Did anyone hear about the Mars Rover - i'm sure most of you did. I just want to know what everyone thinks about the mission to mars. Is is reasonable? I think it's a waste of money, but maybe some of you have other opinions.

I think that the Mars rover is great!  We don't know enough about outer space yet and we need to learn more about it.  Thesun is becoming ared giant I am sure some of you are aware and in probally several hundred to milloins of years it will continue to expant and engulf Murcury and venus.  The Earth will remain but will become to hot for life as we know it.  That is why I believe that it is important to explore the universe.  I believe it is Jupiter thats they believe is still forming and whos atmosphere is a lot like Earths early one.

I agree with DarkMagic, it's not like it's going to affect us or our children or our grand-children or our great-grand-children...boy I'm starting to feel old :lol:

Anyway we should concetrate more on reality then our fantasies.

I don't have any particular complaints about unmanned space missions, especially carried out by a lot of different countries together.  But the idea of manned missions to Mars seems like a bad joke to me.  I mean, not that there's anything inherently wrong with it, but how the heck is anyone going to pay for such a thing?

[quote="JBL"]I wouldn't mind if I thought there were money available.  But right now, we could use a tax increase just to close up the defecit.[/quote]
Ah, but it's not like the deficit is a new thing... I read something once about the reason we keep a deficit, and it seems that we do it intentionally for some economic reason other than not having the money to get rid of it; unfortunately, I don't remember the details or the source.

Nonono -- that's a debt.  Debt isn't always bad -- every country in the world has it.  For a country to have (some) debt isn't necessarily a problem.  What is a problem is if debt gets to large -- namely, when you spend more money then you have for a very, very long time.  Right now, we are spending much more money then we have.  A reasonable attempt at getting to Mars would be really, really expensive.  There are some people (on both sides of the political spectrum) who worry that we're approaching the stage at which we have so much debt that people start to worry about whether we'll ever pay it back.  That's bad.  If we had the money sitting around, i.e. we weren't running (much) defecit otherwise, I would be all for research leading to a Mars mission.  (I'm much more skeptical of a moon base -- who really needs a moon base?)  But the cost is too prohibative for me when we're running big defecits.

I heard once that they were considering using water on the moon as a fueling station for mars missions. Water obviously doesn't work as a fuel to get off the Earth and get to Mars (weight), but if fueled in space, it could work.
(H2O + energy ---> H2 + O2)

[quote="confuted"]I kind of like the idea, even of manned missions. Sure, it'll be expensive, but think about the benefits to education from the space race. We don't have the technology or equipment to send men to Mars now; those things will have to be invented and built. People will have to learn more things in order to build them; math and science will likely get pushed more in schools. Sure, the government will be spending a lot of money, which will come out of us in the form of taxes, but hopefully it will have enough sense to spend that money supporting US industry and invention, in which case it's all in the country and essentially will just jump start the economy, as well as the benefits to education.[/quote]
*ahem* Do you realize that in the short run educatin will sharply decrease in quality [b]because[/b] of the budget cuts they'll receive if we were to send men to Mars? You're talking in the extreme long run and who knows whether the U.S. will hold out until actual good comes out of this or not?

It won't support the entire industry, only those space-related or president/vice-president/etc.-related

I think one of the reason for this expedition planning is to check if mars really have Martians living on it or not :lol:

[quote="Tare"]*ahem* Do you realize that in the short run educatin will sharply decrease in quality [b]because[/b] of the budget cuts they'll receive if we were to send men to Mars? You're talking in the extreme long run and who knows whether the U.S. will hold out until actual good comes out of this or not?

It won't support the entire industry, only those space-related or president/vice-president/etc.-related

I think one of the reason for this expedition planning is to check if mars really have Martians living on it or not :lol:[/quote]
Evidently you know nothing of what happened in the last space race. Math and science education were greatly promoted so that we would beat the Soviets. Also, although the money will initially go to the industries with a direct interest in space, those industries will pay their new employees, who will then go buy groceries and cars, etc. Any time there is a major increase in one industry, money goes around. As for the Martians... that's not worthy of a response.

[quote="bluemoose"]My ultimate concern is about people moving their dumping ground to another planet. That is not what we need to be doing.[/quote]
Why the heck not? It's another planet, and not ours. Furthermore, that's not even close to being economical at this point in time, or in the even remotely near future.

Does anyone have a compelling reason why these missions must be manned (outside of inspirational, and questionable, intangibles, which are far less valuable now than during the space race)?  Considering the dramatic cost increase involved in making a mission manned instead of unmanned, it seems that much more could be done if we scratched the manned portion.

As for the economic effect you cite of government putting money in the hands of engineers, who then spend on whatever - wouldn't this same argument justify a massive federal health program (or any one of a zillion other big-government programs)?  I would have thought you were a natural enemy of big government & massive spending programs.

The effect on education will likely be minimal this time around.  Last time, the space race was about competition and fear - both great motivators to overhaul education.  It's not clear what it's about this time.

As a side note, Sputnik may have been more responsible for getting evolution back in the schools than anything else.  I'll wager against the mission to Mars having a similar effect on evolution education this time around.

I'd rather see a Manhattan Project to get off oil than Man on Mars.  If successful, it would be a huge boon geopolitically and environmentally.

I am an enemy of big government. I would rather see NASA become privitized; more would likely get done, anyway, and there would probably be more funding. Perhaps that will be considered in the plan to get to Mars. Meanwhile, since it doesn't seem that the size of the government will be decreasing in the near future, and since the program already exists, I'll take whatever advances in science and math we can get.

I think that a primary motivator this time around is the fact that China is shooting men into space and vows to put them on the moon. India and Japan are also making technological advances at a rapid rate. The Soviet Union (and Russia) are defunct and aren't feasible rivals in anything technologically related. I think that the administration realizes that if we don't keep up with these other nations, we will be surpassed; maintaining our status as a superpower is a large part of what makes the USA what it is. Right now, we don't have the capacity to put a man in the moon, and haven't put men in orbit for months. I think fear and competition are both motivators this time around too.

I highly doubt Japan will get anywhere seeing we can count the number of times we've launched something into space without screwing something up in the past couple of years with one hand. We're pretty much terrible at space stuff which is ironic because we are one of the more meticulous people. I don't know about the other two so I'll leave it to the other experts.

Doesn't Japan launch a lot of satellites?

Heh yes it does, and I apologize for the long sentences. Let me highlight the words you missed:
[quote="Tare"]I highly doubt Japan will get anywhere seeing we can count the number of times we've launched something into space [b]without screwing something up[/b] in the past couple of years with one hand. We're pretty much terrible at space stuff which is ironic because we are one of the more meticulous people. I don't know about the other two so I'll leave it to the other experts.[/quote]
It's like launching missiles :lol:

To the extent fear and competition are motivators this time around, I think it's fair for a cynic to ask 'fear of what and competition from whom' and answer that with domestic politics rather than foreign threats.  Missile defense might be defensible with the response 'fear and competition', but I don't buy it for the space race at this point.  Besides, it still begs the question 'Why manned?'  Strategically speaking, you get much more bang for your buck unmanned.

Agreed on privatized space exploration.  I'd rather see human space exploration done [url=http://bulletin.ninemsn.com.au/bulletin/eddesk.nsf/All/827F6EAFF27B6DFBCA256D1E00079135]like this[/url].

I just don't get why we are so bent on discovering bacterial life - yes imagine the benefits<jk><jk> on mars when according to recent survey, the earth will lose 1 million species in a little over half a century at present rates of destruction. I am now really glad I am going to be dead in 80 some years - then again they probably will invent a life preservative pill. i won't take it though because I can't stand earth looking like a junkyard. ugh!

I think the Mars Rover is a great way to find out about what else is in the universe. It could one day help us discover things about our planet as well as ones that are farther away and undiscovered ones.

Welcome to reality - the earth will be too trashed to preserve by the time we find anything useful about the universe and extra terrestrial life - how can we preserve what little we have now? Why do we worry about other things when the first thing we should be worrying about is our own world. Mr. ruscyzk is right about it just being about some space race. Knowing mankind, nothing useful will result out of this project before we blow our world to smitherines(however you spell that).

I feel that your last statement was rather misinformed. We have not yet blown our world to smitherines, and the end of all life as we know it is not near. Species have been going extinct since they began to exist; they will continue to do so. If you believe in Darwin's work, it's necessary that inferior species become extinct and superior ones take their place. If species A prays on species B until species B is extinct and then species A has nothing to eat, species A either adapts or also becomes extinct; in reality, A prays on B, B's population decreases, A's population decreases, B's population grows due to less predation, A's population grows, and the cycle repeats, in general of course. No need to give specific counter examples, they exist and I don't deny it. Anyway, are you going to tell me that the original space race had no benefits to man kind? If you will, please don't tell me over the internet, and don't tell me while you're watching satellite TV. The things we have in space, made possible by technology that was developed for the space race and subsequent interest in business applications of space, play a larger role in your life than you seem to think.

[quote="confuted"]If you believe in Darwin's work, it's necessary that inferior species become extinct and superior ones take their place.[/quote]

That's not true at all.  Darwin's work applied to individuals within species, not species within the biosphere.  "Weaker" species [i]can't[/i] evolve -- they are the members of another species who get eaten.
Species have definitely been going extinct at a high rate since people have begun to have major effects on ecosystems.  Whether or not that is entirely due to human action is not necessarily clear -- a lot of species went extinct during the last ice age, too.  It is certain that we are seriously altering the environment on much of the face of the planet in ways that will have effects that we cannot possibly predict.

I don't think the internet was developed as part of or relating to the space race -- by the military in part, but not in relation to the space race.  Although you certainly do have a point with satelite TV ;).

I don't believe that the internet was developed as a direct consequence of the space race either, but a lot of the data being transmitted (between continents especially) is done via satellite.

An argument against a manned mission to Mars isn't necessarily an argument against a space program in general. You could rightly argue that there have been many benefits from the Space Program, but what specifically can you say have been the benefits of the moon landings? And the tremendous resources needed to create a manned mission to Mars would probably kill a lot of the other space programs - including the Hubble Telescope, which has been one of the most beneficial space projects of the last 15 years.

It seems to me that the unmanned projects should continue until more specific identifiable benefits can be found for manned missions. Rather than put so much of our resources into such risky project, we should diversify our space program to accomplish a lot of smaller, more productive, more immediate tasks. The original space race (and much of the Cold War in general) was quite irrational. I don't think we should be looking back to that model for how we want to do things today. We should be learning from the past to make better, smarter decisions.

[quote]That's not true at all. Darwin's work applied to individuals within species, not species within the biosphere. "Weaker" species can't evolve -- they are the members of another species who get eaten.
[/quote]

Actually, by Darwin's work, species do go extinct, completely naturally. There have been 5 great extinctions (known at least) since the beggining of life. Each of these eliminated a majority of the species on earth. This is all natural. Also, the average species has a lifetime of 4 million years before they go extinct.

If you believe that an asteroid crashed into the earth, killing the dinosaurs and other species, I ask this question: Was that natural?  Did it not destroy a majority of species at a huge huge rate? (at least much faster than all humans have been doing?)

I believe in conservation. We go to mars to explore new land... maybe to even discover the long coveted signs that life may have once existed. We are is search of water, yet all NASA can offer are vague hopes ( hopes disguised by the superficial research they have presented to the public). It may seem like such a great and momentous feat, but in fact it is another exploration that is likely to reveal little.

What are a consequences even if we do discover remnents of life or microorganisms and fossils. Can we make any use of these discoveries? That is the question.

Now scientists themselves want to investigate the exotic red planet covered with dust and rocks. They plan to ship supplies and build artificial greenhouses so the scientists can live off the oxygen supplied by those plants. It take 40 acres of foliated vegetation to sustain human life, and it takes two years of provisions - refering to the fact that mars and the earth line up  every two years. That is hundreds of thousands of gallons of water, not to mention the amount of food the has to be brought by every transport. I can assert that even our children won't be seeing the consequences of this endeavor. However, right now we eagerly pay our tax money to support a cause which doesn't have visible results. What if this is another useless space race to occupy mars? What if this isn't about extra-terrestrial life.

Bush has suggested that the whole idea of a permanent colony on mars and the moon would be feasable - who is he kidding? This is a hapless project that will cost us billions upon billions of dollars and world turmoil as we enter another phase of the space race. It disgusts me to imagine the consequences of such an endeavor as this.

Perhaps all this is...a very expensive form of entertainment. After all some of us want to live on the Moon for no apparent reason (besides it being "cool") and others pay millions to go to space (like that guy in Russia). Maybe this will become a new form of business or something, I don't know, humans are weird...

[quote="tetrahedr0n"]Actually, by Darwin's work, species do go extinct, completely naturally. There have been 5 great extinctions (known at least) since the beggining of life. Each of these eliminated a majority of the species on earth. This is all natural. Also, the average species has a lifetime of 4 million years before they go extinct.

If you believe that an asteroid crashed into the earth, killing the dinosaurs and other species, I ask this question: Was that natural?  Did it not destroy a majority of species at a huge huge rate? (at least much faster than all humans have been doing?)[/quote]

I didn't deny anything that you just said.  I did write that species go extinct naturally -- you may have seen my reference to an ice age in there.  However, this has nothing to do with evolution.  Also, you'll note that I didn't say that this wasn't natural.  What I did say is that we are making (and have been making) massive alterations in the ecosystems of the world, with consequences we cannot possibly understand as of now.  They might be minor, or they might be horrifically disasterous -- but we can't say for sure.

Seperately, I would not call the asteroid natural in the same sense as evolution.  Our world extends as far as our atmosphere.  Ignoring space travel temporarily, that's all we have to work with.  I think comparing human activities to something quite as random, purposeless, and extra-terrestrial as an asteroid hit is rather silly.  Better to talk about ice ages or volcanoes.  But space isn't going to solve the problems that we're (potentially) creating for ourselves here on earth.  We have enough difficulty keeping people alive in the Artic and Antartic to make the prospect of maintaining human life on another planet seem just silly.

I do like unmanned space things, though -- Voyager may start giving us some really interesting information as it passes out of the Solar System, Hubble is great, and satelites are very useful for a variety of things.

It's really much too expensive to move things into space to make that an economically feasable idea, scrambled.  Also, the scientific community would probably not approve.  There are better partial solutions to the waste problem -- they just aren't being implemented.  For example, limiting use of plastics and things like styrofoam, mandating recycling (and, ideally but less feasably, composting as well), requiring companies to limit the packaging that products come in, and punative sales taxes could all be used to help limit consumption.  Also, not all garbage goes to waste -- significant portions of Manhattan and Boston are built on top of landfill.  Now, I readily admit that none of these solutions fixes the problem entirely, but they all can do something about the problem.  And they're all cheaper than Mars :)

Scrambled wrote:I believe we should dump all our trash on mars...

[hide]the reason being is that on earth, our landfills make things completely toxic in the areas around it.  Filling landfills is a huge problem, and more and more land is being turned into landfill.  So in a few centuries, our earth is filled with trash.[/hide]

That would cost waaay too much money.  Do you realize how much it costs just to get an ounce of weight off of the Earth?

this is in the future... and people are too lazy to do the stuff you said jbl.  Guess why not everyone is doing it?  Besides, eventually, the trash will still pile up under all of your solutions, so we have to export it into space sometime(maybe throw it into the sun).  But out of space exporting, mars and moon are the most feasible.

About trash...my parents are really into recylcing and composting, and saving energy.  I think it's interesting to see how much we can recycle (my weekly job), compared to how much trash we have each week.

We compost tons of stuff.  My science fair project was to get a pile to turn in the dead of Nebraska winter using a heat lamp.  Pretty hard...

There's lots people can do to conserve, & I suppose some may get on me about us having 6 kids in my family, but I think it's interesting to see how much you can do when you really try to have an eye out for conserving.

Americans waste a ton of resources.

[quote="white_horse_king88"]Why should trash deposit be even viewed as not economically feasible - I know I know. However, it is just sad that money is what holds society together, and it is that which will tear society in half. Money is so big an issue - it's kinda sad how we are all servants of that green piece of paper. Sad....[/quote]

That is why economics is called "the dismal science"--because it confronts the reality of human nature and human society more than almost any other science. The reason money is important is that money is useful--almost every worthwhile thing that people want done, from curing AIDS to housing the homeless to monitoring election advertising for fairness and accuracy to relaxing after a hard day's work all costs money.

But the happy, nondismal aspect of sound economic theory is that economics teaches how to set up incentives so that people get worthwhile things done--and people respond to incentives. One of the reasons that immigrants (like many of the participants on this forum, or their parents) continue to stream into the United States is that the United States was founded by persons who had a profoundly correct understanding of economics--certainly much better than the leaders of China, for example, have had any time up till now. Policies that let people work in conditions of freedom, exchanging things of value with one another so that each person has what he most values, are policies that reduce infant mortality, improve public order, increase civil liberty, and generally make the world a better place. I think the biggest lack in the education of most young people in the United States government-operated school system is a proper appreciation of economics--and I think that stems from the largely economically irrational organization of that system. Learn more about economics, and you can learn more about how to make the world a better place.

Mark Twain (to whom many great lines are falsely attributed, and indeed this quotation can't be verified by the authoritative online source) is said to have noted that "there are two things money can't buy: happiness, and the love of a good woman." I have what money can't buy, and I don't mind giving away for free this advice, and the volunteer labor I put into local math contests and the like. But I have to find housing and food for my children just like all other parents, and I much appreciate an economic system like the one of the country in which I now live (I have lived elsewhere) in which I can choose freely what work to pursue, how to allocate my time, and what to spend my income on.

Technically, even though recycalling is a good idea, i am still in support of dumping our trash on mars.

a) do you know how many people are too lazy to recycle? i dont have statistics, but the amount of recycleable things in the trash is enourmous.  Besides, there is much more unrecycalable trash in the dump.  Toilet paper is about 70 percent of trash in landfills.  Nobody composts toilet paper (who would want to handle that?)

b) i suppose it will get more feasible in the future.  However, i may have my expectations too high...

[quote="JBL"]Listen, recycling is (or has been -- this is changing) viewed as not economically feasable -- and it's certainly a lot cheaper than Mars travel.  You can even put a lower bound on carting stuff to Mars if you can put a lower bound on energy prices, since it will always take a certain amount of energy to get off the surface of the earth, and that number doesn't change.  (At least not by any meaningful amount.)[/quote]
In megajoules of energy, no, that number does not change significantly. However, there are numerous systems under development to make shooting things into space much much cheaper. Ideas that I know of that are under consideration are a rail-gun type propulsion system up an Alaskan mountain, some sort of Gene Rodenbury-esque tether system, and an ion propulsion drive (I believe that was only for once you were in space though)

If you're in favor of dumping trash on Mars, are you in favor of paying to dump your trash on Mars? I have no idea how much it would cost, but I do know that it would be outrageously expensive. I have other objections to turning our neighbor planet into a garbage dump, but they won't be necessary right now.

Do you pay to have your trash moved to a landfill? Im quite uninformed...

It's in your tax.

I say it's not [b]our [/b]problem.


Set R is the set of all possible rectangles such that (1) only the grid points shown here are used as vertices, (2) all sides are vertical or horizontal and (3) no two rectangles in the set are congruent. What fraction of the rectangles in set R are squares? Express your answer as a common fraction.

[asy]import graph;
size(50);
for(int i = 0; i<5; ++i)
for(int j = 0; j<5; ++j)
fill(Circle((i,j),.1));[/asy]

There are 4 squares
(
width 1, height 1
width 2, height 2
width 3, height 3
width 4, height 4
)
There are a total of 10 rectangles
(
width 1, height 1,2,3,4
width 2, height 1,2,3,4
width 3, height 1,2,3,4
width 4, height 1,2,3,4

And of those 16, all but the 4 squares can be rotated 90 degrees to form another rectangle.
So there are $ \frac{16\minus{}4}{2}\equal{}6$ nonsquare rectangles, or 10 rectangles.
)

4/10=2/5


inequalities
calculus
inequalities unsolved

$0<\alpha<\pi$
Prove that:
$\cot{\dfrac{1}{4}\alpha}-\cot{\alpha}>2$

Hey, your puzzle is very nice and I also have a sol for it too.  I have 2 sol for your puzzle. The first one , I only use AM-GM and it is very easy to solve your puzzle. But the second sol is more hard than the first one and also more value than the first. And as I do the second one, I found that I can find a min of  $f(\alpha)=\cot{\dfrac{1}{4}\alpha}-\cot{\alpha}$

  Ok , here is mine.
 Put $x=cos (\dfrac{\alpha}{4})$ so $x$ is in $[\dfrac{\sqrt{2}}{2}..1]$
  Expand all of  $f(\alpha)$ and you will find that
$f(\alpha)=w(x)= \dfrac{4x^2-1}{4x(2x^2-1)\sqrt{1-x^2}}$
I will use caculus in here.
So $w'(x)= \dfrac{-16x^4+16x^6+4x^2-1}{4(2x^2-1)^2 \sqrt[3]{(1-x^2)^2}}$
Put $h(x)= -16x^4+16x^6+4x^2-1$.
We can solve $h(x)$ in this case. In fact $h(x)=(4x^3-2x-1)(4x^3-2+1)$
I have solved $r1(x)=4x^3-2x-1$ and $r2(x)=4x^3-2+1$
$r1(x)$ has only one real root  that satisfy $xr1=\dfrac{1}{6}\sqrt[3]{27+3\sqrt{57}}+\dfrac{1}{\sqrt[3]{27+3\sqrt{57}}}$
$r2(x)$ has only one root too but it is negative
So  $f(x)\geq w(xr1)$  in which $xr1=\dfrac{1}{6}\sqrt[3]{27+3\sqrt{57}}+\dfrac{1}{\sqrt[3]{27+3\sqrt{57}}}$
 :)


A driver approaches a toll booth and randomly selects two coins from his pocket. If the pocket contains 2 quarters,2 dimes, and 2 nickels, what is the probability that the value of the two coins he selects will be at least enough to pay the 30-cent toll? Express your answer as a common fraction.

is the answer[hide]$\frac{1}{2}$[/hide]?
jorian

[hide="not sure"]Is it $\frac{3}{5}$?[/hide]

i get

[hide="my solution"]$\frac{3}{5}$.  the number of ways he can select the coins is 6 choose 2 or $15$ the combinations that will work are:

quarter a and quarter b
quarter a and dime a
q a and dime b
q b and d a
q b and d b
q a and n a
q a and n b
q b and n a
q b and n b

thats $9$ so $\frac{9}{15}=\frac{3}{5}$[/hide]

I think Walk Around The River and laughinghead505 are right.  :maybe:

Walk Around The River and laughinghead505 are both right. The answer is $\frac{3}{5}$.

Yay! :lol: Here's how I got my answer:

[hide]We know that for the value of the two coins to be at least 30 cents, at least one of the coins has to be a quarter. We also know that if one of the coins is a quarter, the value of the two coins will be at least 30 cents no matter what. So we just need to find the probability of him choosing a quarter for at least one of the coins.

To do this, I just subtracted the probability of none of the coins being a quarter from one:

$1-\frac{4}{6}*\frac{3}{5}=\frac{3}{5}$[/hide]


let $ a_1, a_2, ... , a_{2n \plus{} 1}$ be $ 2n \plus{} 1$ integers. let property $ P$ be the following: for any $ 2n$ numbers out of $ a_1, a_2, ... , a_{2n \plus{} 1}$ there exists a way to divide the $ 2n$ numbers into two groups of $ n$ numbers whose sum of elements are equal.
prove that $ a_1, a_2, ... , a_{2n \plus{} 1}$ are all equal if and only if $ a_1, a_2, ... , a_{2n \plus{} 1}$ has property $ P$.

We only show $ P$ property implies equality. The other half is trivial.
We can assume all $ a_i\ge0$, otherwise replace $ a_i$ with $ a_i\minus{}\min a_i$.
We see $ a_i$ has same parity as $ \sum a_i$. If we replace $ a_i$ with $ b_i\equal{}[\frac{a_i}2]$, we see the property still holds with $ b_i\le a_i$ and thus the new series still have same parity. In other words, $ a_i\equiv a_j\mod 4$. Keep doing this operation, we will show $ a_i\equiv a_j\mod 2^k$, for all natural $ k$. Hence, all number have to be equal 'cause we can choose $ 2^k>\max a_i$.
Q.E.D.

Let $ S\equal{}a_1\plus{}a_2\plus{}...\plus{}a_{2n\plus{}1}$.
It's obviously that for all $ k$: $ S\minus{}a_k$ is even. So, all numbers are even, or all numbers are odd.
Let $ b_1\equal{}a_1\minus{}a_1\equal{}0$, $ b_2\equal{}a_2\minus{}a_1$, ..., $ b_{2n\plus{}1}\equal{}a_{2n\plus{}1}\minus{}a_1$.
It's easy to see that new sequence $ b_k$, $ k\equal{}1,2,...,2n\plus{}1$ has the same property, and $ b_1\equal{}0$, $ b_2$, ..., $ b_{2n\plus{}1}$ are even all.
Suppose that $ b_2\neq 0$ and $ b_2\equal{}2^{l}c$, and $ b_1$, $ b_3$, ..., $ b_{2n\plus{}1}$ are divisible by $ 2^l$ (without loss of generality, we can assume that $ b_2$ has the minimum power of 2 in the sequence).
If we replacing $ c_k\equal{}\frac{b_k}{2^l}$ ($ k\equal{}1,2, ..., 2n\plus{}1$), then new sequence has the same property, but $ c_1\equal{}0$, and $ c_2$ is odd. Contradiction (all numbers must be odd or all numbers must be even). So, $ b_1\equal{}b_2\equal{}...\equal{}b_{2n\plus{}1}\equal{}0$ and $ a_1\equal{}a_2\equal{}...\equal{}a_{2n\plus{}1}$.

If all numbers are equal, it's obviously, that the sequence has property $ P$.


[quote="G-UNIT"][quote="4everwise"]852 digits are used to number a book consecutively from page 1. How many pages are there in the book?[/quote]

[hide]9 + 90 = 99, 852-99= 753, 753/3= 251

99 + 251= [b] 350 pages [/b][/hide][/quote]

woops i made an error with the 2 digit it should be 180.  So its 99 + 221 or 320

Good job guys! I have a new question regarding pages and books.

You are trying to number as many pages as you possibly can in a book. You have plenty of 1's, 2's, 4's, 5's, 6's, 7's, 8's, 9's and 0's. There are only twenty-five 3's. How many pages can you number?

I think you got it right. I didn't do this question myself yet.  :huh:

[quote="ruofanisonfire"][hide]every ten pages needs a three. Then you need a three for 30, 130, and 230. you can number 232 pages. 22 threes for the ones digits and another three for each 30. [/hide][/quote]
[hide]Do you need a "3" for page number 31?[/hide]

oops! :blush:
[hide]I forgot to put in page numbers for 31,32,131,etc
hmmm...
I can't figure out a formula for this, but the answer is 132 pages.[/hide]

Good Job!  :surf:

There is no formula. You just use a single 3 for every 10 pages and an extra 10 3's when going from 30-39

[quote="Devendra Patel"]isnt it 320[/quote]
that was the first problem, but in the 6th post 4everwise made a new problem

Yeah, I thought that so many people were visiting the first post that I might as well post a continuation.


quadratics
ceiling function
pigeonhole principle
AMC
AIME

The members of a distinguished committee were choosing a president, and each member gave one vote to one of the $27$ candidates. For each candidate, the exact percentage of votes the candidate got was smaller by at least $1$ than the number of votes for that candidate. What is the smallest possible number of members of the committee?

[hide]
Say that there are $m$ members of the committee.  If a particular candidate received $x$ votes, the exact percentage of the votes received by that candidate is $\frac{100x}{m}$.  So for each candidate we have

$\frac{100x}{m}+1 \leq x$.

Summing this over all 27 candidates, we see that

$\frac{100m}{m}+27 \leq m\Rightarrow m^{2}-27m-100 \geq 0$.

Solving the quadratic, we see that $m \geq 31$.
[/hide]

Sorry for bumping an old thread; I was just studying and wondered if that was really the correct solution. When I took, I got:
$\frac{x}{100-x}\geq max(\left [\frac{x}{27}\right ])\implies min(x)=81$

where $[]$ denotes cieling. (Edit: I thought I knew the tag but forgot, what is the code again?

thx. now i know  what i did... I put lciel ... I can't smell ceiling...
thats what I did...

anyway any idea about whose solution is right.

Well, I don't think that first solution is right... because

by Pigeonhole, there's going to be at least one candidate with two votes,

and so $\frac{2(100)}{31}+1 \leq 2$ is clearly false...

Rather we see that $4 \geq  1+\frac{100(4)}{x}$, where $x$ is our minimum... so $x > 133$... so $134$ is our answer.

Notice $5 \geq  1+\frac{100(5)}{x}$ produces $x \geq  135$... so $134$ is our answer.

Wow, talk about a stupid mistake . . .

$\frac{100m}{m}$ is obviously 100, so my solution actually gives $m\geq 127$.  That still doesn't agree with white_horse_king88, though.

Hmmm... I can't see what went wrong in my solution but does anyone have the answer key so we know the right answer at least?

According to Kalva, it is indeed 134. This was from the AIME II 2003.

Yes, the problem with my solution is that it never takes into account the fact that the numbers of votes must be integers.

I agree with #6, the right answer is 134

[quote=mathman523]Can someone please explain post #6. I'm having a bit of trouble understanding it.

1.26aime#122003[/quote]

[hide=Different Solution]When optimized, each candidate will get about $1/27=37/999=3.7037037\dots\%=3\%+(7/10+1/270)\%=3\%+(19/27)\%$ of the vote.  From the last part of the equation, 8 one candidates could have 4 votes and the other 19 could have 5 votes.  $8*4+19*5=32+95=\boxed{127}$[/hide]

Our answers disagree.  I'm probably wrong though.

[hide=edit]Yeah, I was wrong.
It's 4 [i]percent[/i] and five [i]percent[/i] of the vote.  However, the vote doesn't have to be integers...[/hide]

Votes are generally integers.... after all, who as heard of a 4.5 votes? :P

find the minimum $k$ such that for all $p \leq k$, $$p - \frac{p}{k} \cdot 100 \geq 1$$

$$\frac{pk - 100p}{k} \geq 1$$ $$\frac{p(k - 100)}{k} \geq 1$$

Why did I just get 101?

Why doesn't this work?

[hide=incorrect solution]
Nobody can get $0$ or $1$ votes since $0< 0 + 1$ and $1 < 1 + \frac{1}{m}$ where $m$ is amount of people in committee/

So the "difference" between vote and percentage for each person is at least $2$.

We can write this as

\[ \frac{100c}{m} + 2 \le c \]

where $c$ is the number of votes a particular committee member got.

Summing these inequalities up for all committee members ( and multiplying by $m$, we get

\[100m + 54m \le m^2 \]

which means $m \ge 154$
[/hide]

Thank you!

[quote=hOLObESUTO]
So the "difference" between vote and percentage for each person is at least $2$.
[/quote]

The problem said that the difference between votes and percentage is at least $1.$ Am I missing something in your logic?

For example, we could have some guy get $3$ votes and $4\%$ of the total. Consider a committee with $75$ people where $3$ guys vote for this one dude, and everyone else votes for another. Sure, this is impossible, but I don't see how the preevious bit of logic or anything else in the problem rules this out as a possibility.

Okay, your point makes sense, but even if you replace $2$ with $1$ it still doesn't work?

I make an erroneous leap in logic/mixed up the concepts when I realized that everyone got at least $2$ votes.

[quote=white_horse_king88]The problem with that number is that if we try to maximize all the votes for each candidate, by pigeonhole, at least one will have $4$ votes. If you plug $4$ in to the equation with $127$ total votes, it doesn't work. So the total number of committee members has to satisfy the original statement... much like plugging solutions back into equations just to make sure they work.[/quote]

I believe that's the issue. Your solution is now almost identical to post #2's (last step differs, but you get the idea).

Let $v$ be the number of votes a candidate receives, and $m$ be the number of members. We have that
$$\frac{100v}{m} \leq v-1 \implies \frac{m}{100} \geq 1 + \frac{1}{v-1}$$
To minimize $m$, we need to maximize $v$. To do this, we let some $v = \left\lceil \frac{m}{27} \right\rceil$. Now we have
$$(m-100)\left(\left\lceil \frac{m}{27} \right\rceil - 1\right) \geq 100$$
We ignore the ceiling function for now to find a value that is close to an actual solution. This gives us that
$$(m-100)\left(\frac{m}{27} - 1\right) \geq 100 \implies m \geq 127$$
However, if we use $127$ in the original equation, we have
$$(m-100)\left(\left\lceil \frac{m}{27} \right\rceil - 1\right) = 27\cdot3 < 100$$
We note that for all $m \in [108, 134]$ that $\left(\left\lceil \frac{m}{27} \right\rceil - 1\right) = 4-1 = 3$. Thus we can look for an $m$ in that range such that $(m-100)\left(\left\lceil \frac{m}{27} \right\rceil - 1\right) = 3(m-100) \geq 100$. This gives that $m-100 \geq 34$, or that $m \geq \boxed{134}$.

[hide="Solution"]

Let's just assume for a while that we will have n people voting. If one of the 27 candidates receives 1 vote, then they will have $\frac{1}{n}$ percent of the vote. Well what is that value?

I know that this percentage must be at least 1 less than the number of votes. Well this is impossible, since $\frac{1}{n}$ is not 0. Hence none of the candidates can ever have 1 vote.

For 2, I know that I must have 1% or less of the vote. This is by the problem statement. Well 1% with 2 votes is just 2/200, therefore $n \geq 200$.

For 3, I see that this must be 2% or less of the vote. I have $n \geq 150$

For 4, $n \geq 134$

For 5, $n \geq 125$.

Now I see that n is steadily decreasing, meaning that if I have 125 or more, then I can have 6, 7 8 and so on number of votes and this would work. n is decreasing, because as I have increasing values for the number of votes received, the percentage is increasing. So I'll get eventually 50 votes is 49 percent or less, which means there will be about 100 votes. You can see that it is getting smaller.

Now I look at the 5 restriction. If I have 125 votes, I can give 5+ votes to each candidate. Well I can min give 5 votes to each candidate, but I have 27 and 27*5 is already 135. This will not satisfy, I can only get up to 133 like this using the restriction of 4 and 5.

If I look at the 4 restriction, I see that I can get up to 4*27=108, and I can get to 134 easily by adding some numbers that are not always 4. So the answer is $\boxed{134}$.

[/hide]

My explanation isn't really the best, but the idea is somewhat clear!


the font size of the forum is too small.
If I zoom in , the latex picture is terrible.

Can you increase you font size? I feel it is easy in technology.

Thanks

[quote="PowerOfPi"]There usually is a way to increase text size with your browser. What do you use?[/quote]

firefox , thanks

[quote="JBL"][quote="water2011"]If I zoom in , the latex picture is terrible.[/quote]  Select the option "zoom text only."   :roll:[/quote]

Thanks

[quote="JBL"][quote="water2011"]If I zoom in , the latex picture is terrible.[/quote]  Select the option "zoom text only."   :roll:[/quote]

Where is that option??? I am sorry, I just can't find it...

Look under "view", and then "zoom". It is the last one on the list.


What is the least number of coins you can have and still pay exactly the cost of any purchase less than one dollar?

[quote=AIME15]We definitely need $ 4$ pennies to be able to reach $ 0,1,2,3,4 \pmod{5}$.  We need at least one nickel so we can make $ 0,1,2,\ldots,9\pmod{10}$.  We need $ 2$ dimes by similar reasoning, one quarter, and a half dollar.  This adds up to $ 4\plus{}1\plus{}2\plus{}1\plus{}1\equal{}\boxed{9}$.  Not totally confident of this answer, someone please confirm.[/quote]

Yes you're right. But another way to think about it is in an extreme case(0.99). We need at least 9 coins(half dollars don't count).


algebra
polynomial
LaTeX
algebra proposed

[hide="Latex"]The code for $ x^{2007}$ is:
[code]$x^{2007}$[/code][/hide]


How do you do this one without doing too much work? (2003 National Sprint # 14)

For what value of a is there a right triangle with sides $a+1$, $6a$, and $6a+1$? I have the answer but I want the solution.

[hide]I got 10 as the answer. (a+1)^2+(6a)^2=(6a+1)^2  Now simplify as much as you can, and you should end up with a^2=10a. Well i know that 10*10 = 100 so that works. I dont know if this is what you call a fast solution, it took me about a 1 or 2 mins to do.[/hide]

Yeah, when you're doing a national, you don't want to bust your time all on one problem, as the first ten aren't as easy as they are in the Chapters, etc, making it harder to manage time and stuff... That is, unless you are amazingly fast at algebra and stuff... Thanks tho...

After I thought about the problem, I realized that the Algebraic way was byfar the easiest, and that i just added wrong  :blush:

to do these kind of problems fast, you need to do these problems a lto. and i mean a lot since it'll have to just jump to your head. I did the problem in about 30 sec since i did so many of these problems in 8th grade math (i skipped grades in math)

The algebra really isn't that bad.  For mathcounts I would advise being able to use $\((a+b)^{2}=a^{2}+2ab+b^{2}$ very, very quickly since it will really increase your speed for algebra problems.

[hide]6a+1 is obviously the hypotenuse.  a+1 and 6a must then, be the legs.
a^2+b^2=c^2.
a^2+2a+1+36a^2=36a^2+12a+1
a^2=10a so obviously, a=10.
The Pythagorian triple is 11,60,61.
That's as fast as I can get.[/hide]

I dont know if this solution is any quicker but it avoids some messy foiling.  As everyone said 6a+1 is obviously the hypotenuse so (a+1)^2 + (6a)^2=(6a+1)^2, so subtracting the (6a)^2 from both sides and doing difference of two squares you can get a slightly easier polynomial to factor.  BTW, This seems like a very easy Nationals question.


There is a Q charged circular loop with radius R.From center of the loop to the distance L,there exists magnetic field 2B,and from L to the distance R there is magnetic field with strength B.Both magnetic fields are going to zero(0) value slowly.Find the velocity of the loop when magnetic field in the system is zero.

Due to the change of the magnetic flux, an electric field will be generated along the loop, causing it to spin. Suppose the total flux decreases with the rate $\text d\Phi /\text dt \equiv \mathcal E$, which yields an electric field $\mathcal E/2R\pi$. The total torque of this field on the loop is $\tau = EQR = \mathcal EQ/2\pi = \text d\Phi Q/2\pi\text dt$, and the corresponding change of the loop's angular momentum is $\text dL = \tau \text dt = \text d\Phi Q/2\pi$. Since $L_{t = 0}= 0$ and $\Phi_{t \rightarrow \infty}= 0$, we get
\[\Delta L = MR^{2}\omega = \frac{\Phi_{0}Q}{2\pi}= \frac{BQ}{2}(R^{2}+L^{2}) \Rightarrow \omega = \frac{BQ}{2M}\frac{R^{2}+L^{2}}{R^{2}}.\]

So, you need to specify the mass of the ring, as well.


In the development of first law of thermodynamics,Sears and Salinger proceeds in the following way:

They have referred to a PV diagram where a gas has been taken from one eqlbm point a to another point b.They have considered an adiabatic free expansion from a to c,followed by an adiabatic reversible expansion from c to b.
Again they have taken another path where the gas undergoes an adiabatic reversible expansion from a to d,followed by an adiabatic free expansion from d to b.

Ultimately, they have said that the work in all adiabatic paths are equal between two states.

In this formulation they have neglected any dissipative work in the adiabatic free expansion process.(That is the process from a to c,from d to b).They have mentioned this,but did not justify.

Why should we neglect the dissipative work in that path?Not referring to 1st law of thermodynamics please clarify what would be the problem if we do not neglect the work.All my friends are saying that since first law is correct,hence...
I do not want this type of answer.I need to understand in physical terms why in the adiabatic free expansion process dissipative work is not changing the configuration.

Oh!!!Possibly I got the way:

Dissipative work<=> heat flow.
But we are dealing with adiabatic process.

Hence,contradiction!

Hence,assuming no dissipative work is justified.


geometry
circumcircle
parallelogram
trigonometry
inequalities
function
geometric transformation

Let $A_{1}A_{2}A_{3}$ be an acute triangle, and let $O$ and $H$ be its circumcenter and orthocenter, respectively. For $1\leq i \leq 3$, points $P_{i}$ and $Q_{i}$ lie on lines $OA_{i}$ and $A_{i+1}A_{i+2}$ (where $A_{i+3}=A_{i}$), respectively, such that $OP_{i}HQ_{i}$ is a parallelogram. Prove that
\[\frac{OQ_{1}}{OP_{1}}+\frac{OQ_{2}}{OP_{2}}+\frac{OQ_{3}}{OP_{3}}\geq 3.\]

Change notation: triangle $ABC$ circumcenter $(O)$ orthocenter $H,A_{1}\in OA,A_{2}\in BC$ parallelogram $OA_{1}HA_{2}$ let line $HA$ cut $BC$ at $A'$ and $A''$ be mid point of $BC$ then
$\angle (HA_{2},BC)=90-B+C\Rightarrow HA_{2}=\frac{HA'}{\sin(90-B+C)}$ and $HA'= HB\cos C=2R\cos B\cos C$ therefore $HA_{2}=\frac{2R\cos B\cos C}{\cos(B-C)}$ and $OA_{2}\ge OA''=R\cos A$ thus
$\sum_{a,b,c}\frac{OA_{2}}{OA_{1}}=\sum\frac{OA_{2}}{HA_{2}}\ge\sum\frac{R\cos A}{\frac{2R\cos B\cos C}{\cos(B-C)}}=\sum\frac{\cos A\cos(B-C)}{2\cos B\cos C}\ge\frac{3}{2}\sqrt[3]{\frac{\prod\cos(B-C)}{\cos A}}\ge 3$
because we have
$\prod\cos(B-C)=\prod\frac{\sin(B+C)\cos(B-C)}{\sin A}=\frac{\prod(\sin 2B+\sin 2C)}{8\prod\sin A}\ge\frac{8\prod\sin 2A}{8\prod\sin A}=8\prod\cos A$
We are done.

Here's a slightly different approach: (use GeMath's notation).
Considering triangle $ HBA_{2}$, which has angles $ \angle HBA_{2} \equal{} 90 \minus{} C$, $ \angle HA_{2}B \equal{} C \minus{} B \plus{} 90$, $ \angle BHA_{2} \equal{} B$, and using the fact that $ BH \equal{} 2R cos B$ (which can be derived from Sin Rule in AHB), gives us, using Sin Rule:
$ HA_{2} \equal{} \frac {2R cos B cos C}{cos (B \minus{} C)}$
$ BA_{2} \equal{} \frac {R sin 2B} {cos (B \minus{} C)}$
Using Cosine Rule in $ OBA_{2}$, where $ OB \equal{} R$, using value of $ BA_{2}$ found above, $ \angle OBA_{2} \equal{} 90 \minus{} A$, we have:
$ OA_{2} \equal{} \sqrt {R^2 \plus{} R^2 * (\frac {sin 2B}{cos (B \minus{} C)})^2 \minus{} 2R^2 \frac {sin 2B sin A} {cos (B \minus{} C)}} \equal{} R \frac {\sqrt {(cos (B \minus{} C))^2 \plus{} (sin 2B)^2 \minus{} 2 sin 2B sin A cos (B \minus{} C) }}{ cos (B \minus{} C)}$
Note that:
$ (cos(B \minus{} C))^2 \plus{} (sin 2B)^2 \minus{} 2 sin 2B sin A cos (B \minus{} C) \equal{} (cos(B \minus{} C))^2 \plus{} (sin 2B)^2 \minus{} 2 sin 2B sin (B \plus{} C) cos (B \minus{} C) \equal{} (cos(B \minus{} C))^2 \plus{} (sin 2B)^2 \minus{} sin 2B (sin 2B \plus{} sin 2C) \equal{} (cos(B \minus{} C))^2 \minus{} sin 2B sin 2C \equal{} (cos(B \minus{} C))^2 \minus{} \frac { cos (2B \minus{} 2C) \minus{} cos (2B \plus{} 2C)} {2} \equal{} (cos (B \minus{} C))^2 \minus{} \frac { (2 (cos (B \minus{} C))^2 \minus{} 1) \minus{} (2 (cos (B \plus{} C))^2 \minus{} 1) }{2} \equal{} (cos (B \minus{} C))^2 \minus{} [ (cos (B \minus{} C))^2 \minus{} (cos (B \plus{} C))^2] \equal{} (cos (B \plus{} C))^2 \equal{} (cos A)^2$,
from which we have:
$ OA_{2} \equal{} R \frac {cos A}{cos (B \minus{} C)}$,
so $ \frac {OA_{2}}{OA_{1}} \equal{} \frac {OA_{2}}{HA_{2}} \equal{} \frac {cos A}{2 cos B cos C}$.
We can prove a stronger inequality: $ \prod OA_{2} \geq \prod HA_{2}$ , from which our required inequality follows from AM-GM. This inequality reduces to the following:
$ cos A cos B cos C \leq 1/8$, which is a well-known inequality that follows from Jensen applied to the concave function $ f(x) \equal{} In (cos x)$.

- The fact that $ OA_{2} \equal{} \frac {R cos A}{cos (B\minus{}C)}$ is equivalant to proving that $ \angle OBA_{2} \equal{} 90 \minus{} |B\minus{}C|$, by considering Sin Rule in $ OBA_{2}$. This is equivalant to proving that $ \angle A_{2}A'' \equal{} \angle OA_{2}A''$. Perhaps there is a nice, Euclidean method of doing this, but I'm not sure how.

[hide="Solution"]
Let $ N$ be the midpoint of $ OH$ and $ M$ be the midpoint of $ BC$.  Reflect $ A$, $ B$, and $ C$ into $ N$ to form $ A'$, $ B'$, and $ C'$ respectively.  Now, if $ X$ is the midpoint of $ AH$, then since $ AH \equal{} 2R\cos A \equal{} 2OM$, we have that $ HX \equal{} OM$, so $ XHMO$ is a parallelogram, and so $ X$ and $ M$ are reflections into the midpoint of $ OH$.  It follows that $ X$ lies on $ B'C'$.  Now, $ B'C'\parallel{}BC\perp AH$, so $ HA$ is perpendicular to $ B'C'$.  Furthermore, $ M$ is the midpoint of $ BC$, so $ X$ is the midpoint of $ B'C'$, so $ H$ is on the perpendicular bisector of $ B'C'$.  Similarly, it is on the perpendicular bisector of $ A'B'$, so $ H$ is the circumcenter of $ \triangle A'B'C'$.  Now, the reflection into $ N$ maps $ H$ to $ O$, so $ O$ is the orthocenter of $ \triangle A'B'C'$ since $ H$ was that of $ \triangle ABC$.  Noitce that $ \frac {OQ_1}{OP_1} \equal{} \frac {HP_1}{OP_1}$ by the parallelogram condition.  Since $ P_1$ and $ Q_1$ are reflections into $ N$, $ P_1$ lies on $ B'C'$.  Thus, $ P_1$ lies on the perpendicular bisector of $ AH$, so $ AP_1 \equal{} HP_1$.  Now, it suffices to show that $ \sum \frac {AP_1}{OP_1} \ge 3$.  Let the altitudes of $ \triangle A'B'C'$ be $ A'J$, $ B'K$, and $ C'L$.  We have that $ \frac {XH}{JO} \equal{} \frac {AX}{JO} \equal{} \frac {AP_1}{OP_1}$.  It then suffices to have that $ \sum \frac {XH}{JO}\ge 3$.  However, since $ A'O \equal{} AH \equal{} 2OM \equal{} 2HX$, we have that we want for\[ \sum \frac {A'O}{JO}\ge 6\iff \sum \frac {A'J}{JO}\ge 9\]Now, notice that\[ \sum \frac {JO}{A'J} \equal{} \frac {\sum [B'OC']}{[A'B'C']} \equal{} 1\]so\[ \sum \frac {JO}{AJ} \sum \frac {AJ}{JO}\ge (1 \plus{} 1 \plus{} 1)^2 \equal{} 9\]by Cauchy and we are done.  [/hide]

By simple complex numbers with $a_1,a_2,a_3$ on the unit circle,
\[p_1=\frac{a_1(a_1+a_2)(a_1+a_3)}{a_1^2+a_2a_3},\; q_1=\frac{a_2a_3(a_2+a_3)}{a_1^2+a_2a_3}.\]Thus
\[\frac{OQ_1^2}{OP_1^2}=\left|\frac{OQ_1^2}{OP_1^2}\right|=\left|\frac{q_1\overline{q_1}}{p_1\overline{p_1}}\right|=\left|\frac{a_1^2(a_2+a_3)^2}{(a_1+a_2)^2(a_1+a_3)^2}\right|=\frac{R^2\cdot OM_1^2}{4OM_2^2\cdot OM_3^2},\]where $M_1,M_2,M_3$ are the midpoints of $BC,CA,AB$, respectively. Since $OM_i=R\cos{A_i}$,
\[\sum\frac{OQ_i}{OP_i}=\sum\frac{\cos{A_i}}{2\cos{A_{i+1}}\cos{A_{i+2}}}\ge\frac{3}{2}\sqrt[3]{\frac{1}{\cos{A_1}\cos{A_2}\cos{A_3}}}\ge3,\]where the last inequality follows from the fact that $\log\cos{x}$ is concave for $x\in(0,\pi/2)$.

[hide="Solution"]
A few definitions:
$H_1$ is the reflection of $H$ over $A_2A_3$
$H^*_1$ is the foot of the altitude of $A_1$ onto $A_2A_3$ -- trivially, also the midpoint of $HH_1$.
$M_1$ is the midpoint of $A_2A_3$

First, we show that $H_1$, $Q_1$, and $O$ are collinear. We know that $A_1O \parallel HQ_1$, so since $H_1$ lies on $A_1H$, $\angle H_1HQ_1 = \angle H_1A_1O$. Now, it is well known that $H_1$ lies on the circumcircle of $\triangle A_1A_2A_3$, so $\angle H_1A_1O = \angle A_1H_1O$. In addition, since $H_1$ is simply the reflection of $H$ over $A_2A_3$, and $Q_1$ lies on $A_2A_3$, $\angle HH_1Q_1 = \angle H_1HQ_1 = \angle H_1A_1O$; thus, by AA similarity, $\triangle H_1HQ_1 \sim \triangle H_1A_1O$. Finally, since $A_1O \parallel HQ_1$ and trivially $H_1H \parallel H_1A_1$, we must have that $OH_1 \parallel Q_1H_1$, which implies that $O$, $Q_1$, and $H_1$ are collinear.

Now, let $R$ be the circumradius of $\triangle A_1A_2A_3$. Note that by properties of parallelograms, $OP_1 = HQ_1$, and also note that $HQ_1+OQ_1 = R$. Finally, note that $\angle HA_1O$ is the difference between $\angle OA_1A_2 = \frac{\pi}{2} - \angle A_3$ and $\angle HA_1A_2 = \frac{\pi}{2} - \angle A_2$, so it is either $A_2-A_3$ or $A_3-A_2$. Since $\cos(-x) = \cos(x)$, it won't end up mattering which one it is. Using the well-known fact that $\angle A_2OA_3 = 2 \angle A_1$, we can derive that
$$OQ_1 = \frac{OM_1}{\cos(\angle A_2 - \angle A_3)} = \frac{OA_2 \cos(\angle A_1)}{\cos(\angle A_2 - \angle A_3)} = \frac{R \cos(\angle A_1)}{\cos(\angle A_2 - \angle A_3)}$$
Now, using this we can derive that
$$\frac{OQ_1}{OP_1} = \frac{OQ_1}{R-OQ_1} = \frac{R*\frac{\cos(\angle A_1)}{\cos(\angle A_2 - \angle A_3)}}{R * \frac{\cos(\angle A_2 - \angle A_3) - \cos(\angle A_1)}{\cos(\angle A_2 - \angle A_3)}} = \frac{\cos(\angle A_1)}{\cos(\angle A_2 - \angle A_3) - \cos(\angle A_1)} = \frac{\cos(\angle A_1)}{\cos(\angle A_2 - \angle A_3) + \cos(\angle A_2 + \angle A_3)} = \frac{\cos(\angle A_1)}{2\cos(\angle A_2)\cos(\angle A_3)}$$
By symmetry, we have that $\frac{OQ_1}{OP_1} + \frac{OQ_2}{OP_2} + \frac{OQ_3}{OP_3} = \sum \frac{\cos(\angle A_i)}{2\cos(\angle A_{i+1})\cos(\angle A_{i+2})} \ge 3\sqrt[3]{\frac{\cos(\angle A_1)\cos(\angle A_2)\cos(\angle A_3)}{8\cos^2(\angle A_1)\cos^2(\angle A_2)\cos^2(\angle A_3)}}$

Now we prove the fact that everyone seems to be brushing aside with an unjustified "log cos is convex" -- namely, the fact that $\cos(\angle A_1)\cos(\angle A_2)\cos(\angle A_3) \le \frac{1}{8}$. To show this, first note that for $A, B \in [0, \frac{\pi}{2}]$, $\cos(2A)\cos(2B) = (\cos^2(A)-\sin^2(A))(\cos^2(B)-\sin^2(B)) = \cos^2(A)\cos^2(B)-\sin^2(A)\cos^2(B)-\cos^2(A)\sin^2(B)+\sin^2(A)\sin^2(B)$ $\le \cos^2(A)\cos^2(B)-2\sin(A)\sin(B)\cos(A)\cos(B)+\sin^2(A)\sin^2(B) = (\cos(A)\cos(B)-\sin(A)\sin(B))^2 = \cos^2(A+B)$. Using this, we have that $\cos(\angle A_1)\cos(\angle A_2)\cos(\angle A_3) \le  \cos^2\left(\frac{\angle A_1+\angle A_2}{2}\right)\cos(\angle A_3) = \cos^2\left(90-\frac{\angle A_3}{2}\right)\cos(\angle A_3) = \sin^2\left(\frac{\angle A_3}{2}\right)\cos(\angle A_3)$ $ = \left(\frac{1-\cos(\angle A_3)}{2}\right)\cos(\angle A_3) = \frac{1}{8} - \frac{(\cos(\angle A_3)-\frac{1}{2})^2}{2} \le \frac{1}{8}$, as desired.

With this fact, the proof is practically done. Just note that
$$3\sqrt[3]{\frac{\cos(\angle A_1)\cos(\angle A_2)\cos(\angle A_3)}{8\cos^2(\angle A_1)\cos^2(\angle A_2)\cos^2(\angle A_3)}} = 3\sqrt[3]{\frac{1}{8\cos(\angle A_1)\cos(\angle A_2)\cos(\angle A_3)}} \ge 3\sqrt[3]{\frac{1}{8*\frac{1}{8}}} = 3$$
as desired.
[/hide]


Will the following reactions be primarily substitution or elimination?

$ CH_3CH(Br)CH_3 \plus{} OH^\minus{}(H_2O) \minus{}>$

[hide]$ OH^\minus{}$ is stabilized in water so cant go Sn2,it can go Sn1 though and maybe some elimination.
[/hide]

$ CH_3CH(Br)CH_3 \plus{} OH^\minus{}(Ethanol) \minus{}>$

[hide]Isnt this case almost the same as above?[/hide]

Aq. KOH favours substitution and alcoholic KOH favours elimination. (Think Why?)


AMC
USA(J)MO
USAMO
monovariant
graph theory

How was the usamo? I expect to get $ 28\pm2$, unless I just did something completely stupid (which I do not foresee at all).

I expect to get $ 14\pm2$ after solving 1 and 4, so hopefully, red mop (likewise, unless I did something incredibly stupid that I didn't catch).

[hide="I FAILED SO BADLY... AARGH :wallbash_red: "][size=59]~10[/size] :blush: [/hide]

I expect 6+. I hope for 8~

I made good progress on 5-6 and solved 3 but made a careless mistake on 4 (almost solved) :wallbash_red:

Maybe something like

0
0
6
3
1
1

Which would be pretty cool for a first USAMO. More likely,

0
0
6
1
1
1

Which would still be pretty cool for a first USAMO.

[quote="jb05"]Seriously? The one problem you solved is the second hardest? That, my friends, involves some skilled strategizing. Really, though, that is very impressive.[/quote]

Nick me and calvin all got that one. We failed at 1/2 though  :mad:

I think the hardness level (least to greatest) was

3,4,5,6,2,1

well, atleast in my case

I thought 4 was easy, I got that one too, but took too long...
However, I halfway got #3...

ok I was being optomistic

A more realistic score:

0
0
3
1
1
1

I think I got #1 and #4, and I was one step away from #3.

You must absolutely pwn graph theory/combinatorics and suck at NT if you had 3,4,5,6,2,1 as your difficulty... :P

Anyway since SC doesn't have enough USAMO qualifiers on the board, I think I got $ 14\pm 2$ also. My solution to 1 and 4 were essentially the same as the official ones and I wrote some non-trivial stuff for 2&5, so I'm hoping that I get that score...

GAH

I failed number 2 because I subtracted 1 from one side and added one on the other side.

 :maybe:

I got x1x2...x(k+1)-1=a^4+a^2+2

so i faild

That was #1. And don't feel bad, I failed it too. I don't quite know how, except that that day was just a bad day for me. Pwned #4 though.

I got $ \emptyset$.  Or rather, to use the superior surreal notation, I got fuzzy.

Okay, I deserve to be smacked for that pun.  Sorry, guys.

Edit:  So I would have gotten #1 in about 30 mins, and I don't know about #3.  #2 I did invert around A, but then proceeded to go nowhere.  I probably would have gotten about 9 points on Day 1, for the inversion and a solution to #1.

For Day 2, I think I would have gotten #4, but I'm not sure.  #5 I wouldn't have gotten, as I didn't see the obvious monovariant and focused on rewriting one of the $ r_i$ as a linear combination over $ \mathbb{Q}$ of the other two, which did no good.  #6 I might have gotten, as graph theory is one of my stronger points.  But probably not.  So probably 8 points on Day 2, for a total of 16-17.  Ah well - I will just have to practice more.  Next year...we'll see.
[/hide]

OH MAN, IT LOOKS LIKE NC IS PWNING MOP, WITH 3 MOPPERS!

I got 14, so I made Red Mop.

hm it seems I got 21
One point away from HM  :(... I'm depressed

4.

It doesn't sound very impressive, but I'm really happy about it.

 :)


Idon't think that any product of 3 consecutive integers equals 341.

[quote="math92"]Let (a* b * c) ¸ (a + b + c) = 341 be an equation where a, b and c are consecutive positive integers. What is the least possible value of a?[/quote]

I'm not sure what this wording means. Does a*b*c=341 or does a+b+c=341 or both?

yea, i can't figure it out either, please clarify.

well 341 has factors of 11 and 31 right? and consecutive could mean 11,31,and 51.  The thing i don't get is that comma.

[quote="math92"]i didn't write the problem. i didn't get the comma, and i thought you guys would.[/quote]

who wrote the problem?

We're not all Einsteins here, although you can go to his grave and ask this.
what does the comma mean?

it should read (a*b*c)/(a+b+c) = 341. I think this was from national mathcounts this year....problem 21 sprint if memory serves.

It's very easy with the factorization of 341

With that added bit of knowledge, it's much easier.
[hide="Solution"]
WLOG, let a<b<c

Factor 341 and replace a with b-1 and c with b+1:

$\frac{(b-1)b(b+1)}{3b}=11\cdot 31$

Then cancel the b in the numberator and denominator and multiply by 3:

$(b-1)(b+1)=3\cdot 11\cdot 31$

And we immediatley see that b is 32, so a's smallest value is 31.

Note that a can be 31,32,or 33.[/hide]

It's actually quite simpler than that assuming that $\frac{abc}{a+b+c} \text{ can equal } 341$

[hide="heres how I did it"]
$341 = 11 \cdot 31$

the smallest (and only) set of 3 consecutive integers which has an 11 and a 31 which are required for 341, is 31, 32, 33. any higher and you're in 62, 63, 64 which has no 11. Thus, a can be either 31, 32, or 33 and the smallest is $\boxed{31}$

If you have time when doing this, check and you'll find that it indeed works.

[/hide]


[quote="pardesi"]$Let$ $(u_{jk})$ , $1\leq j\leq 3$, $1\leq k \leq 2$ $be$ $real$ $numbers$
$and$ $let$ $N$ $be$ $an$ $integer$ $such$ $that$

           $\max_{1\leq k \leq 2}\sum_{j = 1}^{3}|u_{jk}| \leq N$

$suppose$ $M$ $and$ $l$ are $positive$ $integers$ $such$ $that$ $l^{2}< (M+1)^{3}.$
$prove$ $that$ $there$ $exists$ $integers$ $\zeta_{1},\zeta_{2},\zeta_{3}$ $not$ $all$ $0$, $such$ $that$

 $\max_{1\leq j \leq 3}|\zeta_{j}| \leq M$ $and$

$|\sum_{j = 1}^{3}u_{jk}\zeta_{j}| \leq \frac{MN}{l}$ $,$ $for$ $k = 1, 2$[/quote]
I think this should go in number theory. Also it is difficult to read from too much latex.

Let the a's and b's be real numbers, and N an integer with

\[\max(|a_{1}|+|a_{2}|+|a_{3}|,|b_{1}|+|b_{2}|+|b_{3}|)\le N\]

suppose M and L are positive integers such that $L^{2}< (M+1)^{3}$.

prove that there exists integers $c_{1},c_{2},c_{3}$ not all 0, such that
\[\max(c_{1},c_{2},c_{3})\le M\] and
\[|a_{1}c_{1}+a_{2}c_{2}+a_{3}c_{3}|, |b_{1}c_{1}+b_{2}c_{2}+b_{3}c_{3}|\]
are both \[\leq \frac{MN}{l}\]

thanks me@home for restatimg the problem but i think this has much to do with 'inequalities' than with 'number theory' :lol: .please someone do give  this a  :blush: try


P = 4n+1 and p is the prime
just prove n^n ≡ 1 mod p


An investor buys stock worth 10,000 (dollars) and holds it for n business days.
Each day he has an equal chance of either gaining 20% or losing 10%. However
in the case he gains every day (i.e. n gains of 20%), he is deemed to have
lost all his money, because he must have been involved with insider trading.
Find a (simple) formula, with proof, of the amount of money he will have
on average at the end of the n days.

I think,

Solution:
[hide]
The possibilities can be written as a pretty straightforward probability tree terminating at the nth branch (where the head of the tree is branch 0), and so the expected earnings holdings H after n days with a starting principle P (P=10,000) is the binomial sum:

$H=\binom n0 0+\binom n1 .5^{n}(1.2^{n-1}.9)P+\binom n2 .5^{n}(1.2^{n-2}.9^{2})P+\dots+\binom nn .5^{n}(1.2^{0}.9^{n})P$

$=.5^{n}P(\binom n1 1.2^{n-1}.9+\binom n2 1.2^{n-2}.9^{2}+\dots+1.2^{0}.9^{n})$
$=.5^{n}P[(1.2+.9)^{n}-1.2^{n}]$

$\boxed{=.5^{n}P[2.1^{n}-1.2^{n}]}$
[/hide]

[hide]On average, each day, he gains $(-10+20)/2 = 5 \text{percent}$.

Thus, after n business days, his average total will be $10000(1.05)^{n}$

Now, we subtract out the case of each day being a 20% gain.

This has value $P_{\text{event}}10000 (1.2)^{n}= \frac{1.2^{n}}{2^{n}}10000 = 10000 (0.6)^{n}$

Thus, $E = 10000(1.05)^{n}-10000(0.6)^{n}= \boxed{10000(1.05^{n}-0.6^{n})}$[/hide]

[b]Edit:[/b] Same as above.

Aha! My answer and Zuton's are the same after you take into account the $.5^{n}$ in mine.

And interestingly, substituting some integer days reveals he only benifits from holding the investment for at least 4 days-- for instance, if he cashes out on the first day, he must lose because if he got the 20% increase then he got it on all n days (n=1).


limit
integration
function
real analysis
real analysis unsolved

This is from an exam and I can't deal with it.

Let $ f_n : (0, 1) \longrightarrow \mathbb{R}$ satisfy Lipschitz condition with constant 1 and for every measurable set $ A \subset (0, 1)$ there exists $ \lim_{n \longrightarrow \infty} \int_A f_n$ and it is finite. Prove that a sequence $ (f_n)$ is convergent in $ L_1 (0, 1)$ space.

We divide the proof into several steps.
[b]Step 1.[/b] Let us show the set $ \{f_n\}$ has at most one limit point in $ L^1([0,1])$. Indeed, if $ (g_m)$ and $ (h_m)$ are two subsequences of $ (f_n)$ that converge in $ L^1([0,1])$ to $ g$ and $ h$, respectively, then
\[ \int_E g = \lim_{m\to\infty} \int_E g_m = \lim_{n\to\infty} \int_E f_n = \lim_{m\to\infty} \int_E h_m = \int_E h
\]
for any measurable set $ E \subset (0,1)$. So
\[ \int_E (g - h) = 0 \quad \text{ for any measurable set } E \subset (0,1),
\]
which implies $ g \equiv h$.

[b]Step 2.[/b] Let us show that the set $ \{f_n\}$ is relatively compact in $ L^1([0,1])$. First, note that the sequence $ (f_n(x_0))$ is bounded for any $ x_0\in(0,1)$. Indeed, taking $ E$ so that $ x_0 \in E$, we obtain
\[ f_n(x_0) \mu(E) = \int_E f_n(x) dx - \int_E \left( f_n(x) - f_n(x_0) \right) dx,
\]
where the first term in the RHD is bounded by data, and the second terms is bounded because functions $ f_n$ are Lipschitz continuous with the same Lipschitz constant.
Now, fixing arbitrary $ x_0\in(0,1)$, we deduce that
\begin{align*} \sup_{n\in\mathbf N} \|f_n\|_{L^1([0,1])} & \le \sup_{n\in\mathbf N} |f_n(x_0)| + \sup_{n\in\mathbf N} \int_0^1 |f_n(x) - f_n(x_0)| dx \\
& \le \sup_{n\in\mathbf N} |f_n(x_0)| + 1 < \infty. \end{align*}
Since
\[ \int_0^1 |f_n(x + h) - f_n(x)| dx \le |h| \quad \text{ for any } n\in\mathbf N \text{ and } h\in\mathbf R,
\]
we at last conclude that the set $ \{f_n\}$ is relatively compact in $ L^1([0,1])$.

[b]Step 3.[/b] Let us prove our initial statement. By relative compactness of the sequence $ (f_n)$ in $ L^1([0,1])$, it contains a [i]subsequence[/i] that converges to some $ f$ in $ L^1([0,1])$. Now suppose, contrary to our claim, that [i]the whole[/i] sequence $ (f_n)$ doesn't converge to $ f$ in $ L^1([0,1])$, then there exist a constant $ C > 0$ and a subsequence $ (f_{n_k})$ such that
\[ C \le \|f_{n_k} - f\| \quad \text{ for any } k\in\mathbf N.
\]
But by Steps 1-2, the sequence $ (f_{n_k})$ must contain a subsequence that converges to $ f$ in $ L^1([0,1])$, contradicting the above inequality. This contradiction proves our claim, and we are done.

[b]P.S.[/b] This problem is easily generalized to the case where the interval $ (0,1)$ is replaced with an open, bounded set $ \Omega \subset \mathbf R^N$, with the same proof.


trigonometry
geometry
trig identities
Law of Sines

Is the law of tangents any more useful than the law of cosine/sines? Or is it just a pointless variation?

actually the law of sines/cossines is the law of tangents

see http://mathworld.wolfram.com/LawofTangents.html

Law of tangents, I believe, is above the level of classroom math.  This should be moved.

this formula is useless...except at ictm...haha

The Tangent Law is hilariously pointless:

$\frac{\tan\frac{1}{2}{(\alpha-\beta)}}{\tan\frac{1}{2}{(\alpha+\beta)}}=\frac{a-b}{a+b}$

Where $a$ and $b$ are sides of a triangle, and $\alpha$ and $\beta$ are angles opposite to $a$ and $b$ respectively.

The Sine Law uses the same variables but is much simpler and more applicable than the Law of Tangets:

$Law.of.Sines>Law.of.Tangents$

The Tangent Law is derived from the Law of Sines:

From ${b}\sin{\alpha}={a}\sin{\beta}$ (the Law of Sines) we see

$(a+b)(\sin{\alpha}-\sin{\beta})=(a-b)(\sin{\alpha}+\sin{\beta})$ or $\frac{\sin{\alpha}-\sin{\beta}}{\sin{\alpha}+\sin{\beta}}=\frac{a-b}{a+b}$

Using the sum and difference formulas (which are $\sin{\alpha}+\sin{\beta}=2\sin(\frac{\alpha+\beta}{2})\cos(\frac{\alpha-\beta}{2})$ and $\sin{\alpha}-\sin{\beta}=2\cos(\frac{\alpha+\beta}{2})\sin(\frac{\alpha-\beta}{2})$) we get:

$\frac{2\cos\frac{1}{2}(\alpha+\beta)\sin\frac{1}{2}(\alpha-\beta)}{2\sin\frac{1}{2}(\alpha+\beta)\cos\frac{1}{2}(\alpha-\beta)}=\frac{a-b}{a+b}$

And since $\frac{\sin\alpha}{\cos\alpha}=\tan\alpha$ we get

$\frac{\tan\frac{1}{2}{(\alpha-\beta)}}{\tan\frac{1}{2}{(\alpha+\beta)}}=\frac{a-b}{a+b}$

The Law of Tangents was useful before the invention of calculators.


Find all functions $ u: \mathbb{R} \rightarrow \mathbb{R}$ for which there is a strictly increasing function $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that:

$ f(x\plus{}y)\equal{}f(x)u(y)\plus{}f(y)$ for all $ x,y \in \mathbb{R}$.

[quote="moldovan"]Find all functions $ u: \mathbb{R} \rightarrow \mathbb{R}$ for which there is a strictly increasing function $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that:

$ f(x \plus{} y) \equal{} f(x)u(y) \plus{} f(y)$ for all $ x,y \in \mathbb{R}$.[/quote]
Result $ u(x) \equal{} a^x\forall x\in R$ where $ a\in \mathbb R^{ \plus{} }$.
Solution:Firsly , it is easy to see $ u(0) \equal{} 1$;$ f(0) \equal{} 0$.
For all $ x;y;z\in R$ we have $ f(x \plus{} y \plus{} z) \equal{} f(x \plus{} y)u(z) \plus{} f(z) \equal{} f(x)u(y)u(z) \plus{} f(y)u(z) \plus{} f(z)$.
But $ f(x \plus{} y \plus{} z) \equal{} f(x)u(y \plus{} z) \plus{} f(y \plus{} z) \equal{} f(x)u(y \plus{} z) \plus{} f(y)u(z) \plus{} f(z)$Thus $ u(y)u(z) \equal{} u(y \plus{} z)$ (1)for all $ y;z\in R$.
Thus $ u(x)\ge 0$ for all $ x\in R$.
In the initial equation , when $ y \equal{} \minus{} z$ we have $ f(x)u( \minus{} x) \plus{} f( \minus{} x) \equal{} 0$.Because $ f(x)$ is a strictly increasing function so that $ f(x) > 0 \forall x > 0$ and $ f(x) < 0\forall x < 0$.
In the initial equation,again, when $ x \equal{} y;y \equal{} x$ we have $ (u(y) \minus{} 1)f(x) \equal{} (u(x) \minus{} 1)f(y)$.
If $ u(x) \equal{} 1$ thus $ f(x) \equal{} x$ satisfies the equation.
If $ u(x)$ is not a constant.Hence either $ u(x) > 1 \forall x > 0$ or $ u(x) < 1 \forall x > 0$.
[b]Case 1[/b]: $ u(x) > 1 \forall x > 0$: Thus when $ x > y > 0$ we have immediately $ u(x) > u(y)$. Since (1) we must have $ u(x) \equal{} a^x\forall x\in R; a > 1$ ( a classical equation).Then $ f(x) \equal{} a^x \minus{} 1$ satisfies.
[b]Case 2[/b] $ u(x) < 1 \forall x > 0$ thus $ u(x) \equal{} a^x\forall x\in R; a < 1$.
In conclusion, $ u(x) \equal{} a^x; a > 0 \forall x\in R$ [/u]

[quote=Allnames]
If $ u(x)$ is not a constant.Hence either $ u(x) > 1 \forall x > 0$ or $ u(x) < 1 \forall x > 0$[/quote]

Why is that?


calculus
integration
calculus computations

Calculate:

\begin{align}
  & Calcular: \\
 & \int_{0}^{1}{\left| x(2x-1) \right|}dx \\
\end{align}

TKX

hello, we have $ \int_{0}^1|x(2x\minus{}1)|\,dx\equal{}\frac{1}{4}$.
Sonnhard.

Alô Antonio 1965, acho melhor v. postar em inglês...

What are the items (1)  (2)  (3) ?

\[ for\;\;  0<x<{1\over 2} \equal{}> |x(2x\minus{}1)|\equal{}x(1\minus{}2x) \;\; and\;\; for\;\; x>{1\over 2} \equal{}> |x(2x\minus{}1)|\equal{}x(2x\minus{}1)\]
so , divide the interval of integration and it's done


calculus
integration
function
algebra unsolved
algebra

$ a_0, a_1, ..., a_n\in\mathbb{R}$ such that
$ a_0\plus{}\dfrac{a_1}{2}\plus{}...\plus{}\dfrac{a_n}{n\plus{}1}\equal{}0$.
Pove that the equation
$ a_0\plus{}a_1x\plus{}...\plus{}a_nx^n\equal{}0$ have at least a real root.

I have just solved in my head, but the solution by Integration.

I would like to see the solution without Integration.

It should be said $ n\equal{}1,\ 2\,\ \cdots$ and $ a_n\neq 0$.

Yes a solution without integration is worth seeing :
but here it is best to integrate :
consider :
$ g(x) \equal{} a_{0}x\plus{}\frac{a_{1}}{2}x^{2}\plus{}....\plus{}\frac{a_{n}}{n\plus{}1}x^{n\plus{}1} \plus{}C$
so, $ g(1)\equal{}g(0)$
so, by $ Rolle's$ $ theorem$ :
$ f(x) \equal{} g'(x)$ has a root in $ (0,1)$

[quote="pqrs"]Thanks :)

But, what is the Rolle's theorem :huh:[/quote]

If we have a continous function f on integral $ <a,b>$ and $ f(a)\equal{}f(b)$ then $ \exists_{c\in <a,b>}f'(c)\equal{}0$ :wink:


Hi !

How to find the real solution of this equation : $ x^5\plus{}5x^3\plus{}5x\minus{}1\equal{}0$ ?

The solutions are in the form (Maple):
$ \dfrac{\alpha\delta}{5}\minus{}5\dfrac{\bar{\alpha}}{\delta}$ where $ \alpha^5\equal{}1$ and $ 41139\plus{}30923\delta^5\plus{}16\delta^{10}\equal{}0$.
The real solution ($ \approx{0.193}$) is obtained with $ \alpha\equal{}e^{4i\pi/5}$  and $ \delta\approx{3.67\plus{}2.67 I}$.

A correction:
In my preceding post, the minimal polynomial of $ \delta$ is false; you must read: $ \delta^{10}\minus{}3125\delta^5\minus{}9765625$.
The real solution ($ \approx{0.193}$) is obtained with $ \alpha\equal{}e^{4i\pi/5}$ and $ \delta\approx{3.67\plus{}2.67 I}$; more precisely this last $ \delta$ satisfies $ \delta^5\equal{}\minus{}6250cos(2\pi/5)$.

You can have a look on:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=261079

[quote="blang"]Hi !

How to find the real solution of this equation : $ x^5 \plus{} 5x^3 \plus{} 5x \minus{} 1 \equal{} 0$ ?[/quote]

You can give soft Maple?

[quote="blang"]Hi !

How to find the real solution of this equation : $ x^5 \plus{} 5x^3 \plus{} 5x \minus{} 1 \equal{} 0$ ?[/quote]

Let put  $ x \equal{} t \minus{} 1/t$ then the equation becomes  $ t^5 \minus{} 1/t^5 \minus{} 1 \equal{} 0$. Now everything is OK.

I think we get Sonhard solutions then.

Yes it works ! Then the real solution is $ x \equal{} \phi^{\frac {1}{5}} \minus{} \frac {1}{\phi^{\frac {1}{5}}}$.

But how do you find $ x \equal{} t \minus{} \frac {1}{t}$ ?

Yes I have not seen that.
But the only polynomials of degree 5 that are transformed in a polynomial in the form $ t^{10}\plus{}at^5\plus{}b$ by setting $ x\equal{}t\minus{}1/t$ are the polynomials:
$ x^5\plus{}5x^3\plus{}5x\plus{}\alpha$ (up to a constant).
Namdung did you know the solution ? If not, you are a genius.

[quote="loup blanc"]Yes I have not seen that.
But the only polynomials of degree 5 that are transformed in a polynomial in the form $ t^{10} \plus{} at^5 \plus{} b$ by setting $ x \equal{} t \minus{} 1/t$ are the polynomials:
$ x^5 \plus{} 5x^3 \plus{} 5x \plus{} \alpha$ (up to a constant).
Namdung did you know the solution ? If not, you are a genius.[/quote]

Of course not. But I did know the trick. So I put $ x \equal{} t \plus{} a/t$and find adequate $ a$. Actually I did by following way

1) $ f(x) \equal{} x^5 \plus{} 5x^3 \plus{} 5x \minus{} 1$
2) I find: $ expand(f(t\plus{}1/t))$ ---> No thing special
3) I find: $ expand(f(t\minus{}1/t))$ ---> Luckily I get result = $ t^5 \minus{} 1/t^5$.

So, I did not know the solution. But I'm not genius. I just act by analogy. In fact, the substitution x = a(t+b/t) solves much og equations.

Have fun.

Hi Namdung,
the change $ x\equal{}t\plus{}a/t$ works for equations in the form $ x^5\plus{}bx^3\plus{}\dfrac{b^2}{5}x\plus{}e\equal{}0$ and for equations in form $ x^3\plus{}px\plus{}q\equal{}0$.
    You knew the trick and it's very well.

Of course, you can always substitute $ x\equal{}u\plus{}v$ and write $ u$ and $ v$ in terms of $ x$ such that all but some set of desired terms cancels. That's the motivation I've seen for the solution to the depressed cubic.

[quote="loup blanc"]Hi Namdung,
the change $ x \equal{} t \plus{} a/t$ works for equations in the form $ x^5 \plus{} bx^3 \plus{} \dfrac{b^2}{5}x \plus{} e \equal{} 0$ and for equations in form $ x^3 \plus{} px \plus{} q \equal{} 0$.
    You knew the trick and it's very well.[/quote]

Yes, thank you Loup Blanc. Actually I learnt this trick from the following equation:
        $ x^5 \plus{} 10x^3 \plus{} 20x \minus{} 18 \equal{} 0$


let P,A,B be three points wich arnt colinear.and let M be a point wich is colinear with A and B. prove that:

the angle betwin the circumcircles of the triangles PAM and PBM ,is fixed.(i mean stable )

When you talk about the angle between the circles, do you mean the angle formed by the tangents at the intersecting points?

yes ure right.in spesial form we have orthogonal circles.


[quote=GameBot]Probability with Repeated Events Part 1

We learn how to find the probability a repeated event occurs a certain number of times out of a given number of trials.  We do so by determining if a weaker player is better off playing a single winner-take-all point, or playing a best-of-five series of points.[/quote]
COOOL

I think you get an award for
Longest Bump That Didn't Need To Happen

LOL.

Okay this forum makes me laugh so much; I don't know it I can stand it :rotfl:


analytic geometry
combinatorics open
combinatorics

consider a square in a cartesian plane, with an angle in the origin and with a rational side length. for each point with rational coordinate inside the given square, can we find a finite integer n such if dividing the square in n*n square table, the point is in the center of one of the little square divided??

It is not even true for a square parallel to the axes.  In the unit square with vertices $(0, 0)$, $(0, 1)$, etc., you cannot put ${(\frac{1}{3}, \frac{1}{3}}$ at the center of a square in usch a tiling.


I was "Watching" the Intermediate IS B forum when I was enrolled in it, but now that mine has been moved to the archives and a new forum has replaced it, I would like to unsubscribe to the messages that I am still getting. How can I do this?

Good question :P For now there's nothing we can do about it. Will be dealt with in the future though ..  :lol:

[quote="mel"]Do you think it'd be possible for me to get temporary access to it so that I can click the "Stop Watching..." link?[/quote]Sure.


geometry perimeter trigonometry parameterization

Hey guys,
anyone remember the answers to the 2008 nysml competition?
if you do, could you please post them? thank you

um, well anyone remember the questions? :(

Below are 6 of the individual questions.  (I am missing the first pair.)  Feel free to post your solutions.

[hide="Questions 3 and 4"]
[b]I3.[/b] An isosceles triangle has perimeter 20 and two sides each of length 8.  Compute the area enclosed by the triangle.

[b]I4.[/b] Compute the sum of the reciprocals of the [u][b]real[/b][/u] solutions to
\[ x^5 \minus{} 4 x^4 \plus{} 3 x^3 \minus{} 8 x^2 \plus{} 32 x \minus{} 24 \equal{} 0.
\]
[/hide]

[hide="Questions 5 and 6"]
[b]I5.[/b] Compute $ \sin(2008^\circ) \cos(28^\circ) \minus{} \sin(28^\circ) \cos(2008^\circ)$.

[b]I6.[/b]  A "[i]Primitive Pythagorean Triangle[/i]" is a right triangle with side-lengths that are three positive integers such that no two side-lengths have a common factor other than 1.  Compute the number of primes $ p < 2008$ for which there are Primitive Pythagorean Triangles such that the difference between the length of the hypotenuse and the length of one of the legs equals $ p$.
[/hide]

[hide="Questions 7 and 8"]
[b]I7.[/b] $ \frac{1 \plus{} 8i}{7 \plus{} 4i} \equal{} a \plus{} bi$ where $ a$ and $ b$ are real numbers and $ i^2 \equal{} \minus{}1$.  When the complex number $ a \plus{} bi$ is plotted in the complex plane, compute its distance from the origin.

[b]I8.[/b]  Compute the number of $ n \in S \equal{} \{ 1, 2, 3, \dots, 2008 \}$ for which $ x^2 \minus{} y^2 \equal{} n$ has a solution with $ x$ and $ y$ integers.
[/hide]

First two:

[hide="Questions 1 and 2"]
[b]I1.[/b] Compute $ x$ where $ 2^8\plus{}2^8\plus{}2^8\plus{}2^8\equal{}8^x$.

[b]I2.[/b] The number 2008 is a four-digit number consisting of four even digits whose sum is 10. Compute the number of integers in $ \{1000,1001,1001,\ldots,9999\}$ that consist of four even digits whose sum is 10.
[/hide]

Ideas for all solutions:
[hide="1 and 2"]
1. Express as 2 to some power and then use $ 8\equal{}2^3$.
2. Case work.
[/hide]

[hide="3 and 4"]
3. I used Herons. You could also drop the perpendicular to the base.
4. Notice $ x^5\minus{}4x^4\plus{}3x^3\equal{}\frac {\minus{}x^3}8 (\minus{}8x^2\plus{}32x\minus{}24)$.
[/hide]

[hide="5 and 6"]
5. $ \sin$ addition formula, and $ \cos{\minus{}x}\equal{}\cos{x}$
6. Most people used the parameterization of primitive Pythagorean triples and went on that. I instead started with $ x^2\plus{}y^2\equal{}(x\plus{}p)^2$ and showed that $ p|x$ or $ p|2$ as well as $ p|y$ (and since its primitive, $ p|2$).
[/hide]

[hide="7 and 8"]
7. Express in polar form or just know that $ \frac{|z|}{|w|}\equal{}\left|\frac zw\right|$. In the end I multiplied by conjugate etc and found the quotient to check, though
8. $ (x\plus{}1)^2\minus{}x^2\equal{}2x\plus{}1$, so you can get all odds. Then $ (x\plus{}2)^2\minus{}x^2\equal{}4x\plus{}4$, so you can get all multiples of 4. Now use squares mod 4 to show you can't get the rest.
[/hide]

Okay these are my answers... can someone plz check them?
I'm not sure about #6 -- whats the answer to that?

1. 10/3
2. 3/4
3. 4* sqrt (15)
4. 11/6
5. 0
6. ??
7. 1
8. 1506

#2 is 34.

Here's a full solution to #6:

First, note that 2008 is a total red herring. In other words, it has nothing to do with the problem.
Start with $ x^2\plus{}y^2\equal{}(x\plus{}p)^2$.
$ y^2\equal{}2xp\plus{}p^2$, so $ p|y^2$ and we have $ p|y$. Let $ y\equal{}kp$ for some integer $ k$. Now,
$ x^2\plus{}(kp)^2\equal{}x^2\plus{}2xp\plus{}p^2$
$ 2xp\equal{}(k^2\minus{}1)p^2$
$ 2x\equal{}(k^2\minus{}1)p$

So $ p$ divides the left hand side, and so $ p|2$ or $ p|x$. But $ p$ already divided $ y$ and we want the triple to be primitive, therefore $ p|2$ so 2 is the only prime that works. (3-4-5 triangle). Then correct answer is 1.

Oh, yeah that makes sense. Sorry, 3/4 was a typo, i meant 34.

thanks guys

[quote="caffeineboy"]...so 2 is the only prime that works. (3-4-5 triangle). Then correct answer is 1.[/quote]

Yes, but 3-4-5 isn't the only way to get 2.  There's also 8-15-17, 12-35-37, etc.  In fact there should be an infinite number of triples that work: for coprime integers $ m, n$, we can generate Pythagorean triples as $ (m^2\minus{}n^2, 2mn, m^2\plus{}n^2)$.  If $ n\equal{}1$ then $ m^2\minus{}n^2\equal{}m^2\minus{}1$ and $ m^2\plus{}n^2\equal{}m^2\plus{}1$ which differ by 2, as needed.  If $ m$ is odd then all numbers in the triple will be even so it won't be primitive.

But for even $ m$ it will be primitive.  $ 2mn\equal{}2m$, so for it to share a factor with $ m^2\minus{}1$, which is odd, $ m$ would have to divide $ m^2\minus{}1$, but it already divides $ m^2$ so that's impossible.  Similarly $ 2m$ and $ m^2\plus{}1$ are coprime.  Finally, $ m^2\minus{}1$ and $ m^2\plus{}1$ differ by 2, yet 2 divides neither of them (both odd), making these coprime as well.  None of them share common factors so this is a primitive triple, and we can generate infinitely many by using $ n\equal{}1$ and any even $ m$.

The question asked "how many primes $ p$...", not "how many Pythagorean triples...".  So it doesn't matter how many times $ 2$ is generated, the answer is still just $ \boxed{1}$ (the number of primes that work).


Suppose Y1....Yn are independently and identically distributed, with E(Yi) = u and var(Yi) = sigma^2.

Also, X = g(sample mean of Y)

Show that E(X) = g(u) + O(1/n) and Var(X) = (g'(u))^2/n + O(1/n^2)

try with \[ g(Y_i)\equal{}g(u\plus{}(Y_i\minus{}u)) \equal{}g(u) \plus{} g'(u) \cdot (Y_i\minus{}u) \plus{} \frac{g''(u)}{2}(Y_i\minus{}u)^2 \plus{} (\text{error term})\]
Keyword: delta method

Thanks for your reply.

But I still don't know how you should proceed...


If $a*b = b\sqrt{a^b}$, what is the value of $2*(3*2)$?


I know but few who have not gotten in. This probably has more to do with who I know than the general acceptance rate.

Not incredibly hard, but hard enough that you should definitely put significant effort into the application.  If you are passionate about and generally good at math, that will shine through if you put in the time, and you will get in.

^^Agreed
The amount of people I knew that got in was about the same as the amount that didn't. Wasn't the acceptance rate about 30% last year? Seems to be getting harder every year, But E^(pi*i)=-1 is right. :)


It seems that quoting isn't working for me.

[quote]this[/quote]

the "this" has quotes around it in the post, but it doesn't show up.


calculus
integration
trigonometry
complex analysis

yes, but I think that it's hard to continue.
Hint: use the residue at infinity.

Ok.
It is generally used $z=4e^{it}$, in this case the result becomes $\frac{7i\pi}{180}$.


algebra
polynomial
real analysis
real analysis unsolved

Define polynomials $f_{n}(x)$ for $n \ge 0$ by $f_{0}(x) = 1$, $f_{n}(0) = 0$ for $n \ge 1$, and

$\frac{d}{dx}f_{n+1}(x) = (n+1)f_{n}(x+1)$

for $n \ge 0$. Find, with proof, the explicit factorization of $f_{100}(1)$ into powers of distinct primes.


According to the standard convention for exponentiation,
\[2^{2^{2^2}} \equal{} 2^{\left(2^{\left(2^2\right)}\right)} \equal{} 2^{16} \equal{} 65,\!536.\] If the order in which the exponentiations are performed is changed, how many [u]other[/u] values are possible?

$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

I think it is 1.  We can write $ (2^2)^4\equal{}4^4\equal{}256$.  I believe that is the only other way to do it.

[hide]Through experimentation, $ 2^{2^2}$ is always 16 (just check 2 cases) and so the values are either 256 or 65536. So B.[/hide]


Find all rational angles $0\leq x \leq90$ such that $sinx$ is also rational

I think we 've already discussed about the following pb:
find all rational value of x(radian) such that cos(x*Pi) is also rational. Make a research,
there were many ways to prove it.


Hi,

As the resources you have on your hands to study a topic increase, it becomes much more important and critical to take note of them (and write them down) with a good method. For this matter my story at university was tragic. I gathered a lot of resources (books, PDFs, papers, and etc.) relatively for all the courses I took at the university. But I have not been so successful in them (but to be honest I had not been too bad though, teachers consider me as someone who is at top of class :oops: !!).

I thought a lot about the reason why I can't do good, and I came to these conclusions:
1.When you study a topic, you have to, somehow, get an overal and comprehensive view of that. You can't do this without writing it down in a systematic, tidy and classified way. A PILE of small facts about sth is worthless, because they can be easily forgotton. You need a general view after all. I was unable to find a way to systematically take note of a lot of things I read until today!!!

2.Having a lot of books is more harmful than helpful.

These days I finally decided to remedy my bad method; My way didn't work. So to solve my problem, I need your help most!  :| Assist me to find good method(s), by telling me your way to study.

thx

For one thing keep your notes nice and neat, usually in a binder of some sort.

Have a title on each paper so you know what subject, part, etc.

Write side-notes, those usually clear up things really nicely.

Keep your homework, sometimes homework are the best notes.

Thats what I do.

The key to studying is having good notes. If you don't take good notes in class, you will need to compile them yourself before you study. This isn't a huge problem, since the process of creating good notes helps re-enforce your understanding, too, but it takes time.

Use rough pictures and graphs liberally, and use formatting (levels of indentation and bullets) to help you organize your information. [hide="cont"]I've found that capitalization doesn't help as much, but that's just me. This way, when you read, you can just look at the left-most part of the page if you only want to catch the main ideas, but can shift towards the center of the page if you want to go into detail. This doesn't mean that you have to have 5 different colored pens, as some people do, but if you can switch quickly, it helps to use them in pictures.[/hide]

Make sure that everything you write is in context and near like content [hide="cont"]If the teacher uses a vocabulary word and defines it, don't just throw it serially into your notes, but keep a running list of vocabulary definitions on the side, and make a cross reference from your main notes. This just makes it easier for you to find what you're looking for, and to focus on certain things when you study. It also prevents it from being a blob of text.[/hide]

Of course, you always want to put in the derivations of any formulas or statements you find. [hide="cont"]History may be boring at times, but knowing the experimental procedures by which a fact became known can help you retain it when you're going through your notes; I find it easier to remember sequences of facts than advanced conclusions (even if the sequence leads to such a conclusion).[/hide]

If you ever need an example of good notes, or even a copy of good notes, check out [b]MIT's OpenCourseWare (ocw.mit.edu)[/b], which has course materials for almost all of MIT's classes. If you find that you just can't keep up with the lecturer, or that there's just too much information to sift through, looking at someone else's notes on the same topic can help.

Lastly, never hesitate to ask to borrow someone else's notes! But it's best to do this as the class is progressing, since right before the exam, people will be less willing to give them to you.

[hide="On fine memory"]At some point in time, you may need to end up plowing through lots of small facts. If you want to improve your recollection of small facts (which is important), exercise in the mornings and make sure you're fresh before class.

Also, a fright response in context tends to help you remember things; a friend of mine said that medical school they yell at you if you do something wrong and tell you the correct answer, and the increased adrenaline and norepinephrine flow ensures that you will remember it.[/hide]

Thanks A M Y for the ocw, I never looked at it that way!  :)
I thank you too, hunter34. :wink:

But I have to add sth here. I had used to writing my notes on A4 papers in the recent years, but I soon found that it became too much cumbersome to handle them and as mentioned before I knew a lot small things about the matter, but not a little whole view, even for God's sakes! :maybe:  :( I guessed that it should be because I can't manage these a lot of single papers and so decided to switch back to the "notebook method." :maybe: Then another problem occured and it was that:

Part of my personal problem is that when you write your notes in a notebook, let you're reading sth on statistics for example test of hypothesis. You take the notes after studying and put your notebook down. Then perhaps a week later you may be reading conditional probability and bayes rule from another book and solve some fine problems or may be you find an interesting thing in a paper related to this. Now you want to jot this down too. Here is where I am stuck!!! :| Where should I write it to keep the logical and practical flow of the material?!

At the end by rereading your replies: I wonder why I didn't have these problems when I was in school! :o  Perhaps the problem is not the methods, be it A4s or notebooks, but it is ME who has changed and has lost the habit of good focused studying. I don't know. But whatever it is, I am indomitably trying to recover.

When I first started college, I took notes writing backwards on the page (I was bored).  This seemed like fun until it came time to study.  Then I had a problem - I couldn't read backwards nearly as easily as I could write backwards, so I'd have to hold a mirror by my notes to read them.

So, I scrapped that system and found a much better one.  I didn't take notes: I paid attention in class and thought about what was going on instead.  Meanwhile, I'd cut a deal with someone in the class whereby I'd help them with their studying if I could borrow their notes.  Pick a person who takes good notes, and it's a winning strategy.  The key, though, to making this work is to get ahead in the class early and stay ahead.  Hard concepts need to rattle around in your head for a while before they make sense.  If you mindlessly take notes and hope to cram for tests, you're sunk.

I used to take quite good notes when I was a student though I also always tried to understand what was said before writing it down. Fortunately, my professors went just at the right speed for me to do both. I found that relying on other people's notes is generally a bad idea especially on the notes of somebody who needs "help with their studies". One technique I found useful before the exams was to read the notes and to write down a short "summary" of them omitting all the "obvious" things and leaving only the things that require effort to grasp. Then I studied this summary instead of the notes.

[quote="A M Y"]

Also, a fright response in context tends to help you remember things; a friend of mine said that medical school they yell at you if you do something wrong and tell you the correct answer, and the increased adrenaline and norepinephrine flow ensures that you will remember it.[/quote]

 :rotfl: That is so true. I usually cannot memorize all the small facts for my AP US History class. Sometimes, my teacher would ask a "detailed" question and direct it at me... Our exchange would went like this:

Teacher: So [name], what's significant about Herbert Hoover?
Me: Um...[adrenaline] I don't know.
Teacher: Well, did you read the chapter?
Me: [adrenaline] Yes.
Teacher: Then, what was he known for?
Me: [stammers and shaking]
Teacher: I will get back to you ok?
Me: [turn the pages and trembling]

Until now, I still remember that he is associated with Food Administration during the World War (but was it I or II? :D ).

I agree with hunter34 and A M Y.  Some of this is a repeat of with they have mentioned  (repetition is a valid learning technique   :P ). Some things that work for me:

Keep a journal of the most important things of the day.
[hide="continued"] I write down ideas, concepts, questions, theorems, sketches of proofs, connections and associations, references to other notes, ect.  [/hide]

Use bold colors or highlighters to help catch your attention to the main points.
[hide="continued"]Come up with an association to the colors.  For me, red is reserved for the main points, blue for questions to be answered, green for side comments, orange for connections to other problems, ect.  When I see those colors, I know what they reference.  [/hide]

Keep several sets of notes running on any given topic.
[hide="continued"]As hunter 34 and A M Y said, it is vital to give your notes meaningful titles, include the title of book/paper/talk, author/lecturer, date, and some key words that tell you what the notes are about.  Typically, I have one set for general question; things that occur to me while reading or listening that I would like to explore when time allows; always referencing a page number or general context in/on which the thought occurred.  I leave space for referencing where these questions are answered.  If the question is "easy" I just leave the space there on the page.  Another set of notes will be on how the material is being presented; keeping the general flow of the argument helps keep me on track, focused, and less confused if many tangents are taken.  Another set of notes may just be for tangents.  [/hide]

Make a one page summary (or two pages, but try to keep it short) of your notes to put at the beginning of a particular set.
[hide="continued"]Most of the time, this ends up actually reversing the order of how the material is presented in the notes. For example, if the goal of the notes is to get to one or two proofs, but several lemmas or other theorems are needed, as well as examples, the summary may start with the main theorems, and mention that the proof depends on several lemmas, but then go one to briefly discuss what the theorems are "good" for, or some other personal take.  After studying, I find that the summary is really the important part.  It gives you an aerial view of what's happening below, and if you need to fill in the details, it will tell you where to go in your notes.  [/hide]

You probably already do this, but bullet your notes so that it is easier to read.

Also, write headings so that when you go back and reread your notes you know what is where.

Another tip is to rewrite your notes afterward. THis way, not only will your notes be neater, you will also have read throught your notes while rewriting (which might help you remember them).

Do not take notes on everything. If you already know it, why write it down? It wastes time and paper.

Use abreviations, especially if the teacher is giving to much informatino in a short amount of time.

[url=http://www.clt.cornell.edu/campus/learn/LSC%20Resources/cornellsystem.pdf]Cornell notes[/url] are pretty good. It depends on the teacher's requirement though.

My old History teacher wanted our Cornell notes to be done [i]her[/i] way (with questions, massive conclusions, and inferences). My AP bio teacher however, said that as long as they are in the Cornell format, he doesn't care about how you take notes because it is your own way of learning.

Also: No matter what note taking system you are using, it is helpful scanning the reading first, identifying key topics, and then going over (and actually reading it) while taking notes. This helps minimize note size and you won't waste time. (And also don't take notes on what you already know - that's the whole key point of note taking).

@Mr. Rusczyk: Wow  :D you had a Da Vinci thing going on there.

i am unable to take notes. I just can't do it. It hinders me greatly...
 :(  anybody with note taking hints?


limit
geometry
similar triangles
calculus
calculus computations

Please don't type responses or delete stuff in here. I'm studying with a friend, and we're using this to view our work.

Thanks.

Find a number $a$ such that $\lim_{x\to -2}\frac{3x^2+ax+a+3}{x^2+x-2}$ exists.

Let's work with the fraction and change the denominator so that it doesn't equal $0$. $\frac{3x^2+ax+a+3}{x^2+x-2}=\frac{3x^2+ax+a+3}{(x+2)(x-1)}$ .

Letting $a=15$, we get $\frac{3x^2+15x+18}{(x+2)(x-1)}=\frac{3(x+2)(x+3)}{(x+2)(x-1)}$. The limit here exists.

It also exists for $a=-1$.

The volume of the trough is determined by $V=\frac{1}{2}bhl$, where $b$ and $h$ are the base and height of the isosceles triangle, and $l$ is the length of the trough.

Using similar triangles, we can tell that $b=3h$, so the new formula is $V=\frac 32h^2l$.

We know that $\frac{dV}{dt}=12$, and $\frac{dl}{dt}=0$, and want to find $\frac{dh}{dt}$.

Differentiating implicitly, $\frac{dV}{dt}=3hl\cdot\frac{dh}{dt}$. We can now plug in the variables.

$12=(3)(10)\left(\frac12\right)\frac{dh}{dt}$ and $\frac{dh}{dt}=\frac 45$.


If j = 0.012 (012 repetitive) k = 0.012 (with the 12 repetitive) and m = 0.012 (with 2 repetitive), what is the maximum difference between any two of j, k, and m?

Convert them to fractions.

You should readily see that:
\[ j \equal{} \frac {12}{999} \\
k \equal{} \frac {12}{990} \\
m \equal{} \frac {11}{900}
\]
So the differences are:
\[ |j \minus{} k| \equal{} \frac {2}{18315} \approx 0.0001092 \\
|k \minus{} m| \equal{} \frac {1}{9900} \approx 0.000101 \\
|j \minus{} m| \equal{} \frac {7}{33300} \approx 0.00021
\]
So the maximum difference between any two of them is $ |j \minus{} m| \equal{} \frac {7}{33300}$.