ARML
quadratics
trigonometry
modular arithmetic
geometry
rectangle
AMC

Since many of the longer threads seem to get off-topic every now and then, I'd like to pre-emptively set up a thread to discuss the actual problems from the 2008 ARML contest, which will take place this weekend. Obviously,

DO NOT DISCUSS THE PROBLEMS UNTIL SATURDAY AFTER 500 P.M. EASTERN.

The 500 pm is there in case things run late at some site -- the relay round is supposed to be over by 200 pm Eastern. In the past, certain sites have had slight delays for various reasons.

I'd like to know which problems you liked, which ones you thought were too easy or too hard, which were too tedious, etc. Maybe you saw a clever solution that betters the one in the solutions handout, or were able to trivialize a problem on your calculator (if so, take pride -- this is the last year calcs are allowed  :P ).

...what did y'all think? Easier than expected? Harder than expected? Orthogonal to your expectations?

Just one comment about difficulty level: while it may have gotten there by a slightly different route, the overall difficulty was very similar to 2007. In 2007, the top 5 Division A teams scored 171-169-163-162-161 and the top 5 Division B teams scored 116-115-113-112-112. In 2008, the top 5 Division A teams scored 170-168-162-161-160 and the top 5 Division B teams scored 122-115-114-109-108. To a first approximation, that looks the same across the two years.

I'll add that, against that background, all of Southern California's teams improved by 14-18 points from 2007 to 2008.

hmm.. the relays was harder than expected, the team questions were easier..

for indiv. the order from ezest to hardest was

1 2 3 4 8 5 6 7 in my opinion.

Team round was harder. Power round was easier ( problem 9 on power is a specialization of a problem from The Art and Craft of Problem Solving). Individual round was easy. Number 8 was much easier than previous years. In my opinion, indiv. round easiest to hardest:  23415876. 1st tiebraker was crazy though. Super relay was over in half a minute because of guessing.

Did [i]anyone[/i] at any of the four sites get tiebreaker question #1 in under 10 minutes? No one did at Las Vegas, so what about the other three sites?

(This was generating's own, and favorite, question.)

Kent: The winner was at Penn State, but he didn't get the first one.  However, at least one person who got a 7 at Penn State did.
My team's average individual was a 4, usually tends to be a bit higher (around a 5).  We did great on team (missed #9 only) and power (34/40 while not getting anywhere on #10) but did poorly on the rest (0/4 on relays).  That's basically what always happens for us.

I found that the difficulty curve on individual was much steeper but none of the questions were anywhere near as hard as the last two years' #8's.  None of the questions were quite as easy as a lot of ARML individuals tend to be.

I think one person at Penn State (the site winner, no one with an 8) did.

Also, did any team get more than 16 pts on the relays? We (NYC A) thought we had bombed them with that score (and one relay team contributing half of it), but it seems to have turned out alright  :wink:

[b]Team:[/b]
Didn't really look at them, since I only focused on mine.

[b]Power:[/b]
I liked the theme of the power question.

[b]Individuals:[/b]
Pretty rough.  Here's what I thought of each of the questions:
1. Nice question, but cruel (I got this one right, but I almost said 107).
2. Nice.
3. Nice, but the $ \frac{1}{2008}$ made things a bit annoying.
4. I found myself doing a lot of work on this one.
5. Nice, but tricky.
6. Looked harder than it was.
7. A bit intimidating unless you knew what to do.
8. Not so terrible for a #8.

[b]Relays:[/b]
These were tougher than I thought.

[b]Super Relay:[/b]
#8 on the super relay was guessable, but I wasn't sure what kinds of numbers I was going to get.

[b]Tiebreakers:[/b]
I immediately thought of Ceva's Theorem for the first one.  I didn't even attempt the second one because I couldn't read the equation from where I was sitting. (I was at Penn State.)  The third one wasn't too bad.

I liked the power question, except for the official answer sheet on some of the problems- I wound up grading #4 because I was annoyed at it.
How many people recognized #9 from the more general version in Art and Craft of Problem Solving?

The highest scores on the relays at the western site were coming from the Taiwan teams- we joked that they must have been easier in Chinese from the grading room. I think the stronger Taiwan team matched that 16.

[quote="yjneb"]Also, did any team get more than 16 pts on the relays? We (NYC A) thought we had bombed them with that score (and one relay team contributing half of it), but it seems to have turned out alright  :wink:[/quote]
That's what won you the championship. That's 60% of your advantage over us. Best teams, relay only:

NYC A: 8 + 8 = 16
Exeter Red: 8 + 6 = 14
Wash. Gold: 2 + 10 = 12
Col. Granite: 6 + 6 = 12
AAST Mu A: 4 + 8 = 12
NC A: 0 + 10 = 10
SFBA A: 2 + 8 = 10
SoCal A: 2 + 8 = 10
Chicago A: 6 + 4 = 10
Texas Gold: 6 + 4 = 10
Minn. Gold: 6 + 4 = 10
NYC N: 2 + 8 = 10
W. Mass A: 2 + 8 = 10

I don't see any Division B teams with more than 6.

[quote="Kent Merryfield"]cognos599, that doesn't come anywhere close to matching the actual scores. By score, the two hardest questions were clearly 8 and 6; 7 was considerably easier than either of those. And question #1 was the hardest of the first four - the easiest was probably #2, but #4 also had high scores.[/quote]

#1 was not hard, it was just really tricky. once ppl establish that $ n<20$ for the step size of the arithmetic sequence being $ n$, many ppl said that $ n \equal{} 19$ and therefore the answer was $ 50 \plus{} 19 \times 3 \equal{} 107$, when it required one to do:

$ n<20$
$ 3n<60$
$ 50 \plus{} 3n <110$
$ n \equal{} \fbox{109}$

on #3, everyone on my team (Indiana Gold) got it right. and #2 was missed by a few people. #8, in my opinion was really easy to brute force, i would just do:

$ \frac{k(3k \minus{} 1)}{2}$ is a perfect square, so just brute force it from here,

you know that $ k$ and $ 3k \equal{} 1$ are relatively prime for any $ n>1$ so they cannot share any common factors therefore attempt to make $ k$ a perfect square and $ 3k\minus{}1$  twice one. now just brute force

k            3k-1
4             11
9             26
16           47
25           74
36           107
49           146
64           191
81           242 this is $ 9^2 \times 11^2 \times 2$ which is twice a perfect square so our answer is $ \fbox{81}$

yeah #6 may have been harder than #7, but besides that i think my guess is right for me.  :P

[quote="Kent Merryfield"]cognos599, that doesn't come anywhere close to matching the actual scores. By score, the two hardest questions were clearly 8 and 6; 7 was considerably easier than either of those. And question #1 was the hardest of the first four - the easiest was probably #2, but #4 also had high scores.[/quote]

Well, you should also consider that there are many biases for why 6 and 8 will be the two lowest scoring problems.  First, they often times are the two hardest problems.  This gives everyone the impression that they'll be hard and many people start off the problem already intimidated.  Also, they come after 5 and 7, respectively, which means you'll usually be left with less time to solve them already.

Top relay scores:
[quote="Kent Merryfield"]NYC A: 8 + 8 = 16
Exeter Red: 8 + 6 = 14
Wash. Gold: 2 + 10 = 12
Col. Granite: 6 + 6 = 12
AAST Mu A: 4 + 8 = 12
NC A: 0 + 10 = 10
SFBA A: 2 + 8 = 10
SoCal A: 2 + 8 = 10
Chicago A: 6 + 4 = 10
Texas Gold: 6 + 4 = 10
Minn. Gold: 6 + 4 = 10
NYC N: 2 + 8 = 10
W. Mass A: 2 + 8 = 10 [/quote]Add Taiwan B with 10 + 12 = 22. They had the highest score anywhere on each question.

[quote="junggi"][quote="DataBox"]
I was on the SFBA(A) Team and 5 on power was done by Brian Zhang and Anthony Lu, while 10 on power was done by Evan O Dorney.  I think 9 was done by Yangpin Chen and 10 was done by Evan O Dorney and me.[/quote]
what
no the #9 was me
i pretty much spent the entire time doing this problem and rechecking it
hmm i thought bowei did #10
well i thought he said something along the lines of "#10 is trivial with the result of #8"
hmm i dont remember what #5 was, but i remember reading the rest of the power and thinking that it was way too easy
(especially in comparison to power rounds of the other years i went to (06,07=origami and continued fractions))
[/quote]

hmmm...my bad, i couldn't remember who did 9 for team
bowei didn't do 10, he talked about it being trivial for like 10 minutes and then moved onto something else

[quote="DataBox"][quote]
Kent Merryfield wrote:
Ted, I'd think that if the bus ride were to affect you negatively, it would be worse on Friday (closer to the ride) and not as bad Saturday after you had a chance to rest in bed. There were probably more of our guys up and around after 11:00 than yours, although of course I don't know what went on in people's rooms.

Here's a specific question that only one of the 15 members of the SFBA A team could answer: How did you do that on Friday evening? Who was responsible for answering #9 and #10 on the team round and #5 and #10 on the power round?

DiscreetFourierTransform wrote:
I'm not on the A team, but I talked to the kids who did these problems.
Evan O'Dorney did #10 power, and Brian Zhang did #5 power (which struck me as being ridiculously difficult for only a #5 on power.)[/quote]

I was on the SFBA(A) Team and 5 on power was done by Brian Zhang and Anthony Lu, while 10 on power was done by Evan O Dorney.  I think 9 was done by Yangpin Chen and 10 was done by Evan O Dorney and me.[/quote]
Hardly...
A more precise way to say it: 5 on power was done by Brian, and I assisted him with the calculations. And I'm assuming here, but I believe 10 was also done by Evan, with you assisting.

Not to discredit you, but I think it's a bit too much to say Brian and I co-solved 5 (which is what it sounds like from your post).

Also, 10 on team wasn't that difficult. Evan and someone else solved it in the last couple of minutes, and I came up with the same answer semi-independently (glanced at their scratch paper) ~30 seconds later.

:lol: yeah, that seems a little more correct than my previous assertion

To answer the OP, I liked the individual problems in general even though I did not do so well.
#1 - Strange problem that seemed to have little purpose other than to confuse people (including me)
#2 - Nice problem; it was easy, but good for a #2
#3 - Fairly routine problem - nothing out of the ordinary
#4 - Cool problem - but easily brute-forced with Ptolemy's
#5 - Fairly simple but I missed it due to a reading error
#6 - Interesting idea - I did not see the trick
#7 - Not too hard but took awhile, taking valuable time away from working on the next problem
#8 - Not nearly as hard as the standard number eight. I figured out that $ p$ and $ \frac{3p\minus{}1}{2}$ were perfect squares with plenty of time to spare. I recognized that it was a Pell's equation but forgot how to solve them - for some reason, I figured that brute force wouldn't work - that $ p$ would end up being somewhere in the thousands - had I tried the first few squares, I would have gotten this one.

The team problems were fine albeit easier than usual. The power question was fun. The relay questions may have been a little bit too hard.

My absolute favorite way to do #9 on Power was not mentioned in the solutions packet -- I wonder if any teams did it this way...

[hide="calculusy!"]
Suppose the region $ R$ has opposite corners at $ (0,0)$ and $ (w,h)$. Integrate $ f(x,y)\equal{}\sin(2\pi x)\sin(2\pi y) \,dx\,dy$ over the region. Notice that if we integrate $ f$ over any region with integer width or length, we get zero. Therefore,
\[ \int_0^h\int_0^w f(x,y)\,dx\,dy \equal{} 0 \quad \Rightarrow \quad
\frac{1}{4\pi^2}(\cos(2\pi w)\minus{}1)(\cos(2\pi h)\minus{}1) \equal{} 0,
\]
which means either $ \cos(2\pi w)\equal{}1$ or $ \cos(2\pi h)\equal{}1$. That is, at least one of $ w$ and $ h$ is an integer!
[/hide]

The solution suggested by the checkerboard tiling is a discretized version of that- we're integrating $ f(x)f(y)$, where $ f(x)\equal{}\begin{cases}1&0\le x<\frac12\\\minus{}1&\frac12\le x<1\end{cases}$, extended to be periodic of period 1. The discretized version lets you get away without explicitly writing down integrals.

[quote="jmerry"]
Incidentally, I almost quibbled SFBA A's perfect power round down to a 39- they were just so terse in #4.[/quote]

Hear that, SFBA?  That's why we had that marathon practice where i had you rewrite PQ solutions over and over until they were long enough - looks like it paid off!  Thanks for taking that advice to heart..   :)

As for the team/individual mismatch, i'll take full responsibility for that as the coach.  My biggest priority with the A team was working on team and power, which totally paid off on those rounds but at the expense of individuals apparently.  i just assumed with 14 USAMO qualifiers on the team i didn't have to worry about content, but we probably could have used some work on speed (as borne out by the relay scores as well).  That's at the top of my agenda for next year content-wise..

does anyone know the rank of the Indiana gold team?

Tim, the more I look at trying to explain the wild shape-of-score difference between our A teams, the less [i]any[/i] explanation based on training methods makes any sense to me.

For instance: did your A team really train for this any differently than your B team? In our case, I can't see that there was much difference - we were doing the same problems in practices with the two teams. Your A team was much better than our A team on team and power and our A team was much better on individual - but the same pattern did not hold for the B team. Our B team did 4 points better than your B team on team, the same on power, 8 points better on individual, and the same on relays. That's just not the same divergence of results - and if it were really the training methods, wouldn't you see something?

Our group was a little older than yours (there are 8 people graduating from the the A and B teams combined) and probably quite a bit more experienced. In making marginal A team vs. B team decisions, I did consider experience and I was thinking about team and power when I did that. So why did my more experienced guys excel in individual rather than team/power? I have no particularly good explanation. It just happened.

----

cognos599: Indiana Gold was 20th in Division A (and ahead of everyone in Division B) with a score of 134. Among other things, you beat TJ A by 3 points.

[quote="Kent Merryfield"]
For instance: did your A team really train for this any differently than your B team?[/quote]

Actually, yes.  i had assistant coaches working with each of our four groups separately, and our B team was the one i actually never spent any time with personally.  The work i did with A versus S/F was very different, and the time i spent with the A team was exclusively focused on PQ writing strategies and division of labor on Team.  So the results from my teams make perfect sense to me - except for the F team's PQ of 35, but several of those students had read ACoPS and said that did a lot of the work for them.. :)

Here's the difficulty table for all teams on the individual round. (Well, almost all teams - I don't yet have the five international teams from the Las Vegas site.) Compare this to my earlier post for top-12 teams only.

$ \begin{tabular}{cccccccccc} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & Total \\
2008 & 5.31 & 10.84 & 7.31 & 8.18 & 5.32 & 1.59 & 3.74 & 0.60 & 42.20\\
2007 & 8.91 & 9.89 & 6.32 & 4.53 & 2.32 & 4.44 & 8.79 & 0.05 & 45.25 \\
2006 & 6.24 & 6.77 & 8.54 & 7.91 & 8.30 & 1.70 & 9.49 & 0.03 & 48.97 \\
2005 & 10.31 & 7.62 & 9.33 & 3.58 & 11.51 & 0.53 & 4.49 & 0.27 & 47.63 \\
2004 & 10.17 & 6.14 & 1.68 & 4.20 & 1.84 & 4.99 & 1.94 & 0.31 & 31.27 \\
2003 & 5.65 & 4.64 & 8.34 & 0.30 & 1.36 & 2.13 & 1.04 & 0.23 & 23.69\end{tabular}$

Problems #1 and #7 were particularly interesting: not seen as particularly difficult (team averages $ >10$) for top-12 teams but the bane of Division B teams (team averages of about 4 for #1 and about 2 for #7).

Here's a comparison of the scoring from 2007 to 2008:

$ \begin{tabular}{cccccc}&Team&Power&Indiv.&Relay&Total\\
\minus{}&\minus{}&\minus{}&\minus{}&\minus{}&\minus{}\\
2008 top 12&34.3&34.5&78.9&8.8&156.6\\
2007 top 12&28.3&30.6&80.3&20.0&159.2\\
\minus{}&\minus{}&\minus{}&\minus{}&\minus{}&\minus{}\\
2008 lowA/highB&29.0&29.5&60.1&6.5&125.0\\
2007 lowA/highB&20.4&24.5&62.9&10.5&118.3\\
\minus{}&\minus{}&\minus{}&\minus{}&\minus{}&\minus{}\\
2008 others&19.9&19.2&32.2&1.6&72.0\\
2007 others&12.5&14.2&32.2&2.9&61.7\\
\minus{}&\minus{}&\minus{}&\minus{}&\minus{}&\minus{}\\
2008 all&23.1&22.7&42.2&3.3&90.6\\
2007 all&16.2&18.6&45.2&6.6&86.7\end{tabular}$

The verdict here? The team and power rounds were both easier than those in 2007. The individual round was very, very slightly more difficult, despite there being no impossible questions - that's mostly about sneaky questions like #1. The relays were much harder in 2008. However, the relay being harder made little difference to mid-B and low-B teams, since those teams don't get points on the relays anyway. The result was to leave the top-end scores relatively unchanged, but bring up the average mid-B and low-B score.

[quote="Ignite168"]TJA was missing some of their more prominent members though[/quote]
Specifically, Hamster, Jacob, and Haitao were all not there, though Daniel Li was.

[quote="archimedes1"][quote="TMA"][quote="paladin8"]one reason that the coaches hypothesized is that all of the practice individual rounds we gave them were from other ML's, which were considerably easier. I'm not sure the team had enough exposure to ARML level individual questions.[/quote]
(1) it's not true that ALL the individual rounds in practice were from other MLs, I remember we did at least the 2005 individual round in one of the first practices
[/quote]
Not quite...  That was the only ARML Individual we did, and only #1-6 at that.  But I don't think that is the issue.  Maybe it was just a bad day.[/quote]
If each member has a bad day in every other day, then a bad day for the whole team would be only $ \frac{1}{2^{15}}$ in probability.
I think sleep-deprivation might be the main reason responsible for SFBA A team's poor performance in second day events.


calculus
LaTeX
limit
trigonometry
trig identities
tangent double angle

A sequence is defined as follows:
a0 = 0; an = an-1 + sqrt(1+an-1^2) for n>0
Find limit of an/2^n-1 as n goes to inf

Here n and n-1 should be in subscript. I am still familiarising myself with Latex)

Calculus belongs in the calculus forum.

Here is the problem in LaTeX:

$ a_{0} \equal{} 0$
$ a_{n} \equal{} a_{n\minus{}1} \plus{} \sqrt{1 \plus{} a_{n\minus{}1}^{2}}$ for $ n > 0$

find $ \displaystyle\lim_{n\to\infty}\frac{a_{n}}{2^{n\minus{}1}}$

There's not that much calculus:

Let $ a_n\equal{}\cot \theta_n$ where $ \theta_n \in (0,\frac{\pi}{2}]$ (we can see each term in the sequence will be nonnegative). Then $ \theta_0\equal{}\frac{\pi}{2}$ and we have
\[ \cot \theta_{n\plus{}1}\equal{}\cot \theta_n \plus{}\csc \theta_n\equal{}\frac{1\plus{}\cos \theta_n}{\sin \theta_n}\equal{}\cot \frac{\theta_n}{2}\]
giving $ \theta_{n\plus{}1}\equal{}\frac{\theta_n}{2}$. So we end up with
\[ \theta_n\equal{}\frac{\pi}{2^{n\plus{}1}},a_n\equal{}\cot \frac{\pi}{2^{n\plus{}1}}\]
As $ x\rightarrow 0$, we have $ \cos x\rightarrow 1$ and $ \frac{\sin x}{x}\rightarrow 1$ which yields an answer of $ \boxed{\frac{4}{\pi}}$.

May I ask, what was the motivation for the solution above?

Sorry, I'm kinda out of it...  :oops:

The first few terms lead to a mess of nested radicals, so if there's going to be a closed form, it'll have to be trigonometric instead of algebraic.

In this case, we want $ \sqrt{1\plus{}a_n^2}$ to be nice, suggesting either a substitution involving $ \tan \theta$ or $ \cot \theta$. But we also see that $ a_n$ tends to $ \infty$ as $ n$ tends to $ \infty$. It would be convenient to have $ \theta$ tend to 0, so contangent is a better choice.

The radicals themselves aren't actually a problem; writing the recursion backwards gives

$ \frac{a_n^2 \minus{} 1}{2a_n} \equal{} a_{n\minus{}1}$

which is strongly suggestive of the tangent double-angle formula.


function
Support
integration
algebra
domain
real analysis
real analysis unsolved

Is the following correct?

I have a simple series RL circuit with a square wave input (periodic) defined by:

[code]
e(t) = {10 0 <= t < 2
       { 0 2 <= t < 4
With a period of 4

E(s) = 1/(1-exp(-2as) * int (0 -> 2) exp(-st) f(t) dt
[/code]

Here the period = 2a thus a = 2

E(s) = 1/(1-exp(-4s) * int (0->2) exp(-st) (10) dt
E(s) = 1/(1-exp(-4s) * 10 * exp(-st)/-s | 0 -> 2
E(s) = 10/s * (1-exp(-2s)/(1-exp(-4s)

Since this is an RL circuit, the equation is

E(s) = I(s) (R + sL)
thus,
I(s) = E(s) / (R+sL)
= 10/s * (1-exp(-2s)/(1-exp(-4s)(R+sL)
= 10 / (s * (R+sL) * (1+exp(-2s))

Now, I don't have a transform where the exponent is on the bottom - how do I get the inverse laplace transform?

Suppose $f$ is a well behaved function supported on $[0,\infty).$ (Integrable would be nice. Compact support would be a convenience). Define

$g(t)=\sum_{k=0}^{\infty}f(t-kP)$

If $f$ is compactly supported, then $g$ is periodic of period $P$ on $[x,\infty)$ for some $x.$

Then $\mathcal{L}(g)(s)=\int_0^{\infty}f(t)e^{-st}\,dt$

${=\int_0^{\infty}\sum_{k=0}^{\infty}f(t-kP}e^{-st}\,dt$

$=\sum_{k=0}^{\infty}\int_0^{\infty}f(t)e^{-s(t+kP)}\,dt$

$=\sum_{k=0}^{\infty}e^{-skP}int_0^{\infty}f(t)e^{-st}\,dt$

$=\mathcal{L}(f)(s)\sum_{k=0}^{\infty}e^{-skP}=\frac{\mathcal{L}(f)(s)}{1- e^{-Ps}}.$

So that's what the $1-e^{-Ps}$ in the denominator means - that you have the Laplace transform of a function put together by this kind of periodic duplication.

I have a 1+exp in the denomator, and yes it was derived from a input periodic function.

But, how do i convert it to the time domain?


geometry
angle bisector
geometry unsolved

$ \triangle ABC$ is an acute triangle. $ AD$ is its altitude. $ D$ is on $ BC$

$ P$ is a point on $ AD(P\neq A,D)$, $ BP$ meets $ AC$ at $ M$, $ CP$ meets $ AB$ at $ N$.

(Obviously, $ \angle NDA \equal{} \angle MDA$(According to CMO))

Let $ MN$ meets $ AD$ at $ Q$,  $ EF$ is a line through $ Q$ and meet $ NC$, $ AC$ at $ E$ and $ F$

Prove that:

$ \angle EDA \equal{} \angle FDA$

I think it's very simple:

Let EF meet BC at K. We have that the pencil C(ANQB) is harmonic so (EFQK) is an harmonic quatern. But the pencil D(EFQK) is harmonic with $ \angle QDK \equal{} \frac{\pi}{2}$ then DQ is the internal angle bisector of $ \angle EDF$.

[quote="¬[ƒ(Gabriel)³²¹º]¼"]I think it's very simple:

Let EF meet BC at K. We have that the pencil C(ANQB) is harmonic so (EFQK) is an harmonic quatern. But the pencil D(EFQK) is harmonic with $ \angle QDK \equal{} \frac {\pi}{2}$ then DQ is the internal angle bisector of $ \angle EDF$.[/quote]

Yes, very nice proof. I'm being silly. :oops:

If do it in tradational way, I mean no projective methods, it will be quite a hard problem I think :blush:


Hi guys, i need help on these 2 questions. I'm taking algerbra through corrsopondents and i don't have anyone to go to for help so i was wondering if someone could please help out with these questions, thank you very much. :(

Questions

1) Find N: A/N - M + 5K/X = Y

5) Solve: Y/7 - 3/4 = 2Y/5

Theres no examples in my textbook so i'm having one heck of time figuring these out

[hide="Solution to Problem 1"] All you're trying to do is isolate N (so that your final equation is N = blah blah blah). To do so, you just get N on one side.

Start with $\frac{A} {N}-M+\frac{5K} {X}=Y$
Multiply both sides by $N$ to get $A-MN+\frac{5KN} {X}=YN$
Move everything with an $N$ over to one side, resulting in $A=YN+MN+\frac{5KN} {X}$
Factor an $N$ out: $A=N(Y+M+\frac{5K} {X})$
Finally, isolate $N$:

${{N=\frac{A} {(Y+M+\frac{5K} {X})}}}$[/hide]

[hide="Solution to Problem 2"] Find the Least Common Multiple of the denominators, which is 140, so multiply it all by 140.

$140(Y/7 - 3/4) = 140(2Y/5)$, which when distributed gets: $20Y-105=56Y$

Get all the Y's on one side to get $-105=36Y$ and divide by 36: $Y=\frac{-105} {36}$ which can be reduced to $\boxed {Y=\frac{-35} {12}}$[/hide]


Solve the equation $ (1 \plus{} 2x)^3 \equal{} 27(2x^2 \plus{} x(1 \minus{} 2\sqrt {x}))$

Hello
well  x=1  is d solution but
if we put x=1/4 in $ (1 \plus{} 2x)^3 \equal{} 27(2x^2 \plus{} x(1 \minus{} 2\sqrt {x}))$
 it should also  be the solution.
Thank u


[b]
Can you please tell me how i can put my work of TeXnicCenter here?
I just don't know how you put the symbols here!
Thanks![/b]

When you put dollar signs around \int^1_0 \sum^{\infty}_{i=0} \vert f_n(x)-f(x)\vert dx

it becomes

$\int^1_0 \sum^{\infty}_{i=0} \vert f_n(x)-f(x)\vert dx$

Voila!

Yes well try not to put everything between $ ... $ signs :) Just the formulaes, not the text ;)


Prove that:
(cot A+tan B)/(cot B+tan A)= cot A .tan B

Please show ur all steps.
thanks. :)

[hide="Solution"]Let $ tan A\equal{}a, tan B\equal{}b$. It follows that $ cot A\equal{}\frac{1}{a}, cot B\equal{}\frac{1}{b}$ because $ tan A\cdot cot A\equal{}1$
$ \frac{cot A\plus{}tan B}{cot B\plus{}tan A}\equal{}\frac{\frac{1}{a}\plus{}b}{\frac{1}{b}\plus{}a}\equal{}\frac{\frac{ab\plus{}1}{a}}{\frac{ab\plus{}1}{b}}\equal{}\frac{b}{a}\equal{}b\cdot\frac{1}{a}\equal{}tan B\cdot cot A$[/hide]

hello, we have
${ \frac{\cot(A)+\tan(B)}{\cot(B)+\tan(B)}}-\cot(B)\tan(B)=$
$ \frac{\cot(A)+\tan(B)-\cot(A)\tan(B)(\cot(B)+\tan(A))}{\cot(B)+\tan(A)}=$
$ \frac{\cot(A)+\tan(B)-\cot(A)-\tan(B)}{\cot(B)+\tan(A)}=0$
Sonnhard.

[quote="Poincare"]i don't quite understand dr gauber but you stole the words off the tip of of my tongue![/quote]He worked backwards from the identity and arriving at the conclusion that it is equal to zero because the steps are 'algebraically reversible'.

Looking over Dr's steps, the solution is correct, but the first step should have been

$ \frac{\cot(A)\plus{}\tan(B)}{\cot(B)\plus{}\tan(A)}\minus{}\cot(A)\tan(B)$

as this was just the original problem with $ \cot(A)\tan(B)$ subtracted from both sides.

He wrote

$ \frac{\cot(B)\plus{}\tan(B)}{\cot(B)\plus{}\tan(A)}\minus{}\cot(B)\tan(B)$ (By mistake, I'm sure, as it went correctly after that.)

Alternative solution:

$ \frac{\cot{A}\plus{}\tan{B}}{\cot{B}\plus{}\tan{A}}$

Take the numerator first:

$ \cot{A}\plus{}\tan{B}\equal{}\frac{1}{\tan{A}}\plus{}\frac{\tan{A}\tan{B}}{\tan{A}}\equal{}\frac{1\plus{}\tan{A}\tan{B}}{\tan{A}}$

Now the denominator:

$ \cot{B}\plus{}\tan{A}\equal{}\frac{1}{\tan{B}}\plus{}\frac{\tan{B}\tan{A}}{\tan{B}}\equal{}\frac{1\plus{}\tan{A}\tan{B}}{\tan{B}}$

Divide them:

$ \left(\frac{1\plus{}\tan{A}\tan{B}}{\tan{A}}\right)\left(\frac{1\plus{}\tan{A}\tan{B}}{\tan{B}}\right)\equal{}\left(\frac{1}{\tan{A}}\right)\tan{B}\equal{}\boxed{\cot{A}\tan{B}}$

There are dozens of ways to do this, I guess.

Yes. It's just part of trigonometry, which I learned as part of Alg II. In my school, the "typical" student would take this in 10th grade. But as I, like most students, are advanced, I took this in 7th.

although you didn't cotangent, and probably neither csc or sec, do you know it what it means?

They're just reciprocals.

sec=1/cos
csc=1/sin
cot=1/tan.

Here in the philippines, we do not really focus on $ \sec x, \csc x$ and $ \cot x$. We just focus on the major trigonometric ratios. But of course when asked to find the six functions, we know how to deal with it.

It's the same for most americans as well. We learn the reciprocal functions, but we don't spend as much time doing things with them as we do with the basic 3.

[quote="Bovy11"]In your countries,you study cotangent at high school??
Here in Japan,No... :([/quote]
in my country all students study $ \cot x$ in 7th grade...not necesarily at a certain level


modular arithmetic
algorithm
number theory
Euclidean algorithm

solve for $ 0\leq z \leq 1147$ such that:
$ z^{329}\equiv452\pmod{1147}$

[hide="outline"]$ GCD(329,\phi(1147)) \equal{} GCD(452,1147)$
Then use the euclidean algorithm to solve for $ u$ in:
$ 329u\minus{}\phi{1147}v \equal{} 1, \therefore b^u \equiv x$[/hide]

Lots of calculation ....

$ 1147=31 \times 37$
The question
$ \Leftrightarrow \begin{cases} z^{29}=18 (mod 31)\\
z^5=33(mod 37) \end{cases}$ by Fermat's little theorem.

$ \Leftrightarrow \begin{cases}1=18z(mod 31) \\
z^{5 \times 7+1}=33^7z(mod 37) \end{cases}$

$ \Leftrightarrow \begin{cases}18z=1(mod 31) \\
1=7z(mod 37) \end{cases}$

$ \Leftrightarrow \begin{cases}z=19(mod 31) \\
z=16(mod 37) \end{cases}$

$ \Leftrightarrow z=608(mod 1147)$


Post some interesting anagrams.

For example:

The Wii was very popular when it came out:

WII GAME CONSOLE = I'M NEW SOCIAL EGO

one of my favourites:
ELEVEN PLUS TWO = TWELVE PLUS ONE :)

a parody of that:
TWENTYONE PLUS TWO = TWENTYTWO PLUS ONE  :rotfl:

this in fact gives us a family of anagrams. :roll:

You don't have to discover them.  Just search the internet and post funny ones.

MOTHER-IN-LAW = WOMAN HITLER

How about this one:

THE ART OF PROBLEM SOLVING = BE FOR SPORT! LONG LIVE MATH!

[quote="randomdragoon"]DORMITORY = DIRTY ROOM
<<my name>> = THE NEXUS P = THE SEX PUN[/quote]
wow
i finally figured out what your sig meant

The guns n' roses lead singer goes under the name Axl Rose.
[hide="Look here if you are above 18"]It's an anagram of Oral Sex.[/hide]

Thank you, bubka, for educating us as to why that is. :omighty:

KYOTO = TOKYO

One was the capital of Japan until 1868, and the other is the capital of Japan today.  Guess which is which.

Justin Timberlake = I'm a jerk, but listen.
Public relations = Crap built on lies.

[quote="morphling"]President Clinton, of the USA -- To copulate, he finds interns[/quote]

I like it.

FUNERAL = REAL FUN

[quote="1=2"]delete = de leet

alt f4=4falt

enter = entre

page up = age pup

If there was a d in page up, I would say "Aged pup"[/quote]

revolting. :(

A rather interesting one... sorry to anyone if they get insulted by this.

The Asian tsunami = It's nausea in math.

[quote="bubka"][quote="mathnerd314"]Truly remarkable anagram:

To be or not to be, that is the question. Whether 'tis nobler in the mind to suffer the slings and arrows of outrageous fortune...

=

In one of the Bard's best-thought-of tragedies, our insistent hero, Hamlet, queries on two fronts about how life turns rotten.[/quote]

[size=150]WOW.[/size][/quote]

Apparently it was created by someone named Cory Calhoun.

Steve Krakowski came up with this one:

"That's one small step for a man, one giant leap for mankind." -Neil Armstrong

=

A thin man ran; makes a large stride, left planet, pins flag on moon! On to Mars!


Show that for any positive integer $ m$, there exist integers $ a,b$ such that

$ |a|,|b|\leq m$, and

$ 0<a\plus{}b\sqrt{2}\leq \frac{1\plus{}\sqrt{2}}{m\plus{}2}$.

Consider when $ a$, $ b$ are $ 0,1,2,...,m$. This is $ (m\plus{}1)^2$ values. $ a \plus{} b\sqrt {2}\le m*(1 \plus{} \sqrt {2})$. Break this up into $ (m \plus{} 1)^2 \minus{} 1$ intervals. By the pidgeonhole principle, one of these intervals has two of these numbers. (and they cannot be equal). If we subtract these, we get a number in the form $ a \plus{} b\sqrt {2}$ where $ a,b$ satisfy the given hypothesis. Hence there exists $ a,b$ so

$ 0 < a \plus{} b\sqrt {2}\le (1 \plus{} \sqrt {2})\cdot \frac {m}{(m \plus{} 1)^2 \minus{} 1}\equal{}(1 \plus{} \sqrt {2})\cdot \frac{1}{m\plus{}2}$

But $ \frac {m}{m^2 \plus{} 2m}\equal{} \frac {1}{m \plus{} 2}$ is true. So the result is true.

the intervals are [0,a],(a,2a],(2a,3a], etc.


calculus
integration
ratio
induction
number theory
prime factorization

Find all pairs $ (x,y)$ such that $ x^y = y^{x - y}$

 $ x,y \in \mathbb{Z}$

[hide]If $ x \equal{} y$, then $ x^x \equal{} 1$, so $ x \equal{} y \equal{} 1$.

Otherwise, we may assume WLOG $ x < y$. In this case, $ x^x | x^y$. Dividing both sides by $ x^x$, we get $ x^{y \minus{} x} \equal{} (y/x)^x \times y^{\minus{}y}$. Thus, $ y/x$ must be an integer, so we let $ k \equal{} y/x$. Substituting, $ x^{x(k \minus{} 1)} \equal{} k^x  (kx)^{\minus{}kx}$. Taking the $ x$-th root of both sides, we get $ x^{k \minus{} 1} \equal{} k k^{\minus{}k} x^{\minus{}k}$. Multiplying both sides by $ x^k$, $ x^{2k \minus{} 1} \equal{} \frac{k}{k^k}$. Note that the LHS is always an integer, where as the RHS is not when $ k > 1$. Therefore, it is necessary that $ k \equal{} 1$, so $ x \equal{} y$, contradicting our assumption that $ x < y$.

Therefore, $ (x, y) \equal{} (1, 1)$ is the only solution.

I think I'm missing something... like, say, proving this is the only nonnegative solution? Not sure.
[/hide]

[quote="Zhero"]
Otherwise, we may assume WLOG $ x < y$. [/quote]

Why?

[hide]
It is clear that both x and y are nonzero. If $ x < 0$ and $ y > 0$, then the LHS is integral while the RHS is not. Similarly, if $ x > 0$ and $ y < 0$, the LHS is not integral while the RHS is. If $ x < 0$ and $ y < 0$, then we can write $ x \equal{} \minus{} x_1, y \equal{} \minus{} y_1$ and get that $ ( \minus{} 1)^{x1} x_1^{y_1} \equal{} y_1^{x_1 \minus{} y_1}$, so every solution in this case will correspond to a solution with $ x > 0$ and $ y > 0$ and x even. Thus, assume $ x > 0$ and $ y > 0$. Clearly $ x \ge y$ because otherwise the RHS would not be integral while the left is. Also, note that x and y decompose into the same set of primes. For any prime p, let $ a_p$ be the exponent of p in the prime factorization of x, and define $ b_p$ analogously for y. For $ p | y$, $ a_py \equal{} b_p(x \minus{} y)$, or $ \frac {a_p}{b_p} \equal{} \frac {x \minus{} y}{y}$. Thus, the ratio $ r \equal{} \frac {a_p}{b_p}$ is constant over all p, so since $ x \ge y$ then $ r \ge 1$ and $ y | x$. If we write $ x \equal{} ky$, then the equation becomes $ k \equal{} y^{k \minus{} 2}$. Manually checking $ k \equal{} 1, 2, 3, 4$ gives us the solutions $ (x, y) \equal{} (1, 1), (9, 3), (8, 2)$ and if $ k > 4$ then it can be easily proved by induction that $ y^{k \minus{} 2} \ge 2^{k \minus{} 2} > k$, so no other solutions exist. Hence, our solutions are $ (1, 1), (9, 3), (8, 2)$, and $ ( \minus{} 8, \minus{} 2)$.
[/hide]

[quote="Temperal"]Why?[/quote]
Woops. <_<

At least I have half of it. :/

EDIT: And even then, my proof could have been condensed to half a sentence. I shall meditate in silence now...

It is obviuosly that $ x\equal{}y^k$ for any $ k\in \mathbb N$ and the result follow ...


S20

The inscribed circle of an equilateral triangle has radius 1 unit. A smaller circle is tangent to this circle and to two sides of the triangle. What is the radius of the smaller circle?

[hide]Draw a line enclosing the smaller circle into a small equilateral triangle of its own.  Now, since the incenter of an equilateral triangle is coincident with its centroid, and the centroid is two-thirds of the way from the vertex to the opposite side, we can conclude that the height of the smaller triangle is one-third the height of the larger triangle.  This means that the large triangle is similar to the smaller triangle with a factor of three.  All lengths preserve this ratio, so the radius of the small circle is just one-third of the radius of the larger circle.[/hide]


Hey guys, what's the trick behind the pathwalking problems requiring you to find the number of ways that you can go from one point to another only going up or right?

For example, from DUKE MATH MEET:

1. How many ways can one move from (0,0) to (3,5) without going through points (1,1) and (1,2) and moving only right and up?

Thanks!

There are no tricks, but only simple recurrence:
[i]Number of paths to (m,n)=Number of paths to (m-1,n) + Number of paths to (m,n-1)[/i].
Using this formula you can step by step calculate [i]Number of paths to (m,n)[/i] for pairs (m,n) s.t. m+n=1, m+n=2, m+n=3, and so on

If you are not allowed to go to (1,1) and (1,2) then you should assume [i]Number of paths to (1,1)=Number of paths to (1,2)=0[/i]

So (if I didn't miscalculate) answer for your example is 16.

There are only two separate routes you can take not passing (1,1), and (1,2), and they are through (0,3) and (2,0). That is C(5,2) + C(6,1) = 10 + 6 = 16.

We could also use the points (0,4), (1,3), (3,1).
So the answer would be 1*4+2*6+2*1=18

[quote="white_horse_king88"][quote="TripleM"]There most definitely is a trick! Well, at least for rectangles. For other shaped grids, then you have to do it that way. Its easy enough to find an explicit formula for, say, the number of paths from (0,0) to (m,n) not passing through (a,b) or (c,d).[/quote]

Do you mean applying inclusive-exclusive formula? I think that works. It would be paths to (m,n) - paths from (0,0) to (a,b) times paths from (a,b) to (m,n) - paths from (0,0) to (c,d) times paths from (c,d) to (m,n) + paths from (0,0) to (a,b) times paths from (a,b) to (c,d) times paths from (c,d) to (m,n)? Would that work?[/quote]

Yup. Assuming that a<=c and b<=d of course, which is what I meant. And of course those individual paths are just (something choose something).


What's the command line so that the batch window will remain open in case of an error in Asymptote? (or alternatively is there any way to produce an error log?).

I do not know of any Asymptote command that allows you to do that, but you can follow stevem's suggestion in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148557[/url] and create a bat file that turns the echo off, calls the asy.exe and then pauses. Then you run this bat file on your code file instead of running asy.exe directly and everything works just as one would wish: asy.exe does its job, shows the errors but the pause command keeps the window open. At any rate, it works on my PC and I am quite happy with this solution :).

Also you can use redir.exe* (second suggestion in the same thread and now also in the [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration#Showing_Asymptote_error_messages_in_TeXnicCenter]AoPSWiki[/url]) to put the error messages in a log which, if it contains any errors, is opened in TeXnicCenter or can be changed to open in any editor you wish.

*redir.exe is a very old but very effective DOS program which redirects standard error output to a file, something that I expect is trivial in Linux but difficult to do in Windows.

Thanks fedja, I didn't notice the other topic first  :blush:


factorial
function
complex analysis
complex analysis unsolved

Why is Mclaurin series true for all values of x? Can any one prove it? Thanks.

It isn't in general.  If it is true for all $x$, it's because the terms in the series decay fast enough at $\infty$ (usually due to the presence of factorials or something else) to make everything work.

But surely the coefficients are found by working on the origin of the original function, so it's supposed to be true only at the origin?


geometry
circumcircle
geometric transformation
reflection
rotation
parallelogram
dilation

Prove the existence of the 9-point circle.  That is, prove that the same circle passes through the midpoints of the sides, the feet of the altitudes, and the midpoints of the segments joining the orthocenter and vertices.

Given the existence of such a circle, prove that it's center is the midpoint of the segment joining the orthocenter and circumcenter of the triangle.

Where on that line does the centroid of the triangle lie?

BUMPing up my sad, unanswered geometry problems.

In triangle ABC, the altitudes AD, BE and CF intersect at the orthocenter, H. From right triangle BEC, <HBC = 90-C, and from right triangle CFB, <HCB = 90-B. Therefore, <BHC = 180-(90-B)-(90-C) = 180-A.

Take H and reflect it across BC to Ha1. Since <BHa1C + <BAC = (180-A)+A = 180, quadrilateral BACHa1 is cyclic, so Ha1 lies on the circumcircle.

Next, rotate H 180 degrees around the midpoint of BC (which we'll call Ma) to Ha2. Since BHCHa2 is a parallelogram, <BHa2C = <BHC = 180-A, and then we find that quadrilateral BACHa2 is also cyclic.

We can do the same stuff with the other sides of the triangle to find that the nine points A, B, C, Ha1, Ha2, Hb1, Hb2, Hc1, and Hc2 all lie on the circumcircle of triangle ABC. Now, dilate the circumcircle by a factor of 1/2 around the orthocenter. Ha1, Hb1, and Hc1 go to D, E, and F respectively, Ha2, Hb2, and Hc2 go to Ma, Mb, and Mc respectively, and A, B, and C go to the midpoints of HA, HB, and HC respectively. So, we've shown that these nine points lie on the same circle. Furthermore, we've shown that the center of the nine point circle is midway between the orthocenter and the circumcenter.

To locate the centroid, consider the dilation around G with ratio of -2. This takes Ma, Mb, and Mc to A, B, and C repsectively, which means it takes the nine-point circle to the circumcircle. Thus, it takes the nine point center to the circumcenter, which shows that the centroid lies on the trisector of NO closer to N (N=nine point center and O=circumcenter), or equivalently, G lies on the trisector of OH closer to O.

Dear Mathlinkers,
a new approach...
see
http://jl.ayme.pagesperso-orange.fr/  vol. 2, Les cercles de Morley, Euler p. 3-5

Sincerely
Jean-Louis


algebra
polynomial
vector
analytic geometry
superior algebra
superior algebra unsolved

Hi, got this problem about Krawtchouk polynomials. Have no idea how to approach, so need some hints. (am new to these polynomials too and hope that this is in the correct section)

Let $q$, $n$ be integers greater than 1. Let $k$ be an integer such that $0\le k\le n$.

Define the Krawtchouk polynomial to be $K_{k}(x)=\sum_{j=0}^{k}(-1)^{j}{x\choose j}{n-x\choose k-j}(q-1)^{k-j}$. Note: let $j$ denote any integer. ${x\choose j}=\frac{1}{j!}x(x-1)\ldots (x-j+1)$.

Let $Z_{q}$ denote the ring of integers modulo $q$ and $u$ be a vector of length $n$ over $Z_{q}$, that is, $u\in Z_{q}^{n}$. Let $wt(u)$ be the weight of $u$, that is, the number of non-zero coordinates in $u$. Let $v\in Z_{q}^{n}$. $u\cdot v$ denote the scalar product, that is, $u\cdot v=u_{1}v_{1}+u_{2}v_{2}+\ldots+u_{n}v_{n}$.

Suppose $wt(u)=i$ and let $\xi$ be the primitive $q$th root of unity in the complex field. Then prove $\sum_{wt(v)=k}\xi^{u\cdot v}=K_{k}(i)$.

Thanks!

hi,
use inclusion exclusion principle and the fact that for fixed $u \in \mathbb{Z}^{k}$ the map $v \to \zeta^{(u.v)}$ is an additive character so its sum over non zero $v$ is $-1$.

thanks buddha. got it.

anyone has any idea for the motivation behind krawtchuok polynomials?


How can I make a column for this one? I want to save space.^^

Please align this or make this into two columns:

\begin{align*}
1 - \cos \gamma &= 1 - \dfrac{a^2 + b^2 - c^2}{2ab} \\
&= \dfrac{2ab - a^2 - b^2 + c^2}{2ab} \\
&= \dfrac{c^2 -(a^2 - 2ab + b^2)}{2ab} \\
&= \dfrac{c^2 - (a - b)^2}{2ab} \\
&= \dfrac{(c - a + b)(c + a - b)}{2ab} \\
&= \dfrac{(2s - 2a)(2s - 2b)}{2ab} \\
\dfrac{1 - \cos \gamma}{2} &= \dfrac{(s - a)(s - b)}{ab} \\
\sin^2 \dfrac{\gamma}{2} &= \dfrac{(s - a)(s - b)}{ab} \\
\sin \dfrac{\gamma}{2} &= \sqrt{\dfrac{(s - a)(s - b)}{ab}}
\end{align*}

\begin{align*}
1 + \cos \gamma &= 1 + \dfrac{a^2 + b^2 - c^2}{2ab} \\
&= \dfrac{2ab + a^2 + b^2 - c^2}{2ab} \\
&= \dfrac{(a + b)^2 - c^2}{2ab} \\
&= \dfrac{(a + b + c)(a + b - c)}{2ab} \\
&= \dfrac{2s(2s - 2c)}{2ab} \\
\dfrac{1 + \cos \gamma}{2} &= \dfrac{s(s - c)}{ab} \\
\cos^2 \dfrac{\gamma}{2} &= \dfrac{s(s - c)}{ab} \\
\cos \dfrac{\gamma}{2} &= \sqrt{\dfrac{s(s - c)}{ab}} \\
\end{align*}
where $ s \equal{} \dfrac{a\plus{}b\plus{}c}{2}$

There are plenty of ways. Here is the most trivial one:
\begin{tabular}{ccc}
\begin{aligned}
1 - \cos \gamma &= 1 - \dfrac{a^2 + b^2 - c^2}{2ab} \\
&= \dfrac{2ab - a^2 - b^2 + c^2}{2ab} \\
&= \dfrac{c^2 -(a^2 - 2ab + b^2)}{2ab} \\
&= \dfrac{c^2 - (a - b)^2}{2ab} \\
&= \dfrac{(c - a + b)(c + a - b)}{2ab} \\
&= \dfrac{(2s - 2a)(2s - 2b)}{2ab} \\
\dfrac{1 - \cos \gamma}{2} &= \dfrac{(s - a)(s - b)}{ab} \\
\sin^2 \dfrac{\gamma}{2} &= \dfrac{(s - a)(s - b)}{ab} \\
\sin \dfrac{\gamma}{2} &= \sqrt{\dfrac{(s - a)(s - b)}{ab}}
\end{aligned}
&\phantom{.\hspace{50pt}.}&
\begin{aligned}
1 + \cos \gamma &= 1 + \dfrac{a^2 + b^2 - c^2}{2ab} \\
&= \dfrac{2ab + a^2 + b^2 - c^2}{2ab} \\
&= \dfrac{(a + b)^2 - c^2}{2ab} \\
&= \dfrac{(a + b + c)(a + b - c)}{2ab} \\
&= \dfrac{2s(2s - 2c)}{2ab} \\
\dfrac{1 + \cos \gamma}{2} &= \dfrac{s(s - c)}{ab} \\
\cos^2 \dfrac{\gamma}{2} &= \dfrac{s(s - c)}{ab} \\
\cos \dfrac{\gamma}{2} &= \sqrt{\dfrac{s(s - c)}{ab}} \\
\end{aligned}
\end{tabular}
(I spread the columns a bit so that they do not look too ugly together).

Thanks a lot! That was very nice. Please accept my friend invitation.  :D


calculus
derivative
function
geometry
linear algebra
matrix
vector

Given two functions, namely, f(x,y) and g(x,y). Can anyone tell me why
$ \frac {dfdg}{dxdy} \equal{} \left| \frac {\partial f}{\partial x} \frac {\partial g}{\partial y} \minus{} \frac {\partial g}{\partial x} \frac {\partial f}{\partial y} \right|$ ?

Hi,

I don't have an answer to that question, but this is what I got in its original form:

$ df \cdot dg \equal{} \left| \frac {\partial f}{\partial x} \frac {\partial g}{\partial y} \minus{} \frac {\partial g}{\partial x} \frac {\partial f}{\partial y} \right| dx \cdot dy$,

where the dot is an ordinary multiplication.

Does that make sense to you know? If it is still not, then I will provide the whole problem in its entirety.

Your notation still makes no sense to me, but I think I can see what's going on here- this is about area. Specifically, we want the area covered by $ h(x,y)\equal{}(f(x,y),g(x,y))$, and the RHS is the determinant of the Jacobian matrix.

Yes, you are right about the area thing. Could you write down the proper way? And how does determinant of the Jacobian matrix appear? Is there a way by derivation to show this?

I found the notation to write this equation:

$ \frac {\partial(f,g)}{\partial(x,y)} \equal{} \left| \frac {\partial f}{\partial x} \frac {\partial g}{\partial y} \minus{} \frac {\partial g}{\partial x} \frac {\partial f}{\partial y} \right|$

However, I don't see how $ \partial(f,g)$ fit to the problem instance I encountered which requires the area which should be in form of $ df \cdot dg$.

Can anyone shed some light?

References:

[url=http://en.wikipedia.org/wiki/Jacobian]Jacobian[/url] and [url=http://en.wikipedia.org/wiki/Change_of_variable]change of variable[/url] and [url=http://planetmath.org/encyclopedia/JacobiDeterminant.html]the same at Planetmath[/url].

Ultimately it comes back to this geometry/linear algebra fact: take vectors $ v_1,v_2,\dots,v_n$ in $ \mathbb{R}^n.$ Form the paralellepiped spanned by these vectors. That is, let $ S \equal{} \{t_1v_1 \plus{} t_2v_2 \plus{} \cdots \plus{} t_nv_n: 0\le t_1\le 1,\dots,0\le t_n\le 1\}.$ Then the $ n$-dimensional measure of $ S$ (area for $ n \equal{} 2,$ volume for $ n \equal{} 3$ and so on ) is the absolute value of the determinant of the matrix created by letting the columns (or rows - it doesn't matter) of the matrix be the vectors $ v_1,\dots,v_n.$

That determinant could be negative, positive, or zero. If it's zero, the alleged paralellepiped is squashed flat into something lower dimensional and has "volume" zero. Whether it is positive or negative indicates orientation; we can change the orientation just by listing the same vectors in a different order. That's why we take the absolute value to be the "volume."

Thanks, Kent. I found a book which illustratively shows how the Jacobian determinant comes about for that case -- and yes, it is like what you said.


$ N$ is a $ 1990$-digit number divisible by $ 9$.The sum of its digits is $ N_1$. The sum of the digits in $ N_1$ is $ N_2$. The sum of the digits in $ N_2$ is $ N_3$. Find $ N_3$.

[hide]All numbers divisible by 9 (except 0) have a sum of digits equal to 9 after adding the sum of the digits enough (example:846-->18-->9).  The largest sum of digits from a 1990 digit number is one where all digits are 9.  9*1990 = 17910. The sum of digits is 18, and the sum of this is 9. Since the sum of digits is reduced to 9  in three steps for the greatest sum of digits possible, the answer will be 9 for all other 1990 digit numbers.[/hide]

I am not sure.
$ \sum$$ %Error. "periodN" is a bad command.
_3 = N_4$ :maybe:

The answer given was for a specific $ N_1$.
You must generalize for all $ N_1$.

how can I write the sum of digits $ N_3$ with Latex

Perhaps this may be of some help.

http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols

Let me know if you have any more questions.

Like vamathletes said, $ N_{3}$ always equal 9.

If the original N is divisible by 9, its sum of the digits (no matter how many times they are taken) must also be divisible by 9, so its 0 mod 9.

since the biggest N has its sum of digits as 1990*9=17910,

N1 must be between 0 to 17910 with its sum of its digits 0 mod 9.
(so it can N1=0,9,18,36...,17910)

So N2 has to be 0,9,18,27,36, with 36=sum digits of 9999.

so N3 has to be 0 or 9. (9=1+8=2+7=3+6=9)

Then we look at the fact that if N3 was 0, that would make N2=0, N1=0, so N must be 0, but that would not make a 1990-digit number, so the answer is

Always $ 9$

[quote="SnipedYou"][hide]Notice that 108, 1008, 10008, etc are all divisible by 9. So $ 1 \plus{} 8 \equal{} 9$ and $ 9 \plus{} 0 \equal{} 9$[/hide][/quote]

Or you can use $ 999, \ 9999, \cdots$ like vamathletes did.

Well, this topic was very helpful.

Thanks, Fraenkel.

[quote="the future"][quote="Fraenkel"]Simply stating that the sum of $ N_2$ is either a two-digit number less than $ 90$ or $ 9$ will suffice to show that $ N_3$ is 9.[/quote]

No, N2 is a 2-digit number bigger or equal to 9, less than or equal to 36 (not 90) and it has to be divisible by 9, to show that N3 is 9.

Your statement is not true.. 85 fits your case, but if 85=N2, N3=13..[/quote]

If $ N_2\equal{}85$ then $ N_1$ is not divisible by 9.

[quote="Fraenkel"][quote="the future"][quote="Fraenkel"]Simply stating that the sum of $ N_2$ is either a two-digit number less than $ 90$ or $ 9$ will suffice to show that $ N_3$ is 9.[/quote]

No, N2 is a 2-digit number bigger or equal to 9, less than or equal to 36 (not 90) and it has to be divisible by 9, to show that N3 is 9.

Your statement is not true.. 85 fits your case, but if 85=N2, N3=13..[/quote]

If $ N_2 \equal{} 85$ then $ N_1$ is not divisible by 9.[/quote]

You never said N1 was divisible by 9

[quote="Fraenkel"]Simply stating that the sum of $ N_2$ is either a two-digit number less than $ 90$ or $ 9$ will suffice to show that $ N_3$ is 9.[/quote]

you said N was divisible by 9, which means N2 is divislble by 9, but you have to prove that, you cant just skip to that second step.  Futhernore, even if I said N1=81 thats wrong because N2 cannot equal 81, or 72 or 63....

anyways this has been answered..

[quote=the future]you said N was divisible by 9, which means N2 is divislble by 9, but you have to prove that, you cant just skip to that second step.[/quote]

I think we all know how to prove this, because the sum of the digits of a multiple of 9 must be divisible by 9.  Thus, if $ N_1$ is divisible by 9, then the sum of its digits can be expressed in the form $ 9k$, where $ k$ is a nonnegative integer.  So $ N_2$ is divisible by 9.  Just use induction if you wish to prove in general.


function
integration
calculus
trigonometry
calculus computations

hope i haven't asked this one already...   :|

show

$ \sum_{k=0}^\infty\;\frac{(k!)^{2}}{(2k)!}\;=\;\boxed{\frac{2}{9}\left(\frac{\pi}{\sqrt 3}+6\right)}$

[quote="perfect_radio"]I don't know if you asked this, but I can tell you that didilica already posted a solution: http://www.mathlinks.ro/Forum/viewtopic.php?t=4364 (the 6th post)[/quote]

nice.   :coolspeak:  pretty [b]unconventional [/b]method i must say. anyhow, i will post
my solution later.  :)

now to find $ \sum_{k=1}^\infty\;\frac{(-1)^{k-1}\cdot (k!)^{2}}{(2k)!}$

$ \text{answer}: \;\;\;\quad\boxed{\frac{1}{5}\left(1+\frac{4}{\sqrt 5}\cdot\text{cosech}^{-1}\;2\right)}$

If I'm not mistaken, the identical method works but with the final substitution of $ x = \frac{i}{2}$.

Another , standard method, for calculating this sum is to use the beta function as follows:

$ \sum_{k=0}^{\infty}\frac{(k!)^{2}}{(2k)!}=\sum_{k\geq 0}\frac{\Gamma(k+1)^{2}}{\Gamma(2k+2)}(2k+1)+$

$ =\sum_{k\geq 0}\int_{0}^{1}(2k+1)(x(1-x))^{k}dx=2\int_{0}^{1}\sum_{k\geq 0}k(x(1-x))^{k}dx$

$ +\int_{0}^{1}\sum_{k\geq 0}(x(1-x))^{k}dx=2\int_{0}^{1}\frac{x(1-x)}{(1-x+x^{2})^{2}}dx+\int_{0}^{1}\frac{1}{1-x+x^{2}}dx$

$ =\int_{0}^{1}\frac{1+x-x^{2}}{(1-x+x^{2})^{2}}dx=\frac{2\sqrt{3}\pi}{27}+\frac{4}{3}$.

Of course the Riemann integral can be calculated by using the standard techniques.

[quote="didilica"]Another , standard method, for calculating this sum is to use the beta function as follows:

$ \sum_{k=0}^{\infty}\frac{(k!)^{2}}{(2k)!}=\sum_{k\geq 0}\frac{\Gamma(k+1)^{2}}{\Gamma(2k+2)}(2k+1)$

$ =\sum_{k\geq 0}\int_{0}^{1}(2k+1)(x(1-x))^{k}dx=2\int_{0}^{1}\sum_{k\geq 0}k(x(1-x))^{k}dx$

$ +\int_{0}^{1}\sum_{k\geq 0}(x(1-x))^{k}dx=2\int_{0}^{1}\frac{x(1-x)}{(1-x+x^{2})^{2}}dx+\int_{0}^{1}\frac{1}{1-x+x^{2}}dx$

$ =\int_{0}^{1}\frac{1+x-x^{2}}{(1-x+x^{2})^{2}}dx=\frac{2\sqrt{3}\pi}{27}+\frac{4}{3}$.

Of course the Riemann integral can be calculated by using the standard techniques.[/quote]

sweet !  :thumbup:

big fat bummer...  :( i was hoping since
$ \int_{0}^{\frac{\pi}{2}}\;\sin^{2k+1}\;x\;\;\;\textbf dx\;=\;\frac{(2^{k}\cdot k!)^{2}}{(2k+1)!}$

then manipulating slightly, we have
$ \sum_{k=0}^\infty\;\frac{(k!)^{2}}{(2k)!}\;=\;\sum_{k=0}^\infty\;\frac{(2k+1)}{2^{2k}}\cdot \int_{0}^{\frac{\pi}{2}}\;\sin^{2k+1}\;x\;\;\;\textbf dx$
$ =\;2\;\int_{0}^{\frac{\pi}{2}}\;\sin x\cdot\left(\;\sum_{k=0}^\infty\;\frac{(2k+1)}{2^{2k+1}}\cdot \sin^{2k}\;x\;\right)\;\;\;\textbf dx\;\;\;\quad\;\;\;\quad\ldots\;\;\;\ldots\;\;\;(A)$

and so, if we can express
$ \sum_{k=0}^\infty\;\frac{\sin^{2k+1}\;x}{2^{2k+1}}\;=\;\textbf f\;(x)$

then differentiating w.r.t. $ x$, we will get
$ \sum_{k=0}^\infty\;(2k+1)\cdot\frac{\sin^{2k}\;x\cdot\cos x}{2^{2k+1}}\;=\;\textbf f\;'(x)$

and so, finally, from $ (A)$,
$ \sum_{k=0}^\infty\;\frac{(k!)^{2}}{(2k)!}\;=\;2\;\int_{0}^{\frac{\pi}{2}}\;\left(\;\textbf f\;'(x)\cdot\tan x\;\right)\;\;\;\textbf dx\;\;\;\quad\;\;\;\quad\ldots\;\;\;\ldots\;\;\;(B)$

which of-course means, [i]first[/i], i have to exactly know what $ \textbf f\;(x)$ is,
and [i]second[/i], that the integral in $ (B)$ is actually going to be [u]doable[/u] !   :|

[quote="didilica"]Your $ f(x)=\frac{2\sin(x)}{4-\sin^{2}(x)}$, since that is a series of the form $ \sum_{n=0}^{\infty}a^{2n}$, $ |a|<1/2$.[/quote]

how did you get that ?   :(

anyhow, plugging it directly in $ (B)$ gives me $ 2\;\int_{0}^{\frac{\pi}{2}}\;\;\frac{8\sin x+2\sin^{3}x}{(4-\sin^{2}x)^{2}}\;\;\textbf dx$

$ =\;\left[\;\frac{4\left(\;\sqrt 3\cdot(7+\cos \;(2x)\; )\cdot\left(\tan^{-1}\left(\frac{2\tan\left(\frac{x}{2}\right)-1}{\sqrt 3}\right)\;-\;\tan^{-1}\left(\frac{2\tan\left(\frac{x}{2}\right)+1}{\sqrt 3}\right)\right)-24\cdot\cos x\;\right)}{(7+9\cdot\cos \;(2x)\; )}\;\right]_{0}^{\frac{\pi}{2}}$

i don't think i like my solution in $ (B)$ anymore !  :dry: will i get the
right answer eventually, even that i do [u]not[/u] know...  :|

misan, you know at least twice as much as I do but i'd like to point out the sum of a geometric series is a standard result :P

Of course you get the right answer.

The substitution $ \cos x=y$ shows that

$ 2\int_{0}^{\pi/2}\frac{8\sin x+2\sin^{3}x}{(4-\sin^{2}x)^{2}}dx=4\int_{0}^{1}\frac{5-y^{2}}{(3+y^{2})^{2}}dy=\frac{4}{3}+\frac{2\sqrt{3}\pi}{27}$.


LaTeX
complex analysis
complex analysis unsolved

download file to see the problem...  :arrow:

[quote="cezar lupu"]Jeane,why don't you post it here on the forum?Do you know $Latex$? ;)[/quote]
[quote="cezar lupu"]Cezareee...i  am a new member and i don't know yet $Latex$. I promise i'll learn it. OPEN YOUR MOBILE PHONE pls....  :D [/quote]


algebra
polynomial
superior algebra
superior algebra unsolved

This might be obvious but could someone explain to me the difference between
Q[sqrt(2),sqrt(3)] and Q(sqrt(2),sqrt(3)) ie the parentheses and the brackets.
The first should be the field of polynomials over Q but what is the representation exactly for both. I am getting confused here.

Thank you

The first object is the ring generated by rational numbers, $ \sqrt {2}$, and $ \sqrt {3}$. Its elements are $ P(\sqrt {2},\sqrt {3})$, where $ P$ is a polynomial with rational coefficients.
The second one is the field generated by rational numbers, $ \sqrt {2}$, and $ \sqrt {3}$. Its elements are $ R(\sqrt {2},\sqrt {3})$, where $ R \equal{} P/Q$, and $ P$ and $ Q$ are polynomials with rational coefficients.

And unless I completely forgot my algebra, they are actually the same thing, because you can rationalize the denominator.


abstract algebra
superior algebra
superior algebra unsolved

Let $ R$ be a ring and $ r\in R$ a central element.
Let $ M,N$ be two left $ R$-modules. The element $ r$ defines $ R$-module homomorphisms $ \alpha_r: M\to M, \alpha_r(m)\equal{}rm$ and $ \beta_r: N\to N, \beta_r(n) \equal{} rn$.
If $ \alpha_r$ is an isomorphism and $ \beta_r\equiv 0$ prove that $ Ext_R^1(N,M)\equal{}0$.

$ 0$ is an iso on $ \mathrm{Ext}$, so $ \mathrm{Ext}$ is $ 0$.


On a standard 12-hour clock, through how many positive degrees does the hour hand move from 9:15 a.m. to 12 noon of the same day? Express your answer as a decimal to the nearest tenth.

Measure the angle between the hour hand at $ 9: 30$ and the hour hand at $ 12: 00$.
$ 30(3)\minus{}\frac{15}{60}(30)\equal{}\boxed{82.5^{\circ}}$


linear algebra
matrix
algebra
polynomial
complex numbers
linear algebra unsolved

If a matrix $ A$, with all entries all complex numbers, statisfies $ A^3\equal{}\minus{}A$, prove or disprove that $ A$ is diagonalizable over $ C$, where $ C$ is the set of complex number. Thanks in advance.

More general question: suppose that $ P(x) \equal{} (x \minus{} r_1)(x \minus{} r_2)\cdots(x \minus{} r_k)$ where $ r_i\ne r_j.$ (That is, suppose $ P$ can be factored completely into [i]distinct[/i] linear factors.) Suppose $ P(A) \equal{} 0.$ Is $ A$ necessarily diagonalizble? (In this particular case, $ P(x) \equal{} x^3 \plus{} x \equal{} x(x \minus{} i)(x \plus{} i).$)

A start: what does this tell you about the minimal polynomial for $ A?$

Is $ x^3\plus{}x$ is the minimal polynomial for the matrices ? Thanks in advance.

The minimal polynomial divides $ x^3\plus{}x.$ It might - or might not  - have lower degree.


(Problem 6 on page 36) Prove that, for any real numbers [tex]x,y, z[/tex],

[tex] 3(x^2-x+1)(y^2-y+1)(z^2-z+1)\geq (xyz)^2+xyz+1.[/tex]

My inequalities skills are somewhat lacking, but I believe that:

(x^3+1)/(x+1) = x^2 - x + 1 would be of help here.

I have a brute force solution, expressing the difference as a sum of squares.

2*(LHS-RHS) = (x-1) :^2: (y-1) :^2: (z-1) :^2:  + (x :^2: +1)((y-1) :^2: (z-1) :^2: +(y-z) :^2: ) + (y :^2: +1)((z-1) :^2: (x-1) :^2: +(z-x) :^2: ) + (z :^2: +1)((x-1) :^2: (y-1) :^2: +(x-y) :^2: )

The only equality case is x=y=z=1.
It took two pages of algebra to get there- I hope there's a nicer solution, but this may be the best; it's a symmetric polynomial of degree 2 in each variable, so it has to be a sum of squares for the inequality to work.

$3(x^2-x+1)(y^2-y+1)(z^2-z+1)\geq (xyz)^2+xyz+1$ ------- (1)

(1) is weaker than (2):

$\displaystyle
((xyz)^2 + xyz +1)^2 \leq \frac{9}{8}(x^4+1)(y^4+1)(z^4+1) $ ------- (2)

since $\displaystyle \frac{x^4+1}{2}\leq (x^2-x+1)^2$ [See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13532&highlight=simple+inequality]here[/url]].

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6366]here[/url] for a discussion of (2).


[quote="Yuriy Solovyov"]If $ a,b,c,d > 0$ prove, that
$ \frac{bcd}{a^{3}+3bcd}+\frac{acd}{b^{3}+3acd}+\frac{abd}{c^{3}+3abd}+\frac{abc}{d^{3}+3abc}\leq 1$[/quote]

The inequality is equivalent to:
$ \sum_{sym}\frac{a^{3}}{a^{3}+3bcd}\ge 1$
We have:
$ \sum_{sym}\frac{a^{3}}{a^{3}+3bcd}\ge \sum_{sym}\frac{a^{3}}{a^{3}+b^{3}+c^{3}+d^{3}}=1$


Prove $1\cdot 2 \cdot 3 \cdots (p-2) \equiv 1 \mod p$ where p is prime.

[hide]$(p-2)! \equiv 1 \bmod p$. By Wilson's theorem $(p-2)!(p-1) \equiv-1 \bmod p$, from which the thesis.[/hide]

for each $x \in \mathfrak{T}=[2;p-2]$ there is only one $\sigma(x)=x^{-1}(\mod p)$ ($\sigma: \mathfrak{T}\to \mathfrak{T}$) (it follows from the Bezout's theorem for linear cominations and continue divisions); so let's take two factors such that their product is $\equiv 1 (\mod p)$ and multiply by 1: it will give $\equiv 1^{\frac{p-3}{2}+1}\equiv 1 (\mod p)$


How many pairs of positive integers $ (a,b)$ with $ a\le b$ satisfy $ \frac{1}{a}\plus{}\frac{1}{b}\equal{}\frac{1}{6}$

[hide]

Multiply by $ 6ab$ to clear the fractions.

$ 6a \plus{} 6b \equal{} ab$

Rearranging and factoring using SFFT

$ (a \minus{} 6)(b \minus{} 6) \equal{} 36$

We have (7, 42); (8, 24); (9, 18); (10, 15); (12, 12). 5 pairs.

[/hide]

well it is equivalent to (a+b)/(ab)=1/6
so 6a+6b-ab=0

factoring that so that you get the SSFT form, you get (a-6)(b-6)=36

36 is as products of two numbers

(1,36), (2,18), (3,12), (4,9), (6,6)

then (a,b) become (7,42), (8,24), (9,18), (10,15), (12,12)

so thats 5 total.

there are also 4 others in which b<a ((6,6) is redundant), and 8 negative ones

[hide]
$ ab\equal{}6a\plus{}6b$

$ (a\minus{}6)(b\minus{}6)\equal{}36$

$ (a\minus{}6,b\minus{}6)\equal{}(1,36),(2,18),(3,12),(4,9),(6,6)$

Answer is $ 5$.
[/hide]

EDIT: Too late.  But here is another (harder) way:
[hide]
Clearly $ 6< a\leq 12$.  So find $ \frac{1}{6}\minus{}\frac{1}{7}$, $ \frac{1}{6}\minus{}\frac{1}{8}$, and so on.  Of the $ 6$ possibilities, only $ \frac{1}{6}\minus{}\frac{1}{11}$ fails to reduce to a unit fraction.  So the answer is $ 5$.
[/hide]

thanks everyone,
now can someone please respond to #10, its really hard...
note: there aren't solutions on the hmmt website so you can't cheat ha!


[quote="zaya_yc"]$ \frac {a}{x} \plus{} \frac {b}{x} \plus{} \frac {c}{z} \equal{} 0$[/quote]

I hope you mean $ \frac {a}{x} \plus{} \frac {b}{y} \plus{} \frac {c}{z} \equal{} 0$

[hide="if so..."] We have $ \frac {a}{x} \plus{} \frac {b}{y} \plus{} \frac {c}{z} \equal{} 0$.  Multiplying this equation by $ \frac{xyz}{abc}$ gives

$ \frac {yz}{bc} \plus{} \frac {xz}{ac} \plus{} \frac {xy}{ab} \equal{} 0$

Also, squaring the first equation yields

$ \frac{x^2}{a^2}\plus{}\frac{y^2}{b^2}\plus{}\frac{z^2}{c^2}\plus{}2\left(\frac{xy}{ab}\plus{}\frac{yz}{bc}\plus{}\frac{xz}{ac}\right)\equal{}1$

Since $ \frac {yz}{bc} \plus{} \frac {xz}{ac} \plus{} \frac {xy}{ab} \equal{} 0$, we have

$ \frac{x^2}{a^2}\plus{}\frac{y^2}{b^2}\plus{}\frac{z^2}{c^2}\equal{}1$[/hide]

[hide]Let $ \frac{x}a\equal{}A,\frac{y}b\equal{}B,\frac{z}c\equal{}C$. Hence $ A\plus{}B\plus{}C\equal{}1$. $ \frac1A\plus{}\frac1B\plus{}\frac1C\equal{}\frac{AB\plus{}BC\plus{}CA}{ABC}\equal{}0$, which gives $ AB\plus{}BC\plus{}CA\equal{}0$. Thus $ A^2\plus{}B^2\plus{}C^2\equal{}(A\plus{}B\plus{}C)^2\minus{}2(AB\plus{}BC\plus{}CA)\equal{}1$. Q.E.D.[/hide]


$ R$ - commutative ring  with $ 1\neq 0$.
let  $ h(x)\equal{}r_nx^n\plus{}\dots \plus{} r_1x\plus{}r_0$ be a zero-divisor in $ R[X]$.
prove that there exists nonzero $ s\in R$ s.t. $ sh\equiv 0$

[quote="knoppix"]$ R$ - commutative ring  with $ 1\neq 0$.
let  $ h(x) \equal{} r_nx^n \plus{} \dots \plus{} r_1x \plus{} r_0$ be a zero-divisor in $ R[X]$.
prove that there exists nonzero $ s\in R$ s.t. $ sh\equiv 0$[/quote]
Suppose $ f(x) \equal{} a_nx^n \plus{} ... \plus{} a_0$ and $ g(x) \equal{} b_kx^k \plus{} ... \plus{} b_0$ so that $ f(x),g(x)$ are non-zero polynomials.
And that $ f(x)g(x) \equal{} 0$ so multiplying out coefficients we get $ a_nb_k^{x \plus{} k} \plus{} ... \equal{} 0$ so it means $ a_nb_k \equal{} 0$ but $ a_n$ and $ b_k$ were non-zero. So $ R$ has zero divisors.

EDIT: Yeah, what I wrote is a different problem. Sorry.

[quote="sylow_theory"][quote="knoppix"]$ R$ - commutative ring  with $ 1\neq 0$.
let  $ h(x) \equal{} r_nx^n \plus{} \dots \plus{} r_1x \plus{} r_0$ be a zero-divisor in $ R[X]$.
prove that there exists nonzero $ s\in R$ s.t. $ sh\equiv 0$[/quote]
Suppose $ f(x) \equal{} a_nx^n \plus{} ... \plus{} a_0$ and $ g(x) \equal{} b_kx^k \plus{} ... \plus{} b_0$ so that $ f(x),g(x)$ are non-zero polynomials.
And that $ f(x)g(x) \equal{} 0$ so multiplying out coefficients we get $ a_nb_k^{x \plus{} k} \plus{} ... \equal{} 0$ so it means $ a_nb_k \equal{} 0$ but $ a_n$ and $ b_k$ were non-zero. So $ R$ has zero divisors.[/quote]

Read the problem again. ;)

*thinking about a proof on less than 10 pages at this moment*

  darij

let $ g\equal{}b_m x^m \plus{} ... \plus{} b_0$ some non-zero polynomial of smallest degree $ m$ that annihilates $ h\equal{}r_n x^n \plus{} ... \plus{} r_0$. if $ m\equal{}0$, we're done. otherwise $ m \geq 1$. this yields $ b_m h \neq 0$, thus $ b_m r_i \neq 0$ for some $ i$ and therefore $ r_i g \neq 0$. choose some maximal $ j$ with $ r_j g \neq 0$. now we have $ 0\equal{}hg\equal{}(r_j x^j \plus{} ... \plus{} r_0)g$, so $ r_j b_m \equal{} 0$. but then $ r_j g$ has degree smaller than $ m$ and annihilates $ h$, a contradiction.

Can someone construct an example of $ R[x]$?

[quote="ZetaX"]Example for what¿[/quote]
A ring that has property like [b]knoppix[/b] describes.

He only required it to be a commutative ring with unity, right¿ Or do I miss something¿

this is one of the first exercises in "Introduction to commutative algebra" by M.F. Atiyah I.G. Macdonald.
It is true in $ R[x_1,x_2,...,x_n]$ for the same reasons.

[quote="knoppix"][quote="-oo-"][...]and therefore $ r_i g \neq 0$. [...][/quote]
maybe it's trivial, but could you explain why?[/quote]
be careful! $ r_i g\equal{}0$ would mean $ b_mr_i\equal{}0$ contradiction :wink:

$ r_ig\equal{}0$ means $ r_ib_mx^m\plus{}r_ib_{m\minus{}1}x^{m\minus{}1}\plus{}\dots\plus{}r_ib_0\equal{}0$
you mean that then all coef's need to be 0's ?
I think this isn't necessary, if our ring would be f.e. $ \mathbb{Z}_2$ then for example poly $ X^2\plus{}X\equal{} 0$ ,although it has nonzero coef's

You mistake polynomials for the corresponding function. $ x^2\plus{}x$ is not zero over $ \mathbb F_2$.

oops :blush:
sorry for the last question :wink:

everything clear, thanks !

eagle's remark for $ R[x_1,...,x_n]$ follows by an easy induction.
@sylow: $ 2X^2$ is a zero divisor in $ \mathbb{Z}/4[x]$, or what are you lookin' for?


geometry
geometric transformation
homothety
geometry unsolved

the circles of centers O and A are tangent at B, the circles O and F are tangent at G. A chord CD of the circle O is tangent to the circle A at E and tangent to the circle F at H. BE and DG meet at a point P and BC and GH meet at Q. Prove that PQ and CD are parallel. Can I have your nice short solutions ?  :)

Here's one way.

The angle between $ DG$ and $ DB$ is $ 90^{\circ}$.  The angle between $ BD$ and $ BE$ is the sum of arcs $ DB$ and $ BE$.  Similarly, the angle between $ BC$ and $ GH$ is $ 90^{\circ}$ plus the sum of arcs $ CG$ and $ GH$.  Let circle $ A$ meet the line of centers again at $ K$.  By homothecy, arcs $ GH$ and $ KE$ are congruent.  We also have $ \widehat{BE}\minus{}\widehat{EK}\equal{}\widehat{GD}\minus{}\widehat{CB}$.  Thus, it follows $ m\angle{GPB}\equal{}m\angle{GQB}$, and so $ GB$ and $ PQ$ are antiparallel.  This completes the proof.


I have this really nasty virus right now and I can't seem to get rid of it. Here are some of my symptoms. Utterly random websites open at a rate of about 10 minues. My keyboard randomly decides not to work. I get a whole bunch of pop-up ads that are ironically for anti-virus programs. Finally this happened. I decided to make a youtube video of it:
[youtube]VExNOjztgVA[/youtube]

http://www.youtube.com/watch?v=VExNOjztgVA

Hi, please download Trend Micro's HijackThis v2.0.2, run it, and post a log.

Well if this occurred fairly recently, then I would suggest running a system restore. (check in your acessories)
If this doesnt work then get some good antivirus and run it in safe mode.

For future reference dont use IE7 because it is the most targetted browser for hacks and exploits.

-I'm assuming you are on Windows.

this happened to me before. what i did was i nuked my hard drive worked perfectly
used my backup os cd to reinstall xp

Sounds like a serious [url=http://en.wikipedia.org/wiki/Trojan_horse_(computing)]trojan[/url]. I'd get Spyhunter

You have gotten a bad, bad case of trojans, adware and possibly, spyware. Remove anything important and personal. If there is a credit card number, or stuff like that, remove it. The best thing to do is format your computer and reinstall your OS again. If that isn't possible, get Spyware Doctor or Adaware. It will help remove quite a lot of the infections. Dont even download the anti-viruses from the ads. They are viruses themselves. I did that and I paid for it.


For [i]a[/i]and [i]n[/i]natural numbers, let [i]k[/i]be the greatest natural number such that $n>ka$. If we define $n \Rightarrow a = n(n-a)(n-2a)...(n-ka)$, then what is $\frac {72\Rightarrow8}{18\Rightarrow2}$?

A) $4^5$
B) $4^7$
C) $4^8$
D) $4^9$
E) None of above


I dont know if this question sould be in MATHCOUNTS or somethng,so mods,do ur stuff if u feel like it....

If a^2+b^2+c^2=ab+bc+ca,FInd what kind of triangle can a,b,c be sides of?

Use the inequality $a^2+b^2+c^2 \geq ab+bc+ac$.

HOw about not usin that and just usin simple 8th grade maths?Just use whats given,its simple...no inequality stuff...

[quote="shobber"]The proof of the inequality is also of middle school math.

$2a^2+2b^2+2c^2=2ab+2ac+2bc$
$\Longleftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=0$
$\Longleftrightarrow (a-b)^2+(b-c)^2+(a-c)^2=0$

Do you see it now?  :)[/quote]
$(a-b)^2\geq0, (b-c)^2\geq0, (c-a)^2\geq0$ and $(a-b)^2+(b-c)^2+(c-a)^2=0$

$\Leftrightarrow (a-b)^2=(b-c)^2=(c-a)^2=0$

$\Leftrightarrow a=b=c$

Do you understand his solution now?

[quote="riddler"]Can't you the sum like this?
$=a^2+b^2+c^2=ab+ac+bc$
which imlies either that $a^2=ab$ or $a^2=ac$ or $a^2=bc$
So $a=b=c$ on simlification???? :agent:[/quote]

Not nessecary , but the idea behind is if $x^2+y^2+z^2=0$ then we can say that $x=y=z=0$ . (or else say $x>0$ will result in $x^2+y^2+z^2>0$ which contradict with the equation .)

 :)

[quote="riddler"]what do you do now? :(  :?[/quote]

C'mon riddler,its too simple from there...Same method i used shobber...I didnt understyand what u meant by use the inequality stuff,i just multiplied both the sides by 2 and ta-dah,there the answer was!! :lol:


function
induction
algebra unsolved
algebra

Please find all functions $f: \mathbb{R}\to \mathbb{R}$ with $f(x)+f(x-1)=x^{2}$ and $f(1)=1$.

Thanks a lot

for all $t$ such that $0<t<1$ let $f(t)$ be arbitrary real value. then by induction calculate $f(t+n)$ for any integer $n$.since you know that $f(1)=1$ easy induction gives you $f(n)=\frac{1}{2}n(n+1)$.
any such function satisfies.many solutions.


trigonometry
inequalities
geometry proposed
geometry

Prove that

$ \sum\cos \frac{A}{2}\geq2\sum\frac{\sqrt{s(s\minus{}a)}}{b\plus{}c}$

.

use $ cosA/2 \equal{} \sqrt {\frac {s(s \minus{} a)}{bc}}$ . By AM-GM $ cosA/2 \ge 2 \frac {\sqrt {s(s \minus{} a)}}{b \plus{} c}$

The cyclic summation gives the inequality  :)


geometry
circumcircle
trigonometry
parallelogram
rectangle
cyclic quadrilateral
power of a point

$ABCD$ is a cyclic quadrilateral with $AC \perp BD$; $AC$ meets $BD$ at $E$. Prove that \[ EA^2 + EB^2 + EC^2 + ED^2 = 4 R^2 \]
where $R$ is the radius of the circumscribing circle.

Let O be the quadrilateral circumcenter. The power of the point E to the circumcircle (O) is

$EA \cdot EC = EB \cdot ED = (R - EO)(R + EO) = R^2 - EO^2$

Let M, N be the midpoints of AC, BD. WLOG, assume $EA \le EC, EB \le ED$, otherwise relabel the quadrilateral.

$EO^2 = EN^2 + EM^2 = (MA - EA)^2 + (NB - EB)^2 =$

$= \left(\frac{AC}{2} - EA\right)^2 + \left(\frac{BD}{2} - EB\right)^2 = \left(\frac{EA + EC}{2} - EC\right)^2 + \left(\frac{EB + ED}{2} - EB\right)^2 =$

$= \frac{(EC - EA)^2}{4} + \frac{(ED - EB)^2}{4} = \frac{EA^2 + EB^2 + EC^2 + ED^2}{4} - \frac{EA \cdot EC + EB \cdot ED}{2} =$

$= \frac{EA^2 + EB^2 + EC^2 + ED^2}{4} - (R^2 - EO^2)$

$EA^2 + EB^2 + EC^2 + ED^2 = 4R^2$

Another soln

Let a,b,c,d = EA,EB,EC,ED

we can calculate the circumradius of ABD as  $\frac{\sqrt{(a^2 + b^2)(a^2 + d^2)(b+d)^2}}{2(b+d)a}$ using the classic $R = \frac{abc}{4\Delta}$ formula.

Then $4R^2 = \frac{(a^2 + b^2)(a^2 + d^2)}{a^2} = a^2 + b^2 + d^2 + \frac{b^2d^2}{a^2}$.  From power of a point, bd = ac so this becomes $4R^2 = a^2 + b^2 + c^2 + d^2$.

$SUM = (EA^2+EB^2)+(EC^2+ED^2) = AB^2+CD^2$

Let $w$ the circumcircle of $ABCD$

$L \in w, AL \parallel BD\Rightarrow$

$AB = LD$

$\widehat{CAL}=90^0 \Rightarrow CL$ is a diameter $\Rightarrow$

$\widehat{CDL}=90^0$

So the triangle $CDL$ is right

Finally,

$SUM = AB^2+CD^2 = LD^2+CD^2 = CL^2 = (2R)^2$

By the way, if the line $AL$ is tangent, then all above is correct if we replace $L$ by $A$.

($OA \perp BD , AC \perp BD \Rightarrow  AC$ is diameter
Now we have $AB = AD \Rightarrow$ $SUM = AB^2+CD^2 = AD^2+CD^2 =AC^2$)

This is just idea.
From pithagora theorem we obtain AE^2+BE^2+CE^2+DE^2=(AB^2+BC^2+CD^2+AD^2)/2
Then if we apply cosine rule to triangles AOB,BOC,COD,AOD and then sum it we get:
AB^2+BC^2+CD^2+AD^2=2R^2(4-cosAOB-cosBOC-cosCOD-cosDOA)
I think that its not hard to prove that cosAOB+cosBOC+cosCOD+cosDOA=0 where AOB+BOC+COD+DOA=360(these are angles)

Here is another simple solution:

We have: \[EA^2+EB^2+EC^2+ED^2=\frac{a^2+b^2+c^2+d^2}{2}\] so we want to prove that: $a^2+b^2+c^2+d^2=8R^2$.  We have from the Law of Sines:\[ \{\begin{array}{ccc
}a=2R\sin{\angle{ACB}}\\\\b=2R\sin{\angle{BCD}}\\\\c=2R\sin{\angle{DBC}}=2R\cos{\angle{ACB}}\\\\d=2R\sin{\angle{ACD}}=2R\cos{\angle{BCD}} \end{array}\] thus: \[a^2+b^2+c^2+d^2=4R^2(\sin^2{\angle{ACB}}+\cos^2{\angle{ACB}}+\sin^2{\angle{BCD}}+\cos^2{\angle{BCD}})=4R^2*2=8R^2\] as desired.

Indeed, the [u]shortest proof[/u] of this simple problem is [b]basketball9[/b]'s. For $x=m(\angle ADB)$ obtain  $AB=2R\sin x$ $CD=2R\cos x$ .

Thus, $EA^2+EB^2+EC^2+ED^2=$ $\left(EA^2+EB^2\right)+\left(EC^2+ED^2\right)=$ $AB^2+CD^2=$ $4R^2\left(\sin^2x+\cos^2x\right)=4R^2$ .

However, I"ll present here some remarkable properties of the cyclical ortodiagonal quadrilaterals. Denote :

$\left|\begin{array}{cccc}
AB=a & BC=b & CD=c\\\
DA=d & AC=e & BD=f\end{array}\right|\ ;$ the midpoints $\{M,N,P,Q,U,V\}$ of the sides and the diagonals

$\left\{[AB],[BC],[CD],[DA],[AC],[BD]\right\}$ respectively ; the projections $\{X,Y,Z,T\}$ of the point $E$ on the sides

 $\left\{[AB],[BC],[CD],[DA]\right\}$ respectively ; $G\in MP\cap NQ$ ; the circumcircle $w=C(O,R)$

of the given cyclical quadrilateral $ABCD$ for which $AC\perp BD$ . Prove easily that the following remarkable properties :

$\blacktriangleright\ EA^2+EB^2+EC^2+ED^2=AB^2+CD^2=AD^2+BC^2=4R^2\ .$

$\blacktriangleright\ AB^2+BC^2+CD^2+DA^2=AC^2+BD^2+4\cdot OE^2=8R^2\ .$

$\blacktriangleright\ \left\{\begin{array}{ccc}
X\in PE & OM=PE=\frac c2\\\\
Y\in QE & ON=QE=\frac d2\\\\
Z\in ME & OP=ME=\frac a2\\\\
T\in NE & OQ=NE=\frac b2\end{array}\right|\implies$ the points $\{M,N,P,Q,X,Y,Z,T\}$ are concyclically ([b][u]circle of eight points[/u][/b] - $\beta$).

$\blacktriangleright\ OMEP$ and $ONEQ$ are parallelograms $\implies$ the centroid $G$ of the quadrilateral $ABCD$  is midpoint of $[OE]$ and the center of $\beta$ .

With other words, the $E$-median in $\triangle AEB$ is the $E$-altitude in $\triangle CED$ and the $E$-median in  $\triangle CED$ is the $E$-altitude in $\triangle AEB$ a.s.o.

Thus, the points $X$ , $Z$ belong to the circle with diameter $MP$ and the point $Y$ , $T$ belong analogously to the circle with diameter $NQ$ .

The quadrilateral $MNPQ$ is a rectangle with the center in $G\in MP\cap NQ$ . In conclusion, the points $\{X,Y,Z,T\}$ belong to the

circumcircle of the rectangle $MNPQ$ .

By pythagorus theorem we get to prove-$ AB^2+CD^2=4R^2$
now, in $\Delta ABC$ $\frac{AB}{sin BCA}$=$2R$=$\frac{CD}{sin CAD}$ in $\Delta ACD$ but cad=cbd=90°-bca so $sin CAD =cos BCA$
 squaring and adding the above equations we get
$AB^2+CD^2=4R^2(sin^2CAD+cos^2CAD)$
which proves it.

Let $O$ be the centre of the circle, Let $\angle AOB =x $ and Let $\angle COD =y$
Now, since, $\angle BCE=\frac{x}{2}$ and $\angle CBE = \frac{y}{2} \implies x+y=180^{\circ}$
Therefore, $$AB^2=2R^2-2R^2\cos x  \text{ and } CD^2=2R^2-2R^2\cos y= 2R^2+2R^2\cos x$$
Hence, $$EA^2+EB^2+EC^2+ED^2=AB^2+CD^2= 4R^2$$
Proved!!

[quote=Math.Is.Beautiful]https://youtu.be/mZaeingxjh4
I think this can now be easily formalized[/quote]

Elegant application of the constant inscribed angle given a fixed chord length.


inequalities
function
inequalities unsolved

Let $x_1,\ldots,x_n\geq0$. For $m\leq n$ suppose $a_1,\ldots,a_m\geq0$ are reals with $\sum a_i=k$. For this partition $\alpha = (a_1,\ldots,a_m)$ of $k$, recall the Muirhead definition of the symmetric function $[\alpha]$ of $x_1,\ldots,x_n$: Put $a_{m+1}=\cdots=a_n=0$ and define
$[\alpha]: = [a_1,\ldots,a_m] : = [a_1,\ldots,a_n]: = \sum_{\sigma\in S_n} x_1^{a_{\sigma(1)}} \ldots x_n^{a_{\sigma(n)}}$.
For $0\leq r\leq \frac k2$ consider the function $f(r): = [k-r,r]$. Muirhead tells us that $f$ is decreasing. Prove that moreover $f$ is [i]concave[/i], i.e. for $0\leq r< s\leq \frac k2$ and $0< h<s-r$ the ineq $f(r)+f(s)\geq f(r+h)+f(s-h)$ holds.

Nobody?
Well OK it is really straightforward. Just factorize, $f(r)+f(s)\geq f(r+h)+f(s-h)$ is equivalent to
$(x^h-y^h)(x^{k-s-r}-y^{k-s-r})(x^{s-r-h}+y^{s-r-h})\geq0$ as the two first factors have the same sign.
In between I have found a more general concavity, see http://www.mathlinks.ro/Forum/viewtopic.php?p=384471#p384471 :)


For positive integers $ n$, $ S(n)$ denotes the sum of its digits of $ n$,for instance $ S(611) \equal{} 6 \plus{} 1 \plus{} 1 \equal{} 8$.
Let $ M$ be the maximum value of $ S(a) \plus{} S(b) \plus{} S(c)$ for three-digits positive integers $ a,b,c$ such that $ a \plus{} b \plus{} c \equal{} 2005$.

How many triplets $ (a,b,c)$ such $ S(a) \plus{} S(b) \plus{} S(c) \equal{} M$ are there?

(However ,for instance,$ (900,900,205)$ and $ (900,205,900)$ are treated as [b]distinct[/b] triplets)

[hide]
Let the three integers be $ 100a_1 \plus{} 10a_2 \plus{} a_3$, $ 100b_1 \plus{} 10b_2 \plus{} b_3$, and $ 100c_1 \plus{} 10c_2 \plus{} c_3.$

Since their sum is 2005, we have $ 100(a_1\plus{}b_1\plus{}c_1) \plus{} 10(a_2\plus{}b_2\plus{}c_2) \plus{} (a_3\plus{}b_3\plus{}c_3) \equal{} 2005$.

We let $ d_i \equal{} a_i \plus{} b_i \plus{} c_i.$ Then we have $ 100d_1 \plus{} 10d_2 \plus{} d_3 \equal{} 2005$ and we want to maximize $ d_1\plus{}d_2\plus{}d_3$.

My solution is a little ugly here, it consists of listing out all possible ordered triples $ (d_1, d_2, d_3)$ using that since each digit must be less than or equal to 9, $ d_i \le 27.$

Using that, the possible ordered triples $ (d_1, d_2, d_3)$ are:

(18,20,5)
(18,19,15)
(18,18,25)
(19,10,5)
(19,9,15)
(19,8,25)
(20,0,5)

and the ordered triple that minimizes d_1 + d_2 + d_3 is (18,18,25). We now use balls in urns several times to get:

the number of ways to choose the hundreds digits is $ \binom{17}{2}$ (hundreds digit can't be 0)
the number of ways to choose the tens digit is $ \binom{20}{2}$
the number of ways to choose the hundred digit is $ \binom{27}{2}$

and we get the desired answer by multiplying these numbers.

[/hide]

Are you sure? I get the final answer as 110 after brute force.

[quote="pythag011"][hide]
Let the three integers be $ 100a_1 \plus{} 10a_2 \plus{} a_3$, $ 100b_1 \plus{} 10b_2 \plus{} b_3$, and $ 100c_1 \plus{} 10c_2 \plus{} c_3.$

Since their sum is 2005, we have $ 100(a_1 \plus{} b_1 \plus{} c_1) \plus{} 10(a_2 \plus{} b_2 \plus{} c_2) \plus{} (a_3 \plus{} b_3 \plus{} c_3) \equal{} 2005$.

We let $ d_i \equal{} a_i \plus{} b_i \plus{} c_i.$ Then we have $ 100d_1 \plus{} 10d_2 \plus{} d_3 \equal{} 2005$ and we want to maximize $ d_1 \plus{} d_2 \plus{} d_3$.

My solution is a little ugly here, it consists of listing out all possible ordered triples $ (d_1, d_2, d_3)$ using that since each digit must be less than or equal to 9, $ d_i \le 27.$

Using that, the possible ordered triples $ (d_1, d_2, d_3)$ are:

(18,20,5)
(18,19,15)
(18,18,25)
(19,10,5)
(19,9,15)
(19,8,25)
(20,0,5)

and the ordered triple that minimizes d_1 + d_2 + d_3 is (18,18,25). We now use balls in urns several times to get:

the number of ways to choose the hundreds digits is $ \binom{17}{2}$ (hundreds digit can't be 0)
the number of ways to choose the tens digit is $ \binom{20}{2}$
the number of ways to choose the hundred digit is $ \binom{27}{2}$

and we get the desired answer by multiplying these numbers.

[/hide][/quote]

Are there really $ \binom{27}{2}$  triples $ (c_1,c_2,c_3)$ such $ c_1\plus{}c_2\plus{}c_3\equal{}25$ and $ 0\le c_1,c_2,c_3\le 9$?

For $ a_i,b_i$ as well.


ba dar nazar gereftane yek graphe jahatdare ghaviaaan hamband $D$ farz konid $f(D)$ toole kootahtarin gashte basteye molaghatkonandeye har raas bashad.. sabet konid rooye hameye graphhaye jahatdare ghaviaan hamband ba $n$ ras $max f(D)=[ \frac{(n+1)^{2}}{4}]$ $(n \geq 2)$([ ]>>> kaf)

be nazaram ba esteghra betoonim ye balayi saresh biyarim :

chon graph ghavian hambandeh pas ye dore jahat dar dare oon doro ye ras begirim va har rasi khareje in dor ke be rashaye in dor vasle ro ba yek yal ba hamoon jahat be oon rase jadid vasl konim va baraks be vozooh ke ghavian hamban mimoneh. bad esteghra bezanim . nazare to chiye sepehr.

roo chi mikhay esteghra bezani??be tedade yalha tavajooh konaa age yal ezafe mikoni!!

khob chon har dor hadeaghal 3 ras dare va ma be jaye oon rasha ye ras mizarim pas tedade rasha kam mishe pas rooye n esteghra mizanim

khube bad niss beshin fekr kon vali rahe rahat tararam dare!!!

khob harki soale (+)dare west(makhsoosan age akharin soale fasle khodesh bashe) ro bebineh roosh sakht fekr mikoneh!!!

are khob manam didam soal ghashange gozashtam roosh fekr konid!!! :!:


geometry
circumcircle
parallelogram
geometry proposed

A result about mixtilinear incircle:
http://www.mathlinks.ro/viewtopic.php?t=186117

[quote]Let $ M$ be an arbitrary point on the circumcircle of $ ABC$ and let the tangents from this point to the incircle of the triangle meet the side $ BC$ at $ X_{1}$, and $ X_{2}$. Prove that the second intersection of the circumcircle of triangle $ MX_{1}X_{2}$ with the circumcircle of $ ABC$ (different from $ M$) is fixed and it coincides with the tangency point of the mixtilinear incircle in angle $ A$ with the circumcircle.

Note: The [b]mixtilinear incircle[/b] in angle $ A$ is the circle tangent to the sides $ AB$, $ AC$ and to the circumcircle of $ ABC$ internally.[/quote]

[i]Proof[/i]

Let $ (K_a)$ be the [i]mixtilinear circle[/i] of $ \triangle ABC$ wrt $ A$. Denote $ B',C',A'$ respectively by the tangency points of $ (K_a)$ with $ AB,AC$ and the circumcircle $ (O)$ of $ \triangle ABC$. Let $ A_1,B_1,C_1$ respectively be the tangency points of $ (I)$ with $ BC,CA,AB$. Consider the inversion through pole $ I$, power $ k \equal{} r^2$, we have $ \mathcal {I}(I,k):$ $ B\mapsto M_b$, $ C\mapsto M_c$ and $ A\mapsto M_a$, where $ M_a,M_b,M_c$ respectively are the midpoints of $ B_1C_1,C_1A_1,A_1B_1$, thus $ (O)\mapsto (M_aM_bM_c)$- which is the $ 9 \minus{}$ point circle wrt $ \triangle A_1B_1C_1$. Therefore if $ A''$ is the image of $ A'$ through $ \mathcal {I}(I,k)$, $ A''\in (M_aM_bM_c)$ $ (*)$. We will prove that $ A''$ will be the orthocenter of $ \triangle A_1M_bM_c$ $ (1)$.

Indeed, let $ B''$ and $ C''$ respectively be the image of $ B'$ and $ C'$ through $ \mathcal {I}(I,k)$. Since $ B'IA'B$ is concylcic quadrilateral (Can be proved easily by angle chasing). Therefore, we conclude $ B'',M_b,A''$ are collinear. Thus $ \angle IM_bA'' \equal{} \angle IC_1B''$ $ \equal{} \angle C_1IA$ $ \equal{} 180^{\circ} \minus{} \angle M_cIM_b$, which implies that $ M_bA''\|IM_c$ $ \Longrightarrow$ $ M_bA''\perp A_1M_c$ $ (**)$. Combine $ (*)$ and $ (**)$, we obtain the result $ (1)$.

Now, denote $ B_2,C_2$ by the tangency points of $ MX_1,MX_2$ with $ (I)$. The inversion through pole $ I$, power $ k$, maps $ B_2, A_2$ and $ C_2$ into themselves. Then if $ \mathcal {I}(I,k):$ $ X_1\mapsto M_1$, $ X_2\mapsto M_2$, then $ M_1,M_2$ respectively will be the midpoints of $ A_1B_2$, $ A_1C_2$, furthermore,  both $ M_1,M_2\in ([IA_1])$. $ \mathcal {I}(I,k):$ $ M\mapsto M_3$. Therefore $ (MX_1X_2)\mapsto (M_1M_2M_3)$ through $ \mathcal {I}(I,k)$. Note that $ M_3$ is also the midpont of $ B_2C_2$. Hence, $ (M_1M_2M_3)$ will be also the $ 9 \minus{}$ point circle wrt $ \triangle A_1B_2C_2$. We will prove that $ A''\in (M_1M_2M_3)$ $ (2)$.

Indeed, Let $ O_1,O_2$ repsectively be the $ 9 \minus{}$ point circle center of $ \triangle A_1B_1C_1$ and $ \triangle A_1B_2C_2$. Denote $ H_c,H_b,P_c,P_b$ respectively by the projections of $ C_1,B_1,C_2,B_2$ onto $ A_1B_1,A_1C_1,$ $ A_1B_2,A_1C_2$. Then $ H_bH_c\|P_cP_b$ (since both of them $ \perp A_1I$). But since $ A'',O_1,M_a$ are collinear and $ A''M_a\perp H_bH_c$ then $ O_1M_a\equiv O_1A''\perp H_bH_c$. Similarly, $ O_2M_3\perp P_bP_c$. Therefore, $ O_1A''\| O_2M_3$ $ (3)$. In the other hand, it is easy to prove that $ R_{([AO])} \equal{} R_1$, where $ R_i$ is the radii of $ (O_i)$ $ \Longrightarrow$ $ R_1 \equal{} R_2$, which implies $ O_1A'' \equal{} O_2M_3$. Combine with $ (3)$, we conclude $ O_1A''O_2M_3$ are parallelogram. Therefore, $ O_2A'' \equal{} O_1M_3 \equal{} R_1 \equal{} R_2$ $ \Longrightarrow$ $ A''\in (M_1M_2M_3)$, we obtain the result $ (2)$.

As the result, $ A'\in (MX_1X_2)$ or we can say that $ (MX_1X_2)$ always passes $ A'$ which is fixed. Our proof is completed then. $ \square$.

Best regard
[i][b]mathVNpro[/b][/i]

Dear livetolove212 and MathVnpro
Can you help me to solve the original problem I asked.
Maybe they are not the same.

I'm sure that you didn't read that link carefully. See two topics below:
http://www.mathlinks.ro/viewtopic.php?t=186117
http://www.mathlinks.ro/viewtopic.php?t=275799

[quote="gb2124"]Given a triangle ABC.M is a point on its circumference and its tangents to the incircle of ABC meets BC at M1 and M2.
Prove that there is a fixed circle which is always tangent to the circumferece of MM1M2 when M varies.[/quote]

[b]Synthetical proof[/b] (related to [b]jayme[/b]):
We can show that $ (MM_1M_2)$ passes through a tangency of $ (O)$ and A-Mixtilinear incircle of triangle $ ABC$.
Let $ GK,LK$ be the tangents of $ (I) (G,L\in (O)$)
It's easy to show that $ GL$ is the tangent of $ (I)$ therefore $ (I)$ is the incircle of triangle $ GLK$.
Thus $ KI$ passes through the midpoint of arc $ GL$. But $ KI$ passes through the midpoint of arc $ BC$ then $ BC//GL$.
Let $ J \equal{} (MM_1M_2)\cap GK, H \equal{} (MM_1M_2)\cap LK$.
Because $ (I)$ is the excircle of $ \Delta MM_1M_2$ hence $ JH$ is tangent to $ (I)$.
Then $ (I)$ is the excircle of $ \Delta KJH$. On the other hand, $ (I)$ is the incircle of triangle $ KGL$ then from the construction of the Mixtilinear incircle and excircle, the K-Mixtilinear excircle of triangle $ KJH$ is also a K-Mixtilinear incircle of triangle $ GKL$. Or $ (MM_1M_2)$ is tangent to K-Mixtilinear incircle of triangle $ KGL$ (a fixed circle)

Nice with inversion [b]mathVNpro[/b], I will complete your sol :

Consider the inversion through pole $ I$, power $ k \equal{} r^2$, we have $ \mathcal {I}(I,k)$:

$ (MX_1X_2) \mapsto (M_1M_2M_3)$ and $ (M_1M_2M_3)$ always passes $ H$ which is fixed.
[geogebra]94bb1d945cd7011c67bc5a30aed868b7b8fb781d[/geogebra]

$ \triangle DHH' \equal{} \triangle IJM_3 \Longrightarrow HH'M_3J$ is a parallelogram $ \Longrightarrow HU \equal{} H'M_3 \equal{} HJ \Longrightarrow (M_1M_2M_3)$  is always tangent to $ (H,HJ)$ which is a fixed circle $ \Longrightarrow (MX_1X_2)$ is always tangent to a fixed circle.

[geogebra]22708c943312de2b540eb66765ce745cd6de66ee[/geogebra]


Given that \[ f(x) \equal{} \frac{1}{1 \minus{} \frac{1}{1 \minus{}  \frac{1}{1\minus{}x}}}.\] Compute $ \left(f\left(f(\minus{}2)\right)\right)^{\minus{}2}$. Express your answer as a common fraction.

after doing a lot of math, we find that f(-2)=-2. Therefore, f(f(-2)) also equals -2. So the answer is -2^-2, or 1/4.


inequalities
symmetry
inequalities unsolved

In fact I think this holds:
\[\sum x \sum \frac{1}{x}> 6(\frac{(x+y)(y+z)(z+x)}{8xyz})^{3}\ge (x+y+z)\max\{ \left(\frac{1}{x}+\frac{1}{y}\right)_{cyc}\}\]
Which is very tight for large $x,y,z$ since $RHS-LHS=\frac{1}{z}$... the middle quantity is quite sandwiched!

And I don't have proof yet :(

[quote="gemath"]Given $x,y,z>0$ prove that
$6(\frac{(x+y)(y+z)(z+x)}{8xyz})^{3}\ge (x+y+z)\max\{\frac{x+y}{xy},\frac{y+z}{yz},\frac{z+x}{zx}\}$
 :)[/quote]
I think, it's wrong: $x=1.7052$ and $y=z=1.$ :wink:

That's pity for this result, but I still hope about an inequality had the form $...\ge max\{...\}$(it needs symmetry) do you know any inequality similar it ? arqady

[quote="gemath"]... but I still hope about an inequality had the form $...\ge max\{...\}$(it needs symmetry) do you know any inequality similar it ? arqady[/quote]
For non-negative $a,$ $b$ and $c$ holds:
$a^{3}+b^{3}+c^{3}-3abc\geq\frac{\max\{a,b,c\}}{2}\cdot\sum_{cyc}(a-b)^{2}.$ :D

$WLOG$ $\sum{x}=xyz$, $\sum(x=tg{a})$ $\Rightarrow$ $\sum{a}=\pi$. $WLOG$ $a\leq b\leq c$ => our inequality <=>

$6(\frac{(tga+tgb)(tgb+tgc)(tgc+tga)}{8.tg{a}.tg{b}.tg{c}})^{3}\geq (\sum{tg{a}})\frac{tga+tgb}{tg{a}.tg{b}}$ <=> $6(\frac{1}{8.cosa.cosb.cosc})^{3}\geq \frac{sin^{2}c}{cosa.cosb.cosc}$

<=> $3\geq 4^{4}.(sinc.cosc.cosb.cosa)^{2}=4^{2}.(sin{2c}.(cos(a+b)+cos(a-b)))^{2}=$
$=4^{2}(sin2(a+b).(cos(a+b)+cos(a-b)))^{2}$. $a+b$ is 'independent' from $a-b$ => we must take $cos(a-b)=1$

=> our inequality is true iff $3\geq 4^{2}(sin2c.(1-cosc))^{2}$


probability
conditional probability
probability and stats

$ n$ points are choosen on the segment $ [0;1]$ independently and uniformly.
The segment is therefore cut onto $ (n\plus{}1)$ smaller segments.
Let's denote their lengths as $ S_{1}$, $ S_{2}$, ..., $ S_{n\plus{}1}$ (from left to right).

I know that $ S_{i}$ are identically distributed (but not independent). So $ E(S_{i})\equal{}\frac{1}{n\plus{}1}$, as $ E(S_{1}\plus{}...\plus{}S_{n\plus{}1})\equal{}1$.

I wish to find $ Var(S_{i})$ (or just $ Var(S_{1})$) [u]without[/u] explicit use of the pdf. Any help is appreciated!

Just consider what happens when you add one point to an already chosen configuration of $ n$ points and notice that $ E S_1(n\plus{}1)\equal{}ES_1(n)\minus{}\frac 12 E [S_1(n)^2]$ (Note that the conditional probability that the $ n\plus{}1$-st point will change the left interval is $ S_1$ and the conditional expectation of the change under the condition that the change happens is $ S_1(n)/2$; then pass from conditional expectations to the full ones. All claims I made do not formally need pdf but a clear understanding of what the uniform distribution is is a must).

Thanks for the idea  :idea: !!!!

I was searching for some (geometric?/combinatorial?) argument that $ E(S_{i}^{k}(n))\equal{}\frac{1}{C_{n\plus{}k}^{k}}$. This can be done with the explicit use of the pdf of $ S_{1}$, but the result is so nice... It asks for a non-pdf proof...


Anybody with a good TI-84 game?

I suggest Pheonix, and Skate.

If u have a REALLY good game FOR THE TI-83, 84, etc......, PLZ POST!!!

If you can, plz let us downlaod it!

do u have it?

can u let us download it?

uhhhh block dude is cool

its in puzzle pack


Uh, $ 1\minus{}\dfrac 1{2^2} \equal{} \dfrac 34 > \dfrac 12$. In fact, any one term of that sum is greater than $ \dfrac 12$.

[quote="t0rajir0u"]He means product.[/quote]
yes.you are true. It means product.


how many rectangles can be in a 6x6 square?

How do you do this, i know how to find how many squares, but what is the formula for the rectangles?

[hide]A rectangle is defined by two horizontal lines and two vertical lines. There are $ \binom{x \plus{} 1}{2}$ ways to choose two distinct lines out of $ x \plus{} 1$ lines.

In an $ m\times n$ grid, the number of rectangles is $ \binom{m \plus{} 1}{2}\binom{n \plus{} 1}{2}$.

In this particular case: $ \binom{7}{2}^2 \equal{} 21^2 \equal{} \boxed{441}$.

EDIT: Little mistake in computation.[/hide]

actually, the answer key stated that the answer was 441??

I think that's what i_like_pie meant.

$ \binom{7}{2}\equal{}21$, not $ 15$
$ 21^2\equal{}441$


Let n be a positive integer. Give a combinatorial proof that

C(2n, n+1) + C(2n, n) = 1/2 * C (2n+2, n+1)

i.e. to obtain the identity by counting the objects of a set in two different ways and set the results equal to one another.

This is a proof by substitution. The problem requires a combinatorial proof, which means proving the identity by counting the number of objects of some set in two different ways and setting the results equal to one another.

It also didn't actually finish with the desired result ;)

It's difficult to give hints for this type of question, but I'll give it a try: [hide="hint"] you're counting something having to do with splitting up a group of $ 2n \plus{} 2$ people.  Two of these people are Paul and Sally, so on one side you want to try splitting it up from their perspective.[/hide]

As with any binomial coefficient identity, there are many possible proofs.  The simplest one of this particular identity is just using Pascal's Identity and $ \binom{n}{r} \equal{} \binom{n}{n \minus{} r}$: [hide]$ 2\left(\binom{2n}{n} \plus{} \binom{2n}{n \plus{} 1}\right) \equal{} 2\binom{2n \plus{} 1}{n \plus{} 1} \equal{} \binom{2n \plus{} 1}{n \plus{} 1} \plus{} \binom{2n \plus{} 1}{n} \equal{} \binom{2n \plus{} 2}{n \plus{} 1}$[/hide]


function
topology
analytic geometry
real analysis
advanced fields
advanced fields unsolved

Consider the space $\Omega$ of all functions from $\mathbb{R}\to \mathbb{R}$. This space is just $\mathbb{R}^\mathbb{R}$ so endow it with the product topology. Then let $\mathcal{B}$ be the Borel $\sigma$-algebra which is generated by those open sets.

The set of all continuous functions is a subset $C \subset \Omega$. Is it true that $C \not \in \mathcal{B}$?

I think that $C$ is too small to be Borel. The base of topology is formed by all sets of the form $\{f\in\Omega\colon a_{k}<f(x_{k})<b_{k},\ k=1,\dots,n\}$ for various choices of $n$ and $x_{k},a_{k},b_{k}\in\mathbb R$. Note that the membership in any of such sets is determined by the values of $f$ on a finite subset of $\mathbb R$. In other words, if $U\subset\Omega$ is open and $f\in U$, then there is a finite set $E\subset \mathbb R$ such that $\{g\in \Omega\colon f=g\text{ on }E\}\subset U$. From this the following statement should follow: if $B\subset\Omega$ is Borel and $f\in B$, then there is a countable set $E\subset \mathbb R$ such that $\{g\in \Omega\colon f=g\text{ on }E\}\subset B$.

[quote="mlok"] From this the following statement should follow: if $B\subset\Omega$ is Borel and $f\in B$, then there is a countable set $E\subset \mathbb R$ such that $\{g\in \Omega\colon f=g\text{ on }E\}\subset B$.[/quote]
This isn't the case as, for example, $B = \{f \mid (\forall x\in\mathbb R)\, f(x)\in [0,1]\}$ is closed in the given topology.

[quote="djmycroft"]
This isn't the case as, for example, $B = \{f \mid (\forall x\in\mathbb R)\, f(x)\in [0,1]\}$ is closed in the given topology.[/quote]

No, this set is not closed. What mlok wrote is correct. Every Borel set is defined by a predicate that depends only on the value of the function at countably many points.

Update: What I wrote here is wrong  :( .

Um.. But now I think that I was wrong and [b]djmycroft[/b] is right. The complement of $B$ is the union of open sets $U_{a}=\{f\colon f(a)>1\}$ and $V_{a}=\{f\colon f(a)<0\}$. The problem with my argument is that the countable set is allowed to depend on $f$. This property passes to unions and countable intersections, but not complements. So I only proved that the set of all continuous functions is not $G_{\delta}$. (Naturally, I want to get partial credit.)

$C$ is a Borel set. The idea of the proof is to use the fact that f is a continuous function if and only if it's uniformly continuous on each compact interval [-n,n].

Let $UC(X, \delta, \epsilon)$, where $X \subset \mathbb{R}$, be the set of all $f \in \Omega$ such that $| f(x)-f(y) | \le \epsilon$ holds for each $x,y \in X$ with $|x-y| \le \delta$. This set is a closed set. The set $UC(X)$ of all $f \in \Omega$ that is uniformly continuous on $X$ is an $F_{\sigma\delta}$ set : take $\delta = 1/m$ and $\epsilon = 1/n$. Since $\mathbb{R}$ is a countable union of compact intervals, $C$ is an $F_{\sigma\delta}$ set, too.

$ C$ is known to be [b]not[/b] measurable w.r.t. the product $ \sigma$-algebra of $ \mathbb{R}^\mathbb{R}$. but anyone know its proof?

The product $ \sigma$-algebra of $ \mathbb{R}^\mathbb{R}$ is the $ \sigma$-algebra generated by { $ \pi_x^{\minus{}1}(E_x)$ : $ E_x$ a Borel subset of $ \mathbb{R}$ and $ x \in \mathbb{R}$ } where $ \pi_x$ are coordinate projections : $ \pi_x(f) \equal{} f(x)$. It is smaller than the Borel $ \sigma$ algebra of $ \mathbb{R}^\mathbb{R}$


how do you do this problem

The four vertices of a rectangle coincide with four vertices of a regular hexagon. If each side of the hexagon is 2, find the area of the square.

please include an explanation

 :whistling:  :spider:  :moose:

Hehe... Random smilies.

[hide="hint"]
Draw the altitudes to the rectangle from a vertex of the hexagon not a vertex of the rectangle, making 30-60-90 triangles.
[/hide]

[hide]
er what square?

For the rectangle, draw it out.
It shouldn't be that hard to locate the rectangle.
Two of the sides of the rectangle are 2.
For the other (long) sides, drop altitudes fro the vertex across from the side of the rectangle. This forms 30-60-90 triangles. The hypenuse is 2, thus the shortest side is 1 and the longer side (half that of the rectangle) is $\sqrt 3$
Since that is half of the rectangle's other side, then the dimensions of the rectangles are $2\sqrt 3$ and $2$.
The area is $4\sqrt 3$[/hide]


In the sequence 0, 1, 1, 3, 6, 9, 27, ..., the first term is 0.
Subsequent terms are produced by alternately adding and
multiplying by each successive integer beginning with 1.  For
instance, the second term is produced by adding
1 to the first term; the third term is produced by
multiplying the second term by 1; the fourth term
is produced by adding 2 to the third term; and so
on.  What is the value of the first term that is greater than 125?

Can someone explain how the pattern works? I don't understand the sequence works.

Continuing this sequence from the 27, we add four to make 31, then multiply 31 by four to make 124, then add five to 124 to make 129. Thus, $\boxed{129}$ is the first term that is greater than [code]125[/code].