calculus geometry inequalities AMC USA(J)MO USAMO

Hey guys, I'm preparing for CMO right now and I want to go to the IMO during my high school year (preferably next year :P). I want to ask all of the better people (like yufei, elyot,.... ) what do you do to improve your math skills? I just realized how important it is or me to do past contests and think about them without peeking at the solutions. So I just wanna kinda get an idea how you guys do to improve and probably I could add them to my schedule.  ;)

[quote="Mathgod"]the better people (like yufei, elyot,.... )[/quote]

I really don't belong that close to Yufei in a sentence.  His level of mathematical skill just pwns mine.  Ask him about how to improve.  I really didn't work at it very much at all (unlike CS, which I practiced for hours and hours for).  I really just got lucky and had a good time.

crazy...
ur only gr10 next year..
no need to hurry.
sweep cayley pascal fermat first.

It's hard for me to summarize how to prepare for Olympiads, but a couple things come to mind

1) Practice, practice, practice. There is no substitute. After a while, you'll feel that you've seen all types of problems, and if you can get that far before your senior year, then IMO is in your hands.

2) Let your interests carry you. However, losing interest (burnout, etc), I find, is one of the worst things that can happen to a math olympian. It does happen, inevidably, though it is usually temporary. On the other hand, if you really want to do well on Olympiads, don't get too carried away with higher math theory (adv. calculus, linear/abstract algebra, analysis, etc). Know some for recreation and exposure, but they will not help as much in Olympiads as simply practicing Olympiad problems. I personally think that it's a lot easier to learn a whole bunch of university math than to do well on Olympiads. Though some may disagree.

The first steps are hard ... it gets easier after that. Trust me.

Thanks for Yufei's comment, that will help me a lot :P, but it's just some times when i'm solving cmo problems for practice (that's what I'm doing now), sometimes I encouter questions I don't know how to do. My friend said it's better to leave them there and think about it rather than peeking at the solutions, but then I would be only able to solve a few of the problem in a day and I'm sometimes worried about if i'm going to have enough time. what do you guys think? I'm kinda learning experiences from the older people and that could improve me doing problems.   ;)

lowest_reynolds: Most Olympiad books on the market are quite decent. What's important is not what books you have, rather, how you use them. It's easy to a have a bookshelf full of the best Olympiad books, but unless you spend the time and effort to work through the problems, the books won't be of much use.

Personally, my first "Olympiad" book was Problem Solving Through Problems. In that summer (after grade 9), I worked through Mathematical Olympiad Challenges and a bit of Engel. They're nice books in my opinion.

Mathgod: That's a tough question. I would personally go for a mix. There are some extremely novel ideas in some problems that you'll probably never think of yourself, so it would definitely help to know it. What I do is that I training through at least two different sources simultaneously: one which I have access to the solutions to (e.g., books, Olympiads), and one which I don't have the solutions to, but the problems are hard enough (for my current level) to worth the investigation and writing up (e.g., Olymon, Crux). The latter also helps you to learn to write up polished solutions, which most of us would not bother with the former route; plus, you get to see your solutions published under your name, which is always fun.

[quote="lowest_reynolds"]Just a question Yufei,

What summer reads would you recommend?  I got "Mathematical Olympiad Challenges," but it doesn't seem to be progressive at all (the topics in the book seem random imo).

I've heard good reviews about Engel's book, and I guess what's left for me to do now is to make the transition from "Problem Solving/Olympiad" Math to "Olympiad" Math.

I've also heard good things about nsato's "NT Packet", Kedlaya's geometry e-book, Mildorf's Inequality packet, and all those other packets out there XD.  Any remarks about them?

Thanks![/quote]

How can we obtain these packets? :D

[quote="lowest_reynolds"][quote="Elyot"]I've never read an olympiad book in my life.  You don't need that... all you need to do is try problems.  Just think about them long enough that you understand what the problem is about.  It takes some time.  Then when you read the solution, you'll realize why the technique employed was necessary and you'll know when to apply it next.[/quote]

Just out of curiosity Elyot, where did you turn to for sources of Olympiad problems?[/quote]

Really, I didn't do many at all.  Other than math camps, I really didn't do much math during high school.

So should I like work on a CMO problem for like 2-3 days, if I still don't get it, then I look at the solution?( I know i sound a bit stupid, but then I"m really tring to improve myself those days :P)

Russian olympiad books  and textbooks are the best :D

[quote="carpo"]Where is Mildorf's Inequality packet?[/quote]

http://web.mit.edu/~tmildorf/www/Inequalities.pdf

Eloyt, just wondering, I didn't see your name appear in 2005 USAMO qualifying list. did u write it unofficially?

[quote="Mathgod"] sometimes I encouter questions I don't know how to do. My friend said it's better to leave them there and think about it rather than peeking at the solutions, but then I would be only able to solve a few of the problem in a day and I'm sometimes worried about if i'm going to have enough time.  ;)[/quote]

I find it useful and time-efficient to peek at the solution for a hint after some time of being stuck, and sometimes the hint is enough for me to finish the problem. Sometimes I really don't see when the problem is going, and then I read the solution until I get it. I know that's really cheap, but otherwise I spend way too much time being stuck on the problem and getting nowhere. If I do, o pillar of cheapness, read the whole  solution, I roll over the steps in my mind and write it up for myself, or try to solve it several days later when it faded from my recollection.


Can you write numbers in the fields of a 5 * 5 grid so that in any 2 * 2 part of the grid the sum of the four numbers should be negative, while in any 3 * 3 part of the grid the sum of the nine numbers should be positive?

I saw this problem and I gave up looking at it.

You can set up a bunch of inequalities and see if they work with eachother but that would take a while ...

its a problem for 11 and 12 year old children......

the answer is: no, because I am too lazy.

Seriously the answer is either yes or no, but I don't know how it would work so i'm guessing no...


Find $ m$ such that $ (m^2\minus{}1)x^2\minus{}(m\plus{}2)x\plus{}1$ have one and one only of it zeros intern to the interval $ ]\minus{}1,1[.$

[quote="123456789"][hide]$ \minus{} 1 \le x < 2$.[/hide][/quote]

i know the answer. And your are not correct.

Am I close?

Also, is that supposed to be $ \left[\minus{}1,1\right]$?

[quote="LoboSolitario"]Find $ m$ such that $ (m^2 \minus{} 1)x^2 \minus{} (m \plus{} 2)x \plus{} 1$ have one and one only of it zeros intern to the interval $ ] \minus{} 1,1[.$[/quote]
Is your interval [-1,1]?(I don't spam, seriously)

[b]123456789[/b] i don't know if you are close because i don't see your work.

[b]ghjk[/b] statement is correct.

[hide="Answer"]

Edit: $ \minus{} 1 < m < 1$ or $ 1 < m < 2.$[/hide]

Why is $ m\equal{}2$ in the solution set? When $ m\equal{}2$, doesn't the quadratic become
\[ 3x^2\minus{}4x\plus{}1?\]Then, the roots are $ \frac{1}{3}$ and $ 1$, and that's more than one zero in the given interval. Am I missing something?
And when $ m\equal{}\minus{}1$, don't we get $ x\equal{}1$? Then only one zero lies in the interval.

Already saw the mistake. $ m$ can't be $ 2$.

You are right.

When $ m\equal{}\minus{}1,$ $ m^2\minus{}1\equal{}0$ and we don't have a quadratic.

[quote="123456789"]The problem didn't state that it must be a quadratic.[/quote]

I don't think so.

[quote="LoboSolitario"]have one and one only of it [b]zeros[/b] [/quote]


Prove that there are no positive integers a, b, n > 1 such that
a^n - b^n divides a^n + b^n.

Does it help to write a^n + b^n as a^n - b^n + 2b^n and then show that 2b^n is not divisible by a^n - b^n.

Anyway does anyone know how to type superscripts? ^ is all very well when it comes to understanding, but it certainly is messy.

So any efficient of typing superscripts other than  22.

I wonder if this appears: 2n.

an-bn|an+bn =>
an-bn|2an => there is k \in N such as 2an=k(an-bn) =>
(k-2)an=kbn. [b][1][/b]

If gcd(a,b)=d then a=a'd and b=b'd where gcd(a',b')=1. So [1] comes to
(k-2)a'n=kb'n.
So we can assume gcd(a,b)=1.

Now, [1] implies the existance of an integer p such as (k-2)=pbn and k=pan.

If k is odd, then gcd(k,k-2)=1 so p=1, so bn+2=an (NOT POSSIBLE)

If k is even then k=2k', so 2|p or else 2|gcd(a,b)=1. So p=2p'.
So k'-1=p'bn and k'=p'an. Since gcd(k',k'-1)=1, we get an=bn+1 (NOT POSSIBLE EITHER)
-------------------------------------------------------------------------------------
I don't know why this problem remained unsolved for so long in here


linear algebra
matrix
superior algebra
superior algebra solved

A\in M_n(R) with the property that det(A+X)=detA+detX for any X\in M_n(R)...
Prove that A=O_n.

We consider n>1
For X=A we get
det(2A)=2detA
2^n.detA=2detA =>detA=0
the equation becomes det(A+X)=detX for any X \in Mn(R)

If A =/=0, let r= rankA \geq 1
A is equivalent to matrix define by blocks

T=
(I_r o)
(0  0)

A=
PTQ where P,Q invertible consider

X=
PWQ where

W=
(0 0)
(0 I_(n-r))

then

det(A+X)=det(PQ) =/=0
det(X)=detP.detW.detQ =0  since detW=0

this contradict the equation det(A+X)=detX

So r=rankA=0 and A=0

Kuba if you have a different proof let me know, because I  always
learn something with several proofs of a problem.

µoubi
10x

lets take X such that
A+X=
a_11 0 0 ... 0
a_21 1 0 .... 0
a_31 0 1......0
..............
a_n1 0.........1
we can see that det(X)=0... results that det(A+X)=0=>a_11=0
results easyly that a_ii=0, i=1,n
now for the other elements we just permute some lines and columns to have it on the main diagonal(we don't change the value of the determinant).
In this way we obtain a_ij=0,i,j=1,..,n
so A=O_n
is not my proof.. is in the book...

In dimension 2 we have a geometrically aspect
We have seen detA=0
for any X \in M_2(R) det(A+X)=detX (1)

A=
(a  b)
(c  d)  detA=0 <=> ad-bc=0

X=
(x  y)
(z  t)

the equation (1) gives  equation of hyperplan

at + xd - bz - yc = 0 for any 4-uple (x,y,z,t) \in R^4

a= b = c = d = 0

A = 0_2

Could be a nice problem for the first course on determinant in 10th grade

yes indeed it could be if we would make in the 10th grade the determinant and the matrices :D :D :D.. we make them only in 11th grade

Well, it seems that Dan Seclaman is a very tough teacher, since he puts his name just on tough problems, published earlier by other people. My respect! By the way, this problem was published long time ago by one of our best problemists, Constantin Cocea. But maybe he copied this problem from Dan Seclaman!  :)

Is Dan Seclaman a university  or high school professor ?

i don't know.. but i think that he's a high school teacher since in the book at the author rubtic writes prof. and not dr. or lector
this book is from 1993 and it has some realy tough problems :D
i'll put some more here :D


inequalities
algebra
polynomial
three variable inequality

[quote="silouan"]Vasc in his book has a proof of
$3(a^{4}+b^{4}+c^{4})+4(a^{3}b+b^{3}c+c^{3}a)\geq0.$[/quote]But the above is stronger. :)

[quote="silouan"]Vasc in his book has a proof of
$3(a^{4}+b^{4}+c^{4})+4(a^{3}b+b^{3}c+c^{3}a)\geq0.$[/quote]
So Vasc inequality only occurs when $a=b=c=0$ ??
If pvthuan's inequality is true, I think I can solve it.  :)

[quote="spanferkel"][quote="arqady"][quote="zhaobin"]
I don't know why don't you write it as: \[7(a^{4}+b^{4}+c^{4})+10(a^{3}b+b^{3}c+c^{3}a)\geq0.\] :D[/quote]
Maybe because, $21=4^{2}+2^{2}+1^{2}$ $?$[/quote]or because some of his keys are stuck?  :lol:[/quote]

or maybe "21"="2"+"1" and "30"="3"+"0", but "2"+"1"="3+"0"="3".  :lol:

[quote="pvthuan"]OK, Can you solve it? Note that it is quite sharp. The following is wrong. \[2(a^{4}+b^{4}+c^{4})+3(a^{3}b+b^{3}c+c^{3}a)\geq0.\] Can you disprove it?[/quote](a,b,c)=(-10,9,-4)

Very nice proof Arqady .But how did you come up with such an indetity ?
Vasc's proof :
It suffices to consider the case when one of x,y,z is negative .
Let z<0 and it suffices to prove that
$3(x^{4}+y^{4}+z^{4})+4x^{3}y\geq 4(y^{3}z+z^{3}x)$ or it is enough to show that
$3(x^{4}+y^{4}+x^{3}y)\geq 4(y^{3}z+z^{3}x)$
and he considers two cases $x\leq y$  and $x\geq y$
Both cases are solved with AM-GM  :)

Yes, mine is true and solved, Anh Cuong. Now you can solve it!

It seems that my mistake has given you a lot of fun.  :lol: Anyway, that is not a big one. Mine is still true. (I did not notice that $21$ and $30$ has a common integer factor) :P

Arqady did you solved Thuan's one ?
If yes please post your solution.

[quote="arqady"][quote="silouan"]But how did you come up with such an indetity ?
[/quote]
Using that $21=4^{2}+2^{2}+1^{2}$ I tried to prove the pvthuan's inequality ,
and the vasc's inequality  has turned out. :D
In fact $4^{2}+(-2)^{2}+(-1)^{2}=21.$ :wink:[/quote]
But if we use this way to prove pvthuan's inequality we must have the denominator is $51$ .
And the 21 you write is not just 21,it should be $51\cdot 21$ or $51\cdot 7$

[quote="Anh Cuong"]I tried to follow Arqady way by expand:
$\sum (4a^{2}+2b^{2}+c^{2}+mab+nac)^{2}$
but no success  :( , maybe must wait for him.
Oke, i will post my solution. The technique you see below can be used for all polynomial cyclic inequality degree less than or equal to $5$.
Oke, take $A=a+b+c=1,B=ab+bc+ca,C=abc$
In fact I just need to prove that:
$(7(a^{4}+b^{4}+c^{4})+10(a^{3}b+b^{3}+c^{3}))(7(a^{4}+b^{4}+c^{4})+10(ab^{3}+bc^{3}+ca^{3})) \geq 0$
This equal to:
$f(C) =$ $2604C^{2}+(2004B^{2}-2964B+1502)C+576B^{4}-1232B^{3}+1232B^{2}-616B+119$
For every value of $B \in [-\frac{1}{3},\frac{1}{3}]$, consider:
$f'(C)=5208C+2004B^{2}-2964B+1502$
There are three case here, for the case $f'(C) \geq 0$ or $f'(C) \leq 0$ refer to $ABC$ theorem, we just need to consider two case:1 variable is equal to 0 or two variables are equal. But this is not difficult
The remain case, consider
$f(\frac{-2004B^{2}+2964B-1502}{5208})=$ $\frac{8300}{651}-\frac{3750}{217}B-\frac{3525}{217}B^{2}-\frac{19850}{217}B^{3}+\frac{41325}{217}B^{4}\geq 0$
for $B \in [\frac{-1}{3},\frac{1}{3}]$.
This is not difficult as well.
You will know more about this technique when coming to VietNam to join IMO 2007.   :wink:
I think we also can solve the below problem by this technique, find the best constant $k$ such that this inequality holds:
$a^{4}+b^{4}+c^{4}+k(a^{3}b+b^{3}c+c^{3}a) \geq 0$
Also this technique can be applied for Vasc famous inequality:
$(a^{2}+b^{2}+c^{2})^{2}\geq 3(a^{3}b+b^{3}c+c^{2}a)$  :D
Enjoy  :) .[/quote]

What are you doing? How can I imagine such computation?

[quote="arqady"][quote="silouan"]Vasc in his book has a proof of
$3(a^{4}+b^{4}+c^{4})+4(a^{3}b+b^{3}c+c^{3}a)\geq0.$[/quote]
$3(a^{4}+b^{4}+c^{4})+4(a^{3}b+b^{3}c+c^{3}a)=\frac{1}{7}\cdot\sum_{cyc}(4a^{2}-2b^{2}-c^{2}+2ac+4ab)^{2}.$ :)
silouan, what vasc's proof? Thank you.[/quote]
arqady, what a nice proof! How do you think the original inequality can be expressed as your inequality?

[quote="Mathsniper"]
How do you think the original inequality can be expressed as your inequality?[/quote]
I think, yes. Maybe even stronger with a very ugly numbers. :wink:

Very nice, Arqady! :first: So $t_{max}$ has minimal algebraic degree 8.  :P Which may be the algebraic degree of $k_{best}$?

Very nice!But how can you show that the $k_{best}$ is really the best?


superior algebra
superior algebra unsolved

prove that if G is isomorphic to H and H is isomorphic to K, then G
is isomorphic to K.

If there exists $i : G \to H$ and $j : H \to K$, then $i(j)$ is an isomorphism $G \to K$.  It's not too hard to go through the axioms to prove that a composition of isomorphisms is also an isomorphism.


Will they be allowed for the competition?

According to the Official Competition Rules page:

"Rulers (straight edges), protractors, and compasses will not be allowed for any test in the competition."

Oh oops didn't see that  :blush: .

Thanks.

I don't think you should need them. The geometry test will have pictures.

[quote="joml88"]Is duck tape allowed? :idea:[/quote]
 :!: I have checked and it seems like our rules don't expressly forbid it. If this causes some kind of problem we may ban duck tape under the "no straightedge" rule, and if you disrupt the test with your duck tape antics we may eject you, but right now I can't see how duck tape could do any harm, so I'm not going to forbid it. (What if you snap your pen in half or rip your answer sheet? :P)

[quote="joml88"]Is duck tape allowed? :idea:[/quote]

lol. is that to keep people form talking when your testing or what?

I dunno, I just wanted to see what they'd say :D  You can do just about anything with duck tape...so use your imagination!

dude what is duck tape

[i]People usually mean [url=http://en.wikipedia.org/wiki/Duct_tape]this[/url] when they say "duck tape."[/i]

[quote="chess64"]huh isn't it duct tape[/quote]
For a long time I have wondered about the distinction myself. I have come to the conclusion that they're the same thing. The Internet, by the way, is your friend on matters like this (I mean: "Just *** google it!")

True soldiers know the difference between Duck Tape and plain old duct tape: http://www.octanecreative.com/ducttape/duckvsduct.html

ooo rah!


What is the smallest whole number greater than
\[ \sqrt{30} + \sqrt{56} + \sqrt{110}? \]

Beautiful problem, Mr. Crawford!

Because $ \, 5 < \sqrt {\,30} < 6 \, $  (why?) and ... it is easy to see that the answer is 23, 24 or 25...

The answer is in fact [hide] 24 [/hide]

Hint: use the "geometric/arithmetic means inequality" to approximate each root!

(Part of the) Answer:
[hide] $ \sqrt {\,30} = \sqrt {5 \cdot 6} \le \frac{{5 + 6}}{2} = 5.5 $
This is a very-good approximation because 5 and 6 are near each other!  ;)
 [/hide]

[quote="math92"][hide]
square everything and we get:
30+56+110=196.
the square root of 196 is 14. 14+1=$ \boxed{15} $[/hide][/quote]

You can't square this expression to get what you got!

AM-GM isn't the only way, but you do know that it's anywhere between 22 and 26. Do you notice that somehow, 30 is almost between 25 and 36, 56 is almost between 49 and 64, and 110 is almost between 100 and 121, with it just one closer to the smaller square root?

sorry, i didn't know how to do the problem

[quote="fskilnik"]This is a very-good approximation because 5 and 6 are near each other!  ;)
[/quote]

I'm not quite sure what you mean by that... what's your perspective of "near"?

Good point.

AM = GM if and only if all terms are equal and their difference is related to the "nearness" of the terms considered. To put it precisely, I guess we need Calculus. I think Taylor expansions for the square roots would suffice, by the way.

Do you have any suggestions? How would you approach this "nearness" ?   :)

[hide]
$5^2+2^2=29$
$7^2+2^2=53$
$10^2+4^2-108$
So, the original answer would be 27, but since there are 3 4s, the $2^2$s, you subtract 3 and get 24??[/hide]
I don't know a solid way to do this problem, but fskilnik may be right, I think you'd have to use calculus to have a good way to solve it. But, since it is in the Basics, it means that there is probably an easier way.


geometry
circumcircle
parallelogram
symmetry
angle bisector
power of a point
geometry unsolved

[b]Problem 1.[/b] Points $D$ and $E$ are chosen on the sides $AB$ and $AC$, respectively, of a triangle $\triangle ABC$ such that $DE\parallel BC$. The circumcircle $k$ of triangle $\triangle ADE$ intersects the lines $BE$ and $CD$ at the points $M$ and $N$ (different from $E$ and $D$). The lines $AM$ and $AN$ intersect the side $BC$ at points $P$ and $Q$ such that $BC=2\cdot PQ$ and the point $P$ lies between $B$ and $Q$. Prove that the circle $k$ passes through the point of intersection of the side $BC$ and the angle bisector of $\angle BAC$.

[i]Nikolai Nikolov[/i]

I got $QC = \sqrt{(QA)(QN)}$ and $PB = \sqrt{(PA)(PM)}$

What next?

Following the post of Jumbler:
Let $F$ be the point on $BC$ between $Q$ and $P$ such that $BP=PF$ and $CQ=QF$.
We have, $PF^{2}=BP^{2}=PM.PA$, $QF^{2}=CQ^{2}=QN.QA$.
Then, $<PFM=<PAF$, $<QFN=<QAF$ and $<MAN+<MFN=180^{0}$.
So, $F\in k$.
Since $NADF$ is cyclic, then $<FAD=<FND=<QCN+<QFN=<CAQ+<FAQ=<CAF$.
So $AF$ is the angle bisector of $<ABC$.

Let A' be the midpoint of BC, so that AA' is  the A-median. Since $DE \parallel BC,$ by Ceva's theorem, BE, CD meet at $X \in AA'.$ Let AA' meet the circumcircle (O) of the $\triangle ABC$ again at Y and let $X_{0}\in AA'$ (on the opposite side of BC than Y) be a point such that $A'X_{0}= A'Y.$ Let $CX_{0},\ BX_{0}$ meet AB, AC at $D_{0},\ E_{0}.$ Again, by Ceva's theorem, $D_{0}E_{0}\parallel BC.$

[hide="extra calculation"]Even though not necessary to solve the problem, we can find the exact positions of $D_{0},\ E_{0}$ on AB, AC. Power of A' to (O) is $\frac{a^{2}}{4}= A'B \cdot A'C = A'Y \cdot A'A = A'X_{0}\cdot A'A = X_{0}A' \cdot m_{a},$ hence $X_{0}A' = \frac{a^{2}}{4m_{a}}$ and $\frac{AX_{0}}{X_{0}A'}= \frac{AA'-X_{0}A'}{X_{0}A'}= \frac{4m_{a}^{2}}{a^{2}}-1.$ By van Aubel's theorem,

$2\ \frac{AD_{0}}{D_{0}B}= \frac{AD_{0}}{D_{0}B}+\frac{AE_{0}}{E_{0}C}= \frac{AX_{0}}{X_{0}A'}= \frac{4m_{a}}{a^{2}}-1$

$\frac{AE_{0}}{AC}= \frac{AD_{0}}{AB}= \frac{AD_{0}}{AD_{0}+D_{0}B}= \frac{4m_{a}^{2}-a^{2}}{4m_{a}^{2}+a^{2}}= \frac{b^{2}+c^{2}-a^{2}}{b^{2}+c^{2}}= 1-\frac{a^{2}}{b^{2}+c^{2}}$
[/hide]
$BX_{0}CY$ is a parallelogram, its diagonals $BC,\ X_{0}Y$ cutting each other in half at A'. The angles

$\angle D_{0}AX_{0}\equiv \angle BAY = \angle BCY = \angle CBE_{0}= \angle BE_{0}D_{0}\equiv \angle X_{0}E_{0}D_{0},$

are then equal and the quadrilateral $AD_{0}X_{0}E_{0}$ is cyclic. Consequently, $M_{0}\equiv X_{0}\equiv N_{0}$ are identical, $P_{0}\equiv A' \equiv Q_{0}$ are also identical, so that $P_{0}Q_{0}= 0.$ By the problem conditions (point P lies between B and Q), the intersection X of BE, CD lies on the segment $X_{0}A'.$ As the line $DE \parallel BC$ moves from $D_{0}E_{0}$ to BC, PQ monotonically increases from 0 to BC. Hence, a unique position of $X \in X_{0}A'$ exists, for which $PQ = \frac{BC}{2}.$ Assume now that the position of $X \in X_{0}A'$ is such that the circumcircle k of the triangle $\triangle ADE$ is tangent to BC at a point T. Then T is the midpoint of the arc DE of the circle k opposite to A, so that AT bisects the angle $\angle EAD \equiv \angle CAB \equiv \angle A.$ Let $AM \equiv AP$ meet the circumcircle of the triangle $\triangle ABT$ again at U. Then

$\angle BTU = \angle BAU \equiv \angle DAM = \angle DEM \equiv \angle DEB = \angle CBE \equiv \angle TBM$

which implies $TU \parallel BM.$ Similarly,

$\angle TBU = \angle TAU \equiv \angle TAM = \angle BMT$

which implies $BU \parallel MT.$ Thus BMTU is a parallelogram, its diagonals BT, MU cut each other in half at P and $PT = \frac{BT}{2}.$  In exactly the same way, we can find $QT = \frac{CT}{2}.$ Combining, $PQ = PT+QT = \frac{BT+CT}{2}= \frac{BC}{2}.$ But this is true for a unique position of $X \in X_{0}A',$ of the corresponding triangle $\triangle ADE,$ and of its circumcircle k.

[b]Could anybody verify if my solution is correct? (sorry for no picture)[/b]

Let $Y$, $Z$, be the midpoints of $AB$ and $AC$, respectively. $YZ=\frac{1}{2}*BC=PQ$, also $YZ//PQ$ by similarity, so $YZQP$ is a parallelogram, which gives $PY//QZ$.

Let $l$ be a line such that $PY//l//QZ$ that passes through $A$, let $l\cap BC=X$, by similarity, we have that $BP=PX$ and $QC=QX$.

Then $\angle BAP=\angle DEB=\angle MBP$ (by inscribed angles and parallel lines), so $\triangle PAB\sim\triangle PBM$, so $BP^{2}=PM\cdot PA=PX^{2}$,

By the converse of power of a point, the circumcircle of $AMX$ is tangent to $BC$ at $X$, by symmetry, the circumcircle of $\triangle ANX$ is tangent to $BC$ at $X$,

since the circle through a point, and tangent to a line at a point defines a unique circle, $AMXN$ is cyclic. Hence, the circumcircle of $AENXMD$ is tangent to $BC$ at $X$.

Since $DE//BC$, $X$ is the midpoint of arc $DXE$, then $\angle DAX=\angle EAX$ by inscribed angles, Conclusion follows.


What schools have the top math programs and where are seniors going?

Also keep in mind... http://en.wikipedia.org/wiki/Gourman_Report

The Gourman Report seems fairly unreliable.
do you guys concur with it? (princeton > everyone  :D )

[quote="mathemonster"](princeton > everyone  :D )[/quote]

What's wrong with that?

The Gourman Report is idiosyncratic and, yes, quite unreliable.

I don't like any of these ranking methods because they don't take into account what's really important about a University. In my opinion the best way to judge a University, is with some metric that shows how the college changes its students from the time they enter to the time they graduate. It should not be based on endowment size or faculty research or number of putnam fellows - but on individual and personal stories or growth and development.

But since this is too hard, they settle on these statistics that don't really correlate well to the actual academic experiences of the students. I acknowledge that reputation is a very important factor in a school, because it's the first thing that will get noticed on your resume as a young person. But a better reputation doesn't always mean a better fit (educationally, socially, or financially) for every individual.

yes I agree...that doesnt mean i dont want to go to an ivy though :D


I will be posting a few questions within the next couple of min. Please help with whichever you  can.  :lol:

A truck traveling at a constant rate of 45  miles per hour leaves Albany. One hour later a car traveling at a rate of 60 miles per hour also leaves albany traveling the same direction. How long will it take for the the two viechles meet?

Rate of the truck- 45 mph.
Time: x+1 [+1 because it leaves Albany one hour later]

Distance= Rate*Time
Therefore, 45(x+1)

Rate of car- 60 mph
Time: x
Therefore, 60x

When they meet, the distance they traveled must be equal, 45(x+1)= 60x =>
Bla bla bla, X= 3.

3 hours.

[quote="PiLuvah"]thnxxx, also another question, what is the value of 1001C999... bigggggg numbers.....[/quote]

Remember, nCr= nCn-r.

Applying this, 1001C999= 1001C2= 500500

Note that you can also do this on your calculator. (I'm not trying to encourage anyone to be lazy or anything, but both methods involve "bigggggg numbers"...)


I also found the location to be a lake in Darfur.

Hmm, my deepest voice tells me that the author of the original question had another Harazi in mind...

  darij

Ah, now I know: he meant [b][url=http://www.al-bab.com/yemen/cole8.htm]Mohammed Omar Al-Harazi[/url][/b], suspect of preparing the attack on the USS Cole in 2000. In fact, maybe not only Raf (Rote Armee Fraktion?) is interested where he is  :D

  dg

BTW, Raf, do [i]you[/i] happen to know where he hides?  :D

  dg

And don't forget it is not the first time a terrorist hides as a mathematician. I guess he is now preparing a bomb attack on MathLinks.

  Darij

Even if I'm not Raf, the right sense of his question was : Does someone know if Harazi is now in France for his studies, as it was an opportunity for him?
In the affirmative, Harazi and Raf may be in the same class.

Pierre.

I can confirm the rumours :D Harazi has indeed won that scholarship that I have announced on the site a long time ago, and he is going to join a preparation class for ecole normale superieure this fall.

If I am not mistakening he is already there :)

We await him to find a internet connection there, so share with us his opinion about this :)

[quote="grobber"][quote="RC-7th"]Longitude:26,81667
Latitude:10,75

We're looking for the lake in sudan, right? The location I found is also in Darfur.[/quote]

I really thought you were joking :D:D.[/quote]

I kinda was joking, the majority of my search results were about the person.

Congratulations, Harazi!!

Now I guess it was a rhetoric question, Raf  :cool:

But some of us, me for instance, indeed were wondering why he hadn't posted anything for more than a week!

  Darij

Congratulations, harazi! :)

And I'll join Darij in saying that I was also wondering why I didn't see him around for quite some time.

Raf, or harazi, tell us if the two of you see one another, okay? :)

Bonjour mes amis,
 These are the only words I know in French. There is a great rule in life: bad things never dissapear, and I'm not an exception. So, I won't leave the site, don't worry. The bad point is that only today I found the informatique club, so here I am. I am not in Sudan, I am in Louis le Grand in Paris, near Sorbonne. And yes, I am studying in MPSI 1, but I didn't meet raf till now. Probably he is not in the same class as me, but it would be funny to meet him. Anyway, I am such an insignifiant person that I really do not deserve a whole topic. That is why I end here.


Let $ ABC$ be an $ A$-isosceles triangle. Let $ D\in (AB)$ , $ E\in (AC)$ be any points so that $ DE$

is tangent to the circle with the center in the midpoint $ O$ of $ [BC]$ and which is tangent

to $ AB$ , $ AC$. Prove that the values of $ m(\widehat {DOE})$ , $ BD\cdot CE$ and $ \frac {OD\cdot OE}{DE}$ are constant.

$ \widehat{ DOE}$ is constant because O becomes the A-exincenter of $ \Delta ADE$ , thus $ \widehat{DOE} \equal{} 90 \minus{} \frac {1}{2}A$

In the triangle $ ODE$ we have $ OD.OEsin(\widehat{DOE}) \equal{} r.DE$ , if $ \widehat{DOE} , r$ are constant  then $ \frac {OD.OE}{DE}$ is constant as well.

[hide]I cannot prove that BD.CE is constant , is this value constant Virgil?[/hide]

Yes, this is the essential conclusion. Try, is interesting relation.

Im looking forward to seeing that proof because i dont see it  :oops: .

But on particular cases verify you that is it "working" ?! Try to prove that $ \triangle BOD\sim\triangle CEO$ ...

Thanks, now i see it.

The circle ODE goes through the incenter of ADE , thus it is tangente to BC, so we have the angle equality <DOB=<DEO=<OEC . the same way we have <BDO= <EOC. The triangles BOD and EOC are similar, then EC/BO=BD/CO ...EC.BD=OB^2 =constant

Cute problem


How many positive perfect cube factors does $2^4 * 3^6 * 5^{10} * 7^9$ have?

Is there any fast way to do this?  That's besides counting the ones that have 3 powers, 6 powers, 9 powers, and so on.

[hide]Is it 96 cuz if so then 4/3 rounded down =1 6=2 10=3 and 9=3. Add one to each to get 2/3/4/4. Multily  them to get 96[/hide]

I'm glad i'm asking all the problems I have BEFORE my states tomorrow.  That way I can get help.

[quote="biffanddoc"][hide]
ooo cubes.
for each of the prime factors, the cube must have a multiple of 3 for each exponent.
2^0,3
3^0,3,6
5^0,3,6,9
7^0,3,6,9

Thats 2*3*4*4=96[/hide][/quote]

That's exactly how I did it.  Note that there are solutions in the back of the handbook as well.


quadratics
integration
real analysis
real analysis unsolved

Hi guys,

I am trying to calculate the class number of the imaginary quadratic field Q(sqrt(-5)). I've reduced it to calculating the sum 1/5 - 1/15 + 1/25 - 1/35 + ... (sum 1/n, n is odd and divisible by 5). How do I calculate this sum? Thanks.

That's $\frac15\cdot(1-\frac13+\frac15-\frac17+\cdots)=\frac15\sum_{i=0}^\infty\int_0^1 (-x^2)^n\,dx=\frac15\int_0^1\frac1{1+x^2}\,dx=\frac{\pi}{20}$.

That's not the only way to see it, but it does generalize nicely. Any series $\sum_n\frac{c_n}{n}$ with $c_n$ periodic (and $c_n$ having average value zero) can be calculated by integrating a rational function.


let $a,b,c\in\mathbb{R}^{+}$ prove:

$\sqrt{\frac{a(b+c)}{2}}+\sqrt{\frac{b(a+b)}{2}}+\sqrt{\frac{c(c+a)}{2}}\le a+b+c$

$\sum_{cyc}\sqrt{\frac{a(b+c)}{2}}\le \sum_{cyc}a$
$\Leftrightarrow \sum_{cyc}\frac{a}{a+b+c}\sqrt{\frac{b+c}{2a}}\le 1$
Cause $f(x)=\sqrt{x}$ is concave for $x>0$ so from Jensen:
$\sum_{cyc}\frac{a}{a+b+c}f\left(\frac{b+c}{2a}\right)\le f\left(\sum_{cyc}{\frac{a}{a+b+c}}\left({\frac{b+c}{2a}\right)}\right)=1$

Using the inequality $\ AM-GM$ , we have
          $\sum_{cyc}\sqrt{\frac{a(b+c)}{2}}$ $\leq$   $\sum_{cyc}\frac{a+\frac{b+c}{2}}{2}$ $\leq$ $\sum_{cyc}a$
                So the inequality is stric ! :rotfl:  :D
*) Comprise this inequality.
   Let $\ x_{1},x_{2},....,x_{n}$ be positive real number such that $\sum_{}x_{i}=1$
Prove that   $\sum_{cyc}\sqrt[n]{\frac{x_{i}(2006-x_{i})}{2}}$ $\leq2006$.
                which is true ? :D

[quote="2006"]Using the inequality $\ AM-GM$ , we have
          $\sum_{cyc}\sqrt{\frac{a(b+c)}{2}}$ $\leq$   $\sum_{cyc}\frac{a+\frac{b+c}{2}}{2}$ $\leq$ $\sum_{cyc}a$
                So the inequality is stric ! :rotfl:  :D
[/quote]
The inequality is not strict, anyway :) [But your answer is true]

It should be $\sum_{cyc}\frac{a+\frac{b+c}{2}}{2}=\sum_{cyc}a$

[quote="2006"]
*) Comprise this inequality.
Let $\ x_{1},x_{2},....,x_{n}$ be positive real number such that $\sum_{}x_{i}=1$
Prove that   $\sum_{cyc}\sqrt[n]{\frac{x_{i}(2006-x_{i})}{2}}$ $\leq2006$.
                which is true ? :D[/quote]

Are you sure? It is false, set $n=10000$, and $x_{1}=x_{2}=...=x_{10000}=\frac{1}{10000}$

$\sum_{cyc}\sqrt[n]{\frac{x_{i}(2006-x_{i})}{2}}>10000 \cdot \sqrt[10000]{\frac{1}{10}}> 5000 > 2006$


Prove that for any triple $ (a,b,c)$ of primes, the following inequality holds: $ a^2b\plus{}a^2\plus{}ac^2\plus{}115a\plus{}b^2c\plus{}b^2\plus{}c^2\plus{}27c\plus{}176>6ab\plus{}22ac\plus{}14bc\plus{}5b$.


Let S=(1,2,3,4....,24,25).  Commpute the number of elements in the largest subset of S such that no two elements in the subset differ by the square of an integer.


Find all real solutions of the equation

$\begin{equation*}
\sqrt{x^{2}+2px-p^{2}}-\sqrt{x^{2}-2px-p^{2}}=1,
\end{equation*}$

where $p$ is a positive real number.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]


The polynomial

$x^5+ax^4+40x^3+bx^2+cx-32$

has positive real roots. Find $a, b,$ and $c$.

I'm curious as to how you're going to prove this without an application of AM-GM...

[quote="eryaman"][hide]The $x^{5}$ and $32$ remind us of $(x-2)^{5}$, which has positive real roots.  Expanding this gives us
$x^{5}-10x^{4}+40x^{3}-80x^{2}+80x-32$, which fits the given polynomial.

So $a=-10$, $b=-80$, and $c=80$.[/hide][/quote]

Yeah, I think more importantly we want to prove that it has to be that. But it is the only one, in case you were wondering.

His solution is 95% of the legwork.

It's a trivial application of AM-GM after that.

But really, is AM-GM appropriate for this forum (just out of curiosity)

[quote="blahblahblah"]But really, is AM-GM appropriate for this forum (just out of curiosity)[/quote]

I've seen some threads on AM-GM... that's why I put it here, because I think it would be too simple for Pre-Olympiad.

IMO, AM GM in most instances is appropriate for Intermediate...I haven't attempted this problem so I'm not totally sure it is at "intermediate" level. But more advanced inequalities belong in Pre Olympiad and above (even Cauchy).


search
absolute value
algebra
polynomial
Rational Root Theorem

1st root is 3. And it has no more R roots. Why do you want 3 different solutions... if it has one very trivial??

Yes, the answer is correct.But I solve it in a beutiful way.Look!
We can rewrite (x^2-4)+(x^3-8)=24=>(x-2)(x+2)+(x-2)(x^2+2x+4)=24=>(x-2)(x+2+x^2+2x+4)=24=>
(x-2)(x^2+3x+6)=24, which has the solution x=3.A
   Another way is that:we can suppose that x isn't 3.We take x>3.Then x^2+x^3>36.Also we take x<3.Then x^2+x^3<36.So, the solution is x=3. ;)  ;)

[quote="stancioiu sorin"]Solve in R the equation x^2+x^3=36.Please find at least 3 ways to solve it.[/quote]

[hide="Yet another"]
The LHS is monotonically increasing for positive $x$, so only one solution.  For $-1 \le x \le 0$ the LHS has absolute value $<1$.  For $x < -1$ the LHS is negative.

:)
[/hide]

[quote="ArcticMonkey"]One more solution,

[hide]
$x^2+x^3=36$
$x^2 - 9 = 27 - x^3$
$(x-3)(x+3)=(3-x)(x^2+3x+9)$

From here we get the solution $x=3$. Lets say $x\neq 3$, so we can divide both sides of equation by $(x-3)$,

$x+3=-(x^2+3x+9)$
$x^2+4x+10=0$

$D=4^2-40 <0$, thus the only solution is $\boxed{x=3}$
[/hide][/quote]
   Yes,it's a beautiful solution!Now I think BOMBON understood that it's an interesting problem for its different ways to solve it.

by decartes, there is one real, positive root (looking at sign variation), rational root theorem limits our search, and we find that x=3 is a solution


Hello everyone, I am just curious, I cannot find anywhere this notation being used, but I like it. So I am going to put down the notation without explanation of its meaning, and I would really appreciate it if someone would translate what they think it means. I am doing this because if people cannot interpret what this notation means, obviously I shouldn't be using it

here it is

" Let $ f(x)\in{C^{k\geq{n}}}$"

Any response would be appreciated.

[quote="t0rajir0u"]If you use a notation that you can't find being used, you should always define it anyway.  It looks like you're talking about the space of functions with derivatives of at least $ n$ orders.[/quote]
 :D ...Your right...but at least you understood it

[quote="MysticTerminator"]Doesn't the notation C^n already exist for that?[/quote]

I have never seen someone say $ C^{k\geq{n}}$, but then again, I am only in eleventh grade...so I haven't seen that much. So are you saying that it is a common usage of this notation?

[quote="MysticTerminator"]I'm saying that it's a pointless notation since the shorter and commonly used alternative C^n already exists[/quote]

But don't they have different meanings?

$ C^{n}$ implies that something is n-times differentiable
$ C^{k\geq{n}}$ implies that something is [u]at least[/u] n times differentiable.

$ C^n$ does not mean [i]exactly[/i] $ n$ times differentiable - the set of functions with derivatives of $ n^{th}$ but not $ (n\plus{}1)^{th}$ order do not form a vector space.  The spaces $ C^n$ are defined so that $ C^{\infty} \subset ... \subset C^n \subset C^{n\minus{}1} \subset ... \subset C^1$, so (and I'm not sure why I didn't catch it before) the standard notation already has the meaning you want to capture.

So maybe I am misunderstanding the notation, here is what my understanding is

$ f(x)\in{C^{8}(a,b)}$ says that on the interval $ (a,b)$ $ f(x)$ maybe differentiated 8 times.

So lets say for a trivial example I want to say

"f(x) is at least eight times differentiable across the reals"

Are you saying that it wouldnt be "$ f(x)\in{C^{k\geq{8}}\left(\mathbb{R}\right)}$" it would just be "$ f(x)\in{C^{8}\left(\mathbb{R}\right)}$"?

EDIT:

I understand now, since for a function to be say 10th order differentiable it must be eight order differentiable, so this notation does encompass all the higher order derivatives.

$ C^n$ means that the functions is $ n\minus{}$times differentiable [i]and[/i] that the $ n\minus{}$th derivative is continuous!


So to produce doublespacing in my document, I used \linespread{2}. However, I want singlespacing in a few parts of my document. For example, when I have a list or enumeration, I want singlespacing for the items since I think doublespacing in those areas does not look very pretty. Is there a way to accomplish this? Thanks.

Have a look at section 6.3.1 of [url=http://www.ctan.org/tex-archive/info/lshort/english/lshort.pdf]The Not So Short Introduction to LATEX2e[/url] where it says:
[quote]Note that the effect of the \linespread command is rather drastic and
not appropriate for published work. So if you have a good reason for changing
the line spacing you might want to use the command:
\setlength{\baselineskip}{1.5\baselineskip}[/quote]

Hmm, do you know of any way to accomplish the opposite of that? That is, be able to set the spacing to doublespaced by default and use \setlength{\baselineskip}{1.0\baselineskip} whenever I want to single space a body of text. Since almost all of my document will be doublespaced, adding that code to every paragraph would be on the cumbersome side.

Have a look at the [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=linespace]TeX FAQ[/url] which (grudgingly) recommends the setspace package. There's more at [url=http://web.mit.edu/answers/latex/formatting/latex_spacing.html]How to change the SPACING inside documents[/url]

Details of the setspace package are in the [url=http://www.csit.fsu.edu/~mimi/tex/doc/html/catalogue/entries/setspace.html]TeX Catalogue Online[/url]

*sigh* I was afraid that I'd have to use setspace. Last time I tried to get that installed I ran across a bunch of problems.

Oh well, thanks for the help.

If you use \renewcommand{\baselinestretch}{2} in the preamble you will get double spacing by default


[quote="ehsan2004"]Let $ a,b,c \in \mathbb{R}^n$. Prove that $ |a \plus{} b| \plus{} |b \plus{} c| \plus{} |c \plus{} a| \leq |a \plus{} b \plus{} c| \plus{} |a| \plus{} |b| \plus{} |c|$.[/quote]
Let $ a\equal{}x, b\equal{}y, c\equal{}z$. See solution 3.

[quote="manlio"]Sorry, but I cannot open this file, it is all black :maybe:

[quote="NguyenDungTN"][quote="ehsan2004"]Let $ a,b,c \in \mathbb{R}^n$. Prove that $ |a \plus{} b| \plus{} |b \plus{} c| \plus{} |c \plus{} a| \leq |a \plus{} b \plus{} c| \plus{} |a| \plus{} |b| \plus{} |c|$.[/quote]
Let $ a \equal{} x, b \equal{} y, c \equal{} z$. See solution 3.[/quote][/quote]
Here is it :)


What is the sum of the reciprocals of the roots of the equation
\[ \frac {2003}{2004}x \plus{} 1 \plus{} \frac {1}{x} \equal{} 0?
\]
$ \textbf{(A)}\ \minus{}\! \frac {2004}{2003} \qquad \textbf{(B)}\ \minus{} \!1 \qquad \textbf{(C)}\ \frac {2003}{2004} \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ \frac {2004}{2003}$

Is it -1   ?
I didn't understand the answer quite correct but I know that x have no solution  :wink:

[hide="Solution"]
First, multiply by $ x$ on both sides to get
$ \frac {2003}{2004}x^2 \plus{} x \plus{} 1 \equal{} 0$

By Vieta, we have the following:
$ a \plus{} b \equal{} \minus{} \frac {2004}{2003} \\
ab \equal{} \frac {2004}{2003}$

We are looking for $ \frac {1}{a} \plus{} \frac {1}{b} \equal{} \frac {a \plus{} b}{ab} \equal{} \frac { \minus{} \frac {2004}{2003}}{\frac {2004}{2004}} \equal{} \minus{} 1 (B)$.
[/hide]

[hide] Consider a polynomial $ P(x)\equal{}a_nx^n\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}....\plus{}a_1x\plus{}a_0$. The polynomial that has the roots that are reciprocals of P(x) is found by substituting $ \frac{1}{x}$ in for x and then multiplying by $ x^n$ to get a polynomial with the coefficients of P(x) in the reverse order.

In this problem, the polynomial that has the desired roots is $ x^2\plus{}x\plus{}\frac{2003}{2004}$. The sum of the roots, by Vieta's, is $ \textbf{(B)}\ \minus{} \!1$.[/hide]

$$\text{full and clear sol}$$

$$\implies \frac{2003}{2004}x+1+\frac{1}{x}= \frac{2003x^2 + 2004x + 2004}{2004x} = 0$$
Multiply both sides by $x$ to get that:
$$2003x^2 + 2004x  + 2004 = 0$$

Let $a$ and $b$ the roots of this equation.
$$\implies \frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab}$$

By Vieta's Formula,
$$a + b = -2004/2003$$
$$ab = 2004/2003$$

$$\implies \frac{a + b}{ab} = \boxed{-1}$$

[quote=IAmTheHazard][quote=Possible][quote=Dr Sonnhard Graubner]hello, the solutions of your equation $ \frac{2003}{2004}x^2\plus{}x\plus{}1\equal{}0$ are
$ x_1\equal{}\minus{}\frac{1002}{2003}\plus{}\frac{2}{2003}\sqrt{752502}i$
$ x_2\equal{}\minus{}\frac{1002}{2003}\minus{}\frac{2}{2003}\sqrt{752502}i$
A algebraic computation gives $ \frac{1}{x_1}\plus{}\frac{1}{x_2}\equal{}\minus{}1$
So the right answer is $ \boxed{B}$.
Sonnhard.[/quote]

Thats fairly hard to bash without a calculator[/quote]

Lol that's how he does things

[quote=yofro]Wow, Sonhard actually posted a half-solution...[/quote]

yeah I'm really surprised[/quote]
Friend  you can do by the sum of roots and product of roots that is alpha + beta is  -b/a and alpha beta is c/a

[quote=HumanCalculator9]bruh use Vieta's bash is bad[/quote]

Nah. 1/a+1/b = (a+b)/ab.
Then, use vietas

Really, the simplest solution is to just plug in 1/x for x and then use vietas.


geometry
trapezoid
trigonometry
trig identities
Law of Sines
geometry unsolved

ABC is an isosceles triangle. AB=CB and D is a point on ABand  <DCB = 30 degree. Find the measure of  <B.

sorry the first picture is picture. second is true

[color=darkblue][b]A well-known similar problem.[/b][/color]

[color=darkred] Let $ABC$ be an $A$- isosceles triangle for which $A=20^{\circ}$. There is a point $D\in (AB)$ so that $AD=BC$. Ascertain $m(\widehat{BDC})\ .$[/color]

[color=darkblue][b]Remark.[/b] In the my book of geometry ([i]Publishing house GIL[/i]), for this problem appear six synthetical methods.[/color]

Yesterday I tried to solve the problem(synthetical solution) in the way that Virgil said , but I failed.I thought that the problem was the same  with $angleA =20$, but it is not.  I'll try again later.

[u] Babis[/u]

[quote="ayhanozbekk"] $ABC$ is an isosceles triangle, $AB=CB$ and $D$ is a point on side $AB$ [color=blue]such that $BD=AC$[/color] and  $\angle DCB = 30$ degrees. Find the measure of  $\angle B$.

sorry the first [color=red]picture is picture[/color]. second is true[/quote]
  1. I do not see any figure.
  2. Please write the problem again.
  3. I have tried to guess the problem. Is the new wording right?
  4. I please V. Nicula to clear the wording. I can not understand his solution because my figure does not work.
  5. With my wording I tried the problem with trig.but the answer $30$ is not right.
   [u]Babis[/u]

dont mention the picture. I again write the question.   ABC is an isosceles  triangle ( AB = CB )  . Given point D which on the side AB  and AC = DB . if the angle m(DCB) = 30 degree  then find angle m(DCB) ?
is it clear?

[quote="Virgil Nicula"][color=darkblue][b]Proof.[/b] Construct the equilateral $\triangle BCE$ so that the line $BC$ don't separate the points $B$, $E$ $\Longrightarrow$ $ACED$- isosceles trapezoid $(ED\parallel AC)$ $\Longrightarrow$
$AE\perp BC$ and $m(\widehat{EDC})=m(\widehat{EAC})=10^{\circ}$ $\Longrightarrow$ $m(\widehat{BDC})=m(\widehat{BDE})+m(\widehat{EDC})$ $\Longrightarrow$ $m(\widehat{BDC})=30^{\circ}\ .$[/color][/quote]

Please, state WHICH problem this solves and make an exact statement about which point is which. It got all messed up.

[b]Farenhajt[/b], you did not read the all my previous messages. I solved above the my proposed problem from the mentioned book.

[color=darkred][b]An extension.[/b] Let $ABC$ be an isosceles triangle $(AB=AC)$. Suppose that $AB>BC$.

Then exists the point $D\in (AB)$ so that $AD=BC$.Denote $m(\widehat{BDC})=y>A$.

Prove that there is the following trigonometrical relation : $\boxed{\ 2\sin\frac{A}{2}\sin y=\sin (y-A)\ }\ \ (*)\ .$

[b]A particular case.[/b] $A=20^{\circ}$ $\Longleftrightarrow$ $y=30^{\circ}\ .$[/color]

[color=darkblue][b]Proof.[/b] Suppose w.l.o.g. $AB=AC=1$. Denote  $AD=BC=l<1$ and $A=2x$. Apply the Sinus' theorem in the triangles $ADC$ and $BCD$ :
$\frac{l}{\sin (y-2x)}=\frac{1}{\sin y}\ \ ;\ \ \frac{l}{\sin y}=\frac{1-l}{\cos (y-x)}=\frac{1}{\sin y+\cos (y-x)}$ $\Longrightarrow$ $\sin^{2}y=\sin y\sin (y-2x)+\sin (y-2x)\cos (y-x)$ $\Longrightarrow$
$1-\cos 2y=\cos 2x-\cos (2y-2x)+2\cos (y-x)\sin (y-2x)$ $\Longrightarrow$ $1-\cos 2x=\cos 2y-\cos (2y-2x)+2\cos (y-x)\sin (y-2x)$ $\Longrightarrow$ $\sin^{2}x+\sin (2y-x)\sin x=\cos (y-x)\sin (y-2x)$ $\Longrightarrow$ $\sin x[\sin x+\sin (2y-x)]==\cos (y-x)\sin (y-2x)$ $\Longrightarrow$
$2\sin x\sin y\cos (y-x)=\cos (y-x)\sin (y-2x)$ $\Longrightarrow$ $2\sin x\sin y=\sin (y-2x)\ .$

[b]Remark.[/b] The relation $(*)$ is equivalently with the relation $\tan y=\frac{\sin A}{\cos A-2\sin \frac{A}{2}}\ .$[/color]

[quote="ayhanozbekk"]   ABC is an isosceles  triangle ( AB = CB )  . Given point D which on the side AB  and AC = DB . if the angle m(DCB) = 30 degree  then find angle [color=red]m(DCB) [/color]?
is it clear?[/quote]
  Do you want to say $\angle DBC$ and not $\angle DCB$ , as you write?

 If so , then the problem has not solution. If we put $\angle B = x$ , using Law of sines we find $4\sin \frac{x}{2}\sin(x+30) =1$. But this equation has not positive root on $(0, \frac{pi}{2})$. ML Silouan told me that programm MATEMATICA gives as solution $x=-60$ degrees.If this is true , then the problem has ... problem.Please , correct , if possible, because it seems nice.

 Babis

 [color=blue]Mmm ! Now I think to take angle $B$ obtuse and not acute and the point $D$ on the extension of side $BA$ in the direction to $A$. I'll try!!! [/color]

[quote="ayhanozbekk"]ABC is an isosceles triangle. AB=CB and D is a point on ABand  $\boxed{\angle DCB =30^{\circ}\ }$. Find the measure of  <B.[/quote]
[size=150][color=red]I think that $m(\widehat{CDA})=30^{\circ}\ !$ In this case, the answer is $B=20^{\circ}$.[/color][/size] See the my last message.


inequalities
Cauchy Inequality
inequalities unsolved

Let $ a,b,c >0$ s.t $ ab\plus{}bc\plus{}ac\plus{}2abc\equal{}1$ .Prove that
$ \frac{1}{a\plus{}b\plus{}2}\plus{}\frac{1}{c\plus{}b\plus{}2}\plus{}\frac{1}{a\plus{}c\plus{}2} \le 1$

First substitute $ a \equal{} \frac{x}{2}$, then notice the condition became $ xy\plus{}yz\plus{}zx\plus{}xyz \equal{} 4$. This is equivalent to $ \sum \frac{1}{x\plus{}2} \equal{} 1$. Then by Cauchy inequality,
$ \frac{4}{x\plus{}y\plus{}4} \leq \frac{1}{x\plus{}2} \plus{} \frac{1}{y\plus{}2}$
Sum up and done.


Do I need to install a pdf file creation util in order to be able to create pdf(s).  As part of the of the LaTeX installation.

If so what are some recommendations.   :?:

[url=http://oku.edu.mie-u.ac.jp/~okumura/texwiki/?PDF%E3%81%AE%E4%BD%9C%E3%82%8A%E6%96%B9]How to make of the PDF[/url]
This site is Japanese.
However, I enter if very careful explanation is various.
You use the translation sites, and please read.

You actually don't need a PDF creator, just a PDF reader. TexInc (If that's the one you are using.) builds, and prints it on PDF.


exists is 67%, proven by Stephen Unwin.  anyone read the book?

im 100% sure he exists

and how can you get such a precise percentage (67%) by using science?

I believe I speak for a great many people on the forums when I ask:

"Who sez?"

[hide]1. Because God made it for us to live on it(not cuz of some big bang!)

2. Wasn't he lonely (or because he loved us, the same reason ur mom gave birth to u) go check in the bible!

3. God is the Alpha, Omega, Beginning and End, Savior, Messiah, Redeemer and friend, Saving Grace, Living God,... there's a song called Prince of Peace.

Our pastor gave a chain sermon on how he's or Shield, Support, Shepherd, and Soverein.  He's mostly, our (heavenly) Father.

4. GOD was always here.

5. See above answer.[/hide]

are you sure sunnyray? let me ask you another question: man hates his creation, does God hate his creation too?  and for your number 1, He probably was lonely.  he couldn't love us until he creates us.

1. [hide]Most likely because 4.6 billion years ago, globs of chemicals combined in a reaction that produced the first prokaryotic cells. Some of this first life then evolved over billions of years to become us.[/hide]

2. [hide]See first sentence of above answer.[/hide]

3. [hide]How the heck am I supposed to know? (Guess my religious orientation)[/hide]

4. [hide]See above answer.[/hide]

5. [hide]See answer to #3.[/hide]

that doesn't explain it.  i read that 6 billion years ago, the universe was a cloud of dust.  then 4.5 billion years ago, well, you know what happened.

hmm...i only thought that the world existed 6 bil. years ago...my brain must be out of date.

God, gods, or both?

I think there is [b]a[/b] God. There aren't gods, in my opinion.

[quote="LynnelleYe"]What I've read says the current universe formed 12 or 13 billion years ago. The commonly accepted theory is that the Big Bang exploded out of nowhere and for no particular reason, but there are a couple others floating around, like that the Big Bang's been happening and re-happening in a long cycle infinitely into the past and will happen and re-happen infinitely into the future. Are you referring to what's now the solar system?[/quote]

There was a "particular reason," as you choose to put it, but it is unknown to us and irrelevant; as the laws of physics and science break down at such scales, we have no way of knowing.

ok one simple answer to your question sunnyray.  there might not be a God because...let me think...I AM GOD! MWAHAHAHAHA! sorry. i was just joking.  um...there isn't a God because who created God? that's the answer to your question.

sunnyray wrote:gameworld... to ur question

[hide]God LOVES us... so why would he HATE us... u kno how in the bible, God disciplines ppl.... but lyk the way our parents disciplines US... even if ur dad shouts at u, its bcuz he wants u to know what u did wrong, and point u in the rite direction

man dont create

man aren't perfect, lyk God.  how can we compare to Him?

also... he knew us even before he was born... even before we could talk he knew everyword we said.... its pretty complicated, hard to realize and understand....[/hide]

besides... how could you prove there ISNT a God.

because, have u ever created something and blew it up just for the fun of it?

if god was so perfect, he wouldnt create imperfect beings (like us). Even the bible says god isnt perfect (or at least in jewish theology)

does god believe in another god?

why shouldnt he? If i created a simulation, would i b perfect?

i would definietly be evil=)

and for the original topic, it is completely bogus.  you cant do probability like that!!!!

And i believe that it is 85% because 85% of ppl believe in god and its an 85% that ur one of them

Prove that they aren't. I've already asserted the fact that I don't believe there is any other rational solution to a non-evolution world other than creationism. Now it is your job to show me that there is. That is how an argument/debate/discussion works :)

I never said there had to be another alternative. I only said it isn't guaranteed that there isn't, so if evolution is falsified, God hasn't been proven. You asserted that evolution and creationism are the only possibilities, you prove it.
If you absolutely must have a counterexample (though the burden of proof is on you) what about the belief that aliens created/seeded/etc. the earth? Evolution will not have taken place but God won't necessarily exist either (though he still could). The aliens could have come about through evolution or something else--falsifying evolution on earth doesn't falsify it through the whole universe. It's a wacky belief with no supporting evidence but no falsifying evidence either.

I didn't say those were the only two options. I just said that if evolution was proven to be false, then creationism would be the only [b]reasonable[/b] option left.

I'm sorry but that alien idea really is wacky...

Well, djimenez how can u say [quote]No, evolution is a scientifical fact, while creationism is a myth![/quote] when it is called the [b]Theory[/b] of Evolution, rather than the [b]Theorem [/b]of Evolution?

(sry i dont know how to quote passages with the name included)

[quote="djimenez"][quote="LynnelleYe"]Prove that evolution and creationism are the only two alternatives.[/quote]

No, evolution is a scientifical fact, while creationism is a myth![/quote]

Stating that an idea is wacky doesn't make it unreasonable. It has no evidence falsifying it, just like creationism (meaning some types of creationism. Young-Earth Creationism and the like do have falsifying evidence.) How wacky an idea is depends on a person's perspective, and obviously reality doesn't. It's the supporting evidence plus the lack of falsifying evidence that's important, and so far only evolution has both.
And, like I said before, there may be another option that we have no clue about. We can't know that all the possibilities have been brought up already.

And evolution hasn't been proven. Nothing in science is ever proven. 'Theory' is as good as it gets in science--theory of gravity, germ theory, etc. etc. Also the definition of a 'theory' in science is an explanation for some observations in the natural world, very well-supported by evidence. Nothing about it being tentative or not yet proven, which it can't be anyway. However, some types of creationism have been falsified.

And don't forget, just because we haven't thought of them yet doesn't mean there are more reasonable solutions out there. There are some things beyond human comprehension. Then again, isn't that how the myth of creationism started anyway?...

The probability of god existing is x where

0 <= x <= 1

let's make a comprimise and let x= 1/2

There's a big difference between knowing and [i]believing[/i], all people seem to be saying in this post is what they believe because of their religion. Okay, assume God doesn't exist, then try to prove that he/she does. Now, assume God does exist and try to prove that he/she does not. What are you left with? Nothing. You shouldn't mix God with math, you're just going to be left with nothing, or be even more confused than you were when you started.

There are few exceptions...

If you're looking for a book that will challenge your perspective about the world (no matter what you believe), read "In the Begninning was Information" by Werner Gitt. It is the most logical book I've ever read. It's like one huge 10000-step proof.

The main points:

There are three entities that comprise the universe arround us: Matter, Energy, and Information.

First he goes into a huge description of the laws of the universe and the significance there-of.

For instance, matter and energy cannot be created or destroyed (given that E = mcc). So therefore the laws of the universe must have been put into place after matter and energy were created, or they must have always existed.

He then goes on to prove (basing a lot of his argument on the second law of thermodynamics) that they cannot have always existed because of information.

Information HAS TO have a sender with intelligence. When you consider the simplest bacteria, which has billions of bits (the standard unit of infomation) of coded information in its DNA, all the way up to the complexity human brain which can store 10 Tbites of information and process several hundred terabites per second, it is obvious that this infomation had to have an information giver. Information cannot create itself.* Therefore someone had to have put this information into existance (creating trillions of trillions of trillions of terebites of information) when matter and energy were created--otherwise we'd still be a big blob of mass or a tighly swirling dot.

His conclusion: Someone (God) must have created matter, energy, and information before the laws of the universe were in place; or created matter, energy, and information simultaneous to creating the laws of the universe.

My conclusion: There is a 100% chance of God.

*If you had monkeys trying to type "Encyclopedia Britannica" (just the title) randomly, you would need monkeys covering the entire surface of the earth, stacked one mile deep, each residing in 1 cubic foot of space, typing for 20 billion years to get it right once. (This is assuming 45 keys on the standard typewriter).

Interesting debate...  it wasnt exactly my intention to start a debate about whether God exist or not... (I just found it funny that someone said god had 67% chance of existing)

What can God do besides creating matter/life?  Can he do anything he pleases?

i agree with sunny ray on this whole subject. God does love us, but he gives us a choice. that choice is to follow him or not to follow him. we can either love him back and accept his free gift of salvation, or we can completely ignore him and live our lives our own way.  he wants you to love him and be with him in eternity, but he isnt going to force us to follow him.

I've always wondered how hard it would be for us to create our own universes. I'm sure that any day now someone is going to find out SOME way to manipulate other universes... please  :(

Come on, it can't be that hard.

What if other universes don't exist? We can't manipulate things that don't exist, can't we?


Now this entire page of threads is spammy!

Ah well. BTW I got 100 posts yesterday, mostly in MO forum!

Wow mew...

Spammer.

Spammer.

Spammer.

:rotfl:

Spam is also a noun.

Spammer=one who spams.

lol, someone from indiana is a hoosier...

Oh so thats why its called the Hoosier State....

Oh.

Hoosier (pronounced /ˈhuːʒɚ/) is the official demonym for a resident of the U.S. State of Indiana.

Hoosier=One who hooses. Isnt that when you spray water on your plants?

Ah yes you are correct. then what does hoose mean?

DUCK DUCK GOOSE! HA! So people from Indiana have to run a lot in that game.


Prove that the Euler-Totient function i.e. phi(n) is even for n>2.

If $(a,n) = 1$ then $(n-a,n) = 1$ so we can pair up all the relatively prime numbers.

Or, for n>2, n has at least one odd prime power factor (or it's 2^k, in which case it's even as well) and thus p-1 is even for that part, and the multiplicative nature of phi makes it a product, thus even.

[quote="Hokkage"]and the multiplicative nature of phi.[/quote]

This is worth discussing.  The original problem is too easy.

[b]Follow-up.[/b]  For a natural number $n$, we define $\varphi(n)$ to be the number of positive integers $a$ less than $n$ such that $gcd(a, n) = 1$.

Show that if $gcd(a, b) = 1$ then $\varphi(ab) = \varphi(a) \varphi(b)$.

[quote="Farenhajt"]This follows directly from $\phi(n)=n\left(1-{1\over p_{1}\right)\left(1-{1\over p_{2}}\right)\dots\left(1-{1\over p_{k}}\right)}$ where $p_{i}$ are all distinct prime factors of $n$.[/quote]

So prove this identity.   :wink:

So, we have to prove $\phi(n)=n\left(1-{1\over p_{1}}\right)\left(1-{1\over p_{2}}\right)\dots\left(1-{1\over p_{k}}\right)$

Well then, let's count the opposite, i.e. the numbers which AREN'T coprime with $n$.

If $n=p_{1}^{a_{1}}p_{2}^{a_{2}}\dots p_{k}^{a_{k}}$, then the counting goes by the inclusion-exclusion principle as follows:

We must include all numbers which have at least one common prime factor with $n$, and we have ${n\over p_{1}}+{n\over p_{2}}+\dots+{n\over p_{k}}$ of those, then subtract those which have at least two common prime factors with $n$, and we have $\sum_{i<j}{n\over{p_{i}p_{j}}}$ of those, then add back those which have at least three common prime factors with $n$ etc, ending with numbers which have $k$ common prime factors with $n$, and we have ${n\over{p_{1}\dots p_{k}}}$ of those. Subtracting the obtained sum from $n$ and factoring, we get the above identity.


function
induction
set theory
advanced fields
advanced fields theorems

Is Omega^Omega uncountable??

According to my book (Elementary Set Theory), it says that it is uncountable, as Omega^Omega >= 2^Omega >= Aleph(1), which is known as uncountable. (According to part of the proof / analysis of Cantor's Contium Hypothesis But for another book, it says that when counting ordinal numbers, it is like this:
Omega < Omega + 1 < Omega + 2 < ... < Omega + Omega = Omega * 2 < Omega * 3 < Omega * 4 < ... < Omega * Omega = Omega ^ 2 < Omega ^ 3 < Omega ^ 4 < ... < Omega ^ Omega < ...
And it says that as this can be counted out, it is coutable, with the smallest uncountable ordinal as Epsilon(0).

Omega = the set N, which is the smallest infinite ordinal and cardinal number.

$ \omega^\omega$ is uncountable.  $ \epsilon_{0}= \omega^{\omega^{\omega^\cdots}}$ is the smallest ordinal which cannot be reached by a finite number of steps of addition, multiplication and exponentiation, starting with $ \omega$.  It's much much larger than $ 2^\omega$ or $ \aleph_{1}$.

no.
this is an ambiguity, when you're talking of ordinals (which is one of the reasons why they introduced aleph notation, i think): exponentiation of ordinals does not match with exponentiation of cardinals.
so $ \omega^\omega$ is a countable ordinal, while $ 2^{\aleph_{0}}$ is an uncountable cardinal.

My apologies.  It's been a little while since I took set theory  :oops:

[quote="r2sputin"]If you view $ \omega$ as the set of natural numbers, then $ \omega^{\omega}$ is the set of infinite sequences of natural numbers, which is in fact uncountable. This can easily be seen by identifying the decimal expansion of any number $ \alpha\in(0,1)$ with an infinite sequence of naturals (but you would have to be careful with numbers that have two expansions.)

But as pointed out above, that is not how ordinal exponentiation is defined. If you have two ordinals $ \alpha,\beta$, then $ \alpha^{\beta}$ is a subset of the set of functions from $ \beta$ to $ \alpha$: it is those functions which are nonzero for only finitely many points $ \gamma\in\alpha$. So $ \omega^{\omega}$ is the set of all infinite sequences of natural numbers, which eventually terminate at $ 0$. This is in fact a countable set. It can be identified with the union $ \mathbb{N}\cup(\mathbb{N}\times\mathbb{N})\cup(\mathbb{N}\times\mathbb{N}\times\mathbb{N})\cup\ldots$.

Also, it turns out that $ \epsilon_{0}$ is countable.[/quote]

Shouldn't that $ \alpha\times\beta$ be the set of functions from $ \beta$ to $ \alpha$? It shouldn't be the exponential for $ \alpha^{\beta}$.

Also, why does the infinite sequences have to "eventually terminate at $ 0$"?

And for the union $ \mathbb{N}\cup(\mathbb{N}\times\mathbb{N})\cup(\mathbb{N}\times\mathbb{N}\times\mathbb{N})\cup\ldots$, are you meaning that the family for the sets in the union are having a natural number set as the index set?

And why is this union countable?

No, $ \alpha\times\beta$ would be the Cartesian product of those two sets. In general, if $ A$ and $ B$ are sets, then $ A^{B}$ is the set of functions from $ B$ to $ A$. For example, when people write $ 2^{\aleph_{0}}$, they mean the set of functions from $ \aleph_{0}$, which can be viewed as $ \mathbb{N}$, to $ \{0,1\}$, which represents $ 2$ (any number can be represented as a set.) That is, the set of binary sequences.

For your second question, it is a convention made to ensure that the resulting set is well ordered. If you view $ \omega^{\omega}$ as the set of infinite sequences of natural numbers, it won't be well ordered in any reasonable way. If the sequences terminate at $ 0$, you have the lexicographical order, which is a well ordering.

For your third question, I mean that this is a countable union. If you have any sequence of naturals which terminates at $ 0$, you can write it as $ (n_{1},\ldots,n_{j},0,\ldots)$. Then you can identify this with the ordered $ j$-tuple $ (n_{1},\ldots,n_{j})$. This is an element of $ \prod_{1}^{j}\mathbb{N}$. Any such sequence of naturals can be represented in this way for some $ j$, so you know that the set of all these sequences can be embedded into the union $ \bigcup_{j=1}^{\infty}\left(\prod_{1}^{j}\mathbb{N}\right)$ (this is just shorthand for the union I wrote in the first post.)

The reason that this union is countable is because for any $ j$, $ \prod_{1}^{j}\mathbb{N}$ is countable. This can be shown by induction once you have that $ \mathbb{N}\times\mathbb{N}$ is countable. The result then follows from the fact that a countable union of countable sets is countable.

But \omega^{\omega} can be defined as the mapping from \omega to \omega, and then can be transformed into a set of natural sequence in natural number, in which the terms are respective natural number that the functions map to.

But for this set of natural sequence of natural numbers, the sequences are infinite, which will not "EVENTUALLY" terminate at 0. For example, the sequence 0, 1, 2, 3, ... is  not included in the union of the sequences.

Also, \omega^{\omega} >~ 2^{\omega} ~ 2^{\aleph_{0}} >~ \aleph_{1}, which then is uncountable.

Then is \omega^{\omega} countable??

[quote]Then is \omega^{\omega} countable??[/quote]
The answer is yes and no. [b]r2sputin[/b] made it very clear a few posts ago that $ \omega^{\omega}$ can mean two different things: an uncountable cardinal or a countable ordinal.

Ok. Let me get it clear:

The countability of a cardinal, is simply by whether it is equipotent to natural number set, but for ordinal, what is the definition of countability?

Is it to check whether the ordinal is ISOMORPHIC to the natural umber set?

no, then difference is not a matter of definitions of countability/uncountability.
it's the exponentiation that is defined in a different way.

[b]darken[/b]'s third question could be simply answered as follows: $ \bigcup_{k \equal{} 1}^{\infty} \mathbb{N}^k \equal{} \mathbb{N}\cup (\mathbb{N} \times \mathbb{N}) \cup (\mathbb{N} \times \mathbb{N} \times \mathbb{N}) \cup \, ...$ is countable, because it is countable union of countable sets.


In a magic square, the sum of the three entries in any row, column, or diagonal is the same value.  The figure shows four of the entries of a magic square.  Find $x.$

[asy]
size(100);defaultpen(linewidth(0.7));
int i;
for(i=0; i<4; i=i+1) {
draw((0,2*i)--(6,2*i)^^(2*i,0)--(2*i,6));
}
label("$x$", (1,5));
label("$1$", (1,3));
label("$19$", (3,5));
label("$96$", (5,5));[/asy]

If its

[hide]$219$[/hide]

then I'll post my solution...

[quote="rjt"][hide]
Hm, I'm getting $200$. Likewise, I'll post my solution if this is correct.
[/hide][/quote]
That's also what I got.

[hide="a solution"]I also got 200, and here is my solution. Let the box be in the form

x  19  96
1  a    b
c  d    e

Because each line has to add up to the same thing, $19+96+x=1+c+x$ and $c=114$. Then, $114+96+a=19+a+d$, and $d=191$. Using the same logic, it is easy to find $b=209$. Now, the magic square looks like this

x    19    96
1    a    209
114 191  e

Because $x+a+e=114+a+96$, $x+e=210$. Because $114+1+x=96+209+e$, $x-e=190$, using a system of equations, $2x=400$, and $x=200$  [/hide]

[quote=Altheman]i suspect there is a solution where you don't compute any other number...while it does not really matter, in my opinion, it is best to find the most elegant solution ...[/quote]

The center square has to be $x-95$ from the diagonal. You can compute the other squares.

(sorry for the long bump)

[hide=Solution]Note that the bottom left square is $114$, making the bottom middle square $191$. Then the middle square in the final column is $209$. Let the remaining two be $a,b$ and we have $x+a+b=114+a+96$ and $x+b=210$. So then $x-b=190$ and the value is $2x=400 \implies x=\boxed{200}$.[/hide]

i got $\boxed{200}$ :)

[img]https://i.postimg.cc/Dwnn22pb/image.png[/img]

i apologize for bad handwriting :(

[hide=Solution]The bottom left square is $114$, the center square is $x-95$, the bottom middle square is $191$, the bottom right square is $x-190$, so $(x-190)+(x-95)+x=x+115$, meaning $x=\boxed{200}$.[/hide]

Common sum is $115+x$, so bottom left corner is $114$. So center is $x-95$. Thus, bottom right corner is $210-x$. So $x-94=306-x\implies x=\boxed{200}$.


I would like to know how our Washington kids did on AIME I or II(not yet taken, but afterwards, please)... so i have the general idea of how our state did, and how that will affect my chance... thanks.

P.S. although I already know some of the scores of kids, i would certainly like to know others...

AIME I: 11

...assuming I filled out the form correctly :P

I got somewhere between 6 and 9 (forgot my exact score), but no USAMO for me...

i didn't do well this year...... :(
hopefully next year :)

I didn't do well either, though I didn't get 0. ;) I tired with joml88's sophomore score so that doesn't feel so bad.

i got 3. but i am not in Washington anymore...

[quote="xxreddevilzxx"]i got 10. Tied with someone else from our school. I cant believe I actually beat my AIME score with USAMO score. :P[/quote]
i'm guessing that would be Zach?
nice job. :)

I got an 8 on the AIME I.....but I made some dumb mistakes (I read one question as $a+b+c$ instead of $a\times b\times c$ and so I missed it) and I probably should have had at least a 10.

I'm in 8th grade so that was a good enough score to move on to the USAMO. Yay!

this thread was for last year ;)

i guess still works

i got 10 this year...
im happy with my score, but i could have gotten better, as i missed questions 8,10,11,14,15
out of those 5 i could have gotten all of them (except maybe 14)...

good job to both 8 grader qualifiers
and also the 9th graders.

hopefully some  of you guys will make MOP this summer so Adam wouldn't be alone  :D

[quote="samath"]I got an 8 on the AIME I.....but I made some dumb mistakes (I read one question as $a+b+c$ instead of $a\times b\times c$ and so I missed it) and I probably should have had at least a 10[/quote]

Lol I did the exact same thing. (Among other silly mistakes such as miscounting on #13)

Yeah my score isn't that good so no need to talk about it here.


algebra
polynomial
geometry
3D geometry
calculus
integration
trigonometry

(a) Find a polynomial with integer coefficients of smallest degree  having  $\sqrt{2}+\sqrt[3]{3}$ as a root.
(b) Solve $1+\sqrt{1+x^2}(\sqrt{(1+x)^3}-\sqrt{(1-x)^3})=2\sqrt{1-x^2}$

I obtained the same result but I am not able to explain why such a polynomial is of lowest degree to satisfy the given condition.
Any idea :?:

I haven't got any idea of how to solve this type of question like finding a polynomial
maintaining a certain condition. Can someone help me please how do you derive the equation?
And what is the problem if we take a polynomial
P(x)=x=sqrt(2)+cubrt(3)?
Where does it fail to maintain the condition?
And for the problem no (b) I got a equation
4x^10+32x^8+152x^6+110x^4-68x^2+5=0
But then I can't proceed. Please give me a hint.

[quote="Oni"]I haven't got any idea of how to solve this type of question like finding a polynomial
maintaining a certain condition. Can someone help me please how do you derive the equation?
And what is the problem if we take a polynomial
P(x)=x=sqrt(2)+cubrt(3)?
Where does it fail to maintain the condition?
[/quote]

$P(x)=x-\sqrt{2}-\sqrt[3]{3}$ does not have integer coefficients. But I don't understand how to derive that polynomial either.

Hi! Oni,
Can you explain how you obtained the equation?
I have got an answer x = 0.292899242086 by numerical method but this value does not satisfy your equation.
Michael.

Thank you for telling me how to derive the equation.
But what does it mean by ' integer coefficients'
Does it mean that the equation must have a integer number which contains no 'x'
Or it means that all the elements of the equation must  have a  integer before 'x'
If the first one is meant then my equation x= sqrt(2)+cubrt(3) is certainly incorrect
But if the second one is meant then what is the problem with my equation.
Look there is a condition there that the equation must have smallest degree
But x^6-6x^4-6x^3+12x^2-36x+1=0 have a degree 6
whereas x=sqrt(2)+cubrt(3) have a degree 1

And what about the second problem?
What will I do after getting the equation
4x^10+32x^8+152x^6+100x^4-68x^2+5=0

And Micheal, for the second problem, if you put x=0.292899242086 in the given equation thendoes it satisfy? I have a calculator which contains only ten digits. So I can't take the full value of x. I worked without the last three digits of x and found that the equation does not hold with your value.
Oni

Dear Oni,
I checked with my calculator ( Texas TI 89 ) again.
The value is a very good approximation to the solution of  the original equation.
Michael.

In a degree $n$ polynomial $P(x) = a_0x^n + a_1 x^{n-1} + \ldots + a_{n-1}x^1 + a_n$, the [i]coefficients[/i] are the numbers $a_0, a_1, \ldots, a_{n-1}, a_n$ where $a_0$ is said to be the coefficient of $x^n$, etc., and $a_n$ is the constant coefficient.  A polynomial with integral coefficients is one in which the coefficients are all integers.  Your polynomial fails because the constant coefficient is not an integer.  (Note that a polynomial is not an equation; you keep acting like your polynomial is $x = \sqrt{2}+\sqrt[3]{3}$, but that's not a polynomial, it's an equation.  The polynomial associated to that equation is $x - (\sqrt{2}+\sqrt[3]{3})$, which doesn't have integral coefficients.)

I realize you probably know most of this, but I always think thorough is better.

For the second question, your polynomial has 4 real and 6 complex roots.  Of these, one of the four real roots is a root of the original question (the one luimichael named) and the other 3 are extraneous.  What you do when you get to a tenth degree polynomial you have to solve is to give up and go try a different method.  (I haven't found a suitable method yet; my first attempt ended up with a fifth degree polynomial, which is in fact no better than where you've gotten yourself (since your polynomial is actually fifth degree in $x^2$).  Right now, I'm experimenting with some trig substitutions, so perhaps that will get me where I need to go.)


to what extent is it possible to solve problems purely based on logic than formulae??? Especially if you have options for the answers? I fnd this aspect actually interesting and would love to use this thread to explore the possibilities further..

here is what made it interesting....

How many matches would be played in a knockout tennis tournament with 512 players???

i solved it... 256+128+....+1 = 511

Now Surprsingly, there was a simpler approach... for 1 to win out of 512, 511 players have to loose....and in a knockout in every match there is one looser...so there have to be 511 matches for 511 loosers...

Are such approaches possible for most problems.... i believe they must be....what are your views??? any more examples...especially in geometry would help...

Mods,
is it ok to post this in this thread???

Of course, most problems tend to lend themselves towards a simple elegant solution and a solution where it's just bruteforce. It's all about growing up, mathematically. It's kinda like how when I was a freshman, I did AIME counting problems by listing and counting  every single possible element that fit the requirements.

Nowadays, I usually like to group them, so I only have a couple hundred groups to count. It makes the work easier.

Elemennop couldn't say this any better. It's just like speaking language. When you're new to the language, say when you're baby, all you can say is "blah, blah blah?" or "AHHHH blah!"  :lol: But as you experience more, you'll be able to communicate better and find ways to express your feelings more clearly.

Same goes with math. In many times, there are elegant cases. It's just that not many people know how to find it.

Good example is mathematician Gauss. I remember reading that when he was young in one class, he was given this problem in class:

What is sum of 1,2,3,.....100?

While other students added this by hand and got wrong answer, Gauss used "grouping" method (100+1,99+2,etc..) to get the correct answer quickly. It's just how experienced you are. There is no magic about it.

Many problems will require you to do at least a little brute force work or a formula or two (this one didn't).  However, there is usually a couple of ways to do the problem - one that is elegant and others that require brute forcing.

Just make sure to remember not to spend too much time searching for an elegeant solution in timed contests like the AIME where a brute force solution is solvable in less time.


Well we all listen to music. some listen to crap music some listen to good lol well i had quite a heated argument with my friend who is a music student about music and he told me that a series of sounds could only be considered music if they followed some pattern nd some scale( he blabbered some technical terms i couldn't understand  :D ) while i said nething pleasing to my ear is music. Does that mean if someone breaks some glass bottles and its pleasing to my ear its music?

i know musical taste mostly depends on self perception but to what extent? is there a definition for music and noise?

[quote]When speaking of noise in relation to sound, what is commonly meant is meaningless sound of greater than usual volume[/quote]

That's Wikipedia's definition of noise... As far as music is concerned, there are many definitions([url]http://en.wikipedia.org/wiki/Definition_of_music[/url]), but this paragragh from Wikipedia seems as an interesting definition (which is pretty much related to the subject of the thread):
[quote]An oft cited definition of music, made by Wynton Marsalis among others, is that it is "sound organized in time." The fifteenth edition of the Encyclopaedia Britannica describes that "while there are no sounds that can be described as inherently unmusical, musicians in each culture have tended to restrict the range of sounds they will admit." Michael Linton, took the definition a step further to add that the form in which music is organized is an important element of the music itself. His definition of music is "the organization of sound and silence into forms that carry culturally derived meanings, cultivated for aesthetic or utilitarian purposes".[/quote]

Well I'm pretty sure lots of people would consider what I listen to (read: Killswitch Engage, Metallica, Necrophagist, Emperor, Guns N' Roses, Immortal) noise >_>

but there's some psycho artists like Merzbow (is that how you spell it? not quite sure) who actually make noise as music.

My music teacher said that music has taken on a remarkably new form since the 20th Century and now music is really just 'organised sound.' I mean when we say music we think of people like Mozart, Beethoven, Billie Joe,.....etc. But they are really stereotypical perceptions of music.

Yet, there are people like Cage who decided to incoporate a different meaning into 'music' and created music based more on process rather than outcome (e.g. Chromaticism, Retrograde, Inversion etc.).

Music is like a language in a sense. You can scribble anything you want on a piece of paper but we have conventions and rules that form what we call "writing". Similarly, you can make all the noise in the world but it's only music if you have some level of synchronization between instruments and if there is some melody (ie a repetition of a sequence of notes with minor variations) being produced. I'm no music expert but that's basically how I would distinguish between a car muffler and an orchestra.

Music is much less like writing than you think. At one point, at the height of Classicism it was, but now, dissonance, polytonal music, strange rhythms, and other things are allowed. On the other hand, language has strict rules of grammar, capitalization, etc.

Music is actually more like drawing/painting than writing, but this analogy doesn't help anyone at all.

[quote="Temperal"]language has strict rules of grammar, capitalization, etc.[/quote]
... all of which are frequently violated for effect, especially in poetry but also in prose fiction (e.g. stream-of-consciousness writing).  (Not that this has anything to do with the larger discussion.)

The top part here: http://www.google.com/search?hl=en&client=firefox-a&rls=com.google:en-US:official&hs=MCE&defl=en&q=define:music&sa=X&oi=glossary_definition&ct=title
is pretty interesting, it uses "music" in a way we would call figuratively, (e.g. "the music of the wind through the grass"), so it's defined as different for everyone, which makes sense.

And JBL: But they still hold [b]generally[/b], unlike the laws of Classicism, which are routinely shattered in most modern music.

Again, I think not in poetry.  Which makes sense, because (in some sense) poetry is much more about sound and style (relative to nonpoetry writing) -- that is, poetry is a more "musical" form of writing.

I should say that I know very little about music, certainly very little about music theory, so I'm not sure when you refer to "modern music" whether that includes (for example) the various forms of popular music.  It seems to me (again as an extreme layman) that these areas of music may violate classical norms, they certainly have their own sets of grammars and rules (albeit those which have likely not been formalized).  It seems to me more like the case of dialects or pidgins than of having no rules whatsoever.

For me the sound I enjoy is music.I listen mostly rock music.


Show that the nth root of a rational (for n a positive integer) cannot be a root of the polynomial x^5 - x^4 - 4x^3 + 4x^2 + 2.

A bit more:
This polynomial has the Galois group $S_5$, thus cannot be solved by radicals.

Sure about that ZetaX? Some can, for example:

$\frac{x^6-1}{x-1}$

I would like a answer without the use of Galois group.


For each positive integer $n$ , define $a_n = 20 + n^2$ and $d_n = gcd(a_n, a_{n+1})$. Find the set of all values that are taken by $d_n$ and show by examples that each of these values is attained.

I don't think your answer is completely correct, Andreas.

And you (should) know that you should actually post a proof, not just an answer.

[quote="inteluser"]Since 243 is prime[/quote]

Hm, strange, $243 = 3^5$ :D

Bi subtraction we obtain $d_n|2n+1$. So $2n\equiv -1$ mod $d_n$, or rather $4n^2\equiv 1$ mod $d_n$ and $n^2\equiv 20$ mod $d_n$. Using the Chinese theorem for the residuum we can say $n^2\equiv \frac{1}{4}\equiv -20$ mod $d_n$, so $d_n|81$
($d_n\equiv 1$ mod $2$)

Your proofs are incomplete. The problem requires examples for the values of $d$ attained.

$(20+(n+1)^2, 20+n^2) = (20 + n^2, 2n+1) = (2n+1, \frac{n}{2} - 20) = (2n+1, 81)$
From this we can solve $2n+1 = d$ to get the appropriate values of $n$.
Indeed:
$(a_{40}, a_{41}) = 81$
$(a_{13}, a_{14}) = 27$
$(a_4, a_5) = 9$
$(a_1, a_2) = 3$ and
$(a_2, a_3) = 1$ can easily be obtained by trial an error.

I think d| 81 does not imply that d would be the solution if and only if it is a divisor of 81

dn = gcd(20+n^2,  2n + 1)
Since 4 is not the gcd so
dn = gcd(80+4n^2, 2n + 1)
Implies
dn = gcd(80 + 4n^2,4n^2-1)
dn divides 81

[b][u]Bonus fact:[/u][/b] For each positive integer $n$ , define $a_n = x + n^2$ and $d_n = gcd(a_n, a_{n+1})$. Then $d_n \vert 4x+1$.
[b][u]Proof:[/u][/b] $d_n \vert x+n^2$, $d_n \vert x+(n+1)^2$.
$\Rightarrow d_n \vert x+(n+1)^2-(x+n^2)$
$\Rightarrow d_n \vert 2n+1 \Rightarrow d_n \vert 2n^2+n$.
Also, $d_n \vert x+n^2 \Rightarrow d_n \vert 2x+2n^2$
$\Rightarrow d_n \vert 2x+2n^2-(2n^2+n)$
$\Rightarrow d_n \vert 2x-n \Rightarrow d_n \vert 4x-2n$
Also, $\Rightarrow d_n \vert 2n+1$
$\Rightarrow d_n \vert 2n+1+ 4x-2n$
$\Rightarrow d_n \vert 4x+1$, which proves our claim :).

Just put $x=20$ and we get the result that $d_n \vert 81 \Rightarrow$ $d_n \in \{1,3,9,27,81\}$.


timhlu wrote:Prove that if a, b, and c are positive integers, then

a^3+b^3+c^3 :ge: 3abc

[hide]Divide by three and rewrite abc as (a^3*b^3*c^3)^(1/3). We get

(a^3+b^3+c^3)/3 :ge: (a^3*b^3*c^3)^(1/3)

But this is just the AM-GM (Arithmetic Mean-Geometric Mean) inequality for a^3, b^3, and c^3.  (AM-GM states that the arithmetic mean is always greater than the geometric mean.)

[/hide]

pshhhh... i think my proof is better. it deserves 8/7 for brevity :)

More inequality fun:

a3 + b3 + c3 + 3abc :ge: ab(a+b) + bc(b+c) + ac(a+c).

[hide]This is a Schur's inequality :

http://planetmath.org/encyclopedia/SchursInequality.html

Mmm, spoiler doesn't hide URLs ....

[/hide]


I'd like to include graphic files in my LaTex document using TeXnicCenter.  DVI documents require EPS graphics which are too large and slow down development.  The option "draft" I'm finding is unsatisfactory; I'm just left with a bunch of holes in my document, and the flow of information is interrupted.  So I'm attempting to use the package "vpe" for source specials so that I can create PDF files with smaller graphic files and have the option of going back and forth from PDF document to source code easily with the mouse.

I installed the package vpe on my system using the MiKTex package manager, then refreshed the FNDB database via the MiKTex settings manager.  I next included the line:

\usepackage{vpe}

in my main document then switched to LaTeX=PDF output.  When I attempt to build the file I receive the message:

Package vpe Warning:  Option 'system' is not specified and the automatic detection failed, therefore package loading is aborted.

The vpe instructions describe:

F.2 PdfTeX and dvips
--------------------
a) Load the package `vpe.sty', eg:
     \usepackage{vpe}
b) Produce the pdf file, but with enabled \write18{} feature
   for system commands.  This can be achieved by setting the
   command line option `--shell-escape' for pdfTeX or TeX.
   An alternative is the boolean variable `shell_escape'
   in the configuration file `texmf.cnf'.  For security
   reasons I recommend the command line option.

Can anybody help me?
Thanks,
Shaw

In Build -> Define Output Profiles -> LaTeX=>PDF, Command line arguments to pass to the compiler add --shell-escape after -interaction=nonstopmode "%pm" leaving a space between them. I haven't tried this though and I hope this works for pdflatex instead of the pdftex they quote.

Hello Steve.  The command-line argument I have is:

-interaction=nonstopmode "%pm"  --shell-escape

but I get the same error message.  I tried it with just one dash before shell, got the same message, and then I tried in desperation the command:

\beginpackage[system]{vpe}

and still got the same error message.  Not sure where to go next.

Have you configured the perl script as in the instructions? I presume you have installed perl to run it.

Hi Steve.  Thanks, I'll look into it.  Don't you love Google?  This is what I did:

Google:  What is perl and how do I download it? :)


I haved used LaTeX previously, but is not installed on my current computer.

I tried to install MiKTeX but I get an error message,

"Check sum-up mismatch" which says that something is wrong with babel.cab.

Does someone know an answer to this problem?

I am currently using WinXP Pro Korean w/ SP1.

I have done some research, but have not found any answer.

Does anyone know an alternate program for MiKTeX?

The TeX Users group keeps a list of TeX distributions:
[url]http://www.tug.org/interest.html#free[/url]

Unfortunately, I can't help you with your MikTeX installation problem -- I've never tried to install it on a Korean system.  You don't seem to be the only person with this problem; see: [url]http://sourceforge.net/tracker/?group_id=10783&atid=110783&func=detail&aid=547948[/url], although it doesn't seem to be too helpful.  :(

I reinstalled Windows XP Pro (English Version).  Still doensn't work.  I will try different methods....

I have spent more than 16 hours trying to solve this problem.  i have been searching through TeX user archives, etc.

I reinstalled windows a couple of times....

oK, so that babel.cab problem is gone (good!) but when every I tried to install MiKTeX,  "initexmf.exe" encounters a runtime error and "could not configure the computer" for MiKTeX....does anyone know anything for this problem?(The whole installation is complete except this step.  TeXnic center does not recognize that MiKTeX is actually installed.)  Help!

In the worst case, I will submit the USAMTS problems with handwritten workouts.

Problem solved.

There were two causes of the problem

1. downloading files were corrupted.

2. MikTeX doesn't like foreign language based Windows.

I had to download MiKTeX korean version from Korean TUG.


geometry
circumcircle
trigonometry
Euler
geometric transformation
homothety
incenter

let $ ABC$ be a triangle($ \angle A>\angle B>\angle C$) with circumcircle $ (O)$ and incircle $ (I)$.$ AI$ cuts $ (O)$ at $ D$.$ DN \perp BC(N \in BC)$.Let $ E$ is a point such that $ N$ is midpoint of $ EI$.$ DE$ cuts $ (O)$ at $ Q$ .Prove that $ QB\equal{}QA\plus{}QC$

Trig solves it fairly easily:
Since Ptolemy gives us $ QA*BC \equal{} QC*AB \plus{} QB*AC$, $ QC \equal{} QA \plus{} QB$ is equivalant to $ BC(QC \plus{} QB) \equal{} QC*AB \plus{} QB*AC$, or $ \frac {QB}{QC} \equal{} \frac {a \minus{} b}{a \minus{} c}$.
Let $ \angle A \equal{} 2A, \angle B \equal{} 2B, \angle C \equal{} 2C$. A quick angle chase gives us $ BD \equal{} DC$, so $ N$ is the midpoint of $ BC$, so triangles $ BEC, BIC$ are congruent, $ \angle EBC \equal{} C, \angle ECB \equal{} B, \angle DBC \equal{} \angle DCB \equal{} A$. Then $ \frac {QB}{QC} \equal{} \frac {sin \angle EDC}{sin \angle EDA} \equal{} \frac {sin \angle ECD * sin \angle EBC} {sin \angle ECB * sin \angle EBD} \equal{} \frac { sin (A \minus{} B) sin C}{ sin (A \minus{} C) sin B} \equal{} \frac {cos (2A \minus{} 90) \minus{} cos (90 \minus{} 2B)}{cos (2A \minus{} 90) \minus{} cos (90 \minus{} 2C)} \equal{} \frac {a \minus{} b}{a \minus{} c}$, as required.
Here we have used trig-Ceva in triangle $ DBC$, with point $ E$, and the fact that $ sin x * sin y \equal{} cos (x \minus{} y) \minus{} cos (x \plus{} y)$.

See the following post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122759[/url]

If I look at this problem again after all this time, I see a different solution now, that shows the connection with our beloved Feuerbach point.

The statement immediately suggests a connection to this point, because the desired relation is satisfied by the Feuerbach point of a triangle ABC, and it's medial triangle A'B'C'. In other words, $ \left|FB'\right| \equal{} \left|FA'\right|\plus{}\left|FC'\right|$.

In Hyacinthos message #10485, Antreas Hatzipolakis conjectures the following result (which was first proven by Darij Grinberg):
[quote="Antreas Hatzipolakis"]Let ABC be a triangle and A'B'C' the cevian
triangle of I.

Ab := the Orthogonal projection of A on BB'
Ac := the Orthogonal projection of A on CC'

Similarly Bc,Ba; Ca,Cb

The Euler Lines of AAbAc, BBcBa, CCaCb are concurrent at F, the Feuerbach point.[/quote]
We will use all these facts to give an alternative solution. Note that this solution is in no way easier or more direct than the ones given in this topic, or the topic Naoki Sato kindly points at. It's just a nice way to apply the properties of the Feuerbach point.

[b]Proof:[/b] Call $ UVW$ the antimedial triangle of $ ABC$. If we could show that $ Q$ is the Feuerbach point of $ UVW$, the proof would be complete. We will show this using the theorem on the Euler lines, as stated above:
[list]
[*] $ D$ is clearly mid arc point of $ BC$ in the nine-point circle of $ UVW$. The point $ D'$, which denotes the antipodal point in the incircle of $ UVW$ (centered at $ I$) of the point of tangency of the incircle and $ VW$ is known to lie on the Euler line of $ UU_bU_c$ (For definition of this triangle, see the above quotation of A. Hatzipolakis). However, you can prove this fact yourself, it's not that hard. The homothecy with center $ F$, the Feuerbach point of $ UVW$, mapping the incircle of $ UVW$ to the NPC of $ UVW$, maps $ D'$ to $ D$. Therefore, $ D$ lies on the Euler line of $ UU_bU_c$.

[*] $ E$ is clearly the incenter of $ UBC$, and therefore the midpoint of $ UI$. This is obviously the circumcenter of $ UU_bU_c$, which lies on the desired Euler line.[/list]
Conclusion: $ Q$ must be the Feuerbach point of $ UVW$, due to Hatzipolakis' theorem.

I hope you enjoyed it...  :wink:

I used to think that any time there was a condition  a = b + c  for segments
  it was a legacy of distances from triangle vertices to a line through
  centroid.

  A proof of this fundamental fact of modern geometry was given in

http://www.mathlinks.ro/Forum/viewtopic.php?t=32412.

  Maybe there is a way to somehow apply this identity to the

  problem above.

   Thank you.
   M.T.


Let $ ABC$ a triangle, the base $ BC$ is fixed and the vertex $ A$ variable on a half-plane determined by $ BC$, where the angle $ A$ is constant.
Find the locus of the incenter that triangle.

[hide="Hint"]We have $ \angle{BIC} \equal{} 90^0 \plus{} \frac {1}{2}\angle{A}$[/hide]

[hide="Hint 2"]
The locus of points that form a fixed angle with a segment is an arc of a circle.

[hide="Further Hint"]Do that with $ \angle BIC$.[/hide]
[/hide]

[hide="Solution"]
look at the hints and combine them
[/hide]


Given the sequence $ 10^{\frac {1}{11}},10^{\frac {2}{11}},10^{\frac {3}{11}},\ldots,10^{\frac {n}{11}}$, the smallest value of $ n$ such that the product of the first $ n$ members of this sequence exceeds $ 100000$ is:

$ \textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 11$

[hide]

$ 100000\equal{}10^5$

Find the smallest integer value of n such that
$ \frac {1}{11} \plus{} \frac {2}{11} \plus{} \frac {3}{11} \plus{} .... \plus{} \frac {n}{11} > \frac {55}{11}$

$ 1 \plus{} 2 \plus{} 3 \plus{} ... \plus{} n > 55$

$ n \equal{} \boxed{11, (E)}$

[/hide]

[hide="Solution"]Recall that $ \sum_{k \equal{} 1}^{n}k \equal{} \frac {n(n \plus{} 1)}{2}$.

$ 10^5 < 10^{\frac {1}{11}\cdot\frac {n(n \plus{} 1)}{2}}\implies 110 < n^2 \plus{} n\implies 0 < n^2 \plus{} n \minus{} 110$.
We have equality at $ n \equal{} 10$ by the quadratic formula.
So the answer is clearly $ \boxed{\textbf{(E)}\ 11}$[/hide]


An Exponential Equation
Find unequal positive, rational numbers, a, b,
 (other than 2 and 4)
 such that

$a^{b}=b^{a}$

It has been posted many times before... search for it (but I think it was in a beginner's forum)

[quote="Altheman"][quote="rnwang2"]You can prove the only solutions are $(2,4)$ and $(4,2)$.[/quote]

there are other solutions...[negative...][/quote]
[quote]An Exponential Equation
Find unequal [b]positive[/b], rational numbers, a, b,
(other than 2 and 4)
such that

a^{b}=b^{a}
[/quote]

consdier the function $f(x)=\frac{\ln{x}}{x}$
$f$ has a max at x=e, any two numbers a, b satisfying the above must lie on two sides of line x=e. :wink: