MATHCOUNTS
AMC
AIME
probability
calculus
AMC 10
AMC 8

I hear many people say they take multiple calculators into tests - but some prefer scientific or graphing or different brands like casio or texas -- what are the differences, and does it matter which brand? How can I figure out which is best for tests like AMC 10 and Mathcounts and AIME (if calcs are allowed)??

there is no calculator allowed on the AIME.  However, by far the best calculator to use on mathcounts or AMC 10 is the TI-89 Titanium.  its good for mathcounts because it keeps things in simplified form, not decimals, it can solve equations and it can factor

same reasons for AMC 10.  If you get to know how to use it, it can be really really helpful

you should use a calculator you are most used to, not just use some calculator because its hi tech. even a ti-30 may be better than a ti 89 if you are used to it.

thanks - right now I'm just using a 15 dollar TI-34 roman numeral two (can't type that) that my school mandated every student had - it has a 2 line display and is blue. Anyone familar with that?  Is this too simple a calculator to use? Am I at a disadvantage in not using a higher tech one? I'm good at math and spatial relations, and stuff, so I'm sure whatever calculator I get, I can figure out -- how do you guys who have very high tech ones figure out how to best use? Did anyone show you or you just played around with??? I really need your advice.

TI89s are banned on most math contests including the AMC10.

I would recommend a calculator from the TI-83+ series.

If you don't want to spend much money, then you can get probably get a really cheap TI-83+ (my mom actually got paid $\$9$ to get one with teacher discount & rebates).

However, if you want the programming aspects of the calculators, then go for the 84+SE instead (its not too much more, but it is faster & has more memory).

You can get programs to do useful things like find prime factorizations of numbers, etc, but calculators really aren't useful because all of the problems can be done by hand.

I think the 89 series simplifies problems TOO much and makes you lazy.

GACK! I CANNOT READ

nvm this post
like the 10th wasted post in two days

i think ti-83-84s are best to use, becuase for most other calculators i use, keys get stuck once a while, and you cannot press the buttons fast. for ti-83-84s the keys jus pop right back up after you press it, so you can simultaneously enter 3-4 digits and expect it to be in the correct order

[quote="7h3.D3m0n.117"]TI-89 if u can afford it
TI-83+ if not...the silver edition for 84 seems "laggy" to me[/quote]
The silver edition is probably slow because you probably still have all those unnecessary programs (like probability simulator and other language programs).

I would recommend a calculator that can do a lot, but you are comfortable with. (For me, I like my TI-84 plus silver edition. I tried a TI-89, but I am horrible at using it...)

[quote="7h3.D3m0n.117"]TI-89 if u can afford it
TI-83+ if not...the silver edition for 84 seems "laggy" to me[/quote]

What do you mean "laggy?"  The 84+SE is about one and a half times faster (plus is has 4 times the RAM, but not all of it is available to the user :()

However, lots of APPS do make it slower.  A simple solution is to delete a lot of the useless apps in multiple languages.  An easier solution is to get Omnicalc, which has the ability to make your apps screen instantaneous, no matter how many apps you have.

If you look hard enough, a TI-89 won't be so much more expensive than an 84. And I think it's a lot more convenient navigating through the TI-89. I love 89s, but then again, I also read the thousand page manual....

I never use calculators on [u]any[/u] contest that I have taken and seriously, I see no advantage from using a calculator during the AMC's imo. But, having a calculator handy for calc class or science classes, with graphing capabilities, is certainly a plus.

[quote="10000th User"]I never use calculators on [u]any[/u] contest [/quote]

so you mentally calculate items like

$\sqrt{5567}$?

[quote="Treething"]So what major contest requires you to calculate $\sqrt{5567}$?[/quote]

well sometimes u need to calculate items i jst thought of that randomly

I use a Texas Instruments Math Explorer.
I got it for $\$$10.
It is by far the BEST calculator in the world.

LOLJK

I actually use a TI-84 but I sometimes leave it somewhere and can't find it...

As far as im aware, im pretty sur that ti-89s are allowed on all contests.  i used it on the amc and sat.

[quote="Ravi B"]I believe the ACT forbids the TI-89.  Not sure about the AP calculus exam.[/quote]
The ACTR forbids all calculators with a QWERTY keyboard.

That's true, but the ACT also forbids some non-Qwerty calculators, such as the TI-89.  See
[url]http://www.actstudent.org/faq/answers/calculator.html[/url].

Also, I believe the AP calculus exam does allow the TI-89, at least on those parts of the test where a calculator is permitted.  See
[url]http://www.collegeboard.com/student/testing/ap/calculus_ab/calc.html?calcab[/url].

I rarely use my calculator to solve problems (except for finding approximations, etc and occasionally to factor large numbers).  However, I program mine to make various programs (and games, of course).

Its quite fun to try and program decent programs in languages as simple as TI's built in Basic  :D

I believe Ravi is correct about calculator use:

Both ACT and SAT forbid QWERTY calculators, Palm Pilots, cell phones, and similar calculating devices.

The ACT family of tests FORBID the TI-89.

The SAT family of tests, including AP Calc, PERMIT the TI-89.  The AP Calc rules specifically state that the calculator memory will NOT be cleared, and that student may bring whatever programs they want to use in their calculators.

AMC-10/12 permits any calculator allowed for the SAT.

Obviously, this is confusing to everybody.  My daughter's school denied students the use of the TI-89 for the PSAT.  They also denied use of the TI-89 for the first half of the AMC-A contest.  [A proctor looked up the rules, then returned the confiscated TI-89's to the students part way through the contest.]

Sorry about your daughter being denied use of an allowed calculator.

My daughter was forbidden from using a TI-84 on the AMC 8, because the proctor mistakenly thought graphing calculators were not allowed.  Fortunately, Meena had a backup calculator (TI-34) with her.

When Meena took the AMC 12, she brought in a copy of the calculator rules, just in case.  Fortunately, the second proctor knew the rules already.


In this game, we post rules like the previous rule games, except that we are also allowed to modify existing rules.

How it will work is:

The first poster posts [b]two[/b] rules, the second one abiding by the first, gaining 3 points.

Every turn, the next poster may make changes to the rule set, subject to the condition that the modified rule must abide by all the previous rules and all the rules following the modified rule must abide by it. These rule changes must be made one at a time, for obvious reasons. Somewhere within that operation, they must post a new rule, gaining "n" points, where n is the number of rules at that time. The poster must list all the rule changes in the order that they are to be applied, and the poster cannot modify the last rule before his/hers. (Thus if the first poster were to only post one rule, the next poster would have nothing to change.)

There is no limit to the number of changes that can be made; however, if any rule is found to infringe a previous rule at any time during the process, that poster loses all the points (s)he gained in that round, plus 1 point for every such infringement throughout the process. It is possible for a rule to infringe itself, in which case 1 point is still deducted. (In this way, an inept enough user can lose many more than n points.) Then, a new round starts.

Posters may [i]not[/i] post rules saying, "Rules strictly after this one", to evade the self-infringement rule. If that happens, that poster will lose 1 point for every rule strictly before his/hers, gaining only one point for the rule that he/she posted.

If a newly posted rule makes it impossible to post new rules, the next poster will have to find a way to change the existing rules to allow him/her to post a rule.

For example: The following ruleset:

1. No rule may contain a word more than seven letters long.
2. No rule may contain words less than two letters long.
3. No rule may have words with more than two vowels.

A player might modify the first rule to say, "No rule may contain a word more than twelve letters long."
Then, he might add: "Eschew monosyllabic letter combinations."

: making the final ruleset:

1. No rule may contain a word more than twelve letters long.
2. No rule may contain words less than two letters long.
3. No rule may have words with more than two vowels.
4. Eschew monosyllabic letter combinations.

But that user forgot about rule 3, so he loses one point instead of gaining four.

So: I'll start:

1. No sentence may contain the words "all", "rules", or "must" in any singular, plural, or conjugated form except in quotes, nor may those words be in quotes where they would normally exist without.
2. Use "quotes" around the word "quotes".

[hide="Scores"]
sdkudrgn88 - 3
[/hide]

To clarify: You can't edit rule 2 until you make rule 4.

1. You must repeat a word twice, and only twice.
2. Use "quotes" around the word "quotes".
3. Ah, being the amicable person I am, every (English) letter should always exist as a component in at least one word, for this ensures pangrammatic quality is in such a proclamation conforming future rules. Kazoos are awesome.

[hide="Scores"]
sdkudrgn88 - 3
james4l - 3?
[/hide]

What do you mean by "one change"? Like this?

[quote="james4l"]1. You must repeat a word twice, and only twice.
2. Use "quotes" around the word "quotes".
3. Ah, being the amicable person I am, every (English) letter should always exist as a component in at least one word, for this ensures pangrammatic quality is in such a proclamation conforming future rules. Kazoos are awesome.

[hide="Scores"]
sdkudrgn88 - 3
james4l - 3
[/hide]

What do you mean by "one change"? Like this?[/quote]

That's correct.

Well, you should list them like this, listing all rule changes and showing the final rule list for the next person to modify:

[quote="Properly done"]
Change rule 1: You must repeat a word twice, and only twice.
New rule: Ah, being the amicable person I am, every (English) letter should always exist as a component in at least one word, for this ensures pangrammatic quality is in such a proclamation conforming future rules. Kazoos are awesome.

1. You must repeat a word twice, and only twice.
2. Use "quotes" around the word "quotes".
3. Ah, being the amicable person I am, every (English) letter should always exist as a component in at least one word, for this ensures pangrammatic quality is in such a proclamation conforming future rules. Kazoos are awesome.
[/quote]

Anyway, my turn!

New rule: The quick brown fox jumps over the lazy dog. No rule shall contain a word with exactly six letters.

1. You must repeat a word twice, and only twice.
2. Use "quotes" around the word "quotes".
3. Ah, being the amicable person I am, every (English) letter should always exist as a component in at least one word, for this ensures pangrammatic quality is in such a proclamation conforming future rules. Kazoos are awesome.
4. The quick brown fox jumps over the lazy dog. No rule shall contain a word with exactly six letters.

(It is entirely permissible to not modify any rules at all.)

[hide="Scores"]
sdkudrgn88 - 7
james4l - 3
[/hide]


If a positive integer $n$ makes $4^n+2^n+1$ a prime number, prove that $n$ must be a power of $3$


geometry
circumcircle
analytic geometry
geometry solved

Hello,

Let $\triangle ABC$ be a given triangle. Let $P\in (BC), Q\in (CA)$ and $R\in (AB)$ be three points such that:
\[
{{BP}\over{PC}}={{CQ}\over{QA}}={{AR}\over{RB}}={{1-c}\over{c}}
\]
where $c\in {\Bbb R}$ is a constant.

Determine, as $c$ varies, the locus of the circumcentre of $\triangle PQR$.

Looks definitely like a higher degree curve...

  Darij

yes, something like the folium of Descartes. Can you find the equation of the curve?

Hardly... anyway, if I could, then only in trilinear or barycentric coordinates, and even this only if I find too much time today.

  Darij

Darij: did you find some time to write down the computations? ;)

Since the calculations turned out to be very hard, I have given up the problem and forwarded it to the Hyacinthos newsgroup ([url=http://groups.yahoo.com/group/Hyacinthos/message/10687]Hyacinthos message #10687[/url]). Now you can read [url=http://groups.yahoo.com/group/Hyacinthos/messages/10668?expand=1]the resulting thread[/url], and [url=http://perso.wanadoo.fr/bernard.gibert/Exemples/k271.html]Bernard Gibert's website for this curve[/url].

  Darij


In  the exterior  of  an  equilateral  triangle  $ABC$  of side  $\alpha$  we  construct  an isosceles  right-angled  triangle  $ACD$  with  $\angle CAD=90^0.$The lines  $DA$  and  $CB$  meet  at  point  $E$.

(a) Find  the  angle  $\angle DBC.$

(b) Express  the  area  of triangle  $CDE$  in  terms  of  $\alpha.$

(c) Find the  length  of  $BD.$

[hide](a) $\Delta DAB$ is isosceles, so $\angle DBA = 15^o$ and $\angle DBC = 45^o$.

(b) $\angle DCE = 105^o$, $\angle CDE = 45^o$ and $\angle DEC = 30^o$.

So $\Delta ABE$ is isosceles and $AE = \sqrt{3}\alpha$.

So $\left[DCE\right] = \frac{\alpha^2}{2} + 2 \cdot \frac{\sqrt{3}\alpha^2}{4} = \frac{\alpha^2(\sqrt{3} + 1)}{2}$.

 [/hide]

For part $c)$
$\angle BAD=150^\circ$, and $BA=AD$, so $BD=\sqrt {a^2+a^2+\sqrt 3a^2}=\sqrt{\sqrt 3+2}a$.


We denote with $m_a$, $m_g$ the arithmetic mean and geometrical mean, respectively, of the positive numbers $x,y$.

a) If $m_a+m_g=y-x$, determine the value of $\dfrac xy$;

b) Prove that there exists exactly one pair of different positive integers $(x,y)$ for which $m_a+m_g=40$.

[hide]
a) Writing this in terms of $x$ and $y$ we get:

\[\frac{x+y}{2}+\sqrt{xy}=y-x.\]

Get radical by itself, square, and collect like terms to get:

\[y^2-10xy+9x^2=(y-9x)(y-x)=0.\]

So $y=9x$ and $y=x$.  But checking answers we see that the first works whereas the second does not.  So $\frac xy =\frac 19.$[/hide]

[quote="mna851"]These problems aren't very difficult to work with, but they seem hard for a seventh grade test.[/quote]Not at all. Wait to see the finals :) ;)

[hide="solution1"]

$\frac{x + y}{2} + \sqrt{xy} = y - x$
$y - x - \frac{x + y}{2} = \sqrt{xy}$
$\frac{y}{2} - \frac{3x}{2} = \sqrt{xy}$
$\frac{y - 3x}{2} = \sqrt{xy}$
$\frac{y^2 - 6xy + 9x^2}{4} = xy$
$y^2 - 10xy + 9x^2 = 0$
$\left(x - y \right) \left(9x - y \right) = 0$
$y = x or y = 9x$, but $y = x$ does not work.
hence, $\frac{x}{y} = \frac{1}{9}$
[/hide]

[hide="solution2"]
well.. i used the above solution to help me find out x and y.
since $\frac{x}{y} = \frac{1}{9}$,
$y = 9x$
since the result is 40, $y - x = 40$
$9x - x = 40$, $x = 5, y =45$

that is the only solution i found... i think i am doing something wrong in going about this.... Valentin tell me if i am making a wrong aproach, because i tried to use a long equation, but it didnt work.[/hide]

[b]Question a[/b]
[hide]$\frac{x + y}{2} + \sqrt{xy} = y - x$
$x+2\sqrt{xy}+y=y-x$
$(\sqrt{x}+\sqrt{y})^2=2(\sqrt{x}+\sqrt{y})(\sqrt{y}-\sqrt{x})$
$(\sqrt{x}+\sqrt{y})=2(\sqrt{y}-\sqrt{x})$
$\sqrt{y}=3\sqrt{x}$
$y=9x$

So we obtain $\frac{x}{y}=\frac{1}{9}$[/hide]

[b]Question b[/b]
[hide]
$y-x=8x=40$
$x=5$
$y=45$[/hide]

I don't get it, what if $m_a + m_g = 40 \neq y - x$.. we can't use the previous theorem?

[hide="Answer a"]x/y = 1/9[/hide]

[hide="Answer b"]x=5 , y=45[/hide]

[quote="perfect_radio"]$m_a+m_g=40$ yields $\sqrt{x}+\sqrt{y}=4\sqrt{5}$, to which there are two solns: $5,40$ and $20,20$. (the last one - not good)[/quote]

isnt $x,y$ different  :)

[hide="A"]$\frac{x+y}{2}+\sqrt{xy}=y-x\Rightarrow \sqrt{xy}=\frac{y-3x}{2}\Rightarrow 4xy=y^2-6xy+9x^2\Rightarrow$
$y^2-10xy+9x^2=0\Rightarrow (y-9x)(y-x)=0$, so either $y=x$ or $y=9x$.
If the former, then we have $\frac{y+y}{2}+\sqrt{y^2}=0\Rightarrow y+y=0$, which is possible only if $y=0$, which contradicts a given. Therefore, we must have $y=9x$, so $\frac{x}{y}=\boxed{\frac{1}{9}}$.[/hide]

I don't understand why you can assume in the second case that $M_a+M_g=y-x$. :huh: Everything I do gives me $x-y=80$.

[quote="JesusFreak197"]I don't understand why you can assume in the second case that $M_a+M_g=y-x$. :huh: Everything I do gives me $x-y=80$.[/quote]

I didn't assume that ....

Well, speaking to the people who did. For you, is there a way to solve that equation other than intuitively?


linear algebra
matrix
linear algebra unsolved

Let M be an 8 x 8 matrix such that each element of the matrix belong to the set in GF(2).
How do I find the permutuation matrix such that the even row of the matrix M become the first four rows of a P and odd row of the matrix M become the last four row of matrix P?


calculus integration function trigonometry derivative algebra polynomial

[quote="Killer"]Find all continuous ,bounded functions $f: R\to R$,which verify $2f(x)=\int_{x-1}^{x+1}f(t)dt$.[/quote]
All constant functions, and no others.

Proof (unfortunately not quite elementary): Given a function with this property, the Fourier transform (as a distribution, so it's defined even if $f$ is not integrable) satisfies
$\hat{f}(s) = \frac{\sin s}{s} \hat{f}(s)$. Iterating, we see that $\hat{f}(s) = \left( \frac{\sin s}{s}\right)^n \hat{f}(s)$ for all $n$. Send $n$ to infinity, thus $\hat{f}(s) = 0$ for $s \neq 0$. Thus $\hat{f}$ is a linear combination of derivatives of $\delta_0$, and $f$ is a bounded polynomial, hence constant.

I spent some time this evening trying to find a more elementary proof than hpe's, but I never got there. I had some interesting intermediate results along the way:

$f$ must be differentiable. In fact, $f$ must be $C^{\infty}.$ There's a formula for the $n$-th derivative:

$f^{[n]}(x)=\frac1{2^n}\sum_{k=0}^n{n\choose k}(-1)^kf(x+2k-n)$

From that, we see that $|f^n(x)|$ is bounded by the same bound we have for $|f(x)|.$ As a consequence of that, by Taylor's Theorem, $f(x)$ is given everywhere by a power series with an infinite radius of convergence. (That is, $f$ is entire.)

But I never could get any mileage out of that power series.

On the other hand, hpe has actually proved a more general theorem. The same proof works on the following:

Suppose $h$ is a nonnegative measurable function such that $\int_{-\infty}^{\infty}h(x)\,dx =1.$ Suppose $f$ is a bounded continuous function on $(-\infty,\infty)$ such that for all $x, \,f(x)=\int_{-\infty}^{\infty}h(x-t)f(t)\,dt.$ (That is, $f=h*f$.)

Then $f$ is constant.

The reason hpe's proof works on this theorem is that for $s\ne0,\, |\hat{h}(s)|<\hat{h}(0)=1.$

[quote="Kent Merryfield"]I spent some time this evening trying to find a more elementary proof than hpe's, but I never got there. I had some interesting intermediate results along the way:

$f$ must be differentiable. In fact, $f$ must be $C^{\infty}.$ There's a formula for the $n$-th derivative:

$f^{[n]}(x)=\frac1{2^n}\sum_{k=0}^n{n\choose k}(-1)^kf(x+2k-n)$

From that, we see that $|f^n(x)|$ is bounded by the same bound we have for $|f(x)|.$ As a consequence of that, by Taylor's Theorem, $f(x)$ is given everywhere by a power series with an infinite radius of convergence. (That is, $f$ is entire.)
[/quote]
There are nontrivial entire solutions of the problem (try $f(z) = e^{\lambda z}$; the resulting transcendental equation in $\lambda$ is $2 \lambda = e^\lambda - e^{-\lambda}$ and has infinitely many solutions), but none are bounded on the reals.


Please only answer if you are competing in MATHCOUNTS this year.

I'm in 6th grade.

[color=orange][size=75][solafidefarms edited the 9th grade option so he could vote][/size][/color]

I'm in 7th grade.

And, BTW, why did you put 9th grade as an option? Ninth graders no longer compete in MATHCOUNTS.

They don't? Thanks for telling me that.

(Whew! :D )

Huh. I'm in 8th--
but can't 5th graders compete, too?

I don't think they can officially compete.

why is 9th grade part of vote?
6, 7,8 , no 9
im 8th.

I enjoy AMC/AIME/USAMO tons more than MathCounts (harder problems, more thinking)   =)

[quote="hunter34"]8th.

I disagree with cheese, I think AMC$ \le$MATHCOUNTS.[/quote]
...I think you agreed with him.

I like AMC (and further) type of problems better.

7th. owned at chapter, and then got owned...

[quote="hedgehog12"]I'm in seventh grade.
Mathcounts is for only 6,7 and 8 grade.[/quote]

no in my class we have to compete against high schools and universities^^ me pwn them to !!!!! 17 or more trophy out of 20 usually

[quote="1=2"][quote="CheeseIsGood"]I'm in 8th, but I wish I wasn't... MathCounts is more fun than competitions like the AMC.[/quote]

You seem a lot older... especially in Mafia games. :)

7th, owned at school, chapter is 3.65 days from when I post this.

MathCounts has gotten a bit boring. AIME owns MathCounts.[/quote]
I say go 6th graders  :D


How would you use AM-GM to find the minimum value of something like $ a^6\plus{}b^6\plus{}c^6\minus{}6abc$?

[url=http://www.artofproblemsolving.com/Wiki/index.php/RMS-AM-GM-HM]RMS-AM-GM-HM[/url] is a more "expanded" version of AM-GM. So basically you could state using AM-GM,
\[ \frac {a^6 \plus{} b^6 \plus{} c^6 \minus{} 6abc}{4} \geq \sqrt[4]{(a^6)(b^6)(c^6)( \minus{} 6abc)}
\]
and then solve from there.

I'm not good with inequalities, but try reducing the 6th powers with cauchy or something like that

[quote="RisingStar"][quote="#H34N1"][url=http://www.artofproblemsolving.com/Wiki/index.php/RMS-AM-GM-HM]RMS-AM-GM-HM[/url] is a more "expanded" version of AM-GM. So basically you could state using AM-GM,
\[ \frac {a^6 \plus{} b^6 \plus{} c^6 \minus{} 6abc}{4} \geq \sqrt {(a^6)(b^6)(c^6)( \minus{} 6abc)}
\]
and then solve from there.[/quote]

That should be a 4th root, not a square root, and AM-GM isn't valid for negative numbers.[/quote]

Yeah, I just noticed that. And I just noticed that $ \minus{}6abc$ is negative.  :wink:

@Original problem: You can't do that with AM-GM, since there's a negative.

When I try to solve it I get $ a^6 \plus{} b^6 \plus{} c^6 \minus{} 6abc\ge4abc\sqrt [4]{ \minus{} 6abc}$

I'm stuck from here.

@ CircleSquared:  umm... whats a cauchy?

[i]Clarification: a,b,c are all real numbers.[/i]

The [url=http://mathworld.wolfram.com/CauchysInequality.html]Cauchy Schwarz Inequaltity[/url] is an inequality that basically states:
\[ \left( \sum_{i \equal{} 1}^{n}a_ib_i \right)^2 \le \left (\sum_{i \equal{} 1}^{n}a_i^2 \right )\left (\sum_{i \equal{} 1}^{n}b_i^2 \right ).
\]

You can use Cauchy to normally simplify large inequalities.

I'm not sure, but is it possible to do the following ($ m$ denotes the minimum value)?

$ a^6 \plus{} b^6 \plus{} c^6 \equal{} 6abc \plus{} m$

And then use the AM-GM inequality to say:

$ \frac {a^6}{3} \plus{} \frac {b^6}{3} \plus{} \frac {c^6}{3} \geq 3\sqrt [3]{\frac {a^6b^6c^6}{27}}$

$ \frac {a^6 \plus{} b^6 \plus{} c^6}{3} \geq a^2b^2c^2$

$ 6abc \plus{} m \equal{} 3a^2b^2c^2$
$ m \equal{} 3abc( 2 \minus{} abc )$

Let $ abc$ $ \equal{}$ $ x$.

$ m \equal{} 3x(2 \minus{} x)$
$ m \equal{} \minus{} x^2 \plus{} 6x$
$ m \equal{} \minus{} 3$.

I think you're forgetting what inequalities means. By the reasoning you use,

$ 3 \ge 2$ and $ 3 \ge 1 \implies 2=1$, which is false.

Where did you get the first inequality anyway?

[quote="Begoner"]I'm not sure, but is it possible to do the following?

$ a^6 \plus{} b^6 \plus{} c^6 \geq 6abc$
[/quote]

You can't assume this fact. The problem asks for the [b]minimum value[/b] of the entire inequality, and therefore this assumption is false.

I think you can get it simply with AM-GM, if this is correct:
[hide="Solution with AM-GM"]$ \frac {a^6 \plus{} b^6 \plus{} c^6}{3} \ge \sqrt [3]{a^6 b^6 c^6}$ by AM-GM (note: all terms are positive), so we have
$ a^6 \plus{} b^6 \plus{} c^6 \ge 3a^2 b^2 c^2$.

Thus,
$ a^6 \plus{} b^6 \plus{} c^6 \minus{} 6abc \ge 3a^2 b^2 c^2 \minus{} 6abc$, and by completing the square on the RHS we have
$ a^6 \plus{} b^6 \plus{} c^6 \minus{} 6abc \ge 3(abc \minus{} 1)^2 \minus{} 3 \ge \minus{} 3$.

Now we simply show that $ \minus{} 3$ is attainable: indeed, for $ a \equal{} b \equal{} c \equal{} 1$, we have $ a^6 \plus{} b^6 \plus{} c^6 \minus{} 6abc \equal{} \minus{} 3$.[/hide]

If any of the variables are negative, replace them with the positive equivalent, and this does not increase the sum, so it suffices to consider when $ a,b,c$ are non-negative. \[ a^6\plus{}b^6\plus{}c^6\plus{}1\plus{}1\plus{}1\ge6\cdot \sqrt[6]{a^6\cdot b^6\cdot c^6\cdot 1\cdot 1\cdot 1}\]
\[ a^6\plus{}b^6\plus{}c^6\minus{}6abc\ge \minus{}3\]
equality occurs (for example) when $ a\equal{}b\equal{}c\equal{}1$.

[quote="#H34N1"]You can't assume this fact. The problem asks for the [b]minimum value[/b] of the entire inequality, and therefore this assumption is false.[/quote]

My bad -- that was completely wrong.  Is the edited version correct, though?

What does #H34N1 mean by splitting the equation?

[quote="archimedes1"][quote="#H34N1"][quote="archimedes1"]$ \frac {a^6 \plus{} b^6 \plus{} c^6}{3} \ge \sqrt [3]{a^6 b^6 c^6}$ by AM-GM (note: all terms are positive), so we have
$ a^6 \plus{} b^6 \plus{} c^6 \ge 3a^2 b^2 c^2$.

Thus,
$ a^6 \plus{} b^6 \plus{} c^6 \minus{} 6abc \ge 3a^2 b^2 c^2 \minus{} 6abc$, and by completing the square on the RHS we have
$ a^6 \plus{} b^6 \plus{} c^6 \minus{} 6abc \ge 3(abc \minus{} 1)^2 \minus{} 3 \ge \minus{} 3$.

Now we simply show that $ \minus{} 3$ is attainable: indeed, for $ a \equal{} b \equal{} c \equal{} 1$, we have $ a^6 \plus{} b^6 \plus{} c^6 \minus{} 6abc \equal{} \minus{} 3$.[/quote]

If you are going to use AM-GM for your solution, you cannot "split" an equation. You must use the entire equation when using AM-GM. For example, we do not know if $ a,b,c$ are directly or inversely proportional, thereby affecting the value of the entire equation.[/quote]
Wait, I'm still not sure why my solution is invalid.  Why can I not "split" an equation if we are not given any special constraints on $ a,b,$ and $ c$?[/quote]

Never mind, I read your solution wrong the first time.  :( Must be because of the 3 AIME's I did today.

Archimedes' solution is correct. The thing to notice is that when you do a series of inequalities, you have to check that equality occurs.

[hide="hint"]apply AM-GM to a^6+b^6+c^6[/hide]

EDIT: oops, didn't see that archimedes already posted the whole solution

Altheman also posted a nice extension [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=197106]here[/url].


Consider those functions $ f$ that satisfy $ f(x + 4) + f(x - 4) = f(x)$ for all real
$ x$. Any such function is periodic, and there is a least common positive period
$ p$ for all of them. Find $ p$.

(A) $ 8$ (B) $ 12$ (C) $ 16$ (D) $ 24$ (E) $ 32$

Thanks in advance!

The majority of AMC problems are already in the archives, so try a search or the Contests tab before you post AMC problems.


From a point interior to an equilateral triangle, the distances to the sides are 4, 6, and 8, respectively. What is the length of a side of the traingle?

[hide]Let $ s$ be the side of the triangle so the area is $ \frac{s^2\sqrt{3}}{4}$

The area can also be calculated by treating 4, 6, and 8 as altitudes so $ \frac{1}{2}4\cdot s\plus{}\frac{1}{2}6\cdot s\plus{}\frac{1}{2}8\cdot s\equal{}9s$

$ \frac{s^2\sqrt{3}}{4}\equal{}9s$

$ s\equal{}\boxed{12\sqrt{3}}$[/hide]

heres a more interesting question.

From a point interior to an equilateral triangle, the distances to the [b]vertices[/b] are 4, 6, and 8, respectively. What is the length of a side of the triangle?


For a class trip to a museum, 461 students and 20 teachers will be taking buses.  Each bus can seat a maximum of 52 persons.  What is the LEAST NUMBER of buses needed for the trip?

MY WORK:

Total number of people is 461 + 20 = 481, right?

I then divided 481 by 52 persons and got 9.25.

I said the LEAST NUMBER of buses is 9 based on the division work above.

The book's answer is 10.  Why is that?

your work is right, but if you only had 9 buses, there would still be .25 left, so you have to round up in this case. At most, 9 buses can fit 9*52=468 people. This won't seat all everyone, so they need another bus.

[quote="bpms"][quote="Interval"]If I round 9.25 to the nearest unit, I will not get 10.

Can you explain another way why the answer is [color=blue]AT LEAST 10 buses?[/color][/quote]
I don't understand why.
It would make perfect sense to cut a bus into four and sell 3 parts of it on ebay, no?[/quote]
:rotfl: :rotfl:

You must use the 10th bus to seat the extra 13 people.

If we had 9, we would only have 462 people accounted for, so that is why we couldn't have only 9.

No where in the problem does it say each bus must be filled.

So we use the tenth bus as a way to supplement the original 9, and hence 10 is our answer.

mathgeniuse^ln(x) made is very clear why the answer is 10 buses.

Thanks!

You are welcome.

I also mean t468, not 462.


1) Given $ a^2b \plus{} b^2c \plus{} c^2a \equal{} 0.$ Prove that $ a^2 \plus{} b^2 \plus{} c^2\geq6(ab \plus{} ac \plus{} bc).$
2) Given $ a^3b \plus{} b^3c \plus{} c^3a \equal{} 0.$ Prove that $ a^2 \plus{} b^2 \plus{} c^2\geq3(ab \plus{} ac \plus{} bc).$

[hide="Really cheap solution to #1"]
This is equivalent to
$ \frac{1}{3!}\cdot\sum_{\text{sym}}a^2b^0c^0\geq\sum_{\text{sym}}a^1b^1c^0$

Because $ (2,0,0)\succ(1,1,0)$, this holds by Muirhead's inequality, and we are done.
[/hide]

[quote="grn_trtle"]
This is equivalent to
$ \frac {1}{3!}\cdot\sum_{\text{sym}}a^2b^0c^0\geq\sum_{\text{sym}}a^1b^1c^0$

Because $ (2,0,0)\succ(1,1,0)$, this holds by Muirhead's inequality, and we are done.
[/quote]
Explain please, why "we are done"? Thank you!

[b]@ grn_trtle:[/b] That is not a symmetric sum -- it is cyclic.

Ah, you're right... I knew that was too easy  :P
Does the fact that it's a cyclic sum help with the proof? I'm a little unsure on the properties of symmetric/cyclic sums.

Can we say that  $ a, b, c \neq 0$  after considering the case of $ a\equal{}b\equal{}c\equal{}0$  ?
Please help   :blush:

[quote="Potla"]Can we say that  $ a, b, c \neq 0$  after considering the case of $ a \equal{} b \equal{} c \equal{} 0$  ?
[/quote]
Yes, of cause! We can say this.

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=457573


Let $c_1,c_2,\ldots ,c_n,b_1,b_2,\ldots ,b_n$  $(n\geq 2)$ be positive real numbers. Prove that the equation
\[ \sum_{i=1}^nc_i\sqrt{x_i-b_i}=\frac{1}{2}\sum_{i=1}^nx_i\]
has a unique solution $(x_1,\ldots ,x_n)$ if and only  if $\sum_{i=1}^nc_i^2=\sum_{i=1}^nb_i$.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48981[/url]

[color=#FF0000][moderator edit: Here too: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193776. Thanks WakeUp for the link!][/color]

Set $x_i = \theta_i^2+b_i$, $1\le i\le n$. We obtain
\[
\sum_{i\le n}c_i \theta_i = \frac12 \sum_{i\le n}\bigl(\theta_i^2 + b_i\bigr) \iff \sum_{i\le n} \bigl(\theta_i-c_i\bigr)^2 = \sum_i \bigl(c_i^2-b_i\bigr)\triangleq S.
\]
Note that if $S=0$, then we indeed have a unique solution: $x_i = c_i^2+b_i$. If $S<0$, we obtain no solutions, whereas for $S>0$, it is not hard to see the solutions are not unique: (a) $x_1=b_1+(c_1+\sqrt{S})^2$, $x_i=b_i+c_i^2$, $i\ge 2$ and (b) $x_i = b_i+(c_i+\sqrt{S/2})^2, i\in\{1,2\}$ and (b) $x_i = b_i+c_i^2$ for $3\le i\le n$ are distinct solutions.


parameterization
probability
function
probability and stats

Let $ Y$ be a bernoulli random variable with two possible values $ a$ and $ b$ where $ P(Y\equal{}a) \equal{} p$ and $ P(Y\equal{}b) \equal{} 1\minus{}p$.  If we condition on $ Y$, then a random variable $ X$ will have a normal distribution with parameters $ Y$ and $ \sigma^{2}$, or Conditional $ X$~Normal($ Y$,$ \sigma^{2}$).  Determine the unconditional proabability density of $ X$.

if you want the distribution function for example:
\[ P[X > x] \equal{} P[X>x | Y\equal{}a] \cdot P[Y\equal{}a] \plus{} P[X>x|Y\equal{}b] \cdot P[Y\equal{}b] \equal{} \ldots\]


HMMT
AMC
AIME
MATHCOUNTS
geometry
AMC 12
AMC 10

Preparing for SAT Is, HMMT, and the AIME  :D

i'm not really gonna do anything about teh AMC B cuz i'm not taking it.  :D but i guess i'm sorta studying for AMc's by studying for the mATHCOUNTS states(3/3 for me). i'm working my tail off to make nationals.  :weightlift:  :weightlift:

I keep getting headaches when trying to study AoPS vol 1 or 2, and studying other math, so I figure I'll do nothing.

Not in the mood at all

I'm having difficulty sitting up straight. Got a flu and a cold at the same time... Didn't take the A, so I'd better be decent on the B--if I can stop sleeping for about 20hrs/day.

when do they post top scores in different States on the AMC 10A? Or do they wait for the AMc 10b to finish and post them together?

Is there anything we are supposed to be doing? I sucked up the AMC 10, probably because of stupidity but until I confirm that with the 12 I'll blame it on not enough sleep + flu.

But seriously, we have ten days. If you're screwed, you're screwed.

Absolutely nothing since my school isn't offering it. But even if it did, I wouldn't do anything besides sharpening pencils.

sharpening pencils - u only need 2 or 3.

It takes seconds to do that.  :lol:

Now it's time to do WOOT and study for HMMT/AIME/MAML. Maybe I'll do like a practice AMC 10 or something in the next week or so just to warm up, though.

[quote="now a ranger"]what is that?

I haven't heard about Pokemon since I watched cartoons in 2nd grade and collected cards in kindergarten.[/quote]
DUDE, POKEMON KICKS BUTT.
I have leafgreen, it's > than firered.

[quote="nat mc"]are u kiddiing me? Classic Red is the best

I'll probably be doing a lot of cubing[/quote]
Yeah yeah yeah classic red>blue
but firered<leafgreen

I'm not taking it because my school isn't offering it and I have no way of transporting myself to the area university that is offering it.

[quote="me@home"][quote="solafidefarms"]I am doing absolutely nothing. I've already qualified for the AIME twice, so once more isn't that important. A perfect would be good, but I've got a long time to study/make sure it's possible.[/quote]
I was going to reply asking how you qualifying twice seeing as we haven't gotten back our round three scores, when I realized that I hadn't actually verified that and now I must thank you for reminding me to do so...
wow, they lowered their standards! a bad 70 got me in... wasn't it 73 last year?

Also, is it just me or are they docking points for stupid things this year... I have lost all my points but 1 on typos.

Well, who else is in the two way qualifier round besides us?[/quote]
Last year it was like 69 as well.
matt276eagles said he got in with 68

I'll read aops vol 2 and 1 , chapters 6 and 7 in 2, and 21 in 1

at home, I will just do the past AMC Bs from 2003-2006 slowly

Studying for AIMEs/w00t/usamo. All that fun stuff.

im actually going to be studying a little. i had the flu, so i couldnt take 10a.  so this is my only shot at making aime.  i dont actually care about making aime, but i dont want to get some terrible score and look stupid.

I'm studying all the tests and reading Aops volume 1.

Man, I wish my school offered the AMC B contests.  :(

So, what exactly is the HMMT?

Right now I'm cramming for the AMC12B... I'm finally reading the transcripts of my PS class which I neglected to attend because it was too hard... Maybe I'll get lucky and get higher than a 100.

I was being a lazyhead and didn't do any practice tests....ugh
now I'm gonna give myself 20-30min for first 20 problems of every amc12 test, and then I'll do the rest after I'm done with this.


calculus
integration
number theory unsolved
number theory

Consider the equation

\[x^2 + y^2 + z^2 + t^2 - N \cdot x \cdot y \cdot z \cdot t - N = 0\]

where $N$ is a given positive integer.

a) Prove that for an infinite number of values of $N$, this equation has positive integral solutions (each such solution consists of four positive integers $x, y, z, t$),

b) Let $N = 4 \cdot k \cdot (8 \cdot m + 7)$ where $k,m$ are no-negative integers. Prove that the considered equation has no positive integral solutions.

Firsly, i think  the number $N$ should be in the form $N=4^{k}(8t+7)$

This one can be solved by this way:

 a)
   For each positive integer $n$, let $N=a^2+2$ then the equation have root $(x,y,z,t)=(a(a^2+2),a,1,1)$. QED

  b)
    We easily prove  (by $mod 4$) that if 4 numbers $a,b,c,d$ have the property $a^2+b^2+c^2+d^2\vdots\ 4$ then each number $\vdots\ 2$.
    Now, assume that for some $N=4^k(8t+7)$ such that  the equation have some roots like $(a,b,c,d)$.Then after $k$ steps we deduce to $(a')^2+(b')^2+(c')^2+(d')^2=(8t+7)(4^{4k}a'b'c'd'+1)$ for some positive numbers $(a',b',c',d')$ . We easily see that $(8t+7)(2^{4k}a'b'c'd'+1)\equiv\ 7 [mod 8]$ for any non-negative integer $k$ which is impossible.QED

There question left find all $n$ such that the equation have solution

oh i'm sorry my solution is totally stupid (part b) i'm still thinking about it

If $z=t=1$ then equal $\leq > N=\frac{x^2+y^2+2}{xy+1}=\frac{(x-y)^2}{xy+1}+2$ .since $N$ is integer $\Rightarrow x=y$.Because that I don't understand your solution

Interested in the second question .. quite hard !

[quote=t_c_lhp]Firsly, i think  the number $N$ should be in the form $N=4^{k}(8t+7)$

This one can be solved by this way:

  b)
    We easily prove  (by $mod 4$) that if 4 numbers $a,b,c,d$ have the property $a^2+b^2+c^2+d^2\vdots\ 4$ then each number $\vdots\ 2$.
    Now, assume that for some $N=4^k(8t+7)$ such that  the equation have some roots like $(a,b,c,d)$.Then after $k$ steps we deduce to $(a')^2+(b')^2+(c')^2+(d')^2=(8t+7)(4^{4k}a'b'c'd'+1)$ for some positive numbers $(a',b',c',d')$ . We easily see that $(8t+7)(2^{4k}a'b'c'd'+1)\equiv\ 7 [mod 8]$ for any non-negative integer $k$ which is impossible.QED

There question left find all $n$ such that the equation have solution[/quote]

I think your answer for b) is wrong


topology
function
Support
real analysis
real analysis unsolved

Let X be a locally compact Hausdorff
$ C_{c}(X)$ be the set of all continuous real -valued function on $ X$ with compact support
\[ C_{0}(X) \equal{} \{f\in C_{c}(X) : \mbox{st : there exists a sequence} : (g_{n})\subset C_{c}(X) : g_{n}\rightarrow f \,\, \mbox{uniformly}\}
\]
let [b]$ f\in C(X)$ st[/b] : for any $ \alpha > 0$, $ \{ x\in X : |f(x)|\geq \alpha\}$ is compact, prove that $ f$ is in $ C_{0}(X)$

I used the contraspositve proof but getting stuck

any idea will be appreciated

This problem makes no sense no matter how you look at it. Typo aside, it's still obvious  :?:

Kalle
why it looks obvious? could you please show me?

You are not allowed to use other definition of $ C_{0}(X)$.

many thanks

Your extra condition makes no sense, because it's automatically true. $ \{x: |f(x)|\ge\alpha\}$ is closed, and a closed subset of a compact set is compact. Your "definition" of $ C_0$ makes no sense, because it boils down to "$ C_0\equal{}C_c$", and the standard definition has $ C_0$ strictly containing $ C_c$.

First of all, replace two occurrences of $ C_c(X)$ with $ C(X)$ (continuous functions on $ X$) for this to make sense!

Now then, pick a small $ \alpha$ and choose a neighborhood $ U$ of $ K_\alpha \equal{} \{x \, : \, |f(x)| \geq \alpha \}$ with compact closure. For each point $ x \in K_\alpha$ choose a continuous function $ h_x: X \minus{} > [0,1]$ that maps every point of $ U^c$ to 0, and $ h_x(x) \equal{} 1$ (see the proof of Urysohns lemma and use that $ X$ is completely regular). By compactness there are finitely many points $ x_n$ such that the $ h_{x_n}^{ \minus{} 1}[(1 \minus{} \alpha, 1]]$ cover $ K$. Let $ g_\alpha \equal{} \max(h_{x_1}, \ldots, h_{x_m}) \cdot f$. Note that $ g_\alpha \in C_c(X)$ as it is supported in $ \overline{U}$. Noting that $ f$ is bounded (since $ K_1$ is compact), it follows that $ g_\alpha \to f$ uniformly as $ \alpha \to 0$.

edit: at the time of editing the definition of $ C_0(X)$ is still wrong. This is what it should be:
$ C_{0}(X) \equal{} \{f\in C(X) : \mbox{st : there exists a sequence} : (g_{n})\subset C_{c}(X) : g_{n}\rightarrow f \,\, \mbox{uniformly}\}$


superior algebra
superior algebra unsolved

Hi,
 the prrof of the following statement is at this book pg 81

Show that if (p^n) | |G| for every n>=1

then |{H<G : |H|=p^n }| = 1 (mod p).

Does aynone have the book and can tell us the proof. Thanks

[quote]Let $K = \{ H < G : |H| = p^{n}\}$.

Show that $\forall n \in \mathbb{N}$, that if $(p^{n}) | |G|$, then $|K| = 1 \, (\bmod \, p)$.[/quote]


analytic geometry
trigonometry
algebra
function
domain
trig identities
Law of Cosines

Write a program that gets as input the coordinates of three points and shows whether they are in a horizontal line,vertikal line or in an angle of 35' with ox.

You mean as in:

horizontal line = all three points have the same Y cooridnate
vertical line = all three points have the same X cooridnate?

So the only problem is to detect the 35° angle?

Given three points $ A(x_a,y_a)$, $ B(x_b,y_b)$, $ C(x_c,y_c)$, we wish to find $ \angle BAC$, for instance (do this for all three angles, I suppose?).

One could use the Law of Cosines on the angle and the coordinates. Although it takes longer to code, it should work most efficiently. For these purposes, however, it is safe to assume some margin of error caused by rounding issues in C.

Another way, much easier to code, (although it may not account completely for the domain and range of $ \tan X$) would be to use right triangles using $ AB$ and $ AC$ as hypotenuses. In other words,

$ \angle BAC\equal{}\left|{\tan^{\minus{}1}{\frac{y_b\minus{}y_a}{x_b\minus{}x_a}}\minus{}\tan^{\minus{}1}{\frac{y_c\minus{}y_a}{x_c\minus{}x_a}}}\right|$

You would have to work around, testing different cases to determine the general range for the angle (is $ \angle BAC$ acute? right? obtuse? straight? reflex?). Other than that, this should work.


integration
logarithms
calculus
Euler
calculus computations

show

$\;\int_{0}^{1}\;\ln{ \left( \frac{1}{\ln{ ( \frac{1}{x}) }}\right) }\;dx\;$

i.e.

$\;\int_{0}^{1}\;-\ln\; ( \;-\ln{ ( \;x\; ) }\; ) \;dx\;=\;\gamma$

where $\gamma$ is 0.577 215 665 (Euler's constant)

PS: i solved it using contour integration, but i wanted to
know other solutions as well...  :pilot:

${x=e^{-u}}$
${dx=-e^{-u}du}$

${\int_{0}^{1}\log \left(\frac{1}{\log \left(\frac{1}{x}\right)}\right) \, dx=\int_{\infty }^{0}e^{-u}\log (u) \, du}$

See http://en.wikipedia.org/wiki/Euler-Mascheroni_constant

this is one of the definition of the Euler Mascheroni constant

we can only make a substituition, dont we?

How do you find that integral with contour integration?


Two brothers sold their flock of sheep. They sold each sheep for as many pieces of gold as the number of sheep in the original flock. They shared the money in the following way: The first 10 pieces of gold were given to the elder brother, the next 10 pieces to the younger one, then the next 10 pieces went to the elder brother again, and so on. At the last turn, the younger brother’s share was less than 10 pieces of gold, therefore the elder one gave him his pocket-knife to make their profits equal. How many pieces of gold was the pocket-knife worth?
PS: I can't solve it! Maybe I miss the problem. Anybody helps me is welcome!

I bet the problem is that each sheep sells for a number of gold equal to the number of sheep currently in the flock:

ie if there are 30 sheep, the first sells for 30 gold, the next for 29 gold, etc.

I actually took that one, and I believe Zuton Force is correct.

Well, I just can go on the solution like this:
Since as The Zuton's Force stated above, we will have the total money is: n(n+1)/2 with n is the number of sheep
Now, since when 2 brothers divide their money equally, we will have:
n(n+1)=4k(k is a positive integer)
If we set x is the value of the knife-pocket,....I'm stuck here :blush: .
Please post the full solution if you want to help me! :wink:

You're all misinterpreting this. There is a single flock of $ n$ sheep, which sells for $ n$ pieces of gold each, so there are $ n^2$ pieces of gold to be distributed between the two brothers.

Because of the way the brothers take turns splitting the gold, we know that $ n^2\equiv k\pmod{20},$ where $ 10 < k < 20.$ But the only squares $ \mod 20$ are $ 0,1,4,5,9,16,$ and the only one in our range is $ 16.$ The first brother got $ 10$ of the last $ 16$ pieces of gold and the second brother only $ 6.$ Hence if the first brother gives a knife worth $ 2$ pieces of gold to the second brother, they'll be even.

By the way, ghjk has slightly mislabeled the source: this is from CSULB Math Day at the Beach, 2006, not 2007. It was a multiple choice problem, with possible answers $ 1,2,3,4,5.$ The correct answer was $ 2.$ When we gave this exam, out of $ 157$ participants, $ 4$ said the answer was $ 1,$ $ 11$ gave the correct answer of $ 2,$ $ 4$ said the answer was $ 3,$ $ 39$ said the answer was $ 4,$ $ 9$ said the answer was $ 5,$ and $ 90$ left the problem blank. That made it the hardest of the fifteen multiple choice problems. Tedrick and Macky both left it blank, but Jeff and Djordjo got the correct answer. And Zellex gave the common incorrect answer of $ 4.$ (Yes, I have records.)

I got this question from a colleague with a penchant for the history of mathematics. I'll have to ask him whether it's an actual historical question (from someone like Fibonacci or Robert Recorde) or whether it's just in the style of a historical question.


$f: \mathbb{R}^n\to \mathbb{R}^n$ a $C^1$ function. Suppose there exists
a dense subset $E$ in $\mathbb{R}^n$ such that the range of $f$ contains $E$
and such that $df(x)$ is invertible for any $x \in E$. Is $f$ necessary surjective.

You need the local inversion theorem, slightly modified to suit your needs (actually you need to use it on the compact ball $Bc$ with center $0$ and radius $R$; since $df$ is uniformly continuous on $B$, looking at the proof of the theorem shows that there exists $r>0$ fixed such that for every element $e$ of $E$, the open ball $B(e,r)$ is the image of a local diffeomorphism induced by $f$. Combined with the density of $E$, this is enough to prove the surjectivity on $Bc(0,R)$. You then can take $R$ as big as you wish.)


probability
function
parameterization
combinatorics unsolved
combinatorics

there is a class with m girls and n^2-m guys and they are arranged in an n x n square.

If the men and women are randomly placed what is the probability that a guy will be next to a girl as a function of m and n?

It is slightly unclear: do you mean, "what is the probability that a particular guy will end up next to at least one girl?"  In this case, there is a $ \frac{4}{n^{2}}$ likelyhood that he is placed in the corner, with probability $ 1-\frac{n^{2}-m-1}{n^{2}-1}\cdot\frac{n^{2}-m-2}{n^{2}-2}$ of being next to a girl, and then make similar calculations for if he is along an edge or in the center.  The result is something quite nasty which doesn't simplify.

If instead the question means, "what is the probability that a particular guy will end up next to a particular girl?" then it is a little cleaner because we get to dispose of the parameter $ m$ entirely, and it ends up working out to $ \frac{4}{n+n^{2}}$.

If instead the question means, "what is the probability that any guy will end up next to any girl?" then the answer is 1 if $ 1 \leq m \leq n^{2}-1$ and 0 otherwise.

I meant "what is the probability that a particular guy will end up next to at least one girl?". Could you show the logic behind your result if he is placed in the corner?

Suppose he is placed in the upper-left corner.  Then the only way he could be [i]not[/i] placed next to a girl (that's the "$ 1-$") is if he's placed next to two boys.  The probability the person below him is male is $ \frac{n^{2}-m-1}{n^{2}-1}$ because there are $ n^{2}-1$ people who could go there, and $ n^{2}-m-1$ of them are male.  Then the second factor is equivalent, except that we must already take as given that the person below him is male, so there are $ n^{2}-2$ people who could go to his right of whom $ n^{2}-m-2$ are male.

The other cases are essentially the same: if he's placed along an edge, he has three possible neighbors.  If he's placed in the middle somewhere, he has four possible neighbors.


geometry
analytic geometry
graphing lines
slope
AMC
AIME
geometric transformation

Three circles, each of radius 3, are drawn with centers at $(14,92)$, $(17,76)$, and $(19,84)$.  A line passing through $(17,76)$ is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it.  What is the absolute value of the slope of this line?

[hide]The line will cut exactly halfway between the two centers of the circles (i.e. through the midpoint)
The midpoint is $(\frac{33}{2}, 88)$.
$|m| = \left|\frac{88-76}{\frac{33}{2}-17}\right| = \left|\frac{12}{-\frac 12}\right| = |-24| = \boxed{24}$[/hide]

[quote="Pineappleperson"]why must it cut through the 2 circles.?[/quote]
Don't revive old threads. Please.

[quote="4everwise"][quote="Pineappleperson"]why must it cut through the 2 circles.?[/quote]Draw a diagram and you'll see. :)

About reviving this thread, it's ok. I mean, where else could you post a question or comment? Creating a new topic? No, that would be bad. :P[/quote]
Wait, so we aren't supposed to make new threads for AIME problems when we don't understand the solution?

It can be shown by the distance formula that none of the circles overlap. Since the line passes through the center of one circle, it divides that circle in half. Hence we can ignore that circle for the rest of the problem. We need a point for which the line divides the area of the other two circles equally. Consider the midpoint of the centers of the other two circles. It can be shown that a $180$ degree rotation about that point maps one circle to another. Hence, the figure formed by the two circles is symmetrical about that midpoint. Hence the area is divided into two equal halves. A simple slop calculation reveals that the answer is $\boxed{33}$.

Wait, so is the answer 24 or 33?  I got 24 too...

I still don't understand why the line has to pass through the midpoint of the centers of the circle, as long as it goes in between the two circles wouldn't the area of the parts on both sides still be equal?

But the line can't go Entirely in between the two circles. No matter what you do, due to where the 1776 circle is placed, you cannot make a line that has one circle completely on one side and the other completely on the other side. Go check it out by graphing it.

Oh I see now.  But usually I don't try graphing AIME problems, is this something you should do?

So the kalva solution outline goes like this "Line will always divide circle center Q evenly. Suppose it passes through (17-x, 92). Then it passes through (17-x/2, 84). We require 3 - x = 2 + x/2, so x = 2/3 and gradient = 16/(-2/3)."  .  I was trying to prove that the distance from the center of the (14,92) circle to the chord (which is 3-x) is equal to the distance from the center of (19,84) circle to the chord (which is 2 + x/2), but I couldn't do it.   :oops: Can anyone post a hint/solution?

Sorry if this was a bad revive, but note that 1492 was the year Columbus found America, 1776 was the year that the Declaration of Independence was signed, and 1984 was a monumental novel written by George Orwell.

7 year revive, but hey, cool fact to be stated :P


geometry
incenter
circumcircle
parallelogram
geometry proposed

Let a triangle ABC. AA',BB',CC' are its altitudes. D,E,F are incenters of triangles AB'C',BC'A',CA'B'. The incircle of triangle ABC touches BC,CA,AB at M,N,P. Prove that the circumcenter of triangle DEF is the orthocenter of triangle MNP.

Denote H’  the orthocenter of  △MNP
EI=BI(1-cosB)  IF=CI(1-cosC)
EI*BI=IF*IC   〈=〉 BIsin(B/2)=CIsin(C/2)  〈=〉 IM=IM
So E.B.C.F are concyclic  =〉∠IEF=∠ICB=∠H’NP=C/2  we have IB∥H’N(both ⊥PM)
=〉NP∥EF
PN=2r*cos(A/2)  ,where r is the radius of the incircle
EF=2BC* sin(B/2)* sin(C/2)
BC=MB+CM=BIcos(B/2)+CIcos(C/2)=r(cot(B/2)+cot(C/2))=r*sin(B/2+C/2)/(sin(B/2)*sin(C/2))
So PN=EF    =〉  △DEF △MNP are congruence    =〉 I is the orthocenter of  △DEF
Quadrilateral IDH’M , IPH’F , INH’E are parallelogram
H’D=H’E=H’F=r
=〉 H’ is the circumcenter of △DEF


I am looking for an extreme principle solution to Eisenstein's criterion, which states that if $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{0}$ is a polynomial with integer coefficients and a prime $q$ divides $a_{n-1}, a_{n-2},...a_{0}$ but not $a_{n}$ and $q^{2}$ does not divide $a_{0}$, then $P$ is irreducible over the rationals.

[quote="Phelpedo"]... and $q$ does not divide $a_{0}$...[/quote]
um im not pretending to be an expert on this topic, but i think you have a small error, according to MathWorld(the $q$ is supposed to be squared): [url=http://mathworld.wolfram.com/EisensteinsIrreducibilityCriterion.html]EisensteinsIrreducibilityCriterion[/url]

[i]Extended Eisenstein's Criterion:[/i] Let $f(x)=\sum_{k=0}^{n}a_{k}x^{k}\in\mathbb{Z}[x]$ be a polynomial.  If there exists $k\in\mathbb{Z}_{n}$ and prime $p$ such that $p|a_{0},...a_{k},p\not| a_{k+1},p^{2}\not| a_{0}$, then there exists irreducible factor $g(x)$ of $f(x)$ such that $\deg g\geq k$.


Let a,b,c >0 such that abc=1. Find the biggest K such that
           1/a(1+b) +1/b(1+c) +1/c(1+a) >=K. :D

THe answer is 3/2. We should prove that the expression is at least 3/2. This is easy, since we can put a=x/y, b=/z and c=z/x and reduce it to  \sum xy/(xz+yz) \geq 3/2.
  Observation: one can prove the more general inequality:
   \sum _cyclic 1/a(1+b) \geq 3/(1+abc).,


I'm curious to know about possible shortcuts for this common type of contest problem:

A lady has an unlimited supply of quarters, dimes, nickels and pennies.  Considering only the value of the coins involved, find the number of distinct ways the lady can make exactly 50 cents.  (Note:  Since we are considering only the value of the coins involved, quarter 1, dimes 1 and 2, and nickel 1 would be the same way as quarter 2, dimes 2 and 3, and nickel 4, for example.)

I don't know about how 'short' my method is, but I don't think it's that bad, assuming it's right...
[hide]We want the number of possible quadruples a, b, c, d such that 25a+10b+5c+d=50. We can tell that for a, b, c, and d to all be integers, d must be a multiple of 5, so we can write it as 5e for some nonnegative integer e. Dividing both sides by 5, we get
5a+2b+c+e=10.
We only need to solve this for when a=0, 1, and 2.
When a=0, b can be anywhere from 0 to 5, and for every b there are 10-2b+1 ordered pairs (c,e). So that's 11+9+7+...+1=36 possibilities.
When a=1, the equation is 2b+c+e=5, and b can be 0, 1, or 2. When b is 0, there are 6 pairs (c,e), when b is 1, there are 4 pairs, and when b is 2, there are 2 pairs. So that's another 12 possibilities.
When a=2, the equation is 2b+c+e=0, and the only solution in nonnegative integers is b=c=e=0. That's one more possibility.
So, 36+12+1=49.[/hide]

hmm, altough I don't yet understand, I thought I heard before that if you get page 96 of [url=http://www.math.upenn.edu/~wilf/DownldGF.html]generatingfunctionology[/url] then problems like this become really easy.   maybe?

[quote="SJS"](Note:  Since we are considering only the value of the coins involved, quarter 1, dimes 1 and 2, and nickel 1 would be the same way as quarter 2, dimes 2 and 3, and nickel 4, for example.)[/quote]

What the heck does that mean?  :?

Basically it means we don't distinguish among different coins of the same type.

I seem to recall a student answering a question like this one very quickly a few years ago while watching a MathCounts countdown round at the state or national level.   I can remember being impressed with the speed of the response and felt there must have been some shortcut available.  Now I've run across another problem like it and would like to figure out the shortcut if one exists.

i'm not quite sure about this, but i think it's the coefficient of $x^{50}$ term of the expanded polynomial

$(1+x+x^2+x^3+...)(1+x^5+x^{10}+x^{15}...)(1+x^{10}+x^{20}+x^{30}...)(1+x^{25}+x^{50}+x^{75}+...)$

i dunno how you go about finding the coefficient though....any ideas?

if you look in section 17-4 and 17-5 of AoPS volume two, they have something kinda like what i said in my previous post, but i didn't quite understand that chapter  :(

Your suggestion about the use of generating functions may well be on target --but I still don't see how it leads to a fast solution to the problem.   I, too, need to study this topic in greater detail.

If this question appeared in a MathCounts round of any type, I doubt generating functions were intended, though I suppose that wouldn't stop them from being helpful.

Yeah, on a MC test, my suggestion would be to just make a neat, organized, systematic table.  I also don't think they'd give you questions whose answers were abnormally large. So, a table is probably your best bet.

For a question of this sort of reasonable size, a systematic counting method is best.  Generating functions can be useful for larger similar problems by combining them as geometric sums and then expanding out again.

If my memory serves me correctly, I thought the student during the MathCounts countdown round knew the answer almost instantaneously -- he could not have generated a table in that short period of time.   On the other hand, this was a few years ago -- so I suppose it's possible I could have forgot about some aspect of the MathCounts problem that might have made it much easier than this one.  Having said this, I'm still curious if a generating function could somehow be manipulated mentally to arrive at the correct answer in just a second or two.

Or he might have done the question before and remembered the answer.

are you sure it was 50 cents and it used all four of those coins??

If there are only two coins, the problem is simplified dramatically.  If there are only 3 it is also not bad.  If there were all four and the sum was a significant number like 50, then I don't see how he did it.

I'm sure it was this general type of problem.   I don't recall the exact amount or the specific coins involved in the MathCounts competition, but I initially thought there might be a quick way to solve any problem like this one.   Now I'm wondering if no such technique actually exists and that such problems can be made easier (or harder) depending on the amount and the coins.


I know four IMO's have passed since, but...

why were two members of the Israeli team awarded a bronze medal and an honorable mention, respectively, despite their scores?  :o

[quote="xpmath"]I suppose you already know this, but [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1585848#1585848]here[/url] it states "Contestant was awarded by the decision of the Jury due to special circumstances."

Is it really that important what these "special circumstances" were?[/quote]

important? not exactly... but I already found out, at least the gist of it.  :oops:


function
number theory unsolved
number theory

Given functions $f,g:N^*\rightarrow N^*$, $g$ is surjective and $2f(n)^2=n^2+g(n)^2$, $\forall n>0$. Prove that if $|f(n)-n|\le2005\sqrt n$, $\forall n>0$, then $f(n)=n$ for infinitely many $n$.

By the way, how many people solved this problem? When I found it, I considered it quite difficult, but who knows, maybe it has a very simple solution.  :?

I think I have got it:

Since $f^2(n)=\left(\frac{n+g(n)}{2}\right)^2+\left(\frac{n-g(n)}{2}\right)^2$ we have that there exist integers $k,u,v$ such that:

$f(n)=k(u^2+v^2)$, $n=k((u-v)^2-2v^2)$, $g(n)=k((u+v)^2-2v^2)$. Now take $n$ to be a big enough prime of the form $8k+3$ or $8k+5$, we have k=n, or that $k=n$, or $a^2-2b^2=n$, but the last option implies that 2 is a square modulo $n$, contradiction!, so we must have: $k=n$, and then $f(n)=n(u^2+v^2)$ and here we have: $f(n)=n$, or $f(n)\geq 2n$, but then $f(n)-n=n>2005\sqrt{n}$, for big enough $n$...

Is it right??? :blush:

Something like that I used, too. Very short solution, as usual!  :D  :D . But why did you put  :blush: ?

sorry for offtopic but I cannot hold my curiousity  :D :

[b]Harazi[/b]: how many problems have you proposed this year ?

I do not remember...  :D ... many...  :D , but they are all simple... :D

[quote="harazi"] but they are all simple... :D[/quote]
 :)  :)  :)  ;)

What if without the condition "$ g$ is surjective"?

Hope it is correct,
We take primes $p_n \equiv 3 $ mod $8$ and then let $(x_n)_{n\ge1}$ be a sequence such that
                                                                   $g(x_n)=p_n \forall n$
Now we take note that $2$ is a quadratic non-residue mod $p_n$ and so
$p_n \mid 2f(x_n)^2-x_n^2$ implies that
                                                                    $p_n \mid f(x_n)$ and $p_n \mid x_n$
so let $x_n=b_np_n$ and $f(x_n)=a_np_n$
Now it is pretty clear from the last condition that for sufficiently large $n$, $a_n=1$ and we conclude.

EDIT: 150th post. :)

Did we use the condition " g is surjective" ?

Yes, we did other wise we can't say if the sequence $x_n$ exists.

No. In Pascual's solution we didn't use it. :(

I think that solution is not fully complete and the surjectivity is kind of necessary to fill in the details.

Why is it not fully complete. Can you tell me why? :huh:

In that solution basically the choice to have 8k+3 primes as images invokes surjectivity.

Oh, I see. Sorry for being such ignorant  :blush:

My solution:
Let $p$ be a prime such that $(\frac{2}{p})=-1$. Since $g$ is surjective we have that there exists $n$ such that
$g(n)=p$.
$2f(n)^2=n^2+g(n)^2$ is $2f(n)^2\equiv n^2(mod p)$. If $(p,f(n))=1$ there exist $f(n)^{-1}$ which implies $(\frac{2}{p})=1$ contradiction. So we have $p|f(n),p|n$. Let $f(n)=kp$ and $n=mp$. Equation reduces to
$m^2-2k^2=-1$. So $m,k$ are solutions to negative Pell equation so $m=u_n+2v_n$ and $k=u_n+v_n$ where $u_n,v_n$ are solutions to $u^2-2v^2=1$.
$|f(n)-n|-2005\sqrt{mp}=p|k-m|-2005\sqrt{mp}\le 0$. But since $k-m$ can be arbitrary large we have $f(n)=n$ for infinitely many $n$.

[quote=PepsiCola]My question: If $|f(n)-n|\le2005\sqrt n$, then if we divide by $n$ on both sides, $|\frac {f(n)} {n} - 1| \le \frac {2005} {\sqrt n}$ so taking the limit as $n$ approaches infinity means that $|\frac {f(n)} {n} - 1| = 0$ for sufficiently large $n$, so doesn't that mean $\frac {f(n)} {n} - 1 = 0$ or $\frac {f(n)} {n} = 1$ or $f(n) = n$ for sufficiently large $n$?

I do not know where I am making a mistake, because this does not use any of the other information?[/quote]

For $n \rightarrow \infty$ let $q$ be the rational such that $0<q< \frac {2005} {\sqrt n}$ and such that $q=\frac{r}{n}$ for $r \in \mathbb N$ ([b]It's existence is the main flaw in your solution[/b]) then we might have $|\frac{f(n)}{n}-1| = q$ $\implies$ $|f(n)-n|=qn=r$ so we might have $f(n)=n+r>n.$

So " as $n$ approaches infinity means that $|\frac {f(n)} {n} - 1| = 0$ " is the problem in your solution.

This seems very confusing. Is surjectivity necessary or not? There seems to have been a debate about this before but there is still no answer.
IMO, looking just at primes of the form 8k+3 and taking n to be one of those primes instead of g(n)  we can bypass the surjectivity issue.
Sorry to revive

Same as above$,$ if we let $n=p\equiv 3 \pmod 8$ for a large prime $p,$ and assuming $p\nmid f(n),$ we would get $$2f(n)^2\equiv g(n)^2\pmod p$$ implying $2$ is a quadratic residue $\pmod p$ $,$ a clear contradiction$.$ Therefore if we assume that for all large $n\geq N$ we have $f(n)\neq n,$ by our choice of $n=p$ we see that $p\mid f(p)$ and since $p>N$ is large$,$ it should be true that $f(p)\geq 2p.$ But $f(p)\leq p+2005\sqrt{p},$ which is certainly not possible for large $p.$

We actually don$'$t need surjectivity of $g.$


[hide] 240 /5 = [b] 48 miles per hour [/b] [/hide]

note its not the average of the two because more time is spent at 40 mph

Exactly, the whole trip took 5 hours, so...
[hide]$\frac{240}{5}=48$[/hide]

easier way[hide], if it were 4 and 6 mph and the same way both directions 1/4+1/6=10/24
So 48/10 or 4.8 mph then becomes 48 mph with 40 and 60
48mph[/hide]

lol nat mc I don't think thats particularly the easiest because you still have to know that 3/5 and 2/5 are the involved ratios which takes another step (you probably just saw that instantly in your head but to people like me it doesn't come as quickly  :) ).

The harmonic mean of 40 and 60 is
$\frac{2}{\frac{1}{40}+\frac{1}{60}}$ = $\frac{2}{\frac{1}{24}}$ or 48.

Not exactly easy...

[quote="math92"]If a car goes 40 miles per hour (mph) for 120 miles, then increases its speed to 60 mph for the next 120 miles, what is the car’s average speed over the whole 240 miles?[/quote]

He travels for a total of 240 miles and a total of 5 hours, so the average should be

$\frac{240}{5}=48$

[quote="math92"]If a car goes 40 miles per hour (mph) for 120 miles, then increases its speed to 60 mph for the next 120 miles, what is the car’s average speed over the whole 240 miles?[/quote]

[hide]
math92, you can stop your spamming spree.

3hrs+2hrs=5hrs.

240=(rate)(5)
rate=48 miles an hour.

[/hide]

[quote="Xantos C. Guin"]Another one.

The speed record for waddling from A to B to C is an average of 60 fpm. B is halfway between A and C. A penguin imposter is trying to break the speed record, miscalculates and waddles at 30 fpm till he reaches B. What is the average speed the penguin imposter needs to waddle from B to C to break the record?[/quote]

Uhhh... I don't think he can do it, the time's already up when he reaches B
What's with penguins again?

[hide=answer] First of all, we need to find out how long it took them to do the first 120 miles, and to do that we need to use the formula time=distance/velocity. time= 120/40=3 hours.

Next, we need to use our formula again, and do time=120/60 which is equal to time = 2 hours.

Next, we add the two times together, and get 3+2=5 hours, so then we use our formula, velocity=distance/time, velocity= 240/5 , which is equal to 48 mph.

yeah i wonder how you found this thread.

why have there been so many useless bumps today?

yes this thread is useless and should be locked

[quote=Xantos C. Guin]Another one.

The speed record for waddling from A to B to C is an average of 60 fpm. B is halfway between A and C. A penguin imposter is trying to break the speed record, miscalculates and waddles at 30 fpm till he reaches B. What is the average speed the penguin imposter needs to waddle from B to C to break the record?[/quote]

answer is 90+fpm as $${90+30\over 2}=60$$

[quote=razhder]isnt average speed just total distance/total time[/quote]

yes it is.


inequalities
algebra
polynomial
integration
abstract algebra
vector
calculus

This is quite a beautiful problem.

Let $ f(x) \equal{} \sum_{k \equal{} 1}^n c_k e^{a_k x}$ ($ c_k,a_k\in\mathbb C$) be an exponential polynomial of order $ n$. Prove that
\[ |f(0)|^2\le n\int_{ \minus{} 1}^1 |f(x)|^2\,dx\]
The proof is 3 lines long. Can you find it? ;) (I failed and felt rather miserable about it :(  :blush: ).

I'm a long way from finding the proof - so far, just fishing around for examples. I do note that if $ f$ is a Dirichlet kernel (with the appropriate period), then the right hand side is exactly $ 2$ times the left hand side.

Also, if $ n\equal{}1,$ this boils down to showing that $ \frac{\sinh(2\text{Re}\,a)}{\text{Re}\, a}\ge 1,$ and again there's a factor of $ 2$ worth of "slack" in that particular estimate.

I didn't think about the problem yet, but I'll make one little observation.  I believe we can quickly get a weaker bound, replacing the factor of $ n$ with some factor $ g(n)$.  Namely, we can use the fact that any two norms on a finite-dimensional vector space are equivalent.

Ravi: I'm not sure that's all that much help. Also, to nitpick: $ |f(0)|$ would be a seminorm, not a norm.

[quote="leshik"]Is this inequality sharp?[/quote]
No, $ n$ is not the best constant. But, as Kent already noticed, it is sharp up to a factor of $ 2$ and that is completely satisfactory for all imaginable applications.
[quote="RaviB"]
I believe we can quickly get a weaker bound, replacing the factor of $ n$ with some factor $ g(n)$.
[/quote]
Certainly, but the whole game here is to get a bound linear in $ n$ ;).

I guess that it's time for a hint: it is enough to prove that for some (fixed!) point $ a\in [ \minus{} 0,5,0,5]$  the inequality $ |f^2 (a)|\le n \int |f|^2$ holds, where the integral is taken over the segment $ [ \minus{} 0,5,0,5]$ and then use the fact that $ span(e^{\lambda_i t})$ is invariant under shifts :wink:
And you are right,fedja, the problem is brilliant!!!
P.S This idea is not mine ,I'm just working on the problems of such type now and found Borwein-Erdelyi article. So I  failed too :(   :blush:

[quote="leshik"]I guess that it's time for a hint: it is enough to prove that for some (fixed!) point $ a\in [ \minus{} 0,5,0,5]$  the inequality $ |f^2 (a)|\le n \int |f|^2$ holds, where the integral is taken over the segment $ [ \minus{} 0,5,0,5]$ and then use the fact that $ span(e^{\lambda_i t})$ is invariant under shifts :wink:
[/quote]

Please continue  :) , I cannot see  :( .


$\forall x,y,z>0$
prove that \[xyz+2\ge x+y+z\]

I think it is not true.let $x=10,y=10,z=\frac{1}{100}$


linear algebra
matrix
algebra
polynomial
algebra proposed

We use the characteristic polynomial of a matrix of order $3$: $A^3-Tr(A)\cdot A^2+Tr(A^*)\cdot A-\det A\cdot  I_3=O_3$. Since $A^3=A^2=O_3$ and $\det A=0$, we get $Tr (A^*)=0$, and this is exactly the conclusion.


inequalities
LaTeX
inequalities proposed

If x,y,z>0 and x+y+z=1, prove that {[(1-y)(1-z)/x]^1/2}+ the last 2 radiquals (it's a simetric sum)>=2*3^1/2.

I'm sorry! I don't know to write math simboles.[/quote][/u]

I think you mean
if $x,y,z>0$ and $x+y+z=1$  prove that

$\sqrt{\frac{(1-y)(1-z)}{x}}+ \sqrt{\frac{(1-x)(1-z)}{y}}+ \sqrt{\frac{(1-x)(1-y)}{z}}\geq 2\sqrt 3$

BTW, to write Latex take only the symbols of the dollar sign $\$$ in the beginning and the end of the mathematical expression

Take the substitutions $\displaystyle x=\frac{a}{a+b+c},y=\frac{b}{a+b+c},z=\frac{c}{a+b+c}$ and we need to prove $\displaystyle \sum \sqrt{\frac{(a+b)(a+c)}{a}}\geq 2\sqrt{3(a+b+c)}$\leftrightarow ${\displaystyle \sum\sqrt{\frac{bc}{a}+a+b+c}}\geq 2\sqrt{3(a+b+c)}$.We have ${\displaystyle\sqrt{\frac{bc}{a}+a+b+c}}\geq \frac{\sqrt{3(a+b+c)}+\sqrt{\frac{bc}{a}}}{2}$.Summing these we get $\sum \sqrt{\frac{bc}{a}}\geq \sqrt{3(a+b+c)}$,which becomes $\sum ab\geq \sqrt{3abc(a+b+c)}$,well known. :)


how do I find the homomorphisms Z/5-->G ?

What's $ G$¿
In general, you can take some element $ g$ of order $ 5$ (or the trivial=neutral one of order $ 1$), then map $ n \mod 5$ to $ g^n$.


The vertices of a triangle have coordinates $(-3,1)$, $(-2,-3)$, and $(x,0)$. Find $x$, given $x$ is positive adn the area of the triangle is $7.5$ square units.


geometry
geometric transformation
reflection
geometry proposed

A rectangular piece of paper has top edge $AD$. A line $L$ from $A$ to the bottom edge makes an angle $x$ with the line $AD$. We want to trisect $x$. We take $B$ and $C$ on the vertical ege through $A$ such that $AB = BC$. We then fold the paper so that $C$ goes to a point $C'$ on the line $L$ and $A$ goes to a point $A'$ on the horizontal line through $B$. The fold takes $B$ to $B'$. Show that $AA'$ and $AB'$ are the required trisectors.

This is a very nice way to trisect an angle in origamic geometry (which in Euclidean geometry, we can't always do). The solution is an application of easy reflection and some computation with ratios. I will post the solution later. Sorry!!!  :!:

Here's the solution. Let $\angle{C'AA'}=\alpha$, then $\angle{BAA'}= 90-x+\alpha$, $BA'$ is horizontal, so $\angle{ABB'}=90^{o}$, hence $\angle{BB'A}=90^{o}-\angle{A'AB}=x-\alpha$
By reflection properties, $AB=A'B', AB'=BA'$, $AA'$ is common so $\triangle{AA'B}$ is congruent to $\triangle{A'AB}$ hence $\angle{B'AA'}=\angle{BA'A}=x-\alpha$, also $\angle{AB'A'}=\angle{ABA'}=90^{o}$
Now in $\triangle{C'AA'}$ $C'B'=B'C'$ and $AB'\perp C'A'$ hence $\angle{C'AB'}=\angle{B'AA'}=x-\alpha$
So $\angle{C'AA'}=2\angle{C'AB'}=2(x-\alpha)$, but $\angle{C'AA'}=\alpha$, so:
$2(x-\alpha)=\alpha$
$\alpha=\frac{2x}{3}$
So $\angle{C'AB'}=\angle{B'AA'}=\frac{x}{3}$, $\angle{A'AD}=x-\frac{2x}{3}=\frac{x}{3}$ hence $\angle{C'AB'}=\angle{B'AA'}=\angle{A'AD}$ so $AA'$ and $AB'$ are valid trisectors. QED


Team:
Michael Blaisse
Greg Brockman
Andrew Freddo
Alex Zozula

First Alternate:
Taylor Yi

Second Alternate:
Sydney Creutz

This year had a lot of competition... congrats to everyone (Taylor and Alex are also AoPSers, so that makes three out of the top six.  And you thought AoPS was only good for math :P ).


geometry circumcircle incenter rectangle parallelogram ratio trapezoid

Let $P$ be a variable point on arc $BC$ of the circumcircle of triangle $ABC$ not containing $A$. Let $I_1$ and $I_2$ be the incenters of the triangles $PAB$ and $PAC$, respectively. Prove that:

[b](a)[/b] The circumcircle of $?PI_1I_2$ passes through a fixed point.

[b](b)[/b] The circle with diameter $I_1I_2$ passes through a fixed point.

[b](c)[/b] The midpoint of $I_1I_2$ lies on a fixed circle.

In addition to the incenters $I_1, I_2$ of triangles $\triangle PAB, \triangle PAC,$ let $I_0, I_3$ be incenetrs of triangles $\triangle ABC, \triangle PCB.$ Let $(O)$ be circumcircle of $\triangle ABC$ and $L, M, N$ midpoints of the arcs $\stackrel \frown {BC}, \stackrel \frown {AC}, \stackrel \frown {AB}$ opposite to $A, B, C.$ Quadrilaterals $ABI_1I_0, ACI_2I_0$ are cyclic with circumcircles $(N), (M).$ Internal bisectors $BI_0, PI_2$ of $\angle ABC, \angle APC$ both pass through $M$ and internal bisectors $CI_0, PI_1$ of $\angle ACB, \angle APB$ both pass through $N.$  Perpendicular bisectors of $I_0I_2, I_0I_1,$ intersecting at $Q,$ also pass through $M, N,$ because $\triangle I_0MI_2, \triangle I_0NI_1$ are isosceles, $MI_0 = MI_2, NI_0 = NI_2$ being radii of the circles $(M), (N),$ hence $\angle QMI_0 + \angle QNI_0 = \frac{\angle CAB}{2}.$ Since $\angle NI_0M = \angle CI_0B= \frac{\angle CAB}{2} + 90^\circ,$ then from concave quadrilateral $QMI_0N,$ $\angle NQM = \angle  NI_0M - (\angle QMI_0 + \angle QNI_0) = \frac{\angle CAB}{2} + 90^\circ - \frac{\angle CAB}{2} = 90^\circ.$  But then the angle $\angle I_2I_0I_1 = 90^\circ$ is also right, because $I_0I_1 \perp QN, I_0I_2 \perp QM$.  Similarly, the remaining angles of the quadrilateral $I_0I_1I_3I_2$ are also right, i.e. it is a rectangle. Perpendicular bisectors of the segments $I_0I_1, I_0I_2$ are the midlines of this rectangle and their intersection $Q$ is the intersection of the rectangle diagonals $I_0I_3, I_1I_2$ and their common midpoint. As the point $P$ moves on the arc $\stackrel \frown {BC}$ opposite to $A,$ midpoint $Q$ of the segment $I_1I_2$ moves on a fixed circle $(K)$ with diameter $MN.$ Circumircle $(Q)$ of the rectangle $I_0I_1I_3I_2$ has diameter $I_1I_2$ and passes through the fixed incenter $I_0$. This proves the propositions (b) and (c).

Let $KI_0$ cuts $(O)$ at $S, T,$ such that the point $S$ is on the same side of the line $MN$ as the point $P.$ Since $I_0$ is orthocenter of $\triangle LMN,$ quadrilateral $ANI_0M$ is a kite with $AM = I_0M, AN = I_0N.$ In addition, $K$ is midpoint of $MN,$ hence quadrilateral $TNI_0M$ is a parallelogram and $KI_0 = KT.$ Power of $K$ with respect to the circle $(O)$ is $KS \cdot KT = KS \cdot KI_0 = KN \cdot KM = KQ^2.$ This means that the point $S$ is an inversion image of the point $I_0$ in the circle $(K).$ Consequently, $SN = I_0N \cdot \frac{KQ^2}{KN \cdot KI_0},$ $SM = I_0M \cdot \frac{KQ^2}{KM \cdot KI_0}$ and $\frac{SN}{SM} = \frac{I_0N}{I_0M} = \frac{I_1N}{I_2M}.$ Since the angles $\angle I_1NS \equiv \angle PNS = \angle PMS \equiv \angle I_2MS$ are equal, the triangles $\triangle I_1SN \sim \triangle I_2SM$ are similar by SAS and $\frac{SN}{SM} = \frac{SI_1}{SI_2}$. From inversion of the point $I_0$ into the point $S,$ $QI_0 = QS \cdot \frac{KQ^2}{KS \cdot KQ},$ hence $\frac{I_1I_2}{NM} = \frac{2 QI_0}{2 KQ} = \frac{QS}{KS}.$ Thus the triangles $\triangle NSM \sim \triangle I_1SI_2$ are similar, having the 2 corresponding sides $\frac{SN}{SM} = \frac{I_1N}{I_2M}$ in the same ratio and the corresponding side and median $\frac{I_1I_2}{NM} = \frac{QS}{KS}$ in the same ratio. Therefore, the angles $\angle I_1SI_2 = \angle NSM$ are equal. But since the angles $\angle NSM = \angle NPM \equiv \angle I_1PI_2$ are equal, it follows that the angles $\angle I_1SI_2 = \angle I_1PI_2$ are equal, regardless of the position of the point $P$ on the given arc $\stackrel \frown {BC}$ of the circumcircle $(O).$ Thus the quadrilateral $PSI_1I_2$ is always cyclic, which means that the circumcircle of the triangle $\triangle PI_1I_2$ always passes through the fixed point $S \in KI_0 \cap (O)$. This proves the proposition (a).

[color=#0000FF][b]Comment:[/b][/color] Quadrilateral $ATNM$ is an isosceles trapezoid with $MN \parallel AT.$ Since $MN \perp AL,$ it follows that $AT \perp AL$ is external bisector of the angle $\angle CAB$ and $TL$ is a diameter of the circle $(O).$ As a result the angle $\angle LSI_0 \equiv \angle LST = 90^\circ$ is right, which means that the fixed point $S$ is the tangency point of A-mixtilinear incircle of $\triangle ABC$ with its circumcircle $(O).$

For proposition a, see here: http://www.mathlinks.ro/viewtopic.php?t=313870
b, Denote $ I$ the incenter of triangle $ ABC, E, F$ the midpoint of arc $ AB$ and $ AC$, respectively $ \Rightarrow I,I_1\in (E,EA), I,I_2\in (F,FA)$
$ \angle BII_1\equal{}\angle BAI_1, \angle CII_2\equal{}\angle CAI_2$.
$ \Rightarrow \angle I_1II_2\equal{}\angle BIC\minus{}\angle BII_1\minus{}\angle CII_2\equal{}90^o\plus{}\angle BAC/2\minus{}(\angle BAI_1\plus{}\angle CAI_2)\equal{}90^o$
c, Denote $ L,M$ the midpoint of $ II_1,II_2$, respectively.
Since $ EL\perp II_1, I_2I\perp I_1I \Rightarrow EL//II_2 \Rightarrow EL$ passes through the midpoint $ K$ of $ I_1I_2$. Similarly we get $ ILKM$ is a rectangle
$ \Rightarrow \angle EKF\equal{}90^o \Rightarrow K$ lies on the circle with diameter $ EF$.

Hello!

The point (a) of the proposed problem can be generalized:
Let $ ABC$ be a triangle, $ \Omega$ is a circle through the points $ B$, $ C$ and $ P\in\Omega$ (the points $ P$ and $ A$ are on different sides of $ BC$).
If $ \left\{X,B\right\} \equal{} AB\cap\Omega$, $ \left\{Y,C\right\} \equal{} AC\cap\Omega$ and $ I_{1}$, $ I_{2}$ are the incenters of $ \Delta PXB$, $ \Delta PYC$ then prove that the circumcircle
of $ \Delta PI_{1}I_{2}$ passes through a fixed point.

Solution:
Let $ \Gamma$ be the circle that tangent to the lines $ AB$, $ AC$ and to the circle $ \Omega$ (the circles $ \Gamma$ and $ \Omega$ are internally tangent).
$ \left\{S\right\} \equal{} \Gamma\cap\Omega$ (the points $ S$ and $ A$ are on different sides of $ BC$), $ \left\{K\right\} \equal{} AB\cap\Gamma$ and $ \left\{L\right\} \equal{} AC\cap\Gamma$.
From here [url]http://forumgeom.fau.edu/FG2003volume3/FG200325.pdf[/url] please read carefully the lemma1 + the solution
$ \left\{N,S\right\} \equal{} SK\cap\Omega$ and $ \left\{M,S\right\} \equal{} SL\cap\Omega$.
$ M$, $ N$ are the midpoints of the arcs $ \widehat{CY}$ and $ \widehat{BX}\Rightarrow I_{1}\in PN$ and $ I_{2}\in PM$.

Let $ V\in\Gamma$ such that $ PV$ is tangent to the circle $ \Gamma$.
$ \Omega$ is the circumcircle of $ \Delta PXB$, $ I_{1}$ is the incenter of $ \Delta PXB$, $ \left\{K\right\} \equal{} XB\cap\Gamma$ and $ \left\{V\right\} \equal{} PV\cap\Gamma\Rightarrow$
$ \Rightarrow I_{1}\in KV$ and the points $ P,S,I_{1},V$ are concyclic (see the lemma1)
Similarly, $ \Omega$ is the circumcircle of $ \Delta PYC$, $ I_{2}$ is the incenter of $ \Delta PYC$, $ \left\{L\right\} \equal{} YC\cap\Gamma$ and $ \left\{V\right\} \equal{} PV\cap\Gamma\Rightarrow$
$ \Rightarrow I_{2}\in LV$ and the points $ P,S,I_{2},V$ are concyclic (see the lemma1)

So, the points $ P,S,I_{1},I_{2},V$ are concyclic $ \Rightarrow$ the circumcircle of $ \Delta PI_{1}I_{2}$ passes through the fixed point $ S$.

Best regards, Petrisor Neagoe :)


calculus
integration
combinatorics unsolved
combinatorics

Determine the number of integral solutions to

\[x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=28 \]

With conditions
a) $x_{i}\geq 0$
b) $x_{i}> 0$
c) $x_{i}> i$ where ($i=1,2,3,4,5$)
$\hline$

Please help me with the specific solution above, but you are also encouraged to solve the general problem $\sum_{i}x_{i}= N$, with corresponding conditions.
I really appreciate a clear solution to this.

[hide]a) and b) both using

[b]Theorem[/b]. Let $m$ and $n$ be positive integers.
i)There are $C_{n-1}^{m-1}$ ordered $m$-tuples $(x_{1},x_{2},...,x_{m})$ of positive integers satisfying the equation $x_{1}+x_{2}+...+x_{m}=n$.
ii)There are $C_{n+m-1}^{m-1}$ ordered $m$-tuples $(x_{1},x_{2},...,x_{m})$ of nonnegative integers satisfying the equation $x_{1}+x_{2}+...+x_{m}=n$.

c) You can let $y_{i}=x_{i}-i$.[/hide]


pigeonhole principle
combinatorics unsolved
combinatorics

In a round-rubin chess tourment, every player has won at least once. Prove that there are two players with exactly the same number of wins.

Isn't it clear? :?

The number of wins of a player ranges between $0$ and $n-1$. However, each player has won at least once, so this number actually ranges from $1$ to $n-1$, i.e. it can take $n-1$ values. Since there are $n>n-1$ values..


geometry
trigonometry
inequalities
geometry unsolved

All of the angles of the triangle ABC are acute. $ AB \equal{} c, BC \equal{} a, CA \equal{} b$, $ a \ne b \ne c \ne a$. $ AA_m, BB_m, CC_m$ are medians. $ AA_l, BB_l, CC_l$ are angle bisectors. $ AA_h, BB_h, CC_h$ are altitudes.

a) Is it true that we can find the areas of the triangles $ A_iB_iC_i$, $ i \equal{} m, l, h$, using only a, b and c?
b) Is it true that we can compare the areas of those three triangles?
c) Prove or disprove that $ S_{A_mB_mC_m} \plus{} S_{A_hB_hC_h} \plus{} S_{A_lB_lC_l} \ge S_{ABC}$.

$ S_{ABC}$ is the area of the triangle ABC.

part c is clearly false: take an equilateral triangle

and for a... of course, since the sides determine everything else. finding the expressions is a different matter :) though for $ i \equal{} m$ it's definitely easy.

Yes

d) Find the areas of $ S_{A_iB_iC_i}, i \equal{} l, h$ using only a, b and c  :D

[quote="CDP100"] [color=darkred]Let $ ABC$ be an acute triangle. Denote $ S = [ABC]$ and the lines

$ \begin{array}{c} AA_m, BB_m, CC_m\mathrm {\ - \ the\ medians} \\
 \\
AA_l, BB_l, CC_l\mathrm {\ - \ the\ angle\ bisectors} \\
 \\
AA_h, BB_h, CC_h\mathrm {\ - \ the\ altitudes}\end{array}$ where $ \begin{array}{c} \{A_m, A_l, A_h\}\subset (BC)\ . \\
 \\
\{B_m, B_l, B_h\}\subset (CA)\ . \\
 \\
\{C_m, C_l, C_h\}\subset (AB)\ .\end{array}$

$ 1\blacktriangleright$ Can find we the areas of $ \triangle\ A_kB_kC_k$ , $ k\in \{m, l, h\}$, using only $ a,b,c$ ?

$ 2\blacktriangleright$ Can compare we the areas of those three triangles ?

$ 3\blacktriangleright$ Prove that $ [A_mB_mC_m] + [A_hB_hC_h] + [A_lB_lC_l]\le \frac {3S}{4}\ .$[/color] [/quote]
[hide="Answer."]
$ \begin{array}{cc} 1\blacktriangleright & \mathrm {yes\ .} \\
 \\
2\blacktriangleright & \mathrm {only\ (m,l)\ and\ (m,h)\ !} \\
 \\
3\blacktriangleright & \mathrm {yes\ ( = \frac S4\ ,\ \le\frac S4\ ,\ \le\frac S4)\ .}\end{array}$[/hide]
[hide="Proof."]
[color=darkred][b][u]Lemma.[/u][/b] For an interior point $ P$ of the acute triangle $ ABC$ denote :

$ \{\begin{array}{cc} X\in AP\cap (BC)\ ,\ \frac {XB}{XC} = x \\
 \\
Y\in BP\cap CA\ ,\ \frac {CB}{CA} = y \\
 \\
Z\in CP\cap AB\ ,\ \frac {ZA}{ZB} = z \\
 \\
\{P\} = AX\cap BY\cap CZ\implies xyz = 1\end{array}\|$ $ \implies$ $ [XYZ] = \frac {2S}{(1 + x)(1 + y)(1 + z)}\ .$[/color]

[color=darkblue][b][u]Particular cases :[/u][/b]

$ \{\begin{array}{cccc} (m)\blacktriangleright & P: = G & \implies & [A_mB_mC_m] = \frac S4\ . \\
 \\
(l)\blacktriangleright & P: = I & \implies & [A_lB_lC_l] = \frac {2abcS}{(a + b)(a + c)(b + c)}\le \frac S4\ . \\
 \\
(h)\blacktriangleright & P: = H & \implies & [A_hB_hC_h] = 2S\cos A\cos B\cos C\le \frac S4\ .\end{array}$

I used the well-known inequalities $ \|\begin{array}{c} \frac {abc}{(a + b)(b + c)(c + a)}\le \frac 18 \\
 \\
\cos A\cos B\cos C\le\frac 18\end{array}\ .$[/color]

[b][u][color=darkred]An open problem :[/u][/b] $ \frac {abc}{(a + b)(b + c)(c + a)}\ge\cos A\cos B\cos C\ ?!$[/color][/hide]


algebra
function
domain
real analysis
real analysis unsolved

Suppose f:R^m ->R^m is continuous and satisfies
|f(x)-f(y)|>=c|x-y|
for all x,y and some constant c>0, prove that f is a bijection with continuous inverse

It has been discussed before. I can't find the thread so I am going to pirate the proof of someone else: Clearly the range of $ f$ is closed, but by the invariance of domain theorem $ f$ is also an open map. Hence, since $ R^n$ is connected, $ f$ is surjective.

Is this from a prelim test for undergraduates?  :maybe:


vector
combinatorics solved
combinatorics

P is a polyhedral which all its faces are traingle and has odd edges, we name its vertices 1,2,...,n. show the number of face which its vertices is clockwise increasing ,is odd.

It's probably the clocwise direction in the plane of that particular face when we look perpendicularly to it from outside the polyhedron. It seems natural, but maybe we need sam-n's confirmation on this one.

He didn't really say convex polyhedron so "outside" can be a tricky notion. What if the polyhedron has a "gap" inside ?

Seems like you're right. I didn't exactly pay attention to the text: I was just trying to help you, since I don't really enjoy combinatorics and I wasn't planning to think hard about this problem or anything :D. Ok, we'll just wait for him to explain then.

What means [quote]are traingle and has odd edges[/quote] ? Odd edges?
Anyway, it reminds me of a result called Sperner's Lemma?
Does anyone heard of it?

I think he means that the polyhedron has an odd number of edges.

odd edge =odd number of edges  :D as Grobber mentioned :) and clockwise increasing in each face is'nt clear?maybe it's due to my poor english that I can not convey the concept :(

I have a solution.
For an edge AB, if the number written in A is bigger than the one written in B construct an arrow perpendicular to vector AB and lying in the face containing AB.
There are two such faces, but we choose the one which lies more at the right for direction AB( imagin the polyhedral like a stone and a man moving from A to B. Then he draws an arrow on the face which is on his right).
Then a face satisfies the condition if and oly if there is just one arrow inside of it, and doesn't if and pnly if there are two arrows inside of it( we cannot have 0 or 3 arrows). As the number of edges, so the number of arrows is odd, the conclusion follows.


I'd like to take a few moments to thank all the people who made this spam possible.

I see that you are trying to rise up from the ranks of the Navier-Stokes Equations and attempting to become a... something.  Seems sort of cheap.  Although I'm only posting to become a... something.

Post number 10.  I have now evolved from a Charmander into a...

er...I'm not spamming to increase my post count. If I wanted to do that, I could go answer easy questions in Intermediate Topics / Pre-Olympiad.

Still Yang-Mill Theory, about 100 more posts to go to become Navier-Stroke Equations! Maybe I can use this forum to reach goal :)

Just kidding lol....

Yeah, I'm getting kind of close. I just need to go on a spamming rampage here or in F and G...

[quote="tetrahedr0n"]I just need to go on a spamming rampage...in F and G...[/quote]

Yea you go ahead and try that.

MysticTerminator appears to be a hermaphrodite.

[quote="JBL"]Do you think I'll get something special when I reach 10,000? ;)  (At lifetime rate of posting, I'll be there within 3 years.  And I'll be back to 1st in total posts within a 2 centuries ...)[/quote]

At 10,000 you'll get nothing special except spam.

This is my best spam of the day.

Spam.

if you want to see REAL spam:
noml.org > Message board> Infamous Word Association Game, What If... Game, anything that has modan posting in it, and the like.

But most of us want a good post count here

...? I still don't understand how post count matters...nobody respects tare more than simon, for example (actually, tare left a while ago, so simon may have the higher post count by now ... but still)


function
trigonometry
quadratics
algebra
quadratic formula

Find the value of $\sin x$, given that $\cos x = \tan x$

[hide]Rewrite to get $\cos{x}=\frac{\sin{x}}{\cos{x}}$.
$\sin{x}=\cos^2{x}$.
$\sin{x}=1-\sin^2{x}$.
$\sin^2{x}+\sin{x}-1=0$.

Using the quadratic formula, we have that $\sin{x}=\frac{-1\pm\sqrt{5}}{2}$. [/hide]

yes, but
$\frac{-1-\sqrt{5}}{2}$
is less than $-1$

[quote="236factorial"][quote="shobber"]$\sin{x} \in [-1, 1]$.[/quote]
x can be anything?  :huh:[/quote]
Yes. $\sin{x} \in [-1, 1]$ for all $x \in R$.

[quote="chess64"][quote="236factorial"][quote="shobber"]$\sin{x} \in [-1, 1]$.[/quote]

x can be anything?  :huh:[/quote]

He means that the range of $\sin x$ is $[-1,1]$, so only one of those answers is correct.[/quote]

oh duh  :roll:  I thought shobber was giving an answer, sorry for the suspicion  :blush:

That way, the only answer is $\frac{\sqrt5-1}{2}$.


symmetry
search
inequalities unsolved
inequalities

Let $a,b,c$ positive real numbers prove the following ineq

$\frac{2a^n-b^n-c^n}{b^2-bc+c^2}+\frac{2b^n-a^n-c^n}{c^2-ac+a^2}+\frac{2c^n-b^n-a^n}{a^2-ba+b^2}\geq 0$   :D

I think this appeared in some GM during 2003.
It is not easy, for sure...

My first thought is Cebasev's ineq...
Probably useful.... :)  :)

Yes it was posted in Gazeta Mathematica by Vasc ,but I have no issues from GM so for me this problem is unsolved .

If posted before please give the link because I can't find it with the search button.
Thank you

[quote="Soarer"][quote="socrates"]Due to the symmetry of the ineq we may suppose WLOG that $a\geq b\geq c$
Then $a^n\geq b^n\geq c^n$

and $\frac{1}{b^2-bc+c^2}\geq\frac{1}{a^2-ac+c^2}\geq\frac{1}{a^2-ab+b^2}$  as well.

Applying Cebasev's ineq we easily reach the conclusion... :)

Am I right?? :lol:[/quote]

you are wrong, because $\frac{1}{b^2-bc+c^2}\geq\frac{1}{a^2-ac+c^2}\geq\frac{1}{a^2-ab+b^2}$ is not true. BTW, this is posted before, and vasc gave a solution. Please search.[/quote]

You are right Soarer.....
My bad.....

Anyway, thanks for noticing :)

the link is [url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=%2A2a%5En-b%5En-c%5En%2A&t=25551[/url]

See there a nice solution  :)


This is what I conjecture:
There is a constant (lets call it amf  :D ) and an integer N (another constant) such for all n>N, there is at least one prime p such that
n \leq p \leq n+(amf)*ln(n).

I saw many conjectures like: between n 2 and (n+1) 2 is at least a prime, but mine is the best  :D.

Actually the constant can be increased such that we get N=1, but it would be nice to get a large N and obtain a small amf. I think if we tend N to  \infty, amf tends to 1. Or I hope it does  :D .

A little history:
  The conjecture with n^2 and (n+1)^2 is due to Schinzel and Sierpinski. Legendre made another conjecture, but with x and x+sqrt(x). Schinzel conjectured that for x>=8 we can find a prime between x and x+(lnx)^2. And now...your conjecture. I don't know why I feel that this conjecture is true, but it is allmost impossible to prove. I hope I'm wrong.

Rankin proved that the gap between consecutive primes is infinitely often larger than any multiple of log(n): see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48789 .

Yes, $\log(n)$ is too small for the asymptotics of the gap between primes.  I believe it was conjectured by Cramer to be at most $O({\log(n)}^2)$ and it is unknown whether it can be as large as ${\log(n)}^{a}$  for $a$ between 1 and 2.


trigonometry
AMC
USA(J)MO
USAMO
geometry
angle bisector
exterior angle

Let $ P$ be a point interior to the triangle $ ABC$ such that $ \angle PAB\equal{}\angle PAC\equal{}20^\circ$, $ \angle PCA\equal{}10^\circ$, $ \angle PCB\equal{}30^\circ$. Prove that $ PA\equal{}PB$.

An idea:
1) Use Sine law for triangle APC to express AC in terms of AP.
2) It is easy to express AB in terms of AC (AB=BC) from 1) it may
   be expressed in terms of AP.
3) Use Cosine law for triangle APB.

I think the problem is in some way similar to USAMO 1996-1-2.

[quote="Leon"]Let $ P$ be a point interior to the triangle $ ABC$ such that $ \angle PAB \equal{} \angle PAC \equal{} 20^\circ$, $ \angle PCA \equal{} 10^\circ$, $ \angle PCB \equal{} 30^\circ$. Prove that $ PA \equal{} PB$.[/quote]
[hide="Solution"]
Notice that $ \angle BAC \equal{} \angle BAP \plus{} \angle PAC \equal{} 20 \plus{} 20 \equal{} 40$ and $ \angle ACB \equal{} \angle ACP \plus{} \angle PCB \equal{} 10 \plus{} 30 \equal{} 40$.  Hence, $ \angle ABC \equal{} 180 \minus{} \angle BAC \minus{} \angle ACB \equal{} 180 \minus{} 40 \minus{} 40 \equal{} 100$.  Let $ \angle PBA \equal{} \alpha\implies \angle PBC \equal{} \angle ABC \minus{} \angle PBA \equal{} 100 \minus{} \alpha$.  Allow $ AP$ to meet $ BC$ at $ Q$.  Thus, $ \angle QPC \equal{} \angle PAC \plus{} \angle PCA \equal{} 10 \plus{} 20 \equal{} 30 \equal{} \angle QCP\implies QP \equal{} CQ$.  Therefore, $ \angle BPQ \equal{} 180 \minus{} \angle BCP \minus{} \angle PBC \minus{} \angle QPC \equal{} 20 \plus{} \alpha$.  Now, notice that $ \frac {BQ}{PQ} \equal{} \frac {BQ}{CQ} \equal{} \frac {AB}{AC}$ since $ AQ$ is an angle bisector of $ \angle BAC$.  Notice that
\[ \frac {AB}{AC} \equal{} \frac {\sin \angle ACB}{\sin \angle ABC} \equal{} \frac {\sin 40}{\sin 100} \equal{} \frac {\sin 40}{\sin 80}
\]
Also, notice that
\[ \frac {\sin 40}{\sin 80} \equal{} \frac {BQ}{PQ} \equal{} \frac {\sin \angle BPQ}{\sin \angle PBQ} \equal{} \frac {\sin (20 \plus{} \alpha)}{\sin (100 \minus{} \alpha)}
\]
Thus, we let $ 40 \equal{} 20 \plus{} \alpha\implies \alpha \equal{} 20$ and $ 100 \minus{} \alpha \equal{} 80\implies \alpha \equal{} 20$.  Thus, $ \alpha \equal{} 20$ either way, so $ \angle PBA \equal{} \alpha \equal{} 20 \equal{} \angle PAB\implies PA \equal{} PB$. [/hide]

This is a solution without trigonometry.

Denote $ Q\equal{}BC \cap AP$.  We have:

$ \angle QPC\equal{}20^\circ\plus{}10^\circ\equal{}30^\circ$ (by exterior angle theorem)

$ \angle QPC\equal{} \angle QCP \Longrightarrow PQ\equal{}CQ$

Let $ R$ be the intersection point of $ AC$ with the perpendicular bisector of $ PC$. We have $ \angle PQC\equal{}120^\circ$ and

$ \angle PQR\equal{}\angle RQC\equal{}60^\circ$

The triangles $ \triangle PQR$ and $ \triangle CQR$ are congruent by S.A.S. , so

$ PR\equal{}RC  \Longrightarrow \angle RPC\equal{}\angle RCP\equal{}10^\circ$

$ \angle PRA\equal{}10^\circ\plus{}10^\circ\equal{}20^\circ$ (by exterior angle theorem)

$ \angle PRA\equal{}\angle PAR \Longrightarrow PA\equal{}PR$

The triangles $ \triangle AQB$ and $ \triangle AQR$ are congruent by A.S.A. , so $ AB\equal{}AR$

The triangles $ \triangle APB$ and $ \triangle APR$ are congruent by S.A.S. , so $ PB\equal{}PR$

Thus $ PB\equal{}PA$ and we are done.

[color=darkblue][b][u]Proof.[/u][/b] Denote $ \boxed {m(\widehat {PBA}) \equal{} x}$ . Thus, $ m(\widehat {PBC}) \equal{} 100 \minus{} x$ .  Apply the [b]Ceva's theorem[/b]

([u]trigonometrical form[/u]) : $ \sin\widehat {PBA}\cdot\sin\widehat {PCB}\cdot\sin\widehat {PAC} \equal{} \sin\widehat {PAB}\cdot\sin \widehat {PBC}\cdot\widehat {PCA}$ $ \Longleftrightarrow$

$ \sin x\sin 30\sin 20 \equal{} \sin 20\sin (100 \minus{} x)\sin 10$ $ \Longleftrightarrow$ $ \sin x \equal{} 2\cos (x \minus{} 10)\sin 10$ $ \Longleftrightarrow$

$ \sin x \equal{} \sin x \plus{} \sin (20 \minus{} x)$ $ \Longleftrightarrow$ $ \sin (20 \minus{} x) \equal{} 0$ $ \Longleftrightarrow$ $ \boxed {x \equal{} 20}$ $ \Longleftrightarrow$ $ PA \equal{} PB$ .[/color]


calculus
integration
real analysis
real analysis solved

$\lambda$ real number find the solutions of the system of equation integrals

$f(x)=e^x+\lambda\int_{0}^{x}e^{x-y}f(y)dy$ if $0\leq x\leq 1$
$g(x)=e^x+\lambda\int_{0}^{1}e^{x-y}g(y)dy$ if $0\leq x\leq 1$

We rewrite the given integral equations as follows.

\[e^{-x}f(x)=1+\lambda \int_0^x e^{-y}f(y)dy\ \cdots\ [1]\]

\[e^{-x}g(x)=1+\lambda \int_0^1 e^{-y}g(y)dy\ \cdots\ [2]\]

Differentiating for $[1]$ with respect to $x$ ,

$-e^{-x}f(x)+e^{-x}f'(x)=\lambda e^{-x}f(x)$
$\Longleftrightarrow f'(x)=(1+\lambda)f(x)\Longleftrightarrow f'(x)=Ae^{(1+\lambda)x}$,  where $A$ is a constant number.

Plugging $x=0$  into $[1]$, we get $A=1$,yielding $f(x)=e^{(1+\lambda)x}$

Now let $\int_0^1 e^{-y}g(y)dy=C\ \cdots\ [3]$, we have $g(x)=(1+\lambda C)e^x$, where $C$ is a constant number.

From $[3]$, we have

$C=\int_0^1 e^{-y}(1+\lambda C)e^y dy=\int_0^1 (1+\lambda C)dy=[(1+\lambda C)y]_0^1\ =C$

$\Longleftrightarrow (1-\lambda )C=1$ ,since $\lambda \neq 1$, we have $C=\frac{1}{1-\lambda}$, yielding

$g(x)=\frac{e^x}{1-\lambda}$

Consequently, the answer is

\[f(x)=e^{(1+\lambda )x}, \ g(x)=\frac{e^x}{1-{\lambda}}\ \ (0\leqq x \leqq 1)\]

kunny's first answer is correct. For an alternate method, use Laplace transforms, noting the convolution integral. If $F(s)$ is the Laplace transform of $f(x),$

$F(s)=\frac1{s-1}+\lambda\cdot\frac1{s-1}\cdot F(s),$ from which

$F(s)=\frac1{s-1-\lambda},$ which implies $f(x)=e^{(1+\lambda)x}.$

[quote="Kent Merryfield"]On the second one, that should be $g(x)=e^x(1+\lambda C).$ Integrating, we finde that $C=\frac1{1-\lambda}.$ Thus, if $\lambda\ne1$, the solution is $g(x)=\frac{e^x}{1-\lambda}.$ (If $\lambda=1$ there is no solution.)[/quote]

Yes, you are right. Thank you for pointing out my mistake, Kent Merryfield.

I have just edited.

kunny