Two circles $ w_1,w_2$ in the plane centered at $ A,B$ intersect at $ C,D$. Suppose the circumcircle of $ ABC$ intersect $ w_1$ at $ E$ and $ w_2$ at $ F$, where the arc $ EF$ not containing $ C$ lies outside $ w_1,w_2$. Prove that this arc $ EF$ is bisected by line $ CD$.

I've seen somewhere this problem. As I remember it is from Moscow Olympiad.

[b][color=darkblue]$ AEBC$ - cyclic, so $ \angle{CBE} = 180^\circ - \angle{EAC} = 180^\circ - \alpha$.

$ \left\| \begin{array}{c} \angle{EDC} = \180^\circ - \frac {1}{2}\angle{EAC} = 180^\circ - \frac {1}{2}\alpha \\
 \\
\angle{CDB} = 90^\circ - \frac {1}{2}\angle{CBD} = \frac {1}{2}\alpha. \end{array} \right\| \Rightarrow \angle{EDB} = 180^\circ \Rightarrow E,D,B$ are collinear.

By analogy $ A,D,F$ are collinear.

So $ \angle{EDA} = \angle{FDA} \Rightarrow \triangle{EDA} \sim \triangle{FDB} \Rightarrow \angle{EAD} = \angle{DBF}$.

$ \left\| \begin{array}{c} \angle{ECD} = \frac {1}{2}\angle{EAD} \\
 \\
\angle{DCF} = \frac {1}{2}\angle{DBF} \end{array} \right\| \Rightarrow \angle{ACD} = \angle{DCF} \Rightarrow \widehat{EH} = \widehat{HF}$, q.e.d.

[/color][/b]
[img]http://i014.radikal.ru/0805/4c/c0053b9cd713.jpg[/img]

[quote="Ahiles"]
$ \angle{CDB} \equal{} 90^\circ \minus{} \frac {1}{2}\angle{CBD} \equal{} \frac {1}{2}\alpha.$
[/quote]

I think you're wrong. How do you know it's the same $ \alpha$?

Yes, I understood... Very stupid mistake...

[quote="Ahiles"]No, I'm true. Look:
$ \angle{CBD} \equal{} 180^\circ \minus{} \alpha$, where $ \alpha \equal{} \angle{EAC}$.
[/quote]

How do you know this? You have not known yet that $ E,D,B$ are collinear. You are going to prove it!!

Probably is better to denote the points $ E\equiv (A)\cap BD$ and $ F\equiv (B)\cap AD,$ where $ (A),\ (B),$ are the circles centered at points $ A,\ B,$ respectively.

It is easier now, to prove that the quadrilateral $ AEBC$ is cyclic, because of $ \angle AEB \equal{} \angle ADE \equal{} 180^{o} \minus{} \angle ADB \equal{} 180^{o} \minus{} \angle ACB.$

Similarly the quadrilateral $ AFBC$ is cyclic and then we have $ \angle ECD \equal{} \frac {\angle EAD}{2} \equal{} \frac {\angle FBD}{2} \equal{} \angle FCD$ and we are done.

Kostas Vittas.

a good problem for inversion practicing!
I choose point $ C$ as the center of inversion and an arbitrary radius $ k$.
the problem turnes into this one:
[color=blue]problem:$ \triangle CA'B'$ is given.points $ F',E'$ are the intersections of the prependicular bisectors of the segments$ CB',CA'$,with $ A'B'$,respectively,and meet eachother at $ D'$(infact is the circumcenter of$ \triangle CA'B')$
prove that $ CD'$ is the bisector of $ \angle E'CF'$[/color]
[color=red]solution: it is clear and dont need to prove! [/color]

[quote="vittasko"]Probably is better to denote the points $ E\equiv (A)\cap BD$ and $ E\equiv (B)\cap AD,$ where $ (A),\ (B),$ are the circles centerd at points $ A,\ B,$ respectively.

It is easier now, to prove that the quadrilateral $ AEBC$ is cyclic, because of $ \angle AEB \equal{} \angle ADE \equal{} 180^{o} \minus{} \angle ADB \equal{} 180^{o} \minus{} \angle ACB.$

Similarly the quadrilateral $ AFBC$ is cyclic and then we have $ \angle ECD \equal{} \frac {\angle EAD}{2} \equal{} \frac {\angle FBD}{2} \equal{} \angle FCD$ and we are done.

Kostas Vittas.[/quote]

That's a superb solution, vittasko. Maybe you should try the one I named "equal angle" as well...


Given acute triangle $\triangle ABC$ in plane $P$, a point $Q$ in space is defined such that $\angle AQB = \angle BQC = \angle CQA = 90^\circ.$  Point $X$ is the point in plane $P$ such that $QX$ is perpendicular to plane $P$. Given $\angle ABC = 40^\circ$ and $\angle ACB = 75^\circ,$ find $\angle AXC.$

This one is kind of interesting

[hide="i love these emoticons :starwars:  :juggle:   :jump:  ]Let $L, M$, and $N$ be the midpoints of $AB, BC$, and $AC$, respectively. Well BQC and AQC are right triangles, we have $CM=MQ$, and $QN=NC$. Hence triangle CQM is congruent to triangle CQN, so $\angle QNM = \angle MNC$. Now drop a perpendicular from C to MN, let the foot be H. triangle NHC is congruent to triangle NHQ by SAS, thus $\angle NHC = \angle NHQ$, guess what, QH is an altitude of triangle QMN. Now let $R$ be a plane through H perpendicular to line MN. Let $l$ be a line through H perpendicular to plane P. $l$ is parallel to QX. We know Q lies on the plane $R$, and $l$ lies on plane $R$, hence XQ lies on plane $R$ and XH is perpendicular to MN.   hence C, H, X are collinear so CX is perpendicular to MN. But MN is parallel to AB so CX is perpendicular to AB. By symmetry BX, AX are perpendicular to BC, AC, respectively, yeah X is the orthocenter. Now we just angle chase, angle BAC = 65, angle XCA= 90-65 = 25. angle XAC = 90-75=15. So AXC = 180-25-15=140. [/hide]

Edit- apparently orthogonal projection does not preserve perpendicularity, so I revised the proof...

[hide=Short Solution][b]Construct spheres with centers at the midpoints $AB, AC, BC$(their $2$ intersection points are two possible locations of $Q$ but just pick $1$ since it doesn't matter). Stick to the plane though since that's what we care about and $2$-D cross sections are easier to visualize. As a result $X$ is just the orthocenter thanks to the fact that the spheres intersect the triangles at the feet of the altitudes and if you don't already know this $\angle{AXC}=180^{\circ}-\angle{ABC}$(this is a standard Olympiad Geometry fact) so the answer is $\boxed{40}$.[/b][/hide]


AMC
AIME
USA(J)MO
USAMO
percent
AMC 12 A

about how many people participated in this year's AMC 12.
from last year it was about 90000 on A and 30000 on B. adding up to 120000
but how many of these are people who took the test twice.

[quote]Please note that we were able to score and report on about 128,000 A-date answer sheets from about 2,500 schools in 17 days. Scoring and reporting on approximately 100,000 B-date answer sheets from about 2,000 schools is underway as I write this. I expect that it will take about the same amount of time. The entire AMC staff is well aware of the AIME I dealine and the entire staff is currently working overtime to get the reports out to you as fast as possible.

Steve Dunbar
MAA Director, American Mathematics Competitions [/quote]

So, 128000 for A and 100000 for B. Now, to find the ones that took both, find it out by yourself (it won't be easy..).

It's difficult and time-consuming to tell how many people took both an A-date contest and a B-date contest.  There are several reasons it is difficult and time consuming.
1.  There are four possible combinations of A-date and B-date taking.  That means four times the amount of tracking.
2.  Students have no unique identifier, we only have the name and to some degree the city, state or region to go by.   If Al Smith from New York City takes the 10 A and Alan Smith from New York City takes the 12 B, is it the same person or two different individuals?
3. Most records are recorded by school, for the puposes of determining school winners, team scores, and where to send an AIME.  If a student takes the A-date contest at the school they attend regularly, but takes the B-date contest at (say) a Math Circle or district-wide Math Club, then it is harder to determine  that the same student took the contests twice.  Multiply that by 30 students in a Math Circle attending 20 different high schools and it could be quite time-consuming.  Add to that the fact that it is quite difficult to distinguish (say) a Math Circle registration from private schools and academy registration.

Because the number of AIME qualifiers is smaller, it is less time-consuming to consider that pool of contest-takers.  FOr AIME 2004, I did study the number of AIME qualifiers who took the both an A-date and a B-date contest.  Out of 15,323 qualifiers, 719 qualified on both an A-date and a B-date contest.  That's 4.6%.  Some quick surveys that suggest that about 5% of participating schools take both the A-date and B-date contest confirm that 5% is a reasonable estimate.

We could figure all this out in greater detail, but I believe it is more important to get the A-date, B-date, AIME, USAMO, web-posted statistics and the Summary Book results back to you as quickly as possible.

Steve Dunbar
MAA Director, American Mathematics Competitions

We don't know yet how many students took the AMC 12, A-date and B-date.
As mentioned previously, we are still scoring, and until we finish scoring the answer sheets, we can't know.  Even then, late mailings will trickle until about the end of February, and we get scores for adminstrations in a few foreign countires by a different means.  There is some inevitable clean-up and editing of the score reports which slightly alters the number of contests considered.  Altogether, we won't know the exact number until sometime in mid-March at soonest.  Until, then the estimates I posted yesterday are the best available.

Steve Dunbar
MAA Director, American Mathematics Competitions

Thanks for the estimates, and best wishes in your continued scoring work. I would have guessed a higher percentage of multiple AMC 10/12 takers, so it's interesting that you have two lines of evidence that both suggest about a 5 percent overlap.

So, will the state scores for the A be released soon, or not until after all the scores are compiled?


AMC
AMC 12
number theory
relatively prime

How did you guys do on Fermat/Cayley/Pascal?

i woke up late, was super-sick, missing good-calculator, subway had failure, ran 3 blocks to contest, started contest late and immediately, then i had to go to the bathroom but I didnt, rushed and missed easy part B and couldnt get last one so I guessed, then contest-guy lost the contest answer bubble sheets, then i went home and there was a black cat on my porch.

as to actual questions, i solved the first 23 in like 14 mins, but thats because i didnt have any time..  and i missed the hexagon, my reasoning was like "not 5 therefore 4, go gogogogo"

24 was tricky but i got it i think, 25 i just sat there.

I had "performance anxiety". I'm expecting a horribly low score.

Done Part A in 4 mins, done Part B in 11 mins. Had 45 mins to do Part C and my brain didn't feel like working so I got two of them only. Add to that a possible missed question in Part B and yeah... (I missed the hexagon in the circle, I put 3, I believe the answer is 2 though I'm NOT SURE. Can someone clarify?)

Afterwards, without the pressure of the test, I solved 24 quite quickly. (The answer is C, by the way)

Yuck. Not happy about my score.

ya its 2..

not 3, suppose its 3, clearly 3 adjacent sides.

Now take A B C to be the 3 points of tangency in that order, O center of circle.

The vertices of the hexagon between A and B and between B and C, call them X and Y.  Clearly XA = XB and YB = YC.  If XB=XY then the origin of the circle is on the perp. bisectors through A B C, and we are done.  But this is clear, else perp bisectors through A and C meet in different places.

Haha, I guess everyone was scared..

I slept in as well, I was dreaming about being locked up with some weird math problems/symbols, woke up at 7:30 am (I usually wake up at 6:30 am), and when I got to school, I had:
(1) this girl beside me sniffing every 5 seconds
(2) announcement (10 mins long) in the midst of the contest; and
(3) horrible image of Elyot, who screwed up in part B to score 144.
(sorry elyot if you ever see this, but it really scared me.. :blush: )

Personally I think Fermat is the most stressful one, the last 2 problems could qualify for Euclid, but you only have to do it within 1 hour.

The last problems of these contests have always been quite time consuming. Though, I think having done AMC12 the week before really helped my performance.

Speaking of dreams, I had an interesting one the night before the contest: I dreamt that I received a photocopied version of the contest (since more people showed up than signed up, something that happens each year), and the problems were copied and pasted in some random order (insteaded of questions 1, 2, 3, ..., it was like 5, 17, 3, 21, ...). I did not notice this and proceeded as usual, then found out after the contest that I bubbled every question in the wrong place... Weird dream.

Anyways, for the hexagon one, you only need to make the observation that 3 lines determine a unique circle (that lies with in the hexagon). This fact comes from the existence of incircle/excircles.

Lol, interesting dream Yufei..

I've been dreaming about Seminar quite a lot for past 2 years, i think.. And in every one of my dream, I was the shortest one in there with Sarah.. (no offense to you sarah..  :D )

So how did you do on the last one? I seem to have most of them right, I think I got the last one, but I'm to scared to post it just in case it's wrong.. If anyone thinks he/she got it, please post the answer here.. I was going to be the one posting since I had math circle tonight, but no one really wanted to take the problems up, and Dr. Rice couldn't even remember the answer, let alone the solution.

And I agree with the hexagon problem, I also put 2.

[quote="alpha123"]I got all the problems except I am not sure on the last one though. I put D) 61 as the answer. Can someone confirm this and tell me if I did it right[/quote]

see http://www.mathlinks.ro/Forum/viewtopic.php?t=27914

[quote="andytyk"]I didn't write the Fermat, wrote the Cayley but I thought that 1-24 on both contests were quite simple.  On the Cayley, the last question seemed a bit more difficult than last year's but the rest of the contest seemed easier.  I would have to say last year's Fermat was more difficult as well.[/quote]
Well, the last question on Cayley this year did bug me for a while (btw, i took Fermat not Cayley).
but infact, it is not hard.
The key is we can convert any Pythagorean triple into the riciprocal format by dividing by their smallest common devisor.
like 3 4 5 converts into
1/20  1/15   1/12 which should work.
5 12 13 into
1/156   1/65   1/60 etc.
And thus, after some checkings, we are done  :cool:

I couldn't make the observation so I spent 25 minutes mulling around with Pythagorean triplets until I hit upon the idea.  Had c = (ab)^2  / sqrt(a^2 + b ^2), but knowing that the Pythagorean triplets we learn have relatively prime sides, each side must have been multiplied by k.  That lead to c = k(ab)^2 / sqrt(a^2 + b^2) which meant k = n*sqrt(a^2+b^2).

It's not exceedingly difficult but I would say it was harder than last year, then again,  I only had 2-3 minutes for question 25 last year.

[quote="Iam1337"]i'm being really stupid but i cant get this one again cause im trying to calc my score on fermat.
20. In trianlge ABC, if AB=AC = x + 1 and BC = 2x-2, where x > 1,then the are of the triangle is always equal to

thanks guys[/quote]

Use Phythagorean to find height, then we are done
Answer: E
2(x-1) :sqrt: x

Thanks got it after just had a mind blank.  So what kind of scores do you think you guys got on fermat?

[quote="bugzpodder"][quote="jpark"]I've been dreaming about Seminar quite a lot for past 2 years, i think.. And in every one of my dream, I was the shortest one in there with Sarah.. (no offense to you sarah..  :D )
[/quote]  As I recall Helen Alexender was about the same height as you.  Oleg dont strike me as a tall person also  :D

So whos mak anyways?[/quote]

Hey, you don't think I'm THAT short,  do you?  :D

My friends tell me I'm petite, not short..  :blush:

of course not, Jenn :P  actually I think you are about the same height as me which means I am shorter as a guy (but as you know its hard to judge ones own height)  :blush:

Hey I'm not that short!  ...could grow a couple inches though, lol.

*sigh* everyone's taller than me!

I actually think you're taller than me, Jack..


MIT
college
Stanford
CMU
Carnegie Mellon

I am interested in majoring in computer science. I have a few colleges in mind: MIT, Stanford, and Carnegie Mellon. Can anyone let me know the differences between these schools regarding their computer science departments?

Also, I live in Washington State and it offers a program called Running Start in which high school juniors and seniors can take college level courses; thus, getting their AA degree and high school diploma at the same time. I have viewed the transfer policy of many colleges and emailed a few of them, and none seems to offer a good response as to weather my AA degree or credits will transfer. Can anyone give his or her input to this question?

I can help you with the question about Running Start. Minnesota has the first program of that kind in the country, known here as PSEO. In general, any program that lets you take college credits as part of a state-funded high school program has some HUGE advantages, such as

1) taking some challenging, interesting courses at low expense,

2) getting used to the college experience before you have to choose a college to enroll in for your undergraduate degree,

3) being able to show a challenging high school program, which will help you get into a more competitive undergraduate degree program,

4) meeting friends with similar academic interests who may not go to your local high school,

5) building confidence through all of the above.

Some participants in PSEO and Running Start programs are surprised to find out that many top colleges accept NO credits used as part of a high school program for transfer credit. (MIT has its own system for evaluating previous college-level studies for credit, but for several other colleges the general answer is no credit, period. Some other colleges offer advanced standing for AP scores or other test scores, but again no transferable credits for previous college study.) But that is not really a disadvantage of such a program, because your strong "high school" program (taken partly at a local college) is what helped you get into the top school in the first place.

A high school counselor will nearly always tell you that it is better to stay in your local high school, because the high school has a better social atmosphere for you. If you disagree, you'll be agreeing with me--I think high school is a royal waste of time for many able students. It's great to be FORMALLY in high school as long as possible, especially for eligibility for the AMC tests, but it is ridiculous to go to regular high school for high school if you are eager to do Running Start college for high school.

What other questions do you have about the trade-offs involved in Running Start study?

[quote="The Sigma"]I am interested in majoring in computer science. I have a few colleges in mind: MIT, Stanford, and Carnegie Mellon. Can anyone let me know the differences between these schools regarding their computer science departments?  [/quote]

They are equivalent for technical education. Stanford also has nontechnical education.  Apart from that and any financial differences in attending the schools, the real difference is location.

Carnegie Mellon is located in... Pittsburgh.
Stanford is located in paradise, though next to a ghetto.  It is more Asian than the other two schools.  It being California, an automobile is essential.

MIT is located in Boston, a city filled with college students.  As a result, the social (and dating) pool of both smart and average young people is enormous compared to the other places.


What is the coefficient of $ x^4$ in the expansion of $ (1\minus{}2x^2)^5$?

By the Binomial Theorem, $ (1\minus{}2x^2)^5\equal{}\cdots\plus{}\dbinom52(1)^3(\minus{}2x^2)^2\plus{}\cdots\equal{}\cdots\plus{}\boxed{40}x^4\plus{}\cdots$.


ratio
search
combinatorics unsolved
combinatorics

There are 2006 students and 14 teachers in a school. Each student knows
at least one teacher (knowing is a symmetric relation). Suppose that, for
each pair of a student and a teacher who know each other, the ratio of the
number of the students whom the teacher knows to that of the teachers
whom the student knows is at least t. Find the maximum possible value
of t.


The square ABCD is inscribed in the circle (C). Let M be a point on the short arc $AB$.
Prove that:
\[ MC\cdot MD > 3\sqrt{3} MA \cdot MB.  \]

It seems trivial ....maybe i make some mistake i don't see...

From sine law $S_{MCD} = \frac{1}{2} \cdot MC \cdot MD \cdot \sin( \angle CMD ) =\frac{\sqrt{2}}{4} \cdot MC \cdot MD$

and $S_{MAB} = \frac{1}{2} \cdot MA \cdot MB \cdot \sin( \angle AMB ) =\frac{\sqrt{2}}{4} \cdot MA \cdot MB$ .

But if $a$ is the side of the square  $S_{MAB} = \frac{1}{2} \cdot a \cdot h$ therefore $S_{MCD}= \frac{1}{2} \cdot a \cdot (a+h)$ .

Now $\frac{MC \cdot MD }{MA \cdot MB } = \frac{ S_{MCD} }{S_{MAB} } = \frac{a+h}{h} = 1+ \frac{a}{h} \geq 1+ \frac{a}{ a \cdot \frac{\sqrt{2}-1}{2}} =3+2 \sqrt{2} > 3\sqrt{3}$

because $h$ is maximum if $M$ is in the middle of the arc and there $h = a \cdot \frac{\sqrt{2}-1}{2}$

 :?


The digits $ 1,2,3 \ldots 9$ are arranged, one per circle, in the
triangle shown so that the sum of the numbers on each side of the triangle arc the same. What is positive difference between the greatest and least sums possible?

[asy]pair A,B,C;
B = dir(60); C = dir(0);
for(int i = 0; i < 3; ++i){
draw(Circle((i*A+(3-i)*B)/3,.10));
draw(Circle((i*B+(3-i)*C)/3,.10));
draw(Circle((i*C+(3-i)*A)/3,.10));
}[/asy]

If s is the sum, then 3s=1+2+3+4+5+6+7+8+9+x+y+z =45+x+y+z where x,y and z are the numbers in the vertexes of the triangle. So, the biggest sum is when x,y,z are 7,8,9 and smallest when they are 1,2,3.
Difference: 23-17=6


function
calculus
derivative
algebra
polynomial
calculus computations

Since $ f$ and $ g$ are differentiable at $ x\in\mathbb{R}^n$, we have

$ f(\mathbf{x \plus{} h}) \equal{} f(\mathbf{x}) \plus{} \nabla f(\mathbf{x}) (\mathbf{h}) \plus{} \| \mathbf{h}\| E_f (\mathbf{h})$,
and $ E_f(\mathbf{h})\rightarrow 0$.

What could $ E_f$ be here? The statement is talking about the product rule. Is it the Jacobian of f or something?

Thanks!

It might be clearer if you consider the single-variable equivalent: $ f(x \plus{} h) \equal{} f(x) \plus{} f'(x) \cdot (h) \plus{} \| h \| E_f (h)$

[hide="Hint"]It says that the function can be written as the sum of a linear approximation and some error term. What properties can we infer on the function $ E_f$?[/hide]
By the way, you know it's not the Jacobian because all the other terms in the equation are scalars :P

First of all, this is a calculus question and should probably be in the Calculus forum.

As for the question itself, there doesn't necessarily exist a "nice" expression for $ E_f(h)$. It's a statement about the error of the best possible linear approximation to f at x (which comes from the derivative). All that is garaunteed about it is that if f is differentiable at x then this error is sufficiently small if h is sufficiently small.

If f has continuous second partial derivatives you can use the mean value theorem to get a "Lagrange type" error for $ E_f(h)$ in terms of the second derivatives of f at some point in the neighborhood (of size h) of x. This is exactly the statement of Taylor's Theorem for first order polynomial approximation.


just wondering...

is there a pattern to 1+1/2+1/3+......+1/n?

thanks!

There is, but it's a pretty long formula, I believe.
When you take $\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots$, the sum tends to infinity, it diverges. This is known as the harmonic series.

how does it go to infinity?

1/1 through 1/10 is just under 3

-jorian

If you keep adding, you'll notice that it does not converge to a value. BTW, if you add the first $11$ terms, you get something like $3.019877344...$. So it is over $3$!

[url=http://mathworld.wolfram.com/HarmonicSeries.html]A website[/url]

In this website, look at Oresme's proof. That looks like it is the most elementary proof that the series goes to infinity.

i researched some more on the harmonic series and i get it now

thanks guys

I do know that $e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}...$, but this is a little different.

Quevvy, that makes no sense at all!
$\sqrt[4]{\frac{1}{1^{4}}+\frac{1}{2^{4}}+\frac{1}{3^{4}}...}\neq \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots$

Besides, the sum diverges!


modular arithmetic
number theory unsolved
number theory

[quote="who"]find all natural nos. n s.t. n^2+n+1 is a prime and
 2n+1Cn=1(mod n^2+n+1)[/quote]
Find all $n\in\mathbb{N}$ such that $n^{2}+n+1\in\mathbb{P}$ and $\binom{2n+1}{n}\equiv 1\bmod n^{2}+n+1$.

Is this what you mean?

dude, the biggest challenge now  is to understand what you mean! use tex man is easy and let the stuff much more clear! :D

Just a note: \[2n+1\equiv \sqrt{-3}\pmod{n^{2}+n+1}.\]

am repeating q. any progress?
 i came to know this is open pro.

The only solutions below $10^{5}$ are $n=8$ and $n=24.$

thanks for help maxal,any theoretical progress?

The next post containing "any progress" will be deleted without any warning ;)
And you didn't answer amcavoy's question (the answer is very likely to be yes).


If a infinite set of points in the plane determines only integer distances, then all the points lie on a straight line.

The statement is incorrect, the set of points has to be infinite! And this is a well-known result due to Erds and Anning.

In another hand, you may find arbitrary large finite sets of points not contained in a line such that the distance between any two is an integer. I think that all known such finite sets with not too few elements (probably at least 5, I think) are contained in a circle and that is an open question to find others.

In the same idea you may prove that if E is an infinite plane set of points such that for every A,B,C in E pairwise distinct, the distance from A to the line (BC) is an integer then E is contained in a line.

Pierre.


advanced fields
advanced fields unsolved

If C is a circle embedded smoothly in R4, show that there exists a 3-dimensional hyperplane H such that the orthogonal projection of C to H is an embedding.

I totaly have no idea on how to start this question. I have problem with even the definitions. What does it mean by "a circle embedded smoothly in R4"????

A $ C^k$-smooth embedding is a map that is a $ C^k$-diffeomorphism onto its image. So you are given a map $ f\colon C\to\mathbb R^4$ such that $ f'(x)\ne 0$ for any $ x\in C$. The problem is to find $ H$ such that $ \pi_H\circ f$ is an embedding too, where $ \pi_H\colon\mathbb R^4\to H$ is the orthogonal projection. By the chain rule, all you need is $ f'(x)\notin H^{\perp}$. Here $ H^{\perp}$ is simply the line through the origin that is perpendicular to $ H$.

I don't know what is the degree of smoothness ($ k$) that is assumed in the problem. If it's merely $ C^1$, that may not be enough.  :maybe: If it's $ C^2$ or higher, then the map $ x\mapsto\frac{f'(x)}{|f'(x)|}$ is at least $ C^1$, and therefore its image does not cover $ S^3$, which implies the existence of $ H$.

Or does it? $ f'(x)\notin H^{\perp}$ only implies that $ \pi_H\circ f$ is an immersion.  :blush:  We also need $ f(x)\minus{}f(y)\notin H^{\perp}$ for $ x\ne y$. Therefore, we should consider the map $ F(x,y)\equal{}\frac{f(x)\minus{}f(y)}{x\minus{}y}$ defined as $ F(x,x)\equal{}f'(x)$ on the diagonal. It is $ C^{k\minus{}1}$ smooth from $ C\times C$ into $ \mathbb R^4\setminus\{0\}$. Once again, the radial projection of its image does not cover $ S^3$ -- but this time, we really need those four dimensions.


topology
vector
real analysis
real analysis unsolved

Usual metric on R: The usual metric for two elements 'x' and 'y' in R is defined to be the absolute difference between 'x' and 'y'.

Usual metric on R^2: For elements x=(x1,x2), and y=(y1,y2) in R^2, the usual metric on R^2 for these two elements is defined as (x1-y1)^2 + (x2-y2)^2.

Can anyone find three distinct elements in R^2 that have the same distance (as defined by the usual metric on R^2), and show that there can be no 3 distinct elements in R that can be mapped to these points, via through an isometric mapping.

Equilateral triangles do exist.
If $ a,b,c$ are distinct real numbers, then (after relabeling) $ a<b<c$, which implies $ b\minus{}a<c\minus{}a$.

Thank you, mlok.

By definition,

Topology: A topology on a set 'X' is a collection 'i' of subsets of 'X' called the open sets, satifying:

1) Any union of elements of 'i' belongs to 'i'

2) Any finite intersection of elements of 'i' belongs to 'i'

3) Both the empty set and 'X' belong to 'i'.

What does it mean when a problem states that a subset of R^2 is 'radially open' if and only if it contains an open line segment in each direction about each of its points?
Can you describe mathematically what the definition of a 'radially open' set is?

The description you have is mathematical. "Mathematical" does not mean "using math symbols".

A description with symbols can go like this: $ V\subset\mathbb R^{2}$ is radially open if for any $ a\in V$ and any unit vector $ \vec u$ there is $ r > 0$ (depending on $ a$ and $ \vec u$) such that $ a\plus{}t\vec u\in V$ whenever $ |t| < r$. Does this help? I did not think so either.


The $ 27$-element sets $ A_{1}$, $ A_{2}$, ..., $ A_{35}$ have the property that every three of them have exactly one element in common. Show that all $ 35$ sets have an element in common.

[hide="hint"]propose that there isn't exist an element common to all the sets. [/hide]

I wonder if u can call that a hint :wink:

Yeah, please give something meaningful  :wink:

Can u please post your solution mychrom?

Ok, I see that nobody solved it. I tried some things but nothing... Is the problem correct? And BTW, mychrom, if u have a solution why don't u post it?

I found out that Furedi proved in an article (Z. Furedi, On a problem of Deza and Frankl, Ars Combinatortia, 13 (1982),221-222) that that the optimal l-wise t-intersecting families(i.e. the intersection of every l sets has cardinal t) for $l\geq3$ are trivially intersecting. However, I can not figure out why this configuration is optimal ( we do not know the total number of elements...maybe the fact that the sets are 27-uniform helps :maybe: ) and I can not find that article either.
If anyone has some ideas pls help (maybe the problem is much easier) :)

I'll keep reviving this topic  :P because I am very interested in a solution.
I discussed the problem with some friends ( for the moment I have very little time as the national olympiad is close) and one of them proved that the maximal chain containing a fix set is at least 8 long ( by chain I mean a sequence of sets that have all an element in common). I also proved ( I do not recall if in the general case or in the case where all the sets were not intersecting) that the minimum number of elements in the reunion of the 35 sets is at least 112. I'll think a little more on it today but I have the impression that i am still far from the solution. So , again, if u have any ideas please post. :)

Well, I don't have much idea at the moment, and I don't know about those l-wise t-intersecting stuff you talk about. Anyway, I can tell you this problem is only from Iran 2005, 2nd Round. Maybe it has an easy solution, and you're thinking too hard...

[quote="ciprian"]@dragonfire: do you mean the second Iran TST from 2005 or the National OLympiad ? :maybe: I doubt that this problem was used in Iran...[/quote]

Well, I mean the 2nd Round of the National Olympiad

We have triples $34 \choose 2$ with first set. So $\frac{{34 \choose 2}}{27}$ of them have this same element. Because ${7 \choose 2}>{34 \choose 2}/27>{6 \choose 2}$. So we have $8$ sets with this same element (we add first set). We can assume that this are $A_{1},...A_{8}$ sets and $x$ is shared element.  Suppose that exist another set $B$ which doesn't have $x$ element. Exist ${8 \choose 2}> 27$ triples which every have set $B$ and is build by first 8 sets. So two diffrent triples have this same element $y$. So at least 3 sets have this same element $x$ and $y$. Contradiction with condition that every three of them have exactly one element in common.


integration
limit
function
calculus
parameterization
derivative
real analysis

hi,

i will be happy if somebody helps to compute the limit via the

result below:

The result : $x\geq-2 \Rightarrow 1-x =< e^{-x}=< 1-x+(\frac{e^{x}}{2})^{2}$

$f(x) = \int_{0}^{1}\frac{e^{-x(1+t^{2})}}{1+t^{2}}dt$
The limit : $\ds\lim_{x\rightarrow x_{0}}\frac{f(x)-f(x_{0})}{x-x_{0}}$
thanks in advance.

the function [img]http://www.mathlinks.ro/Forum/latexrender/pictures/c/8/6/c862de0fe379021732dfc36b3449523bccfccffc.gif[/img]  is constant

then f'(x)=0

then[img]http://www.mathlinks.ro/Forum/latexrender/pictures/4/b/2/4b2b985aba0a72fe27bf8fcdf3567312d0598ed1.gif[/img] is equal to 0 cause its f'(x_0)

Is the x in the numerator supposed to be x^2? Not that it matters but. Forget the f for a moment and think of the integrand as an integral depending on parameter. Your book should have something on how to differentiate those (you take a partial).

goldbach how can the function be constant? The integral depends on a parameter. Take a look at the graph of the integrand.

$f'(x_{0}) = \int_{0}^{1}{-e^{-x_{0}(1+t^{2})}dt}$

This can be evaluated and I think there was a recent post on that. Not sure if you care for that though.

P.S. This no consecutive posts in the span of 10 minutes is enormously stupid.  :mad:

sorry i am so sorry

it isn't constant there is the parametre x into

Hate to say this, but isn't what you're trying to compute the definition of a derivative. Just take the derivative of the integral. ROFL. cancel out for the cool.

i am sorry ,i have made mistake in the result .the condition is that x>=-2 not x>=2.for the other data,i guess they are correct.

It's just wrong- that's how you handle a changing limit, not a parameter.
The actual derivative can't be computed in terms of elementary functions; it depends on definite integrals of $e^{-t^{2}}$.

first off you must calculate the integral that's f(x) and then calculate the derivative

You don't want to do that, goldbach. Look again at jmerry's post, right above yours.

----

$\frac{d}{dx}\int_{a}^{b}f(x,t)\,dt$

$=\lim_{h\to0}\frac1h\left[\int_{a}^{b}f(x+h,t)\,dt-\int_{a}^{b}f(x,t)\,dt\right]$

$=\lim_{h\to0}\int_{a}^{b}\frac{f(x+h,t)-f(x,t)}h\,dt$

*** and now you need appropriate hypotheses to be able to invoke one of the theorems that permits the interchange of limit and integral ***

$=\int_{a}^{b}\int_{h\to0}\frac{f(x+h,t)-f(x,t)}h\,dt= \int_{a}^{b}\frac{\partial}{\partial x}f(x,t)\,dt.$

A sufficient condition for the interchange would be that $\frac{\partial}{\partial x}f(x,t)$ be continuous on $[a,b]\times$ (a bounded interval). That would make the partial derivative uniformly bounded, from which a Mean Value Theorem argument would allow us to conclude that the difference quotient converges uniformly to the partial derivitive, which in turn allows us to invoke the Riemann integal theorem that uniform convergence on a bounded interval justifies the intechange of limit and integral. Other, somewhat weaker, conditions would allow the use of the Lebesgue tools for interchanging limit and integral.

thank you for your answers .what i would to clarifiy
[quote]
Kent Merryfield said :
A sufficient condition for the interchange would be that  be continuous on  (a bounded interval). That would make the partial derivative uniformly bounded, from which a Mean Value Theorem argument would allow us to conclude that the difference quotient converges uniformly to the partial derivitive, which in turn allows us to invoke the Riemann integal theorem that uniform convergence on a bounded interval justifies the intechange of limit and integral. Other, somewhat weaker, conditions would allow the use of the Lebesgue tools for interchanging limit and integral.
[/quote]
I guess you missunderstood what i want and how.in my level(last year in high school) ,we are not suppposed to know neither functions with two variables nor  Riemann integal theorem.We are obliged to use the result to find the derivative.An other thing ,This two question was proposed in a test in 2002.[/quote]

[quote="toetoe"]
I guess you missunderstood what i want and how.in my level(last year in high school) ,we are not suppposed to know neither functions with two variables nor  Riemann integal theorem.We are obliged to use the result to find the derivative.An other thing ,This two question was proposed in a test in 2002.[/quote]

then you must calculate the integral that's f(x) and then calculate the derivative

[quote="goldbach"]then you must calculate the integral that's f(x) and then calculate the derivative[/quote]
Did you miss the point that you [b]can't[/b] do that? The integral has no elementary representation.

The best you can do at the level you're talking about is to completely ignore the part of my post that toetoe quoted and just [i]assume[/i] that you can interchance the limit and the integral and hence differentiate under the integral sign. In other words, calculate first, ask questions later. Very Eulerian, and an attitude I assume fairly often.

Another way to look at it: you know that the derivative of a sum is the sum of the derivatives. What's an integral but a smeared out sum?

The given inequality is quite awkward to use, and I'm not even sure it is true - are you sure that's supposed to be $\left(\frac{e^{x}}2\right)^{2}$ on the right?

yes,i am sure.you can even prove it using $g(x) = 1-x+\left(\frac{e^{x}}{2}\right)^{2}-e^{-x}$.

i tried to use the fact of $0=<t=<1 \Rightarrow -(t^{2}+1)>-2$but it seems that it doesn't work .

hi,
i guess the problem couldn't be solved here .so ,i would like to   ask the administrators if i should move this topic to  "Unsolved problems".

There's no answer because we can't tell what the question is supposed to be. If it's computing the derivative, we can write down an integral for it, but no more; the integral we get is impossible in terms of elementary functions.

in fact,i have asked my maths teacher today and he told me

that there was an error in that question ,he said that the

derivative can't be solved in our level .however i still would like

to see how you can compute the derivative using the way you

want even if your method isn't in the level of high school.

The derivative is $f'(x_{0})=\int_{0}^{1}-e^{-x_{0}(1+t^{2})}\,dt$ as §outh§tar wrote. We can transform this slightly;
$f'(x_{0})=-\int_{0}^{1}e^{-x_{0}}e^{-x_{0}t^{2}}\,dt=-e^{-x_{0}}\int_{0}^{\sqrt{x_{0}}}e^{-s^{2}}\,\frac{ds}{\sqrt{x_{0}}}= \frac{-e^{-x_{0}}}{\sqrt{x_{0}}}\int_{0}^{\sqrt{x_{0}}}e^{-s^{2}}\,ds$

That's it. The integral can't be evaluated exactly- it's well-known that the antiderivative of $e^{-x^{2}}$ is not an elementary function, and we would need that antiderivative.


calculus
integration
trigonometry
calculus computations

Evaluate $ \int_0^{\frac {\pi}{2}} \frac {(\sin x)^{\cos x}}{(\cos x)^{\sin x} \plus{} (\sin x)^{\cos x}} dx$.

Using substitution $ \boxed{x\equal{}\frac{\pi}{2} \minus{} t}$ we get, that $ \int_0^{\frac {\pi}{2}} \frac {(\sin x)^{\cos x}}{(\cos x)^{\sin x} \plus{} (\sin x)^{\cos x}} dx\equal{} \int_0^{\frac {\pi}{2}} \frac {(\cos t)^{\sin t}}{(\sin t)^{\cos t} \plus{} (\cos t)^{\sin t}} dt$

Therefore $ 2\int_0^{\frac {\pi}{2}} \frac {(\sin x)^{\cos x}}{(\cos x)^{\sin x} \plus{} (\sin x)^{\cos x}} dx\equal{}\int_0^{\frac {\pi}{2}} \frac {(\sin x)^{\cos x}\plus{}(\cos x)^{\sin x}}{(\cos x)^{\sin x} \plus{} (\sin x)^{\cos x}} dx\equal{} \frac{\pi}{2}.$

So $ \int_0^{\frac {\pi}{2}} \frac {(\sin x)^{\cos x}}{(\cos x)^{\sin x} \plus{} (\sin x)^{\cos x}} dx\equal{}\frac{\pi}{4}.$


Can somebody prove the sum of squares formula using finite differences?

what do you mean by sum of squares formula?
there are many you could be referring to...

I also don't know what he means when he says 'finite differences'.

probably the nicest proof for this is shown with the little flash application on the side of the main AoPS website. It sums the square into a little box.


Yup, who do you think it's gonna win this IMO? I have included in the options the first 10 countries in order of total number of points at IMO 2004. I will reserve my vote for a later time, but you can vote starting now :)

Moldova! :D ... ur...  :? I mean China :P

If nothing untoward happens, it will go to China. But Russia, Bulgaria, Vietnam and the United States are great contenders too.

Hey Valentin: what about complete optimists ? I mean, could you add option "Other" to the poll ?  :D

Myself, I think China or Russia will win

If there's no accident, it goes to China.

[b]Russia[/b] will(must) win. Go Russia!!!!  :D  :D  :D

Of course many gues from (С?Г - по - ру??ки) vote for Russia.
Realists vote for China. But don't you think VietnaM  team can have best rezult,
because after 2 years they probably will have IMO in their country
and now they are preparing very strong ?
Remember that Vietnam TST is very hard.

[quote="Ragingg"]Why doesn't anyone think USA will have a chance? :([/quote]Actually 5 people seem to think so :) Probably because China is the en-titre world champion, because it's been always either first place or second place since the '90s and probably because most of the last year's US team  were seniors, thus there will be many un-experienced people on the IMO US Team this year (un-experienced in the sense that they didn't take a previous IMO)... :?

[quote="Ragingg"]Why doesn't anyone think USA will have a chance? :([/quote]
It is a fate  :P

France will win!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! :D

:? Well.;. maybe we will do better than last year...

My prognosis:
1. China
2. Russia
3. USA
4. Moldova
5. Bulgaria
6. Romania
7. Vietnam

yeah, i agree with Xixas on the countries for the top 3.

[quote="lego"][quote="shobber"]Problems are easy. So any predictions of the cut line?  :)[/quote]

the same it was in 1987  :evilgrin:[/quote]
Huh?  :?
Is the problems of 1987 as easy as this time?

I don't know but I'm rooting for Bulgaria; somehow I don't think they will have messed up on the first day's problems since they seemed particularly easy.

I think Vietnamese will win. Because Union, Union.
 :roll:

[quote="Xixas"]I think 36 for gold, 27 for silver and 17 for bronze.[/quote]I think you are giving some of the students too much credit :)

[quote="Xixas"]I think 36 for gold, 27 for silver and 17 for bronze.[/quote]

14,23,31?
Just a wild guess. The problems are weird this year.

Think Vietnam did well for the hard questions. Just hearsay. And maybe not so good for an easy question. Also completely hearsay.

[quote="Valentin Vornicu"]My opinion for the top 3 countries is

1. China
2. Russia
3. US[/quote]Well I was wrong. US got 1 point more than Russia, so there is a switch :)

How come India did so terribly
Abhishek Dang was the only silver medallist.


ratio
conics
ellipse
geometry proposed
geometry

In $xy$ plane,denote the distance betwen a line $l$ and a point $A$ by $d(l,A)$.Given that three distinct points $A,B,C$, let $f(l)=d(l,A)^2+d(l,B)^2+d(l,C)^2$.
Determine the shape of the triangle such that three distinct lines minimize $f(l)$.

It is well known - Huyghens - that if L has a fixed direction, f(L) will be minimal when L goes through the centroid g of ABC.
Now if we consider the following facts :
1) if three points lie on a circle with radius r, the sum of the squares of the distances from the three points to any diameter of the circle is constant (= 3/2r^2)
2) if L0 is the focal axis of the Steiner circumellipse of ABC, a and b the corresponding semimajor and semiminor axis, an orthogonal affinity with axis L0, ratio b/a maps the circle (G,a) to the ellipse
Hence, f(L) is minimal when L = L0
It follows that, if TWO distinct lines minimize f(L), the ellipse is a circle and ABC is equilateral
Kind regards. Jean-Pierre

Thank you for your reply, jpe.

You are right.

kunny


function
algebra
polynomial
integration
calculus
real analysis
real analysis unsolved

[i found this problem-additional point b added, not quite difficult, but beautiful and useful, related to some already posted problems, i hope you'll enjoy it :)]
Let f be a continuous function on [a,b] s.t.
∫ab xk f(x) dx = 0, for each k in 0,1,...,n where n ≥ 2.
a) prove that f has n + 1 zeros in (a,b)
b) give an example of such function

If $f$ has $n$ (or less) zeros it would be possible to build a polynomial P of degree at most n being zero at the same place as $f$ and of same sign (Lagrange polynomial with an adequate sign).

But then $\int_a^b P(x) f(x) dx = 0$ and $P(x) f(x) \geq 0$

$f = 0$ is such function  ;)

[hide]I'll look for a less stupid example  :rotfl: [/hide]

The "less stupid" example (tµtµ's words) would be a Legendre polynomial.

Let $f(x)=\left(\frac{d}{dx}\right)^{n+1}\left([(x-a)(b-x)]^{n+1}\right).$

This $f(x)$ happens to be a polynomial of degree $n+1.$

Calculate the integral $\int_a^bx^kf(x)\,dx$ by integration by parts, integrating $f$ and differentiating the power of $x.$ This will show that the integral is zero.

Slight variation needed in tµtµ's proof: for your interpolating polynomial, only use zeros of $f$ at which $f$ changes sign. That way, your polynomial $P$ will have only simple zeros, and you can do what you want (the sign changes) without exceeding the prescribed degree.

The Lagrange approach is the expected example indeed.

About the first part here is another solution, more technically.

We denote by F(x) = ∫ax f(t) dt.

remark:

For k ≥ 1, integrating by parts, 0 = ∫ab xk f(x) dx = ∫ab xk F'(x) dx =
bk F(b) - ak F(a) - k ∫ab xk-1 F(x) dx, and because F(a) = F(b) = 0 we conclude that
∫ab xk-1 F(x) dx = 0, for k ≥ 1, where F is a continuos function on [a,b].

Now it's easy to apply the induction.

Suppose that ∫ab f(x) dx = 0, by the mean value theorem there exists a point x1 in (a,b) s.t. f(x1) = 0.
If ∫ab f(x) dx = 0 and ∫abx f(x) dx = 0, by the above remark, ∫ab  F(x) dx = 0, so there exists a point y1 in (a,b) s.t. F(y1) = 0.
That means [1] ∫ay1 f(x) dx = 0. But ∫ab f(x) dx = 0, so [2] ∫y1b f(x) dx = 0.
From [1], [2] and the mean value theorem, there are 2 points x1 in (a,y1), x2 in (y1, b) which are zeros for f.
If the proprerty holds for n, using the same remark, there exists points y1, y2,... yn in (a,b), all in this order which are zeros for F.
That means :
[1] ∫ay1 f(x) dx = 0
[2] ∫ay2 f(x) dx = 0
...
[n] ∫ayn f(x) dx = 0

And
[n+1] ∫ab f(x) dx = 0

From [i] and [i+1] results ∫yiyi+1 f(x) dx = 0, i = 0,...,n [y0 = a, yn+1 = b].

Using the mean value theorem we conclude the induction.

[quote="Kent Merryfield"]Slight variation needed in tµtµ's proof: for your interpolating polynomial, only use zeros of $f$ at which $f$ changes sign. That way, your polynomial $P$ will have only simple zeros, and you can do what you want (the sign changes) without exceeding the prescribed degree.[/quote]

I don't understand how can we guarantee that the polynomial changes the sign as we want.  :blush:

[quote="mareleG"][i found this problem-additional ... beautiful and useful , related to some already posted problems ..
[b] Let $f$  be a continuous function on $[a,b]$ such that
\[ (*)\; \; \; \; \; \; \; \int\limits_{a}^{b} x^k f(x)\; dx = 0\; \; , \; \; \; \forall k \in \{0,1,...,n\} \; ,\; (n \ge 2).  \]
a) prove that $f$  has $n + 1$ zeros in $(a,b)$
b) give an example of such function [/b] [/quote]
[size=75][color=blue][b]First let us note that the question is well-know in the theory of orthogonal polynomials. [/b][/color] [/size]
[size=150] [b] a) [/b] [/size] For $k=0$ in (*)  one finds  $\int\limits_{a}^{b}f(x)\; dx =0 \; \; ,\; \; \; f\in C[a,b] .$ It follows that $f$ has at least a zero in $(a,b) .$
 Denote by $x_1,x_2,...,x_j$ [b]all distinct zeros which are located   in [/b] $(a,b) .$ According to the first  remark we have $\begin{array}{|c|} \hline j\ge 1\\ \hline \end{array} .$
 Let us assume that   $\begin{array}{|c|}\hline j \le n\\ \hline\end{array}\; .$   In case $f\not \equiv 0$,  consider the  $F(x): = (x-x_1)(x-x_2)\cdots (x-x_j)f(x) : =h(x)f(x)$
    where $h(x): = (x-x_1)(x-x_2)\cdots (x-x_j) .$ From (1)  observe that \[ (**)\; \; \; \; \int\limits_{a}^{b}p(x)f(x)\; dx = 0 \; ,  \]  for  all polynomials of degree $\le n$ .    Because   $F(x) \ge 0 \; \; ,\; \; F\in C[a,b] \; , F\not \equiv 0$ on $[a,b] ,$    one obtain $(***)\; \; \; \int\limits_{a}^{b}F(x)\; dx >0 .$
But (**), together with the fact that [b]degree(h)[/b]$\le n$ , enables us to write    $\int\limits_{a}^{b}F(x)\; dx =\int\limits_{a}^{b}h(x)f(x)\; dx =0$  which contradicts (***).
Therefore $\begin{array}{|c|} \hline k \ge n+1 \\ \hline \end{array}$  which completes the proof. In case $f\equiv 0$ on $[a,b]$ we have nothing to prove.
[size=150][b]b)[/b][/size] A non-trivial example is
\[ f(x): = \left[\left(x-a\right)^{n+1+\alpha}\left(b-x\right)^{n+1+\beta}\right]^{(n+1)} \; \; ,\; \; x\in [a,b], \]
where  $\alpha>-1 \; ,\; \beta >-1$ (conditions in order to assure the convergence of some integrals).

After a good night's sleep, I have finally understood Kent's remark. The proof indeed works.

[quote="flip2004"] Denote by $x_1,x_2,...,x_j$ [b]all distinct zeros which are located   in [/b] $(a,b) .$ According to the first  remark we have $\begin{array}{|c|} \hline j\ge 1\\ \hline \end{array} .$[/quote]

Here distinct zeros has no meaning. I'm sure you meant what Kent said above.

[quote="perfect_radio"] I have finally understood Kent's remark. The proof indeed works. Here distinct zeros has no meaning. I'm sure you meant what Kent said above.[/quote]
Please follow my solution: the roots $x_1,x_2,...,x_j$ are [b]mutually distinct ,i.e.$x_i\ne x_{\alpha}$ for $i\ne \alpha$.[/b]


Let $x$ and $y$ be positive integers such that $xy|x^2 +y^2 +1$. Show that $\frac{x^2+y^2+1}{xy}=3$.

Masoud Zargar


analytic geometry
conics
ellipse
parabola
function
geometry
geometric transformation

If you have a curve on a coordinate plane, does that mean that at least two of the coordinates have to have the same 'x' or 'y' coordinate. Please help.

I don't think so...

$ y \equal{} x^3$???

EDIT: I think this is in ellipses, parabolas, etc.

Not necessarily, it depends on how "complete" the curve is. Usually if the curve resembles half (or more) of a circle/ellipse then it will have two points that have the same $ x$ or $ y$-coordinate.

If the curve is the graph of some function, then for each object "x" there can be only one image "y": in other words, if you draw a vertical line through the curve there can be only one intersection. Of course, there are several curves that are not functions but have considerable importance: circles, ellipses, and other curves described by polar coordinates.
If your curve is the graph of some function, then it can happen that two (or more) objects "x" are mapped into the same image "y": in this case, if you draw an horizontal line through the curve, it will intersect it more than one time. When this line intersects the graph only one time, the function is called "injective" or "one to one": different objects are mapped into different images.

Alright thanks. Thanks for the information. I think I was only imagining curves that go from one point, and then curve down to one point, and then curve in the opposite direction, as if it were the reflection of the curve.

Ummm, not really. I think I was thinking of something along the lines of;

x = z/y

$ x \equal{} \frac{z}{y}$ is three dimensions.  Is that what you mean, or am I misunderstanding you?

When I said 'z', I meant z being any real number. Sorry for not including that.

x = z/y, for any real number 'z'.

Thats the same as y=z/x which is a hyperbola. For example, 1/x. That also is a one-to-one function.


I have a list of 200 books and authors I have to memorize in about 2 weeks.  How do I go about doing something like that?  Any tips from you study masters out there?

Examples:

Little Women - Louisa May Alcott
A Death in the Family - James Agee
Dandelion Wine - Ray Bradbury
Tobacco Road - Erskine Caldwell
A Farewell to Arms - Ernest Hemingway

Obviously, some of these I already know, but I am unfamiliar with about 85% of them, being a math person rather than a literature person.  Advice please!

You say them aloud until they are imprinted in your mind. I am a memorisation person! But prob. is if you don't go through it alot & have no connection to them....
BTW I like Ray Bradbury books. I dream of them.

You have 14 days.

The first 10 days, take 20 each day. Do reviews every day of the ones you have already done.

Last 4 days, get some friends and quiz one another and circle the ones you don't remember.

It also helps that you only remember 199 rather than 200 since you will know that the one you did not remember is not in your memory bank!

I was thinking of maybe reading two of the books that I can never memorize, but I don't know if I'll have the time, and 2 out of 200 is not a big impact.  Of course, it could be the difference between a C and a D.  What do you guys think?

Maybe flashcards is not a bad idea. :maybe:

Flash cards is what I would do. I would also use a lot of mnemonics. What is the purpose of memorizing all of these books and authors?

I have a CRAZY English teacher.  I already made flashcards the day he gave us the assignment, but it's still a lot...

mneumonics will make it a whole lot easier then. What is the format? Will you get 200 book titles and have to write the authors, 100 book titles and 100 authors, or just 200 blank spaces?

It helps me if I'm writing something down as I go, and not just staring a paper. Quizzing friends also helps (face to face or via IM)


inequalities
geometry
trigonometry
geometric transformation
reflection
cyclic quadrilateral
geometry unsolved

Let Q be a quadrilateral whose side lengths are a, b, c, d, in that order. Show that the area of Q does not exceed (ac + bd)/2.

The trigonometry solution is not difficult, but I was looking for a different solution. My idea was splitting the quadrilateral into this: (image attached).

Eventually, I got the inequality down to this:
\[ ac \plus{} bd\ge x(y \plus{} z) \plus{} (x \minus{} w)(z \minus{} t)
\]
But I am very bad at inequalities. So, before I proceed, could I get a hint?

EDIT: Also, the RHS of that inequality is the area of the quadrilateral.

Okay I found a far, far easier solution involving a reflection. Here's a hint how to solve the inequality I posted anyway (I think...)

[hide="Hint"]
Ptolemy's Inequality
[/hide]

[hide]It's well known that given side lengths of a quadrilateral, the cyclic quadrilateral has greatest area.  So make this guy cyclic.  Now the area of the quadrilateral is the sum of the areas of the two triangles into which it is divided by a diagonal.   Thus, $ [ABCD] \equal{} \frac{ab \sin A}{2} \plus{} \frac{cd \sin C}{2} \equal{} \frac{ab\plus{}cd}{2} \sin A$, because $ \sin A \equal{} \sin (\pi \minus{} A) \equal{} \sin C$.  Then $ \sin A \le 1$ so we are done.[/hide]

Nice :). I hadn't thought of that.

Anyway my beast reflection proof is to reflect D around the perp. bisector of AC. to X. ABCX has same area as ABCD, but the opposite sides now paired (i.e. the order is now a-b-d-c of the sides). Now, just use area rule for signs and you are done.

Well thats very similar to yours anyway i guess ;)


Page 342, art of problem solving vol 2. I am understanding all of the processes of the derivation of fermat's theorem until we get to what should be the easiest part. (using x as multiplication), we have
a x 2a x 3a...x (p-1)a is congruent to 1 x 2 x 3.... x (p-1) (mod p). I factored out the a and then made it into
a(p-1)! is congruent to (p-1)! (mod p), and then divided by (p-1)!. I really dont see how one can obtain a^(p-1) is congruent to 1 from this.

Also, page 204 problem 210. A circular coin of radius 1 cm is dropped near the center of a chessboard (8X8 of course) comprised of squares of side length 2 cm. What is the probability that the coin lands so that it is entirely within one square? (lines have zero width).
I do not understand how the answer to this is zero.
I appreciate any help given forth. I may post more of these at a later time, for there are a few more that I am unsure about.

That's not the simplest proof of Fermat's Theorem I think. Here's one that I came up with that I think is much simpler. We work with induction on a. Assume it works for some a. We shall prove it also works for a+1. We expand out (a+1)^p=a^p+p*a^(p-1)+p(p-1)/2*a^(p-2)+...+p*(p-1)/2*a^2+p*a+1. So all the terms that have p in them are divisible by p, and then the others are just a^p and 1, which sum to a+1 (mod p) since a^p:equiv:a(mod p).

For the second question, of all the places the coin could land, for it to land inside a square, there is exactly one place per square for it to land. But there are infinitely many places it could land. (Consider its center: we need it to have odd integers for its x- and y-coordinates, which are picked from real numbers between 0 and 16.) That's sixty-four cases out of infinitely many, which means probability zero.

[quote="ComplexZeta"]
Here's one that I came up with that I think is much simpler. We work with induction on a. Assume it works for some a.[/quote]

It may be obvious, but to fill out your inductive hypothesis you should point out the a for which it works. :wink: This method is nice and fast, but I still like the AoPS method because it can easily be extended to prove the more general Euler's Theorem, that a^phi(n) = 1 (mod n), when (a,n) = 1. And then Fermat's Theorem just comes straight from there.

Jelyman, I think you have a typo in your post that is very important. It should be that [b]a^(p-1)*(p-1)![/b] is congruent to (p-1)! (mod p). You are multiplying (p-1) a's together. So then dividing by (p-1)! gives you a^(p-1) is congruent to 1 (mod p).

Yes, you are right. It clearly works for a=1 since a^p=1:equiv:1(mod p). Furthermore, we can go backwards replacing a by a-1 instead, and the same argument will apply. But yes, the other argument can be extended further to prove other results.


The area of a circle inscribed in an equilateral triangle is $ 48\pi$. The perimeter of this triangle is:

$ \textbf{(A)}\ 72\sqrt{3} \qquad
\textbf{(B)}\ 48\sqrt{3}\qquad
\textbf{(C)}\ 36\qquad
\textbf{(D)}\ 24\qquad
\textbf{(E)}\ 72$

[hide="solution"]
First calculate the radius of the circle, which is $ \sqrt {48} \equal{} 4\sqrt {3}$. Now make a $ 30 \minus{} 60 \minus{} 90$ degree triangle with the radius being the 30 degree side, and from one vertex of the triangle to the center of the circle being the hypotenuse and the 60 degree side is half of a side of the triangle. So that half of a side in the equilateral triangle is $ 4\sqrt {3}\cdot \sqrt {3} \equal{} 12$ making the perimeter $ (12\cdot 2)\cdot 3\equal{}\boxed{72}$. Choice $ E$. If I am wrong sorry I was in a hurry.
[/hide]

let r is radius of inscribed circle!
So r=4√3. let a is side of the triangle. a=2√3r. so a=24 then 3a=p p=72

[hide="alternate solution"]First, we let the side length of the triangle be $ a.$
We have the area formula $ [ABC]\equal{}rs.$ The inradius is $ \sqrt{48}\equal{}4\sqrt{3}$ and the semiperimeter is $ \frac{3a}{2}.$ Since $ \triangle ABC$ is equilateral, the area is $ \frac{a^2\sqrt{3}}{4}.$ Now we can form an equation. \[ \frac{a^2\sqrt{3}}{4}\equal{}4\sqrt{3}\times\frac{3a}{2}\Rightarrow a\equal{}24\] which means the perimeter is $ 3\times24\equal{}72\equal{}\boxed{\text{(E)}}.$[/hide]


prove that for any value of [i]x[/i], the equation $2x^4 - 4x^2 - 1$ always has a greater value than the equation $x^4 - 2x^2 - 4$

nvm got it.

[hide]
for $\left[ {2x^4 - 4x^2 - 1} \right]$ to be greater than $\left[ {x^4 - 2x^2 - 4} \right]$, the following inequality must be true:
\[ \left[ {2x^4 - 4x^2 - 1} \right] - \left[ {x^4 - 2x^2 - 4} \right] > 0 \]
rearreanging this, we get:
\[ \left( {2x^4 - x^4 } \right) + \left( { - 4x^2 + 2x^2 } \right) + \left( { - 1 + 4} \right) > 0 \]
\[ x^4 - 2x^2 + 1 + 2 > 0 \]
\[ \left( {x^2 - 1} \right)^2 + 2 > 0 \]

since $\left( {x^2 - 1} \right)^2$ will always be positive (greater than 0) or equal to 0.
when $\left( {x^2 - 1} \right)^2=0$, 2>0
[/hide]


Find all positive integers  $n>3$  such that  $n$  divides  $(n-2)!.$

I suppose, that it's true for all composite integers greater that 5

4 wouldn't work.

neither does 9, etc. so your method doesn't work.  You need to exclude squares of primes as well.

[quote="dondigo"]I suppose, that it's true for all composite integers greater that 3[/quote]

Except $4$ and square of primes ;)

Well, tµtµ forgot "odd" before primes, that's all :p  j/k

It will work for squares of primes greater than 4. For a prime $p$, $p$ will appear $p-1$ times, one for each of $p$, $2p$, $3p...$ $p(p-1)$ ($p(p-1)\leq p-2$ since $p\geq 2$). $p^2$ will divide $(p-2)!$ as long as $p-1\geq 2$, that is, for any odd $p$.

I will give the solution I gave at this contest as I took part.
I will use the theorem of Wilson.This theorem says: The positive integer $p>2$  is a prime number if only if:$(p-1)! \equiv -1(mod p)$. So $n$ is not a prime number.Now if

$n=ab$ the $a,b$ will appear at the $(n-2)!$ so if $n=ab$ the  $n$ divides $(n-2)!$

Finally if $n=q^2$ with  $q>2$ we have that $q=\sqrt n$. In addition $2q<n-1$ <=>

$2\sqrt n<n-1$    <=> $n^2-6n+1>0$ which holds for $n\geq 6$ . So $2q^2$ divides

$(n-2)!$ and $q^2$ divides $(n-2)!$

After all these we conclude that $n$ divides $(n-2)!$ when $n$ complex number $\geq 6$.


induction
combinatorics proposed
combinatorics

$2^{n-1}$ subsets are choosen from a set with $n$ elements, such that every three of these subsets have an element in common. Show that all subsets have an element in common.

This is cute :).

Note that we must have chosen precisely one set from each pair of complementary subsets of $[1,n]=\{1,2,\ldots,n\}$. Let $k\le n$ be the smallest positive integer $m$ with the property that all sets with $k$ elements were chosen. If we manage to prove that $k=n$ we'e done, for then one of the sets with $n-1$ elements must not have been chosen, and hence we have a chosen set with one element; this element will be contained in all the sets.

Assume $k<n$. If $A$ is a subset of $[1,n]$ with $n-k+1$ elements, we can find two sets with $k$ elements (which are necessarily chosen) whose intersection is contained in the complement $[1,n]\setminus A$ of $A$. Since all chosen sets must cut this intersection, $A$ cannot be chosen, so its complement must be. $A$ was an arbitrary set of size $n-k+1$, so all sets with $k-1$ elements are chosen. This contradicts the minimality of $k$.

So I assume the problem means distinct subsets or else we would have a counterexample or does the phrasing imply that already?...

[hide="hmm this is a lot more convoluted than the previous solution?"]
We proceed by induction on $ k$, the number of elements in our set $ S$.  We can verify $ k \equal{} 3$ directly with some casework, so assume that the statement is true for $ k \equal{} n \minus{} 1$.

Suppose there isn't an element in all the subsets.  We first show that there is some element in at least $ 2^{n \minus{} 2}$ of the subsets.  If that is false, then we can find $ 2^{n \minus{} 2}$ subsets of $ S$, none of which containing some fixed element $ e\in S$.  Then, we have $ 2^{n \minus{} 2}$ subsets of $ S\setminus\{e\}$, so by the induction hypothesis there is some element in all these sets, a contradiction.

If the greatest number of subsets that an element is in is exactly $ 2^{n \minus{} 2}$, choose $ 2^{n \minus{} 2}$ subsets of $ S$ and an element $ e$ for which this is true.  Then, the remaining $ 2^{n \minus{} 2}$ don't contain $ e$, and we have $ 2^{n \minus{} 2}$ subsets of $ S\setminus\{e\}$, so by the induction hypothesis there is some element $ f$ in all these sets.  However, $ f$ can't be in the original $ 2^{n \minus{} 2}$ sets, so since the number of ways to choose subsets of $ S$ both containing $ e$ and not containing $ f$ is $ 2^{n \minus{} 2}$, the first collection of subsets contains all of these sets.  Thus in particular, we have the subset $ \{e\}$, and since any 3 subsets have one element in common, every subset must contain $ e$, a contradiction.

Suppose that we have an element $ e$ in $ j > 2^{n \minus{} 2}$ subsets $ s_1,\cdots,s_j$.  Then, the sets $ t_i \equal{} s_i\setminus \{e\}$ are subsets of $ S\setminus \{e\}$, and since there are $ 2^{n \minus{} 1}$ subsets of $ S\setminus \{e\}$, for some $ k,l$, $ t_k \equal{} S\setminus t_l$, so that $ s_k\cap s_l \equal{} \{e\}$.  however, since every three subsets contain a common element, we must have that every subset contains $ e$.  This completes the inductive step.
[/hide]


A dragon has 100 heads.  A knight can cut off 15, 17, 20, or 5 heads, respectively, with one blow of his sword.  In each of these cases, 24, 2 ,14, or 17 new heads grow on its shoulders.  If all heads are blown off, the dragon dies.  Can the dragon ever die?

[hide]If the knight has to keep cutting the heads off in that order, then no, the dragon cannot die.[/hide]

[hide]after each sequence of hits/regaining of heads, the dragon will not have lost any number of heads and the sequence will continue again, so the dragon will never die.[/hide]

Oh, I didn't know the knight had to cut off heads in that order...oops :blush:

I'm not sure I understand the problem.

If the knight has to use his sequence of attacks, then after each series, the dragon will not have lost any heads.

If the knight is free to use whichever attack he wishes, repeatedly cutting off either 17 or 20 heads will do the trick, assuming he is allowed to cut off say...20 heads if only 16 remain.

If the knight can only use attacks if there are that many number of dragon heads left, he will never be able to kill the dragon, since the number of dragon heads (mod 3) is never changed by an attack.

I probably am overcomplicating things again though...

It is a bit confusing, so I don't know if I'm interpretting it right, but I see nowhere that it has to be in that sequence and I think it would be too simple if he could cut off more heads than are present.  I think the part that throws people is the placement of the word "respectively," but I think it means if 15 then 24, if 17 then 2, etc in which case towersfreak2006's (mod 3) argument should be it.  Cuz otherwise you could just think of repeatedly cutting off 15 and growing back 14, in which case any number headed dragon would be killable.

nr1337, what's your answer/interpretation?

[quote="djshowdown2"][quote="Scrambled"][quote="eryaman"][hide]Obviously if the knight cuts of 20 heads each time, and the dragon only grows 2 heads each time, the dragon will run out of heads after 6 hits.[/hide][/quote]
but thats subtracting 18 every time and 100 isnt divisible by 18[/quote]

He was just think since 20 > 2 then the dragon's number of heads would decrease each "round" and he'd eventually run out.  Like 10 heads left, if you can cut off 20... you know?  Which I know isn't the right way to look at the problem, I'm just saying  :lol:[/quote]Yep, that was my logic.

by curring off the heads in the order of: 15,17,20,5, etc., etc., etc. new heads start reappearing: in order of: 24,2,14,17, etc etc etc

so let us say:
if he cuts in this sequence, he cuts of 57 heads, then 57 grow back .: our knight is a complete idiot for trying to kill an immortal dragon

a better answer would be [hide]the[/hide] dragon [hide]can't be killed[/hide] by [hide]this head-blowing technique[/hide] thing. [hide]Consider the problem mod 3[/hide] .But there are [hide]probably other ways[/hide] to kill [hide]the dragon[/hide] like lighting [hide]a fire inside[/hide] of its [hide]trachea[/hide].

Now if the dragon had 101 heads, it could be killed.  But since our idiot of a knight crushed one head earlier today by rolling a boulder off a cliff, the dragon is now immortal.


function
analytic geometry
graphing lines
slope
ratio
calculus
integration

I was reviewing my mechanics textbook when I came across this:

[quote]$ v_{xf} \equal{} v_{xi} \plus{} a_{x}t \, ( \, \text{for constant } a_{x} \, )$

Because velocity at constant acceleration varies linearly in time according to the above equation, we can express the average velocity in any time interval as the arithmetic mean of the initial velocity and the final velocity.

$ v_{x, \text{ avg }} \equal{} \frac { v_{xi} \plus{} v_{xf} }{2} \, ( \, \text{for constant } a_{x} \, )$[/quote]

I don't understand the arithmetic mean part. How is that different from integrating the first equation to get the second?

To get velocity (as a time function) one can integrate acceleration with respect to time. Doesn't seems that is the case in your "Quote". It cames only to the average of velocity, in the case of constant acceleration...

It's not different, I don't think... Because integrating acceleration at constant accelration should give average velocity.

Perhaps I'm not being clear. Why can we assert the second equation without doing any algebra? The arithmetic mean is basically the sum of entires divided by the number of entries, but I don't see how this carries meaning when we talk about linear functions.

try drawing your velocity line

for constant acceleration your velocity is line with a slope equal to acceleration

isn't your average velocity just the average of the velocity initial and final?

just the definition of average velocity

The misunderstanding seems to lie in the fact that Prime factorization is taking as a statistical mean velocity, and here mean velocity is the ratio of displacement and corresponding time interval (say, a constant velocity that would yield the same displacement in a given time interval).

Integral of a function divided by the difference in x values is the average of the function.

Graphically, if you think about a velocity-time graph with constant acceleration, then it's a linear function, so it's like a triangle. The average velocity will be a horizontal line in the middle of that function. The integration of the function, or the area of the triangle, divided by the time, or the base of the triangle, will give the average value of the function.

The average velocity is defined as space(s)/t(time).
v=u+at and the space is s=at^2/2+ut.So average velocity=(at^2/2+ut)/t=(at+2u)/2=(at+u+u)/2=(v+u)/2

Thank you for the solution costigioni. Here's another one I thought of:

Using the mean value theorem of integrals, we have:

$ V_{av} \equal{} \frac{ \int_{t_{1}}^{t_{2}} V(t) \, dt }{\triangle t } \equal{} \frac{V_{0} \triangle t \plus{} \frac{1}{2} a \triangle t^{2} }{\triangle t} \equal{} V_{0} \plus{} \frac{1}{2} a \triangle t$

From there costigioni's solution follows.


both A and n are natural number greater than 1,show that the number of
natural number which is fewer than An-1 and conprime with An-1 is the
multiple of n.(note:An-1 means A's nth power minus 1)

PLS give a complete solution.thanks

You mean, show that $n | \varphi(a^{n}-1)$?

Well, since $a^{n}\equiv 1 \bmod (a^{n}-1)$ we conclude that the order of $a \bmod (a^{n}-1)$ must be $n$.

But this order must divide the size of the multiplicative group $\varphi(a^{n}-1)$.


My answer is not one of the choices, but I'm pretty sure I'm correct, but just in case...

Segment AB is both a diameter of a circle of radius 2 and a side of an equilateral triangle ABC.  The circle also intersects AC and BC at points D and E, respectively.  The length of AE is

a) 1.5
b) 5/3
c) (radical 3)/2
d) radical 3
e) (2 + radical 3)/2

(I don't have latex)

thanks to anyone who posts

[quote="firecricket91"][quote="indianamath"]i have very likely overlooked something obvious...

but if AB is the diameter of a circle and a side of an equilateral triangle, the circle would be tangent to BC at B...so point E=Point B.

so AE i think is equal 4...[/quote]

that made no sense, i think it should be $ 2\sqrt{3}$, just do some angle chasing[/quote]

oh yeah sorry. i drew my picture wrong :( . i think you're right.

[hide="solution"]We label the center of the circle F. Like firecricket said, we do some angle chasing and we find out that FEB is equilateral with side length 2. Then, with some more angle chasing, we arrive at FAD. We draw an altitude from F, and get two 30-60-90s. We apply trig(I mean memorizing 30-60-90 ratios) and get $ \overline{AE}\equal{}2\sqrt{3}$[/hide]

yeah, I had 2 radical 3

but obviously it's not one of the choices.

...yay i'm right

Here's a simple way to solve it:
[hide]Since $ AB$ is a diameter, $ \angle AEB$ is right. Thus $ AE$ is an altitude and it has length $ 2\sqrt{3}$.[/hide]

Nice one, Neal!  I like that one the best, for its lack of angle chasing.  I solved it by making the center O and noting that $ AO\equal{}OE\equal{}2$ and the angle between them is 120 degrees, so it's a $ 1\minus{}1\minus{}\sqrt{3}$ triangle.  I find it useful to have that triangle memorized, even though it's a near-trivial extension of a standard one.

What answer did the book give?  Was there a solution?


Slove this equation: $ x\sqrt {x^2 \plus{} x \plus{} 1} \plus{} \sqrt {x^2 \minus{} x \plus{} 1} \equal{} x \plus{} \sqrt {x^4 \plus{} x^2 \plus{} 1}$

[hide="Solution"]We notice that $ (x^2+x+1)(x^2-x+1)=x^4+x^2+1$.

Rearranging a bit, ${ x(\sqrt{x^2+x+1}-1)=\sqrt{x^2-x+1}(\sqrt{x^2+x+1}-1}$
Putting all the terms on one side gives us $ (x-\sqrt{x^2-x+1})(\sqrt{x^2+x+1}-1)=0$

Thus, either $ x=\sqrt{x^2-x+1}$ or $ \sqrt{x^2+x+1}=1$.

$ x^2=x^2-x+1\Longrightarrow x=1$ or $ x^2+x=0\Longrightarrow x=0,-1$.

Thus, we have three solutions, $ x=-1,0,1$[/hide]


algebra polynomial function calculus derivative trigonometry inequalities

Find positive real numbers a and b such that f (x) = ax−bx^3 has four extrema on [−1,1], at each of which |f(x)| = 1.

A polynomial of degree $ n$ can have no more than $ n\minus{}1$ points with derivation equal to $ 0$. Therefor, your question has no answer.

Yes, but the extrema of a differentiable function on a closed interval need not happen only at points of derivative 0 ....

Preview
Two of your extrema are going to be at your end points, so for x={1,-1},|ax-bx^3|=1 implies that |a-b|=1 and b=a+/-1. The other two extrema will be at the zeros of the derivative, a-3bx^2=-3b(x+a/3b)(x-a/3b), which are {a/3b,-a/3b}. For the case that a-b=1, b=a-1, the equation is ax-bx^3=1. Then substitute a/3b for x and a-1 for b, and solve for a, which will give you B as well. The equation is a(a/3(a-1))-(a-1)a^3/27(a-1)^3=1. One graph is the negative of the other (a and b, and -a and -b are both solutions), and one extrema is the negative of the other, so you only need to solve for this case and derive the others from it.

Yes.  As it happens, one of the two resulting polynomials satisfies $ f(\cos x) \equal{} \cos 3x$, which seems like a rather odd coincidence -- is there any way to show it directly?  Also, are there any solutions to this problem if we allow the $ x^0$ and $ x^2$ coefficients to be nonzero?

JBL, if we make the x^0 and x^2 coefficients nonzero, we're shifting the polynomial's single point of inflection.  If it has two relative extrema, one must be to the right of the point of inflection and the other must be left of it, and one must be above it and the other must be below it.  I expect that if the point of inflection is on the x interval (-1,1) and the y interval (-1,1), there will be a solution.  Otherwise, there won't be.  If there is one, you can use a method similar to the one I used here, but it'll be a bit more complicated.  I'm not sure what you mean when you say that one of the resulting polynomials satisfies f(cos(x))=cos(3x).  Isn't that a periodic function?  Do you mean that the polynomial satisfies it on the interval [-1,1]?  I'm pretty new to this and I've never seen a function defined like that, that's interesting.  f(cos(x))=cos(3x) is equivalent to f(x)=cos(3cos^-1(x)) on the interval [-1,1], but the latter is only defined on that interval and the former seems to be defined for all of x.  Does that kind of function definition have a name?

JBL is talking about the [url=http://en.wikipedia.org/wiki/Chebyshev_polynomials]Chebyshev polynomials[/url].  The condition that $ f$ is a polynomial satisfying $ f(\cos x) \equal{} \cos nx$ is enough to uniquely identify $ f$ as a polynomial on $ [\minus{}\infty, \infty]$ because its behavior is specified at an infinite number of points (all of $ [\minus{}1, 1]$).

One way to think about them (especially within the context of this problem) is that they translate the interval $ [ \minus{} 1, 1]$ to the interval $ [ \minus{} 1, 1]$ in a certain unique way; that is, JBL appears to be suggesting the result [url=http://en.wikipedia.org/wiki/Chebyshev_polynomials#Minimal_.E2.88.9E-norm]here[/url].


respond in spoiler, extra pts for generalizing, due date is sep. 28 6 PM EST

1. An isosceles triangle with base 10 cm has area 60 cm.:^2:.What is its perimeter?

2. 3 circles of radii 2, 4, and 6 are tangent to each other (a threesome!) externally. What is the area of the triangle formed by connecting their centers?

3. In a unit square ABCD, the point G is on AB and b units away from A, and the point F is on CD and a units away from D. The point E is the intersection of AF and DG. What is the distance from E to AD?

4.  Tree A stands 5 meters from Tree B. A map indicates that a treasure is buried 2 meters from Tree A and 4 meters from Tree B. What is the greatest number of places you are required to dig to be sure of finding the treasure?

5. The lengths of two sides of a parallelogram are 5 and 12. Find the sum of the squares of the diagonals of the parallelogram.

[hide]

1) 36 cm (use pythagorean triplets)

2) Triangle would have legs 6 and 8 giving it an area of 24.

3) Don't feel like drawing it out

4) 2

5) Again, drawing laziness

[/hide]

[hide]

1.h = 60/(10/2) = 12

side  =  :sqrt: (5 :^2:  + 12 :^2: ) = 13

p = 2*13 + 10 = 36

generalize:

b=base

A = Area

h = A/(b/2)

side =  :sqrt: (h :^2:  + (b/2) :^2: )

p = 2 :sqrt: ((A/(b/2)) :^2:  + (b/2) :^2: ) + b

2. 6-8-10 triangle... right... A = 24

generalize: r_1, r_2, r_3 are the radii

herron's

A =  :sqrt: ((r_1+r_2+r_3)(r_1)(r_2)(r_3))

3.

AG = b

DF = a

EH is perpedicular to AD at H

let's say EH = x

and DH = m and AH 1-m

So then porportions...

x:b = m:1

x:a = (1-m):1

m = (x/b)

m = 1- (x/a)

(x/b)  =  1 - (x/a)

ab - bx = ax

(a+b)x = ab

x = ab/(a+b)

and doesn't that generalize the prob?

4. You have two points A and B representing the trees. they are 5 meters away... then you draw circle of radius 2 and 4 from A and B respectively... they intersect in two points... so

the answer is 2???

generalize:

draw a locus...and see how many points they intersect in???

5. for parallelogram ABCD

the formula is 2(BC:^2: + DC :^2:) = AC :^2:  + BD :^2:

so the answer is 338 units

how you get that is:

you have parallelogram ABCD.

You draw BD parallel to AC that meets DC at F.

You draw BE perpendicular to DF at E

BE :^2:  + (DC + CE):^2: = DB:^2:   (since DC+CE = DE)

so then

BE:^2: + DC:^2: + 2DC*CE + CE:^2: = DB:^2:

since BE:^2: + CE:^2: = BC:^2:

then DC:^2: + 2DC*CE + BC:^2: = DB:^2: (we'll say this is equation #1)

BE:^2: + (CD - CE):^2: = AC:^2:

ok... now we have BE squared plus (CD - CE):^2: which is actually (CF - CE):^2: but since DC = AB = CF we just used CD since it's actually a side... and BF = AC we used AC

you expand that to get

BE:^2: + CD:^2: - 2CD*CE + CE:^2: = AC:^2:

and since again BE:^2: + CE:^2: = BC:^2:... we have

CD:^2: - 2CD*CE + BC:^2: = AC:^2: (equation #2

now we have equations #1 and #2... we add them together...

DC:^2: + 2DC*CE + BC:^2: = DB:^2:

CD:^2: - 2CD*CE + BC:^2: = AC:^2:

we cancel the 2DC*CE

and we get

2(BC:^2: + DC :^2:) = AC :^2:  + BD :^2:

and to generalize...

use the formula... and if you forgot through my ramblings...

2(BC:^2: + DC :^2:) = AC :^2:  + BD :^2:

[/hide]


I'm given these 'description' of a graph:

Consider the function p(x), -b<x<b, which is made up of the two straight line segments which connect (-b,0) to (0,c) and (0,c) to (b,0), where b, c are both positive

I need to graph p(x) and determine a formula for p(x) - i.e. I need ONE formula

But using the equation of a line I actually came up with two formulas:

1. y=(c/b)(x+b)

2. y=(-c/b)(x-b)

So now I need a way to combine these two formulas and end up with a single function that will still give me the graph I started out with...

Any ideas?

The absolute value function might be useful.


Coin $ A$ is flipped three times and coin $ B$ is flipped four times. What is the probability that the number of heads obtained from flipping the two fair coins is the same?

$ \textbf{(A)}\ \frac {19}{128}\qquad
\textbf{(B)}\ \frac {23}{128}\qquad
\textbf{(C)}\ \frac {1}{4}\qquad
\textbf{(D)}\ \frac {35}{128}\qquad
\textbf{(E)}\ \frac {1}{2}$


function
calculus
derivative
algebra proposed
algebra

Solve of the equation:
$\large \sqrt{x}+\sqrt[3]{x+7}=\sqrt[4]{x+80}$

Obviosly$\large \sqrt{x}+\sqrt[3]{x+7}-\sqrt[4]{x+80}$ increase, therefore x=1 unique solution.

How can you say that the function is increasing ? Is it so obvious ?  :maybe:

Of course I know that it's the way to prove it (by derivative), but how can you deduce ?  :(


Let $ S(n)$ be the sum of digits of positive integer $ n$.Show that the sequence
$ S(2008), S(2008^2), S(2008^3),\cdots$ isn't bounded below.

More than :
Let a is an even number and $ 5\not |a$ then the sequence : $ S(a^n)$ is not bound.

[quote="ZetaX"]You mean bounded above, right¿ (It's surely bounded below by $ 0$).

In fact for any $ a \neq 10^k$, the sequence $ S(a^n)$ is not bound as we can get $ S(a^n)$ to start with any sequence of digit's (especially an arbitrary number of $ 9$'s).[/quote]
It is the second time at this week,when i see that someone use this fact in proving another problem.This very useful fact(or maybe a theorem?),indeed.

Dear [b]Erken[/b] could you post some refference or a prof of this theorem :?:  Thank you in advance.


In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded 1 point, the loser got 0 points, and each of the two players earned 1/2 point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned in games against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points, against the other nine of the ten). What was the total number of players in the tournament?

[hide="Hint to get you started"]
This is AIME number 14???

We know of at least 10 players, so lets say n+10 players play.

The general formula for the number of games played is n(n-1)/2, so the bottom 10 played 45 games amongst themselves. This is half then number of games played by the bottom 10, so they played 90 games in total.

So by our formula, the 10+n players in the tourney played (10+n)(9+n)/2 games.

Note the number of points in the tournement equals the number of games.
Find a way to combine what you know into a slove-able expression.

[/hide]


Show that: \[3xyz(x+y+z) \leq (xy+yz+zx)^{2}\]

[hide]Upon expansion, we find that the given inequality is equivalent to
$x^{2}{(y-z)}^{2}+y^{2}{(z-x)}^{2}+z^{2}{(x-y)}^{2}\geq 0.$[/hide]

[quote="rem"]$x^{2}yz+y^{2}xz+z^{2}xy\le x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}$
This is true by Muirhead(or AM-GM, too)[/quote]
Also true by Cauchy!  :lol:

Denoting $a=xy,b=xz,c=yz$ gives us that the ineq is eqiuvalent to
$(a+b+c)^{2}\geq 3(ab+bc+ac)$,which is equivalent to $a^{2}+b^{2}+c^{2}\geq ab+bc+ac$.Which is well-known.[Proof for it:$LHS-RHS=\frac12((a-b)^{2}+(b-c)^{2}+(c-a)^{2})$]

We've got the inequality: $(a+b+c)^{2}\geq 3(ab+bc+ca)$;$a,b,c \in \mathbb{R}$

Substitute: $a=xy,b=yz,c=zx$


Groucho, Zippo, and Harpo have $ \$65$ altogether. Groucho and Zippo together have $ \$1$ more than Harpo has, and Groucho has $ \$15$ more than Zippo. How many dollars does Groucho have?

This is just three equations in three unknowns:
$ G\plus{}H\plus{}Z\equal{}65$
$ G\plus{}Z\equal{}H\plus{}1$
$ G\equal{}Z\plus{}15$
Now all one has to do is solve these three equations, which is straightforward.

Uh...it said the answer was $ 21$, but I got $ \boxed{24}$..

Call Zippo $ x$ money.

Thus $ x\plus{}x\plus{}15\plus{}2x\plus{}14\equal{}65$

$ 4x\equal{}36$
$ x\equal{}9$.

Thus Groucho has $ 9\plus{}15\equal{}\boxed{24}$....

$ 24$ is correct. The answer key is incorrect. Report an error.


algebra
polynomial
abstract algebra
algebra proposed

Let $f$ be an irreducible(in $Z[x]$) monic polynomial with integer coefficients and of odd degree greater than $1$.Suppose that the modules of the roots of $f$ are greater than $1$ and that $f(0)$ is a square-free number.Prove that the polynomial $h(x)=f(x^{3})$ is also irreducible.  :)  :)

It is very beautiful problem , Same as romanian TST 2002. :roll:

Can you post your solution?As you say,i think you have the offical solution. :)  :)

If you solved the same problem at romanian`s TST 2002 ,you can easily solve this problem too. :?

It's also appears in a RMO,for 11th grade(if i remember exactly)
Maybe Valentin will be kind to post the offical solution?  ;)


LaTeX
calculus
integration
logarithms
calculus computations

integrate
log(x+sqrt(x^2 +a^2))dx
srry for latex

Via integration by parts,

$ \int \ln \left( x\plus{}\sqrt{x^{2}\plus{}a^{2}}\right) dx$

$ \equal{}\allowbreak x\ln \left( x\plus{}\sqrt{\left( x^{2}\plus{}a^{2}\right) }\right) \minus{}\int \frac{x}{\sqrt{x^{2}\plus{}a^{2}}}dx$

$ \equal{}\allowbreak x\ln \left( x\plus{}\sqrt{x^{2}\plus{}a^{2}}\right) \minus{}\sqrt{x^{2}\plus{}a^{2}}\plus{}C$


function
complex analysis
advanced fields
advanced fields unsolved

$f(z)$ is analytic in $C-{0}$

$|f(z)|\leq \frac{1}{\sqrt|z|}$ for all $z\in C-{0}$

Prove that f is identically zero

With that slower-than-$1/z$ growth rate, $f$ must have a removable singularity at the origin. Filling that in, it is a bounded entire function, hence constant, and the only possible constant value is zero.


$ \partial E$ is the boundery of E.

For any A and B, $ (\partial A)\cup (\partial B) \supset (\partial (A\cup B))$.

but for countable set $ E_i, E_i \cap E_j = \emptyset (i \neq j)$,
$ \cup (\partial E_i) \supset \partial (\cup E_i)$.

Above fomular is right?

Thanks

[quote="water2011"]but for countable set $ E_i, E_i \cap E_j \equal{} \emptyset (i \neq j)$,
$ \cup (\partial E_i) \subset \partial (\cup E_i)$. [/quote]
I don't think so. In $ \mathbb{R},$ let $ E_i\equal{}[i,i\plus{}1).$

Then $ \cup (\partial E_i) \equal{}\mathbb{Z}$ while $ \partial (\cup E_i)\equal{}\emptyset.$

[quote="water2011"]$ \cup (\partial E_i) \supset \partial (\cup E_i)$.

[/quote]
I had revised the formular.


algebra
polynomial
group theory
abstract algebra
induction
Galois Theory
superior algebra

let $\phi_m(x)$ be the mth cyclotomic polynomial. Suppose $p$ doesnt divide $m$. Prove that it has $\phi(m)$ integer roots in $Z_p$ if and only if $p=1$ (mod m).

[hide]Hint: Consider a splitting field for it over $Z_p$, call it $L$ and prove $G(L: Z_p)$ is isomporphic to the multiplicative subgroup of $Z_m$ generated by the powers of $p$.[/hide]

Thats a direct consequence of/equivalent to the known theorem:
Let $q$ be prime divisor of $\Phi_n(m)$, then $q|n$ or $q\equiv 1 \mod n$.

Prove:
By the well known formula $\prod_{d|n} \Phi_d(x) = x^n-1$, we get that $m^n \equiv 1 \mod q$, thus $ord_q(m) | n$.
When $ord_q(m)=n$, we get because of fermats theorem that $n|q-1$, so $q\equiv 1\mod n$.
So assume that $ord_q(m)$ is a true divisor of $n$ from now.
Then $m$ is also a zero of some $\Phi_k(x)$ with $k|ord_q(m)|n \neq k$. So $x^n-1$ has a multiple root $\mod q$. But looking at it's derivation gives that this is just possible for $q|n$, so we are done.

Now this gives that there are no zeros modulo primes not $\equiv 1 \mod n$ and not dividing $n$.
So let $p \equiv 1 \mod n$ be a prime. Then any primitive root $\mod p$ will do.

I know, but please read the hint I gave, Your proof is the usual proof I would also give, but today I found a wonderful one! and also a more general statement.

Here's what I did:

If $p\not|m$, then $x^m-1$ has $m$ distinct roots in a finite extension of $\mathbb Z_p$, and these roots must be of the form $1,\zeta,\zeta^2,\ldots,\zeta^{m-1}$ for some $\zeta\ne 1$, because every multiplicative subgroup of a finite field is cyclic. Now, the roots of $\phi_m$ are precisely $\zeta^n,\ 1\le n<m,\ (n,m)=1$ (we can see this by induction on $m$, using $x^m-1=\prod_{d|m}\phi_d$). Now, if $\phi_m$ has even [i]one[/i] root in $\mathbb Z_p$, then $x^m-1$ has all its roots in $\mathbb Z_p$, because they are all powers of that one root of $\phi_m$. Since the number of roots of $x^m-1$ in $\mathbb Z_p$ is $(m,p-1)$, we get $m=(m,p-1)$, i.e. $m|p-1$.

P.S.

This is just one implication; the converse is obvious :).

I rememeber teh problem has beeen posted before, in the olympiad section.

[quote]Hint: Consider a splitting field for it over $Z_p$, call it $L$ and prove $G(L: Z_p)$ is isomporphic to the multiplicative subgroup of $Z_m$ generated by the powers of $p$.[/quote]

"Notational" question:
Is $G(L: Z_p)$, the galois group of the galois extension $L/Z_p$ ?
(I guess yes :))


find real triples(x,y,z)that satisfy x^4+y^4+z^4-4xyz=1

First, are you sure this is the correct statement. Secondly, this subforum is dedicated to very difficult problems that you cannot solve and you think are very interesting and difficult.

I think harazi means there should be integer triples, and the problem is easy because of AM-GM :)


let $ a,b,c$ is nonegative . prove that:
$ a^6\plus{}b^6\plus{}c^6\minus{}3a^2b^2c^2 \ge 16(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2$

Could somebody point out where I have gone wrong?

My solution:

$ a^6\plus{}b^6\plus{}c^6\minus{}3a^2b^2c^2\equal{}\frac{1}{2}(a^2\plus{}b^2\plus{}c^2)(a^2\minus{}b^2)^2(b^2\minus{}c^2)^2(c^2\minus{}a^2)^2$.

$ (a^2\minus{}b^2)^2(b^2\minus{}c^2)^2(c^2\minus{}a^2)^2\equal{}(a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2$.

Thus, we are to prove that $ (a^2\plus{}b^2\plus{}c^2)(a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2\ge 32$. (Assuming $ a\neq b\neq c$, since if two or more of them are equal, RHS becomes 0 and LHS gets automatically greater.)

However, above inequality is not true.

We have also

$ a^6 \plus{} b^6 \plus{} c^6 \minus{} abc(a^3\plus{}b^3\plus{}c^3) \ge 16(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2$.

[quote="FantasyLover"].......

Thus, we are to prove that $ (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2\ge 32$. (Assuming $ a\neq b\neq c$, since if two or more of them are equal, RHS becomes 0 and LHS gets automatically greater.)

However, above inequality is not true.[/quote]
If you can show counterexamples, i.e. values for which the inequality does not hold, then it $ \implies$ the inequality originally proposed is not correct. Your method is all right.

Although tuantam1lan's original inequality and my inequality are exactly same (I haven't changed a bound or anything), $ 1, \frac{1}{2}, \frac{1}{3}$ satisfies tuantam1lan's inequality and it doesn't satisfy mine :(

[quote="spanferkel"][quote="FantasyLover"]Could somebody point out where I have gone wrong?

My solution:

$ a^6 \plus{} b^6 \plus{} c^6 \minus{} 3a^2b^2c^2 \equal{} \frac {1}{2}(a^2 \plus{} b^2 \plus{} c^2)(a^2 \minus{} b^2)^2(b^2 \minus{} c^2)^2(c^2 \minus{} a^2)^2$.

[/quote]
WHat about the degrees of your identity?[/quote]
Whoops...

It was $ \frac{1}{2}(a^2\plus{}b^2\plus{}c^2)((a^2\minus{}b^2)^2\plus{}(b^2\minus{}c^2)^2\plus{}(c^2\minus{}a^2)^2)$...

[quote="Vasc"]We have also

$ a^6 \plus{} b^6 \plus{} c^6 \minus{} abc(a^3 \plus{} b^3 \plus{} c^3) \ge 16(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2$.[/quote]

Dear Vasc, this inequality can also be true
$ a^6 \plus{} b^6 \plus{} c^6 \minus{} abc(a^3 \plus{} b^3 \plus{} c^3) \ge 16.2 (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2$.

It seems that the general case $ a^6 \plus{} b^6 \plus{} c^6 \minus{} abc(a^3 \plus{} b^3 \plus{} c^3) \ge k(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2$.[/quote]
Best constant $ k$ lies between $ 16.2$ and $ 16.3$. I don't know if my answer is correct ?


what was the formula for the
perimeter of a rectangle?

I need it for the national competition.
Thank you.


:( The problem of proving a sets question is a big headache for me:

for example:
Prove that A-B=A <->B-A=B.

can anybody teach me abt how 2 tackle this such of questions,pls???

Try drawing makeshift sets representing $ A\B$ and then try doing it again for some sets such that $ A\B = A$. What can you notice about the location of $ B$ relative to $ A$?
This should give you an idea of how to approach this.

A and B are mutually exclusive ,right??? :maybe:
Then how i going to prove B-A ⊆ B and B ⊆ B-A so that B-A=B if i go frm left hand side to right???


[quote=harazi]Let $x,y>1$ different integers, coprime. Prove that for infinitely many primes $p$, the exponent of $p$ in $ x^{p-1}-y^{p-1}$ is odd.[/quote]
See [url=https://artofproblemsolving.com/community/c6h196188]here[/url], [url=https://artofproblemsolving.com/community/c6h272718]here [/url]and [url=https://artofproblemsolving.com/community/c6h606020]here[/url] for a complete solution.


The AMC/AIME class that's supposed to be going now -- did you all get thrown out of the room and find yourselves unable to log back on?

no, were just sitting here with nobody teaching us, as a result theres nothing on the screen and nobody has posted anything except for a classroom tip.

Drat.  Both Mathew and I have been kicked off, I don't know why.  Sorry about this.

It's okay.. I'll do an AIME while I wait.. lol

Well, here's a problem while you wait (from the Mandelbrot Problem Corner, of roughly AIME difficulty, I'd say):

Given points A, B, C and D on a circle such that the measure of arc AB = the measure of arc BC, with D on minor arc AB.  Let E be the foot of the perpendicular from B to (segment) CD.  If (the length of segment) AB = 13, BD = 9, and BE = 5, compute the length of segment AD.


linear algebra
matrix
calculus
derivative

Ok, this is a really dumb question, but I have to understand. The question deals with a minor detail in the proof "A square matrix $A$ is invertible iff the equation $AX = Y$ has a unique solution for all $Y$". I'm not sure the question should be put here, since the question could be applied on real numbers or any other structure.

"Assume that $A$ has an inverse $B$. It is easy to show that $X = BY$ is one solution to the equation. Since $BA = E$ the solution is also unique, because $AX = Y \Rightarrow X = BY$".

My problem: Nowhere in this proof is it mentioned that inverses are unique. Is this not required for the proof? It seems like it should be, but I can't see where. Therefore, I don't think I fully understand this. Can anyone enlighten me?

edit: To make my question a bit clearer: Is the uniqueness of inverses a part of the statement $AX = Y \Rightarrow X = BY$ or is this not a necessary condition? Is it necessary somewhere else?

The uniqueness of inverses holds in all grup ..
  :cool:

[quote="Kalle"]I know, but is it necessary to know that?[/quote]
I am not sure to understand your question ??!!  Because of the element $A$ is invertible then its inverse $B$ is unic , so the solution of $AX=Y$ is unic ..

  :cool:

If B and B' are both inverse matrix of A, you can compute the product BAB' as
BAB'=(BA)B'=IB'=B'
BAB'=B(AB')=BI=B

Thus B=B'

[quote="Diogene"][quote="Kalle"]I know, but is it necessary to know that?[/quote]
I am not sure to understand your question ??!!  Because of the element $A$ is invertible then its inverse $B$ is unic , so the solution of $AX=Y$ is unic ..
[/quote]
This is my question.  I thought it was clear enough...  My problem is that how do we arrive at that $AX = Y$ is equivalent to $X = A^{-1}Y$? My book's derivation does not mention that it is necessary for there to be only one inverse. So I asking if it is, and why. Is it for the implication $AX = Y \Rightarrow X = A^{-1}Y$?

Thanks bizz, but that's not what I meant to ask. I know my question is a trivial one, since it is pretty much equivalent to knowing how we solve $5x = y$ in real numbers, but I'm still not sure.

We can derive the uniqueness of inverses from this; we could also prove uniqueness separately.

Suppose that $B$ and $C$ are both inverses of $A$.

$AX=Y$ implies $X=BAX=BY$, and $A(BY)=ABY=Y$, so $X=BY$ is the unique solution to $AX=Y$. Similarly, $X=CY$ is the unique solution to $AX=Y$. This implies that $BY=CY$ for all $Y$, so $B=C$ as linear transformations.

Thank you jmerry!

I know I ask really dumb questions sometimes (like this, and the other one you answered about differentiation), but I have a tendency to get stuck with formalism sometimes and not trust my own conclusions . Thanks again.


I don;t know whether this should be posted in this section. What I am looking for is a computation formula for the Feigenbaum constants $\delta=4.669201660910299\ldots$. Could we know that they are transcenental numbers?

Maybe this will help you :
http://mathworld.wolfram.com/FeigenbaumConstantApproximations.html
http://mathworld.wolfram.com/FeigenbaumConstant.html


So the tournament's September 27 at Alpharetta High School. If you haven't registered, you can still make the late registration I believe.

Also, who's going?

Walton's coming. And no, I haven't actually registered yet. I did tell Maja that we were coming and that counts right?! :-)

My team won't be there. It's National Hunting & Fishing Day, when I don't have to have a license to go hunting and fishing; therefore I'll be busy. My [i]team[/i] has the annual club-party then; between huntin', fishin', and flippin' squirrel burgers, I'll be downright occupied.

I assume that sweet tea will be involved?  :D

I'll probably be there. What time does it start?

I can't participate until after around 1:30pm because of SAT prep.  :(


To the nearest whole number, find the number of revolutions a 27-inch diameter bicycle wheel makes in a five-mile race. Use 3.14 as an approximation for $ \pi$.

A 27 inch diameter wheel would have circumference $ 27 \cdot 3.14 \equal{} 84.78$ inches. It travels for $ 5$ miles, or $ 5 \cdot 5280 \cdot 12 \equal{} 316800$ inches. Dividing, we get the number of revolutions made. $ \frac{316800}{84.78} \equal{} 3736.73036... \rightarrow 3737$.