The bottom, side, and front areas of a rectangular box are known. The product of these areas is equal to:

$ \textbf{(A)}\ \text{the volume of the box} \qquad\textbf{(B)}\ \text{the square root of the volume} \qquad\textbf{(C)}\ \text{twice the volume}$
$ \textbf{(D)}\ \text{the square of the volume} \qquad\textbf{(E)}\ \text{the cube of the volume}$

[hide]
Let the length be denoted as $ l$, the width as $ w$, and the height $ h$. The area of the bottom is $ lw$, the area of the side is $ lh$ and the area of the front is $ wh$. The product of these is $ l^2w^2h^2$, which is the square of $ lwh$, the volume.  Thus, $ \boxed{D}$ is the correct answer.
[/hide]

[quote="Discrete_Math"][hide]
Let the length be denoted as $ l$, the width as $ w$, and the height $ h$. The area of the bottom is $ lw$, the area of the side is $ lh$ and the area of the front is $ wh$. The product of these is $ l^2w^2h^2$, which is the square of $ lwh$, the volume.  Thus, $ \boxed{B}$ is the correct answer.
[/hide][/quote]
Almost right, except check which one you picked at the end...  :wink:


Three cubes are stacked as shown. If the cubes have edges lengths in $1$, $2$, $3$, and shown, what is the length of the portion of segment $AB$ that is contained in the center cube?

umm that is a sort of bad image with no AB

[quote="me@home"]Hint: 2d side views :wink:[/quote]

How would I do that.

$\sqrt{6^{2}+3^{2}}=3\sqrt{5}$.

my problem is i don't know where A or B is so it looks like that old problem...

Sorry. I didn't get the old solution so I posted it again? I couldn't revive it.

[quote="funcia"]does that mean sqrt6 isn't right  :maybe:[/quote]
Actually it is correct, I'm 100 percent sure.
If you have aops vol. 1's solution booklet look in there, it has the solutions.

i know i just did that problem in preparation for chapters

[hide] Okay, I can't do the fancy equations that some of you are doing, but here is a good way to do it.
Say that the bottom left most corner of the bottom square is point C. Then the distance from point C to point B is sqrt18 (Yes that is a right angle, even though it isn't even an angle). Now imagine we construct a triangle, which is triangle ABC. One leg of triangle ABC is AC with length 6 units. The second leg is CB which is sqrt9. This makes the hypotenuse of the triangle, and length of AB sqrt54, which is the same has 3*sqrt6. Since the middle cube has frac1/3 of the total height of leg AC. That makes the length of the segment that is the proportion that is in the middle cube frac1/3*3*sqrt6, or 6. Sorry for my non cool symbolness, and lack of periods.  [/hide]


geometry
geometric transformation
reflection
ratio
vector
perpendicular bisector
geometry solved

A circle with center on perpendicular bisector of side $BC$ of triangle $ABC$
intesect $AB$ and $AC$ at $X,X'$ and $Y,Y'$.
 $XY$ and $X'Y'$ intersect $BC$ at $P,P'$ prove that $PM=P'M$ ($M$ is midpoint of $BC$)

[color=blue][b]Problem.[/b] Given a triangle ABC. Let a circle with center on the perpendicular bisector of the segment BC intersect the line AB at the points X and X' and the line AC at the points Y and Y'. Let the lines XY and X'Y' intersect the line BC at two points P and P'. Prove that the points P and P' are symmetric to each other with respect to the midpoint of the segment BC.[/color]

[i]Solution.[/i] We will use directed segments.

Since our circle has its center on the perpendicular bisector of the segment BC, it is symmetric with respect to this perpendicular bisector (in fact, a circle is symmetric with respect to any line through its center). In other words, the reflection in the perpendicular bisector of the segment BC leaves our circle fixed. On the other hand, this reflection maps the point B to the point C. Since reflection in a line leaves powers of points with respect to circles fixed, it thus follows that the power of the point C with respect to our circle equals to the power of the point B with respect to our circle. Now, the power of the point C with respect to our circle is $CY\cdot CY^{\prime}$, and the power of the point B with respect to our circle is $BX\cdot BX^{\prime}$. Hence, $CY\cdot CY^{\prime}=BX\cdot BX^{\prime}$. Also, $AX\cdot AX^{\prime}=AY\cdot AY^{\prime}$ by the intersecting chords theorem.

By the Menelaos theorem, applied to triangle ABC and the collinear points X, Y and P, we have

$\frac{BP}{PC}\cdot\frac{CY}{YA}\cdot\frac{AX}{XB}=-1$,

so that $\frac{BP}{PC}=-\frac{XB}{AX}\cdot\frac{YA}{CY}=-\frac{BX}{AX}\cdot\frac{AY}{CY}$.

Similarly, $\frac{BP^{\prime}}{P^{\prime}C}=-\frac{BX^{\prime}}{AX^{\prime}}\cdot\frac{AY^{\prime}}{CY^{\prime}}$. Thus,

$\frac{BP}{PC}\cdot\frac{BP^{\prime}}{P^{\prime}C}=\left(-\frac{BX}{AX}\cdot\frac{AY}{CY}\right)\cdot\left(-\frac{BX^{\prime}}{AX^{\prime}}\cdot\frac{AY^{\prime}}{CY^{\prime}}\right)$
$=\frac{BX\cdot BX^{\prime}}{CY\cdot CY^{\prime}}\cdot\frac{AY\cdot AY^{\prime}}{AX\cdot AX^{\prime}}=\frac{BX\cdot BX^{\prime}}{BX\cdot BX^{\prime}}\cdot\frac{AY\cdot AY^{\prime}}{AY\cdot AY^{\prime}}=1$,

so that $\frac{BP}{PC}=\frac{P^{\prime}C}{BP^{\prime}}$. In other words, $\frac{BP}{PC}=\frac{CP^{\prime}}{P^{\prime}B}$. Hence, the points P and P' divide the segment BC in the same ratio but "from opposite sides". In other words, the points P and P' are symmetric each other with respect to the midpoint of the segment BC. Problem solved.

  darij

Three questions, what are directed segments,are they vectors or...?
How to write letters in diferent colours?
How to write tha "click here to reveal hidden ..." thing?

Well, I can answer the last two questions. :D

When you are posting, there is something above your postbox labelled: [color=red]FONT COLOR[/color]. Also, there's something called the [hide]hide[/hide]. :D

Vector is a [b]quantity[/b] which has both size and direction.
Directed segment is a [b]segment[/b] which contains three elements: initial point, direction and length. The terminal point can be found whenever the three elements are known.

We use directed segments to represent vectors.

[quote="M@re"]I still don't get it, but never mind.... Thanx[/quote]
Well, it means: Vector is a quantity, directed segment is a segment. They are not the same.

Well I understanded that they are not the same :D :D :D


Problem:
Show that the product of four consecutive positive integers can never be a perfect square.

Thanks, both answers work well.
The answer to Rep's last problem can be solved along pretty much the same method he outlined in his proof:
For the sake of cleanliness, let x=28
then :sqrt: 28*29*30*31+1 =  :sqrt:(x)(x+3)*(x+1)(x+2)+1
=:sqrt:(x^2+3x)(x^2+3x+2)+1
now, by letting x^2+3x=y it is easy to see that the middle part is one away from a perfect square.  Adding one, it looks like this:
(y)(y+2)+1 = y^2+2y+1 = (y+1)^2
therefore, substituting the numbers back in, :sqrt:(28^2+3*28+1)^2 = 28^2+3*28+1 = 869

By the way, what do you do to get your text to look so good (like the square root roof to cover the whole thing)?

Now that this has been moved to "Intermediate", I'll give a response in spoiler, although the LaTeX reveals a lot... [hide]

If the algebra in the previous posts doesn't just come to you by divine inspiration, or if you are just better at seeing patterns, you might just try multiplying out the first few cases:

Since we are seeking perfect squares, it makes sense to ask "How close are these products to a perfect square?"

In each case, our product seems to be 1 less than a perfect square.  [Try a few more cases to convince yourself.]

If this is always true, the only perfect square our sequence could produce is zero.  But we were told the sequence must consist of positive integers.

If you want to "prove" our observation is correct, we proceed...

I happened to recall the formula [tex]n^2-1=(n-1)\times(n+1)[/tex].  Let's try to apply that, here.

The product of the middle two terms of our sequence always gives the larger of the two factors.  The product of the outer two terms of our sequence always gives the smaller of the two factors.

Now I think I know the formula!  We use a little more algebra to see if we are correct.  We will consider the general sequence [tex]\{x, x+1, x+2, x+3\}[/tex]

Let [tex]y=x^2+3x+1[/tex].  Then

[/hide]

[quote="tokenadult"]How long have each of you been involved in mathematical problem-solving?[/quote]

um, a bit more than one year.
Sorry for my *stupid* solution, i missed the fact that its always one less than a perfect square  :D  :D

[quote="tokenadult"]How long have each of you been involved in mathematical problem-solving?[/quote]

Let's see.. the AoPS  books arrived at my house at the beginnning of spring break, so for me it's a few months.


function
induction
limit
algebra unsolved
algebra

[quote="Chen241290"]$f(x)=\pm x$[/quote] how could you do ? type it clearly,ok,it's easy to talk,let's do it

$f(f(xy)+x+y)=f(x+y)+xy$ $\forall x,y\in R \ldots (*)$
Let $a=f(0)$
In $(*)$: y=0    $f(a+x)=f(x)$
$\Rightarrow f(a)=f(0)=a$
In $(*)$: x=a,y=1  $f(f(a)+a+1)=f(a+1)+a \Rightarrow f(a+1)+a=f(a+(a+1))=f(a+1) \Rightarrow a=0$
In $(*)$: x=$\sqrt{-t}$,y=$-\sqrt{-t}$ ; t<0
$f(f(t))=a+t=t; t<0$ $(f(f(t)=t; t\le0)$
[b]If f(m)=f(n):[/b]
   Let p such that $p^2>max(4m,4n)$
  $\Rightarrow p^2>4m$ and  $p^2>4n$
   In $(*)$: $x=\frac{p+\sqrt{p^2-4m}}{2}$, y=$\frac{p-\sqrt{p^2-4m}}{2}$
  $f(f(m)+p)=f(p)+m$
  In $(*)$: $x=\frac{p+\sqrt{p^2-4n}}{2}$, y=$\frac{p-\sqrt{p^2-4n}}{2}$
   $f(f(n)+p)=f(p)+n$
$f(m)=f(n) \Rightarrow f(p)+n=f(f(n)+p)=f(f(m)+p)=f(p)+m \Rightarrow m=n$
Therefore f is inyective

f is continous and inyective, hence f is strictly monotonous
  [b] I.-  If f is increasing:[/b]
   f(x)<0; x<0
   In $(*)$: $x=\frac{z+\sqrt{z^2-4f(t)}}{2}$,
$y=\frac{z+\sqrt{z^2-4f(t)}}{2}$ $(f(t)\le0, since \ t\le0)$
$f(f(f(t))+z)=f(z)+f(t)$
$f(t+z)=f(t)+f(z)$; t$\le0$
$0=f(0)=f(q+(-q))=f(q)+f(-q)$("q" o "-q" is non positive)
If $u\le0$ $f(u+v)=f(u)+f(v)$
If $u>0$ $f(u+v)=-f(-(u+v))=-f((-u)+(-v))$
$=-(f(-u)+f(-v))=-(-f(u)-f(v))=f(u)+f(v)$
Hence $f(u+v)=f(u)+f(v); u,v \in \Re$ and f is continuous (Cauchy's
equation)
$\Rightarrow f(x)=cx+d$ In $(*)$ we have $f(x)=\pm x$, but f is increasing,
therefore f(x)=x

[b]II.- f is decreasing[/b]
Let $k=f(-1)>f(0)=0$
f is continuous and dom(f) is an interval($\Re$)
Hence f($\Re$)\is an interval
$0=f(0)\in$ f($\Re$)
$k=f(-1)\in$ f($\Re$)
$\Rightarrow [0,k] \in$ f($\Re$) (interval)
If $h\in[0,k] \exists$ s $tq \ f(s)=h\ge0=f(0)\Rightarrow s\le0$
$f(h)=f(f(s)=s;s\le0 \Rightarrow f(f(h))=f(s)=h$
In $(*)$: $x=\frac{w+\sqrt{w^2-4f(h)}}{2}$,
$y=\frac{w+\sqrt{w^2-4f(h)}}{2}$
$f(f(f(h))+w)=f(w)+f(h)$
$f(h+w)=f(h)+f(w)$  $;h\in[0,k]$
$f(h+mh)=f(h)+f(mh)$
by two inductions:
f(mh)=mf(h); m is an integer
$\frac{h}{n}\in[0,k]$ n is an integer
Analogously:
$f(m.\frac{h}{n})=mf(\frac{h}{n})$
For m=n: $f(h)=nf(\frac{h}{n})$
Hence: $f(\frac{m}{n}.h)=\frac{m}{n}f(h)$
Is equiv. to: f(rh)=rf(h); r is a rational
If j\in$\Re$, $\exists \ a \ succession (r_n)\subset \ rationals \ such \ that \lim_{n\to\infty}r_n = j$
f is continuous, therefore:
$f(jh)= \lim_{n\to\infty}f(r_n .h)= \lim_{n\to\infty}r_nf(h)=jf(h)$
$f(x)=x.\frac{f(h)}{h}$
In particular: $f(x)=x.\frac{f(k)}{k} \ (lineal)$
In $(*)$: $f(x)=\pm x$ ,but f is decreasing \Rightarrow f(x)=-x
Hence $f(x)=\pm x$

[quote="tranthanhnam"]find all continuous function $f(x): R\to R$ such that :
     $f(f(xy)+x+y)=f(x+y)+xy$ $\forall x,y\in R$[/quote]

so,what if we delete the requirment that "f(x) is continuous"?will there be any more suitable F(x)? :huh:
I think it would be more difficult.


[quote="thatwasafail"]Which polygon has the same number of diagonals as it does sides?[/quote]
im such an idiot....
I solved this like 10 sec after i posted it!
Ans:pentagon

Solution, for anyone who wants it:

Number of diagonals is $ \binom{n}{2}\minus{}n\equal{}\frac{n^2\minus{}3n}{2}\equal{}n\implies n^2\minus{}3n\equal{}2n\implies n^2\equal{}5n\implies \boxed{n\equal{}5}$


What is the range of the rational function $f(x)=\frac{3x^{2}+2x-1}{x^{2}-4}$

Look at the domain first.

You aren't taking any even roots, but you are dividing.
You need to make sure you aren't dividing by zero.
Look for what values of X make a division by zero.

The denominator factors as $(x-2)(x+2)$
The numerator factors as$(x+1)(3x-1)$
Therefore, you are dividing by zero when one of the denominator's factors is zero.
Therefore, the domain of the function is $(-\infty,-2)\cup(-2, 2)\cup(2, \infty)$

Look at what we get out when we evaluate all elements of the domain, we see that we can evalute everything but$\pm2$, so look at the function at these numbers, and conclude that they are the only numbers not in the range.

While actually evaluating for these, you end up dividing by zero, so use synthetic division to find the remainder.

If you graphed the function, you'll see that the range doesn't include some other intervals as well

[quote="NeverOddOrEven"]you also gots an asymptote at $y=3$[/quote]

Yes, I know that. But have you looked at the graph yet?


ARML quadratics function geometry rectangle calculus derivative

so what did you guys think of ARML this year? i had a lot of fun but was disappointed at how i did on the actual contest, especially that this is my last one :(

You guys did fine.  The teams were really exemplary this year.

And hey, you could always do what I did and come back as a coach...

Overall, ARML was a lot of fun.  Last year I was on Chicago's D-Team and had a blast.  This year was even better.  I made Chicago A, which was totally unexpected, plus I went from 2 to 6 on individual over since last year.  Plus, I really enjoyed the power this year ... probably the best power I've done in two years of practice and competition.

Thank you to all the organizers and coaches for doing a wonderful job with everything.  Good job to all the teams and students, and if you're not a senior, I'll see you next year!

JB

ARML was really fun. I got my score from 3 to 5! :lol:  and since I have 2 more years of ARML left in me, I will shoot for National competition next yr (7 or 8). I should have had 6 or 7 (hitting myself after not reading "2" in front of sin and cos75 on #7), but since I was bad last year, I am pretty content. Plus, I got Team Medal despite having two people with Site or National Award-caliber.

The only thing that disappointed me was not being able to talk to Mr. Ruscyzk. I had nothing to talk to him about.... or I was just afraid to talk to him... I guess  :oops:

Overall, this year was fun. The Song contest was great too! Macky should have won it.... that would have marked the first ARML Song Contest winner not to sing. :D

Apparently we (chicago A) got clobbered on the last relay... I think EU was probably the only second seat to hand back a good answer, since I can't imagine anyone getting the third wrong (although "least upper bound" could be confusing.)

Anyhow, I didn't think our A-team would be good (since we were missing so many people) but it was just really solid down the line.

yeah, all three texas teams were slaughtered on the relay rounds as well. thats what made texas silver lose to wisconsin in div. B...... by one point! but overall, arml was a pretty darn fun trip.  :)

[quote](although "least upper bound" could be confusing.)[/quote]
One of the relays on our B team had trouble with that- it cost us a tie for the B division site award.

There were far more individual scores of 4 and 5 than last year, since problems 1,3, and 5 were all nearly as easy as the easiest problem last year. On the other hand, problems 6 and 8 were both about as hard as the hardest problem last year, and nobody got everything.

The Power round was probably too easy, and not really interesting enough- the average for all teams (excluding the three guest teams) was 31. In 2004, only two teams got more than 31.

For several reasons, the scores were closer together than most years, and that made it a year for upsets. Our A team didn't quite beat SFBA A, but we came as close as any western site team ever has.

The "least upper bound" was very confusing, especially since there wasn't anything "least" about it... but I was the only one on my whole team to get it right. :P

I'm extremely surprised at how easy #8 was, aside from one problem... you just need to express it as a quadratic function in terms of $a$, rather than a cubic function in terms of $x$. Once I did that (this morning), I got it in about five minutes. :roll: I guess nobody but the very small number of people who got it were able to see that in time?

As for my individual score, I got a 3... but considering that this is my first year (9th grade), and I made two extremely dumb mistakes... I think I did pretty well. :) I can't wait for next year, to see if I can get a 6 or something. :P

Yeah, I loved ARML this year.  Getting to go to tiebreakers (site tiebreak with a 6) was really exciting.  I was really close to getting a 7--I just misread #7 and switched a sin and a cos.  Of course, I was also very close to getting a 5, as I spent over 9 minutes solving #4 (did #3 in my head), got an answer I didn't like, then came up with a completely different solution in the last 30 seconds or so.

Here's my solution for #4:
Let D be a right angle.  Drop the perpendicular BE from B to CD.  Then BD is a diagonal of rectangle ABED.  F coincides with D, and so EPGF is similar to ABED.  Thus ABED is a square, and BE=height=8.

Also, I solved #8 by factoring, not by making it a quadratic in a.  Once I saw the $a^2-1$ constant term, it got pretty easy.  I did have to spend 5 minutes guessing and checking and playing with signs, but I got it.

Minnesota was in an interesting situation this year in that beyond a few top students, there didn't seem to be much differentiation in ability, so it was really up in the air who would be on the Gold (A) team and who would be on Maroon (B).  The results of this were bad--at the time of the lunch break, we were seriously considering the possibility that Maroon would win division B but get bumped up to division A by beating Gold.  But they didn't do well enough on relays for that to happen.

woohoo, the team i was on got 50th. Go south carolina B team.  and even better, our A team got tied for 9th. this is all in the B division.

Well, here's my strategy for #8...

Put the equation in terms of $a$, so $a^2-(x^2+2x)a+(x^3-1)=0$. Use the quadratic formula to find the roots $a=x-1$ and $a=x^2+x+1$. Subtract $a$ from both sides of both terms, and you have the original equation factored to $(x-(a+1))(x^2+x+(1-a))$. To have only one real factor, both factors of $x^2+x+(1-a)$ must be complex, so $b^2<4ac \Rightarrow 1<4(1-a)$, and $a<\frac{3}{4}$. That would actually only take about three minutes, I think, leaving over six minutes extra, since #7 was so easy. I'm surprised that more people didn't get this right.

I also made the same mistake on #7... and on #2 I lettered counter-clockwise instead of clockwise, which was specified. >_<

There was a 10th grader on my team who had a 5, made a dumb mistake on #7, and almost got #6, which would have gotten him a 7. :what?: We also had a 10th grader on our A Team (Southern California) who I think made two dumb mistakes and got a 5.

[quote="JesusFreak197"]
I also made the same mistake on #7... and on #2 I lettered counter-clockwise instead of clockwise, which was specified. >_<
[/quote]

I made the same mistake on #2. Pradeep, sitting to the left of me, also did the same thing...causing us to falsely think that we got the problem right for a whole minute after the answers were collected. So I get a 5. Three more years to go. I'll get somewhere. :)

Southern California A was 2 points away from SF A!! :mad:  You know what we did wrong? We said from the start that we couldn't win. Next year, we'll be #1 at San Jose.  :blush:

ARML is seriously fun, not just because of the math part, but also because of all the laughing. :rotfl: I'm seriously going to miss some of the seniors.

Thanks for an awesome competition.

[quote="jmerry"]
The Power round was probably too easy, and not really interesting enough- the average for all teams (excluding the three guest teams) was 31. In 2004, only two teams got more than 31.
[/quote]

The Power round was intentionally written to be a bit easier this year, and a head grader was heard to say (early on Sat. a.m.) that average scores were predicted to be in the 30s.

I thought the questions were sufficiently interesting, and laughed aloud several times while reading them.

[quote="codeblue87"]i thought it was a pretty fun competition. the sad part is we(florida) would have won division B nationally if our sponsors had actually determined alternates by their skill level. instead, they put names in a hat and randomly drew out two (17 people went).

so it turns out one of the alternates got a 5 while two of the people whose scores counted got 1s(and these people were predicted to do poorly).

haha, we did really bad on relay too, which cost us.

at least we were site winners(pennsylvania) and i had a good time.  next year we should have a better group going, which we'll need for division A :)[/quote]

what kind of a team with oleg rudenko on it enters div. B?   :huh::huh::huh:

[quote="SilentStealth"]Haha it was fun. I'm not a very diehard math type of person. I just stay after school to take a competition once in a fortnight.

I did quite well though.. It's good to know that I'm not the only one making 2 stupid mistakes and ending up with a 5.

On number 2 I also lettered incorrectly, and answered 2 instead of 1/2 ><
I didn't know how to do the circle within ellipse question, but I should have gotten #8, which is just some algebra.

The highlite of the actual competition for me was the trapezoid question when I totally missed the similar triangles but got it right by drawing a diagram (to scale) and estimating. ^^
I used the grains on my pencil as a ruler--it was total skill XD

I'm gonna have to say doing the pi rap thing was pretty stupid--my friend Kevin signed us up for fun, but hey, we got free candy =)[/quote]

dude. that is so incredibly awesome. I knew ppl could get AIME geometry by drawing to scale, but they have actual rulers...

well, we had oleg but the team didn't have that much depth. the people sponsoring the FL team made very bad decisions about who to let on, so there were a lot of good people (keenan, yannik, jason larue, others) who we didn't have.

if we would have had 10 or so USAMO qualifiers and a couple others i'm sure we would have entered under A.

are you from FL arnav? i know someone has told me about you but i can't quite remember..

[quote="SilentStealth"]#8, which is just some algebra.[/quote]
That was the only question I could not figure out how to do.

[quote="codeblue87"]if we would have had 10 or so USAMO qualifiers and a couple others i'm sure we would have entered under A.[/quote]
Not true. All new teams, regardless of quality, are in B division their first year. Those new teams that are of A quality get there by winning a national B award.

Yes indeed.

As for me, I worked in the scoring room all meet, and thus got to see the scores as they piled in.  I was surprised at some of the teams, and when the grader read scores of 12 and 14 for one of the relays the whole room fell silent almost, trying to figure out who went nuts on the relay.

Obviously, number 6 and 8 were killers, and I've been still trying to understand the method in number 8 using calculus...

Wait... a method of solving #8 involving Calc? Would that be derivatives? My dad suggested that, but that's extremely complicated... there doesn't seem to be much point to it, when it can be solved so much more quickly.

I tried setting the derivative equal to 0, solving, plugging the smaller root into the original thingy and setting its value there negative or the larger root in and showing it's positive but that was just a heck of a lot of work

no I'm not floridian  :rotfl: ok so that wasn't actually funny but I like that emoticon

apologies for the misunderstanding about the divisions

I just posted an outline of a solution using calculus in the similar thread in the Intermediate forum.  Mystic -- was it necesarily obvious that the root with the - sign would be the smaller root?  I'm skeptical.  In any event, that solution should turn out the same as mine.  (You have to compare absolute values, which is equivalent to comparing squares, which is equivalent to examining the difference of the squares, which is equivalent to what I did.)

There was a number 8 three or four years ago that looked like it required calculus.  Any ARML problem that looks like it can be solved with calculus, shouldn't be -- there will definitely be a nicer solution.  I imagine they work hard to keep the calculus methods too long or computationally nasty to be really doable in the 7 or 8 minutes that top competitors will usually have for number 8.

I love you Joel!

oh man., we almost had to pay out our bet!

#8 from 2003 looks like a calculus problem, but I challenge anyone to come up with a calculus solution in under half an hour. On the other hand, a non-calculus solution is pretty easy.

ha, no one on our team got #6, and we did awful in the relays. our entire team had two 6 minutes and one 3 minutes, but we were first up until that.


I'm looking for a proof of the identity

$ \sum_{n \equal{} 0}^{\infty}f(x\plus{}n) \equal{}\sum_{n \equal{} 0}^{\infty}\frac{B_{n}D^{n}F}{n!}$

where D is the differential operator d/dx and dF/dx = f(x)

Sorry, I got this off a physics research paper(typical), though the form I gave is invalid, the argument goes like this,

$ \sum_{n \equal{} 0}^{\infty}f(x\plus{}n) \equal{}\sum_{n \equal{} 0}^{\infty}\sum_{k \equal{} 0}^{\infty}\frac{n^{k}f^{(k)}(x)}{k!}$

By Taylor's theorem(when it be valid of course), exchanging order of summation we have,

$ \sum_{k \equal{} 0}^{\infty}{\zeta}(\minus{}k)\frac{f^{(k)}(x)}{k!}$

since for negative integers,
$ {\zeta}(\minus{}k) \equal{}\frac{(\minus{}1)^{k}B_{k\plus{}1}}{(k\plus{}1)!}$

We can obtain the desired result!!!

Very useful in physics I'm sure!

[quote="yasiru89"]Sorry, I got this off a physics research paper(typical), though the form I gave is invalid, the argument goes like this,

$ \sum_{n \equal{} 0}^{\infty}f(x \plus{} n) \equal{} \sum_{n \equal{} 0}^{\infty}\sum_{k \equal{} 0}^{\infty}\frac {n^{k}f^{(k)}(x)}{k!}$

By Taylor's theorem(when it be valid of course), exchanging order of summation we have,

$ \sum_{k \equal{} 0}^{\infty}{\zeta}( \minus{} k)\frac {f^{(k)}(x)}{k!}$

since for negative integers,
$ {\zeta}( \minus{} k) \equal{} \frac {( \minus{} 1)^{k}B_{k \plus{} 1}}{(k \plus{} 1)}$

We can obtain the desired result!!!

Very useful in physics I'm sure![/quote]

sorry for the additional factorial in the $ {\zeta}(\minus{}k)$ closed form!!

So thus I came across the connection to Taylor's series, imagine my surprise when I learnt that one can think of $ f(t\plus{}n)$ as the result of the operator $ {\exp}{n{\delta}}$ applied on $ f(t)$
Then we have,
$ \sum_{n \equal{} 0}^{\infty} f(n) \equal{} \sum_{n \equal{} 0}^{\infty} {\exp}{n{\delta}} f(0)$

When t is limited to 0.
We can sum the exponential series as a geometric progression(this works even for the operator)and,
$ \sum_{n \equal{} 0}^{\infty} f(n) \equal{} (1 \minus{} {\exp}{{\delta}})^{\minus{}1} f(0)$

since,
$ \frac{{\delta}}{{\exp}{{\delta}} \minus{} 1} \equal{} \sum_{n \equal{} 0}^{\infty}
\frac{B_{n}}{n!} {\delta}^{n}$
we substitute and we have,

$ \sum_{n \equal{} 0}^{\infty} f(n) \equal{} \minus{}\sum_{n \equal{} 0}^{\infty} \frac{B_{n}}{n!} {\delta}^{n\minus{}1}f(0)$

A form more similar to that expected(the operations on the operator are valid)


pigeonhole principle
AMC
USA(J)MO
USAMO
induction

Let $n=2^k$ for some positive integer k. Prove that given a set of $2n-1$ distinct arbitrary integers, it is always possible to find $n$ integers within this set such that their sum is divisible by n

Call the $2n-1$ numbers $a_1,a_2 \cdots a_{2n-1}$

Consider the $2n-1$ numbers:

$a_1$
$a_1+a_2$
$\vdots$
$a_1+a_2+ \cdots + a_{2n-1}$

since $2n-1 > n$, we have that two of the above numbers a equivelent (mod n), so the difference of these is divisable by n. QED

By the way, there is no need for the $2^k$ condition. Also, i think we could have done it with n numbers just as easily...

But the problem says "find $n$ numbers", and you haven't covered that part. For instance, if the first and the last sums are equal modulo $n$, you'll have $2n-2$ numbers, not $n$.

[quote="seamusoboyle"]Then use induction. Its not too hard.[/quote]

That would be great for the USAMO!!!

For any problem I can't figure out, just write "Use induction, it's not that hard".
 :rotfl:

hmm i think this works...

we note that every integer can be expressed in the form $A(2^m)$, for odd integer $A$ and positive integer $m \ge 0$

here we also see that $2n-1 - n +1 = n$, and since $n=2^k$, we know that regardless of how we choose our $2n-1$ integers, we will always have $n$ integers that take on the form $A(2^j)$ for $j \ge k$, and hence, adding these up, it is clear that the sum will be divisible by $2^k$

[quote="gotztahbeazn"]we know that regardless of how we choose our $2n-1$ integers, we will always have $n$ integers that take on the form $A(2^j)$ for $j \ge k$[/quote]

why

the only integer in that form is $n = 2^k$

yea disregard my solution its wrong :blush:

[quote="Bictor717"]I thought the solution was straightforward when seamusoboyle suggested induction.

[hide="This is what you would induct on..."]You can test $k=0,1$ cases pretty easily.
Now assume it works for $k$ and show it leads to $k+1$. Our assumption is that given a set of $2n-1$ arbitrary positive integers, we can find a $n$-element subset with the property that its elements sum to a multiple of $n$.
The $k+1$ case we need to prove is that (since $2^{k+1}=2n$) given a set of $4n-1$ arbitrary positive integers, we can find a $2n$-element subset with the property that its elements sum to a multiple of $2n$.[/hide]
[hide="This is how I reasoned it out."]Pick any $2n-1$ elements from the $4n-1$-element set, apply the assumption, and set those $n$ aside. You are left with a $3n-1$-element set. Apply the assumption again twice to get two more $n$-element sets. So we can find 3 disjoint $n$-element sets of integers with the property that its elements' sum is divisible by $n$.
Consider the sums $\mod 2n$. They can only be $0$ or $n$. Pigeonhole says at least two must be equivalent. Take the union of those two. It's a $2n$-element subset of the the $4n-1$ set of arbitrary numbers with the property that its elements sum to a multiple of $2n$.[/hide]
Sorry if you got lost.[/quote]

Same solution as mine! I think this is a good problem, as you think it needs pigeonhole, but its so easy using induction! The thing that gave it away to me was the $2^k$ in the problem. Whenever you see something like this in a problem, think induction!


function
limit
calculus
inequalities open
inequalities

[b]Conjecture:[/b] Suppose $m_{r}$ is the $r$--th degree power mean, \[m_{r}=\left(\frac{x_{1}^{r}+\cdots+x_{n}^{r}}{n}\right)^{1/r}.\] Suppose $r,s,t$ are real numbers such that $r\geq s\geq t.$ Prove or disprove \[m_{r}^{s-t}m_{s}^{t-r}m_{t}^{r-s}\leq 1,\] or, equivalently, \[\left(\frac{m_{r}}{m_{s}}\right)^{s-t}\leq \left(\frac{m_{s}}{m_{t}}\right)^{r-s}.\] EDIT: Reversed the sign

What are you doing bodan? :wink:

Ok. Give me solutions please!!!

[quote="bodan"]
or, equivalently, \[\left(\frac{m_{r}}{m_{s}}\right)^{s-t}\geq \left(\frac{m_{s}}{m_{t}}\right)^{r-s}.\] [/quote]

or, equivalently, $\left(\frac{m_{r}}{m_{s}}\right)^{\frac{1}{r-s}}\geq \left(\frac{m_{t}}{m_{s}}\right)^{\frac{1}{t-s}}.$

Assume $t>0$. (for $t<0$ it seems to be false.) Then we can assume wlog $s=1$ (replace $x_{i}$ by $x_{i}^{s}$). Put $m_{1}=: c$.  This leaves us with $\left(\frac{m_{r}}c\right)^{\frac{1}{r-1}}\geq \left(\frac{m_{t}}c\right)^{\frac{1}{t-1}}.$ for $r>1> t$.
This is just a function of one variable: $f: r\mapsto \left(\frac{m_{r}}c\right)^{\frac{1}{r-1}}$.
Obviously, we have  $\lim_{r\to0}f(r)=\frac{m_{1}}{m_{0}}\ge 1$ and $\lim_{r\to\infty}f(r)=1.$ Note as well $\lim_{r\to1}f(r)=\exp(\frac{m^\prime_{1}}{m_{1}})\ge 1$.  And in fact it seems that the function $f(r)$ might always be [i][b]strictly decreasing [/b][/i]for $r>0$, so the [i][b]converse [/b][/i]of this beautiful conjecture would be true.
Lots of calculus ahead.  :maybe: I wouldn't do it. (Has Mildorf an idea?  :wink: )


number theory
greatest common divisor
relatively prime

Let $ k$ be a positive integer such that $ \frac{k(k+1)}{3}$ is a perfect square.
Show that $ \frac{k}{3}$ y $ (k+1)$ are perfect squares.

Like here in my solution: [hide="Link"][url]http://www.mathlinks.ro/Forum/viewtopic.php?p=895928[/url][/hide]

[hide="Solution"]
$ k(k+1)=3n^{2}$ for $ n\in\mathbb{Z}$
We have $ \textrm{gcd}(k,k+1)=1$, so it must be:
$ k=a^{2}\wedge k+1=3b^{2}$ or $ k=3a^{2}\wedge k+1=b^{2}$ for some $ a,b\in\mathbb{Z}$.
But 1st case is impossible, becouse:
$ a^{2}+1=3b^{2}$ hasn't got any solution (look at $ mod\ 3$ for example)[/hide]

Exactly, so it can't be divisible by $ 3$.

Couldn't you just say $ 3$ divides one of $ k, k+1$, so we divide that one by $ 3$, and then that and the remaining number are relatively prime, giving the desired result?  I suppose it's no simpler, but just an alternate way to do it...


probability geometry perimeter LaTeX videos

A math problem says to add A and B.  By mistake, I subtracted B from A, you got 66 or 1/3 of A+B.  What is B?

abcd is a four digit number more than 3000 and less than 4000.  abcd=$ a^a\plus{}b^b\plus{}c^c\plus{}d^d$.  What is abcd?

The jayhawks (jayhawks pwn final four) have 100 females and 80 males on band.  The jayhawks have 80 females and 100 males in strings.  60 female are both in strings and band.  there are 230 students, how many males are in band?

2 six sided dice are rolled.  what is the probability that the product of the 2 numbers will be more than ten?  express as common fraction

an equilateral triangle (nonzero area) has the property where the perimeter and the area of the triangle is the same.  what is the side lenght of the triangle?

any help is appreciated.  thanks

if 66 is 1/3 of a+b, then a+b=198(check if my mental math is wrong at any point)so we have two equations, namely:
a+b=198
a-b=66
adding these two you get
2a=264, or a=132...just solve for b now
132+b=198
b=198-132=56 ans.

for the second problem, note that a=3, and that no digit can be larger than 5.
im not entirely sure how to do it..i might figure it out later...

for number 3, instead of finding the probability that the product is greater than 10, find the probability of getting atmost 10 and subtract it from 1.

so if the 1st die is a 1, then there are 6"succeses" for the second die, since the product in each case is less than 10
if the 1st die is  a 2, there are 5 succeses
" " 1st die is a 3, there are 3 succeses
1st die is a 4, 2 succes
1st die is a 5, 2 succeses
1st die is a 6, 1 success
adding these, you get 6+5+3+2+2+1=18 out of 36 possibilities, so 18/36 or 1/2 is the probability of getting at most 10
subtracting from 1, you get that the product being greater than 10 has a probability of 1/2...i found that kinda surprising....

for the =lateral triangle problem, just set 3s^2*sqrt3/4=3s and solve
so s^2*sqrt3=4
s^2=4/sqrt3, or s=2/(4throot)3
simplify if you wish

[quote="ajain"][quote="mathking123"]if 66 is 1/3 of a+b, then a+b=198(check if my mental math is wrong at any point)so we have two equations, namely:
a+b=198
a-b=66
adding these two you get
2a=264, or a=132...just solve for b now
132+b=198
b=198-132=56 ans.

[/quote]

u do mean =66 ans.  right?[/quote]

yeah...oops.... :oops:  :rotfl:

[quote]2. abcd is a four digit number more than 3000 and less than 4000.  abcd=$ a^a \plus{} b^b \plus{} c^c \plus{} d^d$.  What is abcd?

3. The jayhawks (jayhawks pwn final four) have 100 females and 80 males on band.  The jayhawks have 80 females and 100 males in strings.  60 female are both in strings and band.  there are 230 students, how many males are in band?[/quote]

2. Note that there has to be one 5 somewhere. Thus, there must be a three, and then trial and error.

3. Uh, it gives you 80 males in band, unless you were asking how many were in strings and band, in which you would use Inclusin-Exclusion, for 180+180-60-n = 230, thus n is 70.

uh...it does give you how many males are in the band, but if you even [i]tried[/i] to do this problem, you would see that there are some males that will be in [b]both[/b] strings and band.

and about #2, i think the answer is 3435, i tried it and  think it works.

@uldivad9:  in the PIE, its going to be (180-60)+(180-60) + 60, or simply 120 + 180

[quote="mathking123"]
so if the 1st die is a 1, then there are 6"succeses" for the second die, since the product in each case is less than 10
if the 1st die is  a 2, there are 5 succeses
" " 1st die is a 3, there are 3 succeses
1st die is a 4, 2 succes
1st die is a 5, 2 succeses
1st die is a 6, 1 success
adding these, you get 6+5+3+2+2+1=18 out of 36 possibilities, so 18/36 or 1/2 is the probability of getting at most 10
subtracting from 1, you get that the product being greater than 10 has a probability of 1/2...i found that kinda surprising....

for the =lateral triangle problem, just set 3s^2*sqrt3/4=3s and solve
so s^2*sqrt3=4
s^2=4/sqrt3, or s=2/(4throot)3
simplify if you wish[/quote]

for the die problem, i get 19/36...when the problem says more than ten, would getting ten count as not getting more than ten??

and for the lateral triangle: is $ 3s^2\frac{\sqrt{3}}{4}$ the area of a equilateral triangle, because i only know thsi formulas: $ \frac{s^2}{4}\sqrt{3}$  could you explain this problem a bit more, and possibly write it in latex?

[hide="For the Triangle Problem"]
\begin{align*}
\frac{s^2\sqrt{3}}{4} &= 3s \\
\frac{s\sqrt{3}}{4} &= 3 \\
s &= 3 \cdot \frac{4}{\sqrt{3}} \\
&= \frac{12}{\sqrt{3}} \\
&= \boxed{4\sqrt{3}}
\end{align*}
[/hide]

[quote="Yongyi781"][hide="For the Triangle Problem"]
\begin{align*} \frac {s^2\sqrt {3}}{4} & = 3s \\
\frac {s\sqrt {3}}{4} & = 3 \\
s & = 3 \cdot \frac {4}{\sqrt {3}} \\
& = \frac {12}{\sqrt {3}} \\
& = \boxed{4\sqrt {3}} \end{align*}
[/hide][/quote]

Thanks a lot yongyi781!!! :D

By the way guys, is #2 3435 and #3 10 and #4 19/36???  Some other person got 1/2 for #4, but on # 4 can you have a product =ing exactly ten and count it as more than ten?

for the triangle problem i screwed up with the mental algebra and i admit it, so stop bothering me about getting the wrong answer! :P

for the dice problem, i counted the ways of not getting more than 10, and subtracted it from 1, so getting 10 counts as not getting more than 10...

i probably messed up somewhere else though! :P

[quote="mathking123"]for the dice problem, i counted the ways of not getting more than 10, and subtracted it from 1, so getting 10 counts as not getting more than 10...[/quote]

Thanks for clearing up the confusion.  I was also not bothering you on that triangle problem.  So if ten counts then...is the answer $ frac{17}{36}$.

RANDOM FACT FOR PEOPLE WITH WIIS: If you have the "Internet Channel" you can play Aops For The Win, but I highly reccamend that you join a response game cause it's hard to get your answer in time and first!!!

ehh, i still think im right....can anyone explain what is wrong with my logic? i checked it over(second post i think) and i still think its right...

thanks for any help

OKay...I am going to post how I did the dice problem.  First I found all the possible answers ten or below, listed in this table:
__1_2_3_4_5_6_
1|1  2 3  4  5  6
2|2  4  6  8  10
3|3  6  9
4|4  8
5|5  10
6|6

As you can see, we have [b]19[/b] ways to NOT get more than ten.  So:
$ 1\minus{}\frac{19}{36}\equal{}\frac{17}{36}$

oops...i added wrong :oops:

btw, has anyone here seen the video on youtube called "what you know about math?"?


I was browsing Wikisource today and I found this:

[url=http://en.wikisource.org/wiki/Industrial_Society_and_Its_Future]http://en.wikisource.org/wiki/Industrial_Society_and_Its_Future[/url]

It's really interesting, the way it puts things...

And it was written by a [url=http://en.wikipedia.org/wiki/Theodore_Kaczynski]convicted murderer[/url]..

I've got 3 questions:
1) Did you read ALL of that? :what?:
2) Do you think the industrial society will break down?
3) Do you think the world would be better off or worse off without the "system", as they call it in the article?

I read only parts of it.

[quote] If the system succeeds in acquiring sufficient control over human behavior quickly enough, it will probably survive. Otherwise it will break down. We think the issue will most likely be resolved within the next several decades, say 40 to 100 years.
[/quote]
I don't think the "system" will "fail". At least, not in the near future.
This article goes against itself in many places, but I'm too lazy to quote them now. For instance, in the introduction they say this will be so and that will be so, but later on in the article they give so many conditions for it to happen that it most likely would not happen.

Anyway, this whole article was just weird... but I think we should not form our opinions solely by looking at one source. We should gather different views on the subject and then decide which one to believe :)

Hmm, I did not realise immediately this is the unabomber manifesto

the unabomber is perhaps for most people the most notorious mathematician of the last century

as a child he suffered from some kind of allergic reaction, marking him psychologically

he appeared very gifted as a mathematician, and obtained a phd at a university in michigan
he was a specialist in complex analysis
he also worked at berkeley, where he was again considered gifted but students rated him poorly because of his social behaviour

he then later went on to target scientist, universities etc.... with bombs
He made people lose fingers or eyes, he killed a computer store owner, maimed a geneticist

His own brother recognised his manifesto and turned him in, he even used his own money to compensate the families of his victims

You can read more here http://en.wikipedia.org/wiki/Theodore_Kaczynski

My first conclusion would be that this is an asocial gifted person who wanted to be an anarchist fighting for a better world, but who was actually angry about his own social life

That's why i do not feel like reading his manifesto right now

[quote="fredbel6"] My first conclusion would be that this is an asocial gifted person who wanted to be an anarchist fighting for a better world, but who was actually angry about his own social life[/quote]
Hmm, that's exactely the impression you get when you read it.

All I can say is that is a huge wiki. Pretty wierd mathematicain.

[quote="fredbel6"] as a child he suffered from some kind of allergic reaction, marking him psychologically [/quote]
Whether that had anything to do with the Unabombing is unclear, to say the least.  There was some speculation that his participation in electrical-shock psychiatric experiments to earn money during college, had a more direct effect.

[quote] he appeared very gifted as a mathematician, and obtained a phd at a university in michigan [/quote]
He appeared talented, but not necessarily gifted or amazing as a mathematician.  The head of department at Berkeley said that Kaczynski probably would have been tenured eventually had he stayed there, so apparently he was solid if not a superstar.

[quote]He also worked at berkeley, where he was again considered gifted but students rated him poorly because of his social behaviour  [/quote]
Not necessarily unusual for a mathematics graduate student of his generation and background.

[quote]His own brother recognised his manifesto and turned him in, he even used his own money to compensate the families of his victims  [/quote]

Not exactly.  His brother was eligible to receive the USD 1000000 reward for information leading to capture of the Unabomber, and said that he would accept the funds and use them mostly for compensating the victims, and partly for his brother's legal defense and related expenses.  I never heard of the brother using his own money (separate from the reward) for payments to the victims.

Wasn't he also kicked from another job because of a harassment complaint of a female coworker?


[quote="xxreddevilzxx"]how does the ranking work... because it was fast for me to go from Nvs NP to Conjecture, but it is taking  long time for me to go from POintcare to Riemman.

also, is it 100% dependent on how many posts i make?[/quote]

Yes, it is. It has been discussed several times on the forum. For example,  [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=26876]here[/url].

Best regards,

[quote="xxreddevilzxx"]thanx... i want more starss soon!!!!!!111111[/quote]Be sure not to spam too much ;) We can render your post count not to increase when you post if you try to spam in order to get more stars :P

[quote="filletwho"]What's the definition of spam? Is it just a post that has no meaning, post with too many words, too many posts? I used to think it was a post with swear words or something.[/quote]Meaningless posts, made just to increase your post count. They will have the opposite effect ... ;)


A regular octagon, ABCDEFGH, has side length 12. What is the area of the trapezoid BCDE? Express your answer in simplest radical form.

[hide="answer, but no solution :ewpu:"] $ 72\plus{}72\sqrt{2}$[/hide]

[hide]
You can easily find that one angle of the octagon is 135. So, making the trapezoid, the top two angles are 135º, and since the bottom two are equal, they must both be 45º. So drop a perpendicular from one of the top corners and you get a 45-45-90 triangle. So the height must be 12/sqrt2 or 6sqrt2.

Then, the big base must be 2(6sqrt2)+12 or 12+12sqrt2. This averaged out with the top base is 12+6sqrt2. Multiply the two to get 72+72sqrt2.[/hide]


I'd appreciate some assistance with the following problem:

Given a real valued function f(x) = sqrt ((3x+k)/(w+2x)) where k and w are integers and 0<k<w.  If the largest integer NOT in the natural domain of f is -6 and the smallest integer NOT in the natural domain of f is -10, find the maximum possible value of k+w.

I reasoned as follows:  If w were an even integer greater than or equal to 22, then f would have a discontinuity at an integer less than or equal to -11 which contradicts -10 being the smallest integer not in the domain of f.  Thus I concluded that w was probably less than 22.  Since we want to maximize k+w, I decided to let w=21.   For w=21, the denominator of the radicand is positive for x = -6 and x=-5, so f will be undefined if the numerator of the radicand is negative and defined when the numerator of the radicand is positive or zero.   Since we want f to be undefined when x=-6 and f to be defined when x=-5, we note that 3(-6)+k<0 and 3(-5)+k>=0, which implies 15<=k<18.   Since we're maximizing k+w,  I decided to let k =17 and arrived at the (correct) answer of 17+21 = 38.

Although my final answer is correct, I am unsatisfied with my solution since it involved making a leap  (based on intuition) which turned out to be correct.   I'd be very interested in a more rigorous explanation -- Please help if you can.

[quote="SJS"]I'd appreciate some assistance with the following problem:

Given a real valued function f(x) = $\sqrt{\frac{3x+k}{w+2x}}$ where k and w are integers and 0<k<w.  If the largest integer NOT in the natural domain of f is -6 and the smallest integer NOT in the natural domain of f is -10, find the maximum possible value of k+w.[/quote]

Evaluating at the given points yields non-real values, so we must have

$\frac{-18+k}{w-12} < 0$ and $\frac{-30+k}{w-20} < 0$

Furthermore, since those points are the extreme values, we have -5 and -11 in the domain, so

$\frac{-15+k}{w-10}  > 0$ and $\frac{-33+k}{w-22} > 0$

Consider if in the two left inequalities, the denominator changed signs.  Then w = 11, but then on the right inequalities, for them to change sign, we must have 30 < k < 33, contradicting k < w.  Therefore, in the two left inequalities, the numerator changes sign, so we have 15 < k < 18.  Then it follows that w = 21 from the right inequalities.  I don't see why you can't have k = 16 though.


You don't have to answer all the questions. I would really appreciate your answers. Thanks

1. Portia has waited a long time for her ship to come in, and she has concluded that it will
arrive today with probability 1/4. If it does come, she will receive $16. If it doesn't come
in today, it never will and she will have zero wealth. She has a Bernoulli utility function
equal to the square root of her total income. What is the minimum price at which she
would sell the rights to her ship?
2. Dr. Parkin's car is worth $5,000. But he is careless and leaves the door open and the
keys in the ignition. Consequently his car will be stolen with probability .4. If it is stolen,
he will never get it back. Dr. Parkin also earns $5,000 from the university for which he is
working, and his Bernoulli utility function for wealth is u(w) = ln(w): Suppose that he
can buy $K worth of insurance for his car at a price of $.4K. Let w1 and w2 denote his
wealth level when the car is stolen and not stolen, respectively.
(a) What is Dr. Parkin's utility function considering the uncertainty involved in his car?
(b) If Dr. Parkin buys $K worth of insurance, how much of w1 and w2 does he end up
with?

(c) What is Dr. Parkin's optimal level of wealths, w1 and w2? How much does he pay for
the insurance he buys at the optimum? How much on average will the insurance company
pay back to Dr. Parkin?

3. Quasimodo consumes cereals and other things. His utility function for cereals x and
money to spend on other goods y is given by
u(x; y) = 100x - x2 + y:

(a) Let p denote the price for cereals and y be a numeraire whose price is set equal to
1. What is his demand curve for cereals?

(b) If the price of cereals is $20, how many cereals will he consume? What if the price
of cereals is $30?

(c) Suppose that Quasimodo has $2000 in total to spend and the price of cereals has
increased from $20 to $30. How much does his income have to increase to make him not
complain about the price increase? (In other words, what is the compensating variation?)
(d) What is the change in consumer's surplus when the price changes from $20 to $30?
Compare this number to the answer to (d).

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=271325[/url]


PROBLEM 1-2
Suppose we have a line segment with end points $ P$ and $ Q$. What is the difference between the distance $ PQ$ and the distance $ QP$?

SOLUTION 1-2
This is an interesting question. If we consider distance without paying attention to the direction in which it is measured, then $ PQ \equal{} QP$. But if direction is important, we define $ PQ \equal{}  \minus{} QP$. In basic plane geometry, direction is sometimes specified in diagrams in order to get viewers to move their eyes from right to left instead of from left to right, or from bottom to top rather than from top to bottom.

--------------------------------------------------------
What I didn't understand is $ PQ \equal{}  \minus{} QP$  :huh: ?

The way this problem is worded is a bit confusing. When it says PQ, it means the distance, as measured from P to Q. When it says QP, it's the distance from Q to P (But in the opposite direction of PQ). While the lengths of the two are the same, it is important to note in plane geometry the direction (positive or negative) as well as the length.

[quote="hasan4444"]
[b]If we consider distance without paying attention to the direction[/b] in which it is measured, then $ PQ \equal{} QP$.
[/quote]
This is called ''scalar".

[quote="hasan4444"]
But [b]if direction is important[/b], we define $ PQ \equal{} \minus{} QP$.
[/quote]
This is called "vector".

Do you mean the direction is important in some cases?

Yes in some cases direction is really important.It seems you didn't learn yet about Vectors at your school. :o


[quote="Buffalo"]Let $F_{n}$ be the Fibonacci numbers with $F_{1}$=$F_{2}$=$1.\ $ For natural numbers $a$ and $k$, define $n\left(  a,k\right)  $ as the smallest natural number $n$ $s.t.$ $F_{n}$ is divisible by $a^{k}.$

  Prove that there exists two numbers $c\left(  a\right)  $ and $k_{0}\left(
a\right)  $ $s.t.\ n\left(  a,k\right)  $=$c\left(  a\right)  a^{k}$ for all $k$>$k_{0}\left(  a\right)  .$

  Try to make an estimation of $c(a)$ and $k_{0}\left(  a\right)  $.[/quote]
Well, by definition, if $n(a,k)$ is the minimum value verifying the property using ths fact that \[\gcd(F_n,F_m)=F_{\gcd(m,n)}\] we get that $n(a,k+1)=\lambda n(a,k)$, i am trying to solve in generak but it's a hard problem.

Best Regards and good luck.

RobertuX the Crazy math

[quote="Buffalo"]Let $F_{n}$ be the Fibonacci numbers with $F_{1}$=$F_{2}$=$1.\ $ For natural numbers $a$ and $k$, define $n\left(  a,k\right)  $ as the smallest natural number $n$ $s.t.$ $F_{n}$ is divisible by $a^{k}.$

  Prove that there exists two numbers $c\left(  a\right)  $ and $k_{0}\left(
a\right)  $ $s.t.\ n\left(  a,k\right)  $=$c\left(  a\right)  a^{k}$ for all $k$>$k_{0}\left(  a\right)  .$

  Try to make an estimation of $c(a)$ and $k_{0}\left(  a\right)  $.[/quote]
Are you sure it is true?

Not quite, but I can't find a counterexample with my computer, in fact, it's impossible to find one by computer because of $k_{0}\left(a\right)$. Up to now, $k_{0}\left(a\right)$ seems to be small, maybe bounded w.r.t.  $a$.


induction
combinatorics proposed
combinatorics

A positive integer is written in each square of an m x n board. The allowed move is to add an integer k to each of two adjacent numbers (whose squares have a common side). Find a necessary and sufficient condition for it to be possible to get all numbers zero by a finite sequence of moves.

Consider instead a net of $m\times n$ lattice points. We are given $mn$ positive integers as the labels of the points, and our goal is to label the edges belonging to the lattice in such a way that the sum of the numbers around each point equals the label of that point. Call this last condition $C$. As you can imagine, the points represent the cells, while the edges represent the connections between cells.

Name the labels of the points $a_{i}^j$, just like in any matrix. An obvious necessary condition is $\sum_{i+j=2k}a_{i}^ j=\sum_{i+j=2k+1}a_{i}^ j$.

Conversely, let's assume this condition holds. We use induction. Let $x_i^ j$ be the horizontal edge connecting $a_i^ j,a_i^{j+1}$, and let $y_i^ j$ be the vertical edge connecting $a_i^ j,a_{i+1}^ j$. I don't think I'll get into all the necessary details, but you can check for yourself that if we label $x_1^1,\ x_2^ 1,\ldots,\ x_m^1$ s.t. $\sum_{i+1=2k}x_i^1-\sum_{i+1=2k+1}x_i^1=\sum_{i+1=2k}a_i^1-\sum_{i+1=2k+1}a_i^1$, then we can label $y_1^1,\ y_2^1,\ldots,\ y_{m-1}^1$ in such a way that $a_i^1$ (the points on the first column) satisfy condition $C$.

We can now use the induction hypothesis on the $m\times (n-1)$ table which is left after eliminating the first column, by replacing the labels $a_{i}^2$ with $a_i^2-x_i^1$.

Maybe it would be helpful to draw a picture.


Well, I believe that this game is very cool, so I want to make a topic about it.

The rules are:

1) We will begin with 3 sudokus
2) If you have the answer to anyone of them, post the solution and the sudoku number, and you will earn 1 point. If you are wrong, you don't earnany point and we continue with the same sudoku.
NOTE: If you don't know how to make the sudoku board, copy and paste [hide="this code:"]
[code]$\begin{tabular}{|c|c|c||c|c|c||c|c|c|}
\hline & & & & & & & & \\
\hline & & & & & & & & \\
\hline & & & & & & & & \\
\hline
\hline & & & & & & & & \\
\hline & & & & & & & & \\
\hline & & & & & & & & \\
\hline
\hline & & & & & & & & \\
\hline & & & & & & & & \\
\hline & & & & & & & & \\
\hline \end{tabular}$[/code][/hide]and place the right numbers between the &'s (the last number of the row between the & and the \, and the first between hline and the first &).
If you can't do it, just attach an image or a document (with Excel it's very easy).
3) After from posting the solution, post another sudoku. If you do, you earn another point. If not, I will post another sudoku and you will not earn the other point.
4) If you don't know how to play sudoku, look [url=http://www.sudoku.name/rules/en]this url[/url] and you will know all what you need.
5) Please post if you want to participate in this tournament. I will start if there are 15 people in the tournament.
6) Don't post unsolvable sudokus. if you do, you lose 3 points.
7) GOOD LUCK!!!!!!!!!!!!!!

EXAMPLE (very easy):
$ \begin{tabular}{|c|c|c\parallel{}c|c|c\parallel{}c|c|c|} \hline & 4 & 1 & & 9 & & 7 & & 5 \\
\hline & 5 & 2 & 8 & & 4 & & 1 & \\
\hline & & & & & & & & 3 \\
\hline \hline 7 & & 3 & 1 & & & & & \\
\hline & 2 & 5 & & 4 & & & & 1 \\
\hline 1 & & 4 & & & & 5 & & 2 \\
\hline \hline & & & 4 & 5 & 7 & & 6 & \\
\hline & & 7 & & & 6 & 2 & & 8 \\
\hline 4 & 1 & & 9 & & & & & \\
\hline \end{tabular}$

[b]The completely useless registered players list:[/b]
1) BOGTRO
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
12)
13)
14)
15)

[hide="NOTE:"]
yes, this was a BUMP.
A BUMP is when... blah blah blah, look [url=http://www.mathlinks.ro/viewtopic.php?t=196860]this[/url] and you will know what a BUMP is[/hide]

lol

(This message is a BUMP. Please make it longer before

BUMPING).

Does anyone more want to register? Does anyone want to shame me because I created a completely failed game?

come on!
(The BUMP is too small. Please makSHUT UP!)

[quote="BOGTRO"]
This is the FINAL BUMP.

A BUMP is when someone makes a useless post so that the topic goes to the top

of the Forum, when viewed.

A BUMP is entirely useless.

A BUMP is when you want people to see the topic.

You would want someone to see your topic if

1. Its a game where you sign up, and you need people to see this.

2. You just like being on top of other people.

3. You're bored

The first option applies to this game

and many more.

The first option is because applications may be a little slow

Or it may be people just don't scroll down.

The second option is a problem. You are obsessed with power.

The third option applies to me. I am making a long post because I'm bored.

Yay.

Apparently, not many people are interested.

If this BUMP does not work I call for a LOCK.
[/quote]

taken from Survivor...

 :(

sure, ill join

im too lazy to do the problem right now so ill do it later


calculus
derivative
algebra
polynomial
real analysis
real analysis unsolved

Suppose that f(x) and g(x) have n + 1 derivatives on [ a - k, a + k] and that f^(k)_(a) = g^(k) = 0 for 0<=k<n and g^(n)_(a) is not = 0. Use Taylor polynomials to show that lim {f(a)/g(a)}=f^(n)_(a)/g^(n)_(a) as x->a

does this have to do with L'Hospitals rule

this looks like l'hopiltals rule:

f ' (x) =  lim h->0 [ f(x+h) - f(x) ] / h =

let u = x + h
and a = x
as h approaches 0, u approaches a

f ' (u) = lim u ->a [ f(u) - f(a) ] / (u - a)

---

if f(a) = 0:

f ' (x) = lim x ->a f(x) / (x - a)
g ' (x) = lim x ->a g(x) / (x - a)

so f(x) / g(x) = f ' (x) / g ' (x) as x approaches c

------

i've noticed i can look at the syntax of other peoples expression, but what do i write first so its recoginized as an expression and not text?

[quote="oooooh ya"]i've noticed i can look at the syntax of other peoples expression, but what do i write first so its recoginized as an expression and not text?[/quote]
Dollar signs. (dollar)\frac{x^n}{52+z_1}(dollar) becomes $\frac{x^n}{52+z_1}$.


Let a,b,c >0 such that    $ ab\plus{}bc\plus{}ca\equal{}1$     .Find min:

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

It's an old problem of can_hang.
We can solve it by the lenma:
$ \frac{a^2}{b} \plus{} \frac{b^2}{c} \plus{} \frac{c^2}{a} \geq\ \frac{(a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c)}{ab\plus{}bc\plus{}ca}$
 :)

[quote="nguoivn"]It's an old problem of can_hang.
We can solve it by the lenma:
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \geq\ \frac {(a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c)}{ab \plus{} bc \plus{} ca}$
 :)[/quote]
Suppose that $ c\equal{}mic\{a;b;c\}$
<=>$ \sum_{cyc} \frac {a^2}{b} \minus{} \sum_{cyc} a \geq \frac {(\sum_{cyc} a^2 \minus{} \sum_{cyc} ab)(a \plus{} b \plus{} c)}{ab \plus{} bc \plus{} ca}$
<=>$ (a \minus{} b)^2[\frac {a \plus{} b}{ab} \minus{} \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca}] \minus{} (a \minus{} c)(b \minus{} c)[\frac {b \plus{} c}{ac} \minus{} \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca}] \geq 0$
Easy to prove that $ \frac {a \plus{} b}{ab} \minus{} \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \geq 0$
$ \frac {b \plus{} c}{ac} \minus{} \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \geq 0$
Done!


Syntax Error wrote:The sum of all five terms of an arithmetic sequence is 55.  Find the sum of the first three terms of the sequence if the sum of the last three is 48.

[hide]Middle term: 11

1, 6, 11, 16, 21

18[/hide]

...so the sum would be 18. you didn't offer any justification for the 4th number being 16, but I assume you just did 48/3, which is correct.

Here's how I did this problem with variables. It's much slower, but is more justified.

First 5 terms of the sequence:
$a+(a+r)+(a+2r)+(a+3r)+(a+4r)=55$
$5a+10r=55$

Last 3 terms of the sequence:
$(a+2r)+(a+3r)+(a+4r)=48$
$3a+9r=48$

Now, just solve the systems of equations to find the variables.
$3a+9r=48$
$a=16-3r$

$5(16-3r)+10r=55$
$16-3r+2r=11$
$r=5$

Now that we have this, the rest is obvious.


:maybe: Normally its around 5 each year... I just came to know of 3 yet-
Akriti Gaur(Delhi)
Jaswanth Madhavan(Chennai)
Smitha Radhakrishnan(NRI)
Maybe Some Archit NRI too...
Are they aopsians?
Edit: Disclaimer: All information stated above is based on sensible but unconfirmed reports. But yeah, don't beat me up if its wrong  :blush:

I'm rejected.

Some guy on that "unofficial" MIT decisions IRC channel said they MIT had 400 Indian applicants. 400 applicants and only 5 admitted? That's scary.


Does anyone have 1997-1998 solutions for warm up 20? can you even buy this book anymore?
If anyone has the solutions could you please post/pm me?
Thanks so much! :)

I believe this post goes in this appropriate thread:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133189&start=540

I do.  :)

But this isn't the appropriate forum and I'm not giving it out for free. Also, I am not giving out handbooks in my free tests thread.

akhil0422 could you just give answers for warmup 20?
thanks.

Even if someone pays me, I won't give it out because I got it from a SPECIAL SOURCE and that SPECIAL SOURCE does not let me give stuff out and I won't break my promise to my SPECIAL SOURCE or tell you who that SPECIAL SOURCE is.

And that SPECIAL SOURCE is in all caps because his username is too? AIME15 clearly? Does it matter?


induction
number theory open
number theory

It is known that $a^2 + b^2 = 3(c^2 + d^2)$ has no solutions $\in \mathbb{N}$.

Generalizing the problem, we wish to find the smallest integer $c$ such that $\sum_{k=1}^{n} a_n^2 = c \sum_{k=1}^{n} b_n^2$ has no solutions for two sets of naturals $a_n, b_n$.  For $n = 4$, I have tested $c = 2..11$ and always been able to find a solution, and I have not been able to find a counterexample for $n = 3$ either.  So my question is:

Prove or disprove:  $\sum_{k=1}^{n} a_n^2 = c \sum_{k=1}^{n} b_n^2$ always has a solution $\in \mathbb{N}$ for $n > 2$, regardless of the value of $c$.

(And if you disprove, describe $c$ in relation to $n$.)

Edit: I first forgot that none is allowed to be $0$. Now it should be correct.

Case $n=1$: trivial.

Case $n=2$: exactly those $c$ work that can already be written as sum of two squares (see case $n=3$ for the reason).

Case $n=3$:
Exactly those $m$ that are not of the form $4^s u$ with $u \equiv 7 \mod 8$ can be written as sum of three squares.
By factoring out some factors $4$ we can assume that $4 \nmid c$.
To avoid that $m$ or $cm$ can be written already as sum of two or less squares, use the characterisation of the sums of two squares: all their prime factors $\equiv 3 \mod 4$ occure an even number of times each.
So take (e.g., there are a lot more possibilities) a prime $p \equiv 3 \mod 8$ that doesn't divide $3c$. Then $p$ and $3p$ can be written as sum of three but not of two squares. But then at least one of the numbers $pc$ or $3pc$ is not $\equiv 0;4;7 \mod 8$, so can be written as sum of three squares and by the factor $p$ not as sum of a lesser number of squares.

Case $n \geq 4$:
By induction on $n$, take $a_1,a_2,...,a_{n-1},b_1,b_2,...,b_{n-1}$ that fulfill the condition for $c$ and $n-1$. Now take any integer that is a square and can be written as sum of two nonzero squares, e.g. $5^2=3^2+4^2$, so in general $m^2=k^2+l^2$.
Then the sets $a_1 m , a_2 m , ... , a_{n-2} m , a_{n-1} k , a_{n-1} l$ and $b_1 m , b_2 m , ... , b_{n-2} m , b_{n-1} k , b_{n-1} l$ contain no zeros but fulfill the property.


The midpoints of square $ ABCD$ are connected to form square $ EFGH$. If $ EF \equal{} 10$ units, what is the area, in square units, of square $ ABCD$?

[asy]size(101); pair A, B, C, D, E, F, G, H;
A = (0,2); B = (2,2); C=(2,0); D=(0,0); E=(1,2); F=(2,1); G=(1,0); H=(0,1);
draw(A--B--C--D--cycle);
draw(E--F--G--H--cycle);
label("$A$", A, NW);
label("$B$", B, NE);
label("$C$", C, SE);
label("$D$", D, SW);
label("$E$", E, N);
label("$F$", F,  ENE);
label("$G$", G, S);
label("$H$", H, W);[/asy]

Side length of ABCD=$ 2*\frac{10}{\sqrt{2}}\equal{}10\sqrt{2}$
Area of ABCD=$ (10\sqrt{2})^2\equal{}100*2\equal{}\boxed{200}$ square units.

Another way you can do it (a quicker way) is notice that ABCD's area must be twice the area of EFGH (since EFGH come from ABCD's four midpoints), so EFGH's area is $ 10^2\equal{}100$, making us conclude that $ [ABCD]\equal{}100*2\equal{}\boxed{200}$.

EDIT: Beaten.


I recieved an email notification that a reply had been added to the topic AoPS Kings test POSTED(all levels) at 7:44 EST, yet the last post on it was me at 2:50 EST (both times PM).

I thought earlier that I got a false notification, but I just figured it was one I hadn't deleted.

Mail might have been late. Not a confirmed bug (my notifications seem ok). Keep a watch on it and report if other emails get late / missing.

Its not all of them, just that one so far.

Also, I got a notification for the post before mine, so I don't think it could have just been late...

Forward me both emails please (valentin at mathlinks.ro )

Ok, I sent them.

Note that the second one was grouped as a conversation by gmail (same day, same message).

I suspect that the same bug that affected the links of the topics affected this issue also. I think it is solved now, but if anything like this pops up again, please let me know.

I got another message about a reply had been posted in AoPS Kings test POSTED(all levels) last night, after I got off, and there still hasn't been a reply.

You may have changed something after that, but I don't know (I got off a little before 10 PM EST.

Yeah, not sure, let's just wait this one out see what happens. I have no clues as to what (else) can be wrong.

Whoa, I don't think this is a bug.  Arnav (MysticTerminator) had repeatedly been spamming in that thread, and I just kept deleting it until he stopped.  So perhaps the TRN went out for his post, but the post was deleted by me, so nothing showed up.

Yes, that probably would cause it to set off.  When was the last time you deleted stuff?

Thanks!

I'm not sure of the times, but that's approximately correct.  Looking back, I'm relatively sure that's the problem. $\boxed{QED}$


Given a regular hexagon ABCDEF with sides of length 1, find the area of triangle ACE.

[hide="Answer"]
Note that $\triangle ACE$ is an equilateral triangle since $\triangle ABC$ ~ $\triangle AFE$ ~ $\triangle CDE$ by SAS (an angle and two sides of hexagon).

Take $\triangle ABC$ for example. We know $\overline{AB} = 1$ and $\overline{BC} = 1$ with $\measuredangle ABC = 120^\circ$ (since it's part of hexagon). Thus, we have $\measuredangle BAC = 30^\circ$. Make an altitude from $\measuredangle ABC$ and we can easily figure out that the half of $\overline{AC} = \frac{\sqrt 3}{2}$ and thus, $\overline{AC} = \sqrt 3$.

$A = \frac{s^2 \sqrt 3}{4} = \frac{3 \sqrt 3}{4}$.

QED.
[/hide]

Thanks for pointing out.

Hide your answers silver so other people can try it before looking.

How I did it:
[hide]First of all, the 6 congruent triangles formed in a regular hexagon by the diagonals are all equilateral. The side length is also 1. $\triangle ABC$ is 3 of these triangles.

Use the formula for the area of an equilateral triangle: $\frac{s^2\sqrt{3}}{4}$

$3*(\frac{1^2\sqrt{3}}{4})$ = $\frac{3\sqrt{3}}{4}$[/hide]   :lol:

Well, if you draw the diagram, you'll see that a regular hexagon has 6 equalateral triangles with sides of 1. If you draw the triangle ACE, you'll realize that it goes through half of all the triangles or 3 equalateral triangles. By knowing the equalateral triangle formula you can tell that it is equal to[b] (3x3^1/2)/4[/b].

Sorry if it's hard to read my answer, I don't know how to do that latex stuff.  :oops:

[hide]
ACE is equilateral. AC is sqrt3 because of 30-60-90 rt triangles.
using sqrt3 in the formula, we get: 3sqrt3/4
[/hide]

[quote="Iversonfan2005"][quote="chess64"][quote="Iversonfan2005"]...I don't know how to do that latex stuff.  :oops:[/quote]

It's easy (and I'm in 8th grade!) Just click LaTeX on the sidebar and start learning!  :)[/quote]

Hey, where's the sidebar?[/quote]

Just click this: http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php

Thanks Chess, I'll use it if I start getting a lot of exponents in my answers.


MATHCOUNTS counting distinguishability search

On the refrigerator, MATHCOUNTS is spelled out with 10 magnets, one letter
per magnet. Two vowels and three consonants fall off and are put away in a
bag. If the Ts are indistinguishable, how many distinct possible collections of
letters could be in the bag?


Find $ z\equal{}x\plus{}2i, x\in \mathbb{R} , z\in \mathbb{C}$ such that the number $ |z\minus{}3\minus{}4i|$ attains it's minimum value .

[hide="Solution"]$ z\minus{}3\minus{}4i \equal{} (x\minus{}3) \minus{} 2i$. We want to minimize $ \sqrt{(x\minus{}3)^2 \plus{} 4}$, which is minimized when $ (x\minus{}3)^2$ is minimized, which occurs when $ \boxed{x\equal{}3}$.[/hide]


If a grating is ruled 580 lines per millimeter and the screen is 65 centimeters from the grating, how far from the axis will the violet line of a spectrum appear


Wow this is creepy... I hope everything turns out ok, and the man will be found...

[quote="lightrhee"]I was watching local news yesterday, and I found out that Robert, who won two bronze in IMO for Canada, is missing.
Read [url=http://www.canada.com/globaltv/edmonton/story.html?id=3fcebc8b-db36-4933-b7e9-014a93bbf26c]this[/url].[/quote]
That is creepy. I hope he's found safe and ok.

Thanks for letting us know.  Please keep us posted.

Oh my gosh, he's missing?!!!

Him and Alex taught me Mao. Hehehehe.  They also had some of the scariest Mao rules I've ever seen.

he must be proving riemann hypothesis inside the closet.

Thank you for lightning the mood killmesoft. :roll:
This is serious though.

Today was the first day of the Alberta math camp, so I went there to meet the kids(I met Mathgod. First thing he said to me was that he was told by Yufei that I am crazy at Rubik's cube :blush: ).

On the way there and back, I saw four [url=http://grad.econ.ubc.ca/cpbl/robert/poster16Aug.pdf]posters[/url]. There are some information on [url]http://findrobert.com[/url].

a friend of mine from UT told me about this yesterday. Best wishes to robert

Me, Dr. Andy Liu, and Jerry Lo(some of you know him from the National Camp) were interviewed by CBC. We told them that Robert is a very nice guy, that this is something unexpected, and that he is the one of the smartest people we know.

finerobert.com is updated and now it looks more organized.

er, I think it is findrobert.com, not finerobert!

[quote="weileaf"][quote="lightrhee"]Me, Dr. Andy Liu, and Jerry Lo(some of you know him from the National Camp) were interviewed by CBC.[/quote]
Are you going to be on CBC tonight? I feel the media should cover this incident a little more, there were huge media coverage when Cecilia Zhang was missing. (Maybe this isn't an appropriate comparision)[/quote]
yes it isn't. Cecilia Zhang was special to her family as Robert is to his. You can't really compare to incidences like this.

I hope they find him soon..

Keep us updated.

I was so sorry to hear the news.  I remember Robert as always happy, and an all-around good guy.  It is truly tragic that this happened to someone so young and bright.

[quote="bugzpodder"]This is my last post at AOPS[/quote]

[quote="bugzpodder"]I do not want to hijack this thread from being Robert to about Putnam[/quote]

Then don't.  wtf?  There are so many better places to be a dramawhore than this thread.

I met him at the CUMC this year; seemed like a really neat, undoubtedly brilliant guy.  My condolences to his family and friends.

I also saw Robert at CUMC this year, never got a chance to talk to him. What makes me more sad is the picture on findrobert.ca is a picture of him at CUMC in Montreal.

I gave my best wishes to Robert's family and friends.

First of all, my sincere condolences to Robert's family and friends. Even though I've never met him, his being double-winner three years ago inspired me to work hard in mathematics. So this is really dreadful news. Rest in peace, Robert.

p.s. I agree with the blahblahblah, it's analogous to me saying right now: I do not want to hijack this thread from being Robert to being bugzpodder, but I do not think you (bugz) should have said what you said.

It's just unreal... and unfair...

Although I have never met him... RIP Robert

My condolences to the family and friends.

I met him at the IMO in Japan... I'm SHOCKED.

I've been following this in the news for a few days. I didn't know what to say about it on the boards, since I never had the pleasure of meeting him. But I saw another picture of his on the net today, and I relized something:
He had a great smile.   :)

[quote="José"]What was officially? murder, tragedy or suicide?[/quote]

Suicide... but probably they won't have the conclusive investigation finished for a while.  I know it makes no sense... but it makes no sense to his family either.

Any further information?? Maybe a witness?

A very bright young mathematician (who, in particular, you do not know personally) killed himself for reasons unbeknownst to us.  Can't you leave it at that, at least for the purposes of this thread?

It is ok to be curious. I will admit though, suicide doesn't sound right to me. But, i guess as blahblahblah says, lets leave it at this.

a year has passed since Robert went away.  I just want to say that you are not forgotten Robert.  Your friends will always remember you.  Rest in peace.

Which one is it? http://www.student.math.uwaterloo.ca/~rfurmani//rbl/guestbook.cgi

 :(  Poor guy. I have never met him but I wish I did. He sounds like an absolutely brilliant person. It is so unfair that he is snatched from the world just like that.


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

Suppose that R is a commutative ring without zero divisors. Show that the characteristic of R is 0 or prime..

Note: It doesn't say that R is an integral domain so we do not know whether R has unity.

[quote="ZetaX"]How do you define characteristic¿[/quote]

A characteristic is the smallest postive integer n such that for every element [i]a[/i] in a ring, [i]an[/i] = 0

You can adapt the prove for the case with a unit:
First, assume that the characteristic $c$ is not $0$, since then there is nothing to show.
When $c=pq$ with nontrivial $p,q$, look at $(px)(qy)=(pq)(xy)=0$. So we know that $px=0$ or $qy=0$ for all $x,y$. Assume that there is some $x$ such that $px \neq 0$ (otherwise, the characteristic would be $\leq p<pq$). Then $qy=0$ for all $y$, also contradicting since $q<pq$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

A group $ G$ of order $ 22$ contains elements $ x$ and $ y$, where $ x$ is not $ 1$ and $ y$ is not a power of $ x$. Prove that the subgroup generated by these elements is the whole group $ G$.

What is the size of $ \langle x\rangle$?  What is the size of $ \langle x,y\rangle$?

Suppose $ x \neq 1$.
We know that $ |x|$ divides $ 22 \implies |x| \equal{} 2$ or $ |x| \equal{} 11$.
$ |x| \equal{} 2 \implies |y| \equal{} 11$ while $ |x| \equal{} 11 \implies |y| \equal{}2$. In either case we find that the two elements of G generates 22 elements.

[quote="akech"]$ |x| \equal{} 2 \implies |y| \equal{} 11$ while $ |x| \equal{} 11 \implies |y| \equal{} 2$.[/quote]
This is wrong.  It is quite possible for both $ x$ and $ y$ to have the same order.  Think about my hint a little more; this is really a question about subgroups, not elements.

In fact, it is not hard to see there are only two groups of order 22, one is cyclic and other is dihedral.


The two common external tangent segments between two non intersecting circles cut off a segment along one of the common internal tangents. Show that all three segments have the same length.

This problem is really not that hard. You just need to use the "equal-tangents" theorem a whole bunch of times.


What is the seventh term in the arithmetic sequence $ 4,\frac72,3,\frac52,\ldots$?

Okay, there are $ 6$ term decreases, each of $ \frac12$.  So we have $ 4 \minus{} \frac12 \times 6$, or $ 4\minus{}3$, which equals $ \boxed{1}$.


Solve in the set of natural numbers the equation :

$ n!\plus{}n\equal{}n^3$

Divide the equation by $ n$
$ (n\minus{}1)! \plus{} 1 \equal{} n^2$
$ (n\minus{}1)(n\minus{}2)! \equal{} (n\plus{}1)(n\minus{}1)$
$ (n\minus{}2)! \equal{} n\plus{}1$
$ n \equal{} 5$

$ n! < n^3 \implies n < 6$
By inspection, $ n\equal{}5$ is the only solution.


geometry
geometric transformation
reflection
geometry unsolved

Triangle $ ABC$ satisfy $ \widehat{C}\equal{}\frac{\pi\plus{}\widehat{B}}{2}$. Denote $ M$ be midpoint of $ BC$, $ D$ be symetric point of $ M$ by altitude $ AH$ of triangle $ ABC$.
a) Prove that $ MD\equal{}AB$.
b) Exbisector of $ \widehat{A}$ cut $ BC$ at $ E$, incircle tagent $ AB$ at $ F$. Prove that $ CE\equal{}2AF$.

Denote by $ N$ the middle point of the side $ AB$. We have $ \angle MHN\equal{}\angle B$, and $ \angle HMN\equal{}\angle HCA\equal{}90^{\circ}\minus{}\frac{\angle B}{2}$. Hence we conclude that $ HM\equal{}HN$. Notice that $ \triangle HAB$ is right-angled at $ H$, hence we have $ AB\equal{}2\cdot HN\equal{}2\cdot HM\equal{}DM$

Denote by $ P$ the reflection of $ C$ across $ AH$, and by some calculations, we can easy to see that our problem is equivalent to prove that $ AP\equal{}PE$.
Indeed, we have $ \angle PEA\equal{}\angle PAE\equal{}\frac{\angle C\minus{}\angle B}{2}$

I'm sorry, I don't have enough time to explain my solution, but I think now it's very easy and you can do it by yourselt.


irrational number
number theory unsolved
number theory

Prove that:$\forall$ two integers m,n, $\exists{}x\in\mathbb{R}$ such that both $\{mx\}$ and $\{nx\}\in (\frac{1}{3},\frac{2}{3})$

EDIT: I'm sorry I've read something like for all $x$ irrational there is an integer $m,n$ such that ...
Anyway I promess I'll solve this problem (ZetaX now knows why :))

This is probably the kind of proof you would love to avoid...
For any $x \in \mathbb{R}-\mathbb{Q}$ the sequence $a_n = \left\{ nx \right\}$ is dense in $[ 0 ; 1 ]$.
Choose an irrational number $0 < x < \frac{1}{6}$.
By density of $a_n$, there is an positive integer $m$ such that $\left\{ mx \right\} \in \left( \frac{1}{3} ; \frac{1}{2} \right)$.
Since $0 < x < \frac{1}{6}$, we have $\left\{ \left( m + 1 \right) x \right\} \in \left(0 + \frac{1}{3} ; \frac{1}{2} + \frac{1}{6} \right) = \left( \frac{1}{3} ; \frac{2}{3} \right)$.

How does this solve the problem¿ :?

PS: {x} answering problem {x} ... nice  :D

Oh, this is a problem of a National Olym pic 96 or 97...


Μου φαίνεται ότι είναι εύκολη αλλά σαν να μου έχει κάτσει στραβά και δεν έχω βρει κάποια λύση.... :mad:

Αν $ 0<\delta<\frac{2}{\pi}$ να δειχθει τότε ότι:

$ \left|\int^{\frac{2}{\pi}}_{\delta}sin(\frac{1}{x})dx\right|\leq\frac{4}{\pi^{2}}$

Μια λύση εδώ [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1129294#1129294[/url]


I wonder whether the following inequality holds for $ k \geq 3$?

For any $ a^1_1,\ldots,a^1_n,\ldots,a^k_1,\ldots,a^k_n$ then
\[ \prod_{i\equal{}1}^k \sum_{j\equal{}1}^n (a^i_j)^2 \geq n^{k\minus{}2}(\sum_{j\equal{}1}^n \prod_{i\equal{}1}^k a^i_j)^2.
\]

[quote="officemax"]I wonder whether the following inequality holds for $ k \geq 3$?

For any $ a^1_1,\ldots,a^1_n,\ldots,a^k_1,\ldots,a^k_n$ then
\[ \prod_{i \equal{} 1}^k \sum_{j \equal{} 1}^n (a^i_j)^2 \geq n^{k \minus{} 2}(\sum_{j \equal{} 1}^n \prod_{i \equal{} 1}^k a^i_j)^2.
\]
[/quote]

No - in the particular case when $ a^1_j \equal{} a^2_j \equal{} ... \equal{} a^k_j$ for every $ j$ (so that $ a^i_j$ is actually independent of $ i$) you get the power mean inequality with opposite sign, what must obviously be wrong in general. Even for $ k \equal{} 3$ it's wrong.

An assertion which is somewhat similar but true is the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=102983&search_id=1536997288]generalized Hölder inequality[/url] (see also http://www.mathlinks.ro/Forum/viewtopic.php?t=3574 and http://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality#Generalization for something more general).

  darij


The sum of the last four digits of Marian’s phone number is 30. How many such four-digit sequences are there?

[hide]We first need quadrouples (x,y,z,w) that will make x+y+z+w=30.

(3,9,9,9) (4,8,9,9) (5,7,9,9) (5,8,8,9) (6,6,9,9) (6,8,8,8) (6,7,8,9) (7,7,7,9) (7,7,8,8)
(I think that's all)

All of them can be arranged into more because order matters.

In the order I listed them, $4+12+12+12+6+4+24+4+6=84$.[/hide]


Let $ E$ and $ F$ be the respective midpoints of $ BC$ and $ CD$ on unit square $ ABCD$. If lines $ DE$ and $ BF$ intersect at $ G$, what is the area of triangle $ BDG$?

[asy]size(200);
pair A=(0,0), B=(0,1), C=(1,1), D=(1,0), E=waypoint(B--C,0.5), F=waypoint(C--D,0.5), G=IP(B--F,D--E);
D(unitsquare);
D(E--D--B--F);
MP("A",D(A),SW);
MP("B",D(B),NW);
MP("C",D(C),NE);
MP("D",D(D),SE);
MP("E",D(E),N);
MP("F",D(F),E);
MP("G",D(G),NE);[/asy]
[hide="Solution"]The equation of line $ BF$ is $ y \equal{} \minus{} \tfrac{x}{2} \plus{} 1$, and of line $ DE$ is $ y \equal{} \minus{} 2x \plus{} 2$. Solving, we get $ G\left(\tfrac{2}{3},\tfrac{2}{3}\right)$.

The distance from $ \left(\tfrac{2}{3},\tfrac{2}{3}\right)$ to $ \left(\tfrac{1}{2},\tfrac{1}{2}\right)$ is $ \tfrac{\sqrt {2}}{6}$, and the area of $ BDG$ is $ \sqrt {2}\cdot\tfrac{\sqrt {2}}{6}\div2 \equal{} \boxed{\tfrac{1}{6}}$.[/hide]

I find that another way to approach a problem like this is to do side ratios, and find out what the ratio of a one part of a line is to the other part of it that was intersected. Since ratios wont change the area, it doesn't matter what they are, as long as every thing is matched up.

Sorry if this isn't clear. If its not, I will post a picture.

Also, it is a well known fact that the medians of a triangle trisect each other. So the required area is 2/3 of one quarter of the area of the square.

[hide]part of code copied from i_like_pie's:
[asy]size(200);
pair A=(0,0), B=(0,1), C=(1,1), D=(1,0), E=waypoint(B--C,0.5), F=waypoint(C--D,0.5), G=IP(B--F,D--E);
D(unitsquare);
D(E--D--B--F);
MP("A",D(A),SW);
MP("B",D(B),NW);
MP("C",D(C),NE);
MP("D",D(D),SE);
MP("E",D(E),N);
MP("F",D(F),E);
MP("G",D(G),NE);

pair H=midpoint(B--G), I=midpoint(G--D), J=midpoint(B--D), K=midpoint(E--F);
D(E--F,gray);
D(H--I--J--cycle,gray);
D(C--J,dashed+gray);
MP("H",D(H),NNE,blue);
MP("I",D(I),ENE,blue);
MP("J",D(J),SW,blue);
MP("K",D(IP(E--F,C--J)),E,blue);[/asy]

$ H$ and $ I$ are midpoints of $ \overline{BG}$ and $ \overline{GD}$, respectively.

$ \triangle BGD : \triangle EGF : \triangle CEF \equal{} 4: 1: 3$
$ \triangle CEF \equal{} \frac {1}{8}$

which gives us $ \triangle BGD \equal{} \frac {1}{8}\cdot\frac {4}{3} \equal{} \boxed{\frac {1}{6}}$.
[/hide]

Post your solution , I Tried one but couldn't able to complete.
Can anyone plz tell how these ratio come up [b]@undefined117[/b]

[hide="yep"]
Note that $ \overline{EF}\parallel{}\overline{BD}$. That means $ m\angle FED \equal{} m\angle EDB \equal{} m\angle EFB \equal{} m\angle FBD$ because they're alternate interior angles. Additionally, $ m\angle EGF \equal{} m\angle BGD$ since they're vertical angles. That makes $ \triangle BGD \sim \triangle EGF$, and since $ \overline{BD}: \overline{EF} \equal{} 2: 1$, $ \triangle BGD: \triangle EGF \equal{} 4: 1$.

And of course, you could always just count the triangles I drew..
[/hide]


If $ a,b,c,d$ are positive real numbers,prove the inequality

$ \frac{a}{b\plus{}2c\plus{}3d} \plus{}\frac{b}{c\plus{}2d\plus{}3a} \plus{}\frac{c}{d\plus{}2a\plus{}3c} \plus{}\frac{d}{a\plus{}2b\plus{}3c} \geq \frac{2}{3}$

sorry if it had been posted before

thanks


geometry
3D geometry
sphere
calculus
integration
vector
Gauss

Hi, could anybody explain what electric flux is? And also how to do this question:

What is the electric flux through a sphere that has a radius of 1.00m and carries a charge of +1.00 MC at its center?

Electric flux through a surface $ S$ is simply the surface integral of electric field, something written as the following:
$ \Phi = \int\int_{S}\vec{E}\cdot d\vec{A}$, here $ d\vec{A}$ is an infinitesimal area vector that is perpendicular to the surface.
The Gauss' Law tells us that $ \Phi = \frac{q}{\epsilon_{0}}$.
In case of your problem the flux through the surface does not really depend on the radius of the sphere it only depends on the amount of charge in the surface....

[quote="aidan"]So the radius is completely irrelevant? But i thought the larger the radius of the sphere, the larger the surface?[/quote]

If you expand the integral for Gauss's Law (since it's a general relation between the net flux and the charge enclosed) for a sphere with radius $ r$,

\begin{eqnarray*}\Phi&=&\oint\vec{E}\cdot\vec{dA}\\ &=& \vec{E}\cdot\oint\vec{dA}\\ &=& \frac{kq}{r^{2}}\cdot{A}\\ &=& (\frac{q}{4{\pi}r^{2}\epsilon_{o}})({4{\pi}r^{2}})\\ &=& \frac{q_{in}}{\epsilon_{0}}\end{eqnarray*}

Since $ r$ cancels out, you can see that radius is irrelevant.

Also, the electric flux does not depend on the shape of the surface that it is considered... as long as all surfaces encompass the same charge and no other extra charges the flux through all of them will be the same...


inequalities
function
inequalities solved

Let a,b,c be positive real numbers. Prove that

 3^(a+b+c) \geq (1+(a+b)/c)^c(1+(b+c)/a)^a(1+(c+a)/b)^b

The inquality is homogenous, thus we may suppose that  a+b+c = 1., and may be rewritten as
a a b b c c   \geq 1/3
That is
aLn(a) + bLn(b) + cLn(c)  \geq Ln(1/3).
Since the function f(x) = xLn(x) is convex, from Jensen's inequality, we have :
aLn(a) + bLn(b) + cLn(c)  \geq 3[(a+b+c)/3][Ln[(a+b+c)/3]] = Ln(1/3), as desired.
Note that equality holds iff a = b = c.

Pierre.


Prove that some multiple of $ \sqrt2$ lies within $ \frac{1}{1000000}$ of an integer

[hide]
$ \{ \sqrt{2}\cdot k \}$ is dense in the set $ [0,1)$.
Apply kronecker's, since it is a continuous function to get the desired result.[/hide]


The first term of a sequence of positive integers is any two-digit integer.  Each subsequent term is the sum of the tens digit and the square of the ones digit of the previous term.  One possible sequence is $ 14$, $ 17$, $ 50$, $ 5$, $ 25$,... .  If the first term of one such sequence is $ 97$, what is the $ 2008^{th}$ term of the sequence?

After a few terms, the sequence becomes $ 9, 81, 9, 81, 9, ...$, with $ 9$ as the odd-numbered terms.  Thus, the answer is $ \boxed{81}$.


integration
geometry
3D geometry
sphere
inequalities
calculus
parallelogram

In some finite-dimensional Banach spaces $X$ the norm $\parallel x\parallel $ can be represented as \[\parallel x\parallel =\int_{S(X^{*})}|x^{*}(x)|d\mu(x^{*}),\quad (*)\] where $S(X^{*})$ is the unit sphere of the dual space, and $\mu$ is some positive measure on $S(X^{*})$.

For example, $\mathbb R^{n}=\ell_{2}^{n}$ has property (*), with $\mu$ being the surface measure, appropriately normalized. Also, $\ell_{1}^{n}$ (the $n$-dimensional space with $\ell_{1}$ norm) satisfies (*), with $\mu$ being a sum of point masses. But $\ell_{\infty}^{3}$ fails (*), as I realized [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=118752]last night[/url]. Note that $\ell_{\infty}^{2}$ is isometric to $\ell_{1}^{2}$ and therefore satisfies (*).

I conjecture that (*) holds for $\ell_{p}^{n}$ (for all $n$) if and only if $1\le p\le 2$. Any suggestions/pointers to literature are much appreciated.

just a question : what makes you thinks it works for $1<p<2$ ? It works for $p=2$ because $l^{2}$ is the same that is dual so the unit sphere of the dual is "the same" that the sphere of $l^{2}$, and it is easy for $p=1$ because the $l^{1}$ norm is well adapted to this problem ($\int |x^{*}(x)|$ is quite similar to $\sum |x_{i}|$). But I can't see what could be better for the case $1<p<2$ that for the case $2<p<\infty$ ? (sorry for spoiling  :) )

The cases $p=1$ and $p=2$ appear as extremes: $\mu$ concentrated at $\pm e_{i}^{*}$ versus $\mu$ spread out evenly. The intermediate values of $p$ may have $\mu$ of variable density which is higher near $\pm e_{i}^{*}$.  :huh:

Since (*) implies [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=118752]Hlawka's inequality[/url] on $X$, one can try to find a counterexample to the inequality on $\ell_{p}$, $1<p<2$...

Integrating over the unit sphere of the dual norm is a red herring; an integral over any ball or sphere in any norm suffices. The real question is which norms are convex combinations of absolute values of linear functionals.

[hide="Erroneous"]Here's why we can't have this for the $p$-norm when $p$ is greater than 2: any such combination $N$ satisfies the parallelogram inequality $N(u+v)^{2}+N(u-v)^{2}\le 2N(u)^{2}+2N(v)^{2}$ for all $u$ and $v$ (at least over the reals; I haven't tested the complex case).
A counterexample to this inequality for $p>2$ in dimension $\ge2$: $u=(1,1),v=(1,-1)$. $\|u+v\|_{p}=\|u-v\|_{p}=2$ and $\|u\|_{p}=\|v\|_{p}=2^{\frac1p}$. We have $4>2\cdot 2^{\frac2p}$.
Incidentally, you're wrong about the $\infty$ norm in 2 dimensions.[/hide]

My previous post was in error- I'll see what I can salvage.

Out of curiosity, I tried to extend this representation for $\ell_{\infty}$ norm to three dimensions (knowing it would fail). The norm
\[\parallel x\parallel =|x_{1}+x_{2}+x_{3}|+|x_{1}+x_{2}-x_{3}|+|x_{1}-x_{2}+x_{3}|+|x_{1}-x_{2}-x_{3}| \]
produced a funky unit ball: I attached its Maple plot. Does it have a name?

Edit: Sorry about the extra space around the plot; this is what Maple does.  :(

That's one of the Archimedean or semi-regular solids. You can get it by truncating either a cube or an octahedron. Its symmetry group should be the same as the symmetry group for either the cube or the octahedron. I'll check tomorrow with someone who's familiar with the semi-regular solids for the conventional name of this particular one.

http://en.wikipedia.org/wiki/Cuboctahedron

cheers,
-Q

Thanks, qleak. Back to my original question: a necessary condition for (*) to hold is that $X$ be isometric to a subspace of $L^{1}(\Omega,\mu)$ for some measure space $(\Omega,\mu)$. Is this also sufficient?

The answer is yes, and is (unsurprisingly) well-known. First, there is an easy equivalence between two properties:
[list](A) The norm of $X$ can be approximated by the sums of absolute values of linear functionals
(B) The unit ball of $X^{*}$ is a [u]zonoid[/u], which is a limit of [url=http://mathworld.wolfram.com/Zonotope.html]zonotopes[/url] in the Hausdorff metric
[/list]
It is a classical result (Bolker 1969) that $X$ embeds into $L_{1}$ isometrically if and only if the unit ball of $X^{*}$ is a zonoid. The $\ell_{p}$ spaces of dimension $3$ or higher admit such an embedding precisely when $1\le p\le 2$. However, all $2$-dimensional Banach spaces are $L_{1}$-embeddable.