calculus
integration
logarithms
trigonometry
inequalities
calculus computations

For $ 0\leq x <\frac{\pi}{2}$, prove the following inequality.

$ x\plus{}\ln (\cos x)\plus{}\int_0^1 \frac{t}{1\plus{}t^2}\ dt\leq \frac{\pi}{4}$

[quote="kunny"]For $ 0\leq x < \frac {\pi}{2}$, prove the following inequality.

$ x \plus{} \ln (\cos x) \plus{} \int_0^1 \frac {t}{1 \plus{} t^2}\ dt\leq \frac {\pi}{4}$[/quote]

$ \int_0^1 \frac {t}{1 \plus{} t^2}\ dt \equal{} \frac{1}{2}\log{2}$.

consider $ f(x) \equal{} x \plus{} \log (\cos x)$.

$ f'(x)\equal{} 1 \minus{} \tan x$, which means $ f$ has maximum at $ x \equal{} \frac{\pi}{4}$.

therefore, $ LHS \le \frac{\pi}{4}$.

$ \int_{0}^{1}\frac {t}{1 \plus{} t^{2}}\ dt \equal{} \frac {1}{2}\int_{1}^{2} \frac {1}{u}\ du \equal{} \frac {1}{2} \ln 2$

Differentiating and setting to zero, $ x \plus{} \ln (\cos x)$ attains its maximum when $ x \equal{} \frac {\pi}{4}$ and is $ \frac {\pi}{4} \minus{} \ln \sqrt {2}$ which shows that the inequality holds.


Prove that the polynomial x200*y200+1 cant be expressed in the form P(x)*Q(y) where P(x) and Q(y) are polynomials of x and y respectively.

Solution: Suppose  x200*y200+1≡ P(x) *Q(y).Then by the equalities:
P(0)*Q(0)=1
P(0)*Q(y)=1
P(x)*Q(0)=1
We get P(x)*Q(y)=1 which is not possible.

How does one get those three equalities???

uff :( I don't know how to superscript the 200-s,they are the powers of x and y.


calculus
integration
trigonometry
geometry
calculus computations

Consider a parameterized curve $ C: x \equal{} e^{ \minus{} t}\cos t,\ y \equal{} e^{ \minus{} t}\sin t\ \left(0\leq t\leq \frac {\pi}{2}\right).$

(1) Find the length $ L$ of $ C$.

(2) Find the area $ S$ of the region bounded by $ C$, the $ x$ axis and $ y$ axis.

You may not use the formula $ \boxed{\int_a^b \frac {1}{2}r(\theta)^2d\theta }$ here.

Hi kunny. I make these:

($ 1$) $ \sqrt{2}(1\minus{}e^{\frac{\minus{}\pi}{2}})$

and

(2) $ \frac{1\minus{}e^{\minus{}\pi}}{4}$.

Thanks, kunny. I'm afraid I just used the formula you gave for the second part.


wait a minute-is this the formula for the sum of the triangular numbers up to n?
if not, what is it?

It's for the sum of the first n square numbers. Sorry, I should have mentioned that.  :blush:

Thanks! Some of that is a little above me, but I see how you proved it.  :)

Just wondering-how would you yourself be able to derive this...not just prove it...?

That is under the hockeystick identity.  It directly manipulates the summation into the closed form.

Hockeystick identity can be proven, but I won't unless you make me.

This can be proved more simply if you first assume that it's in the form an^3+bn^2+cn+d. A summary is taking that S(n+1)=S(n)+n^2, applying the above form, expanding, and solving the system for a, b, c, and d.

Can you prove it is a cubic?  I am not opposing you, but it seems interesting.

And solving a four variable system is just as tedious as using the hockeystick identity.  (I think).

uh...its a cubic because its a linear amount of squares? would that work as a proof?

Oops, sorry.  I think you could just make the guess that the solution is a cubic and then by proving that S(n+1)=S(n)+(n+1)^2 and S(1)=1, that the closed form is right.  So you wouldn't need to prove the answer is a cubic.

Yeah-and, it acually turns into a 3 variable system, as d is clearly 0. This can be shown as S(0)=0.


isnt there a pin for being in the top 1% of amc 12 takers?

[url]http://www.unl.edu/amc/d-publication/d1-pubarchive/2003-4pub/04tm12/1012html/12Atm04-text.html[/url]

Hey hommie, when am i gonna see the top scorers on yo site.

Peace.

[quote="JBL"]They don't calculate it, although he might know it, too.

:edit:  I take it back, they do calculate it.  Who'da thunk it?[/quote]

Whoa Whoa! You said "who'da thunk it" for a while I thought only my friend used it. And by the way if you score more than 120 do you automacticly get in?

[quote="A+MATH"]And by the way if you score more than 120 do you automacticly get in?[/quote]

If you score more than 120 on the 10 or 100 on the 12, you are automatically in.

[quote="A+MATH"]Whoa Whoa! You said "who'da thunk it" for a while I thought only my friend used it.[/quote]

I think the phrase "who'da thunk it" was popularized by some comedian or other celebrity. It's used all over the United States for humorous effect in conversation.

50,145 scored exactly a 72.5?

[quote="AMCDirector"]Scanning for the 2004 AMC 10 A and 2004 AMC 12 A ended yesterday.  That
means we have all data entered, and we have some raw statistics.  More
complete statistics should be available later this week.

We had 77,371 answer sheets for the 2004 AMC 10 A.  This compares to
77,421 in 2003.

There were 5 perfect papers.  The 1% level is 110.0 with 829 students,
which is 1.071% (There were 703 students at 110.5.)  The median is between
69.0 with 38,214 students and 68.5 with 39,390 students.

We had 99,439 answer sheets for the 2004 AMC 12 A.  This compares to
99,322 in 2003.

There were 12 perfect papers.  The 5% level is 103.0 with 5,211 students,
or 5.24%.  (There were 4674 students at 103.5, or 4.70%.) There were 7104
students at 100.0 or above, for 7.14%.  The median is between 72.5 with
50,145 students and 73.0 with 48,942 students.[/quote]

I think he means at least 72.5.  Similarly for the other numbers.


induction
combinatorics proposed
combinatorics

There are 2000 cities in Graphland; some of them are connected by roads.
For every city the number of roads going from it is counted. It is known that there are exactly two equal numbers among all the numbers obtained. What can be these numbers?

The answer is that 1000 is the only possibility.  Here is an ugly way of writing the proof of this:

We have a list of 2000 numbers which contains every number from 0 to 1999 once, except that one number is missing and one number is included twice.  It is inconsistent to have both a 0 and a 1999, so one of these must be missing.  Suppose first that 0 is absent and that 1999 is represented.  We will construct the graph recursively: choose a vertex to be the vertex of degree 1999.  (There cannot be two vertices of degree 1999, or there would be no vertices of degree 1, and that would be bad.)  Then among the other 1999 vertices, one must be the vertex of degree 1, and it has its edge already.  (If there were two vertices of degree 1, the maximal possible degree would have been 1997, i.e. we would have no vertex of degree 1998, and that would be bad.)  So among the remaining 1998 vertices, the maximal possible degree is 1998, and it must be connected to each of the not-yet-considered vertices.  (And again, there can only be one of these.)  We must have a vertex of degree 2 among these, so we have exhausted its edges.  (And it must be unique.)  Then among the remaining 1996 vertices, the maximal possible degree is 1997, and that vertex must be unique, and so on.  Finally, we have found a vertex of degree 1, 2, 3, ..., 998, 1001, 1002, ..., 1999.  We have not yet considered 3 vertices, and they currently have degree 999.  Only one of them must be the vertex of degree 999 (and there cannot be more than 1, or else there would be no vertex of degree 1000).  So the other two vertices must be connected to each other, and they must have degree 1000.

Has this question appeared on the forum or some other contest before?  It feels vaguely familiar.

Sorry to revive this, but I'm going through some old Tuymaada problems, and the only solver here is unhappy about his effort, [b]and[/b] he has missed a finer point of the problem.

Of course it is a familiar problem - its usual formulation uses handshakes. Let us introduce induction where induction seems not to be, and consider a graph on $2n$ vertices, such that exactly two degrees of vertices are equal. We claim the common value of those two equal degrees can only be $n$ (when there is no $0$ degree vertex, but there is a $2n-1$ degree vertex) or $n-1$ (when there is no $2n-1$ degree vertex, but there is a $0$ degree vertex). Clearly, each of these two cases on a graph $G$ corresponds to the other on the complementary graph $\overline{G}$.

For $n=1$, this is trivial; either we have the one possible edge, or we don't.
For $n>1$, notice, as above, that one cannot have both a $0$ and a $2n-1$ degree vertex.
Say there is a $2n-1$ degree vertex, so there also is a $1$ degree vertex. Notice, as above, that the $2n-1$ degree cannot be duplicated, and neither the $1$ degree. Remove these two vertices; we are left with a graph on $2(n-1)$ vertices, each of degree one less than before, so having the same property of exactly one degree duplicated. Apply the induction hypothesis; since this graph has no $0$ degree vertex, its duplicated degree must be $n-1$, so that of the original graph was $n$.
Say there is a $0$ vertex. The complementary graph has a $2n-1$ degree vertex, so we may apply the above. Then the duplicated degree in the original graph is $(2n-1) - n = n-1$, and we are completely done.


Find the monotonically increasing interval of $f(x)=2\sin{(\frac{\pi}{6}-2x)}$ if $x \in [0, \pi]$.

Note: I spent more than half an hour on this stupid problem......dunno what i was doing...... So guys, how long does it take for you to kill it?

Well, the $2$ in front, and the $\frac{\pi}{6}$ inside just play with the graph. I think what really affects the increasing/decreasing parts is just the $-2x$. $\sin(-x)$ monotonically increases from $x=-\frac{3\pi}{2}$ to $x=-\frac{\pi}{2}$. Then:
$\frac{\pi}{6}-2x=-\frac{\pi}{2}\Rightarrow x=\frac{\pi}{3}$
$\frac{\pi}{6}-2x=-\frac{3\pi}{2}\Rightarrow x=\frac{5\pi}{6}$

So the interval is $\left(\frac{\pi}{3},\frac{5\pi}{6}\right)$.

That took about ten minutes.

We can find $f^{\prime}(x)$ then solve $f^{\prime}(x) > 0$ for $0 \leq x \leq \pi$...

$f^{\prime}(x) = -4\cos{(\frac{\pi}{6}-2x)}$
$-4\cos{(\frac{\pi}{6}-2x)} > 0$
$\cos{(\frac{\pi}{6}-2x)} < 0$

Solving the inequality

$\frac{\pi}{2} + k.2\pi < \frac{\pi}{6}-2x < \frac{3\pi}{2} + k.2\pi \quad \forall{k}\in \mathbb{Z}$
$\frac{\pi}{3} + k.2\pi < -2x < \frac{4\pi}{3} + k.2\pi $
$-\frac{2\pi}{3} + k.\pi < x < -\frac{\pi}{6} + k.\pi $

$k = 1 \quad \Rightarrow \quad \frac{\pi}{3} < x < \frac{5\pi}{6}$

what about $[\frac{\pi}{3}, \frac{5 \pi}{6}]$

[quote="towersfreak2006"]At $\frac{\pi}{3}$ and $\frac{5\pi}{6}$, the function is at its minimum and maximum, respectively, so it is not monotonically increasing, right?[/quote]
But if $x_1= \frac{\pi}{3}$ and $x_2= \frac{\pi}{3} + dx$. Then $f(x_2) > f(x_1)$. So it is still monotonically increasing.

:huuh: By monotonically increasing, do you mean the interval where it is an arithmetic progression? :?


Hi everyone,

I'm working on a project dealing with some graph problems. I have N nodes in the graph, many of which are connected to each other but some of which are not. The task is to divide the N nodes into M  fully connected graphs of equal length, so that each node is included in exactly one of the M sub-graphs.

I'm trying to do this as efficiently (i.e. quickly) as possible. Can anyone give me any ideas on how to go about this, or links to some online literature that may help?

Thanks,
~Gorkk

Hey!
Will this help:  http://www.mathlinks.ro/Forum/viewtopic.php?highlight=country+splitting&t=18611

?


Let $ a,b \in R$ such that:
$ 0<b \le a\le 2, ab^2 \le 2$
Prove that:
$ a\plus{}2b \le 4$

[quote="NguyenDungTN"]Let $ a,b \in R$ such that:
$ 0 < b \le a\le 2, ab^2 \le 2$
Prove that:
$ a \plus{} 2b \le 4$[/quote]
[hide="Solution"]
Firstly, allow us to set $ ab^2 \equal{} 2$ instead of $ ab^2\le 2$.  We know that it works when $ a \equal{} 2$ and $ b \equal{} 1$.  Now, we allow $ b \equal{} 1 \plus{} \epsilon$ and thus $ a \equal{} \frac {2}{(1 \plus{} \epsilon)^2}$.  We realize that $ \epsilon$ is positive otherwise $ a > 2$.  We also realize that $ 1 \plus{} \epsilon\le \sqrt [3]{2}\implies \epsilon\le \sqrt [3]{3} \minus{} 1\implies 0\le \epsilon\le \sqrt [3]{2} \minus{} 1 < .3$ since otherwise $ b > a$.  Now, we have that we wish to show that
\[ \frac {2}{(1 \plus{} \epsilon)^2} \plus{} 2 \plus{} 2\epsilon\le 4
\]\[ \frac {1}{(1 \plus{} \epsilon)^2} \plus{} 1 \plus{} \epsilon\le 2
\]\[ 1\le (1 \minus{} \epsilon)(1 \plus{} \epsilon)^2 \equal{} (1 \minus{} \epsilon^2)(1 \plus{} \epsilon) \equal{} 1 \plus{} \epsilon \minus{} \epsilon^2 \minus{} \epsilon^3
\]\[ \epsilon^3 \plus{} \epsilon^2\le \epsilon
\]\[ \epsilon^2 \plus{} \epsilon \minus{} 1\le 0
\]\[ \frac { \minus{} 1 \minus{} \sqrt {5}}{2}\le \epsilon\le \frac { \minus{} 1 \plus{} \sqrt {5}}{2}
\]
Which is true since $ \frac { \minus{} 1 \minus{} \sqrt {5}}{2} < 0$ and $ \frac { \minus{} 1 \plus{} \sqrt {5}}{2} > \frac { \minus{} 1 \plus{} 2}{2} > .3 > \sqrt [3]{2} \minus{} 1$. [/hide]

[quote="NguyenDungTN"] [color=darkred]Let $ x\ ,\ y\in R$ such that $ 0 < y\le x\le 2\ ,\ xy^2 \le 2$ . Prove that $ x + 2y\le 4$ .[/color] [/quote]
[b][u]Proof.[/u][/b] In the first quadrant ($ x\ge 0\ ,\ y\ge 0$) of an analytical system $ xOy$  consider

the plane surface $ \mathcal S$ bounded by the $ Ox$ - axis, the lines $ x - y = 0$ , $ x = 2$

and the graph $ \mathcal G_f$ of the function $ f(x) = \sqrt {\frac 2x}\ ,\ x> 0$ . Therefore,

$ (x,y)\in\mathcal S$ $ \Longleftrightarrow$ $ \{\begin{array}{c} 0\le x\le 2\ ,\ y\ge 0 \\
 \\
x - y\ge 0\ ,\ xy^2\le 2\end{array}$. Thus we must find $ \max_{(x,y)\in\mathcal S} (x + 2y)\ .$

Prove easily that the maximum point is $ T(2,1)\in\mathcal G_f$ , i.e. $ x + 2y\le 4\ .$

[b][u]Remark.[/u][/b] The tangent line in the point $ T$ to the graph $ \mathcal G_f$ is $ x + 2y - 4 = 0\ .$

How about using Abel summation?? I think just 30 secnds to kill it with Abel! :lol:

Could someone demonstrate the use of Abel Summation here to me please?


calculus
integration
inequalities
function
real analysis
real analysis unsolved

Let $ f: [0,a]\minus{}R$ a derivable function with $ f'$ continous and $ f(a)\equal{}0$ . Prove that $ (\int_0^a f(t)\ dt )^2 \le \frac{a^3}{3}\int_0^a [f'(t)]^2\ dt$

Note that $ f(t)\equal{}\minus{}\int_t^af'(s)\,ds.$ That means that

$ \minus{}\int_0^af(t)\,dt\equal{}\int_0^a\int_t^af'(s)\,ds\,dt\equal{}\int_0^a\int_0^sf'(s)\,dt\,ds\equal{} \int_0^asf'(s)\,ds$

Now apply the Schwarz inequality to that:

$ \left(\int_0^af(t)\,dt\right)^2\equal{}\left(\int_0^atf'(t)\,dt\right)^2\le \left(\int_0^at^2\,dt\right)\left(\int_0^a[f'(t)]^2\,dt\right).$

That's our inequality. The general criterion for equality in the Cauchy-Schwarz inequality leads to this inequality being sharp for $ f(t)\equal{}c(t^2\minus{}a^2).$


Let $ A \equal{} (0,9)$ and $ B \equal{} (0,12)$. Points $ A'$ and $ B'$ are on the line $ y \equal{} x$, and $ \overline{AA'}$ and $ \overline{BB'}$ intersect at $ C \equal{} (2,8)$. What is the length of $ \overline{A'B'}$?

$ \textbf{(A)}\  2 \qquad \textbf{(B)}\  2\sqrt2 \qquad \textbf{(C)}\  3 \qquad \textbf{(D)}\  2 \plus{} \sqrt 2\qquad \textbf{(E)}\ 3\sqrt 2$

[hide]

shove it on ur calc and u get $A' = (6,6)$ and $B' = (4,4)$, hence $2\sqrt{2}$ is the answer (B)

[/hide]

[hide]You know point C is on lines AA' and BB', so you can find the equation of these lines using points A, B, and C. The equation of line AA' is y=-1/2(x)+9 and the equation of line BB' is y=-2x+12. As for both points A' and B' y=x, substitute in x for every y in both equations and you get A'(6,6) and B'(4,4). Using the distance formula for these two points gets 2*rt2, which is answer B.[/hide]


this kind is popular and quite difficult

this problem is very useful in may hard inequalities


How to find the area between the curves, $ r\equal{}asin \theta$ and $ r^2\equal{}a^2 cos 2 \theta$.

I have the answer, but I don't know how to get it.

ans: $ \frac{1}{2}a^2(1\minus{}\sqrt{2}\plus{}cot^{\minus{}1}\sqrt{2})$.

A good start would be to draw a picture (or have a graphing calculator do it).  Then find the angles and radii of intersection so you set up your limits of integration.  Then it's just an exercise in polar integration.


AMC
USA(J)MO
USAMO
AIME
geometry
cyclic quadrilateral

How long have you studied the USAMO (only for USAMO, not for AMC and AIME)? Does you score match your study days?

To me, I got 9, but I only studied for 7 days. So 9/7 scores improvement per day. Quite good!

Let's see if scores are high when you study for really long time.
Or just personal talent, never study, but can get high.
Also give us some suggestion on how long should we study to prepare for next year to make higher score.

Thanks!

Hah, I pwn you. :lol:
Study: (near) none
Score: 14  :D

[quote="probability1.01"]I don't think this "Study: (near) none" constitutes "pwnage." I don't mean to single out pieterminate--this is to all red MOPpers. If you think you can just sail your way through olympiad material on talent, or you can just try to get lucky, be forewarned that while fun, MOP is ultimately meant to train potential IMO team members. I don't blame any 9th graders for not preparing, especially considering what resources/experience you have, but there's no point in being proud of it.[/quote]

Well...there was this USAMO that I didn't [i]quite[/i] make (dismal aime+amc scores), so........ :(
yeah.......i must have gotten lucky....... :roll:
[size=59]
[color=white]Hah, Im luckier than you!!! Pwned!!![/color][/size]   :D

[quote="pieterminate"]

Well...there was this USAMO that I didn't [i]quite[/i] make (dismal aime+amc scores), so........ :(
yeah.......i must have gotten lucky....... :roll:
[size=59]
[color=white]Hah, Im luckier than you!!! Pwned!!![/color][/size]   :D[/quote]

Huh? If you didn't qualify how did you get scored :?

He was one of those 57 people who got an 8 on the AIME, but less than a 212 index.

Ha. I studied less than a week. No wonder i got majorly owned. Next time i'll study for about 30 days. ;) You just wait. :rotfl:

I didn't do too much studying because I thought I was hopelessly doomed to getting a 0.  Thankfully that didn't happen :D

BUT next year I'll study for 7 months straight because of WOOT!  w00t!

I only studied for a week or two seeing as I wasn't so sure I would make USAMO  ;) I was also focusing on preparing for AMC/AIME so...

Anyway, next year I'll take some olympiad courses and hopefully get some good olympiad practice, so I'll do at least reasonably well. As well, AIME Problem Series so I do actually qualify for USAMO  :)

I studied for a few nights before the USAMO, I spent all my time practicing AIME problems.

[quote="SamE"][quote="probability1.01"]I don't think this "Study: (near) none" constitutes "pwnage." I don't mean to single out pieterminate--this is to all red MOPpers. If you think you can just sail your way through olympiad material on talent, or you can just try to get lucky, be forewarned that while fun, MOP is ultimately meant to train potential IMO team members. I don't blame any 9th graders for not preparing, especially considering what resources/experience you have, but there's no point in being proud of it.[/quote] I just want to second this comment. So often have I seen smart kids just throw it all away by not studying. They seem to think that if they get the same grade as someone else without studying, it somehow makes them better than that other person. This is far from true. While it may help you with beginner's luck (as it did me), you need to seriously sit down and study. [b]Just because you made blue MOP without knowing what a cyclic quadrilateral was, doesn't mean you're even going to make blue next year.[/b] Yeah, that's what happened to me. It's never something to brag about, because it just means you got lucky. If you're going to do well, you really have to buckle down and put in the hours of practice it takes to get better. I didn't, and I'm paying for it this year.[/quote]

Heh, I know I was extremely lucky...4 NT?  :D


If $d(n)$ is the number of divisors of n and $\sigma(n)$ is the sum of the divisors of n, prove that $\frac{\sigma(n)}{d(n)} \geq \sqrt{n}$.

Use AM- GM inequality to prove that $1+p+p^2+....+p^k \geq (k+1).p^\frac{k}{2}$.
  Then the result follows by applying the formula of $\sigma{(n)}$ and $d(n)$.

We discussed that here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23974


Fie $\ f: [0,\infty) \rightarrow R$ o functie astfel incat sa avem inegalitatea
$\ f\left(\frac{x+y}{2}\right)\leq\ f\left(\sqrt{\frac{x^2+y^2}{2}}\right)$.Sa se arate ca $\forall x,y,z$ pozitive, atunci avem inegalitatea
$\ f\left(\frac{x+y+z}{3}\right)\leq\ f\left(\sqrt{\frac{x^2+y^2+z^2}{3}}\right)$.

Pai din ipoteza functia este crescatoare si din inegalitatea mediilor obtinem ultima inegalitate.

Da.Asta e raspunsul corect.E prima data cand fac si eu un exercitiu cu functii,..asta e cam usor, nivel de clasa... :blush:


Let k be an integer . The sequence a_n is defined by:
a_0=0 ,a_1=1, $a_{n}=2ka_{n-1}-(k^{2}+1)a_{n-2}$ n>=2

If p is a prime number  of the  for 4m+3 prove that:

$a_{n+p^{2}-1}=a_{n}{mod p}$ and $a_{n+p^{3}-p}=a_{n}(mod p^{2})$


Cevians $AD$ and $CF$ are drawn in $\triangle ABC$ such that $\frac{AF}{FB}=\frac{1}{2}$ and $\frac{CD}{DB}=\frac{1}{3}$ . Denote the intersection of $AD$ and $CF$ by $P$. Line $BP$ is drawn, intersecting side $AC$ at $E$. Find $\frac{CE}{EA}$.

Nice.  :lol: I first did the question with mass points. Then I looked at it and I was like "duh!" [hide]Ceva's Thm.[/hide]

[hide]With Ceva's Theorem is straightforward. $\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1$.
Hence $\frac{1}{2} \cdot 3 \cdot \frac{CE}{EA} = 1$ $\Longrightarrow$ $\frac{CE}{EA} = \frac{2}{3}$.[/hide]


linear algebra
matrix
linear algebra solved

$A,B,C \in M_n(R)$ suppose that $A\neq B,\; A^4=B^4,\; C^3A=C^3B,\; B^3A=A^3B$

Does $A^3+B^3+C^3$ is an ivertible matrix ?

It can't be invertible.

$(A^3+B^3+C^3)(A-B)=A^4-A^3B+B^3A-B^4+C^3A-C^3B=0.$

Since $A-B$ is not the zero matrix, then $A^3+B^3+C^3$ must be singular.


geometry
trapezoid
circumcircle
trigonometry
parallelogram
geometry proposed

Given regular 17-gon $ A_1\cdots A_{17}$. Prove that two triangles formed by lines $ A_1A_4$, $ A_2A_{10}$, $ A_{13}A_{14}$ and $ A_2A_3$, $ A_4A_6$, $ A_{14}A_{15}$ are equal.

[i]Author: N.Beluhov[/i]

Let $ X \equiv A_1A_4 \cap A_2A_{10},$ $ Y \equiv A_2A_{10} \cap A_{13}A_{14},$ $ Z \equiv A_{13}A_{14} \cap A_1A_4.$
Let $ D \equiv A_2A_3 \cap A_{14}A_{15},$ $ E \equiv A_{14}A_{15} \cap A_{4}A_{6},$ $ F \equiv A_{4}A_{6} \cap A_2A_3.$

$ A_{13}A_{14}A_4A_6$ is isosceles trapezoid $ \Longrightarrow$ $ YZ  \equiv A_{13}A_{14}\parallel A_4A_6 \equiv EF,$
$ A_{1}A_{4}A_3A_2$ is isosceles trapezoid $ \Longrightarrow$ $ ZX \equiv A_{1}A_{4} \parallel A_2A_3 \equiv FD,$
$ A_{2}A_{10}A_{15}A_{14}$ is isosceles trapezoid $ \Longrightarrow$ $ XY \equiv A_{2}A_{10} \parallel A_{14}A_{15} \equiv DE$

$ \Longrightarrow$ $ \triangle XYZ \sim \triangle DEF$ are centrally similar, having parallel sides.

Let $ P \equiv XY \cap EF \equiv A_4A_6 \cap A_2A_{10}$ and $ Q \equiv YZ \cap FD \equiv A_{13}A_{14} \cap A_2A_3.$
Let $ O$ be circumcenter and $ R$ circumradius of the regular 17-gon. Let $ \phi \equal{} \angle A_1OA_2 \equal{} \frac{2\pi}{17}.$

$ \angle A_4A_{10}P \equal{} \angle A_4A_{10}A_2 \equal{} \frac{_1}{^2} \angle A_4OA_2 \equal{} \phi,$
$ \angle PA_4A_{10} \equal{} \pi \minus{} \angle A_{10}A_4A_6 \equal{} \frac{_1}{^2}A_{10}OA_6 \equal{} \pi \minus{} 2 \phi$

$ \Longrightarrow$ $ \triangle A_4A_{10}P$ is isosceles with $ A_4P \equal{} A_4A_{10} \equal{} 2R \sin \frac{_1}{^2} \widehat{A_4OA_{10}} \equal{} 2R \sin 3 \phi.$

$ \angle QA_2A_{14} \equal{} \angle A_{2}A_3A_{13} \equal{} \frac{_1}{^2}\angle A_{2}OA_{13} \equal{} 3 \phi,$
$ \angle A_2A_{14}Q \equal{} \angle A_{3}A_{13}A_{14} \equal{} \frac{_1}{^2}\angle A_{3}OA_{14} \equal{} 3 \phi$

$ \Longrightarrow$ $ \triangle QA_2A_{14}$ is isosceles with $ QA_2 \equal{} QA_{14} \equal{} \frac{A_2A_{14}}{2 \cos 3 \phi} \equal{} \frac{2R \sin \frac{_1}{^2} \widehat{A_2OA_{14}}}{2 \cos 3 \phi} \equal{} \frac{R \sin 6 \phi}{\cos 3 \phi} \equal{} 2R \sin 3 \phi.$

$ \Longrightarrow$ $ QA_{14} \equal{} A_4P.$

$ EPYA_{14}$ is parallelogram $ \Longrightarrow$ $ EP \equal{} YA_{14}.$
$ ZQFA_4$ is parallelogram $ \Longrightarrow$ $ FA_4 \equal{} ZQ.$

$ EF \equal{} EP \plus{} PA_4 \minus{} FA_4 \equal{} YA_{14} \plus{} A_{14}Q \minus{} ZQ \equal{} YZ$

$ \Longrightarrow$ $ \triangle XYZ \cong \triangle DEF$ are congruent.


Who is everyone's favorite chess player, I think Kramnik is awesome!

Anand, cuz Indians are awesome :D

Plus he has the highest ELO at the moment!

[quote="shaggy75"]Anand FTW! :lol:[/quote]

Anand rules !!!!!!!!!!!! :first:

ANAND FTW!!!!

We have the same name!!!

Whenever I goto a chess tournament, ppl are like "Darn I have to play Anand now"

my favorite GM is karpov butu did not post him. oh well. GO KARPOV!!!!! :thumbup:

topalov is pretty sicknasty, I like his play against the english attack---owned Anand.

who won the championship??????/

kramnik or anand??

i believe it is kramnik????

or is it anand?

Alright, revival?
Anyway, I think that the 19 year old guy in TIME was pretty good. (I don't know his name)

Of course, the new #1 rated Magnus Carlsen is missing from this poll.

Although he isn't in the top echelon of players, my favorite player would probably be Larry Christiansen. I like his ruthless attacking style.
Plus, I'm slightly biased towards him because he was my chess coach and he was really funny.

I like Magnus Carlsen, Levon Aronian and Garry Kasparov. They are awesome :)

i like the old dead guy bobby fisher and kasparov

Can you honestly not read the date of the original post?

Darn. Would have picked Vassily Ivanchuk, but he's not there!

I like how well Kramnik plays his games.
http://www.chessgames.com/perl/chessgame?gid=1440796
(Troll)

I'd go with Aronian since Caruana is not there.

Why is Leko or Morozevich in this poll if they hardly ever play top tournaments anymore?

Viswanathan Anand is simply the best player you all have seen . There shouldn't be any more confusion in this issue .
Other players are useless. :starwars:  :yup:

Even thought he is not very famous... he is one of the greatest players that sacrifices all his pieces to win.Paul Morphy.[url]http://en.wikipedia.org/wiki/Morphy_versus_the_Duke_of_Brunswick_and_Count_Isouard[/url]

scroll towards the bottom

EDIT: OOPS didn't see that it was current GM... but something worth including


integration
logarithms
limit
function
abstract algebra
Support
real analysis

f is defined on $ \int^{\infty}_{ \minus{} \infty} f(t) dt$

If $ f$ is non-negative and $ \int^{\infty}_{ \minus{} \infty} f(t) dt$ is FINITE,

then $ f$ is bounded on $ ( \minus{} \infty,\infty)$ ?

If so, Why?

If not, what condition has to be added?

conssider $ f(x) \equal{} \minus{} 1_{[0,1]} \ln x$
then $ \int_{ \minus{} \infty}^{\infty} f < \infty$ and $ f \geq 0$ but   $ f$ is not bounded at point $ 0$

[quote="Extremal"]conssider $ f(x) = - 1_{[0,1]} \ln x$
then $ \int_{ - \infty}^{\infty} f < \infty$ and $ f \geq 0$ but   $ f$ is not bounded at point $ 0$[/quote]

I think this is divergent.

if f(x) is well defined on ${ ( - \infty},\infty)$

then bounded ?

[quote="Extremal"]it's not divergent :
it's equal to ${ - \int_{0}^{1}\ln x dx = - x\ln x|_{0}^1} + \int_{0}^{1} dx = 1 < \infty$ (by parts)[/quote]

$ \lim_{x \rightarrow 0} x\ln x = 0$ ???

I don't think so...

even ln 0 is not defined.

let's $ x>0$ is small and  then $ \ln x \equal{} \minus{} y$ where $ y > 0$ if $ x \to 0\plus{}0^{\plus{}}$ then $ y \to \infty$
 also $ x \equal{} e^{ \minus{} y}$ hence our limit can be stated in this way:
$ \lim_{y \to \infty} e^{ \minus{} y} y \equal{} 0$
try to show  : $ \lim_{y \to \infty}\frac {y}{e^{y}} \equal{} 0$.
use this fact $ e^{y} \equal{} \sum \frac {y^{n}}{n!} > Cy^{2}$ for some $ C > 0$ provided $ y > 0$

[quote="Extremal"]let's $ x > 0$ is small and  then $ \ln x \equal{} \minus{} y$ where $ y > 0$ if $ x \to 0 \plus{} 0^{ \plus{} }$ then $ y \to \infty$
 also $ x \equal{} e^{ \minus{} y}$ hence our limit can be stated in this way:
$ \lim_{y \to \infty} e^{ \minus{} y} y \equal{} 0$
try to show  : $ \lim_{y \to \infty}\frac {y}{e^{y}} \equal{} 0$.
use this fact $ e^{y} \equal{} \sum \frac {y^{n}}{n!} > Cy^{2}$ for some $ C > 0$ provided $ y > 0$[/quote]

WOW.. Thanks. Anyway, is a summable function that is well-defined on R bounded?

I want to know.. please help me, and thanks again.

[quote="Extremal"]I'm not familiar  with such word as  : "well-defined"
what is  " well-defined" ? is it continuous?
if you explain me, then I try to help you  :)[/quote]

sorry for ambiguous word..

I mean

"for all x in R, 0 $ \leq$ f(x) < $ \infty$"

or " f:R->R+ " (R+ is non negative real number)

sorry for poor english..

[quote="lysangc"]

then $ f$ is bounded on $ ( \minus{} \infty,\infty)$ ?

[/quote]

[quote="lysangc"]...
"for all x in R, 0 $ \leq$ f(x) < $ \infty$"

... [/quote]
thus by definition of "well-defined" $ f$ is bounded  :) .

No, most emphatically, that is not what bounded means. "Finite everywhere" does not imply "bounded."

We can make extremal's example be defined and finite everywhere by doing this:
\[ f(x)\equal{}\begin{cases}\minus{}\ln x,&0<x\le 1\\0&\text{otherwise}\end{cases}\]
(The only thing extremal did that I'm changing is that he said $ \mathbf{1}_{[0,1]}$ instead of $ \mathbf{1}_{(0,1]}.$)

That particular example is not continuous, but it is possible to have a continuous unbounded integrable function.

Define $ T(x)\equal{}\max(1\minus{}|x|,0)\equal{}\begin{cases}1\minus{}x,&0\le x\le 1\\ 1\plus{}x,&\minus{}1\le x\le 0\\ 0,&|x|\ge 1\end{cases}.$

Let $ f(x)\equal{}\sum_{k\equal{}1}^{\infty}kT(k^3(x\minus{}k)).$

$ \int_{\mathbb{R}}f(x)\,dx\equal{}\sum_{k\equal{}1}^{\infty}\frac1{k^2},$ which is finite.

$ f$ is a continuous function. It has an upward spike centered at each integer; as you go outward, the spikes get higher and narrower.

:oops:  I'm sorry,  I just was thinking by bounded - $ f(x)<\infty$ for all $ x$.
I absolutely agree with Kent's argument.
In some sense  $ T(x)$ is positive kernel with compact support and unit norm in $ L_{1}$, it's  also equal to $ \frac {1_{[ \minus{} 1,1]}*1_{[ \minus{} 1,1]}}{2}$ or square of convolutions of  characteristic function. This kernel also is used  as approximation unit, to prove theorem about: to inverse Fourier transform.
in some sense in Kent's example  $ f(x)$ is discrete convolutions of $ T(x)$ with $ 1/x^{2}$, but not exactly.  Kent has made small the support of $ T(x)$ but save the norm in $ L_{1}$ . It's looks like this operation $ T_{A}(x) \equal{} AT(Ax)$, also such operation as $ f_{s}(x) \equal{} f(x \minus{} s)$ don't change the norm in $ L_{1}$. thus he put : $ f(x) \equal{} \sum_{1}^{\infty} \frac {1}{k^{2}} k^{3} T(k^{3}(x \minus{} k))$ therefore $ \parallel{}f\parallel{}_{1} \equal{} \sum \frac {1}{k^{2}}$.
it's great idea  :) .
You also can choose another positive kernel,  another $ A$ and another function instead $ \frac {1}{x^{2}}$
for example if you choose $ \frac {1}{x^{4}}$ you will get $ \parallel{}f\parallel{}_{1} \equal{} \zeta(4)$


Euler
search
number theory
number theory open

Hope  the moderators allow this unusual question here .. but something started troubling me...

If N=2^32+1

It is easy to see  that 3^N is NOT 3 (moodulo N).  and this requires only 32 squaring of the number 3 which is not hard. even without a calculator..

Yet why didn't Fermat know (after all it was his 'little' theorem and his prime) that this number is NOT prime.

Why it has too wait till Euler?
Something is strange here... how come Fermat did not notice it????

(PS- Move if this is not appropiate here in this form)

Don't blame Fermat too hardly. The test of primality you are referring to is known as Pepin's test, and appeared in 1877. A little late for Fermat to know it...

There are many 'easy' results which have been discovered later after the tools needed to prove them have been elaborated.

Pierre.

No ,  This is known much earllier then Pepin's test.

Pepin's test  fails because  [(3)^(2^16)  == (-1) mod (F(5)]   is untrue.

What I said was simply Fermat's little theorem eg:

3^N  is NOT equal to 3 . (mod N)
And that means N is not prime

That  'little theoeem' was certainnly popularized by him and i assume known , by him., AND as said before, SIMPLE  (I mean only 32 squarings are needed) enough for him to do.

Hope you see that... Whar am I missing?

That's what bothers me.

I sympathize with Fermat with respect to the $641|2^{32} + 1$ oversight.

The fact is that even without $ \it relatively$ modern primality tests or even Fermat's Little Theorem, which presumably was popularized by Fermat, hence the name, there are simple ways of proving $641|2^{32}+1$.  After viewing this thread, I decided to search for one, and found this doozie:

$641 = 5*2^7 + 1 = 2^4 + 5^4$.  Therefore, $641|5^4 2^{28} + 2^{32}, 5^4 2^{28}-1$, hence, $641|2^{32}+1$, their difference.

On the other hand, in Fermat's defence, he was a brilliant mathematician.  If he were posed the problem of proving that $641|2^{32} + 1$, I don't doubt that he would have proven it elegantly.  But $641$ isn't exactly a prime that comes to mind immediately (at least it wasn't at the time  ;) ) when testing for primality; it took a calculation savant like Euler to spot it, only so that we all can take cheap shots at Fermat for missing it.  And even if Fermat searched for primes as large as $641$, which we'll never know for certain, there are so many small primes to check that he probably didn't exert a whole-hearted effort to prove divisibility, or non-divisilbity, by most of them.

Michael -

I think yoiu are missing the point. It is NOT hindsight/20-20 type of reasoning ...  and no one is  taking a cheap shot  at Fermat that he missed 641.

The point was, if  applied his own FLT, it was  sure, and  [b]he would know for sure that he did NOT  have to do  MORE THAN 32 squaring..[/b] which I  believe is not too much not to expect from him...(After all simplest Primality tests  depend on this so  one can easily rule out  a given large number as not prime by this  test without having to  find a factor)

It is NOT a case of him NOT trying all little primes etc... it is a case of him not doing a 32 step process which he  ought to have known that it would  not take long. .

******* New Thought ****

Now going of on a tangent .. and it is not directly related to the above, but Euler (who understood FLT) not only found out the number is not prime, but its factor also. He did not have to search through all the  little primes though .. it can be easily proven that if F(5) had factors all have to be of the form:
128*k + 1  or  129,  257, 385, 513,..641.. etc  and if you  disallow 129,385,513 .. etc (They are not prime) .. to find 641 was not that hard. .It was the next number after 257...

Note that I am not blaming Fermat for not  deducing that he only has to try  prime factors of the type 128K+1  (though it is not hard to prove it now.. and that is what helped Euler).

PS - BTW the FLT  test for prime is not new..  Specially afeter Euler.. mathematicians routinely used it to  show   and know some  numbers (2^n-1 types)   as non-prime  though they did not know their factors...


well, another problem:
Two condenser of capacity C(1)= 10 pF and C(2)= 20pF are linked by a switch T and a resistor of resistence R(1)= 100 ohm.
Initially the switch is off and both the condenser are uncharged. The plates of C(1) are linked to a battery of electromotive force V(0)=4.5 V, so that the positve pole of the batery is oriented towards the upper plate of C(1).
The battery is removed. The switch T is closed.

a) Find the values of the electric charges Q(1) and Q(2) on the upper plates of the 2 condensers at the equilibrium.

At the equilibrium, at the istant t(0), the plates of C(2) are linked to a resistor of resistence R(2)= 30 ohm.

b)Find the values of the electric current in R(1) and R(2) immediately after t(0)

C) Find the values of the electric charges on the plates of C(1) and C(2) at the new state of equilibrium and calculate the total work done in the resistors during the transient state after the istant t(0).

I've solved the point a) using integrals, even if this kind of problem shouldn't require them...
in the point b) shouldn't R(1) be 0?
enjoy

ok..the part a) can be solved without integrals.
We use the conservation of the charge at the equilibrium, when the potential of C(1) is the same of C(2)'s one.
we have:
C(1)*V(0)= V(1)*C(1) + V(2)C(2);
V(2) = V(1)
then, after few passages: Q(1)=V(0)*C(1)^2/(C1+C2)
                                     Q(2)= C(1)*C(2)*V(0)/(C1+C2)

for the part b)...well immediately after t0, there's no current through R(1), while the current in R(2) is V(2)/R(2)

c) if i consider again the conservation of the charge and V(1)=V(2) after an infinite time, i have exactly the same charges of the point a), so the total work should be 0.
I'm not sure of what i wrote, in particular about the point c) :lol: ...correct me!

For part c), you cannot use charge conservation as you did for part a) as charge is not conserved in the upper half of the circuit. It seems that c) will be complicated (integrating currents etc.) though I am trying to find a simple way.

Actually it's very simple. As current through both resistors is zero at equilibrium, charge on both capacitors is zero. Hence work done can be easily calculated.

For c), I think that the resistance 2 will be sure to leave $ C_{2}$ without any charge, because as long as there is charge on $ C_{2}$, there is a potential difference across the resistance, which is the same to say there is a current. And if $ V_{2}$ is zero, also $ V_{1}$ is zero. Thus, we may conclude that in the end $ Q_{1}\equal{}Q_{2}\equal{}0$. But, where is the initial charge?

In the end we have got $ Q_{1}\equal{}Q_{2}\equal{}0$ and the work is equal to the energy of the capacitors at $ t_{0}$.
[b]@pedroomagico[/b] the charge is there but it has flew through the circuit so that in every point there is the same amount of positive and negative charge...

let's see if i've understood e if i can write a solution:

well, at the equilibrium there'll be no current flowing through the circuit. That means C(1) and C(2) are at the same potential.
Now if if V(2) is different from 0, in R(2) must be flow a current. This is impossible at the equilibrium, so V(1)=V(2)=0. Then we have Q(1)=Q(2)=0.

We use the conservation of energy to calculate the work. W=0.5C(1)V(1)^2 + 0.5C(2)V(2)^2
Ok..do you confirm my answer for part b)?


geometry
perpendicular bisector
geometry proposed

Two cirlces $ C_1$ and $ C_2$, with center $ O_1$ and $ O_2$ respectively, intersect at $ A$ and $ B$. Let $ O_1$ lies on $ C_2$. A line $ l$ passes through $ O_1$ but does not pass through $ O_2$. Let $ P$ and $ Q$ be the projection of $ A$ and $ B$ respectively on the line $ l$ and let $ M$ be the midpoint of $ \overline{AB}$. Prove that $ MPQ$ is an isoceles triangle.

It is enough to prove that $ M$ lies on the perpendicular bisector of $ PQ$. Let the midpoint of $ PQ$ be $ X$. Since $ AP\parallel BQ$, $ MX$ must be parallel to them as well. So $ MX$ is perpendicular to $ PQ$ as desired.

almost obvious but I spent about an hour to solve it :lol:


[quote="kunny"]For distinct real numbers $ a,\ b,\ c,\ d$, prove that $ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \plus{} 4\geq 6(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)$.
When does the equality hold?[/quote]
try $ a \equal{} 1; 1 < b,c,d < \sqrt {6}$

Yes, you are right! Sorry for you to take time, everyone. :oops:

I have just edited.

kunny

[quote="kunny"]For real numbers $ a,\ b,\ c,\ d$ such that $ a < b < c < d$ and $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 12$.

Prove the following inequality.
\[ a^8 \plus{} b^8 \plus{} c^8 \plus{} d^8 \plus{} 38(a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4)\geq 12(a^6 \plus{} b^6 \plus{} c^6 \plus{} d^6) \plus{} 140.
\]
When does the equality hold?[/quote]

Is the problem easy?

My question
Why dont you let $ 0\le a < b < c < d$,$ a \plus{} b \plus{} c \plus{} d \equal{} 12$
and
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \plus{} 38(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)\geq 12(a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3) \plus{} 140$
And now,(if it is true),SMV will work well :)

What's SMV?
Could you show me your solution, Allnames?


inequalities
number theory
prime numbers
national olympiad

[b]First day[/b]

1) Prove that the equation $ x^{5} \plus{} 31 \equal{} y^{2}$ has no integer solution.

2) Let $ r$ be a real number such that the sequence $ {a_{n}}$ of positive real numbers satisfies the equation $ a_{1} \plus{} a_{2} \plus{} .... \plus{} a_{m \plus{} 1} < \equal{} ra_{m}$ for each positive integer $ m$. prove that $ r > \equal{} 4$.

3) Let SABC be a regular triangular pyramid. find the set of all points $ D (D! \equal{} S)$ in the space satisfing the equation $ |cos ASD \minus{} 2cosBSD \minus{} 2 cos CSD| \equal{} 3$.

[b]Second day[/b]

1) For the positive real numbers $ a,b,c$ prove that $ c/{(a \plus{} 2b)} \plus{} d/{(b \plus{} 2c)} \plus{} a/{(c \plus{} 2d)} \plus{} b/{(d \plus{} 2a)} > \equal{} 4/3$

2) The inner point $ X$ of a quadrilater is [i]observable[/i] from the side YZ if the perpendicular to the line YZ meet it in the colosed interval [YZ]. The inner point of a quadrilater is a k-point if it is observable from the exactly k sides of the quadrilater.
 Prove that if a convex quadrilater has a 1-point then it has a k-point for each k=2,3,4.

3) Find all prime numbers $ p,q < 2005$ such that $ q | p^{2} \plus{} 8$  and $ p|q^{2} \plus{} 8$.

[quote="mon"][b]First day[/b]

1) Prove that the equation $ x^{5} \plus{} 31 \equal{} y^{2}$ has no integer solution.

2) Let $ r$ be a real number such that the sequence $ {a_{n}}$ of positive real numbers satisfies the equation $ a_{1} \plus{} a_{2} \plus{} .... \plus{} a_{m \plus{} 1} < \equal{} ra_{m}$ for each positive integer $ m$. prove that $ r > \equal{} 4$.

3) Let SABC be a regular triangular pyramid. find the set of all points $ D (D! \equal{} S)$ in the space satisfing the equation $ |cos ASD \minus{} 2cosBSD \minus{} 2 cos CSD| \equal{} 3$.

[b]Second day[/b]

1) For the positive real numbers $ a,b,c$ prove that $ c/{(a \plus{} 2b)} \plus{} d/{(b \plus{} 2c)} \plus{} a/{(c \plus{} 2d)} \plus{} b/{(d \plus{} 2a)} > \equal{} 4/3$

2) The inner point $ X$ of a quadrilater is [i]observable[/i] from the side YZ if the perpendicular to the line YZ meet it in the colosed interval [YZ]. The inner point of a quadrilater is a k-point if it is observable from the exactly k sides of the quadrilater.
 Prove that if a convex quadrilater has a 1-point then it has a k-point for each k=2,3,4. (I personally find this problem a little strange since I can't see a 1-point in a convex quadrilater but maybe it's just my impresion...)

3) Find all prime numbers $ p,q < 2005$ such that $ q | p^{2} \plus{} 8$  and $ p|q^{2} \plus{} 8$.[/quote]
Second day
1. By Cauchy-Schwarz inequality:
$ \frac{c}{(a \plus{} 2b)} \plus{}\frac{d}{(b \plus{} 2c)} \plus{}\frac{a}{(c \plus{} 2d)} \plus{}\frac{b}{(d \plus{} 2a)} \ge \frac{(a\plus{}b\plus{}c\plus{}d)^2}{2(ab\plus{}bc\plus{}cd\plus{}da\plus{}ac\plus{}bd)} \ge \frac{4}{3}$
Because $ 3(a\plus{}b\plus{}c\plus{}d)^2 \ge 8(ab\plus{}bc\plus{}cd\plus{}da\plus{}ac\plus{}bd)$
$ \Leftrightarrow (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}d)^2\plus{}(d\minus{}a)^2\plus{}(a\minus{}c)^2\plus{}(b\minus{}d)^2 \ge 0$
 which is clearly true!

yes, this was the easiest, toghether with 1 from the first day which works with an easy reasoning modulo 4( the "hardest" part would be when $ x\equal{}1(mod 4)$ but if we rewrite the equation as $ x^{5}\plus{}2^{5}\equal{}y^{2}\plus{}2^{2}$ and decompose the first part of the equality we obtain that $ x\plus{}2$ should not be of the form $ 4k\plus{}3 \equal{}> x$ could not be of the form $ 4t\plus{}1$  and we are done).
But I think 2 from the second day is quite strange...I can't see how a 1-point looks like :maybe: i could only draw a 2-point in the interior of a convex quadrilater...

ok, that's true, my typing mistake  :P   :wink:

Has somebody already been posting all these problems to their own threads?

I don't think so, but you've posted it right? :)

Anyway, does anybody know the medal cut-off of this olympiad? Who's going to Zhautykov Olympiad 2009?


Given $ artanh {x} \plus{} artanh {y}\equal{}{(1/2)}ln({5}^2)$, show that $ y\equal{}{(2\minus{}3x)}/{(3\minus{}2x)}$

[hide="hint"]
$ \arctan{x} \equal{} \dfrac{1}{2} \log_e \dfrac{1\plus{}x}{1\minus{}x}$
[/hide]


This is the place I store somehow meaningful 1s and 0s.

I never thought I'd see the day when all my 2GB RAM was used, much less my 10GB swap too...

cule were you running a program or something

Uhhhh... yes? I was running my science fair project encryption, but I didn't check the computational intensity of this particular key.

complete with cowboy hat, in true Joe style.

My encryption, the one I filed a provisional patent for recently? It's a new method of encryption that my science fair project is about.

So we were playing around with genetic engineering: we were trying to breed a steer with 6 legs so hopefully we'd get more rear leg meat. Unfortunately they turned out to be a set of front legs. Ah well.

ummm isnt there another cow behind it? :D  :ninja:

'Tis _still_ [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=156754]not a cow[/url] :P... The group he is in may be cattle, or beeves, but they are certainly not cows! He would be most offended; he has no offspring.

OH NO! BILLY AND LAURA ARE GOING ON A RAMPAGE!

How 'bout the rest of the gunmen in the picture? :P


abstract algebra
superior algebra
superior algebra solved

Prove that $S^{-1}A$ is a free $A$-mod iff $S\subseteq U(A)$.

If $S\subset U(A)$ then it is clear that $S^{-1}A\simeq A$ as rings and $A$-modules.

Assume now that $S^{-1}A$ is a free $A$-module.

Let $\frac{x}{a}$ and $\frac{y}{b}$ be two [b]different[/b] elements of $S^{-1}A$ with $x,y\in A$ and $a,b\in S$.
Then $(ya)\frac{x}{a}-(xb)\frac{y}{b}=0$. If $ya=xb=0$, then by definition $\frac{x}{a}=\frac{y}{b}$ which contradicts our choice. This means that not both $ya$ and $xb$ are $0$ in $A$, so $\frac{x}{a}$ and $\frac{y}{b}$ are linearly dependent over $A$.
So any two (distinct - this extra assumption can be obviously removed, but not with the same proof) elements of $S^{-1}A$ are linearly dependent over $A$. This means that if $S^{-1}A$ is free, then it must be of rank $1$. I assume that we exclude $S^{-1}A=0$ which is a theoretical possibility if $S$ is allowed to contain $0$.

$S^{-1}A$ is torsion free as an $A$-module because it is free. Let $s\in S$. If $s$ is a zero-divisor in $A$, then there exists $r\in A\neq 0$ such that $rs=0$. But then $\frac{s}{1}=0$ in $S^{-1}A$ by definition of the localization. Hence $s\cdot\frac{1}{1}=0$ in $S^{-1}A$. The later is torsion free over $A$, so $\frac{1}{1}=0$ which is a contradiction. This proves that all the elements of $S$ are not zero-divisors in $A$

Let $\frac{x}{a}$ be a basis for $S^{-1}A$ as $A$-module. There exists $x'\in A$ such that $\frac{xx'}a=\frac{1}{a}$. Since the elements of $S$ are non-zero divisors in $A$, we get $xx'=1$ in $A$, so $x$ is invertible in $A$. Multiplication by an invertible element of $A$ does not change a basis, so $\frac{1}{a}$ is also a basis for $S^{-1}A$ as $A$-module.
Let $s\in S$. Then $as\in S$ because $a\in S$ and $S$ is closed under multiplication. Since $\frac{1}{a}$ is a basis, there exists $r\in A$ such that $\frac{r}{a}=\frac 1{as}\Leftrightarrow rs=1$. This means that $s\in U(A)$.
$s$ is chosen at random in $S$, so we are done.


There are nine unit triangles in the three rows of this triangle. If you add three more rows to the bottom of this triangle so that each row has 2 more unit triangles than the previous row, how many unit triangles will there be?

problem no 31  please see the figure in  the following pdf link

http://www.mnspe.org/mathcounts/2007tests/510-07_C_Countdown.pdf

answer is 36 triangles

You just add the number of triangles per row, which is
\[ 1\plus{}3\plus{}5\plus{}7\plus{}9\plus{}11 .\]
(Note how we have 6 numbers for 6 rows.) We can calculate this manually and get $ 36$, but in a CD round, this method could be very slow.

Instead, we can notice that this is the sum of the first $ 6$ odd numbers. We also have a formula for the sum of consecutive odd numbers, which states that the sum of the first $ n$ odd numbers is $ n^2$. A proof and further explanation can be found [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=291245]here[/url]. Applying this to our problem, we see that there are $ 6^2\equal{}\boxed{36}$ small triangles because this quantity is equal to the sum of the first 6 odd integers.

powerofpi u accidentally included the row of 11

I purposely included the row of 11  :) . That is the number of triangles in the sixth row, or the third added on.


linear algebra
matrix
linear algebra unsolved

This one is really cute: is it true that given any unitary ring we can find two square $nxn$ ($n$ is fixed) matrices such that the only matrices which commute with both of them are the scalar matrices?

Indeed it is :).

Take one matrix to have $1$ right under the main diagonal and zero everywhere else, and the other one to have $1$ right above the main diagonal and zero everywhere else.

Just the type of problem you want to solve before going to sleel, as sam-n once said.  :D


Here is a classic problem that appeared in many past math competitions.

But in this problem, I want you to find 3 different ways to do this.

Post as many ways as possible you can think of. I only know 3 although I'm not sure whether there are anymore.

Calculate:

$\frac {1}{1 \cdot 2} + \frac {1}{2 \cdot 3} + \frac {1}{3 \cdot 4} + .... \frac {1}{256 \cdot 257}$.

Find answer and as many ways you can do to this problem on solving it. :)

Bleh - I'm lazy so easy way:

[hide]$\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3} + \ldots + \frac{1}{256}-\frac{1}{257} = 1 - \frac{1}{257} = \frac{256}{257}$

But ofcourse, this appears in a millisecond to the trained telescoping eye.[/hide]

That's one way with correct answer.

What whitehorseking used was something called telescoping. This is *very* useful way. :D

Here's an easier way. Notice theres a pattern. The sum up to the first number is 1/2. The sum up to the second term is 2/3. The sum up to the third term is 3/4. The pattern is- the sum of n terms is n/(n+1). So the answer is 256/256+1= [b]256/257[/b]

Great! Two ways are introduced. I guess I'll introduce the third way.

I give the credit to Mathematical Circles by Fomin, Genkin, and Itenberg.

The answer is just:
[hide]
$\frac {n-1}{n}$[/hide]

Ok. Here is the proof on why this is the answer. This is proven by Mathematical Induction.

[hide="Proof"]
The question we're trying to prove is....

$\frac {1}{1 \cdot 2} + \frac {1}{2 \cdot 3}... + \frac {1}{(n-1)n} = \frac {n-1}{n}$.

To start, we should plug in 1 and examine the result. However, this doesn't work because one side gives $\frac {1}{(1-1)1} = \text {undefined}$ and other sides gives $\frac {0}{1} = 0$. So, I'm not sure why the major step of induction doesn't work out. But if we plug in 2, it works.

Anyway, the major idea of induction is that if we assum $n$ works, and prove $n+1$ works also, this would prove that it works in all cases.

$\displaystyle \frac {1}{1 \cdot 2} + \frac {1}{2 \cdot 3}... + \frac {1}{(n-1)n} = \frac {n-1}{n}\\
\displaystyle [\frac {1}{1 \cdot 2} + \frac {1}{2 \cdot 3}... + \frac {1}{(n-1)n}] + \frac {1}{n(n+1)} = \frac {(n+1)-1}{n+1}\\
\text {Before simplifying the left side, when we find the right side, it gives us} \frac {n}{n+1}\\
\text {Left:} \frac {(n-1)}{n} + \frac {1}{n(n+1)} = \frac {(n-1)(n+1)+1}{n(n+1)}\\
=\frac {n^2-1+1}{n(n+1)} = \frac {n^2}{n(n+1)}\\
=\frac {n}{n+1} = \frac {n}{n+1} \text {a result in right side}$.

Therefore, this method works.[/hide]

This is not well-known I think. Telescoping and pattern is more known. I may be wrong but.

Finding a pattern and using induction are very nearly the same thing. To use induction you have to know the answer, which you usually do by finding a pattern :)
Finding a pattern by itself is very useful, especially at the getting started level, but remember if you want to explain WHY the pattern works, you will need to do something more. Eg induction :D


Threads related to Chemistry Olympiads or similar sources.

Chemistry Marathon @ AoPS is located [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=229844]here[/url].


Find the polynomial $ f(x)$ with deg $ < 3$ so that with $ \alpha$ is given $ \in [ \minus{} 1;1]$ we have $ |f(1)|\leq 1,|f( \minus{} 1)|\leq 1,|f(\alpha)|\leq 1$ and $ max_{[\alpha,1]} f(x) \equal{} \frac {\alpha ^2 \plus{} 2\alpha \plus{} 5}{4(\alpha \plus{} 1)}$

Nobody like this problem?
[hide="Hint"]Use Lagrange identity for $ f(x)$ at $ \minus{}1,\alpha,1$[/hide]


In isosceles triangle ABC, E is the midpoint of base BC.  Points F and G are on line segments AB and AC, respectively, and EFG is an equilateral triangle.  If BC = 6 and AB = 5, then FG = (-24/11)(a+b*sqrt3).  Find the ordered pair (a,b) of rational numbers.  THanks.

Important Edit: BC=6, not 5.

So ABC is also an equilateral triangle and FGE is an equilateral triangle inscribed in ABC but its sides are not parallel to ABC?

-interesting_move

Assuming that the problem statement is correct, we have two equilateral triangles ABC and FGE.  I believe that their corresponding sides [i]must[/i] be parallel to each other.  Then FG is half of BC; it is 2.5.  I can't find a pretty way of turning this into a number of the form (-24/11)(a+b*sqrt3)

Is is possible that you meant AB = AC = 5?

I don't think that the sides of EFG must be parallel to ABC.  You can see why by trying to construct some equilateral triangles within an equilateral triangle.

-interesting_move

but E is the mid point of BC
in any other triangles E would be somewhere else along BC

Oh yes, I forgot about that given!

-interesting_move

Indeed, they are parallel.

-interesting_move

[quote="hgmm76"]Sorry for the confusion, there was a typo and i didnt proofread.  base BC=6, the legs =5.  Sorry.[/quote]

Okay good. Cause this is a distinctly different problem than the one you posted originally.

[quote="hgmm76"]Anyways, my main question was, when I arrived at FG=3, i wasnt sure how to write it as (-24/11)(a+b*sqrt3). Since (a+b*sqrt3) is rational, i assumed b=sqrt 3, then solved for a. But if b=-sqrt3, i'll get another answer for a, therefore 2 ordered pairs would work, but from the wording of the question there is only supposed to be 1. So is one of the cases invalid, if so, why? Thanks.[/quote]

What makes you think FG = 3? You should look at that again.

im really not sure.  But doing it again i get FG=2(sqrt15-3), but im still unsure.  The first time i got FG=3 because i thought that EFG would make an equilateral triangle with all points bisecting their respective sides and FG parallel to the base.  Which of those assumptions were wrong?

[quote="darkquantum"]the way i see it, the sides don't necesarily have to be parallel, and thus the vertices of EFG don't necesarily have to bisect their respective sides[/quote]

Yes, we NOW know that the sides don't have to be parallel.  But at the time, hgmm76 said that BC was 5, and so we were all referring to that (if BC = 5, then both ABC and FGE are equilateral triangles, and thus the sides must be parallel).  Now that we know BC is 6, we know they aren't parallel.

Actually in the original problem he had AB = BC = 5, and the length of AC was unspecified. Which I think left the problem unsolvable. But even if you were given a value for AC, it makes the problem a little different than the one that's given.

For example, suppose you have an isosceles triangle ABC with AB = BC = 5 and AC = 6. Let E be the midpoint of BC and let points F and G be on line segments AB and AC, respectively, such that EFG be an equilateral triangle. Find the length of FG.

can someone tell me how they posted my edited problem?

Okay, I think I solved the actual problem (base = 6, legs = 5) I will just make one remark and put the rest in spoiler.

First, I think FG will be parallel to BC. The other two sides (EF and EG) will not be parallel to AB & AC, but FG will be parallel to BC.

[hide]Okay, let the side of the equilateral triangle be s (how original!) Dropping the altitude from A to BC, we see its length must be 4 since we have that nice 3-4-5 right triangle. Okay, now let the point where the altitude from A intersects FG be called P. EP=s:rt3:/2. Now, since FG is parallel to BC, triangle AFG is similar to ABC. Therefore, the ratio of the altitudes from A will be equal to the ratio of the bases of the triangles, so (4-(s:rt3:/2))/4=s/6. Solving, we get s=24/(4+3:rt3:), or 24(3:rt3:-4)/11[/hide]

Sorry, I didn't put it in the form. [hide]So: a would be 4 and b would be -3.[/hide] Wow, my last post looks weird with little :rt3: marks all over the place. I think I'll do that again.[hide]This is random stuff that I put here just to make the  emoticons look  weird.  This is :rt2: still just  :Sigma: a lot  :pi:  of random  :equiv:  stuff. Okay :ge: , I think  :approx:  that's enough.[/hide]

Because it's an altitude of equilateral triangle EFG with sides length s.


geometry
analytic geometry
rectangle
trigonometry
perimeter
ratio
Pythagorean Theorem

how do you find the area of a hexagon? a pentagon?

[url=http://en.wikipedia.org/wiki/Pentagon#Regular_pentagons]Pentagon[/url], [url=http://en.wikipedia.org/wiki/Hexagon#Regular_hexagon]hexagon[/url], [url=http://en.wikipedia.org/wiki/Polygon#Area_and_centroid]general regular polygon[/url].

The common method to find the area of a regular $ n$-gon is to break it up into $ n$ small isosceles triangles with side lengths $ R,R,s$, where $ s$ is the side length and $ R$ is the circumradius. Then the problem is finding the area of this isosceles triangle.

Then it really depends on what is given. Even if one has all the side lengths, a unique $ n\ge4$-gon can't be determined. Given coordinates, one can always use something like the [url=http://en.wikipedia.org/wiki/Shoelace_formula]shoelace formula[/url] or [url=http://en.wikipedia.org/wiki/Pick%27s_Theorem]Pick's Theorem[/url], though.

How do you find the area of a regular octagon? Is there a formula?

Yup. [url=http://letmegooglethatforyou.com/?q=area+of+a+regular+octagon]Right here[/url]. :rotfl:

whoa!
that was awesome!
how do you do that? :)

Go to [url=http://letmegooglethatforyou.com]here.[/url] :P

The apothem is the key to all regular polygons, and their areas.  It's a matter of finding the length of the apothem when trying to find the polygons's areas.

Oh so the area of an octagon is basically the area of the square in the middle + 4 x (the area of the triangle) + 4 x (the area of the rectangle).

The area of the square is $ s^2$ where $ s$ is the side length of the octagon.
To find the area of the triangles, you need to first find the side length. You do this by dividing the side length of the octagon by rad2. (pythagorean theorum). Then you should get srad2/2 as the side length of the triangle. In order to get the area of the triangle you square the side length and divide by 2, so the area of the triangle is $ s^2/4$. That means that the area of four triangles is $ s^2$. The area of the rectangle is just (s^2)rad2/2. Multiplying this by four, you get (2s^2)rad2, so the area of the octagon given the side length is $ s$ is just $ 2s^2 \plus{} (2s^2)\sqrt{2}$. But of course you can factor $ 2s^2$ to get;

$ 2s^2 (1 \plus{} \sqrt{2})$

[quote="dragon96"]Unless it's like a regular 7-gon or 1982306401-gon (or something like that), I usually find it helpful to draw some diagonals. For the regular hexagon, connecting all the opposite points split it up to 6 equilateral triangles with side length equal to the hexagon's side length. For the regualr octagon, I drew it like this:

[geogebra]67236edc465cd636230949a3fab308649473b47a[/geogebra]

That breaks the octagon down into a workable rectangles and triangles[/quote]

I can't see the picture.  :?:  :?:

Divide any polygon into triangles, use come up with some formula's based on what you found and then try to prove them; it should work for many of them (if not all).

[quote="007math"][quote="Wickedestjr"]Well, I think it gets kind of tricky working with a 7-gon polygon and an 11-gon polygon. :wink:[/quote]

You can assign points coordinates and use shoelace to find area.[/quote]

Still tricky to assign the points.  Also, you could split into $ s$ amount of triangles of the same area (let $ s$ equal the number of sides in the polygon) and then use trig (:D) to solve the triangle and ultimately, find the area.

Which is exactly what Z-coli and akhil0422 said :wink:

$ \frac {3s^{2}\sqrt{3}}{2}$

^For regular hexagons.

How you derive
(3s^2*sqrt3)/2 is by cutting a hexagon into 6 equal equilateral triangle, and multiply it by six

When working with the area of any shape it is important to list the given information then try to use your prior knowledge of shapes.  The answer about drawing diagonals and what not is a good response.  I would have divided the regular hexagon into 6 equilateral triangles.  Obviously, this is assuming I know the length of 1 side.  To find the area of one triangle, I would split the length of one side in half then use the pythagorean theorem to get the height.  Multiply base by height to get the area of one of the six triangles.  Then multiply this by six to get the total area of the regular hexagon.  Formulas mean absolutely nothing if you can't see the concrete actions behind the formula.  They're great in the short run but not for the understanding one needs to continue to solve problems of the like in the future.

(1/2) a P
or (1/2) apothem*Perimeter

This is an equation for REGULAR Polygons. If irregular, do as others have said, divide into shapes you know.

The area of a regular n-gon is the apothem times the perimeter divided by 2. To find the apothem you make one side of the regular n-gon the base of an isosceles triangle. Next take the length of the side and divide it by 2. Since the isosceles triangle is a 30-60-90 and the short side the base divided by 2. So the apothem is half the base radical 3. The definition of an apothem is the distance from the center of regular n-gon to the perpendicular bisector of one side.

If you bothered to read the thread, everything that you said has already been stated.

If you want to find the area of a regular hexagon, then you must split it along the opposite diagonals to form 6 equilateral triangles. Then you just find the area of one of the triangles and multiply it by 6.

Continuing from last time, if you divide one of the equilateral triangles in two with the altitude, (perpendicular to the midpoint of the base), then you get two 30 60 90 triangles. You can use the ratio 1, sqrt(3), and 2 to find any missing sides if you have to.


3/2.

You can get it by using Maclaurin's series, or by using the L'Hospital rule (only once!).

LHR solution:

L=lim (6sinx(cosx)^5)/(2xcos2x+sin2x)=lim (6sinx(cosx)^5)/(2xcos2x+2sinxcosx)=lim (6(cosx)^5)/(2cos2x(x/sinx)+2cosx).

As x/sinx->1 and cosx->1, L=3/2, qed.

Or you can do it without LHR...

1-(cosx)^6=(1-(cosx)^2)(1+(cosx)^(2+cosx)^4)=((sinx)^2)(1+(cosx)^(2+cosx)^4)

Hence, L=lim (((sinx)^2)(1+(cosx)^2+(cosx)^4))/2xsinxcosx=lim sinx(1+(cosx)^2+(cosx)^4)/2xcosx

As sinx/x->1, L=lim (1+(cosx)^2+(cosx)^4)/2cosx=3/2, qed.

thank you
i prefer without LHR rule is very simple ^^


calculus
integration
logarithms
calculus computations

Hi all,

i've always wondered about the resolution of integrals of this sort:
\[ I \equal{} \int \frac{x^2 \,dx}{1\plus{}x^4} \]
It is tabulated in almost all books on integrals, like Schaum, etc. but have never been able to reproduce it nor to find it anywhere. Hope you can give a clue.

Cheers,
Smm.

:welcomeani:

[hide="Method 1"]Divide by $ x^2$: $ I \equal{} \int \frac {x^2}{1 \plus{} x^4}\,dx \equal{} \int\frac {1}{x^2 \plus{} \frac {1}{x^2}}\,dx$

$ \equal{} \frac {1}{2}\left(\int\frac {1 \minus{} \frac {1}{x^2}}{x^2 \plus{} \frac {1}{x^2} \plus{} 2 \minus{} 2}\,dx \plus{} \int\frac {1 \plus{} \frac {1}{x^2}}{x^2 \plus{} \frac {1}{x^2} \plus{} 2 \minus{} 2}\,dx\right)$

$ \equal{} \frac {1}{2}\left(\int\frac {\left(x \plus{} \frac {1}{x}\right)'}{\left(x \plus{} \frac {1}{x}\right)^2 \minus{} 2}\,dx \plus{} \int\frac {\left(x \minus{} \frac {1}{x}\right)'}{\left(x \minus{} \frac {1}{x}\right)^2 \plus{} 2}\,dx\right)$

and now make $ x \plus{} \frac {1}{x} \equal{} t$, $ x \minus{} \frac {1}{x} \equal{} u$, to get

$ I \equal{} \frac {1}{2}\left(\int\frac {1}{t^2 \minus{} 2}\,dt \plus{} \int\frac {1}{u^2 \plus{} 2}\,du\right)$

$ \equal{} \frac {1}{4\sqrt {2}}\ln\left|\frac {t \minus{} \sqrt {2}}{t \plus{} \sqrt {2}}\right| \plus{} \frac {1}{2\sqrt {2}}\arctan\left(\frac {u}{\sqrt {2}}\right) \plus{} C$

and now you just have to substitute back.[/hide]

[hide="Method 2"]$ 1\plus{}x^4 \equal{} (1\plus{}x^2)^2 \minus{} 2x^2 \equal{} [1\plus{}x^2 \plus{} \sqrt{2}x][1\plus{}x^2 \minus{} \sqrt{2}x]$ and so

$ \frac{x^2}{1\plus{}x^4} \equal{} \frac{x^2}{[x^2 \plus{} \sqrt{2}x \plus{} 1][x^2 \minus{} \sqrt{2}x \plus{} 1]}$

and now you must split this into partial fractions.
[/hide]

Carcul, simply excellent!! Thanks so very much for your help!

Cheers,
Smm.


geometry
function
inequalities
geometry unsolved

The points $A$, $B$, $C$, $D$ lie on a straight line, in this order. A circle $k$ passes through the points $B$ and $C$. The two tangents from the point $A$ to the circle $k$ touch $k$ at the points $M$ and $N$; the two tangents from the point $D$ to the circle $k$ touch $k$ at the points $K$ and $L$. The lines $MN$ and $KL$ intersect the line $BC$ at $P$ and $Q$, respectively.
[b]1)[/b] Prove that the points $P$ and $Q$ do not depend on $k$.
[b]2)[/b] If $AD=a$, $BC=b$, and the segment $BC$ moves along $AD$, find the minimum length of $PQ$.

Proof of (1):

Let's focus on P (the soln is obviously the same for Q). MN is the polar of A with respect to k, so AB/AC=PB/PC, and since a is outside [BC] it means that P is inside [BC], so P is a fixed point, because AB/AC is a fixed value.

I'm sure the relations AB/AC=PB/PC and DB/DC=QB/QD can successfully be used to give a soln to (2), but I'm kind of in a lazy move right now :D

In order to close the topic...
It is a piece of cake now (after 1 year from last post on this problem), like grobber says.

Let $x$ be $AB$, $y$ be $CD$, then $BP=\frac{bx}{2x+b}$, $CQ=\frac{by}{2y+b}$, and we are to maximize $BP+CQ$, but $f(t)=\frac{bt}{2t+b}$ is a concave function, $x+y=a-b$, so applying Jensen's inequality for $f(x)+f(y)$ we obtain $f(x)+f(y)\leq 2f\left(\frac{x+y}{2}\right)=\frac{b(a-b)}{a}$.


limit
LaTeX
calculus
integration
calculus computations

$ \lim_{x \to 0}{\frac{e^x-1-x}{x^3}}$

[hide]
I used L'Hopital's Rule, and got 1/0.  Is that correct?[/hide]

this is too easy. just use L`Hospital  3 times. u get answer as 1/6

Hi, [b]math92[/b]

I think that you meant

$ \lim_{x \to \infty} \frac {e^{x} \minus{} 1 \minus{} x}{x^{3}}$.

Am I right? ;)

[quote="coquitao"]I think that you meant

$ \lim_{x \to \infty} \frac {e^{x} \minus{} 1 \minus{} x}{x^{3}}$.

Am I right? ;)[/quote]

if above post is rite, then watever i said is rubbish

[quote="coquitao"]To the contrary, [b]b555[/b]. You are right unless I'm wrong! ;)[/quote]

hehe. thats  obvious.  lets wait for  math92 to confirm

$ e^x\equal{}1\plus{}x\plus{}\frac{x^2}{2}\plus{}\frac{x^3}{6}\plus{}O(x^4)$.
So, $ \lim_{x \to 0}\frac{e^x\minus{}1\minus{}x}{x^3}\equal{}\lim_{x \to 0}\frac{\frac{x^2}{2}\plus{}\frac{x^3}{6}\plus{}O(x^4)}{x^3}\equal{}\left(\lim_{x \to 0}\frac{1}{2x}\right)\plus{}\frac{1}{6}$.

Thus, $ \lim_{x \to 0}\frac{e^x\minus{}1\minus{}x}{x^3} \neq \frac{1}{6}$.

[hide="I think..."]$ \left(\lim_{x \to 0}\frac{e^x\minus{}1\minus{}x}{x^3} \equal{} \infty \right)$[/hide]

Its the limit as x approaches 0.  NOT infinity.

The limit as $ x$ approaches $ 0$ as well as $ \infty$ is $ \infty$ (positive or negative depending on whether x is positive or negative), and one cannot use L'Hopital's Rule because it is not an indeterminate form - it is of the form $ \frac{1}{0}$, which means that it diverges to infinity.

You can use L'hopital's rule twice.  The problem is using it a third time.

[quote="b555"]but what is wrong  in using lhospital rule  here, and answer seems quite obvious with it.  :?:[/quote]

Simple exercise for you... "Under what condition/s can you apply L'H's rule?"

Note that it is not in the spirit of L'H's rule that you can differentiate the numerator and denominator until you see something that you like to see. :wink:

[quote]Simple exercise for you... "Under what condition/s can you apply L'H's rule?"

Note that it is not in the spirit of L'H's rule that you can differentiate the numerator and denominator until you see something that you like to see. :wink:[/quote]

ya  sorry, saw that later.

[quote="IndoChina"]Note that it is not in the spirit of L'H's rule that you can differentiate the numerator and denominator until you see something that you like to see. :wink:[/quote]  I don't know, I think it's in the spirit but just not in the word :).  "The spirit of the theorem" versus "the word of the theorem" -- I think this is a distinction lots of mathematicians recognize :)

Applying l'Hopital's rule, I get $ \infty$.

I think:

As $ x$ goes to $ \minus{}\infty$, the limit is $ 0$.

As $ x$ goes to $ 0$ from the right, the limit is $ \plus{}\infty$ and $ \minus{}\infty$ as $ x$ goes to $ 0$ from the left.

How does one write a right arrow in $ LaTeX$?  and "The limit as x goes to infinity"?

hello, the limit $ \lim_{x \to 0}\frac{e^x\minus{}1\minus{}x}{x^3}$ doesn't exist.
Sonnhard.

I agree: there is an infinite jump discontinuity or saltus (interesting Latin cognates) at the origin.

hello, yes, that is right, we have
$ \lim_{x \to 0\minus{}0}\frac{e^x\minus{}1\minus{}x}{x^3}\equal{}\minus{}\infty$
$ \lim_{x \to 0\plus{}0}\frac{e^x\minus{}1\minus{}x}{x^3}\equal{}\plus{}\infty$
Sonnhard.

Yes, I've checked it with l'Hopital's rule, sketched the graph and checked that against the graph given by a gaph plotter site on the internet.

(By the way, Sonnhard, do you have any ideas on my "y = ln (cos x), "integration by substitution" and "integration" posts?)

I've just seen how to write $ \lim_{n\to\infty}$.  What about a dollar sign or a percentage sign?


superior algebra
superior algebra unsolved

1. If p|(q^p-1)/(q-1) (or p|(q^p+1)/(q+1)),then whether or not p|q-1 (or p|q+1)? where p prime number.
2. Let p and p' be primes. The equations

     5^p - 3^{p'}=2;
     3*5^p - 2^{p'+2}=7;
     2^{p'+2} - 5^p =3;

have no solution?

I think it will be more proper to post this topic in the "number theory" subforum.


[color=darkblue]Hi every body. Can you help me?

Let $ A\equal{}(x\plus{}y\plus{}z)^3\minus{}(x\plus{}y\minus{}z)^3\minus{}(x\minus{}y\minus{}z)^3$. I want factor $ A$. Can you help me?[/color]

[quote="hien"]Are you sure typing corect? I have used Maple 11 but don't get any solution[/quote]

[color=darkblue]Yes. I tried to factor it, but I don't have any solution. I will see my problem. [/color]

[color=darkblue]I see it in a book and tried done it but I can't. Hi every one! What do you think about this problem?[/color]

[quote="Euclid's Turtle"]A=(x+y+z)^3-(x+y-z)^3-(x-y-z)^3 + (-x-y-z)^3 factors. Perhaps it was a mistake and you require the 4th bracket?

Well if so, the answer is 2(z-x)(z^2-2xz+3y^2+x^2)

Cheers[/quote]
Thank you!  :P


Given x,y,z >0 .Prove that
$\sqrt{2x} + \sqrt{2y} +\sqrt{2z}$ <= $\sqrt{x+y} +\sqrt{y+z}+\sqrt{z+x}$

[quote="NC7"]Given x,y,z >0 .Prove that
$\sqrt(2x) + \sqrt(2y) +\sqrt(2z)$ >= $\sqrt(x+y) +\sqrt(x+y)+\sqrt(x+y)$[/quote]

You mean: $\sqrt{2x}+\sqrt{2y}+\sqrt{2z}\leq\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}$, right?

I think $\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\geq\sqrt{2x}+\sqrt{2y}+\sqrt{2z}$.
$2(x+y)\geq x+y+2\sqrt{xy}$.
Therefore, $\sqrt{2(x+y)}\geq\sqrt x+\sqrt y$
Similarly, $\sqrt{2(y+z)}\geq\sqrt y + \sqrt z$, $\sqrt{2(z+x)}\geq\sqrt z + \sqrt x$.
Then, sum three inequalities.


Prove that [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/6f5e7a7f03e748bcdc45686ce4f2e8f2.gif[/img]
where a,b,c are reals s.t [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/2d1ed4ba97d969b231d8a32c1815fdd7.gif[/img]

I just wonder how many times and in how many forms should we discuss great Vasc's inequality $ (a^2+b^2+c^2)^2\geq 3(a^3b+b^3c+c^3a)$?  :?

Where can i find a proof of $(S_2)^2 \geq 3(\sum_{cyc} a^3b)$ ?

See http://www.mathlinks.ro/Forum/viewtopic.php?t=6026 .

  Darij

Thank you very much,Darij! That's just what I want.


A compound containing only xenon and fluorine can be obtained by shining light on a flask that contains .526g of xenon and excess fluorine gas. When all the xenon was consumed, 0.678g of the new compound was obtained.

Determine the empirical formula of the compound.

Molar Masses:
Mg: 24.31g/mol
O: 16.00g/mol
Xe: 131.29g/mol
F: 18.99g/mol

Does anyone know how to determine the empirical formula given all of this information?

Try to answer the following questions:

1) Why do you give the molar masses of Magnesium and Oxygen?

2) You know the mass of compound that forms and you also know the mass of Xenon it contains. So what is the mass of fluorine?

3) Given those masses, and the molar masses of Xe and F, what is the proportion of both elements in the compound (the empirical formula)?


geometry
projective geometry
geometry unsolved

Given a triangle $ ABC$ with $ AC \not\equal{} BC$. Let $ w$ be the incircle of $ ABC$ with center $ I$.$ w$ touch $ AB,BC$ and $ AC$ at $ D,E$ and $ F$ ,respectively. Let $ M$ be the second intersecting point of $ w$ and $ CD$.the tangent to $ w$ at $ M$ intersects the line $ AB$ at point $ G$ and $ CI \cap EF \equal{} S$.Prove that $ GS \perp CI$

apply Newton-Line theorem AL,BH,DM,EF are concurrent
A,D,B,G are harmonic division So AD/DB=AG/BG
CF/FA*AG/BG*BE/EC=CF/FA*AD/BD*BE/EC=1  so G,E,F are colinear
Thus we are done

Alternatively...

$ G$ is the pole of $ CM$, so $ G$ lies on the polar of $ C$, that is, $ EF$, and we are done.
This is a pretty well-known fact, and point $ S$ seemed pretty unnecessary...


calculus
derivative
function
limit
calculus computations

hello

the question i am stating below isnt much as the concept related which i need to clear up :blush:  -:

$ f(x)=  \{ \begin{array}{rcl}-x^{3}+c & \mbox{for}& \ x\leq0 \\ 3x-2 &\mbox{for}& \ x>0 \end{array}$ then -:

[b](Choose the correct option)[/b]

[i]a)f(x) is increasing in [0,∞)
b)f(x) is decreasing in [0,∞)
c)f(x) is increasing in (-∞,0]
d)none of these[/i]

Now the problem is , if i differentiate, i get -3x^2 for x<=0, 3 for x>0
So the function is increasing only in (0,∞). Hence option (d). But the correct answer marked is a). Why?
At 0, function has derivative =0, the function is neither decreasing  nor increasing. However, i need to consider a range to find whether the func. is increasing or not, so do i choose x-dx to x or x to x+dx  ;dx here showing infinitesimal change ?
Could someone plz help by providing the general method to decide whether the func is increasing in an interval when its derivative is 0 at a point  :maybe:

Thank You

It's a poorly-posed question.  (a) is true if and only if $ c \leq-2$.  (b) and (c) are false for any value of $ c$.  Your function does not have a derivative at $ x= 0$, regardless of the value of $ c$.  (If $ c =-2$, your function has a cusp there; if $ c \neq-2$, your function isn't even continuous at 0.)

A function $ f$ is (strictly) increasing of an interval $ I$ if and only if, for all $ x < y \in I$ we have $ f(x) < f(y)$.  So, for example, the function $ f(x) = x^{3}$ is increasing on $ \bf R$, even though it has a point of 0 derivative.

[quote="JBL"] (b) and (c) are false for any value of $ c$.  Your function does not have a derivative at $ x= 0$, regardless of the value of $ c$.[/quote]

So how does one determine how the function behaves at $ x=0$?

We look at the value and various one-sided limits. In this case, $ f(0)=c$, $ \lim_{x\to 0^{-}}f(x)=c$, and $ \lim_{x\to 0^{+}}f(x)=-2$. We can make sense of one-sided derivatives as well; if $ c=-2$ so $ f$ is continuous, $ \lim_{h\to 0^{+}}\frac{f(h)-f(0)}{h}=3$ and $ \lim_{h\to 0^{-}}\frac{f(h)-f(0)}{h}=0$ (If $ c\neq-2$, the first limit doesn't exist). For $ f$ to be differentiable, these two one-sided derivatives would have to agree.

It does generally make sense to talk about a function being increasing on a closed interval; by the definition JBL quoted, the largest possible interval on which a function is increasing is always closed unless that function is discontinuous.


calculus
integration
calculus computations


very nice problem.
For the begining,I guess the answer is $-\frac 7{10}$
right? :)

First of all the numbers are positive ?

Secondly the condition $a+b+c=1$  smells weighted Jensen, but i don't try it yet.

[quote="tµtµ"]Don't have ideas for now but Zhaobin, try with a=b=-1.2  ;)[/quote]

Don't expect to have "nice" numbers in the solution, because
Mathematica says that the minimum occurs, for example, if $b=c$, and
$b$ is the root of  $b^4-b^3-b^2+b-1=0$, $b\approx -1.178724176105222...$.
This would mean that the minimal value of the expression $P$ is approx.
$-0.71305576355891849206...$

Really, I forgot Karamata, please post your solution  :) .

[quote="leepakhin"][quote="zhaobin"][quote="Anh Cuong"]Well, this is a nicer problem if you want to work with pleasant numbers  :)
Give $a,b,c \geq 0$ satisfying $a+b+c=1$. Prove that:
$\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2} \geq \frac{1}{2}$  :)[/quote]
but that's easy by karamate ;)[/quote]
But $\frac{a}{1+a^2}$ is concave for $|a|\leq\sqrt{3}$... Are you sure Karamata works? :?

Anyway, my solution goes as follows.
Since $a\leq 1$, we have
$\frac{a}{1+a^2}\geq\frac{a}{1+1^2}=\frac{a}{2}$.
Summing up analogous inequalities yield the desired result. :)[/quote]
sure,karamate does,
let $f(x)=\frac{x}{1+x^2}$
wlog $a\ge b \be c$
we have:$1\ge a$,$1+0 \ge a+b$,$1=a+b+c$.
by karamate we have$f(1)+f(0)+f(0) \le f(a)+f(b)+f(c)$...
but your proof is nicer :)

to AC:
Karamate's inequality:
you can see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14975]www.mathlinks.ro/Forum/viewtopic.php?t=14975[/url]

[quote="Anh Cuong"]Give three real numbers $a,b,c$ satisfying $a+b+c=1$. Find the minimum value of
$P=\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}$  :)[/quote]
Hi [b]AnhCuong[/b]
Have you looked for the maximum value yet ? It is actually much nicer
For real $x,y,z$ satisfying $x+y+z=1$
\[\sum_{cyc} \frac{x}{1+x^2} \leqslant \frac{9}{10} \]
or real $x,y,z$ satisfying $xyz=1$
\[\sum_{cyc} \frac{x}{1+x^2} \leqslant \frac{3}{2} \]

[quote="urun"]
Hi [b]AnhCuong[/b]
Have you looked for the maximum value yet ? It is actually much nicer
For real $x,y,z$ satisfying $x+y+z=1$
\[\sum_{cyc} \frac{x}{1+x^2} \leqslant \frac{9}{10} \]
or real $x,y,z$ satisfying $xyz=1$
\[\sum_{cyc} \frac{x}{1+x^2} \leqslant \frac{3}{2} \][/quote]
They are old problems.
For the first inequality Asume $ x\geq y\geq z.$
If $ z \geq \dfrac{-3}{4}$ then
\[ LHS \leq \sum \dfrac{36x+3}{50}=\dfrac{9}{10}\]

If $ c<\dfrac{-3}{4}$
Now, for $ 2>x,y>\dfrac{1}{2}$ then $ -3<z<\dfrac{-3}{4}$.So
\[ \dfrac{x}{x^2+1}+\dfrac{y}{y^2+1}+\dfrac{z}{z^2+1}<\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{-3}{10}<\dfrac{9}{10}\]

In other cases we have \[\dfrac{x}{x^2+1}+\dfrac{y}{y^2+1}<\dfrac{1}{2}+\dfrac{2}{5}\]

Since $\dfrac{z}{z^2+1}<0.$

For the second inequality it follows from obvious inequality $ \frac{x}{1+x^2}\leq \dfrac{1}{2}$
We are done!  :P  :wink:


This is a very simple problem, but I cannot solve it.

Given a finite string, show an efficient algorithm to calculate the least number of insertions needed to transform it in a palindrome.

Example: if input=aabaa, the answer is 0; if the input is caabaa, then the answer is 1 (caabaa[c]); if the input is aabcaa, the answer is 1 too (aa[c]bcaa;aabc[b]aa).
If possible, show also the modified string (all of them).

Consider this:

Let the function, method, or operation be called insertPal
Return 0 if the string has length 0 or 1 or points to null
Store the first number of the string (first)
Starting at the end of the string, look at the letters until char=first or index = 0
* If you do not find this string (index = 0), return 1 + insertPal(original string minus the first letter)
* If you do find the string (char = first):
Let 0 be the position of the first letter, x be the position of the last letter, length-1 be the position of the last letter:  return length -1 - x + insertPal ( substring from 1 to x-1 )

Hopefully it works.


By what percent must the radius of a circle decrease in order to decrease the area of the circle by $ 75\%$?

Since area essentially equals radius squared, the square root of the new area, 1/4, equals the new radius, 1/2 or [u][b]50%[/b][/u]


Solve $\sqrt[3]{6x+28} - \sqrt[3]{6x-28} = 2$.

I moved the term $\sqrt[3]{6x-28}$ over to the right side of the equation and cubed both sides.

I ended up with $8 = 2\sqrt[3]{6x-28} + \sqrt[3]{(6x-28)^2}$.

Setting $\lambda = \sqrt[3]{6x-28}$, I got $\lambda^2 + 2\lambda - 8 = 0$.

Are there in fact only two solutions to this cubic root question?

OK, using your method I obtained the same [b]two[/b] answers as I did.

I guess there are only two solutions...

if you are not convinced those are the only 2 answers, graph it, but i am not sure if there are any complex solutions

Every positive number has a positive and negative square root, because both positive and negative numbers have positive squares.

However, every number (doesn't have to be positive) has only one cube root, because positive numbers have positive cubes and negative numbers have negative cubes (and $\sqrt[3]{0}=0$.  So because every number has exactly one cube root, there are no restrictions on values under a cube root, and a cube root won't have more than one value.

So you don't have to worry about there being any more answers.

Yes, there are only 2 solutions (including complex). The first method (the math is correct) gives 2 solutions for $\sqrt[3]{6x-28}$; cubing and solving for $x$ produces one $x$ for each $\sqrt[3]{6x-28}$.

Different idea:

[hide]     $\sqrt[3]{6x+28}-\sqrt[3]{6x-28}-2=0$

Notice that when $x+y+z=0$, we have

     $x^{3}+y^{3}+z^{3}=3xyz$

Now we have

     $-8=\sqrt[3]{36x^{2}-784}$

Cubing, solving for $x$ gives:

\fbox{6}[/hide]

[quote="erik-the-red"]Solve $\sqrt[3]{6x+28}-\sqrt[3]{6x-28}= 2$.

I moved the term $\sqrt[3]{6x-28}$ over to the right side of the equation and cubed both sides.

I ended up with $8 = 2\sqrt[3]{6x-28}+\sqrt[3]{(6x-28)^{2}}$.

Setting $\lambda = \sqrt[3]{6x-28}$, I got $\lambda^{2}+2\lambda-8 = 0$.

Are there in fact only two solutions to this cubic root question?[/quote]

This cube isn't technically a cubic equation right?

[quote="4everwise"][quote="jhcreinhardt"]By definition it has three roots, if it has one complex, then it must have another (its conjugate).  it could have a double root, but it has to have three.[/quote]That's only if it's a polynomial with rational coefficients.[/quote]

Actually it only has to be a polynomial with real coefficients.


Well there have been many times that I've searched for past problems, and I find the topic that I was looking for. Yet, the topic is about 20 pages long and it would take too long to look through all of it. Would it be possible with multi-paged topics (when you click on it) to have it go straight to the post itself instead of just showing the whole topic?
Sorry if I'm not clear.

Thanks.

Search by posts instead of topics, that should work.


In a bag, there are 3 red marbles and B blue marbles. Two marbles are randomly selected from the bag without replacement. The probability that the two marbles are the same color is 0.5.  What is the sum of all the possible values of B?

[hide="Solution"]Either both marbles are red or both marbles are blue.

The probability that both are red is (3/(B+3))*(2/(B+2)).
The probability that both are blue is (B/(B+3))*((B-1)/(B+2)).
The probability that both are the same color is just the sum of these probabilities.

(6/(B^2+5B+6)) + ((B^2-B)/(B^2+5B+6)) = 1/2

Simplifying everything, we get
B^2-7B+6=0.

Looking at the second term, we see that the sum of all possible values of B is [b]7[/b].[/hide]

Oh I just messed up on the signs of the equations


If $x,y$ are integers and $17$ divides both $x^2 -2xy + y^2 -5x + 7y$ and $x^2 - 3xy + 2y^2 + x - y$ , then prove that $17$ divides $xy - 12x + 15y$.

[hide]$x^2-3xy+2y^2+x-y=(x-y)(x-2y+1)$. Since 17 is prime, it must divide either of these factors.

[b]Case 1:[/b] 17 divides $x-y$. Writing $x^2-2xy+y^2-5x+7y$ in terms of $d=x-y$ gives $d^2-5d+2y$. So $2y$, and therefore $y$, is divisible by 17. But $x\equiv y$ so $x$ is divisible as well. From there the result clearly follows.

[b]Case 2:[/b] 17 divides $x-2y+1$. Here $x\equiv 2y-1\Rightarrow d\equiv y-1$.

Now $d^2-5d+2y\equiv d^2-3d+2\equiv (d-1)(d-2)$. So $d\equiv 1$ or $d\equiv 2$.

The first case gives $y\equiv 2$ and $x\equiv 3$ so $xy-12x+15y$, which reduces to $xy+5x-2y$, is congruent to $6+15-4=17$ which works. The second case gives $y\equiv 3$ and $x\equiv 5$ so $xy+5x-2y\equiv 15+25-6\equiv 34\equiv 0$. So this also works.

This covers all of the cases, so $xy-12x+15y$ is always divisible by 17.[/hide]

could any kind person pleeaasseee help me with scorpiuss's solutionm in case 2.(all the congruencies are mod 17,right?)

I thought about this for an hour and this is what I came up with.

17 divides x^2 - 3xy + 2y^2 - 5x + 7y

or 17 divides (x-y)(x - 2y + 1)

=> 17 divides (x-y) or 17 divides (x - 2y + 1).

Case 1: 17 divides x-y

Also, 17 divides x^2 - 2xy + y^2 -5x + 7y

          17 divides (x-y)^2 + 7y - 2x.

    => 17 divides 7y - 5x

   or 17 divides 7(y-x) + 2x

  => 17 divides 2x

 => 17 divides x.

But 17 divides x-y.

So, 17 divides y.

Therefore, 17 divides xy - 2x + 15y :jump:

Case 2 : 17 divides x - 2y + 1

Also       17 divides x^2 - 2xy + y^2 - 5x + 7y

Or          17 divides x(x - 2y + 1) + y^2 - 6x + 7y

=>         17 divides y^2 - 6x + 7y...................(2)

Also       17 divides x - 2y + 1

=>          17 divides -2y^2 + xy + y..................(1)

(2) =>     17 divides 2y^2 - 12x + 14y.................(3)

adding (1) and (3) =>  17 divides xy - 12x + 15y. :jump:

[quote=Pork_Chop8][quote="scorpius119"]....$0\equiv (d-1)(d-2)$. So $d\equiv 1$ or $d\equiv 2$.....[/quote]
Unfortunately, mod's don't work quite like regular algebra. This assumption can't stand by itself. It must be accompanied with something along the line of since 17 is prime, one of d-1 or d-2 must be divisible by 17 because nothing else contributes to the product being divisible by 17. Though 1 is a factor of 17, even a million ones doesn't make the product any more divisible by 17. etc.[/quote]

 this works itself. because of Lagrange's theorem


function
complex analysis
complex analysis unsolved

Supposed $ f$ is an analytic function from $ D$ to $ D$, where $ D$ is a unit disk, so that there are two different values $ a$ and $ b$ so that $ f(a)\equal{}a$ and $ f(b)\equal{}b$. Prove by Schwarz Lemma that $ f(z)\equal{}z$ for all $ z$ in $ D$. Thank you for the solution before.

Suppose $ M$ is the Moebius automorphism of the unit disk mapping $ a$ to the origin and $ b$  to some $ \beta\ne{0}$.  Consider  $ g(z)\equal{}M(f(M^{\minus{}1}(z)))$.

We have $ g(0)\equal{}M(f(M^{\minus{}1}(0)))\equal{}M(f(a))\equal{}M(a)\equal{}0$, so $ g(z)$ satisfies the conditions of Schwarz lemma and $ |g(z)|\leq{|z|}$ with equality if and only if $ g(z)\equal{}\exp(i\phi)z$.  But
 $ g(\beta)\equal{}M(f(M^{\minus{}1}(\beta)))\equal{}M(f(b))\equal{}M(b)\equal{}\beta$ and hence
$ g(z)\equal{}\exp(i\phi)z$, but again since $ g(\beta)\equal{}\beta$ we must have $ \phi\equal{}0$ and $ g(z)\equal{}z$. Then $ f(w)\equal{}M^{\minus{}1}(g(M(w))\equal{}M^{\minus{}1}(M(w))\equal{}w$.


An athlete's target heart rate, in beats per minute, is $ 80\%$ of the
theoretical maximum heart rate. The maximum heart rate is found by
subtracting the athlete's age, in years, from 220. To the nearest whole
number, what is the target heart rate of an athlete who is 26 years old?

according to the problem, the answer should be (200-26)*4/5.

...evaluate it.

194*4/5=776/5=155.2

answer : 155


Of all the mathletes at Wantagh Middle School, 80% own
computers and 40% are in band. However, 10% of all the
mathletes neither own computers nor are in band. What
percent of all the mathletes both own computers and are in
band?

[hide]  .80-x  +   .40-x   +   x= .9

1.2-x=.9

x=.3

Thus 30% percent of them are in both.[/hide]

100% - 10% = 90%
so 90% are in a band, own a computer or both

suppose x% are both in a band and have a computer
40% - x% = in a band but don't have a computer
80% - x% = have a computer but not in a band
40+80-x= 90
120-x=90
120-90=x
30=x

so 30% of the mathletes are in a band and have a computer  :lol:


algebra
polynomial
Rational Root Theorem
algebra unsolved

Let $p(x) = x^{3}+x+1$. Since $p^\prime (x) = 3 x^{2}+1$, it follows that $p$ has exactly one real root. Thus, $x_{1}\in \left(-1, 0 \right)$ and $\overline{x_{2}}= x_{3}$. Also, by the rational root theorem, the only rational roots of $p$ can be $\pm 1$, so $x_{1}\in \mathbb R \setminus \mathbb Q$.
By Viete we have that $x_{1}x_{2}x_{3}=-1$, i.e. $x_{1}|x_{2}|^{2}=-1$, so
\[\sum |x_{1}|^{2}= |x_{1}|^{2}-\frac2{x_{1}}= \frac{x_{1}^{3}-2}{x_{1}}= \frac{-x_{1}-3}{x_{1}}=-1-\frac3{x_{1}}, \]
which is in $\mathbb R \setminus \mathbb Q$, since $x_{1}\in \mathbb R \setminus \mathbb Q$.

Let's find [b]an[/b] integer polynomial that has $|x_{1}|^{2}, |x_{2}|^{2}, |x_{3}|^{2}$ among its roots. Since $|x_{2}|^{2}= |x_{3}|^{2}= x_{2}x_{3}$, it suffices to find two integer polynomials $g,h$ such that $g \left( \left| x_{1}\right|^{2}\right) = 0$ and $h \left( x_{2}x_{3}\right) = 0$.
We take:
- $g(x) = p \left( \sqrt x \right) p \left(-\sqrt x \right) =-x^{3}-2 x^{2}-x+1$;
- $h(x) = \prod \left( x-x_{1}x_{2}\right) = x^{3}-x^{2}-1$.


function
vector
linear algebra
matrix
Online Math Open
linear algebra unsolved

Characterize the functions $f$ from $R^n$ into $M_n(R)$ such that for any vector $X$ and any orthogonal matrix $O$ one has $f(OX)=Of(X)O^{-1}$.

Let's note $(e_1,e_2,..,e_n)$ the canonical basis of $\mathbb{R}^n$.
First, one will study the conditions imposed on $M=f(e_1)$:
Take $X \in \mathbb{R}^n$ an arbitrary vector of unit length, and $O \in O_n(\mathbb{R})$ an orthogonal transformation that maps $e_1$ to $X$. Then, $f(X)=f(Oe_1)=Of(e_1)O^t=OMO^t$. But $M$ cannot be arbitrary because this relation must hold for any $O \in O_n(\mathbb{R})$ such that $Oe_1=X$. Hence, this shows that if $U,V \in O_n(\mathbb{R})$ satisfies $Ue_1=Ve_1$ then $UMU^t=VMV^t$, which means that $M$ commutes with $V^tU$.

Let's take for example:
$U= \left( \begin{array}{c c } 1 & 0 \\ 0 & U_1 \end{array} \right)$ and $V= \left( \begin{array}{c c } 1 & 0 \\ 0 & V_1 \end{array} \right)$ and $M= \left( \begin{array}{c c } a & v^t \\ w & M_1 \end{array} \right)$.
Then $W=V^tU= \left( \begin{array}{c c } 1 & 0 \\ 0 & O_1 \end{array} \right)$ where $O_1$ can be any $O_1 \in O_{n-1}(\mathbb{R})$. So straitforward computations show that this imposes that $v=w=0$ and that $M_1$ commutes with any matrix of $O_{n-1}(\mathbb{R})$, that is that $M_1$ is a scalar matrix.
$M= \left( \begin{array}{c c } a & 0 \\ 0 & \lambda Id \end{array} \right)$.

Now, I think one can caracterise $f$ as follows: There exists two function $p,q : \mathbb{R}_+ \to \mathbb{R}$ such that $f(X)$ is the linear transformation described by:
if $v \in \mathbb{R}^n$ is a vector, write $v=a+b$ where $a$ is colinear to $X$ and $b$ orthogonal to $X$. Then $f(X)v=p(|X|)a + g(|X|)b$. I think it works because $f$ has to be of this form, and these transformations seem to work because of their nice geometric interpretation.

Hope this is correct, which I begin to doubt ..


algebra
polynomial
geometry
algebra unsolved

Two friends, Marco and Ian, are talking about their ages.
Ian says, "My age is a zero of a polynomial with integer coefficients."
Having seen the polynomialp(x) Ian was talking about, Marco exclaims, "You mean, you are seven years old?
Oppps, soory I miscalculated! p(7) = 77 and not zero."
"Yes, I am older than that," Ian's agreeing reply.
The Marco mentioned a certain number, but realizes after a while that he was wrong again because the value
of the polynomial at that number is 85.
Ian sighs, "I am even older than that number."
Determine Ian's age.

Let $ t$ be Ian's age. Also, in "Marco mentioned a certain number...", let $ l$ be that number. Also, $ 7 < l < t$.
Obviously, $ P(7) \equal{} 77, P(l) \equal{} 85, P(t) \equal{} 0$.
Now, we have $ t \minus{} 7|77$, $ t \minus{} l|85$ and $ l \minus{} 7|8$. From the last one we gather four cases: $ l \equal{} 8$, $ l \equal{} 9$, $ l \equal{} 11$ and $ l \equal{} 15$. Obviously, $ t > l\geq8$, so from the first one we gather $ t \equal{} 14$, $ t \equal{} 18$ and $ t \equal{} 84$.
Now, from the second one we have $ t \minus{} l \equal{} 1$, $ t \minus{} l \equal{} 5$, $ t \minus{} l \equal{} 17$ and $ t \minus{} l \equal{} 85$. We check and easily see that the only possibility is $ l \equal{} 9$, $ t \equal{} 14$, $ t \minus{} l \equal{} 5$.
So, Ian's age is $ 14$. We can also find, using Lagrange, that $ P(x)$ can equal $ \minus{}3x^2\plus{}52x\minus{}140$ (which is the only solution for a second degree polynomial).

A very nice little problem, although it seems rather straightforward. Where's it from?

That question was from the area level of the Philippine Mathematical Olympiad 2007.


Connecting the vertices of a regular $ n$-gon we obtain a closed (not necessarily convex) $ n$-gon. Show that if $ n$ is even, then there are two parallel segments among the connecting segments and if $ n$ is odd then there cannot be exactly two parallel segments.


This is going to sound really dumb.

Electrons in a beam of density $ \rho \equal{} 10^{15}$ electrons/m are accelerated through a potential of 100 V. Stuff about other stuff....

I can do everything about this problem such as transmission coefficient and things, but the answer is supposed to be in current not beam density. How do I go from a beam density of electrons with a certain potential to a current?


geometry
3D geometry
algebra
polynomial
quadratics
number theory
prime numbers

Can it be proven that infinitely many primes are the difference of 2 consecutive cubes?

This means that there are infinetly many primes of the form $(x+1)^3-x^3 = 3x^2+3x+1$. But till now there is no polynomial with degree $\geq 2$ for that it's been proven that it gets infinetly many prime values.

Okay. That's a pity - I was working on an IMO proposal but got unexpectedly stuck in the final line because of this one small case - which turns out to be an open problem.

So, goodbye, IMO problem... :(

Dirichlet did manage to prove that a quadratic in two variables x,y represents infinitely many primes. Maybe this can be extended to i variable...

Bomb

[quote="bomb"]Dirichlet did manage to prove that a quadratic in two variables x,y represents infinitely many primes.[/quote]

Are you sure about that? That contradicts what ZetaX said.

Just my curiosity, but.. what was this IMO problem? :D

Arne, you have high ambitions... don't forget that each year 1,2,3 problems are proposed by HL. And no, it does not contradict the statement, since polynomials in many variables have completely different behaviour. For example, there is a polynomial in 22 variables such that all positive values that it takes are prime numbers. A simple example of polynomial in 2 variables which represents infinitely many prime numbers is $X^2+Y^2$.

[quote="harazi"]Arne, you have high ambitions... don't forget that each year 1,2,3 problems are proposed by HL.[/quote] Yeah, Gabriel!  :D

Subtle irony is felt... no good reason is...  :P

To be honest, I don't think this problem (the one this thread originated from) has a good chance to become an IMO problem, but I have another problem, which is very beautiful, and some people (the few people who know it, and one of them ís Hojoo Lee, since we know each other) say it has a very good chance. Obviously, the chances are still not big, since I probably need a lot of luck, but we will see. :)


inequalities
geometry
geometric transformation
reflection
number theory
inequalities proposed

are you sure about the ineq?
[hide]$ LHS \leq \sum \frac{\sqrt{a(a\plus{}bc)}}{2\sqrt{abc}}$
but $ \sum \sqrt{a(a\plus{}bc)}\leq \sqrt{(a\plus{}b\plus{}c)(a\plus{}b\plus{}c\plus{}ab\plus{}bc\plus{}ca)}\leq 2$(B-C-s)
so i get $ \frac{\sqrt{a(a\plus{}bc)}}{2\sqrt{abc}}\leq\frac{1}{\sqrt{abc}}$ [/hide] :huh:

[hide="hint"] by AM-GM  we have :

$ \sum\frac {\sqrt {a(a \plus{} bc)}}{b \plus{} ca}\leq\frac {1}{2}\sum\frac {a \plus{} a \plus{} bc}{b \plus{} ca}$

and we have :

$ \sum\frac {a \plus{} a \plus{} bc}{b \plus{} ca} \equal{} \sum\frac {2a \plus{} bc}{b(a \plus{} b \plus{} c) \plus{} ca} \equal{} \sum\frac {2a \plus{} bc}{(a \plus{} b)(b \plus{} c)}$

now the inequality become :

$ \sum(2a \plus{} bc)(a \plus{} c)\leq\frac {(a \plus{} b)(a \plus{} c)(b \plus{} c)}{2\sqrt {abc}}$

wich i think easy to prove ;)[/hide]

[quote="Guest"]Mr/Ms mto:  If you do not like inequality problems, why come the inequality section?
Perhaps it is too hard for you to prove any nontrivial inequality?
"Modesty helps one to go forward, whereas conceit makes one lag behind."[/quote]

Could u prove 2>1??? :D

To mto
I have an elegant problem for you. Solve the "3x+1" problem, as you disturb us with yours cxonsiderations. Your appeal in mathematic isn't the same for everybody. You like algebra, combinatorics and number theory; other may prefer inequalities, geometry or calculus. Mathematics is a whole.

[quote="mto"][quote="can_hang2007"]Let $ a,b,c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
\[ \frac {\sqrt {a(a \plus{} bc)}}{b \plus{} ca} \plus{} \frac {\sqrt {b(b \plus{} ca)}}{c \plus{} ab} \plus{} \frac {\sqrt {c(c \plus{} ab)}}{a \plus{} bc} \le \frac {1}{2\sqrt {abc}}
\]
:)[/quote]

   Frankly, i prefer the elegant questions on algebra, number theory or combinatorics than those ugly question on inequality. It 's hard in term of skill and not in term of reflection[/quote]
Dear mto, each person has their favorite. Maybe with you, inequalities are not interesting but with other people they are. So If you don't like inequalities, you can post your reply on the other forum. Why did you post your reply here, was it a form of spam?