MIT
college
Harvard
MATHCOUNTS
Support
geometry
geometric transformation

Does anyone else here ever get the idea that a large portion of adults' college talk is a mere ploy? I don't mean to say that college lacks importance or that early planning is superfluous. What I mean is this:

From an early age, they ask us what college we want to go to. They are pleased if we answer with "MIT" or "Caltech," but they never tell us exactly what college is, other than that it is a place where you go to "further your education."

Then, they goad us to participate in many activities, solely for the purpose of college applications. Sports and participation in student government are necessities that must not be forgotten. Previously-mandatory science fair projects continue to be mandatory in high school.

In our freshman year, they tell us to go take the SAT. They are not pleased with a 2200 score. Slowly, college applications become the reason behind mostly everything we do. We hardly ever do things for fun anymore.

Any progress we make is downplayed and merely increases their expectations. They come to expect "A" grades on every test and in every last AP class. Heck, they even expect us to get perfect/near-perfect PSAT and SAT scores. They never took as many AP classes and have no clue as to the format of the SAT, aside from the 2400 maximum score.

"Do you want to go to _____ (insert name of "bad" college)?"

"You don't have to do anything in life. You don't have to go to college or get a job."

The worst part? They went to that so-called "bad" college. Their jobs earn six-figure salaries.

They attempt to motivate us by telling us that there are kids in India and China who forgo lunch in order to afford school supplies. We will be competing with them for college entrance. Unsurprisingly, it does not make us feel any better.

They regard all of our classmates as competitors. They are "cohorts," not "friends." They ceaselessly seek ways to prove that we are superior to them.

Aside from our muddled brains (late-night studying for AP classes) and our broken resolve (darned SAT), it seems that we don't have much left. We learn to do what we do tolerably well, but is it worth doing?

So in conclusion, it seems to me that lots of this hype is phony. Phony like Holden Caulfield would say. I might be disgruntled and disillusioned, but I can't help to think that I am at least partly correct.

I think a lot of people share your opinion and experience, I certainly know at least a couple people that went through this in high school. But you should really consider the alternative, parents that care too much vs parents that care too little.

Not to turn this into a sob story, but my parents on the other hand dont know my sat scores, or even what a good sat score would be. I had to fill out my college apps, sat and gre registrations, and drive myself to most of these. I rather liked the freedom and responsibility of choosing my own fate, but frankly, if I had someone pushing me to do better, I would have probably been much better off.

My best advice if this is troubling you is to find your own reason to go to college.

No offense mark, and I feel sorry for you, but I know for a fact your situation isn't anywhere near typical. If I had to make a guess, it would be about 3 kids in my grade took the SATs in freshman year. Most don't take AP classes, and most would think there was a mistake on the grading if they got a 2200.

But I do agree. Throughout all of high school, parents and teachers say that you have to do this or that to get into a  good college. I don't personally see the reason to get all hyped up about it. People can get great jobs as long as they are intelligent and handy, and the difference between Harvard and a slightly worse college like Caltech won't make much of a difference. You're only in highschool once, and it's better in my opinion to have fun than to work yourself out just to get to a slightly better college.

Plus, (here's something that lots of people miss) even if going to Harvard could make your salary 10,000 dollars higher, who cares? Last time I checked, money doesn't define happiness, and if you think it does, then you aren't happy.

I just noticed that you're from hawaii. I have two friends from college that were from hawaii, one is now in med school, the other is a swim instructor and they're both pretty relaxed people. Not to generalize or anything, but given the hawaiian atmosphere, you should already know how to wind down and relax, probably just need to find some time to do so.

With me, those activities that we are "goaded to participate in" are actually done because I enjoy them, e.g. math. I don't do math solely because I was like "Oh hey if I study for MATHCOUNTS I might get into a good college."

[b]@ shallow:[/b] I find it somewhat ironic that someone with your username would make such a comment about Hawaiians.  :roll:

[b]@ OP:[/b] It's probably a combination of your parents:
(1) Having good intentions/realizing your potential but not knowing how to motivate you or direct their enthusiasm
(2) Living vicariously through your academic success

You should try to determine which things your parents say that belong into which category, and to what extent. Try to do so as impartially as possible (rather than out of convenience or confirmation bias), and interpret what they say accordingly.

You'll find that while your parent's attitude is terrible (especially regarding your friends), some of the things they suggest are worth considering. We can stipulate a few things:

(1) Some schools suit your needs better than other schools. The higher your SAT scores are, the more likely you are to gain admission to a school that's good for you (all else equal), because you're more likely to get admitted into [i]any[/i] school. The same argument applies to grades and quality/variety of EC's. Sometimes, spending time on things that you aren't interested in (like studying for the SAT's) is worth the cost if the benefit is that it will help you go to a place where you can further your interests.

(2) At the same time, you should focus on identifying and developing your interests. Taking part in a wide variety of experiences can help you identify your interests, directly or indirectly. You may be surprised! At the same time, if you really dislike something, it's probably not worth spending your time on, because you could be spending that time doing something else of interest.

(3) It takes people a while to figure out what college they want to attend. MIT and Caltech may be obvious choices for people of certain interests, but it's worth taking the time to explore those (and other) colleges in more depth. This is one of the advantages of starting to look at colleges early. Chances are that in the end, your top choice will be an upper tier school anyway (perhaps even MIT or Caltech), but you'll have a better idea of why you want to go there (this will also be apparent in your overall application and your interview, which would increase your chance of ultimately gaining admission). Also, because [i]noone[/i] can count on getting admitted to a particular school, it's good to have well thought-out alternate choices.

[quote="TZF"]Living vicariously through your academic success [/quote]

That's a biggie.  Of course, they're your parents and they want to make sure you're successful because you're probably their pride and joy, and also because they want to make sure they can live easily after a few years of tough parenting.  However, another comment I wanted to elaborate on was...
[quote="Ihatepie"]difference between Harvard and a slightly worse college like Caltech[/quote]
Not endorsing any one college over another, but this statement supports the fact that the applicant is the one who matters.  The college are there to give you options.  One applicant could easily say that Caltech is better than Harvard and viceversa.  That means you can make the same argument with, say MIT and a relatively unknown school that may have the same resources in your future field.  It just depends on what you make with what you have.
Make sure you like whatever your parents tell you to do, whether it be going to a college, etc.  Give yourself some good reasons, then check out the campus.  My math teacher has a story where some graduate from my high school went to UPenn, without taking a campus visit, and hated the weather there as soon as the person got there.  You can imagine the agony, four years of living somewhere you hate, then finally getting out of there.  Just make sure you make the right decisions, not your parents.

Sometimes I have trouble reminding my parents that I'm the high school senior and that they're not.

On a related note:
A 2200 SAT in freshman year and your parents are angry and you live in Hawaii?  I think most Hawaiians are pretty laid-back people.  But parents have the drive to make the lives of their children better than their lives.

My sister graduated from Caltech and is now working on cancer research. To me, it seems strange that she was accepted because she only took one AP class (chem; she got a 3 on the AP exam). She didn't participate in any sports or the student government.

Unrelated: Hawaii really isn't as laid back as most people think.

My dad went to the local college/medical school and is now a radiologist. I asked him why I would need to go to MIT to become a radiologist like him. I could easily go to the same college he did and get the same job, which is more stable and pays better than engineering, astronomy, physics, etc. (if you believe online government job statistics). I can't remember exactly what he said in response.

Find your own reasons for wanting to go to a high-tier college.

You're on this site. This probably means you like talking to intelligent people. High-tier colleges have higher percentages of intelligent people to collaborate with. Odds are you'd be bored and unhappy in the local college.

Yes, parents are sometimes annoying, hypocritical, illogical, phony, and generally full of ____.

That said, try to cut them some slack.  The fact is that they care about you and want you to succeed, and in the grand scheme of high school life, getting into a good college is probably the most obvious step you can take toward a "successful" life (whatever that means).  Of course, it is true that many parents want to send their kids to ivys because it serves *their* notions of what it means to be a "successful" parent.  (Even the most noble of parents probably can't help but feel smug at having children who attend Harvard.)  But in general, you should give them the benefit of the doubt regarding their intentions.

You should also take a moment to reflect on your privileged existence.  To paraphrase what shallow was saying, maybe your parents are annoying, but would you rather have parents who are hell-bent on SAT prep and ivy league, or parents who have low expectations and dare not dream of sending their children to the ivy league?  I assure you that the majority of American parents fall in the latter category.

You also mentioned that your parents are doing just fine with their no-name degrees and yet demand that you get a fancy degree.  I should point out that many parents who have no-name degrees and are successful DO take that lesson to heart and *refuse* to shell out the money for private colleges, essentially denying that option to their children.  Would you prefer that your parents had this attitude?

I don't mean to be overly harsh, because I do sympathize with you and totally disagree with your parents' tactics.  But try to put it in perspective.  Besides, in a couple years (if your parents have their way) you'll be in college a thousand miles away and you'll be free.

I'd like to comment and say that it isn't unfair for parents to have expectations of their children, especially if they put in so much effort into raising them. While it's not right for them to dictate your life or force you down an undesirable path, there's nothing wrong with them encouraging you in any particular direction (so long as that direction is ethical and perceived to be in your best interest).

i agree that parents can be a little too serious about college and how it is the center of a highschooler's existence. my parents give some pressure, but i know kids who are in worse families that actually beat them if they get B+'s. i find though that what i used to have to be forced to do i do well now, and i have my own reasons for doing them. because my parents motivated me to start certain things, i love doing those things a lot. so though they nag sometimes, i guess i have to be thankful.


In a closed container, 36.03 grams of Carbon combines with 5.05 grams of H2 to produce liquid hexane. At 38.000 degrees Celsius, the liquid hexane and hexane vapor reach equilibrium. A sample of the vapor has a density of 0.000803g/mL. Assuming that the liquid hexane has surface area of 1 square inch, calculate the average force (in pounds) acting upon one mol of liquid hexane. (Given: R = 62.364 Torr mol$^-1$ K$^-1$)

Can someone verify that this can actually be calculated?

It's possible to determine the number of moles (using limiting reactants), as well as the vapor pressure. However, is it legitimate to convert vapor pressure from Torr to Psi (pounds per square inch), then since the surface area is 1, to convert Psi to pounds, then divide pounds by moles?

I don't need the answer, that's my homework, I'm just wondering if the question makes sense.

Pressure is all you need, number of moles and masses are here just to confuse you. Seems like they worked ;)


inequalities
trigonometry
inequalities open

$ For\; xy \plus{} yz \plus{} zx \equal{} 1\; and\; x,y,z > 0,\; prove\; \frac {x}{\sqrt {x^{2} \plus{} 1}} \plus{} \frac {y}{\sqrt {y^{2} \plus{} 1}} \plus{} \frac {z}{\sqrt {z^{2} \plus{} 1}}\leq \frac {3}{2}$

Thanks.

Don't know if this is any help but:

$ (x-1)^{2} \ge 0 \Rightarrow\, x^{2} + 1 \ge 2x$

$ \Rightarrow\, \frac{x}{\sqrt{x^{2}+1}} + \frac{y}{\sqrt{y^{2}+1}} + \frac{z}{\sqrt{z^{2}+1}}$

$ \le \sqrt{\frac{x}{2}} + \sqrt{\frac{y}{2}} + \sqrt{\frac{z}{2}}$

${ \le \frac{3}2}(x+y+z)$, by Cauchy-Schwarz with $ (\sqrt{x},\sqrt{y},\sqrt{z})$ and $ (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$.

This would do it if we can show that $ x+y+z \le 1$ for $ xy+yz+zx \le 1$ which I can't see.  Hope this is useful.

[quote="AndrewTom"]Don't know if this is any help but:

$ (x - 1)^{2} \ge 0 \Rightarrow\, x^{2} + 1 \ge 2x$

$ \Rightarrow\, \frac {x}{\sqrt {x^{2} + 1}} + \frac {y}{\sqrt {y^{2} + 1}} + \frac {z}{\sqrt {z^{2} + 1}}$

$ \le \sqrt {\frac {x}{2}} + \sqrt {\frac {y}{2}} + \sqrt {\frac {z}{2}}$

${ \le \frac {3}2}(x + y + z)$, by Cauchy-Schwarz with $ (\sqrt {x},\sqrt {y},\sqrt {z})$ and $ (\frac {1}{\sqrt {2}},\frac {1}{\sqrt {2}}, \frac {1}{\sqrt {2}})$.

This would do it if we can show that $ x + y + z \le 1$ for $ xy + yz + zx \le 1$ which I can't see.  Hope this is useful.[/quote]

The last inequality isn't correct. Try $ x = y = 1, z = 0$
Here is my solution:

Since $ xy + yz + zx = 1$ we have
$ \frac{x}{\sqrt{x^2 + 1}} = \frac{x}{\sqrt{(x + y)(x + z)}} \le \frac{2x}{2x + y + z} \le \frac{1}{2}(\frac{x}{x + y} + \frac{x}{x + z})$

Similarly, we have
$ \frac{y}{\sqrt{y^2 + 1}} \le \frac{1}{2}(\frac{y}{y + z} + \frac{y}{y + x})$
$ \frac{z}{\sqrt{z^2 + 1}} \le \frac{1}{2}(\frac{z}{z + x} + \frac{z}{z + y})$

Adding three inequalities above completes our proof./.

Nice solution, II93110. (I put the $ \frac{3}{2}$ on the wrong side, so it should have been $ \frac{2}{3}$ and then I should have started again.)

Thanks a lot.

Just one thing.
$ \frac{x}{\sqrt{(x\plus{}y)(x\plus{}z)}} \leq \frac{1}{2}\left( \frac{x}{x\plus{}y}\plus{}\frac{x}{x\plus{}z}  \right )$
is true but
$ \frac{x}{\sqrt{(x\plus{}y)(x\plus{}z)}}\le\frac{2x}{2x\plus{}y\plus{}z}$
is not true. (direction is reversed.)

So we can just erase that step.

if we put x=1/a , y=1/b.... we get the inequality
1/√1+a^2+.....≤3/2 with abc=a+b+c so we can put a=tanA.... with A+B+C=180°
the inequality becomes cosA+cosB...≤3/2 wich is well known [/list][/code]

just put $ x\equal{}\tan \frac{A}{2}$ cyclically, where $ A,B,C<\frac{\pi}{2}$ and $ A\plus{}B\plus{}C\equal{}\pi$ and use jensen or sth..

[quote="calm"]Thanks a lot.

Just one thing.
$ \frac {x}{\sqrt {(x \plus{} y)(x \plus{} z)}} \leq \frac {1}{2}\left( \frac {x}{x \plus{} y} \plus{} \frac {x}{x \plus{} z} \right )$
is true but
$ \frac {x}{\sqrt {(x \plus{} y)(x \plus{} z)}}\le\frac {2x}{2x \plus{} y \plus{} z}$
is not true. (direction is reversed.)

So we can just erase that step.[/quote]
Forget about that step, $ \frac{x}{\sqrt{(x\plus{}y)(x\plus{}z)}}\leq\frac{1}{2}\left(\frac{x}{x\plus{}y}\plus{}\frac{x}{x\plus{}z}\right )$  is direct consequence of AM-GM.


The numbers $ a_n$ are such that :

$ \frac{a_n}{a_{n\minus{}1}} \equal{} \frac{(n\minus{}1)(2n\minus{}1)}{(n\plus{}2)(2n\minus{}3)}$

Find an expression for $ \frac{a_n}{a_1}$ and hence, or otherwise, evaluate $ \sum_{n\equal{}1}^{\infty}a_n$ when $ a_1 \equal{} \frac{2}{9}$ .

$ \frac {a_n}{a_{n \minus{} 1}} \equal{} \frac {(n \minus{} 1)(2n \minus{} 1)}{(n \plus{} 2)(2n \minus{} 3)}$

$ \Longleftrightarrow  \frac {(n \plus{} 2)(n \plus{} 1)na_n}{2n \minus{} 1} \equal{} \frac {(n \plus{} 1)n(n \minus{} 1)a_{n \minus{} 1}}{2n \minus{} 3}$

$ \equal{} \cdots \equal{} \frac {(1 \plus{} 2)(1 \plus{} 1)\cdot 1\cdot a_1}{2\cdot 1 \minus{} 1} \equal{} 6a_1$, yielding

$ \frac {a_n}{a_1} \equal{} \frac {6(2n \minus{} 1)}{n(n \plus{} 1)(n \plus{} 2)}\ (n\geq 1).$

The rest is left for you.

As kunny showed, we have $ \frac {a_n}{a_1} \equal{} \frac {6(2n \minus{} 1)}{n(n \plus{} 1)(n \plus{} 2)}\ (n\geq 1).$
So,
$ \frac {a_n}{a_1} \equal{} \frac 6 {n \plus{} 2} \left( \frac 2 {n \plus{} 1} \minus{} \frac 1 n \right)$

$ \equal{} \frac {12}{(n \plus{} 1)(n \plus{} 2)} \minus{} \frac 6 {(n \plus{} 2)n}$

$ \equal{} \frac {12}{n \plus{} 1} \minus{} \frac {12}{n \plus{} 2} \minus{} \frac 3 n \plus{} \frac 3 {n \plus{} 2}$

$ \implies a_n \equal{} 3a_1 \left( \frac 4{n \plus{} 1} \minus{} \frac 3{n \plus{} 2} \minus{} \frac 1 n \right)$

So, $ \sum_{n \equal{} 1}^{\infty}a_n \equal{} \frac 1 3$


I am currently reading "Common Sense on Mutual Funds" by John Bogle and I think it is an interesting book, relatively simply written even though there are some things that I dont understand.

Ok, several things I wanted to discuss in this first post. First of all, Bogle suggests long term investing, do you agree, I too see this as superior. But in this case, isnt what people like traders and QA's do quite meaningless, because they are not the ones who create the earnings of the company, I see Bogle as critiqiuing all of this. Of course, he critiques mutual fund managers (and the industry) because he says that they have high costs and it is better to simply invest in one of the market index which is both simpler and brings on average better returns. I see this as interesting but was wondering about your opinions.

Also, some things I just did not completely understand. I didnt really understand the part where Bogle explains that there are four ways to look at the costs of mutual funds. I am refering to annual cost as percentage of assets managed, annual costs as percentage of totoal equity, etc, etc. Does this mean that different funds charge in these four different ways, because otherwise I dont understand how you can have a different perspective, because it will give completely different results, which makes it seem impossible that you can simply have different views of an absolute truth. So how does this work.

I have other questions that I will ask later because this post will be too big to post everything an once. Thanks!

Anyways, does anyone have any opinions on all this, I am still not done the book but Bogle is critiquing mutual funds that are actively managed very heavily and believes in the index. It seems to be the repetitive idea of the book. Has anyone read this book?


Let $ ABCD$ be a quadrangle with $ \angle A\equal{}\angle B\equal{}\angle C$
$ I;J$ are center of incircle of triangles $ ABC;ACD$,respectively.
Prove that $ IJ$ through center of circumcircle of $ ABCD$

[quote="chien than"][color=brown]Let $ ABCD$ be a quadrangle with $ \angle A = \angle B = \angle C$
$ I;J$ are center of incircle of triangles $ ABC;ACD$,respectively.
Prove that $ IJ$ through center of circumcircle of $ ABCD$[/color][/quote]

[hide="proof"]$ \angle A = \angle B = \angle C = \alpha$ $ \implies$ $ 360 - 3\alpha = 180 - \alpha$(because $ ABCD$ is inscriptible) $ \implies$ $ \alpha = 90$ $ \implies$ $ ABCD$ rectangle.
$ \Delta ADC \equiv \Delta ABC$ from where $ r_{ADC} = r_{ABC} = r$.
 $ \{\begin{array}{cccc} O\in IJ \cap AC,\angle JOM = \angle NOI \\
 \\
JM \bot AC,M\in AC, JM = r \\
 \\
IN \bot AC,N\in AC, IN = r \end{array}$ $ \|\begin{array}{cccc} \\
 \\
 \\
 \\
 \\
\end{array}$ $ \implies$ $ \Delta ION \equiv \Delta JOM$ $ \implies$ $ NO = OM$

$ \{\begin{array}{cccc} IP \bot AD,P\in AD \\
 \\
JQ \bot BC,Q\in BC \end{array}$ $ \implies$ $ PD = BQ$ $ \implies$ $ PA = QC$ $ \implies$ $ AN = MC$.But $ NO = MO$ therefore $ O$ is the middle of $ AC$,the center of the circumcircle of $ ABCD$.[/hide]


geometry
ratio
number theory
relatively prime

A chord of a circle is perpendicular to a radius at the midpoint of the radius. The ratio of the area of the larger of the two regions into which the chord divides the circle to the smaller can be expressed in the form $\frac{a\pi+b\sqrt{c}}{d\pi-e\sqrt{f}}$, where $a$, $b$, $c$, $d$, $e$, and $f$ are positive integers, $a$ and $e$ are relatively prime, and neither $c$ nor $f$ is divisible by the square of any prime. Find the remainder when the product $abcdef$ is divided by 1000.

[hide="Answer"]Drawing radii to the endpoints of the chord and labeling those as $OA$ and $OB$, we have two 30-60-90 triangles, with arc $AB$ having a measurement of $120^\circ$. Therefore, the larger region has area $\frac{2}{3}\pi r^2+\frac{r^2\sqrt{3}}{4}$, while the smaller has a measurement of $\frac{1}{3}\pi r^2-\frac{r^2\sqrt{3}}{4}$. Factoring out an $r^2$ that will cancel in the ratio, we are left with $\frac{\frac{2}{3}\pi +\frac{1}{4}\sqrt{3}}{\frac{1}{3}\pi-\frac{1}{4}\sqrt{3}}=\frac{8\pi+3\sqrt{3}}{4\pi-3\sqrt{3}}$, and we have $abcdef=8(3)(3)(4)(3)(3)\equiv \boxed{592}\bmod{1000}$.[/hide]


logarithms
LaTeX
calculus
integration
ratio
number theory
prime factorization

For any positive $ n$, let
\[ f(n)\equal{}\begin{cases}\log_{8}n\text{,}&\text{if}\ \log_{8}n \ \text{is rational,} \\0 \text{,}&\text{otherwise.}\end{cases}
\]
What is $ \displaystyle \sum_{n\equal{}1}^{1997}f(n)$?

$ \textbf{(A)}\ \log_{8}2047 \qquad
\textbf{(B)}\ 6 \qquad
\textbf{(C)}\ \frac{55}{3} \qquad
\textbf{(D)}\ \frac{58}{3} \qquad
\textbf{(E)}\ 585$

Thanks in advance!

[hide]$ \log_{8}n$ is rational only if n is a power of 2.  The summation includes the powers of 2 from 2 to 1024.  $ \log_{8}{2^1 2^2 2^3\ldots 2^8 2^9 2^{10}} \equal{} \log_{8}{2^{55}} \equal{} 55(\log_{8}{2}) \equal{} 55*\frac {1}{3} \equal{} \frac {55}{3}$

The answer is (c)[/hide]

Edit: Latex mistakes fixed

[hide]$ \log_{8}n \equal{} \frac {1}{3}\log_{2}n$ is rational iff $ \log_{2}n$ is rational.
If $ n$ is an integer, then $ \log_{2}n$ is rational iff $ n \equal{} 2^k$ for some integer $ k$.
For $ n \equal{} \overline{1,1997}$, $ n \equal{} 2^{0},2^{1},\ldots,2^{10}$ are the such values for $ n$.
Since $ f(n) \equal{} 0$ for all other values of $ n$:

$ \displaystyle \sum_{n \equal{} 1}^{1997}f(n) \equal{} f(2^0) \plus{} f(2^1) \plus{} \ldots \plus{} f(2^{10})$

$ \equal{} \log_{8}(2^0) \plus{} \log_{8}(2^1) \plus{} \ldots \plus{} \log_{8}(2^{10})$

$ \equal{} \frac {1}{3}\log_{2}(2^{0 \plus{} 1 \plus{} \ldots \plus{} 11})$

$ \equal{} \frac {1}{3}(0 \plus{} 1 \plus{} \ldots \plus{} 10)$

$ \equal{} \frac {55}{3}\Rightarrow C$[/hide]

[quote="314.math"]For any positive $ n$, let
\[ f(n) \equal{} \begin{cases}\log_{8}n\text{,} & \text{if}\ \log_{8}n \ \text{is rational,} \\
0 \text{,} & \text{otherwise.}\end{cases}
\]
What is $ \displaystyle \sum_{n \equal{} 1}^{1997}f(n)$?

$ \textbf{(A)}\ \log_{8}2047 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ \frac {55}{3} \qquad \textbf{(D)}\ \frac {58}{3} \qquad \textbf{(E)}\ 585$

Thanks in advance![/quote]

[hide]We only care about numbers $ n$ of the form $ \log_{8} n \equal{} p/q$, where $ n,p,q$ are integers. This means $ n \equal{} 8^{p/q}$, so $ n$ is $ 8$ to the power of a rational. Since $ n$ is an integer, we need to consider which integers are powers of 8 to a rational.

Notice that if $ n$ is not a power of 2, then it's of the form $ n \equal{} 2^{r} \cdot s$, where $ s$ is odd. But this means $ \log_8 {n} \equal{} \log_8 (2^{r} \cdot s) \equal{} \log_8 2^{r} \plus{} \log_8 s \equal{} \log_8 8^{r/3} \plus{} \log_8 s \equal{} r/3 \plus{} \log_8 s$. But since $ s$ is odd, $ \log_8 s$ can't be rational (why? well, if it were, $ 8^{p/q} \equal{} s$, meaning $ 8^{p} \equal{} s^{q}$, so $ 2^{3p} \equal{} s^{q}$, but $ s^{q}$ is not even, while $ 2^{3p}$ is).

So $ n$ must be a power of 2. Now we simply compute:
$ \log_8 2 \plus{} \log_8 2^{2} \plus{} \log _8 2^{3} \plus{} \cdots \plus{} \log_8 2^{10} \equal{} \log_8 2^{1 \plus{} 2 \plus{} \cdot \plus{} 10} \equal{} \log_8 2^{55} \equal{} \log_8 8^{55/3} \equal{} 55/3 \implies \textbf{C}$[/hide]

[quote="314.math"][quote="Quickster94"]$ \log_{8}n$ is rational only if n is a power of 2.[/quote]

Why is $ \log_{8}n$ rational only if $ n$ is a power of $ 2$?  Why isn't it when $ n$ is a power of $ 8$?  Thanks![/quote]

Powers of 8 are powers of 2... $ 8 \equal{} 2^{3}$, $ 8^{2} \equal{} 2^{6}$, etc...

EDIT: You can read my post to see why $ \log_{8} n$ is rational iff $ n$ is a power of $ 2$. Try to generalize this for any base.

[quote="314.math"][quote="Quickster94"]$ \log_{8}n$ is rational only if n is a power of 2.[/quote]

Why is $ \log_{8}n$ rational only if $ n$ is a power of $ 2$?  Why isn't it when $ n$ is a power of $ 8$?  Thanks![/quote]

When n is a power of 8, we get integers, which is a more precise term...

for that matter why is $ log_2{n}$ rational iff n is a power of 2, it is $ integral$ iff n is a power of 2. anyone care to prove it is rational iff $ n \equal{} 2^k$? Like couldnt $ log_2{5}$ for example be rational (expressable as the ratio of 2 integers) although it is not integral?

If $ \log_2 n \equal{} \frac{p}{q}$ then $ n^q \equal{} 2^p$ so the only prime appearing in the prime factorization of $ n$ is $ 2$.

[hide="Initial Thoughts"]We see that if $ \log_{8}n$ is irrational then we don't have to include it in the sum.  So we are really adding all $ \log_{8}n$ such that it is rational and $ n\leq1997$[/hide]
[hide="Solution"]We can pull a 1/3 out of $ \log_{8}n$ to get $ \frac{1}{3}\log_{2}n$.
All the possible n are in the form $ 2^{(p/q)}\equal{}n$ but because n is an integer between 1 and 1997, $ q\equal{}1$ and $ p\equal{}0,1,2,...,10$
Taking the sum of these and dividing by 3 you get
$ \frac{0\plus{}1\plus{}...\plus{}10}{3}\equal{}\frac{55}{3}\implies\boxed{C}$[/hide]


[hide]Just find the length of the longest diagonal $\sqrt{a^2 + b^2 +c^2}  = \sqrt{2^2 + 3^2 +6^2} = \sqrt{49} = (A) 7 ft$[/hide]

First, use the pythagorean theorem for the triangle with the sides 6 and 3, so you get the hypotenuse as the  :sqrt: 45. Then use the pythagorean theorem again to find the diagonal of the triangle of sides 2 and  :sqrt: 45, so the hypotenuse would be the  :sqrt: 49, or [b]7[/b].

[hide]Use the distance formula, except in three dimensions.
$\sqrt{4+9+36}=\sqrt{49}=7$[/hide]

[hide]because right triangles are everyone's best friend, look for one from one point to the opposite. One leg (in my case) is 2. The other is 3root5 (3^2 + 6^2, square root of that.) Then i use the pythagorean theorem with 2 and 3root5 as the legs of my triangle. Then you get 7. (A)[/hide]

If it flies from one corner to the other than just find the space diagnol (2^2)+(3^2)+(6^2)=49 and square root 7(A)

[hide]This problem requires us to use the pythageoran therom twice. First let us use the pythageoran theorem to find a diagonal length of the box's base.

3^2+6^2=45, thus the diagonal is sqrt(45).

Using that base diagonal, let us now use the pythag theorem to the hypotinuse of the triangle made by the diagonal found on the base, the altitudal corner line, and the fly's path(hypotinuse).

2^2+sqrt(45)^2=49, thus the diagonal is sqrt(49), or 7.

The fly must go 7 feet to reach the opposite corner.[/hide]

In actualy, wouldn't the answer be dependent on the size of the fly?

[i]In actualy, wouldn't the answer be dependent on the size of the fly?[/i]

No, because no matter how big the fly is, it still would be traveling that much of a distance. The only difference is how fast it will get there.  :P

I have a follow-up question to this -

What is the shortest distance an ant would have to crawl to go from one corner to the opposite one?  And it can crawl up a wall.

[hide="hint"]
Try unfolding the box into a two-dimensional figure[/hide]

[hide="My Answer"]
$\sqrt{66}$

I basically just used the hint, and found a 6 by 5 rectangle that goes from one corner to another, when the whole figure is folded up into a box.  You just use pythagorean theorom to find the diagnol, which is $\sqrt{66}$[/hide]

[quote]In actualy, wouldn't the answer be dependent on the size of the fly?[/quote]
we assume the fly is like a particle (no mass or volume)

[quote="MCrawford"]A fly is inside an empty, closed, rectangular box that measures two feet by three feet by six feet.  If he flies from one corner to the opposite corner, what is the minimum distance he must travel?
\[ \text{(A)}\ 7 \text{ft} \qquad \text{(B)}\ 8 \text{ft} \qquad \text{(C)}\ 9 \text{ft} \qquad \text{(D)}\ 10 \text{ft} \qquad \text{(E)}\ 11 \text{ft}  \][/quote]

[hide]the distence is just the square root of the sum of the squares.

$\sqrt{2^2+3^2+6+2}=\sqrt{49}=7$[/hide]

[quote="michaeln"]Indeed, but if the fly is a big fat round fly with a 1ft. diameter, it could start with its posterior end touching one corner and end with its anterior end touching the opposite corner. That way, would the fly not travel only 6ft?  :P[/quote]

Then it wouldn't be able to touch a corner.

You lose.


algebra
polynomial
superior algebra
superior algebra unsolved

Let A be a commutative ring with 1 and let A[x] be the ring of polynomials in a indeterminate x, with coefficients in A. Let $ f\equal{}a_{0}\plus{}a_{1}x\plus{}...\plus{}a_{n}x^{n} \in A[x]$.
 $ f$ is said to be primitive if $ (a_{0},....a_{n})\equal{}A[x]$. Prove that if $ f,g \in A[x]$then
$ fg$ is primitive if and only if  $ f$ and $ g$ are primitive

HINT Replace "prime" by "maximal ideal" in the proof of Gauss's Lemma for $ \mathbb{Z}[x]$.


pigeonhole principle
number theory solved
number theory

The integers a1,...,an give at least k+1 different remainders when they are divided by n+k. Prove that some of them add up to an integer divisible by n+k.

WLOG, suppose a_1, a_2, ..., a_k+1 are all distinct mod n+k. Consider the following n+k numbers:

Type 1 (2k+1 numbers):
a_1
a_2
a_1 + a_2
a_1 + a_3
a_1 + a_2 + a_3
a_1 + a_2 + a_3 + a_4
...
a_1 + ... + a_k-1
a_1 + ... + a_k-1 + a_k
a_1 + ... + a_k-1 + a_k+1
a_1 + ... + a_k-1 + a_k + a_k+1

Type 2 (n-k-1 numbers):
a_1 + ... + a_k+2
a_1 + ... + a_k+3
...
a_1 + ... + a_n

If all of them are distinct mod n+k, then one of them = 0 (mod n+k) & we are done. Suppose not, then by Pigeonhole Principle, there exist two numbers x & y among the n+k of them s.t. x = y (mod n+k)

Note that x & y cannot have the same number of terms since otherwise x & y each has the same terms except for the last term, which are distinct mod n+k as a_1, a_2, ..., a_k+1 are all distinct mod n+k. WLOG, assume the number of terms in the sum for x is larger than that for y, then all the terms of y are also the terms of x, which implies that x-y is the sum of elements of a subset of {a_1, a_2, ..., a_n}.

Since x-y = 0 (mod n+k), we are done. :)

I don't understand: if x=a1+a3 and y=a2 then x-y is not a sum of ai (?)

WLOG, suppose a1, a2, ..., ak+1 are all distinct mod k.

Consider 3 groups of numbers:

Type 1 (k+1 numbers): a1, a2, ..., ak+1
Type 2 (k+1 numbers): S - a1, S - a2, ..., S - ak+1; where S = a1 + a2 + ... + ak+1
Type 3 (n-k) numbers: S, S + ak+2, S + ak+2 + ak+3, ..., S + ak+2 + ... + an

By Pigeonhole Principle, there exists 2 numbers x and y such that x = y (mod n+k).  Since a1, a2, ..., ak+1 are all distinct mod k, x and y cannot be both from Type 1 or both from Type 2.  If at least one of x or y is from Type 3, then we are done (assume all terms of x are terms of y, then y-x is a sum of some ai).

The only case left to consider is WLOG if x is from Type 1 and y is from Type 2.  If x = ap and y = S - aq where p \neq q then we are also done, by a similar argument.  Assume x = ap and y = S - ap.  Note that there are actually 3 such pairs (ap1, S - ap1), (ap2, S - ap2), (ap3, S - ap3) since we have n+k+2 numbers considered modulo n+k.

Then 2ap1 = S (mod n+k),
2ap2 = S (mod n+k),
2ap3 = S (mod n+k)

But this is clearly a contradiction as ap1, ap2, ap3 are distinct.  So x and y must fall into one of the categories mentioned above.

Actually dblues' argument is for a weaker version of the problem, where there are at least 2k different remainders mod n+k.  In that case direct construction would work.

Type 1 (3k numbers):
a1
a2
a1 + a2
a1 + a2 + a3
a1 + a2 + a4
a1 + a2 + a3 + a4
...
a1 + ... + a2k-2 + a2k-1
a1 + ... + a2k-2 + a2k
a1 + ... + a2k-2 + a2k-1 + a2k

Type 2 (n-2k numbers):

a1 + ... + a2k+1
a1 + ... + a2k+2
...
a1 + ... + an

I can understand the a1 + a3 mistake...I did exactly the same thing the first time I tried to solve the easier version of the problem during training. :blush:


AMC
AIME
floor function
geometry
rectangle
geometric transformation
rotation

Here is the second Mock AIME of they year!

The rules are:
1. No calculators
2. Answers are integers 0-999 inclusive
3. 3 hour time limit
4. Answers must be submitted within 15 days of 7/25/2006. (I will allow 15 extra minutes for the procrastinators :))

We are going by the honor system. When you have finished the test, send me your answers via private message. Please refrain from discussing the problems with other people until the test is over. The problems are posted below and they are also in PDF format.

For nice looking images, please use a $100\%$ viewing window. :yup:

If you don't have Adobe Acrobat, the problems are available [url=http://www.artofproblemsolving.com/Wiki/index.php/Mock_AIME_2_2006-2007/Problems]here[/url] on the AoPS Wiki.

For clarification on some problems,

-If $n$ is partitioned into integers $a,b,c,d,$ then $n=a+b+c+d.$

-$\lfloor ~ \rfloor$ refers to the greatest integer function. For example, $\lfloor 2.5\rfloor =2$ and $\lfloor \pi \rfloor =3.$

- Whenever it appears in a problem, $i=\sqrt{-1}.$

Hope I didn't miss anything. :)

Just a clarification, $O$ was the intersection of the diagonals of the rectangle on #12 right?

[hide]
on question 9, shouldn't the answer in question be negative?
[/hide]

[quote="junggi"]on #11, by sum of squares of the roots, do you mean the sum of squares of x y and z ?[/quote]Yeah. If the solutions are: $(x_{1},y_{1},z_{1}),$ and $(x_{2},y_{2},z_{2},)$ find $x_{1}^{2}+y_{1}^{2}+z_{1}^{2}+x_{2}^{2}+y_{2}^{2}+z_{2}^{2}.$

To vvr1590: Thanks for catching that. I'll fix that in the PDF.

[quote="arjunlandes"]In #7, do you mean to say that $n$ is not divisible by the square of any prime?[/quote]Aw come on, you guys know what I mean. $m\sqrt{n}$ in simplest form, find $m+n.$  :P

Just trying to be precise. The first way you said changes the problem significantly. :wink:

[quote="calc rulz"]So is the PDF always more up-to-date than the AoPS Wiki version? Because I see various inconsistencies between the two...[/quote]They were the same until yesterday. I'll fix things up in a moment.  :)

[quote="computer_nuke"]So when is the latest date we can turn in answers to you?[/quote]

It's in his signature!!!

Yeah, sure.

I won't be able to post the solution document for a while, since i'm on vacation. You guys can discuss when the test is over though.  :)

So it's August 10, when can we start discussion?

Wait, I thought the answers were due sometime today, meaning we can only start discussing tomorrow?

someone pm me with the answers? I want to check my results.

ummmm 4everwise, did you get my pm about my answers? because it still is in my "outbox" even though I sent it on the 8th... I"m confused...

[quote="mysmartmouth"][quote="rzsolt"]Any one pwn this thing, and want to post there answers? Or PM it to me...?[/quote]

Wait until tommorrow please.  I just turned in my answers a few hours ago, but some people have not (it's not due until later, I think).[/quote]
No problemo.  :lol:

[quote="Pakman2012"]I tried some of these problems.  It was kinda hard.
The later ones were a little harder than I'd hoped.[/quote]
Nothing there seemed 'trivial' to me, but I'm inexperienced.

I didn't get any geo. Geo is hard!

Currently connected to net with my cell phone. (Bluetooth stuff) It's really slow, so I won't be replying to messages till Friday.

[hide="Answers"]
1. 050
2. 000
3. 464
4. 003
5. 100
6. 005
7. 054
8. 456
9. 351
10. 045
11. 003
12. 186
13. 164
14. 277
15. 576
[/hide]

I'm sure you guys will correct any typos I make...  :P

Hm...in his PM, 4everwise told me the answer to 15 was 4!*3!*3, which comes out to 432...might have to wait for him to get back from vacation for clarification

I said 432 and 4everwise said that was wrong and the answer is 576.

all problems have been posted in separate threads in HSM,  I just created the related post collection

[url=https://artofproblemsolving.com/community/c3751275_2007_mock_aime_2]here[/url]

enjoy  / start solving

USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more AIME mocks [url=https://artofproblemsolving.com/community/c2439872_aime_mocks]here[/url]
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


trigonometry
integration
real analysis
calculus
function
limit
real analysis unsolved

Prove that trigonometric series $\displaystyle \sum_{n=0}^{\infty}a_ncos(nx)+b_nsin(nx)$ absolutely convegres for all  $x\in[a,b]$ iff the series $\displaystyle\sum_{n=0}^{\infty}|a_n|+|b_n|$ converges

The "if" part is trivial by the comparison test. For the "only if" part let us write $a_n\cos nx+b_n\sin nx=\rho_n \sin (nx+\theta_n)$ where $\rho_n=\sqrt{a_n^2+b_n^2}$ and $\theta_n\in [0,2\pi]$. Since, $\sum_{n=1}^\infty \rho_n |\sin(nx+\theta_n)|$ converges for all $x\in [a,b]$ there exist $M\in \mathbb N$ and measurable subset $E$ of $[a,b]$ such that $m(E)>0$ and $\sum_{n=1}^\infty \rho_n|\sin (nx+\theta_n)|\, dx\leq M$ for all $x\in E$. Thus, by Beppo-Levi theorem we obtain:
$\sum_{n=1}^\infty \rho_n \int_E \sin^2(nx+\theta_n)\, dx \leq \sum_{n=1}^\infty  \rho_n \int_E|\sin (nx+\theta_n)|\, dx = \int_E \sum_{n=1}^\infty \rho_n |\sin(nx +\theta_n)|\, dx\leq M m(E) <\infty$.
On the other hand we have: $\int_E \sin^2(nx+\theta_n)\, dx\to \frac{m(E)}{2}$ by Riemann-Lebesgue lemma and the identity $\sin^2\alpha=\frac{1-\cos 2\alpha}{2}$, where $m(\cdot)$ denotes the Lebesgue measure on $\mathbb R$. We conclude that $\sum_{n=1}^\infty \rho_n$ converges, which gives the result.

I think this is the Denjoy-Luzin theorem:

1. Denjoy, A., [i]"Sur l'absolue convergence des séries trigonométriques"[/i], C.R. Acad. Sci. 155: (1912) 135–136.
2. Luzin, N., [i]"On the convergence of trigonometric series"[/i], Moskau Math. Samml. (in Russian) 28: (1912) 461–472.

You proved a stronger result: The conclusion follows if an interval is replaced by any set of positive measure.

We can handle the interval version without measure theory. The sets \[E_k = \{ x\in [a,b]: \sum_{n=1}^\infty\rho_n|\sin (nx+\theta_n)|\, dx\leq k\}\] are closed, and their union is $[a,b].$ By Baire, some $E_k$ contains an interval $[c,d].$ Continuing without measure theory, the integration of any partial sum of $\sum_{n=1}^\infty\rho_n\sin^2 (nx+\theta_n)$ over $[c,d]$ can be carried out explicitly, and the result follows.

I apologize in prior if i do not see something here, but why $E_k$ are closed? You don't know if the function $x\stackrel{f}\mapsto \sum_{n=1}^\infty \rho_n|\sin(nx+\theta_n)|$ is continuous in $[a,b]$.

Actually, if you knew that, you need nothing: The conclusion follows by uniform convergence.

But that function is lower-semicontinuous, and that's enough.

We don't need to be so fancy: Suppose $x_m\to x_0,$ where $x_m$ is a sequence in $E_k.$ Then for any $N,  \sum_{n=1}^N\rho_n|\sin (nx_0+\theta_n)|=\lim_{m\to \infty}\sum_{n=1}^N\rho_n|\sin (nx_m+\theta_n)|\le k$ by continuity. Since $N$ is arbitrary,$\sum_{n=1}^\infty\rho_n|\sin (nx_0+\theta_n)|\le k$ as desired,

PS: 2222 posts! Cosmic.


Let a>b and n be arbitrary positive integers. Prove that n divides $\phi(a^{n}-b^{n})$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=99900 .

  darij

[quote="Rust"]Exist prime $p|(a^{n}-b^{n})$, suth that $p\not |(a^{d}-b^{d})$ for all divisor d<n. It mean $n=ord_{p}(a/b)|(p-1)|\phi(a^{n}-b^{n})$.[/quote]

With $a^{n}-b^{n}=2^{6}-1$, this is false.

There is already a topic to discuss it -> locked.


inequalities
number theory solved
number theory

Prove that if$n\in \mathbb{N}$then
$2n\leq \varphi (n)+\sigma (n)\leq nd(n)$

The RHS inequality fails with $n = 1$, but anyway... Letting $n > 1$, everything works. So let's prove it!

1. For any integer $n > 1$, $\varphi(n) + \sigma(n) \leq nd(n)$, where the equality holds iff $n$ is a prime.

Proof: RHS is multiplicative. Moreover, for any $x, y \in \mathbb{Z}^+$: $\varphi(xy) + \sigma(xy) \leq (\varphi(x) + \sigma(x))(\varphi(y) + \sigma(y))$, if $\gcd(x,y) = 1$. So we're done if we can skew that 1. is actually true just for perfect prime powers. Now this is trivial, since $\varphi(p^k) + \sigma(p^k) = (p^k - p^{k-1}) + (1 + \ldots + p^{k-1} + p^k) \leq (k+1)p^k = p^k\cdot d(p^k)$, where the equality holds iff $k = 1$. Game over!

2. For any integer $n > 0$, $2n \leq \varphi(n) + \sigma(n)$, where equality holds iff $n = 1$ or $n$ is a prime.

Proof: $2 \cdot 1 = \varphi(1) + \sigma(1)$, so suppose $n > 1$. According to Euclid, let $n = \prod_{i=1}^r p_i^{\alpha_i}$, where $r \in \mathbb{Z}^+$; $p_1, p_2, \ldots, p_r\in\mathfrak{P}$ are pairwise distinct and $\alpha_i > 0$ is an integer, for any $i = 1, 2, \ldots, r$. You have $2n \leq \varphi(n) + \sigma(n)$ iff $\displaystyle\frac{\varphi(n)}{n} + \frac{\sigma(n)}{n} \geq 2$, which is almost trivial. In fact, by well-known formulas for $\varphi(\cdot)$ and $\sigma(\cdot)$ and also employing an useful extension of Bernoulli's inequality, you argue

$\qquad LHS = \prod_{i=1}^r \left(1 - \frac{1}{p_i}\right) + \prod_{i=1}^r \left(1 + \frac{1}{p_i} +\ldots + \frac{1}{p_i^{\alpha_i}}\right) \geq$ $\left(1 - \sum_{i=1}^r \frac{1}{p_i}\right) + \left(1 + \sum_{i=1}^r \frac{1}{p_i} \right) =2$

where equality holds iff $r = \alpha_1 = 1$. Yeah, nice, you see...


Let  a circle $C(O, R)$  and a diameter $AB$  . A circle (K) touches internally circle (C) at D and radius $OA$  at point  $C$ . A chord  $EF$ of the greate circle $(C)$ wich is perpendicular to AB  touches the circle $(K)$ at  $Z$  .Prove that points  $D , Z , B (or D, Z , A)$ are collinear.

 [u]Babis[/u]

I suppose that the line $OD$ separates the points $A,\ Z$.
$(K)=C(I,r),\ IZ\parallel OB,\ \dfrac{DI}{DO}=\frac rR =\dfrac{IZ}{OB}\Longrightarrow Z\in BD.$.

This is an immediate consequence of similarity right?


Simplify, where a, b, c are real numbers such that the following expression will never be undefined.

$\frac{a^4-(ab-ac)^2}{b^2-(a+c)^2}+\frac{a^2b^2-(ac-a^2)^2}{c^2-(a+b)^2}+\frac{a^2c^2-(a^2-ab)^2}{a^2-(b+c)^2}$

Nope, Your answer is correct.  Well, it *could* be solved by knowing basic algebra simplifying concepts learned in middle school.

I get the same answer,
Why is this too hard for this forum if it requires only a concept related to this forum? :roll:

Never mind.  That was only a guess.  Forget what I wrote.  Maybe I'll even change it.  :lol:

[quote="tiredepartment"]Well, it *could* be solved by knowing basic algebra simplifying concepts learned in middle school.[/quote]

All right, I think I missed something in algebra and I don't think I want to go through what robinhe did...could someone explain the simplifying concept?(pls and thank you.)

factor out a^2, use difference of squares, let a+b+c=P

Well, I used basically difference of squares.  Because $a^2-b^2=(a+b)(a-b)$, $a^2-(b+c)^2=(a+b+c)(a-(b+c)) \Rightarrow (a+b+c)(a-b-c)$.  :)


probability
abstract algebra
inequalities
algebra
system of equations

True or false:
It is possible to bias two dice such that when rolled, the probability of any sum between $2$ and $12$ coming up is equal.

[hide="WRONG  :oops:  "]
Including the $2$ and $12$ as sums, True, Leave a six sided dice unbiased and those sums remain in equal probability.
If not including those two sums, False, Since the sums of $3$ and $11$ need a $1$ and a $6$ making a bias against those two sides will affect their probably.
[/hide]

a thought...

[hide]
Lets try it if both dice are biased the same way:

(P_1) ^2 = 1/11 = (p_6)^2
P_1 = P_6 = sqrt(1/11)

1/11 = 2 x P_1 x P_2
P_2 = 1/2 sqrt(1/11) = P_5

1/11 = (P_2)^2 +2(P_1 x P_3)
P_3 = 3/8 sqrt(1/11) = P_4

Alas this fails as they do not sum to 1, too bad.[/hide]

[quote="AstroPhys"][hide="Possibly   :maybe:  "]
Including the $2$ and $12$ as sums, True, Leave a six sided dice unbiased and those sums remain in equal probability.
If not including those two sums, False, Since the sums of $3$ and $11$ need a $1$ and a $6$ making a bias against those two sides will affect their probably.
[/hide][/quote]I'm not sure you understand the question.  I want it so that $P(2) = P(3) = \cdots = P(12)$.

Well, so far I have that the dice must be biased differently:

[hide]
Let $p_{N}$ be the probability of rolling $N$ on a single dice:
$P(2)=P(7)$ $\implies p_{1}^{2}=2 p_{1}p_{6}+2p_{2}p_{5}+2 p_{3}p_{4}> 2 p_{1}p_{6}$ $\implies p_{1}> 2p_{6}$ $\implies p_{1}\not = p_{6}$ $\implies P(2) \not = P(12)$

[/hide]

Still working on the case in which the dice are different, but from the way it looks, it might force some of the probabilities to be negative...

[hide="A possible approach"]Let $n_{i}$ and $k_{i}$ denote the probability of rolling an $i$ on dice $n$ and $k$, for $i=1,2,3,4,5,6$, and let $P(x)$ denote the probability of rolling a sum of $x$ on dice $n$ and $k$, for $x=2,3,4,...,12$.  Set up a long system of equations, and hope for a quick inconsistency.
$P(2) = n_{1}k_{1}$
$P(12) = n_{6}k_{6}\implies n_{1}k_{1}= n_{6}k_{6}$.
$P(11) = n_{5}k_{6}+n_{6}k_{5}\implies n_{5}k_{6}+n_{6}k_{5}= n_{6}k_{6}= n_{1}k_{1}$.
And so forth. [/hide]

Well yeah, but its not a particularly easy system to solve...

I also put it into my computer just on the side:
[code]No more memory available.
Mathematica kernel has shut down.
Try quitting other applications and then retry.[/code]
2GB just disappears with these things, doesn't it

Hmm...

[hide]
(a_1)(b_1) = 1/11 = P(2)
(a_6)(b_6) = 1/11 = P(12)

Note that if a_1 > a_6 and b_1 > b_6  this is impossible, so either a_1 = a_6 and b_1 = b_6 or one of those inequalities must be switched...

so lets assume one of the inequalities is switched:
P(7) = stuff + (a_1)(b_6) + (a_6)(b_1)

so (a_1)(b_6) + (a_6)(b_1) < 1/11

but rearrangement gives us that this is greater than (a_1)(b_1) + (a_6)(b_6) = 2/11, a clear contradiction.

so a_1 = a_6 and b_1 = b_6, but again we look at P(7)

P(7) = stuff + (a_1)(b_6) + (a_6)(b_1)
=stuff + (a_1)(b_1) + (a_6)(b_6)
= stuff + 2/11 > 1/11
contradiction.

So its impossible.[/hide]


trigonometry
calculus
integration
trig identities
Law of Cosines

In scalene, but nonright triangle ABC tan(A)= 5/12. If the length of each of the sides of Triangle ABC is an integer, find the smallest possible perimeter.

This may help, but is nowhere near a solution... :(

[hide]$\tan A=\frac{5}{12}\Rightarrow \cos A=\pm \frac{12}{13}$. Use the Law of Cosines to get $a^2=b^2+c^2\mp \frac{24}{13}bc$, where $a=BC$ and $b$ and $c$ are the other side lengths. In order for both sides to be equal, they both must be integral, so one of the sides must be divisible by 13. Let $c=13d$, $d$ integral, so that the equation becomes $a^2=b^2+169d^2\mp 24bd$.[/hide]

but the correct answer is 80 according to the answer key I have.....

I think I may have a solution but I will post it later. Basically I just drew an altitude and used the sum of tangents formula to express the tangent and then I tried to minimize the sides by plugging in small values.


function
combinatorics unsolved
combinatorics

Let $A_{1}, A_{2}, \cdots, B_{1}, B_{2}, \cdots$ be sets such that $A_{1}= \oslash , B_{1}= \{0\}. A_{n+1}= \{x+1|x \in B_{n}\}$ and $B_{n+1}= A_{n}\cup B_{n}/ A_{n}\cap B_{n}$ for all positive integers $n$. Determine all $n$ such that $B_{n}= \{0\}$.

My guess is all power of $2$, but I can't prove that.. It's better if one uses generating function to solve this, thanks..  :)

do you mean $B_{n+1}= (A_{n}\cup B_{n})\backslash (A_{n}\cap B_{n})$?

The "slash" means "without" and yup, the rest is true..  :)


function
number theory
totient function
Galois Theory
number theory solved

let p be a prime and n be n>=1

let phi be the usual totient function

show that n divides phi(p^n-1)

i have a proof (i think) but it uses galois theory??  if anyone is interested i'll give it, but i'm sure many people here can give straightforward proofs?

thx

frederic

$p$ has order $n$ in the multiplicative group mod $p^n-1$.

That is the proof.


linear algebra
matrix
limit
calculus
calculus computations

Can someone explain to me how to evaluate this kind of problem? I don't understand. I know the answer to the first two parts but I am not entirely sure how I got them correct and I have no idea for the 3rd one.

Use algebra to evaluate the limits if they exist:

${ f(x) = \left\{\begin{matrix} x^2 - 2, & 0 < x < 3 \\
2, & x = 3 \\
2x + 1, & 3 < x \end{matrix}\right.}$

$ \lim_{x\rightarrow 3^ + }{f(x) = }$
$ \lim_{x\rightarrow 3^ - }{f(x) = }$
$ \lim_{x\rightarrow 3}{f(x) = }$

$ \lim_{x\to 3}f(x)\equal{}a$ if and only if both $ \lim_{x\to 3^\minus{}}f(x)\equal{}a$ and $ \lim_{x\to 3^\plus{}}f(x)\equal{}a$. This is a property of limits on the real line, and should be easy to see from the definition. Flipping that around, $ \lim_{x\to 3}f(x)$ fails to exist if either one-sided limit fails to exist or if the two one-sided limits disagree.

Do the one-sided limits you have calculated already agree?

Yes they both agree.

[hide]I got 7 for each of the first two[/hide]


trigonometry
function
inequalities
logarithms
calculus
calculus computations

If  $0\leq x \leq\frac{\pi}{2}$

determine which is larger cos(x)   or   $e^{\frac{-x^2} {2}}$

Something cheap and quick, albeit incomplete:

$\cos(x)=1-\frac{x^2}2+\frac{x^4}{24}+O(x^6)$

$e^{-x^2/2}=1-\frac{x^2}2+\frac{x^4}8+O(x^6).$

From this it is clear that there exists an interval $(0,\delta)$ on which $\cos(x)<e^{-x^2/2}.$

Note that that's also clearly the direction of the inequality for $x=\frac{\pi}2.$ I would try to see if I could prove that direction of the inequality for the whole interval.

Just take logarithms. You need to prove that $\ln \cos x\le -\frac{x^2}2$. The values at $x=0$ coincide, so it suffices to check the inequality for the derivatives. But they are $-\tan x$ and $-x$ and the inequality $\tan x\ge x$ is very well known and easy to prove.  :P


AMC
USA(J)MO
USAMO
probability
AIME
AMC 12
trigonometry

Did anybody think this year's AMC 10 was easier than previous years'?

I unofficially got a 132, but I really think I could've gotten a 150. I knew how to do all the problems (I checked after the test), but I spent too much time double-checking my work. I'm so angry I didn't get a 150.

How did everybody else do?

How about the AMC 12? I head it was hard this year...

I did worse than last year.

I got 144 this year.   mistake on #24.

i thought the amc 12 was harder than previous years..

but it could have been just me since i did 4 of the questions i didnt do right  after the test.

however, its because there was like a graph problem with abosolute values (it was like number 14) that took forever to do and i didnt even get the right answer to.  also its a lot easier to do it after the test when there is no pressure so i cant really say i would have got those 4 problems right during the test even if i knew how to do them (for like the one with a1+a2 < a4+a5, during science class i realized that you dont have to calculate them out since the chances of that is the same chances of a1+a2> a4+a5, so you just have to do 120-(ways of a1+a2=a4+a5)/2-to get the answer), but that doenst mean that i could have thought of that way during the test.

i think i shouldve taken the amc 10 a.  number 15 on the 12 was number 24 on the 10 and i could do it in 2 mins, and i unofficially took the 10 and i got a 138.

i wouldnt really care about this but its that they also use your amc score for the usamo index and i need to get a higher amc score on the 10b that im taking

Can anyone please post the answers to the AMC 12?
Thank you.

It isn't already up?

Okay, well here they are.

1. D
2. A
3. C
4. B
5. B
6. A
7. D
8. C
9. D
10. D
11. C
12. B
13. B
14. A
15. D
16. D
17. D
18. C
19. C
20. E
21. D
22. C
23. D
24. D
25. D

There were 11 Ds as the answer on this test.

Yes 17 is D, and that is final I think.

- On number 25 of AMC 10, I thought of using Power of a Point, but I didn't notice how to apply it until a minute after the contest.

- On numbers 22 and 23, I skipped over them, because when I skimmed over them, they seemed to be about probability and combinations, so I thought they would be hard for me, but after the test, I did them within a minute or two.

I'm so mad.

And our school didn't sign up for AMC 10 B so I have to take the AMC 12 B.

And about the Ds on the AMC 12, somebody in my school guessed on 5 problems, and got 4 of them right.

And do most people know their unofficial scores already?

Could someone post the problems for the AMC 12?

[quote="rem"]Hmm, I think 17 is C because 4 does not work. :maybe:[/quote]

4, 2, 1, 4.

4 isn't less than 2 and 1 unfortunately.

I put C too on the test.

When and where do you guys think the problems will be released? I didn't take the A test, so i really want to see what the problems will look like for the B.

Are we legally allowed to scan them and post them in this forum?

And oops, I left out the [b]140-150[/b] option on the poll for AMC 12. Can a moderator fix it?

I don't think they can be released for there is a copyright law, right?

I actually thought this year's 10 was harder than last year...
or maybe I just messed up really bad.
Either way, I bombed it.

No but like you get $ a_1\equal{}a_4$ and should be $ a_1<a_4$...

Grrr. If you guys got the full time to take the test, that frustrates me.
This was my first year in taking both the AMC 10A and AMC 12A.
We got roughly 80 to 90 mins to do both.... :mad:
So...I couldn't compare it to any of the year's before.
However, My question is: Do some of the problems from the AMC 10 always appear on the AMC 12?
For me, the AMC 10&12 were O.K. I probably did pretty badly, though. :(
And what's the difference between the A & B tests?

The issue was already brought up countless times two years ago and last year (I think? I'm kind of blanking out). I don't know why the AMC does not appear to have changed their policy, but I think they are aware of the problem.

There's already a [b]sticky[/b] thread to complain or whatever...

Everyone was posting histories a while back, so I will.

[u]4th grade:[/u]
AMC 8: forgot

--skipped 5th--

[u]6th grade:[/u]
AMC 8: 18
AMC 12A: 99.5(REALLY ANNOYING!)
AMC 12B: 92.5

[u]7th grade:[/u]
AMC 8: 21
AMC 12A: 108
AMC 12B: 111
AIME: 3

--rejoined my age peers--

[u]second 7th grade:[/u]
AMC 8: 25
AMC 12A: 126
AMC 12B: Taking Wednesday after HMMT.

[quote="AMCDirector"]In response to the future's scenario:

(1) Yes, we are well aware that it could happen.

(2)Yes, as probability1.01 pointed out it has been thoroughly discussed each of the last several years.  It would be good to review the discussions then before renewing the same discussion.

(3) Yes, we have considered alternative selection rules, most of them coming from the membership of this forum.  Each selection scheme had objections to it at least as great as the current set of rules.  There was no compelling reason to change.

I remain willing to consider alternative USAMO selection rules provided they are written as an algorithm.  See my post from a year ago for the rules.  I suggest that altenratives appear in a new thread, not here.

Steve Dunbar
MAA Director, American Mathematics Competitions[/quote]
Note that the old thread from last year is still up and stickied.

You know, I really appreciate what Mr. Dunbar and the AMC do for contestants, especially in terms of feedback. They actually listen. And make changes if it is necessary. They are definitely under-appreciated. There is definitely a reason that the AMC is the most popular math competition in the US.

What Pakman2012 said. Just being on this forum (the AMC one) for the past few days, and I'd already seen how helpful they've been. :)

[size=42]My History:
My school district was stupid and wouldn't EVER let me skip grades--even my parents didn't want me too, and I was withheld the experience that is AOPS until a few months ago. My school district never let us take any advanced math tests, either. -prods-
Therefore, my current record sucks. And is very short and incomplete....

2008
AMC 8-->20? Don't remember. My teacher placed extra emphasis on that "it doesn't matter". Pft.
AMC 10 & 12-->Still waiting for results. Forgot to circle the questions on my test packet. :oops: [/size]

Eh. Forgive that. Just a bit of ranting to get out of my system. Just ignore it.

wait you don't need to be in a certain math class to be able to take the amc's anyone can. you certainly don't need to skip any grades to be able to take them. but i am not sure. So did you take the 10 or the 12 cuz u wrote AMC  10 & 12 --> Still waiting

Looks like the "averages" of the AMC are misleading compared to that of AoPSers. I got 126  :( .It's supposed to be over top 1%, but then it's not much in here.

I think BOTH A tests were harder than the previous years. I know that the 12 was harer my many testimonies, although the difficulty of the 10 was just my personal feeling. Maybe because I got overly excited. However, the lack of calculators didn't really affect me.

I think that for the average (below top 1%) student, 120 was almost impossible to obtain this year, as more of the problems were moderately difficult.  The AMC 10A seemed easier to AoPS'ers because we're used to consistently getting moderately difficult questions.  Therefore, I think the cutoff will be pretty low (115.5)

this year, 23 on amc 8,
120 on AMC 10A...
barely...

lol i barely beat you i got
AMC 8: 23
AMC 10A: 121.5

[quote="MustItAlwaysBeMe"]And I was thrown in a land far, far, away to the 73.5s on AMC 10As and 19s on AMC 8s...

I'm terrible, but I honestly don't care...I take math competitions for fun.[/quote]

This is excellent!  Any competition whether it's math, science, music, sports is about having fun.  In the process of having fun hopefully you learn something about math and about yourself, too.  If you're having fun, you'll continue to pursue the activity and more than likely improve.  A mid-70's score is nothing to sneeze at especially if it's one of the first times you've seen AMC10-type questions.  Using that knowledge the next time, the B contest next week or even if it's not until next year, will help to improve the score.

It's good to hear you had fun.  We do like knowing that people [i]enjoy[/i] taking the tests.

Wishing you all the best,

Elizabeth Claassen
AMC Accounting Clerk
  and former AHSME taker ;)

*sighs*...

I screwed my AMC 8s last two years, and i screwed them up again..
oh well...
[size=59]Here was MY situation for being really really aggravated with the school. Of course, an 8th grader taking the AMC 10/12 is abnormal. I asked the math counselor, she's like. sure you can come. so i went. i asked for a paper and she gives me the AMC 12. i said, "i'm here for the AMC 10" and she's like, "oh. i ran out. here's a copy if you want" so i ended up having to take the AMC 12. i was really upset. I haven't gotten my scores b/c 1. i didnt' mark my answers. 2. our school didn't grade them before they sent them.[/size]


integration
logarithms
calculus
calculus computations

Prove that, $ \int_0^\infty{x\ln x \over 1\plus{}x^3}\equal{}{2\pi^2 \over 27}$.

Consider integral :
$ I(\alpha)\equal{}\int_{0}^{\infty} \frac{x^{\alpha}}{1\plus{}x^{3}}dx$
1) step : Try to note that it's beta function. ( substitute $ \frac{1}{1\plus{}x^{3}}\equal{} z$)
2) note that $ \frac{d I(\alpha)}{d \alpha}|_{\alpha\equal{}1} \equal{} your integral$
3) calculate answer.
 :roll:


This is something I came up with walking home from school today. I'm not sure how good the wording is, but here goes...

Given a minor arc in a plane, prove it lies on a unique circle.

[hide]Only one circle will have that arc because all the others will be either smaller or bigger. I don't know how to put that into symbols. :oops:[/hide]

First, define "arc". If it's parabolic or elliptic or hyperbolic or semicubic or God knows what arc, then it certainly doesn't belong to any circle. And if it's a circular arc, then it already IS a part of a circle :) just pick three points on it and use the fact that any three non-collinear points define a unique circle, i.e. the circumcircle of the corresponding triangle.

I agree with Farenhajt. You need to specify what kind of arc. If you do know that it is a circular arc it is, pick 3 points, and construct the three perpendicular bisectors of the 3 segments formed by picking 2 of the segments. This will be a unique point (it is the center of the circle, and so it is an unique circle) since the three perpendicular bisectors of the sides of a [i]triangle [/i] intersect at the same point. I pondered about this problem about a year ago, so I know.  :)


Since $\binom{p}{i}=\frac{p!}{i!(p-i)!}$, both of the terms in the denominator are made up of factors less than $p$ and thus do not divide $p$ (since any prime cannot be divided by any number less than itself other than 1). Thus, none of the factors in the bottom will divide $p$ out from the top, and thus it is divisible by $p$.

[quote="dondigo"]Prove that if $p$ is a prime number, then $p$ divides $\binom{p}{i}$, $1\leq{i}\leq{p-1}$[/quote]

[hide]
rewrite as :
$\frac{p!}{(p-i)!i!}$. $p$ is relatively prime to all integers in the interval $[1, p - 1]$ thus it won't be divisible by either $(p-i)!$ or $i!$. So we can say $\displaystyle{p \choose i} = p * x$ for any integer $1 \le i \le p -1$ where $x$ is some integer. thus $p \mid \displaystyle{ p \choose i}$[/hide]


linear algebra
matrix
algebra
polynomial
linear algebra solved

Let $A$,$B\in {\mathbb R}^{2x2}$ with $detA$,$detB\in {\mathbb R}_+$. Show that
$det(A^n+B^n)+det(A^n-B^n)\geq 4[det(AB)]^{\frac{n}{2}}$ holds for any $n\in {\mathbb N}$.

Nice and easy. Just use the identity $ det(X+Y)+det(X-Y)=2(det(X)+det(Y)) $ and AM-GM.The conclusion follows. ;)

[quote="cezar lupu"]$ det(X+Y)+det(X-Y)=2(det(X)+det(Y)) $ [/quote]

this identity is funny. It is straightforward in the case of $\mathbb{R}$ or $\mathbb{C}$ with density but what is the usual algebraic proof ?

You write the matrices and make the computations.  :D  Anyway, 2x2 matrix computations are always easy. :P

[quote="cezar lupu"] $ det(X+Y)+det(X-Y)=2(det(X)+det(Y)) $ [/quote]

Is there an identity like this one for matrices in $M_3(R)$ ?

I have almost found one for matrices of order $3$ but I can't post it on the forum.I want to propose this problem for the National Olympiad in 2006. ;)

This is just incredible!! First, do you really think it is so difficult to find such identities and even more to prove them after the formula was given, after tons of problems using polynomials and matrices? Secondly, do you really think that is good to announce the problems you are going to propose for next year olympiad one year before, Cezar? Excuse me, but I just can't understand this... :?

Ok,sorry! You are right.This was a mistake from my part to say that.I promiss it won't happen again. :oops:  :blush:


let $ y \equal{} z \equal{} 1$ then we have $ 12(x\plus{}1)^{4}(x\plus{}2)\geq 16(x^{2}\plus{}2x\plus{}3)^{2}x$ while $ x\rightarrow \infty$ it is wrong

[quote="chien than"]Let $a;\,b;\,c$be positive real numbers.Prove that:
\[3(x+y)^{2}(y+z)^{2}(z+x)^{2}(x+y+z)\ge 16(x^{2}+y^{2}+z^{2}+xy+yz+zx)^{2}xyz \]
[/quote]
It's not of you :mad:  I was posted it at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=135464]here[/url]


ARML Practice will be this weekend, May 10th, at Hammond School in Columbia, and every following Saturday until the competition. An e-mail should be sent about it soon (if not already). Bring \$6 for some pizza and drinks, or you can bring your own food for lunch. Air conditioning will be working this weekend!!

Sure, that's possible. We'll ask what people want before we order.

Awesome :)

Also what if they don't let us in? Will there be a sign saying "all state math team" so they know we aren't intruders?

Also, what are the times? :)

(i didn't attend last time)


limit
integration
calculus
induction
calculus computations

Evaluate the limit:

$ \lim_{n \to \infty} \sum_{k \equal{} 1}^{n}{\frac {1 \cdot 3 \cdot 5 \cdot \dots \cdot (2k \minus{} 1)}{2 \cdot 4 \cdot 6 \cdot \dots \cdot 2k \cdot (2k \plus{} 2) }}.$

Atention : On the denominator are : k+1 therms, and on the numerator are only k therms (on each member of the sum).

Note that $ \prod_{k\equal{}1}^{n} \frac{2k\minus{}1}{2k} \equal{} (\minus{}1)^n \binom{\minus{}1/2}{n}$, hence generalized binomial series yields $ \frac{1}{\sqrt{1\minus{}x}} \equal{} 1 \plus{} \sum_{n\equal{}1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots (2n\minus{}1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \, x^n$.

Then the given limit is equal to $ \frac{1}{2}\int_{0}^{1} \left( \frac{1}{\sqrt{1\minus{}x}} \minus{} 1 \right) \, dx \equal{} \frac{1}{2}$.

Thank you :D.

Anybody have a solution without integrals? :D

It's an hypergeometric series: $ \frac{a_{k\plus{}1}}{a_k} \equal{} \frac{2k\plus{}1}{2k\plus{}4}$, with $ \alpha \equal{} 2$, $ \beta \equal{} 1$, and $ \gamma \equal{} 4$. Therefore, its sum is $ \frac{\minus{} \gamma a_1}{\alpha \plus{} \beta \minus{} \gamma} \equal{} \frac{\minus{}4 \frac18}{2\plus{}1\minus{}4} \equal{} \frac12$.

Finally, I came up with a solution:

So, we have that:
\[ \sum_{k \equal{} 1}^n \frac {1 \cdot 3 \cdot 5 \cdot \dots \cdot (2k \minus{} 1) }{2 \cdot 4 \cdot 6 \cdot \dots \cdot (2k \plus{} 2)} \equal{}
\]

\[ \equal{} \sum_{k \equal{} 1}^n \left( \frac {1 \cdot 3 \cdot \dots \cdot (2k \minus{} 1)}{2 \cdot 4 \cdot \dots \cdot 2k} \minus{} \frac {1 \cdot 3 \cdot \dots \cdot (2k \plus{} 1)}{2 \cdot 4 \cdot \dots (2k \plus{} 2)} \right) \equal{} \frac {1}{2} \minus{} \frac {1 \cdot 3 \cdot \dots \cdot (2n \plus{} 1)}{2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n \plus{} 2)}.
\]
But it is easy to show (induction works also, i think) that :

$ \frac {1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n \plus{} 1)}{2 \cdot 4 \cdot 6 \cdot \dots (2n \plus{} 2)} < \sqrt {\frac {1}{2n \plus{} 3}} \to 0$, for all $ n \in \mathbb{N*}$.

So, the desired limit is $ \lim_{n \to \infty} \sum_{k \equal{} 1}^{n}{\frac {1 \cdot 3 \cdot 5 \cdot \dots \cdot (2k \minus{} 1)}{2 \cdot 4 \cdot 6 \cdot \dots \cdot 2k \cdot (2k \plus{} 2) }} \equal{} \frac {1}{2}$.


Prove that for all $k \in \mathbb{N}_0$ there is an infinity of perfect squares of the form $- 7 + n \cdot 2^k$.

With what numbers can we replace $-7$ s.t. the above remains (eventually) true?

We can take every integer $m \equiv 1 \mod 8$ instead of $-7$ and only those when we avoid that $m$ is divisible by $4$:
When $m$ is divisible by $4$ divide through it often enough to reach that $m$ is not divisible by $4$, the 'being square'-property will not change. When $m \equiv 2 \mod 4$, then $m$ is clearly no square, so we are left with odd $m$, for which also only $m \equiv 1 \mod 8$ remains possible.

Let $x,y$ be odd with $x^2 \equiv y^2 \mod 2^k$ or equivalently $2^k|(x+y)(x-y)$.
Now $\gcd(x+y,x-y)=\gcd(x+y,2y)$ is divisible by $2$, but not by $4$, so at least one factor is just divisible by $2$, but not by $4$.
This let us remain with $x \equiv \pm y \mod 2^{k-1}$, thus there are for given $x$ exactly $4$ such $y$ that are different $\mod 2^k$.
Now all squares of odd numbers are $\equiv 1\mod 8$, so at most, and by the previous said also at least (there are $2^{k-1}$ 'odd' residue classes and $2^{k-3}$ are $\equiv 1 \mod 8$), the residues $\mod 2^k$ that are squares are that $\equiv 1 \mod 8$.

And the existence of one $x$ with $x^2 \equiv m \mod 2^k$ implies trivially that of infinety many.


Find the area of triangle $ ABC$ give altitudes lengths $ 12, 15, and 16.$

Use the law of cosines to get the angle and then use $ A\equal{}\frac{1}{2}ab\sin{C}$.


[hide="1"]
$3a+3b+3c+3d=807\implies a+b+c+d=269$.

The four numbers are 89, 72, 61, and 47, the largest of which is $\boxed{89}$.[/hide]

[quote="jeez123"]Help please...

#1
Four whole numbers, when added three at a time, give the sums 180, 197, 208 and 222. What is the largest of the four numbers?

#2
In a certain three-digit number, if the sum of the digits is doubled, the result is 16 more than 3 times the tens digit. One more than twice the hundreds digit is the units digit. The hundreds digit is 11 decreased by twice the tens digit. Find the three-digit number.[/quote]

[hide="2"]since two times the hundreds digit is less than 10,  the hundreds digit is from 1-4. since its 11 minus an even number, it must be odd.  this means the hundreds digit is 1 or 3.  so the tens digit is either 5 or 4.  and the units is 3 or 7.

so the sum of the digits *2=3*(tens digit)+16, which is 5 or 4.
so if its 5, the sum of the digits is 31/2.  that doesnt exactly work... :P

if its 4.  2s=12+16=28.  so the sum of the digits is 14.  and the tens is 4, meaning the hundreds is 3.  14-3-4=7.
so the number is 347.[/hide]


[quote="shadysaysurspammed"]1)$x^{\frac{1}{3}}+(2x-3)^{\frac{1}{3}}={12x(x-1)}^{\frac{1}{3}}$

2)Find the greatest value of:-

$\frac{x+2}{2x^2+3x+6}$

3)IF a,b,c,x,y,z are real numbers and
$(a+b+c)^3=3(bc+ca+ab-x^2-y^2-z^2)$
show that a=b=c and x=y=z=0.[/quote]
#3 is a proof. Off to intermidiate we go!

as for number 3 , are u sure u give the right question?

if it is

$(a+b+c)^2$ instead of $(a+b+c)^3$ then it is true that $a=b=c$ and $x=y=z=0$ . Or else I dont think it is true.

Shyong,In the book the question is the same,Even then could you xplain a bit.
And i need confirmation for the 1st one

well , if a=b=c=2 and x=y=z=0 , we still cant get (a+b+c)^3=3(ab+bc+ac-x^2-y^2-z^2)

i think it is $(a+b+c)^2$ because we have $(a+b+c)^2\geq 3(ab+ac+bc)$ by am-gm so

$3(x^2+y^2+z^2)=3(ab+ac+bc)-(a+b+c)^2\leq 0$ and so $x=y=z=0$ and $a=b=c$

[quote="shadysaysurspammed"][quote="shyong"]
i think it is $(a+b+c)^2$ because we have $(a+b+c)^2\geq 3(ab+ac+bc)$ by am-gm so

[/quote]

How did we get that by am-gm?Won't we get $(a+b+c)^2\geq9abc$?[/quote]

That inequality isn't even true; you are looking for $(a+b+c)^2\geq 9(abc)^{\frac 23}$

[quote="mathmanman"][quote="shadysaysurspammed"][quote="shyong"]
i think it is $(a+b+c)^2$ because we have $(a+b+c)^2\geq 3(ab+ac+bc)$ by am-gm so

[/quote]

How did we get that by am-gm?Won't we get $(a+b+c)^2\geq9abc$?[/quote]

Well, shyong's inequality is equivalent to : $a^2 + b^2 + c^2 \ge ab + bc + ca$, which directly comes from AM-GM :)[/quote]
yup,I was wrong.sorry(Not the question but about the inequality thing,I still don't know if the question is right but if u guys are saying so,then it has to be $(a+b+c)^2$ :) )


if  [x]  is the greatest integer which is not greater than x,
For any non-negative integer $n$, prove that $[(1 + \sqrt{3})^{2n+1}]$can be divided by $2^{n+1}$

It isn't too hard if you've seen problems of this type before.  I have no time to post a complete solution.

[hide="Hint 1"] An alternate way to write the Fibonacci sequence (edit: for even terms) is as $F_{2n} = \lfloor \frac{\phi^{2n}}{ \sqrt{5} } \rfloor$. [/hide]
[hide="Hint 2"] $\left( 1 - \sqrt{3} \right)^{2n+1}$ gets very, very small.  :) [/hide]

toetoe, I think your solution might be a little confusing unless someone reading it already knew what the inevitable method would be (are you sure your $\LaTeX$ in the second part is correct?), so...

[hide="Solution"]
Let $\alpha = 1 + \sqrt{3}, \beta = 1 - \sqrt{3}$.

Let $\omega =\alpha^2 = 4 + 2 \sqrt{3}, \zeta = \beta^2 = 4 - 2 \sqrt{3}$.

Let $a_n = \alpha^{2n+1} + \beta^{2n+1} = \alpha \omega^n + \beta \zeta^n$, which is integer for all $n$ by binomial theorem.

Then $a_n$ is a second order linear recurrence with characteristic equation $x^2 = (\omega + \zeta) x - (\omega \zeta) = 8x - 4$.  Moreover, because $| \beta | < 1$ we have

$\lfloor (1 + \sqrt{3})^{2n+1} \rfloor = \lfloor \alpha \omega^n \rfloor = a_n$

The recurrence takes the form

$a_{n+1} = 8 a_n - 4 a_{n-1}$

With $a_0 = 2, a_1 = 20$.  And then the result is obvious by induction.  QED.  :) [/hide]

Same as above but  perhaps  a cuter  way:

Can you prove $(\frac{\sqrt 3 + 1} 2 )^ {2n+1} - (\frac{\sqrt 3 - 1} 2 )^ {2n+1}$ is an integer?
(Same as saying $(\sqrt 3 + 1 )^ {2n+1} - ( \sqrt 3 - 1 )^ {2n+1}$ is divisible by $2^{n+1}$ )

We do know that $0<( {\sqrt 3 - 1} )^ {2n+1}<1$ :)

Will be edited later (when I get some time) :  Error as pointed out below...

[quote="Gyan"]Another, or perhaps cuter  way:

Can you prove $(\frac{\sqrt 3 + 1} 2 )^ {2n+1} - (\frac{\sqrt 3 - 1} 2 )^ {2n+1}$ is an integer?
(Same as saying $(\sqrt 3 + 1 )^ {2n+1} - ( \sqrt 3 - 1 )^ {2n+1}$ is divisible by $2{n+1}$ )

We do know that $0<( {\sqrt 3 - 1} )^ {2n+1}<1$ :)[/quote]

... it's not.   :huh:  The problem asks for divisibility by $2^{n+1}$, not $2^{2n+1}$.

Although I see what you mean.  Applying this logic to my solution would make it a little faster  :oops:

Yep! a little sloppy  a little too quick. :! You need to use $(1 \pm \sqrt 3)^2 = 2(2 \pm \sqrt 3)$


superior algebra
superior algebra solved

Let $ I,J$ be ideals of a commutative ring $ R$ such that $ I\plus{}J\equal{}R$. Prove that $ IJ\equal{}I\cap J$.

I have that $ IJ\subseteq I\cap J$.

[quote="NcG"]Let $ I,J$ be ideals of a commutative ring $ R$ such that $ I \plus{} J \equal{} R$. Prove that $ IJ \equal{} I\cap J$.

I have that $ IJ\subseteq I\cap J$.[/quote]

Hello.
Assume $ x$ is in $ I\cap J$ and let us prove that $ x$ is in $ IJ$.
Since  $ I \plus{} J \equal{} R$, there exist $ i$ in $ I$ and $ j$ in $ J$ such that $ i \plus{} j \equal{} 1$.
Multiply by $ x$ : $ xi \plus{} xj \equal{} x$.
Each term of the left member is in $ IJ$, thus the left member is in $ IJ$, thus the right member is in $ IJ$.

Friendly,
P.


vector
analytic geometry
geometry unsolved
geometry

Let triangle ABC, H is orthocentric. prove that:
$\Large tgA.\vec{HA}+tgB.\vec{HB}+tgC.\vec{HC}=\vec{0}$

It follows from the definition of baricentric coordinates of a point $P$ wrt $\triangle ABC$.

Given $\triangle ABC$ and a point $P$ on it's plane we have $(x+y+z)OP=xOA+yOB+zOC$ for any point $O$ then the baricentric coordinates of $P$ are $(x: y: z)$.

As a corolari we get that if $P=O$ then $0=xPA+yPB+zPC$

With some calculations we get that $x: y: z=[PBC]: [PCA]: [PAB]$

So as the baricentric coordinates of $H$ are $(tgA: tgB: tgC)$ teh result is easy  :wink:

José Carlos Chavez Sandoval


In triangle $ABC,$ the medians to the sides $\overline{AB}$ and $\overline{AC}$ are perpendicular. Prove that $\cot B+\cot C\ge \frac23.$

[hide="Solution"]Let the median from $B$ meet $\overline{AC}$ at $E$ and the median from $C$ meet $\overline{AB}$ at $F.$ Denote the centroid as $G.$ It follows that $BG=2GE$ and $CG=2GF.$ For the sake of convenience, let $GE=x$ and $GF=y.$

Now we use the cotangent addition formula (can be derived on the spot):
\[\cot (a+b)=\frac{\cot a \cot b-1}{\cot a+\cot b}. \]It follows that
\begin{eqnarray*}\cot B &=& \cot (\angle ABE+\angle CBE) \\ &=& \frac{\left(\frac{2x}{y}\right)\left(\frac{2x}{2y}\right)-1}{\frac{2x}{y}+\frac{2x}{2y}}\\ &=& \frac{2x^{2}-y^{2}}{3xy}. \end{eqnarray*}Similarly (by symmetry or calculation),
\[\cot C =\frac{2y^{2}-x^{2}}{3xy}. \]Adding these two expressions,
\begin{eqnarray*}\cot B+\cot C &=& \frac{2x^{2}-y^{2}}{3xy}+\frac{2y^{2}-x^{2}}{3xy}\\ &=& \frac{x^{2}+y^{2}}{3xy}\\ &=& \frac{x}{3y}+\frac{y}{3x}. \end{eqnarray*}Hence, by AM-GM,
\[\cot B+\cot C =\frac{x}{3y}+\frac{y}{3x}\ge 2\sqrt{\left(\frac{x}{3y}\right)\left(\frac{y}{3x}\right)}=2/3. \]$\mathbb{QED}$
[/hide]

Let $AX$ be the altitude with $X$ on $BC$.
Let $G$ be the centroid and $D$ the midpoint of $BC$.

$CotB=\frac{BX}{AX}$ and $CotC=\frac{CX}{AX}$
Thus $CotB+CotC=\frac{BC}{AX}$
Since $BGC$ is a right triangle,
$GD=\frac{AD}{3}=\frac{BC}{2}$

Also $AX \le AD$
Thus $CotB+CotC=\frac{BC}{AX} \ge \frac{BC}{AD} = \frac{2}{3}$

Thus the result.


calculus
integration
combinatorics unsolved
combinatorics

a,b,c,d ,n is positive integral,1<=a,b,c,d<=n,1<=a-b+c<=n,1<=a-b+d<=n,1<=a-c+d<=n,1<=b-c+d<=n,then how many solutions of (a,b,c,d)? and why ?


Assume that $ a,b,c$ are positive numbers such that $ a\plus{}b\plus{}c\equal{}3$.Prove that
$ \frac{a}{b^2\plus{}c^2}\plus{}\frac{b}{c^2\plus{}a^2}\plus{}\frac{c}{b^2\plus{}a^2}\geq\frac{3}{2}$

[quote="Ibrogimov"]Assume that $ a,b,c$ are positive numbers such that $ a \plus{} b \plus{} c \equal{} 3$.Prove that
$ \frac {a}{b^2 \plus{} c^2} \plus{} \frac {b}{c^2 \plus{} a^2} \plus{} \frac {c}{b^2 \plus{} a^2}\geq\frac {3}{2}$[/quote]
Let $ x \equal{} y \equal{} 1.4, z \equal{} 0.2 , LHS \minus{} RHS \equal{} \minus{} 0.0489... < 0$ :wink:


geometry
rotation
calculus
integration
perpendicular bisector
angle bisector

Someone has drawn a line on a piece of paper. You have a ruler; however, the ruler one straight side and no angles (assume the other three sides have been chewed off by a dog) and on this good side there are 1000 equal scale increments (in or cm, no difference). Give a proof of how you can draw a perpendicular line to the given line using only the ruler and a pencil.

P.S. You can draw lines with measurements; no compass tricks are allowed (this is not trivial).

[quote="Zratonx"]Hint
draw a pair of diagonal lines on one side of the line with the paper... and another one on the other side than you have two pts and a perpendicular line

Don't sound like a math problem to me[/quote]

It's still a math problem.. a construction problem

what do u mean? it doesn't sound correct to me only using a ruler (NO COMPASS ALLOWED).

Do you have a solution?

Okay, I think I have a solution:

[hide]At an angle (not right) to the line, attached at one end to the line, draw a line segment of some length, say, 3.  Now, at the other end of that line segment, place the end of the ruler, and rotate the rest of the ruler around that point until the ruler's 3 measurement point reaches the line.  Now, connect the "vertex" with that 3 measurement.  This gives us an isosceles triangle, with part of the given line as a base.  Now, using the measurements of the ruler, find the midpoint of the base, and connect that midpoint with the vertex, and you have an altitude, and, thus, a line perpendicular to the given line.[/hide]

The only problem with that is you don't know whether the base will be exactly measurable using the ruler.  Other than that, it works.

[quote="dts"][quote="Mikey"]from the same two points on the line you can repeat the two lines construction on the other side of the line and join the two tips of the triangles[/quote]

No, you can't.  You'd have to get the angle exactly correct, or the resulting angle wouldn't be right.  Just because two isosceles triangles have the same "equal sides length" doesn't mean they're congruent.[/quote]

I see

[hide]To do this you use the perpendicular bisector. Find the center of the line with the ruler (you can do that) than just draw a line as perpendicular as you can get it (this is not the line you will end up with) than using that line as an approxamation of where the vertex should fall comstruct a equilateral triangle with sides congruent to the line and using the line as a side. Now the vertex shouldn't be lying on the perpendicular that you just drew on there this is good. The perpendicular bisector is the line drawn through the vertex into the midpoint of the line. [/hide] Sorry if it's a bit confusing.

So you can only use the ruler to measure integer length..... right?       Which makes this very hard........ Can you determine an irrational length with the ruler?
Be good if you could.... then you can just draw a triangle(with one of the sides be a segment on the given line), measure the length of each side of the triangle....... use Pathygorean Theorem to find the point where the altitude intersect the given line, and connect it to the vertex of the triangle

JBL wrote:Well, I can construct a right angle, but I don't know how to rotate it onto our line.  I can give a hint as to how to accomplish this: [hide]try to construct the angle bisector of two given lines.[/hide]  But I'm not convinced I know how to go on from there.  It's certainly possible to construct irrational lengths, but it's not at all possible to construct an arbitrary irrational length, which is what you'd need.  Also, you can't even measure except for integral lengths.

Let point A and B be on the given line such that AB=2. Draw line AC such that AC=2, and let M be the midpoint of AC so AM=1. Extend AB to point D, and BC intersect MD at E. Draw AE and let it intersect CD at F. BF is parallel to AC (proof: by ceva's, AM/MC*CF/FD*DB/AB=1, DF/FC=DB/BA...... BF//AC). Let G be a point on line BF such that BG=2 (F is on segment BG), angle BGA=angle GAC, angle BGA=angle BAG, so angle CAG=angle GAB, so AG is the angle bisector of angle BAC, and AG perpendicular to BC.

[quote="JBL"]dts, the trick you are using would work, you'd just need two different isosceles triangles and the line connecting their vertices would be perpendicular to the other line.  However, what you're doing is exactly what dpopov was refering to when he said "no compass tricks!"  So, it works, but it doesn't satisfy the conditions of hte problem.

A+, how do you construct the equilateral triangle?  (Also, why did you even need that first line?)

(Also, you should all stop talking about the "endpoints of the line."  Lines don't have endpoints.)

Zratonx, what you said doesn't make any sense to me at all.  You draw two pairs of lines ... and end up with two points ... and suddenly you have a perpendicular line?[/quote]

The first line was there just to help you find the 3rd vertice of the equilateral triangle. Your right about the lines having no endpoint, I knew that but I was lazy and line is easier to write than segment.I thought about using two isolese triangles but if you just use one equilater and connect the vertice to the mispoint that makes a perpendicular bisector. Plus it's not a compass trick because I used that extra line as a guide to where to but my two segment.

[quote="A+MATH"]
The first line was there just to help you find the 3rd vertice of the equilateral triangle. Your right about the lines having no endpoint, I knew that but I was lazy and line is easier to write than segment.I thought about using two isolese triangles but if you just use one equilater and connect the vertice to the mispoint that makes a perpendicular bisector. Plus it's not a compass trick because I used that extra line as a guide to where to but my two segment.[/quote]

So HOW did you CONSTRUCT the equilateral triangle (you didn't just use your eyes to approximate, did you?)? How did you use your first line to help(appoximated with your eyes?)?


A point $ (x,y)$ is randomly selected such that $ 0 \le x \le 3$ and $ 0 \le y \le 6$. What is the probability that $ x\plus{}y \le 4$? Express your answer as a common fraction.

I didnt really get that. I mean the solution, i didnt get how to make it into a rectangle and then shade it off.

The rectangle comes from the region such that $ 0\le x \le 3$ and $ 0 le y \le 6$, try graphing it yourself.  Same thing with the shading it off.

[hide=Solution (from Alcumus)]We draw the region and mark off the area where $x+y \le 4$:

[asy]
draw((0,0)--(3,0)--(3,6)--(0,6)--cycle);
fill((0,0)--(0,4)--(3,1)--(3,0)--cycle, gray(.7));
dot((0,0));
dot((3,0));
dot((0,6));
dot((3,6));
dot((0,4));
dot((3,1));
label("(0,0)", (0,0), W);
label("(0,6)", (0,6), W);
label("(0,4)", (0,4), W);
label("(3,1)", (3,1), E);
label("(3,0)", (3,0), E);
label("(3,6)", (3,6), E);
[/asy] The area of the rectangle is 18. The area of the shaded region, a trapezoid, is $\frac{1}{2}(1+4)\cdot3=\frac{15}{2}$. The probability that the point ends up in the shaded region is then $\boxed{\frac{5}{12}}$.[/hide]

A point $ (x,y)$ is randomly selected such that $ 0 \le x \le 8$ and $ 0 \le y \le 4$. What is the probability that $ x+y \le 4$? Express your answer as a common fraction.

I took the bottom and top bases, instead of left and right. With a height of $4$, and bases $3$ and $3\sqrt2$, I got a different answer. :huh: I dunno why I'm wrong.


In chess, a knight moves in an L-shaped manner -- two spaces in one direction and one space in a direction perpendicular to the first direction -- as shown. Beginning in and including the upper left corner (marked *), what is the most number of squares in a $4\times 4$ chessboard that a knight can visit without visiting any square more than once?

[img]http://i96.photobucket.com/albums/l167/adasarathy/chessboard.gif[/img]

[quote="anirudh"]Is there a way you could show that? :maybe:[/quote]
yes, there is a way, that was meant to be a challenge.

[quote="jhredsox"]can someone show how you get 15?

and can you move the one space, then turn first?

-jorian[/quote]
Read my post...
I know the proof, but I've seen it before so if no one gets it in a few days I'll show it.
Here is a hint:
[hide="hint"]Consider parity[/hide]

[quote="mathgeniuse^ln(x)"][quote="bpms"]15 is the answer, can you prove that 16 isn't possible?[/quote]

Well, if you reached 16 then you had to hit the square you started off with again, which is not allowed.

So if 15 is possible than 15 is the answer.[/quote]
But you include the corner you start at when you count out of 16...

[hide]i got 15, i just tried to waste as little space as possible.  I dont think there is any systemiatic way except guess and check.[/hide]

[quote="nutz_for2.718281828"][hide="proof of 16"]down right right; down left left; down right right; right up up; up left left[/hide][/quote]
:roll: you only count the ones you land on for each move.
If no one posts the proof in a couple days I'll do it.

Oh, you do...

I thought otherwise...that's why i didn't get 15


Sorry, I've only included the most common ones.

Edit:
HUH????! How did Japanese drop off the list?

I learn no languages in school until high school..... but I am going to take French.

meh. i WISH my school taught Japanese. i'd take it FOR SURE!!!!!!

i'm not a nativeenglish speaker, tho, so i shouldnt be talking.

I am learning Japanese. And attempting Korean.

Picked italian, dropped, picked up chinese, dropped, picked up Russian, and holding it.

Umm last year just French, this year (well I suppose this year is over, but w/e) French and Latin (I'd say Latin warrants a spot in the most common ones), next year Latin and Greek.

Weird. None of Latin, Greek, Russian, Italian, or Korean are offered in my school district.

umm i live in the US and speak english more fluently... but chinese was my first language... does that still count?? anyway... me takes french in skool bonjour!! me wishes there was chinese... EASY A+!!!!!!!!!!!!!!!!!!!!!!!!!! :rotfl:

There's Chinese at my school... but no Chinese people take it because of our wonderful Chinese parents who think we need to be trilingual (sorry parents)... although that is kind of cool...

All the Spanish people take Spanish.  :|

And why in the world to we celebrate Cinco de Mayo when it is not even celebrated in the majority of the Spanish-speaking world... and it isn't even an obligatory Mexican holiday?

i would want to take chinese if my school offered it. easy to make the grade  :wink:


Find all positive integers $n$ that are quadratic residues modulo all primes greater than $n$.

Let $n=2^k\prod _1 ^m q_i \cdot a^2$

There exists infinitely many primes $p$ for which $2^k\prod _1 ^m q_i$ is non-residue mod $p$ (by Chinese remainder theorem and Dirichlet's theorem), hence $n$ has to be a square and obviously all such numbers satisfy conditions of the problem.

[quote="ZetaX"]We can avoid Dirichlet (the only analytical step) by using quadratic reciprocity for Jacobi symbols (the rest of the proof being the same).

Here an adapted copy of more or less the same from a post of mine on MathLinks:

Assume that $n$ isn't a square, but a square $\mod $ all primes $>n$.

Let $n=2^s b = 2^s p_1^{v_1} p_2^{v_k} ... p_k^{v_k}$ be the prime factorisation and $b$ it's greatest odd factor. Since $n$ is not a square, either (w.l.o.g.) $v_1$ is odd or $s$ is odd.

1. case: all $v_i$ are even, so $s$ is odd.
Then simply taking an integer $q$ (coprime to all primes $\leq n$) not dividing $n$ and $\equiv 3\mod 8$ gives $\left( \frac{n}{q} \right) = \left( \frac{2^s}{q} \right) \left( \frac{b}{q} \right) = (-1)^s = -1$.

2. case: $v_1$ is odd.
Now let $c$ be a quadratic non-residue $\mod p_1$ and take an integer $q$ (coprime to all primes $\leq n$) such that $q \equiv 1 \mod 8$, $q \equiv c \mod p_1$ and $q \equiv 1 \mod p_i$ for all other $p_i$.
Then we get:
$\left( \frac{n}{q} \right) = \left( \frac{2}{q} \right)^s \left( \frac{p_1}{q} \right)^{v_1} \left( \frac{p_2}{q} \right)^{v_2} ... \left( \frac{p_k}{q} \right)^{v_n} = \left( \frac{q}{p_1} \right)^{v_1} \left( \frac{q}{p_2} \right)^{v_2} ... \left( \frac{q}{p_k} \right)^{v_k}=(-1)^{v_1}=-1$.

In both cases, $n$ is not a quadratic residue $\mod q$, thus not for at least one of it's prime divisors. This contradicts the assumption of $n$ being a square $\mod$ all primes $>n$.

There are some tough ones:

1. If $n$ is a $k$-th power $\mod $ all primes $p>n$, then:
- $8 \nmid k$ ; then $n$ is a $k$-th power of an integer.
- $8 | k$ ; then $n$ is a $\frac{k}{2}$-th power of an integer.

2. If $n$ is a power of $2$ $\mod$ all primes $p>n$, then $n$ is also a power of $2$ seen as an integer.[/quote]

Thank you thank you thank you!

You wouldn't believe how long i have been trying to find the answer to this!

Thank you so much ZetaX,

Laura


A server has an availability of 97%. I can cluster it with an identical node and bring the availability up to 99.91% (97% + 97% * (1-97%). The second node has to cover the 3% unavailibility of the first node and to a maximum of its availability 97%

What if I have 3 active nodes with 97% availability each. I have one standby node (97% availability). what would be the availability of this cluster?

I assume that you mean that your claster of $4$ nodes is available if and only if at least $3$ nodes in it are available and the availabilities of nodes are independent. Then the answer is
\[4\cdot 0.97^{3}-3\cdot 0.97^{4}\approx 0.9948\]
The first term represents the sum of probabilities of availability of $3$ given nodes (there are $4$ combinations) and the second term takes care of overcounting the situation when all $4$ nodes are available (it was counted $4$ times in the first term and you need it to be counted just once; hence $-3$).


inequalities
blogs
inequalities proposed

Let $ a,b,c$ be positive real numbers such that $ abc\equal{}1$. Prove that
\[ \frac{8}{a\plus{}b\plus{}c}\plus{}1 \ge \frac{11}{ab\plus{}bc\plus{}ca}.\]
:)

This is a silly inequality as we can assume wlog that $ \sum a\geq \sum ab$ and vice versa, so the problem (generalize 8,11 to any k,k+3) is just $ \sum a\geq 3$...

[quote="me@home"]This is a silly inequality as we can assume wlog that $ \sum a\geq \sum ab$ and vice versa.[/quote]
No you can't because replacing a,b,c by their reciprocals yields a different inequality.  :(

[quote="arqady"][quote="can_hang2007"]Let $ a,b,c$ be positive real numbers such that $ abc \equal{} 1$. Prove that
\[ \frac {8}{a \plus{} b \plus{} c} \plus{} 1 \ge \frac {11}{ab \plus{} bc \plus{} ca}.
\]
:)[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159857[/quote]
Sorry, arqady. I dont know that you have posted it before.  :blush:
So, let me post, my full proof.
Without loss of generality, we may assume that $ a \equal{} \max\{a,b,c\}$, then $ a \ge 1$. Now, putting $ t \equal{} \sqrt {bc} \le 1\Rightarrow a \equal{} \frac {1}{t^2}$. Now, rewrite our inequality as
\[ 1 \minus{} \frac {3}{a \plus{} b \plus{} c} \ge \frac {11}{ab \plus{} bc \plus{} ca} \minus{} \frac {11}{a \plus{} b \plus{} c}
\]

\[ \Leftrightarrow (ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c \minus{} 3) \ge 11(a \minus{} 1)(b \minus{} 1)(c \minus{} 1)
\]
Now, by AM-GM Inequality, we see that
\[ ab \plus{} bc \plus{} ca \ge 2a\sqrt {bc} \plus{} bc \equal{} \frac {2}{t} \plus{} t^2 \equal{} \frac {t^3 \plus{} 2}{t}
\]

\[ a \plus{} b \plus{} c \minus{} 3 \ge a \plus{} 2\sqrt {bc} \minus{} 3 \equal{} \frac {1}{t^2} \plus{} 2t \minus{} 3 \equal{} \frac {(t \minus{} 1)^2(2t \plus{} 1)}{t^2}
\]

\[ (a \minus{} 1)(b \minus{} 1)(c \minus{} 1) \equal{} (a \minus{} 1)(bc \minus{} b \minus{} c \plus{} 1) \le (a \minus{} 1) \left( bc \minus{} 2\sqrt {bc} \plus{} 1\right) \equal{} \frac {(1 \minus{} t^2)(t \minus{} 1)^2}{t^2}
\]
Hence, it suffices to prove that
\[ \frac {(t^3 \plus{} 2)(2t \plus{} 1)(t \minus{} 1)^2}{t^3} \ge \frac {11(1 \minus{} t^2)(t \minus{} 1)^2}{t^2}
\]

\[ \Leftrightarrow 2t^4 \plus{} 12t^3 \minus{} 7t \plus{} 2 \ge 0
\]
With this inequality, see here: http://www.mathlinks.ro/viewtopic.php?t=237625 :)

See also here: [url=http://canhang2007.wordpress.com/]My Inequality blog[/url][/url]


My father's age 5 years ago plus twice my age now gives 65, while my age 5 years ago plus three times my father's age now gives 130.  What is my father's age?

Let my age be $ x$, and my father's, $ y$.
We get the equations:
$ y \minus{} 5 \plus{} 2x \equal{} 65$
$ 3y \plus{} x \minus{} 5 \equal{} 130$
Simplifying,
$ y \plus{} 2x \equal{} 70$
$ 3y \plus{} x \equal{} 135$
Multiplying the second equation by 2, we get:
$ y \plus{} 2x \equal{} 70$
$ 6y \plus{} 2x \equal{} 270$

Subtracting the first equation from the second,
$ 5y \equal{} 200$
$ y \equal{} 40$
So my father's age is $ \boxed{40}$


algebra
polynomial
trigonometry
function
algebra proposed

[quote="N.T.TUAN"][color=darkred]Let $P (x) = x^{3}-3x+1.$ Find the polynomial $Q$ whose roots are the ﬁfth powers of the roots of $P$.[/color][/quote]
[b][u]Proof.[/u][/b] $\left\{\begin{array}{c}x^{3}=3x-1\\\ y=x^{5}\end{array}\right\|$ $\Longrightarrow$ $y=x^{2}(3x-1)=3(3x-1)-x^{2}$ $\Longrightarrow$ $\left\{\begin{array}{c}x^{3}-3x+1=0\\\ x^{2}-9x+(3+y)=0\end{array}\right\|$ $\Longrightarrow$

$\left\{\begin{array}{c}x^{2}-9x+(3+y)=0\\\ 9x^{2}-(6+y)x+1=0\end{array}\right\|$ $\Longrightarrow$ $\left|\begin{array}{cc}1 & 3+y\\\ 9 & 1\end{array}\right|^{2}=\left|\begin{array}{cc}1 & 9\\\ 9 & 6+y\end{array}\right|\cdot \left|\begin{array}{cc}9 & 3+y\\\ 6+y & 1\end{array}\right|$ $\Longrightarrow$

$(9y+26)^{2}+(y-75)\left(y^{2}+9y+9\right)=0$ $\Longrightarrow$ $\boxed{y^{3}+15y^{2}-198y+1=0}$.

Here is an interesting solution using tirogonometry:
We now that,
$\sin{3\phi}=3\sin{\phi}-4\sin^{3}{\phi}$
now lets multipliy both sides by $2$ and rearrange it:
$(2\sin{\phi})^{3}-3(2\sin \phi)+2\sin 3\phi=0$   $(1)$
now this looks quite similar to the original one: ($x^{3}-3x+1$)
We can read this as , if every time that for $\phi$ the last part of $(1)$ is equal to the last part of $P$
then for this angle $2\sin \phi$ is the solution for $(1)$.
So :
$2\sin 3\phi =1$
The solution for this equation:
$\phi =10^{\circ}+k\cdot 120^{\circ}$   and $\phi = 50^{\circ}+k \cdot 120^{\circ}$
using that the sine function is symmetrink we get three differant sine values, that means three differant solutions:
The solutions for $(1)$ are:
$2\sin 10^{\circ}$
$2\sin 130^{\circ}$
$2\sin 250^{\circ}$
So if the $Q$ polynomial has the following form:
$Q=x^{3}+ax^{2}+bx+c$
Then:
$a=-2^{5}(\sin^{5}10^{\circ}+\sin^{5}130^{\circ}+\sin^{5}250^{\circ})=15$
$b=2^{10}(\sin^{5}10^{\circ}\cdot \sin^{5}130^{\circ}+\sin^{5}130^{\circ}\cdot \sin^{5}250^{\circ}+\sin^{5}250^{\circ}\cdot \sin^{5}10^{\circ})=-198$
$c=-2^{15}\cdot \sin^{5}10^{\circ}\cdot \sin^{5}130^{\circ}\cdot \sin^{5}250^{\circ}=1$
So the polynomial we are looking for:
$Q=x^{3}+15x^{2}-198x+1$