[quote="rrusczyk"]The Getting Started board is for students who are just getting started in problem solving.  Problems and questions that appear here should be those that might be on MATHCOUNTS or beginning AMC tests or other beginning problem solving programs.[/quote]

I think there are getting to be WAY too many intermediate-level (or higher) problems and questions posted on this Getting Started forum, perhaps because it doesn't take long for someone to get good enough at problem-solving to forget how hard much easier problems are for someone without experience. As a rule of thumb, I would expect that anyone who is taking the AMC 12 for the second time, or who is already too old to participate in Mathcounts after having been involved in Mathcounts when younger, is probably really at an intermediate level (or perhaps an advanced level) and should be posting questions on the Intermediate forum (or the Advanced forum).

[quote="rrusczyk"]Generally, the Getting Started board will include topics of middle school and high school math such as word problems, linear and quadratic equations, basic geometry, divisibility, fundamental counting techniques, etc.[/quote]

Mr. Rusczyk here is really referring to United States grade levels when he speaks of "high school math" as leading to getting started level problems. In most parts of the world, the math topics he indicated are all junior high topics. Interestingly, he probably overestimated how easy most participants here think geometry problems are when he posted the paragraph quoted above. There is a level of geometry that I would call "middle school geometry" that is pretty easy--or at least my son and I had learned most of that at fifth-grade age--but which can make for nontrivial math contest problems. That, I suppose, is what Mr. Rusczyk meant by "basic geometry." Any geometry that shows up on the last half of an AMC 10 or AMC 12 is probably better deemed intermediate level geometry for most participants on this board.

[quote="rrusczyk"]Moderators will occasionally post Challenge Problems (which will be designated as such) - if you feel you are already sufficiently advanced that most of the problems in this board are pretty easy for you, [b]please leave the Challenge Problems for less experienced students.[/b]  Anything not marked a challenge should be considered a question - we welcome our more experienced members to answer these questions.  Also, if you are not a moderator but have a good Challenge Problem, please post it![/quote]

Note the emphasis added to one part of the quoted passage. The younger guys are not going to get practice in problem-solving here unless at least one forum, which logically should be the Getting Started forum, is a place where problems stay posted to entice the problem-solving beginners without the experts posting answers--worse still, without even using spoiler text. There is plenty of intellectual challenge for the experienced problem-solvers in the intermediate or advanced forums--or could be, if more of the problems at those levels that appear on Getting Started were put at their proper level.

I am not a moderator of this forum, but I would like to respond to Mr. Rusczyk's invitation and start posting challenge problems for all comers at a true [b]getting started[/b] level, the only level at which I can reliably solve math problems. I have an abundant collection of Chinese-language problem-solving books and many of the best English-language books with problems at the appropriate level for this forum, from which I could adapt a great many problems suitable for the people who really are just getting started. But I'd rather do that over on Brand X if the participants here don't THINK before they post, putting problems at unfitting levels or posting solutions without spoiler.

Can you tell I'm feeling frustrated  :wink:  ? I think there are a lot of really smart guys and gals here, who can help one another--and help the younger, less experienced participants--learn a lot of math. But the basic netiquette of "read your fine instructions" and "keep posts on topic" needs more practice here. Capable people get even more capable if they practice, and one thing that would be good to practice on these forums is [b]reading[/b] carefully to put posts in the most fitting forum, and to make sure that posted solutions use spoiler text enough not to spoil the fun of problem-solving. Especially, I would caution the people who have been involved in problem-solving for years now to remember their first few months of problem-solving (if they still can remember): that is the getting started level, not the level of anything that looks easy or trivial now, after gaining experience.

End of vent mode  :)  Yes, I would like the people who are looking for intermediate-level problems to be able to get one-stop shopping by looking for those on the Intermediate forum, while people who are new to problem-solving find problems here on the Getting Started forum that are approachable for beginners without a lot of mathematical techniques in their problem-solving toolchest. Because there is no upper bound to the difficulty level on the Advanced forum, it's not like anyone should run out of problems to solve while the AoPS site becomes more welcoming for the many lurkers who are, er, probably just getting started in problem-solving.

One problem is, I don't think there is a lot of "Getting Started" students participating this forum frequently, but I totally agree with you.

But there might be more if there are lots of problems for them to solve.

[quote="tokenadult"]
I am not a moderator of this forum, but I would like to respond to Mr. Rusczyk's invitation and start posting challenge problems for all comers at a true [b]getting started[/b] level, the only level at which I can reliably solve math problems. I have an abundant collection of Chinese-language problem-solving books and many of the best English-language books with problems at the appropriate level for this forum, from which I could adapt a great many problems suitable for the people who really are just getting started. But I'd rather do that over on Brand X if the participants here don't THINK before they post, putting problems at unfitting levels or posting solutions without spoiler.

[/quote]

I agree whole-heartedly that you should become a moderator, so we could have an example.

[quote="Chinaboy"]I agree whole-heartedly that you should become a moderator, so we could have an example.[/quote]

Well, I'll defer volunteering to be a moderator for a while, but I would be open to doing that, and meanwhile I'll start preparing the first few posts from my projected Getting Started Challenge Problem Series. Meanwhile, there are some good examples of the right level in some of the Mathcounts Challenge posts recently put up on the [url=http://www.artofproblemsolving.com/Forum/viewforum.php?f=4]Mathcounts Forum[/url] by Syntax Error. So the getting started folks can try those problems out while I hunt up some others for posting here.


How many people qualify for USAMO and AIME each year?
I heard that AIME is like around 200 or so.

Uh...no...It's on AoPS like 10 times, and the AMC site.

Last year, there was 514 people who took the USAMO (I think).  And then between 8,000 and 9,000 people who took the AIME.

The statistics are all on the site. Even some people's scores are on the site

Detailed statistics for the 2007 AIMEs, including information allowing you to determine the total number of participants is at:
http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2007-aa/07-AIMEstats.shtml

You can find similar information for the years 2001-2006 by searching on :
http://www.unl.edu/amc/e-exams/e7-aime/archiveaime.shtml


An isosceles triangle is inscribed in a circle so that one of its sides is a diameter. The ration of the area of the triangle to the are of the circle is 1:a. Express a in terms of pi.

If the triangle is an isosceles triangle, then the height must be equal to the radius of the circle. The base is 2 times the radius. The area of the triangle is (1/2)base*height=(1/2)*2*r*r=[b]r^2[/b] The area of the circle is [b]r^2*pi[/b]. The ratio of the area of the triangle to the area of the circle is (r^2)/(r^2*pi) which is equal to 1/pi. So a must be just one pi or

[b]pi[/b]

PowerOfPi


topology
geometry
3D geometry
sphere
real analysis
real analysis unsolved

Let $ B\left(0,1\right)\subset\mathbb{R}^{n}$ be the open ball centered at $ 0$ of radius $ 1$.  Show that the one-point compactification of $ B\left(0,1\right)\subset\mathbb{R}^{n}$ is homeomorphic to the $ n$-sphere $ S^{n}$.

Essentially, this is the same as showing $ B\left(0,1\right)\sim S^{n}\setminus\left\{\left(1,0,\dots ,0\right)\right\}$, right?  How do we construct such a homeomorphism?  Thank you :)


inequalities
analytic geometry
LaTeX
real analysis
real analysis theorems

Suppose that there is two family of cardinal numbers, (a_i), (b_i), i are elements of an index set I, with a_i < b_i for all i in I.

Then we will have the sum of the family (a_i) smaller than the product of (b_i) (with the arithmetics in cardinal numbers.)

I've read through several book for the proof, and they are all in the same form - two parts:
(1) Prove that the sum of the family (a_i) smaller than or equipotent to the product of (b_i)
(2) Prove that the equality cannot hold.

For the second part of the proof, it says that whatever subset P of the product of (b_i) that is equipotent to the sum of the family (a_i), P can be divided into the union of a family (P_i), in which for all i in I, P_i is eqipotent to a_i. Then the i-th coordinate of every point in P_i consistutes a subset, which is smaller than or equal to the equipotential of a_i, and thus (b_i) \ (a_i) is non empty, and by the Axiom of Choice, the cartesian product of (b_i) \ (a_i) for all i in I is non-empty.

Then, there always exist a point in which P cannot be mapped to but is in the product of b_i, and thus the equality cannot hold.

I have two questions, unsolved in this two weeks:
(1) The division of the subset P, is restricted, and is only took once, which does not seem to be arbitary enough.
(2) Although there is always one extra point in the product of b_i, but for sets which is large enough, adding one will still be equipotent to itself. (By the definition of infinite sets)

This post probably belongs more naturally in the Advanced Topics forum, where you placed your other one.  Perhaps a moderator will move it.
Also, the forums have LaTeX enabled; see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690
for more.

[quote="darken"]Then, there always exist a point in which P cannot be mapped to but is in the product of b_i, and thus the equality cannot hold.
...
(2) Although there is always one extra point in the product of b_i, but for sets which is large enough, adding one will still be equipotent to itself. (By the definition of infinite sets)[/quote]

I believe you've misunderstood how this proof is working in your second objection.  It's a proof by contradiction.  Assume the two cardinals are the same.  Then there exists a bijection between them (or possibly a surjection or injection from one to the other).  Then you find an element of one side that cannot possibly have been mapped to by the bijection.  But this is a contradiction, so the claimed bijection cannot exist and the two cardinals are not equal.  (This is the classic structure of a diagonal argument.)  I strongly suspect that this is actually what the structure of the proof is.


$ x^t\plus{}x^n\equal{} y^n$

Find all solutions, with proof, that satisfy the above equation, where $ t$ is even and $ n$ is prime.

A nearly same problem appeared in [b]INMO 2008(Prob. no. 2)[/b]:
Find all triples($ p,x,y$) such that $ p^x\equal{}y^4\plus{}4$, where $ p$ is a prime and $ x,y$ are natural numbers.

I solved the problem(at home, of course!) using arguments & cross-arguments. I am eager to see if the solution of the given problem also follows the same way! :)


In right triangle ABC, let AC be the hypotenuse and angle CAB have measure alpha.  Place point D on the opposite side of AC as B so that angle DCA is right.  Let angle DAC have measure beta. Let the foot of the perpendicular from D to AB be E.  Quadrilateral AECD is cyclic, and I suspect that an application of Ptolemy's theorem on this quadrilateral can be used to prove the sine sum-angle formula (sin(alpha + beta)), but I can't make it work.  Help requested.

yes, you can use ptolemy to prove sin addition. So keep working on it.


Hi!

Given a $ 2$ by $ n$ rectangle, find the number of modes in which this rectangle can be totally covered  with $ 1$x$ 1$ or $ 1$x$ 2$ pieces. The pieces can be layed both vertically and horizontally. The pices can't overlap and can't get out the rectangle.

Examples:

$ n \equal{} 1$ => 2
$ n \equal{} 2$ => 7 (see the figure)

So, let the number of ways to tile the 2x$ n$ be $ f(n)$. From a tiling of f(n-2), we can add any of those 7 to the end to make a tiling of a rectangle of length n, or from one of f(n-1), we can add any of 2, for $ f(n) \equal{} 7f(n \minus{} 2) \plus{} 2f(n \minus{} 1)$. However, this approach is flawed.

[hide="overcounting rules"]
for a 2x3 rectangle, I just counted:

[code]
+--+--+--+
|  |  |  |
+--+--+--+
|  |  |  |
+--+--+--+
[/code]
[i]twice[/i].

[/hide]

In English, "rectangle."

This is one of the 8 tiling problems posted (but only 2 solved) [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9736]here[/url], lo those many years ago.  Like most such problems, the best approach is to hit it with the right recursions.  Let $ a_n$ be the number of ways to tile a $ 2 \times n$ rectangle with $ 1 \times 1$ and $ 2\times 1$ tiles and let $ b_n$ be the number of ways to tile a $ 2 \times n$ rectangle with one corner missing with $ 1 \times 1$ and $ 2 \times 1$ tiles.  Start with your $ 2 \times n$ rectangle and consider the left edge: we can tile it with a single $ 2 \times 1$ tile (one way, with $ a_{n \minus{} 1}$ ways to finish), with two $ 1 \times 1$ tiles (one way, with $ a_{n \minus{} 1}$ ways to finish), with two $ 2 \times 1$ tiles (one way, with $ a_{n \minus{} 2}$ ways to finish), or with a $ 2 \times 1$ and $ 1 \times 1$ tile (two ways, with $ b_{n \minus{} 1}$ ways to finish).  Thus $ a_n \equal{} 2a_{n \minus{} 1} \plus{} a_{n \minus{} 2} \plus{} 2b_{n \minus{} 1}$.  Now, how many ways are there to tile a $ 2 \times n$ rectangle with one corner missing?  We can fill the "extra" square with a $ 1 \times 1$ tiles (one way, with $ a_{n \minus{} 1}$ ways to finish) or with a $ 2 \times 1$ tile (one way, with $ b_{n \minus{} 1}$ ways to finish).  Thus $ b_n \equal{} a_{n \minus{} 1} \plus{} b_{n \minus{} 1}$.

Now we have two recursive equations with two unknown sequences.  We can convert them into one recursive equation in the following way: $ 2b_{n \minus{} 1} \equal{} a_n \minus{} 2a_{n \minus{} 1} \minus{} a_{n \minus{} 2}$, so $ 2(b_{n \minus{} 1} \minus{} b_{n \minus{} 2}) \equal{} (a_n \minus{} 2a_{n \minus{} 1} \minus{} a_{n \minus{} 2}) \minus{} (a_{n \minus{} 1} \minus{} 2a_{n \minus{} 2} \minus{} a_{n \minus{} 3})$, but also (from our other equation) $ b_{n \minus{} 1} \minus{} b_{n \minus{} 2} \equal{} a_{n \minus{} 2}$, so $ 2a_{n \minus{} 2} \equal{} a_n \minus{} 3a_{n \minus{} 1} \plus{} a_{n \minus{} 2} \plus{} a_{n \minus{} 3}$ or $ a_n \equal{} 3a_{n \minus{} 1} \plus{} a_{n \minus{} 2} \minus{} a_{n \minus{} 3}$.  With initial values $ a_0 \equal{} 1, a_1 \equal{} 2, a_2 \equal{} 7$ we can compute as much of the sequence as we like.

The sequence is [url=http://www.research.att.com/~njas/sequences/A030186]A030186[/url] in the [url=http://www.research.att.com/~njas/sequences/]OEIS[/url].


Let $BD$ and $CE$ be the medians of triangle $ABC$ and $P$ and $Q$ be the midpoints of $BE$ and $CD$ respectively. Prove that $PQ$ is divided into $3$ equal parts by the segments $BD$ and $CE$.

[hide="Solution"]By Thales' Theorem $DE \parallel QP \parallel CB$. Let $QP \cap CE = R$ and $QP \cap BD = S$. So $QR = 2DE = PS$. On the other hand $\Delta ADE \sim \Delta ESP$. Hence $DESQ$ is a parallelogram, and so the diagonals bisect each other. Thus $S$ is the centroid of $\Delta BEQ$. Then $QS = 2SP$ and finally $PS = SR = RQ$. [/hide]
btw I don't think the source is correct  :maybe:

[quote="Sepp"]
btw I don't think the source is correct  :maybe:[/quote]

You're right, and nice solution... :)

[hide="Two words"] Midsegment Theorem[/hide]

[hide="short explanation"] Let the medians cut $PQ$ so that the points $P, R, S, Q$ are on the segment in that order.  Then $PR$ is the midsegment of $\triangle{BED}$, making it half of $ED$, as is $QS$, and because $PQ$ is $\frac{3}{2}$ of $ED$, the result follows.  You could instead use the midsegments of $\triangle{BDC}, \triangle{BEC}$ instead if you want.[/hide]


If $ x$ varies as the cube of $ y$, and $ y$ varies as the fifth root of $ z$, then $ x$ varies as the $ n$th power of $ z$, where $ n$ is:

$ \textbf{(A)}\ \frac{1}{15} \qquad
\textbf{(B)}\ \frac{5}{3} \qquad
\textbf{(C)}\ \frac{3}{5} \qquad
\textbf{(D)}\ 15 \qquad
\textbf{(E)}\  8$

[hide="Click for solution"]
$ x\equal{}k_1y^3, y\equal{}k_2z^{\frac{1}{5}}$, so $ x\equal{}k_1{k_2}^3z^\frac{3}{5}\equal{}k_3z^\frac{3}{5}$ so the answer is $ \boxed{\textbf{(C)}}$.
[/hide]

There is a typo in this problem, the last relation should be "then $ x$ varies as the $ n$th power of $ z$."

Can you please elaborate? I do not really understand this...

EDIT never mind i got it...


AMC AIME factorial trigonometry geometry rhombus USAMTS

If you have any questions such as way to attack some problems, you may ask here. There will be (hopefully) all 15 problems posted in AoPS but till then, this thread will serve that role to answer any questions.

could some one put complete solution for #15?i had a discussion with Paladin8 on that problem, he said the answer was 27 ;)

[hide="my answers"]
the bold ones are theones that i missed, the ones with ? are the ones im not sure about

1. 84
2. 901
3. 725
4. 124
5. 936
6. 360
7. 408
[b]8. 89 ?[/b]
9.46
[b]10. 65[/b]
[b]11. 458[/b]
12.906
13. 899
[b]14. 183 ?[/b]
[b]15. ?? [/b] i put 3, dunno if its right or not[/hide]

EDIT: just saw [url=http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2006-aa/06answers-aime1.html]this[/url], BLAH!!!
looks like paladin8 was right :)

I'm not all too excellent at math... I only got a 4 on this AIME. I was wondering why the solution to #2 is 901 and not 946 :\. What I did was I figured that there is a minimum sum and a maximum sum, and  (I assumed, perhaps incorrectly) using 90 elements of the 100 in that set you can form any sum in-between the maximum and the minimum, therefore the number of sums you can get is the maximum, minus the minimum, plus one. I'm obviously missing something :\. Is there a sum in-between the max and the min that you can't form using 90 elements?

I also made a GLARING, ULTRA-STUPID mistake on number 4. I figured out how many zeros in 5!, then 10!, ... , then 100!, and for 100! I got 24 zeros, so I (stupidly) put 976 as my answer. I forgot that all the factorials were being multiplied together >_<. Wow.

I liked number 5. It was simple. I thought number 6 was the question from hell... is there any elegant way of going about that? It just seems like a bunch of ugly counting. I figured out (again, I'm not sure if this is right) that you need to find numbers that have three distinct digits and are not divisible by 3 or 111. Is this at least a correct start?

I only answered 1 - 7. :( I really should read those AoPS books I have stashed away on my bookshelf.

-BLt

I got a 7, which I am pretty happy with.  I also made a stupid mistake on #4 - I realized that you had to double-count 25's, but somehow forgot that 100 is a multiple of 25. :roll:

To thebrokenlight:  your method is correct.  The minimum is 4095 and the max is 4995.

I missed #8 too. I don't know, I just assumed that the maximum came when $T$ was a square. Apparently not. I guess an algebraic solution would have been best...

i didn't do 8, but the solution for 8 is with trig.
let the small angle of the samll rhombous be theta then work from there...
its not that hard actually
but i never even tried trig in the test, i was trying to get a diophantic equation, it never happened

[quote="thebrokenlight"]I thought number 6 was the question from -... is there any elegant way of going about that? It just seems like a bunch of ugly counting. I figured out (again, I'm not sure if this is right) that you need to find numbers that have three distinct digits and are not divisible by 3 or 111. Is this at least a correct start?
-BLt[/quote]

There's an elegant solution.  I'll post it later if no one beats me to it.

Hmmm... Is this right?

[hide="Problem 6"]For each .abc there is a .(10-a)(10-b)(10-c) and those two add up to one.  There are a total of 9*8*7 numbers, so divide that by two, what was wrong?[/hide]

On 6, you express it as a fraction, so each element of S is something/999.  It's just finding the sum of the ones where a, b, and c are different from there.

[quote="thebrokenlight"]I thought number 6 was the question from -... is there any elegant way of going about that? It just seems like a bunch of ugly counting. I figured out (again, I'm not sure if this is right) that you need to find numbers that have three distinct digits and are not divisible by 3 or 111. Is this at least a correct start?[/quote]

Well, it doesn't matter if the numbers are divisible by 3 (or 111) as long as they've got different digits... Essentially, the trick was finding that there are 10 * 9 * 8 distinct numbers, and that each of 0-9 occurs 72 times in each of the three places. Then just add'em all up over 999.

yes and to find the sum there is an easy way

you count the number of times each digit appears in the hundreds, tens, and units place. so, in the number abc, if a is fixed then there are 9 choices for b and 8 choices for c
so each digit appears 72 times in the H place
Similarly, each digit appears 72 times in T snd the U place.

so sum = 7200(1+2+..+9) + 720(1+2+..+9) + 72(1+2+..+9)
 and the answer is that divided by 999.

sorry, didnt realize that sky had the same thing as me.

[quote="daveed"]Can any-one post a solution for 8?  I was stumped on that one... or maybe my drawing was mocking me.  One or the other.[/quote]
[hide="Using area of rhombus and sin double-angle formulae"]
The illustration shows part of P, Q, and T.

The area of rhombus P is $A_P=a^2\sin(\alpha)=\sqrt{2006}$
The area of rhombus T is $A_T=a^2\sin(\gamma)$

Simplifiying:
\begin{eqnarray*} A_T &=&a^2\sin(\gamma)\\ \gamma&=&\pi-2\alpha\\ A_T&=&a^2\sin(\pi-2\alpha)\\ A_T&=&a^2\sin(2\alpha)\\ A_T&=&a^2\times 2\sin(\alpha)\cos(\alpha)\\ A_T&=&\sqrt{2006}\times 2\cos(\alpha)\\ A_T&\approx&89.58\times\cos(\alpha)\\ \end{eqnarray*}
[/hide]

Or you can just say there are 720 numbers whose average is 1/2, so their sum is 360...

[quote="thebrokenlight"]
I liked number 5. It was simple. I thought number 6 was the question from -... is there any elegant way of going about that? It just seems like a bunch of ugly counting. I figured out (again, I'm not sure if this is right) that you need to find numbers that have three distinct digits and are not divisible by 3 or 111. Is this at least a correct start?
[/quote]
What you're doing is adding up all numbers of the form $abc/999$, where $a,b,c$ are distinct.  So, for any choice of $a$ (and there are ten of them), you have $9 \cdot 8 = 72$ choices for $b$ and $c$, and similarly for the other two digits.  That means that when you add up all possible choices, you have $72(0 + 1 + ... + 9) \cdot 10^2 + 72 (0 + 1 + ... + 9) \cdot 10 + 72 (0 + 1 + ... + 9)$.  If you sum that (not hard to do), and divide by $999$, you end up with 360.

Just to let you all know, AoPS has permission from AMC to reproduce their contests on the website. We did the same thing with the AMC's.

based on what I was able to work through, I should have had a 5.

but I misread the first question completely and answered 900 for number two.

At least I got 3,4, and 10 right.

I got #8 wrong, because i didn't count right. And I think i got another wrong, but i'm not sure because I don't have any of my scratch paper or the test.  If I'm lucky, I'll get a 10.  I know i got 1-5 and 13 right, so I feel a bit better.

Btw, does the line for the one with 8 circles pass through tangent points?

I think that I missed the answer for the trig problem (number 12, I think) by a 100 because of making an addition error in the very end after finding all the solutions. :wallbash:
But my score is still at least 9. :)

for #7, how can u set up a ratio of triangle areas by squaring the height?  isnt it base times height divided by 2?  what happened to the base and over 2 part...i can see the canceling out of the 2, but each similar triangle has a different base....right?

(speaking of solution in AIME Math Jam)

[quote]for #7, how can u set up a ratio of triangle areas by squaring the height?  isnt it base times height divided by 2?  what happened to the base and over 2 part...i can see the canceling out of the 2, but each similar triangle has a different base....right?[/quote]

Because since they're similar triangles, the ratio of bases will equal the ratio of the heights.[/quote]

In similar figures, the ratio of areas is the square of the ratio of linear components.

Is there any way of finding out who proposed these problems? And as an answer to people who felt it was easier, I will say I am torn. I scored lower this year than I have in the past and I usually do, but after seeing all of the solutions I find the test unsatisfying, if not easier than usual.

Our whole school is taking the AIME II because of some SAT statewide practice thing for juniors that was on the same day as the AIME I. They sent us the test anyway though, so we could use them as practice. Anyway, I took the AIME I yesterday for about an hour before giving up. I checked the answers I had up to that point, and got 5/15. Is that good for a freshman?

[quote="paladin8"][quote="pieterminate"][quote="nirvanatear"]argh.................
i should've got 12...
since the only hard questions are #14,#12...
i havent learnt much about the trigonometric function thingy yet so i left #11 blank[/quote]

12 was one of the easiest!
[hide]unless one makes a stupid mistake :(  :(  :(  :(  [/hide][/quote]

I'd have to argue that #12 was not one of the easiest problems... I think it was either the hardest or the second hardest. The other mildly difficult one would be #15, which was really not that hard (to prove somewhat, but to guess, no). #14 had a simple solution that required a lot of computation, #13 was just too easy, #11 was an easily provable recursion, so yeah.[/quote]Hmm, I personally think #12 was the easiest out of the last 5, it was the only one which I found the solution to during the test (made a dumb mistake though).

[quote="eryaman"][/quote][quote="paladin8"][quote="pieterminate"][quote="nirvanatear"]argh.................
i should've got 12...
since the only hard questions are #14,#12...
i havent learnt much about the trigonometric function thingy yet so i left #11 blank[/quote]

12 was one of the easiest!
[hide]unless one makes a stupid mistake :(  :(  :(  :(  [/hide][/quote]

I'd have to argue that #12 was not one of the easiest problems... I think it was either the hardest or the second hardest. The other mildly difficult one would be #15, which was really not that hard (to prove somewhat, but to guess, no). #14 had a simple solution that required a lot of computation, #13 was just too easy, #11 was an easily provable recursion, so yeah.[/quote][quote="eryaman"]Hmm, I personally think #12 was the easiest out of the last 5, it was the only one which I found the solution to during the test (made a dumb mistake though).[/quote]

Amen! sum identity................divide.............find factors......done?
[hide]though stupid mistakes still plague me :(  :( [/hide]

[quote="eryaman"][quote="paladin8"]

I'd have to argue that #12 was not one of the easiest problems... I think it was either the hardest or the second hardest. The other mildly difficult one would be #15, which was really not that hard (to prove somewhat, but to guess, no). #14 had a simple solution that required a lot of computation, #13 was just too easy, #11 was an easily provable recursion, so yeah.[/quote]Hmm, I personally think #12 was the easiest out of the last 5, it was the only one which I found the solution to during the test (made a dumb mistake though).[/quote]

I see. Well I thought it was much harder than #11 or #13 for sure, because for one thing trig is always intimidating. And seeing as how almost everyone said they made a stupid mistake on it, which usually means they missed a case, I would say a considerably smaller number of people got #12 than #11 or #13. But I guess it depends on your strengths and weaknesses as well, so we could debate for a long time and not arrive at any real conclusion.


algebra
binomial theorem
combinatorics unsolved
combinatorics

The result is immediate from the binomial theorem if you just consider the coefficent of $ x^{n}$ for the LHS and RHS of the following equality:

$ (x+1)^{2n}=(x+1)^{n}(x+1)^{n}$.

This is the most standard special case of Vandermonde's identity (also known as Vandermonde's convolution).  It has appeared many, many times on the forum before.


Just wondering, which do you think is better for Black as a response to e4,
French Defence, Sicilian Defence, or other. If other, please post and say what.

Sicilian defense, personally -> easier to manipulate

Sicillian because the French depends on white making e4 and then d4. If the second move is different you might be playing a whole different game. However, sicillian you can immediately change your opponents gameplay to Sicilian with 1...c5...you don't have to depend on white's moves as much.

Sicilian, b/c keeps steady pawn structure

french, its solider and safer, which is mainly blacks objective, and if he makes a mistake trying to attack like crazy, youve got an easy win

[quote="dbkarp"]Play caro-kann (1.e4 c6 2._ d5) it's super defensive though, depends on how you like to play. Sicilian is way better than French.[/quote]

Totally agreed, Caro-Kann is the best. I can't play Sicilian for my life, though it's pretty good, I would think.

Sicilian is dynamic and versatile, I feel French leaves you powerless early-mid in the game. But then, I'm not a defensive player.


function
calculus
derivative
integration
probability and stats

Let $ X_1, X_2$ and $ X_3$ be random variables, either continuous or discrete. The joint moment-generating function of $ X_1, X_2$ and $ X_3$  is defined by

$ m(t_1, t_2, t_3) \equal{} E(e^{t_1X_1 \plus{} t_2X_2 \plus{} t_3X_3})$

It is known that $ m(t,t,t)$ gives the moment generating function of $ X_1 \plus{}X_2 \plus{} X_3$ while
$ m(t,t,0)$ gives the moment generating function of $ X_1 \plus{} X_2$

Show that

 $ \frac {\partial^{k_1 \plus{} k_2 \plus{} k_3}m(t_1, t_2, t_3)} {\partial{t_1^{k_1}}\partial{t_2^{k_2}} \partial{t_3^{k_3}}} \equal{} E(X_1^{k_1}X_2^{k_2}X_3^{k_3})$ when you set $ t_1 \equal{} t_2 \equal{} t_3 \equal{} 0$

It's just a joke: take the  derivative in your integral, considering $ t_1,t_2,t_3$ to be parameters.The question is : why can we do so ? But it follows from Calculus  :)


Find the Laurent series that converges for $ 0 < | z \minus{} i| < R$ of

$ \frac {1}{1 \plus{} z^2}$.

I have been given the hint to break it up as

$ \frac {1}{1 \plus{} z^2} \equal{} (\frac {1}{z \minus{} i})(\frac {1}{z \plus{} i})$ and then expand $ \frac {1}{z \plus{} i}$. I am kind of confused about this, because the series is centered at $ i$. I'm not exactly sure how to do it because the center isn't 0.

The solution is -$ \sum_{n \equal{} 0}^{\infty}(\frac {i}{2})^{2n \plus{} 1}(z \minus{} i)^{n \minus{} 1}$

Substitute $ w\equal{}z\minus{}i$. It looks a lot nicer when we're expanding with respect to a variable. You should recognize a geometric series.


So everyone just lists one thing. They have to provide a website that verifies their prices (stocks are a different matter, you can easily look up how much a stock was worth on any date, so please provide how much it is worth accurately!)

Once we get up to a million dollars, we just compiile it up in a spreadsheet or whatever.

RULES:

1) You cannot suggest something someone else has already suggested.
2) You cannot suggest a house
3) You cannot suggest anything in excess of $ \$$500,000.
4) Profits in stocks can be calculated by a person but they must be done on a single turn.
5) Each post is considered as 1 turn. Once we get up to a million, the person who suggested the [u]most [/u] expensive things (added up) gets to take that and a new game begins. In teh second game, anything mentioned in the first game cannot be used. I'll start: Maybach 62S -$ \$$426,000 - [url=http://autos.yahoo.com/maybach_62_s/]Click to view[/url]

[url=http://charisgames.com/product/WMR55?meta=FRG]5x5 Rubik's Cube[/url] and put it on display and say "I've solved that [i]hundreds[/i] of times.
$ \$ 28.98$

 :first:  :winner_first:

[quote="#H34N1"]So everyone just lists one thing. They have to provide a website that verifies their prices (stocks are a different matter, you can easily look up how much a stock was worth on any date, so please provide how much it is worth accurately!)

Once we get up to a million dollars, we just compiile it up in a spreadsheet or whatever.

RULES:

1) You cannot suggest something someone else has already suggested.
2) You cannot suggest a house
3) You cannot suggest anything in excess of $ 500,000. 4) Profits in stocks can be calculated by a person but they must be done on a single turn. 5) Each post is considered as 1 turn. Once we get up to a million, the person who suggested the [u: 8a9cc51d68]most[/u: 8a9cc51d68] expensive things (added up) gets to take that and a new game begins. In teh second game, anything mentioned in the first game cannot be used. I'll start: Maybach 62S\minus{}$426,000 - [url=http://autos.yahoo.com/maybach_62_s/]Click to view[/url][/quote]

you might wanna fix the latex

do:

[code]$\$$[/code] for the dollar sign.

get MLB2K7 for Xbox

59.99

[url]http://www.amazon.com/2K-Sports-Major-League-Baseball/dp/B000M5KPYY[/url]

Go buy a dell precision 690; check every option; 43,445 at this time.

http://dell.com

can't really prove this one :P

Go hire some nice prostitutes with VD. LOL jk

I'd give most of it to charity.

A (16GB RAM) maxed out (with 2x 750GB HDs) Mac Pro:

$ \$$8,595.00 LOL

http://store.apple.com/1-800-MY-APPLE/WebObjects/AppleStore.woa/9294000/wo/S476yJcT07uu2Sz1IEF2IZQSDgt/3.?p=0\
-or-
http://www.apple.com/macpro/

http://configure.us.dell.com/dellstore/config.aspx?c=us&cs=04&kc=6W300&l=en&oc=2400mp&s=bsd

Dell Projector - kinda stinks though


let x^2+y^2+z^2=1 then prove that
 x+ y+ z+ 1/(xyz) >=48^1/2
 i have solve this inequality by SMV theorem and i am looking for another method?

[quote="M.A"]and i am looking for another method?[/quote]
The Mixing Variables Method  kill's it easy enough.

what is the SMV theorem and the mixing variable method???
please answer me =D i want to learn  :D  :D  :D


Let $ a,b,c$ be the positive numbers whose sum is not larger than $ 3$. Find the minimum possible value of the expression
  $ S\equal{}\frac{a\plus{}1}{a(a\plus{}2)}\plus{}\frac{b\plus{}1}{b(b\plus{}2)}\plus{}\frac{c\plus{}1}{c(c\plus{}2)}$.

[quote="lasha"]Let $ a,b,c$ be the positive numbers whose sum is not larger than $ 3$. Find the minimum possible value of the expression
  $ S \equal{} \frac {a \plus{} 1}{a(a \plus{} 2)} \plus{} \frac {b \plus{} 1}{b(b \plus{} 2)} \plus{} \frac {c \plus{} 1}{c(c \plus{} 2)}$.[/quote]
Posted before.. Use that $ \frac {a \plus{} 1}{a(a \plus{} 2)} \equal{} \frac {1}{2} \left ( \frac{1}{a} \plus{} \frac {1}{a \plus{} 2} \right )$ and AM-HM..


limit
integration
algebra unsolved
algebra

[color=darkblue]Hi every One. I need your solution about these  limmits problem:

1) Find $ \lim_{n \to \infty} \left ( \frac {1}{n \plus{} 1} \plus{} \frac {1}{n \plus{} 1} \plus{} ... \plus{} \frac {1}{n \plus{} n} \right)$

2) Find $ \lim_{n \to \infty} \left ( \frac {|sin1|}{2} \plus{} \frac {|sin2|}{2^2} \plus{} ... \plus{} \frac {|sinn|}{2^n} \right)$

3) Find $ \lim_{n \to \infty} \left (1 \plus{} \frac {1}{2^2} \plus{} \frac {1}{3^2} \plus{} ... \plus{} \frac {1}{n^2} \right)$[/color]

hello, to 3) it is well-known that $ \sum_{i\equal{}0}^\infty\frac{1}{i^2}\equal{}\frac{\pi^2}{6}$.
Sonnhard.

1) $ \int_1^2\frac{dx}{x}\equal{}ln2$;
2) $ Im\sum_{n\equal{}1}^{\infty}x^n$, were $ x\equal{}\frac{e^i}{2}$. Therefore
 $ lim\equal{}Im\frac{x}{1\minus{}x}\equal{}Im\frac{cos1\plus{}isin1}{2\minus{}cos1\minus{}isin1}\equal{}\frac{2sin1}{5\minus{}4cjs1}.$

Thank Rust and Dr Sonnhard Graubner. Can you give me how to done it?


The numbers from $ 1$ to $ n^2$ are written in the cells of a $ n\times n$ board. When placing a new number in some empty cell, we write on the blackboard the sum of all the numbers already written in cells of the same row or column with this cell. After all numbers are inserted, we calculate the sum of all written numbers on the blackboard. Provide a way of placing the numbers for which the calculated sum is minimal.

[i]K. Kohasi (К. Кохась)[/i]

[hide="Note"]It is not important whether we write the sum before or after a new number is inserted, as this will just modify the final sum by a constant value: $ 1\plus{}2\plus{}\ldots\plus{}n^2$.[/hide]

Let our last sum be $ S$. After we have finished filling numbers, for all rows and columns we will be summed one of the $ n$ numbers in that row or column $ n-1$ times, one of them $ n-2$ times,..., one of them $ 1$ time, one of them never. Therefore,
$ S=(n-1)(a_1+...+a_n)+(n-2)(a_{n+1}+...+a_{2n})+...+0.(a_{n^2-n+1}+...+a_{n^2})+(n-1)(b_1+...+b_n)+(n-2)(b_{n+1}+...+b_{2n})+...+0.(b_{n^2-n+1}+...+b_{n^2})$; such that $ (a_1,...,a_{n^2})$ and $ (b_1,...,b_{n^2})$ are permutations of $ (1,...,n^2)$. This sum takes its minimum value if $ ( a_{k.n+1},...,a_{(k+1).n} )=( {b_{k.n+1},...,b_{(k+1).n} )=( k.n+1,...,(k+1).n )}$ (they are sets but i couldn't type the {} signs) for every $ k=0,1,...,n-1$. To prove that take an arbitrary $ a_i$ permutation. If we got a $ (i,j)$ pair such that $ i<j$ and $ a_i>a_j$, change the places of $ a_i$ and $ a_j$. The new value of $ S$ is less then or equal to the initial value of $ S$. After finitely many such operations we will reach the desired idle permutation. Do the same for $ b_i$.
Also this value can be achieved. Write the numbers in $ 1,2,...,n^2$ order to make the belov table.
1              n+1   2n+1 ..................  n^2-n+1
n^2-n+2    2       n+2   2n+2 .....     n^2-2n+2
.
.
.
2n             3n      4n ........     n^2    n

Example for 3 is
147
825
693


If $ r$ and $ s$ are the roots of the equation $ ax^2 \plus{} bx \plus{} c \equal{} 0$, the value of $ \frac {1}{r^2} \plus{} \frac {1}{s^2}$ is:

$ \textbf{(A)}\ b^2 \minus{} 4ac \qquad\textbf{(B)}\ \frac {b^2 \minus{} 4ac}{2a} \qquad\textbf{(C)}\ \frac {b^2 \minus{} 4ac}{c^2} \qquad\textbf{(D)}\ \frac {b^2 \minus{} 2ac}{c^2}$
$ \textbf{(E)}\ \text{none of these}$

We have $ (r^2 \plus{} s^2)/(r^2s^2) \equal{} (r^2 \plus{} s^2)/(c^2/a^2)$
From Newton sum, we have
$ aS_1 \plus{} b \equal{} 0$
$ aS_2 \plus{} bS_1 \plus{} 2c \equal{} 0$

So $ S_1 \equal{} \minus{} b/a$ and so we have
$ aS_2 \minus{} b^2/a \plus{} 2c \equal{} 0 \implies S_2 \equal{} \frac {b^2/a \minus{} 2c}a \equal{} \frac {b^2 \minus{} 2ac}{a^2} \equal{} r^2 \plus{} s^2$.

So we have $ \frac1{r^2} \plus{} \frac1{s^2} \equal{} \frac {b^2 \minus{} 2ac}{c^2}$.

EDIT: Another way is to note that $ 1/s$ and $ 1/r$ are both roots of $ cx^2 \plus{} bx \plus{} a$.

Newton sums: http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums
To apply it to this problem you can substitute $ x \equal{} 1/x$ and use newton sums from there.


MIT college Princeton probability search

Hi!
Like the title of the post say, I am looking for admission exams from universities.
I just want to see and try to do exams just to know if I can do it.
Does someone know where can I get it?? I have try in the websites of various universities but there aren't admission exams.
thanks!

Well, at least in the US, universities don't tend to have their own admission exams. They rely on stuff like the SAT. However, I did see this MIT math assessment thing one time...

the MIT math assessment is for already admitted MIT pre-frosh, to aid in placement. It has nothing to do with admissions. I don't know of any US schools with admissions exams, although I think it used to be common in the pre-College Board era.

thanks for your reply!
but, Are there SAT exams online?Can I get SAT exams or stuff like this on some website???
thanks again!! :lol:

well, the best way to get SAT prep is to go to a bookstore and buy the princeton review kaplans, or barron's SAT guide (any SAT book will work, but those are the better ones)

[url=http://www.collegeboard.com/student/testing/sat/prep_one/prep_one.html]SAT Preparation Center[/url]

[url=http://www.actstudent.org/testprep/index.html]ACT Test Prep[/url]

These are the official sources for sample tests made the same designers as the real tests.

Thanks a lot tokenAdult
by the way
 Does someone know if it is important the ACT test???
the samples in the website are just so easy.
Is it useful this test to apply to a good university??...

Caltech has admissions exams for transfer students.

Here in the United States, some regions of the country prefer the ACT admission test, and some prefer the SAT, but most colleges will accept either test. The most selective colleges generally require one of those (and tend to strongly prefer the SAT) and frequently require multiple [url=http://www.collegeboard.com/student/testing/sat/about/SATII.html]SAT II tests[/url] as well. But unlike the system in many other countries, in the United States admission is not based strictly on test scores--colleges often admit applicants with lower test scores while rejecting some applicants with higher test scores. High test scores increase the probability of admission, but do not guarantee it.

If you are applying as a citizen of another country (as I'm guessing from the flag on your user profile), then admission standards can be even more demanding than for United States applicants. Successful international applicants generally have to have very high test scores, grades, and demonstrated accomplishments in extracurricular activities to get into the top schools in the United States--as many international applicants do have wonderful admission criteria to mention on their applications.

Good luck in your college search and applications.

thanks for the information
I am not applying now, I am just searching information for a possible future.
thanks again! Your help was so useful for me!!! :lol:


find all n,p such that:

N^8-P^5=P^3+N^6  :)

hallo,
see here
http://www.mathlinks.ro/Forum/topic-79907.html  :)


use a linear approximation L(x) to an appropriate function f(x), with an appropriate value of a, to estimate the given number.

Square root of 80


Let $\mathbb{N}^*=\{1,2,\ldots\}$. Find al the functions $f: \mathbb{N}^*\rightarrow \mathbb{N}^*$ such that:

(1) If $x<y$ then $f(x)<f(y)$.
(2) $f\left(yf(x)\right)=x^2f(xy)$ for all $x,y \in\mathbb{N}^*$.

a) Take $x=1$ then $f(yf(1))=f(y)$ with (1) gives $f(1)=1$
b) Take $y=1$ then $f(f(x))=x^2f(x)$
c) Take $x=y$ then $f(xf(x))=x^2f(x^2)$
d) Take $y=f(x)$ then $f(f(x)^2)=x^2f(xf(x))=x^4f(x^2)=f(f(x^2))$ so $f(x^2)=f(x)^2$
We have $f(xf(x))=x^2f(x^2)=x^2f(x)^2=(xf(x))^2$
Take $y=xf(x)$ then $f(y)=y^2$ and its indeed solution.

Unless I am mistaken, you have shown that $ f(y)\equal{}y^{2}$ for all $ y$ that are of the form $ xf(x)$ (please correct me if I am wrong).  In order to show that $ f(y)\equal{}y^{2}$ for ANY $ y$, we may proceed as follows:

Assume $ f(y)>y^{2}$.  Then, $ f(f(y))>f(y^{2})\equal{}f(y)^{2}$, because of (1), and where d) has been used.  Then,  using b), $ f(y)^{2}<f(f(y))\equal{}y^{2}f(y)$, or $ f(y)<y^{2}$, absurd.  Similarly, assume $ f(y)<y^{2}$, and in the same way find that $ f(y)>y^{2}$, absurd again.  Or $ f(y)\equal{}y^{2}$.

Substitution in the original equation yields $ x^{4}y^{2}$ at both sides, or $ f(y)\equal{}y^{2}$ is indeed the only solution.


[quote="Potla"]I could not solve this INMO problem: :blush:
Define a sequence $ < a_n > _{n\ge 1}$ be $ a_1 \equal{} 1, a_2 \equal{} 2,$ and $ a_{n \plus{} 2} \equal{} 2a_{n \plus{} 1} \minus{} a_n \plus{} 2$ for $ n\ge1$. Prove that for any $ m$, $ a_ma_{m \plus{} 1}$ is also a term in the sequence.

Please help. :([/quote]

With the given recursive relations. inductively  prove that:
$ a_n \equal{} (n\minus{}1)^2 \plus{} 1$ .
Now we get that $ a_ma_{m \plus{} 1}\equal{} ((m\minus{}1)^2 \plus{}1)(m^2\plus{}1) \equal{}(m(m\minus{}1)\plus{}1)^2 \plus{}1 \equal{} a_{m(m\minus{}1)} \plus{} 2$.
$ QED$.

We can solve the recursive formula directively as follows.

$ a_{n\plus{}2}\equal{}2a_{n\plus{}1}\minus{}a_n\plus{}2\Longleftrightarrow a_{n\plus{}2}\minus{}a_{n\plus{}1}\equal{}a_{n\plus{}1}\minus{}a_{n}\plus{}2$.

Let $ b_n\equal{}a_{n\plus{}1}\minus{}a_n,\ b_1\equal{}a_2\minus{}a_1\equal{}1$, we have $ b_{n\plus{}1}\equal{}b_n\plus{}2$. Thus $ b_n$ is an arithmetic sequence with a common difference 2, we have $ b_n\equal{}b_1\plus{}2(n\minus{}1)\equal{}1\plus{}2(n\minus{}1)\equal{}2n\minus{}1$.

$ \therefore a_n\equal{}a_1\plus{}\sum_{k\equal{}1}^{n\minus{}1} b_k\ (n\geq 2)$

$ \equal{}1\plus{}\sum_{k\equal{}1}^{n\minus{}1} (2k\minus{}1)\equal{}(n\minus{}1)^2\plus{}1\ (n\geq 1)$.


I've seen many solutions for this problem, and I want to see more.

Given: $ F_{n\plus{}2} \equal{} F_{n\plus{}1} \plus{} F_{n}, F_1 \equal{} F_2 \equal{} 1$

Prove: $ F_{n}^{3} \plus{} F_{n\plus{}1}^{3} \ne \frac{3}{7}(F_n)(F_{n\plus{}1})(F_{n\plus{}2})$

[hide="an idea"]
We shall prove this by contradiction.  Assume there are integers $ F_{n}, F_{n \plus{} 1},F_{n \plus{} 2}$ in the Fibonacci series such that they satisfy this condition.

$ F_{n}^3 \plus{} F_{n \plus{} 1}^3 \equal{} \frac {3}{7}(F_{n})(F_{n \plus{} 1})(F_{n \plus{} 2})$

Factoring the left side, and noting that $ F_{n} \plus{} F_{n \plus{} 1} \equal{} F_{n \plus{} 2}$, we simplify to get $ 7F_{n}^2 \minus{} 10F_{n}F_{n \plus{} 1} \plus{} 7F_{n \plus{} 1}^2 \equal{} 0$.  However, by the discriminant, there are no integral values $ F_{n},F_{n \plus{} 1}$ that make this expression true.  Thus, we have a contradiction.  They cannot be equal.
[/hide]

[hide="Solution 1"]
$ F_{n}^{3} \plus{} F_{n \plus{} 1}^{3} \equal{} \frac {3}{7}(F_n)(F_{n \plus{} 1})(F_{n \plus{} 2})$
$ \iff (F_n \plus{} F_{n \plus{} 1}) \cdot [ (F_{n \plus{} 1} \minus{} F_{n})^2 \plus{} (F_{n})(F_{n \plus{} 1}) ] \equal{} \frac {3}{7}(F_n)(F_{n \plus{} 1})(F_{n \plus{} 2})$
$ \iff F_{n \minus{} 1}^2 \plus{} (F_n) (F_{n \plus{} 1}) \equal{} \frac {3}{7}(F_n)(F_{n \plus{} 1})$
$ \iff F_{n \minus{} 1}^2 \equal{} \minus{} \frac {4}{7} (F_n) (F_{n \plus{} 1})$, a clear sign contradiction.[/hide]

[hide="Solution 2"]
$ F_{n}^{3} \plus{} F_{n \plus{} 1}^{3} > F_{n \plus{} 1}^{3} > (F_n)(F_{n \plus{} 1})^2$

It suffices to show that $ F_{n \plus{} 1} > \frac {3}{7} F_{n \plus{} 2}$, which is straightforward (use Binet's or anything else)[/hide]

Hmmm... are you sure that you wrote your problem correctly? It seems to be a very loose inequality...

[hide="Another"] $ F_n^3 \plus{} F_{n \plus{} 1}^3 \equal{} F_{n \plus{} 2}( F_{n \minus{} 1}^2 \plus{} F_n F_{n \plus{} 1}) > F_n F_{n \plus{} 1} F_{n \plus{} 2} > \frac {3}{7} F_n F_{n \plus{} 1} F_{n \plus{} 2}$. [/hide]

This is a strange problem.

One may also note that $ F_{n \minus{} 1} F_{n \plus{} 1} \equal{} F_n ^2 \plus{} ( \minus{} 1)^n$, which leads to yet another painless solution.

[i](this particular relation is more interesting than the solution itself)[/i]

1,1,2

the product of those is not divisible by 7..so the LHS is an integer, but the RHS is not

yay?

I think by $ \ne$ he means "is never equal to"

Yes, it is a very loose inequality, but so many interesting solutions have come from it that I decided to post it.
[hide="Another"]By Fermat's Last Theorem,
$ (2F_n)^3 \plus{} (2F_{n\plus{}1})^3 \ne (F_{n\plus{}2})^3$

$ (2F_n)^3 \plus{} (2F_{n\plus{}1})^3 \ne (F_n \plus{} F_{n\plus{}1})^3$

$ 8F_n^3 \plus{} 8F_{n\plus{}1}^3 \ne F_n^3 \plus{} 3F_n^2F_{n\plus{}1} \plus{} 3F_nF_{n\plus{}1}^2 \plus{} F_{n\plus{}1}^3$

$ 7F_n^3 \plus{} 7F_{n\plus{}1}^3 \ne 3F_nF_{n\plus{}1}^2 \plus{} 3F_n^2F_{n\plus{}1}$

$ 7F_n^3 \plus{} 7F_{n\plus{}1}^3 \ne 3(F_n)(F_{n\plus{}1})(F_n \plus{} F_{n\plus{}1})$

$ F_n^3 \plus{} F_{n\plus{}1}^3 \ne \frac{3}{7}(F_n)(F_{n\plus{}1})(F_{n\plus{}2})$[/hide]


trigonometry
modular arithmetic
algebra
polynomial
Vieta
complex numbers
absolute value

If the six solutions of $x^6 = -64$ are written in the form $a+bi$ where $a$ and $b$ are real, then the product of those solutions with $a > 0$ is

$\text{(A)} \ -2 \qquad \text{(B)} \ 0 \qquad \text{(C)} \ 2i \qquad \text{(D)} \ 4 \qquad \text{(E)} \ 16$

Sorry, I'm not too familiar with cis....so....is there another way to deal with this one?

Notation: $\text{cis }\theta=\cos\theta+i\sin\theta=e^{i\theta}.$

alan was essentially using DeMoivre's Therorem, which can be phrased variously as:

$(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta)$
$(\text{cis }\theta)^n=\text{cis }(n\theta)$
$\left(e^{i\theta}\right)^n=e^{in\theta}.$

Even more generally (and alan used this as well):

$\text{cis }\alpha\cdot\text{cis }\beta=\text{cis }(\alpha+\beta)$

or, in more easily remembered notation, $e^{i\alpha}e^{i\beta}=e^{i(\alpha+\beta)}.$

When you multiply two complex numbers you multiply their absolute values and add their arguments. (The argument of a complex number is the angle needed to represent it in polar coordinates. There's an inherent ambiguity here, as the angle is only determined $\pmod{2\pi}$ and when we say to add the angles, the addition must be performed $\pmod{2\pi},$ or $\pmod{360^{\circ}},$ if we're working in degrees.)

While I suppose there could be other ways to work this problem, by far the simplest would involve knowing how to find the $n$th roots of any complex number. Which is what alan did.

Except there is an easier way to work this problem, taking advantage of it being a multiple choice problem.

All six of the roots of $z^6=-64$ have absolute value $|z|=\sqrt[6]{64}=2.$ They are placed around the circle $|z|=2$ in such a fashion that they form the vertices of a regular hexagon. How many of those vertices lie in the right-hand half plane? "about half" of them, but we need to be more specific. Do any of them lie on the imaginary axis? The points at the intesections of the imaginary axis and that circle are $\pm2i;$ a quick inspection shows that these two numbers are both roots of the equation.

So of the six roots of the equation, two lie on the vertical axis, two lie in the right hand half plane, and two lie in the left hand half plane. The answer is the product of two of these roots. The product of two complex numbers of absolute value 2 must have absolute value 4. Of the possible answers given only one has absolute value 4, so it must be the solution.

(One of those explanations that takes far less time to think than to write out. The problem writers could have slowed down this train of thought by including $-4$ or $4i$ among the choices for answers.)

[hide]Divide both sides by $-64$ to get $-\frac{x^6}{2^6} = 1$, or $-(x/2)^6=1$. Let $y=x/2$ so $y^6 = -1$. Therefore $y=x/2=e^{k\pi i/6}\forall k \in \{ 1,3,5,7,9,11\}$, so $x=2e^{k\pi i/6}$. But $a=2\cos (k\pi/6)$ is only positive when $k=1,11$, so our answer is $4e^{2\pi i}=\boxed {4}$.[/hide]

Is there any way to do this without DeMoivre's/Geometric? Any way to do it with Vieta's/Newton's Sums?

Here is one way.

[hide="Solution"]
Note that if $r$ is a root of the polynomial $x^6 + 64 = 0$, then so is $-r$. Therefore if $r'$ is a root of the polynomial with non-zero real part, exactly one of $r', -r'$ has a positive real part.

Thus it suffices to find which roots have zero real part and are a root of the polynomial. Its easy to see that only $\pm 2i$ satisfy this so by Vieta the product of the roots of the polynomial which are not $\pm 2i$ is $16$. If $a,-a,b,-b$ are the rest of the roots of the polynomial (WLOG $a,b$ have positive real parts), then we have $a^2b^2 = 16 \implies ab = \pm 4$. Since $-4$ is not an answer choice, the answer must be $\boxed{\textbf{(D) } 4}$.
[/hide]


geometry
3D geometry
arithmetic sequence
number theory unsolved
number theory

Prove that the cube roots of three odd prime numbers cannot be three terms of an A.P.(not necessarily consecutive).

EDIT : Oh, I'm sorry.. I thought they were perfect cubes...  :blush:

i said cube roots. if i'm mistaken please explain your solution.

what? you mean $\sqrt[3]{101},\sqrt[3]{5}$ and $\sqrt[3]{3}$ are terms of an AP? what is the first term and common difference of the AP? can you explain?

i think your solution has many problems. why should $q^{3}-r^{3}|p^{3}-r^{3}$?
they can be terms of an AP even if they doesn't satisfy your proposition.

Ok, I want to ask something..
If $p = 2, q = 3, r = 5$, trivially..
How can we prove that they are not in arithmetic progression ? Thx..

my approach is like this: prove that for $p>q>r$, the equations $\sqrt[3]{p}=\sqrt[3]{q}+md$ and $\sqrt[3]{p}=\sqrt[3]{r}+nd$ cannot both satisfy at the same time where $m,n\in \mathbb N$ and $d\in \mathbb R^$.
but of course there may be other nice approaches. and i'm not sure if my approach will work since i could not get anything out of it.

Ok, I can get something for you..
Our task is to prove that $\frac{\sqrt[3]{p}-\sqrt[3]{q}}{\sqrt[3]{q}-\sqrt[3]{r}}$ is not rational. Otherwise, we can find $m,n \in \mathbb{N}$ so that they will satisfy the property...  :)


Let a triangle ABC and a point P in plane.Denote its altitudes are $ h_a,h_b,h_c$ and A',B',C' are midpoint of BC,CA,AB correspondence.Prove that
$ PA^2 \plus{} PB^2 \plus{} PC^2\ge\frac {4S}{3}\max\lbrace \frac {PA \plus{} PA'}{h_a},\frac {PB \plus{} PB'}{h_b},\frac {PC \plus{} PC'}{h_c}\rbrace$

I think the inequality is wrong.Here is my correction :
 $ PA^{2}\plus{}PB^{2}\plus{}PC^{2}\ge\frac{4\sqrt{3}S}{3}\max\lbrace\frac{PA\plus{}PA'}{h_{a}},\frac{PB\plus{}PB'}{h_{b}},\frac{PC\plus{}PC'}{h_{c}}\rbrace$

I think the source problem of this inequality is :
$ PA^2 \plus{} PB^2 \plus{} PC^2\ge 2\sqrt {3}S\max\lbrace\frac {PA}{h_a},\frac {PB}{h_b},\frac {PC}{h_c}\rbrace$
You can see it and find your solution ! :D   :rotfl:


The arithmetic mean of three numbers $ x,y,z$ is 24. The arithmetic mean of $ x,2y,z\minus{}7$ is 34. What is the arithmetic mean of $ x,z$? Express your answer as a decimal to the nearest tenth.

$ x \plus{} y \plus{} z \equal{} 72\implies2x \plus{} 2y \plus{} 2z \equal{} 144$ (1)
$ x \plus{} 2y \plus{} z \minus{} 7 \equal{} 102\implies x \plus{} 2y \plus{} z \equal{} 109$ (2)
(2)-(1)  $ x \plus{} z \equal{} 35\implies \frac {x \plus{} z}2 \equal{} \boxed{17.5}$


logarithms
real analysis
real analysis unsolved

prove or disprove:
$ \sum_{i\equal{}1}^{\plus{}infinity}(n^{1/n}\minus{}1)$ diverges

$ n^{1/n}\minus{}1\equal{}e^{\frac{\ln n}n}\minus{}1>\frac{\ln n}n>\frac1n$ for large enough $ n.$

The series diverges by comparison to the harmonic series.

I think it's looks like this
http://www.mathlinks.ro/viewtopic.php?t=181459
 :wink:


$f:\mathbb{R} \longrightarrow \mathbb{R}$
$|f(a)-f(b)| \leq |a-b|$
and $f(f(f(0)))=0$

Prove that $f(0)=0$

hint: prove that $|f(f(0))| \geq |f(0)|$ and $|f(f(0))| \leq |f(0)|$ and $|f(f(0))-f(0)| \leq |f(0)|$ using few times the first condition in the problem, then it's easy.


Τελικά στην παλιά άσκηση που τέθηκε στους φοιτητές:

ν.δ.ό. $ a^b\plus{}b^a > 1$ για $ a,b \in (0,1)$

ποια είναι η λύση; Αν είναι είδη λυμένη στο topic εκείνο, μπορείτε να το βάλετε σαν link εδώ;

Την επανέφερε πρόσφατα μια κουβέντα με τον πατέρα μου, και θα ήθελα να ξέρω τη λύση της.

(Αν γίνεται, παραθέστε τη λύση ολοκληρωμένα.)

Ευχαριστώ πάρα πολύ.


analytic geometry
geometry unsolved
geometry

I am learning about constructible numbers, and I got stuck on this problem. Would anyone give some help?
In the real Cartesian plane, given an angle $ \alpha$ of radian measure $ 2 \pi a/b$ with a,b are integers and 3 does not divide b.  Show that it is possible to construct an angle $ \frac{\alpha}{3}$ starting from the given angle.
I think I need to show that $ x\equal{}cos \frac {\alpha}{3}$ is constructible, which means degree of $ Q(x)/Q$ is a power of two, but I don't really know where to start.

Lemma: Let $ m,n$ be 2 integers pairwise prime. If $ 2\pi/m$ and $ 2\pi/n$ are constructible or given then $ \dfrac{2\pi}{mn}$ is constructible.
Proof: use Bezout Theorem.
We chose $ m\equal{}b$ and $ n\equal{}3$. Then $ \dfrac{2\pi}{3b}$ is constructible. Obviously $ \dfrac{2\pi{a}}{3b}$ is also constructible.

More precisely:
$ \dfrac{2\pi{a}}{b}$ is given and $ \dfrac{2\pi{a}}{3}$ is constructible. There exist integers $ u,v$ s.t. $ bv\plus{}3u\equal{}1$. Thus $ \dfrac{2\pi{a}}{3b}\equal{}u\dfrac{2\pi{a}}{b}\plus{}v\dfrac{2\pi{a}}{3}$ is constructible.


superior algebra
superior algebra unsolved

Here is a proof that I am struggling with:

"Prove that the sum of any two ideals A and B, where A+B={a+b|a is an element of A and b is an element of B} is an ideal."

It's one of those problems that makes me think "duh" of course it is, but proving it?

Thanks in advance.

MM

put the definitions togehter, nothing else has to be done

Hi -oo-

That is what I thought it was; it just looked as if there was something more to it.  My thought was to apply the ideal test to A + B but then I thought, as you did, that all I needed to do was combine the definitions.

Thanks,
--MM

Hi -oo-

You already did; I was going to make this proof far more difficult than it needed to be, so I am grateful for the time you have saved me!   :blush:

--MM


Prove that if $a,b,c$ are nonnegative real numbers, we have
\[\sqrt[3]{\frac{a^{2}+bc}{b+c}}+\sqrt[3]{\frac{b^{2}+ca}{c+a}}+\sqrt[3]{\frac{c^{2}+ab}{a+b}}\ge \sqrt[3]{9(a+b+c)}\]
:)

Dear can_hang2007, can you please post your solution if possible. :wink:

Thank you very much.

[quote="manlio"]Dear can_hang2007, when does  you allowed to post this proof ?

I hope not too much late  :wink:

Thank you very much for this very nice inequality.[/quote]
My hint:
If $ a,b,c$ be the side lengths of a triangle, then
\[ \frac{ab}{c}\plus{}\frac{bc}{a}\plus{}\frac{ca}{b} \ge \sqrt[3]{9(a^3\plus{}b^3\plus{}c^3)}\]
:)

[quote="tranquoc"]it not trus[/quote]
What is your counter example?

[quote="manlio"][quote="can_hang2007"][quote="manlio"]Dear can_hang2007, when does  you allowed to post this proof ?

I hope not too much late  :wink:

Thank you very much for this very nice inequality.[/quote]
My hint:
If $ a,b,c$ be the side lengths of a triangle, then
\[ \frac {ab}{c} \plus{} \frac {bc}{a} \plus{} \frac {ca}{b} \ge \sqrt [3]{9(a^3 \plus{} b^3 \plus{} c^3)}
\]
:)[/quote]

Thank you very much, can_hang2007. I will try.  :)[/quote]

How do you prove this hint, can_hang2007?

Thank you very much.


inequalities trigonometry function logarithms limit integration

Prove or disprove the following inequalities:
1.$a^{a}+b^{b}\geq  a^{b}+b^{a}$

2.$a^{2a}+b^{2b}\geq  a^{2b}+b^{2a}$

3.$a^{a}+b^{b}+c^{c}\geq  a^{b}+b^{c}+c^{a}$
Where $0<a, b, c< 1$

I think 1 and 3 is true.I don't know if 2 is true.
I think it is hard to prove them by elemetry method.

A hint for (1).
Assume that $a \geq b$ and denote $a^{b}=c$, $b^{b}=d$ and $\frac a{b}=k$ ($c \geq d$, $k \geq 1$).
We have to show that $c^{k}-d^{k}\geq c-d$.

My computer shown that the largest $k$ such that

$a^{ka}+b^{kb}\geq a^{kb}+b^{ka}$

for any positive $a$ and $b$ is $k=2.71828182...=e$.

Hm, Vasc. I've followed your hint. But Vasc, did you mean
$c^{k}-d^{k}\geq c-d$ for all $c\geq d, k\geq 1$?
I don't think so.
Ah. I need your help for http://www.mathlinks.ro/Forum/viewtopic.php?t=116189

[quote="zeikii"]I have a solution for (1), it's different to the Vasc's solution. It uses the weighted AM-GM inequality.[/quote]Please post this solution. :!:

[quote="zeikii"]But Vasc, did you mean
$c^{k}-d^{k}\geq c-d$ for all $c\geq d, k\geq 1$?
I don't think so.[/quote] I didn't mean this inequality for all positive $c\geq d, k\geq 1$, but only for $c=a^{b}$, $d=b^{b}$ and $k=a/b$.  :P

Vasc, here is my solution:
Rewrite the inequality:
$\frac{b^{a}}{a^{b}+b^{a}}a^{a}+\frac{a^{b}}{a^{b}+b^{a}}b^{b}\geq  1$
Now, note that $\frac{b^{a}}{a^{b}+b^{a}}+\frac{a^{b}}{a^{b}+b^{a}}= 1$, use weighted AM-GM inequality, we only have to prove that
$(\frac{a}{b})^{ab^{a}-ba^{b}}\geq  1$ which is easy by using function.Done.
Vasc, can you post your full solution? I think your method can prove the $3^{rd}$ part.

[quote="zeikii"]I have a solution for (1), it's different to the Vasc's solution. It uses the weighted AM-GM inequality.
And the largest posible value of k, i think, it's $e$.[/quote]

I proved this same exact problem exactly this way when working on this problem (from Crux Mathematicorum magazine):\[m,n,\in \mathbb{Z}\ \ \sin^{2m}\cos^{2n}\leq \frac{m^{m}n^{n}}{(m+n)^{m+n}}\]
any ideas?

[quote="zeikii"]Vasc, can you post your full solution? I think your method can prove the $3^{rd}$ part.[/quote]
Use the known inequality
$f(c)-f(d) \geq (c-d)f'(d)$
for convex function $f(x)=x^{k}$. We still have to prove that $f'(d) \geq 1$, which is not hard (you may use Bernoulli's Inequality).
We can prove (3) using (1).

[quote="zeikii"]Vasc, here is my solution:
Rewrite the inequality:
$\frac{b^{a}}{a^{b}+b^{a}}a^{a}+\frac{a^{b}}{a^{b}+b^{a}}b^{b}\geq 1$
Now, note that $\frac{b^{a}}{a^{b}+b^{a}}+\frac{a^{b}}{a^{b}+b^{a}}= 1$, use weighted AM-GM inequality, we only have to prove that
$(\frac{a}{b})^{ab^{a}-ba^{b}}\geq 1$ which is easy by using function.Done.[/quote]
This is wrong. First, you "rewrite" your ineq as $(ab)^{a}+(ab)^{b}\ge a^{b}+b^{a}$ which is stronger than the original and wrong. And in fact, $(\frac{a}{b})^{ab^{a}-ba^{b}}\leq 1$  :(
But I think we [i]can [/i]use weighted AM-GM as follows:
 ${a^{a}+b^{b}}=a^{b}a^{a-b}+b^{a}b^{b-a}\ge (a^{b}+b^{a})\left(\frac{a}{b}\right)^\frac{a-b}{a^{b}+b^{a}}\ge a^{b}+b^{a}$. Am i right?  :P

And I think the same method with weighted AM-GM works for (3). We just need to show $a^{a-b}b^{b-c}c^{c-a}\ge1$, equivalently $(\frac{a}{b})^{a-b}(\frac{b}{c})^{a-c}\ge1$, equivalently  $(\frac{b}{a})^{b-a}(\frac{c}{b})^{c-a}\ge1$. Assuming wlog that $b$ is the medium one, we have either $a\ge b \ge c$ or $a\le b\le c$, in both cases that holds. The restriction $a,b,c\le1$ isn't needed.

On the other hand, I don't think this can be extended for more than three numbers.

[quote="spanferkel"]  ${a^{a}+b^{b}}=a^{b}a^{a-b}+b^{a}b^{b-a}\ge (a^{b}+b^{a})\left(\frac{a}{b}\right)^\frac{a-b}{a^{b}+b^{a}}\ge a^{b}+b^{a}$. Am i right?  :P[/quote]
$a^{b}a^{a-b}+b^{a}b^{b-a}\ge (a^{b}+b^{a})\left(\frac{a}{b}\right)^\frac{a-b}{a^{b}+b^{a}}$ ?
Why?

[quote="spanferkel"]And I think the same method with weighted AM-GM works for (3). We just need to show $a^{a-b}b^{b-c}c^{c-a}\ge1$[/quote]Why?

Oops. Good question :blush: The way I've used AM-GM rather yields ${a^{a}+b^{b}}\ge(a^{b}+b^{a})\left(\frac{a^{a^{b}}}{b^{b^{a}}}\right)^\frac{a-b}{a^{b}+b^{a}}$. And if we suppose $a\ge b$, the fraction $\frac{a^{a^{b}}}{b^{b^{a}}}$ isn't always $\ge1$.
Even worse (=uglier) for 3 variables.

So how can it be solved with weighted AM-GM?
And Vasc, how can you prove (3) using (1)? By the way, do you use the $a,b,c<1$ condition?

[quote="Vasc"]  A solution will appear in Special Issue of The Journal of Nonlinear Sciences and its Application (TJNSA), dedicated to the 60th Anniversary of Professor Themistocles M. Rassias. [/quote]
See here
http://www.tjnsa.com/

hi there, i'm zeikii of 5 years ago. after graduating high school in 2007, i stopped logging in to mathlinks so it seems that my old id "zeikii" has been deleted. to day i googled this name and found out that this inequality has been solved. thanks you all.

ops i found it (by your notice)
[code]http://www.jstor.org/pss/2324012[/code]
so the only thing encouraged me was that i proposed it and solved it by myself after several days ^^ (well, i'm just kidding).
but i really really happy when i heared about the last one's solution ^^. just want you to know :)
(actually it takes me alot of time to understand the solution since i don't study about math anymore ^^, but i'll take my time, try to figure out what i desire).

on the other hand, at that time (2006) i proposed myself the inequality with n numbers and their permutation, but i prefered the inequality of 2 numbers so that i forgot it. and i thought it was non sense :|

[i]Lemma. If $r\in (0,e]$ and $a$ and $b$ are positive real numbers,

 then[/i] $a^{ra}+b^{rb}\ge a^{\frac{r(a+b)}{2}}+b^{\frac{r(a+b)}{2}}.$

The proof. $2(a^{ra}+b^{rb})$

$\ge a^{ra}+a^{rb}+b^{ra}+b^{rb}$

$\ge 2(a^{\frac{r(a+b)}{2}}+b^{\frac{r(a+b)}{2}}). \blacksquare$

I found the following inequality.

[i]Proposition. If $r\in (0,e]$ and $f:[0,t]\to \mathbf{R^+}$ is a continuous function,

 then[/i] $\int_0^t f(x)^{rf(x)}dx\ge \int_0^t f(x)^{\frac{r(f(x)+f(t-x))}{2}}dx.$

The proof.  For $x\in [0,t]$,

$f(x)^{rf(x)}+f(t-x)^{rf(t-x)}\ge f(x)^{\frac{r(f(x)+f(t-x))}{2}}+f(t-x)^{\frac{r(f(x)+f(t-x))}{2}}.$

Therefore,

$2\int_0^t f(x)^{rf(x)}dx$

$=\int_0^t f(x)^{rf(x)}dx+\int_0^t f(t-x)^{rf(t-x)}dx$

$\ge \int_0^t f(x)^{\frac{r(f(x)+f(t-x))}{2}}dx+\int_0^t f(t-x)^{\frac{r(f(x)+f(t-x))}{2}}dx$

$=2\int_0^t f(x)^{\frac{r(f(x)+f(t-x))}{2}}dx. \blacksquare$

[quote="SeiichiManyama"][i]If $r\in (0,e]$ and $a$ and $b$ are positive real numbers,

 then[/i] $a^{ra}+b^{rb}\ge a^{\frac{r(a+b)}{2}}+b^{\frac{r(a+b)}{2}}.$ [/quote]
I think the following inequality also holds:
$a^{4a}+b^{4b}\ge a^{2(a+b)}+b^{2(a+b)}$.

I think the following inequality also holds:
$a^{4.3a}+b^{4.3b}\ge a^{2.15(a+b)}+b^{2.15(a+b)}$.

Yes, my computer tells me that $r$ can be as big as \[r\approx 4.33690094\approx(\sqrt{2}+1)e^{2-\sqrt{2}}.\] The extremal case is situated around $a=b\approx .5566679  \approx e^{\sqrt{2}-2}$, and there we have $a^{ra}\approx e^{-\sqrt{2}}$. All that seems to be more than a mere coincidence. (Remember post #5 which started it all!) But I suppose the proof of this conjecture is not for tomorrow. :huh:

A. Coronel and F. Huancas proved the following inequality:

$a^{2a}+b^{2b}+c^{2c}\geq  a^{2b}+b^{2c}+c^{2a}.$

http://ajmaa.org/cgi-bin/paper.pl?string=v9n1/V9I1P3.tex

–rb ln b ≤ 1 ≤ 1 / a < (ln a – ln b) / (a – b),
which implies γ > g(s).
( 1 ≥ a > b > 0, γ = b ^ (rb),  s = a / b, g(t) = e ^ (– ln t / (t – 1)) )
http://www.journalofinequalitiesandapplications.com/content/pdf/1029-242X-2014-509.pdf

Why?

[quote="SeiichiManyama"]A. Coronel and F. Huancas proved the following inequality:

$a^{ra}+b^{rb}+c^{rc}\geq  a^{rb}+b^{rc}+c^{ra} $  for all r $  \in $ [0, e].

http://www.journalofinequalitiesandapplications.com/content/pdf/1029-242X-2014-509.pdf[/quote]

After reading hungkhtn's book on Inequalities, one can say these type of inequalities are doable by the method of integration. :)

[quote=Vasc]Professor Ahmet Yasar Ozban from Ankara (Turkey) recently said me that the inequality
$a^a+b^b\ge a^b+b^a$
was proposed by M.Laub (Israel) in 1985 (problem E 3116) and solved by I.Ilani (Israel) in 1990 in JSTOR,AMM,Vol.97,No.1,1990.
In addition, problem E 3116 contains also the following generalization
$x_1^{x_1}+x_2^{x_2}+...+x_n^{x_n}\ge x_1^{y_1}+x_2^{y_2}+x_n^{y_n}$,
where $(y_1,y_2,...,y_n)$ is any permutation of $(x_1,x_2,...,x_n)$.[/quote]


ratio
rotation
geometry unsolved
geometry

Let quadrilateral convex ABCD. Construct outside ABCD  equidrilateral triangle ABX;BCY;CDZ;DAT; Let E,F is midpoint of AT;BY respectively; and O is center of triangle CDZ.
Prove that XO is perpendicular to EF and calculate ratio XO/EF

My solution is based on rotation.

 We have $\overrightarrow{EF} = \frac{1}{2}(\overrightarrow{AB} + \overrightarrow{TY})$.

 Let $I$, $J$, $M$ and $N$ be the midpoints of $BC$, $AD$, $AB$ and $CD$. Then

 $\overrightarrow{TY} = \overrightarrow{TJ} +\overrightarrow{JI} + \overrightarrow{IY} = \overrightarrow{TJ}+\overrightarrow{IY} + \frac{1}{2}(\overrightarrow{AB} + \overrightarrow{CD})$

 Rotate $R^{90^{\circ}}$ the figure $90^{\circ}$, we obtain

 $R^{90^{\circ}}(\overrightarrow{EF}) = R^{90^{\circ}}(\frac{1}{2}(\overrightarrow{AB} + \overrightarrow{TY}))$

 $= \frac{2}{\sqrt{3}}\overrightarrow{XM} + R^{90^{\circ}}(\overrightarrow{TJ}+\overrightarrow{IY} + \frac{1}{2}(\overrightarrow{AB} + \overrightarrow{CD}))$

 $= \frac{2}{\sqrt{3}}\overrightarrow{XM} + \frac{\sqrt{3}}{2}\overrightarrow{AD} + \frac{\sqrt{3}}{2}\overrightarrow{CB} + \frac{1}{\sqrt{3}}\overrightarrow{XM} + \sqrt{3} \overrightarrow{NO}$

 $=\sqrt{3} \overrightarrow{XM} + \sqrt{3}\overrightarrow{MN} + \sqrt{3} \overrightarrow{NO} = \sqrt{3} \overrightarrow{XO}$

 Hence $XO$ is perpendicular to $EF$ and $\frac{EF}{XO} = \sqrt{3}$.


geometry
3D geometry
sphere
integration
calculus
calculus computations

Can someone please give me a formula to find the surface area of a spherical cap.
I already know that :
Let
r = radius of sphere
h = height of cap
Volume=((pi)h²/3)(3r - h)

Thanks!

$ f(x)\equal{}r(x)\equal{}\sqrt{r^2\minus{}x^2}$
$ \ell_c : y\equal{}c: c\le r$

$ x_a\equal{}\sqrt{r^2\minus{}c^2}$
$ x_b\equal{}r$

\[ S\equal{}2\pi \int^{x_b}_{x_a} r(x)\sqrt{1\plus{}(f^{\prime}(x))^2}\,dx\]
\[ S\equal{}2\pi r(x_b\minus{}x_a)\]
Then by $ x_b\minus{}x_a\equal{}h$,
\[ S\equal{}2\pi rh\]
where $ h$ is the height of the cap, and $ r$ is the sphere's radius. I'm kinda uneasy about that answer, if someone could check it that would be nice. What I did find is that the area of the inscribed cone's lateral side is always less than S, which is promising.
$ \pi x_a\sqrt{h^2\plus{}c^2}< S$

Ah! Surface area of a sphere is $ S\equal{}4\pi r^2$ and if we consider a cap of height $ r$ then we find the cap's surface area is $ 2\pi r^2$, or one half that of an entire sphere. Exactly as it should be.


geometry
circumcircle
incenter
inradius
3D geometry
prism
absolute value

A circle $i$ with center $I$ lies inside a circle $o$ with center $O$. Find the locus of the circumcenters of all triangles $AIB$, where $AB$ is a (variable) chord of the circle $o$ such that the line $AB$ touches the circle $i$.

It IS a great problem. It generalizes Euler's identity $OI=R^2-2Rr$, where O, I, R and r are the circumcenter, the incenter, the circumradius and the inradius of a triangle, and it yields a very easy proof of Poncelet's porism for triangles (I leave this as an exercise). Wonderful!

Here is my solution of the problem.

Let R be the radius of the circle o with center O, and let r be the radius of the circle i with center I. Let M be the circumcenter of triangle AIB. Then, the locus of the point M when the chord AB varies is the circle with center O and radius $\frac{R^2-OI^2}{2r}$.

[i]Proof.[/i] It is enough to show that $OM = \frac{R^2-OI^2}{2r}$ for any position of the chord AB.

We will use directed angles modulo 180° throughout the proof. First, note a lemma:

[b]Lemma 1.[/b] Let ABC be a triangle, R its circumradius, and $h_c$ its altitude to the side AB. Then, $2Rh_c=ab$.

[i]Proof.[/i] Let C' be the point diametrically opposite to the point C on the circumcircle of triangle ABC. Then, since the segment CC' is a diameter of the circumcircle, we have CC' = 2R and < CBC' = 90°. Also, since the point C' lies on the circumcircle, we have < CC'B = < CAB. On the other hand, let N be the foot of the altitude to the side AB of triangle ABC. Then, < CNA = 90° and < CAN = < CAB. Hence, < CNA = < CBC' and < CAN = < CC'B. Thus, the triangles CNA and CBC' are directly similar, so that CN : CA = CB : CC'. Hence, $CC^{\prime}\cdot CN = CB\cdot CA$. In other words, $2R\cdot h_c = a\cdot b$. This proves Lemma 1.

Consider the triangle AIB in our configuration. Its circumradius is AM. Its altitude to the side AB is r (since the distance from the point I to the line AB is r, as the circle i touches the line AB). Hence, applying Lemma 1 to triangle AIB, we get $2\cdot AM\cdot r = AI\cdot BI$, i. e. $2r\cdot AM = AI\cdot BI$.

We want to show $OM = \frac{R^2-OI^2}{2r}$. In other words, we want to show $2r\cdot OM = R^2-OI^2$. But $R^2-OI^2$ is the absolute value of the power of the point I with respect to the circle o; hence, if the line AI meets the circle o at a point G (apart from A), then $R^2-OI^2 = AI\cdot IG$. Hence, we have to prove $2r\cdot OM = AI\cdot IG$. Equivalently, we have to show 2r : AI = IG : OM.

The point M is the center of the circle through the points A, I, B. Hence, after the central angle theorem for directed angles modulo 180°, we have < MAB = 90° - < BIA. On the other hand, since the point O is the center of the circle through the points A, B, G, the same theorem yields < BAO = 90° - < AGB. Hence,

  < MAO = < MAB + < BAO  = (90° - < BIA) + (90° - < AGB)
        = 180° - < BIA - < AGB = - < BIA - < AGB
        = - < (BI; AI) - < (AI; BG) = - < (BI; BG) = - < IBG.

Also, being the circumcenters of triangles ABG and AIB, the points O and M both lie on the perpendicular bisector of the segment AB. Hence, $OM\perp AB$ and < (OM; AB) = 90°. Thus,

  < MOA = < (OM; OA) = < (OM; AB) + < (AB; OA)
        = 90° + < BAO = 90° + (90° - < AGB) = 180° - < AGB
        = - < AGB = - < IGB.

This, together with < MAO = - < IBG, shows that the triangles MOA and IGB are inversely similar. Hence, IG : MO = BI : AM. In other words, IG : OM = BI : AM. But we want to show that 2r : AI = IG : OM. Hence, it is enough to show that 2r : AI = BI : AM. But this follows from $2r\cdot AM = AI\cdot BI$. And it's done.

  Darij

Well Darij you have very profound solution.
But I think you have no need to demonstrate such obvious lemma as 2R* Hc = ab.

Here is my solution:
Let X, Y are two points on a small circle. Draw tangents AB and CD, T = {AB∩CD}.
AB, CD are radical axes of circles (AIB)∩(ABCD) and (CID)∩(ABCD).
It means T is a radical center of circles (AIB), (CID), (ABCD).
As a result T is on a radical axe of (AIB) and (CID). I also belongs to it.
Hence IT is perpendicular to UV (where U, V are centers of (AIB) and (CID))
But IT is also perpendicular to XY, it means XY || UV.

So we have for line joining two points on a small circle is parallel to the line joining
(AIB) centers. We can choose three points X, Y, Z on a small one and corresponding to them U, V, W - (AIB) centers and for every point K on a small corcle we have K’ corresponding to it on (AIB) one.
By using parallelism of lines which we deduced it can be easy shown that if X, Y, Z, K are on a circle than U, V, W, K’ are on a circle too. So the geometrical locus is a circle.

Now I think I can say that this problem was proposed for Tournament of the Towns 2004 (Autumn competition) and his price was 6 points out of 8 – maximum.

Very nice! What I am missing is, however, a demonstration that the locus circle has center O. Unfortunately, it is not so clear what exactly was asked in the problem.

Anyway, could you post a topic with links to all TOT Autumn problems? I intend to try all non-combi ones.

Thanks!

  Darij


Where is he?  What does he do now? I'm just curious.

[quote="sandor"]He is located in New York and is a financial trader.  He still sees Rich as often as possible, and snoops around this site once in a while, although he's forgotten a significant fraction of the math he once knew.[/quote]

Sandor, are you somehow related to the CMU Stats Professor John P. Lehoczky?

I asked him this once too (I went to CMU).  The answer is no.


vector
integration
logarithms
quadratics
limit
calculus
calculus computations

Q. Two ships stand separated on a straight coastline by a distance $ \lambda$. Ship 1 moves $ \bot$ to the coastline at a constant speed, while Ship 2 moves with that same constant speed such that its velocity vector is always aligned along the position of Ship 1. After sufficient time, both ships will move along a straight line, at a constant separation $ x$. Find $ x$

Forgot to add the answer  :oops:
[hide="Answer"]$ x\equal{}\frac{\lambda}{2}$[/hide]

ah a chase problem

suppose ship one is at $ (\lambda, 0)$
ship two is at the origin
lets trace the path of ship two:

first, ship one is obviously at $ (\lambda, ct)$ at any given t

and ship two is headed at this point, so diff wrt x

$ y'(x)\equal{}\frac{ct\minus{}y}{\lambda\minus{}x}$

using arclength we know that $ ct\equal{}\int_0^x \sqrt{1\plus{}(\frac{dy}{dw})^2}dw$

so $ \int_0^x \sqrt{1\plus{}(\frac{dy}{dw})^2}dw\equal{}y\minus{}(x\minus{}\lambda)\frac{dy}{dx}$ let $ q(x)\equal{}y'(x)$

then $ \int_0^x \sqrt{1\plus{}q^2(w)}dw\equal{}y\minus{}(x\minus{}\lambda)q(x)$ now differentiate

$ \sqrt{1\plus{}q^2(x)}\equal{}q(x)\minus{}(x\minus{}\lambda)q'(x)\minus{}q(x)\equal{}(\lambda\minus{}x)q'(x)$

$ \frac{dq}{\sqrt{1\plus{}q^2}}\equal{}\frac{dx}{\lambda\minus{}x} \to \ln ( q \plus{} \sqrt{1\plus{}q^2})\equal{}\ln \lambda \minus{} \ln (\lambda\minus{}x)$

$ \ln \left[ (q\plus{}\sqrt{1\plus{}q^2})(1\minus{}\frac{x}{\lambda})\right] \equal{}0 \to q\plus{}\sqrt q \equal{} \frac{1}{1\minus{}\frac{x}{\lambda}}$

solving the quadratic gives us $ q\equal{} \frac{\frac{1}{1\minus{}\frac{x}{\lambda}}\minus{}(1\minus{}\frac{x}{\lambda})}{2}\equal{}y'(x)$

from here we can solve the diff eq and get $ y(x)\plus{}C\equal{}\frac{1}{2}\left[ \int \frac{dx}{1\minus{}\frac{x}{\lambda}}\minus{}\int (1\minus{}\frac{x}{\lambda} )dx\right]$

solving with $ y(0)\equal{}0$ we get $ y(x)\equal{}\frac{\lambda}{2}\left[\frac{1}{2}(1\minus{}\frac{x}{\lambda})^2 \minus{} \ln(1\minus{}\frac{x}{\lambda})\right] \minus{}\frac{\lambda}{4}$

so as $ t\to \infty$ we have $ x \to \lambda$ and so $ \lim_{x\to \lambda}ct\minus{}y\equal{}\lim_{x\to \lambda}(\lambda\minus{}x)y'(x)$

$ \equal{}\lim_{x\to \lambda}(\lambda\minus{}x)\frac{\frac{\lambda}{\lambda\minus{}x}\minus{}(\frac{\lambda\minus{}x}{\lambda})}{2}\equal{}\lim_{x\to \lambda} \frac{\lambda^2\minus{} (\lambda\minus{}x)^2}{2\lambda}\equal{}\frac{\lambda}{2}$


[img]http://img232.imageshack.us/img232/5530/fcdvxc.jpg[/img]
I used the conjugate two times, but it seems that it doesn't work.


geometry
3D geometry
octahedron
pyramid
trigonometry
AMC

(From BAMM Test of Ingenuity 1995)

15. How many "friday the 13ths" are possible in a normal 365-day year?

(A) Zero, one, or two
(B) Only one
(C) One or two
(D) One or three
(E) One, two, or three

17. Suppose it is true that exactly two of the following three statements are true.  Also, statement 1 is false or it is not true that statements 2 and 3 are both true or both false.

1. Statement 2 is false and my dog is not named "Euclid".
2. My dog is white and is named Euclid.
3. Statement 1 is false and my dog is white.

We can conclude that my dog is:
(A) white and named Euclid
(B) named Euclid, but not white
(C) white, but not named Euclid
(D) neither white nor named Euclid
(E) none of the above

18.  The tetrakaidecahedron is a polyhedron composed of 14 faces: 6 squares and 8 regular hexagons.  At each vertex, two hexagons and one square meet.  Let :theta: be the angle between one of the hexagonal faces and the adjacent square face.  Find cos :theta:.

(A) -1/:rt3:
(B) 1/:rt3:
(C) -1/sqrt(5)
(D) -:rt3:/4
(E) -1/3

Please include your solutions.  Enjoy!

17. [hide]A. We have that two of the statements are true, and one is false. Suppose Statement 1 is true. Then statement 2 must be false. If statement 2 is false, the dog is not white, which makes statement three false. This doesn't fit the restrictions. So statement 1 is false, and statements 2 and 3 are true, which we can easily verify works.[/hide]

Ah, gotta love number 18.  Yeah for 3D geometry!

15.

[hide]

Whichever day of the week is the 13th in January, the 13th in February will be

31 (mod 7) = 3 days later in the week.  So we find the residues (mod 7) of the

number of days in each month: 3, 0, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3.

Then we compute how many days later in the week the 13th of month i will be

compared with the 13th of January: 0, 3, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5.

Since there are three 3s, we see that it is possible to have Friday the 13ths in

February, March, and November.  It is also possible to have 1 or 2 Friday the

13ths since there are modulo classes of size 1 and 2.  Hence (E).

[/hide]

17.

[hide]

I went with a truth table with the four possibilities for whether the dog is White and named Euclid (T/F for each) down the side and statements 1, 2, and 3 across the top.

The only instance where exactly two statements are true is when the dog is both white and named Euclid.  In that case, statement 1 is false and 2 and 3 are true.

So we get (A).

[/hide]

18.

[hide]

Say we have a vertex at the origin, and we have a unit square

that is axis aligned with its top right corner at the origin.  Let A = (-1,0) and

B = (0,0). Let H_1 be the hexagon that share edge AB with the square.  Let C

be the next vertex (proceeding counterclockwise from B) of H_1.

The key is that if we project edge BC into the plane, we get y = x (easy to see by symmetry).  Now add point G in the interior of H_1, such that GB projects onto

the y-axis and CBG is a 30/60/90 triangle.  Then we have GC = GB cos:theta:.

This is 1/2 = cos(:pi: - :theta:)  :rt3: /2.  So cos ( :pi: - :theta:) = 1/:rt3:.

As a result, cos(:theta:)=-1/:rt3:.

[/hide]

--Dan

My method for number 18:  [hide]We recognize the described figure as a truncated octahedron.  If we look only at the top plane cut, we see it will be parallel to the plane which divides the octahedron into two right square pyramids.  Thus, the angle between those two faces of our figure with the long name is the supliment of the angle between a triangular face of a right square pyramid and its base (if the sides are equilateral).  We can calculate cosine of that angle with very little difficulty by dropping a perpendicular from the top vertex and forming a right triangle, and we also get cos :theta: = -1/:rt3: (the minus coming because we have the supliment of the angle.)[/hide]


A repunit is a number that can be represented by only 1's for some integer base. For example, 11, 7, and 4 are repunits since 11=11 base 10, 7=111 base 2, and 4 = 11 base 3. What is the largest repunit that is less than 2009?

Uh I'm assuming you mean integers. It has to be less than 2009, meaning 2008 is the maximum possible, so if 2008 works, we win. Which it does.


Calculate $ \frac{3 \cdot 5! \plus{} 15\cdot 4!}{6!}$

\begin{align*}
\frac{3\cdot 5!+15\cdot 4!}{6!} &= \frac{3\cdot 5+15}{6\cdot 5} \\
&= \frac{30}{30} \\
&= \boxed{1} \\
\end{align*}

It's not too bad just to evaluate the expression without dividing by $ 4!$ first though.

[hide]We factor and rearrange the terms in the factorials:

\begin{align*}
\frac{3\cdot 5! + 15\cdot 4!}{6!} &= \frac{3\cdot 5! + 3\cdot 5!}{6!} \\
&= \frac{2\cdot 3 \cdot 5!}{6!} \\
&= \frac{6!}{6!} \\
&= \boxed{1}
\end{align*}[/hide]

[quote=NathanBikeSki][quote=GameBot]Calculate $ \frac{3 \cdot 5! + 15\cdot 4!}{6!}$[/quote]

[hide=My Solution]
\begin{align*}
\frac{3 \cdot 5! + 15\cdot 4!}{6!}&=\frac{3 \cancel{\cdot 5!}  + 15\cdot 4!}{\cancelto{6}{6!}}\\&=\dfrac{3+15\cdot (4\cdot (3\cdot 2)\cdot 1)}{6}\\&=\dfrac{3+15\cdot 4\cancel{\cdot 6}\cdot 1}{\cancel{6}}\\
&=3+60\\&=\boxed{\frac{3 \cdot 5! + 15\cdot 4!}{6!}=\boxed{63}}
\end{align*}
[/hide][/quote]

Blud u cant only divide two numbers in that division thing and ignore the others.

The answer is 1. Here is how you may check your answer:

$ \frac{3 \cdot 5! + 15\cdot 4!}{6!} = \frac{3 \cdot 120 + 15 \cdot 24}{720} = \frac{360 + 360}{720} = \frac{720}{720} = \boxed{1}$


geometry
area of a triangle
geometry unsolved

Consider an acute triangle $ABC$ scribe in a circle $(O;R)$.Let $M$ is in the plane which contains $A,B,C$.$X,Y,Z$ are the foots of perpendicular from$M$ to $BC,CA,AB$,respectively.$Q\in(OM)$and$Q\neq M;OM=OQ$.The lines $x,y,z$ are the images of $BC,CA,AB$respectively in a hothometic $V_Q^{\frac{1}{2}}$.$D,E<F$ are the poles of $x,y,z$ with $(O)$.Prove that:
             $\frac{S(DEF)}{S(ABC)}\ge \frac{256}{R^2}.|P_{M/(XYZ)}|$
[i][u]Notes:[/u][/i]$S(IJK)$is the area of a triangle $IJK$
                           $P_{P/(I;r)}=r^2-IP^2$


Oh no, where do you get such an idea (and even making it into a thread)? :shock:

Hm... however I think statistically it can sometimes happen :wink: [Depending on your definition of after :wink:]

[quote="sinajackson"]Is it True that after any person dies :( ,then somewhere in this world, a baby is born... :lol:[/quote]
Maybe sometimes, but definitely not always. If a catastrophe happens and many people die in an instant, I don't think that many babies can be born at the same time.

Alfred Tennyson once wrote in some poem:
"Every second a man dies, every second one is born."

Charles Babbage reportedly wrote to him, suggesting him to edit this to:
"Evey second a man dies, evey second one and one-tenth is born."

 :rotfl:


let $ ABC$ be a  triangle and $ D,E,F$  the points where the incircle touches $ BC,CA,AB$ respectively.the parallel to $ AB$ trough $ E$ meets $ DF$ at $ Q$ and the parallel to $ AB$ through $ D$ meets $ EF$ at $ T$ prove that $ CF,DE,QT$ are concurrent

Denote K be the intersection of FE and BC. From (BCDK)=-1, we have the harmonic pencil F(BCDK). By intersecting of this pencil with QE, DT, we have that FC passes through the midpoint of QE, DT (since QE || DT || AB). Hence result.


function
integration
inequalities
real analysis
real analysis unsolved

Let $ a > 0, \alpha \geq 0$.Find the integrable function $ f: [0,a] \minus{} > [0, \infty)$ with the property that $ \forall x \in [0,a]$ we have  $ f(x) \leq \int_{0}^{x}t^{\alpha}f(t)dt$.

_______________________________________________
   $ SHORTLIST 2004$

If $ f$ satisfies the inequality then $ g: [0,a]->[0,\infty)$, $ g(x)=\int^x_0 t^\alpha f(t)\,dt$ also does it:
$ f(x)\le g(x)\Leftrightarrow \int^x_0 t^\alpha f(t)\,dt\le \int^x_0 t^\alpha g(t)\,dt \Leftrightarrow g(x)\le \int^x_0 t^\alpha g(t)\,dt$
$ g$ is continous $ \Rightarrow x^\alpha g(x)$ primitivable $ \Rightarrow \exists  F: [0,a]->\mathbb{R}$, $ F'(x)=x^\alpha g(x)  \forall  x \in [0,a]$
$ x^\alpha g(x) \le x^\alpha \int^x_0 F'(t) \,dt \Rightarrow F'(x) \le x^\alpha [F(x) -F(0)]$, multiplying by $ e^{- \dfrac{x^{\alpha+1}}{\alpha+1}}$, one gets
$ [F'(x)-x^\alpha (F(x)-F(0))] e^{- \dfrac{x^{\alpha+1}}{\alpha+1}} \le 0$
$ \Rightarrow \left( \dfrac{F(x)-F(0)}{e^{ \dfrac{x^{\alpha+1}}{\alpha+1}}} \right) ' \le 0$, so the function $ G(x)= \dfrac{F(x)-F(0)}{e^{ \dfrac{x^{\alpha+1}}{\alpha+1}}}$  is decreasing $ \Rightarrow G(x) \le G(0)  \forall x>0$, but $ G(0)=0 \Rightarrow G(x) \le 0$. Also from $ F'(x)=x^\alpha g(x)>0$ we get F is increasing, so $ F(x)-F(0) \ge 0  \forall x>0 \Rightarrow G(x) \ge 0  \forall x>0 \Rightarrow G(x)=0 \Rightarrow F(x)=F(0)=0,g(x)=0,f(x)=0 .$


function
calculus
derivative
parameterization
calculus computations

Consider the function given by the equations $ x=\theta+\frac{1}{2}sin2\theta$ and $ y=\theta-\frac{1}{2}sin2\theta$ where $ \theta$is a real number

(i)Show that $ \frac{dy}{dx}=tan^{2}\theta$

(ii)Show that $ \frac{d^{2}y}{dx^{2}}=tan\theta\sec^{4}\theta$

(i) $ \frac{dy}{dx}=\frac{1-cos2\theta}{1+cos2\theta}=\frac{2sin^{2}\theta}{2cos^{2}\theta}=tan^{2}\theta$

(ii) $ \frac{d^{2}y}{dx^{2}}=\frac{2tan\theta}{2cos^{4}\theta}=tan\theta\sec^{4}\theta$

Just in case you didn't see what Ognjen09 did, they used the formula for taking derivatives of parametric equations which is dy/dx = (dy/dt)/(dx/dt) where t is the parameter. However in your case, theta is the parameter but it makes no difference in the formula.


ratio
geometry
circumcircle
power of a point
radical axis
perpendicular bisector
geometry unsolved

What is a "pencil" ?

Also, why you proved $\frac{p_{a}}{p_{b}}$ is constant? Not $\frac{AK}{BK}$. I'm confused..

Pencil of circles is a set of circles with a common radical axis. You can have a pencil of intersecting circles (all circles passing through 2 given points P, Q), a pencil of tangent circles (all circles tangent at a given point P), or a pencil of non-intersecting circles with 2 base points P, Q (see the figure). The radical axis of a pencil of intersecting circles is the line PQ. The radical axis of a pencil of tangent circles is their common tangent at P. The radical axis of a pencil of non-intersecting circles is the perpendicular bisector of PQ. In all cases, the centers of all circles in the same pencil are collinear. Also in all cases, the radical axis belongs to the pencil as a circle of infinite radius. For a pencil of tangent circles, the tangency point P belongs to the pencil as a circle of zero radius. Likewise, for a pencil of non-intersecting circles, the 2 base points P, Q belong to the pencil as circles of zero radius. If (A), (B), (O) are 3 circles in the same pencil (any type), K an arbitrary point on (O) and $p_{a},\ p_{b}$ powers of K to (A), (B), then $\frac{p_{a}}{p_{b}}= \frac{AO}{BO}.$ This is a constant independent on the position of K on (O). For a pencil of non-intersecting circles, this holds even if we choose the circles (A), (B) with zero radius (identical with the base points P, Q). But power of a point K to the point $A \equiv P$ is square of the distance KA, same for the point $B \equiv Q.$ So we get $\frac{KA^{2}}{KB^{2}}= \frac{AO}{BO},$ which is constant. If we square-root it, $\frac{KA}{KB}= \sqrt{\frac{AO}{BO}}$ is also constant.

[quote="yetti"]Thus the locus of points with the constant ratio of distances from 2 given points A, B is a circle, which belongs to the same pencil with the points A, B. [/quote]

Then, how do you know no other points are on the locus?

The points A, B are different from each other, otherwise any point would belong to the locus and the problem would not make sense. We can also assume that $\frac{AK}{BK}\neq 1,$ otherwise we would have AK = BK and the locus would trivially be the perpendicular bisector of the segment AB. Assume that some other point K' not on the circle (O) belongs to the locus, i.e. $\frac{AK'}{BK'}= \frac{AK}{BK}= \sqrt{\frac{AO}{BO}}\neq 1,$ i.e., K' is not on the perpendicular bisector of the segment AB either. A unique circle (O') from the pencil with the base points A, B and with the circle (O) passes through K', its center O' different from O, i.e., $OO' \neq 0,$ and $\frac{AK'}{BK'}= \sqrt{\frac{AO'}{BO'}}.$ Then

$\frac{AO}{BO}= \frac{AK'^{2}}{BK'^{2}}= \frac{AO'}{BO'}= \frac{AO \pm OO'}{BO \pm OO'}$

$AO \cdot BO \pm AO \cdot OO' = BO \cdot AO \pm BO \cdot OO'$

$AO \cdot OO' = BO \cdot OO'$

(either all signs are plus, or all signs are minus). Since $OO' \neq 0,$ it follows that $AO = BO.$ Since no circle of the non-intersecting pencil with the base points A, B is centered on the segment AB, O cannot be the midpoint of the segment AB. Consequently, either the points $A \equiv B$ are identical, which is a contradiction. Or O is at infinity, (O) is a circle of infinite radius in our pencil, identical with the radical axis of the pencil, identical with the perpendicular bisector of the segment AB, hence AK = BK, which is also a contradiction.

Is the circumcircle of triangle XYZ Apollonius's circle?

What differences have between pencil of circles wrt two points A,B and Apollonius's circle wrt two points A,B?

Sorry for not answering earlier. The circumcircle $ \mathcal K$ of the $ \triangle XYZ$ is the K-Apollonius circle of any $ \triangle ABK$ with $ A, B$ fixed and $ K \in \mathcal K.$ The points $ A, B,$ considered as circles of zero radius, and the circle $ \mathcal K$ form a pencil of non-intersectiong circles. The posted problem is a special case of a general problem for 3 coaxal circles $ \mathcal A, \mathcal B, \mathcal K,$ forming a pencil of any type. The solution applies to the general problem.

Ha I find it amusing that I glanced over this page when I was younger and I was like wow thats an intimidating solution, I'm not going to read it.

Anyway, I guess my question is:

How does one come up with such a solution? It is so remarkable. I suppose that the foundation is this lemma:

Let $ \omega_1$ and $ \omega_2$ be circles center $ O_1$ and $ O_2$ and $ P$ is a point, $ L$ is the projection of $ P$ onto $ O_1O_2$, $ M$ is the midpoint of $ O_1O_2$. Then
\[ \mathcal{P}_{\omega_1}(P) \minus{} \mathcal{P}_{\omega_2}(P) \plus{} R_1^2 \minus{} R_2^2 \equal{} 2LM \cdot O_1O_2
\]
The rest of the solution follows naturally I supppose...Take convenient points to cancel things and such. But I just have no idea how one would come up with this.

[quote="maggot"]Let $X,Y,Z$ be three points on a circle $O$ and $A,B$ be two points such that $\frac{AX}{BX}=\frac{AY}{BY}=\frac{AZ}{BZ}$. Show that $\frac{AK}{BK}$ is constant for all points $K$ on the circle.[/quote]

[quote="yetti"]Let (A), (B) be 2 arbitrary non-concentric circles with radii a, b (intersecting or not) and let K be a point not on them. Let $p_{a}= KA^{2}-a^{2},\ p_{b}= KB^{2}-b^{2}$ be powers of K to (A), (B), respectively. Let M be the midpoint of the segment AB, let the radical axis of (A), (B) meet their center line AB at R, and let L be foot of the normal from K to the center line.

$a^{2}-b^{2}+p_{a}-p_{b}= KA^{2}-KB^{2}=$

$= LA^{2}-LB^{2}= (LA+LB)(LA-LB) = 2LM \cdot AB$

If K is on the radical axis, then $p_{a}= p_{b},$ so that

$a^{2}-b^{2}= 2RM \cdot AB$

Substituting this to the above equation with $p_{a},\ p_{b},$

$p_{a}-p_{b}= 2(LM-RM) \cdot AB = 2LR \cdot AB$

Let (O) be a circle passing through K, which belongs to the same pencil with the circles (A), (B). The 3 circles have a common radical axis and their centers are collinear, hence the point R is common. Since $K \in (O),$ power p of K to (O) is zero. Applying the above formula to the circle pairs (A), (O) and (B), (O),

$p_{a}= 2LR \cdot AO,\ \ \ p_{b}= 2LR \cdot BO$

which yields $\frac{p_{a}}{p_{b}}= \frac{AO}{BO}= \text{const},$ regardless of the position of K on (O). As a special case, let (A), (B) be circles of zero radius (a = b = 0), i.e., points A, B. Powers of K to the points A, B are just $p_{a}= KA^{2},\ p_{b}= KB^{2},$ so that $\frac{AK}{BK}= \sqrt{\frac{AO}{BO}}= \text{const},$ regardless of the position of K on (O). Thus the locus of points with the constant ratio of distances from 2 given points A, B is a circle, which belongs to the same pencil with the points A, B. The 3 points X, Y, Z, such that $\frac{AX}{BX}= \frac{AY}{BY}= \frac{AZ}{BZ}$ define this circle (the circumcircle of the $\triangle XYZ$). Given the points A, B, one of the 3 points X, Y, or Z would be sufficient to define this circle, because a unique circle of a given pencil of non-intersecting circles (defined by the points A, B) passes through every point of a plane.[/quote]


I'm at the point where:

Enter the full path of the directory, where the executables (latex, tex, etc.) of your TeX-distribution are located:

I typed in:

C:/texmf/miktex/bin/yap.exe

then NEXT

I got a big white X saying

"The application 'LaTex' has not been found in the direcotry of your TeX-distriubtion.

Please enter the directory where executable files (tex.exe, latex.exe, etc.) are located, as your distriubiton's directory.

I'm so frustrated! Help please!

[quote]Enter the full path of the [size=117][color=red][u]directory[/u][/color][/size], where the executables (latex, tex, etc.) of your TeX-distribution are located:[/quote] Yet you've put the name of the DVI viewer called YAP.
Is C:\texmf\miktex\bin the directory on your system where latex.exe, tex.exe etc are located?

You need also to say what you are doing, where the instruction comes from, what program you are installing, which operating system etc. There are some very intelligent people on this forum but I doubt that they are capable of reading your mind.

No, but it was something else very similar...

Gosh, after 5 hours of figuring out, I finally got it through!!! I'm so excited. Thanks for replying!!


quadratics
algebra
number theory
relatively prime
Diophantine equation

Find the least integer x>1 such that [tex]\frac{3x^{2}}{2} - \frac{x}{2}[/tex] is a perfect square.

Use Pell's equation: if 3x^2/2-x/2 = a^2, then rearrange to find

(6x-1)^2 - 24a^2 = 1.

The Fundamental Solution to the pells equation m^2-24a^2=1 is (5,1), and from there all solutions can be found. The next few solutions are (49,10), (485,99), .... Obviously 49 cant equal 6x-1 because that would make x not an integer. But, 6x-1 can equal 485, giving [b]x=81[/b].

i understand all of it except how you find the solutions (49,10) (485,99).. etc

could you please explain it?

Since the fundamental solution is (5,1), all solutions (x_n,y_n) are given by

x_n + y_n sqrt24 = (5 + 1 sqrt24)^n. (Dont ask why... I dont remember :))

So, we easily find that (5+sqrt24)^2 gives 49+10 sqrt24, so the second solution is (49,10). Likewise, (5+sqrt24)^3 gives the third solution of 485 + 99 sqrt24. In general, this is not the best way to characterize the solution set, since there are recursions and even explicit formulas that you can use. But by hand, it is easiest to just use this method. See if you can use this method to find the recursion and even the explicit formula!

Pell's equations to come up often in questions involving solving equations for integer solutions with squares.  A lot of problems of these types can be solved by noting that if you have a quadratic equation in the integers, then the discriminant in the quadratic equation must also be the square of an integer.  Here are three problems I stumbled upon this weekend, which can be solved by Pell's equation.

1.  Prove that there exist infinitely many triples (a, b, c) of positive integers s.t. a, b, c are pairwise relatively prime and a :^2: *b :^2: + b :^2: * c :^2: + a :^2: *c :^2: is the square of an integer.

2.  A right triangle is called "nearly isosceles" if its two legs are consecutive positive integers, and its hypotenuse is a positive integer.  The smallest such triangle is (3, 4, 5).  Find the next four smallest "nearly isosceles" right triangles.

3.  There exist infinitely many positive integer solutions (verify?) (m,n) to m + (m+1) + (m+2) + ... + (n-2) + (n-1) + n = m*n.  Find five such solutions to this equation.

zscool, for your second problem, do you mean that the hypotenuse must be an integer as well?

1, 2,  :rt5: technically satisfies the question.

[quote="zscool"]2.  A right triangle is called "nearly isosceles" if its two legs are consecutive positive [b]integers.[/b]  The smallest such triangle is (3, 4, 5).  Find the next four smallest "nearly isosceles" right triangles.[/quote][/b]

Heh.

ok, i understand now.. i read up on pell equations some too..

just curiously, is there another way to solve the problem i provided? or is the pell equation the way to go?


Alicia is 150 centimeters tall. On a sunny day, she casts a
200-centimeter shadow at the same time that a flagpole casts a
24-meter shadow. How many meters tall is the flagpole?

The height:shadow ratio is 150/200=3/4, so the flag's height is 3/4 * 24=18 meters

Use a ratio.
If 150 cm person creates a 200 cm shadow,
then every 3 cm produces 4 cm shadow.
3/4*24=18.