articles Columbia percent videos geometry Harvard college

To readers of this biology forum,

I have decided to initiate a BIO-NEWS thread. There are no rules, except not to spam a certain topic, and to cite your sources

Of course, this thread is open to all news article you or I wish to post.

All that is left, is I hope you enjoy future articles, from me, or different users.

-arb95

[img]http://www.agtcenter.com/images/bovine.gif[/img]

"The new tool, called the Illumina BovineSNP50 BeadChip, is a glass slide containing thousands of DNA markers, some of which may be associated with important production traits such as disease resistance.

ARS researchers at the Roman L. Hruska U.S. Meat Animal Research Center (USMARC) in Clay Center, Neb., and the National Animal Disease Center in Ames, Iowa, are using the BeadChip to research bovine respiratory disease (BRD). Cattle can carry BRD without showing any symptoms, so infected animals can be difficult to identify. Tools like the BeadChip may facilitate identification of genetic markers for traits like BRD resistance.

Another project employing the BeadChip technology is a USMARC investigation into the influence of genetics on feed efficiency. Research leader Cal Ferrell, geneticist Mark Allan and their colleagues are identifying phenotypes--visible characteristics--that relate to post-weaning feed efficiency and lifetime productivity in beef cattle.

The researchers are also using the BeadChip to find relationships between DNA markers and phenotypes that can be used to enhance genetic selection and management in beef cattle. These studies could eventually allow researchers to develop tools that can be used to guide animal breeding selection and management decisions.

The BeadChip has research applications for both beef and dairy cattle. Design of the BeadChip was led by ARS researchers at Beltsville, Md., in collaboration with ARS scientists at Clay Center and colleagues from the University of Missouri and the University of Alberta in Canada. The chip is being used at all those locations and many others--in at least 23 locations in 11 countries."

Source: [url]http://www.sciencedaily.com/releases/2008/08/080807081926.htm[/url]
IMG of Cow with Illumina chip : [url]http://www.agtcenter.com/images/bovine.gif[/url]

Pretty interesting news article, I thought:

[color=green][size=117]'Sputnik' Virus Orbits, Hijacks Other Viruses
Eric Bland, Discovery News

Aug. 13, 2008 -- A virus that takes advantage of other viruses for its own survival has been found by scientists in France.

"This opens up whole new avenues of scientific exploration in terms of how viruses interact," said Curtis Suttle, a scientist at the University of British Columbia who was not involved in the research, which was recently published in the journal Nature.

In 2003 scientists at the Mediterranean Institute of Microbiology in Marseilles, France, found the world's largest virus in a U.K. cooling tower, a behemoth so large scientists initially discarded it as a bacteria. (Bacteria are typically many times larger than viruses.)

The French scientists continued to study the large virus, called mimivirus, and recently found an even larger one, which they named mamavirus.

Then they found something no one had seen before -- a third virus orbiting, and actually parasitizing, the mamavirus. The scientists named it Sputnik.

Viruses are a unique form of life. Unlike most other forms, they don't have the cellular equipment to reproduce. Instead they have to infect other cells, hijacking their equipment to produce thousands of little replicas.

Unfortunately for the infected cells, those copies typically pile up inside and eventually explode. The host cells are typically killed in the process.

Sputnik, in essence, hijacks a hijacker. About 100 times smaller than mimivirus or mamavirus, it can't directly infect the other viruses. Instead, it co-infects the same cells that mimivirus and mamavirus infect, usually amoebas.

Sputnik then uses its genetic material to hijack the cellular machinery churning out copies of mimivirus and mamavirus, forcing them to make copies of Sputnik instead. As a result, the spread of mimivirus and mamavirus in the amoeba is slowed.

Two organisms crammed into another creates a lot of free-floating genetic material. When scientists sequenced Sputnik's genome, they found traces of many different organisms, said Eugene Koonin, a researcher at the National Institutes for Health who worked with the French team.

"Sputnik is a chimera. It has several genes that are distantly but clearly related to more familiar viruses, like pox viruses or the mimivirus or mamavirus," said Koonin.

"Then again, it has some genes that are strange to see," he added. "It's very peculiar."

Some of those peculiar genes come from archaea, a different branch of life altogether, but about 50 percent of Sputnik's genome was similar to marine viruses, leading scientists to speculate that Sputnik is not the only virus of its kind.

"There are on average 50 million viruses in every milliliter of sea water," said Suttle, who studies marine viruses. "It's certainly possible that these viruses are very widespread."

The viruses don't have any particular medical value yet, according to the scientists. They don't infect humans, and it is highly unlikely that doctors could modify Sputnik to parasitize viruses that attack humans, say both Koonin and Suttle.

The scientists also dismiss discussions about whether the discovery is proof positive that viruses are alive, a debate that itself has refused to die.

"It's a matter of semantics," said Koonin. "They are part of the biological world. They have their own genome. It depends on what your definition of alive is and what your preferences are."

Alive or not, the discovery "shows just how little we know about the diversity of life," said Suttle. "It's an intriguing discovery."

Source: [url]http://dsc.discovery.com/news/2008/08/13/mimivirus-sputnik.html[/url]
ATTACHED IMG: [url]http://dsc.discovery.com/news/2008/08/13/virus-zoom.html[/url][/size][/color]

"Evidently we weren’t the only ones to wonder how maggots thrive in some of the most disgusting conditions imaginable.

Why don’t they get sick? Ask Norman Ratcliffe, a researcher on the immune systems of invertebrates, who has been figuring out the defenses that maggots have going for them.

Turns out the little buggers secrete chemicals that can halt the growth of bacteria or even kill them. He and colleagues at Swansea University in Wales have derived an antibiotic from those secretions that is active against a range of microbes, including strains of MRSA, the drug-resistant staph that’s a particularly nasty superbug.

It even has a trademarked name, Seraticin, and the researchers are hopeful it can be developed as a medicine, though it has yet to be tested in humans. Check out the Reuters video featuring Ratcliffe (if your aren’t eating at the moment) for more info on maggots and microbes.

For more technical details on Ratcliffe’s work, there’s a recent paper in the journal Microbes and Infection.

Maggots’ flesh-eating ways have long been used to cleaning nasty wounds. Just a few years back the FDA even decided maggots could be regulated as medical devices for prescription wound care.

If you don’t believe us, take a look at the package insert for Medical Maggots. Scroll down a bit and you’ll come across a passage that makes a pill sound like an appealing alternative:

Beware: Medical Maggots may try to escape from the dressing, especially when they have debrided all necrotic tissue or when they are satiated (finished eating). Medical Maggots should not be used on more than one patient, nor allowed to wander away from their host patient."

Source: [url]http://blogs.wsj.com/health/2008/08/08/maggots-are-enough-to-gag-superbugs/[/url]

"[b]OTHER OPTIONS FOR SINUSITIS

Some nonantibiotic treatments that may ease sinus symptoms:
• Decongestant sprays (use for up to three days)
• Saline irrigation
• Topical steroids
• Over-the-counter pain medicines, such as Advil, Tylenol[/b]

Sinus infections are among the most common and aggravating medical conditions, diagnosed in around 31 million Americans each year. And for most patients, treatment has long included an antibiotic.

Now, medical experts are pushing for a less-aggressive approach.  Around 21% of antibiotic prescriptions for adults in the U.S. are for sinusitis, even though studies show the drugs often do little or no good. Short-term sinusitis can be caused by either a bacterial infection, which may respond to treatment with antibiotics, or a viral infection, which doesn't. Doctors believe the overwhelming majority of cases begin as viral infections, and less than 2% of those turn into bacterial infections.

The American Academy of Otolaryngology -- Head and Neck Surgery, which represents ear, nose and throat specialists, issued new guidelines last year for treating sinusitis. Among these: Doctors generally should wait 10 days before even considering an antibiotic. If the sinus sufferer is feeling better by then, the bug is almost surely a virus, and antibiotics would be of no use.

But if the problem lasts more than 10 days, chances are good it is bacterial. At this point, if the patient's illness is relatively mild, the guidelines leave it up to the doctor to decide between prescribing an antibiotic or monitoring the patient for as much as another week.

Some patients, and their doctors, might not be willing to wait. Keisha Herbin Smith knew what she wanted when she called her doctor's office last month about symptoms including congestion, a headache and a runny nose. Ms. Herbin Smith, a 30-year-old research coordinator from Woodbridge, Va., asked the nurse for an antibiotic to head off a full-blown sinus infection, so it wouldn't ruin a long-planned weekend at the beach. An antibiotic "just makes you feel better sooner," says Ms. Herbin Smith, who got her prescription and enjoyed her vacation, and later went to see her doctor when she returned.

Studies have shown that antibiotics may help at least some people with bacterial infections feel better faster. An analysis of 13 previous studies, performed for the academy to gather evidence for the new guidelines, found that 87% of sinusitis patients who took antibiotics felt better after seven to 12 days. But among patients who didn't take the drugs, 73% felt better by that point.

Getty Images
And the drugs have drawbacks. The analysis's lead author, Richard Rosenfeld, found that 25% of sinusitis patients who got antibiotics had side effect such as gastrointestinal problems, skin rashes or headaches. By contrast, among sinusitis sufferers who didn't take the drugs, 14% also had such problems. Widespread use of antibiotics is also blamed for the emergence of resistant bacteria that don't respond to traditional drugs.

"Certainly, the path of least resistance is to write an antibiotic," says Bradley Marple, a professor at the University of Texas Southwestern Medical Center in Dallas and a co-author of the new guidelines. Dr. Marple says he has gotten more comfortable in recent years with treating relatively mild sinus infections without the drugs. He says he sits down to discuss options with each patient and lays out the evidence behind his advice.

Some patients are adjusting to drug-free treatments. Wendi Woolley, a graduate student at the University of Alabama, says her doctor recently gave her a saline rinse rather than an antibiotic for her sinus problem. "I don't want to take a chance" on possible drug side effects, she says.

Other medical authorities want to go even further than the academy's guidelines to minimize the use of antibiotics in treating sinusitis. An analysis by international researchers published in the Lancet medical journal in March said the drugs "are not justified even if a patient reports symptoms for longer than seven to 10 days." That's because the way doctors typically diagnose sinusitis, based on symptoms, isn't typically enough to determine which patients will benefit from antibiotics, the authors concluded.

Still, habits are hard to break. Daniel Merenstein, director of research programs for family medicine at Georgetown University, knows the arguments against antibiotics as well as anyone; he's a co-author of the Lancet analysis that recommended against their use. Dr. Merenstein says he nevertheless often prescribes an antibiotic for a patient after a week of difficult and worsening sinus symptoms, rather than waiting 10 to 17 days. "I think it's a patient expectation," he says. "I'm willing to risk it and say, 'maybe you're that subset that will get better faster.'"

Source: [url]http://online.wsj.com/article_email/SB121848024875230551-lMyQjAxMDI4MTE4MzQxODMwWj.html[/url]

[img]http://www.fossilsonline.com/images/stock/gw3_sm.jpg[/img]

Scared of sharks? An interesting encounter between a greenland shark and polar bear. Polar bears are generally not considered of having any natural enemies...

[i][b]In Shark Vs. Polar Bear Smackdown, Shark Wins
Jennifer Viegas, Discovery News [/b]

"Aug. 15, 2008 -- Adult polar bears, among the largest and most powerful carnivores, are thought to have no natural enemies, but that assessment might change in light of a polar bear jawbone recently found in the stomach of a Greenland shark.

While Greenland sharks, which can grow up to 24 feet long, are known to have eaten large seals, porpoises, an entire reindeer and parts of a horse, the discovery is rare evidence that they may also feast upon polar bears.

"These sharks eat anything they find dead, and do some active hunting as well," Kit Kovacs, who made the find, told Discovery News.

"We cannot determine whether the young bear was carrion or not -- nobody can," added Kovacs, who is the leader of the Biodiversity Research Program at the Norwegian Polar Institute in Tromso.

Kovacs made the unusual discovery while investigating the deaths of multiple harbor seals at Svalbard, an Arctic archipelago. She suspected that Greenland sharks, one of two species of sleeper sharks there, might be the culprits. Prevalent in the region, they are one of few species that might be capable of taking down polar bears, which grow to around 10 feet and weigh up to 1,500 pounds.

Kovacs and her team performed autopsies on collected Greenland shark specimens. They were shocked to find the polar bear bone, but in hindsight, she said, "this finding is not likely to be anything new."

"Greenland sharks have been in the Arctic for millions of years," she explained, suggesting that fatal encounters between marine and terrestrial predators would be inevitable over such an extended period.

"Habitat loss due to global warming is so much more important in terms of a threat to polar bears," added Kovacs.

It's even possible that climate change could have set the stage for the fatal encounter.

In recent years, scientists have reported that sea ice is melting in the Arctic at unprecedented rates. The International Union for Conservation of Nature and Natural Resources classifies polar bears as a vulnerable species, with global warming mentioned as the bear's most significant threat.

Seals make up the bulk of the polar bear diet, and the predators do most of their seal hunting on the ice.

At least one study, published in journal Polar Biology, found that the bears have become so desperate for food that they are now resorting to cannibalism.

Another climate change factor that could have made the shark-meets-bear encounter more likely is a documented growth of the Arctic sleeper shark population.

Aerial surveys have spotted hundreds of fins circling regions such as Prince William Sound off the south coast of Alaska.

Vince Gallucci, a University of Washington professor of fisheries and aquatic sciences, has studied shark populations for more than a decade. He believes the aerial count "would be a high number of sharks in one spot for any place in the world."

Climate change, regular regional climate shifts, and commercial fishing appear to have altered local ecosystems. The change appears to be hurting some animals, such as sea lions, but benefiting others.

"Sharks, being the more efficient eaters, just may be able to take greater advantage of changes in the food that's available," Gallucci explained.

Greenland sharks can descend to almost 1.5 miles underwater, so it is unclear if the Svalbard polar bear died, sunk and was snatched, or if the shark scavenged or killed it close to the water's surface.

Kovacs joked, "I won't be going swimming there again."[/i]

Source: [url]http://dsc.discovery.com/news/2008/08/15/polar-bear-shark-02.html[/url]
IMG: [url]http://www.fossilsonline.com/images/stock/gw3_sm.jpg[/url]

If the shark meets the polar bear in water, then definitely shark wins......This is basically the same with letting a shark fights a polar bear on ice.

"Scientists at the Smithsonian Institution have discovered a new species of bird in Gabon, Africa, that was, until now, unknown to the scientific community. Their findings were published in the international science journal Zootaxa today, Aug. 15.

The newly found olive-backed forest robin(Stiphrornis pyrrholaemus) was named by the scientists for its distinctive olive back and rump. Adult birds measure 4.5 inches in length and average 18 grams in weight. Males exhibit a fiery orange throat and breast, yellow belly, olive back and black feathers on the head. Females are similar, but less vibrant. Both sexes have a distinctive white dot on their face in front of each eye.

The bird was first observed by Smithsonian scientists in 2001 during a field expedition of the National Zoo's Monitoring and Assessment of Biodiversity Program in southwest Gabon. It was initially thought, however, to be an immature individual of an already-recognized species. Brian Schmidt, a research ornithologist at the Smithsonian's National Museum of Natural History and a member of the MAB program's team, returned to Washington, D.C., from Gabon in 2003 with several specimens to enter into the museum's bird collection. When he compared them with other forest robins of the genus Stiphrornis in the collection, Schmidt immediately noticed differences in color and plumage, and realized the newly collected birds might be unique.

"I suspected something when I found the first bird in Gabon since it didn't exactly match any of the species descriptions in the field guides," Schmidt said. "Once I was able to compare them side by side to other specimens in our collections it was clear that these birds were special. You, of course, have to be cautious, but I was still very excited at the prospect of possibly having found a new species of bird."

To ensure that the specimens Schmidt collected were a new species, geneticists at the Smithsonian's National Zoo compared the DNA of the new specimens to that of the four known forest robin species. The results clearly showed that these birds were in fact a separate and distinct species.

Discovering an unknown mammal or bird species is far from a common event. Before the 20th century, the rate of discoveries was great—several hundred new species were being described each decade. Since then, however, the pace has slowed and new species of vertebrates are generally only found in isolated areas.

Now officially recognized, the olive-backed forest robin brings Gabon's number of known bird species to 753. Other than its existence, however, little is known to science about this newcomer.

There is some knowledge about the species' habitat choice since all of the birds seen and heard in the wild were found in dense forest undergrowth. Other facts such as specific diet, mating and nesting habits, and the species' complete habitat range are all things that still need research.

"This discovery is very exciting for us," said Alfonso Alonso, who directs the Biodiversity Program in Gabon. The opportunity to study areas the tropics of Gabon allows scientists to learn about the organisms that live there and in turn develop plans to protect them in the future. "Finding the olive-backed forest robin strongly underscores the importance of our research. This helps us show the conservation importance of the area."

The MAB program is part of the Center for Conservation Education and Sustainability at the National Zoo. This particular study in the program is being conducted in the Gamba Complex of Protected Areas, a coastal region in southwestern Gabon containing the Loango and Moukalaba-Doudou National Parks with a restricted-access industrial corridor between them.

Scientists in the program are assessing the species diversity of the region, conducting applied research on the impact of management and development and providing biodiversity education programs locally to guide the regional conservation strategy. The program has partnered with the Gabonese government and Shell Gabon to integrate biodiversity conservation into energy development. The partnership has produced the first in-depth study of rainforest biodiversity in this area of Central Africa, provided relevant scientific advancements on the effects of development on biodiversity and identified conservation strategies for the long-term management of the area.

"Although finding an unknown species like the olive-backed forest robin was not the goal of the MAB project," Schmidt said, "it is definitely a reminder that the world still holds surprises for us."

Source, IMG: [url]http://www.biologynews.net/archives/2008/08/16/smithsonian_scientists_discover_new_bird_species.html[/url]

A similar article yet shorter: [url]http://news.yahoo.com/s/nm/20080815/sc_nm/bird_gabon_dc[/url]

"Cocoa flavanols, the unique compounds found naturally in cocoa, may increase blood flow to the brain, according to new research published in the Neuropsychiatric Disease and Treatment journal. The researchers suggest that long-term improvements in brain blood flow could impact cognitive behavior, offering future potential for debilitating brain conditions including dementia and stroke.

In a scientific study of healthy, older adults ages 59 to 83, Harvard medical scientists found that study participants who regularly drank a cocoa flavanol-rich beverage made using the Mars, Incorporated Cocoapro® process had an eight percent increase in brain blood flow after one week, and 10 percent increase after two weeks.

In this first-of-its-kind study, the researchers found both short and long-term benefits of cocoa flavanols for brain blood flow, offering future potential for the one in seven older Americans currently living with dementia. When the flow of blood to the brain slows over time, the result may be structural damage and dementia. Scientists speculate that maintaining an increased blood flow to the brain could slow this cognitive decline.

"The totality of the research on cocoa flavanols is impressive. This is just one more study adding to an increasing body of literature connecting regular cocoa flavanol consumption to blood flow and vascular health improvements throughout the body," said Harold Schmitz, Ph.D., chief science officer at Mars, Incorporated, which has supported research on cocoa flavanols for more than 15 years. "Though more research is needed, these findings raise the possibility that flavanol-rich cocoa products could be developed to help slow brain decline in older age."

The Body of Evidence
Contrary to statements often made in the popular media, the collective research demonstrates that the vascular effects of cocoa flavanols are independent of general "antioxidant" effects that cocoa flavanols exhibit in a test tube, outside of the body. While research aimed at studying the potential role of cocoa flavanols in the context of blood vessel and circulatory function continues, a number of previously published studies already suggest that the consumption of cocoa flavanols can have important beneficial effects on the function of the body's network of blood vessels. The body of research not only suggests that cocoa flavanols may provide a dietary approach to maintaining cardiovascular function and health, but also points at new possibilities for cocoa flavanol-based interventions for vascular complications associated with cognitive performance, skin health and age-related blood vessel dysfunction.

[b]Future Cocoa Flavanol Research Directions[/b]
For more than 15 years, Mars, Incorporated has conducted and/or supported a significant portion of the research undertaken in the field of cocoa flavanols and reported new insights in peer-reviewed scientific literature. Working in collaboration with top research institutions around the world, Mars, Incorporated continues to lead the way in exploring the nutritional and medical potential of cocoa flavanols. Mars' commitment to rigorous scientific research of cocoa and flavanols is evidenced by more than 100 peer-reviewed research publications and more than 80 patents held by the company. Mars also developed and patented the breakthrough process called Cocoapro® that helps retain more of the naturally occurring flavanols in cocoa. The high-flavanol cocoa powders made using the Mars Cocoapro® process are thoroughly characterized in terms of nutrient content, as well as standardized with respect to flavanol level and flavanol profile. Through the newly created Mars Botanical division, Mars will continue to develop and apply industry-leading analytical techniques and standards to further investigate the biomedical potential of cocoa flavanols. For more information on the many research studies on cocoa flavanols, visit [url]http://www.healthycocoa.com/[/url]"

Source: [url]http://www.biologynews.net/archives/2008/08/18/researchers_link_cocoa_flavanols_to_improved_brain_blood_flow.html[/url]


inequalities function limit quadratics calculus algebra polynomial

Not sure if anyone has posted these before, but for reference...

All sums and products in this post are evaluated with i = 1 to n, unless specified otherwise.

[b]AM-GM inequality[/b]

Given $a_1,a_2,\ldots,a_n$ non-negative reals, we have
\[{{\sum a_i}\over{n}}\ge{\sqrt[n]{\prod a_i}}\]

Note the restriction. This is because there is no geometric mean for a set of data with $2k$ elements, in which an odd number of elements are negative.

[b]General Mean Inequality[/b]

Define a set of functions $m_j(a) = \left({{\sum a_i^j}\over{n}}\right)^{1/j}$.

Note that by definition, $m_0$ does not exist; we introduce the geometric mean as $m_0 = \lim_{k \rightarrow 0} m_k$.
Then for non-negative reals $a_1,a_2,\ldots,a_n$ we have
\[m_x \le m_y  \quad \text{for arbitrary reals} \  \ x<y.\]

In particular, $m_2$ is the quadratic mean, $m_1$ the arithmetic mean, $m_0$ the geometric mean, and $m_{-1}$ the harmonic mean.

[b]Cauchy-Schwarz inequality[/b]

Given reals $a_1,a_2,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$, we have
\[(\sum a_i^2)(\sum b_i^2) \ge (\sum a_ib_i)^2.\]

[b]Cauchy-Schwarz inequality: variant[/b]

Given reals $a_1,a_2,\ldots,a_n$ and positive reals $b_1,b_2,\ldots,b_n$, we have
\[\sum\left({{a_i^2}\over{b_i}}\right) \ge {{\sum a_i^2}\over{\sum b_i}}.\]

[b]Chebyshev's inequality[/b]

Given reals $a_1 \ge a_2 \ge \ldots \ge a_n \ge 0$ and $b_1 \ge b_2 \ge \ldots \ge b_n$, we have
\[{{\sum a_ib_i}\over{n}} \ge {{\sum a_i}\over{n}}{{\sum b_i}\over{n}}.\]

[b]Minkowsky's inequality[/b]

Given reals $a_1,a_2,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$, we have
\[\sqrt{\sum a_i^2} + \sqrt{\sum b_i^2} \ge \sqrt{\sum (a_i+b_i)^2}\]

[b]Nesbitt's inequality[/b]

For all positive reals $a$, $b$ and $c$, we have
\[{{a}\over{b+c}} + {{b}\over{c+a}} + {{c}\over{a+b}} \ge {{3}\over{2}}.\]

[b]Schur's inequality[/b]

Given positive reals $a,b,c$ and real $r$, we have
\[a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b)\ge 0.\]

What about the following :

Jensen (!)
Power mean (!)
Hardy - Littlewood
Schur (!)
Muirhead (!)
Rearrangement
Minkowski
Holder
Newton
Maclaurin

I've got that sum(1/x_i) = sum(x_i)/prod(x_i)

and then

sum(x_i)/(sum(x_i)/prod(x_i)) = prod(x_i)

So we are proving that prod(x_i)>=n^2... that is as far as I can get :\

Spoon wrote:I've got that sum(1/x_i) = sum(x_i)/prod(x_i)

That's not exactly true.

Hint: [hide]make squares.[/hide] End of hint.

ok, i can see why it's not true. that is good... hmm, what do you mean by make squares?

Well, the Cauchy-Schwarz inequality mainly concerns squares, but this question has none. How do we go about making some?

A substitution sounds good.

Huh? Maybe I'm just confused or something, but the proof looks pretty trivial to me. The LHS of Cauchy-Schwarz is right there.

sigh :(, i just don't see the relationship to

"( (i from 1 to n) (a_i)^2) ( (i from 1 to n) (b_i)^2)  ( (i from 1 to n) (a_i * b_i)) "

I need to get sleep though, I will look at this tomorrow... if you could, post the solution in spoiler, thx.

[quote="tetrahedr0n"]There is almost no point in knowing anything but AMGM for aime. for usamo, last year it would have helped very much to Schur. (and obviously AMGM). Also, you shouldnt forget inequality techniques such as bunching for solving olympiad problems.[/quote]

actually, if you are keen on your basic skills, and with help of homogenuity, you could have solved that inequality easily.  or you can bash using calculus is which what I did ;)  I regret going on to the last one now -- i knew i probably couldnt have get it, and should have spent time on the inequality one more.

There are other inequalities not that used as the ones posted but still:
Bernoulli,
Nesbitt,
Popoviciu,
Klakmin,
Carlson,
and the following lemmas:
1) O consider this one extremely useful:
x1,x2,....,xn,a1,a2,...an>0 n,m natural numbers  n,m>=0

x1^m/a1+...+xn^m/an>= (x1+...xn)^m/(n^(m-2)  * (a1+...+an))
2) x1,....,xn>0
(x1+1/x1)(x2+1/x2).......(xn+1/xn)>=(x1+1/x2)......(xn+1/x1)
3) a1,...,an>0
 :sqrt: (a1+...an)+ :sqrt: (a2+...+an)+.....+ :sqrt: (an-1+an)+ :sqrt: an >=  :sqrt: (a1+4a2+...+n^2*an)

[quote="MysticTerminator"]no, trivial inequality is actually the name for the inequality x:^2::ge:0. It's not just a description of it.

Ummm...maybe a stupid question, but how do you [i][b]prove[/b][/i] x:^2::ge:0?[/quote]
Well by trichotomy or Axiom of Comparison you get: x<0, x=0, or x>0.
If x=0, then you get 0:^2::ge:0 :arrow: 0:ge:0 (Multiplicative Property of 0) which stands true
If x<0, you set x' as the additive inverse of x (note: x' is always positive since x<0). You get -x':^2>0:arrow:-x'*-x'>0 (Product of Exponent) :arrow: (-1*x')(-1*x')>0 (Multiplicative Property of -1) :arrow: (-1*-1)(x'*x')>0 (Associativity, Commutativity, Asoociativity) :arrow: 1(x'*x')>0 (I forgot the name of the Theorem) :arrow: x'*x'>0 (Multiplicative Identity) :arrow: x':^2:>0 (Product of Exponent) which is true since x' is a positive number.
If x>0, x:^2:>0 by Multiplicative Property of Order.

I know, I know, don't laugh.

Why don't we include geom ineq? Some are kind of useful in algebra especially the trigonometric ones. Someone should make a post with useful Theorems,Inequalities,Lemmas...etc. I think I'll make that post. If it's borring to someone sorry! But I think the idea is not that bad.

[quote="bilot4"][quote="Ezbakhe Yassin"]$a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\geq abc(a+b+c)$[/quote]
Could someone post a proof for this inequality :blush: Could we use rearrangement or Schur?[/quote]

let $x=ab,y=bc,z=ca$ and its equivalent to $x^{2}+y^{2}+z^{2}\ge xy+yz+zx$. :lol:

Brenoulli inequality

[img]http://upload.wikimedia.org/math/2/d/3/2d31027230ed6534aad225f85e1f10bd.png[/img]
for x>-1  and n£N

regards

Is chebyshev's inequality true for two monotone sequence with real variables? Or one of the sequences must be with positive terms?

[quote="mudok"]Is chebyshev's inequality true for two monotone sequence with real variables? [/quote]
Yes!

Then in the first post of this topic condition of chebyshev's inequality is incomplete.

Given a condition $xy+yz+xz=1$ and how can we use SMV method?

[quote="Soarer"][quote="cezar lupu"]Arne,can you post Newton's inequality? I'm really interested in it.[/quote]
$S_i S_{i+2} \ge S_{i+1}^2$ where $S_i$ equals the i-term expressions of $\prod (1+x_i)$[/quote]
I think that this is not newton's inequality.Firstly he define $ a_i $ as follwing:
$ a_k=\displaystyle\frac{n_k}{\binom{n}{k}} $ where $ n_k $ is the $ s_k $ from above.After this,the inequality is : $a_{k-1}a_{k+1}\le a_k^2 $

[quote="Mildorf"]Let $a$ and $b$ be positive reals. Then
\[ \frac{(1+k)(a-b)^2+8ab}{4(a+b)} \ge \sqrt[k]{\frac{a^k+b^k}{2}}  \]
for $k \in \{-1,0,1\} \cup [2,\infty)$ and
\[ \frac{(1+k)(a-b)^2+8ab}{4(a+b)} \le \sqrt[k]{\frac{a^k+b^k}{2}}  \]
for $k \in (-\infty,-1] \cup [1,3/2]$. Equality holds in either whenever $a=b$ or $k=\pm 1$. (For $k > 0$ we allow $a, b$ to be zero.)[/quote]

Could you give a proof?

[quote="smy2012"]
Could you give a proof?[/quote]
i have ask for a [b]elementary[/b] proof on AoPS,but cannot found.Usually we use this inequality for $k=0,2$,and it is easy to check(just expanding).
If you want to see the complete proof, see(good luck to you! :D )

I really don't mean to be rude, but I don't think this thread has done crap. Is it possible to create a PDF of identities, inequalities, and substitutions common to inequality solving? Eg. $\frac{1}{x+y}\le\frac{1}{4}(\frac{1}{x}+\frac{1}{y})$

[quote="bilot4"][quote="Ezbakhe Yassin"]$a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\geq abc(a+b+c)$[/quote]
Could someone post a proof for this inequality :blush: Could we use rearrangement or Schur?[/quote]
It is simple AM-GM Inequality as
\[ a^{2}b^{2}+b^{2}c^{2}\ge 2ab^2c \]
\[ b^{2}c^{2}+c^{2}a^{2}\ge 2abc^2 \]
\[ c^{2}a^{2}+a^{2}b^{2}\ge 2a^2bc \]
Summing up all of them, we get $2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})\ge 2(a^2bc+ab^2c+abc^2)=2abc(a+b+c)$, from which we get the desired result.
This is also equivalent to $a^2(b-c)^2+b^2(c-a)^2+c^2(a-b)^2\ge 0$. :)

This was written 2 day before i was born and here now i am studying what you wrote


number theory
greatest common divisor
algebra
polynomial
function
domain
inequalities

Hmm... does anyone have the solution to that......

and DBR is ignoring me... Has anybody solved "Determine all primes p for which there exists a positive integer k such that the base-p representation of 4k^2 consists of digits only equal to 1?"

DBR doesn't know the solution as far as I know. I think its a problem from Alex Zhai...(and I think he knows a solution)

@ pythag011: TOTALLY TRIVIAL!!!
I enjoyed this problem-- it combines a lot of number theory concepts :)

[hide]
We want to solve $ 4k^2 = 1 + p + \cdots + p^{d - 1}$ in positive integers. Note that since the LHS is even, d must be even.
Claim: $ d$ is a power of 2.
Proof: Suppose otherwise. Then we can write $ d = 2^io$, where $ o$ is odd. Then we can factor the RHS as
$ (1 + p + \cdots + p^{o - 1})(1 + p^{o} + \cdots + p^{(2^i - 1)o})$
But note that the gcd of these 2 factors divides $ 2^i$ by simply by simply looking at the remainder upon polynomial division. But since the first factor is odd, the gcd is 1. Then the first factor is a perfect square. However this is impossible as it is sandwiched between $ (p^{\frac {o - 1}{2}})^2$ and $ (p^{\frac {o - 1}{2}} + 1)^2$. This proves the claim.
Now if $ d = 2$ then $ p + 1 = a^2,p = (a - 1)(a + 1)$ for some positive integer $ a$. So $ p = 3$. ($ 4 \cdot 1^2 = 11_3$) If $ d = 2^i > 2$ then
$ 4k^2 = (1 + p)(1 + p^2)(1 + p^4) \cdots (1 + p^{2^{i - 1}})$
The gcd of any 2 factors is 2. So each is 2 times a perfect square (as none can be a perfect square). We need to solve
We need to solve (*)
\begin{eqnarray*} 1 + p & = & 2a^2 \\
1 + p^2 & = & 2b^2\end{eqnarray*}
Working in $ Z\left[\sqrt {2}\right]$ (which is an Euclidean domain) we get
$ p = (a\sqrt {2} - 1)(a\sqrt {2} + 1)$
$ p^2 = (b\sqrt {2} - 1)(b\sqrt {2} + 1)$
The 2 factors of $ p$ above are prime because they have norm $ p$. Let
$ \pi = a\sqrt {2} + 1$
$ \pi ' = a\sqrt {2} - 1$
Since $ b\sqrt {2} - 1$ and $ b\sqrt {2} + 1$ are relatively prime (their common factor
divides their difference, 2), and neither is a unit, we must have
$ \pi ^ 2 | b\sqrt {2} - 1$
or
$ \pi '^2 | b\sqrt {2} - 1$
We consider the first case.
Note
$ \pi ^ 2 = (2a^2 + 1) + 2a \sqrt {2}$
By computation the constant term of $ \frac {b\sqrt {2} - 1}{\pi ^ 2}$ is
$ \frac { - (2a^2 + 1 + 2ab)}{4a^4 - 4a^2 + 1}$
which must be an integer so
$ 2a^2 + 1 + 2ab \geq 4a^4 - 4a^2 + 1$
and rearranging,
$ b \geq a^3 - \frac {3}{2} a^2$
However $ b < \sqrt {2}a^2$ from (*). So
$ \sqrt {2} a^2 > a^3 - \frac {3}{2} a^2$
from which it follows $ a < 3$. The other case $ \pi '^2 | b\sqrt {2} - 1$ is similar:
We get the inequality $ \sqrt {2} a^2 > a^3 - \frac {a}{2} + \frac {1}{2a}$, from which it follows that $ a < 3$ as well.
Checking $ a = 1,2$ we see that $ p = 3,7$ which are indeed solutions.
(For $ p = 7$, $ 4 \cdot 10^2 = 1111_7$.)
[/hide]

Actually the first part is probably more than necessary, as we only really need $ 4|d$. But it seems elegant.

I never claimed to have a solution.  But silver-sheep's seems nice....

I can just hear your voice speaking that.  :P

Nice, Holden.

And the solution to the functional equation?

Mieh... I seem to have lost the team contest problems...

[quote="silversheep"]Suppose otherwise. Then we can write $ d \equal{} 2^io$, where $ o$ is odd. Then we can factor the RHS as
$ (1 \plus{} p \plus{} \cdots \plus{} p^{o \minus{} 1})(1 \plus{} p^{o} \plus{} \cdots \plus{} p^{(2^i \minus{} 1)o})$
But note that the gcd of these 2 factors divides $ 2^i$ by simply by simply looking at the remainder upon polynomial division. But since the first factor is odd, the gcd is 1. Then the first factor is a perfect square. However this is impossible as it is sandwiched between $ (p^{\frac {o \minus{} 1}{2}})^2$ and $ (p^{\frac {o \minus{} 1}{2}} \plus{} 1)^2$.[/quote]
I don't think it is sandwiched in between for sufficiently large $o$, but as you noted only $4|d$ is necessary.

Here's a slightly different method, which takes a bit more effort since it does not really use the fact that $p$ is prime (only that it is odd, but it's pretty clear that it can't be even anyway).

[hide="Solution"]For $p=3$ and $p=7$, we can find solutions easily, so assume that $p\not\in\{3,7\}$.

Take the minimal $n$ such that
\[4k^2=\frac{p^n-1}{p-1}=p^{n-1}+\cdots+p+1\]has a solution $k$, so $\pmod{2}$ gives us $n=2m$ for some $m$. Then
\[4k^2=(p^m+1)\left(\frac{p^m-1}{p-1}\right),\]where $\gcd(p^m+1,p^m-1)|2$. Note that $p^m+1=a^2$ means that $p^m=(a-1)(a+1)$, whence $\{a-1,a+1\}=\{1,p^m\}$ from $\gcd(a-1,a+1)|2$. But this is clearly impossible unless $p^m=3$ (and we've assumed $p\ne3$), so instead we must have
\[p^m+1=2a^2,\qquad 2b^2=\frac{p^m-1}{p-1}=p^{m-1}+\cdots+p+1.\]Taking this $\pmod{2}$ as above yields $m=2l$ for some $l$. Thus
\[b^2=\left(\frac{p^l-1}{p-1}\right)\left(\frac{p^l+1}{2}\right),\]where again we note that $\gcd(p^l-1,p^l+1)|2$. If $l$ is even, then $p^l+1\equiv2\pmod{4}$, so
\[2|l\implies 2|1+p+\cdots+p^{l-1}=\frac{p^l-1}{p-1}\implies \frac{p^l-1}{p-1}=4c^2\]for some $c$, contradicting the minimality of $n$. So $l$ must instead be odd, which implies that $(p^l-1)/(p-1)$ is odd as well. Hence $p^l+1=2d^2$ for some $d$. However, this means that
\[2a^2=p^m+1=p^{2l}+1=(2d^2-1)^2+1\implies a^2=d^4+(d^2-1)^2.\]Because $\gcd(d^2,d^2-1)=1$, we can set $\{d^2,d^2-1\}=\{u^2-v^2,2uv\}$ where $\gcd(u,v)=1$.

[b]Case 1:[/b] $d^2=u^2-v^2$ and $d^2-1=2uv$. Since $\gcd(u,v)=1$ and $d$ is odd (i.e. $u^2-v^2=d^2\equiv1\pmod{4}$), this forces $u=r^2+s^2$ and $v=2rs$, i.e.
\[1=(r^2-s^2)^2-4rs(r^2+s^2)=r^4-4r^3s-2r^2s^2-4rs^3+s^4=(r-s)^4-8r^2s^2,\]or equivalently,
\[(rs)^2=\left(\frac{t-1}{2}\right)\left(\frac{t+1}{2}\right)\left(\frac{t^2+1}{2}\right)\]where $t=r-s$. But these three terms are easily seen to be pairwise coprime, so $(t-1)/2$ and $(t+1)/2$ are two consecutive squares. This is impossible, however, since $t=1\implies d^2=1\implies p^l=1$.

[b]Case 2:[/b] $d^2=2uv$ and $d^2-1=u^2-v^2$. Since $\gcd(u,v)=1$ and $d$ is even (i.e. $u^2-v^2=d^2-1\equiv-1\pmod{4}$), we have $u=2r^2$ and $v=s^2$ for some $\gcd(r,s)=1$ and $s$ odd, so
\[1=2uv-u^2+v^2=s^4+4r^2s^2-4r^4=(s^2+2r^2)^2-8r^4.\]Let $s^2+2r^2=2x+1$ so that
\[r^4=\frac{x(x+1)}{2}.\]Depending on the parity of $x$, we either have $x=2y^4$ and $x+1=z^4$ (if $x$ is even) or $x=y^4$ and $x+1=2z^4$ if ($x$ is odd) for some $\gcd(y,z)=1$. In the first situation,
\[2y^4=x=z^4-1\implies\left(\frac{y^2}{2}\right)^2=\left(\frac{z-1}{2}\right)\left(\frac{z+1}{2}\right)\left(\frac{z^2+1}{2}\right),\]so as in the first case, $z=1\implies d=0\implies p^l=-1$. In the second situation,
\[2z^4=x+1=y^4+1\implies \left(\frac{z^2}{2}\right)^2=\left(\frac{y^2+1}{2}\right)^2+\left(\frac{y^2-1}{2}\right)^2.\]But we also have
\[y^2=\left(\frac{y^2+1}{2}\right)^2-\left(\frac{y^2-1}{2}\right)^2,\]whence
\[\left(\frac{yz^2}{2}\right)^2=\left(\frac{y^2+1}{2}\right)^4-\left(\frac{y^2-1}{2}\right)^4,\]It's well-known (i.e. Fermat) and easy to establish by infinite descent that this means $y^2=1\implies x=1\implies s=r=1\implies d=2\implies p^l=7$. However, we've assumed that $p\ne7$, so we're done.[/hide]

[b]Edit:[/b] I meant that the proof can be adapted to show that solutions only exist when $p=4k^2-1$ for some $k$ (where the primality restriction is lifted).


let $a,b,c$ be the three sides of a triangle.
show that $3+\sum_{cyc}\frac{a(a+b-c)}{c(c+b-a)}\geq \sum_{cyc}\frac{a^2+b^2}{ab}$

I hope someone will give a very nice solution to this problem. ;)

[quote="Albanian Eagle"]let $a,b,c$ be the three sides of a triangle.
show that $3+\sum_{cyc}\frac{a(a+b-c)}{c(c+b-a)}\geq \sum_{cyc}\frac{a^{2}+b^{2}}{ab}$
I hope someone will give a very nice solution to this problem. ;)[/quote]
First of all, ugly proof.
Let $b+c-a=x,$ $a+c-b-y$ and $a+b-c=z.$
Then our inequality equivalent to the following inequality
$\sum_{cyc}(x^{5}z+2x^{4}z^{2}+x^{3}y^{3}-x^{4}yz-x^{3}y^{2}z-2x^{2}y^{2}z^{2})\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(x^{5}z-x^{4}yz)+\sum_{cyc}(x^{4}z^{2}-x^{3}y^{2}z)+\sum_{cyc}(x^{4}z^{2}+x^{3}y^{3}-2x^{2}y^{2}z^{2})\geq0\Leftrightarrow$
$\Leftrightarrow\frac{1}{21}\cdot\sum_{cyc}(11x^{5}z+8y^{5}x+2z^{5}y-21x^{3}y^{2}z)+\frac{1}{2}\cdot\sum_{cyc}(x^{4}z^{2}+y^{4}x^{2}-2x^{3}y^{2}z)+$
$+\sum_{cyc}(x^{4}z^{2}+x^{3}y^{3}-2x^{2}y^{2}z^{2})\geq0,$ which true by Cauchy.

Second of all, ugly proof. :P
We notice after substitution by $x' = s-a, y'=s-b, z'=s-c$ where $2s = a+b+c$   and then further $x = \frac{x'}{y'+z'}$ and so forth, the inequality is ${\frac{x}{y}+\frac{y}{z}+\frac{z}{y}\ge 2(x+y+z}$
This is easy, since $2\frac{x}{y}+\frac{y}{z}\ge 3\sqrt[3]{\frac{x^{2}}{yz}}\ge 3\sqrt[3]{8x^{3}}= 6x$, noticing that $xyz \leq \frac{1}{8}$


C is the center of two concentric circles.  CB is the radius of the smaller circle and it happens to be 5.  AB = 8 and B is the location of 90-degrees.  The bigger circle is the shaded region.

I must find length AC using the Pythagorean Theorem.
C is a point on the bigger circle connected to B.

How do I find the area of the shaded region given the information above?

so is $\angle{ABC}=90^\circ$ if it is then
[hide]
$\overline{AC}^{2}=25+64=89\Rightarrow\overline{AC}=\sqrt{89}$

so i think thats the radius of the bigger circle..(if i'm reading the question right..it seems u have a typo)..then u can compute the area of the bigger circle and subtract it from the smaller circle
[/hide]

The textbook answer is 64pi square units.

Who is right?

umm lol if u continued with my solution u'd get

$89\pi-25\pi=\boxed{64\pi}$

[quote="ckck"][hide]$AB$ is a tangent to circle C so the area of the annulus would be that squared, so it would be $64\pi$[/hide][/quote]

Well, AB isn't really a Tangent, it is half a tangent from what you are saying.


probability
geometry
real analysis
real analysis unsolved

I need to solve a "game of darts" problem, where the dart isn't punctual, but had an area itself.

Imagine a big circular area (Ra radius) and a total of "N" little circles (Rb radius and N very big) INSIDE the big circle. The little circles are distributed uniformly into the big circle. Then you "throw" the dart (thats a circle too, with Rc radius). What is the probability of reaching some of the Ra radius circles? (If the dart reaches partially one of the little circles Rb, we assumed a success.)

Thanks again… I hope my english is good enough to be understood  :blush:

Two circles overlap if and only if the distance between their centers is no more than the sum of their radii. With that adjustment, this is equivalent to the version where the dart is a point.

Thanks, you are right  :lol: I'll try that  :)


What is the area of a semicircular region tangent to two sides of unit square, with endpoints of its diameter on the other two sides?

I think that's "what is the biggest semicircle that can be drawn in a unit square"
[hide="solution"]
(too lazy to sketch on geogebra)
we have a unit square $ ABCD$, first notice that the semicircle is symetrical around the diagonal $ AC$
Draw two lines, length r, from the tangent points to the center of the circle. They meet at $ 90^o$ to one another and the the angle between that radius and the semicircle's diameter is $ 45^o$
If we continue the tangent lines we can make a right angle triangle with side lengths $ (1\minus{}r), (1\minus{}r)$ and $ r$

so apply pythagoras and find $ r\equal{}2 \pm \sqrt{2}$, but we can't have $ r \ge 1$ so discard the positive.

$ A\equal{} \dfrac{1}{2} \pi (2\minus{} \sqrt{2})^2 \equal{} \pi (3\minus{}2\sqrt{2})$[/hide]


geometry
probability
3D geometry
sphere
MATHCOUNTS

[quote="jogaman323"][b]13.[/b] A bowling ball cannot weigh more than 16 pounds and must have a diameter of $ 8 \frac12$  inches.  How many square inches are in the surface area of a bowling ball before finger holes are drillled?  Express your answer as a common fraction in terms of pi.

[b]21.[/b]  There is a combination of exactly nine pennies, nickels, dimes and quarters in Lisa's pocket.  The coins have total value of 79 cents.  If one coin is randomly selected, what is the probability that it is a dime?  Express your answer as a common fractoin.

[b]27.[/b]  When starting a unit, a teacher announced that each day after the first, the class must do twice the total number of problems that had been assigned in all previous days.  At the end of six days, the students had completed one-third of the unit's problems.  How many days did it take to complete the entire unit?

[b]28.[/b]  Phoenix hiked the Rocky Path Trail last week.  It took four days to complete the trip.  The first two days she hiked a total of 22 miles.  The second and third days she averaged 13 miles per day.  The last two days she hiked a total of 30 miles.  The total hike for the first and third days was 26 miles.  How many miles long was the trail?[/quote]

[hide="my answers"]
13) too lazy to actually do, but the surface area of a sphere is $ 4\pi r^2$.

21) um, 4 pennies, 2 quarters, 2 dimes, 1 nickel, so $ \frac{2}{9}$

27) the seventh day, they do 2/3 of the problems in the book, and (1/3)+(2/3)=1, so $ 7$

28) after some annoying algebra, I got $ 78$
[/hide]

[quote="alanchou"][quote="jogaman323"][b]13.[/b] A bowling ball cannot weigh more than 16 pounds and must have a diameter of $ 8 \frac12$  inches.  How many square inches are in the surface area of a bowling ball before finger holes are drillled?  Express your answer as a common fraction in terms of pi.

[b]21.[/b]  There is a combination of exactly nine pennies, nickels, dimes and quarters in Lisa's pocket.  The coins have total value of 79 cents.  If one coin is randomly selected, what is the probability that it is a dime?  Express your answer as a common fractoin.

[b]27.[/b]  When starting a unit, a teacher announced that each day after the first, the class must do twice the total number of problems that had been assigned in all previous days.  At the end of six days, the students had completed one-third of the unit's problems.  How many days did it take to complete the entire unit?

[b]28.[/b]  Phoenix hiked the Rocky Path Trail last week.  It took four days to complete the trip.  The first two days she hiked a total of 22 miles.  The second and third days she averaged 13 miles per day.  The last two days she hiked a total of 30 miles.  The total hike for the first and third days was 26 miles.  How many miles long was the trail?[/quote]

[hide="my answers"]
13) too lazy to actually do, but the surface area of a sphere is $ 4\pi r^2$.

21) um, 4 pennies, 2 quarters, 2 dimes, 1 nickel, so $ \frac {2}{9}$

27) the seventh day, they do 2/3 of the problems in the book, and (1/3)+(2/3)=1, so $ 7$

28) after some annoying algebra, I got $ 78$
[/hide][/quote]

Thanks... but 28's wrong ( the rest are correct, though.. :) )

And can you be kind and post $ how$ you got the answers on some of 'em?          That's what I'm intrested in..
And by the way, on #21, wouldn't there be other combinations?

Oh wait, woops, I think
[hide="28 is"]
$ \boxed{52}$
[/hide]

[hide]For #21, you know there has to be 4 or 9 pennies.  Clearly there won't be 9, so there are 4 pennies, and you just have to find the combinations with only nickels, dimes, and quarters so that there's 5 coins and 75 cents.  You'll find that only 1 combination works. :) [/hide]

My extremely short solution to #27 (which was an old mathcounts problem) was basically how I did it.

I'll do the solution for #28 later unless someone already gets it.

If someone reads this topic that's actually from Oklahoma, go ahead and answer the questions.

[hide]13. 18 and $ \frac {1}{16}$ pi
21. $ \frac {2}{9}$
27. 7 days
28. 52[/hide]

Note:  Show HOW you did it!  (Mr. B)

13.Solution:
The surface area of a sphere with radius $r$ is \[4\pi r^2.\] The sphere in question has diameter $8\frac{1}{2}=\frac{17}{2}$, radius $\frac{17}{4}$, and surface area \[4\pi\left(\frac{17}{4}\right)^2 = \frac{17^2}{4}\pi = \boxed{\frac{289\pi}{4}}.\]


Could someone please recommend a "classic" book (written in the 20th century) as independent reading for me? Examples would be Fahrenheit 451, The Great Gatsby, The Things They Carried, One Flew Over the Cuckoo's Nest, Their Eyes Were Watching God, etc.

[quote="max_tm"]Camus' "L'Etranger" is an amazing 20th century classic IMO. I'm not sure if there are any good translations in English however (I've never bothered to look, there may very well be).[/quote]
I might decide to read that in French.  :wink:

Any other suggestions?

[quote="max_tm"]Camus' "L'Etranger" is an amazing 20th century classic IMO. I'm not sure if there are any good translations in English however (I've never bothered to look, there may very well be).[/quote]

Yeah, there is. The Stranger.

Some good books are:

Flannery O’Connor, The Habit of Being
George Orwell, Homage to Catalonia
Walker Percy, Lost in the Cosmos
Malcolm X and Alex Haley, The autobiography of Malcolm X

[quote="Ignite168"]How would stuff written in the 20th century be considered a classic?[/quote]
Well, people generally refer to them as classics.

[quote="iamagenius"]i suggest fahrenheit 451. it's a very good book, we read it[/quote]

yes, it is.


The ratio $ \dfrac{10^{2000}\plus{}10^{2002}}{10^{2001}\plus{}10^{2001}}$ is closest to which of the following numbers?

$ \text{(A)}\ 0.1\qquad
\text{(B)}\ 0.2\qquad
\text{(C)}\ 1\qquad
\text{(D)}\ 5\qquad
\text{(E)}\ 10$

[quote="worthawholebean"]The ratio $ \dfrac{10^{2000} \plus{} 10^{2002}}{10^{2001} \plus{} 10^{2001}}$ is closest to which of the following numbers?

$ \text{(A)}\ 0.1\qquad \text{(B)}\ 0.2\qquad \text{(C)}\ 1\qquad \text{(D)}\ 5\qquad \text{(E)}\ 10$[/quote]
[hide]This fraction equals to $ \frac{10^{2000}(101)}{10^{2001}(2)}$. Canceling out results in $ \frac{101}{20}$ which equals closest to 5 or $ D$. [/hide]

Factor out the $10^{2000}$ from the numerator and denominator and you are left with
$\dfrac {101}{20}$ which is closest to 5 so the answer is $\boxed {D}$

[hide="Solution"]
Let $x = 10^{2000}.$ Then the expression becomes \[\dfrac{x + 100x}{10x + 10x} = \dfrac{101x}{20x} = \dfrac{101}{20}\] which is about $\boxed{5}.$
[/hide]

[hide="Pro"]

Divide everything by $10^{2000}$.  Then $\dfrac{1+100}{10+10} \implies \mathbf{D}$[/hide]

[hide="Really easy solution"]We want to find what $\dfrac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}$ approximates to. This is $\dfrac{10^{2000}+10^{2002}}{2\times 10^{2001}}$, obviously.

Now note that $10^{2000}$ has little significance in the overall value of the number; it is far eclipsed by $10^{2002}$ by two orders of magnitude. Therefore, we can just take out that term. We now have $\dfrac{10^{2002}}{2\times 10^{2001}}=\boxed{\textbf{(D) }5}$ and we are done.

This concept of "little significance" actually pops up in math; for example, one intuitive way to find $\displaystyle\lim_{x\to \infty}\dfrac{x^3-5x^2+3}{5x^3-4x+1}$ is to look at which terms have the most significance; which term is most important in the change of the function itself.[/hide]


geometry
circumcircle
ratio
function
trigonometry
geometry solved

On each side of the triangle $ABC$ a regular $n-$gon is constructed in the exterior on the triangle and sharing the side with the triangle. Find the values of $n$ for which the centers of the three $n-$gons are the vertices of an equilateral triangle.

This may not be relevant, but is there an explicit formula for R(n)/S(n), where R(n) is the circumradius of a regular n-gon whose side length is S(n)?

It is a not nice problem, in my opinion ;)

Let me get this straight: the question is for which $n$ is that triangle equilateral no matter what $ABC$ is?

Yes, you right (otherwise take regular triangle :) )
That is why problem isn't nice.

But in this case the problem is quite nice, unless I'm making some huge mistake:

Consider the limit case $A\to$ the midpoint of $BC$. The triangle $MNP$ (where $M,N,P$ are the centers of the $n-$gons) is equilateral when $n=3$, and the ratio $\frac{NP}{\hbox{the altitude from M of MNP}}$ is clearly an injective function of $n$, so $n=3$ is a necessity. It's well-known (Napoleon's problem) that it's also sufficient.

[quote="Myth"]Yes, you right (otherwise take regular triangle :) )[/quote]

Actually, it is enough that the triangle formed by the centers of the regular n-gons erected outwardly on the sides BC, CA, AB of triangle ABC is equilateral for [i]one non-equilateral[/i] triangle ABC to conclude that n = 3. In fact, there is the following theorem:

[b]Theorem 1.[/b] Let ABC be a triangle with sidelengths a = BC, b = CA, c = AB. On the sides BC, CA, AB of this triangle ABC, we outwardly erect isosceles triangles BCA', CAB' and ABC' with bases BC, CA, AB and equal base angles

  < A'BC = < A'CB = < B'CA = < B'AC = < C'AB = < C'BA = t,

where, of course, 0 < t < 90. Then,

[b](a)[/b] we have C'A' = A'B' if and only if b = c or t = 30;
[b](b)[/b] the triangle A'B'C' is equilateral if and only if the triangle ABC is equilateral or t = 30.

From Theorem 1 [b](b)[/b], the problem easily follows. In fact, if we call A', B', C' the centers of the regular n-gons erected outwardly on the sides BC, CA, AB of triangle ABC, then, because of the similarlity of the regular n-gons, we have < A'BC = < A'CB = < B'CA = < B'AC = < C'AB = < C'BA; in other words, the triangles BCA', CAB' and ABC' are isosceles triangles with bases BC, CA, AB and equal base angles. Now, applying Theorem 1 [b](b)[/b], since the triangle A'B'C' is equilateral, we see that either the triangle ABC is equilateral or t = 30. But the triangle ABC is not equilateral by assertion; hence, t = 30. This clearly corresponds to the case when the regular n-gons are equilateral triangles, and thus n = 3.

Remains to establish Theorem 1.

[i]Proof of Theorem 1:[/i] First let's prove Theorem 1 [b](a)[/b].

In fact, in http://www.mathlinks.ro/Forum/viewtopic.php?t=16912 , I have showed the formula

$C^{\prime}A^{\prime}\ ^2-A^{\prime}B^{\prime}\ ^2=u^{2}\cdot \left( 1-2\cos \left( 2t\right) \right) \cdot \left( c^{2}-b^{2}\right) $,

where $u=\frac{1}{2\cos t}$. Since $u\neq 0$, we have $C^{\prime}A^{\prime}\ ^2-A^{\prime}B^{\prime}\ ^2=0$ if and only if either 1 - 2 cos (2t) = 0 or $c^2-b^2=0$. In other words, we have C'A' = A'B' if and only if either 2 cos(2t) = 1 or b = c. But 2 cos (2t) = 1 is equivalent to $\cos\left(2t\right) = \frac12$, i. e. to 2t = 60 (in fact, since 0 < t < 90, we have 0 < 2t < 180, and the only solution of $\cos\left(2t\right) = \frac12$ for 0 < 2t < 180 is 2t = 60), and therefore to t = 30. Hence, we have C'A' = A'B' if and only if either t = 30 or b = c. This proves Theorem 1 [b](a)[/b].

After Theorem 1 [b](a)[/b], we have C'A' = A'B' if and only if either t = 30 or b = c. Similarly, A'B' = B'C' if and only if either t = 30 or c = a. Hence, the triangle A'B'C' is equilateral (i. e., both equations C'A' = A'B' and A'B' = B'C' hold) if and only if either t = 30 or the triangle ABC is equilateral (i. e., both equations b = c and c = a hold). This proves Theorem 1 [b](b)[/b].

Thus, Theorem 1 is proven.

Note that Theorem 1 can be extended to isosceles triangles erected inwardly on the sides of triangle ABC:

[b]Theorem 2.[/b] Let ABC be a triangle with sidelengths a = BC, b = CA, c = AB. On the sides BC, CA, AB of this triangle ABC, we erect isosceles triangles BCA', CAB' and ABC' with bases BC, CA, AB and equal base angles

  < A'BC = < A'CB = < B'CA = < B'AC = < C'AB = < C'BA = t;

hereby, -90 < t < 90, and a positive t means that the isosceles triangles BCA', CAB' and ABC' are erected outwardly on the sides of triangle ABC, while a negative t means that the isosceles triangles BCA', CAB' and ABC' are erected inwardly. Then,

[b](a)[/b] we have C'A' = A'B' if and only if b = c or t = 30 or t = - 30;
[b](b)[/b] the triangle A'B'C' is equilateral if and only if the triangle ABC is equilateral or t = 30 or t = -30.

Using Theorem 2, we can show that, if we erect n-gons internally on the sides of a non-equilateral triangle ABC and their centers happen to form an equilateral triangle, then n = 3.

  Darij

Let $\omega =\cos \frac{2\pi }{n}+i\sin \frac{2\pi }{n},$ and let $a,b,c$ be the affixes of the triangle's vertices oriented clockwise. If $z$ is the affix of the center of the $n$-gon constructed on the side $AB,$ then we have $b-z=\omega \left( a-z\right) ,$ therefore $z=\frac{\omega a-b}{\omega -1}.$ We deduce that the three centers are the vertices of an equilateral triangle if and only if $\frac{\omega a-b}{\omega -1}+\varepsilon \frac{\omega b-c}{\omega -1}+\varepsilon ^{2}\frac{\omega c-a}{\omega -1}=0,$ where $\varepsilon $ is either one of the nonreal cubic roots of the unity.
Using $\varepsilon ^{3}=1,$ the latter becomes $\left( a+b\varepsilon +c\varepsilon ^{2}\right) \left( \omega -\varepsilon^{2}\right) =0.$
It follows that either $ABC$ is an equilateral triangle and $n\geq 3$ is an arbitrary integer or $n=3.$


circumcircle
combinatorics unsolved
combinatorics

Let $ n\ge2$ be an integer. A regular $ (2n\plus{}1)$-gon is divided into $ 2n\minus{}1$ triangle by diagonals which do not meet except at the vertices. Prove that at least three of these triangles are isosceles.

Consider the triangle which contains the center of the circumscribed circle of the polygon. It divides the polygon into three parts (exterior of the triangle). Then, prove (using the same tools as in IMO 2006#2 :http://www.mathlinks.ro/Forum/viewtopic.php?t=101306) that there is an isoscele triangle in each of the three regions (or that the initial triangle is isoscele and that there are two isoscele triangles in the two non-trivial regions).

Pierre.


FORTY +  TEN  +TEN= SIXTY

Each letter means number
Find it
(I don't know what section this problem belongs, ...)

　２９７８６
　　　８５０
＋　　８５０
－－－－－－
　３１４８６

N = 0, I = 1, F = 2, S = 3, X = 4, E = 5, Y = 6, R = 7, T = 8, O = 9

Probably narrowing the possibilities, for example from the first column $ 2N$ ends in zero then $ N\equal{}0$ or $ 5$.

Another example is $ 1 \le Mod(O\minus{}I,10) \le 2$ because $ R\plus{}2T \le 8\plus{} 2 \cdot 9 \plus{} 2 \equal{}28$
or $ F\minus{}S\equal{}1$


See http://www.mathlinks.ro/Forum/viewtopic.php?t=26104 for the first inequality and http://www.mathlinks.ro/Forum/viewtopic.php?t=35036 for the second one.

  darij

Silouan, apparently you have seen the inequalities before on THIS forum since you give Vasc as a reference. What's the use of posting them again here?

I have seen these at a post of Soarer and he told that ''I use these Vasc ineqs''.
But i have not seen these ineqs again in the forum so i posted this topic


integration
calculus
logarithms
trigonometry
calculus computations

$ \int cosec^3x$

How can you identify if an integration is to be integrated by parts?

but $ \int tan^3$ can be done without parts :S

[hide="Hint"]$ P \equal{} \int \csc^3x\,dx \equal{} \int \csc^2x \cdot \csc x\,dx$[/hide]

[hide="Answer"]$ P \equal{} \int \csc^3x\,dx \equal{} \int \csc^2x \cdot \csc x\,dx \equal{} \minus{} \cot x \cdot \csc x \minus{} \int \cot^2x \cdot \csc x\,dx$

$ \equal{} \minus{} \cot x \cdot \csc x \minus{} \int (\csc^2x \minus{} 1) \cdot \csc x\,dx \equal{} \minus{} \cot x \cdot \csc x \minus{} P \plus{} \int \csc x\,dx$

$ \equal{} > P \equal{} \frac {1}{2}\left(\ln\left|\tan \frac {x}{2}\right| \minus{} \cot x \cdot \csc x\right) \plus{} C.$
[/hide]

[quote="Bos1234"]How can you identify if an integration is to be integrated by parts?[/quote]

In most cases, we don't know if we are to integrate by parts or not: we have to try it. However, there are some standard cases in which we should know, based on the experience of others, that we can/must integrate by parts. Your case (an odd power of csc x or sec x) is one of them.


let $a,b,c,d$ be positive real numbers. Prove that
\[\left(\underset{cyc}{\sum}\frac{1}{a}\right)\left(\underset{cyc}{\sum}\frac{1}{a+b}\right) \ge \frac{16}{abcd+1}\]
:)

Nice cefer. My solution is longer than yours. :)


number theory
least common multiple
probability

I thought I understood odds, but I don't see how this can be possible:

A bag contains several chips which are red, white or blue.  The odds in favor of selecting a blue chip are 1 to 2, a red chip are 1 to 4, and a white chip and 1 to 14.  What is the least number of chips the bag must contain?

I'm assumming the word after white chip is supposed to be are... but I keep ending up with the equation $\frac{3}{5}x=x$

yeah i think it's are, otherwise it doesn't mean anything.
to solve the problem, solve for the LCM of 2,4,14. that would be the min number of chips.

 [hide]
$answer=28$ [/hide]

So how many of each color chip would be in this bag?

so you will have at laest
14 blue chips, 7 red, and 2 white chips.

well odds mean blue: not blue and red: not red and white: not white, so the probability of getting a blue is 1/3

...but 1/3+1/5+1/15 is not equal to 1

Thats what I thought.  but even had it mean the other way and you said the answer is 28, 14+7+2 = 23

odds are different from probability.  Probability is successes over total.  Odds are successes to failures.  So If the odds are $1 \text{ to } 4$ the probability is $\displaystyle \frac{1}{5}$, $1 \text{ to } 2$ is $\displaystyle \frac{1}{3}$, and $1 \text{ to } 14$ is $\displaystyle \frac{1}{15}$.  So it is $\text{lcm(3,5,15)}=15$

Oooh I learned something new today!

$3 + 5 + 1 = 15$

Obviously.


Show Show that the substitution z=1/(y^2) reduces the differential equation dy/dx+y=xy^3 to dz/dx-2z=-2x.

Hence find the general soulution of the differential equation dy/dx+y=xy^3

hello, you need the fact that
$ 2yy^'=-\frac{1}{z^2}z^'$.
Sonnhard.

Can someone show me the approach to this problem?

[quote="zhou_qiang"]find the general soulution of the differential equation dy/dx+y=xy^3[/quote]
This is a Bernoulli's differential equation.

Rewrite it as follows: $ \frac{y'}{y^3}\plus{}\frac1{y^2}\equal{}x.$

Now this "suggests" to substitute $ z\equal{}\frac1{y^2}\implies z'\equal{}\minus{}\frac{2y'}{y^3},$ hence $ \frac{y'}{y^3}\equal{}\minus{}\frac12z'.$

Now you just replace this into your equation and we're done.


linear algebra
matrix
vector
analytic geometry

The following problem was confusing me, so any help would be really appreciated!

Let L: R^3 -> R^3 be given by L([x, y, z]) = [-4y - 13z, -6x + 5y + 6z, 2x - 2y - 3z].

What is the matrix for L with respect to the basis
B = ([-1, 6, 2], [3, 4, -1], [-1, -3, 1])?

Take each of the basis vectors and substitute its coordinates in L([x, y, z]) = [-4y - 13z, -6x + 5y + 6z, 2x - 2y - 3z]. The first one will be [-24-26, 6+30+12, -2-12-6]= [-50, 48, -20]
So you get three such vectors. Then arrange their transposes in a matrix to get the matrix of the operator wrt B.

The matrix should be similar to the matrix for $L$ in the standard basis. In the standard basis, it has (more or less by inspection) the matrix

$A=\begin{bmatrix}0&-4&-13\\-6&5&6\\2&-2&-3\end{bmatrix}.$

Now, we have a basis $B.$ Stack this together to make a matrix:

$P=\begin{bmatrix}-1&3&-1\\6&4&-3\\2&-1&1\end{bmatrix}$

Now suppose that vector $x$ has coordinates $(y_{1},y_{2},y_{3})$ in basis $B.$ What that means is that if $B=\{b_{1},b_{2},b_{3}\},$ then $x=y_{1}b_{1}+y_{2}b_{2}+y_{3},b_{3}.$ But we have a better way of writing that: write $y$ as a column vector and we have $x=Py.$ If we have the standard coordinates $x,$ then we could find the base-$B$ coordinates $y$ by $y=P^{-1}x.$

So the standard coordinates of $L(x)$ are $Ax=APy.$ To find the base-$B$ coordinates of this, left-multiply by $P^{-1}$ to get $P^{-1}APy.$

Hence the matrix for $L$ in basis $B$ is $P^{-1}AP.$ To finish the computation, you'll have to compute $P^{-1}$ and then do that matrix multiplication.


How many inches are in the perimeter of a square with a diagonal six inches long? Express your answer in simplified radical form.

If the side length of the square is $ s$, then $ s\sqrt{2}\equal{}6 \implies s\equal{}3\sqrt{2}$. Thus, $ 4s\equal{}\boxed{12\sqrt{2}}$.


[quote="mdk"]Hi,

A uniform steel beam has a mass m. On it is resting half of an identical beam (L/2). What is the vertical support force at each end?[/quote]

mdk,

First, if the beams are identical then the first beam supports the entire weight of the second beam despite the fact that only half of the second rests on the first.

Second, the two beams together extend to a total length of $\frac32 L$. Their center-of-mass is at $\frac34 L$.

To ensure that the beams do not rotate the sum of clockwise moments must equal the sum of counter-clockwise moments.

$\Rightarrow T_{1}(\frac34 L-0) = T_{2}(L-\frac34 L)$

I assume that $T_{2}$ is applied at the end of the first beam.

$\Rightarrow T_{1}\cdot\frac{3L}{4}= T_{2}\cdot\frac{L}{4}$

$\Rightarrow \frac{T_{2}}{T_{1}}= 3$

To ensure the beams do not fall the sum of upward tensions has to equal the total weight pulling down.

$T_{1}+T_{2}= 2mg$

$T_{2}= 3T_{1}$ from balancing the moments.

$T_{1}+3T_{1}= 2mg$

$4T_{1}= 2 mg$

$T_{1}= \frac{mg}{2}\implies T_{2}= \frac{3mg}{2}$

$T_{1}$ is applied at the left end of beam 1 and $T_{2}$ is applied at the right end of beam 1.


trigonometry
real analysis
real analysis unsolved

Prove that :
a) $ A = $ { $sin n / n \in N $ } is dense in $ [-1,1] $
b) $ B= $ { {$ \sqrt n $ } $ / n \in N* $} is dense in $ [0,1] $

[quote="romano"]Prove that :
a) $ A = $ { $sin n / n \in N $ } is dense in $ [-1,1] $
[/quote]

Actually, I don't think this one is true, because $\frac{\sin x}x$ tends to $0$ as $x\to\infty$. Unless by $\sin n/n$ you meant something else than $\frac{\sin n}n$? :?

Oh , sorry for my stupid writing !
 I mean A = {sin n with n \in N} is dense in [-1,1]

Then it follows from the fact that $1$ is an irrational multiple of $\pi$, meaning that $\{n\}$ is dense modulo $\pi$. It's basicly the same result as the one you posted before, asking to show that $\{n\alpha\}$ is dense modulo $1$ when $\alpha$ is irrational.

As for the second one, it follows from this:

Given a sequence $(a_n)_n$ s.t. $a_n\to\infty$ and $a_{n+1}-a_n\to 0$, the set $\{\{a_n\}|n\in\mathbb Z\}$ is dense in $(0,1)$. It's almost obvious :). Moreover, this too has been discussed before.

There is too the result: $E= \{ \sin{P} \  | \  P$  is primer $\}$ is dense in $[0,1]$ but I won't play to prove that because it is very hard for me  :blush: !
If someone is interested to give a proof...

Man, this is a DIFFICULT PROBLEM! The only solution I saw uses extremely strong results. If you have found another solution, please show it to us.

I suppose that if this were easy, an answer would have been already given ;) . I wanted to go up the post with the case or somebody would have passed to side. Such an amount of worse.

I apologise, I did not want to be rude, but this problem is far too difficult for almost all members of mathlinks comunity. In fact, we have already discussed it on advanced forum and Moubi indicated a source with a solution. Don't get my words as an ofense.

No, that's nonsense. The Prime Number Theorem is much more likely to be useful.

Well jmerry how will you apply the prime no. theorem.of all the things this is something i am pretty familiar about.I am very interested if you give a solution using the above theorem.

I never said that I had any idea how to prove this. I don't see any reason that it would be false, but that's not a proof.


Find all sequences $x_{1},x_{2},...,x_{n}$ of distinct positive integers such that
 $\frac{1}{2}=\sum_{i=1}^{n}\frac{1}{x_{i}^{2}}$.

Very interesting question.
Let's see...
Let
$P = x_{1}* x_{2}* \ldots * x_{n}$
$X_{i}= \frac{P}{x_{i}}= \prod_{k \neq i}^{n}x_{k}$
Now:
$\frac{1}{2}= \sum_{i=1}^{n}\frac{1}{x_{i}^{2}}= \frac{\sum_{i=1}^{n}X_{i}^{2}}{P^{2}}$
That means:
$P^{2}= 2 \sum_{i=1}^{n}X_{i}^{2}$
So $P$ must be [u]even[/u].
Because $2 \sum_{i=1}^{n}X_{i}^{2}$ is an even square we must have:
$S = \sum_{i=1}^{n}X_{i}^{2}\equiv 2 \mod 4$
or
$S = 2^{2k-1}* G^{2}$, where $G$ is odd.
I. $S = \sum_{i=1}^{n}X_{i}^{2}\equiv 2 \mod 4$
If all $X_{i}$ are even, then $S$ will be divisible by 4.
So there exist $X_{t}$, which is odd.
But we have:
$X_{t}= \prod_{k \neq t}^{n}x_{k}$, so all $x_{i}$ except $x_{t}$ must be odd.
We have 2 cases:
1. $x_{t}$ is even.
Then, because $x_{t}| X_{i}$ for $i \neq t$ $X_{i}, i \neq t$ will be even and $X_{t}$ will be odd, which means:
$S \equiv 1 \mod 4$. Contradiction.
2. $x_{t}$ is odd.
Now all $x_{i}$ will be odd, so as their product $P$. But $P$ must be even. Contradiction.

II.
A)
$S = 2^{2k-1}$
Now $P = 2^{2k}= \prod x_{i}$
So $x_{i}= 2^{y_{i}}$
But $x_{i}\neq 1$, because otherwise $\sum_{i=1}^{n}\frac{1}{x_{i}^{2}}>1$
This means that $\{x_{i}\}_{i=1}^{n}\subset \{2^{i}\}_{i=2}^{\infty}$
So...
$\sum_{i=1}^{n}\frac{1}{x_{i}^{2}}< \sum_{i=2}^{\infty}\frac{1}{2^{i}}= \frac{1}{2}$

[EDITED]
B)
Now, if $S = 2^{2k-1}*G^{2}$, then $x_{i}= z_{i}2^{y_{i}}$ where
$\sum y_{i}= 2^{k}$
$\prod z_{i}= G^{2}$
I don't know how to proceed.....  :(

[quote="krassi_holmz"]
Because $2 \sum_{i=1}^{n}X_{i}^{2}$ is an even square we must have:
$S = \sum_{i=1}^{n}X_{i}^{2}\equiv 2 \mod 4$
or
$S = 2^{2k-1}$
[/quote]
I not think so. Are you sure? Example $S=8\cdot 9$.

What is your solution? Please post concrete!

[quote="wojto111"]Observe that $\frac{\pi^{2}}{6}-1-\frac{1}{2^{2}}> \frac{1}{2}$[/quote]

Nice hint  :D

But I think that it's <, not > (RHS it's about 0,39) :P

wojto, I had the same observation, but this leads only to the fact, that $\frac{1}{4}$ [u]must[/u] be in the sequence. (otherwise the sum will be less than $\frac{1}{9}+\frac{1}{16}+...<\frac{1}{2}$)

This seems to me to be an exceptionally difficult problem.  At least I can find some solutions.  Consider $S= a_1a_2...a_k.$  Then $\sum_ {x\mid S} \frac{1}{x^2} = (1+\frac{1}{a_1^2})(1+\frac{1}{a_2^2})...(1+\frac{1}{a_k^2}).$

So, for example, if $S=2\cdot3\cdot4\cdot5 = 120$, then $\sum_ {x\mid S} \frac{1}{x^2} =22100/14400 = 1.5 + 500/14400 = 1.5 + \frac{(10^2)+(20^2)}{120^2}= 1.5 + \frac{1}{6^2}+\frac{1}{12^2}$.

Thus if T is the set of all divisors of S, except 1, 6 and 12, $\sum_ {x\in T} \frac{1}{x^2} =\frac{1}{2}$.  In particular $T=(2,3,4,5,8,10,15,20,24,30,40,60,120)$.

Of course we note that it is also true that $500=400+64+36$, which generates solution $(2,3,4,5,8,10,12,24,30,40,60,120)$, and T equal to all the divisors of $2\cdot3\cdot4\cdot7=168$, except 1, 14,42, and 84, works as well.  I have no idea how to find all the other solutions.


If $ \frac{3}{x}\plus{}\frac{5}{x}\equal{}1\frac{1}{3}$ what is the value of $ x$ ?

[quote="GameBot"]If $ \frac {3}{x} \plus{} \frac {5}{x} \equal{} 1\frac {1}{3}$ what is the value of $ x$ ?[/quote]
$ \frac{8}{x} \equal{} \frac{4}{3}$
$ x \equal{} 2(3) \equal{} \boxed{6}$.


function
analytic geometry
graphing lines
slope

Find the value of x such that function $ f(x) \equal{} |x\minus{}1| \plus{} |x\minus{}9| \plus{} |x\minus{}8| \plus{} |x\minus{}2|$ attains the least value.

It's better if you say, find the "value(s)" of x such that...

[hide="Answer"]$ \boxed{\begin{array}{cc} \textbf{Slope of f} & \textbf{Interval} & - 4 & x < 1 & - 2 & 1 < x < 2 & 0 & 2 < x < 8 & 2 & 8 < x < 9 & 4 & 9 < x \end{array} }$

$ f$ decreases until $ x = 2$, then stays level until $ x = 8$, and then starts increasing.

That means the minimum value occurs at all $ x \in (2,8)$, and this value is $ 14$. The same value of $ f$ is attained at the endpoints of this segment, so we can extend our interval to $ \boxed{x \in [2,8]}$[/hide]

[quote="vinskman"]Find the value of x such that function $ f(x) \equal{} |x \minus{} 1| \plus{} |x \minus{} 9| \plus{} |x \minus{} 8| \plus{} |x \minus{} 2|$ attains the least value.[/quote]
[hide]Here's my solution:
We have: $ /x\minus{}1/\plus{}/8\minus{}x/ \geq7$
              $ /x\minus{}2/\plus{}/9\minus{}x/ \geq7$
Therefore, the minimum value is 14.
The equality occurs when x=5(note that there is more than 1 equality).
PS: This problem can also be generalized and solved exactly the same as above.[/hide]


Consider a triangle ABC where AB<BC are fixed lengths but angle ABC varies. Prove, using Euclidean geometry, that as angle ABC increases from 0 to 180, angle BAC is monotonically decreasing.

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=315275[/url]


$ \text{Find the marked angle}$

[img]http://img374.imageshack.us/img374/1531/m001yr3.jpg[/img]

by easy trigonometry calculation, the answer is 30 deg. After that if one needs a "quick" proof, I have made one:

1. AB=AC.
2. Angle A = 20deg. So CE=EA.
3. Angle CDB=50deg. So BC=CD.
4. AB/CE=AC/CE=2CEcos20deg/CE = 2cos20deg.
5. AD/CD=AC/BC -1 = (1/2sin10deg)-1 =(1-2sin10deg) / 2sin10deg
6. 4cos20deg * sin10deg = 2sin30deg - 2sin10deg = 1-2sin10deg, therefore AD/CD=AB/CE. Triangle ADB is similar to triangle CDE. Angle CED=Angle ABD = 30deg.

it's an old problem:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=179738[/url]

Dear Mathlinkers,
a nice technic for this adventitous angle, can be seen on my website:
http://perso.orange.fr/jl.ayme    vol. 2   In Memorian "Juan Carlos Salazar"
Sincerely
Jean-Louis


limit
real analysis
real analysis unsolved

Let $ \{ x_n \}_{n\equal{}1}^{\infty}$ and $ \{ y_n \}_{n\equal{}1}^{\infty}$ be sequence of non-negative real numbers satisfying
\[ x_{n\plus{}1} \leq x_n \plus{} \frac{1}{n^2} , y_{n\plus{}1} \leq y_n \plus{} \frac{1}{n}, \forall n \in \mathbb{N}
\]
Can the limit $ \lim_{n\to\infty} x_n$ and $ \lim_{n\to\infty} y_n$ always exist,respectively?

thxxxxx  :D

In Case 2: Setting $ y_{n\plus{}1} \equal{} y_{n} \plus{} \frac {1}{n}$ yields $ \lim_{k \to \infty} y_{k} \equal{} y_{1} \plus{} \lim_{k \to \infty} \sum_{n\equal{}1}^{k} \frac {1}{n} \to \infty$
For Case 1: $ 0 < x_{n} < x_{1} \plus{} \lim_{k \to \infty} \sum_{n\equal{}1}^{k} \frac {1}{n^2} \equal{} x_{1} \plus{} \frac {\pi^2}{6}$ This implies, the sequence is bounded. How do we proceed after this?  :maybe:

[quote="madness"]In Case 2: Setting $ y_{n + 1} = y_{n} + \frac {1}{n}$ yields $ \lim_{k \to \infty} y_{k} = y_{1} + \lim_{k \to \infty} \sum_{n = 1}^{k} \frac {1}{n} \to \infty$
For Case 1: $ 0 < x_{n} < x_{1} + \lim_{k \to \infty} \sum_{n = 1}^{k} \frac {1}{n^2} = x_{1} + \frac {\pi^2}{6}$ This implies, the sequence is bounded. How do we proceed after this?  :maybe:[/quote]

$ \{x_n\}$ has a subsequence $ \{x{}_n{}_k \}$ converges to some point $ L_1$

If $ {x_n}$ does not conveges...

Does it have another subsequence $ \{ x{}_m{}_k \}$ converges to $ L_2$ ?

If there is..

take some $ n_0 \in \mathbb{N}$ such that $ \frac{1}{n_{0}^2} \leq |L_2-L_1|$

so we can get a contradiction


geometry
trapezoid
ratio
quadratics
trigonometry
algebra
quadratic formula

In trapezoid $ABCD,$ leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD},$ and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular.  Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001},$ find $BC^2.$

[hide="Answer"]Label the intersection of the diagonals $E$. $AE^2+BE^2=11$, $BE^2+CE^2=BC^2$, $AE^2+DE^2=1001$, so we have $11-BE^2=1001-DE^2\Rightarrow BE^2=DE^2-990$. Substituting, $CE^2+DE^2-990=BC^2=CD^2-990$. Drawing the perpendicular $AF$ to $DC$, we have $CF=\sqrt{11}$. $AF^2=BC^2=1001-(CD-\sqrt{11})^2=990-CD^2+2\sqrt{11}=CD^2-990$, so we have $CD^2-CD\sqrt{11}-990=0\Rightarrow CD=10\sqrt{11}$. Therefore, we have $BC^2=CD^2-990=\boxed{110}$.[/hide]

[hide]Let the perpendicular from $A$ to $CD$ intersect $BD$ at $F$.  Let the two diagonals intersect at $E$, and let $\angle{EAF}=\theta$.  Then \[ \angle{EAF} \cong \angle{EBA} \cong \angle{ECB} \cong \angle{EDC} = \theta \]  Thus, with all four of these triangles having an angle of theta and a right angle, we have four similar triangles.  Let $AE=a$, $BE=b$, $CE=c$ and $DE=d$.  Note that \[ \frac{EF}{a}=\frac{a}{b}=\frac{b}{c}=\frac{c}{d} \]  So these triangles are in a constant ratio (it can be observed that the ratio of one to the next in the counter clockwise direction is $\cot{\theta}$; let this ratio, also equal to $\frac{b}{a}=x$).  So,
\[ a^2+d^2=a^2+a^2x^6=a^2(1+x^6)=a^2(1+x^2)(1-x^2+x^4)=1001 \]
\[ a^2+b^2=a^2+a^2x^2=a^2(1+x^2)=11 \]
Dividing the top by the bottom,
\[ 1-x^2+x^4=91 \implies (x^2)^2+(x^2)-90=0 \]
Using the quadratic formula,
\[ x^2=\frac{1 \pm \sqrt{1+360}}{2}=10, -9 \]
Hence, $x^2=10$.  And because $BC^2=AB^2x^2$, $BC^2=11(10)=\boxed{110}$[/hide]

[hide][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=699&sid=0410aaa1bd30b9a182e2c716eee02a44[/img]

Let $BC=x$. This means that $AC=\sqrt{11+x^2}$. Now notice that $\triangle ABC \sim \triangle BCD$. Therefore, $\frac{BD}{BC}=\frac{AC}{AB} \implies BD=\frac{x\sqrt{x^2+11}}{\sqrt{11}}$. Another valuable piece of info we get from these similar triangles is that $%Error. "cosACB" is a bad command.
=\cos ABD=\frac{x}{\sqrt{x^2+11}}$.

Now, it seems like applying the law of cosines on $\angle ABD$ will come out nicely. $AD^2=AB^2+BD^2+2(AB)(BD)(\cos ABD)=11-\frac{x^2(x^2+11)}{11}-4=1001$. Simplifying further yields $(x^2+99)(x^2-110)=0$, so $BC=x=\boxed{110}$. [/hide]

[quote="JesusFreak197"][hide="Answer"]Label the intersection of the diagonals $ E$. $ AE^2 \plus{} BE^2 \equal{} 11$, $ BE^2 \plus{} CE^2 \equal{} BC^2$, $ AE^2 \plus{} DE^2 \equal{} 1001$, so we have $ 11 \minus{} BE^2 \equal{} 1001 \minus{} DE^2\Rightarrow BE^2 \equal{} DE^2 \minus{} 990$. Substituting, $ CE^2 \plus{} DE^2 \minus{} 990 \equal{} BC^2 \equal{} CD^2 \minus{} 990$. Drawing the perpendicular $ AF$ to $ DC$, we have $ CF \equal{} \sqrt {11}$. $ AF^2 \equal{} BC^2 \equal{} 1001 \minus{} (CD \minus{} \sqrt {11})^2 \equal{} 990 \minus{} CD^2 \plus{} 2\sqrt {11} \equal{} CD^2 \minus{} 990$, so we have $ CD^2 \minus{} CD\sqrt {11} \minus{} 990 \equal{} 0\Rightarrow CD \equal{} 10\sqrt {11}$. Therefore, we have $ BC^2 \equal{} CD^2 \minus{} 990 \equal{} \boxed{110}$.[/hide][/quote]

Wait, where did you get the substitution of $ \minus{}990$ in the equation $ CE^2 \plus{} DE^2 \minus{} 990 \equal{} BC^2 \equal{} CD^2 \minus{} 990$?

[hide]Let $ E$ be the foot of the perpendicular from $ A$ to $ CD$.
Note that since $ AC \perp BD$, $ \triangle ABC \sim \triangle BCD \implies \frac {AB}{BC} \equal{} \frac {BC}{CD} \implies BC^2 \equal{} CD \cdot \sqrt {11}$. Also, by the Pythagorean theorem, $ AE^2 \plus{} ED^2 \equal{} AD^2$, but $ AE \equal{} BC$, $ ED \equal{} CD \minus{} \sqrt {11}$, so this simplifies to $ CD^2 \minus{} \sqrt {11} \cdot CD \minus{} 990 \equal{} 0$. Solving ($ CD$ must be positive), $ CD \equal{} \frac {\sqrt {11} \plus{} \sqrt {11 \cdot 19^2}}{2}$, so $ BC^2 \equal{} CD \cdot \sqrt {11} \equal{} \frac {11 \plus{} 11 \cdot 19}{2} \equal{} 110$. [/hide]

As for mr.k_74's question, JeseusFreak197 had earlier that $ BE^2 \plus{} CE^2 \equal{} BC^2$, so he substituted for $ BE^2$ in this equation.

[hide=quick]
Set the trapezoid on the coordinate plane with $C$ as the origin and $CD$ lying on the $x$ axis. Furthemore denote $CD=e$ and $BC=a$
Then looking at slopes we have $(\frac{a}{e})(\frac{a}{\sqrt{11}})=1$. Thus $a^2=e\sqrt{11}$. Now doing distance formula on
$AD$ gives $a^2+(e-\sqrt{11})^2=1001$. Now just substitute and you get a quadratic in $a^2$ which an be easily found to have positive solution
$\boxed{110}$[/hide]

[quote=lifeismathematics][hide=quick from this lemma] lemma: if $AB \perp BC$ and $CD\perp BC$, such that $AB=h_{1}  , DC=h_{2}$ and $BC=d$ then $d^2=h_{1}h_{2}$
proof of lemma is easy from similarity, then we have from this lemma $x^2=y\sqrt{11}$ and from orthogonal diagonal theorem we have $x^2+1001=y^2+11$, solving for  $x^2$ we get $x^2=\boxed{110}$[/hide][/quote]

Wow, this is the simplest solution. :o

I literally had to use 3 similar triangles and 2 pythagorean theorems to bash out the answer  :P


Sam has two ostrich eggs and lives in a building with 36 floors. He is told that if he drops the ostrich egg from a certain floor or higher, the egg will break. If he drops an egg from any lower floor, the egg will not break. His job is to determine the lowest floor at which an egg will break. What is the minimum number of drops required that will guarantee that Sam will know the distinct floor as soon as the second egg breaks?

[hide="Solution"]
12

You just take the square root of the number of floors, and double it. (since there are two eggs) If there were n eggs and m floors, you would just take the nth root of m and multiply it by n. This works because the maximum number of floors each egg is being dropped off is equal.
[/hide]

i have a strategy with less than 12 drops
drop the first egg at the following floors (in order) UNTIL the first egg breaks (if it breaks move on to the next egg)
8
15
21
26
30
33
35
36

if it breaks when you drop the first egg on the 8th floor, test 1-7 using the second egg (1 first, 7 last) and when it breaks on one floor but not the one below, you know the floor
if it breaks on 15, test on 9-14
if it breaks on 21 test 16-20

continue this way and each way needs a maximum of 8 tests
(i know this is confusing... i might try to reword it)

bottom line--- only needs maximum of 8 drops

[quote="mewto55555"]My solution:

Drop from 18: Depeding on what happens, move up/down halfway (to 9 or 27)

BY repeatedly halving, you mminimize the time. Since 36 is greater than $ 2^5$ but less than $ 2^6$ it will take 6 drops maximum.[/quote]
What happens if you break it on the 18th floor, then break it on the 19th floor?
Stevenmeow's is correct...

If you break it on the 18th floor, move down to the 9th floor.

In general, the maximum amount of necessary drops is $ n$, when the number of floors is the $ n^{th}$ triangular number, and you are given two or more eggs.

[b]mewto55555[/b]: Your method doesn't work when the eggs break either in floors lower than the 9th, or higher than the 27th.


I had my first personal pentium computer,presently.
I'm at a difficulty in applying '[b] mathematical[/b] science[b]'[/b] latex[u] technique[b]. Can our mathematics communities
help by showing how such latex application can ber more useful to me in my personal[/b] mathematics research? How about using same technique in '
[/u] mathematical[u] logic[/u].[u]'[/u]

                                           Respectfully,
                                          A.Eddington

what do you mean by Mathematical science latex "technique?


superior algebra
superior algebra unsolved

Let $G$ be a finite group of odd order. Prove that there is a $g \in G$ such that $g^{-1}$ is not conjugated to $g$.

Actually, $\forall g \neq 1$, $g$ and $g^{-1}$ will never be conjugated ... this can be shown by assuming the contrary, playing a little with the expressions and showing that eventually $g$ must have order $2$, which is absurd.


inequalities
geometry
circumcircle
inradius
Euler
trigonometry
geometry proposed

Let R and r be the circumradius and the inradius of a triangle ABC, respectively. Also, let $r_1$ be the inradius of triangle XYZ, where X, Y, Z are the points where the incircle of triangle ABC touches the sides BC, CA, AB.

Prove that $2r_1 \leq r \leq \sqrt{r_1R}$.

[b]PS.[/b] Proposed solution (in Russian) on http://kvant.mccme.ru/1995/03/gif/95_03-20.gif .

  Darij

First inequality is easy to deduce from Euler inequality in triangle XYZ:
  r ≥ 2r’.   For the second we need some calculations:

YZ = (p-a)*2sin(a/2) = 2(p-a)*√(p-b)(p-c)/bc = √a(p-a) *√r/R

(we use fact that p(p-a)(p-b)(p-c)= s2 and abc = 4prR)

  r’ = XY*YZ*XZ/2r(XY+YZ+XZ)  or we need to prove  R*r’ ≥ r2.

Using first statement it can be deduce:

  √abc(p-a)(p-b)(p-c) ≥ 2r2∑√a(p-a)

 √a(p-a) = √2* √a/2*(p-a)≤ [a/2+(p-a)]/√2  => ∑√a(p-a)  ≤ p√2  =>

 √abc(p-a)(p-b)(p-c) ≥ 2√2p*r2

   abc*(p-a)(p-b)(p-c) ≥ 8p2*r^4

   abc*p(p-a)(p-b)(p-c) ≥ 8p^3*r^4

   4prR *p2r2 ≥ 8p^3*r^4    or  R ≥ 2r.

   which again turns Euler inequality.

[quote="darij grinberg"][b]PS.[/b] Proposed solution (in Russian) on http://kvant.mccme.ru/1995/03/gif/95_03-20.gif .[/quote]
Trigonometry :(


summer program
MathPath
Mathcamp
Ross Mathematics Program

[quote="jli"]Is anybody here going to MathPath during the summer?[/quote]

I am.  :)

[quote="felmo"]no i cant go to america cos its kind of stupid[/quote]

Yup america does kinda suck

America only sucks to other countries because they think everybody here is either Bush or Rice.

Everyone is bush or rice where I live.  Well, besides me.  I still hear debates to this day (literally, this day) of how much kerry sucks and why he should be killed.  It's kind of scary actually.

On topic, I got some brochure about some math camp.  I don't know if it was mathpath or not.  Where is mathcamp and how much it is?

Is it mathcamp or mathpath? I got the MathCamp brochure, but it's too expensive, and I have a conflict.

Mathcamp and mathpath are two different camps, but are heded by the same people. Mathpath is for younger students, and mathcamp is for older students.  You can find out more in the "Other US Contests & Programs" forum.

I am!!!!!!!!

There's another thread called "MathPath 2005 Information Now Posted" asking the same question, and you'll probably find more usernames there. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=mathpath&t=22358]Click here[/url]

[quote="mathlover"]i wanted to soooooooooo bad, but my parents said it cost too much. and finishing those 5 problems for NOTHING?!?!?!?! and boy were they hard!!![/quote]

same thing with me! I finished the problems, got accepted, raised some money, and the day I'm about to send my registration package, [u]they change their mind[/u]!  :wallbash_red:  :wallbash_red:  :wallbash_red:

[quote="tarquin"]I'm not because it's too expensive and I'm going to some other math camp here in Texas.[/quote]Which camp is that?  San Marcos?

[quote="tarquin"]Yes, San Marcos TX Mathworks Honors Camp ..[/quote]woah cool, I'm going to that too.  I guess I'll see you in 8 days.  :D

I think next summer I will probably try to go to MathCamp or Ross or something.  Hopefully my parents will help pay for it.  What do camps like these cost?

My camp, for some reason gave my family a huge break and only made us pay 500 dollars out of the regular price of 1750...

It really depends on your financial status and if you've received any scholarships for the camp etc.

My family really gets more than average so I have no idea why I'm paying so little.

I am going to mathcamp this summer in Oregon. Is there anyone going?

Check the MathCamp forum in the Other US Contests and Programs forum.  ;)

[quote="apusey"]I am doing to many other summer camps[/quote]

which ones?

I want to get a head start on my foreign language requirements so I won't be able to go.


inequalities
calculus
derivative
inequalities unsolved

If $a$, $b$, $c$ are sides of an triangle, then
\[3(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-1)\geq2(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}).\]
Try it ?

This one is really interesting.If I don't make any mistakes then it is equal to:
 [tex](5a-5b+3c)(a-b)^2+(5b-5c+3a)(b-c)^2+(5c+5a-3b)(c-a)^2 \geq 0 [/tex]
Let $ a=x+y;b=y+z;c=x+z $.
If $ x \geq y \geq z $ ,since
[tex] (a-b)(b-c)(c-a) \geq 0 \Leftrightarrow (a-b)^3+(b-c)^3+(c-a)^3 \geq 0 [/tex] so it is clear.
If $ z \geq y\geq x  $ then our inequality is:
[tex] (4x-z)(x-z)^2 +(4z-y)(y-z)^2+(4y-x)(x-y)^2 \geq 0 [/tex]
We just need prove it when  $ z>4x $ and it is not too difficult now.... :)

First part of your solution is beautiful and correct.
It remains to prove the case "not too difficult"
              z>4x (!).

Is it about this problem that was refined or another one, but looks like this Kvant problem?

http://www.mathlinks.ro/viewtopic.php?t=5842

thank you very much!

After some computations and the classical substitution I found that this is the same as the following beautiful inequality:
  For any $ x,y,z\geq 0 $ we have $ x^3+y^3+z^3+2(x^2y+y^2z+z^2x)\geq
 3(xy^2+yz^2+zx^2) $. But I couldn't find a smart solution for this one. I think it can be solved using derivatives and some computations, but I don't like this approach. Anyone could help me to prove this beautiful inquality?

O.K. Anh Cuong. Your solution is now complete in my opinion. :)
I have a shorter solution and I will post it the next week.

Lagrangia, if I well understand your question, this inequality $A\geq0$
refines the Kvant inequality $B\geq0$ since the first involves the second:
    $3B=2A+\frac{a}{c}+\frac{c}{b}+\frac{b}{a}-3\geq2A$

I think I managed to find a solution for the other inequality that solves this problem. Please, anyone, check the computations! First of all, if $ x^2y+y^2z+z^2x\geq xy^2+yz^2+zx^2 $ we are done since $ x^3+y^3+z^3\geq xy^2+yz^2+zx^2 $. Now, we can obviously take $ x=max(x,y,z) $ and suppose that $ x^2y+y^2z+z^2x\leq xy^2+yz^2+zx^2 $, which means $ (x-y)
(x-z)(y-z)\leq 0 $. This implies $ y\leq z$. Thus, we can take $ y=a, z=a+b, x=a+b+c $ with $ a,b,c\geq 0$ and introduce those in the inequality. If my computations are write the inequality will reduce to $ b^3+c^3+2a(b^2+bc+c^2)\geq b^2c $ and this is obvious.

I have treated the case $ z>4x$ with the following substitutions :

$ a=z-x , b=z-y$   and  $ c=z-4x $  so that $a,b,c \geq 0$.

After some computations and using $c \geq a+b$ you have $a^3+b^3\geq ab^2$ (obvious)

[quote="Vasc"]O.K. Anh Cuong. Your solution is now complete in my opinion. :)
I have a shorter solution and I will post it the next week.[/quote]

Can you please post your solution, Vasc?

Thank you very much.

I am sure I posted my solution somewhere in this forum.
I copy now the solution given in the book Old and New Inequalities.
We may assume that [i]c[/i] is the smallest among a[i], b, c[/i]. Then let $x=b-\frac{a+c}{2}$.
After some computations, the inequality becomes
$(3a-2c)x^2+(x+c-a/4)(a-c)^2\geq 0$,
that is
$(3a-2c)(2b-a-c)^2+(4b+2c-3a)(a-c)^2\geq0$.
But, $3a-2c>0$ and $4b+2c-3a=3(b+c-a)+b-c>0$.
This completes the proof.


How many distinct real roots does the equation $ \sqrt{x\plus{}4\sqrt{x\minus{}4} } \minus{}\sqrt{x\plus{}2\sqrt{x\minus{}1} } \equal{}1$ have?

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


In the equation $ 7\plus{}x\plus{}y\equal{}19$, $ x$ and $ y$ are positive one-digit whole numbers. What is the least value possible for $ x$?

To minimize x, we maximize y. The largest one digit whole number is 9, so substitute this in for y.

$ 7 \plus{} x \plus{} y \equal{} 19$

$ 7 \plus{} x \plus{} 9 \equal{} 19$

$ 16 \plus{} x \equal{} 19$

$ x \equal{} 3$


Determine all positive integers $n$ such that the number $\sum_{i=1}^{n} \frac{n}{i!}$ is integer.

It was year ago in Slovak math olympiad. Here's my solution:
Clear the denominators and write in this form:
$\sum_{i=1}^{n} \frac{n}{i!}=\frac{2\cdot3\cdot4\ldots n+3\cdot4\cdot5\ldots n+\ldots+(n-1)\cdot n+ n +1}{(n-1)!}$
NUMERATOR is in form $k\cdot (n-1)+2$ and
DENOMINATOR is divisible by $(n-1)$

From this flows that $n-1$ can be just $0,1,2$ (or $n=1,2,3$). For these it realy is an integer.


geometry
rectangle
combinatorics unsolved
combinatorics

It is given a 1001*1001 board divided in 1*1 squares. We want to amrk m squares in such a way that:
1: if 2 squares are adjacent then one of them is marked.
2: if 6 squares lie consecutively in a row or column then two adjacent squares from them are marked.

Find the minimun number of squares we most mark.

The answer is $601200=\frac35(1001^2-1)$

First, we show that among any five consecutive squares in a row or column, at least three are marked.
Case 1: Two consecutive squares in the block of five are marked. Then, since at least two of the other three are adjacent to each other, at least one of them is marked.
Case 2: No two consecutive squares in the block are marked. Then the markings form one of these two patterns: M_M_M or _M_M_. The first pattern has three marked squares, so we are done in that case. In the second case, we can form a block of six including those five, which then has the pattern _M_M_?. Whether the ? is marked or not, this block has no adjacent marked squares, contradicting the rules of the construction and making this case impossible.

Now, we can partition all but one  corner square of the array into $5\times 1$ or $1\times 5$ blocks by first dividing into rectangles $1001\times 1000$, $1000\times 1$, and $1\times 1$. This establishes $\frac35(1001^2-1)$ as a minimum. To achieve this minimum, we need an explicit construction:
Let the squares be represented by ordered pairs of integers $(m,n)$ with $0\le m,n \le 1000$. Mark $(m,n)$ if $m+n \equiv 1$ or $3$ or $4$ mod $5$. This obviously is a valid marking; the sum $m+n$ ranges over consecutive integers in any $k\times 1$ or $1\times k$ block. Also, every $5\times 1$ or $1\times 5$ block has exactly 3 marked squares.
To count the number of marked squares, we divide all but the square $(0,0)$ into nonoverlapping $5\times 1$ or $1\times 5$ blocks; each small block has 3 marked squares for a total of $601200$ and the corner is not marked.


probability counting derangement AoPSwiki

8 people put their ID cards together and randomly shuffle them. Then each person picks an ID card randomly from the pile. What's the probability that nobody gets their own ID?

$\frac{!8}{8!}$

Look up derangements on AoPSWiki or Mathworld

[quote="neelnal"]7/8 x 6/7 x 5/6 x 4/5 x 3/4 x 2/3 x 1/2 x 1

= 1/8[/quote]

That's not right, if the second person picks the ID that belonged to the first person, then the third person have exactly 6 allowable ID to pick, not 5.

I have never seen the derangement notation before so I wasn't expecting the answer like

 $!8/8!$. If you have seen the derangement notation, try to find out how the method works.

We can count the complement - the number of arrangements where at least one person gets his/her ID, and subtract that from the total number of arrangements - 8!

There are $\binom{8}{1}$ ways to choose which person gets his/her ID, and $7!$ ways to arrange the rest. But this double counts the cases where two people get their ID, so you subtract $\binom{8}{2}6!$. Using PIE to correctly count the rest, you get
\[\binom{8}{1}7!-\binom{8}{2}6!+\cdots+\binom{8}{7}1! \binom{8}{8}=25537 \]

The desired probability is then $\frac{8!-25537}{8!}$ which equals
\[\frac{14783}{40320}\]

[quote="calculuslover800"][quote="mysmartmouth"]Derangements:

$\boxed{!n = n! \sum_{k=0}^{n}\frac{(-1)^{k}}{k!}}$

Use it wisely![/quote]

It's useful to note that the summation approaches $e$ as $n$ gets large, which is why $!8/8! \approx 1/e$[/quote]

This is easily derived if we note that:

$\sum_{k=0}^{\infty}\frac{1}{k!}= e$


Given a convex quadrilateral $ABCD$ with any $M$ inside it, called $G$ is centroid of $\triangle MCD$ prove that
$S_{DAB}+S_{MAB}+S_{CAB}=3S_{GAB}$   :P

[quote]Given a convex quadrilateral  $ABCD$with any $M$ inside it, called $G$ is centroid of $\triangle MCD$ prove that
 $S_{DAB}+S_{MAB}+S_{CAB}=3S_{MAB}$
[/quote]

please write the correct form of the problem :wink:

okay I have edited it I type error sorry  :blush: