modular arithmetic
algebra
polynomial
trigonometry
inequalities
ratio
complex numbers

[hide="4"]The polynomial is $\frac{x^{n+1}-1}{x-1}$, so the roots of the polynomial are given by $e^{\theta i}$ where $\theta=\frac{2k\pi}{n+1}$ and $k$ varies from 1 to $n$.

$\frac{1}{e^{\theta i}-1}=\frac{1}{\cos \theta -1+i\sin \theta}$. Multiplying by conjugate, this is $\frac{\cos \theta-1-i\sin \theta}{2-2\cos \theta}$.

The real part is $\frac{-1}{2}$ since $\cos \theta \neq 1$.

The imaginary part will be 0 if $k=\frac{n+1}{2}$. Since $\sin (2\pi -\theta)=-\sin \theta$ and $\cos (2\pi -\theta)=\cos \theta$, then for each $k<\frac{n+1}{2}$, add the imaginary part for $k$ with that for $n+1-k$. The imaginary parts cancel.

This means that the sum of all of the imaginary parts is 0, so the sum is $\boxed{-\frac{n}{2}}$.[/hide]

[hide="3"]$x^8-x^5=\left(x^4-\frac 12 x\right)^2-\frac 14 x^2$

$\Rightarrow x^8-x^5+x^2-x+1=\left(x^4-\frac 12 x\right)^2+\frac 34 x^2-x+1$

$=\left(x^4-\frac 12 x\right)^2+\left( \frac{\sqrt{3}}{2}x-\frac{1}{\sqrt{3}} \right)^2+\frac 23$

By trivial inequality, this is always greater than or equal to 2/3, so the solution is $\boxed{\text{all reals}}$.[/hide]

[quote="riddler"][quote="Farenhajt"][hide="1"]${{2x+30}\over{x+7}}=2+{16\over{x+7}}\Longrightarrow(x+7)\in\{\pm 1,\pm 2,\pm 4,\pm 8,\pm 16\}$ - 10 different values.[/hide]
[/quote]

How can $\pm1$ be a value of $x$?[/quote]

$\pm 1$ is a value for $x+7$, not $x$.

[quote="riddler"][quote="Farenhajt"][hide="1"]${{2x+30}\over{x+7}}=2+{16\over{x+7}}\Longrightarrow(x+7)\in\{\pm 1,\pm 2,\pm 4,\pm 8,\pm 16\}$ - 10 different values.[/hide]
[/quote]

How can $\pm1$ be a value of $x$?[/quote]

Read more carefully: $\pm 1$ is value of $x+7$ (I didn't bother to calculate all $x$'s since the question asks only how many of them there is)

The first exercise is more interesting if the ratio is ${\frac{5x+12}{3x+5}}$ or $\frac{2x^2+3x+1}{3x-1}$.

[quote="Farenhajt"][quote="felipesa"]Thank you  all !!!!:)

[quote="Farenhajt"]
[hide="2"]\begin{eqnarray*} 74892^{359}6379^{207}9538^{179}3756^{723} &\equiv& 2^{359}(-1)^{207}(-2)^{179}1^{723}\pmod{5}\\ &\equiv& 2^{359}\cdot(-1)\cdot(-2^{179})\pmod{5}\\ &\equiv& 2^{359}2^{179}\pmod{5}\\ &\equiv& 2^{538}\pmod{5}\\ &\equiv& 4^{269}\pmod{5}\\ &\equiv& (-1)^{269}\pmod{5}\\ &\equiv& -1\pmod{5}\\ &\equiv& 4\pmod{5}\end{eqnarray*}[/hide][/quote]

Farenhajt ,
[hide]
Could you explain how do get $74892^{359}6379^{207}9538^{179}3756^{723} \&\equiv\& 2^{359}(-1)^{207}(-2)^{179}1^{723}\pmod{5}$?
Thanks[/hide][/quote]

[hide="Explanation"]Well, for congruences in general the following is valid:
\begin{eqnarray*}a\equiv b\pmod{m}\land c\equiv d\pmod{m} &\Longrightarrow& ac\equiv bd\pmod{m}\end{eqnarray*}

From this directly follows also:
\begin{eqnarray*}a\equiv b\pmod{m}\Longrightarrow a^k\equiv b^k\pmod{m}\end{eqnarray*}

Now we have:
\begin{eqnarray*}74892\equiv 2\pmod{5} &\Longrightarrow& 74892^{359}\equiv 2^{359}\pmod{5}\\ 6379\equiv -1\pmod{5} &\Longrightarrow& 6379^{207}\equiv (-1)^{207}\pmod{5}\\ 9538\equiv -2\pmod{5} &\Longrightarrow& 9538^{179}\equiv (-2)^{179}\pmod{5}\\ 3756\equiv 1\pmod{5} &\Longrightarrow& 3756^{723}\equiv 1^{723}\pmod{5}\end{eqnarray*}

Now multiply right-hand sides and you'll get the identity.[/hide][/quote]
Thanks again ,Now I got it.


How many factors does $12$ have?

Did you list them?

There is a better way if you did. We know that the prime factorization of 12 is $2^2 * 3$. Notice that every factor of $12$ is in the form $2^n * 3^m$, where $n$ and $m$ are non-negative integers less than or equal to the power in the prime factorization.

So, for the prime factorization of $12$, there are $3$ choices for $n$, they are ${0,1,2}$. For $m$, there are two choices, ${0,1}$. So, there are thus $3$ chioces for $n$ times $2$ choices for $m$, which is $3 * 2 = 6$ total factors of $12$.

Now, how many factors does $1296$ have?

Challenge: How many factors of $1296$ aren't multiples of $9$?

pretty simple..

[hide]$1296=2^4\cdot3^4$, so there are $(4+1)(4+1)=25$ factors. If a factor is not to be divisible by $9$ then the exponent of $3$ must be less than $2$, so there are $(1+4)(1+4-3)=10$ such factors.[/hide]

[hide]1296=2^4*3^4, so there are 5*5=[b]25 [/b]factors.  If it cannot be a factor of 9, then the choices for 3 can only be 0 or 1.  So there are 5*2=[b]10[/b] choices if it can't be a factor of 9.[/hide]


Support
college
MIT
geometry
Harvard
Gauss
geometric transformation

Is it true that affirmative action actually hurts Asian-Americans when applying to college? Also, why did I see Asian-Am activists campaigning FOR affirmative action in the UMichigan case? On the front page of the Boston Globe, there was a picture of an Asian girl hugging a black girl after the Supreme Court decision. Should Asian-Am support affirmative action?

I support it...hey, anything to raise my chances of going to a good college :) Although I seriously doubt it concerns me I still support it because...um...go Asians! :)

Well as far as I know, asians do not get increased admission chances like African Americans.  Asians are really smart and some of my friends have said that when they look at applications, the adcoms think it is "just another Asian."  Seariously though that is totally unfair.  Being a straight A student, getting high test scores, etc. is hard.  Supposedly all adcoms think that Asians are all alike.

In conclusion college admissions is unfair.

Like for example, UC schools do not have affirmative action.  Therefore Berkely is 41% asian and UCLA is 37%.  Most other top schools are from 22% (Carnegie Mellon) to 6% (Vanderbilt).  (note: these are schools I looked at so I got this from admissions info).  So therefore being asian really isnt helping too much...

I actually don't know too much about it b/c Im not asian, so feel free to correct me....

[quote]Should Asian-Am support affirmative action?[/quote]

Everyone should be entitled to their own choice; all Asian-Americans need not all support it or all oppose it (and it there will always be Asian-Americans on both sides, presumably).

[quote="mathfanatic"]Everyone should be entitled to their own choice; all Asian-Americans need not all support it or all oppose it (and it there will always be Asian-Americans on both sides, presumably).[/quote]

There will be nobody on either side when the barriers to education melt into nothing.

My friends always why do you want to be so smart. I say that competition is the fundamentals of capitalism, and I call some of them commi-symps. For affirmative action. But the asian american thing is racist. But what it has done is make alot of Asians wore harder and harder. Letting them know that they have tougher competition and univerities accepting only a limited number minorities. This happened in Utah. The people who don't make med school keep on complaing about admission process. And blaming that on minorities, But minorites only make up 15% of the med school enrollment. It is just like racist commercials senator Jeesie Helms aired on TV for reelections in the last decade. Stuff like you wanted that job, but it was taken by a black guy, and you know you were better qualified (this is egotism).

i totally disagree with all forms of affirmative action. if you take a step back, it's ridiculous; colleges should not know the race or ethnicity of their applicant, they should not strive to meet certain quotas or make attempts at seeming 'totally fair.' MIT boasts of their student body, calling 50% of it 'people of color.' why would one brag about being unfair to more qualified applicants. i'm not racist, sexist, or anything, but the whole concept just makes me sick.

I completely agree with codeblue.

Affirmative action makes college application more unfair than fair.

You all are taking a very short-sighted view of "fairness."  What is fairness?  Why do you seem to imagine that only considering one's individual achievements would be more "fair?"  As a quick (and obviously extreme) counterexample, if someone were locked in a closet after school until the next morning, is it "fair" to that student for colleges to not accept him on the basis of a lack of extracurricular activities?  I think not: what would be far fairer in that circumstance would be for the college to make allowances for that student in some way.

Individuals don't make themselves, nor do they make their resumes.  They are influenced by thousands of factors including their socio-economic status, their neighborhood, how connected their families are, how good their school is, and on and on.  To pretend that all that is relevant is what you yourself have done is a bad joke: what you have had the chance to do is at least as significant.

The standard terminology for this is the "level playing-field."  There is no level playing field in this country.  As such, there is no fairness.  Affirmative action is an attempt to introduce fairness.

As for the original question, I think it's rather silly.  But as for its intent: if an Asian American in this country were to make up all of his decisions in a purely self-centered fashion, he would be opposed to affirmative action.  He would also favor a government in which he was supreme dictator.

Hmmm, seeing I want to take over a third world country and take complete control over that country to conquer neighboring countries, maybe I should oppose AA...I thought AA was for the AAs too and not just for the AAs if you know what I'm sayin :wink:

if it's good for asians, i'll support it. if it isn't, i won't. in other words, i only like it if it's does something good for me.

[quote="JBL"]

As for the original question, I think it's rather silly.  But as for its intent: if an Asian American in this country were to make up all of his decisions in a purely self-centered fashion, he would be opposed to affirmative action.  He would also favor a government in which he was supreme dictator.[/quote]

There are other reasons to oppose affirmative action. Furthermore, does affirmative action really help those that it claims it helps? Moreover, if affirmative action is going to hurt me, then why should I support it? Am I being too "self-centered" when I refuse to support something that may hurt me?

Also, I think my question was perfectly valid considering that many Asian-Americans do not know everything to the issue, e.g. tare. Consider this: there is actually a lot of support for the Republican Party among Asian-Americans. I would have thought that Asian-Americans-- being a minority and all-- would support the Democratic Party. However, there are two reasons-- according to my AP US Gov teacher-- why Asian-Americans support the Republican Party: (1) the Democratice Party does not support affirmative action for Asians; (2) many Asian-Americans are small business owners --> more likely to go Republican. Who knew?

I don't agree with affirmative action at all because it doesn't seem like its very fair. I don't believe colleges should know the race and gender of an applicant until after they make the admissions decision. For a long time, people have been wanting equal rights and opportunities, and I think affirmative action doesn't give them to everyone. Colleges accept a certain number of minority students no matter what their qualifications. Admissions should be based only on things like grades, activities, and the like.

Being a female and 2 different races, many people have told me that I should agree with affirmative action because it could help me out. I don't want to get into a college because I'm a certain gender or race, no matter how qualified I may be. Think about how many people are rejected in favor of accepting a diverse group of students who may be less qualified. I don't know about most of you, but I want to get into a college because I earned it, not because I'm a certain race or gender that will help meet their quotas.

edit: grammatical and spelling errors

how would we mask people's ethnicity?  With a name like Gonzalez, what are you to assume about their race?  And I wouldn't be surprised if some people use their race or culture as a topic for essays.

I think there are much more 'unfair decisions' stemming from legacies, geographic diversity, celebrity, etc.

[quote="JBL"]I personally think you're overestimating the combined importance of the two characteristics, 1600 and being a minority.  Admissions officers aren't going to treat a 1600 incredibly different than they will a 1580.  Although you are right that they will probably get in to a lot of good schools, the entire class of people with 1600's, barring other stupidities like not going to class, are likely to be cross-admits no matter what.  The fact that, under affirmative action, such students would get chosen above equally qualified students is pretty immaterial -- at that level all the equally qualified students get chosen, too.[/quote]

As far as I know, both Harvard and Stanford (no clue about other institutions) reject close to 50% of their 1600 applicants.  SATs alone do not get students into the best universities.  But I suspect that a 1600-scoring minority, or one in the high 1500s, is extremely rare, both absolutely and also relatively to the rest of their minority cohort.

You mentioned it before - Harvard just doesn't have enough beds for all the 1600 scorers.

That's fine, but Harvard doesn't have enough beds for all of anybody.  The point is that, barring some massive other mistake, students with 1600's all get in somewhere good.  There aren't many 1600's at community colleges.  Of course not everyone gets into their first-choice school, but 1600's all get into schools that are a lot of people's first choice school.

Okay, I found some interesting hardcore statistics on SAT Scores. Sorry to post the whole article in-thread, but the link to it is now dead. There were also some graphs ect. in the original article that I can't reproduce here.

------------------------------------------------------------------------------------------------
[u]The Expanding Racial Scoring Gap Between Black and White SAT Test Takers[/u]

In the 12-year period between 1976 and 1988, the black-white scoring gap on the Scholastic Assessment Test closed significantly. The improvement in black scores was so strong that some educators predicted that within a generation the black-white gap would disappear altogether.

Unfortunately, this was not to be. In fact, since 1988 the racial gap in SAT scores has become wider and there is no compelling evidence that any improvement is in the offing.

In 1976 The College Board published an analysis of the racial differences in scores of the Scholastic Assessment Test (SAT). At that time the average black score was about 240 points, or 20 percent, below the average white score. When The College Board next examined the racial scoring gap in the early 1980s, the gap had shrunk to 200 points. Black scores were then 17 percent lower than white scores. By 1988 the black-white SAT scoring gap was down to 189 points. The trend was encouraging. Many people in the educational community came to believe that in time the racial scoring gap would disappear altogether. But progress in closing the SAT gap stopped abruptly and now it has begun to open up. For each of the past three years the gap between black and white scores on the SAT test has expanded.

In 2002 the average black score on the combined math and verbal portions of the SAT test was 857. The mean white score on the combined math and verbal SAT was 17 percent higher at 1060.

Over the past 15 years there has been only a very small improvement in African-American SAT scores. In 1988 the combined mean score for blacks on both the math and verbal portions of the SAT was 847. By 2002 the average black score had risen only 10 points, or about one percent, to 857. In 2002 the average combined score on the SAT for black students actually dropped by two points from last year.

Despite the small overall improvement of black SAT scores over the past 14 years, the gap between black and white scores has actually increased. In 1988 the average combined score for whites of 1036 was 189 points higher than the average score for blacks. In 2002 the gap between the average white score and the average black score had grown to 203 points. In the past year alone the black-white scoring gap on the SAT increased by two points.

Not only are African-American scores on the SAT far below the scores of whites and Asian Americans, but they also trail the scores of every other major ethnic group in the United States including Puerto Ricans and Mexican Americans. In fact, American-Indian and Alaskan native students on average score more than 100 points higher than black students.

[i]Explaining the Black-White SAT Gap[/i]

There are a number of reasons explaining the continuing and growing black-white SAT scoring gap. A major factor in the SAT racial scoring gap is family income. There is a direct correlation between family income and SAT scores. For both blacks and whites, as income goes up, so do test scores. Some 28 percent of all black SAT test takers came from families with annual incomes below $20,000. Only 5 percent of white test takers came from low-income families. At the other extreme, 5 percent of all black test takers came from families with incomes of more than $100,000. The comparable figure for white test takers is 24 percent. But income alone does not explain the racial scoring gap. Consider these facts:

 Whites from families with incomes of less than $10,000 had a mean SAT score of 980. This is 123 points higher than the national mean for all blacks.
 Whites from families with incomes below $10,000 had a mean SAT test score that was 46 points higher than blacks whose families had incomes of between $80,000 and $100,000.
 Blacks from families with incomes of more than $100,000 had a mean SAT score that was 142 points below the mean score for whites from families at the same income level.

Clearly, one of the main factors is that black students across the board are not being adequately schooled to take these tests. Public schools in many neighborhoods with large black populations are underfunded, inadequately staffed, and ill equipped to provide the same quality of secondary education as is the case in predominantly white suburban school districts. Many of the black students who graduate from suburban high schools are recent newcomers to the suburbs and received their elementary school education in lower-quality inner-city schools. These students have been placed in a position from which they have been forced to play catch up.

Data from The College Board shows that 52 percent of white students who take the SAT are ranked in the top 20 percent of their high school class. This compares to 32 percent of black test takers. Some 46 percent of white students who take the SAT report that their high school grade point average is in the A range. This compares to only 22 percent of black test takers. These figures alone can explain the large racial scoring gap on the SAT.

Furthermore, data from The College Board confirms that black students who take the SAT have not followed the same academic track as white students. Nearly all blacks and whites who take the SAT have studied algebra in high school. But white SAT test takers are slightly more likely than black SAT test takers to have completed courses in geometry.

In higher level mathematics courses such as trigonometry and calculus, whites hold a large lead. In 2002, 47 percent of white SAT test takers had taken trigonometry in high school compared to 38 percent of black test takers. More than one quarter of white test takers had taken calculus in high school. Only 14 percent of black students had taken calculus, about one half as many as whites. Thirty-three percent of white SAT test takers had taken honors courses in mathematics compared to 20 percent of black SAT test takers.

Similar discrepancies appear in the level of instruction in English, the other major component of the SAT. Some 85 percent of white test takers had completed coursework in American literature compared to 73 percent of black test takers. Nearly three quarters of whites, but only 55 percent of blacks, had taken high school courses in composition. Some 75 percent of whites and 62 percent of blacks had completed coursework in grammar. A full 39 percent of all white test takers had completed honors courses in English compared to 28 percent of black test takers.

Whites were also far more likely than blacks to have taken honors courses in science and social studies. Given the huge differences in course study between black and white high school students, it comes as no surprise that white SAT scores are significantly higher than black SAT scores. Whites, who are more likely to attend high-quality schools, have simply achieved a greater mastery of the subject matter than have blacks.

There are other reasons that contribute to the large scoring gap between blacks and whites on the SAT. These include:

 In many cases black schoolchildren are taught by white teachers who have low opinions of the abilities of black kids from the moment they enter the classroom. These teachers immediately write off black students as academic inferiors and do not challenge them sufficiently to achieve the skills necessary to perform well on standardized tests.
 Black students who study hard are often the subject of peer ridicule. They are accused of "acting white" by other blacks. This so-called "ghetto chic" in the form of peer pressure to shun academic pursuits undoubtedly has a dragging effect on average black SAT scores.
 Black students may be subject to what Stanford psychology professor Claude Steele calls "stereotype vulnerability." Steele contends that black students are aware of the fact that society expects them to perform poorly on standardized tests. This added pressure put upon black students to perform well in order to rebut the racial stereotype in fact makes it more difficult for them to perform well on these tests.
 Black students in some urban schools may be taught an Afrocentric curriculum that may serve to increase black pride and foster an awareness of black culture, but this form of education pays little attention to the subject matters that are covered on the SAT.
 Even middle-class blacks tend to be brought up in basically segregated surroundings. They are not taught the pathways and modes of thinking that are embedded in white culture and reflected in standardized tests. Black families that urge their children to go to college are often first-generation college graduates who grew up in households without the systems that support first-rate academic achievement.
 School administrators and guidance counselors often believe that black students are less capable and less able to learn. They routinely track black students at an early age into vocational training or into a curriculum that is not college preparatory. Black students are rarely recommended for inclusion in gifted education, honors, or advanced placement programs. Once placed on the slow academic track, most black kids can never escape. By the time black students are juniors and seniors in high school, they are typically so far behind their white counterparts in the critical subject areas necessary to perform well on standardized tests that they have little hope of ever matching the scores of whites on the SAT.

[i]Almost No Blacks Among the Top Scorers on the Scholastic Assessment Test[/i]

Before we conclude our report on the black-white SAT scoring gap, it is important to note how these test scores will impact African-American higher education in the event that the current effort to ban race-sensitive admissions at colleges and universities becomes standard practice at all institutions of higher education. Under an admissions system in which race can no longer be used as a positive factor in the admissions process, standardized test scores will almost certainly become a more important component in deciding who is admitted and who is rejected at our leading colleges and universities.

The latest statistics on standardized test scores for college admissions show clearly that if the race-neutral admissions policies now in place in California, Texas, Florida, Georgia, and Washington State are applied nationwide, blacks will be almost totally excluded from admission to the nation's highest-ranked colleges and universities. The reason for this is that only a very tiny percentage of college-bound black students score at the top of the SAT scale.

Under the SAT scoring system, students hoping to qualify for admission to any of the nation's 25 highest-ranked universities and 25 highest-ranked liberal arts colleges need to score at least 700 on each portion of the SAT.

For admission to the very highest ranked, brand-name schools such as Princeton or MIT, applicants realistically need scores of 750 to be considered for admission. Yet, as we shall see, only a minute percentage of black test takers score at these levels. Thus, in a race-neutral admissions environment, high-ranking colleges and universities will choose their first-year students from a pool in which there will be very few blacks.

Let's be more specific about the SAT racial gap among high-scoring applicants. In 2002, 122,684 African Americans took the SAT test. They made up 9.2 percent of all SAT test takers. But only 838 African-American college-bound students scored 700 or above on the math SAT and only 822 scored at least 700 on the verbal SAT. Nationally, 83,689 students of all races scored 700 or above on the math SAT and 59,662 students scored 700 or above on the verbal SAT. Thus, in this top-scoring category of all SAT test takers, blacks make up only 1 percent of the students scoring 700 or higher on the math test and only 1.4 percent of the students scoring 700 or higher on the verbal SAT.

If we eliminate Asians and other minorities from the statistics and compare just white and black students, we find that 5.1 percent of all white SAT test takers scored 700 or above on the verbal portion of the test. But only 0.7 percent of all black SAT test takers scored at this level. Therefore, whites were more than seven times as likely as blacks to score 700 or above on the verbal SAT. Overall, there are more than 43 times as many whites as blacks who scored at least 700 on the verbal SAT.

On the math SAT, only 0.7 percent of all black test takers scored at least 700 compared to 6.2 percent of all white test takers. Thus, whites were nearly nine times as likely as blacks to score 700 or above on the math SAT. Overall, there were 51 times as many whites as blacks who scored 700 or above on the math SAT.

If we raise the top-scoring threshold to students scoring 750 or above on both the math and verbal SAT  a level equal to the mean score of students entering the nation's most selective colleges such as Harvard, Princeton, and CalTech  we find that in the entire country 195 blacks scored 750 or above on the math SAT and 218 black students scored 750 or above on the verbal portion of the test. Nationwide, 26,838 students scored at least 750 on the math test and 20,160 scored at least 750 on the verbal SAT. Therefore, black students make up 0.7 percent of the test takers who scored 750 or above on the math test and 1 percent of all test takers who scored 750 or above on the verbal section.

Once again, if we eliminate Asians and other minorities from the calculations and compare only blacks and whites, we find that 0.18 percent of all black test takers scored 750 or above on the verbal SAT compared to 1.7 percent of all white test takers. Thus, whites were nearly 10 times as likely as blacks to score 750 or above on the verbal portion of the test. Overall, there were 54 times as many whites as blacks who scored at or above the 750 level.

On the math SAT, only 0.16 percent of all black test takers scored 750 or above compared to 1.8 percent of white test takers. Thus, whites were more than 11 times as likely as blacks to score 750 or above on the math SAT. Overall, there were 65 times as many whites as blacks who scored 750 or above on the math section of the SAT.

In a race-neutral competition for the approximately 50,000 places for first-year students at the nation's 25 highest-ranked universities, high-scoring blacks will be buried by a huge mountain of high-scoring nonblack students. Today, under prevailing affirmative action admissions policies, there are about 3,000 black first-year students matriculating at these 25 high-ranking universities, about 6 percent of all first-year students at these institutions. But if these schools operated under a strict race-neutral admissions policy where SAT scores were the most important qualifying yardstick, these universities could fill their freshman classes almost exclusively with students who score at the very top of the SAT scoring scale. As shown previously, black students make up at best between 1 and 2 percent of these high-scoring groups.

If the nation now insists on race-blind college admissions, it must face the near certainty that the percentage of black students at the nation's highest-ranked colleges and universities will drop from the present average of about 6 percent to 2 percent or less.

[i]Black Students Are Also Losing Ground on the ACT Standardized Admission Test[/i]

Nearly as many black students now take the ACT college admission test as sign up for the competing SAT examination. But, in common with the more prestigious SAT, the scoring gap between blacks and whites is widening. In addition, almost no blacks score at the very highest level of the ACT performance scale which is generally necessary to win admission to the nation's most prestigious colleges and universities.

The scores of black students on this year's college entrance examination of the American College Testing Program are cause for increased concern. Many students in the Midwest, the Deep South, and the Rocky Mountain states take the ACT test for college admission rather than the Scholastic Assessment Test. And the ACT test is becoming increasingly important to college-bound blacks. The number of black students taking the ACT is up 20 percent from 1998. The rise in the number of African-American students taking the ACT is the result, at least in part, of the fact that a greater percentage of high school students are now preparing for college in the heavily black Deep South states of Alabama, Tennessee, Arkansas, Mississippi, and Louisiana. In these states, almost all college-bound students take the ACT test.

Nationwide, 120,311 black students sat for the ACT test in 2002. This is only slightly fewer than the 122,684 black students who took the SAT test this year. If present trends continue, only a few years will pass before more black students nationwide will be taking the ACT test than sitting for the SAT. In 2002 blacks made up 10.8 percent of all ACT test takers. Blacks made up 11.4 percent of all students who took the SAT test.

[i]The Racial Trend in ACT Scores[/i]

In 2002 the median score for whites on the ACT was 21.7. (The ACT test is scored on a scale of 0 to 36.) For blacks, the median score was 16.8. Thus, on average, blacks scored 14 percent lower on the ACT than did whites.

The serious fact is that the racial gap on the ACT test has been expanding over recent years. The median scores for both blacks and whites had remained the same for the three years from 1997 to 1999, at 21.7 and 17.1, respectively. But since that time the black score had dropped 0.1 point each year while the white score has held steady at 21.7.

Don Carstensen, vice president for educational services at the American College Testing Program, told JBHE, "The drop in average ACT composite scores for blacks can be attributed to the fact that the number of students who have taken the ACT has increased significantly, creating a more heterogeneous group of test takers. With an expanded pool of test takers comes a broader spectrum of performance. The good news, of course, is that black students are considering going to college."

[i]Few Blacks at the Top of the ACT Scoring Grid[/i]

The nation's highest-ranked colleges and universities seek students who score 28 or above on their ACT test. Nationwide, only 1,180 black students scored 28 or above on the ACT test. They made up slightly less than 1 percent of all black ACT test takers. In contrast, 86,831 white students scored 28 or above on the ACT test this year. They made up 11.2 percent of all white ACT test takers. Thus whites were 11 times more likely than blacks to score at a level equal to the mean score of students admitted to the nation's most prestigious colleges and universities. This data tends to show that if colleges and universities were unable to take race into account during the college admissions process  such as is the case today for state-chartered institutions in California, Florida, Georgia, Texas, and Washington  blacks would be placed at a huge disadvantage for winning any places at the nation's leading institutions.

If we examine ACT scores at the highest scoring levels, we find an even larger disparity. Of the 120,311 blacks who took the ACT test this year, not one scored a perfect score of 36. On the other hand, there were 96 white students who received the highest score of 36.

But here is the most discouraging statistic in this year's ACT report: In 2002 more than 87 percent of all white test takers scored at or above the median score for blacks.

Copyright  2002. The Journal of Blacks in Higher Education. All rights reserved.

Thank you for finding that.  It was interesting, though a little unnerving.

It is downright sad to see the failure of public education shown so clearly.

[quote="gauss202"]I don't think its a failure of public education as much as it is a failure of the nation as a whole. Just think of how much untapped economic potential there is here.[/quote]

I always feel a little uncomfortable blaming a failure on "the nation" as if everyone's score is the responsibility of the system.  The nation doesn't control the system.  At least, I don't feel that I have any control over it.

I think the failure is a cultural one.  We live in a culture where people say "education is not my job" and everyone washes their hands of it while government succeeds to the minimum level necessary to retain political support.  Or perhaps they fail because political support is easier to exact from the uneducated.  Regardless...

Students are people.  People are animals.  Animals follow incentives.

Failure is a construct we use to explain the result of "irrational" incentives.  Can we offer a better system of incentives?  Not until our culture changes from one where education is more important than a nice car, an episode of Friends, or the Super Bowl (or just the ads during the Super Bowl).

Changing incentives is complicated by the fact that the United States is [i]not[/i] one culture.  Anybody who has lived in two different geographical regions knows this.  Incentives react on a local level (as local as the individual) and it will take a national effort to make a national change.

Unfortunately any national effort must butt heads with the entirety of the current system.  Is that an impossible task?

Until we see a major systemic change, the best we can do is target individuals.  I think this is exactly why black students are at such a disadvantage -- who is going to step into the ghetto to target a student for excellence??

I think the lack of an answer to that question means that black students above all are held captive by our system of public schools.

I just read an interesting statistic.  In general, black students preform better when taught by black teachers and white students when taught by white teachers.  But when class sizes get small enough, that gap disappears.  In fact, smaller class sizes are related to increased performance, pretty much across the board.  And that's within this current system with which you find so much fault.  I came back and read over some of the things you've said, and one thing popped out at me:
[quote="MCrawford"]Until we see a major systemic change, the best we can do is target individuals.[/quote]
I don't think you're right.  I think our system, with a little bit of clever management, has the capacity to do much better than it is now.  Imagine elementary school classes of fifteen students, growing to 20 or 25 by high school.  There's no aspect of the current system that prevents it, except cost.  Everything gets better: teachers will be happier workers under better conditions; the abilities of both students and teachers can be more accurately assessed; the specifics of a curriculum matter less; and most importantly, teachers can spend more time with individual students -- targetting of individuals -- and avoid the stereotypes of students that Gauss's article talks about.  I'm not suggesting that everyone in America will ace the SAT's overnight, but there is overwhelming evidence to support the idea that smaller class sizes lead to good things for students and for school systems.

So, I guess what this boils down to is just a question: does shrinking class sizes count as "major systemic change," and if not, why are you discarding or downplaying it as a possible solution?  (This isn't meant to be an attack at all -- that's just the most straightforward way to phrase the question.)

[quote="JBL"]I just read an interesting statistic.[/quote]

What's the full citation of the statistic, and the full citation of the (probably differing) underlying primary source study where the methodology used to reach that conclusion would be explained?

[url=http://www.nytimes.com/2004/05/26/education/26memo.html]This[/url] New York Times article (subscription required) says:

[quote] Academic studies show small class size carries many benefits, even mitigating racial problems that interfere with learning. A recent study by Tom Dee, a Swarthmore professor, in "The Review of Economics and Statistics" concluded that both white and black children achieved more when they were taught by teachers of their own race. This is bad news for black children since the vast majority of teachers, even in big cities, are white and the vast majority of urban children - 85 percent in New York City - are minority.

But there is a hopeful exception. If classes are small, Dr. Dee found, black children do equally well with a white or black teacher. "It may be because there's more personal interaction, less chance for stereotyping," Dr. Dee said.[/quote]

The abstract for that journal article is:[quote]Abstract: In previous research, a substantial gap in test scores between white and black students persists, even after controlling for a wide range of observable characteristics. Using a newly available data set (the Early Childhood Longitudinal Study), we demonstrate that in stark contrast to earlier studies, the black-white test score gap among incoming kindergartners disappears when we control for a small number of covariates. Real gains by black children in recent cohorts appear to play an important role in explaining the differences between our findings and earlier research. The availability of better covariates also contributes. Over the first two years of school, however, blacks lose substantial ground relative to other races. There is suggestive evidence that differences in school quality may be an important part of the explanation. None of the other hypotheses we test to explain why blacks are losing ground receive any empirical backing.[/quote]

The Times article also mentions a variety of other benefits, some of which I mentioned as well, which are non-empirical.  For example, the claim that it is easier to teach smaller classes than larger classes.  I have seen it repeatedly asserted that smaller class size is associated with better results across the board, but I do not have a reference for that.

Actually, I'm wrong.  That paper tricked me.  The real abstract is this:

[quote]Abstract: Recommendations for the aggressive recruitment of minority teachers are based on hypothesized role-model effects for minority students as well as evidence of racial biases among nonminority teachers. However, prior empirical studies have found little or no association between exposure to an own-race teacher and student achievement. This paper presents new evidence on this question by examining the test score data from Tennessee's Project STAR class-size experiment, which randomly matched students and teachers within participating schools. Specification checks confirm that the racial pairings of students and teachers in this experiment were unrelated to other student traits. Models of student achievement indicate that assignment to an own-race teacher significantly increased the math and reading achievement of both black and white students.[/quote]

(It's in the February issue of [u]Review of Economics and Statistics[/u].)

I support affirmative action.

Hopefully it gives me more of a chance to get into college considering I am black?

The issue is discussed in the Thursday 24 June 2004 New York Times article "[url=http://www.nytimes.com/2004/06/24/education/24AFFI.final.html?hp]Top Colleges Take More Blacks, but Which Ones?[/url]"

That's an excellent article. It talks about something that I've been saying for quite a while; and that I've even mentioned in this thread.

This is the statement that really stood out to me though: "Professor Massey said that the admissions officials he talked to at these colleges seemed surprised by the findings about the black students. 'They really didn't have a good idea of what they're getting,' he said."

How is that possible? It seems to me that there's a disconnect between the admissions officials and the students that they admit. Perhaps Universities should start collecting more information from their applicants. I don't just mean information on ethnicity - but more information in general. That way they can at least know more about the students who are applying.

The other thing that stood out to me was the difference of opinion as to what the sociological causes are of differential achievement. As Lee Bollinger puts it, "The issue is not origin, but social practices. It matters in American society whether you grow up black or white. It's that differential effect that really is the basis for affirmative action." But is that really true? How does that explain the differential representation between black students who are descendents of recent immigrants, and those who are descendents of American slaves?

I think the Bakke decision, by making the discussion less about social justice and more about diversity, cut off a lot of the discussion about what's really going on sociologically here.

For those who haven't heard of him, Henry Louis Gates is a really great sociologist, who always brings up these interesting types of questions. He's written some good books, including one with Cornell West called [u]Race Matters[/u]. But before you read that I would recommend [u]The Souls of Black Folks[/u] by W.E.B. DuBois. It's a book that I feel every American should read.

I think there's a lot of evidence for an "immigrant advantage," regardless of race.  I'm personally convinced that that is why Asians do so well here, by and large.

I also think Bollinger misses the point, somewhat.  It seems to me that really serious problem isn't just discrimination and social effects, it's the collected social (and perhaps personal) burden of 350 years of it.  That's what sets African-Americans (and American Indians) apart, in my mind, from other groups who have suffered at some point in time.


Can someone help me by explaining what is motivic unity? I googled but couldn't find anything understandable. Thanks!

Sometimes a change of search terms helps.

For example try searching under "Motivic development", Musical unity" ,"Motive(s) in Music".

Maybe these sites will help you:

http://jan.ucc.nau.edu/~tas3/inv.html

http://jan.ucc.nau.edu/~tas3/intrmz.html

http://www.flashmusicgames.com/compose/introduction.html


calculus
integration
number theory proposed
number theory

Prove that if $ x\neq 1$ then the equation $ x^{2000}\minus{}1\equal{}y^2(x\minus{}1)$ has no integral solutions.

This is not correct. (0,1) and (0,-1) are solutions.

But, apart from that, I believe there are no other solutions. I think I have a proof, but that uses a nice result (with a complicated proof).

The result is due to Wilhelm Ljunggren and states

if $ aX^2 \minus{} bY^2 \equal{} 1$ is solvable, then $ aX^2 \minus{} bY^4 \equal{} 1$ has at most one solution. ($ a,b$ are positive integers and $ a$ is non-square).

We use this result for $ a \equal{} 2\,b \equal{} 1$ for which (1,1) is the only solution.

Back to original problem.

We have

$ (x^{1000} \plus{} 1)(\frac{x^{1000} \minus{} 1}{x\minus{}1}) \equal{} y^2$. If $ x$ is even, then we see that $ x^{1000} \plus{} 1$ must be a perfect square. So assume $ x$ is odd.

$ (x^{1000} \plus{} 1)(x^{500} \plus{} 1)(x^{250} \plus{} 1)(x^{125} \plus{} 1)(\frac {x^{125} \minus{} 1}{x \minus{} 1}) \equal{} y^2$.

Notice that apart from a possible common factor of 2, all the terms are co-prime as

$ 2 \equal{} (x^{1000} \plus{} 1) \minus{} (x^{500} \plus{} 1)(x^{250} \plus{} 1)(x^{125} \plus{} 1)(x^{125} \minus{} 1)$

Thus we see that

one of $ x^{1000} \plus{} 1$ or $ x^{500} \plus{} 1$ must of be of the form $ 2M^2$, which contradicts the above theorem by Wilhem Ljunggren, if $ x \neq 1$

I hope I didn't do my usual goof up.

Sorry , the problem is as follows
[quote="silouan"]Prove that if $ x\neq 1$ then the equation $ x^{2000} \minus{} 1 \equal{} y^2(x \minus{} 1)$ has no positive integral solutions.[/quote]

Using the theorem I stated seems overkill and perhaps out of scope for the Olympiad (I have no clue, though).

What solution did you have in mind silouan?


induction
number theory unsolved
number theory

let n >= 2 be positive integer. find the least possible number of element of the set A={1,2,...,2n-1,2n} that should be deleted in order to the sum of any two different elements remained be a composite number.

We have L, the least possible number removed, is greater than or equal to n (because we can always remove all odd numbers and this will guarantee all composites) Claim: the answer is always n
[hide="My first stab:"]
Assume there is a possible way with n-1 numbers crossed out. By induction, we have the result for the first 2n-2 numbers, so that we know there are n-1 numbers crossed out among the first 2n-2. Therefore, neither of 2n or 2n-1 are crossed out. Now what would be really cool would be if the following worked: Add 2n to each of the n-1 non crossed out numbers, yielding n-1 new numbers which all must be not prime. Add 2n-1 to each uncrossed, giving another n-1 new numbers which can't be prime. But all these numbers mentioned are between n and 2n, and we have listed 2n-2 therefore exhausting all numbers in this range. But by Chebychev's Theorem there is always a prime in this range, so contradiction. Of course this argument doesn't work if there exists a consecutive pair of numbers that aren't crossed off (this is true, but proving it proves the problem first :D ). So no really nice way like this.[/hide]
Here is my real solution:
By Chebychev's Theorem, there is some prime between 2n and 4n. There is also one between 2n-1 and 4n-2, 2n-k to 4n-2k, etc... n to 2n. Each prime tells us (p-1)/2 pairs that sum to it. The problem is solved when one realizes the impossiblity by considering first the largest prime in n, 2n, then considering the largest prime in the group starting one above this prime, then considering the largest in the group starting one above that prime,... until you've reached 4n, and considering the number of pairs completely within (1,...2n) that sum to each specific prime. I'm not quite sure how to write up a rigorous proof of this, but I know why it must be true. If someone understands this, please try to help!


Point $P$ is inside triangle $ABC$.  When line segments $APD$, $BPE$, and $CPF$ are drawn, with $D$, $E$, and $F$ on the triangle's perimeter, $AP=6$, $BP=9$, $PD=6$, $PE=3$, and $CF=20$.  Find the area of $ABC$.

this same problem was in another thread:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10954


The 20 members of a local tennis club have scheduled exactly 14 two-person games among themselves, with each member playing in at least one game. Prove that within this schedule there must be a set of 6 games with 12 distinct players.

[hide]
$2(14)$ means 28 players and since each player plays atleast once, there are atmost 8 people who play more than one game. Call a game a "repeated game" if it has somebody who has already played in it. There are atmost 8 repeated games. In the other 6 games, all the players must be distinct.
[/hide]


1/Let $ a,b,c>0$. Prove that:
$ \sqrt{\frac{a^3}{a^3\plus{}(b\plus{}c)^3}}\plus{}\sqrt{\frac{b^3}{b^3\plus{}(a\plus{}c)^3}}\plus{}\sqrt{\frac{c^3}{c^3\plus{}(b\plus{}a)^3}}\geq 1$
2/Let $ x,y,z\geq 0, x\plus{}y\plus{}z\geq 6$. Find the min value of
$ P\equal{}\sqrt{1\plus{}x^3}\plus{}\sqrt{1\plus{}y^3}\plus{}\sqrt{1\plus{}z^3}$
3/Find the min value of: $ A\equal{}cos^4x\plus{}\sqrt{sin^4x\plus{}\frac{cos^4x}{2}\plus{}\frac{sin^2x\cdot cos^2x}{2}}$ $ \plus{}\sqrt{\frac{cos^4x}{4}\plus{}2sin^4x\plus{}\frac{sin^2x\cdot cos^2x}{2}}$
4/Find $ n\in N$ such that:
$ (\sum^n_{k\equal{}1}a^2_k)(\sum^n_{k\equal{}1}a_k)\minus{}(\sum^n_{k\equal{}1}a^3_k)\geq 6\prod^n_{k\equal{}1}a_k$
which is always true with all values of $ a_i>0(i\equal{}1,2,3,...,n)$

[quote="Kyoshiro"]1/Let $ a,b,c > 0$. Prove that:
$ \sqrt {\frac {a^3}{a^3 \plus{} (b \plus{} c)^3}} \plus{} \sqrt {\frac {b^3}{b^3 \plus{} (a \plus{} c)^3}} \plus{} \sqrt {\frac {c^3}{c^3 \plus{} (b \plus{} a)^3}}\geq 1$
2/Let $ x,y,z\geq 0, x \plus{} y \plus{} z\geq 6$. Find the min value of
$ P \equal{} \sqrt {1 \plus{} x^3} \plus{} \sqrt {1 \plus{} y^3} \plus{} \sqrt {1 \plus{} z^3}$
3/Find the min value of: $ A \equal{} cos^4x \plus{} \sqrt {sin^4x \plus{} \frac {cos^4x}{2} \plus{} \frac {sin^2x\cdot cos^2x}{2}}$ $ \plus{} \sqrt {\frac {cos^4x}{4} \plus{} 2sin^4x \plus{} \frac {sin^2x\cdot cos^2x}{2}}$
4/Find $ n\in N$ such that:
$ (\sum^n_{k \equal{} 1}a^2_k)(\sum^n_{k \equal{} 1}a_k) \minus{} (\sum^n_{k \equal{} 1}a^3_k)\geq 6\prod^n_{k \equal{} 1}a_k$
which is always true with all values of $ a_i > 0(i \equal{} 1,2,3,...,n)$[/quote]
1) You can see it in Ineq Unsolved problems
2)$ f(x)\equal{}\sqrt{1\plus{}x^3},f'(x)\ge0,f"(x)\ge 0$ so by Jensen $ f(x)\plus{}f(y)\plus{}f(z)\ge 3f(\frac{x\plus{}y\plus{}z}{3})$ and using $ f'(x)\ge 0$ we get $ f(x)\plus{}f(y)\plus{}f(z)\ge 3f(\frac{x\plus{}y\plus{}z}{3})\ge 3f(2)\equal{}9$
4) N<4 because if N>4 $ a_1\equal{}a_2\equal{}...\equal{}a$ a very big number and degree of LHS bigger than RHS N=1,2 no,if N=3 its true because $ (a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)\minus{}a^3\minus{}b^3\minus{}c^3\equal{}a^2(b\plus{}c)\plus{}b^2(a\plus{}c)\plus{}c^2(b\plus{}a)\ge 6abc$ (AM-GM)

[quote="dima ukraine"]
1) You can see it in Ineq Unsolved problems
[/quote]
Can you show me the link, please? :)

for 3 , using $ \sum \sqrt{a^2\plus{}2b^2\plus{}ab} \geq 2\sum a$ and a little bit substitution  :wink:

1 is also Jensen, let $ a\plus{}b\plus{}c\equal{}1$, then use Jensen on $ f(x)\equal{} \sqrt{\frac{x^3}{x^3\plus{}(1\minus{}x)^3}}$


superior algebra
superior algebra unsolved

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

It's clear that $ IJ \subset I \cap J$. For the second inclusion note that $ I \cap J \equal{} (I \plus{} J)(I \cap J) \subset I (I \cap J) \plus{} (I \cap J) J \subset IJ \plus{} IJ \equal{} IJ$.

How do you prove $ IJ \plus{} IJ \equal{} IJ$?

It follows from the fact that $ IJ$ is an ideal (see the definition of $ IJ$).


category theory
superior algebra
superior algebra theorems

Show that every Boolean ring without identity can be embedded into a Boolean ring with identity.

Let $A$ be a $K$-algebra (not with identity). We define another $K$-algebra $B$ with identity as follows:

Elements of $B$ are pairs $(k,a) \in K \times A$, which we denote by $k+a$. Addition is defined pointwise, and multiplication is motivated by the notation:

$(k+a)*(k'+a')=kk'+(ka'+k'a+aa')$
$l(k+a)=lk+la$

We can embed $K$ and $A$ into $B$. The image of $1 \in K$is the identity of$B$.

Now if$A,K$are boolean rings (hence of characteristic $2$), this holds also for $B$:

$(k+a)^{2}=k^{2}+(2ka+a^{2})=k^{2}+a^{2}=k+a$

If we take $K = \mathbb{Z}/2$, we can embed every boolean ring in one with identity.

I hope this works :-). By the way, $A \to B$ yields a functor, which is left-adjoint to the forgetful functor from unitary $K$-algebras to $K$-algebras.


inequalities
modular arithmetic
number theory proposed
number theory

Prove that for every positive integer $n$ we have:

$[\sqrt{n}+\sqrt{n+1}]=[\sqrt{4n+2}]$

where $[t]$ denote the integer part of real $t$..

Letting $f(n)=\sqrt{4n+2}-\sqrt{n}-\sqrt{n+1}$ ,$n\geq 1$

We shall first claim that $0<f(n)\leq f(1)<1$

But $f(n)>0$ $\Longleftrightarrow \sqrt{4n+2}>\sqrt{n}+\sqrt{n+1}$ $\implies (2n+1)^2>4n(n+1)$ which is true .

For the left hand side , we shall prove $f(n)>f(n+1)$ $\Longleftrightarrow \sqrt{4n+2}+\sqrt{n+2}>\sqrt{n}+\sqrt{4n+6}$ $\implies (4n+3)^2>4n(4n+6)$ which is also true after squaring and expanding .

Since $f(n)$ is decreasing , hence $f(n)\leq f(1)=\sqrt{6}-\sqrt{2}-1<3-1-1=1$

Since $0<f(n)<1$ , hence $[f(n)]=0$ .....(i)

But we know that if $0<x-y<1$ ,then it must be either $[x]=[y]$ or $[x]=[y]+1$

So from (i) , it is either $[\sqrt{4n+2}]=[\sqrt{n}+\sqrt{n+1}]$ or $[\sqrt{4n+2}]=[\sqrt{n}+\sqrt{n+1}]+1$

we shall now prove that the latter case wont work .

In fact , the latter case only works iff there exist an integer $k$ such that

$\sqrt{4n+2}>k>\sqrt{n}+\sqrt{n+1}$ (equal sign wont occur since both $\sqrt{4n+2}$ and $\sqrt{n}+\sqrt{n+1}$ cannot be integer ) .

but above is equivalent to $(k^2-1)^2>4nk^2>(k^2-1)^2-1$ which cannot happen for any $k$ .

Hence we must have $[\sqrt{4n+2}]=[\sqrt{n}+\sqrt{n+1}]$

Remark : We can also have $[\sqrt{4n+1}]=[\sqrt{4n+2}]=[\sqrt{4n+3}]=[\sqrt{n}+\sqrt{n+1}]$  :) anyone wanna prove it . :P

[quote]Prove that for every positive integer n we have:

[\sqrt{n}+\sqrt{n+1}]=[\sqrt{4n+2}]

where [t] denote the integer part of real t..[/quote]

First, $ \sqrt{n}+\sqrt{n+1}<\sqrt{4n+2}. $ This follows from:
$\sqrt{n}+\sqrt{n+1}=\sqrt{(\sqrt{n}+\sqrt{n+1})^{2}}=\sqrt{2n+1+\sqrt{4n^{2}+4n}}<\sqrt{2n+1+\sqrt{4n^{2}+4n+1}}=\sqrt{4n+2}$
Now, it suffices to prove that there is no integer $m$ such that: $\sqrt{n}+\sqrt{n+1}<m\leq\sqrt{4n+2}$.
Our last inequality is equivalent to $4(n^{2}+n)<(m^{2}-2n-1)^{2}\leq(2n+1)^{2}$
And from this..$m^{2}-2n-1=2n+1, m^{2}=4n+2$. Last equation is impossible, because $m^{2}\not\equiv 2 \pmod 4$
So, $[\sqrt{n}+\sqrt{n+1}]=[\sqrt{4n+2}]$


Ilian will randomly choose a value for $ p$ from the integers 0
through 13, inclusive, and will make $ \frac{p}{
13}$ the coordinate of
point $ A$. The coordinate of point $ B$ is $ \frac{2}
{3}$ . Points $ A$ and $ B$ will be
on the same number line. What is the probability that point $ A$
and point $ B$ will be less than $ \frac{1}{5}$
of a unit from each other?
Express your answer as a common fraction.

Point $ A$ will then have to be between $ \frac{2}{3}\minus{}\frac{1}{5}\equal{}\frac{7}{15}$ and $ \frac{2}{3}\plus{}\frac{1}{5}\equal{}\frac{13}{15}$.

Therefore, $ \frac{p}{13}$ needs to be between $ \frac{7}{15}$ and $ \frac{13}{15}$.
p could then only be one of the values $ 7$ through $ 11$.

So the answer is $ \frac{11\minus{}7\plus{}1}{13\minus{}0\plus{}1}\equal{}\frac{5}{14}$

Note:The denominator is 14, not 13, because the question states :"0 through 13 INCLUSIVE"


Mel constructs a pyramid by gluing 25 cubes in a 5*5*1 layer1 and then making a 3*3*1 layer on top of the 5*5*1 and a single cube on top.  Then Mel wants to paint the outside red.  What fraction of the cubes have exactly 3 sides of paint when the cubes are seperated?

On the bottom layer, that's any cube on the edge but not on the corner, which makes to be 12 cubes. On the next layer, only the corners will have 3 sides of paint, and there are 4.

16/35?

when i counted, i forgot that she could paint the bottom of the pyramid too  :oops:


if f(2x)= 2 over 2+x for all x>0, then 2f(x) equals what?

[hide="Solution?"]
$2f(x)=\frac{4}{2+\frac{x}{2}}$

EDIT: w00, go careless mistakes![/hide]

i dunno if it's right or not that's why I'm askin you for the souliton.

[hide="Solution"]$f(2x) = \displaystyle\frac{2}{2+x}$
$f(x) = \displaystyle\frac{2}{2+\frac{x}{2}} = \displaystyle\frac{4}{4 + x}$
$2f(x) = \boxed{\displaystyle\frac{8}{4+x}}$[/hide]

[hide]
If $f(2x) = \frac{2}{2+x}$, then in the function f() x is being halved, so $f(x)$ must be $\frac{2}{2+.5x}$. From here just do the multiplication.
[/hide]


I'm bored, entertain me.
Say something interesting or something.
Oh I have an idea.
Let's play the "I make you laugh game"
Everyone posts something and if someone laughs, they lose.
BE HONEST!
I start:
[img]http://www.jasperfforde.com/specops/images/cheese_alert.jpg[/img]

Fine then :lol:
[img]http://i30.photobucket.com/albums/c340/davemarkowitz/pirates/pirate_keyboard.jpg[/img]

I just lost the game (not the "I make you laugh game").

[img]http://www.blogiseverything.com/files/pics/funny_math.jpg[/img]

You know how in sc both Zeratul and the zealots go like, "My life for Auir!"
Funniest misinterpretation:
"My wife's a liar!"

Aw daern it I lost this game four times and [i]the[/i] game twice!

[img]http://joi.ito.com/images2/mathjoke.gif[/img]

Lawl, 1=2 and Cheeseisgood's posts were funny... seen 'em before, though.

... I lost... THE game...

And, ummm..... I like chocolate mlik?

I haven't lost yet... I know something funny, but I don't know how to post YouTube videos.


[quote="kunny"]$ \frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}$?[/quote]
Do you understand what you write ? I don't understand
  :)


search
combinatorics unsolved
combinatorics

Let n be a positive integer with n>1.Fix 2n points in space in such way no four of them are in the same plane,and select any n^2+1 segments determined by the

given points . prove that these segment form at least n triangles.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1151946802&t=6323

Pierre.


geometry
inequalities
geometric transformation
homothety
circumcircle
conics
ellipse

It posted here
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=457344#p457344

$3-2a=a+b+c-2a=-a+b+c$ etc., so we have to prove:
\[ (-a+b+c)(a-b+c)(a+b-c)\leq (abc)^2\iff \] \[ \iff(a+b+c)(-a+b+c)(a-b+c)(a+b-c)\leq 3(abc)^2\iff \]
\[ \iff \frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}{16}\leq\frac{3}{16}(abc)^2 \iff \] \[ \iff\sqrt{\frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}{16}}\leq \frac{\sqrt{3}}{4}abc. \]
Let $P$ be the area of triangle whose sides are $a,b,c$ and $R$ be the radius of corresponding circumcircle. Then $LHS=P$ and using $R=\frac{abc}{4P}$ we get:
\[ P\leq \frac{\sqrt{3}}{4}abc\iff \]
\[ \iff \frac{\sqrt{3}}{3}\leq R. \]
We got that your inequality is equivalent to: in triangle whose sides are $a,b,c$ holds $R\geq \frac{\sqrt{3}}{3}$. Let's transform the triangle using homothety with scale $\frac{k}{3}$. Now we have to prove that in any triangle whose sum of sides is $k$ we have $R\geq\frac{\sqrt{3}}{9}k$. In other words, in any triangle whose radius  of circumcircle is $R$ and sum of sides is $k$ we have $k\leq 3\sqrt{3}R$. It's easy to see that for constant $R$ we can increase $k$ making the triangle isosceles (just draw ellipse and you will see it). It's easy to count that sum of sides of isosceles triangle whose basis defines central angle $\alpha$ is $2R(\sin(\alpha/2)+2\cos(\alpha/2))$, where $\alpha\in(0,2\pi)$. We have to prove that $f(\beta)=\sin(\beta)+2\cos(\beta)\leq \frac{3\sqrt{3}}{2}$, where $\beta=\alpha/2\in(0,\pi)$. We have only to prove this for $\beta\in(0,\pi/2)$, because for $\beta\in(\pi/2,\pi)$ we have $f(\beta)\leq f(\pi-\beta)$. For $\beta\in(0,\pi/2)$ we can substitute $t=\sin(\beta)$ and then $\cos(\beta)=\sqrt{1-t^2}$ (without any $\pm$). We have to prove that for any $t\in [0,1]$ hlods
\[ t+2\sqrt{1-t^2}\leq\frac{3\sqrt{3}}{2} \]
(and this is equivalent to the beginning inequality). $\frac{3\sqrt{3}}{2}>1$ so we can sqare
\[ 2\sqrt{1-t^2}\leq \frac{3\sqrt{3}}{2}-t \]
and get
\[ 4(1-t^2)\leq \frac{27}{4}-3\sqrt{3}t+t^2. \]
After some transformations it becomes
\[ 5t^2-3\sqrt{3}t+\frac{27}{4}-4\geq 0, \]
which clearly holds.

Oh no, you were faster! The proof you posetd link is at all more elegant.

Let $b+c-a=x,\ c+a-b=y,\ a+b-c=z$, then we have $x+y+z=3$ and $a=\frac{y+z}{2},\ b=\frac{z+x}{2},\ c=\frac{x+y}{2}$. The inequality becomes: \[(x+y)^{2}(y+z)^{2}(z+x)^{2}\ge 64xyz.\] Wlog assume $x \ge y \ge z$. Define $f(x,y,z)=(x+y)^{2}(y+z)^{2}(z+x)^{2}-64xyz$. Consider: \[f(x,y,z)-f\left(x,\frac{y+z}{2},\frac{y+z}{2}\right) = \left(\frac{y-z}{2}\right)^{2}\left[64x-54(y+z)^{2}\left(a+{b+c \over 2}\right)^{2}\right] \ge 0\] We have to show, with $t=\frac{y+z}{2}$ that: \[f\left(x,\frac{y+z}{2},\frac{y+z}{2}\right) \ge 0 \\ \iff (x+t)^{4}(2t)^{2}-64xt^{2}\ge 0\\ \iff \left(\frac{x-3}{2}\right)^{2}\left(\frac{(x+3)^{4}}{4}-64x\right) \ge 0.\] That’s obvious since AM-GM, $(x+3)^{4}\ge 256x$.
Done.


A pyramid has a square base $ ABCD$ and vertex $ E$.  The area of square $ ABCD$ is $ 196$, and the areas of $ \triangle ABE$ and $ \triangle CDE$ are $ 105$ and $ 91$, respectively.  What is the volume of the pyramid?

$ \textbf{(A)}\ 392 \qquad \textbf{(B)}\ 196\sqrt {6} \qquad \textbf{(C)}\ 392\sqrt {2} \qquad \textbf{(D)}\ 392\sqrt {3} \qquad \textbf{(E)}\ 784$

[hide]
Since square $ ABCD$ has area $ 196$, it has side length $ 14$.  Drop perpendiculars from $ E$ to $ AB$ and $ CD$ at $ M$ and $ N$, respectively.  Since $ [ABE]\equal{}105$, $ \frac{1}{2}14EM\equal{}105$, so $ EM\equal{}15$.  Since $ [CDE]\equal{}91$, $ \frac{1}{2}14EN\equal{}91$, so $ EN\equal{}13$.  Thus $ \triangle EMN$ is a $ 13\minus{}14\minus{}15$ triangle, with area $ 84$.

So the perpendicular $ EX$ to side $ MN$, which is also the height of the pyramid, satisfies $ 84\equal{}\frac{1}{2}14EX$, so $ EX\equal{}12$.  Thus the volume of the pyramid is $ \frac{1}{3}12*196\equal{}784$.  Answer is $ E$.
[/hide]


linear algebra
matrix
vector
linear algebra unsolved

[quote="Fierytycoon"]T is a linear map from [tex]F^n[/tex] to [tex]F^m[/tex] and M(T) is a m by n matrix with the value [tex]a_{i, j}[/tex] at row i and column j. If we are using the standard bases, then show that

[tex]T(x_1, ..., x_n) = (a_{1,1}x_1+...+a_{1,n}x_n,...,a_{m,1}x_1+...+a_{m,n}x_n[/tex]

for each [tex](x_1, ..., x_n)[/tex] in [tex]F^n[/tex].

Any help would be appreciated.[/quote] :huh:
I know it cannot be true what I'm saying.. it's too simple but...

isn't this the definition of matrix multiplication? :|  :|

[quote="Peter VDD"]:huh:
I know it cannot be true what I'm saying.. it's too simple but...

isn't this the definition of matrix multiplication? :|  :|[/quote]

That is what I thought too, but this is an exercise from Axler's textbook, so I don't think it would be that obvious.

From looking at a margin note (refering to an assertion elsewhere in the text), I think perhaps that the intended problem is to prove that all linear maps from [tex]F^n[/tex] to [tex]F^m[/tex] have the form [tex]T(x_1, ..., x_n) = (a_{1,1}x_1+...+a_{1,n}x_n,...,a_{m,1}x_1+...+a_{m,n}x_n)[/tex], regardless of what [tex]a_{i,j}[/tex] is.

Of course. That is the definition of matrix multiplication.

But if that's in a serious college book it can't be the solution. Probably it's a generalisation of it somehow?

Yeah, it's quite odd. I am posting the problem just as it is stated in the book (well, the wording is not the exact same, but it is the same problem).

Oh well, guess I'll just skip this exercise.  ;)

[quote="Peter VDD"]Perhaps that book just uses another definition of matrix multiplication and you have to show they're equal?[/quote]

No, I'm pretty sure that they defined matrix multiplication in virtually the same way...

Oh well, nevermind about this exercise. No need to lose any sleep over it.  :D

[quote="Fierytycoon"]Hold on...this is not the definition of matrix multiplication...

We are asked to prove that every linear map from [tex]F^n[/tex] to [tex]F^m[/tex] is given by matrix multiplication.[/quote]

 :?:  :?:  :?:

What is the confusion? The problem is to prove that any linear map from [tex]F^n[/tex] to [tex]F^m[/tex] can be represented by a matrix. It is a standard result I believe, but I can't seem to find a proof of it.

[quote="Fierytycoon"][tex]T(x_1, ..., x_n) = (a_{1,1}x_1+...+a_{1,n}x_n,...,a_{m,1}x_1+...+a_{m,n}x_n[/tex][/quote]
=> this follows straight ahead from the linearity, and this is the definition of matrix multiplication?

Hmm...I'm sorry, I think I had better take a closer look at some of the definitions and proofs in my textbook. I think that I am missing something obvious (for some reason, I think I forgot that the value of Tv is defined by the values of T on the basis vectors of v) but I think I can sort through the details myself.

Thank you for your replies though.


trigonometry
ratio
geometry
perimeter
similar triangles
Pythagorean Theorem

Consider a $\triangle {ABC}$, with $AB \perp AC$. We know that $BC=24$. Consider a point $P$ in $AB$, another point $Q$ in $BC$ and another point $R$ in $AC$, such that $APQR$ is a square of side $7$. Find $AB+AC+BC$.

[hide="hint"]
Use similar triangles.
[/hide]

No one??... [hide="Well, my hint"]Form a cuadratic....[/hide]

[hide] Using similar triangles, we get $7x=(x-7)\sqrt{24^2-x^2}$. And I don't feel like solving that. [/hide]

I'm pretty sure my equation is correct, since I got that same equation with both similar triangles and solving with sum of the areas.

I got an answer...but only with the help of a graphing calculator. [hide=":("] Legs are $16\pm4\sqrt2$. Answer: $\boxed{56}$ [/hide]

[hide="Solution (not the best)"]
Let $\angle ACB = \theta$.  Then

$24 = 7 \sec \theta + 7 \csc \theta$

And

$AB + BC + CA = 38 + 7 \tan \theta + 7 \cot \theta$

Let $x = \tan \theta + \cot \theta$.  Now, $\frac{ \cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} = \frac{1}{\cos \theta \sin \theta}$.  Moreover,

$\sec \theta + \csc \theta = \frac{\cos \theta + \sin \theta}{\cos \theta \sin \theta} = \frac{24}{7}$

Squaring,

$\frac{1 + 2 \sin \theta \cos \theta}{\sin^2 \theta \cos^2 \theta} = \frac{24^2}{7^2}$, or

$x^2 + 2x - \frac{24^2}{7^2} = 0$
$(x + \frac{32}{7})(x - \frac{18}{7}) = 0$

We find our positive solution at $x = \frac{18}{7}$, and therefore

$AB + BC + CA = \boxed{ 56 }$. [/hide]

[quote="calculuslover800"]Would you explain in more detail how you got that equation?

The largest of the roots that my computer could find is 21.6569, which is less than the length of the longest side, so it can't be correct.  Here I'm assuming x is the length of the sum of the sides, do you mean that x is the length of one of the shorter sides?  If so, which one, because the other positive root is 10.3431.[/quote]

x is one of the legs. Both of the numbers you gave are correct. [hide] I drew a right triangle with 24 as the hypotenuse, x as a leg and $\sqrt{24^2-x^2}$ as the other leg. Then drawing in the square, I used similar triangles with one of the smaller triangles (the one touching the leg of length x) and the large right triangle. The small right triangle has legs $x-7$ and $7$ and the large right triangle has legs $x$ and $\sqrt{24^2-x^2}$. I put these in a ratio and cross-multiplied. :D [/hide]

Uh....this solution was pretty lucky...

[hide]
Looking at the triangle, we can compare two similiar triangles, QPB and CQR.

Setting $CR = x$ and $BP=y$,

We see that $\frac{y}{7}=\frac7x \rightarrow xy=49$.

Now, using Pythagorean Theorem, we get

$(x+7)^2+(y+7)^2=24^2 \rightarrow x^2+14x+y^2+14y+98=576$

$2xy=98$ So, $x^2+14x+y^2+14y+98=576 \rightarrow x^2+14x+y^2+14y+2xy=576$

$(x+y)^2+14(x+y) = 576 \rightarrow (x+y)(x+y+14)=576 \rightarrow x+y=18$

So, the perimeter is $18+7+7+24\rightarrow \boxed{56}$

[/hide]

BTW, a silly question....how do I box answers again?

[quote="Zellex"]BTW, a silly question....how do I box answers again?[/quote]
\boxed{56} :D

[quote="quantabe"][quote="José"][quote="quantabe"] How i did it is my secret.[/quote]

:spam:

Then, if you don't post your solution, the post is nonsense, because we can't see why you got that......[/quote]

ok ok i edited my post  :D[/quote]

Come on man!!!! Post your answer! do not be silly!

let the legs be $x$,$y$, then by similar triangles, we have

$\frac{x-7}{7}=\frac{7}{y-7}$
$xy-7(x+y)=0$
also by pythagorus, $(x+y)^2-2xy=24^2$
$(x+y)^2-14(x+y)-24^2=0$
$x+y=32$, it follows the answer is $56$


della works at the fish store. Today she is filling 3 fish tanks with water. Together the 3 tanks hold 29 1'4 gallons. Tank b holds half as much water as tank a, and tank c holds 8 quarts less that tank a. How many gallons of water does each tank hold?   please help i need it pretty quickly if possible thank you so much

put everything is terms of a

$ a\plus{}\frac {1} {2} a\plus{}a\minus{}8\equal{}29.25$ i dont know what you meant when you said 29 1'4
$ \frac{5} {2} a\minus{}8\equal{}29.25$
$ \frac{5} {2}a\equal{}37.25$
$ a\equal{}14.9$
then put the other variables in to get your answer
$ a\equal{}14.9$
$ b\equal{}7.45$
$ c\equal{}6.9$

He probably meant to say $ 29 \frac {1}{4}$ like abacadaea has in his solution.


Putnam
inequalities
number theory unsolved
number theory

Find the smallest positive integer $n$ with the following property:

For any integer $m$ with $0 < m < 2004$, there exists an integer $k$ such that

\[\frac{m}{2004}<\frac{k}{n}<\frac{m+1}{2005}.\]

He, He! :)
This looks like the Putnam 2005 A1 (I think :D )
I think that $\frac{4}{5}$ would do :?
[Edit: Sorry, I didn't read the question carefully :blush: , but it is still similar tu the Putnam Question....]

I think [tex]2004\frac{1}{2}[/tex] will do too!  :D

1) I see why n=4009 is a minimum possible n for m=2003
2) and I can show that n=4009 will do for any other 0 < m < 2004
therefore n=4009

But my way of proving it is quite long, I probably miss some obvious thing. Can anybody show quick solution?

To show that $4009$ suffices, just use that $\frac{a+c}{b+d}$ lies between $\frac{a}{b}$ and $\frac{c}{d}$ for any positive integers $a,b,c,d$.
To show that it is the smallest: take $m=2003$ and do some boring work.

Well, it's no surprise I could not solve this question without doing "boring work", because even after you've showed me $\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}$ I could not realize that the inequality is true until I did some boring work with it :)
I have to train myself on it...

Thanks, ZetaX

Also you can use this lemma:
Let $ a,b,c,d,p,q$ positive integers such that $ \frac{a}{b}<\frac{p}{q}<\frac{c}{d}$ and $ bc\minus{}ad\equal{}1$. Then $ p\geq{a\plus{}c}$ and $ q\geq{b\plus{}d}$.
Proof: It's easy for the reader

Put $ m\equal{}2003$, as $ 2004.2004\minus{}2003.2005\equal{}1$ thus $ n\geq{2004\plus{}2005}$. Hence $ n\equal{}4009$.


geometry
geometric transformation
reflection
incenter
geometry unsolved

Prove that the triangle formed by the points of contact of the sides of a given triangle with the ex circles corresponding to these sides is equivalent to the triangle formed by the points of contact of the sides of the triangle with the inscribed circle.

There is a more general result, which states that the cevian triangles of two isotomic points with respect to a given triangle have the same area. Note that the Nagel point and the Gergonne point are obviously isotomic conjugates. Thus, the intouch and extouch triangle have equal area. For a proof of the referred property see the topic [url=http://www.artofproblemsolving.com/viewtopic.php?t=295353]collinear-shariguin[/url].

P.S. Alternatively, note that the extouch triangle is the pedal triangle of the Bevan point, which is the reflection of the incenter about the circumcenter. Then use Euler's theorem.


Prove that \[ e^{it}+e^{is}=2\cos\left(\frac{t-s}{2}\right)e^{i\left(\textstyle\frac{s+t}{2}\right)}.  \]

$s=\frac{s+t}{2}+\frac{s-t}{2},\ t=\frac{s+t}{2}-\frac{s-t}{2}.$

Euler' formula; $e^{ix}=\cos x+i\sin x.$


I used the following lemma (generalised AM-GM):                         ((a^p+b^p+c^p)/3)^1/p >= ((a^q+b^q+c^q)/3)^1/q when p>=q. What we must prove is E=(x+y+z)(a^3/x+b^3/y+c^3/z) >= (a+b+c)^3/3. By applying Cauchy we get E >= E1= (a^(3/2)+b^(3/2)+c^(3/2))^2. Let (s, t, v) = (a^3, b^3, c^3). Then apply the lemma for s, t, v, p=1/2, q=1/3 and we get E1 >= ((a+b+c)^3)/3, which is exactly what we want.

Notice that by the same manner we can obtain the generalisation a1^m/x1+..+an^m/xn >= (a1+..+an)^m/n^(m-2)*(x1+..+xn). For n=3 and m=3 we get this problem.

Use Holder's inequality we have:
$ (\frac{a^3}{x}\plus{}\frac{b^3}{y}\plus{}\frac{c^3}{z})^{1/3} (1\plus{}1\plus{}1)^{1/3}(x\plus{}y\plus{}z)^{1/3} \geq a\plus{}b\plus{}c$
Done!

[quote="Moubinool"]Prove that

a^3 /x + b^3/y + c^3/z  \geq  (a+b+c)^3/(3(x+y+z))

for all positive real numbers a,b,c,x,y,z[/quote]
Prove that \[\frac{a^3 }{x} +\frac{ b^3}{y }+ \frac{c^3}{z } \geq  \frac{(a+b+c)^3}{3(x+y+z)}.\]
for all positive real numbers   $ a,b,c,x,y,z$.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=419938&p=3753812#p3753812]2009 NST[/url]


$ Problem$

Enumerate the number of $ 2$-regular simple graphs.

(i) Vertices are numbered.
(ii) Up to Isomorphism.

A graph $ G(v,e)$with vertices is called $ k$-regular if degree of every vertex is $ k$.

[b]A approach:[/b] Every $ 2$-regular graph can be written as union of disjoint vertex cycles of length $ \geq 3$. From this [url=http://www.mathlinks.ro/viewtopic.php?t=196533]post[/url] we know a closed form to express a finite set as union of disjoint cylces but that includes singletons and transpositions. Do we have a way of counting the cycles excluding singletons and transpositions ?

Let us count the possible cycles including a vertex say vertex 1. Let $ f(n)$ denote the number $ 2$-regular graphs of order $ n$.

Let $ P_k$ be a permutation of $ k$ vertices from $ \{1,\dots,n\}$. We construct a cycle starting vertex 1, then first element of $ P_k$, then the second element of $ P_k$ and so on upto the $ k$ th element of $ P_k$ element connects back to vertex1. (The reverse of the $ P_k$ also gives rise  to same cycle). There exists a bijection between a $ P_k$ (and the reverse of $ P_k$) and cycles of length $ k\plus{}1$.

$ \frac{(n\minus{}1)(n\minus{}2)}{2} f(n\minus{}3) \plus{} \frac{(n\minus{}1)(n\minus{}2)(n\minus{}3)}{2} f(n\minus{}4) \plus{}\dots \plus{} \frac{(n\minus{}1)\dots 4}{2} f(3) \plus{} \frac{(n\minus{}1)\dots 1}{2} \equal{} f(n)$

$ f(n)\equal{} \frac{(n\minus{}1)!}{2} \sum_{k\equal{}3}^{n} \frac{f(n\minus{}k)}{(n\minus{}k)!}$

How can we proceed from here ?


algebra
polynomial
function
vector
Ring Theory
superior algebra
superior algebra unsolved

(a) Prove that for every $ q$ prime power, one has that for all $ n$ there exists an irreducible polynomial $ P(x)\in\mathbb{F}_q[x]$ such that $ \deg P\geq n$.
(b) Is (a) true if we change the condition $ \deg P\geq n$ by $ \deg P \equal{} n$?

[Moderator edit: this proof is wrong, and see one of Jmerry's posts at http://www.mathlinks.ro/viewtopic.php?t=247222 for corrections.]

I will prove that irreducible polynomials exist by counting them. The number of monic irreducible polynomials of degree $ n$ is $ \frac1n\sum_{k|n}q^k\mu\left(\frac nk\right).$ This is a multiplicative function which takes on the value $ \frac {q^{r^j} \minus{} q^{r^{j \minus{} 1}}}{r^j}$ at a prime power $ r^j,$ so it is never zero.

Let $ I(k)$ be the number of monic irreducible polynomials of degree $ k.$ Since every monic polynomial factors uniquely as a product of irreducible polynomials, we have
$ \prod_{k \equal{} 1}^\infty \frac1{\left(1 \minus{} x^k\right)^{I(k)}} \equal{} \sum_{n \equal{} 1}^\infty q^nx^n$.
The right side counts the polynomials of degree $ n$ directly; the left asks how many copies of each irreducible polynomial are in the factorization, so an irreducible polynomial of degree $ k$ contributes a factor of $ \sum_{j \equal{} 0}^\infty x^{jk}.$

Now, take derivatives. The left side gives
$ \left(\sum_{k \equal{} 0}^\infty kI(k)\frac {x^{k \minus{} 1}}{1 \minus{} x^k}\right)\cdot\left(\prod_{k \equal{} 1}^\infty \frac1{\left(1 \minus{} x^k\right)^{I(k)}}\right)$
and the right side gives $ \sum_{n \equal{} 0}^\infty nq^nx^{n \minus{} 1} \equal{} \frac q{(1 \minus{} qx)^2}.$
Multiplying by $ x$ and dividing by the original equation, we have
$ \sum_{k \equal{} 0}^\infty kI(k)\frac {x^k}{1 \minus{} x^k} \equal{} \frac {qx}{1 \minus{} qx} \equal{} \sum_{n \equal{} 1}^\infty q^nx^n.$
Equating coefficients then gives $ \sum_{k|n}kI(k) \equal{} q^n$ and the Mobius inversion formula gives $ I(n) \equal{} \frac1n\sum_{k|n}q^k\mu\left(\frac nk\right).$

The other interpretation here is that the field with $ q^n$ elements is the universal splitting field for irreducible polynomials of degree dividing $ n$- the computations here can actually be used to prove that every such polynomial splits completely in that field and that every element of the field is a root of such a polynomial (including the base field as roots of irreducible linear polynomials).

[quote="djimenez"](a) Prove that for every $q$ prime power, one has that for all $n$ there exists and irreducible polynomial $P(x)\in\mathbb{F}_q[x]$ such that $\deg P\geq n$.
(b) Is (a) true if we change the condition $\deg P\geq n$ by $\deg P = n$?[/quote]

How do you talk about $F_q$ without already knowing existence of such $P$???

In order to construct $F_q$, we use the existence of irreducible polynomials over $F_p$; that doesn't say anything about irreducible polynomials over $F_q$.

Start with the field of integers mod $p$. We prove the existence of irreducible polynomials over this field, and use construct a field with $q$ elements as the field generated by a root of such a polynomial. That's enough to start the problem.
The universal splitting field property, which follows from my computation, allows us to say that the field is unique up to isomorphism- and this computation also establishes the property for $F_{q^n}/F_q$.

I can prove the only algebraic fact I need here by running this argument at primes first.

[quote="fleeting_guest"][quote="jmerry"]In order to construct $F_q$, we use the existence of irreducible polynomials over $F_p$; that doesn't say anything about irreducible polynomials over $F_q$.[/quote]

Yes, but any argument for constructing irreducible polynomials of varying degree $n$ over $F_p$ is already a construction for $F_q$.   As far as I can tell, there is no way of constructing $F_q$ without having already solved the same problem up to a change of notation $p \to q$.[/quote]

Could you please explain what are you talking about.  That is, please, rephrase what you wanted to say, because, I think it is not clear (at least to me).

It is true that in order to prove that $F_q$ exists for every $q=p^n$, where $p$ is prime and $n$ is a positive integer, you have to prove that there are primitive polynomials of degree $n$ (at least that is how it goes the only proof that I have seen).

An easy way to construct $F_{q^n}$ from $F_q$ (regardless if $q$ is a prime or a prime power) is just to take a primitive polynomial $P_n(x)\in F_q[x]$ of degree $n$, and $F_{q^n}\equiv F_q[x]/(P(x))$. Equivalently (if you need a nice computational model) you can represent $F_{q^n}$ as $F_q^n$ (the $n$-dimentional vector space over $F_q$) where the addition in the field is the addition of the vector space and the multiplication is basically the multiplication of polynomials, and after that you take mod $P_n(x)$. What I mean as multiplication of polynomials is that you may think about the $n$-tuple $(a_0,a_1,\dots,a_{n-1})$ as $a_0+a_1x+\cdots+a_{n-1}x^{n-1}$.

Best,

[quote="djimenez"][quote="fleeting_guest"] any argument for constructing irreducible polynomials of varying degree $n$ over $F_p$ is already a construction [over] $F_q$.   As far as I can tell, there is no way of constructing $F_q$ without having already solved the same problem up to a change of notation $p \to q$.[/quote]

Could you please explain ....  It is true that in order to prove that $F_q$ exists for every $q=p^n$, where $p$ is prime and $n$ is a positive integer, you have to prove that there are primitive polynomials of degree $n$ [/quote]

My point was this:
take any construction of finite fields $F_q$.  This requires finding irreducible polynomials of degree $n$ over $F_p$ (for an infinite set of $n$, such as the set of all positive integers).  For any such construction, what needs to be a prime (rather than a prime power) $p$ is the characteristic of the field, not its number of elements. One can substitute $q = p^k$ in the construction with no problems, and this immediately produces irreducible polynomials of the same degrees over $F_q$.

The set of irreducible polynomials is an algebraic variety, which acquires more points for larger $q$.
Also the structure of the finite fields -- cyclic multiplicative group, Frobenius automorphism, Galois groups -- is uniform in $q$;  it's hard to find proofs that work for prime fields but not all finite fields of the same characteristic.

Sorry if I misunderstood something, but why do you need an irreducible polynomial to construct $\mathbb{F}_{p^n}$¿
(Because you can construct $\mathbb{F}_{p^n}$ as spliting field of the polynomial $x^{p^n}-x$ in $\mathbb{F}_{p}$)

(1) I think that irreducible polynomials are not enough, at least not of a construction of the field that is worth to do computations. I think you need primitive polynomials!

(2) Yes, $\mathbf{F}_{p^n}$ is the spliting field of $x^{p^n}=x$ over $\mathbf{F}_{p}$, but, can you explicitely construct this field just with that polynomial, with the sum and multiplication tables? Probably not! On the other hand, there is a theorem that says that if $R$ is a commutative ring and $I$ is a maximal ideal (not prime, but maximal, that is the difference between irreducible and primitive polynomials), then $R/I$ is a field. It is not hard but tedious to prove that $\mathbf{F}_{p^n}$ may be constructible as $\mathbf{F}_{p}[x]/\langle P_n(x)\rangle$ for any primitive polynomial $P_n$  of degree $n$. Of course, if you use different primitive polynomials, you are going to get different tables of addition and multiplication, but at the end, they are isomorphic.

Best,

1.: In a field, I don't see the difference between primitive and irreducible...  :?
2.: My answer was more to this (I think with constructing in this case proving the existence is meant ¿):

[quote="jmerry"]In order to construct $F_q$, we use the existence of irreducible polynomials over $F_p$; that doesn't say anything about irreducible polynomials over $F_q$.[/quote]

[quote="fleeting_guest"]As far as I can tell, there is no way of constructing $F_q$ without having already solved the same problem up to a change of notation $p \to q$.[/quote]

[quote="djimenez"]
It is true that in order to prove that $F_q$ exists for every $q=p^n$, where $p$ is prime and $n$ is a positive integer, you have to prove that there are primitive polynomials of degree $n$ (at least that is how it goes the only proof that I have seen).[/quote]

3. can you construct it explicitly (giving the tables) with only knowing that an irreducible polynomial exists¿

[quote="ZetaX"]1.: In a field, I don't see the difference between primitive and irreducible...  :?
[/quote]

Sorry, may be I messed up the definition.  Basically a primitive polynomial of degree $n$ over $\mathbb{F}_q$ is an irreducible factor of $x^{q^n-1}-1$. I don't remember the details and I am a little lazy to look for them right now, specially because my algebraic code theory is in other place, but the fact is important.

[quote="ZetaX"]3. can you construct it explicitly (giving the tables) with only knowing that an irreducible polynomial exists¿[/quote]

Well, I don't think so, but you may try! The construction I know don't just need the existence of a primitive polynomial, it really need the polynomial itself!


function
geometry
circumcircle
national olympiad

Hi all !
These are the problems from Bulgarian National Olympiad 2006. Enjoy!

[b]Problem 1.[/b] Consider the set $A=\{1,2,3,4,\dots ,2^n\}$, $n\geq 2$. Find the number of its subsets $B \subseteq A$, such that if $x$ and $y$ are different elements of $A$, with sum a power of two, then exactly one of $x$ and $y$ is in $B$.

[i]Aleksandar Ivanov[/i]

 [b]Problem 2.[/b] Let $R^+$ be the set of the positive real numbers and $f: R^+\to R^+$ is functions, such that $f(x+y)-f(x-y)=4\sqrt{f(x)f(y)}$ for all $x>y>0.$
 a) Proove that $f(2x)=4f(x)$ for all $x\in R^+$.
 b) Find all such functions.

[i]Nikolai Nikolov, Oleg Mushkarov[/i]

[b] Problem 3.[/b] The natural numbers are written in sequence, in increasing order, and by this we get an infinite sequence of digits. Find the least natural $k$, for whom among the first $k$ elements of this sequence, any two nonzero digits have been written different times.

[i]Aleksandar Ivanov, Emil Kolev[/i]

[b]  Problem 4.[/b] Let $p$ be prime, such that $p^2$ divides $2^{p-1}-1$. Proove that for any natyral $n$ the number $(p-1)(p!+2^n)$ has at least three different prime divisors.

[i]Aleksandar Ivanov[/i]

 [b]Probem 5.[/b] Given is $\t ABC$, and $ a BAC=30^\circ$ and $ a ABC=45^\circ$. Consider all couples of points $X$ and $Y$, such that $X$ and $Y$ are respectively on the rays $AC^{\rightarrow}$ and $BC^{\rightarrow}$, and $OX=BY$, where $O$ is the circumcenter of $\t ABC$. Proove that the bisectors of the segments $XY$ pass through fixed point.

[i]Emil Kolev[/i]

 [b]Problem 6.[/b] Point $O$ is given in the plane. Find all sets $S$ with more than one point in $S$, such that for any $A\in S$, $A\neq O$, the circle with diameter $OA$ is in $S$.

[i]Nikolai Nikolov, Slavomir Dinev[/i]

Here are some links for the topics where they are posted:
1. http://www.mathlinks.ro/Forum/viewtopic.php?p=530892#530892
2. http://www.mathlinks.ro/Forum/viewtopic.php?p=530894#530894
3. http://www.mathlinks.ro/Forum/viewtopic.php?p=530895#530895
4. http://www.mathlinks.ro/Forum/viewtopic.php?p=530898#530898
5. http://www.mathlinks.ro/Forum/viewtopic.php?p=530900#p530900
6. http://www.mathlinks.ro/Forum/viewtopic.php?p=530904#p530904

Dima was this problems proposed for a day or ther were two days,with 3 problems every day?
And how much time have you had to solve this problems?

The problems were given in IMO format.
This means: two days, each day 3 problems, each day 4h 30min for the problems

Hey! the sums are really good ,I will try the second one and get back to u in a day or 2


prove that every planar graph is 5-colorable.

This is exercice #27 here  :D

http://www.animath.fr/cours/graphes.pdf

Pierre.

Pierre one day you have to translate some of your papers to english!! especially the graph one,
 :D  everybody says its good, but i can't read it, though my first language is spanish...

I think you should better learn french whereas waiting for me to translate it... :P

Pierre.

Pierre ,I can see ,you are like Myth too :D [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=covering&t=1896[/url]

From post 7.

Who said that I have an heart??? :P

Pierre.

I do believe ,you have Mr.bornsztein.
I by myself learned a lot from your solutions and elegant ideas on the forum.

I can read it, but it's a full page :? but I indeed wonder why people don't translate their work into english if it's good enough for world level. :)


Treething bangs his head on another tree when he brings home a Tree-C from TreeSchool one day. A Tree-C lets you draw numbers on a board. Then, he draws a word SPAMMYISBEST. How many words using the letters from $SPAMMYISBEST$ can be made that are before $SPAMMYISBEST$ ( assuming all the words formed from those $12$ letters are all words ) in the alphabet?

I assume that only words alphabetically before SPAMMYISBEST count?

[quote="Treething"]$\frac{12!}{3!2!}= 39916800$

'cept i can only draw numbers, and spammyisbest is made of letters[/quote]

Hm, how did you get that?

He didn't read the problem.

His solution is the number of ways to order 12 letters where letters are viewed as indistinguishable.

There are 12! ways to order all the letters, but there are three S's and two M's, so we need to divide by 2!3!.

However, the problem has an alphabetical stipulation on it.


geometry
trigonometry
arithmetic sequence

(a) The sides of the triangle $ ABC$ have lengths $ BC \equal{} a, AC \equal{} b$ and $ AB \equal{} c$, where $ a,b,c$. The sizes of the angles $ A, B$ and $ C$ form an arithmetic sequence.

(i) Show that the area of triangle $ ABC$ is $ ac\frac{\sqrt{3}}{4}$.

Given that $ a\equal{}2$ and $ \sin A \equal{} \frac{\sqrt{15}}{5}$, find

(ii) the value of $ b$,

(iii) the value of $ c$.

(b) The internal angles of an $ n$-sided polygon form an arithmetic sequence with first term $ 143$ degrees and common difference $ 2$ degrees.

Given that all of the internal angles are less that $ 180$ degrees, find the value of $ n$.

a)
 i) If A, B, C are in arithmetic progression then $ B\equal{}\frac{A\plus{}C}{2}$.
But $ A\plus{}B\plus{}C\equal{}\pi$. Then $ B\equal{}\frac{\pi}{3}$
Now use the formula $ A_{ABC}\equal{}\frac{ac\sin B}{2}$

ii), iii)
Use $ \frac{a}{\sin A}\equal{}\frac{b}{\sin B}\equal{}\frac{c}{\sin C}$ and $ \sin C\equal{}\sin(A\plus{}B)$

Summing the interior angles in two ways:

$ n(143\cdot2\plus{}2n\minus{}2)/2\equal{}180(n\minus{}2)$

$ n^2\minus{}38n\plus{}360\equal{}0\implies n\equal{}19\pm1$

but $ 143\plus{}2(18\minus{}1)<180,143\plus{}2(20\minus{}1)>180$ so $ n\equal{}18$


quadratics
number theory proposed
number theory

[quote="Crux"]Can someone give me a proof of the fact that any prime $ p$ dividing $ 4n^2 \plus{} 1$ for $ n$ an integer verifies $ p \equiv 1 \mod 4$
using only basic notions, I mean without quadratic residues ?[/quote]
We will prove that if $ p\in\mathbb{P}$,and $ p|x^2 \plus{} 1$ for some $ x\in\mathbb{Z}$,then $ p \equal{} 4k \plus{} 1$.
Proof:
Since $ p$ is odd,then it can be only $ 4k \plus{} 1$ or $ p \equal{} 4k \plus{} 3$.
Let $ p \equal{} 4k \plus{} 3$,then
$ x^2\equiv \minus{} 1$ $ (mod p)$,hence $ x^{p \minus{} 1} \equal{} (x^{2})^{\frac {p \minus{} 1}{2}}\equiv ( \minus{} 1)^{\frac {p \minus{} 1}{2}} \equal{} \minus{} 1$ $ (mod p)$,
contradiction to the Fermat's theorem.So $ p$ can be only of the form $ 4k \plus{} 1$.

$ (\minus{}1)^{\frac{p\minus{}1}{2}}\equiv (4n^2)^{\frac{p\minus{}1}{2}}(mod\ p)\equiv (2n)^{p\minus{}1}(mod\ p)\equiv 1(mod\ p)$ gives $ \frac{p\minus{}1}{2}$ even


search
quadratics
trigonometry
geometry
circumcircle
3D geometry
prism

Construct a quadrilateral which is inscribed and circumscribed, given the radii of the respective circles and the angle between the diagonals of quadrilateral.

[color=darkblue] Let $ O$ be the center of the circumcircle. From $ OI \equal{} \sqrt {R \plus{} r^2 \minus{} r\sqrt {4R^2 \plus{} r^2}}$, we can construct the point $ I$ - the incenter.

 If  $ P$ is the intersection of the diagonals,  then $ OP \equal{} \dfrac{2R^{2}d}{R^{2} \plus{} d^{2}}$ (proved  [url=http://www.mathlinks.ro/viewtopic.php?search_id=1422695076&t=101127]here[/url]}). So, the point $ P$ is also constructible. Points $ P$, $ I$, $ O$ lie on a line in this order.

 If $ K$ and $ L$ are midpoints of the diagonals $ AC$ and $ BD$, then $ OL \perp BD$ and $ OK \perp AC$, and we deduce, that quadrilateral $ OLKP$ is inscribed in a circle $ \omega$ with diameter $ PO$ (this circle can be constructed). Points $ K$, $ L$ and $ I$ lie on the Newton's line. We need to construct a line through $ I$, which cuts $ \omega$ at points  $ K$ and $ L$, such that $ \angle{KPL} \equal{} \alpha$, where $ \alpha$  - the given angle betwwen diagonals. If $ M$ is midpoint of $ PO$, then $ \angle{KML} \equal{} 2\alpha$, and our problem is to construct the isosceles triangle $ MKL$ with known vertex $ M$, angle $ \angle{KML}$, length of the side $ KM$ and point $ I$ on the side $ KL$.

  Apply Stewart in  $ \triangle{KML}$ for $ MI$ and get

 $ \small{MI^2 \equal{} \dfrac{MK^2\cdot IL \plus{} ML^2\cdot IK \minus{} IK\cdot IL \cdot KL}{KL} \equal{} MK^2 \minus{} IK\cdot IL \equal{} MK^2 \minus{} IK(KL \minus{} IK) \equal{} }$

  $ \small{ \equal{} MK^2 \plus{} IK^2 \minus{} IK\cdot KL \Longleftrightarrow IK^2 \minus{} IK\cdot KL \minus{} MI^2 \equal{} 0}$

We solve the quadratic $ IK$. Note that $ KL \equal{} 2MK\sin \alpha$. Because the construction is possible, the equation has two roots. But just one satisfies us because $ x_1\cdot x_2 \equal{} \minus{} MI^2 < 0$. So we can find $ IK$. We construct the circle $ \Omega$ with the center $ I$ and radius $ IK$. Clearly, circles $ \omega$ and $ \Omega$ intersect at $ K$ (there exist two such $ K$s, we''l continue with one of them). The line $ KI$ meets $ \omega$ at $ L$.

Line $ PL$ meets the circumcircle at $ B$ and $ D$, but line $ PK$ at points $ A$ and $ C$.

The construction is finished.[/color]

Let $A,B,C,D$ be the vertices of the wanted quadrilateral with incircle $ (I)$ and circumcircle $ (O).$ According to [i]Fuss relation[/i] (mentioned by Ahiles), distance $ IO$ is constructible, then we can place the incircle and the circumcircle. According to Poncelet porism, $O,I$ and diagonal intersection $P \equiv AC \cap BD$ are fixed for all quadrilaterals with incircle $(I)$ and circumcircle $(O)$ $\Longrightarrow$ $P$ is constructible. Let $ M,N$ the orthogonal projections of  $ O$ on $AC,BD,$ it follows that $ OM$ and $ ON$ are perpendicular bisectors of $AC,AB$ and $ MN \equiv n $ is the Newton line of the quadrangle $ ABCD$ passing through its incenter $I.$ Hence it suffices to draw the line $n$ passing through $ I$ such that $ \angle MPN$ is given. If $K$ is center of the circle with diameter $ OP,$ then central angle $ \angle MKN$  is known as well $\Longrightarrow$ Isosceles triangle $ \triangle MKN$ is constructible. Placing the constructed chord $ MN$ in the circle $ (K)$  passing through $ I$ completes the quadrilateral $ABCD$ easily.

Ok let \(\angle A, \angle B\) be the two adjacent angles of the quadrilateral, draw a \(\Delta ABC\) with the given two angle measures. Let \(A'B'C'\) be the intouch triangle of \(\Delta ABC\), let \(X \in A'B'\)\(: C'X \perp B'A'\), let \(AX \cap BC =D, BX \cap CA =E, \square ACDE\) is the required .


A $4*13$ board is given, and $52$ usual playing  cards are also given (4 swits, and 13 numbers on each swit). We have to put this cards in the board one card in each square in such a way that if 2 squares share a side, their cards have either the same swit, or the same number. Find all ways to do it.

If I remember correctly, each row must contain a swit (as you call it :); is that the correct word?).

yes, and i dont know if it is swit or suite ;)  ;)
but Grobber, can you prove it?

[quote="Pascual2005"]Grobber, i am waiting for your solution![/quote]

And I am waiting that you are stating full sources ! TOT have a year, round (spring/autumn), problem number etc. I think you guys don't need to wonder in the case of reposting.


trigonometry
AMC
AIME
function
algebra
domain
trig identities

[quote="vamathletes"]I understand kohjhsd's solution, but what is Stewart's Theorem?[/quote]

Stewart's Theorem is a relation between the sides of a triangle and the length of a segment from a vertex to a point on the opposite side.

Let a, b, and c be the sides of a triangle, and Let p be a segment from vertex A to a point on the opposite side dividing it into x and y. Then,

[img]http://upload.wikimedia.org/math/0/c/4/0c4494e6b6f8b99efe18352144f70161.png[/img]

[img]http://upload.wikimedia.org/wikipedia/commons/thumb/7/77/Stewart%27s_Theorem.svg/407px-Stewart%27s_Theorem.svg.png[/img]

A better way to memorize this is

$dad+man=bmb+cnc$  :rotfl:

when (based on frostmourne's diagram)

$m=x$, $n=y$, $d=p$

And Stewart's Theorem saves you a lot of trouble when your teacher gives you three pages of Law of Cosines homework.  :wink:

$dad+man=bmb+cnc$

By the way you pronounce this

"Dad plus man equals bomb plus sink(cnc)"

Not only is this formula hilarious, it can be useful in any occasion where you have a triangle and a line connecting a vertex with the opposite side. You can try to prove this easily using the Law of Cosines.

That Stewart guy must be one heck of a character.  :lol:

how does the law of cosines help with this? stewarts has nothing to do with angles  :huh:

[quote="frostmourne"]
[img]http://upload.wikimedia.org/math/0/c/4/0c4494e6b6f8b99efe18352144f70161.png[/img]

[img]http://upload.wikimedia.org/wikipedia/commons/thumb/7/77/Stewart%27s_Theorem.svg/407px-Stewart%27s_Theorem.svg.png[/img][/quote]

Let $M$ be the point the line segment from $A$ intersects $BC$. Let $\theta = \angle AMB$ then $180^{\circ}-\theta = \angle AMC$. Now use the fact that $\cos \theta+\cos (180^{\circ}-\theta) = 0$.

[quote="Andrew Kim"]$dad+man=bmb+cnc$

By the way you pronounce this

"Dad plus man equals bomb plus sink(cnc)"[/quote]

The improved version that I learned was "command 'n conquer plus bomb = dead man."

I always learned

"A bomb in the sink = a dead man"

[quote="cincodemayo5590"][quote="Treething"]A man and his dad built a bomb in a sink.

there.[/quote]

Crazy terrorists...[/quote]

Or brilliant geometers?
Is this also called Apollonius' Theorem? I think I've seen it somewhere else too...

And good call on the law of cosines applications. I don't think I've ever blitzed through so much useless trig so quickly.

Off topic,
I'm getting bored in Algebra II.... My AIME score and practices make my school class look like a fifth grade math class.

True class excerpt today. We were doing elementary trig, and our teacher was telling us that the sin function had a domain over all reals, which was, of course, incomplete information, as usual... He neglected to tell us/them that it could take complex values too, but... whatever...

Teacher: You can put anything into the $\sin$ function, even the kitchen sink!
Student: (types intently into TI-84)
Student: That doesn't work.
Student: (holds up calculator screen, I see it)
Student: (has typed the following:)
$\sin(\text{TheKitchenSink})$
$\text{DOMAINERROR}$

This is serious. It wasn't meant to be funny. He was actually confused why it didn't work.

Yep, that's all we do in Algebra Two Honors. I thought they were supposed to WEED out the people like that...

[quote="DiscreetFourierTransform"][quote="cincodemayo5590"][quote="Treething"]A man and his dad built a bomb in a sink.

there.[/quote]

Crazy terrorists...[/quote]

Or brilliant geometers?
Is this also called Apollonius' Theorem? I think I've seen it somewhere else too...[/quote]
I think that Apollonius is when m=n

Lol DiscreetFourierTransform, sounds like my Precalculus Honors class to me. Ah well, this is what AoPS is for, isn't it?

Yep.
That's why we have the AMC, AIME, Mandelbrot, CAML, ARML...
Need I go on? Just let me say, thank goodness for AoPS.

Apollonius also has a circle which belongs to him... it's called Apollonius's Circle. It's a good thing he decided to share it.