inequalities
Cauchy Inequality
inequalities proposed

If $a_{1} \geq a_{2} \geq \ldots \geq a_{n} \geq 0,$ $b_{1}, b_{2}, \ldots, b_{n} \in R$ and $\displaystyle \sum_{i=1}^{k} a_{i} \leq \sum_{i=1}^{k} b_{i},$ $\forall k=1,n$ then we have: $\displaystyle \sum_{i=1}^{n} a_{i}^{2} \leq \sum_{i=1}^{n} b_{i}^{2}.$

cheers! :D :D


inequalities
algebra
polynomial
inequalities proposed

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

$ \frac {2a^2 \plus{} 3bc}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {2b^2 \plus{} 3ca}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {2c^2 \plus{} 3ab}{a^2 \plus{} ab \plus{} b^2}\ge5$[/quote]

----------------
Solution is found in the book of Dr. Vasile Cirtoaje (GIL Romania) at pg. 322

I found a nice Theorem for proving almost all symmetric homogeneous inequalities of degree six.

[b]Theorem.[/b] [i]Any symmetric homogeneous polynomial of degree six can be written in the form
$ f(a,b,c) \equal{} Ax^2 \plus{} (Bp^3 \plus{} Cpq)x \plus{} Dp^6 \plus{} Ep^4q \plus{} Fp^2q^2 \plus{} Gq^3$,
where $ A,B,C,D,E,F,G$ are real numbers and $ p \equal{} a \plus{} b \plus{} c$, $ q \equal{} ab \plus{} bc \plus{} ca$, $ x \equal{} abc$.
If
$ f(a,0,0)\ge 0$, $ f(a,1,0) \ge 0$, $ f(a,1,1)\ge 0$ for all $ a\ge 0$,
and
$ f(a,1,1)\ge \frac {A(a \minus{} 1)^4}{27}$ for all $ a\ge 1$,
then $ f(a,b,c) \ge 0$  for any nonnegative real numbers $ a,b,c$.[/i]

For our inequality we have $ A \equal{} 0$. Therefore, it suffices to prove it for $ c \equal{} 0$ and for $ b \equal{} c \equal{} 1$.

Actually, this theorem fails for some strong inequalities $ f(a,b,c) \ge 0$  with $ A > 0$ and with equality for $ a \equal{} b \equal{} c$ and again for $ \frac a{k} \equal{} b \equal{} c$  ($ k > 1$) and any cyclic permutation thereof.
The following theorem kills also this inequality type.

[b]Theorem 2.[/b] [i]Any symmetric homogeneous polynomial of degree six can be written in the form
$ f(a,b,c) \equal{} Ax^2 \plus{} (Bp^3 \plus{} Cpq)x \plus{} Dp^6 \plus{} Ep^4q \plus{} Fp^2q^2 \plus{} Gq^3$,
where $ A,B,C,D,E,F,G$ are real numbers and $ p \equal{} a \plus{} b \plus{} c$, $ q \equal{} ab \plus{} bc \plus{} ca$, $ x \equal{} abc$.
If
$ f(a,0,0)\ge 0$, $ f(a,1,0) \ge 0$, $ f(a,1,1)\ge 0$ for all $ a\ge 0$,
and there is a real number $ \delta \neq \minus{} 2$ such that
$ f(a,1,1)\ge \frac {4A(a \minus{} \delta)^2(a \minus{} 1)^4}{81(\delta \plus{} 2)^2}$ for all $ a\ge 1$,
then $ f(a,b,c) \ge 0$  for any nonnegative real numbers $ a,b,c$.[/i]

To prove this strong inequality above we need choose $ \delta \equal{} k$.
For ordinary inequalities (as Iran 1996), we can choose $ \delta \equal{} 0$ or $ \delta \equal{}\infty$. Also, we can use Theorem 1, which is stronger than Theorem 2 for $ \delta \equal{}\infty$.

I have found a solution
[hide="solution"]
By AM-GM, and Nesbitt's Inequality
$ \sum \frac {2a^2 \plus{} 3bc}{b^2 \plus{} bc \plus{} c^2} \ge \sum \frac {4a^2 \plus{} 6bc}{3(b^2 \plus{} c^2)} \equal{} \frac {4}{3} \sum \frac {a^2}{b^2 \plus{} c^2} \plus{} 2\sum \frac {1}{\frac {b}{c} \plus{} \frac {c}{b}} \ge \frac {4}{3}\cdot \frac {3}{2} \plus{} 2\cdot \frac {1}{2}\cdot 3 \equal{} 5\qquad Q.E.D$  :) [/hide]

EDIT:  :oops: sorry this is wrong

Why does?

[quote="ocha"]
$ \frac {4}{3} \sum \frac {a^2}{b^2 \plus{} c^2} \plus{} 2\sum \frac {1}{\frac {b}{c} \plus{} \frac {c}{b}} \ge \frac {4}{3}\cdot \frac {3}{2} \plus{} 2\cdot \frac {1}{2}\cdot 3$  :) [/quote]

hey, if i'm not mistaken :
$ \frac{1}{\frac{b}{c}\plus{}\frac{c}{b}} \le 2$

[quote="Vasc"]Actually, this theorem fails for some strong inequalities $ f(a,b,c) \ge 0$  with $ A > 0$ and with equality for $ a \equal{} b \equal{} c$ and again for $ \frac a{k} \equal{} b \equal{} c$  ($ k > 1$) and any cyclic permutation thereof.
The following theorem kills also this inequality type.

[b]Theorem 2.[/b] [i]Any symmetric homogeneous polynomial of degree six can be written in the form
$ f(a,b,c) \equal{} Ax^2 \plus{} (Bp^3 \plus{} Cpq)x \plus{} Dp^6 \plus{} Ep^4q \plus{} Fp^2q^2 \plus{} Gq^3$,
where $ A,B,C,D,E,F,G$ are real numbers and $ p \equal{} a \plus{} b \plus{} c$, $ q \equal{} ab \plus{} bc \plus{} ca$, $ x \equal{} abc$.
If
$ f(a,0,0)\ge 0$, $ f(a,1,0) \ge 0$, $ f(a,1,1)\ge 0$ for all $ a\ge 0$,
and there is a real number $ \delta \neq \minus{} 2$ such that
$ f(a,1,1)\ge \frac {4A(a \minus{} \delta)^2(a \minus{} 1)^4}{81(\delta \plus{} 2)^2}$ for all $ a\ge 1$,
then $ f(a,b,c) \ge 0$  for any nonnegative real numbers $ a,b,c$.[/i]

To prove this strong inequality above we need choose $ \delta \equal{} k$.
For ordinary inequalities (as Iran 1996), we can choose $ \delta \equal{} 0$ or $ \delta \equal{} \infty$. Also, we can use Theorem 1, which is stronger than Theorem 2 for $ \delta \equal{} \infty$.[/quote]

Of course, I only like classic ineq but indeed, they are very nice and useful lenmas, Vasc. Congratulation!  :)

Below are some "strong" inequalities of degree 6 with nonnegative $ a,b,c$.

$ (1) \ \ \ \ \ \ \ \ \ \sum \frac {9a^2 \minus{} 4bc}{3a^2 \plus{} 2b^2 \plus{} 2c^2} \ge \frac {15}{7}$;

$ (2) \ \ \ \ \ \ \ \ \ \sum \frac {a(b \plus{} c)}{ka^2 \plus{} b^2 \plus{} c^2} \le \frac {6}{k \plus{} 2} \ \ \ \ for \ \ 5 \minus{} 2\sqrt 6 \le k \le 5 \plus{} 2\sqrt 6$;

$ (3) \ \ \ \ \ \ \ \sum \frac {a}{ka^2 \plus{} b^2 \plus{} c^2} \le \frac {9}{(k \plus{} 2)(a \plus{} b \plus{} c)} \ \ \ \ for \ \ 7 \minus{} 2\sqrt {11} \le k \le 7 \plus{} 2\sqrt {11}$;

$ (4) \ \ \sum \frac {b \plus{} c}{ka^2 \plus{} b^2 \plus{} c^2} \le \frac {18}{(k \plus{} 2)(a \plus{} b \plus{} c)} \ \ \ \ for \ \ 25 \minus{} 4\sqrt {38} \le 17k \le 25 \plus{} 4\sqrt {38}$;

$ (5) \ \ \ \ \ \ \ \ \ \sum \frac {6a(b \plus{} c) \minus{} bc}{12a^2 \plus{} b^2 \plus{} c^2} \le \frac {33}{14}$;

$ (6) \ \ \ \ \ \ \ \ \ \sum \frac {a(b \plus{} c) \minus{} bc}{a^2 \plus{} 3b^2 \plus{} 3c^2} \le \frac {3}{7}$.

Equality holds for $ a \equal{} b \equal{} c$ and again for $ \frac a {w} \equal{} b \equal{} c$ or any cyclic permutation. We have
$ (1) \ \ \ w \equal{} \frac 4{3}$;
$ (2) \ \ \ w \equal{} 2 \plus{} \frac {\sqrt 6}{3}$ for $ k \equal{} 5 \minus{} 2\sqrt 6$, and $ w \equal{} 2 \minus{} \frac {\sqrt 6}{3}$ for $ k \equal{} 5 \plus{} 2\sqrt 6$;
$ (3) \ \ \ w \equal{} \frac {9 \minus{} \sqrt {11}}{5}$ for $ k \equal{} 7 \plus{} 2\sqrt {11}$;
$ (4) \ \ \ w \equal{} \frac {42 \minus{} 5\sqrt {38}}{11}$ for $ k \equal{} \frac {25 \plus{} 4\sqrt {38}}{17}$;
$ (5) \ \ \ w \equal{} \frac 4{3}$;
$ (6) \ \ \ w \equal{} \frac 4{3}$.

$ \frac {2a^2 \plus{} 3bc}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {2b^2 \plus{} 3ca}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {2c^2 \plus{} 3ab}{a^2 \plus{} ab \plus{} b^2}\ge5$
$ \iff \sum (2a^2 \plus{} 3bc)(c^2 \plus{} ca \plus{} a^2)(a^2 \plus{} ab \plus{} b^2)\ge 5\prod (a^2 \plus{} ab \plus{} b^2)$
$ \iff \sum_{sym}a^6 \plus{} 2\sum_{sym}a^5b \plus{} \sum_{sym}a^3b^2c\ge3\sum_{sym}a^4b^2 \plus{} \sum_{sym}a^3c^3$
By Miurhead inequality,we have:
$ \sum_{sym}a^2b^2c^2\le \sum_{sym}a^3b^2c.........(*)$
$ \sum_{sym}a^4b^2 \plus{} \sum_{sym}a^3b^3\le2\sum_{sym}a^5b.......(1)$
By Schur inequality, we have:
$ 2\sum_{sym}a^4b^2\le \sum_{sym}a^2b^2c^2 \plus{} \sum_{sym}a^6$
$ 2\sum_{sym}a^4b^2\le \sum_{sym}a^3b^2c \plus{} \sum_{sym}a^6 .......(2)$
adding up $ (1),(2)$ ,we have,
$ \sum_{sym}a^6 \plus{} 2\sum_{sym}a^5b \plus{} \sum_{sym}a^3b^2c\ge3\sum_{sym}a^4b^2 \plus{} \sum_{sym}a^3c^3$
$ QED$

[quote="arqady"]Vasc, I think we can obtain similar theorems for degree 7 and 8. :wink:[/quote]
Indeed, Arqady. If we write these polynomoials as
$ f_7(a,b,c)\equal{}Apx^2\plus{}g(p,q)x\plus{}h(p,q)$,
and
$ f_8(a,b,c)\equal{}(Ap^2\plus{}Bq)x^2\plus{}g(p,q)x\plus{}h(p,q)$,
then the both theorems above are true by replacing $ A$ with $ Ap$ for $ f_7$, and with $ Ap^2\plus{}Bq$ for $ f_8$.

[quote="nguoivn"]Thanks Vasc for nice theorems.
I think your theorems is very useful for the symetric ineqs which degree 6.  :)
Althought I think with the ineq which degree 7 and 8, that's so ugly and hard to expand (by hand) :oops:[/quote]

It is not necessary to expand. :wink:

[quote="Vasc"][quote="arqady"]Vasc, I think we can obtain similar theorems for degree 7 and 8. :wink:[/quote]
Indeed, Arqady. If we write these polynomials as
$ f_7(a,b,c) \equal{} Apx^2 \plus{} g(p,q)x \plus{} h(p,q)$,
and
$ f_8(a,b,c) \equal{} (Ap^2 \plus{} Bq)x^2 \plus{} g(p,q)x \plus{} h(p,q)$,
then the both theorems above are true by replacing $ A$ with $ Ap$ for $ f_7$, and with $ Ap^2 \plus{} Bq$ for $ f_8$.[/quote]
The main condition in Theorem 2 becomes
$ f(a,1,1)\ge \frac {4A(a \plus{} 2)(a \minus{} \delta)^2(a \minus{} 1)^4}{81(\delta \plus{} 2)^2}$ for all $ a\ge 1$,
and
$ f(a,1,1)\ge \frac {4[A(a \plus{} 2)^2 \plus{} B(2a \plus{} 1)](a \minus{} \delta)^2(a \minus{} 1)^4}{81(\delta \plus{} 2)^2}$ for all $ a\ge 1$,
respectively.


An airplane flies west at 150m/s at an altitude of 1000m. A searchlight oon the ground in the same vertical plane must be kept on the airplane. What is the rate of revolution of the searchlight when the airplane is 500m due east of the searchlight?

let z(t) be the angle of elevation, searchlight be at (0, 0), and the airplane coming from the positive x values towards (0, 1000). so v = dx/dt = -150 m/s.
let the distance of the airplane from (0, 1000) be x(t).
note that x = 1000 cot z, so dx/dt = -1000 csc^2 (z) (dy/dt).
Thus dz/dt = (dx/dt) / (-1000 csc^2 (z)) = -150 / (-1000*(5/4)) = 600/5000 = 3/25 radians/second.


Find all functions $ f$ on the reals such that:
$ [f(x)\plus{}f(y)][f(u)\plus{}f(z)]\equal{}[f(xu\minus{}yv)\plus{}f(xv\minus{}yu)$ for all $ x, y, u, v$

Please help ASAP.

a very good problem, see here:
http://www.mathlinks.ro/viewtopic.php?p=118703#118703

[i]Locked by mod. Please use existing topics instead of reposting problems.[/i]


No. What's the need of having one if there's already an AMC 8, 10, and 12?

Those are the only AMC levels that exist. If you're in middle school or below, your only easiest choice is AMC 8.

why, are you in 6th grade cgyao?

No there isnt. For one thing, the AMC people would definitely not get enough participants under 7th grade to make it worthwhile for them.

Um, most people under grade 6 are not serious about competing yet.

Or AMC 8/10/12 :roll:

I agree with Thunder365 and AwesomeToad - there would not be enough participants in an AMC 6 for the AMC committee to spend time on it.

[quote="$ LaTeX$"]I know a 6th grader that made AIME.....

If you're in 6th grade, you could try Math League.[/quote]

I heard about that too.

Though Math League doesn't matter to most people at all...

except the people at your school who are noobs heehee I've been there

Most 6th graders can't do that well. Come to think about it, most people can't do that.

[quote="$ LaTeX$"]I know a 6th grader that made AIME.....[/quote]

4th grader that made USAMO was on my ARML team.

AMC 8 is for the people in elementary and middle school. An AMC 6 would be pointless, because there are almost no problems that require creativity at that level.

Oh I know about that too. Did he make MOP?

But as I said, most people can't do that sort of thing.

AMC 6 wouldn't make much sense as those problems would be the same level as many other novice competitions out there.

[quote="Maybach"]According to the AMC website, people as young as [b]8[/b] take the AMC 8.

And if there WAS an AMC 6 it would probably be too easy.  It would probably be like all fractions and division.

And sixth graders can always do MC(assuming your school isn't a "junior high" style) and school rounds are kinda around that level.

In our school, only like 3 kids take the AMC 8 each year..........and the top score is usually about 18.  :rotfl:[/quote]

Usually kids under grade 7 tend to take Math League, as $ \text{\LaTeX}$ said.

I couldn't do MC to 7th grade :( so I had to settle for Math League in Grades 4-6

Of course, for the very young ones on AoPS, we could make a mock AMC6 under "Middle School Tournaments"

But of course there are other options:
Math League
Bergen County Mathematics Competition
Mathcounts(if you're lucky)
MOEMS

Yeah, we get the point.  We don't need 10 people saying "Nope, there is no AMC 6, but you can take Math League!"

Locked.


Find  "x" if:
[2Sin18*Sin18] / [2Cos42º - Sin36º]  =  Tan x

hello, we have $ \sin(36^{\circ})=\sqrt{\frac{5}{8}-\frac{\sqrt{5}}{8}}$ and ${ \cos(42^{\circ})=-\frac{1}{8}\sqrt{3}(1-\sqrt{5})+\frac{1}{4}\sqrt{\frac{1}{2}(5+\sqrt{5})}}$ so we have $ \tan(x)=\frac{1}{2}\left(\sqrt{5}+\sqrt{15+6\sqrt{5}}\right)$ and we found $ x=\arctan\left(\frac{1}{2}\left(\sqrt{5}+\sqrt{3(5+2\cdot\sqrt{5})\right)+2k\pi}$ with $ k\in \mathbb{Z}$.
Sonnhard.


algebra
function
domain
geometry theorems
geometry

[quote="The soul of rock"]Hello all Mathlinkers
As the big of our forum, many theorems and wonderful results has been prove and this work may be never end. So, I wonder that why we don't combine our results in one e-book? I think it is easy for us to learn, because not all member have condition to access internet everyday.
What do you think about my idea?[/quote]

Long ago, I proposed to Valentin to make an offline version of MathLinks accessible (i. e. a possibility to download the whole forum or at least its mathematical parts). He said this would lead to some problems, and I tend to believe these are of legal nature - if I understand it correctly (which I am anything but sure of), the posts are property of their respective authors, and it would require the permission of every of them to distribute the discussions as a whole.

Such problems can be actually circumvented if we get an agreement on the part of the content providers. Herewith I am [url=http://creativecommons.org/licenses/publicdomain/]placing all my posts on MathLinks in the public domain, excluding the contents of quote tags, attachments and posts beginning with "The author of this posting is:"[/url]. If we find more people to agree on this, this could make possible a pretty complete downloadable version of the contents of MathLinks discussions - we mainly need the people who really contributed some stuff of their own, not necessarily the ones who were just posting questions (which, upon being rewritten - what is often necessary in any case - become not copyrightable at all, but even without being rewritten should fall under fair use [b]in most cases[/b]).

What is the opinion of others on this? Should an extra topic be started?

  darij

It'll definitely be useful. I'm all for it - assuming we don't face any legal issues. It'll be a rather large task though. If you need any further hands on deck I'd like to help.


geometry
3D geometry
calculus
real analysis
real analysis unsolved

Prove that if $S\subseteq \mathbb{R}^{n}$ is uncountable then there exists $x\in S$ such that $x$ is a condensation point -that is to say, that every neighborhood of $x$ contains uncountably many elements of $S$. I tried to show the counterreciprocal using that $\mathbb{R}^{n}$ is separable, but couldn't work it out.. =(

[quote="Emis"]counterreciprocal[/quote]
The English is "contrapositive".

Trying the contrapositive: for each $x\in S$, let $d(x)=\inf_{y\in S,y\neq x}\|x-y\|$, the distance from $x$ to the rest of $S$. This distance is zero if $x$ is an accumulation point and positive if $x$ is an isolated point.
Suppose all points of $S$ are isolated. For each $x\in S$, let $B_{x}$ be the open ball centered at $x$ with radius $\frac{d(x)}{2}$. These balls are disjoint; if $z\in B_{x}$, $x$ is the closest member of $S$ to $z$. Each $B(x)$ contains some rational point $q(x)$ (this is separability); by disjointness, there are at most countably many balls and $S$ is countable.

Cover $\mathbb{R}^{n}$ by nonoverlapping cubes of side length 1. (Make the vertices lattice points.) There are countably many such cubes. If there were only countably many members of $S$ in each cube, then $S$ would be countable - hence there are uncountably many members of $S$ in at least one of the cubes.

Now, subdivide that cube into cubes of side length $\frac12,$ and repeat the argument - at least one of the smaller cubes contains uncountably many members of $S.$ Continue to repeat the argument. We have a nested sequence of closed cubes of diameters tending to zero, so there is a unique intersection point $x$ of all of those cubes. But any neighborhood of $x$ contains one of the cubes that appeared in our argument, hence contains uncountably many members of $S.$

In going for a generalization, instead of focusing on "separable," how about using "$\sigma$-compact"?

[quote="jmerry"]
The English is "contrapositive".
[/quote]
Oh.. Thanks for that =P (and of course for the rest of your reply)

I had tried something similar to what you wrote, but what happens if there exist some points which are accumulation points but not condensation points?

[quote="Kent Merryfield"]
In going for a generalization, instead of focusing on "separable," how about using "$\sigma$-compact"?
[/quote]
I'm afraid we haven't talked about that yet in my Calculus class.. Thanks for your reply :)

$\sigma$-compact merely means a countable union of compact sets - something I used quite explicitly in my proof.


This one was WAY to easy. The cutoff is going to be high...

[quote="amd2357"]Does anyone know when the AMC website will post the answers for AIME I or does anyone else know another source?[/quote]We will be solving all the problems in tonight's Math Jam at 7:30 PM ET / 4:30 PM PT.  If you can't make it then, you can always read the transcript.

I took it unofficially and checked my answers over with a couple of friends. Looks like I got a 6, with 4 dumb mistakes and a misinterpretation on the coins problem. I am definitely not gonna listen to music as I check my answers on the Alt. AIME.

I would like to know that too...
Plezz if any one knows it:D

I agree. Very easy AIME. Wouldnt be surpized if you need a 9 to get in even in 10th or below. *waits for answers to come up so he can see how many he got right*... should be somewhere between 8-10.Unless of course, i bubbled some wrong  :?

Well, the answers my friends and I pretty much agreed on were:

1. 942
2. 12
3. 109
4. 294
5. 630 (if you could tell whether a coin in the stack was face up or down)
6. 45
7. 150
8. 113
9. 74
10. 47
11. 544
12. 25

[quote="NightFlarer"]yif man, on the coin problem did you also think that having two adjacent coins with the bottom coin showing blank on top and the top coin showing blank on bottom was illegal?  I did!  Luckily I took the AIME I for practice so I'm not upset or anything...[/quote]

I just thought that you could only tell if the top coin was face up or face down, so I just put 140. Yeah, I also just took it for practice. I wonder how they'll interpret that problem. Probably that you can tell if something's face up or down in the middle, but I thought that was a little too confusing.

[quote="1234567890"]3. I believe it was 214, because we have 15 primes under 50, so 15 *14 + 4, because we have 2 3 , 3^ 3 , 5 3 ,7 3 all have factors less than 50
5. I think it was 140, because we have 2(8!/(4!4!))[/quote]

Number three. 2*17=17*2  :)
Number five: Call it U if the coin is face up and D if it is face down
you can have D on D, U on U and U on D. You just cant have D on U.

Thus you can have 8 Ds, 7D's then a U, 6D's then 2 U's... 8 U's for a total of 9, not 2, ways to arrang ethe coins regardless of color.

I agree with yif man on all of those but 6,8,10,11 (I didnt sovle those ones). I then got 182 on 13-but am not 100% sure on that one. I was solving 10 when time ran out... had I had 5 more minutes I might have gotten that one as well.

edit: yeah, I found the coin problem slightly confuisng as well. But I figgure you can tell the differnces becaue if you couldnt the "no coins face to face" wouldnt matter  :) . As it gave you that information, I assumed that the coins were arranged such that you could see what face was up on each coin.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=182820#p182820]Here are the answers[/url] that many people have confirmed.

the official answers are up on the AMC website
http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2005-aa/05answersaime1.html

I also misinterpreted the coins problem! ARGH! I forgot that you could have the back to back non-heads.. man I'm mad that was probably the easist problem. And I missed the band one b/c I factored out 294 wrong... man I was dumb. But I have four more chances at the AIMES. Anyone whose taking the alternates, you ran out of luck, bc it's probably not going to be this easy, and that's coming from an 8th grader

Darn...so close on number 14.  I found the square somehow (Thank you graph paper), and that the sides were longer than 13.9, so I assumed they were 14: 960. 13.91...:(

I hate stupid mistakes:
#3: Not fair! 1*2*2 2 is not three divisors, (same for 3, 5, 7)
#4: Don't know how I missed this one. Somehow my answer passed the check initially.
#5: For some reason I only found 4 different combos of up/down, instead of 9 (Forgetting that you can have, say two down, or three down, etc.)
#7:I used 12 as the leg of a triangle when it was the hypotenuse, so I had $7+\sqrt{144\bold{+}3}$ instead of $9+\sqrt{144\bold{-}3}$=>154 instead of 150
#10: Didn't realize that the median could be above the line or below...
#14: So close, as I said before.

I answered all 15, got 7 of them, but whittled me down to a 2...:(

Most of those random guesses were pretty close... (I actually was going to guess 074 for #9, but then changed it :()

My stupid mistakes:

#3:   I forgot 4^3 and 6^3 had 6 and 15 proper divisors respectively.  :mad:
#4:   Found only all heads up and all heads down. :oops:
#9:   I don't know how I got that wrong, my answer was 77. :?
#13: I thought there were C(10,5) lines, but I ended up overcounting all distinct possible paths.


Let $ a$, $ b$, $ c$ be nonnegative reals such that $ a+b+c=1$.

Prove that

$ \sum_{\text{cyc}}\sqrt{a^{2}+abc}\geq\frac23\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)$.

$3 \sum \sqrt{a}\sqrt{1-b}\sqrt{1-c}   \ge 2 \sum \sqrt{a} $

Is there a nice trigo substituition? I've tried $a = sin x, b = sin y, c = sin z$, and the LHS becomes $3 (sin (x + y + z) + sin x sin y sin z)$ which lends itself to Jensen's/AM-GM, but unfortunately, we have a weak term on the LHS? Or what?

Equivalent to

\[3 \sum a\sqrt{a^2 + b^2}\sqrt{a^2 + c^2} \geq 2\sum a \sum a^2\]

for any three positive reals $a, b, c.$ How now?

But

\[\sqrt{1 + 9abc} \geq \frac{2}{3}\sum \sqrt{a}\] is not true.

$\sqrt{a^2+abc}+\sqrt{b^2+bca}+\sqrt{c^2+cab}\geqq \frac{2}{3}(\sqrt{a}+\sqrt{b}+\sqrt{c})$

$\Longleftrightarrow \sqrt{\frac{a}{bc}+1}+\sqrt{\frac{b}{ca}+1}+\sqrt{\frac{c}{ab}+1}\geqq \frac{2}{3} \left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)$

Proof:$\sqrt{\frac{a}{bc}+1}+\sqrt{\frac{b}{ca}+1}+\sqrt{\frac{c}{ab}+1}$

$\geqq \sqrt{\left(\sqrt{\frac{a}{bc}}+\sqrt{\frac{b}{ca}}+\sqrt{\frac{c}{ab}\right)^2+(1+1+1)^2}=\sqrt{\frac{c}{ab}+\frac{a}{bc}+\frac{b}{ca}+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+9}}$

$=\sqrt{\frac{(a+b+c)^2}{abc}+9}=\sqrt{\frac{1}{abc}+9}=\sqrt{\frac{a+b+c}{abc}+9}=\sqrt{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+9}$

${\geqq \sqrt{\frac{1}{a\sqrt{bc}}+\frac{1}{b\sqrt{ca}}+\frac{1}{c\sqrt{ab}}}+9}\ \cdots\ (E)$

Since ${\frac{1}{a\sqrt{bc}}=\frac{a+b+c}{a\sqrt{bc}}\geqq \frac{1}{\sqrt{bc}}+\frac{2}{a}}$ , similarly

${{\frac{1}{b\sqrt{ca}}\geqq \frac{1}{\sqrt{ca}}+\frac{2}{b}},\ \frac{1}{c\sqrt{ab}}\geqq \frac{1}{\sqrt{ab}}+\frac{2}{c}}$

$(E)\geqq  \sqrt{\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+9}\geqq \sqrt{3\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)+9}$

By the way, for $0<x\leqq 9$, we have $\sqrt{3x+9}\geqq \frac{2}{3}x$.

Therefore $(E)\geqq \sqrt{3\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)+9}\geqq
\frac{2}{3}\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)$ Q.E.D.

kunny

[b]kunny[/b] dear ,
I don't understand your solution !

[quote="kunny"]By the way, for $0<x\leqq 9$, we have $\sqrt{3x+9}\geqq \frac{2}{3}x$.

Therefore $(E)\geqq \sqrt{3\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)+9}\geqq
\frac{2}{3}\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)$[/quote]

but $\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\geq 9$...

  darij

I think that :
We have :if $x_1\geq x_2\geq x_3;y_1\leq y_2\leq y_3 $ then
$(x_1+x_2+x_3)(y_1+y_2+y_3)\geq 3(x_1y_1+x_2y_2+x_3y_3) (*)$
$\sum\sqrt{\frac{a}{bc}+1}=\sum\sqrt{\frac{a}{bc}+\frac{1}{a+b+c}}=\sum\sqrt{\frac{(a+b)(a+c)}{bc}}$

Let $a\geq b \geq c$ ,then:
$\frac{(a+b)(a+c)}{bc}\geq\frac{(b+c)(b+a)}{ac}\geq\frac{(c+a)(c+b)}{ab}$
And $\frac{1}{\sqrt{(a+b)(a+c)}} \leq \frac{1}{\sqrt{(b+c)(b+a)}}\leq\frac{1}{\sqrt{(c+a)(c+b)}}$.

Use (*) we have :
$(\sum\sqrt{\frac{a}{bc}+1}).(\sum\frac{1}{\sqrt{(a+b)(a+c)}})\geq 3.(\sum\frac{1}{\sqrt{ab}}).$
Last inequality :Prove that :
$\sum\frac{1}{\sqrt{(a+b)(a+c)}} \leq \frac{9}{2}$.
Can you help me to finish my solution ?

Your last inequality is, obviously, wrong. Just approach $ a,b\to 0$ and $ c\to 1$, then $ LHS\to\infty$.

I spent 15 minutes for it yesterday. I didn't prove it. BUT... If I correctly remember, it was solved on the forum before. Unfortunately, I couldn't find the link.

[quote="kunny"][quote="Reyes"]I think that :
We have :if $x_1\geq x_2\geq x_3;y_1\leq y_2\leq y_3 $ then
$(x_1+x_2+x_3)(y_1+y_2+y_3)\geq 3(x_1y_1+x_2y_2+x_3y_3) (*)$

You seemed to use Tchevichev's inequality,I think, you have a wrong applying.[/quote][/quote]
No, it is correct!

I understand what you said,Myth.

Sorry for,[b]Reyes[/b].

Please help me to prove that
\[ \sqrt{9abc+1}\geq \frac{2}{3}(\sqrt{a}+\sqrt{b}+\sqrt{c}) \]
is not true.

We have reduced the inequality to proving
$3\sum_{cyc} x\sqrt{(x^2+y^2)(x^2+z^2)}\geq 2(x+y+z)(x^2+y^2+z^2)$
Lemma:
$2(x^3+y^3+z^3)-(\sum x^2y)=\sum (x+y)(x-y)^2$
proof: immediately from $x^3+y^3-xy(x+y)=(x+y)(x-y)^2$
Now we see that $3 \sum_{cyc} x \frac {(x^2+y^2)+(x^2+z^2)}{2} -RHS=\frac 12 \sum (x+y)(x-y)^2$ using Lemma.
However $3 \sum_{cyc} x \frac {(x^2+y^2)+(x^2+z^2)}{2}-LHS= \frac 12 3 \sum (\sqrt{x^2+y^2}-\sqrt {x^2+z^2})^2$
So we are left to prove that
$\sum (x+y)(x-y)^2 \geq 3 \sum  x(\sqrt{x^2+y^2}-\sqrt {x^2+z^2})^2$.
We prove that $(x+y)(x-y)^2 \geq 3 z (\sqrt{z^2+y^2}-\sqrt {z^2+x^2})^2=3 \frac {(x-y)^2(x+y)^2}{(\sqrt {z^2+y^2}+\sqrt {z^2+x^2})^2}$.
Thus we have to prove $(\sqrt {z^2+y^2}+\sqrt {z^2+x^2})^2 \geq 3z(x+y)$ or
$2z^2+x^2+y^2+2 \sqrt{z^2+x^2} \sqrt {z^2+y^2} \geq 3z(x+y)$.
However $\sqrt{z^2+x^2} \sqrt {z^2+y^2} \geq z^2+xy$ by CBS.
Thus we reduce it to $4z^2+x^2+y^2+2xy \geq 3z(x+y)$ or $4z^2+(x+y)^2 \geq 3z(x+y)$.
However $4z^2+(x+y)^2 \geq 4z(x+y)$ by AM-GM and so we are done.

It's a nice solution! Congratulations ! Strange ,in your solution :
$...4z^2+(x+y)^2\geq 3z(x+y)$ .When do we have equality ?.

Is this inequality true for $(1/2;1/3;1/6)$ :?

If you mean $\sqrt{9abc+1}\geq \frac{2}{3}(\sqrt a+\sqrt b+\sqrt c)$, then NO.

I mean to the $\sum_{sym}\sqrt{a^2+abc}\ge 2/3(\sqrt{a}+\sqrt{b}+\sqrt{c})$

I checked your values and the inequality holds.

It is strange, the inequality you have discussed here I found in an old number of AMM, proposed by Walther Janous as a geometrical inequality. A really tough problem. I remember spending one hour on it with no results. Now, finally, I see a solution. Thanks, iura.

[quote="Chang Woo-JIn"] Let $a, b, c>0$ and $a+b+c=1$. Prove that
$\sum\sqrt{a^{2}+abc}\geq\frac{2}{3}\sum\sqrt{a}$.[/quote]
Put $a=x^{2},b=y^{2},c=z^{2}$.
[quote="iura"]We have reduced the inequality to proving
$3\sum x\sqrt{(x^{2}+y^{2})(x^{2}+z^{2})}\geq 2\sum x\sum x^{2}$[/quote]
Squaring, and using the inequality $\sqrt{(x^{2}+y^{2})(x^{2}+z^{2})}\geq x^{2}+yz$ etc, this in turn is implied by

$9\sum x^{2}(x^{2}+y^{2})(x^{2}+z^{2})+18\sum yz(y^{2}+z^{2})(x^{2}+yz)-4\left(\sum x\right)^{2}\left(\sum x^{2}\right)^{2}$

$=\frac{1}{6}\sum(y-z)^{4}(x^{2}+15y^{2}+15z^{2}+4yz+8zx+8xy)\geq0$.

It's a problem was proposed by Walther Janous, see [url=http://journals.cms.math.ca/CRUX/]CRUX Mathematicorum[/url], Vol. 17 (1991), p. 207, Problem $1663^{*}$.

The above solution by Ji Chen, see Vol. 18 (1992) , pp. 188 - 189.

This problem also solved by Marcin E. Kuczma.
[quote="harazi"]The inequality you have discussed here I found in an old number of AMM, proposed by Walther Janous as a geometrical inequality. A really tough problem.[/quote]
Walther Janous and Magda P. Nakasis, Problem E $3146^{*}$, Amer. Math. Monthly, Vol. 93 (1986), p. 299 and Vol. 95 (1988), pp. 767 - 769.


[hide]If we add them up, we find that he has a total of 108 gallons of punch.  We know that there must be an amount of blue punch divisible by three, because otherwise the red punch would not be divisible by three.  Testing 15 (for blue), we find that there is no way to get $\frac{108-15}{3}=31$ without opening the remaining bottles.  Thus he keeps 21 bottles of blue punch for himself.[/hide]

[color=blue]Moderator kyyuanmathcount's note: Please use hide tags when posting answers or solutions, so others can do the problems without having to look at answers, thanks![/color]

[quote="ch1n353ch3s54a1l"]A vendor has six jugs of punch with capacities of 13, 15, 16, 21, 21, and 22 gallons. One jug, which he keeps for himself, is full of blue punch. The other five jugs are full of red punch. Without opening any of the jugs, he sells 1/3 of the red punch to Randy and the remaining 2/3 of the red punch to Lisa. How many gallons of the blue punch did he keep for himself?[/quote]

[hide]21 gallons

Add together $15$, $16$, $21$, $21$, and $22$ to get $95$. $95$ isn't divisible by $3$.
Add together $13$, $16$, $21$, $21$, and $22$ to get $93$. It is divisible by $3$ but there's no way to get $\frac{1}{3}$ or $\frac{2}{3}$ of $93$ with the given jugs.
The only way it would work is if you had $13$, $15$, $16$, $21$, and $22$. Therefore, the answer's $21$[/hide]


At the national curling championships, there are three teams of four players each. After the championships are over, the very courteous participants each shake hands three times with every member of the opposing teams, and once with each member of their own team.

How many handshakes are there in total?

So we have 3 teams with 4 people in them. Let's say one of the teams has members $ A$, $ B$, and $ C$.

$ A$ would shake hands with two other people -- $ B$ and $ C$.

$ A$ would also shake hands with $ 8$ other people -- the people in other teams.

That means $ A$ shakes hands with $ 10$ people in total. We have 12 people, and $ 12\times10 \equal{} 120$

However, we are counting each handshake twice($ A$ with $ B$, and $ B$ with $ A$ are being counted as separate handshakes.) So we divide by two to get $ 60$.

$ \boxed{60}$

(I'm not sure if I did this right.)

ok i think you did everything right but forgot to account for the number of handshakes. So we have 12 people and first of all they are going to shake hands with everyone not on their team 3 times so

12*8*3 and we will divide by 2 to avoid duplicates, thus yielding our answer as 144.

Now each person shakes hands with each person on his/her team once so 12*3 (and divide by 2) or 18. Summing together gives 162.


[quote="Rzeszut"][quote="parinaz.sa"]let $a,b,c,d,r$ be natural numbers and we have $ab=cd$ ,$T=a^{r}+b^{r}+c^{r}+d^{r}$ then prove that $2^{T}-1$ is composite number.[/quote]
I hope you mean $a,b,c,d\in\mathbb{N}^{+}$, because if $a=b=c=d=0$, then $2^{T}-1=0$, and for $b=d=0$ we can put $a=c=r=1$ and it gives $T=2$ and $2^{T}-1=3$.[/quote]
yeah i mean  $a,b,c,d\in\mathbb{N}^{+}$

[quote="parinaz.sa"]let $a,b,c,d,r$ be natural numbers and we have $ab=cd$ ,$T=a^{r}+b^{r}+c^{r}+d^{r}$ then prove that $2^{T}-1$ is composite number.[/quote]
Borrowing the idea from the posts found here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33417
we proceed as follows.

$a=\frac{cd}{b}.$ So, $c=hm, d=kn, b=hk.$

$T ={(mn)}^{r}+{(hk)}^{r}+{(hm)}^{r}+{(kn)}^{r}$
$\Rightarrow T = (m^{r}+k^{r})(n^{r}+h^{r})$

Note that both the factors on the $RHS$ above are greater than 2.

Now, we know that $(a^{q}-1)=(a-1)(a^{q-1}+a^{q-2}+...+1)$ for any two naturals $a$ and $q.$

Just put $a=2^{m^{r}+k^{r}}$ and $q=n^{r}+h^{r}$ and the result follows.


Three hundred students took a test and all three hundred test
scores are shown in the table below. What is the least possible
integral score a student could earn and still have a higher score
than 75 percent of the students taking the test?
$ \begin{tabular}[t]{c|c|c|c|c|c|c|c|c|c|c|c|c|c}
Score &0 &1 &2 &3 &4 &5 &6 &7 &8 &9 &10 &11 &12\\\hline
\#Students &3 &6 &12 &13 &16 &23 &28 &43 &51 &45 &37 &16 &7\end{tabular}$

To get a 75%, you must get at least a 9.  But since you want higher than a 75%, the answer is 10.

Let's go a bit further than simply stating the solution :)

If there are 300 people taking the test, 75 students constitute 25% of the class, and 225 students are 75% of the class.

Adding from the right, 10, 11 and 12 you get 60.  So far, you have 240 students beneath you.
Adding 9, 10, 11 and 12 you get 105:  There are now 195 students beneath you: Not enough!

So 10 is as low as you can go: it puts 240 students beneath you.


There are eight points in the plane no three of them collinear. Find the maximum number of triangle formed out of these points such that no two triangles have more than one vertex in common.

bye

[hide]
8 triangles are possible, and no more.

Label the vertices 1,2,3,4,5,6,7,8.

Then we can have eight triangles:

1,2,3
1,4,5
1,6,7
8,2,4
8,5,6
8,3,7
2,5,7
3,4,6

Since there are 7 potential edges off a vertex, and each triangle uses two of these, there is no way we can have more than 3 triangles on a vertex.

Now suppose we can have 9 triangles. Then these gives us 9 unordered triples, or 27 "uses" of vertices. But each vertex can only be used 3 times, meaning we can have at most 3*8 = 24 "uses", a contradiction.

So 8 is the best we can do.

[/hide]


[quote="ckck"]In triangle ABC there is a point P on BC such that AP bisects angle BAC. If AB=9, AC=6, and BP=6, find AP.[/quote]
[hide]With the ABT we can find that PC=4. With stewart's, we can find that $324+216=10x^{2}+240$
$x=\sqrt{30}$[/hide]


At the end of the recent Linguistics Math Jam, I promised that I would provide a list of resources for people interested in studying linguistics.  This is that list.

I thought that I would begin by providing information about the two linguistics competitions that people would be likely to hear about, namely the North American Computational Linguistics Olympiad (NACLO) and International Linguistics Olympiad (ILO).

North American Computational Linguistics Olympiad (NACLO)  [url]http://www.naclo.cs.cmu.edu/[/url]
Past problems:  [url]http://www.naclo.cs.cmu.edu/practice.html[/url]

International Linguistics Olympiads
The Wikipedia pages has compiled the links to sites of the past olympiads, many of which have problems posted.  [url]http://en.wikipedia.org/wiki/International_Linguistic_Olympiad[/url]

Linguistics Challenge (an archive of problems from the NACLO's predecessor):  [url]http://webscript.princeton.edu/~ahesterb/puzzles.php[/url]

The Linguistics Math Jam was great!  I've never studied linguistics before and I'd like to know if anyone has any recommendations for instructional materials (books, or anything else) for real [i]beginners[/i].

What are the recommended or established steps in learning this material?   Beginner's Step One...Step Two...Step Three...   :D

Thank you.

Thanks!   :D

I'm not sure there really [i]is[/i] an established way for learning linguistics, but this is what I'd say...

First, there is a deep split in linguistics between studying the sound systems of languages (phonetics/phonology) and the way words are put together to form sentences (syntax).  You could study either one more or less independently of the other.

If you're interested in phonetics/phonology, the first thing to learn is what the different speech sounds are, and how they are made in the vocal tract.  For example, here is a chart of the symbols in the International Phonetic Alphabet, which has symbols to represent all speech sounds in all languages-- [url]http://www.phonetics.ucla.edu/course/chapter1/chapter1.html[/url]  You would want to understand how you would pronounce all the symbols, and what the words at the side of the chart refer to.  For instance, the box for [p] is in the column labeled "bilabial," and the row labeled "plosive," which means that  that sound is made by stopping airflow out of your nose and mouth (plosive) by closing off the flow of air through your mouth by closing your lips against each other (bilabial).  This is called "articulatory phonetics."

Most phonology textbooks have a chapter on it at the beginning of the book, but a much more fun and interesting way to learn it would be to get an introductory phonetics book, like [u]Vowels and Consonants[/u] by Peter Ladefoged.  It will also tell you how these sounds appear in sound spectrograms.  The software most linguists use for recording and analyzing sound spectorgrams is free, if you want to play around with it and see what your speech looks like.  It's called "Praat," and you can download it here [url]http://fonsg3.hum.uva.nl/praat/[/url].

Beyond that, you would want to learn phonological feature theory, which is the notation system used for writing out rules for how languages handle sound patterns.  For example, the plural suffix in English is /s/, but it can be pronounced as an [s] after certain sounds, like the [t] in "cats," but [z] after other sounds, like the [g] in "dogs."  One way of writing this rule would be [s] --> [+voice]  /[+voice] ___, which reads "[s] becomes voiced after a voiced segment.  (The real rule is more complicated than this example.)  The introductory phonology textbook I have is David Odden's [i]Introducing Phonology[/i], but there are probably others out there.  I haven't read it cover to cover, but it looks like a pretty good book, and it has lots of problems of manageable size for a beginner.

The other thing is what you look for in a set of data when you go to describe the sound system of a language, but working a lot of phonology problems, like the ones in Odden's book, should help with that.

If you're interested in syntax, the book that I learned from was Andrew Carnie's [i]Syntax:  A Generative Introduction[/i].  The "big ideas" that help you solve problems are constituency, movement operations, and the basic setup of X' theory.

On the other hand, I got started learning linguistics by learning a lot of dead languages that were related to each other, and picking up on the similarities and differences between them.  If you like ancient Indo-European languages, this is a good site to start with:  [url]http://www.utexas.edu/cola/centers/lrc/eieol/index.html[/url]  There are introductory lessons and glossed and parsed examples of real texts.

I think the most important thing is to find something you're interested in playing around with, and then play around with it.

Best of luck!

This is [i]tremendously[/i] helpful to me.  I will be checking out every one of your leads and I greatly appreciate your help.

Thank you so much!

Mischa


inequalities
real analysis
real analysis solved

Prove that for every \alpha \in [1,oo)U{oo} thee exists a sequence (k_n)_n of natural numbers s.t.
lim{n->oo}(1+1/2+1/3+...+1/k_n)^(1/n) -> \alpha

We use the fact that lnn<1+1/2+...+1/n<lnn+1 (or something like this, I do not remember exactly). Let  \alpha real. Take the sequence k_n=[e^x_n] where x_n= \alpha ^n. Using the above inequality we obtain that the limit is  \alpha . For  \alpha = \infty  we can take k_n=[e^n^n].


I added an entry for Ohio in the [url=http://www.artofproblemsolving.com/Wiki/index.php/How_to_join_an_ARML_team]How to join an ARML team[/url] wiki.  If anyone knows more than I do (which is likely) about the Ohio team selection process, please edit the section as you deem appropriate.

Has anyone gotten an invitation before soley based on their AMC scores?  Because I only got that in 8th grade, but not last year.  And then I probably wouldn't have gotten an invitation based on that this year, since I was on ARML last year.


function
inequalities
calculus
calculus computations

I am guessing the answer is C because of the Mean Value Theorem, but I don't feel 100% confident.  What is holding me up from being confident about it is that the MVT states there exists a $ c_1$ in $ (0,x)$ such that $ \frac{f(x)\minus{}f(0)}{x}\equal{}f'(c_1)$.  However, the $ c_1$ that is used for $ f$ and the $ c_2$ that is used for $ g$ might be very different, and hence, $ g'(c_2)$ might be bigger.  If someone could give some explanation, that would be awesome :].  Thanks!

19) Let $ f$ and $ g$ be twice differentiable real-valued functions defined on $ R$.  If $ f'(x) > g'(x)$ for all $ x > 0$, which of the following inequalities MUST be true for all $ x > 0$?

A) $ f(x) > g(x)$.
B) $ f''(x) > g''(x)$.
C) $ f(x)\minus{}f(0) > g(x)\minus{}g(0)$.
D) $ f'(x)\minus{}f'(0) > g'(x)\minus{}g'(0)$.
E) $ f''(x)\minus{}f''(0) > g''(x)\minus{}g''(0)$.

Consider the function $ h(x) = f(x) - g(x)$.  Differentiating we find that $ h'(x) > 0$ for all $ x > 0$.  What does this imply about $ h(x) - h(0) = f(x) - f(0) - (g(x) - g(0))$ for $ x > 0$?

That makes it very obvious.  Thanks so much!

Alternatively, it's easy to eliminate the other choices by choosing $ f$ and $ g$ to be equations of lines.


Let $n \in \mathbb{N}$ ........ :)
Then,

find all $n$ such that:

                     $\left [ \frac{n^2}{5} \right] = p,$

where $p$ is a prime number........

Carlos Bravo :)
Lima -PERU

The given equality may be rewritten as $5p \leq n^2 < 5p+5.$
Note that a square is $0,1,4$ modulo $5$.
Thus :
- If $n^2 = 0 \mod [5]$ then $n^2 = 5p^$ which forces $p=5,n=5.$
- If $n^2 = 1 \mod [5]$ then $n^2 = 5p+1$, which is $5p = (n-1)(n+1)$ where $n+1 \geq n-1 \geq 2.$
Thus ($n-1 = 5$ and $n+1 = p$) or ($n-1=p$ ad $n+1 = 5$). This leads to the solutions $n=6,p=7$ and $n=4,p=3.$
- If $n^2 = 4 \mod [4]$ then $n^2 = 5p+4$ which is $5p = (n-2)(n+2)$,  where $n+2 \geq n-2 \geq 2.$
Thus ($n-2 = 5$ and $n+2 = p$) or ($n-2=p$ ad $n+2 = 5$). Since $p$ is prime, it gives no more solutions.

Pierre.


search
geometry
incenter
circumcircle
inradius
function
linear algebra

IMO 2003 ShortList:

[url=http://140.247.141.165/~ajorza/math/mathfiles/imoshortlist/doehring/imo2003s_sol.pdf]IMO ShortList 2003 with solutions[/url] [b][size: 5.23 MB][/b]

[b]The links to the single problems on the forum can be found below. These links also indicate where you should discuss the solutions to the problems.[/b]

Orlando Döhring

Thanks for posting the SL, I really appreciate it.

Sorry if I seemed impatient, it's just that I have been waiting a long time to see it.

There is a problem proposed by Reid Barton, which is it?

Well, please register or at least say your name then I will provide the answer. Otherwise you will have to search the forum...  :D

[quote="darij grinberg"]Thanks Orl!

  darij[/quote]

You are welcome, jhaussmann and darij.  :)

I am not the guest, but I am curious as to the problem that was proposed by Reid Barton.

Oh well, it was already posted here on MathLinks...

  http://www.mathlinks.ro/Forum/viewtopic.php?t=6062

  Darij

Reid Barton proposed that last Algebra problem, the inequality. Look it up in my ISL03 file above.  :)

Thank you very, very, very, very  much, orl. :P

[quote="orl"]IMO 2003 ShortList:

[url=http://www.artofproblemsolving.com/Forum/orl/IMO_2003_shortlist.pdf]IMO ShortList 2003[/url][/quote]
Thank you very much,orl! Would you have the solution?Thanks!

Hereby I want to announce that the full IMO ShortList 2003 is available now. I know it is not a very sensational message for my 1000th postings.  :)  But I think you appreciated it having me on your site.  :D

PS: But somehow it was obvious that THIS special posting refers to ISLs/ILLs.  :)

[b][size=150]Problems on the board: [/size][/b]

[u][size=150]By category:[/size][/u]

 [b]GEOMETRY[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=264#264]IMO ShortList 2003, geometry problem 1[/url]

Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$, $AB$, respectively. Show that $PQ=QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$.

[url=http://mathlinks.ro/Forum/viewtopic.php?p=19089#19089]IMO ShortList 2003, geometry problem 2[/url]

Three distinct points $A$, $B$, $C$ are fixed on a line in this order. Let $\Gamma$ be a circle passing through $A$ and $C$  whose centre does not lie on the line $AC$. Denote by $P$ the intersection of the tangents to $\Gamma$ at $A$ and $C$. Suppose $\Gamma$ meets the segment $PB$ at $Q$. Prove that the intersection of the bisector of $\angle AQC$ and the line $AC$ does not depend on the choice of $\Gamma$.

[url=http://mathlinks.ro/Forum/viewtopic.php?p=18550#18550]IMO ShortList 2003, geometry problem 3[/url]

Let ABC be a triangle, and P a point in the interior of the triangle. Let D, E, F be the feet of the perpendiculars from P to the sides BC, CA, AB. Assume that

  $AP^{2}+PD^{2}=BP^{2}+PE^{2}=CP^{2}+PF^{2}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=119988#119988]IMO ShortList 2003, geometry problem 4[/url]

Let $\Gamma_1$, $\Gamma_2$, $\Gamma_3$, $\Gamma_4$ be distinct circles such that $\Gamma_1$, $\Gamma_3$ are externally tangent at $P$, and $\Gamma_2$, $\Gamma_4$ are externally tangent at the same point $P$. Suppose that $\Gamma_1$ and $\Gamma_2$; $\Gamma_2$ and $\Gamma_3$; $\Gamma_3$ and $\Gamma_4$; $\Gamma_4$ and $\Gamma_1$ meet at $A$, $B$, $C$, $D$, respectively, and that all these points are different from $P$. Prove that

 \[
  \frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}.
 \]

[url=http://mathlinks.ro/Forum/viewtopic.php?t=5642]IMO ShortList 2003, geometry problem 5[/url]

Let $ABC$ be an isosceles triangle with $AC=BC$, whose incentre is $I$. Let $P$ be a point on the circumcircle of the triangle $AIB$ lying inside the triangle $ABC$. The lines through $P$ parallel to $CA$ and $CB$ meet $AB$ at $D$ and $E$, respectively. The line through $P$ parallel to $AB$ meets $CA$ and $CB$ at $F$ and $G$, respectively. Prove that the lines $DF$ and $EG$ intersect on the circumcircle of the triangle $ABC$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=263#263]IMO ShortList 2003, geometry problem 6[/url]

Each pair of opposite sides of a convex hexagon has the following property: the distance between their midpoints is equal to  $\sqrt{3}/2$ times the sum of their lengths. Prove that all the angles of the hexagon are equal.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=119989#119989]IMO ShortList 2003, geometry problem 7[/url]

Let $ABC$ be a triangle with semiperimeter $s$ and inradius $r$. The semicircles with diameters $BC$, $CA$, $AB$ are drawn on the outside of the triangle $ABC$. The circle tangent to all three semicircles has radius $t$. Prove that

 \[
  \frac{s}{2}<t\le\frac{s}{2}+\biggl(1-\frac{\sqrt{3}}{2}\biggr)r.
 \]

 [b]NUMBER THEORY[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=119990#119990]IMO ShortList 2003, number theory problem 1[/url]

Let $m$ be a fixed integer greater than $1$. The sequence $x_0$, $x_1$, $x_2$, $\ldots$ is defined as follows:

$x_i=  2^i$ if $0 \leq i\leq m-1$ and $x_i = \sum_{j=1}^{m}x_{i-j},$ if $i\geq m$.

[url=http://mathlinks.ro/Forum/viewtopic.php?p=19088#19088]IMO ShortList 2003, number theory problem 2[/url]

Each positive integer a is subjected to the following procedure, yielding the number d = d(a):

[b](a)[/b] The last digit of a is moved to the first position. The resulting number is called b.
[b](b)[/b] The number b is squared. The resulting number is called c.
[b](c)[/b] The first digit of c is moved to the last position. The resulting number is called d.

(All numbers are considered in the decimal system.) For instance, a = 2003 gives b = 3200, c = 10240000 and d = 02400001 = 2400001 = d(2003).

Find all integers a such that $d\left( a\right) =a^{2}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=262#262]IMO ShortList 2003, number theory problem 3[/url]

Determine all pairs of positive integers $(a,b)$ such that

 \[
  \frac{a^2}{2ab^2-b^3+1}
 \]

 is a positive integer.

[url=http://mathlinks.ro/Forum/viewtopic.php?p=19129#19129]IMO ShortList 2003, number theory problem 4[/url]

Let $b$ be an integer greater than $5$. For each positive integer $n$, consider the number

 \[
  x_n=\underbrace{11\cdots1}_{n-1}\underbrace{22\cdots2}_{n}5,
 \]
 written in base $b$.

Prove that the following condition holds if and only if $b=10$:  [i]there exists a positive integer $M$ such that for any integer $n$ greater than $M$,  the number $x_n$ is a perfect square.[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=18558#18558]IMO ShortList 2003, number theory problem 5[/url]

An integer $n$ is said to be [i]good[/i] if $|n|$ is not the square of an integer. Determine all integers $m$ with the following property: $m$ can be represented, in infinitely many ways, as a sum of three distinct good integers whose product is the square of an odd integer.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=266#266]IMO ShortList 2003, number theory problem 6[/url]

Let $p$ be a prime number. Prove that there exists a prime number $q$ such that for every integer $n$, the number $n^p-p$ is not divisible by $q$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15464]IMO ShortList 2003, number theory problem 7[/url]

The sequence $a_0$, $a_1$, $a_2,$ $\ldots $ is defined as follows: $ a_0=2, \quad a_{k+1}=2a_k^2-1 \quad $ for $k \geq 0$. Prove that if an odd prime $p$ divides $a_n$, then $2^{n+3}$ divides $p^2-1$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=119994#119994]IMO ShortList 2003, number theory problem 8[/url]

Let $p$ be a prime number and let $A$ be a set of positive integers that satisfies the following conditions: [b](1)[/b] the set of prime divisors of the elements in $A$ consists of $p-1$ elements; [b](2)[/b] for any nonempty subset of $A$, the product of its elements is not a perfect $p$-th power. What is the largest possible number of elements in $A$ ?

[b]ALGEBRA[/b]

[url=http://mathlinks.ro/Forum/viewtopic.php?p=19085#19085]IMO ShortList 2003, algebra problem 1[/url]

Let $a_{ij}$ (i, j = 1, 2, 3) be real numbers such that

  $a_{ij}>0$ for i = j;
  $a_{ij}<0$ for $i\neq j$.

Prove the existence of positive real numbers $c_{1}$, $c_{2}$, $c_{3}$ such that the numbers

  $a_{11}c_{1}+a_{12}c_{2}+a_{13}c_{3},$
  $a_{21}c_{1}+a_{22}c_{2}+a_{23}c_{3},$
  $a_{31}c_{1}+a_{32}c_{2}+a_{33}c_{3}$

are either all negative, or all zero, or all positive.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=17477]IMO ShortList 2003, algebra problem 2[/url]

Find all nondecreasing functions $f: \mathbb{R}\rightarrow\mathbb{R}$ such that
(i) $f(0) = 0, f(1) = 1;$
(ii) $f(a) + f(b) = f(a)f(b) + f(a + b - ab)$ for all real numbers $a, b$ such that $a < 1 < b$.

[url=http://mathlinks.ro/Forum/viewtopic.php?p=100351#100351]IMO ShortList 2003, algebra problem 3[/url]

Consider two monotonically decreasing sequences $\left( a_k\right)$ and $\left( b_k\right)$, where $k \geq 1$, and $a_k$ and $b_k$ are positive real numbers for every k. Now, define the sequences

  $c_k = \min \left( a_k, b_k \right)$;
  $A_k = a_1 + a_2 + ... + a_k$;
  $B_k = b_1 + b_2 + ... + b_k$;
  $C_k = c_1 + c_2 + ... + c_k$

for all natural numbers k.

[b](a)[/b] Do there exist two monotonically decreasing sequences $\left( a_k\right)$ and $\left( b_k\right)$ of positive real numbers such that the sequences $\left( A_k\right)$ and $\left( B_k\right)$ are not bounded, while the sequence $\left( C_k\right)$ is bounded?

[b](b)[/b] Does the answer to problem [b](a)[/b] change if we stipulate that the sequence $\left( b_k\right)$ must be $\displaystyle b_k = \frac{1}{k}$ for all k ?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=265#265]IMO ShortList 2003, algebra problem 4[/url]

Let $n$ be a positive integer and let $x_1\le x_2\le\cdots\le x_n$ be real numbers.
Prove that

             \[
          \Biggl(\sum_{i,j=1}^{n}|x_i-x_j|\Biggr)^2\le\frac{2(n^2-1)}{3}\sum_{i,j=1}^{n}(x_i-x_j)^2.
             \]
Show that the equality holds if and only if $x_1, \ldots, x_n$ is an arithmetic sequence.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=119999]IMO ShortList 2003, algebra problem 5[/url]

Let $\mathbb{R}^+$ be the set of all positive real numbers. Find all functions $f: \mathbb{R}^+ \longrightarrow \mathbb{R}^+$ that satisfy the following conditions:

- $f(xyz)+f(x)+f(y)+f(z)=f(\sqrt{xy})f(\sqrt{yz})f(\sqrt{zx})$ for all $x,y,z\in\mathbb{R}^+$;

- $f(x)<f(y)$ for all $1\le x<y$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=20653#20653]IMO ShortList 2003, algebra problem 6[/url]

Let $n$ be a positive integer and let $(x_1,\ldots,x_n)$, $(y_1,\ldots,y_n)$ be two sequences of positive real numbers.  Suppose $(z_2,\ldots,z_{2n})$ is a sequence of positive real numbers such that $z_{i+j}^2 \geq x_iy_j \qquad$ for all $1\le i,j \leq n$.

Let $M=\max\{z_2,\ldots,z_{2n}\}$.  Prove that

 \[
  \biggl(\frac{M+z_2+\cdots+z_{2n}}{2n}\biggr)^2\ge
  \biggl(\frac{x_1+\cdots+x_n}{n}\biggr)\biggl(\frac{y_1+\cdots+y_n}{n}\biggr).
 \]

[b]COMBINATORICS[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=261#261]IMO ShortList 2003, combinatorics problem 1[/url]

Let $A$ be a $101$-element subset of the set $S=\{1,2,\ldots,1000000\}$. Prove that there exist numbers $t_1$, $t_2, \ldots, t_{100}$ in $S$ such that the sets

 \[
  A_j=\{x+t_j\mid x\in A\},\qquad j=1,2,\ldots,100
 \]

 are pairwise disjoint.

[url=http://mathlinks.ro/Forum/viewtopic.php?p=18977#18977]IMO ShortList 2003, combinatorics problem 2[/url]

Let $D_1$, $D_2$, ..., $D_n$ be closed discs in the plane. (A closed disc is the region limited by a circle, taken jointly with this circle.) Suppose that every point in the plane is contained in at most 2003 discs $D_i$. Prove that there exists a disc $D_k$ which intersects at most $7\cdot 2003 - 1 = 14020$ other discs $D_i$.

[url=http://mathlinks.ro/Forum/viewtopic.php?t=5785]IMO ShortList 2003, combinatorics problem 3[/url]

Let $n \geq 5$ be an integer. Find the maximal integer $k$ such that there exists a polygon with $n$ vertices (convex or not, but not self-intersecting!) having $k$ internal $90$ angles.

[url=http://mathlinks.ro/Forum/viewtopic.php?p=18993#18993]IMO ShortList 2003, combinatorics problem 4[/url]

Given n real numbers $x_1$, $x_2$, ..., $x_n$, and n further real numbers $y_1$, $y_2$, ..., $y_n$. The elements $a_{ij}$ (with $1\leq i,\;j\leq n$) of an $n\times n$ matrix are defined as follows:

$a_{ij}=\left\{
\begin{array}{c}
1\text{\ \ \ \ \ \ if\ \ \ \ \ \ }x_{i}+y_{j}\geq 0; \\
0\text{\ \ \ \ \ \ if\ \ \ \ \ \ }x_{i}+y_{j}<0.
\end{array}
\right.$

Further, let B be an $n\times n$ matrix whose elements are numbers from the set {0; 1} satisfying the following condition: The sum of all elements of each row of B equals the sum of all elements of the corresponding row of A; the sum of all elements of each column of B equals the sum of all elements of the corresponding column of A. Show that in this case, A = B.

[url=http://mathlinks.ro/Forum/viewtopic.php?p=19090#19090]IMO ShortList 2003, combinatorics problem 5[/url]

Regard a plane with a Cartesian coordinate system; for each point with integer coordinates, draw a circular disk centered at this point and having the radius 1/1000.

[b](a)[/b] Prove the existence of an equilateral triangle whose vertices lie in the interior of different disks.
[b](b)[/b] Show that every equilateral triangle whose vertices lie in the interior of different disks has a sidelength > 96.

[url=http://mathlinks.ro/Forum/viewtopic.php?p=100355#100355]IMO ShortList 2003, combinatorics problem 6[/url]

Let $f(k)$ be the number of all non-negative integers $n$ satisfying the following conditions:

[b](1)[/b] The integer $n$ has exactly $k$ digits in the decimal representation (where the first digit is not necessarily non-zero!), i. e. we have $0 \leq n \leq 10^k$.

[b](2)[/b] These $k$ digits of n can be permuted in such a way that the resulting number is divisible by 11.

Show that for any positive integer number $m,$ we have $f(2m) = 10 f(2m - 1).$

[b]Remark: ISL problems A4, G1, G6, C1, N3, N6 were chosen for IMO.[/b]

[quote="Anonymous"]You did a great job orlando, thank a lot for posting all shorlist IMO.
Moubinool[/quote]

Thanks a lot for your appreciation.  :)

BTW the IMO 2003 ShortList file is available in reasonable file size now.

oh
I am sorry I made  mistake
I have already download it

Thank you so much orl you helped me so much.Thanks! :D


Hey, What's this whole hubhub over this math search? How hard is it?

usamts.org

Look at the problems, that would be a pretty good indicator.


geometry
3D geometry
counting
distinguishability
combinatorics open
combinatorics

In how many ways can 5 red cubes, 5 white cubes, and 5 blue cubes be placed into 3 (distinguisable)

kangaroos so that no kangaroo has more than 8 cubes and no kangaroo has more than 3 cubes of

anyone color?


Let $A_1, A_2, ..., A_{63}$ be the nonempty subsets of {1, 2, 3, 4, 5, 6}. For each of these sets $A_i$, let $\pi(A_i)$ denote the product of all the elements in $A_1$. Then what is the value of $\pi(A_1)+\pi(A_2)+...+\pi(A_{63})$?

I noticed a pattern, which took forever to spot.

[hide]Let $S_n$ be the set of all subsets of $\{ 1,2,3,4,5,6\}$ with exactly $n$ elements.

The polynomial $\prod_{r=1}^6 (rx+1)=1+\sum_{n=1}^6 \left( \sum_{A_k \in S_n} \pi(A_k) \right) x^n$ because the $x$ term contains the sum of 1 through 6, the $x^2$ term is the sum of all possible products of two of these integers, etc.

Substitute $x=1$ to get $1+\sum_{k=1}^{63} \pi(A_k)=\prod_{r=1}^6 (r+1)=7!=5040 \Rightarrow \boxed{5039}$.[/hide]

[hide]let $P(x)=(x+1)(x+2)(x+3)(x+4)(x+5)(x+6)$, the expansion gives us every possible product in one of the coefficients of the polynomial, but it over counts slightly because we do not want the empty set, say we want the $\pi\{4,5,6\}$, then we choose 6,5,4 and multiply them by x's in the other terms, so the sum of the coefficients will be our answer

then if we do $P(1)$, we get the sum of the coefficients of our polynomial, but we do not want the coefficient of x^6 because that corresponds to choosing none of them, so we want $P(1)-1=7!-1=\boxed{5039}$[/hide]

EDIT: my solution is pretty much the same as scorpius' oh well...


Two equal parallel chords are drawn $ 8$ inches apart in a circle of radius $ 8$ inches. The area of that part of the circle that lies between the chords is:

$ \textbf{(A)}\ 21\frac {1}{3}\pi \minus{} 32\sqrt {3} \qquad\textbf{(B)}\ 32\sqrt {3} \plus{} 21\frac {1}{3}\pi \qquad\textbf{(C)}\ 32\sqrt {3} \plus{} 42\frac {2}{3}\pi$
$ \textbf{(D)}\ 16\sqrt {3} \plus{} 42\frac {2}{3}\pi \qquad\textbf{(E)}\ 42\frac {2}{3}\pi$

[hide="Solution"]
The chords must both be $ 4$ inches from the center of the circle, so we get some $ 30^\circ\minus{}60^\circ\minus{}90^\circ$ triangles. It is now easy to see that the area is $ 2\left(\frac12\cdot8\sqrt3\cdot4\right)\plus{}\frac{120^\circ}{360^\circ}\cdot\pi\cdot8^2\equal{}\boxed{\textbf{(B)}\ 32\sqrt3\plus{}21\frac13\pi}$.
[/hide]


real analysis
geometry
real analysis unsolved

Hi,
Could anyone help me with this problem:

Suppose f:[a,b]->R^2 is a path of finite arc length. Prove that it has zero lebesgue area.

thanks

Let $ L$ be the length, understood as the total variation of $ f$ on $ [a,b]$. For any integer $ N\ge 1$ you can partition $ [a,b]$ by points $ a_i$ in such a way that the total variation of $ f$ on each subinterval is $ L/N$. The image of $ f$ is contained in the union of $ N$ disks centered at $ f(a_i)$ of radius $ L/N$. The total area of the disks is $ \pi L^2/N$.


function
real analysis
real analysis solved

Let $X\subset \mathbb{R}$ posses the following property: any function $f:\mathbb{R}\to \mathbb{R}$, satisfying the condition that its restriction to any segment $[a,b]\subset X$ is a continuous fucntion, is continuous on $X$.

Prove that $X$ is union of at most countable number of intervals and points. (Note: point $\{x\}$ is a segment $[x,x]$.)

The interior of the set is a union of disjoint open intervals, so we only need to look at the rest of $X$. There are countably many points of $X$ which are endpoints of those countably many open intervals forming the interior, and there might be points of $X$ which are not endpoints of intervals contained in $X$. Let's cal these points "lonely" :).

If a lonely point $x\in X$ is an accumulation point, then we can define a function to be $0$ on $X\setminus\{x\}$, and $1$ in $x$. This function has a continuous restriction on each $[a,b]\subset X$, but it's not continuous on $X$. This means that there are no lonely accumulation points, so the subset of lonely points of $X$ is discrete and thus countable.


modular arithmetic
number theory proposed
number theory

Find all natural numbers $ a$ and $ b$ satisfying $ a^3\minus{}b^3 \equal{} 2009$.

[quote="MeKnowsNothing"]Find all natural numbers $ a$ and $ b$ satisfying $ a^3 \minus{} b^3 \equal{} 2009$.[/quote]

Let $ a\equal{}b\plus{}x$ (and so $ b,x>0$) and we get $ 3b^2x\plus{}3bx^2\plus{}x^3\equal{}2009$ and so $ x$ is a positive divisor of $ 2009$ and $ x\equal{}2\pmod 3$ and so $ x\in\{41,287,2009\}\ge 41$ but then $ LHS>41^3>2009\equal{}RHS$

And so no solutions in natural numbers for this equation.


If $ x$, $ y$ and $ z$ are positive reals prove that:
$ \sum \sqrt{(x\plus{}y)(x\plus{}z)} \geq x\plus{}y\plus{}z\plus{}\sqrt{3(xy\plus{}yz\plus{}zx)}$

[quote="canada"]If $ x$, $ y$ and $ z$ are positive reals prove that:
$ \sum \sqrt {(x \plus{} y)(x \plus{} z)} \geq x \plus{} y \plus{} z \plus{} \sqrt {3(xy \plus{} yz \plus{} zx)}$[/quote]
Put $ a \equal{} \sqrt {x \plus{} y} \plus{} \sqrt {x \plus{} z} \minus{} \sqrt {y \plus{} z} , b \equal{} \sqrt {x \plus{} y} \plus{} \sqrt {y \plus{} z} \minus{} \sqrt {x \plus{} z} ,$
$ c \equal{} \sqrt {y \plus{} z} \plus{} \sqrt {x \plus{} z} \minus{} \sqrt {x \plus{} y}$
We have $ ab \plus{} bc \plus{} ca \equal{} 2\sum \sqrt {(x \plus{} y)(x \plus{} z)} \minus{} 2(x \plus{} y \plus{} z)$
So we must to prove that $ ab \plus{} bc \plus{} ca \ge 2\sqrt {3(xy \plus{} yz \plus{} zx)}$
But $ abc(a \plus{} b \plus{} c) \equal{} 2\sum(\sqrt {(x \plus{} y)(x \plus{} z)})^2 \minus{} \sum(x \plus{} y)^2 \equal{} 4(xy \plus{} yz \plus{} zx)$
Since we rewrite inequality in this form
$ ab \plus{} bc \plus{} ca \ge\sqrt {3abc(a \plus{} b \plus{} c)}$
which is true by AM-GM.

u can also see here:
http://www.mathlinks.ro/viewtopic.php?t=297110


blogs
algebra
polynomial
vector
number theory

[quote="srulikbd"]how to generalise the following: $ \sqrt{a}+\sqrt{b}+\sqrt{c}$ is irrational a,b,c natural not squares. this case is easy, but larger ones? also, easy is $ a^{1/3}+b^{1/3}$ irrational, what about larger?[/quote]

What about $ (a,b,c)=(4,9,2)$? That is irrational but two of the numbers are squares.

[quote="srulikbd"]$ \sqrt{a}+\sqrt{b}+\sqrt{c}$ is irrational a,b,c natural not squares. [/quote]

It's not clear what the direction of implication is, but since Altheman showed that the right direction is clearly false, you seem to mean the other direction; that is,

$ a, b, c \in \mathbb{N}, \sqrt{a}, \sqrt{b}, \sqrt{c}\not \in \mathbb{N}\implies \sqrt{a}+\sqrt{b}+\sqrt{c}\not \in \mathbb{Q}$

Algebraic number theory provides a massive generalization of this kind of principle because it looks at this problem in terms of algebraic field extensions, but that's overkill.

You might want to refer to my blog post about [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=134466]minimal polynomials[/url], and try to prove something about the minimal polynomial of $ x+y$ given the minimal polynomials of $ x, y$.

[quote="t0rajir0u"]
Algebraic number theory provides a massive generalization of this kind of principle because it looks at this problem in terms of algebraic field extensions, but that's overkill.
[/quote]
What do you mean by overkill? I think its relatively easy. Consider $ E=\mathbb{Q}(\sqrt{a},\sqrt{b},\sqrt{c})$ if $ \sqrt{a},\sqrt{b},\sqrt{c}\not \in N$ then $ \mathbb{Q}< E$ and since $ E$ is a vector space over $ Q$ the rationals $ q+r\sqrt{a}+s\sqrt{b}+t\sqrt{c}\in \mathbb{Q}$ are the trivial ones, i.e. $ q\in \mathbb{Q}$ and $ r=s=t=0$. Pretty much like that.

[quote="t0rajir0u"]Algebraic number theory provides a massive generalization of this kind of principle because it looks at this problem in terms of algebraic field extensions, but that's overkill.
[/quote]

Overkill is good if it kills a lot of the surrounding stuff as well.

Ie, killing a fly with a cannon is good if that fly is part of a swarm of flies.

Using a method which generalizes the problem is not bad here, because it's insightful...

Well, yes, but it's slightly beyond the scope of this particular forum and problem, although I do agree that it's an insightful generalization.  Although, sylow_theory, it might be better to adjoin each of $ \sqrt{a}, \sqrt{b}, \sqrt{c}$ at a time in case one is contained in the field extensions generated by the others, i.e. $ \sqrt{2}, \sqrt{3}, \sqrt{6}$.

The easiet case can immediately be generalized for $ n$ primes $ p_{1},p_{2},..p_{n}$ because it immediately follows from the fact that $ [\mathbb{Q}(\sqrt{p_{1}},...,\sqrt{p_{n}}): \mathbb{Q}] = 2^{n}$ which can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=153532]here.[/url].


Hi.

I can do well on AMC's(average 138-144) and AIME's(average 6-8), but I couldn't solve a single problem on USAMO.

What should I do?

Please help.

How long have you tried on the USAMO problems?

I don't even know where to start...

that's my problem...

hmm...interesting...

are you unfamiliar with "proof" questions? Also, which USAMO problems have you tried?

1s tend to be the easiest. Try to solve some of the older USAMO problems. They tend to be to easier...

Sometimes it's just a lack of exposure to difficult problems...maybe try really hard to do the AIME problems that you miss (and don't give up)? Do problems slightly above your normal "level"?

Practice proofing? Its a different set of skills, I think.

Also, try to solve USAMO problems for a long time... many people can't solve them because they give up after 5 minutes...

What are perfect scores on AIME and USAMO?

[quote="Poincare"]i would suggest doing REALLY well on the AIME then try the USAMO problems if you understand the USAMO solutions or if you don't get engel or something.[/quote]

Really? I know people who can solve ~4 AIME problems but a lot of USAMO problems with enough time...USAMO #1s don't require much background knowledge... mostly they require problem solving skill and time...

Note: Do not be scared by some USAMO problems. I found that the #1s on USAMOs that look the hardest are actually the easiest...

Do easier olympiads like Canada and early USAMOs. Don't get frustrated; Zeitz' Art and Craft of Problem Solving has the best section on psychological strategies.


geometry
logarithms
MATHCOUNTS
geometric series

how would u do this problem?

franky's mom wants him to mow the lawn. unfortunately, franky is lazy and only mows 70% of the unmowed lawn at a time. so his mom settles from him to mow a total of 95% of the lawn. how many times must franky mow the lawn in order to have mowed at least 95% of it.

I hope its right

[hide="solution"]Lets say the area of the lawn is 100ft^2

the area remainig after each he mows the lawn is 100(.3)^n
We are to find how many times he needs to mow 95% of the the lawn if evey time he goes out to mow it he mows 70% of the remainig lawn

so I set up the equation as follows

$100(.3)^{x}>or = 5$ which is the remaining unmowed lawn after 95% is mowed

$(.3)^{x}=.05$ we solve for x

$x=\frac{\log .05}{\log .3}$ about 2.46 Now since x must be an integer the answer is x=3.[/hide]

after one lawn-mowing 70% of the lawn is mown, after two 91% is mown, and after [b]three[/b] 97.3% is mown.

go go mathcounts method

The general method is done by using logarithms.

Suppose he only does $a\%$ of the time, and his mom wants him to do atleast $b\%$. Every time, he does $a\%$ more:

1st time $a\%$
2nd time: $(a+\frac{a^{2}}{100})\%$
3rd time: $(a+\frac{a^{2}}{100}+\frac{a^{3}}{10000})\%$
Etc...
nth time: $(a+...+\frac{a^{n}}{100^{n-1}})\%$
Using the formula for the sum of a geometric series,
nth time: $\frac{a(1-(\frac{a}{100})^{n}}{1-\frac{a}{100}}$

Now solve for $n$.

Can anyone correct me if I am wrong?

Your method is technically correct, but unbelievably tedious.  Jhonny gives the preferred method.

Rather than look at the amount he has mowed, we look at what he [b]hasn't[/b] mowed.  The progression of this value is quite simple; it becomes $30$% of itself every time Franky mows the lawn.  In other words, after the $n^{th}$ time Franky mows the lawn,

$100(.3)^{n}$%

Of the lawn remains to be mowed, as Jhonny gave earlier.  Then we solve with logs.


Greetings,

Currently with the contest to be held in mid December this later year, there are only two problem reviewers (Erin J. Schram for USAMTS and MellowMelon). They have already been really awesome help but I feel like I need more reviewers to make sure that the level of problems is right. I haven't decided whether I want this to be harder than real AMCs but still, I don't want to make it too different than AMC.

If anyone's interested, please send me private message. Problem reviewer has to graduate this year (That is, upcoming June or May) or graduated in past years. I do not want high school students as reviewers as I rather have them as the contestants.

Thanks.

Addition to the Topic:

If a high student submits a problem or wants to make a problem, he/she is free to do that.  It will appear as Your Name - Round Name Proposal (This is also allowed for any other problem not directly used in the contest).


Find all function $ f: R \rightarrow R$ such that
\[ \sum_{i\equal{}1}^{2549}f(x_{i}\plus{}x_{i\plus{}1})\plus{}f(\sum_{i\equal{}1}^{2550}x_{i})
\leq \sum_{i\equal{}1}^{2550}f(2x_{i})\]
for all real $ x_1,x_2,...x_{2550}$

Now I found that $ f(x) \equal{} c$ is only solution  :jump:

[hide="solution"]
[u]Solution[/u]$ x_1\equal{}x_2\equal{}x,x_3\equal{}x_4\equal{}...\equal{}x_{2550}\equal{}0$
$ \therefore  f(x) \leq f(0)$
$ x_1\equal{}x_3\equal{}...\equal{}x_{2549}\equal{}k$
$ x_2\equal{}x_4\equal{}...\equal{}x_{2550}\equal{}\minus{}k$
$ \therefore  2f(0) \leq f(x)\plus{}f(\minus{}x)$
$ \therefore  f(x)\equal{}c$ is only solution. :lol:[/hide]


Calculate the current flowing through the circuit in the diagram given below(or rather attached here).
What answer did you get?

MissSmartyPants, there is no file attached.

I did attach it..dunno why it hasn't cum...but must have looked pretty silly...
 :lol:
Sorry..here it is..(I hope)

First step. Find the equivalent resistance to the 3, 4, 5, and 10 resistors. As the first three are in parallel, then their equivalent resistance is 1/Req = 1/3 + 1/4 + 1/5 = 47/60, and so Req = 1.3 Ohm. The total resistance is then R = 1.3 + 10 = 11.3 Ohm.

Second step. Write Ohm's law for this equivalent resistance: U = RI.

Third step. Write Ohm's law for the voltage source: E = U + rI, where E is the e.m.f. and r the internal resistance (2 Ohm).

Fourth step. Solve the two equations for I: E = I(R + r) => I = 6.4/(11.3 + 2) = 0.48 A.


Maybe known...

Let $p, q$ and $r$ be prime numbers. It is given that $p$ divides $qr - 1$, $q$ divides $rp - 1$, and $r$ divides $pq - 1$.
Determine all possible values of $pqr$.

Here you go:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=pr-1&t=36538
http://www.mathlinks.ro/Forum/viewtopic.php?t=792


Are you sure that this question is correct?  I cannot get an answer for this.
you mean that
$(2000000+a*100000+b*10000+c*1000+d*100+e*10+f)*3=a*1000000+b*100000+c*10000+d*1000+ex100+f*10+2$
right?

Yeah, I don't get one, but I see that

285714*3=857142

Note that $f=4$, since nothing else results in a units digit of 2 upon multiplication by 3. Then substitute and use the same logic for all the other digits.

This is exactly like USAMTS 18.1.1.

[quote="Altheman"]huh if abcdef=x, then $(2*10^{6}+x)(3)=(10)(x)+2$[/quote]

OMG, you are right

So the answer is $\frac{5999998}{7}$

[quote="VIPeter"]Let $2abcdef\times3=abcdef2$, find abcdef.[/quote]
Sorry! :blush:
The equation should be $2abcde\times3=abcde2$


Prove that for every integer number n, 60 devise P=n^5+5n^4-15n^3-65n^2+74n+120.

Trivially $P \equiv n^5 + n^4 + n^3 - n^2 + 2n \equiv n^3 + n^2 + n^3 - n^2 + 2n \equiv 2n(n^2+1) \equiv 0 \bmod 4$. Moreover, by Fermat's little theorem: $P \equiv n^5 + 2n^4 + n^2 + 2n \equiv n^3 + 2n^2 + n^2 + 2n \equiv n(n^2 + 2) \equiv 0 \bmod 3$ and $P \equiv n^5 - n \equiv n - n \equiv 0 \bmod 5$. So $60 \mid P$, q.e.d.


AMC
AIME
geometry
3D geometry
rectangle
number theory
greatest common divisor

A $150\times 324\times 375$ rectangular solid is made by gluing together $1\times 1\times 1$ cubes. An internal diagonal of this solid passes through the interiors of how many of the $1\times 1\times 1$cubes?

I know that it has something to do with PIE, but I don't know how to apply it on this... :huh:

And it is hard to visulize..

Can anyone post a solution? Thanks,

[hide]If we use the inclusion-exclusion principle, the answer  \begin{eqnarray*} &=& 150+324+375-\gcd(150,324)-\gcd(150,375)-\gcd(324,375)+\gcd(150,324,375) \\ &=& 150+324+375-6-75-3+3 \\ &=& \boxed{768} \end{eqnarray*}
You really just have to know that to do the problem
[/hide]

Umm question: how do you "just know how to do a problem" (not including intuition)?   ;)   As in why would you use the inclusion-exclusion principle... :roll:

[quote="t0rajir0u"]I would do something of a mental dialogue.

"Hmm.  I have no idea how to count this value directly."
"Well, let's use PIE then!"
"Ok!  :D "

:D[/quote]

wow.....you have to teach me how to do that!  :rotfl:

Sorry, I should've stated in this way ( as pieterminate puts):

"how do you know that u have to use PIE in this problem?"

Thanks,

Does anyone know how to generalize this? Especially because there is a really simple equation in the two dimension sense of the problem (given an $m X n$ rectangle find the number of unit square the diagonal passes thru). It is something like m+n-1 for relatively prime m,n.

For a*b*c, a+b+c-gcd(a,b)-gcd(b,c)-gcd(c,a)+gcd(a,b,c) for 3 dimensional(i believe).

And for 2 dimensional, i believe it's just for a*b, a+b-gcd(a,b).

So it's just the inclusion thingy... ;)

I guess what you're saying is right, the reason I remember a+b-1 is because, being an 8th geometry class, we only had to look at the problem when the two were relatively prime, simplifying to the 1. By the way, how [i][b]do[/b][/i] you get into the gcd? I can understand why it goes there, but having trouble coming up with it from scratch.[/i]

Well, say you have a rectangle that is $a$ by $b$, and they are relatively prime, it is easy to see that the diagoinal goes through $a+b-1$ unit squares. Say you have a rectangle that is $x$ by $y$. with $\gcd(x,y)=d$. Then you can break up the part where the diagonal goes through into $d$ rectangles of dimensions $\frac xd$ by $\frac yd$ where it passes through the endpoints of these rectangles. We know $\gcd\left(\frac xd,\frac yd\right)=1$, so the diagonal of each of these little rectangles passes through $\frac xd+\frac yd-1$ unit squares. there are $d$ of these little rectangles, so the number of unit squaresit goes through altogether is $d\left(\frac xd+\frac yd-1\right)=x+y-d$

Yep!  :)

And now prove by induction for the next dimensions... :D

yea the idea behind this is pretty cool

first we can think of it in 2d cuz its ezier... but the 3d version is essentially the same thing

each time the internal diagonal hits a "gridline" on the coordinate plane, it will be entering a new square, hence $a+b$

but, if this line hits the gridline exactly at the corner of a square, then we will be counting this case twice -- once when we count how many vertical lines our diagonal crosses, and once when we count how many horizontal lines our diagonal crosses

hence, the number of corners that this diagonal will hit is $\gcd(a,b)$ and so our answer is $a+b - \gcd(a,b)$

now we can just extend this into 3 dimensions basing it off the same concept


Proove that $ \lim_{x\to\plus{}\infty}sin(x\plus{}1)^{1/3}\minus{}sin(x)^{1/3}\equal{}0$

Is that

a) $ \sqrt [3]{\sin (x+1)}-\sqrt [3]{\sin x}$

or

b) $ \sin\sqrt [3]{x+1}-\sin\sqrt [3]{x}$?

[hide="Assuming it is b"]since the limit doesn't exist in a:

\begin{eqnarray*}\left|\sin\sqrt [3]{x+1}-\sin\sqrt [3]{x}\right| & = & 2\left|\cos\frac{\sqrt [3]{x+1}+\sqrt [3]{x}}{2}\sin\frac{\sqrt [3]{x+1}-\sqrt [3]{x}}{2}\right|\\ &\leqslant & 2\left|\sin\frac{\sqrt [3]{x+1}-\sqrt [3]{x}}{2}\right|\\ & = & 2\left|\sin\frac{1}{2\left(\sqrt [3]{(x+1)^{2}}+\sqrt [3]{x(x+1)}+\sqrt [3]{x^{2}}\right)}\right|\\ & < &\varepsilon\end{eqnarray*}

for arbitrarily small positive $ \varepsilon$ and sufficiently large $ x$[/hide]

Of course b was assumed.  :lol:
[hide="Assuming it is a"] Taking $ x \equal{} 2k\pi$ we obtain $ \lim \equal{}\sqrt [3]{\sin{1}}$ whereas taking $ x \equal{} 2k\pi\plus{}\pi$ we have $ \lim \equal{}\minus{}\sqrt [3]{\sin{1}}$. This means that the limit does not exist [/hide]


i need help

there are $2^{2^{3}}$ labelled families of subsets of a 3-set. How many unlabelled families are there??


Write $ 80\%$ of $ \frac34$ of $ 1.25$ as a common fraction.

One solution:
Simplify by converting to decimals.
.8*.75*1.25=.75=3/4.

I like this solution.

$ 1.25$ and $ 0.8$ are reciprocals, so multiply those to get $ 1$.  :lol:

It's pretty straight forward from there.

Good job ernie.  But Jongao, I think it's easier to convert to fractions:

$ \frac{4}{5} %Error. "TIMES" is a bad command.
\frac{3}{4} %Error. "TIMES" is a bad command.
\frac{5}{4}$

Hmm now you can use common sense.


geometry
angle bisector
similar triangles

In triangle $ ABC$, internal bisector of angle $ A$ intersects with $ BC$ at $ D$. Let $ E$ be a point on $ \left[CB\right.$ such that $ \left|DE\right|\equal{}\left|DB\right|\plus{}\left|BE\right|$. The circle through $ A$, $ D$, $ E$ intersects $ AB$ at $ F$, again. If  $ \left|BE\right|\equal{}\left|AC\right|\equal{}7$, $ \left|AD\right|\equal{}2\sqrt{7}$ and $ \left|AB\right|\equal{}5$, then $ \left|BF\right|$ is

$\textbf{(A)}\  \frac {7\sqrt {5} }{5} \qquad\textbf{(B)}\  \sqrt {7} \qquad\textbf{(C)}\  2\sqrt {2} \qquad\textbf{(D)}\  3 \qquad\textbf{(E)}\  \sqrt {10}$

I used Stewart's. How do you do it without it?

There is a nice way to find the length of the angle bisector without Stewart's Theorem.

Consider the following diagram. By similar triangles, $ \frac cd\equal{}\frac{d\plus{}\frac{mn}d}b\implies bc\equal{}d^2\plus{}mn\implies d^2\equal{}bc\minus{}mn$.


This is slightly annoying but not serious: When I click on the "Delete Attachment" button for some PM in my inbox which has an attachment, all I receive is this error:

[quote]General Error

No post ID was specified[/quote]

The only thing I can do is delete the whole PM.

  darij

Another one, even less critical. (I don't know for sure how it is triggered, but probably it has to do with somebody sending me a PM and revoking it thereafter. Or old PM's being automatically deleted from the inbox.)

  darij

Yea if you could reproduce that it would be great, obviously seems like a subtraction that shouldn't happen at some point.

Did you have $ 16$ PM's in your inbox?  It looks pretty ironic to me, because $ 65535 \equal{} 2^{16} \minus{} 1$...


induction
real analysis
real analysis unsolved

Prove that euclidian plane isn't the union of disjoint open circles(circles cannot be reduced to the point).
Sorry if it has already been posted.

There is no "open circle". I guess you simply mean "circles".

Assume the contrary. Now pick a circle $C_{1}$ in your covering, and denote by $r_{1}$ its radius. Denote by $c_{1}$ its center. $c_{1}$ must belong to a circle $C_{2}$ of the covering, and clearly this circle must have radius $r_{2} < \frac{r_{1}}{2}$. Now repeat the same process on $C_{2}$. By induction you'll get a sequence of disjoints circles $C_{n}$ of radii $r_{n}$ such that ${r_{k} < \frac{r_{1}}{2^{k}}}$. This means the centers $(c_{n})$ of the circles form a Cauchy sequence in the plane, whence $c_{n} \rightarrow c$, and $c \in \cap D_{n}$, where $D_{n}$ is the open disc with frontier $C_{n}$. But $c$ must lie on a circle $C$ of the covering with radius $r$, and since all those circles must be disjoint, any circle $C_{n}$ containing $c$ must have radius $>r$, which is absurd since $r_{n} \rightarrow 0$.


any formula for finding area for pentagons?

No. Any shape with 5 sides is a pentagon.

The formula for any polygon uses trig, which is too advanced for

this forum. However, If you know the side length and the apothem,

(The apothem is a perpendicular from the center to one of the sides),

the formula is $ \frac{apothem \cdot perimeter}{2}$.

The trig one is:

$ \frac{S^2N}{4\tan\frac{\pi}{N}}$, where $ S$ is the length of

a side, $ N$ is the number of sides, and $ \tan$ is trig.

That was in radians. In degrees,

$ \frac{S^2N}{4\tan\frac{180}{N}}$.

The trig is advanced.

[quote="BOGTRO"]No. Any shape with 5 sides is a pentagon.

The formula for any polygon uses trig, which is too advanced for

this forum. However, If you know the side length and the apothem,

(The apothem is a perpendicular from the center to one of the sides),

the formula is $ \frac {apothem \cdot perimeter}{2}$.

The trig one is:

$ \frac {S^2N}{4\tan\frac {\pi}{N}}$, where $ S$ is the length of

a side, $ N$ is the number of sides, and $ \tan$ is trig.

That was in radians. In degrees,

$ \frac {S^2N}{4\tan\frac {180}{N}}$.

The trig is advanced.[/quote]

Assuming the figures are regular :wink:

Of course, if you knew $ \tan{\frac{180^\circ}{108^\circ}}$ beforehand or by typing it into a scientific calculator, you could find the area of a regular pentagon.


probability
factorial
modular arithmetic

[hide="Hint for 1"] At some point, a factorial will be divisible by $18$.  When is this point?  (Every subsequent factorial will also be divisible by $18$, since they are all multiples of the previous ones.) [/hide]

[quote="moogra"]1. If you divided the sum of the expression 1! + 2! + 3! + … + 98! + 99! + 100! by 18, what would be the remainder?[/quote]
[hide]$6!=720,$ which is divisible by 18. Every factorial after $6!$ will also be divisble by 18, so focus on 1-5.

$1!+2!+3!+4!+5!=1+2+6+24+120=153$

$153\equiv\boxed{9}\pmod{18}$[/hide]

[quote="moogra"]1. If you divided the sum of the expression 1! + 2! + 3! + … + 98! + 99! + 100! by 18, what would be the remainder?

2. Katelyn and Carly are playing a game that consists of ten different numbered cards laying face down on a table.  The faces of the cards are labeled using the numbers 1 – 10 (each card is numbered differently).  If Katelyn and Carly each turn over a different card, what is the probability that the sum of the two cards is even?
By the way, it's only 1 set of cards, so it is like only 1,2,3,4,5,6,7,8,9,10 for them 2.[/quote]

[hide="#2"]This is the probability that both cards are even plus the probability that both cards are odd (since the events are disjoint).

The first probability is $(\frac{1}{2})(\frac{4}{9})$.
The second probability is the same.

The total probability is thus $\frac{4}{9}$.[/hide]


You roll a die, and I roll a die. You win if the number showing on your die is strictly greater than the one on mine. If we play the game five times, what is the chance you win at least 4 times?

[hide]
The probability of winning a game is 5/12

the probabilty of winning 4 out of 5 is
5*(5/12)^4*(7/12)
the probililty of winning 5 out of 5 is
1*(5^12)^5*1

add them together and you get
3125/20736
[/hide]

[hide="Hint"]
Start with the probability of winning one game.
[hide="More of a Hint"]
Then consider the number of ways you can win 4 and 5 games, using combinations.
[hide="Even more of a Hint"]
Just kidding :D
[/hide][/hide][/hide]

well Combination(5 4) = 5, and Combination(5 5) = 1 , i thought the probability of winning would be .5

Am I right that the probability of winning 1 game is 15/36?

[hide="probability of winning one game"]
Yeah, I think it's $\frac{15}{36}$.
[/hide]

Probability of winning one game:
There are $6 \cdot 6 = 36$ possible outcomes.  If your roll is $1$, there are $5$ rolls I could make that make me win.  Similarly, if you roll $2$, I have $4$ rolls that make me win.  The sum of all possible outcomes where I win is $5+4+3+2+1+0 = 15$.  Therefore the probability is $\frac{15}{36}$

Winning four games:
The number of ways to do this is $\binom 5 4 \left( \frac{15}{36} \right)^4 \left( \frac{21}{36} \right)^1$

Winning all five games:
The number of ways to do this is $\binom 5 5 \left( \frac{15}{36} \right)^5 \left( \frac{21}{36} \right)^0$

Summing them, we get $\left( \frac{15}{36} \right)^4 \left( \frac{10}{3} \right)$

say P is when you have strictly greater, think of switching the dice, then P is also when the other has strictly greater, also we need to count the 2 people having equal rolls, so $2P+\frac{6}{36}=1 \implies P=\frac{5}{12}$, then winning 4 times is just $\binom{5}{4}*\left(\frac{5}{12}\right)^4*\left(\frac{7}{12}\right)+\left(\frac{5}{12}\right)^5$

Wow... yeah $\frac{15}{36}$ doesn't reduce... I just haven't been with it lately.

I think it reduces.  Maybe it's a matter of opinion.

Ok, yeah, this makes me feel really stupid, but I keep seeing this notation everywhere, and I know I should know what it means by now, but for some reason, I don't.

So, someone, please help!  What does $\binom{5}{4}$ mean?

$\binom{n}{k}$ is the number of ways of picking k [i]unordered[/i] outcomes from n possibilities.  See http://mathworld.wolfram.com/Combination.html.

[quote]So basically it's a notation for combinations?[/quote]

Yep.

Sweet, now I went from completely clueless about the explanation to understanding it perfectly...  I need to hang around here more...


real analysis
calculus
derivative
real analysis unsolved

For $ t \geq 0$, put

$ \phi(x,t) \equal{} \left\{\begin{array}{ll} x\; & \mbox{if }\, 0 \leq x \leq \sqrt t \\
\minus{} x \plus{} 2\sqrt {t}\; & \mbox{if }\, \sqrt t \leq x \leq 2 \sqrt t \\
0\; & \mbox{otherwise} \end{array}\right.$

and put $ \phi(x,t) \equal{} \minus{} \phi(x,|t|)$ if $ t < 0$.

Show that $ \phi$ is continuous on R^2 and $ \frac {\partial\phi}{\partial t}(x,0) \equal{} 0$.

I don't think the partial derivative even exists at (0,0) but someone please correct me.

Never mind. The partial exists. I am an idiot.


I am having some trouble determining the products for that chemical reaction. I think there will be H2O?, CO2?, and what else? Thanks.

hmm it seems like you would get C6H6O6, and then the H2 would combine with CO3 to make carbonic acid which decomposes into H2O and CO2, and then you would have some sodium ions?

im not sure though


Delete this if this is spam, please.

News Flash: Valentin Vornicu passes harazi for 4th on the post count list, with 5019 posts! Congrats, I guess!

And I'm almost going to break the top 100, yay

:o
what are you doing spamming a forum instead of paying attention to AMC12 class?
oh wait i need to pay attention too :o

[quote="junggi"]
what are you doing spamming a forum instead of paying attention to AMC12 class?
oh wait i need to pay attention too :o[/quote]

but i don't  :P
oh wait my 1st sems start this monday  :o

I'm in 115th place if you count posts in the member list. The person above me has 962, the person below me has 940. I'm in heavy traffic!

grobber is in first place with a whopping 7541!

The post counting system doesn't include posts in G&FF. Besides. Would you like it if post count rose easier? You could be in 2225th place! The place you ARE in is:(about 13 thousand people have 0 posts) 2389!!!!!

edit: on page 48 if you go to the member list and list it in post descending order.

hey this is a weird topic how about ur nailpolish?

Koda, in the suggestions and announcements forum, there is a topic about shortening your signature. It MAY NOT be distracting!

P.S. I'm in 114th place. The person below me has 986 posts, the person above me has 997, the person above him has 999 posts. The person above HIM has 1000.

Happy NSE to me, happy NSE to me... Happy NSE to nutzy boy, happy NSE to me...

P.S. I'm in 112nd place, and 3 posts ahead #113...

P.P.S. I'm not spamming...

Here is the link to my name, I'm #112.

[url]http://www.artofproblemsolving.com/Forum/memberlist.php?start=100[/url]

[quote="asmallgoat"]SPAM! I LOVE SPAM! I LOVE SPAM!!! SPAM SOME MORE Y'ALL!!!!!spamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspams[/quote]

Go away!
(mumbles something rude about treething)

I'm at 223. I'm closer than I though I'd be.

If you're on mathlinks skins, click on memberlist on the list of icons in the top right. If you're on aops skins, then I don't know, so you'd better just change to mathlinks and look there.

Or I could look for you- you're currently at 591.

wow AoPs is mathlinks how why when where who?????what???damn


Two squares $ ABCD$ and $ CDEF$ are in perpendicular planes. If $ \ell$ is the side of the squares, find the distance between the lines $ BD$ and $ CE$ in terms of $ \ell$.

Edit: planes not plans

[b]firstly what are are 'perpendicular plans'[/b]
this is the solution based on my understanding of the question.
JOIN [i]CE[/i] AND [i]BD[/i] AND MIDPOINT [i]'O'[/i] OF [i]CE [/i]TO [i]'D'[/i]
IT IS CLEARLY VISIBLE THAT [i]BD [/i]AND [i]CE [/i]ARE PARALLEL AND [i]OD [/i]IS THE PERPENDICULAR DISTANCE BETWEEN THEM.
[i]OD [/i]IS HALF DIAGONAL [i]CE[/i]
= (l *sqrt2)/2

Sorry. I meant perpendicular planes.

$ BD$ and $ CE$ aren't parallel.

if you see extreme points of bd and ce, distance bc=de=l
and corresponding distances b/w points on bd and ce are =l
so, distance=l
or i misunderstood again...


geometry
circumcircle
AwesomeMath
summer program
trigonometry
Pythagorean Theorem
cyclic quadrilateral

Let $ ABC$ be a triangle with $ AB\equal{}AC\equal{}20$ and $ BC\equal{}24$. Point $ D$ lies on the small arc $ AB$ of its circumcircle such that $ AD\equal{}15$ and $ BD\equal{}7$. Prove that $ CD$ is a diameter of this circle.

Cheap way:

Use Ptolemy's Theorem to immediately find that $ CD \equal{} 25$, which satisfies those darn Pythagorean triples, which implies that the angle that subtends the arc $ CD$ is equal to $ 90$, which implies that $ CD$ is a diameter.

EDIT: Oh apologies. I just realized that post about writing good solutions. As I'm not an AwesomeMath student, I'll stop posting here unless I could write a nice and decent solution.

Without Ptolemy's:

[hide]
Assume that $ CD$ is not the diameter. Then angles $ CAD$ and $ CBD$ are not right. Thus by the Pythagorean theorem $ CD$ is not $ 25$. Therefore $ CD$ is either greater than or less than $ 25$. If it were greater than $ 25$, each of the angles $ CAD$ and $ CBD$ would be greater than 90. If it were less than $ 25$, each of the angles $ CAD$ and $ CBD$ would be less than 90. However, being a cyclic quadrilateral, ACBD necessarily has opposite angles that sum to 180. Therefore our original assumption that $ CD$ was not the diameter was false, and $ CD$ is the diameter.
[/hide]

I think that the ptolemy solution is the cleanest...

@leoxnlin: Your argument isn't very rigirous...Like if CD>25, then blah about the angles...where is the justification?

I believe that if you find the justifications for these statements, you could just as well directly argue that CD=25, so by the converse of the pythagorean theorem, CAD, CBD are right, so CD is a diameter.

We could always justify using the Law of Cosines...

Okay. But if you use the law of cosines and the fact that the opposite angle sum to 180, you might as well just solve and say that CD=25... (remember that $ \cos \pi \minus{} A \equal{} \minus{} \cos A$).

Also this isn't the best problem because it is not clear that point $ D$ exists. The problem stipulates that $ D$ has three properties:
i) $ D$ is on small arc $ AB$
ii) $ AD \equal{} 15$
iii) $ BD \equal{} 7$

If you think about it, each condition on it's own is a circle or a part of a circle. It is not obvious that these three circles should have a common point.

In my opinion, a full solution would show that the diagram is even possible.


$ a,b,c> 0$,and $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \equal{} 1$,
Prove that
$ \frac {1}{\sqrt {a^2 \plus{} b^3}} \plus{} \frac {1}{\sqrt {b^2 \plus{} c^3}} \plus{} \frac {1}{\sqrt {c^2 \plus{} a^3}} \plus{} \frac {1}{\sqrt {a^2 \plus{} c^3}} \plus{} \frac {1}{\sqrt {b^2 \plus{} a^3}} \plus{} \frac {1}{\sqrt {c^2 \plus{} b^3}} \le 1$

I've no idea.Please help me to prove it. Thank you.

There is another hard inequality,Please prove it,too:
$ a,b,c > 0$,and,$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$
Prove that
$ \frac {a}{ab \plus{} 3} \plus{} \frac {b}{bc \plus{} 3} \plus{} \frac {c}{ca \plus{} 3} \le \frac{3}{4}$

Thank you.

[quote="zjlywjh001"]There is another hard inequality,Please prove it,too:
$ a,b,c > 0$,and,$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$
Prove that
$ \frac {a}{ab \plus{} 3} \plus{} \frac {b}{bc \plus{} 3} \plus{} \frac {c}{ca \plus{} 3} \le \frac {3}{4}$

Thank you.[/quote]
Use $ (a^2\plus{}b^2\plus{}c^2)^2\geq3(a^3c\plus{}b^3a\plus{}c^3b).$ :wink:

zjlywjh001, I have mistake in my proof.  :(  I am sorry.

Do you think the two probelms are a bit hard,aren't they?

[quote="tronghieu"]See here: http://www.mathlinks.ro/viewtopic.php?t=274135[/quote]

Very good,but who can solve my first problem?

I've solve the secend problem within two lines by using Vasc's inequality.
Now,Who can solve my first problem?


I looked around a bit and I didn't see any real logic to the forum numbering system? Like some related forums are in a row but others are kind of random. Between forums 300-320, there are only 6 actual forums and the rest are unused, and they range from classroom math to canada. So given the spotty forum numbering system and the already amusing phone number for mathcamp, was forum number 314 a request or just a coincidence?

Actually, there most likely are a full 21 forums in that range. It's just that the others are hidden for various reasons, most likely that they're linked to a class from around that time.

This whole site is on math, I don't think Mathcamp has any more influence/say than anyone else.

I would say this is heavy proof Mathcamp has been Chosen for some higher purpose, but the people in forum 3141 would laugh at me in 10 years.

The owners of forums 3141 and 3142 can fight that out amongst themselves.

Besides, 3.1415 only rounds to 3.142 by convention. There's something called "banker's rounding," where .5s round to the nearest even value (to avoid the upward bias created by conventional rounding), and you could just as easily make a "round to odd" system.

[quote="Darmani"]
Besides, 3.1415 only rounds to 3.142 by convention. There's something called "banker's rounding," where .5s round to the nearest even value (to avoid the upward bias created by conventional rounding), and you could just as easily make a "round to odd" system.[/quote]
However, 3.1415[i]9[/i]... rounds to 3.142 under any convention :wink: (and it rounds to 3.1416, not 3.1415)

Yeah, I was just about to come and post a retraction, but...too late.


I've been doing the problems from the South African Mathematics Olympiad, and was wondering, what's considered a good score on the 2nd round of the senior division?  Like what would normally be the average score, and what score would you normally need to qualify for the 3rd round.  I couldn't find any of this information on their website.

Do you mean the South African Mathematical Olympiad?

I don't really have an idea about the qualifying score for the final round.

Strange, since I should actually know - I even proposed a problem for the final round...

Yea, South African  :blush:

Sorry about that.

The best 100 in the country qualify for 3rd round. I don't know the average score, but this year anybody with less than around 75% didn't qualify.


geometry
geometric transformation
reflection
rhombus

Given a rectangular pool table and a cue ball of unsubstantial radius in the middle, construct a path such that it hits cushion AB, BC, then CD, such that A,B,C,D are the vertices of the table and AB>BC, then passes over where it started.

are we given the dimensions of the table?

See the attached diagram. We consecutively reflect $ X$ over $ AB$, $ BC$ and $ CD$ to get $ X_{3}$, and then we easily identify contact points as the respective intersections. The proof of the construction, and the proof of the fact that the path is actually a rhombus, I'll leave to someone else as an exercise.


geometry
circumcircle
geometric transformation
reflection

The triangle ABC has CA = CB. P is a point on the circumcircle between A and B (and on the opposite side of the line AB to C). D is the foot of the perpendicular from C to PB. Show that PA + PB = 2PD.

Solution

Take Q on the ray PB so that PD = DQ. Then CA = CB, CP = CQ, ∠CAP = ∠CBQ, so triangles CAP and CBQ are congruent, so AP = BQ. Hence 2PD = PQ = PB + PA.

By the solution it seems to be an easy problem,but I cant understand where point P is. Could you please send a figure, including the helping construction?

What exactly don't you understand? P is just an arbitrary point on the arc AB of the circumcircle of triangle ABC. Maybe, you don't understand how to construct Q ? The point Q is the reflection of the point P in the point D.

But there is an incorrect step in the solution:

[quote="Iris Aliaj"]Then CA = CB, CP = CQ, ∠CAP = ∠CBQ, so triangles CAP and CBQ are congruent,[/quote]

In fact, if two triangles XYZ and X'Y'Z' satisfy XY = X'Y', XZ = X'Z' and < XYZ = < X'Y'Z', then in general you cannot conclude that the triangles XYZ and X'Y'Z' are congruent; you can conclude it only if you know that XZ > XY.

Here is a better proof of the congruence of triangles CAP and CBQ: We have CA = CB and CP = CQ.

In the isosceles triangle PCQ, we have < PCQ = 180 - 2 < CPQ = 180 - 2 < CPB = 180 - 2 < CAB (cyclic). Now, < CAB = A, and since triangle ABC is isosceles, A = B. Thus, < PCQ = 180 - 2 A = 180 - (A + B) = C = < ACB. Thus,

    < BCQ = < PCQ - < PCB = < ACB - < PCB = < ACP.

Therefore, we have CA = CB, CP = CQ and < ACP = < BCQ. This allows us directly to infer the congruence of triangles CAP and CBQ.

  Darij


calculus
integration
trigonometry
logarithms
calculus computations


[quote="mathgeek2006"][quote="math92"]After 3 games Pete is 4 for 12. Give a realistic ratio(number of hits: times at bat) in the fourth game, assuming that he raises his average to .500.[/quote]
What exactly does "realistic" mean?[/quote]

Haha, realistic as in not over 10 at bats.

[quote="DarkKnight"][quote="math92"]After 3 games Pete is 4 for 12. Give a realistic ratio(number of hits: times at bat) in the fourth game, assuming that he raises his average to .500.[/quote]

[hide]
He could go 2 for 3 or 4 for 6.
[/hide][/quote]

howd u get that?