AMC
AIME
geometry
trigonometry
trig identities
Law of Cosines

[quote="beta"]Here's a pretty easy #14

14.  O is the center of the circle. AC = BD = 78, OA = 42, OX = 18. Find the area of the shaded area.[/quote]

It's not THAT easy... but I solved it. So that's saying something for a #14.

[hide]
First, extend OX to touch circle O at two points. Since r = OA = 42, we apply PoP to X to get (DX)(XB) = (42+18)(42-18) = (60)(24) = (30)(48). Since DX+XB = 78, we know XB = 30 and DX = 48. Similarly, we can do that for AC to get AX = 38, XC = 30.

Now draw in OD and OC. Let the intersection of OC and DX be E. We note that <XOC = <BDC because <BDC inscribes an angle twice as big as <XOC (the extention of OX touches the circle at the midpoint of arc BC). Then DCE ~ OXE. So by similarity, we have DE/OE = CE/XE = DC/18. We also know DE+XE = 48 and CE+OE = r = 42. Finally, from another PoP at E we have CE(84-CE) = DE(78-DE). Using these equations, we can go through a bunch of algebra to solve for DC = 42. That means DCO is equilateral => <COD = 60.

The area of the shaded region is [sector COD] - [triangle COD] + [triangle CXD]. Time to TeX.

The area of the sector is $\frac{1}{6}42^2\pi = 294\pi$. The area of triangle COD is $\frac{42^2\sqrt{3}}{4} = 441\sqrt{3}$. We don't know the area of triangle CXD, but we know its side lengths... Heron's! We have $DX = 48$, $CX = 30$, and $DC = 42$. So $s = \frac{48+30+42}{2} = 60$. Then the area is $\sqrt{60(60-48)(60-30)(60-42)} = 360\sqrt{3}$. Finally, the area of the shaded region is $294\pi-441\sqrt{3}+360\sqrt{3} = 294\pi-81\sqrt{3}$ which is what kalva says. Thankfully. :)

EDIT: wrote 40 a couple times instead of 42[/hide]

I did it a similar way. I used the law of cosines to solve for an angle and (1/2)ABsin(c) to calculate some areas. I got the right answer, and it wasn't that hard, but I feel like there should be an easier way to do it. Anyone else have any ideas? There might not be an easier way, but this one definitely was easy for a 14.

Yeah, it was like a huge list of basic steps; so not difficult, but somewhat time-consuming. Then again, a #14 is almost always time-consuming.

[quote="beta"]Ah I did it slightly easier way.
[hide]
Drop perpendicular from O to BD, say the foot is E(the midpoint of BD). Pythagoras on triangle DEO we get $OE=9\sqrt{3}$. How nice OX=18. This implies angle DXO=60 degrees. DX=39+9=48, BX=30=CX. Since arc CD=arc AB, angle DXC=2*arcCD=angleDOC. Thus DCXO is cyclic quad => angle OCD=angleDXO=60 => OCD is equilateral => angle DOC=60= angle DXC. Now use absin\theta/2 formula we can find area of triangle DOC and DXC. we can use  sector DOC - [DOC] + [DXC] to get our desired area. [/hide][/quote]

Nice. I'm no good at connecting points that aren't given to me =P. I wouldn't have thought of dropping that perpendicular. That's a nice solution though.

[quote="paladin8"]Nice. I'm no good at connecting points that aren't given to me =P. I wouldn't have thought of dropping that perpendicular.[/quote]

Most of the time, auxiliary lines are not arbitrary; they usually come from some natural relationship between various parts of your diagram. Well, that and experience. In this case, a chord relates very well with the center of the circle through this perpendicular line. In fact, knowing either the length of the chord or the distance from the center will immediately give you the other one, because of the right triangle: r^2=d^2+(l/2)^2, r=radius, d=distance to chord, l=length of chord. Knowing that relationship between chord and center helps to motivate the perpendicular that beta drew. So keep you eyes and your mind open, and these auxiliary lines will jump out at you.

BTW, that was a nice solution, beta. The helpful perpendicular and the cyclic quad were very good observations!

Yeah, I should've seen it. If it were just one chord I would've, but there were two! :P


[quote="anirudh"]you could just visit an easy to navigate forum [url=http://4aunlimited.co.nr]here[/url] or if that link doesn't work, then click [url=http://dasarathy0.tripod.com]here[/url].[/quote]

can you please stop posting links to your website?  it's getting really irritating when i click on them.  thanks.

[quote="Farazy"]http://groups.myspace.com/mathdorksunited

Join if you can... Lets make this a huge kick butt myspace community!

 :D

MathCounts portion of my site is down...  :mad:  My site will be up in september again... I am waiting for my uncle to bring the server down from Massachusetts!  If someone would be willing to temporarily host all of this stuff i have I would really appreciate it.[/quote]

Is this really MathCounts?!

i'mm willing to host. it. pm me the stuff as a zip file

Farazy:

DONT SPAM!!!!

you have already posted this topic in the getting started forum

it was moved here.  :D . It was not farazy's fault. a mod moved it.

I didnt mean to post it twice... something must've happened... Sheesh!  :blush:


integration
inequalities
function
calculus
real analysis
real analysis unsolved

Given $f: [0,1] \rightarrow %Error. "mathdd" is a bad command.
{R}$ absolutely continuous such that $f(0)=0$ and $\int_{0}^{1}{|f '(x)|^{2}dx} < + \infty$, find $lim_{x \rightarrow 0^{+}} (x^{\frac{-1}{2}}f(x))$.

I think that limit is $0$.

Since $f$ is absolutely continuous, it's differentiable a.e., and the equality $f(x)=\int_0^xf'dt$ holds. We thus have $\frac{f^2(x)}x=\frac 1x\cdot\left(\int_0^xf'dt\right)^2\le\frac 1x\cdot\int_0^x|f'|^2dt\cdot\int_0^x1^2dt=\frac 12\int_0^x|f'|^2dt$ (the inequality is just Cauchy-Schwartz).

In order to show that $\frac{f(x)}{\sqrt x}\to 0$ as $x\to 0$ it thus suffices to prove that $\int_0^x|f'|^2dt\to 0$ as $x\to 0$. This follows, for example, from the Dominated Convergence Theorem:

Set $g_n(t)=\chi_n(t)\cdot|f'(t)|^2$, where $\chi_n$ is the characteristic function of $\left[0,\frac 1n\right]$. Now, the sequence $g_n$ is dominated by the $L^1([0,1])$ function $|f'|$ and converges pointwisely to $0$ on $(0,1]$, so we get $\int_0^1g_ndt\to 0$ as $n\to\infty$, and since $\int_0^1g_ndt=\int_0^{\frac 1n}|f'|^2dt$, we're done.

Hi

[quote="grobber"]This follows, for example, from the Dominated Convergence Theorem:[/quote]

I agree, this follows from the absolute continuity of the integral anyhow.

Another approach(...or more exactly question):
$2\int_x^{1}\frac{f(t)}{\sqrt t}=f(x)(\sqrt(x)-1)|_x^1-\int_x^1 f'(t)(\sqrt(t)-1)$.Since $f(0)=0$we have  $2|\int_x^{1}\frac{f(t)}{\sqrt t}|=|\int_x^1 f'(t)(\sqrt(t)-1)dt|\leq (\int_x^1 |f'(t)|^2)*k<\infty$(Cauchy-Schwarz inequality).Can we deduce from here that $\frac{f(x)}{\sqrt(x)}$ has limit 0 by "saying"(i didnt succed to prove this one) that $\sqrt(x)$ should dominate $f(x)$ in 0?I dont wanna bother u with this one after Groober posted an answer,but I wonder if someone can continue with this idea and get the answer.


logarithms
function
algebra
domain
calculus
integration
complex analysis

Harazi we have a complex analysis forum now. Please use it! :P

Oh, sorry, I didn't see it.  Anyway, the complex analysis that I know reduces to multiplying complex numbers...  :blush:

First of all, let's show that given a continuous map $f:\mathbb C\to\mathbb C^*$, it has a lifting $g$, i.e. there is a continuous map $g:\mathbb C\to\mathbb C$ s.t. $f=\exp(g)$.

Take some $\tilde x_0=\in\mathbb C$ s.t. $\exp(\tilde x_0)=f(0)$. Each point in $y\in f(\mathbb C)$ has a neighbourhood $U_y$ s.t. $\exp^{-1}(U_y)$ is a disjoint union of open subsets of $\mathbb C$, each one mapped homeomorphically onto $U_y$ by $\exp$ (this holds because $f(x)\ne 0,\ \forall x\in\mathbb C$). It's clear that we can choose the neighbourhoods $U_y$ to be open disks of a certain fixed radius.

Now partition the complex plane into square regions which contain their bottom and left edge, but not their top and right edge (so that we actually have a partition) so small that each one of them is mapped by $f$ completely into one of the open sets $U_y$ (this is possible because of the continuity of $f$ and the fact that the open balls $U_y$ all have the same radius). If such a square region is contained in some $U_y$, once we choose the component of $\exp^{-1}(y)$ where we want to lift the restriction of $f$ to it, the lifting is unique, because the logarithm is a well defined and bijective function from $U_y$ to the desired component. We can now start lifting the restrictions of $f$ to our regions one by one (there are countably many such regions): first we lift the restriction of $f$ to the square region which contains $0$ to the component of $\exp^{-1}(f(0))$ which contains $\tilde x_0$, and then, at each step, we lift the restriction of $f$ to a region which is adjacent to one of the regions where we have already lifted $f$. In this way we can cover the entire complex plane with the regions where we lift $f$, thus getting a lifting of $f$.

As for analyticity in a certain point $\tilde x\in\mathbb C$ for $g$, it's enough to observe that in a small enough neighbourhood of $\tilde x$, $g$ can be considered the composition between the analytic function $f$ from $\mathbb C$ to some small enough neighbourhood of $f(x)$, where $f(x)=\exp(g(\tilde x))$, and the logarithm function, which is well defined, analytic and bijective if the involved mentioned neighbourhoods are small enough, so $g$ is analytic in all the points of the complex plane, and thus entire.

I hope it's correct. I'm not very familiar with this stuff either, but at least it looks like the right idea, in case I did something wrong.

Another approach :

 We know that, if $f \not= 0$ everywhere then $\log|f(z)| = \text{Re} \log f(z)$  is a harmonic function, i.e, satisties Laplace equation. Since $\mathbb{C}$ is a simply connected domain, there exists an entire function $g$ such that $\text{Re}(g(z)) \equiv \log|f(z)|$. Now
\[|e^g|= e^{\text{Re}g} = e^{\log|f|} = |f| \] we must have $f = \alpha e^g$, where $|\alpha |=1$, which is equivalent to $f= e^{g*}$ for some $g*$

[quote="harazi"]Let $f$ be a complex entire series that does not have zeroes and converges in all complex plane. Prove that  there exists an entire series $g$ which converges ever the entire plane and such that $f=e^g$.[/quote]

$f'/f$ is entire (as $f \neq 0$ everywhere).  Its integral $g$ is therefore also entire, and we can check that $g + C$ solves the problem for some constant(s) $C$.


I was doing this one question, i was able to get the answer, but it was different from the back of the textbook. Heres the question, all i need is your answer, no need for work to be shown.

If x+3 is a factor of kx^3 + 4x^2 + 2kx - 1, what is the value of k?

basically by plugging in $x=-3$ into the equations you need to find k so that it equals 0.  So in this case $k=-\frac{35}{33}$

WROOOOOONNNGGGG!!!!!!

Right answer:

[hide]
He probably had the right idea, because the sign is just off.

Use synthetic division on the proposed root -3 (or long division). The remainder is -33K+35, so for the remainder to be 0, $k=\frac{35}{33}$[/hide]

oh right my bad, eh 36-1 doesn't equal -35...  i got to stop making such mistakes

$k=\frac{35}{33}$

also i reccommend just plugging in the -3 since it makes the calculations much quicker.


[quote="silouan"]If $a,b,c>0$  prove the following inequality

$\frac{a^6}{b^2+c^2}+\frac{b^6}{a^2+c^2}+\frac{c^6}{a^2+b^2}\geq\frac{abc(a+b+c)}{2}$[/quote]

 :D  Yeah Really it's very easy using Chebusev inequality twice and combine with Nesbit inequality......


By FLT, $ 3^4 \equal{} 1 \pmod 5$, or in other words, the powers of 3 cycle every four powers. Thus to compute $ 3^{2007} \pmod 5$, we see that that is the same as $ 3^3 \pmod 5$, which is just 2.

So the possible values of x are $ 3$ and $ 8$.

[quote="williams"]given that x is a +ve 1 digit number,if 3^(2007) +x is a multiple of 5,find d +ve value of x.[/quote]
[hide="This may be a solution"]

If $ 3^{2007} \plus{} x\equiv 0 \pmod 5$ , find $ x$.

By Fermat's little thereom :   $ a^{p \minus{} 1}\equiv 1\pmod p$

$ \therefore 3^4\equiv 1\pmod 5 \implies 3^{4k}\equiv 1\pmod 5$ ,where $ k\in \mathbb N$

Thus,

$ 3^{2007} \plus{} x \equal{} 3^{2004}.3^3 \plus{} x\equiv 0 \pmod 5$

$ \implies 3^3 \plus{} x\equiv 0\pmod 5$

$ \implies 2 \plus{} x\equiv 0\pmod 5 \implies x\in\{3,8,13,18....\}$

But, $ x$ is a single digit , $ \therefore x \equal{} 3 ,8 .$

[/hide]


A goat is tethered to the outside of a $ 30$ ft. by $ 10$ ft. barn, $ 10$ ft. from the corner on one of the longer sides of the barn. If the rope is $ 30$ ft. long, what is the total area of the land the goat can graze, assuming the goat does not eat the rope?

[hide="Hint"]You're going to be dealing with circles here.  But not quite.  The fact that the barn is in the way means that some of the circle is off-limits.  So the available land is $ 900\pi\minus{}x$ square feet.  Your job is to find $ x$.[/hide]

The area that the goat can graze on without the rope folding around any corners of the barn is a semicircle with area $ 450\pi ft^2$.
There are two areas where the rope folds around one corner. When the rope folds around the corner closest to the beginning($ 10 ft$), the area created is a quarter of a circle with area $ 100\pi ft^2$. When the rope folds around the other corner, the region created is another quarter of a circle with area $ 25\pi ft^2$.
One more region is created where the rope folds around two corners, the area of this region is $ 25\pi ft^2$.
So the total area that the goat can graze on is :
$ 450\pi ft^2 \plus{} 100\pi ft^2 \plus{} 25\pi ft^2 \plus{} 25\pi ft^2 \equal{} 600\pi ft^2$.


inequalities
function
inequalities proposed

If $a, b, c$ are nonnegative numbers satisfying $a+b+c=3$ and $\lambda$ is a real number in $[0;1]$; then we have
$\lambda(abc)^{k}+(ab)^{k}+(bc)^{k}+(ca)^{k}\le max(\lambda+3;(\frac{3}{2})^{2k})$
(When $\lambda =0$, we get a famous result, of Harazi ?  :lol: )
I use mixing variables.
Can you solve if $\lambda$ is not in $[0;1]$ ? We can mix variables, too, but after that, I can prove it. I think it not true.

OK, it is a difficult problem. I will give a hint :
Assume $a \ge b \ge c \ge 0$ and put $a=t+x,b=t-x \ \ (t \ge x \ge 0)$. Consider the following function :
$f(x)=\lambda(t^{2}-x^{2})^{k}c^{k}+(t^{2}-x^{2})^{k}+c^{k}(t+x)^{k}+c^{k}(t-x)^{k}$
 We can prove $f'(x) \le 0 \forall x \ge 0$, then we have $f(x) \le f(0)$. We only need prove the given inequality when $a=b$.
Continue !


What do you think?

Do you think MATHCOUNTS State sprint round is harder or AMC 8 is harder?

Honestly, there is no competition. School MC=AMC8 level.

[quote="A+MATH"]I do much better at AMC-8 than MC so obviously MC is harder  :lol:  :P .[/quote]

Same.

MC is harder, b/c AMC8 is m/c, but MC is fill in the blank. However, AMC8 problems are (technically) on a higher level of difficulty than MC problems, though just barely (2.3 points harder, scores adjusted and averaged for 10 years of competitions, on a sale of 100)

I think that the AMC is hard but not as hard as MC

I think the reason for this is the audience.  For the AMC 8, anyone can take it, so the makers want to make it so that almost everyone can answer one question making some questions very easy.  Also, there isn't anything after AMC 8.  But, Mathcounts state is different, because these are the best people from each chapter, which has the best people from each school, making the questions harder, or else everyone will get perfects.  And the porpuse of the state round is to decide the national contestants, so if too many people get too many points, then it will be harder to distinguish between the contestants.

yeah... most definitely mc is much harder..... funner too 4 me cuz you're actually competing with ppl out of your school and in higher grades

MATHCounts, because I do a lot better on AMC 8, and the questions I don't get on the AMC, I understand easily where I went wrong. On the sprint round, I don't do as well, and it takes some working out to find the correct answer and how I went wrong there.
 :lol:

I haven't taken the AMC 8 yet, but MathCounts is pretty easy for me.

Those scores are do not look like they were made by the same person! Did you get tons better?

[quote="d34gl3r"]Geez i was only an alternate for mathcounts chapter competition,a nd i only got 17 on the amc8 but i got a 133.5 on the amc10?? lol
  ;)[/quote]

I had the same kind of jump in my scores.  But hey, I'm not complaining! :D

I'd say school mathcounts sprint round and amc8 are on the same level, but all the higher mathcounts rounds are much harder than the amc8. I aced the amc8 this year and obviously didn't do so at states mathcounts and won't do so when i go to nationals!


I am posting this in here because I think it is an advanced problem because it relies so much on your intuitive problem solving ability.  If you know the solution already because it is famous I think, please don't post it.  Anyways, here it is:
A census-taker knocks on a door, and asks the man inside how many children they have and how old they are.
"I have three daughters, their ages are integers, and the product of the ages is 36," says the father.
"That's not enough information," responds the census-taker.
"I'd tell you the sum of their ages, but you'd still be stumped."
"I wish you'd tell me something more."
"Okay, my oldest daughter Annie is mathematically inclined."
What are the ages of the three daughters?

What is
[quote="problem"]mathematically inclined[/quote]?

It means good at mathematics. Do you think that's relevant to the problem though?

Well, it probably means that she is over, say, 3 years of age.

ok tell me if I'm wrong but: (Spoiler)

[hide]The ages they could be are:

(two 1s) 1,1,3 and the sum is 38(one 1)1,2,18 and the sum is 21
1,3,12 and the sum is 16
1,4,9 and the sum is 14
1,6,6 and the sum is 13(no 1s)2,2,9 and the sum is 13
3,3,4 and the sum is 10
2,6,3 and the sum is 11

But notice the problem said: I'd tell you the sum of their ages, but you'd still be stumped.

So that means the ages either have to be 1,6,6 or 2,2,9 because their sums are the same...

And according to the last thing the father said: Okay, my oldest daughter Annie is mathematically inclined.

That means he only has one oldest daughter, so 1,6,6 will not work.

Therefore the ages are 2,2, and 9.[/hide]

Ok.. please tell me if I'm right or wrong

Correct, great job.  It took me a long time to finally get that problem.


geometry
trigonometry
analytic geometry
ratio
Pythagorean Theorem
perpendicular bisector
similar triangles

$\triangle ABC$ has $AC=7$, $BC=24$, $\angle C = 90^\circ$. $M$ is the midpoint of $AB$, $D$ lies on the same side of $AB$ as $C$ and has $DA=DB=15$. Find the area of $\triangle CDM$.

[hide="Somewhat-convoluted solution"]$AM=BM=\frac{25}{2}$, so $DM=\sqrt{225-\frac{625}{4}}=\frac{5\sqrt{11}}{2}$. $CM=AM$ because $M$ is the midpoint of $AB$, so $CM=\frac{25}{2}$. Draw the altitude from $D$ to $CM$ and call it $DH=x$. Now, say $\angle ABC=\alpha$. By definition, $\angle AMC=2\alpha$, so $\angle CMD=90-2\alpha$. From there, we have $\sin CMD=\sin (90-2\alpha)=\cos 2\alpha=2\cos ^2\alpha -1=\frac{527}{625}=\frac{x}{\frac{5\sqrt{11}}{2}}$. Solving, we find $x=\frac{527\sqrt{11}}{250}$. Now, multiplying out for area, we have $\frac{1}{2}bh=\boxed{\frac{527\sqrt{11}}{40}}$.[/hide]

[b]EDIT:[/b] [hide="By the hint"]$CM=\frac{25}{2}$, $DM=\frac{5\sqrt{11}}{2}$, $\sin AMD=\cos 2\alpha =\frac{527}{625}$, $\frac{1}{2}ab\sin AMD=\boxed{\frac{527\sqrt{11}}{40}}$.[/hide]

[hide="Coordinate Geometry Solution"]Put the figure on a coordinate plane such that $M$ is at the origin, $A=\left(0,\frac{25}{2}\right)$, and $B=\left(0,-\frac{25}{2}\right)$.  We know that $D$ must lie on the $x$-axis.  By Pythagorean Theorem, we have $DM=\frac{5\sqrt{11}}{2}$ and $D=(\frac{5\sqrt{11}}{2},0)$.  We must find the $y$-coordinate of $C$.  Since $\angle ACB$ is right, we have
\[ x^2+y^2=\frac{625}{4} \]
Since $C$ is $7$ units away from $A$, we have
\[ x^2+\left(y+\frac{25}{2}\right)^2=49 \]
Expanding the second equation, we get
\[ x^2+y^2+25y+\frac{625}{4}=49 \]
Substituting,
\[ 25y+\frac{625}{2}=49\Rightarrow y=-\frac{527}{50} \]
The area is $\left(\frac{1}{2}\right)\left(\frac{5\sqrt{11}}{2}\right)\left(\frac{527}{50}\right)=\boxed{\frac{527\sqrt{11}}{40}}$.[/hide]

A different solution:

[hide]We know $DM$ is the perpendicular bisector of $AB$. So $DM = \sqrt{15^2 - \left(\frac{25}{2}\right)^2} = \frac{5\sqrt{11}}{2}$.

Using similar triangles we establish some well known ratios which tell us that if the perpendicular from $C$ is extended to $AB$ at $E$, then $AE = (AC^2)/AB$, and so pluggin numbers in $AE = \frac{49}{25}$.

Then $EM$, the height of triangle $CDM$ with base $DM$ is $25/2 - 49/25 = \frac{527}{50}$.

By $\frac{1}{2}bh$, $\frac{1}{2} * \frac{527}{50} * \frac{5\sqrt{11}}{4} = \frac{527\sqrt{11}}{40}$, which isn't an integer between 0 - 999. Oh well, I'm guessing it was asking the sum of those integers there or something.[/hide]

And a comment of JesusFreak's solution:

[hide]Trig is always the way to go.  :)

I actually thought that since $\angle{HDM} = 2\alpha$, then we want $\cos{2\alpha}$, which led to Law of Cosines on $AMC$, LOL.

So I guess $\cos{2\alpha} = \displaystyle\frac{7^2 - 12.5^2 - 12.5^2}{-2(12.5)(12.5)} = \frac{527}{625}$.[/hide]

I think it's easier to just say $\cos 2\alpha =2\cos ^2\alpha -1$, and $\cos \alpha =\frac{24}{25}$, since it's a right triangle...

I also found the similar triangles solution. You can find the height in almost exactly the same way.

Well, here's my solution:

[hide][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=667[/img]

Let $\theta = \angle CMD$ and $\phi = \angle AMC$. Notice that $\theta + \phi = 90^\circ$ since $\triangle ADB$ is isosceles. We need to find \[ [CMD]=\frac{1}{2}\sin \theta (CM)(DM).  \] It is well known that the median to the hypotenuse of a right triangle is equal to half of the hypotenuse, so $CM=AM=MB=25/2$. Also, by the pythagorean theorem, $MB^2+DM^2=BD^2$, so $DM=\frac{5}{2}\sqrt{11}$. By the law of cosines, $AC^2=AM^2+MC^2-2(AM)(MC)\cos \phi\implies\cos\phi=\frac{527}{625}$. Recall that $\sin \theta = \cos (90-\theta)=\cos\phi=527/625$, so our answer is \[ [CMD]=\frac{1}{2}\left( \frac{527}{625}\right) \left(\frac{25}{2}\right)\left(\frac{5}{2}\sqrt{11}\right)=\boxed {\frac{527}{40}\sqrt{11}}.  \][/hide]


floor function
AMC
USA(J)MO
USAMO
number theory unsolved
number theory

Show that if $ x$ is a real number and $ n$ is a positive integer, then $ \sum_{k\equal{}1}^n \frac{\lfloor kx \rfloor}{k} \le \lfloor nx \rfloor$.

[hide="what I have so far"]

$ \sum_{k\equal{}1}^n \frac{\lfloor kx \rfloor}{k} \equal{} \sum_{k\equal{}1}^n \frac{kx \minus{} \{kx\}}{k}$
where $ \{kx\}$ is the fractional part of $ kx$.

$ \sum_{k\equal{}1}^n \frac{kx \minus{} \{kx\}}{k} \equal{} nx \minus{} \sum_{k\equal{}1}^n \frac{\{kx\}}{k}$

Since we know $ nx\minus{}1 \le \lfloor nx \rfloor \le nx$, we need to show $ \sum_{k\equal{}1}^n \frac{\{kx\}}{k} \ge 1$ and we will be done.

That's all I have, I couldn't show $ \sum_{k\equal{}1}^n \frac{\{kx\}}{k} \ge 1$.

[/hide]

[quote="ark_xm"]
[hide="what I have so far"]

That's all I have, I couldn't show $ \sum_{k \equal{} 1}^n \frac {\{kx\}}{k} \ge 1$.

[/hide][/quote]

For sure you cannot solve this because it's not true for all $ x$. Take a look at the case when $ x \in \mathbb{Z}$.

[hide="A suggestion"]

Consider all the cases $ \frac{t}{n} \leq \{ x \} < \frac{t\plus{}1}{n}$, $ t\equal{}\overline{0,n\minus{}1}$.

[/hide]

Hmmm, I still can't solve it. Anyone else?

It's from USAMO, 1981
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101711[/url]

Ah nice, yeah my TA solved it and this is a very beautiful problem :)
Very fun solution to write up.


conics
geometry
geometric transformation
rotation
trigonometry

Find the equation for the conic $4x^2 - 8xy + 4y^2 - 5x - 5y = 0$ when it's rotated such that one of its axes/axis is parallel to the x-axis.  However, you may not use trig in your solution.

[hide="hint"]
Consider the substitution:

 $u = x+y$
$v = x-y$
[/hide]

Please, please, hide your original work, to show respect to the other people on this forum.

[quote="#20002 pwns"][hide]So you want $\frac{u+v}{2}=x,\frac{u-v}{2}=y$.
Now we have: $4x^2 - 8xy + 4y^2 - 5x - 5y = 0 = 4(\frac{u+v}{2})^2-8(\frac{u+v}{2})$ $(\frac{u-v}{2}) +4(\frac{u-v}{2})^2 - 5(x+y)$ $=4(\frac{u+v}{2})^2-8(\frac{u+v}{2})(\frac{u-v}{2}) + 4(\frac{u-v}{2})^2-5u$
$=(u+v)^2-2(u^2-v^2)+(u-v)^2-5u=$ $2(u^2+v^2)-2(u^2-v^2)-5u=2(2v^2)-5u=4v^2-5u=0$
This mapping of $(x,y) \implies (u,v) = (x,y) \implies (x+y,x-y)$ is equivalent to a 90 degree rotation, right?[/hide][/quote]

[hide="almost"]
It's not 90 degrees, and there's a slightly easier way to do it.

Even though you've done all the work you haven't actually answered the question... what is the conic in the xy-plane when rotated?  Although this is trivial now.

[/hide]


linear algebra
matrix
group theory
abstract algebra
quadratics
topology
number theory

The invertable element of $End(R^{n})$ is $GL(R^{n})$, for any fixed $a \in GL(R^{n})$,

the set of elements of $GL(R^{n})$ satisfying: $x^{T}a x=a$ is a subgroup of $GL(R^{n})$.

Is this group orthogonal group $O(n)$?

1) This is interesting only if $a$ is symmetic invertible. Then $a$ is associated to a non degenerate form with signature $(p+,q-)$ where $p+q=n$. $\{x|x^{T}ax=a\}$ is the so called "indefinite orthogonal group" $O(p,q)$ (cf. Wikipedia). It's an algebraic Lie group of dimension $\frac{n(n-1)}{2}$. $(p,q)$ determines this group up to isomorphism with $O(p,q)\approx{O(q,p)}$.
For example $O(3,1)$ is the orthogonal group (of dimension 6) for the space-time. The fundamental group of $SO^{+}(3,1)$ , the identity connected component of $O(3,1)$ is $\mathbb{Z}_{2}$; $SO^{+}(3,1)$ has the "plate property" like $SO^{+}(3)$.
2) Another point of view:
The Diophantine equation $ux^{2}+vy^{2}+2wxy=k$ (with $u,v,w,k\in\mathbb{Z}$) can be rewrited:
\[\begin{pmatrix}x\\y\end{pmatrix}^{T}U\begin{pmatrix}x\\y\end{pmatrix}=k, with: U=\begin{pmatrix}u&w\\w&v\end{pmatrix}\]
If there exists $A\in{GL_{n}(\mathbb{Z})}$ s.t. $A^{T}UA=U$ then $[x,y]^{T}$ is a solution implies that the $A^{m}[x,y]^{T}$ are also  solutions.
Unfortunately, in the general case, $A$ doesn't exist; an exception is the Pell equation: $x^{2}-ny^{2}=1$ where $n$ isn't a square (if not the solution is obvious): if $[p,q]^{T}$ is a solution then we can choose \[A=\begin{pmatrix}p&nq\\q&p\end{pmatrix}\]
The problem is the $(p,q)$ choice to obtain all the solutions (ie. a fundamental solution); it's linked to the continued fraction for $\sqrt{n}$ (cf. Wikipedia).

I can't see why all orthogonal groups of signature (n,0) are isomorphic to $O(n,R)$?

It there a simple anwser?

Hi kalazealot.
Let $A,Q$ be an invertible symmetric matrix with signature $(p,q)$ and its associated quadratic form. Let $D=diag(I_{p},-I_{q})$ and let $Q_{0}$ be its associated quadratic form. Let $O(Q),O(Q_{0})=O(p,q)$ be the associated orthogonal groups. According to the Sylvester's inertia theorem, there exists an invertible matrix $P$ s.t. $P^{T}AP=D$. We deduce the required isomorphism:$U\in{O(Q)}\rightarrow{P^{-1}UP\in{O(p,q)}}$.

Remark about this affirmation in part 2) of my precedent post:
"Unfortunately, in the general case,  $A$ doesn't exist"
Now I am not sure!


Solve equation

$ \lfloor{\frac{5\plus{}6x}{8}}\rfloor\equal{}\frac{15x\minus{}7}{5}$

$ x \equal{} k \plus{} f$ where $ k$ is the integer part of $ x$ and $ f$ is the fractional part.

$ \lfloor{\frac {5 \plus{} 6k \plus{} 6f}{8}}\rfloor \equal{} \frac {15k \plus{} 15f \minus{} 7}{5}$

$ \lfloor{\frac {5 \plus{} 6k \plus{} 6f}{8}}\rfloor \equal{} 3k \plus{} 3f \minus{} 1 \minus{} \frac {2}{5}$

$ \lfloor{\frac {5}{8} \plus{} \frac {3k \plus{} 3f}{4}}\rfloor \equal{} 3k \plus{} 3f \minus{} 1 \minus{} \frac {2}{5}$

$ \lfloor{\frac {3k \plus{} 3f}{4}}\rfloor \equal{} 3k \plus{} 3f \minus{} 1 \minus{} \frac {2}{5}$

This is how far I can get it..

You can't assume $ \left\lfloor {5\over 8}\plus{}{3k\plus{}3f\over 4}\right\rfloor\equal{}\left\lfloor{3k\plus{}3f\over 4}\right\rfloor$ (e.g. what happens if $ k\plus{}f\equal{}1$?).

[hide="Solution"]$ {15x\minus{}7\over 5}$ must be integer, hence $ 15x\equal{}5n\plus{}7\iff x\equal{}{5n\plus{}7\over 15}$. That yields $ {5\plus{}6x\over 8}\equal{}{10n\plus{}39\over 40}$, so we get $ \left\lfloor{10n\plus{}39\over 40}\right\rfloor\equal{}n$. That is equivalent to

$ n\leqslant {10n\plus{}39\over 40}<n\plus{}1$

$ 40n\leqslant 10n\plus{}39<40n\plus{}40$

$ 30n\leqslant 39<30n\plus{}40$

The LHS yields $ n\leqslant {13\over 10}\implies n\leqslant 1$, and the RHS yields $ n>\minus{}{1\over 30}\implies n\geqslant 0$. Hence the only possible values for $ n$ are $ n\equal{}0$ and $ n\equal{}1$. Those yield $ x\equal{}{7\over 15}\lor x\equal{}{4\over 5}$[/hide]


$x,y,z$ are all positive real such that

$x+[y]+\{z\}=13.2$

$[x]+\{y\}+z=14.3$

$\{x\}+y+[z]=15.1$

Find $x,y,z$ .

*$[x] =$integer part of $x$ . $\{x\}$ = fraction part of $x$

[quote="Farenhajt"]As $\{a\}+[a]=a$, adding up the equations we get

$2(x+y+z)=42.6\iff x+y+z=21.3$

Subtracting the first equation we get

$\{y\}+[z]=8.1$

Since $0\leqslant \{y\}<1$ and $[z]$ is integer, the only possibility is $\{y\}=0.1,[z]=8$

In the same manner we find $\{x\}=0,[y]=7$ and $[x]=6,\{z\}=0.2$

Now $x=[x]+\{x\}=6, y=7.1, z=8.2$[/quote]

 :first:


In triangle $ABC$, $AD$ is angle bisector ($D$ is on $BC$) if $AB+AD=CD$ and $AC+AD=BC$, what are the angles of $ABC$?

this is my solution
first for a triangle ABC
we have B=2C <=> b 2 =c 2 +ca.
(it can be proved easily)
return to our problem
AB+AD=CD and AC+AD=BC
<=> b 2 + bc + 2bccos(A/2) = ab+ac (1)
and bc + c 2 +2bccos(A/2) = ab (2)
(1) - (2) and we have b 2 = c 2 + ca
<=> B = 2C
we can write (1) in another way
(1) <=> 2bccos(A/2) = (a - b)(b + c)
<=> sinAsinBsinC = (sinA - sinB)(sinA + sinB)sin(A/2)
<=> sinAsinBsinC = 4cos((A+B)/2)sin((A-B)/2)sin((B+C)/2)cos((B-C)/2)sin(A/2)
 = 4sin(C/2)sin((A-B)/2)cos(A/2)cos(C/2)sin(A/2)   (because B = 2C)
<=> sin((A-B)/2) = sinB
<=> A = 3B
we also have B = 2C. So
A = 120o
B = 40o
C = 20o  :D

Are there any other solutions?(Not using trigonometry)? :|

Let $B'$ be the reflection of $B$ in $AD$, then easily we have $B'C=AC-AB=BC-CD=BD=B'D$, so $\angle B=\angle AB'D=2\angle C$.
Meanwhile, let $E$ be the intersection of the perpendicular bisector of $AC$ and $BC$. Then $\angle AEB=2\angle C=\angle B$, thus $AB=AE=CE$, and hence $DE=CD-CE=CD-AB=AD$, so we have $\angle DAE=\angle DEA=2\angle C$, hence $\angle CAD=3\angle C$, and consequently we know that $\angle A=6\angle C$.
 Finally we have $\angle C=20^\circ$, $\angle B=40^\circ$,$\angle C=120^\circ$.


geometry perimeter inradius calculus integration

Find all $\triangle{ABC}$ such that:
1) the lengths of $\overline{AB},\overline{BC},\overline{CA}$ are all integers.
2) the area of $\triangle{ABC}$ is numerically eaqual to the perimeter of $\triangle{ABC}$.

For example: A triangle with three sides (6,8,10).

[hide="hint"]Heron[/hide]
[hide="hint 2"]s(s-a)(s-b)(s-c) must be a perfect square, otherwise the sides could not be integers.[/hide]

Triangle with sides (5,12, 13).

The perimeter would be 30 and the area would be 30. :ddr:

There are 5 different triangles satisfying the situation... how to find them all ?

[quote="jelly119"]...circumference...[/quote]
thats only for circle, no? i dont understand...

[quote="tarquin"]The lengths of the sides satisfy

$\frac {(a+b-c)(a-b+c)(-a+b+c)}{a+b+c} = 16$[/quote]

Why?

Just equate the area of the triangle using Heron's to the perimeter of the triangle.

I figured you had done that.  I was trying to do the same thing, but I didn't get very far.  Did you change $s$ to $\frac {a+b+c}{2}$ at the beginning?

$\frac {\sqrt {(a + b + c)(a + b - c)(a - b + c)(-a + b + c)}}{4} = a + b + c$

Square both sides and kill one of the $a + b + c$'s.

Yeh...I mean "perimeter"! Forgive my poor English... :blush:

[quote="tarquin"]The lengths of the sides satisfy

$\frac {(a+b-c)(a-b+c)(-a+b+c)}{a+b+c} = 16$[/quote]

And how do we go further more ?

Let a+b-c=x, a-b+c=y, b+c-a=z
so xyz/x+y+z=16
x+y+z=circumference
so find all such possibilties by this equation but....
thats the same as tarquin.....
 :(

Sorry, but I got lost on that tarquin, what is $r$ supposed to represent, what is the "inradius" and where did those formulas come from?  Sorry if that is a basic question.

[quote="tarquin"]One of my fellow campers told me that since $rs=A$ and $A=2s$ then $rs = 2s \Rightarrow r = 2$.

I never even thought of using the formula for the inradius.
Now it's just a matter of finding all triangles that have an inradius of 2.[/quote]
In fact, the original problem is: Find all triangles with sides of integral lengths and having a inradius of 2. ;)

Hmm...

The remaining triangles cannot be right (the remaining being the ones that aren't {5,12,13} or {6,8,10}.

How you get this is by drawing a right triangle, a circle inscribed in it, and drawing all the tangents. Then, you label the two parts of the hypoteneuse x and y, and the unknown parts of the legs will also be x and y by two-tangent theorem. The rest is just simplifying it.

This only applies to right triangles

$(x+2)^2+(y+2)^2=(x+y)^2$
$x^2+4x+4+y^2+4x+4=x^2+2xy+y^2$
$4(x+y+2)=2xy$
$xy-2x-2y-4=0$
$xy-2x-2y+4=8$
$(x-2)(y-2)=8$
$(x,y)=(3,10),(4,6)$

and also (10,3) and (6,4) if you really want to, but that's pointless...

very nice solution bob123, we all got hypnotized by Hexaditidom's hint  :D

Ha...I thought that I could sovle this problem soon.... :)


linear algebra
matrix
linear algebra unsolved

Find the matrices $ A\in\mathcal{M}_n(\mathbb{C})$ s.t. $ \forall{X}\in\mathcal{M}_n(\mathbb{C}),rank(AX\minus{}XA)\not\equal{}1$.

$ cI$, where $ c \in \mathbb{C}$.
WLOG, let us assume that $ A$ is jordan form, since we can conjugate $ AX\minus{}XA$ without affecting the rank.

1. We show that A is diagonal.
Proof: Let us do one example, say there is a 1 at the $ (n\minus{}1,n)$th position (right above the diagonal)
Suppose $ A$ is expressed as the matrix of a linear transformation with respect to the basis $ v_1,...,v_n$.
Consider the matrix representation with respect to $ v_1,...,v_{n\minus{}1},2v_n$. The matrix representation will then change by dividing 2 in the last row and multiply 2 in the last column. We see that for our jordan matrix $ A$, these two operations would only change the 1 at $ (n\minus{}1,n)$th position and make it become 2. That means, if $ P$ is the change of basis matrix, we have $ rank(PAP^{\minus{}1} \minus{} A) \equal{} 1$, thus the $ (n\minus{}1,n)$th entry of A must be 0. By similar reason for other positions, we conclude that $ A$ is diagonal.

2. A = cI for some c.
Proof: It suffices to show that every eigenvalue of A is equal. If the (i,i)th entry and the (j,j)th entry of A are different, we set A to be $ E_ij$, where every entry is 0 except the (i,j)th position, which is 1. In this way, we can see that $ AX$ = i-th column of A in j-th column, 0 otherwise. $ XA$ = j-th row of A in i-th row, 0 otherwise. Thus, $ AX\minus{}XA$ = 0 except the (i,j)-th entry, which is the difference of the (i,i)th entry and (j,j)th entry, which makes $ rank(AX\minus{}XA) \equal{} 1$ if they aren't equal.


number theory
greatest common divisor
articles
algorithm
prime factorization
Euclidean algorithm

First off, I apologize if I posted this in the wrong forum.

Secondly, I am a high school drop out and it's been a few years since I was in school, so I may be asking middle school questions. :)

Anyways, hopefully my questions aren't silly, but I never paid attention in all of my school years, so pardon my ignorance :)

1) Adding and subtracting mixed fractions is it nessecary (ever?) to first convert to improper fractions?

2) [b]GCD and LCD[/b]
I understand them (i think). GCD is the [b]largest [/b]number which evenly divides 2 or more numbers? LCD on the other hand is the [b]lowest common[/b] number which evenly divides 2 or more numbers (Integers?)

GCD is used when reducing fractions:

64/128 would yield a GCD of 64, correct? which would reduce the fraction to 1/2

LCD is used when performing an operation on fractions like addition - becuz we need to share a common denominator before an operation can continue???

What gets me is GCD is said to be best determined (or what I understand anyways - maybe the book is saying something different) by finding the prime factorization of each number and mupliying the common factors. This I understand.

So for example:

12 and 30
12 = 2 (2)(3)
30 = 2 (3)(5)

Which yields a GCD of 6

But I can just as easily in my head determine that 6 is the GCD, so I figured GCD was probably handy for large numbers where the math can't be easily done in one's head. Then I read this article on some other math site which says GCD isn't handy for large numbers, so instead use Euclidian Theorm or something to that effect.

I remember LCD from middle school, but don't remember anything of GCD.

Knowing this, can someone explain to me when I would use GCD and LCD, cuz for the time being I can't really see a difference between them???

Regards,
Alex

LCD is the [b]lowest common denominator[/b], you can only substract and add fractions if they have a common denominator.

For example: $\dfrac{1}{3}$ and $\dfrac{3}{5}$ have LCD of $15$.

GCD is the [b]greatest common divisor[/b]. It is very different from LCD. For example, you have two integer say $12$ and $26$, then the $GCD(12,24)=6$

(In order to find GCD, you can use Euclidean algorithm)

I think that in the Euclidean algorithm, you take the numbers to find the GCD of. Subtract the smaller from the larger, and in your numbers in the GCD, substitute the difference for the larger. Keep doing this until you have GCD(something, 0). The remaining number is the GCD. Basically it uses the principle that GCD(A,B)=GCD(A-B,B).

Hello, Alex. Welcome to this forum. Your question is very welcome and don't be afraid to post because some people might consider "silly."

First of all, ignore Euclidean's algorithrm. That's a complicated concept and I didn't learn it till the Russian math text (translated to English).

Second, use of GCD and LCD are very limited by hand (without using calculators) because in order to find GCD or LCD, we have to look at the number's prime factors like you did with 12 and 30. Imagine you're doing with two numbers like 151489248634 and 7256786431878. That's a lot of work by hand.

Third, GCD doesn't really have "specific-time" to use it. Other than the time you're using it in algebra for sometime, I don't think I ever really used GCD until I came across math competitions. LCD is something you'll use often. This is because in order to add or subtract (multiply or divide does not matter) the fractions which have different denominators, you have to conver them into same number. This number is LCD.

I hope I answered all your questions. If you have anymore, feel free to ask.

Silverfalcon


modular arithmetic
algebra unsolved
algebra

For every positive integer $ n$ show that: $ [\sqrt{n}\plus{}\sqrt{n\plus{}1}]\equal{}[\sqrt{4n\plus{}1}]\equal{}[\sqrt{4n\plus{}2}]\equal{}[\sqrt{4n\plus{}3}]$.

[hide]Since $ n^2 \equiv 0,1 \pmod{4}$, for all $ n \in \mathbb{N}$ we have: $ K^2 \le 4n \plus{} 1 < 4n \plus{} 2 < 4n \plus{} 3 < (K \plus{} 1)^2$ for some $ K \in \mathbb{N}$.
Therefore: $ [\sqrt {4n \plus{} 1}] \equal{} [\sqrt {4n \plus{} 2}] \equal{} [\sqrt {4n \plus{} 3}] \equal{} K$.

Note that: $ (\sqrt {n} \plus{} \sqrt {n \plus{} 1})^2 \equal{} 2n \plus{} 1 \plus{} 2\sqrt {n^2 \plus{} n} > 2n \plus{} 1 \plus{} 2(n) \equal{} 4n \plus{} 1$

And that: $ (\sqrt {n} \plus{} \sqrt {n \plus{} 1})^2 \equal{} 2n \plus{} 1 \plus{} 2\sqrt {n^2 \plus{} n} < 2n \plus{} 1 \plus{} 2(n \plus{} \frac{1}{2}) \equal{} 4n \plus{} 2$

Therefore: $ \sqrt {4n \plus{} 1} < \sqrt {n} \plus{} \sqrt {n \plus{} 1} < \sqrt {4n \plus{} 2}$.
And thus: $ [\sqrt {n} \plus{} \sqrt {n \plus{} 1}] \equal{} [\sqrt {4n \plus{} 1}] \equal{} [\sqrt {4n \plus{} 2}] \equal{} [\sqrt {4n \plus{} 3}]$ as desired.[/hide]


superior algebra
superior algebra unsolved

Let $ A$ be a ring and $ S$ a subset of $ A$. Let $ A[S^{-1}]$ be an extension of $ A$ to a ring such that every $ s$ in $ S$ has a multiplicative inverse. Show that a canonical homomorphism $ e: A\to A[S^{-1}]$ is injective if for all $ s \in S$ we have $ sx = 0 \implies x=0$.

a ring extension consists - per definition - of an injective ring homomorphism.

perhaps you just mean [i]the[/i] localization?

MeKnows, just be careful that you cannot "invert" some subset of a commutative ring arbitrarily: if you want the definition of a localized ring $ S^{-1}A$ to be consistent, you have to take a multiplicative subset.


Demonstrati ca oricare ar fi numerele reale $a,b,c$,urmatoarea relatie are loc:
$6+a^{2}+b^{2}+c^{2}+ab+bc+ac\geq4(a+b+c)$
 [hide]
PS:Eu unul nu am mai vazut inegalitatea asta.Daca apartine altcuiva,scuzele mele :blush: [/hide]
PPS:Inegalitatea este simpla[cel putzin solutia asa e]
Have fun!

intr-adevar e super simpla.inmultesti cu 2 toata inegalitatea si se rescrie:
$12+(a+b)^{2}+(b+c)^{2}+(c+a)^{2}\ge 8(a+b+c)$.notand $x=a+b,y=b+c,z=c+a$ se reduce la $(x-2)^{2}+(y-2)^{2}+(z-2)^{2}\ge 0$ care e evident adevarata.

Corect.Sau
1.Membru stang-Membru drept=$\frac12((a-1)^{2}+(b-1)^{2}+(c-1)^{2}+(a+b+c-3)^{2})$
2.De asemenea,se mai poate rezolva si prin AM-GM.[Postez mai incolo]
3.$4+(a+b)^{2}\geq4(a+b)$ si se aduna analoagele.
Mai sunt alte solutii? :lol:

Folosim principiul trinomului si punem inegalitatea sub forma:
$a^{2}+a(b+c-4)+b^{2}+c^{2}+bc-4b-4c+6$
Iese $\Delta=-3b^{2}+-3c^{2}+8b+8c-2bc-8$ Atunci avem de demonstrat
$3b^{2}+3c^{2}-8b-8c+2bc+8\geq\ 0$
Mai aplicam o data principiul trinomului si obtinem
$\Delta=-32c^{2}+64c-32=-(4\sqrt{2}c-4\sqrt{2})^{2}\leq\ 0$, de unde rezulta concluzia.

Of, of Sorine, ai scris faza aia inainte sa apuc eu sa redactez rezolvarea in Latex  :D

Banuiesc ca ar merge o spargere dupa perechile de indici...

Eu unul n-am stiut inegalitatea asta, dar am vazut o rezolvare cu trinomul de gradul II.. :roll:

Eu ma gandesc ca e cam greoaie cu trinomul...Daca vrei, incearca...

Nu stiu mai Tudor, si mie mi se pare cam greu, dar n-am nicio alta idee.. :(

Ti-am spus, sparge-o, rupe-o, n-o lasa... :D


inequalities
function
triangle inequality

how can I prove that
$2y^2+x^2 \geq xz + 2xy$ where x,y,z >0

You can't, just take $z \rightarrow \infty$.

okay, I guess you have a point there, here is the original question and part of my solution

Let a,b,c be the lengths of the sides of a triangle. Prove that

\[ a^2b+(a-b) + b^2c(b-c) + c^2a (c-a) \geq 0  \]

and determine when equality occurs.

Let $x = \frac{b+c-a}{2}$, $y = \frac{a+c-b}{2}$,$z = \frac{a+b-c}{2}$

So, $a = y+z$, $b =x +z$, $c =x+y$

Use the properly of triangle inequality, $x,y,z > 0$

Use variable substitution, the original inequality becomes

\[ (y+z)^2(x+z)(y-x) + (x+z)^2(x+y)(z-y) + (x+y)^2(y+z)(x-z) \geq 0 \]

expand:

\[ (y^2+z^2+2yz)(x+z)(y-x) + (x^2+z^2+2xz)(x+y)(z-y) + (x^2+y^2+2xy)(y+z)(x-z) \geq 0 \]

\[ (y^2+z^2+2yz)(xy+yz-x^2-xz) + (x^2+z^2+2xz)(xz+yz-xy-y^2) + (x^2+y^2+2xy)(xy+xz-yz-z^2) \geq 0 \]

\[ \sum_{cyc}2xy^3 + x^3y -x^2yz-2x^2y^2 \geq 0 \]

\[ \sum_{cyc}2xy^3 + x^3y -x^2yz-2x^2y^2 \geq 0 \]

I am not sure about the last two steps.

Cuz paladin8 said it is impossible to prove what I have on my first poste.

I got it. All I have to do is to expand the cyc function. Then I will get the answer.

Can anybody tell me the use of cyc functions though?

Is it there just to save time or is there any property that goes with cyclic functions?

$2y^2+x^2\geq xz+2xy$ is not true, but it is possible that

$2y^2+x^2\geq xz+2xy$ where $x,y,z$ are sides of a triangle is true.  See the difference?


I just downloaded flash like twice, but when i try to get into the classroom, it still says i need flash installed. I followed all the procedures on the website but its still not working.

I see that you got in -- what was the trick?

I used my dad's laptop, which does have flash installed.

For whatever reason, i was able to download adobe reader, but not flash. Whenever i downloaded reader, it said something like "agree and download", but when i installed reader, it said "agree and install". I dont know if that made a difference.

If you have firefox your settings may disable some plug-ins, and by settings I mean it may be set not to allow things that it isn't 100% certain are safe.
If not, then I don't know what you're doing wrong.


geometry
rectangle
combinatorics proposed
combinatorics

On a plane given three rectangulars whose any two rectangulars have parallel sides mutually.Find the possible maximum value for which the plane can be divided by these rectangulars. Note that you may count as one for the part which aren't involved to any rectangulars.For example given one rectangular, plane will be divided into two parts.

Is the following restatement correct?

Three rectangles are drawn in the plane with their sides parallel.  What is the maximum number of regions into which the plane may be divided as a result?

Thank you for your correction to my poor English.
Your statement of this problem is correct.
Japan Mathematical Olympiad preliminary was held on Jan. 8
But I have been difficulty in translating the context. :(


function
inequalities
algebra
functional equation
algebra proposed

Find all functions from the positive reals to reals with the property
$f(x^{2})+f(y^{2})=2f(xy)$  for each positive reals x,y and $f(1)=2000$

[hide="big hint"]Let f(x)=g(lnx), or f(e^x)=g(x)

we have  g(2lnx)+g(2lny)=2g(lnx+lny)

letting 2lnx=u,2lny=v

[u]g(x)+g(y)[/u]=g((x+y)/2)
          2

This is jensens functional equation, working backwards, setting f(1)=2000 gives the solution[/hide]

But there is no way to solve Jensen's functional equation, is there?

jensens functional equation has solution kx+n, see here:[url]http://eqworld.ipmnet.ru/en/solutions/fe/fe3103.pdf[/url]

How do you prove kx+n is the only solution?

Generally with functional equations, there is a basis of knowledge of simpler funtional equations, which you reduce the problem into. Personally, I don't know how to prove the equation, but it is a famous equation, so the proof is still valid. :)


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Edit: I know what's going on already, but it seems that I cannot delete the post, so..

---
Let $ A$ be a ring such that
1. for each maximal ideal $ m$ of A, the local ring $ A_m$ is Noetherian.
2. for each $ x \neq 0$ in $ A$, the set of maximal ideals of $ A$ which contain $ x$ is finite.
Show that $ A$ is Noetherian.

I have some idea but I didn't make use of condition 2. Is this correct?

Consider an ascending chain of proper ideals $ a_1 \subset a_2 \subset \cdots$. Then the union $ \cup a_i$ is also a proper ideal, which lies in some maximal ideal $ m$. Localize with respect to $ m$ and use the fact that $ A_m$ is Noetherian shows that the chain of ideals stabilize.

I think there are commutative non-Noetherian ring $ A$ with an identity such that for each maximal ideal $ m$ of $ A$, the local ring $ A_m$ is Noetherian. Therefore you have to use the condition 2. too.

you cannot say much about ideals in $ A_\mathfrak{m}$, the correspondence does not work as in the case of prime ideals. in fact, (2) is important: $ \prod_{i \in \mathbb{N}} \mathbb{Z}/2$ is not noetherian, but every localization is a local boolean ring, i.e. $ \mathbb{Z}/2$ and thus noetherian.

atiyah gives a very detailled hint to this exercise. just follow it.


geometry
combinatorics solved
combinatorics
combinatorial geometry

Consider a circle with a radius of 16 cm. 650 points inside this circle are given.
A ring is the part of the plane that's included between two concentric circles of radius 2 cm and 3 cm respectively.
Show that a ring can be placed such that at least 10 of the 650 given points are covered by this ring.

Nobody seems to be interested ...
For me it's an interesting question but I don't manage to solve it :?

[quote="Arne"]Nobody seems to be interested [/quote]

Perhaps a way to the solution :

The area of the ring is 5 \pi cm2

The sum of the area of the 650 ways of putting the ring = 3250 \pi  cm2 is > 9 x area of circle of radius 16+3=19 cm = 3249 \pi  cm2 (*) so there is a point P in the circle of radius 19 cm which is covered 10+ times when the ring is centered on (wolog) P1, .., P10.
Then putting the ring centered on P will cover P1,.. P10 ( 2 <= |PMi| <= 3)

It seems to be too simple to be correct but where is the mistake ?

(*) 3249 < 3250 : it can't be a coincidence  :) , with this we can't be too far from the solution  :D  :D

Seems to be correct ... didn't think of that.
Thanks :)


Source: GA Varsity Math Tournament 2001

50. The parametric equations for two lines are $(2t+3, 5t+6, -t+6)$ and $(t+2, 4t+3, -3t+20)$

Find the shortest distance between the two lines to the nearest hundredth unit.

[hide]By the distance formula, $d(t)=\sqrt{[2t+3-(t+2)]^2+[5t+6-(4t+3)]^2+[-t+6-(-3t+20)]^2}$. Simplifying, we have $d(t)=\sqrt{6t^2-48t+206}$. The numerator of the derivative is then $12t-48$. The absolute minimum of $d(t)$ occurs when $12t-48=0 \Rightarrow t=4$. At this time, the corresponding points on the lines are $(11,26,2)$ and $(6,19,8)$, and by the distance formula, the shortest distance between the two lines is $\sqrt{110}\approx\framebox{10.49 units}$. [/hide]

[hide]

The question is intentionally misleading you to believe that the value of $t$ for the first line must be the same as the value of $t$ for the second line.  The question is better written as: $(2t_1+3, 5t_1+6, -t_1+6)$ and $(t_2+2, 4t_2+3, -3t_2+20)$.  For instance, it's easy to check that when $t_1 = 1$ and $t_2 = 3$, the two points are $(5,11,5)$ and $(5,15,11)$ and are only $\sqrt{0^2 + 4^2 + 6^2} = \sqrt{52}$ apart, which is less than your answer.  However I'm not quite sure how to find the actual minimum.  I'm sure there's a formula for it that might be worth memorizing.

[/hide]

[hide] Write the equation of the plain $\alpha$ containing the first line and parallel to the second. If I made no mistake it is $\alpha: -11x+5y+3z-15=0$ and its normal equation is $\alpha: \frac{ -11x+5y+3z-15}{\sqrt 155}$.
The distance from the second line to $\alpha$ is the distance from any point of this line, say $P(2,3,20)$ or $d=\frac{38}{\sqrt 155}$[/hide]

A few more approaches:

[hide="Solution 1"]
Let $P = (2t + 3, 5t + 6, -t + 6)$ and $Q = (s + 2, 4s + 3, -3s + 20)$.  Then
\begin{eqnarray*} PQ^2 &=& (2t - s + 1)^2 + (5t - 4s + 3)^2 + (-t + 3s - 14)^2 \\ &=& 30t^2 - 50ts + 26s^2 + 62t - 110s + 206 \\ &=& 30t^2 - (50s - 62)t + 26s^2 - 110s + 206 \\ &=& 26s^2 - (50t + 110)s + 30t^2 + 62t + 206. \end{eqnarray*}

As a quadratic in $t$, this expression is minimized when
\[ t = \frac{50s - 62}{60}. \]
As a quadratic in $s$, this expression is minimized when
\[ s = \frac{50t + 110}{52}. \]

Solving this system of linear equations, we get
\[ t = \frac{569}{155}, \quad s = \frac{175}{31}. \]

Plugging these into the formula above gives
\[ PQ = \frac{38}{\sqrt{155}}. \]
[/hide]

[hide="Solution 2"]
When $PQ$ is minimized, the vector $\overrightarrow{QP}$ is perpendicular to both lines.  The direction vector of the first line is $\overrightarrow{u} = (2,5,-1)$, and the direction vector of the second line is $\overrightarrow{v} = (1,4,-3)$.  So, we want $\overrightarrow{QP} \cdot \vec{u} = \overrightarrow{QP} \cdot \vec{v} = 0$, which gives the equations
\begin{eqnarray*} 2(2t - s + 1) + 5(5t - 4s + 3) - (-t + 3s - 14) &=& 0, \\ 1(2t - s + 1) + 4(5t - 4s + 3) - 3(-t + 3s - 14) &=& 0. \end{eqnarray*}

This gives us two linear equations in $t$ and $s$, and their solution is the same as in solution 1.
[/hide]

[hide="Solution 3"]
A slight variation on solution 2.  Once we know that $\overrightarrow{QP}$ is perpendicular to $\overrightarrow{u}$ and $\overrightarrow{v}$, then $\overrightarrow{QP}$ is parallel to $\overrightarrow{u} \times \overrightarrow{v} = (-11, 5, 3)$.  Therefore,
\[ \frac{2t - s + 1}{-11} = \frac{5t - 4s + 3}{5} = \frac{-t + 3s - 14}{3}. \]
This again gives two linear equations in $t$ and $s$.
[/hide]

All this may seem like overkill, but it's good to view a problem from different angles.


[quote="Wolfram MathWorld"]In a game of 40 moves, the number of possible board positions P(40) is at least 10^(120) according to Peterson (1996).  However, this value does not agree with the 10^(40) possible positions given by Beeler et al. (1972)...

(http://mathworld.wolfram.com/Chess.html)[/quote]

It seems rather obvious that Peterson's number is incorrect... Consider the fact that any square can contain one of 6 distinct white pieces, 6 distinct black pieces, or it can be empty.  This makes for 13 possible conditions for each square.  There are 64, squares, so we get $ 13^{64}$.  Of course, this is a gross overestimate, because we can never have less than 32 empty squares, and we could never fill all 64 squares with kings, for example.  This [b]overcount[/b] looks to be less than Peterson's number...

$ 13^{64} < 10^{120}$
$ 64 \log{13} < 120 \log{10}$
$ 71.2923745 < 120$

Did I do something wrong?

I didn't thouroughly look at your post, but the reason this scenario is so difficult is the conditions.  For example, not only to you have to consider the factors you stated, but you also need to account for any pieces that were taken, any pawn promotions, etc.  Your argument is still a huge overestimate, but I don't think you're wrong.  After all, you're quote says that Peterson's number disagrees with that one given by Beeler, so it maybe the wrong one.  And, you're overestimate is still greater than Beeler's number.

[quote="seamusoboyle"]Mysmartmouth is saying is that there must be less than $ 13^{64}$ posible board configurations.

Peterson's saying that its more than $ 10^{120}$

Peterson could have been talking about the possible number of games up to 40 moves perhaps?[/quote]

I'm inclined to believe that's what Peterson was actually talking about as well.


an acid (Mr = 163) of % purity 99.1% is to be investigated using a titration. the aim is to check the purity of the acid by titration with sodium hydroxide.

7.1g of acid added to a beaker and dissolved in de-ionised water. Solution then poured into a 250cm^3 graduated flask and filled to the mark with de-ionised water.

A burette is filled with the acid solution. The average titre found when 25.00cm^3 portions of 0.150M NaOH solution were titrated with the acid solution was 23.66cm^-3 (average titre from range of results)

1) calculate the molarity of the acid solution

2) using this molarity determine the concentration of the acid in grams per dm^3

3) calculate the % purity of the acid

yes the molar ratio is 1:1

(if that's your question)

No.  He was giving you tools to solve the problem.

molarity of solution 1 times the volume of solution 1 equals the molarity of solution 2 times the volume of solution 2.  This is true for 1:1 mole reactants, as occurs in your case.

The purity is determined as in all cases by dividing by the molarity measured by reactivity by the molarity measured by measurement of reactant (weight or volume of acid, in this case).

If the prior response wasn't enought, this supplement should be enough for you to easily solve this question.

There is one, simple answer to every titration question: write down the reaction equation. Than all you have to do is simple stoichiometry :)


$\frac {1 \cdot 2 \cdot 4 \cdot + 2 \cdot 4 \cdot 8 \cdot + ... 10 \cdot 20 \cdot 40}{1 \cdot 3 \cdot 9 + 2 \cdot 6 \cdot 18 .... 10 \cdot 30 \cdot 90}$

Simplify the above.

Glad to see some South African problems!  More problems from this olympiad can be found [url=http://science.up.ac.za/samo]here[/url] There are two sections
juniors : grades 8 and 9
seniors : grades 10 to 12
Each section has 3 rounds round one and two are multiply choice and round 3 is written.  So there is a good range of problems there. enjoy!

Yes.

Problems are very good. This competition is also listed in Art of Problem Solving Resources Page also.

Please do the problems. :)

Especially this one. :D

h_s: [hide]$2\cdot4\cdot8 = 2\cdot(1\cdot2\cdot4)$?[/hide]

Yeah, H_S did make and error, but it doesn't affect the final result, since it cancels out anyways.

I noticed that as well -- if I were making a short-answer contest, this is a question to avoid, I think, since there are so many wrong ways to get the correct answer.


The length of the common internal tgt. to 2 circles is 7 and a common external tangent is 11; then the product of their radii is how much??????

[hide]
Distance between the two circles' centers=$k$
We draw the radii to the points where the tangents meet the circles, then we connect the circles' centers, to get two right triangles. For the internal tangent, $(r_{1}+r_{2})^{2}+7^{2}=k^{2}$. For the external tangent, $(r_{1}-r_{2})^{2}+11^{2}=k^{2}$ (WLOG $r_{1}>r_{2}$). Setting these two equal, we get $r_{1}r_{2}=\boxed{18}$.
[/hide]

Wasn't this a question on the 1983 Individual portion of ARML?

I can envision an external tangent, but what is an internal tangent line? As in what would it look like?

The common internal tangents of two circles are the lines tangent to both circles such that the circles are on different sides of the tangent lines.

here, i will try out asymptote and try to answer your question.

[asy]size(100); pair A,B,C,D,E,F; A=(0,1); B=(-1,2); C=(-2,1); D=(0,-1); E=(1,-2); F=(2,-1); draw(circumcircle(A,B,C)); draw(circumcircle(D,E,F)); draw((0,3)--(0,-3)); draw((-3,0)--(3,0));[/asy]

these are the common internal tangents to two circles.

Ah! I get it now, as well as where the two equations he used come from, thanks!


function
vector
geometry
3D geometry
sphere
advanced fields
advanced fields unsolved

what do you mean by $f(x)=-f(x)$? can you clraify that please? i think you need to change something

I mean an odd function but I made a misspelling.

[quote="puuhikki"]Let $f: \mathbb{D}^{2}=\{(x,y)\in\mathbb{R}^{2},x^{2}+y^{2}\leq 1\}$ be a continuous odd function such that if $x\in \mathbb{S}^{1}$, then $f(-x)=-f(x)$. Prove that for some $x_{0}\in\mathbb{D}^{2}$ we have $f(x_{0})=0$.

Is this a correct idea: Let $A=\{x\in \mathbb{D}^{2}: f(x)<0\}$ and $B=\{x\in \mathbb{D}^{2}: f(x)>0\}$. Take two points $x_{1}\in A$ and $x_{2}\in B$ and join them by line segment $\gamma$. Then how can I show that $f(\gamma \cap \partial A)=0$?[/quote]Is $f$ real valued? In that case, how about considering $g: [-1,1] \to \mathbb{R}$, defined by $g(x) = f(x,0)$ and looking for a zero of $g$ via the intermediate value theorem? On the other hand, if $f$ has values in $\mathbb{R}^{2}$, then your idea will probably not work.

The function is $f: \mathbb{D}^{2}\to \mathbb{R}^{2}$.

Aha, you mean a vector field on D2.

This is a classical problem in algebraic topology.

Let me change notation a little bit. Let X be the vector field you call f. X is continuous and we have X(-p) = -X(p) when p is on S1.

X induces a continuous tangent vector field on the closed northern hemisphere N of the sphere S2.

Let's define a vector field Y on the closed southern hemisphere S too by setting Y(p) = -X(-p). Y is continuous. On the equator Y(p) = -X(-p) = -(-X(p)) = X(p) so X and Y agree on the equator.

Therefore we can join them to a global continuous vector field on S2. It is well known that every such field vanishes somewhere. If this happens on N then we are done. If it happens on S at a point p, then X(-p) = -Y(p) = 0 and again we are done.


induction
number theory
relatively prime

$ 6\equal{}3\cdot 2\plus{}4\cdot 0$
$ 7\equal{}3\cdot 1\plus{}4\cdot 1$
$ 8\equal{}3\cdot 0\plus{}4\cdot 2$

Now prove by induction that if it is true for $ k$ then it is true for $ k\plus{}3$  :)

[hide]We can write the argument above as a non-inductive argument, as well. Consider the generation scheme:

$ 3k \equal{} 3(k)\plus{}4(0)$
$ 3k\plus{}1 \equal{} 3(k\minus{}1) \plus{} 4(1)$
$ 3k\plus{}2 \equal{} 3(k\minus{}2)\plus{}4(2)$

Where the minimum values on $ k$ are $ 0, \; 1, \; 2$ for those 3 cases, respectively. This scheme creates all non-negative integers except $ 1,2,5$.[/hide]

[quote="azn6021023"]This isn't exactly a proof, but given $ a$ and $ b$, where $ a$ and $ b$ are relatively prime, the largest number that you cannot make with multiples of $ a$ and $ b$ is expressed as $ ab \minus{} a \minus{} b$, or $ (a \minus{} 1)(b \minus{} 1) \minus{} 1$.  For this case, it is $ 5$.[/quote]

See http://en.wikipedia.org/wiki/Chicken_McNuggets_problem


geometry
superior algebra
superior algebra unsolved

Let $ \Gamma$ be a variety of monoids such that not all monoids of $ \Gamma$ are groups. Prove that if $ A \in \Gamma$ and $ B$ is a submonoid of $ A$, there exist monoids $ S \in \Gamma$ and $ C$ and epimorphisms $ \varphi : S \rightarrow A, \;\varphi_1 : S \rightarrow C$ such that $ ((e)\varphi_1^{\minus{}1})\varphi\equal{}B$ ($ e$ is the identity element of $ C$).

[i]L. Marki[/i]

[quote="ehsan2004"]a variety of monoids[/quote]
Could you define that term? I have read about monoids and seen varieties in algebraic geometry but I don't know what is a variety of monoids.


Prove that the digit six appears an infinite number of times in the sequence.

2,7,1,4,7,4,2,8...

What is the sequence? It could be anything! Including a sequence that has no sixes in it...

$a_1=2, a_2=7$, to get $a_3$ and $a_4$ multiply$a_1 \times a_2= 2 \times 7=14$ giving $a_3=1,a_4=4$ to get your next digit/s multiply $a_2 \times a_3=7\times 1 \rightarrow a_5=7$, etc.

2, 7, 1, 4, 7, 4, 2, 8, 2, 8, 8, 1, 6, 1, 6, 1, 6, 6, 4, 8, 6, 6, 6, 6, 6, 3, 6, 2, 4, 3, 2, 4, 8, 3, 6, 3, 6, 3, 6, 3, 6, 1, 8, 1, 8, 1, 2, 8, 1, 2, 6, 8, 3, 2, 2, 4, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 6, 8, 8, 8, 8, 2, 1, 6, 8, 2, 1, 2, 4, 8, 2, 4, 6, 4, 8, 4, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 4,...........

Consider the 11 consecutive 8's; somewhere after them in the sequence they correspond to  alternating  6's and 4's, which correspond somewhere later in the sequence to  alternating 2's and 4's which correspond later yet to a run of  8's  later in the sequence, which in turn reproduce the alternating 6's and 4's ......this proccess goes on indefinitely hence we have infinitely many 6's. We also have infinitely many 8's, 2's, and 4's.


In an isosceles triangle, the base equals 24cm and the lateral sides equal 15cm. Find the lengths of the inradius and the circumradius.

Use Pithagoras to find the area and then use:

$ (ABC)\equal{}\frac{abc}{4R}\equal{}(a\plus{}b\plus{}c)\frac{r}{2}$

 :)

Base = 24, sides = 15.

We can drop the altitude, forming two congruent right triangles.
The length of this altitude is $ \sqrt{24^2-15^2}=\sqrt{351}=3\sqrt{39}$.
The area is obviously $ (1/2)(24)(3\sqrt{39)=36\sqrt{39}}$ but also (using the formula K = rs),
$ r\frac{24+15+15}{2}=27r$
So we can equate the two expressions and divide by 27 so $ r=\frac{4\sqrt{39}}{3}$.
The area is also $ \frac{abc}{4R}$.
$ \frac{24\cdot 15 \cdot 15}{4R}=36\sqrt{39}$
$ \frac{15 \cdot 15}{R}=6\sqrt{39}$
$ \frac{75}{R}=2\sqrt{39}$
$ R = \frac{75}{2\sqrt{39}}$

[quote="IntrepidMath"]Base = 24, sides = 15.

We can drop the altitude, forming two congruent right triangles.
The length of this altitude is $ \sqrt {24^2 - 15^2} = \sqrt {351} = 3\sqrt {39}$.
The area is obviously $ (1/2)(24)(3\sqrt {39) = 36\sqrt {39}}$ but also (using the formula K = rs),
$ r\frac {24 + 15 + 15}{2} = 27r$
So we can equate the two expressions and divide by 27 so $ r = \frac {4\sqrt {39}}{3}$.
The area is also $ \frac {abc}{4R}$.
$ \frac {24\cdot 15 \cdot 15}{4R} = 36\sqrt {39}$
$ \frac {15 \cdot 15}{R} = 6\sqrt {39}$
$ \frac {75}{R} = 2\sqrt {39}$
$ R = \frac {75}{2\sqrt {39}}$[/quote]

Umm what?
No offense but your area is wrong screwing up the whole problem
Im pretty sure the altitude isnt $ \sqrt {24^2 - 15^2} = \sqrt {351} = 3\sqrt {39}$

The altitude to the base is actually $ \sqrt{15^2\minus{}12^2}\equal{}9$  :wink:


Marcia has a total of 65 game pieces worth 89 points. Each game
piece is worth one or two points. How many one-point game pieces
does she have?


There is an interesting five-digit number. With a 1 after it, it is three times as lardge as with a 1 before it. What is it ?

[i]from book "The Moscow puzzles" by Boris A. Kordemsky

Please hide Your solution . Write inside Hide ... /Hide
[/i]
 :wink:

[quote="gaussintraining"][hide] $ 42857$   :D [/hide][/quote]
Solution ?
 :juggle:

[hide] $ 142857$ or the repeating part of $ \frac{1}{7}\equal{}\overline{.142857}$, has many interesting properties relating to the fact that each of its first $ 6$ multiples has the same digits as the others, just in a different order.

$ 142857$

$ 285714$

$ 428571$

$ 571428$

$ 714285$

$ 857142$

When the question stated that it was a five digit number, which when adding a digit to the front or back of it resulting in $ 2$ numbers, with the property that one was $ 3$ times the other (and of course noting that a $ 5$ digit number with another digit makes a $ 6$ digit number), I thought of this sequence of numbers, since it is often the solution to puzzles of that kind with $ 6$ digit numbers and multiples that can be rearranged. A quick check confirmed that $ 142857\times{3}\equal{}428571$. So the original $ 5$ digit number was $ \boxed{42857}$ [/hide]

I found a different way.Here it is.

[hide="Click here to reveal hidden content"]

Call the number abcde, with the variables the digits. Therefore, with what is given, you can conclude that abcde1=3(1abcde). You can write abcde1 and 1abcde in expanded form. This is what abcde1 would be like in expanded form: 100,000a + 10,000b + 1,000c + 100d +10e + 1. 1abcde would be: 100,000 + 10,000a + 1,000b + 100c + 10d + e.
By substitution you can form the conclusion that 100,000a + 10,000b + 1,000c + 100d + 10e + 1=3(100,000 + 10,000a + 1,000b + 100c + 10d + e.) Distributing gives 100,000a + 10,000b + 1,000c + 100d + 10e + 1=300,000 + 30,000a + 3,000b + 300c + 30d + 3e. Subtracting from both sides forms 70,000a + 7,000b + 700c + 70d + 7e= 299,999. Dividing by 7 yields 10,000a + 1,000b + 100c + 10d + e=42857. Taking this and putting it together makes abcde. Therefore, abcde=[b]42857[/b]

[/hide]

My solution started similarly to Steve's:

[hide]

Let x = the 5-digit number

10x + 1 = 3(100000 + x)
10x + 1 = 300000 + 3x
       7x = 29999
        x = 42857

[/hide]

I just found this forum, and I'm liking it a lot!


abstract algebra
superior algebra
superior algebra unsolved

give an example of a family $ \{M_i\}_{\in I}$ and $ N$
      $ R$ _modules such that

$ HOM_R(\Pi_{i\in I}M_i,N) \ncong_Z   \Pi_{i\in I}HOM_R(M,N_i)$

Try $ I\equal{}\mathbb N$ and $ R\equal{}M_i\equal{}N$ and see what happens to the family $ \{id: M_i\to N\}_i$.


The radius of Earth at the equator is approximately 4000 miles. Suppose a jet flies once around Earth at a speed of 500 miles per hour relative to Earth. If the flight path is a neglibile height above the equator, then, among the following choices, the best estimate of the number of hours of flight is:

$\textbf{(A)}\ 25 \qquad
\textbf{(B)}\ 8 \qquad
\textbf{(C)}\ 75 \qquad
\textbf{(D)}\ 50 \qquad
\textbf{(E)}\ 100$

Wow... I always thought the AHSME's were harder than [b]that[/b].

[hide]
Answer

\[\frac{2 \cdot 4000 \cdot \pi}{500}=16\pi=50.26\]

Thus, the answer is $D$!

[/hide]

Exactly right.

To whitehorseking88, good one. :)

The AHSME I'm posting on the MC are pretty neat ones. You'll see some nasty ones in future. :?

What about this one

Given that function f is represented as f(x) = x/10 is x is divisible by 10, and f(x) = x+3 if x is not divisible by 10. Let a_0 = 1993 and a_{n+1} = f(a_{n}). What is the smallest n such that a_n = 1?

[color=darkred]ModThing: Read what Silverfalcon said :P .[/color]

[quote]The radius of Earth at the equator is approximately 4000 miles. Suppose a jet flies once around Earth at a speed of 500 miles per hour relative to Earth. If the flight path is a neglibile height above the equator, then, among the following choices, the best estimate of the number of hours of flight is:

A) 25
B) 8
C) 75
D) 50
E) 100[/quote]

[hide]the equatoer of the earth is $2\pi4000.$

$h=\displaystyle\frac{2\pi4000}{500}=16\pi$

$h\approx50 hours$[/hide]

Rawr :P . *abuses moderator powers*

Does the plane travel in a great circle around the earth or does it circle once at the north pole :P ?

Guys, let's calm down here.

Silverfalcon- Let's not get too worked up. Yes, the problems were too difficult for the Mathcounts forum, but I don't think you had to say it that way to communicate the fact over. C'mon, I've talked to you before and usually you're much more good natured than that.  ;)

DPopov and White_horse_king- Let's try to remember that this is the Mathcounts forum. Those problems are a little (just a tad ;) ) too difficult for the contest. Next time, let's try to stick with problems that are more suitable for Mathcounts.

In general, this forum is suitable for Mathcounts level problems, but more difficult follow-up problems are also welcome, as well as challenges that are a bit tougher than Mathcounts level (i.e. easy AHSMEs). So, let's not get any problems that are too tough (i.e. any AMC problems after #15).

Thanks, and happy new year!

True, true.

But AHSME got really harder in early 80's before AIME was established. They're hard, I'm telling you. I tried them. ;)

[hide]4000 mile radius or 8000 mile diameter.  So D(pi) gives us the cirucmference of approxiamately 25120 and traveling at 500mph, we need about 50 hours or D.[/hide]

[quote="white_horse_king88"]LOL. Mentioning messier problems. Here's a harder one.

In triangle ABC, M is the midpoint of side BC, AB = 12 and AC = 16. Points E and F are taken on AC and AB, respectively, and lines EF and AM intersect at G. If AE = 2AF then EG/GF equals.

AHSME 1975.

[color=darkred]ModThing: What Silverfalcon said :P .[/color][/quote]

what's the answer to that question?

when airplanes flies, it usually flies like 10000meters above the ground level... so I think we should include that to calculate the amount of time it flied... it will be more than 50 hours..

and

this is a fact...

it takes only about 30 hours to circle the earth with a plane  :D

[quote=white_horse_king88]LOL. Mentioning messier problems. Here's a harder one.

In triangle ABC, M is the midpoint of side BC, AB = 12 and AC = 16. Points E and F are taken on AC and AB, respectively, and lines EF and AM intersect at G. If AE = 2AF then EG/GF equals.

AHSME 1975.

[color=darkred]ModThing: What Silverfalcon said :P .[/color][/quote]

Solution with [b]stroller[/b], [b]mathscienceclass[/b], [b]mathleticguyyy[/b], [b]brianzjk[/b], [b]kevinmathz[/b], [b]awang11[/b], [b]dchenmathcounts[/b], and [b]RadiantCheddar[/b].

Let $D$ be the point on line $AB$ with $AD = 8,BD=4$. We note that by SAS similarity, $\triangle ACD\sim \triangle AEF$. Define $CD\cap AM = K$. Then, let $\mathcal{YUH}$ be the homothety about $A$ that takes $\triangle ACD$ to $\triangle AEF$. As $\mathcal{YUH}$ takes $AM$ to itself and $CD$ to $EF$, we must have that $\mathcal{YUH}$ takes $K$ to $G$. This implies that $\frac{EG}{GF} = \frac{CK}{KD}$.

Now, we use barycentric coordinates with regard to $\triangle ABC$. We have
\[M=(0:1:1), D = (1:2:0).\]
Note that $AM$ is parametrized by $(t:1:1)$ and $CD$ is parametrized by $(1:2:t)$, so we get that $K=(1:2:2)$. Homogenize $K$ to get
\[K = \left(\frac{1}{5},\frac{2}{5},\frac{2}{5}\right) = \frac{3}{5} \cdot \left(\frac{1}{3},\frac{2}{3},0\right) + \frac{2}{5} \cdot (0,0,1).\]
Now, we see that
\[\frac{EG}{GF}=\frac{CK}{KD} = \frac{3}{2}.\]


Prove that : $ \frac{a}{a\plus{}2b} \plus{} \frac{b}{b\plus{}2c}\plus{}\frac{c}{c\plus{}2a} \ge  \frac{a}{2a\plus{}b}\plus{}\frac{b}{2b\plus{}c}\plus{}\frac{c}{2c\plus{}a}$ , where $ a,b,c>0$ .

[quote="linker"]Prove that : $ \frac {a}{a \plus{} 2b} \plus{} \frac {b}{b \plus{} 2c} \plus{} \frac {c}{c \plus{} 2a} \ge \frac {a}{2a \plus{} b} \plus{} \frac {b}{2b \plus{} c} \plus{} \frac {c}{2c \plus{} a}$ , where $ a,b,c > 0$ .[/quote]
$ \frac {a}{a \plus{} 2b} \plus{} \frac {b}{b \plus{} 2c} \plus{} \frac {c}{c \plus{} 2a} \ge1\geq \frac {a}{2a \plus{} b} \plus{} \frac {b}{2b \plus{} c} \plus{} \frac {c}{2c \plus{} a}.$


Find the volume of the region in space defined by
\[ |x + y + z| + |x + y - z| \le 8 \qquad \text{and} \qquad x, y, z \ge 0. \]


could you help me with any idea to find this?or is this a divergent one?
could anyone explain how to check the convergence of a series?

$ \sum_{k\equal{}1}^{\infty}\frac{1}{(3k\minus{}1)(3k\minus{}2)}$

thanks before :)

It definitely converges by comparison to $ \sum \frac{1}{k^2}$.  We can write $ f(x) \equal{} \sum_{k\equal{}0}^{\infty} x^{3k} \equal{} \frac{1}{1 \minus{} x^3}$ and integrate twice.

oh,thank you for your reply..
but,,actually I'm a beginner in calculus..how can we compare it with $ \sum \frac{1}{k^2}$ and make $ f(x)$ as you wrote before?
so, in case $ \sum_{k \equal{} 1}^{\infty}\frac {1}{(3k \minus{} 1)(3k \plus{} 1)}$,, is it a convergent one?

you can just look at the $ 9k^2$ term. E.g. $ \sum \frac {1}{9k^2} < \sum \frac {1}{k^2}$.


maybe this seems easy for u ppl.. but i just dun get the answer....
out lectur didnt talk abt it in calss.. now i am in trouble....

prove that is A is an infinite set then so is the set A*B for any nonempty set B. what happenes is B is the empty set?

What is A*B? I mean, what does the * stand for?

a*b= a times b
this is a simble for computer science.. sorry i didnt clear it out b4...
*=times
/=deivde .....
^=power.....
and so on...

by the way.. how u ppl type all the math simbles?


This is an Epithalamium Verso. Made by Cassanova on Wincustomize.


Let $ A \in \mathcal{M}_{n}(\mathbb{Z})$ with unit determinant such that for every positive integer $ k$, there exists some matrix $ X \in \mathcal{M}_{n}(\mathbb{Z})$ such that $ X^{k} \equal{} A$. Prove that $ A \equal{} I_{n}$.

This is question 11401 from the AMM Dec 2008.
As someone famous on this forum wrote once: "I will not post the proof of that result, because it is not fair, but take it as a hint: that problem is very easy!"
At the end of the month I'll post of 4 line-solution if there is none.

I've managed to show that the eigenvalues of $ A$ are all $ 1$, but I can't see any further path.

PS. Actually, the problem is from a Romanian NMO Shortlist from 2007.

The condition that $ A$ has unit determinant is not needed (this condition does not appear in the AMM problem either), as long as we assume that $ \det A\neq 0$.
Indeed, this problem has a very short solution. Please remind me to post it when the deadline is over (which is the last of April, I believe).

darij, from NoName\s notebook

Ah, ok. I didn't know the problem was still open (as I said, this statement with "unit determinant" appears like this in the RMO SL from 2007).

Thanks to Moubinool for reminding me to post my solution, which goes as follows:

Let $ p$ be an arbitrary prime not dividing $ \det A$, and let $ k \equal{} |\mathcal{GL}_n(\mathbb{Z}/(p))|$. By our premise, there exists an $ X \in \mathcal{M}_n(\mathbb{Z})$ such that $ X^k \equal{} A$. Since $ \det A \not\equiv 0 \mod p$, we know that $ A \mod p$, and hence $ X \mod p$, lies in $ \mathcal{GL}_n(\mathbb{Z}/(p))$. By Lagrange, $ A \equal{} X^k \equiv I_n \mod p$. Now, since $ p$ was chosen arbitrarily, $ A$ must equal $ I_n$ globally.

darij, from NoName\s notebook

I did not find any short solution.
The NoName's proof is great !

There could be a solution without LAgrange:
First we prove that the eigenvalues of A are 1 .After that we use the Jordan form of A :$ A \equal{} TJ_A T^{ \minus{} 1}$ where $ J_A$ is the matrix where the diagonal are the eigenvalues and T is the matrix formed by the eigenvectors of A.


Let $ a$, $ b$, $ c$ and $ k\le 1$ be positive reals such that $ abc\le 1$.
Prove that
\[ \dfrac{a}{b}\plus{}\dfrac{b}{c}\plus{}\dfrac{c}{a}\ge a^k\plus{}b^k\plus{}c^k\]

:)

[quote="ElChapin"]Let $ a$, $ b$, $ c$ and $ k\le 1$ be positive reals such that $ abc\le 1$.
Prove that
\[ \dfrac{a}{b} \plus{} \dfrac{b}{c} \plus{} \dfrac{c}{a}\ge a^k \plus{} b^k \plus{} c^k
\]
:)[/quote]
We have:$ LHS \ge \frac{a\plus{}b\plus{}c}{\sqrt[3]{abc}} \ge a\plus{}b\plus{}c \ge a^k\plus{}b^k\plus{}c^k$ :)

Thanks :) , but $ a\plus{}b\plus{}c\ge a^k\plus{}b^k\plus{}c^k$ is not always true :maybe: . Try $ k\equal{}0.1$, $ a\equal{}1.7$, $ b\equal{}0.6$ and $ c\equal{}0.5$.

Also, I know that $ \dfrac{a}{b}\plus{}\dfrac{b}{c}\plus{}\dfrac{c}{a}\ge a\plus{}b\plus{}c$, but how did you see it inmediately like $ \dfrac{a}{b}\plus{}\dfrac{b}{c}\plus{}\dfrac{c}{a}\ge \dfrac{a\plus{}b\plus{}c}{\sqrt[3]{abc}}$?

:)

[quote="ElChapin"]Thanks :) , but $ a \plus{} b \plus{} c\ge a^k \plus{} b^k \plus{} c^k$ is not always true :maybe: . Try $ k \equal{} 0.1$, $ a \equal{} 1.7$, $ b \equal{} 0.6$ and $ c \equal{} 0.5$.

Also, I know that $ \dfrac{a}{b} \plus{} \dfrac{b}{c} \plus{} \dfrac{c}{a}\ge a \plus{} b \plus{} c$, but how did you see it inmediately like $ \dfrac{a}{b} \plus{} \dfrac{b}{c} \plus{} \dfrac{c}{a}\ge \dfrac{a \plus{} b \plus{} c}{\sqrt [3]{abc}}$?

:)[/quote]
Sorry,I think $ abc\equal{}1$ :) .we can prove that: $ LHS \ge \frac{a\plus{}b\plus{}c}{\sqrt[3]{abc}}$very easy
WE have: $ \frac{a}{b}\plus{}\frac{a}{b}\plus{}\frac{b}{c} \ge 3\sqrt{\frac{a^2}{bc}}\equal{}\frac{a}{\sqrt[3]{abc}}$
Similar,we have Q.E.D :)

:lol:  Yes, that is exactly my proof. With slight modifications, using am gm, this problem is solved.

:)


algebra
function
domain
induction
system of equations
algebra unsolved

slove the system of equations:
$\frac{1}{xy}=\frac{x}{z}+1,\frac{1}{yz}=\frac{y}{x}+1,\frac{1}{zx}=\frac{z}{y}+1$
in the domain of real number.

1) prove that $x,y,z$ have same sign
2) you can assume $x \geq y \geq z > 0$ and easily get that $\frac{1}{xy} \geq 2$ and $\frac{1}{zx} \geq 2$
3) Going back again to our equations you get $x=y$ and then similary prove that $y=z$
4) So there are two solutions: for $x,y,z >0$ and $x,y,z < 0$

Hope this helps  :)

Yes, its exactly this solution- http://www.om.edu.pl/zadania/zwardon/zwardon2004r.pdf (page 12, prb.4)
The problem comes from Czech-Polish-Slovak Maths Olympiad 2004.

here is my solutions:
Put $a=\frac{y}{x}$, $b=\frac{z}{y}$, $c=\frac{x}{z}$
we have: $a+1=\frac{y}{x}+1=\frac{1}{yz}=\frac{x}{z}\cdot{\frac{1}{xy}}=c\cdot({c+1})$
as like that we have $b+1=a\cdot({a+1})$ and $c+1=b\cdot({b+1})$
so $a^2+b^2+c^2=3$
Now, applying AM-GM, we obtain $a=b=c=1$ and we have 2 solutions
$(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$ and $(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$
okie?
@filip: i can't dowload this page?

you solution are both very nice.thank you :)  :)

nguyenquockhanh: method with substition which you presented to the original problem would work well here too, I guess  :) nice one, but I've got feeling I've seen it before - where did you get it from?

my teacher gave me this problem and i need 1 minute to solve it
and i've never seen it before-okie?
@megus: sorry for my poor english

[quote="nguyenquockhanh"]okie, now, solve newpro:
$\frac{y}{x}+\frac{z}{y}+\frac{t}{z}+1=\frac{1}{tv}$
$\frac{z}{y}+\frac{t}{z}+\frac{v}{t}+1=\frac{1}{vx}$
$\frac{t}{z}+\frac{v}{t}+\frac{x}{v}+1=\frac{1}{xy}$
$\frac{v}{t}+\frac{x}{v}+\frac{y}{x}+1=\frac{1}{yz}$
$\frac{x}{v}+\frac{y}{x}+\frac{z}{y}+1=\frac{1}{zt}$[/quote]
Well, I just solved the general MEGA-equation:
\[
1+\sum_{j=1}^{n-3} \frac{x_{j+k}}{x_{j-1+k}}=\frac{1}{x_{n-2+k}x_{n-1+k}}, \quad k=1,2,\dots,n.
\]
 [HINT: consider the difference of consecutive equations and use a nice argument as before comments to show that  every variable is equal to the others.]
obtaining that $x_1=x_2=\dots =x_n$ [use induction in some sense] then $x_i=\frac{\pm 1}{\sqrt{n-2}}$ for every $i$.
I don't have space in the margin of the forum to write the solution so I hope anyone try to do it in a simple WAY!.
GOOD LUCK

robetux :D , i think MEGA-equation not easy
because when n=4 then system has infinite roots, so check your result, please.okie?

[quote="nguyenquockhanh"]okie :D
and now, solve new pro:
$\frac{y}{z}+\frac{z}{y}+1=\frac{1}{tx}$
$\frac{z}{t}+\frac{t}{z}+1=\frac{1}{xy}$
$\frac{t}{x}+\frac{x}{t}+1=\frac{1}{yz}$
$\frac{x}{y}+\frac{y}{x}+1=\frac{1}{zt}$[/quote]

okie: the system has infinite roots:
$x=z=t$, $y=t=\frac{1-t}{1+t}$ in here $t\ne{-1,0,1}$


I'm having trouble completing this proof.

Proove that for any positive integer, the sum of the reciprocals of all of the integer's factors is equal to:

$\frac{\text{the sum of all of the factors}}{\text{the integer}}$

Here is the progress I've made thus far:

WLOG let $k$ be a positive integer with 4 distinct prime factors, {a, b, c, d}. Then

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{1+a+b+c+d+ab+ac+ad+bc+bd+cd+abd+acd+bcd+abcd}{abcd}$

Where the long numerator represents every factor of the integer. This form works for any integer that does not contain powers of a prime factor, but I am unsure of how to deal with powers of primes; for example, the above proof works for the integer $(7)(11)(23)$ but not $(7^2)(11)(23)$.

I have tried creating new variables for each power, for instance $a=3$, $b=3^2$ but this does not work because then $ab=3^3$ which is not a factor. I also tried creating a new variable for each $2^n$ power, for instance $a=3$, $b=3^2$, $c=3^4$, $d=3^8$. But this still results in over-counting because if an integer contains, for instance, the factor $3^9$,  you would need $2^n$ powers up to $3^8$, but you would end up multiplying everything through $3^{15}$ or in general $3^{2^{n+1}-1}$.

Please help me complete this! How do I include powers of prime factors without over-counting?

What i did was just realize that to get a common denominator you want to LCM of all the factors (or the number itself.)
when you do this you multiply the numerator of each fraction by $\frac{\#}{factor}$ where the factor is the factor in the denominator.
when you do this to all the fractions you will have the $\#$ in the denominator and one of each factor in the numerator so you will have the desired result.

[quote="bthings_2000"]$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{1+a+b+c+d+ab+ac+ad+bc+bd+cd+abd+acd+bcd+abcd}{abcd}$[/quote]

This is not true -- the RHS what you want, but the LHS is not, and the two sides are not equal.  Fixing the LHS may give you a better idea of how to do the other cases.

[quote="bthings_2000"]
WLOG let $k$ be a positive integer with 4 distinct prime factors, {a, b, c, d}. [/quote]

You lose rather a lot of generality with this step, I think  :|

Edit:  [hide="Rather large hint"] Multiply the entire problem by $k$. [/hide]

Ahh... yes

When you multiply by $k$ you get:

$\frac k{f_0}+\frac k{f_1}+...+\frac k{f_n}$

where $f_0$ to $f_n$ are the factors in order of size. But since $f_i\cdot f_{n-i}=k$,

$\frac k{f_0}+\frac k{f_1}+...+\frac k{f_n}=f_n+f_{n-1}+...+f_0$ which is the sum of all the factors

[quote="nat mc"]But since $f_i\cdot f_{n-i}=k$[/quote]

Yes, exactly!  :)  This is the central point to the question, and gets rid of all that nasty business of dealing with powers and variables.

[hide]Let N be the number with factors $d_1,d_2,\cdots,d_{k-1},d_k$. Note that $d_i\times d_{k-i+1}=N$.
The sum of the reciprocals of the factors is
$\frac{1}{d_1}+\frac{1}{d_2}+\cdots+\frac{1}{d_{k-1}}+\frac{1}{d_k}$,
and we let our common denominator be N. So multiply $\frac{1}{d_i}\times\frac{d_{k-i+1}}{d_{k-i+1}}=\frac{d_{k-i+1}}{N}$,and we obtain
$\frac{d_k+d_{k-1}+\cdots+d_2+d_1}{N}$, as desired.[/hide]


trigonometry
geometry
trig identities
Law of Sines
geometry proposed

In acute triangle $ ABC$ points $ A_1,B_1,C_1$ are midpoints of the sides $ BC,AC,AB$ respectively. It is known, that
$ AA_1 \equal{} d(A_1,AB) \plus{} d(A_1,AC)$
$ BB_1 \equal{} d(B_1,AB) \plus{} d(A_1,BC)$
$ CC_1 \equal{} d(C_1,AC) \plus{} d(C_1,BC)$,
where $ d(X,YZ)$ is distance from point $ X$ to the line $ YZ$. Prove, that triangle $ ABC$ is equilateral.

According to the Law of Sines, the first relation is equivalent with $ m_{a} = \frac {a}{2}(\sin{B} + \sin{C}) = a\cos{\frac {A}{2}}\cos{\frac {B - C}{2}}$. On other hand, $ \cos{\frac {B - C}{2}} = \frac {b + c}{a} \cdot \sin{\frac {A}{2}}$, and thus the above equality rewrites as $ 2m_{a} = (b + c) \sin{A} = 2S \left(\frac {1}{b} + \frac {1}{c}\right) = h_{b} + h_{c}$. Hence, $ m_{a} + m_{b} + m_{c} = h_{a} + h_{b} + h_{c}$, and according to the fact that $ m_{a} \geq h_{a}$ (and the other analogues), we conclude that $ ABC$ is equilateral.


Number is called palindrome if it's digits can be reversed without changing the number (value).

$ (i)$ Given that a number is four-digit palindrome, what is the probability that the number is a multiple of 99?

$ (ii)$ Given that four-digit number is a multiple of 99, what is the probability that the number is palindrome?

Well there are 90 four digit palindromes and 90 four digit multiples of 99.  So I guess the probability of both is $ \frac{90}{9000}\cdot\frac{90}{9000}\equal{}\frac{1}{10000}$, yes?

$ {1001a\plus{}110b\over 99}\equal{}{91a\plus{}10b\over 9}\equal{}10a\plus{}b\plus{}{a\plus{}b\over 9}$

Hence we have $ 9$ possible palindromic multiples of $ 99$, from $ a\equal{}1$ to $ a\equal{}9$.

1. There are $ 9\cdot 10$ four-digit palindromes, so the probability is $ {9\over 9\cdot 10}\equal{}{1\over 10}$

2. There are $ 91$ four-digit multiples of $ 99$ (from $ 99\cdot 11$ to $ 99\cdot 101$), hence the probability is $ {9\over 91}$

JRav, you have two mistakes.  First, you tried to calculate the wrong thing.  (The question doesn't ask for the likelyhood that any 4-digit number is both divisible by 99 and a palindrome, it asks for the probability of one of these events, conditioned on the other.)  Second, your calculation wouldn't be valid unless you could demonstrate the independence of the events "being a 4-digit palindrome" and "being divisible by 99."


[quote="ekolost"]ok this was a practice question in a paper of mine.

a empty jar weighs 260g. to the nearest 10g.

A)write down the maximum weight of the jar
B) write down the minimum weight of the jar.

i been told its either 264g or 265g but i dont know which ones right or if theyre even right!

i will be gratefull if you can explain it.

i'll tell you interesting facts if you can help me to show my gratitude.

Thank You For Your Time.[/quote]
Each of these numbers will make 260 if rounded to nearest tenth.
${255, 256,257,258,259,260,261,262,263,264}$.
Clearly, $264$ is the largest in the set while $255$ is the smallest in the set.

[quote="buzzer11"]hm, i cant edit your post, akech, but its not TENTH its TENS.
im sure it was just a typo, but there's a big difference between .1 and 10.[/quote]

Thanks!!
It is due to my limited knowledge of the English language. It is my fourth language

haha, quadlingual?
wow, i wish i was that talented...  :lol:

thank you as promised heres your facts.

Horses can't vomit.

If you sneeze too hard, you can fracture a rib.
If you try to suppress a sneeze,
you can rupture a blood vessel
in your head or neck and die.
If you keep your eyes open by force,
they can pop out.

Wearing headphones for just an hour
will increase the bacteria in your ear by 700
times.

Mosquito repellents don't repel. They hide you. The spray blocks the mosquito's sensors so they don't know you're there.

Donkeys kill more people annually than plane crashes.

[quote="akech"]Those are very interesting facts!!!![/quote]
Wow, quadlingual. I'm sort of quadlingual, but not very fluently.

The fourth fact is false. If you try to keep your eyes open, you'll automatically close them; you can't fight it.

I don't think that 'quadlingual' is a real English word - at that point you become multilingual and a polyglot.

Yeah, well I'm not an English major, so I guess I can't argue with you.


calculus
integration
quadratics
number theory
greatest common divisor
induction
algebra proposed

$ m$ is an odd integer.sorry not to have mentioned :oops:

[quote="aadil"]if $ \alpha$ and$ \beta$ are the roots of $ x^2 \plus{} mx \minus{} 1 \equal{} 0$.and $ \lambda_n \equal{} \alpha^n \plus{} \beta^n$.then prove that:
1.$ \lambda_n$ is an integer.
2.$ gcd(\lambda_n,\lambda_{n \plus{} 1}) \equal{} 1$
$ m$ is an odd integer[/quote]

$ \lambda_n$ is the sequence :
$ \lambda_0\equal{}2$
$ \lambda_1\equal{}\minus{}m$
$ \lambda_{n\plus{}2}\equal{}\lambda_n\minus{}m\lambda_{n\plus{}1}$

From this, the fact that $ \lambda_n$ is an integer is an immediate consequence.
And since $ \gcd(\lambda_0,\lambda_1)\equal{}\gcd(2,\minus{}m)\equal{}1$ since $ m$ is odd, an immediate induction gives the second result.

[quote="aadil"]i also proved it using induction.
can it be done in some other way?[/quote]

For the first part, you can use the direct computation of $ \lambda_n$
For the second, I think that any direct solution would be rather ugly.

And for both, I wonder what could be the interest of any other solution since the one I gave is very short and simple and use only elementary knowledge (IMHO)


function
induction
algebra proposed
algebra

find all $ f: R \to R$ such that :
\[ f(x\plus{}y)\plus{}f(x\minus{}y)\equal{}2f(x)\plus{}2F(y)\]

$ y\equal{}0 \Longrightarrow f(0)\equal{}0$
$ x\equal{}y \Longrightarrow f(2x)\equal{}4f(x)$
By induction we can show that $ f(nx)\equal{}n^2f(x) (*)$
Replacing $ x\equal{}1$ in (*) we have $ f(n)\equal{}n^2f(1)$ for every $ n \in \mathbb{N}.$
$ x\equal{}\frac{m}{n}$ in (*) $ \Longrightarrow f(m)\equal{}m^2f(1)\equal{}n^2f(\frac{m}{n})$
Implying that $ f(\frac{m}{n})\equal{}(\frac{m}{n})^2f(1)$
So for all $ x\in \mathbb{Q^\plus{}}$ we have $ f(x)\equal{}cx^2$
Replacing $ x\equal{}0$ in the original equation we get that $ f(y)\equal{}f(\minus{}y)$
So for all $ x\in \mathbb{Q}$ we have $ f(x)\equal{}cx^2.$
For extension to reals we need monotonicity or continuity.


The Early-Bird enrollment deadline for [url=http://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php]Worldwide Online Olympiad Training[/url] is approaching!  The enrollment fee increases on July 15, so enroll soon to receive the discounted price.

If you have any questions about WOOT, read the Math Jam transcripts [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=93]here[/url] or [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=115]here[/url], and/or post your questions in this thread.

The link to last night's transcript isn't working for me. (Well, the page doesn't show the transcript?)


[quote="towersfreak2006"]1. How many integers $n$ in the interval $1\le n\le 1000000$ are such that the equation \[ n=x^y  \] has some solution in integers $x$ and $y$ with $x\neq n$?

10. Four distinct planes pass through a common point, but no three contain the same line. Find the number of disjoint regions into which three-dimensional space is partitioned by these four planes.[/quote]

[hide="1"]$y\leq 19$.
There is 1 for 19, 1 for 18, 1 for 17, 1 for 16, 1 for 15, 1 for 14, 1 for 13, 2 for 12, 2 for 11, 2 for 10, (3-1)=2 for 9, (4-1)=3 for 8, (6-1)=5 for 7, (8-2-1)=5 for 6, (14-2-1)=11 for 5, (30-3-2-1)=24 for 4, (98-5-2-2-1-1)=87 for 3, (998-24-5-3-2-2-1-1-1)=959 for 2, and since $x \neq n$, $y\neq 1$ and that's the end. Is the answer $1+1+1+1+1+1+1+2+2+2+2+3+5+5+11+24+87+959=\boxed{1109}$?[/hide]

[hide="10"]$16$?[/hide]

[hide="10"]Three planes can partition into 8 regions. The 4th plane passes through 6 of these regions. 14?[/hide]


If $ a,b,c \in [0,\infty)$ then prove the inequality:

$ a^3\plus{}b^3\plus{}c^3\plus{}2(ab^2\plus{}bc^2\plus{}ca^2) \ge 3(a^2 b\plus{}b^2c\plus{}c^2a).$

[quote="moldovan"]If $ a,b,c \in [0,\infty)$ then prove the inequality:

$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 2(ab^2 \plus{} bc^2 \plus{} ca^2) \ge 3(a^2 b \plus{} b^2c \plus{} c^2a).$[/quote]

Assume : $ c\equal{} min \{a;b;c\}$

$ a\equal{}c\plus{}x \ ; \ b\equal{}c\plus{}y \ ; \ x,y \geq 0$

It is equivalet to : $ \left( 2\,{y}^{2}\minus{}2\,xy\plus{}2\,{x}^{2} \right) c\plus{}2\,x{y}^{2}\minus{}3\,{x}^{2}y\plus{}
{y}^{3}\plus{}{x}^{3} \geq 0$

Done !


Does any body know how to change the location of the title of any chapter? I need to center it for my thesis and I have no idea how to do it! Also, is it possible to put the 'chapter 1 - 'title of the chapter' ' somehow???

Thanks

Hi,

Visit [url]http://zoonek.free.fr/LaTeX/LaTeX_samples_chapter/0.html[/url]
(French)

i hope that's will help you.

You can also use predefined chapter styles provided by the [url=http://www.tug.org/tex-archive/help/Catalogue/entries/fancyhdr.html]fancyhd[/url]package and the [url=http://www.tug.org/tex-archive/macros/latex/contrib/titlesec/titlesec.pdf]titlesec[/url] package.


Anyone heard of a Romanian guy called Slash`?

Well that is my nick on undernet :D So why are you looking for me ?

I randomly decided to post on the last unlocked thread on the fun factory.

Please, just please, don't EVER revive old threads. Thank you.

[size=89][color=#FF0000]Please, just please, don't EVER backseat moderate. Thank you. —PhireKaLk6781[/color][/size]

[quote="yrushi"]Wow, > 9 year necropost![/quote]

You mean a >9 year [i]revival.[/i]

[color=green]chao: Do not feed the trolls.[/color]


[quote=ALGEBRA]Let $a_1,a_2,...,a_n,b_1,b_2,...,b_n$ are $2n$ positive real numbers. Prove that
$$\prod_{1\leq i,j\leq n}{\min (a_i+a_j,b_i+b_j)} \leq \prod_{1\leq i,j\leq n}{\min (a_i+b_j,a_j+b_i)}.$$[/quote]
Anyone? iirc when I try this months ago, I can show that the first one implies the second.


calculus
derivative
limit
analytic geometry
graphing lines
slope
real analysis

Given 5 symmetric points ...

Here's something inspired by the contents of a numerical analysis course:

$\frac{-f(x+2h)+8f(x+h)-8f(x-h)+f(x-2h)}{12h}=f'(x)+O(h^5).$

To verify this, expand out the Taylor polynomials.

(Without bothering to consult a numerical analysis text, I derived this formula by Richardson extrapolation.)

I'm sorry but your formula is too advanced for me.  What exactly does it mean?

Look at it this way:

$P'(1992)=\lim_{h\to\0}\frac{P(1992+h)-P(1992)}h$

But you can't take that limit, because you only have finitely many values of for $P(t).$ All you can do is find particular values of that quotient.

$\frac{P(1994)-P(1992)}2$ is an approximation of $P'(1992),$ but it's biased - it's the average slope on one side of the point you're interested it.

$\frac{P(1992)-P(1990)}2$ is just as good an approximation, an since it comes from the other side, it's biased the other way.

So we can do better than either of the two values above by averaging them. This would give us the estimate

$P'(1992)\approx \frac{P(1994)-P(1990)}4.$

You can see that this is the average slope over an interval with 1992 in the middle.

We have another estimate: $P'(1992)\approx\frac{P(1996)-P(1988)}8.$

This isn't as good an estimate as the previous one, becuase it involves a larger $h.$ There should be a way to get an even better estimate by combining these last two values somehow - actually, by extrapolating. That's where the formula in my previous post came from.

Much better.  Thank for taking time to type all that work.


Dissect an equilateral triangle with side length 1 into n small triangles. Show that the sum of the squares of the 3 side lengths of all these small triangles >= 3.  :D

(Thanks grobber, I have changed lengths to side lengths)

Are the small triangles also equilateral? You sum the squares of all three sides of each triangle? What does "the length of a triangle" mean? :)

[Of course, if the answer to the first question is "yes", the rest becomes clear]

One more problem, which have come to my mind:
One has 2005 equilateral triangles with sides $1$, $\frac{1}{2}$, ..., $\frac{1}{2005}$. What is the smallest number $a$ s.t. it possible to place all small triangles into equilateral triangle with side $a$?
Source: [i]Russian TST, 1984[/i]

[quote]Are the small triangles also equilateral? You sum the squares of all three sides of each triangle? What does "the length of a triangle" mean?  [/quote]

No, grobber, the n small triangles could be in any form (as long as they are still triangles... :D ) And for each triangle, suppose they have side length a,b,c, let s=a^2+b^2+c^2. Now sum s over ALL n small triangles, (of course each triangle has its own s) and show that the sum (of s) >= 3.  :lol: (hope it's clearer now...)

(I have edited my first post to make the problem clearer)

:D  :D  :D
No problems!
We just need to mention that $a^2+b^2+c^2\geq 4\sqrt 3\cdot S$.
So, in fact, we have $\sum\geq 4\sqrt{3}\cdot \frac{\sqrt{3}}{4}=3$ !!!

Yeah, you got it Myth, sorry for my computational error.... :blush:


Let g(x)= x^2 - 3x

Simplify completely:  (g(2+h) - g(2-h)) / (2h)

kind of confused. :maybe:


algebra
polynomial
quadratics
superior algebra
superior algebra unsolved

Let $\mathcal K$ be a field with $2^{n}$ elements, $n \in \mathbb N^\ast$, and $f$ be the polynomial $X^{4}+X+1$. Prove that:
(a) if $n$ is even, then $f$ is reducible in $\mathcal K[X]$;
(b) if $n$ is odd, then $f$ is irreducible in $\mathcal K[X]$.

[hide="Remark."]I saw the official solution and it wasn't that difficult, but I just couldn't solve this bloody problem.[/hide]

Nobody? :( I can not spoil the fun since I have also read the official solution...As perfect_radio said, the problem is not that hard but  I was not able to find a solution in the contest either :(

Is this problem so "bloody" that nobody is interested? :maybe:  I think not...
Hint: (2^n-1,3)=?

Don't know what the hint has to do with the problem...

The polynomial is irreducible over $K=\mathbb F_{2}$ (easy checking). Let $\alpha$ be a root of the polynomial. $[K[\alpha]: K]=4$, thus if the polynomial factors in some finite extension $L$ of $K$, then $[L: K]$ has to be even (a linear factor implies $L$ to contain $K[\alpha]$, two quadratic factors give $[L[\alpha]: L]=2$ and again the result by multiplicity of degrees).
Thus for $n$ odd, it is irreducible.
Conversely, it factors over any field containing $\mathbb F_{4}$, which happens for $n$ even.

If you make a few computations, then you'll see that finding a third root of unity is the key.

PS: I was pretty sure a solution using the $[: ]$ notation existed, but I bearly know what it means...

I found the third root of unity in the contest but couldn't solve it even afterwards...how sick is that ?!?!...i think I understood ZetaX's solution, but it would have never crossed my mind ....nice one....I would however appreciate something more..."elementary" :blush: