Rudolph bikes at a constant rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin biking at the same time and arrive at the $ 50$-mile mark at exactly the same time. How many minutes has it taken them?

Not that sure of this, but I'll post it anyway.

[hide]

Rudolph takes 49 breaks, so his total time (hours) in breaks is $ \frac{49\cdot5}{60}\equal{}\frac{49}{12}$

Jennifer takes 24 breaks, so her total break time is $ \frac{24\cdot5}{12}\equal{}2$

Let $ r$ be Rudolph's rate (mph). Since $ 50\equal{}r\cdot t$, the time that it takes him to bike 50 miles is $ \frac{50}{r}$.

Similarly, the time that it takes Jennifer to bike 50 miles is $ \frac{50}{\frac{3}{4}r}$

Adding and equating Rudolph's total time and Jennifer's total time, we have

$ \frac{49}{12}\plus{}\frac{50}{r}\equal{}2\plus{}\frac{50}{\frac{3}{4}r}$

Solving, we get $ r\equal{}8$, so Rudolph's time is $ \frac{49}{12}\plus{}\frac{50}{8}\equal{}\frac{124}{12}\equal{}\frac{31}{3}$

His time in minutes is therefore $ \frac{31}{3}\cdot 60\equal{}\boxed{620}$

[/hide]

Apparently Rudolph and Jennifer must be slow and out-of-shape, although they were going 50 miles. Six mph on a bike? I had to check my work three times because I was sure I had something wrong.  :maybe:

[quote="shake9991"]sucks for the people who thought that they rested at the 50 mile mark and got an answer of 625...[/quote]

totally agree with u there.....cus i made that mistake... :(

This problem was a killer.  I put 625 too.

The problem said "arrived at the 50 mile mark at the same time" so it's kinda obvious they were cutting off the timer when they arrived  :roll: ...

[hide]In 50 miles, R rests 49 times or 245 minutes, while J rests 24 times for 120 minutes.

Let r be the rate of R, thus 3r/4 is the rate of J.

d=rt

Thus, $ 50/r \plus{} 245 \equal{} 50/(3r/4) \plus{} 120$ solve $ r \equal{} 2/15$.

Then, R took $ 50*15/2 \equal{} 375$ minutes RIDING to make it, plus the 245 minutes he rested, gives $ \boxed{620}$ - works out for J.[/hide]

Considering last year's #2 was also a word problem, which also tricked a bunch of people, ya shoulda been more careful  :P .

Did anyone think Rudolph had 48 breaks instead of 49 and got 600.
 :mad: i could've got 8 on AIME but because of this i got 7

Hehe, I used an hour to check my solutions, and I caught this, hehehe *evil laughter* hahahaHAHAHAHAHA....(but I still missed 2 other problems without catching the mistakes...sigh... :( )

[hide="Used this quick method on the test even though I put 625"]
Say it takes them $ t$ minutes to arrive, and Rudolph and Jennifer bike at rates of $ r$ and $ \frac {3}{4}r$ miles per minute, respectively. Rudolph takes $ 49$ breaks so he wastes $ 245$ minutes and thus is actually biking for $ t \minus{} 245$ minutes. In the same way, Jennifer bikes for $ t \minus{} 120$ minutes. Since they both bike the same distance, we have that:
\[ \frac {3}{4}r(t \minus{} 120) \equal{} r(t \minus{} 245)
\]\[ \frac {3}{4}(t \minus{} 120) \equal{} t \minus{} 245
\]\[ 3t \minus{} 360 \equal{} 4t \minus{} 980
\]\[ t \equal{} \boxed{620}
\]
[/hide]

[hide="Answer"]
We put everything in terms of $ r$ being Rudolph's rate and $ t$ being time it takes. So, we have Rudoplh taking 245 minutes of breaks, and Jennifer taking 120 minutes of break. Setting up our equation, we have $ r(t\minus{}245)\equal{}\frac{3}{4}r(t\minus{}120)$. Solving, we have $ r$ cancelling out, and $ t$ being $ \boxed{620}$.[/hide]

[hide=Solution]
We let $3x$ and $4x$ be the time it take Rudolph and Jennifer to go $1$ mile, respectively.

We then know that $150x+5\cdot49=200x+5\cdot24$.

We see that this simplifies to $5(49-24)=50x$.

We see that this simplifies to $25=10x$ so $x=2.5$.

We see that our answer is $150\cdot2.5+5\cdot49=200\cdot2.5+5\cdot24=\boxed{620}$.
[/hide]

[hide= Solution]
I got $625$ until I saw that they stopped the timer right when they arrived...

So we can name Rudolph's speed to be $x$, and Jennifer's speed to be $\frac{3x}{4}$. We can also name the amount of time that it takes for Ruldpoh to complete one mile to be d. The time that it takes Jennifer to complete $2$ miles can be $t$. So if we put these values into an equation we get...
$$2 = (\frac{3x}{4})(t)$$
$$1=xd $$
Since they arrived at the same time and arrived at the same time, this means that they took the same amount of time. Rudolph has taken $245$ minutes of a break and  Jennifer has taken $120$ minutes of a break. So since d and t are the values for $1$ and $2$ miles respectively, we can write the equation...
$$50d + 245 = 25t + 120. $$
So if we solve for $t$ we get $2d + 5$ and we can plug this into one of the above equations. We get...
$2 = (2d+5)(\frac{3x}{4})$ and we know $xd=1$ so we can put that in a solve for $d$...
$\frac{8}{3} = (2d+5)(\frac{1}{d}) \Rightarrow \frac{8}{3} = 2 + 5(\frac{1}{d}) \Rightarrow d = \frac{15}{2}.$
We know that the time is $50d + 245$ so we can put that in and we get $50( \frac{15}{2}) + 245 = \boxed{620}$

[/hide]

[hide=Solution]Let's say Rudolph goes $x$ miles per hour and Jennifer gives $\frac{3}{4}x$ miles per hour. The total amount of five-minute breaks Rudolph takes is $49$ and the total amount of five-minute breaks Jennifer takes is $24$. This means we have $\frac{50}{x} \cdot 60 + 49 \cdot 5 = \frac{50}{\frac{3}{4}x} \cdot 60 + 24 \cdot 5$. Solving gives $x=8$, so our answer is $\frac{50}{8} \cdot 60 + 49 \cdot 5 = \boxed{620}$.[/hide]

Whew, made it through the monster (RATE PROBLEM) alive!


Let $ a,b,c,d\geq0$ and $ ab\plus{}bc\plus{}cd\plus{}da\equal{}1$. Prove that \[ \displaystyle\frac{a^3}{b\plus{}c\plus{}d}\plus{}\displaystyle\frac{b^3}{a\plus{}b\plus{}d}\plus{}\displaystyle\frac{c^3}{a\plus{}b\plus{}d}\plus{}\displaystyle\frac{d^3}{a\plus{}b\plus{}c} \geq \frac{1}{3}\][/code]

Note: all variables are positive.

$ (*)$ $ \sum \frac {a^3}{x_1}\geq \frac {(a \plus{} b \plus{} c \plus{} d)^3}{4(x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4)}$
See [url=http://www.mathlinks.ro/viewtopic.php?t=138785]here[/url] for the $ (*)$ inequality.

Using this we only have to prove that $ (a \plus{} b \plus{} c \plus{} d)^2 \geq 4$ $ \Leftrightarrow$ $ (a \plus{} b \plus{} c \plus{} d)^2\geq 4(ab \plus{} bc \plus{} cd \plus{} da)$  $ \Leftrightarrow$ $ (a\plus{}b\plus{}d\minus{}c)^2 \geq 0$

Let S be our left member of the inequality, and $ x\equal{}b\plus{}c\plus{}d$, $ y\equal{}c\plus{}d\plus{}a$, $ z\equal{}d\plus{}a\plus{}b$, $ t\equal{}a\plus{}b\plus{}c$. we can supose that $ a\geq b\geq c \geq d$ and for sure that $ a^n \geq b^n \geq c^n \geq d^n$ for any natural n diffrent from 0, and $ \frac{1}{x} \geq \frac{1}{y} \geq \frac{1}{z} \geq \frac{1}{t}$. with the rearengements inequality we have $ a^2\plus{}b^2\plus{}c^2\plus{}d^2 \geq ab\plus{}bc\plus{}cd\plus{}da\equal{}1$. With cebasev's we have $ S \geq\frac{1}{4} \cdot (a^3\plus{}b^3\plus{}c^3\plus{}d^3)(\frac{1}{x}\plus{}\frac{1}{y}\plus{} \frac{1}{z}\plus{}\frac{1}{t})$. Alsaw $ a^3\plus{}b^3\plus{}c^3\plus{}d^3 \geq \frac{1}{4} (a^2\plus{}b^2\plus{}c^2\plus{}d^2)(a\plus{}b\plus{}C\plus{}d)$ or $ a^2\plus{}b^2\plus{}c^2\plus{}d^2 \geq1$ and $ 3(a\plus{}b\plus{}c\plus{}d)\equal{}x\plus{}y\plus{}z\plus{}t$ .Therefore \[ S\geq\frac{1}{48}(x\plus{}y\plus{}z\plus{}t)(\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\plus{}\frac{1}{t})\geq \frac{16}{48}\equal{}\frac{1}{3}\]


$ P_{n}Q_{n} \equal{} 1, Q_{n} \equal{} Q_{0} \plus{} \sum _{k \equal{} 0} ^{n \minus{} 1} P_{k}$ for all $ n \in N_{0}, Q_{0} \equal{} \alpha$
Following properties are given.
1. $ P_{n}$ is decreasing and positive sequence
2. $ Q_{n} \succ \sqrt {\alpha^2 \plus{} 2n}$
3. $ Q_{n} \prec \sqrt {\alpha^2 \plus{} 2n \plus{} \alpha^{ \minus{} 2} \plus{} \frac {1}{2}ln(\frac {2n}{\alpha^2} \plus{} 1)}$
Then, find minimum value of n that
$ P_{n} \prec \frac {1}{2e}$

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=318117[/url]


Find the polynomial $P(x)$ such that :
 $P(x^2-2)=P(x)[P(x)-2]$

I think that the solutions are $P(x)=0$, $P(x)=3$ and $P(x)=Q_n(x), n \in \mathbb{N}$ , where $Q_1(x)=x+1$ ,$Q_{n+1}(x)=Q_n(x^2-2)$


Hey, I just thought that some people may find interesting this. Some people is filming independently (and free) the [url=http://www.newvoyages.com/]New Voyages[/url] of the original generation of Star Trek. I haven't seen them yet (I am downloading them) there are two so far! Let's see how it goes.

Best,

I love StarTrek, however I can't cope with the original series. They are ... out of sync with the others - they look just ... ugly :)
I've just taken a short look on this site, and it seems that they wanted to change this aspect, which is great. However, I could not understand what they are trying to do.

And isn't StarTrek a registred trademark of the Paramount Pictures? Or at least I thought so :? :?

[quote="Valentin Vornicu"]I love StarTrek, however I can't cope with the original series. They are ... out of sync with the others - they look just ... ugly :) [/quote]
What are original series in your opinion? Before Jean-Luck?
I like Star Trek. And I like Enterprise. But I don't know whether I will see the continuation (the last series from Enterprise saga I saw was about the attack to the Earth; and I don't know what happened next :( )

No, the original series were made in the 60's, staring Cpt. James T. Kirk and Enterprise 1701-[b]A[/b] (as opposed to Picards' NCC 1701-[b]D[/b], and later on, NCC 1701-[b]E[/b]).

Picard and his crew were the start of a new Age in StarTrek, when the show, I belive, brought many many inovations in its script: (ie small laptops, PDAs, Mobile Communications Devices, LCDs, Touchscreens - things that didn't exist back in '87, but do exist now :) ).

Enterprise show has been canceled :( (season 4 is the last season). You are probably reffering to the 3rd season (the Xindy threat). Season 4 is much nicer. The Federation is almost formed as an alliance thought up by Archer to discover clocked Romulan ships that were attacking random vessels in the system ;) We finally see some Remans in a StarTrek series :D

What they are trying to do is basically to film "New Voyages" from the Kirk's generation. Nevertheless, they can't profit from that, because it is not part of Paramount.

Basically they are trying to say: "Well, we liked it that much that we want [b]it[/b] back, not [i]the Next Generation[/i], not [i]Deep Space 9[/i], but the original one." And that is what they try to do, they write their own scripts, but based on the same set of caracters than Kirk's generation, and with the same political situation of the federation than it, basically, like if it would be continue running.

Best,

But my question is [b]can[/b] they do that? Can they use the Star Trek name and the names of other characters and such to do this? :? Also this kind of work requires huge funds ... how do they plan to finance themselves, if not by selling the movies? ...

Actually they are doing it very efficiently. They do it by love to the art. The actors spend their free time doing it, but it is not their job, they don't receive any money for that. On the other hand, the rest of the expenses, well, they could low it to $15K per episode. It is still high, but is like two or three of the actors and technicians are the ones who give the money.

On the other hand, aparently they can use whatever they want as far as they don't get any profit of it (at least directly), and proof of that is that they are doing it in public.

Now, that is what I heard in a CNN article that was shown last night. That is how I found the site.

Best,

[quote="Gyan"][quote]I love StarTrek, however I can't cope with the original series. They are ... out of sync with the others - they look just ... ugly Smile [/quote]

Now .. now .. don't diss that. It just not nice.[/quote]Well, it's my opinion. It might be influenced by the fact that I first watched TNG (not knowing that TOS existed in the first place), then DS9, then Voyager, and then a couple of episodes from TOS. You can't put them together. I admit TOS was nice for the 60's, but in the 3rd millenium it can't pass along. Even StarWars were modificated in 2000 to make them look nicer and similar with the new series. This is what this new show tries to do with TOS probably ;)

Well Valentin, it happens with everything. Just thing about this, most of us consider very slow the movies of the 60's or before, kind of boring, but that is because we have been exposed too much to the contemporary movies.

The same thing happens with Star Trek. TOS, as you call it, it was a sensation on his era. I really didn't watch them until very recently, but I had watched the movies. Somehow it is may be to slow, or to naive in many aspects for our "contemporary" vision.

I remember when I took a course in movie apreciation and they told me that some experts considered "The Citizen Kane" one of the two best movies of the century. I saw the movie and I was disapointed. Of course, when you read all the innovations in that movie, it was great for its time.

Anyway, I honestly prefer to watch TNG than TOS. Well, that was just a comment.

Best regards,


[quote="XAKKER"]x,y,z>0 xy+yz+zx+2xyz=1 prove that: 32xyz+4(x+y+z)=>10[/quote]

Try $ z\equal{}0,x\equal{}y\equal{}1$  :wink:

[quote="akai"][quote="XAKKER"]x,y,z>0 xy+yz+zx+2xyz=1 prove that: 32xyz+4(x+y+z)=>10[/quote]

Try $ z \equal{} 0,x \equal{} y \equal{} 1$  :wink:[/quote]

in terms given  x,y,z>0


factorial
calculus
ratio
calculus computations

Hello.  This may not be a calculus problem perse, but it is giving me problems in calculus.

Say you are dealing with the Power Series.  O.k. you want to know if the series An  (from 1 to infinity) converges.  An= (X^n)/(n!)  .  O.k. you apply the ratio test, and you've got:

lim n --> infinity   (X^(n+1))/(n + 1)!/X^n/n!

which you multiply out as (X^(n+1))/(n+1)!      *     n!/X^n

and you get lim n --> infinity  |X|/(n+1)  which equals 0

O.k. I understand everything in that problem except how the 1/(n+1)! multiplied by a n!  leaves you with a 1/(n + 1).

I know a facorial is like 3! = 3*2*1, but I don't get what they are doing.  I'm assuming it is something like you can do with an exponent where you have 1/3^(n + 1) multiplied by a 3^n, and you are left with a 1/3.  I guess we just didn't do factorials much when I was in precal.   :blush:

This is a consequence of the recursive definition of factorials. $(n+1)!=(n+1)\cdot n\cdot(n-1)\cdots1$, so $(n+1)!=(n+1)\cdot n!$. This is the most important thing to know about factorials.

Alright thanks.  I didn't see that identity anywhere in the book.

You're also likely to run into situations like this (as part of a ratio test for series convergence): can we reasonably simplify

$\frac{(2n)!}{(2n+2)!}$ ?

Be a little bit careful with this one!


function
topology
real analysis
real analysis solved

For what sets $A$ can we find a $C^{1}$ class function $f$ with real values and defined on the reals such that $ f^{-1}(0)=A$?

Belief: $f^{-1}(0)$ can be any closed set.
justification: It has to be closed by continuity (i think).
And conversely, take any closed set, let $A$ be its complement. Then $A = \cup_i I_i$ where $I_i$ are open disjoint intervals. Then plase the famous $C^\infty$ function that's non-zero in $I$ and zero eslewhere. Then add those function on $I_i$'s.

Yes, it has to be closed. By the definition of continuity, the preimage of any open set is open. Taking complements, the preimage of any closed set is closed.

That construction works just fine, and we are done.

Also for $R^n$ any closed set can be realized as zero locus of a smooth function.  It was proved by H. Whitney in the 1940's, not an easy theorem if I remember.

Hope that the following makes sense:

Here's an idea how to prove the result for $\mathbb [0,1]^n$ for merely $C^1$ functions. The idea is almost as before: The complement of our closed set is $A$, and its a countable union of not necessarily disjoint open balls $B_j$. So $A = \bigcup_j B_j$. We would like to put a bump function on each $B_j$ and to add them up. We could multiply each by a very small factor, $c_j$ to make sure everything converges. However, we'll need to verify that the result is in $C^1$ and not merely continuous.

And to do so, all we need is to find a norm on $C^1$ that makes it into a banach space. Then we'll choose our constants $c_j$ so that the functions will be absolutely convergant with respect to that norm. Then by completeness, the sum will be in $C^1$.

And I think that this is the norm we need $\|f \| = \|f\|_\infty + \sum_j  \left\|\frac {\partial f}{\partial x_j}\right\|_\infty$.


topology
abstract algebra
vector
limit
Functional Analysis
combinatorial geometry
absolute value

Let $\mathcal M$ be a closed linear subspace of a reflexive Banach space $\mathcal B$, and let $x_0\in\mathcal B\setminus\mathcal M$. Prove that there is a $y\in\mathcal M$ such that $\|x_0-y\|=d(x_0,\mathcal M)$.

the set $\{y\in M, d(x,y)\leq 2d(x,M)\}$ si weakly compact. there exists $y_n$ in $M$ so that $d(x,y_n)\to d(x,M)$ and $y_n$ weakly convergent. By Krein (?) theorem one can find $z_n$ in the convex hull of $y_n$ such that $z_n$ convergent (norm topology) and $d(x,z_n)\to d(x,M)$.

The converse is true as well. In other words, this is a characterization of reflexive Banach spaces. Here's what I did (for the converse, I mean):

Suppose for every $\mathcal M,\ x_0\not\in\mathcal M$ as above we can find $y\in\mathcal M$ such that $\|x_0-y\|=d(x_0,\mathcal M)$. In particular, this holds when $\mathcal M$ is a hyperplane (the kernel of a continuous linear functional). Let $f$ be any linear functional of norm $1$, and let $x_0$ be such that $f(x_0)>0$. By an application of the Hahn-Banach Theorem on extensions of linear functionals, it's easy to see that we can find a linear functional $\tilde f$ of norm $1$ whose kernel is $\mathcal M$ and such that $\tilde f(x_0)=d=d(x_0,\mathcal M)$. Since a linear functional is uniquely determined by its kernel, it's norm, and the fact that it has a positive value in a certain point outside its kernel, we get $\tilde f=f$, so $f(x_0)=d$.

Now take $y\in\mathcal M$ according to the hypothesis (i.e. such that $\|x_0-y\|=d$). We have $\|x_0-y\|=d=f(x_0)=f(x_0-y)\le\|x_0-y\|$, so every $\le$ is actually an equality. This means that $|f|$ reaches the value $1=\|f\|$ on the closed unit ball of $\mathcal B$. By a theorem of James, if every continuous linear functional reaches the maximum of its absolute value on the closed unit ball, then the Banach space is reflexive.

i think you prove that $f(x)=d(x, ker(f))$ for any functional of norm one in any banach space, and i'm not sure if that is correct.

It's true in any normed space if you also know that $f(x_0)>0$, and I pretty much outlined the proof. Now I'm just repeating myself.

You take the functional $f$ of norm $1$ and kernel $\mathcal M$, and a point $x_0$ outside $\mathcal M$. Let $d=d(x_0,\mathcal M)$. You don't even need the Hahn-Banach Theorem: that's required for smaller spaces; in our case, we already know that $\mathcal M$ is a hyperplane, so every vector in the space is of the form $ax_0+y$, where $y\in\mathcal M$ and $a$ is a scalar. Now define $\tilde f(ax_0+y)=ad$. Obviously, $\tilde f$ has norm $1$, and we also have $\tilde f(x_0)=d>0$. If $0<d'=f(x_0)$, then $\tilde f(x)=\frac d{d'}f(x),\ \forall x$ in the space, so $f,\tilde f$ cannot possibly have norm $1$ at the same time unless $d=d'$, which is equivalent to $\tilde f=f$.

Is there anything wrong with that? :?

i see now that it's true but i still dont see why you would need hahn-banach, but maybe i'm just confused.

$d(x,ker(l))=\inf_{y\in ker(l)}\|x-y\|\geq\inf_{y\in ker(l)}|l(x-y)|=|l(x)|$

on the other hand let $\|a_n\|=1$ such that $|l(a_n)|\to 1$. Then $y_n=x-\frac{l(x)}{l(a_n)}a_n\in ker(l)$ and

$d(x, ker(l))\leq\lim_n\|x-y_n\|=\lim_n\frac{|l(x)|}{|l(a_n)|}=|l(x)|$

Didn't I just say in my previous post that you don't need Hahn-Banach? :) Here's how I encountered the problem before, in a form which required the theorem (which is what confused me): prove that given a subspace $\mathcal M$ of a normed space and a point $x_0$ with positive distance $d$ to $\mathcal M$, there is a linear functional of norm $1$ which takes the value $0$ on $\mathcal M$ and $d$ in $x_0$. As I said, here we don't need the theorem, since we can simply define the functional by $f(ax_0+y)=ad$ (here $\mathcal M$ is a hyperplane, not just any subspace, as in the problem I had encountered).

ok, but it's not "obvious" that $\tilde f$ has norm one, in fact it's not true.

As far as I can tell, yes, it is true.

You have $|\tilde f(ax_0+y)|=|ad|$, which is $\le |a|\cdot\left\|x_0+\frac 1ay\right\|$ if $a\ne 0$ (by the definition of $d$ as being $d=\inf_{y\in\mathcal M}\|x_0+y\|$) and $0$ if $a=0$, so the norm of $\tilde f$ is at most $1$. On the other hand, take $y\in\mathcal M$ such that $\frac d{\|x_0+y\|}$ is within $\varepsilon>0$ from $1$. Then, since $\tilde f(x_0+y)=\tilde f(x_0)=d,\ \frac{|\tilde f(x_0+y)|}{\|x_0+y\|}\ge 1-\varepsilon$, so the norm of $\tilde f$ is at least $1$.

this is my last post on this subject :)

although it now makes sense (to me), for me the obvious part was that

$x=y+ax_0=(x-\frac{l(x)}{l(x_0)}x_0)+\frac{l(x)}{l(x_0)}x_0$ so

$\tilde f(x)=ad=\frac{d}{l(x_0)}l(x)$, hence $\|\tilde f\|=\frac{d}{|l(x_0)|}$

that's all i had to say.


Humanitarian organization has to deliver 18 tons of humanitarian aid which is distributed in n containers, where n≥35. They have 7 trucks each that can carry infinite number of containers whose total mass does not exceed 3 tons. It is known that these trucks can carry any 35 containers, from given n containers, in one tour. Prove that these trucks can carry all n containers in one tour.

Do not double post!
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=224776[/url]


trigonometry
logarithms
complex analysis

[quote="color"]Please solve $\sin^{2}{z}=-1$.[/quote]
Hint: Eliminate the square and do $\sin z = \pm i$

Now we will find all $z=x+iy$ such that $\sin z = i$ and the case $-i$ I leave to you. Using identities we find that,
$\sin (x+iy) = \sin x \cosh y+i\cos x\sinh y = i$
Equating real and imaginary parts we find that,
$\sin x \cosh y = 0 \ \ (*)$
$\cos x \sinh y = 1 \ \ (**)$

Look at (*) since $\cosh y \not = 0$ for real $y$ it must be that $\sin x = 0$ that means $x=2\pi n$ for $n\in \mathbb{Z}$.

If $x=2\pi n$ that means $\cos x = (-1)^{n}$ and so by (**) we have,
$(-1)^{n}\sinh y = 1$
Rewrite this,
$\sinh y = (-1)^{n+1}$
So,
$e^{y}-e^{-y}= 2(-1)^{n+1}$
This leads to the quadradic equation,
$e^{2y}+2(-1)^{n}e^{y}-1 = 0$
Solve,
$e^{y}= \frac{2(-1)^{n+1}\pm \sqrt{4+4}}{2}= (-1)^{n+1}\pm \sqrt{2}$
We [b]cant[/b] choose the negative sign so,
$e^{y}= (-1)^{n+1}+\sqrt{2}\implies y = \ln \left(\sqrt{2}+(-1)^{n+1}\right)$

With all of that we finally have all the solutions,
$z = 2\pi n+i\cdot \ln \left( \sqrt{2}+(-1)^{n+1}\right) \ \ \ n\in \mathbb{Z}$

Now it remains to find the completely anagolous equation,
$\sin z =-i$
Which I think is real easy cuz $\sin (-z) =-\sin z$.
So you dont need to go through all of that mess.


limit
logarithms
calculus
integration
function
calculus computations

Quick question:

The well-known limit $ \lim_{n\to\infty}\left( {\frac{1}
{{n \plus{} 1}} \plus{} \frac{1}
{{n \plus{} 2}} \plus{}  \cdots  \plus{} \frac{1}
{{n \plus{} n}}} \right) \equal{} \ln 2,$ it's representated by the integral $ \int_0^1\frac1{x\plus{}1}\,dx.$

Why the integration limits are defined from zero to $ 1\,?$

The choice of function and of limits are inextricably tied together. We rewrite a typical term of the series as $ \frac1n\left(\frac1{1\plus{}\frac in}\right)$, then let $ x_i\equal{}\frac in$. If we do this, $ \frac1n$ is the length of the interval between $ x_{i\minus{}1}$ and $ x_i$, and $ \frac1{1\plus{}\frac in}$ is the value of the function $ f(x)\equal{}\frac1{1\plus{}x}$ at $ x_i$. The limits come from the range of the $ x_i$ (and the intervals they sample from); $ x_0\equal{}0$ and $ x_n\equal{}1$, so we integrate from zero to $ 1$.

We could have chosen $ x_n\equal{}1\plus{}\frac in$ instead; that would have made the integral $ \int_1^2 \frac1x\,dx$. We could have factored the term as $ \frac2n\left(\frac1{2\plus{}\frac{2i}{n}}\right)$ and set $ x_i\equal{}\frac{2i}{n}$; that leads to $ \int_0^2 \frac1{2\plus{}x}\,dx$.


Did anyone take it? What did you guys think?

I took it.  I left 2 blank, and am fairly confident about the other 23.  Blah, worse than last year, oh well . . .

I took it. #19 blank, #25 wrong, very confident about the other 23. I almost got #19, but I had 30 seconds left and I wasn't sure enough about the answer. Oh well.

Yeah I had trouble with 19 too.
I put 11 as my answer, but now I'm fairly sure I'm wrong. :(

Really? I proved that it had to be odd... but maybe I'm wrong:

$ d\equal{}2006$, obviously.

Using $ P(1)$ we can find $ a\plus{}b\plus{}c \equal{} 0$. Still pretty straightforward.

Then using $ P(2)$ I found that $ 3a\plus{}b \equal{} \minus{}7$.

Then I plugged into $ P(3)$.

$ P(3) \equal{} 81 \plus{} 27a \plus{} 9b \plus{} 3c \plus{} d$
$ \equal{}2087 \plus{} 24a \plus{} 6b$
$ \equal{}2087 \minus{}56\minus{}8b \plus{} 6b$ (Here I used the equation i found using $ P(2)$)
$ \equal{}2031 \minus{} 2b$

Since $ b$ is an integer, this $ P(3)$ has to be odd. (?)

The answer is 5. Doing a few algebraic manipulations, you can show that the value must be 5 mod 6. On the whole, I did about as well as I did last year. I think Blair has 2 perfect scores on part I this year though: one a junior, and one a sophomore. Good luck to everyone on part 2! Maybe people will do better even better than usual now that scholarships are an incentive...


linear algebra
matrix
induction
algebra
polynomial
Putnam
quadratics

If I'm given 2 matrices, E and F (n X n) and I know that E, E+F, E+2F, ... E+4F all have  inverses...

Can I show that E+5F has an inverse for n=2?  I'm thinking it is an induction problem, but don't know how to go about it.

The determinant of $ E\plus{}xF$ is a polynomial in $ x$ of degree at most 2. What do you know about its values?

This is Putnam 1994 A4.

(I know, because this week's homework for my "Putnam problems" class is the 1994 A problems.)

I think that the word "integer" should appear somewhere in the problem.

perfect_radio is right. The word "integer" should appear.

[b]Putnam 1994 A4[/b]

Let $ A$ and $ B$ be $ 2\times 2$ matrices with integer entries such that $ A,A\plus{}B,A\plus{}2B,A\plus{}3B,$ and $ A\plus{}4B$ are all invertible matrices whose inverses have integer entries. Show that $ A\plus{}5B$ is invertible and that its inverse has integer entries.

[hide]
I used a brute force approach and found that $ Det(B) \equal{} 0$.
So B is singular.
Since it is a $ 2 \times 2$ the matrix $ B$ has a row that is a multiple of the other.
I don't know how to use that.
[/hide]

A little slicker:

[hide]The determinant of a $ 2\times 2$ matrix is a quadratic polynomial in the entries of that matrix. This means that $ f(k)\equal{}\det(A\plus{}kB)$ is a quadratic (or lower degree) polynomial in the single variable $ k.$

A matrix is invertible over $ \mathbb{Z}$ iff its determinant is $ \pm1.$ Hence the five values $ f(0),f(1),f(2),f(3),$ and $ f(4)$ all lie in the set $ \{\minus{}1,1\}.$ By pigeonhole, three of those five values are equal. Any quadratic or lower degree polynomial which takes on the same value three times must be constant. Hence $ f(k)$ is a constant function, either $ f(k)\equal{}1$ or $ f(k)\equal{}\minus{}1.$

We conclude that $ A\plus{}kB$ is invertible over $ \mathbb{Z}$ for all $ k\in\mathbb{Z}.$

[Of course, this is all merely elaboration of what jmerry said in the second post on this topic.][/hide]

A matrix is invertible over $ \mathbb{Z}$ iff its determinant is $ \pm1.$

I gave some thought to it and can't figure out why that is true exactly.

If the product of two integer matrices is the identity, both have to have unit determinants.
Conversely, we can express inverses in terms of minors- for an integer matrices, the entries of the inverse are integers divided by the determinant.

Ahh I see.
So,
$ 1\equal{}Det(I)\equal{}Det(AA^{\minus{}1})\equal{}Det(A)Det(A^{\minus{}1})$
thus
$ Det(A^{\minus{}1})\equal{}\frac{1}{Det(A)}$
Since $ A^{\minus{}1}$ and $ A$ have elements in $ \mathbb{Z}$, their determinant is also in $ \mathbb{Z}$.

Thus $ \frac{1}{Det(A)}$ is an integer and $ Det(A)\equal{}Det(A^{\minus{}1})\equal{} \pm1$

I'll add one trivial point to Kent's solution: If we first square the polynomial, then we don't need the Pigeonhole Principle.

But we need the five data points either way. Squaring the polynomial makes it a fourth degree polynomial. If we only had the hypothesis for the four points $ A,A\plus{}B,A\plus{}2B,A\plus{}3B,$ then this proof (either your version or mine) wouldn't work.


trigonometry
algebra
system of equations

Solve system of equations: $ \begin {cases} 2^x \plus{} 2^y \equal{} x^2 \plus{} y^2 \plus{} 2 \\
x^4 \plus{} y^4 \equal{} 1\end {cases}$ , with $ x,y \in R$ and $ x, y > 0$.

Let $ x\equal{}\cos^{1/2}z$
Let $ y\equal{}\sin^{1/2}z$
Then, $ \sin^{2}z\plus{}\cos^{2}z\equal{}1$
And
$ 2^{\cos^{1/2}z}\plus{}2^{\sin^{1/2} z}\equal{}\cos z \plus{} \sin z \plus{} 2$

I do not know how to solve this by hand, so, I found the zeroes using my calculator.
$ z\equal{} 0\plus{}2\pi k, \frac{\pi}{2}\plus{}2\pi k$

So, the solutions are $ (0,1)$ and $ (1,0)$


What is the greatest proper factor of 6! ?

A proper factor of a number $ n$ means that it is a factor of $ n$, but is not $ 1$ or $ n$ itself.  Therefore, to maximize the proper factor of $ 6!$, we divide out the smallest factor of $ 6!$ (besides $ 1$ of course), which happens to be $ 2$.  So we have $ \frac {6!}{2} \equal{} \frac {720}{2} \equal{} \boxed{360}$

Actually, 1 is a proper factor of any number (except 1 of course).

According to[url=http://mathworld.wolfram.com/ProperFactor.html]Wolfram Mathworld[/url], $ 1$ does not count as a proper factor, but does count as a proper [i]divisor[/i].  However, [url=http://wiki.answers.com/Q/What_is_a_proper_factor_in_math]WikiAnswers[/url] says that there is debate on it...  I'm really not that sure.


analytic geometry
geometry
trapezoid
LaTeX

When plotted in the standard rectangular coordinate system,
trapezoid $ ABCD$ has vertices $ A(1, \minus{}2)$, $ B(1, 1)$, $ C(5, 7)$ and
$ D(5, 1)$. What is the area of trapezoid $ ABCD$?

[hide="Solution"]Trapezoid ABCD has vertices
A(1, -2), B(1, 1), C(5, 7) and D(5, 1).
To find the area of the trapezoid,
draw it and draw the line BD(I don't know how to draw with latex, Sorry!). Two right
triangles have been formed. The
area of triangle BDC is ½ × 6 × 4 =
12. The area of triangle BAD is
½ × 4 × 3 = 6.
12 + 6=[b]18[/b][/hide]

Or just solve it using the method to solve trapezoids: Add the bases, multiply by the height, then divide by 2. The height is $5-1=4$, $b_1=1-(-2)$, and $b_2=7-1=6$. Therefore, our answer is $\dfrac{4(3+6)}{2}=2(3+6)=\boxed{18}$.


geometry
rectangle
arithmetic sequence
puzzles

I have to organise a Maths club for some 9-year-old kids, can someone suggests some simple puzzles or brainteasers which is interesting. :wink:

Thank you very much! :P

Have them count all of the possible rectangles found in this diagram if all vertices of each rectangle must coincide with a set of 4 dots of this array.  (An asterisk here represents a dot.)
The dots are to be one unit apart horizontally and vertically from
their nearest neighbors.

* ... * ... *
* ... * ... *
* ... * ... *
* ... * ... *

Things for your students (or other readers) to consider:
[hide]1) Squares are a type of rectangle.  2) Orientation:
you may have non-vertical and/or non-horizontal shapes [/hide]

I think this is a good puzzle for the kids, thank you very much box of rocks! :P
I would still like to have more suggestions of puzzles and brainteasers, can I please have more suggestions?

Thank you very much! :P

[quote="moogra"]You can teach them how to add 1+2+3+4...+n really fast. I thought that was fun in 4th grade[/quote]

I was thinking of that as well, I can tell them about the formula for the sum of arithmetic sequence by then. :) Can I please have more suggestions?

Thank you very much! :P

[quote="OHO"]What's about the Tower of Hanoi?

[url]http://en.wikipedia.org/wiki/Tower_of_Hanoi[/url][/quote]

What a good suggestion! OHO, thank you very much! :P
Tower of Hanoi is a very good puzzle indeed, I used to play this puzzle when I was a kid, I still find this puzzle really fun.
Can I please have more suggestions?

Thank you very much! :P


A few questions:

1. What are the best priced hotels in the area?
2. How is it possible to get on-university dorm lodging? I;ve heard that people are planning to do that, but I was not sure how.

[url=http://web.mit.edu/hmmt/www/faq.shtml#boston]This[/url] should help you find hotels in the area.

As far as staying in the dorms, I think this is a great option!  Email hmmt-request@mit.edu telling them you want to stay on campus.  If you don't get a response in a couple days after emailing them, let me know and I'll get things moving.  The only thing about this is that your team will be split up a lot.  Staying on campus means that you'll be the guest in someone else's room.  So your team will be split up, at least a little bit.  This shouldn't be a problem though.

I think we'll opt for the hotels  :wink: .


Δίνεται η συνάρτηση $f(x)=\sqrt{(x^2+1)(x^2+2)(x^2+3)(x^2+4)+1}$  ,,me χ ανήκει στους πραγματικούς.

Αφού δείξετε ότι η $f$ είναι πολυωνυμική να υπολογίσετε το

$\int^{1}_0f(x)dx$


Let $ R(x)$ be the remainder of the division of a given polynomial $ P(x)$ by the trinomial $ T(x) \equal{} x^2 \minus{} (a \plus{} b)x \plus{} ab$, where $ a \neq b$.

(a) Determine the coefficients $ m$ and $ n$ of $ R(x)$, in terms of $ a,b,P(a)$ and $ P(b)$.

(b) Determine $ m$ and $ n$ in the particular case where $ P(x) \equal{} x^{200}, a \equal{} \minus{} 1$ and $ b \equal{} 2$.

(c) Demonstrate that, in the particular case of (b), the two coefficients are integers.

[hide=" a"]Let $ R(x) \equal{} mx \plus{} n$. Notice that $ T(x) \equal{} (x \minus{} a)(x \minus{} b)$, so $ P(a) \equal{} ma \plus{} n$ and $ P(b) \equal{} mb \plus{} n$. So we find that $ m \equal{} \frac {P(a) \minus{} P(b) }{a \minus{} b}$ and $ n \equal{} \frac {bP(a) \minus{} aP(b) }{b \minus{} a}$.[/hide]

[hide="b"]Just plug in the values.
$ m \equal{} \frac {1 \minus{} 2^{200} }{ \minus{} 3} \equal{} \frac {2^{200} \minus{} 1 }{3}$

$ n \equal{} \frac {2(1) \minus{} ( \minus{} 1)(2^{200})}{2 \minus{} ( \minus{} 1)} \equal{} \frac {2^{200} \plus{} 2}{3}$[/hide]

[hide="c"]We must show that $ 3| 2^{200} \minus{} 1$ and $ 3 | 2^{200} \plus{} 2$. We see that

$ 2^{200} \minus{} 1 \equiv (\minus{}1)^{200} \minus{} 1 \equiv 0 \mod 3$ so $ m$ is an integer.

$ 2^{200} \plus{} 2 \equiv 2^{200} \minus{} 1 \equiv 0 \mod 3$ so $ n$ is an integer.[/hide]


inequalities
trigonometry
vector
triangle inequality
trig identities
Law of Cosines
Pythagorean Theorem

Can you give some proofs of the Triangle inequality theorem

Hmm... cauchy is the standard proof...

[hide]
\[ \sqrt{a_1^2 \plus{} a_2^2 \plus{} \ldots \plus{} a_n^2} \plus{} \sqrt{b_1^2 \plus{} b_2^2 \plus{} \ldots \plus{} b_n^2} \ge \sqrt{(a_1 \plus{} b_1)^2 \plus{} (a_2 \plus{} b_2)^2 \plus{} \ldots \plus{} (a_n\plus{}b_n)^2}\]
\[ \leftrightarrow a_1^2 \plus{} b_1^2 \plus{} \ldots \plus{} a_2^2 \plus{} b_2^2 \plus{} \ldots \plus{} a_n^2 \plus{} b_n^2 \plus{} 2\sqrt{(a_1^2 \plus{} a_2^2 \plus{} \ldots \plus{} a_n^2)(b_1^2 \plus{} b_2^2 \plus{} \ldots \plus{} b_n^2)} \ge\] \[ (a_1\plus{}b_1)^2 \plus{} (a_2\plus{}b_2^2) \plus{} \ldots \plus{} (a_n \plus{} b_n)^2\]
\[ \leftrightarrow \sqrt{(a_1^2 \plus{} a_2^2 \plus{} \ldots \plus{} a_n^2)(b_1^2 \plus{} b_2^2 \plus{} \ldots \plus{} b_n^2)} \ge a_1b_1 \plus{} a_2b_2 \plus{} \ldots \plus{} a_nb_n\]

which is true by Cauchy.
[/hide]

[hide="A different perspective"]The triangle inequality is a direct consequence of the law of cosines
$ c^2 \equal{} a^2 \plus{} b^2 \minus{} 2ab \cos \theta$
since $ | \cos \theta | \leq 1$.

Proving the law of cosines with vectors uses the formula
$ \cos \theta \equal{} \frac{u \cdot v}{|u| \, |v|}$
and dots $ u\minus{}v$ with itself.

That the right hand side of that formula always produces values in $ [\minus{}1,1]$ is true by Cauchy.[/hide]

Can't we just say that the shortest path between two points is along a straight line?

Or drop a perpendicular from a vertex to the opposite side and use the Pythagorean theorem?


Prove that $x^{3}+y^{3}+z^{3}= 3xyz$ if and only if $x+y+z = 0$.

That's not necessarily true. The equality holds for $x=y=z$.


Here is some Armenian team results,wiche are already known.

[b]Tigran Sloyan-      7 1 * * 3 *[/b]
[b]Vardan Verdyan-  7 0 * * 0 *[/b]
[b]Narek Tovmasyan- 7 0 * * 0 *[/b]
[b]Nerses Aramyan-  7 0 * * 0 *[/b]
[b]Tigran Hakobyan-  7 0 * * 0 *[/b]
[b]David Karagulyan- 7 0 * * 0 *[/b]

It is not correct according to the official website, found here:
[url]http://imo2006.dmfa.si/results.html[/url]

[quote="vardanch"]$TEAM RESULTS$

Tigran Sloyan       7 1 7 6 3 0  sum 24
Vardan Verdiyan   7 0 7 0 0 0  sum 14
Nerses Aramyan   7 0 1 7 0 0  sum 15
David Karagulyan  7 0 1 1 0 0  sum 9
Narek Tovmasyan 7 1 7 6 3 0  sum 14
Tigran Hakobyan   7 0 1 6 0 0  sum 14

SUM 90[/quote]
Guys he had a little typo,here is the correct results :wink: .

Tigran Sloyan       7 1 7 6 3 0  sum 24
Vardan Verdiyan   7 0 0 7 0 0  sum 14
Nerses Aramyan   7 0 1 7 0 0  sum 15
David Karagulyan  7 0 1 1 0 0  sum 9
Narek Tovmasyan 7 1 7 6 3 0  sum 14
Tigran Hakobyan   7 0 1 6 0 0  sum 14

So we have a silver,a bronze and four honors!

[quote="nik"]Georgian's have much better results :)[/quote]
Not so much but a little,BUT in GEO's team ther isn't anybody hwo got more then 20 point,in our team I have 24 :P.

Georgian's have much better results

Not so much but a little,BUT in GEO's team ther isn't anybody hwo got more then 20 point,in our team I have 24 .

That actually does not help,

The point is that they have a better total and they have better awards then the armenians.

Congrutulations Tigran, you did a good job

P.S Nick do i know you???

[quote="Soarer"]why should one start this kind of meaningless comparison....
every country should try to compare with China  :D
then most of them will stop comparing. :D  :D[/quote]

I don't agree with you.These two countries and also Azerbaijan
are all Caucasus countries and all have almost(with at most 3 million difference) same population.
And population is very small compared with China.So I think it is natural that these countries compete and compare the results of their countries.
And China has a very huge population which is very great advantage for the country, so actually you can't(must not) compare China with these countries.

I don't think there's anything wrong with comparing results with nations similar to yours (population, training program etc.)

The UK regularly watches the progress of our Western European neighbours and also the top Commonwealth countries. It gives a far better indication into how you are collectively doing that say comparing yourself to China and knowing you will probably always being doing worse.

I know the IMO is an individual competition but's its sometimes good to know just how good your team were this year.

[quote="nik"]In 2001, 2002, 2003 when I was contestant georgians result was more than much beter.
I think 4  medals are much better than 2 too.

Last 13 years: Georgia-Armenia 9:4

irakli
you know me.[/quote]
Do not forget that  your team had 94point,and our had 90,this year,and  that's why I said that your result isn't much better.
Also I think that Armenia and Georgia is brother counterys,so it isn't important for us to have batter results than you,but I can't say the same words about  [size=75]a[/size]zerbaijan,and this year our result was much and much more batter than [size=75]a[/size]zerbaijan's result :P  and this is most important for us :wink: .

armenians are the worst country in the world.   cindirdilar, peyser usagi peyserler.       !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!


Suppose that $ x$, $ y$, $ z$ are positive real numbers such that $ xyz\ge 1$.
Prove that
$ (1\plus{}x\plus{}y)^2\plus{}(1\plus{}y\plus{}z)^2\plus{}(1\plus{}z\plus{}x)^2\le3(x\plus{}y\plus{}z)^2$,
with equality if and only if $ x\equal{}y\equal{}z\equal{}1$.
 :?:

Generally, I do love Irish Inequality problems very much.
They always make me happy.  :)


Let $ ABC$ be an isosceles triangle with $ AB\equal{}AC$ and $ \angle BAC\equal{}100$. Let $ D$ be the point on the production of $ BA$ (order of points is $ B\minus{}A\minus{}D$) such that $ AD\equal{}BC$. Find $ \angle ADC$.

Sorry if I post a mistake solution but my result isn't nice!
Use $ \frac {AC}{AD} \equal{} \frac {AC}{BC}$ we have:
$ \frac {sinD}{sin(100^0 \minus{} D)} \equal{} \frac {sin40^0}{sin100^0}$
This condition give me a result $ D \equal{} 35,94040439...$.
May be I wrong!
 :blush:  :blush:  :blush:

[b]tdl[/b] have you checked this result by some computer programs or there is maybe standard method for solving this type of equations?

Originally problem states that point $ D$ is taken on production of $ AB$  and it is not defined is it beyond $ B$ or $ C$ but normally it should be beyond $ B$ and in this case problem is easy.

Second case is difficult one and most probably it is not necessarily to consider it.

Thank you [b]tdl[/b]

[color=darkred]I think that the [b]correct enunciation[/b] is following lower :[/color]

[quote="delegat"][color=darkred]Let $ ABC$ be an $ A$- isosceles triangle with $ \boxed {m(\widehat {BAC}) = 108^{\circ}}$.

Let $ D\in (AB)$ so that $ \{\begin{array}{c} A\in (BD) \\
\ AD = BC\end{array}$. Find $ x\equiv m(\widehat {ADC})$.[/color][/quote]
[color=darkblue][b][u]Proof I (synthetical)[/u][/b] Denote $ E\in (BC)$ for which $ EA = EB$.

Observe that $ \|\begin{array}{c} AB = CA = CE \\
\ \triangle\ ABE\sim\triangle\ BCA\end{array}$. Thus, $ \frac {BA}{BE} = \frac {CB}{CA} = \frac {AD}{CE}\Longrightarrow$

$ \frac {AB}{AD} = \frac {EB}{EC}$ $ \Longrightarrow$ $ AE\parallel CD$ $ \Longrightarrow$ $ m(\widehat {ADC}) = 36^{\circ}\ .$

[b][u]Proof II (trigonometric).[/u][/b] Apply the [b]Sinus' theorem[/b] in the triangles :

$ \|\begin{array}{cc} \triangle\ ABC\ : & \frac {AB}{BC} = \frac {\sin 36^{\circ}}{\sin 108^{\circ}} \\
 \\
\triangle\ ACD\ : & \frac {AC}{AD} = \frac {\sin x}{\sin (108^{\circ} - x)}\end{array}\|\implies$

$ \sin x\sin 72^{\circ} = \sin (72^{\circ} + x)\sin 36^{\circ}$ $ \Longleftrightarrow$

$ \cos (72^{\circ} - x) - \cos (72^{\circ} + x) = \cos (36^{\circ} + x) - \cos (108^{\circ} + x)$.

Since $ \|\begin{array}{c} \cos (72^{\circ} - x) = - \cos (108^{\circ} + x) \\
\ \cos (72^{\circ} + x) = - \cos (108^{\circ} - x)\end{array}$ obtain

$ \cos (108^{\circ} - x) = \cos (36^{\circ} + x)$, i.e $ \boxed {\ x = 36^{\circ}\ }\ .$
[/color]


Find all positive integers $ n$ and $ x_1,x_2,...,x_n$ such that
$ x_1\plus{}x_2\plus{}....\plus{}x_n\equal{}3n$ and $ \frac{1}{x_1}\plus{}\frac{1}{x_2}\plus{}...\plus{}\frac{1}{x_n}\equal{}1\plus{}\frac{1}{4n}$

$ x_{1},x_{2},...,x_{n}$ must be positive integers, too?

We have that $ 3n*(1 \plus{} \frac {1}{4n}) \equal{} (x_{1} \plus{} x_{2} \plus{} .... \plus{} x_{n})*(\frac {1}{x_{1}} \plus{} \frac {1}{x_{2}} \plus{} ... \plus{} \frac {1}{x_{n}})\geq n^2$.
So, $ 3 \plus{} \frac {3}{4n}\geq n$, $ n \equal{} 1, 2, 3$.
For $ n \equal{} 1, 2$ we don't have any positive integer solutions.
For $ n \equal{} 3$:
$ x_1 \plus{} x_2 \plus{} x_3 \equal{} 9$ and $ \frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} \frac {1}{x_3} \equal{} \frac {13}{12}$.
It's can be easily proved, that only $ (x_1, x_2, x_3) \equal{} (2, 3, 4)$ and all permutations are solutions. If you want, I can show it.


I'm going back for second year this summer. I'm a junior from from MA, live about 8 minnutes away from the dorms lol...

Where are you from lemondemon?

I'm from Lexington. I'm also in 8th grade, so I have to live with a counselor or something like that.


trigonometry
calculus
derivative
integration
geometry
geometric transformation
rotation

Hey,

Is there a third way of deriving the Trigonometric Angle Addition Identities (AAIs),
$ {\cos{\left({\alpha} \plus{} {\beta}\right)}} \equal{} {{\cos{\alpha}\cos{\beta}} \minus{} {\sin{\alpha}\sin{\beta}}}$
$ {\sin{\left({\alpha} \plus{} {\beta}\right)}} \equal{} {{\sin{\alpha}\cos{\beta}} \plus{} {\cos{\alpha}\sin{\beta}}}$
$ {\tan{\left({\alpha} \plus{} {\beta}\right)}} \equal{} {\frac {{\tan{\alpha}} \plus{} {\tan{\beta}}}{{1} \minus{} {\tan{\alpha}\tan{\beta}}}}$
Besides deriving them geometrically from a figure,
[hide="Click to See Figure."]
[img]http://img137.imageshack.us/img137/2944/trigsumformulauf2.png[/img]
[/hide]
Or from Euler's Formula,
$ {{e}^{i\theta}} \equal{} {{{\cos{\theta}}} \plus{} {i\sin{\theta}}}$

Is there a third way?

Thanks,

-APSStudent

There is a clever way to derive them using the derivatives of $ \sin$ and $ \cos$.  Instead of defining the derivative in the normal way, you define them based on integrals, and then you use the solutions to the differential equation
$ f''\plus{}f\equal{}0$.

I can't remember the details but it is doable, although it uses calculus.

Hey,
[quote="JRav"]
There is a clever way to derive them using the derivatives of $ \sin$ and $ \cos$.  Instead of defining the derivative in the normal way, you define them based on integrals, and then you use the solutions to the differential equation
$ f'' \plus{} f \equal{} 0$.

I can't remember the details but it is doable, although it uses calculus.
[/quote]
Hmm, I wasn't able to figure out how to derive the Angle Addition Identities this way, does anyone else know how?

Thanks,

-APSStudent

Hey,
[quote="t0rajir0u"]
Depending on how you define the sine and cosine, the proof of what their derivatives are depends on the angle addition formulas.
[/quote]
Hmm, I'll keep that in mind and try again, although a bit more information would be appreciated.
In addition, thanks for the link the entry was interesting.
[quote="JRav"]
See the attached document. The first and second page develop sine and cosine. Then head to Lemma 2.4 and go from there.
[/quote]
Awesome, this helps explain the derivation alot more. Thanks. In addition, I noticed the attachment was titled "Lecture Notes 15," where might I find the other 14 lecture notes?

Thanks,

-APSStudent

Actually, there are 26 others.  That is number 15 in my collection of 27 based on the Honors Single Variable Calculus course at my university.  I posted the others some weeks ago so look under my old posts, for "Lecture Notes".  If you can't find them, tell me and I'll get them to you.

[quote="t0rajir0u"]Depending on how you define the sine and cosine, the proof of what their derivatives are depends on the angle addition formulas.

The angle addition identities hide a basic underlying principle which is the way [url=http://mathworld.wolfram.com/RotationMatrix.html]rotation matrices[/url] compose (and this is in turn the underlying principle behind Euler's formula):  see Formulation 2 [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=216343]here[/url].  From a sufficiently high-concept point of view, all of trigonometry is the [url=http://en.wikipedia.org/wiki/Representation_theory]representation theory[/url] of the [url=http://en.wikipedia.org/wiki/Circle_group]circle group[/url].[/quote]Nice read and 'undergrad' level math topics in a typical US college but that may be too hard for him. Perhaps he is looking for a [b]geometric[/b] or a [b]vectorial[/b] demonstration.... :ninja:


question:determine the ratio in which the line 3x-y = 9 divides the segment joining the points (1,3) and (2,7)
             if u can represent the situation graphically , it wud be nice.

The line doesn't intersect that line segment.  On the line 3x-y=9, y is negative when x is between 1 and 2.

yes, i know that too. so, try externally dividing it.
and afterwards explain what that ratio means.

[quote="analytic"]question:determine the ratio in which the line 3x-y = 9 divides the segment joining the points (1,3) and (2,7)[/quote]

the line through $ A(1,3)$and$ B(2,7)$ is $ y\equal{}4x\minus{}1$.
so crossing point is calculated as: solve "$ y\equal{}4x\minus{}1,3x\minus{}y\equal{}9$"
then it is$ C(\minus{}8.\minus{}33)$.
So,$ AB: BC\equal{}\sqrt{17}: 9\sqrt{17}\equal{}1: 9$

Then $ 3x\minus{}y\equal{}9$ divedes externally $ 1: 10$ into $ AB$.

what was the original problem? i mean what was your actual question?

i've psted the actual problem in the first post in this topic by me.


induction
geometry
number theory
modular arithmetic
power of a point

It's been since I did the math chat and since it's vacation, I decide to revive it. The first one will be next Friday, 6/18/04.  The topic we're going to decide will be:

1. Modular arithmetic
2. Power of a Point Theorem
3. Explanation of what Mathematical Indution and Contradiction is using for
4. Effect of the Pythagorem Theorem on deciding the type of the triangle
5. Basic Proofs on geometry

If you're interested, please sign up below:

Sincerely,

Silvefalcon

Hi, I'll do it, my AIM is js1527k.

Question:
*Are you going to be teaching us these topics, or do you want to have us explain it, or are you just going to be giving problems about it?

i'll join if my baseball team doesn't make it to our league's finals

Using my brother's AIM account since AOL won't let me make one: rb79mobilepod.  I'll join as long as there is no math jam that day.  What time will this be?

mine is hmmmmmbbq, ill be there. what time?

I'd like to do it, but i don't have AIM. What should I do?

[url=http://www.aim.com/get_aim/win/latest_win.adp?aolp=]The newest version of AIM.  Click the "New Users Click Here" button[/url]

Well there is not much that you can do.  Aim is the easiest thing for us to do this on since we aren't highly organized.  I believe that you can use Aim express w/out downloading aim.  You still need a sn though.  Here's a link [url]http://www.aim.com/get_aim/express/aim_expr.adp?aolp=1[/url] Someone will probably post the transcript of it afterwards so you can always look at that.

Silverfalcon, will you just send a chat invitiation?

We usually just use the chatroom called "MathChat".  You can invite yourself in their but one of the other people including me will invite you in if you want us to.  It doesn't matter either way.

Just want to remind everyone about the Mathchat.

Date: June 18
Time: 5 PM EST-whenever everyone leaves
Where: Aol Instant Messenger Chatroom( Mathchat )

If you have any questions my aim sn is BaseballJoe2002...I really need to get a new one because I don't play baseball much anymore... or you can talk to silverfalcon since he is hosting the mathchat( his sn is jwew05 if he hasn't changed it since last time).
I hope to see everyone there.

bah, i guess i can't come. i'm going deep-sea fishing from 5-11, so i won't be able to make it unless weather's bad and the trip's cancelled

Just to make sure, BobofBobs, do you have your sn now or should I exclude you from today's discussion? Please post as soon as possible!

Sorry to tell all of you this but due to my personal business, today's discussion has been cancelled. I'm very sorry about it.

Um, can we still have a discussion w/o you?  I will set up the chatroom in your absence if that is okay...

Unless Silverfalcon has some problem w/it, the chatroom is MathChat and you can invite yourself.

Is this okay?


where are the results of rmo 2006 for  mumbai region put up?

Why do you need it now?Even the INMO has got over by now.....As far as I knw ,there is no website which puts up the results for mumbai region though there is one for maharashtra region


induction
combinatorics unsolved
combinatorics

On a $2^n$ by $2^n$ chessboard, one square is cut out. An "L-piece" covers three squares like an L. Show that the board can be tiled with L-pieces.

By induction (just use the $n=2$ pattern)  it is clear immediately that a "big" L-piece of sidelength $2^n$, i.e. the board with a whole quarter of it cut out, can be covered by L-pieces.
So if we deal first with the big L-piece formed by the three quarters that don't contain the hole, the problem is reduced to $2^{n-1}$. etc...  Finally we arrive at $n=1$.

Now a harder one: same problem for a square board of side $n>5$ not divisible by 3, with a 1x1 hole. :)

Thanks for such a nice generalization, spanferkel.  :)

First we prove that for $n>5$, an $n$ by $6$ board is "tileable". Indeed, it follows from the fact that $n$ can be written in form of $2a+3b$ where $a,b$ are nonnegative.

So from an $n$ by $n$ board, we can gradually remove an $n$ by $6$ board (of course we choose those with no holes), and we will end up with a $7$ by $n$, an $8$ by $n$, a $10$ by $n$, or an $11$ by $n$. Now we remove from the other side, and we will get a square with side $7$,$8$,$10$, or $11$. Now we only need to check these four cases, which can be done by hand.

Is this the same as your solution?

[quote="spanferkel"]Yeah, only I had hoped for something more formal than 'doing by hand' the 7,10,11 cases.  There are quite a bit of cases to distinguish, according to where the hole is  :([/quote]

Eh, it takes quite some work, doesn't it?

Anyway I'll post it soon  :)

Well, first we can reduce the cases by a quarter, since the board is symmetric (or less, for odd $n$)

Then, for example, if we want to prove for the hole at position $(1,1)$, we can try to tile the rest of the board, placing one L at $(1,2),(2,1),(2,2)$. Then, if the hole is at $(1,2)$, we can tile at $(1,1),(2,1),(2,2)$, and the rest is the same. By this way, we can reduce a lot of cases.

But still a lot remains.  :(

I can't see any better way than this.[/code]


function
induction
inequalities
algebra unsolved
algebra

A function $f: [0,1]\rightarrow R$ satisfies $f(x)\geq 0$ for $0<x<1$, $f(1)=1$, and $f(x+y)\geq f(x)+f(y)$ for $x,y,x+y\in [0,1]$.

Show that $f(x)\leq 2x$ for all $x\in [0,1]$

We can easily find that $f(\frac{1}{2^{n}})\le \frac{1}{2^{n}}$ $\forall n\in \mathbb{N}$
next notice that $x\in [\frac{1}{2},1]\rightarrow f(x)\le 1\le 2x$
with induction it is simple to show that the inequality holds for all intervals $[\frac{1}{2^{n}},\frac{1}{2^{n-1}}]$

It mean, that f(x) increase and $f(\frac{1}{2^{k}})\le \frac{1}{2^{k}}$.
Therefore $f(x)<2x \ \forall x>0.$


A swimmer plans to swim at least 100 laps during a 6 day perio. during this period the swimmer will increase the number of laps completed each day by one lap. What is the least number of laps the  swimmer must complete on the first day

This was on my test today and i just couldn't get it:
A cargo plane leaves an airport travling 180mph. two hours later a jet plane leaves the same airport travling at 900mph in the same direction how long will it take for the cargo plane to catch up to the jet.

and one last question

The dimensions of a brick in inches, are 2 by 4 by 8. How may  such bricks are needed to have a total volume exactly three cubic feet?

any and all help will be greatly appreciated

1.
x + x +1 + x +2 + x +3 + x +4 +x +5=100
6x+15=100
6x=85
doesnt divide evenly it is 14 remainder 1.. so least possible is [b]15[/b]

[quote="FrankieBlueEyes"]
This was on my test today and i just couldn't get it:
A cargo plane leaves an airport travling 180mph. two hours later a jet plane leaves the same airport travling at 900mph in the same direction how long will it take for the cargo plane to catch up to the jet.
[/quote]

the jet plane starts 180 times 2=360 miles behind the cargo plane.
Expressing their distance from the airport as a function of time, t, we have

distance of cargo plane from airport =180t
distance of jet plane from airport =900(t-2)
                                                       =900t - 1800

When cargo plane catches up to the jet, their distances are equal:
180t=900t - 1800
720t=1800
     t=2.5 hours

the amount of time it takes is after the cargo plane has flown 2.5 hours
this means that the jet plane has flown 2.5-2=0.5 hours

Sorry FrankieBlueEyes it [i]is[/i] less than 1 , I made some mistakes in calculations  :blush:

It's corrected now.

[quote="NoSoupForYou"][quote="chess64"][quote="FrankieBlueEyes"]
The dimensions of a brick in inches, are 2 by 4 by 8. How may  such bricks are needed to have a total volume exactly three cubic feet?
[/quote]

1 cubic feet = 144 cubic inches
3 cubic ft. = 432 cb. in.
The volume of a brick is 64.
$\frac{432}{64}=6.75$, I think.[/quote]

$1$ cubic ft. = $12^3=1728$ cubic in.
$\frac{1728\cdot 3}{64}=\boxed{81}$[/quote]

oops  :blush: , *hits head against table* *sighs * *repeats "$\text {squares } \neq \text {cubes}$ 50 times*


superior algebra
superior algebra unsolved

Let $ G$ be a group and $ x,y\in G$ and $ z \equal{} xyx^{ \minus{} 1}y^{ \minus{} 1}$.Prove that if $ z$ commutes with both $ x,y$ then:
$ (yx)^n \equal{} z^{{n}\choose {2}}y^nx^n$

$ z$ is typically called the commutator of $ x$ and $ y$, abbreviated $ [x,y]$.  Notice that:

$ xy \equal{} zyx$

Now using this fact, we just move things around, knowing that each time we do, we get an extra $ z$ in there.  Since $ z$ commutes with $ x$ and $ y$ (and itself), we need not pay attention to where the $ z$'s are.  This just boils down to a counting arugument, what is the number of times you need to switch $ xy$ with $ zyx$.


I dont get this question.. is there something wrong with the sign or am i just stupid..

[quote="Treething"]A triangle has side lengths of $x$, $y$, and $z$.

Prove that $\sqrt{x}+\sqrt{y}+\sqrt{z}\geq 2\sqrt{x}$.[/quote]
[hide]
$\sqrt{y}+\sqrt{z}\geq \sqrt{x}$
$y+z+2\sqrt{yz}\geq x$
$y+z \geq 2\sqrt{yz}$
$y+z>x$
[/hide]
Grrrr.
I'm so close I think

$\sqrt{x}+\sqrt{y}+\sqrt{z}\geq 2\sqrt{x}$
$\sqrt{y}+\sqrt{z}\geq \sqrt{x}$
since they're sides of a triangle, both sides are positive so the $\geq$ doesn't change when we square
$y+z+2\sqrt{yz}\geq x$
we know from triangle inequality that
$y+z>x$
and adding the positive $2\sqrt{yz}$ to the greater side doesn't change the inequality.


Let $ \frac{1}{\sqrt{2}}\leq a,b,c \leq \sqrt{2}$.Prove that:
$ \frac{3}{a\plus{}2b}\plus{}\frac{3}{b\plus{}2c}\plus{}\frac{3}{c\plus{}2a}\geq \frac{2}{a\plus{}b}\plus{}\frac{2}{b\plus{}c}\plus{}\frac{2}{c\plus{}a}$

Put a+b+c=1

By CS, $ LHS /ge 9 /ge RHS$ obvious in the given bounds for a,b,c.

hell_ever wrote:
[quote]Put a+b+c=1

By CS, $ LHS \geq 9 \geq RHS$ obvious in the given bounds for a,b,c.[/quote]
wrong.
Let $ a\plus{}b\plus{}c\equal{}1$
By CS $ LHS\equal{}\frac{3}{a\plus{}2b}\plus{}\frac{3}{b\plus{}2c}\plus{}\frac{3}{c\plus{}2a}\geq \frac{27}{3(a\plus{}b\plus{}c)}\equal{}9$ and
$ RHS\equal{}\frac{2}{a\plus{}b}\plus{}\frac{2}{b\plus{}c}\plus{}\frac{2}{c\plus{}a}\geq \frac{18}{2(a\plus{}b\plus{}c)}\equal{}9$
I see no proof here.
This inequality is indeed easy, but not as easy as you pretend to be.

[quote="Inequalities Master"]Let $ \frac {1}{\sqrt {2}}\leq a,b,c \leq \sqrt {2}$.Prove that:
$ \frac {3}{a \plus{} 2b} \plus{} \frac {3}{b \plus{} 2c} \plus{} \frac {3}{c \plus{} 2a}\geq \frac {2}{a \plus{} b} \plus{} \frac {2}{b \plus{} c} \plus{} \frac {2}{c \plus{} a}$[/quote]
Let b=x_2 and $ c \equal{} x_3$
$ A \equal{} LHS \minus{} RHS \equal{} \sum_{i \equal{} 1}^{3} \frac {x_i \minus{} x_{i \plus{} 1}}{(x_i \plus{} 2x_{i \plus{} 1})(x_i \plus{} x_{i \plus{} 1})} \geq 0$
 Note that: $ \sum \frac {x_i \minus{} x_{i \plus{} 1}}{x_ix_{i \plus{} 1}} \equal{} 0$
hence :$ 6A \equal{} 6 \sum_{i \equal{} 1}^{3} \frac {x_i \minus{} x_{i \plus{} 1}}{(x_i \plus{} 2x_{i \plus{} 1})(x_i \plus{} x_{i \plus{} 1})} \minus{} \sum \frac {x_i \minus{} x_{i \plus{} 1}}{x_ix_{i \plus{} 1}}$
$ \equal{} \sum_{i \equal{} 1}^{3} \frac {(x_i \minus{} x_{i \plus{} 1})^2(2x_{i \plus{} 1} \minus{} x_i)}{(x_i \plus{} 2x_{i \plus{} 1})(x_i \plus{} x_{i \plus{} 1})x_ix_{i \plus{} 1}}$


Let a,b,c be positive real numbers with a+b+c=1.Prove that
(1/a+1/b+1/c)$\geq$25/(1+48abc)


[quote="anonymous1173"]$(i+1)^{i}=?$[/quote]
$e^{-\frac{\pi}{4}}(cos(ln\sqrt{2})+isin(ln\sqrt{2}))$
one branch of it:)


calculus
integration
logarithms
trigonometry
limit
calculus computations

Calculus:
\[ I = \int\limits_{0}^{\pi/4}xtan(x)dx\]

The first step you should be able to do it yourself: that is the integration by parts, the result you will get is the one where you may need help on.

According to Mathematica:
[code]In[3]:= \[Integral]x Tan[x] \[DifferentialD]x

Out[3]= (I x^2)/2 - x Log[1 + E^(2 I x)] +
 1/2 I PolyLog[2, -E^(2 I x)][/code]

So I don't think integration by parts is going to do very much. Looks like this can't be expressed in elementary terms.

My solution is quite long, but it is the first thing I came up with, so it's rough.

First, let

$ S \equal{} \int_{x \equal{} 0}^{\pi/2} \log \sin x \, dx \equal{} \int_{x \equal{} \pi/2}^0 \log \sin(\pi/2 \minus{} x) ( \minus{} dx) \equal{} \int_{x \equal{} 0}^{\pi/2} \log \cos x \, dx$.

So

$ 2S \equal{} \int_{x \equal{} 0}^{\pi/2} \log \frac {\sin 2x}{2} \, dx \equal{} \int_{x \equal{} 0}^{\pi/2} (\log \sin 2x \minus{} \log 2) \, dx$

$ \equal{} \frac {1}{2}\int_{x \equal{} 0}^\pi \log \sin x \, dx \minus{} \frac {\pi}{2} \log 2$.

But

$ \int_{x \equal{} 0}^\pi \log \sin x \, dx \equal{} S \plus{} \int_{x \equal{} \pi/2}^\pi \log \sin x \, dx \equal{} S \plus{} \int_{x \equal{} 0}^{\pi/2} \log \sin(\pi/2 \plus{} x) \, dx \equal{} 2S$,

and combining these results gives $ S \equal{} \minus{} \frac {\pi}{2} \log 2$.

Now consider the integral

$ I \equal{} \int_{x \equal{} 0}^{\pi/4} x \tan x \, dx \equal{} \left[ \minus{} x \log \cos x \right]_{x \equal{} 0}^{\pi/4} \plus{} \int_{x \equal{} 0}^{\pi/4} \log \cos x \, dx$
$ \equal{} \frac {\pi}{8} \log 2 \plus{} \frac {1}{2} \int_{x \equal{} 0}^{\pi/2} \log \cos \frac {x}{2} \, dx \equal{} \frac {\pi}{8} \log 2 \plus{} \frac {1}{4} \int_{x \equal{} 0}^{\pi/2} \log \left(\frac {1 \plus{} \cos x}{2}\right) \, dx$
$ \equal{} \frac {1}{4} \int_{x \equal{} 0}^{\pi/2} \log (1 \plus{} \cos x) \, dx$.

Combining this with S gives

$ 4I \minus{} S \equal{} \int_{x \equal{} 0}^{\pi/2} \log \frac {1 \plus{} \cos x}{\sin x} \, dx \equal{} \int_{x \equal{} 0}^{\pi/2} \log \cot \frac {x}{2} \, dx \equal{} \minus{} 2 \int_{x \equal{} 0}^{\pi/4} \log \tan x \, dx$.

With the substitution $ x \equal{} \tan^{ \minus{} 1} t$, $ dx \equal{} dt/(1 \plus{} t^2)$, we obtain

$ 4I \minus{} S \equal{} \minus{} 2 \int_{t \equal{} 0}^1 \frac {\log t}{1 \plus{} t^2} \, dt$.

Next, we integrate by parts with the choice $ u \equal{} \log t$, $ du \equal{} dt/t$, $ dv \equal{} dt/(1 \plus{} t^2)$, $ v \equal{} \tan^{ \minus{} 1} t$:

$ 4I \minus{} S \equal{} \minus{} 2 \left(\lim_{a \rightarrow 0^ \plus{} } \left[(\log t)(\tan^{ \minus{} 1}t)\right]_{t \equal{} a}^1 \minus{} \int_{t \equal{} 0}^1 \frac {\tan^{ \minus{} 1} t}{t} \, dt\right) \equal{} 2 \int_{t \equal{} 0}^1 \frac {\tan^{ \minus{} 1} t}{t} \, dt$.

This final integral is equal to Catalan's constant, which we see by expanding the arctangent as a power series and integrating term by term:

$ \int_{t \equal{} 0}^1 \frac {\tan^{ \minus{} 1} t}{t} \, dt \equal{} \int_{t \equal{} 0}^1 \sum_{k \equal{} 0}^\infty \frac {( \minus{} 1)^k t^{2k}}{2k \plus{} 1} \, dt$
$ \equal{} \sum_{k \equal{} 0}^\infty \frac {( \minus{} 1)^k}{2k \plus{} 1} \int_{t \equal{} 0}^1 t^{2k} \, dt \equal{} \sum_{k \equal{} 0}^\infty \frac {( \minus{} 1)^k}{(2k \plus{} 1)^2} \equiv C$.

Thus

$ I \equal{} \frac {S \plus{} 2C}{4} \equal{} \frac {C}{2} \minus{} \frac {\pi}{8} \log 2$, where $ C$ is Catalan's constant.  Of course, there are some convergence issues that I did not address, so the above is not completely rigorous, but it is basically correct.

atomicwedgie: I thought about doing something like that, but I like making a substitution on the first step and making it
\[ \int_0^1\frac{y\arctan y}{1\plus{}y^2}\,dy\]
That might be another way of getting into the power series.

The substitution you mentioned was actually the very first thing I tried.  I was unable to get it to work, because both $ \tan^{ \minus{} 1} x$ and $ x/(1 \plus{} x^2)$ have similar power series and I ran into convergence problems when I expanded both.  Expanding only one or the other did not enable me to find an easy formula for the term-by-term integral.  If there's a way to proceed from this form, I'm not smart enough to see it... :(


algorithm
number theory
Euclidean algorithm
Diophantine equation

Hi,

I just dont understand a particular thing with this, i would really appreciate all help =)

We shall ﬁnd all solutions of

$ 77x \plus{} 42y \equal{} 35.$

First we calculate $ gcd(77, 42)$ using the Euclidean Algorithm:

$ gcd(77, 42) \equal{} 7$

Since 7 does divide 35, this means that the linear Diophantine equation
does have integer solutions. To actually ﬁnd the solutions we ﬁrst reverse
the steps in the Euclidean Algorithm:

$ 7 \equal{} 42 \minus{} 35$

$ \equal{} 42 \minus{} (77 \minus{} 42)$
$ \equal{} (\minus{}1) \cdot 77 \plus{} 2 \cdot 42$

So we take $ u \equal{} \minus{}1$ and $ v \equal{} 2$ One solution is then

$ x_{0} \equal{} (\minus{}1) \cdot \frac {35}{7} \equal{} \minus{}5, y_{0} \equal{} 2 \cdot \frac {35}{7} \equal{} 10.$

The main thing i dont get is "Euclidean Algorithm in reverse". Could someone please help me =S

Thankyou

[quote="t0rajir0u"]I gave a long example of this type of computation [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?ppage=5&w=634&t=217361&spp=5&sst=85]here[/url].[/quote]

thankyou =)


Suppose the function $ f$ satisfies the following:

(a) $ f(n) > 0$ for all integers $ n$

(b) $ f(n) \equal{} [f(n\minus{}1)]^2$, and

(c) $ f(4) \equal{} 81$.

Find $ f(1)$.

We use fact c and work backwards, plugging into b and keeping a in mind so we get positive square roots.

$ 81 \equal{} f(4) \equal{} f(3)^2$

$ 9 \equal{} f(3) \equal{} f(2)^2$

$ 3 \equal{} f(2) \equal{} f(1)^2$

$ f(1) \equal{} \sqrt {3}$


This is the problem:

Let f(n) be a function defined on the set of all positive integers and with all its values in the same set. Prove that if f(n+1) > f[f(n)] for each positive integer n, then f(n) = n for each n.

Well my problem is that I don't understand why f(1) is the smallest value and then the next smallest is f(2).

In general why is it that f(1) < f(2) < f(3) < ... < f(n), or monotonically increasing?

Also a solution from a book states that the smallest element cannot be of the form f(f(n)). Why?

Remember : Let S be a set of natural numbers. Always S has a smallest value and this value is in S .

Let S(1) be the sequence {f(1),f(2),f(3),..}.
For all m>=2 there is f(m) > f(f(m-1)) , so f(m) cant be the smallest value of S(1), so this one must be f(1), and f(m) > f(1) for all m >=2.
Now, let S(p) be the sequence {f(p),f(p+1),f(p+2),..}.
For all m>=p+1 there is f(m) > f(f(m-1)) , so f(m) cant be the smallest value of S(p), so this one must be f(p), and f(m) > f(p) for all m >=p+1.

So there is the sequence :  f(1) < f(2) < ... < f(p) < f(p+1) < ... .
Manifestly p < m  ===> f(p) < f(m) , increasing function
Also f(1) >=1 , f(2)>f(1) so f(2) >=2 , f(3)>f(2) so f(3) >=3, …., f(p) >f(p-1) so f(p) >=p , …..
We have f(f(p)) < f(p+1) so f(p) < p+1 , because f is an increasing function.
 (f(p) >=p) & (f(p)<p+1)     ===> f(p)=p  (q.e.d)

Do you follow me ? .. :)

We made the same thing at the training weekend in Belgium.
We also suposed there is a minimum etc. But we suposed the minimum was 0 and not 1...  :?


geometry
circumcircle
incenter
geometry proposed

Let $ ABC$ be a triangle and $ A'$, $ B'$, $ C'$ be the feet of the altitudes dropped from $ A$, $ B$, $ C$ to $ BC$, $ CA$, $ AB$, respectively. Call $ M$, $ N$, $ P$ the tangent points of the incircle $ \mathcal{C}(I)$ of triangle $ ABC$ with its sides $ BC$, $ CA$, $ AB$, respectively. Let $ I_1$, $ I_2$, $ I_3$ be respectively the incenters of triangles $ AB'C'$, $ BC'A'$, $ CA'B'$. Prove that the circumcenter of triangle $ I_1I_2I_3$ is the orthocentre of triangle $ MNP$ and determine the circumradius of triangle $ I_1I_2I_3$ in term of $ BC\equal{}a$, $ CA\equal{}b$, $ AB\equal{}c$.

[quote="April"]Let $ ABC$ be a triangle and $ A'$, $ B'$, $ C'$ be the feet of the altitudes dropped from $ A$, $ B$, $ C$ to $ BC$, $ CA$, $ AB$, respectively. Call $ M$, $ N$, $ P$ the tangent points of the incircle $ \mathcal{C}(I)$ of triangle $ ABC$ with its sides $ BC$, $ CA$, $ AB$, respectively. Let $ I_1$, $ I_2$, $ I_3$ be respectively the incenters of triangles $ AB'C'$, $ BC'A'$, $ CA'B'$. Prove that the circumcenter of triangle $ I_1I_2I_3$ is the orthocentre of triangle $ MNP$ and determine the circumradius of triangle $ I_1I_2I_3$ in term of $ BC \equal{} a$, $ CA \equal{} b$, $ AB \equal{} c$.[/quote]

We have $ I_1$ ,$ I_2$ ,$ I_3$ be the orthorcenters of triangles $ ANP$ ,$ BMP$ ,$ CMN$ ,respectively (see theorem 3.8 in file )
so this problem was very easy :D and $ R(I_1I_2I_3) \equal{} R(MNP)$

I think this is not true when $ \triangle ABC$ is obtuse!


Can you create a planar graph with $8$vertices whose complement is still a planar graph?
Do you have a general idea about this event? :)

[quote="The Captain"]I found it! :lol: It can be shown that the statement does not hold for $n \geq 9$
Unfortunately the proof is not elementary.Anyway for this case consider $Q_3$ and its complement.[/quote]

 :lol: what a pitty.
I have an idea but it seems that I arrive late to answer your question.
still I want to say my idea for someones who are like to find other solutions.
you can make a graph with 8 vertices and draw the maximum edges till your graph stay planar.
it seems the compliment of this graph must be planar.


I just think that this is a funny problem :)


Does the solution of the following differential equation continues untill $ x \equal{} a$?

$ y' \equal{} y^{2} \plus{} x^{2}$
$ y(0) \equal{} 0$
$ a \equal{} 1$

The answer is: yes, when $ 0 \le x \le 1$ , $ 0 \le y' \le y^{2} \plus{} 1 \equal{} > 0 \le y \le tan(x) < \equal{} \plus{} oo$

Can someone explain what am I looking in the solutions in order to answer this type of questions?

Being unable to find an explicit solution of $ y'\equal{}y^2\plus{}x^2$, you should consider comparing it with a solution of a simpler differential equation. "Comparison" means that you can prove that one solution is smaller than the other. For example, if $ u'\equal{}u^2\plus{}1$, then on the interval $ [0,1]$ you have $ y\le u$ (under the initial condition $ y(0)\equal{}u(0)\equal{}0$). This requires some proof: begin it by introducing $ x_0: \equal{}\inf\{x\colon y(x)>u(x)\}$.


geometry
inequalities
similar triangles
geometry solved

The incircle W of triangle ABC touches its
sides at the points D, E, F. For any point P on W,
let d be the sum of the distances from P to
the sides of the triangle DEF.
-> prove that PA+PB+PC > 2d

Has someone an idea??

All I could do is this: $2d=\frac{PE\cdot PF+PE\cdot PD+PF\cdot PD}{r}$, where r is the inradius. That's not much, but maybe it helps?

Erdos-Mordell Theorem tells us that for any point P inside a triangle ABC, the following inequality holds:

PA + PB + PC $\geq$ 2d where d is the distance from P to the sides of the triangle.

Equality takes place iff and only iff P is the center of a equilateral triangle ABC.

From this theorem your problem derives very easy.

Sure, but here d = sum of the distances from P to the sides of DEF, not ABC... :D

This must beat all records: An absolutely easy problem standing unsolved for one and a half year! Let's stop this misery.

[quote="alekk"]The incircle W of triangle ABC touches its
sides at the points D, E, F. For any point P on W,
let d be the sum of the distances from P to
the sides of the triangle DEF.
-> prove that PA+PB+PC > 2d[/quote]

Looks like Valentin's hint about the Erdös-Mordell inequality has not really been understood, since Erdös-Mordell is really an important step in the solution. Probably most important is the following lemma:

[b]Lemma 1.[/b] If GH is a chord of a circle k, and g and h are the tangents to the circle k at the points G and H, and if P is an arbitrary point on k, then $d(P;\;GH)^2 = d(P;\;g)\cdot d(P;\;h)$. Hereby, for every point $P_1$ and every line $g_1$, the abbreviation $d(P_1;\;g_1)$ means the distance from the point $P_1$ to the line $g_1$.

Lemma 1 is really well-known and one can easily prove it either using similar triangles or using simple trigonometry. Now, applied to the incircle of triangle ABC and the chord GH being one of the chords EF, FD, DE, Lemma 1 yields $u^2 = yz$, $v^2 = zx$, $w^2 = xy$, where u, v, w are the distances of the point P to the sides EF, FD, DE of triangle DEF, and x, y, z are the distances of the point P to the sides BC, CA, AB of triangle ABC. In other words, $u=\sqrt{yz}$, $v=\sqrt{zx}$, $w=\sqrt{xy}$.

But the Erdös-Mordell inequality, applied to the triangle ABC and the point P (of course, the point P lies inside the triangle ABC, since it lies on the incircle), yields $PA+PB+PC\geq 2\cdot (x+y+z)$. Note that we even have the stronger > sign here: $PA+PB+PC > 2\cdot (x+y+z)$ (this is because equality in the Erdös-Mordell inequality holds only if the triangle ABC is equilateral and the point P is its center, what is not possible for a point P lying on the incircle of triangle ABC).

Now, AM-GM gives $y+z\geq 2\sqrt{yz}$, what, together with $u=\sqrt{yz}$, implies $y+z\geq 2u$. Similarly, $z+x\geq 2v$ and $x+y\geq 2w$. Thus,

$2\cdot (x+y+z) = (y+z)+(z+x)+(x+y) \geq 2u+2v+2w = 2\cdot (u+v+w)$.

Since $PA+PB+PC > 2\cdot (x+y+z)$, we thus get

$PA+PB+PC > 2\cdot (u+v+w) = 2d$

(since d = u + v + w by definition), and the proof is done.

  Darij


quadratics
abstract algebra
number theory
number theory unsolved

Can you prove that with every prime p exit n<=p such that n is not a quadratic residue mod p

It's not true for $p = 2$.

$x^{p-1}-1 \bmod p$ has roots $1, 2, 3, ... p-1$ by Fermat's Little Theorem, and those are its only roots by Lagrange's Theorem.

For $p > 2$ of course $p-1$ is even; hence,

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

The left factor contains all of the quadratic residues; by Lagrange's Theorem there must be exactly $\frac{p-1}{2}$ of them, hence, there are exactly $\frac{p-1}{2}$ quadratic non-residues.

what a disaster!!(I typed casually). The really thing is :
n<=sqrt(p) :blush:

Still not true for $p = 3$.

It suffices to show that there exist quadratic residues $m = n^{2}\le p$ which are not quartic residues.

For $p \equiv 1 \bmod 4$ we have

$x^{ \frac{p-1}{2}}-1 = \left( x^{ \frac{p-1}{4}}-1 \right) \left( x^{ \frac{p-1}{4}}+1 \right)$

And then the conclusion is obvious.

For $p \equiv 3 \bmod 4$...

"With every prime p, exist a nonquadratic residue n mod p such that n<sqrt(p) +1"
Why do you say that it is wrong. It is in "PROBLEMS IN ELEMENTARY NUMBER THEORY" of Hoojoo Lee :?:

[quote="luu"]"With every prime p, exist a nonquadratic residue n mod p such that n<sqrt(p) +1"
Why do you say that it is wrong.[/quote]

I didn't say it was wrong, but the problem as he stated it didn't hold for $p = 3$.  It holds now though, thanks.


logarithms
calculus
integration
analytic geometry
geometry
calculus computations

If a shaded region is bounded by the graphs $ y\equal{}2x$ and $ y\equal{}xe^x$, what is the volume of the solid generated when the region is revolved around the x-axis?

I'm not sure when to use the washer method and when to use disk method.

In explaining these to my class, "washer method or shell method?" is not question I ever ask, not in that form. We choose our method based on other considerations. (And "washer" and "disk" are really the same method; the alternate method is "shell".)

1. Draw the flat, 2-dimensional graph of the object before rotating. Figure out which curve is right/left/top/bottom. Find points of intersection. In this case, we have points of intersection at $ (0,0)$ and $ (\ln 2,2\ln 2),$ with $ y\equal{}2x$ as the top curve and $ y\equal{}xe^x$ as the bottom curve.

2. Now: which way would we like to slice that - vertically (which will lead to a $ dx$ integral) or horizontally (which will lead to a $ dy$ integral). Note that our equations are given in $ y\equal{}f(x)$ form. We'll use that directly if we slice vertically, while if we want to slice horizontally, we'll have to solve for $ x$ in terms of $ y.$ Do we want to solve $ y\equal{}xe^x$ for $ x$ in terms of $ y$ - uh, no, we don't. So for computational algebraic reasons, we're pretty much forced to slice vertically.

3. Draw a typical slice. It's vertical, and it's located approximately at the horizontal coordinate $ x$ for some $ x$ between $ 0$ and $ \ln 2.$ It has width $ dx.$ It's height extends from a bottom of $ y\equal{}xe^x$ to a top of $ y\equal{}2x.$

4. Now revolve that slice around ... (taking the time to re-read the problem with extra care) ... the $ x$-axis. We get a three dimensional shape - skinny and flat, so that its thickness is $ dx.$ What is it? It's a "washer." Find the cross-sectional area, which will be the area of an annulus - you get that by taking the area of an outer circle and subtraction off the area of an inner circle. Multiply that by the thickness $ dx$ and you have $ dV.$

5. Now integrate.

Our decision as to whether this was a washer method or a shell method wasn't confirmed until the middle of my step 4. We didn't set out to used the washer method - we just did what we could and arrived at whatever method that led us to.


There are n boxes, each having a numbered lock. Now each box has a numbered key, ie box i has key j. Now u have to open all the boxes. Initially u hv no keys, so u hv to break open one box. Now, once u open d box, u get they key inside it, nd then go on opening boxes as long as possible. Then u break open another box. In how many arrangements of keys can u open all d n boxes without breaking more than k locks.

The question is a bit vague as is: are you asking for the number of initial configurations, or number of ways you can open the boxes?

In the former, the answer should just be $ \sum_{j \equal{} 1}^k S(n,j)$, where $ S(n,j)$ is the [url=http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind]Stirling number of the first kind[/url].

ya rite, i had got that, it shud be summation of mod (S(n,k)) i guess. Is there a formula for this summation. i have modified the first post


[hide][quote="kunny"]
$=(x-1)\{(b-c)x^2-(a+b-2c)x+2(a-b)\}$[/quote]

the quadratic can be factored using the quadratic formula:

$x=\frac{a+b-2c\pm\sqrt{(a+b-2c)^2-4(b-c)(2)(a-b)}}{2(b-c)}$

the discriminant simplifies to $a^2-6ab+4ac+9b^2-12bc+4c^2$, which conveniently factors to $(a-3b+2c)^2$.  so the 2 roots are

$\frac{a+b-2c+a-3b+2c}{2(b-c)}=\frac{2a-2b}{2(b-c)}=\frac{a-b}{b-c}$
and
$\frac{a+b-2c-a+3b-2c}{2(b-c)}=\frac{4b-4c}{2(b-c)}=2$

so the factors are $(x-1)(x-2)((b-c)x-a+b)$[/hide]

That's right perfect number :)

Try this one now,

$E=x^3(b-c)-x(b^3-c^3)+(b^3c-bc^3)$

Find the factors of $E$.

Yup,right on kunny :)

While we are at it,why don't we try another one :D

$\text{Factorise} \ (x+y+z)^5-x^5-y^5-z^5$ without expandiong $(x+y+z)^5$.Show your method as well
 :)

I got it! But this problem is left for anyone. :D


topology
complex analysis
function
limit
complex analysis unsolved

1. (a) Let $ A$ and $ B$ be open subsets of $ \mathbb{C}$, let $ \varphi: A\to B$ be a homeomorphism of $ A$ onto $ B$. Assume that $ z_j$ is a sequence in $ A$ which converges to a point $ a\in\partial A$, and that $ \varphi(z_j)$ converges to a point $ b$, prove that $ b\in\partial B$.
(b) Let $ A=D\setminus\{0\}=\{z\in\mathbb{C}: 0<|z|<1\}$ be a puntured unit disc, and let $ B$ be the annulus $ B=\{z\in\mathbb{C}: 1<|z|<2\}$. prove that there is no holomorphic isomorphism of $ A$ onto $ B$.

(c) Define explicitly a diffeomorphism of $ A$ onto $ B$, with $ A$, $ B$ as in $ (b)$.

2. Consider the sequence $ f_n(z)=\frac{n}{z+n}$. These functions are all homomorphic in $ \mathbb{C}\setminus\{-\mathbb{N}\}$. Determine the limit on $ \mathbb{C}\setminus\{-\mathbb{N}\}$ and prove that on this open set the sequence converges compactly to this limit.

3. Take $ E\subset\mathbb{C}$, $ \varphi: E\to\mathbb{C}$  is continuous. If $ C(D,E)=\{f\in C(D\mathbb{C}): f(D)\subset E\}$, then $ \varphi\circ  (_): C(D,E)\to C(D,\mathbb{C})$, $ f\mapsto\varphi\circ f$.
Prove that this map is continuous in the compact-open topology.

Thanks very much in advance.

I think you have a typo in your first question: Let $ \{z_n\}\subset A$ be a sequence converging to $ a\in\partial A$ and let $ f(z_n)$ converge to $ b$. Assume by way of contradiction $ b\in B$. Then there exists $ r > 0$ such that $ B_r(b)\subset B$, and since $ f$ is a homeomorphism $ f^{ \minus{} 1}(B_r(b))$ is open and contained in $ A$. However by continuity we have $ b \equal{} \lim f(z_n) \equal{} f(\lim z_n) \equal{} f(a)$ and so $ f^{ \minus{} 1}(b) \equal{} a\in\partial A$ contrary to $ f^{ \minus{} 1}(B_r(b))\subset A$. Thus $ b\in\partial B$.

b) Lemma: Let $ G$ be a region, let $ a\in G$ and suppose that $ f: (G\setminus\{a\})\to\mathbb{C}$ is an analytic function such that $ f(G\setminus\{a\}) \equal{} \Omega$ is bounded. Then $ f$ has a removable singularity at $ z \equal{} a$ and if $ f$ is one-one $ f(a)\in\partial \Omega$.

Proof of Lemma: There exists $ r > 0$ such that $ \Omega\subset B_r(0)$, so let $ \{z_n\}$ be a sequence of points in $ G\setminus\{a\}$ converging to $ a$. Then
\[ |(z_n \minus{} a)f(z_n)|\le|(z_n \minus{} a)|R
\]
and $ (z_n \minus{} a)\to 0$. Hence $ \lim_{z\to a}(z \minus{} a)f(z) \equal{} 0$, and $ a$ is a removable singularity of $ f$. Now assume $ f$ is one-one so that $ f: (G\setminus\{a\})\to\mathbb{C}$ is a homeomorphism. Note that $ G\setminus\{a\}$ and $ \Omega$ are both open. So choose a sequence of points $ \{z_n\}\subset G\setminus\{a\}$ converging to $ a\in\partial G$. Then $ \lim f(z_n) \equal{} f(\lim z_n) \equal{} f(a) \in\partial \Omega$ by part (a).

You should see where to go from here.


geometry unsolved
geometry
U sviti matematyky

Let $ ABCD$ be a convex quadrilateral. $ M$ is the centroid of triangle $ ABC$, and $ N$ is on segment $ MD$ such that $ MN: ND \equal{} 1: 3$. The points $ E,F$ are on $ AN,CN$ respectively such that $ ME\parallel{}AD,MF\parallel{}CD$. Prove that $ AF,CE,BD$ concur.

[quote="TomciO"]$ M$ is a centroid of the which triangle?[/quote]

Sorry... edited  :blush:

Just prove E,F are centroid of triangle BDC,ABD , respectively
Denote F',E'are centroid of the two triangles respectively
then you have MF'∥CD MF'=1/3CD   ME'∥AD  ME'=1/3AD
So F'=F  E'=E


LaTeX
probability
trigonometry
geometry
3D geometry
tetrahedron
dodecahedron

This the NM Marathon.

An easier version of the other one. Let's do easy questions.
Suprisingly, there are lots of NMers on AoPS, and I know a lot of them.
Obviously not as many as other states, but still.
I'll tell some people to contribute and make this just the second NM thread to have double digit replies (behind UNM PNM competition).

Post solutions.
Post new problems.
Post from MOEMS/easy Chapter MC to easy MC Nats/mid AMC10/12
Use $ \text{\LaTeX}$ if you feel like it.
Have fun.

Prob- There is a party at the ABQ Convention Center. Each person who attends shakes hands with everybody else twice. There is an exception, however. Silly Sylvester is feeling grumpy and crabby, so he shakes hands with nobody. There are 182 total handshakes. How many people were at the party?

 :mad: Yep silly sylvester, that's him....

th other party goers- :|  :)  :blush:  :roll:

looks like the ans is five, but it isn't....

Let n=# of partygoers.  Since Sylvester doesn't shake hands with anybody, then the number of handshakes is:

2(n-1)(n-2)/2=182
(n-1)(n-2)=182
n=15

Therefore, there are $ 15$ people (INCLUDING SYLVESTER!!!!!)

NP:
Each of the 10 students in a certain class has a $ 1/10$ chance of falling asleep during the entire class time (wakes up after class is dismissed).  If two students are chosen at random, what is the probability that one is asleep and the other is awake???

Two people:    :sleeping:  :rotfl:

You want the prob Student A is sleeping, and Student B is awake, multiplied by two, since it could be the other way around.

$ 2*(9/10)*(1/10)$
$ 9/50$

NP-Since these are easier problems, I'll give a really easy trig one.

Given sin 2x= 2 sin x cos x (double angle formula)= 7/10, find

$ sin^2x$ + 2 sin x cos x + $ cos^2x$

The middle term is given already (it's $ 7/10$).  To find the other two terms, let $ a$ be the opposite leg of angle $ x$, $ b$ be the adjacent leg of $ x$, and $ h$ be the hypotenuse.  Then:

$ (sin x)^2 \plus{} (cos x)^2$
$ (a^2/h^2) \plus{} (b^2/h^2)$

$ a^2 \plus{} b^2/h^2$

From the Pythagorean Theorem:

$ a^2 \plus{} b^2 \equal{} h^2$
and...
$ h^2/h^2 \equal{} 1$

Therefore, the answer is $ 1 \plus{} 7/10$, or $ 1.7$

NP:Frank is taking a true/false test.  He flips a coin for each of the 10 questions.  If it lands heads, then he puts true for that question.  If it lands tails, then he puts false.  After finishing the test, he double-checks each question by flipping the coin again and putting [i]that[/i] answer down instead, whether it's different from the original answer or not.  What is the probability that he gets all 10 questions??
 Frank:  :read:

[quote="prezcoin"]The middle term is given already (it's $ 7/10$).  To find the other two terms, let $ a$ be the opposite leg of angle $ x$, $ b$ be the adjacent leg of $ x$, and $ h$ be the hypotenuse.  Then:

$ sin^2x \plus{} cos^2x$= $ 1$
$ (a^2/h^2) \plus{} (b^2/h^2)$

$ a^2 \plus{} b^2/h^2$

From the Pythagorean Theorem:

$ a^2 \plus{} b^2 \equal{} h^2$
and...
$ h^2/h^2 \equal{} 1$

Therefore, the answer is $ 1 \plus{} 7/10$, or $ 1.7$
[/quote]

Yeah, you can do that, but if you wanted to, you could have just said- By the identity, $ sin^2x \plus{} cos^2x$= $ 1$, so my plugging everything in, we get the answer 17/10, or 1.7. BTW, fixed your sin and cos squared notation.

You find it doesn't matter what he even wrote down the first time, since he is replacing all of the answers while he is "double-checking". So, we just pretend that the first time did not happen. Each answer has 2 options and 1 correct, so Frank will get a perfect paper with probability

$ (1/2)^{10}$
$ 1/1024$

NP- What is the surface area, to the nearest hundredth, of a tetrahedron with side length 4.2*sqrt(65)?

[quote="pinkmuskrat"]NP- What is the surface area, to the nearest hundredth, of a tetrahedron with side length 4.2*sqrt(65)?[/quote]

I assume you mean a regular tetrahedron?

If so, then there are 4 triangular faces that are equilateral.  The formula would be

$ 4s^2\sqrt{3}/4$
or
$ s^2\sqrt{3}$

Therefore, the answer is:
$ (65*17.64)\sqrt{3}$
$ 1146.6\sqrt{3}$

$ 1985.97$

NP:Everyday, the mice of a certain house hold a drawing for who fights the cat of the house.  $ 7/12$ of the mice can beat the cat, $ 1/3$ lose, and $ 1/12$ doesn't win or lose.  If the mice win, they get 3 oz. of cheese. If they lose, then they get -4 oz. of cheese.  If they tie, then they get no cheese.  What is the expected value of the amount of cheese the mice get after one fight?

Mouse: :help:    Cat:   :furious:

well, I still like my topic... these problems are so trivial...

[hide]Let $ E$ be the expected value. Then
\begin{eqnarray*}
E=\frac{7 \times 3+4 \times (-4) + 1 \times 0}{12}
=\frac{21-16}{12}
=\frac{5}{12}
\end{eqnarray*}

So, our answer is $ \boxed{\frac{5}{12}}$ oz. of cheese. YAY FOR CHEESE!![/hide]

NP: For all integer values $ n\ge 2$, $ k$ will divide $ n^3-n$. What is the greatest possible value of $ k$?

We can either just try numbers, in which will work for contests, (but we'll never know for SURE it works) or prove it.
A. Trying integers from from 2 up, we get $ 6, 24, 60, 120, 210...$, and we find the ans is $ 6$.
B. Factor $ n^3\minus{}n$
n(n^2-1)
n(n+1)(n-1)
(n-1)(n)(n+1)
We see two things that occur, for sure. This expression divides 2 and 3, since any three consecutive integers include an even, and a multiple of 3, respectively.
To make sure nothing else is here, we just try the smallest value of n, 2, and get 6. We already have k atleast 6, so we are done. $ 6$

NP: Find the surface area, in inches squared, of dodecahedron with the apothem of one face length 3 feet. Express in simplest radical form.

An apothem of a regular polygon is the line segment drawn from the center of the polygon perpendicular to one of its edges.

 from wiki

[quote="pinkmuskrat"]NP: Find the surface area, in inches squared, of dodecahedron with the apothem of one face length 3 feet. Express in simplest radical form.[/quote]

Ummm...
You realize that this requires trig, right?

EDIT: I think you misunderstood "dodecahedron". Let's change that to icosahedron.(Apothem 3 feet)

I think that's why he posted it... So neither you or I could do it, since we don't know anything about trig (idk about u, but this is my guess)

Yeah, sorry, change to icosahedron. Was having mind blank about polyhedra. This is easy marathon, so I agree.
And, I didn't know you didn't know trig.... besides its like the most basic form.... normal nice friendly right triangles   :)  :)  :trampoline:

ermm...

NP: I like MATHCOUNTS. If i randomly make each letter in the word "mathcounts" capitalized, or undercase, what is the probability that it is not MathCOUNTS, MATHCOUNTS, mathCOUNTS, MATHcounts, MATHCounts, or MathCounts? express your answer as a percent to the nearest tenth.

The total number of ways is $ 3^{10}\equal{}59049$.  The probability is:

$ 59043/59049\equal{}99.98983 percent$

NP: I put $ \$$1000 in the bank.  The annual rate is 13% interest.  What is the total interest  after 5 years (compund)?

with some arithmetic, we see that $ 1000 \times (1.13)^5 \equal{} \boxed{1842.44}$ dollars

NP: evaluate $ \sqrt {19 \plus{} 6\sqrt {10}} \plus{} \sqrt {19 \minus{} 6\sqrt {10}}$.

Square the expression:

$ x^2\equal{}19\plus{}6\sqrt{10}\plus{}19\minus{}6\sqrt{10}\plus{}2\sqrt{361\minus{}360}$
$ x^2\equal{}38\plus{}2$
$ x^2\equal{}40$
$ x\equal{}\sqrt{40}$
$ \boxed{\sqrt{40}}$

NP: I had mowed my lawn for 10 minutes when two friends decided to help.  Without my friends, I mow the lawn in 2 hours.  Each of my friends takes 4 hours.  How long will it take us to mow the lawn, including the first 10 minutes?

You do $ \frac{1}{12}$ of the work in 10 minutes, and both of them together do $ \frac{1}{12}$ of the work in 10 minutes. After the first 10 minutes, there is $ \frac{11}{12}$ left to do. That takes the 3 of you 5.5-10minute periods to do, which equals 55 minutes. So, it takes you 65 minutes in all.

NP: 2 people are playing a game. They take turns rolling 2 dice, and the first person to get an odd sum from those two numbers wins. What is the probability that the person that rolls first wins?

no... it's that he wins if the rolls go on forever until the first person to get an odd sum.

[hide="Hint"]It is an infinite geometric sequence.[/hide]

The probability for the first person is:
$ P_{win}\equal{}\frac{1}{2}\plus{}\frac{1}{2}\left(\frac{1}{2}\left(\frac{1}{2}\plus{}\frac{1}{2}\left( ... \right) \right)\right)$
Which we can write recursively as:
$ f(x)\equal{}\frac{2\plus{}f(x\minus{}1)}{4}$
For x turns, where $ f(0)\equal{}0$
However, we can't work with it like that, so we need to change it to:
$ f(x)\equal{}\frac{2\left(\sum^{x\minus{}1}_{i\equal{}0} 4^i\right) \plus{} 1}{2\cdot 4^x}\\
\equal{}\frac{2\left(\frac{4^x \minus{}1}{3}\right) \plus{} 1}{2 \cdot 4^x}\\
\equal{}\frac{2\left(4^x \minus{} 1\right) \plus{} 1}{6 \cdot 4^x}\\
\equal{}\frac{2\cdot 4^x \minus{} 1}{6 \cdot 4^x}$
Slap it in a limit:
$ \lim_{x\to\infty} f(x)\\
\equal{}\lim_{x\to\infty} \frac{2\cdot 4^x \minus{} 1}{6 \cdot 4^x}$
Using our friend, the l'hospital rule:
$ \equal{}\lim_{x\to\infty} \frac{2\cdot 4^x \cdot \ln 4}{6\cdot 4^x \cdot \ln 4}\\
\equal{}\lim_{x\to\infty} \frac{1}{3}$
We can throw away the limit and get the last answer:
$ \boxed{P_{win}\equal{}\frac{1}{3}}$

I probably did something wrong, but oh well. :lol:

NP
How many 5-digit numbers are there such that the first three digits add up to the last two?
i.e. 12306 -> 1+2+3 = 06

Very beautiful and complicated, but... Look at your first statement. This is greater then $ \frac {1}{2}$ already, and you claim the answer is $ \frac {1}{3}$.

The probability is the sum of an infinite geometric sequence. The probability the first guy, call him quantumaccountant, rolls an odd sum and wins is $ \frac {1}{2}$. He'll roll even the other half of the time. In order for quantum to win, he needs the other guy, call him connaissance, to roll even. This occurs with $ \frac {1}{2} \cdot \frac {1}{2} \equal{} \frac {1}{4}$. Now quantum returns with the 3rd roll, with $ \frac {1}{4} \cdot \frac {1}{2}$ additional chance of winning.

Continue the geometric sequence reasoning and sum it.

$ P_{win} \equal{} \frac {1}{2} \plus{} \frac {1}{8} \plus{} \frac {1}{32} \plus{} \frac {1}{128} \plus{} ...$

$ P_{win} \equal{} \sum^{n}_{i \equal{} 0} \frac {1}{2^{2n \plus{} 1}}$

$ P_{win} \equal{} \frac {\frac {1}{2}}{1 \minus{} \frac {1}{4}} \equal{} \boxed{\frac {2}{3}}$

NP= same

same new problem? hm, i bet i can come up with a good one.

Prove they Pythagorean Theorem in 5 different ways.

I love this problem. We can also do this with other well known theorems

Hey, I found out what I did wrong. I changed the function from the recursive form wrong; it should be:
$ f(x)\equal{}\frac{2\left(\sum^{x\minus{}1}_{i\equal{}0} 4^i\right) \plus{} 1}{2\cdot 4^{\boxed{x\minus{}1}}}$
Working it from there gives 2/3.  :blush:

comment: that was way too much (as in the solution). The simplest methods will always be best

I am looking for the New Mexico HS Mathematics Contest (UNM / PNM Statewide) / Round 2: 2001 problem set in order to fill the gaps in my collection [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url]


function
trigonometry
induction
real analysis
real analysis unsolved

There is such an identity I want to prove:

${{\frac{1}{\Gamma(n-s+1)}\sum_{j=l}^{n}\frac{(-1)^{l+j}(n-l)!}{\Gamma(s-j+1)(j-l)!}= (-1)^{l}\frac{\sin(\pi s)}{\pi (s-l)}}}$

for all integers $n\geq l\geq 0$ and real numbers $s\in{\mathbb R}$

Hint:  By Mathematical Induction on the integer $n$  :wink:


The cost of a pair of sunglasses and a case for them is 10 dollars.
If the sunglasses cost 9 dollars more than the case, what is the cost of the case?

[quote="tcjy"]The cost of sunglasses and a case for them is $ \$10$. If the sunglasses cost $ \$9$ more than the case, what is the cost of the case?[/quote]

Let $ c$ be the cost of the case.  Then, $ c\plus{}9$ is the cost of the sunglasses.  We then get:
\[ c \plus{} c \plus{} 9 \equal{} 10
\]
\[ 2c \equal{} 1
\]
\[ \boxed{c \equal{} \$0.50}
\]

Ok, so this is how you do this.

The sunglasses cost 9 more than the case. So look at this algebraic equation.

S= 9+c
c+c+9=10
2c+9=10
2c=1
c= 50 cents

So then you know the case, and if you want, you can find out how expensive sunglasses are!

easy way to do it-just think, if a case was free, the glasses would be 9 dollars. we need to get to 10 dollars, so lets increase everything by $ \frac{10\minus{}9}{2}$ (do you see why?), getting 9.50 for glasses and .50 for the case.

this cuold be a really eassy cd problem.

Well it doesn't say if it has tax. So no.

ok so no tax but what if it comes in with a mail in rebate

Ok. Stop trying to be a smart guy. Ok? It's not even funny.

this is how I solved it

I guess then I check.

and make sure there is no mail or rebate announced in the problem.

Are you referring to smartytarty or saszs?


function
probability
algebra
binomial theorem
probability and stats

I'm not really sure how to go about this, so any help would be appreciated!

Suppose you throw a six-sided die five times.  Find the probability that the sum of the outcomes of the throws is 15 using generating functions.

Let $ Z$ be the total number of points and set
$ Z \equal{} X_1 \plus{} ... \plus{} X_5$
where $ X_i$ is the number of points shown on the ith die.
The probability generating function of each $ X_i$ is
$ G_{X_i}(s) \equal{} \frac {s \plus{} ... \plus{} s^6}{6}$.
Since the $ X's$ are independent:
$ G_Z(t) \equal{} \frac {(s \plus{} ... \plus{} s^6)^5}{6^5} \equal{} \frac {s^5(1 \minus{} s^6)^5}{6^5(1 \minus{} s)^5}$. (*)
Expand in (*) $ (1 \minus{} s^6)^5$ with the binomial theorem and $ (1 \minus{} s)^{ \minus{} 5}$ with the binomial series ( http://mathworld.wolfram.com/BinomialSeries.html ).
In (*) the coefficient of $ s^{15}$ is the required probability.

Thanks!  I get that

1/(1-s)^5 = 1 + (5 choose 1) s + (6 choose 2) s^2 + (7 choose 3) s^3 + (8 choose 3) s^4 + (9 choose 5) s^5

and that

(1-s^6)^5 = 1 - (5 choose 1) s^6 + (5 choose 2) s^12 - (5 choose 3) s^18 + (5 choose 4) s^24 - (5 choose 5) s^30

However, I'm confused about how I use these to figure out the coefficient of s^15.

$ \frac {s^5(1 \minus{} s^6)^5}{6^5(1 \minus{} s)^5}\equal{}\frac{1}{6^5}s^5\left({5 \choose 0}\minus{}{5 \choose 1}s^6\plus{}...\right)\left({\minus{}5 \choose 0}\minus{}{\minus{}5 \choose 1}s\plus{}...\right)$.
The coefficient of $ s^{15}$ is
$ \frac{1}{6^5}\left({5 \choose 0}{\minus{}5 \choose 10}\minus{}{5 \choose 1}{\minus{}5 \choose 4}\right)$.


You are at a party and you have several sheets of paper with dimensions $m \times n$ centimeters where $m$ and $n$ are prime numbers. You distribute these sheets of paper among the people present and give them a pair of scissors too.

Their job is to reduce the given sheet of paper to pieces exactly $1cm \times 1cm$ in size.

All they are allowed to do is - at any stage pick up any one piece of paper with them and cut it along a straight line parallel to any of the edges.

When they have accomplished this task, they report the number of cuts made to you.

Prove that the difference between the maximum and the minimum numbers reported is less than or equal to $\frac{m+n}{m-n}\times \frac{m+n}{m-n}$.

This seems so simple, so I'm inclined to say that this is WRONG.

[hide="solution?"]Well, there's only one way to do this. Make m+n-2 cuts total.

m+n-2-m-n+2=0.

$\left(\frac{m+n}{m-n}\right)^{2}\geq0 \leq \left(\frac{m+n}{m-n}\right)^{2}$

Tada[/hide]

Why exactly is that the only way? I'm not good at math.

EDIT: Oh I get it.

So, I'm not familiar with what's acceptable on this forum, but I'm guessing that I shouldn't reveal the response in the previous post.

[hide="My attempt at figuring out what you just said"]I'm just trying to reason out your answer. Are you saying that because there's only one way of getting 1 by 1 squares (m+n-2), subtracting it from itself (to find the difference in max and min) is 0, which is always greater than or equal to 0 when squared?[/hide]