function
superior algebra
superior algebra theorems

The author writes that a field $E$ is algebraically closed if and only if any finite extension of $E$ is $E$. Ok, but why not directly say "if and only if any algebraic extension of $E$ is $E$" ? Am I wrong ?

Consider $F=\mathbb{C}$. It is algebraically closed. But the function field $\mathbb{C}(X)\neq \mathbb{C}$, because $\mathbb{C}(X)$ is not a finite extension of $\mathbb{C}$.

Thanks Arne for pointing out my mistake. I was not reading the question carefully.
julien_santini's assertion is correct indeed. For an algebraically closed field $E$, there is no proper algebraic extension. Suppose $\alpha$ is algebraic over $E$, then by algebraic closedness irr$(\alpha, E)=x-\alpha$. Thus $\alpha\in E$.


Let  ABCDEF  be  a  convex hexagon such that angles  B + D + F = 360    and  AB*CD* EF= BC*DE*FA   .  PROVE THAT  BC*AE*FD = CA*F*DE

  please solve with  inversion

This was posted a lot of times, cf. the links at http://www.mathlinks.ro/Forum/viewtopic.php?t=55892 . I have two solutions of my own to this beautiful problem, but at the moment time doesn't permit me to post either of them...

  Darij


[quote="Quevvy"]If Xavier Ezekiel Manchella has 40 minutes to do 30 sprint round problems, how much time would he get for 40 sprint round questions? (assuming that he gets the same amount of time per problem)[/quote]

set up a proportion

[hide]
$\frac{30}{40}= \frac{40}{x}$

cross multiply to get

$1600=30x$

$53.\overline{3}= x$
[/hide]


Let
$ u_k = \int_{0}^{\pi}e^{iv}\{\frac{1}{|v + 2k\pi|} - \frac{1}{|v + \pi + 2k\pi|}\} dv$
I want to show
$ \sum\limits_{k\geq 1} |u_k|$ convergent.
Can anyone help me? thanks a lot!

$ \dfrac {1} {|v \plus{} 2\pi k|} \minus{} \dfrac {1} {|v \plus{} \pi \plus{} 2\pi k|} \equal{} \dfrac {\pi} {(v \plus{} 2\pi k)(v \plus{} \pi \plus{} 2\pi k)}, v\geq 0$
than you can majorize your row as the following($ v \in [0,\pi]$): $ |u_k | \leq \dfrac {\pi^2} {(2\pi k)(\pi \plus{} 2\pi k)} \equal{} \dfrac 1 {2k(2k \plus{} 1)}$

it looks so easy for you :-)
Thanks a lot!

but what if here $ |k|\geq1$ instead of $ k \geq 1$?

Locked for double-posting. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=311559]here[/url].


inequalities
algebra
polynomial
Vieta
trigonometry
absolute value

If $ x$ and $ y$ are real and $ x^2\plus{}y^2\equal{}1$, compute the maximum value of $ (x\plus{}y)^2$.

I get that $ (x\plus{}y)^2\equal{}x^2\plus{}2xy\plus{}y^2\equal{}1\plus{}2xy$.  Then, the solution says that $ \frac{x^2\plus{}y^2}{2}\geq\sqrt{x^2y^2}\equal{}xy$ by the Trivial Inequality.  How is this the Trivial Inequality?  I thought the Trivial Inequality was $ x^2\geq 0$.

We also see that if we call
s1 as x+y.
s2 as xy.
where x and y are real
by vietta we can construct a polynomial z^2+/-(x+y)z+xy.
we see that (x+y)^2-(x-y)^2=4s2.
so we see that the roots of this polynomial are real.
So all we care about is the discriminant must be positive.
so b^2-4ac=s1^2-4s2. So this is why this inequality is trivial.
You should learn about symmetric polynomials.

[quote="314.math"]If $ x$ and $ y$ are real ... Then, the solution says that $ \sqrt {x^2y^2} \equal{} xy$[/quote]This part of the solution is not entirely correct, there should be absolute value signs if they are working with reals, in general.

I'd make trigonometric substitution $ x\equal{}\sin \varphi, \ y\equal{}\cos \varphi$.

$ (\sin \varphi \plus{} \cos \varphi)^2\equal{}1\plus{}\sin 2\varphi \implies max\ (x\plus{}y)^2\equal{}2$

[quote="dgreenb801"]Yes, $ (x \minus{} y)^2 \ge 0 \implies\frac {x^2 \plus{} y^2}{2} \ge xy$[/quote]

How did you get this?

[quote="Flame"]I'd make trigonometric substitution $ x \equal{} \sin \varphi, \ y \equal{} \cos \varphi$.

$ (\sin \varphi \plus{} \cos \varphi)^2 \equal{} 1 \plus{} \sin 2\varphi \implies max\ (x \plus{} y)^2 \equal{} 2$[/quote]

Did you get that by using the fact that the sin of anything has to be either equal to or less than 1?

Yes. In the same manner you'd normally get minimum (though it's obvious in this case).

Sorry to respond before you, dgreenb801, but while I'm typing here...

$ (x\minus{}y)^2 \geq 0 \implies x^2\minus{}2xy\plus{}y^2 \geq 0 \implies x^2\plus{}y^2\geq 2xy \implies \frac{x^2\plus{}y^2}{2}\geq xy$

[quote="3.14.math"]Then, the solution says that $ \frac{x^2\plus{}y^2}{2}\geq\sqrt{x^2y^2}\equal{}xy$ by the Trivial Inequality. How is this the Trivial Inequality? I thought the Trivial Inequality was $ x^2\geq 0$.[/quote]

$ \frac{x^{2}\plus{}y^{2}}{2}\geq\sqrt{x^{2}y^{2}}\equal{}xy$
$ x^2\plus{}y^2 \ge 2xy$
$ x^2\plus{}y^2\minus{}2xy \ge 0$
$ (x\minus{}y)^2 \ge 0$
we can let $ x\minus{}y\equal{}a$ so $ a^2 \ge 0$ which is the trivial inequality.

@3.14 math it is actually supposed to be 2sinxcosx which has max of 1.
this is because 2sinxcosx=sin2x=which is maximized when x=45 which makes this equal to 1.[/hide]

You can use Cauchy–Schwarz for this problem.

[list]$ (x \plus{} y)^2\ \equal{} \ (1\cdot x \plus{} 1\cdot y)^2\ \leqslant\ (1^2 \plus{} 1^2)(x^2 \plus{} y^2)\ \equal{} \ 2$[/list]
Equality is attained when $ x \equal{} y \equal{} \frac {\sqrt2}2.$ Hence the maximum value is $ 2.$


I am typing up a report-like thingy that have sections and chapters, with theorems, lemmas, problems, etc.  The command \begin{theorem}, etc. from amsthm will automatically give me a number for the theorems.  However, it doesn't reset at a new section/chapter.  How can I do that??

I don't know about the details, since I just take the inequality packet from the AMC website by Kiran Kedlaya, and then delete the content and used the skeleton...

Put different sections in different files?

Define a custom command?

I assume you have a command like \newtheorem{theorem}{Theorem} in the preamble of your document.  Try changing it to \newtheorem{theorem}{Theorem}[section].

\setcounter{theorem}{0} resets the counter to 0 whenever you wish.


ok here it goes.
What is the remainder when 17+177+1777+...+177777777777777777777 is divided by 8. note that the last number has 21 digits; 1 1 and 20 7s.
i know i should use mod 8 but any solutions/strategies?

[hide="Solution"]
(Excuse me if I have some wrong notation, but you will hopefully understand what I mean).
\[ 17 + 177 + 1777 + \dots + 177777777777777777777\pmod{8} \\
\equiv \\
(17\pmod{8} + 177\pmod{8} + 1777\pmod{8} + \dots + 177777777777777777777\pmod{8})\pmod{8}\]
Now $ 17\pmod{8}$, $ 177\pmod{8}$, and $ 1777\pmod{8}$ are relatively easy to find:
\[ 17\equiv 1\pmod{8} \\
177\equiv 177 - 160\equiv 17\equiv 1\pmod{8} \\
1777\equiv 1777 - 1600\equiv 177\equiv 1\pmod{8}\]
For a number to be divisible by $ 8$, the final $ 3$ digits must be divisible by $ 8$.  Since $ 1777, 17777, \dots, 177777777777777777777$ all have the same final three digits $ (777)$, they will all have the same remainder.  Thus
\begin{align*} 17 + 177 + 1777 + \dots + 177777777777777777777 & \equiv \underbrace{1 + 1 + 1+ \ldots + 1}_{20} \\
& \equiv 20 \\
& \equiv\boxed{4}\pmod{8}\end{align*}
[/hide]

[hide="Solution"] Alright, we want mod 8, so that means we only care about the last 3 digits. We need to add:
$ 017 \plus{} 177 \plus{} 777 \plus{} 777\plus{} \dots \plus{} 777$

How many $ 777$'s is that? Well the last number ends in $ 20$ $ 7$'s, so since you dont count the first 2 numbers, there are $ 20 \minus{} 2 \equal{} 18$ $ 777$'s.
$ 777 \cdot 18 \plus{} 177 \plus{} 17 \equal{} 14180$. since $ 14176\equiv 0 \mod (8)$, $ 14180 \equiv 4 \mod (8)$

But dont take my word for it. Seriously, I am probably wrong.[/hide]

EDIT: Darnit beaten to the solution. But im right? Wow.

Edit: jxl's solution is pretty interesting

[hide="Alternatively"] Each number is the previous number times $ 10$ and then with a $ 7$ added.
We know that $ 17\equiv 1\pmod{8}$. So multiplying one times $ 10$ and then adding $ 7$ also gives $ 1\pmod{8}$. Since the same thing is done every time, all of the numbers are $ 1\pmod{8}$ so the answer is $ 20*1\equiv 4\pmod{8}$ or $ 4$.
[/hide]

Aha! i see!.
Thank you guys.
Mods u guys can lock.

lol Whenever I see your posts I think its cgyao with a new avatar.

woah!
lolz
i just realized my post rating hehe.
Make this not spammy, so here.
Thunder365, you messed up the latex.
it should be [code] \pmod 8[/code]


inequalities
geometry
incenter
circumcircle
geometry proposed

Prove that in any $\triangle  ABC$ at least one of the following two inequalities is true:

\[ \frac {_{9}}{^{4}}R^2 \plus{} 4r^2 \plus{} 7Rr\ge p^2 \ \ , \ \    9R^2 \minus{} 5r^2 \minus{} 2Rr\ge p^2 \]

Let $ I,O,G,N$ be the incenter, circumcenter, centroid and nine-point center of $\triangle ABC.$ $\overline{ IG}$ is an internal cevian of $ \triangle ION,$ since $G$ is between $ N$ and $O.$ Hence, either  $ IN \ge IG$ or $ IO \ge IG.$ Now, we use the well-known formulas

$IN \equal{} \frac {_1}{^2}R \minus{} r$

$IO \equal{} \sqrt {R^2 \minus{} 2Rr}$

$IG \equal{} \sqrt {\frac {_1}{^9}(p^2 \plus{} 5r^2 \minus{} 16Rr)}$

$IN \ge IG \Longrightarrow \  \frac {_1}{^2}R \minus{} r \ge \sqrt {\frac {_1}{^9}(p^2 \plus{} 5r^2 \minus{} 16Rr)} \Longrightarrow \ \frac {_9}{^4}R^2 \plus{} 4r^2 \plus{} 7Rr\ge p^2 $

$ IO \ge IG \Longrightarrow \ R^2 \minus{} 2Rr \ge \frac {_1}{^9}(p^2 \plus{} 5r^2 \minus{} 16Rr) \Longrightarrow \ 9R^2 \minus{} 5r^2 \minus{} 2Rr\ge p^2 $


Ok,i post my solution :)
From the condition we get $(a+b+c)abc \geq ab+bc+ca \Longrightarrow (a+b+c)(ab+bc+ca) \geq \frac{(ab+bc+ca)^2}{abc}=2(a+b+c)+\frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a} \geq 2(a+b+c)+3\sqrt[3]{abc}$(*)
we have two cases:
If $abc \geq 1$ then $\frac{3}{a+b+c}+2{abc} \leq 3$(because $abc \geq 1$ and $a+b+c \geq 3$)
So that $a+b+c \geq \frac{3}{a+b+c}+\frac{2}{abc}$
If $abc \leq 1$.From (*) $\Longrightarrow (a+b+c)(ab+bc+ca) \geq 3abc+2(a+b+c)$ (1)(because $abc \leq 1 \Longrightarrow \sqrt[3]{abc} \geq abc$)
$a+b+c \geq \frac{3}{a+b+c}+\frac{2}{abc} \Longleftrightarrow (a+b+c0^2abc \geq 3abc+2(a+b+c)$
Because $(a+b+c)abc \geq (ab+bc+ca) \Longrightarrow (a+b+c)^2abc \geq (a+b+c)(ab+bc+ca) \geq 3abc+2(a+b+c)$ (from (1))
So that the problem is proved :D  :D

My proof:
Since $\frac{abc(a+b+c)}{ab+ac+bc}\geq1$ then remain to prove that
$a+b+c\geq\frac{3}{a+b+c}\cdot\frac{abc(a+b+c)}{ab+ac+bc}+\frac{2}{abc}\cdot\left(\frac{abc(a+b+c)}{ab+ac+bc}\right)^2.$
But the last inequality equivalent to $\sum_{cyc}(a^3b^2+a^3c^2-2a^2b^2c)\geq0,$ which obviously true.

[quote="Cezar Lupu"]Let $a, b, c$ three positive real numbers such that $a+b+c\geq\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.
  Prove the following inequality

                                                  \[ a+b+c\geq\frac{3}{a+b+c}+\frac{2}{abc}. \]

  [i] Cezar Lupu, Valentin Vornicu [/i][/quote]
Let $a, b, c,x,y$  positive real numbers such that $a+b+c\geq\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.
  Prove the following inequality

                                                  \[ \frac{x+3y}{9}(a+b+c)\geq\frac{x}{a+b+c}+\frac{y}{abc}. \]


[quote="Ignite168"]Last year CT had a total of 0 perfect scores on both the 10 and the 12, hopefully we have some perfect ones this year.

I got a 138/144 on the 10 so no perfect here but I might win the State :)[/quote]

I got 144 on the 12.  Made mistake on #24.

I got 150 on the 10 last year. Does that count?

Only people taking it in CT count so no to Phelpedo and mathgeniuse^ln(x).

dkushnir: I'm a 9th grader.

[quote="spiderboy95"]I'm an eigth grader and got a 24 on the AMC 8. Boohoo, missed that last question!  :([/quote]

Was there seriously only one perfect AMC 8 in the whole state?!?!?!?  I couldn't take it. :(

Could you guys stop reviving old topics and instead make new ones?


trigonometry
geometry
incenter
trapezoid
geometric transformation
reflection
geometry solved

Let $ABC$ be a triangle. Let $A^{\prime}$ be the point at which the excircle of triangle $ABC$ across from $A$ touches the side $BC$; similarly define the points $B^{\prime}$ and $C^{\prime}$. Prove that the line through the point $A^{\prime}$ parallel to the bisector of the angle $CAB$, the line through the point $B^{\prime}$ parallel to the bisector of the angle $ABC$, and the line through the point $C^{\prime}$ parallel to the bisector of the angle $BCA$ are concurrent.

I think I have a proof, but it's rather computational. I'll post it a bit later. I just wanted to say that it's more efficient (I think) to look at it from this point of view:

Let $ABC$ be a triangle, and let $H_a,H_b,H_c$ be the feet of the altitudes from $A,B,C$ respectively. Let $X,Y,Z$ be the projections of $A,B,C$ on $H_bH_c,H_cH_a,H_aH_b$ respectively. Then the perpendiculars from $X,Y,Z$ to $BC,CA,AB$ (respectively) concur.

If $M,N,P$ are the projections of $X,Y,Z$ on $BC,CA,AB$ respectively, then the problem is equivalent to $f(M,N,P)=MB^2-MC^2+NC^2-NA^2+PA^2-PB^2=0$. All we need to do is show this, and since $f(H_a,H_b,H_c)=0$, after replacing $MB=H_aB\pm H_aM$ and the others it's not that hard. I'm trying to avoid writing it in detail because there are different cases according to the choice of signs (notice that $\pm$), but I think that in the end we get something like $\sum_{\hbox{cyc}}\sin(B-C)\cos A=0$, and this is clear after expanding $\sin(B-C)$.

I'll try to be more specific if nobody finds a nice geometric solution.

Once again an op02 problem. This time it's geo problem 164 = Russia 92, by L. Emelyanov. Here is my solution:

Let I be the incenter of triangle ABC; then we have to show that the parallels to the lines AI, BI, CI (i. e. to the angle bisectors of triangle ABC) through the points A', B', C' concur.

Let the incircle of triangle ABC touch the sides BC, CA, AB at X, Y, Z, and let D, E, F be the midpoints of these sides. If a, b, c are the sidelengths and s = (a+b+c)/2 is the semiperimeter of triangle ABC, then BX = s-b and CA' = s-b, so that BX = CA', and, consequently, the points X and A' are symmetric to each other with respect to the midpoint D of the side BC. In other words, the point D is the midpoint of the segment XA'.

Now, if H is the orthocenter of triangle XYZ, then XH is perpendicular to YZ. But YZ is perpendicular to AI (since the points Y and Z are symmetric to each other with respect to the line AI); so, XH is parallel to AI.

Since the points D, E, F are the midpoints of the sides BC, CA, AB of triangle ABC, the triangles DEF and ABC are homothetic. Thus, the angle bisectors of these triangles are parallel. Consequently, if S is the incenter of triangle DEF, then DS || AI. Now, if the parallel to AI through A' meets the line HS at V, then the quadrilateral XHVA' is a trapezoid with parallel bases XH and A'V (because both of these bases are parallel to AI). The line DS passes through the midpoint D of the leg XA' of this trapezoid and is parallel to the bases XH and A'V (because it is also parallel to AI); hence, this line is the midparallel of the trapezoid and must also pass through the midpoint of the other leg HV. In other words, the point S is the midpoint of HV; hence, V is the reflection of H in S. Hence, the parallel to AI through A' passes through the reflection of H in S. Similarly, the parallel to BI through B' and the parallel to CI through C' also pass through the reflection of H in S. Therefore, the three parallels concur at one point - namely, at the reflection of H in S. Problem solved.

  Darij


Determine the maximum possible real value of the number $ k$, such that
\[ (a \plus{} b \plus{} c)\left (\frac {1}{a \plus{} b} \plus{} \frac {1}{c \plus{} b} \plus{} \frac {1}{a \plus{} c} \minus{} k \right )\ge k\]
for all real numbers $ a,b,c\ge 0$ with $ a \plus{} b \plus{} c \equal{} ab \plus{} bc \plus{} ca$.

[quote="Ahiles"][color=brown]Determine the maximum possible real value of the number $ k$, such that
\[ (a \plus{} b \plus{} c)(\frac {1}{a \plus{} b} \plus{} \frac {1}{c \plus{} b} \plus{} \frac {1}{a \plus{} c} \minus{} k )\ge k
\]
for all real numbers $ a,b,c\ge 0$ with $ a \plus{} b \plus{} c \equal{} ab \plus{} bc \plus{} ca$[/color][/quote]
$ (a \plus{} b \plus{} c)\left(\frac {1}{a \plus{} b} \plus{} \frac {1}{c \plus{} b} \plus{} \frac {1}{a \plus{} c} \minus{} k \right)\ge k\Leftrightarrow\frac{(ab\plus{}ac\plus{}bc)(a\plus{}b\plus{}c)}{(a\plus{}b)(a\plus{}c)(b\plus{}c)}\geq k$
Thus, the answer is $ 1.$

let $ a\equal{}0,b\equal{}c\equal{}2$ so $ k \leq 1$.
$ a\plus{}b\plus{}c\equal{}ab\plus{}bc\plus{}ca$ so $ 1\equal{}\frac {ab\plus{}bc\plus{}ca}{a\plus{}b\plus{}c}$
$ \sum a .\sum \frac {1}{a\plus{}b} \geq k.(\sum a \plus{} \frac {\sum ab}{\sum a})$.
so
$ \sum ab.\frac {\sum a^2\plus{}3\sum ab}{\prod (a\plus{}b)} \geq k.\frac {\sum a^2\plus{}3\sum ab}{\sum a}$
so
$ \sum a.\sum ab \geq k.\prod (a\plus{}b)$
let k=1 so that
$ abc \geq 0$.
when $ a\equal{}0$ or $ b\equal{}0$ or $ c\equal{}0$
$ \sum a .(\sum \frac {1}{a\plus{}b} \minus{}1)\equal{}1$


Point $E$ is on side $\overline{AB}$ of square $ABCD$. If $\overline{EB}$ has length one and $\overline{EC}$ has length two, then the area of the square is

$\text{(A)}\ \sqrt{3}\qquad \text{(B)}\ \sqrt{5}\qquad \text{(C)}\ 3 \qquad \text{(D)}\ 2\sqrt{3}\qquad \text{(E)}\ 5$

$(C)$ :sleeping:  :sleep2: --Super easy.

[hide]
well first draw the square and label the points. You get a right triangle with hypotenuse 2 and 1 leg side length 3. Use the pyth theorem to get sqrt 3 as a side. That squared = 3. So $c$[/hide]


Use Newton's summation formulae or Wronski's forumula.

Bomb

[quote="sagar1331"]:( What is the sum of the sequence 1^4 + 2^4 ... n^4[/quote]
$1^4+2^4+...+n^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$.
You can prove this, using, for example, induction.

A generalization:
Let $m, n$ be two positive integers. Then
\[ \sum_{k = 0}^{n - 1} k^m = \frac{1}{m + 1} \sum_{k = 0}^m \left( \left(\begin{array}{c} m + 1\\ k \end{array}\right) \cdot B_k \cdot n^{m + 1 - k} \right)  \]
Where $B_k$ is the $k$-th Bernoulli number defined by the recurrence relation.
\[ \left\{\begin{array}{l} B_0 = 1\\ \sum^m_{j = 0} \left(\begin{array}{c} m + 1\\ j \end{array}\right) B_j = 0, %Error. "tmop" is a bad command.
{for} m \geqslant 1 \end{array}\right.  \]
The first $12$ values of Bernoulli numbers are
\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12\\ \hline B_n & 1 & - \frac{1}{2} & \frac{1}{6} & 0 & - \frac{1}{30} & 0 & \frac{1}{42} & 0 & - \frac{1}{30} & 0 & \frac{5}{66} & 0 & - \frac{691}{2730}\\ \hline \end{array}  \]
For more about Bernoulli numbers see: http://mathworld.wolfram.com/BernoulliNumber.html

There is the reccurence relation between $\sum_{i=1}^{j}i^j$, but there isn't common type of it(or all these formulas are very long and uncomfortable).


Are there any integer solutions to $x^{2}+y^{4}=z^{4}$. If there are any, they are clearly all even.

Pherma prove, that these equation had not nontrivial ($xy\not =0$) solutions.

Sorry, but can you post his prove? I know only equation $x^{4}+y^{4}=z^{2}$.

[quote="Rust"]Edvards. Last Fermat theorem.[/quote]
I don't undertand. What is ''Edvards''? I used dictionary but i can't understand it!

Can you explain better how Fermat's Last Theorem implies this fact?

N.T. Tuan right. Ferma prove, that equation $x^{4}+y^{4}=z^{2}$ had not nontrivial solutions. In book Edvards "Last Fermat's theorem" you can see about it.
But in book Ribenboum "Last Fermat's theorem" you can find, that these equation $x^{4}+y^{4}=z^{2}$ had not solutions in Gaus's integer numbers $a+bi$.

It had not nontrivial solution.
If (x,y,z) solution and $(x,y)=1$, then $z=a^{2}+b^{2}$, $x=Re(a+bi)^{4},y^{2}=Im(a+bi)^{4}$ or $x=Im(a+bi)^{4},y^{2}=Re(a+bi)^{4}$ were $(a,b)=1$.
In first case $y^{2}=4ab(a^{2}-b^{2})\to a=c^{2},b=d^{2}$ (because (a,b)=1), therefore we get another solution $y^{2}+d^{4}=c^{4}$ with $c<z=c^{4}+d^{4}$. It give contradion with minimal solutions.
Second case $y^{2}=(a^{2}-b^{2})^{2}-4a^{2}b^{2}$ give $\frac{k^{2}-1}{k^{2}+1}=r^{2}, k\in N, k>1, r\in Q$ and had not solution too.


function
integration
limit
real analysis
real analysis unsolved

Assume that the function $ f$ is defined on $ {(\minus{}\infty, \plus{}\infty)}$, measurable and satisfies following condition: $ \displaystyle\int^{\plus{}\infty}_{\minus{}\infty} |f(x)|^2dx<\plus{}\infty$. Show that \[ \displaystyle\lim_{\alpha\to \plus{}\infty}{\alpha}^2\mu\{x\in \mathbb{R}: |f(x)|>\alpha\}\equal{}0,\]
where $ \mu\{A\}$ - is $ Lebesque$ measure of the set $ A$.

Hint: Consider the sequence of functions $ g_{\alpha}(x)\equal{}\begin{cases}\alpha^2&|f(x)|>\alpha\\0&\text{otherwise}\end{cases}.$

(And to make it a sequence, replace $ \alpha$ by an arbitrary sequence $ \alpha_k\to\infty.$)

Note that $ 0\le g_{\alpha}\le |f|^2.$

Is there any convergence theorem that would help?


calculus
logarithms
complex numbers
imaginary numbers

I dont know where this goes, but it does require that you know somethings about calculus.

Find $x$ such that $x^i = i$

Hint: such an $x$ is real!!!

hmm, maybe this should be moved to a more serious subforum

If it needs calculus, the calculus computations and tutorials forum might be better.

very nice,  :D I  just find it amazing that you can have a real number $x$ such that $x^i = i$ Also, I think from that answer, you can figure out any other solutions. Care to post solution?

\[x^i = e^{i(\frac{\pi}{2}+2\pi k)}\]
\[{i \ln{x} = i (\frac{\pi}{2} + 2\pi k)}\]
\[\ln{x} = \frac{\pi}{2} + 2\pi k\]
\[x = e^{\frac{\pi}{2} + 2\pi k}\]
I'm not sure if it's okay to take the log of an imaginary number, but this is what I did.


A cop is patrolling a highway from a plane flying directly above the road, altitude 1 mile, on a course parallel to the road, at a constant speed of 120 mph. Radar aboard shows an oncoming car approaching the plane along the line of sight at 136 mph at a distance of 1.5 miles. Does the cop arrest the driver?

No calculus in High School forums!   Especially not Getting Started!  :(


I've seen this problem from IMTS (I forgot which round) and I wondered how would you do this.

Arrange the number from 1 through 97 such that the number adjacent to each other has absolute value of difference of 7 or 9 (another word, $|a_j - a_{j+1}| = |a_{j+1} - a_j| = 7 \text{or} 9$.

If the amount of numbers I have to arrange is small, I could use guess and trial but arranging 97 numbers is bit long for guess-and-trial.

How would you do this kind of problem? Thank you!

[quote="jhaussmann5"]Interesting problem.  It was round 21, by the way: [url=http://www.cms.math.ca/Competitions/IMTS/imts21.html]www.cms.math.ca/Competitions/IMTS/imts21.html[/url]

I first observed that the terms of the sequence must alternate with respect to parity.  So, I wrote the numbers 1, 3, 5, 7, ... in one row, and the numbers 2, 4, 6, 8, ... in another row below the first, and connected numbers which had a difference of 7 or 9.  Yeah, the graph gets a bit messy, but it helps to visualize what's going on.

I then notice that the number 1 only connects to two other numbers, namely 8 and 10.  Some other small numbers have the same property, so this helps to nail down the numbers at the left end of the graph.  For example, the sub-sequence 8, 1, 10 must appear, as well as 9, 2, 11, and so on.  It's possible that 1 or 2 is the first or last term of the sequence, but it's a start.  Playing around a little then gives you an answer.[/quote]

That's same problem I found them.

Hmm.. I got on the seeing "one number is connected to only two numbers" but I have little bit different thing. Like if the number is 21, there are actually four numbers that are connected: 28,30,14,12. So, this makes it harder.

 :(

That's why you have to play around; it's not obvious.

Here's another hint: does $a_1$ have to be odd or even?


show $x^{3}+3y^{3}=9z^{3}$ has no nontrivial solutions.

[hide="So do you mean nontrivial integer solutions?"]
If so x must be divisible by 3 so we can let x = 3a.

Then $27a^{3}+3y^{3}= 9z^{3}.$  Dividing each side by 3 we get:

$9a^{3}+y^{3}= 3z^{3}$

For the same reason as earlier, y must be divisible by 3, so we let y = 3b.

$9a^{3}+27b^{3}= 3z^{3}$ Dividing again by 3:

$3a^{3}+9b^{3}= z^{3}$

So z must be divisible by 3, so z = 3c.

$3a^{3}+9b^{3}= 27c^{3}$ Finally, dividing by 3 again, we get:

$a^{3}+3b^{3}= 9c^{3}$, which is like our previous equation. And in order to find nontrivial solutions, we will not reach a definite solution. So, there are no nontrivial integer solutions.[/hide]

Is this the method called "infinite descent"?


he diagonal of square I is a+b. The perimeter of square II wih twice the are of I is:

a. $ (a+b)^{2}$
b. $ \sqrt{2}$$ (a+b)^{2}$
c. $ 2(a+b)$
d. $ 4(a+b)$

i got:
[hide][b]D[/b]
so, for the first square, since the diagonal is lenght $ a+b$, a side lenght of the square is $ \frac{(a+b)\sqrt{2}}{2}$. so the area of this square is $ \frac{(a+b)^{2}}{2}$. the area of the other square is twice this, so its area is $ (a+b)^{2}$. a side length of this square is thus $ a+b$. so its perimeter is $ 4(a+b)$[/hide]

[hide="Answer"]The perimeter of square II is 4(a+b). To see this, follow the following steps:

1. From the general relation between the side of a square and its diagonal, write the relation between the side of square I and its diagonal.

2. From the relation between the areas, find the relation between the side of square II and the side of I.

3. Finally, the perimeter is 4 times the side of a square.
[/hide]

When using LaTeX, put curly braces around numbers.

[hide="Simplier Solution"]The diagonal is $ (a+b)$ so the area is $ \frac{1}{2}(a+b)^{2}$ so the area of the second square is $ (a+b)^{2}$ so the side is $ (a+b)$ so the perimeter is $ 4(a+b)$. $ \boxed{D}$[/hide]

[quote="abacadaea"]mihail, you dont need to find the areas, you just need the ratios of the sides.

[hide]length of a side of a square is $ /sqrt2(a+b)/2$ after we rationalize it. if the ratio of a squares side length is 1:x, then the ratio of the area is 1:x^2. in this case, x^2=2 so $ x=/sqrt2$. multiply $ /sqrt2(a+b)/2$ by $ /sqrt2$ and get a+b for the sidelength of square II. the perimater is 4 times the sidelength so [b]d[/b] is the answer[/hide][/quote]
You should use backslashes instead of slashes.
Corrected version:
[quote="abacadaea"]mihail, you dont need to find the areas, you just need the ratios of the sides.

[hide]length of a side of a square is $ \sqrt2(a+b)/2$ after we rationalize it. if the ratio of a squares side length is 1:x, then the ratio of the area is 1:x^2. in this case, $ x^{2}=2$ so $ x=\sqrt2$. multiply $ \sqrt2(a+b)/2$ by $ \sqrt2$ and get a+b for the sidelength of square II. the perimater is 4 times the sidelength so [b]d[/b] is the answer[/hide][/quote]

oh right and the 1:x^2 should be $ 1: x^{2}$


Let a,b,c be positive numbers such that: $ a \plus{} b \plus{} c \equal{} \sqrt [8]{2}$
Prove that:
$ \sqrt [8]{\frac {a^ 13}{b^3(a^2 \plus{} b^2)}} \plus{} \sqrt [8]{\frac {b^ 13}{c^3(b^2 \plus{} c^2)}} \plus{} \sqrt [8]{\frac {c^ 13}{a^3(c^2 \plus{} a^2)}}\geq\ 1$

It's obviously trues because by Cauchy Schwarts, we only need to prove:
$ \sum\ \sqrt [8]{a^3b^3(a^2 \plus{} b^2)} \leq\ \sqrt [8]{2} (a \plus{} b \plus{} c)$
(always trues by Am-Gm)

Also using Am-Gm, you can try with my old and harder problem:
Given $ a, b, c \geq\ 0$. Prove that:
$ \sqrt [9]{\frac {a^3}{b \plus{} c}} \plus{} \sqrt [9]{\frac {b^3}{c \plus{} a}} \plus{} \sqrt [9]{\frac {c^3}{a \plus{} b}} \geq\ \frac {1}{\sqrt [9]{2}}(\sqrt [9]{ab} \plus{} \sqrt [9]{bc} \plus{} \sqrt [9]{ca})$
 :)


floor function
calculus
integration
function

How many of the first 1000 positive integers can be expressed in the form
\[ \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor, \]
where $x$ is a real number, and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$?

[hide]To start with, let's let

If  is an integer, then , and we have that

for any integer . Even better, if one notices that each floor has an even number in it, then we have

this shows that any values we obtain in the range between  and  , then any other value in the range is just an integral multiple of 10 more than the values in that original range. What this basically says is that we need to check to see which values the function takes on in the interval between 0 and 10; all of these occur as  ranges from  to . Note also that a floor changes value whenever it increased by an integer; the occurs at all multiples of  for . All multiples of  covers multiples of  and , so we only need to test multiples of  and  that are less than ; these multiples are, in order:

which correspond to function values of

Thus, in every group of ten integers, six of them can be obtained by the function. Noting that  is made up of  groups of , then our answer is .[/hide]

[hide]
Note that every time $x$ reaches a multiple of $\frac{1}{24}$, the function's value possibly changes. So if we let $x = \frac{n}{24}$, where $n$ is an integer, the function will still have the same range. So we have

$f(n) = \lfloor \frac{n}{12} \rfloor + \lfloor \frac{n}{6} \rfloor + \lfloor \frac{n}{4} \rfloor + \lfloor \frac{n}{3} \rfloor$

This function increases by $1$ at every multiple of $3$, $4$, $6$, and $12$. When these multiples coincide, $f(n)$ skips some numbers. We can check to see that at every multiple of $6$, one number is skipped, and at every multiple of $12$, three numbers are skipped. So, every $12$ numbers, the function skips $4$ numbers. For the first $1000$ positive integers, we have $3 \leq n \leq 1200$. The function skips $1200 \cdot \frac{4}{12} = 400$ values. So it does not skip $1000 - 400 = 600$ values.
[/hide]

Alternatively, using Hermite's Identity,...

[hide="Solution"]
Recall that $ \sum_{k\equal{}0}^{n\minus{}1}\left\lfloor x\plus{}\frac kn\right\rfloor\equal{}\lfloor nx\rfloor$. Then we can rewrite $ \lfloor 2x \rfloor \plus{} \lfloor 4x \rfloor \plus{} \lfloor 6x \rfloor \plus{} \lfloor 8x \rfloor\equal{}4\lfloor x\rfloor\plus{}\left\lfloor x\plus{}\frac18\right\rfloor\plus{}\left\lfloor x\plus{}\frac16\right\rfloor\plus{}2\left\lfloor x\plus{}\frac14\right\rfloor\plus{}\left\lfloor x\plus{}\frac13\right\rfloor$
$ \plus{}\left\lfloor x\plus{}\frac38\right\rfloor\plus{}4\left\lfloor x\plus{}\frac12\right\rfloor\plus{}\left\lfloor x\plus{}\frac58\right\rfloor\plus{}\left\lfloor x\plus{}\frac23\right\rfloor\plus{}2\left\lfloor x\plus{}\frac34\right\rfloor\plus{}\left\lfloor x\plus{}\frac56\right\rfloor\plus{}\left\lfloor x\plus{}\frac78\right\rfloor$. There are $ 12$ terms here (we don't actually have to write all of it out; we can just see where there will be duplicates and subtract accordingly from $ 20$). Starting from every integer $ x$, we can keep adding to achieve one higher value for each of these terms, but after raising the last term, we will have raised the whole sum by $ 20$ while only achieving $ 12$ of those $ 20$ values. We can conveniently shift the $ 1000$ (since it can be achieved) to the position of the $ 0$ so that there are only complete cycles of $ 20$, and the answer is $ \frac{12}{20}\cdot1000\equal{}\boxed{600}$.
[/hide]

[hide="alternate solution"]
The function can change only when x is a multiple of $\frac{1}{24}$.  So we can write $x = \frac{n}{24}$ There would be 24 distinct values if $0 < x \le 1$ if it wasn't for "repeats".  We need to subtract the number of times the function stays the same.  This is when $\frac{n}{24}$ is not a multiple of $\frac{1}{6}$ or $\frac{1}{8}$  There are 6 times $\frac{n}{24}$ is not a multiple of $\frac{1}{6}$, 8 times $\frac{n}{24}$ is not a multiple of $\frac{1}{8}$, and 2 times $\frac{n}{24}$ is not a multiple of either.  So there are $6 + 8 -2 = 12$ times $\frac{n}{24}$ is not a multiple of $\frac{1}{6}$ or $\frac{1}{8}$ for $0 < x \le 1$.  Thus 12 distinct integers can be expressed with $0 < x \le 1$.

Since we are only looking for how many of the first 1000 positive integers can be expressed this way, we have $x \le 50$.  There must be 12 values of x in every range $k - 1 < x \le k$ where $k \in \mathbb{N}$  So $50 \cdot 12 = \boxed{600}$ positive integers less than 1000 can be expressed this way.[/hide]

Sorry for revive.

[quote="TheAmazingOne"]For the first $1000$ positive integers, we have $3 \leq n \leq 1200$. [/quote]

I still don't understand, why $3 \leq n \leq 1200$ ?

[quote=HookLengthFormula]Sorry for revive.

Does anyone know what is wrong in [hide = this]In this solution, $\{x\}$ will denote the fractional part of $x$, or $x - \lfloor x \rfloor$.

We can rewrite this expression as $$2x - \{2x\} + 4x - \{4x\} + 6x - \{6x\} + 8x - \{8x\} = 20x - \{2x\} - \{4x\} - \{6x\} - \{8x\}.$$Let $n = \{2x\} + \{4x\} + \{6x\} + \{8x\}.$ For each multiple of $20$, there are $5$ possible values, when $n = 0, 1, 2, 3, 4$. Thus, the answer should be around 200, but the AoPS Wiki says that it is $600$, which is way off. What is wrong with this reasoning? [/hide] solution? Thanks[/quote]

edit: oops sry misunderstanding

Can somebody tell me what's incorrect though?

[quote=HookLengthFormula]Sorry for revive.

Does anyone know what is wrong in [hide = this]In this solution, $\{x\}$ will denote the fractional part of $x$, or $x - \lfloor x \rfloor$.

We can rewrite this expression as $$2x - \{2x\} + 4x - \{4x\} + 6x - \{6x\} + 8x - \{8x\} = 20x - \{2x\} - \{4x\} - \{6x\} - \{8x\}.$$Let $n = \{2x\} + \{4x\} + \{6x\} + \{8x\}.$ For each multiple of $20$, there are $5$ possible values, when $n = 0, 1, 2, 3, 4$. Thus, the answer should be around 200, but the AoPS Wiki says that it is $600$, which is way off. What is wrong with this reasoning? [/hide] solution? Thanks

By the way, this is from itised. I got the permission from him to post this since I couldn't solve it.[/quote]

If you just plug in some values of $x$ you will see why

[quote=HookLengthFormula] For each multiple of $20$, there are $5$ possible values, when $n = 0, 1, 2, 3, 4$. [/quote] I think you should check this, there are more than just those. You have to look at the number of different fractions with denominators $2,4,6,8$. Each time you consider an $x$ between two of these fractions the value of $\lfloor 2x \rfloor+\lfloor 4x \rfloor+\lfloor 6x \rfloor+\lfloor 8x \rfloor.$ [hide=...]Why would the official AoPS solution be wrong? The solution is perfectly fine[hide=.]I've done this problem before so I know. :) [/hide][/hide]

SOLUTION 1
We can experiment with all numbers less than or equal to one in the form $\frac{m}{24}$: $\frac{1}{24}, \frac{2}{24} \to 0$, $\frac{3}{24} \to 1, \frac{4}{24}$, $\frac{5}{24} \to 2$, $\frac{6}{24}$, $\frac{7}{24} \to 4$, $\frac{8}{24} \to 5$, $\frac{9}{24}$, $\frac{10}{24}$, $\frac{11}{24} \to 6$, $\frac{12}{24}$, $\frac{13}{24}$, $\frac{14}{24} \to 10$, $\frac{15}{24} \to 11$, $\frac{16}{24},\frac{17}{24} \to 12$,$\frac{18}{24}, \frac{19}{24} \to 14$, $\frac{20}{24}\to 15$, $\frac{21}{24}, \frac{22}{24}, \frac{23}{24} \to16$, $\frac{24}{24} \to 20$. We can easily that that $12$ of $20$ numbers can be achieved. Extending this, of the first $1000$ numbers, there are $\frac{12}{20} \cdot 1000 = \boxed{600}$ possible values.

SOLUTION 2
Imagine that we increase $x$ from $0$ to $1$ as the values goes from $0$ to $2+4+6+8=20$. We seek to find how many integers between $1$ and $20$ we skipped. We skip whenever $2x$, $4x$, $6x$, and $8x$ become integers at simultaneously.

When $x=1/2$ and $x=1$, all four values become integers at the same time, so skip three integers, and when $x=1/4$ and $x=3/4$, the values $4x$ and $8x$ become integers at the same time, so we skip one integer for both.

In total we skip $3+3+1+1=8$ of the first $20$ integers. Extending this, of the first $1000$ numbers, there are $\frac{12}{20} \cdot 1000 = \boxed{600}$ values.

I found that all numbers $1 - 16, 20 - 36, 40 - 56, \ldots, 980-996, 1000$ could be made, giving $16+17 \cdot 49 + 1 = 850$. Why is this wrong? I got this by noting if $p \leq x \leq p+1$, then the minimum is at $20p$ and the maximum is at $2(p+1)-1+4(p+1)-1+6(p+1)-1+8(p+1) - 1 = 20p+16$.

For any $x \in [k, k+1), 2x \in [2k, 2k+2), 4x \in [4k, 4k+4), 6x \in [6k, 6k+6), 8x \in [8k, 8k+8)$.
Obviously $20k \le \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor \le (2k+1)+(4k+3)+(6k+5)+(8k+7)=20k+16$
We can evenly divide $[k, k+1)$ into $2$ subsets, $4$ subsets, $6$ subsets, and $8$ subsets, respectively.
We can also evenly divide $[k, k+1)$ into $\text{gcd}(2, 4, 6, 8)=24$ subsets.
When $x \in [k+\frac{5}{24}, k+\frac{6}{24}) \rightarrow [k+\frac{6}{24}, k+\frac{7}{24})$, both $\lfloor 4x \rfloor$ and $\lfloor 8x \rfloor$ increase by $1$, thus $3$ is skipped.
When $x \in [k+\frac{11}{24}, k+\frac{12}{24}) \rightarrow [k+\frac{12}{24}, k+\frac{13}{24})$, all of $\lfloor 2x \rfloor$, $\lfloor 4x \rfloor$ and $\lfloor 8x \rfloor$ increase by $1$, thus $7, 8, 9$ are skipped.
When $x \in [k+\frac{17}{24}, k+\frac{18}{24}) \rightarrow [k+\frac{18}{24}, k+\frac{19}{24})$, both $\lfloor 4x \rfloor$ and $\lfloor 8x \rfloor$ increase by $1$, thus $13$ is skipped.
Thus, \[ \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor = 20k + \ell, \ell \in \{0,1, 2, 4, 5, 6, 10, 11, 12, 14, 15, 16\}\]
The total counts shall be $\frac{1000}{20}\cdot {12} = \boxed {600}$.

p6: Every time one of these floors increases from one number to the next, we get a new number. Hence we want to count the amount of total "jumps." There are $8 \cdot 50 = 400$ for the final floor, $300$ for the third, $200$ for the second, and $100$ for the first, which are in the form $\frac{n}{8}, \frac{n}{6}, \frac{n}{4}, \frac{n}{2}$ respectively. But $\frac{n}{2}$ and $\frac{n}{4}$ are already counted, so exclude them. For the ones common to our two remaining options, we must have common denominator $2$, so the amount of ways is $400 \cdot \frac{2}{8} = 300 \cdot \frac{2}{6} = 100$. The answer is $400+300-100=\boxed{600}$.

My solution, was that for the last floor we have $400$ jumps, and for the third, $300$, second $200$, first $100$, so this is just $\boxed{600}.$ By some counting.


abstract algebra
function
group theory
superior algebra
superior algebra solved

Consider the additive group Z sub 12 and the mapping f: Zsub12->Zsub 12 given by f([x])=[4x]. Prove that f is a group homomorphism and describe ker(f).

Looks like a homework problem. liltubbs25, can you tell us what you've already tried?

I've got one very simple-minded suggestion. Since this is a very small finite group, how about listing the full table of all of the values of $f$ for all possible arguments. And just what is the definition of $\ker(f)$, anyway?

The kernel of f is the set of all elements of G that map to the identity in H. The set is denoted Ker(f) and ker(f)={x :varepsilon: |1 sub H}

We're dealing in a case of $f: G\mapsto G$ where $G=\mathbb{Z}_{12}.$ We're writing the group operation as addition. When we do that, we call the identity $0,$ and, in fact, we write the identity in $\mathbb{Z}_{12}$ as $0$ (or as $[0]$ if you would prefer - I wouldn't prefer.)

That definition of kernel was a little mechanical. How about adapting it to the case at hand and defining it this way:

$\ker(f)=\{x\in G : f(x)=0\}.$

Now, you tell me what the kernel is in this case.

[quote="liltubbs25"]-4x?[/quote]
The kernel is a set. Your answer is nonsense.

where does f([x])=[4x]comeinto the problem?

Since this is a small group why dont you use the group table.It will not only help you in this problem but also a variety of problems you will solve later on as it helps you to visualise the the situation at hand.....

... such that f(x)=0. After all, 0 is the identity of this group.


as far as I know, brian = 32, sherry and alex = 28 (alex should have gotten like 35 but whatev)

who got the 26? (is it TEDDYYYYYYYYYYYYYYYYYYYYY)

me = 25

two other 25s, two 24s, three 23s

five 22s - any chance of them being invited?

Eh I believe that Teddyyyyyyyyyyyyyyyyyyyyyyyyyyyy got 25 also. Not sure who got the 26. Oh well, I don't really have much to say, except that: The race is very close@@

there is also a 28 you are not accounting for.

I got a $23$. And for some reason Haitao wanted me to edit what was here before, so I'll just say...

THE RACE IS VERY CLOSE

Very good Jacob. The race is very close.

OH MAN GUYS GUYS IF WE HAVE A 3-DAY TST FOR TIEBREAKERS SINCE THE RACE IS VERY CLOSE IT WILL TOTALLY MESS UP THE FLIGHT SCHEDULE SO DON'T TIE OK

[quote="diophantient"][quote="mysmartmouth"]So the TST is like the next level up from the USAMO?  When was it given, and to whom?[/quote]
The TST is given at the end of MOP to the winners (and maybe HM?).
To decide who goes to IMO, the top six scores of TST+USAMO are taken.[/quote]

This year, the TST will be held much earlier. It will be held right after the USAMO ceremony in Washington, in about a week or so. Good luck to those who are taking it.  :lol:

[quote="billzhao"]This year, the TST will be held much earlier. It will be held right after the USAMO ceremony in Washington, in about a week or so. [/quote]

... So what's the point of going to MOP?  I was under the impression that it was to train for the TST, but if the TST is being held before MOP, then couldn't people who aren't going to MOP but may be able to take the TST take it?

The majority of MOP has always been people with no chance at the IMO team- that year. Having a TST at all is actually a recent development.
For those with a chance at the team, MOP is preparation for the IMO itself. For everyone else, it's preparation for future contests.

Under the system from my years, this year's team would consist of the four students with scores $\ge 26$, and two of the three with scores of 25. If one of those 7 declined the offer, the team would be set; I don't know what the tiebreaker procedure would be otherwise. One or both of the students with scores of 24 might also have been invited to MOP as alternates.

One issue with the early TST- what does a senior who doesn't make the team do?

I kinda doubt that they'd kick them out of MOP.

they will be invited to MOP, not sure if they'd feel like going

and I don't think anyone particularly wanted the early TST, but the Vietnam IMO people want the team members' names early

I don't think MOP has ever been designed to train for the TST... it's usually held right before MOP, even if it wasn't held at the early date it's being given on this year.

MOP instead develops problem-solving and proof-writing skills for the IMO and future USAMOs

[quote="Phelpedo"]Also, isn't MOP (for Black Group) a chance to learn last-minute "brute force" techniques like dumbassing and Lagrange Multipliers?[/quote]
Hmmm darn if that is the case then I sort of wasn't paying attention last year... however this is not currently very relevant as the race is very close.

I would hope blacks already know standard dumbassing techniques and Lagrange Multipliers... Wow, there are such huge misconceptions about MOSP out there...

The TST is usually at the beginning. It is not about preparing for anything, except for the six IMO team members.


Prove that a graph is bipartite if and only if every inducted cycle has a pair number for length.

What do you mean by inducted cycle and pair number?  Do you mean cycle and even number?

yes I mean cycle and even number. English is not my native language. :blush:

Taking any vertex in the cycle as a starting point. Suppose that this vertex is in set A, then the next vertex in the cycle is in set B and the following one in set A. In general every alternate vertex in the cycle is in set A. A cycle is a closed path so, eventually, the path leads back to the starting vertex. This vertex is in set A so the path length must be 2, 4, 6, 8, etc, ie the cycle has an even number of edges.

You correctly proved that a bipartite graph can have only cycles of even length, but you haven't proved the converse (that graphs with cycles of even length are bipartite).  This is a classical result of graph theory and it has doubtless appeared on the forum before.


function
inequalities proposed
inequalities

Let $\Delta$ denote the set of all triangles in a plane. Consider the function $f: \Delta\to(0,\infty)$ defined by $f(ABC) = \min \left( \dfrac ba, \dfrac cb \right)$, for any triangle $ABC$ with $BC=a\leq CA=b\leq AB = c$. Find the set of values of $f$.

is it $[1,\frac{\sqrt{5}+1}{2}]$? I just find its maximun and minimum.

$\frac{b}{a}=m \; , \; \frac{c}{b}=n$

$a \leq b \leq c \; \Longrightarrow \; m,n \geq 1$. Equality holds when triangle [b]ABC[/b] is equilateral.

$a+b>c \; \Longleftrightarrow \; m+1>mn$

$i) \; m \leq n \; \Longrightarrow \; m \leq n < 1+\frac{1}{m} \; \Longrightarrow \; min\{m,n\}=m<\frac{\sqrt{5}+1}{2}$

$ii) \; n \leq m \; \Longrightarrow \; n \leq m < \frac{1}{n-1} \; \Longrightarrow \; min\{m,n\}=n< \frac{\sqrt{5}+1}{2}$

We can easily show that there are positive reals $m,n$ such that $min\{m,n\}=\frac{\sqrt{5}+1}{2}-\epsilon$ for all $0<\epsilon<\frac{\sqrt{5}-1}{2}$ and hence the answer is $\left[1,\frac{\sqrt{5}+1}{2}\right)$

hi
Solution is equal to (1,(1+√5)/2) because
Let a triangle in a plane with a≥b≥c arbitrary. now we fix a and c and change the value of b. so we get a continues function of b that set of value of that determine with min{a/b,b/c}.clearly this function is bounded so it have suprimum  wherever a/b=b/c or equivalent b^2=ac.
Hence we find the state that satisfying 1. b+c>a (triangle rule),  2. b^2=ac ,3. a≥b≥c>0
Equivalent
bc+c^2>ac=b^2
Equivalent
(b/c)^2-b/c-1<0
So ,(1-√5)/2<b/c=a/b<(1+√5)/2 .but by 3 we get b/c>1 and a/b>1


inequalities
number theory
relatively prime

Determine all positive intergers $ n$ such that $ n$ is divisible by any positive integer $ m$ which satisfies $ m^2\plus{}4<\equal{}n$.

*bump*

no one seems to be getting it.... So here is the answer, please post the solution.  Thanks.  :blush:

[hide="answers"]$ 1,2,3,4,5,6,7,8,10,12,18,24$[/hide]

Seven hours doesn't constitute a bump.  The AoPS forums don't work that quickly; be patient.

I will show that any such $ n$ is at most 39.

[hide="Solution"]Let $ n$ be any integer that is at least 13 and that satisfies the given condition.  Let $ M$ be the largest integer such that $ M^2 \plus{} 4 \le n$.  Because $ n \ge 13$, we have $ M \ge 3$.  By the given conditions, we know that $ n$ is divisible by $ M \minus{} 2$, $ M \minus{} 1$, and $ M$.  We know that $ M \minus{} 1$ is relatively prime to $ M \minus{} 2$ and $ M$; also $ M \minus{} 2$ and $ M$ share at most a factor of 2.  So $ n$ is divisible by $ \frac{M(M\minus{}1)(M\minus{}2)}{2}$.  In particular, $ n$ is at least $ \frac{M(M\minus{}1)(M\minus{}2)}{2}$.

On the other hand, because $ M$ is largest, we know that $ (M \plus{} 1)^2 \plus{} 4 > n$.  Combining that inequality with what we learned from the previous paragraph, we get
\[ \frac{M(M\minus{}1)(M\minus{}2)}{2} < (M \plus{} 1)^2 \plus{} 4.
\]
Solving that inequality, we find that $ M \le 5$.  Thus $ n$ is less than $ 6^2 \plus{} 4$, which is 40.[/hide]

Out of curiosity, what is AHSMC?

t0rajir0u- sorry.  :oops:

Ravi B- Alberta High School Mathematics Competition.


trigonometry
logarithms
quadratics
algebra
quadratic formula

(1) Solve the system of equatioin.
$\begin{cases}\log_{2x}y+\log_{x}{2y}=1\\ \log_{2}{xy}=1 \end{cases}$

(2) Find the values of $\sin \theta ,\ \cos \theta$ such that $\sin^{3}\theta+\cos^{3}\theta =\frac{13}{27}\ (90^\circ <\theta <180^\circ).$

[hide="Hint for 1"] Substitute $a = \log_{2}x, b = \log_{2}y$ and do some base-changing. [/hide]

I was thinking about that, but I got nowhere :/

EDIT
[hide]
Let $a = \log_{2}y$, and $b=\log_{2}x$

Using t0rajir0u's hint, we get $\frac{a}{b+1}+\frac{a+1}{b}= 1$

Replacing the 1's in the fraction with $a+b$, we get:

$\frac{a}{a+2b}+\frac{2a+b}{b}= 1 \rightarrow \frac{a}{a+2b}+\frac{2a}{b}= 0 \rightarrow ab+2a^{2}+4ab = 0 \rightarrow (a)(2a+5b)=0$.

When $a=0$, $y=1$. So, $x=2$.

$a=-\frac{5}{2}b$.

Solving for $x$ and $y$ gives us $\boxed{(2^{\frac{-2}{3}}, 2^{\frac{5}{3}}), (2,1)}$.[/hide]

I see now.  I originally defined my $a$ and $b$ different from your $a$ and $b$ :P

Well, here goes nothing for #2

I brute forced the end; I didn't see any elegant way of solving it :(

[hide="2"]
Let $x = \sin \theta$, and $y = \cos \theta$

$x^{3}+y^{3}= \frac{13}{27}$
$x^{2}+y^{2}= 1$

Let $x+y = a$

Notice, $-1\le a\le 1$

$(x^{2}+y^{2})(x+y) = a$
$x^{3}+y^{3}+xy^{2}+yx^{2}= a$
$\frac{13}{27}+\left(\frac{a^{2}-1}{2}\right)(a) = a$
$a^{3}-3a+\frac{26}{27}= 0$

We notice that $a=\frac{1}{3}$ is a root, so divide and use quadratic formula to get $a=\frac{\sqrt{105}-1}{6}, \frac{-\sqrt{105}-1}{6}$

However, we notice that the last two solutions are too big in the  interval for $90\le \theta\le 180$.

Next, to solve for $x$ and $y$, we need to find $x-y$, so $(x-y)^{2}= x^{2}-2xy+y^{2}= 1-2\cdot \frac{a^{2}-1}{2}\rightarrow 2-a^{2}$

So, $\sin \theta = \frac{\sqrt{17}+1}{6}, \cos \theta = \frac{1-\sqrt{17}}{6}$[/hide]


analytic geometry
graphing lines
slope
function
vector
real analysis
real analysis unsolved

A surface is given by $ z\equal{}f(x,y)$, where $ f(x,y)\equal{}xy$. A person stands at the point $ (3,\minus{}4)$ and wishes to walk in the direction of the greatest descent. What is the direction $ \vec{v}$ in the plane in which he will start walking and what is the slope of the surface in this direction?

Steepest descent is given by the gradient of the function.

What function? $ F(x,y,z): \equal{}f(x,y)\minus{}z$ ?

Well the gradient is $ \nabla f\equal{}(y,x)$.

But the person stands in the plane. The direction he wishes to walk should obviously be 3-dimensional, not 2-dimensional.

No, we want a direction "in the plane". That means the horizontal projection of the actual path.

Well, that's just me copying the question wrong. The original formulation asks for a vector $ \vec{v} \in \mathbb{R}^3$.


(i)Let S be the universal set and n(X)=k.Find the probability of selecting

    two subsets A and B of the set X such that B=A'.

(ii)Two subsets A and B of a set S consisting of 'n' elements are constructed

    randomly.Find the probability that A and B are mutually exclusive and

    exhaustive.

(iii)Two subsets A and B of a set S consisting of 'n' elements are constructed

    randomly.Find the probability that A is a subset of B.

(iv)A five digit number is written down at random.Find the probability that the

      number is divisible by 5 and no two consecutive digits are identical

1)Let A have 0 elements.This means that B has k elements
For n(A)=1,no of ways=k
For n(A)=2,no of ways=kC2
and so on........
We need'nt select anything for B,coz everything apart from A is automatically part of B

So P=2^k/(2^k)^2=1/2^k

2)I think this one is the same as 1)

3)Let A have 1 element.B can now have 0,1..........n-1 elements from those apartf from the one chosen for A.So no of ways=nC1*2^n-1
 So P=sum(i=1 to n)(nCi*2^n-i)/2^2k
Don't know bout summin up the series :D

4)No is divisible by 5=>last digit=5 or 0
Second last can have 9 values,ditto for the next two from the last.
First digit can have 8 values only,discounting 0
So P=81/625

Hi ashmant_wiz07,

1)Since we are selecting two subsets,shouldn't the number of

ways to do this be (2^k)C(2)=(2^k)((2^k)-1)/2 ,also the

favourable number of ways should be (2^k)/2=2^(k-1).

This appears to be so  because if you select any one set from

2^k then the other is automatically selected,so we need to

divide by 2 to avoid double counting.therefore the correct

answer should be (2^(k-1))/((2^k)((2^k)-1)/2)=1/(2^(k-1)-1)

4)you have missed the obvious situation that if the digit in the

middle is same as that at the unit's place then there will be 9

possibilities for the digit at the tens place i.e for the second

 digit from the left.Now this will give rise to more favourable

 cases than what you have calculated.


Kind of like basic counting except it is counting happy numbers.

Happy numbers:A number where you square each digit of the number add them up, and keep doing it, it is going to get to 1. For example: 13=1^2+3^2=10=1^2+0^2=1. Thus, 19 is a happy number. I will start.

1

The only way you know for sure it isn't a happy number is if it cycles, like 2 and 4.

Next one is:

7  :)

[quote="PowerOfPi"]Next is 46.[/quote]
Sorry, $ 46$ is not happy. :noo:

Next happy number after $ 49$ is $ 68.$

Happy numbers from 10,000 to 15,000  :) :
10000 is a Happy number
10003 is a Happy number
10009 is a Happy number
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14494 is a Happy number
14495 is a Happy number
14507 is a Happy number
14511 is a Happy number
14515 is a Happy number
14516 is a Happy number
14526 is a Happy number
14537 is a Happy number
14546 is a Happy number
14549 is a Happy number
14551 is a Happy number
14556 is a Happy number
14561 is a Happy number
14562 is a Happy number
14564 is a Happy number
14565 is a Happy number
14570 is a Happy number
14573 is a Happy number
14594 is a Happy number
14614 is a Happy number
14615 is a Happy number
14617 is a Happy number
14625 is a Happy number
14641 is a Happy number
14645 is a Happy number
14648 is a Happy number
14651 is a Happy number
14652 is a Happy number
14654 is a Happy number
14655 is a Happy number
14671 is a Happy number
14684 is a Happy number
14702 is a Happy number
14704 is a Happy number
14705 is a Happy number
14708 is a Happy number
14711 is a Happy number
14716 is a Happy number
14720 is a Happy number
14723 is a Happy number
14724 is a Happy number
14732 is a Happy number
14734 is a Happy number
14735 is a Happy number
14738 is a Happy number
14740 is a Happy number
14742 is a Happy number
14743 is a Happy number
14750 is a Happy number
14753 is a Happy number
14761 is a Happy number
14780 is a Happy number
14783 is a Happy number
14801 is a Happy number
14804 is a Happy number
14807 is a Happy number
14810 is a Happy number
14812 is a Happy number
14813 is a Happy number
14821 is a Happy number
14823 is a Happy number
14831 is a Happy number
14832 is a Happy number
14837 is a Happy number
14840 is a Happy number
14846 is a Happy number
14864 is a Happy number
14870 is a Happy number
14873 is a Happy number
14889 is a Happy number
14898 is a Happy number
14911 is a Happy number
14912 is a Happy number
14921 is a Happy number
14939 is a Happy number
14944 is a Happy number
14945 is a Happy number
14954 is a Happy number
14988 is a Happy number
14993 is a Happy number

I,ve created a java program:
// Java program to check a number is a Happy
// number or not
import java.util.Scanner;
class GFG {

// Utility method to return sum of square of
// digit of n
static int numSquareSum(int n)
{
	int squareSum = 0;
	while (n!= 0)
	{
		squareSum += (n % 10) * (n % 10);
		n /= 10;
	}
	return squareSum;
}

// method return true if n is Happy number
static boolean isHappynumber(int n)
{
	int slow, fast;

	// initialize slow and fast by n
	slow = fast = n;
	do
	{
		// move slow number
		// by one iteration
		slow = numSquareSum(slow);

		// move fast number
		// by two iteration
		fast = numSquareSum(numSquareSum(fast));

	}
	while (slow != fast);

	// if both number meet at 1,
	// then return true
	return (slow == 1);
}

// Driver code to test above methods
public static void main(String[] args)
{
	Scanner in=new Scanner(System.in);
	System.out.println("Strating and ending NO.");
	int s=in.nextInt();
	int e=in.nextInt();
	for(int i=s;i<=e;i++){
	if (isHappynumber(i))
		System.out.println(i +" is a Happy number");
}
}
}

No more?
[img] https://cdn.fansshare.com/photo/xwallpapers/xerneas-pokemon-and-872210946.jpg[/img]


Prove that a prime of the form $2^{2^n}+1$ cannot be the difference of two fifth powers of two   positive integers. :)

By $x^5-y^5=(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)$ you get that this two integers must have difference one (otherwise it's not prime).
But then $x^5-y^5 \equiv x-y \equiv 1 \mod 5$ which can't be, because it would imply that $2^n \equiv 0 \mod 5$ which is absurd.


Do you guys intend to offer an Intermediate Geometry class in Fall 2005?

Currently, we do plan to offer Intermediate Geometry at that time.  It might even be ready by the summer time, though in either case, we can't make any guarantees since we never know how we will need to spend our time in the upcoming months.


In the sequence $ 2001, 2002, 2003, \ldots$, each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is $ 2001 \plus{} 2002 \minus{} 2003 \equal{} 2000$. What is the $ 2004^\text{th}$ term in this sequence?

$ \textbf{(A)} \minus{} \! 2004 \qquad \textbf{(B)} \minus{} \! 2 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 4003 \qquad \textbf{(E)}\ 6007$


Determine the least possible value of the largest term in an arithmetic progression of seven distinct primes.

It's Jugoslavia 96, I remember that the largest term was less than $1000$ and also remember using divisibility $mod 7$.


inequalities function integration calculus limit geometry logarithms

Could anyone please help me prove Young's inequality?

Problem: If $ f: \left[0,A\right]\longrightarrow\mathbb{R}$ is a continuous and a strictly increasing function satisfying $ f(0) \equal{} 0$, then for every positive $ 0 < a\leq A$ and $ 0 < b\leq f(A)$ holds that $ \int^{a}_{0}f(t)dt \plus{} \int^{b}_{0}f^{ \minus{} 1}(t)dt\geq ab$

[quote="JRav"][url=http://en.wikipedia.org/wiki/Young%27s_inequality#Proof_of_the_standard_form]Proof of Young's Inequality[/url].[/quote]

The above link proves the standard inequality, using the transformation in the integrals for $ f(t)\equal{}t^{p\minus{}1}$. But what happens when $ f(t)\neq t^{p\minus{}1}$ :?:

Actually,for $ f: [a,b] \to [c,d]$ continuous and strictly increasing we have $ \int_a^b f(x)dx\plus{}\int _c^d f^{\minus{}1}(x)dx\equal{}bd\minus{}ac$.

iasu Γιώργος .
I never have seen the solution for this problem (I know only the case, when $ f = t^{p - 1}$  ) but I think I have a idea
If you will  have any question please ask me and I'll explain you.
I assume that $ A = 1$ it's not hard to verify for the case $ A\neq 1$, I believe you can do it yourself.
we need to show that
\[ \int_{0}^{a}f(t)dt + \int_{0}^{b}f^{ - 1}(t)dt \geq ab
\]
where $ 0\leq b \leq f(1)$ and $ 0 \leq a \leq 1$ and $ f$ is a continuous and a strictly increasing function satisfying $ f(0) = 0$
 there exist integers $ M(n), N(n)$ such that $ \lim \frac {M(n)}{n} = a$ and $ \lim f( \frac {N(n)}{n} ) = b$
since $ \mathbb{Q}$ is dense on $ \mathbb{R}$.
It's well-known  that $ f$ can be approximated sufficiently good via the function $ f_{n}(t) = \sum_{k = 1}^{n} f( \frac {k - 1}{n}) \chi_{[(k - 1)/n, k/n]}(t)$  as $ n \to \infty$ then note that $ f^{ - 1}$ can be approximated sufficiently good via the function $ f^{ - 1}_{n}(t) = \sum_{k = 1}^{n} \frac {k - 1}{n} \chi_{[f((k - 1)/n), f(k/n)]}(t)$  as $ n \to \infty$ since $ f$ is strictly increasing continuous.
(remark: "sufficiently good" here means $ f_{n}$ converge to $ f$ in $ C([0,1])$)
indeed by Cantor theorem $ f$ is uniform continuous on $ [0,1]$ therefore  if $ |x - y|\leq \frac {1}{n}$ then $ |f(x) - f(y)|\leq \varepsilon_{n}$ where $ \varepsilon_{n} \to 0$ as $ n \to \infty$ hence $ |f_{n}(t) - f(t)| \leq \varepsilon_{n}$ for all $ t \in [0,1]$
therefore $ |f_{n}(t) - f(t) |_{C[0,1]} \leq \varepsilon_{n} \to 0$ as $ n \to \infty$

Thus :
\[ \int_{0}^{a}f(t)dt + \int_{0}^{b}f^{ - 1}(t)dt = \lim \int_{0}^{M(n)/n}\sum_{k = 1}^{n} f( \frac {k - 1}{n}) \chi_{[(k - 1)/n, k/n]}(t) dt + \int_{0}^{f(M(n)/n)}\sum_{k = 1}^{n} \frac {k - 1}{n} \chi_{[f((k - 1)/n), f(k/n)]}(t) dt = \lim \sum_{k = 1}^{M(n)} f( \frac {k - 1}{n}) \frac {1}{n} + \sum_{k = 1}^{N(n)} \frac {k - 1}{n} (f( \frac {k}{n}) - f( \frac {k - 1}{n} ) ) = \lim \sum_{k = 1}^{M(n)} f( \frac {k - 1}{n}) \frac {1}{n} + \frac {k - 1}{n}f( \frac {k}{n} )|_{1}^{N(n)}  - \sum_{k = 1}^{N(n)}\frac {1}{n}f(\frac {k - 1}{n} )
\]
Now suppose $ M(n) > N(n)$ case ($ M(n) < N(n)$ considered similary)
then we have :
\[ = \lim \sum_{k = N(n) + 1}^{M(n)}\frac {1}{n} f( \frac {k - 1}{n} ) + \frac {N(n) -1}{n} f( \frac {N(n)}{n} ) \geq \lim \sum_{k = N(n) + 1}^{M(n)}\frac {1}{n} f( \frac {N(n)}{n} ) + \frac {N(n)-1}{n} f( \frac {N(n) }{n} ) = \lim \frac {M(n) - N(n) - 1}{n} f( \frac {N(n)}{n} ) + \frac {N(n) - 1}{n} f( \frac {N(n) }{n} ) = \lim \frac {M(n)}{n}f( \frac {N(n)}{n} ) + \mathcal{O}( \frac {f(N(n)/n)}{n} ) = ab
\]
since $ b$ is bounded and $ f$ is increasing .
if you understood everything well, then try to consider case $ N(n) > M(n)$

Comparison of the area of some figure, $ ab$ means the area of a rectangular.

Yes, of course , this problem has geometrical interpretation.
For example.
$ \int_{0}^{b}f(t)dt$ this is green area on the picture.
$ \int_{0}^{b} f^{\minus{}1}(t)dt$ this is yellow area on the picture.
and $ ab$ this is area of the rectangular $ OaFb$.
we see that $ S_{green}\plus{} S_{yellow} \geq S_{OaFb}$

No questions Extremal, thank you very much :)

Let $ a,b$ be positive real numbers.Prove that following inequality.
\[ a\ln (1 \plus{} a) \plus{} e^b > 1 \plus{} ab \plus{} b
\]
1986 Nagoya Institute of Technology


I'm trying to find out if the non-palindrome language is context-free or not. I'd use the pumping lemma, if only could find the generating language of non-palindrome grammar.

:huh: If you had the non-palindrome grammar, you would already know that it was context-free.  One uses the pumping lemma to show that a language is not context free.  :maybe:  (At least, I'm sure that one of us is very confused about what you're trying to do.)  Perhaps you should try to decide whether you think the language is or is not context-free, and then follow the approach that is suggested by such a belief.  (I, for one, don't think I want to try to pump this particular language -- where are you going to find a string which, when you break it up to pump it in a particular way, [i]must[/i] give a palindrome?  So perhaps you should think about characterizing "non-palindromes" and translating such a characterization into a grammar.)

You're right. And after the characterization of non-palindromes and the translation into a grammar?

As for the grammar, I would say something like $ S \equal{} 1S0 | 0S1, A \equal{} 1B0 | 0B1, B \equal{} 1B0 | 0B1 | 1 | 0 | \epsilon$ (I may have written rubbish)

Well, if you write a context-free grammar for the language, that means by definition that the language is context-free.  So then you'd be done.

The grammar you've written has a few of the right elements of the grammar you are looking for, but is largely nonsensical: you can't ever get from the variable $ S$ to any other variable or to a terminal, so it doesn't generate any strings at all.  Again, you should try (in natural language) to characterize what it means for a string to not be a palindrome.  Then you should try to formalize this characterization as a grammar.


Let a=b=c=1 then :
The left side = 3
The right side = 126/32 >3

[quote="Ligouras"]Let a, b, c positive real naumbers. Prove that
$ \frac {a^2 \plus{} 2bc \plus{} a}{(b \plus{} c)^2} \plus{} \frac {b^2 \plus{} 2ca \plus{} b}{(c \plus{} a)^2} \plus{} \frac {c^2 \plus{} 2ab \plus{} c}{(a \plus{} b)^2} \ge \frac {9(a \plus{} b)(b \plus{} c)(c \plus{} a) \plus{} 6(a \plus{} b \plus{} c)^2}{4(a \plus{} b)(b \plus{} c)(c \plus{} a)}$ .

Thanks

Panagiote Ligouras[/quote]

And the next?

Let a, b, c positive real numbers. Prove that
$ \frac {a^2 \plus{} 2bc \plus{} a}{(b \plus{} c)^2} \plus{} \frac {b^2 \plus{} 2ca \plus{} b}{(c \plus{} a)^2} \plus{} \frac {c^2 \plus{} 2ab \plus{} c}{(a \plus{} b)^2} \ge \frac {27(a \plus{} b)(b \plus{} c)(c \plus{} a) \plus{} 8(a \plus{} b \plus{} c)^2}{12(a \plus{} b)(b \plus{} c)(c \plus{} a)}$ .


inequalities
geometry
inradius
circumcircle
inequalities unsolved

Let $\triangle ABC$ be a triangle and denote with $r,R,p$ the inradius, the circumradius, the semiperimeter respectively. Show that

\[
D \le \max \left\{ {a,b,c} \right\} - \min \left\{ {a,b,c} \right\} \le \frac{{2D}}{{\sqrt 3 }}
\]

where $D= \sqrt {p^2  - 3r\left( {4R + r} \right)} $.

Let $a=x+y$, $b=y+z,z+x$.
WLOG, assume $a\ge b \ge c$, then $y \ge x \ge z$
Simplifying, $D^2 = x^2+y^2+z^2-yz-zx$
Now assume $m=y-x, n=x-z$, then $m+n=y-z$
we have to prove
$\frac{m^2+n^2+(m+n)^2}{2} \le (m+n)^2 \le \frac{2}{3}(m^2+n^2+(m+n)^2)$
The first one is equivalent to
$2mn \ge 0$
the second one is equivalent to
$2(m^2+n^2) \ge (m+n)^2 $
first one is obvious and second one is cauchy.


in a circle, the length of the diameter is 2cm more than twice the length of a chord. The sum of the lengths of the diameter and the chord is 50. Find the distance from the midpoint of the chord through the center of the circle to a point on the circle.

A. $ 15$
B. $ 32$
C. $ 33$
D. $ 18\sqrt3$
E. not here

[hide]Let $ x$ be the length of the chord, $ d$ the diameter. We see that $ 2x \plus{} 2 \equal{} d$, $ x \plus{} d \equal{} 50$. Solving, $ x \equal{} 16,d \equal{} 34$. Thus, the length is $ \sqrt {17^2 \minus{} 8^2} \plus{} 17 \equal{} 32\Rightarrow B$.[/hide]

In general, if a chord of length $ a$ and a diameter of length $ b$ are given, then the distance between them is $ \sqrt {\left(\frac {b}{2}\right)^{2}\minus{}\left(\frac {a}{2}\right)^{2}}$. This can be seen by dividing the diameter into two radii and drawing the radii from the diameter to the endpoints of the chord, then applying Pythagorean's. This holds for chords nonparallel to the diameter as well, since all chords equidistant from the diameter are congruent.

[quote="alanchou"]All you have to do is rotate the diameter to make it parallel to the chord, so yeah.[/quote]

Well, yeah, that's almost equivalent to what I stated if you think about it.


Find the remainder when $x^{n}$ is divided by $x^{2}-x-1$

[hide="hint"]
Suppose $x^{n}= Q(x)(x^{2}-x-1)+R(x)$ where $\deg R \le 1$. Can you find two points to define $R$?
[hide="further hint"]
Try $x = \frac{1 \pm \sqrt{5}}{2}$.[/hide][/hide]

x^n=(x^2-x-1)*Q(x)+R(x)     (1)
then we know that deg(R)<=1, so R(x)=ax+b
x^2-x-1 has two roots so by putting that roots into (1), we get two equations for R(x), solve the system and there you go :roll:

And the answer is $R_{n}(x)=F_{n}x+F_{n-1}$, where $F_{k},k\geqslant 0$ is the Fibonacci sequence ($F_{0}=0,F_{1}=1$) - which can be proven even inductively, without the explicit solving.


I would like to know more about the problem culture of the strongest IMO countries. Russian and Viet Nam have opposite paradigms, both do well at IMO but their problems are so different! Is it true ? Though I wonder if a contestant in one coutry would be selected in the other one. USA, Romania and Bulgaria are somehow in a middle path if compared with these two "extremes"." Does China have a problem culture or do they only put very hard, technical and boring problems without any sense ?

I think Russia is famous for its Combinatoric questions. Especially the Leningrad Mathematical Olympiad is one of the few competitions using graph problems regularly. I feel Hungary is one of the rare countries using solid geometry. Most countries restrict to Euclidean geometry problems. What are your experiences ?

Romania also uses a lot of solid geometry (that is geometry in space) in the Olympiad. I proposed of the problems of solid geometry last year :)

I had one in 2001 at the TST, proposed by Mihai Baluna :)

I think in most places, solid geometry tends to "die out", because nobody cars about this subject in the IMO anymore. In Romania, however, solid geometry problems are used regularly in the final round of the NMO, for 8'th and 10'th grade.

as another thread on this forum pointed out, China really has millions of students studying Solid geo, cause that's a must for college entrance
I'll bump up that thread soon.
P.S.those Chinese students don't study Number theory, which is an essential for many of you.


inequalities
function
logarithms
inequalities open

First, I will introduce a notation that I call "function mean."  The function mean $f_{mean} (a_k)$ of a monotonic, one-to-one function $f(x)$ and a sequence of nonnegative reals $a_1, a_2, ... a_n$ is given by $f^{-1} \left( \sum_{k=1}^{n} \frac{ f(a_k) }{n} \right)$.

Then the Power Mean inequality states, for $f(x) = x^p$ and $g(x) = x^q$ and $p > q$,

$f_{mean} (a_k) \ge g_{mean} (a_k)$.

Notice that it is possible to express many other inequalities this way - for example, AM-GM is expressible by setting $f(x) = x, g(x) = \ln x$ and AM-HM is expressible by setting $f(x) = x, g(x) = \frac{1}{x}$.

I suspect that it is possible to pose conditions for any two general functions $f(x), g(x)$ in order that the above inequality is true.

So, my question:

For a non-negative sequence of real numbers $a_1, a_2, ... a_n$ and two functions $f(x), g(x)$, under what conditions is it guaranteed that

$f_{mean} (a_k) \ge g_{mean} (a_k)$

?

what a nice problem :)

but first,you must add the condition that $f^{-1},g^{-1}$ are exist.
my answer is if that $f(g^{-1})$ is convex on $I$,then for every $a_k \in I$
we have:$f_{mean} (a_k) \ge g_{mean} (a_k)$

proof:

$f_{mean} (a_k) \ge g_{mean} (a_k) \iff f^{-1} \left( \sum_{k=1}^{n} \frac{ f(a_k) }{n} \right) \ge g^{-1} \left( \sum_{k=1}^{n} \frac{ g(a_k) }{n} \right)$
$\iff \left( \sum_{k=1}^{n} \frac{ f(a_k) }{n} \right) \ge f \left(g^{-1} \left( \sum_{k=1}^{n} \frac{ g(a_k) }{n} \right) \right)$
$\iff \left( \sum_{k=1}^{n} \frac{ f(g^{-1} g(a_k) )}{n} \right) \ge f \left(g^{-1} \left( \sum_{k=1}^{n} \frac{ g(a_k) }{n} \right) \right)$
then with $f(g^{-1})$ is convex  apply to jesen. :)

Such an elegant proof, thank you, I had been wondering about this for some time  :)


Two is $ 10 \%$ of $ x$ and $ 20 \%$ of $ y$. What is $ x \minus{} y$?

$ \textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 2 \qquad
\textbf{(C)}\ 5 \qquad
\textbf{(D)}\ 10 \qquad
\textbf{(E)}\ 20$

[hide="Click for solution"]
$ 0.1x\equal{}2$ and $ 0.2y\equal{}2$, so $ x\equal{}20$ and $ y\equal{}10$, so $ x\minus{}y\equal{}10$, or $ \boxed{\textbf{(D)}}$.
[/hide]


function
quadratics
algebra
polynomial
calculus
derivative
calculus computations

Suppose that $f''$ is continous on $[a, b]$ and that $f$ has three zeros in the interval. Prove that $f''$ has at least one zero in $(a, b)$. Generalize this result.

How would you prove this statement?

Here's a useful quick paraphrase of Rolle's Theorem: "Between any two zeros of a function lies at least one zero of its derivative."

So for this problem, if $f$ has three distinct zeros, then that leaves two intervals each of which must have a zero of $f'.$ Then with two zeros of $f',$ by Rolle's Theorem again, there's a zero of $f''$ between them.

A consequence of this: a quadratic polynomial (and by that, I mean that the coefficient of the $x^{2}$ term is nonzero) cannot have three zeros. From the argument above, that would imply that its second derivative has a zero, but its second derivative is a nonzero constant.

The problem asked you to generalize; one direction to go is toward higher order derivatives (and, for the followup, higher degree polynomials.)


A bag contains one marble which is known to be either green or red. A green marble is put into the bag, and the bag is shaken well. A marble is drawn randomly, and it is green.  This marble is returned, the bag shaken, and a second draw made - also green. What is the probability that the marble still in the bag is also green?

To make sense of that, you need a prior probability that the original marble was green. There is not enough information as stated, although I'd guess that a 50% prior probability was intended.


Corinne is a basketball player who makes 75% of her free throws over the course of a season. In a key game, she shoots 12 free throws and makes only 7 of them. The fans think that she failed because she was nervous. Is it unusual for her to perform this poorly? The solution in my book is to find out the probability of making a basket on at most 7 free throws. Why we need to find out at most 7 free throws instead of exactly 7 free throws? So for P (x<=7) =16%, therefore my book concludes that she will make at most 7 of her 12 free throws about 16% of the time, or roughly in one of every six games. While below her average level, this performance is well within the range of the usual chance variation in her shooting. How do I know if it's within the usual range of her shooting by looking at 16%?  Thank you.

This all relates to the convention in statistics to consider as "improbable" the events of the kind $ X > EX \plus{} c$ or $ X < EX \minus{} c$ whose probability is less than $ 5\%$. If you do not get into that event, you are in the "normal range". Various statistical tests are in the same spirit. This is all extremely voluntaristic, of course, but nobody has ever come with a substantially better idea.


geometry
circumcircle
inequalities
analytic geometry
similar triangles
triangle inequality
geometry solved

We are given triangle $ABC$ with $AB = 3$, $BC = 4$, $CA = 5$.  Find the minimum value, over all points $D$ inside the triangle, of $12AD + 5BD + 13CD$.

Hmmm, no one has replied, so I'll add an observation:

In the plane, let A = (0,3), B = (0,0), C = (4,0).  The minimal point seems to be
\[D = \left( \frac{126}{85},  \frac{108}{85} \right).\]
For this point, we have $\angle ADB = 90^\circ$.  Proof?

I thought you had a solution, since you posted the problem into the "Proposed & Own Problems" section, and hence I didn't post my solution...

In fact, the problem is a special case of the following theorem (С. И. Зетель, [i]Задачи на максимум и минимум[/i], Москва / Ленинград 1948):

[color=blue][b]Theorem 1.[/b] Let ABC be a triangle, and l, m, n three positive numbers such that there exists a triangle XYZ with sidelengths YZ = l, ZX = m and XY = n.

Erect three triangles A'BC, AB'C, ABC' outwardly on the sides BC, CA, AB of triangle ABC which are all similar to triangle XYZ; this means that

  BC : CA' : A'B = B'C : CA : AB' = BC' : C'A : AB = YZ : ZX : XY = l : m : n.

Then, the circumcircles of triangles A'BC, AB'C, ABC' and the lines AA', BB', CC' all concur at one point P.

Moreover, [b]if this point P lies inside triangle ABC[/b], then the following two facts hold:

- It satisfies < BPC = 180° - < YXZ, < CPA = 180° - < ZYX and < APB = 180° - < XZY.
- If D is a variable point in the plane of triangle ABC, then the sum $l\cdot AD+m\cdot BD+n\cdot CD$ is minimal if and only if D = P.[/color]

[i]Proof.[/i] Call P the point of intersection (different from C) of the circumcircles of triangles A'BC and AB'C. Then we will first show that this point P also lies on the circumcircle of triangle ABC', and that the equations < BPC = 180° - < YXZ, < CPA = 180° - < ZYX and < APB = 180° - < XZY are satisfied, and then prove that the point P lies on the lines AA', BB', CC'.

Since the point P lies on the circumcircle of triangle A'BC, the quadrilateral BPCA' is cyclic, and thus < BPC = 180° - < BA'C. Since the triangle A'BC is similar to triangle XYZ, we have < BA'C = < YXZ, and thus < BPC = 180° - < YXZ. Similarly, < CPA = 180° - < ZYX. Thus,

  < APB = 360° - < BPC - < CPA = 360° - (180° - < YXZ) - (180° - < ZYX)
            = < YXZ + < ZYX
            = 180° - < XZY        (sum of angles in triangle XYZ)
            = 180° - < AC'B          (since the triangles XYZ and ABC' are similar).

And this yields that the quadrilateral AC'BP is cyclic, and thus the point P lies on the circumcircle of triangle ABC'. Hence, the point P is a common point of the circumcircles of triangles A'BC, AB'C, ABC'. Now, as we have seen before, < APB = 180° - < XZY; but < XZY = < A'CB because of the similar triangles XYZ and A'BC, and thus < APB = 180° - < A'CB. But < A'CB = < A'PB, since the point P lies on the circumcircle of triangle A'BC; therefore, < APB = 180° - < A'PB. This yields that the points A, P, A' lie on one line, i. e. the point P lies on the line AA'. Similarly, the point P lies on the lines BB' and CC'.

In order to complete the proof of Theorem 1, it remains to show that the sum $l\cdot AD+m\cdot BD+n\cdot CD$ is minimal if and only if D = P.

Well, after the Ptolemy inequality applied to the quadrilateral BA'CD, we have $CA^{\prime }\cdot BD+A^{\prime }B\cdot CD \geq BC\cdot A^{\prime }D$, with equality only if the quadrilateral BA'CD is convex and cyclic. Since BC : CA' : A'B = l : m : n, the condition $CA^{\prime }\cdot BD+A^{\prime }B\cdot CD \geq BC\cdot A^{\prime }D$ is equivalent to $m\cdot BD+n\cdot CD \geq l\cdot A^{\prime }D$. Hence, we have $m\cdot BD+n\cdot CD \geq l\cdot A^{\prime }D$ with equality if and only if the quadrilateral BA'CD is convex and cyclic. Also, by the triangle inequality, $AD+A^{\prime }D\geq AA^{\prime }$, with equality if and only if the point D lies on the segment AA'.

Thus,

$l\cdot AD+m\cdot BD+n\cdot CD\geq l\cdot AD+l\cdot A^{\prime }D=l\cdot \left( AD+A^{\prime }D\right) \geq l\cdot AA^{\prime }$,

with complete equality only if both the quadrilateral BA'CD is convex and cyclic and the point D lies on the segment AA'. Since AA' is a constant value, it follows that the sum $l\cdot AD+m\cdot BD+n\cdot CD$ has the minimum $l\cdot AA^{\prime }$ which is achieved if and only if both the quadrilateral BA'CD is convex and cyclic and the point D lies on the segment AA'. Now, the latter happens if and only if D = P (in fact, the quadrilateral BA'CD being cyclic implies that the point D lies on the circumcircle of triangle A'BC, and since the point D also lies on the segment AA', the point D must be the meet of the circumcircle of triangle A'BC with the segment AA', what means that D = P). Hence, the expression $l\cdot AD+m\cdot BD+n\cdot CD$ is indeed minimized for D = P, and Theorem 1 is proven.

Now, applying Theorem 1 in your case, with l = 12, m = 5, n = 13 leads to the construction of the minimum point. Note that in this case, l, m, n are not just the sidelengths of a triangle; they are even the sidelengths of a right-angled triangle, i. e. we have < XZY = 90°. And thus, from Theorem 1, it follows that the required minimum point P satisfies the equation < APB = 180° - < XZY = 180° - 90° = 90°.

I have not checked your coordinates, but they look right. I am not used to solving such problems with coordinate geometry anyway.

Note that Theorem 1 partly solves a famous problem: For a given triangle ABC and three given reals l, m, n, construct the point D such that the sum $l\cdot AD+m\cdot BD+n\cdot CD$ is minimal. In fact, Theorem 1 works if the reals l, m, n are positive and are the sidelengths of a triangle, and the point P constructed above lies inside triangle ABC. I am quite interested in what happens if the first or the third condition is not satisfied (i. e. the point P lies outside triangle ABC or some of the reals l, m, n are negative). If the second condition is not satisfied - i. e. if the numbers l, m, n are positive but not sidelengths of a triangle; for instance, say $l \geq m + n$ -, then the answer is quite easy:

[color=blue][b]Theorem 2.[/b] Let ABC be a triangle, and l, m, n three positive numbers such that $l \geq m + n$.

If D is a variable point in the plane of triangle ABC, then the sum $l\cdot AD+m\cdot BD+n\cdot CD$ is minimal if and only if D = A.[/color]

This is quite easy to prove; in fact, for every point D in the plane, $AD+BD\geq BA$ and $AD+CD\geq CA$ by the triangle inequality, and thus

$l\cdot AD+m\cdot BD+n\cdot CD\geq\left(m+n\right)\cdot AD+m\cdot BD+n\cdot CD$
$=m\cdot\left(AD+BD\right)+n\cdot\left(AD+CD\right)\geq m\cdot BA+n\cdot CA$
$=0+m\cdot BA+n\cdot CA=l\cdot AA+m\cdot BA+n\cdot CA$,

and equality holds if and only if both triangle inequalities $AD+BD\geq BA$ and $AD+CD\geq CA$ become equalities, what forces the point D to lie on both of the lines AB and AC, and thus to coincide with the point A; thus, equality holds if and only if D = A. This shows that the minimum value of the sum $l\cdot AD+m\cdot BD+n\cdot CD$ is obtained for D = A. This proves Theorem 2.

  Darij

Darij, thanks for the solution, and for the note about the section, one day I will learn...


$x(n)=\frac{1}{(a+b)(2a+b)...(na+b)}$ where a,b>0

Give an equivalent of $x(n)$ with two terms

So far, I have

$x(n)\sim \frac1{n!a^nn^{b/a}}+?$


[quote="IOO"]Let $ f(x) \equal{} x^{ \minus{} 1/2}$  if $ 0 < x < 1$ and  $ f(x) \equal{} 0$ otherwise. Let$ \{r_n \}_n$ be an enumeration of rationals, and set $ g(x) \equal{} \sum_{n \equal{} 1}^{\infty} 2^{ \minus{} n}f(x \minus{} r_n).$ Prove that $ g \in L^1(\mu)$,  $ g$ is discontinuous at every point, unbounded on every interval, $ g^2 < \infty$ a.e., but $ g^2$ is not integrable on any interval.[/quote]

This is an exercise in using the theorems you learned recently. Why don't you tell us where you're stuck?