LaTeX calculus geometry parallelogram ratio algebra polynomial

give yourself 40 minutes to complete the problems in the attachment

Yes, but hide them so people can work them first.

[hide="1"]don't know how to do all that sigma notation with latex, but it simplifies out to $\frac{n(n+1)}{2}=210 \longrightarrow n=20$[/hide]
[hide="8"]
$3+\frac{3}{2}+\frac{3}{4}....=6$[/hide]

[quote="prybarczyk"]Those are all right.  Number 2 is easy with Calculus, but more tricky without it.[/quote]

don't you simply have to find the minimum  :huh:

[hide="number 6"]calling each term $a$ and $b$, we have $a-b=3$ and $(a-b)^3=27=a^3-3a^2b+3ab^2-b^3 \Longrightarrow ab(a-b)=3$, so $ab=x^2-9^2=1 \Longrightarrow x=\sqrt{82} \Longrightarrow x^2=\boxed{82}$[/hide]

[hide="number 3"]isn't it just the number of divisors of $-1420$?[/hide]

Do you already have files on your computer that you will be posting or are you going to type them up along the way?  If you're going to be typing them up you might as well take Shobhit's suggestion and learn $\LaTeX$ so that you can type them straight into your post.  $\LaTeX$ is so much better than MathType!  It's not all that hard to learn enough to do the basics (though admittedly harder than MathType).

[hide="About number 3"]
I think $p=7$ is the only answer unless I'm terribly mistaken.[/hide]

I split this post from the other thread because these problems were a bit off topic (trying to keep organization in the forum ;) )

Btw, I forgot to to tell you Shobhit, but to do sigma notation you do

\sum_{i=0}^n i

\[ \sum_{i=0}^n i \]

[hide="2"]i used calculus on this one and got the min at $x=\frac12$ with a minimum value of $\boxed{2}$.[/hide]

[hide="the one about alcohol"]let $a$ be the amount taken out each time. Ater the first "move", there will be $A$ liters of alcohol. We then remove $A$ liters of this solution, or $\frac{A^2}{250}$ liters of alcohol. We then add another $A$ liters of alcohol. B/c the ratio of water to alcohol is 16:9, there are $90$ liters of alcohol in the final solution.
Solving $2A-\frac{A^2}{250}=90$ for $A$ yields the answer, $\boxed{50}$.[/hide]

[hide="Non Calc Number 2"]
Notice that we can write this as

\begin{eqnarray*} \frac{4x^2+8x+13}{6(x+1)} &=& \frac{4(x+1)^2+9}{6(x+1)}\\ &=& \frac 23(x+1)+\frac 3{2(x+1)} \end{eqnarray*}

Noting that $x+1$ is positive, we can use AM GM which yields

\[ \frac 23(x+1)+\frac 3{2(x+1)}\ge 2\sqrt{\frac 23(x+1)\frac 3{2(x+1)}}. \]

Things cancel under that radical leaving 2...which must be the minimum.[/hide]

[quote="joml88"][hide="Number 3"]

If $p$'s a root then \[ p^6+ap^5+bp^4+cp^3+dp^2=1470=2\cdot 3\cdot 5\cdot 7^2.  \]

Factor out a $p^2$ on the LHS....the only $p^2$ on the RHS is $7^2$.  Right?[/hide][/quote]

nice solution for number 2

[hide="but for number 3"] joe, take for instance this polynomial:

\[ P(x)=(x+1)\cdot(x-2)\cdot(x-3)\cdot(x-5)\cdot(x-7)^2  \]

This will have as its constant term $-1470$, but will have roots at $2$, $3$, $5$, $7$...note that they say that $p$ is [i]root[/i] of the polynomial, not that its a divisor...

Also note that the product of the roots of the polynomial are 1470...finding the prime factorization of 1470 gives you all the values of $p$.[/hide]

[hide="much ado about number three..."]
But your polynomial has an $x$ term.  Give me a polynomial like that which doesn't have an $x$ term and I'll be satisfied.  (don't actually try to...you could be there for a while :P )

My first explanation might not have been very good so here's another one:

Our polynomial is $x^6+ax^5+bx^4+cx^3+dx^2-1470$.  If this has a root at $p$ then we know that $p^6+ap^5+bp^4+cp^3+dp^2-1470=0$ or $p^6+ap^5+bp^4+cp^3+dp^2=1470$.  Now, we can factor a $p^2$ out on the LHS.  So we will have $p^2$ times some integer equals 1470.  But since $1470=2\cdot 3\cdot 5\cdot7^2$ the only way this is possible is if $p=7$.[/hide]

[quote="Iversonfan2005"][quote="arkmastermind"][hide="5"]$\boxed{35}$ - pretend ABCD is a parallelogram[/hide][/quote]

yeah, that's what I did, but I didn't think it was a good way to do it :P

How would you do it if you don't consider it to be a parallelogram, like Ptolemy mixed with Brahmagupta, etc? but then again, you'd be assuming its cyclic  :|[/quote]

[hide]
Let $x=DP, y=BD, w=CP,$ and $z=AP$.  So then the area of the quadrilateral is (note that $\sin \theta=\sin(180-\theta)$

\begin{eqnarray*} \frac 12\sin 30 yz+\frac 12\sin 30xw+\frac 12\sin 150 yw+\frac 12\sin 150 xz &=& \frac 12\sin 30(yz+xw+yw+xz)\\ &=& \frac 12\sin 30(x+y)(w+z)\\ &=& \frac{14\cdot 10}{2\cdot 2}\\ &=& 35 \end{eqnarray*}
[/hide]

[quote="Iversonfan2005"]^^oh okay, cool, so i was completely headed in the wrong direction with my thinking  :blush:

I actually tried to do that, but I didn't realize that it factored to $(x+y)(w+z)$[/quote]

I used grouping to factor it...probably should have said that before :P

lol about number 3, yeah i realized that $x$ term was 0 a little after posting but i had already left the comp,

sooo

yeah you're right, my bad  :lol:

[quote="arkmastermind"]iversonfan, whats brahmagupta?

i dont think ive heard of it before..[/quote]

For any cyclic quadrilateral, the area$=\sqrt{(s-a)(s-b)(s-c)(s-d)}$, where $s$ is the semiperimeter and all the others are the sides.

It's really similar to Herons and it's really good to know :)

[quote="arkmastermind"]hm. yeah i should probably learn that...ive never seen it before. thanks for the information!  :)[/quote]

Have you heard of Shoelace theorem Asif? I just learned it about a month or two back on this board and I think it is also very useful since we're on the topic of geometry.

For any polygon, say a quadrilateral with vertex points at $(a,b), (c,d), (e,f), (g,h)$ you list the points out in clockwise/counterclockwise order, so let's say that I already did, then you list them as:

a b
c d
e f
g h
a b

(note that you have to include $(a,b)$ twice.)

Now simply find the determinant $(ad+cf+eh+gb)-(bc+de+fg+ha)$ (hence the name shoelace) and divide the result by $2$ to get the area. It's very useful and any $n$ gon can be solved this way.

In fact, on the USC contest this year, I think number 20 or something (the one with the coordinate parallelogram) took 2 seconds to solve with this trick  :lol:

[quote="Iversonfan2005"]guess all eight problems were solved...so i'll include them in the document with all the problems. BTW Joe, were you able to get the solutions on your website? I already have them done, so if you want me to email them to you that's fine with me...[/quote]

No, I don't think I'll be able to do that.  I don't know how to upload files to my site so that they can be downloaded by other people :(

yeah i was reading geometry unbound (kiran kedlaya) and came across shoelace there...very useful.


[quote="hohoho11"]Two circles of radius 1 lie in a plane and are tangent to each other. How many circles of radius 2 are in this same plane and tangent to both circles?

A. 3
B. 4
C. 5
D. 6
E. 7[/quote]

[hide="Hint"]Consider all the possible ways for them to be tangent (sounds slightly stupid) such as Internally to one, Externally to another.

I saw this one in the MathCounts section a long time ago.

[/hide]

[hide="Solution"]1 large circle is internally tangent to both small circles, 2 large circles are externally tangent to both small circles, and 2 large circles are externally tangent to one small circle and internally tangent to the other, giving 5 circles in total. $ \Rightarrow\boxed{C}$[/hide]


I have been having some trouble figuring out the remaining side of acute scalene triangles when given two sides...nothing seems to work for me...any suggestions/solutions?

Think about the meaning of a triangle being acute. Given two sides $ a, b,$ and a third unknown side $ c,$ we have $ a^2\plus{}b^2>c^2.$ We also have the restriction of $ a\neq b\neq c.$ This should limit the choices when only whole numbers are involved.

The answer may not be a whole number...i can find out some possible ones but dont know how to find the angle cause tangent only works on right triangles if i am correct...

If you know the angle between the sides, you could use the Law of Cosines but that is too advanced for Middle School.

nah i know the law of cosines
but i just know two sides...so if i could find an angle that would make it loads easier

Well, then there is a range of uncertainty. You can change the length of the third side without making it so that it isn't acute. What are the first two side lengths?

i found it...the answer is 40<x<41
the two given sides were 9 and 41 and given that 9 was smallest side and 41 was largest, using the c^2<a^2+b^2 and a is 9...41^2-9^2=1600 or 40^2 therefore the remaining side must be greater than 40 but less than 41 cause 41 is the longest side length

Also, you could notice that 9-40-41 is a Pythagorean triple so anything smaller than that would have an angle greater than 90. So it $ x>40$. And it can't be 41 or bigger because then it would be the biggest side. So you got the right answer.

Some mistakes during posting:
first of all this is not right level
second of all you should put the ? sticky if you need help just remember that :D


inequalities
induction
inequalities unsolved

The following one was proposed by my teacher for a test. It's not difficult but kinda beautiful...
Let $ x,y,z\in (0,\plus{}\infty)$ and $ n\in \mathbb{N^{*}}$. Prove that $ \left(x^{n\plus{}3}\minus{}x^{n}\plus{}3\right)\left(y^{n\plus{}3}\minus{}y^{n}\plus{}3\right)\left(z^{n\plus{}3}\minus{}z^{n}\plus{}3\right)\geq \left(x\plus{}y\plus{}z\right)^{3}$

[quote="Γιώργος"]The following one was proposed by my teacher for a test. It's not difficult but kinda beautiful...
Let $ x,y,z\in (0, \plus{} \infty)$ and $ n\in \mathbb{N^{*}}$. Prove that $ \left(x^{n \plus{} 3} \minus{} x^{n} \plus{} 3\right)\left(y^{n \plus{} 3} \minus{} y^{n} \plus{} 3\right)\left(z^{n \plus{} 3} \minus{} z^{n} \plus{} 3\right)\geq \left(x \plus{} y \plus{} z\right)^{3}$[/quote]
The inequality is true for all $ n \in \mathbb{ N }$ . ( I consider that $ 0 \in \mathbb{ N }$ )

Use induction ! For $ n\equal{}0$,we obtain an immediate consequence of Holder's Inequality.Then,use the trivial inequality $ x^{n \plus{} 4} \minus{} x^{n\plus{}1} \plus{} 3 \geq x^{n \plus{} 3} \minus{} x^{n} \plus{} 3$ which holds for every $ x \geq 0$ and $ n \in \mathbb {N}$ , ...

easy but nice one..  :) Just show that $ x^n^ + ^3 - x^n + 3 \geq x^3 + 1 + 1$ which is equivalent to $ (x^3 - 1)(x^n - 1)\geq0$ which is obvious...!Then apply Holder's inequality.... :P OOPS!I think I am late.. :(


percent
geometry
geometric transformation

"March 7 issue - The name—Cambridge School of Languages—conjures images of spires and Anglo-Saxon aristocrats conversing in the Queen's English. But this Cambridge is composed of a few dank rooms with rickety chairs at the edge of a congested Delhi suburb. Its rival is not stately Oxford but the nearby Euro Languages School, where a three-month English course costs 16 dollars ([i]replaced dollar symbol[/i]). "We tell students you need two things to succeed: English and computers," says Chetan Kumar, a Euro Languages manager. "We teach one. For the other"—he points to a nearby Internet stall—"you can go next door."

The professors back in Cambridge, England, would no doubt question the schools' pedagogy. There are few books or tapes. Their teachers pronounce "we" as "ve" and "primary" as "primmry." And yet such storefront shops aren't merely the ragged edge of the massive English-learning industry, which in India alone is a $100 million-per-year business. They are the front lines of a global revolution in which hundreds of millions of people are learning English, the planet's language for commerce, technology—and, increasingly, empowerment. Within a decade, 2 billion people will be studying English and about half the world—some 3 billion people—will speak it, according to a recent report from the British Council.

From Caracas to Karachi, parents keen for their children to achieve are forking over tuition for English-language schools. China's English fever— elevated to epidemic proportions by the country's recent accession to the World Trade Organization and the coming 2008 Olympics—even has its own Mandarin term, Yingwen re. And governments from Tunisia to Turkey are pushing English, recognizing that along with computers and mass migration, the language is the turbine engine of globalization. As one 12-year-old self-taught Eng-lish-speaker from China's southwestern Sichuan province says, "If you can't speak English, it's like you're deaf and dumb."

Linguistically speaking, it's a whole new world. Non-native speakers of English now outnumber native speakers 3 to 1, according to English-language expert David Crystal, whose numerous books include "English as a Global Language." "There's never before been a language that's been spoken by more people as a second than a first," he says. In Asia alone, the number of English-users has topped 350 million—roughly the combined populations of the United States, Britain and Canada. There are more Chinese children studying English—about 100 million—than there are Britons.

The new English-speakers aren't just passively absorbing the language—they're shaping it. New Englishes are mushrooming the globe over, ranging from "Englog," the Tagalog-infused English spoken in the Philippines, to "Japlish," the cryptic English poetry beloved of Japanese copywriters ("Your health and loveliness is our best wish," reads a candy wrapper. "Give us a chance to realize it"), to "Hinglish," the mix of Hindi and English that now crops up everywhere from fast-food ads to South Asian college campuses. "Hungry kya?" ("Are you hungry?"), queried a recent Indian ad for Domino's pizza. In post-apartheid South Africa, many blacks have adopted their own version of English, laced with indigenous words, as a sign of freedom—in contrast to Afrikaans, the language of oppression. "We speak English with a Xhosa accent and a Xhosa attitude," veteran actor John Kani recently told the BBC.

All languages are works in progress. But English's globalization, unprecedented in the history of languages, will revolutionize it in ways we can only begin to imagine. In the future, suggests Crystal, there could be a tri-English world, one in which you could speak a local English-based dialect at home, a national variety at work or school, and international Standard English to talk to foreigners. With native speakers a shrinking minority of the world's Anglophones, there's a growing sense that students should stop trying to emulate Brighton or Boston English, and embrace their own local versions. Researchers are starting to study non-native speakers' "mistakes"—"She look very sad," for example—as structured grammars. In a generation's time, teachers might no longer be correcting students for saying "a book who" or "a person which." Linguist Jennifer Jenkins, an expert in world Englishes at King's College London, asks why someAsians, who have trouble pronouncing the "th" sound, should spend hours trying to say "thing" instead of "sing" or "ting." International pilots, she points out, already pronounce the word "three" as "tree" in radio dispatches, since "tree" is more widely comprehensible.

Not everyone is as open-minded about English, or its advance. The Web site of the Association for the Defence of the French Language displays a "museum of horrors"—a series of digital pictures of English-language signs on Parisian streets. But others say such defensiveness misses the point. "This is not about English swamping and eroding local identities," says David Graddol, author of the British Council report. "It's about creating new identities—and about making everyone bilingual."

Indeed, English has become the common linguistic denominator. Whether you're a Korean executive on business in Shanghai, a German Eurocrat hammering out laws in Brussels or a Brazilian biochemist at a conference in Sweden, you're probably speaking English. And as the world adopts an international brand of English, it's native speakers who have the most to lose. Cambridge dons who insist on speaking the Queen's English could be met with giggles—or blank stares. British or American business execs who jabber on in their own idiomatic patois, without understanding how English is used by non-natives, might lose out on deals.

To achieve fluency, non-native speakers are learning English at an ever-younger age. Last year primary schools in major Chinese cities began offering English in the third grade, rather than middle school. A growing number of parents are enrolling their preschoolers in the new crop of local English courses. For some mothers-to-be, even that's not early enough; Zhou Min, who hosts several English programs at the Beijing Broadcasting Station, says some pregnant women speak English to their fetuses. At Prague's Lamea children's English-language school, 3-year-olds sing songs about snowmen and chant colors in English. Now 2-year-olds have a class of their own, too.

For the traditional custodians of English—the British and, more recently, the Americans—this means money. The demand for native English-speakers is so huge that there aren't enough to go around; China and the Middle East are starting to import English teachers from India. The average price of a four-day business-English course in London for a French executive runs 2,240 euro. Despite—or perhaps because of—all the new Englishes cropping up, it's the American and British versions that still carry prestige, particularly with tuition-paying parents. Australia and Britain, in particular, have invested heavily in branding themselves as destinations for learning English. More than 400 foreign English-teaching companies are trying to break into China. On a visit to Beijing last week, British Chancellor Gordon Brown said the Chinese thirst to acquire the language was "a huge opportunity for Britain," which already boasts a 1.3 billion pound English-teaching industry. Says Jenkins, "Owning English is very big business."

To see big business in action, one need only walk down London's busy Oxford Street, where ads hawk instant access to the language of success: DOES YOUR ENGLISH EMBARRASS YOU? BUSINESS ENGLISH FOR BEGINNERS; LEARN ENGLISH IN JUST 10 WEEKS! Above clothing stores, bustling English-language schools are packed with eager twentysomethings from around the world. Ben Beaumont, a buoyant 28-year-old Briton, presides over a class that includes a South Korean business manager, a nurse from rural Japan and an Italian law student. "Do you want a lot of homework or a little?" he asks. The class is unequivocal: "A lot!"

Volunteers like Steve Restains, who teaches English in Phuket, are helping rebuild countries hit by last year's tsunami Volunteers like Steve Restains, who teaches English in Phuket, are helping rebuild countries hit by last year's tsunami Why such enthusiasm? In a word, jobs. A generation ago, only elites like diplomats and CEOs needed English for work. "The ante on what's needed is going up year by year," says Graddol. "Throughout organizations, more people need more English." In China, the Beijing Organizing Committee for the 2008 Olympics is pushing English among staff, guides, taxi drivers and ordinary citizens. For lower-middle classes in India, English can mean a ticket to a prized call-center job. "With call centers, no longer is speaking English one of the important skills to get a good job," says Raghu Prakash, who runs an English-language school in Jaipur. "It is the skill." At the new Toyota and Peugeot plant in the Czech Republic, English is the working language of the Japanese, French and Czech staff. Says Jitka Prikrylova, director of a Prague English-language school: "The world has opened up for us, and English is its language."

Governments, even linguistically protectionist ones, are starting to agree. Last year Malaysia decided to start teaching school-level math and science in English. In France, home of the Academie Francaise, whose members are given swords and charged with defending the sanctity of the French language, a commission recommended last fall that basic English be treated like basic math: as part of the mandatory core curriculum beginning in primary school. As it turns out, the minister of Education didn't agree. No matter; French schoolchildren are ahead of their government: 96 percent of them are already studying the language as an elective in school.

Technology also plays a huge role in English's global triumph. Eighty percent of the electronically stored information in the world is in English; 66 percent of the world's scientists read in it, according to the British Council. "It's very important to learn English because [computer] books are only in English," says Umberto Duirte, an Uruguayan IT student learning English in London. New technologies are helping people pick up the language, too: Chinese and Japanese students can get English-usage —tips on their mobile phones. English-language teachers point to the rise of Microsoft English, where computer users are drafting letters advised by the Windows spell check and pop-up style guides. In the temple town of Varanasi, India, Sanjukta Chaterjee says she's astonished by the way her 7-year-old son learns the language, through CDs and video. "Our teachers were strict that we should practice, and speak the language till we were near-perfect," she says. "Now there's an additional technological finesse to learning English."

Schools are becoming more and more creative. Last August, South Korea set up its first English immersion camp. The Gyeonggi English Village, built on a small island in the Yellow Sea and subsidized by the provincial government, comes complete with a Hollywood-style fake bank and airport, where students must conduct all transactions in English. "Through the camp, we want to train capable global citizens, who can help Korea win international competition in this age of globalization," says Sohn Hak Kyu, governor of Gyeonggi province, who started the program. In one class, eighth grader Chun Ho Sung, wearing a long black wig and posing as British heartthrob Orlando Bloom, sweats under the lights of a mock television studio as he prepares to be interviewed. "Do you think you are handsome?" asks the anchorwoman. Shyly, in broken English, Chun responds: "Yes, I do. I am very handsome." The audience of other students collapses in giggles.

While courses like Gyeonggi's sound simple, English and its teaching are inexorably becoming more complex. Ilan Stavans, an Amherst College professor, recently finished a translation of Cervantes's "Don Quixote" into Spanglish, the English-Spanish hybrid spoken in the United States and Mexico. Writing in the journal English Today last spring, Hu Xiaoqiong argued for reorientating China's English curriculum toward China English, incorporating Chinese phrases like "pay New Year calls," a Spring Festival tradition, and "no face," to be ashamed—as Standard English. In countries like Germany, where most kids begin English as early as the second or third grade, the market for English studies is already shrinking. German language schools no longer target English beginners but those pursuing more-expert niches: business English, phone manners or English for presentations. Beginning-English classes are filled with immigrants from places like Turkey and Russia, eager to catch up with the natives. As with migrants the world over, they're finding that their newfound land is an English-speaking one."

[b] Non-Native Speakers are transforming the global language. What do think about it and the way it will change we communicate ?[/b]

Editor note: Here's the link: http://www.msnbc.msn.com/id/7038031/site/newsweek/; I took it out of the source field to make this forum's top level render more clearly.

All that is false. English is "yours", both British and American, because is your BIG BUSINESS.

Moreover, it's hypocrit that the authors flaunt "Non-Standard" Englishes, like Scots or Ghanian English, but afterwards they are not valid for business or science. I don't imagine myself talking to my boss in Chinglish

English is the language of science because the US dominates this field. English is the language of Internet because it was invented in the US
English is the language of business because the US is the economic superpower.

So English dominates the world because the US is the superpower.

As somebody said "language is always companion of empire"

Take care


conics
parabola
calculus
integration
algebra
system of equations
calculus computations

[quote="Ikerous"]I'm trying to find the equation of a unique parabola for a program

I'm given two points, (0,0) and (c,d) as well as the arch length between the two points (L).  The general equation is obviously going to be y=ax^2 + bx.  I tried to solve by plugging in c and d and then using the arc length integral but it became so gaudy I'm not even sure it'd be possible to isolate one variable in terms of knowns.

Is there any equation in terms of c, d and L for the parabola?  Or even a series or loop/boolean combinations i could use to get a and b?

tia[/quote]

From $y=ax^{2}+bx$ know that a and b satisfy this system of equations

$d=ac^{2}+bc$
$L=\frac{-b\sqrt{1+b^{2}}+(b+2ac)\sqrt{1+(b+2ac)^{2}}-\mbox{arcsinh}(b)+\mbox{arcsinh}(b+2ac)}{4a}$


How many diagonalizable matrices are there in $\mathcal{M}_{n}(\mathbb{F}_{q})$?

P.S.: I couldn't find it on this forum, sorry if it's already there.

I think the answer is
$q^{n^{2}}\left(\sum_{m=1}^{n}\sum_{a_{1}+...+a_{n}= n \atop a_{1}\leq \cdots \leq a_{m}}\frac{q^{m}}{q^{a_{1}^{2}+\cdots+a_{m}^{2}}}\right)$

this is obtained by considering the centralizer of non-similar diagonalizable matrices. I guess it can be simplified but I don't know how to do so.

soarer,
Do you  have a little bit more details about the proof?


algebra
function
domain
geometry proposed
geometry

Dear All My Friends,
There are 1 point, 1 line and 1 circle on our plane. How we can do to get interesting simple and new geometrical fact?
Best regards,
Bui Quang Tuan

The circle will divide a plane into 2 part (not obvious this is Jordal theorem with three papers proof :D ) and the line must be through not bounded domain and the point must belong to one of two domains (we consider circle as a part of bounded domain)! :P


geometry
3D geometry
prism
analytic geometry
trigonometry
trapezoid
parallelogram

http://mcis.jsu.edu/mathcontest/geometryans07.pdf

$Numbers$

2
 4
 8
10
11
12
14
15
17
18
21
25
26
32
39
44
45
49

These are the ones I missed.

Can someone post the solutions for the numbered ones? THANKS!  :lol:

Make the side lengths be a,b, and c, where a and b are the legs and c is the hypotenuse.

[hide="14"]So we have that a+b+c=26, and $a^{2}+b^{2}+c^{2}=288$. We also know that $a^{2}+b^{2}=c^{2}$. So plug this in to get that c=12. So we know that $a^{2}+b^{2}=144$. We also know that a+b=14. We square this to get that $a^{2}+2ab+b^{2}=196$. Substitute to get $2ab=52$. Divided by four to get the area which is 13.[/hide]
[hide="21"]Make the side lengths of the rectungular prism sides a,b and c. We know that ab=9, bc=16, and ac=25. So we want the volume which is abc. Multiply all of these equations to get $a^{2}b^{2}c^{2}=9(16)(25)$. Square root all of this to get $abc=3(4)(5)$ which is 60.[/hide]
[hide="44"]Just use the formula $\frac{Ax_{1}+By_{1}+C}{\sqrt{A^{2}+B^{2}}}$
Where $(x,y)$ is the coordinates and the standard form is $Ax+By+c=0$. When you plug this in you get the answer is $frac{6}{5}$.[/hide]
I'll do more later.

[hide="10"]Change the equations to $x^{2}+y^{2}=5$ and $2x^{2}-y^{2}=4$. Add them together to get $3x^{2}=9\rightarrow x=\pm \sqrt{3}$. Plug this back in to get $x=\pm \sqrt{2}$. So the points are $(-\sqrt{3}, \sqrt{2}), (-\sqrt{3},-\sqrt{2}), (\sqrt{3},-\sqrt{2}), (\sqrt{3}, \sqrt{2})$. You can use shoelace to find the area is $4\sqrt{6}$.[/hide]

[hide="11"]Ok to have the midpoint as the origin obviously the coordinates have to be additive inverses. So plug them into the equations to get
\[y=4x^{2}+8x-4 \]
and
\[-y=4x^{2}-8x-4 \]
so add them together to get $0=8x^{2}-8\rightarrow x=\pm 1$ so plug this back in to find $y=\pm 8$ The points are $(-1,-8)$ and $(1,8)$ distance formula says that the answer is $2\sqrt{65}$[/hide]

[hide="12"]Let $CD=a$. Knowing that $[CBD]+[ACD]=[ABC]$ and the sin area formula, we have \[\frac{4a\sin 60}{2}+\frac{8a\sin 60}{2}=\frac{(8)(4)\sin 120}{2}\] Solving for a results in a=8/3[/hide]

[quote="now a ranger"]http://mcis.jsu.edu/mathcontest/geometryans07.pdf
[/quote]

cool, I am preparing for ICTM and in need of extra algebra II problems... too bad they don't have full solutions  :(

[hide="32"]midpoint of AB=O midpoint of CF=P

and radius of semicircle P is r

draw OP, this is 9+r

PB=6-r

and OB=9. then use pythagorean theorem to solve for r and then subtract from 6.

[/hide]

[hide="44"]just use the formula $\frac{Ax+By+C}{\sqrt{A^{2}+B^{2}}}$ where the line is $Ax+By+C=0$ and the point is $(x,y)$.[/hide]

ill post solutions to the other two problems later.

btw 49 is like some problem from a previous vestavia and was on the grissom this year i posted a solution in bama forum

[quote="diophantient"][hide="outline of 4"]
$\sin 18=\frac{\sqrt{5}-1}{4}$
Use that and law of cosines a couple times to find the sidelengths and whatnot. [/hide]
[hide="number 8"]
Note that all triples are expressible as $2rs$, $r^{2}-s^{2}$, and $r^{2}+s^{2}$
If the smallest side was $r^{2}-s^{2}$ then r isn't an integer.
So $r+s=9$.
Let r=4, s=5.
40-9-41
nope
Let r=3, s=6
36-27-45
nope
r=2, s=7
28-45-53
So far it's all we got.
r=1, s=8
16-63-65
So our answer is $14 \cdot 45=630$[/hide]
[hide="15"]
We see that AC $4\sqrt{5}$
The area is 40.
$BF=4\sqrt{5}$
$AF=2\sqrt{5}$
Now note that FEA~DEB
Let AE=x
$\frac{\sqrt{x^{2}-20}{8-x}=\frac{\sqrt{5}}{3}$
$\frac{x^{2}-20}{x^{2}-16x+64}=\frac{5}{9}$
$9x^{2}-180=5x^{2}-80x+320$
$4x^{2}+80x-500=0$
$x^{2}+20x-125=0$
$x=-10+15$
$x=5$
So AF=$8-5=3$[/hide]
[hide="17"]
$y^{4}=27y$
$y=0$
$y^{3}-27=0$
$(y-3)(y^{2}+3y+9)$
$y=3$ is the only real.
So our two points are (0,0) and (9,3)
Distance formula gives $\sqrt{90}=3\sqrt{10}$[/hide]
[hide="25, one liner"]
Note that AD is the altitude.[/hide]
[hide="26"]
Let DE=x, and the altitude of ADE=$\frac{40}{x}$.
The Altitude of ABC is $\frac{160}{x}$
The altitude of DECB is $\frac{120}{x}$
The altitude of CDE= the altitude of DECB, and the base is x.
$\frac{x \cdot \frac{120}{x}}{2}=60$[/hide][/quote]
Wait why is AD is altitude for number 25?

[hide="45"]  The two quarter circles added together minus the big shaded region plus the smaller shaded region is equal to the area of the square so 1^2pi*1/4*2-B+S=1, so -B+S=1-1/2pi, so S-B=1/2pi-1[/hide]

[hide="14"] Say the lengths of the sides are a, b, and c, where c is the hypotenuse.  We have that a+b+c=26 and $a^{2}+b^{2}+c^{2}=288$.  We know that $a^{2}+b^{2}=c^{2}$, so $a^{2}+b^{2}+c^{2}=c^{2}+c^{2}=2c^{2}=288$.  Thus, $2c^{2}=288\Rightarrow c=12$.  So, we have that 12+a+b=26, so a+b=14.  Since $a^{2}+b^{2}=144$, we have that $b=\sqrt{144-a^{2}}$.  Thus, \[a+\sqrt{144-a^{2}}=14\Rightarrow 14-a=\sqrt{144-a^{2}}\Rightarrow 196+a^{2}-28a=144-a^{2}\Rightarrow 2a^{2}-28a+52=0\Rightarrow a^{2}-14a+26=0\]  Now, we know that one root of this equation is $a$ and the other is $b$.  Since we are looking for $\frac{ab}{2}$, we want the product two roots for this equation, divided by two.  The constant is the product of the two roots, so $ab=26\Rightarrow \frac{ab}{2}=\boxed{13}$ [/hide]
[hide="45"]
Say that the two arcs intersect at Y.  Say that the area of the bigger shaded region is $x$.  We can easily find that the area of circular sectors DAB and DAC are both $\frac{\pi}{4}$.  Thus, the areas of the regions contained by D, C, and Y, and A, B, and Y are both $\frac{\pi}{4}-x$.  Thus, the area of everything except the small region is $\frac{\pi}{4}-x+\frac{\pi}{4}-x+x=\frac{\pi}{2}-x$.  Thus, the area of the small shaded region is ${1-(\frac{\pi}{2}-x)=x+1-\frac{\pi}{2}}$.  So, the difference is $x-(x+1-\frac{\pi}{2})=\boxed{\frac{\pi}{2}-1}$[/hide]
[hide="49"] Say that the midpoint of AC is M and that of BC is N.  Also, say that AM=a and BN=b.  Thus, we have that AC=2a and BC=2b.  Thus, the radius of the big semicircle a+b.  So, the area of the shaded region is the (area of the big semicircle-area of the two semicircles).  This is $\frac{\pi(a+b)^{2}}{2}-\frac{a^{2}\pi}{2}-\frac{b^{2}\pi}{2}=\frac{2ab\pi}{2}$. We have that \[DA=\sqrt{AC^{2}+DC^{2}}=\sqrt{4a^{2}+1}\] and \[DB=\sqrt{BC^{2}+DC^{2}}=\sqrt{4b^{2}+1}\]  We also have that AB=2(a+b).  Since angle ADB is right, we have that \[BD^{2}+AD^{2}=AB^{2}\Rightarrow 4a^{2}+4b^{2}=4a^{2}+8ab+4b^{2}\Rightarrow 2=8ab\Rightarrow ab=\frac{1}{4}\]  Now, since we want $\frac{2ab\pi}{2}$, we can plug in our value for $ab$ to find that the area of the shaded region is $\boxed{\frac{\pi}{4}}$[/hide]

[hide="39"]Call the point where Ef and Ad meet O.  So we have the height of trapezoid EFCB as h and the height of triangle AEF as g.  We know frmo similar triangles that (BD+DC)/(g+h)=(DC)/(g)-note DC is the same as EF.  Cross multiplying gives us the fact that BD(g)=DC(h), but DC(h) is the area of the parallelogram-28.  We also have similar triangels BDA and EOA.  So then (BD)/(g+h)=(EO)/(g), cross multiplying gives us BD(g)=EO(g+h), but BD(g)=28 from earlier so EO(g+h)=28.  Now note the area of taht shaded traingle is percisely 1/2*EO*(g+h), so it is 1/2*28=14.[/hide]
umm can somennoe explain 25..

[hide="Q4"]sum of all angles in a dodecagon is $1800$
each angle is $150$
we can work out $\angle P_{1}P_{3}P_{5}=120$
use cosine in triangle $P_{1}P_{2}P_{3}$
$P_{1}P_{3}^{2}=4+4-2*2*2*cos(150)=8+4\sqrt{3}$

area of triangle $P_{1}P_{3}P_{5}=\frac{1}{2}P_{1}P_{3}*P_{3}P_{5}*sin(\angle P_{1}P_{3}P_{5})=\frac{1}{2}P_{1}P_{3}^{2}sin(120)=3+2\sqrt3$

i don't understand the $sin18$ way. can someone please explain thanks  :blush:  [/hide]

[hide="8"]
We have that \[\sqrt{a^{2}+b^{2}}=c\Rightarrow a+\sqrt{a^{2}+b^{2}}=81\Rightarrow 81-a=\sqrt{a^{2}+b^{2}}\Rightarrow a^{2}-162a+6561=a^{2}+b^{2}\Rightarrow b^{2}=6561-162a\Rightarrow \sqrt{6561-162a}=b\]  Thus, the area of the triangle is $\frac{a\sqrt{6561-162a}}{2}$, and we can say that the value of a that would maximize this equation is 27, but if a=27, then c=54, so $b=27\sqrt{3}$, which is not an integer.  So, a must be as close to 27 as possible.  Now, since b is an integer, we know that $\sqrt{6561-162a}$ is also an integer.  Thus, $6561-162a$ is a perfect square.  Thus, $\frac{6561-162a}{9^{2}}=81-2a$ is a perfect square.  Now, we can do some quick guessing and checking to get that $a=40, a=36, a=28, a=16$.  Since 28 is closest to 27, we know that $a=28$ and $c=53$ since a+c=81.  Now, using the Pythogorean Theorem, we have that $b=45$, so $\frac{ab}{2}=\frac{28*45}{2}=14*45=\boxed{630}$.  (Note that figuring out that a must be close to 27 needs knowledge of caculus, but that was just a shortcut.  It also would have worked to test out all of the values of $a$ and see which one would make the largest area). [/hide]

[quote="Art of Owna"]There is nothing to do with sin 18..all I did was law of cosines twice and the 1/2absinC area formula..

and nvm about that earlier post..I can do 25..all you need to know is the tangent properties

[hide="25"]first note that AD is NOT the height (even if it is drawn like it)  Call the point where one of the circle meets AB X.  Call the point where both circles meet Y.  Call the point where the smaller circle meets BD M and where the bigger circle meets DC N.  Call the point where the other circles meets AC Z.  So  AY=AX=AZ from tangnet properties.  ALso, BX=BM, NC=CZ and MD=DY=DN  So AX+XB=260, AX+ZC=400 (substitue AX for AZ) and BX+MD+MD+CZ (substituting in the values)=520, so that means that ZC-XB=140.  Also, XB+ZC+2MY=520, so substituing ZC=140+XB into that second equation we have 2XB+2MD=520-140=380, so XB+MD=190, well since XB=BM so that means BM+MD=190 meaning BD=190.[/hide][/quote]
oops I thought that since the circles were tangent their radii were perpendicular to the line, and that the radii were parallele to the base.


induction
number theory
AMC
relatively prime

A triangular array of numbers has a first row consisting of the odd integers $ 1,3,5,\ldots,99$ in increasing order.  Each row below the first has one fewer entry than the row above it, and the bottom row has a single entry.  Each entry in any row after the top row equals the sum of the two entries diagonally above it in the row immediately above it.  How many entries in the array are multiples of $ 67$?
[asy]size(200);
defaultpen(fontsize(10));
label("1", origin);
label("3", (2,0));
label("5", (4,0));
label("$\cdots$", (6,0));
label("97", (8,0));
label("99", (10,0));

label("4", (1,-1));
label("8", (3,-1));
label("12", (5,-1));
label("196", (9,-1));
label(rotate(90)*"$\cdots$", (6,-2));[/asy]

[quote="E^(pi*i)=-1"]A triangular array of numbers has a first row consisting of the odd integers $ 1,3,5,\ldots,99$ in increasing order.  Each row below the first has one fewer entry than the row above it, and the bottom row has a single entry.  Each entry in any row after the top row equals the sum of the two entries diagonally above it in the row immediately above it.  How many entries in the array are multiples of $ 67$?[/quote]
[hide]We can see that numbers directly below numbers in the first row always come in the pattern (let the number in the first row be n) n, 4n, 16n, etc.
Since 67 is prime, the number of numbers in the array in the same vertical column as it will be the number of multiples of 67. We can easily see that since the array "shrinks", the last multiple of 67 occurs on the edge, at the 17th row, and so there are 17.[/hide]

[hide="Solution"]
Let the $ k$th number in the $ n$th row be $ a(n,k)$. We claim by [[induction]] on $ n$ or inspection that $ a(n,k) = 2^{n-1}(n+2k-2)$. Indeed, note that $ a(1,k) = 2^{1-1}(1+2k-2)=2k-1$, which is the correct formula for the first row. By definition of the array, $ a(n,k) = a(n-1,k)+a(n-1,k+1)$, and by our inductive hypothesis,
\begin{align*}a(n,k) &= a(n-1,k)+a(n-1,k+1)\\ &= 2^{n-2}(n-1+2k-2)+2^{n-2}(n-1+2(k+1)-2)\\&=2^{n-2}(2n+4k-4)\\&=2^{n-1}(n+2k-2)\end{align*}
thereby completing our induction.

Since $ 2^{n-1}$ and $ 67$ are [[relatively prime]], we just have to consider the $ n+2k-2$ term. Since every row has one less element than the previous row, $ k \le 51-n$. Hence $ n+2k-2 \le 100-n$, and the only multiple of $ 67$ which fits this range is $ 67$ itself. For $ n+2k-2 = 67$, note that we need $ n$ to be odd, and also that $ n+2k-2 = 67 \le 100-n \Longrightarrow n \le 33$. We can check that all odd $ n$ satisfying $ 1 \le n \le 33$ indeed contains one entry that is a multiple of $ 67$, and so the answer is $ \frac{33+1}{2} = \boxed{017}$.
[/hide]

[hide]
Probably not the most foolproof way, but I found a pattern for the triangle

On line 1, the 67 appears as the 34th term out of 50.
Multiples of 67 cannot appear on the even lines simply because there isn't that many terms
Then on line 3, a multiple of 67 appears as the 33rd term out of 48.
So I concluded that the term the multiple of 67 appears on decreases for every two terms lost per row. Therefore, the 18th term of the row with 18 terms is the last multiple of 67 to appear.  So 34 - 18+ 1 = 17
[/hide]

[hide=Solution]
Call the row with $n$ numbers, row $n$. Thus, the first row is row $50$ and the row below is $49$, going all the way to $1$. Now, after a while, you see that the only rows that have a chance of containing a multiple of $67$ are when $n$ is even. To make our life easier, factor out all the unnecessary multiples of $4$. In these rows, you get consecutive numbers as such:

Row $50$ - $1,3,5,\dots, 99$
Row $48$ - $3,5,7,\dots, 97$
Row $46$ - $5, 7, 9, \dots 95$

At this point, it's obvious that each row has its bounds shifted by $2$, on each side. Thus, we want $99-2n \geq 67$, to which $n=0,1,2,\dots, 16$, for an answer of $\boxed{017}$

Nice solution @above. But even better with the almost 10 year bump!   :-D

It's okay to bump a thread if they have a solution / discussion , though

The numbers in the leftmost diagonal are $1, 4, 12, 32, 80 \ldots$ and thus the $n$th number in said diagonal is $n \cdot 2^{n-1}$. We can see that in the $n$th row, the common difference is $2^n$, so the $x$th term in the $n$th row is $n \cdot 2^{n-1} + (x)2^n = 2^{n-1}(n+2(x)) = 2^{n-1}(n+2x)$. Clearly we will have $50$ rows that range from having $1$ term to having $50$ terms. Obviously $2^{n-1}$ will always be coprime to $67$, so $n+2x$ must be a multiple of $67$, meaning that $n+2x \equiv 0 \pmod {67}$. Now, we take cases. If $n = 1$, there is one solution; $x = 33$. If $n = 2$, there are no solutions. If $n = 3$, there is one solution; $x = 32$. If $n = 4$, there are no solutions, and so on. The last working value is $n = 33$, which gives $x = 17$. Anything after that when $n$ is odd gives $x+n > 50$, and $n$ is even never gives solutions. There are $\boxed{17}$ numbers in the list $1, 3, 5, \ldots, 33$.

I’m actually getting something slightly different @above, the explicit formula for numbers in the $n$th row is $K_n = n 2^{n-1} + k2^n$, where $0 \le k \le 50-n$, so we have $n+2k \equiv 0 \mod 67$. Clearly if $n$ is odd by inspection there exists no $k$ that satisfies the condition, but in order to find the maximum odd $n$ that works we have $n+2(50-n) \ge 67 \implies n \le 33$. Therefore there are $\boxed{017}$ solutions

@below $k=0$ is the first term, not $k=1$

@above The first term in the first row is 1, but plugging in $n = k = 1$ to your formula gives $3$? I understand we're indexing differently but I'm not sure what I've done wrong.


Of the students attending a school party, $ 60\%$ of the students are girls, and $ 40\%$ of the students like to dance.  After these students are joined by $ 20$ more boy students, all of whom like to dance, the party is now $ 58\%$ girls.  How many students now at the party like to dance?

[hide] Say that there were $ 3k$ girls and $ 2k$ boys at the party originally. $ 2k$ like to dance. Then, there are $ 3k$ girls and $ 2k\plus{}20$ boys, and $ 2k\plus{}20$ like to dance.

Thus, $ \dfrac{3k}{5k\plus{}20}\equal{}\dfrac{29}{50}$, solving gives $ k\equal{}116$. Thus, the number of people that like to dance is $ 2k\plus{}20\equal{}\boxed{252}$.[/hide]

Thats EXACTLY what I did, but I forgot that only 40% of the people initially present wanted to dance, so I put my answer as 600 (the final number of people), instead of 600*.42=252, which was the real answer.
 :mad:

yay! one of the few problems i got right =D

[quote="123456789"][quote="dingzhou"]hm...i thought it was a trick question, i think this is the easiest aime problem i ever seen, i mean i can do it with ease...

well, i know that i can get at least one right one aime I  :D[/quote]
This problem might have been even easier:
http://www.artofproblemsolving.com/Wiki/index.php/2006_AIME_I_Problems/Problem_1[/quote]

Another close one:
http://www.artofproblemsolving.com/Wiki/index.php/2005_AIME_I_Problems#Problem_2

Let $3x = \text{girls}$ and let $2x = \text{boys}$.

$20$ more boys come.
$$\frac{3x}{5x + 20} = 58/100$$.

Hence $x = 116$.

Hence our answer is just:
$$\frac{40}{100}\cdot580 + 20 = 252$$

[hide=Solution with no (hard) algebra, only logic]
We notice that $\frac53$ of the number of girls is the total number of people.

We notice that after $20$ more people join, $\frac{50}{29}$ of the girls is the total number of people.

We notice that the number of girls always stays the same, so $\frac{50}{29}-\frac{5}{3}=\frac{5}{87}$ of the number of girls must be $20$.

We calculate there to be $20\cdot\frac{87}{5}=348$ girls.

We notice that $40$ is $\frac23$ of $60$ so $\frac23\cdot348+20=\boxed{252}$.
[/hide]

[hide=Solution]Let the amount of students attending be $x$. Then $0.6x$ are girls and $0.4x$ are boys. After the boys join, there are $x+20$ people attending. We now have $\frac{0.6x}{x+20}=\frac{29}{50}$, or $30x=29x+29 \cdot 20 \implies x=580$. The amount who originally like dancing is $580 \cdot \frac{2}{5}=232$, and adding $20$ gives $\boxed{252}$.[/hide]

50 dollars and an ice cream cake I sillied this

OOPS I didn't forget about what I just said

[hide=Solution]Let $p$ be the number of people at the party before the 20 boys join. Therefore, we have the equation
$$\frac{3}{5}(p)=\frac{29}{50}(p+20)$$
Proceeding to solve this equation finds that $p=580$. The amount of people that like to dance after the 20 boys join is just
$$\frac{2}{5}(p)+20=\frac{2}{5}\cdot 580+20=232+2=\boxed{252}$$[/hide]

[size=50][hide=.]susususususususususususuusususus[/hide][/size]

Let, the total amount of students, be $10x,$ then there are $4x,$ boys, and $6x,$ girls. Hence, we have $\frac{6x}{10x+20}=\frac{29}{50}$, hence, $290x+580=300x$, hence, $10x=580$, hence, we have $580\cdot \frac{2}{5}+20=\boxed{252}.$


probability
calculus
integration
inequalities
search
real analysis
real analysis unsolved

Hi all

I have been working on the following problem for a while, and am quite stuck. Hopefully someone can help me out. This question is from Rudin's Real and Complex Analysis.

Given a probability space (X, m), assume f is in L^r for some r (And hence for all p<r by a previous result). Show that the limit as p -> 0 of the L^p norm of f is equal to the exp{ integral( log(|f|) ) }

Now I know how to use Jensen's inequality to show that the limit has to be greater then or equal to the RHS. I can also show the reverse inequality when the measure of the set of points where x=0 is non 0. So I have reduced the problem a bit, but now I have been stuck here for a few days. I thought about applying log to the LHS sides and applying L'hopital's rule and this gives a nice (albeit still not solved) picture, but of course we have to show that we can actually differentiate with respect to p, which is not clear, especially if we don't assume that integral( log(|f|) ) is not -infinity. Hence I have tried to work on the two different cases, where the integral is finite and when it is -infinity, but have not been able to make any further progress on either case.

Any advice on how to proceed on either of these two cases would be greatly appreciated.

Thanks!

There are two solutions [url=http://www.mathlinks.ro/viewtopic.php?t=75936]here[/url].

Awesome thanks. I tried looking for this problem by search but couldn't find it. Also, I jsut realized I posted in the wrong section lol. Thanks again.


Let a, b, c, d are unequaled real numbers satisifying a+1/b = b+1/c = c+1/d = d+1/a = x.
Find the value of x.

The answer is $x=\pm \sqrt{2}.$

Express $b,c,d,$ in terms of $a,x$ we have $b=\frac{1}{x-a},c=\frac{x-a}{x^2-ax-1},d=\frac{x^2-ax-1}{x^3-ax^2-2x+a},$ thus

Plugging these into $d+\frac{1}{a}=x,$ and rearranging, we have $x(x^2-2)(a^2-xa+1)=0.$...... :D


Prove or do not prove that the equation:
$ \sqrt[7]{x}\minus{}\sqrt[5]{x}\equal{}\sqrt[3]{x}\minus{}\sqrt{x}$ -  has exactly one solution ?

[quote="Obel1x"]Prove or do not prove that the equation:
$ \sqrt [7]{x} \minus{} \sqrt [5]{x} \equal{} \sqrt [3]{x} \minus{} \sqrt {x}$ -  has exactly one solution ?[/quote]

$ x \equal{} 1$ and $ x\equal{}\minus{}1$ and $ x \equal{} 0$ :?:

How do u explain that there aren't any other solution ? :!:

Well, if we found more than one solution then we have successfully disproved the statement, so we are done.

what if the problem was for exactly three solutions ?


Question:
According to my calculation, $ \left( {\frac {8}{{15}}} \right) \equal{} \minus{} 1$ and $ \left( {\frac {{ \minus{} 8}}{{15}}}\right) \equal{} \minus{} 1$. But we have$ \left( {\frac {{ \minus{} 8}}{{15}}} \right) \equal{} \left( {\frac {{ \minus{} 1}}{{15}}} \right).\left( {\frac {8}{{15}}} \right) \equal{} ( \minus{} 1).( \minus{} 1) \equal{} 1$. Is it wrong? Maybe I stupid but can anybody answer my question?

I think that $ \left({\frac{8}{{15}}}\right) \equal{}\left({\frac{8}{{5}}}\right)\left({\frac{8}{{3}}}\right)\equal{}(\minus{}1)(\minus{}1)\equal{}1$


In right Triangle ABC, AB is the hypotenuse and is 2 cm more than the longest leg.  The other leg is half of AB.  Give the length of the shortest leg, in cm.

I can get to the point where I get a quadratic equation, which I shouldn't have to do to solve the problem though.  Any clues?

You don't need to use the quadratic at all. Just remember the Pythagorean Triple:
3:4:5

[quote="kysuke"]The other leg is half of AB.[/quote]

It is not a $ 3\minus{}4\minus{}5$ triangle, hunter34.

[hide]It's a $ 30^{\circ}\minus{}60^{\circ}\minus{}90^{\circ}$ triangle, as the sides are in the ratio of $ 1: \sqrt{3}: 2$. From here it should be fairly simple to solve.[/hide]

[hide]I think you mean 30-60-90.
3:4:5 can't have one leg half of the hypotenuse...

So the triangle is a 30-60-90 triangle.
The length of the hypotenuse is $ 2x$ and the length of the longer leg is $ x\sqrt{3}$
$ (2\minus{}\sqrt{3})x\equal{}2 \implies x \equal{} \boxed{4\plus{}2\sqrt{3}}$
[/hide]

I had thought of one leg being half of the hypotenuse and it could be a 30-60-90 triangle but is the other leg was hypotenuse minus 2.  I wasn't sure if there was a complete way to determine that before doing the work and I'm still not sure.


Does anyone know what kind of scores are needed on the usnco national exams to qualify to be a finalist?

Nothing is etched in stone, but if you can get 55+ on the MC and 85+ on the FR you're probably going to get in.

I refer you to a post of Scott Rabin, an IChO medalist [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=978118620&t=33817&start=60](original thread).[/url]

[quote]
For everyone interested in how they grade the national exam, lets run through the basics.

1) If you're not making above a 55/60 on the Multiple Choice, you may have some problems qualifying
2) If you're answering none of the higher-level (d through g) questions in the Free Response, you may have some problems qualifying
3) If you're absolutely terrible at labs, you'll have NO PROBLEM getting in.

[...]

For everyone interested in going to the camp next year, I STRONGLY recommend taking all previous tests that are posted online.  That's how I prepared my first year, and how I helped inteluser prepare (who qualified but chose MOP over USNCO :P).  Also, if you think you have an exceptionally good chance at making it in, I recommend taking an organic class and lab at a local college.  That will prepare you for the absolute MINDJOB that is the organic lectures at camp.
[/quote]

what kind of "problem solving" do you have to do in the labs (based on past years)? would you have to like make up a protocol or something?


Dear Forum

Let X be a n by p matrix of which rank is p and be it's hatmatrix H = X(X'X)^(-1)X'.

Then Let Partition X into X = [X1| X2] where X1 n by q and X2 n by (p-q)

H1 = X1(X1'X1)^(-1)X1'

Define X2* as (I-H1)X2 and its hatmatrix H2

i.e. H2 = X2*(X2*'X2)^(-1)X2*'

Then show that H = H1 + H2

I am Sorry for poor notation. If you have any question about the problem, let me know that.

Thanks in advance.

I'll assume your matrices are real and your prime notation is for the transpose.

We have $ X \equal{} \begin{bmatrix}X_1 & X_2\end{bmatrix}$ where $ X_1$ is $ n\times q$ and $ X_2$ is $ n\times(p \minus{} q).$ Assume $ n\ge p,$ so certainly $ n > q$ and $ n > p \minus{} q.$ We're assuming that $ X$ is of full rank; that implies that $ X_1$ and $ X_2$ are of full rank.

$ H \equal{} X(X^TX)^{ \minus{} 1}X^T \equal{} \begin{bmatrix}X_1 & X_2\end{bmatrix} \left(\begin{bmatrix} X_1^TX_1 & X_1^TX_2 \\
X_2^TX_1 & X_2^TX^2\end{bmatrix}\right)^{ \minus{} 1} \begin{bmatrix}X_1^T \\
X_2^T\end{bmatrix}.$

We define $ H_1 \equal{} X_1(X_1^TX_1)^{ \minus{} 1}X_1^T$ and $ X_2^* \equal{} (I \minus{} H_1)X_2.$

Then $ H_2 \equal{} X_2^*(X_2^{*T}X_2^*)^{ \minus{} 1}X_2^{*T}$

$ \equal{} (I \minus{} H_1)X_2(X_2^T(I \minus{} H_1)^2X_2)^{ \minus{} 1}X_2^T(I \minus{} H_1)$

(where I have noted that $ H_1$ is symmetric).

We're asked to show that $ H_1 \plus{} H_2 \equal{} H.$

(I haven't done anything yet - I'm just clarifying the notation.)

One comment: $ H$ is idempotent and $ Hy$ is the orthogonal projection of a vector $ y$ into the column space of $ X.$

We can prove this by employing the Gram-Schmidt process on the columns of the original matrix $ X.$ This amounts to writing $ X\equal{}QR$ where $ Q$ is an $ n\times p$ matrix whose columns are orthonormal - that is, $ Q^TQ\equal{}I$ - and $ R$ is an invertible upper triangular $ p\times p$ matrix. That $ R$ is invertible follows from $ X$ being of full rank.

Then $ H\equal{}QR(R^TQ^TQR)^{\minus{}1}R^TQ^T\equal{} QR(R^TR)^{\minus{}1}R^TQ^T$

$ \equal{} QRR^{\minus{}1}(R^T)^{\minus{}1}R^TQ^T\equal{}QQ^T.$

In block form, $ X\equal{}\begin{bmatrix}X_1&X_2\end{bmatrix}\equal{} \begin{bmatrix}Q_1&Q_2\end{bmatrix} \begin{bmatrix}R_1&A\\0&R_2\end{bmatrix}.$

It follows from above that

$ H\equal{}\begin{bmatrix}Q_1&Q_2\end{bmatrix} \begin{bmatrix}Q_1^T\\Q_2^T\end{bmatrix}\equal{} Q_1Q_1^T\plus{}Q_2Q_2^T.$

Since $ H_1\equal{}Q_1Q_1^T,$ our task is to prove that $ H_2\equal{}Q_2Q_2^T.$

$ X_2^*\equal{}(I\minus{}Q_1Q_1^T)X_2\equal{}(I\minus{}Q_1Q_1^T)(Q_1A\plus{}Q_2R_2)$

$ \equal{}Q_1A\plus{}Q_2R\minus{}Q_1Q_1^TQ_1A\minus{}Q_1Q_1^TQ_2R_2\equal{}Q_2R_2$

where we use that $ Q_1^TQ_1\equal{}I$ and $ Q_1^TQ_2\equal{}0$ from the fact that the columns of $ Q$ are orthonormal.

But then by a repetition of the computation that gave us $ H,$ we get that

$ H_2\equal{}Q_2R_2(R_2^TQ_2^TQ_2R_2)^{\minus{}1}R_2^TQ_2^T\equal{}Q_2Q_2^T.$ [i]q.e.d.[/i]

Generalization: omit the hypothesis that $ X$ is of full rank.

That requires us to redefine the hat operation. Given $ X$, define $ \widehat{X}$ to be the matrix that gives the orthogonal projection of $ \mathbb{R}^n$ onto the column space of $ X.$

Let $ H \equal{} \widehat{X}.$ Suppose $ X \equal{} \begin{bmatrix}X_1 & X_2\end{bmatrix}.$ Let $ H_1 \equal{} \widehat{X_1}.$

Define $ X_2^* \equal{} (I \minus{} H_1)X_2.$ Define $ H_2 \equal{} \widehat{X_2^*}.$

Prove that $ H \equal{} H_1 \plus{} H_2.$

Second generalization: Suppose $ X \equal{} \begin{bmatrix}X_1 & X_2 & X_3\end{bmatrix}.$ Define $ H_1$ and $ H_2$ as above. Now construct the proper definition of $ H_3$ depending on $ X_3$ such that $ H \equal{} H_1 \plus{} H_2 \plus{} H_3.$

As far as I understand,  $ H_1$ projects $ \mathbb R^n$ onto $ A \equal{} \text{Im} (X_1)$ and $ H_2$ projects $ \mathbb R^n$ onto the projection $ B$ of $ \text{Im} (X_2)$ on $ \text{Im} (X_1)^\perp$. Since $ \text{Im} (X) \equal{} A \oplus B$ and $ A \perp B$, $ H \equal{} H_1 \plus{} H_2$. However, I think that Gram-Schmidt is required for a step above (the $ \oplus$ one), so maybe it's a not that different argument...


calculus
derivative
calculus computations

Thanks, i'll try to find the correct statement. Meanwhile, could you help me with the similar problem:

Let $u=\ln(x+\sqrt{1+x^2})$ and $v=\ln(y+\sqrt{1+y^2})$. Transform the following equation:
\[(1+x^2)\frac{\partial^2 z}{\partial x^2}+(1+y^2)\frac{\partial^2 z}{\partial y^2}+x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=0\].
This one looks easier, but I still can't figure it out  :( .

Danica, please post each problem in a separate topic. I've done this for you this time, but please do it yourself the next time ;)

PS :welcomeani: on AoPS-MathLinks :D

Note that $u$ depends on $x$ only and $v$ depends on $y$ only. This simplifies things a lot. Let $z(x,y)=Z(u,v)$. Then, using the chain rule, we get
$z_x=Z_uu_x=Z_u \frac 1{\sqrt {1+x^2}}$. Now, using the chain and the product rules, we obtain $z_{xx}=Z_{uu} u_x^2+Z_u u_{xx} =Z_{uu} \frac 1{1+x^2} - Z_u \frac x{(1+x^2)^{3/2}}$. From here it follows that $(1+x^2)z_{xx}+x z_x= Z_{uu}$. Similarly, the part with partial derivatives with respect to $y$ transforms into $Z_{vv}$. So, the final answer is $Z_{uu}+Z_{vv}=0$. Hope this helps :).

I would write that transformation in the opposite direction: $x=\sinh u$ and $y=\sinh v$. After that, it's a whole lot of chain rule.

Differentiating that transformation, $1=\cosh u\cdot\frac{\partial u}{\partial x}$, so $\frac{\partial u}{\partial x}=\frac1{\cosh u}\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\cdot\frac{\partial u}{\partial x}= \frac{\partial z}{\partial u}\cdot\frac1{\cosh u}$, so $x\frac{\partial z}{\partial x}=\tanh u\frac{\partial z}{\partial u}.$

$\frac{\partial^2 z}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\right)=\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial u}\cdot\frac1{\cosh u}\right)=\frac{\partial z}{\partial u}\cdot\frac{-\sinh u}{\cosh^2 u}\cdot\frac{\partial u}{\partial x} +\frac{\partial^2 z}{\partial u^2}\cdot\frac{\partial u}{\partial x}\cdot\frac1{\cosh u}=\frac{\partial^2 z}{\partial u^2}\cdot\frac1{\cosh^2 u}-\frac{\partial z}{\partial u}\cdot\frac{\sinh u}{\cosh^3 u},$ so $(1+x^2)\frac{\partial^2 z}{\partial x^2}=\frac{\partial^2 z}{\partial u^2}-\tanh u\frac{\partial z}{\partial u}.$

The terms involving $y$ and $v$ are similar. Adding the terms, the equation transforms to the Laplace equation $\frac{\partial^2 z}{\partial u^2}+\frac{\partial^2 z}{\partial v^2}=0$.


geometry
circumcircle
trigonometry
angle bisector
geometry proposed

$ABC$ is a scalene triangle. $D$ is a point on the interior angle bisector of $\angle{A}$. $E$ is the intersection (other than $A$) of the circumcircle of triangle $ABD$ and the line containing $\overline{AC}$. $F$ is defined analogously with respect to triangle $ACD$ and $\overline{AB}$. Show that $D$ is on $\overline{BC}$ or the circumcircle of $ABC$ if and only if

$AB \cdot FD \cdot EC = AC \cdot ED \cdot FB.$

Again, simple angle chasing yields DF=DC,DE=DB,BF=CE. Therefore $AB\cdot FD\cdot EC=AC\cdot ED\cdot FB\Leftrightarrow AB\cdot CD=AC\cdot BD\text{ or }EC=BF=0\Leftrightarrow\sin\angle ADB=\sin\angle ADC\text{ or }EC=BF=0\Leftrightarrow D\in BC\text { or }D\text{ lies on the circumcircle of }ABC$.

[quote="Mildorf"]I'm a little confused - how did you get $BF = CE$ from angle chasing? Unless I am mistaken, it seems that if this is always true, then there are issues when $D$ approaches $A$ with $E$ and $F$.[/quote]
Well...not actually angle chasing, but I proved the 2 triangles are congruent.

Ok, I think I see it. I didn't mean for this to be a boring problem. I solved it by inversion about $A$, Menelaus on $\triangle{AC'F'}$, substituting the angle bisector condition $\frac{E'D'}{E'A} = \frac{B'D'}{B'A}$, and noting that the given becomes equivalent to $\frac{C'D'}{D'B'} = 1$ for non-degenerate cases; leading to the conclusion that $D' \in \mbox{Circumcircle}(\triangle{AB'C'})$.


geometry
inequalities
inequalities proposed

Ponder This Geometry Inequality: $\frac{1}{2}a(a+b+c)\geq m_b h_b + m_c h_c$

What a tough one it is,I have tried all myself out but I still can't tackle it.
And I think you'd better off posting it at the geometry section.

I found it stronger than:

$(\frac{x+y}{2})^2(\frac{y+z}{2})^2(\frac{z+x}{2})^2\geq xyz(\frac{x+y+z}{3})^3$


[quote="soumik"]Device a formula to caclulate how many integers lie between m^n and m^(n+1)?
m,n are naturals.[/quote]

maybe $ m^{n\plus{}1}\minus{}m^n\minus{}1$ ?

[quote="soumik"]No idea.....BTW show how u got it.[/quote]

Hemmm, are you joking ?
Or maybe I misunderstand something.

The number of integers strictly between two integers $ a > b$ is $ a \minus{} b \minus{} 1$ : $ \{b \plus{} 1, b \plus{} 2, ..., b \plus{} (a \minus{} b \minus{} 1)\}$

@ pco, this is not what I mean,
(PLS TELL ME WHERE FROM U USE LATEX....)

I MEANT NUMBER OF INTEGERS BET, - TAKE AN EXAMPLE.....
2^2 and 2^3, n=2, m=2 here.....
Get it?

Number of integers lying between 2 consecutive powers of an integer.....I wanted a formula for it.....
Got it?

[quote]I MEANT NUMBER OF INTEGERS BET, - TAKE AN EXAMPLE.....
2^2 and 2^3, n=2, m=2 here.....
Get it?[/quote]

The number of integers strictly between $ b$ and $ a$ with $ a > b$ is $ a \minus{} b \minus{} 1$

So the number of integers strictly between $ m^n$ and $ m^{n \plus{} 1}$ is $ m^{n \plus{} 1} \minus{} m^n \minus{} 1$


abstract algebra
algebra
polynomial
superior algebra
superior algebra unsolved

$ A$ is Noetherian, prove $ A[[x]]$, the ring of power series Noetherian

Translate Hilberts (¿) proof for $ A[x]$; to do this, don't work by the degree, but by the $ x$-adic valuation (that is: a power series is "bigger" if it has more zeros at the beginning).

Suppose $ J$ is an ideal of $ A[[x]]$.

Define $ d(f)\equal{}\mathrm{min}\{i|a_i\neq 0\}$, for all $ f(x)\equal{}\sum_{i\equal{}1}^\infty a_i x^i \in A[[x]]$ with $ f(x)\neq 0$.
and $ l(f)\equal{}a_{d(f)}$.

Let $ I\equal{}\{l(f)|f\in J, f\neq 0\}\cup\{0\}$ be the collection of the leading coefficient of the polynomials in $ J$, then $ I$ is an ideal of $ A$.

$ A$ is Noetherian implies $ I$ is finitely generated, so
$ I\equal{}Al(f_1)\plus{}Al(f_2)\plus{}\cdots\plus{}Al(f_m).$

Let $ d_i\equal{}d(f_i)$ and $ N\equal{}\mathrm{max}_{1\leq i\leq m} {d_i}$

[color=darkred]HELP me check the following statement!! Thanks [/color]

Define $ q_N: A[[x]]\rightarrow A[x]$ by $ \sum_{i\equal{}0}^\infty a_i x^i\mapsto \sum_{i\equal{}0}^N a_i x^i$.

[IS THIS MAP FROM $ A[[x]]$ to $ A$ or from $ A[[x]]$ to $ A[x]$? last time my prof wrote this map is from $ A[[x]]$ to $ A$, but I dun think the map define this way is in $ A$.]

Then clearly, $ q_N$ is a surjective ring homomorphism.

Let $ J_N\equal{}\{q_N(f)|f\in J\}\subseteq A\plus{}Ax\plus{}Ax^2\plus{}\cdots \plus{}Ax^N$,
Then $ A[x]$ is Noetherian implies $ J_N$ is finitely generated,
i.e. $ J_N\equal{}A q_N(g_1)\plus{}\cdots\plus{}A q_N(g_n)$

Claim: $ f_1,\cdots f_m, g_1,\cdots, g_n$ generates $ J$ as an $ A[[x]]$-modlule.
Take $ f\in J$,
$ J_N$ finitely generated implies $ q_N(f)\equal{}a_1 q_N(g_1)\plus{}\cdots a_n q_N(g_n)$ for some $ a_1,\cdots, a_n\in A$.
let $ f'\equal{} f\minus{}a_1 q_N(g_1)\plus{}\cdots\plus{}a_n q_N(g_n)\equal{}\sum_{i\equal{}N}^\infty b_i x^i$.

[Is $ f'\in J$?????]

Then $ l(f')\equal{}c_1 l(f_1)\plus{}\cdots\plus{}c_m l(f_m)$ for some $ c_1,\cdots, c_m$.

I dunno how to continue the proof, can anyone help ? Once i proved the claim, then I am done. Thank you


modular arithmetic
ratio
number theory
relatively prime

Each side of an arbitrarly triangle is divided into $ 2002$ congruent segments. After that, each vertex is joined with all "division" points on the opposite side.
Prove that the number of the regions formed, in which the triangle is divided, is divisible by $ 6$.

[i]Proposer[/i]: [b]Dorian Croitoru[/b]

[hide]Let the triangle be $ ABC$.  It is clear that the number of regions into which the triangle is divided into by the lines from $ A,B$ is $ 2002^2$.  As for the lines from $ C$, each line adds another region (below) for each line that it intersects.  The number of lines that it intersects is $ 2001\plus{}2001\plus{}1\equal{}4003$ where the $ 1$ takes account for line $ AB$.  But, this overcounts since when a line from $ C$ intersects a line from $ A$ and $ B$ at a single point, the region we counted does not exist.  Thus, if $ r$ is the number of points of intersection of three lines (excluding of course the points $ A,B,C$), then the total number of regions is

$ 2002^2\plus{}4003\cdot 2001\minus{}r\equiv 1\minus{}r\pmod{6}$

Therefore, we are to prove that $ r\equiv 1\pmod{6}$.  Well, suppose we have $ P$ an intersection of 3 distinct lines.  Let the line from $ A$ cut $ BC$ in a ratio $ 2002\minus{}x: x$ and similarly define $ y,z$.  Then by Ceva's Theorem we have

$ \frac{2002\minus{}x}{x}\cdot\frac{2002\minus{}z}{z}\cdot\frac{2002\minus{}z}{z}\equal{}1$

If $ x,y,z$ are pairwise distinct then each permutation of $ x,y,z$ works, giving $ 6$ different solutions.  Since we are only concerned with the remainder of $ r$ upon division by $ 6$, we disregard this case.  If two of the variables are equal, without loss of generality $ y\equal{}z$, then we have $ \frac{2002\minus{}x}{x}\equal{}\frac{p^2}{q^2}$ for relatively prime $ p,q$.  Thus, let $ 2002\minus{}x\equal{}tp^2,x\equal{}tq^2$ where $ t\equal{}(2002\minus{}x,x)\equal{}(2002,x)$.  Then we have

$ 2002\minus{}tq^2\equal{}tp^2\Rightarrow 2002/t\equal{}p^2\plus{}q^2$

Notice that $ 2002\equal{}2\cdot7\cdot11\cdot13$.  It is well known that if $ a\equiv3\pmod{4}$ is a prime for which $ a|p^2\plus{}q^2$ then $ a|p,a|q$.  Thus, since $ (p,q)\equal{}1$ we must have $ 2002/t\equal{}2$ which gives only one valid solution for $ p,q$: $ p\equal{}q\equal{}1$.  So, we deduce that $ x\equal{}1001$ and so going back to the original equation and remembering that $ y\equal{}z$ yields the only solution $ y\equal{}z\equal{}1001$.  Thus, there is only one solution for $ x,y,z$ not pairwise distinct and so $ r\equiv1\pmod{6}$.[/hide]


Ένα ωραίο που βρήκα :

Να υπολογιστεί το:

$ \int \sin e^x dx$

Θέτουμε $ u\equal{}e^x$ , οπότε $ x\equal{}lnu$ ή $ dx\equal{}1/u du$
[hide]Αν δεν κάνω κάπου λάθος :maybe:  η λύση είναι:
Άρα το ολοκλήρωμα γίνεται $ \int sinu/u du$ που είναι ισοδύναμο με $ \int u'sinu/u du$ και από παραγοντική ολοκλήρωση παίρνουμε αυτό που ζητάμε[/hide]

Παιδιά , δεν το προσπάθησα , αλλά αυτό το ολοκλήρωμα ( $ I\equal{} \int \frac {sinx}{x} dx$ )   δεν το έχω ξαναδεί .Φαίνεται ελκυστικό από πρώτη ματιά .

 Θα περιμένω να δω πώς θα το υπολογίσετε , για να μη σας πάρω.... και τη μπουκιά από το στόμα!

 Μπάμπης

[quote="stergiu"]Παιδιά , δεν το προσπάθησα , αλλά αυτό το ολοκλήρωμα ( $ I \equal{} \int \frac {sinx}{x} dx$ )   δεν το έχω ξαναδεί .Φαίνεται ελκυστικό από πρώτη ματιά .

 Θα περιμένω να δω πώς θα το υπολογίσετε , για να μη σας πάρω.... και τη μπουκιά από το στόμα!

 Μπάμπης[/quote]
Είναι από αυτά που δεν υπολογίζονται στοιχειωδώς όπως και τα $ e^{x^2},\frac{1}{lnx},sin(x^2),\frac{lnx}{x\minus{}1},\sqrt{1\plus{}x^3},\frac{e^x}{x}$

Ummm guys, ο r_boris έχει απόλυτο δίκιο, αυτά τα ολοκληρώματα είναι εκτός του διαφορικού δακτυλίου των στοιχειωδών συναρτήσεων.

Mostel, για το αρχικό ολοκλήρωμα, δεν μπορείς να πεις τίποτα περισσότερο εκτός από το ότι είναι ίσο με $ \int \sin(e^x) dx \equal{} \textrm{Si} (e^x)$, όπου $ \textrm{Si}(x) \equal{} \int \frac{\sin x}{x} dx$ η ολοκληρωτική συνάρτηση του ημιτόνου.

Τι ακριβώς είχες στο μυαλό σου? Επίσης, gregg, δε νομίζω ότι η παραγοντική ολοκλήρωση θα σε οδηγήσει κάπου ;)

Cheerio,

Durandal 1707

[quote="Durandal"]Ummm guys, ο r_boris έχει απόλυτο δίκιο, αυτά τα ολοκληρώματα είναι εκτός του διαφορικού δακτυλίου των στοιχειωδών συναρτήσεων.

Mostel, για το αρχικό ολοκλήρωμα, δεν μπορείς να πεις τίποτα περισσότερο εκτός από το ότι είναι ίσο με $ \int \sin(e^x) dx \equal{} \textrm{Si} (e^x)$, όπου $ \textrm{Si}(x) \equal{} \int \frac {\sin x}{x} dx$ η ολοκληρωτική συνάρτηση του ημιτόνου.

Τι ακριβώς είχες στο μυαλό σου? Επίσης, gregg, δε νομίζω ότι η παραγοντική ολοκλήρωση θα σε οδηγήσει κάπου ;)

Cheerio,

Durandal 1707[/quote]

Durandel , ακριβώς για αυτό δεν είπα στο Στέλιο τίποτα , παρά τους άφησα να .... παλαίψουν ! Διότι το ολοκλήρωμα δεν υπολογιζεται  σχολικώς , παρά αν το το κάνουν (πανεπιστημιακώς) ολοκλήρωμα σειράς και πάλι δεν ξέρω  τι ακριβώς θα προκύψει. Απορώ που το βρήκε ο Στέλιος , γιατί σίγουρα θα του πήρε πολύ ώρα παιδευόμενος !
 Μπάμπης


limit
floor function
trigonometry
function
algebra
real analysis
real analysis unsolved

Evaluate the limit $ \lim_{n \to \infty}{\lfloor n^2 \cdot \sin^{2}{\frac {1}{n}} \rfloor}$, where $ n \in \mathbb{N}^*$ and $ \lfloor x \rfloor$ is the value of the floor function for the real number $ x$.

[hide="solution"]$ \sin^2\left(\frac {1}{n}\right) < \frac {1}{n^2}$ for all $ n > 0$, so $ 0 < n^2\sin^2\left(\frac {1}{n}\right) < 1$ for all $ n > 0$, and the limit is $ 0$.[/hide]


Please do not discuss any aspect of Round 4 of the USAMTS until Monday, March 21st.  This is to make sure that we have received all the submissions before solutions are publicly discussed.

When will you be finished logging solutions?  I sent mine a few weeks ago, and it's still unlogged.
Thanks.

As it says on your "My USAMTS" page: [quote]Most Round 4 Papers will not be marked received until March 24.[/quote]

I thought the solutions had to be in by March 14.

They did.  March 14 was the postmark deadline.

It is now OK to begin discussing Round 4.

As a reminder, we will be having a Math Jam on Wednesday 3/23 at 7:30 PM Eastern to discuss the solutions to the Round 4 problems.


induction
modular arithmetic
LaTeX
number theory proposed
number theory

Solve the equation $3^x=2^xy+1$ in positive integers.

Are you sure?  :?
If $x=2^kr$ then we obtain $k+2\geq 2^kr$ for $k>0$ and $1\geq r$ for $k=0$... Too easy (?)
Am I missing something?

yes, it is very easy problem, showing by induction on the max power of $2$ dividing $x$, what is the highest power of $2$ dividing $3^x-1$ easy problem...

[quote="Myth"]Am I missing something?[/quote]I said easy diophantine equation ... :)

i dont get the solution??' details for not so good mexikan aspiring olympic please :)[hide][/hide]

by induction $3^{2^n}\cong 1+2^{n+2}\pmod{2^{n+3}}$ so for $m$ odd $3^{m2^n}\cong 1+2^{n+2}\pmod{2^{n+3}}$. This is easy if you have seen other problems like it. Otherwise, not really.

Andrei

i heard that this equation can be solved using fields. can anyone show me how to do that?

what is $\| \cdot \|_2$ in your solution?

That curious symbol $||n||_2$ for the highest (exponent of a) power of $2$ dividing $n$ cannot be rotted out... Where does it come from¿ Any book¿ I#M strongly against it, since it means something else in my opinion!
But why not using $\LaTeX$ and using $m,n$ instead of $x,y$¿ Any reason¿

It's simple I don't like using Latex . :D

Using $\LaTeX$ is a moral :!:

You should use it! E.g., the above isn't very good readable. I just don't want to read such things in the knowledge that the writer could have done better by using $\LaTeX$ and hope other people think so, too. But are you posting here for posting only or with the intention that somebody reads and understands it¿

$\LaTeX$ is standard here, and most Mathlinkers use it. There is as good no reason for not using it, especially not something as 'I don't want to learn/use it'.

Most agree with the use of $\LaTeX$, it's accepted widely here.
So use it please.

Sorry,guys.My teacher have just teached me how to write 3^n as above and something which is similar to that.So I only want to try write it here..

could some of you guys explain your solutions a little better, i really don't understand,

myth, you use the substitution $x=2^k*r$

i suppose the most important question is:

[b]What made you think of that substitution?[/b]

it really seems kind of random to me...

also i don't see how this:

(putting in the substitution)

$3^{2^{k}r}=2^{2^{k}r}y+1$ implies $k+2\ge 2^{k}r$

huh? where did the +2 come from?

sorry if these are stupid questions, but i completely do not understand what you did and i want to learn... :)

[quote="perfect_radio"]Maybe you understand what is written in the following PDF.

I can't be sure of what made Myth think of that substitution, but I think it is experience. Many people seem to know that trick. I didn't know it until this problem.[/quote]

thanks, that was a very good solution, i understand it perfectly, i will still put some thought into the $3^{2^n}-1$ stuff, that seems to be a useful technique, and has a few nice results associated with it

[quote="perfect_radio"]Mind you, that's not my creation. (I wrote that PDF, but I think you understand what I mean) Many people solved it at that TST and used the exact same method. It seems that idea Myth used is unusually well-known.[/quote]

i realize, but you took the time to type it up  :)

If x=1 then (x,y)=(1,1)
If x>1 then $ 3 \equiv -1 (mod 4)$ and $2^x y +1 \equiv 1 (mod 4) $ so x is even.
If x>6 then we have repetition of $ x \not\equiv 0 (mod 2^x) $
Thus we have solutions (x,y)= (2,2), (4,5), (1,1)

if x is odd we get $3^x - 1\equiv 2 (mod 4)$ thus $x\leq1$ and we get $(x,y)=(1,1)$
if x is even , from $LTE$ we have $v_2(3^x-1^x)=v_2(2)+v_2(4)+v_2(x)-1=v_2(x)+2$
thus , $v_2(x)=x-2$ and let $x=2^t*i$ where i is an odd positive integer
we get $t=2^t*i-2$ which is impossible for $3\leq i$ , thus $i=1$
we see inductively that $t<2^t-2$ for $3\leq t$
since we were in the case where x is even , we only have to check $t=1$ and $t=2$
both of them work , thus we get the solutions $(x,y)=(2,2)$ and $(x,y)=(4,5)$


linear algebra
matrix
combinatorics unsolved
combinatorics

Any bipartite Eulerian graph has an even number of spanning trees.

i wanted to try out an inductive proof-i think i had it some way through but its so large now that i can't guarantee its validity.

instead what one can do is to try to evaluate the number of spanning trees more directly using the Laplacian(recall the laplacian matrix of a graph is defined as $L_G=((l_{i,j})), l_{i,i}=d(v_i),l_{i,j}=\delta_{\{i,j\}\in E(G)}$). Then the matrix tree theorem tells us that any cofactor of the Laplacian counts the number of spanning trees of $G$ (note that all of this holds on a graph on a labelled set $V$). WLOG we can choose the deleted vertex $u$(to calculate the cofactor) to be such that if the bipartition is $(U,V), u\in U,|U|\leq|V|$. having deleted the corresponding row and column we see that $|U'|<|V|$ where $U'=U\setminus\{u\}$. reading all the terms modulo 2 we get a matrix of the form
 $\begin{array}{cc} 0&A\\A^{T}&0\end{array}$
where $A$ is the incidence matrix between $U'$ and $V$. Since $|U'|<|V|$ every term in the expansion of the determinant will necessarily contain a term from the main diagonal blocks and so in zero and that finishes the proof.

Here is a proof.I must remind the Matrix-Tree Theorem first:
Let $A$ be the adjacency matrix of graph$G$ and $D$ be a matrix that its diagonal is the degrees of vertices of $G$ and other elements are zero. If we delet an arbitrary row and column of $D-A$ the determinant of the remaining matrix is equal to number of spanning trees of $G$.
For the solution arrange vertices of $G$ such that vertices of the first part of $G$ are first. We just use $0,1$ in matrix $D-A$ because parity is important here. If $a$ vertices are in firtst part of $G$ and $b$ in the other vertice, these elements of $D-A$ are zero:$x_{i,j}| 1\leq i,j \leq a$ or $a<i,j \leq a+b$. we can delete first row and column or last row and column of $D-A$ such that these two zero square matrixes become not equal. Then the determinant of this matrix is zero because you can not find a set of nonzero elements in different rows and columns. So from the Matrix-Tree Theorem it results directly to the solution.

Thank you all. The original question was asking us to deduce it from the following statement:

A graph is even if and only if each edge lies in an odd number of circuits.


Through a point inside a triangle, three lines are drawn from the vertices to the opposite sides forming six triangle sections. Then:
(A) the triangles are similar in opposite pairs
(B) the triangles are congruent in opposite pairs
(C) the triangles are equal in area in opposite pairs
(D) three similar quadrilaterals are formed
(E) none of the above relations is true

[hide]Answers B and C means the same thing and are true if the point of intersection is the center of gravity.
Answer A is true if the point is the orthocenter.
Answer D is obviously false.
So the correct answer is E.[/hide]


calculus
derivative
integration
logarithms
trigonometry
parameterization
calculus computations

$ \int_{0}^\pi \log\left( 1-2\alpha \cos x+\alpha^{2}\right) dx$

The preferred method is to differentiate it with some parameter and form a differential equation.

You can find it also [url=http://www.physicsforums.com/showthread.php?t=173543&page=4]here[/url].
(Page 4).

[quote="sos440"]For $ |r|<1$,

$ \ln (1-2r \cos x+r^{2}) =-2 \sum_{n=1}^{\infty}\frac{r^{n}}{n}\cos nx.$
[/quote]

How do you get that?

First, $ 1+r^{2}-2r\cos x=(1-z)(1-\overline{z})$, where $ z=re^{ix}$. We apply the standard power series for $ \ln$: $ \ln(1-z)=\sum_{n=1}^{\infty}\frac{z^{n}}{n}= \sum_{n=1}^{\infty}\frac{r^{n}\cos nx}{n}+i\frac{r^{n}\sin nx}{n}$.
Similarly, $ \ln(1-\overline{z})=\sum_{n=1}^{\infty}\frac{r^{n}\cos nx}{n}-i\frac{r^{n}\sin nx}{n}=\overline{\ln(1-z)}$. When we add the two series, the imaginary parts cancel and the real parts are as sos440 wrote.

Of course, the power series only converges inside the unit disk.


I need to find a mystery book for an english book report. I need somthing that is over 150 pg - 300 pg and in the 7th grade level. I would like it if it wasn't too complicated (complicated example is the war of the worlds) and dosn't have much horror. I like funny books.

Anything from Agatha Christie is good.

The best one in my opinion is Poirot's Last Case.

or Elephants Can Remember!!! Miss Marple: The Complete Short Stories is really good but I think that is a bit long.

[quote="funcia"]how about And Then There Were None by Agatha Christie?[/quote]
I recommend this book too.

Don't know if you're into Criminal Justice stuff but:

Presumed Innocent - Scott Turow

The Blessing Way - Tony Hillerman

I like Cards on the Table by Agatha Christie too.

[quote="Kent Merryfield"][quote]Presumed Innocent - Scott Turow [/quote]
Remember that he said he was in the 7th grade. I'd have to characterize that book as "R-rated." Sexual situations, adult themes, adult motivations, and failing his test of "not too complicated."

And great as Hillerman's stories are, that's a lot of American Indian culture, religion, and philosophy to deal with for a 7th grade book report.[/quote]

Those were both on possible books for the required summer reading  so I'd think they were both appropriate.

Plus, he doesn't have to get into the themes and such for his book report, he can just do the plot which should get him what he wants, and then maybe even throw in some analysis for that extra push.

ashpotter is a she i think

try "chasing vermeer".  its kind of a mystery

"the curious incident of a dog in the night time"-- you cant be immature to read this

then there was this awesome book i read, i forget the title.  if any one know the cover, it had a red curtain with a rabbit in a hat.
something about the ""rabbit hole", like the alice in wonderland rabbit hole.

face on the milk carton

and then you have nancy drew and the hardy boys.  so overrated.

how could i forget about [size=200][b]the westing game[/b][/size]

its such an awesome book

I think the best choices to read would be any Sherlock Holmes novel, such as [i]A Study in Scarlet[/i] or [i]The Sign of the Four[/i].

Or for something quite different,

Hardy Boys!!!!!!

Oh man, I've read so many of them...

[quote="star99"]ashpotter is a she i think

try "[u]chasing vermeer"[/u].  its kind of a mystery

"the curious incident of a dog in the night time"-- you cant be immature to read this

then there was this awesome book i read, i forget the title.  if any one know the cover, it had a red curtain with a rabbit in a hat.
something about the ""rabbit hole", like the alice in wonderland rabbit hole.

face on the milk carton

and then you have nancy drew and the hardy boys.  so overrated.

how could i forget about [u][size=200][b]the westing game[/b][/size][/u]

its such an awesome book[/quote]
the  underlined are books i read and think they are actually pretty good(not up to awesome agatha christie ones though)

I've always liked Raymond Chandler. [i]The Big Sleep[/i] would be a good place to start with him - although if "not too complicated" is any kind of issue, I've got to warn you that after you read that you'll still not entirely know who killed whom - in fact, one legend is that Chandler himself, consulting for the screenplay for the movie based on this book, was asked who had killed a particular character and answered that he didn't know.

Walter Mosely ([i]Devil in a Blue Dress[/i], among others) follows from Chandler in some ways, but in his own setting (African Americans in Los Angeles). But Mosely is also rather an adult taste - some of his books get into some pretty rough material.

i remember the title now!

down the rabbit hole: an echo falls mystery

[quote="star99"]"the curious incident of a dog in the night time"-- you cant be immature to read this[/quote]
I've never read this book, but I hear it's good.

i was accually trying to stay away from hardy boys and Nancy drew (I think i did book reports on them in 4/5th grade) Anyway
[quote]Try The Da Vinci Code, and Angels and Demons, both by Dan Brown if those are mysteries. I love those books. They're definitely over 300 pages, but both are adult level.[/quote] by Jenn7

I have read The Da Vinci Code. I think About 600 pages are a little too big for the book report.

I asked my high school friend about And then there were none by Agatha Christie. He read it in 7th grade and thought it was pretty cool. I think i'll give it a shot. Would it really be mystery though? And is it horrific or gory?

Agatha Christie novels are generally pretty tame in terms of violence and gore.  I've always enjoyed them for an easy, entertaining read.  Raymond Chandler is good, too -- a little rougher, in general.

[quote="rnwang2"][quote="ashpotter"]I asked my high school friend about And then there were none by Agatha Christie. He read it in 7th grade and thought it was pretty cool. I think i'll give it a shot. Would it really be mystery though? And is it horrific or gory?[/quote]
Yes, it is a mystery, because it is mysterious. It's also a little creepy at times (especially the poem *shiver*).[/quote]
dude i just knew the answer by the time the there was 6 little indians
it is overall one of the best agatha christie books i have ever read, and i got it for easter one year. :lol: i read it in one day or something it was so good. i recommend it over all father brown, sherlock holmes, and agatha christie books that i might tell you to read...basically if you have to pick one pick this one.

[quote="rnwang2"][quote="funcia"]dude i just knew the answer by the time the there was 6 little indians[/quote]
Wait, you're saying you knew who the killer was when there were 6 people left?  :o[/quote]
yeah his/her character in the beginning tells you. it was interesting when i was thrown for a loop after *something*, of course i was right all along when i realised something else. but dude we can't give out plots on math forums.
that's what wiki is for  :rotfl:
http://en.wikipedia.org/wiki/And_Then_There_Were_None
i think...(for those interested in plot however i didn't really read article)

i know you picked your book, but Remote Man, by Elizabeth Honey is a really good book.  Its not 100% mystery though.  It's similar in format to the CyberKidz series, if youve heard of those.

and its not gilda jones, its gilda joyce (my previous post)

[quote="Ignite168"]We were assigned the Westing Game in 3rd grade, I doubt that's appropriate for a 7th grade book report.[/quote]

3rd grade? Are you serious? wow! I think I read it in 6th.... I'm not sure how an 8 year old got motivation to read a book that seemed unending

my class also read the book in sixth grade

I certainly am the odd one here, I never liked Agatha Christie books, there I've said it!
I'll take Sir Conan Doyle's Holmes books over them anyday.
How about one of the shorter Chichton or Ludlum books? The plots are well worth the read.

All Poirot books by Agatha Christie are pretty good.

Also, I actually did like Sherlock Holmes, but it might be adult level or too long.

Why not an H.G.Wells classic? Or Asimov?

and as for Poirot vs. Holmes, I dunno - I'm inclined to back Holmes..

what other Poirot books are there? He's cool.

This is a wierd true story about how much immagination could get into you after you read mystery books about murders

One day me and my friends were having lunch at some chinnese restraunt. as wer were talking about our mystery books (for the book report) and the murders in them, we looked out the window and saw the perfect murder scene (unfourtenently, no blood anywhere)
There was a rack with "Wine Seller" on it. On the bottom left of it, lay some kind of a kinfe (you know those chinese ones that look like this?)
 ________
|____|

There was also a plate, a shovel and a broken bottle on the ground and a wierd looking tree (we were certain there was some secret entrance on the tree)

Ah, finally someone has recommended Conan Doyle's work.

[quote="alkjash"]Also, I actually did like Sherlock Holmes, but it might be adult level or too long.[/quote]

Adult level?  Questionable.  Too long? Almost certainly not.

Or, for an even more antique taste of mystery, try "The Murders in the Rue Morgue" by the great Edgar Allan Poe.

[quote="ashpotter"]I ended up not liking And there there was none. Thanks for all the help! My book report book is ABC Murders (also by Agatha). It's ending is really sweet[/quote]

That's strange....I really like And Then There Were None.
The ABC Murders is excellent though(can't see how an Agatha Christie book wouldn't be less than great.) Poirot was always rather good.

Oh and iff you haven't done your book report yet I'd search how to pronounce Poirot if you aren't the pronunciation savant. :ninja:

i suggest the first book [i]Framed[/i] by Malcolm Rose....


[b]An interesting identity in a triangle.[/b] $ (a\minus{}b)^{2}\plus{}(a\minus{}c)^{2}\minus{}2(a\minus{}b)(a\minus{}c)\cos A\equal{} a^{2}\cdot\left(1\minus{}\frac{2r}{R}\right)\ge 0$.

This is definitely not true; I have checked a few cases.

[quote="Virgil Nicula"]$ (a\minus{}b)^{2}\plus{}(a\minus{}c)^{2}\equal{} a^{2}\cdot\left(1\minus{}\frac{2r}{R}\right)\ge 0$.[/quote]

[b]Sorry, Nsato,[/b] I forgot a term !
Correctly is such : $ (a\minus{}b)^{2}\plus{}(a\minus{}c)^{2}\minus{}2(a\minus{}b)(a\minus{}c)\cos A \equal{} a^{2}\cdot\left(1\minus{}\frac{2r}{R}\right)\ge 0$.
I corrected. Thank you !

I don't know if this helps, but...

Construct $ E, F$ on rays $ BA, CA$ such that $ BE\equal{}CF\equal{}a$.  Then (all signs work out because the cosine of the supplement of an angle is the opposite of the cosine of the angle) the expression on the RHS is $ EF^{2}$.

Yes, this is the "spring" of this identity. It is very usually in some remarkable inequalities. I"ll come back soon with some examples.


algebra
polynomial
calculus
integration
function
real analysis
real analysis unsolved

Let $ (a_n)$ be increasing sequence of positive integers. Is there (any?), (nice?) condition(s) (using measure?) allowing us to state convergence or divergence of the series: \[ \sum_{n \equal{} 1}^{\infty }{1\over{a_n}}\]

There are plenty of necessary conditions and there are plenty of sufficient conditions.  Are you looking for something like [url=http://en.wikipedia.org/wiki/Natural_density]natural density[/url]?  Clearly every sequence with positive natural density can be compared to the harmonic series.  On the other hand, the primes have zero natural density but diverge.

This question is too general.  A vacuous answer is to define the "measure" of a sequence to be $ \sum_n \frac{1}{a_n}$.

I asked myself this question and started searching. I'd like some key-words that will help me to find some material on that. Natural density is my first clue - thanks. Could you give some condition please? Where can I find them if there are so many?

One thing that is always worth trying is the Cauchy condesation trick: $ \sum_n \frac 1{a_n}<\plus{}\infty$ if and only if $ \sum_k \frac{2^k}{a_{2^k}}<\plus{}\infty$. This trick often makes the convergence/divergence much more transparent. Often you can apply it several times (more precisely, you can do it as long as the terms of the series form a non-increasing sequence).

Standard tools include direct or limit comparison (for example, to a polynomial), the integral test (if $ a_n$ is given by or can be bounded by an integrable function), or [url=http://en.wikipedia.org/wiki/Summation_by_parts]summation by parts[/url].

Depending on how $ a_n$ is defined, you may find it conceptually more useful to investigate the growth of the counting function $ c_n \equal{} | \{ a_1, .... \} \cap \{ 1, 2, ... n \} |$ directly.  If $ a_n$ grows slowly, then $ c_n$ will grow quickly and you should try to prove divergence.  On the other hand, if $ a_n$ grows quickly, then $ c_n$ will grow slowly and you should try to prove convergence.  Sometimes you can find bounds on $ c_n$ using [url=http://en.wikipedia.org/wiki/Sieve_theory]sieve methods[/url] (again, depending on how $ a_n$ is defined).


[quote="Halsafar"]I have an equation which can be factored to (2x-3)(4x+1)
I am suppose to show use of the sign char method.
I have no idea how to use the sign chart method.  I've re-read the lesson several times and it just won't make sense to me.[/quote]
Do you mean an inequaity?
When $x=\frac{3}{2}$ or $-\frac{1}{4}$, the expression becomes zero. So we consider the ranges $x\leq -\frac{1}{4}$, $-\frac{1}{4}<x<\frac{3}{2}$ and $x\geq\frac{3}{2}$.
Next time please put this type of questions in the high school forum. :)


conics
ellipse
inequalities
quadratics
triangle inequality
algebra
quadratic formula

Find all numbers x,y such that:

(1 - x)² + (x - y)² + y² = 1/3

[hide]
$ 1\minus{}2x\plus{}x^2 \plus{}x^2\minus{}2xy\plus{}y^2\plus{}y^2\equal{}1/3$
$ 2/3\equal{}\minus{}2x^2\plus{}2x\plus{}2xy\minus{}2y^2$
$ 3(x^2\minus{}x\minus{}xy\plus{}y^2\plus{}1)\equal{}0$
$ (x\minus{}0.5)^2\plus{}(y)^2\minus{}xy\equal{}\minus{}1.25$
I detect an ellipse, but I forgot how to get rid of that annoying xy term (is it shifting axes?).
[/hide]

[hide="Like this?"]

$ x^2 \minus{} x \minus{} xy \plus{} y^2 \plus{} 1 \equal{} 0$

$ y^2 \minus{} xy \plus{} x^2 \minus{} x \plus{} 1 \equal{} 0$

$ D \equal{} x^2 \minus{} 4x^2 \plus{} 4x \minus{} 4 \equal{} \minus{} 3x^2 \plus{} 4x \minus{} 4$

As long as $ D\ge 0$, ordered pairs $ (x,y) \equal{} (x,\frac {x\pm\sqrt D}{2}$ are solutions for $ x\in\mathbb{R}$

[/hide]

Now I think it has no solutions.

[hide="On x"]

$ \minus{}3x^2\plus{}4x\minus{}4\ge 0$

First, we see when $ \minus{}3x^2\plus{}4x\minus{}4\equal{}0$

That's nowhere, because discriminant is less than zero. And, for any other value of x, the inequality isn't true, so no solutions?

[/hide]

Well ... I don't know the answer correct. But I think your answer is correct.  :wink:

Try AM-QM  :wink:

edit: given $ \ge \frac {|1 \minus{} x| \plus{} |x \minus{} y| \plus{} |y|}{3} \ge \frac 13$ by triangle inequality, [edit] quick cases shows all positive, $ 1 \minus{} x \equal{} x \minus{} y \equal{} y \Longrightarrow x \equal{} 2y \equal{} 1 \minus{} y$, $ (x,y) \equal{} \left(\frac 23, \frac 13\right)$

Here is one of the routes for work and the answer:
[hide]
It becomes:

$ 1 \minus{} 2x \plus{} x^2 \plus{} x^2 \minus{} 2xy \plus{} y^2 \plus{} y^2 \equal{}$ 1/3

Subtract 1/3 from each side and combine like terms:

$ 2y^2 \minus{} 2xy \plus{} 2x^2 \minus{} 2x \plus{} \frac23 \equal{} 0$

Divide both sides by 2:

$ y^2 \minus{} xy \plus{} x^2 \minus{} x \plus{} \frac13 \equal{} 0$

Using the quadratic formula:

D stands for the discriminant in the quadratic formula here:

D = $ \minus{} 3x^2 \plus{} 4x \minus{} \frac43$

EDIT:  This can be expressed as $ \frac{\minus{}1}{3}(3x \minus{} 2)^2$.

$ y \equal{} \frac { x \pm \sqrt {D}}{2}$

For the discriminant, the only solution is x = 2/3.

Substituting this back into y, you get y = 1/3.

The answer is $ (\frac23, \frac13).$

[/hide]

I continued on what Caelestor done, he had a mistake, so it would turn out that it has no solutions. Arrange showed it nicely.

And QM-AM solution is fascinating.

Thanks man. Your solution is very good and thanks for everybody help. But it has more solutions ?

Wow, where did I go wrong?
I need to review my conics.

Is that the only solution?

Hi, but anyone knows how to find the quadratic formula: y² -xy+x²-x+1/3 = 0     to    $ \minus{}3x^{2}\plus{}4x\minus{}\frac{4}3$, what's the a,b and c of equation, please ?

a = 1 because it's the coefficient of $ y^2$
b = -x because it's the coefficient of $ y$
c = the whole trinomial $ (x^2 \minus{} x \plus{} \frac13)$ and it is a constant relative to the $ y$ variable

a=1, b=-x, c=(x^2-x+1/3)

We're taking the quadratic according to y.

[quote="luiseduardo"] But it has more solutions ?[/quote]

No, it does not have more (REAL) solutions.

I refer to an edit I had made in the solution post earlier.

Because the discriminant can be factored as $ \frac{\minus{}1}{3}(3x \minus{} 2)^2$,

the only value of x that makes the discriminant equal to 0 is x = $ \frac23$.

With a 2 for an exponent, otherwise $ (3x \minus{} 2)$ will be a positive value
when x is not equal to 2/3.  And, because the leading multiplier of this
factored discriminant form is a $ \frac {\minus{}1}{3}$, then this discriminant
will be negative under the square root symbol giving all values that are
not real, EXCEPT for when x = $ \frac23$, for which the value of the
discriminant is zero.

Ah ok, now I understand all. Thanks Arrange your tan. And Thanks to everybody here.  :)

By Cauchy Schwarz $ 3 [(1\minus{}x)^2 \plus{} (x\minus{}y)^2 \plus{} y^2] \ge (1\minus{}x\plus{}x\minus{}y\plus{}y)^2 \equal{} 1$

Equality occurs when 1-x = x-y = y which gives $ (x,y) \equal{} \left (\frac{2}{3}, \frac{1}{3} \right)$


logarithms
combinatorics unsolved
combinatorics

We are to measure $n$ marbles with the help of a one-armed balance. We are allowed to measure one or two marbles at a time. It may happen, however, that one of the results is misread. Denote by $f(n)$ the minimum number of weighings neccessary to determine the weight of each marble. Prove that $f(n) < n+3+\log_{3}n$.

[i]Komal, A. 256, January 2001[/i]

[i]Edited by Myth[/i]

Very nice! Move it from Algebra to Combinatorics ;)

I don't see how to get this $\log_{3}n$ :( , but the estimate $f(n)<n+3+\log_{2}n$ is quite clear.
Step 1) Weight first marble two times.
1.a. If weighs are equal then it is real weight of first marble.
1.b. Otherwise one of the weightings is false. Then weight first marble once again, now you can weight all other marbles - it will take $3+(n-1)$ weightings.
Step 2) (after 1.a.) So we know real weight of first marble. Weight marbles in pairs: $(1,2)$, $(2,3)$, ..., $(n-1,n)$. Using the weight of first marble we will calculate weight of all other marbles.
3) Weight marble $n$.
3.a. If it is the same weight as for step 2) we conlude all weightings are true.
3.b. If it is not the same then weight $n$ once again. Therefore we will know where mistake took place: weightings in pairs or weighting of $n$-th marble. The last case is clear.
4) So mistake take place in weightings in pairs. Weight now the central marble, it will give its true weight, and compare with previous weight. It will give us an informations whether mistake took place  in left or right half. And so on (binary search). It will take $\log_{2}n$  operations.


ABCD is a cyclic quadrilateral. The tangent at A meets the ray CB at K, and the tangent at B meets the ray DA at M, so that BK=BC and AM=AD. Show that the quadrilateral has two sides parallel.

[quote="adey_007"]You need to prove AC/AK = a/b (where a = AB, b = BC and so on) and BD/BM = a/d

Then use the median formula for AB. You get two equations in AC and BD which are equal.

We can then easily deduce that if these two equations are equal, then b=d or in other words AB is paralled to CD as required.

[/hide][/code][/i][/quote]
thank you for your answer


Denote that [x](x is a real number) is the largest integer number $\le$ x.
Let be given a sequence of integer numbers $x_1,x_2,x_3...$ satisfying:
$x_1=1$ and $x_n=[\sqrt{x_1+...+x_n}]\forall n\ge 2$
$x_{2005}=?$

James are you sure about this problem?:?
Mybe must be
$x_{n+1}=[\sqrt{x_1+...+x_n}]$? :?

yes you are right [b]Tiks[/b].Sorry everybody?
so....o someone helps me? :oops:

[quote="vietnamesegauss89"]I think$x_{2005}=997$[/quote]
I am not think so,but mybe... :(.
I just prove that $950<x_{2005}<998$ but I don't know what then :( .
 [b]vietnamesegauss89 [/b] can you post your solution?,of cours if you have that :lol: .

[quote="vietnamesegauss89"]By using induction,we have:$x_{2^k+k-1}=2^{k-1}$if $[(2^k+k-n)/2]=y\Rightarrow x_n=2^{k-1}-y\Rightarrow x_{2058}=1024\Rightarrow x_{2005}=997$
 P/s:I'm sorry,I don't good English[/quote]
Guys who can explain how to prove by induction this fact$x_{2^k+k-1}=2^{k-1}$? :?

[quote="vietnamesegauss89"]By induction,we have$\sqrt{x_1+x_2+...+x_{2^k+k-2}}=2^{k-1}$[/quote]
Are you sure?becaus if I calculy well,then
$x_1+x_2+...+x_{35}=258$ :( .


Express in simplest form:
$(x+\frac{1}{x})^{2}+(y+\frac{1}{y})^{2}+(xy+\frac{1}{xy})^{2}-(x+\frac{1}{x})(y+\frac{1}{y})(z+\frac{1}{z})$

The answer is supposed to be $4$ but I can't see how that could be ... maybe $z=xy$??

Yeah, as it stands the expression is clearly a function of $z$...


Let x,y,z are three real numbers such as $ x \geq y \geq z >0$ and $ 1\plus{}4\sqrt{2}\minus{}2\sqrt{2}x^2\equal{}z^2\equal{}5\minus{}4y^2$

find the maximum of $ P\equal{}xy\plus{}2yz\plus{}zx$

Sorry for my bad English

Nobody helps me ?

This problem is hard, isn't it ?

It's just a problem of a contest for students in grade 9 in my district :D

I've waited too long for the solution, where are u can_hang and nguoivn, help me to do this problem plz :D

It isn't a nice ineq (too many conditions). Verry sory my friend, but I think nobody want to waste time for it.  :oops:
And another reason: we are rather busy because MIC (Mathsvn Inequality Contest) is now taking place.
For registering, please click this link:
http://www.maths.vn/forums/showthread.php?t=17758
Have fun, my friend.  :)

Of course it isn't a nice problem because of the ugly condition, also non-cyclic and non-symetric, but this is only for student in grade 9 and it must have a nice hint :D , plz help me to find it out :D

P/S:I'm not relly interseted in the MIC ( my skill is not good  :oops: )


Hey all,

Just wondering... any techers here? Would like to get to know as many of my fellow classmates as possible.  :D

I'm assuming you're on the new forum since you post at CC as well.

If there are any other 2010 Caltech students around here, go here ASAP: http://caltech2010.houseofcheesecake.com/  :lol: