inequalities
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inequalities proposed

[b]Suppose that a,b,c,d are arbitrary real numbers. Then[/b]
\[ \begin{array}{|c|} \hline \\ a(a-b)(a-c)(a-d) +b(b-c)(b-d)(b-a) + c(c-d)(c-a)(c-b) +d(d-a)(d-b)(d-c) + abcd \ge 0\\ \\ \hline \end{array}\; .  \]
[[i]Case $d=0 , a>0,b>0,c>0$ is well known.[/i] ]

This is the special case $e=0$ of the inequality

$\sum (a-b)(a-c)(a-d)(a-e)\geq 0$

in five variables $a,b,c,d,e$. This is one part of the old IMO problem IMO 71/1.

Peter

[quote="flip2004"][quote="Peter Scholze"]...  This is one part of the old IMO problem IMO 71/1... Peter[/quote]
Thank you very much !   Today I observed that the inequality (IMO 71/1)  is a special case of an earlier result  estabilshed by
[b]Oppenheim A. and Roy O.Davies[/b] , [i]Inequalities of Schur's type[/i] ,Math.Gaz., 48 [b] (1964)[/b] 25--27.[/quote]

May I know the full name of Math.Gaz.? (Is it an abbreviation actually?)

Oddly enough, in my trolling of sci.math over the last night and a bit, I found:

http://groups.google.com/group/sci.math/browse_frm/thread/6caeb7e1a97c0fd2/62a49911c389f63f#62a49911c389f63f

[quote="blahblahblah"]Oddly enough, in my trolling of sci.math over the last night and a bit, I found:....[/quote]
Thanks, I visit the posted question.A remark: Robert Israel  uses Maple. The question was proved by  a computer ? Not bad, but ... ?

[quote="Soarer"][quote="flip2004"][quote="Soarer"][b]Oppenheim A. and Roy O.Davies[/b] , [i]Inequalities of Schur's type[/i] ,Math.Gaz., 48 [b] (1964)[/b] 25--27. May I know the full name of Math.Gaz.? (Is it an abbreviation actually?)[/quote]
[[b]Math.Gaz.[/b] ] $\equiv$[b]The MATHEMATICAL GAZETTE[/b]= published weekly,  by the Worcester Polytechnic Institute Mathematical Sciences Department.
 (Indeed it's an abbreviation, also used  in [i] Mathematical Reviews[/i] )[/quote]
May I know where I can find the article? I searched some database and can't get any issue as early as 1964... :([/quote]
Unfortunatelly, I hav'nt find it  on Internet. Only as a resource in [i]The Mathematical Gazette[/i] site. See:
         =====================
            http://www.wpr3.co.uk/gazette/
            http://www.wpr3.co.uk/gazette/
           http://www.wpr3.co.uk/gazette/1960-69.html
         ==========================
Try on Google ! However, better   search the paper   in a Mathematical library at a certain University.

A recent issue from [i]Math.Gaz.[/i] is here (to download):

http://www.mathsyear2000.org/resources/ma/pdfs/Maths_Gazette.pdf


trigonometry
induction
Recursive Sequences

The sequence $ \{x_{n}\}_{n \ge 1}$ is defined by
\[ x_{1} \equal{} 2, x_{n \plus{} 1} \equal{} \frac {2 \plus{} x_{n}}{1 \minus{} 2x_{n}}\;\; (n \in \mathbb{N}).
\] Prove that

a) $ x_{n}\not \equal{} 0$ for all $ n \in \mathbb{N}$,

b) $ \{x_{n}\}_{n \ge 1}$ is not periodic.

[quote="Peter"]The sequence $ \{x_{n}\}_{n \ge 1}$ is defined by
\[ x_{1} \equal{} 2, x_{n \plus{} 1} \equal{} \frac {2 \plus{} x_{n}}{1 \minus{} 2x_{n}}\;\; (n \in \mathbb{N}).
\]
Prove that [list=a][*] $ x_{n}\not \equal{} 0$ for all $ n \in \mathbb{N}$, [*] $ \{x_{n}\}_{n \ge 1}$ is not periodic.[/list][/quote]
$ a \equal{} \arctan {2}$
$ x_1 \equal{} \tan{a}$
$ x_{n \plus{} 1} \equal{} \frac {x_n \plus{} \tan{a}}{1 \minus{} x_n\tan{a}}$
We prove by induction that:
$ x_{n} \equal{} \tan{na}$
Suppose $ x_n$ is period T.
$ \tan{(n \plus{} T)a} \equal{} \tan{na}$
$ \rightarrow Ta\equiv 0(\mod \pi)$
So $ a \equal{} \frac {mk\pi}{T}$
We has a result :
$ k\in Q$ then $ \tan{k\pi}$ is not equal 2.

[hide="Another approach"]
As above let $ \alpha \equal{} \arctan 2$ so that $ x_n \equal{} \tan n\alpha$. For [b](a)[/b] suppose for the sake of contradiction that $ x_n \equal{} 0$ for some $ n$. Let $ m \equal{} n/2$ so that \[ x_n \equal{} x_{2m} \equal{} \frac{2\tan m\alpha}{1\minus{}\tan^2 m\alpha} \equal{} \frac{2x_m}{1\minus{}x_m^2}.\] If $ n$ is even, $ x_n \equal{} 0 \iff x_m \equal{} 0$. Otherwise let $ n \equal{} 2^k (2j\plus{}1)$ for some nonnegative integers $ j,k$. We can divide out powers of 2 until we are left with \[ x_{n/2^k} \equal{} x_{2j\plus{}1} \equal{} \frac{2\plus{}x_{2j}}{1\minus{}x_{2j}} \equal{} 0,\] so $ x_{2j} \equal{} \minus{}2$. But \[ x_{2j} \equal{} \frac{2x_j}{1\minus{}x_j^2} \equal{} \minus{}2.\] This equation has no rational roots so we have a contradiction (inductively, $ x_n$ must be rational since $ x_1 \equal{} 2$ is rational). Therefore there are no zero values of $ \{x_n\}$.

For [b](b)[/b] suppose that $ x_{k\plus{}t} \equal{} x_{k}$ for some positive integers $ k,t$ with $ t\ge 1$. Then \[ x_{k\plus{}t} \minus{} x_{k} \equal{} \tan (k\plus{}t)\alpha \minus{} \tan k\alpha \equal{} \frac{\tan t\alpha}{\cos(k\plus{}t)\alpha} \equal{} \frac{x_t}{\cos(k\plus{}t)\alpha} \equal{} 0,\] so $ x_t \equal{} 0$. However, this is impossible by (a). Thus all terms of $ \{x_n\}$ are distinct.
[/hide]


since it takes tim 6 hours to paint the fence, he can do 1/6 of it in an hour. Lisa can to 1/5 of it and Tivanka a 1/3. since they are working together, just by simply adding the times each can do in an hour up gives how much of the fence they can do, thus giving us the formula:
(1/6 + 1/5 + 1/3) x = 1
simplifying gives x = 10/7 hours

[quote]Tim, Lisa, and Tivanka are painting a fence black. It would take Tim 6 hours to paint the fence alone. It would take Lisa 5 hours to paint the fence alone. It would take Tivanka, a talented fence painter, only 3 hours to paint the fence alone. How long will it take all three of them working together to paint the fence?[/quote]

yeah, i quoted that so i can refer back easier. anyways to the solution!

in one hour Tim can paint 1/6 of the fence. in one hour Lisa can paint 1/5 of the fence. in one hour Tivanka can paint 1/3 of the fence. then you add this up to to find how much of the fence they can paint in one hour.
so: 1/6+1/5+1/3=1/x=7/10
solving for x, then x=10/7 of an hours

pretty much the same thing. fractions of what people can do in an hour

[hide]$\frac{1}{6}+\frac{1}{5}+\frac{1}{3}=\frac{5+6+10}{30}=\frac{7}{10}$

Flip the fraction to get 10/7ths of a hour.[/hide]

Billy


Find the largest integer that is a factor of
$(a-b)(b-c)(c-d)(d-a)(a-c)(b-d)$
for all integers $a,b,c,d$.

$12|(a-b)(b-c)(c-d)(d-a)(a-c)(b-d)$ (this was posted many times, even recently so I won't be proving this, anyway it's straightforward), so we're need to find $a,b,c,d$ such that :

$(a-b)(b-c)(c-d)(d-a)(a-c)(b-d)=12$

take: $a=b+1$,  $b-c=2$, $b-d=1$ and it is easy to check that in this case indeed $(a-b)(b-c)(c-d)(d-a)(a-c)(b-d)=12$ so we're done.

Do You have any idea how to kill this problem in a quick way?

Yes. For every integer $ n$ greater than $ 4$ we may find a set of numbers $ \{a,b,c,d\}$ such that every one of them has different residue $ mod \ n$ :wink:
The conclusion follows...


I know you can block others from your blog, but can you set blogs that won't appear in the newly-updated list? I really don't want to see certain blogs, but I can't block them right now, except getting the owner to block me, which they aren't likely to do and isn't anonymous anyway.

A good idea. I will add it to the todo list, but can't guarantee a date for it.

Uh oh, am I on the list of blogs you don't wanna see?

Might be useful, although I don't think there are any blogs I want to block at the moment


[quote="firecricket91"]What is the smallest positive integer a such that for any x, 5x^11+11x^5+ax is divisible by 55.[/quote]
[hide="Solution"]
Look at it $ \bmod 5$ first.  Thus, $ 5x^{11} \plus{} 11x^5 \plus{} ax\equiv 11x^5 \plus{} ax \equal{} x(11x^4 \plus{} a)\equiv x(11 \plus{} a)\bmod 5$.  Now, since this number is divisible by $ 55$ for all $ x$, we have that $ 11 \plus{} a\equiv 0\bmod 5\implies a\equiv 4\bmod 5$ (since it is for all $ x$, we can have where $ 5$ is not a factor of $ x$.)

Now, look at the equation $ \bmod 11$.  Thus, $ 5x^{11} \plus{} 11x^5 \plus{} ax\equiv x(5x^{10} \plus{} a)\equiv x(5 \plus{} a)\equiv 0\bmod 11\implies 5 \plus{} a\equiv 0\bmod 11\implies a\equiv 6\bmod 11$.

Hence, $ a\equiv 6\bmod 11$ and $ a\equiv 4\bmod 5$.  From here, we conclude that $ a\equiv 39\bmod 55$ (we can do this by guessing and checking or by setting variables and creating a system of equations.)  Therefore, the lowest value of $ a$ is $ \boxed{39}$. [/hide]

[b][hide="generalization"][/b]
$ a,b \in odd \ prime \ \Rightarrow \ ab|ax^b \plus{} bx^a \plus{} (ab \minus{} a \minus{} b)x \ (\forall x \in \mathbb{N})$[/hide]


Could someone help me with this one? Im not quite sure how to do it.

What is the number of digits in 416525 when written in the usual base 10 form?

[hide]
Maybe it would be more helpful to think of [tex]4^{16}[/tex] as [tex]2^{32}[/tex].
[/hide]

I think it is a quite complicated explanation for such easy problem ;)

I didn't think to do it that way at first. Nice solution Danbert!  :D

[hide] $2^7 \cdot 10^{25}$ is $128$ followed by $25$ zeros so there are $25+3=28$ digits[/hide]


[color=blue]He aquí un problema interesante que resolvimos en un grupo de discusión paraguayo. Es sencillo, pero después nos planteamos varias preguntas adicionales. Los invito a que no se limiten con el enunciado, que está adjunto debido al apuro.
¡Éxitos a todos para la Iberoamericana![/color]

No se adjuntó, después, cuando puedas, tratá de nuevo

[color=darkblue]   Discúlpenme. Creí que se había adjuntado el archivo. Por lo visto aún me falta mucho para dominar este sistema. Ahora que me fijo… hay un botón llamado "Add Attachment". El otro día estaba demasiado apurado como para verlo. Voy a probar nuevamente.

   Gracias Por la observación.[/color]


Thank you very much, Khue, for your very nice and hard inequality,

I hope someone can solve it.
I will post my solution after 1 week.

Yes, nice. This one has something similar to the one by Zhao Bin and another one by Le Vi and me. For yours, I have solution for $k=10$, Can has a solution for $k=7$, which is the best constant for the left hand side exceeds $3/\sqrt{2+k}$.

[quote="ehku"][color=blue]Prove the following inequality for the non-negative real numbers $a,b,c$ and the constant $k \ge-2$ :
$\sqrt{\frac{a^{2}}{a^{2}+kab+b^{2}}}+\sqrt{\frac{b^{2}}{b^{2}+kbc+c^{2}}}+\sqrt{\frac{c^{2}}{c^{2}+kca+a^{2}}}\ge \min \left(1;\frac{3}{\sqrt{k+2}}\right)$
When does the equality hold ?[/color]

This is a problem I posted in 2006 in this forum, but I didn't receive any solution.[/quote]

I solved it by mixing variable, much it is really complicate. As a matter of fact, we can change the inequality to $a=b$ by a lot of technique, but the rest when $a=b$ is hard.

[quote="ehku"][color=blue]Prove the following inequality for the non-negative real numbers $a,b,c$ and the constant $k \ge-2$ :
$\sqrt{\frac{a^{2}}{a^{2}+kab+b^{2}}}+\sqrt{\frac{b^{2}}{b^{2}+kbc+c^{2}}}+\sqrt{\frac{c^{2}}{c^{2}+kca+a^{2}}}\ge \min \left(1;\frac{3}{\sqrt{k+2}}\right)$
When does the equality hold ?[/color]

This is a problem I posted in 2006 in this forum, but I didn't receive any solution.[/quote]
It's not very hard, my solution is only using classical inequalities, let I post it (note that it suffices to consider the case $k \ge 7$,)
Settting $x=\frac{b}{a},y=\frac{c}{b},z=\frac{a}{c}$ then $xyz=1$, we will show
\[\sum\frac{1}{\sqrt{x^{2}+kx+1}}\ge \frac{3}{\sqrt{k+2}}\]
Since $x,y,z>0$ and $xyz=1$ then there exist $m,n,p>0$ such that $x=\frac{n^{2}p^{2}}{m^{4}},y=\frac{p^{2}m^{2}}{n^{4}},z=\frac{m^{2}n^{2}}{p^{4}}$, the inequality becomes
\[\sum\frac{m^{4}}{\sqrt{m^{8}+km^{4}n^{2}p^{2}+n^{4}p^{4}}}\ge \frac{3}{\sqrt{k+2}}\]
By Holder inequality, we have
\[LHS^{2}\cdot \sum m(m^{8}+km^{4}n^{2}p^{2}+n^{4}p^{4}) \ge (m^{3}+n^{3}+p^{3})^{3}\]
It suffices to prove that
\[(k+2)(m^{3}+n^{3}+p^{3})^{3}\ge 9\sum m(m^{8}+km^{4}n^{2}p^{2}+n^{4}p^{4}) \]
or
\[k(m^{3}+n^{3}+p^{3})((m^{3}+n^{3}+p^{3})^{2}-9m^{2}n^{2}p^{2})+2(m^{3}+n^{3}+p^{3})^{3}-9\sum m(m^{8}+n^{4}p^{4}) \ge 0 \]
Since $k \ge 7$ and $(m^{3}+n^{3}+p^{3})^{2}-9m^{2}n^{2}p^{2}\ge 0$, then it suffices to consider the case $k=7$, then we have to show
\[(m^{3}+n^{3}+p^{3})^{3}\ge \sum m(m^{8}+7m^{4}n^{2}p^{2}+n^{4}p^{4}) \]
Hay
\[\sum_{\text{sym}}( 5m^{6}n^{3}+2m^{3}n^{3}p^{3}-7m^{5}n^{2}p^{2})+\sum_{\text{sym}}(m^{6}n^{3}-m^{4}n^{4}p) \ge 0 \]
which is true by Muirhead theorem. :)

Very nice proof,

[hide="can"]@Can. Hi, so why you say your solution is so complicated? [/hide][/list]

OK, the old technique has now been improved, Can

http://www.mathlinks.ro/Forum/viewtopic.php?t=146081

Congratulation.  :lol:

Thank you, Can. Nice proof. If you visited this forum last year, I wouldn't post my inequality in here  :lol:
My solution bases on the following inequality (for all non-negative real numbers $a,b,c$):
$\frac{a^{2}+2ab}{a^{2}+kab+b^{2}}+\frac{b^{2}+2bc}{b^{2}+kbc+c^{2}}+\frac{c^{2}+2ca}{c^{2}+kca+a^{2}}\ge \min \left(1;\frac{9}{k+2}\right)$

[quote=ehku][color=blue]Prove the following inequality for the non-negative real numbers $a,b,c$ and the constant $k \ge-2$ :
$\sqrt{\frac{a^{2}}{a^{2}+kab+b^{2}}}+\sqrt{\frac{b^{2}}{b^{2}+kbc+c^{2}}}+\sqrt{\frac{c^{2}}{c^{2}+kca+a^{2}}}\ge \min \left(1;\frac{3}{\sqrt{k+2}}\right)$
When does the equality hold ?[/color]
This is a problem I posted in 2006 in this forum, but I didn't receive any solution.[/quote]
Prove the following inequality for the non-negative real numbers $a,b,c$ and the constant $k \in[4,8]$ :
$$\sqrt{\frac{a^{2}}{a^{2}+kab+(8-k)b^{2}}}+\sqrt{\frac{b^{2}}{b^{2}+kbc+(8-k)c^{2}}}+\sqrt{\frac{c^{2}}{c^{2}+kca+(8-k)a^{2}}}\ge 1$$


[quote="Rep123max"]I know that there is a floor function for finding the number of digits in any number but I was trying to find a way for finding the digits in any factorial. As far as I know the floor function ([logx]+1) is useless for this because you either need to know the factorial or add tons of logs.

Any help?[/quote]

But... floor isn't for finding the number of digits, floor is for finding the greatest integer less than or equal to the number...

As a side note, the math emoticons don't show up *at all* on the white background...

I think Rep123Max is refering to the formula stating that the number of digits in a number x is equal to [log x]+1

You can estimate ln n! fairly easily by approximating it with an integral.

ln n! = sum (k=1 to n) [ln k]
:approx: integral (1 to n) ln x dx
= x ln x - x | (1 to n)
= n ln n - n

This is not a [i]great[/i] approximation, but it gets you somewhere in the ballpark.

To get a better approximation you need to use Stirling's Formula:
n! :approx: (n/e)^n*sqrt(2*pi*n).

This gives, ln n! :approx: n ln n - n + 1/2*ln(2*pi*n).

This approximation is still not perfect, but it's pretty good. The Stirling Formula comes from the first few terms of the more general Stirling Series. By taking more terms of the Stirling Series you can get better and better approximations to n!.

I really doubt they might "change it up". The reason being: it's easy to find the number of zeroes at the end (look through the mathjam transcripts; it's somewhere in there), and it's hard to approximate n!.

Most TI calculators nowadays can handle exponents up to 10^999. Like my TI-86 can do factorials up to 449! So overflow shouldn't be a problem until you start dealing with ones larger than that or try to manipulate them in someway that makes them to large.

But I doubt that they would ever ask you a question like this without some additional information. Because, for one, it's not a terribly interesting problem; and secondly, no one would expect high school students to know Stirling's Approximation.

Uh, if you are willing to use a calculator, why not just add up the logs???

Just use the summation function on your calculator- I know that 83+s have it.

How many times do they ask questions like that? I've never seen it...

You push (2nd) then (stat) which takes you to the list menu. What you have to do is simply define lists and the calculator can sum them as long as they have...i think the limit is 999 or so elements.

Anyway...once in the list menu, under Math you find "sum(" and under Ops you find "seq(". And then you define a list. It should be in the format:

sum(seq(expression, variable, lower bound, upper bound, increment))

the increment part is optional; it automatically assumes the increment is 1, but you can set a different one.


geometry
Triangle
incenter
IMO Shortlist
geometry solved
midpoint
congruent triangles

A point $D$ is chosen on the side $AC$ of a triangle $ABC$ with $\angle C < \angle A < 90^\circ$ in such a way that $BD=BA$.  The incircle of $ABC$ is tangent to $AB$ and $AC$ at points $K$ and $L$, respectively.  Let $J$ be the incenter of triangle $BCD$.  Prove that the line $KL$ intersects the line segment $AJ$ at its midpoint.

Denote $Z \in BJ \cap AC$.

We have that $\frac{BJ}{JZ}=\frac{a+c}{CD}$, becuase $BJ$ is the bisector of $\angle{CBD}$.

Thus $ZJ=\frac{BJ \cdot CD}{a+c}$, and $BJ=\frac{BJ \cdot (CD+a+c)}{a+c}$.

Now the main idea is to use the transversal theorem in $\triangle{ABZ}$, for $KL$ and the cevian $AJ$.

So let's denote $T \in AJ \cap KL$.
Therefore $TA=TJ \Longleftrightarrow \frac{BK}{KA}\cdot ZJ+\frac{ZL}{LA}\cdot BJ = BZ$

$\iff \frac{a+c+CD}{a+c}-\frac{p-b}{p-a}\cdot \frac{CD}{a+c}= \frac{ZL}{LA}$. ($\star$)

Now from the bisector theorem $\frac{ZD}{ZC}=\frac{c}{a}\iff ZD=\frac{c \cdot CD}{a+c}$.

And from $AD=b-CD$, we have that $AZ = b-CD+\frac{c \cdot CD}{a+c}$, so:

$ZL = \frac{(a+b-p-CD)(a+c)+c \cdot CD}{a+c}$, and $AL=p-a$.

Thus $(\star)$ is equivalent with:

$\frac{a+c+CD}{a+c}-\frac{p-b}{p-a}\cdot \frac{CD}{a+c}= \frac{(a+b-p-CD)(a+c)+c \cdot CD}{(a+c)(p-a)}$, which after clearing the denominator is true.

[EDIT:typing errors]

[color=darkred][b][u]PP13[/u].[/b] Let $\triangle ABC$ such that $C<A<\frac{\pi}{2}$. Let $D\in [AC]$ such that $BD=BA$. The incircle $w=C(I,r)$ of $\triangle ABC$ touches its sides in $K\in [AB]$ and $L\in [AC]$. Let $w_{1}=C(J,r_{1})$ be the incircle of $\triangle BCD$. Denote $T\in KL\cap AJ$. Prove that $TA=TJ$.[/color]

[color=darkblue][b][u]Proof.[/u].[/b] The incircle $w_{1}$ touches  $\triangle BCD$ in $U\in BD$, $V\in BC$ and $W\in BC$. Observe that $DC=AC-AD=b-2c\cdot\cos A=$ $\frac{b^{2}-(b^{2}+c^{2}-a^{2})}{b}$ $\implies$ $DC=\frac{a^{2}-c^{2}}{b}$. Thus, $VC=\frac{1}{2}\cdot (BC+CD-BD)=\frac{1}{2}\cdot(a+\frac{a^{2}-c^{2}}{b}-c)\implies VC=\frac{p(a-c)}{b}$ . Since $VC=JC\cdot\cos \frac{C}{2}$ obtain that $\boxed{JC=\frac{p(a-c)}{b\cdot\cos\frac{C}2}}$. Denote $P\in KL\cap CI$. The property $PC\perp PB$ is well-known (see the lower remark). Thus,  $APJW$ is cyclically. Analogously $P\in UV\cap CJ$, i.e. $\boxed{\ P\in KL\cap UV\ }$. Observe that $\boxed{PC=a\cdot \cos \frac{C}{2}}\implies$ $\frac{PC}{JC}=\frac{ab\cdot\cos^{2}\frac{C}{2}}{p(a-c)}=\frac{p-c}{a-c}$ $\implies$ $\frac{PC}{p-c}=\frac{JC}{a-c}=\frac{PJ}{p-a}$ $\implies$ $\boxed{\ \frac{PJ}{PC}=\frac{p-a}{p-c}\ }$. Apply the [u][b]Menelaus[/b]' theorem[/u] to transversal $\overline{PTL}$ in $\triangle AJC$ : $\frac{PJ}{PC}\cdot\frac{LC}{LA}\cdot\frac{TA}{TJ}=1$ $\implies$ $\frac{p-a}{p-c}\cdot\frac{p-c}{p-a}\cdot\frac{TA}{TJ}=1$ $\implies TA=TJ$. So $PC\perp PB$ $\Longleftrightarrow$ the quad. $PKIB$ is cyclically $\Longleftrightarrow$ $m(\widehat{PKB})=m(\widehat{PIB})=\frac{B+C}{2}$ or $m(\widehat{PKA})=m(\widehat{PIB})=\frac{B+C}{2}$[/color]

[img]8922[/img]

I have a synthetic solution:
Let S be the foot of the perpendicular from B, M the point where the incircle of BCD touches AC, F the midpoint of SL.
Let Q be the point where the perpendicular bisector of SL meets KL.

Now, the triangle SQL is isosceles, and by angle chasing we get $\angle QSL = \angle SLQ = 90-\frac{\alpha}2$.
With angle chasing you can also find that $\angle JDM = 90-\frac{\alpha}2$, therefore $SQ \parallel JD$.

Now we calculate $2LM = 2CL-2CM = (AC+BC-AB)-(DC+BC-BD) = AC-DC = AD$.
Therefore $AS = SD = LM$, $SL = DM$, $2SF = DM$.

Consider the homotety with centre A and factor 2:
- it sends S to D
- therefore it sends the line QS to the line JD
- it sends F to M
- therefore it sends the line QF to the line JM
- therefore it sends Q to J

Finally, Q is on the lines AJ and KL, then Q is P! But the homotety had a factor 2; we conclude $JP = PA$.

This problema was in the first Brazilian TST!

[quote="Arne"]That's not so suprising, this is a Shortlist problem.[/quote]

by the way I would like to know if you have the last IMO short list.

http://www.mathlinks.ro/Forum/viewtopic.php?t=106306

How did this succeed to get disclosed in August 2006?

  Darij

Let $ S$ be the midpoint of $ AJ$ and let $ T$ be the intersection of lines $ JD$ and $ KL$. Moreover, let $ M$ be the foot of the perpendicular of $ J$ on $ CD$. It's our aim to show that $ K$, $ S$ and $ T$ are collinear.
First note that triangles $ \triangle TLD$ and $ \triangle AKL$ are similar. This follows from a short angle-chase. Hence $ DT/DL \equal{} LA/LK$. Moreover, $ LA/LK \equal{} 1/(2\sin\angle CAB/2)$  $ (*)$. This follows from the sine law and the addition theorem. Another consequence of the similarity is that $ AKDT$ is cyclic.
Additionally, $ AL \minus{} LD \equal{} 2MD$  $ (**)$. This is easily seen, as $ AL \minus{} LD \equal{} 2AL \minus{} AD \equal{} AB \plus{} AC \minus{} BC \minus{} AD \equal{} AB \plus{} CD \minus{} BC$ and $ 2MD \equal{} BD \plus{} CD \minus{} BC$, hence the conclusion.
We have $ MD/JD \equal{} \sin \angle CAB/2$  $ (***)$.
Now, applying Menelaos' theorem in triangle $ \triangle AJD$ yields:
\[ \frac {AS}{SJ} \cdot \frac {JT}{DT} \cdot \frac {DL}{LA} \equal{} 1 \\
\Leftrightarrow JT \equal{} LA \cdot \frac { DT}{DL} \equal{} \frac{LA^2}{LK}.\]
This is equivalent to $ JT \cdot LK \equal{} (JD \plus{} DT)LK \equal{} LA^2 \Leftrightarrow JD \cdot LK \equal{} LA^2 \minus{} DT \cdot LK \equal{} LA(LA \minus{} DL)$, where $ DT \cdot LK \equal{} LA \cdot DL$ follows from the power of a point wrt quadrilateral $ AKDT$. The conclusion follows now from combining $ (*)$, $ (**)$ and $ (***)$.

Dear Mathlinkers,
can we imagine a proof with trigonometry, ratio, angle, homothetie, Menelaüs and case of egality of triangles?
I think so
A new challenge...
Sincerely
Jean-Louis

I have another idea.

let $m(JAD)=b,m(JAI)=a$ $\implies$  $m(IAK)=a+b$

$m(DCJ)=m(JCB)=c$  $\implies$

$m(JBC)=m(JBD)=a+b-c,$

$m(DBI)=90-3(a+b)-c,m(IBK)=90-a-b-c$

$AJ\cap KL=\{P\}$

$AP=BJ{\frac{cos(a+b+c).cos(a+b)}{cosa}}$

$AJ=BJ{\frac{cos(2a+2b)sin(a+b+c)}{cos(a+b)sina}}$,

in $\triangle ABJ$  $sin(b+c)cos(2a+2b)=cos(a+b+c)sina\cdots (1)$

$2cos(a+b+c)cos(a+b)cos(a+b)sina=cosacos(2a+2b)sin(a+b+c)$  $\iff$  $2cos^{2}(a+b)sin(b+c)=cosasin(a+b+c)$

$\iff$  $(cos(2a+2b)+1)sin(b+c)=cosasin(a+b+c)$  $\iff$  $(cos(2a+2b)+1)sin(b+c)=sin(2a+b+c)-sin(b+c)cos(2a+2b)$  $\iff$

$sin(b+c)(2cos(2a+2b)+1)=sin(2a+b+c)$ $\iff$ $2cos(2a+2b)sin(b+c)=sin(2a+b+c)-sin(b+c)=2sinasin(a+b+c)$  Which is obviously true Q.E.D

[quote="Yimin Ge"]http://www.artofproblemsolving.com/Forum/download/file.php?id=9462&mode=view
[/quote]
How can you draw geometries shapes?

Dear Mathlinkers,
I come back with the same idea where I made a typo:
can we imagine a proof without trigonometry, ratio, angle, homothetie, Menelaüs and case of egality of triangles?
I think so.
Sincerely
Jean-Louis

Found the following solution as mentioned by @jayme

Let $I_A$ be the incenter of $\triangle ABD$, then $D,J, I_A,L$ are concyclic with diameter $I_AJ$ (for example use moving point to prove it). Let $H_A$ be the orthocenter of $AI_AL$, then $\measuredangle I_AH_AL = \measuredangle LAI_A = \measuredangle I_ADL$. Hence $H_A$ also lies on $\odot(I_AJ)$. Therefore $AH_A \parallel JL$ (both perpendicular to $I_AL$) and $H_AJ \parallel AL$ (both perpendicular to $I_AH$), implying that $AH_AJL$ is a parallelogram. Finally we conclude as $H_A$ lies on $KL$.

[quote=Yimin Ge]Let $P$ be the point of intersection of $JD$ and $KL$, $I$ be the incentre of $\triangle ABC$, $Q$ be the excentre of $\triangle BAD$ opposite to $B$ and $M$ be the midpoint of $AD$.

A little angle-chasing shows that $\angle PDL=\angle DAQ =\angle ALK=\angle AKL = 90^\circ-\alpha/2$, $\angle AQM=\alpha/2$ and $\angle JBQ=\beta/2$.

From $\angle ALK=\angle QAD$ we conclude that $KL\parallel AQ$, so it is enough to prove that $QJ=2\cdot QP$.

It follows from $\angle PDQ=\angle PDL=\angle AKL=\angle AKP$ that $AKDP$ is cyclic, hence $\angle QPA=\angle DKA$, so the triangles $QPA$ and $AKD$ are similar. Hence
\[\frac{QP}{AK}=\frac{QA}{AD}= \frac{1}{2}\cdot\frac{QA}{AM}=\frac{1}{2\sin\frac{\alpha}{2}},\ \ \ \ \ \ \ \ (1) \]
the latter following from $\QMA=90^\circ$.

Also, since $\angle JBQ=\angle IBA=\beta/2$ and $\angle JQB=\alpha/2=\angle IAB$, the triangles $QJB$ and $AIB$ are similar, so
\[\frac{QJ}{AI}=\frac{BQ}{BA}=\frac{\sin(90^\circ-\frac{\alpha}{2})}{\sin\frac{\alpha}{2}}= \frac{\cos\frac{\alpha}{2}}{\sin\frac{\alpha}{2}}.\ \ \ \ \ \ \ \ (2) \]
From (1) and (2), we conclude that
\[\frac{QJ}{QP}=2\cdot\sin\frac{\alpha}{2}\cdot\frac{\cos\frac{\alpha}{2}}{\sin\frac{\alpha}{2}}\cdot \frac{AI}{AK}= 2\cdot\cos\frac{\alpha}{2}\cdot\frac{AI}{AK}.\ \ \ \ \ \ \ \ (3) \]
However, since $AKI=90^\circ$, we have
\[\frac{AI}{AK}=\frac{1}{\cos\frac{\alpha}{2}}\ \ \ \ \ \ \ \ (4) \]
so combining (3) and (4), we obtain
\[\frac{QJ}{QP}=2, \]
as desired. $\Box$[/quote]

It was incredibly beautiful

define X as the point on AC such that JX is parallel to KL, now we want to show that AX=2AL
 By simple angle chasing we get JXD is isosceles, so if we define H as the projection of J on AC we have DX=2DH
s is the semi perimeter of ABC and   s' is the semi perimeter of BDC, so we have s=(a+b+c)/2 and s'=(a+b+c-AD)/2
so we conclude by writing 2AL=2(s-a)=2s-2a=2s'+2(AD/2)-2a=2(s'-a)+AD=2DH+AD=DX+DA=AX
and we are done.

[img width=75]https://cdn.discordapp.com/attachments/872490629714808853/947215925109723146/unknown.png[/img]

Fantastic problem!

Let the incircle intersect $\overline{BC}$ at $M$. Let $X$ be the intersection of $\overline{LM}$ and the angle bisector of $BAC$. Let $N$ be the midpoint of $\overline{AJ}$. By the Iran Lemma, $\angle AXB=90^\circ$. Consider the homothety at $A$ of scale factor $2$ that takes $X$ to $A'$ and $L$ to $L'$. By symmetry, $BA=BA'$ and $A'$ is on $\overline{BC}$. By symmetry, \[\angle ALN=90-\frac{\angle A}{2}=\angle BDJ=\angle BA'J=\angle AL'J,\] so $\overline{KL} \parallel \overline{JL'} \parallel \overline{LN}$, so $N$ is on $\overline{KL}$.

Edit: 100th post :D

Why did I try to length chase/PoP the first claim for so long...

Let $K'$ and $L'$ be the reflections of $A$ in $K$ and $L,$ respectively.

[b][color=lightblue]Claim:[/color][/b] $BIJK'$ is cyclic.
[i]Proof.[/i] \begin{align*}\angle IBJ&=180-\angle BJC=180-(90+\tfrac{1}{2}\angle BDC)\\&=90-\tfrac{1}{2}(180-\angle ADB)=\tfrac{1}{2}\angle A=\angle K'AI=\angle IK'B\end{align*} since $AI=IK'.$ $\blacksquare$

Then, $$\angle JK'A=180-\angle BIC=180-(90+\tfrac{1}{2}\angle A)=90-\tfrac{1}{2}\angle A=\angle LKA=\angle L'K'A$$ so $K',J,$ and $L'$ are collinear. By homothety at $A$ with scale factor $\tfrac{1}{2},$ $\overline{AJ}\cap\overline{KL}$ is the midpoint of $\overline{AJ}.$ $\square$

Let $KL\cap JD=P$ and $AJ\cap KL=E$ and $\angle BAC=a$. Note that $\angle LDP=90^{\circ}-\frac{a}{2}=\angle DLP$. By Menelaus's theorem on $\triangle AJD$ with collinear points $E,L,P$, we have that $$\frac{AE}{EJ}\cdot \frac{JP}{DP}\cdot\frac{DL}{LA}=1 \implies \frac{AE}{EJ}=\frac{DP}{LD}\cdot\frac{AL}{JP}$$ Note that by Law of sines in $\triangle LDP$, we have that $\frac{DP}{LD}=\frac{\sin{(90^{\circ}-a/2)}}{\sin{a}}=\frac{1}{2\sin(a/2)}$. Now $$JP=DP+JD=\frac{DL}{2\sin{(a/2)}}+\frac{BD+DC-BC}{2\sin{(a/2)}}=\frac{DL+BA+DC-BC}{2\sin{(a/2)}}=\frac{BA+AC-BC-AL}{2\sin{(a/2)}}=\frac{2AL-AL}{2\sin{(a/2)}}=\frac{AL}{2\sin{(a/2)}}\implies \frac{AL}{JP}=2\sin{(a/2)}$$ Hence $\frac{AE}{EJ}=\frac{1}{2\sin{(a/2)}}\cdot 2\sin{(a/2)}=1$ as desired.

Let $X$ be the projection of $B$ onto $AC$, the incenter of $ABC$ be $I$, $T = BX \cap KL$, $M = AI \cap KL$, the incircle of $BCD$ touch $\overline{ADC}$ at $P$, and $S_{\delta}$ denote the semiperimeter of triangle $\delta$. It's easy to see $XA = XD$ and $AI \perp KL$.

Clearly, we have $$S_{ABC} - S_{BCD} = \frac{CA - CD}{2} = \frac{AD}{2} = XA$$ so $$LP = CL - CP = (S_{ABC} - BA) - (S_{BCD} - BD) = XA.$$ Thus, basic length chasing yields $AL = XP$ and $XL = DP$. Now, because $$\angle TLX = 90^{\circ} - \angle MAL = 90^{\circ} - \frac{\angle A}{2} = 90^{\circ} - \frac{\angle BDA}{2} = \angle JDP$$ follows from properties of external bisectors, we know $TLX \cong JDP$. This implies that $JPXT$ is a rectangle, so $$\overrightarrow{TJ} = \overrightarrow{XP} = \overrightarrow{AL}$$ which means $ALJT$ is a parallelogram. Hence, we know $\overline{KTL}$ bisects $AJ$, as required. $\blacksquare$

[b]Remark:[/b] I almost missed the parallelogram argument completely. In fact, I was prepared to perform a full out length bash.

[hide=Diagram][img width=34]https://media.discordapp.net/attachments/986049049834713139/1076716673186418739/Screenshot_2023-02-18_at_21.07.33.png[/img][/hide]

Let $M$ be the midpoint of $AD$, $N$ be the perpendicular from $J$ to $CD$, and $P$ be the point such that $L$ is the midpoint of $AP$.

Note that $2AL-2ND=(AB+AC-BC)-(BD+CD-BC)=AD$ so we have \[ND=2AL-ND-AD=2AL-AN=AL-(AN-AL)=PL-NL=PN\]
Since $JN\perp PD$, \[\angle JPA=\angle JDC=\frac12 (180^\circ-\angle BDA)=90^\circ-\frac12\angle BAC=\angle KLA\] so $JP\parallel KL$. Since $KL$ bisects $AP$, it must also bisect $AJ$.

[hide = Merry Christmas with a very natural trig solution of mine from 8 years ago]

Our strategy is as follows: $KL$ bisects $AJ$ if and only if the areas of $AKL$ and $JKL$ are equal -- the best way to view this is by showing that the distances $JS$ $(S\in KL$) and $AN$ (where $N = AI \cap KL$, with $AI \perp KL$) from $A$ and $J$ to $KL$ are equal in length. Computing $AN$ seems straightforward, while $JS$ shall be computed from the trapezoid $IJSN$ with $\angle JIN = \angle AIC = 90^{\circ} + \frac{\beta}{2}$ after we compute the base $IN$ and the leg $IJ$.

From triangle $ANK$ we have $AN = AK\cos\angle IAK = (p-a)\cos \frac{\alpha}{2}$. From triangles $INK$ and $AIK$ we obtain $IN = IK\sin \frac{\alpha}{2} = AK \sin \frac{\alpha}{2} \tan \frac{\alpha}{2} = (p-a)\sin \frac{\alpha}{2} \tan \frac{\alpha}{2}$.

To compute $IJ = CI - CJ$, let $X$ and $Y$ be the feet of the perpendiculars from $I$ and $J$ to $BC$. The right triangles $CIX$ and $CJY$ give $CI = \frac{CX}{\cos \frac{\gamma}{2}} = \frac{a+b-c}{2\cos \frac{\gamma}{2}}$ and $CJ = \frac{CY}{\cos \frac{\gamma}{2}} = \frac{BC+CD-BD}{2\cos \frac{\gamma}{2}} = \frac{a + b - c - AD}{2\cos \frac{\gamma}{2}}$ (we used $BD = AB = c$ from the problem conditions). Since $AD = 2AT = 2c\cos\alpha$ from the isosceles triangle $ABD$ (here $T$ is the midpoint of $AD$), we obtain
\[ IJ = \frac{AD}{2\cos \frac{\gamma}{2}} = \frac{c\cos\alpha}{\cos \frac{\gamma}{2}}.\]
Now focus on the trapezoid $JINS$. If $Z$ is the foot of the perpendicular from $I$ to $JS$, then
\[ JS = JZ + ZS = IJ\sin\frac{\beta}{2} + IN = \frac{c\cos\alpha\sin\frac{\beta}{2}}{\cos \frac{\gamma}{2}} + \frac{b+c-a}{2}\frac{\sin^2 \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} \]
and now it suffices to verify that the latter equals $AN = \frac{b+c-a}{2}\cos \frac{\alpha}{2}$. The difference of these is (using $c = 2R\sin \gamma = 4R\sin \frac{\gamma}{2}\cos\frac{\gamma}{2}$ and $\cos^2\frac{\alpha}{2} - \sin^2\frac{\alpha}{2} = \cos\alpha$)
\[ 4R\sin\frac{\gamma}{2}\sin\frac{\beta}{2}\cos\alpha - \frac{b+c-a}{2\cos \frac{\alpha}{2}}\cos\alpha = R\cos\alpha\left(4\sin\frac{\beta}{2}\sin \frac{\gamma}{2} - \frac{\sin \beta + \sin \gamma - \sin\alpha}{\cos \frac{\alpha}{2}}\right)  \]
and so it remains to show $\sin\beta + \sin \gamma - \sin \alpha = 4\cos \frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}$. For this, compute
\[ \sin\beta + \sin \gamma - \sin \alpha = \sin \beta + \sin \gamma - \sin(\beta + \gamma) = \sin\beta + \sin \gamma - \sin\beta\cos\gamma - \cos\beta\sin \gamma \]
\[ = \sin\beta(1-\cos \gamma) + \sin\gamma(1-\cos \beta) = 4\sin\frac{\beta}{2}\cos\frac{\beta}{2} \sin^2\frac{\gamma}{2} + 4\sin\frac{\gamma}{2}\cos\frac{\gamma}{2} \sin^2\frac{\beta}{2} \]
\[ = 4\sin\frac{\beta}{2}\sin\frac{\gamma}{2}\left(\sin\frac{\gamma}{2}\cos\frac{\beta}{2} + \cos\frac{\gamma}{2}\sin\frac{\beta}{2}\right) =  4\sin\frac{\beta}{2}\sin\frac{\gamma}{2}\sin \frac{\beta+\gamma}{2} =  4\cos \frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}. \]

[/hide]

[hide = Remark] Some of the final steps could be quicker if one instead expresses $AN$ and $IN$ via the inradius $r$ and angles of $ABC$ and uses directly the well known formula $r = 4R\sin \frac{\alpha}{2}\sin \frac{\beta}{2}\sin \frac{\gamma}{2}$ (whose proof in fact goes through an expression such as $\sin \beta + \sin \gamma - \sin \alpha$, which happened to us at the end of the illustrated proof). [/hide]

Let $CJ$ meet $KL$ at $X$ and the external $A$-bisector at $Y$. Since $KL\parallel AY$, we have the condition is equivalent to $X$ being the midpoint of $JY$.

Now $X$ is the iran point, so $BX\perp CX$ and we now need to show that $\angle JBX=\angle XBY$. But since $Y$ is the $C$-excentre it is not hard to chase that both angles equal $90^\circ-\frac{A}{2}$. So we are done.

easy solve without diagram

We use barycentric coordinates with $A=(1,0,0)$, $B=(0,1,0)$, and $C=(0,0,1)$. Let $D=:(d,0,1-d)$ so
\[
db=AD=2c\cos A=\frac{b^2+c^2-a^2}{b}.
\]Also $K=(s-b:s-a:0)$ and $L=(s-c:0:s-a)$. The angle bisector of $\angle BCD$ is given by $(a:b:t)$ where $t$ varies. Since $BD=c$, by the Angle Bisector Theorem we have $CD=\frac{a}{a+c}(b-db)=\frac{ab(1-d)}{a+c}$. Thus the angle bisector of $\angle CBD$ is given by $(a(1-d):t:c+ad)$ where $t$ varies. It follows that
\[
J=(a(1-d):b(1-d):c+ad)=\left(\frac{a(1-d)}{a+b+c-db},\frac{b(1-d)}{a+b+c-db},\frac{c+ad}{a+b+c-db}\right).
\]
Line $KL$ is given by
\[
\begin{vmatrix}
x & y & x\\
s-b & s-a & 0\\
s-c & 0 & s-a
\end{vmatrix}
=0
\]
which simplifies to $(s-a)x-(s-b)y-(s-c)z=0$. The midpoint of $\overline{AJ}$ is
\[
\frac{1}{2}\left(1+\frac{a(1-d)}{a+b+c-db},\frac{b(1-d)}{a+b+c-db},\frac{c+ad}{a+b+c-db}\right).
\]
Now we just need to check that it lies on line $KL$. This is easy:
\begin{align*}
&(s-a)(2a+b+c-ad-db)-(s-b)b(1-d)-(s-c)(c+ad)\\
=\ &2as+sb+sc-sad-sdb-2a^2-ab-ac+a^2d+adb-sb+sdb+b^2-b^2d-sc-sad+c^2+cad\\
=\ &a(2s-b-c-2a)-ad(2s-a-b-c)+b^2+c^2-b^2d\\
=\ &-a^2+b^2+c^2-(b^2+c^2-a^2)\\
=\ &0.
\end{align*}
We are done. $\square$


Let $ x,y,z\in \mathbb{R}^\plus{}$, Find the minimum of

(a) $ \frac{x^2\plus{}y^2\plus{}z^2}{xy\plus{}yz}$

(b) $ \frac{x^2\plus{}y^2\plus{}2z^2}{xy\plus{}yz}$.

a) $ \frac {x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} yz} \equal{} \frac {(x^2 \plus{} \frac {y^2}{2}) \plus{} (\frac {y^2}{2} \plus{} z^2)}{xy \plus{} yz}$, Apply $ AM \minus{} GM$ inequality.
 $ \frac {(x^2 \plus{} \frac {y^2}{2}) \plus{} (\frac {y^2}{2} \plus{} z^2)}{xy \plus{} yz}\geq \frac {\frac {2}{\sqrt {2}}(xy \plus{} yz)}{xy \plus{} yz} \equal{} \sqrt {2}$.
b)  $ \frac {x^2 \plus{} y^2 \plus{} 2z^2}{xy \plus{} yz} \equal{} \frac {(x^2 \plus{} \frac {2y^2}{3}) \plus{} (\frac {y^2}{3} \plus{} 2z^2)}{xy \plus{} yz}$, Apply $ AM \minus{} GM$ inequality.
$ \frac {(x^2 \plus{} \frac {2y^2}{3}) \plus{} (\frac {y^2}{3} \plus{} 2z^2)}{xy \plus{} yz}\geq \frac {2\frac {\sqrt {2}}{\sqrt{3}}(xy \plus{} yz)}{xy \plus{} yz} \equal{} 2\sqrt{\frac {2}{3}}$.

(a) The minimum value is $ \sqrt{2}$.  Expanding
\[ \left( x \minus{} \frac{\sqrt{2}}{2} y \right)^2 \plus{} \left( \frac{\sqrt{2}}{2} y \minus{} z \right)^2 \ge 0,\]
we get $ x^2 \plus{} y^2 \plus{} z^2 \minus{} \sqrt{2} xy \minus{} \sqrt{2} yz \ge 0$, so
\[ \frac{x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} yz} \ge \sqrt{2}.\]
Equality occurs, for example, if $ x \equal{} 1$, $ y \equal{} \sqrt{2}$, and $ z \equal{} 1$.

(b) The minimum value is $ \sqrt{8/3}$.  Expanding
\[ \left( x \minus{} \sqrt{\frac{2}{3}} y \right)^2 \plus{} \frac{1}{3} \left( y \minus{} \sqrt{6} z \right)^2 \ge 0,\]
we get $ x^2 \plus{} y^2 \plus{} 2z^2 \minus{} \sqrt{8/3} xy \minus{} \sqrt{8/3} yz \ge 0$, so
\[ \frac{x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} yz} \ge \sqrt{\frac{8}{3}}.\]
Equality occurs, for example, if $ x \equal{} 2$, $ y \equal{} \sqrt{6}$, and $ z \equal{} 1$.

[quote="nsato"](a) The minimum value is $ \sqrt {2}$.  Expanding
\[ \left( x \minus{} \frac {\sqrt {2}}{2} y \right)^2 \plus{} \left( \frac {\sqrt {2}}{2} y \minus{} z \right)^2 \ge 0,
\]
we get $ x^2 \plus{} y^2 \plus{} z^2 \minus{} \sqrt {2} xy \minus{} \sqrt {2} yz \ge 0$, so
\[ \frac {x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} yz} \ge \sqrt {2}.
\]
Equality occurs, for example, if $ x \equal{} 1$, $ y \equal{} \sqrt {2}$, and $ z \equal{} 1$.
[/quote]
Can we always assume the minimum value this way and prove the inequality? I mean, for instance I could assume that the minimum is $ \frac{3}{2}$ and then prove it and show that, equality when $ x\equal{}y\equal{}z\equal{}1$ :?:

Or prove in the way that Achiles followed?

[quote="math_07"][quote="nsato"](a) The minimum value is $ \sqrt {2}$.  Expanding
\[ \left( x \minus{} \frac {\sqrt {2}}{2} y \right)^2 \plus{} \left( \frac {\sqrt {2}}{2} y \minus{} z \right)^2 \ge 0,
\]
we get $ x^2 \plus{} y^2 \plus{} z^2 \minus{} \sqrt {2} xy \minus{} \sqrt {2} yz \ge 0$, so
\[ \frac {x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} yz} \ge \sqrt {2}.
\]
Equality occurs, for example, if $ x \equal{} 1$, $ y \equal{} \sqrt {2}$, and $ z \equal{} 1$.
[/quote]
Can we always assume the minimum value this way and prove the inequality? I mean, for instance I could assume that the minimum is $ \frac {3}{2}$ and then prove it and show that, equality when $ x \equal{} y \equal{} z \equal{} 1$ :?:

Or prove in the way that Achiles followed?[/quote]
Of course you can't prove that the minimum value is $ \frac{3}{2}$. There is a counter-example! (Namely $ x\equal{}z\equal{}1, y\equal{}\sqrt{2}$)

$(\text a)$ We have, by AM-GM:
$$x^2+y^2+z^2=x^2+\frac12y^2+\frac12y^2+z^2\ge xy\sqrt 2+yz\sqrt 2$$
so $\frac{x^2+y^2+z^2}{xy+yz}\ge\sqrt2$, with equality when $(x,y,z)=\left(\sqrt2,2,\sqrt2\right)$.
$(\text b)$ We have, by AM-GM:
$$x^2+y^2+2z^2=x^2+\frac23y^2+\frac13y^2+2z^2\ge\frac{2\sqrt6}3xy+\frac{2\sqrt6}3yz$$
so $\frac{x^2+y^2+2z^2}{xy+yz}\ge\frac{2\sqrt6}3$, with equality when $(x,y,z)=\left(2\sqrt6,6,\sqrt6\right)$.


geometry
geometric transformation
dilation
ratio
area of a triangle
Heron\u0027s formula

Find all triangles $ \triangle ABC$ such that $ \overline{AB},\overline{BC},\overline{AC},$ and the shortest altitude are consecutive integers.

[hide="Hidden by mod"]shortest altitude goes to the longest side. let $ x$ be the shortest side. altitudes are never longer than sides, so $ x \minus{} 1$ is the altitude.

so like let's solve $ \sqrt {2x \minus{} 1} \plus{} \sqrt {4x} \equal{} x \plus{} 2$

$ 2x \minus{} 1 \plus{} 4x \plus{} 2\sqrt {8x^2 \minus{} 4x} \equal{} x^2 \plus{} 4x \plus{} 4$

$ x^2 \minus{} 2x \plus{} 5 \equal{} 2\sqrt {8x^2 \minus{} 4x}$

$ x^4 \minus{} 4x^3 \plus{} 14x^2 \minus{} 20x \plus{} 25 \equal{} 32x^2 \minus{} 16x$

$ x^4 \minus{} 4x^3 \minus{} 18x^2 \minus{} 4x \plus{} 25 \equal{} 0$

$ (x \minus{} 1)(x^3 \minus{} 3x^2 \minus{} 21x \minus{} 25) \equal{} 0$

the first thingy gives us $ x \equal{} 1$. the second thingy is horrifying and you can verify that it doesn't give off anything integral.

since we can't really have an altitude of length 0, nor can we have a 1-2-3 triangle, this leads me to believe that there's no triangles that fit the given conditions, unless i'm misinterpreting something[/hide]

[quote="Lazarus"]altitudes are never longer than sides[/quote]
What do you mean by that? Of course it's possible for some altitude to be longer than the shortest side. In the 3-4-5 right triangle, 4 is an altitude...

BTW it's not entirely clear if the original problem asks about cases where all 4 lengths are different. But regardless of that, I don't think you can assume the altitude is $ x\minus{}1$.

err, my bad, i meant that shortest altitude $ \leq$ shortest side, middle altitude $ \leq$ middle side, and longest altitude $ \leq$ longest side

and i'm pretty sure that you can assume that.

draw the altitude to the longest side and pythag theorem it with the other two sides.

Right. So I think the point here is that the 4 numbers $ \{a,b,c,h\}$ can be just 3 consecutive integers, as in the triangle with sides $ (5,5,6)$ whose short altitude is 4.

[quote="Lazarus"]
<snip>
so like let's solve $ \sqrt {2x \minus{} 1} \plus{} \sqrt {4x} \equal{} x \plus{} 2$
<snip>
since we can't really have an altitude of length 0, nor can we have a 1-2-3 triangle, this leads me to believe that there's no triangles that fit the given conditions, unless i'm misinterpreting something[/quote]

You're correct in that there are no triangles that fit the given conditions. I compared the results of Heron's formula and the product of the longest side and shortest altitude to obtain the same polynomial. Edit: okay, I get your solution now, I hadn't seen your second post above.

@randomgraph: I meant four [b]distinct[/b] consecutive integers--sorry if that was unclear.

Extension: Find all triangles  such that $ \overline{AB}$,$ \overline{BC}$,$ \overline{AC}$, and the shortest altitude $ h_s$ are consecutive numbers in an arithmetic progression.

[quote="akessler"]
I meant three [b]distinct[/b] consecutive integers--sorry if that was unclear.
[/quote]

4.  :wink: (Lazarus's equation came from two applications of Pythagorean on the altitude, as he noted above)

The extension yields the same result; if the common difference is $ d$ then dilate the triangle by ratio $ 1/d$.


calculus
integration
trigonometry
logarithms
calculus computations

Evaluate $ \int_0^1 \frac{(1\minus{}x\plus{}x^2)\cos \ln (x\plus{}\sqrt{1\plus{}x^2})\minus{}\sqrt{1\plus{}x^2}\sin \ln (x\plus{}\sqrt{1\plus{}x^2})}{(1\plus{}x^2)^{\frac{3}{2}}}\ dx$.

put $ x\equal{}\text{sinh}y$ and then its straightforward...

hello, i don't think it is so easy.
Sonnhard.

hello kunny, he meant $ x\equal{}\sinh(y)$, the hyberbolic function.
Sonnhard.

Thank you! I should have looked up by Wikipedia etc.

answer $ \equal{} sin (\log(1\plus{} \sqrt{2} )) \plus{} \frac{cos(\log(1\plus{}\sqrt{2} ) }{\sqrt{2}} \minus{} 1$ ,actually its late night here and i have to go to bed so will post my entire solution tomorrow in the evening sorry for the inconvienience .

[quote]Today's calculation of Integral 462
created by kunny
Evaluate $ \int_0^1 \frac {(1 - x + x^2)\cos \ln (x + \sqrt {1 + x^2}) - \sqrt {1 + x^2}\sin \ln (x + \sqrt {1 + x^2})}{(1 + x^2)^{\frac {3}{2}}}\ dx.$
[/quote]

 $ \int_0^1 \frac {(1 - x + x^2)\cos \ln (x + \sqrt {1 + x^2}) - \sqrt {1 + x^2}\sin \ln (x + \sqrt {1 + x^2})}{(1 + x^2)^{\frac {3}{2}}}\ dx.$

Now separate the integral as :

${ \int_{0}^{1}\frac {(1 + x^2)(cos\ln (x + \sqrt {1 + x^2}))}{(1 + x^2)^{\frac {3}{2}}}}$  $ - \int_{0}^{1}\frac {xcos(\ln(x + \sqrt {1 + x^2})}{(1 + x^2)^{\frac {3}{2}}}$  $ - \int_{0}^{1}\frac {(\sqrt {1 + x^2})(sin\ln(x + \sqrt {1 + x^2})}{(1 + x^2)^{\frac {3}{2}}}$

in the first integral put $ \ln (x + \sqrt {1 + x^2})) = t$  and evaluate .

in the 2nd integral use parts .when we use parts we see that a part of this and the third integral cancels out .

leaving $ (sin\ln(x + \sqrt {1 + x^2})$ from 0 to 1 and  $ \frac {(cos\ln (x + \sqrt {1 + x^2}))}{(1 + x^2)^{\frac {1}{2}}}$ from o to 1

after evaluating i get the answer as mentioned above.

i hope this is correct .

thankyou  :)

Thanks KUNNY people like u are great source of encouragement for me


geometry
rectangle
combinatorics unsolved
combinatorics

Initially,the number $ 1$ or $ \minus{}1$ is arbitrarily assigned to each cell of a $ 9X9$ board.For each cell $ C$,compute the product of all numbers in cells sharing exactly one edge with $ C$.Simultaneously,all numbers are replaced by the corresponding products.Is it always possible,by performing the above operation a finite number of times,to change all $ 81$ numbers to $ 1$'s?

[quote="KDS"]Initially,the number $ 1$ or $ \minus{} 1$ is arbitrarily assigned to each cell of a $ 9X9$ board.For each cell $ C$,compute the product of all numbers in cells sharing exactly one edge with $ C$.Simultaneously,all numbers are replaced by the corresponding products.Is it always possible,by performing the above operation a finite number of times,to change all $ 81$ numbers to $ 1$'s?[/quote]
No. Take the following pattern ($ o$ means $ \minus{} 1$)

$ \texttt{ \plus{} \ o\ o\ \plus{} \ \plus{} \ \plus{} \ o\ o\ \plus{} \ }$
$ \texttt{o\ \plus{} \ \plus{} \ o\ \plus{} \ o\ \plus{} \ \plus{} \ o\ }$
$ \texttt{o\ \plus{} \ \plus{} \ o\ \plus{} \ o\ \plus{} \ \plus{} \ o\ }$
$ \texttt{ \plus{} \ o\ o\ \plus{} \ \plus{} \ \plus{} \ o\ o\ \plus{} \ }$
$ \texttt{ \plus{} \ \plus{} \ \plus{} \ \plus{} \ \plus{} \ \plus{} \ \plus{} \ \plus{} \ \plus{} \ }$
$ \texttt{ \plus{} \ o\ o\ \plus{} \ \plus{} \ \plus{} \ o\ o\ \plus{} \ }$
$ \texttt{o\ \plus{} \ \plus{} \ o\ \plus{} \ o\ \plus{} \ \plus{} \ o\ }$
$ \texttt{o\ \plus{} \ \plus{} \ o\ \plus{} \ o\ \plus{} \ \plus{} \ o\ }$
$ \texttt{ \plus{} \ o\ o\ \plus{} \ \plus{} \ \plus{} \ o\ o\ \plus{} \ }$

 or

$ \texttt{o\ \plus{} \ \plus{} \ o\ \plus{} \ o\ \plus{} \ \plus{} \ o\ }$
$ \texttt{o\ o\ o\ o\ \plus{} \ o\ o\ o\ o\ }$
$ \texttt{o\ o\ o\ o\ \plus{} \ o\ o\ o\ o\ }$
$ \texttt{o\ \plus{} \ \plus{} \ o\ \plus{} \ o\ \plus{} \ \plus{} \ o\ }$
$ \texttt{ \plus{} \ \plus{} \ \plus{} \ \plus{} \ \plus{} \ \plus{} \ \plus{} \ \plus{} \ \plus{} \ }$
$ \texttt{o\ \plus{} \ \plus{} \ o\ \plus{} \ o\ \plus{} \ \plus{} \ o\ }$
$ \texttt{o\ o\ o\ o\ \plus{} \ o\ o\ o\ o\ }$
$ \texttt{o\ o\ o\ o\ \plus{} \ o\ o\ o\ o\ }$
$ \texttt{o\ \plus{} \ \plus{} \ o\ \plus{} \ o\ \plus{} \ \plus{} \ o\ }$

Both constellations are stationary, as each o is neighbor of an odd number of o's, and each + is neighbor of an even number of o's. Iterating the patterns, it is easy to see that there are stationary constellations for any $ a\times b$-board where $ a,b\in \{4,9,14,...\}$.
Likewise, the pattern

$ \texttt{o\ o\ \plus{} \ o\ o\ \plus{} \ ...\ \plus{} \ o\ o\ }$
$ \texttt{ \plus{} \ \plus{} \ \plus{} \ \plus{} \ \plus{} \ \plus{} \ ...\ \plus{} \ \plus{} \ \plus{} \ }$
$ \texttt{o\ o\ \plus{} \ o\ o\ \plus{} \ ...\ \plus{} \ o\ o\ }$
$ \texttt{ \plus{} \ \plus{} \ \plus{} \ \plus{} \ \plus{} \ \plus{} \ ...\ \plus{} \ \plus{} \ \plus{} \ }$
.
.
.
$ \texttt{ \plus{} \ \plus{} \ \plus{} \ \plus{} \ \plus{} \ \plus{} \ ...\ \plus{} \ \plus{} \ \plus{} \ }$
$ \texttt{o\ o\ \plus{} \ o\ o\ \plus{} \ ...\ \plus{} \ o\ o\ }$

shows that this is also the case for any $ a\times b$-board where $ a\in\{3,5,7,...\}$ and $ b\in\{2,5,8,11,...\}$.

For some of these rectangles there also exist other stationary constellations, but I haven't been able to find any for another rectangle. I would further conjecture that if there is no stationary constellation, the process will sooner or later result in all "+" for any constellation, but I have no idea how to prove this  :maybe:


calculus
integration
trigonometry
logarithms
parameterization
calculus computations

question 1:
$\begin{document}
\displaystyle
\mathop \int_0^{1}\frac{1}{x(1-x)^\frac{1}{2}}dx
\end{document}$

question 2:
$\begin{document}
\displaystyle
\mathop \int_0^{1}\frac{x-1}{lnx}dx
\end{document}$

For question 1 i get 0 as the answer any opinions?

[quote="Kent Merryfield"]Unless eplag meant to say $\int_0^1\frac{dx}{\sqrt{x(1-x)}}$ which would be a different problem and which would converge.[/quote]
For this integral, let $x=\sin^2 t$ and it becomes $\int_0^\frac{\pi}{2} 2dt = \pi$

[quote="liyi"]the first integral diverges because that $\frac{1}{x(1-x)^\frac{1}{2}}>\frac{1}{x}$ and $\int_0^a \frac{dx}{x}$ diverges.[/quote]

$
\begin{document}
\displaystyle
\mathop \int_0^{1}\frac{1}{x(1-x)^ \frac{1}{2}}=x(1-x)^\frac{-1}{2}=\frac{x(1-x)^\frac{1}{2}}{\frac{1}{2}}\longrightarrow f(1)-f(0) = 0
\end{document}$

Thanks guys for the info, eplag, this is what i get as my answer for q1  :? i think we both have it wrong

[quote="Jimmy"]$\int \frac{1}{x(1-x)^{1/2}}\  dx = 2\ln(\sqrt{1-x}-1)-\ln(x)$
$\left [ 2\ln(\sqrt{1-x}-1)-\ln(x) \right]_0^1$ diverges.
When I typed this into my calculator it gave me $\infty +2\pi i$ (clearly not a number):D
fotia, can you explain your steps? :?[/quote]
actually jimmy, i was afraid of this. i ignored the ln rule for integrals and just re arranged the fraction so that it can be easier to calculate. so instead of it being 1 / x(1-x)^1/2 i just made the 1/2 negative to remove the fraction. from there I did the anti derivative. I can see my mistake :) thanks for your input.

(i have made this post into a new topic) topic name: converge/diverge

[quote="blahblahblah"]It IS $\ln 2$.  This is a classic example of Leibniz's Rule.

Let $I(t)=\int^1_0 \frac {x^t-1}{\ln x}$.  Then formally, (bear with me here)

$I'(t)=\int^1_0 x^tdx=\frac {1}{t+1}$.  [/quote]
Quite a nice idea, indeed.  :)  I was trying how to add a parameter to the integral, but I never tried $x^t$.  :blush:

[quote="fotia"]here's one more question that caught my eye while flipping though my mathematics analysis book.

Given that F(x)=$
\begin{document}
\displaystyle
\mathop \int_0^{x}f(s)ds$ and $F(\infty)=1,$ prove that for any $a\geq1$
$
\begin{document}
\displaystyle
\mathop\int_0^{\infty}x^{a}dF(x) = a \int_0^{\infty}x^{a-1}(1-F(x))dx
\end{document}$, where the two sides either converge or diverge together.

its a bit tougher for my taste ;) , but check it out[/quote]

thanks guys for the help, fotia, i've seen a question like that im my book too, maybe we have the same one :P any ideas?

[quote="blahblahblah"][quote="Jimmy"]I think the second one relates to the Li integral, which doesn't have a closed form.
$\int_0^1 \frac{x-1}{\ln(x)}\approx .69314717965363$
When I saw this I thought the answer was $\ln(2)$, but it's slightly off. :([/quote]

It IS $\ln 2$.  This is a classic example of Leibniz's Rule.

Let $I(t)=\int^1_0 \frac {x^t-1}{\ln x}$.  Then formally, (bear with me here)

$I'(t)=\int^1_0 x^tdx=\frac {1}{t+1}$.  Integrating WRT $t$, we get

$I(t)=\ln (t+1)+C$.  Since $I(0)=0$, we must have $C=0$, and setting $t=1$, we get that $I(1)=2$, which is the desired result.[/quote]
That's a really cool integrating method, I guess my TI-89 isn't acurate past a few decimal places with integral approximations. :(


$ \bullet$For which $ n$ is true:
If $ A \subset \mathbb{R}^n$ konvex closed then $ A\plus{}A$ is konvex closed?
$ \bullet$For which $ n$ is true:
If $ A \subset \mathbb{R}^n$ konvex closed then $ A\minus{}A$ is konvex closed?

For all $ n$, for
if $ x\pm y \in A\pm A$ and $ u\pm v \in A\pm A$, then
$ \lambda (x\pm y) \plus{} (1\minus{}\lambda)(u\pm v) \equal{} (\lambda x \plus{} (1\minus{}\lambda)u)  \pm  (\lambda y \plus{} (1\minus{}\lambda)v) \in A \pm A$

Convexity is automatic. The interesting part is whether these sets are closed.
(English note: "convex" is spelled with a "c")

Yes I Know convex.
 $ A$ convex closed then $ A\plus{}A$ is convex. Then $ \frac{A\plus{}A}{2}\equal{}A$. Then $ A\plus{}A$ is convex closed.
The second part: $ A\minus{}A$ convex.For $ n\equal{}1$ is true that is cloed. But for $ n\equal{}2$ let $ f(x)\equal{}\frac{1}{1\minus{}x^2}$. Then graph of $ f$ and the interior is konvex closed but A-A is not.


Given that $ \frac1a\plus{}\frac1b\equal{}\frac7{24}$ and $ a\plus{}b\equal{}14$, what is the product of $ a$ and $ b$?

Remember that $ \frac{1}{a} \plus{} \frac{1}{b} \equal{} \frac{a \plus{} b}{ab}$. Since we know that $ \frac{1}{a} \plus{} \frac{1}{b} \equal{} \frac{7}{24}$, and $ a \plus{} b \equal{} 14$, we can say that \[ \frac{7}{24} \equal{} \frac{14}{ab}\] This means the product is simply $ 2 \cdot 24 \equal{} \boxed{48}$.


$ a_{1}> 0$
${ a_{n+1}= log(1+a_{n}})$
a)Prove that $ a_{n}$ converge to 0
Find lim
$ \lim\frac{n.(a_{n}-2)}{log(n)}$

$ \lim\frac{n.(a_{n}\minus{}2)}{log(n)}\equal{}\minus{}\infty$, because $ a_{n}\to 0$

I'm sorry
lim $ \frac{n.a_{n}\minus{}2}{log(n)}$

I don't undestant, why you need $ log(n)$.
Let $ b_{n}\equal{} na_{n}$, then $ b_{n\plus{}1}\equal{} f_{n}(b_{n}),\ f_{n}(x) \equal{} (n\plus{}1)ln(1\plus{}\frac{x}{n})$.
Let $ x_{n}$ is fixed point for $ f_{n}(x)$. It is easy to chek, that $ 2 < x_{n}< 2\plus{}\frac{1}{n}$.
Therefore $ \lim_{n\to\infty}b_{n}\equal{} 2$.


How many different positive 10 digit integers can be formed using the digits in the following set only once in each number?

{0, 0, 0, 1, 2, 3, 5, 7, 8, 8}

[hide]There's 10! different ways to arrange the numbers, but there's three zeroes and 2 eights, so the total number is $\frac{10!}{3!2!}=403200$.[/hide]

[hide="I got"]403200[/hide].  Same method.

[hide]There are $\frac{10!}{3!2!}=302400$ arrangements, but there are some sequences with zeroes at the beginning. The number of arrangements with zero at the beginning (there are three zeroes) is $3\displaystyle\left(\frac{9!}{2!2!}\right)=272160$. So $302400-272160=\boxed{30240}$.[/hide]
$\frac{10!}{3!2!}\neq403200$ according to my calculator...

Discarding the 0s theres 7!/2! = 2520 ways of ordering the other digits. Theres 7 places to insert a 0, then 8 places after the first zero is inserted, then 9. 7*8*9/3! = 84. 84 * 2520 = 211680? Can someone explain whats wrong with my reasoning.

[quote="VenDiddy"][hide]There are $10!$ arrangements when not considering the duplicates. Because 2 is repeated twice and 0 is repeated three times, the number of unique arrangements of the 10 digits is $\frac{10!}{2!3!}$. All of the numbers starting with 0 aren't 10 digit numbers, and there are $\frac{9!}{2!2!}$ of these (fixing one of the 0's as the first digit).

The total number of digits is $\frac{10!}{2!3!}-\frac{9!}{2!2!}=\boxed{211680}$[/hide][/quote]

[hide]Don't you need three of the $\frac{9!}{2!2!}$ because there are three zeroes that don't make the sequence a 10 digit integer? [/hide]

Wait, now I'm confused.  I guess I agree with chinesecheckers...

[hide="answer"]$\frac{7*9!}{3!*2!}$

7 possibilities for the first digit, then 9,8,7,..1 possibilities for the rest, thus 9!, then we over counted since we considered 0,0,0 as distinct digits, there are exactly 3! times the same thing is counted for 0's, same with 8's[/hide]

EDIT: forgot to hide  :blush:

dunno what took so long.  maybe there wouldn't be so many mistakes if the question were harder ;)

but yeah, i agree with altheman

LOL.  I always agree with the last answer I see.  It's kinda like the Animal Farm debate...


For any $x_i\geq 0$  $(i=1,2,3,\cdots ,n)$ , $m\in \mathbb{R^+}$ prove or disprove that

$\frac{\prod^n_{i=1}(x_i+m)}{\left(\sum^n_{i=1}(x_i+m)\right)^n}\geq \frac{\prod^n_{i=1}x_i}{\left(\sum^n_{i=1}x_i\right)^n}$

No one out there giving this question a shoot ?

 :P

[quote="silouan"]I think that for a very small $x_i$ it is wrong[/quote]why?

[quote="silouan"]I think that for a very small $x_i$ it is wrong[/quote]

Any counter example ?

[quote="Vasc"]For $m\rightarrow\infty$ we get $0 \ge RHS$, which is clearly wrong. :blush:[/quote]Don't we have $\frac{1}{n^n}$ for LHS?

[quote="Vasc"]For $m\rightarrow\infty$ we get $0 \ge RHS$, which is clearly wrong. :blush:[/quote]

Oh , thats weird . I have a proof for it so i guess i need someone to point out what is wrong .

[hide="my solution"]

The inequality is also equivalent to

$\frac{\sum x_i}{\sum (x_i+m)}\geq \left(\prod\frac{x_i}{(x_i+m)}\right)^{\frac{1}{n}}$

since $\{\frac{x_i}{x_i+m}\}$ and $\{x_i+m\}$ is in same order , applying chebyserv

$\frac{\sum x_i}{\sum (x_i+m)}\geq \frac{1}{n}\sum \frac{x_i}{x_i+m}$

Lastly , it follows AM-GM

$\frac{1}{n}\sum \frac{x_i}{x_i+m}\geq \left(\prod\frac{x_i}{(x_i+m)}\right)^{\frac{1}{n}}$

Kindly point out anything wrong  :)
[/hide]

Nice solution, shyong! :)
Indeed, I was wrong when I said:
"For $m\rightarrow\infty$ we get $0 \ge RHS$, which is clearly wrong".  :o


[hide]Factor out $7^{110}$ on the left side, simplify the left, and the cancel out the 41's to get $7^{2n}=7^110$
n=55[/hide]


[quote="fleeting_guest"][quote="vinxenz87"]Can you give more informations about the Z.T.'s  application to the chess game?[/quote]

Which theorem of Zermelo?

The fact that either black or white has a strategy to defeat or match the other player can be thought of as an argument using Zermelo's axiom of choice (looking forward within a potentially infinite game tree), but this is not really essential.[/quote]

I exactely don't know which theorem is.
I read that one of two players can force only one  of the three ending status of the game starting from the first position.
If exists a strategy(the best) that gives to white victory in n moves than does a different strategy that gives to white victory in k>n moves exist?


Given $q$ is prime number.
For each prime number $p$ let $1+\frac{1}{2}+...+\frac{1}{p}=\frac{a_p}{b_p} ($ where $gcd(a_p;b_p)=1)$.

Prove that there exist infinitely many prime number $p$ such that $q\not |a_p$

Can you give a hint? Is it useful to try to appply Dirichlet th which states that in an arithmetic progresion there is an infinite number of primes?

Yes,  using Dirichlet's theorem is fruitful:

Take any prime $p=qm+1$. Then ($\lambda$ is primitive root modulo $q$):

$1+...+\frac{1}{p}\equiv 1+2^{q-2}+...+p^{q-2}\equiv m \frac{\lambda ^{(q-1)(q-2)}-1}{\lambda ^{q-2}-1}+1 \equiv 1 \mod q$ so $q \not | a_p$

I think your solution have some bugs: In the sumation multiples of q appears. You cannot write them as $(smth)^{q-2}$. I also tried this way, but no results came. Can you fix your solution? Or explain why it works? :?

I don't use Drichle theoem but I must use the result : among $n$ and $2n$ there is at least one prime number.

And two results:
a/ if $q\not |a_n$ then $q\not| a_{qn+k} \forall k=0,1,2,...,q-1$
b/ $q\not|a_m\forall m\in[q^k,2q^k-1] \forall k$

You're right Xirti, but for now I don't know how to fix it  :?


Question:
What are the time restrictions on the comprehensive competition at the state finals level?

Thanks!

uh they refuse to say that officially for some reason

looks like it'll be 90 minutes from the schedule on the website?

thanks.

hmm...how does scoring work again?
i saw your other post regarding scoring...but then it seems really unlikely that berman only got a 113...

oh wait maybe it is only 5 points for short answer, 4 for multiple choice

I was probably deluded


Peng Chu = x-22 dollars
Li Juan = x-3 dollars

so 2x-25<2x.

i think the problem is wrong.

it should say 'same 1 cd' instead of 'same 2 cds'

If $ 2x \minus{} 25 < 2x$, then wouldn't $ x$ have infinitely many solutions?

exactly.

i dont think you should put $ \infty$ so i think the problem is wrong.

Hmm, that's what I thought, so that's why I asked this problem.

According to the review, the answer is 24.

Or maybe the money is like this.

Peng Chu- $ 2x \minus{} 22$

Li Juan- $ 2x \minus{} 3$

$ 4x \minus{} 25 < 2x$

$ 2x \minus{} 25 \equal{} 0$

$ x \equal{} 12.5?$

However, note that we want to find the [b]greatest[/b] amount of money those 2 CD's could cost. I think this has some impact on the problem.

[quote="Problem"]Peng Chu and Li Juan are each saving to buy the same two CD's. Peng Chu is lacking $ \$22$ and Li Juan is lacking $ \$3$. When they combine their money they still do not have enough to buy the two CD's. What is the greatest number of whole dollars the two CD's could cost?[/quote]

Say x is the cost of the 2 CD's

Peng Chu has x-22
Li Juan has x-3
Together they have $ 2x\minus{}25<x$

So $ x<25$

Therefore maximum integral value is 24, agreeing with the review.

James got it right, there was no reason to assume that they both wanted 2x as opposed to x.


Fifteen men sit in $ 15$ seats in a line. All of the $ 15$ get up and then reseat themselves using all $ 15$ seats, each sitting in the seat he was in before or a seat next to the one he occupied before. In how many ways can the men be reseated?

987
if im not mistaken....

[hide]
it is a recursive solution...
leads to fibonacci series..
[/hide]

Yes Keshav this is Fibonacci recurrence.

can anyone explain the solution.....  :maybe:

@anurags92
Consider the people as 1,2,3...
Let them be seated originally in  the order 1,2,3...
Let n(k) denote the number of ways k people can be reseated.
Now, 1 can either remain in his seat(Case 1) or go to the second seat(case 2).
Case1
It now boils down to 14[k-1] people being reseated
So no of ways in this case=n(14)[n(k-10)]

Case 2
If 1 goes to seat 2,2 has to go to seat 1.
Therefore it boils down to reseating 13 [k-2] people.
Therefore in this case no of ways =n(13).[n(k-2)]

Therefore total no of ways =n(15)=n(14)+n(13).
n(k)=n(k-1) +n(k-2)
this is fibonacci series,with n(1)=1,n(2)=2.
find n(15).

In a derangement, no person goes back to the seat he was in before. So you'll have to add those extra cases separately-tedious.


inequalities
trigonometry
inequalities unsolved

In triangle ABC,prove or disprove:
$\frac{1-cosA}{1-cosBcosC}+\frac{1-cosB}{1-cosCcosA}+\frac{1-cosC}{1-cosAcosB} \geq 2$

What a beautiful inequality! I have a rather long proof, but I hope you like it.

[color=blue][b]Theorem 1.[/b] For any triangle ABC, we have

$\frac{1-\cos A}{1-\cos B\cos C}+\frac{1-\cos B}{1-\cos C\cos A}+\frac{1-\cos C}{1-\cos A\cos B}\geq 2$.[/color]

[i]Proof of Theorem 1.[/i] Let $s= \frac{a+b+c}{2}$ be the semiperimeter of triangle ABC. Then, it is well-known (see, e. g., [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=208379#p208379]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22198 post #3[/url]) that the numbers x = s - a, y = s - b, z = s - c are positive and satisfy y + z = a, z + x = b, x + y = c and x + y + z = s. Now, by the half-angle formulas, $\sin^2\frac{A}{2}=\frac{\left(s-b\right)\left(s-c\right)}{bc}=\frac{yz}{\left(z+x\right)\left(x+y\right)}$. Thus, $1-\cos A=2\sin^2\frac{A}{2}=2\frac{yz}{\left(z+x\right)\left(x+y\right)}$. Similarly, $1-\cos B=2\frac{zx}{\left(x+y\right)\left(y+z\right)}$ and $1-\cos C=2\frac{xy}{\left(y+z\right)\left(z+x\right)}$, so that

$1-\cos B\cos C=\left(1-\cos B\right)+\left(1-\cos C\right)-\left(1-\cos B\right)\cdot\left(1-\cos C\right)$
$=2\frac{zx}{\left(x+y\right)\left(y+z\right)}+2\frac{xy}{\left(y+z\right)\left(z+x\right)}-2\frac{zx}{\left(x+y\right)\left(y+z\right)}\cdot 2\frac{xy}{\left(y+z\right)\left(z+x\right)}$
$=2x\frac{z\left(y+z\right)\left(z+x\right)+y\left(x+y\right)\left(y+z\right)-2xyz}{\left(y+z\right)^2\left(z+x\right)\left(x+y\right)}=2x\frac{\left(x+y+z\right)\left(y^2+z^2\right)}{\left(y+z\right)^2\left(z+x\right)\left(x+y\right)}$.

Hence,

$\frac{1-\cos A}{1-\cos B\cos C}=\frac{2\dfrac{yz}{\left(z+x\right)\left(x+y\right)}}{2x\dfrac{\left(x+y+z\right)\left(y^2+z^2\right)}{\left(y+z\right)^2\left(z+x\right)\left(x+y\right)}}=\frac{\left(y+z\right)^2}{x^2\left(y^2+z^2\right)}\cdot\frac{xyz}{x+y+z}$.

We will now use the notation $\sum$ for cyclic sums. Then, as a result of this computation, the inequality of Theorem 1, namely $\sum\frac{1-\cos A}{1-\cos B\cos C}\geq 2$, rewrites as $\sum\frac{\left(y+z\right)^2}{x^2\left(y^2+z^2\right)}\cdot\frac{xyz}{x+y+z}\geq 2$. Upon multiplication by $\frac{x+y+z}{xyz}$, this inequality becomes $\sum\frac{\left(y+z\right)^2}{x^2\left(y^2+z^2\right)}\geq 2\frac{x+y+z}{xyz}$, or, equivalently, $\sum\frac{\left(y+z\right)^2}{x^2\left(y^2+z^2\right)}\geq 2\sum\frac{1}{yz}$.

Now, let's forget what the variables a, b, c stood for in the above part of the proof (namely, for the sidelengths of triangle ABC), and define three new variables a, b, c as follows: $a=\frac{1}{x}$, $b=\frac{1}{y}$, $c=\frac{1}{z}$. Then, of course, $x=\frac{1}{a}$, $y=\frac{1}{b}$, $z=\frac{1}{c}$. Now,

$\frac{\left(y+z\right)^2}{x^2\left(y^2+z^2\right)}=\frac{\left(\frac{1}{b}+\frac{1}{c}\right)^2}{\left(\frac{1}{a}\right)^2\left(\left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2\right)}=\frac{a^2\left(b+c\right)^2}{b^2+c^2}$

and

$\frac{1}{yz}=\frac{1}{\frac{1}{b}\cdot\frac{1}{c}}=bc$.

Thus, the inequality in question, $\sum\frac{\left(y+z\right)^2}{x^2\left(y^2+z^2\right)}\geq 2\sum\frac{1}{yz}$, becomes $\sum\frac{a^2\left(b+c\right)^2}{b^2+c^2}\geq 2\sum bc$. Now, since $2\sum bc=\sum\left(ca+ab\right)$, we can equivalently transform this inequality as follows:

$\sum\frac{a^2\left(b+c\right)^2}{b^2+c^2}\geq 2\sum bc\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ \sum\frac{a^2\left(b+c\right)^2}{b^2+c^2}\geq\sum\left(ca+ab\right)$
$\Longleftrightarrow\ \ \ \ \ \sum\left(\frac{a^2\left(b+c\right)^2}{b^2+c^2}-\left(ca+ab\right)\right)\geq 0$
$\Longleftrightarrow\ \ \ \ \ \sum\left(ab\left(a-b\right)\frac{b+c}{b^2+c^2}-ca\left(c-a\right)\frac{b+c}{b^2+c^2}\right)\geq 0$
$\Longleftrightarrow\ \ \ \ \ \sum ab\left(a-b\right)\frac{b+c}{b^2+c^2}-\sum ca\left(c-a\right)\frac{b+c}{b^2+c^2}\geq 0$
$\Longleftrightarrow\ \ \ \ \ \sum bc\left(b-c\right)\frac{c+a}{c^2+a^2}-\sum bc\left(b-c\right)\frac{a+b}{a^2+b^2}\geq 0$
$\Longleftrightarrow\ \ \ \ \ \sum bc\left(b-c\right)\left(\frac{c+a}{c^2+a^2}-\frac{a+b}{a^2+b^2}\right)\geq 0$
$\Longleftrightarrow\ \ \ \ \ \sum bc\left(b-c\right)\frac{\left(bc+ca+ab-a^2\right)\left(b-c\right)}{\left(c^2+a^2\right)\left(a^2+b^2\right)}\geq 0$
$\Longleftrightarrow\ \ \ \ \ \sum bc\left(b-c\right)^2\frac{bc+ca+ab-a^2}{\left(c^2+a^2\right)\left(a^2+b^2\right)}\geq 0$.

Upon multiplication with $\left(b^2+c^2\right)\left(c^2+a^2\right)\left(a^2+b^2\right)$, this inequality becomes

$\sum bc\left(b^2+c^2\right)\left(b-c\right)^2\left(bc+ca+ab-a^2\right)\geq 0$.

Now, WLOG assume that $a\leq b\leq c$. (This assumption is legitimate, since the inequality $\sum bc\left(b^2+c^2\right)\left(b-c\right)^2\left(bc+ca+ab-a^2\right)\geq 0$ is symmetric in $a$, $b$, $c$.) Define the numbers

$a_1=bc\left(b^2+c^2\right)$; $b_1=ca\left(c^2+a^2\right)$; $c_1=ab\left(a^2+b^2\right)$;
$u_1=\left(b-c\right)^2\left(bc+ca+ab-a^2\right)$;
$v_1=\left(c-a\right)^2\left(bc+ca+ab-b^2\right)$;
$w_1=\left(a-b\right)^2\left(bc+ca+ab-c^2\right)$.

Since $c\geq a$, we have $bc\left(b^2+c^2\right) \geq ba\left(b^2+a^2\right) = ab\left(a^2+b^2\right)$; in other words, $a_1\geq c_1$.
Since $c\geq b$, we have $ca\left(c^2+a^2\right) \geq ba\left(b^2+a^2\right) = ab\left(a^2+b^2\right)$; in other words, $b_1\geq c_1$.

Since $b\geq a$, we have $ab\geq a^2$ and thus $bc+ca+ab-a^2\geq bc+ca+a^2-a^2=bc+ca\geq 0$, so that $u_1=\left(b-c\right)^2\left(bc+ca+ab-a^2\right)\geq 0$. Since $c\geq b$, we have $bc\geq b^2$ and thus $bc+ca+ab-b^2\geq b^2+ca+ab-b^2=ca+ab\geq 0$, so that $v_1=\left(c-a\right)^2\left(bc+ca+ab-b^2\right)\geq 0$.

Finally,

$u_1+v_1+w_1=\sum\left(b-c\right)^2\left(bc+ca+ab-a^2\right)$
$=\sum\left(b-c\right)^2\left(2bc-\left(a-b\right)\left(c-a\right)\right)$ (since $bc+ca+ab-a^2=2bc-\left(a-b\right)\left(c-a\right)$)
$=\sum 2bc\left(b-c\right)^2-\underbrace{\sum\left(b-c\right)^2\left(a-b\right)\left(c-a\right)}_{=\left(\sum\left(b-c\right)\right)\cdot \left(b-c\right)\left(a-b\right)\left(c-a\right)}$
$=\sum 2bc\left(b-c\right)^2-\underbrace{\sum\left(b-c\right)}_{=0}\cdot\left(b-c\right)\left(a-b\right)\left(c-a\right)$
$=\sum 2bc\left(b-c\right)^2\geq 0$.

Hence, we can apply Theorem 3 from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=387639#p387639]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=65385 post #2[/url] to the numbers $a_1$, $b_1$, $c_1$, $u_1$, $v_1$, $w_1$ instead of $a$, $b$, $c$, $u$, $v$, $w$, and get $a_1u_1+b_1v_1+c_1w_1\geq 0$. In other words, $\sum bc\left(b^2+c^2\right)\left(b-c\right)^2\left(bc+ca+ab-a^2\right)\geq 0$. Thus, Theorem 1 is proven.

  Darij

after this:
$\sum\frac{\left(y+z\right)^2}{x^2\left(y^2+z^2\right)}\geq 2\sum\frac{1}{yz}$.
my proof is too longer than you,your solution is nice.during the proof ,I use the software "bottema" to prove it is true which was programed by prof. Lu Yang.

now we gave a simple prove of the above inequality:
Thus, the inequality in question, $\sum\frac{\left(y+z\right)^2}{x^2\left(y^2+z^2\right)}\geq 2\sum\frac{1}{yz}$, becomes $\sum\frac{a^2\left(b+c\right)^2}{b^2+c^2}\geq 2\sum bc$. Now, since $2\sum bc=\sum\left(ca+ab\right)$, we can equivalently transform this inequality as follows:
$\sum \frac{a^2(b+c)^2}{b^2+c^2}=\sum \frac{[(a^2+b^2+c^2)-(b^2+c^2)](b+c)^2}{b^2+c^2}$
$=\sum a^2\sum\frac{(b+c)^2}{b^2+c^2}-\sum(b+c)^2$
$\geq \sum a^2\cdot\frac{(2\sum a)^2}{2\sum a^2}-\sum(b+c)^2$
$=2(\sum a)^2-\sum (b+c)^2=2\sum bc$
[color=red]By the way:What about the upperbound of the inequality?[/color]


A number’s decimal place is shifted one place to the right, producing a new number
that is 36 greater than the original. What is the original number?

Let the number be $ 10a\plus{}b$. Thus we have $ 100a\plus{}10b\equal{}10a\plus{}b\plus{}36$. Thus $ 90a\plus{}9b\equal{}36$, and $ 10a\plus{}b\equal{}\boxed{4}$.


Let {an} thuộc R+; $a^{2}(n+1)$ $\geq$ a1+a2+...+an (n$\geq$ 1)
   Prove that there exist c>0 such that an$\geq$ nc (all n in N*)

Let me see if I understand:

[quote="bui_haiha2000"]Let $\{ a_n \}_{1}^{\infty} \in \mathbb{R}^{+}$ such that $a^{2}_{n+1} \ge a_1+a_2+...+a_n \forall n \geq 1$

Prove that there exist $c>0$ such that $a_n \geq cn \forall n \in \mathbb{N}^*$[/quote]

:)

Yes exactly.Anybody help me solve this problem


donbaleye namotenahi az adade $0$ va $1$ ra dar nazar begirid ke jomleye avale an $0$ ast va har jomle barabar ast ba kenare ham gharar dadane jomleye ghabl va gharineye jomleye ghabl ( yani har $0$ ra be $1$ va har $1$ ra be $0$ tabdil kardan ) yani  $a_{1}= 0$ va $a_{2}=01$ , $a_{3}=0110$ , $a_{4}=01101001$ , ....
sabet konid $a_{2k}=$gharineye $a_{k}$ va sabet konid donbale dore tanavob nadarad!!

[quote="sepehr ahmadzadeh"]... sabet konid $a_{2k}=$gharineye $a_{k}$...[/quote]

gharine yani.....

$a_{4}$ =gharineye $a_{2}$??

fek konam har jomle barabar bashe ba gharineie jomleie ghabl va khode jomleie ghabl iani avval gharine bade khode jomle.
doroste?

aval khode danblaye ghabl miad badesh gharinash yani agar jomleve avali $f_{1}=0$ dar in sooorat $f_{2}=01$ ok???

ghesmate avvale soal ke be rahati ba esteghra sabet mishe,inke dore tanavob nadare ro nemifahmam ,manzooretoon oon donbale ie ke  az gharar dadane f(i) ha poshte sare ham be vojood miad ?

are khob ghesmate aval ke asooone ... are manzooram donbale namotenahieee $f_{k}$ dore tanavob nadare


analytic geometry
combinatorics proposed
combinatorics

Each point of a plane with both coordinates being integers has been colored black or white. Show that there exists an infinite subset of colored points, whose points are in the same color, having a center of symmetry.

[EDIT: added condition about being infinite - now it makes sense]

This is a very weak problem, but I think it's still nice and appropriate for problem 4.

The solution:

Let's take a look at a square grid of $ n\times n$ points, with one diagonal having endpoints $ (1,1)$ and $ (n,n)$. There are $ n^2$ points, so at least $ \frac {n^2}{2}$ points of the same colour, say black. So there are $ K \equal{} \binom{\frac {n^2}{2}}{2}$ pairs of black points, and thus $ K$ midpoints of these pairs (Each midpoint can be counted more than once). However, the points that can be midpoints of these $ n^2$ points are (at most) points of the form $ (\frac {a}{2},\frac {b}{2})$, where $ a,b$ are integers such that $ 2\leq a,b\leq 2n$. There are $ M \equal{} (2n \minus{} 1)^2$ such points. As a result, at least one point serves as a midpoint of $ P \equal{} \frac {K}{M}$ pairs. Taking $ n$ to infinity, we see that $ P$ tends to infinity as well. That is, we can increase $ P$ as much as we can. We are done.

Edit: Well, if this is still unclear, then suppose what we want is not true. Then there must be a point which is the midpoint of a maximum number of pairs of points. Let this point be $ L$, and let it be the midpoint of $ J$ pairs of points. From the above proof, we can see that we can increase $ P$ to be more than $ J$. This is a contradiction.

You have proven only that there exists a set with as large as one can imagine number of points, but still finite. The problem is about existence of an infinite one.

I have edited, michaj. I hope it's clear now.

[quote="JBL"][quote="swensen"]Edit: Well, if this is still unclear, then suppose what we want is not true. Then there must be a point which is the midpoint of a maximum number of pairs of points. Let this point be $ L$, and let it be the midpoint of $ J$ pairs of points. From the above proof, we can see that we can increase $ P$ to be more than $ J$. This is a contradiction.[/quote]

No, you don't understand: consider the family of sets $ \{1\}, \{1, 2\}, \{1, 2, 3\}, \ldots, \{1, 2, 3, \ldots, n\}, \ldots$.  This family contains sets of unbounded size, but it doesn't contain any infinite sets.[/quote]

Thanks. It was my misunderstanding.

Anyway, I've come up with another solution which I am sure is correct, but I am too lazy to type it right now, so I'll just type the outline.

Again, we consider the lattice points. Suppose there are no infinite set of points of the same color symmetric with respect to $ (0,0)$, and also none with respect to $ (1,0)$. For large enough $ |a|$ or large enough $ |b|$, we can prove that points $ (a,b)$ and $ (a \plus{} 2,b)$ have the same color. Now, we'll prove that there must be an infinite set of points of the same color symmetric with respect to $ (1,1)$. Suppose there is none. Then again, for large enough $ |a|$ or large enough $ |b|$, we can prove that points $ (a,b)$ and $ (a \plus{} 2,b\plus{}2)$ are of the same color. Now try to find an infinite pairs of points symmetric with respect to $ (1,1)$  :)

Note, by the way, that this problem is far from being original  :wink:

I think it was a hard one I gave a solution by working on lines and rotating them!!! It's really large it's not possible to submit it!!!


What is the simplified value of $ 55^2$ - $ 45^2$?


calculus
integration
logarithms
function
algebra
rational function
complex numbers

$\int_{0}^\infty \frac{ln(x)}{(x+1)^{2}}dx$

I was following examples in my book and would solve it by integrating
(A) $\int_{see\ the\ pic}\frac{log(z)}{(z+1)^{2}}dz$
$log(z) = Log|z|+i\theta,\ 0<\theta < 2\pi$ (so a branch cut at positive real axis)
and then using the residue theorem I could solve it by finding the answer of the integral along the line 1. The integral along the circle 2 and 4 -> 0. So i'm left with 1 and 3.
I define at
1: (*)$z = x+\epsilon+i*\epsilon$ dz = dx. If we let R->\infty and epsilon -> 0 I get
$\int_{0}^\infty \frac{ln(x)}{(x+1)^{2}}dx$

and at 3 by letting $z = x+\epsilon-i*\epsilon$ I get the integral
$\int_\infty^{0}\frac{ln(x)+i*2\pi}{(x+1)^{2}}dx$
But 1 and three take eachother out and I'm left with
$\int_{0}^\infty \frac{-i*2\pi}{(x+1)^{2}}dx = i*2*\pi*Res(...)$

And don't know my misstake. Is it the figure? Should (A) be along another one? Or is my misstake at (*)?

Thank you.

You must integrate $\frac{(\log z)^{2}}{(z+1)^{2}}$ over the contour instead. Note that your method gives a good way to integrate any rational function over $(0,\infty)$.

[quote="kahsi"]
But 1 and three take eachother out and I'm left with
$\int_{0}^\infty \frac{-i*2\pi}{(x+1)^{2}}dx = i*2*\pi*Res(...)$

And don't know my misstake. Is it the figure? Should (A) be along another one? Or is my misstake at (*)?

Thank you.[/quote]

The contours over 1 and 3 do not cancel each other out.  This is because they are being integrated along the edge of a branch cut.  To see this, plot parametrically the imaginary part of Log[z] between 0 and 2pi.  Like this in Mathematica:

ParametricPlot3D[{r Cos[t],r Sin[t],t},{t,0,2 PI},{r,0.001,1},BoxRatios->{1,1,1}]

Look at that plot and convince yourself that an integral along the edge at t=0 is not the same as an integral along the edge at t=2pi.

So on 1 let $z=re^{0i}$ and on 3, let $z=re^{2\pi i}$

We get:

$\int_{c_{1}}\frac{Log[z]}{(z+1)^{2}}dz=\int_{0}^{\infty}\frac{ln(r)}{(r+1)^{2}}dr$

$\int_{c_{3}}\frac{Log[z]}{(z+1)^{2}}dz=\int_{\infty}^{0}\frac{ln(r)+2\pi i}{(r+1)^{2}}dr$

Add those together and doing the integration results in $-2\pi i$ which is the value of the integral along the entire contour as R goes to infinity and $\epsilon$ to zero.  This is the same as calculating:

$\int_{C}\frac{Log[z]}{(z+1)^{2}}dz=2\pi i \mathop\text{Res}_{z=-1}\left\{\frac{Log[z]}{(1+z)^{2}}\right\}=-2\pi i$

However, the lights in my stars are all out so maybe someone should verify this.

I misunderstood Kashi above when he stated "1 and 3 take each other out".  Now I understand he meant the:

$\int_{0}^{\infty}\frac{ln(r)}{(1+r)^{2}}dr$

components cancel and that's the reason Kalle suggested integrating:

$\int\frac{Log^{2}(z)}{(1+z)^{2}}dz$

around the keyhole contour above.  I'd like to work it through just to keep things straight then:  I'll write it as:

$\int_{K}\frac{Log^{2}(z)}{(1+z)^{2}}dz=2\pi i \mathop\text{Res}_{z\,=\,-1}\frac{Log^{2}(z)}{(1+z)^{2}}=-(2\pi i)^{2}$

where K is the contour above.

First consider the contour over $C_{2}$ as $R\to\infty$

$\left|\int_{C_{2}}\frac{Log^{2}(z)}{(1+z)^{2}}dz\right| \leq\int_{C_{2}}\left|\frac{Log^{2}(z)}{(1+z)^{2}}\right|dz \leq\frac{(ln(R)+2\pi)^{2}}{(R-1)^{2}}2\pi R$

and $ln^{2}(R)/R$ goes to zero as $R\to\infty$.

For the contour over $C_{4}$:

$\left|\int_{C_{4}}\frac{Log^{2}(z)}{(1+z)^{2}}dz\right| \leq\int_{C_{4}}\left|\frac{Log^{2}(z)}{(1+z)^{2}}\right|dz \leq\frac{-ln(\epsilon)+2\pi}{(1-\epsilon)^{2}}2\pi\epsilon= 2\pi\left[\frac{2\pi\epsilon-\epsilon ln(\epsilon)}{(1-\epsilon)^{2}}\right]$

Now, as $\epsilon\to 0$ the denominator goes to 1 and by L'Hopital's rule the product $\epsilon ln(\epsilon)$ goes to zero thus the quotient goes to zero and therefore the integral does as well.

We're left with:

$\int_{0}^{\infty}\frac{ln(r)}{(1+r)^{2}}dr=\pi i\left(1-\int_{0}^{\infty}\frac{dr}{(1+r)^{2}}\right)$

$\int_{0}^{\infty}\frac{ln(r)}{(1+r)^{2}}dr=\pi i(1-1)=0$

Sorry Kashi.

Also, I think it's interesting that this technique can be generalized for any positive integer n. That is, express the following integral as a function of n (using contour integration):

$\int_{0}^{\infty}\frac{ln^{n}(x)}{(1+x)^{2}}dx$


Assume that a quadrangle $ ABCD$ have $ AD \equal{} BC$, $ AC\cap BD \equal{} O$, bisector of $ \widehat{DAB}$ and $ \widehat{CBA}$ intersec at $ I$. Prove that: midpoint of $ AB, CD, OI$ are collinear.

Let $ R$, $ S$ be the intersections of $ AI$ with $ BD$ and respectively $ BI$ with $ CA$. Denote by $ M$, $ N$, $ P$ and $ Q$ the midpoints of the segments $ AB$, $ CD$, $ OI$, and $ RS$, respectively.

From the complete quadrilateral $ ORISBA$, we conclude that $ P$, $ Q$ and $ M$ are collinear. Hence we have to prove that $ M$, $ Q$ and $ N$ are also collinear.

Indeed, let $ \frac{RD}{RB}\equal{}\frac{AD}{AB}\equal{}\frac{BC}{BA}\equal{}\frac{SC}{SA}\equal{}k$, we have \[ \overrightarrow{QN}\equal{}\frac{1}{2}\left(\overrightarrow{RD}\plus{}\overrightarrow{SC}\right)\equal{}\frac{k}{2}\left(\overrightarrow{BR}\plus{}\overrightarrow{AS}\right)\equal{}\frac{k}{2}\cdot\overrightarrow{MQ}\]
Hence $ M$, $ N$ and $ Q$ are collinear and our solution is completed.


vector
analytic geometry
geometry
linear algebra
linear algebra theorems

I was studying these days and I realized that I didn't know what was a vector. Does anyone know its precise definition? What does it mean in terms of physics? Does anyone know a good book talking about the essencial structures of mathematics [like philosophical issues]?

thanks

The key thing about vectors, from a physical standpoint, is that they transform in a particular way under linear coordinate transformations.

There is a good discussion of vectors, especially their physical meaning and how they should behave under linear transformations, at the beginning of a book called [u]Classical Dynamics[/u] by Marion and Thornton.

There is a more abstract interpretation of vectors from a Differential Geometry perspective. It can be read about in the book [u]Introduction to Differential Geometry[/u] by Spivak.


MATHCOUNTS
geometry
3D geometry
Pythagorean Theorem

[hide]For the 2nd one the formula is d:^3:/3. You arrive to it by first figuring out the diagonal of a square (2*a^2), then use the Pythagorean Theorem again on the triangle standing up. You'll get 2*a:^2:+a:^2:=d:^2:  3*a:^2:=d:^2:  a=d/:rt3:

Since a volume of a cube is a:^3:, the volume can also write it as d:^3:/3

Substituting 12 for d:^3:/3 gives us 576.[/hide]

Hope I didn't mess up on the algebra...that would be embarrasing

MCrawford wrote:(a) If the diagonal of a sqare is 12 cm, what is the area of the square in cm :^2: ?

(b) If the diagonal of a cube is 12 cm, what is the volume of the cube in cm :^3: ?

[hide]

a)Formula is d^2/2

So it's 72

b) The diagnol of a cube =  :rt3: * s

So (12/ :rt3: )^3 = 192 :rt3:

[/hide]

[12/sqrt(2)]^2 = 72
b) is solved similarly except diagonal = s*sqrt(3). volume = 192sqrt(3)

What you're doing should be a good lesson to many students -- hunting for old problems to solve on the forum is a great way to learn more about mathematical problem solving.


In the triangle $ABC,\angle A=\frac{\pi}{3},D,M$ are points on the line $AC$ and $E,N$ are points on the line $AB$ such that $DN$ and $EM$ are the perpendicular bisectors of $AC$ and $AB$ respectively. Let $L$ be the midpoint of $MN$. Prove that $\angle EDL=\angle ELD$

Clearly, $M,D,E,N$ are concyclic points, on a circle with midpoint $L$.
So $\angle ELD = 2\angle EMD = \frac{\pi}{3}$.
$\Delta DEL$ is equilateral.


There are $100$ students in the class. Every student know at least $67$ other students. Prove that there exist four students who all know each other.

Consider the graph of acquiantances. The number of edges is at least $\frac{1}{2}\times 6700 > \frac{100^{2}}3$ thus, using Turan's theorem, this graph contains a $4$-clique, and we are done.

Pierre.

Could you please post some details ?How did you use Turan's theorem ?

[quote="silouan"]How did you use Turan's theorem ?[/quote]

In a pretty straight way...


superior algebra
superior algebra unsolved

Show that $\mathbb{C}\times \mathbb{C}$ and $\mathbb{C}\otimes_{\mathbb{R}}\mathbb{C}$ are isomorphic rings.

I used the map $(\alpha, \beta)\mapsto \alpha\otimes \beta$  to try to prove injectivity and surjectivity, but did not succeed. Is this even the correct map?

What is the definition of $\mathbb{C}\otimes_{\mathbb{R}}\mathbb{C}$?

@adbul: It's the tensorproduct.

@pouging:
Your map isn't additive, so it can't be an isomorphism. (And it isn't surjective, but this should be well known...)

Use the universal property of the tensor product  :wink:

an easy exercise: if $L$ is a $K$-algebra and $f \in K[x]$, then $K[x]/(f) \otimes_{K}L \cong L[x]/(f)$ as $K$-algebras. this yields

$\mathbb{C}\otimes_{\mathbb{R}}\mathbb{C}\cong \mathbb{C}[x] / (x^{2}+1) \cong \mathbb{C}[x]/(x-i) \times \mathbb{C}[x]/(x+i) \cong \mathbb{C}\times \mathbb{C}$


Keisha's basketball team must decide on a new uniform.  The seventh-graders will pick the color of the shorts (black or gold) and the eighth-graders will pick the color fo the jersey (black, white, or gold), but the two groups of players will not confer together.  If, for both garments, each possible color is equally likely to be chosen, what is the probability that the shorts will be a different color than the jersey? Express your answer as a common fraction.

There are $2$ ways for the colors to be the same (both gold, both black).
There are $6$ total ways to choose the colors ($2*3$)
$\frac{2}{6} = \boxed{\frac{1}{3}}$

^you have to take the complement of that @AKAL3
it asks for DIFFERENT colors.


Let $ S$ be a set with $ 2002$ elements, and let N be an integer with $ 0 \leq N \leq 2^{2002}$. Prove that it is possible to color every subset of $ S$ either black or white so that the following conditions hold:

(a) the union of any two white subsets is white;
(b) the union of any two black subsets is black;
(c) there are exactly $ N$ white subsets.

I think you have to generalize to a non-unique n and prove by induction?  Right?  This is my first time so I could be totally wrong, as I barely qualifed for AIME.

Yep you are right. Ill give this one a shot too, but please correct me if I am wrong or I am missing some steps.

Lets consider when x=2.
So we have 0 <= N <= 2^1.
We can choose N=2 and say its elements are (x,y).

Let the empty set be white, {x} be white, {y} be black, and {x,y} to be black.
Cleary (a), (b), and (c) can be satisfied. (a) with the empty set and {x}, (b) with {y} and {x,y}, and (c) with the empty set and {x}.
Similarily, we can prove this for N=0 and N=1...for N=0, consider when every subset is colored black. For N=1, cosider when the empty set is white, {x} is black, {y} is black, and {x,y} is black.

So we can assume that x is true, and we need to prove this for x+1 (if we prove this to be true, x+1 will cover 2002).

So now this new set has x+1 elements with 0<= N <= 2^(x+1).
If N <= 2^x, then we can use the fact that the case for the set with size x is satisfiable (assumption step). Thus, this follows that for N <= 2^x, just use the corresponding subset of x elements (out of the size x+1 set) to choose which subsets will be white/black. We let the last remaining element be black (every subset with this element is black when combined) so that N is still within the limits.

Now, consider when 2^x < N <= 2^(x+1). In this case, make every black element a white element and every white element a black element. Then, from the set with x elements, we can use the assumption step to choose white/black subsets to satisfy this condition. Furthermore, N <= 2^x for the black subsets in the size x set because 2^x < N <= 2^(x+1) for the size x+1 set. Let the last element be white (so every subset with that element is white when combined).


[hide]only units digits contribute to units digits, so we have 7*9*1*3*5*7*9*...*1*3

1*3*5*7*9 has units digit 5. 5 raised to any power still has a units digit of 5. since 7*9*1*3 is odd, multiplying it by 5 still has a units digit of 5

5[/hide]

[quote="rnwang2"]Find the units digit of the product $7\cdot 9\cdot 11\cdot\ldots\cdot1999\cdot 2001\cdot 2003$, where each successive term is two more than the previous term.[/quote]
[hide]Since there will be many 5s in the unit digit of such a sequence (you need only one of them), and since the product of 5 and any odd number is 5, the unit digit of the above sequence will be 5.[/hide]

[hide]5 raised to any odd power is still 5, so the answer is $\boxed{5}$[/hide]

[hide]we want to find the units digit so basically: 7*9*1*3*5*7*1... We notice there's a 5, 5 times any odd # has a units digit of 5, so the units digit is 5. :D[/hide]

[hide="solution"]15 will be in that list, so now the units digit can only be 5 or 0. Since there is no multiple of 2 in the list, the units digit has to be 5.[/hide]

[quote="rnwang2"]Find the units digit of the product $7\cdot 9\cdot 11\cdot\ldots\cdot1999\cdot 2001\cdot 2003$, where each successive term is two more than the previous term.[/quote]

[hide]there is at least one multiple of 5.  however, there arent any multiples of 2 so the units digit must be [b]5[/b][/hide]


AwesomeMath
summer program
Mathcamp
MATHCOUNTS
MathPath
AMC
USA(J)MO

Who all is going to the AwesomeMath camp this summer down in Texas (besides me)?

I might go.
The test you had to take to go there, was it like a qualification test? Cuz if it was i'm screwed.

it turned out to be that the 6 I answered are probably right. (as i checked over and over and over...)

is there a restriction on what grade you have to be in?

Since, I'm in 7th grade this year, I'm going to mathpath. Next year, I'm shooting for awesomemath.
Mr RR Pibb 2 (Nathan Benjamin): The only thing I don't like about awesomemath is the name. they're trying to make math sound cool.
pocketaces2838 (me): I like Mathcamp though. Straightforward and to the point.
Mr RR Pibb 2: lol

Anyway, I wanna make MOP in all of my high school years, although its alot easier to make it in freshman/junior than soph/senior, since the soph/junior cutoff is significantly higher than freshman (and I will be more prepared in junior) and to make black (senior) MOP, you have to be one of the winners.
This year the cutoffs (i believe) were 25, 19, 9.

Lol, the Indiana team (- Sachin and Lauren... crap that's only 1/2 the team... well, John and Nathan make up 7/8 of the personality anyways) were poking fun at the awesomemath name for a while... and I don't blame them. Awesome math...?

...Well as long as the camp's good...I doubt anyone wouldn't go because the name is bad  :huh: :dry:
Anyways I'm going

[quote="asdf4"]Actually I havent turned it in yet. They said I have to turn it in by the end of the week.
Right now, I have the first 6 done. (but probably some of them are wrong)
and i doubt that i will magically figure out the last 4.
Anyways, if I go, I'll be the only stupid person there.[/quote]

Don't worry, you'll do fine. I only got 4 right and breezed in.

yay!!!! Me happy!!! Now that I have #7. (after spending hours and hours on it)

[quote="Ignite168"][quote="kyyuanmathcount"]Give him a break, the coach is chinese... Zuming Feng?  It makes the name sound Asian, and since the majority of the people there are asian, it might sound appealing to some.[/quote]

Wait, how does AwesomeMath = Asian?  :?[/quote]

He used to be the coach of the USA IMO team right?Together with Titu Andreescu...

And to be on topic: I'm going to MathCamp, I went last year too and it was [b]awesome[/b]

wow I feel stupid, 8/20 on red...

[quote="samath"]One note for ALL people going to awesomemath who play tennis, bring a tennis racquet and hopefully some balls. i know that ch1n353ch3s54a1l and i do...[/quote]

i don't play tennis but i'll bring a couple anyways

 :roll:

[quote="Phelpedo"][quote="nebula42"]They said (about the homework) "This will help us determine the group you will be initially placed on. There will be flexibility regarding moving from a group to another during the program."
Do you think this might mean that when we get there, we'll take a test, and then they'll combine the results from that and our homework to determine our groups?
Just a random guess.  :wink:  :P[/quote]

If they said anything else in the email, could you post that too for those of us who didn't get it :maybe:[/quote]
All right... here's the original e-mail:

[quote="Dr. Titu Andreescu"]Dear AMSP participant,

Attached you will find four HW problem sets that reflect the level of the four groups we intend to have at the program this summer. Please solve the problems from the set that both challenges you and yet you feel comfortable working on. Please have the solutions ready when you come to Dallas. This will help us determine the group you will be initially placed on. There will be flexibility regarding moving from a group to another during the program.

You will also find attached the list of the AMSP 2006 participants, listed in the alphabetical order.

Best regards,

Dr. Titu Andreescu
[/quote]

[quote="samath"]I now have all but 1, 5, 6, 11, and 19 on the red--but I've got a pretty good start on one of them.

EDIT: Now I've kinda got 19 and I got 11 (I think), so I have 17 solutions.[/quote]

trade your answer for 19 for my answer for 6 :D ?

just got another one, now I have:

2, 3, 4, 6, 8, 9, 10, 12, and 16

Well, got all the yellow down...

On red #6, I got no solution... do they mean lateral surface area?

[quote="Phelpedo"][quote="nebula42"]
If they said anything else in the email, could you post that too for those of us who didn't get it :maybe:[/quote]
All right... here's the original e-mail:

[quote="Dr. Titu Andreescu"]Dear AMSP participant,

Attached you will find four HW problem sets that reflect the level of the four groups we intend to have at the program this summer. Please solve the problems from the set that both challenges you and yet you feel comfortable working on. Please have the solutions ready when you come to Dallas. This will help us determine the group you will be initially placed on. There will be flexibility regarding moving from a group to another during the program.

You will also find attached the list of the AMSP 2006 participants, listed in the alphabetical order.

Best regards,

Dr. Titu Andreescu
[/quote]

Ah, thanks a bunch. Now any chance you could forward/post the list of participants?

p.s. Does anybody get #13 on the blue test?[/quote]
Someone told me I'm not allowed to attach this list, so I've taken it out (but now I can't delete this post).  :wink:

[quote="not_trig"]On red #6, I got no solution... do they mean lateral surface area?[/quote]

I agree, the answer with the interpretation of surface area=lateral area + base area has to be 0 (or 4 other complex ones). But interpreting it as lateral surface area yields a real answer (or 0). But with it being 0, the arithmetic progression isn't true.

I think they mean lateral surface area - it gives a nice answer >_>

eh...
I'm too lazy to write out a solution for #6, and I already printed out all the other solutions when I found out the it should be lateral area was wrong, so I'll just skip it.

[quote="bpms"]Can someone put up the problems for white? I'm thinking about applying to awesomemath next year, but I want to see if it is too advanced for me...[/quote]
You could also just try the admission test problems here:
[url]http://www.awesomemath.org/applicationinfo.html[/url]
(Look at the section entitled "Admission Test.")

im going 2 china this summer, but i'm considering the program 2.
how is it? from the way it looks, it should be very challenging... maybe next year.

This thread is over 6 months old... and you didn't even post a relevant response.

There is an awesomemath forum now, here: http://www.artofproblemsolving.com/Forum/index.php?f=442


trigonometry
inequalities proposed
inequalities

we have:${f(n)=\underbrace{\sin\sin...sin1}}$
prove that:$\frac{1}{\sqrt{3n+1}}<f(n)<{\sqrt{\frac{5}{5+n}}}$

见张景中 李浩 《实迭代》（湖南教育出版社，1991，12）一书，第46-48页例4.1

The left part is not true. The right one can be proved by introduction

sorry,Lovasz .the question isn't mine.
so i don't know it is right or not. but ,if u have a proof on right side ,please write it down.
thank u :ninja:


number theory
relatively prime
number theory proposed

Let $ x,y$ be relatively prime different natural numbers. Prove that for infinitely many primes $ p$ the exponent of $ p$ in $ x^{p\minus{}1}\minus{}y^{p\minus{}1}$ is odd.

[quote="ehsan2004"]Suppose that $ x \equal{} p^{2k}(p\plus{}1)\plus{}1$ and $ y \equal{} p^{2k}\plus{}1$ where $ p\geq3$ then $ gcd(x,y) \equal{} 1$ and we have $ e_{p}(x^{p\minus{}1}\minus{}y^{p\minus{}1}) \equal{} e_{p}(x\minus{}y) \equal{} e_{p}(p^{2k\plus{}1}) \equal{} 2k\plus{}1$. :D[/quote]
I think you misunderstood N.T.TUAN's question, since actually $ x$ and $ y$ are supposed to be fixed.

sorry ,I am construct infinit $ x,y$ but I don't know it's true for every $ x,y$ where $ gcd(x,y)\equal{}1$ :maybe:

Let r be an odd prime divisor of x^3 -y^3.
 Let q be a prime such that (q/r)=-1.
 We have ([(x^3q -y^3q) / x^3- y^3]/r)=-1.
 Let p be a prime divisor of [(x^3q -y^3q) / x^3- y^3] with odd exponent. So p works.

I have a idea.it's for some x then there is infinite many p (prime number) such that $ x^{p\minus{}1}$ # $ 1$ mod $ p^{2}$

I think that would be a much more difficult problem.
As far as I know, it is unknown whether there are infinitely many such p when x=2.


I'm bringin this qustion up already posted here,but i need the solution to it,not just the answers.

Find all possible sides of a triangle whose area is equal to tits perimeter.

This problem is asked many times.

Search the forum. Locked.


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

Solve the equation :

$ X^3 \plus{} X \plus{} 2 I_{2} \equal{} O_{2}$ ,$ X\in \mathcal M_{2}$

edit: can someone move this to an appropriate section ? I m sorry I thought I was in liniar algebra.I cannot delete it.

One does not "solve" an equation like this. There are infinitely many answers - a great profusion of answers. There are some things that might be said about those answers.

$ x^3\plus{}x\plus{}2\equal{}(x\plus{}1)(x^2\minus{}x\plus{}2)$ which has roots $ \minus{}1,\frac{1\pm i\sqrt{7}}2.$

The eigenvalues of $ X$ must be among those roots, and the minimal polynomial of $ X$ must divide this polynomial. One possibility is $ X\equal{}\minus{}I.$

You didn't specify a field for the entries of this matrix. If the field is $ \mathbb{C},$ we have great freedom in writing answers. For instance, we could have

$ X\equal{}A\equal{}
\begin{pmatrix}\frac{1\plus{}i\sqrt{7}}2&0\\0&\minus{}1\end{pmatrix}.$

Or anything similar to that - in other words, let $ P$ be a completely arbitrary invertible matrix and let $ X\equal{}PAP^{\minus{}1}.$

If the field is $ \mathbb{R},$ then our choices are narrowed. We can only have the complex eigenvalues if we have both of them. That means that the only way we could include the $ \minus{}1$ eigenvalue is to have the minimal polynomial be $ x\plus{}1$ and that would give us only $ X\equal{}\minus{}I.$ But we'd still have infinitely many choices with the complex eigenvalues.

If the field is $ \mathbb{Q},$ our choices are narrower, but still infinite. Besides $ \minus{}I,$ we would have the companion matrix

$ X\equal{}C\equal{}\begin{pmatrix}0&\minus{}2\\1&1\end{pmatrix}$

and anything similar to that ($ X\equal{}PCP^{\minus{}1}$).


algebra
difference of squares
special factorizations

In a right  triangle $ABC$ ,with $\angle A = 90$  we have $a = 58641980875 58641980875$ and $b= 56296294152 56296294152$ . Find  side $c$ of this triangle.

 [i]( This problem was posed  at a  school test for class 9 !) [/i]

 [u]Babis[/u]

$c=\text{ A really large number}$.

[hide="No, really..."]
Umm... try doing the difference of squares, and trying to find some kind of factor that's squared, or to the fourth, etc.... but wow, how long did the kids have to calculate this?[/hide]

[quote="Farenhajt"]Both numbers are divisible by $10^{11}+1$[/quote]

How did you figure that out?

[quote="robinhe"]Note that both of those really large numbers are two smaller numbers put together (e.g 456*1001=456456).[/quote]

Ooooohh. That's cool.  :)  I completely didn't notice that.

[quote="Farenhajt"]Both numbers are divisible by $10^{11}+1$, hence the simplest form is

$c=(10^{11}+1)\sqrt{2345686723\cdot 114938275027}$[/quote]

 Exactly my solution ! Did you try to simplify the sqrt, or this is impossible ?

               Babis

 [i]The problem was given to me by a mathematician. His son had to solve it in the school , without calculator. When he has the solution of the teacher , I'll post it. Anyway , it seems strange.And the problem is that in our middle school we do not teach change of base ![/i]

It appears it's impossible. Prime factorization (by computer!) is

$(79\cdot 29692237)\cdot(461\cdot 249323807)$


modular arithmetic
number theory unsolved
number theory

let $ k\geq 0$ be an integer and $ p$ be a prime number. Prove that

$ 1^{k}+2^{k}+\cdots+(p-1)^{k}\equiv 0\pmod{p}$

What about $ k=0,p-1,2p-2,...$¿
And see http://www.mathlinks.ro/Forum/viewtopic.php?t=40171 .