I need help on school round # 28.

For a set of five position integers, none greater than 100, the mean is 1.5 times the mode.  If 31, 58, 98, x and x are the five integers, what is the value of x. this should be easy im just kind of tired.

We quickly see that $ x$ is the mode. This means $ \frac{31\plus{}58\plus{}98\plus{}2x}{5}\equal{}1.5x$. Cross multiplying, we get $ 5.5x\equal{}187$. This means $ x\equal{}37$. Plugging back in, we see that $ /boxed{x\equal{}37}$ is indeed correct.

I don't think there is a point posting here if you're just sort of lazy :lol:  and don't feel like solving it.

Do similar triangles fall under "proportional reasoning?"

[quote="cgyao15"]??
that has nothing to do with this...[/quote]

Well I didn't want to waste getting a new thread for this. I was just curious, and my teacher didn't say.


Hi folks, I have a query regarding the shape of the universe.

What is the shape of the universe? Is it - (a) A Sphere or  (b) An Ellipsoid? If it is either (a) or (b) , can you justify your answer? I know, most people including myself, think it is a sphere, but why can't it be an ellipsoid?

Secondly, Another interesting (might be) question, I thought of this problem (actually it's of my own creation).

[b] Problem [/b]

How long does it take for a body traveling with almost (or near) the speed of light to traverse the entire volume of the universe?

I think the volume of the universe can't be assumed as "infinite" because since the universe is spherical (generally thought of as) a sphere can have only finite volume.

[b] P.S.: [/b] Don't use overly complex physics equations (or whatever) to convey or explain your answers, because I am not so good at physics. Maybe try answering them through logic.

Thanks!

i think it is a ellipsoid since universe is cooling it means its expanding so its ellipsoid ok????
and a body wont possess mass if it'd travel with a speed of light... this is in acc to Einstien's equation E=mc^2

notify me ur thoughts

[quote="Gen8"]
P.S.: Don't use overly complex physics equations (or whatever) to convey or explain your answers, because I am not so good at physics. Maybe try answering them through logic.

[/quote]

Your question itself is overtly complex physics.

You'd like to read [url=http://www.space.com/scienceastronomy/mystery_monday_040524.html]this[/url].

One tenant of modern cosmology is that the Universe has no boundary (or edges) -- basically, you couldn't come up to the "end" of the Universe. That would mean that whatever shape the Universe has, it cannot be as simple as the inside of a sphere or ellipsoid. It must have a shape that we (we meaning most laymen) are unfamiliar with and probably cannot imagine.

You might wonder, "if it has no boundary, doesn't that mean that it is infinite?" The answer to that question is not necessarily. Imagine the 2-dimensional surface of a balloon. It is finite in size (surface area), yet it has no boundary. The universe could have a shape that is a 3-dimensional analog to that. Like a 3-dimensional coating of a 4-dimensional sphere. Impossible to imagine right?

Yet, things can exist, even if we cannot imagine them! That is the power of mathematics -- to describe and study things logically, even when we cannot draw or imagine them.

Hmmm... I can imagine. Nice one, are you some research bloke?

How do you traverse the entire "volume"?

And that recent Scientific American Article conjectured that the universe was expanding nonuniformly... So it wouldn't be a sphere or ellipsoid, it would be an irregular blob. (However you define the "Shape" of the universe is up to you... )

Additionally, considering how we are able to "see" stuff near the beginning of the Big Bang, if my understanding is correct that would imply a universe size greater than or equal to some number near 13 billion light years across, at its longest axis. (Which also implies that the expansion of the universe occurs at, near, or faster than c. Which throws a wrench into your plans.)

I think Gauss202 has the correct path..., but i dont think its unimaginable what the universe looks like..
I have even heard some amazing theories that it is cylindrical...
But i dont think that we can make any correct models using the modern theory, also the Big-Bang theory is in my opinion "bullsh..".

And also, i dont think anyone has seen as far as the "end of the universe" or even "near it" so even if we know that Galaxies are moving away and other things, we cannot make a prediction of the whole universe itself expanding..

So you cannot basically give an answer to problems like this, you can at most speculate, just maybe by the use of some untested theory(GR has not been tested on every "scale"..) or maybe use your imagination and think of a solution yourself.

[quote="yeahnigga"]we cannot make a prediction of the whole universe itself expanding..[/quote]

Redshift?

P.S. http://en.wikipedia.org/wiki/3-sphere

[quote="yeahnigga"]
also the Big-Bang theory is in my opinion "bullsh.."...
[/quote]

You have to be careful, because there is a lot of misconception (due to its misrepresentation in popular culture) of what the Big-Bang theory actually says. All it says, really, is that the Universe was once small and hot, and has since expanded and cooled. [i]Exactly[/i] how small and how hot, and what happened before that time, is not technically part of the Big-Bang theory. Those details depend on the theories of the laws of physics at extremely high energy, high gravity, and small scale.

But there is [i]lots[/i] of [url=http://en.wikipedia.org/wiki/Big_Bang#Observational_evidence]physical evidence[/url] to cooberate the basic thesis of the Big Bang model, which is that the Universe expanded from a state that was once very small and hot and that this expansion and cooling led to certain features of the Universe that we see today.

You should read about the [url=http://en.wikipedia.org/wiki/Big_Bang]details[/url] and the [url=http://en.wikipedia.org/wiki/History_of_the_Big_Bang_theory]history[/url] of the Big Bang model, and also some of the [url=http://en.wikipedia.org/wiki/Non-standard_cosmology]alternative models to the Big-Bang model[/url] before you completely make up your mind.

Well you know Gauss202 I already did that, i have read a lot about the big-bang theory, and not the wide public writings, but technical discreptions, i also have taken courses on GR etc.

and AFTER this i concluded what i said. And still will maintain my opinion until i get some non-speculative evidence that it may be true.

That's fair enough. That's the thing about science, it's based on inductive reasoning. So you can never be [i]absolutely[/i] sure about the truth of any model. You can only say things like "I like this model because, when coupled with my other beliefs in models, it fits more observations than any of these other models".

Yes Gauss202, you are absolutely right, we will never know if something is completely right, only upto experiments...
so whatever the theory will be i am sure there will be scientists who wont accept it... Thats the mystery of science i guess :roll:

Well, I really appreciate all your replies. So, is the universe "finite"?

[b] Digression [/b]

I personally feel (based upon my observance of many of his posts) that[b] "gauss202" [/b] possess one of the most (if not the most) brilliant analytical minds that I've come across. (Be it in reality or virtual reality like in here).

Yeah, the universe is finite, but constantly expanding, and I guess there's infinite room for it to expand.

The tenet proposed by Einstein -" Any body in the universe can travel at most at the speed of light." Is this tenet applicable for all the bodies in the universe? What if, some spaceship build by "aliens" traveled at a speed more than the speed of light?

It is due to this tenet that we aren't able to travel to other galaxies (also, it needs lots of fuel) but there have been instances of UFOs being reported in the U.S., so, clearly if they were true then for those UFOs to travel to Earth, they must have traveled at a speed more than the speed of light. (Since Earth is millions of light years away from its nearest galaxy), so this leads to a violation of Einstein's theory, right?

[b] It might be possible that those UFOs might have reduced their 'speed' after reaching Earth, because if they didn't, we can't spot them in the sky. [/b]

Can anyone throw some light on this particular view of mine (I know it might be incorrect, do correct my reasoning if it's fallible).

[quote="Gen8"]The tenet proposed by Einstein -" Any body in the universe can travel at most at the speed of light."[/quote]

I don't think Einstein ever said that. I think what Special Relativity says is much more complicated than that. Special relativity tells us (indirectly) that nothing with rest mass can be moved from rest to a speed greater than the speed of light in vaccum in an inertial frame due to a force acting on the object.

This leaves the door open to several options in which an object could "travel at a speed greater than the speed of light". For one, if the inertial frame itself is doing something funny -- like its scale is expanding, then the rate of separation betwen two objects could be greater than the speed of light. This is actually believed to be the case with the Expansion of the Universe -- where some galaxies are moving apart at a rate faster than the speed of light because of expansion. Not due to any physical force that is pushing them apart. This is also hypothesized to have happened in an early stage of the Universe called the [url=http://en.wikipedia.org/wiki/Inflationary_epoch]Inflationary Epoch[/url].

Secondly, relativity only says that an object with a rest mass cannot be accelerated from [i]rest[/i] to a speed above the speed of light. It doesn't say anything about an object (if such a thing exists) that has no "rest state" or rest mass. Such objects (if they exist) are called [url=http://en.wikipedia.org/wiki/Tachyons]Tachyons[/url]. (This is not science imo though, because there has been no direct detection of such things. Their existence is purely speculative).

Thirdly, relativity only says that an object cannot exceed the speed of light [i]in vacuum[/i]. Light travels more slowly through matter. And in fact, it has been observed, in nuclear reactors for instance, cases where [url=http://en.wikipedia.org/wiki/%C4%8Cerenkov_radiation]radiation particles exceed the speed of light inside of the reactor[/url].

I agree with you about the aliens thing though. The vastness of separation between galaxies makes it implausible (in my eyes) for anything to make a purposeful expedition to our planet from another galaxy, considering the time it would have taken to get here.

It doesn't rule out, however, the possibility of visitation from star systems within our galaxy (most of which are still pretty freaking far away), or of automated explorers that were sent out from somewhere a long time ago.


Calculate:
$S=\tan a.\tan2a+\tan2a.\tan3a+...+\tan(n-1)a.\tan na$

We have:
\begin{eqnarray*}n+S&=&\sum_{k=1}^{n}\left[1+\tan ka\cdot\tan(k+1)a\right]\\ &=&\sum_{k=1}^{n}\frac{1}{\tan a}\left[\tan(k+1)a-\tan ka\right]\\ &=&\left[\tan (n+1)a-\tan a\right]\cdot\cot a\\ \Longrightarrow S&=&\cot a\tan(n+1)a-(n+1) \end{eqnarray*}


Consider n students. In each group of 4 students, at least a student is friend of the other 3.  Show that there's a student who is friend of all the others. If A is a friend of B's, then B is a friend of A's.

My solution is too simple to be correct

I consider a group A, B, C, D where A is friend of B,C,D but B is not a friend of C's. Now A has 3 friends. Then i consider all the groups composed by A, B, C and the n-4 remaining students. In this situation, the fourth student (E,F...N) is friend of A,B,C or A is friend of B,C and E,F...N. In the second case we showed it. In the first case A will have:
3 + (n-4) friends, hence n-1. Finish.

Other solutions? Is mine correct?
enjoy!

I would begin, "Either everyone is friends with everyone else (in which case we're done) or there exist two people, call them B and C, who are not friends.  In this later case, choose any two others, A and D, and WLOG A is friends with B, C, D."  Otherwise, it looks perfect.


function
algebra
system of equations
algebra unsolved

Solve this system of equations with $ x, y \in R$:

$ \left\{ {\begin{array}{*{20}c}
   {(1 \plus{} 4^{x \minus{} y} ).5^{1 \minus{} x \plus{} y}  \equal{} 1 \plus{} 3^{x \minus{} y \plus{} 2} }  \\
   {x^2  \minus{} 3y\sqrt {y \minus{} \frac{1}{x}}  \equal{} 1 \minus{} 2y}  \\
\end{array}} \right.\begin{array}{*{20}c}
   {}
\end{array}$

hello, setting $ x\minus{}y\equal{}t$ and let $ f(t)\equal{}(1\plus{}4^t)5^{1\minus{}t}\minus{}1\minus{}3^{t\plus{}2}$ so our function is strictly monotonoulsy
decreasing and our equation $ f(t)\equal{}0$ has only the solution $ t\equal{}0$ so it must be $ x\equal{}y$ and from the second equation we get in this case $ x\equal{}y\equal{}\frac{1}{2}(1\minus{}\sqrt{5})$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, setting $ x \minus{} y \equal{} t$ and let $ f(t) \equal{} (1 \plus{} 4^t)5^{1 \minus{} t} \minus{} 1 \minus{} 3^{t \plus{} 2}$ so our function is strictly monotonoulsy
decreasing and our equation $ f(t) \equal{} 0$ has only the solution $ t \equal{} 0$ so it must be $ x \equal{} y$ and from the second equation we get in this case $ x \equal{} y \equal{} \frac {1}{2}(1 \minus{} \sqrt {5})$.
Sonnhard.[/quote]

Hi, Mr Sonnhard Graubner! How do you solve $ f(t) \equal{} (1 \plus{} 4^t)5^{1 \minus{} t} \minus{} 1 \minus{} 3^{t \plus{} 2}$  is strictly monotonoulsydecreasing?


Yeah, the first digit would somehow represent 567 using a symbol or something.

Yeah, why not?
To have base 1000, you'd have to come up with 1000 different symbols...
so 567_1000'd just be one digit of whatever symbol it corresponds to.

[quote="Jimmy"]Can you write a number like 567 in base 1000?[/quote]

I think so, wouldn't it just be:

5(1000^2) + 6(1000^1) + 7(1000^0) = 5006007

[quote="tarquin"]dangvh, that's 567 base 1000 to base 10 heh.[/quote]

oops sorry lol  :blush: lol my bad

[quote="sky9073"]
To have base 1000, you'd have to come up with 1000 different symbols...
[/quote]
Thanks, that's what I thought.

Heh, that would be past the capabilities of ASCII...that would be pretty painful to do as well.

567 in base 1000 would be:
567*(1000)^0

good luck creating 531 new symbols for this.

If you hold alt and press a # on the #pad you get a symbol.

1=☺
2=☻
3=♥
.
.
.
567=7

there fore 567 would be 7 in this thing.

Ther you go seems reasonable to me lol.

err... don't know how that "quote" got there

Actually, those are ASCII symbols, and there are only 256 of them. Oh well, now we need 10-bit bytes. :D
PS: ▐▀

[quote="xdl615x"]Can someone explain to me throughly what change of base is? and what you are all talking about? Sorry.. I'm new..[/quote]

THe normal numbers that you are used to seeing are called base ten. This is because they are expressed as sums of powers of ten:

$423=4\cdot 10^2+2\cdot 10^1+3\cdot 10^0$

If you just change the ten to another number, you have the number in another base. For example, $1321_{four}=1\cdot 4^3+3\cdot 4^2+2\cdot 4^1+1\cdot 4^0=121_{ten}$

WOW! Thanks! that was an AWESOME explanation I got it in like 4 seconds!


Let S denote the sum of all of the three digit positive integers with three distinct digits.
Compute the remainder when S is divided by 1000.

[hide="Answer"]For this problem, it would be easiest to consider each digit distinct. So, each digit in the set 1,2,3,4,5,6,7,8,9 would appear 9*8 = 72 times (10 choices for A, 9 for B, 8 for C). Likewise, we can have B and C appear indistinctivey, or 8*8 = 64 for tens digit.

Therefore, our final answer would just be:

$ 45(100)(72)\plus{}45(10)(64)\plus{}45(1)(64)\equal{}355680$. Remainder over 1000 is $ \boxed{680}$.[/hide]

thanks for your help

but please answet the others


A few days ago, I came across a software for Chemistry named Chemsketch. It's very useful for organic chemistry. However, it is just a 30-day trial version. So anyone have full version of it, could you please share it with me? Thank you very much in advance.
[url]http://www.freechemsketch.com[/url]

Actually, you can (legally) obtain Chemsketch for free.  From the site you linked: ``Please note, this software is for personal, home and educational use only, but can be downloaded for a 30-day trial period by users in commercial and for-profit organizations.''

By the way, let me know if you figure out how to make 3D pictures of orbitals.  I have tried to do so, without success.


integration
combinatorics unsolved
combinatorics

Let $k$ be a positive integer. Show that for all $n>k$ there exist convex figures $F_{1},\ldots, F_{n}$ and $F$ such that there doesn't exist a subset of $k$ elements from $F_{1},..., F_{n}$ and $F$ is covered for this elements, but $F$ is covered for every subset of $k+1$ elements from $F_{1}, F_{2},....., F_{n}$.

Maybe this works.

Take a convex poligon $P$ with $\binom{n}{k}$ vertices and consider a set $T$ of $\binom{n}{k}$ pairwise disjoint triangles each of which has a vertex in common with $P$, the 2 sides adjacent to it being contained in the two sides of $P$ it belongs to. Make a correspondence between each $S\subset\{1,2...n\}$, $|S|=k$ and a triangle $t_{S}\in T$ so that no triangle is assigned to two different sets. Now define the sets $T_{i}\subset T$, $1\leq i\leq n$, in the following way $t_{S}\in T_{i}\leftrightarrow i\in S$. Notice that for each set $X\subset\{1,2...n\}$ with $|X|=k$,    ${\bigcap \{ T_{i}| i\in X \}=\{t_{X}}\}$, since $\forall i\in X ( t_{S}\in T_{i}) \leftrightarrow \forall i\in X(i \in S)$. For the same reason, for each $Y\subseteq \{1,2...n\}$ with $|Y|=k+1$,  $\bigcap \{ T_{i}| i \in X \}=\emptyset$. Now consider the sets of points (indeed, polygons) in the plane $Q_{i}=P-\bigcup T_{i}$, $1\leq i\leq n$. Clearly each of them is convex, because of the way we chose the elements of $T$. On the other hand, $\forall X\subset \{ 1,2...n\}$ with $|X|=k$
$\bigcup\{T_{i}|i\in X\}=P-\bigcup (\bigcap \{ T_{i}|i \in X \}) \}= P-\bigcup \{ t_{X}\}\subset P$, while  $\forall Y \subset \{ 1,2...n \}$ with $|Y|=k+1$,
$\bigcup\{T_{i}|i \in Y \}=P-\bigcup (\bigcap \{ T_{i}|i\in Y \}) \}= P-\bigcup\emptyset=P$.  And we are done.

Sorry for my english.

I can't understand how is taken the set $ T $.
can you be more clear or give an example,please?


geometry
circumcircle
projective geometry
geometry proposed

Here's my solution... calculate only :(
$B_{1}C_{1}= \frac{|pow_{(ABC)}K|.BC}{KB.KC}\\ \text{And }\overrightarrow{KA_{1}}= \overrightarrow{KA}. \frac{pow_{(ABC)}K}{KA^{2}}\\ \text{So }B_{1}C_{1}^{2}.\overrightarrow{KA_{1}}+C_{1}A_{1}^{2}.\overrightarrow{KB_{1}}+A_{1}B_{1}^{2}.\overrightarrow{KC_{1}}\\ = \frac{(pow_{(ABC)}K)^{3}}{(KA.KB.KC)^{2}}(BC^{2}.\overrightarrow{KA}+CA^{2}.\overrightarrow{KB}+AB^{2}.\overrightarrow{KC}) = \overrightarrow{0}$
It means that $K$ is the Lemoine point of $\triangle A_{1}B_{1}C_{1}$
Let $H$ projection of $A$ to $BC$, $M, A'$ is the midpoint of $BC, AH$
$\overrightarrow{AH}= \overrightarrow{AB}+\frac{AB.cosB}{BC}.\overrightarrow{BC}= \overrightarrow{AB}+\frac{c^{2}+a^{2}-b^{2}}{2a^{2}}.\overrightarrow{BC}\\ \Rightarrow 2\overrightarrow{AH}= \overrightarrow{AB}+\frac{c^{2}+a^{2}-b^{2}}{2a^{2}}.\overrightarrow{BC}+\overrightarrow{AC}+\frac{a^{2}+b^{2}-c^{2}}{2a^{2}}.\overrightarrow{CB}\\ =\overrightarrow{AB}+\overrightarrow{AC}+\frac{c^{2}-b^{2}}{a^{2}}\overrightarrow{BC}= \frac{a^{2}+b^{2}-c^{2}}{a^{2}}.\overrightarrow{AB}+\frac{c^{2}+a^{2}-b^{2}}{a^{2}}.\overrightarrow{AC}\\ \Rightarrow 2a^{2}.\overrightarrow{AH}+2(b^{2}+c^{2}-a^{2})\overrightarrow{AM}= 2b^{2}\overrightarrow{AB}+2c^{2}\overrightarrow{AC}\\ \Rightarrow a^{2}.\overrightarrow{AA'}+\frac{1}{2}(b^{2}+c^{2}-a^{2})\overrightarrow{AM}=\frac{1}{2}(a^{2}+b^{2}+c^{2})\overrightarrow{AK}$
It means that $A'M$ passes through $K$ ...

[quote="April"][color=darkred]The intersection point of simedians of a triangle is called [i]Lemoine’s point[/i].
$1.$ Lines $AK,\, BK$ and $CK,$ where $K$ is [i]Lemoine’s point[/i] of triangle $ABC,$ intersect the circumscribed circle at points $A_{1},\, B_{1}$ and $C_{1},$ respectively. Prove that $K$ is [i]Lemoine’s point[/i] of triangle $A_{1}B_{1}C_{1}.$
$2.$ Prove that lines that connect the midpoints of the sides of a triangle with the midpoints of the corresponding heights intersect at [i]Lemoine’s point[/i].[/color][/quote]
[color=darkred][b]Lemma[/b] (well-known). In $\triangle ABC$ denote $T\in BB\cap CC$ and $S\in (BC)$.

Then $S\in AT$ $\Longleftrightarrow$ the ray $[AS$ is the $A$- symmedian of $\triangle ABC$.[/color]

 [color=darkblue]$1.\blacktriangleright$ Denote $M\in BB\cap CC$, $D\in BC_{1}\cap B_{1}C$ and $\{\begin{array}{c}X\in B_{1}B_{1}\cap C_{1}C_{1}\\\ Y\in C_{1}C_{1}\cap A_{1}A_{1}\\\ Z\in A_{1}A_{1}\cap B_{1}B_{1}\end{array}$.
Apply the Pascal's theorem to the degenerate hexagons :

$\{\begin{array}{ccc}BC_{1}C_{1}CB_{1}B_{1}& \Longrightarrow & X\in KD\\\ BBB_{1}CCC_{1}& \Longrightarrow & M\in KD\end{array}\|$ $\Longrightarrow$ $\overline{XAKA_{1}MD}$ $\Longrightarrow$ $K\in A_{1}X$.

Analogously prove that $K\in B_{1}Y\cap C_{1}Z$. Using the above lemma,

$K\in A_{1}X\cap B_{1}Y\cap C_{1}Z$ $\Longrightarrow\ K$ is the Lemoine's point of $\triangle A_{1}B_{1}C_{1}$.

$2.\blacktriangleright$ Denote : $V\in BC\ ,\ AV\perp BC$; $S\in AK\cap BC$;

the middlepoints $U$, $W$ of the segments $[BC]$, $[AV]$ respectively.

Prove easily that $\frac{KA}{KS}=\frac{b^{2}+c^{2}}{a^{2}}$, $US=\frac{a|b^{2}-c^{2}|}{2(b^{2}+c^{2})}$, $UV=\frac{|b^{2}-c^{2}|}{2a}$.

Therefore, $\frac{KA}{KS}\cdot\frac{US}{UV}\cdot\frac{WV}{WA}=1$. From the Menelaus' theorem obtain $K\in UW$.[/color]


trigonometry
limit
function
calculus
derivative
calculus computations

Ascertain $\mathrm{Im}f$ , where $f(x)=\frac{\sin x-x\cos x}{x-\sin x}\ ,\ x\in \left(0,\frac {\pi}{2}\right]\ .$

I'm not sure what you're asking, but

$\lim_{x\to\frac{\pi}{2}} f(x) = \frac{2}{\pi-2}$  and   $\lim_{x\to 0} f(x) = 2$

What is the monotony of the function ? You seek Im f  !.
[b]Definition[/b]. For $f\ :\ A\longrightarrow B,\ \mathrm{Im}f=f(A)=\{y\in B\ |\ (\exists)\ x\in A, f(x)=y\ \}.$

Sorry, for some reason I thought you meant limit, but I still have no idea what you're talking about, so I'll let someone else answer.

Well i guess one way you could to it is find min and max of that function on $(0,\frac{pi}{2})$ and then say that is the range.....

Well, the function is monotonically decreasing, so the range would just be $\left(2, \frac{2}{\pi-2}\right)$ right?

$\mathrm{Im(f)}=\left[\frac{2}{\pi -2},2\right)$. You prove that the function is decreasing. this is the problem !!

[quote="Kent Merryfield"]He means the image, also known as the range, of this function. Since the function is a continuous function on an interval, its range must be an interval (by the Intermediate Value Theorem). The problem is to find that interval.[/quote]

Thanks, I understood after he gave the definition, but your explanation is much simpler.  Now I have to try to finish the problem :) .

O.K.,Jimmy. Good luck and much love for and from ...... :rotfl: ......math !

The geometrical interpretation of this problem is
http://www.mathlinks.ro/Forum/viewtopic.php?t=45414


inequalities
calculus
logarithms
real analysis
real analysis unsolved

[quote="valica"]Prove that:

$\frac{18*19}{ln19}+\frac{1973*1974}{ln1974} < \frac{16*17}{ln17}+\frac{1975*1976}{ln1976}$[/quote]
Here's why this belongs in the Calculus section.[hide="***"]
Set $f(x) = \frac{(x-1)x}{\ln x}$, then the goal is to prove  $f(19)-f(17) < f(1976)-f(1974)$.

This will follow if $f''>0$ on $[10,\infty)$, since $f(19)-f(17) = 2 f'(c_1)$ and $f(1976) - f(1974) = 2f'(c_2)$ with  $c_1 < 19, \, c_2 > 1974$.

Now
$f''(x) = \frac{2x (\ln x)^2 - 3x \ln x}{x (\ln x)^3} + \frac{\ln x + 2x - 2}{x(\ln x)^3}$, and both fractions are clearly positive for $x>10$.
[/hide]

[quote="valica"]You're right,but more simple:

It's easy to see that

$19^x + 1974^x \leq 17^x + 1976^x \forall x\geq 1$
We simply integrate this in [1,2] and obtain the inequality.[/quote]That's miraculous. I prefer a method :)


Another problem
   p is primes and $p=2q+1$
  prove that  $(q!)^2 + (-1)^q$ divisible to p

[quote="Kuba"]Can you please explain what is a "pride" number? thank you[/quote]

I think this is the correct definition of pride numbers [just put "numbers" instead of people"]:

http://dictionary.cambridge.org/define.asp?key=62818&dict=CALD

  :P  :P  :P

[quote="mashimaro"]Another problem
   p is pride number and $p=2q+1$
  prove that  $(q!)^2 + (-1)^q$ divisible to p[/quote]
And it is a basic fact when we elementary prove that each $p=4k+1$ is a sum of perfect squares.
 :D  :D  :D

[quote=Pascual2005]for the nsecound "pride" nukmber problkem:

$q!^2=(2q)!(-1)^q=(p-1)!(-1)^q$ and then claim follows by wilsons theorem for pride numbers

 :D  :D  :D[/quote]

Could you explain the first step?

I'll explain [b]Pascual2005's[/b] solution:
First observe the following congruences :
$(p-1) \equiv -1 \pmod{p}$
$(p-2) \equiv -2 \pmod{p}$
etc.. the last one being $\frac{p+1}{2} \equiv -\frac{p-1}{2} \pmod{p}$
Observe also that  $\frac{p-1}{2}=q$.
Now if you multiply all those congruences you get :
$(q+1)(q+2) \cdots (2q)\equiv (-1)^q q! \pmod{p}$ (remember that $p-1=2q$)
Now multiply this congruence with $q!$ and then apply Wilson's theorem :
$$(-1)^q (q!)^2 \equiv (2q)! \equiv (p-1)! \equiv -1 \pmod{p}$$
So $$(q!)^2+(-1)^q \equiv (-1)^{q-1}+(-1)^q \equiv 0 \pmod{p}$$ as desired :) .

How about this one?

Let ${p}$ be a prime number such that dividing ${p}$ by 4 leaves the remainder 1. Show that there is an integer ${n}$ such that $n^2 + 1$ is divisible by ${p}$.

I want to solve it with Wilsons theorem. A little advise on how to get started would be nice, have been trying many things, but it feels as if i'm missing something obvious.

Figured it out. Used Wilsons theorem to write it as n^2 congruent to (4k)!, then realised(4k)! is congruent to( (2k)!)^2

Cheers

$(p-1)!$ is congruent to $-1\pmod{p}$ by Wilson's theorem.  We are given that $q=\frac{p-1}{2}$ so  $(p-1)!=1(p-1)2(p-2)... (\frac{p-1}{2})(\frac{p-1}{2}+1)\equiv 1(-1)2(-2)...q(-q) \pmod{p}$, we have $1(-1)2(-2)...q(-q)=q!^2$ if $q$ is even. Additionally, if $q$ is even, $-1^q=1$. $(p-1)!\equiv q!^2\equiv -1\pmod{p} \Longrightarrow q!^2+1\equiv 0\pmod{p}$ If $q$ is odd, the expression can be written as $-q!^2$. Similarly, $-1^q= -1$. $-q!^2-1$ can be written as $-(q!^2+1)$, which we know is congruent to $0\pmod{p}$ from above.

[quote=Kuba]Can you please explain what is a "pride" number? thank you[/quote]
What they said.

I believe pride actually means prime, based on the solutions.

The solution for the second problem:
$(p-1)!=1(p-1)2(p-2)...(\frac{p-1}{2})(\frac{p-1}{2}+1)$. This simplifies to $1(-1)2(-2)...(\frac{p-1}{2})(-\frac{p-1}{2}) \pmod{p}$. Notice that $p\equiv 1\pmod{4}$, so $p-1\equiv 0\pmod{4}$ meaning $\frac{p-1}{2}$ is even. Since there will be an even number of terms, the expression is positive and can be written as $(\frac{p-1}{2})!^2\equiv -1\pmod{p}$ by Wilson's identity which means $n=(\frac{p-1}{2})!$ works.

The first post was for the first problem


I feel kind of guilty as a comoderater that I'm not putting any effort into gathering people to this forum. I'd tried posting interesting questions/question series, but most of them are left unsolved. What am I doing wrong? :cry:

Any ideas would be greatly appreciated.

Tare,

Honestly I think that much of the content in this forum is slightly above the heads of the average MATHCOUNTS student.  I am mostly to blame for this as most of my Challenge Problems have been on the high side of MATHCOUNTS-level problems (or a little higher), but part of the problem is that most of the students who come here are not challenged by the ordinary MATHCOUNTS problem, so they want to see the harder stuff.

I plan to post a few easier problems and along with the harder problems there should be a good mix to help bring students in and keep them.  I certainly don't want to discourage the harder stuff that students can springboard into, but we should keep in mind that MATHCOUNTS covers a certain set of problem areas that students will be looking to master.

Also, AoPS is still young.  It will take some time to get to the point where we have a few dozen interested MATHCOUNTS students who post regularly.  Give it time and eventually the forum will be self-sustaining.  I believe that.


Two rectangles of different size have the same side ratio. They are covering each other in a way that on the interior of each side of the bigger triangle there is one vertex of the smaller rectangle. Find those side ratios which makes that construction possible.

So far so good? But what do you conclude about the side ratio a/b and c/d as asked. And what values can you assume to make this construction possible, e.g. the ratio a/c?

[hide]The rectangles have the same ratio, $\frac{d}{c}=\frac{b}{a}=k$ and: $b=m+n$, $a=m'+n'$ such that:

$cos\ {\alpha}=\frac{m}{d}=\frac{m}{kc} \implies m=kc.cos\ {\alpha}$,  $sin\ {\alpha}=\frac{n}{c} \implies n=c.sin\ {\alpha}$

$cos\ {\alpha}=\frac{m'}{c} \implies m'=c.cos\ {\alpha}$, $sin\ {\alpha}=\frac{n'}{d}=\frac{n'}{kc} \implies n'=kc.sin\ {\alpha}$ consequently:

$\frac{b}{a}=\frac{m+n}{m'+n'}=\frac{k.cos\ {\alpha}+sin\ {\alpha}}{k.sin\ {\alpha}+cos\ {\alpha}}=k \implies k^2.sin\ {\alpha}= k.sin\ {\alpha} \implies k=1$

[/hide]

I got the same answer as you, but It looks simple rather than very natural, so I didn't think this answer was correct.

Here is my solution.

Since $a=c\cos \alpha+d\sin \alpha,\ b=d\cos \alpha+c\sin \alpha,$ we have $\frac{a}{b}=\frac{c+d\tan \alpha}{d+c\tan \alpha}.$Rewrite this equation as $\frac{a}{b}=\frac{\frac{c}{d}+\tan \alpha}{1+\frac{c}{d}\tan \alpha}.$ If $\frac{a}{b}=\frac{c}{d}=r,$ then we have $r=\frac{r+\tan \alpha}{1+r\tan \alpha}.$ Solve for $r,$ we get $r=1.$ On the other hand  if $\frac{a}{b}=\frac{d}{c}=r',$ then we obtain $r'=1.$

And what is the range for the ratio $\frac{a}{c}$?

[hide="A elementary solution."]Given are two rectangles $ABCD$, $MNPQ$ so that $M\in (AB)$, $N\in (BC)$, $P\in (CD)$, $Q\in (DA)$ and $\frac{AB}{AD}=\frac{MN}{MQ}=k$. Prove that $k=1$.

[b]Proof.[/b] Denote $AB=a$, $AD=b$, $MN=c$, $MQ=d$, $AM=x$, $BN=y$.
$\triangle BMN\sim \triangle CNP\Longrightarrow \frac cd=\frac yx=\frac{a-x}{b-y}\Longrightarrow$
$k=\frac ab=\boxed {\frac cd=\frac yx=\frac{a-x}{b-y}}=\frac{a-y}{b-x}=\frac{a+y}{b+x}\Longrightarrow\frac{a-x}{b-y}=\frac{a+y}{b+x}\Longrightarrow$
$(a-b)(x+y)=(x^2-y^2)\Longrightarrow a-x=b-y\Longrightarrow k=1$
and in this case $x=y$ and $\frac ac\in (1,\sqrt 2]\ .$[/hide]


Is there some formal notion of distribution? For example, the size of the set of natural numbers and the size of the squares are the same, but the squares are sort of more sparsely distributed than the natural numbers (e.g., 1/n diverges, but 1/n^2 converges).

You can define 'densities' for subsets of the (positive) integers.
Let $A$ be a subset of those and $A(x)$ be it's counting function, so giving the number of elements of $A$ that are $\leq x$. Then one can define:

lower asymptotic density: $d_L(A) : = \liminf_{n \to \infty} \frac{A(n)}{n}$
upper asymptotic density: $d_U(A) : = \limsup_{n \to \infty} \frac{A(n)}{n}$
asymptotic density: if the limit exists, take $d(A) : = \lim_{n \to \infty} \frac{A(n)}{n}$
Schnirel'man density: $\sigma(A) : = \inf_{n \in \mathbb{N}} \frac{A(n)}{n}$

The primes and the squares have density $0$ for all those, but differences occure often.

Another way would be to find lower or upper (asymptotic) bounderies or even a asymptotic formula itself.
E.g. when $A$ is the set of all primes, the asymptotic value of $A(x)$ is $\frac{x}{ln(x)}$, and when $A$ is the set of all squares it's $\sqrt x$.


trigonometry
calculus
inequalities
invariant
inequalities proposed

Find the lower bound of $\cos(A-B)+\cos(B-C)+\cos(C-A)$

Is it supposed that $A,B,C$ should be the angles of a triangle  :?:

If this is NOT the case, let me try to solve. Introduce $x=A-B$, $y=B-C$, $z=C-A$, so that $x+y+z=0$. We have to minimize

$\cos x+ \cos y+ \cos z = \cos x + \cos y + \cos (x+y) =: f(x,y)$.

The arguments $x,y\in\mathbb{R}$ of $f()$ are free variables. Therefore, we can use calculus to look for critical points, for which

$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0$.

In this case, we must have $-\sin x - \sin(x+y)=0$ and $\sin y - \sin(x+y)=0$, which means

$\sin x = \sin y = -\sin (x+y)\;\;(=\sin z)$

From $\sin x=\sin y$ we deduce that either (i) $x=y\;[2\pi]$ or (ii) $x+y=\pi\;[2\pi]$.

The second case (ii) implies $\sin (x+y)= 0= \sin x=\sin y$, so that $x=k\pi$, $y=\ell\pi$, $z=-(k+\ell)\pi$. Therefore, $f(x,y)=(-1)^k + (-1)^\ell + (-1)^{k+\ell}$. The minimum of such an expression is obviously $-1$.

The first case (i) implies $\sin x + \sin 2x=0$, and excluding $\sin x=0$ (already examined), we have $\cos x=-\frac{1}{2}$, i.e. $x=\pm\frac{2\pi}{3}\;[2\pi]$. Similar equality for $y$ (with the same sign). The corresponding value for $f$ is $-\frac{3}{2}$.

Since $-\frac{3}{2}<-1$, the minimum is equal to $-\frac{3}{2}$.

[i]edit: I corrected a calculation mistake.[/i]

Are there non-calculus solutions?  :?

And, what's the lower bound if $A,B,C$ are angles in a triangle?

I'm not sure how to deal with the case in which A,B,C are angles of a triangle, but for free A,B,C,
recall the identity $cos x + cos y + cos z + cos (x+y+z) = 4 cos \frac{x+y}{2} cos \frac{y+z}{2} cos \frac{z+x}{2}$
Then it is easy to find the result.

Well, at the moment I can't see how to minimize the product...

oops, it seems that i have made a mistake..  :oops:

Let's solve all variants of the problem at once:

[color=blue][b]Theorem 1.[/b] If A, B, C are three arbitrary angles, then $\cos\left(B-C\right)+\cos\left(C-A\right)+\cos\left(A-B\right)\geq -\frac32$.
[b]Theorem 2.[/b] If A, B, C are three angles from the interval ]0°; 180°[, then $\cos\left(B-C\right)+\cos\left(C-A\right)+\cos\left(A-B\right)>-1$.
[b]Theorem 3.[/b] If A, B, C are three acute angles, then $\cos\left(B-C\right)+\cos\left(C-A\right)+\cos\left(A-B\right)>1$.[/color]

Note that all these three inequalities are sharp (and Theorem 2 is also sharp for A, B, C being the angles of a triangle, and Theorem 3 is also sharp for A, B, C being the angles of an acute-angled triangle).

[i]Proof of Theorem 1.[/i] Denote $x=\pi+\left(B-C\right)$, $y=\pi+\left(C-A\right)$ and $z=-\pi+\left(A-B\right)$. Then, cos x = - cos (B - C), cos y = - cos (C - A) and cos z = - cos (A - B). But since

$x+y+z=\left(\pi+\left(B-C\right)\right)+\left(\pi+\left(C-A\right)\right)+\left(-\pi+\left(A-B\right)\right)=\pi$,

we can apply http://www.mathlinks.ro/Forum/viewtopic.php?t=30785 post #6 Lemma 1 to the three reals x, y, z and get $\cos x+\cos y+\cos z\leq \frac32$. In other words, $\left(-\cos\left(B-C\right)\right)+\left(-\cos\left(C-A\right)\right)+\left(-\cos\left(A-B\right)\right)\leq \frac32$. Hence, $\cos\left(B-C\right)+\cos\left(C-A\right)+\cos\left(A-B\right)\geq -\frac32$, and Theorem 1 is proven.

[i]Proof of Theorem 2.[/i] Since the angles A, B, C are from the interval ]0°; 180°[, their cosines all lie between -1 and 1. In other words, 1 > cos A > -1, 1 > cos B > -1 and 1 > cos C > -1.

Assume that $\cos B+\cos C\geq 0$. Then, cos A > -1 yields $\cos A \left(\cos B + \cos C\right) \geq -\left(\cos B + \cos C\right)$, so that

$\cos B\cos C+\cos C\cos A+\cos A\cos B=\cos B\cos C+\cos A\left(\cos B+\cos C\right)$
$\geq \cos B\cos C-\left(\cos B + \cos C\right)=\left(1-\cos B\right)\left(1-\cos C\right)-1>-1$

(since 1 - cos B > 0 and 1 - cos C > 0, as 1 > cos B and 1 > cos C).

Now assume that cos B + cos C < 0. Then, cos A < 1 yields $\cos A \left(\cos B + \cos C\right) > \cos B + \cos C$, so that

$\cos B\cos C+\cos C\cos A+\cos A\cos B=\cos B\cos C+\cos A\left(\cos B+\cos C\right)$
$>\cos B\cos C+\left(\cos B + \cos C\right)=\left(1+\cos B\right)\left(1+\cos C\right)-1>-1$

(since 1 + cos B > 0 and 1 + cos C > 0, as cos B > -1 and cos C > -1).

Thus, in every case, we have $\cos B\cos C+\cos C\cos A+\cos A\cos B>-1$.

Now, since the angles A, B, C are from the interval ]0°; 180°[, their sines are positive: sin A > 0, sin B > 0, sin C > 0. Thus,

$\cos\left(B-C\right)+\cos\left(C-A\right)+\cos\left(A-B\right)$
$=\left(\cos B\cos C+\sin B\sin C\right)+\left(\cos C\cos A+\sin C\sin A\right)+\left(\cos A\cos B+\sin A\sin B\right)$
$>\cos B\cos C+\cos C\cos A+\cos A\cos B>-1$,

and Theorem 2 is proven.

[i]Proof of Theorem 3.[/i] Since the inequality $\cos\left(B-C\right)+\cos\left(C-A\right)+\cos\left(A-B\right)>1$ is symmetric in A, B, C (in fact, because of cos (-x) = cos x it is invariant under a transposition of any two of the angles A, B, C), we can WLOG assume that $A\geq B\geq C$. Then, the angles A - B, B - C and A - C are positive. Also, these angles are all $<\frac{\pi}{2}$ (since A, B, C are acute angles), so that their cosines are positive: cos (A - B) > 0, cos (B - C) > 0 and cos (A - C) > 0.

Since $A-C<\frac{\pi}{2}<\pi$, we have $\pi-\left(A-C\right)>0$. Thus, the three angles A - B, B - C and $\pi-\left(A-C\right)$ are all positive. The sum of these three angles is $\left(A-B\right)+\left(B-C\right)+\left(\pi-\left(A-C\right)\right)=\pi$. Hence, these angles A - B, B - C and $\pi-\left(A-C\right)$ are the three angles of a triangle. Hence, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=21188 , we have $1<\cos\left(A-B\right)+\cos\left(B-C\right)+\cos\left(\pi-\left(A-C\right)\right)$. In other words, $1<\cos\left(A-B\right)+\cos\left(B-C\right)-\cos\left(A-C\right)$. Since cos (A - C) > 0, this yields $1<\cos\left(A-B\right)+\cos\left(B-C\right)+\cos\left(A-C\right)$. Since cos (-x) = cos x for every angle x, this becomes $1<\cos\left(A-B\right)+\cos\left(B-C\right)+\cos\left(C-A\right)$. In other words, $\cos\left(B-C\right)+\cos\left(C-A\right)+\cos\left(A-B\right)>1$, what proves Theorem 3.

  Darij


trigonometry
inequalities
inequalities proposed

Given three real positive numbers $x,y,z$ such that $x^2+y^2+z^2+2xyz=1$, prove that
\[ 1+x+y+z+xy+yz+zx+xyz\geq\frac23(x+y+z)^4.  \]

my solution is very ugly.

i use $x = cosA$, $y = cosB$, $z = cosC$.

so, i use $cosA = \frac{b^2+c^2-a^2}{2bc}$ and analogous for $cosB$ and $cosC$.

now, let $a = x+y$, $b = x+z$ and $c = y+z$.

opening the inequality, we prove with bunching (the computations are very ugly).

Let $x=\sin A$,$y=\sin B$,$z=\sin C$.
Now if $x^2+y^2+z^2+2xyz=1$ we can say:$A+B+C=\frac{\pi}{2}$

[quote="pvthuan"][quote="zhaobin"]here is my prove,i think it is not nice.
let $x=\cos A,y=\cos B,z=\cos C$
we want to prove  $(1+\cos A)(1+\cos B)(1+\cos C) \geq \frac{2}{3}(\cos A+\cos B+\cos C)^4$
let $f(A,B,C)=(1+\cos A)(1+\cos B)(1+\cos C) - \frac{2}{3}(\cos A+\cos B+\cos C)^4$
try to prove $f(A,B,C) \geq f(\frac{A+B}{2},\frac{A+B}{2},C)$ with $\cos A+\cos B+\cos C>1$
sorry i haven't explain it so clear.[/quote]

So I shoudl write my inequaliy in a more readable form as follows:

Given three positive real numbers $x,y,z$ such that $x^2+y^2+z^2+2xyz=1$, prove that
\[ (1+x)(1+y)(1+z)\geq{\tfrac23}(x+y+z)^4.  \]
The mixing variable method Zhaobin used for this inequality works well for a large number of other inequalties. The most difficult step in the method is to show, for example, that $f(x,y,z)\geq f(x,t,t)$, where $t$ is a means of $y$ and $z$. Unfortunately, Zhaobin missed to explain this step. :D :roll:[/quote]
that's because I can't explain it well in english. :D


calculus
function
algebra
polynomial
integration
logarithms
partial fractions

How would I decompose this?

$ \frac{t^{2}}{1\plus{}t}$

Hi;
I would break it down into:

$ \frac{1}{t\plus{}1}\plus{}(t\minus{}1)$

that's right MrMagician..bobbym, you can solve it by simply using long division..BTW, partial fraction decomposition is usually used for decomposing improper rational expression so long as the denominator term is factorable...

A couple of lines I use in teaching the subject as part of Calculus II:

Partial fractions is not a technique of integration. It is a technique of algebra that happens to be useful in integration.

And:

Partial fractions is a technique to be used when you have at least two [i]different[/i] factors in the denominator.

So, what does that make MrMagician's problem here? There aren't two different factors, so partial fractions is inappropriate. In the larger picture, which is that we're dealing with rational functions, there's always a first step: use long division to get a polynomial plus a proper fraction. That suffices here.

Now, what if we had $ \frac{x}{(x\plus{}1)^2}\ ?$

There still aren't two different factors, so it's still true that partial fractions is inappropriate. This is already a proper fraction, so long division isn't needed. What can we do?

Let's suppose this is an integration problem. I like asking my integration problems as definite integrals, so we'll do that.

$ \int_0^1\frac{x}{(x\plus{}1)^2}\,dx.$

One thing we can do to make our life simpler: make the obvious "little" substitution, which in this case is $ u\equal{}x\plus{}1.$ (That would also have been useful in the original problem.) We have $ du\equal{}dx$ and $ x\equal{}u\minus{}1.$ The integral becomes

$ \int_1^2\frac{u\minus{}1}{u^2}\,du\equal{}\int_1^2\left(\frac1u\minus{}\frac1{u^2}\right)du\equal{}\left.\left(\ln u\plus{}\frac1u\right)\right|_1^2\equal{}\ln 2\minus{}\frac12.$

Hi;

I am aware that it must be a proper fraction and that the denom must be factorable, so I said break it down instead of saying a partial fraction decomposition.

I think that:

$ \frac{1}{t\plus{}1}\plus{}(t\minus{}1)$

is easier to integrate. That is the usual reason for a partial fraction decomposition.

oh I forgot..what I've said in my last post was a mistake..when a rational function is improper, we simply use long division to obtain the proper form..Now to Kent Merryfield's example, $ \frac{x}{(x\plus{}1)^2}$, we're aware it's already proper fraction...of course, we can always decompose this..the denominator  $ (x\plus{}1)$ is repeating twice so we expect 2 partial fractions...So we have
$ \frac{x}{(x\plus{}1)^2}\equal{}\frac{1}{x\plus{}1}\plus{}\frac{\minus{}1}{(x\plus{}1)^2}$,
which we can now easily integrate..A remark to this is that, we can also decompose a proper fraction even if the denominator terms are not distinct..

[color=red]This is a rule one can find in many texts:[/color]
[color=blue]For each factor of the denominator of the form $ (px\plus{}q)^m$ with $ m\geq 1$, the partial fraction decomposition contains the sum of $ m$ partial fractions of the form
$ \frac{A_{1}}{px\plus{}q}\plus{}\frac{A_2}{(px\plus{}q)^2}\plus{}\cdots \plus{}\frac{A_m}{(px\plus{}q)^m}$
where each number $ A_k$ is a constant.[/color]

I think, what we mean now by "decompose" a fraction to its partial fractions is to express the fraction (which is proper) as a sum of those partial fractions..For example, for the partial fraction decomposition of, say, $ f(x)/g(x)$ to be found, [b][i]it is essential that[/i][/b] $ f(x)$ [b][i]have lower degree than[/i][/b] $ g(x)$..If this is not the case, we can use [i][b]long division[/b][/i] to obtain such expression.


calculus
integration
function
calculus computations

Calculate
\[ \iiint\limits_{x^2  \plus{} y^2  \plus{} z^2  \leqslant 1} {\frac{{x^4  \plus{} 2y^4 }}{{x^4  \plus{} 4y^4  \plus{} z^4 }}}.
\]
I hope that someone can solve it. Very easy and its solution is very short.

Call the original integral $ I.$ Make the volume-preserving and ball-preserving substitution that sends $ (x,y,z)$ to $ (z,y,x).$ The integral would then become

$ I\equal{}\iiint_{x^2\plus{}y^2\plus{}z^2\le1}
\frac{2y^4\plus{}z^4}{x^4\plus{}4y^4\plus{}z^4}\,dx\,dy\,dz$

Add this to the original expression to find that

$ 2I\equal{}\iiint_{x^2\plus{}y^2\plus{}z^2\le1}1\,dx\,dy\,dz\equal{}\frac{4\pi}3.$

Hence $ I\equal{}\frac{2\pi}3.$

Incidentally, this shows that the integral converges and isn't even improper. The integrand is nonnegative and bounded (since we have to add another nonnegative function just to get to $ 1$) and continuous except at a single point (the origin).


superior algebra
superior algebra unsolved

Prove that every element in $S_{n}$ has a unique cycle decomposition, using group actions.

It's an elegant alternative of the usual proof :)

Hi.

The proof is not very different from the usual one, it just uses the language of group actions.

Let $\sigma$ be an fixed element of $S_\Omega$ ($\Omega$ a finite, nonempty set) and consider the canonical action of $\langle\sigma\rangle$ on $\Omega$. The orbits $\Omega_{1}, \Omega_{2}, ..., \Omega_{r}$ define a partition of $\Omega$.
Now let $\omega_{i}\in\Omega_{i}$ and $L_{i}=|\Omega_{i}|$, then the cycle decomposition of $\sigma$ is obviously given by $\sigma=\prod_{i=1}^{r}{(\omega_{i}, \sigma(\omega_{i}), \sigma^{2}(\omega_{i}), ..., \sigma^{L_{i}-1}(\omega_{i}))}$


I'm writing data to a file, and I want to get the numeric data from the file, perform calculations on it, delete the original file, and re-write the data to a new file.

More compactly, I want to put numbers into my program, save them, and then access them again later to perform calculations, and then be able to access them again later to do the same thing.

In C:

[code]
#include <cstdio>

...

FILE *f = fopen("file","r");
fscanf(f, ...
...
freopen("file","w",fin);
fprintf(f, ...
...
fclose( f );

That should solve your problem. [/code]


geometry
3D geometry
geometric transformation
reflection

A cube of glass is placed on a horizontal plane which perfectly absorbs lights. The walls are transparent except on (the upper), completely black. A parallel beam of light hits the cube and the plane with an angle of 45.
Using reflection and rifraction's rules, calculate the area not reached by any ray. Assume the refractive index is 2^(0.5)

[hide]a(2-3^(-.5)), where a is the edge length[/hide]

Can you show it? You must consider also the rays hitting the couple of vertical faces with an angle of 90 relative to the perpendicular

Is light reflected off the lower surface or something?

No, the floor absorbs the light, but we have to consider the ray hitting the close-up face (and the face behind) parallel to the surface...sorry for my english

Is there anyone who can explain the solution of this problem?
thanks a lot

I have still not got the intricacy in the problem. I got the answer by simple goemetry and ray optics.

well my problem is picturing rays in 3d, i don't know where the shadow appears, which form it has..please, can you be so kind to help me?

I have made a picture but can't post it. can we add paint files to a post? How did you add your picture?

thank you very much! you can't post .bmp files, so save the picture as a jpeg file

Thanks.
Consider rays at either extreme of the darkened surface. All rays to left of one extreme and right of other extreme are not absorbed. Angle of incidence of refracted ray is 45 degree and refracted angle is 30 degree. Diagram is in side view and not to scale.

well..i still have a different result. The lenght of the darkened side you showed is a*(1/(3)^.5). The other side should be 5, so why do you have a*(2-(3)^.5)?
thanks

Sory, my mistake.
One length is a/(3^.5) and other is a. Hence area=a^2*(3^-.5)


Finals are coming and I need help with the study guide.  Here is the question:  HOw many bonds does...
1. Hydroxyl group form?
2. Methyl group form?
3. Amino group form?
4. Carboxyl group form?

For 1 I got one, and 4 i got four, but I'm not sure.

*pulls out chemistry notes**dusts off*... wonderful subject... just wonderful...
1. hydroxyl- that would be one.
2. methyl is CH3... so thats one extra.
3. amino... AH... i always hated nitrogen...
4.  GR. now i [i]really[/i] have to look at my notes...

ok i couldnt find my notes. im really pretty sure on those first two, though. that help at all?


geometry
analytic geometry
ratio
area of a triangle
combinatorics unsolved
combinatorics

Manuel, do you allow collinear points?

I think some points on a line located at distances powers of a trascendental number, and a point outside the line, will give the result...

I also think something like that should work, but Pascual pay attention to the fact that if four points are collinear then you have two triangles with equal (to $0$) areas.... ;)

Pierre.

Yes Pierre's right. Let's  try with no collinear points.

But can't you simply add points one by one so that the condition is satisfied? :?

Maybe manuel meant points with integer coordinates?  :?  ;)

[quote="manuel"]Does there exists a configuration of an infinite amount of points in the plane, such that no area of a triangle is a multiple of another area of another triangle in the set?[/quote]

you mean rational multiple,of course?
yes,i had the same idea as grobber of choosing inductively(then you would have a countable number of points)

yes, forbidden parts are a union of sets of equidistan paralel lines in finitely many directions, that doesn't cover the whole plane.

That's it.
An you may 'generalize' it to rational ratios of the areas.

Pierre.


inequalities
AMC
USA(J)MO
USAMO
algebra unsolved
algebra

Let a,b,c>0 be real numbers such that
                      a^2+b^2+c^2+abc=4
prove that:
             0<=ab+ac+bc-abc<=2
i have seen one solution at kalva demon,but it is not elegant..
can anyone solve it by applying trigonometric substutition?

It has been discussed  before and I think blahblahblah sent this question  ;)

A trigonometric substitution is this one: $a=2 cosA ,b=2 cos B ,c=2 cosC$ and the inequality reduces to the following:
                              $ 0\leq\sum{ cos A cos B }-\prod{cos A}\leq2$ and from here we will see. ;)


number theory
greatest common divisor
modular arithmetic
number theory unsolved

If $ p$ is a prime such that $ p\equiv 3\bmod 4$, and if there exist integers $ a$, $ b$, $ c$, $ d$ such that
\[ a^{2p} \plus{} b^{2p} \plus{} c^{2p} \equal{} d^{2p}
\]
then, show that $ p$ divides $ abc$.

[i]Alternative formulation.[/i] We have a prime number $ p$ such that $ 4\mid p \minus{} 3$, and four integers $ a$, $ b$, $ c$, $ d$ such that $ a^{2p} \plus{} b^{2p} \plus{} c^{2p} \equal{} d^{2p}$. Prove that $ p\mid abc$.

Sketch. Suppose it is not true. Reduce the equation so that no prime divides everyone. One of a, b, and c is odd, WLOG c. We have

$d^{2p}-c^{2p}= (d^{p}+c^{p})(d^{p}-c^{p})$
$= (d+c)(d-c)(d^{p-1}-d^{p-2}c+d^{p-3}c^{2}-\ldots+c^{p-1})(d^{p-1}+d^{p-2}c+\ldots+c^{p-1})$.

We can show by casework that at least one of these terms is 3 mod 4, so that there exists a prime q = 3 mod 4 occuring with odd multiplicity dividing the RHS of $a^{2p}+b^{2p}= d^{2p}-c^{2p}$.

So........ it is usually pretty strange to have a prime 3 mod 4 dividing a sum of two squares, as q must now divide both a and b. Err ok so this appears to be pretty awesome, since if q is not p and divides neither c nor d (true since we've reduced), then by evaluating gcds, q can not divide two of

$d+c, d-c, d^{p-1}-d^{p-2}c+d^{p-3}c^{2}-\ldots+c^{p-1}, d^{p-1}+d^{p-2}c+\ldots+c^{p-1}$

so q divides only one of these factors. However q must divide $a^{2p}+b^{2p}$ with even multiplicity, so consequently it also divides $d^{2p}-c^{2p}$ with even multiplicity, a contradiction. Hence we must have q = p and so p | abc.

BTW I'm reconstructing the solution having seen this problem before... don't think I could have figured it out myself..

[quote="wojto111"]I don't understand why we have that $d^{2p}-c^{2p}$ can't divides $q$ in even multiplicity.[/quote]
We picked q earlier to divide one of the factors with odd multiplicity... since q divides only one of the factors, it must divide $d^{2p}-c^{2p}$ with odd multiplicity, while we also showed that it divided the LHS with even multiplicity.

I don't understand your solution but I invent similar prove which I understand :).

Reduce the equation so that no prime divides everyone. Because $x^{2}\equiv 0 \vee 1 (mod 4)$. So we can suppose that $c^{2}, d^{2}\vee 1 (mod 4)$. $d^{2p}-c^{2p}=(d^{2}-c^{2})(d^{2*(p-1)}....+c^{2*(p-1)})$. So $d^{2*(p-1)}....+c^{2*(p-1)}\equiv 3 (mod 4)$. So we have prime number $q \equiv 3 (mod 4)$ dividing  $d^{2*(p-1)}....+c^{2*(p-1)}$ in odd multiply. If we suppose that $q \not = p$. We have that $q \not |d^{2}-c^{2}$. Because $q | a^{2p}+b^{2p}$. We have that $q$ divide $a,b$.  So $q$ divides $a^{2p}+b^{2p}$ in even multiply. We have contradiction with $q \not = p$. If $q=p$ we have that $p | a^{2p}+b^{2p}$ and $p|a,b$ so $p|abc$. Maybe ours proves are isomorfic  :D .

[quote="wojto111"] We have that $q \not |d^{2}-c^{2}$. [/quote]

Why is this?  :maybe:

[quote="wojto111"]If $ p$ is a prime such that $ p\equiv 3\bmod 4$, and if there exist integers $ a$, $ b$, $ c$, $ d$ such that
\[ a^{2p} \plus{} b^{2p} \plus{} c^{2p} \equal{} d^{2p}
\]
then, show that $ p\mid abc$.[/quote]
If $ a,b,$ or $ c$ is 0 then we are done.  Suppose $ abc\neq 0$.  WLOG $ \gcd(a,b,c) \equal{} 1$.
Now, in what follows we suppose that $ p\not| abc$ and seek to contradict (1).  If $ 2\not| abc$, then use Euler's Theorem modulo 8.  (The odd squares mod 8 are $ \equiv 1$.)  We contradict (1).
If $ 2\mid d$ then WLOG $ 2\mid a$.  So, $ b^{2p} \plus{} c^{2p}\equiv 0\pmod{2^{2p}}$.  This forces both $ b$ and $ c$ to be odd, then again modulo 8 gives a contradiction to (1).
Lastly, it is possible for $ 2\mid a$ and $ 2\mid b$ but $ 2\not|cd$, so $ c^{2p}\equiv d^{2p}\pmod{2^{2p}}$. (I am stuck here...thought I had it but I don't...)
I am tired and going to bed.  I hope this helps, I will check in the morning to see what others think.

Let's fill the small gaps:

Lemma:
If $ q \equiv 3 \mod 4$ is prime and $ q|x^2\plus{}y^2$, then $ q|x,y$.
Proof:
Assume that $ q \nmid y$, so there is an inverse $ y^{\minus{}1} \mod q$. This gives $ (xy^{\minus{}1})^2 \equiv \minus{}1 \mod q$. But $ z^2 \equiv \minus{}1 \mod q$ has no solution, so we have $ q|y$. $ q|x$ follows immediatly.

Corollary:
A prime $ q \equiv 3 \mod 4$ divides $ x^2\plus{}y^2$ an even number of times.
Proof:
Let $ q^m$ be the highest power of $ q$ dividing both $ x\equal{}q^m s,y\equal{}q^mt$. Then $ x^2\plus{}y^2 \equal{} q^{2m} (s^2\plus{}t^2)$ and $ q$ doesn't divide at least one of $ s,t$, so by the lemma doesn't divide $ s^2\plus{}t^2$. So $ q^{2m}$ is the exact power if $ q$ dividing $ x^2\plus{}y^2$, being an even power.

We may assume $ \gcd(a,b,c,d)\equal{}1$.
Looking $ \mod 4$, we get that exactly two of $ a,b,c$ are even. W.l.o.g. assume that $ a,b$ are even.
We have $ a^{2p}\plus{}b^{2p}\equal{}d^{2p}\minus{}c^{2p}\equal{}(d^2\minus{}c^2)(d^{2p\minus{}2}\plus{}d^{2p\minus{}4}c^2\plus{}...\plus{}c^{2p\minus{}2})$. Now  $ c,d$ are odd, thus $ c^2 \equiv d^2 \equiv 1 \mod 4$, giving $ d^{2p\minus{}2}\plus{}d^{2p\minus{}4}c^2\plus{}...\plus{}c^{2p\minus{}2} \equiv p \equiv 3 \mod 4$. Thus there is some prime $ q \equiv 3 \mod 4$ dividing $ d^{2p\minus{}2}\plus{}d^{2p\minus{}4}c^2\plus{}...\plus{}c^{2p\minus{}2}$ an odd number of times.
Now $ q$ divides $ a^{2p}\plus{}b^{2p}$ an even number of times, thus $ q|d^2\minus{}c^2$.

From $ q|d^2\minus{}c^2$, we have $ c^2 \equiv d^2 \mod q$, giving $ 0 \equiv d^{2p\minus{}2}\plus{}d^{2p\minus{}4}c^2\plus{}...\plus{}c^{2p\minus{}2} \equiv p d^{2p\minus{}2}  \mod q$. Thus $ q|p d^{2p\minus{}2}$.

1. case: $ q|d$.
Then $ q|c$. As $ q|a^{2p}\plus{}b^{2p}$, by the lemma $ q|a,b$. But now $ q|a,b,c,d$, contradicting $ \gcd(a,b,c,d)\equal{}1$.

2. case: $ q|p$.
Then $ q\equal{}p$ and thus $ p|a^{2p}\plus{}b^{2p}$, so by the lemma $ p|a,b$, especially $ p|abc$.


geometry
circumcircle
search
geometry unsolved

Let ABC be a triangle with $ AA_1,BB_1,CC_1$ as their medians and $ G$ be the centroid.If the points $ A,C_1,G ,B_1$ be concyclic then prove that

$ 2a^2\equal{}b^2\plus{}c^2$ ,where $ a\equal{}BC,b\equal{}CA,c\equal{}AB$

Call O the circumcenter of ABC; on the circumcircle of $ AB_1C_1$ O is the antipode of A so $ OG \perp GA$ then

$ AO^2\equal{} GO^2\plus{}AG^2$ $ \Longleftrightarrow \ R^2 \equal{} R^2 \minus{} \frac{a^2\plus{}b^2\plus{}c^2}{9} \plus{} \frac{2b^2\plus{}2c^2\minus{}a^2}{9}$ $ \Longleftrightarrow \ 2a^2\equal{}b^2\plus{}c^2$

Search in Amazon the book College Geometry, by Nathan Altshiller Court, has been recently reedited.

[quote="pankajsinha"]Can it be proved that if circumcircle of $ AB_1C_1$ passes through $ G$ then $ O$ must lie on it and thus $ OA$ will be the diameter. Also,can you tell me about any book which deals with Geometry at this level.
Thanks in advance.[/quote]

hem, the circumcircle of $ AB_1C_1$ pass always through O and O is the antipode of A because $ \angle OB_1A \equal{} \angle OC_1A \equal{} 90^\circ$


trigonometry
geometry
parallelogram
geometry proposed

In a right triangle  $ABC$ with $\angle A = 90$  we draw the bisector $AD$ .
Let    $DK \perp AC , DL \perp AB$ .
     Lines $BK , CL$ meet each other at  point $H$ .
  Prove that $AH \perp BC$.

[u]Babis[/u]

Indeed, nice and easy problem !

[hide="Here is a short and fast metrical solution and a generalization of this problem."]
$S\in AH\cap BC\ ;\ l_a\equiv AD=\frac{2bc}{b+c}\cdot \cos \frac A2\Longrightarrow AD=\frac{bc\sqrt 2}{b+c}\Longrightarrow$

$AL=LD=DK=KA=\frac{bc}{b+c}\ ;\ LB=\frac{c^2}{b+c}\ ,\ KC=\frac{b^2}{b+c}\ .$

The [u]Van Aubel's relation[/u]: $\frac{HA}{HS}=\frac{LA}{LB}+\frac{KA}{KC}\Longrightarrow \boxed {\ \frac{HA}{HS}=\frac{a^2}{bc}\ }\ .$

The [u]Ceva's theorem[/u]: $\frac{SB}{SC}\cdot \frac{KC}{KA}\cdot \frac{LA}{LB}=1\Longrightarrow \frac{SB}{SC}=\frac{LB}{KC}\Longrightarrow$

$\boxed {\ \frac{SB}{SC}=\frac{c^2}{b^2}\ }=\frac{c^2}{a}\ : \ \frac{b^2}{a}\Longrightarrow AS\perp BC\Longrightarrow \boxed {\ AH\perp BC\ }\ .$

[color=red][b]Extension 1.[/b] $A\in (0,180^{\circ})\Longrightarrow\frac{HA}{HS}=\frac{b^2+c^2}{bc}\ \ \wedge\ \ \frac{SB}{SC}=\left(\frac cb\right)^2$,

[u]Remark.[/u] The point $S$ is the foot of the [u]symedian-line[/u] from the vertex $A\ .$

[b]Extension 2.[/b] $D\in (BC)\ ,\ \frac{DB}{DC}=k\Longrightarrow \frac{HA}{HS}=k+\frac 1k\ \ \wedge\ \ \frac{SB}{SC}=k^2\ .$

[u]Consequence.[/u] Giving a segment $[BC]$ and a positive number $k$, we can construct

the positions of the points $\{S_1,S_2\}\subset BC$ for which $\frac{S_1B}{S_1C}=\frac{S_2B}{S_2C}=k^2\ .$[/color][/hide]
[hide="A equivalent enunciation with the proposed problem (from a old Romanian geometric book)."]
[quote][color=blue]" Let $A$, $B$ be two fixed points and let $\widehat {BAX}$ be a right angle.
Given is a mobile point $M\in (AX$. Define: $N\in BM\ ,\ \widehat {BAN}\equiv \widehat {MAN}$;
the projections $P$, $R$ of the point $N$ on the rays $[AB$, $[AM$ respectively;
the intersection $L\in BR\cap MP$. Ascertain the geometrical locus of the point $L\ .$"[/color][/quote][hide="Answer."]The circle with the diameter $AB$ without the points $A$ and $B\ .$[/hide][/hide]

here a little bit more general result, from which easily follows the Virgil's generalization of the original problem.

LemmaRL16022006

Let ABCD be a parallelogram and p be a parallel to diagonal AC. Let's call E, F and G the intersection of p with AB, AD and DC respectively. And finally let's call X and Y the intersection between EC and BF, BG respectively.
Then it holds that XE/XC*YE/YC = (BE/BA)^2.

[quote="Virgil Nicula"][b]Very nice your generalization, Sprmn21 ![/b] The points $X$, $Y$ are the feet of two isigonal-lines from the vertex $B$ in the triangle $BEC$. [u]Very nice again ![/u] It is very usefully in the some construct problems. [color=green]  Viva  [/color]        [color=red]  Italia  [/color] ![/quote]

Thank you very much Virgil, for your compliments.

I have to point out that, for the picture I have in mind,  BX, BY are isogonal only if ABCD is a rombus. Do you agree?

In my opinion, I think you aren't right, Sprmnt21 !
The Steiner's theorem is "[u] if and only if [/u]":

[b]Steiner's theorem:[/b] Given are $\triangle ABC,\ \{M,N\}\subset BC$.
Then $\frac{MB}{MC}\cdot \frac{NB}{NC}=\left(\frac{AB}{AC}\right)^2\Longleftrightarrow \widehat {MAB}\equiv\widehat {NAC}\ .$

[quote="Virgil Nicula"]In my opinion, I think you aren't right, Sprmnt21 !
The Steiner's theorem is "[u] if and only if [/u]":
[/quote]

Yes. I know this, but is not this the point.

perhaps giving a picture it will be more clear, what I would mean.

The Claimis that HE/HC*JE/JC = (BA/BE)^2.

Using two times Menelaos to the triangle ECK and the transversal BF and BD we have the thesys.

But only in case FACK is a rombus KH is the symmedian of KEC.

From another view point:
Drawn the circle with diameter $AD$ such that $P$ is the second cut point with $BC$.Now: $AP$ is the altitude from $A$.Also: $<LPA=<APK=45^0$. Then $CL$ and $BK$ concur on the altitude $AP$ in $H$.(Converse of Blanchet's theorem).

In what follow P is the foot of A on BC. From secants theorem we have CA*CK = CD*CP and BA*BL = BD*BP. Using those relations and that deiriving from bisector teorem, i.e. AC/AB = DC/DB, we deduce that CK/BL = CP/BP.  Then as AK = AL, from Ceva we have the claim.

PS

What is Blanchet's theorem?

You can see this Theorem 327-II pag.471-472 F.G.-M book:If $AD$ is the altitude of $ABC$ and $BM, CN$ are two cevians concurrent at point on $AD$ then $AD$ is the bisector of $<NDM$.

Construct squares on AB, AC  and on LB,KC   and  use all those nice properties

   of Vecten configuration together with homotheties of the squares.

   T.Y.

   M.T.


algebra
function
domain
algebra unsolved

i would like to ask some terms if possible,relations
to one another and also where could i find free sources online
to read about them,also some solved problems with them
would also be nice if possible:

1)GCD-Domains
2)Bezout ring
3)fractions ring
4)Q(R) (where R is a ring but i dont know what Q(R) is)
5)multiplicative system(i guess i would know what this
is but im not sure)

Most of those can be found in a first abstract algebra text. Some of the typical ones are Dummit and Foote, Herstein, Fraleigh, although I personally use Rotman for everything since it goes so far beyond undergraduate algebra. Also wikipedia or mathworld.wolfram is always a good source for straight-up definitions.


See all the problems from 5-th Kyiv math festival [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=506789#p506789]here[/url]

Let $a, b, c, d$ be positive integers and $p$ be prime number such that $a^2+b^2=p$ and $c^2+d^2$ is divisible by $p.$ Prove that there exist positive integers $e$ and $f$ such that $e^2+f^2=\frac{c^2+d^2}{p}.$

See for a more general one: http://www.mathlinks.ro/Forum/viewtopic.php?t=66248
But this special case follows for example directly from the Two-Squares-Theorem: a positive integer can be written as sum of two squares iff every prime divisior $\equiv 3 \mod 4$ occurs an even number of times in the decomposition.

There is a mistake in the problem. :)  Put a=1, b=2, p=5, c=3, d=9. But there are not such e and f for 9 .

[quote="ZetaX"]$ e \equal{} f \equal{} 3$¿[/quote]
I've mixed up, $ d \equal{} 6$

Then it depends on the definition of positive  :P


Hey guys,

Northview is thinking of hosting a math tournament early next year (around October or so). Would you guys be interested in coming?

Cheers,

Karthik

Depending on when, Walton would probably come.

Karthik, some other guys and me are writing the test questions. btw, Karthik, did you take a look at some of my test questions yet?

When, if at all, are we having this tournament, anyway?


I think we should solve another problem
Find all $n\in N^*$such that$\binom{2n}{n}\vdots (n+1)^2$

Dosnt turn it into much of an easier problem, though...

[quote="nthd"]prove or disprove that
$\forall k\in N^*,$always$\exists n\in N^*$such that
 $\frac{(2n)!}{[(n+k)!]^2}\in Z$
[u][b]NOTE[/b][/u]:+)$k=1$,we have $n=5$
            $\frac{10!}{(6!)^2}=7$
         +)$k>1$...... :([/quote]

I think (but not sure) the below problem is possible help us to solve nthd's one.

[b]Problem[/b] prove that for all $k \in N^*$ there exist infinity many positive integers $n$ such that :

$\frac{(2n)!}{n!(n+k)!}$ is integer


Let $a_{n}$ be a nonnegative sequence. Prove that if $\sum a_{n}$ converges, then so does $\sum \frac{ \sqrt{a_{n}}}{n}$

Also AM  \ge GM works here,too.
$2\frac{\sqrt{a_{n}}}{n}\le \frac{1}{n^{2}}+a_{n}$ :)


How many triples of natural numbers (a,b,c) are there such that a+b+c=111 and a,b,and c are in geometric progression.

EDIT: oh yea also I tried doing a+ar+ar^2=111, and then a(1+r+r^2)=111

and I tried a=1,3,37, and 111, and then found if r(r+1) was a factor of the remaining answers and I got (37,37,37) and (1,10,100) and (100,10,1)-since they can be switched, but apparently that doesnt give the right answer....so I was also wondering what was wrong with my method.

EDIT2: oh and I think this problem might be poorly worded.  a,b,c in that exact order must form a geometric progression.  So you cant have like (10,1,100)-I initially just copied down the prboelm from the old AMC

Thanks in advance

Let $ a\leq b\leq c,$ we have $ b^2 \equal{} ac$ and $ a \plus{} b \plus{} c \equal{} 111.$
Then $ 0\leq (a\minus{}c)^2\Longleftrightarrow (a\plus{}c)^2\geq 4ac$
$ \Longleftrightarrow (111\minus{}b)^2\geq 4b^2$
$ \Longleftrightarrow 3(b\plus{}111)(b\minus{}37)\leq 0.$
Since $ b>0,$ we have $ b\leq 37.$

Let $ d \equal{} (a,c)$, then $ d^2|b^2$ so $ d|b$. Then let $ a \equal{} da'$, etc. So $ b'^2 \equal{} a'c'$, but then $ a'$ and $ c'$ are perfect squares. Let $ a' \equal{} x^2$ and $ c' \equal{} z^2$ and $ (x,z) \equal{} 1$. So $ b' \equal{} xz$.

So $ d\cdot (x^2 \plus{} xz \plus{} z^2) \equal{} 111 \equal{} 3\cdot 37$

Since a,b,c are natural numbers, $ x^2 \plus{} xz \plus{} z^2\ge 3$ with equality when $ x \equal{} z \equal{} 1$. So we have to check cases $ x^2 \plus{} xz \plus{} z^2 \equal{} 3,37,111$...$ \text{max}\, (x,z)\le 10$...

The problem with your method is that $ r$ is not necessarily an integer.

Edit:  [b]N.B.:[/b]  The primes of the form $ a^2 \plus{} ab \plus{} b^2$ for [i]integer[/i] (not necessarily positive) $ a, b$ are precisely $ 3$ and the primes congruent to $ 1 \bmod 3$.  In other words, they are the norms of the non-real [url=http://mathworld.wolfram.com/EisensteinPrime.html]Eisenstein primes[/url].

I think that I had a similar problem in one of my mock tests where you had a finite geometric sequences of positive integers where the ratio was not an integer...I got a good number of complaints that the problem was faulty :D.


A man jogs to the next town at 6 mph and walks back at 2mph.
what was his average speed for both journeys?

suppose he goes $d$ miles, then it takes $\frac{d}{6}$ and $\frac{d}{2}$ hours to go out then back, respectively, the average speed is distance over time; he traveled $2d$ and it took him $\frac{d}{6}+\frac{d}{2}$ hours, hence the average speed is $\frac{2d}{\frac{d}{6}+\frac{d}{2}}=\frac{2}{\frac{1}{6}+\frac{1}{2}}$

note the form of this answer; it is the harmonic mean of the speeds, 2 and 6

so what is the answr? 3 or (2 and 6)? or 12

It is 3, because the harmonic mean (2ab/a+b)  is 24/8=3.
Or you can use the way Altheman said.

i'm sorry, i probably misunderstood the question.  :oops:


1) Compare But-1-yne, But-2-Yne. 1yne has a dipole moment while it is zero for the 2nd compound due to symmetry. But yet 1yne has a lower BP than 2yne. Why?

2) We usually prepare alkynes from Vicinal dihalides. Consider 1,2-DihaloButane. And adding alc.KOH we get 3 products and infact to prepare a proper alkyne (the only possible one), we need to react the less stable product with NaNH2. Does this occur spontaneously on reacting with alc.KOH and NaNH2.

Also we use NaNH2 cause it is a strong base to extract the proton from Vinyl Halide (assuming), why do we need a stronger base incase of vinyl halide?

3) Predict the product(s) with proper mechanism.

[b]2,3-Dibromo,2-MethylButane --- (excess of NaNH2) ------>[/b]

Another random doubt, not related to alkynes. Is Cyclohepta1,3,5-triene planar??

[quote="pardesi"]2. we normally use alc.KOH first followed by $ \ce{NaNH2}$ yes that's because vinylic Hydrogen are difficult to extract like hydrogens directly attached to benzene....
i couldn't get the second part...what do u mean
no idon't think cyclo hepta 1,3,5 triene is planar :)[/quote]

Why is vinylic C-H bond more stable? Don't tell me its because of sp2.

no it's not stable....that will  be non sense because we don't know what are we comparing the stability to :)
but the Carbon hydrogen bond in case of vinclic hydrogen is much stronger simply because of the hybridisation the Carbon assumes...well that's just apart of reason because then one could argue why is the carbon hydrogen bond in case of alkynes easily cleaved ....the reason is the electrons are more towards the carbon hence a litlle effort by a strong base can cleave it to form a stable intermediate than itself...that is negative charge on ths sp hybrdized carbon

vinylic H is stable because it is a bond fomed by sp2 - s overlap which has higher stability compared to sp3 - s of alkyl halides.

and by that reasoning how will u explain the high acidity of 1-alkynes ...because the sp-s overlap must be even more stable

In case of that we need to account for lowry bronsted theory. The alkynes you are talking will create cabanions and these will create carbocations.

I've added one more question, anybody wants to try?

[quote="Cyber-Shreyas"]
Another random doubt, not related to alkynes. Is Cyclohepta1,3,5-triene planar??[/quote]

Its bcos there is a sp3 carbon  :D

[quote="Carcul"]Cycloheptatriene is not planar: its structure is similar to cyclooctatetraene.[/quote]

You mean tub shaped? Can you describe the structure a little more?  :maybe:


A, B, C and D are four positive whole numbers with the following properties:

(i) each is less than the sum of the other three, and (ii) each is a factor of the sum of the other three.  Prove that at least two of the numbers must be equal.  (An example of four such numbers: A=4, B=4, C=2, D=2.)

Thanks

[hide="Proof"]By the second condition, there must exist positive integers $x,y,z,t$ such that

\begin{eqnarray*}b+c+d &=& xa\\ a+c+d &=& yb\\ a+b+d &=& zc\\ a+b+c &=& td\end{eqnarray*}

and also, all of them must be at least $2$ (if, for example, $x=1$, then $b+c+d=a$ and we must have $b+c+d>a$ by the first condition).

Assume two of the numbers are equal - WLOG we'll take $a=b$. Then $a+c+d=b+c+d\implies xa=yb\implies{x\over y}={b\over a}=1\implies x=y$. Now assume that two of $x,y,z,t$ are equal - WLOG we'll take $x=y$. Then $a+b+c+d=xa+a=a(x+1)$ and $a+b+c+d=yb+b=b(y+1)$, hence $a(x+1)=b(y+1)\implies{a\over b}={y+1\over x+1}=1\implies a=b$.

Therefore, we've proven that $a=b\iff x=y$, hence the problem is equivalent to proving that two of $x,y,z,t$ must be equal.

Take the initial four equations and add $a$ to the first, $b$ to the second, $c$ to the third and $d$ to the fourth. If we denote $s=a+b+c+d$, then we get

\begin{eqnarray*}s &=& (x+1)a\\ s &=& (y+1)b\\ s &=& (z+1)c\\ s &=& (t+1)d\end{eqnarray*}

which gives

$a={s\over x+1}, b={s\over y+1}, c={s\over z+1}, d={s\over t+1}$

Therefore

$s=a+b+c+d={s\over x+1}+{s\over y+1}+{s\over z+1}+{s\over t+1}$

which gives

${1\over x+1}+{1\over y+1}+{1\over z+1}+{1\over t+1}=1$

Now assume all $x,y,z,t$ are different. Since they are all at least $2$, the RHS is at most ${1\over 3}+{1\over 4}+{1\over 5}+{1\over 6}={19\over 20}<1$, hence can never be equal to $1$. Therefore, at least two of $x,y,z,t$ must be equal. QED[/hide]

Where would I be without AOPS community?

Thanks for the solutions...


quadratics
trigonometry
search
complex numbers
algebra
quadratic formula

[hide] $z$ and $z-1$ have the same magnitude and complex part, so their real parts must have the same absolute value. It follows that $z = \frac{1}{2}\pm \frac{\sqrt{3}}{2}i = e^{\pm \frac{\pi i}{3}}$.[/hide]

[quote="Altheman"]Prove: If $z\in\mathbb{C}$, $|z|=1$ and $|z-1|=1$, then $z=e^{\pm \frac{\pi i}{3}}$.

I have two solutions so far.[/quote]

[hide="Solution 1?"]$\sqrt{(a+1)^{2}+b^{2}}=1$ and $\sqrt{a^{2}+b^{2}}=1$, solving, you get $a=-1/2$, $b=\pm\sqrt{3}/2$, so $z=-1/2\pm \frac{\sqrt{3}i}{2}$ which is equivalent to $e^{\pm \frac{\pi i }{3}}$[/hide]

[hide="Solution 2?"]Geometrically, $z$ and $z-1$ are on a unit circle, so then draw from the center of the circle two radii that intersect at two points on the circle in the same direction that have a distance of 1 from each other which is horizontal...the points can be either in the positive imaginary part or the negative imaginary part of the graph...so then an equilateral triangle is formed by the chord and two radii...so the angle between the radii on the positive real axis is $\pm \frac{\pi}{3}$...so $z=e^{\pm \frac{i\pi}{3}}$.[/hide]

[hide]
\[|x|=1\Rightarrow \overline{x}=\frac{1}{x}\]
As $\overline{z-1}=\overline{z}-1$, we get
\[\frac{1}{z-1}=\frac{1}{z}-1\]
which simplifies to $z^{2}-z+1=0$, and quadratic formula gives us
\[z=\frac{1\pm i\sqrt{3}}{2}\]
which happens to equal $e^\frac{\pm \pi i}{3}$.
[/hide]

$|z|^{2}=z\overline z=1\wedge |z-1|^{2}=(z-1)(\overline{z}-1)=z\overline{z}-z-\overline{z}+1=1\implies z+\overline{z}=1$
$\Re(z) =\frac{z+\overline{z}}{2}=\frac{1}{2}\implies \cos\theta =\frac{1}{2}\implies \theta=\pm\frac{\pi}{3}\implies z=e^{\pm i\frac{\pi}{3}}\blacksquare$

[quote="Altheman"]i was thinking about the z*conjugate(z)=1, and a geometric interpretation...basically we have that z lies on two unit circles centered at 1,0 and 0,0, a intersection point makes an equilateral triangle with the centers, and we get arg(z)=\pm pi/3 immediately[/quote]

nice, very nice.

Wow... way overposted problem :!:

Please try doing a search (maybe not in this forum)


Let $ p, q$ be two consecutive odd prime numbers. Prove that $ p\plus{}q$ is a product of at least $ 3$ natural numbers $ > 1$ (not necessarily different).

$ p, q\equal{}1,3(mod \;\; 4)$, so $ 4|(p\plus{}q)$, but $ p\plus{}q>4$, so there must be another factor of $ p\plus{}q$, done.

Another way...

Suppose $ p,q$ exist (wlog with $ q>p$) such that the hypothesis is not verified, thus $ p\plus{}q\equal{}2r$, where $ r$ is a prime. Now, we have $ 2p<2r<2q$, namely $ p<r<q$. But this is impossible since $ p$ and $ q$ are consecutive primes.

[quote="stephencheng"]$ p, q \equal{} 1,3(mod \;\; 4)$, so $ 4|(p \plus{} q)$, but $ p \plus{} q > 4$, so there must be another factor of $ p \plus{} q$, done.[/quote]

 :?:   Take, for example, $ p\equal{}13$, $ q\equal{}17$. They are consecutive primes but $ p\equiv q\equiv 1\mod 4$. Am I wrong?

[quote="geda"][quote="stephencheng"]$ p, q \equal{} 1,3(mod \;\; 4)$, so $ 4|(p \plus{} q)$, but $ p \plus{} q > 4$, so there must be another factor of $ p \plus{} q$, done.[/quote]

 :?:   Take, for example, $ p \equal{} 13$, $ q \equal{} 17$. They are consecutive primes but $ p\equiv q\equiv 1\mod 4$. Am I wrong?[/quote]
No, probably, stephencheng thought of consequtive odd numbers which are primes

geda wrote:
[quote]Take, for example, p=13,q=17 . They are consecutive primes but p=q=1(mod 4). Am I wrong?[/quote]

ith_power wrote:
[quote]No, probably, stephencheng thought of consequtive odd numbers which are primes[/quote]

Umm ,I understand the problem just like what ith_power said, but no big deal, both of them are easy.


linear algebra
matrix
function
linear algebra unsolved

Let A be real square matrix such that $A+A^{T}= I$. Prove that $\det A > 0$.

That's a problem from 2007 Vojtech Jarnik Competition.

We use Schur's decomposition: for every square A there exists a complex matrix U such that U is unitary (UU*=I) and $UAU^{*}=T$, where T is upper triangular (an elementary proof can be found here - [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=61229[/url]

Let U be the unitary matrix from Schur's lemma, $UAU^{*}=T$. Conjugate of both sides yields $UA^{*}U^{*}=T^{*}$, but since A is real, $A^{*}=A^{T}$. Adding both equalities and applying $A+A^{T}=I$ and $UU^{*}=I$ gives $T+T^{*}=I$. Now it's obvious that T is diagonal and its every diagonal entry is of the form $\frac{1}{2}+a_{k}i$. All entries are eigenvalues of T, and if z is an eigenvalue, so is z* (conjugate) - easy conclusion form T's characteristic polynomial. We are almost done - it's simple to check that detT>0, and T is similar to A, so detA=detT>0.

[size=75]BTW, wiesz Karol, że za to zadanie można czekoladę wygrać? ;)[/size]

Ahh, similar to my solution - Schur's lemma is very useful:D

[size=59]Yes, I know;P Tak tez je znalazlem i mi sie spodobalo.[/size]

Very closely related to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=143797]this problem[/url] posted right next to it.

Note in this case that $A-\frac12I$ is real and skew-symmetric.

Another solution, from my friend, beautiful:

- first, we prove that the set of solutions of $A+A^{T}=I$ is convex, i.e. if A and B are solutions, so is $\lambda A+(1-\lambda) B$ (easy)
- A is nonsingular; if detA=0, then Ax=0 for some nonzero x, and then $x^{T}Ax=0$ and $x^{T}Ax=0$; so $x^{T}(A+A^{T})x=x^{T}x=0)$ but $x^{T}x=0$ iff x=0 - contradiction;
- since A is convex, it is connected (in topological sense); determinant function is continuous, so the image of a connected set is also connected; hence, if X is the set of solutions, det(X) is connected and $0 \notin det(X)$ - this implies that either det(X)=R+ or det(X)=R-; to prove that it's R+, we notice that $A=\frac{1}{2}I$ satisfies the equation, QED


AMC
AIME
USA(J)MO
USAMO
ARML
AIME I
AIME II

Here is a list of dates for math contests that I know of.  Feel free to add any other dates of contests that you know of, especially local ones.   These dates are for 2005 only.

AMC 10/12 A - February 1, 2005
AMC 10/12 B - February 16, 2005
AIME I - March 8, 2005
AIME II - March 22, 2005
USAMO -  April 19 & 20, 2005

[b][u]2007[/u][/b]
AIME--March 12
Alternate AIME--March 28
OHMIO--March 31
USAMO--April 24-25
ARML--June 1

[b][u]2008--tentative[/u][/b]
AMC A--February 12
AMC B--February 27
AIME--March 18
Alternate AIME--April 2
USAMO--April 29-30

[b][u]2009--tentative[/u][/b]
AMC A--February 10
AMC B--February 24
AIME--March 17
Alternate AIME--April 1
USAMO--April 28-29

If anyone knows about any other local or national math contests, please post them here.  Specifically, it would be nice to know the dates for the next Mandelbrot competition, Mu Alpha Theta, and Tri-C Math and science fair (a math contest at Cuyahoga Community College--the last one was this past Autumn).

Sorry, that was a typo on my part :oops: .

Chapter is on February 9, States is in April

2008 MATUCOUNTS State: March 8th
Each Chapter has it's own date.
2008 ARML: May 31st. Leave for Penn State on May 30th, and return on June 1st.


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

[quote="prester"]Find all function $ f$ which are defined for all $ x \in \mathbb{R}$ and, for any $ x,y$ satisfy
$ xf(y) \plus{} yf(x) \equal{} (x \plus{} y)f(x)f(y)$[/quote]
Let $ P(x,y)$ be the assertion $ xf(y)\plus{}yf(x)\equal{}(x\plus{}y)f(x)f(y)$

If $ \exists u\ne 0$ such that $ f(u)\equal{}0$, then $ P(x,u)$ $ \implies$ $ f(x)\equal{}0$ $ \forall x$. So from now, we'll consider $ f(x)\ne 0$ $ \forall x\ne 0$

The equation may then be written $ x(f(x)\minus{}1)f(y)\plus{}yf(x)(f(y)\minus{}1)\equal{}0$ and, considering $ x,y\ne 0$ : $ \frac{x(f(x)\minus{}1)}{f(x)}\plus{}\frac{y(f(y)\minus{}1)}{f(y)}\equal{}0$

And so $ g(x)\plus{}g(y)\equal{}0$ $ \forall x,y\ne 0$ where $ g(x)\equal{}\frac{x(f(x)\minus{}1)}{f(x)}$ and so $ g(x)\equal{}0$ and so $ f(x)\equal{}1$ $ \forall x\ne 0$ and we got a second family of solutions.

Hence the solutions :
=============
$ f(x)\equal{}0$ $ \forall x$

$ f(x)\equal{}1$ $ \forall x\ne 0$ and $ f(0)\equal{}a$ for any real $ a$

[quote="prester"][quote="pco"]So from now, we'll consider $ f(x)\ne 0$ $ \forall x\ne 0$

[/quote]

Dear pco, why do you exclude function that are not zero only for some $ x \in \mathbb{R}$ and not for any? I mean what about discontinous functions?[/quote]

Because I have shown that if $ \exists u\ne 0$ such that $ f(u)\equal{}0$, then $ f(x)\equal{}0$ $ \forall x$

So, either $ f(x)\equal{}0$ $ \forall x$, either $ f(x)\ne 0$ $ \forall x\ne 0$

[quote="prester"]Find all function $ f$ which are defined for all $ x \in \mathbb{R}$ and, for any $ x,y$ satisfy
$ xf(y) \plus{} yf(x) \equal{} (x \plus{} y)f(x)f(y)$[/quote]
Let $ P(x,y) \iff xf(y) \plus{} yf(x) \equal{} (x \plus{} y)f(x)f(y)$
$ P(x,0) \iff xf(0) \equal{} xf(x)f(0)$
$ x \neq 0, P(x,x) \iff 2xf(x) \equal{} 2xf(x)^2 \iff f(x) \equal{} 0$ or $ f(x) \equal{} 1$, for all $ x \neq 0$
Assume that $ a,b \neq 0, f(a) \equal{} 0, f(b) \equal{} 1$. Then:
$ P(a,b) \iff a \equal{} 0$ contradiction. So either $ f(x) \equal{} 0 \forall x \neq 0$ or $ f(x) \equal{} 1 \forall x \neq 0$.
If $ f(x) \equal{} 0 \forall x \neq 0$, then $ P(0,1) \iff f(0) \equal{} 0$ and hence $ f(x) \equal{} 0 \forall x \in \mathbb{R}$, which is seen to be a solution.
If $ f(x) \equal{} 1 \forall x \neq 0$, then $ f(x) \equal{} 1, \forall x \neq 0, f(0) \equal{} c$ is a solution for all $ c \in \mathbb{R}$

Yes, ok. Thank you guys for your solutions. My solution is the same as mathias

As pco and mathias have proved, there are no discontinuous solutions of type $ f(x)\equal{}0,\ \forall x \in \mathbb{A}$ and $ f(x)\equal{}1\ \forall x \in \mathbb{R}\setminus A$ where $ \mathbb{A}$ is any subset of $ \mathbb{R}$ as indicated in the book from which I got this problem. Obvioulsy the second solution $ f(x)\equal{}1\ \forall x \ne 0$ can be discontinuous at $ x\equal{}0$ when $ f(0) \ne 1$


Here are all of the Logarithmic identities that mathmaticians currently know of. If I missed any, please let me know. Also, for those less aquainted with logs, it would help if you could prove the identities that have been posted without proofs.

$\log_{a}b+\log_{a}c=\log_{a}b\times c$.

$\log_{a}b-\log_{a}c=\log_{a}\frac{b}{c}$.

$\log_{a}b=\frac{1}{\log_{b}a}$.

$\log_{a}b=x$ iff $a^{x}=b$.(obvious). :D

$\ln{z}=\frac{log z}{log e}$. Actually, I think this might be wrong. Someone please correct me if this is wrong.

EDITS: Sorry, bad at LaTeX. :blush:

[quote="bigboy82892"]Here are all of the Logarithmic identities that mathmaticians currently know of.[/quote]

This is a [b]rather arrogant claim.[/b]  Semantically speaking, every true statement about logarithms is an identity.  For example, you've forgotten

$\log_{a}1 = 0$
$\log_{a}b^{c}= c \log_{a}b$
$a^{x}= e^{x \ln a}$
$\log_{a^{n}}b^{m}= \frac{m}{n}\log_{a}b$
$\log_{a}b \log_{b}c = \log_{a}c$

In fact, it's trivial to generate an infinite list of identities by repeatedly combining a small, compact list of identities in various ways.

By the way, you've provided two special cases of the general identity

$\log_{a}b = \frac{\log_{c}b}{\log_{c}a}$

[quote="bigboy82892"]$\log_{a}b=x$ iff $a^{x}=b$.(obvious). :D [/quote]

I fail to see how a definition is ever obvious.  Definitions are definitions.

Is it best to just memorize them to become familiar with them? Is there a good way to memorize these? Or should I prove these first like I've heard a person do, then memorize.  :maybe:

[quote="now a ranger"]prove these first[/quote]

Always the best way to go.  Once you've proven them to yourself you don't need to memorize them, since you can quickly retrace your mental steps through the proof instead.  Understanding makes remembering easier.

An especially good way is to associate with every logarithmic identity a corresponding exponential identity (this is assuming you know those backwards and forwards already; they are a little easier to learn because they make more sense).

[quote="t0rajir0u"]This is a [b]rather arrogant claim.[/b][/quote]
But it got someone's attention [b]quickly[/b] instead of having to wait until tomorrow when I'll be cramming all day. Then having to check instead the day after when I'm still cramming for my stupid bio test. Then I'll have to check the day after, by which time I'd have forgotten I'd ever posted, so it wouldn't have been for several weeks that I checked it, and when I finally responded, everyone would have said "Why are you bringing up an ancient topic?" and everyone would have been totally inconvenienced.

Sorry.

Personally, I was interested in learning some new log id's, particularly those involving $\ln$. Also, when someone posts the ln id, I would desire that he posts the [b]beginning[/b] or a [b]hint[/b] as to how to prove the id. Thank you.

You could've just asked for a run-through of standard identities with proof.  It doesn't take long.

[b]Definition:[/b]  $\log_{a}b = c \Leftrightarrow b = a^{c}$

Corollary:  $a^{\log_{a}b}= b$

Corollary:  $\log_{a}1 = 0$

Corollary:  $\log_{a}a = 1$

[b]Identity:[/b]  $\log_{a}(bc) = \log_{a}b+\log_{a}c$

Proof:  $a^{\log_{a}b+\log_{a}c}= a^{\log_{a}b}\cdot a^{\log_{a}c}= bc$.

[b]Identity:[/b]  For reals $r$, we have $\log_{a}b^{r}= r \log_{a}b$.

Proof:  $a^{r \log_{a}b}= \left( a^{\log_{a}b}\right)^{r}= b^{r}$

Corollary:  $\log_{a}\frac{1}{b}= \log_{a}b^{-1}=-\log_{a}b$

Corollary:  $\log_{a}\frac{b}{c}= \log_{a}b-\log_{a}c$

[b]Identity:[/b]  $\log_{a}b = \frac{\log_{c}b}{\log_{c}a}$

Proof:  $c^{\log_{a}b \log_{c}a}= \left( c^{\log_{c}a}\right)^{\log_{a}b}= a^{\log_{a}b}= b$

Corollary:  $\ln z = \frac{\log z}{\log e}$

Corollary:  $\log_{a^{n}}b^{m}= \frac{m}{n}\log_{a}b$

Corollary:  $\log_{a}b = \frac{1}{\log_{b}a}$

For some fun, prove

$a^{\log{b}}=b^{\log{a}}$.

[hide="Hint"] A good way to do this is to write both sides in the form
[quote="t0rajir0u"]$a^{x}= e^{x \ln a}$[/quote] [/hide]

$e=\lim_{n\to\infty}(1+\frac{1}{n})^{n}$.
Which is also, incidentally, equal to the following expression:
$\sum_{n=0}^{\infty}\frac{1}{n!}$.

That's true.
$e$ is also known as the base of the natural logarithm, and is close to $2.7182818$...
There's lots of cool things you can do with $e$ in calculus, but for dealing with logs, the reason it is important is because $\ln x$ actually means $\log_{e}x$.


What fraction in the interval $\frac{7}{10}\leq x\leq \frac{8}{11}$ has the least denominator?

I think you reversed the less-than signs, in which case the answer would be $\frac{1}{2}$.

but then ubemaya would be right.. or am i missing something (unless the problem was written wrong on purpose)

If a fraction is greater than $\frac{7}{10}$, it is greater than $\frac{3}{10}$.

The answer book says the answer is

[hide="..."]
$\frac{5}{7}$  :| [/hide]

the answer would be 5/7 if u didnt write the problem wrong, as i see u have edited it...

Thanks for getting mad at people when you wrote the problem wrong.

Anyway, convert the fractions into decimal form and test small denominators.

[quote="#H34N1"]Sorry Guys. I wrongly wrote it :blush:[/quote]

I incorrectly wrote it.

[hide]Multiply both sides by 110q.

We then get

77q<=110p<=80q.

Then start going and get p=5, q=7[/hide]

Technically, oyu didn't specify integers...

he's saying that there could be decimals in teh numerator or the denominator


geometry
inradius
incenter
trigonometry
LaTeX
MATHCOUNTS
area of a triangle

One of the sides of a triangle is divided into segments of 6 and 8 units by point of tangency of the inscribed circle. If the radius of the circle is 4, then what is the length of the shortest side?

Don't give me hints, I just want the solution

The solution won't help any if you don't learn how to solve similar problems on your own.

For convenience, we can label the vertices of the triangle, $A$, $B$, and $C$ with side $a$ opposite $A$, etc.

If, $c$ is divided into $6$ and $8$, then The portion of, say, $a$ nearest $c$ must be $6$ and the portion of $b$ nearest $c$ must be $8$.  (You could reverse those without loss of symmetry).  The other segments of the sides must measure $d$ for some $d$.  Now, the area of a triangle is the inradius times the semiperimeter, which is $6+8+d$, thus the area is $56+4d$.  The area is also
$\sqrt{s(s-a)(s-b)(s-c)}={}\&\sqrt{(56+4d)(50+3d)(48+3d)(42)}$  You would set this equal to $56+4d$, and probably square both sides, but I'm too lazy to do this just now.

100th post! :wow:

ABC is the triangle, and I is the incenter

\tan AIB/2+BIC/2+CIA/2=tan 180=0=tanAIB/2+tanBIC/2+CIA/2

x/4=-(8/4+6/4)/(1-8/4*6/4)
x=7, the sides are 6+7,7+8,8+6, the smallest being 13

That works, but will Derek be able to use the solution to work out similar problems on his own?

And speaking of "so 1337," there is such a thing as $\LaTeX$.
 :yankchain:  :yankchain:  :yankchain:  :yankchain:

I'm not that bright so I won't get hints

Thanks guys

roadnottaken - I used your method before posting here but I realized it gets too messy so yeh.

[quote="roadnottaken"]I wasn't sure how messy it would get, but I didn't have the energy to work it out at the time, so I left it as a method.  Obviously, someone else knew an easier way to solve it (even if he doesn't believe in $\LaTeX$ :wink: )[/quote]

i don't believe in latex when i am doing a history paper, but getting off task and doing math

There is actually an old mathcounts problem. This problem has a not so messy solution without trig:

[hide]Since the tangents are congruent the sides are 14, 6+x, and 8+x. The semiperimeter is 14+x. So the area is $\sqrt{(14+x)(x)(8)(6)}$ and $4(14+x)$. Equating them we get that $(14+x)(x)(6)(8)=16(14+x)^{2}$ which is the same as $3x=14+x$ so x is 7. The sides are 13, 14, and 15.[/hide]


i don't know why, but i keep askgin problems from the same category each day...

its weird... :roll:

One base of a given isosceeles trapezoid is 1/2 the length of the other, each of the congruent side is 5 inches long, and the area of the trapezoid is 36 square inches. What is the sum of the least pair integers that can be used as measures of the bases :?:

is the answer by any chance 18?
\my solution is based on the fact that, in an incosceles trapeziod, you can "cut off" right triangles at both ends, both of which would have hypot. = to 5. since it's integers the other 2 sides must be 3 and 4. so 4 is the height, and the formula for area is av(b1 and b2)*h. so 36/4=9 which must be the average of the bases. let the one base be  x and the other be x+6(since you have those 3's in the right traingles that we "cut off". so if average of x, x+6 is 9 then the sum mus t be 18 (and x=6)

the answer is for me to know and you to find out.

[hide="Answer"] i actually don't know the answer, but i got 18 too.  :P [/hide]

Yeah the answer is 18. I quadruple-checked.


linear algebra
matrix
factorial
algorithm
combinatorics proposed
combinatorics

Consider some matrix $ n\times n$ and with vertices $ A,B,C,D$ ordered in clockwise order.Define the notation of "house" as some column with basement on $ AD$ and with width equal to $ 1$ and with height equal to some non-negative integer.The side $ AD$ is completely fulfilled with these houses and this figure is called "city".
"City" is named "good" if diagonal $ AC$ doesn't intersect neither of these houses [geogebra]93b6dde2b536feaadd8edc2ca0a6804114ad1738[/geogebra]
a)Determine the value of different possilbe "good" cities for the given $ n$?
b)Prove that "cities" consisting from the same set of "houses" form equivalence classes.
c)Find the number of equivalence classes
d)Find some explicit formula of number of elements in equivalence class.

I can't view the applet, and I'm not 100% certain that I understand the question correctly.  Is it equivalent to the following?

Compute the number of sequences $ (a_0, \ldots, a_{n \minus{} 1})$ of integers such that $ 0 \leq a_i \leq i$ for each $ i$.

If so, the answer is clearly $ (n \minus{} 1)!$.  Part (b) is (I think) obvious in any case.  If my understanding is correct, part (c) is the $ (n\minus{}1)$th Catalan number -- for each such city, sort the blocks in increasing order.  This preserves the property that $ a_i \leq i$, and the resulting objects are in an immediate bijection with Dyck paths of length $ n \minus{} 1$.  For the last question, it's not immediately clear to me, but I strongly suspect the answer is known.

you got it correctly,I am pretty sure.
yup,my explanation was just a poor try to make the problem a little bit harder,nevertheless it finds application rather often.
part a) and b) are quite obvious,but what does Dyck path mean?
I guess it is the shortest way to reach one corner of the matrix from another not intersect the main diagonal,right?
and this is pretty obvious too...the last part seems to be the most interesting :wink:

http://mathworld.wolfram.com/DyckPath.html -- it's one of the standard/most frequently appearing objects enumerated by the Catalan numbers.  Hopefully I'll have some time to devote to (d) in the near future.

That factorial for (a) looks off by 1; I see $ n!$ from the same argument. The Catalan number should also be the $ n$th rather than $ n\minus{}1$th.

I'll think about (d) as well.

jmerry is right that I fouled up the indexing.

[quote="Erken"]need a hint?or you haven't devoted any time to the problem yet?...[/quote]  So, I forgot about this a long time ago and then just stumbled on it again, and I would like a hint, if you had one ;)  I can give an efficient recursive algorithm for calculating the number of good cities in the equivalence class with a given city, but that's not really an explicit formula, exactly.