geometry
analytic geometry
geometry proposed

Given triangle $ ABC$ with centroid $ G$, orthocenter $ H$ and first and second Brocard points is $ \Omega_{1},\Omega_{2}$ prove that
\[ G\Omega_{1}^{2}+\Omega_{1}H^{2}+G\Omega_{2}^{2}+\Omega_{2}H^{2}=2GH^{2}\]

If $U$ is the midpoint of $\overline{GH},$ then it suffices to show that $\Omega_1U^2+\Omega_2U^2=\frac{_1}{^2}GH^2.$ We resort to barycentric coordinates with respect to $\triangle ABC.$ Equation of orthocentroidal circle $\mathcal{G}$ on diameter $\overline{GH}$ is given by

$\mathcal{G} \equiv a^2yz+b^2xz+c^2xy-\frac{_2}{^3}(x+y+z)(S_Ax+S_By+S_Cz)=0.$

Powers $p_1,p_2$ of $\Omega_1 \left ( \frac{1}{b^2}:\frac{1}{c^2}:\frac{1}{a^2} \right)$ and $\Omega_2 \left ( \frac{1}{c^2}:\frac{1}{a^2}:\frac{1}{b^2} \right)$ WRT $\mathcal{G}$ are then

$p_1=\frac{ \frac{2}{3} \left (\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \right) \left ( \frac{S_A}{b^2}+\frac{S_B}{c^2}+\frac{S_C}{a^2} \right)- \left ( \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \right)}{\left ( \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \right)^2}$

$p_2=\frac{\frac{2}{3} \left (\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \right) \left ( \frac{S_A}{c^2}+\frac{S_B}{a^2}+\frac{S_C}{b^2} \right)- \left ( \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \right)}{\left ( \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \right)^2}$

Keeping in mind that $S_B+S_C=a^2,$ $S_C+S_A=b^2$ and $S_A+S_B=c^2,$ we get

${p_1+p_2=\frac{1}{\left ( \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \right)^2} \left [ 3 \cdot \frac{2}{3} \left (\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \right) -2 \left ( \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \right)\right]}=0$

$\Longrightarrow \Omega_1U^2-\frac{_1}{^4}GH^2+\Omega_2U^2-\frac{_1}{^4}GH^2=0\Longrightarrow \Omega_1U^2+\Omega_2U^2=\frac{_1}{^2}GH^2.$


find all $ (a,b,c) \in Z^3$ such that:
i) $ gcd(a,b) \equal{} 1$
ii) $ a^2 \equal{} b^5 \plus{} c^{10}$

Can you explain what $ \mathbb{Z}^3$ is (is it the set of integers raised to the third power)? Or do you mean $ \mathbb{Z}_3$?

Well, $ \mathbb{Z}^3 \equal{} \mathbb{Z}\times \mathbb{Z} \times \mathbb{Z}$, denoting Descartes' product. It simply means that $ (a,b,c)$ is an ordered triple of integers.

[quote="hsiljak"]Descartes' product[/quote]  In English we say "Cartesian product."


whats your favotite comic? tell me to add more polls if your favorite's not on there.  my favorite is calvin and hobbes or garfield :lol: also, tell us the funniest comic you could find and post it here.

sorry i can't add anymore.  the post won't let  me :(

I've never read Calvin and Hobbes or Blonde and I don't like The Lockhorns much, but I really, really like all the others, and a lot that isn't on the poll as well. For example...Sherman's Lagoon, Overboard, Cathy, Rhymes with Orange, and some whose names I can't think of off the top of my head. I read the entire comics page in the news every day lol...

oh yeah! i put those on the poll...but it says "you had filled too many polls." i actually put all those down, but it wouldn't let me :( .  could you post some of your favorite strips on here?

simpsons. they do make comics of the simpsons.

i already said.  i put every single comic but it won't let me put more. sorry about that.  they make simpson comics? i never knew that.

oh yeah by the way, is there any site that has tons of comics and then we could post those strips on this forum? i'm not sure where to start...

I know they've got Simpsons comic books, but I didn't think they had a comic strip.

[quote="gameworld7"]yeah i know. i'm asking for comic strips in this topic.[/quote]

He was responding to Athena  :maybe:

As for comics, there's a lot at http://www.ucomics.com but don't go posting them all around the forum, unless there's an actual demand for them.

umm...it doesn't work for me. does it work for u guys?

How could you forget Peanuts!?  That's my favorite strip!

He omitted it because the poll could fit no more options.

thank you. someone read the posts on here.  i have one i gues sort of funny comic i saw in my garfield book today. i don't know how to post it (the strip) here but i could tell you guys.

Jon: Hello? Yes this is Jon Arbuckle. The police? You say vegetation is dying? Crops are ruined? People are dropping in the streets?

Jon (still):Well yes, I did.  yes sir, i will...right away.

Jon talking to Garfield: I have to put my shoes back on.

Garfield: That explains the peeling wallpaper.

ok here's another one i found...it wasn't that funny.

Alarm Clock: BRRING!

Alarm Clock: tick tick tick tick tick tick (Garfield is getting fustrated)

Garfield puts alarm clock in a box sending to tierra del fuego and mailman comes: mailman says:YAAAHHH!

the mailman thought that there was a bomb inside the box. :lol: hehe...that wasn't that funny. only when you read it.

[quote="gameworld7"][quote="yif man12"]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11547

*sigh*[/quote]

sorry! i didn't notice it![/quote]

Heh, no problem. A change in scenery (or in topic  :) ) can't do any harm.


let $ a,b,c\geq 0$. Prove that
$ \sqrt{(a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})}\geq 1\plus{}\sqrt{1\plus{}\sqrt{(a^2\plus{}b^2\plus{}c^2)(\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}}}$
help me

[quote="VIMF"]let $ a,b,c\geq 0$. Prove that
$ \sqrt {(a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})}\geq 1 \plus{} \sqrt {1 \plus{} \sqrt {(a^2 \plus{} b^2 \plus{} c^2)(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}}}$
help me[/quote]
It is well-know inequality:
By Cauchy-Schwarz Inequality,we have:
$ (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})\equal{}\sqrt{\left({a}^{2}\plus{}{b}^{2}\plus{}{c}^{2}\plus{}2ab\plus{}2bc\plus{}2ca\right)\left(\frac{1}{{a}^{2}}\plus{}\frac{1}{{b}^{2}}\plus{}\frac{1}{{c}^{2}}\plus{}2.\frac{1}{ab}\plus{}2.\frac{1}{bc}\plus{}2.\frac{1}{ca} \right)}$
$ \geq \sqrt{\left({a}^{2}\plus{}{b}^{2}\plus{}{c}^{2} \right)\left(\frac{1}{{a}^{2}}\plus{}\frac{1}{{b}^{2}}\plus{} \frac{1} {{c}^{2}} \right)}\plus{}2\sqrt{\left(ab\plus{}bc\plus{}ca \right)\left(\frac{1}{ab}\plus{}\frac{1}{bc}\plus{}\frac{1}{ca} \right)}\equal{}\sqrt{\left({a}^{2}\plus{}{b}^{2}\plus{}{c}^{2} \right)\left(\frac{1}{{a}^{2}}\plus{}\frac{1}{{b}^{2}}\plus{} \frac{1} {{c}^{2}} \right)}\plus{}2\sqrt{\left(a\plus{}b\plus{}c \right)\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c} \right)}$
$ \Rightarrow {\left(\sqrt{(a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})}\minus{}1 \right)}^{2}\geq 1\plus{}\sqrt{\left({a}^{2}\plus{}{b}^{2}\plus{}{c}^{2} \right)\left(\frac{1}{{a}^{2}}\plus{}\frac{1}{{b}^{2}}\plus{} \frac{1} {{c}^{2}} \right)}$
So we have Q.E.D
@VIMF:
[hide]Hello anh,bít em là ai hông???:D[/hide]

[quote="trungk42sp"][quote="VIMF"]let $ a,b,c\geq 0$. Prove that
$ \sqrt {(a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})}\geq 1 \plus{} \sqrt {1 \plus{} \sqrt {(a^2 \plus{} b^2 \plus{} c^2)(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}}}$
help me[/quote]
It is well-know inequality:
By Cauchy-Schwarz Inequality,we have:
$ (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \equal{} \sqrt {\left({a}^{2} \plus{} {b}^{2} \plus{} {c}^{2} \plus{} 2ab \plus{} 2bc \plus{} 2ca\right)\left(\frac {1}{{a}^{2}} \plus{} \frac {1}{{b}^{2}} \plus{} \frac {1}{{c}^{2}} \plus{} 2.\frac {1}{ab} \plus{} 2.\frac {1}{bc} \plus{} 2.\frac {1}{ca} \right)}$
$ \geq \sqrt {\left({a}^{2} \plus{} {b}^{2} \plus{} {c}^{2} \right)\left(\frac {1}{{a}^{2}} \plus{} \frac {1}{{b}^{2}} \plus{} \frac {1} {{c}^{2}} \right)} \plus{} 2\sqrt {\left(ab \plus{} bc \plus{} ca \right)\left(\frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca} \right)} \equal{} \sqrt {\left({a}^{2} \plus{} {b}^{2} \plus{} {c}^{2} \right)\left(\frac {1}{{a}^{2}} \plus{} \frac {1}{{b}^{2}} \plus{} \frac {1} {{c}^{2}} \right)} \plus{} 2\sqrt {\left(a \plus{} b \plus{} c \right)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \right)}$
$ \Rightarrow {\left(\sqrt {(a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})} \minus{} 1 \right)}^{2}\geq 1 \plus{} \sqrt {\left({a}^{2} \plus{} {b}^{2} \plus{} {c}^{2} \right)\left(\frac {1}{{a}^{2}} \plus{} \frac {1}{{b}^{2}} \plus{} \frac {1} {{c}^{2}} \right)}$
So we have Q.E.D
@VIMF:
[hide]Hello anh,bít em là ai hông???:D[/hide][/quote]
[hide="i know"]thấy tên có chữ trung đoán thử có phải là trungdeptrai (Lương Quang Trung không nhỉ) - sai thôi nha[/hide]

[quote=sqing]Crux problem 3486:
Let $a,b,c$ be positive reals. Prove that:
$$\sqrt[3]{3(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\geq 2\left(\frac{bc}{a^2+bc}+\frac{ca}{b^2+ca}+\frac{ab}{c^2+ab}\right)$$[/quote]
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $f(w^3)\geq0$, where
$f(w^3)=\sqrt[3]{\frac{uv^2}{w^3}}-\frac{2(4w^6+18u^3w^3-27uv^2w^3+9v^6)}{8w^6+27u^3w^3-54uv^2w^3+27v^6}$.
$f'(w^3)=-\frac{\sqrt[3]{uv^2}}{3w^4}-\frac{18(27u^3v^6-27uv^8+8v^6w^3-4u^3w^6)}{(8w^6+27u^3w^3-54uv^2w^3+27v^6)^2}\leq0$
because we'll prove now that $27u^3v^6-27uv^8+4v^6w^3-4u^3w^6\geq0$.
Indeed, let $g(w^3)=27u^3v^6-27uv^8+4v^6w^3-4u^3w^6$.
Hence, $g$ is a concave function, which says that it's enough to prove the last inequality for an extremal value of $w^3$,
which happens in the following cases.
1. $w^3\rightarrow0^+$ and we see that $g\geq0$ because even $27u^3v^6-27uv^8\geq0$;
2. $b=c=1$, which gives $(a-1)^2(8a^4+24a^3+30a^2+17a+2)\geq0$.
Id est, $f$ is a decreasing function, which says that it's enough to prove the original inequality for $b=c=1$, which gives
$(a-1)^2(6a^9-19a^8+43a^7-15a^6+89a^5+73a^4+81a^3+83a^2+37a+6)\geq0$, which is obvious.
Done!

[quote=sqing]Crux problem 3486:
Let $a,b,c$ be positive reals. Prove that:
$$\sqrt[3]{3(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\geq 2\left(\frac{bc}{a^2+bc}+\frac{ca}{b^2+ca}+\frac{ab}{c^2+ab}\right)$$[/quote]

[img]http://s1.sinaimg.cn/middle/006ptkjAzy75UdLZe8g20&690[/img]

[quote=sqing]Crux problem 3486:
Let $a,b,c$ be positive reals. Prove that:
$$\sqrt[3]{3(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\geq 2\left(\frac{bc}{a^2+bc}+\frac{ca}{b^2+ca}+\frac{ab}{c^2+ab}\right)$$[/quote]
$$\iff \frac{1}{2}\sqrt[3]{3(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}+\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ca}+\frac{c^2}{c^2+ab}\geq 3.$$
$$ \Leftarrow  \frac{1}{2}\sqrt[3]{3(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}+\frac{(a+b+c)^2}{(a+b+c)^2-ab-bc-ca}\geq 3.$$


Help to Find the Limit

$\mathop{lim}_{x \to \infty }\left({\sqrt[3]{{x^{3}-2x^{2}}}-x-1}\right)$

Um calculus in the calculus forum please?

DUDE
that's my job..JK
http://www.artofproblemsolving.com/Forum/index.php?f=296
is probably the calc forum you are looking for.

This is a double post.
See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142832


(If this topic or question has been posted in another "topic", please notify me and I will go there)

At one point in time - I will not get too specific, since I honestly do not remember the day - the AoPS website closed down for maintenence.

To keep all visitors occupied, a problem was proposed for our entertainment.

The question was,

"If the number 123456789101112131415161718... is continued indefinitely, will it ever be palindromic?"

I thought this question to be very interesting, one of the most interesting questions I had heard in a long time.

Immediately a way to prove this came to me, but like all of my proofs, it is [i]very[/i] strange. Please give me some comments - i.e. "you can't do that" or "that's completely wrong" - and I would be most grateful.

Answer:

No. The first instance of n consecutive zeros that appears if you list the natural numbers in order starting from 1 is 10^n. What this means is that assuming there is a point when this number becomes palindromic, then there cannot be a power of 10 to the left of the halfway line. This is because if there is, then that means that there are n consecutive zeros to the right of the halfway line, but at most n-1 consecutive zeros to the left, and this is a contradiction because the number should be a palindrome.

Note: The halfway line would be the imaginary vertical line drawn that separates the number into two equal sections. It can be drawn through a number (for example the halfway line for 12321 would be a vertical line through the 3).

So now we have established that a power of ten cannot be to the right of the halfway line. Ok, so the natural argument that follows would be that there can be a palindrome without any power of 10 to the right of the halfway line. But even so, there cannot be a palindrome, for this reason:

Since the numbers 1-10 are placed on the left of the halfway point (we know this because 1234567891011121314151617181920 is not palindromic, and neither are any of the "previous" numbers), and the numbers get longer as you go on (soon you will be adding numbers like 100, and 1000, and so on to the end), what it means is that the whole long 1234567891011121314... thing is in a sense shifted to the left. The point of this is that if the last power of 10 placed in the number is say 10^n, then 10^(n-1), 2*(10^(n-1)), 3*(10^(n-1)), 4*(10^(n-1)), and 5*(10^(n-1)) will all be to the LEFT of the halfway line. This is another contradiction because there are more consecutive n-1 zeros on the left than on the right.

Therefore the answer is no.

If you could comment, I would appreciate it, especially one of the moderators. What would I receive point wise if this was an official solution to an offical USAMTS problem? What could I do to make this more clear?

Thanks

[quote="somedude"]just think of it like this:
1, 2, 3, 4, 5, 6, its really just all the counting numbers.  the thing would have to end with ...4131211101987654321

there are no counting numbers in this pattern, unless the number has more digits than this which would be HIGHLY unlikely.  so the answer is no[/quote]

I'm not sure I understand your point, but the integer after (one more than) 4131211101987654320 does work.

I don't understand your HIGHLY unlikely argument.

Thanks for replying!

[quote]I'm not sure I understand your point, but the integer after (one more than) 4131211101987654320 does work.

I don't understand your HIGHLY unlikely argument.[/quote]
but that would meant the preceding number would be 41312111019876543[b]19[/b] which would mess up the pattern on the other side of the number (cause its a palindrome).  There is no way this could work.

that doesnt work, because what if if u continue, u get that again?

Yes, TripleM and scrambled basically pointed out what I was trying to get across. I know what you mean somedude, that's what I thought first, but you can't convince someone like that unless you further elaborate.

So, what do you think of my proof? Any comments? ;)

Adding this message so that this post will be listed near the top; where I hope it is more likely to get spotted by someone.

[i]Someone[/i] who is willing to give comments!

 :lol:  Just kidding, but seriously guys, I need feedback.  :(

If there is no feedback, consider it good. No need to post and make the thread up top.
There are more problems in this site that are [b]unsolved[/b]. Please give them a try.

[quote="TripleM"]And what if it was ..(a million digits)..4131211101987654321 ? How can you PROVE that it will never work? You can't just say it probably won't, thats the whole question :P[/quote]

You have to imagine an infinate spectrum of numbers.  No matter haw long the string of numbers there will be a middle... right?  This is where the symmetry will go aroung. ex.12321 or 524425.  I don't know how to explain it, but if I just think about the numbers that would have to lie before and after this wouldn't provide a pallyindrome.  I can't think of a way you could possible prove this, other than just reasoning.

hmm imaging that the middle number's center is in the middle of the sequence.
ex 437293848392734.  it would follow in the sequence
43729384839273[b]3[/b]    4372938[color=blue]4[/color]8392734   43729384839273[b]5[/b]
no matter how long this number is it simply can't work!  Oh I finally though of a proof.  Yes I am thinking while I'm typing.  Anyways look at the sequence I just typed.  Imagine these three numbers.  THe end of the first would have to match up with the biginning of the third one...  that would mean you would have to totally skip around and mess up the pallindrome.  That of course is assuming that the center of the pallindrome is directly in the center.  This same reasoning can be proved if the center is of the middle by one two three... etc  except the second to last corresponds with the second one and so on.  Yes it's not much of a proof, but it's the best I can come up with, and one of the best fathomable probably.


Hi

you have $ n$ digits "9" and you have to write as big number as it possible.

you cannot use any other signs, only 9's.

so you can only concatenate or power them.

fot $ n \equal{} 1$ the answer is obvious: 9.

for $ n \equal{} 2$ there are 2 possibilities: $ 99$ (concatenation) and $ 9^9$ (power)
the bigger one is $ 9^9$, so the answer is $ 9^9$

for $ n \equal{} 3$ there are:
$ 999,\ 99^9,\ 9^{99},\ 9^{9^9}$.
the biggest is the last one.

how can I prove (I mean the formal proof) that for every $ n\ge 2$ the answer will be
like $ 9^{9^9}$ but containing $ n$ 9's (I don't know how to type this in TeX).

In TeX, just put { } braces around each term, 9^{a}=9^a, for one digit integers, but when TeX sees a^10, it will write $ a^10$.

Use induction. Suppose it's true up to n-1.

Now for n:

Let there be x nines in the base of the number, and n-x nines in the exponent. Using the assumed, the number is biggest when those n-x nines are in a tower ($ 9^{9^{9^...}}$).

Now we will prove that $ m\equal{}{\underbrace{99\ldots9}_{x}} ^{9^y}\ge{ \underbrace{99\ldots9}_{x\plus{}1}} ^y$ That's easy, as the second number is less than 11 times the first number, so it's up to $ m^{9^x\minus{}x}\ge 11^x$, which is easy and true.

Now the conclusion.

First we put $ x$ numbers in the base of that big number. We proved that the exponent is tower-like.  At the end, we proved that with tower-like exponents, the biggest is the one which has only 9 as a base. The end.


[quote="tranthanhnam"]let $a,b,c$ be positive reals
 prove that :
$\sqrt{ab(a+b)} \sqrt{cb(c+b)} \sqrt{ac(a+c)} \geq \sqrt{4abc+(a+b)(b+c)(c+a)}$[/quote]

Maybe
$\sqrt{ab(a+b)}+ \sqrt{cb(c+b)} +\sqrt{ac(a+c)} \geq \sqrt{4abc+(a+b)(b+c)(c+a)}$ ??  :)

Then known....

[quote="tranthanhnam"]let $a,b,c$ be positive reals, prove that:

$\sqrt{ab(a+b)}+ \sqrt{cb(c+b)} +\sqrt{ac(a+c)} \geq \sqrt{4abc+(a+b)(b+c)(c+a)}$[/quote]

I think I have an elemental prove.

Squaring both sides, we have:

$ab(a+b)  +  bc(b+c)  +  ca(c+a)  +  2\sqrt{a^2bc(a+b)(a+c)}  +  2\sqrt{ab^2c(b+a)(b+c)}  +  2\sqrt{abc^2(c+a)(c+b)}\geq 4abc  +  (a+b)(b+c)(c+a)$

Doing some simplifications, we get:

$\sqrt{a^2bc(a+b)(a+c)}  +  \sqrt{ab^2c(b+a)(b+c)}  +  \sqrt{abc^2(c+a)(c+b)}\geq 3abc$

which is true since:

$\sqrt{a^2bc(a+b)(a+c)}\geq abc$, $\sqrt{ab^2c(b+a)(b+c)}\geq abc$ and $\sqrt{abc^2(c+a)(c+b)}\geq abc$


Let $ S$ be the set $ {1,\ 2,\ 3,\ \cdots ,\ 19}$. Consider the subset of $ S$ such that:

(a) $ A$ has five elements.
(b) the difference of any two elements of $ A$ is greater than one.

Find the possibilities of A satisfying this conditions.

We can find the number of possibilities of $ A$ with a simple counting argument

We want five numbers for the set $ A$, let them be $ o$. Let the remaining numbers be $ |$ which seperate the $ o$'s. Since each $ o$ must be atleast two away from any other $ o$, we are given;
$ o\parallel{}o\parallel{}o\parallel{}o\parallel{}o$ with $ 4$ dividors remaining to place inbetween the $ o$'s.

So our answer is $ \binom{5\plus{}4}{4} \equal{} \boxed{126}$

[quote="ocha"]We can find the number of possibilities of $ A$ with a simple counting argument

We want five numbers for the set $ A$, let them be $ o$. Let the remaining numbers be $ |$ which seperate the $ o$'s. Since each $ o$ must be atleast two away from any other $ o$, we are given;
$ o\parallel{}o\parallel{}o\parallel{}o\parallel{}o$ with $ 4$ dividors remaining to place inbetween the $ o$'s.

So our answer is $ \binom{5 \plus{} 4}{4} \equal{} \boxed{126}$[/quote]

I think you made a slight error as there are 19 numbers, so I think you meant to say $ \binom{15}{5}$

All I did was 5-element sets so that no two numbers are consecutive.  This is just $ \binom{19\minus{}5\plus{}1}{5}$.

Or did I mess up?

mathgeniuse^ln(x), your solution is right, or the answer is $ \binom{19 \minus{} 5 \plus{} 1}{5} \equal{} 3003$.

[quote="mathgeniuse^ln(x)"]

I think you made a slight error as there are 19 numbers, so I think you meant to say $ \binom{15}{5}$

[/quote]

But Kunny said that the difference between any two elements is "[b]greater than 1[/b]". Which implies It must be atleast $ 2$.


Prove that the closed form for: \[\sum_{k=0}^{m}k^{n}\]
is a polynomial in $m$ of degree $\leq n+1$.

http://mathworld.wolfram.com/FaulhabersFormula.html

[hide="However, an Intermediate-level solution"] Let $P(m) = \sum_{k=0}^{m}k^{n}$.  Then the first difference is

$P(m+1)-P(m) = (m+1)^{n}$

Which is a polynomial of degree $n$.  It follows by some well-known lemmas that $P(m)$ is a polynomial of degree $n+1$.  In fact, because the first difference is monic it follows by some other lemmas that the leading term of $P(m)$ is $\frac{1}{n+1}m^{n+1}$. [/hide]


algebra
polynomial
function
complex numbers

Prove:
Given that f(x) = Anx^n + An-1X^n-1+ ....A1X + A0, and An does not = 0

=> if f(x) has a root of the form (A+Bi), then it must have a root of the form (A-Bi). (complex roots)

All I've figured out so far is the conjugate roots theorem, but our professor gave us a hint that we should utilize the remainder theorem

The Remainder theorem is not actually necessary.  Instead, you should investigate some properties of the conjugation function.

[hide="Hint"] What is the conjugate of $f(a+bi)$? [/hide]

Thanks so much for the fast reply, Although the remainder Theorem is not necessary how would you use the theorem?? It is somewhat peculiar that if a polynomial is divided by x-c then the remainder is f(c) would be utilized to prove such a problem.

Please clarify your hint for me. f(a+bi)? Would I just plug (x+bi) into the polynomial function?

Regard, Chris

$f(x) = a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}$.  Hence

$f(a+bi) = a_{n}(a+bi)^{n}+a_{n-1}(a+bi)^{n-1}+...+a_{1}(a+bi)+a_{0}$

What is the conjugate of this?

Or, more specifically:  What is the conjugate of $(a+bi)^{k}$?  And what is the conjugate of $(z+w)$, where $z, w$ are some complex numbers?

The Remainder Theorem proof actually works along the same lines as this one, but it doesn't actually need to be invoked.  Basically, think about $f(a-bi)$.


Please tell me, why you find $ x \equal{}\minus{}0.90441$?

hello, i think you can only use a numerical method to find the roots.
Sonnhard.

[quote="thanhnam2902"]Find $ x \in R$ of equation: $ x^5 \plus{} 3x^4 \plus{} 6x^3 \plus{} 4x^2 \plus{} 8x \plus{} 7 \equal{} 0$[/quote]

$ x^5 \plus{} 3x^4 \plus{} 6x^3 \plus{} 4x^2 \plus{} 8x \plus{} 7 \equal{}(x\plus{}0.904410010167919...)$ $ (x^2\plus{}2.93555466937901...x\plus{}5.03252306934488...)$ $ (x^2\minus{}0.839964679546928...x\plus{}1.53796660380073...)$

and none of these numbers seem to be a rational number or of the form $ p\plus{}q\sqrt r$ (looking at continuous fractions).

Where is this problem from ?


quadratics
MATHCOUNTS
algebra
quadratic formula
special factorizations

could someone please post alot!! of practice quadratic equations and a neat way of solving them without trial and errors ( something my teacher doesn't know how to do....) :?

factoring becomes less trial and error and more strategy when you notice that:

$(x-r)(x-s) = x^2 - (r+s)x + rs$ where r and s are roots of the quadratic.

Say you have the equation:

$x^2 + 24x + 143 = 0$

notice that $143 = 13 \cdot 11$ and $24 = 13 + 11$

Thus, the factored form is:

$(x+11)(x+13) = 0$ so the roots are $-11, -13$

also, completing the square works too.

say you have:

$x^2 + 6x = 16$

$x^2 + 6x + 9$ is a perfect square ($(x+3)^2$) so add 9 to each side

$x^2 + 6x + 9 = 25$

take the square root:

$x+3 = \pm 5$

$x = 2, -8$

here's some for factoring:
$x^2 + 30x + 81 = 0$
$x^2 + 25x + 150 = 0$
$x^2 - 2x - 15 = 0$
$x^2 - x - 420 = 0$

have fun!

[quote="shobber"][quote="GoBraves"]I dont feel like posting any problems now, but do you know the quadratic formula? If you dont, then here it is-

$\frac{-b \pm \sqrt {b^2 - 4ac}}{2a}$[/quote]
I think finding this formula is also a good practice.[/quote]

You mean proving it??? That's not very easy for middle school students!!

But... before you leave middle school (or for Mathcounts students, 5th grade), this formula must be memorized!!!!!!!!

To find the quadratic formula, simply complete the square. Completing the square is also a great method to solve equations.

$ax^2+bx+c$
$x^2+\frac{b}{a}x=\frac{-c}{a}$
$x^2+\frac{b}{a}x+(\frac{b}{2a})^2=\frac{-c}{a}+(\frac{b}{2a})^2$
$(x+\frac{b}{2a})^2=\frac{-4ac}{4a^2}+\frac{b^2}{4a^2}$
$x+\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}$
$x=-\frac{b}{2a}\pm\sqrt{\frac{b^2-4ac}{4a^2}}$
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

i use hamster's method.....it works relatively well unless c in $ax^2+bx+c$ has many factors and/or has more than (in my opinion) 3 digits, in which case i would use the quadratic formula
and of course, you can always complete the square

guys, thanks alot

and could someone help me how to solve quadratic equations with coeficcients rather than just coefficient of 1?

thats where the quadratic formula and completing the square come in

but factoring is possible too :D

I always attempt factoring for a bit before going to formula (i never complete the square) unless it has fractional coefficients.

$2x^2 + 9x + 9 = 0$

in this case we want something in the form of $(2x+r)(x+s)$

finding factors of 9, we have 3, 3 and 9,1. 3 + 6 = 9, so put them in:

$(2x+3)(x+3) = 0$

$x = -\frac 32, -3$

try foiling out $(2x+3)(x+3)$ to verify the factorization.

If a is prime, it is easy like this. If a is not prime, only attempt to factorize for about 30 seconds, then go on to the formula. For example:

$12x^2 + 22x + 10 = 0$

In this case, we can divide by 2, so do that first.

$6x^2 + 11x + 5 = 0$

now we either want it in the form of $(6x + r)(x + s)$ or $(2x+r)(3x+s)$

finding factors of 5, we get only 5, 1.

becasue $5 + 6\cdot1 = 11$, we can put them in for r and s in the first form to get $(6x+5)(x+1) = 0$

$x = -\frac65, -1$

once again, you should try foiling it out to check the factorization.

I'm too lazy to think these up (they're a lot harder becasue you have to multiply the factors then add them....)

In fact, factoring can be the fastest way. Do you really want to solve the equation
$x^2 + 2x + 1 = 0$ when you can factor it into $(x+1)^2$?
 x = -1.

Of course, it can be impossible to factor nicely sometimes.

Meh, you could make a graphing calculator program that would generate random numbers and you could solve them.

$ax^2+bx+c=0$
$3x^2-5x+2$
$x^2+2x+1=0$
$y^2-y+1$

A special problem:
what special number is a root of:
$x^2+x+1$

A trick question:
WHat is $i^2+4i-3$

think about that last one...

[hide="hint about it"]
imaginary numbers

[/hide] :P

[quote="math92"][quote="nat mc"]All you do is complete the square[/quote]
does that work all the time?[/quote]
$a \neq 0$

[quote="IntrepidMath"]A special problem:
what special number is a root of:
$x^2+x+1$

A trick question:
WHat is $i^2+4i-3$[/quote]

For the trick question:

[hide="Answer"]$i=\sqrt{-1} \Rightarrow i^2=-1 \Rightarrow -1+4i-3=4i-4$[/hide]

As for the other problem, are you sure you typed it correctly? Because I know of a special number that is a root of $x^2-x-1$... as for the equation given, the roots are just $\frac{-1\pm \sqrt{-3}}{2}$.

[quote="anirudh"]factor using the quadratic formulae

$5^2+5-49$[/quote]
 :huh:  But what you have to do? $5^2+5-49=25+5-49=-19$ cannot factor. 19 is prime   :huuh:

I've found the easiest way to solve quadratics is to get really good at the quadratic formula and to factor only when it is obvious it can be factored. But in Mathcounts you will probably be able to factor easily on most things unless it tells you answer involves a square root ("express your answer in simplest radical form").

The quadratic formula is really short, so it shouldn't be difficult to apply it. Only factor the quadratic if you already know it by heart or the variables are very compatible.

In the case of a quadratic such as $2x^2+7x+29$, use the quadratic formula.

[quote="djshowdown2"][quote="JesusFreak197"]$\binom{i}{u}$, Pikachu![/quote]
Wow, never saw this before, but gotta say I love it  :lol:

Anyhow, chances are IntrepidMath meant $x^2-x-1$ and anirudh meant $5x^2+5x-49$[/quote]

your right. it was just a typographical err

You should really only use the quadratic equation if you really need to.

I always try to factor it, as it really is much quicker.
I really don't like completing the square......

Completing the square is useless. The formula comes from completing the square. SO, if you are thinkingabout completing the square because you can't factor, DONT. Use the formula

I know I should probably already know this but what  do you mean by 'completing the square'?

Completing the square puts an $(x-k)^2$ on one side and a $c$ on the other side. You do this by taking half of the coefficient of the x term, and add its square and subtract this will give us the form above.
Example:
$x^2-2x+7=0$
$x^2-2x+1-1-7=0$
$(x-1)^2-8=0$
$(x-1)^2=8$


limit
function
algebra
domain
abstract algebra
calculus
calculus computations

i just need perhaps a hint or pointer about proving the next theorem:
if the sums a_n,b_n,d_n when n=1 to infinity (and d_n=sum(a_ib_j) where i+j=n) converge to A,B,D respectively then AB=D.

can i use the method of the proof of marten's (i think you type it this way) theorem here?

This is a corollary of the following theorem also by Abel: if $\sum_{n=0}^\infty c_nz^n=f(z)$ for $|z|<1$ and $\sum_{n=0}^\infty c_n$ converges, $\sum_{n=0}^\infty c_n=\lim_{x\to 1^-}f(x)$

Prove or quote this (the proof uses summation by parts), and then apply to $f(z)=\sum a_n z^n$, $g(z)=\sum b_nz^n$, $h(z)=f(z)g(z)=\sum d_nz^n$. We have $h(z)=f(z)g(z)$ for $|z|<1$ because the defining series converge absolutely.

i can use it also when |z|<=1 right? (i'm just affirming it).

That's not a very clear question- what part are you trying to use? We don't have absolute convergence when $z=1$, and we might not have convergence at all. The basic argument includes a limit as $z\to 1$; this can be adapted to any other point on the unit circle, but we still need all of the other conditions of the problem at that point.
Inside the circle, everything converges absolutely and it's easy.

i'm asking if also |z|=1 because my original question doesn't have z^n in the sums, that's the only thing that bothers me from this method.

It's a very common problem-solving trick to introduce a new variable so you can apply new methods and then specialize to the original problem. For this problem, we get at the sums by defining new functions which "should" have the desired values at a particular point, but that point isn't in the domain of the functions. It's on the boundary, and an Abelian theorem lets us get at that value with a limit of the function from inside the domain.

Our functions $f(z),g(z),h(z)$ are only defined for $|z|<1$; we're working with the limits as $z\to1^-$ here. Since $a_n\to 0$, $b_n\to 0$, the defining series for these functions converge absolutely inside the unit circle.


search
number theory proposed
number theory

Determine all pair of $ (a,n)$ be possitive integer such that : all prime divisor os $ a^n\plus{}1$ is also prime divisor of $ a\plus{}1$


solve
$ 3 tan^3 x \minus{} tanx\plus{}3.\frac{1\plus{}sinx}{cos^2 x}\minus{}8cos^2 (\frac{pi}{4}\minus{}\frac{x}{2})\equal{}0$

hello, do you mean
$ 3\tan^3(x)\minus{}\tan(x)\plus{}3\left(\frac{1\plus{}\sin(x)}{\cos^2(x)}\right)\minus{}8\cos^2\left(\frac{\pi}{4}\minus{}\frac{x}{2}\right)\equal{}0$
Sonnhard.

Can you tell me what was the point of rewriting the previous post? It's obvious that this is what he means..

hello, ok, it looks a little bit better, but nothing more.
Sonnhard.


search
inequalities
inequalities proposed

It was posted and it was posted in a Romanian TST 1988 or 1989. I cannot find it in the search button but I had the solution from this post. It is not my solution. [quote="Megus"]I will prove it by induction. It is true for $n=1$, so suppose it is true for $n$. We have:
\[ \sum\limits_{i=1}^n a_i ^3 \geq \sum\limits_{i=1}^n a_i ^2 + 2\sum\limits_{i<j}^n a_ia_j  \]
Adding $a_{n+1}^3$ to both sides we have to prove that:
\[ \sum\limits_{i=1}^n a_i ^2 + 2\sum\limits_{i<j}^{n} a_ia_j +a_{n+1}^3 \geq \sum\limits_{i=1}^{n+1} a_i ^2 + 2\sum\limits_{i<j}^{n+1} a_ia_j  \]
equivalently:
\[ a_{n+1}^2+2a_{n+1}\sum\limits_{i=1}^n a_i \leq a_{n+1}^3  \]
\[ a_{n+1}+2\sum\limits_{i=1}^n a_i \leq a_{n+1}^2  \]
I will prove this by induction as well - suppose it is true for $n+1$, adding to both sides $a_{n+1}+a_{n+2}$ we are left to prove:
\[ a_{n+1}^2 +a_{n+1}+a_{n+2} \leq a_{n+2}^2  \]

\[ a_{n+1}(a_{n+1}+1)\leq a_{n+2}(a_{n+2}-1)  \] which is true due to condition $a_{n+1}<a_{n+2}$

The equality holds when the numbers are consecutive
Hence the result[/quote]

[quote="cezar lupu"]Actually, this inequality is an old Romanian Olympiad. I think it has been proposed by V. Vajaitu. ;)[/quote]
I saw only this:
let $a_i$ are various naturals numbers. Then:
$a_1^3+a_2^3+...+a_n^3+3(a_1^5+a_2^5+...+a_n^5)\geq4(a_1+a_2+...+a_n)^3$
(one of the old Russian olympiad) ;)

Well, maybe you saw what you saw, but i'm preety sure of what i'm saying.
I'm not drunk now, so... :P


[color=blue][size=150]Comentario:[/size][/color] :idea:
Este es un conjunto de listas de preguntas de diversos niveles que he ido editando,
como veran, cada lista tiene un conjunto de preguntas, las cuales cuentan con su refencia de origen,
cada lista pretende agrupar preguntas de naturaleza similar, pero asu vez, cada lista de recursos variados,
podremos encontrar fuentes como: [color=blue]IMO ShortLists, Brazilian Training Programs,  NZimo Training Program,
Bulgarian Competitions, etc...[/color]

Esta primera entrega es Algebra, y engloba preguntas referidas de IMOShortLists....
aqui podremos encontrar Temas como: [color=blue]Algebra Funcional, Seuencias, Ecuaciones Algebraicas, Inecuaciones, etc..
[/color]
PD: Adjunto la version en PDF. :lol:
Intentenlos y disfrutenlos !!!

[color=blue][size=150] Algebra Training - Test 01 [/size][/color]
[color=blue] Date: May 30, 2005.[/color]

[i]Spanish Version[/i]

$\blacktriangleright$ [color=red]Problema 1.[IMO 98 ShortList][/color]
Sean $r_{1},r_{2},\ldots ,r_{n}$ numeros reales mayores o
iguales que 1. Probar que:
\[ \frac{1}{r_{1}+1}+\frac{1}{r_{2}+1}+\cdots +\frac{1}{r_{n}+1}\geq \frac{n}{ \sqrt[n]{r_{1}r_{2}\cdots r_{n}}+1}.  \]

$\blacktriangleright$ [color=red]Problema 2.[IMO 99 ShortList][/color]
Encontrar todas las funciones $f: \mathbb{R} \rightarrow \mathbb{R}$, tal que:
\[ f(x - f(y)) = f(f(y)) + xf(y) + f(x) - 1.  \]

$\blacktriangleright$ [color=red]Problema 3.[IMO 2001 ShortList][/color]
Encontrar todos los enteros positivos $a_1, a_2,...,a_n$, tal que:
\[ \frac{99}{100} = \frac{a_0}{a_1} + \frac{a_1}{a2} + ... + \frac{a_{n-1}}{a_n}.  \]

donde $a_0 = 1$ y $(a_{k+1}-1)a_{k-1} \geq a_k^2(a_k - 1)$ para $k = 1,2,\ldots,n-1.$

$\blacktriangleright$ [color=red]Problema 4.[IMO 95 ShortList][/color]
Sean $a, b$ y $c$ numeros reales positivos.Determine todos los
reales positivos, tal que:
$i) x + y + z = a + b + c;$
$ii) 4xyz - (a^2x + b^2y + c^2z) = abc.$

$\blacktriangleright$ [color=red]Problema 5.[IMO 97 ShortList][/color]
Sea una progresion aritmetica cuyos terminos son enteros
positivos.Esta PA contiene el cuadrado de un entero y el cubo de
un entero. Mostrar que esta progresion contiene la potencia
sexta de un entero.

$\blacktriangleright$ [color=red]Problema 6.[IMO 96 ShortList][/color]
Sean $a, b, c$ reales positivos tal que $abc = 1.$ Probar que:
\[ \frac{ab}{a^5 + b^5 + ab} + \frac{bc}{b^5 + c^5 + bc} + \frac{ac}{a^5 + c^5 + ac} \leq 1.  \]
Cuando se cumple la igualdad?

Carlos Bravo :)
Lima- PERU

En el problema 3 faltan condiciones!!! Aqui esta el link con el problema original.

http://www.mathlinks.ro/Forum/viewtopic.php?t=15624

Sip .. cierto!!!

 Gracias por la observacion....

ya he hecho los cambios....

PD: La idea es que ... intenten resolver las preguntas un tiempo prudente, buscar soluciones alternativas y/o comparalas
con las que podemos encontrar en la seccion Resources/IMO ShortLists....

Bueno, a intentarlos. ...

Carlos Bravo :)
Lima -PERU

Carlos en el problema $4$ no deberia ser $a^{2}x + b^{2}y + c^{2}z$?

SI..

siempre existen errores de tipeo ..

gracias nuevamente por la observacion ..

espero ideas de sus soluciones...

Carlos Bravo :)
Lima -PERU


Let $ a,b,c$ be positive real numbers. Prove that
$ \sum \sqrt{\frac{a^3}{a^3\plus{}\frac{1}{4}abc\plus{}b^3}} \le 2.$

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers. Prove that
$ \sum \sqrt {\frac {a^3}{a^3 \plus{} \frac {1}{4}abc \plus{} b^3}} \le 2.$[/quote]
Actually, this inequality follows from a result of Vasc: If $ x,$ $ y,$ $ z$ are positive real numbers, then
$ \sqrt{\frac{x^2}{4x^2\plus{}5yz}}\plus{}\sqrt{\frac{y^2}{4y^2\plus{}5zx}}\plus{}\sqrt{\frac{z^2}{4z^2\plus{}5xy}} \le 1.$ (*)

Let me explain why:

Applying the AM-GM Inequality, we have
$ \frac{1}{4}abc\plus{}b^3\equal{}\frac{1}{4}(abc\plus{}4b^3) \ge \frac{5}{4}\sqrt[5]{ab^{13}c}.$
Therefore, it suffices to prove that
$ \sqrt{\frac{a^3}{4a^3\plus{}5\sqrt[5]{ab^{13}c}}} \plus{}\sqrt{\frac{b^3}{4b^3\plus{}5\sqrt[5]{bc^{13}a}}}\plus{}\sqrt{\frac{c^3}{4c^3\plus{}5\sqrt[5]{ca^{13}b}}} \le 1.$
Setting $ m \equal{}\frac{\sqrt[5]{ab^{13}c}}{a^3},$ $ n\equal{}\frac{\sqrt[5]{bc^{13}a}}{b^3}$ and $ p\equal{}\frac{\sqrt[5]{ca^{13}b}}{c^3}.$ We have to prove that
$ \frac{1}{\sqrt{4\plus{}5m}}\plus{}\frac{1}{\sqrt{4\plus{}5n}}\plus{}\frac{1}{\sqrt{4\plus{}5p}} \le 1$
for $ m,$ $ n,$ $ p>0$ and $ mnp\equal{}1.$

Since $ m,$ $ n,$ $ p>0$ and $ mnp\equal{}1,$ there exist positive real numbers $ x,$ $ y,$ $ z$ such that $ m\equal{}\frac{yz}{x^2},$ $ n\equal{}\frac{zx}{y^2}$ and $ p\equal{}\frac{xy}{z^2}.$ The above inequality becomes
$ \sqrt{\frac{x^2}{4x^2\plus{}5yz}}\plus{}\sqrt{\frac{y^2}{4y^2\plus{}5zx}}\plus{}\sqrt{\frac{z^2}{4z^2\plus{}5xy}} \le 1,$
which is just (*).

And the proof of (*), see here: http://can-hang2007.blogspot.com/2010/01/inequality-33-v-cirtoaje.html


I was just wondering if anyone had a good method to find the reciprocal of an infinite series;

for example take

F(x) = a + bx + cx^2 + ..... to infinity;

then how would I find the coefficients (A, B, C .....) of

1/F(x) = A + Bx + Cx^2 + ...... to infinity;

If anyone knew a good way of doing this, could you please let me know, thanks

In general, there's only a method when $ a\neq 0$.  And it's just a method; there's no guarantee that there will be  "nice" way to write $ A$, $ B$, etc.

What you probably want to do is write out the product and equate coefficients.

$ 1 \equal{} (a \plus{} bx \plus{} cx^2 \plus{} ...)(A\plus{}Bx\plus{}Cx^2 \plus{} ...)$

So $ 1 \equal{} aA$, $ 0 \equal{} aB \plus{} Ab$, $ 0 \equal{} aC \plus{} bB \plus{} cA$, etc.

Then you can solve for $ A, B, C, ...$.  These are the first three:

$ A \equal{} \frac{1}{a}$, $ B\equal{}\minus{}\frac{b}{a^2}$, $ C\equal{}\frac{b^2}{a^3}\minus{}\frac{c}{a^2}$

Of course, when doing this you need to be very careful about convergence!


inequalities AMC AIME USA(J)MO USAMO USAMTS geometry \/closed

Is there a difference between WOOT and the Olympiad Classes? Are the Olympiad Classes (i.e. Inequalities) going to be offered again?

Nope, they won't be offered again any time soon.  :(

The Olympiad classes go into greater detail, but WOOT has a greater emphasis on proofs.

[quote="4everwise"]
The Olympiad classes go into greater detail, but WOOT has a greater emphasis on proofs.[/quote]
This is why I would have prefered, instead of WOOT, a full range of olympiad classes. I don't care whether my instructor has a PhD or won the IMO. I don't care about proofs; those can be learned through practice on the Olympiad boards here. What I care about is that the Olympiad classes presumably offered great teaching and problem sets, focused on one subject, at a very low price. If it's anything like AIME Problem Series versus Intermediate Counting, Olympiad classes would have given me more problems and teaching time.

I really don't care about the teaching. I just want the dang problems. And when the Olympiad classes (presumably) gave more problems on one subject than WOOT, I'd want the regular Olympiad classes.

Now, If someone were to wrap all of last year's WOOT problems up into a .pdf and sell that for $\$20$...

If problems are all you want, you might as well look at the IMO shortlist and buy (or even better, borrow) 102 combinatorics problems and other books of that sort. Most WOOT problems come from sources like those anyway. Meanwhile, the olympiad fora have some interesting problems (although I think a lot of them are ill-suited for USAMO practice), but you definitely won't learn how to write good proofs there, since it is not a formal contest setting. I don't know how helpful the teaching is, but if it's the problems, they are out there, and it's not that hard to find good ones.

I like the idea that the olympiad classes are shorter length. So that I can actualy fit them into a summer or something. I still want to take woot in fall of 2007  :D

Yeah, I would like the Olympiad classes better, it's easier to learn proofs on your own than the actual concepts.

Anyway, I'm probably taking WOOT in 2007 fall.

Rather than paying the money and doing woot, I'm just doing olympiad/pre-olympiad problems on this forum, and using two prize books from USAMTS, namely the geometry and 360 problems books.  Also, whatever other sources of practice I can find.  I find that I can learn more efficiently by actually attempting problems and exploring these forums instead of taking a class.

As far as I know, the class basically [i]is[/i] doing problems. But with some guidance and discussion and teamwork. I'd do WOOT if I could afford it. :P

I like WOOT. One can study Olympiad topics on his own; there are plenty of good books out there. WOOT allows you to actually get feedback and comments on your proofs. It's no use being able to solve a problem if you can't write a rigorous proof for it, and if you've never taken the USAMO you'd be surprised (I learned the hard way...gah). As they say, everybody overestimates their score on the USAMO the first time they take it.

I really enjoy working problems on my own, and certainly believe that doing practice olympiads in a competition setting can help your proof writing greatly. That said, there are definitely some significant advantages involved. First off, there are a lot of techniques involved in motivating solutions which are not at all obvious in the solutions themselves--a lot of the time when you look at a solution and think it would be impossible to get, there is in reality some kind of common approach underlying it. Second, there is potentially a lot to be gained by getting constructive feedback on your proof-writing, especially if you are new to the whole business of competition math.

I am not sure whether or not WOOT is worth it, but its not necessarily immediately dismissable. Of course, other people's milage may vary.


In less than 30 seconds, solve the following problems without a calculator:
[hide]104*109=
93*99=
Express as a common fraction: 0.123456456456456456...=[/hide]

ACTUALLY... your fraction is close... but the answer is 41111/333000
*about to $ \text{\LaTeX}$ it*
$ \frac{41111}{333000}$
and try to use LaTeX
the code is:

[code]$\frac{x}{y}$[/code]
for x/y

Thanks. How do you make square roots and squares?

square root

[code]$\sqrt{x}$[/code]

for $ \sqrt{x}$

and

[code]$x^2$[/code]

$ x^2$

also check out the $ LATEX$(dont know how to do cool latex symbol) tutorial in the menu to the left!

[quote="zolojetto"]square root

[code]$\sqrt{x}$[/code]

for $ \sqrt {x}$

and

[code]$x^2$[/code]

$ x^2$

also check out the $ LATEX$(dont know how to do cool latex symbol) tutorial in the menu to the left![/quote]
Actually it's
[code]
x^{2}
[/code]
Because if it's $ 2\plus{}$ digits, and I use your code, it turns into
$ 2^13$
Whereas if you use the brackets, it's
$ 2^{13}$

Hey stevenmeow, do you do Number Sense? NS pwns!!! (I'm from TX but I moved)

the first two are just basic number sense... ~10 sec total
the last one is just annoying fraction crap oh well

i guess it is doable with pencil/paper if you can write/calculate fast enough

[quote="stevenmeow"]actually the fraction isnt THAT hard
this is a NS rule:
take the non repeating digits and add one group of the repeated digits then subtract the nonrepeated section
if this is confusing, the number we end with for the proposed decimal is 123456
then you subtrace the non repeated section
123456-123=[b]123333[/b]
now you count off backwards on the first number saying 9 if the digit is in the repeated section and saying 0 if it is not
again easier done than said this is just [b]999[/b](for 654)[b]000[/b](for321)
now the first bolded number is the numerator and the second bolded number is the denominator
$ \frac {123333}{999000}$ then simplify if possible

$ \frac {41111}{333000}$
while you may not be able to do it in the proposed 30 seconds, this is a sometimes useful shortcut in sprint

side notes:
LaTeX in $ \text{\LaTeX}$ is [code]$\text{\LaTeX}$[/code]

and NS is something you have to practice at, just like mathcounts[/quote]
Faster? I found that out when i was bored after chap last year. Then I learned the orthodox way and it's way easier.


inequalities
function
three variable inequality

Let $x,y,z$ be non-negative numbers such that $x+y+z=3$. Then,

$\frac1{2+x^2+y^2}+\frac1{2+y^2+z^2}+\frac1{2+z^2+x^2}\leq \frac3{4}$.

We have $\displaystyle \frac{1}{2+x^2+y^2}\leq \frac{2}{4+(x+y)^2}$ so we have $\displaystyle \frac{1}{2+x^2+y^2}\leq\frac{2}{4+(3-z)^2}$ and since $\displaystyle f(z)=\frac{1}{4+(3-z)^2}$ is concave we are done.

[quote="mateivld"]$\displaystyle f(z)=\frac{1}{4+(3-z)^2}$ is concave we are done.[/quote]
it's neither convex nor concave on [0,3]

In fact, we have

$\sum_{cycl}\frac{9(a^2+b^2)}{2(a+b+c)^2+9(a^2+b^2)} \ge \frac{3}{2}$

But my proof is too ugly!!

let
$ f(x,y,z)\equal{} \frac{1}{2\plus{}x^{2}\plus{}y^{2}}\plus{}\frac{1}{2\plus{}y^{2}\plus{}z^{2}}\plus{}\frac{1}{2\plus{}z^{2}\plus{}x^{2}}$
assume that z=max(x,y,z)
we have
$ f(x,y,z)\minus{}f(x,\frac{y\plus{}z}{2},\frac{y\plus{}z}{2})\equal{}(y\minus{}z)^2(\frac{z^2\plus{}y^2\minus{}2x^2\plus{}4yz\minus{}4}{(2\plus{}x^2\plus{}z^2)(2\plus{}x^2\plus{}y^2)(8\plus{}4x^2\plus{}(y\plus{}z)^2)}\minus{}\frac{1}{(y^2\plus{}z^2\plus{}2)(4\plus{}(y\plus{}z)^2)})\le 0$
so we will prove that
$ f(x,\frac{y\plus{}z}{2},\frac{y\plus{}z}{2})\le \frac{3}{4}$ ***
let y+z=a so x=3-a
*** which is equivalent to
$ \frac{\minus{}3(5a^2\minus{}4a\plus{}4)(a\minus{}2)^2}{4(5a^2\minus{}24a\plus{}44)(4\plus{}a^2)}\le 0$  (which is obvious true)
q.e.d

[quote="FantasyLover"][quote="mateivld"]We have $ \displaystyle \frac {1}{2 \plus{} x^2 \plus{} y^2}\leq \frac {2}{4 \plus{} (x \plus{} y)^2}$ so we have $ \displaystyle \frac {1}{2 \plus{} x^2 \plus{} y^2}\leq\frac {2}{4 \plus{} (3 \minus{} z)^2}$ and since $ \displaystyle f(z) \equal{} \frac {1}{4 \plus{} (3 \minus{} z)^2}$ is concave we are done.[/quote]
Can someone explain how we are simply 'done' by noticing that the function $ f(z)$ is concave?

Thanks. :)[/quote]
If $ f(z)\equal{}\frac {1}{4 \plus{} (3 \minus{} z)^2}$ is concave,indeed the problem is done....
Because:
$ f(x)\plus{}f(y)\plus{}f(z) \le 3f\left(\frac{x\plus{}y\plus{}z}{3}\right)\equal{}3f(1)$
So $ LHS \le 3\frac{2}{4\plus{}(3\minus{}1)^2}\equal{}\frac{6}{4\plus{}4}\equal{}\frac{3}{4}$
Did you understand it know?

[quote="Vasc"]Let $ x,y,z$ be non-negative numbers such that $ x \plus{} y \plus{} z \equal{} 3$. Then,

$ \frac1{2 \plus{} x^2 \plus{} y^2} \plus{} \frac1{2 \plus{} y^2 \plus{} z^2} \plus{} \frac1{2 \plus{} z^2 \plus{} x^2}\leq \frac3{4}$.[/quote]
This inequality is true for all reals $ x,$ $ y$ and $ z$ such that $ x\plus{}y\plus{}z\equal{}3.$

[quote="Vasc"]Let $ x,y,z$ be non-negative numbers such that $ x \plus{} y \plus{} z \equal{} 3$. Then,

$ \frac1{2 \plus{} x^2 \plus{} y^2} \plus{} \frac1{2 \plus{} y^2 \plus{} z^2} \plus{} \frac1{2 \plus{} z^2 \plus{} x^2}\leq \frac3{4}$.[/quote]
This ineq is in Iran TST 2009.

In Iran 2009 inequality, the condition is $ a^2\plus{}b^2\plus{}c^2\equal{}3$

I dont have a solution but this problem is equivalent to -
Set $ 2a^2 \equal{} x^2 \plus{} y^2 \hdots$

The inequality changes to If $ a \plus{} b \plus{} c \ge 3$ are positive reals then -
$ \frac {3}{2} \ge \sum \frac {1}{1 \plus{} a^2}$

[quote="Ronald Widjojo"]In Iran 2009 inequality, the condition is $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$[/quote]
Dear my friend, the condition is $ a\plus{}b\plus{}c \equal{}3$
See : [url]http://www.mathlinks.ro/viewtopic.php?p=1492633#1492633[/url]

[quote="Ronald Widjojo"]In Iran 2009 inequality, the condition is $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$[/quote]
the condition is a+b+c=3.Vasc post this problem in 2006.

[quote="Vasc"]Let $ x,y,z$ be non-negative numbers such that $ x + y + z = 3$. Then,

$ \frac1{2 + y^2 + z^2} + \frac1{2 + z^2 + x^2} + \frac1{2 + x^2 + y^2}\leq \frac3{4}.$[/quote][quote="arqady"]
This inequality is true for all reals $ x,$ $ y$ and $ z$ such that $ x+y+z=3.$[/quote]$288\left(\frac{3}{4}-\sum{\frac{1}{2+y^2+z^2}}\right)\prod{\left(2+y^2+z^2\right)}$

$=\sum{\Big\{3(1-x)\left[\left(1+2\sqrt{19}\right)x^2+\left(1+4\sqrt{2}+2\sqrt{19}+\sqrt{38}\right)x-10-\sqrt{38}\right]}$

${+\left(x^2-yz\right)\left[\left(3-2\sqrt{2}+6\sqrt{19}+\sqrt{38}\right)x-11+14\sqrt{2}-2\sqrt{19}-\sqrt{38}\right]\Big\}^2}$

$+\left(85+4\sqrt{19}\right)\prod{(y-z)^2}\geq 0.$

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=282926

See also here
[url=http://www.artofproblemsolving.com/community/c6h275815p1492633]Iran TST 2009[/url]
[quote=sandumarin] The following inequality is true:
If $ a,b,c>0,a+b+c=3,k\ge 1 $ then:
$ \frac{1}{2k+a^2+b^2}+\frac{1}{2k+b^2+c^2}+\frac{1}{2k+c^2+a^2}\le\frac{3}{2(1+k)} $
____
Marin Sandu[/quote]

[quote=omegatheo][quote="Soarer"][quote="mateivld"]$ \displaystyle f(z) \equal{} \frac {1}{4 \plus{} (3 \minus{} z)^2}$ is concave we are done.[/quote]
it's neither convex nor concave on [0,3][/quote]
If we study the sign of f'' it results that f''<0 so is concave.[/quote]

Dear friend, f'' is not always negative on [0, 3] !!


prove that if f continuous and f' exist on (a,b) then exist c that a<c<b and
f'(c)<(f(b)-f(a))/(b-a)

Yes, Because he forgot that f is not linear(ax+b)
I know this because bluestar is my classmate and this problem is our homrwork.Please help us!!!! :!:

Let $\ell(x)$ denote the (graph of) the line joining $(a,f(a))$ and $(b,f(b))$.  Say that $f(x)<\ell(x)$ for some $x\in (a,b)$.  Then apply the mean value theorem on $[a,x]$.  Say that $f(x)>\ell(x)$ for some $y\in (a,b)$.  Then apply the mean value theorem on $[x,b]$.  Since $f$ is not linear at least one of the previous conditions is satisfied and we're done.

to tell you the truth I and my classmates don't understand anything of your solution!Can you write it more clearly ?

If I wrote it any more clearly I would be doing your homework for you, which I am loathe to do.  Figure out the details yourself.

North American conditions and semantics don't apply worldwide.

Maybe you misunderstood us.if you don't like to write out.it's OK.I know you is a"5 stars" member, is different from us ( little stars).We will ask others.don't care about us

Don't be surprised if you don't get any answers at all in the future with an attitude like that.


[hide] Let $ y \equal{} x^{2010}$. Then,

$ y \plus{} \frac {1}{y} \equal{} 1 \plus{} y^2$

$ \implies y^2 \plus{} 1 \equal{} y \plus{} y^3$

$ \implies y^3 \minus{} y^2 \plus{} y \minus{} 1 \equal{} 0 \implies (y \minus{} 1)(y^2 \plus{} 1)$.

Hence, $ y \equal{} 1$, $ y \equal{} \pm i$.

However, upon inspection, we find that $ x^{2010} \equal{} \pm i$ does not work with the given equation.

Hence, the only solution is $ x^{2010} \equal{} \boxed{1}$.[/hide]

Edit: Ignore (go a few posts down).

Gaussintraining, I think that is wrong as $ x^{2011} \neq y^2$, as $ y^2 \equal{} x^{4020}$. My solution:

It is equivalent to
$ x^{4020} \plus{} 1 \equal{} x^{2010} \plus{} x^{4021}$.

Case 1: $ x \ge 1$.
Then $ x^{4021} \ge x^{4020}$ and $ x^{2010} \ge 1$. We thus find that $ x \equal{} 1$ is the only solution.

Case 2: $ 0 \le x \le 1$
Then $ x^{4020} \ge x^{4021}$ and $ 1 \ge x^{2010}$. We thus find that $ x \equal{} 1$ is the only solution.

Case 3: $ x < 0$
Then the equation is equivalent to $ x^{4020} \minus{} x^{2010} \plus{} 1 \equal{} x^{4021}$, or $ (x^{2010} \minus{} \frac {1}{2})^2 \plus{} \frac {3}{4} \equal{} x^{4021}$. But then the left side is positive and the right side is negative, so there is no solution in this case.


Here is an inequality I came up with,but I`m preety certain it`s known.and anyway,it`s very easy:
Let a,b,c be positive real numbers with $a^{2}+b^{2}+c^{2}=3$.Prove that the following inequality occurs: \[4(a^{3}+b^{3}+c^{3})\geq3(a+b+c+abc)\] I have two nice solutions for this ineq..maybe there are more :)

$4(a^{3}+b^{3}+c^{3}) \geq (a+b+c)(a^{2}+b^{2}+c^{2})+3abc$

Which Muirhead kills very quickly :).

Letting sums be symmetric,
$\frac{3}{2}\sum a^{3}\geq \sum a^{2}b+\frac{1}{2}\sum abc$

Chebyshev: \[4(a^{3}+b^{3}+c^{3}) \geq 4(a+b+c)\] So we have to prove that \[(a+b+c) \geq 3abc \iff (a^{2}+b^{2}+c^{2})(a+b+c) \geq 9abc\] which is true by rearrangement.

I have a solution which applies some identitys.These are \[a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)\] and \[(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc\] First we multiply the first identity by 4: \[4(a^{3}+b^{3}+c^{3})=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)+12abc \Rightarrow \]  \[4(a^{3}+b^{3}+c^{3})=(a+b+c)(3-ab-bc-ca)+12abc \Rightarrow \]  \[4(a^{3}+b^{3}+c^{3})=3(a+b+c+abc)-(a+b+c)(ab+bc+ca)+9abc\] if we prove that $9abc-(a+b+c)(ab+bc+ca)\geq0$ the problem is solved.
Now we apply the second identity.We have: \[(a+b+c)(ab+bc+ca)=(a+b)(b+c)(c+a)+abc\]  \[(a+b)(b+c)(c+a)+abc\geq9abc\leq >\]  \[(a+b)(b+c)(c+a)\geq8abc\] which is true applying AM-GM for a,b;b,c;c,a;
What do you think? :)

Very cool proofs :D
The nice thing is that my other two proofs are different :)  :)
Here is my first solution:
By Cebisev,we have \[3(a^{3}+b^{3}+c^{3})\geq(a+b+c)(a^{2}+b^{2}+c^{2})\] ,which means that \[3(a^{3}+b^{3}+c^{3})\geq3(a+b+c)\] ,so we only have to prove \[a^{3}+b^{3}+c^{3}\geq 3abc\] Which is just AM-GM :)

One more nice solution:
Let the power mean of order $t$ of numbers $a,b,c$ be $\mu_{t}$, i.e. $\mu_{t}= \left(\frac{a^{t}+b^{t}+c^{t}}{3}\right)^{\frac1t}$ for $t\neq 0$ and $\mu_{0}= \sqrt[3]{abc}$. Then the ineqality rewrites as \[4\cdot 3\mu_{3}^{3}\geqslant 3\cdot \left( 3\mu_{1}+\mu_{0}^{3}\right)\] for $\mu_{2}=1$. Then cancelling $3$ at both sides and homogenising, we get \[4\mu_{3}^{3}\geqslant 3\mu_{1}\cdot \mu_{2}^{2}+\mu_{0}^{3}.\] But from the power mean inequality we have $3\mu_{3}^{3}\geqslant 3\mu_{1}\mu_{2}^{2}$ and $\mu_{3}^{3}\geqslant \mu_{0}^{3}$ which ends the proof. :)


[quote="rboyal"]my last question guys.  I have this word question, just having trouble with it.  Thank you for the help, i really apericate it. Thank you

4) Milton ran north at 8 kilometers per hour.  Four hours later Harret set out in pursuit at 16 kilometers per hour.  How long did it take harret to catch milton?[/quote]

Unfortunately, I'm illiterate.

[hide="Solution"]Basically, Milton gets a 4 hour head start on Harret, which for him, is a $(4*8)=32 kilometer$ head start. Harret runs 8 kilometers faster than Milton, so every hour Harret will gain 8 kilometers on Milton. All you do now is $\frac{32} {8}=\boxed {4 hours}$[/hide]

thats what i was thinking, but i don't think that this question could be that straight forward, but thanks for you help anyways!!


[hide="My sorution"]

so just divide the hexagon into 6 equilateral triangles. so then the radius of the circle is equal to the height of one of those equilateral triangles, which is $\sqrt{3}$. So we find the area of the circle is
\[\pi(\sqrt{3})^{2}=\boxed{3\pi}\]
[/hide]

[quote="ckck"]A circle is inscribed in a regular hexaogn with side length 2. Find the area of the circle.[/quote]

[hide]Split the hexagon into 6 equilateral triangles.  The height of each triangle is $\sqrt{3}$.  This is also the radius of the circle so the area $3\pi$.[/hide]


Where could i find previous national mathcounts tests to take for practice?

try your MC coach if he or she is experienced then they might have past nats tests, you can also try the MC store at mathcounts.org.

thanks, are there any websites that have these tests or do you have to buy books with them inside

I'm pretty sure umm, this site might not be the most legit site in da world.

I was lucky enough to download that zip file and will work on uploading that so everyone can download it.
:)

Ohh... then yashar.biz is violating tons of stuff too.
Think I should just abort the uploading mathcrazed?

I don't think it does, I mean isn't post- 2001 tests all over the internet? But it's like someone went and deleted all pre2000, and all year national tests from the web.

um for some reason your link isnt up anymore it said that it violated terms of service.  Is there another one or is that the only one.

go here for more information on past tests

http://www.mathlinks.ro/Forum/viewtopic.php?t=133189


I used to live there. Anyone go to stillwater middle school/????

I used to live there too! Went 2-5 at Westwood, then left for Jenks.


Let $a$, $b$, $c$ be positive real numbers. Find all real numbers $x$, $y$, $z$ such that:

$x + y + z = a + b + c$
$4xyz - (a2x + b2y + c2z) = abc$

Not sure if this is correct.

====================

Well, if x,y,z are positive reals:
------

We have $4xyz - abc = \sum a^2x = \sum \frac{a^2}{1/x} \ge \frac{(\sum a)^2}{\sum 1/x} = \frac{xyz(x+y+z)^2}{xy+yz+zx}$

So $(\frac{x+y+z}{3})^3 \ge abc = \frac{4xyz(xy+yz+zx) - xyz(x+y+z)^2}{xy+yz+zx}$,

whence $(\sum xy)(\sum x)^3 \ge 108xyz(\sum xy) - 27xyz(\sum x)^2$.

Now, let $p,q,r = \sum x, \sum xy, xyz$.  We have $108rq = p^3q + 27p^2r = p^2 ( pq + 27r ) \ge (3q)(pq + 27r) \ge (3q)(36r) = 108rq$ with equality when $p^2 = 3q$, $pq \ge 9r$, so $x=y=z$.  Plugging it back in and doing a similar method yields that $a,b,c$ must be equal for it to work.


there is no error in my latex file but gives a warn AS Latex Warning: 'h' float specifier changed to 'ht'. ..what does this warn mean? any effect on my file?

It is a reminder that 'h' does not mean exactly [i]here[/i] for a float. It is placed [i]here[/i] only if there is enough room for it, otherwise it appears at the top of the next page ie 'ht'. This warning is reminding you of this. So you shouldn't use \begin{figure}[h] but tell LaTeX where you want the figure if it doesn't fit with eg \begin{figure}[ht]. The warning is just telling you this and that it has replaced [h] by [ht].
See any LaTeX book or [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=lshort-english]The Not So Short Introduction to LaTeX2e[/url]


Do you all have any tips on how to improve speed on the AMCs?  It seems like that's my biggest problem and also it's why I don't get higher scores :-(.

I think AMC/AIME/USAMO is more about not wasting time rather than pure speed, by wasting time I mean spending a long time on a problem because of not seeing something "obvious"

i think that drawing accurate diagrams is probably a big waste of time

also, if you don't recognize a method to do a problem in about 30 seconds, its probably best to look further down

[quote="krustyteklown"]i think that drawing accurate diagrams is probably a big waste of time[/quote]

lol, yeah I saw your question booklet with your scribblings...but hey, you got a 150 so it works for you :)

On AIME/Olympiads though it would probably be much wiser to draw neat and accurate drawings since you'll probably be using them for a large amount of time and it may even help you to make conjectures about stuff in the diagram.

See this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=73652]thread[/url].

Edit: Oops, wrong topic.  Sorry.

MCrawford has a blog entry on this subject......

Yeah dude well it'd help if you could link me.  =)

http://www.artofproblemsolving.com/Forum/weblog.php?w=7

Well, not exactly related..... but close.

I have the same problems as mentioned in the blog entry, and am aiming to do improve my score signifiantly on B with better speed.


probability
MATHCOUNTS
LaTeX
ratio
geometric series

Suppose there is a wheel with thirty equal spaces to land on. You have an equal chance to land on each space.

There are 16 blue spaces, 8 red spaces, and 6 white spaces.

Landing on a blue space means you win.
Landing on a red space means you lose.
And landing on a white space means you must spin again, and you have neither won nor lost.

The question is, what are your chances of winning? and of losing?
They should add to 1.

I thought of this question when we had an open house night at my school, with games such as these. I think the best way to solve it is to take 6 out of the denominator and then put the winning or losing number over that. Is this method correct?

I just want to know the answer for my own benefit. My stats teacher isn't the most intelligent man alive, and he could not help me.

for the first time you spit i think it must be still 16/30 for winning and 8/30 because when you spit it there is a chance to turn into white again... and you can never win or lose by spitting to white forever. And if you will go for 2nd spit 6/30*16/30

Well, there's a problem like this in the Mathcounts section, and yes, the chance of winning is $\frac{16}{24}=\frac{2}{3}$.

Look at how you can win:
You win on your first spin OR your second OR your third OR your fourth .......
You either win or spin again. So we a win looks like: W, SW, SSW, SSSW, SSSSW, SSSSSW,...........This is just a simple geometric sequence.

$P(win)=  \sum_{k=0} \displaystyle \left(\frac{6}{30} \right)^k  \cdot \frac{16}{30}= \frac{16}{30} \displaystyle \left( \frac{1}{1- \frac{6}{30}} \right)= \frac{16}{30} \cdot \frac{30}{24}= \frac{16}{24} = \frac{2}{3}$.

The probability of losing is:  $1-P(win)= \frac{1}{3}$. $\clubsuit$

Yeah I set it up like thejerk did with the infinite series.  I am not so eloquent as he is with Latex but if you realize you have a 16/30 chance of winning each spin and your chances of getting more and more spins will be a geometric series with ratio 6/30.

You should get W = 16/30( 1 / (1 - 6/30 ) ) = 16 / 30 ( 5 / 4) = 2 / 3.

Obviously in this game you win or you lose and since your chances must total to 1:

L = 1 - W = 1 / 3.


the polynomial
$4 x^{4}-a x^{3}+b x^{2}-c x+5$
 has four positive (real)roots
such that
$\frac{r_{1}}{2}+\frac{r_{2}}{4}+\frac{r_{3}}{5}+\frac{r_{4}}{8}=1$
find them

[quote="10000th User"][hide]By Vieta, we have $r_{1}r_{2}r_{3}r_{4}=\frac54$

On the other hand, by AM-GM
\[\begin{aligned}\frac{r_{1}}2+\frac{r_{2}}4+\frac{r_{3}}5+\frac{r_{4}}8&\ge4\sqrt[4]{\frac{r_{1}}2\frac{r_{2}}4\frac{r_{3}}5\frac{r_{4}}8}\\ 1&\ge2^{2}\sqrt[4]{\frac{r_{1}r_{2}r_{3}r_{4}}{2^{6}\cdot5}}\\ \frac1{2^{8}}&\ge\frac{r_{1}r_{2}r_{3}r_{4}}{2^{6}\cdot5}\\ \frac54&\ge r_{1}r_{2}r_{3}r_{4}\end{aligned}\]
Since equality is achieved, the quantities $\frac{r_{1}}2,\frac{r_{2}}4,\frac{r_{3}}5,\frac{r_{4}}8$ are equal and so the roots are $2r,4r,5r,8r$. Using the same Vieta equation, we find $r^{4}=\frac1{2^{8}}$. Since the roots are positive real, $r=\frac14$, and so the roots are $\frac12,1,\frac54,2$[/hide][/quote]

wow that's a nice solution
how did you think of using AM-GM?

[quote="tjhance"]wow that's a nice solution
how did you think of using AM-GM?[/quote]
When working in a problem, it's more like analyzing it and testing out stuff. But, there are a few things that suggested me to try AM-GM: the roots are positive real, a sum is given related to roots, the product of the roots is given... see what I mean?  :surf:


without using the calculator, show that
$cos(\frac{2\pi}{5})+cos(\frac{2\pi}{5})=-\frac{1}{2}$

[quote="yrhc@eht, corrected"]without using the calculator, show that
$cos(\frac{2\pi}{5})+cos(\frac{4\pi}{5})=-\frac{1}{2}$[/quote]

Let $z\in \mathbb C$ ,$z=cos \left(\frac{2\pi}{5}\right)+i\cdot sin \left(\frac{2\pi}{5}\right)$
So we have:
cos $\left(\frac{2\pi}{5}\right)=\frac{z^{2}+1}{2z}$ and
$cos \left(\frac{4\pi}{5}\right)=\frac{z^{4}+1}{2z^{2}}$
therefore we have to prove that
$\frac{z^{3}+z+z^{4}+1}{2z^{2}}=-\frac{1}{2}$ $\Leftrightarrow$ $1+z+z^{2}+z^{3}+z^{4}=0$ $\Leftrightarrow$ $\frac{z^{5}-1}{z-1}=0$ ,wich is true because $z^{5}=cos(2\pi)+i\cdot sin(2\pi)=1$

x=18 degree.Then 5x=90 or 2x=90-3x then
sin2x=sin(90-3x)
2sinx.cosx=cos3x=4cos^3(x)-3cosx
we divide this equation by cosx
2sinx=4cos^2(x)-3
2 sinx=4.(1-sin^2(x)-3
if sinx=t then 2t=4.(1-t^2)-3
t=(-1+root5)/4
sin18=cos72=(-1+root5)/4
sin54=cos36=1-2sin^2(18)=(1+root5)/4
cos144=-sin54==-(1+root5)/4

cos72+cos144=-1/2


please explain in details

ThanX

what is the probability that five card hand has:

a) at leat one of each of the four values: ace, king, queen, and jack?
b) the same number of hearts and spades?


trigonometry
integration
calculus
function
geometry
3D geometry
floor function

The game is simple, count numbers using creative ways to express your numbers. Let's see how far we get.

Ill start:

$ \sin^2{\alpha}\plus{}\cos^2{\alpha}$

This has been done before, but oh well.

$ 2\equal{}\sqrt{2\plus{}\sqrt{2\plus{}\sqrt{2\plus{}\cdots}}}$

The old topic was locked.

To avoid this, you should stipulate that spam posts be not allowed. In other words, no "the post above me+1" or anything like that.

[b]The number of little pigs' houses that the Big Bad Wolf tried to blow down.[/b]

[b]the amount of piggies in the piggy song

This little piggy went to town...[/b]

$ \int_{\sqrt[3]{9}}^{3}x^2\hspace{1 mm}dx$ (more calculus-like)

$ \pi(18)$

(where $ \pi(n)$ is the [url=http://en.wikipedia.org/wiki/Prime-counting_function]prime counting function[/url])

The largest perfect cube in the Fibonacci sequence.

[quote="ZhangPeijin"][quote="alanchou"]$ \int_{\sqrt [3]{9}}^{3}x^2\hspace{1 mm}dx$ (more calculus-like)[/quote]
Stop insulting my inability to do integrals that aren't linear.
:(
\[ \left(\sqrt {\frac {\sqrt [3]{\sqrt {4824612 \plus{} \sqrt {4824612 \plus{} \sqrt {4824612 \plus{} ...}}}} \plus{} 1337}{2^{2}5^{2} \plus{} 3B_{13}}}\right)^{\dfrac{2^{10}}{3(4\cdot43691 \minus{} 1) \minus{} 1}}
\]
[/quote]

you might want to recheck that, I don't think it's even rational

$ \frac{d}{dx}(11x)$

that's 11 for you common folks

[quote="mz94"][u][i][b]X[/b][/i][/u][/quote]
I think that underlining it may affect the value.

$ 12i^{2349205286976}$

[quote="alanchou"][quote="ZhangPeijin"][quote="alanchou"]$ \int_{\sqrt [3]{9}}^{3}x^2\hspace{1 mm}dx$ (more calculus-like)[/quote]
Stop insulting my inability to do integrals that aren't linear.
:(
\[ \left(\sqrt {\frac {\sqrt [3]{\sqrt {4824612 \plus{} \sqrt {4824612 \plus{} \sqrt {4824612 \plus{} ...}}}} \plus{} 1337}{2^{2}5^{2} \plus{} 3B_{13}}}\right)^{\dfrac{2^{10}}{3(4\cdot43691 \minus{} 1) \minus{} 1}}
\]
[/quote]

you might want to recheck that, I don't think it's even rational

$ \frac {d}{dx}(11x)$

that's 11 for you common folks[/quote]
It simplifies to $ 3^{2}$
Unless I can't calculate using a calculator...
[b]WHEN _ DINE TOGETHER, THE FIRST TO RISE IS THE FIRST TO DIE![/b]
-Some crazy old lady in Harry Potter

[quote="33286"][quote="mz94"][u][i][b]X[/b][/i][/u][/quote]
I think that underlining it may affect the value.
[/quote]

No, only overlining.

The lucky # for a party of travelers in Tolkien's [i]Hobbit[/i].

404:  One of the most famous computer error message numbers, also the area code of Atlanta, Georgia (I think).

$ \left \lfloor \frac {5^{12}}{10^2} \right \rfloor \mod {1000}$

$ 406\equal{}406/sin{90}$ meh, i need to do integration...

$ 410 \equal{} \frac {111110}{271}$

Another rule: your method of "counting" MUST uniquely identify your number.

413: the second number over 400, whose digits add up to 8.

415; the third positive multiple of 5 greater than 400.

The (((1st prime)th prime)th prime)th power of two times the square root of ((3^2)+(4^2)+((3*4)^2))


geometry
circumcircle
Triangle
reflection
IMO Shortlist
concurrency
IMO Longlist

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

[quote="orl"]A line $l$ is drawn through the intersection point $H$ of altitudes of acute-angle triangles. Prove that symmetric images $l_a, l_b, l_c$ of $l$ with respect to the sides $BC,CA,AB$ have one point in common, which lies on the circumcircle of $ABC.$[/quote]

See the note "Anti-Steiner points with respect to a triangle" on my website http://de.geocities.com/darij_grinberg/ .

  Darij

[quote="vickymano"]Hey, the link is broken !![/quote]

I guess I can rectify that situation :) Here's my solution:

Let $AD, BE, CF$ be altitudes of $\triangle ABC$. Also, let $l$ meet sides $BC, CA, AB$ of $\triangle ABC$ at points $K, L, M$ respectively. WLOG we may assume that $K$ lies on the extension of $CB$ beyond $B$, and points $L$ and $M$ lie on the boundary of the triangle. It suffices to show that $l_b, l_c$ concur on the circumcircle of $\triangle ABC$. Then, by analogous arguments $l_a, l_b$ also concur on the circumcircle of $\triangle ABC$, hence $l_a, l_b, l_c$ concur on the circumcircle of $\triangle ABC$.

Now, it is well-known that the reflections of $H$ with respect to the sides of $\triangle ABC$ all lie on the circumcircle of $\triangle ABC$. So let $Q, R$ be the reflections of $H$ in sides $AC, AB$ respectively. Also, note that $RM \equiv l_c$ and $QL \equiv l_b$. So, our problem reduces to proving that $RM$ and $QL$ concur on the circumcircle of $\triangle ABC$, which is also the circumcircle of $\triangle RAQ$. Also let $RM$ intersect $QL$ at $O$. We have:

$\angle ARM+\angle AQL=\angle AHM+\angle AHL=180^0$

Also, the angle sum in pentagon $RAQLM$ is $540^0$, hence we have that:

$\angle RAQ+\angle RML+\angle QLM=360^0 \leftrightarrow \angle RAQ+\angle ROQ=180^0$

Hence $R, A, Q, O$ are concyclic, and therefore $O$ does indeed lie on the circumcircle of $\triangle ABC$, as desired.

Dear Mathlinkers,
see also

http://jl.ayme.pagesperso-orange.fr/Docs/Droite%20de%20Simson%20de%20Fe.pdf  p. 7-12.

Sincerely
Jean-Louis


:? $1^*.$ If a,b,c > 0 and a+b+c=3, then $9(ab+bc+ca)\le 22+5abc.$
 :mad: $2^*.$ In the triangle ABC: If s=3, then $18Rr\le 11+3r^2$ :rotfl:

[quote="levi"]$1^*.\ \ \ If\ \ \ a,b,c\ >\ 0\ and\ a+b+c=3\ \ \ then\ \ \ 27(ab+bc+ca)\le 9+15abc.$
$2^*.\ \ In\ the\ triangle\ ABC:\ \ If\ \ \ s=3\ \ \ then\ \ \ 18Rr\le 11+3r^2$[/quote]

 Levi,I think the first one is obtained from Schur.Here is her sister: if $a+b+c=1$ then we have

   $7(ab+bc+ca)\leq2+9abc$. ;)

Anyway,I will make calculations when I get to Bucharest,because I have to leave in an hour,but I'm sure it is Schur. ;)

The second ineq. is obtaining from the first ineq. through the substitution a:=s-a, b:=s-b, c:=s-c, where the new a, b, c are the sides of the triangle ABC. :roll:  :o  :idea:  :!:  :!:

Levi,you stated that if there are $a,b,c>0$ such that $a+b+c=3$ then the following inequality holds:

                   $9(ab+bc+ca)\leq{22+5abc}$

 By Schur,we have $(a+b+c)^3+9abc\geq4(a+b+c)(ab+bc+ca)$.But, $a+b+c=3$ so the inequality becomes
 $12(ab+bc+ca)\leq9(abc+1)$.I hope I haven't made some stupid mistakes. :?

 Anyway,Levi I think this works.Do you agree with me? :huh:


With 48 m of fence, Matt enclosed a square corral for his horse.  With his 48 m of fence, Nick built a rectangular corral which was twice as long as wide.  What is the ratio of the area of Matt’s corral to the area of Nick’s corral?

A.1:1	B.2:1	C.3:2	D.7:6	E.9:8

[hide=" click to reveal hidden content"]

I think this is $ 9: 8$:

Matt's corral is a $ 12 \times 12$ square with area $ 144$, Nick's is a $ 16 \times 8$ rectangle with area $ 128$.

Therefore the ratio is $ 144: 128 \equal{} 18: 16\equal{}9: 8$.

[/hide]


trigonometry
algebra
polynomial
complex numbers

Find the sum

$\sum_{k=1}^{89}\tan^2{k}$

i know i posted this in the intermediate forum, but it was too difficult, and i still have not gotten an answer, so hopefully, i can get a solution here...

Let's find a polynomial such that this $89$ numbers are the roots of it, then the coefficients will give the sum.
We have $(cos(x)+i \cdot sin(x))^n = cos(nx) + i \cdot sin (nx)$
$\implies (1+i \cdot tan(x))^n = \frac{1}{cos(x)^n} (cos(nx) + i \cdot sin (nx))$.
Write $z: = tan(x)$.
Thus, $\sum_{k=0}^n \binom{n}{k} i^k z^k = \frac{1}{cos(x)^n} (cos(nx) + i \cdot sin (nx))$.
Now let $n=180$ and let $x$ having 'integer-valued degree', so $\sum_{k=0}^{180} \binom{180}{k} i^k z^k = \frac{1}{cos(x)^n} (cos(nx) + i \cdot sin (nx))= \frac{(-1)^x}{cos(x)^n}$.
Now look at the imaginary part, giving:
$z \sum_{k=0}^{89} \binom{180}{2k+1} (-1)^k (z^2)^k=0$.
But this is the polynomial we wanted, since its roots are $tan(k^\circ)^2$ (we also counted $tan(0)=0$, which can be neglected).
So $\sum_{k=1}^{89} tan(k^\circ)^2 =\frac{ \binom{180}{177}}{\binom{180}{179}} = \frac{15931}{3}$.

nice solution, this problem has been bugging me...

[quote="ZetaX"]
Now let $n=180$ and let $x$ having 'integer-valued degree', so $\sum_{k=0}^{180}\binom{180}{k}i^{k}z^{k}= \frac{1}{cos(x)^{n}}(cos(nx)+i \cdot sin (nx))= \frac{(-1)^{x}}{cos(x)^{n}}$.
Now look at the imaginary part, giving:
$z \sum_{k=0}^{89}\binom{180}{2k+1}(-1)^{k}(z^{2})^{k}=0$.
.[/quote]

Could anyone explain these lines ?
In the sum he put n=180 but next the <  =  > he left it as n

Ah, that commes from copying ^^
Just see $n$ as $180$ from that point on it is said to set on hat value.

[quote="mathson"][quote="ZetaX"]since its roots are $ tan(k^\circ)^2$ [/quote]
Why?
I can't understand this.[/quote]

I'm still waiting.  :D

*finds ancient thread and feels compulsed to answer*

Well, the polynomial should be $ z \sum_{k \equal{} 0}^{89} \binom{180}{2k \plus{} 1} ( \minus{} 1)^k (z)^k$ (without the square). In more generallity, if $ \alpha$ is a root of $ p(x^2)$, then $ \alpha^2$ is one of $ p(x)$.


I haven't read his books, but my math team coach always talks about him.
His logic problems are really great.

[i]What is the name of this book[/i] is really good.

[quote="Bob"]Has anyone read any of Raymond Smullyan's books? He writes mainly about mathematical logic and philosophy. Some of the titles include: [i]The Riddle of Scheherazade, The Tao is Silent, To mock a Mockingbird, and What is the Name of this Book?[/i] I haven't read any of his books and would like to know what other people think of them.[/quote]

  I've read both "The Riddle of Scheherazade" and "The Lady or The Tiger" and I can personally tell you that they are both excellent books.  Mr. Smullyan has the unique ability to present logic problems in the most intuitive and fascinating manner.  His books are wonderfully organized and talk-in-depth about logic problems that have puzzled mathematicians and logicians for centuries.  This book lays a solid foundation for begginers and ametuers alike, who are interested in the world of logic.

Have I read Smullyan? Several of them. One that is often ignored but should be of interest to anyone seriously interested in Logic (real Logic, not just puzzles) is his "Forever Undecided". Of course, his best from the point of view of literature is his "The Tao is silent", with the best chapter being "Is God a Taoist", although you can ignore the part about angels.


calculus
integration
trigonometry
function
limit
calculus computations

Decide whether the improper integral is convergent or divergent:

(x=0 to x=1) ∫1 / (sin x) dx

Any ideas?

Sure. The only end that matters is the $x\to0^+$ end. As $x\to 0,\sin x$ "looks like" $x.$ Exploit that.

Thanks once again for your help Kent... I feel kind of stupid, but I'm not sure how to calculate the integral of 1 / sin(x)... if somebody could help I'd be very grateful!

That misses the point. You can decide whether this integral converges without finding the antiderivative.

Here's a version of the limit comparison test for improper integrals: Supose $f$ and $g$ are positive functions on $(a,b)$ and suppose that $\lim_{x\to a^+}f(x)=\lim_{x\to a^+}g(x)=\infty$ and that $\lim_{x\to a^+}\frac{f(x)}{g(x)}=1.$

Then (assuming there are no other reasons for these integrals to be improper)

$\int_a^bf(x)\,dx$ converges if and only if $\int_a^bg(x)\,dx$ converges.

Note that $\lim_{x\to0}\frac{\frac1{\sin x}}{\frac1x}=\lim_{x\to0}\frac{x}{\sin x}=1.$

Hence the convergence of $\int_0^{\frac{\pi}2}\frac1{\sin x}\,dx$ is equivalent to the convergence of $\int_0^{\frac{\pi}2}\frac1x\,dx.$ And that diverges.

As an aside, it isn't to hard to guess what the antiderivative of $\frac1{\sin x}$ is, if you know what the antiderivative of $\sec x$ is and if you note that $\frac1{\sin x}=\csc x.$ But from my point of view, that's a waste of time. Think about the sizes of functions, and see that $\frac1{\sin x}$ near zero as something that might as well be $\frac1x$ near zero - too big for a convergent integral.

Ahh totally makes sense... I was basically just overcomplicating the problem... Thanks so much for your help!

A way to compute the integral is to notice that $\frac{1}{\sin x} = \frac{\sin x}{\sin^2 x} = \frac{\sin x}{1 - \cos^2 x}$.


induction
inequalities
floor function
geometric sequence
algebra unsolved
algebra

Professor Piraldo takes part in soccer matches with a lot of goals and judges a match in his own peculiar way. A match with score of $m$ goals to $n$ goals, $m\geq n$, is [i]tough[/i] when $m\leq f(n)$, where $f(n)$ is defined by $f(0) = 0$ and, for $n \geq 1$, $f(n) = 2n-f(r)+r$, where $r$ is the largest integer such that $r < n$ and $f(r) \leq n$.

Let $\phi ={1+\sqrt 5\over 2}$. Prove that a match with score of $m$ goals to $n$, $m\geq n$, is tough if $m\leq \phi n$ and is not tough if $m \geq \phi n+1$.

Sketch of the solution:

1. If n is written in the fibonacci basis (as a sum of distinct fibonacci numbers containing no neighbors), then the representation of f(n) is just that of n with a 0 in the end.
Proof: induction (r will be such that f(r) = n or f(r) = n - 1 depending on the last digit of the fibonacci representation of n).

2. f(n) = phi*n + b, where 0 < b < 1.
Proof: You get f(n) > phi*n with an easy induction.
 For the other inequality, express n and f(n) by using the closed form for fibonacci numbers, and majorate b by an infinite geometric progression which adds up to 1.

Best Regards,
Marcio

Actually, one can prove that $f(n) = \lfloor \phi n+\phi^{-1}\rfloor$. Or, equivalently, $f(n)+1 = \lfloor(n+1)\phi)\rfloor$. My proof was an induction on $n$, albeit a really ugly one (lots and lots of ugly inequalities with integer/fraction parts).

I prefer the Fibonacci base solution (Zeckendorf's theorem, isn't it?).

[quote=HansV]Itˋs not difficult to show that f(2i) = 3i and f(2i -1) = 3i - 1.
(i being any positive integer).

Then just make the two  comparisons for each of the two statements.[/quote]

Sorry for  the bump, but $f(2n) = 3n$ doesn't hold, though... Because:
$f(1) = 2\cdot 1 - f(0) + 0 = 2$
$f(2) = 2\cdot 2 - f(1) + 1 = 3$
$f(3) = 2\cdot 3 - f(2) + 2 = 5$
$f(4) = 2\cdot 4 - f(2) + 2 = 7$

So $f(2\cdot 2) = f(4) = 7 > 3 \cdot 2$.

$f(2n-1) = 3n-1$ doesn't seem to hold either, because:

$f(2\cdot 5 - 1) = f(9) = 15 > 14 = 3\cdot 5 - 1$

Through small cases it [i]seems[/i] that $f(2n) \geq 3n$ and $f(2n-1) \geq 3n-1$, is that what you meant?


geometry
circumcircle
perimeter
inequalities
trigonometry
geometry proposed

Let $ R,r,s$ be the circumcenter, radius of the incircle, and half perimeter of $ \triangle ABC$. Show that $ \triangle ABC$ is a right triangle iff $ 2R\plus{}r\equal{}s$ :)

hello Mathias,setting $ R\equal{}\frac{abc}{4A}$ and $ r\equal{}\frac{A}{s}$ in your inequality then we get
$ abc\plus{}2(s\minus{}a)(s\minus{}b)(s\minus{}c)\equal{}2s\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$
after squaring and factoring we get
$ (a^2\plus{}b^2\minus{}c^2)(a^2\plus{}c^2\minus{}b^2)(b^2\plus{}c^2\minus{}a^2)\equal{}0$.
A nice problem.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello Mathias,setting $ R \equal{} \frac {abc}{4A}$ and $ r \equal{} \frac {A}{s}$ in your inequality then we get
$ abc \plus{} 2(s \minus{} a)(s \minus{} b)(s \minus{} c) \equal{} 2s\sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$
after squaring and factoring we get
$ (a^2 \plus{} b^2 \minus{} c^2)(a^2 \plus{} c^2 \minus{} b^2)(b^2 \plus{} c^2 \minus{} a^2) \equal{} 0$.
A nice problem.
Sonnhard.[/quote]
WOW! I can barely manage to multiply it out after squaring ! :P Did you use a computer?

[quote="Dr Sonnhard Graubner"]hello Mathias, i'm sitting in front of my PC, but this was a known theorem and so was the result clear and i used my PC only for the control of the result, or with Polya, i worked backwards.
Sonnhard.[/quote]
Ok. I had a feeling that it was well-known, but I had never seen it before :)

This was my solution:

$ 2R \plus{} r \equal{} s \iff$

$ 2(2R \plus{} r)^2 \equal{} 2s^2 \iff$

$ 8R^2 \equal{} (2s)^2 \minus{} 2(s^2\plus{}r^2\plus{}4Rr) \equal{} a^2\plus{}b^2\plus{}c^2 \iff$

$ \sin^2 A \plus{} \sin^2 B \plus{} \sin^2 C \equal{} 2$

I have seen this one a couple of times before, though :) The idea is just to see that $ \sin^2 C \equal{} \sin^2 A\plus{}B \equal{} (\sin A \cos B \plus{} \sin B \cos A)^2$. After some random manipulation, it was equivalent to $ \cos A \equal{} 0 \iff A \equal{} 90$ or $ \cos B \equal{} 0 \iff B \equal{} 90$ or $ \tan A \equal{} \tan 90 \minus{} B \iff A \equal{} 90 \minus{} B \iff C \equal{} 90$. :)

[quote="Mathias_DK"] [color=darkred]Let $ R,r,s$ be the circumradius, inradius and semiperimeter of $ \triangle ABC$. Show that $ \triangle ABC$ is right iff $ 2R + r = s$ .[/color] [/quote]
[color=darkblue][b][u]Proof 1[/u].[/b]

Prove easily that the equation $ f(x)\equiv sx^3 - (4R + r)x^2 + sx - r = 0$ has the roots $ \tan\frac A2$ , $ \tan\frac B2$ , $ \tan\frac C2$ , i.e.

$ \sum\tan\frac A2 = \frac {4R + r}{s}$ , $ \sum\tan\frac B2\tan\frac C2 = 1$ , $ \prod\tan\frac A2 = \frac rs$ . Observe that $ f(1) = 2(s - 2R - r)$ . Therefore,

$ 2R + r = s$ $ \Longleftrightarrow$ $ f(1) = 0$ $ \Longleftrightarrow$ $ 1\in\left\{\tan\frac A2,\tan\frac B2,\tan\frac C2\right\}$ $ \Longleftrightarrow\ 90^{\circ}\in \{A,B,C\}$ $ \Longleftrightarrow$ $ \triangle ABC$ is right.

[b][u]Remark[/u].[/b] Show analogously that $ \underline{\overline{\left\|\ \begin{array}{ccc} s = \sqrt 3\cdot (R + r) & \Longleftrightarrow &{ 60^{\circ}\in \{A,B,C\}\ } \\
 \\
s\sqrt 3 = 3R + r & \Longleftrightarrow & 120^{\circ}\in \{A,B,C\}\end{array}\ \right\|}}$ .[/color]

[color=darkblue][b][u]Proof 2[/u].[/b]

 $ 2R + r = s$ $ \Longleftrightarrow$ $ 2 + \frac rR = \frac sR$ $ \Longleftrightarrow$ $ 1 + \cos A + \cos B + \cos C = \sin A + \sin B + \sin C$ $ \Longleftrightarrow$

$ \cos^2\frac A2 + \sin\frac A2\cos\frac {B - C}{2} = \sin \frac A2\cos\frac A2 + \cos\frac A2\cos\frac {B - C}{2}$ $ \Longleftrightarrow$

$ \left(\cos\frac A2 - \cos\frac {B - C}{2}\right)\left(\cos\frac A2 - \sin\frac A2\right) = 0$ $ \Longleftrightarrow$ $ \cos\frac A2 = \cos\frac {B - C}{2}\ \ \vee\ \ \cos\frac A2 = \sin\frac A2$ $ \Longleftrightarrow$

$ A = B - C\ \ \vee\ \ A = C - B\ \ \vee\ \ \tan\frac A2 = 0$ $ \Longleftrightarrow$ $ B = 90^{\circ}\ \ \vee\ \ C = 90^{\circ}\ \ \vee\ \ A = 90^{\circ}$ .

[b][u]Remark[/u].[/b] I used the well-known relations $ \left\|\begin{array}{c} \sin A + \sin B + \sin C = \frac sR \\
 \\
\cos A + \cos B + \cos C = 1 + \frac rR\end{array}\right\|$ .

[/color][color=darkblue][b][u]Proof 3[/u].[/b]

Suppose w.l.o.g. $ A\ne 90^{\circ}$ . Therefore, $ 2R + r = s$ $ \Longleftrightarrow$ $ 2R(1 - \sin A) + (s - a)\tan\frac A2 = s - a$ $ \Longleftrightarrow$

$ 2R\left(\sin\frac A2 - \cos\frac A2\right)^2 + (s - a)\left(\tan\frac A2 - 1\right) = 0$ $ \Longleftrightarrow$ $ 2R\left(\sin\frac A2 - \cos\frac A2\right)\cos\frac A2 + (s - a) = 0$ $ \Longleftrightarrow$

$ R(\sin A - 1 - \cos A) + (s - a) = 0$ $ \Longleftrightarrow$ $ a - 2R(1 + \cos A) + (b + c - a) = 0$ $ \Longleftrightarrow$ $ 2R(1 + \cos A) = b + c$ $ \Longleftrightarrow$

$ 1 + \cos A = \sin B + \sin C$ $ \Longleftrightarrow$ $ 2\cos^2\frac A2 = 2\cos\frac A2\cos\frac {B - C}{2}$ $ \Longleftrightarrow$ $ \cos\frac A2 = \cos\frac {B - C}{2}$ $ \Longleftrightarrow$

$ A = B - C\ \ \vee\ \ A = C - B$ $ \Longleftrightarrow$ $ B = 90^{\circ}\ \ \vee\ \ C = 90^{\circ}$ . In conclusion, $ 90^{\circ}\in \{A,B,C\}$ .

[b][u]Remark[/u].[/b] I used  the well-known relations $ r = (s - a)\tan\frac A2$ , $ a = 2R\sin A$ and usual trigonometrical formulas.[/color]

This follows from the identity
\[ \cos A \cos B \cos C \equal{} \frac{s^2 \minus{} (r \plus{} 2R)^2}{4R^2}.\]

Dear Mathlinkers,
is it possible to imagine a synthetical proof ?
Sincerely

Let me give a  geometry method right now：
[b]Lemma[/b]：
let $ r,r0,r1,r2$ be the incircle radius of Δ$ ABC$，Δ$ BCD$，Δ$ ABD$，Δ$ ACD$，then $ r \plus{} r0 \equal{} r1 \plus{} r2$ .

[b]The Lemma's Briefly Proof[/b]：
First of all，The inecenters $ IIoI1I2$ frame a rectangle，so $ IIo \equal{} I1I2 \equal{} m$，$ (r \plus{} r0)^2 \equal{} m^2\minus{}EIo^2\equal{}m^2 \minus{} (AB \plus{} CD \minus{} AC \minus{} BD)^2 \equal{} m^2 \minus{} n^2$
Empathy，we can get $ (r1 \plus{} r2) \equal{} m^2 \minus{} n^2$，then $ r \plus{} r0 \equal{} r1 \plus{} r2$ .

[b]Now[/b]，come back to this problem：
According to the [b]Lemma[/b]，let ∠$ BAC$ be the max angel of Δ$ ABC$，$ AD$ be the diameter .
then $ r1 \plus{} r2 \equal{} 1/2(AB \plus{} BD \plus{} CD \plus{} CA) \minus{} 2R$，so we have：$ r \plus{} r0 \equal{} 1/2(AB \plus{} BD \plus{} CD \plus{} CA) \minus{} 2R$ .
hence  $ r \plus{} 2R \minus{} S \equal{} 1/2(BD \plus{} CD \minus{} BC \minus{} 2r0)$，so if $ r \plus{} 2R \equal{} S$：then $ BD \plus{} CD \minus{} BC \equal{} 2r0$ .
hence $ DG \equal{} GIo$，so ∠$ IoDG \equal{} 45$，∠$ BDC \equal{} 90$，then we get ∠$ BAC \equal{} 90$，It means Δ$ ABC$ is a right triangle .
And we can found A triangle $ ABC$ is acute,right or obtuse according to $ r \plus{} 2R$ is than $ S$.

[quote="Virgil Nicula"][quote="Mathias_DK"] [color=darkred]Let $ R,r,s$ be the circumradius, inradius and semiperimeter of $ \triangle ABC$. Show that $ \triangle ABC$ is right iff $ 2R + r = s$ .[/color] [/quote]
[color=darkblue][b][u]Proof 1[/u].[/b]

Prove easily that the equation $ f(x)\equiv sx^3 - (4R + r)x^2 + sx - r = 0$ has the roots $ \tan\frac A2$ , $ \tan\frac B2$ , $ \tan\frac C2$ , i.e.

$ \sum\tan\frac A2 = \frac {4R + r}{s}$ , $ \sum\tan\frac B2\tan\frac C2 = 1$ , $ \prod\tan\frac A2 = \frac rs$ . Observe that $ f(1) = 2(s - 2R - r)$ . Therefore,

$ 2R + r = s$ $ \Longleftrightarrow$ $ f(1) = 0$ $ \Longleftrightarrow$ $ 1\in\left\{\tan\frac A2,\tan\frac B2,\tan\frac C2\right\}$ $ \Longleftrightarrow\ 90^{\circ}\in \{A,B,C\}$ $ \Longleftrightarrow$ $ \triangle ABC$ is right.

[b][u]Remark[/u].[/b] Show analogously that $ \underline{\overline{\left\|\ \begin{array}{ccc} s = \sqrt 3\cdot (R + r) & \Longleftrightarrow &{ 60^{\circ}\in \{A,B,C\}\ } \\
 \\
s\sqrt 3 = 3R + r & \Longleftrightarrow & 120^{\circ}\in \{A,B,C\}\end{array}\ \right\|}}$ .[/color]
[/quote]
Great!
Just faboulous... I love the proof ! :) It works too with $ \cos A$ or $ \sin A$, instead of $ \tan \frac{A}{2}$, but the part with $ \cos A$, is a bit complicated :) Brilliant! It seems like magic to me..


a,b,c,d \geq 0 , a+b+c+d=4  ===>  \sqrt (a+b+c) +  \sqrt (b+c+d) +  \sqrt (c+d+a) +  \sqrt (d+a+b)  \geq 6
Thank  you

Do you remember the topic: The easies inequality? Well, this is a variant of that, puting a=4x and so on. We have  \sum  \sqrt b+c+d= 2 \sum  \sqrt 1-x \geq 2 \sum 1-x=6.