calculus Asymptote algebra polynomial function limit conics

I am taking a summer calculus course and it goes very fast and I seem to be falling behind.  Anyway, I have a few questions (more to come):

(1) Does the way that you find slant asymptotes guarentee that there is only one slant asymptote?  Why does polynomial long division and ignoring the remainder give you a slant asymptote?

(2) A car starts at a high speed and its speed then decreases slowly.  Sketch a graph of the distance the car has traveled as a function of time.
---> I drew a horizontal line going across that then curved and fell to the right

(3) When you are doing limits w/trig functions, why does it make a difference if you use radians or degrees?  Why is [tex]\displaystyle \lim_{\theta\to0}{\frac{sin\theta}{\theta}}[/tex] equal to  :pi: /180 when  :theta:  is in radians, and 0 when  :theta:  is in degrees?

(4) If you have an graph of distance v. time and it looks like an upside down parabola, how do you tell which is greater: the average velocity between t=1 and t=3 or the instantneous velocity at t=2?

(5) A particle moving at varying velocity along a line and s=f(t) represents the particle's distance from a point as a function of time, t.  Sketch a possible graph for f if the average velocity of the particle between t=2 and t=6 is the same as the instantaneous velocity at t=5.
-----> For this I drew a straight line, because that implies constant velocity since slope of a line is constant, but I am not sure.

(6) Find [tex]\displaystyle \lim_{x\to0^+}{x^x}[/tex].  I tried graphing this on my calc but I got this really strange graph that had all of these dots left of the y-axis.  I really don't understand this graph.

(7) Can someone please explain to me the delta-epsilon definition of a limit?

(8) Find [tex]\displaystyle\lim_{x\to\infty}{\frac{3e^{x}+2}{2e^x+3}}[/tex].

(9) Find [tex]\displaystyle\lim_{x\to\infty}{\frac{2e^{-x}+3}{3e^{-x}+2}}[/tex].

(10) Find [tex]\displaystyle\lim_{h\to0}{\frac{\sqrt{4+h}-2}{h}}[/tex].

(11) Find a value of k so that the limit exists: [tex]\displaystyle\lim_{x\to-\infty}{\frac{e^{2x}-5}{e^{kx}+3}}[/tex].

Okay, these are all of my questions for the first two sections.  Sorry, I know this is a lot.  Thanks for your help!

Shoot, I meant to press "preview" instead of enter.  It will take me a few minutes to finish editing the top message.

EDIT: Done.

[quote="JS1527"](2) A car starts at a high speed and its speed then decreases slowly.  Sketch a graph of the distance the car has traveled as a function of time.
---> I drew a horizontal line going across that then curved and fell to the right[/quote]

The "distance that the car has traveled" keeps increasing as long as the car is moving at all. So that graph shouldn't be falling to the right. It rises rapidly at first, and then rises more slowly.

[quote="JS1527"](7) Can someone please explain to me the delta-epsilon definition of a limit?[/quote]

Hmmm, this is a classic question about a rigorous calculus course. I THINK Michael Spivak's Hitchhiker's Guide to Calculus is what helped me most to "get" this, but it's been a couple years, and I may be misremembering which of the several books I read about this did the most to get my thinking unstuck.

I'll look forward to seeing responses to your other questions from other participants, and to seeing more tips on how to wrap your mind around the epsilon-delta definition of limits.

Thanks Tokenadult.

[quote]The "distance that the car has traveled" keeps increasing as long as the car is moving at all. So that graph shouldn't be falling to the right. It rises rapidly at first, and then rises more slowly.[/quote]

Okay I drew a graph that looks like the graph of [tex] f(x)=\sqrt{x}[/tex]Is this correct?

I know that these may seem really easy to some of you but this class is really fast and I am having trouble w/simple stuff.  Also, you don't have to answer all, if you could just help with one or two I would greatly appreciate it.

2) I think that works.
3)Unless I'm seriosly mistaken the limit is equal to 1. Do you know L'Hopitals rule?
4)Average velocity between a and b is (f(b)-f(a))/(b-a)
that should help i think...i can't answer your question without knowing more about the graph
5) My hint from 4 should help.
6)think about why the dots are there, for example: what is (-2)^-2?
that's why it's a right sided limit.
7)this is tough, someone else should explain this, but it basically says a limit is based on tolerance ranges instead of "motion." That was very unclear. Sorry.
8+)for the limits remember that the constants have nothing to do with the limit as x goes to infinity. (except for 10, which is trickier, think of how you derived the derivative formulas)

Thanks Jeffery,

[quote]Unless I'm seriosly mistaken the limit is equal to 1. Do you know L'Hopitals rule?[/quote]

No, I don't know L'Hopital's rule.  I used my TI-89 to find them.  When it is in radian mode, the limit is 1.  When it is in degree mode, it says the limit is  :pi: /180.  I really don't understand why there would be a different limit for radians and degrees.

[quote]Average velocity between a and b is (f(b)-f(a))/(b-a)
that should help i think...i can't answer your question without knowing more about the graph
[/quote]

Are you implying that you have to actually find the derivative, and you can't estimate accurately?  The graph is a parabola-shape concave downward in the first quadrant.  They don't ask this question specifically -- It just occured to me.

[quote]5) My hint from 4 should help.[/quote]

So a straight line would be wrong for #5?

[quote]7)this is tough, someone else should explain this, but it basically says a limit is based on tolerance ranges instead of "motion." That was very unclear. Sorry. [/quote]

It's okay, but it is unclear at least to me.  What do "tolerance range" and "motion" mean here?

[quote]8+)for the limits remember that the constants have nothing to do with the limit as x goes to infinity. (except for 10, which is trickier, think of how you derived the derivative formulas)[/quote]

(8) 3/2
(9) 2/3
(10)

Take the expression and manipulate it to get h out of denominator:
[tex]\frac{\sqrt{4+h}-2}{h}*\frac{\sqrt{4+h}-2}{\sqrt{4+h}-2} = \frac{h}{h\sqrt{4+h} + 2h}[/tex] so the limit is 0?  Is this a valid substitution?

(11) k=2

for 1, think of the hints i gave you about the other limits and see if you can extend it to this problem.
for 4, the expression i gave is the exact value of the average velocity over the interval.
for 5, since the velocity varies you can't have a straight line...

for 10, think of the definition of a derivative. remember that 2 is 4^(1/2). Your multiplication trick doesn't work (denomintor is still 0).

remember that f'(h)= lim (x approaches 0) of (f(h+x)+f(h))/x.

for 7 i must say that im not totally sure myself. let a be the point we are taking the limit to. my understanding is this:
we conceptually think of limits in terms of motion. i think of a point moving along the curve to the number it approaches. however, the formal definition is actually different. it says for a certain range around a (a +/- delta) there is a range around the limit (L+/- epsilon) such that f(a+/- delta)=L+/- epsilon.

Someone tell me if I'm wrong or give a clearer explanation.

[quote]for 1, think of the hints i gave you about the other limits and see if you can extend it to this problem. [/quote]

Okay, I get why the limit = 1.  When you look at the graph it's a "dampened" trigonometic function.  As x->0 from either side, graph approaches (but is undefined at) 1.  However, this is in BOTH degrees and radians.  Why does the TI89 say the limit in each mode are different?

[quote]for 5, since the velocity varies you can't have a straight line... [/quote]

Okay how about this: a graph that looks like part of y=-1/x.  Starts at (2,1) then has a vertex at (5,2) and ends at (4,6).

[quote]for 10, think of the definition of a derivative. remember that 2 is 4^(1/2). Your multiplication trick doesn't work (denomintor is still 0). [/quote]

[tex]\displaystyle\lim_{h\to0}{\frac{\sqrt{4+h}-2}{h}}=\displaystyle\lim_{h\to0}{\frac{\sqrt{4+h}-\sqrt{4}}{h}}[/tex].  I see how this is the definition of a derivative.  So you want to find [tex]f'(4)[/tex] for the function [tex]f(x)=\sqrt{x}[/tex].  So that is a power function, so you apply power rule to get [tex]\displaystyle f'(x)=\frac{1}{2}*x^{-\frac{1}{2}}[/tex].  Now find f'(4).  [tex]\displaystyle f'(4)=\frac{1}{2}*4^{-\frac{1}{2}}=\frac{1}{4}[/tex].  Is this correct?

[quote]for 7 i must say that im not totally sure myself. let a be the point we are taking the limit to. my understanding is this:
we conceptually think of limits in terms of motion. i think of a point moving along the curve to the number it approaches. however, the formal definition is actually different. it says for a certain range around a (a +/- delta) there is a range around the limit (L+/- epsilon) such that f(a+/- delta)=L+/- epsilon.

Someone tell me if I'm wrong or give a clearer explanation.[/quote]

To be honest -- I don't really get it.  I will try to read about it in my textbook.

Your answers look ok to me...

sorry about the limit definition...the textbook probably is the way to go.

[quote]Quote:
for 1, think of the hints i gave you about the other limits and see if you can extend it to this problem.

Okay, I get why the limit = 1. When you look at the graph it's a "dampened" trigonometic function. As x->0 from either side, graph approaches (but is undefined at) 1. However, this is in BOTH degrees and radians. Why does the TI89 say the limit in each mode are different?
[/quote]

Any thoughts?  Glad my other answers are correct, thanks for the help.

Take a look at my reply and other posts on another question (on Intermediate, I think).

I still don't get the radians/degrees thing, and for that limit problem, it is h->0 not h->inf for the definition of a derivative, so I am not sure my answer is correct...

Thanks for all the help so far.

EDIT: nvm, it actually is h->0 but I made a typo.  My bad.  I still am wondering about the radians/degrees thing though.

[quote="JS1527"]

(3) When you are doing limits w/trig functions, why does it make a difference if you use radians or degrees?  Why is [tex]\displaystyle \lim_{\theta\to0}{\frac{sin\theta}{\theta}}[/tex] equal to  :pi: /180 when  :theta:  is in radians, and 0 when  :theta:  is in degrees?

[/quote]

I don't know calc, but are you sure the answer isn't :pi:/180 for degrees and 1 for radians?

just as a heads-up...I was told to always remember that:

lim x->0  of sin(x)/x = 1

my class covered that one when we learned about the squeeze theorem...

as for why it is different in radian vs. degrees...for one thing there is a difference in the scale with radians and degrees.  Unless the problem specifically says to use degrees you should always use radians in calculus.

hope that helps!

[quote]I don't know calc, but are you sure the answer isn't pi/180 for degrees and 1 for radians?[/quote]

That is correct, my bad.

Now, I know that the scale is different, but the graph is the same.  The RANGE for the graphs of f(x)=sin(x) and f(x)=cos(x) are STILL on the interval [-1,1]!  So, either way, as the values for x->0, the values for f(x) will approach 1.  Where does the pi/180 come from???  This is what is confusing me.

Because you're dividing by theta.  For the same angle (in practice), your theta for degrees will be greater (180/:pi: times, to be exact) than your theta for radians.  So sin :theta:/:theta: doesn't remain the same.

[quote]3)Unless I'm seriosly mistaken the limit is equal to 1. Do you know L'Hopitals rule?
[/quote]

You are right, but you can't prove that limit using L'Hopital.

Would an angle drawn on an unit circle suffice?  You would see that the leg eventually becomes essentially the arc of the circle.

You can't use L'hopital? its a 0/0 indeterminant form. is there another condition required for L'hopital that i don't know about?

You can find the value of that limit using L'Hopital, but she doesn't [i]know [/i] L'hopital, so it doesn't matter.

As for the degree/radian question, draw the graphs of y = sin(x) and y = x, where x is in both degrees and radians.  Notice that when x is in degrees, the sin curve is much flatter -- x is coming in with a much larger slope.  In radians, though, they come in at almost exactly the same speed.  In other words, look at the space between the two graphs near zero.  That is, informally, L'hopitale's rule, and is a relatively intuitive way of thinking about that limit.

[quote="Mathworld"]A function f(z) is said to have a limit [tex]\lim_{z\to a} f(z)=c[/tex] if, for all [tex]\epsilon>0[/tex], there exists a [tex]\delta>0[/tex] such that [tex]\left\vert{f(z)-c}\right\vert<\epsilon[/tex] whenever [tex]0<\left\vert{z-a}\right\vert<\delta[/tex].[/quote]

Okay, it's a little bit hard to break it down, but let me restate it and see if it helps you:

"We say that an expression has a limit, [tex]\lim_{z\to a} f(z)=c[/tex], if, for any region near c, there is a different region near a such that if z is in that region, f(z) is in the first."

This guarantees that f won't "jump" suddenly and move away from c while z stays close to a.

OR:

"There is some small interval such that the values of f over that interval are arbitrarily close to the limit value."

Please ask for clarification if there are things that aren't clear.

1. Suppose you want to find the slant asymptote of P(x)/Q(x), where P and Q are polynomials. By the division algorithm for polynomials (which can be proved rigorously),

P(x)/Q(x) = S(x) + R(x)/Q(x), where the degree of S(x) is
degree of P - degree of Q.

and the degree of R(x) is less than the degree of Q(x).

So the highest degree term in R is less than the highest degree term in Q and the limit R/Q as x--> infinity is zero. This means that when x is very large, R/Q will be negligible relative to S(x), which shoots off to infinity.

2. Let s(t) be the position of the car at time t. The speed of the car is then defined as |s'(t)|, but since the car moves in one direction only, we can measure distance in the direction in which the car is going, then s'(t) will be positive for all t, and |s'(t)| = s'(t). The car starts at a high speed initially ---> s'(t) >0 initially, so initially the graph of s(t) vs. t is sloping upward. Then the car's speed slowly decreases, so the slope gets less and less steep with time and more and more shallow.

Graphically, what you described (s(t) = sqrt(t)) is good.

3. First let's show that the limit is 1 when we measure the angle in radians. The advantage of working in radians is that the arclength of a section of the unit circle is given by the central angle subtended. So visually, we can see from a picture of the unit circle that

sin(theta) < theta < tan(theta) when theta > 0.

Follow this link for a good presentation of this: http://www.ies.co.jp/math/java/calc/LimSinX/LimSinX.html.

If we work in degrees, instead of writing arclength = theta we have to write arclength = (theta)(180 degrees/pi radians). If you continue the argument as is done in the link, you find that the limit of sin(theta)/(theta) is in fact pi/180 if theta is measured in degrees.

4. Average velocity is the slope of the line joining (1, s(1)) and (3, s(3)). Instantaneous velocity at t = 2 is the slope of the function at that point. Judge which of these is greater based on how the graph looks.

5. Your answer is good

6. The limit is 1. It's hard to see this without having had logs and exponentials in calculus because you haven't defined x^x rigorously. Since you want the limit from the RIGHT, you don't need to worry about the dots to the left the y axis that your calculator gives you.

7. Epsilon delta definition of a limit...like the other people have said, this is hard to understand the first time around. I certainly didn't understand it the first time that I saw it. Only a year later did it start to make more sense.

Suppose lim x--> c f(x) = L. The epsilon delta definition says that for any e >0 we can find a d > 0 such that 0<|x - c|<d --> |f(x) - L| < e.

We require that 0 < |x - c| because we don't want the value of the function right at the point of interest to matter with reference to the limit. Consider
f(x)  = 2x when x is not 2
       = 3 when x is 2

This function has a point discontinuity at x = 2, but we still feel that intuitively, the function approaches 4 when x approaches 2. If we made our definition of limit say that |x - c| < d ---> |f(x) - L| < e then in situations such as the one given above, the limit would not exist, contrary to our intuition about limits.

In words, we say that the limit = L if we can make the distance between the L and the function get as small as we want by making the distance between x and c small enough. To understand this, it helps to think of some functions such that the limit does NOT exist and think about why they fail to meet the definition. For example, why is

f(x) = 0 if x >0
f(x) = 1 if x < = 0 discontinuous at the origin?

Why is f(x) = 0 if x is rational, 1 if x is rational discontinuous EVERYWHERE.

 It'll make more sense with time.

9) As x --> 0, e^(-x) --> 0, so you can substitute 0 in for e^(-x) and get that the limit is  3/2.

8) We get 8) from 9) by dividing the numerator and denominator in the fraction considered in 9) by e^(x).

Using L'hospital implies knowing the derivative of sin x. And to find the derivative you need to use the limit. So its kind of circular.

Thanks so much for all of the responses.  (I just found out I got 100/100 and 10/10 on first test and quiz! Yay!)


calculus
integration
trigonometry
conics
calculus computations

Let substitute $u=sin\theta \Longrightarrow du=cos\theta d\theta \Longrightarrow d\theta =\frac{du}{\sqrt{1-u^2}}$ then the integral changs in the form $\int_{0}^{1} \frac{du}{(1-e^2u^2)\sqrt{1-u^2}}$ and again substitute $t=1-e^2u^2 \Longrightarrow dt=-2e^2udu \Longrightarrow du=\frac{dt}{-2e\sqrt{1-t}}$ and $u=\frac{\sqrt{1-t}}{e}$ then the integral changs in the form: \[ \frac12 \int_{1-e^2}^{1} \frac{dt}{\sqrt{\frac{e^4}{4}-(t-1+\frac{e^2}{2})^2}}=\frac12 [ Arcsin \frac{t-1+\frac{e^2}{2}}{\frac{e^2}{2}}]_{1-e^2}^{1}=\frac{\pi}{2}  \]

it does not matter it can be solved same way[code]app[/code]

[quote="kelvin911"][quote="kunny"]Evaluate $\int_0^{\frac{\pi}{2}} \frac{d\theta }{(1-e^2\sin ^ 2 \theta)^3}\ (e<1\ \text{is a constant number}).$[/quote]

It is the most bizzare letter for a constant o_O[/quote]

Not really. It probably has some relation with conic section, so e stand for eccentricity. (don`t know if this is the really reason)


algebra
polynomial
superior algebra
superior algebra unsolved

Let a>0 such that all $ Q[a^{\frac{1}{n}}]$ are equal for all n. Prove that $a=1$.

Sorry if this sounds stupid, but what is $Q[a]$? :?

The set of $ f(a)$ where f is a polynomial with rational coefficients. Sorry for not writing that.

Oh, I should have imagined it, but I was expecting $\mathbb Q$ instead of $Q$ :). I'm sorry.

$Q[a^{1/n}]$ is an extension of $Q$ degree $n$

All  $Q[a^{1/n}]$ are equal, then  they are all isomorph to each other

then they have the same degree $[Q[a^{1/n}:Q]=n$  this imply $a=1$

See the slightly more general version [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=640097#640097]here[/url].


Poate cineva sa-mi explice si mie uramtoarele denumiri:
-functie bijectiva
-functie inejectiva
-functie surjectiva
-functie crescatoare
-functie descrescatoare?
    Va multumesc anticipat!

$\boxed{1}$ Fie $f: A\to B$ o functie. Spunem ca $f$ este o [b]functie injectiva[/b],  daca pentru oricare $x,y\in A$ avem $f(x)=f(y)$. (se mai poate exrima shi astfel: $\forall x,y\in A$ : $f(x)=f(y)\Rightarrow x=y$).

$\boxed{2}$ Fie $f: A\to B$ o functie. Spunem ca $f$ este o [b]functie surjectiva[/b],  daca pentru orice $y\in B$ exista cel putin un element $x\in A$ astfel incat $y=f(x)$. (In caz contrar $f$ nu e surjectiva daca $\exists y\in B$ pt care $y\neq f(x)$ pentru orice $x\in A$).

$\boxed{3}$ O functie $f: A\to B$ se numeste [b]bijectiva[/b] daca este simultan [b]injectiva si surjectiva[/b]. (Ex. $f: \mathbb{R_{+}^{*}}\to\mathbb{R_{+}^{*}}, \ f(x)=x^{2}$ este bijectiva)

O functie $f$ este [b]inversabila[/b] (si o notam cu $f^{-1}$) daca si numai daca este bijectiva.

Monotonia functiilor: Fie doua multimi reale $A,B$ si functia $f: A\to B$, atunci:

(a) $f$ se numeste [b]crescatoare[/b] daca $\forall x_{1},x_{2}\in A$ daca $x_{1}<x_{2}$ rezulta ca $f(x_{1})\leq f(x_{2})$

(a') $f$ se numeste [b]strict crescatoare[/b] daca $\forall x_{1},x_{2}\in A$ daca $x_{1}<x_{2}$ rezulta ca $f(x_{1})< f(x_{2})$

(b) $f$ se numeste [b]descrescatoare[/b] daca $\forall x_{1},x_{2}\in A$ daca $x_{1}<x_{2}$ rezulta ca $f(x_{1})\geq f(x_{2})$

(b') $f$ se numeste [b]strict descrescatoare[/b] daca $\forall x_{1},x_{2}\in A$ daca $x_{1}<x_{2}$ rezulta ca $f(x_{1})>f(x_{2})$

[i][b]Sau:[/b][/i]

- $f$ este [b]crescatoare[/b] daca $\forall x_{1}\neq x_{2}$ din $A$ avem $\frac{f(x_{1})-f(x_{2})}{x_{1}-x_{2}}\geq 0$.

- $f$ este [b]strict crescatoare[/b] daca $\forall x_{1}\neq x_{2}$ din $A$ avem $\frac{f(x_{1})-f(x_{2})}{x_{1}-x_{2}}> 0$.

- $f$ este [b]descrescatoare[/b] daca $\forall x_{1}\neq x_{2}$ din $A$ avem $\frac{f(x_{1})-f(x_{2})}{x_{1}-x_{2}}\leq 0$.

- $f$ este [b]strict descrescatoare[/b] daca $\forall x_{1}\neq x_{2}$ din $A$ avem $\frac{f(x_{1})-f(x_{2})}{x_{1}-x_{2}}< 0$.

Ultimile exprimari sunt de fapt consecintele primei derivate (notata cu $f^{'}$). Dar pentru a-ti explica ce inseamna shi derivata trebuie sa mai scriu destul de mult dar nu mai scriu(din teama ca nu vei intelege) :( Poate cineva mai experimentat te va la muri daca esti curios. :D! Spor la lucru!

Multumesc mult Slizzel!Ai dreptate, o sa o intreb pe doamna mea profesoara ce nu inteleg.Si tie spor la lucru!


How do I find the domain of logs?  I have two questions below.

Find the domain and the intercepts, IF ANY, of each function.

1) h(x) = log_1/2 (x^2 - 2x + 1)

2) f(x) = In (3 - x)

Thank you.

Bruno,

Great job as always.  However, I am a baby math person who has not seen questions in a while.  I am preparing for a state test and need step by step help, you see?  I don't understand the math symbols that you always use to answer my questions.
Symbols like a GIANT R and strange-looking curved capital letter E.  What is that?

What do these symbols mean?  Is there a site that explains math symbols and terms easily?

Thanks

$\mathbb{R}$ means the set of all real numbers.  Therefore, $\mathbb{R}-\{1\}$ means all reals except for 1.

I don't see where he used the curved capital E.  $\Sigma$ means 'the sum of' but that wasn't used in his answer.

[quote="joml88"]I don't see where he used the curved capital E.  $\Sigma$ means 'the sum of' but that wasn't used in his answer.[/quote]
I think he meant $\in$ ;)

Ah, ok  :blush: .  $\in$ means "is an element of" or "is in".  Therefore $x\in\{0,1,2,3\}$ means that $x$ is an element of $\{0,1,2,3\}$ or $x$ is in the set $\{0,1,2,3\}.$

Yes, this is exactly what I was looking for.

Thank you very much.


So absolutely unsure about this so I would need some experts to correct me and tell me where I can improve in Number Theory (brilliant subject, I am just hopeless at it).

Since $ (n, n+1)= (n+1,n+2)=1$ we have $ n+1= m^{k}$ where $ k$ is a positive integer and $ k<r$. Which implies that $ n(n+2)= n^{2}+2n = (n+1)^{2}-1 = m^{2k}-1 = m^{r-k}\Rightarrow m^{3k}= m^{r}+m^{k}$.

Clearly, $ n(n+1)(n+2)$ is even so let $ m= 2L$ where $ L$ is a positive integer. So we have $ 2^{3k}L^{3k}= 2^{r}L^{r}+2^{k}L^{k}\Rightarrow 2^{2k}L^{2k}= 2^{r-k}L^{r-k}+1$.

$ LHS$ is always even while $ RHS$ is always odd. A contradiction.

[quote="mdk"]$ n$ is a positive integer. Prove that we cannot find an integer $ m$ and an integer $ r > 1$ such that    $ n(n+1)(n+2) = m^{r}$[/quote]

Note that $ (n+1)^{2}-n(n+2)=1$ hence $ \gcd(n(n+2),n+1)=1$. so if $ n(n+1)(n+2)$ were a perfect $ r$th power then $ n(n+2)$ and $ n+1$ must be perfect $ r$th powers as well. Let $ n+1=a^{r}$. so $ n(n+2)=(a^{r}-1)(a^{r}+1)=a^{2r}-1=(a^{2})^{r}-1$, which is one less than a perfect $ r$th power and thus can never be one. This is the contradiction we intended.


How would you show that for any real number x > 1/4, x is not in the Mandelbrot Set?

Thanks.

It doesn't really belong in Intermediate.  :wink:

Let $f(z)=z^{2}+c$, where $c>1/4$ is real. Must show $f(f(f\dots (0)\dots))\to \infty$. For any real number $z$ we have $f(z)-z=(z-1/2)^{2}+(c-1/4)$, hence $f(z)\ge z+(c-1/4)$. The claim follows.


function
real analysis
real analysis unsolved

We say that the functions $ p,q: \mathbb{R}\rightarrow\mathbb{R}$ have property $ (P)$ if $ p(q(x))\plus{}q(p(x))\equal{}x$.

Prove that if $ f,g: \mathbb{R}\rightarrow\mathbb{R}$ have property $ (P)$ and they are strictly increasing, then $ f,g$ are continuous and invertible, and the inverse functions of $ f,g$ [b]do not [/b]have property $ (P)$.


A group of 6 friends went to the movies. Colin and Megan insist on sitting next to each other and Abe wants to sit next to them as well. Lee, Dani, and Libbey also want to sit together. How many ways can they do this?

I solved this problem, but I don't know if I'm correct:
Could someone check my solution and answer?

[hide] You have 6C3 + 6C3  = 40 ways to organize the two groups of 3 people who want to sit together. Then you add 6 more possibilities for moving around Lee, Dani and Libbey. Finally, you add two more possbilities for moving around Colin and Megan within their group of three.

40 + 6 + 2 = 48 total ways.

[/hide]

Assuming that they sit in a line...

First we can assign Colin, Megan, and Abe to one group, and Lee, Dani, and Libbey to another group, so that there are first $ 2$ ways to arrange these groups.

Then, notice that in the second group, there are $ 3!\equal{}6$ ways to arrange them.

In the first group, we again subdivide: Colin is alone and Megan and Abe are in a group. There are $ 2$ ways to do this, and then $ 2$ more ways to arrange Megan and Abe within their group.

Finally, our answer is $ 2\cdot6\cdot2\cdot2\equal{}\boxed{48}$.

i think you have the right answer but the wrong solution :D
[hide="hint"]
when you are working out a counting problem, the cases in independent events must be multiplied to derive the final answer
(for dependent possibilities-- in the same event, you add)[/hide]


geometry
geometric transformation
reflection
3D geometry
tetrahedron
AMC

--Scoreboard--
(Updated 12/14)

Congratulations to everyone who submitted solutions so far!

1. TripleM - 150
1. MysticTerminator - 150
3. JBL - 144
3. interesting_move - 144
3. zscool - 144
6. Syntax Error - 143
6. tetrahedr0n - 143
8. The SouthPaw - 133.5
9. paladin8 - 132
9. Chinaboy - 132
11. bugzpodder - 126
12. Spoon - 121.5
13. confuted - 120.5
14. WarpedKlown1335 - 119
15. Tare - 108
16. jeffrey - 100
17. Rep123max - 96.5

Look at those scores!

This deserves a post of its own.

We have our very first perfect score!  Congratulations to TripleM for getting all of the problems right.  (Or at least the same answers as the answer sheet.)

I have updated the scoreboard to reflect recent submissions.  Thanks everyone for sending in your answers.  Please continue to do so.

Uh, it doesn't look like the scoreboard's been updated...

Sorry, forgot to push submit after I switched tabs.  Better now?

Why is Paladin8's 132 higher than The SouthPaw's 133.5?

And what's with the double first and second places?

Oops.  All better.  I think.

This is embarrassing.  :oops::oops::oops::oops::oops:

Um, actually it was me that got the 133.5

Urgh.  How much longer can this humiliation continue?

On a side note, the more mistakes I make, the more the post count increases and the more attention is paid to AMC.  Hm.  This is almost a good technique, except for the humiliation part.

I know what'll happen next: paladin8 will say he got a 138.  I know what happened.  In the PM, I somehow counted 23 correct instead of 22, and 23*6 = 138...but 22*6 = 132.  Anyway, I hope it's all fixed now.  I don't think I can stand much more of this humiliation.

First I called Fierytycoon a she accidentally.  Then I didn't update it.  Then I put a lower number higher than another number.  Then I went 1 2 1 2 3 4 5 6....

Oh wait, it makes sense now why TheSouthpaw was higher than paladin8.

but how bout the 2.5 points for ommited questions

what? i got a 138!

...

just kidding :D, i notcied that I missed three but got a 23 and thought it very strange :)

Hm.  Both Neal and tetrahedr0n answered the same 23 problems, and got them all right.  Congrats!

[hide]Do I sense a conspiracy here?[/hide]

hm, thanks, but i asked tetrahedr0n and he omitted a different question than i did.

Oh.  Oops.  Ya, tetrahedr0n omitted 21 and 22, whereas you omitted 21 and 23.

can you post the solutions after awhile?

no don't i promise i'll send in answers eventually probably after midterms but that tetrahedron question is evil grr

make sure you guys dont take too much time

I shall try to submit it asap.

-interesting_move

*Updates*

Whoa, look at MysticTerminator's score.

MysticTerminator wrote:yay go Neal numba 3 (or 4 if i ever stop being so lazy and e-mail my answers to matfan)

exactly

[hide]this is in spoiler[/hide]

w00t i'm as good as MMM i deserve to go to IMO. or not

Mystic's score has been changed to reflect an error.  Now we have two perfects.

congrats grads!

[hide]oooh, spoiler, i just like to say "congrats grads" a lot. um. oooh, spoiler[/hide]

Updated again.

Alrighty, I know everyone is anxious to discuss the solutions and stuff, so I've decided that Tuesday, December 16th will be the last day to submit answers.  After that, I'll open up the problems to discussion on this board.  At that point, everyone is free to post their solutions, compare notes, etc.  This way, we'll be able to hear different people's methods of solving each problem.

[Cross-posted to general announcement about AMC Mock Contest #1]

why does paladin8 rank higher than chinaboi?

determine all the scores one can achieve on an AMC grading scale (with 25 problems) (or with n problems)

grrrr, i should have gotten that stick problem with a diameter, sigh. i need to think more logically next time. but blah! so many people with 144 points! not cool! well cool! but blah!

Alright, I fixed that paladin8-chinaboy thing, and updated for (probably) the last time.  We'll now be discussing the solutions to the problems.


function LaTeX algebra polynomial trigonometry

Demonstrate that if f(x) is an even function, then g(x)=x * f(x) is an odd function

[hide]
EDITTED $ \text{\LaTeX}$ typo
$ f(-x) = f(x)$
$ g(-x) =-x\cdot f(-x) =-x\cdot f(x)\implies\text{g(x) is an odd function}$.
[/hide]

In general:

[b]even times even gives even[/b]
h=f*g means h(-x) = f(-x)g(-x)=f(x)g(x)=h(x)
[b]
even times odd gives odd[/b]
h=f*g, f even and g odd, produces h(-x)=f(-x)g(-x)=f(x)[-g(x)]=-f(x)g(x)=-h(x)

[b]odd times odd gives even[/b]
h=f*g means h(-x)=f(-x)g(-x)=[-f(x)][-g(x)]=f(x)g(x)=h(x)

[hide]Let $ h(x)\equal{}x$.  Since $ h(x)\equal{}\minus{}x$, $ h$ is an odd function.  So, odd times even equals odd, which is our answer.[/hide]

[quote="archimedes1"][hide]Let $ h(x) \equal{} x$.  Since $ h(x) \equal{}\minus{}x$, $ h$ is an odd function.  So, odd times even equals odd, which is our answer.[/hide][/quote]

this makes a lot of sense :D . I mean, there's a reason the functions are called "even" and "odd" right? This must be it.

I don't think so: even and odd functions don't behave much like even and odd integers under either addition or multiplication.  (In particular, the rule you're quoting isn't one which is followed by even and odd integers.)  In fact, we have the rules:

$ o\plus{}o \equal{} o$
$ e\plus{}e \equal{} e$
$ e\plus{}o \equal{} ?$
$ o\cdot o \equal{} e$
$ e\cdot e \equal{} e$
$ o\cdot e \equal{} o$

They actually behave much more like positive and negative numbers (which is totally unsurprising in hindsight because of the definition).  If I had to guess, the reason it's "even" and "odd" has to do with polynomials.

I believe JBL's comment is right.

An even polynomial is one where x is raised to only even powers, where an odd polynomial is where x is raised to only odd powers:

Examples of even polynomials:
$ y\equal{}x^{2}$, $ y\equal{}5x^{4}\plus{}8$, $ y\equal{}383x^{578298}\minus{}283x^{678}\plus{}38297387x^{2}$

Examples of odd polynomials:
$ y\equal{}238x^{7}\minus{}398209x^{5}$, $ y\equal{}x^{2473097375}\plus{}3458x$, etc.

In fact, we can add negative exponents to either and it won't matter.  Odd functions can have $ x^{\minus{}(2k\plus{}1)}$ terms in them, while even ones can have $ x^{\minus{}2k}$ terms for k is a natural number.

the reason mentioned by krsattack is absolutely the correct one...as i found out in my reading of aops vol 1

[quote="JBL"] If I had to guess, the reason it's "even" and "odd" has to do with polynomials.[/quote]

To add, it also bothered me why we call them even and odd when they dont exactly behave like intergers. However, I beleive even and odd comes from not just from polynomial but from "infinite" polynomials.
For example,
$ \sin x \equal{} x\minus{}\frac{x^{3}}{3!}\plus{}\frac{x^{5}}{5!}\minus{}...$
You see this "infinite" polynomial has all coefficients odd. So its called an "odd function".


Karel is a frustrated painter who works by day in an electrical repair shop. Inspired by the color-coded bands on resistors, he is painting a series of long, narrow canvases with bold colored stripes. However, Karel is lazy and wants to minimize the number of brush strokes it takes to paint each canvas.

Abbreviating each color to a single uppercase letter, Karel would write the stripe pattern red-green-blue-green-red as "RGBGR" (quotes added for clarity). It would take him three brush strokes to paint this pattern. The first stroke would cover the entire canvas in red (RRRRR). The second stroke would leave a band of red on either side and fill in the rest with green (RGGGR). The final brush stroke would fill in the blue stripe in the center (RGBGR).

Given a stripe pattern stripes as a String, calculate the minimum number of brush strokes required to paint that pattern.

Definition

Class:	StripePainter
Method:	minStrokes
Parameters:	String
Returns:	int
Method signature:	int minStrokes(String stripes)
(be sure your methodis public)

Notes
-	The blank canvas is an ugly color and must not show through.

Constraints
-	stripes will contain only uppercase letters ('A'-'Z', inclusive).
-	stripes will contain between 1 and 50 characters, inclusive.

Examples
0)

"RGBGR"

Returns: 3

Example from introduction.
1)

"RGRG"

Returns: 3

This example cannot be done in two strokes, even though there are only two colors. Suppose you tried to paint both red stripes in one stroke, followed by both green stripes in one stroke. Then the green stroke would cover up the second red stripe. If you tried to paint both green stripes first, followed the red stripes, then the red stroke would cover up the first green stripe.
2)

"ABACADA"

Returns: 4

One long stroke in color 'A', followed by one stroke each in colors 'B', 'C', and 'D'.
3)

"AABBCCDDCCBBAABBCCDD"

Returns: 7

4)

"BECBBDDEEBABDCADEAAEABCACBDBEECDEDEACACCBEDABEDADD"

Returns: 26

relate to parathesis correct form and to parity of strokes inside one such kind of
paranthesis...

see the base cases are
AABB
BBAA

ABAB
BABA

ABBA
BAAB

think of the final mixture of stripes as beeing an onion.
you just have to unfold it :)

[quote="mahbub"]can't get it :([/quote]

It's similar to the "Edit Distance" problem I believe.

edit distance works by letter by letter DP. how can u this prb do by that?


Let a,b,c,d are real numbers.Prove that:
$ \frac {5}{2}\sum_{cyc} a^4 \plus{} \sum_{sym} a^2b^2 \geq (a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3)(a \plus{} b \plus{} c \plus{} d) \plus{} (a \minus{} b)(b\minus{}c)(c\minus{}d)(d\minus{}a)$

[quote="No Reason"]Let a,b,c,d are real numbers.Prove that:
$ \frac {5}{2}\sum_{cyc} a^4 \plus{} \sum_{sym} a^2b^2 \geq (a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3)(a \plus{} b \plus{} c \plus{} d) \plus{} (a \minus{} b)(b \minus{} c)(c \minus{} d)(d \minus{} a)$[/quote]
My solution :)


Points $ A$, $ B$, $ C$, and $ D$ lie on a line, in that order. If $ AB\equal{}2$ units, $ BC\equal{}5$ units and $ AD\equal{}14$ units, what is the ratio of $ AC$ to $ BD$? Express your answer as a common fraction.

Draw a line with those points. Since you're given that AB = 2, BC = 5, and AD = 14, you can find CD, which equals 7.
So, the ratio of AC: BD is (2+5) : (5+7) --> 7:12 or 7/12


In $ \triangle ABC$, $ \angle B$ $ >$ $ \angle C$.$ T$ is the midpoint of arc $ BAC$ and $ I$ is the incenter of $ \triangle ABC$.$ E$ is a point such that $ AEI\equal{}90$ and $ AE$ and $ BC$ are parallel.$ TE$ intersect the circumcircle of $ \triangle ABC$ at $ P$ for the second time.$ \angle IPB\equal{}\angle ABC$.find $ \angle BAC$.

$ (O)$ is circumcircle of $ \triangle ABC.$ $ AI$ cuts $ (O)$ again at $ X$ diametrically opposite of $ T.$
$ (S)$ is circle with diameter $ AI,$ $ E \in (S),$ and $ IE \parallel TX,$ both $ \perp (AE \parallel BC).$
$ \measuredangle IEP \equal{} \measuredangle XTP \equal{} \measuredangle XAP \equal{} \measuredangle IAP$ $ \Longrightarrow$ $ APIE$ is cyclic, $ P$ is intersection of $ (O), (S)$ other than $ A.$ Let $ XP$ cut $ BC$ at $ D$ and $ (S)$ again at $ Q.$ From inversion in circle $ (X)$ with radius $ XB \equal{} XC \equal{} XI,$ it follows that $ ID \perp BC,$ $ \overline{XD} \cdot \overline {XP} \equal{} XI^2$ $ \Longrightarrow$ $ \measuredangle IAQ \equal{} \measuredangle IPQ \equal{} \measuredangle IPX \equal{} \measuredangle XID$ $ \Longrightarrow$ $ AQ \perp BC.$ Let $ Z \equiv PI \cap AO.$ Then $ \measuredangle ZAX \equal{} \measuredangle OAX \equal{} \measuredangle XAQ \equal{} \measuredangle IAQ \equal{} \measuredangle IPQ \equal{} \measuredangle ZPX$ $ \Longrightarrow$ $ APXZ$ is cyclic and $ Z \in (O)$ is diametrically opposite to $ A.$ Then $ \measuredangle IPB \equal{} \measuredangle ZPB \equal{} \measuredangle ZAB \equal{} \measuredangle CAB \minus{} (90^\circ \minus{} \measuredangle ABC).$

Assuming $ \measuredangle IPB \equal{} \measuredangle ABC$ $ \Longleftrightarrow$ $ \measuredangle CAB \equal{} 90^\circ.$

Assuming $ \measuredangle IPB \equal{} \minus{} \measuredangle ABC$ $ \Longleftrightarrow$ $ \measuredangle CAB \plus{} 2 \measuredangle ABC \equal{} 90^\circ$ $ \Longleftrightarrow$ $ \measuredangle BCA \minus{} \measuredangle ABC \equal{} 90^\circ,$ which is possible only for $ \measuredangle BCA > \measuredangle ABC.$

Digging deep in Aops history.

Let $AI$ meet $(ABC)$ at $M$. With $IE\parallel MT$ by Reim $APIE$ - cyclic. Hence $API=90^\circ$. Let $PI$ meet $(ABC)$ at $D$. Angle condition say that $AC=BD$. This gives $AD=BC$. But $AD$ is diameter and so is $BC$. Answer is $\angle A=90^\circ$.


LaTeX
number theory proposed
number theory

prove that there is infinty composite numbers like N such that
n devides
  n-1        n-1
3      -    2

Well he wrote a "-" and no "+" ;)
But nonetheless he should rewrite it using $\LaTeX$.
Read Valentin's introduction when necessary: http://www.mathlinks.ro/Forum/viewtopic.php?t=5261 .


integration
logarithms
trigonometry
limit
calculus
calculus computations

\begin{eqnarray*}\int_{0}^{\infty} \frac {\sin^2 x}{x} \, e^{ - 3x} \, dx & = & \int_{0}^{\infty} \sin^2 x \, e^{ - 3x} \left( \int_{0}^{\infty} e^{ - xt} \, dt \right) \, dx \\
& = & \int_{0}^{\infty} \int_{0}^{\infty} \sin^2 x \, e^{ - (3 + t)x} \, dt dx \\
& = & \int_{0}^{\infty} \int_{0}^{\infty} \sin^2 x \, e^{ - (3 + t)x} \, dx dt \quad \cdots \cdots (*) \\
& = & \int_{0}^{\infty} \int_{0}^{\infty} \frac {1 - \cos 2x}{2} \, e^{ - (3 + t)x} \, dx dt \\
& = & \int_{0}^{\infty} \frac {1}{2} \left( \frac {1}{3 + t} - \frac {3 + t}{4 + (3 + t)^2} \right) \, dt \\
& = & \int_{3}^{\infty} \frac {1}{2} \left( \frac {1}{u} - \frac {u}{4 + u^2} \right) \, du \qquad (u = t + 3) \\
& = & \lim_{R\to\infty} \int_{3}^{R} \frac {1}{2} \left( \frac {1}{u} - \frac {u}{4 + u^2} \right) \, du \\
& = & \lim_{R\to\infty} \frac {1}{2} \left[ \log \left( \frac {u}{\sqrt {4 + u^2}} \right) \right]_{3}^{R} \\
& = & \lim_{R\to\infty} \frac {1}{2} \left[ \log \left( \frac {R}{\sqrt {4 + R^2}} \right) - \log \frac {3}{\sqrt {13}} \right] \\
& = & - \frac {1}{2} \log \frac {3}{\sqrt {13}} \\
& = & \frac {1}{4} \log \frac {13}{9} \end{eqnarray*}

Note that step (*) needs to be justified. You may rely on some advanced theories (for example, monotone convergence theorem) or make some direct approaches restricted to this particular step.

p.s. IndoChina, you're right. I made a slight mistake at 5th line but now I fixed it.

sos440 :
You have made a slight arithmetic error. The answer given by \$math\$ is right.


Trying to find all real and imaginary zeroes of the following polynomial function:

2x^(3)+4x^(2)+18x+36

I know how to find all of the possible solutions, but don't know how to find out the exact solutions except for plugging every possible solution in synthetic division or substitution, but this is obviously the long way. Any better way to do this?

Group terms and factor:

$2(x^3+9x)+4(x^2+9)$
$(2x+4)(x^2+9)$

And you're done.

Can you explain how you grouped the terms like that?


If expanded, 3^2007 will have d more digits that 2^2007. Find d.

Thank you.

The number of digits in any integer $ n$ is $ \lfloor1\plus{}\log n\rfloor$, so $ 3^{2007}$ has

$ \lfloor1\plus{}\log 3^{2007}\rfloor\equal{}\lfloor1\plus{}2007\log3\rfloor$ digits. $ \log3\approx0.477$, which gives us $ \lfloor958.339\rfloor\equal{}958$ digits, and $ 2^{2007}$ has

$ \lfloor1\plus{}\log 2^{2007}\rfloor\equal{}\lfloor1\plus{}2007\log2\rfloor$ digits. $ \log2\approx0.301$, which gives us $ \lfloor605.107\rfloor\equal{}605$ digits.

Thus, our answer is $ 958\minus{}605\equal{}\boxed{353}$.


quadratics
articles
AoPSwiki
algebra
polynomial
AoPS Books
sum of roots

I was working on a random grab-bag of questions. This one came up (I'm not entirely sure what difficulty section it should be in):

Solve for x and y:
$ 4xy \equal{}\minus{}15$
$ 4x^{3}\plus{}4y^{3}\equal{} 49$

The substitution $ x \equal{}\frac{\minus{}15}{4y}$ is plain, but the ensuing algebra is so ugly (e.g. $ 4(\frac{3375}{64y^{3}})\plus{}4y^{3}\equal{} 49$) it makes me think I'm missing something. I don't think you are expected to be a calculator whiz or whatever, so is there some easier substitution or trick that could stop me from having to force a less than elegant method through to get a solution for this? I played with the equation a little but I couldn't see anything that leapt at me. I don't think this question is [i]meant[/i] to be ugly, and for all I know the answer is right in front of me.

Any hints would be appreciated.

consider the second equation

$ 4(x^{3}\plus{}y^{3})\equal{}4(x\plus{}y)(x^{2}\minus{}xy\plus{}y^{2})$

$ \equal{}4(x\plus{}y)(x^{2}\plus{}y^{2}\minus{}\frac{15}{4})$

so, $ 4(x\plus{}y)(x^{2}\plus{}y^{2}\minus{}\frac{15}{4})\equal{} 49$

maybe you can play around with this

there might be an easier way still

Well...we have:

$ xy\equal{}\frac{\minus{}15}{4}$
and
$ x^{3}\plus{}y^{3}\equal{}\frac{49}{4}$

$ (x\plus{}y)^{3}\equal{}x^{3}\plus{}3x^{2}y\plus{}3xy^{2}\plus{}y^{3}$. Since we know $ x^{3}\plus{}y^{3}$ and $ xy$,
this can be rewritten as $ (x\plus{}y)^{3}\equal{}\frac{49}{4}\plus{}\frac{\minus{}45}{4}(x\plus{}y)$. To find $ x\plus{}y$...I have to go, the bell just rang...but I think you get the idea.

[quote="Icy"]Well...we have:

$ xy \equal{}\frac{\minus{}15}{4}$
and
$ x^{3}\plus{}y^{3}\equal{}\frac{49}{4}$

$ (x\plus{}y)^{3}\equal{} x^{3}\plus{}3x^{2}y\plus{}3xy^{2}\plus{}y^{3}$. Since we know $ x^{3}\plus{}y^{3}$ and $ xy$,
this can be rewritten as $ (x\plus{}y)^{3}\equal{}\frac{49}{4}\plus{}\frac{\minus{}45}{4}(x\plus{}y)$. To find $ x\plus{}y$...I have to go, the bell just rang...but I think you get the idea.[/quote]

its school for you and its 3:30 in the morning for me :D

after you find x+y you can make x or y subject and sub back into eqn 1 and find the other variable

im off to bed
good morning

[quote="Icy"]this can be rewritten as $ (x\plus{}y)^{3}\equal{}\frac{49}{4}\plus{}\frac{\minus{}45}{4}(x\plus{}y)$. To find $ x\plus{}y$...I have to go, the bell just rang...but I think you get the idea.[/quote]

This is a general cubic, which is usually bad.  The OP's method was quadratic in $ y^{3}$, which is usually a much better alternative.

[hide="However..."] Observe that $ x\plus{}y \equal{} 1$ is a root.  Knowing this makes our lives easier.  Substitute $ u \equal{} x\plus{}y$; then

$ 4u^{3}\plus{}45u\minus{}49 \equal{} 0\Leftrightarrow$
$ (u\minus{}1)(4u^{2}\plus{}4u\plus{}49) \equal{} 0\Leftrightarrow$
$ (u\minus{}1)( (2u\plus{}1)^{2}\plus{}48 ) \equal{} 0$

$ u \equal{} 1$ is the only real solution, so if that's all we're interested in then we have a simple system

$ xy \equal{}\minus{}\frac{15}{4}$
$ x\plus{}y \equal{} 1$

$ x, y$ are therefore the roots of the quadratic equation

$ a^{2}\minus{}a\minus{}\frac{15}{4}\equal{} 0\Leftrightarrow$
$ 4a^{2}\minus{}4a\minus{}15 \equal{} 0\Leftrightarrow$
$ (2a\minus{}1)^{2}\minus{}4^{2}\equal{} 0\Leftrightarrow$
$ 2a\minus{}1 \equal{}\pm 4\Leftrightarrow$
$ a \equal{}\boxed{\frac{5}{2},\minus{}\frac{3}{2}}$. [/hide]

That last line...where did you get that quadratic equation involving $ a$ from? I remember reading an article on the AoPSWiki about some sort of property, but I don't remember what it was called, and I think you used it here. Gah! It would have been so much easier to find $ \left(\frac{3375}{16}\right)\plus{}4y^{6}\equal{}49y^{3}$! I thought that there would be a simpler method...like the ones in the AoPS books, but I was wrong, and made it harder in the process...eventually requiring knowledge I lack to solve....Wow, that sucks. :(

Hi, how did you get this line?

[quote="t0rajir0u"]
Observe that $ x+y = 1$ is a root.  Knowing this makes our lives easier.  Substitute $ u = x+y$; then
[/quote]

----------------------------------------------------------------------------------------------------
[quote="Icy"]That last line...where did you get that quadratic equation involving $ a$ from? I remember reading an article on the AoPSWiki about some sort of property, but I don't remember what it was called, and I think you used it here. . :([/quote]

His using polynomial technique $ \alpha+\beta$=$ {-b}{a}$ and $ \alpha\beta$=$ \frac{c}{a}$

Using this fact and the simultaneous equation you can form a quadratic and solve for $ a$

But tor, how do you which value takes $ \frac{5}{2}$ or $ \frac{-3}{2}$

---------------------------------------------------------------------------------------------------

I was thinking along the lines like this??

consider the second equation

$ 4(x^{3}+y^{3}) = 4(x+y)(x^{2}-xy+y^{2})$

$ = 4(x+y)(x^{2}+y^{2}-\frac{15}{4})$

so, $ 4(x+y)(x^{2}+y^{2}-\frac{15}{4}) = 49$

$ 4(x+y){(x+y)^{2}-2xy-\frac{15}{4}) = 49}$

$ 4(x+y)[(x+y)^{2}-\frac{15}{2}-\frac{15}{4}] = 49$

Maybe it doesn't lead to anywhere :oops:

[quote="bos1234"]Hi, how did you get this line?

[quote="t0rajir0u"]
Observe that $ x\plus{}y \equal{} 1$ is a root.  Knowing this makes our lives easier.  Substitute $ u \equal{} x\plus{}y$; then
[/quote][/quote]

He said in his post:

[quote="t0rajir0u"][quote="Icy"]this can be rewritten as $ (x\plus{}y)^{3}\equal{}\frac{49}{4}\plus{}\frac{\minus{}45}{4}(x\plus{}y)$. To find $ x\plus{}y$...I have to go, the bell just rang...but I think you get the idea.[/quote][/quote]

Just eyeball it anyway it's easy to get $ x\plus{}y \equal{} 1$ as a root. If you really want to, you could use this to find any other roots.

[quote="bos1234"]Hi, how did you get this line?

[quote="t0rajir0u"]
Observe that $ x\plus{}y \equal{} 1$ is a root.  Knowing this makes our lives easier.  Substitute $ u \equal{} x\plus{}y$; then
[/quote]

----------------------------------------------------------------------------------------------------
[quote="Icy"]That last line...where did you get that quadratic equation involving $ a$ from? I remember reading an article on the AoPSWiki about some sort of property, but I don't remember what it was called, and I think you used it here. . :([/quote]

His using polynomial technique $ \alpha\plus{}\beta$=$ {\minus{}b}{a}$ and $ \alpha\beta$=$ \frac{c}{a}$

Using this fact and the simultaneous equation you can form a quadratic and solve for $ a$

But tor, how do you which value takes $ \frac{5}{2}$ or $ \frac{\minus{}3}{2}$

---------------------------------------------------------------------------------------------------
[/quote]

What do $ \alpha$ and $ \beta$ mean? The sum of roots? And what does that have to do with $ xy\equal{}\minus{}\frac{15}{4}$ and $ x\plus{}y\equal{}1$

I think he's just referring to the Viete's formula(the letters doesn't matter), it's used in [b]@t0rajir0u[/b]'s solution when he get the quadratic of a.

Indeed; in a quadratic $ ax^{2}\plus{}bx\plus{}c$, the sum of the roots is $ \frac{\minus{}b}{a}$ and the sum is $ \frac{c}{a}$, which you can see by using roots from the quadratic formula.


The AMC website says:

"The USAMO index cutoff for participants was 233 if you took the AIME I, or
220.5 if you took the AIME II.
In addition, students in grades 7th through 10th achieving an AIME score of 9 or above
were also selected to write the USAMO."

Wait.  But if they've adjusted the index over 10 points (1 AIME-point) for the difference between the I and II exams, why is the floor value a constant?

(Okay, so I got an 8 on the AIME-II and that's my motivation for bringing this up.  But it still seems objectively unfair.)

I'm just guessing, but maybe of all the people who made the USAMO via AIME II, the lowest AIME score is a 9.

Hmm...I guess that clause about porportional qualifiers kicked in.


ratio geometry MATHCOUNTS inequalities LaTeX search algebra

POST YOUR ANSWERS UP SO EVERYBODY WILL KNOW HOW TO SOLVE IT

[b]PROBLEM 1[/b]
 Audrey's 300 mile trip took 6 hours to complete. She spent 2/3 of the total time of her trip going the first 1/3 of the distance. What is the ratio of her average speed during that first 2/3 of the time to her average speed during the remaining 1/3 of her time? Express your answer as a common fraction

[quote="Brut3Forc3"][hide="Problem 1"]She spends her first 4 hours going 100 miles, so the initial part is 25 mph. She spends the rest going 200 miles in 2 hours, which is 100 mph. 25/100 = 1/4[/hide]
Uhh, do I add a problem now?[/quote]

yes, you can do that

[b]problem 2[/b]

what is the area of the quadrilateral with vertices at (1,1),(5,2),(4,4),(2,3)?

post your answer up everybody! solve the daily problem!

You could at least explain that it's Shoelace...

[b]Problem 3[/b]

Fabric is run through a machine that fastens sequins in rows and columns such that each sequin is four inches vertically and horizontally from the next closest sequins, forming a grid. There is always a four inch space with no sequins at the beginning, end and side margins of the fabric. if a piece of fabric 60 inches by 72 inches is fed through the machine, how many sequins will be attached?

ps: this one is a little harder than the others

[b]problem 4[/b]

The Moisture Co. produces wipes that are sold 70 to a rectangular carton measuring 6'' by 4'' by 2''. The carton is completely filled. The company wants to sell a carton with dimensions 6'' by 4'' by y'' containing 100 wipes. What is the least possible value of y? Express your answer as a decimal to the nearest tenth.

ps: good job, guys! you have done a great job! keep going!

the new problem is up! what are you guys waiting for?

These are not Intermediate level problems; they are MathCounts problems at best. See the marathon in the MathCounts forum.

these are not mathcounts problems
i got it from a math challenge problems book

Just because you get problems from a non-mathcounts book doesn't mean they aren't mathcounts level, and unfortunately, they are about the same difficulty as problems that would appear on a mathcounts chapter competition.  No offense or anything, but did you get them from math book targeted at the general population that enjoys math class, or from a math book targeted at those who excel at [b]problem solving[/b]?

I wouldn't even go that far. Try classroom math for these types of problems.

ok! if you say so. i mean like i didn't know it was the mathcounts level

Might as well continue this thread with a real problem...
[b]
Problem 5[/b]

For $ x,y,z\in \mathbb R^ \plus{}$, given that $ xyz \equal{} \frac{1}{3}$, show that $ \sum\frac {z}{xy}\ge 1$

Non-specified sums are cyclic.

AM-GM:
$ \sum_{cyc}\frac{x}{yz}\ge 3\frac{xyz}{x^2y^2z^2}\equal{}1$

[quote="madness"]Put 1/xy =3z and so on
So the inequality reduces to 3(x*x+y*y+z*z)>=1
i.e. its enough if we prove that (x^2+y^2+z^2)>=1/3=xyz
But clearly x^2+y^2+z^2>=xy+yz+zx>=xyz
Hence proved
:)[/quote]

1. $ \text{\LaTeX}$ is your friend ;)

2. $ xy+yz+zx\ge xyz$ is generally not true $ (x=y=z=2),$ but $ xy+yz+zx\ge(xyz)^\frac{2}{3}$ is ;)

x*y+y*z+z*x>=3*(xyz)^2/3 = 3^1/3>1/3 (AM-GM)
Hence proved.

P.S. I don't know how to use latex as the files don't seem to download on my PC.
 :)

[quote="madness"]
P.S. I don't know how to use latex as the files don't seem to download on my PC.
 :)[/quote]
There is no need to download $ \text{\LaTeX}$. All you need to do is put dollar signs $ \$$ around the formula in the forums when you post. But you do need to know how to use the simple typesetting codes. Search around this forum, there are links in this forum/AoPSWiki that can guide you and get you started.

I will.

Given a quartic $ c_4x^4\plus{}c_3x^3\plus{}c_2x^2\plus{}c_1x\plus{}c_0\equal{}0$ with 4 positive real roots, determine the minimum possible value of $ \frac{c_1c_3}{c_0c_4}$.

Hmmm, it seems to be nicer when the $ c_{4} \equal{} 1$, so I'll assume that. If not, you end up with zero factors of $ c_{4}$ on top and two on the bottom, and that's a little weird. Alternatively, you could put $ c_{4}$ on top...

[hide]Let the roots be $ a,b,c,d$. $ c_{1} \equal{} \minus{} (bcd \plus{} abd \plus{} acd \plus{} abc),c_{3} \equal{} \minus{} (a \plus{} b \plus{} c \plus{} d)$, $ c_{0} \equal{} abcd$, so the quotient is equal to $ \dfrac{(bcd \plus{} abd \plus{} acd \plus{} abc)(a \plus{} b \plus{} c \plus{} d)}{abcd} \equal{} (a \plus{} b \plus{} c \plus{} d)(\dfrac{1}{a} \plus{} \dfrac{1}{b} \plus{} \dfrac{1}{c} \plus{} \dfrac{1}{d})\ge16$, which follows from Cauchy (or multiplying out and AM-GM), equality iff $ a \equal{} b \equal{} c \equal{} d \equal{} 1$, which means the polynomial is $ (x \minus{} 1)^{4}$, and $ c_{3} \equal{} \minus{} 4,c_{2} \equal{} 6,c_{1} \equal{} \minus{} 4,c_{0} \equal{} 1$. So yeah $ \boxed{16}$.[/hide]

[quote="Temperal"]This is a marathon. It's definition is that when someone solves a problem, a person different from the person whom posted the last problem posts a new problem. Generally, not posting a new problem isn't considered posting a new problem.

Anyone up to posting a new problem?[/quote]

wtf was the point of that?

New problem:

$ x^4\minus{}ax^3\plus{}2x^2\minus{}bx\plus{}1\equal{}0$ has at least one real root. Prove $ a^2\plus{}b^2\ge 8.$

[quote="CatalystOfNostalgia"]Hmmm, it seems to be nicer when the $ c_{4} \equal{} 1$, so I'll assume that. If not, you end up with zero factors of $ c_{4}$ on top and two on the bottom, and that's a little weird. Alternatively, you could put $ c_{4}$ on top...[/quote]

Well, $ a \plus{} b \plus{} c \plus{} d \equal{} \minus{} \frac {c_3}{c_4}$, $ abc \plus{} bcd \plus{} cda \plus{} dba \equal{} \minus{} \frac {c_1}{c_4}$, and $ abcd \equal{} \frac {c_0}{c_4}$, so the quotient with a, b, c, and d is simply $ \frac {( \minus{} \frac {c_3}{c_4})( \minus{} \frac {c_1}{c_4})}{\frac {c_0}{c_4}} \equal{} \frac {c_3c_1c_4}{c_0c_4^2} \equal{} \frac {c_1c_3}{c_0c_4}$.

yeah i know but my point is then the minimum is like 16/c_4 then you can just make c_4 really big, so it's kinda weird.

In terms of $ n$, find the minimum value of $ f(x)\equal{}\displaystyle\sum_{i\equal{}1}^{n}|x\minus{}i|$.


[hide="Chinese Remainder Theorem"]$1155$ is divided by $2$, the remainder is $1$, and minimum of a common multiple of $3, 5, 7, 11$
$770$ is divided by $3$, the remainder is $2$, and minimum of a common multiple of $2, 5, 7, 11$
$924$ is divided by $5$, the remainder is $4$, and minimum of a common multiple of $2, 3, 7, 11$
$1320$ is divided by $7$, the remainder is $4$, and minimum of a common multiple of $2, 3, 5, 11$
$1680$ is divided by $11$, the remainder is $8$, and minimum of a common multiple of $2, 3, 5, 7$

$1155+770+924+1320+1680=5849 \equiv 1229 \pmod{2\times 3\times 5\times 7\times 11}$

Therefore, $\boxed{1229}$[/hide]

[quote="Art of Owna"]WHen n is divided by 2, the remainder is 1
when it is divided by 3, the remainder is 2
when it is divided by 5 the remainder is 4
when it is divided by 7 the remainder is 4
when it is divided by 11 the remainder is 8

what this the smallest posivitive value of n?[/quote]

[hide="Um..."]
\[n\equiv 1\pmod{2}\]

\[n\equiv 2\pmod{3}\]

\[n\equiv 4\pmod{5}\]

\[n\equiv 4\pmod{7}\]

\[n\equiv 8\pmod{11}\]
Thus, $n+1$ is divisible by $2,3,5$, whereas $n+3$ is divisible by $7,11$.

From the first, $n=30p-1$ for some integer $p$.
From the second, $n=77q-3$ for some integer $q$.
\[30p-1 = 77q-3 \]

\[30p = 77q-2 \]

\[77q\equiv 2\pmod{30}\]
From guess and check, I get $q=16$ and $p=41$. Thus, $n=1229$.[/hide]


Putnam
trigonometry
symmetry
inequalities
geometry
3D geometry
sphere

What is the largest possible radius of a circle contained in a 4-dimensional hypercube of side length 1?

The answer is $ \boxed{\frac {\sqrt {2}}2}.$ In order to take advantage of symmetries, let's scale it up to a hypercube of side length 2, and center it at the origin as $ [ \minus{} 1,1]^4.$ I claim that any circle in $ \mathbb{R}^n$ can be parameterized as $ \mathbf{x}(t) \equal{} r\left[(\cos t)\mathbf{a} \plus{} (\sin t)\mathbf{b}\right] \plus{} \mathbf{c}$ where $ \{\mathbf{a},\mathbf{b}\}$ is an orthonormal set in $ \mathbb{R}^n.$ Here, $ r$ is the radius of the circle and $ \mathbf{c}$ is the center of the circle. (For any given circle, this representation is far from unique.) Symmetry considerations allow us to assume in our case that the center is at the origin - that is, $ \mathbf{c} \equal{} \mathbf{0}.$

We want to maximize $ r$ such that $ r\left[(\cos t)\mathbf{a} \plus{} (\sin t)\mathbf{b}\right]$ is contained in the hypercube for all real $ t,$ for any orthonormal $ \{\mathbf{a},\mathbf{b}\}.$

$ \mathbf{x}$ is contained in the hypercube if each component is contained in $ [ \minus{} 1,1].$ The $ j$th component is $ r(a_j\cos t \plus{} b_j\sin t).$ It is well known that the maximum of this simple harmonic motion - the amplitude - is $ r\sqrt {a_j^2 \plus{} b_j^2}.$ Hence we need $ r\sqrt {a_j^2 \plus{} b_j^2}\le 1,$ which we rewrite as $ a_j^2 \plus{} b_j^2\le \frac1{r^2}.$ Adding up these inequalities, we have $ \sum_{j \equal{} 1}^4(a_j^2 \plus{} b_j^2)\le\frac4{r^2}.$ But $ \mathbf{a}$ and $ \mathbf{b}$ were components of an orthonormal set, so in particular, $ \sum_{j \equal{} 1}^4a_j^2 \equal{} \sum_{j \equal{} 1}^4b_j^2 \equal{} 1.$ Hence, $ \sum_{j \equal{} 1}^4(a_j^2 \plus{} b_j^2) \equal{} 2,$ which gives us $ 2\le\frac {4}{r^2}$ from which we get $ r\le\sqrt {2}.$

Letting $ \mathbf{a} \equal{} \frac1{\sqrt {2}}(1,1,0,0)$ and $ \mathbf{b} \equal{} \frac1{\sqrt {2}}(0,0,1,1)$ and $ r \equal{} \sqrt {2}$ gives us $ \mathbf{x}(t) \equal{} (\cos t,\cos t,\sin t,\sin t),$ which does stay within the hypercube. Hence the maximum value of $ r \equal{} \sqrt {2}$ is acheived.

The final answer of $ r \equal{} \frac {\sqrt {2}}{2}$ is obtained by scaling back from side length 2 to the original side length 1.

How many points do you think proving that you can find a circle of radius $ \frac{\sqrt{2}}{2}$ would be worth? I'm hoping for 2...

I got the equation of a sphere, intersected it with two planes that run along the diagonals of the hypercube, and then found the maximum value for r that forced x,y,z,t<1/2 and got the answer. Is that any good?

My proof shows that a circle in a hypercube of side length $ 2$ in $ \mathbb{R}^n$ must have a radius no larger than $ \sqrt {\frac {n}2}.$ If $ n$ is even, we can reach that radius by letting $ r\mathbf{a} \equal{} (1,1,\dots,1,0,0,\dots,0)$ and $ r\mathbf{b} \equal{} (0,0,\dots,0,1,1,\dots,1)$ as we did for $ n \equal{} 4.$

What is the situation when the dimension $ n$ is odd?

aushani, I don't see that that shows that that is the largest possible $ r.$ What is the proof that the maximal solution lies within your restriction?

[quote="Kent Merryfield"]
What is the situation when the dimension $ n$ is odd?[/quote]

I tried to do this problem by looking at smaller cases. I was excited when it was trivial in $ R^2$. Then I drew the cube in $ R^3$...and after moving the circle around the cube in my mind, not being able to find a "natural" way to make the radius maximal nor having any idea how to translate the picture into a constraint optimization problem, I promptly gave up.

[quote="zeb"]How many points do you think proving that you can find a circle of radius $ \frac {\sqrt {2}}{2}$ would be worth? I'm hoping for 2...[/quote]

Me too.  I had such a circle, plus a proof that you could assume the circle to be centered at the center of the cube (basically just symmetry considerations), and an observation that if the radius is greater than that value, there's a circle centered at the cube center lying completely outside the cube.  I thought for a while that that would lead to a proof, but it didn't.

Orient the hypercube so that it's the boundary of $ [0,1]^n$. For the even case the corners of the planar cut are $ \vec{0}$ the origin, the vector $ \vec{1}$ of all ones, $ \vec{x} \equal{} (0,...,0,1,...,1)$ half-zeros and half-ones, and the complement of that ($ \vec{1} \minus{} \vec{x}$).

In the odd case you get $ \vec{0},\vec{1}$ and the third intersection point is $ (0,...,0,\frac {1}{2},1,...,1)$ half-zeros, half-ones and the odd one out is $ \frac {1}{2}$.

I hand-waved a bit on the proof that you need $ \vec{0}$ and $ \vec{1}$ in general position (if no corner is in the cut pull the closest intersection point towards it, all edges of your cut surface lengthen; if only one corner is in it pull the furthest intersection point from that towards the diametrically opposite corner and all edges lengthen again).

Supposing that, the third point which will define the planar cut is some vector of "$ \infty \minus{}$norm 1" (shifted), call it $ \vec{x}$, that is at least one coordinate is 0 or 1 (say 0). Similarly by extending the edge $ (\vec{0},\vec{x})$ to an edge of the hypercube you can show one coordinate of 1 and one of 0 is attained. The last point of intersection is simply $ \vec{1} \minus{} \vec{x}$, and your cut is actually a rhombus, with largest incircle radius being the min distance from the center of the rhombus, obviously $ \vec{\frac {1}{2}}$, to the center of one of the edges.

That is, $ \min\{\sqrt {2\left(\frac {1}{2}\right)^2 \plus{} \sum_{i \equal{} 1}^{n \minus{} 2}\left(\frac {x_i}{2}\right)^2},\sqrt {2\left(\frac {1}{2}\right)^2 \plus{} \sum_{i \equal{} 1}^{n \minus{} 2}\left(\frac {1 \minus{} x_i}{2}\right)^2}\}$.

Those numbers are squares less than one, so to maximize pick exactly half of the $ x_i$s to be 1 and the other half 0. With an odd number, pick that last one out to be $ \frac {1}{2}$, for a radius of $ \sqrt\frac{4n\plus{}5}{16}$ in $ H_{2n\plus{}1}$.

So, in the cube, pick two diametrically opposite corners, then the other two intersection points are going to be the mid-points of two diametrically opposite edges both not incident to the corners chosen. (That's the best I can write without drawing a picture ;) )

Putnam grading is really harsh, right ? I remember last year for the probability question I had basically a correct proof but forgot to multiply by a factor (it was 2k+1 / 3k+1 or something) that came from an easy observation and I think I got a 1 for the problem. For this problem, I would be hoping for one point at most. Of course, I'm not an expert on Putnam grading by any means.

The answer is $ \sqrt{\frac{n}{8}}$ for all dimensions $ n$.

Are you sure of that for odd $ n$?

The existence question is equivalent to the existence of a pair of orthonormal vectors $ (x_1,x_2,\cdots,x_n)^T$ and $ (y_1,y_2,\cdots,y_n)^T$ such that $ x_1^2 \plus{} y_1^2 \equal{} x_2^2 \plus{} y_2^2 \equal{} \cdots \equal{} x_n^2 \plus{} y_n^2$.

You managed to hit the bound in say the cube? I knew I wasn't rigorous in picking two corners... so how did you do it?

I was wondering if there was a group-theoretic way to attack this problem.  It seems to me that the maximal circle should be contained in some regular $ n$-gon whose points are points on the surface of the hypercube for some $ n$, and if that can be proven then the problem becomes describing certain finite subgroups of $ SO_4$...?

That would be $ \sqrt{\frac {n}2}$ if we use $ [ \minus{} 1,1]^n$ as our hypercube. I've got the proof above that it is no greater than that, but can you really reach that for odd $ n?$

My first try at an example: $ n \equal{} 3.$ I'm guessing that our best chance would be to put the circle in a plane orthogonal to $ (1,1,1).$ Completing that to an orthogonal set and normalizing, I get $ \mathbf{a} \equal{} \frac1{\sqrt {2}}(1, \minus{} 1,0)$ and $ \mathbf{b} \equal{} \frac1{\sqrt {6}}(1,1, \minus{} 2).$ Multiplying through by $ r \equal{} \sqrt {\frac32}$ I get $ r\mathbf{a} \equal{} \left(\frac {\sqrt {3}}2, \minus{} \frac {\sqrt {3}}2,0\right)$ and $ r\mathbf{b} \equal{} \left(\frac12,\frac12, \minus{} 1\right).$

Then my parametric equations are

$ x \equal{} \frac {\sqrt {3}}2\cos t \plus{} \frac12\sin t$

$ y \equal{} \minus{} \frac {\sqrt {3}}2\cos t \plus{} \frac12\sin t$

$ z \equal{} \minus{} \sin t$

Yes, that does stay inside the cube and is a tight fit in all three dimensions.

So while I'm still skeptical of Albanian Eagle's claim, it's at least true for $ n \equal{} 3.$

Once we have the $ n\equal{}3$ example, we can add more dimensions two at a time- each belongs to only one of the orthogonal pair. It does work.

Well, I can pretty much assume I zeroed this question on false assumptions  :roll:

[quote="Kent Merryfield"]I claim that any circle in $ \mathbb{R}^n$ can be parameterized as $ \mathbf{x}(t) \equal{} r\left[(\cos t)\mathbf{a} \plus{} (\sin t)\mathbf{b}\right] \plus{} \mathbf{c}$ where $ \{\mathbf{a},\mathbf{b}\}$ is an orthonormal set in $ \mathbb{R}^n.$[/quote]
And proof?

Basically, all you need to do is pick two points on the circle $ 90^{\circ}$ apart, subtract the center, and normalize. Let's put it this way: how would you describe a circle in $ \mathbb{R}^n?$

It would be nice to prove that the largest radius of an $ m$-dimensional sphere that fits inside the $ n$-dimensional cube ($ n\geq m$) of side $ 1$ is $ \sqrt{\frac{n}{4m}}$.  :)

I think my argument gets one side of it - that the radius is less than equal to what you just said. (I'd prefer to make it a hypercube of side length 2 and quote your number as $ \sqrt{\frac{n}{m}},$ if you don't mind.)

Take an orthonormal set $ \{\mathbf{a}_1,\mathbf{a}_2,\dots\mathbf{a}_m\}$ and look at the set $ \left\{\sum_{i\equal{}1}^mt_j\mathbf{a}_j\ : \ \sum_{j\equal{}1}^mt_j^2\equal{}1\right\}.$ That's a sphere. Now go through my proof above; it still works in this situation.

So the challenge is to prove that you can achieve this maximum (if it's even true). We know we can have it for $ m\equal{}1$ and all $ n$ (that's just the long diagonal.) We know we can have it for $ m\equal{}2$ and all $ n;$ that's what we've already argued here. We have it trivially for $ m\equal{}n.$

Also I think it is easy to prove it can be reached for $ m|n$. And by something similar to an argument above we only need to check it for $ n\le 2m\minus{}1$

Albanian Eagle's conjecture, in language that resembles functional analysis:

Fix $ n.$ Let $ \Lambda_m$ be the set of all $ m$-dimensional subspaces of $ \mathbb{R}^n.$ We claim that
\[ \sup_{S\in \Lambda_m}\,\sup_{x\in S\setminus\{0\}}\,\frac{\|x\|_2}{\|x\|_{\infty}}\equal{}\sqrt{\frac{n}{m}}\]

I claim we have already proved $ \le$ in this inequality; we're trying to prove $ \equal{}.$ The re-phrasing is just to see if the alternate language brings any more ideas to mind. Or perhaps it could even be re-phrased again as something about $ n\times m$ real matrices.

Let's try one more way to re-express this.

What is the largest value of $ r$ such that there exists an $ n\times m$ real matrix $ A$ (with $ m\le n$) such that $ A^TA\equal{}r^2I$ and every element of the diagonal of $ AA^T$ is $ \le 1?$

We must have $ r^2\le\frac{n}{m}.$ The proof of this: If every element of the diagonal of $ AA^T$ is $ \le 1,$ then $ \operatorname{trace}(AA^T)\le n.$ But $ \operatorname{trace}(AA^T)\equal{}\operatorname{trace}(A^TA)\equal{} \operatorname{trace}(r^2I)\equal{}r^2m.$ Hence, $ r^2m\le n$ or $ r^2\le\frac{m}{n}.$

Conjecture: for all $ m$ and $ n,$ there exists a matrix $ A$ such that $ A^TA\equal{}\frac{n}{m}\,I$ and every element of the diagonal of $ AA^T$ is $ 1.$

Somehow solved the problem lying in bed last night when I thought it wasn't possible for me to tamper with any geometry past 3 dimensions. Solution is basically the same as prof. Kent's save for a different basis and me backing my steps with some not-too-hard-to-follow intuition.

[hide="Solution"]We start by trying to parameterize a circle in $\mathbb{R}^n$. Since a circle lies in a plane, and a plane is the set of linear combinations of its basis vectors $\vec{v_1}, \vec{v_2} \in \mathbb{R}^n$, a circle centered at the origin on the plane is the set of linear combinations of $\vec{v_1}, \vec{v_2}$ such that $a\vec{v_1} + b\vec{v_2}$ has constant magnitude. Now let $\vec{v_1}, \vec{v_2}$ be an orthonormal basis in $\mathbb{R}^n$. Then $\|a\vec{v_1} + b\vec{v_2}\| = \sqrt{(a\vec{v_1} + b\vec{v_2}) \cdot (a\vec{v_1} + b\vec{v_2})} = \sqrt{a^2 + b^2} = r$, with $r$ being the radius of the circle, and setting $a = r\cos{\theta}, b = r\sin{\theta}$ gives us a simpler expression for $a, b$. Hence, the circle can be parameterized as $\vec{s} + r\cos{\theta}\vec{v_1} + r\sin{\theta}\vec{v_2}$ with $\vec{s}$ the center of the circle.

Let the hypercube have vertices $\left(\pm\tfrac{1}{2}, \pm\tfrac{1}{2}, \pm\tfrac{1}{2}, \pm\tfrac{1}{2}\right)$. For a circle to be contained in the hypercube, we must have that each component of $\vec{s} + r\cos{\theta}\vec{v_1} + r\sin{\theta}\vec{v_2}$ is in the interval $\left[-\tfrac{1}{2}, \tfrac{1}{2}\right]$. Clearly, setting $\vec{s} = \vec{0}$ allots for the maximum radius. If we set $\vec{v_1} = (a_1, a_2, a_3, a_4)^T, \vec{v_2} = (b_1, b_2, b_3, b_4)^T$, we have that $$-\tfrac{1}{2} \le r\cos{\theta}a_i + r\sin{\theta}b_i \le \tfrac{1}{2}, 1 \le i \le 4.$$ Then, differentiating each component yields that each component attains a maximum when $-r\sin{\theta}a_i = r\cos{\theta}b_i \rightarrow \tan{\theta} = \tfrac{b_i}{a_i}$. Then $\cos{\theta} = \tfrac{a_i}{\sqrt{a^2_i + b^2_i}}$ and $\sin{\theta} = \tfrac{b_i}{\sqrt{a^2_i + b^2_i}}$ and the maximum of each component is $r\sqrt{a^2_i + b^2_i}$. It follows that $$r = \frac{\frac{1}{2}}{\max_{1 \le i \le 4}(\sqrt{a^2_i + b^2_i})}.$$

Now we claim that the lowest that $M = \max_{1 \le i \le 4}\left(\sqrt{a^2_i + b^2_i}\right)$ can be is $\sqrt{\tfrac{1}{2}}$, which occurs when the squares of all components of $\vec{v_1}$ and $\vec{v_2}$ are equal to $\tfrac{1}{4}$ though this isn't necessarily the only equality case. To prove this, suppose for the sake of contradiction that $M < \sqrt{\tfrac{1}{2}}$. Then we must have that $a^2_i + b^2_i < \tfrac{1}{2}$ for three values of $i$. WLOG assume that these three values of $i$ are $1, 2, 3$ so $$a^2_1 + a^2_2 + a^2_3 + b^2_1 + b^2_2 + b^2_3 < \tfrac{3}{2}.$$ But since $\vec{v_1}, \vec{v_2}$ is orthonormal, we have $$a^2_1 + a^2_2 + a^2_3 + a^2_4 + b^2_1 + b^2_2 + b^2_3 + b^2_4 = 2.$$ Then $a^2_4 + b^2_4 > \tfrac{1}{2}$, establishing the contradiction. Thus, the largest $r$ is equal to $\boxed{\tfrac{\sqrt{2}}{2}}$, attained with basis $$v_1 = \left(\tfrac{1}{2}, \tfrac{1}{2}, \tfrac{1}{2}, \tfrac{1}{2}\right)^T, v_2 = \left(\tfrac{1}{2}, \tfrac{1}{2}, -\tfrac{1}{2}, -\tfrac{1}{2}\right)^T.$$[/hide]


algebra
system of equations
algebra unsolved

Solve in real numbers the system of equations: \begin{align*}
\frac{1}{xy}&=\frac{x}{z}+1 \\
\frac{1}{yz}&=\frac{y}{x}+1 \\
\frac{1}{zx}&=\frac{z}{y}+1 \\
\end{align*}

If x=y=z then $x=y=x=\pm \frac{1}{\sqrt{2}}$. If not all three of them are equal then x>y or y>z or z>x. Suppose x>y (other cases similar).
Divide equation 2 by eq 3:
$\frac{\frac{1}{yz}}{\frac{1}{zx}}=\frac{\frac{y}{x}+1}{\frac{z}{y}+1}$ which is equivalent to $\frac{x^{2}}{y^{2}}=\frac{y+x}{y+z}$ Well, x>y and therefore y+x>y+z, yielding x>z. Now repeat same reasoning with the first and second equation. This results in z<y. Same reasoning again with first and third equation gives us x<y which contradicts our assumption. Done.

[quote="Beat"]Solve in real numbers the system of equations:
$\frac{1}{xy}=\frac{x}{z}+1$ (1)
$\frac{1}{yz}=\frac{y}{x}+1$ (2)
$\frac{1}{zx}=\frac{z}{y}+1$ (3)[/quote]
If x<0.
If y>0 .(1)$\rightarrow{z>0}$;(3)$\rightarrow$ Left hand <0<Right Hand
So y<0.(2)$\rightarrow{z<0}$.Thus x,y,z<0
If x>0,similarly ,we have x,y,z>0
We consider x,y,z>0.Suppose $x\geq{y}$
 From (2),(3) we have :$\frac{x^{2}}{y^{2}}=\frac{x+y}{y+z}$
thus$x\geq{z}$
On the other hand ,from(1),(2),similarly,we have
$\frac{z^{2}}{x^{2}}=\frac{x+z}{y+x}$ thus $y\geq{z}$
So $x\geq{y}\geq{z}$
From(1) and (3),we have $\frac{z^{2}}{y^{2}}=\frac{x+z}{y+z}$ Thus $x\leq{y}$.So x=y=z .We easily have :$x=y=z=\pm{\frac{1}{\sqrt{2}}}$ (Consider x,y,z<0) :)


Prove that: there are unlimited group of 6 real numbers that satisfy:
 a+b+c=d+e+f
a^2+b^2+c^2=d^2+e^2+f^2

Let $ a\equal{}b\equal{}c\equal{}d\equal{}e\equal{}f$ then $ a^2 \plus{} b^2 \plus{} c^2 \equal{} d^2 \plus{} e^2 \plus{} f^2$ for all numbers a.

What does
[quote="trangiaquan"]unlimited group of 6 real numbers [/quote]
mean?

It means an infinite amount of groups of 6 real numbers.


Let $ J$ be any set, and for each $ j \in J$, let $ A_j$ be contained in $ X$. Show that $ \mathcal{C}\left(\bigcap \{A_i: j \in J \}\right) \equal{} \bigcup\{ \mathcal{C}(A_j): j \in J \}$.

Here $ \mathcal{C}(A_i) \equal{} E \backslash A_j$.

Consider an arbitrary element $ a\in J$, we have:

$ a \in \mathcal{C}\left(\bigcap A_j: \right) \Leftrightarrow \exists A_j: a\not\in A_j \Leftrightarrow \exists j: a\in\mathcal{C}(A_j) \Leftrightarrow a\in\bigcup\mathcal{C}(A_j)$

This implies what we want to prove.

I think this problem is also a corollary of the Principle of Inclusion and Exclusion :oops:


group theory
abstract algebra
linear algebra
matrix
superior algebra
superior algebra theorems

If H is an infinite subgroup of a group G=<X>, and xHx^(-1) is a subset of H for all x in X, does it imply H is normal in G?  :roll:

but i cant show that x^(-1)Hx is a subset of H, how do you get the first relation?

What does "for all x" mean? Apply the one thing we know with a different value.

But X is just a set of generators, it may not include its inverses, right?

So you've shot down the argument, which was based on $ X\equal{}G$.

OK, just build something. Generators $ x,y$, relation $ xyx^{\minus{}1}\equal{}y^2$. The subgroup generated by $ y$ is not normal.

I found it:
$ x\equal{}\begin{bmatrix}
2 & 1\\
0 & 1
\end{bmatrix},y\equal{}\begin{bmatrix}
1 & 1\\
0 & 1
\end{bmatrix}$

It makes me feel better. :D

Anyway, thank you jmerry.

I have one question of low level...
When xyx^-1 = y^2, Why <y> is not normal in <{x,y}> ?


Twice four and twenty blackbirds
Sitting in the rain
I shot and killed a quarter of them
How many do remain?

[hide="Psh"] None of them.  They've all flown away ;) [/hide]


[b]In a math class, 12 out of 15 girls are freshmen and 11 out of 15 boys are freshmen. What is the probability that in a randomly selected group of five studenst from the class, there will be two freshmen girls and three freshmen boys?[/b]

[hide="answer"]1306800/17100720=0.076[/hide]

[hide="where did i go wrong"] there are 12C15 ways to choose 12 girls and 2C12 to choose two freshmen girls. there are 11C15 ways to choose 11 boys and 3C11 ways to choose 3 freshmen boys.
the total probability is:
2C12/12C15*3C11/11C15. which is wrong. their solution used permutations but how would you do this problem with combinations?[/hide]

I think you got the right concept, but the denominator part was wrong.
I know this for sure, because I missed this one, too.

[hide="I think"][b]Total[/b]: It should be $ \binom{30}{5}$, because there are 30 students and 5 to choose.

[b]2 girls[/b]: $ \binom{12}{2}$

[b]3 boys[/b]: $ \binom{11}{3}$

So the answer is :$ \frac{\binom{12}{2}\cdot\binom{11}{3}}{\binom{30}{5}}\approx{0.076 }$[/hide]

[hide]Mihail911, the denominator is supposed to be the number of possible group combinations. Since there are 30 people in this class, there are $ \binom{30}{5}$ or $ 142506$ possible group combinations. The numerator is supposed to be the amount of groups that meet the restrictions. Now we calculate the amount of groups with 2 freshmen girls and three freshmen boys. This is simply $ \binom{12}{2}\cdot\binom{11}{3}$ or $ 10,890$ possible combinations. Now we calculate $ \frac {10,890}{142,506}$ and get $ \boxed{.076}$. [/hide]

What does (30)
               ( 5 )

mean?

[quote="Ihatepie"]What does (30)
               ( 5 )

mean?[/quote]

It's binomial. It's [u]combination[/u]. It's derives from $ \frac{n!}{(n\minus{}k)!k!}$ It's the same thing. It's just  different way to say $ _{30}C_5$ or 30 choose 5.

$ \binom{n}{r}$ is sometimes called a binomial coefficient due to it's use in the binomial theorem.  Also, it's the number of ways to choose $ r$ items from $ n$.  Finally, $ \binom{n}{r}\equal{}\frac{n!}{(n\minus{}r)!r!}$

Furthermore, n!, or any (number/variable)! is (n)(n-1)(n-2)...(1). It is called "n factorial."


Consider a box containing $ n$ balls numbered $ 1$ to $ n$. If $ 455$ is the number of ways to get three balls in this box that are not part of the same two or three balls called consecutive numbers. Determine the value of $ n$.

[hide]Answer: 15

Fix some three balls, do not number them, then distribute balls around them which corresponds to non negative integer solutions of $ x_1\plus{}x_2 \plus{}x_3\plus{}x_4\equal{}n\minus{}3$ which are $ n\minus{}3\plus{}4\minus{}1\choose 4\minus{}1$ in number.[/hide]

The answer is [hide]$ n\equal{}17$.[/hide]

hi bros

 if there are n different things and are lined up . one is required to pick up  m  things such that no two are consecutive.

(n-m+1)Cm

[hide="Complimentary Counting"]
Clearly there are $ n \choose 3$ number of ways to choose any three balls from the box. Now, if we subtract the number of ways where you choose two or three consecutive balls, then we will be left with only the number of ways to choose nonconsecutive balls.
By PIE, we know that the number of ways to choose two or three consecutive balls equals the number of ways to choose two consecutive balls minus the number of ways three consecutive balls.
The number of ways to choose two consecutive balls is $ (n \minus{} 1)(n \minus{} 2)$ because there are $ n \minus{} 1$ ways to choose the first of two consecutive balls, the second ball is the one immediately after it so there is only one choice for this one, and then $ n \minus{} 2$ ways to choose any remaining third ball.
The number of ways to choose three consecutive balls is $ n \minus{} 2$, because you can choose all but the last 2 balls to be the first of a set of three consecutive balls.
Hence the equation to solve is:
$ 455 \equal{}$ $ n \choose 3$ $ \minus{} [(n \minus{} 1)(n \minus{} 2) \plus{} (n \minus{} 2)]$, which makes
$ n \equal{} \boxed{17}$
[/hide]

[quote="Talegari"][hide]Answer: 15

Fix some three balls, do not number them, then distribute balls around them which corresponds to non negative integer solutions of $ x_1 + x_2 + x_3 + x_4 = n - 3$ which are $ n - 3 + 4 - 1\choose 4 - 1$ in number.[/hide][/quote]

If we fix three balls, then we have the arrangement $ x_{1}  \| ball  \| x_{2}  \| ball \| x_{3}  \| ball  \| x_{4}$, where $ x_{1},x_{2},x_{3},x_{4},x_{5}$ represent the number of balls between the fixed balls. So, we have $ x_1 + x_2 + x_3 + x_4 = n - 3$

Since we cannot have consecutive numbers, $ x_{2}$ and $ x_{3}$ must be integers $ \geq 1$, while $ x_{1}$ and $ x_{4}$ can be integers $ \geq 0$. Let us set $ x_{5}=x_{2}+1$ and $ x_{6}=x_{3}+1$.

Thus, we want the number of non-negative solutions to $ x_1 + x_5 + x_6 + x_4 = n - 5 \Rightarrow \binom{n - 5 + 4 - 1}{4-1}=455$$ \Rightarrow \boxed{n=17}$.

[i][b]Consider a box containing $ n$ balls numbered $ 1$  to $ n$ . If $ 455$ is the number of ways to get three balls in this box that are not part of the same two or three balls called consecutive numbers. Determine the value of $ n$ .[/b][/i]

[hide="Check it - I bet it is the coolest solution provided"]It is simply $ ^{n\minus{}2}C_{3} \equal{} 455$
=> $ (n \minus{} 2)(n \minus{} 3)(n \minus{} 4) \equal{} (13)(14)(15)$
=> $ n \equal{} 17$
[hide="Explanation:"] First place $ n \minus{} 3$ balls any where un-numbered. Now we need to select $ 3$ places from $ n \minus{} 2$ places between them including the end places. Now these $ 3$ places can be used to put the balls.[/hide][/hide]


What do you think will be the cutoff qualifying score for AMC12B ?

hmm...i heard it was harder than the A from some people and easier from others...I didnt personally take it but im guessing that if you scored anywhere from 90+ you should be ok...but dont take it from me!

any idea for 10B???is 110 ok!

In response to zolojetto I don't think the amc 10B cutoff has ever been as low as 110..
In my opinion, it was easy to pass, but EXTREMELY hard to get perfect, because of confusing and tricky problems, like 24 and 25

I think amc 12B was easy.. I actually UNDERSTOOD all the solutions, so the cutoff is probably 100

ok fine let me rephrase that: I could actually do all of them by myself and understand other peoples' solutions

I understood all the solutions for most contests I fail at.
I really doubt the cutoff is 100.

I'd hazard something around 97 for the 12B.
And the 10B was easy, but hard to get perfect, as mentioned earlier. 117, perhaps?

Wow I'm hoping for the cutoff for 12B to be 90+ (but I doubt it). My school just mentioned they supported AMC the day before so I didn't study  :( . Apparently I did as well as the seniors and juniors there.

The only thing that made the 10B hard to get perfect on, imo, was #25, and only if you didn't draw a picture.


AoPS has one bug on wednesday.  It has 2 bugs on thursday.  On friday, it has 6 bugs!  On sunday, it has 120 bugs :o !  After a bug invasion on Wednesday, it has 40320 bugs.  After another wave of bugs, on sunday AoPS has a whopping 479002600 bugs!   After 13 days(starting at the first wednesday), the Administrators destroy 1/5 of all the bugs. After those bugs are destroyed,  how many bugs will AoPS have?   :D [b]no calculators[/b]!!!! otherwise this problem will be dramatically easy.  Good luck.  If nobody can answer this problem I will post a very helpful hint on thurday.

If we could zap 1/5 of our bugs on one day, that would be a most excellent day!

[hide="HiNt"]
Read the following BACKWARDS! :D :D :D :diablo:

lairotcaf
...
[/hide]

would anyone like me to post a solution?  236 if you calculated your number, it would be wrong.  I want answers simplified to a number, not like what 236 gave me.  Who dares to solve the bugs of AoPS :D

Okay, then, it's $\frac{13!\cdot4}{5}$, which is equal to 4981616640.  :D


How much knowledge does one need to get to nationals for the Physics Olympiad? If I take AP Physics C and know AP Physics B material, can I get there? (rising junior so I have another year after this)

I don't know much about the USAPhO, but I assume it is similar to other olympiads in more way than one.  Presumably, if you learn the material taught in an introductory physics course (Physics C), you could make it to nationals.  However, it seems likely that the USAPhO is based more on how you synthesize and apply that material than how you memorize certain methods.

Take for example the USAMO.  Each question may be solved using only concepts learned through precalculus, but not everyone who knows precalculus well can answer the questions.  Something of the sort must be true for the USAPhO as well.


[hide]Factor out (2m-3)(n-1).  Let's call that b.  m-n-4=a so m-n-2=a+2
We have b(x^2+ax-2a-4)=1 so b and that quadratic equation in the center are reciprocals.  Since m and n are integers, there's no way b is a fraction; b is a integer too.  The only possibility for a fraction is in the quadratic term.  But since we want integer solutions for x and m and n are integers, how can there be a fraction anywhere in that quadratic?  There can't so there's no solution.[/hide]

Oops, there was a mistake on my previous proof.  [hide]They are only reciprocals when b and the quadratic are plus or minus one.  Let's check the possibilities.  Can (2m-3)(n-1) equal 1?  Since both m and n are integers, their terms have to both equal either plus one or minus one.  A quick check tells that (2,2) and (1,0) are solutions.  If we plug (2,2) into the quadratic equation, we get x^2-4x+4=1 or x^2-4x+3=0.  We factor to get the solution (x-3)(x-1) so (2,2) works.  Plugging (1,0), we get x^2-3x+2=1 or x^2-3x+1=0 which has irrational roots.  Now, we set (2m-3)(n-1)=-1 so we get (1,2) and (2,0).  Trying (1,2) for the quadratic, we get -x^2+5x-6=-1 or x^2-5x+6=1 or x^2-5x+5 which has irrational solutions.  Trying (2,0) for the quadratic, we get -x^2+2x=-1 or x^2-2x=1 or x^2-2x-1 which also has irrational solutions.  So (2,2) is the only solution.[/hide]


The number $ 273*49*5$ is an eight-digit number where two of the digits have been erased and replaced by asterisks. The number is divisible by $ 9$ and $ 11$. What are the two missing digits?

[quote="grafitti123"]The required number is $ 27374985$.
You can get it with direct application of divisibility rules.[/quote]
that is what I did, but its full of listing numbers and trial and error. I smell Chinese Remainder Theorem... FI FI FO FUM! Sorry. Just had to do that.

It's not necessarily complete brute-force.

$ \minus{}2\plus{}7\minus{}3\plus{}a\minus{}4\plus{}9\minus{}b\plus{}5\equal{}a\minus{}b\plus{}12\equiv a\minus{}b\plus{}1\equiv 0\bmod 11\implies a\minus{}b\equiv \minus{}1\bmod 11$

$ 2\plus{}7\plus{}3\plus{}a\plus{}4\plus{}9\plus{}b\plus{}5\equal{}a\plus{}b\plus{}30\equiv a\plus{}b\plus{}3\equiv 0\bmod 9\implies a\plus{}b\equiv \minus{}3\bmod 9$.

$ a\minus{}b\equal{}11k\minus{}1$
$ a\plus{}b\equal{}9j\minus{}3$

Now you can force $ k$ and $ j$ to be even by adding these and using mod 2, and then bound them because your digits must be between 0 and 9, so there's less cases to check

This is only a little bit faster...

[hide]An equivalent rule for divisibility by 99 is to partition the number into blocks of two digits and see if those sum to a multiple of 99. Note that this rule also means that we can switch a digit with any other one that is two away. So it suffices to find digits such that $ 273549ba$ where $ b$ is the second $ *$ and $ a$ is the first $ *$. Thus $ 10b \plus{} a \equal{} \minus{}(27 \plus{} 35 \plus{} 49) \pmod{99}$ and it's arithmetic from here.[/hide]


What determines which LaTeX sequences can be clicked to display the source and which ones can't?  Length?

Examples:

$x$

$x^2 - 3x + 2$

$\displaystyle \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$

[quote="mathfanatic"]What determines which LaTeX sequences can be clicked to display the source and which ones can't? [b] Length? [/b]
[/quote]Yes.


modular arithmetic
calculus
integration
algebra
polynomial
number theory

I was thinking to made with the rest of a number of the following form $7k+a$ at a perfect square, but i like your solution ,too.

Nice.

Just btw, that's assuming that $x$ isn't divisible by $7$, but it happens that if it were the LHS would be $7k+1$, and they would still be non-congruent.

[quote="towersfreak2006"][quote="mychrom"]show that the equation $x^{4}-x^{2}+1=7y$ has no integer solution.[/quote][hide]The equation above implies $x^{6}+1=7y(x^{2}+1)$. By Fermat's Little Theorem, $x^{6}\equiv 1\pmod{7}$, so the left hand side is always congruent to $2\pmod{7}$ while the right hand side is congruent to $0\pmod{7}$ for integral values of $x$ and $y$. Hence, there are no solutions.[/hide][/quote]
I concur - that would be my solution too; factorizations of polynomials of the form $x^{2}-x+1$ are useful to notice.


Merhabalar.
Turkiyenin fizik olimpiatin son yillarin sorulari bulmak istiyorum .Yardimci olabilir misiniz

Tübitak, 1990-2003 yılları arasında Ulusal Fizik Olimpiyatları 1. aşama, 2. aşama, takım seçme ve kamplarda sorulan soruları iki ciltlik bir eserde topladı. Bunları kitapçılardan ya da TÜBİTAK yayınlarından bulabilirsin. Ayrıca sadece 1. aşamaya yönelik istersen Altın Nokta Yayınları'nın kitabını önerebilirim. Bu kitap da 1993-2006 yılları arasında sorulan 1. aşama soruları ve çözümlerini içeriyor.


quadratics
algebra
polynomial
Vieta
product of roots
sum of roots

The teacher has written a quadratic on the board, but Heather can't read the linear term.  She can see that the quadratic term is $ 4x^2$ and that the constant is $ \minus{}24$.  She asks her neighbor, Noel, what the linear term is.  He decides to tease her and just says, `One of the roots is 4.'  Heather then says, `Oh, thanks!'  She then correctly writes down the linear term.  What was the linear term (include entire term, not simply the coefficient.)

By Vieta the product of roots is -6, and one of the roots is 4, so the other root is -1.5, so the sum of roots is 2.5. Un-vieta it and the linear term is -10x.

Alternatively: plug in x=4.


function
analytic geometry
vector
real analysis
real analysis unsolved

We proved that $\mathbb{N}^{2}$ is countable using the bijection: $\varphi: \left\{\begin{array}{rll}\mathbb{N}^{2}&\longrightarrow&\mathbb{N}^{*}\\(p,q)&\longmapsto&2^{p}(2q+1)\end{array}\right.$.
My question is: Can we generalize this proof for $\mathbb{N}^{q}~(q\in\mathbb{N}^{*})$ using the function $\psi: \left\{\begin{array}{rll}\mathbb{N}^{q}&\longrightarrow&\mathbb{N}^{*}\\(n_{1},\dots,n_{q})&\longmapsto&q^{n_{1}}(qn_{2}+1)\dots(qn_{q}+q-1)\end{array}\right.$  ?
[b]Or simply[/b] is $\psi$ a bijective function?
Thanks in advance.

No, that's not the way to generalize the method. It's not injective and it's not surjective. For $q=3$, we can't get $1,3,4,7,9,10,12,13,\dots$, and $(0,1,0)$ and $(0,0,2)$ both map to 8.

The way to generalize it is to use $\varphi$ on the first two "coordinates" of the vector to map $\mathbb{N}^{n}$ to $\mathbb{N}^{n-1}$, and so on.

Edit: the above works if you have have a "pairing function" $\varphi$
 that send $\mathbb{N}^{2}$ to the whole of $\mathbb{N}$.


search
combinatorics proposed
combinatorics

For which $ n\in \{3900,3901,...,3909\}$ can the set $ \{1,2,...,n\}$ be partitioned into disjoint triples in such a way that in each triple one of the number equals the sum of the other two?


In the diagram, the circle has radius 6, DE is a chord, angle AOB is a right angle, and AO=BO=$ 3\sqrt{2}$. What is the area of the shaded region?

[hide="Hint"]Find the area of the shaded region plus the triangle (using sectors), and then subtract the area of the triangle.[/hide]

Oh! [b]123456789[/b] you can post your solution?

[hide="Solution"]There is another way to do this, that uses two $ 15^\circ - 30^\circ 135^\circ$ triangles, but I prefer this way:
The area of the shaded region is a semicircle plus two $ 30^\circ$ sectors, plus a $ 120^\circ - 30^\circ - 30^\circ$ triangle, minus the area of $ \triangle ABO$. Hence, the area is:
\begin{align*} & \frac {1}{2}\cdot \pi \cdot 6^2 + 2 \left(\frac {1}{12} \cdot \pi \cdot 6^2\right) + \frac {1}{2} \cdot 6^2 \cdot \sin 120^\circ - \frac {{\left(3\sqrt {2}\right)}^2}{2} \\
& = 18\pi + 6\pi + 9\sqrt {3} - 9 \\
& = \boxed{24\pi + 9\sqrt {3} - 9} \end{align*}
[/hide]

its basically finding the area of the major segment(120/360)*2*3.14*6 then add to two smaller tringles formed by drawing the radiusi.e(1/2*3*square root of two *6)


algebra
polynomial
complex numbers
algebra unsolved

Let $ P(x)$ be a nonzero polynomial such that, for all real numbers $ x$, $ P(x^2 \minus{} 1) \equal{} P(x)P(\minus{}x)$. Determine the maximum possible number of real roots of $ P(x)$.

I have a solution for this problem but it requires advance theory in algebra.

Denote $ f(x) \equal{} x^2 \minus{} 1$ and $ f^{k} (x)$ is defined recursively as: $ f^{0} (x) \equal{} x$ and $ f^{k}(x) \equal{} f(f^{k\minus{}1}(x))$.

First, $ P(x)$ must be monic. And its factorization from the [b] Fundamental Theorem of Algebra [/b] is of form:
\[ P(x) \equal{} \prod_{i\equal{}1}^{n} (x\minus{}c_i)\]
where $ c_i \in \mathbb{C}$ and $ n \equal{} \deg P$.

The hypothesis $ P(x^2 \minus{} 1) \equal{} P(x) P(\minus{}x)$ gives:
\[ \prod_{i\equal{}1}^{n} (x^2 \minus{} 1 \minus{}c_i) \equal{} \prod_{i\equal{}1}^{n} (x\minus{}c_i) \prod_{i\equal{}1}^{n} (\minus{}x\minus{}c_i) \equal{} (\minus{}1)^n \prod_{i\equal{}1}^{n} (x^2\minus{}c_i)\] (1)

Then $ n$ is even. Let $ y \equal{} x^2$, (1) implies
\[ \prod_{i\equal{}1}^{n} (y \minus{} 1 \minus{}c_i) \equal{} (\minus{}1)^n \prod_{i\equal{}1}^{n} (y\minus{}c_i^2)\]

Due to the [b] Unique Factorization Theorem [/b], the complex numbers $ c_i^2$ are a permutation of the numbers $ 1\minus{}c_i$. Since a permutation could be written as product of disjoint cyclic permutations, we could partition the set $ c_i$ to sets $ c_{i,j}$ where $ c_{i,j\plus{}1}^2 \equal{} 1\plus{}c_{i,j}$ and $ c_{i,k}^2 \equal{} 1 \plus{} c_{i,1}$.

This means we could sort the roots of $ P(x)$ into roots of $ f^{k}(x) \equal{} 0$. And hence, any such $ P(x)$ satisfying the condition of the problem must have the form:
\[ P(x) \equal{} \prod_{t\equal{}1}^{k} f^{i_t}(x)\]

The remaining task is proving that for any $ k$, $ f^{k}$ has no more than 4 real roots. And the final conclusion is:

$ P(x)$ has at most 4 (real) roots.

[b] I really wish to know the proof without such high level result.  :wink: [/b]

$ (x^2\plus{}x)^{n_1} (x^2\minus{}x\minus{}1)^{n_2}$ fits the bill

[quote="t0rajir0u"]Can you demonstrate $ P(x)$ such that it has exactly $ 4$ real roots?  And as far as facts about polynomials unique factorization can hardly be considered "advanced."[/quote]

The advanced stuff I mentioned comes from "The Fundamental Theorem of Algebra" not the Unique Factorization. Anyway, thank you for your comment. I realize that we can modify the proof so as to eliminate the use of such a theorem.

Also, I made a big mistake :D  . It should be: "All the real roots of $ P(x)$ are roots of $ f^{k}(x)$ for some $ k$", not $ P(x) \equal{} \prod_{t \equal{} 1}^{k} f^{i_t}(x)$. (It means we can only conclude that $ P(x) | \prod_{t \equal{} 1}^{k} f^{i_t}(x)$.)

[quote="hsbhatt"]$ (x^2 \plus{} x)^{n_1} (x^2 \minus{} x \minus{} 1)^{n_2}$ fits the bill[/quote]
If we find $ P(x)$ we can solve this problem with easy solution
Let $ P(x) \equal{} (x^2 \plus{} x)^{n_1} (x^2 \minus{} x \minus{} 1)^{n_2} \plus{} Q(x)$
in there $ degQ(x) \equal{} k <n\equal{}degP(x)$ or $ Q(x) \equal{} 0$ with all $ x$
$ P(x^2 \minus{} 1) \equal{} P(x).P( \minus{} x)$ so $ 2k\equal{} k\plus{}n$ contradiction
So $ P(x) \equal{} (x^2 \plus{} x)^{n_1} (x^2 \minus{} x \minus{} 1)^{n_2}$

P/s:How do you find $ P(x)$,thanks :D

So sad it seems that you did not see my post earlier. :(  If you had seen it, maybe you realize that the polynomial

$ P(x) \equal{} ((x^2 \minus{} 1)^2 \minus{} 1)^2 \minus{} 1 \minus{} x$

satisfies the condition. Yet it fails to have the form.


I see some topics on different countries but none on Canada

[quote="fatehi"]wated to congratulate the Canadian team on their performance. This question is for any one the team  members: quite frankly are you happy with your results?[/quote]

Absolutely... this was my first and only IMO, so I'm actually pretty thrilled with a bronze.  Our 6th member who didn't get a medal was pretty bummed, but he's got 2 years to come back and do better.  The other 4 are all very strong and will return next year to do even better!


42 is the arithmetic mean of a group of 30 numbers. If two numbers, 84 and 44 are removed, what is the arithmetic mean of the remaining numbers.

Brute force
[hide]old total=1260
new total=1132
1132/28=283/7[/hide]

Really, close you were [hide]1/14[/hide] off[/hide]


function
real analysis
calculus
integration
real analysis unsolved

An analysis problem created by me for once!

Suppose $f: \mathbb{R}^{d}\rightarrow \mathbb{R}^{+}$ is not (Lebesgue) measurable. So we can't integrate it, but we can make do with a weaker definition:

The lower integral of $f$ is defined to be
$LI(f) = \sup_{g}\int g(x) dx$
where $g$ ranges over all measurable functions st $g \le f$ everywhere.

The upper integral is defined as
$UI(f) = \inf_{h}\int h(x) dx$
where $h$ ranges over all measurable functions st $h \ge f$ everywhere.

Given some fixed $\epsilon > 0$, exhibit (such as it is, since you have to use the Axiom of Choice) a function $f$ with
$LI(f)+\epsilon \le UI(f) < \infty$

Well, it may be picky, but taking $g$ measurable doesn't guarantee the existence of $\int{g}$. If $g$ is positive or negative, then it can always be defined. Otherwise, how shall we define $\int{g}$ ?

Edit: Ok, we shall assume $g \geq 0$ also, I guess  :D

Pick any non-Lebesgue measurable set, for instance one of the usual Vitali sets $V \subseteq [0,1]$. We know that $\mu^{*}(V)>0$, by completeness of the Lebesgue measure. Now consider two measurable subsets $E$ and $F$ such that $E \subseteq V \subseteq F$. It is clear that $\mu(F-E)>\alpha>0$, where $\alpha$ does not depend on $E$ nor $F$ but only on $V$ (otherwise, by unions and intersections, one can respectively construct two measurable sets $E$ and $F$ such that $\mu(F-E)=0$, which implies the measurability of $V$ -absurd-).

Now we can apply the above by defining $f(x)=0$ outside $[0,1]$ and on $V$, $f(x)=1$ everywhere else.

Every measurable function less than $f$ satisfies $UI(f) \leq 1$, and $UI(f)-LI(f) >\alpha$, by what precedes. Dilating $f$ allows us to adjust the result.

Following all of this, I am now wondering something: can we exhibit two sets $E,F$ (we can assume $F$ measurable if necessary) such that $E \subseteq F$, $m^{*}(E)=m^{*}(F)$, and $m^{*}(F-E)>0$ ?
Instead of reasoning on $E$ and $F$ above, this would allow us to think independently on $E$ and on $F$.