How to solve this: $\lim_{t \to \infty} -t^{x-1} e^{-t}$; x is real.

You can mess around with some other methods, but as far as I know at this moment L'Hospital cures this patient with ease.
$\lim_{t\to\infty} \frac{-t^{x-1}}{e^t} = \lim_{t\to\infty} \frac{-(x-1)t^{x-2}}{e^t} = ... = \lim_{t\to\infty} \frac{product of finite number of finite terms assuming x is finite}{e^t} = 0, x \not= \infty$
As x is real, then the limit is always $0$.
By the way what I said in squashed italics is, "product of finite number of finite terms assuming x is finite".


function
algebra
rational function
algebra unsolved

[quote="hipatitusb"]Call me brain dead, but I cannot think of a real world application for a second degree or higher polynomial. Or a rational function, 1/x, 1/(x^2), etc. What I really need is a function whose real-world usefulness only applies to a certain domain. Like, it generally makes no sense to have run negative miles. Anyway, if someone thinks of a good application, post it!
Thanks, hipatitusb.[/quote]

Cubic functions can be applied to problems involving volume, so minimizing usage of material for a given volume. Negative numbers also have use. Ask you bank manager whether he thinks -10 in your account means something,


A biased coin is such that the probability of a head in each toss is 1/3. What is the probability of having k heads in 10 tosses?

[hide="is it..."]the binomial theorem?

As in,
$ (\frac{1}{3}+\frac{2}{3})^{10}$

So the probability of getting $ k$ heads is
$ \binom{10}{k}(\frac{1}{3})^{k}(\frac{2}{3})^{10-k}$[/hide]


A five digit positive integer is a "mountain number" if the first three digits are in ascending order and the last three digits are in descending order. For example, 35 761 is a mountain number, but 32 323 and 35 655 are not. How many 5 digit numbers greater than 70 000 are mountain numbers?

How do you do this without writing them all out?

[hide="hint"]The number must start with 789.[/hide]

[hide="another hint"]the last two digits are distinct digits from the set 0~8, ordered.[/hide]

[hide="yet another hint"]There are 9 choose 2 = 36 ways to do that. And that's the answer.[/hide]

great last hint james. you almost gave it away (oh wait, you did give it away)

I'd rather have him do what he did than just give the solution outright.

Thank you James. That was helpful. :clap2:  :D


calculus
integration
trigonometry
logarithms
LaTeX
calculus computations

1. integral of (x^2)/2(1+x^4) dx

2. integral of (sinx)^2 dx

[hide=" hint for 1"]

You want the denominator to match a well known trigonometric identity. [/hide]

[hide=" hint for 2"]

See 1, it's another trig identity.
[/hide]

I still can't get the answer for #2, any more hint?

#2

Note:

$ \sin^2x \equal{} \frac{1\minus{}\cos{2x}}{2}$

That was what I did, but I can't solve the problem further. :(

Remember that $ \int \cos{u}\,du\equal{}\sin{u}\plus{}C$. I think you're looking for a $ \sin{x}$ attached to something it shouldn't be.

$ \int \sin(x^2)\,dx$ is not an elementary function. $ \int (\sin(x))^2\,dx$ is. Can you integrate a constant? Can you integrate $ \cos(cx)$?

I think for the first:
Multiply and divide by 2:
$ \int \frac {2\times x^{2}}{4\times (1 \plus{} x^{4})}\,dx \equal{} \frac {1}{4} ( \int \frac {x^{2} \plus{} 1}{x^{4} \plus{} 1}\,dx \plus{} \int \frac {x^{2} \minus{} 1}{x^{4} \plus{} 1}\,dx$)
Then divide both Nr. and Dr. of both the integrals by $ x^{2}$ and then substitute $ x \minus{} \frac {1}{x} \equal{} y$ for the first and $ x \plus{} \frac {1}{x} \equal{} z$ for the second to get the answer.

that's the perfect method for the first one....

For the second one, milin's method is working...

$ \int \frac{x^{3n\minus{}1}}{1 \plus{} x^{4n}}\,dx \equal{} \frac{1}{4n \sqrt{2}} \left[2 \arctan\left(\frac{x^{2n}\minus{}1}{\sqrt{2} x^n}\right) \plus{} \ln \left(\frac{x^{2n}\minus{} \sqrt{2}x^n \plus{} 1}{x^{2n} \plus{} \sqrt{2}x^n \plus{} 1}\right)\right] \plus{} C$

[quote="cdymdcool"]
I know that. But the integral of cosx doesn't apply for the integral of sin(x^2)?[/quote]

This is not the thing you wrote in your first post!

[quote="cdymdcool"]
2. integral of (sinx)^2 dx[/quote]

$ \text{\LaTeX}$ note: using latex isn't that hard. There is a good tutorial to this way of symbol writing on this forum!!


Wow. I haven't been here since the "Elder Days" of AoPS... but now that there's a Biology forum, I'm back. =D Hey everyone... *points to title* I guess you can tell what part of bio I'm crazy about, lol. If you have any questions about stuff in that area, feel free to post or PM me. ^^ *waves*

--Hann

yeah im the biomaster...lol, but yeah im really glad biology is a forum now. my favorite areas of biology are ecology/evolutionary biology and molec./cell/developmental bio

anatomy and physiology is cool, shobber. heres the thing that ive noticed about biology classea round the world:

many other countries spend less time on anatomy/physiology and spend more time on other topics in the broad scope of biology like plants and zoology and ecology. in the US, most of the biology classes ive seen that are normal cover anatomy, genetics and DNA, and thats about it. they really restrict their stuff (im sure there are exceptions, and most AP bio classes cover it all well). what do they emphasize more in china?

personally, i think the anatomy stuff is not only conceptually easier, but a little less interesting than theoretical and unifying stuff like evolution and endosymbiotic hypotheses and stuff...but true, anatomy/physiology IS one of the ultimately cool applications for all of biology. one thing that draws me to ecology is how life is structured in the world and how life interacts with its abiotic and biotic environments at all different scales. (one of the things that makes biology cool is its super organization from the chemistry of life with molecules and atoms all the way through ethology and the behavior of living things and how everything fits together real well). pretty much all biology rules.

yeah, one of the amazing things about bio is the vast range of applications it has to everything around us. even technology is based largely off biology - engineers tend to begin by imitating nature - or strive to do so, as in SciAm's latest issue on neuromorphic chips.

personally I'm a microbiologist through and through. Bio I love because of how it can help people, microbio because of the immense complexity that can be contained in a tiny cell. I've always had a frightening (so I'm told) interest in diseases and learning more about them... if I were a US citizen, my dream would be to work in CDC's NCID someday. microbes are to bio as black holes are to physics. Seduisante. =D

i will try to tell you as much as i can about the things we learn in China, but it takes time......it is very difficult for a Chinese student to understand biology in English...

This is semi-relevant to the topic.

I am going to be starting at university this coming September. I'll be an Molecular and Cell Biology (MCB) major, with emphasis in Genetics, at UC Berkeley. I'm thinking about changing my emphasis to immunology, but I am not yet sure what I want to do with biology. I am thinking pre-med, or possibly research. As of now, I just wanna leave my options open....

Hi all.

Ankur- way to go! i myself am looking into the MCB program at UC Berkeley as an option...it is notorious for being very good. as an MCB major, your horizons are really broadened. you gan get your MD if you want to do physical medicine, or Ph.D. for research. theres really a lot you can do!

Is anyone else as fascinated by prions and viroids as I am? :lol:

[quote="4everwise"]I am really interested medicine and cures for certain diseases, although this isn't related to biology at school. One day, I saw this cool NOVA program about a possible cure for cancer that involved the use of cutting of blood to a tumor.

For more information, you can watch the whole movie online at the PBS website.[/quote]
High school biology contains the most basic knowledges, but has little about medicine. At least it is in China.

i like the genetics the most and physiology and systematics
Ripple your microbiology has so small creatueres that my eyes really find it difficult to figure them out   :?:  can you help me a bit to understand the concepts in immunology :?:  :?:  :?:  how it works. but the PRIONS always attract me


1. Given that $ a,b,c$ are positive such that $ abc \equal{} 1$, prove that
$ \frac {ab}{c^5(a \plus{} b)^2}\plus{}\frac {bc}{a^5(b \plus{} c)^2}\plus{}\frac {ca}{b^5(c \plus{} a)^2}\geq\frac34$.

2. Given that $ a,b,c$ are positive and $ a\plus{}b\plus{}c\equal{}3$, find the minimum value of $ \frac{a^2}{b^2\plus{}c^2\minus{}bc}\plus{}\frac{b^2}{a^2\plus{}c^2\minus{}ac}\plus{}\frac{c^2}{a^2\plus{}b^2\minus{}ab}$.

For 1

Multiply both sides by $ a^{5}b^{5}c^{5}$ to get $ \frac{a^{6}b^{6}}{(a\plus{}b)^{2}}\plus{}\frac{b^{6}c^{6}}{(b\plus{}c)^{2}}\plus{}\frac{c^{6}a^{6}}{(c\plus{}a)^{2}} \ge \frac{3(abc)^{5}}{4}\equal{}\frac{3}{4}$.

Then, by Power-Mean, we need to prove that $ \frac{a^{3}b^{3}}{a\plus{}b}\plus{}\frac{b^{3}c^{3}}{b\plus{}c}\plus{}\frac{c^{3}a^{3}}{c\plus{}a} \ge \frac{3}{2}$.

This is equivalent to $ \frac{1}{c^{3}(a\plus{}b)}\plus{}\frac{1}{a^{3}(b\plus{}c)}\plus{}\frac{1}{b^{3}(c\plus{}a)} \ge \frac{3}{2}$, given that $ abc\equal{}1$.

Letting $ a\equal{}\frac{1}{x}$, $ b\equal{}\frac{1}{y}$, $ c\equal{}\frac{1}{z}$.

This gives the equivalent inequality of
$ \frac{x^{2}}{y\plus{}z}\plus{}\frac{y^{2}}{z\plus{}x}\plus{}\frac{z^{2}}{x\plus{}y} \ge \frac{3}{2}$.

Then, by Cauchy-Scwarz, all we have to do is prove that if $ xyz\equal{}1$, then $ \frac{x\plus{}y\plus{}z}{2} \ge \frac{3}{2}$.

This is just true by AM-GM, as $ x\plus{}y\plus{}z \ge 3(xyz)^{\frac{1}{3}}\equal{}3$.

[quote="mathgeniuse^ln(x)"]
This is equivalent to $ \frac {1}{c^{3}(a \plus{} b)} \plus{} \frac {1}{a^{3}(b \plus{} c)} \plus{} \frac {1}{b^{3}(c \plus{} a)} \ge \frac {3}{2}$, given that $ abc \equal{} 1$.
[/quote]

That's an [url=http://www.mathlinks.ro/viewtopic.php?p=365178#p365178]IMO Problem[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=25778]Generalization of the same[/url]


search
number theory unsolved
number theory

Hi Everyone!
First of all i would like to know what TST is called in Iran?
All i'm sure of is that its not iran's national olympiad. Like for example the first question of iran's national MO in the second round was this:
$n$ is a natural number bigger than one and $p$ is a prime number. Also $\frac {p-1}{n}$ and $\frac{n^3-1}{p}$ are natural numbers. Prove that $4p-3$ is a square.
I solved this halfway since i'm not really experinced in the MO :(. If any one solved it please post it. Thanks! :lol:
If you are interested in the other questions tell me and i'll post it for you ;)

I think this question is posted at least 3 times in the number theory section. Maybe you can search about it.


geometry
analytic geometry
graphing lines
slope
calculus
conics
parabola

Let point A be at (1,4). If a line passes through A and both the x and y axis, what is the area of the smallest triangle that can be formed in the first quadrant with its sides being the x-axis, y-axis, and the line.

[hide="big hint"]let the slope be m, then get an equation for the area in terms of m. Then use either calculus or your knowledge of parabolas to find the min.[/hide]

If it passes through $ (1,4)$ then it has equation $ \minus{}m(x\minus{}1)\plus{}4\equal{}y$

$ x\equal{}0$ gives the $ y$-intercept to be at $ 4\plus{}m$

$ y\equal{}0$ gives the $ x$ intercept to be $ 1\plus{}4/m$

for the triangle to be in the first quadrant, we have m is positive.

Hence, we must minimize $ (4\plus{}m)(1\plus{}4/m)$ which is $ \ge (2\plus{}2)^2\equal{}16$ by cauchy-schwartz w/ equality when $ m/(4/m)\equal{}4/1$ <--> $ m^2\equal{}16$ <--> m=4.

Hence the min. area is 16.

dont you have to divide that by 2 then? (for the area of the triangle formula)
 :D

then the answer is $ \boxed{8}$


if $ x \bigcirc y \equal{} (x^y)^x$, what is the units digit of $ 4 \bigcirc 3$?

This is equivalent to $ (4^3)^4\equal{}4^{12}\equal{}2^{24}$. Units digits of powers of $ 2$ cycle through every four digits, so the units of $ 2^{24}$ is equivalent to the units digit of $ 2^4$, which is $ \boxed{6}$.


search
number theory
combinatorics solved
combinatorics

show that at least one subset of four elements from any seven integers has a sum that is divisible by 4.  Is there a particularly elegant/short solution without going through every case?  is there a generalization?

It is a well_known result from Erds and someone else (Graham?) that for each positive integer n and each set S of 2n-1 integers, there exists a subset A of S, with |A| = n such that the sum of the elements of A is divisible by n.

The proof is not so difficult and use unduction on n, starting with a proof of the result in the case where n is a prime. I have the feeling that it may have been proposed before on that forum...

Pierre.

I had a feeling that this question might have been posted before so i did a search on this forum however I did not find it.

See, for example, http://www.mathlinks.ro/Forum/viewtopic.php?highlight=nullstellensatz&t=19496#

[quote="pbornsztein"]It is a well_known result from Erds and someone else (Graham?)
[/quote]

No, not Graham  :) ; actually, his co-authors are A. Ginzburg and A. Ziv [Theorem in additive number theory, Bull Research Council Israel 10 (1961), 41--43.]


1)Let {a_1,a_2,a_3,a_4,a_5}={1,2,3,4,5}.Find the maximum and minumum value of a_1+2a_2+3a_3+4a_4+5a_5=?

2)For positive integers x,y,z satisfies
x+y+z=90  and
x/n=y/n+1=z/n+2 for n element of positive integer.How many positive (x,y,z) integers can be found that satisfy these conditions?

3)A is a set that all elements are smaller than 102.A does not include the sum of any elements.What can be the maximum number of element of A?

1 is just the rearrangement inequality. Maximum is 55, minimum is 35.

2 is just simple algebra: x = y+n = z+2n, so the numbers are 30-n, 30, 30+n, so we have 29 solutions.

I've seen 3 around the forums recently. Have a search for it.

(And if possible, change your source field to the actual name of the books, so that it is useful.)


Slove this equation: $ 2cos2x\minus{}sin2x\equal{}2(sinx\plus{}cosx)$

It is equivalent to

$ 2(1 \minus{} 2\cos^2 x) \minus{} 2\sin x \cos x \equal{} 2(\sin x \plus{} \cos x)$

Divide by 2 to get

$ 1 \minus{} 2\cos^2 x \minus{} \sin x \cos x \equal{} \sin x \plus{} \cos x$

or

$ 1 \minus{} \cos^2 x \equal{} \sin x (1 \plus{} \cos x) \plus{} \cos x(1 \plus{} \cos x)$

$ (1 \minus{} \cos x)(1 \plus{} \cos x) \equal{} (\sin x \plus{} \cos x)(1 \plus{} \cos x)$

$ 1) 1 \plus{} \cos x \equal{} 0, x \equal{} (2k\plus{}1) \pi$

$ 2) 1 \plus{} \cos x \neq 0,$

then

$ 1 \minus{} \cos x \equal{} \sin x \plus{} \cos x$,

$ x \equal{} 2k\pi \plus{} \frac {\pi}{2}$


superior algebra
superior algebra theorems

I am trying to show that every element of the quotient group G = Q/Z has finite order, but I'm not entirely sure where to start. Hints?


function
algorithm
algebra unsolved
algebra

Are there rational functions f, g and h of rational coefficients such that
$(f(x))^{3}+(g(x))^{3}+(h(x))^{3}=x$
i find somthing but i wait for an admirable Algorithme or généralization.
good luck. :D


email
trigonometry
geometry
geometry unsolved

email:
             edson_laura_galvez@hotmail.com

Add me I am of Perú

[quote="EDSON_"][size=134][color=red][b]Geometers of the world, help me to resolve ![/b] I am of Peru.[/color][/size][/quote]
$0<m(\widehat{ABE})=180-10x$ $\Longrightarrow$ $\boxed{\ 0<x<18\ }\ .$ Apply the Sinus' theorem in the triangles $ABC$, $BCE$, $ABE$ :

$\frac{BA}{BC}=\frac{\sin 2x}{\sin 7x}\ ;\ \frac{BC}{EC}=\frac{\sin E}{\sin x}\ ;\frac{EA}{BA}=\frac{\sin 10x}{\sin E}\ .$ From the product of these relations obtain : $\sin 2x\cdot\sin 10x=\sin 7x\cdot\sin x$ $\Longleftrightarrow$ .....................................................

Exactly it is what lacking me;  solves the trigonometrical equation :(

If someone could for pure geometry?

i think virgil didnt meant that, did he?

he is one of the good geometers.but not the best geometer of the

world :!:

SOME TIME AGO THAT DOES NOT GO OUT FOR ME THIS PROBLEMITA :(

[quote="Amir.S"][quote="Virgil Nicula"][quote="EDSON_"][size=134][color=red][b]Geometer of the world, help me to resolve ![/b] I am of Peru.[/color][/size][/quote]
$0<m(\widehat{ABE})=180-10x$ $\Longrightarrow$ $\boxed{\ 0<x<18\ }\ .$ Apply the Sinus' theorem in the triangles $ABC$, $BCE$, $ABE$ :

$\frac{BA}{BC}=\frac{\sin 2x}{\sin 7x}\ ;\ \frac{BC}{EC}=\frac{\sin E}{\sin x}\ ;\frac{EA}{BA}=\frac{\sin 10x}{\sin E}\ .$ From the product of these relations obtain : $\sin 2x\cdot\sin 10x=\sin 7x\cdot\sin x$ $\Longleftrightarrow$ .....................................................[/quote]

you remove "s" from Geometers and you think only you are a geometer ,ha?:rotfl:

hhmmmm,wait,let's see . do you think you are  a geometer?? :rotfl:[/quote]

Edson is so sorry that was not what he want to say

$0<m(\widehat{ABE})=180-10x$ $\Longrightarrow$ $\boxed{\ 0<x<18\ }\ .$ Apply the Sinus' theorem in the triangles $ABC$, $BCE$, $ABE$ :

$\frac{BA}{BC}=\frac{\sin 2x}{\sin 7x}\ ;\ \frac{BC}{EC}=\frac{\sin E}{\sin x}\ ;\frac{EA}{BA}=\frac{\sin 10x}{\sin E}\ .$ From the product of these relations obtain : $\sin 2x\cdot\sin 10x=\sin 7x\cdot\sin x$ $\Longleftrightarrow$ $2\cos x\sin 10x=\sin 7x$ $\Longleftrightarrow$

$\sin 9x+\sin 11x=\sin 7x$ $\Longleftrightarrow$ $\sin 9x=-2\sin 2x\cos 9x$ $\Longleftrightarrow$ $\tan 9x=-2\sin 2x<0$ $\Longrightarrow$ $10<x<18\ .$ Observe a solution $x_{1}=15\ldots\ldots$

holla guyz i hope you can understand what i do. i send the synthetic solution.


The two triangles shown are similar.  What is the perimeter of the larger triangle?
[metapost=0a09fbbb47847f1803caa944ec131504] save u,A,B,C; pair A,B,C;

 % Set the value of u to scale the entire image
 u:=.25cm;

 A:=origin;
 B:=5u*dir(-20);
 C:=4u*dir(-20-78.46);

 draw A--B--C--cycle;
 draw A--(2B)--(2C)--cycle shifted (10u*dir(20));

label(btex $ 4$ cm etex,origin) rotated (160-78.46) shifted (2.3u*dir(-130));
label(btex $ 5$ cm etex,origin) rotated (-20) shifted (2.5u*dir(0));
label(btex $ 7$ cm etex,origin) rotated (20) shifted (4.7u*dir(-60));

label(btex $ 8$ cm etex,origin) rotated (160-78.46) shifted (8u*dir(0));
label(btex $ 10$ cm etex,origin) rotated (-20) shifted (15u*dir(10));
[/metapost]

Could you please take a screenshot and post it, as we can not see the diagram?

Presumably, one triangle has lengths 4, 5, 7 and the other has lengths 8, 10. Because the ratios of corresponding sides must be equal, the last side is 14. So the answer is 32 (8+10+14).


I don't know if this is really factoring but you can try this:

$ \displaystyle\frac {(x(x\plus{}k)(x\minus{}k))\minus{}1}{x}$

If you just multiply through using the order of operations, then simplify it, you will get the same expression. However, I'm not sure whether this is really considered factoring or not. Sorry I can't help much more.

When all else fails, use the quadratic formula. Something like this I believe?

$ \left(x \minus{} \frac{k^2 \plus{} \sqrt{k^4 \plus{} 4}}{2}\right)\left(x \minus{} \frac{k^2 \minus{} \sqrt{k^4 \plus{} 4}}{2}\right)$

Hmm... my first post outside of other topics & P&B.


Let G be a graph on n points with at least (n^2) /4 + 1 edges. Can one color G using two colors ?

No, we cannot color $G$ in 2 colors (i.e. it is not bipartite). Suppose it was bipartite, then we can divide its vertices into 2 groups of sizes $p$ and $q$, such that no two vertices of the same group are connected. Then the maximum number of edges there can be are $pq$. But, $p+q=n$, so max value of $pq$ is $\frac{n^2}{4}$.

Turan's theorem (for the triangles) states that if a graph has $n$ vertices and no triangle then it has at most $frac {n^2} 4 $ edges.

Pierre.

The graph $G$ of $2n$ vertices has at least $n^2+1$ edges. Prove that $G$ contains at least $n$ triangles. ;)


Let f(x) be a differentiable function in (0,2), then what is the value
2
∫ f(x) dx  ?
0

It could be anything. Is that really the question you wanted to ask?

is it 2f(c) where c belongs to (0,2)? if it is so , prove it

I guess the answer is yes as this is an immediate consequence of the First Mean Value Theorem for Integrals.


logarithms
integration
calculus
calculus computations

It's clear to me that for reals the normal definition of exponentiation is unsuitable, and so it becomes necessary to define

$a^{b}= e^{b \ln a}$

Where $\ln x = \int_{1}^{x}\frac{dt}{t}$ and $e^{x}= \ln^{-1}x$

Now, something that's always bothered me is how one goes about proving logarithmic (and therefore exponential) properties in this way.  For example, how does one go about showing that

$\ln (ab) = \ln a+\ln b$

?  This is equivalent to showing that

$\int_{1}^{ab}\frac{dt}{t}= \int_{1}^{a}\frac{dt}{t}+\int_{1}^{b}\frac{dt}{t}$

Which is to me far from obvious.

A simple proof of that nonobvious fact:

$\int_{a}^{ab}\frac1x\,dx=\int_{1}^{b}\frac1{at}\cdot a\,dt=\int_{1}^{b}\frac1x\,dx$, using the substitution $x=at$.

It's also reasonably nice working with those integrals straight from the definition; just use geometric partitions.


Among any four people in a room, there is one who knows the other three. Show that there is a person who knows everybody in the room.

Assuming we have at least four people...

[hide]
Suppose, for every person, there exists another person in the room that they do not know. There are two cases:

(1) There exist $A, B, C, D$ such that $A$ does not know $B$ and $C$ does not know $D$. Then none of these four people know the other three. Contradiction.

(2) There exist $A, B, C, D$ such that $A$ does not know $B, C, D$. Again, none of these four people know the other three. Contradiction.

So for some person in the room, there doesn't exist another person that they do not know (i.e. they know everyone).[/hide]


Can you recommend a internet site where there are good stuff on real analysis? thanks!


Point P lies inside ABC, and is such that PAC = 18°, PCA = 57°, PAB = 24°, and PBA = 27, calculate ise m(PBC)=?

[hide="Answer"]$ 15^{\circ}$.[/hide]
[hide="Hint"]Use Ceva's theorem in trigonometric form.[/hide]
Similar problems:
http://www.mathlinks.ro/viewtopic.php?t=259144
http://www.mathlinks.ro/viewtopic.php?t=258638


Of the following statements, the only one that is incorrect is:

$ \textbf{(A)}$ An inequality will remain true after each side is increased, decreased, multiplied or divided (zero excluded) by the same positive quantity.
$ \textbf{(B)}$ The arithmetic mean of two unequal positive quantities is greater than their geometric mean.
$ \textbf{(C)}$ If the sum of two positive quantities is given, ther product is largest when they are equal.
$ \textbf{(D)}$ If $ a$ and $ b$ are positive and unequal, $ \frac {1}{2}(a^2 \plus{} b^2)$ is greater than $ [\frac {1}{2}(a \plus{} b)]^2$.
$ \textbf{(E)}$ If the product of two positive quantities is given, their sum is greatest when they are equal.

E is incorrect. Suppose the number is $ 100$; this can be written as $ 10*10$ or $ 5*20$ (among other possibilites). But clearly, $ 10\plus{}10<20\plus{}5$.

[quote="MCTP"]E is incorrect. Suppose the number is $ 100$; this can be written as $ 10*10$ or $ 5*20$ (among other possibilites). But clearly, $ 10 \plus{} 10 < 20 \plus{} 5$.[/quote]
Yes, in fact, you can have an infinite sum... for example, if the number is 1, pick a very very huge (positive) number and its reciprocal to get the greatest possible sum. And having the numbers equal is the lowest possible sum. :)


function
induction
algebra unsolved
algebra

Find all functions $f:$ $\mathbb R \rightarrow \mathbb R$ such that
$f(x+y)+f(x)f(y)$ $=$ $f(x)(1+y)+(1+x)f(y)+f(xy)$

[hide="Hint"]There are three solutions.
constant,linear,quadratic.[/hide]

I think I found the the solutions hinted by pardesi, but I still can't complete the proof.

[hide]Let's call
$E(x,y)$: $f(x+y)+f(x)f(y)=f(x)(1+y)+(1+x)f(y)+f(xy)$.

First of all, $E(x,0)$ $\to$ $f(x)f(0)=(2+x)f(0)$ $\to$ $f(0)=0$,
because we can check $f(x)=2+x$ is not a solution.

Now, $f(0)=0$, $E(1,-1)$ $\to$ $f(1)f(-1)=3f(-1)$.

Case 1) $f(0)=0$, $f(1)=3$,
$E(x,1)$ $\to$ $f(x)=3x$, we can check it's a solution!

Case 2) $f(0)=0$, $f(-1)=0$,
$E(x,-1)$ $\to$ $f(x-1)=f(-x)$, so $f$ is symmetric around $x=\frac{-1}{2}$.
In particular, $f(1)=f(-2)=a$, and
$E(-2,1)$ $\to$ $a^{2}=2a$, so $a=0$ or $a=2$.
Either way,
$E(x-1,1)$ $\to$ $f(x)=(3-a)f(x-1)+ax$.

Case 2.1) $f(1)=a=0$, $f(x)=3f(x-1)$, and by induction $f(n)=0$ for integer $n$.
Case 2.2) $f(1)=a=2$, $f(x)=f(x-1)+2x$, and by induction $f(n)=n^{2}+n$ for integer $n$.[/hide]

yes $f(x)=3x,x^{2}+x,0$
are the solutio :) ns but the last part of yours needs some more ...

first case
[hide]
assume $f(n)=0$ for integers and $f(x+n)=3^{n}f(x)$
plug $y=y+n$
$3^{n}(f(x+y)+f(x)f(y))=f(x)(y+n+1)+3^{n}f(y)(1+x)+f(x(y+n))$
$f(x(y+n))+f(x)(y+n+1)=3^{n}(f(x)(y+1)+f(xy))$.take $x=2$
$f(2y+2n)=3^{n}f(2y)=f(2y+2n)=3^{2n}f(2y)$ so $f(y)=0$ for all real $y$
[/hide]
second case
[hide]
i will use $f(n)=n^{2}+n$ for integers to prove $f(x)=x^{2}+x$ for reals
set $f(x)=g(x)+x^{2}+x$.now $g(n)=0$ for integers.
from the condition we have
$g(x+y)+g(x)g(y)+y^{2}g(x)+x^{2}g(y)=g(x)+g(y)+g(xy)$ for all $x,y$ (*)
$y=1$ gives us $g(x+1)=g(x)\to g(x+n)=g(x)$ where $n$ integer,$x$ real.
now plug in $y+n$ instead of $y$.
$g(x+y)+g(x)g(y)+(y+n)^{2}g(x)+x^{2}g(y)=g(x)+g(y)+g(x(y+n))$ and now with (*)
$g(x(y+n))-g(xy)=(n^{2}+2ny)g(x)$for $n$ integer and real $x,y$. take $n=2$ and $y=-1$
$g(x)=g(-x)$.put $y=-y$ in (*), combine with (*) and you get $g(x-y)=g(x+y)$ and now take $x=y=a/2$ and you have $g(a)=g(0)=0$ for all real $a$
[/hide]
very nice :)


complex numbers
number theory
relatively prime

As proposed by t0rajir0u

Determine all triplets $ (a,b,c)$ of natural numbers such that $ a^2 \plus{} b^2 \equal{} 5^c$

But we have other examples like $ 7^2\plus{}24^2\equal{}25^2\equal{}5^4$, so it has to be more than that...

Lets look in $ \mathbb{Z}[i]$.

$ (a \plus{} bi)(a \minus{} bi) \equal{} [(1 \plus{} 2i)^n(1 \minus{} 2i)]^n$

Let $ x \equal{} a \plus{} bi$ and $ y \equal{} a \minus{} bi$. $ x$ and $ y$ are conjugates. By unique factorization in $ \mathbb{Z}[i]$, we have that $ x$ can be written as a product of the primes $ (1 \plus{} 2i)$ and $ (1 \minus{} 2i)$ up to units which are $ 1, \minus{} 1,i, \minus{} i$ which just switch around $ a$ and $ b$. So we can ignore the units and write $ u \equal{} (1 \plus{} 2i)^e(1 \minus{} 2i)^f$. Then $ v$ being the conjugate of $ u$ is $ (1 \plus{} 2i)^f(1 \minus{} 2i)^e$. Multiplying these gives $ e \plus{} f \equal{} n$. So

$ u \equal{} a \plus{} bi \equal{} (1 \plus{} 2i)^e(1 \minus{} 2i)^{n \minus{} f}$

for $ f \equal{} 0,1,2,...,n$. It is easy to see that all of these solutions work so we're done. If you want $ a,b$ explicitly, then note that $ a$ is the real part of $ u$ which is $ \frac {1}{2}(u \plus{} \overline{u})$.

A comment: a nicer problem in my opinion is find all triples given that $ (a,b)\equal{}1$...

Both cases are fairly easy to describe; if, WLOG, $ e \ge n \minus{} f$ then $ a \plus{} bi \equal{} 5^{n\minus{}f} (1 \plus{} 2i)^{e\minus{}n\plus{}f}$ and the relatively prime case just corresponds to $ n \equal{} f$.

Anyway, I wasn't too serious about proposing this generalization - the really interesting generalization is to $ a^2 \plus{} b^2 \equal{} n$, where it is in fact possible to describe all solutions; see, for example, the discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=219017]here[/url].

I thought that the Gaussian integers didn't have unique factorization... $ (1 \plus{} i)\cdot(1 \plus{} i) \equal{} 2\cdot i$, all of which are primes...

edit: Never mind, did some research, 2 is not prime... which is obvious from the above. I need some sleep.


function
vector
calculus
integration
real analysis
probability
limit

Let $ X_1, X_2,...,X_n$, $ n>2$ be i.i.d. random variables with pdf $ f(x)\equal{}2x$ for $ x \in (0,1)$ and $ M\equal{}max(X_1,...,X_n)$. Find $ E(M|X_1\equal{}z)$.
I tried to find density function of random vector $ (M,X_1)$ and I got $ g(m,x)\equal{}2(2n\minus{}2)xm^{2n\minus{}3}$ for $ 0<x<m<1$ but i think it's wrong because the integral from this density function on set $ \{(m,x): 0<x<m<1\}$ is not $ 1$.  :(
Can anyone help me?  :D

for $ a<b$, it seems very easy to compute the quantity:
\[ P[ X_1 < a; M<b] \equal{} P[X_1<a] P[X_2<b] P[X_3<b] \ldots P[X_n<b]\]
and you can then find the joint density of $ X_1, M$, isn't it ?

Yes, and I get the joint density $ g(m,x)\equal{}2(2n\minus{}2)xm^{2n\minus{}3}$ for $ 0<x<m<1$ but, as I wrote before, the integral from this density is $ \neq 1$  :mad:, maybe I make a mistake in my calculation. Can anyone write how it should be done?  :blush:

Take a look at your density function don't you think there is a problem there ?

this is because the density is not absolutely continuous with respect to the Lebesgue measure  :)  If there was a density $ f(x,y)$, then
\[ P[X_1 \equal{} M] \equal{} \frac{1}{n} \equal{} \int_{(x,m) \in [0;1]^2} 1_{ \{x\equal{}m\} } f(x,m) dx dm \equal{} 0\]
which is absurd! What you've got is part that is absolutely continuous with respect to the Lebesgue measure.

hmm ok, so what is correct density of random vector $ (M,X_1)$? Or there is another way to find $ E(M|X_1\equal{}z)$? :)

Well the direct way should be ok I think

you have
$ E[M|X_1\equal{}z]\equal{}E[z1_{M\equal{}X_1}|X_1\equal{}z]\plus{}E[M1_{M>X_1\equal{}z}|X_1\equal{}z]$
so
$ E[M|X_1\equal{}z]\equal{}z/n\plus{}(2n\minus{}2).\int_{z}^{1}m.m^{2n\minus{}3}.dx\equal{}z/n\plus{}\frac{2n\minus{}2}{2n\minus{}1}(1\minus{}z^{2n\minus{}1})$

(using the fact that the density of maximum of n-1 variable $ X_i$ is $ (2n\minus{}2).m^{2n\minus{}3}dm$)

PS 1:
what do you think of it? (making errors in calculations is one of my favorite occupation so double check this)

PS 2:
Calculating conditionnal measure of probability is almost always a difficult task.

Something is wrong in Your solution, because the correct answer is $ \frac{2n\minus{}2\plus{}z^{2n\minus{}1}}{2n\minus{}1}$, but I don't know why :/ Any other ideas how to get it?

and what if you say that
\[ P[M < m|X_1 \equal{} x] \equal{} \lim_{h \to 0} \frac {P[(x < X_1 < x \plus{} h) \cap (M < m)]}{P[x < X_1 < x \plus{} h]} \equal{} m^{2n \minus{} 2} 1_{m > x}
\]
then it gives you
\[ E[M|X_1 \equal{} x] \equal{} \int_{m \equal{} 0}^{m \equal{} 1} m d P[M < m|X_1 \equal{} x]
\]

\[ E[M|X_1 \equal{} x] \equal{} x.x^{2n \minus{} 2} \plus{} \int_{m \equal{} x}^1 m. (2n \minus{} 2)m^{2n \minus{} 3} dm \equal{} \frac {2n \minus{} 2 \plus{} x^{2n \minus{} 1}}{2n \minus{} 1}
\]
isn't it ?
Rq: what is exactly your definition of the conditional probability $ P[M \in A|X]$ ? this concept is a little bit tricky for non-discrete random variables. There are theorems that ensure that such an object exists (because it does not always exist). But of course, for nice spaces (polish, with borel $ \sigma$-algebra), it always exists. Anyway, I am just curious about the definition you are using ..

Hi Alekk

First how do you get the $ m.m^{2n\minus{}2}$ term ?
It's like a derivative in the distribution sense of $ m^{2n\minus{}3}.1_{m>x}dm$ which must be a dirac distribution in $ x$ but I'm not sure about the details.

Second I realize that what went wrong in my calulation is the fact :
$ P(X_1\equal{}M|X_1\equal{}z)$ and $ P(X_1\equal{}M)$ are not equal.

Definition of conditional probability that I use is: $ (\Omega ,\Sigma ,P)$ - probability space, $ \Sigma \supset \mathcal{S} \minus{} \sigma \minus{}$ algebra. The probability of some event $ A \in \Sigma$ given $ \sigma$-algebra $ \mathcal{S}$ is $ P(A|\mathcal{S})(\cdot): \Omega \rightarrow \mathbb{R}$ that
1) $ \forall B \in \mathcal{B}: P^{\minus{}1}(A|\mathcal{S})(B) \in S$, ($ \mathcal{B}$ is borel sigma algebra)
2) $ \forall C \in \mathcal{S} : P(A \cap C)\equal{}\int_{C} P(A|\mathcal{S})(\omega)dP(\omega)$
This is the formal form of conditional probability   :P

[quote="TheBridgeTwo"]
 how do you get the $ m.m^{2n \minus{} 2}$ term ?
[/quote]
as you sais, the "derivative" of $ m \mapsto 1_{\{m>x\}} m^{2n\minus{}2}$ is $ x^{2n\minus{}2} \delta_x \plus{} (2n\minus{}2)m^{2n\minus{}3}1_{m>x}.dm$ where $ dm$ is the Lebesgue measure and $ \delta_x$ is a dirac. Does it make sense ?


What rank are you in?

Pointcare Conjecture is where I am in. I am almost in Riemann hypothesis.

i'm Riemann Hypothesis. i think 11 people are at the top rank.

I'm posting this to see what my rank is.

Edit: I'm in Navier Stokes equations

I'm in Birch and Swinnerton Dyer (just today!).
I noticed, I'm the only one in this rank that joined in 2005  :?

[quote="jhollenbeck"]Im a  riemman hypothosis :) they have weird rank names[/quote]

they are millenium prizes names. There are 6 or 7 of them

[quote="236factorial"][quote="h78reg"]how long do you think it will take you to be number 1 in posts? :)[/quote]

forever, grobber has too much of an advantage.  :([/quote]

It shouldn't be too hard, since you're packing in a whopping 15 posts a day. Yeesh, so many relatively new members with astronomical posts per day...

Im a Pointcare Conjecture as of rightnow.

[quote="236factorial"]lately, 80-120 posts per day  :o[/quote]
Whoa, that's [i]a lot[/i]. :D

I'm a Poincare Conjecture right now.

I'm Poincare now but I'm gonna get to the next pretty soon. I average around 3 posts total per day, though on active days I get about 20. I try not to post if it really wouldnt contribute, else I'd have a higher post count.

[quote="frt"][quote="236factorial"]lately, 80-120 posts per day  :o[/quote]
Whoa, that's [i]a lot[/i]. :D
[/quote]

It helps when you're on several hours a day *sigh*

[quote="h_s_potter2002"][quote="Spencer"]I thought that a lot of people would be in New Member.
'cause I think that this site is becoming very popular.[/quote]

It's just a lot of the New Members aren't active, and therefore wouldn't be voting in this poll.[/quote]

[hide="click here for an embarrasing fact"]There are 6769 people with no posts  :( [/hide]

After all, there isn't too many users that are actually [i]active[/i]

just out of curiosity, how many of those 6500 users have become users over 3 weeks ago? ;)

Out of the 9000+ new members, only about 600 have joined in the last 4 weeks.  :( That means most of our new members aren't really flourishing. Maybe math too boring for them? It sure is for 90% of the people.

the new members probably havn't discovered this forum and if they did would be too embarressed to vote on being a new person.

I know of one person who doesn't have any posts, but is still active (or at least was) active.  Like he/she did some of the Mock AMC's and AIME's and stuff, and probably just lurks, and participates in the classes.

P and something...

why didn't u put that in your poll?

[quote="h78reg"][quote]anybody know the post count to pass hodge conjecture?

[edit] isnt there also a P vs NP rank?[/quote]

see the second post ;)[/quote]

errrr...there's nothing on the second post that tells us this?

Pointcare Conjecture  sometimes post counts dont matter its their contribution to the site and a total spammer could get a high post count and who is Tare

[quote="white_horse_king88"]It takes time for new people to get used to the forums. I wasn't very active for the first few months of my registration either.[/quote]

I think I mentioned this too, but the only forum i thought existed at first was Mathcounts  :blush: Therefore, I wasn't postinig a lot in the beginning.

[quote="jhollenbeck"]i like this forum because its all fun and games and polls :)[/quote]

yeah, but it's nothing about math, so don't get hooked only on this forum.  :)
But i do see you posting elsewhere, that's good.

that's the problem. What i do, is taht i aswer about 10 questions, and then i come here for a break.  :)

last time i checked this poll, there were no posters below Rimmean Hypothosis.  Now there is a majority below Rimmean Hypothosis :lol:  :o


geometry
geometric transformation
reflection
circumcircle
angle bisector
geometry proposed

Let $ABC$ is a triangle inscribed in $(O,R)$. $D$ is the objection of $A$ on $BC$. $I$ is the incebter of $ABC$. $(J,Ra)$ is the excribed circle of angle $BAC$ of triangle $ABC$.
Prove that $D, I, O$ are collinear $\iff R = Ra$.

Let $J'$ be the reflection of $J$ in $BC$. Then it's not hard to show that $D,I,J'$ are colinear. This means that $D,I,O$ are collinear iff $I,O,J'$ are collinear. If $K,L$ are the excenters corresponding to $B,C$ respectively, then $D,I,J'$ are collinear iff $J'$ is the reflection of $I$ in $O$, which is the circumcenter of $J,K,L$ iff $JJ'=2R_a$ is equal to the circumradius of $JKL$, which is $2R$, because $\mathcal C(O,R)$ is the nine-point circle of $JKL$.

Let G be the point whwre the excircle touche BC, M be the midpoint of BC, N be the point where the incircle touche BC and P the antipode of N on the incircle (I, IN).

It is well known that P lies on AGand that GM=NM, then, as  PI=NI, it holds that IM//AG. Then, from similaty of involved triangles, we have that GM/MK=AI/IK=DN/NK or GM/DN=MK/NK. Now if L=OI^AD, it results that OI/IL=MN/DN=GM/DN=MK/NK or OI/IL=GK/KD.
As AI is the angle bisector of <BAC then AI and OM meet on (O,R) let's say at H. Then it easy to see that <GJA = <OHA = <OAJ.

This is true in general. To go to the problem,

let's suppose that L==D. Then, as OI/IL = GK/KD, OG//IK and then, as <GJA=<OAJ, R=Ra;

Going back through the latter argument one can prove the "only if" part and this close the proof.

if R=Ra, as


calculus
integration
trigonometry
real analysis
real analysis unsolved

Hi Guys,

I have an integration problem which i do not understand. I'd be very appreciative if anybody could solve this and show (briefly) how it is solved?

I've uploaded a screenshot of the problem - Click the following link..

[url]http://putfile.com/pic.php?pic=main/8/21711573480.jpg&s=f10[/url]
Many Thanks

Substitution.

[hide="Step 1"] Let $u = 5x^{2}+2$. [/hide][hide="Step 2"] Then $du = 10x \, dx$. [/hide][hide="Step 3"] We can rewrite the integral as $\int \sin u \, (x \, dx)$. [/hide][hide="Step 4"] We can further rewrite this as $\frac{1}{10}\int \sin u \, du$. [/hide][hide="Step 5"] This integrates as $-\frac{1}{10}\cos u =-\frac{1}{10}\cos (5x^{2}+2)$. [/hide]


trigonometry
inequalities
calculus
calculus computations

Find the maximum $2^{\sin x}+2^{\cos x}$

The maximum occurs when $\sin{x}=\cos{x}$, which is when $x=\frac{\pi}{4}$

So the maximum$=2^{\frac{2+\sqrt{2}}{2}}$

to obtain maximum , we only need to consider the case where both $\sin x$ and $\cos x$ are non-negative .Suppose $0\leq x\leq \frac{\pi}{2}$ , we see that for any $\frac{\pi}{4}<x\leq \frac{\pi}{2}$ , we can always change it into $0\leq x\leq \frac{\pi}{4}$ by using ${\sin x=\cos (\frac{\pi}{2}-2}$ or $\cos x=\sin (\frac{\pi}{2}-x)$ . Now the question is equivalent to find the maximum of

$2^{\sin x}+2^{\sin (\frac{\pi}{2}-x)}$ in $x\in [0,\frac{\pi}{4}]$

But $f(x)=2^{\sin x}$ is convex over the range (we can somehow approx $\sin x=x$ for small $x$ and we know that $2^{x}$ is convex .)

So by Jensen , $(\frac{\pi}{4},\frac{\pi}{4})$ majorise $(x,\frac{\pi}{2}-x)$ and hence the maximum occurs when $x=\frac{\pi}{4}$


modular arithmetic
number theory unsolved
number theory

I have a trouble in solving the problem in baltic way 2001, I need some help!

Let $ a$ be an odd interger number. Prove that $ gcd(a^{2^m}\plus{}2^{2^m},a^{2^n}\plus{}2^{2^n})\equal{}1$ $ \forall m,n\in \mathbb {N}$, $ m,n>0$ and $ m,n$ are distinct.

[u]Solution[/u]. Let $ p$ be a prime.  Suppose $ p|\gcd(a^{2^m} + 2^{2^m},a^{2^n} + 2^{2^n})$.  Assume without loss of generality that $ n > m$.  Then $ a^{2^m}\equiv - 2^{2^m}\pmod p$.  Raise both sides to the power of $ 2^{n - m}$.  Then we have $ - 2^{2^n}\equiv a^{2^n}\equiv (a^{2^m})^{2^{n - m}}\equiv( - 2^{2^m})^{2^{n - m}}\equiv 2^{2^n}\pmod p$.  As a result, $ p|2^{2^n + 1}\implies p = 2$.  Thus $ a$ is even, a contradiction. So no prime $ p$ divides $ \gcd(a^{2^m} + 2^{2^m},a^{2^n} + 2^{2^n})$, and so $ \gcd(a^{2^m} + 2^{2^m},a^{2^n} + 2^{2^n}) = 1$ for odd $ a$. $ \Box$


Have any of you ever taken the IQ test at http://www.highiqsociety.com? It's a very accurate site, and it's free too. Tell me some of your scores, we can discuss the questions too.

[quote="banblink182"]It's a very accurate site.[/quote]

Wrong. It's not an accurate site, and if we can discuss the questions here, the test has no construct validity, because there is no test security.

Please read [url=http://learninfreedom.org/iqbooks.html]something intelligent about IQ tests[/url] before assuming that everything you see on the Web is validated or meaningful.

Do you think we can trust mensa iq tests?

mensa you cannot trust. their test is too easy, the only mistakes you can make is being distracted for a minute. But that site highiqsociety.com is really good. The tests have enormous level (which I really appreciate :)) and the analysis seems careful and accurate.

But I agree with tokenadult, if we start discussing questions here, it'll make the test valueless. I suggest discussing the results and the analysis the tests give instead :)
Can be really interesting to see what the tests say to other people, so you can verify it doesn't just echoes the same thing over and over again :D

Judging by how much you know about the questions, I assume you have all taken it already. So there should be no harm in discussing the problems. Taking the same test twice is almost as bad as discussing the solutions - to someone smart. The only thing we have to watch out for are people who have not yet taken the test and are planning to. Hopefully that does not apply to anyone.

I got a 144, with quite a bit of stupid mistakes. After looking it over another time I got every question besides the one with the A, B, C, D, E, and F things all connected with circles and lines.

Mensa is a little bit too easy...but there are still word scrambles that I cannot do. The insatiable thingy.

[quote="banblink182"]Judging by how much you know about the questions, I assume you have all taken it already. So there should be no harm in discussing the problems. Taking the same test twice is almost as bad as discussing the solutions - to someone smart. The only thing we have to watch out for are people who have not yet taken the test and are planning to. Hopefully that does not apply to anyone.[/quote]

You may have noticed there are multiple tests ;)

And there are other people on the forum besides you and me :P

:-) I know. That was kind of my point.

What did you guys get?

132 (but didnt bother with the polyhedron foldings because i did this at lunch at my school)

you did the polyhedron foldings? :P

I had no clue where to start... it seems I _really_ suck at geometry :?

I had a 145 on the first test. Got yet to try their super test (which is the most accurate for most people here)

I got 125... probably it's due to the fact that on the stereometry questions I was simply guessing since my stereometric imagination equals 0... and the vocabulary I hardly got right...

  Darij

I found the folding questions the easiest, but I found the one regarding the BCDEF thing the hardest - the one with the letters connected by lines (some double, some single), and I also thought some were kind of strange - like the monkey one.

Do you know what the highest score on that test is?
The highest I've been able to get is 151.


Reminder to everyone to visit the Missouri Forum!

Meh. There is no need for a president anymore.

Wait, which white person is Obama running against?
Is Obama an democrat or republican? Or is he a Federalist? Gah, its too confusing!!!!!!!!!

Isn't Obama a democratic-republican? I'm not sure, because he has both characteristics of a mule and an elephant. And isn't Obama running against Biden? My logic is that if Biden murders Obama, he will be president, so they are enemies.

Is "them" another one of them...what do you call them...verbs???

How many electorate votes does Missouri have?????? :huh:

I'm not sure, but I know California is 1 or 2 and Rhode Island is 60 or so.
Wait. Do electoral votes have to do with those really good electoral colleges?

I think it has something to do with the average IQ of the people in the state. Not sure though...

Dammit. Now that you ask, I can only cipher it as Indignant Questions. So, the more angry a place is, the more electoral votes it has?

Hm wow. North Korea/Iran must be crucial-what do you call them-slide states? GAH I CANT REMEMBER!


geometry
circumcircle
inradius
search
combinatorics unsolved
combinatorics

The Wonder Island is inhabited by Hedgehogs. Each Hedgehog consists of three segments of unit length having a common endpoint, with all three angles between them $120^{\circ}$. Given that all Hedgehogs are lying flat on the island and no two of them touch each other, prove that there is a finite number of Hedgehogs on Wonder Island.

Cute problem.

Suppose that we had an infinite number of hedgehogs. Then for any arbitrarily small real $r$, we could find two hedgehogs such that the distance between their centers was $<r$. This is easy to prove: if you assume a minimal distance $r$ between any two centers, then we can draw non-intersecting circles of radius $r/2$ and centered at each hedgehog center... but then the Wonder Island contains an infinite area, contradiction (...well, I guess it could be the Wonder Island for a reason, but...).

So take two hedgehogs whose centers $A, B$ are very close (at distance $r$, say). It is obvious that they touch. For proof: let $a, b$ be the triangles with vertices as the vertices of $A, B$ respectively (that is, the triangles which contain the hedgehogs). Note that $a, b$ are equilateral with circumradius $1$. Since $AB=r$ and the inradius of $a, b$ is $1/2$, $B$ is in the incircle of $a$ and vice-versa. Now break up each of $a, b$ into three pieces: $a_1, a_2, a_3, b_1, b_2, b_3$, where each piece is bounded by two sides of the hedgehog and the corresponding side of the triangle. Since $A$ is in $b$, it's in some $b_i$, and similarly $B$ is in some $a_j$. Now, taking the sides of $b_i, a_j$, we get our intersection.

A search for hedgehog gives me...: http://www.mathlinks.ro/Forum/viewtopic.php?t=27872


inequalities
trigonometry
geometry
inequalities proposed

Can anyone prove this inequality:
$\alpha+\beta+\gamma+\delta=360$
$2\sqrt{2}(\sum sin{\alpha/2}) \ge \sum sin{\alpha}$

Do you mean $\alpha, \beta, \gamma, \delta>0$? (otherwise take $\alpha=3\pi$, $\beta=-\pi$, $\gamma=\delta=0$).
Suppose these are positive numbers. Then $\alpha\leq 2\pi$, so $\frac{\alpha}{2}\leq \pi$ $\Rightarrow$ $\sin\frac{\alpha}{2}>0$ and $\sin\alpha=2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}\leq 2\sin\frac{\alpha}{2}$.
Maybe you mean $\sqrt{2}$ instead $2\sqrt{2}$?

Yes Myth, you're right $\alpha ,\beta ,\gamma ,\delta >0$. It was a geometry problem ($\alpha ,\beta ,\gamma ,\delta$ were angles of a quadrilateral) which i reduced to this inequality. And also there should be $\sqrt{2}$ instead of $2\sqrt{2}$ , sorry  :blush: .
In rest I guess it's right  ;)

Just for interest, maybe you can post the genuine, geometrical problem(hopfully without many mistakes  :lol: )


Three points of a 5 x 5 square array are connected. How many square units are in the area of the largest possible triangle that can be formed?

[quote="Brut3Forc3"][quote="ghjk"][quote="Brut3Forc3"]The largest triangle will have base and height 4, and have area 8.[/quote]
I think it must be a square with base and height 5 and have area 25/2 :wink: .[/quote]
I think not. :D
[asy]unitsize(20);
for (int i=1; i<6; ++i){
for (int j=1; j<6; ++j){
dot((i,j));
}
}
pair A=(1,1), B=(5,1), C=(4,5);
draw (A--B); draw (B--C); draw (C--A);[/asy][/quote]
My figure(just to compare):
[asy]unitsize(20);
for (int i=1; i<6; ++i){
for (int j=1; j<6; ++j){
dot((i,j));
}
}
pair A=(1,1), B=(5,1), C=(1,5);
draw (A--B); draw (B--C); draw (C--A);[/asy]

Notice how its 5 dots, but only a 4 unit length. :wink:


geometry
circumcircle
radical axis
angle bisector
projective geometry
power of a point
perpendicular bisector

A member of the greek comunity has given me the following problem. He has a trig.solution and I'd like a synthetic solution based on radical center and without trig/try ,if possible.This is the reason that I put the problem in this section. Thanks in advance.

[b]Problem[/b]

In the image, $ MN$ is a diameter , agnles $ MKA,NKD$ are equal and angles $ BKA , CKD$ are equal. Prove that lines $ BC,DA,MN$ concur .

 Babis

Let $ P: MA \cap ND$, then cause the perpendicolar line from K to MN is the angle bisector of $ \angle AKD$ K is the feet of the altitude from P in triangle PMN. Then Da pass allways trought the harmonic conjugate of K wrt MN when A moves on the circle.

[color=darkblue] [b]Nice and easy problem ![/b] I"ll present an equivalent enunciation with a bit supplement in conclusion.[/color]

[quote="Virgil Nicula"] [color=darkred][b][u]An equivalent enunciation.[/u][/b] Let $ s$ be a semicircle with the diameter $ [AB]$ and let $ F\in (AB)$

be a given point. Consider two mobile points $ \{M,N\}\subset s$ so that $ \widehat {AFM}\equiv\widehat {BFN}$ . Prove that

 the line $ MN$ passes through a fixed point and ascertain the geometrical locus of $ L\in AM\cap BN$ .[/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote $ R\in AB\cap MN$ and $ P\in (MN)$ for which $ PF\perp AB$ . Observe that the rays $ [FP$ , $ [FR$

are the interior and exterior bisectors of the angle $ \widehat {MFN}$ . Thus, the point $ P$ is harmonical conjugate with

the point $ R$ w.r.t. the pair of points $ \{M,N\}$ , i.e. the point $ P$ belongs to the polar line $ d$ of the point $ R$

w.r.t. the circle $ w$ with the diameter $ [AB]$ . Since $ d\perp AB$ obtain $ d\equiv PF$ , i.e. and [b]the point $ R$ is harmonical

conjugate with the fixed point $ F$ w.r.t. the pair of points $ \{A,B\}$[/b] . In conclusion, and [b]the point $ R$ is fixed[/b].

Prove easily that the points $ A$ , $ B$ are the $ N$ - exincenter and the $ M$ - exincenter respectively for $ \triangle MFN$ .

Thus, the lines $ AM$ , $ BN$ , $ d\equiv PF$ are concurrently in the point $ L$ which is the $ F$ - exincenter of $ \triangle MFN$ .

In conclusion, $ L\in d$ and [b]the geometrical locus of $ L$ is the polar line $ d$ of the fixed point $ R$ w.r.t. the circle $ w$[/b] .[/color]

If you want a solution with radical axis:
[hide="Solution"]
Let $ O$ be the center of the circle and let $ O'$ be the intersection of the circumcircle of $ \triangle KAD$ with $ MN$. Now, we have that \[ \angle O'DA \equal{} \angle AKM \equal{} \angle DKN \equal{} \angle O'AD\] so $ O'A \equal{} O'D$, so $ O'$ is the intersection of the perpendicular bisector of $ AD$ with $ MN$, which is a diameter.  It follows that $ O \equal{} O'$, so $ OKAD$ is cyclic.  With similar reasoning, $ BKOC$ is cyclic.  Applying the Radical Axis Theorem with $ ABCD$, $ OKBC$, and $ OKAD$, we have that $ MN$, $ AD$, and $ BC$ concur. [/hide]


Triangles $ ABC$ and $ A_1B_1C_1$ have the same area. Using compass and ruler, can we always construct triangle $ A_2B_2C_2$ equal to triangle $ A_1B_1C_1$ so that the lines $ AA_2$, $ BB_2$, and $ CC_2$ are parallel?

it's possible

First, we can assume that $ A_2\equal{}A$

if $ AB_2C_2$ exists,
then consider triangles $ ABC$ and $ AB_2C_2$ with $ BB_2//CC_2$
1. if $ BC//B_2C_2$, using the fact $ [ABC]\equal{}[AB_2C_2]$, we can show B=B_2, C=C2
2. let $ P\equal{}BC \cup B_2C_2$, also using $ [ABC]\equal{}[AB_2C_2]$, we'll have $ AP//BB_2//CC_2$, thus $ \frac{BP}{CP}\equal{}\frac{B_2P}{C_2P}$

so we want to find point $ P$ in the line $ BC$ and point $ P_1$ in the line $ B_1C_1$ so that $ AP\equal{}A_1P_1$ and  $ \frac{BP}{CP}\equal{}\frac{B_1P_1}{C_1P_1}$.

we construct another triangle $ A_3BC \sim A_1B_1C_1$, the point $ P$ on $ BC$ satisfies $ {{AP}\over{A_3P}} \equal{} {{A_1P_1}\over{A_3P}} \equal{} {{B_1C_1}\over{BC}}$.

all the points $ Q$ such that $ {AQ \over A_3Q} \equal{} {B_1C_1\over BC}$ is the circle of Apollonius which can be drawed by compass and ruler.

the existence can also be proved without much difficult

[quote="ltsin"]it's possible

2. let $ P \equal{} BC \cup B_2C_2$, also using $ [ABC] \equal{} [AB_2C_2]$, we'll have $ AP//BB_2//CC_2$

[/quote]

I don't  understand! :blush:   Would you explain why it is ?


$ A$ is $ n\times m$ matrix, $ B$ is $ m\times n$ matrix, $ m<n$. $ I_n$ is diagonal matrix.
If $ AB\equal{}I_n$, prove : the colume vectors of $ B$ is linear independence.

but I think $ B$ is $ m\times n, m<n$ , colume vectors of $ B$ is linear dependence.

thanks

If $ m<n$, the product can't be the identity.


I see this appearing in some posts... an algebraic real of dimensions 2,3,...

What is this? Can someone explain me? It appears not to be in our algebra course as far as I can see... [unless it's some other word in dutch]

Degree of algebraic number $a$ is the minimal $n$ s.t. $a$ is a root of the equation of degree $n$ with integer coefficients.


number theory
relatively prime
algebra unsolved
algebra

Let x, y positive integers. Si xy/ x^2 + y^2 - x
prove that x is a perfect square.

Please, i think this is not algebra, but i need your help!!!
Thanks

First we set $k: =gcd(x,y)$, i.e. $x=ku,y=kv$ with $u,v$ relatively prime.
Now from $k^{2}uv|k^{2}u^{2}+k^{2}v^{2}-ku$ we deduce $k|u$ and thus $u=kw$. Now we must have $k^{3}wv|k^{4}w^{2}+k^{2}v^{2}-k^{2}w$, that is $kwv|k^{2}w^{2}+v^{2}-w$ where still $u=kw$ and $v$ are relatively prime.
Since now $w|v^{2}$ and $w,v$ are relatively prime, we must have $w=1$, proving that $x=k^{2}$ is a perfect square. The remaining condition is $kv|k^{2}+v^{2}-1\Longleftrightarrow k|v^{2}-1\wedge v|k^{2}-1$, which has for instance the nontrivial solution $v=3,k=2$ giving $x=4,y=6$.

Thanks columbus, youre solution is great, but....
can you explain me the nontrivial part please??

 :lol:


geometry
perimeter
AMC
AIME
AIME I
Pythagorean Theorem

$\triangle{APM}$ has $\angle{A}=90^\circ$ and perimeter $152$. A circle center $O$ (on $AP$) has radius $19$ and touches $AM$ at $A$ and $PM$ at $T$. Find $OP$.
simple.

The topic title says 192 but the post says 152  :?:

does it? :huh: you sure?.....okay it was a mistake. :blush:

Well, although the problem "looks" simple, it's not easy. In fact, this was #14 on AIME!

I think we can start with letting $a=OP$. Then by Pythagorean theorem, $PT=\sqrt{a^2-361}$, and the perimeter of $\triangle POT=a+19+\sqrt{a^2-361}$. Then, since $\triangle POT$ and $\triangle PMA$ are similar, we can set up the equation regarding the perimeters of two triangles.

What I have is this:
$(\text{Perimeter of } \triangle POT)\cdot \frac{PA}{PT}=(\text{Perimeter of } \triangle PAM)=152$

$\frac{(a+19+\sqrt{a^2-361})(a+19)}{\sqrt{a^2-361}}=152$

Then all we have to do is to somehow solve this equation...


yes, typo! :blush:  10x! :D

now I have corrected it!

[quote="Lagrangia"]This inequality is not very hard, it just has a nice idea(i think :) ).

Prove that: $ (\frac{x}{12})^{n}+(\frac{y}{28})^{n}+(\frac{z}{105})^{n}+(\frac{t}{70})^{n} \geq \frac{4}{420^{n}}$ if $\frac{(2x+3y-5z)^{2}+(3x+3z-7t)^{2}+11y^{2}+1}{2}=x(12y+9z-24t)-y(15z-4y)$.
/quote]

An old problem, becomes new...

 :)


is it possible to teach yourself the BC Calc curriculum in two months?

Have you been taking calc to this point (probably the AB due to the nature of the question)? If so, I'd say it is definitely $\emph{possible}$. It really isn't too much more material than you'd cover in a general calc class, so two months should be sufficient.

Now if you haven't been in calc, and are starting from scratch in attempts to learn $\emph{everything}$. Then you'd be pushing it.

I took AP Calculus (covering the AB material) last year, and in the two months or so before the AP Exam, I tought myself the BC material by doing sample problems my teacher gave me.  I ended up getting a 5 on the AP Calculus BC Exam and on its AB Subscore.

So, depending on if you have learned the AB material, on how much time you can spend on learning, and on how motivated you are, it is possible to learn the BC material in about two months.


Given that $ \log_{12} {4}\equal{}t$. Find $ \log_{12}{108}$ in terms of t.

A similar way to find it is to notice that $ 108 \equal{} 4 \cdot 27 \equal{} 4 \cdot 3^3$, so:
\[ \log_{12} 108 \equal{} \log_{12} 4 \plus{} 3 \log_{12} 3
\]
Since $ \log_{12} 4 \equal{} t$ and $ \log_{12} 3 \equal{} 1 \minus{} t$, we have that $ \log_{12} 108 \equal{} t \plus{} 3(1 \minus{} t) \equal{} 3 \minus{} 2t$, as math154 said.


inequalities
trigonometry
inequalities proposed

[quote="Anh Cuong"]Give $a,b,c \geq 0$, prove that:
   $\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{6(a+b+c)}\leq a+b+c-3\sqrt{abc}$ [/quote]
It is not homogeneous. Should it be $\sqrt[3]{abc}$?

[quote="Anh Cuong"]Also, try the inequality:
Give $a_{1},a_{2},...,a_{n}\geq 0$. Find the best $k$ for this inequality holds :
$\frac{\sum_{1 \leq i<j \leq n}(a_{i}-a_{j})^{2}}{2n(a_{1}+a_{2}+..+a_{n})}\leq a_{1}+a_{2}+...+a_{n}-n\sqrt[n]{a_{1}a_{2}...a_{n}}$  :)[/quote]
Where is $k$? ;)

Opps, sorry, i was too sleepy last night :| . Thank you for the comment, i fixed it already.  :)

I find that for $n=3$, the best $k$ is $k=\frac{8\cos^{2}(20)}{1+2\cos(20)}\approx 2.45336$ and equality holds if $a: b: c=1: 1: (1+6\cos(20))=1: 1: 8\cos^{3}20$ or a permutation. :D  :D

I doubt that $k_{max}$ can be written in closed form for bigger $n$...

Note that it we can write the ineq (for general $n$) in terms of power sums $p_{k}$ as \[\frac{n\sum a_{i}^{2}-(\sum a_{i})^{2}}{k\sum a_{i}}\le \sum a_{i}+np_{0}\ \ \Longleftrightarrow\ \ \frac{p_{2}^{2}}{p_{1}^{2}}+k\frac{p_{0}}{p_{1}}\le k+1.\] As $p_{0}\le p_{1}$, this is in a sense slightly stronger than (7) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110149]here[/url] which states \[\frac{p_{2}^{2}}{p_{1}^{2}}+(n-1) \left(\frac{p_{0}}{p_{1}}\right)^{\frac{\scriptstyle n}{\scriptstyle{n-1}}}\le n.\] :)

[quote="Anh Cuong"]Give $a,b,c \geq 0$, prove that:
   $\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{6(a+b+c)}\leq a+b+c-3\sqrt[3]{abc}$
Also, try the inequality:
Give $a_{1},a_{2},...,a_{n}\geq 0$. Find the best $k$ for this inequality holds :
$\frac{\sum_{1 \leq i<j \leq n}(a_{i}-a_{j})^{2}}{k(a_{1}+a_{2}+..+a_{n})}\leq a_{1}+a_{2}+...+a_{n}-n\sqrt[n]{a_{1}a_{2}...a_{n}}$  :)[/quote]

It's easy to show that we only need to consider the above inequality incase there're $n-1$ numbers equal. We can use EV-method or SMV-method. But I dont think I should try anymore. Problem is including only one-variable now.

[quote="hungkhtn"]
It's easy to show that we only need to consider the above inequality incase there're $n-1$ numbers equal. We can use EV-method...[/quote]Yes, that's exactly what I did. ( :blush: )
 [quote="hungkhtn"]But I dont think I should try anymore. Problem is including only one-variable now.[/quote]And that's of course why we cannot probably expect a "nice solution" (in Arqady's sense :) ) for each $n$  :(

I have no doubt about EV strength, I also refer to one variable case based on my ABC. Can you post the solution using SMV, hungkhtn ?
To spanferkel: your new problem also can be done by Vasc'EV or my ABC  :D


Q1 In how many ways can the letter of the English alphabet be arranged so that there are 7 letters between a and b
[hide]I am getting (18!)(7!)(2) but answer is (18)(2)(24!)[/hide]

Q2 How many 3 digit numbers xyz with x,z<y can be formed ?
[hide]240[/hide]

Hi jackie-chan01.

Q1.

Pick $ 7$ letters to go between $ a$ and $ b$ in $ \binom{24}{7}$ ways and arrange them in $ 7!$ ways. That is, $ \frac{24!}{17!}$

Then we can have none of the $ 17$ letters to the left of $ a$ and $ 17$ to the right or $ 1$ to the left and $ 16$ to the right and so on: $ 18$ choices.

Arrange the $ 17$ letters in $ 17!$ ways and split them in two to get one of the $ 18$ possibilities above.

This gives $ (24)!(18)(2)$.

Thank you andrewtom

but do we have some more elegant or a shorter method for Q2 ?

[hide="Q2"]
There are three cases in this problem.

[b]CASE 1[/b]
$ x>z$
If you choose any three distinct integers from $ 0$ to $ 9$, inclusive, you will have one that is the largest, one that is the smallest, and one that is in the middle.  The largest will be the $ y$ value, the smallest will be the $ z$ value, and the one in the middle will be the $ x$ value.  So for this case, there is a total of $ \binom{10}{3}\equal{}120$ different numbers.
[b]
CASE 2[/b]
$ x<z$
This case is basically the same as CASE 1, except this time the smaller number goes to the $ x$ value, and the middle value goes to the $ z$ value.  However, $ x$ cannot be $ 0$, because the number would not be a three-digit one if it was.  So we can only choose from the integers that are from $ 1$ to $ 9$, inclusive.  Thus, the number of different numbers in this case is $ \binom{9}{3}\equal{}84$.
[b]
CASE 3[/b]
$ x\equal{}z$
Here, we only need to choose two numbers, because $ x$ and $ z$ will be the same.  Once again, we cannot choose $ 0$, because the number would not be a three-digit one.  So the number of different numbers in this case is $ \binom{9}{2}\equal{}36$.

Therefore, our final answer is $ 120\plus{}84\plus{}36\equal{}\boxed{240}$.  I apologize if this is somewhat confusing  :( [/hide]


The outside edge of a fountain is a the locus of pts 2m from the centre. The outside edge of a flower bed is the locus of pts 3m from the centre of the fountain. There is no gap between the flower bed and the fountain. Sketch the flower bed and calculate its area.

Isn't it just two concentric circles? The area between them is $9\pi - 4\pi = 5\pi$ (in m²).


A fair coin is to be tossed $10$ times. Let $i/j$, in lowest terms, be the probability that heads never occur on consecutive tosses. Find $i+j$.

[hide="The force of brute force"]
Clearly the number of tails is at least 5. Let's consider each case:

- 5 T's: We have _T_T_T_T_T_, where we need to put H's in five of the six _'s. This gives us $\binom{6}{5}$ ways.

- 6 T's: We have _T_T_T_T_T_T_. This gives us $\binom{7}{4}$.

- 7 T's: _T_T_T_T_T_T_T_. $\binom{8}{3}$.

- 8 T's: _T_T_T_T_T_T_T_T_. $\binom{9}{2}$.

- 9 T's: _T_T_T_T_T_T_T_T_T_. $\binom{10}{1}$.

- 10 T's: _T_T_T_T_T_T_T_T_T_T_. $\binom{11}{0}$

Thus there are $\binom{6}{5}+\binom{7}{4}+\binom{8}{3}+\binom{9}{2}+\binom{10}{1}+\binom{11}{0}=144=3^2\cdot 2^4$ ways, so our answer is $\frac{3^2\cdot 2^4}{2^{10}}=\frac{9}{64}$.
[/hide]

[hide="another idea"] I broke this up into simpler cases of 2 tosses, 3 tosses, and 4 tosses initially.  The possible outcomes of flipping 2 times are $HH, HT, TH, TT.$  In this case, heads do not occur on consecutive tosses in $3 \text{ out of the } 4 \text{ outcomes}.$  3 tosses: the possible outcomes of flipping 3 times are $HHH, HHT, HTH, THH, TTH, THT, HTT, TTT.$  In this case, heads do not occur on consecutive tosses in $5 \text { out of the }8\text{ outcomes}.$  4 tosses: the possible outcomes of flipping 4 times are $HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THTH,$   $TTHH, HTTH, THHT, TTTH, TTHT, THTT, HTTT, TTTT.$  In this case, heads do no occur on consecutive tosses in $8 \text{ out of the }16\text{ outcomes}.$  From this we might notice that the successful outcomes for 2 tosses, 3 tosses, and 4 tosses make up the first few terms in the $\text{ Fibonacci sequence.}$ (I am not quite sure how to prove this).  Listing out the terms from 3 onward in the Fibonacci sequence gives us $3, 5, 8, 13, 21, 34, 55, 89, 144,...$  The 144 corresponds with the successful outcomes of tossing a coin 10 times.  There are $2^{10}=1024$ ways of flipping a coin 10 times.  Our probability would be $\displaystyle\frac{144}{1024}=\displaystyle\frac{9}{64}.$  So $i+j=9+64=73.$ [/hide]

To show that the Fibonacci approach works, let $T_n$ be the number of good tosses of length $n$. We can see that any sequence of $T_{n-1}$ may still be made good by simply appending a tails after it, and any sequence of $T_{n-2}$ may be made good by adding a TH to it (this covers the possibilities of when we can or cannot add H to the $T_{n-1}$ sequences). Thus, we have our relation of $T_{n}=T_{n-1}+T_{n-2}$ for all $n\ge3$. Observing that $T_1 = 2$, $T_2=3$, then we can recursively find that $T_{10}=144$, and our answer is \[ \frac{T_{10}}{2^{10}}=\frac{144}{1024}=\frac{9}{64}\Longrightarrow 9+64=73. \]

Does Principle of Inclusion and Exclusion work for this?

[hide="55% of a PIE solution"]You could do PIE, but you have to do casework along with it.

We count the number of combos with the first pair of consecutive heads starting at $n$, going from $n=1$ to $n=9$.

1)  We start with HH.  All other $2^8 = 256$ combos work.
2)  We have xHH.  Clearly x cannot be H; if so, it's covered in case $1$.  All other $2^7 = 128$ combos work.
3)  We have xTHH - I'm putting the T in for the same reason as in case $2$.  In this case, the remaining x can be anything, as can the other $6$ coin flips.  $2^7 = 128$ combos work.
4)  We have xxTHH.  We have $2^5$ choices for the last five flips.  We have $3$ choices for the first two - we can't have HH, because then the HH in positions 4/5 won't be the first HH.  This gives $3 \cdot 2^5 = 96$ combos.
5)  We have xxxTHH.  We have $2^4$ choices for the last four flips.  If the middle x is T, we're fine - so we have at least $4$ good combos for xxx.  If the middle x is H, that forces the other x's to be T, for one more possibility for the x's, giving a total of $5$.  Overall, we have $5 \cdot 2^4 = 80$ combos.

You can continue, but it gets a little tedious...and you can almost see some recursion going on.[/hide]

[quote=itised]What's wrong with this?

[hide = Wrong Solution]
Note that the probability that heads never occur on consecutive tosses is equivalent to the probability that tails never occur on consecutive tosses. The only other possibility is if heads and tails alternate. There are $2$ possibilities that can happen, when it starts heads and when it starts tails. There are $2^{10} = 1024$ total possible coin flip combinations. Thus, the probability that heads never occur on consecutive tosses plus the probability that tails never occur on consecutive tosses is $\frac{511}{512}.$ They are equal probability, so the probability that heads never occur on consecutive tosses is $\boxed{\frac{511}{1024}}$, which is obviously incorrect.
[/hide][/quote]

It doesn't account for the fact that there are possible sequences in which consecutive heads and consecutive tails occur.


geometry
geometric transformation
reflection
circumcircle
projective geometry
similar triangles

Let $ABC$ be a triangle and $H$ its orthocenter. We draw two lines through the point $H$ which are perpendicular to each other. The first line intersects the lines $BC$, $CA$, $AB$ at the points $A_{1}$, $B_{1}$, $C_{1}$, respectively; the second line intersects the lines $BC$, $CA$, $AB$ at the points $A_{2}$, $B_{2}$, $C_{2}$, respectively. Let $A^{\prime}$, $B^{\prime}$, $C^{\prime}$ be the midpoints of the segments $A_{1}A_{2}$, $B_{1}B_{2}$, $C_{1}C_{2}$, respectively. Prove that these points $A^{\prime}$, $B^{\prime}$, $C^{\prime}$ lie on one line.

I didn't complete it, but the main idea is here:

First of all, let the 2 lines through H be named d and d'. Let the circle of diameter C1C2 be named (C). Define (A) and (B) similarly. Circle (A) passes through the reflection of H in BC and, of course, analogous properties take place for (B) and (C).

We can show pretty easily that the angle between (B) and (C) is either A or \pi-A; the circles pass, as I've already said, through the reflection of H in CA and in AB respectively, points which are on the circumcircle of ABC.

If we make d parallel to BC we can see that the other intersection point of (B) and (C) is on the cicumcircle of ABC, and from the angle argument in the previous paragraph we find that this intersection point is always on the circumcircle of ABC. From here we derive the conclusion.

If you're not convinced by the argument concerning the angle between circles, just apply an inversion of pole H and this will turn circles into lines and everything will be Ok then.

Grobber, it's a very interesting idea but I don't understand the solution completely. How do you conclude that the point of intersection of the circles always lies on the circumcircle?

BTW, the result sam-n quoted is known as the Droz-Farny Theorem (in fact, one of many Droz-Farny theorems). I had posted a proof on [url=http://www.cip.ifi.lmu.de/~grinberg/]my triangle geometry website[/url] under the title "From the Complete Quadrilateral to the Droz-Farny Theorem". But it's quite a difficult proof!

  Darij

I think I've actually downloaded your article, but I have to admit I didn't have the time to read it yet.

Let me put it in another way:

Assume you use an inversion of pole H. This would turn the circles (B) and (C) into 2 lines having angles A or \pi-A (because inversion preserves angles). These lines pass through 2 certain fixed points on the image of the circumcircle, which are the images of the reflections of H in CA, AB by the given inversion (call these fixed images Xb and Xc). If for a particular position of d (namely d||BC; this case is very easy to analyze) the intesection point of the images of (B) and (C) lies on the circumcircle of ABC then this happens for all the positions because this means that the chord XbXc of the image of the circumcircle of ABC has a corresponding angle to the center of size 2A (or 2\pi-2A, whichever is <=180).

From here I derived the fact that circles (B), (C) and the circumcircle always intersect. The same is, of course, true for (A) and (B) on one hand and for (A) and (C) on the other.

Now asume (A), (B) and (C) aren't concurrent. Take points A", B", C" s.t. A" is the other interesction point of (B) and (C). We have shown that A", B", C" lie on the circumcircle of ABC (call it (ABC)). Let X, Y, Z be the reflections of H in BC, CA, AB respectively. We have the following situation: (A) does not pass through A" (assumption), but it intersects (B) and (ABC), so it must do so in Y. In the same way we conclude that it passes through Z. We already knew that it passes through X, so it's the circle passing through X, Y, Z and that's (ABC). This is a very particular case and it's easy to dismiss (it's the case when AB and AC are perpendicular and d and d' are AB and AC or AC and AB respectively).

Grobber, thanks a lot! I think the proof can be still simplified as follows: The inversion with pole H maps the reflections of H in BC, CA, AB to three points Xa, Xb, Xc such that triangle XaXbXc is directly similar to triangle ABC (this is easy to prove by a simple angle chase). Now, the images of the circles (A), (B), (C) in this inversion are three lines through the points Xa, Xb, Xc, respectively, and intersecting each other at the angles A, B, C. Hence, these three lines meet at one point on the circumcircle of triangle XaXbXc. Inverting back, we obtain that the circles (A), (B), (C) meet at one point on the circumcircle of the triangle formed by the reflections of H in BC, CA, AB, i. e. on the circumcircle of triangle ABC. This does it. Thank you!

  Darij

I see. Much shorter. I was sort of lost in all the points and circles and so on and I didn't see the nicer, simpler picture :). Anyway, I read (some of ) your article. Looks like the proofs here follow a completely different line of thought.

Bye! :)

For your possible interest, here are two messages I have sent to Hyacinthos. The first contains a modification of Grobber's proof of the Droz-Farny theorem; the second gives a very simple proof, just by angle chase.

[quote][b]Hyacinthos message #9845 (edited to update the link to website)[/b]

[b]From:[/b] Darij Grinberg
[b]Subject:[/b] A synthetic proof of the Droz-Farny theorem

In the past of the Hyacinthos newsgroup, several proofs were presented for the so-called Droz-Farny theorem. Here is one possible formulation of this theorem:

[i]  If H is the orthocenter of a triangle ABC, and g and g' are two [b][here I omitted the word "perpendicular"][/b] lines through H, such that g meets BC, CA, AB at X, Y, Z, and g' meets BC, CA, AB at X', Y', Z', then

  (a) The circles with diameters XX', YY', ZZ' are coaxal; they meet at H and at another point which lies on the circumcircle of triangle ABC.

  (b) The midpoints of the segments XX', YY', ZZ' are collinear.[/i]

In the note "From the Complete Quadrilateral to the Droz-Farny Theorem" on my website

  http://www.cip.ifi.lmu.de/~grinberg/

I gave a rather complicated proof of the above. Meanwhile, Grobber at MathLinks (= "ben_goss_ro" at Hyacinthos) found a much simpler derivation at the MathLinks thread Advanced Section -> Geometry -> Proposed & Own Problems -> nice and hard as usual. Here I will repeat this derivation with some modifications and additions:

Since (b) obviously follows from (a), we will prove (a) only.

We will use directed angles modulo 180; all angles are measured in degrees.

The circle with diameter XX' passes through the point H and through the reflection D of this point H in the line BC. [Actually, the fact that our circle pass through H is an obvious consequence of < XHX' = 90; now, since the center of the circle lies on BC, the circle also passes through the reflection of H in BC.] Similarly, the circle with diameter YY' passes through H and the reflection E of H in CA, and the circle with diameter ZZ' passes through H and the reflection F of H in AB.

In the following, we will show that the circles with diameters XX', YY' and ZZ' have, apart from H, another common point, which lies on the circumcircle of triangle ABC.

At first, we remember the following fact:

[i](*) The reflections D, E, F of the orthocenter H in the sidelines BC, CA, AB lie on the circumcircle of triangle ABC.[/i]

We will use this fact later. At first, we calculate the angle between the circles with diameters YY' and ZZ'.

If ty and tz are the tangents to the circles with diameters YY' and ZZ' at their common point H, then

  < (ty; tz) = < (ty; g') - < (tz; g') = < (ty; HY') - < (tz; HZ').

Now, as a secant-chord angle, < (ty; HY') = < HYY' = < (g; CA), and similarly, < (tz; HZ') = < (g; AB). Hence,

  < (ty; tz) = < (g; CA) - < (g; AB) = < (AB; CA) = < BAC.

In other words, the angle between the circles with diameters YY' and ZZ' at their common point H is < BAC.

Now perform an inversion with center H. The points D, E, F are mapped to three new points D', E', F'. The circles with diameters XX', YY', ZZ' (all passing through the center H of inversion) are mapped to three lines d, e, f. Since inversion preserves angles, the angle between the two lines e and f must equal the angle between their preimages - i. e. the circles with diameters YY' and ZZ' -, i. e. it must equal < BAC.

[Actually, we have let out some steps here. Strictly speaking, inversion preserves angles, but changes their sign. But strictly speaking, we must also mention the fact that the directed angle between two circles is defined uniquely only if we say at which of their two common points we measure the angle. If k and k' are two circles, then their angle at one common point equals minus their angle at the other common point! Now, if we have two circles k and k' through a point H, and we invert both of them with center H, then we get two lines g and g'. If we call H' the common point of the circles k and k' different from H, then we see that the angle between the lines g and g' equals minus the angle between the circles k and k' at their common point H', and hence it equals plus the angle between the circles k and k' at their common point H. Hence, we are allowed to say that the angle between the lines e and f is < BAC.]

Now, since the circles with diameters XX', YY', ZZ' pass through the points D, E, F, their images - i. e., the lines d, e, f - pass through the images of these points - i. e. the points D', E', F'. On the other hand, by the definition of the inversion we have HD * HD' = HE * HE', thus HD' / HE' = HE / HD. However, since the points D, E, F lie on the circumcircle of triangle ABC (see (*)), we have HA * HD = HB * HE, or HE / HD = HA / HB. Hence, HD' / HE' = HA / HB, and thus D'E' || AB. Similarly, E'F' || BC and F'D' || CA. Therefore,

  < E'D'F' = < (D'E'; F'D') = < (AB; CA) = < BAC = < (e; f).

Now we can quickly show that the lines d, e, f meet on the circumcircle of triangle D'E'F':

Let the line e meet the circumcircle of triangle D'E'F' at the point T (apart from E'). Then,

  < (e; TF') = < E'TF' = < E'D'F'      (cyclic)
             = < (e; f).

Hence, the lines TF' and f are parallel, and since both of them pass through F', they coincide, i. e. the line f passes through T. Similarly, the line d passes through T. Hence, the lines d, e and f concur at one point T on the circumcircle of triangle D'E'F'. Now, inverting back, we see that the circles with diameters XX', YY' and ZZ' concur (apart from H) at one point on the circumcircle of triangle DEF, i. e. of triangle ABC (because of (*)). Therefore, they have two common points (namely the point H and their second point of concurrence on the circumcircle of triangle ABC), meaning that they are coaxal. This completes our proof of (a).

  Darij Grinberg[/quote]

[quote][b]Hyacinthos message #9851[/b]

[b]From:[/b] Darij Grinberg
[b]Subject:[/b] The Droz-Farny theorem is a piece of cake

It regularly happens in mathematics that someone poses a conjecture, it remains open for a long time, until someone else finds a long and elaborate proof, then after some time another one succeeds to simplify the proof to some extent, the proof gets simplified more and more, and finally it becomes half a page long, and people wonder why nobody found such a proof before. Something of this kind can happen in elementary mathematics. For instance, take the "Droz-Farny theorem". Here is a somewhat extended formulation of this theorem:

[i]Let g and g' be two mutually perpendicular lines through the orthocenter H of a triangle ABC. The line g meets the sidelines BC, CA, AB at A', B', C'; the line g' meets the sidelines BC, CA, AB at A", B", C". Then:

(a) The circles with diameters A'A", B'B", C'C" pass through the point H.
(b) These circles have a common point Q different from H; this point Q lies on the circumcircle of triangle ABC.
(c) The circles with diameters A'A", B'B", C'C" are coaxal.
(d) The midpoints of segments A'A", B'B", C'C" lie on one line.
(e) The relation
    B'C' : C'A' : A'B' = B"C" : C"A" : A"B"
holds.[/i]

The part (c) is the actual theorem of Arnold Droz-Farny; it was given without proof in Honsberger's "Episodes in Nineteenth and Twentieth Century Euclidean Geometry". (b) is a stronger variant of (c), while (e) is a result by Floor van Lamoen.

You see that (a) is trivial (the angles < A'HA", < B'HB" and < C'HC" are right angles), while (c) and (d) immediately follow from (b). Once (b) is proven, (e) becoms quite simple, too. Hence, the main issue is to prove (b).

In November 2002, I began searching for a synthetic proof, could not find any and therefore posted the question to Hyacinthos in December 2002 (Hyacinthos message #6128). Floor van Lamoen gave a trigonometric proof using pivot point theory (Hyacinthos messages #6144 and #6245). I searched for a rather elementary proof and found one in 2003 with the assistance of Nikolaos Dergiades (Hyacinthos message #7381, see also the note "From the Complete Quadrilateral to the Droz-Farny Theorem" on my website). Nick Reingold found a proof of (c) only using the polar relationship (Hyacinthos message #7384). Milorad Stevanovic proved (c) using vectors (Hyacinthos message #9130). All these proofs were quite interesting but far from being easy. Later, the Droz-Farny theorem was posted on the MathLinks forum (as a "hard" problem!). Suddenly, one of the forum members - Grobber - found an elementary proof of (b) using some angle chase and the fundamental properties of inversion (Hyacinthos message #9845). This proof was easier than most of the former ones, but it still was a bit tricky. Now, this morning, I have rather accidentally come up with a very simple proof, a proof using nothing but trivial angle chase.

In fact, we begin with the well-known fact that the reflections D, E, F of the orthocenter H of triangle ABC in the sides BC, CA, AB lie on the circumcircle of triangle ABC. Of course, these reflections D, E, F also lie on the altitudes AH, BH, CH.

In the following, we will measure angles in degrees and use directed angles modulo 180.

As we have already seen, the point H lies on the circle with diameter A'A". The reflection D of the point H in the line BC also lies on this circle (in fact, the circle with diameter A'A" is symmetric with respect to the line BC, since its center lies on BC). Hence, the circle with diameter A'A" meets the circumcircle of triangle ABC at the point D and at one new point, which we will call Q. We will show that this point Q also lies on the circles with diameters B'B" and C'C".

In fact,

  < HA'D = < HA'C + < CA'D
         = < HA'C + < HA'C
      (since < HA'C = < CA'D from the symmetry of
      the points H and D with respect to BC)
         = 2 < HA'C = 2 < (g; BC),

and similarly < HC'F = 2 < (g; AB). On the other hand,

  < FQD = < FAD         (cyclic)
        = < FAB + < BAD
        = < FCB + < BAD        (cyclic)
        = < (CH; BC) + < (AB; AH)
        = < (CH; AB) + < (AB; BC) + < (AB; BC) + < (BC; AH)
        = 90 + < (AB; BC) + < (AB; BC) + 90
        = 180 + 2 < (AB; BC) = 2 < (AB; BC),

and < HQD = < HA'D (cyclic), so that

  < HQF = < HQD - < FQD = < HA'D - < FQD
        = 2 < (g; BC) - 2 < (AB; BC)
        = 2 < (g; AB) = < HC'F,

and it follows that the point Q lies on the circle through the points H, F and C', i. e. on the circle with diameter C'C". Similarly, Q lies on the circle with diameter B'B". Altogether, our point Q lies on the circumcircle of triangle ABC and on the circles with diameters A'A", B'B", C'C". This proves (b). As I have already said, (c) and (d) follow directly from (b).

Now, the proof of (e) goes as follows: Since Q and H lie on the circle with diameter C'C", we have < QC'H = < QC"H, hence < QC'A' = < QC"A", and similarly < QA'C' = < QA"C". Thus, the triangles QC'A' and QC"A" are (directly) similar, and we can conclude

  C'A' / C"A" = QC' / QC".

Similarly, B'C' / B"C" = QC' / QC". Thus, B'C' / B"C" = C'A' / C"A". The same reasoning shows C'A' / C"A" = A'B' / A"B", and combining this, we arrive at

  B'C' / B"C" = C'A' / C"A" = A'B' / A"B",

what is nothing but an equivalent restatement of (e).

Finally a question. The Droz-Farny theorem, at least its part (c), was discovered by Arnold Droz-Farny. He surely had a proof. Does anybody know what his proof was like?

  Darij Grinberg[/quote]

I solved this 3 months ago I don't have the time to write all the details but my methode was like this:

I tried to prove that the circumcircles of HA1A2 and HB1B2 and HC1C2 have another common point beyond H.( its obvious that this is the same as what the problem mentions) .Then i inverted through H with the power HA.HA''( A'' is the foot of the altitude from A). then I reached three diameters of three different circles, and i had to prove that they are concurrent, ther i used spiral similarity. :)

And in the book: Problems in Geometry ( I.F.Sharigin) there is an analytical solution: it takes the two lines through H as the x axis and y axis... :)

I think the following more general problem holds:

Given a complete quadrilateral $ABCD$ (in our case $D=H$) and a line $\ell$ (in our case this is the line at infinity), by Desargue's theorem $\ell$ cuts the pairs of opposite sides of the quadrilateral in points which form an involution. Now consider two lines through $D$ s.t. they cut $\ell$ in points which correspond to each-other in the mentioned involution. These lines cut $BC, CA, AB$ in $A_1,A_2,B_1,B_2,C_1,C_2$, just like in our problem. Take $A'$ to be the harmonic conjugate of $A_1,A_2$ wrt $\ell\cap BC$. Define $B'$, $C'$ similarly. Then $A',B',C'$ are collinear.

I don't know for sure if it's true, but it can be transformed through projective transformations in a much simpler configuration. For example, I examined the case when $AD$ is the line at infinity with Euklides, and it seemed to hold. I didn't bother to prove it, but I'm sure it can't be that hard, because in this case there are lots of pairs of parallel lines and similar triangles and so on.

[quote="grobber"]I think the following more general problem holds:

Given a complete quadrilateral $ABCD$ (in our case $D=H$) and a line $\ell$ (in our case this is the line at infinity), by Desargue's theorem $\ell$ cuts the pairs of opposite sides of the quadrilateral in points which form an involution. Now consider two lines through $D$ s.t. they cut $\ell$ in points which correspond to each-other in the mentioned involution. These lines cut $BC, CA, AB$ in $A_1,A_2,B_1,B_2,C_1,C_2$, just like in our problem. Take $A'$ to be the harmonic conjugate of $A_1,A_2$ wrt $\ell\cap BC$. Define $B'$, $C'$ similarly. Then $A',B',C'$ are collinear.[/quote]

I believe this is just the D-$\ell$-perpendicular variant of the Droz-Farny theorem. I encourage you to look at [url=http://forumgeom.fau.edu/FG2001volume1/FG200122index.html]Floor van Lamoen's paper about $P\ell$-Perpendicularity[/url] (I have read only the first two pages and they were very instructive).

  Darij

You recommended this paper to me before, and I read it then :). "Instructive" is a light word! :)

Dear mathlinkers,
see for example Forum Geometricorum
http://forumgeom.fau.edu/FG2004volume4/index.html  A purely synthetic proof ...
Sincerely
Jean-Louis

Here is [b]Cosmain Pohoatza and Ta Hong Son (April)[/b]'s article about this problem, I will try to find some different proof :( (Hope so).

Let me restate the problem so that it can fit my solution:
[b]"Let $ ABC$ be a triangle and $ H$ its orthocenter. We draw two lines through the point $ H$ which are perpendicular to each other. The first line intersects the lines $ BC$, $ CA$, $ AB$ at the points $ A_{1}$, $ B_{1}$, $ C_{1}$, respectively; the second line intersects the lines $ BC$, $ CA$, $ AB$ at the points $ A_{2}$, $ B_{2}$, $ C_{2}$, respectively. Let $ A_3$, $ B_3$, $ C_3$ be the midpoints of the segments $ A_{1}A_{2}$, $ B_{1}B_{2}$, $ C_{1}C_{2}$, respectively. Prove that these points $ A_3$, $ B_3$, $ C_3$ lie on one line"[/b]
[i]Proof[/i]
[b]Lemma $ 1$[/b]:[color=darkblue]Let $ ABC$ be triangle with orthocenter $ H$. A line passing through $ H$ intersects $ AB$, $ AC$, respectively, at the point $ P,Q$. The line through $ H$ which is perpendicular to $ PQ$ intersects $ BC$ at $ M$. Prove that: $ \frac {\overline {HP}}{\overline {HQ}} \equal{} \frac {\overline {MB}}{\overline {MC}}$.[/color]
[i]Proof lemma $ 1$[/i]:
Through $ A$, let $ Ax$ be the parallel line to $ PQ$. Through $ H$, let $ Hx$ be the parrallel line to $ BC$. We have: $ \frac {\overline {HP}}{\overline {HQ}} \equal{} A(PQHx)$, $ \frac {\overline {MB}}{\overline {MC}} \equal{} H(BCMx)$. But $ AQ\bot HB$, $ AP\bot HC$, $ AH\bot Hx$, $ AD\bot HM$. Hence, $ A(PQHx) \equal{} H(BCMx)$. Thus, $ \frac {\overline {HP}}{\overline {HQ}} \equal{} \frac {\overline {MB}}{\overline {MC}}$.
[b]Lemma $ 2$[/b]: [color=darkblue]Given $ 3$ lines $ (d_1),(d_2)$. Let $ \{A_1,B_1,C_1\} \in (d_1)$, $ \{A_2,B_2,C_2\} \in (d_2)$ and $ A_0\in A_1A_2$, $ B_0\in B_1B_2$, $ C_0\in C_1C_2$ such that:
$ i)$ $ \frac {\overline {A_1B_1}}{\overline {A_1C_1}} \equal{} \frac {\overline {A_2B_2}}{\overline {A_2C_2}}$[/color]
$ ii)$ $ \frac {\overline {A_0A_1}}{\overline {A_0A_2}} \equal{} \frac {\overline {B_0B_1}}{\overline {B_0B_2}} \equal{} \frac {\overline {C_0C_1}}{\overline {C_0C_2}}$.
[color=darkblue]Then $ A_0,B_0,C_0$ are collinear.[/color]

[i]Proof lemma $ 2$[/i]:

Due to $ \frac {\overline {A_0A_1}}{\overline {A_0A_2}} \equal{} \frac {\overline {B_0B_1}}{\overline {B_0B_2}}$. Therefore:
$ \vec {A_0B_0} \equal{} \frac {\overline {A_0A_2}}{\overline {A_0A_1} \plus{} \overline {A_0A_2}}.\vec {A_1B_1} \plus{} \frac {\overline {A_0A_1}}{\overline {A_0A_1} \plus{} \overline {A_0A_2}}.\vec {A_2B_2}$. Argue the same, we will get $ \vec {A_0C_0} \equal{} \frac {\overline {A_1C_1}}{\overline {A_1B_1}}.\vec {A_0B_0}\Longrightarrow A_0,B_0,C_0$ are collinear.
Back to our problem.
According to the [b]lemma $ 2$[/b]. In order to prove $ A_3,B_3,C_3$ are collinear, it is enough to prove: $ \frac {\overline {A_1B_1}}{\overline {A_1C_1}} \equal{} \frac {\overline {A_2B_2}}{\overline {A_2C_2}}$ or $ A(B_1C_1A_1H).\frac {\overline {HC_1}}{\overline {HB_1}} \equal{} A(B_2C_2A_2H).\frac {\overline {HC_2}}{\overline {HB_2}}$ or $ (CBA_1D).\frac {\overline {HC_1}}{\overline {HB_1}} \equal{} (CBA_2D).\frac {\overline {HC_2}}{\overline {HB_2}}$ ($ D$ is the projection of $ A$ onto $ BC$) or $ (CBA_1D).\frac {\overline {A_2C}}{\overline {A_2B}} \equal{} (CBA_2D).\frac {\overline {A_1C}}{\overline {A_1B}}$ (Thanks to [b]lemma $ 1$[/b]) or $ \frac {\overline {DC}}{\overline {DB}} \equal{} \frac {\overline {DC}}{\overline {DB}}$ but this is obviously [b]true[/b].
Therefore $ A_3,B_3,C_3$ are collinear.
Our proof is completed then.

Yes,that is also Nguyen Minh Ha's proof of this theorem. :lol: Is it a strange proof? :P


Let $ a,b,c$ be nonnegative real numbers. Prove that

$ \frac{a\plus{}2b\plus{}3}{c\plus{}2b\plus{}3}\plus{}\frac{b\plus{}2c\plus{}3}{a\plus{}2c\plus{}3}\plus{}\frac{c\plus{}2a\plus{}3}{b\plus{}2a\plus{}3}\ge 3$

[quote="dduclam"]Let $ a,b,c$ be nonnegative real numbers. Prove that

$ \frac {a \plus{} 2b \plus{} 3}{c \plus{} 2b \plus{} 3} \plus{} \frac {b \plus{} 2c \plus{} 3}{a \plus{} 2c \plus{} 3} \plus{} \frac {c \plus{} 2a \plus{} 3}{b \plus{} 2a \plus{} 3}\ge 3$[/quote]
It's equivalent to $ \sum_{cyc}(2a^3\minus{}3a^2b\plus{}abc\plus{}3a^2\minus{}3ab)\geq0,$ which obviously.

[quote="fjwxcsl"]See:
http://www.irgoc.org/bbs/dispbbs.asp?boardid=12&id=2980&star=1#2980[/quote]

I don't see where it's ?

Solution :
Do note $ A\equal{}a\plus{}1$ and       $ B\equal{}b\plus{}1$  and     $ C\equal{}c\plus{}1$
Now we haves to prove :
$ \displaystyle \frac{A\minus{}C}{C\plus{}2B}\plus{}\frac{B\minus{}A}{A\plus{}2C}\plus{}\frac{C\minus{}B}{B\plus{}2C}\geq 0\,\,(**)$
Since the inequality is symmetric in the variables, and without loss of generality , we may assume that $ a\,\geq\,b\,\geq\,c$ so  :      $ A\,\geq\,B\,\geq\,C$

$ (**)\,\longleftrightarrow\,\, \displaystyle (A\minus{}B)^{2}(2A\plus{}2B\minus{}C)\plus{} (A\minus{}C)(B\minus{}C)(2A\minus{}B\plus{}2C)\geq 0$

wich is true  :wink:

Sorry,you made misstake here  :wink:

[quote="mehdi cherif"]
Since the inequality is symmetric in the variables, and without loss of generality , we may assume that $ a\,\geq\,b\,\geq\,c$ so  :      $ A\,\geq\,B\,\geq\,C$
 [/quote]

Unfortunately it is not true. Try to switch places A with B and you will see that it's not symmetric :wink:

Mîndrilă Claudiu, it's easy enough. See my post upward.  :wink:

[quote="Mîndrilă Claudiu"]Congratiulations [b]dduclam[/b]. It`s a very nice inequality. I`ll try to solve it, but maybe after few days you will post the solution...[/quote]

Thank you,Mîndrilă Claudiu. My solution assume $ c\equal{}min(a,b,c)$ and set $ c\plus{}1\equal{}z$,$ a\plus{}1\equal{}z\plus{}m$,$ b\plus{}1\equal{}z\plus{}n$ with $ m,n\ge0$

[quote="dduclam"]Let $ a,b,c$ be nonnegative real numbers. Prove that

$ \frac {a \plus{} 2b \plus{} 3}{c \plus{} 2b \plus{} 3} \plus{} \frac {b \plus{} 2c \plus{} 3}{a \plus{} 2c \plus{} 3} \plus{} \frac {c \plus{} 2a \plus{} 3}{b \plus{} 2a \plus{} 3}\ge 3$[/quote]
Setting $ x\equal{}a\plus{}1,y\equal{}b\plus{}1,z\equal{}c\plus{}1$, then the inequality becomes
\[ \sum \frac{x\plus{}2y}{z\plus{}2y} \ge 3\]
\[ \Leftrightarrow \sum (x\plus{}2y)(2z\plus{}x)(2x\plus{}y) \ge 3(2x\plus{}y)(2y\plus{}z)(2z\plus{}x)\]
\[ \Leftrightarrow \sum (x\plus{}2z)(2x^2\plus{}5xy\plus{}2y^2) \ge 3(9xyz\plus{}4\sum x^2y\plus{}2\sum xy^2)\]
\[ \Leftrightarrow 2\sum x^3\plus{} 9\sum x^2y \plus{}6\sum xy^2 \plus{}30xyz \ge  3(9xyz\plus{}4\sum x^2y\plus{}2\sum xy^2)\]
\[ \Leftrightarrow 2\sum x^3\plus{}3xyz \ge 3\sum x^2y\]
By Schur's Inequality, we get
\[ \sum x^3\plus{}3xyz \ge \sum x^2y\plus{}\sum xy^2\]
It suffices to prove that
\[ \sum x^3\plus{}\sum xy^2 \ge 2\sum x^2y\]
\[ \Leftrightarrow \sum x(x\minus{}y)^2 \ge 0\]
which is true. :)

[quote="can_hang2007"][quote="dduclam"]Let $ a,b,c$ be nonnegative real numbers. Prove that

$ \frac {a \plus{} 2b \plus{} 3}{c \plus{} 2b \plus{} 3} \plus{} \frac {b \plus{} 2c \plus{} 3}{a \plus{} 2c \plus{} 3} \plus{} \frac {c \plus{} 2a \plus{} 3}{b \plus{} 2a \plus{} 3}\ge 3$[/quote]
Setting $ x \equal{} a \plus{} 1,y \equal{} b \plus{} 1,z \equal{} c \plus{} 1$, then the inequality becomes
\[ \sum \frac {x \plus{} 2y}{z \plus{} 2y} \ge 3
\]

\[ \Leftrightarrow \sum (x \plus{} 2y)(2z \plus{} x)(2x \plus{} y) \ge 3(2x \plus{} y)(2y \plus{} z)(2z \plus{} x)
\]

\[ \Leftrightarrow \sum (x \plus{} 2z)(2x^2 \plus{} 5xy \plus{} 2y^2) \ge 3(9xyz \plus{} 4\sum x^2y \plus{} 2\sum xy^2)
\]

\[ \Leftrightarrow 2\sum x^3 \plus{} 9\sum x^2y \plus{} 6\sum xy^2 \plus{} 30xyz \ge 3(9xyz \plus{} 4\sum x^2y \plus{} 2\sum xy^2)
\]

\[ \Leftrightarrow 2\sum x^3 \plus{} 3xyz \ge 3\sum x^2y
\]
By Schur's Inequality, we get
\[ \sum x^3 \plus{} 3xyz \ge \sum x^2y \plus{} \sum xy^2
\]
It suffices to prove that
\[ \sum x^3 \plus{} \sum xy^2 \ge 2\sum x^2y
\]

\[ \Leftrightarrow \sum x(x \minus{} y)^2 \ge 0
\]
which is true. :)[/quote]

Nice,Can  :)
But how will we do with following inequality?

$ \frac {a \plus{} 2b \plus{} 1}{c \plus{} 2b \plus{} 1} \plus{} \frac {b \plus{} 2c \plus{} 1}{a \plus{} 2c \plus{} 1} \plus{} \frac {c \plus{} 2a \plus{} 1}{b \plus{} 2a \plus{} 1}\ge 3$

And general problems:

1/ $ \frac {a \plus{} 2b \plus{} k}{c \plus{} 2b \plus{} k} \plus{} \frac {b \plus{} 2c \plus{} k}{a \plus{} 2c \plus{} k} \plus{} \frac {c \plus{} 2a \plus{} k}{b \plus{} 2a \plus{} k}\ge 3$ for $ k\ge0$

2/  $ \frac {a \plus{} mb \plus{} n}{c \plus{} mb \plus{} n} \plus{} \frac {b \plus{} mc \plus{} n}{a \plus{} mc \plus{} n} \plus{} \frac {c \plus{} ma \plus{} n}{b \plus{} ma \plus{} n}\ge 3$ for $ m,n\ge0$

[quote="dduclam"]And general problems:

1/ $ \frac {a \plus{} 2b \plus{} k}{c \plus{} 2b \plus{} k} \plus{} \frac {b \plus{} 2c \plus{} k}{a \plus{} 2c \plus{} k} \plus{} \frac {c \plus{} 2a \plus{} k}{b \plus{} 2a \plus{} k}\ge 3$ for $ k\ge0$

2/  $ \frac {a \plus{} mb \plus{} n}{c \plus{} mb \plus{} n} \plus{} \frac {b \plus{} mc \plus{} n}{a \plus{} mc \plus{} n} \plus{} \frac {c \plus{} ma \plus{} n}{b \plus{} ma \plus{} n}\ge 3$ for $ m,n\ge0$[/quote]
1/ Setting $ x \equal{} a \plus{} \frac {k}{3},y \equal{} b \plus{} \frac {k}{3},z \equal{} c \plus{} \frac {k}{3}$, the inequality becomes
\[ \sum \frac {x \plus{} 2y}{z \plus{} 2y} \ge 3
\]
It is the previous.
2/ Setting $ x \equal{} a \plus{} \frac {n}{m \plus{} 1},y \equal{} b \plus{} \frac {n}{m \plus{} 1},z \equal{} c \plus{} \frac {n}{m \plus{} 1}$, the inequality beomces
\[ \sum \frac {x \plus{} my}{z \plus{} my} \ge 3
\]
This inequality is not always true. :)

[quote="can_hang2007"][quote="dduclam"]And general problems:

1/ $ \frac {a \plus{} 2b \plus{} k}{c \plus{} 2b \plus{} k} \plus{} \frac {b \plus{} 2c \plus{} k}{a \plus{} 2c \plus{} k} \plus{} \frac {c \plus{} 2a \plus{} k}{b \plus{} 2a \plus{} k}\ge 3$ for $ k\ge0$

2/  $ \frac {a \plus{} mb \plus{} n}{c \plus{} mb \plus{} n} \plus{} \frac {b \plus{} mc \plus{} n}{a \plus{} mc \plus{} n} \plus{} \frac {c \plus{} ma \plus{} n}{b \plus{} ma \plus{} n}\ge 3$ for $ m,n\ge0$[/quote]
1/ Setting $ x \equal{} a \plus{} \frac {k}{3},y \equal{} b \plus{} \frac {k}{3},z \equal{} c \plus{} \frac {k}{3}$, the inequality becomes
\[ \sum \frac {x \plus{} 3y}{z \plus{} 2y} \ge 3
\]
It is the previous.
2/ Setting $ x \equal{} a \plus{} \frac {n}{m \plus{} 1},y \equal{} b \plus{} \frac {n}{m \plus{} 1},z \equal{} c \plus{} \frac {n}{m \plus{} 1}$, the inequality beomces
\[ P\equal{}\sum \frac {x \plus{} my}{z \plus{} my} \ge 3
\]
This inequality is not always true. :)[/quote]

You're right,it's not true when $ m\ge4$. But problem: Find the min value of P is a hard problem.

Of course, the following is also holds:
\[ \sqrt{\frac{a\plus{}2b}{a\plus{}2c}}\plus{}\sqrt{\frac{b\plus{}2c}{b\plus{}2a}}\plus{}\sqrt{\frac{c\plus{}2a}{c\plus{}2b}}\ge 3,\forall a,b,c\ge 0, a\plus{}b\plus{}c\equal{}1\]
There is a very nice solution exist  :wink:

Setting :$ x\equal{}a\plus{}2b;y\equal{}b\plus{}2c;z\equal{}c\plus{}2a$ ;$ x,y,z \in [0;\frac{3}{2} ]$
Assume that :$ a\plus{}b\plus{}c\equal{}1$=>$ x\plus{}y\plus{}z\equal{}3$
<=>$ \sum \sqrt{\frac{x}{2\minus{}x} } \ge 3$
$ [\sum x^2(2\minus{}x) ]LHS^2 \ge (x\plus{}y\plus{}z)^3$.
We need to prove:
$ (x\plus{}y\plus{}z)^3 \ge 3[ 2(x\plus{}y\plus{}z)(x^2\plus{}y^2\plus{}z^2)\minus{}3x^3\minus{}3y^3\minus{}3z^3]$
<=>$ 4\sum x^3\minus{}3\sum xy(x\plus{}y)\plus{}6xyz \ge 0$
It's easier than schur ;)

yes, your solution is correct but not good  :D I don't know what is Schur's inequality  :D
We only need prove that:
\[ x^3\plus{}y^3\plus{}z^3\plus{}3\ge 2(x^2\plus{}y^2\plus{}z^2)\]
\[ \iff (1\plus{}x^3\minus{}2x^2\plus{}x\minus{}1)\plus{}(1\plus{}y^3\minus{}2y^2\plus{}y\minus{}1)\plus{}(1\plus{}z^3\minus{}2z^2\plus{}z\minus{}1)\ge 0\]
\[ \iff x(x\minus{}1)^2\plus{}y(y\minus{}1)^2\plus{}z(z\minus{}1)^2\ge 0\]

[quote="zaizai-hoang"]Of course, the following is also holds:
\[ \sqrt {\frac {a \plus{} 2b}{a \plus{} 2c}} \plus{} \sqrt {\frac {b \plus{} 2c}{b \plus{} 2a}} \plus{} \sqrt {\frac {c \plus{} 2a}{c \plus{} 2b}}\ge 3,\forall a,b,c\ge 0, a \plus{} b \plus{} c \equal{} 1
\]
There is a very nice solution exist  :wink:[/quote]

$ \bigstar$ The following inequality is true (don't need $ a\plus{}b\plus{}c\equal{}1$  :wink: ):

$ \sqrt {\frac {a \plus{} 2b}{a \plus{} 2c}} \plus{} \sqrt {\frac {b \plus{} 2c}{b \plus{} 2a}} \plus{} \sqrt {\frac {c \plus{} 2a}{c \plus{} 2b}}\ge 3 ; \forall a,b,c\ge 0$

 $ \bigstar$ A stronger and harder one:

$ \sqrt {\frac {a \plus{} 2b}{c \plus{} 2b}} \plus{} \sqrt {\frac {b \plus{} 2c}{a \plus{} 2c}} \plus{} \sqrt {\frac {c \plus{} 2a}{b \plus{} 2a}}\ge 3 ; \forall a,b,c\ge 0$


We have $ a,b,c \in R^\plus{}$ and $ a \plus{} b \plus{} c \equal{} 1$, prove that \[ 9abc\left( {a^2  \plus{} b^2  \plus{} c^2 } \right) \plus{} 24abc \le 1
\]

i think it was posted in here before .... anyway , here's my solution
$ 9abc(a^2\plus{}b^2\plus{}c^2)\plus{}24abc \leq 1$
$ \leftrightarrow 9[1\minus{}2(ab\plus{}bc\plus{}ac)] \plus{}24 \leq \frac{1}{abc}$
$ \leftrightarrow 33 \leq \frac{1}{abc} \plus{} 18 (ab\plus{}bc\plus{}ac)$
$ \leftrightarrow 33 \leq \frac{1}{ab}\plus{}\frac{1}{ac}\plus{}\frac{1}{bc} \plus{} 18 (ab\plus{}bc\plus{}ac)$
$ \leftrightarrow 33 \leq (\frac{1}{4,5ab} \plus{} 18 ab)\plus{} (\frac{1}{4,5bc} \plus{} 18 bc)\plus{} (\frac{1}{4,5ac} \plus{} 18 ac)\plus{} \frac{7}{9}(\frac{1}{ab}\plus{}\frac{1}{bc}\plus{}\frac{1}{ac})$
which is true since
$ RHS \geq 4\plus{}4\plus{}4\plus{}\frac{7}{9}(\frac{1}{ab\plus{}bc\plus{}ac}) \geq  33$

[quote="zkw"]Well, can anyone gives a link to the old post of this problem...thanks[/quote]
The stronger inequality see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87919