quadratics
algebra proposed
algebra
quadratic equation

Given a quadratic trinomial $f\left(x\right)=x^2+ax+b$. Assume that the equation $f\left(f\left(x\right)\right)=0$ has four different real solutions, and that the sum of two of these solutions is $-1$. Prove that $b\leq -\frac14$.

The equation $f(x)=0$ must have two real roots, say $x_1,x_2.$ The roots of the equation $f(f(x))=0$ are all real iff $x_{1,2}\ge -\frac{\Delta}{4},$ where $\Delta=a^2-4b.$ It follows that $-a=x_1+x_2\ge -\frac{1}{2}\Delta$ hence $4b\le a^2-2a (*).$
Now let $y_1,y_2,y_3,y_4$ be the roots of $f(f(x))=0.$ Two of them, say $y_1,y_2$ are the roots of the equation $x^2+ax+b-x_1=0,$ and the other two satisfy the equation $x^2+ax+b-x_2=0.$
If, for instance, $y_1+y_2=-1,$ then $a=1$ and plugging in $(*)$ gives the desired result.
Suppose that $y_1+y_3=-1.$ It is easy to check that this implies $y_1^2+y_3^2\ge \frac12.$
We have
\[ y_1^2+ay_1+b-x_1=0, \]
and
\[ y_3^2+ay_3+b-x_2=0. \]
Adding up yields
\[ y_1^2+y_3^2+2b=0, \]
hence
\[ b= -\frac{y_1^2+y_3^2}{2}\le -\frac14 , \]
as desired.

this is basically the same, but in a slightly different approach...

note that $ f(f(x)) \equal{} x^4 \plus{} 2ax^3 \plus{} x^2(a^2 \plus{} a \plus{} 2b) \plus{} x(a^2 \plus{} 2ab) \plus{} (b^2 \plus{} ab \plus{} b)$...

we have that two of the roots of $ f(f(x))$ sum up to $ \minus{} 1$, then, $ f(f(x))$ factorizes as

$ f(f(x)) \equal{} (x^2 \plus{} x \plus{} k)\left(x^2 \plus{} (2a \minus{} 1)x \plus{} \dfrac{b^2 \plus{} ab \plus{} b}{k}\right)$, for some real $ k$.

comparing the coefficients of $ x^2$ and $ x$ it follows that

$ \dfrac{b^2 \plus{} ab \plus{} b}{k} \plus{} 2a \minus{} 1 \plus{} k \equal{} a^2 \plus{} a \plus{} 2b$ and $ \dfrac{b^2 \plus{} ab \plus{} b}{k} \plus{} k(2a \minus{} 1) \equal{} a^2 \plus{} 2ab$

substracting the equation it follows that $ 2k(a \minus{} 1) \equal{} (2b \plus{} 1)(a \minus{} 1)$...

suppose $ a \equal{} 1$. then $ \dfrac{b^2 \plus{} 2b}{k} \plus{} k \equal{} 2b \plus{} 1$. note that the sum of the roots in both factors of $ f(f(x))$ is $ \minus{}1$, so its independent term is at most $ \frac {1}{4}$... it follows that $ 2b \plus{} 1\leq \dfrac{1}{4} \plus{} \dfrac{1}{4}$ from where it follows easily that $ b\leq \dfrac{ \minus{} 1}{4}$

suppose $ 2k \equal{} 2b \plus{} 1$. from the same argument as before $ k\leq\dfrac{1}{4}$ from where we conclude that $ b\leq\dfrac{ \minus{} 1}{4}$, and we're done.  :D


inequalities
logarithms
Bosnia
algebra
High school olympiad
circumcircle

[hide="Hidden by mod"]After multiplying everything out it is equivalent to:
$ 4(\sum_{cyc}a^{3}) \plus{} 23abc > 4(\sum_{cyc}a^{2}(b \plus{} c))$
which is obvious, because by Schur we have:
$ (\sum_{cyc}a^{3}) \plus{} 3abc\geq \sum_{cyc}a^{2}(b \plus{} c)$
So finally we have:
$ 4(\sum_{cyc}a^{3}) \plus{} 23abc > 4(\sum_{cyc}a^{3}) \plus{} 12abc\geq 4\sum_{cyc}a^{2}(b \plus{} c)$ Q.E.D. :wink:[/hide]

[hide="Hidden by mod"]Since the inequality is homogeneous, we could normalize: $ a + b + c = 1$ and the inequality turns out to be

$ \sum_{cyc}{\ln{(1 + \frac {4a}{1 - a}})} > \ln{25}$.

Then analyze the function $ f(x) = \ln{(1 + \frac {4a}{1 - a})}$.

I conjecture that : For all function $ f: \mathbb{R}^0\rightarrow\mathbb{R}$ whose second derivative has only one solution. Let $ x_1, x_2, ...,x_n$ be non-negative variable such that $ x_1 + x_2 + ... + x_n = s = const$. Then $ f(x_1) + f(x_2) + ... + f(x_n)$ get its minimum values when some of the variables are 0 and the others are all 0. (Actually, we just need the condition that $ f(x_1) + ... + f(x_n) \geq \min\{ f(x_1 + x_2) + f(0), 2f(\dfrac{x_1 + x_2}{2} \} + f(x_3) + ... + f(x_n)$).

Transform the problem using algebra, I got the following geometric inequality: Suppose that we have a triangle with sides $ a,b,c$ and $ m_a, m_b, m_c$ be its median. We have: $ \frac {m_a}{a}\frac {m_b}{b}\frac {m_c}{c} \geq \dfrac{5}{8}$. The equality occurs only when it is a right isoscele triangle. I wonder if there is a beautiful method to solve this problem without refering to the original one.[/hide]

This is really interesting Namely, before 2 weeks we had Balkan M.O. preparation test and question equivalent to your inequality (related to medians and sidelengths) was proposed and as I remember we could not solve it by using some standard methods but in Sharygin's book there is non/standard solution to this inequality. I will try to find a book and scan that part.

Thanks [b]delegat[/b] for the information.   :| I 'll try it before looking at such an elegant solution.

[quote="delegat"]If $ a$, $ b$ and $ c$ are positive reals prove inequality:

$ (1 \plus{} \frac {4a}{b \plus{} c})(1 \plus{} \frac {4b}{a \plus{} c})(1 \plus{} \frac {4c}{a \plus{} b}) > 25$[/quote]
It's a privet case of my inequality. :D
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59512

[b]mathangel[/b] Here I attached scanned pages with solution to geometric inequality written above.

Thanks for sharing.

While looking for the solution, I transformed the problem to another one:

Given triangle $ ABC$ whose sides are $ a,b,c$ and whose circumradius is $ R$. Denote by $ R_m$ the radius of the circumcircle of the triangle formed by the 3 medians of $ \triangle ABC$. Then we have the inequality:

$ R_m \geq \frac{5}{6} R$.

It seems that this is the original problem in the book, isn't it ? Anyway, the solution is elegant.

You are absolutely right [b]mathangel[/b] The same question as you wrote is asked in that book.

[quote="arshakus"]this the problem of the gyumri olymiad in 2008, when I was in 7class, in our class no one solved, and after seeing the solution we didn't understant anything, but now it is very easy after opening the brackets we get
$a^3+b^3+c^3+12abc>a^2b+b^2a+c^2a+a^2c+c^2b+b^2c$ which is true by chur's ineq, which is $a^3+b^3+c^3+3abc\ge a^2b+b^2a+c^2a+a^2c+c^2b+b^2c=>$ [b]PROOVED[/b][/quote]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=2913000]Bosnia Herzegovina Regional Olympiad 2008:[/url]
$ \left(1\plus{}\frac{4a}{b\plus{}c}\right)\left(1\plus{}\frac{4b}{a\plus{}c}\right)\left(1\plus{}\frac{4c}{a\plus{}b}\right) > 25$

$\Leftrightarrow a^3+b^3+c^3+7abc>a^2b+b^2a+c^2a+a^2c+c^2b+b^2c$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=54&t=3165&hilit=ineq][b]Book: Inegalitati; Authors:L.Panaitopol,V. Bandila,M.Lascu[/b][/url]
http://www.artofproblemsolving.com/Forum/posting.php?mode=quote&f=151&t=569686&p=3344462:
 Let $a,b,c$ be  positive real numbers . Prove that\[(1+\frac{6a}{b+c})(1+\frac{6b}{c+a})(1+\frac{6c}{a+b})> 43.\]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=496088&p=2785707#p2785707
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2685748#p2685748
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&p=3344407#p3344407

Regional Olympiad - Federation Of [b]Bosnia And Herzegovina 2008[/b]
[quote=delegat]If $ a$, $ b$ and $ c$ are positive reals prove inequality:

\[ \left(1\plus{}\frac{4a}{b\plus{}c}\right)\left(1\plus{}\frac{4b}{a\plus{}c}\right)\left(1\plus{}\frac{4c}{a\plus{}b}\right) > 25.\][/quote]

If $ a$, $ b$ and $ c$ are positive reals [url=https://artofproblemsolving.com/community/c6h3165p10538523]prove inequality:[/url]
$$ ( \frac{4a}{b+c}+1)( \frac{4b}{c+a}+1)( \frac{4c}{a+b}+1)\geq 25+\frac{16abc}{(a+b)(b+c)(c+a)}. $$

[quote=sqing]If $ a$, $ b$ and $ c$ are positive reals [url=https://artofproblemsolving.com/community/c6h3165p10538523]prove inequality[/url]:
$$ ( \frac{4a}{b+c}+1)( \frac{4b}{c+a}+1)( \frac{4c}{a+b}+1)\geq 25+\frac{16abc}{(a+b)(b+c)(c+a)}. $$[/quote]

[url=http://blog.sina.cn/dpool/blog/s/blog_5618e6650102zvcu.html]7436[/url].


counting
distinguishability
geometry
3D geometry
octahedron
rotation

Source: AMC-12 2000, #25

Eight congruent equilateral triangles, each of a different color, are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron (two colored octahedrons are distinguishable if neither can be rotated to look just like the other).

Please don't double post. ;)
[hide="Let me have a go at this problem..."]
Without restrictions, we can color the octagon, i.e. the triangles, in $8!$ ways because each is a different color. Now we have to divide by the number of ways we can rotate it so that it's the same. We can rotate right, left, up, or down, from each point, so the answer should be $8!/4!=8 \cdot 7 \cdot 6 \cdot 5 = 1680$
[/hide]

[quote="Iversonfan2005"]Did I do it right?

[hide]I said that there are $8!$ ways the triangles can be colored, but since the bottom $4$ can't be the same as the top $4$, I divided it by $4!$ because each side would have $4$ different colors, so $\frac{8!}{4!}=1680$. Don't know, but it doesn't seem like the right way to do it :o[/hide][/quote]

I don't really understand your reasoning.

[quote="chess64"]
[hide="Let me have a go at this problem..."]
Without restrictions, we can color the octagon, i.e. the triangles, in $8!$ ways because each is a different color. Now we have to divide by the number of ways we can rotate it so that it's the same. We can rotate right, left, up, or down, from each point, so the answer should be $8!/4!=8 \cdot 7 \cdot 6 \cdot 5 = 1680$
[/hide][/quote]

You're reasoning doesn't make sense either.  From what you are saying I would think we would divide by $8\cdot 4$ since there are, according to you, 4 ways to rotate each vertex and there are 8 vertices.

On the other hand, correct reasoning would be to say that if you fix a vertex then there are 6 other orientations of the cube with this point fixed (consider which face is facing 'forward'...there are 6 faces).  But then consider the point on the other end of the long diagonal of the cube.  The orientations around this point are the same as the first point we looked at.  Thus there are $\frac{8\cdot 6}{2}=24$ orientations of one color scheme of the cube.  The rest of the solution is the same as the other solutions given.

Ah...this one...haven't tried it yet...
[hide]Here's what I would say.  Just like you said, break it into two sets of 4.  Then, find the number of ways to distribute.  That is $\binom{8}{4}=70$.  Then, take the number of ways to order each set of 4 (considering them circular, so that 1234 is the same as 2341).  The number of ways to order each set will be 4!/4=6.  Then, the number of ways to fit each set together is 4 (think two halves of the octahedron coming together).  Thus, we have 70*6*6*4=10080.[/hide]


In my class today, we learned how to find the area of a rectangle by using the formula. My teacher asked us to prove the formula (or the rectangle area postulate) though in the book, it is mentioned as a postulate which it is so obvious and doesn't need to be proved. Can any one do the proof? :help:

This isn't so much as a rigorous proof but more of an intuitive understanding, which I think your teacher may be looking for.

Remember that area is counting the number of unit squares. So consider a rectangle that is 6 by 4. That means there are 6 squares on the long side, and 4 on the short. Therefore, it is easy to see the number of unit squares is just $ 6 \times 4 \equal{} 24$.

We can now apply this with variables. Imagine a rectangle that's $ a$ by $ b$. There are $ a$ squares on one side and $ b$ on the other. So, we have an area of $ ab$.

Incidentally, my geometry textbook postulated that a square's area is $ s^2$, and derived the rectangle's area from that. Basically, consider the square $ (a\plus{}b)^2$. Take away two squares of $ a^2$ and $ b^2$; two $ a \times b$ rectangles are left.

The other two area postulates: "congruent figures have the same area" and "the area of a figure is the same as the sum of the areas of its nonoverlapping parts".

You can't prove anything until you know what your axioms are, and in the reasonable constructions of area I know of, the area of a rectangle is taken as the basis for the [b]definition[/b] of area.


6 boys and 6 girls are seated randomly in a row of 12 chairs. What is the probability that no two boys are seated next to one another and no two girls are seated next to one another?  Express your answer as a common fraction.

[hide] Answer: 1/462[/hide]

[quote="darkmatter47"]So there are two ways, out of 924, equals $ \frac {2}{924}$ = $ \frac {1}{472}$[/quote]
Heh. You messed up your division at the very end. It's $ \frac2{924} \equal{} \frac1{462}.$

Whoops :blush:  I actually didn't divide wrong, I typed wrongly.

[quote="darkmatter47"][hide="Solution"]First we figure out the total number of possibilities for to arrange the six boys.

$ \frac {12!}{6!6!}$

Since it is obvious that we can meet the conditions only if we start only with girl on the left and alternate boy, girl, boy girl, etc. or if we start with a boy and alternate accordingly, there are two ways, out of 924, equals $ \frac {2}{924}$ = $ \frac {1}{462}$[/hide][/quote]
Where did you get 12!/(6!6!)? was it 12c6? how come?

[hide]It's 2(6*6*5*5*4*4*3*3*2*2*1*1)=2(36*25*16*9*4) ways, whatever that is. I'm leaving it to you to do the computation.[/hide]

To andrew liu,

It is 12c6. You can either find the number of ways you can seat the boys, or the number of wars you can seat the girls. The opposite gender will have to fill in the extra spaces, so that does not need to be taken into account.

[hide="My Solution"]
Since the only way to seat them as such is either
BGBGBGBGBGBG
or
GBGBGBGBGBGB
So that's two possible ways.
Then you figure out how many possible ways there are to seat all of them.
Since there are $ 12!$ possible way to seat them all IF the girls and boys are distinct, which I am assuming that they are not.
So you do
$ \frac{12!}{6!6!}$
You have the two $ 6!$s because there 6 boys and 6 girls, therefore, seeing as they are not distinct, allowing repeats, which can be removed by dividing my the 6 factorials.
So then you get
$ \frac{2}{\frac{12!}{6!6!}}$
Which simplifies down to
$ \frac{1}{462}$
[/hide]

Yes. Its not 12p6 because the actual order of the individual boys/girls does not matter, only the general placement.

thank you ppl who submitted. I get it now  :)

Slightly different way to do this.
[hide]As an alternate solution :

Look at the probabilities of filling the seats in order in acceptable ways.

Probability that first seat is filled in an acceptable way : $ \frac{12}{12}$
Since 1st seat can be boy or girl.

Since next seat must be filled by the opposite gender of 1st seat, probability that second seat is filled in an acceptable way :  $ \frac{6}{11}$

Next seat :  $ \frac{5}{10}$

Continuing that logic :

$ \frac{12}{12} \cdot \frac{6}{11} \cdot \frac{5}{10} \cdot \frac{5}{9} \cdot \frac{4}{8} \cdot \frac{4}{7} \cdot \frac{3}{6} \cdot \frac{3}{5} \cdot \frac{2}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} \cdot \frac{1}{1} \equal{} \frac{1}{462}$

The other solutions are maybe nicer, but it helps to have different ways to approach problems.

[/hide]

I remembered this problem from last year, I got it wrong :( the correct answer was 1/924.


function
integration
limit
calculus
calculus computations

Hi, I'm having some difficulty to prove the question related to measure. Wonder if anyone can help.

Let $ (X, B, \mu)$ be a measure space and $ h$ a nonnegative measurable function on $ X$. Let $ uE\equal{}\int_E h d\mu$. Show that $ u$ is measure on $ B$.

$ u(\emptyset)\equal{}\int_{\emptyset}hd\mu\equal{}0$ since $ h\chi_{\emptyset}$ vanishes everywhere.
Since $ h\geq 0$, then $ u(E)\geq 0$.
Let $ (E_n)$ be a disjoint sequence in $ X$ with union $ E$. Define $ h_n\equal{}\sum_{k\equal{}1}^nh\chi_{E_k}$. Since $ h_{n\plus{}1}\minus{}h_n\equal{}h\chi_{E_{n\plus{}1}}\geq 0$ and $ h_n$ converges to $ h\chi_{E}$, apply MCT gives
$ u(E)\equal{}\int h\chi_{E}d\mu\equal{}\lim_{n\rightarrow\infty}\int h_nd\mu\equal{}
\lim_{n\rightarrow\infty}\int\sum_{k\equal{}1}^nh\chi_{E_k}d\mu\equal{}\lim_{n\rightarrow\infty}\sum_{k\equal{}1}^n\int h\chi_{E_k}d\mu\equal{}\sum_{k\equal{}1}^{\infty}\int h\chi_{E_k}d\mu\equal{}\sum_{k\equal{}1}^{\infty}u(E_k)$.


[quote="RDeepMath91"]
$\geq \sum_{cyc}log_{a}\sqrt{bc}$
[/quote]

There is a small typo ...  :)


limit
real analysis
real analysis unsolved

Hey!

[quote]The function

$ f: \mathbb{R} \longrightarrow \mathbb{R} \\
f(x) \equal{} \sum\limits_{n \equal{} 1}^\infty \frac {\sin((n \plus{} 1)! \cdot x)}{n!}$

is continuous and nowhere differentiable.[/quote]
OK,

$ \frac {\sin((n \plus{} 1)! \cdot x)}{n!}$ is continuous for every $ n \in \mathbb{N}$ and $ f_N$ converges uniformly

$ \Rightarrow f$ is continuous.

But at the moment I don't see how to prove that the limit $ \lim_{g \to 0} \frac {f(x \plus{} g) \minus{} f(x)}{g}$ does not exist.


[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \frac {1}{2a \plus{} b} \plus{} \frac {1}{2b \plus{} c} \plus{} \frac {1}{2c \plus{} a} \geq\ \frac {2}{\sqrt [3]{(a \plus{} b)(b \plus{} c)(c \plus{} a)}}$
 :)
PS: From it, we have a weaker result but also nice:
$ \frac {1}{2a \plus{} b} \plus{} \frac {1}{2b \plus{} c} \plus{} \frac {1}{2c \plus{} a} \geq\ \frac {\sqrt {3}}{\sqrt [6]{(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)^2}}$[/quote]
Multiplying both sides for $ a\plus{}b\plus{}c$, we can rewrite the inequality as
$ \sum \frac{a\plus{}b}{2a\plus{}b} \plus{}\sum \frac{c}{2a\plus{}b} \ge \frac{2(a\plus{}b\plus{}c)}{\sqrt[3]{(a\plus{}b)(b\plus{}c)(c\plus{}a)}}.$
According to the Cauchy Schwarz Inequality, we have
$ \sum \frac{a\plus{}b}{2a\plus{}b} \equal{}\frac{3}{2}\plus{}\frac{1}{2} \sum \frac{b}{2a\plus{}b} \ge \frac{3}{2}\plus{}\frac{1}{2}\cdot \frac{(a\plus{}b\plus{}c)^2}{\sum b(2a\plus{}b)} \equal{}2,$
and
$ \sum \frac{c}{2a\plus{}b} \ge \frac{(a\plus{}b\plus{}c)^2}{3(ab\plus{}bc\plus{}ca)}.$
So, it suffices to prove that
$ \frac{(a\plus{}b\plus{}c)^2}{3(ab\plus{}bc\plus{}ca)} \plus{}2 \ge  \frac{2(a\plus{}b\plus{}c)}{\sqrt[3]{(a\plus{}b)(b\plus{}c)(c\plus{}a)}}.$
Now, the AM-GM Inequality yields that
$ \frac{(a\plus{}b\plus{}c)^2}{3(ab\plus{}bc\plus{}ca)}\plus{}1\plus{}1 \ge 3\sqrt[3]{\frac{(a\plus{}b\plus{}c)^2}{3(ab\plus{}bc\plus{}ca)}}.$
And we come up with the inequality
$ 3\sqrt[3]{\frac{(a\plus{}b\plus{}c)^2}{3(ab\plus{}bc\plus{}ca)}} \ge \frac{2(a\plus{}b\plus{}c)}{\sqrt[3]{(a\plus{}b)(b\plus{}c)(c\plus{}a)}},$
or
$ (a\plus{}b)(b\plus{}c)(c\plus{}a) \ge \frac{8}{9}(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca).$
The last one is obvious by the AM-GM and we are done.

Yes, it's also the way I created it  :lol:


30. Let $ABCD$ be an isosceles trapezoid with bases $AB = 92$ and $CD = 19$ Suppose $AD = BC = x$ and a circle with center on $AB$ is tangent to segments $AD$ and $BC$. If $m$ is the smallest possible value of $x$ then $m^2 =$
(A) $1369$
(B) $1679$
(C) $1748$
(D) $2109$
(E) $8825$

Great Problem
[hide]
Extend $AD$ and $BC$ so that they meet at a point $E$. Let the center of the circle be $O$. Extend $O$ so it is perpendicular to $BC$ at a point $F$. Let the point where $EO$ intersects $AB$ be $G$. Segment $EF$ will be equal to the radius of circle $O$, because if $BC$ is tangent to circle $O$, it must pass through the center. We need to set an inequality so that $EO-EG\geq OF$.

To find $OF$, we see that triangles $ECO$ and $EFO$ are similar. Given $OC=46$,$EO=92h$,
given that $OF=a$:

$\frac{EC}{OC}=\frac{EO}{OF} \Rightarrow \frac{\sqrt{92^2*h^2+46^2}}{46}=\frac{92h}{a} \Rightarrow a=46\sqrt{\frac{92^2*h^2}{92^2*h^2+46^2}}$

Now, we want to find the maximum value of OG.

$\sqrt{\frac{92^2*h^2}{92^2*h^2+46^2}}=73h$

Solve for $h$, then:

$BC=\sqrt{(46-19/2)^2+(73h)^2}$.

$\boxed{BC^2=m^2=2109}$

The answer is $\boxed{D}$
[/hide]


Daniel and Scott are playing a game where a player wins as soon as he has two points more than his opponent.  Both players start at zero, and points are earned one at a time. If Daniel has a 60% chance of winning each point, what is the probability he will win the game?

~Solved by x biz kit l and andy17null in an hour~

-Isaac

Oui, this was in an interesting problem.

Once you see the "gimmick" its pretty easy to solve, but the thing is, can anyone give additional cases of the same phenomenon?

It confused me for a while, for I was getting into infinite series and trying a bunch of things that I was afraid I couldn't solve without some insane manipulating skills.

Probability, probability...
My bane...

Consider the possible results after 2 throws: either Daniel has won, Scott has won, or we are still exactly even.
So if Daniel wins with probability x, then x = 0.6*0.6 + 2*0.6*0.4*x
Solving this gives x = 9/13.

This idea is usually found in examples similar to this with a person winning after a string of coin tosses. A similar idea is, for example, in this problem:
A drunk person starts out n units from point A and n units from point B (in a straight line). He then takes one step towards A or towards B (50/50 chance). Whats the expected number of steps until he reaches one of A or B?
OK maybe its not really the same idea but it sort of is. Or not. ah try solving it anyway..

$ \documentclass[11pt]{article} \usepackage{amsmath} \pdfpagewidth 8.5in \pdfpageheight 11in \newcounter{prob_num} \setcounter{prob_num}{1} \newcommand{\prob}[5]{\bigskip \bigskip\arabic{prob_num}.\stepcounter{prob_num} #1 \par\nopagebreak[4]\medskip A.\ #2\hfill B.\ #3\hfill C.\ #4\hfill D.\ #5\hfill E.\ NOTA} \begin{document} \prob{What is$2+2$ ?}{4}{5}{6}{7} \prob{What is$\sqrt{100}$ ?}{81}{10}{9}{1} \prob{Evaluate$\displaystyle\sum_{n=1}^\infty \frac{1}{n^2}$ .} {$\displaystyle\frac{1}{e}$ } {$\displaystyle\frac{2}{\pi}$ } {$\displaystyle\frac{\pi^3}{8}$ } {$\displaystyle\frac{\pi^2}{6}${ } \end{document}$

/edit

really old ppost


A circle of radius 3 is tangent at the midpoint of one side of an equilateral triangle and passing through the vertex opposite this side. What is the area of the region that is outside of the circle and inside of the equilateral triangle?

EDIT: missing word added

[quote="10000th User"]A circle of radius 3 is tangent at the midpoint of one side of an equilateral and passing through the vertex opposite this side. What is the area of the region that is outside of the circle and inside of the equilateral triangle?[/quote]

[color=brown][b]my answer is :  9.641528612   :blush: [/b][/color]
 :wink:

Triangle $ ABC$, $ AB$ tangent, $ C$ is the opposite vertex, $ O$ centre of circle, circle cuts triangle at $ D$ and $ E$, $ N$ is midpoint of base (tangent) $ AB$, $ D$ is between $ A$ and $ C$, $ E$ is between $ C$ and $ B$. Part of triangle outside circle is twice triangle ($ ANC$, - (triangle $ DOC$ + sector $ ODN$)). Triangle $ ANC$ is $ 30 \minus{} 60 \minus{} 90$, triangle $ DOC$ is $ 30 \minus{} 30 \minus{} 120$. Hope this helps as I have to rush out.

[quote="10000th User"]A circle of radius 3 is tangent at the midpoint of one side of an equilateral and passing through the vertex opposite this side. What is the area of the region that is outside of the circle and inside of the equilateral triangle?[/quote]
[color=darkred]
[b]Arent u mean the area with wellow colure ??!!!
wait to send my complet solution
 :play_ball: [/b][/color]

I had to add a missing word: 'triangle'.

[quote="almonkez"][quote="10000th User"]A circle of radius 3 is tangent at the midpoint of one side of an equilateral [b]triangle[/b] and passing through the vertex opposite this side. What is the area of the region that is outside of the circle and inside of the equilateral triangle?[/quote]

[color=brown][b]my answer is :  9.641528612   :blush: [/b][/color]
 :wink:[/quote]Not quite correct...

@castigioni: please post a solution :)

@AndrewTom: I would like to see your alternative solution to finish :)

I have edited my post above, putting in another pair of brackets. (I'm afraid I was in a hurry.)

My diagram is the same as that of almonkez except that the point he calls $ D$, I called $ N$; I did not join $ D$ and $ E$ to $ N$, but joined $ O$ to $ D$.

Area of triangle $ ABN$ is $ \frac{1}{2} \times \frac{6}{\sqrt{3}} \times 6 \equal{} 6 \sqrt{3}$.

Sector $ ODN$ has area $ \frac{1}{2} \times 3^{2} \times \frac{\pi}{3} \equal{} \frac{3 \pi}{2}$.

The area of triangle $ AOD$ is $ \frac{1}{2} \times 3^{2} \times \sin \frac{\pi}{3} \equal{} \frac{9\sqrt{3}}{4}$

Therefore required area is $ 2(6\sqrt{3} \minus{} \frac{3\pi}{2} \minus{} \frac{9\sqrt{3}}{4}) \equal{} \frac{15\sqrt{3} \minus{} 3\pi}{2}$, which is approximately $ 8.277992076$.

Hope this helps (and is correct).

[quote="AndrewTom"]Therefore required area is $ 2(6\sqrt {3} \minus{} \frac {3\pi}{2} \minus{} \frac {9\sqrt {3}}{4}) \equal{} \frac {15\sqrt {3} \minus{} 3\pi}{2}$, which is approximately $ 8.277992076$.

Hope this helps (and is correct).[/quote]Everything else is right, just a slight computational mistake: the area is $ \frac {15\sqrt {3}}2\minus{}3\pi$.

oh , im sorry i used scr 2 in stead of scr 3  :wallbash_red:


[quote="JL"]Sorry if my question seems out of place here, but i was just wondering which universities would provide a solid undergrad math course. There has been a lot of discussion on this forum about the top US universities such as MIT, CalTech, Harvard, Princeton, etc. How do the top UK universities Cambridge, Oxford and Imperial College compare? Any advice is much appreciated  :)[/quote]

Existence of tutorial system, better teaching, comparatively lesser research/faculty at the top levels (improving, but UK still in recovery
from Thatcherite lean years).  Less competitive more social.   Better pubs. More questionable foods but many superior ones.  Greater ease of travel to Europe and conferences.  More left-wing, Euro-ish, Yank-resenting polity of the sort that takes Noam Chomsky seriously;  Oxford Union are radicals compared to the politics at or around Harvard and MIT, much less Princeton.

Quality and funding differences may equilibrate as US dollar declines.


I think you made a mistake in typing.....
Your Question should have been like this, for $ a,b,c>0$ prove that:
$ \frac{{x^6 z^3 }}{{x^2 y^6 z \plus{} y^4 z^5 }} \plus{} \frac{{y^6 x^3 }}{{y^2 z^6 x \plus{} z^4 x^5 }} \plus{} \frac{{z^6 y^3 }}{{z^2 x^6 y \plus{} x^4 y^5 }} \ge \frac{3}{2}$, else I cannot solve it...

I have found a solution which is trivial and childish:::

[hide="solution"]Observe that
$ \frac {{x^6 z^3 }}{{x^2 y^6 z \plus{} y^4 z^5 }} \plus{} \frac {{y^6 x^3 }}{{y^2 z^6 x \plus{} z^4 x^5 }} \plus{} \frac {{z^6 y^3 }}{{z^2 x^6 y \plus{} x^4 y^5 }}$
$ \equal{} \frac {{x^6 z^3 }}{{y^4 z (x^2y^2 \plus{} z^4) }} \plus{} \frac {{y^6 x^3 }}{{ z^4x(y^2z^2 \plus{} x^4) }} \plus{} \frac {{z^6 y^3 }}{{x^4y(z^2 x^2 \plus{} y^4) }}$
$ \geq \frac {{x^6 z^2 }}{{y^4.2xyz^2 }} \plus{} \frac {{y^6 x^2 }}{{ z^4.2yzx^2 }} \plus{} \frac {{z^6 y^2 }}{{x^4.2zxy^2 }}$  By AM GM inequality
$ \geq \frac {1}{2}(\frac {x^5}{y^5} \plus{} \frac {z^5}{x^5} \plus{} \frac {y^5}{z^5})\geq\frac {3}{2}$ which is true, again, from AM GM
Is my solution correct? :) [/hide]

It's because
if $ z(x^2y^2\plus{}z^4)\ge2xyz^2$, then $ \frac{1}{z(x^2y^2\plus{}z^4)}\le\frac{1}{2xyz^2}$

Oh, again the reverse sign mistake!!!!! :blush:
Sorry for the inconvenience in post#3...... :(
But post #2 is completely correct..........  :rotfl:

Here is a stronger one !!

$ LHS \geq \sum_{cyc}\frac{x^3z}{y^3x\plus{}y^2z^2}\geq \frac32$

[quote="Albanian Eagle"]Here is a stronger one !!

$ LHS \geq \sum_{cyc}\frac {x^3z}{y^3x \plus{} y^2z^2}\geq \frac32$[/quote]
It's quite samilar with my inequality just use holder to show it. :)


What exactly is atomic stability? Does stability refer to the enthalpy of the atom?

If so, why would a full and half filled energy level be most stable?

No it is not the enthalpy of the atom.  That would be related to the atomic decay or half-life of the atom.

What you are talking about is the relative energy states of the various orbitals for the electrons contained in the orbitals.

When a whole level if full, there is an interaction among the electrons that make the addition of the last electron to fill the level result in a lower energy position than one would otherwise expect.  Same for the half full orbitals.  You can think of this (althought the physics is different) as equivalent to the duck flying just behind and to one side of another duck -- it's easier because of the first duck's passage.  Same with following someone's track in deep snow.  The placement of the last electron to either fill or half fill the orbitals allows all of the electrons in that orbital have a lower energy position than they would if the orbital was not full or half full.

The stability related to ease of getting the electron(s) to move out of their full or half full orbitals and react with other atoms.


Find all real numbers $\alpha$ with the property that every member of the sequence

$cos \alpha, cos2 \alpha,........., cos2^{n}\alpha$

is negative.

[hide]uh...assuming that n ranges to infinity,

360-2x must be cylic with all terms greater than 90 and less than 270

I know 120 is a solution...[/hide]

[quote="FieryHydra"]If $k \in \mathbb{N}$ and $m \in \mathbb{Z},$ then
$\cos{2^{k}\alpha}< 0$
$\Rightarrow \frac{\pi}{2}+2m\pi < 2^{k}\alpha < \frac{3\pi}{2}+2m\pi$
$\Rightarrow \frac{\pi}{2^{k+1}}+\frac{m\pi}{2^{k-1}}< \alpha < \frac{3\pi}{2^{k+1}}+\frac{m\pi}{2^{k-1}}$

I guess it shouldn't be too hard to proceed from here.[/quote]

Would you mind showing us how you had proceed?


inequalities
analytic geometry
inequalities unsolved

[quote="bibobeo"]$a,b,c,d satisfy$
$a^{2}+b^{2}=1$
$c+d = 3$
Prove that $ac+bd+cd \leq \frac{9+6\sqrt{2}}{4}$[/quote]
The hint to ugly proof:
Let $f(a,b,c,d)=m(c+d)^{2}+n(a^{2}+b^{2})-ac-bd-cd$
and in $(a,b,c,d)$ the $\min f$ holds.
Hence, $f'_{a}=f'_{b}=f'_{c}=f'_{d}=0$ and $a^{2}+b^{2}\neq0,$ $c^{2}+d^{2}\neq0.$

Clearly, to get a maximum value of the expression, $a$ and $b$ must be positive.
Then let $a = cos \alpha, b = sin \alpha, 0 \leq \alpha \leq \frac{\pi}{2}$.
Then, $ac+bd = kcos(\alpha-\beta) \leq |k|$ where $tan \beta = \frac{d}{c}$ and $|k| = \sqrt{c^{2}+d^{2}}$.

$P = ac+bd+cd \leq \sqrt{c^{2}+(3-c)^{2}}+c(3-c) = \sqrt{2c^{2}-6c+9}+(3c-c^{2})$
Now let $f(x) = \sqrt{2x^{2}-6x+9}+(3x-x^{2})$. Since the expression under the root is positive, then by AM-GM, we have
$f(c) \leq \frac{ \frac{\sqrt{2}}{3}\cdot (2c^{2}-6c+9)+\frac{3}{\sqrt{2}}}{2}+(3c-c^{2})$
$=-(c^{2}\cdot \frac{3-\sqrt{2}}{2}-c(3-2\sqrt{2})-\frac{9\sqrt{2}}{4})$
$\leq-(-\frac{(3-\sqrt{2})^{2}+9\sqrt{2}\cdot \frac{3-\sqrt{2}}{3}}{4 \cdot \frac{3-\sqrt{2}}{3}})$
$= \frac{9+6\sqrt{2}}{4}$.

Therefore, $P \leq \frac{9+6\sqrt{2}}{4}$ as desired..  :)

have no idea?
we can see that the inequality holds when $a=b=\frac{1}{\sqrt{2}}$ and $c=d=\frac{3}{2}$
then applying Cauchy ...
I remember someone on Mathlinks teach me this. (Maybe Sylow_theory)  :P
Try it. (The numbers are not very nice  :wink: )

I haven't figured it out.. Can you post yours, bibobeo ? The one with CS ?  :wink:

[quote="kunny"]RDeepMath91, in your soltion, when does equality holds?  :maybe:

I think that this problem is wrong. :maybe:

Use Cauchy-schwarz's inequality, we have that $ac+bd\leq |ac+bd|\leq \sqrt{(a^{2}+b^{2})(c^{2}+d^{2})}$,

thus $ac+bd+cd\leq \sqrt{c^{2}+d^{2}}+cd\leq \sqrt{c^{2}+d^{2}}+\frac{c^{2}+d^{2}}{2}$.

Now $\sqrt{c^{2}+d^{2}}$ means that the distance of the origin and points $(c,\ d)$ on the line $c+d=3$ in $c-d$ cordinate plane.

So $ac+bd+cd<\infty.$[/quote]

When $2c^{2}-6c+9 = \frac{9}{2}\Rightarrow c = \frac{3}{2}= d$ and $\alpha = \beta = tan^{-1}(\frac{d}{c}) = \frac{\pi}{4}\Rightarrow a = b$.

Sorry, I seem to have made a misunderstnding. This problem is valid. :oops:

[quote="RDeepMath91"]
Now let $f(x) = \sqrt{2x^{2}-6x+9}+(3x-x^{2})$. [/quote]

$f'(x)=\frac{4x-6}{2\sqrt{2x^{2}-6x+9}}+3-2x=\frac{2(3-2x)(x^{2}-3x+4)}{(1+\sqrt{2x^{2}-6x+9})\sqrt{2x^{2}-6x+9}}.$

Since we have $2x^{2}-6x+9=2\left(x-\frac{3}{2}\right)^{2}+\frac{9}{2}>0,\ x^{2}-3x+4=\left(x-\frac{3}{2}\right)^{2}+\frac{7}{4}>0$,

The sign of $f'(x)$ can be detemined by the sign of $3-2x$. Thus $f(x)$ has a local maximum at $x=\frac{3}{2}$.

My friend, he said he had a solution with "coordinate" (I don't know what that in English, but he use circle and line)


Find the least $ k\in N$ such that
$ \frac{k(2n)!}{n!(n\plus{}2)!}\in N$ for all n
2)Find the least k such that
$ \frac{k(3n)!}{n!(n\plus{}1)!(n\plus{}2)!}$ is integer.

[hide="hint"]
Use Poligac lemma
$ e_{p}(n!) \equal{}\sum_{i \equal{} 1}^{\plus{}\infty}[\frac{n}{p^{i}}]$
[/hide]

There are trivial problems - k=2 for 1) and 2).


Find all infinite bounded sequences $ a_1,a_2,...$ of positive integers such that for all $ n > 2$
$ \displaystyle a_n\equal{}\frac{a_{n\minus{}1}\plus{}a_{n\minus{}2}}{gcd(a_{n\minus{}1},a_{n\minus{}2})}$
 :P  :roll:

Since the sequence is bounded, we have finite possible combination of $ (a_n,a_{n\plus{}1})$.
Hence, after a few steps, the sequence has to be periodic.
In other words, we can see $ a_1,\cdots,a_m,a_{m\plus{}1},\cdots,x,a_m,a_{m\plus{}1}$.

Assume $ t\equal{}gcd(a_m,a_{m\plus{}1})$, we have $ t|a_{m\plus{}1}|a_m\plus{}x\Rightarrow t|x$.
Going backwards, we see $ t$ divides all the number in between the two $ a_m$.

Let $ a_m\equal{}at,a_{m\plus{}1}\equal{}bt$, then $ a_{m\plus{}2}\equal{}a\plus{}b$.
Since $ gcd(a\plus{}b,b)\equal{}1$, $ gcd(a\plus{}b,bt)\equal{}gcd(a\plus{}b,t)\le t$.
But we already have $ t|a_{m\plus{}2}$, so $ gcd(a_{m\plus{}2},a_{m\plus{}1})\equal{}t$.
Going forward, $ gcd(a_{n\plus{}2},a_{n\plus{}1})\equal{}t,\forall n\ge m$.
In other words, we have
$ a_{n\plus{}2}\equal{}\frac{a_n\plus{}a_{n\plus{}1}}t$ for all $ n\ge m$.
Let $ r_1<r_2$ be the two roots to $ x^2\minus{}\frac1tx\minus{}\frac1t\equal{}0$. To make sure the sequence bounded, $ |r_1|,|r_2|\le1$.
Then we can derive $ t\ge2$ and
Now $ a_{n\plus{}2}\minus{}r_2 a_{n\plus{}1}\equal{}r_1(a_{n\plus{}1}\minus{}r_2a_n),\forall n\ge m$ with $ \minus{}1<r_1<0$.
To make sure the sequence have only integers, $ a_{n\plus{}1}\equal{}r_2a_n$ with $ r_2\equal{}1$.
Hence $ t\equal{}2$.

Going backwards,it is easy to see the whole sequence has to be constant.
Then we can verify $ a_1\equal{}a_2\equal{}2$ is the only solution.


trigonometry
logarithms
calculus
calculus computations

hello, we have $ \arcsin(i\minus{}1)\equal{}iarsinh(1\plus{}i)$
Sonnhard.

$ e^{iz} \equal{} \cos z\plus{}i \sin z$
$ e^{\minus{}iz} \equal{} \cos(\minus{}z) \plus{} i \sin(\minus{}z) \equal{} \cos z \minus{} i \sin z$

$ \sin z \equal{} \frac{e^{iz} \minus{} e^{\minus{}iz}}{2i}$.

Let $ w \equal{} \sin z$ or equivalently, $ z \equal{} \sin^{\minus{}1} w$.  Then

$ 2iw \equal{} e^{iz} \minus{} e^{\minus{}iz}$

$ (e^{iz})^2 \minus{} 2iw e^{iz} \minus{} 1 \equal{} 0$

$ e^{iz} \equal{} \frac{2iw \pm \sqrt{(2iw)^2 \plus{} 4}}{2} \equal{} iw \pm \sqrt{1 \minus{} w^2}$

$ z \equal{} \minus{}i \log\left(iw \pm \sqrt{1\minus{}w^2}\right) \equal{} \sin^{\minus{}1} w$.

Substituting $ w \equal{} i\minus{}1$ gives the desired result.

However, it should be noted that

$ \sqrt{1\plus{}2i} \equal{} \pm\left(5^{1/2} e^{i \tan^{\minus{}1} 2}\right)^{1/2} \equal{} \pm 5^{1/4} e^{i (\tan^{\minus{}1} 2)/2}$

$ \equal{} \pm 5^{1/4} \left(\cos \frac{\tan^{\minus{}1} 2}{2} \plus{} i \sin \frac{\tan^{\minus{}1} 2}{2}\right)$

$ \equal{} \pm (\varphi^{1/2} \plus{} i \varphi^{\minus{}1/2})$,

where $ \varphi \equal{} (1\plus{}\sqrt{5})/2$ is the golden ratio.  Thus

$ \sin^{\minus{}1} (i\minus{}1) \equal{} \minus{}i \log \left((\pm \varphi^{1/2} \minus{} 1) \plus{} i(\pm \varphi^{\minus{}1/2} \minus{} 1)\right)$.

Since we know that

$ \log z \equal{} \log |z| \plus{} i (\arg z \plus{} 2 \pi k), \quad k \in \mathbb{Z}$,

we obtain

$ \sin^{\minus{}1} (i\minus{}1) \equal{} \minus{}i \left( \log (\varphi \minus{} \varphi^{1/2}) \minus{} i (\tan^{\minus{}1} \varphi^{\minus{}1/2} \plus{} 2\pi k)\right)$
$ \equal{} 2\pi k \minus{} \tan^{\minus{}1} \varphi^{\minus{}1/2} \minus{} i \log (\varphi \minus{} \varphi^{1/2})$.

There is another solution set, however, since there is the other root:

$ \sin^{\minus{}1} (i\minus{}1) \equal{} \pi(2k\plus{}1) \plus{} \tan^{\minus{}1} \varphi^{\minus{}1/2} \minus{} i \log (\varphi \plus{} \varphi^{1/2})$.


geometry
geometric transformation
reflection
circumcircle
analytic geometry
symmetry
parallelogram

By generalizing a problem from our IMO training, I have found the following triangle geometry problem:

[b]Problem, version 1:[/b]

Let P and Q be two points in the plane of a triangle ABC such that the lines CP and CQ are symmetric with respect to the angle bisector of the angle BCA. The lines AQ and BQ meet the lines BP and AP at the points N and M, respectively. Show that the lines CM and CN are symmetric with respect to the angle bisector of the angle BCA.

[i]Remarks.[/i]

[b]1.[/b] The above problem also has an equivalent version using the notion of isogonal conjugates:

[b]Problem, version 2:[/b]

Let P and Q be two mutually isogonal conjugate points with respect to a triangle ABC, and let E be an arbitrary point on the line CQ. The lines AE and BE meet the lines BP and AP at the points N and M, respectively. Show that < ACN = < MCB, i. e. that the lines CM and CN are symmetric with respect to the angle bisector of the angle BCA.

[Note that the notations in versions 1 and 2 are different: The point E of version 2 corresponds to the point Q of version 1.]

[b]2.[/b] For those who don't know, [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=18472]the notion of isogonal conjugate points[/url] is defined as follows: If P is a point in the plane of triangle ABC, then the reflections of the lines AP, BP, CP in the angle bisectors of the angles CAB, ABC, BCA concur at a new point Q, which is called the [i]isogonal conjugate[/i] of P with respect to triangle ABC. Then, of course, P is again the isogonal conjugate of Q with respect to triangle ABC; hence, the relation of the points P and Q is mutual, and we can say the points P and Q are mutually isogonal conjugate points.

[b]3.[/b] As I have already said, the problem above is a generalization of a problem Arend Bayer discussed with us during the IMO training session at Oberwolfach:

Let H be the orthocenter and O the circumcenter of a triangle ABC. Let E be an arbitrary point on the line CO. The lines AE and BE meet the altitudes BH and AH at the points N and M, respectively. Show that < ACN = < MCB, i. e. that the lines CM and CN are symmetric with respect to the angle bisector of the angle BCA.

(In fact, this follows from version 2 of the problem above, since the points H and O are always mutually isogonal conjugate points - prove this for yourself!)

[b]4.[/b] There is a solution of the above problem which uses trilinear coordinates; however, it can be easily made synthetic by replacing trilinear coordinates by distance ratios.

  Darij

[quote="darij grinberg"]Let P and Q be two points in the plane of a triangle ABC such that the lines CP and CQ are symmetric with respect to the angle bisector of the angle BCA. The lines AQ and BQ meet the lines BP and AP at the points N and M, respectively. Show that the lines CM and CN are symmetric with respect to the angle bisector of the angle BCA.
[/quote]

Let me try my hands at projective geometry.

Denote the angle bisector of $\angle ACB$ as line $\ell$

Consider a projective transformation $\mathcal{P}_1$ ($X \to X'$) that takes the line perpendicular to $\ell$ at $A$ to infinity, and preserves the symmetry about $\ell$. (see end for the existance of such transformation)

Then $\ell'$, $(AC)'$, $(BC)'$, $(PC)'$, $(QC)'$ are parallel. Also $(AC)'$ and $(BC)'$ are on opposite sides of $\ell'$ and are equidistant to it (since $\mathcal{P}_1$) preserves symmetry). Similarly with $(PC)'$, $(QC)'$.

Now, consider an affine transformation $\mathcal{A}$ ($X' \to X''$) with the property that $A''B'' \bot \ell''$. The lines $\ell''$, $(AC)''$, $(BC)''$, $(PC)''$, $(QC)''$ are still parallel and still satisfy the properties in the previous paragraph.

Now, consider another projective transformation $\mathcal{P}_2$ ($X'' \to X'''$) with the property that $(AB)'''$ is the line at infinity and the symmetry about $\ell'''$ is preserved.

Then $(PN)''' \parallel (QM)'''$ and $(PM)''' \parallel (QN)'''$, so $P'''M'''Q'''N'''$ is a parallelogram with centre on $\ell'''$ (since the midpoint of $P'''Q'''$ is on $\ell'$). It follows that $M'''$ and $N'''$ are equidistant from $\ell'''$ and on opposite sides of it.

Let $m'''$ denote the line parallel to $\ell'''$ through $M'''$. Similarly with $n'''$, then tracing back the transformations, we see that $m = MC$ and $n = NC$. Since symmetry is preserved throughout, the result follows.

To see that the two projective transformations exist, consider the following. For $\mathcal{P}_1$ (and similarly with $\mathcal{P}_2$), consider a point $P$ in space (not on the plane $(ABC)$) such that $PC \bot (ABC)$. Now consider a plane $\Pi$ that does not contain $P$ and is perpendicular to $\ell$. Then projecting the points on plant $(ABC)$ onto $\Pi$ with perspective $P$ yields the desired transformation.

I wonder if the solution can be compressed by combining the three transformations into two or one  :?

Half a year ago I promised to repost my solution of this problem (which was lost because of the hacker attack on MathLinks in december 2003). Sorry for forgetting this promise for a long time - here it goes.

I will solve the problem in its version 1 and with slightly modified notations (sorry for that, but I am copying the formulas from one of my earlier, German writeups where the notations are different):

[color=blue][b]Theorem 1.[/b] Let P and Q be two points in the plane of a triangle ABC such that the lines AP and AQ are symmetric to each other with respect to the angle bisector of the angle CAB. The lines BP and CQ meet each other at a point X, and the lines CP and BQ meet each other at a point Y. Then, the lines AX and AY are symmetric to each other with respect to the angle bisector of the angle CAB.[/color]

[i]First proof of Theorem 1.[/i] We will first introduce some notations. For any arbitrary point P in the plane of triangle ABC, let x(P) be the signed distance from the point P to the side BC of the triangle ABC; hereby, "signed" means that it is the distance from the point P to the line BC with a sign, and the sign is $ \plus{}$ if the point P lies in the same halfplane with respect to the line BC as the point A, and $ \minus{}$ if the points P and A lie in different halfplanes with respect to the line BC. Similarly, define the signed distances y(P) and z(P) from the point P to the sides CA and AB. The first important property of these signed distances is:

[color=blue][b]Lemma 2.[/b] Let P and Q be two arbitrary points in the plane of triangle ABC. Then, the points A, P, Q are collinear if and only if $ \frac{y\left( P\right) }{z\left( P\right) }\equal{}\frac{y\left( Q\right) }{z\left( Q\right) }$.[/color]

Of course, similar criteria hold for the collinearity of B, P, Q or of C, P, Q.

[hide="Proof"]
[i]Proof of Lemma 2.[/i] Let Y and Z be the orthogonal projections of the point P on the lines CA and AB, and let Y' and Z' be the orthogonal projections of the point Q on the lines CA and AB. Since $ PY\perp CA$ and $ QY^{\prime}\perp CA$, the lines PY and QY' are parallel.

The length PY equals to the non-signed distance from the point P to the line CA; but if we use directed segments and direct the line PY properly, then it will equal to the signed distance from the point P to the line CA. In other words, if we direct the line PY properly, we have PY = y(P). Similarly, we can direct the line QY' in such a way that QY' = y(Q), and in this way, the lines PY and QY' are equally directed. Similarly, we direct the lines PZ and QZ' in such a way that PZ = z(P) and QZ' = z(Q).

Now, assume that the points A, P, Q are collinear; then, since PY || QY', Thales yields $ \frac{PY}{QY^{\prime }}\equal{}\frac{AP}{AQ}$, so that $ \frac{y\left( P\right) }{y\left( Q\right) }\equal{}\frac{AP}{AQ}$. Similarly, $ \frac{z\left( P\right) }{z\left( Q\right) }\equal{}\frac{AP}{AQ}$, and thus $ \frac{y\left( P\right) }{y\left( Q\right) }\equal{}\frac{z\left( P\right) }{z\left( Q\right) }$, or, equivalently, $ \frac{y\left( P\right) }{z\left( P\right) }\equal{}\frac{y\left( Q\right) }{z\left( Q\right) }$.

Conversely, if $ \frac{y\left( P\right) }{z\left( P\right) }\equal{}\frac{y\left( Q\right) }{z\left( Q\right) }$, then we have $ \frac{y\left( P\right) }{y\left( Q\right) }\equal{}\frac{z\left( P\right) }{z\left( Q\right) }$. Denote $ k \equal{}\frac{y\left( P\right) }{y\left( Q\right) }\equal{}\frac{z\left( P\right) }{z\left( Q\right) }$, so that $ y\left( P\right) \equal{} k\cdot y\left( Q\right)$ and $ z\left( P\right) \equal{} k\cdot z\left( Q\right)$, and let P' be the point on the line AQ such that $ AP^{\prime }\equal{} k\cdot AQ$. Then, similarly to the above, we can show that $ \frac{y\left( P^{\prime }\right) }{y\left( Q\right) }\equal{}\frac{AP^{\prime }}{AQ}$ and $ \frac{z\left( P^{\prime }\right) }{z\left( Q\right) }\equal{}\frac{AP^{\prime }}{AQ}$. Since $ \frac{AP^{\prime }}{AQ}\equal{}\frac{k\cdot AQ}{AQ}\equal{} k$, this becomes $ \frac{y\left( P^{\prime }\right) }{y\left( Q\right) }\equal{} k$ and $ \frac{z\left( P^{\prime }\right) }{z\left( Q\right) }\equal{} k$. Thus, $ y\left( P^{\prime }\right) \equal{} k\cdot y\left( Q\right) \equal{} y\left( P\right)$ and $ z\left( P^{\prime }\right) \equal{} k\cdot z\left( Q\right) \equal{} z\left( P\right)$. Since y(P') = y(P), the points P and P' are equidistant from the line CA, and thus, if they were distinct, this would yield PP' || CA. Similarly, the equation z(P') = z(P) would yield PP' || AB. But this combined together would lead to CA || AB, what is absurd. Thus, the points P and P' cannot be distinct; hence, P = P', and since we know that the point P' lies on the line AQ, it follows that the point P lies on the line AQ, so that the points A, P, Q are collinear.

Thus, Lemma 2 is proven.[/hide]

One can use Lemma 2 to prove a number of facts, e. g. the Ceva theorem. The above proof of Lemma 2 was simple, but imagine that a number of authors use calculus to establish the direction $ \frac{y\left( P\right) }{z\left( P\right) }\equal{}\frac{y\left( Q\right) }{z\left( Q\right) }\Longrightarrow A,P,Q\text{ collinear}$ (in fact, they show the injectivity of the function $ f\left( x\right) \equal{}\frac{\sin x}{\sin\left(\varphi\minus{}x\right) }$ for fixed $ \varphi$ by differentiating)...

Now let's show a corollary of Lemma 2:

[color=blue][b]Lemma 3.[/b] Let P and Q be two arbitrary points in the plane of triangle ABC. Then, the lines AP and AQ are symmetric to each other with respect to the angle bisector of the angle CAB if and only if $ y\left( P\right)\cdot y\left( Q\right) \equal{} z\left( P\right)\cdot z\left( Q\right)$.[/color]

Of course, similar criteria can be stated for the lines BP and BQ to be symmetric with respect to the angle bisector of the angle ABC or for the lines CP and CQ to be symmetric with respect to the angle bisector of the angle BCA.

[hide="Proof"]
[i]Proof of Lemma 3.[/i] Let Q' be the reflection of the point Q in the angle bisector of the angle CAB. Then, the line AQ' is the reflection of the line AQ in the angle bisector of the angle CAB. Thus, the lines AP and AQ are symmetric to each other with respect to the angle bisector of the angle CAB if and only if the line AP coincides with the line AQ', i. e. if and only if the points A, P, Q' are collinear.

Now, reflections with respect to lines are congruence transformations, i. e. they leave distances fixed. Consider the reflection with respect to the angle bisector of the angle CAB; this reflection maps the line CA to the line AB and the point Q to the point Q'. Thus, the distance from the point Q' to the line AB equals to the distance from the point Q to the line CA. In other words, z(Q') = y(Q) (in fact, these are signed distances, but the equality of the signs can be easily seen). Similarly, y(Q') = z(Q).

Now, as we have seen above, the lines AP and AQ are symmetric to each other with respect to the angle bisector of the angle CAB if and only if the points A, P, Q' are collinear. According to Lemma 2, this holds if and only if $ \frac{y\left( P\right) }{z\left( P\right) }\equal{}\frac{y\left( Q^{\prime }\right) }{z\left( Q^{\prime }\right) }$. But in view of z(Q') = y(Q) and y(Q') = z(Q), we can rewrite this condition as $ \frac{y\left( P\right) }{z\left( P\right) }\equal{}\frac{z\left( Q\right) }{y\left( Q\right) }$, or, equivalently, as $ y\left( P\right)\cdot y\left( Q\right) \equal{} z\left( P\right)\cdot z\left( Q\right)$. Thus, Lemma 3 is proven.[/hide]

Now we are ready to prove Theorem 1:

Since the lines AP and AQ are symmetric to each other with respect to the angle bisector of the angle CAB, Lemma 3 yields $ y\left( P\right)\cdot y\left( Q\right) \equal{} z\left( P\right)\cdot z\left( Q\right)$.

Since the points B, X, P are collinear, Lemma 2 shows that $ \frac{z\left( X\right) }{x\left( X\right) }\equal{}\frac{z\left( P\right) }{x\left( P\right) }$. Similarly, the collinearity of the points C, X, Q yields after Lemma 2 that $ \frac{x\left( X\right) }{y\left( X\right) }\equal{}\frac{x\left( Q\right) }{y\left( Q\right) }$. Thus, $ \frac{z\left( X\right) }{x\left( X\right) }\cdot\frac{x\left( X\right) }{y\left( X\right) }\equal{}\frac{z\left( P\right) }{x\left( P\right) }\cdot\frac{x\left( Q\right) }{y\left( Q\right) }$, what simplifies to

$ \frac{z\left( X\right) }{y\left( X\right) }\equal{}\frac{z\left( P\right) }{x\left( P\right) }\cdot\frac{x\left( Q\right) }{y\left( Q\right) }$.

Similarly,

$ \frac{y\left( Y\right) }{z\left( Y\right) }\equal{}\frac{y\left( P\right) }{x\left( P\right) }\cdot\frac{x\left( Q\right) }{z\left( Q\right) }$.

Dividing these two equations, we obtain

$ \frac{y\left( Y\right) }{z\left( Y\right) }:\frac{z\left( X\right) }{y\left( X\right) }\equal{}\left(\frac{y\left( P\right) }{x\left( P\right) }\cdot\frac{x\left( Q\right) }{z\left( Q\right) }\right) :\left(\frac{z\left( P\right) }{x\left( P\right) }\cdot\frac{x\left( Q\right) }{y\left( Q\right) }\right)$,

thus

$ \frac{y\left( Y\right) }{z\left( Y\right) }\cdot\frac{y\left( X\right) }{z\left( X\right) }\equal{}\left(\frac{y\left( P\right) }{x\left( P\right) }\cdot\frac{x\left( Q\right) }{z\left( Q\right) }\right)\cdot\left(\frac{x\left( P\right) }{z\left( P\right) }\cdot\frac{y\left( Q\right) }{x\left( Q\right) }\right) \equal{}\frac{z\left( P\right)\cdot z\left( Q\right) }{z\left( Q\right)\cdot z\left( P\right) }\equal{} 1$,

so that $ \frac{y\left( X\right)\cdot y\left( Y\right) }{z\left( X\right)\cdot z\left( Y\right) }\equal{} 1$, and thus $ y\left( X\right)\cdot y\left( Y\right) \equal{} z\left( X\right)\cdot z\left( Y\right)$. According to Lemma 3, this shows that the lines AX and AY are symmetric to each other with respect to the angle bisector of the angle CAB, and Theorem 1 is proven.

[i]Second proof of Theorem 1.[/i] This proof will use projective involutions. Recall the Desargues involution theorem:

[color=blue][b]Desargues involution theorem.[/b] If A, B, C, D are four points and g is a line, and if we define the lines a = AB, b = BC, c = CD, d = DA, e = AC, f = BD, then the pairs of points $ \left(a\cap g;\;c\cap g\right)$, $ \left(b\cap g;\;d\cap g\right)$ and $ \left(e\cap g;\;f\cap g\right)$ are conjugate pairs of points under one involution of the line g.[/color]

This theorem has a projective dual version:

[color=blue][b]Dual Desargues involution theorem.[/b] If a, b, c, d are four lines and G is a point, and if we define the points $ A \equal{} a\cap b$, $ B \equal{} b\cap c$, $ C \equal{} c\cap d$, $ D \equal{} d\cap a$, $ E \equal{} a\cap c$, $ F \equal{} b\cap d$, then the pairs of lines (AG; CG), (BG; DG) and (EG; FG) are conjugate pairs of lines under one involution of the pencil of lines through the point G.[/color]

Apply this theorem to the four lines a = BP, b = BQ, c = CQ, d = CP and to the point G = A. Since $ BP\cap BQ \equal{} B$, $ BQ\cap CQ \equal{} Q$, $ CQ\cap CP \equal{} C$, $ CP\cap BP \equal{} P$, $ BP\cap CQ \equal{} X$, $ BQ\cap CP \equal{} Y$, it will follow that the pairs of lines (BA; CA), (QA; PA) and (XA; YA) are conjugate pairs of lines under one involution of the pencil of lines through the point A. But an involution of a pencil of lines is uniquely determined by two conjugate pairs of lines; hence, considering the pairs of lines (BA; CA) and (QA; PA), which are both symmetric with respect to the angle bisector of the angle CAB, we conclude that our involution of the pencil of lines through the point A must be the reflection with respect to the angle bisector of the angle CAB. Hence, as the pair of lines (XA; YA) is a conjugate pair under this involution, it follows that the lines XA and YA are symmetric to each other with respect to the angle bisector of the angle CAB. And Theorem 1 is established again.

[b]EDIT:[/b] I just noted that Theorem 1 can be used to give an elementary proof of a beautiful fact about isogonal conjugates which is usually proven with the help of conics or even cubics:

[color=blue][b]Theorem 4.[/b] Let ABC be a triangle, let P and P' be a pair of isogonal conjugate points with respect to triangle ABC, and let Q and Q' be a pair of isogonal conjugate points with respect to triangle ABC. Then, the points $ PQ\cap P^{\prime}Q^{\prime}$ and $ PQ^{\prime}\cap P^{\prime}Q$ are a pair of isogonal conjugate points with respect to triangle ABC as well.[/color]

[i]Proof.[/i] Denote $ U \equal{} PQ\cap P^{\prime}Q^{\prime}$ and $ U^{\prime}\equal{} PQ^{\prime}\cap P^{\prime}Q$. We have to prove that these points U and U' are isogonal conjugates with respect to triangle ABC.

Since the points P and P' are isogonal conjugates with respect to triangle ABC, the lines AP and AP' are symmetric with respect to the angle bisector of the angle CAB. Thus, the angle bisector of the angle CAB is also the angle bisector of the angle PAP' (to be more precise, either the internal or the external angle bisector of this angle). On the other hand, as the points Q and Q' are isogonal conjugates with respect to triangle ABC, the lines AQ and AQ' are symmetric with respect to the angle bisector of the angle CAB. Hence, we can say that the lines AQ and AQ' are symmetric with respect to the angle bisector of the angle PAP'. Thus, we can apply Theorem 1 to the points Q and Q' in the plane of triangle AP'P. The lines P'Q and PQ' meet each other at the point U', and the lines P'Q' and PQ meet each other at the point U. Theorem 1 then yields that the lines AU' and AU are symmetric with respect to the angle bisector of the angle PAP'. In other words, the lines AU' and AU are symmetric with respect to the angle bisector of the angle CAB. Similarly, the lines BU' and BU are symmetric with respect to the angle bisector of the angle ABC, and the lines CU' and CU are symmetric with respect to the angle bisector of the angle BCA. Therefore, the points U and U' are isogonal conjugate points with respect to triangle ABC, and Theorem 4 is proven.

  Darij


For all positive integers a and x, let [AlX be defined as the sum of the integers from a to x, inclusive. where a<x. Which of the following equals [11lX]-[9l(x-2)]

A. 0
B. x-19
C. x-20
D. 2x-19
E. 2x-20

And also, to all the Indian members of AoPSS, happy Diwali.  :)

Thanks in advance for the question.  :D [/i]

[quote="AIME15"]Hint:  Write [9|(x-2)] as [11-2|(x-2)-2][/quote]

Ugh, after a dozen tries. Would the answer happen to be E? 2x-20? If it isn't there goes 20 minutes of my life.  :maybe:

Happy Diwali to you too!! :)

For those of you who are like "What?", Diwali is an Indian Festival of Light.  You light firecrackers and fireworks, and in India, i believe there is no school today.

[quote="Forte"][quote="math154"]$ [11|x] \equal{} 11 \plus{} 12 \plus{} \cdots \plus{} (x \minus{} 2) \plus{} (x \minus{} 1) \plus{} x$, and

$ [9|(x \minus{} 2)] \equal{} 9 \plus{} 10 \plus{} 11 \plus{} 12 \plus{} \cdots \plus{} (x \minus{} 2)$.

From this, we see that
$ [11|x] \minus{} [9|(x \minus{} 2)]$

$ \equal{} (11 \plus{} 12 \plus{} \cdots \plus{} (x \minus{} 2) \plus{} (x \minus{} 1) \plus{} x) \minus{} (9 \plus{} 10 \plus{} 11 \plus{} 12 \plus{} \cdots \plus{} (x \minus{} 2))$

$ \equal{} \minus{} (9 \plus{} 10) \plus{} (11 \plus{} 12 \plus{} \cdots \plus{} (x \minus{} 2)) \minus{} (11 \plus{} 12 \plus{} \cdots \plus{} (x \minus{} 2)) \plus{} (x \minus{} 1) \plus{} x$

$ \equal{} \minus{} 19 \plus{} 2x \minus{} 1 \equal{} 2x \minus{} 10\implies\boxed{E}$.

So you are correct. :D[/quote]

Thanks for the hint. It made all the difference.  :) Really appreciate it. ^^[/quote]

No problem.

Is this from that SAT book of yours, by any chance :wink:

Learn to solve the *new operation* problems, those get people sometimes because they're intimated by $ 3 \star 4$, when it says $ a \star b\equal{}ab$...

[quote="AIME15"][quote="Forte"][quote="math154"]$ [11|x] \equal{} 11 \plus{} 12 \plus{} \cdots \plus{} (x \minus{} 2) \plus{} (x \minus{} 1) \plus{} x$, and

$ [9|(x \minus{} 2)] \equal{} 9 \plus{} 10 \plus{} 11 \plus{} 12 \plus{} \cdots \plus{} (x \minus{} 2)$.

From this, we see that
$ [11|x] \minus{} [9|(x \minus{} 2)]$

$ \equal{} (11 \plus{} 12 \plus{} \cdots \plus{} (x \minus{} 2) \plus{} (x \minus{} 1) \plus{} x) \minus{} (9 \plus{} 10 \plus{} 11 \plus{} 12 \plus{} \cdots \plus{} (x \minus{} 2))$

$ \equal{} \minus{} (9 \plus{} 10) \plus{} (11 \plus{} 12 \plus{} \cdots \plus{} (x \minus{} 2)) \minus{} (11 \plus{} 12 \plus{} \cdots \plus{} (x \minus{} 2)) \plus{} (x \minus{} 1) \plus{} x$

$ \equal{} \minus{} 19 \plus{} 2x \minus{} 1 \equal{} 2x \minus{} 10\implies\boxed{E}$.

So you are correct. :D[/quote]

Thanks for the hint. It made all the difference.  :) Really appreciate it. ^^[/quote]

No problem.

Is this from that SAT book of yours, by any chance :wink:

Learn to solve the *new operation* problems, those get people sometimes because they're intimated by $ 3 \star 4$, when it says $ a \star b \equal{} ab$...[/quote]

Heh. Yes it was by the way.  :P

[quote="mewto55555"]Or you could just pick a random number for x like 12:

[11|12]-[9|10]=23-19=4.

Substituting in 12 for x in all of our choices we find that 2x-20=4. So it is E.[/quote]

That sure beats the heck out of all that algebra work. None the less, both methods are good, thank you.  :)


[quote="Julia Robet"]Find the least value of the expression: $H=\sum \frac{x^2}{y+z}$
where $x,y,z$ are positive numbers satisfying $\sum \sqrt{x^2+y^2}=2006$.[/quote]
$H=\sum_{cyc}\frac{x^4}{x^2y+x^2z}\geq\frac{(x^2+y^2+z^2)^2}{x^2y+x^2z+y^2x+y^2z+z^2x+z^2y}\geq\frac{1}{2}\cdot\sqrt{3(x^2+y^2+z^2)}=$
$=\frac{1}{2\sqrt2}\cdot\sqrt{3\cdot\sum_{cyc}(x^2+y^2)}\geq\frac{1}{2\sqrt2}\cdot\sum_{cyc}\sqrt{x^2+y^2}=\frac{1003}{\sqrt2}.$ :)

[quote="arqady"]$\frac{(x^2+y^2+z^2)^2}{x^2y+x^2z+y^2x+y^2z+z^2x+z^2y}\geq\frac{1}{2}\cdot\sqrt{3(x^2+y^2+z^2)}=$[/quote]
Why??
[quote="arqady"]$=\frac{1}{2\sqrt2}\cdot\sqrt{3\cdot\sum_{cyc}(x^2+y^2)}\geq\frac{1}{2\sqrt2}\cdot\sum_{cyc}\sqrt{x^2+y^2}=\frac{1003}{\sqrt2}.$ :)[/quote]
And why?
Sory :blush: .

[quote="Julia Robet"][quote="arqady"]$\frac{(x^2+y^2+z^2)^2}{x^2y+x^2z+y^2x+y^2z+z^2x+z^2y}\geq\frac{1}{2}\cdot\sqrt{3(x^2+y^2+z^2)}=$[/quote]
Why??[/quote]
$\frac{(x^2+y^2+z^2)^2}{x^2y+x^2z+y^2x+y^2z+z^2x+z^2y}\geq\frac{1}{2}\cdot\sqrt{3(x^2+y^2+z^2)}\Leftrightarrow$
$\Leftrightarrow4(x^2+y^2+z^2)^3\geq3(x^2y+x^2z+y^2x+y^2z+z^2x+z^2y)^2\Leftrightarrow$
$\sum_{cyc}(4x^6+9x^4y^2+9x^4z^2+2x^2y^2z^2-6x^4yz-6x^3y^3-6x^3y^2z-6x^3z^2y)\geq0\Leftrightarrow$
$\Leftrightarrow2\cdot\sum_{cyc}(x^6-x^4y^2-x^4z^2+x^2y^2z^2)+$
$+\sum_{cyc}(2x^6+11x^4y^2+11x^4z^2-6x^4yz-6x^3y^3-6x^3y^2z-6x^3z^2y)\geq0$
The last inequality is true by Schur and Muirhead.
[quote="Julia Robet"]
[quote="arqady"]$=\frac{1}{2\sqrt2}\cdot\sqrt{3\cdot\sum_{cyc}(x^2+y^2)}\geq\frac{1}{2\sqrt2}\cdot\sum_{cyc}\sqrt{x^2+y^2}=\frac{1003}{\sqrt2}.$ :)[/quote]
And why?
[/quote]
Let $x^2+y^2=a^2,$ $x^2+z^2=b^2$ and $y^2+z^2=c^2,$ where $a\geq0,$ $b\geq0$ and $c\geq0.$ Then
$\frac{1}{2\sqrt2}\cdot\sqrt{3\cdot\sum_{cyc}(x^2+y^2)}\geq\frac{1}{2\sqrt2}\cdot\sum_{cyc}\sqrt{x^2+y^2}\Leftrightarrow3(a^2+b^2+c^2)\geq(a+b+c)^2.$
Other way to the proof that $H\geq\frac{1}{2}\sqrt{3(x^2+y^2+z^2)}:$
$\frac{x}{y+z}\geq\frac{8x-y-z}{4(x+y+z)}\Leftrightarrow(2x-y-z)^2\geq0.$
Hence, $H\geq\frac{1}{4(x+y+z)}\cdot\sum_{cyc}x(8x-y-z)=\frac{4(x^2+y^2+z^2)-xy-xz-yz}{2(x+y+z)}.$
Remain to prove that $\frac{4(x^2+y^2+z^2)-xy-xz-yz}{x+y+z}\geq\sqrt{3(x^2+y^2+z^2)}.$
But   $\frac{4(x^2+y^2+z^2)-xy-xz-yz}{x+y+z}\geq\sqrt{3(x^2+y^2+z^2)}\Leftrightarrow$
$\Leftrightarrow(4(x^2+y^2+z^2)-xy-xz-yz)^2\geq3(x^2+y^2+z^2)(x+y+z)^2\Leftrightarrow$
$\Leftrightarrow16(x^2+y^2+z^2)^2-8(x^2+y^2+z^2)(xy+xz+yz)+(xy+xz+yz)^2\geq$
$\geq3(x^2+y^2+z^2)^2+6(x^2+y^2+z^2)(xy+xz+yz)\Leftrightarrow(13u-v)(u-v)\geq0,$
where $u=x^2+y^2+y^2$ and $v=xy+xz+yz.$
The last inequality is obviously true. And our inequality is proved.
More way is SOS.

[quote="arqady"][quote="Julia Robet"][quote="arqady"]$\frac{(x^2+y^2+z^2)^2}{x^2y+x^2z+y^2x+y^2z+z^2x+z^2y}\geq\frac{1}{2}\cdot\sqrt{3(x^2+y^2+z^2)}=$[/quote]
Why??[/quote]
$\frac{(x^2+y^2+z^2)^2}{x^2y+x^2z+y^2x+y^2z+z^2x+z^2y}\geq\frac{1}{2}\cdot\sqrt{3(x^2+y^2+z^2)}\Leftrightarrow$
$\Leftrightarrow4(x^2+y^2+z^2)^3\geq3(x^2y+x^2z+y^2x+y^2z+z^2x+z^2y)^2\Leftrightarrow$
$\sum_{cyc}(4x^6+9x^4y^2+9x^4z^2+2x^2y^2z^2-6x^4yz-6x^3y^3-6x^3y^2z-6x^3z^2y)\geq0\Leftrightarrow$
$\Leftrightarrow2\cdot\sum_{cyc}(x^6-x^4y^2-x^4z^2+x^2y^2z^2)+$
$+\sum_{cyc}(2x^6+11x^4y^2+11x^4z^2-6x^4yz-6x^3y^3-6x^3y^2z-6x^3z^2y)\geq0$
The last inequality is true by Schur and Muirhead.[/quote]I don't like Muihear because it' s very strong and no popular :( .
[quote="arqady"]
Let $x^2+y^2=a^2,$ $x^2+z^2=b^2$ and $y^2+z^2=c^2,$ where $a\geq0,$ $b\geq0$ and $c\geq0.$ Then
$\frac{1}{2\sqrt2}\cdot\sqrt{3\cdot\sum_{cyc}(x^2+y^2)}\geq\frac{1}{2\sqrt2}\cdot\sum_{cyc}\sqrt{x^2+y^2}\Leftrightarrow3(a^2+b^2+c^2)\geq(a+b+c)^2.$
Other way to the proof that $H\geq\frac{1}{2}\sqrt{3(x^2+y^2+z^2)}:$
$\frac{x}{y+z}\geq\frac{8x-y-z}{4(x+y+z)}\Leftrightarrow(2x-y-z)^2\geq0.$
Hence, $H\geq\frac{1}{4(x+y+z)}\cdot\sum_{cyc}x(8x-y-z)=\frac{4(x^2+y^2+z^2)-xy-xz-yz}{2(x+y+z)}.$
Remain to prove that $\frac{4(x^2+y^2+z^2)-xy-xz-yz}{x+y+z}\geq\sqrt{3(x^2+y^2+z^2)}.$
But   $\frac{4(x^2+y^2+z^2)-xy-xz-yz}{x+y+z}\geq\sqrt{3(x^2+y^2+z^2)}\Leftrightarrow$
$\Leftrightarrow(4(x^2+y^2+z^2)-xy-xz-yz)^2\geq3(x^2+y^2+z^2)(x+y+z)^2\Leftrightarrow$
$\Leftrightarrow16(x^2+y^2+z^2)^2-8(x^2+y^2+z^2)(xy+xz+yz)+(xy+xz+yz)^2\geq$
$\geq3(x^2+y^2+z^2)^2+6(x^2+y^2+z^2)(xy+xz+yz)\Leftrightarrow(13u-v)(u-v)\geq0,$
where $u=x^2+y^2+y^2$ and $v=xy+xz+yz.$
The last inequality is obviously true. And our inequality is proved.
More way is SOS.[/quote]
Hi, it's only : $\sqrt{3(x^2+y^2+z^2}\ge x+y+z$.
It's easy  ;) .

Julia Robet, you don't permit to post and discuss this problem now. Because it was publised in the last issue of a Maths magazine in Vietnam. And this problem was the same: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88009
Please Admin lock these topics

Chebyshev works also.
Edit: I don't see Nameless's post.

We have this inequality:
$\ H\geq\sqrt{3(x^2+y^2+z^2)}$
And $\sqrt{6(x^2+y^2+z^2)}\geq\sum\sqrt{x^2+y^2}$
=>$\ H\geq\frac{1003\sqrt{2}}{2}$ :lol:


What I do when I want to start a new $\text{\TeX}$ file is to open an existing one and simply delete all the stuff in the middle.

[quote="Rep123max"]Is there a way to have all new projects on Texnic Center to begin with all the formatting stuff without having to type it there? By formatting stuff [/quote]Yes, you can do this by using Templates.

Inside your TeXnicCenter program directory you'll find a Template directory and inside that another folder called Documents. [list][*]Make a new folder inside that called 'Pre-formatted' or anything else you fancy
[*]In TeXnicCenter create a document with the formatting you want
[*]Save the document in the tempate directory you created.[/list]Then in future, when creating a new document, go to File, New from Template (or press Ctrl+N or put an icon on the toolbar) and choose your template.

You may wish to have lots of different templates for different occasions.


I try to prove that the equality

$ \frac{1}{2n} (1+\frac{1}{3}+\frac{1}{5}+\ldots+\frac{1}{4n-1}) -
\frac{1}{2n+1} (1+\frac{1}{3}+\frac{1}{5}+\ldots+\frac{1}{4n+1})-
\frac{1}{(2n+1)^2}\\
=2(\frac{1}{1\cdot (4n+1)}+\frac{1}{3\cdot (4n-1)}+\frac{1}{5\cdot (4n-3)}+\ldots\frac{1}{(2n-1) \cdot (2n+3)}\\
-\frac{1}{1\cdot (4n-1)}-\frac{1}{3\cdot (4n-3)}-\frac{1}{5\cdot (4n-5)}-\ldots-\frac{1}{(2n-1)\cdot (2n+1)})$

holds for every $ n\in \mathbb{N}$ and even do not know how to start... Could you please give me some advice?

[quote="solenoid"]I try to prove that the equality

$ \frac {1}{2n} (1 \plus{} \frac {1}{3} \plus{} \frac {1}{5} \plus{} \ldots \plus{} \frac {1}{4n \minus{} 1}) \minus{} \frac {1}{2n \plus{} 1} (1 \plus{} \frac {1}{3} \plus{} \frac {1}{5} \plus{} \ldots \plus{} \frac {1}{4n \plus{} 1}) \minus{} \frac {1}{(2n \plus{} 1)^2} \\
\equal{} 2(\frac {1}{1\cdot (4n \plus{} 1)} \plus{} \frac {1}{3\cdot (4n \minus{} 1)} \plus{} \frac {1}{5\cdot (4n \minus{} 3)} \plus{} \ldots\frac {1}{(2n \minus{} 1) \cdot (2n \plus{} 3)} \\
\minus{} \frac {1}{1\cdot (4n \minus{} 1)} \minus{} \frac {1}{3\cdot (4n \minus{} 3)} \minus{} \frac {1}{5\cdot (4n \minus{} 5)} \minus{} \ldots \minus{} \frac {1}{(2n \minus{} 1)\cdot (2n \plus{} 1)})$

holds for every $ n\in \mathbb{N}$ and even do not know how to start... Could you please give me some advice?[/quote]
My first thought was to use this identity:
$ \frac {1}{ab} \equal{} \frac {1}{b \minus{} a} \left ( \frac1{a} \minus{} \frac1{b} \right )$

E.g. $ 2\frac {1}{1\cdot (4n \plus{} 1)} \equal{} \frac {1}{2n} \left ( \frac {1}{1} \minus{} \frac1{4n \plus{} 1} \right )$.

I'm going to bed now, so I can't help you, but this may get you started :)

Mathias, thanks for hint!
Indeed, if we apply the equality $ \frac {1}{ab} \equal{} \frac {1}{b\plus{}a} \left ( \frac1{a} \plus{} \frac1{b} \right )$, we obtain

$ \frac {1}{2n} (1 \plus{} \frac {1}{3} \plus{} \frac {1}{5} \plus{} \ldots \plus{} \frac {1}{4n \minus{} 1})\\\equal{}
2[\frac1{4n} (1\plus{}\frac1{4n\minus{}1})\plus{}\frac1{4n}(\frac1{3}\plus{}\frac1{4n\minus{}3})\plus{}\ldots\plus{}\frac1{4n}(\frac1{2n\minus{}1}\plus{}\frac1{2n\plus{}1})]\\\equal{}
2[\frac {1}{1\cdot (4n \minus{} 1)} \plus{} \frac {1}{3\cdot (4n \minus{} 3)} \plus{} \ldots\frac {1}{(2n \minus{} 1) \cdot (2n \plus{} 1)}]$

and

$ \frac {1}{2n \plus{} 1} (1 \plus{} \frac {1}{3} \plus{} \frac {1}{5} \plus{} \ldots \plus{}\frac{1}{2n\plus{}1}\plus{}\ldots\plus{} \frac {1}{4n \plus{} 1}) \minus{} \frac {1}{(2n \plus{} 1)^2}\\ \equal{}
2[\frac1{2(2n\plus{}1)} (1\plus{}\frac1{4n\plus{}1})\plus{}\frac1{2(2n\plus{}1)}(\frac1{3}\plus{}\frac1{4n\minus{}1})\plus{}\ldots\plus{}\frac1{2(2n\plus{}1)}(\frac1{2n\minus{}1}\plus{}\frac1{2n\plus{}3})]\\\equal{}
2[\frac {1}{1\cdot (4n \plus{} 1)} \plus{} \frac {1}{3\cdot (4n \minus{} 1)} \plus{} \ldots \plus{} \frac {1}{(2n \minus{} 1)\cdot (2n \plus{} 3)}]$.

After taking the difference we obtain what we wanted to prove.


and me. :)
We pwned.  We have won for 20 years.

*Sorry.  This is a reply to wz118's post.

unneeded revival...
Yea, sure, but Jongao contributed to the team and I didnt because of 1. ONE tiebreaker question I didnt bother doing.

Its all my fault.


algebra
polynomial
superior algebra
superior algebra unsolved

is it possible that in a Neother Ring or more specifically, $ k[x_1,\cdots k_n]$, there exist $ f_1,\cdots f_n$ that $ \sqrt{(f_i)}\equal{}\sqrt{(f_1,\cdots f_n)}$?
I guess not.
this question come up to me when trying to prove, for a proj variety $ Y$, O(Y)=k.
 if there exist n polynomials $ f_i$ that $ \frac{f_i}{x_i^{d_i}}\equal{}\frac{f_j}{x_j^{d_j}}$ on k[X]/I, then this means $ (f_i)(x_j^{d_j})\equal{}(f_j)(x_i^{d_i})$, take root and assume $ \sqrt{(x_i)}\equal{}{all}$ on k[X]/I,

Of course there are. For example: all $ f_i$'s are equal. Less trivially, we can have $ f_r \equal{} x^r$ for some fixed variable $ x \equal{} x_i$.


what is

[(3x+2y)^2(3x-2y)^2]^3

yeah..
if u cant understand it

find 29 on nats sprint 2007

[quote="stevenmeow"]isnt the question how many terms are in that expression

the expression is $ \left(\left(3x \plus{} 2y\right)^2\left(3x \minus{} 2y\right)^2\right)^3 \equal{} \left(\left((3x \plus{} 2y)(2x \minus{} 2y)\right)^2\right)^3 \equal{} \left(9x^2 \minus{} 4y^2\right)^6$
and this has 7 terms (dont know how to explain this but i hope i helped)[/quote]

shouldn't it be $ (6x^2 \minus{} 4y^2)^6$ at the end?

[quote="stevenmeow"]isnt the question how many terms are in that expression

the expression is $ \left(\left(3x \plus{} 2y\right)^2\left(3x \minus{} 2y\right)^2\right)^3 \equal{} \left(\left((3x \plus{} 2y)(3x \minus{} 2y)\right)^2\right)^3 \equal{} \left(9x^2 \minus{} 4y^2\right)^6$
and this has 7 terms (dont know how to explain this but i hope i helped)
EDIT: fixed typo[/quote]

Helping you explain: The exact expressions $ 9x^2$ and $ \minus{} 4y^2$ don't matter anymore, so just replace this with $ a$ and $ b$. So we get $ (a \plus{} b)^6$... look familiar? Yes, the binomial expansion!

$ (a \plus{} b)^6 \equal{} \sum_{i \equal{} 0}^6 \binom 6i a^ib^{6 \minus{} i}$ which clearly has 7 terms, each corresponding to when $ i \equal{} 0, 1, 2, 3, 4, 5,$ and $ 6$.

ohhh, i figured out what i did wrong
i just messed up on the binomial theorem
i thought i had the wrong method.


geometry
geometric transformation
reflection
analytic geometry

Using the grid below, follow the clues to find the location of the treasure.
Start at $ (\minus{}1, \minus{}1)$.  Follow the line $ y\equal{}x$ until $ x\equal{}4$.
Reflect this new point over the $ x$-axis.
From this new point add seven to the $ y$-coordinate
	and subtract five from the $ x$-coordinate.
This is the location of the treasure.
How far is the treasure point from the starting point?

[asy]for(int i = 0; i < 11; ++i){
    draw((i,0)--(i,10)); draw((0,i)--(10,i));
    }

draw((5,0)--(5,10),linewidth(2bp));
draw((0,5)--(10,5),linewidth(2bp));

draw((11,11)--(-1,-1),Arrows);

label("$y$",(5,10),N);
label("$x$",(10,5),E);
label("$y=x$",(11,11),E);[/asy]

[asy]for(int i = 0; i < 11; ++i){
    draw((i,0)--(i,10)); draw((0,i)--(10,i));
    }

draw((5,0)--(5,10),linewidth(2bp));
draw((0,5)--(10,5),linewidth(2bp));

draw((11,11)--(-1,-1),Arrows);

label("$y$",(5,10),N);
label("$x$",(10,5),E);
label("$y=x$",(11,11),E);

label("$S$",(4,4));
label("$1$",(9,9));
label("$2$",(9,1));
label("$3$",(9,8));
label("$G$",(4,8));[/asy]

$ S$:Start at $ (\minus{}1, \minus{}1)$.
$ 1$:Follow the line $ y\equal{}x$ until $ x\equal{}4$.
$ 2$:Reflect this new point over the $ x$-axis.
$ 3$:From this new point add seven to the $ y$-coordinate.
$ G$:From this new point subtract five from the $ x$-coordinate. This is the location of the treasure.

Thus, a position of the treasure is the place that from start point $ \boxed{\text{add four to the y\minus{}coordinate}}$.


If a1,a2,.., an>0, n>=3 then the following inequality holds:
(a1-a3)/(a2+a3)+(a2-a4)/(a3+a4)+...+(an-a2)/(a1+a2)>=0.

[i]unknown source[/i]

[quote="Lagrangia"]If a1,a2,.., an>0, n>=3 then the following inequality holds:
(a1-a3)/(a2+a3)+(a2-a4)/(a3+a4)+...+(an-a2)/(a1+a2)>=0.[/quote]

Add n to both sides.

We must prove

(a1 + a2)/(a2 + a3) + ... + (an + a1)/(a1 + a2) \geq n.

Let b1 = a1 + a2, b2 = a2 + a3, ..., bn = an + a1.

Then we must prove b1 / b2 + ... + bn / b1 \geq n.

But that is trivial by rearrangement inequality.
That is, suppose we have A = {b_c1, b_c2, ... , b_cn} in nondecreasing order.
Then B = {1/b_c1, 1/b_c2, ..., 1/b_cn} is in nonincreasing order.
Then the minimal sum obtained by taking each element of A and multiplying it with an element of B is \sum b_ci * 1/b_ci, which is n.


The point of this game is to name reasons for why 1337 is important. 1 point per reason, and a max of 3 reasons per post.

The person with the most points wins!

1337 is important b/c...lolz... this is hard... b/c its like probably part of a zip code somewhere... :ninja:  :rotfl:

[quote="batteredbutnotdefeated"]wats so good about it? wats so good about 42?[/quote]

you could always try wikipedia...

$ 42$ folds of a sheet of paper of normal thickness is almost exactly the distance from the earth to the moon.

[quote="tcjy"][quote="batteredbutnotdefeated"]wats so good about it? wats so good about 42?[/quote]

42=6*7

4+2=6

42 was featured in "The hitchiker's guide to the galaxy" (a book)[/quote]

"The meaning of life is... 42...  :huh:"  most people don't even know where that's from... i already know without wanting to... i read the whole increasingly innaccurately named trilogy...

1337:
The reversal of the digits of $ 1337^{1000}$ is a prime number!!!

42:
The sum of the least consecutive primes ending in "1" (42 = 11 + 31).

In the first book of his Hitchhiker "Trilogy" Douglas Adams writes, "42 is the answer to life, the universe, and everything." Note that 42 lies between a twin prime pair {41, 43}.

 42 = (pi(4)*pi(2)) * (prime(4)*prime(2)). Note that 42 is the only number less than 20000000 with this property.

$ 42^{42} \plus{} 42\pi$ is prime. It is interesting that all numbers less than 42 with this property are are all prime factors of 42.

All five numbers 1 + 42, 1 + 4242, 1 + 424242, 1 + 42424242 and 1 + 4242424242 are prime.

[quote="Thunder365"]$ 42^{42} \plus{} 42\pi$ is prime.[/quote]

Because an irrational number is certainly prime, right?

the word 1337 is the only word that is a noun that is compromised completely of numbers. i think.

6337 = beet your wrong

1337 is important because...um...its leetspeak!!(lol)

1337=13 and 37
ok
does not make sense
 :lol:

[quote="meewhee009"]42 is 2 plus 40 (:[/quote]

where'd you get 42?

1337 is special because people bother to make a thread about it

[quote="batteredbutnotdefeated"]wats so good about it? wats so good about 42?[/quote]
@Chompy Here ^


linear algebra
matrix
algebra
polynomial
vector

Is the only linear transformation/ matrix T: V->V where A = A^2 the identity transformation/matrix?

 thanks

Key word: idempotent.

Absolutely not! There are many, many idempotent matrices. (You didn't say that $ V$ was finite dimensional, but I'll start with that assumption.)

Fact: if $ P(A)\equal{}0$ for some polynomial $ P,$ then every eigenvalue $ \lambda$ of $ A$ satisfies $ P(\lambda)\equal{}0.$

Therefore if $ A$ is idempotent - that is, if $ A^2\equal{}A$ - then the only possible eigenvalues of $ A$ are $ 0$ and $ 1.$

If all of the eigenvalues of $ A$ are $ 1$ and $ A$ is idempotent, then $ A\equal{}I.$ That is, the identity is the only invertible idempotent matrix.

If all of the eigenvalues of $ A$ are $ 0$ and $ A$ is idempotent, then $ A$ is both nilpotent and idempotent - and the only possibility that that that leaves is for $ A$ to be the zero matrix.

But in between those two extremes - for $ A$ that have both some eigenvalues of $ 1$ and some eigenvalues of $ 0$ - there is a wealth of examples.

Some ideas: The simplest is the block matrix $ \begin{pmatrix}I_k&0\\0&0\end{pmatrix}$ that has the first $ k$ entries on the main diagonal equal to $ 1$ and all other entries equal to zero.

Or you could let $ J_n$ be the $ n\times n$ matrix of all $ 1$'s. Then $ A\equal{}\frac1nJ$ is an idempotent of rank $ 1.$

(Fun fact: for an idempotent matrix, the rank equals the trace.)

If $ u$ is a unit vector (written as a column vector), meaning that $ u^Tu\equal{}1,$ then $ A\equal{}I\minus{}uu^T$ is an idempotent of rank $ n\minus{}1.$ It's the orthogonal projection onto the hyperplane perpendicular to $ u.$

So far, my examples have been real symmetric matrices. But there are non-symmetric idempotents. For instance, $ A\equal{}\begin{bmatrix}1&17\\0&0\end{bmatrix}$ is idempotent.

thanks a lot for the reply :)

 I have a question that asks me to determine all of such transformations, how would I go about doing this? :S thanks again

The one thing you do know is that all idempotents have to be diagonalizable. Why? There's the identity, the zero, and then all of the other (nontrivial) idempotents. Every non-trivial idempotent has as its [b]minimal polynomial[/b] the polynomial $ x(x\minus{}1).$ A matrix is diagonalizable precisely when its minimal polynomial can be completely factored but contains no repeated factors.

So that's a way to characterize them - work backwards from the diagonal case.

[quote="Kent Merryfield"]The one thing you do know is that all idempotents have to be diagonalizable. Why? There's the identity, the zero, and then all of the other (nontrivial) idempotents. Every non-trivial idempotent has as its [b]minimal polynomial[/b] the polynomial $ x(x \minus{} 1).$ A matrix is diagonalizable precisely when its minimal polynomial can be completely factored but contains no repeated factors.

So that's a way to characterize them - work backwards from the diagonal case.[/quote]

 Sorry, I don't know what a minimal polynomial is supposed to be, neither do I know about diagonalizibility.. It's because my class hasn't gone over these concepts yet.. Is there a way to show this using more elementary linear algebra? thanks

[quote="Carcul"][quote]Is the only linear transformation/ matrix T: V->V where A = A^2 the identity transformation/matrix?[/quote]

No. Any linear transformation whose matrix representation, relative to some base, is $ \begin{bmatrix}I_r & 0 \\
0 & 0 \end{bmatrix}$ is idempotent and is not equal to the identity. Is this enough?[/quote]
 Yes, thanks, I realized this after Kent Merryfield's post. How can I "determine all such transformations"?

Thanks


$ a, b, c$ are distinct rational roots of $ x^3\plus{}ax^2\plus{}bx\plus{}c\equal{}0$. Find the values of $ a,b,c$.

[quote="ertanrock"]$ a, b, c$ are distinct rational roots of $ x^3 \plus{} ax^2 \plus{} bx \plus{} c \equal{} 0$. Find the values of $ a,b,c$.[/quote]
Replacing $ a,b,c$
$ 1) 2a^3\plus{}ba\plus{}c\equal{}0$
$ 2) b^3\plus{}ab^2\plus{}b^2\plus{}c\equal{}0$
$ 3) c(c^2\plus{}ac\plus{}b\plus{}1)\equal{}0$

From $ 3)$ $ c\equal{}0$
Replacing in $ 2)$ $ 0\equal{}b^2 (b\plus{}a\plus{}1)$,  $ b\plus{}a\plus{}1\equal{}0$, $ b\equal{}\minus{}1\minus{}a$

Replacing in $ 1)$ $ 2a^3\plus{}(\minus{}1\minus{}a)a\equal{}0$, $ 2a^2\equal{}1\plus{}a$
$ a\equal{}1$ for $ b\equal{}\minus{}2$ or
$ a\equal{}\minus{}\frac{1}{2}$ for $ b\equal{}a$
so $ (a,b,c)\equal{}(1,\minus{}2,0)$

notice that $ \minus{}abc\equal{}c$ then $ ab\equal{}\minus{}1$

From $ 2)$, $ b^3\plus{}b^2\minus{}b\plus{} c \equal{} 0$
if $ b\equal{}0$ then $ c\equal{}0$ contradiction then $ a\equal{}\minus{}\frac{1}{b}$

From $ 1)$, $ 2a^3\plus{}c\minus{}1\equal{}0$
$ c\equal{}\frac{2}{b^3}\plus{}1$

For $ b^3\plus{}b^2\minus{}b\plus{}\frac{2}{b^3}\plus{}1\equal{} 0$, $ b^6\plus{}b^5\minus{}b^4\plus{}2\plus{}b^3\equal{} 0$ , $ b^6\plus{}b^5\minus{}b^4\plus{}b^3\plus{}2\equal{} 0$
only possible rational roots $ \minus{}2,\minus{}1,1,2$
Checking those cases, only solution for $ b\equal{}\minus{}1$ but $ c\equal{}b$

$ f(x)\equal{}x^3\plus{}ax^2\plus{}bx\plus{}c$ have roots $ a,b,c$.
We have
$ a\plus{}b\plus{}c\equal{}a (1)$
$ ab\plus{}bc\plus{}ca\equal{}\minus{}b (2)$
$ abc\equal{}c (3)$
$ b\equal{}\minus{}c$ (1)

$ 1)$If $ c<>0$ and $ b<>0$
 $ a\equal{}1/b$ (3)

 $ ab\plus{}bc\plus{}ca\equal{}\minus{}b$
 $ 1\minus{}b^2\minus{}1\equal{}\minus{}b$
 $ b\equal{}0$ or $ b\equal{}1$
 But  $ b<>0$
 We have $ (a,b,c)\equal{}(1,1,\minus{}1)$
 But $ f(1)\equal{}2<>0$

$ 2)$$ b\equal{}0,c\equal{}0$
$ f(x)\equal{}x^3\plus{}ax^2\equal{}x^2(x\plus{}a)$
$ a$ is root of $ f(x)$
$ 0\equal{}f(a)\equal{}2a^3$
$ a\equal{}0$
We have one solution $ (a,b,c)\equal{}(0,0,0)$

if $ f(x)\equal{}x^3\plus{}ax^2\plus{}bx\plus{}c$ and $ a,b,c$ were roots then
$ a\plus{}b\plus{}c \equal{} \minus{}a ...(1)$
$ ab\plus{}bc\plus{}ca \equal{} b...(2)$
$ abc\equal{}\minus{}c ...(3)$
from (3) we have, $ c(ab\plus{}1)\equal{} 0 \Rightarrow c\equal{}0$ or $ ab\plus{}1\equal{}0$
now suppose that $ c\equal{}0$, we will get from (2)  $ ab\equal{}b \Rightarrow a\equal{}1$
and from (1) $ 1\plus{}b \equal{}\minus{}1 \Rightarrow b\equal{}\minus{}2$
$ \therefore a\equal{}1, b\equal{}\minus{}2$ and $ c\equal{}0$ is a solution

But i cant reach to anything when i put $ ab\plus{}1\equal{}0$
anyone can?


geometry
circumcircle
Euler
trigonometry
geometry proposed

Let $ ABC$ be a tiragnle. Let $ X,Y,Z$ be a point on $ BC,CA,AB$ such that triangle $ XYZ$ is similar to $ ABC$.Let $ H,O$ be orthocenter,circumcenter of triangle $ ABC$ and $ O^{'}$ be a circumcenter of trinalge $ XYZ$. Prove that $ O^{'}H \equal{} O^{'}O$.

and What is source of this problem?

[hide="hint"]
I think,it will be useful:
$ O$ is orthocenter of $ \triangle XYZ$.[/hide]

Please Erken, if you have a solution, then share it with us; this is an Olympiad forum, and no Intermediate section.
If you have no solution, then there's no point giving hints.

Very nice problem Leonhard Euler!  :lol:
One can immediately see the starting point for this proof, inspired by the following theorem:
It is well known and easy to prove that, given a point $ P$ inside a triangle, the triangles $ XYZ$ and $ DEF$ are similar, where $ X,Y,Z$ are on $ BC,AC,AB$ such that $ \angle PXC \equal{}\angle PYA \equal{}\angle PZB \equal{}\alpha$, and $ DEF$ is the pedal triangle of $ P$. (We call such a triangle $ XYZ$ an $ \alpha$-projection triangle)

Unfortunately, I can't help you as far as the source is concerned, but I can present [b][u]my solution:[/u][/b]

I claim that the inverse result is in this case also true:
If a triangle $ XYZ$ is as described in Leonhard Euler's problem, then it must be true that $ \angle OXC \equal{}\angle OYA \equal{}\angle OZB$ (i.e.: then $ XYZ$ is an $ \alpha$-projection triangle.)
[hide="Why?"]Well, draw an arbitrary line $ UV$ with $ U,V$ on $ AC,AB$ respectively. Now draw point $ W$ such that $ \Delta UVW$ is similar to $ \Delta BCA$.
If we repeat this construction with new segments, all parallel to $ UV$, we obtain that the geometrical locus of points $ W$ is a line through $ A$. This line as a unique intersection with $ BC$, so given a point $ X$ on $ BC$, there is exactly one triangle $ XYZ$ such that $ XYZ$ and $ ABC$ are similar. Because we can also find a $ \alpha$-projection triangle $ XYZ$ of $ O$ for each $ X$ on $ BC$, we come to the important conclusion:

[b]If $ XYZ$ satisfies the conditions as stated in the problem, then $ XYZ$ has to be an $ \alpha$-projection triangle.[/b] [/hide]
Because $ XYZ$ is an $ \alpha$-projection triangle, then clearly $ AZOY, BXOZ, CYOX$ are cyclic. This gives us:
$ \angle YOX \equal{} 180^{\circ}\minus{}\angle C$ and $ \angle ZYO \equal{}\angle ZAO \equal{} 90^{\circ}\minus{}\angle C$, yielding that $ XO\bot YZ$.
Analogue arguments for $ YO$ and $ ZO$ show us that $ O$ is the orthocenter of $ XYZ$.

Call $ M$ the midpoint of $ BC$.
We now know that $ \angle O'OH$ is the angle between the Euler lines of $ \Delta XYZ$ and $ \Delta ABC$. Because $ \Delta ABC$ has the same Euler line as its medial triangle, we conclude that $ \angle O'OH \equal{}\angle XOM$.
This makes us conclude that $ \frac{OO'}{OH}\equal{}\frac{2OM}{OX}\equal{} 2\cos{XOM}$

We deduce that $ \left|OO'\right| \equal{}\left|O'H\right|$.

[img]http://img166.imageshack.us/img166/9995/leonhardeulermathlinkseoz8.jpg[/img]

Nice solution, Jan. Another two proofs for the preliminary result which states that $ O$ is the orthocenter of $ \triangle{XYZ}$ can be found at: [url]http://www.mathlinks.ro/viewtopic.php?p=810795#810795[/url].


Too lazy too think hard,but do u do somethin like keep x+y=a and x/y=b,substitue them and u can then solve it. :huh:

[quote="eminem"]hey guys, a really stupid IIT question.......

x<0, y<0

x+y+x/y= 1/2
and
(x+y)x/y = -1/2

find x,y

this is from IIT JEE 1990 , I couldnt believe my eyes when i saw it! wish IIT JEE was that easy nowadays!![/quote]
Are you sure????
You should latex your posts!
[hide="Sol"]
$x+y=a$
$\frac{x}{y}=b$
$a+b=\frac{1}{2}$
$ab=\frac{-1}{2}$
$b=\frac{-1}{2a}$
$a+\frac{-1}{2a}=\frac{1}{2}$
$\frac{2a^2-1}{a}=1$
$2a^2-a-1=0$
$2a^2-2a+a-1=0$
$2a(a-1)+1(a-1)=0$
$a=\frac{-1}{2}$
$a=1$
$x+y=1$
$\frac{x}{y}=\frac{-1}{2}$
$2x=-y$
$x-2x=1$
$x=-1$
why isnt it coming? :?  :(
[/hide]

[quote="eminem"]hey i see the mistake ya made riddler, you got x/y = 1 or -1/2 now u took it to be -1/2 but thats wrong as x is negative , y is  negative , so x/y has to be positive , hence x/y=1 and not -1/2

now do it , it should come  ;)[/quote]
sorry maybe, I was feeling sleepy. :blush:


Find all f(x) functions R to R such that f(2007)=2008 and for all x,y real numbers

f(4xy) = 2y(f(x+y) + f(x-y))


probability
real analysis
real analysis theorems

What is an intuitive explanation of the result: "given $(A_{n})$ independent events of a probability space $(\Omega,T,P)$ such that $\sum{P(A_{n})}=+\infty$, $P(\limsup A_{n})=1$" ? Intuitively, why wouldn't there be a "hole" with positive probability such that the hypotheses still hold ?

Note that I know the proof and I know what the result means but I'd like some intuitive sense of it ...

Excluding the trivial case $\exists n\ P(A_{n})=1$, one can restate the assumption $\sum{P(A_{n})}=+\infty$ as $\prod (1-P(A_{n}))=0$. This makes it clear what is going on: the product of probabilities of independent events "not $A_{n}$" is zero.

Well, the point I was trying to make is that I cannot understand why it works intuitively ...


how to prove that intersection point of heights,
circumcenter and centroid within triangle are collinear?

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=84848&start=61]this post[/url] and the next one.


algebra polynomial function calculus derivative induction search

Prove that for any polynomial $ P\left(x\right)$ with degree $ <n$, we have
 $ \sum_{k \equal{} 0}^{n} \left( \minus{} 1\right)^n \frac {n!}{k!\left(n \minus{} k\right)!}P\left(k\right) \equal{} 0$.

Exactly
 $ \sum_{k \equal{} 0}^{n} ( \minus{} 1)^k \frac {n!}{k!(n \minus{} k)!}P(k) \equal{} 0$, were $ n> deg(P(x)).$
Let S is operator $ P(x)\to P(x\plus{}1)\minus{}P(x).$ Then it followed from $ S^n(P(x))|_{x\equal{}0}\equal{}\sum_{k\equal{}0}^n(\minus{}1)^kC_n^kP(k)$ and $ S^n(P(x))\equiv 0$.

i  study  university tpu  - tomsk city- sibip

[quote="Yustas"]Indeed, a very beautiful problem. I'll give you a hint: first, it is enough to prove for $ P(x) \equal{} x^p$.
Next, consider the function $ f(x) \equal{} \sum_{k \equal{} 0}^{n}( \minus{} 1)^{n \minus{} k}\binom{n}{k}e^{kx} \equal{} (e^x \minus{} 1)^n$. Now look at the derivatives when $ x \equal{} 0$ and use induction on $ deg(P)$.
Where did you learn russian? :lol:[/quote]
We can prove with Lagrange law at the points :$ \{0,...,n\}$
Are you Nguyen Hong Quy at Vinh University?

This theorem has numerous applications in calculating combinatorics.For example:
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1400603014&t=41800

  Also it would be useful for some problems in AMM and Maths & Young magazine (in my country!)...

But in my opinion,it's not so strong ,or I dont know how to "exhaust" it  :?:


arithmetic sequence
number theory proposed
number theory

If $ a,b,c$ are positive real numbers, do $ a,b,c,a\plus{}b\minus{}c,b\plus{}c\minus{}a,c\plus{}a\minus{}b,a\plus{}b\plus{}c$ form a seven term arithmetic progression in some order?

[quote="hell_ever"]If $ a,b,c$ are positive real numbers, do $ a,b,c,a \plus{} b \minus{} c,b \plus{} c \minus{} a,c \plus{} a \minus{} b,a \plus{} b \plus{} c$ form a seven term arithmetic progression in some order?[/quote]

Answer is no.

Wlog say $ a\geq b\geq c$
Let $ S\equal{}a\plus{}b\plus{}c$
The seven numbers are $ S\minus{}(b\plus{}c)$, $ S\minus{}(a\plus{}c)$, $ S\minus{}(a\plus{}b)$, $ S\minus{}2c$, $ S\minus{}2a$, $ S\minus{}2b$, $ S$
Obvously, six of these numbers may be easyly ordered :

$ S\minus{}2a$, $ S\minus{}(a\plus{}b)$, $ S\minus{}2b$, $ S\minus{}(b\plus{}c)$, $ S\minus{}2c$ and $ S$
The seventh, $ S\minus{}(a\plus{}c)$ is somewhere between $ S\minus{}(a\plus{}b)$ and $ S\minus{}(b\plus{}c)$

So the commun difference is $ d\equal{}a\minus{}b\equal{}2c\equal{}b\minus{}c$ and we have $ b\equal{}3c$, $ a\equal{}5c$, $ S\equal{}9c$ and the six ordered numbers are :
$ \minus{}c$, $ c$, $ 3c$, $ 5c$, $ 7c$ and $ 9c$
The seventh is $ 3c$

And so, the seven numbers are the terms of a six-term arithmetic progression and not seven-term arithmetic progression (we know that $ c\neq 0$ since the problem says "positive" real numbers).


The cost of a bus trip was 180 dollars. the people who signed up for the trip agreed to split the cost equally .however ,six people did not show up , so that those who did go each had to pay 1.50 dollars more. How many people actually went on the trip

Let $ x$ be the number of people who were supposed to show up. If they had all showed up, each person's fee would be $ \frac{180}{x}$. However, because $ 6$ people did not show up, each person's fee was $ \frac{180}{x\minus{}6}$. Thus we can set up an equation: $ \frac{180}{x} \plus{} 1.5 \equal{} \frac{180}{x\minus{}6}$. Solving this equation, we have $ x \equal{} 30$ and $ x \minus{} 6 \equal{} \boxed{24}$.


geometry
geometric transformation
rotation

In a circular arrangement of 9 light bulbs, each light bulb can either be ON or OFF. How many different ways can the lights be turned on or off if rotations of the same arrangement are considered to be the same?

Generally, when rotations are not distinct, you should find the what the answer would be if rotations were distinct, and then divide by the number of original objects.

[hide]This problem has a twist though ... There are $2^{9}$ ways to turn the lights on (rotations distinct), but if you try to divide by 9, the thing explodes. There is something else going on here.[/hide]
[hide="hint2"]If the lights are 100000000, for instance, there are 9 rotations of it that all count as the same combination (100000000, 010000000, 001000000, etc.).

However, if the lights are turned on 001001001, are there really 9 rotations that you need to account for, or just 3?[/hide]


inequalities
function
inequalities solved

Let A1,A2,...,A[2n] be a set of positive real numbers and let
B1,B2,...,B[2n] be this same set of numbers arranged in  order of magnitude so that

                         B1 \geq B2 \geq ... \geq B[2n]

Prove that

B1B2...Bn + B[n+1]B[n+2]...B[2n] \geq A1A2...An + A[n+1]A[n+2]...A[2n]

This old inequality is known as "LeCointe's inequality".

It is not very difficult, the most complicated is to have a clear proof (!)
We have A(k) = B(p(k)) , where p is a permutation(bijective function) . p : {1,2,...,n} --> {1,2,...,n}
The proof is trivial if all A(k)=B(k) (i.e. p = identity).
Suppose now p is not the identity function, and note :
Gc = {1,..,n}&{p(1),..,p(n)} , Gd = {1,..,n}-Uc ; Lc = {n+1,..,2*n}&{p(n+1),..,p(2*n)} , Ld = {n+1,..,2*n}-Lc
Then : {1,..,n} = (Gc)u(Gd) ,  {n+1),..,2*n} = (Lc)u(Ld)
{p(1),..,p(n)} = (Gc)u(Ld) ,  {p(n+1),..,p(2*n)} = (Lc)u(Gd)
Xc = Product(k in Uc){B(k)} , Xd = Product(k in Ud){B(k)}
Yc = Product(k in Lc){B(k)}, Yd = Product(k in Ld){B(k)}.
Manifestly Gd,Ld have the same number of elements, so it is easy to see : Xd>= Yd, Xc >= Yc
So the initial inequality holds :
Xc*Xd + Yc*Yd >=  Xc*Yd + Yc*Xd <===> (Xc - Yc)*(Xd-Yd) >= 0

Of course, in my precedent message the permutation must be  p:{1,2,...,2*n}-->{1,2,...,2*n}


For 0<a,b,c<1 real numbers prove that

a^2+b^2 +c^2  \leq a^2*b+b^2*c +c^2*a +1

Posted before. Source:

[i]Italy 1993[/i]

Can you please tell me the post name or title ?

Thank you very much.

Consider it as a function in a,b,c. You will find it is convex and from now on everything is clear ( to be rigourous, take 0<=a,b,c<=1).

Strange - I can't find it anymore but I'm sure it was posted.

So, a solution:

Fix b, c - we want to prove f(a) = a(1 - b) - a(c) + (b + c - bc) \leq 1.

Now notice that f is a convex function of a, so it suffices to check a = 0, a = 1.

So the maximum can occur only if each of a, b, c equals 0 or 1.
This gives an easy case check.

Another solution:
  sum a^2-sum a^2b=sum a^2(1-b)<=sum a(1-b). Thus, we have to prove that a+b+c-ab-bc-ca<=1, which is simple: the function in three variables is linear.


how do you know when you should use power of a point concepts?

[hide]
In a circle, if two chords intersect, the products of the lengths of the pieces of each cord will be equal

[/hide]

that's correct C^2, but that's only one of the power of a point theorems. usually power of a point is useful when you have a circle with segments intersecting, and ur looking for the length of a particular segment.

So you should basically use it when you have a circle, then with tangents or secants, or chords.

[quote="abacadaea"]anyone have a solution to
[url]http://www.mathlinks.ro/Wiki/index.php/1971_Canadian_MO_Problems/Problem_1[/url][/quote]

The solution is right underneath the problem.


$ n \equiv 1 \pmod{8} \implies n^2 \equiv 1 \pmod{8}$

$ n \equiv 3 \pmod{8} \implies n^2 \equiv 9 \pmod{8} \implies n^2 \equiv 1 \pmod{8}$

$ n \equiv 5 \pmod{8} \implies n^2 \equiv 25 \pmod{8} \implies n^2 \equiv 1 \pmod{8}$

$ n \equiv 7 \pmod{8} \implies n^2 \equiv 49 \pmod{8} \implies n^2 \equiv 1 \pmod{8}$

This takes care of all odd numbers, so if $ x$ and $ y$ are odd, then $ x^2 \equiv y^2$, and thusly, $ (x^2 \minus{} y^2) \equiv 0 \pmod{8}$.

[quote="Lizzie91"]Let x,y be two odd integers.
Then, is $ (x^{2} \minus{} y^{2})$ a multiple of 8 always?[/quote]
If $ x$ is odd then
$ x\equal{}2k\plus{}1 \rightarrow x^2\equal{}4k^2\plus{}4k\plus{}1\equal{}4k(k\plus{}1)\plus{}1$
But $ k(k\plus{}1)$ is even so $ k(k\plus{}1)\equal{}2m$
So $ x\equal{}8m\plus{}1$
And respectively $ y\equal{}8n\plus{}1$
So $ x\minus{}y\equal{}8(m\minus{}n)$


[quote="7h3.D3m0n.117"][quote="ckck"][quote="7h3.D3m0n.117"]The equation MATH-COU=NTS can be made true if each of the letters is replaced by a different digit and the letter groups are read as 4-digit and 3-digit integers. Both Ts have the same value. If C=3, what is the value of the letter 0?

This seems waaayyy too easy it makes sense no matter how I check it but it is a Workout problem...[/quote]
[hide]Because t-o=t, the only digit that can do this is 9 so o=9.[/hide][/quote]

what??

[hide]
isn't the answer 0 because t-0=t?

[/hide][/quote]
My bad. I think you're right.

if U > H then you must borrow 1 from T.

then O could be 9.

however, if U<H then there is no need to borrow and O = 0.

[quote="7h3.D3m0n.117"][quote="7h3.D3m0n.117"][quote="ProtestanT"]what if U > H? then o would be 9[/quote]

impossible because both Ts have the same value...therefore it MUST be 0[/quote][/quote]
I disagree. Just because you borrow from T doesn't mean you actually change its value.
Look at this for example:
 ATH - OU = TS.
 151 - 92 = 59.
 157 - 99 = 56.If the 5s are "T"s, then they stay the same because they are the same when O = 9!  Borrowing doesn't change "T"s actual value.

[quote="archimedes1"][/quote][quote="7h3.D3m0n.117"][quote="7h3.D3m0n.117"][quote="ProtestanT"]what if U > H? then o would be 9[/quote]

impossible because both Ts have the same value...therefore it MUST be 0[/quote][/quote][quote="archimedes1"]
I disagree. Just because you borrow from T doesn't mean you actually change its value.
Look at this for example:
 ATH - OU = TS.
 151 - 92 = 59.
 157 - 99 = 56.If the 5s are "T"s, then they stay the same because they are the same when O = 9!  Borrowing doesn't change "T"s actual value.[/quote]

Your reasoning is very well thought out except for: "...different digit" as said in the original question

[quote="7h3.D3m0n.117"][/quote][quote="archimedes1"][/quote][quote="7h3.D3m0n.117"][quote="7h3.D3m0n.117"][quote="ProtestanT"]what if U > H? then o would be 9[/quote]

impossible because both Ts have the same value...therefore it MUST be 0[/quote][/quote][quote="archimedes1"]
I disagree. Just because you borrow from T doesn't mean you actually change its value.
Look at this for example:
 ATH - OU = TS.
 151 - 92 = 59.
 157 - 99 = 56.If the 5s are "T"s, then they stay the same because they are the same when O = 9!  Borrowing doesn't change "T"s actual value.[/quote][quote="7h3.D3m0n.117"]

Your reasoning is very well thought out except for: "...different digit" as said in the original question[/quote]
He gave an example of where o=9, not the actual solution

[quote="ajai"]Look at it this way:

MATH
- COU
_____
  NTS

which is:

MA[b]T[/b]H   If T remains the same, wouldn't O have to be 0?
- 3[b]O[/b]U
_____
   N[b]T[/b]S[/quote]
No, that is not correct. If H+U>9, then o must equal 9 for the two "t"s to be equal.


What happened to Missouri's super awesome jedi force predictions skizillz? I was very disappointed.

 :starwars:  :starwars:  :starwars:

Nope. All the states' results are official. At least we were the closest state this year with  49.5% Mccain and 49.3% Obama. (5886 vote difference)

2000 was 576 in Florida, right? Was that the closest ever?

The official result was around 900 vote Bush lead after recounts. Some sources say that Bush won only by 154 votes. Anyway, a 900 vote lead in Florida is $ .015%
$ percent. I think thats the closest.... :rotfl:

Wait, actually the officail result was 193 vote Bush lead according to

[url]http://www.cnn.com/ELECTION/2000/results/[/url]

which is a $ 0.003$ percent lead.  :o  :o

Uh...another reason (the?) constitution (I really don't think it should be capitalized) sucks. I mean, how do people spend time debating something stupid like abortion or gay rights? Seriously.

How many times has Al Gore's son been arrested?

No, I read in a book that it was 154 vote lead but the official result according to CNN is 193 Bush vote lead

http://www.cnn.com/ELECTION/2000/results/

Yeah, sorry! So close to discovering one of the hundreds of reasons my usernames have 154.

That is one of the great mysteries of the world.

OMG! Upon further research, apparently there was a dispute over whether 215 or 176 votes from Palm Beach County should be added for Gore. They would give Bush a 154 or 193 vote lead respectively. However, the Florida Supreme Court ruled that Bush's lead should be 193 votes. :(

BUT WHY IS IT MATH154? is that the reason?

Sorry, no. You still have lots of time to figure out.

No, I'm afraid not. :wink: Too bad it's not so good for procrastinators.

Fine, fine,... the stupidest reason why it's 154--Runescape and recommendations. Figure out yourself.

[quote="AIME15"]Why do you all play runescape? thats a stupid game.[/quote]

I don't play. I was referring to the olden days.

I was sick of 20 hour days, skipping bathroom breaks, breakfast lunch and dinner, all so I could get another Firemaking level.

Im serious. I woudl stay up until my parents had gone to bed, then run and get in another 6 hours.

my dad kills our internet at 10:30 so i cant do that crap :(

Ah, AIME, don't feel sorry for yourself. Without suffering, there would be no improvement. Asian motto, Benjamin Franklin's quote.

Stupid idiots.

Benjamin franklin was 100% asisisisisaan.