b+c)+\left(\frac{b^3}{a^2}+\frac{c^3}{b^2}+\frac{a^3}{c^2}\right)\geq
(2+k)\left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right)$

I don't know which $k$ is maximal, but here's one problem with concrete $k$ (if finding maximum is very hard):
$17(a+b+c)+10\left(\frac{b^3}{a^2}+\frac{c^3}{b^2}+\frac{a^3}{c^2}\right)\geq
27\left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right)$

Thanx for any ideas or solutions to these (for me) very hard problems.

[quote="Yu-Dong Wu"]Using the software that called BOTTEMA which was invented by Prof. Lu Yang(a Chinese mathematician), we obtain the following result:

`The best possible maximal const `.k.` is a root of the following polynomial :`
16*k^7+83*k^6-207*k^5-1440*k^4+3537*k^3+23220*k^2+14229*k-32535
`which is between (`, 1/2, `，`, 225/253, `)`

And we can find the following equation has only one real root.

16k^7+83k^6-207k^5-1440k^4+3537k^3+23220k^2+14229k-32535=0

It's approximation is 0.8892597769.[/quote]

hi, do you mind sending me the software?

It's wonderful result, that maximal $k$. But in other way I'd like to see a proof for some specific $k$ without using computer. For example if $k=0$ it's quite easy ;) What about slightly bigger $k$? If $k=\frac{1}{2}$ it comes to:

$3(a+b+c)+2\left(\frac{b^3}{a^2}+\frac{c^3}{b^2}+\frac{a^3}{c^2}\right)\geq
5\left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right)$

[quote="ondrob"]

I don't know which $k$ is maximal, but here's one problem with concrete $k$ (if finding maximum is very hard):
$17(a+b+c)+10\left(\frac{b^3}{a^2}+\frac{c^3}{b^2}+\frac{a^3}{c^2}\right)\geq
27\left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right)$

[/quote]

this is so hard   :roll:

ondrob, do you have a solution for it?

No, I haven't got it... I find it very hard. After all estimates I tried, sooner or later I found that these estimates were too bad... I'd like to see a solution too...

I would like to ask two questions :

1) Could [b]Yu-Dong Wu[/b] briefly explain the ideas he used to derive the polynomial equation for the optimal $k$ ?

2) Is the two-variable version of the problem, i.e., find the optimal $k$ so that, for $a,b>0$,

$(1+k)(a+b) + \left(\frac{a^3}{b^2}+\frac{b^3}{a^2}\right) = (2+k)\left (\frac{a^2}{b}+\frac{b^2}{a}\right)$

easier to cope with ? I am aware that the 2-variable problem is symmetrical, while the original 3-variable problem is not, but perhaps it might provide us with some insights :idea:


$ \triangle ABC$, the projections of $ A,B,C$ to sides $ BC,CA,AB$ are $ A_1,B_1,C_1$ respectively. Prove that the Euler's lines of $ \triangle AB_1C_1, \triangle A_1BC_1, \triangle A_1B_1C$ have a common point.

It has been posted before http://www.mathlinks.ro/Forum/viewtopic.php?t=113311

Kostas Vittas

Oh thanks~ :lol:
I just couldn't find that...

There is more general problem. See the Theorem 1.8 of note "Generalization of the Feuerbach point" on darij grinberg website http://de.geocities.com/darij_grinberg/


I just want to get a sense of the overall scores so please be honest.

7- i was really close on #6... two steps away from the answer when the guy said 1 minute... and i suddenly couldn't do math! blah... guessed 6.. the answer was 66

7: long division somehow never occurred to me on #6, I reduced it to a congruence and made an educated guess, having no time left to calculate.

6:actually on number 7 long division was the first idea i had. i tried to break up the problem into three minature problems however like AllanZ i ran out of time before i could complete it!

6--on #6, my faulty thought process was lets multiply 729 by numbers with successively more digits that get more and more 0's in the result--I guess that technically would have worked eventually, but it would have taken a 64 or so digit number. The long division solution was nice, though. On #5, I knew even digits couldn't be in the middle, and neither could five. 9731 works, so now we have to choose between 5 and 2 at the beginning. Somehow, I didn't quite realize that 8 was a digit.


hmmm, With Cauchy and Chebychev I can get it down to $ \frac{\sqrt{3}}{3}$

[quote="orl"]Let $ a_1, a_2, \ldots, a_{100}$ be nonnegative real numbers such that $ a^2_1 \plus{} a^2_2 \plus{} \ldots \plus{} a^2_{100} \equal{} 1.$ Prove that
\[ a^2_1 \cdot a_2 \plus{} a^2_2 \cdot a_3 \plus{} \ldots \plus{} a^2_{100} \cdot a_1 < \frac {12}{25}.
\]
[i]Author: Marcin Kuzma, Poland[/i][/quote]

Anybody has idea ?

[quote="Allnames"]use
Lagrangre Multiplier method,I found The Maximum of S (our RHS) is $ 0,458879$[/quote]

would you post the full solutions?

 :roll:

Can someone please post a solution properly for this question? I am in desperate neeed of it.

[quote="Allnames"]use
Lagrangre Multiplier method,I found The Maximum of S (our RHS) is $ 0,458879$[/quote]

Are you sure? :wink:

[quote="tin_124"]who can show me When does Equalities hold ?[/quote]
It doesn't hold. :lol:

Could anyone post the solution or even a sketch of it because it's rather hard inequality and if I'm right this problem has been rejected by IMO Commete. I can't prove it myself but I'm very interested in the method which kills it.

Without knowing Lagrange multipliers this problems is nearly impossible. Let $\sum a_i^2a_{i+1} =S$. Using Lagrange multipliers, one finds that the maximum of $S$ is attained if $a_{i-1}^2+2a_ia_{i+1}=-2\lambda a_i$  for all $i$. Since $\lambda$ is constant, the sequences $(a_{i-1}^2+2a_ia_{i+1})$ and $(a_i)$ are the same up to some constant factor. Hence one can use Cauchy-Schwarz as follows:
\[(3S)^2=\left(\sum a_i(a_{i-1}^2+2a_ia_{i+1}) \right)^2 \le \left( \sum a_i^2 \right) \left( \sum (a_{i-1}^2+2a_ia_{i+1})^2 \right) \mbox{.}\] From here it's not hard to obtain the bound $\sqrt{2}/3$.

Let $x_1,x_2,\cdots,x_n (n\ge 3) $ be non-negative real numbers satisfying $x_1+x_2+\cdots+x_n=1$ ,[url=https://artofproblemsolving.com/community/c6h507933p2853319]Show that[/url]\[x^2_1x_2+x^2_2x_3+\cdots+x^2_{n-1}x_n+x^2_nx_1\leq \frac {4}{27}.\]

[quote=msceok][quote="orl"]Let $ a_1, a_2, \ldots, a_{100}$ be nonnegative real numbers such that $ a^2_1 \plus{} a^2_2 \plus{} \ldots \plus{} a^2_{100} \equal{} 1.$ Prove that
\[ a^2_1 \cdot a_2 \plus{} a^2_2 \cdot a_3 \plus{} \ldots \plus{} a^2_{100} \cdot a_1 < \frac {12}{25}.
\]
[/quote]
This is not a neat problem. In fact, the official solution proves the stronger inequality with the constant $ \frac{\sqrt{2}}{3}$.[/quote]

please enter ur solution. its right. because 12/25>ur solution.

[quote=orl]
[i]Author: Marcin Kuzma, Poland[/i][/quote]

PhD Marcin KUCZMA

I hope this is correct  :)
For the sake of a clearer and more legible solution, let $a_1^2=x_1\dots a_{100}^2=x_{100}, \text{ and, }a_1=y_1\dots a_{100}=y_{100}$, thus the condition becomes $\sum_{i=1}^{100}x_i=1 \text{, and } \sum_{i=1}^{100}x_iy_{i+1}<\frac{12}{25}$
Furthermore notice that: $\sum_{i=1}^{100}x_iy_{i+1}\overset{\text{Chebyshev}}{\le}100\bigg(\frac{1}{100}\sum_{i=1}^{100}x_i\bigg)\bigg(\frac{1}{100}\sum_{i=1}^{100}y_i\bigg)=\sum_{i=1}^{100}\frac{1}{100}y_i$
Now, $\sum_{i=1}^{100}\frac{1}{100}y_i\overset{\text{C-S}}{\le}\sqrt{\sum_{i=1}^{100}y_i^2}\sqrt{\frac{1}{100}}$, however notice that $\sum_{i=1}^{100}y_i^2=\sum_{i=1}^{100}x_i=1$, thus the inequality is equivalent to $\frac{1}{10}<\frac{12}{25}$, which is clearly true  $\blacksquare$


Prove that $ \sqrt{2}$ is irrational by using mathematical induction.

Scroll down to [url=http://en.wikipedia.org/wiki/Square_root_of_2#Proofs_of_irrationality]Analytic Proof[/url].  It's the second proof.


AMC
USA(J)MO
USAMO
trigonometry
LaTeX
linear algebra
matrix

[size=200]Problem 3 should have been[/size]:
[hide]Prove that $\sqrt{1-\cos{A}}\sqrt{1-\cos{B}}+\sqrt{1-\cos{C}}\sqrt{1-\cos{D}}=\sqrt{1-\cos{(A+C)}}\sqrt{1-\cos{(A+D)}}$
[/hide]

Rules: 4.5 hours, no calculators, no computers, no brains (jk), etc

Submitting Solutions: Either LaTeX it or write it up and scan it. Then pm me with your solutions. You could even LaTeX it into the pm.

Scoring: Each problem is out of 7 points. I and possibly other graders will grade your solutions and send feedback.

In a few days or so, I'll post the solutions (or maybe in a couple days, I'm not sure).

There should be a second day after the USAMO comes out. After then, we'll have scores, solutions, and eventually results.

[hide="Comments"]#1 is sort of well-known, but not too much so and it makes a good problem.

#2 is...pretty cool?

#3 is my favorite problem[/hide]

Nice work with those questions!
I think they're probably beyond me, but I can tell they're nice, and I'll try them out.  :lol:

[hide="hrmdiddyderm"]
Looks a bit easier than an actual USAMO, but nice nonetheless.
[/hide]

Umm grr...there's an error in #3 so it's not quite what I wanted it to be.

Hmm sorry what should I do? Should I post the problem I wanted or leave it as is? (I guess the current problem is valid but it's not what I wanted it to be).

Ummm... if there is still a nice solution to the current problem, then you should probably just leave it, but otherwise just post what you meant it to be. Actually w/e you might as well post your intended problem anyway.

[quote="calc rulz"]Rules: 4.5 hours, no calculators, no computers, no brains, etc[/quote]

So, if we can't use our brain, how are we supposed to solve the problems?

Man, me and my brain don't get along. so i don't use it anyway.

and for an alternate interpretation: doesn't that put like white people at a huge disadvantage?

#1 seems to have a slight problem...

[hide]The last column should have $n-1$ in the exponent, not $n$; took me a while to realize why my approach wasn't working...  :huh: [/hide]

Hmm okay that's what it should be...
But otherwise it's the same, it's still a geometric series of length $n$.

Umm hmm should I? Well yeah I guess if you use the result of my intended #3 there is a nice solution to the older problem. The old #3 just a bit of extra and random/stupid (but not too messy) work in it.

So I guess I should say use the new one, but if you did it with the old one then I'll accept it.
[hide]
\item Let $A$ be an $n \times n$ matrix with $n \ge 4$ such that each row is a geometric sequence of length $n$ that consists entirely of nonzero terms, as shown:
\[\begin{pmatrix}a_{1}& a_{1}r_{1}& a_{1}r_{1}^{2}& a_{1}r_{1}^{3}& ... & a_{1}r_{1}^{n-1}\\ a_{2}& a_{2}r_{2}& a_{2}r_{2}^{2}& a_{2}r_{2}^{3}& ... & a_{2}r_{2}^{n-1}\\ ... & ... & ... & ... & ... & ... \\ ... & ... & ... & ... & ... & ... \\ a_{n}& a_{n}r_{n}& a_{n}r_{n}^{2}& a_{n}r_{n}^{3}& ... & a_{n}r_{n}^{n-1}\end{pmatrix}\]
Prove that the determinant of $A$ is $0$ if and only if at least two of the common ratios are equal.

\item Let
\[f(n) : = \frac{\sum^{n}_{k=1}\frac{(2n)! 2^{k}}{ (2k-1)! (2n-2k+1)!}}{\sum^{n}_{k=0}\frac{(2n)! 2^{k}}{2k! (2n-2k)!}}\]
Let $k$ be the smallest real number such that $f(n) \le k$ for all positive integers $n$. With proof, find $k$.

\item Let $A$, $B$, $C$, and $D$ be the angles of a convex quadrilateral. Prove that $\sqrt{1-\cos{A}}\sqrt{1-\cos{B}}+\sqrt{1-\cos{C}}\sqrt{1-\cos{D}}=\sqrt{1-\cos{(A+C)}}\sqrt{1-\cos{(A+D)}}$
.[/hide]

Are you sure the new #3 is true? For example take the angles to be A = 90, B = 60, C = 120, D = 90.

Umm I get 1.93... for both sides so I think it is true

Okay, maybe if a mod could edit the first post to mention the (minor) error in #1?


1.Find the sum upto first n-terms of the series

$ \frac {1}{4} \plus{} \frac {1.3}{4.6} \plus{} \frac {1.3.5}{4.6.8} \plus{} ....$

2.Evaluate

 $ \sum_{n \equal{} 1}^{\infty}\left(\frac {\tan \frac {\theta}{2^n}}{2^{n \minus{} 1}\cos \frac {\theta}{2^{n \minus{} 1}}}\right)$

I believe the answer to the first one is
\[ 1\minus{}\frac{3.5.7\cdots}{4.6.8\cdots}\equal{}1\minus{}\prod_{k\equal{}1}^n\frac{2k\plus{}1}{2k\plus{}2}\equal{}1\minus{}2^{\minus{}2n}{2n\plus{}1\choose n}\]


Not sure if it's too well known or too hard.

Solve in integers:

$2^{n}-7=x^{2}$

well by inspection you see:

$n=3$ and $x=1$ works

EDIT: There are more values btw (obviously)
$n=4$ and $x=3$ & $n=5$ and $x=5$ also works

In general $x^{2}$ must be odd because L.S. is odd..therefore $x$ is odd..but not all odd values work..so..

The trick is subtracting $1$ from both sides; this yields $2^{n}-8=x^{2}-1$ which implies $2^{3}\cdot [2^{n-3}-1]=(x-1)(x+1)$.

Note: I did not realize that $2^{n}-1$ is actually not always prime... duh

[quote="zgorkster"]The trick is subtracting $1$ from both sides; this yields $2^{n}-8=x^{2}-1$ which implies $2^{3}\cdot [2^{n-3}-1]=(x-1)(x+1)$.  Since $x-1$ and $x+1$ are the same parity, there are only four cases for the integer solutions to this equation.  So for the solutions $(n,x)$, we obtain $(3,1),(4,3),(5,5)$ as the only solutions.[/quote]

Why would there be only four cases?

It's very well-known, I've seen three other topics on it.  I posted in one, but my solution misses a case.  Anyway, the largest solution is $(n, x) = (15, 181)$ which is why this is not such an easy question.

Edit:  [hide="Musings"] Obviously $n$ cannot be even for large $n$ because $2^{n}$ will be a square, and the only squares that differ by $7$ are $9, 16$.

If $n$ is large and odd, let $n = 2k+1$ and let $y = 2^{k}$.  Then we are solving

$2y^{2}-x^{2}= 7$

We can use Pell's methods to characterize all integer solutions and probably show that only finitely many of them have $y$ a power of $2$.  I think. [/hide]

All mentioned here n = 3, 4, 5, 7, 15 and no more.

See ITEM 31 here:
http://www.inwap.com/pdp10/hbaker/hakmem/number.html#item31


Let a,b,c positive numbers such that abc=1. Prove that

 $ \sqrt{\frac{b\plus{}c}{a^3\plus{}1}}\plus{}\sqrt{\frac{c\plus{}a}{b^3\plus{}1}}\plus{}\sqrt{\frac{a\plus{}b}{c^3\plus{}1}}\geq\frac{9}{a\plus{}b\plus{}c}$

As usually I'm being unable to post a complete solution.  I'm failing to finish it just due to a single proof. Here's my approach:
\[ \sqrt\frac{a^3+1}{b+c}+\sqrt\frac{b^3+1}{c+a}+\sqrt\frac{c^3+1}{a+b}=\sqrt\frac{a^3+abc}{b+c}+\sqrt\frac{b^3+abc}{c+a}+\sqrt\frac{c^3+abc}{a+b}\]
or,\[ \sqrt\frac{a^3+abc}{b+c}+\sqrt\frac{b^3+abc}{c+a}+\sqrt\frac{c^3+abc}{a+b}=\sqrt\frac{a(a^2+bc)}{b+c}+\sqrt\frac{b(b^2+ca)}{c+a}+\sqrt\frac{c(c^2+ab)}{a+b}\leq(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})(a^2+b^2+c^2+ab+bc+ca)\]
Since,\[ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\leq\frac{3}{2}\]
so,\[ \sqrt\frac{a^3+abc}{b+c}+\sqrt\frac{b^3+abc}{c+a}+\sqrt\frac{c^3+abc}{a+b}\leq\frac{3}{2}\times{2(a^2+b^2+c^2)}={3}{(a^2+b^2+c^2)}\]
Thus,\[ (\sqrt\frac{b+c}{a^3+abc}+\sqrt\frac{c+a}{b^3+abc}+\sqrt\frac{c^3+abc}{a+b})(\sqrt\frac{a^3+abc}{b+c}+\sqrt\frac{b^3+abc}{c+a}+\sqrt\frac{a+b}{c^3+abc})\geq{9}\]
i.e.\[{ (\sqrt\frac{b+c}{a^3+abc}+\sqrt\frac{c+a}{b^3+abc}+\sqrt\frac{c^3+abc}{a+b})\geq\frac{3}{(a^2+b^2+c^2)}}\]
Hereafter I failed to proceed.....

[quote="Marius Mainea"][quote="Sunkern_sunflora"]
Since,
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\leq\frac {3}{2}
\]
so,[/quote]

 Try a=4 ,b=c=1[/quote]
Isn't this Nesbitt's inequality, and shouldn't it be$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\geq\frac {3}{2}$?

[quote="Marius Mainea"]Let a,b,c positive numbers such that abc=1. Prove that

 $ \sqrt {\frac {b \plus{} c}{a^3 \plus{} 1}} \plus{} \sqrt {\frac {c \plus{} a}{b^3 \plus{} 1}} \plus{} \sqrt {\frac {a \plus{} b}{c^3 \plus{} 1}}\geq\frac {9}{a \plus{} b \plus{} c}$[/quote]
$ \sum_{cyc}\sqrt {\frac {b \plus{} c}{a^3 \plus{} 1}}\geq3\sqrt[6]{\prod_{cyc}\frac {b \plus{} c}{a(a^2\plus{}bc)}}\geq\frac {9}{a \plus{} b \plus{} c}.$ :wink:

I'm sorry........I was full of sleep and my proof was full of mistakes. Not only in the Nesbitt's inequality, but also in applying cauchy. :oops:
arqady wrote:
   \[ \sum_{cyc}\sqrt {\frac {b \plus{} c}{a^3 \plus{} 1}}\geq3\sqrt[6]{\prod_{cyc}\frac {b \plus{} c}{a(a^2\plus{}bc)}}\geq\frac {9}{a \plus{} b \plus{} c}\]

I still didn't get why you used the 6th root. And also please clarify the last step.......infact your solution seems too short to guess, at least for me. :huh:

I'm sorry........I was full of sleep and my proof was full of mistakes. Not only in the Nesbitt's inequality, but also in applying cauchy. :oops:
[quote="arqady"]
\[ \sum_{cyc}\sqrt {\frac {b \plus{} c}{a^3 \plus{} 1}}\geq3\sqrt [6]{\prod_{cyc}\frac {b \plus{} c}{a(a^2 \plus{} bc)}}\geq\frac {9}{a \plus{} b \plus{} c}
\][/quote]
I still didn't get why you used the 6th root. And also please clarify the last step.......infact your solution seems too short to guess for me. :huh:

This is my ugly solution  :blush:
Using Holder we got that
$ \sum \sqrt {\frac {b + c}{a^3 + 1} }\geqslant \sqrt {\frac {8(a + b + c)^3}{\sum (b + c)^2(a^3 + abc)}}$
it remains to prove that
$ \sqrt {\frac {8(a + b + c)^3}{\sum (b + c)^2(a^3 + abc)}} \geqslant \frac {9}{a + b + c}$
which is
$ 8(a + b + c)^5\geqslant 81(4\sum a^3bc + \sum_{sym} a^3b^2 + 2\sum a^2b^2c)$
and then boom boom...you will get that $ L.H.S \geq R.H.S$

+Of course it must be 6th root cause mr.arqady used AM-GM for that...

Ooh, I am sorry, Sunkern_sunflora. I didn't see your post. :(
[quote="Sunkern_sunflora"]
[quote="arqady"]
\[ \sum_{cyc}\sqrt {\frac {b \plus{} c}{a^3 \plus{} 1}}\geq3\sqrt [6]{\prod_{cyc}\frac {b \plus{} c}{a(a^2 \plus{} bc)}}\geq\frac {9}{a \plus{} b \plus{} c}
\]
[/quote]
I still didn't get why you used the 6th root. [/quote]
Because by AM-GM $ \sum_{cyc}\sqrt {\frac {b \plus{} c}{a^3 \plus{} 1}}\geq3\sqrt [3]{\prod_{cyc}\sqrt {\frac {b \plus{} c}{a^3 \plus{} 1}}} \equal{} 3\sqrt [6]{\prod_{cyc}\frac {b \plus{} c}{a(a^2 \plus{} bc)}}.$
[quote="Sunkern_sunflora"]And also please clarify the last step......[/quote]
The last inequality is equivalent to
\[ (a \plus{} b \plus{} c)^6(a \plus{} b)(a \plus{} c)(b \plus{} c)\geq729abc(a^2 \plus{} bc)(b^2 \plus{} ac)(c^2 \plus{} ab)
\]
for which I didn't find a nice proof, but it obvious after expanding.

[quote="Marius Mainea"]Let a,b,c positive numbers such that abc=1. Prove that

 $ \sqrt {\frac {b \plus{} c}{a^3 \plus{} 1}} \plus{} \sqrt {\frac {c \plus{} a}{b^3 \plus{} 1}} \plus{} \sqrt {\frac {a \plus{} b}{c^3 \plus{} 1}}\geq\frac {9}{a \plus{} b \plus{} c}$[/quote]

  Follows $ GM\geq HM$ and $ AM\geq HM$ we get

 $ LHS\equal{}\sum {\sqrt{\frac{b\plus{}c}{a^2\plus{}bc}\frac{1}{a}}}\geq\sum {\frac{2}{\frac{a^2\plus{}bc}{b\plus{}c}\plus{}a}}\equal{}\sum {\frac{2(b\plus{}c)}{(a\plus{}b)(a\plus{}c)}}\geq\frac{2\cdot9}{2(a\plus{}b\plus{}c)}\equal{}RHS$


$ a,$ $ b$ and $ c$ are real numbers such that $ \{a,b,c\}\equal{}\{a^4\minus{}2b^2,b^4\minus{}2c^2,c^4\minus{}2a^2\}$ and $ a\plus{}b\plus{}c\equal{}\minus{}3.$
Find the values of $ a,$ $ b$ and $ c.$

we have $ \{a,b,c\}\equal{}\{a^4\minus{}2b^2,b^4\minus{}2c^2,c^4\minus{}2a^2\}$ so $ \{a\plus{}1,b\plus{}1,c\plus{}1\}\equal{}\{a^4\minus{}2b^2\plus{}1,b^4\minus{}2c^2\plus{}1,c^4\minus{}2a^2\plus{}1\}$ so :
$ 0\equal{}(a\plus{}1)\plus{}(b\plus{}1)\plus{}(c\plus{}1)\equal{}(a^4\minus{}2b^2\plus{}1)\plus{}(b^4\minus{}2c^2\plus{}1)\plus{}(c^4\minus{}2a^2\plus{}1)$
$ \equal{}(a^2\minus{}1)^2\plus{}(b^2\minus{}1)^2\plus{}(c^2\minus{}1)^2$ therefore
$ a^2\equal{}b^2\equal{}c^2\equal{}1$ and since $ a\plus{}b\plus{}c\equal{}\minus{}3$ we get $ a\equal{}b\equal{}c\equal{}\minus{}1$.

Nice solution, 	tchebytchev! What about the following problem:
$ a,$ $ b$ and $ c$ are real numbers such that $ \{a,b,c\} \equal{} \{a^6 \minus{} 2b^2,b^6 \minus{} 2c^2,c^6 \minus{} 2a^2\}$ and $ a \plus{} b \plus{} c \equal{} \minus{} 3.$
Find the values of $ a,$ $ b$ and $ c.$ $ ?$ :wink:


limit
trigonometry
real analysis
real analysis unsolved

Let $ f$ be defined on $ (\minus{}1,1)$, $ f'(0)$ exists.  Suppose $ \minus{}1 < \alpha_n < \beta_n < 1$ and $ \alpha_n,\beta_n \rightarrow 0$ as $ n \rightarrow \infty$.  Put
$ D_n : \equal{} \frac{f(\beta_n) \minus{} f(\alpha_n)}{\beta_n\minus{}\alpha_n}.$
Prove that:
(i) if $ \alpha_n < 0 < \beta_n$, then $ \lim D_n \equal{} f'(0)$.
(ii) if $ 0 < \alpha_n < \beta_n$ and $ \{\beta_n/(\beta_n\minus{}\alpha_n)\}$ is bounded,  then $ \lim D_n \equal{} f'(0)$.
(iii)  If $ f'$ is continuous in $ (\minus{}1,1)$, then $ \lim D_n \equal{} f'(0)$.

Find an example for which $ f$ is differentiable in $ (\minus{}1,1)$, $ f'$ not continuous at $ 0$, and $ \lim D_n \neq f'(0)$ but $ \lim D_n$ exists.

The following formula should prove useful:

$ \frac {f(\beta_n) \minus{} f(\alpha_n)}{\beta_n \minus{} \alpha_n} \equal{} \frac {\beta_n \minus{} 0}{\beta_n \minus{} \alpha_n}\cdot\frac {f(\beta_n) \minus{} f(0)}{\beta_n \minus{} 0} \plus{} \frac {0 \minus{} \alpha_n}{\beta_n \minus{} \alpha_n}\cdot\frac {f(\alpha_n) \minus{} f(0)}{\alpha_n \minus{} 0}$

with suitable modifications if either $ \alpha_n \equal{} 0$ or $ \beta_n \equal{} 0.$

For the last item, consider this example:

$ f(x) \equal{} \begin{cases}x^2\cos\left(\frac1x\right), & x\ne0 \\
0, & x \equal{} 0\end{cases}.$

$ \alpha_n \equal{} \frac {1}{2\pi n},\beta_n \equal{} \frac {1}{2\pi(n \minus{} 1)}$

I agree this shows how to solve i,ii,iii . (For (iii) I use MVT to get a sequence $ t_n \rightarrow 0$ such that $ D_n \equal{} f'(t_n) \rightarrow f'(0)$.)

But for your example, I have a question.  Don't we have $ f'(0) \equal{} 0$ and also $ \lim D_n \equal{} 0$ ?

I gave you the wrong $ \beta_n.$ Let $ \alpha_n$ be $ \frac1{2\pi n}$ as before but now let $ \beta_n\equal{}\frac1{\pi(2n\minus{}1)}.$ Try that and see where it leads.


Given positive reals $x,y,z$, prove that

$((x-1)^2+11)((y-1)^2+11)((z-1)^2+11) > 8(xyz+27)^{\frac{3}{2}}$

NOTE: I posted this in Pre-Olympiad, but no one has solved it, so I'm reposting it here.

EDIT: It's been solved now.


I like the following problem very much !
It's maybe my favourite geometry problem together with IMO 1999/5.

ALL INDICES MODULO 5 ! (So the only possible indices are 1, 2, 3, 4, 5.)

Given is a convex pentagon A(1)A(2)A(3)A(4)A(5), no two sides of which are parallel.
Let B(i) be the intersection of A(i+1)A(i+2) and A(i+3)A(i+4) for i = 1, ..., 5.
Let C(i) be the circle trough B(i), A(i+2) and A(i+3) for i = 1, ..., 5.
The circles C(i) and C(i+1) intersect in A(i+3) and D(i+3) for i = 1, ..., 5.
Prove that the five points D(1), D(2), ..., D(5) are concyclic.

ENJOY IT :)

The main idea is that Di is the Miquel point of the quadrilateral A(i-1)AiA(i+1)Bi. This means that the quadrilaterals BiB(i+3)DiD(i+3) are cyclic. Using this and some crazy angle-chasing we get the conclusion. It would take too much to post a complete soln, but here's how I proved that Quadrilateral D2D3D4D5 is cyclic (the others are exactly the same):

Here all 3-letter things mean angles. D1D5D2=D1D5B2-D2D5B2=Pi-D1A1A5+Pi-B2A4A5-(Pi-(Pi-D2A2A3)) (here I used the fact that B5B2D5D2 is cyclic, from that Miquel thing I said before, and D1A1D5A5 and B2D5A5A4 are also cyclic) = 2Pi-(B2A4A5+D1A1A5+D2A2A3) (1)

D1D5D4=Pi-D1A1A5+Pi-D4A4A5=2Pi-(D1A1A5+D4A4A5) (2)

D2D5D4=D1D5D4-D1D5D2 from (1) and (2) we get D2D5D4=D2A2A3+B2A4A5-D4A4A5 (3)

D2D3D4=Pi-D2A2A3+Pi-D4A4A3 (4)

Now we add (3) and (4) and we get D2D5D4+D2D3D4=2Pi+B2A4A5-D4A4A5-D4A4A3. We can see that D4A4A5+D4A4A3=2Pi-A3A4A5, and by replacing this we get D2D5D4+D2D3D4=B2A4A5+A3A4A5, which is clearly Pi, which means that 2 opposite angles in the quadrilateral D2D3D4D5 add up to Pi, so this is a cyclic quadrilateral. We do the same for quadrilateral D1D3D4D5 and we are done.

What's your soln? P.S. My beloved inversion becomes REALLY nasty in this case. We have too much stuff to try it.  :D

I'll post my solution later :D
But what's a Miquel point ? :?

For a convex quadrilateral ABCD let E be the intersection of AB and CD and F the intersection of AD and BC. Then the circumcircles of BEC, AED, AFB, DFC intersect at a point M. This is Miquel's point of the quadrilateral ABCD. It's easy to prove, just mess a bit with the angles. I think there's another thm which states that the orthocenters of the four triangles above are collinear. I'll have to check on this one, I'm not sure I remember right.


function
integration
real analysis
real analysis unsolved

Can we find a function $f$ from $R$ to $R$, infinitely differentiable such that $f,f',...,f^{(n)}$ have no real zeros and $f(x)=o(x^n)$  for $|x|$ big?

Of course, it exists and there are no problems to construct it (say, if $f_0(x)=e^{-|x|}$ and $f_k(x)=\int_{-\infty}^x f_{k-1}(t)dt$, then $f=f_n$ satisfies our conditions).

Make that $f_0(x)=e^{-x^2}.$ We do want $C^{\infty}.$


Let $ p\equal{}4k\plus{}1$ be a prime number. Show that there's an integer solution to $ x^2\minus{}py^2\equal{}\minus{}1$.


calculus
integration
trigonometry
calculus computations

[quote="JRav"]This looks like an elliptic integral.[/quote]

how dcan u say that an integral is elliptic?

[color=blue]
Problem: Integrate the following : $ \int \frac {x\minus{}1}{(x\plus{}1)\sqrt {x^3\plus{}x^2\plus{}x}} dx$

Solution:Let I $ \equal{} \int \frac {x\minus{}1}{(x\plus{}1)\sqrt {x^3\plus{}x^2\plus{}x}} dx$

$ \equal{}\int \frac {x^{2}\minus{}1}{(x\plus{}1)^{2}\sqrt {x^3\plus{}x^2\plus{}x}} dx$

$ \equal{}\int \frac {x^{2}\minus{}1}{(x^{2}\plus{}2x\plus{}1)\sqrt {x^3\plus{}x^2\plus{}x}} dx$

$ \equal{}\int \frac {1\minus{}\frac{1}{x^{2}}}{\left(x\plus{}\frac{1}{x}\plus{}2\right)\sqrt {x \plus{}\frac {1}{x}\plus{}1}} dx$

$ \equal{}\int \frac {1}{\left(x\plus{}\frac{1}{x}\plus{}2\right)\sqrt {x \plus{}\frac {1}{x}\plus{}1}} d\left(x\plus{}\frac{1}{x}\right)$

$ \equal{}\int \frac {1}{\left(x\plus{}\frac{1}{x}\plus{}2\right)\sqrt {x \plus{}\frac {1}{x}\plus{}1}} d\left(x\plus{}\frac{1}{x}\plus{}1\right)$

Let $ x\plus{}\frac{1}{x}\plus{}1 \equal{} t^{2} \implies d\left(x\plus{}\frac{1}{x}\plus{}1\right) \equal{} 2tdt$

$ \implies I\equal{} 2\int \frac {1}{1\plus{}t^{2}} dt$

$ \implies I \equal{} 2 \tan^{\minus{}1}(t) \plus{} c$

$ \implies\boxed{ I \equal{} 2 \tan^{\minus{}1}\sqrt {\left(x\plus{}\frac{1}{x}\plus{}1\right)} \plus{} c}$
[/color]