[quote="Q.E.D.avid"][quote="can_hang2007"][quote="tuantam1lan"]let a,b,c are non-negative such that:$ a \plus{} b \plus{} c \equal{} 1$.prove that:
$ \frac {a^2(b \plus{} c)}{1 \plus{} 4bc} \plus{} \frac {b^2(a \plus{} c)}{1 \plus{} 4ac} \plus{} \frac {c^2(a \plus{} b)}{1 \plus{} 4ab} \leq \frac {1}{4}$[/quote]
Hello, it is clear that
$ LHS \ge \sum a^2(b \plus{} c) \le \frac {1}{4}.$
So your inequality is valid. :)[/quote]

 :D Ooops can hang! You placed the inequality sign the wrong way. It should read:

$ LHS \le \sum a^2(b \plus{} c) \le \frac {1}{4}.$[/quote]
Thank you very much, Q.E.D.avid. I typed it so fast and didn't preview, so I didn't see the mistake in typing.  :blush:

[quote="can_hang2007"][/quote][quote="Q.E.D.avid"][quote="can_hang2007"][quote="tuantam1lan"]let a,b,c are non-negative such that:$ a + b + c = 1$.prove that:
$ \frac {a^2(b + c)}{1 + 4bc} + \frac {b^2(a + c)}{1 + 4ac} + \frac {c^2(a + b)}{1 + 4ab} \leq \frac {1}{4}$[/quote]
Hello, it is clear that
$ LHS \le \sum a^2(b + c) \le \frac {1}{4}.$
So your inequality is valid. :)[/quote][/quote][quote="can_hang2007"][/quote]
Thank you very much,can-hang2007,you really are smarter people.
I created it after read your proof for hungkhtn's inequality in math.vn.it is an innovative simple and not particularly.
another proof of mine:
$ \frac {1 + 4bc}{4} = \frac {(a + b + c)^2}{4} + bc \ge ab + bc + ca$
Consequently:
$ \frac {4a^2(b + c)}{1 + 4bc} \le \frac {a^2(b + c)}{(ab + bc + ca)}$
we have:
$ 4\sum{\frac {a^2(b + c)}{1 + 4bc}} \le \frac {\sum{a^2(b + c)}}{ab + bc + ca} \le \frac {\sum{a^2(b + c) + 3abc}}{ab + bc + ca} = 1$

[quote="phymax"]please dont post the topic with names like "easy" or "hard' , etc.
if its really easy to you then why do you want to post it! that too in "unsolved" section?

please take care of that :)[/quote]
if you have'n idea for problem then you should'n Comment like ditto!

[quote="can_hang2007"][quote="tuantam1lan"]let a,b,c are non-negative such that:$ a \plus{} b \plus{} c \equal{} 1$.prove that:
$ \frac {a^2(b \plus{} c)}{1 \plus{} 4bc} \plus{} \frac {b^2(a \plus{} c)}{1 \plus{} 4ac} \plus{} \frac {c^2(a \plus{} b)}{1 \plus{} 4ab} \leq \frac {1}{4}$[/quote]
Hello, it is clear that
$ LHS \le \sum a^2(b \plus{} c) \le \frac {1}{4}.$
So your inequality is valid. :)[/quote]
 i think we can prove that:$ a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b) \plus{} \frac {3}{4}abc \le \frac {1}{4}$
equality when $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$ or $ a\equal{}b\equal{}\frac{1}{2},c\equal{}0$

[quote="tuantam1lan"][quote="can_hang2007"][quote="tuantam1lan"]let a,b,c are non-negative such that:$ a \plus{} b \plus{} c \equal{} 1$.prove that:
$ \frac {a^2(b \plus{} c)}{1 \plus{} 4bc} \plus{} \frac {b^2(a \plus{} c)}{1 \plus{} 4ac} \plus{} \frac {c^2(a \plus{} b)}{1 \plus{} 4ab} \leq \frac {1}{4}$[/quote]
Hello, it is clear that
$ LHS \le \sum a^2(b \plus{} c) \le \frac {1}{4}.$
So your inequality is valid. :)[/quote]
 i think we can prove that:$ a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b) \plus{} \frac {3}{4}abc \le \frac {1}{4}$
equality when $ a \equal{} b \equal{} c \equal{} \frac {1}{3}$ or $ a \equal{} b \equal{} \frac {1}{2},c \equal{} 0$[/quote]
Well, it is equivalent to Schur's Inequality of third degree, my friend. :)

[quote="can_hang2007"][/quote][quote="tuantam1lan"][quote="can_hang2007"][quote="tuantam1lan"]let a,b,c are non-negative such that:$ a + b + c = 1$.prove that:
$ \frac {a^2(b + c)}{1 + 4bc} + \frac {b^2(a + c)}{1 + 4ac} + \frac {c^2(a + b)}{1 + 4ab} \leq \frac {1}{4}$[/quote]
Hello, it is clear that
$ LHS \le \sum a^2(b + c) \le \frac {1}{4}.$
So your inequality is valid. :)[/quote]
 i think we can prove that:$ a^2(b + c) + b^2(c + a) + c^2(a + b) + \frac {3}{4}abc \le \frac {1}{4}$
equality when $ a = b = c = \frac {1}{3}$ or $ a = b = \frac {1}{2},c = 0$[/quote][quote="can_hang2007"]
Well, it is equivalent to Schur's Inequality of third degree, my friend. :)[/quote]
can we find maximal value of the expression?
$ a^2b + b^2c + c^2a + \frac {3}{8}abc$
with $ a,b,c$ are non-negative such that:$ a + b + c = 1$

[quote="tuantam1lan"][/quote][quote="can_hang2007"][/quote][quote="tuantam1lan"][quote="can_hang2007"][quote="tuantam1lan"]let a,b,c are non-negative such that:$ a + b + c = 1$.prove that:
$ \frac {a^2(b + c)}{1 + 4bc} + \frac {b^2(a + c)}{1 + 4ac} + \frac {c^2(a + b)}{1 + 4ab} \leq \frac {1}{4}$[/quote]
Hello, it is clear that
$ LHS \le \sum a^2(b + c) \le \frac {1}{4}.$
So your inequality is valid. :)[/quote]
 i think we can prove that:$ a^2(b + c) + b^2(c + a) + c^2(a + b) + \frac {3}{4}abc \le \frac {1}{4}$
equality when $ a = b = c = \frac {1}{3}$ or $ a = b = \frac {1}{2},c = 0$[/quote][quote="can_hang2007"]
Well, it is equivalent to Schur's Inequality of third degree, my friend. :)[/quote][quote="tuantam1lan"]
can we find maximal value of the expression?
$ a^2b + b^2c + c^2a + \frac {3}{8}abc$
with $ a,b,c$ are non-negative such that:$ a + b + c = 1$[/quote]
I think that we can, I found thye maximal value, is $ \frac {361}{136}$ but I'm not sur, we must wait for the best solver problem here,

[quote="tuantam1lan"][/quote][quote="can_hang2007"][/quote][quote="tuantam1lan"][quote="can_hang2007"][quote="tuantam1lan"]let a,b,c are non-negative such that:$ a + b + c = 1$.prove that:
$ \frac {a^2(b + c)}{1 + 4bc} + \frac {b^2(a + c)}{1 + 4ac} + \frac {c^2(a + b)}{1 + 4ab} \leq \frac {1}{4}$[/quote]
Hello, it is clear that
$ LHS \le \sum a^2(b + c) \le \frac {1}{4}.$
So your inequality is valid. :)[/quote]
 i think we can prove that:$ a^2(b + c) + b^2(c + a) + c^2(a + b) + \frac {3}{4}abc \le \frac {1}{4}$
equality when $ a = b = c = \frac {1}{3}$ or $ a = b = \frac {1}{2},c = 0$[/quote][quote="can_hang2007"]
Well, it is equivalent to Schur's Inequality of third degree, my friend. :)[/quote][quote="tuantam1lan"]
can we find maximal value of the expression?
$ a^2b + b^2c + c^2a + \frac {3}{8}abc$
with $ a,b,c$ are non-negative such that:$ a + b + c = 1$[/quote]
Hello, we have the known result of Hungkhtn and Vasc is: $ a^2b + b^2c + c^2a + abc \le \frac {4}{27}$ with equality holds for $ a = b = c = 1/3$ and $ a = 2/3,b = 1/3,c = 0$. From this result, we can see that the maximum you asked is $ 4/27.$ :)

$ \frac{3}{8}abc\le abc$. Is that right?
I don't think this nice inequality belongs to hungkhtn, canhang2007. As I know it appeared on Canada Mathematic olympiad 1996 .Sorry If my information source is wrong . :)

[quote="Allnames"]$ \frac {3}{8}abc\le abc$. Is that right?
I don't think this nice inequality belongs to hungkhtn, canhang2007. As I know it appeared on Canada Mathematic olympiad 1996 .Sorry If my information source is wrong . :)[/quote]
Canada's problem is to prove that $ x^2y\plus{}y^2z\plus{}z^2x\leq \frac{4}{27}$ with $ a\plus{}b\plus{}c\equal{}1$ and $ a,b,c>0$ and it's Canada 1999, Problem 5.

[quote="Allnames"]$ \frac {3}{8}abc\le abc$. Is that right?
I don't think this nice inequality belongs to hungkhtn, canhang2007. As I know it appeared on Canada Mathematic olympiad 1996 .Sorry If my information source is wrong . :)[/quote]
It is valid as $ abc \ge 0$ and $ 3/8<1$. Let me explain more, because the inequality of Hungkhtn (you can see it is stronger than Canadian MO 1996 ;) ), we have
$ a^2b\plus{}b^2\plus{}c^2a\plus{}\frac{3}{8}abc \le a^2b\plus{}b^2c\plus{}c^2a\plus{}abc \le \frac{4}{27}.$
Moreover, the equality can attain when $ a\equal{}2/3,b\equal{}1/3,c\equal{}0,$ so this is also our maximum value. :)


The rules are the same in Cap. Sugar's saga.

Problem 1

While I type this, I'm listening to my ipod. Due to difficulties with the computer, only 5 songs of my favorite album were put into it. There are 10 total songs in the whole album. Here are the times for each:

Oceans and Rain: 3 min 45 sec
Ghoom Charakhana: 2 min 54 sec
Transcendence: 3 min 26 sec
Harmon Session Special XI: 2 min 34 sec
Sweet Lassi Dub: 3 min 54 sec
Sirocco: 4 min 11 sec
Der Bauch: 3 min 29 sec
Just You and I: 2 min 59 sec
Caravane: 3 min 1 sec
Shaskin: 3 min 3 sec

It so happens that the average of those songs equals one of the songs +.8 second times 2. Find the average of the minutes and the seconds of that song.

don't you think [b]four[/b] marathons in one forum is a little too much?

@kyyuanmathcount: Yes.
@splashD: Well, I may decide to move it to another forum...

Yeah, I believe people in this forum are really overloaded with work.  I will, however, move it to a forum of your choice, you just have to make the level appropriate :).  Pm me.

Edit: Alright.  Mathcounts -->  MOEMS

Sorry.

#2

My dad needs time to put the other 5 songs onto my ipod. He also needs to finish working on the downstairs bathroom. If he will finish in x-3 days if he works x hours per day, and if he works 8 more hours, he will finish that. He will fix my ipod in one day maximum. If he never fixes my ipod the same day he works, and never starts on my ipod until he is finished in the bathroom, and he will work x hours per day in the bathroom, what is the minimum time needed so that I can have my ten songs onto my ipod?

I have no idea what that question says. Please repeat it clearly.


geometry
perimeter
3D geometry
sphere
incenter

A circle is inscribed in a square. The area of the square divided by  its perimiter is the area of the circle divided by its circumfrence.

1) How big must a triangle be?

2) How big must a pentagon be?

3) Prove the general rule that should be obvious from here.

4) Is the rule also trule for Archimedean solids and spheres?

I'm not sure I understand.  Do you mean "how big must a triangle inscribed in the same circle be"?

Oh, sorry. I meant a triangle that has the property that its area over its perimeter is the same as that as the square and circle. Same for a pentagon.

[hide="the general rule is"]
for [b]any[/b] polygon that has a circle of radius $r$ inscribed in it, perimeter $2s$, and area $K$, $K=rs$. (it can be rewritten as $\frac{K}{2s}=\frac{\pi r^{2}}{2\pi r}$)

To see that this formula works, we can split an $n$-sided polygon with an incircle into $n$ triangles, each formed by the incenter and two consecutive vertices. Let the polygon be $P_{1}P_{2}P_{3}\cdots P_{n}$ with incenter $I$. Then triangle $P_{k}P_{k+1}I$ has a base length of $P_{k}P_{k+1}$ and a height of $r$, therefore an area of $\frac{r}{2}P_{k}P_{k+1}$.

Now add up the areas of the $n$ triangles to get the area of the whole polygon, and note that
\[P_{1}P_{2}+P_{2}P_{3}+\cdots+P_{n}P_{1}=2s\]
so that
\[K=\frac{r}{2}(2s)=rs\]
as desired.
[/hide]

[hide="for three dimensions..."]
if a polyhedron has a sphere of radius $r$ inscribed in it, surface area $S$, and volume $V$, then
\[V=\frac{1}{3}(rS)\]
[/hide]


AMC
AIME
USA(J)MO
USAMO
geometry
induction
pigeonhole principle

I have gotten a lot better at problem solving the past few weeks (i.e., 6's on practice NYSMLs as opposed to 2's; 5's and 6's [out of 6]on practice westchester math league contests). However, I was a wimp last year and decided to only do 12 of the AMC 12, so I got in with a 104.5 and did zilch on AIME. Should I try to work on my AMC skills and do the Art of Problem solving AMC 12 Practice Series, or should I do the AIME series?

Learning to solve AIME problems better will probably improve your AMC score anyway.

It depends on your capabilities. If you are getting under 5s on AIMEs then work on AMC 12... If you are getting over 4s then work on AIME stuff.

You should do AoPS 2. All of it. And do every single AIME/AMC12, ever.

If you do that, the classes will be a waste of time, and you can work through that faster than waiting every week for the next lesson. It also saves you $\$100$.

[quote="Karth"][quote="PenguinIntegral"][quote="OVB"]So you think I can qualify for USAMO doing just practice AIME/AMC 12 and AoPS Volume 2?[/quote]
Yes. Most definitely. What do you think you'd be doing special in WOOT or problem series anyway?[/quote]

Well, I'm not completely sure about problem series, but WOOT is completely a different story. In WOOT, you actually learn how to approach problems (through problems, of course). However, they teach you how to think through certain formulae and show you where to apply what.[/quote]
Also, there's really no harm in trying WOOT. If you decide before the third class that you've changed your mind and don't want to take it anymore, you can drop the class and get a full refund.

[quote="Karth"]Hmm... I had a question about this policy... is this after the third class? As in Building Bridges, Geometry of the Circle, and Induction & Pigeonhole Principle? Or is this Building Bridges A, Building Bridges B, and Geometry of the Circle A?[/quote]
I'm not sure... but you'd be better off asking this question in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=140]Classes Information Forum[/url].  :wink:

[quote="Karth"]Hmm... I had a question about this policy... is this after the third class? As in Building Bridges, Geometry of the Circle, and Induction & Pigeonhole Principle? Or is this Building Bridges A, Building Bridges B, and Geometry of the Circle A?[/quote]

I think Orientation counts as a lesson too

[quote="PenguinIntegral"][quote="OVB"]So you think I can qualify for USAMO doing just practice AIME/AMC 12 and AoPS Volume 2?[/quote]
Yes. Most definitely. What do you think you'd be doing special in WOOT or problem series anyway?[/quote]

Well just doing problems is helpful until you can't solve one of them...then it's kind of pointless unless someone is there to explain how to [i]come up with[/i] the solution, not just solve the problem.

i agree with most comments mentioned above (except somebody talking about W00T... I don't think W00T was part of this thread's purpose :D)

If you are doing [i]okay[/i] on the AIME (around 5 or higher) than you should definitely work on the AIME (unless you're worried about not qualifying; even so, you might think about working AIME problems); by working on AIME, your brain learns more about harder problems, and the mid-range-level problems become easier (So your AMC12 score should also go up).

If you are not quite doing that well on AIME yet, then maybe you should try to do a mix of both. Figure out what any obvious weaknesses you have (for example, if you just skip [i]every[/i] problem that starts out with "A triangle...") and try to cap those holes. Work on AMC12 tests until you feel comfortable working 18 or more problems (just an estimate) and then move on to AIME.

Of course, I shouldn't be talking as I screwed both the AMC12 and AIME last year... and still made it to MOP... dunno how that worked out...

[quote="Mathlete9366"]for example, if you just skip [i]every[/i] problem that starts out with "A triangle..."[/quote]

I'm like that the first year when I really try to study for AIME... :rotfl: (except I also skip every problem with a picture)


[url=http://news.yahoo.com/s/afp/20090528/od_afp/swedeneducationoffbeatiraq_20090528124335]A student in Sweden[/url]

I'm not familiar with Bernoulli's number. Is this a really a huge achievement?

http://news.yahoo.com/s/afp/20090528/wl_mideast_afp/swedeneducationoffbeatreply_20090528174251

So the formula isn't new... and according to http://de.wikipedia.org/wiki/Mohamed_Altoumaimi it's similar to the Worpitzky's formula. Now I don't understand one thing: Worpitzky's formula isn't exactly 'a formula', it's more of an algorithm.

Oh, Sweden.  As if [url=http://plus.maths.org/latestnews/sep-dec03/oxenhielm/]Elin Oxenhielm[/url] wasn't bad enough.

In any case, a single achievement, even if it is real, can always be just a random stroke of luck. If you want to call someone a "genius" or some other word like that, you need some consistent record of many achivements over some reasonable time. The media just cannot wait that long; each reporter needs to be the first to deliver an exciting news, so they often come with premature judgements and announcements, especially when something like "a child prodigy" is concerned. This makes much more harm than good to both the community and the person in question. I only hope that Mohamed and his friends and relatives have enough common sense to just ignore all that media noise around his name.

What "fedja" has said is precisely true. I think, maybe an apt title for this thread should be -- [b] "Another young 'fraudigy'" [/b]  :lol: though, I would be really surprised if he turned out to be a real prodigy. It really sounds pretty absurd (I really can't believe it) how a young person equipped with a limited amount of mathematical tools can solve a problem that has confounded experts for so many years.

[quote="t0rajir0u"]That depends on the nature of the formula.  The [url=http://en.wikipedia.org/wiki/Bernoulli_number]Bernoulli numbers[/url] are rather deep, but any particular formula would have to be evaluated on the basis of its explanatory power or its [url=http://en.wikipedia.org/wiki/Bernoulli_number#Efficient_computation_of_Bernoulli_numbers]complexity[/url].[/quote]
have you looked at his formula yet?

The only picture I could find that includes the formula itself is [url=http://www.thelocal.se/19710/20090528/]here[/url]. This is exactly what is called [url=http://en.wikipedia.org/wiki/Bernoulli_number#Connection_with_the_Worpitzky_number]Worpitzky representation[/url] except $ p$ near the end should be $ (p\plus{}1)$ (I do not know why this error appears on the picture; it certainly results in a different answer, which is incorrect, but it may be just a stupid "misprint"). The Worpitzky formula is far from obvious, of course, but it is rather well-known and I do not see why one should get so excited about its being rediscovered.

It could well be that the picture is just of him in front of something that looks like a math formula. Generally, newspapers do not care much about mathematical accuracy. If the story is some boy-genius, so long as there is something in the background that looks complicated enough to a non-mathematical observer, even if it has nothing to do with what he proved (though this seems to have a bunch to do with what he proved, so it could just have been one of many things he wrote on the board explaining what he proved to someone, and the photographer chose that spot as the most photogenic...)

That said, just as I do not have much faith in photojournalists correctly representing mathematical achievements, I just barely have a little more faith in written journalists correctly representing mathematical achievements. It could be the AP ballooning something not very big to be a big deal. That's not saying this kid isn't incredibly smart, just that his proof may be an incredible achievement for someone his age, with the journalist just jumping to the conclusion that it therefore must be incredible consequential to mathematics as a whole.

But I would have to see the *actual* formula, (and probably ask someone far smarter than me) in order to figure out what is the case.

It seems that professional mathematicians in Sweden are very reluctant to contact news papers. Unfortunately, the only somewhat recent achievement I know of that should have been written about is this: http://www.claymath.org/research_award/Dencker/

The only thing that grabs the average reader's attention is that the kid is Iraqi. I assume this has something to do with why they bothered to publish such a story.

TZF this is random but are you going into pure math or applied?

[b]@ mathmonster [/b]: It would be much better if you had asked [b] TZF [/b] via a PM.

likewise
true, most assume refugees are poor and unresourceful
nationality really has nothing to add to the story


Two particles move along the edges of equilateral triangle $ \triangle ABC$ in the direction
\[ A\rightarrow B\rightarrow C\rightarrow A
\]starting simultaneously and moving at the same speed. One starts at $ A$, and the other starts at the midpoint of $ \overline{BC}$. The midpoint of the line segment joining the two particles traces out a path that encloses a region $ R$. What is the ratio of the area of $ R$ to the area of $ \triangle ABC$?

$ \textbf{(A)}\ \frac {1}{16}\qquad \textbf{(B)}\ \frac {1}{12}\qquad \textbf{(C)}\ \frac {1}{9}\qquad \textbf{(D)}\ \frac {1}{6}\qquad \textbf{(E)}\ \frac {1}{4}$

[hide="Not really a solution"] Draw the perpendicular from $A$ to $BC$.  At the beginning the midpoint of the particles is the midpoint of this perpendicular.  Once they have both moved half a side, the midpoint of the particles is at the midpoint of the perpendicular from $B$.  And so forth symmetrically.

Divide up $\triangle ABC$ into four congruent equilateral triangles, then divide each of those into four more congruent triangles.  The path of the midpoint traces out exactly the boundary of the middle triangle, which has area $\frac{1}{16}$ that of the original.  Hence the answer is $\boxed{ A }$. [/hide]

blah...use vectors to show that the image is a triangle

Could somebody show the method that Altheman is describing with the vectors?

The position vector as a function of time of one point can be written as $ v_1 t\plus{}c_1$. The other is $ v_2 t\plus{}c_2$. Then the midpoint is $ \frac{v_1\plus{}v_2}{2}\cdot t\plus{}\frac{c_1\plus{}c_2}{2}$. It follows that while the velocity (i.e. speed and direction are constant) that the midpoint of those two points also has a constant velocity. Thus in such an interval, the midpoint traces a line segment. The whole path is just the union of 6 line segments.

Each time the particles move one half side the midpoints form a line, because of for example what Altheman said above (you don't need to actually compute anything to see this). Two points form a line and so there are 3 lines which form a triangle that we know must have vertices at the midpoint of the medians of the original triangle. The rest follows from what t0rajir0u said above.

Can someone draw a picture for this since I am having a hard time understanding what the figure looks like.

Thus the answer is $A$, the ratio being $1/16$.

Best regards,
sunken rock

umm this one is rlly easy for a #22
a homothety centered at the centroid and scale factor $\frac{3}{2} \cdot (\frac{2}{3}-\frac{1}{2})=\frac{1}{4}$ gets you the [hide=result]$\boxed{
\textbf{(A)}\ \frac {1}{16}}$[/hide] immediately

I kind of guessed that the locus was an equilateral triangle (after plotting three vertices and noting that a conic section wouldn't work, based on the answer choices), but I was looking at wiki solution #1 for this, and they used the fact that the distance between the two points was linear (at least, it's a square root of a quadratic) in terms of $x$, where $x$ is the distance traversed by the two particles, but I don't understand why that would justify the locus being an equilateral triangle.


inequalities
real analysis
real analysis solved

This is from the county round of the Romanian olympiad, guess what year  :D :

Let b_n be a real sequence s.t. a_n=b_1+b_2+..+b_n is bounded. Let f:[0,1]->R a Riemann integrable function. Prove that

1/n*(\sum {1<=k<=n} f(k/n)*b_k) -> 0 as n->00.

(1/n) \sum f(k/n)*b_k=(1/n)*(a_1*(f(1/n)-f(2/n))+a_2*(f(3/n)-f(2/n))+...+a_(n-1)*(f((n-1)/n)-f(n/n))+a_n*f(n/n)). [b](1)[/b] (I think this is Abel's identity but it is easy anyway)

We can assume a_k>0 for all k. That is because (a_n) is bounded so we can add a b>0 so b_1 such that a_n>0, and if we replace b_1 by b+b_1, we don't change the limit.
So assume 0<m<a_n<M for all n. Then
(1)>(1/n)*m*(f(1/n)-f(2/n)+f(2/n)-...+f((n-1)/n)-f(1)+f(1))=(m/n)*f(1/n).
Because f is Riemann integrable, lim(m/n)f(1/n)=0, so our limit is  \geq 0.
Do the same for M to get our limit  \leq 0 and we are done.

[quote="amfulger (1)>(1/n)*m*(f(1/n)-f(2/n)+f(2/n)-...+f((n-1)/n)-f(1)+f(1))=(m/n)*f(1/n).
Because f is Riemann integrable, lim(m/n)f(1/n)=0, so our limit is  \geq 0.
Do the same for M to get our limit  \leq 0 and we are done.[/quote]

Nobody bothered to see if what I wrote was correct, so nobody noticed the mistake. We can have f(k/n)<f((k+1) so the inequalities are not good.

I can still salvage a solution fortunately. Suppose |ak| \leq M for all k.
|(1)| \leq (1/n)*M*(|f(1/n)-f(2/n)|+|f(2/n)-f(3/n)|+...+|f(n-1/n)-f(n/n)|+|f(1)|.
It is easy to see that after removing the || (write f(a)-f(b) or f(b)-f(a)), in the RightHandSide (RHS), we have M/n*( the difference of two Riemman sums associated to f), so the RHS tends happily to 0  :D , so (1) tends to 0.


Given a triangle $ABC$. Let a circle $\omega$ touch the circumcircle of triangle $ABC$ at the point $A$, intersect the side $AB$ at a point $K$, and intersect the side $BC$. Let $CL$ be a tangent to the circle $\omega$, where the point $L$ lies on $\omega$ and the segment $KL$ intersects the side $BC$ at a point $T$. Show that the segment $BT$ has the same length as the tangent from the point $B$ to the circle $\omega$.

Let $AL$ meet the circumcircle of $\triangle{ABC}$ at $L'$, then obviously $KL||BL'$.
Since $\angle{CTL}=\angle{CBL'}=\angle{L'AC}$ we get that $ATLC$ is cyclic.
Therefore, if $LC$ meets the circumcircle of $\triangle{ABC}$ at $W$, then $\angle{CAL}=\angle{WAL}=\angle{BAT}=\angle{KTB}$.
Thus $BT^2=BK\cdot{BA}$ and the conclusion follows.

More complicated:
Let $\{A,J\}\in\omega\cap AC$; clearly $KJ\parallel BC$. $JAKL$ is cyclic, so $\angle JAL=\angle JKL=\angle CTL\implies CATL$ cyclic and $\angle LCT=\angle LAT\ (\ *\ )$.
If $\{M,N\}\in\omega\cap BC$, order $B-M-N-C$ and $\{A,X\}\in\omega\cap AT$, from $(*)$ we infer $\stackrel{\frown}{MN}-\stackrel{\frown}{NL}=\stackrel{\frown}{LX}$, so $LX\parallel BC\parallel KJ$, thence $\angle LTC=\angle LAC=\angle JAL=\angle KAT$ and $BT$ is tangent to $\odot (KAT)$. Since $AK\equiv AB$ is radical axis of circles $\odot (KAT)$ and $\omega$, we are done.

Best regards,
sunken rock

Join $AT$ and $AL$. Let $\omega\cap AC$ = $M$. Join $KM$.
Since, $\omega$ and circle $ABC$ touch each other,
so, $KM$ is parallel to $BC$.
Thus, $\angle CTL$ = $\angle MKL$ = $\angle MAL$.
∴ $ATLC$ is also cyclic.
So, $\angle ATC$ = $\angle ALC$ = $\angle AKL$ [alternate segment theorem]
∴$BC$ is tangent to circle $AKT$.
By power of a point theorem,
$BT^2$ = $BK\cdot BA$ = $BN^2$.
Hence, $BT$ = $BN$.

Let BS be tangent to $\omega$. BS^2 = BK.BA so we need to prove BT^2 = BK.BA or in fact AKT is tangent to BC at T. ∠AKT = ∠ALC and we want to prove ∠AKT = ∠ATC so we will prove ATLC is cyclic. Let J be random point on Line L which is tangent to $\omega$. ∠ACT = ∠BTJ = ∠KAJ = ∠ALT so ATLC is cyclic as wanted. we're Done.

Let $\omega$ again meet $AC$ at $M.$ Clearly $KM\parallel BC,$ hence $\angle CTL=\angle MKL=\angle CAL$ yields that $ATLC$ is cyclic.
Thus $\angle ATC=\angle ALC=\angle AKL$ and $|BT|^2=|BK|\cdot |AB|,$ so we are done.

Harmonic bundles!
Let $\omega \cap AC=N, \omega \cap BC = D,E$. Also let $U$ be point such a $B$ is midpoint of $UT$ and $KU \cap \omega = V$. From condition we have $KN \parallel BC$.
Note that we want to prove that $(U,T;D,E)=-1$. But $(U,T;D,E)=^K(V,L;D,E)$, so we need to prove that $CV$ is tangent to $\omega$.
But $-1=(U,T;B, \infty_{BC})=^K(V,L;A,N)$, so $VALN$ is harmonic and really $CV$ is tangent to $\omega$. $\blacksquare$


function inequalities calculus integration

SMO 2007, Open #14

Let $N$ be the set of positive integers and let $f: N \rightarrow N$ be such that \[f(m+n) \geq f(m)+f(n)-1\]for any $m, n$ which are elements of $N$ and $f(1)>1$ and $f(8888) < 9000$. Determine the value of $f(2007)$.

if $f(0)=1$
$\forall n\in \mathbb{N}^{*},\ f(n)\ge f(n-1)+f(1)-1\ \Longrightarrow\ f(n)\ge n(f(1)-1)+1\ge n+1\\ we\ take\ g(n)=f(n)-n-1\\ \\ i)\ \forall (n,m)\in \mathbb{N}^{2},\ g(n+m)\ge g(n)+g(m) \\ ii)\ g(8888)<111$
there are infinity of solutions.
if $g$ check i), then $\exists (a,b,h)\in \mathbb{N}\times \mathbb{N}\times \mathbb{N}^{\mathbb{N}}$
such that:
$\forall n\in [|0,a-1|],\ g(n)=0$
$g(a)=b$
$\forall n\in[|a,+\infty[,\ g(n)=Max(\{g(k)+g(n-k)| 0<k<n\})+h(n)$.
for example if we take $a=8888,b=110,$
we have $g(8888)<111$,but $f(2007)=2008+g(2007)$ take a lot of values  who depant of $h$.

[quote="dengmi"]0 isn't a positive integer (natural number).  THus, f(0) is undefined.
Also, all SMO problems have integral solutions.[/quote]
$f(0)=1\ or\ 0$.

I only said myself that aidan can be it A forgets something in the data .if not, where is the fault in what I said

Sorry, I didn't look deep enough.  You're right in saying that there's infinitely many solutions.
x+1 and [102/101x+1] both satisfy the conditions, but x=2008 in equation 1 and 2027 in equation 2 for f(2007).

I didn't know how to do this problem, i thought there wasn't enough information. But i copied the problem verbatim from the problem booklet so i didn't miss any words.


linear algebra
matrix
quadratics
function
linear algebra solved

Let $A=\left(
\begin{array}{cc}
I_{m} & X \\
0 & I_{n} \\
\end{array}
\right)$
$ X \in C^{mxn}, m \geq n$
$ I_{t}$ is a t x t identity matrix.
Express the $\kappa_{2}(A)$ in terms of the singular values of X.

Now, I see that the matrix is reducible. Also I know that the $\kappa_{2}(A)$ is actually $\frac{maximal \hspace{3 mm} singular \hspace{3 mm} value \hspace{3 mm} of \hspace{3 mm} A}{minimal \hspace{3 mm} singular \hspace{3 mm} value \hspace{3 mm} of \hspace{3 mm} A}$

First maybe we can try to find the eigenvalues of  $A^{H}A$ since we know the singular values of A are their roots.
so we get:
$A^{H}A=\left(
\begin{array}{cc}
I_{m} & 0 \\
X^{H} & I_{n} \\
\end{array}
\right)
\left(
\begin{array}{cc}
I_{m} & X \\
0 & I_{n} \\
\end{array}
\right)=
\left(
\begin{array}{cc}
I_{m} & X \\
X^{H} & X^{H}X + I_{n} \\
\end{array}
\right)
$

Any suggestions please?

[quote="jmerry"]$A^*A=\begin{bmatrix}I&0\\X^*&I\end{bmatrix}\cdot \begin{bmatrix}I&X\\0&I\end{bmatrix}= \begin{bmatrix}I&X\\X^*&X^*X+I\end{bmatrix}$

It's really quite easy to multiply matrices in compatible block structures. From here, you just read off what you want to know about the singular values.[/quote]

Sorry.. But I can't see how to extract the information about the singular values of A through $\kappa_2(X)$
And I knew how to multiply these of course...

Right- I'm missing the "unitarily diagonalize" step. Easier said than done- I'll be back later.

All right, here's the eigenvalue (and eigenvector) calculation for $A^*A$:

There are three cases
If $Xv=0$, $\begin{bmatrix}0\\v\end{bmatrix}$ is an eigenvector with eigenvalue 1.
If $X^*v=0$, $\begin{bmatrix}v\\0\end{bmatrix}$ is an eigenvector with eigenvalue 1.
If $X^*Xv=tv$ for some $t\neq0$, $\begin{bmatrix}Xv\\ \lambda v\end{bmatrix}$ is an eigenvector with eigenvalue $\lambda+1$, where $t(\lambda+1)=\lambda^2$.
Let $\lambda+1=x$. Then $tx=(x-1)^2=x^2+1-2x$ and $x=\left(\frac t2+1\right)\pm\sqrt{\frac{t^2}4+t}$.
Both roots are valid, and result in different eigenvalues of $A^*A$. The largest eigenvalue and the smallest eigenvalue are reciprocals, and both correspond to the largest eigenvalue of $X^*X$, which is $||X||_2^2$.

That should do it. I'm not sure exactly what form you want, so I'll leave it here.

Thanx for the eplanation!!
I have several questions though:
The largest eigenvalue and the smallest eigenvalue are reciprocals, and both correspond to the largest eigenvalue of $X^*X$
Why is that?
which is $ ||X||_2^2$ is it?

And also you have to swap the $X$, $X^{H}$ at the beginning... the corresponding eigenvector is just the opposite.

Look more closely at that quadratic for the eigenvalue $x$: the two roots have product 1. Since taking reciprocals reverses order (among positive numbers) and the roots which are not equal to 1 come in reciprocal pairs, the reciprocal of the smallest eigenvalue will be the largest eigenvalue and vice versa.
Also, since $\left(\frac t2+1\right)+\sqrt{\frac{t^2}4+t}$ is clearly an increasing function of $t$, the largest eigenvalue comes from the plus sign and the largest $t$.

Thus the condition number is $\left(\frac {||X||_2^2}2+1\right)+\sqrt{\frac{||X_2||^4}4+||X||_2}$

I'll fix that mistake you noticed. I have everything in the right place where it matters.

But $\kappa(A)$ is defined as the division of the highest singular value of A by the lowest one.
So we have the $\displaystyle \sigma_{max}^2 = \left(\frac {||X||_2^2}2+1\right)+\sqrt{\frac{||X_2||^4}4+||X||_2}$ as the highest eigenvalue of $A^*A$ while the lowest one is:
$\displaystyle \sigma_{min}^2 = \left(\frac {||X||_2^2}2+1\right)-\sqrt{\frac{||X_2||^4}4+||X||_2}$
and hence $\kappa(A) = \frac{\sigma_{max}}{\sigma_{min}}$

Have you tried actually performing that division and rationalizing the denominator?


Duke college Harvard geometry rectangle trigonometry AMC

As many of you (24 exact) waited, here are the problems.

Side note: All rules are posted in the PDF document and all I need is the answers.

Note for Problem 7: That is the problem of sum of many expressions. I got few from other competitions and I give credit to Duke University and Harvard University for this. DO NOT TRY TO FIND THOSE PROBLEMS! This is a honest system so do not please look them up. But problem itself is mainly my work.

Note on Problem 5. The 10 polygons exclude triangle.

Rule for scoring: Top score (6,7) is the qualifying scores. If there is nobody on 6 or 7, it would be 5 and 4 and so on.

Good luck on problem solving

I don't get #1... there aren't any subsets that don't contain $1, 2, 3, \cdots , \text{ or } 22$... Am I misinterpreting the problem or something?

For number 4, $m$ has to be an integer but $n$ doesn't?

#6 has a typo in the third fraction. Also, it doesn't make much sense to me because $M$ is different, depending on whether the number of fractions is even or odd. Infinity is neither even nor odd, correct?

for number 4 i assumed $m$ and $n$ were both positive integers.

I had the same question for number 6 too though

I am not entirely sure i understand #5 though =/

#1 is perfectly worded.

#4, yes $m$ and $n$ are both positive integers.

#6 needs modification. I'll change now and post it as new PDF document.

Answers are due next Monday (extra one day :D )

For number six, must $p,q \in \mathbb{N}$?

Yes, $p,q$ are natural numbers.

But just wondering, I don't think you were one of 24 qualified people, did you? :?

Ok.. There are multiple answer for #6 with wording now.. So, slight change:

After finding $M$, find $a+b+c+d$ if $M$ can be written as $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}$ where $abcd = 33339600$.

#5 makes no sense. The polygons can have any number of sides, and therefore the sums of the angles can change.

[b]EDIT:[/b] I also don't understand the sums in #7. Is the double sum supposed to be the sum as both variables go from 0 to $\infty$, or the sum of $s$ from 0 to $\infty$ of the sum of $t$ from 0 to $\infty$?

[quote="JesusFreak197"]#5 makes no sense. The polygons can have any number of sides, and therefore the sums of the angles can change.[/quote]

I agree. I assumed he meant the smallest possible polygons.

I'm bit confused.

There are only 10 polygons excluding triangle that can be supergon.

The sum of the smallest angles in those 10 polygons mean that take the smallest angle in polygn1, polygon2,polygon3,...polygon 10 and add them up.

Is this answer your question? :?

Isn't it true that $\sum{ab}=\sum{a} \cdot \sum{b}$?

Using the example in the post above yours, it seems to be true.

Well, that doesn't really help me much... this problem is still extremely difficult. I think I'm going to just go for six. :P

I don't know if this has been asked already, but for question 5, can you have the same polygons with different angles?  Like for a triangle with 59,60,61 but also having 58,60,62 as another supergon?

[quote="Press alt f4"]Silverfalcon, I would just like to thank you for the work you did in making these problems and grading and such. I didn't participate in the COP this time but I might next time. So just, thank you.

Oh, Some relations for the sigma notation are
a.$\displaystyle \sum{(a+b)}=\sum{a} + \sum{b}$
b.$\displaystyle \sum{ka}=k\sum{a}$
but I'm not quite sure about your relation. If both a and b are arbitrary variables then how can one sigma define them both?
Maybe you meant
$\displaystyle \displaystyle \sum_{x=a}^{b}\sum_{y=c}^{d}xy=\sum_{x=a}^{b}x\cdot \sum_{y=c}^{d}y$ ?
Where a,b,c, and d are all constants
edit: woohoo 100 posts![/quote]

No, I mean $\sum{ab}=\sum{a}\sum{b}$. If that's not true, then what did Silverfalcon mean by "try expressing the sum as some products" or whatever?

Factoring? There's probably a better way to put it in closed form.

Oh shoot....I managed to procrastinate myself, now I got $1\frac{1}{2}$ days to do all the problems.
 :mad:

YES!!! :idea:  :) I got #6! The answer is...

[hide="Answer!!!..."]yeah right  :P lol[/hide]

Did anyone else get it yet?

Oh, I just saw the change he made to it.

I think I'm just going to submit my answers... if anyone gets #6 or #7, then they either have far too much time, or they deserve to do better than me.

Silverfalcon, thanks for all the work you put into this contest.  :coolspeak: It's really fun  :) I know it's hard to make all these questions! (I once wrote a mock MATHCOUNTS competition in the mc forum, and it was really hard to come up with 38 problems  :wacko: )

Yes, thank you very much.  I didn't get 1 or 5 and 7 is probably wrong, but it was still fun.


[quote="jclarkemathL314159"]Prove that there exist no perfect square with three or more digits comprising of either all ones, all fives, or all nines.[/quote]
[hide]Simply look at the numbers $\bmod{4}$. Any number comprised of either all ones, all fives, or all nines (containing 3 or more digits) will end in either $11$, $55$, or $99$, and thus are always $3 \pmod{4}$. However, perfect squares can only be either $0$ or $1 \pmod{4}$, so it is impossible for these numbers to be perfect squares.
$\text{Q.E.D.}$  :P [/hide]
In fact, the problem would still work even if it had said "2 or more digits" instead.

[quote="towersfreak2006"]Not sure what you mean. $9997^{2}=99940009$.[/quote]
He means
"Prove that no number of the form $\dots 111$, $\dots 555$, or $\dots 999$ can be a perfect square."

[hide="In case anyone wants a hint"]Use $\bmod{4}$.[/hide]

[hide]Modding by 4, all digits before the last two dissapear, since they are all multiples of 100, and thus divisible by 4. Thus the numbers become $11,55,99\equiv 3 \pmod{4}$, but the quadratic residues modulo 4 are 0 and 1.[/hide]

[hide="outline"]Just consider in $modulo 4$.[/hide]

What about [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=107565#p107565]this question[/url] from Belgian (Dutch) MO?  :wink:


Let $A_{0}B_{0}C_{0}$ be a triangle and $P$ a point. Define a new triangle whose vertices $A_{1}B_{1}C_{1}$ as the feet of the perpendiculars from $P$ to $B_{0}C_{0}, C_{0}A_{0}, A_{0}B_{0}$, respectively.
Similarly, define the triangles $A_{2}B_{2}C_{2}$ and $A_{3}B_{3}C_{3}$. Show that $A_{3}B_{3}C_{3}$ is similar to $A_{0}B_{0}C_{0}$.

And if we do the same for a $n$-gon, what do we obtain?

Nobody can solve this?
Think about cyclic quadrilaterals...


inequalities
Ring Theory
ratio
superior algebra
superior algebra unsolved

Don't we have $[K\cdot L : L] \leq [K: L] \leq [K: F]$? (treat this as a question of beginner, any explanation why it is not true are welcome   :) )

I'm a beginner too. Any professionals out there? ;)

Our professionals seem mostly to be analysts...

[quote="Megus"]Don't we have $[K\cdot L : L] \leq [K: L] \leq [K: F]$[/quote]
Not exactly, because $[K: L]$ doesn't make sense. On the other hand, we could use $[K: L']$ where $L'=L\cap K$. This needs some argument.

I don't think those are the right equality cases; if either $L$ or $K$ is a subset of the other, we get inequality trivially.

I meant $K \nsubseteq L$ and $L \nsubseteq K$.

Does anyone have a proof I can understand? :)

I am no professional and certainly no analyst

however

note that $KL$ is also the smallest subring of E containing both K and L

also note that that ring consists of all finite sums of terms of form $kl$ with k in K and l in L

now let $v_{i}$ be a basis of K over F

my claim is that it is also a spanning set (not necessarily a basis, that is where the inequality comes from) of $KL$ over L

keep in mind that you only need to prove that all elements of form $kl$ are linear combinations of those $v_{i}$ with coefficients in L

does this help?

[quote="jmerry"]Not exactly, because $[K: L]$ doesn't make sense. On the other hand, we could use $[K: L']$ where $L'=L\cap K$. This needs some argument.[/quote]

Could you elaborate, jmerry? I'm interested.

The problem has since been edited to fix the equality cases. My post was a musing, not "I have the solution". That idea might work, or it might not.

[quote="Arne"]Let $E$ be a field. Let $F$ be a subfield of $E$, and let $K$ and $L$ be subfields of $E$ containing $F$.
Define $K \cdot L$ as the smallest subfield of $E$ which contains both $K$ and $L$.
Suppose that $[K \cdot L : F]$ is finite.

(a) Prove that $[K \cdot L : L] \leq [K : F]$.
(b) Show that equality occurs if $[K : F] = 2$ and $K \nsubseteq L$.
(c) Show that equality occurs if $[L : F] = 2$ and $L \nsubseteq K$.[/quote]

Let $C = K \cap L$.  We shall prove the stronger statement that $[KL: L] \le [K: C]$ which implies the result as $F \subseteq C$ and $[K: F] = [K: C][C: F]$.
Suppose that $x_1, \ldots, x_d$ is a basis for $K/C$ (finite since $K \le KL$ which is finite over $F$).  I claim that these elements span $KL$ over $L$.  Any element of $KL$ is of the form $\sum l_\iota k_\iota$ where $k_\iota \in K$ and $l_\iota \in L$.  Now each $k_\iota = \sum_{i=1}^n c_i(\iota) x_i$ with the $c_i(\iota) \in C$: but then the $c$ are necessarily in $L$.  So this element of $KL$ is an $L$-linear combination of the $x_i$.

Now suppose $[K: F] = 2$.  Then $[KL: L] = 1$ or $2$.  But if $[KL: L] = 1$ then $KL = L$, which is $K = L$.

Now suppose $[L: F] = 2$.  Then $[L: C] \le 2$.  If $[L: C] = 1$ then $L \subseteq K$.  So $[L: C] = 2$, and so $C = F$.
Now say $L$ generated over $C$ by 1 and $\lambda$.  I claim the $x_i$ which give a basis for $K/C$ are also a basis for $KL/L$: that is, that no $L$-linear combination of the $x_i$ is zero.  If it were, then a non-trivial $C$-linear combination of the $x_i$ and the $\lambda x_i$ is zero.  Collecting terms in $\lambda$ together gives either the coefficient of 1 or of $\lambda$ equal to zero: which is a non-trivial linear rleation in the $x_i$ with coefficients in $C$: or $\lambda$ as a ratio of $C$-linear combinations of the $x_i$: that is, $\lambda$ as an element of $K$.  But $\lambda$ is in $L$, so we have $\lambda \in C$: a contradiction.


[quote="caothujjj"]Let triangle ABC with AB=AC. AH is a altitude of triangle ABC. a circle center O through A, B meet AH at K. From K line a straight line, which parallel with AC and meet AB at L. BK meet CL at E. AE meet (O) at F. Prove that angle AFL = angle BAC[/quote]

  It seems to be hard for everyone!


[quote="chemrock"]What is the major product on performing riemer tiemann reaction to pyrolle.[/quote]
CHO - will be at the 2nd position with respect to the nitrogen of pyrrole

I think the answer will be pyridine as the carbene will insert itself  :D

it's actually chloro Pyridine
and the minor would be Pyrrole 2 aldehyde

[quote="chemrock"]What is the major product on performing riemer tiemann reaction to pyrolle.[/quote]
wekk that could have been stated better....it should have been what happens when pyrrole is treated with alc KOH and $ \ce{CHCl3}$

i think the answer is 3-chloro pyridine.

yes forgot that  it will be $ \beta$ chloro pyridine

[quote="Carcul"]Yes, the answer is 3-chloropyridine. So, how can we formylate pyrrole in position 2?[/quote]

Maybe protect the N with BnBr? And formylate?

You cannot protect the nitrogen, since it is very weakly basic (why?).

you can use houben hoesch(or rather the gattermann modification of this reaction) to formylate pyrrole.

Yes, that's right, because pyrrole is have an electron rich ring. How about my question above?

[quote="Carcul"]Yes, the answer is 3-chloropyridine. So, how can we formylate pyrrole in position 2?[/quote]

Can we use Vilsmeier-Haack ?

bcoz the lone pair on nitrogen is delocalized(its aromatic)

[quote="Carcul"]Yes, Vilsmeier-Haack is even better.

And yes, that's right VARUNARASU.[/quote]

Why is it better?

houben hoesch gives better yields with polyhydroxy phenols(eve better if meta0, it is not generally used for other heterocyclics. vilsmeier haack gives better yield with heterocyclics


$ x,y,z\in R^{\plus{}}$
$ \sum_{x,y,z}\frac{yz}{x (y\plus{}z)^{2}}\geq\frac{9}{4(x\plus{}y\plus{}z)}$

my opinion:
Suppose that
$ x\geq y \geq z$
and$ x\plus{}y\plus{}z\equal{}1$
 :wink: ......

Your inequality is equivalent to
$ \sum_{cyc}\frac {1}{\left(\frac {\sqrt {xy}}{\sqrt {z}} \plus{} \frac {\sqrt {xz}}{\sqrt {y}}\right)^2} \ge \frac {9}{4\sum{\frac {\sqrt {xy}}{\sqrt {z}}.\frac {\sqrt {xz}}{\sqrt {y}}}} \equal{} RHS$

Using  iran 96  :)

[size=167][b]thanks!

beautiful proof![/b][/size]


[color=blue][b]solve this equation: :rotfl:
$ ln(x\minus{}2009)\minus{}lnx\plus{}\frac{1}{x}\equal{}2009$[/b][/color]

[quote="quocbao153"][color=blue][b]solve this equation: :rotfl:
$ ln(x \minus{} 2009) \minus{} lnx \plus{} \frac {1}{x} \equal{} 2009$[/b][/color][/quote]

The function $ f(x)\equal{}ln(x \minus{} 2009) \minus{} lnx \plus{} \frac {1}{x}$ is increasing and

$ limit_{x \to \infty} f(x)\equal{}0$, so no solution.


Three positive reals $x,y,z$ are given so that $xy=\frac{z-x+1}{y}=\frac{z+1}2.$ Prove that one of the numbers is the arithmetic mean of the other two.


Please don't post into the "Solved Problems" section.
And inequalities belong not to Algebra but to inequalities as the name says ;)

Note that your inequality is equivalent with the following:
$\sum_{sym} a^2b \ge \sum_{sym} a^{\frac{5}{3}}b^{\frac{2}{3}}c^{\frac{2}{3}}$
which holds by Muirhead.


Trapezoid $ TRAP$ is a trapezoid inscribed in $ \odot O$ such that $ T$, $ O$, and $ R$ are collinear, $ TR\equal{}8$, and $ \angle PTR\equal{}75$ degrees. The area of trapezoid TRAP, in simplest radical form, can be written in the form $ k\plus{}w\sqrt{p}$. Find the value of $ k\plus{}w\plus{}p$.

[geogebra]9ccbaffe45fdeacbcfc86f10a2f2111b37b78db0[/geogebra]
[hide="solution"]
Since $ TOR$ are collinear, they lie on the diameter of the cirlce and $ TRAP$ is perfectly symetrical.
So $ TP \equal{} 8 \cos(75) \equal{} 8\cos (30\plus{}45) \equal{} 8\left(\frac{\sqrt{6}\minus{}\sqrt{2}}{4}\right)\equal{}2(\sqrt{6}\minus{}\sqrt{2})$
$ AP \equal{} 8\sin(30)\cos(45)\plus{}\cos(30)\sin(45) \equal{} 2(\sqrt{6}\plus{}\sqrt{2})$

By Ptolemy
$ 8\cdot AP \plus{} 4(\sqrt{6}\minus{}\sqrt{2})^2 \equal{} 4(\sqrt{6}\plus{}\sqrt{2})^2$
$ \cdot AP \equal{} 2\sqrt{3}$

$ h\equal{} 2(\sqrt{6}\minus{}\sqrt{2})\cos(75) \equal{} \frac{1}{2}\cdot (\sqrt{6}\minus{}\sqrt{2})^2 \equal{} 4\minus{}2\sqrt{3}$

$ [TRAP] \equal{} \frac{1}{2}\cdot (4\minus{}2\sqrt{3})\cdot 8 \cdot 2\sqrt{3} \equal{} \boxed{16(2\sqrt{3}\minus{}3)}$[/hide]

I did a lot of trig, and I got
\[ 4r^2 \sin^3 \theta \cdot \cos \theta.\]
Is that right?

When I evaluate my formula, I get something around 14.

[hide="Solution"]
So unless I'm being stupid again, the area is just $ \frac12\cdot2\cdot(8 \plus{} 4\sqrt3) \equal{} 8 \plus{} 4\sqrt3\implies8 \plus{} 4 \plus{} 3 \equal{} \boxed{15}$.
[/hide]

[b]123456789:[/b] Your formula is correct.


A vertical line divides a triangle with vertices (0, 0) , (1, 1) , and (9, 1) into two regions of equal area.  What is the equation of this line?

As the line that divides the triangle is vertical line,
so assume that the equation of that line is as follow:

x - a = 0
==> x = a

Asuume triangle is ABC
A=(0,0)
B=(1,1)
C=(9,1)

BC = 8

Area of the triangle : (1/2) . 8 . 1 = 4

Suppose line (x - a = 0) intersects BC at D and AC at E.

then CD = 9 - a

ED = (1/9) . (CD) = (1/9).(9 - a)

Area of triangle CDE = (1/2).(CD).(ED) = (1/2).(9 - a).(1/9).(9 - a)

==> (1/2).(9 - a).(1/9).(9 - a) = 4/2 = 2
==> (9 - a).(9 - a) = 36
==> 9 - a = 6
==> a = 3
So equation of the line is x - 3 = 0

Thanks,
http://www.teacherone.com


Does anyone know the formula for the fibonnacci sequence? in algebratic terms?
like 1, 2, 3, 5, 8, 13, ....

$F(n) = \frac{a^n - b^n}{\sqrt{5}}$

where $a = \frac{1+\sqrt{5}}{2}$ and $b = \frac{1-\sqrt{5}}{2}$.

Hope that helps.

isn't a term of the fibonnacci sequence just the sum of the two terms preceeding it?

chess64 gave you a theorem known as "Binet's Formula"

Another one you can use is the following:

$F_n = \displaystyle\binom{n}{0} + \displaystyle\binom{n-1}{1} + \displaystyle\binom{n-2}{2}  +...$

The sequence above will continue until $n-k$ is less than $k$.

Proofs of these can be found in section 16-3 of AoPS Volume II, but this is much beyond MathCounts, so it's best to remember the pattern as $F_0 = F_1 = 0, F_{n+1} = F_n + F_{n-1}$


Is there a command that makes the whole document double spaced?

[quote="Fierytycoon"]
Use:

\linespread{2}

Or use the setspace package and the \doublespacing command.
[/quote]


The letters $A$ or $B$ are assigned on the points divided equally into $2^{n}\ (n=1,\ 2,\cdots)$ parts of a circumference.If you choose $n$ letters from any succesively arranging points directed clockwise, prove that there exists the way of assignning for which the line of letters are mutually distinct.

Draw a directed graph whose vertices are the sequences of length $ n\minus{}1$,with an edge between two sequences if the last $ n\minus{}2$ letters of the first vertex match the first $ n\minus{}2$ letters of the second.
This graph has two edges into and out of each vertex,so there exists a (directed) path traversing each edge exactly once.We convert this into a cycle of the desired form by starting at any vertex,writting down the sequence corresponding to it,then appendinginturn thelast letter of each sequence we encounter along the path.


I search for *\frac{a^2+bc}{b+c}*, then the search engine gave me a lot of irrelevant links.... How can I search normally? Thanks for your help.

Yes, there is a problem with the search part... I tried $x^y=y^x$, which was post like 10 times, and it appeared no results.... are the admins trying to repear it?

Yes. Read the supplied links :)

Have patience, fixes are on the way, but we are only human.


function
complex analysis
complex analysis unsolved

Show that an even function f(z) can have only even powers in its Taylor expansion about z = 0, and similarly for odd functions.  (Use f(z) - f(-z) = 0 and uniqueness)

Thank you.

So $ f(z)\minus{}f(\minus{}z)$ has a taylor expansion. And this is zero identically, so if

$ f(z)\equal{}\sum a_n x^n$, we get

$ (a_0\plus{}a_1 x\plus{}a_2 x^2\plus{}...)\minus{}(a_0\minus{}a_1 x\plus{}a_2 x^2\plus{}...)\equal{}0$

Collecting terms on the left, we get

$ 2a_1x\plus{}2 a_3 x^3\plus{}...\equal{}0\equal{}0\plus{}0x\plus{}0x^2\plus{}...$

so $ a_1\equal{}a_3\equal{}...\equal{}0$

We can equate coefficients by uniqueness.


geometry
perpendicular bisector
cyclic quadrilateral

Two points $ A$ and $ C$ are chosen on a circle $ \Gamma$. Let $ O$ be an arbitrary point on the perpendicular bisector of $ AC$. The perpendicular bisectors of $ OA$ and $ OC$ meet $ \Gamma$ at points $ B$ and $ D$ respectively such that $ A,B,C,D$ lie on $ \Gamma$ in this order. $ AB$ and $ CD$ when extended meet at $ P$. Show that the segment adjoining the midpoints of $ AB$ and $ CD$ is perpendicular to $ OP$.

$ \textbf{Lemma 1: }$ If $ ABCD$ is a cyclic quadrilateral and $ BA\cap CD = P$ then,
\[ PB^2 - AB\cdot PB = PC^2 - PC\cdot DC
\]
$ \textbf{Proof: }$ From power of point,
\begin{align*} & PA \cdot PB = PD \cdot PC \\
\iff & PB(PB - AB) = PC(PC - DC) \\
\iff & PB^2 - AB\cdot PB = PC^2 - PC\cdot DC \end{align*}

$ \textbf{Lemma 2}:$ If $ AB \perp CD$ then, $ AC^2 - CB^2 = AD^2 - DB^2$
It is quite wellknown.
Now, we need to prove that,
\[ PM^2 - MO^2 = PN^2 - ON^2 \iff PM^2 - PN^2 = OM^2 - ON^2
\]
Here,
\[ PM^2 = \left( PB - \frac {AB}{2}\right)^2 = PB^2 + \frac {AB^2}{4} - AB \cdot PB
\]
And,
\[ PN^2 = PC^2 + \frac {DC^2}{4} - DC.PC
\]
So,
\[ PM^2 - PN^2 = \frac {1}{4}(AB^2 - CD^2)
\]
Agian, from the theorem of Apollonius,
\[ OM^2 = \frac {1}{4}(2AO^2 + 2OB^2 - AB^2)
\\
ON^2 = \frac {1}{4}(2OD^2 + 2OC^2 - DC^2)
\]
Subtracting we get,
\[ OM^2 - ON^2 = \frac {1}{2}(AO^2 - OC^2 + OB^2 - OD^2) - \frac {1}{4}(AB^2 - DC^2)
\]
As, $ AO = OC$, it is left to prove that, $ OB^2 - OD^2 = AB^2 - DC^2$
Which follows from, $ AB = OB$ and $ OD = DC$

EDIT: Eh...Making a lot of typos  :mad:

[quote="nayel"] [color=darkred]Two points $ A$ and $ C$ are chosen on a circle $ w$ . Let $ O$ be an arbitrary point on the perpendicular bisector of the segment $ [AC]$ .

The perpendicular bisectors of the segments $ [OA]$ , $ [OC]$ meet the circle $ w$ at $ B$ , $ D$ respectively such that $ A,B,C,D$ lie on the circle $ w$ in this order.

Denote the intersection $ P\in AB\cap CD$ and the midpoints $ M$ , $ N$ of the segments $ [AB]$ and $ [CD]$ respectively. Prove that $ OP\perp MN$ .[/color] [/quote]

[color=darkblue][b][u]Proof (metrical).[/u][/b]

Observe that $ \left\|\begin{array}{c}
OA\equal{}OC\ \ ;\ \ OB\equal{}AB\ \ ;\ \ OD\equal{}CD\\\\
p_w(P)\equal{}PA\cdot PB\equal{}PC\cdot PD\\\\
2\cdot PM\equal{}PA\plus{}PB\ \ ;\ \ 2\cdot PN\equal{}PC\plus{}PD\end{array}\right\|$ . I"ll show (similarly to [b]Moonmathpi496[/b]) that $ PM^2\minus{}PN^2\equal{}OM^2\minus{}ON^2$ .

$ \blacktriangleright\ 4\cdot\left(PM^2\minus{}PN^2\right)\equal{}(PA\plus{}PB)^2\minus{}(PC\plus{}PD)^2\equal{}$ $ PA^2\plus{}PB^2\minus{}PC^2\minus{}PD^2\equal{}(PA\minus{}PB)^2\minus{}(PC\minus{}PD)^2\equal{}AB^2\minus{}CD^2$ .

$ \blacktriangleright\ 4\cdot\left(OM^2\minus{}ON^2\right)\equal{}$ $ \left[2\left(OA^2\plus{}OB^2\right)\minus{}AB^2\right]\minus{}\left[2\left(OC^2\plus{}OD^2\right)\minus{}CD^2\right]\equal{}$ $ \left(2\cdot OB^2\minus{}AB^2\right)\minus{}\left(2\cdot OD^2\minus{}CD^2\right)\equal{}AB^2\minus{}CD^2$ .

In conclusion, $ PM^2\minus{}PN^2\equal{}OM^2\minus{}ON^2\equal{}\frac 14\cdot\left(AB^2\minus{}CD^2\right)$ , i.e. $ OP\perp MN$ .

[b]Remark.[/b] I denoted the [b]power[/b] $ p_w(P)$ [b]of the point[/b] $ P$ [b]w.r.t. the circle[/b] $ w$ and I used the [b]theorem of median[/b] in $ \triangle ABC\ : \ 4m_a^2\equal{}2\left(b^2\plus{}c^2\right)\minus{}a^2$ . [/color]

Nice proof Mr. Nicula, indeed. :)
But if you just look at [url=http://www.mathlinks.ro/viewtopic.php?t=213117]the solution of yetti[/url] you are going to laugh or cry(like me) :wink:


Find the value of $ K$ for which these two equations will not have a common solution:
\[ 6x \plus{} 4y \equal{} 7\]
\[ Kx \plus{} 8y \equal{} 7\]

hello, multiplying the first equation by $ \minus{}2$ and adding the second equation we get the equation $ x(k\minus{}12)\equal{}\minus{}7$.
So we have for $ k\equal{}12$ a contradiction and it follows there exist no intersection point.
Sonnhard.


Given that $ a, b,$ and $ c$ are the roots of $ x^3\minus{}2x^2\plus{}3x\plus{}4$, find $ (a\plus{}b)(a\plus{}c)(b\plus{}c)$.

Since a+b+c=2 from Vietas theorem, rewrite it as (2-a)(2-b)(2-c).

Then you can use Vietas theorem once more when that term is expanded as you will get like an a+b+c thing, an ab+ac+bc thing, and abc thing, Vietas covers all three.

Even easier, $ (2\minus{}a)(2\minus{}b)(2\minus{}c)$ is just $ P(2) \equal{} 10$.


topology
induction
superior algebra
superior algebra theorems

OK. Maybe this is because I'm not into college algebra yet... but does (the infinite form of ) the axiom of choice imply a well-ordering on the reals? Is it possible to find that well-ordering? and... if not, is it possible to create a new axiom to allow that? :maybe:

(and can we induct on the reals? :D )

The axiom of choice is equivalent to the statement that any set can be well-ordered. On the other hand, the rule of thumb about anything for which you invoked the axiom of choice or any equivalent just to claim it exists is something you're never going to be able to visualize in any concrete way.

As for your last question, try searching for the words "Transfinite Induction." (Also, Zorn's Lemma and the Hausdorff Maximal Principle.)

by the way, a well-ordering on IR implies the existence of a IQ-basis of IR :-)

1. Yes, the axiom of choice implies a well-ordering on the reals. It also implies a well-ordering on any non-empty set. See [url=http://en.wikipedia.org/wiki/Well-ordering_theorem]here[/url].

2. Yes, there is induction, sort of. However, it's a slightly different form called Transfinite Induction that involves ordinals. See [url=http://en.wikipedia.org/wiki/Transfinite_induction]here[/url]

3. I don't think you need to say "infinite form" of the axiom of choice. The 'finite' form is in fact an obvious theorem of standard ZF axioms. The axiom of choice refers to an extension of ZF, and to be an extension it must to be the infinite form. So the distinction probably isn't necessary.

4. One of the reasons for the axiom of choice being controversial is that it allows you state existence of things that can't be, or are really hard to, construct. The canonical example is using the Axiom of Choice to choose an element from each class of $ \mathbb{R}/ \mathbb{Q}$, which I believe is a partition of the reals where for every two sets, there do not exist elements from each with difference being a rational number, and every two elements in one set have their difference rational. (maybe not though, I don't know much about quotients) Anyways, if you try to actually construct that list of elements, you'll find yourself baffled quickly. The Axiom of Choice says you can do this anyways.

The well-ordering of the reals I believe is one of the things that can't be concisely described. Another example of something that is rather difficult to construct is the four pieces of a ball that allow the Banach-Tarski paradox.

EDIT: Der, I got beaten.


I don't get the difference between them. If anybody could help, that would be awesome.


Rules: You have an egg. You choose someone to throw your egg at. Whoever's egg has a "bomb" in it kills the other person. We play until there is only one person left.
I will be taking 14 people to start.
Sign up please, guys!
1. spaceguy524
2. meewhee009
3. eqjj168
4. 15!!!
5. Scamper
6. EggyLv.999
7. Yoshi
8. westiepaw
9. cf249
10. andrewjjiang97
11. lotsofmath
12. bluecarneal
13. Richard_Min
14.

I will join. :)

And don't make the eggs thrown at me the bomb just because I beat you in the previous Egg Toss.

in.

The message is too small. Please make the message longer before submitting.

join.  I hope I don't die early like last time.

In.  Maybe one of these days, I'll win one of these games.

I finally get the rules!

1. spaceguy524
2. meewhee009
3. eqjj168
4. 15!!!
5. Scamper
6.
7.
8.
9.
10.
11.
12.
13.
14.

How do you know if an egg has a bomb?

1. spaceguy524
2. meewhee009
3. eqjj168
4. 15!!!
5. Scamper
6. EggyLv.999
7. Yoshi
8. westiepaw
9. cf249
10. andrewjjiang97
11. lotsofmath
12. bluecarneal
13.
14.

Edit: 100th post!!!  :lol:   :clap2:

All right, he's inactive.
Jjfun1's voice comes over the speaker. "Err... looks like spaceguy is in a coma. Everyone must throw their egg at him. Thank you." The speaker fell silent with an earsplitting squeak. Meewhee shrugged. She held not an egg in her hand, but a smelly sock. BAM! It hit spaceguy. Nothing happened except for the fact that he seemed to wrinkle his nose in disgust while still in his coma. Lotsofmath smashed is egg on spaceguy's head. Spaceguy snored. Yoshi threw a fail egg at lotsofmath, forgetting jjfun1's instructions. The egg seemed to stop in mid-air. It zoomed towards spaceguy and smashed into his face. He exploded.

[b]Deadline: Sunday[/b]
[hide="Player List"]
1.meewhee009
2. Yoshi
3. westiepaw
4. lotsofmath[/hide]

Throw a Tyrannosaurus Rex Egg at westiepaw

[b]Throw a YOSHI egg at lotsofmath[/b].

Yay, I haven't died yet!

[b]Throw a RED Yoshi egg at lotsofmath[/b]

[b]Those who still need to throw: only meeewhee.[/b]

You guys have time zone advantage :o
[b]Throws :rotfl: smiley @ lotsofmath[/b]

Awesome!
Jjfun1 smiled. lotsofmath swung his arm around and threw a Tyrannosaurus Rex Egg at westiepaw. Westiepaw threw a YOSHI egg at lotsofmath. Yoshi threw a RED Yoshi egg at lotsofmath. meewhee threw a  :rotfl: smiley egg at lotsofmath. Everyone looked around, confused. Meewhee stammered, "Ummm. Why is everyone still alive?" Jjfun1 shrugged. "I don't know," she replied, "I guess this poison isn't what it said on the bottle." Suddenly, westiepaw fell to the ground. "I don't feel so good..." she whispered. Rawr.

[b]Deadline: Sunday Night[/b]
Note: Because there are less players, I'm going to make the deadline shorter. :P

[hide="Player List"]

1.meewhee009
2. Yoshi
3. lotsofmath[/hide]

OMG epic luck for two straight rounds!

Throw a  :bomb: at meewhee009

Nice activity guys. :)
Meewhee threw a  :ninja: egg at lotsofmath.
Lotsofmath hurled a  :bomb: egg at meewhee.
Yoshi swung his arm and flung a  :sleeping: egg at lotsofmath.

Suddenly, jjfun1's voice came on.
"Ermm....it looks like there's a glitch with the eggs...they don't seem to be working!" True to her words, the egg that hit lotsofmath broke open, and a bomb with a timer, frozen, was visible. "Err...it seems as if everyone needs to throw another egg at the bomb...to activate it I guess." She flipped through the EGG TOSS GAME HANDBOOK! "Heh. Yeah. So. Everyone throw another egg at the bomb."

The players whispered among one another. One by one, sparkly polka-dotted eggs appeared in their hands.
Meewhee squeezed her eyes shut, tight, and dropped her egg on the bomb.
She opened one eye, then sighed in relief.
Lotsofmath snickered, and smashed his egg on top of the bomb.
BOOM!
An invisible force field had appeared around Meewhee and Yoshi, but not lotsofmath.
He disappeared with a loud [i]pop![/i]

[b]Ok, it's the last round, and this is how it'll work. Meewhee and Yoshi will take turns guessing a number from 1-10 inclusive. When one of them guesses the correct number, they win! Good luck guys, and great game to everyone else who played. Thanks for playing. :) Now....GO![/b]

[color=white]Good luck to both of you![/color]

Wow.
I'm surprised!
[b]Meewhee guessed the number on her very first try. :O[/b]
@Yoshi, nice guess, that's my favorite number. :)
Great game everyone. Thanks for playing!!!!


Coupons in cereal boxes are numbered 1 to 5, and a set of one of each is required for a prize. With one coupon per box, how many boxes on the average are required to make a complete set?

Solution:
[hide]
11.42

if the probability of an event occuring is 1/n, it will take n times before, chances are, u have what u want.
So probabilities for getting coupons:

1 (1st coupon)
4/5 (2nd coupon)
3/5 (3rd coupon)
2/5 (4th coupon)

1/5 (5th coupon)

so each of these can be put into 1/n for by making them each 1/(reciprical).
so, its then 1+5/4+5/3+5/2+5 = 11.42 [/hide]


Find the area of a segment of a circle within a $60^\circ$ sector, with a radius of 9.

[hide]$60^{\circ}$ is $\frac{1}{6}$ of the circle, so the area is $\frac{81\pi}{6}=13.5\pi$.[/hide]

I think he means the circular segment...
[hide]The area of the sector is $\frac{81\pi}{6}$. The area of the triangle is $\frac{81\sqrt{3}}{4}$. Just subtract the area of the triangle from the sector.[/hide]

[hide]$60^\circ$ is $\frac16$ of the entire circle, so the area is $\frac{9^{2}\pi}6=\frac{27\pi}2$[/hide]

Oh I see, I never heard it referred to as the circular segment before.


function
integration
calculus
derivative
real analysis
real analysis unsolved

Let $ f$ be a measurable function of $ {\mathbb R}^n$ to $ {\mathbb C}$ such that $ \sup_{x,y\in{\mathbb R}^n}\frac{|f(x)\minus{}f(y)|}{1\plus{}|x\minus{}y|}<\plus{}\infty$
Prove that $ f$ is sum of a bounded measurable function and of a continously differentiable function with bounded first order derivatives.

Nice to see your proof.  :)

Convolving $ f$ with a positive function $ \phi\in C_c^{\infty}(\mathbb R^n)$ such that $ \int \phi(x)\,dx \equal{} 1$, we get a $ C^{\infty}$ function $ g \equal{} f*\phi$. Clearly,
\[ |g(x) \minus{} f(x)| \equal{} \left|\int (f(y) \minus{} f(x))\phi(x \minus{} y)\,dy\right|\le \int C(|y \minus{} x| \plus{} 1)\phi(x \minus{} y)\,dy \equal{} : C_2
\]
Also, the derivative of $ k$th order $ D^kg(x)$ can be written as $ (\minus{}1)^k\int (f(y) \minus{}f(x))D^k\phi(x\minus{}y)dy$, which is shown to be uniformly bounded as above.

I also obtained a same proof as your. Good work mlok :)

I may be a bit jaded by now, but somehow I am more likely to call this "a standard mollification result" than something "extremely beautiful". It's that damned mechanization of math, again.


calculus integration algebra function domain logarithms limit

I had a bunch of questions like this in my homework and I just don't get the difference. Can someone just help me out with this concept with an example question?

Compute the value of the following improper integral if it converges. If it diverges, enter INF if it diverges to infinity, MINF if it diverges to minus infinity, or DIV otherwise (hint: integrate by parts).

{Integral from 1 to infinity} (9lnx)/(x^6) dx

Determine whether
{Summation from 1 to infinity}(9ln(n))/(n^6)
is a convergent series. Enter C if the series is convergent, or D if it is divergent.

Are they always the same? If one is convergent are they both and vice versa? Thanks =)

[quote="Beachic109"]Are they always the same? If one is convergent are they both and vice versa? Thanks =)[/quote]

This is the [url=http://en.wikipedia.org/wiki/Integral_test_for_convergence]integral test[/url].  You should be able to find it in your textbook.

To determine if an improper integral converges (or its value), replace the problem point(s) in the domain of integration with variables and take limits, e.g.

\[ \int_{1}^{\infty}\frac{9\ln x}{x^6}dx \equal{} \lim_{L\to\infty}\int_{1}^{L}\frac{9\ln x}{x^6}dx\]

now compute the integral as normal: use integration by parts with $ u\equal{}\ln x$ to get

\[ \equal{}\lim_{L\to\infty} \left[ \minus{}\frac{9\ln x }{5x^5}\right]_{1}^{L}\plus{}\int_{1}^{L}\frac{9}{5x^6}dx \equal{} \lim_{L\to\infty} \frac{9}{25}\left[ \frac{5\ln x\plus{}1}{x^5}\right]_{L}^{1}\]
\[ \equal{} \lim_{L\to\infty} \frac{9}{25}\left(1\minus{} \frac{5\ln L\plus{}1}{L^5}\right)\]

Now if the above limit converges, then the improper integral converges and has the same value as this limit.

There is a theorem that states that if $ a_n$ is monotonically decreasing then the series $ \sum_{n\equal{}1}^{\infty}a_n$ and the integral $ \int_{1}^{\infty}a_n\, dn$ converge or diverge together.


hi everyone~

I am Korean...

Maybe someone know the problem which proposed last 2005 in Germany TST.

I try to solve it using a nesbitt's inequality but I failed.

Problem is

1+a/1-a + 1+b/1-b + 1+c/1-c =<2(b/a+c/b+a/c)

Oops! The condition is a+b+c=1, a,b,c>=0

http://www.mathlinks.ro/Forum/viewtopic.php?t=25780

  darij


group theory
abstract algebra
superior algebra
superior algebra solved

any compact subgroup $G$ of $GLn(R)$ is of the following form:
$G=M K M^{-1}$ where:
$K$ is a subgroup of $O(n)$
$M$ an invertible matrices.

Alekk, do you happen to know any normal (meant for mortal human beings) solution for this problem? I asked my teacher and he says it can be done using the Haar measure or something like this, but these tools I imagine cannot be used in a kholle or exam. And anyway, I undersood nothing from his solution.

It is a friend who got this exercice during an oral examination. He told me that he didn't manage to solve it(of course). So I ask on another forum if someone could  solve it with elementary results. It turns out that the usual solution uses Haar stuff you just mentioned but someone gave me a (long) proof with elementary mathematics. I will post a pdf file with the sln.

here is a proof(in fact it is a ps-file):


who can give the solution? please!! :)  :)
thanks for all!!

thanks to you!!
let n be a positive integer, prove that :exist &>0 so that with n positive real x1,x2,...xn arbitrary always exist t>0 so that
&<{tx1},{tx2}...{txn} < 1/2


Let a,b,c be real positive numbers.

Prove that

  \sqrt (a + 1/b) +  \sqrt (1/ab + 1/a) + \sqrt (b + ab)  \geq 3 \sqrt 2

this is a quite straight forward AM-GM application:

first (a+1)(b+1)(ab+1) \geq 8ab (AM-GM for each of the pairs (a,1), (b,1), (ab,1) and multiply) then

\sqrt ( (ab+1)(a+1)(b+1) / ab) \geq 2\sqrt 2 and extracting the cubic root we get

3 ( \sqrt ( (ab+1)(a+1)(b+1) / ab)  )1/3 \geq 3\sqrt 2 and the LHS is exactly the GM of the three terms in LHS of the initial inequality.


The Selective Examination (1) for the National Team of China for IMO 2005
2005年中国数学奥林匹克国家集训队选拨考试（二）

还有几次测试题,费老师能否一并提供  ?     多谢 !


[quote="N.T.TUAN"]For a given integer $d$, let us deﬁne $S = \{m^{2}+dn^{2}| m, n \in\mathbb{Z}\}$. Suppose that $p, q$ are two elements of $S$ , where $p$ is prime and $p | q$. Prove that $r = q/p$ also belongs to $S$ .[/quote]
Let $p=m_{0}^{2}+dn_{0}^{2}, k=\frac{m_{0}}{n_{0}}(mod \ p)=\sqrt{-d}(mod \ p).$
Let $q=pr=m^{2}+dn^{2}$. If $p|m, \ p|n$, then $r=(\frac{mm_{0}-nn_{0}d}{p})^{2}+d(\frac{nm_{0}+n_{0}m}{p})^{2}$.
Else $m=\pm nk(mod \ p)$ and $r=(\frac{mm_{0}\pm nn_{0}d}{p})^{2}+d(\frac{nm_{0}\mp n_{0}m}{p})^{2}.$

following link is easy than what Rust posted http://www.mathlinks.ro/Forum/viewtopic.php?t=66248 but same  :wink: . Sorry, because I don't remember it.


ok the title says it all. i'll start off and the next poster edits it to be funny and so on and so on

"I learned how to count!"

[quote="life=tennis+math"]"I learned female anatomy!!!"
[hide]im a guy by the way[/hide][/quote]

I doubt there's many girls who'd say that anyways.

oh man i learned a lot today
don't remember what though

I learned that math isn't overrated; neither is flushing the toilet in a public restroom.

I learned that nobody learns anything of value in school!

i learned how to make a bomb using only a can in macgyver movie

i learned how to supress the need of going to the bathroom

I learned it's hard to kidnap fat people.  jk.

i learned that fob teachers can be born in brazil while being chinese

[quote="dragongrl6"]well...um, i learned that i suck at basketball?

well in PE, i jammed my finger and now it's painful and swollen.  :stretcher:

i got mad skillz :roll:[/quote]

lol

i learned at school that bowling is a sissy sport (we do it in pe so boring)  :showoff:  :sleep2:  :weightlift:

i learned that substitute teachers are fun to annoy

I learned that when someone is in possession of my Rubik's cube, he/she wouldn't put it in lost and found. He/she would take it and not give it back.


i am trying to submit my solutions but it says the dealine was march 9th even tho today is marh 9th what do i do !!!!!!!!!!!!!!

omg i didnt see that i submitted my last three by mail its 9 o clock and i cant do anything now ><

you could have still submitted it by mail

:o  [quote="peregrinefalcon88"]i just said its 9 o clock i cant submit by mail anymore ><[/quote]
omg what are u going to do  :o
get to one quick.
USAMTS staff: please accept his solns!!! he has proof he finished in time

[quote="mathemonster"]USAMTS staff: please accept his solns!!! he has proof he finished in time[/quote]
How does he have proof?  He posted this topic 4 hours and 48 minutes after the due time.
And even if he did post it before, there is no proof that he finished in time. This is the internet, so you can lie (I don't mean this sarcastically).  It also doesn't even matter if he finished on time provided that he didn't turn in his solutions.  If you don't turn in your solutions on time, then they won't be accepted (except possibly under extraordinary circumstances, such as the situation with Round 2 solutions being lost, but even in this scenario, only the people that turned in their solutions on time were able to resend).

tough luck, read carefully, meh, i'm not being very helpful.

[quote="kcn2rivers"]From the problems sheet: "Web deadline: 3 PM Eastern / Noon Pacific on the due date"[/quote]This was announced in at least four places: on the problem sheet, on the USAMTS website, by email, and on this forum; and on multiple occasions.


suppose A and B are two angles such that
sin A + sin B = 1
cos A + cos B = 0
find the value of 12 cos 2A + 4 cos 2B.

Looks like your computer accidentally double-posted.

Umm...that's the computer problem if his computer accidentally double-posted...Anyhow, do not post a same question twice :D

$\cos A =-\cos B \rightarrow \sin A = \pm \sin B$

If $\sin A =-\sin B$ then $\sin A+\sin B = 0$, so $\sin A = \sin B = 1/2 \rightarrow A=30^{o}$, $B=150^{o}$ or vice-versa.  Either way, $\cos2A = \cos2B = \frac{1}{2}\Rightarrow 12\cos2A+4\cos2B = 8$


geometry
3D geometry
sphere
calculus
derivative
calculus computations

A sphere shrinks at a rate of 1 cm^3 per second. At what rate does the diameter decrease when the diameter is 10 cm? I know it has something to with optimization, and the use of the area formula. The problem is that I do not know the diameter for which it started with and it gave too little info on this problem(at least i think).

You are given diameter (which I'll call $ q$) and rate of change of volume or $ \frac{dV}{dt}$. You should look for a formula that has these relations in them.

$ V=\frac{4}{3}\pi r^{3}$, but you want diameter so: $ V=\frac{4}{3}\pi \left(\frac{q}{2}\right)^{3}=\frac{1}{6}\pi q^{3}$. Now take the derivative to get $ \frac{dV}{dt}=\frac{1}{2}\pi q^{2}\frac{dq}{dt}$ (which is why I used $ q$ to avoid the infamous $ \frac{dd}{dt}$). Your problem asks for $ \frac{dq}{dt}$ at the time of $ q=10$ and $ \frac{dV}{dt}=-1$, so plug in and solve from here.


Prove that there are infinitely many positive integers $ a$ such that the sequence $ (z_{n})$ $ n \ge 1$ ,
$ z_{n} \equal{} n^{4} \plus{} a$ ,does not contain a prime number.

Let a = 4 into n power 4.
I FORGOT THIS IN RMO 2009. GRRR...


MATHCOUNTS
FTW
geometry
3D geometry
LaTeX

Hello,

This is a mock mathcounts sprint round.

Please see the questions [url=http://mersenne.zapto.org/Submission/Sprint.pdf]here[/url].

Please submit your answers [url=http://mersenne.zapto.org/Submission]here[/url]. Do not PM me. PM'ed answers will not count.

Do not disclose any answers via any form of communication.

I will post the target round on Wednesday. I will post all scores and correct answers on Sunday.

I have also attached the questions.

-#H34N1

[b]Clarifications[/b]
[b]#3:[/b] You do not have to express it in simplest radical form.
[b]#6:[/b] Integer Lengths
[b]#9:[/b] $ a$, $ b$, and $ c$ are positive, real numbers. Find the least positive value.
[b]#11[/b] Express your answer in terms of $ n$.
[b]#12.[/b] Express your answer in simplest radical form.
[b]#13[/b] Express your answer in common fraction form.
[b]#19[/b] Either decimal or fraction is permissible.
[b]#20[/b] You need the distance that hte ball travelled.
[b]#29:[/b] There are in reality 99 cards, not 50 cards as mentioned in the contest.

[i]Authentication Issue: If you are misauthenitcated by the Submission form, please wait about 4-5 minutes. This happens when two people are filling out the form. Thank you.[/i]

Answers will not be posted until Sunday, 9:00 PM EST, so you have until then.

[b]Clarification[/b]: There was a typo, and on #3, you do not have to express it in simplest radical form.

the format is exactly like the real ones. nice dude!!

I must remind everyone that they have [b]1[/b] chance before turning in answers. There may be NO corrections.

They are all saved onto a database, and I will take your FIRST submission.

Cool, how did you edit the PDF?
That site is getting better and better!
Edit: It says I am #H34N1 when I go to the submission page, but I am minicon.

[quote="minicon"]Cool, how did you edit the PDF?
That site is getting better and better![/quote]

I made the template. It is not edited off of a previously made MATHCOUNTS pdf.

[quote="cognos599"]#29 is messed up, u might want to check it out. 50 cards numbered 1-99???[/quote]

Yeah I just realized that and have added it to the Clarification section. Forgot to proofread these :oops:

thanks........(fricking message too short)

[quote="mchoi815"]i think there are parts that are missing from some of the problems.  some are listed below

PROBLEMS
6 (not specified whether integers or not)
9 (i believe -infinity works, if only one of the values is neg infinity)
11 (i think it should have, "in terms of n" at the end)
12 ("simplest radical form")
13 ("in common fraction form")
17 (some ppl may not know what perfect numbers are, so define for them maybe)
19 (what form do you want it in? decimal or fraction or mixed numeral?)
20 (clarify what distance you are looking for)[/quote]

6- Integer lengths
9 - Positive value
11 - yes
12 - yes
13 - yes
17 - no
19. when it is not specifried, you can put either
20. i think its pretty clear there

[b]IF YOU ARE MISAUTHENTICATED, THEN PLEASE WAIT 4 OR 5 MINUTES AND RE-AUTHENTICATE. THE AUTHENTICATION BUG HAPPENS WHEN TWO PEOPLE ARE AT THE SITE AT THE SAME TIME[/b]

[quote="minicon"]The bug is caused by deadlock, correct?[/quote]

It should be fixed now. You just need to wait 5 seconds if you are misauthenticated.

This was because you clicked on Submit query when you were supposed to just click on the auto authenticate. :P

[quote="xpmath"]Hmm, I could have gotten another if I had actually felt like doing #30, which was the worst probblem, according to me.[/quote]

i spent 10 minutes multiplying it out  :rotfl:

I put $ \frac{2}{3}\sqrt{3}\minus{}1, \frac{2}{3}\sqrt{3}\plus{}1$ for Target Round #2, and the answer was the same thing, except it was expressed in a different form. Was that counted as incorrect?
because i got 1 and 8 right and i only got 4 points for target, so....

[quote=epic2004_aops]Hey can you post the answers? I finished my sprint[/quote]

why the revive tho?

I don't know I was just surfing and found this and once I finish I found there was no key... can a key be made?

[quote=Kagebaka][quote=green_dog_7983]ask the OP in an pm or something like that[/quote]

The OP hasn't been active in 3 months and this post was made in 2008; there's a high possibility he won't respond[/quote]

true tho

Eh,  I can't access any of the links at all...


limit
real analysis
real analysis unsolved

$ f: \mathbb{Q}\to\mathbb{R}$ is continuous.

prove, that $ f$ has a finite limit in some (at least 1) irrational points.

[hide="Idea"]
Consider metric spaces $ (X, d), \ (Y, \rho)$.
Let $ A \subset X$, and $ g: A \to Y$.
For $ x\in X$ define ${ o_{g}(x) = \inf\{\varepsilon > 0 \ : \exists U - \mbox{open neighbourhood of }x \ : \ \mathrm{diam}\ g(U\cap A}) \leqslant \varepsilon\}$
(we assume that $ \inf \emptyset = + \infty$)
and $ o(g) = \{x \in X \ : \ o_{g}(x) = 0\}$.
Now it's not hard to prove that $ o(g)$ is a countable intersection of open sets, and if $ g$ is continuous in $ x$ then $ o_{g}(x) = 0$.
It is easy to see that if $ x_{0} \in o(f)$ then $ \lim_{x\to x_{0}}f(x)$ exists.
Since $ \mathbb{Q}$ is not a countable intersection of open sets (this follows from Baire Category Theorem), we have $ \mathbb{Q}\subsetneq o(f)$, and $ f$ must have limit in some irrational point.
[/hide]

I didn't work through that idea fully, since my eye was drawn to the word Baire at the bottom. Doesn't Baire Category Theorem only work for complete metric spaces (and $ \mathbb{Q}$ is not)?

That sentence really means "$ \mathbb{Q}$ is not an intersection of open sets in $ \mathbb{R}$".


1: If the entries of a 3 by 3 matrix are chosen randomly between 0 and 1, what are the most likely dimensions of the 4 subspaces? What if the matrix is 3 by 5?

The answer is - it's most likely to be independent. But how can we tell? There are 2^9 possible combinations. But this is where I get stuck...

2: If A and B share the same 4 subspaces, then A is a multiple of B. True or false? Answer is false - involving an invertible matrix. This is talking about subspaces, right? Not just the bases? If they share the same 4 subspace bases, then A MUST be a multiple of B, right? But if they do not share the same 4 subspace bases, but nonetheless span the same subspace, then A doesn't have to be a multiple of B.

I assume that "the four subspaces" are the spaces spanned by the first $k$ columns for $k=0,1,2,3$; I've never heard this terminology before.

For #1, I offer this count:
There are $7\cdot 6\cdot 4=168$ invertible matrices over the field with two elements. All of these are still invertible over the rationals, since they have odd determinant.
In addition, the matrices with determinant $\pm2$ are invertible. Looking at the permutations, we can show that there are three matrices of determinant $2$ and three of determinant $-2$ with elements in $\{0,1\}$, so a total of 174 of the 512 matrices of this form are invertible.
After that, the other possible lists are comparatively easy:
0, 0, ?, ?: 1*8*8=64
0, 1, ?, ?: 512-64=448
0, 0, 0, ?: 1*1*8=8
0, 0, 1, ?: 64-8=56
0, 1, 1, ?: 7*2*8=112
0, 1, 2, ?: 448-128=336
0, 0, 0, 0: 1
0, 0, 0, 1: 7=8-1
0, 0, 1, 1: 14=1*7*2
0, 0, 1, 2: 42=56-14
0, 1, 1, 1: 28=1*7*2*2
0, 1, 1, 2: 84=112-28
0, 1, 2, 2: 162=336-174
0, 1, 2, 3: 174

Invertibility is most likely, but only in a very narrow sense. That's only a third of the time.

For part 2: You've got it right. If the spaces in question are the column spaces I mentioned, $A$ and $B$ share the same spaces whenever $AX=B$ for some upper triangular matrix $X$ with nonzero diagonal entries. (This makes $X$ invertible, so $B=AX^{-1}$, and $X^{-1}$ is also upper triangular)
You should be able to construct plenty of examples.


calculus
derivative
analytic geometry
graphing lines
slope
conics
parabola

2 ways to do it:

[hide]1)the square of the distance from the point (a,1/3 a^2) to the point (15,-3) is (15-a)^2+(-3-1/3 a^2)^2. We wish to minimize this, so set it's derivative equal to 0.

2) the slope of the line tangent to the parabola at the point (a,1/3 a^2) is 2/3 a. the slope of the line perpendicular to this line is -3/(2a). So, the line perpendicular to the parabola at the point (a,1/3 a^2) is (y-1/3 a^2)=-3/(2a) (x-a). We want this line to pass through (15,-3). Solve for a.[/hide]

I used his first way to get the answer. Is it [hide]13.416[/hide]?

Don't use approximations!  Feh.  Who knows what that number is?  Is it (10 :pi: - 18)?  Is it :sqrt:180 (= 6 :rt5:)?  Those two numbers are very close together, but the [i]reasons[/i] that it would be one as opposed to the other are very different!

*Note* edited because I can't subtract.

remmebeer alwasy store your numbers when do the AP CALC exam. and then round at the end to three digits.

Right-o, (10 :pi: - 18).  But it's okay, because now that I'm doing real math, we don't actually have to know how to subtract, we just have to acknowledge the existence of the inverse-additive operator.

And it's worth pointing out that this website is [i]not[/i] the AP calculus exam, and that exact forms really are much nicer than decimal approximations because they contain more information.

Yeah sorry, I was too lazy to do the derivatives and stuff so I just used my calculator and graphed it :).

rlee this probelm is from a non-calculator exam that fubu almost aced.

That isn't really helpful at all.  It might be nice if you gave the full name of the competition and some information (or possibly a link) about it.


Proof that Every second-countable space is first-countable and separable

The first countable fact is obvious; the separability follows from the fact that the union of one point from each basis element forms a countable dense subset.


Determine the position where a parenthese should be put in an exponential tower-
  ${{{{{{{x^{x}}^{x}}^{x}}^{x}}^{.}}^{.}}^{.}}^{x}$there being $x$ exponents odf $x$ such that the resultant number has maximum value($x$ is a natural number)

Is the tower evaluated from the right or the left conventionally?
Is only one pair of parentheses allowed to be added?

the tower is to be evaluated from the left
and only one pair of parentheses is to be added?

I think I have this all clear now.  Solution follows:
[hide]
I am going to invent some notation because I can't currently work Latex.
[X,n] means a stack of n Xs not in parenthesis.
[X, m][X, n] means a stack of height m as the base for a stack of height n adjoined upwards to it.

Write [X, X] as [X, a]([X, b])[X, X - a - b] where 1 <= a, b and a + b <= X
       (1)
Parentheses are inserted around the middle stack of height b.

Evalutating from the left-hand-side:
[X, a] = X^(X^(a-1))
[X, b] = X^(X^(b-1)).
The third stack collapses down the the level of the power of the first.
(1) can therefore be expressed as X raised to the following power, which I then simplify:
X^(a - 1) * X^(X^(b - 1)) * X^(X - a - b).
= X^(a - 1 + X^(b-1) + X - a - b)
= X^(X^(b - 1) + X - b - 1)
Notice how the first and third stacks result in two tier exponentiation whilst the middle stack results in three tier exponentiation.  Note also the independence of this answer upon a!

To maximise [X, X] therefore, we maximise f(X, b) = X^(b-1) + X - b - 1 by varying b.  Since X > 1, f(X, b + 1) = f(X, b) + X - 1 > f(X, b), so
maximise b subject to the inequalities in (1).
This gives a = 1, b = X - 1.

Therefore the maximum value of the stack is [X:1]([X:X-1]).

It has value:  X^(X^(X^(X - 2)))

(N.B. a >= 1 because if a = 0 [X:X] = X^(X^(X-1)).   b >= 1 because the case of a stack vanishing is already covered since the upper stack may have zero height.)
[/hide]


symmetry
geometry
parallelogram
geometry unsolved

$W$ is a polygon which has a center of symmetry $S$ such that if $P$ belongs to $W$, then so does $P'$, where $S$ is the midpoint of $PP'$. Show that there is a parallelogram $V$ containing $W$ such that the midpoint of each side of $V$ lies on the border of $W$.

It's useful to have seen [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35947]this problem[/url] before :).

For compactness reasons, there is a parallelogram $ABCD$ of minimal area containing the polygon $\cal W$ (and which has the center of symmetry $O$ of $\cal W$ as its own center of symmetry). Let's prove that such a polygon satisfies the requirements.

Suppose, for instance, that the midpoints $M,N$ of $AD,BC$ (respectively) do not belong to the border $\partial\cal W$ of $\cal W$. The intersection between $\partial\cal W$ and $AD$ is a closed (maybe degenerate) segment $s$ which lies either in $[AM)$ or $(MD]$. Suppose it's the former. Then, the intersection $\partial\cal W\cap BC$ will be a closed sgement $s'$ which lies in $(NC]$. Now keep $M,N$ and the lines $AB,CD$ fixed, and move $A$ slightly, putting $D=AM\cap CD$, until $AD$ no longer meets $\partial\cal W$, but such that $ABCD$ still contains $\cal W$ (a drawing should make all of this clear). Then $BC$ will not meet $\partial\cal W$ either, and the new parallelogram $ABCD$ will have the same area as the old one. We can slightly shrink it along the direction of $AB$ so that it still contains $\cal W$, and this contradicts the minimality of the area of the initial parallelogram.


geometry
inradius
number theory proposed
number theory

Prove that there are infinite positive integer$ n$ such that $ p\equal{}nr$ , where $ p$ and $ r$ are the semiperimeter and the inradius of a triangle $ ABC$ , where $ ABC$ has integer side lengths.

Has anyone got it?  I've reduced it to finding rational solutions $ x,y$ for $ xy(x\plus{}y\minus{}n)\equal{}n$.  Any ideas?

[quote="cosinator"]Has anyone got it?  I've reduced it to finding rational solutions $ x,y$ for $ xy(x \plus{} y \minus{} n) \equal{} n$.  Any ideas?[/quote]

I don't know english very well...I didn't know I had posted it correctly, but here's my solution.

Let $ p \equal{} 2s$. If $ s \equal{} nr$, $ p \equal{} 2nr$. So if we prove that there are infinite positive intergers $ n$ such that $ s \equal{} nr$, this completes the proof.

$ n \equal{} \frac {s}{r} \equal{} \frac {s^2}{S} \equal{} \frac {s^2}{\sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}} \equal{} \sqrt {\frac {s^3}{(s \minus{} a)(s \minus{} b)(s \minus{} c)}}$

where S is $ ABC$'s area and $ a, b, c$ are $ ABC$'s side length.

If we let $ s \minus{} a \equal{} x, s \minus{} b \equal{} y, s \minus{} c \equal{} z$, than $ x, y, z$ are any possible natural numbers, since $ a, b, c$ are triangle's side lengths.

So we can get $ n \equal{} \sqrt {\frac {(x \plus{} y \plus{} z)^3}{xyz}}$. Now we have to prove that there are infinite many interger $ \frac {(x \plus{} y \plus{} z)^3}{xyz}$ that is a perfect square.

Let's put $ z \equal{} k(x \plus{} y)$. Then the value is $ \frac {(k \plus{} 1)^3(x \plus{} y)^2}{xyk}$. Let's prove that this value can turn into $ 9(k \plus{} 1)^2$ for infinite many interger $ k$.

$ \frac {(k \plus{} 1)^3(x \plus{} y)^2}{xyk} \equal{} 9(k \plus{} 1)^2$.

$ (k \plus{} 1)(x \plus{} y)^2 \equal{} 9xyk$.

If we let $ y \equal{} xt$, $ (k \plus{} 1)(t \plus{} 1)^2 \equal{} 9kt^2$.

So, we can get $ (k \plus{} 1)t^2 \plus{} ( \minus{} 7k \plus{} 2)t \plus{} (k \plus{} 1) \equal{} 0$.

$ D \equal{} ( \minus{} 7k \plus{} 2)^2 \minus{} 4(k \plus{} 1)^2 \equal{} 9k(5k \plus{} 4)$. So we can let $ k(5k \plus{} 4) \equal{} u^2$.

Since $ gcd(k, 5k \plus{} 4)$ is $ 1$ or $ 2$ or $ 4$, we can get $ k \equal{} X^2, 5k \plus{} 4 \equal{} Y^2$ or $ k \equal{} 2X^2, 5k \plus{} 4 \equal{} 2Y^2$.

But either can produce $ Y^2 \minus{} 5X^2 \equal{} 4$.

With mod 8, we can easily prove that $ X, Y$ are not odd. Let's let $ X \equal{} 2v, Y \equal{} 2w$.

So, $ v^2 \minus{} 5w^2 \equal{} 1$. Since this is a pell's equation, it has infinite many solutions.

This completes the proof. :lol:


conics
parabola
Putnam
analytic geometry
graphing lines
slope
symmetry

Given a parabola, construct its focus using the ruler and the compasses.

I liked it because it was so simply stated. This is A5 from Putnam 1955 (I found it on Kalva).

Choose the coordinate system with origin at the (unknown) parabola vertex and the (unknown) parabola axis identical with the y-axis. In this coordinate system, the parabola has an equation $y = ax^2$. An arbitrary line intersecting the parabola at points $P = [x_P, y_P] , Q = [x_Q, y_Q]$ has an equation

$y - y_P = \frac{y_Q - y_P}{x_Q - x_P}\ (x - x_P)$

$y - ax_P^2 = \frac{ax_Q^2 - ax_P^2}{x_Q - x_P}\ (x - x_P) = a(x_Q + x_P) (x - x_P)$

because for the points $P, Q$ on the parabola, $y_P = ax_P^2, y_Q = ax_Q^2$. Similarly, an arbitrary line $RS$ parallel to $PQ$ and intersecting the parabola at points $R = [x_R, y_R] , S = [x_S, y_S]$ has an equation

$y - ax_R^2 = a(x_S + x_R) (x - x_R)$

Since the lines $PQ, RS$ are parallel, their slopes are equal, i.e., $a(x_Q + x_P) = a(x_S + x_R)$. This means that the x-coordinates $\frac 1 2 (x_Q + x_P) = \frac 1 2 (x_S + x_R)$ of the midpoints $M, N$ of the segments $PQ, RS$ are the same, i.e., the line $MN$ connecting the midpoints is parallel to the y-axis, identical with the parabola axis. A line through an arbitrary point on the parabola, for example through the point $P$, and perpandicular to the line $MN$ intersects the parabola at another point $P'$. Because of symmetry of the parabola in our coordinate system, the midpoint $K$ of the segment $PP'$ lies on the parabola axis and the line through $K$ parallel to $MN$ intersects the parabola at its vertex $O$ (the origin). The line perpendicular to $KO$ at $O$ is the tangent to the parabola at the vertex, identical with the x-axis.

The slope $m$ of the parabola at the point $P$ is the derivative of $y = ax^2$ at $x_P$, i. e., $m = 2ax_P$. Let $H$ be the foot of a normal from the point $P$ to the x-axis and $L$ the midpoint of the segment $OH$, i.e., $H = [x_P, 0]$, $L = [\frac{x_P}{2}, 0]$. A line passing through the points $L$ and $P$ has an equation

${y - y_P = \frac{y_L - y_P}{x_L - x_P}\ (x - x_P) = -\frac{ax_P^2}{\frac{x_P}{2} - x_P}}\ (x - x_P) = 2ax_P (x - x_P)$

The line $LP$ passes through the point $P$ and it has the same slope as the parabola at this point, i.e., it is the tangent to the parabola at the point $P$. Now it is possible to proceed in 2 ways. First a parabolic mirror focuses light rays parallel to the mirror axis to the focus. Hence, it would be possible to draw a line through the point $P$ parallel to the axis $OM$ and then reflect this line in the normal to the tangent $LP$ at the point $P$. The reflected line would intersect the parabola axis at the focus $F$. Alternately, it is possible to use the fact that the pedal curve of a parabola corresponding to the focus is the tangent to the parabola at its vertex. The line through the point $L$ perpendicular to $LP$ has an equation

$y - y_L = -\frac{1}{2ax_P}\ ( x - x_L)$

$y = -\frac{1}{2ax_P}\ \left( x - \frac{x_P}{2} \right)$

The y-coordinate of the intersection $F$ of this line with the parabola axis is obtained by putting $x = 0$:

$y_F = \frac{1}{2ax_P} \frac{x_P}{2} = \frac{1}{4a}$

which does not depend on the point $P$. The curvature of the parabola at the vertex $O$ is

$\frac{|y\"|}{(1 + y'^2)^{\frac 3 2}} = \frac{2a}{(1 + 4a^2x^2)^{\frac 3 2}} = 2a$ at $x = 0$

The kissing circle of the parabola at its vertex $O$ has the same curvature and is centered on the axis, i.e., its radius is $r = \frac{1}{2a}$ and its center $C = [0, \frac{1}{2a}]$. The focus of the parabola lies halfway between the points $O$ and $C$, i.e., it has coordinates $[0, \frac{1}{4a}]$. But these are just the coordinates of the constructed point $F$.

Thank you, but I didn't read it :). I'll take your word for it :). There are easier ways (no computations), but there are some things to prove. I'll leave them as exercises :).

Construct a chord of the parabola, and another chord parallel to the first one. Now draw the chord passing through the midpoints of the two initial chords. This last chord is parallel to the axis of the parabola, so it's perpendicular to its directrice. Call its direction $d$.

Now choose any point $P$ outside the parabola, construct the tangents $PT_1,PT_2$ to the parabola, using the fatc that $T_1T_2$ is the polar of $P$ wrt the parabola, and draw a line $d'$ through $T_1$ which is $\|d$. The reflection of $d'$ in $PT_1$ should pass through the focal point. If you do this again for $T_2$, we get two lines intersecting at the focal point.

There might be shorter solutions, but this is all I can think of right now.

I put down the computations because I am not that familiar with conic sections and besides, I am trying to get used to precision. The first computation just showed that the line connecting the midpoints of 2 parallel chords to a parabola is parallel to its axis. Then I very simply constructed the axis and the tangent at the vertex, which is the pedal curve of the focus. The second computation showed a simple way to construct a tangent at an arbitrary point. A normal to this tangent at its intersection with the pedal curve cuts the axis at the focus, which is indeed very simple.

[quote="grobber"]Given a parabola, construct its focus using the ruler and the compasses.[/quote]

I have another solution of this problem, which is based on Grobber's idea, but uses only rational operations (so, for instance, I don't draw tangents to the parabola from a point outside the parabola):

Call $\mathcal{P}$ our parabola.

First, consider a subroutine for our construction:

[b]Subroutine 1.[/b] For any point M on the parabola $\mathcal{P}$, construct the tangent to this parabola $\mathcal{P}$ at the point M.

Indeed, the simplest construction uses the Pascal theorem: Take four arbitrary points A, B, C, D on the parabola $\mathcal{P}$; connect the points $AB\cap DM$ and $CD\cap MA$, and intersect the connecting line with the line BC. Join the point of intersection with the point M; the resulting line is the tangent to the parabola $\mathcal{P}$ at the point M. This is a direct consequence of the fact that, after the Pascal theorem, applied to the points A, B, C, D, M, M on the parabola $\mathcal{P}$, the points $AB\cap DM$, $BC\cap MM$ and $CD\cap MA$ are collinear, where the line MM has to be interpreted as the tangent to the parabola $\mathcal{P}$ at the point M.

Now, apply Subroutine 1 to four different points on the parabola $\mathcal{P}$. You get four tangents to the parabola $\mathcal{P}$. These four tangents bound four triangles, each of these triangles having all its sidelines tangent to the parabola $\mathcal{P}$. But there is a well-known theorem stating that if all sidelines of a triangle are tangent to a parabola, then the circumcircle of this triangle passes through the focus of the parabola. Thus, we have to take our four triangles, draw their circumcircles; since all these circumcircles pass through the focus of our parabola, we can immediately find this focus by intersecting the circumcircles.

  Darij


Given a_i , b_i (i = 1,2,3,4,5) satisfy : $ a_i \leq a_j$ and $ b_i \leq b_j   (i \leq j)$
$ a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} a_5 \equal{} b_1 \plus{} b_2 \plus{} b_3 \plus{} b_4 \plus{} b_5 \equal{} 0$
$ a_2 \plus{} a_3 \plus{} a_4 \plus{} a_5 \leq b_2 \plus{} b_3 \plus{} b_4 \plus{} b_5$
$ a_3 \plus{} a_4 \plus{} a_5 \leq b_3 \plus{} b_4 \plus{} b_5$
Prove that :$ a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_5^2 \leq b_1^2 \plus{} b_2^2 \plus{} ... \plus{} b_5^2$
Plz help me >"<

I don't think the inequality is correct

Put $ a_i \equal{} b_i$ for $ 1\le i\le3$

and $ a_4 \equal{} \minus{}a_5 \equal{} 3$
$ b_4 \equal{} \minus{}b_5 \equal{} 2$

[quote="arpit goel"]I don't think the inequality is correct

Put $ a_i \equal{} b_i$ for $ 1\le i\le3$

and $ a_4 \equal{} \minus{} a_5 \equal{} 3$
$ b_4 \equal{} \minus{} b_5 \equal{} 2$[/quote]
oh sr  :P  I was so sleepy when writing this post so I missed sth ...:P edited :P

I don't understand how does this change anything

Put $ a_4 \equal{} \minus{}a_5 \equal{} \minus{}3$

and $ b_4 \equal{} \minus{}b_5 \equal{} \minus{}2$

[quote="arpit goel"]I don't understand how does this change anything

Put $ a_4 \equal{} \minus{} a_5 \equal{} \minus{} 3$

and $ b_4 \equal{} \minus{} b_5 \equal{} \minus{} 2$[/quote]
ok ... I'll consider again this problem ...>"< anyway, thx:D

[quote="arpit goel"]I don't understand how does this change anything

Put $ a_4 \equal{} \minus{} a_5 \equal{} \minus{} 3$

and $ b_4 \equal{} \minus{} b_5 \equal{} \minus{} 2$[/quote]
I see.... iF $ a_5 \equal{} \minus{}3$, then $ a_5 < a_1 , a_2 , a_3$ , which is wrong :D