This year's National MATHCOUNTS Contest will be on ESPN2 at noon ET on June 3.  Yours truly will be one of the sportscasters for the broadcast.  I hope to see several other AoPSers in the broadcast!

That will be 12 noon ET
11 a.m. CT
9 a.m. PT

[quote="mathgeek2006"]What about MST?  JK, 10AM.[/quote]Actually its not.  MDST is 10:00 AM (Daylight savings not standard time) :P  :lol:  :)  :D  :D

Who decided to put mathcounts broadcast the day before the ARML?!? Now I have to get my neighbor to tape it and everything just gets messed up... I hope it's not on ARML day next year (though next year there won't even be a small chance I'm in it). Oh well...

Do you really think the guys at ESPN are aware of the ARML? :(

I suppose you're right, but of all the times they could do it... I wouldn't even mind if they did it that night, so I can watch it from the hotel. They ought to ask the math people (purposefully dodges writing mathletes) before throwing it on tv.

True, true, you've got a point. You wonder why ESPN chose June 3rd as the exact date when it's a month after the competition. People have said something about the Seplling Bee, but still...it is inconvenient for people in your situation.

Heehee, how ironic, I spelled the word "spelling" wrong, even if it is just a typo. I'll think I'll leave it that way for the humor. :)

Yes, the idea is to put the MATHCOUNTS broadcast during the break on the final day of the Spelling Bee.

I suppose they figure people who like spelling like math also? :P

Wouldn't it be cool if, someday, the situation were reversed: The Spelling Bee was broadcast during the lunch break of MathCounts?  Not likely to happen, I know...

That would be awesome! Hahaha, we mathematicians will quell the noise coming from those nauseating spellers once and for all!

And consider that ESPN is a network dealing primarily with athletic competitions, so they tend to group the academics together. We should be grateful ESPN even does stuff like this.

So now we know why ESPN exists. I don't remember the last time I watched ESPN for more than the time it takes to flip through channels... besides mathcounts last year...

[quote="elvenchamp777"]Are we gonna see you Mr. Ruszcyk?  Be sure to say AOPS RULES somewhere in your broadcast :D

elvenchamp777[/quote]

You'll see me, but I'm pretty sure any AoPS references will get the quick axe.

I think that the whole world should just buy the whole world and give it to me :)

um...

I told my school math teacher about it and he got all excited and wants to see it

is there place where you can see it over again in AoPS?

( and by the way, I was so surprised that day when I saw Mr.Rusczyk. I thought he was much older than that )

[quote="wild octagon"]
Overall I was disappointed with the quality of the ESPN broadcast which portrayed the Countdown Round as a rather dry affair.  The audience never got a chance to see the participants as friendly, fun-loving kids.  There was almost no footage of the early rounds.  We barely caught a glimpse of the winning team.  I would have liked to see solutions for all of the problems, not just a few.[/quote]

Honestly, on stage, the competitors were rather wooden, now had I made it up there, that would have been mighty different :P

[quote="nat mc"][quote="wild octagon"]Someone should make a [i]Spellbound[/i]-type documentary on Mathcounts.  It would do wonders for the world of math enthusiasts.[/quote]

I have actually thought about that myself.  It truly would be awesome like Mr. Rusczyk said.  It would be unbelievable.  Just to see all the practice, the emotions, hard work, andcompetitiveness would be great[/quote]

Yeah, I was thinking about that -- maybe then I would actually get to see it, lol. I don't have cable  :(

[quote="rrusczyk"]...I'll note that the production company more or less had its hands tied...[/quote]
Given the time constraints, I think ESPN did a good job of producing this year's show.  The technical glitches that made last year's show seem amateurish were gone.

Last year, when the bracket format benefitted a Colorado student (my own state), I argued that the ladder format was both more fair and more interesting.  I would make the same argument this year.  Imagine how much more dramatic Neal's victory would have been if he had been forced to burn through 8 opponents on his way to the top.

I would also like to mention that every time I have viewed the bracket-format competitions, somebody always asks "which one is that", when a competitor returns to the stage for later rounds.  There is no continuity -- By the time a student comes back, the audience has forgotten who they are.  With the ladder-format, there is continuity.  You are already familiar with the student who is moving to the next round, and the moderator merely introduces one new student.

I haven't actually timed the show, but it seems each pairing averaged about 3 to 4 minutes.  A 10-person ladder format would give them 6 to 8 minutes more breathing room.  The 2003 broadcast, which used the 10-person ladder-format, was clearly the best at showing students outside the Countdown Round.

Many of the competitors did seem "wooden".  There were few smiles.  [Of course, you don't often see the NFL place-kickers grinning when they're about to try a last-second field goal.]  But not all moments were "wooden".  I was disappointed none of the small talk between moderator and student made it to ESPN.  Karlanna's interractions with the moderator, and the moderator's deferring to the audience were perhaps the lightest moments of the Countdown Round.

[quote="sirorange"]Honestly, on stage, the competitors were rather wooden[/quote]It's difficult for anyone to be relaxed and smiling when they're concentrating so hard on the math problems, with so many eyes and cameras focused on them, with the national championship at stake.

The broadcast was fine for viewers mostly interested in the Countdown problems and results.  I see the annual Mathcounts broadcast as an opportunity to promote the world of math and to attract students to an activity often viewed as nerdy and uncool.  In that sense I found the production disappointing.

After watching the National Countdown Round for a second time, I feel even more strongly that the current format needs revamping.  Mathcounts does a great job of encouraging students to acquire math knowledge and develop problem solving skills.  The Countdown Round, however, is a totally different animal.  Top Mathcounts students who want to succeed at Countdown are forced to spend hours honing very specific non-math-related skills -- skills that have little value in the math world outside of Mathcounts.  I am in awe of Neal Wu and his accomplishments but the Countdown needs to be changed to emphasize math ability while deemphasizing speed and luck.

Here are a few suggestions for improving the current format without totally replacing it:

1.  Eliminate wordy problems.  Do we want to reward speed reading skills or math skills?  Countdown problems should take no more than 10 seconds to read aloud, esp. because the reading time is included in the total solution time.  The longer the problem, the more likely it is that a contestant will get the wrong answer, not because he/she did the math wrong but because he/she misread the problem.  The problem below, for example, taken from this year's Countdown, takes almost 20 seconds to read aloud slowly, which is way too long.

[code]Noah's first four bowling scores were 125, 140, 85 and 150.  All bowling scores are integers.  If the positive difference between his next score and the mean of his first four scores is less than 3, how many possible distinct median values would there be for the five scores?[/code]
2.  Lock out the buzzers until the problem is read aloud.  This is the [i]Jeopardy[/i] approach.  I like it because it reduces the probability that a question will be misread.  It also gives the contestants a couple of seconds to relax and clear their minds without worrying that their opponent will buzz in at any moment.  Anything we can do to help the contestants relax and think clearly would be an improvement.  It's hard enough being on stage in front of a large audience and TV cameras without the added time pressure.

3.  Make the problems harder.  If the contestants can easily arrive at the solution, then speed and luck will be the deciding factors, not math ability.

[quote="wild octagon"]
2.  Lock out the buzzers until the problem is read aloud.  This is the [i]Jeopardy[/i] approach.  I like it because it reduces the probability that a question will be misread.  It also gives the contestants a couple of seconds to relax and clear their minds without worrying that their opponent will buzz in at any moment.  Anything we can do to help the contestants relax and think clearly would be an improvement.  It's hard enough being on stage in front of a large audience and TV cameras without the added time pressure.
[/quote]

Eh... I don't think that one is a good idea. If, for example, there were two contestants, A and B, and the problem was read in 10 seconds... and A did it in 2 seconds and B did it in 7 seconds, then obviously A should get the point. But if the buzzers were locked, both of them could and that's not really fair.

The idea is to make the problems hard enough so that you would need more than 10 seconds to find the solution.

What if it's a long problem that takes 20 seconds to read and A gets it in 11 seconds, while B gets it in 19 seconds?

[quote="wild octagon"]
[code]Noah's first four bowling scores were 125, 140, 85 and 150.  All bowling scores are integers.  If the positive difference between his next score and the mean of his first four scores is less than 3, how many possible distinct median values would there be for the five scores?[/code]
.[/quote]

Hey, I liked that question. Because now I can say I was in the countdown, while still coming in 14th

Is countdown top 16 in your state? Because it is in mine, and if I was 14th, I would just say I made countdown.

That was national countdown, and no, countdown is top 10 in my state. But the weird thing is, in my State State Champion is decided by the countdown, but it doesn't affect your standings. So I say I'm the State Champion but second place. . . .

That's one long countdown. . . . 23 rounds? That's almost as long as our school countdown.


function
search
real analysis
real analysis unsolved

Porve that: if $f:\mathbb{R}\longrightarrow \mathbb{R}$ is diff. then $f'$ cannot be discontinous in the hole $\mathbb{R}$.

This has been discussed before. I think I posted a proof based on the fact that a pointwise limit of continuous real functions has a dense set of continuity points.

I remember the topic was started by liyi, and dickchimney also replied, so these are a few good keywords, if you decide to perform a search in order to find the respective thread :).


geometry
incenter
circumcircle
inequalities
geometry proposed

$I$ and $I_a$ are incenter and excenter opposite $A$ of triangle $ABC$. Suppose $II_a$ and BC meet at $A'$. Also $M$ is midpoint of arc $BC$ not containing $A$. $N$ is midpoint of arc $MBA$. $NI$ and $NI_a$ intersect the circumcircle of $ABC$ at $S$ and $T$.
Prove $S$, $T$ and $A'$ are collinear.

Consider the inversion of pole $N$ and power $NA^2$. This inversion turns the circle $(ABC)$ into the line $AM$. $T$ turns into $I_a$, $S$ turns into $I$, and let the intersection of $NA'$ with the circumcircle be $N'$, so we must show that $NIN'I_a$ is cyclic. This is equivalent to $A'N\cdot A'N'=A'I\cdot A'I_a$. On the other hand, we have $A'N\cdot A'N'=A'B\cdot A'C=A'I\cdot A'I_a$ (the first inequality holds because the value is the power of $A$ wrt the circle $(ABC)$ and the second one holds because the value is the power of $A$ wrt the circle $(IBCI_a)$), so we have the conclusion.

Here is my proof
    It is easy to show that $SITI_a$ is cyclic (by proving the angle $I_aIT=I_aST$.
    We also notice that $IBI_aC$ is cyclic.
    Now consider the three following circles $(ABC), (ISI_aC), (BICI_a)$, their three pairwise radical axises are concurent. Thus S, A' T are collinear. :cool:

I'm sorry. In my diagram S is T, and T is S.
  So in order to prove $TISI_a$ is cyclic, we prove the angle $I_aTS =I_aIS$.

Dear Mathlinkers,
who want to prove this angular equality?
Sincerely
Jean-Louis


Show that if a>=2 and a^n+1 is prime, then a is even and n=2^m for some integer m>=0

and

Show if n>1 and a^n-1 is prime then a=2 and n is prime

Thanks!

If $ n$ has an odd divisor $ d$, then $ a^{n/d} \plus{} 1 | a^n \plus{} 1$; contradiction.  Additionally, if $ a$ is odd then $ a^n \plus{} 1$ is even; contradiction.  When $ a \equal{} 2$ these are the [i]Fermat primes.[/i]

If $ n$ has a nontrivial divisor $ d$ then $ a^{n/d} \minus{} 1 | a^n \minus{} 1$; again, contradiction.  Additionally, if $ a > 2$ then $ a \minus{} 1 | a^n \minus{} 1$; contradiction.  These are the [i]Mersenne primes.[/i]


Prove that if a,b,c>0 then
$ \sum\frac{a^2\plus{}b^2}{a\plus{}b}\leq \frac{3(a^2\plus{}b^2\plus{}c^2)}{a\plus{}b\plus{}c}$

[quote="limes123"]Prove that if a,b,c>0 then
$ \sum\frac {a^2 \plus{} b^2}{a \plus{} b}\leq \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c}$[/quote]
\[ \sum {\frac{{a^2  \plus{} b^2 }}{{a \plus{} b}}}  \equal{} \sum {\left( {a \plus{} b \minus{} \frac{{2ab}}{{a \plus{} b}}} \right)}  \le \frac{{3\left( {a^2  \plus{} b^2  \plus{} c^2 } \right)}}{{a \plus{} b \plus{} c}}
\]
\[ \Leftrightarrow 2\left( {a \plus{} b \plus{} c} \right)^2  \minus{} 2\left( {a \plus{} b \plus{} c} \right)\left( {\sum {\frac{{ab}}{{a \plus{} b}}} } \right) \le 3\sum {a^2 }
\]
\[ \Leftrightarrow \sum {a^2 }  \plus{} 2\left( {\sum a } \right)\left( {\sum {\frac{{ab}}{{a \plus{} b}}} } \right) \ge 4\sum {ab}
\]
\[ \Leftrightarrow \sum {a^2 }  \plus{} 2abc\left( {\sum {\frac{1}{{a \plus{} b}}} } \right) \ge 2\sum {ab}
\]
By Cauchy Schwarz we have
\[ \sum {\frac{1}{{a \plus{} b}}}  \ge \frac{9}{{2\left( {a \plus{} b \plus{} c} \right)}} \Rightarrow 2abc\left( {\sum {\frac{1}{{a \plus{} b}}} } \right) \ge \frac{{9abc}}{{a \plus{} b \plus{} c}}
\]
Hence it suffices to show that
\[ \sum {a^2 }  \plus{} \frac{{9abc}}{{a \plus{} b \plus{} c}} \ge 2\sum {ab}
\]
\[ \Leftrightarrow \left( {\sum {a^2 } } \right)\left( {\sum a } \right) \plus{} 9abc \ge 2\left( {\sum {ab} } \right)\left( {\sum a } \right)
\]
\[ \Leftrightarrow a^3  \plus{} b^3  \plus{} c^3  \plus{} 3abc \ge ab\left( {a \plus{} b} \right) \plus{} bc\left( {b \plus{} c} \right) \plus{} ca\left( {c \plus{} a} \right)
\]
It's Schur inequality.

[quote="limes123"]Prove that if a,b,c>0 then
$ \sum\frac {a^2 \plus{} b^2}{a \plus{} b}\leq \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c}$[/quote]

$ \sum\frac {a^2 \plus{} b^2}{a \plus{} b} \leq \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c}$

$ \leftrightarrow S \equal{} S_a(b \minus{} c)^2 \plus{} S_b(a \minus{} c)^2 \plus{} S_c(a \minus{} b)^2\ge\ 0$

With $ S_a \equal{} \frac {1}{a \plus{} b \plus{} c} \minus{} \frac {1}{2(b \plus{} c)}$

       $ S_b \equal{} \frac {1}{a \plus{} b \plus{} c} \minus{} \frac {1}{2(c \plus{} a)}$

       $ S_c \equal{} \frac {1}{a \plus{} b \plus{} c} \minus{} \frac {1}{2(a \plus{} b)}$

Assum $ a \ge\ b \ge\ c$

We have $ S_b \equal{} \frac {a \plus{} c \minus{} b}{2(a \plus{} b \plus{} c)(c \plus{} a)} \ge\ 0$

Similar we have : $ S_c \ge\ 0$ and $ a^2S_b \plus{} b^2S_a \ge\ 0$

So $ S \ge\ 0$

We have done!

Left - Right =
\[ \sum \left(\frac{a^2\plus{}b^2\plus{}c^2}{a\plus{}b\plus{}c}\minus{}\frac{a^2\plus{}b^2}{a\plus{}b} \right)\equal{} \frac{1}{a\plus{}b\plus{}c}\sum \frac{c}{a\plus{}b}[a(c\minus{}a)\plus{}b(c\minus{}b)]\]
\[ \geq \sum\frac{1}{a\plus{}b}[a(c\minus{}a)\plus{}b(c\minus{}b)] \equal{} \sum (c\minus{}a)\left(\frac{a}{a\plus{}b}\minus{}\frac{c}{b\plus{}c}\right) \\ \equal{} \sum \frac{b(c\minus{}a)^2}{(a\plus{}b)(b\plus{}c)}\geq 0\]


quadratics
algebra
polynomial
number theory proposed
number theory

Quadratic polynomial f(x)=ax^2+bx+c so that (f(x))^3-f(x) has three real roots. Find ordinate of vertices of graph of f(x)=ax^2+bx+c

Solution :
As is known
$ g(x)\equal{}(f(x))^{3}\minus{}f(x)\equal{}f(x)(f(x)\minus{}1)(f(x)\plus{}1)$ .

The  roots of g(x) are roots of the trinomials f(x),f(x)-1 and f(x)+1.We assume that main coefficient of the trinomial f(x) is positive and $ y_{0}$.
Then $ f(x)>0$ and $ f(x)\plus{}1>0$,$ \forall x \in R$ .It is mean that the roots of g(x) are the roots of the $ f(x)\minus{}1$.They number of root not greater than 2.
If $ y_{0}$,then trinomials f(x) and $ f(x)\minus{}1$ has two roots.But is imposible.Then $ y_{0}$.  :!:


calculus
integration
logarithms
function
trigonometry
complex analysis

Prove that for $ n>0$,
\[ \int_{0}^{\infty}{\frac{\cos(nx)}{x^2\plus{}1}\log x\;dx}\plus{}\frac{\pi}{2}\int_{0}^{\infty}{\frac{\sin(nx)}{x^2\plus{}1}\;dx}\equal{}0\]

hmm...
Consider Contour $ \Gamma$ (Picture PI2).
and consider function $ F(z) \equal{} \frac {e^{inz} \ln z}{z^{2} \plus{} 1}$
and integrate $ F(z)$ on this contour.
if my calculation is correct then residue of this function inside this contour at point $ z \equal{} i$ is equal to $ \frac{e^{\minus{}n} \pi}{4}$ thus is  real number.  .
when we integrate $ f(z)$ on the big and small cirlce we get $ 0$ as radii of the big cirlce $ \to \infty$ and radii fo the small cirlce $ \to 0$
when we integrate on the line $ (\epsilon, R)$ we get :
$ \int_{(\epsilon,R)} \frac {\cos nx \ln x \plus{} i \sin nx \ln x }{x^{2} \plus{} 1} dx \equal{} \int_{(\epsilon,R)} \frac {\cos nx \ln x }{x^{2} \plus{} 1} dx \plus{} i \int_{(\epsilon,R)} \frac {\sin nx \ln x }{x^{2} \plus{} 1} dx$
and when we are ont hte line $ ( \minus{} R, \minus{} \epsilon)$ we have $ \ln z \equal{} \ln |z| \plus{} i\pi$ thus :
$ \int_{( \minus{} R, \minus{} \epsilon)}\frac {(\cos nx \plus{} i \sin nx)(\ln|x| \plus{} i\pi)}{x^{2} \plus{} 1}dx \equal{} \int_{(\epsilon,R)}\frac {(\cos nx \minus{} i \sin nx)(\ln|x| \plus{} i\pi)}{x^{2} \plus{} 1}dx \equal{} \int_{(\epsilon,R)}\frac {\cos nx \ln x \plus{} \pi \sin nx }{x^{2} \plus{} 1}dx \plus{} i \int_{(\epsilon,R)}\frac {\pi \cos nx \minus{} \ln x \sin nx }{x^{2} \plus{} 1}dx$
finally we get:
\[ \int_{(0,\infty)} \frac {2\cos nx \ln x \plus{} \pi \sin nx }{x^{2} \plus{} 1} dx \plus{} i * Real \equal{} 2\pi i \frac{e^{\minus{}n} \pi}{4}
\]
compare real parts we get :
$ \int_{0}^{\infty} \frac {\cos nx \ln x \plus{} \pi/2 \sin nx }{x^{2} \plus{} 1} dx \equal{} 0$


there's a square whose sides are in full centimetre . Its area is more than that of a rectangle whose 1 side is 5cm less than double the side of the square and the and the other side is 4 cm less than the side of the square. Also the area of the square is less than that of a rectangle for which one side is smaller by 2 cm and the other greater by 3 cm than the side of the square . Find possible length of side of square

[hide]The first statement means that if x is the side length of the square, then $(2x-5)(x-4)<x^2 \Rightarrow \displaystyle\frac{13-\sqrt{89}}{2}<x<\displaystyle\frac{13+\sqrt{89}}{2}$.  The second statement means that $x^2<(x-2)(x+3) \Rightarrow x>6$.  So $6<x<\displaystyle\frac{13+\sqrt{89}}{2}$.[/hide]


rahe  halhatoon ro vase soale 4 lotfan befrestid. :D

az har adade fard mishe be har addade fard rasid.chera?chon har doadade motavali mitavanand do ozve motavalie donbale bashand.

[quote="fatemeh"][quote="jensen"][quote="fatemeh"]berin az inja  :arrow:  beenahaiat.persianblog.com javaba ro bebinin[/quote]
shoma?[/quote]
ye olampiadi  :)[/quote]
ok i see. :D


HI, I am new here. My name is Arghya. I'm in class 12. :)

u shud hav posted it [url=http://www.mathlinks.ro/viewtopic.php?t=181606&start=320]here[/url]
neways
Welcome
 where r u from???[/url]


quadratics
analytic geometry
graphing lines
slope
inequalities
algebra

Given that $x^{2}+y^{2}=14x+6y+6$, what is the largest possible value that $3x+4y$ can have?

[hide="Solution"]
By Cauchy we know that $3x+4y$ is at its maximum iff $\frac{x}{3}=\frac{y}{4}$, and we are trying to find the intersection of this line with the circle $(x-7)^{2}+(y-3)^{2}=64$.

Substitute $y=\frac{4}{3}x$ to get the quadratic equation $25x^{2}-198x-54=0$ with solutions $x_{1,2}=\frac{99}{25}\pm \frac{2\sqrt{1239}}{25}$. We disregard the smaller solution, as we are looking for the expression's maximum; then $x=\frac{99}{25}+\frac{2\sqrt{1239}}{25}\Rightarrow y= \frac{132}{25}+\frac{8\sqrt{1239}}{75}$.

Plugging this solution back into the initial expression we get that $3x+4y \leq \boxed{33+\frac{2}{3}\sqrt{1239}\approx 56.47}$[/hide]

[quote="paladin8"][quote="13375P34K43V312"]Is there a solution involving the fact that the equation is a circle?[/quote]

Suppose $3x+4y = k$. Then the line formed by this in the plane has slope $-\frac{3}{4}$. Since $k$ is maximized when this line is tangent to the circle, the radius from the circle to the line must be perpendicular, so it has slope $\frac{4}{3}$. Since the center of the circle is $(7,3)$ and it has radius $8$, the tangent point is $\left(7+\frac{24}{5},3+\frac{32}{5}\right)$ (the point that is a distance of $8$ away at a slope of $\frac{4}{3}$), which gives $3x+4y = 73$.

EDIT: cincodemayo5590, how do you get the maximum occurs at $\frac{x}{3}= \frac{y}{4}$ by Cauchy? There is a Cauchy solution [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=418742#p418742]here[/url], but the answer agrees with mine.[/quote]

Doesnt Cauchy's say that $(ax+by)^{2}\leq (a^{2}+b^{2})(x^{2}+y^{2})$, with equality iff $\frac{a}{x}= \frac{b}{y}$ ?  ([url=http://en.wikipedia.org/wiki/Cauchy-Schwarz_inequality]Cauchy's  on Wiki[/url])

I thought about using Cauchy's all the way through, but while i was working it out I forgot a square somewhere and it led nowhere (accidentally). I just used that one part without the rest of it... but I guess that's not valid (?). If I had remembered the square, then I would've ended up with a solution similar to that of the linked thread.

[quote="paladin8"][quote="cincodemayo5590"]

Doesnt Cauchy's say that $(ax+by)^{2}\leq (a^{2}+b^{2})(x^{2}+y^{2})$, with equality iff $\frac{a}{x}= \frac{b}{y}$ ?  ([url=http://en.wikipedia.org/wiki/Cauchy-Schwarz_inequality]Cauchy's  on Wiki[/url])
[/quote]
Yes, but that gives you $(3x+4y)^{2}\le 25(x^{2}+y^{2})$. Since the RHS is not constant, we cannot say that the maximum occurs at $\frac{3}{x}= \frac{4}{y}$ because different $x$ and $y$ may produce a non-equality case that actually gives a larger value for $3x+4y$.[/quote]

Oh.  :oops:  Ok, thanks for pointing that out.  ^_^


Our design team drew all 48 line segments that connect 6 points on one side of rectangle R to 8 points on the opposite side. In at most how many different points inside R can these 48 line segments intersect?

its 8C2*6C2 because every 4 points (2 on both) sides there is exaclty 1 way to create intersecting lines...its 420?


How many $1 \times 10\sqrt 2$ rectangles can be cut from a $50\times 90$ rectangle using cuts parallel to its edges?

WHo can solve this difficult problem?I guess the maxium is 315,but I only can proove it is less than 318
EDIT:By the way ,what is the full mark of each problem?

I don't remember the solution, but there is a document from Andrei Jorza about coloring. There is a solution there. You can find it on the net.

Sorry,I can't find it . :blush:  Could you help me ,or tell me the name of the document?Thank you very much.

How did you get the things you said in the first post?

I think I have solved the whole problem,and the maxium is 315.But,I am not able to post because I felt really hard to express it in English as
 there're some terminologies I don't know how to translate. :blush: I
 am not good at english.On the other hand ,my proof is long and really uniteresting.
So,I'd like to see somebody post a solution here :)

there is anyone that have done this problem?
i solve this in two ways. in one results 312 in the other resullts 315 so i think that 315 its the maximum number of rectangle.

[quote="Omid Hatami"]Consider the lines $x+y=10\sqrt 2$.
It is easy to see that any rectangle $1\times10\sqrt 2$ whose sides are parallel to the axes intersect these lines in one or two segments, but in either case the total length of the intersection is $\sqrt2$ for obvious reasons.[/quote]Just some motivation: our tiling rectangles ($1\times10\sqrt{2}$ and $10\sqrt{2}\times1$) behave nicely $\pmod{1}^2$ and $\pmod{10\sqrt{2}}^2$ (actually, $1/n$ and $10\sqrt{2}/n$ work well for any positive integer $n$, but $n=1$ is the most general option). Take $m\in\{1,10\sqrt{2}\}$; then we hope to find $S\subset(\mathbb{R}/m\mathbb{R})^2$ such that $S\cap R$ has constant area $A$ over all $1\times10\sqrt{2}$ and $10\sqrt{2}\times1$ rectangles, and $S\cap[0,50]\times[0,90]$ has area less than $316A$. So we'll need to take $m=10\sqrt{2}$ to get anything nontrivial. The two obvious choices for $S$ (up to rotation, reflection, and translation) are $\{x+y\equiv0\pmod{m}\}$ and the $2\times2$ chessboard coloring of $(\mathbb{R}/m\mathbb{R})^2$ determined by the lines $x\equiv m/2$, $y\equiv m/2$; both end up working.

Using area we find that $318<\frac{50\cdot{90}}{10\sqrt{2}}<319$ so we can deduce that at most 318 $1\times{10\sqrt{2}}$ rectangles can fit into the $50\times{90}$ rectangle. However this is by no means the maximum because we don't know if they fit exactly or not (as whole rectangles).

[quote=Omid Hatami]I just copied the solution from [url=http://www.ajorza.org/wimo/images/mathfiles/files/resources/tilings.pdf]here[/url]:
It is easy to tile (greedy) this rectangle into $315$ small rectangles. Now, put the big rectangle on the plane $(0, 0), (90, 0), (90, 50), (0, 50)$. Consider the lines $x+y=10\sqrt 2$.
It is easy to see that any rectangle $1\times10\sqrt 2$ whose sides are parallel to the axes intersect these lines in one or two segments, but in either case the total length of the intersection is $\sqrt2$ for obvious reasons. One can manually check that the total length of the intersection of the big rectangle with these lines is $570\sqrt 2-360$, so the biggest number of $1\times10\sqrt 2$ rectangles that fit into the big rectangle is at most \[\left[\frac{570\sqrt2-360}{\sqrt 2}\right]\]So $315$ is the best possible number.[/quote]

Excuse, but I do not understand it at all. Why do we have more than one line such that $x+y=10\sqrt 2$.

Here is a sketch of a different solution.

Throw the $50 \times 90$ rectangle onto the Cartesian plane such that the two sides with side length $90$ are parallel to the $x$ axis and the lower-left corner of the rectangle coincides with the origin. Color a point $(x, y)$ in the Cartesian plane black if $\left \lfloor \frac{x}{50 \sqrt 2} \right \rfloor + \left \lfloor \frac{y}{50 \sqrt 2} \right \rfloor$ is even, and white otherwise. Note that this is essentially just coloring using a standard checkerboard pattern, but with squares of size $5 \sqrt 2 \times 5 \sqrt 2$ rather than $1 \times 1.$ Observe that every $1 \times 10 \sqrt{2}$ rectangle covers an equal area of white as that of black. Hence, since the black area within the $50 \times 90$ is $(90 - 60 \sqrt 2) \cdot (40 \sqrt 2 - 50)$ greater than the white area within the $50 \times 90$, we must "waste" at least $(90 - 60 \sqrt 2) \cdot (40 \sqrt 2 - 50)$ of the board. This gives us that there are at most $\frac{50 \cdot 90 - (90 - 60 \sqrt 2) \cdot (40 \sqrt 2 - 50)}{10 \sqrt 2} \approx 315.807$ rectangles, giving us the desired bound. The construction is shown above.
$\square$

[quote=Kirilbangachev]

Excuse, but I do not understand it at all. Why do we have more than one line such that $x+y=10\sqrt 2$.[/quote]
I think there is a mistake.
 $x+y=k$$10\sqrt 2$


calculus
integration
logarithms
real analysis
real analysis unsolved


if a+b+c+d=1 prove that
abc+bcd+acd+dab <= 1/27+176/27abcd

there is a solution by lagrange method but i need simple way.

We can use the "n-1 Equal Variable Method".

Oh, really it is usefull for making money...

Oh,don't say that,Myth  :lol: .

To payman_pm : Your can find the office solution for it in website http://www.kalva.demon.uk

I said that because many people on the forum are tired with his ....... (I erased the word I wrote here for censored reasons) secrets.

But now all problem he said secret we had solved oursheft.

Myth, you are very subjective and ....
For example, I posted my RCF-Theorem, but you refused to post your IPT-Theorem! ;)

You are very coarse. I didn't say that you are ....  :(

and .......
means
AND ROMANTIC.
That's all. ;)

Here is a nice solution that I know :
$f(a,b,c,d)=abc+bcd+cda+dab-\frac{176}{27}abcd=bc(a+d)+ad(b+c-\frac{176}{27}bc)$. Of course $f$ is symetric. Now let's divide the inequality in 2 cases :
$I$ :
$b+c-\frac{176}{27}bc \leq 0$  then
$f(a,b,c,d) \leq bc(a+d)$ and this is easy to see $\leq \frac{1}{27}$
$II$
$b+c-\frac{176}{27}bc > 0$  then
$f(a,b,c,d) \leq bc(a+d)+(\frac{a+d}{2})^2(b+c-\frac{176}{27}bc) \leq f(\frac{a+d}{2},b,c,\frac{a+d}{2}) =
 f(b,\frac{a+d}{2},\frac{a+d}{2},c) \leq f(\frac{b+c}{2},\frac{a+d}{2},\frac{a+d}{2},\frac{b+c}{2}) =
(\frac{a+d}{2},\frac{b+c}{2},\frac{a+d}{2},\frac{b+c}{2}) \leq f(\frac{1}{4},\frac{b+c}{2},\frac{a+d}{2},\frac{1}{4}) \leq f(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}) = \frac{1}{2}$
and so we are done.


For $ x,y,z \geq 0,$ establish the inequality

\[ x(x\minus{}z)^2 \plus{} y(y\minus{}z)^2 \geq (x\minus{}z)(y\minus{}z)(x\plus{}y\minus{}z)\]

and determine when equality holds.

[hide="Discussion"] WLOG $ x \ge y$.  There is a nice general principle which states that an inequality with lots of terms involving differences behaves nicely when each variable is translated by the same constant.  Hence:

The transformation $ (x, y, z) \mapsto (x \plus{} t, y \plus{} t, z \plus{} t)$ increases the LHS by $ t(x \minus{} z)^2 \plus{} t(y \minus{} z)^2$ while it increases the RHS by $ t(x \minus{} z)(y \minus{} z)$, but clearly $ (x \minus{} z)^2 \plus{} (y \minus{} z)^2 \ge (x \minus{} z)(y \minus{} z)$ (complete the square).  The difference between the LHS and RHS is therefore minimized when $ t$ is negative and as small as possible.

[b]Case:[/b]  $ y \le z$.  We can therefore take $ y \equal{} 0$ and it suffices to prove $ x(x \minus{} z)^2 \ge (x \minus{} z)^2 (\minus{}z)$, a triviality.  Equality holds here when $ x \plus{} z \equal{} 0 \implies x \equal{} z \equal{} 0$.

[b]Case:[/b]  $ y \ge z$.  We can therefore take $ z \equal{} 0$ and it suffices to prove $ x^3 \plus{} y^3 \ge xy(x \plus{} y)$, which follows from Rearrangement or weighted (i.e. repeated applications of) AM-GM.  Equality holds here when $ x \equal{} y$.

Of course, this is the same kind of reasoning that proves Schur's inequality (even the rather hard version). [/hide]

[quote="kunny"]$ L.H.S \minus{} R.H.S. \equal{} x(x \minus{} y)(x \minus{} z) \plus{} y(y \minus{} x)(y \minus{} z) \plus{} z(z \minus{} x)(z \minus{} y)\geq 0$ by Schur's ineqaulity.

The equality holds when $ x \equal{} y \equal{} z$ or some of two of $ x,\ y,\ z$ are equal and the other is 0.[/quote]

Wow, it's interesting how Schur's Inequality basically answers the question! (I just looked it up)

Now, it says on Wikipedia that Schur's Inequality holds true for $ t$ being positive number. So it's not restricted to integer values of $ t$?

$ t$ I mean is:

$ x^t (x\minus{}y)(x\minus{}z) \plus{} y^t (y\minus{}x)(y\minus{}z) \plus{} z^t (z\minus{}x)(z\minus{}y) \geq 0$

Yes.  This is more or less explained by the Vornicu-Schur inequality.


The ratio of the lengths of the radii of two circles is $ 3: 4$. What is the respective ratio of their areas? Express your answer as a common fraction.

Since area involves squaring, we just multiply the {loss of words} factor by itself.

$ \boxed{\frac9{16}}$


geometry
geometric transformation
rotation
geometry proposed

Here's a rather simple problem, but I still like it :). That's because I think there's lots of room for all sorts of solutions (even though I only came up with one):

Show a bounded set in the Euclidean plane which is geometrically congruent to a proper subset of itself.

Hmm.. I guess people just didn't like it. :) I wonder why though. Maybe it was too easy to be interesting?

Here's the example I gave: we take the set to be $\{e^{nix}|n\in\mathbb N,n\ge 1\}$ where $x$ is some irrational multiple of $\pi$, and we take the proper subset to be $\{e^{nix}|n\in\mathbb N,n\ge 2\}$. The proper subset is a rotation of the full set.


This may or may not be relating to MathCounts, but I found a formula a while ago which related to the maximum number of points of intersection between circles and lines...
Does anybody still know what it was?

It had something to do with just circles and then just lines and then something with circles and lines... (the formula)

Feel free to correct

[hide]well i know that the maximum number of points of intersection between $ n$ lines is just the $ n \minus{} 1$ triangular number ($ 1 \plus{} 2 \plus{} 3 \plus{} 4 ... \plus{} n \minus{} 1$)  And then the one circle can intersect another circle at 2 points so im guessing (could be wrong) that the maximum number of intersections between $ m$ circles would be the $ m \minus{} 1$ triangular number times $ 2$.  And then every line can intersect every circle at 2 points so i think n lines intersect m circles at $ 2(n \times{m})$ so one big formula to find the maximum number of intersection points between n lines and m circles would be (ive probably made it alot more complicated than it needs to be)

(the $ n \minus{} 1$ triangular number) + $ 2$(the $ m \minus{} 1$ triangular number) + ($ 2(n \times {m})$)

which now i can see simplifies lol

so your formula would be for $ n$ lines and $ m$ circles

$ \frac {n(n \minus{} 1)}{2}$ + $ m(m\minus{}1)$ + $ 2(n \times{m})$[/hide]

[quote="shentang"]I don't know the formula for circles and lines, but I do know the formula for the maximum number of intersections with x circles. The formula is x^2 - x[/quote]

If you don't know why, you can prove it using induction.


probability counting distinguishability function calculus derivative algebra

Given a bag of $n$ randomly assorted jellybeans and $k$ possible different flavors, what is the probability that one would obtain at least one of each flavor?

In the "randomly assorted" $n$ jellybeans, are we guaranteed to have at least 1 of each flavor?

No, that's what I'm asking the probability of.
View it as having $n$ beans, and each of these is randomly one of $k$ flavors.

Wow, nobody?  :(

I thought this was a [i]very[/i] interesting problem, particularly having worked on it so long only to find I was doing it wrong...

[hide="Possible idea"]Well, you could look at indistinguishable objects and distinguishable urns.

Sort the n jellybeans into k urns by flavor. The number of ways to do this is well-known:

$\binom{n+k-1}{k-1}$

[hide="Explanation"]To divide n objects into k categories, you need k-1 "dividers". Thus, there are n+k-1 spots, and you need to choose k-1 spots for the dividers.[/hide]

Now, you need the number of ways to place them in urns given that each urn already has one. So you "give" each urn one, and distribute the remaining n-k:

$\binom{n-k+k-1}{k-1}=\binom{n-1}{k-1}$

Thus the answer is $\frac{\binom{n-1}{k-1}}{\binom{n+k-1}{k-1}}$

I hope I'm not totally wrong.[/hide]

mathnerd: I think there's an ordered vs. unordered problem here.  In particular, in your solution you count the number of ways to assign the $n$ jellybeans to $k$ colors as $\binom{n+k-1}{k-1}$.  The problem with this is that you are weighting every such coloring equally; for example, if $n = 5$ and we have colors red and blue, you've counted "all five are red" and "two are red, three are blue" as one outcome each.  Given how the colors are assigned (with each jellybean randomly selecting a color for itself), this is not the right way to count: in particular, "two are red, three are blue" should be weighted $10 = \binom{5}{2}$ times as heavily as "all are red."

It is possible to find an ugly summation-based expression with inclusion-exclusion.  Can anyone do anything cleverer?

JBL, what would that ugly summation-based expression be?

We can always bash this with generating functions..

[hide]
If you wanted to weight them, you'd have to consider the generating function $G(x) = \binom{k}{0}+\binom{k}{1}x+\binom{k}{2}x^{2}+\binom{k}{3}x^{3}+\cdots+\binom{k}{k}x^{k}= (1+x)^{k}$, and then consider $(G(x))^{k}= (1+x)^{2k}$. Then, we'd have the simple coefficients of $\sum_{n}\binom{2k}{n}x^{n}$, and the requested probability would be $\boxed{\frac{\binom{2k}{n-k}}{\binom{2k}{n}}}$.
[hide="solving the wrong problem"]
The total number of ways we can choose $n$ jellybeans out of $k$ types can be represented by the generating function $\left(\frac{1}{1-x}\right)^{k}$. Think of each factor as choosing some number of jellybeans from each type. Then, the coefficient of $x^{n}$ is the number of ways we can do this so that we end up with $n$ jellybeans.

I believe that you can get the following result with the extended binomial theorem, but it requires some manipulations that I don't see at the moment. We get that the generating function is equal to $\sum_{n}\binom{n+k-1}{n}x^{n}$.

[hide="calculus derivation"]
Consider taking the derivatives of $\frac{1}{1-x}$. We get that $\frac{d^{k}}{dx^{k}}\frac{1}{1-x}= \frac{(k-1)!}{(1-x)^{k}}$. If we take the derivatives of the generating function form, we get $\sum_{n}\frac{(n+k-1)!}{n!}$. Dividing by $(k-1)!$, we obtain$\frac{1}{(1-x)^{k}}= \sum_{n}\frac{(n+k-1)!}{n!(k-1)!}= \sum_{n}\binom{n+k-1}{n}$.
[/hide]

So then, we see that if we already have one of each type of jellybean, we need to choose $n-k$ more, so the numerator is $\binom{n-1}{n-k}$ and the denominator is the total number of ways we can choose $n$ beans, or $\binom{n+k-1}{n}$, so the requested probability is $\boxed{\frac{\binom{n-1}{n-k}}{\binom{n+k-1}{n}}}$.
This is the same as mathnerd's answer for the wrong problem.
[/hide]

Haha, the actual version of the problem was easier than mathnerd's version :rotfl:
[/hide]

[quote="JBL"]Hamster, you've made a number of mistakes (including some confusion about which variable is which?), which should be clear from your answer when (for example) we take $n = 2k$ or $n = 2k+1$.

For the ugly summation-based expression:[hide]
We do PIE: for a given set of $j \leq k$ colors, there are $j^{n}$ ways to color the jellybeans using only those $j$ colors.  So, what we really want is $\frac{1}{k^{n}}\sum_{i = 0}^{k}(-1)^{i}\binom{k}{i}(k-i)^{n}$.  I don't think this has a nice closed form.[/hide][/quote]

Ah oops. I solved the correct problem as an afterthought, since I arrived at an equivalent to mathnerd's answer, and I simply switched k and n in the beginning which screwed up the rest of the answer. Let me fix this.

[hide="fixes to my solution"]
The correct generating function should be $G(x) = \binom{n}{0}+\binom{n}{1}x+...+\binom{n}{n}x^{n}= (1+x)^{n}$. We then take $G(x)^{k}= (1+x)^{n+k}= \sum_{i}\binom{n+k}{i}x^{i}$. The coefficient of $x^{n}$ is $\binom{n+k}{n}$. When we already have one, we take out the $\binom{n}{0}=1$ because we're not allowed to take zero of one type. Then we have $(1+x)^{n}-1$, which we want to raise to the $k$th power, resulting in a much nastier formula. If anyone would like to continue this, you are welcome to.
[/hide]

You said a number of mistakes. Are there any others I should know about?


geometry logarithms ratio inequalities function AMC

Let
\[ S_1 \equal{} \{ (x,y)\ | \ \log_{10} (1 \plus{} x^2 \plus{} y^2)\le 1 \plus{} \log_{10}(x \plus{} y)\}
\]and
\[ S_2 \equal{} \{ (x,y)\ | \ \log_{10} (2 \plus{} x^2 \plus{} y^2)\le 2 \plus{} \log_{10}(x \plus{} y)\}.
\]What is the ratio of the area of $ S_2$ to the area of $ S_1$?

$ \textbf{(A) } 98\qquad \textbf{(B) } 99\qquad \textbf{(C) } 100\qquad \textbf{(D) } 101\qquad \textbf{(E) } 102$

[hide="solution"]complete the square!
from the first equation you get:$(x-5)^2+(y-5)^2<49$
from the second one you get: $(x-50)^2+(y-50)^2<4998$ so the ratio is $102\Rightarrow \boxed{E}$[/hide]

[hide="Solution"]
Notice that $1=\log_{10} 10$, $2=\log_{10} 100$, and that if $\log a = \log b$, then $a=b$. Hence, \[ 1+x^2+y^2 \le 10x+10y  \] and \[ 2+x^2+y^2 \le 100x+100y.  \] From the first equation, we find that \[ (x-5)^2+(y-5)^2=49  \] is the equation of $S_1$ and \[ (x-50)^2+(y-50)^2 = 4998  \] is the equation of $S_2$, by completing the square. Thus $S_1$ is a circle with radius 7 and $S_2$ is a circle with radius $\sqrt{4998}$, so the ratio of their areas is $(\sqrt{4998}/7)^2=102$, so our answer is $\boxed {(E).}$
[/hide]

Really stupid thinking and I didn't get it on the actual test -.- (realized my messed up thinking later on).

Anyway, here goes:
[hide] Let's tackle the first inequality. $10^x$ is chiefly increasing, so we can put that to both sides and keep the inequality. Putting 10 to both sides of each inequality, we have: \[ 1 + x^2 + y^2 \le 10(x+y)  \] and \[ 2 + x^2 + y^2 \le 100(x+y)  \] These simplify as $x^2 - 10x + y^2 - 10y \le -1$ and $x^2 - 100x + y^2 - 100y \le -2$. Completing the squares, we have \[ (x-5)^2 + (y-5)^2 \le 49  \] and \[ (x-50)^2 + (y-50)^2 \le 4998  \] These are circles. To find the ratio of areas we simply find the square of the ratio of the radii, which is $\frac{4998}{49} = 102$. [/hide]

You all forgot to check an important condition!

[hide]
Remember that the log function can only be applied to positive numbers.  It is pretty clear that
$1 + x^2 + y^2 > 0$ and $2 + x^2 + y^2 > 0$.
We still need that $x + y > 0$.  It turns out, luckily, that the line $x + y = 0$ passes barely under the circles.
So the solution still works, but it could have pretty easily failed.
[/hide]

[hide="Answer"]Set each one equal, then eliminate the log: $1+x^2+y^2=10(x+y)$ and $2+x^2+y^2=100(x+y)$. These are circles: $(x-5)^2+(y-5)^2=49$ and $(x-50)^2+(y-50)^2=4998$. However, we need these for $x+y>0$. Upon checking, however, the circle just barely avoids touching the line $x+y=0$, so we have two complete circles, and the ratio is $\frac{4998\pi}{49\pi}=102\Rightarrow \boxed{E}$.[/hide]


[size=150]The squares BCDE, CAFG and ABHI are constructed outside the triangle ABC. Let GCDQ and EBHP be parallelograms. Prove that APQ is isosceles and rectangular.[/size]

could anyone solve it using complex numbers?

[hide="using complex numbers really isnt fun"]
$ G \equal{} C \plus{} i(C\minus{}A) \equal{} (1\plus{}i)C\minus{}i A$
$ D \equal{} (1\minus{}i)C \plus{} i B$
$ Q \equal{} G \plus{} D \minus{} C \equal{} C \minus{} i A \plus{} i B$
Similarly $ P \equal{} B \minus{} i C  \plus{} i A$
$ Q \minus{} A \equal{} C \minus{} (i\plus{}1)A \plus{} i B \equal{} i (\minus{}i C \plus{} (i\minus{}1)A \plus{} B) \equal{} i (P\minus{}A)$
Q.E.D.
[/hide]

THANKS! Using complex numbers sometimes really needed.


Let $ f$ be an analytic function on some region $ \Omega$ such that $ \left|f(z)\right| < 1$ for all $ z \in \Omega$.

Define (for all $ z$ in $ \Omega$) \[ g(z) \equal{} \sum_{n \equal{} 1}^\infty nf(z)^n.\] Show that $ g$ is a well-defined analytic function on $ \Omega$.

Isn't $ g$ just the composition of $ f$ with $ h(\zeta)\equal{}\sum_{n\equal{}1}^{\infty}n\zeta^n$? The function $ h$ is analytic in the open unit disk (and it's not hard to find an explicit formula for it.)

the basic idea of what's going on here is the maximum modulus principle and the uniform convergence on compact sets of analytic functions is analytic, right ?


Hi, I'm completely stumpped here with this problem

If the total average is 5 what is the missing number ?

8, 7, 6, 2 , ____ . by guessing I know the answer should be 3 which totals to 25/5  then I get 5 as the average. This is pretty easy but what about if it all
ends up with a higher average like (98). It would be pretty hard to guess the answer with  trial and error.

Is there any simple formula I could use to solve this problem ?

Yes there is.  Say that you know the average of $ n$ numbers is $ p$.  Then you know that the sum of all of the numbers has to be equal to $ nq$.  Then it's just simple subtraction.  So in your case, $ n\equal{}5$ and $ p\equal{}5$, so $ np\equal{}25$.  Now let's subtract out: $ 25\minus{}8\minus{}7\minus{}6\minus{}2\equal{}2$.  So the answer is $ \boxed{2}$.  Check your math.

Is there an easy way to remember that formula? I keep forgetting it =(

just remember how average, or the arithmetic mean, is defined.

no, that's normally how I do these problems.  Even with larger numbers.

Oh wait, I was thinking about different problems...
Actually, that's just the "formula" but a different way
Let's take your problem
If your average is 5, then
8 is +3 from 5
7 is +2
6 is +1
2 is -3
Then, the last number is +3+2+1-3+n=0
This gives you n=-3
this means the last number is -3 from 5
5-3=2

[quote="airyork6=)"][quote="zapi2007"]A different way is if you take the distances to the mean and you know those have to add up to 0.
This might make the question faster, but if you have bigger numbers, it might not work as well.  :P[/quote]

bad idea.... :huuh:  :wallbash: might not always work...[/quote]

I dunno why.  It's just taking the definition and then just arithmetic.  It's easily adjustable, depending on what the question's asking

And by distances to the mean, I think he means so that distance is negative.  Probably a better way is the distances from the mean to the numbers, where going backwards is a negative distance.

zapi2007, your method works, but when you have 10+entries each above 50, it becomes much more tedious.  Math is about finding simpler, more elegant ways to do things.

[quote="airyork6=)"]I can barley memorize all the formulas...
Is there a fast way to memorize them all?[/quote]

Basically, your list is

8,7,6,2,n. To find n, we realize that since the average is 5, with 5 numbers, that the sum must be 25. This yeilds n=2.

For any average a, with b elements, the sum of all the elements must be ab.

51,62,36,84,n with average 59, find n.

The average is 59, and there are 5 elements. This means that the sum of the 5 numbers is $ 59 \cdot 5\equal{}295$. Subtracting $ 51,62,36,84$, we get $ n\equal{}62$.

This works the same way for 6 elements (ANY number of elements really)

'Bob has a hat with 6 cards on it. 5 of the cards have 1,5,9,12,3 on them. The final number is smudged. Bob, having a bad memory, cannot remember what the number was. His sister, Anna, recalls that the average is 8. Bob then was able to find the number on the final card. What was the number?'

That question is made up

Since the list has 6 elements, and the average is 8, the sum of the 6 numbers must be 48. Therefore, we can call the missing number n, and solve. $ 1\plus{}5\plus{}9\plus{}12\plus{}3\plus{}n\equal{}48 \rightarrow n\equal{}18$.

The missing number is 18.

Final example (Challenge!):

'Alice has a set of 7 cards, each with a number on them. The numbers on them are 1,3,8,10,11,19,n. The average is 8. When Alice removes a number, the average drops to 7.5. What number did Alice remove?'

[hide="Solution: Do not look until after solved"]Solution: To find n, we know that the sum of all the numbers is 56, so solving for n, we get n=4. After taking out a number, the total sum is 45. Subtracting the sums, we know that she removed 11. Note that we didn't even need to find n!

I hope that this helped you![/hide]

oh easy. Just do sigma x / sigma n that is the total sum of data divide the number of data( it means the number of the numbers) = 5. Simple equation .

23+x/5= 5  . 23+x= 25. x= 25-23 . x= 2
The answer is 2

The above average formula is only for data(s) that are Ungrouped.

The below average formula is for data(s) in a table or graf or grouped.

Sigma fx/ sigma f

You must get the answer for each of the frequency multiply the data then get the total sum of all the frequency multiply by the data in the table then divide by the total number of frequency in the table

All of you are making this way too complicated.  Don't memorize anything.  Understand it.  Then what you don't remember you can re-derive.

Well it's only a bad idea to find the difference from the mean when they tell you all the statistics, but not the mean.
I tried doing that once at CD, since I was trying to be fast...
So I'm like:
93... doesn't work
94... doesn't work
95... doesn't work...
And yeah so I lost but whatever.

[quote="fishythefish"]Um...regarding your solution, you said you didn't need to find n, but didn't you say that n=4?

You have to find n to solve it. Try doing it without finding n![/quote]

You can solve it without solving for n, here is how you do it:

The sum of the first set of numbers is 56 (7*8) and if the average is 7.5, the sum of the numbers is 45 (7.5*6), so 56-45=11.   The number she removed has to be 11.

We did this problem without solving for n, so it is possible.

[quote="fishythefish"]ok, yeah i guess. But how do we end up with people talking about stuff involving sigma, for a simple case of finding averages. It is much easier to understand what u are doing when u memorize a formula, than use the formula, but have no idea what the principle behind it is.[/quote]

Yeah.  All you need to know is how to find the average.  That's it, you don't need to know all this sigma stuff!


function
logarithms
trigonometry
quadratics
complex analysis
complex analysis theorems

I don't know if this is the right place for this, but here it is:

I am trying to show the following:

$\arcsin{x}=-i\ln{(ix+\sqrt{1-x^{2}})}$

It seems that a formula like this would come from the complex definition of the sine, but I am having trouble solving for x in the following equation:

$\sin{x}=m=\frac{e^{ix}-e^{-ix}}{2i}$

Can anyone show me how to prove this?  Am I on the right track?

Thanks for your help.

The key is to treat this as a quadratic in $y=e^{ix}$

We have $m=\frac{y-y^{-1}}{2i}$
$y^2-2imy-1=0$. Solve this for $y$ in terms of $m$, and then take the logarithm.

Of course, any such formula is only one branch of the inverse trig function.

Ok when I do that, I come up with the following:

$\arcsin{x}=\frac{\ln{(ix+\sqrt{1-x^{2}})}}{i}$

and

$\arcsin{x}=\frac{\ln{(ix-\sqrt{1-x^{2}})}}{i}$

Any ideas what went wrong here?  I just solved the quadratic, took the natural logarithm, and divided by i.

Thanks again.

Ahhh, ok.  Sorry about that.

Thanks a lot for the help!


how do i solve this

for what value of x is
abs(3x+5) not positive.

well, if we have the absolute value of that, it always is positive

Basically, since a absolute value is never negative, it has to be 0 for this question.


Find the maximum of $2sin(\theta)+sin(2\theta)+2cos(\theta)+cos(2\theta)$.

[hide]
Cauchy
$(2^{2}+1^{2}+2^{2}+a^{2})(sin^{2}\theta+\sin^{2}2\theta+\cos^{2}\theta+\cos^{2}2\theta)\ge (2\sin\theta+\sin 2\theta+2\cos \theta+2\cos 2\theta)^{2}$
$(10)(2)\ge (2\sin\theta+\sin 2\theta+2\cos \theta+2\cos 2\theta)^{2}$
$\boxed{2\sqrt{5}}\ge 2\sin\theta+\sin 2\theta+2\cos \theta+2\cos 2\theta$
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well I didn't check that it's actually possible for the expression to equal $2\sqrt{5}$. Maybe that's why.

Could you post your solution?

Ya i made a program and it spat out

$\frac{3+3\sqrt{3}}{2}$

Which is when $\theta=30^\circ$

Okay.  My solution was very incorrect, I now see why.  SM4RT, can you prove it without a program?

I don't think it can acutally equal the value I got. My solution deals with the general problem:

$2\sin\theta_{1}+\sin\theta_{2}+2\cos\theta_{1}+\cos\theta_{2}$

My solution does not deal with the restriction $\theta_{2}=2\theta_{1}$

[hide="calculus solution..."]
$f(\theta)=2\sin\theta+\sin 2\theta+2\cos\theta+\cos 2\theta$
$f'(\theta)=2\cos\theta+2\cos 2\theta-2\sin\theta-2\sin2\theta$
Set to 0 to get:
$\cos\theta+\cos 2\theta=\sin\theta+\sin2\theta$
Which has the obvious solution $\theta=30^\circ$ as previously stated.

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[quote="10000th User"]Know well the conditions for Cauchy before you apply it, like AM-GM, know the conditions. Also, no calculus in Intermediate, it's a cheap shot attempt.[/quote]

Calculus is indeed an easy way out, but what else can you do with this problem?


Find all pairs $ (a,b)$ of positive integers such that $ 2a^2\equal{}3b^3$.

[quote="moldovan"]Find all pairs $ (a,b)$ of positive integers such that $ 2a^2 \equal{} 3b^3$.[/quote]

$ 3b^3$ must be even. So $ b\equal{}2y$.

The equation becomes: $ a^2\equal{}12y^3$.

$ a^2$ is divisible by $ 6$, because $ 6$ is square-free we get $ a\equal{}6x$. Substituting in the equation we get: $ 3x^2\equal{}y^3$.

So $ y$ is also divisible by $ 3$. Note $ y\equal{}3t$. Equation becomes: $ x^2\equal{}9t^3$.

So $ t$ is square let it be $ t\equal{}r^2$. We get that the solution are $ a\equal{}18r^3$, $ b\equal{}6r^2$ for any positive integer $ r$.


geometry
geometric transformation
rotation
circumcircle
trigonometry

$P$ is a point inside an equilateral triangle such that the distances from $P$ to the three vertices are $3, 4$ and $5,$ respectively. Find the area of the triangle.

[If I am not wrong, I have a truly beautiful solution. Will wait for others to post their solutions.]

[hide="hint"]This problem, with different distances, is a classical.
A rotation kills it  :D [/hide]

[hide]Let $A,B,C$ be the vertices of the triangle such that $AP=3$, $BP=4$, and $CP=5$.  Rotate the triangle 60 degrees clockwise around $A$.  Let $B', C', P'$ be the images of $B,C,P$, respectively, after rotation.  Consider quadrilateral $APBP'$.  Now $\angle P'AP=\angle P'AB+\angle BAP=\angle PAC+\angle BAP=60$.  Therefore, triangle $\bigtriangleup AP'P$ is equilateral, and $P'P=3$.  Then $P'PB$ is a 3-4-5 right triangle.  We have $[ABP]+[ACP]=[AP'BP]=\frac{9\sqrt{3}}{4}+6$.

Similarly, $[ABP]+[BCP]=4\sqrt{3}+6$ and $[ACP]+[BCP]=\frac{25\sqrt{3}}{4}+6$.  Then $2[ABC]=\frac{9\sqrt{3}}{4}+6+4\sqrt{3}+6+\frac{25\sqrt{3}}{4}+6=\frac{25\sqrt{3}}{2}+18$, and $[ABC]=\boxed{\frac{25\sqrt{3}}{4}+9}$.[/hide]

[color=darkblue][b][u]Generalization.[/u][/b] Let $P$ be a point inside an equilateral triangle such that the distances
from the point $P$ to the three vertices are $x\ ,\ y\ ,\ z$. Find the area of the triangle.

[b][u]Answer.[/u][/b] Denote the lenght $\boxed{a}$ of the side of the equilateral triangle and the area $\boxed{\sigma}$ of the triangle ([b]Pompeiu's triangle[/b]) with the side-lengths $x\ ,\ y\ ,\ z$
(from the [b]Pompeiu's theorem[/b] this triangle exists because the point $P$ doesn't belong to the circumcircle of the given equilateral triangle !). Then $\boxed{a^{2}=\frac{1}{2}\left(x^{2}+y^{2}+z^{2}\right)+2\sigma\sqrt 3\ }\ .$ The proof is identically with the previous classical [b]matt276eagles' proof[/b] in the particular case.[/color]

Euhm, is this wrong? [hide]Let's say $\Delta ABC$ is that triangle. Then $AB=BC=CA=t$ for a positive number $t$. We clearly have \[\frac{1}{2}\cdot t^{2}\cdot \sin\frac{\pi}{3}= [ABC] = [ABP]+[BCP]+[CAP] = \frac{1}{2}\cdot 3\cdot t+\frac{1}{2}\cdot 4\cdot t+\frac{1}{2}\cdot 5\cdot t = 6t.\] This means $t=\frac{24}{\sqrt{3}}$, so $[ABC] = \frac{144}{\sqrt{3}}$. [/hide]

[quote="JBL"][quote="Kurt Gödel"]Euhm, is this wrong? [/quote]

Yes -- distance to vertices $\neq$ distance to sides[/quote]

Woops, I hate misreading problems.  :|

[quote="rjt"][quote="JBL"][quote="Kurt Gödel"]Euhm, is this wrong? [/quote]

Yes -- distance to vertices $\neq$ distance to sides[/quote]

And if the question was distance to sides, Viviandi would make short work of this.[/quote]
Viviandi? What is that?

[quote="PenguinIntegral"]Viviandi? What is that?[/quote]Mathworld says: http://mathworld.wolfram.com/VivianisTheorem.html (note: it's Viviani. Not Vivian[b]d[/b]i.)
In fact, it's the same thing I did, but, yeah, they just gave it a name  :lol:

[quote="Virgil Nicula"]A great man is this Viviani ! [b]Rjt[/b], you know and another thinks about this Viviani ?[/quote]

You can read [url=http://www-history.mcs.st-andrews.ac.uk/Biographies/Viviani.html]here[/url] a brief biography of Vincenzo Viviani, as well as dozens of other mathematicians.

Thanks, [b]JBL[/b] ! I read at your address and is very interestingly.


inequalities
combinatorics unsolved
combinatorics

Find the number of nonnegative integer solutions to the following system of inequalities:
$\sum_{i=1}^k a_i \ge k$ for each $n\ge k\ge 1$ and $a_1+a_2+...+a_n\leq m$ where $m\ge n$
Give the answer in terms of $m,n$

Let $f(n,k)$ be the number of sequences with $a_1+a_2+...+a_n = k$. The original problem is equivalent to computing $F(n,m) = \sum_{k=n}^m f(n,k)$.

It is clear that $f(n,k) = F(n-1,k)$ and, thus, $F(n,m) = \sum_{k=n}^m F(n-1,k)$ implying $F(n,m)=F(n,m-1)+F(n-1,m)$ for $n<m$.
It is also clear that $F(n,n)=C_n$, the $n$-th [url=http://mathworld.wolfram.com/CatalanNumber.html]Catalan number[/url].

Therefore, $F(n,m)=\binom{n+m}{n}\frac{m-n+1}{m+1}$. See [url]http://oeis.org/A009766[/url] for details.


Kelsey drove to school at an average rate of 40 miles per hour and returned home at an average rate of 60 miles per hour. How many miles per hour was her average speed fro the entire trip?

We take the harmonic mean:

$ \frac{2}{\frac{1}{40}\plus{}\frac{1}{60}}\equal{}\frac{2}{\frac{1}{24}}\equal{}\boxed{48}$.