vector
inequalities
function
linear algebra
matrix
topology
real analysis

I am looking for a list of examples of norms in ${\bf R}^n$ apart from those well-known $\| x\|_{r} = {\root r\of {\sum_{i=1}^{n}{x_{i}}^r}}$ and $\sup_{1\leq i\leq n}x_n$
Are there any good sources available on the web?  :roll:

$\mathbb{R}^n$ (over $\mathbb{R}$) is a finite dimensional vector space.  A well-known theorem states that in any finite dimensional vector space, all norms are equivalent.  This means that given two norms $|\cdot|$ and $\|\cdot\|$, there exists $C_1,C_2>0$ such that $C_1\|x\|\leq |x|\leq C_2\|x\|$ for every $x\in \mathbb{R}^n$, and, for all intents and purposes, we can treat equivalent norms as the same.

Here is the outline of a proof.

Say that $X$ is finite dimensional.  Let $\{e_1,e_2,\ldots,e_n\}$ be a basis for $X$, and define

$\|\sum_i a_ie_i\|=\sum_i |a_i|$

Prove that $\|\cdot\|$ is a norm on $X$.

Prove that the map $(a_1,\ldots,a_n)\to \sum_i a_ie_i$ is continuous.

Prove that $\{x\in X:\|x\|=1\}$ is compact.

Conclude that all norms on $X$ are equivalent.

EDIT:  Actually, I tried to do this just for the heck of it (it's an exercise in Folland), and couldn't prove the compactness bit.  However, it's very easy to do this if we use the norm

$\|\sum_i a_ie_i\|=\sqrt{\sum_i a_i^2}$ (the 2-norm) instead.

Use the Cauchy-Schwarz inequality to prove that any norm satisfies a Lipschitz condition w.r.t the 2-norm and hence is continuous.

Finally, take any two norms $|\cdot|_1$ and $|\cdot|_2$.  The function

$f(x)=\frac {|x|_1}{|x|_2}$

is continuous with respect to the 2-norm on the boundary of the unit sphere.  Therefore it has a finite, nonzero maximum and minimum (since $|x|_1$ can't vanish except at zero), and we get the equivalence of $|x|_1$ and $|x|_2$ instantly.

[quote="Kent Merryfield"]Take out your pencil and start drawing.

Let $S$ be a set in $\mathbb{R}^n$ that satisfies the following conditions:

1. $S$ is convex.
2. $S$ is closed, but contains a neighborhood of the origin.
3. $S=-S.$ That is, $x\in S$ iff $-x\in S.$ Call this "balanced."

For $x\ne0$ define $||x||=\frac1{\sup\{r>0: rx\in S\}}.$

This defines a norm. So you can have as many different norms as you can imagine balanced convex sets.

But in $\mathbb{R}^n,$ they will all be equivalent.[/quote]

Well, fair enough. Could we somehow say that: "All norms in ${\bf R}^n$ could be approximated in some sense by those well known norms ${\root r\of{\sum_{i=1}^{n}{x_{i}}^r}}$ or "The set of all those well-known norms differs from all possible norms from only a set of measure zero"  :?:

[quote]Could we somehow say that: "All norms in ${\bf R}^n$ could be approximated in some sense ...[/quote]
Depends on what you mean by "approximate". On the one hand, as blahblahblah has shown, they are all equivalent. That means that if we have two norms, $||\cdot ||$ and $||\cdot ||',$ then there exist positive constants $C_1$ and $C_2$ such that $C_1||x||\le ||x||'\le C_2||x||$ for all $x.$ In turn, that means the same convergent sequences, the same open sets and closed sets, and so on.

If you accept this equivalence as close enough for what you mean by "approximate," then you don't even need a family - you just need one norm.

But if you want more than that, you have to account for different shapes. The standard norms don't give much that's assymetrical or lumpy. Consider these possibilities in $\mathbb{R}^2.$ The first three even give you inner product spaces:

1. $||x||^2=4x_1^2+x_2^2$

2. $||x||^2=7x_1^2+8x_1x_2+5x_2^2$

3. How about the generalization of 1 and 2: Let $A$ be any positive definite symmetric matrix, and let $||x||^2=x^TAx.$

4. Or for some different approaches to shape, how about this:

$||x||=\max(3|x_1|,3|x_2|,2(|x_1|+|x_2|).$

Try drawing the unit ball in each of these cases.

There are two kind of understanding for the concept "approximation"
One is from Topology and the other one from Measure Theory

(1)  "Open sets of the two norms are the same" mean that they have exactly the same topology ${\cal T}$ and subsets such as $S_1=\{x\in {\bf R}^n; \| x\|_1 \leq C_1\}$ and $S_2=\{x\in {\bf R}^n; \| x\|_2 \leq C_1\}$ would have the same interior, exterior and closures...etc. and there is no question further for the word "approximation" unless we further define another norm or metric on the classes of all norms which is a subset of all functions $\{F;F:{\bf R}\mapsto
{\bf R}\}$ or anything alike.

(2)  How about the "measure theoretic approximation"? For example, could we construct a measure $\mu:{\cal T}\mapsto {\bf R}$ such that $\mu (S_{1} - S_2) =0$ where ${\cal T}$ is the class of all Borel measurable sets  :?:

That doesn't make sense, bchui. Take any two of the examples in my last post. Then $S_1$ and $S_2$ are different sets - different interiors, different measures - just different. Saying they generate the same topology (same open sets, same convergent sequences) is not saying that the balls are in any sense the same sets.

Let's make things more clearify. It is well known that every two norms in finite dimensional spaces are always comparable and that $0<s<r\Rightarrow
\| x \|_s\leq\|x\|_r$ How about those other norms such as $\| x\|_S =
\frac{1}{\sup \{r>0|rx\in S\} }$. If comparison is not possible on the entire space ${\bf R}^n$. Could the comparison of the norms $\|x\|_S$ be made according to the sets $S$ only?

[quote="bchui"]It is well known that every two norms in finite dimensional spaces are always comparable ...[/quote]
Comparable in the sense that there exist constants $c$ and $C$ such that $c||x||\le ||x||'\le C||x||.$ That's anything that satisfies the axioms of a norm on a finite dimensional vector space. Period. It makes no difference how you constructed the norms.

[quote="bchui"]... and that $0<s<r\Rightarrow \,\| x \|_s\leq\|x\|_r$ How about ...[/quote]
Now you're reaching for something that just isn't there, and that there's no way to have.

Example: let $\|\cdot\|$ be the usual Euclidean norm in $\mathbb{R}^2$ and let $\|\cdot\|'$ be defined by $\|x\|'^2=2x_1^2+\frac12x_2^2.$

Then $\|x\|\le\|x\|'$ is false and $\|x\|'\le\|x\|$ is also false.


analytic geometry
geometry
AMC
AIME
abstract algebra
Pythagorean Theorem

This problem is in a form of a dialogue:

(A group of AoPSers are hanging out outside in a park during the Christmas holidays, listening as ComplexZeta lectures about the Ax-Kochen Isomorphism Theorem and its applications to the p-adic Norm.)

ComplexZeta: ...and so, if we let the field of p-adic numbers be Z_p(t) and suppose that a set D is a nonprincipal maximal filter on P, the set of primes,  we have Product[1 to k, n in P] Q_p/2(Gamma[Z_p(t)])- Sum(0 to n) 2^(e * T), and it follows that...

Syntax Error: (whispers) Man this stuff is cool beans.

Chinaboy: (whispers) uh...yeah...right...

(rippledance and Eve approach the group, looking as stunning as ever.)

mathfanatic: (rubs his hands) Ah, well Tare, perhaps today is my lucky day. (grins)

s0mp: lolz, good luck.

jojobird: yup, he's gonna need it.

(rippledance and Eve arrive.)

rippledance: Hey guys! I was wondering whether you could help me out with a homework problem I am having trouble on.

mathfanatic: (confronting rippledance) Sure, of course I can help. What is the problem?

rippledance: Well, ABCDEF is a convex hexagon with all sides equal and opposite sides parallel. Angle FAB = 120 degrees. A = (0,0) and B = (b,2), and the y-coordinates of the other vertices are 4, 6, 8, 10 in some order. Find the area of the hexagon.

(Everyone starts solving the problem hurriedly.)

Fierytycoon: Aha! Here is the solution!

(Fierytycoon walks up to rippledance and whispers something in her ear.)

rippledance: Wow, Fierytycoon, you certainly are intelligent.

(rippledance gives Fierytycoon a peck on the cheek.)

Tare: (whispers) Whew, that Fierytycoon certainly is a chickmagnet.

 8-)  8-)  8-)  8-)  8-)

Now, the problem is: what did Fierytycoon whisper into rippledance's ear?

your attempt to mask aime b no. 14 into another problem is clever!

Well think about what type of hexagon this is and how tilted it is compared with the x-axis.

Hahaha...very funny, Fierybeans...you captured everyone's personality quite well there.

Is there some way I can post a diagram?  That is to say, can AoPS host a picture for me for a little while?  This is a question to Mr. Rusczyk.

It's somewhat hard to explain without a diagram...

TheSouthpaw wrote:Ok, call me stupid.

Is what seems to be blatantly obvious true? That [hide]this is a regular hexagon[/hide]??

It is.

And, mathfanatic,  [hide]You know he's not my type[/hide]  </jk>

</jk>, as in "I'm not attracted to male," not "Fiertycoon's my type," FYI.

[quote="mathfanatic"]you captured everyone's personality quite well there. [/quote]

Everyone's but mine...anyway -laughs- that was a weird dialogue. I'm honored that I get to pose the question...=P

And yes, this is a regular hexagon.

The y-coordinates of the vertices should go A=0, B=2, C=6, D=10, E=8 and F=4.

Well, here's what I'd whisper in rippledance's ear:

[hide]Which spice girl are you?

or

You have beautiful eyes.  That might make her kiss me... </jk>[/hide]

But back to the problem.  I can't draw a diagram, so I'll just talk a lot.

Draw A on (0,0), in case that wasn't obvious, and draw B a ways to the left of it, with y-coordinate 2.  F is to the right of A, with y-coordinate 4.  C is to the left of A, with y-coordinate 6.  E is to the right of A, with y-coordinate 8.  And D is slightly to the left of A, with y-coordinate 10.

Now, looking at A, B, F, and the x-axis, we have two nice right triangles.  Drawing X on the x-axis such that AXB is a right angle, and do the same for Y such that AYF is also a right angle.  Now, AXB and AYF have the same hypotenuse, which we'll call x for fun.

Also, BAX + FAY = 60, since FAB = 120 and these two form the remainder of the line.

Now for the reaaally fun part.  Stack AXB on top of AYF such that AF and AB coincide.  You should have a quadrilateral AYFX, with:

YF = 4

FB (or XB) = 2

FYA = 90

BAY = 60 (since BAY = BAX + FAY)

FBA = 90

YFB = 120

Extend AB to point M, forming right triangle AMY.

FBM is also a right triangle with MFB = 60.  Since this is a 30-60-90 triangle, with FB = 2, we know that FM, the hypotenuse, is 2*FB = 4.  Therefore, YM = 4+4=8.  Since AMY is a 30-60-90 triangle, AY = 8/:rt3:.

Now we know FY and AY.  Using the pythagorean theorem, we find that AF = sqrt((8/:rt3:):^2: + 4:^2:) = sqrt(112/3).  Since AF is one of the sides of the hexagon, we can now find the area:

s = sqrt(112/3)

s:^2: = 112/3

A = 3*s:^2:*:rt3:/2 = 3*112/3*:rt3:/2 = 56*sqrt(3) :approx: 97

I do not think that the hexagon by definition is regular.  I can draw several different hexagons with all sides equal and opposite sides parallel.

Wait, the conversation is a contradiction to what Fiertycoon said. He said,
[quote="fiertycoon"]I actually need help with the problem[/quote]
When in the dialogue, he knows the answer...oh, well.

As for whether the hexagon is regular or not, notice that one of the angle is 120 degrees. Along with the other gives, we can conclude that it's a regular hexagon.

Doy!  Oops. Brain slip.  Hex formula is 3s:^2::rt3:/2...

But yes, it must be regular.

It is not a regular hexagon. We know that two angle area 120. What if there are two that are 150 and two that are 90? It is only an equilateral hexagon.

mathfanatic, I am a bit confused. You say to "stack AXB on top of AYF such that AF and AB coincide. " Now I am guessing that that B and F are at the same location. However, you then refer to such angles as YFB and FB. What do you mean by this?


geometry
3D geometry
congruent triangles

Listing the sides, the possible triangles are $ \{(1,1,\sqrt2),(1,\sqrt2,\sqrt3),(\sqrt2,\sqrt2,\sqrt2)\}$ for an answer of $ \boxed3$.

[quote="myyellowducky82"][quote="GameBot"]How many non-congruent  triangles can be drawn such that their vertices lie on the corners of a unit cube?[/quote]
I asked this because wouldn't it be 8 choose 3 for 56?[/quote]

the problem is that some of these 56 trangles are congruent to each other
the problem asks for non-congruent triangles

Oh, okay. so there's 3 different ways, right?

Bump. I got this prob lem on acumen’s. I d9nt get it. Wouldn’t there be 4?

just think about it some more and you will see that there are only 3

Wait why are the distances between a unit cube 1,1, and $\sqrt{2}$?

Because that is all the possible distances between vertices on the unit cube.

[hide=sol]
I just visualized 3 triangles with different side lengths and vertices and I got $\boxed{3}$.
[/hide]


There exists a parallelogram ABCD. There is a point P inside the parallelogram such that angle PCB = angle PAB. Prove that angle ADP = angle ABP.

Use a transformation: $T(\overrightarrow{BA})$, then $P \to P'$, $B \to A$ and $C \to D$. Therefore we have:
\[ \angle{APP'}=\angle{BAP}=\angle{BCP}=\angle{P'DA} \]
Hence $A, P, D, P'$ lie on the same circle, so $\angle{ADP}=\angle{AP'P}=\angle{ABP}$.


[quote="ehsan2004"]$a,b,c\in\ R^+$.Prove that $(a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3)\geq\ (a+b+c)^3$.[/quote]
it's posted for many times... the core idea is $a^5-a^2+3 \ge a^3+2$ then use holder.


inequalities induction algebra function domain polynomial

Source: Bulgarian MO 1980
4. Find all $x$ verifying the inequality
$\sqrt{x+1}>1+\sqrt{\frac{x-1}{x}}$.

[quote="ashwinrk_jain"]answer is $x \varepsilon [1,\infty)$. If the answer is true then i will submit the solution.[/quote]
that's what I got too!
plz do post your solution..it doesn't matter whether u got the right answer or not... :)

BTW, I don't have the answer.. but im pretty sure that u got the correct answer.

well if that is the answer then i assume you could prove it w/ induction (well atleast for $n\in N$)

Ok then then i will submit the whole solution.

we know the term it the square root must be positive.
so ${x+1} \geq  0$
hence we can say x shoulb be always greater than -1.
similarly,
$\frac{x-1}{x} \geq  0$
By this we can say x E (- $\infty$ , 0] U [1 , $\infty$)

Now taking intesection of bot we can say x E (- 1 , 0] U [1 , $\infty$)
we can write the equation as
$\sqrt{x+1} - 1 \geq  \frac{x-1}{x}$
then we can square both the sides(because both sides are positive).
now we get $\frac{x^2 + x + 1}{x} \geq  2\sqrt{x+1}$
hence
$\frac{x^2 + x + 1}{x} \geq  0$
numerator is always > 0 hence 1/x > =0
By taking the intersection we get

$x E [ 1 , \infty)$
hence the answer.

First, use \in instead of E.  You get $x \in S$, which actually looks nice.

Second, your solution goes in the wrong direction.  This is particularly troublesome because one of your steps is squaring, justified by "both sides are positive," when in fact both sides are not positive in the domain you are considering.

Both the sides are positive because

$\sqrt{x+1} - 1 \geq  \sqrt{\frac{x-1}{x}}$
And
$\sqrt{\frac{x-1}{x}}$ is positive hence
$\sqrt{x+1} - 1$ must be also positive because

$\sqrt{x+1} - 1 \geq  \sqrt{\frac{x-1}{x}} \geq  0$
hence $\sqrt{x+1} - 1 \geq  0$ which gives $x\geq 0$ which do not affect the answer. If you find some other mistake then reply.

I think you miss a value... $x = \frac{1+\sqrt{5}}{2}$ wouldn't work - it gives equality.
[hide="my proof"]
square both sides and rearranging gives $\frac{x^2 - x + 1}{x} \ge 2\sqrt{\frac{x-1}{x}}$.  Notice that this is the Am-GM inequality with $x-1$ and $\frac{1}{x}$.  So this will be true as long as both terms are positive, which implies $x > 1$.  Check $x = 1$ and it works too.
So it works for all x from 1 to infinity, except if $x-1 = \frac{1}{x}$.  Solving gives us $x = \frac{1+\sqrt{5}}{2}$.  So the answer is everything from 1 to infinity except the golden ratio.
[/hide]

i don't see how this is an olympiad problem, all you have to do is make it an equality, solve and do interval testing just like any other polynomial-like inequality


Are there math or compsci programs that start in early August and end in late august or September 1? School doesn't start until September 2 for us and most academic programs I've seen end in early August.

Most programs I know of are five weeks or longer and therefore generally start in the middle of July at the latest.  In addition, if you don't live on the east coast you should be aware that many programs on the east coast are geared towards an east-coast school schedule where school starts in August.


analytic geometry
geometry
area of a triangle
Heron\u0027s formula
absolute value

The vertices of a triangle have coordinates $ (\minus{}3,1),(\minus{}2,\minus{}3)$ and $ (x,0)$. Find $ x$, given $ x$ is positive and the area of the triangle is 7.5 square units.

I tried shoestring..but I failed.
[hide="Answer"]$ 1$[/hide]

[quote="alanchou"]you might have done shoestring incorrectly.  The points in clockwise order are
(-3,2)
(x,0)
(-2,-3)[/quote]
Oh, I forgot to repeat the first coordinate.. Silly me.. Alright. Thanks.

What do you mean by 'shoestring'?
 :blush:

Shoestring is a formula derived by matrices to find areas of convex polygons given their coordinates., it is called shoestring because when you finish multiplying everything the crossed numbers pair into what looks like a shoelace.

So would Herron's formula be considered shoestring?

[quote="ZhangPeijin"]Its a way to find the area of any polygon on the carisian plane using matrices.
-fails at trying to sound smart-[/quote]

1) It's called the Cartesian plane
2) it does not use matrices, it's derived from them though

Heron's formula is given in terms of side lengths, not coordinates, so it's not shoestring

I get 1/-26 by using the area of a triangle is equal to 1/2 * the absolute value of

x1(y2-y3)+x2(y3-y1)+x3(y1-y2).

[quote="ZhangPeijin"]Aw cmon, I wanted to sound smart...
:(
Anyways, you basically list the points of the polygon in clockwise order from any point and cross multiply, then add up both sides, then take the absolute value of the difference of the two sums divided by two.[/quote]

first off, that's a rather vague description

and if you do it correctly, the difference is going to be positive, so the absolute value doesn't need to be taken.

[url]http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf[/url]

What is shoelacing? (noob question, I know. I'm stupid.)

If you click on the link that alanchou
posted, it explains it pretty well.

*Doesn't think it has to be in clockwise order.* =P. Actually... it might.

And really, you don't need shoelace for this. Just divide it into two triangles... One with height $ 1$ and base $ x\plus{}\frac{11}{4}$,
The other with height $ 3$ and base $ x\plus{}\frac{11}{4}$.

So $ 4x\plus{}11\equal{}15$
$ x\equal{}1$

The $ \frac{11}{4}$ is found using the x-intercept of the line containing $ (\minus{}3,1)$ and $ (\minus{}2,\minus{}3)$

[quote="ZhangPeijin"]counter clockwise makes the difference negative, which as alan says, is "wrong"[/quote]

yeah, but then again, I know how you do your version of shoestring, and for you, it should be positive

if you do counter clockwise, then you take absolute value

[quote="Yongyi781"][quote="ZhangPeijin"]counter clockwise makes the difference negative, which as alan says, is "wrong"[/quote]

Going counterclockwise and subtracting the \'s from the /'s, is the same as going clockwise and subtracting the /'s from the \'s... and vice versa.[/quote]
Yeah, but since you take the absolute value, it doesn't matter, but generally, going clockwise makes the difference positive and going counterclockwise makes the difference negative, which is as alan says, "wrong".

hmm, I said if [u]you[/u] do it correctly, and just because something isn't correct doesn't always mean it's wrong (although that usually doesn't seem to happen).

I like copying from movies and stuff.

my technique is I take the absolute value whether it is positive or negative.


calculus
integration
calculus computations

this should be an interesting integral to evaluate... i think !   :|

show
$ \int_{0}^{1}\;\;\left( \;\frac{\sum_{k=1}^\infty\;x^{2^{k}-1}}{1+x}\;\right) \;\;\textbf dx\;=\;\boxed{\gamma}$


AMC
AMC 12
AMC 12 A
AMC 12 B
USA(J)MO
USAMO
AIME

My school's math team coach has said repeatedly that for the AMC 10/12s, only the first one you take counts.

So if I take both the AMC12A and AMC12B, then the AMC office will  count only my AMC12A score.

Is this true, and is there writing somewhere that says it?

Thanks

That is not true.  They take the highest score.  So you can qualify with the B test even if you have already taken the A test.

http://www.unl.edu/amc/e-exams/e7-aime/aime.shtml

It says that you need to get above 100 or be in the top five percent.  On either test, even if you have taken both.

This also goes for USAMO qualification.  I know someone who qualified for the AIME but wants a higher score, so he is taking the B test.

Your school's math team coach needs to do what I have done as a coach and read the [url=http://www.unl.edu/amc/]AMC Web site[/url]. There are plenty of informative pages there about all aspects of AMC tests and awards. Your question "Is this true, and is there writing somewhere that says it?" is a very good question to ask a teacher any time the teacher tells you something that sounds strange. In fact, either score of an AMC participant who takes two AMC tests in one year will count. Or both scores will count. Students on my math team have double-qualified for the AIME, for instance. Both their scores have gone into the annual summary of results and awards book published by the AMC and distributed to math team coaches.

He started saying it last year after the AIME, and said that it was part of some new changes that the AMC was incorporating...

[quote]This also goes for USAMO qualification. I know someone who qualified for the AIME but wants a higher score, so he is taking the B test.
[/quote]

While taking both the A and B versions of the AMC 10 or 12 is permitted, one is not to take both versions of the AIME.  The cover letter to the Teacher's Manual on the MAA/AMC site (under AIME Archives) makes this clear in point #3.

[url]http://www.unl.edu/amc/d-publication/d1-pubarchive/2005-6pub/2006AIME-USAMO/TM-%20AIME,2006.pdf[/url]

[quote="gt59"]
While taking both the A and B versions of the AMC 10 or 12 is permitted, one is not to take both versions of the AIME.  The cover letter to the Teacher's Manual on the MAA/AMC site (under AIME Archives) makes this clear in point #3.

[url]http://www.unl.edu/amc/d-publication/d1-pubarchive/2005-6pub/2006AIME-USAMO/TM-%20AIME,2006.pdf[/url][/quote]

I didn't know that, thanks for informing me.  However, I meant that my friend was taking the AMC 12B to improve his index because he did not score as well as he wanted to on the A test.


function
Functional Analysis
advanced fields
advanced fields theorems

In an article, the author affirmed: Then the solution $ u(x,t)$blows up at the finite time $ T^{*}$.

I don't really know what he meant? Blow-up means that the solution tends to infinity or that there are more than one solution at point $ T^{*}$

Would you help me?

Thanks.

I would expect "blows up at time T" to mean "tends to infinity at time T" or (if the situation is appropriate) "exceeds some other kind of bound around time T".

Hence, "blow up" and "decay" are oppsite, aren't they?

Usually what is generally meant by "blow-up" at time $ T$ is the following: the solution exists as some reasonable function of $ x,t$ for all $ t<T$ but it is impossible to extend it beyond $ t=T$ preserving "reasonability". The typical (but by no means the only possible) situation is that "reasonable" means "belonging to some Banach space $ X$ as a function in $ x$ variable and continuous as an $ X$-valued function of $ t$ in the $ t$ variable". In this case, a typical (but, again, by no means the only possible) blow-up scenario is $ \|u(t,\cdot)\|_{X}\to+\infty$ as $ t\to T-$, which is exactly the one described by Xevarion.

Informally speaking, you are right: the notions of "blow-up", "loss of smoothness", "fast growth of norms", etc. are the opposites of "global existence", "smoothing out", "decay of norms", etc.

[quote="fedja"]

Informally speaking, you are right: the notions of "blow-up", "loss of smoothness", "fast growth of norms", etc. are the opposites of "global existence", "smoothing out", "decay of norms", etc.[/quote]

Thank for your interesting comments.

I have a question is that it is safe if we only prove that the solution blows up ( at one T or more than, when the request is to study the blow-up of the solution) without showing the existence? While to study the decay, we must prove the global existence first.

Best wishes.

U.H


geometry
circumcircle
trigonometry
geometry unsolved

Let ABC be a triangle. D is the intersection of the tangent from A (on the circumcircle) and BC. Let E and F lie on the symetrical axis of AB and AC, such that BE and CF are perpendipular to BC. Prove that D, E, F are colinear.

w.l.o.g. $ \gamma > \beta$, for $ \alpha \equal{} \beta$ it's trivial. We have:

$ \tan \angle FDC \equal{} \frac{\overline{FC}}{\overline{CD}}\equal{}\frac{b \sin (\gamma \minus{} \beta)}{2b \sin \gamma \sin \beta} \equal{} \frac{c \sin (\gamma \minus{} \beta)}{2c \sin (\alpha \plus{} \beta) \sin \beta} \equal{} \frac{\overline{EB}}{\overline{BD}}\equal{}\tan \angle EDB$


function
inequalities
calculus
calculus computations

Let f(x,y)=2x+3y.
Let $ \epsilon$ be any positive number.  Show that there is a disk with center (1,1) such that whenever P is in that disk, $ |f(P)\minus{}5|< \epsilon$.  (Give $ \delta$ as a function of $ \epsilon$.)

$ |f(x,y)\minus{}5|\equal{}|2(x\minus{}1)\plus{}3(y\minus{}1)|$

letting $ \delta \equal{} \frac{\epsilon}{5}$
$ (x\minus{}1)^2\plus{}(y\minus{}1)^2 < \delta^2$

$ \to |x\minus{}1|< \delta, |y\minus{}1|<\delta$

$ \to |2(x\minus{}1)\plus{}3(y\minus{}1)| < 5\delta \equal{} \epsilon$

Let $ \textbf{a}\equal{}(2,3),\textbf{o}\equal{}(1,1),\textbf{P}\equal{}(x,y)$. Then $ \textbf{a}\cdot(\textbf{P}\minus{}\textbf{o})\equal{}f(P)\minus{}5$
Let $ |\textbf{P}\minus{}\textbf{o}|\equal{}\delta$. Then the requirement\[ |f(\textbf{P})\minus{}5 |\equal{} |\textbf{a}\cdot(\textbf{P}\minus{}\textbf{o})| \equal{} \sqrt{5} \delta < \epsilon
\iff \delta <\epsilon /\sqrt{5}
\]
as this is the only restriction for a general point $ \textbf{P}$ to satisfy the inequality, the disk center $ \textbf{o}$ radius $ \epsilon/\sqrt{5}$ is minimal


find all $f:\mathbb R\to\mathbb R$ and $g:\mathbb R\to\mathbb R$ such that
$f(x+y)+f(x-y)=2\cdot{f(x)\cdot{f(y)}}$
$(f(x)+g(x))^2=1+2f(2x)$

my solution is a bit boring [recurrence], so I won't post it just answer:

$f(x)=1$ and $g(x)=\pm \sqrt{3} -1$ or $f(x)=0$ and $g(x)=\pm 1$

okie
i think megus'result is exactly
and now, solve newpro:
find all $f:\mathbb R\to\mathbb R$ and $g:\mathbb R\to\mathbb R$ such that
$f(x+y)+f(x-y)=2\cdot{f(x)\cdot{f(y)}}$
$(f(x)+g(x))^2=1+f(2x)$

hm...it's the same - because first equation remains the same, then you have the same possible functions $f$ as in problem above - so just $g$ changes a bit

[quote="Diogene"]No, it's not the same ! Take for  exemple : $f(x)=cos(ax) , g(x)=(\pm\sqrt{2} -1)*cos(ax)$
 :cool:[/quote]

???? why ????   $\sqrt{2}-1$ :D
i agree with megus
the first condition is all we need

Oh man! Do you really think that $f(x) = cos(ax)$ is the same that $f(x) = const.$  :?:
:cool:

Diogene is right - I've must missed one solution, but I wonder how ?  :?

Thanks for finding mistake Diogene  :P

So, does anybody want to post the solution :?:

okie :D
we use::$P=f(x+y+z)+f(x-y+z)+f(x+y-z)+f(x-y-z)$


calculus
integration
function
calculus computations

[quote="kunny"]For a function satisfying $ f'(x) > 0$ for $ a\leq x\leq b$, let $ F(x) \equal{} \int_a^b |f(t) \minus{} f(x)|\ dt$. For what value of $ x$ is $ F(x)$ is minimized?[/quote]

I fixed your typo "dx" -> "dt".

Answer $ x \equal{} \scriptsize{\frac {a \plus{} b}{2}}$.

[hide="Work:"][url=http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign]Differentiating under the integral sign[/url], we have
\[ \frac {dF}{dx} \equal{} \frac {d}{dx}\left\{ \minus{} \int_{a}^{x}\left[f(t) \minus{} f(x)\right]\, dt \minus{} \int_{b}^{x}\left[f(t) \minus{} f(x)\right]\, dt\right\}\]

\[ \equal{} \int_{a}^{x}f^{\prime}(x)\, dt \plus{} \int_{b}^{x}f^{\prime}(x)\, dt\]
\[ \equal{} (2x \minus{} a \minus{} b)f^{\prime}(x): \equal{} 0\]
\[ \Rightarrow x \equal{} \frac {a \plus{} b}{2}\]
[/hide]

[quote="benorin"]

I fixed your typo "dx" -> "dt". [/quote]

Thank you and nice solution! :)

Something more complicated but more elementary
Since f'(x) > 0 , f is strictly increasing so if t < x then f(t) < f(x ) and if t>x then f(x)<f(t), this will help us to get rid the abs

\[ \begin{array}{l}
 \displaystyle\ F(x) \equal{} \int_a^b | f(t) \minus{} f(x)|\;dt \equal{} \int_a^x | f(t) \minus{} f(x)|\;dt \plus{} \int_x^b | f(t) \minus{} f(x)|\;dt \equal{}  \\
  \equal{}  \minus{} \int_a^x {\left( {f(t) \minus{} f(x)} \right)} \;dt \plus{} \int_x^b {\left( {f(t) \minus{} f(x)} \right)} \;dt \equal{}  \\
  \equal{} \left( {x \minus{} a} \right)f\left( x \right) \minus{} \int_a^x {f(t)dt}  \minus{} \left( {b \minus{} x} \right)f\left( x \right) \plus{} \int_x^b {f(t)} \;dt \equal{}  \\
  \equal{} \left( {2x \minus{} a \minus{} b} \right)f\left( x \right) \minus{} \int_a^x {f(t)dt}  \minus{} \int_b^x {f(t)dt}  \\
 \end{array}\]
So
\[ \begin{array}{l}
 \displaystyle\ F'\left( x \right) \equal{} 2f\left( x \right) \plus{} \left( {2x \minus{} a \minus{} b} \right)f'\left( x \right) \minus{} 2f\left( x \right) \equal{} \left( {2x \minus{} a \minus{} b} \right)f'\left( x \right) \\
 F'\left( x \right) > 0 \Leftrightarrow x > \frac{{a \plus{} b}}{2} \wedge F'\left( x \right) < 0 \Leftrightarrow x < \frac{{a \plus{} b}}{2} \wedge F'\left( x \right) \equal{} 0 \Leftrightarrow x \equal{} \frac{{a \plus{} b}}{2} \\
 \end{array}\]
Implies minimum value at x= (a+b)/2, global one


Euler
algorithm
number theory proposed
number theory

$1$.Find the solutions of $3x+5y+4z=6$ where $x,y,z$ are integers.
$2$.Now let $a,b,c,d$ be integers and let $(x_{0},y_{0},z_{0})$ be a solution of the equation $ax+by+cz=d$, Find all the solutions.

hallo,
1,
by eulers algorithm i got: \[\{y-4v+2, y, 3v-2y: y, v \in Z\}\] 2,
it is a linear diophantine equation. from the theoreme of linear diophantine equations we have the solutions \[\{x_{0}+t\frac{bc}{\alpha^{2}},y_{0}+t\frac{ac}{\alpha^{2}},z_{0}-2t\frac{ab}{\alpha^{2}}: t \in Z\},\] where $\alpha = gcd(a,b,c)$.  :)


algebra
polynomial
search
combinatorial geometry
superior algebra
superior algebra open

let $P$ be a polynome and $P'$ he's derivee
show that the root of $P'$ is in the small convexe who conain the root of $P$ in the complex plan?
and if for any root $z$ of $P$ $|z|<1$ then for any root of $P$ ther exist a root w of $P'$ such that
$|z-w|$<1 and sorry for my bad  english

In better English:
Show that the roots of $P'$ are contained in the convex hull of the roots of $P$ for any complex polynomial $P$.

I proved and used this theorem in my very first [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=93984#93984]post[/url] on this forum.

The second question is a well-known open problem (Sendov's conjecture). It looks better if $ <$ is replaced with $ \le$. In recent years there have been at least two purported "proofs" of this conjecture. One of them was even accepted by a journal that will remain unnamed. :roll: (The article was eventually withdrawn.)

I once thought I had a proof (it was a long time ago).  :blush:

Of course, a lot of computer time has been wasted on the search for counterexamples. However, I don't know if anyone tested the following stronger version of the conjecture. Let $ K\subset\mathbb C$ be a closed convex region that is symmetric about $ 0$. Suppose that a polynomial $ p$ has all its zeroes in $ K$. Is it true that $ p(z)\equal{}0\implies\exists w\in K\plus{}z$ such that $ p'(w)\equal{}0$?
In other words, can we replace the Euclidean distance $ |\cdot|$ with an arbitrary norm?


function logarithms algebra domain LaTeX

From what I understand, the logarithmic function can only have positive real numbers.  Is this how I find the domain..... 0 > log?  If so, why are some of them >or = and some of them are just > ?
******************************************

In terms of natural logarithms, my book says:

[color=red]"If the base of a logarithmic function is the number e, then we have the natural logarithm function.  This function occurs so frequently in applications that it is given a special symbol, ln.
Thus, y = log_ex = ln x  if and only if x = e^y."[/color]

I am lost.  I have no idea what the above red statement means.  What does it mean 'if the base is e'?

QUESTION:

log_5(x + 1/x)

I took    0 > x + 1 and x > -1 but don't know why.

I also took x > 0, because of the x in the denominator.

Now my book says: x < -1 or x > 0.  What I don't understand is why x is < -1.  Why isn't it > -1?  See my point?  Help.....

whenever you use logs, you want to get rid of an exponent, for example,
$3^x=27$ to isolate x, you do $\log_3$ and 3 would be considered the base, doing that...
$\log_3 3^x = \log_3 27$ since log and exponents are inverses, we get
$x = \log_327$ or $x = 3$
now having $e$ as the exponent is where the natural logarithm comes in, say we have
$e^x = 45$ and we want to solve for x, so we take the $\log_e$ of both sides to get the exponent
$\log_ee^x = \log_e45$
$x = \log_e45$, since having $e$ as an exponent is so common, they just made an operation $\ln{x}$ instead of writing out the $\log_e$
so for example, the previous $x = \log_e45$ would look like $\ln45$

as for your other question, i am assuming that you want to find the domain of the function $\log_5{x + 1/x}$
so since the domain of logs has to be negative (can you explain [b]why[/b]?) anyway, we have
$x+\frac{1}{x} >0$, now we would be tempted to multiply by x, but if we did that, it creates a problem
[hide="what if x is negative?"]we would have to switch the > to < if x were negative[/hide]
so what if we multiplied by $x^2$, this would NOT give us the problem of having to switch the > to < so doing this...
$x^3+x>0$, factoring this
$x(x^2+1)>0$
and $x >0$ follows from this (do you know why?)
but you may have written the problem wrong, if the plus sign in $\log_5{x + 1/x}$ were a negative, that would change the problem, and make it what your book had

one last thing LEARN TO USE LATEX, you post so many questions, and it would be nicer to use if you would use latex when you ask your questions, just one basic thing, you need to put $\$$ around your math stuff if you want it to look good, also if you want to look at how somebody uses latex, you can just press the quote button on the bottom of their post to see how they did things


Suppose that $ a$, $ b$, $ c$ are positive real numbers. Prove that
\[ \frac{a^{3}}{b^{2}\minus{}bc\plus{}c^{2}}\plus{}\frac{b^{3}}{c^{2}\minus{}ca\plus{}a^{2}}\plus{}\frac{c^{3}}{a^{2}\minus{}ab\plus{}b^{2}}\geq a\plus{}b\plus{}c.\]

This one is stronger than the inequality

$ \frac{a^{3}}{b^{2}\minus{}bc\plus{}c^{2}}\plus{}\frac{b^{3}}{c^{2}\minus{}ca\plus{}a^{2}}\plus{}\frac{c^{3}}{a^{2}\minus{}ab\plus{}b^{2}}\geq\frac{3\left(ab\plus{}bc\plus{}ca\right)}{a\plus{}b\plus{}c}$

which was solved in http://www.mathlinks.ro/Forum/viewtopic.php?t=3221 .

In fact, $ a\plus{}b\plus{}c\geq\frac{3\left(ab\plus{}bc\plus{}ca\right)}{a\plus{}b\plus{}c}$.

to vascs inequality: write
$\displaystyle \sum \frac{a^4}{c^2a+b^2a-abc}\geq \frac{(a^2+b^2+c^2)^2}{a^2b+a^2c+b^2a+b^2c+c^2a+c^2b-3abc}$, do some algebra and get that the rest is equivalent to schur's inequality with a 2 in the exponent: $a^2(a-b)(a-c)+b^2(b-a)(b-c)+c^2(c-a)(c-b)\geq 0$ which can be proved completely similar to the "classical" schur with a 1 in the exponent.

Peter

This is very easy.

We have to prove the inequality

$\frac{a^{3}}{b^{2}-bc+c^{2}}+\frac{b^{3}}{c^{2}-ca+a^{2}}+\frac{c^{3}}{a^{2}-ab+b^{2}}\geq a+b+c$

for positive reals a, b, c.

First note that the denominators of all the fractions in this inequality are positive. This is clear, since $b^{2}-bc+c^{2}=\left(b-c\right)^{2}+bc$, $c^{2}-ca+a^{2}=\left(c-a\right)^{2}+ca$ and $a^{2}-ab+b^{2}=\left(a-b\right)^{2}+ab$. Now, slightly transform the left hand side:

$\frac{a^{3}}{b^{2}-bc+c^{2}}+\frac{b^{3}}{c^{2}-ca+a^{2}}+\frac{c^{3}}{a^{2}-ab+b^{2}}$
$=\frac{a^{4}}{a\left(b^{2}-bc+c^{2}\right)}+\frac{b^{4}}{b\left(c^{2}-ca+a^{2}\right)}+\frac{c^{4}}{c\left(a^{2}-ab+b^{2}\right)}$
$=\frac{\left(a^{2}\right)^{2}}{a\left(b^{2}-bc+c^{2}\right)}+\frac{\left(b^{2}\right)^{2}}{b\left(c^{2}-ca+a^{2}\right)}+\frac{\left(c^{2}\right)^{2}}{c\left(a^{2}-ab+b^{2}\right)}$.

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

$\frac{\left(a^{2}\right)^{2}}{a\left(b^{2}-bc+c^{2}\right)}+\frac{\left(b^{2}\right)^{2}}{b\left(c^{2}-ca+a^{2}\right)}+\frac{\left(c^{2}\right)^{2}}{c\left(a^{2}-ab+b^{2}\right)}$
$\geq\frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{a\left(b^{2}-bc+c^{2}\right)+b\left(c^{2}-ca+a^{2}\right)+c\left(a^{2}-ab+b^{2}\right)}$.

We have

$\left(a^{2}+b^{2}+c^{2}\right)^{2}-\left(a\left(b^{2}-bc+c^{2}\right)+b\left(c^{2}-ca+a^{2}\right)+c\left(a^{2}-ab+b^{2}\right)\right)\left(a+b+c\right)$
$=a^{2}\left(a-b\right)\left(a-c\right)+b^{2}\left(b-c\right)\left(b-a\right)+c^{2}\left(c-a\right)\left(c-b\right)$,

and this is $\geq 0$ by the Schur inequality; thus,

$\left(a^{2}+b^{2}+c^{2}\right)^{2}\geq\left(a\left(b^{2}-bc+c^{2}\right)+b\left(c^{2}-ca+a^{2}\right)+c\left(a^{2}-ab+b^{2}\right)\right)\left(a+b+c\right)$,

and

$\frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{a\left(b^{2}-bc+c^{2}\right)+b\left(c^{2}-ca+a^{2}\right)+c\left(a^{2}-ab+b^{2}\right)}\geq a+b+c$.

Combining our results, we see that

$\frac{a^{3}}{b^{2}-bc+c^{2}}+\frac{b^{3}}{c^{2}-ca+a^{2}}+\frac{c^{3}}{a^{2}-ab+b^{2}}$
$=\frac{\left(a^{2}\right)^{2}}{a\left(b^{2}-bc+c^{2}\right)}+\frac{\left(b^{2}\right)^{2}}{b\left(c^{2}-ca+a^{2}\right)}+\frac{\left(c^{2}\right)^{2}}{c\left(a^{2}-ab+b^{2}\right)}$
$\geq\frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{a\left(b^{2}-bc+c^{2}\right)+b\left(c^{2}-ca+a^{2}\right)+c\left(a^{2}-ab+b^{2}\right)}\geq a+b+c$.

And the inequality is proven.

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=3221 and http://www.mathlinks.ro/Forum/viewtopic.php?t=6551 for proofs.

  Darij

Stop saying problems are easy or people will start to think that they are really easy :P

It looks like this problem may have been motivated by the factoring of the sum of two cubes.

If it was, then perhaps this is a more natural way to begin:

$\frac{a^3+b^3}{a^2-ab+b^2}=a+b\Rightarrow\sum\frac{a^3}{b^2-bc+c^2}\ge\sum a\Leftrightarrow\sum\frac{1}{a^2-ab+b^2}\ge3\frac{\sum a}{\sum a^3}$.

I haven't gotten anywhere with this, but I suspect there's some merit in it.


limit
logarithms
calculus
calculus computations

show that

            $\lim_{n\to \infty}\frac{n}{3^{n}}=0$

I assume you either mean:

$\lim_{n\to \infty} \frac{n}{3^n} = 0$

OR

$\lim_{n\to 0} \frac{n}{3^n} = 0$

The latter is just:

$\frac{0}{3^{0}} = \frac{0}{1} = 0$

The former can be done many ways:

L'hospital:

$\lim_{n\to \infty} \frac{n}{3^n} = \lim_{n\to \infty} \frac{\frac{d}{dn} n}{\frac{d}{dn}3^n} = \lim_{n\to \infty}\frac{1}{\ln 3 \cdot 3^n} = \frac{1}{\infty} = 0$

Rewriting:

$\lim_{n\to \infty} \frac{n}{3^n} = \lim_{n\to \infty} \frac1{\frac{3^n}{n}} = \lim_{n\to \infty} \frac1{3^{(n-\log_3 n)}} = \frac1{3^{\lim_{n\to \infty} (n-\log_3 n)}} = 0$

There are probably more ways to do this too.

i meant the second one n --> infinity

thanks for your solution...


Se considera numere reale $a, b, c, d$ astfel incat \[ a+b+c+d=6 \] si \[ a^{2}+b^{2}+c^{2}+d^{2}=12. \]
 Sa se determine maximul lui $d.$ ;)

Folosind inegalitatea lui Cauchy avem

$3(a^2+b^2+c^2)\geq(a+b+c)^2$ sau
$3(12-d^2)\geq(6-d)^2$......

[quote="cezar lupu"]Yes, indeed. The problem is very easy. Socrates, since when you know romanian language? ;)[/quote]

Thanks.  WE shall talk about it in private...(PM...) ;)  :)

problema a fost data la concursul revistei de matemtica si informatica din constanta si aprtine profesorului Mihai Bogdan.

[quote="cezar lupu"]Yes, indeed. The problem is very easy. Socrates, since when you know romanian language? ;)[/quote]Ar fi o idee sa il intrebi in Romana, daca tot stie.

Sa stii ca ai dreptate, Valentin! :D . referitor la problema: apartine lui sweetconnect=Mihai Bogadn. Este acelasi lucru,nu? :)


function
integration
calculus
calculus computations

Let $f$ be a continuous function on $[a,b]$. Prove that there exists a point $c \in [a,b]$  such that
\[ \int_a^cf(x)dx-\int_c^bf(x)dx=f(c)(a+b-2c) \].
Probably very easy but I can't figure it out..

[quote="xirti"]Any ideas?.. :?[/quote]
This is one of those numerious problems where you apply Rolle's theorem to some cleverly constructed function and get the result in one line. In this particular case the function is
\[ F(x)=\left(\int_a^x f(t)\,dt-\int_x^b f(t)\,dt\right)(2x-a-b)\,. \]
Note that $F(a)=F(b)=(b-a)\int_a^b f(t)\,dt$, so there exists a point $c$ where $F'(c)=0$. This point is what we are looking for.


[img]http://img501.imageshack.us/img501/8063/turkuv4.png[/img]

If $\overline{AB}=\overline{BD}+\overline{ED}$, then compute $\measuredangle\,x$

No trig :whistling:

The answer is 40° if you want to know  :ninja:


Prove that, if $ \frac {x}{a} \equal{} \frac {y}{b}$, then, provided $ a \neq 0, b \neq 0$ and $ a\neq b$,

$ \frac {x^{2} \minus{} y^{2}}{a^{2} \minus{} b^{2}} \equal{} \frac {xy}{ab}$.

$ \frac{x}{a}\equal{}\frac{y}{b}$

$ bx\equal{}ay$

$ bx\minus{}ay\equal{}0$

$ (ax\plus{}by)(bx\minus{}ay)\equal{}0$

$ abx^2\minus{}(a^2\minus{}b^2)xy\minus{}aby^2\equal{}0$

$ abx^2\minus{}aby^2\equal{}a^2xy\minus{}b^2xy$

$ ab(x^2\minus{}y^2)\equal{}(a^2\minus{}b^2)xy$

$ \frac{x^2\minus{}y^2}{a^2\minus{}b^2}\equal{}\frac{xy}{ab}$

(under the assumption $ a \neq b$)

Edit added later:
We can also precede as follows:
$ \frac{x}{a} \equal{} \frac{y}{b}$ implies $ \frac{x^2}{a^2} \equal{} \frac{y^2}{b^2} \equal{} \frac{xy}{ab}$.  Now use that $ \frac{m}{n} \equal{} \frac{p}{q}$ implies $ \frac{m}{n} \equal{} \frac{p}{q} \equal{} \frac{m \minus{} p}{n \minus{} q}$ (provided everything makes sense) to conclude the result.

Thanks. Edited.

If $ \frac{x}{a} \equal{} \frac{y}{b} \equal{} k$, then LHS = RHS = $ k^{2}$.

And when I solved it, I did the first three steps forward, then started from the end to see what I needed to do.


Two curves are given by polar coordinates. r=1+cos a and r=3cos a.

1. Fint the polar coordinates to the intersection between the two curves

2.Find the area over the x-axis and under both of the curves

[hide=" number 1"]
The two graphs form a circle and a cardioid which intersect in two locations. Setting the two equal

$1+cos(a) = 3cos(a)$
$1 = 2cos(a)$
$cos(a) = \frac{1}{2}$
$a = \frac{\pi}{3}or-\frac{\pi}{3}$

r = 3/2 by substituting into either original equation yielding points
$(\frac{3}{2}, \frac{\pi}{3}) (\frac{3}{2},-\frac{\pi}{3})$
[/hide]
[hide=" number 2"]
Break the area into two portions with a line from the pole to the point of intersection
one on 1 + cos(a) from 0 to pi/3
other on 3cos(a) from pi/3 to pi/2
[/hide]

no trig here or polar coordinates

edit: ummm...why isnt this topic moving?

help? :blush:


geometry
perimeter
analytic geometry
Pythagorean Theorem

A pentagon has vertices at (0, 0), (7, 0), (13, 8), (5, 14), and (0, y).
The perimeter of the pentagon is 42 units. What is the value of y?
[hide]So far, I've figured out that the measures of 3 of the 5 side lengths are 7, 10, and 10. So the sum of the last two side lengths should be 15.[/hide]

How many different ways can six people be seated in a row of seven if Jim and John must be in adjacent seats?

Julian and Vino working together can paint a room in 11 hours. Vino and Ted working together can paint a the same room together in 10 hours, and Ted and Julian working together can paint the room in 11 hours. What is the number of hours it will take Julian, Vino and Ted working together to paint 31 identical rooms?

1. Use pythagorean theorem on the sides, so sides are 7, 10, 10. The x-coordinate going from 5 to 0 implies that the next side is 5-12-13, and it is, giving the last side, and y, a value of [u][i][b]2[/b][/i][/u].

[quote="M_M"]A pentagon has vertices at (0, 0), (7, 0), (13, 8), (5, 14), and (0, y).
The perimeter of the pentagon is 42 units. What is the value of y?

[color=red] Solution (Mine): The side lengths are $ 7$, $ \sqrt{(13\minus{}7)^2\plus{}(8\minus{}0)^2}\equal{}10$, $ \sqrt{(13\minus{}5)^2\plus{}(14\minus{}8)^2}\equal{}10$, $ \sqrt{25\plus{}(y\minus{}14)^2}$, and $ y$.

Therefore, $ y\equal{}2$.[/color]
[hide]So far, I've figured out that the measures of 3 of the 5 side lengths are 7, 10, and 10. So the sum of the last two side lengths should be 15.[/hide]

How many different ways can six people be seated in a row of seven if Jim and John must be in adjacent seats?

[color=red] Solution: We can treat Jim and John as a "super person", or a block of 2 people (that take two seats).

There are $ 7*4!*2!\equal{}336$ ways.

Note: the $ 7$ comes from $ 7$ chairs, the $ 4!$ comes from

$ 4$ other people, and the $ 2!$ comes from Jim and John. [/color]

Julian and Vino working together can paint a room in 11 hours. Vino and Ted working together can paint a the same room together in 10 hours, and Ted and Julian working together can paint the room in 11 hours. What is the number of hours it will take Julian, Vino and Ted working together to paint 31 identical rooms?

[color=red] Solution: Let J be Julian's time (to paint one room), V be Vino's time, and T be Ted's time.

Therefore,
J+V=11
V+T=10
J+T=11
J+V+T=16

Therefore, 31(J+V+T)=496. [/color][/quote]

See the above in red.

BOGTRO, I believe your number 3 was incorrect. It should take LESS time for them to finish, working together.

[hide="Solution"]
Let $ j,v,t$ be the 3 people's times. To ease calculations, we can assume that they need to complete "1". So their "distance" is 1. By $ d \equal{} rt\implies1 \equal{} rt$, we have that their rates are $ \frac1j,\frac1v,\frac1t$.
Now it is simply a system of equations.
$ \frac1j \plus{} \frac1v \equal{} \frac1{11}$
$ \frac1v \plus{} \frac1t \equal{} \frac1{10}$
$ \frac1j \plus{} \frac1t \equal{} \frac1{11}$
Solving, we get $ \left(\frac1j,\frac1v,\frac1t\right) \equal{} \left(\frac9{220},\frac1{20},\frac1{20}\right)$. Hence, their combined speed is $ \frac1j\plus{}\frac1v\plus{}\frac1t\equal{}\frac9{220} \plus{} \frac1{20} \plus{} \frac1{20} \equal{} \frac {31}{220}$
Therefore, the time for 1 room is $ \frac1{\frac {31}{220}} \equal{} \frac {220}{31}$, and for 31, we have $ \boxed{220}$ hours.
Note: knowledge of harmonic mean helps here.
[/hide]


I though people might like to take a look at this amazing piece of artwork:

http://www.nemoramjet.com/treeoflife.html

(disclaimer: very large picture, 6768x5651)

in 135 different clades, each described briefly here:

http://www.nemoramjet.com/tolallplates.htm

Maybe it will help somebody learning bio, or just for people can enjoy the artwork.

The CS and biology research that goes into constructing the Tree of Life is really fascinating.  I've been searching without success for something that describes the interesting problems in current research without being too technical.

If you're interested in CS, the best place to start might be the CIPRES (Cyberinfrastructure for Phylogenetics Research) website, [url]http://www.phylo.org/[/url]

The basic idea is to take DNA sequences from a set of organisms, run them through a phylogenetic method which tallies the similarities and differences between the sequences, and then constructs an evolutionary tree.

From the CS standpoint, there are several non-trivial problems:

(1)  Aligning the sequences.  DNA can mutate by transitions and transversions, which don't alter the length of the DNA sequence.  DNA can also mutate by insertions, deletions, and reversals, all of which alter the length of the DNA sequence.  If you have sequences for two genes of different length, how do tell which nucleotide in one sequence corresponds with which nucleotide in the other?  How do you construct an automated method for correctly aligning the sequences?

(2)  Making an accurate model of evolutionary processes so that you can create an accurate method for reconstruct an evolutionary tree from genetic data.

(2)  Finding the best tree.  The best current methods can't be solved exactly in polynomial time.  How do you construct a heuristic that will find a good solution in a reasonable amount of time on a massive data set, with hundreds of taxa and thousands of characters?

(3)  Visualizing an enormous tree like the Tree of Life.

There are also non-trivial problems from the biological standpoint:

(1)  Phylogenetic algorithms assume a strictly tree-like model of evolution, with no transfer of information other than descent.  But, hybridization, recombination, lateral gene transfer, and other processes mean that reticulation happens.  How do you deal with that?

(2)  The 'gene tree,' the evolutionary tree from a single gene, may not be the same as the 'species tree,' the tree showing the evolution of the species, due to things like lateral gene transfer and gene duplication, and incomplete lineage sorting.  What do you do if you have the DNA sequences for several genes, and they don't agree?  How should you even set up a phylogenetic analysis if you have data from several different genes?  Should you combine the data sets before you run, or run each one separately?


P is inside quadrilateralABCD，∠PAB=∠CAD，∠PCB=∠ACD，O1,O2 are circuncenters of  ABC,ACD。prove：∠BO1P=∠DO2P

D and B are isogonal conjugate wrt triangle $ APC$ then $ \angle APD\plus{}\angle BPC\equal{}180^o$
$ \frac{AD}{sin \angle APD}\equal{}\frac{DP}{sin \angle DAP}, \frac{BC}{sin \angle BPC}\equal{}\frac{BP}{sin \angle PCB}$
$ \Rightarrow \frac{DP}{BP}\equal{}\frac{AD.sin \angle DAP}{sin \angle APD}: \frac{BC.sin \angle PCB}{sin \angle BPC}$
$ \equal{}\frac{AD.sin \angle DAP}{BC.sin \angle PCB}\equal{}\frac{AD.sin \angle CAB}{BC.sin \angle ACD}\equal{}\frac{R_2}{R_1}\equal{}\frac{DO_2}{BO_1}$
On the other side, by angle chasing, it's easy to show that $ \angle PDO_2\equal{}\angle PBO_1$
So $ \Delta PDO_2\sim \Delta PBO_1$
Therefore $ \angle PO_1B\equal{}\angle PO_2D$ (QED)

dear [b]livetolove212[/b],thank you for your interesting,I saw the problem on a chinese forum
here is my solution:
By angle chasing, we can conclude ⊿DO2O4∽⊿BAP   BP/R4=AB/R2
⊿PDO4∽⊿BAO1   DP/R4=AB/R1
=>  BP/DP=R1/R2   notice ∠PBO1=∠PDO2  =>  ⊿PBO1∽⊿PDO2

Dear friend,
This figure makes me remember [b]IMO Shortlist 2004, G4 [/b]. Very nice :)


function
invariant
algebra unsolved
algebra

In a sequence: $1,0,1,0,1,0,3,...$ Each term starting from the seventh is the last number of the sum of six previous terms.
Prove, that the configuration : $0,1,0,1,0,1$ cannot appear in this sequence.


function
algebra
polynomial
Functional Equations
pen

Find all functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $n\in \mathbb{N}$: \[f^{(19)}(n)+97f(n)=98n+232.\]

Let us write $ LH\equal{} f^{19}\plus{} 97 f(n) \equal{} 98 n \plus{} 232 \equal{} RH$
One notes that:
1°. The function $ f: N \rightarrow  N$  is strictly increasing.
2°.  $ f^{19}\plus{} 97 f(n)$ needs to be  even number,
since RH is always even.

 Let $ n \longmapsto  k\cdot n$ , so $ f(n) \equal{} k\cdot n, k \in N$ , then  $ f^{19}(n) \equal{} k^{19} \cdot n$,
$ LH\equal{}  k^{19}n \plus{} 97 k\cdot n \equal{} 98 n \plus{} 232$.
For $ n\equal{}1$, we have $ k^{19} \plus{} 97 k  \equal{} 330$,
so  it is not a  solution.
Let $ f(n) \equal{} n \plus{} b, b \in N$, then   $ f^{2}(n)\equal{} n\plus{} 2 b$,
$ f^{19}(n)\equal{} n\plus{} 19 b$.
So we have
$ n \plus{} 19 b \plus{} 97 ( n\plus{}b) \minus{} 98 n \equal{} 232$
Therefore,
$ (19 \plus{} 97) b \equal{} 232$ so $ 116 b \equal{} 232$, and $ b \equal{} 2$.
We have
$ n\plus{} 38 \plus{} 97 ( n\plus{}2) \minus{} 98 n \equal{} 232$
$ 38 \plus{} 194 \equal{} 232$, hence $ f(n) \equal{} n \plus{} 2$ that is  a possible solution.
$ f^{19}(n) \plus{}97 f(n)\equal{} 98n \plus{}232 \equal{}RH$
RH is a 1st degree polynomial.
Hence
$ LH\equal{} f^{19}(n) \plus{}97 f(n)$ must be  1st degree polynomial.That is only possible if
$ LH\equal{}f(n)\equal{} a n\plus{}b$

But if  $ a\in N$
then $ f^{19}(n)\equal{}a^{19}n\plus{}(a^{18}\plus{}...\plus{}a\plus{}1)b$.
So
$ a^{19}n\plus{}(a^{18}\plus{}...\plus{}a\plus{}1)b\plus{}97(an\plus{}b)\minus{}98n\equal{}232$,
$ (a^{19}\plus{}97a\minus{}98)n\equal{}232\minus{}(a^{18}\plus{}...\plus{}a\plus{}1)b\minus{}97b$.
So,
$ (a^{19}\plus{}97a\minus{}98)\equal{}0$
only solution is for $ a\equal{}1$.

[quote="azo"]Suppose $ f(n)^{19} \plus{} 97f(n) \equal{} 98n \plus{} 232$ holds. Picking $ n \equal{} 1$ we see that $ f(1)^{19} \plus{} 97f(1) \equal{} 330 \Rightarrow f(1) [ f(1)^{18} \plus{} 97] \equal{} 330$ We conclude that $ f(1)$ is a divisor of 330. It is easy to see that for $ f(1) \equal{} 1$ equality doesn't hold and, if $ f(1) > 1$ then $ f(1)^{18} \ge 330$. So, no such $ f$ exists![/quote]

I think $ f(n)^{19}\equal{}f(f(...(f(n)...))$ nineteen times. At least that's a common notation, I think. Also, if it meant $ f(n)^{19}$, that would make the problem trivial as you noticed, so I think my interpretation is correct  :P

[quote="Kabor Suk"]Let us write $ LH \equal{} f^{19} \plus{} 97 f(n) \equal{} 98 n \plus{} 232 \equal{} RH$
One notes that:
1°. The function $ f: N \rightarrow N$  is strictly increasing.
2°.  $ f^{19} \plus{} 97 f(n)$ needs to be  even number,
since RH is always even.

 Let $ n \longmapsto k\cdot n$ , so $ f(n) \equal{} k\cdot n, k \in N$ , then  $ f^{19}(n) \equal{} k^{19} \cdot n$,
$ LH \equal{} k^{19}n \plus{} 97 k\cdot n \equal{} 98 n \plus{} 232$.
For $ n \equal{} 1$, we have $ k^{19} \plus{} 97 k \equal{} 330$,
so  it is not a  solution.
Let $ f(n) \equal{} n \plus{} b, b \in N$, then   $ f^{2}(n) \equal{} n \plus{} 2 b$,
$ f^{19}(n) \equal{} n \plus{} 19 b$.
So we have
$ n \plus{} 19 b \plus{} 97 ( n \plus{} b) \minus{} 98 n \equal{} 232$
Therefore,
$ (19 \plus{} 97) b \equal{} 232$ so $ 116 b \equal{} 232$, and $ b \equal{} 2$.
We have
$ n \plus{} 38 \plus{} 97 ( n \plus{} 2) \minus{} 98 n \equal{} 232$
$ 38 \plus{} 194 \equal{} 232$, hence $ f(n) \equal{} n \plus{} 2$ that is  a possible solution.
$ f^{19}(n) \plus{} 97 f(n) \equal{} 98n \plus{} 232 \equal{} RH$
RH is a 1st degree polynomial.
Hence
$ LH \equal{} f^{19}(n) \plus{} 97 f(n)$ must be  1st degree polynomial.That is only possible if
$ LH \equal{} f(n) \equal{} a n \plus{} b$

But if  $ a\in N$
then $ f^{19}(n) \equal{} a^{19}n \plus{} (a^{18} \plus{} ... \plus{} a \plus{} 1)b$.
So
$ a^{19}n \plus{} (a^{18} \plus{} ... \plus{} a \plus{} 1)b \plus{} 97(an \plus{} b) \minus{} 98n \equal{} 232$,
$ (a^{19} \plus{} 97a \minus{} 98)n \equal{} 232 \minus{} (a^{18} \plus{} ... \plus{} a \plus{} 1)b \minus{} 97b$.
So,
$ (a^{19} \plus{} 97a \minus{} 98) \equal{} 0$
only solution is for $ a \equal{} 1$.[/quote]

f(n) need not to be a polynomial
and that the difficulty of this problem

I believe $ f^{(19)}$ means composition, not power. Otherwise azo already shows it is trivial.

We have $ f(n)< \frac{98n\plus{}232}{97}\equal{}n\plus{}2\plus{}\frac{n\plus{}38}{97}$.
So $ f(n)\le n\plus{}2$ when $ n\le59$.
It is not hard to show,
$ f^{(19)}(n)\le n\plus{}38,\forall n\le21$

Then for $ n\le21$,
$ 98n\plus{}232\le97(n\plus{}2)\plus{}n\plus{}38\Rightarrow f(n)\equal{}n\plus{}2,f^{(19)}(n)\equal{}n\plus{}38$.

Actually $ f^{(19)}(20)\equal{}20\plus{}38$ and $ f^{(19)}(21)\equal{}21\plus{}38$ imply
$ f(n)\equal{}n\plus{}2,\forall n\le57$.

Now we will prove $ f(n)\equal{}n\plus{}2,\forall n$ by induction.
Suppose $ f(n)\equal{}n\plus{}2,\forall n<m$.
We have
$ f^{(19)}(m\minus{}36)\equal{}f(m)\equal{}98*(m\minus{}36)\plus{}232\minus{}97*f(m\minus{}36)\\\equal{}98*(m\minus{}36)\plus{}232\minus{}97*(m\minus{}34)\equal{}m\plus{}2$.

Q.E.D.

[quote="xxp2000"]It is not hard to show
$f^{(19)}(n)\le n+38,\forall n\le21 $[/quote]
It is not obvious for me. please show it.

[quote="xxp2000"]So $ f(n)\le n\plus{}2$ when $ n\le59$.[/quote]

It is a silly question to ask to someone who made his post two years ago ... However, then
$ f(f(n))\le f(n)\plus{}2\le n+4$ when $ n\le57$
etc. (step by step, until)
$ f^{(19)}(n)\le  f^{(18)}(n) + 2 \le n\plus{}38$ when $n\le23$ (even more than the $n\le 21$ used).

An easy solution, first checked each side$\mod 97$  then $f^{19}(x) \equiv n+38 \mod 97 \implies $

$\forall n\le 58 : f^{19}(x) \ge n+38 \implies f^{19}(x)+97f(n) =98n+232 \ge n+38 + 97f(n) \implies n+2 \ge f(n) $

and now it will imply $ n+38 \ge ...\ge f^{17}(x)+4\ge f^{18}(x)+2\ge f^{19}(x)$ and thereupon because

$f^{19}(x) \ge n+38, f^{19}(x) \le n+38 \implies f^{19}(x)=n+38 \implies f^{19}(x)+97f(n)=n+38+97f(n)=98n+232$

$\implies \forall n \le 58 : f(n)=n+2 $ and the rest is straightforward using induction. suppose we've proven that for

$n>36$ , $\forall i < n, : f(i)=i+2 \implies f^{19}(n-36)+f(n-36)=98(n-36)+232$

but from induction hypothesis we have: $f(n-36)=n-34$ and

$f^{19}(n-36)=f^{18}(n-34)=f^{17}(n-32)=...=f(n)$ and now substituting the values we see:

$f^{19}(n-36)+f(n-36)=f(n)+97(n-34)=98(n-36)+232 \implies f(n)=n+2$

 and induction step in complete


Suppose $ n$ cuts are made across a pizza. Prove that the maximum number of pieces which can result is $ p_n \equal{} \binom{n}{0} \plus{} \binom{n}{1} \plus{} \binom{n}{2}$

it's the same as the maximum number of regions into which $ n$ lines divide the plane, which is $ 2\plus{}2\plus{}3\plus{}\cdots\plus{}n\equal{}1\plus{}\frac{n(n\plus{}1)}{2}\equal{}\binom{n}{0}\plus{}\binom{n}{1}\plus{}\binom{n}{2}$

If a plane is divided by $ n$ lines into maximum number of regions, there are $ \binom{n}{2}$ pieces that have 'deepest point'. 'Deepest point' means a point of region that is closest to $ \minus{}\infty$ of given direction. Let's draw a line in $ \minus{}\infty$ to count number of regions that don't have the deepest point. The new line passes through $ n$ lines that were originally drawn because in $ \minus{}\infty$, the original $ n$ lines don't pass through each other. So, there are $ n\plus{}1$ regions that don't have the deepest point. $ n\plus{}1$ can be rewritten by $ \binom{n}{0}\plus{}\binom{n}{1}$, so we get the maximum number of region is $ \binom{n}{0}\plus{}\binom{n}{1}\plus{}\binom{n}{2}$.

Let's assume that a plane is divided by $ n$ lines into maximum number of regions, and draw a new line. If this line passes through all $ n$ lines that were originally drawn, these $ n\plus{}1$ lines divide the plane into maximum number of regions. So, we get the inductive form, $ p_{n\plus{}1}\equal{}p_n\plus{}n\plus{}1$.

$ p_n\equal{}p_1\plus{}\sum_{k\equal{}1}^{n\minus{}1}(k\plus{}1)\equal{}\frac{n^2\plus{}n\plus{}2}{2}\equal{}\binom{n}{0}\plus{}\binom{n}{1}\plus{}\binom{n}{2}$.

How about cutting a cake instead of pizza ?
Suppose $n$ cuts are made across a cake. What is the maximum number of pieces which can result?

Generally, how can we prove that the maximum number of regions $x$ $(n - 1)$-dimensional things can "cut" an $n$-dimensional thing into is the sum of the first $(n+1)$ numbers in the the $x$ th row of pascal's triangle. (with the top row being row $0$) ?

Thanks.


function
calculus
integration
calculus computations

1. evaluate: integral from 4 to 6  dx/{(x-3)[(x^2 - 6x + 8)^1/2]}

2. set f(x)=arctan [(a+x)/(1-ax)] , x doesn't equal 1/a

a) show that f'(x)=1/(1+x^2), x doesn't equal 1/a
b) show that there is no constant C such that f(x)=arctan x + C for all x doesn't equal 1/a
c) find constants C_1 and C_2 such that:
    f(x)=arctan x + C_1   for x < 1/a
    f(x)=arctan x + C_2   for x > 1/a

Question 1 has been answered [url=http://www.mathlinks.ro/viewtopic.php?t=261062]here[/url].

Question 2a):
$ f(x) \equal{} \arctan\left(\frac{a\plus{}x}{1\minus{}ax}\right), x \neq \frac{1}{a}$.

$ f'(x) \equal{} \frac{1}{1\plus{}\left(\frac{a\plus{}x}{1\minus{}ax}\right)^2}\cdot \frac{(1\minus{}ax) \plus{} a (a \plus{} x)}{(1\minus{}ax)^2}$

$ \equal{}  \frac{(1\minus{}ax)^2}{(1\minus{}ax)^2 \plus{} (a\plus{}x)^2}\cdot \frac{1\plus{} a^2}{(1\minus{}ax)^2} \equal{} \frac{1 \plus{} a^2}{1 \minus{} 2ax \plus{} a^2x^2 \plus{} a^2 \plus{} 2ax \plus{} x^2}$

$ \equal{}  \frac{1 \plus{} a^2}{1 \plus{} a^2 \plus{} x^2 \plus{} a^2x^2} \equal{} \frac{1}{1 \plus{} x^2}$.

Maybe show us what you have done?

I've done nothing about this yet, because i have not clue what to do. looking for a full solution to the question.

Then probably the first thing you should do is go look at your course notes and textbook, read the text carefully, and pay special attention to worked examples that they include.  Once you have some clue of how to proceed, you'll be in a much better position to begin understanding the question and its solution.

In order to understand question 2, it will really, really help you to draw pictures of the functions in question.


What was with the IMO a long while back? Here is Problem 1 from the IMO 1959:

Prove that (21n+4)/(14n+3) is irreducible for every natural number n.

Now, I know this was a #1, but seriously, this takes about 20 seconds of applying a single technique in a very straightforward manner.

Remember -- number theory was largely unknown back then, at least to high school students.

Yes, that problem is not difficult. Many IMO problems from the early years were not so hard. I think that after more countries joined, they decided to make the contests harder.

Anyone want to write a formal proof just as an example for me to see how proofs are to be written?

A rational number a/b is irreducible if (a,b)=1. (a,b)=1 iff there exist integers x,y so that ax+by=1. Since -2(21n+4)+3(14n+3)=1, (21n+4)/(14n+3) is irreducible for every natural number n.


If Sally can paint a house in 4 hours, and John can paint the same house in 6 hour, how long will it take for both of them to paint the house together?

If any one can help and show how you got the answer. It would be appreciated. Thanks

Sally can paint 1/4 of a house in one hour, while John can paint 1/6 of a house in one hour. Together, they can paint $ \frac{1}{4}\plus{}\frac{1}{6}\equal{}5/12$ of a house in one hour. Thus, it will take them $ 1 \div \frac{5}{12}\equal{}2.4$ hours.

[hide="A Simple Formula"]Note that if we are asked to find how long it will take if $ k$ people with rates $ r_1,r_2,\dots,r_i$ work together to do a job, the answer is simply, $ \dfrac{1}{\frac{1}{r_1}\plus{}\frac{1}{r_2}\plus{}\dots\plus{}\frac{1}{r_i}}$(why?).[/hide]

[quote="pytheagle"][hide="A Simple Formula"]Note that if we are asked to find how long it will take if $ k$ people with rates $ r_1,r_2,\dots,r_i$ work together to do a job, the answer is simply, $ \dfrac{1}{\frac {1}{r_1} \plus{} \frac {1}{r_2} \plus{} \dots \plus{} \frac {1}{r_i}}$(why?).[/hide][/quote]

[hide="Proof of that"]

In one hour, they will do $ \sum_{i\equal{}1}^k \frac{1}{r_i}$. So, if they do x part of (x<1) job in 1 hour, they will do complete in 1/x hours, so we want the reciprocal of the above to express the time [/hide]


Hello.
In this circuit when a $ V_i$ is applied wouldnt the diode be reversed biased because the current leaves the + terminal and enters the - terminal?

But in the soln. they have sketched $ V_0$ as

[img]http://i44.tinypic.com/11i1wy8.jpg[/img]

This is the question

[img]http://i41.tinypic.com/ic5mh3.jpg[/img]

Well, it actually is in reverse polarization:

While the input voltage is positive, the diode is in OFF state, hence the voltage on its ends is simply the voltage on the ends of $ V_i$ source (you have a break in the circuit). That is the half of the sinusoid shown on the sketch. When the input voltage falls under zero (that would be the next half of the sinusoid) diode is on, and the circuit is short-circuited (diode is ON), so the ends of the diode are on the same potential, hence the output voltage is zero, as shown.

What is the problem?

thanks for reply

how did u know that it is in reverse polarization??

Well, diode is working in reverse polarization if the cathode is on a higher potential than the anode (that's the so-called ideal-diode model, in which the current flows through the diode for every positive voltage on it, opposed to a CVD model where the diode gets directly polarized when the difference of potential on its ends grows bigger than 0.6 or 0.7 V).

Intuitive explanation: if the current should go through the diode opposite to the direction diode is pointing, then the diode is OFF /i.e. it is in reverse polarization/.


Let $a,b$ is integer.Prove that there exist $p$ is prime and $n$ is integer such that $p=a+nb+1$

Proofs for the case of $a$ a multiple of $b$ are already on the forum (the easiest case of Dirichlet's theorem on primes in arithmetic progression).


superior algebra
superior algebra unsolved

Let K be a field of characteristic $ p \geqslant 2$ and $ m,n$ two positive integers with $ (m,p)\equal{}1$.           Let$ A,B \in M_{n} ( k )$ with the property $ A^{p^m}\equal{}A$ and $ B^{p}\equal{}B\plus{}I_{n}$.Prove that exist's $ X \in M_{n}( k)$ that $ AX\plus{}XB\equal{}I_{n}$.


Find all n that for each couple of prime divisors  of n a and b, a + b -1 divides n.


vector
complex analysis
complex analysis unsolved

If $ {\cal B}(X,Y)$ is Banach $ \Rightarrow Y$ is Banach.

Fix a nonzero vector $ x\in X$ and a complementary subspace $ Z\subset X$. For $ y\in Y$ let $ L_y\in \mathcal B(X,Y)$ be the operator such that $ Z\equal{}\ker L_y$ and $ L_y(x)\equal{}y$. This shows that $ Y$ is isomorphic to a closed subspace of $ \mathcal B(X,Y)$.


AMC AIME USA(J)MO USAMO Gauss algebra polynomial

okay, for some reason I cant reply to Bill's petition so I am going to do it here.  There is only a few of us Canadians, namely Tom, Bill, kool_dude, and me.  Some of my friends occasionally comes here but they are more interested in contests like AIME and USAMO rather than canadian parts so there is absolutely no point in creating a new section.

anywayz for the Canadian contests, there are a lot of them
Gauss for grade 7,8
Pascal for grade 9
Cayley for grade 10
Fermat for grade 11

all of them are multiple choice (25 questions, like the AMCs)

if you do well on them, you get invited to second contest
I think they are called Fryer, Galois, Hyptia (for grade 9,10,11)
they are full solution, about five questions.  There are room for proof type questions rather than straight calculations

for grade 12 there is the Euclid as well as the COMC (Canadian Open Math Challenge, for all grades), (and when we had grade 13 we have the Descartes)   These are partially answer (an answer will give you full credit but partial marks may be alotted if you get the wrong answer but you show your work) and partially full solutions (olympiad style).

Real olympiad contest/selection includes the CMO (Canadian Math Olympiad)
there is also a contraversial take-home unofficial contest Time Limited Exam - TLE (you are suppose to time yourself in 3 hours times two days... but howmany of them actually does is a question.  our school is the only place where we are proctored in TLE)  it is used to pick a olympiad training camp (15 ppl) as well as USAMO/APMO qualifications (very high stakes for a take-home contest).  you actually get the contest and you are suppose to submit it in 15 days (to those of you naive enough to believe me that the 3hr deadline is followed, not so... several ppl have already told me that they spent more time on it... of course I wont reveal who though)  Also the equilvalent of MOSP is MOCP (math olympiad correspondence program)

for computing contests we have the Canadian Computing Competition as well as a stage 2 (sorta like a camp)

all info can be found either in:
http://www.cemc.uwaterloo.ca
or
http://www.cms.math.ca/Competitions

CMO in its current format is definately easier than USAMO...

but having seen all of the past contests for all the non-olympiad ones, I believe that old Descartes are very interesting (I perfer them over the AIME -- but AIME isnt bad too)
In fact, some topics covered are: pell's equations, characteristic polynomial, generating functions, bijection, pigeon hole, tough geometry problem, locus, conic sections, inequalities, combinatorics.... it almost sounds like an olympiad competitions doesnt it!!  AIME just doesnt have all these

And the CMO is not really hard compared to American contests...

I think it is the same level as an AIME...

that's what the people at this forum say though...

no, AIME is pre-olympiad, CMO is olympiad... there is a significant difference...

the first 20 questions in Pascal and Cayley this year are trivial.  I mean you could just look at the question and figure out the answer.  the part Cs are harder

Are you sure that's not experience speaking? ;)

[quote="Caroline"]Do you know if there is any way for an American to participate legitimately in any of these?[/quote]

In the results booklets, I always see a section for International contestants. So I suppose that you could participate, but I'm not sure about the award structue for that.

The Waterloo contest website is [url]http://cemc.uwaterloo.ca/[/url], check it out yourself.

The website that billzhao pointed out is practically the main website of Canadian Contests since we don't have many here.

Also, I think there are some places that hold Tournament of Towns twice a year.

Not sure about that though.

Yeah, Toronto is one of them.
Four times to be exact. Spring Open, Spring Advanced, Fall Open, and Fall Advanced.

[quote="kool_dudy"]Also, I think there are some places that hold Tournament of Towns twice a year.

Not sure about that though.[/quote]

Tournament of Towns is indeed a contest that happens twice a year.  But every round (one round in fall, the other in spring) there are two levels: the ordinary level and the advanced level.  Also, there are the senior level and the junior level to choose from. (So it's actually four contests in total)  The contest consists of around 7 questions, and they are solution based.  The individual contests last quite long, from 4h to 5h I think.  And only specific universities in specific cities have enrolled in the contest, for example, U of T.

Btw, aren't Fryer, Galois, Hypatia those three contests that are really easy?  I don't think they need any invitation.  Although doing well in Pascal, Cayley and Fermat can get you into regional camps or waterloo seminars...

I would also like to know what other contests are present in Canada?

There are some regional contests, such as
[url=http://www.math.ucalgary.ca/education/JrMath/]Calgary Junior High School Math Contest[/url]
Manitoba Mathematics Contest
[url=http://www.math.ualberta.ca/~ahsmc]Alberta High School Mathematics Competition[/url]
[url=http://ace.acadiau.ca/math/MMC/mmc.htm]Maritime Mathematics Competition[/url]
[url=http://www.math.mun.ca/index.php3?content=includes/outreach/blundon.html]J.B. Blundon Math Contest[/url]

And probably many more.
CNML is also a contest accessible to Canadians.

Do you know anyones that are situated in Toronto?

I wrote the 2002 Fermat.
LMAO Rep, the C section is REALLY REALLY REALLY hard.

i have never written AMC 10 before... but i believe Fermat questions, the first few are much much easier than AMC 10 questions and the last few are harder than AMC 10 questions (judging from recent years)

[quote="Rep123max"]Oh, I forgot to ask. What are the qualifying scores usually when you take Fermat?[/quote]

Fermat doesn't qualify you for anything.

If you're a regional winner (in a metropolis, it usually means a perfect score), I thinkyou would get $200 and invitation to Waterloo Seminar.

Btw, the results and stats for the Fermat contests are on the Waterloo website.

fryer, galois, hypatia r intended for ppl who score top 25% of the multiple choice ones, u have to ask ur teachers about them to see if they r organizing them.

btw if you can get the first 19/20 questions perfectly correct, then you probably have placed in the top 25%

Strangely, in my school, anyone that took an enriched math class could take it.  One student scored 88 on Pascal and still managed to take Fryer.

[quote="Rep123max"]Then how do you take the Fryer and Hypatia and Galois? (what I thought was the Canadian equivalent of AIME).[/quote]
I don't think Fryer and Hypatia and Galois are equivalent of AIME at all, since they are [i]a lot[/i] easier than AIME, or AMC 12, or maybe even easier than AMC 10.
Besides, they don't need any invitation.  Everybody are welcome to write them.

That's what I thought too.  F/G/H aren't any invitational contests.  I think anyone can write them because this sort of case happened at my school.

Fryer and Galois are indeed very easy contests but Hypathia might take a bit more nerves to solve it.

they r recommended to ppl who score 25%, of course other ppl can writed, but they r INTENDED for top 25%


limit
logarithms
real analysis
real analysis solved

yeah...just use l'Hpital to get that it equals $\frac{a_1+...+a_n}{a_1\log{a_1}+...+a_n\log{a_n}}$. since all numbers have to be positive, the numerator is not zero. if the denominator is zero, then there is no limit.

Peter


group theory
abstract algebra
superior algebra
superior algebra unsolved

[quote="MeKnowsNothing"]Let $f: G\to H$ be a surjective group homomorphism between finite groups. Show that the image of a Sylow subgroup of $G$ is a Sylow subgroup of $H$.[/quote]
Here's a rather messy way to approach it.

Let $f: G\mapsto H$ be a monomorphism (surjective homomorphism). Then the monomorphic image of a p-subgroup is a p-subgroup.

Let $P\leq G$ be a p-subgroup (the orders of all the elements are powers of $p$). Then $f[P]$ is indeed a subgroup of $H$ because the homomorphic image of a subgroup is a subgroup. However, is it also a p-subgroup of $H$? We need to show that elements have prime power orders. Say $x\in f[P]$ then, $x = f(y)$ for some $y \in P$. But then, $x^{p^{k}}= f(y)^{p^{k}}$ where $p^{k}$ is order of $y$ (since its a p-subgroup). So $x^{p^{k}}= f(y^{p^{k}})=f(e)=e'$.

Now we need to show that $f[P]$ is not strictly contained in any other p-subgroup. Say $f[P]\leq K$ where $K$ is a p-subgroup of $H$. Then ${f^{-1}[f[P]}=P \leq f^{-1}[K]$  but $f^{-1}[K]$ is a p-subgroup of $G$. Therefore, $P$ is contained in another p-subgroup, since its a ME-subgroup it tells us that (its maximal) $P=K$. So its a Sylow p-subgroup.

[quote="wcdtd"]I just have a small correction in terminology. A monomorphism in a injective homomorphism in the category of groups. I think you meant an epimorphism, which is a surjective homomorphism in the category of groups.[/quote]

Actually that what is what I meant! I began doing the problem with injective homomorphism, then realized it was stupid and deleted everything except for my first sentence which I should have corrected. Thanks.


Prove the area of a convex quadrilateral is:
\[ \frac{1}{4}\sqrt{4e^2f^2-(a^2+c^2-b^2-d^2)^2} \]
where $a,b,c,d$ are the side lengths of the quadrilateral and $e,f$ are the lengths of the diagonals.

Let ABCD be a convex quadrilateral, $\{I\}=AC\cap BD,\ a=AB,\ b=BC,\ c=CD,\ d=DA$,and $\ x=m(\angle AIB)$. Denote S= the area of ABCD.

We have: $S=\frac{ef\sin x}{2}$ (sum of the areas of triangles IAB, IBC, ICD, IDA), hence
(1)  $S^2=\frac{e^2f^2(1-\cos^2 x)}{4}$.
Then express $a^2, \ b^2, \ c^2, \ d^2$ by cos theorem in the same four triangles and obtain:
(2)  $b^2+d^2-a^2-c^2=2ef\cos x$.
From (1) and (2) we easily get the requested relation.


can we say that the sum 1/(c_1-1) +1/(c_2-1)+...+1/(c_n-1) is bounded if 1/c_1+1/c_2+...+1/c_n is bounded where each c_i is greater than 1?


[quote="futurus"]Given a circle (O;R). Let M be a point outside the circle, and let MB, MC be 2 tangent lines to (O;R). Draw a ray Mx between MO and MC. From B, draw a line parallel to Mx and intersects (O;R) at A. Let I be the intersection of Mx and AC.

Prove that M, O, I, C are on the same circle.

Thanks.[/quote]
[hide="Solution"]
Since $ MC$ is a tangent to the circle and $ AB\parallel MI$, we have that $ \angle MCB \equal{} \angle BAC \equal{} \angle MIC$, so $ MICB$ is cyclic.  This  implies that $ I$ lies on the circumcircle of $ \triangle MBC$.  Yet,
\[ \angle OCM \equal{} 90 \equal{} \angle OBM\implies \angle OCM \plus{} \angle OBM \equal{} 180
\]since $ MB$ and $ MC$ are tangents to the circle.  Thus, $ OCMB$ is cyclic, so $ O$ lies on the circumcircle of $ \triangle MCB$.  This means that $ O$ and $ I$ lie on the circumcircle of $ \triangle BMC$, so $ MCIOB$ is cyclic, so $ MCIO$ is cyclic. [/hide]


geometry
cyclic quadrilateral
complex numbers
IMO Shortlist
IMO Longlist

Let $a,b,c,d$ be the complex numbers corresponding to $A,B,C,D$ respectively. Assume the origin of the plane is $O$, the center of $(ABCD)$. Then $|a|=|b|=|c|=|d|\ (*)$. Let $g_a=\frac{b+c+d}3,g_b=\frac{a+c+d}3,g_c=\frac{a+b+d}3,g_d=\frac{a+b+c}3$ be the affixes of the centroids. We can then see that if $p=\frac{a+b+c+d}3$, then $|p-g_a|=|\frac a3|,|p-g_b|=|\frac b3|,|p-g_c|=|\frac c3|,|p-g_d|=|\frac d3|$, and combined with $(*)$ this yields $|p-g_a|=|p-g_b|=|p-g_c|=|p-g_d|$, so $G_a(g_a),G_b(g_b),G_c(g_c),G_d(g_d)$ lie on a circle having $P(p)$ as center.

[quote="orl"]Let $ABCD$ be a quadrilateral inscribed in a circle. Show that the centroids of triangles $ABC,$ $CDA,$ $BCD,$ $DAB$ lie on one circle.[/quote]

The problem is trivial!

In fact, if A', B', C', D' are the centroids of triangles BCD, CDA, DAB, ABC, respectively, then we have to prove that these centroids A', B', C', D' lie on one circle. Now, if G is the centroid of all the four points A, B, C, D, then the mass-regrouping theorem (I guess the name is wrong) says that e. g. the point G lies on the segment AA', and divides this segment in the ratio AG : GA' = 3. Thus, GA' : AG = 1/3, so that GA' : GA = -1/3. This means that the homothety with center G and factor -1/3 maps the point A to the point A'. Similarly, the same homothety maps the points B, C, D to the points B', C', D'. And since the points A, B, C, D lie on one circle, it is clear that their images under this homothety must also lie on one circle, and thus we see that the points A', B', C', D' lie on one circle. Problem solved.

  Darij


I want to apply to colleges for fall 2008 and the deadine for applications is the end of December 2007. I need to apply for the FAFSA but the website says that I need a valid social security no. which I don't have (I have been living outside the US for a very long time). According to the US embassy, it might take me almost [i]6 months [/i]to get a social security no. Will I be eligible for the FAFSA if I state that I have applied for a SSN and would be getting it in a few days? (in any case, I will have it by February).

Best thing to do: call or write the FAFSA. They're the only ones who can give you a definitive answer.


Prove that for any three different rational number $a,b,c$, the quantity \[\sqrt{\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}}\] is always a rational number

The given expression is equal to $\sqrt{\frac{(a^2+b^2+c^2-ab-ac-bc)^2}{((a-b)(b-c)(c-a))^2}}$. The expression inside the square root is a perfect square.


calculus
integration
trigonometry
parameterization
logarithms
real analysis
real analysis unsolved

What's "$coslx$"? Is that a typographical error, and if so, what should it say?

If it is $\cos x$

$\int_{0}^{+\infty}\frac{\sin^3x\cos x}{x^3}dx= \frac{\pi}{4}$

Yes, I've got $\frac{1}{2}\int_0^\infty \frac{\sin t}{t}dt=\frac{\pi}{4}$.

In general case I obtained
1) $\frac{\pi}{16}(k^2+9-6k)$ for $1\leq k\leq 2$;
2) 0 for $k>2$ (though my approximative computations using MathCad don't confirm it with full guarantee).

[quote="Myth"]

In general case I obtained
1) $\frac{\pi}{16}(k^2+9-6k)$ for $1\leq k\leq 2$;
2) 0 for $k>2$ [/quote]

Which expression for general case Myth ?
thx

General case is $\int_{0}^\infty\frac{\sin^3 x\cos kx}{x^3}dx$.

[quote="eugene"]...$\int_{0}^{+\infty}\frac{sin^3xcoslx}{x^3}dx$...[/quote]

I see  "l" is a parameter, I thought it was an error typo  I wonder if eugene wrote
$\cos{\ln x}$

What about the nature of  $\int_{0}^{+\infty}\frac{sin^3xcos\ln x}{x^3}dx$ ?

Do you think it is interesting question?  :?


geometry
circumcircle
projective geometry
power of a point
geometry proposed

Let $ABCD$ be a cyclic quadrilateral. The lines $AB$ and $CD$ meet at $P$, and the lines $AD$ and $BC$ meet at $Q$. Let $E$ and $F$ be the points where the tangents from $Q$ meet the circumcircle of $ABCD$. Prove that points $P,E,F$ are collinear.

Is this not just Pascal's Hexagon with two sides of zero length?

edit: no, it's not  :blush:

I must admit I'm not familiar with Pascal's hexagon. Could you be more clear?

Anyway, this is a trivial problem. Here's my "solution":

$EF$ is the polar of $Q$ with respect to the circle. Because $P$ lies on the polar thanks to a famous lemma, so we are done.

I misread the question, sorry about that. Some stuff about Pascal's Hexagon can be found at http://en.wikipedia.org/wiki/Pascal%27s_theorem

I know the theorem, but I have never seen someone apply it. Thanks tim1234133

Now, how would you prove the lemma I was talking about? It's the following:

Given a circle $\omega$ and a point $P$. Two lines through $P$ intersect $\omega$ in $B$ and $C$, and $D$ and $E$ respectively, as shown in the image. $CD$ and $BE$ intersect in $F$, and $CE$ and $BD$ intersect in $G$.
Prove that $FG$ is the polar of $P$ with respect to $\omega$.

Here's what I would do:

The harmonic conjugate of $P$ wrt $D$ and $E$ lies on the polar of $P$ wrt $\omega$, but also on the polar of $P$ wrt to the two intersecting line $GE$ and $GD$. Similar for the harmonic conjugate of $P$ wrt $B$ and $C$.

The polar of a point wrt a circle is a line, and the polar of a point wrt two intersecting lines is also a line.
Since they have two points in common, they are identic here. We know that $G$ lies on the polar of $P$ wrt the two intersecting lines, so it also lies on the polar of $P$ wrt $\omega$

I don't like this result, could someone give a better proof, perhaps more 'pure'?

Also prove it for $P$ inside the circle.

Using your diagram, let O be the center of the circle. Then let the circumcircles of OBC and ODE meet at X. Since we have $\angle CXE = 360-\angle OXC-\angle OXE = \angle OBC+\angle ODE = \angle CFE$, we have that CXFE is cyclic. Symmetrically, BXFD is cyclic.

Now by radical axes, O, X, and P are collinear. Also by radical axes, X, F, and G are collinear. Furthermore, a fairly simple angle chase reveals that $\angle OXF = 90$, so it would suffice to show that $OX \cdot OP = R^{2}$, where R is the circumradius. This is true, because $\angle OCP = 180-\angle OCB = 180-\angle OBC = 180-\angle CXP = \angle OXC$.

I hope this proof works....

let $I$ be an inversion about the circle, $\omega$, centered at $Q$ passing through $E$.

let $PE\cap \omega=F'$ (where $F'\ne E$), (it is a little sketchy if this exists)

let $I(P)=P'$

note $QE^{2}=QP*QP'$ by definition of inversion, then $QF'=QE$, so $QF'^{2}=QP*QP'$, so $\frac{QF'}{QP'}=\frac{QP}{QF'}$, then $\triangle QP'F\sim\triangle QFP$ by SAS similarity, ($QEP'\sim QP'P$ also by SAS)

hence, $\angle QEP'=\angle QPF\angle QF'P'$, so $QP'FE$ is cyclic

notice that $I(D)=A$ and $I(C)=B$ by power of a point, and $I(P)=P'$ (by definition), since $P$, $C$, and $D$ are collinear, then $ABP'Q$ is cyclic

then $DC$, $EF'$, and $QP'$ are concurrent at $P$, and $QP'FE$, $QP'AB$ are cyclic, so $ABEF'$ is cyclic

since $QE=QF'$, then $QF'$ is also tangent to the circle by the converse of power of a point, so $F\equiv F'$

since $EF'P$ is a line, $EFP$ is a line as desired

[quote="Altheman"]I hope this proof works....

let $I$ be an inversion about the circle, $\omega$, centered at $Q$ passing through $E$.

let $PE\cap \omega=F'$ (where $F'\ne E$), (it is a little sketchy if this exists)

let $I(P)=P'$

note $QE^{2}=QP*QP'$ by definition of inversion, then $QF'=QE$
[/quote]
I think I am missing something, sorry. Why is $QF'=QE$ ?
If this is true you are done, because then $F'=F$

i defined $F'$ as the intersection of [the circle centered at $Q$, radius $QE$], and [$PE$]

i just have to prove that $F'$ lies on $ABCD$, and $QF$ is tangent to that circle

I'm sorry, i misread your definition of $F'$.

Cool proof!  :)


[quote="rd5493"][hide="Answer:"]4
It's just a simple pattern of adding and subtracting.[/hide][/quote]

Why can't the answer be 8?

[quote="AstroPhys"]$1+5= 6$
$6-2=4$
The pattern is like that.[/quote]

Oh, I see... :P

[quote="sharkman"]What number should replace the question mark?

9,7,2,5,7,4,3,2,5,1,4,5,9,8,1,5,6,3,3,4,7,2,[b]5,3,3[/b],8,2,6,3,9,4,5,2,7,5,2,7,9,8,1,5,6,2,?[/quote]

According to your pattern, this is not true.(the bold part is not the same as the pattern)

[quote="AstroPhys"]This answers is : $4$.
Though there appears to be a mistake [quote]....5,3,3,8,....[/quote] ,there should only be one three, otherwise it breaks the pattern.[/quote]
vietkhoa that was already stated.

[hide]the answer is gonna be 4 becuase you just add or subtract the number 2 spaces from that.[/hide]


trigonometry
inequalities proposed
inequalities

[b]Problem 1:[/b]
Let $ a,b,c\ge 0$ such that:
\[ a^2\plus{}c^2\equal{}1,b^2\plus{}2b(a\plus{}c)\equal{}6\]
Prove that: $ b(a\minus{}c)\ge 4$

[b]Problem 2:[/b]
Find the maximum value of:
\[ N\equal{}(\cos{x})^a(\sin{x})^b\]
For $ 0\le x \frac{\pi}{2},a,b\ge 1$

[b]Problem 3:[/b]
Let $ x^3\plus{}ax^2\plus{}bx\plus{}d\equal{}0$ have 3 roots. Prove that:
\[ (27c\plus{}2a^3\minus{}9ab)^2<4(a^2\minus{}3b)^3\]


superior algebra
superior algebra solved

let F be subfield of C.and p(x) be poly. having coeff. from F and is also irreducible in F[x].then show p(x) cant have multiple roots in C. :)
(C-complex nos.)

Fields with characteristic 0 are perfect...


Prove that in a $ ABC$ triangle we have the relation:

$ \boxed{a\tan\frac A2\plus{}b\tan\frac B2\plus{}c\tan\frac C2\equal{}4R\minus{}2r}$.

hello, your equation is equivalent to
$ r\left(\frac{a}{s-a}+\frac{b}{s-b}+\frac{c}{s-c}+2\right)=\frac{abc}{A}$
Setting $ r=\frac{A}{s}$ we get
$ {\frac{a(s-b)(s-c)+b(s-a)(s-c)+c(s-a)(s-b)+2(s-a)(s-b)(s-c)}{A}=\frac{abc}{A}}$
after some algebra.
A nice problem.
Sonnhard.

My solution

Using sin theorem we get:

$ a\tan\frac{A}{2}+b\tan\frac{B}{2}+c\tan\frac{C}{2}=\sum 2R\sin A\cdot \tan\frac A2=4R\sum \sin^2\frac A2=4R(\frac{1-\cos A+1-\cos B+1-\cos C}{2})=2R(3-\sum \cos A)=2R(3-1-\frac rR)=4R-2r$.

[quote]Prove that in $ \triangle ABC$ there is the identity $ \boxed{a\cdot\tan\frac A2 \plus{} b\cdot\tan\frac B2 \plus{} c\cdot\tan\frac C2 \equal{} 4R \minus{} 2r}$ . [/quote]
[b][u]Method 1[/u].[/b] $ \left\|\begin{array}{cc} r_a \equal{} s\cdot\tan\frac A2 & (1) \\
 \\
\ ar_a \equal{} s\left(r_a \minus{} r\right) & (2) \\
 \\
r_a \plus{} r_b \plus{} r_c \equal{} 4R \plus{} r & (3)\end{array}\right\|\implies$ $ \sum a\cdot\tan\frac A2\ \stackrel {(1)}{ \equal{} }\ \sum \frac {ar_a}{s}\ \stackrel{(2)}{ \equal{} }\ \sum\left(r_a \minus{} r\right)\ \stackrel{(3)}{ \equal{} }\ 2(2R \minus{} r)$ .

[b][u]Method 2[/u].[/b] $ \left\|\begin{array}{ccc} S \equal{} s(s \minus{} a)\tan\frac A2 & (4) \\
 \\
S \equal{} sr \equal{} (s \minus{} a)r_a & (5)\end{array}\right\|\implies$ $ \sum a\cdot\tan\frac A2\ \stackrel{(4)}{ \equal{} }\ S\cdot\sum\frac {a}{s(s \minus{} a)} \equal{}$ $ \sum\left(\frac {S}{s \minus{} a} \minus{} \frac {S}{s}\right)\ \stackrel{(5)}{ \equal{} }\ \sum\left(r_a \minus{} r\right) \equal{} 2(2R \minus{} r)$ .

[b][u]Applications[/u].[/b] $ \sum\frac {a}{r_a}\ \ge\ \frac {2s}{2R \minus{} r}$ . Define $ S_n \equal{} \sum a^n\cdot\tan\frac A2\ ,\ n\in \mathrm N$ . Then $ \boxed {\ S_{n \plus{} 1} \equal{} s\cdot S_n \minus{} r\cdot\sum a^n\ ,\ n\in \mathrm N\ }$ .


analytic geometry
geometry
parallelogram

The coordinates of a parallelogram are (5, 3), (6, 8), (7, 4) and
$ (x, y)$ and $ x > 7$. What is the value of $ x \plus{} y$?

5,3 and 7,4 are endpoints of a side
To get from 5,3 to 7,4 -------> (x+2,y+1)
To find the other endpoint of the side containing 6,8 we do the same
So we get 8,9.
We know it isn't 4,7 because x>7 although it also makes it a parallelogram
But our answer isn't 8,9
They asked for the sum, so 17 :D

great answer, i'll just LaTeXify it for you

$ 5,3$ and $ 7,4$ are endpoints of a side
To get from $ 5,3$ to $ 7,4$ -------> $ (x\plus{}2,y\plus{}1)$
To find the other endpoint of the side containing $ 6,8$ we do the same
So we get $ 8,9$.
We know it isn't $ 4,7$ because $ x>7$ although it also makes it a parallelogram
But our answer isn't $ 8,9$
They asked for the sum, so $ 17$ :D


Demonstrati Inegalitatea mediilor(Inegalitatea lui Cauchy pentru medii) in mod geometric.

Doua, trei sau mai multe variabile pozitive ?

Va raspund cam tarziu.Demonstrati pentru doua variabile pozitive.


geometry
trigonometry
Yale
college
analytic geometry
function
number theory

I'm trying to get as many math topics and things to my head. I have geometry and trigonometry now but I think since I'm starting precal next year, I would be pretty busy that time. So, I'm trying to get the topics that older or other people who've been my level think should know on my level.

Good suggestions please!

Thank you :D

u got algebra 2 rite?
and u dont need anymore if u got tat covered

Nonsense; one can always use more math knowledge.

Which field thereof, however, depends on your interests.

I would learn to call precalculus by the term "precalc" (with a 'c' at the end) for short, which is definitely the majority way to say it in American English.

A good review book of all the mathematics a high school student should know is [url=http://www.amazon.com/exec/obidos/tg/detail/-/0387967877/learninfreed]Basic Mathematics[/url], by Yale Professor Serge Lang.

Also very good for learning secondary mathematics from a higher viewpoint is anything by I.M. Gelfand, for example [url=http://www.amazon.com/exec/obidos/tg/detail/-/0486425657/learninfreed]The Method of Coordinates[/url].

The Japanese mathematics textbook series translated into English and published by the American Mathematical Society, for example [url=http://www.amazon.com/exec/obidos/tg/detail/-/0821805827/learninfreed]Mathematics 2: Japanese Grade 11[/url] is also quite interesting.

There is plenty of mathematics to learn before calculus. It is regrettable that United States schools rush through that mathematics so cursorily.

Hope this helps!

P.S. I would never say anyone has geometry understood unless that person can EASILY do every problem in AoPS volume 1 and volume 2.

P.P.S. Moderators, please move this thread to Other Problem
Solving Topics, where it belongs.

No one should be allowed to graduate high school without knowing how to find eigenvalues.

That seems somewhat arbitrary.  Should people, say, be able to graduate high school without being able to calculate the Hessian of a multivariable function?  Or apply the precise definition of a limit?  Or apply the axiom of choice?

Actually, come to think of it, you need to know how to find eigenvalues to actually get anything significant out of the Hessian.

But I don't think you can really say that about any one topic.  Frankly, I would rather have learned stuff like inversion in geometry than a lot of the 'higher' math stuff I learned well I was in high school.

If you know how to calculate eigenvalues, you already know everything you learn in most normal math classes in high school.

[quote="Silverfalcon"]Wait. You guys are like talking not so english  :? I was looking for certain terms of branch of math you know, like complex fraction, number theory, uh.. I don't know.

P.S. I definitely don't agree that you only need algebra I. To graduate and get good grade on SAT, you need to have more than that.[/quote]

First, he said Algebra 2.  I think you can graduate with nothing higher than Algebra as a minimum-you probably wont go to college but it s possible (at least at my school).

For the SAT, the math class you re in does not have as large an effect as you would think.  You can do go even without know alg. 2 or anything higher (except now you will need a little alg 2 knowledge).

True, true. You don't necessarily need anything higher than to pass it. Like some of my teacher said, "You can just take SAT Math now"

:)

Of course, I won't do it yet because I got to study more on literature part.


Suppose $ a_{1}=2$ and $ a_{k+1}= 3a_{k}+1$ for all $ k\geq 1$. Find a general formula for $ a_{1}+a_{2}+...+a_{n}$


I'd just do $a_{n}=-\frac{n+4}{\lfloor n/4 \rfloor+1}$ where $a_{0}= \frac{-4}{1}$

Trig is cool, though... trig is cool...

shaister, the trig one doesnt work, for 2nd term it give -3 instead of -5, it only works for the first term.

You can't divide by 0, so finding a general formula that includes the first four terms is a moot point anyway.

[quote="binomial_4eva"]is it impossible to find one without the [...] ?[/quote]

Yes. Consider $\frac{1+(-1)^{n}+i^{n}+(-i)^{n}}{4}$, which produces the sequence $1,0,0,0,1,0,0,0$

Now you can do an index-shift to make another sequence $0,1,0,0,0,1,0,0...$
Perform another index-shift and multiply by 2 to get $0,0,2,0,0,0,2,0...$
And another, but this time multiplying by 3, to get $0,0,0,3,0,0,0,3...$

Adding these gives the sequence $0,1,2,3,0,1,2,3,0,1,2,3...$

Dividing by $4$ will give a sequence of numbers you will have to subtract from the series $n/4$ to get the desired sequence of denominators.

[hide]

$-\frac{16(n+4)}{4n+10-[(-1)^{n-1}+2(-1)^{n-2}+3(-1)^{n-3}]-[(i)^{n-1}+2(i)^{n-2}+3(i)^{n-3}]-[(-i)^{n-1}+2(-i)^{n-2}+3(-i)^{n-3}]}$

[/hide]

Interesting note:  Shaister's approach and The Zuton Force's second approach are more or less equivalent; consider the identity

$i^{x}= e^{x \ln i}= e^{i \frac{\pi}{2}x }= \cos \frac{\pi}{2}x+i \sin \frac{\pi}{2}x$


Given a usual $8\times 8$ chessboard, we are allowed to interchange any two rows and any two columns. Can we get a board whose left half is white?

:)

I am sorry, maybe I don't understand something, but I think that every time we make an operation, the number of black squares on each row/column is still equal to that of the white ones........
So, considering this, it is impossible? :?

Exactly :). That's what the :) at the end of my previous message stands for. And here's another one: :).


Mathman rides $ \frac{3}{5}$ of a mile in 1$ \frac{1}{2}$ minutes on his bicycle. What is his average speed in miles per hour?

\[ \frac {\frac {3}{5}\text{ miles}}{1\frac {1}{2}\text{ minutes}}\times \frac {60\text{ minutes}}{1\text{ hour}} \equal{} \boxed{24 \text{ mph}}
\]


AMC
AMC 12
AMC 12 B
AIME
USA(J)MO
USAMO
MATHCOUNTS

If you took AMC12 2007 and AMC12 2008 please vote on how you did this year compared to last year

I got a 130.5 last year, and 124.5/126 this year.  I think the lack of calculators really slowed me down this year.

Last year (10th grade), I got a 114 on the 12A.  This year, I got a 135 and a 126 on the A and B respectively.  Huge improvement!  I got so much better from the math camp I went to over the summer...

Last year: $ 144/138$
This year: $ 139.5?/138$

The lack of calculators was probably balanced by my improvement in overall ability between the two years; I didn't perceive the tests to be any easier. I made the same amount of dumb mistakes both years (basically one on each of the four tests), so calculator didn't help there.

I got 150 on 10A and 120 on 12B last year, and I got 138/141 on 12 A/B this year. Big improvement, which I'm happy about.

last year

AMC 10A(the only one i took) 117

This year

AMC10A/B = 121.5/126

heh

last year 132 12A

this year 97.5 12A (assuming no bubbling errors, I haven't seen the official score)

i think i had the worst drop of anyone here

i remember last year's AMC12B. I took AMC12B. (I don't know how well i did yet, though....)
but one thing is definite. THE PROBLEMS WERE A LOT HARDER THIS YEAR!

126 AMC12B 2007 -> 123 AMC12B 2008.  Difference of -3; not too bad.  But unless I get more than 11 on the AIME, my USAMO index will be monotonically decreasing over the past three years.

uhh my AMC 12 score was like the same, but my AMC 10 score improved by 12 points.

I'm cursed to always get a lower score on competitions than the first time I do them seriously.
This happened to me at mathcounts states, nationals, chapters, and now it happened to me on the AMC.

I got the same score on both the A and B, one question less than my score on the AMC last year (which I only took once).

Last year: 138/144
This year: 150/nan

Improvement is always good.

I actually found this year's 12A to be rather enjoyable to take, even without a calculator. The only one that was not easy to do without a calculator on the A was #24.

Last year: 10A = 126 and 12B = 114
This year: 12A = 12B = 132

Last year: 126
This year: 141

I think I got a lot better, though.

Of course, I bombed the B's this year, but I'm willing to attribute that to the fact that I got lost on my way to the host school, barreled in late, forgot my watch, and had very little desk space.  :rotfl:

these were my first AMC's ever (i didn't even consider taking AMC 10 or 12 in 7th grade)
8th grade: 99 on 12A/ 127.5 on 10B

im kinda satisfied with my results even though i made more than a substantial share of mistakes on the 12 (shoulda had 114 :()
i guess this is ok progress after about a year of "real" competition math.
hopefully i can pwn the AIME!

AMC 12B 2007: 102 (or something really close to the cutoff)
AMC 12B 2008: 115.5 (grr 2 stupid mistakes on #2 and #3)

Overall, I improved in math immensely, since last year I didn't have a clue on how to solve some of the problems, and now, all the problems are actually doable if I had 45 minutes more time.

Exactly the same. 2007 AMC 12A: 112.5/2008 AMC 12A: 112.5.
Overall, I think I improved in math, but no calcs slowed me down, and I was really quite ill while I took the AMC this year.
I just hope I do a lot better on the AIME this year. And my AMC 10B score will help my index immensely.

My A score was higher by 3 points; my B score was lower by 3 points. ):

10A/12B:

2007 - 138/109.5 (ugh)
2008 - 138/127.5

yeah... failure at AMC10. Blah. And didn't score as hoped on 12.

150 last year for AMC12A
144 this year for AMC12B
no :stretcher: 16

Does anyone know if there is mock AIME or USAMO on the site. They should be fun

[quote="greentreeroad"]Does anyone know if there is mock AIME or USAMO on the site. They should be fun[/quote]Here are some:
http://www.artofproblemsolving.com/Wiki/index.php/Mock_AIME

does a improvement of 102 to 105 count?

despite my 3 pt improvement, I feel that i really am better at math this year

I have also noticed that I have gotten worse in accuracy at easy problems
so the +hard-easy evens out to the score being about the same :P

lol i deproved, but mostly everyone at my school deproved too (and alot of people here i think)...

But I did improve on my 10 score but thats not as impressive as the 12 (unless you get a perfect).