geometry
3D geometry
calculus
integration

How many different values can $ \angle ABC$ take, where $ A,B,C$ are distinct vertices of a cube?

[hide]There are three types of triangles that one can form with distinct vertices of a cube. They are $ 45\minus{}45\minus{}90$ triangles, $ 30\minus{}60\minus{}90$ triangles, and $ 60\minus{}60\minus{}60$ triangles. The amount of distinct values that angle $ ABC$ can take is $ \boxed{4}$.[/hide]

don't forget that the cube has diagonals.

On a cube with side length s, the shortest possible distance between any two vertices is s, while the largest possible distance is (s)(sqrt(3)). In a 30-60-90 triangle, the hypotenuse has twice the length of the shorter leg. Thus I do not think a 30-60-90 triangle can be formed by connecting 3 vertices of a cube.

I think that 0123456789 (not 123456789  :lol: ) is right.

[hide="Answer"]
If two of the points of an unit cube are connected by an edge, then the triangles formed have side lengths $ (1,1,\sqrt {2})$ and $ (1,\sqrt {2},\sqrt {3})$. Otherwise, we pick an arbitrary point, and we see that we cannot pick the diametrically opposite point, so the only possible triangle is $ (\sqrt {2},\sqrt {2},\sqrt {2})$. In total, these give $ \boxed{5}$ unique possible angles (remembering that there are 2 $ 90^{\circ}$, 2 $ 45^{\circ}$, 3 $ 60^{\circ}$ angles).
[/hide]

Edit: zolojetto, whose solution are you referring to?

i dont see how you got that

for some reason i agree with 0123456789 because the only angles that they can be are 90 and 45 i dont see how you could get a 60 or 30 degree angle using the verticies of a square

well there are 3 different triangles that can be formed
we'll assume its a unit cube (doesnt matter)

1*1*sqrt2
sqrt2*sqrt2*sqrt2
1*sqrt2*sqrt3

the first one we can have 45 and 90
the second one is equilateral so we have 60
the third one is a right triangle but the two non-right angles are not very "good" numbers but they are not 45,60, or 90 so we have 2 more

so we have $ 2\plus{}1\plus{}2\equal{}5$

we do not hvae a 30-60-90 but we havec a 60-60-60 to make up for the 60.  There is no 30, the other triangle has two non integral angles but they are still angles

oh wow i see now i thought it said square and then you could only have 45 and 90.  Wow i need to read the question ( again I phail3)

What? Wait but the angle formed by the space diagonal and face diagonal is $ arctan{\frac{\sqrt2}{2}}$, so we have that angle, its complement, plus the rest.

serialk11r, I think the angle you are refering to is an angle in the 1*sqrt2*sqrt3. This angle has already been counted.

[quote="0123456789"]How can a 30-60-90 triangle be formed from the vertices of a cube?[/quote]Sorry, I was stupid. For some reason, I thought $ 1\minus{}\sqrt{2}\minus{}\sqrt{3}$ triangle was a $ 30\minus{}60\minus{}90$ triangle... lol


function
analytic geometry
graphing lines
slope
absolute value

What is the minimum Value of |x+1|+|x-2|+|5-x| ?

Analyze the given expression:

Case 1: When $ x \ge 5$

$ |x \plus{} 1| \plus{} |x \minus{} 2| \plus{} |5 \minus{} x| \equal{} x \plus{} 1 \plus{} x \minus{} 2 \plus{} x \minus{} 5 \\
 \\
\equal{} 3x \minus{} 6$

Case 2:When $ 2 \le x < 5$

$ |x \plus{} 1| \plus{} |x \minus{} 2| \plus{} |5 \minus{} x| \equal{} x \plus{} 1 \plus{} x \minus{} 2 \plus{} 5 \minus{} x \\
 \\
\equal{} x \plus{} 4$

Case 3: When $ \minus{} 1 \le x < 2$

$ |x \plus{} 1| \plus{} |x \minus{} 2| \plus{} |5 \minus{} x| \equal{} x \plus{} 1 \plus{} 2 \minus{} x \plus{} 5 \minus{} x \\
 \\
\equal{} 8 \minus{} x$

Case 4:When $ x < \minus{} 1$

$ |x \plus{} 1| \plus{} |x \minus{} 2| \plus{} |5 \minus{} x| \equal{} \minus{} x \minus{} 1 \plus{} 2 \minus{} x \plus{} 5 \minus{} x \\
 \\
\equal{} 6 \minus{} 3x$

Now,draw the graph.Its easy to find the minimum value from the graph.

I think the answer is $ \minus{}6$

Do u mean drawing the graph of the 4 cases individually?

The correct answer here is 6  :idea: but how???

hello, the minimum of $ f(x)\equal{}|x\plus{}1|\plus{}|x\minus{}2|\plus{}|x\minus{}5|$ is $ 6$ and will be attained for $ x\equal{}2$, this is clear when we draw the graph of $ f(x)$.
Sonnhard.

Is there a way to know that x=2 w/o graphing? I dont know how to graph three absolute values.

Find the range of each of the expressions given the restriction on $ x$. Like so:

Case 1: $ x \ge 5\implies 3x \ge 15\implies 3x \minus{} 6\ge 9.$ Therefore, the minimum of the first section of the graph is $ 9$.

Case 2: $ 2 \le x < 5\implies 6 \le x < 9.$ Therefore, the minimum of the second section of the graph is $ 6$.

Case 3: $ \minus{}1 \le x < 2\implies \minus{}2 < \minus{}x \le 1\implies 6 < \minus{}x \le 9.$ Therefore, the maximum of the third section of the graph is $ 9$.

Case 4: $ x < \minus{}1\implies \minus{}3x > 3\implies 6\minus{}3x > 9.$

The minimum value throughout all the cases is $ 6$ as you can see above.

Generally speaking, $ |x \minus{} a| \plus{} |x \minus{} b|\ (a<b)$ is minimal when $ a\leq x\leq b$.

[b]Note:[/b] This is a very standard way of viewing sums of such absolute values, so I'll walk through it. It's basically an increasing/decreasing argument. It looks long, but very fast to read through (but, make sure you take the time to understand it if you don't get it right away).

Generally, it is helpful to be able to characterize functions by how they are increasing or decreasing, and by their concavity/convexity, even if you don't know how to (or don't care to) actually compute the exact rate of change (a derivative).

[hide="Walk Through"]You have the sum of three components: $ |x \minus{} ( \minus{} 1)|$, $ |x \minus{} 2|$, and $ |x \minus{} 5|$, which are absolute value-ish functions with vertices at -1, 2, and 5, respectively.

An absolute value function is decreasing with slope -1 until it reaches its vertex, and then increasing with slope 1 after that.

For x < -1:
|x-(-1)| is decreasing with slope -1
|x-2| is decreasing with slope -1
|x-5| is decreasing with slope -1
Thus, their sum is decreasing with slope -3

For -1 < x < 2:
|x-(-1)| is increasing with slope 1
|x-2| is decreasing with slope -1
|x-5| is decreasing with slope -1
Thus, their sum is decreasing with slope -1

For 2 < x < 5:
|x-(-1)| is increasing with slope 1
|x-2| is increasing with slope 1
|x-5| is decreasing with slope -1
Thus, their sum is increasing with slope 1

For 5 < x:
|x-(-1)| is increasing with slope 1
|x-2| is increasing with slope 1
|x-5| is increasing with slope 1
Thus, their sum is increasing with slope 3

We can see that the function in question (the sum of the three absolute values) is decreasing until it gets to x=2, and then increasing after that. So, the minimum occurs at x=2.[/hide]

[b]Follow-up:[/b] Let $ a_n$ be the minimum value achieved by $ \sqrt{\sum_{k\equal{}0}^{n} |x\minus{}k|}$. Find, with justification, the number of elements from $ a_1$ through $ a_{2009}$ that are integers.

You could also think of it on a number line and in terms of distances.
$ |x\plus{}1| \equal{} |x\minus{}(\minus{}1)|$  This is the distance between $ x$ and $ \minus{}1$.
$ |x\minus{}2|$ is the distance between $ x$ and $ 2$.
$ |x\minus{}5|$ is the distance between $ x$ and $ 5$.

Now you can plot the points -1, 2, and 5 on the number line, and you try to find the point $ x$ so that the sum of the distances is the smallest.

Starting with -1 and 5, since they are on the outside:
$ |x\minus{}(\minus{}1)|\plus{}|x\minus{}5|$ is smallest when $ x$ is between -1 and 5, inclusive. So the minimum value of $ |x\minus{}(\minus{}1)|\plus{}|x\minus{}5|$ is the distance between -1 and 5, or 6.
To minimize $ |x\minus{}2|$, you can make $ x\equal{}2$, so that $ |x\minus{}2|\equal{}0$.
Since 2 is between -1 and 5, $ |x\minus{}(\minus{}1)|\plus{}|x\minus{}5|$ is still 6.
$ |x\minus{}2|\equal{}0$, so $ |x\minus{}(\minus{}1)|\plus{}|x\minus{}5|\plus{}|x\minus{}2|\equal{}6$, when $ x\equal{}2$

Let $ y\equal{}|x\plus{}1|\plus{}|x\minus{}2|\plus{}|5\minus{}x|.$

When $ x\geqslant2,$ $ x\plus{}1$ and $ x\minus{}2$ are both non-negative so $ y\equal{}x\plus{}1\plus{}x\minus{}2\plus{}|5\minus{}x|\equal{}2x\minus{}1\plus{}|5\minus{}x|.$ $ 5\minus{}x$ may be negative or positive (depending on whether $ x>5$ or not) so either $ y\equal{}2x\minus{}1\plus{}5\minus{}x\equal{}x\plus{}4$ or $ y\equal{}2x\minus{}1\minus{}5\plus{}x\equal{}3x\minus{}6.$ In either case, the gradient is positive; hence $ y$ is increasing when $ x\geqslant2.$

When $ x<2,$ $ x\minus{}2$ is negative $ 5\minus{}x$ is positive so $ y\equal{}|x\plus{}1|\minus{}x\plus{}2\plus{}5\minus{}x\equal{}|x\plus{}1|\minus{}2x\plus{}7.$ $ x\plus{}1$ may be negative or positive (depending on whether $ x<\minus{}1$ or not) so either $ y\equal{}x\plus{}1\minus{}2x\plus{}7\equal{}\minus{}x\plus{}8$ or $ y\equal{}\minus{}x\minus{}1\minus{}2x\plus{}7\equal{}\minus{}3x\plus{}6.$ In either case, the gradient is negative; hence $ y$ is decreasing when $ x<2.$

It therefore follows that the minimum is attained at $ x\equal{}2.$


algebra
polynomial
number theory proposed
number theory

The only integers that make $ 1\plus{}x\plus{}x^2\plus{}x^3\plus{}x^4\plus{}x^5\plus{}x^6$ a square are $ x\equal{}\minus{}1,0$.

Schmu

$ P(x)\equal{}256(1\plus{}x\plus{}x^2\plus{}x^3\plus{}x^4\plus{}x^5\plus{}x^6)\equal{}P_3^2(x)\plus{}P_2(x),$ were
$ P_3(x)\equal{}16x^3\plus{}8x^2\plus{}6x\plus{}5,P_2(x)\equal{}140x^2\plus{}196x\plus{}231$. Because $ P_2(x)\equal{}0$ had not integer roots and $ \minus{}2|P_3(x)|\plus{}1<P_2(x)<|P_3(x)|\plus{}1$ for $ x\le \minus{}1$ or $ x\ge 2$ P(x) can not be perfect square for integer x.

Rust, thanks for the quick solution!

Me@home: to start, find rational $ a,b,c$ so that $ (x^3 \plus{} ax^2 \plus{} bx \plus{} c)^2$ has leading terms $ x^6 \plus{} x^5 \plus{} x^4 \plus{} x^3$.
There are (only) three free variables so we ought to be able to do this but not more.
We get $ a \equal{} 1/2$, $ b \equal{} 3/8$, and $ c \equal{} 5/16$. Multiply by $ 256 \equal{} 16^2$ to make
it an equation in integers. Now you are on your way....

Schmu

Thanks for the response, I should have thought of that.
What about the following?

Let $ a_{n,k}\equal{}\left(p_{n,k}/q_{n,k}\right)a_{n,k\minus{}1}$. Let $ p_{n,k}\equal{}n\plus{}p_{n,k\minus{}1}$ and $ q_{n,k}\equal{}n\plus{}q_{n,k\minus{}1}$. Let $ a_{n,0}\equal{}1,p_{n,1}\equal{}1,q_{n,1}\equal{}n$. Prove: \[ \left(\sum_{j\equal{}0}^k a_{n,j}x^{k\minus{}j}\right)^n\equal{}\sum_{n(k\minus{}1)\plus{}1}^{nk} x^i\plus{}P(x^{n(k\minus{}1)})\]where $ P(x^d)$ denotes a polynomial of degree $ d$.


[quote="krnboy32"]Determine the angle x in the triangle given below with AB = 1 and BC = 2.

the x is in the angle of <B[/quote]
What triangle below? With the given information it's impossible, perhaps the diagram gives more info?

You need to know the length of the 3rd side.

[quote="8parks11"]most likely

it should be

$cos x=\frac{1}{2}$

so just do inverse cos of $\frac{1}{2}$

if the triangle is different then try using the law of $cos$
(which would require the third side)[/quote]
That's completely ridiculous. You can't assume that.

[quote="8parks11"]im not saying that's the correct answer.
this can't bet that complicated since its the HSB[/quote]
But it could be $\frac{\sqrt{3}}{2}$ or even $\frac{\sqrt{6}+\sqrt{2}}{4}$ without actually being too complicated. It could even be a right triangle.

[quote="krnboy32"]Determine the angle x in the triangle given below with AB = 1 and BC = 2.

the x is in the angle of <B[/quote]

not enought info. Where did you get this problem? Provide the pic

Is $\angle A$ a right angle?  Just a guess.


probability
integration
logarithms
combinatorics unsolved
combinatorics

Case 1 A person dropped a rod which fell into three bits, so whats the probability that these 3 pieces could make up a triangle?

Case 2 The person dropped another rod which fell into 2 parts, and he randomly picked up one of them and smashed it onto the wall and it broke into 2 pieces again. Whats the probability that he could make a triangle with these new 3 pieces?

[quote]Case 1 A person dropped a rod which fell into three bits, so whats the probability that these 3 pieces could make up a triangle?[/quote]

If you view the rod as a ring that is cut 3 times, then the answer is the probability that given a ring with 1 cut, after 2 more cuts the 3 pieces can form a triangle. If the 3 pieces in order from the 1st cut are $a>b>c$, then the probability that they cannot form a triangle is $\frac{1}{4}\because a>b+c \iff a>\frac{a+b+c}{2}$. Since $a>\frac{a+b+c}{2}$, $b>\frac{a+b+c}{2}$, $c>\frac{a+b+c}{2}$ are mutually exclusive, the answer is $1-3\left(\frac{1}{4}\right)=\frac{1}{4}$

[quote]Case 2 The person dropped another rod which fell into 2 parts, and he randomly picked up one of them and smashed it onto the wall and it broke into 2 pieces again. Whats the probability that he could make a triangle with these new 3 pieces?[/quote]

Note that given the rod of length 1, if the first cut is at $x<\frac{1}{2}$ then the probability of the second cut allowing a triangle to be formed is $\frac{1}{2}\frac{x}{1-x}$. So the answer is $\int^\frac{1}{2}_{0}\frac{1}{2}\frac{x}{1-x}\, dx=\frac{1}{2}\ln 2-\frac{1}{4}$

[quote="pleurestique"]a very elegant solution to the first part can be seen here: http://www.cut-the-knot.org/Curriculum/Probability/TriProbability.shtml#Explanation[/quote]

Haha nice... but there's no elegant solution for the second one, or rather the elegant "solution" on cut-the-knot is "slightly" wrong. Actually I also made a similar mistake and obtained $\frac{1}{8}$ at first. Don't ask me how I got it :wink:

If the problem is extended for n-gon, the probability should be $1-\frac{2n}{2^{n}}$, in the same manner as before. But can a similar method to the one used on cut-the-knot work?


function
algebra
domain
special factorizations
complex numbers

[quote="shadysaysurspammed"]$\sqrt{x}+y=11$
$x+\sqrt{y}=3$

Find x,y without hit and trial.[/quote]

What's "without hit and trial"????

[quote="José"]
What's "without hit and trial"????[/quote]

I guess that he means "without using trial and error".. =)

I honestly doubt one would try (and succeed by) trial and error in this problem.
[hide]$\sqrt{x}+y=11$ and $x+\sqrt{y}=3$, so $x+\sqrt{x}+y+\sqrt{y}=14$.  Completing the square, we get $(\sqrt{x}+1/2)^2+(\sqrt{y}+1/2)^2=29/2$ and $\sqrt{x}=-1/2+-\sqrt{29/2-(\sqrt{y}+1/2)^2}$, and so by substitution we can solve for $y$ (or $x$) alone.  I haven't really worked the problem from there, because the numbers aren't really pretty, but it definitely looks doable.  (By the way, $+-$ is "plus or minus"; I'm not sure what the keyword is for the actual symbol...)[/hide]

there is no real solution, is there? if you solve for the x and y in the square root and then combine the equations you find that the real zeros don't work... or that's what I got anyway... is that correct?[/hide]

[quote="kokosai"] (By the way, $+-$ is "plus or minus"; I'm not sure what the keyword is for the actual symbol...)[/quote]
\pm -> $\pm$ :)

I think there are real answers, but I'm not absolutely sure.

[quote="kokosai"]$\sqrt{x}=-1/2\pm\sqrt{27/2-(\sqrt{y}+1/2)^2}$[/quote]

kokosai: This is a bit like bored, because there aren't beautiful numbers, but in this equation you got it, so substitute and you are almost done...

I got the lines $\sqrt{x} = 11-y$ and $\sqrt{y} = 3-x$ then solving for y and x:
$y=x^2-6x+9$ and $x = y^2 -22y+121$ then plugging in y into x we get $x = x^4-12x^3+32x^2+24x+4$ or $\rightarrow 0 = x^4-12x^3+32x^2+23x+4$
the only two real solutions are 5.926691 ad 6.6858794 (using a calculator)
If you plug in x into y, you get $0 = y^4-44y^3+726y^2-5323y+14641$ which has no real solutions... I think the answer may be imaginary.
Then again, I may have some errors...
The only solution that I thought might have worked would be -0.3166247 and 10.9999994 in the second equation, but as soon as you get a negative the answer doesn't work

Right, I can't do arithmetic.  It should be $\sqrt{x}=-1/2\pm\sqrt{29/2-(\sqrt{y}+1/2)^2}$.  As Apprentice wrote, this turns into a quartic, which can be solved exactly by the quartic equation (http://mathworld.wolfram.com/QuarticEquation.html).  There are two distinct pairs of x and y (both real), but they are [i]incredibly[/i] hideous.

What do you mean, Jose?

Thanks, mathmanman.

I tried solving this algebraically, but was not coming up with any values to satisfy both equations.

I then used a graphical approach and came up with the curves for the two equations.

As it turns out, the curves did not intersect each other.

So, it looks like there are no solutions for the system.

OK, there are two distinct pairs of (x,y) that satisfy the equation, but both are complex.  (So I was wrong.)  They are (approximately) $(-0.306285-0.0844712i,10.9244+0.558572i)$ and $(-0.306285+0.0844712i,10.9244-0.558572i)$.  I think those are it though.

[quote="kokosai"][quote="Altheman"]i don't think that we can really have square roots for complex numbers, that introduces ambiguity into the problem because the square root is no longer a function when it has a complex domain, demoivre gives 3 roots, and there is no standard for a "primary" root[/quote]

I'm not sure what you mean; square root is still a function when you extend its domain to complex numbers, and means the same thing.  Check out http://mathworld.wolfram.com/SquareRoot.html .  For example, $\sqrt{1+i}=\frac{\sqrt{2(\sqrt{2}+1)}}{2}+\frac{\sqrt{2(\sqrt{2}-1)}}{2}i$; if you square the number, you will indeed get $1+i$, so the concept of square root for complex numbers makes sense.

By the way, I used the quartic formula to find those two roots.  You can solve for them exactly, but they are extremely ugly, so I've only posted their approximate values.[/quote]

$\sqrt{1+i}=\frac{\sqrt{2(\sqrt{2}+1)}}{2}+\frac{\sqrt{2(\sqrt{2}-1)}}{2}i$ that is true, but $-\frac{\sqrt{2(\sqrt{2}+1)}}{2}-\frac{\sqrt{2(\sqrt{2}-1)}}{2}i$ is also a root (i switched the signs), i suppose that this is ambiguious as to what a "principal" root would be, i made a topic about this earlier, and i got the answer that there was no way to choose a principal root, but it seems that according to

[url]http://mathworld.wolfram.com/PrincipalRootofUnity.html[/url]

you choose the root that has the smallest positive complex argument for roots of unity, but that is "informal" and only for a small subset of complex numbers, so i still don't think there is a standard (eventhough smallest positive argument makes sense)


induction
superior algebra
superior algebra theorems

Suppose that you have a group $G$ and elements $a,b,c,d,e,f$ in $G$. Does associativity guarantee that I could evaluate $abcdef$ by placing parentheses anywhere that I desire? For example, could I just write $a(bc)d(ef)$ or $(a(bc)d)ef$ and start evaluating from there?

I know have another question. I was wondering how you would prove that with induction.

why induction?!

just apply the associative law many times. by the way, you already used this stating these expressions ...

Well, I'm not sure if one doesn't use induction to do that...

Well if you were giving a rigorous proof it would be much better to have proved a lemma that indicates that you can group terms in any way that you want, without having to skip over a large number of intermediate steps where you group terms according to $a(bc)=(ab)c$.

I have a proof.  We did this lemma in abstract algebra this summer.

[u]Proof[/u]: Assume that the lemma is not true.  Then, there exists a smallest natural number $n$, such that for $n$ elements, the arrangement of parantheses matters.

Suppose $k, m \in \mathbb{N}$ such that $k \leq  m < n$.

Then, let $P = (\Pi_{i = 1}^{k}x_{i}) * (\Pi_{i = k+1}^{n}x_{i})$,
and let $Q = (\Pi_{i = 1}^{m}x_{i}) * (\Pi_{i = m+1}^{n}x_{i})$.

But, by the minimality of $n$, and since $m < n$,
$\Pi_{i = 1}^{m}x_{i}= (\Pi_{i = 1}^{k}x_{i}) * (\Pi_{i = k+1}^{m}x_{i})$, for any arbitrary arrangement of parantheses.

Similarly,
$\Pi_{i = k+1}^{n}x_{i}= (\Pi_{i = k+1}^{m}x_{i}) * (\Pi_{i = m+1}^{n}x_{i})$.

Thus, $P = (\Pi_{i = 1}^{k}x_{i}) * [(\Pi_{i = k+1}^{m}x_{i}) * (\Pi_{i = m+1}^{n}x_{i})]$,
and $Q = [(\Pi_{i = 1}^{k}x_{i}) * (\Pi_{i = k+1}^{m}x_{i})] * (\Pi_{i = m+1}^{n}x_{i})$.

And therefore $P = Q$, by the associative law.

Thus, the arrangement of parantheses does not matter for $n$ terms, and therefore our assumption was false.

Therefore, the lemma is true.


AMC
USA(J)MO
USAMO
AIME
number theory
AIME I

If i'm preparing for USAMO, i have only enough money for a book. I have already ACoPS. Do you think i should buy 'Problem solving strategies' or 'the art of problem solving volume 2' that would be more useful for me.

Any help would be appreciated :lol:

Problem Solving strategies gives tons and tons of problems at or above the usamo level. But its slightly terse, although there are a few interesting digressions. It also requires alot of previous experience with the topics, for example the beggining of the number theory chapter simply gives about 20 main results/theorems that are useful in solving olympiad problems but leaves the proofs to the reader. AoPs 2 covers a large spectrum, from mid-high AMC to beggining IMO/USAMO, however the treatment of each topic is very engaging and although doesnt treat it as intensely as Problem solving strategies it does provide a very solid background which in works well with AcoPs (i for instance sometimes read a chapter of AoPs 2 and then go on to an AcoPs chapter on the same topic). Anyway, thats my two cents.

I don't wanna pretend to be an expert on the USAMO or anything, but I have both of those books. Art of problem solving #2 is more AIMEish/low level USAMO where the Problem Solving Strategies is like Mid-High level USAMO, or even alittle higher than that. It all depends on what USAMO score you're aiming at, but I'd definitely say get the Art of problem solving #2, or like check it out of your library at least, because if you don't understand Art of problem solving #2, you'd probably be lost reading Problem Solving Strategies.

PSS is pretty much the perfect book for USAMO prep, but before getting it you should make sure that you know all or most of the topics in AoPS 2. You can look at the Intermediate Forum to get an idea of average AoPS 2 problems. If you can solve most of those, then you should buy PSS. Otherwise buy AoPS 2.

Good luck on USAMO next year! :)

EDIT: Looking at your posts I would say go with AoPS 2. And study it hard. :D

If you can get a good AIME score already, then get PSS. If not, AoPS vol.2 will help you to qualify.

If ACoPS is at the right level for you, I would recommend Problem Solving Strategies.  In my opinion, AoPS V.2 is much more useful for AIME prep than USAMO prep.

[quote="D.P.L"]Is 6 a decent score?[/quote]

Well, what grade are you in?

[quote="D.P.L"]Is 6 a decent score?[/quote]6 is a good score, regardless of what grade you are in.  For example, on the 2006 AIME I, fewer than 10% of the students scored 6 or higher.

However, 6 is not quite enough to qualify for the USAMO.  This generally requires at least an 8, perhaps more depending on your grade and/or your AMC score.

[quote="D.P.L"][quote]Well, what grade are you in?
[/quote]

7th :lol:[/quote]

WOW, that's an incredibly good score for a 7th grader. :D

EDIT: According to the AMC site, there are only 14 people in 7th grade or below who scored a 6 or above. :D


algebra
system of equations
linear algebra
algebra proposed

Solve the following system of equations: \[ \left\{ \begin{aligned} 1 & = x^2 + y^2 + z^2 \\ 8 & = 6x^2 + 6y^2 + 5z^2 - 4xy - 2xz - 2yz \end{aligned} \right. \] for $x,y,z \in \mathbb{R}$.

for begin we have
3=x²+y²-4xy-2xz-2yz
  =(x-y)²-(2xy+2xz+2yz)
  =(x-y)²-(x+y+z)²+1  <=>2=-(2x+z)(2y+z)
and it's finish i think

I developped
2=-(4xy+2xz+2yz+z²)
  =-(4xy+2xz+2yz+z²)-6x²-6y²-4z²+6x²+6y²+4z²
10=6x²+6y²+4z²
5=3x²+3y²+2z²
we have 1=x²+y²+z²
  i havent time for termited this and you can have mistake but it'sfinished i think if it was good

[quote="altusi"]if we substitute the first equality into the second one,we obtain after some computations:

(x+y)²+(x+z)²+(y+z)²+z²+x²+y²=(x-y)²

but:
by cauchy,we have: (1/2+1/2)((x+z)²+(y+z)²)>=(1/sqrt(2)(x+z)-1/sqrt(2)(y+z))²=1/2(x-y)² with equality iff x+z=-z-y

and its not difficult to see that:  (x+y)²+z²+x²+y²>=1/2(x-y)² (by putting z=0) with equality iff x=-y

hence, the equality occur iff z=0, y=1/sqrt(2), x=-1/sqrt(2)

 :rotfl:[/quote]

Other proof:
$8(x^2+y^2+z^2)-(6x^2+6y^2+5z^2-4xy-2yz-2zx)=8-8=0$
Which is equivalent with: $2x^2+2y^2+3z^2+4xy+2yz+2zx=(x+y)^2+(x+y+z)^2+2z^2=0$
Really i don't think this is an interesting problem Arne.. :D

I know, but I actually used some linear algebra techniques to tackle this (we saw this in class as an exercise) and I was wondering what "elementary" solutions people would come up with :)


topology
function
advanced fields
advanced fields unsolved

It is known that any continuous bijection from a compact space to a Hausdorff space is a homeomorphism. As a corollary, any continuous injective map f from a compact space X to a Hausdorff space Y determines a homeomorphism from X to f(X).

Give an example of a continuous injective map $f: (0,1) \rightarrow \mathbb{R}^{2}$ which does not determine a homeomorphism from $(0,1)$ to $f( (0,1) )$.

Try for instance $f(x)=e^{2 i \pi x}$; it's clearly continuous and injective, but $(0,\frac{1}{2}]$ is closed in $(0,1)$ while $f((0,\frac{1}{2}])$ is not closed in $f((0,1))$ (because every open set containing $0$ intersects $f((0,\frac{1}{2}])$. Whence $f$ is not an homomorphism.

I don't quite understand this part. [quote]because every open set containing 0  intersects f((0,1/2))[/quote] Which space are you referring to when you say open sets?

My mistake. This set is closed in $f((0,1))$. I had in mind the counterexample given by $f(x)=(1,e^{\frac{2 \pi x^{2}}{(x^{2}+1)}})$ defined on $]-\infty;0]$ with values on the unit circle of $R^{2}$ ($f$ being given in polar coordinates). In this counterexample $]-\infty;0]$ is closed but not bounded (whence not compact), while you are asking for a case where the set is bounded but not closed.

Ok I have an example but it is kind of tedious to describe analytically so I put it in a picture and I will attach it to this post. The idea (in $R^{2}$) is to start from the point $A=(0,1)$ (excluded), to move along the $y$-axis towards $B=(0,0)$, then to move along the $x$-axis towards $(1,0)$, and then to move back towards the segment $[AB]$, approaching it without touching it. On the picture the red point will keep getting closer and closer to the first segment with the blue point on it, but will never intersect it.

Such a function will be continuous (obvious), in bijection with $(0,1)$ (obvious), but does not describe an homeomorphism since $\forall \epsilon>0, f((0,\epsilon))$ is not open in $f((0,1))$.

The hint given in the book is that the numeral 6 suggests such a map but I can't figure out why.

It's topologically the same as my example ;).


If $n$ is a positive integer such that $1<n<15$, find the sum of all distinct $n$ such that $n!+(n+1)!+(n+2)!$ is an integral multiple of $1024$.

[hide]
factor
$[(n+1)+(n+2)+1)](n!)=(2n+3)(n!)$
since $2n+3$ is always odd, we don't really care
12! has exactly 10 2's, so
$12+13+14=39$
[/hide]

oops  :D
in that case, we add 6 and 10 because they both work
so 39+10+6=55

Hey, can you prove there is no other solution???


I have attached the exercise as a .gif image, because I've made it with MathType. The solution is beautiful, but I am interested in new ideas! Enjoy!  :D

Let $ \alpha = 2 + \sqrt {3}$ and $\beta = 2 - \sqrt {3}$.
Then $0 < \beta < 1$ and for each $n$, using the binomial expansion, we deduce that  $U_n = \alpha^n + \beta^n$ is an integer.
Thus $ U_n - 1 = [ \alpha ^n]$ so that $ a_n = U_n $.

In another hand, it is easy to compute that :
$U_{n+1} = (2+\sqrt{3})\alpha^n + (2-\sqrt{3})\beta^n$
and
$U_{n+2} = (7+4\sqrt{3})\alpha^n + (7-4\sqrt{3})\beta^n$
Thus
$U_{n+2} = 4U_{n+1} - U_n$.
That is $a_{n+2} = 4a_{n+1} - a_n$.
It follows that :
$a_{n+1}^2 - a_na_{n+2} = a_{n+1}^2 - a_n(4a_{n+1} - a_n) = a_n^2 - a_{n+1}(4a_n -a_{n+1}) = a_n^2 - a_{n+1}a_{n-1}$
from which the desired result follows easily by induction.

Pierre.

Well done pal!  :shock: :D
There is another way without the binomial expansion.......  ;)

Yes, you may also use some Galois' theory (special dedicace to Xavier C...), but I think it is a bit too much for this problem  ;)

Pierre.


Just when I thought I could do every problem, I saw that the answer to Target #8 was 21/1024.

I keep getting 5/256 so I would like someone to either tell me where I lost 1/1024 or tell me that there was a mistake on the answer key. Thank you.

btw, it was the problem where you flip n+1 heads in a row to win if you have flipped n tails. What is the probability that you win with 4 consecutive heads?

TTTHHHH

THTTHHHH

TTHTHHHH

TTHHTHHHH

THTHTHHHH

THTHHTHHHH

I think you missed the last one

I thought of that but after 5, it's THTHH which is a winner. Thanks, anyway.

THTHH is not a winner because you flipped 2 heads in a row but you've a already flipped a 2 tails.

Yes, one reason I failed on target was that the wording of the problem ("... you win BY FLIPPING 4 heads in a row") made me think about conditional probability... and I wasted 3 minutes on that and never thought about just simply listing it out... :mad:

I misread t0o and was doing infinite geometric series junk.  :mad:  :mad:  :mad:


induction
number theory unsolved
number theory

We prove for any number $n$ we can reach $9$.
Lemma: For any $n$ with $(n,10)=1$ there is a number $B$ of form $10\cdots01\cdots01\cdots01\cdots12$ divisible by $n$.
Proof: Let $p=\phi(n)$. Take $m>10n$ with $9|m+1$ then $A=1\cdots12$ having $m$ ones is divisible by $n$.
Now take a $1$ at position $lp$ and move it to position $jp+1$ to became the first digit in the number (the remaining 'gap' becomes $0$). Then as we easily see the residue of $A$ mod $n$ changes by $9$ so the residue of $\frac A9$ mod $n$ changes by $1$.
Now performing enough operations we shall bring about a number divisible by $9n$ having the form we wish.
Now to pass to the solution of the problem.

If $n=2^k5^l n'$ where $(n',10)=1$ then multiply $n$ by $2^l5^k$ and eliminate zeros to produce $n'$.
Then take $n'$, multiply it by a suitable constant to become $B$ as in the lemma.
Then remove all zeros from $B$ and multiply it by $9$ producing $10\cdots08$.
Remove again all zeros to produce $18$. Multiply it by $5$ to reach $90$. Finally remove last zero to get $9$.

[Edit: edited according to RobertuX's observation below]

[quote="iura"]We prove for any number $n$ we can reach $9$.
Lemma: For any $n$ there is a number $B$ of form $10\cdots01\cdots01\cdots01\cdots12$ divisible by $n$.
Proof: Let $p=\phi(n)$. Take $m>10n$ with $9|m+1$ then $A=1\cdots12$ having $m$ ones is divisible by $n$. ...
[/quote]
Sorry your proof is NOT correct, if $n=25$ then $A$ is NOT divisible by $5$. So, how you fix it? In fact your lemma is wrong taking $n=25$.

B.R.

RX TCM

[quote="heartwork"]Beginning with any positive integer we can perform the following 2 operations:
1) Multiply it by any positive integer.
2) Delete the 0's in its decimal representation.
Can we always obtain a single-digit number after a finite number of operations?[/quote]
It's TRUE.  Proof by induction:
If $0<n<10$ it's clear, and if $n>9$ then we know that there exists $\lambda\in \mathbb{N}$ such that $10^{\lambda}=1 \ mod \ {9n}$ [I assume that 0 is not in $n$, and if $n$ is divisible by $2$ or $5$ the multiply by $6$ or $2$ and delete the 0, until last digit is not $2$ or $5$],  then exists  a number $m$ such that $mn=\frac{10^\lambda-1}{9}=1\cdots 1$ with $\lambda$ ones, and it's easy to show doing simple calculations by hand that $\times 11$, $\times 5$ and delete 0, $\times 2$ and delete, $\times 9$ and deleeeeeete 0, we arrive to 117, then $\times 6$ and delete, $\times 5$ and delete, $\times 5$ and delete, $\times 5$ and delete, and finaly  $\times 5$ and delete we arrive  to 9.
[NOTE: if last digit is even, then we need to multiply by 5 and delete such times we got an odd digit, ok?  ;)]
So, if $n<10$ then answer is $n$ and if $n>9$ then answer is 9.

Best regards.

RX the Crazy Math.

well iuras argument can work putting $0$'s after his number.

[quote="heartwork"]It is possible to determine for each positive integer the smallest (optimal) number of operations?
From the proof of RobertuX this is at most $8$.[/quote]
I think this quantity depends of the number, I mean, clearly if $n<11$ this number is 1, but for 12 and $14$ are 2, i.e. $x5$ and delete 0, but for $13$, I think optimal is 3, $x8$, delete 0 and $x5$, then an upper bound is $8$ but in general is not well easy to know.

RX


what major schools go to this competition? doesn't east go here for regionals?

After a minute or two of research and a single google search (I'm feeling lucky), the results page shows that some of the top schools that go to Elon are East Chapel Hill (upperclassmen), Durham Academy, Cary Academy, and Enloe.

My advice: if you want to go to state, don't go to Elon.


modular arithmetic
induction
function
algebra
polynomial
number theory proposed
number theory

Let $a,b,c$ be integer numbers such that $(a+b+c) \mid (a^{2}+b^{2}+c^{2})$. Show that there exist infinitely many positive integers $n$ such that $(a+b+c) \mid (a^{n}+b^{n}+c^{n})$.

[i]Laurentiu Panaitopol[/i]

Consider the equation $x^3 - (a+b+c) x^2 + (ab + bc + ca) x - abc = 0$. This equation is the characteristic equation of the recurrence relation $a_n = (a+b+c)a_{n-1} - (ab + bc + ca)a_{n-2} + (abc)a_{n-3}$.
This is equivalent to $a_n = a_1a_{n-1} - \frac{{a_1}^2-a_2}{2} a_{n-2} + (abc) a_{n-3}$.
We note that the formula for $a_n$ is $a_n = a^n + b^n + c^n$. Thus $a_1 = a + b + c, a_2 = a^2 + b^2 + c^2, a_3 = a^3 + b^3 + c^3$.

If $\frac{{a_1}^2-a_2}{2} \equiv 0 \pmod{a_1}$, then $a_n \equiv (abc)a_{n-3} \pmod{a_1} \Rightarrow a_i \equiv 0 \pmod{a_1} \forall i \equiv 1$ or $2 \pmod{3}, i \in \mathbb{Z}^+$, since $a_1 \equiv a_2 \equiv 0 \pmod{a_1}$, and we are done.

Likewise, if $a_1$ is odd, then ${a_1}^2-a_2 \equiv 0 \pmod{a_1} \Rightarrow \frac{{a_1}^2-a_2}{2} \equiv 0 \pmod{a_1}$ (Note that we are guaranteed that $\frac{{a_1}^2-a_2}{2} \in \mathbb{Z}$ since $\frac{{a_1}^2-a_2}{2} = ab + bc + ca \in \mathbb{Z}$)
This means that we once again have the similar case of $a_n \equiv (abc)a_{n-3} \pmod{a_1}$ and we are done.

Thus we can assume that $a_1$ is even and that $\frac{{a_1}^2-a_2}{2} \equiv \frac{a_1}{2} \pmod{a_1} \Rightarrow a_n \equiv \frac{a_1}{2}a_{n-2} + (abc)a_{n-3} \pmod{a_1}$. Note that since $a_n = a^n + b^n + c^n$, we have $a_i$ will have the same parity throughout $\forall i \in \mathbb{Z}^+$. Since we have assumed $a_1$ is even, we have $a_i$ is even $\forall i \in \mathbb{Z}^+$.

Suppose that $\not\exists$ infinitely many $n$ such that $a_1 \mid a_n$, say $M$ is the largest integer such that $a_1 \mid a_M$. Now consider $a_{M+3}$. We have $a_{M+3} \equiv \frac{a_1}{2}a_{M+1} + (abc)a_{M} \equiv 0 \pmod{a_1}$, since $a_{M+1}$ is even. (Contradiction)

$\therefore \exists$ infinitely many $n$ such that $a + b + c \mid a^n + b^n + c^n$.

[quote="ooiler"]i think  n=2^k is a solution :)  :)[/quote] True by induction, but my solution is much stronger, asserting that $\forall \ i \equiv 1$ or $2 \pmod{3}$ we are sure that $a_i \equiv 0 \pmod{a_1}$, and that if we are also given that $a_1 \mid a_3$, then all $a_i$ will be divisible by $a_1$.  :P

There is another super-nice proof of problems like this (the proof is from a mathlinks user and imo gold simo_the_wolf):
For the fundamental theorem of symmetric functions, you can write $a^{n}+b^{n}+c^{n}$ uniquely as a homogenous polynomial with integer coefficients in the variables:
$\sigma_{1}= a+b+c$
$\sigma_{2}= ab+bc+ca$
$\sigma_{3}= abc$.
If in each monomial of this polynomial there is a $\sigma_{1}$ or a $\sigma_{2}$, then a+b+c divides it. If there is only a $\sigma_{3}$, then its degree is a multiple of 3. Therefore, for each n non multiple of 3, we have $a+b+c \mid a^{n}+b^{n}+c^{n}$.

I'm implicitly using symmetric functions, but thanks for pointing out.


linear algebra
matrix
induction
function
algebra unsolved
algebra

I had this problem proposed at the exam today.
please help me solve it.
i tried almost everything today and it didn't work.

\[ A=\left( \begin{array}{ccc}
1 & 2 & 3 \\
0 & 1 & 2 \\
0 & 0 & 1 \end{array} \right)
\]

find the roots of the equation $Y^n=A,n\in N$ with $Y\in M_3$

if you seek only solutions of the form
      1 a b
y=  0 1 c
      0 0 1
andi  think you can show these are the only possible solutions, but i am not sure,
you get that y^n will looc smth like:
1 a*2^(n-1)  "something_else"
0 1 c*2^(n-1)
0 0 1

And making this equal to your matrix you get that a=c=1/(2^(n-2))
Repalacing a and c with these values in Y all you have left to compute is b. Denoting by x(n) the value of the top-right conner of matrix Y^n we have:
x(1)=b
x(n+1)=2*x(n)+1/(2^(2n-4))
And now jut compute x(2), x(3),x(4) and you'll get the idea of how the general term looks like :)

so you want to say that i should make $Y^n$ depending on some recurrent(by induction i guess)
sequences and then how do i calculate the solution ?
i didn't understood that part. :?

i want to say that you should start with y of that form i presented in the beginning and then see how y^n looks like. indeed you will get a formula for the elements on line 1 column 2 and line 2 column 3, a formula which of course depends on a respectively c. you then compute a and c.
so now, in y you only need to know b since you already found out a and c.
so yes when you start computing the powers of y you will get that recursive formula for the upper right element of the matrix y^n .
all i didn't do is start computing x(2), x(3) and so on, which if you compute you will see how x(n) looks like, of course as a function of b. it means you will figure out a formula for x(n), of course a non-recursive one cause a recursive one won't help you compute b. so, just try and compute x(2), x(3) and x(4) and see what you get.

is this an exercice you think we would get at an bacalaureat exam or at politechnic institute in bucharest ?

well, as i know at the politechnic institute in bucharest the exam is grill, therefor i don't see how you could get this subject. for the bacalaureat, well, maybe as one of the hard problems. anyway, you will surelly have som matrixes stuff, and actually triangular matrixes like this one are pretty popular cause they are pretty easy to work with.
since i get it you will have an exam soon, good luck with it!


Euler
geometry
rectangle
integration
calculus
real analysis
real analysis unsolved

Let $ P$ be a rectangle , $ f_{0} : \partial P \rightarrow R)$ continuous and Lipschitz, $ C_{0} \equal{} \{ f \in C^{2}(P) : f \equal{} f_{0} \ on \ \partial P \}$. and finally $ S : C_{0} \rightarrow R$ a functional:

$ S(f) \equal{} \int^b_a (\int^d_c (\frac {\partial f}{\partial x})^{2}\,dy)\,dx \plus{} \int^d_c (\int^a_b (\frac {\partial f}{\partial y})^{2}\,dx)\,dy$.

Write Euler-Lagrange equation for S.

I tried writing: $ S(f) \equal{} \int^b_a (\int^d_c (\frac {\partial f}{\partial x})^{2} \plus{} (\frac {\partial f}{\partial y})^{2}\,dy)\,dx$, so the proper Lagrangian would be $ L(x) \equal{} \int^d_c (\frac {\partial f}{\partial x})^{2} \plus{} (\frac {\partial f}{\partial y})^{2}\,dy$.

Then the Euler-Lagrange equation should be $ \frac {d}{dx}\frac {\partial L}{\partial f^{'}_{x}} \equal{} 0 \leftrightarrow \frac {d}{dx}\frac {\partial }{\partial f^{'}_{x}}\int^d_c (\frac {\partial f}{\partial x})^{2}\,dy \equal{} 0$, ($ L^{'} \equal{} \int^d_c (\frac{\partial f}{\partial x})^{2}\,dy$) and now since $ \frac {dL^{'}}{dx} \equal{} \frac {\partial L^{'}}{\partial f_{x}^{'}}\frac {\partial f_{x}^{'}}{\partial x}$, we can rewrite that as $ \frac {d}{dx}\frac {\frac {d}{dx}\int^d_c (\frac {\partial f}{\partial x})^{2}\,dy}{f_{xx}^{''}} \equal{} 0 \leftrightarrow \frac {d}{dx}\frac {\int^d_c \frac {\partial}{\partial x}(\frac {\partial f}{\partial x})^{2}\,dy}{f_{xx}^{''}} \equal{} 0 \leftrightarrow \frac {d}{dx}\frac {\int^d_c 2f_{x}^{'}f_{xx}^{''}\,dy}{f_{xx}^{''}} \equal{} 0$. but what then?

I didn't read everything because is confused.

But a tip dude.

The lagrangian is $ L(x,y)$ . This is the principal error.

The second is, the lagrangian here don't have any integral, can you figure it out?


Hi, can you please tell me what score you need on the Avogadro Exam (out of 40) to get in the top 95+ percentile?
Thanks! :)

I got 35/40 and was in the 98th, so I would say about 32-33.
P.S. it's not really hard because they always have the same type of questions so just do previous years' problem sets and know some basic gen chem knowledge and you should be set.


calculus HMMT limit function algebra domain logarithms

Hello everyone, I would like to start a marathon for calculus problems, particularly in the level of the HMMT calculus test.  See more at [url=http://web.mit.edu/hmmt/]web.mit.edu/hmmt/[/url].  The way a marathon works is person N solves a problem then proposes a problem, person N+1 solves N's problem and proposes a problem, etc.   I will start by giving a problem.

Source: HMMT 2004 #3
Compute $\lim_{x\to\infty} (\sqrt[3]{x^3+x^2}-\sqrt[3]{x^3-x^2})$

Alternate solutions are encouraged.

Taylor expand.  It's $2/3+O(1/x)$, so the limit is $2/3$.

So now it's your turn to post a problem,blahblahblah.  Anyone could post a problem as long as we can keep up the marathon.

To expand on blahblahblah's answer:
$\lim_{x\to\infty}\sqrt[3]{x^3+x^2} = \lim_{x\to\infty}\frac{1}{3} \frac{3x^2+2x}{(x^3+x^2)^{2/3}} + \lim_{x\to\infty}O(1/x)$
$\lim_{x\to\infty}\sqrt[3]{x^3-x^2} = \lim_{x\to\infty}\frac{1}{3} \frac{3x-2x}{(x^3-x^2)^{2/3}} + \lim_{x\to\infty}O(1/x)$
Thus the sum of those limits is: $\frac{2}{3} + O(1/x)$, because the two RHS limits are both 1.
Answer: $\frac{2}{3}$

y=ln(x^2+y^2)

find y'

Hint: [hide] Use implicit differentiation.  [/hide]

[hide="solution"] $\frac{dy}{dx} = \frac{2x + 2y dy/dx}{x^2+y^2} \rightarrow (x^2+y^2 - 2y) \frac{dy}{dx}= 2x \rightarrow y' = \frac{2x}{x^2+y^2-2y}$[/hide]

Next problem.
Compute $\lim_{x\to 0+} (x^{x^x}-x^x)$

[quote="mli7"]One more:

where is the function f(x)=xln(x) decreasing?

Hint: [hide]This is a tricky one, think about the domain of the function.[/hide][/quote]

$f' (x) =\ lnx + 1 < 0 \rightarrow \ lnx < -1 \rightarrow x < \frac{1}{e} \rightarrow x \in (0,\frac{1}{e})$

Problem III
[hide]
\[ \begin{array}{l} \mathop {\lim }\limits_{x \to 0^ + }^{} (x^{x^x } - x^x ) = \mathop {\lim }\limits_{x \to 0^ + }^{} x^{x^x } - \mathop {\lim }\limits_{x \to 0^ + }^{} x^x \\ a^x = e^{x\ln a} \\ \mathop {\lim }\limits_{x \to 0^ + }^{} x^x = \exp (\mathop {\lim }\limits_{x \to 0^ + }^{} x\ln x) = \exp (\mathop {\lim }\limits_{x \to 0^ + }^{} \frac{{\ln x}}{{\frac{1}{x}}}) \\ \frac{\infty }{\infty }\mathop {\lim }\limits_{x \to 0^ + }^{} x\ln x = \mathop {\lim }\limits_{x \to 0^ + }^{} \frac{{\frac{1}{x}}}{{ - \frac{1}{{x^2 }}}} = \mathop {\lim }\limits_{x \to 0^ + }^{} - x = 0 \\ \mathop {\lim }\limits_{x \to 0^ + }^{} x^x = e^0 = 1 \\ \mathop {\lim }\limits_{x \to 0^ + }^{} x^{x^x } = 0^1 = 0 \\ \mathop {\lim }\limits_{x \to 0^ + }^{} (x^{x^x } - x^x ) = 0 - 1 = - 1 \\ \end{array}  \][/hide]

Problem IV
compute
\[ \int {\frac{1}{{\sqrt[3]{{x + 2}}}}} tg(x)dx  \]

Are you sure that's the right integral (or that it doesn't have limits)? Becaue even Mathematica http://integrals.wolfram.com/ couldn't evaluate it.

How about we make a rule that the person who posts a problem should know how to solve it.  Could someone write a new problem?

[quote="socrates"][quote="Jimmy"]Here's one, but I'm not sure it's up to the level you want:

Compute
\[ \lim_{k\to 0} \frac{1}{k}\int_0^k \left(1+\sin 2x\right)^{1/x} \ dx  \]

Source Putnam 1938 A5.[/quote]

De hospital rule works here.... :)[/quote]

Yep. :yup:

[quote="Beginner"]
For the second part, If $I(t)= \int_{0}^{\infty} \frac{1}{x^2+t} \, dx$ then
$I'(t)= \int_{0}^{\infty} \frac{-1}{(x^2+t)^2} \, dx$ and  $I''(t)= \int_{0}^{\infty} \frac{2}{(x^2+t)^3} \, dx$.
[/quote]

The question asks for when $t = 2$, but Beginner made a function with a variable of $t$, he then takes the derivative of it a few times to get the form we were looking for in the beginning.

[quote="Beginner"]
Now compute I(t) using trig subtitution $x= \sqrt{t} \tan u$
\begin{eqnarray*} I(t) &=& \int_{0}^{\infty} \frac{1}{x^2+t} \, dx \\ &=& \int_{0}^{\infty} \frac{\sqrt{t} \sec^2 u}{t \tan^2 u +t} \, du \\ &=& \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{t}}{t} \, du \\ &=& \frac{ \sqrt{t}}{t} \left( \frac{\pi}{2} \right) \end{eqnarray*}
[/quote]

He then solves for $I(t)$ using a pretty well known Calculus I method.  Note: inside the integrand $t$ is a constant.

[quote="Beginner"]
Differentiate with respect to $t$
$I'(t)= -t^{- \frac{3}{2}} \frac{\pi}{4}$

$I''(t)= t^{- \frac{5}{2}} \frac{3 \pi}{8}$ subtitution $t=2$

$\int_{0}^{\infty} \frac{1}{(x^2+2)^3} \, dx = \frac{I''(2)}{2}= 2^{\frac{-7}{2}} \frac{3 \pi}{8} = \frac{3 \pi \sqrt{2}}{128}$
[/quote]

He then combines the efforts of his part with his second part.

Obviously, Beginner isn't really a beginner.

I hope that helps, Kalle.

Someone should post another question.

Thanks for explained that to Kalle,  Fmako.
Before  i post another problem, i must say that the previous method require some conditions. We can have \[ \frac{d}{dt} \int_{a}^{\infty} f(x,t) \, dx = \int_{a}^{\infty} \frac{\partial f(x,t)}{\partial t} \, dx  \]
if we have three conditions:
1) $f(x,t)$ and $\frac{\partial f(x,t)}{\partial t}$ are continous in $a \leq x< \infty$,  $c \leq t \leq d$.
2)   $\int_{a}^{\infty} f(x,t) \, dx$ convergent.
3)  There exist a continous function $R(x)$, independent from $t$ for which $\left|\frac{\partial f(x,t)}{\partial t}\right| \leq R(x)$, for  $c \leq t \leq d$ and $\int_{a}^{\infty} R(x) \, dx$ convergent.

In advance calculus books (I can't wait to have that course next year :lol: ) there is a similar thing like this due to Leibnitz. \[ \frac{d}{dt} \int_{a}^{b} f(x,t) \, dx = \int_{a}^{b} \frac{\partial f(x,t)}{\partial t} \, dx  \]
this only require no 1) condition for $a \leq x \leq b$.

Back to marathon.
Source: My senior
if $Si(t)= \int_{0}^{t} \frac{\sin u}{u} \, du$. compute $\int_{0}^{\infty} e^{-t} Si(t) \, dt$

$\int_0^{\infty} \int_0^t e^{-t} \frac{\sin u}{u} \, du \, dt = \int_0^{\infty} \int_u^{\infty} e^{-t} \frac{\sin u}{u} \, dt \, du =$

$\int_0^{\infty} \frac{\sin u}{u} e^{-u} \, du = \Im \int_0^{\infty} \frac{e^{(i-1)u}}{u} \, du$

This is the best I could do so far.

$\int \sin u e^{-u} \, du = -\sin u e^{-u} + \int e^{-u} \cos u \, du = -\sin u e^{-u} - \cos u e^{-u} - \int e^{-u} \sin u \, du$

$\int \sin u e^{-u} = \frac{-e^{-u}}{2} (\sin u + \cos u)$

Maybe that will help.

Sorry I can't figuring what you are trying to do Fmako.
This should be useful, $\text{Si} (0) = 0$ and $\text{Si} ' (t) = \frac{\sin t}{t}$.

This is probably similar to what FMako said, but I got the following by integration by parts:

$\int_{0}^{\infty}e^{-t}\textrm{Si}\left(t\right)\,dt=\int_{0}^{\infty}\frac{e^{-t}\sin{t}}{t}\,dt$

I'm not quite sure how to integrate that.

My computer says that the second integral is:

$\int_{0}^{\infty}\frac{e^{-t}\sin{t}}{t}\,dt\approx .785398$

:D  $\frac{\pi}{4} = 0.78539..$. That's the corect answer.
Now I post my solution.
[hide]  We have from Fubini's theorem \[ \int_{0}^{\infty} e^{-y} \, \frac{\sin y}{y} \, dy = \int_{0}^{\infty} \int_{0}^{\infty} e^{-(x+1)y} \sin y \, dx \, dy = \int_{0}^{\infty} \int_{0}^{\infty} e^{-(x+1)y} \sin y \, dy \, dx  \]
Now we try to compute $\int_{0}^{\infty} e^{-(x+1)y} \sin y \, dy$, we can use euler's formula. that is \begin{eqnarray*} \int_{0}^{\infty} e^{-(x+1)y} \sin y \, dy &=& \Im \, \int_{0}^{\infty} e^{-(x+1- i) y} \, dy \\ &=& \Im \, \frac{1}{x+1-i} \\ &=& \Im \, \frac{x+1 +i}{(x+1)^2 +1} \\ &=& \frac{1}{(x+1)^2 +1} \end{eqnarray*}

So we have
\begin{eqnarray*} \int_{0}^{\infty} e^{-y} \, \frac{\sin y}{y} \, dy &=& \int_{0}^{\infty} \frac{1}{(x+1)^2 +1} \, dx \\ &=& \int_{1}^{\infty} \frac{1}{u^2 +1} \, du \\ &=& \lim_{\tau \rightarrow \infty} \arctan \tau - \arctan 1 = \frac{\pi}{2} - \frac{\pi}{4} \\ &=& \frac{\pi}{4} \\ &\approx& 0.78539816339744830961566084581988 \end{eqnarray*}
[/hide]

That is a really nice solution, I suspected that I would have to resort to a double integral to resolve the $\frac1{y}$ term, but I

couldn't figure out how to implement it nicely with the $e^{\text{stuff}}$.

Do you know any other way to do this problem?

I tried:

$\int_0^{\infty} \frac1{u} \frac{d}{du}(\frac{-e^{-u}}{2} (\sin u + \cos u)) \, du$

Among other things, but I couldn't find another way.

I think i have another solution.
Do you know the idea of previous solution inspired by you, FMAKO  :) . After saw your post, I think that we can use double integral to solve this problem.
[hide=" another solution"]
Look at my previous post, amcavoy gave a good respon.  :lol:
From integration by part, the integral is equal to $\int_{0}^{\infty} \frac{e^{-t} \sin t}{t} \, dt$.
Now let $I(s) = \int_{0}^{\infty} \frac{e^{-st} \sin t}{t} \, dt$ . Then \[ I'(s) = \int_{0}^{\infty} - e^{-st} \sin t \, dt = \frac{-1}{s^2+1} .  \]
If you know that $I(0) = \int_{0}^{\infty} \frac{\sin t}{t} \, dt = \frac{ \pi}{2}$ , then you can use fundamental theorem of calculus  \[ \int_0^{1} I'(s) \, ds = I(1) - I(0)= I(1) - \frac{\pi}{2}  \]
and also $\int_0^{1} I'(s) \, ds = \int_0^{1} \frac{-1}{s^2+1} \, ds = - \arctan 1$
So $I(1) - \frac{\pi}{2} = - \frac{\pi}{4}$. And $I(1) = \frac{\pi}{4}$.
Then we are done.
About $I(0)$, you can use search in this forum to find the proof.
[/hide]

I can't find $I(0)$ on the forum right now, although it looks very familiar.

Can you help me find it?

[quote="Beginner"]I think you should search it it was discussed many time[/quote]

I know, but I just couldn't find it...

[quote="Beginner"]
Back to marathon.
Source: My senior
if $ Si(t) \equal{} \int_{0}^{t} \frac {\sin u}{u} \, du$. compute $ \int_{0}^{\infty} e^{ \minus{} t} Si(t) \, dt$[/quote]

Hmm...there is a similar question on the hmmt 2007 test... http://web.mit.edu/hmmt/www/datafiles/solutions/2007/scalc07.pdf

Thanks Athelman. Does anyone knows what justifies introducing the complex variable? I only know about Riemann integrals so there might be something I'm missing. And what justifies taking the Re[] function outside the integral?


How many natural numbers N less than or equal to 12 is 4^N+ 5^N + 6^N + 7^N a multiple of 11?

$ \equal{} (11 \minus{} 7)^N \plus{} (11 \minus{} 6)^N \plus{} 6^N \plus{} 7^N \equal{} \sum_{k \equal{} 0}^{N}\binom{N}{k}11^k( \minus{} 7)^{N \minus{} k} \plus{} \sum_{k \equal{} 0}^{N}\binom{N}{k}11^k( \minus{} 6)^{N \minus{} k} \plus{} 6^N \plus{} 7^N$

$ When$ $ N$ $ is$ $ an$ $ Odd$ $ number,$  $ the$ $ above$ $ number$ $ will$ $ be$ $ a$ $ multiple$ $ of$ $ 11$

$ So$ $ N \equal{} 1,3,5,7,9,11$

There should be a simpler solution, cause no Mathcounts Problems require the knowledge of summation notation

...If summation notation really bothers you so much, you can just write out the sum...

Also, note that $ a^n\plus{}(n\minus{}a)^n$ is always divisible by $ n$ if $ n$ is odd.  So we can group this as $ (4^N\plus{}7^N) \plus{} (5^N \plus{} 6^N)$ and win.

Unfortunately, I do not know how to prove that the even numbers do not work...


function
vector
topology
group theory
abstract algebra
algebra
functional equation

Please solve the problems, I couldn't do them out:

1. Let $ f: \mathbb R\to\mathbb R$ be increasing function. Then does there always exist an interval (a,b) where f is continuous?

2. In $ \mathbb R^2$, a decreasing sequence of connected sets $ \{A_n\}_n$ is given. Is $ \cap A_n$ connected always? (what might be the answer for any topological space?)

3. Let $ f: [0,\infty)\to[0,\infty)$ be a function with $ \forall x,y, f(x \plus{} y) \equal{} f(x) \plus{} f(y)$. Then is f always continuous? (My obs: such a function is conts at 0 iff it is continuous throughout its domain. So is there such a function which is disconts at 0?)

4. Is there an infinite metric space $ X$ such that the dimension of the vector space of continuous functions :$ X\to \mathbb R$ over $ \mathbb R$ is finite?

1. No. You can construct a monotone increasing function that is discontinuous in Q. For example, consider a function on [0,1], and then $ f(x) \equal{} \sum \frac {sgn(x \minus{} r_n)}{2^n}$, where $ r_n$ is an enumeration of rationals in [0,1]. You can of course extend it to a function defined on R.

2. No. Consider a vertical infinite rectangular strip containing 0 with 0 removed.  Shrink the strip to the vertical line containing 0, then the infinite intersection won't be connected.

Remark: I believe that in a Hausdorff space, the intersection of decreasing compact connected set (continuum) is still compact and connected.

3. I'm not sure about this one. From the functional equation one is forced to conclude that $ f(x) \equal{} x$ on $ \mathbb{Q}$, but then we have some freedom on the rest. I think one can extend the function to $ \mathbb{R}$ by Zorn's lemma, but I'm not quite certain.

4. We make a claim, that in an infinite metric space $ X$, there exists a point $ p$ and radius $ r$ so that $ X \minus{} B(p,r)$ is still infinite.
Assuming this claim and keep on using it, we obtain a sequence $ p_i$ and $ r_i$ so that $ B(p_i,r_i)$ miss any other $ p_j$.
Since $ X$ is normal, apply Urysohn's lemma we can find a function $ f_i$ that is 1 at $ p_i$, and 0 outside $ B(p_i,r_i)$. This family of function is linearly independent, so the dimension can't be finite.

To show the claim, assume we can't find such a point. Then for any point $ p$, the set $ E_n \equal{} \{y: d(p,y) > \frac {1}{n}\}$ is finite. Fix $ x$ so that $ d(p,x) > \frac {1}{m}$, and then we note that as there are infinitely many points that is of distance less than $ \frac {1}{2m}$ with $ p$, these points have distance at least $ \frac {1}{2m}$ with $ x$.  So $ x$ violates the assumption that we can't find such a point.

[quote="t0rajir0u"][quote="Soarer"]3. I'm not sure about this one. From the functional equation one is forced to conclude that $ f(x) \equal{} x$ on $ \mathbb{Q}$, but then we have some freedom on the rest. I think one can extend the function to $ \mathbb{R}$ by Zorn's lemma, but I'm not quite certain.[/quote]

This is the [url=http://en.wikipedia.org/wiki/Cauchy_functional_equation]Cauchy functional equation[/url], and there are lots of discontinuous solutions (by the Axiom of Choice - that is, Zorn's lemma).[/quote]

Thanks! I was actually thinking about extending the function to each cyclic subgroup of $ \mathbb{R}/\mathbb{Q}$, and take the "maximal" extension guaranteed by Zorn's lemma.


$ P(x)\equal{}x^{4}\minus{}3x^{2}\plus{}5x^{2}\minus{}9x$.
Find all pairs of distinct integers $ (a,b)$, such that $ W(a)\equal{}W(b)$.

Do you mean $ P(x)\equal{}x^{4}\minus{}3x^{3}\plus{}5x^{2}\minus{}9x$?

I think you meant $ P(a)\equal{}P(b)$, and $ \minus{}3x^3$ instead $ \minus{}3x^2$. Correct me if I'm wrong, please.


geometry
AMC 10
number theory
linear algebra

Given that I have the chance to go to either Exeter or IMSA (Illinois Math and Science Academy), which one should I choose?

Disclaimer: This post might be a bit biased (I go to IMSA :) )

Well, first off, IMSA is pretty darn cheap compared to Exeter (and that's an understatement) -- how often do you find a great, cheap, state-funded residential high school?  IMSA does have quite a lot of math resources, but probably not as much as Exeter.  Sure, it might make a difference, but I don't think it's worth the extra 30k or so each year.

Another thing that makes IMSA stand out is it's mentorship/inquiry program -- on most Wednesdays, we don't have class, and a lot of students pursue independent research with a mentor at a nearby college.  I'm not sure if Exeter has a similar type of program, but opportunity-wise, IMSA definitely has a lot to offer.

There's a lot more about IMSA that makes it great (that I can't think of right now.. :lol: ).  I don't know that much about Exeter, but personally, I'd chose IMSA if you live in IL (a friend of mine, Albert Ni, did).

But of course, it's all up to you -- hope that helped.  If you have any more questions about IMSA in particular, feel free to ask!

What grade are you in, by the way?

I like Exeter a lot, and urged my son to apply to it this year, but if the choice is between IMSA in one's home state (that is your choice, right) and Exeter out of state, I would vote for the in-state cool school. An out-of-state experience no later than college age is a very good idea (I think it was a mistake for me to go to an in-state college as an undergrad), but IMSA is plenty good enough to stay closer to home for while in your teenage years.

Of course I expect to hear from students of Big Red here  ;)  and I will eagerly read what they have to say.

Huh, well Exeter is pretty beastly, but I would definitely say IMSA is the better choice due to money. You would have a lot more pressure over at Exeter from all the money you paid. While you do have Zuming at Exeter, your education is really up to yourself. Since the opportunities are similar in the two schools, the convenience of IMSA definitely wins.

Thanks for the comments.  I applied for 10th grade.

[quote="ThAzN1"]personally, I'd chose IMSA if you live in IL (a friend of mine, Albert Ni, did).[/quote]
Did he choose between IMSA and Exeter?

EDIT:  Money-wise, I think I should be fine; I was more interested in other aspects (workload, opportunities, teachers, classes, etc.)

Yeah, he applied to Exeter his sophomore year (?) at IMSA (which could actually be considered his freshman year -- he came straight from 8th grade) so if he went to Exeter, he'd start as a sophomore...

Cool, so you could be coming next year? I might see you there! (I'll be a senior :) )

[quote]Of course I expect to hear from students of Big Red here Wink and I will eagerly read what they have to say.[/quote]
AoPSer mathfanatic went to Exeter as a tenth grader with a clear math focus and discovered a whole world of extremely bright people who are good at everything including math.  Although he did write a perfect AMC 10 last year and although he does study number theory with Mr. Feng and IMO veterans, mathfanatic will tell you that the reason to go to Exeter is for everything else.

The most striking feature of Exeter is the Harkness teaching method which encourages discourse and discovery among peers.  For math, this means getting a difficult and intelligently designed problem set, figuring it out and presenting your solutions to your peers.  Students from Exeter are exceptionally well prepared for college and graduate level seminar style teaching.

Alohamath (aka, father of mathfanatic)

[quote="ThAzN1"]Disclaimer: This post might be a bit biased (I go to IMSA :) )[/quote]

As will this post; I go to Exeter.

[quote]Well, first off, IMSA is pretty darn cheap compared to Exeter (and that's an understatement) -- how often do you find a great, cheap, state-funded residential high school?  IMSA does have quite a lot of math resources, but probably not as much as Exeter.  Sure, it might make a difference, but I don't think it's worth the extra 30k or so each year.[/quote]

I'm not sure what the difference is between IMSA and Exeter in terms of math resources, but it seems to me that there are two types of resources:
1. good problems
2. good teachers.

Thanks to the internet, everyone can have access to good problems.  As for good teachers, you only really need one; if IMSA has a good math club coach, then I see little difference between IMSA and Exeter.  At Exeter, you work with Mr. Feng, but he can only teach you as much as you are willing to learn.

So I wouldn't choose Exeter over IMSA because one has more resources, or because one will turn you into a math machine.  I think a lot of that has to do with you, not with the school you go to.  If money is truly no object and you're only looking for the better school, then Exeter might make sense, but as far as I have seen, math excellence is much more dependent on devotion to math, rather than on the available resources.

Also, If you plan to focus entirely on math, I think IMSA might make more sense.  At Exeter, one spends a significant amount of time working on the other subjects (at least, I did).  It was beneficial for me, because I think I've broadened my interests a lot, particularly into writing.  But if you want to focus on math and math alone, then IMSA, which, judging by its name and reputation, is completely focused on math/science, might have a workload more amenable to your interests.

When I first came to Exeter, I expected to spend several hours a day working on math by myself.  After a few weeks, I realized that was completely impractical.  There is too much work and too many other things to do.  I have class from 8 to 6 (with some holes in between classes, but probably only an hour or two that could be spent working); dinner and club meetings usually until about 8; and probably about 3 hours of homework a night--although this can vary significantly (as a 10th grader, I could have as little as half an hour, but as an 11th grader things seem to have piled up).  Adding an hour of math work each night might be possible; but then again, one is rarely as efficient as anticipated, and one must spend time with friends.

If you were very serious about math, I know it is possible.  I spend 25-30 hours per week working on the school newspaper, so if I were to devote that to math, that'd be four hours per day...and I would probably be pretty good at math.  I think it would be difficult to maintain that focus, although there are several students here (plus Mr. Feng) who could certainly help push you.

[quote]Another thing that makes IMSA stand out is it's mentorship/inquiry program -- on most Wednesdays, we don't have class, and a lot of students pursue independent research with a mentor at a nearby college.  I'm not sure if Exeter has a similar type of program, but opportunity-wise, IMSA definitely has a lot to offer.[/quote]

We have no such program.

[quote]mathfanatic will tell you that the reason to go to Exeter is for everything else. [/quote]

Pretty much.  We have a great history program, a top-ranked Latin program (at least, that's what I hear--I don't take Latin), solid courses in other liberal arts courses (religion/philosophy is good, and English has some good courses, although they're mostly for seniors), and tons of extracurriculars.  If you are interested in areas outside of math, I would definitely recommend it.

Among clubs, there are a few math/science clubs (although, admittedly, surprisingly fewer than I expected upon arriving).  There's math club, physics club, and I think a recently formed general science club for random projects.  Math club is a pretty big committment; physics club meets once a week and does the USAPhO.

Oh, and the facts on the math program:

- math club meets twice a week (1 hour on Wednesdays, 2 on Sundays) to do pre-olympiad stuff.  We meet every weekday at lunch (half an hour) to work on olympiad problems.
- there are many math offerings, up to linear algebra, plus a special topics course (topic this year: number theory; teacher: Mr. Feng) which most of the top olymiad people take.
- math classes are based on a problem-solving approach, not a textbook-and-formula one.  If you take advantage of it (i.e. do all the problems) then it is very effective.  Basically, at the end of each math class, the teacher assigns some problems from the math "textbook" (basically a 100-page book of problems).  You do them for homework, and the next day in class, each person puts a problem on the board and then the class goes over it.  Repeat.

Thanks for all the information.  It's great to know that Exeter has a lot of other strong areas apart from math.  I'm particularly interested in biology and chemistry, but I haven't heard that much about them at Exeter.  How "strong" are those areas?  Is there much attention/focus or other activities in biology or chemistry?  How are the teachers and resources of those classes?

TJ's better than both of them. (Also biased.)  :P

[quote="probability1.01"]Yeah, but TJ isn't a boarding school.

Edit--In "response" to tokenadult, I'm just saying only people who live near VA can possibly go to TJ.
[/quote]

I take it you consider a school being a boarding school to be a feature rather than a bug.

After edit: Okay, so what you're saying is that TJ is available to a more limited catchment group (people who live in a defined part of northern VA) than, say, IMSA (available to Illinoisans in general) or Exeter (available in principle to anyone in the whole world). Yeah, if that's what you mean, I'd agree with that, and the way I would say it is "Yeah, but TJ is only available to people who live in one part of VA." That statement can either be taken as a reason to move to VA (I know of at least two families that have moved to VA from out of state so that a child could go to TJ) or as a reason why IMSA and (especially) Exeter are better schools: because they can draw from the best of the students (who are in the set {students willing to live away from home as teens}) in a very wide area.

It is maddeningly hard to balance the trade-offs between growing up at home (what I did) and living at a boarding school (what my wife did for six years, in Taiwan). IFF the boarding experience results in a truly superior academic experience, which I acknowledge it does in the cases of IMSA and Exeter, then there is a lot to be said for boarding. I have been testifying before my state's legislature about setting up an IMSA-like school here in Minnesota.

[quote="tokenadult"][quote="probability1.01"]Yeah, but TJ isn't a boarding school.[/quote]

I take it you consider a school being a boarding school to be a feature rather than a bug.[/quote]

I thought the issue was simply that a boarding school is a possibility for someone living in Illinois, whereas TJ is not.


Let $a, b, c, x, y, z$ be positive real numbers such that $cy+bz=a, az+cx=b, bx+ay=c$. Find the minimum value of $f(x, y, z) = \frac{x^2}{x+1}+\frac{y^2}{y+1}+\frac{z^2}{z+1}$.

this is the five solutions to the problem[in chinese]
enjoy it!


[quote="jrrbc"]theorem of S theory

$ (t/s\wedge t/r)\Rightarrow t/(s \plus{} r)$

i want the proof

 :(

theorems not proved from the book [b]Introduction to Mathematical Logic [/b](Elliott Mendelson)

[b]Good Christmas and Happy New Year[/b]:lol:[/quote]
By definition $ a \mid b$ $ \iff$ $ \exists d$ such that $ b \equal{} da$. So $ s \equal{} d_1t$ and $ r \equal{} d_2t$. Thus $ s\plus{}r \equal{} (d_1\plus{}d_2)t$ so $ t \mid s\plus{}r$ since $ \exists d$ such that $ td \equal{} s\plus{}r$ (We get $ d \equal{} d_1\plus{}d_2$).


Triangle $ABC$ is an equilateral triangle with $AC = 3$. Point $D$ and $E$ are selected on $AB$ and $BC$, respectively, such that $BD = BE = 1$. Find the perimeter of the quadrilateral $ACED$.

[hide="Answer"]$BD=BE=1$, so $DE$ is parallel to $AC$ and therefore $DE=1$. $AC+AD+CE+DE=3+2+2+1=\boxed{8}$.[/hide]

[hide]You can use proportions.  BD and BE = 2/6, so since they are parallel it has to be:
1 + 2 + 2 + 3 = 8[/hide]


Do you mean |-6+c| = 8?

I don't think absolute value can be negative.

[quote="symmetry"]So, in other words, there is no solution for c in this case, right?[/quote]

Correct  :)

So, |-6 + c| does not = 6 + c but
|-25| = 25, right?

What about if there is NEGATIVE infront of the absolute value.

Look:

-|25| = -25, right?

Or this one:

-|-25| = -25 NOT 25.

In other words, I do not multiply -1 by -25 like commonly done with parentheses, right?

So, -|-25| does not equal -(-25), right?

lol

[size=150]How to Solve a Linear Absolute Value Equation[/size]

Suppose we need to solve for $c$ in $|ax+b|=c$.

We split into two cases: $ax+b=c$, $ax+b=-c$.

CASE 1
$ax+b=c \Rightarrow x=\frac{c-b}{a}$.

CASE 2
$ax+b=-c \Rightarrow x=\frac{-c-b}{a}$.

$\mathbb{DONE.}$ :D


A number is written with one 1, followed by three 3's, then five 5's, then seven 7's, then nine 9's, then eleven 11's, etc. Thus it begins 13335555577... , and just as it gets to the 13's it has ... 11131313.... What is its 999th digit?

[hide]you can verify ur solution here: http://www.projecteureka.org/problem/question/297[/hide]

Though this is a bit gratuitous, If anyone ever needed some hint, here it is:

[b] Hint [/b]
The above problem demands the application of the arthimetic progression "Sum" formula (to some extent).

Though there maybe more ways of solving the above problem, I had arrived at the correct solution by using the "sum" formula.


Let $ a,b,c \ge 0,ab\plus{}bc\plus{}ca\equal{}1$. Prove that:
$ \sum \frac{ab\plus{}c}{a\plus{}b\plus{}1} \ge \frac{4}{3}$

[quote="NguyenDungTN"]Let $ a,b,c \ge 0,ab \plus{} bc \plus{} ca \equal{} 1$. Prove that:
$ \sum \frac {ab \plus{} c}{a \plus{} b \plus{} 1} \ge \frac {4}{3}$[/quote]
Put $ a\equal{}tanA,b\equal{}tanB,c\equal{}tanC$
$ A\plus{}B\plus{}C\equal{}\pi/2$

I cannot solve it, sorry.  :(

Can you please help me, NguyenDungTN with any idea or trick.

Thank you very much.

manlio, of course, you cann't solve it  :D because above inequality is wrong  :D Try \[ (a,b,c)\equal{}\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)\]
We have:
\[ LHS\minus{}RHS\equal{}\minus{}0,65....\]


limit
trigonometry
quadratics
floor function
algebra
calculus
calculus computations

$ \lim_{k \to 0}2 \tan^{\minus{}1}( \frac{bk\plus{}\sqrt{(bk)^{2}\minus{}tk(l\minus{}mk)}}{k(l\minus{}mk)})$ where $ b,t,l,m$ are constants.

hello, do you meant $ \lim_{k \to 0}2 \arctan\left( \frac{bk\plus{}\sqrt{(bk)^{2}\minus{}tk(l\minus{}mk)}}{k(l\minus{}mk)}\right)$?
Sonnhard.

hello,are your constants $ b,t,l,m$ positive?
Sonnhard.

hello, the term $ \frac{bk+\sqrt{(bk)^2-tk(l-mk)}}{k(l-mk)}$ has the form zero divided by zero for $ k=0$, using the rules of L'Hospital we get
$ \lim_{k=0}-\frac{1}{2}\frac{2b\sqrt{k(b^2k-tl+tkm)}+2b^2k-tl+2tkm}{\sqrt{k(b^2k-tl+tkm}(-l-2mk)}=undef$.
Sonnhard.

Hello,
Hope there havent been computational errors in applying l hopital.
Nevertheless, the following ideas may be useful:

$ \frac{bk \plus{} \sqrt{(bk)^{2} \plus{} tk(l \minus{} mk)}}{k(l \minus{} mk)}$ is a root of the quadratic equation
$ k(l\minus{}mk)x^{2} \minus{} bkx \minus{} t \equal{} 0$

Denoting the above qauntity by $ \alpha$ and its conjugate by $ \beta$
and using the facts $ \sigma \equal{}  \alpha \plus{} \beta \equal{} \frac{2b}{l\minus{}mk}$ and
$ \pi \equal{} \alpha\beta \equal{} \frac{\minus{}t}{k(l\minus{}mk)}$
$ 2 \arctan \alpha \equal{} \arctan \frac{2\alpha}{1 \minus{} \alpha^{2}}$ and writing $ 2\alpha \equal{} \alpha \minus{} \beta \plus{} \sigma$ and $ \alpha^{2} \equal{} \frac{\alpha \pi}{\beta}$
its jus a guess of ideas, dunno if it would work. Stronger ideas may b reqd.

I am pretty sure there are issues taking $ k \to 0$  for instance, for $ 0<k< \frac{tl}{b^2\plus{}tm}\equal{}\gamma$

$ \sqrt{(bk)^{2}\minus{}tk(l\minus{}mk)}\equal{} \sqrt{((b^2\plus{}tm)k\minus{}tl)k}\equal{}\sqrt{b^2\plus{}tm}\sqrt{k(k\minus{}\gamma)}$

is undefined

The limit exists and is equal to $ \pi$ in the sense that for each $ \epsilon > 0$ there exists a $ \delta > 0$ such that for all $ k\in \mathbb{R}$, $ 0 < |k| < \delta$,
\[ \left\vert 2\arctan \left(\frac {b \plus{} \sqrt {b^2 \plus{} tm \minus{} \frac {tl}{k}}}{l \minus{} mk}\right) \minus{} \pi\right\vert < \epsilon
\]
where $ \left\vert \cdot \right\vert$ denotes the [i]complex[/i] norm.  If $ \left\vert \cdot \right\vert$ were to instead denote the real norm, the limit does not exist for the reason ken4000 stated.
[hide="remark"]Actually an even stronger statement holds: $ \lim_{k\to 0}2\arctan \left(\frac {b \plus{} \sqrt {b^2 \plus{} tm \minus{} \frac {tl}{k}}}{l \minus{} mk} \right)\equal{} \pi$ in the complex plane.[/hide]

@joeblow can u post the proof of ur assertion

We clearly have $ \lim_{k\to 0^ - }\frac {b + \sqrt {b^2 + tm - \frac {tl}{k}}}{l - mk} = + \infty$, so $ \lim_{k\to 0^ - }2\arctan \left(\lim_{k\to 0}\frac {b + \sqrt {b^2 + tm - \frac {tl}{k}}}{l - mk}\right) = \pi$.  For $ k > 0$ and sufficiently small (see ken4000's post) the limit becomes
\[ \lim_{k\to 0^ + }2\arctan \left( \frac {b}{l - mk} + i\frac {\sqrt {\left \vert b^2 + tm - \frac {tl}{k}\right \vert }}{l - mk}\right) = \lim_{k\to 0^ + }2\arctan \left(\psi (k) + i\phi (k)\right)
\]
where $ \frac {b}{l} < \psi (k) < c$ for some constant $ c$, and $ \lim_{k\to 0^ + }\phi (k) = + \infty$.  Note that $ \arctan \left(\psi (k)+i\phi (k)\right)=\arctan \left(\frac {1}{\left(\frac {\psi (k) - i\phi(k)}{\psi(k)^2 + \phi(k)^2}\right)}\right)$.  The denominator approaches $ 0$, so the series* for $ \arctan \left(\frac{1}{x}\right)$ implies,
\[ \lim_{k\to 0^ + }2\arctan \left(\psi(k) + i\phi (k)\right) = \lim_{k\to 0^ + }( - 1)^{\left \lfloor \frac {\pi + 2\arg (\psi(k) - i \phi(k))}{2\pi}\right \rfloor}\pi
\]
Since $ \psi(k) > 0$ and $ - \phi(k) < 0$, we have $ \frac {3\pi}{2} < \arg \left( \psi(k) - i\phi (k)\right) < 2\pi$, and therefore $ \left \lfloor \frac {\pi + 2\arg (\psi(k) - i\phi(k))}{2\pi}\right \rfloor = 2$ and the limit in question is $ \pi$.

[hide="*"]Recall that $ \arctan$ is analytic at infinity, and its series is given by (for the usual branch cut):
\[ \arctan \left (\frac {1}{x}\right) = \frac {1}{2}( - 1)^\left \lfloor \frac {\pi + 2\arg (x)}{2\pi}\right\rfloor}\pi - \sum_{n = 0}^\infty ( - 1)^n \frac {x^{2n + 1}}{2n + 1}
\]
In our case $ |x|\to 0$ so $ \sum_{n = 0}^\infty ( - 1)^n \frac {x^{2n + 1}}{2n + 1}$ drops out in the limit.
[/hide]

If $ x \equal{} \frac {bk \plus{} \sqrt {(bk)^{2} \minus{} tk(l \minus{} mk)}}{k(l \minus{} mk)}$,
then $ (l \minus{} mk)x^2 \minus{} 2bx \plus{} \frac {t}{k} \equal{} 0$.
We want to compute $ \lim_{k \to 0}2\arctan(x) \equal{} \lim_{k \to 0}2y$, where
 $ (l \minus{} mk)\tan^2(y) \minus{} 2b\tan(y) \equal{} \frac {t}{k}$.
Taking the limit on both sides shows that $ \lim_{k \downarrow 0}2y \equal{} \pi$ and $ \lim_{k \uparrow 0}2y \equal{} \minus{} \pi$.


trigonometry
calculus
calculus computations

$ y'' \plus{} 2y \equal{} e^{2x} \plus{} x$

I found the characteristic equation: $ m^2 \plus{} 2 \equal{} 0$, but I'm unsure of what to do when I get unreal roots.  :/

Because the roots are imaginary, instead of $ e^{\lambda_1 x}$ and $ e^{\lambda_2 x}$ you have $ \cos (\sqrt {2} x)$ and $ \sin (\sqrt {2}x)$. You can still do variation of coefficients to handle the nonzero righthand side.


find all of the polynomials $p(x,y)$with real coefcients such that :$p(x+y,x-y)=2p(x,y)$


logarithms
limit
integration
calculus
calculus computations

For today's exercise, assume known

[b]Theorem 1[/b]. If a sequence $x_{n}\to\xi ,$ then so does the sequence of arithmetic means

\[\frac{x_{0}+x_{1}+\cdots+x_{n}}{n+1}\to\xi .\]

[b]Theorem 2[/b]. If a sequence $y_{n}\to\xi ,$ then so does the sequence of geometric means

\[\sqrt[n+1]{y_{0}y_{1}\cdots y_{n}}\to\xi .\]

Today's Limits:

Divide the inverval $[a,b]$ (for $0<a<b$) into $n$ equal parts and let the points of division be denoted by the sequence $a=x_{0},x_{1},x_{2},\ldots x_{n}=b$. Prove that the geometric mean
\[\sqrt[n+1]{x_{0}x_{1}\cdots x_{n}}\to\frac{1}{e}\left(\frac{b^{b}}{a^{a}}\right)^{\frac{1}{b-a}}\]
and the harmonic mean
\[\frac{n+1}{\frac{1}{x_{0}}+\frac{1}{x_{1}}\cdots \frac{1}{x_{n}}}\to\frac{b-a}{\log b-\log a}\]
as $n\to\infty$.

Notice that if $0<a<b$, then limit
\[\lim_{n\rightarrow \infty}\left\{ \int_{0}^{1}\left[ bx+a(1-x)\right]^{\frac{1}{n}}dx\right\}^{n}=\frac{1}{e}\left( \frac{b^{b}}{a^{a}}\right)^{\frac{1}{b-a}}\]
holds per [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=115865]this thread[/url], can you think of a similar limit which would $\to \frac{b-a}{\log b-\log a}$ ? Note: I haven't yet, but the conincidence was to great to ignore.

For the geometric mean, we can write it as

$\lim_{n \rightarrow \infty}\prod_{k=0}^{n}\left(a+\frac{k(b-a)}{n}\right)^{\frac{1}{n+1}}= L$.

Taking the natural log, we get

$\lim_{n \rightarrow \infty}\frac{1}{n+1}\sum_{k=0}^{n}\ln{\left(a+\frac{k(b-a)}{n}\right)}= \ln{L}$,

but the limit of this is simply the average value of $\ln{x}$ on the interval $[a,b]$, or

$\ln{L}= \frac{1}{b-a}\int_{a}^{b}\ln{x}dx = \frac{1}{b-a}[x\ln{x}-x]_{a}^{b}= \frac{1}{b-a}\left(b\ln{b}-a\ln{a}+(a-b)\right)$,

simplifying to

$\ln{L}= \ln{\left(\frac{b^{b}}{a^{a}}\right)^{\frac{1}{b-a}}}-1$

$L = \frac{1}{e}\left(\frac{b^{b}}{a^{a}}\right)^{\frac{1}{b-a}}$.

EDIT: The reciprocal of the harmonic mean can also be written as an integral, namely $\frac{1}{b-a}\int_{a}^{b}\frac{dx}{x}$, which immediately gives us the result.


Who can help me with this problem:
Richard is walking up a stair that has 10 steps. With each stride he goes either one step or two steps. In how many different ways can Richard go up the stairs?
And please tell me how to get to the answer

[hide]
To reach the $10th$ step (the top of the stairs), we know the person must have either taken $1$ step from the $9th$ stair, or taken $2$ steps from the $8th$ stair. Since this covers all possible ways to reach the $10$ step, the number of ways to reach the $10th$ step is the sum of the number of ways to reach the $8th$ step and the number of ways to reach the $9th$ step. This is the same for all the steps above and including the second one.

So if we let $d_{n}$ denote the number of ways to reach the $nth$ step, then for all $n \geq 2$
\[d_{n}= d_{n-1}+d_{n-2}\]

$d_{1}$ is obviously equal to $1$; $d_{2}$ is equal to $2$; and from there, you can the above relation to get $89$ ways to climb to the top of the stairs.
[/hide]


Hello there,

[i]I posted the same problem in Classroom Math but I realised that this exercise may be too difficult for that forum, so I am posting it here. (Admins: please feel free to delete original post)[/i]

I am having some trouble with factoring. Could anyone please give me a hint?

1. $\frac{a^{2n}+(a^{n})(b^{n})}{3a^{3n}+(3a^{2n})(b^{n})}$

Thanks.

----
[u] My work: [/u]

1. I tried to factor out a and 3a from both the numerator and the denominator but then the answer said that the denominator had a. If I divide a by 3a, then I will eliminate the a's and thus, my answer will be incorrect.

[hide]

factor a^n and 3a^(2n)

$\frac{a^{n}\left(a^{n}+b^{n}\right)}{3a^{2n}\left(a^{n}+b^{n}\right)}$

$\frac{1}{3a^{n}}$
[/hide]

[quote="catalan"][u] My work: [/u]

1. I tried to factor out a and 3a from both the numerator and the denominator but then the answer said that the denominator had a. If I divide a by 3a, then I will eliminate the a's and thus, my answer will be incorrect.[/quote]

The problem with that is, that you don't have any $3a$'s in your equation. You have some $3a^{2n}$'s, which hints at that you should try dividing by $a^{n}$.


6 people can do 6 things in 6 hours

If So

and all the people work at the same rate

How long does it take 8 people to do 8 things?

Similiar problems to these kinds of ratio problems have confused me forever.

Please Help me

And Show All Your work

[hide]Well, if $6$ people can do $6$ things in $6$ hours, then one person can do one thing in one hour...  I think it would be 8 hours, since $1*8=8$..., but that doesn't sound right.[/hide]

in one hour, the six people do one thing. so in each hour, each of the six people do 1/6 of the thing. if there are 8 hours, they do 8*1/6=4/3 of a thing per hour together. so divide 4/3 into 8 to find how many hours, this is 6.

that helped alot man

another question?

how do you know if something directly varies or indirectly varies in word problems?

like my problem consisted of directly varying variables

Think of it this way:

Direct means
If more x, then more y.

For example, the more people there are, the more food we need.

Indirect means
If more x, then less why.

For example, if we have more people working, the work takes less time.

:) Hope this helped.

yeah it helped

but i think its the other way around though...

like direct is:
the more this, then the more of the other

indirect is:
the more this, then the less of the other

Edit:
SORRY :oops:
I didn't read your explanation correctly
You were absolutely correct


induction
geometry
geometric transformation
IMO Shortlist
binomial coefficients
combinatorics solved
combinatorics

Draw a circle, and place n points on the circumference, where n is a positive integer. Draw lines between each pair of points. What is the maximum of regions that these lines divide the circle into ?

My guess : C(n,2) + C(n,4) + 1

A try with small values gives 2,4,8,16  :D !  So the answer should be 2^(n-2). Alas next is 31  :( ....

. We start with the polygon : 1

. Each time a new diagonal is added a new zone is created : there are C(n,2) - n diagonals

. If no tree diagonals intersect, each time two diagonals crosses a new zone is created : there are C(n,4) diagonal intersections (the ways to choose 4 vertices)

. Not to forget the n small zones between the polygon and the circle : there are n of them.

I think It is something like n 2 - n + 1

Hi Dust,

that's not true. Nevertheless you can write the binomial coefficient as polynom. Define nCr as the quotient n!/(r!(n-r)!). Let the sequence be denoted by u_n. The elements of u_n generate the respective values of the maximum number of circle regions. So we get:

u_n = nC4 + nC2 + 1 = n!/(4!(n-4)!) + n!/(2!(n-2)!) + 1

u_n = [n(n-1)(n-2)(n-3)]/24 + [n(n-1)]/2 + 1 = 1/24 * (n^4-6n^3+23n^2-18n+24)

It is easy to be seduced by the ease of experimentation-conjecture. But this is only part of mathematical investigation. Sometimes a relatively uninformed investigation leads us astray.

It is easy to check as Belenos did that u_1 = 1, u_2 = 2, u_3 = 4, u_4 = 8 and a_5 = 16. The conjecture that u_n = 2^(n-1) is practically inescapable. Yet u_6 = 31 and not 32, so something else is going on.

This is a well-known problem, e.g. it was an early IMO shortlist problem. I provide two sources for the problem:

(1) Richard K. Guy, The strong law of small numbers, American Mathematical Monthly, 95: 697-712, 1988.

This source provides a similar solution to Belenos using binomial coefficients. He also illustrates similar examples as: Check whether f(n) = n^2 + n + 41 is a prime for all positive integers n ?

(2)

Elena Alexandrovna Morozova, Ivan Semenovic Petrakov, V.A. Skvortov:
title: 'International Mathematical Olympiads'
publisher : education, Moscow

Please have a look at http://mech.math.msu.su/~manuilov/morosova.html and you will be capable of reading that altogether there are 4 editions of the book. The problem is covered in all the four editions. This source does not apply binomial coefficients in order to solve the problem but makes use of elementary estimates. Finally the formula is verified by induction on n.

There is also a Romanian translation of this book available:

E.A.Morozova, I.S.Petracov, V.A. Skvortov, Olimpiade internationale de matematica. (probleme, rezolvari, punctaj). Editura Tehnica, Bucuresti, 1978, 367p. (Tradusa din rusa) Lucrarea contine materialele primelor 20 olimpiade internationale de matematica.

For a reference see: http://math.ournet.md/bib.html - Look at item 26.


geometry
inradius
incenter
cyclic quadrilateral

I like this problem!
Assume that the incircle of $\Delta ABC$ meets $BC,AB,AC$ ar $D,E,F$ respectively.
We know the lengths $AB+AC$, $BC$ and the inradius of the triangle $ABC$.
1.We know the inradius but we do not know the position of the incentre still. But, if we draw a line perpendicular to $BC$ and cut the length of the inradius from it and draw a line parallel to $BC$ which passes through it,
we can say that the incentre must line on that line.
2.Note that
 $AB+AC=(AE+BE)+(AF+CF)=(AE+AF)+(BE+CF)=(AE+AF)+(BD+DC)$
$=(AE+AF)+BC$
Hence if we subtract the length $BC$ from $AB+AC$(It can be done by drawing a line of length $AB+AC$ and cutting the length of $BC$ from it and the length of the remaining segment will be the length of $AE+AF$ but as $AE,AF$ are both tangents, $AE=AF$ hence we have got the length of the tangents to the incircle from the vertex $A$, moreover the distance $EI=FI=$inradius is known, also $\angle AEI=\angle AFI=90^{\circ}$ hence we can construct the cyclic quadrilateral $AEFI$.
Hence we have found out the angle $A$.
As we know $\angle A$, we can construct $90^{\circ}+\frac{\angle A}{2}$ which is equal to the angle $\angle BIC$.
3.As we know angle $BIC$, side $BC$ we can construct a circle which passes through $B,C$ and subtends angle $BIC$ on the circumference.(It is very easy to do).
And the intersection of this circle with the line parallel to $BC$ will be the incentre $I$.
Now it is easy to locate the vertex $A$.
Done!

Do we know the lengths AB+BC, BC  ?
we know AC+BC, AB !

but jrrbc, it is the same!
just the names are different but the construction is the same!
For example $f(x,y,z)=f(z,x,y)$ if $f$ is symmetric.
If I do it for triangle $XYZ$ instead of #ABC$ does the construction change? :wink:

[quote="jrrbc"]not just the names: AB+BC, BC  (BC adjacent),  AC+BC, AB (AB oposit)[/quote]
Check properly, I have done the "opposite one ". :wink:


Prove that any k in Z can be written

(*) k = +-1^2 +-2^2+-3^2 +-....+-m^2

where m is an integer, +- means we write + or - depend on m

Prove also the decomposition of k as (*) is infinte way

Application : Find the Erds-Suranyi decomposition of
2002; -2003; -2004

This is well known.

Note that 1 = + 1, 2 = - 1 - 2 - 3 + 4, 3 = - 1 + 2 and 4 = - 1 - 2 + 3.
Now if we can represent a positive integer k this way then we can also represent k+4.
Since for any m, we have m - (m + 1) - (m + 2) + (m + 3) = 4.
So the result easily follows by induction.
If we can represent k > 0 then we can also represent -k (obvious).
So every integer k has such a representation.


How many different 6 letter "words" can be formed from the word THIRST if:

"T" must always immediately precede "H"?

I'm finding the answer to be 120 but the answer on the work sheet says 60.

also as a sidenote, i'm just learning combinatorics and find it really difficult in that it is nearly impossible to check your answer. Often times i question the answer of my teachers as they can sometimes get wrong answers as well. Does anyone else have the same experience?

i mean it's not like geometry where there is a visual representation or algebra where there's numbers, there's just nothing tangible with combinatorics, i mean realistically i can't possibly draw a tree diagram or something for problems that may have thousands of arrangements...

I would say that you're right: Make all the words out of the letters TIRST - there are ${5!\over 2!1!1!1!}=60$ of those - and then just append H after any of the two T's for the total of $120$ words.

[hide]There are ${6\choose 3}$ ways to pick the three places that the Ts and H will go, and the placements of T,T, and H are uniquely determined. There are $3!$ ways to arrange the remaining letters.${6\choose 3}3!=120$.[/hide]

[hide]Just imagine "TH" as one letter.

"THIRST" then has $5$ letters, so there are $5!=120$ "words" that can be formed.[/hide]

I could say something wrong, since the fact that I'm just learning combinatorics, as princevegetam...
Anyway...

I'd consider the group TH as a unique letter, then I'd have $5!=120$ ways to permute the other letters. BUT notice that substituting one T with the other you obtain two indistinguishible anagrams, which count for just one "word" satisfying the problem... Thus you must divide by 2! and get the (I hope so) correct answer $\frac{120}{2!}=60$.

EDIT: Thanks, rnwang2!  :wink:

Oh yeah, that's right. You have to divide by $2!$ since there are $2$ T's.

Then, the answer is $\boxed{60}$.

Right, it's $2!$, not just $2$... I edited for completeness!...  :roll:


les $ ABC$ an acute triangle ,and $ P$ a point inside it,les $ A_{1}$,$ B_{1}$,$ C_{1}$ be the projections of P in $ BC$,$ AC$,and$ AB$
and les $ A_{2},B_{2},C_{2}$ be the projections of P in $ B_{1}C_{1},A_{1}C_{1},A_{1}B_{1}$ ,
and so on........
prove that $ S(A_{n}B_{n}C_{n})\leq S(ABC)/4^{n}$
remark:$ S(XYZ)$ mean the area of triangle $ XYZ$

ok Euler formula \[ S_{pedal}=\frac{1}{4}(1-\frac{OP^{2}}{R^{2}})S_{ABC}\le\frac{S_{ABC}}{4}\]

yes ,that's thru :D ,but you gorget that $ 1-OP^{2}/R^{2}$is not always positive :P

[quote="drapt@"]les $ ABC$ an acute triangle ,and $ P$ a point inside it...[/quote]

[quote="drapt@"]yes ,that's true, but you forget that $ 1-\frac{OP^{2}}{R^{2}}$ is not always positive.[/quote]

You seem to contradict yourself.


[hide]

Well, examining the choices, at first glance it looks like $(D)$ is the only one that will give us our $-4x^2$ term. (because we know that $y^2$ is going to give us a $+2$ to deal with)

So we plug the numbers in to see if they check

$x^2((x+\frac{1}{x})^2 + (x + \frac{1}{x}) - 6)$

$x^2(x^2 + 2(x)(\frac{1}{x}) + (\frac{1}{x})^2 + (x + \frac{1}{x}) - 6)$

$x^2(x^2 + 2 + \frac{1}{x^2} + x + \frac{1}{x} - 6)$

$x^2(x^2 + x - 4 + \frac{1}{x} + \frac{1}{x^2})$

$x^4 + x^3  + -4x^2 + x +1$

So we could conclude the answer is $(E)$

[/hide]


logarithms
calculus
calculus computations

Solve the differential equation

$ x(4y\,dx\plus{}2x\,dy)\plus{}y^3 (3y\,dx\plus{}5x\,dy)\equal{}0.$

ps.: My answer is $ 24x^2 y\plus{}11xy^5 \equal{} C$,
but the answer from the book is $ x^4 y^2\plus{}x^3 y^5\equal{}C$.

The most elementary thing you can do after the fact is differentiate to check.

Your solution doesn't check. It's not correct.

The book solution differentiates to

$ (4x^3y^2\plus{}3x^2y^5)dx\plus{}(3x^4y\plus{}5x^3y^4)dy\equal{}0$

Divide that through by $ x^2y$ and you get the original differential equation. (Which, in retrospect, reveals that the way to have solved it would have been to multiply by $ x^2y.$)

[quote="Kent Merryfield"]The most elementary thing you can do after the fact is differentiate to check.

Your solution doesn't check. It's not correct.

The book solution differentiates to

$ (4x^3y^2 \plus{} 3x^2y^5)dx \plus{} (3x^4y \plus{} 5x^3y^4)dy \equal{} 0$

Divide that through by $ x^2y$ and you get the original differential equation. (Which, in retrospect, reveals that the way to have solved it would have been to multiply by $ x^2y.$)[/quote]
I see. But I would like to know where did I get wrong and how to fix it. The following is my "solution".

$ x(4y\,dx\plus{}2x\,dy)$
$ \equal{}x(2y\,dx\plus{}2y\,dx\plus{}2x\,dy)$
$ \equal{}x[2y\,dx\plus{}2d(xy)]$
$ \equal{}2(xy)\,dx\plus{}2x\,d(xy)$
$ \equal{}d(2x^2 y),$

$ y^3 (3y\,dx\plus{}5x\,dy)$
$ \equal{}y^3[3d(xy)\plus{}2x\,dy]$
$ \equal{}y^2[3y\,d(xy)\plus{}2(xy)\,dy]$
$ \equal{}y^2[2d(xy^2)\plus{}y\,d(xy)]$
$ \equal{}2y^2\,d(xy^2)\plus{}y^3\,d(xy)$
$ \equal{}d\left(\frac23 y^3\cdot xy^2\plus{}\frac14y^4\cdot xy\right)$
$ \equal{}d\left(\frac{11}{12}xy^5\right).$

Thus the original equation can be rewritten as
$ d(2x^2y)\plus{}d\left(\frac{11}{12}xy^5\right)\equal{}0.$
Hence the solution is $ 24x^2y\plus{}11xy^5\equal{}C$.

The issue is somewhere in the second block, since it is true that $ d(2x^2y)\equal{}4xy\,dx\plus{}2x^2\,dy.$

You're making so many leaps from line to line that I'm having trouble following what you were doing. My suggestion is that you rewrite it yourself putting in as many intermediate steps as you can.


[quote="arqady"]$a^2+b^2+c^2=3,a\geq0,b\geq0,c\geq0.$
Prove that:
$(4-abc)(3-a-b-c)\geq3-ab-ac-bc.$[/quote]
Sorry. $4-abc\geq3$ and $3-a-b-c\geq\frac{1}{3}(3-ab-ac-bc)\Leftrightarrow(a+b+c-3)^2\geq0.$

[quote="leepakhin"]Why are you apologizing to yourself and then posting the solution? :huh:[/quote]
Since this is not unsolved problem. :)


Let z be a solution to z^n = 1. Show that
1) for any integer k, z^k is also a solution
2) 1 + z + z^2 + z^3 + ... + z^(n-1) = 0

// Demoivre's formula:  z^k = |z|^k cis(k arg(z))

um....I can do part 1).....cuz it's easy..
but how to do part 2)?????????????

Thanks....

o yea! ...good.....
but then does this method apply to negative k also????

Actually the problem is wrong. $z=1$ is also a solution of $z^n=1$ however $1+z+\cdots + z^{n-1} = n \neq 0$ ;)

actually...now the question is changed...add one more line:
where [b]z doesn't equal to one[/b].....

and just figured out that if using the geometric series, then we have to restrict |z|<1 .....which is not the general case....

any other idea???

[quote="math beginner"]and just figured out that if using the geometric series, then we have to restrict |z|<1 .....which is not the general case....[/quote] That's not true. The geometric series formula works for any finite sequece in geometric progression.

If $n-1$ is an odd number, then $x=-1$ is a solution.

[quote="Rep123max"]$\displaystyle 1+z+z^2+z^3+\ldots+z^{n-1}=\frac{z^n-1}{z-1}=\frac{1-1}{z-1}=0$. All we used was the sum of a geometric series ($\displaystyle a+ar+ar^2+ar^3+\ldots+ar^n=\frac{ar^{n-1}-a}{a-1}$).[/quote]

z^n-1 / z-1. So far you're correct. But if you fill in z=1 you simply get 0/0. So you must take z=1 separatedly.

$\lim_{z\rightarrow 1} \frac{z^n-1}{z-1}=\lim_{z\rightarrow 1} nz^{n-1}=n$


[url]http://arxiv.org/pdf/0812.0930[/url]

Can anyone explain what is stated in the proof above?  :lol:

I now post something arguably not any less worth reading than some of the threads posted by Gen8.

No, I cannot explain a single thing from that paper. But I [i]can[/i] say that the paragraph spacing is horrible and inconsistent; in some places there is an empty line between adjacent paragraphs, in others there is not. And even in the places where there is not an empty line, there is no indent, which is positively (or should I say, negatively) ugly. I would be happy to give a better LaTeX document if someone payed me a reasonable amount to do so.

Can we please stop discussing every new "proof" of some theorems¿ It's kind of annoying, especially as these discussions always have no realcontent!

People are always coming out with attempted proofs at major open problems, and they are always getting shot down. If we open a thread every time someone tries to solve a major open problem, we'd have a lot of threads with disappointing conclusions! This happened recently with a proof of the Riemann Hypothesis (I think?).

Of course, it would be cool if this one worked!

Can you find it?  I would like to see it, even if it is wrong.

I was talking about [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213013]this thread[/url].

What about making a compromise between both views, i.e. a sub-forum of some ML/AoPS section that discusses solutions attempts of notoriously known unsolved problems. This way it does not annoy people who are tired of seeing these threads but those who want are still able to do it. But I agree with Daniel on the disadvantage of these threads on ML/AoPS: We have few experienced math researchers here who can comment on a level like Terence Tao, and those who could decide not to comment.


limit
logarithms
integration
inequalities
trigonometry
real analysis
real analysis unsolved

Give an example of a sequence$ \{x_{n}\}$(if possible) such that $ x_{n+1}-x_{n}\to 0$ but the sequence of its arithmetic mean, $ \{ \frac{x_{1}+\dots+x_{n}}{n}\}$, is not convergent.

Note if $ \{ x_{n}\}$ is convergent then $ \frac{x_{1}+...+x_{n}}{n}$ converges as well (this is a known result) and $ x_{n+1}-x_{n}\to 0$ since $ \lim x_{n+1}= \lim x_{n}$. So we need to find a [b]non-convergent[/b] sequence $ \{x_{n}\}$ such that $ x_{n+1}-x_{n}\to 0$. And I fail even to find that  :blush:

I would have phrased mlok's example as $ x_{n}=\ln n.$ Slightly easier to manipulate that way.

Here's a slightly different and even easier example. Let $ x_{n}=\sqrt{n}.$

It's not hard to show that $ \sqrt{n+1}-\sqrt{n}<\frac1{2\sqrt{n}}$ and thus $ x_{n+1}-x_{n}\to0.$

$ \frac1n\sum_{k=1}^{n}x_{k}=\frac1n\sum_{k=1}^{n}\sqrt{k}\,> \frac1n\int_{0}^{n-1}\sqrt{x}\,dx=\frac23\cdot\frac{(n-1)^{3/2}}n$

And that clearly goes to infinity like a multiple of $ \sqrt{n}.$

---

Can you prove this? If $ \lim_{n\to\infty}x_{n}=\infty,$ then $ \lim_{n\to\infty}\frac1n\sum_{k=1}^{n}x_{k}=\infty\,?$

You could consider that another case of the theorem sylow_theory was quoting when he said "this is a known result".

I once proved this result some time ago, looking through my scrippled notes I found this.

[b]Theorem:[/b] If $ (s_{n})$ is convergent to $ s$ then $ \lim \left( \frac{1}{n}\sum_{k=1}^{n}s_{n}\right) = s$

[b]Proof:[/b] We can find $ N_{1}$ such that $ |s_{n}-s|<\frac{\epsilon}{8}$ for $ n>N_{1}\in \mathbb{N}$. Choose $ N_{2}\in \mathbb{N}$ so that $ \frac{|s_{1}-s|+...+|s_{N_{1}}-s|}{N_{2}}< \frac{\epsilon}{8}$. Choose $ N = \max \{N_{1},N_{2}\}$ then $ |s_{n}-s|<\frac{\epsilon}{2}$ (this inequality is obvious) [b]and[/b] $ \frac{|s_{1}-s|+...+|s_{N}-s|}{N}< \frac{\epsilon}{2}$ (this inequality takes a little work to show but its not so bad).

Now if $ n>N$ we have:
$ \left| \frac{s_{1}+s_{2}+...+s_{n}}{n}-s \right| = \left| \frac{s_{1}+...+s_{n}-ns_{n}}{n}\right| = \left| \frac{(s_{1}-s)+...+(s_{n}-s)}{n}\right| \leq \frac{|s_{1}-s|+...+|s_{n}-s|}{n}= \frac{|s_{1}-s|+...+|s_{N}-s|+|s_{N+1}-s|+...+|s_{n}-s|}{N+(n-N)}\leq \frac{|s_{1}-s|+...+|s_{N}-s|}{N}+\frac{|s_{N+1}-s|+...+|s_{n}-s|}{n-N}< \frac{\epsilon}{2}+\frac{\epsilon/2 (n-N)}{n-N}= \epsilon$ Q.E.D.

---------------------------------------------
My own question I wondered about.

Let $ (s_{n})$ be a sequence of [i]non-negative[/i] numbers so that $ \lim s_{n}= s$ is the following true $ \lim \sqrt[n]{s_{1}...s_{n}}=s$.

My approach: Was to use the AM-GM inequality since $ \sqrt[n]{s_{1}...s_{n}}\leq \frac{1}{n}(s_{1}+...s_{n})$.
This is the UPPER bound. Now if we can find a LOWER bound with the SAME limit then we can use the SqUeZeZe principle. But I cant find a lower limit.

[quote="mlok"]Apply the above result to $ (\log s_{n})$.[/quote]
Thanks, its really is a nice solution, I really like it :)

sylow_theory: notice how similar your proof of this theorem is to the proof I gave [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=156964]here[/url]. Notice also how similar the underlying theorems are, when looked at in the right way.

[quote="mlok"]What if we assume in addition (to the original problem) that $ (x_{n})$ is a bounded sequence?[/quote]
Consider the example of $ x_{n}=\sin(\ln n).$ Then $ |x_{n+1}-x_{n}|<\frac1n,$ satisfying the condition that consecutive differences tend to zero, and the $ x_{n}$ themselves are certainly bounded.

For large $ k,\ \sum_{e^{k\pi}<n<e^{(k+1)\pi}}\sin(\ln n)$ is approximately $ c(-1)^{k}e^{k\pi}$ for some constant $ c.$

Now let $ n$ be approximately $ e^{m\pi}.$ We can approximate

$ \frac1n\sum_{k=1}^{n}x_{n}$ by $ e^{-m\pi}\sum_{k=0}^{m}c(-1)^{k}e^{k\pi}$

and then in turn approximate that by $ \frac{C(-1)^{m}e^{m\pi}}{e^{m\pi}}.$

That subsequence does not converge. It's bounded, of course - any average of the $ x_{n}$ must be contained in any interval that contains all of the $ x_{n}.$ But it doesn't converge.

I left the details a bit sketchy, but everything can be filled in - this is the essence of the argument.