A plank of weight $ W$ and length $ \sqrt{3} R$ lies in a smooth circular trough of radius $ R$. At one end of the plank is a weight $ W/2$. Calculate the angle $ \theta$ at which the plank lies when it is in equilibrium.

This is a problem about static equilibrium of a rigid body. What are the conditions for the equilibrium of such a body?

1) Translational equilibrium: $ \sum_iF_i \equal{} 0$.

2) Rotational equilibrium: $ \sum_iM(F_i) \equal{} 0$.

If the weight can be "reduced" to the extremity...

[hide=" click to see the answer"] \[ \theta \equal{} {\pi \over 6}\][/hide]

Yes, that's it! Can you post the solution? :)

Cyprusad, wouldn't it be more usefull if you find the solution by yourself?

Begin by answering the following question: what are the forces applied to the plank and their direction?

Sure it would; I've tried it a couple of times without much success. There are two normal forces at the point of contact of the plank and the trough, their directions are arbitrary (don't pass through the centre), but we can resolve them into $ x$ and $ y$ components. I've calculated the torques of the weights and the components of the normal forces at the two ends of the plank about the centre of mass of the system (the plank plus the body). The vertical components of both normal forces balance out the weights of the plank and the body, and the horizontal components of the normal forces cancel out each other.

It looks almost correct but I don't get $ \theta$. Perhaps someone can point out what I'm missing out on?

EDIT
Here's what I did, $ N_1$ is the normal on the upper end of the plank and $ N_2$ is the normal at the lower end.
Balancing forces in vertical and horizontal directions..
$ N_{1y} \plus{} N_{2y} \equal{} W \plus{} W/2$
$ N_{1x} \equal{} N_{2x}$
I took the torques about centre of mass of the system which is $ R/ 2 \sqrt{3}$ from the centre of the plank.
Net torque about CM is zero..
$ W( \frac {R} {2 \sqrt{3}} )\sin ( {\frac {\pi} {2} \minus{} \theta} ) \plus{} N_{2y} ( \frac {R} {\sqrt{3}} ) \sin ( {\frac{\pi}{2} \plus{} \theta} ) \equal{} \frac{W} {2} (\frac{R} {\sqrt{3}} ) \sin ({\frac{\pi} {2} \minus{} \theta}) \plus{} N_{2x}( \frac{R} {\sqrt{3}} ) \sin( {\pi \minus{} \theta} )  \plus{} N_{1y} (\frac{2R} {\sqrt{3}} ) \sin({\frac{\pi} {2} \plus{} \theta}) \plus{} N_{1x} (\frac{2R} {\sqrt{3}} ) \sin{\theta}$

Solving which I got, $ \tan{\theta} \equal{} \frac {N_{2x} \plus{} 2N_{1x}} {N_{2y} \minus{} 2N{1y} }$ ; pretty much a dead end.

I think I'm doing something wrong over and over again..
(I hope its not a silly mistake  :blush: )

[quote="Cyprusad"]There are two normal forces at the point of contact of the plank and the trough, their directions are arbitrary (don't pass through the centre)[/quote]

Their directions are not arbitrary. When two surfaces are in contact and one of them is reduced to a point, then the normal force has the direction of the perpendicular to the other surface.

[quote="Cyprusad"]I've calculated the torques of the weights and the components of the normal forces at the two ends of the plank about the centre of mass of the system (the plank plus the body).[/quote]

When calculating torques, we can choose an arbitrary point for that calculation: so, it's reasonable to choose a point about which the biggest number of torques is zero. In this case, choosing the point of the planck where the mass W/2 lies enables us to compute just two torques.

[quote="Cyprusad"]The vertical components of both normal forces balance out the weights of the plank and the body,[/quote]

What is the direction of the force exerted by the mass W/2 on the plank?

[quote="Carcul"]Their directions are not arbitrary. When two surfaces are in contact and one of them is reduced to a point, then the normal force has the direction of the perpendicular to the other surface.[/quote]

Ofcourse that's true but I was talking about the normals at the point of contact of the - point mass, plank and trough - at one end and the - plank and the trough - at the other end. I guess I wasn't clear earlier

[quote="Carcul"] When calculating torques, we can choose an arbitrary point for that calculation: so, it's reasonable to choose a point about which the biggest number of torques is zero. In this case, choosing the point of the planck where the mass W/2 lies enables us to compute just two torques.[/quote]

I thought of that, but then I couldn't figure out the direction of one of the two torques that remain to be calculated, viz. the normal force torque. So that's why I thought of taking the torque about the centre of mass, which was safe albeit wrong and not to mention cumbersome!

[quote="Carcul"]What is the direction of the force exerted by the mass W/2 on the plank?[/quote]

Going by what I've posted in the previous edited post, I took it to be vertically downward. Though all that I posted earlier seems a little incomprehensible, now that I look at it :P.. But, I still haven't got the idea!

Could you elaborate a little more Carcul? I guess I 'm on the verge of sounding silly, thanks for bearing with me..   :rotfl:

[quote="Cyprusad"]I took it to be vertically downward.[/quote]

No, that's not correct. The force exerted by the plank in the body is perpendicular to both bodies. So, what is the direction of the force exerted by the body on the plank?

Its perpendicular to the plank then isn't it?


geometry
rectangle
calculus
calculus computations

rectangle has fixed area eual to 100 units^2 , length L is increasing at 2un/sec. Find L when the width is decreasing at .5 un/sec..................
i know how to do related rates problems, for this u could use the rectangle area formula or a triangle bc rectangle consists of two triangles......
but i dont understand how can u do this, if u dont have any L and W measurements, all u r given is the dW/dt and dLdt???? :?:

$lw=100$
$l'*w+w'*l=0$
$2w-\frac{1}{2}l=0$

use the 1st and 3rd lines to solve for them...

to explain why this would give one answer...we have that l is changing 2 u/s right? so since the area is constant, the change of the width will vary (do you see why?), but there is only one specific moment that we will have the dw/dt being -0.5, so we have to find the dimensions at that moment


superior algebra
superior algebra unsolved

Consider the following commutative diagram

[img]http://upload.wikimedia.org/wikipedia/commons/6/64/FiveLemma.png[/img]

[b]1.[/b] Give examples where $ n$ is not a monomorphism and

(i) $ m$ and $ p$ are monomorphisms;
(ii) $ m$ is a monomorphism and $ l$ is an epimorphism;
(iii) $ p$ is a monomorphism and $ l$ is an epimorphism;

[b]2.[/b] Give examples where $ n$ is not a epimorphism and

(i) $ m$ and $ p$ are epimorphisms;
(ii) $ m$ is an epimorphism and $ q$ is a monomorphism;
(iii) $ p$ is an epimorphism and $ q$ is a monomorphism;

Danken!

are they so hard to find?  I can't even find one

[quote="ZetaX"]For 1. (i) take $ m,p$ as isomorphisms and choose the rest accordingly; you may want to take vactor spaces for simplicity (then you can take $ C,C^\prime$ as anything with high enough dimensions).
All others can be done similarily.[/quote]

I don't know many examples of exact sequences,so I have tried to find examples by taking short exact sequences for the bottom and top sequences but I cannot obtain any examples which fulfills what I want.   In fact, any sequences that I think and makes the diagram commutative doesn't help me because gives me that the middle map is a monomorphism.

I'll appreciate if you can give me an explicit example and then I'll try to find the rest by myself.

Thank you very much in advance

P.S.: this problem is making me crazy  :mad:

Must the top and bottom sequences be exact?  That wasn't specified anywhere.


For a,b,c,d positive real numbers prove that

  39a^3 +15a(b^2+c^2) + 20ad^2 +5bc(b+c+d) \geq 10a^2(b+c) +43a^2d + 39abc +ad(b+c)

hope you won't mind me saying that this is a really ugly inequality manlio :)

:D  :D  That's exactly what i was thinking...

Pierre.

hmm.. I don`t know what do you mean by saying: "ugly inequality",
but I think that a problem is ugly when the solution is ugly..   :play_ball:


b)Solve for all possible values(angles) of x:

1-sin2x-1= cos2x

I can't seem to find an answer for this one!
:)thanks
<3<3

[quote="erma12879"]b)Solve for all possible values of x:

1-sin2x-1= cos2x

I can't seem to find an answer for this one!
:)thanks[/quote]

[hide]The equation simplifies to
\[\cos 2x+\sin 2x = 0. \]
Squaring both sides,
\[\cos^{2}2x+\sin^{2}2x+2 \sin 2x \cos 2x = 1+2 \sin 2x \cos 2x = 0 \]
which gives
\[\sin 4x =-1. \]
The solution is
\[4x = \frac{3 \pi }{2}+2n \pi \]
or
\[x = \boxed{\frac{(3+4n) \pi}{8}}\]
where $n \in \mathbb{Z}$.
[/hide]
Shouldn't this go to High School Basics?

How does $1-sin \left(2x-1\right)$ equal $-sin 2x$ ?
Unless is was meant to be $1-sin \left(2x\right)-1=cos 2x$.

How the problem is written makes me think it is $\sin 2x-1$. :(
And if it is $\sin(2x-1)$, is it in radians or degrees?

no, the question is to solve for all possible angles of sin2x - 1 = cos2x. I know you're supposed to use identities to factor it, but i'm not sure exactly how

So it didn't have 1- at the front. :(

[hide]The equation simplifies to
\[\sin 2x-\cos 2x =1. \]
Squaring both sides,
\[\sin^{2}2x+\cos^{2}2x-2 \sin 2x \cos 2x = 1-\sin 4x = 1 \]
which gives
\[\sin 4x =0. \]
The solution is
\[4x = n \pi \]
or
\[x = \boxed{ \frac{n \pi }{4}}\]
where $n \in \mathbb{Z}$.
[/hide]

[quote]The equation simplifies to $\sin 2x-\cos 2x =1.$ Squaring both sides, $\sin^{2}2x+\cos^{2}2x-2 \sin 2x \cos 2x =$ $1-\sin 4x = 1$ which gives $\sin 4x =0.$ The solution is $4x = n \pi$ or $x = \boxed{ \frac{n \pi }{4}}$ where $n \in \mathbb{Z}$[/quote]
This is wrong : for $n=-1$, $x=-\frac{\pi}{4}$ is no solution !!!!

[quote="Virgil Nicula"][quote="kohjhsd"]So it didn't have 1- at the front. :( [hide]The equation simplifies to $\sin 2x-\cos 2x =1.$
Squaring both sides, $\sin^{2}2x+\cos^{2}2x-2 \sin 2x \cos 2x =$ $1-\sin 4x = 1$ which gives $\sin 4x =0.$
The solution is $4x = n \pi$ or $x = \boxed{ \frac{n \pi }{4}}$ where $n \in \mathbb{Z}$[/hide][/quote]
This is wrong : for $n=-1$, $x=-\frac{\pi}{4}$ is no solution !!!![/quote]
Oops
The extraneous roots should've appeared as I squared them :blush:
If I don't square, then this:
[hide]The equation simplifies to
\[\sin 2x-\cos 2x =1. \]
which is
\[2\sin x \cos x-\cos^{2}x+\sin^{2}x-1=0 \]
or
$2 \sin x \cos x-2 \cos^{2}x = 0$
$\implies 2 \cos x ( \sin x-\cos x ) = 0$
Case 1) $\cos x = 0 \implies x=\frac{\pi}{2}+n \pi$
Case 2) $\sin x = \cos x \implies x = \frac{\pi}{4}+n \pi$
where $n \in \mathbb{Z}$.
[/hide][/quote]


A five digit number, when multiplied by a single digit, gives the result that is the first number in reverse.

What are these numbers?

The single digit is not $ 1$.

[hide="answer"]it's 10989 and 9,
I don't know how to solve this by manipulation
I did it mainly by trial and error
it's easy to see the single digit is 9 because when you consider 2,3,4, you have to keep the last digit of the calculated number very small so that when you put that number back into the original one, it wouldn't carry it to a six digit no, so it leaves just a few choices,
when you consider 5 or above, you need the last digit of the calculated number to be exactly 1.
than you actually know it's 1 _ _ _ 9, the second is either 1 or 0. when you choose 1, it is quickly found that 1 1 _ 9 9 is impossible, and so it would be 1 0 _ _ 9, than the second last digit has to be 8 to make that 0 happen, and to have that 8,
the carrying requires a 9 in the middle.[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose $G$ is a group that if $G\neq H\leq G$ then $H$ is cyclic. What can we say about $G$?
I've proved if $G$ is finite then $G$ is generated with 2 objects (i.e there exist $a,b\in G$ that $G=(a,b)$), and a counterexample for the case $G$is infinite. (The proof is trial)
I want to know more about $G$ that is finite and not cyclic with the previous conditions. I guess if $ab=ba$ then $G$ is isomorph to $\mathbb Z_p\times\mathbb Z_p$

Yes, by using the classification theorem for abelian finite groups your assumption is correct

Let G now be a finite group, nonabelian, with that property

Well , there is this theorem of Sylow : if a prime power divides the order of G, then there is a subgroup with that prime power as order
a well known theorem is that all groups of prime order are cyclic

thus : if the order of G is squarefree (no primes occur twice) it definitely only has cyclic subgroups
nonabelian examples are S3, the direct product of  C7 and C3, and the group of isometries on a regular 7-gon

for the rest, i  am stuck too now


Three listening stations located at $(3300,0)$, $(3300,1100)$, and $(-3300,0)$ monitor an explosion. The last two stations detect the explosion $1$ second and $4$ seconds after the first, respectively. Determine the coordinates of the explosion. Assume that the coordinate system is measured in feet and that sound travels at $1100$ feet per second.

[hide]We apply a new coordinate system, which measures in seconds. So

$A = (3,0)$
$B = (3,1)$
$C = (-3,0)$

Let $E=(x,y)$ be the coordinates of the explosion.

The distance $AB$ is exactly 1 second, so $x=3$ and $y<0$
(because the explosion takes place on the line AB and E is closer to A)

$AE^{2}= y^{2}$
$CE^{2}= (3+3)^{2}+y^{2}$

$\Rightarrow CE^{2}-AE^{2}= 36$

We know that $CE-AE=4$ seconds, so $CE+AE=9$
From the last two equations we get $AE=2.5\Rightarrow \ E(3,\-2.5)$

Now translate back to feet, and we find that  the explosion was placed on the point $(3300,\-2750)$[/hide]


Ben flips a fair nickel four times. What is the probability that Ben flips Head, Tail, Head, Tail, in that order? Express your answer as a common fraction.

Each suuccesive flip has a $ \frac12$ probability, so $ (\frac12)^4 \equal{} \boxed{\frac {1}{16}}$.

Ok. For every flip, you have a 1/2 chance to get a head or a tail. So, head, tail, head, tail, is (1/2)^4. This equals [u][b]1/16[/b][/u].


Can anybody tell me this answer???
P.S please

In any Inorganic Chemistry book you should find the answer.

[quote="valeriummaximum"]I honestly don know the answer for this

C60-buckyballs is cubic close packed structure in its crystalline form.If u insert K atoms into all the tetrahedral and Octahedral holes of the structure,the formula becomes KxC60.

What is value of x?????????
density Of C60=3g/ml

Also find the density of KxC60.[/quote]

CCP = FCC. We have 60 Octahedral voids and 120 Tetrahedral Voids. Does that make your answer any simpler???


analytic geometry
inequalities
trigonometry
geometry proposed
geometry

[quote="A6, Putnam, 1998"][color=darkred]Let $ A$, $ B$, $ C$ be three points with the integer coordinates in $ \mathbb{R}^{2}$. Prove that if

$ (AB \plus{} BC)^{2} < 1 \plus{} 8\cdot [ABC]$ , then  $ A$, $ B$, $ C$ are the vertices of a square.[/color][/quote]

[quote="Virgil Nicula"][color=darkred][b][u]Extension.[/u][/b] Let $ A$, $ B$, $ C$ be three points with the integer coordinates in $ \mathbb{R}^{2}$. Prove that

$ (m\cdot BC \plus{} n\cdot AB)^{2} < 1 \plus{} 8mn\cdot [ABC]\ \Longrightarrow BA\perp BC$ and $ \tan C \equal{} \frac mn\ .$[/color][/quote]
See http://mateforum.ro/viewtopic.php?p=580#580

Since $ A$, $ B$ and $ C$ have integer coordinates we have that $ AB^{2}$, $ BC^{2}$ and $ 2[ABC]$ are all integers so $ AB^{2} + BC^{2}$ and $ 1 + 8[ABC]$ are integers too.
Then inequality $ (AB+BC)^{2}< 1 + 8[ABC]$ implies that:
 $ (AB+BC)^{2}\leq (1 + 8[ABC]) - 1 = 8[ABC] = 4AB.BC.\sin(\angle ABC)\leq$ $ 4AB.BC.1 = 4AB.BC$

On the other hand from $ AM%Error. "geqGM" is a bad command.
$ it follows that: $ (AB + BC)^{2}\geq 4AB.BC$.

Hence: $ 4AB.BC\leq(AB + BC)^{2}\leq 4AB.BC$ so equality must be reached in:

1.	$ (AB + BC)^{2}\geq 4AB.BC$. so $ AB = BC$

2.	$ \sin(\angle ABC) = 1$ so $ \angle ABC = 90$

From 1. and 2. it is obvious that triangle $ ABC$ is right isosceles (right angle at $ B$) so $ A$, $ B$ and $ C$ are vertices of square.


function
real analysis
integration
calculus
Support
topology
real analysis unsolved

Suppose $ f$ is Lebesgue integrable over a set $ E$. I want to show that given $ \varepsilon>0$, there is a step function $ \psi$ such that $ \int_{E}|f\minus{}\psi|<\varepsilon$.
Prior to that, I have shown that (1) there is a simple function $ \varphi$ such that $ \int_{E}|f\minus{}\varphi|<\varepsilon$ and that (2) for any measurable function $ g$ on an interval $ [a,b]$ taking finite values a.e., there is a step function $ h$ such that $ |g\minus{}h|<\varepsilon$.
Thus I see that if I can get a step function $ \psi$ such that $ \int_{E}|\varphi\minus{}\psi|<\varepsilon$, then I'm done. Is there a way I can use (2) to achieve this (maybe by taking $ g$ to be $ \varphi$)?

A simple function is a finite linear combination of characteristic functions of measurable sets.

Can we approximate (in integral) the characteristic function of a measurable set with a step function? We can if the following theorem is true:

Theorem: $ E$ is a measurable subset of $ \mathbb{R}^n$ if and only if for every $ \epsilon>0,$ we can find a set $ S$ which is a finite union of nonoverlapping intervals such that $ E\equal{}(S\cup N_1)\setminus N_2$ where $ N_1$ and $ N_2$ are each sets with outer measure less than $ \epsilon.$

Have you ever seen a theorem like that?

One difficulty to be overcome in the way you're trying to do things is that you don't know that the measure of $ E$ is finite. If it's not, $ |g\minus{}h|<\epsilon$ may not imply that $ \int_E|g\minus{}h|$ is small.

---

The argument would continue. Once you get that you can approximate integrable functions by step functions, then can you approximate (in the same sense) step functions by continuous functions with compact support? Once you have that, can you approximate integrable functions by $ C^{\infty}$ functions of compact support?

What if I first approximate $ E$ by an open set, which is then a countable union of disjoint open intervals? Then I approximate the simple function by step functions in each of these intervals.

Alternatively, is it possible to restrict to a closed interval, say $ [\minus{}n,n]$, by showing that the integral will be small outside the interval and then approximating the function in that interval?

[quote="chungyc"]What if I first approximate $ E$ by an open set, which is then a countable union of disjoint open intervals? Then I approximate the simple function by step functions in each of these intervals.[/quote]
That will work. But just remember the the countable union of disjoint open intervals is an intermediate step, not the final product. We need to get to a finite union of intervals.

[quote]Alternatively, is it possible to restrict to a closed interval, say $ [ \minus{} n,n]$, by showing that the integral will be small outside the interval and then approximating the function in that interval?[/quote]
While true, that only accomplishes a small part of the task. That leaves you with a bounded measurable set, and you still have to approximate that with a finite union of intervals.

[quote="chungyc"]
Prior to that, I have shown that (1) there is a simple function $ \varphi$ such that $ \int_{E}|f \minus{} \varphi| < \varepsilon$ [/quote]
Now the simple function can be approximated as the sums of characteristic functions over open sets (not necessarily disjoint). Any such function can easily be approximated with continuous functions, and continuous functions can be approximated with step functions.

This is the simplest approach imo. The theorem that Kent suggested is so ugly. I dislike that kind of stuff. Makes me uneasy and irritated.


A rectangular floor measures 3.5 meters by 4.5 meters. If 1 inch = 2.54 cm, what is the area of the floor, in square inches? Express your answer as a decimal to the nearest tenth.

Find the new dimensions in inches and multiply for the area: 2.4.

No offense, but I did not think this was a good problem. There was no trick to it, just brute force.


Let $ x,$ $ y$ and $ z$ are non-negative numbers. Prove that:
\[ \sqrt[3]{x^3\plus{}7y^3}\plus{}\sqrt[3]{y^3\plus{}7z^3}\plus{}\sqrt[3]{z^3\plus{}7x^3}\geq2(x\plus{}y\plus{}z)\]

We have $ \sum\sqrt[3]{\frac{x^3\plus{}7y^3}{8}}\ge\sum\frac{x\plus{}7y}{8}\equal{}\sum x$

$ \implies \sum\sqrt[3]{x^3\plus{}7y^3}\ge 2(x\plus{}y\plus{}z)$.

Cool, nayel!
Try it:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=174360


i saw in multiple places that force at impact is $ F\equal{}m*a$ or mass times acceleration
why isnt it $ m*v$ mass times velocity?

Because that is the definition of force.  $ mv\equal{}p$ is a different quantity called momentum and as a matter of fact, $ \frac{dp}{dt}\equal{}\frac{d}{dt}(mv)\equal{}\frac{dm}{dt}v\plus{}m\frac{dv}{dt}\equal{}0\plus{}m\frac{dv}{dt}\equal{}ma\equal{}\vec{F}$.

^^Not quite.  What Newton said was that the change in momentum with respect to time is equal to force, which was expressed as I did above.  This was then re-expressed as $ F\equal{}ma$, which is the second law, not the third.

[quote="JRav"]^^Not quite.  What Newton said was that the change in momentum with respect to time is equal to force, which was expressed as I did above.  This was then re-expressed as $ F \equal{} ma$, which is the second law, not the third.[/quote]

Gah. I even knew it was the 2nd law. I was just thinking about not typing that. Damn my brain!!

[quote]give me 2 more yrs.--and i WILL CERTAINLY proof NEWTON"S 3rd law WRONG   [/quote]
Good luck with that.  It is not possible to prove or derive any of Newton's laws since they are fundamental.  Their acceptance is due to experimentation and verification.  Similarly, we can't derive the Schrodinger Equation.

i heard that schrodinger eq. cud b verified through quantum observation----.... :?:

[quote="rituraj007"][quote="SimonM"]Essentially, because that's what Newton said in his third law, and is what has been verified to an extend that physicists believe it.[/quote]
give me 2 more yrs.--and i WILL CERTAINLY proof NEWTON"S 3rd law WRONG :mad:  :mad:  :mad:[/quote]

I hope you are joking.

[hide]because I have already proved them wrong[/hide]

Newton's 2nd law is supposed to be the main law because using it, you can prove the other laws or atleast understand them. ;)

frankly speaking--i ve always felt a speck of uncertainty in Newton's 3rd law
~~~i m a strict believer of quantum physics :blush:  :D

Schrodinger's Equation can be OBSERVED, not derived.  I don't know why you're doubting Newton's third law.  It's been verified for non-relativistic speeds countless times.  If you're a strict believer of quantum mechanics, then maybe you could enlighten today's particle physicists because they won't admit to fully understanding it.

i never said i do understand it fully--i just know few basics & that's the window 4 further dev.

Well, since science isn't really about belief you might want to look a bit harder at the subject.

On the topic of QT, did anyone else read that article in NS about Quantum Certainty?


geometry
vector
trigonometry
calculus
calculus computations

A disk of radius 1 cm. has a small hole at a point half way between the center and the circumference. The disk is lying inside a circle of radius 2 cm. A pen is put through the hole in the disk, and then the disk is moved once
round the inside of the circle, keeping the disk in contact with the circle without slipping, so the pen draws a curve. What is the area enclosed by the curve

The center of the small disk $ D_{1}$ is moving along a circle of radius $ 1$. Since the circumference of $ D_{1}$ is $ 1/2$ of the circumference of the greater circle ($ D_{0}$), it follows that each revolution of $ D_{1}$ about the center of $ D_{0}$ corresponds to three revolutions of $ D_{1}$ about its own center. (Three instead of two? Why?). Therefore, the vector equation of the curve is $ \vec r(t)=\langle \cos t,\sin t\rangle+\frac{1}{2}\langle\cos 3t,\sin 3t\rangle$.

Caveat emptor.


geometry
3D geometry
inequalities
inequalities unsolved

Find the smallest real number $ x$ such that for arbitrary real numbers $ a, b, c\geq x$ and $ a\plus{}b\plus{}c \equal{} 3$ we have $ a^{3}\plus{}b^{3}\plus{}c^{3}\geq 3$

[color=darkred]Edit by Megus: current Polish MO![/color]

[quote="TomciO"]It's from Polish MO, please delete it![/quote]
I'm very sorry. I found it unsolved in another forum. I had no idea about that.

[quote="TomciO"]Could you inform the people on the mentioned forum? Or just give the link, thanks.[/quote]
I already informed them.


Are there integers $M, N, K$, such that $M+N=K$ and
(I) each of them contains each of the seven digits $1,2,...,6,7$
(II) each of them contains each of the nine digits $1,2,...,8,9$
P.S. Merry Christmas, guys! :)

Is it just the integers from one to nine, or does it just have to contain them?

If its the first one, for the first part, get mod 9 on its ass!

One example: $M= 123456789, N = 321987654, K = 987654321$(Not correct, but you can see what i mean)


geometry
trigonometry
inequalities
angle bisector
exterior angle
geometry unsolved

[quote="sumana saha"]AD the bisector of ang BAC intersects BC at D. BE is perpendicular drawn from B to AC at E, E,D are joined. Ahow that ang CAD > 45 deg[/quote]
I think that the problem statement is wrong. But, I have a result related to the problem statement.So,
[quote="The problem should be"]
In an acute angled triangle $ ABC$, let $ AD$ be the bisector of $ \angle BAC$ which intersects $ BC$ at $ D$. $ BE$ is perpendicular drawn from $ B$ to $ AC$ at $ E$, $ E,D$ are joined. Show that $ \angle CED > \frac {\pi}{4}$
[/quote]As in the case of obtuse angle we get, the opposite result.
Let $ D'$ be the bisector of $ \angle BEC$
Now, it is enough to prove that, $ BD < BD'$
After some calculation we get,
\[ BD < BD' \iff \frac {BD}{BC} < \frac {BD'}{BC} \iff \frac {BD}{DC} < \frac {BD'}{D'C}
\]
Now, by angle bisector theorem,
\[ \frac {BD}{DC} < \frac {BD'}{DC} \iff \frac {BA}{AC} < \frac {BE}{EC} \iff \frac {c}{b} < \tan C \iff c \cos C < b \sin C
\]
Now, from the sin law, $ b \sin C \equal{} c \sin B$. So, we get,
\[ \sin B > \cos C \iff \sin^2 B \minus{} \cos^2 C > 0 \iff \minus{} \cos (B \plus{} C)\cos(B \minus{} C) > 0
\]
In an obtuse angled triangle $ B \plus{} C > \frac\pi2 \Longrightarrow \cos (B \plus{} C) < 0$. And the inequality follows. :)

Here I used a very simple identity: $ \cos (B \plus{} C)\cos(B \minus{} C) \equal{} \cos^2C \minus{} \sin^2 B$
It can be proved easilty by expanding the left hand side.

this is CRMO 07, already solved.
So u could have posted it in the propsed problems section rite?

very sorry for wrong presetation and wrong placement of my problem.

I like to present the following proposed solution of the same problem given by someone to know whether the proof is sufficient
or not

Let the perpendicular BE intersects AD at O.
An intercept EF equal in length of EO is cut from EA. F,O are joined and produced to intersect the produced ED at G.

The triangle EFO is an rt angle isosceles triangle where EO = EF and angle FEO = 90 deg
Hence ang EFG = ang EFO = 45 deg
So the exterior angle of the triangle EFG i.e. angle CEG = angle EFG + angle FGE.
i.e. angle CED is greater than 45 deg.


geometry
analytic geometry
logarithms
limit
Euler
calculus
calculus computations

How good an estimate of 1/1 + 1/2 + 1/3 + 1/4 + ... + 1/n is ln n?

[url=http://en.wikipedia.org/wiki/Taylor%27s_theorem]Taylor's theorem[/url] allows you to estimate the remainder in a Taylor series expansion.

But this isn't a Taylor series expansion, so what JRav said is pretty much useless.

Visualize $ 1 \plus{} \frac12 \plus{} \frac13 \plus{} \cdots \plus{} \frac1n$ as the areas of a set of boxes. On the coordinate plane, draw the box whose base is on the $ x$ axis from $ k$ to $ k \plus{} 1$ and make it of height $ k,$ so that its upper left corner is $ \left(k,\frac1k\right).$

Now draw the curve $ y \equal{} \frac1x$ for $ 1\le x\le n.$ That passes through the upper left corners of all of the boxes, but stops at $ n$ whereas the last box continues to $ n \plus{} 1.$

The area under that curve is $ \ln n.$ Hence:

$ 1 \plus{} \frac12 \plus{} \frac13 \plus{} \cdots \plus{} \frac1n \minus{} \ln n$ is the area that is inside the boxes but either above the curve or in the last box. In your picture, shade in that area.

Now imagine pushing on the shaded area from the right with a ruler held vertically. All of the shaded pieces will neatly stack on top of each other until they all fit, without overlap into the first box. We have proved that:

$ 0 < 1 \plus{} \frac12 \plus{} \frac13 \plus{} \cdots \plus{} \frac1n \minus{} \ln n < 1.$

Furthermore, we can see that the sequence $ 1 \plus{} \frac12 \plus{} \frac13 \plus{} \cdots \plus{} \frac1n \minus{} \ln n$ is decreasing, since with each new step of the process, we cut into the solid last block and leave some of it out. Since the sequence is bounded and decreasing, it must have a limit. On the other hand, since each of the wedges that makes up the shaded area is bigger than a triangle, we have that the shaded area is $ > \frac12.$ In summary, we have the following:

Define $ \gamma \equal{} \lim_{n\to\infty}\left(1 \plus{} \frac12 \plus{} \frac13 \plus{} \cdots \plus{} \frac1n \minus{} \ln n\right).$

The limit exists and $ \frac12 < \gamma < 1.$

This particular limit is quite well known, $ \gamma$ is its conventional name, Euler is associated to it, and you can find approximations for $ \gamma$ in many places.

[quote]$ 1 \plus{} \frac12 \plus{} \frac13 \plus{} \cdots \plus{} \frac1n \minus{} \ln n$ is the area that is [b]inside[/b] the boxes but either above the curve or in the last box. In your picture, shade in that area. [/quote]

Should "inside" be replaced by "outside" ?

I said exactly what I meant. Note that my first and biggest box is $ [1,2]\times [0,1].$ I'm not drawing anything to the left of $ x\equal{}1.$ The curve lies below the top edge of the boxes. Draw the picture for yourself, and draw it carefully.


quadratics
number theory unsolved
number theory

Solve $x^2-92y^2=1$ in positive integers.

Use Dario Alpern's [url=http://www.alpertron.com.ar/QUAD.HTM]Quadratic two integer variable equation solver[/url]. It can give you even step-by-step solution.

all the result can give by
$(1151+120\sqrt92)^n=x_n+y_n\sqrt92$,$(x_n,y_n)\in N^2$
then $(x_n,y_n)$ are all the result.

Proof: standard Pell Equation.
To show that the given one is minimal, don't just test, but use the description of $\sqrt{92}$ as continued fraction.


algebra
polynomial
function
real analysis
real analysis unsolved

Is there a polynomial $ p$ of two variables such that the image of the polynomial $ p(\mathbb{R},\mathbb{R})$ equals to the open interval $ (0, \infty )$?

How about the old example $ (xy\minus{}1)^2\plus{}y^2$?

I had to convince myself that mornik's example works. We can rewrite that as $ p(x,y)\equal{}(x^2\plus{}1)y^2\minus{}2xy\plus{}1.$

We then look along the curve $ y\equal{}\frac{x}{x^2\plus{}1},$ which is where the minimum of $ p(x,y)$ considered as a function of $ y$ for fixed $ x$ lies.

$ p\left(x,\frac{x}{x^2\plus{}1}\right)\equal{}\frac{1}{x^2\plus{}1}.$

So while $ p(x,y)$ is never equal to zero, we do have that the infinum of its range is zero.

(I was trying to prove that this range was impossible, but I couldn't get my proof to work - with good reason, it would seem.)

Much simpler. kent. First it's clear the values are $ \geq 0$, but $ 0$ cannot be reached, since for $ y\equal{}0$ we have $ (xy\minus{}1)^2 \equal{} 1$. However, for $ y\equal{}\varepsilon$ and $ x\equal{}\frac {1} {\varepsilon}$ the value is $ \varepsilon^2$, so $ 0$ is approached at wish. All combined with continuity yields the range to be $ \mathbb{R}_\plus{}^*$.


Find the maximal $ \alpha,$ for which the following inequality
\[ a^3b \plus{} b^3c \plus{} c^3a \plus{} \alpha abc\leq27
\]
holds for all non-negative $ a,$ $ b$ and $ c$ such that $ a \plus{} b \plus{} c \equal{} 4.$

Nice inequality ;) Just let $ a\equal{}b\equal{}c\equal{}\dfrac{4}{3}$ to find the best value. However, I think the following is tight enough

\[ a^3b\plus{}b^3c\plus{}c^3a\plus{}10abc \le 27.\]

I use EMV to solve it. With a little help from Cyclic-inequality-of-degree-3 theorem, the solution is just in 5 lines.

Anyway, this is very nice one! Arqady!

[quote="hungkhtn"]Just let $ a \equal{} b \equal{} c \equal{} \dfrac{4}{3}$ to find the best value. [/quote]
Yes! And for the best value the equality holds in $ 4$ various points. :wink:
[quote="hungkhtn"]However, I think the following is tight enough
\[ a^3b \plus{} b^3c \plus{} c^3a \plus{} 10abc \le 27.
\]
[/quote]
It's wrong for the same $ a \equal{} b \equal{} c \equal{} \dfrac{4}{3}.$
[quote="hungkhtn"]
Anyway, this is very nice one! Arqady![/quote]
Thx, hungkhtn! :lol:

[quote="arqady"][quote="hungkhtn"]Just let $ a = b = c = \dfrac{4}{3}$ to find the best value. [/quote]
Yes! And for the best value the equality holds in $ 4$ various points. :wink:
[quote="hungkhtn"]However, I think the following is tight enough
\[ a^3b + b^3c + c^3a + 10abc \le 27.
\]
[/quote]
It's wrong for the same $ a = b = c = \dfrac{4}{3}.$
[quote="hungkhtn"]
Anyway, this is very nice one! Arqady![/quote]
Thx, hungkhtn! :lol:[/quote]
Yes, Arquady. I just prove that the expression

\[ a^3b + b^3c + c^3a + \alpha abc\]

attains max iff $ abc=0$ or $ a=b=c$. Sorry for wrong computation when $ \al=10$, but the result for your problem is still the same ;)


Solve in $\mathbb R$ the following equation
\[x+2^{x}= \frac{x}{2^{x}}. \]

[hide="Wait, it cant be this easy..."]
Consider $f(x)=x+2^{x}-\frac{x}{2^{x}}$. We want to find the zeroes of $f(x)$.

For $x > 0$ we have $f(x)= x \cdot \left( 1-\frac{1}{2^{x}}\right)+2^{x}$. But $2^{x}> 1 \implies 1-\frac{1}{2^{x}}> 0 \implies f(x) > 0$.

For $x < 0$, put $x=-y$ with $y \in \mathbb{R}^{+}$. We have $f(-y)=-y+\frac{1}{2^{y}}+y \cdot 2^{y}= y \cdot (2^{y}-1)+\frac{1}{2^{y}}$.
But again, $2^{y}> 1 \implies f(-y) > 0$.

To complete the proof, we test $x=0$ and see that $f(0)=1$. Thus, $f(x) > 0 \ \forall x \in \mathbb{R}$ and there are no solutions. [/hide]

That seemed way too easy... did I mess up somewhere?   :P

[i]Follow-up:[/i]
Solve in $\mathbb{C}$.

[hide]
$x2^{x}+2^{2}x = x$

$x(1-2^{x}) = 2^{2x}$

It is obvious that $x \neq 0$ b/c $0 \neq 1$.

If $x > 0, 1-2^{x}< 0$ but then the LHS is negative and the RHS is positive.

If $x < 0, 1-2^{x}> 0$ but then the LHS is negative and the RHS is positive.

Which leads us to the generalization that for $a$ a positive real number with $a > 1$, the following inequality holds for all real $x$:

$x+a^{x}\ge \frac{x}{a^{x}}$

Cool problem.

[/hide]

Actually, cincodemayo5590, it is that easy - I didn't look at your proof till later, but we both did it the same way - i didn't bother with functions :P


The degree of $(x^2+1)^4 (x^3+1)^3$ as a polynomial in $x$ is
$\text{(A)} \ 5 \qquad \text{(B)} \ 7 \qquad \text{(C)} \ 12 \qquad \text{(D)} \ 17 \qquad \text{(E)} \ 72$

[hide]
It becomes $(x^{8}+...)(x^{9}+...)$ with 8 being the degree of the first factor and 9 being the degree of the second factor, making the degree of the whole thing 17, or $\boxed{D}$. NOT 72.
[/hide]

Yeah. Whoops.

-BLt

[hide]Wouldn't $x^8\cdot x^9=x^{17}$? $\boxed{D}$ ;) [/hide]

GAH, you're right. Sorry, had a MAJOR brain fart right there.

-BLt

[quote="4everwise"][hide="Solution"]$\text{Answer: (D)}$

We only care about the variables of highest degree. The expression becomes $(x^2)^4(x^3)^3+\text{something}=x^{14}+\text{something}$[/hide][/quote]

you mean $x^{17}+\text{something}$  ;)


hey I've just finished reading the three musketeers and was wondering if you guys have any recommendations for good books?
thanks

I don't have any good recommendations, but I was wondering if you know if there's any translation for Monte Cristo that's very good, or one that's preferred by most people over others? I plan on reading that book soon, and it'd be nice to know which copy I should keep an eye out for.

[quote="Miss Merrythought"]HARRY POTTER ALL THE WAY![/quote]

That's like asking for fine wine and receiving grape juice.

I read the one with the omninous castle on the front cover.  not sure who translated it.

You may find "The Monk" by Matthew Gregory Lewis interesting ([url=http://www.gutenberg.org/wiki/Main_Page]free online ebook[/url]).  It was first published in 1796.  It has the flavor of the Count, more gothic and the adventure is along the lines of catholic mysticism.  It was suggested to me by an English teacher when I was a freshman in HS, but I have reread it since a few times over.

If you haven't already discovered the Dune series by Frank Herbert, it could potentially be a good waste of the rest of the summer.  However, he has a few other books that very few people talk about, such as "The Dragon in the Sea," "The Eyes of Heisenberg" and  quite a few others.  His witting is passively about ecology, social psychology; basically, technical advances and their impact on humans.

If science fiction is not your thing, you may find what your looking for in Albert Camus, a French author and philosopher from the 1930-50.  I read "A Happy Death" shortly after "The Monk" (suggestion by the same teacher), and from there I went to "The Metamorphosis," Franz Kafka (also at the above linked site).   All of these are classics, but would never be found in a classroom (outside of college and maybe not even there).

All of these are "small" books, but the material contained is deep.


geometry
3D geometry
probability
modular arithmetic

A number cube has the digits 1 through 6 each recorded on a different face. When two such cubes are tossed, what is the probability that the sum of the top two numbers is a multiple of 3? Express your answer as a common fraction.

We need the sum to be either $ 3, 6, 9,$ or $ 12$. The probabilty of rolling a sum of 3 is $ \frac{1}{18}$, a sum of 6 is $ \frac{5}{36}$, a sum of 9 is $ \frac{1}{9}$, and a sum of 12 is $ \frac{1}{36}$. We add these to get $ \boxed{\frac{1}{3}}$.

If you're wondering how I got these probabilities, just count! There are not many and some are just common sense.

Easier way: there's an equal probability for either $ 0,1,2 \pmod{3}$, so rolling a sum of either $ 0,1,2\pmod{3}$ is also equal so the answer is $ \boxed{\frac{1}{3}}$.


The (preliminary) results for all four rounds of Mandelbrot Individual are now on the web at [url]http://www.mandelbrot.org/scores.html[/url].  Yay for playing cards.

(After edit:) Congratulations on your own many math achievements, Mathfanatic, and thanks for sharing the link. I see a lot of the usual names on the list of Mandelbrot leaders.

[quote="tokenadult"]The URL you put on the link has an extra period on the end that results in a "page not found" until I edited the URL. You may want to edit the link.[/quote]

Thanks for pointing that out.  I fixed it.


$a>0,b>0,c>0,d>0,a+b+c+d=4.$
Prove that $5(abc+abd+acd+bcd)\leq12+8abcd.$

Nice problem. :)

Let $f(a,b,c,d) = 5(abc+bcd+cda+dab) - 8abcd = ab(5c+5d-8cd) + 5cd(a+b)$.  Assume, for the sake of contradiction, that $f(a,b,c,d) > 12$.  We must have $5c+5d > 8cd$ or else

$f(a,b,c,d) \leq 5cd(a+b) \leq 5(\frac{(a+b)+c+d}{3})^3 \leq 5(\frac 43)^3 = 11\frac{23}{27} < 12$

Thus, by AM-GM

$f(a,b,c,d) = ab(5c+5d-8cd) + 5cd(a+b)$
${\leq (\frac{a+b}{2})^2(5c+5d-8cd) + 5cd(a+b) = f(\frac{a+b}{2},\frac{a+b}{2},c,d})$

By symmetry, and applying repeatedly, we get

${f(a,b,c,d) \leq f(\frac{a+b}{2},\frac{a+b}{2},c,d})$
${\leq f(\frac{a+b}{2},\frac{a+b}{2},\frac{c+d}{2},\frac{c+d}{2}})$
${\leq f(\frac{a+b}{2}, 1, 1,\frac{c+d}{2}})$
$\leq f(1,1,1,1) = 12$, contradiction... so we are done.


abstract algebra
advanced fields
advanced fields unsolved

Let $ X=Spec(A)$ be an affine variety and let $ I$ be an ideal that is locally principal in the sense that there exist $ f_i$ generating the unit ideal in $ A$ such that $ IA_{f_i}$ is an $ A_{f_i}$-free module of rank 1 for all $ i$.
Show that the complement of the vanishing locus of $ I$ in $ X$ is an affine variety.

Comment: This is a generalization of the known result: If $ f$ is a nonzero divisor in $ A$, then if $ D(f)$ denotes the locus in $ Spec(A)$ where $ f$ does not vanish (primes in $ A$ that do NOT contain $ f$), then $ D(f)$ is affine. In fact $ D(f)=Spec(A_f)$, where $ A$ is the localization of $ A$ at the multiplicative system ${ \{1,f,f^2,\ldots}$. Note that $ A_f\neq 0$ because $ f$ is not a zero divisor in $ A$.
There is also a classical proof for this simple case: $ D(f)$ can be identified to the vanishing locus of $ Tf-1$ inside $ Spec(A[T])=X\times\mathbb A^1$, so $ D(f)$ is actually a closed subvariety of a bigger (than $ X$) affine variety.

Another comment: The ideal $ I$ in the hypothesis has to be a projective $ A$-module of rank one, so it corresponds to a line bundle on $ X$.
So a problem that would imply the previous is: If $ L$ is an algebraic line bundle in the geometric sense ($ p: V\to X$ with fiber $ \mathbb A^1$) over an affine variety and $ s$ is a section ($ s: X\to V$ such that $ p(s(x))=x$ for all $ x\in X$), then the complement of the image of $ s$ inside $ L$ is an affine variety.

Reduce it to this special case using the criterion for affineness given in Hartshorne.

I'm looking for something more constructive because I'm sure it's there, but still... I never trusted that criterion to actually work.

Denote $ U\equal{}X\setminus V(I)$ and $ B\equal{}\mathcal O_X(U)$. How do you show that $ IB\equal{}B$?


induction
geometry
rectangle
3D geometry

Here I have a set of a few problems that I have made.  They are easy, but something about tham spellbounds me.  The problems are as follows:
1.  Prove that the formula for $S=1+2+3+...+n=\frac{n(n+1)}{2}$.
2.  Prove that the formula for $S=1+3+5+...+(2n-1)=n^2$.
3.  Prove that $\sum_{x=1}^{n}F_{x}=F_{n+2}-1$.
4.  Prove that the formula for $S=2+4+6+...+2n=n(n+1)$.
5.  Prove that the formula for $S=1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$.
6.  Prove that the formula for $S=1^3+2^3+3^3+...+n^3=\left(\frac{n(n+1)}{2}\right)^2$.

Good luck in proving these...:)

EDIT:  No induction!!!

Masoud Zargar

why no induction? that is a perfectly valid way to prove something, but anyway

[hide]1. add first and last terms, then second to last, and second, etc

2. draw a square and notice to make a bigger square, you add an odd number, then to make it bigger again, you make an odd number, i.e

[b]0[/b]

0[b]0[/b]
[b]00[/b]

00[b]0[/b]
00[b]0
000[/b]

3. this one is so obvious from induction, but i dont know how to show it another way

4. factor out the 2, and you have the same as problem one

5. ?

6. look on the side bar where they have the visual proofs[/hide]

Here's what I have for number 1...I'll do the rest later; it's 1:06 am and I need to finish an assignment for tomorrow. :P

$2S=\sum_{i=1}^{n}i+\sum_{i=1}^{n}(n+1-i)$
$=\sum_{i=1}^{n}(n+1)$
$=n(n+1)$
$S=\frac{n(n+1)}{2}$

:)

P.S.  I don't like inductions because tehay re aroudn teh lines of...
All balls are red;
I have a ball;
My ball is red...You get the point (It's nto a beautiful method of proof).

Masoud Zargar

The beauty of induction lies in its simplicity, the fact that it's nearly common sense. If you show something is true for a certain case, and you can show that if something is true for a case, and the truth of the successive case follows, then it must be true for all cases following that original case. If you think of your other option of testing every single case individually, induction suddenly looks much nicer.
I see your point about induction being "cheap", as it gives little insight into why a particular pattern holds. Sometimes the cheapness pays off...
These have all been posted before.
[hide="5&6"]For 5, use $\sum^n_{i=1} i^3-(i-1)^3=\sum^n_{i=1} 3i^2-3i+1$.
The LHS is a telescoping series that becomes $n^3$, while RHS is the same sum after its summand is multiplied out. Split up that summand into three sums and solve for $\sum^n_{i=1} i^2$.
For 6, use the analogous sum for $\sum^n_{i=1} i^4$.[/hide]

The fifth question there is also a visual proof at left hand side.

The second one is easy
Let:

1        +  3        +     5      + ... +   2n-1 = S  ( n terms on LHS)
2n-1  +  2n-3   +   2n-5   + .... +1         =S  ( n terms on LHS)

So, 2S=2n+2n+2n+2n+.....2n (altogether n terms)

So, 2S=2n^2

S=n^2

[quote="Kalle"][quote="boxedexe"]
P.S.  I don't like inductions because tehay re aroudn teh lines of...
All balls are red;
I have a ball;
My ball is red...You get the point (It's nto a beautiful method of proof).
[/quote]
Uh, no, that is not what induction is like at all. Your example sounds like some logic fallacy.[/quote]

all balls are red -> one of the balls is red
is perfectly logic but indeed it isnt induction but deduction because it comes from the general case and specifies then. induction is coming from the specific case and then showing the general case ...

Right. I misunderstood. I thought he was trying to show that all balls were red.

[quote="huahe_"]The very first proof of a great theorem is never by induction.

But induction is definitely the first choice for other mathematicians to check what another mathematician has achieved.[/quote]

Wasn't the Last Theorem/Taniyama proved with induction?

Well, it's good to know other people's opinions about induction.

Well, I'm about to pass out because of my 1.5 hour sleep...Here are my ways of proving number 2 and 4...the rest will be posted soon.

2. Prove that the formula for $S=1+3+5+...+(2n-1)=n^2$.
$S=\sum_{i=1}^{n}(2i-1)$
$=2\sum_{i=1}^{n}i-\sum_{i=1}^{n}1$
From my previous proof of $S=1+2+3+...+n=\frac{n(n+1)}{2}$, we knwo that
$2\sum_{i=1}^{n}i-\sum_{i=1}^{n}1$
$=n(n+1)-n$
$=n^2+n-n=\boxed{n^2}$.

For number 4 we have:
4. Prove that the formula for $S=2+4+6+...+2n=n(n+1)$.
$S=\sum_{i=1}^{n}2i=2\sum_{i=1}^{n}i$
$=\frac{2n(n+1)}{2}=\boxed{n(n+1)}$.

According to this, we can say that the formula for $S=a+2a+3a+...+ai$ is
$S=\sum_{i=1}^{n}ai=a\sum_{i=1}^{n}i$
$=\boxed{\frac{an(n+1)}{2}}$

:)

Masoud Zargar

just FYI.

#5 is the number of squares in a $n*n$ square.

#6 is the number of rectangles in a $n*n$ square.

isn't #6 also the number of cubes in a $n*n*n$ cube?


Let $ M$ be a subset of $ \{1,2,3,...,2006\}$ with the following property:
For any $ 3$ elements $ x,y,z$ of $ M$($ x<y<z$), $ x\plus{}y$ does not divide $ z$.
Determine the largest possible size of $ M$.

http://gifted.hkedcity.net/Gifted/Download/notes/IMO-9th-CHKMO-Dec-2006.pdf

There are all of the problems for that year with solutions.

#1 has an easy answer, but a much less easy proof.


inequalities
integration
function
real analysis
real analysis unsolved

Let $ f\in L^{p}(0,\infty)$ and   $ Tf(x) \equal{} \int_{0}^{\infty}\frac {f(y)}{x \plus{} y}dy$

Prove that
\[ (\int_{0}^{\infty}|Tf(x)|^{p}dx)^{\frac {1}{p}}\leq (\int_{0}^{\infty}\frac {1}{x^{\frac {1}{p}}(x \plus{} 1)}dx)(\int_{0}^{\infty}|f(x)|^{p}dx)^{\frac {1}{p}}\]
for $ 1 < p < \infty$

I have tried to use the Generalized Minskowski's inequality but I dont know where does $ (\int_{0}^{\infty}\frac {1}{x^{\frac {1}{p}}(x \plus{} 1)}dx)$  come from?

Any idea will be highly appreciated. thanks

Try the $ L^p$-version of the Schur test with $ x^{\minus{}\alpha}$ as one of the functions ;).

Fedja

I don't know what Schur test is, I have searched for it [url]http://en.wikipedia.org/wiki/Schur_test[/url]
it just talks about $ L^{2}$ functions
If possible, could you please explain more abou the Schur test? thanks[/url]


calculus
integration
trigonometry
logarithms
calculus computations

Compute the integral: $ \int \frac{\sin^2 x\plus{}\sin 2x\plus{}1}{e^{\minus{}x}\plus{}\sin ^2x\plus{}1} \,dx$.

$ \begin{array}{l}
 \smallint \frac{{{{\sin }^2}x \plus{} \sin 2x \plus{} 1}}{{{e^{ \minus{} x}} \plus{} {{\sin }^2}x \plus{} 1}}\,dx \equal{} \smallint \frac{{{e^{ \minus{} x}} \plus{} {{\sin }^2}x \plus{} 1 \plus{} \sin 2x \minus{} {e^{ \minus{} x}}}}{{{e^{ \minus{} x}} \plus{} {{\sin }^2}x \plus{} 1}}\,dx \equal{}  \\
  \equal{} x \plus{} \smallint \frac{{\sin 2x \minus{} {e^{ \minus{} x}}}}{{{e^{ \minus{} x}} \plus{} {{\sin }^2}x \plus{} 1}}\,dx \equal{} x \plus{} \ln \left( {{e^{ \minus{} x}} \plus{} {{\sin }^2}x \plus{} 1} \right) \plus{} c \\
 \end{array}$


source: Griffiths 3rd edition, prob 7.16 & 7.33
Griffiths 4th edition, prob 7.36

An alternating current $I=I_0 \cos(\omega t)$ flows down a long straight wire and returns along a coaxial conducting tube of radius $a$. Both wire and tube have negligible thickness.

a) In what direction does the induced electric field point (radial, circumferential or longitudinal)?

b) Assuming the field goes to zero as s tends to infinity, find $E(s,t)$.
$s$ is the shortest distance from a point in space to the wire.

c) Integrate the displacement current density $J_d$ to get the total displacement current $I_d$.
You may need $\int x \ln x dx=\frac{1}{4} x^2 (2 \ln x -1)$

d) What is the ratio $I_d/I$? If the outer cylinder were 2mm in diameter, how high would the frequency have to be for $I_d$ to be 1% of $I$? (This problem is designed to indicate why Faraday never discovered displacement currents, and why it is ordinarily safe to ignore them unless the frequency is extremely high.)

related:
https://www.physicsforums.com/threads/finding-electric-field-displacement-current-for-time-varying-current-in-coaxial-cable.506183/

a) longitudinal

b) create a rectangle loop of length $f$ (longitudinal) and length ? (radial)
magnetic flux through rectangle = $\Phi_B= \int_s^a \frac{\mu_0 I}{2\pi r} dr \cdot f=\frac{\mu_0 I f}{2\pi}\ln(\frac{a}{s})$
 where $0<s<a$
$E f=-d\Phi_B/dt =  \frac{\mu_0 f}{2\pi} I_0\omega \sin(\omega t) \ln (\frac{a}{s})$
$E(s,t)=\frac{\mu_0 I_0\omega}{2\pi} \sin(\omega t) \ln (\frac{a}{s})$
$E(s,t)=0$ for $s>a$

c) $J_d=\epsilon_0 \frac{dE}{dt} =\frac{\epsilon_0\mu_0 I_0 \omega^2}{2\pi} \cos(\omega t) \ln(\frac{a}{s})$
$I_d=\int_0^a J_d \,dS=\int_0^a \frac{\epsilon_0\mu_0 I_0 \omega^2}{2\pi} \cos(\omega t) \ln(\frac{a}{s}) * 2\pi s \,ds$
$= \frac{1}{4} \epsilon_0\mu_0\omega^2 a^2 \cdot (I_0 cos(\omega t))$

d) $\frac{I_d}{I}= \frac{1}{4} \epsilon_0\mu_0\omega^2 a^2=(\frac{\omega a}{2c})^2$
$\frac{\omega a}{2c}=\frac{1}{10}$
$f=\frac{\omega}{2\pi} = \frac{c}{10\pi a}\approx 10^{10} Hz$


A CONVEX OCTAGON IS INSCRIBED IN A CIRCLE OF RADIUS - 'R'.[u][b]THE OCTAGON HAS FOUR CONSECUTIVE SIDES OF [/b][/u][u][b]LENGTH 3 UNITS & FOUR CONSECUTIVE SIDES OF LENGTH 2 UNITS.[/b][/u]FIND THE AREA OF THE OCTAGON.NOW I'VE SOLVED IT BY SOME LENGTHY GEOMETRY :blush: .BUT THE GIVEN SOLUTION INCLUDES THAT-THE SOLVER HAS DEVIDED THE CIRCLE IN 4 QUADRANTS[u].[b]EACH CONTAINING SIDES HAVING LENGTH 2 & 3 UNITS ALTERNATIVELY[/b],[/u]THEN USED SIMPLE TRIGONOMETRY. :o MY QUESTION IS IF THE REARRANGEMENT IS POSSIBLE OR NOT?????            PLS. HELP....

The area of the octagon is the area of the circle minus the sum of the areas of the segments cut from the circle by the sides. The order in which the segments appear does not matter for the value of this sum, so you can freely rearrange the sides.

[color=red]And, please, post using normal fonts in the future![/color]. Those underlined fat capital letters are plain irritating and, if anything, [i]diminish[/i] your chances to get an answer here.

Thank you fedja for your solution and suggestion  :blush:  :blush:


geometry
circumcircle
trigonometry
inradius
ratio
geometry proposed

[hide="hint"]$ SK$ passes thruogh the midpoint of arc $ BC$ from circle $ \omega_{A}$[/hide]

BaBaK Ghalebi,Could you post your solution,because i know one solution,without using your hint,but with using Menelaus theorem.

[quote="Erken"]Let $ \omega$ be the incircle of $ \triangle ABC$ and $ K$ be the point of tangency with side $ BC$.Denote circle through $ B,C$ as $ \omega_{A}$.If $ \omega_{A}$ is tangent to $ \omega$ at point $ S$,then $ S,K,I_{A}$ are collinear.(where $ I_{A}$ is excenter )[/quote]

Can someone send a figure ? I'm not at home and I have not ... my Cabri with me.
 Thanks !

[quote="Erken"][color=darkred]Let $ \omega(I,r)$, $ \omega_{a}(I_{a},r_{a})$ be the incircle and the $ A$- exincircle respectively of $ \triangle ABC$. Let $ K$ be the point of tangency of the circle $ w$ with the line $ BC$.
Let $ \Omega_{a}((O_{a}, R_{a})$ be the circle which passes through the points $ B$, $ C$ and which is tangent to the circle $ w$ in the point $ S$. Prove that $ I_{a}\in KS$.[/color][/quote]
[color=darkblue][b]Proof.[/b] Denote : the circumcircle $ C(O,R)$ of $ \triangle ABC$ ; the midpoint $ M$ of the side $ [BC]$ ; the point $ A'\in AI\cap OM$ ; the point $ N\in OM\cap \Omega_{a}$ so that the line $ BC$ separates the points $ A$, $ N$. Observe that $ \overline{OO_{a}MA'N}\perp BC$ and $ N\in SK$ because $ \frac{SI}{SO_{a}}=\frac{IK}{O_{a}N}=\frac{r}{R}_{a}$ (see http://www.mathlinks.ro/Forum/viewtopic.php?t=155800 ).  Because $ A'I=A'I_{a}$ is sufficiently to show $ IK=2\cdot NA'$ and apply the [b]Menelaus' theorem[/b] to the points $ S\in IO_{a}$, $ N\in O_{a}A'$, $ I_{a}\in A'I$ and $ \triangle IO_{a}A'$ : $ \frac{SI}{SI_{a}}\cdot\frac{NO_{a}}{NA'}\cdot \frac{I_{a}A'}{I_{a}I}=\frac{r}{R_{a}}\cdot \frac{R_{a}}{\frac{r}{2}}\cdot \frac{1}{2}=1$ $ \implies S\in I_{a}N$. But $ K\in SN$. Thus, $ I_{a}\in SK$.[/color]

[quote]sufficiently to show $ IK=2\cdot NA'$[/quote]
Can somebody show how to prove this?

[color=red]Here is another nice proof from my good friend Bayev Alen(A_L_E_N)  :[/color]
[color=blue]Solution:
If $ AC\equal{}AB$ then the problem is obvious $ \Rightarrow$:
WLOG suppose that $ AC<AB$.Let $ l$ be a tangent line to $ \Omega_{a}$ at point $ S$.Denote $ X\equal{}l\cap BC$.
Let $ XK\equal{}x,AB\equal{}c,AC\equal{}b\Rightarrow KC\equal{}p\minus{}c,KB\equal{}p\minus{}b$.Since $ XS$ is common tangent line to $ \Omega_{a}$ and $ \omega$$ \Rightarrow$ $ [XK]^{2}\equal{}[XS]^{2}\equal{}[XC]\cdot [XA]\equal{}[XK\minus{}KC]\cdot[XK\plus{}KB]$
$ \Rightarrow x^{2}\equal{}(x\minus{}(p\minus{}c))(x\plus{}(p\minus{}b)$,
so we have that $ x\equal{}\frac{(p\minus{}c)(p\minus{}b)}{c\minus{}b}$.Let $ \angle IKS\equal{}\angle IXK\equal{}\alpha$ and $ \angle LI_{a}K\equal{}\alpha_{1}$,where $ L$ is the point of tangency $ \omega_{a}$ with $ [BC]$.So it is sufficient to show that $ \alpha\equal{}\alpha_{1}\Rightarrow\tan{\alpha}\equal{}\frac{r}{x}\equal{}\frac{r(c\minus{}b)}{(p\minus{}c)(p\minus{}b)}$.
But $ \tan{\alpha_{1}}\equal{}\frac{KL}{r_{a}}\equal{}\frac{a\minus{}2(p\minus{}c)}{r_{a}}\equal{}\frac{(c\minus{}b)(p\minus{}a)}{S}\equal{}$
$ \equal{}\frac{(c\minus{}b)(p\minus{}a)pr}{S^{2}}\equal{}\frac{(c\minus{}b)r}{(p\minus{}c)(p\minus{}b)}\equal{}\frac{r}{x}\equal{}\tan{\alpha}$.Hence $ \alpha\equal{}\alpha_{1}\Rightarrow$ points $ S,K,I_{a}$ are collinear.Thus we are done.
[/color]
[color=green]
Are there any other properties of point $ S$?
[/color]

[color=blue][b]Lemma:[/b]  Let two circle $ O_{1}$ and $ O_{2}$ are tangent at $ P$. Line that tangent to $ (O_{1})$ at $ K$ meet $ (O_{2})$ at $ A,B$ .Let $ X$ be the intersection of line $ PK$ with $ (O_{2})$. Then $ XP$ bisect $ \angle APB$[/color]

Let $ T$ be the intersection of circle $ SBC$ with line $ SK$.By lemma, $ TS$ bisect $ \angle BSC$. Therefore we have $ TB = TC$ and $ TB^{2}= TK\cdot TS$. Let $ R = TB$. Then $ TB^{2}= TK\cdot TS$ imply $ R^{2}$ is power of $ T$ wrt $ w$. Let $ U$ be circle with center $ T$, radius $ TB$. Since $ R^{2}$ is power of $ T$ wrt $ w$, circle $ U$ and $ w$ are orthogonal. Let $ H(\neq B)$ be the intersection of circle $ U$ with line $ IB$. Since circle $ U$ and $ w$ are orthogonal, we have $ r^{2}= IH\cdot IB$,where $ r$ is inradius of $ \triangle ABC$.Therefore, $ H$ is inverse of $ B$ wrt $ w$. Let $ L,M$ be the point of tangency of incircle of $ \triangle ABC$ with $ CA,AB$ and $ P,Q$ midpoint of $ KM,KL$. Then it is easy to show $ P$ is inverse of $ B$ wrt $ w$. Therefore, $ P = H$.So circle $ U$ pass $ P$ and similarly also pass $ Q$. Let $ X,Y(\neq K)$ be the point on $ BC$ such $ BX = BK$,$ CY = CK$. Then $ XMLY$ and $ BPQC$ are homothetic where homothetic center is $ K$ and similitude ratio is $ \frac{1}{2}$.  It is  easily to see that midperpendicular of $ XM,YL,ML$ is external bisector of $ \angle B,\angle C$ and internal bisector of $ \angle A$. Therefore, midperpendicular of $ XM,YL,ML$ meet at $ I_{A}$. Therefore $ X,M,L,Y$ lie on circle with center $ I_{A}$. But $ XMLY$ and $ BPQC$ are homothetic where homothetic center is $ K$, similitude ratio is $ \frac{1}{2}$.  So $ B,P,Q,C$ lie on circle with center midpoint of $ KI_{A}$.Therefore, $ T$ is midpoint of $ KI_{A}$.But $ S,K,T$ are collinear. So we are done.


analytic geometry
graphing lines
slope
trigonometry

A worker can lift a box a vertical distance of 1m or roll it up a two meter-long ramp to the same elevation. Either way he does the same amount of work. If he uses the ramp, the applied force required is how much of the force that if he doesn't use the ramp?

A person uses 4.2J of work to push a 0.6kg box on a 30 degree slope, what's the distance that the box travels on the slope?

A Santa Claus looks at a spherical ball, what can you conclude about the image that he sees?

[hide]1. Same amount of work done
$ \Rightarrow F_{1} d_{1} \equal{} F_{2} d_{2}$
$ F_{2} \equal{} F_{1} \cdot \frac{ d_{1}}{ d_{2}} \equal{} F_{1}\cdot \frac{1}{2}$
So the applied force using the ramp is half the force required if he doesn't use the ramp.

2. The height he can raise the box is given by:
$ mgh \equal{} 4.2 \Rightarrow h \equal{} \frac{4.2}{0.6g}$
If $ x$ is the distance the box travels up the slope, then: $ x\sin 30^\circ \equal{} h$ so
$ x \equal{} 2h \equal{} \frac{8.4}{0.6g} \approx 1.43 \text{m}$

Not really sure on the last one...[/hide]

3. Ah, you mean the image on the ball? At first I thought the answer was that he saw a spherical ball...
Assuming that it's reflective, then it would just be a convex mirror with its focus at the center of the sphere. So it would be a virtual, real image.

How can sth be a virtual AND real image?

Force = -dU/dx so giving an energy of distance, just divide change of energy/distance.

Ahahaha, whoops, I didn't mean that. I meant virtual and upright image.


The point M is the mid-point on the side$\overline{BC}= a$ of a triangle ABC. Let $r_{1}, r_{2}, r_{3}$ be the radii of the inscribed circles in the triangles ABC, ABM and ACM respectively. Prove that $\frac{1}{r_{1}} + \frac{1}{r_{2}} \geq 2\left(\frac{1}{4}+\frac{2}{a}\right)$ .

Should the LHS of the inequality be $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}$?  Because according to my sketch is GSP the inequality as written does not always hold.


How do i get the fancyhdr package.  When I try to use it it says i have to install it, but then when i do, it doesn't work.

It's impossible to answer your question since there's no information on which to base an answer. You haven't said which Latex installation you are using, what you did to install fancyhdr and what "it doesn't work" means.

I'm afraid we can't read minds so you will have to tell us more and then we can help.

I also have a question on the Fancyhdr Package. I know this is stupid, but how do I set the text in the header?

[code]\rhead{Right top}
\chead{Middle top}
\lhead{Left top}
\rfoot{Right bottom}
\cfoot{Middle bottom}
\lfoot{Left bottom}[/code]


geometry
3D geometry
algorithm
Rubik\u0027s Cube Solving
search
videos
ratio

53 seconds.  Although now I average around a minute and 15, because I am out of practice. :(

i'm really really really really really really really really really really really really really really really really really really really really really really really really really really really really bad.
i got 1 hour and 4 minutes.

I'm worse.
It would probably take me a day :D

[quote="LynnelleYe"]No, I'm worse.
I've never completely solved a Rubik's cube. Ever.[/quote]

same here =)

Hmm...looks like I'm pretty good. My record is 32 seconds. Although I now average around 50 because I haven't cubed for a while. I just have to replace some stickers and I'll probably get back into it quickly. I was going for the record for the youngest sub-20 solver but I gave up on it. I'm also good at other rubik's like puzzles.

I have found my own solutions for:

Pyraminx
Skewb Cube
Megaminx (12 colored)
5x5x5 cube (based off of 4x4x4 solution which was not mine)

I can also solve:
2x2x2 cube
4x4x4 cube
square-1
sam lloyd's 15 puzzle (I can actually do any size, but the 15 one is most common)

I'm curious about the methods that other people use to solve these. I use the Lars Petrus method, which is one of the faster ones. Does anyone use the fridrich method? If so, do you have any tips on memorizing the algorithms? (i'm not expecting anyone, just in case anyone does)

[quote="Hamster1800"]
I'm curious about the methods that other people use to solve these. I use the Lars Petrus method, which is one of the faster ones. Does anyone use the fridrich method? If so, do you have any tips on memorizing the algorithms? (i'm not expecting anyone, just in case anyone does)[/quote]

Just wait for Macky (litoxe) to post :D But my personal best is 1:17 I think. I average 1:30-1:45 now, though, out of practice and my cube is getting sticky.

EDIT: Thanks for reminding me alan :P Here lies the 1337est post of paladin8. May it be remembered for all of eternity. :yup:

~3 minutes >_<

btw, l33t post count, paladin8 :ninja:

Bleh I'm so slow, my best was 4:31, but I just started learning how to do it like a month ago... Does anyone have any good guides? I used the one [url=http://jjorg.chem.unc.edu/personal/monroe/cube/rubik.html]here[/url] as a base guide, but then made some adjustments. You solve the first layer, 3rd layer corners, 3rd layer edges, then the middle layer. Is there a faster way?

My best is around 34 seconds; average is about a minute.  I use a really inefficient method (~120 moves/solve), so that makes the time a bit higher... :lol:

my record is 1:39

Does anyone else have the anniversary cube? It doesn't turn very well, so i took it apart to adjust the springs, but I didn't see any springs.

First of all, i've never solved a rubiks cube

but.....

I've timed a person consistently solving under 20 seconds (note: he put nail polish on it so it was easy to spin)

also

at ARML 2005 (at San Jose State University, West Coast), one guy went on stage, juggled three rubiks cubes, solved one at a time. o.O

i have to learn

[quote="Priapus"]
at ARML 2005 (at San Jose State University, West Coast), one guy went on stage, juggled three rubiks cubes, solved one at a time. o.O[/quote]

Yeah, that'd be Macky. He has yet to post here! His times would trump us all :juggle: <-- with Rubik's Cubes.

Record time: 54 seconds. Average is about 1:20.

[quote=Aminecraftbear][quote=GreyApatosurus]hi I got A 7 SECOND SOLVE OLL AND PLL SKIP, I WAS DONE ON TH 2ND F2L PAIR(I'M NOT YELLING)[/quote]

nice
lol
ur so lucky(tho i dont usually count solves with skips cuz like mostly they dont happen)[/quote]
ur right

[quote=nr1337]What's your best time for solving a (scrambled :D) 3x3x3 Rubik's  Cube.

My best was 1minute and about 28 seconds.  The problem is whenever I try to solve it as fast as I can I drop the cube a few times  :P[/quote]

45 secs

Just today i got 49.80 sec but i average around 1 min

[quote=itzsea2012][quote=LynnelleYe]No, I'm worse.
I've never completely solved a Rubik's cube. Ever.[/quote]

Same! I scrambled my Rubik's cube by accident once, and had to use a online Rubik's cube solver to unscramble the cube.[/quote]

hmmm
if you want to learn, i highly recommend jperm's videos!  ;)
[url]https://www.youtube.com/@JPerm[/url]

also, my pb is around 12. smth smth sec.
(i forgot, also my cs timer stuff got deleted  :oops: )

My fastest is 53 seconds
I haven't practiced in a year, so I probably can't even solve the cube now.


complex analysis
complex analysis unsolved

show that any four distinct points can be mapped to 1,-1,k,-k by a fractional linear transformation ( k depends on the 4 points). how many solutions are there?


This needs to be updated. When you click view and discuss the problems, you can only see the first 14 problems of the day and the most recent 7, with every problem in between hidden.


geometry
trigonometry
analytic geometry
graphing lines
slope
Pythagorean Theorem
power of a point

I can't draw it. There is the circle; where is E?

NightFlarer wrote:I need help with this problem:

Square ABCD has side length 2.  A semicircle with diameter AB is constructed inside the square, and the tangent to the semicircle from C intersects side AD at E.  What is the length of CE?

Here's what I have so far:

[hide] The midpoint of AB is the center of the circle, so let's call the center of the circle F.  FB is 1, BC is 2, so by Pythagorean Theorem, FC is sqrt5.  Also, let the point of tangency of CE be called G, so that (here I guessed) angle FGC is right.  If it is right, then GC = 2 by Pythagorean Theorem.  Also, I think you could have used Power of a Point, which states that two segments that are tangent to a circle and come from one point are equal, making CG and CB equal.  I can't seem to find the area of GE though.  I know FG is 1, and if only I could find the length of FE I could find GE and finish the problem.  Any help?[/hide]

U r almost there. Note that angles AEC and ECB are supplementary (e.e. their sum is pi radians). Now  angle FEG = 1/2(angle AEC) = 1/2(pi-ECB)

= 1/2(pi -  2*angle FCG)              (since angle  ECB = 2* angle FCG)

= pi/2 - alpha  where alpha = angle FCG

Now from rt. triangle EGF, GE = GF*cot(FEG) = 1* cot(pi/2 - alpha)

=  tan(alpha) = 1/2 ( Because tan(alpha) = tan(FCG)= FG/GC = 1/2

so EC = EG + GC = 1/2 + 2 = 2.5

[quote="NightFlarer"]Thanks for the solution.  Anybody know how to solve this problem without trig or radians?[/quote]

Let the midpoint of AB be G, and let CE intersect the semicircle at point F.  CF=CB=2.  AE=EF=x.  Since AE+ED=2, so ED=2-x,  CE=2+x.
Since EDC is a right triangle, by Pathygorean Theorem,
2^2+(2-x)^2=(2+x)^2

solve for x, we get x=1/2.
EC=2+1/2=5/2

[hide]so label points on the square, A(0,2), B(0,0), C(2,0) and D(2,2)

the circle will have center (0,1) and radius of 1

meaning that its equation will be x 2+(y-1) 2 =1

G will be point (x,y)

subbing into formula for slope

(1-y)/(0-x) and (0-y)/(2-x)

since the two must be perpendicular because tangents are perpendicular to the center of a circle

so -x/(y-1)=-y/(2-x)

-2x+x 2 =-y 2 +y

x 2 -2x+y 2 -y=0

make this equation equal x 2 +(y-1) 2 =1, which equals x 2 +y 2-2y=0

so x 2 +y 2-2y=x 2 -2x+y 2 -y

y=2x

sub that into x 2 +y 2-2y=0

x 2 +4x 2 -4x=0

5x 2 -4x=0

x(5x-4)=0

x=0 or x=4/5

wont use 0 since that would not intersect side AD

sub x=4/5 y=2x

y=2(4/5)

y=8/5

so now we need to find point e by extending CG

m=(8/5-0)/(4/5-2)

=8/5/-6/5

=-4/3

y=mx+b

0=-4/3(2)+b

b=8/3

y=-4/3x+8/3

POI with y=2

2=-4/3x+8/3

-2/3=-4/3x

=.5

sides of triangle CDE are 1.5, 2 and CE

since CE is hypoteneuse of right triangle (DE and CD are sides of square so angle is 90)

c 2 =1.5 2 + 2 2

c 2 =6.25

c=2/5

therefore line CE is 2.5 units long[/hide]

Man, I always miss the obvious solutions..

I did it using Ptolemy's  :?

Wasn't this exact problem (except that the 2 was a 10) on last year's AMC 10?

Yeah it was number 18 on the AMC 12 of either 2004 or 2003.


trigonometry
inequalities
integration
real analysis
real analysis unsolved

Prove that for $ \alpha \in \langle 0,\pi \rangle , n\in\mathbb{N}$

$ \sum_{k\equal{}1}^n\frac{\sin{k\alpha}}{k}>0$.

you can find this sum easy. it's standart problem :
it's equal to $ \frac {\pi \minus{} x}{2}$
hint
$ f(a) \equal{} \sum...$
then
$ f'(a) \equal{} good sum$
we can differentiate it because your sum is "good" sum . :wink:

It's only a finite prefix of the sum, not the entire infinite sum.

[hide="some hints"]This is known as Fejer-Jackson-Gronwall inequality, an elegant proof deu to Turán in which he use two facts:
$ \frac{\sin kx}{k}\equal{}2\int_{x/2}^{\pi/2}\Big(\frac{\sin x/2}{\sin t}\Big)^{2k}\frac{\sin(2k\minus{}1)t}{\sin t}dt$
and $ \sum_{k\equal{}1}^n\sin(2k\minus{}1)x\equal{}\frac{\sin^2nx}{\sin x}$
then sum them we deduce it.[/hide]

See here two proofs by our regreted Alexandru Lupas (flip2004). Although it is written in Romanian, the proofs are understandable: [url]http://www.mathlinks.ro/viewtopic.php?t=114058[/url] and a proof by Myth: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=20812[/url].


limit
real analysis
real analysis unsolved

Let $ (a_n)_n,(b_n)_n,(c_n)_n$ be positive real valued sequences such that $ \lim_{n\to\infty}a_nb_nc_n\equal{}1$ and $ \lim_{n\to\infty}(a_n^3\plus{}b_n^3\plus{}c_n^3)\equal{}3$ .Prove that $ \lim_{n\to\infty}a_n\equal{}\lim_{n\to\infty}b_n\equal{}\lim_{n\to\infty}c_n\equal{}1$

\[ 3=\lim_{n\to \infty}\left(a_{n}^3+b_{n}^3+c_{n}^3\right)\geq\lim_{n\to \infty}{3\sqrt[3]{a_{n}^3b_{n}^3c_{n}^3}}=3\lim_{n\to \infty}{a_{n}b_{n}c_{n}=3}\]
Equality hold only if $ \lim_{n\to \infty}{a_n}=\lim_{n\to \infty}{b_n}=\lim_{n\to \infty}{c_n}=1$

Uhm, I think the idea was to show that ...

What is needed is to study the proof of AM-GM for n=3 so that we can bound away the GM from the AM when the terms are bounded away from being equal, but this sounds like an extremely boring exercise  :wink:

[quote="Yuriy Solovyov"]
\[ 3 = \lim_{n\to \infty}\left(a_{n}^3 + b_{n}^3 + c_{n}^3\right)\geq\lim_{n\to \infty}{3\sqrt [3]{a_{n}^3b_{n}^3c_{n}^3}} = 3\lim_{n\to \infty}{a_{n}b_{n}c_{n} = 3
}\]
Equality hold only if $ \lim_{n\to \infty}{a_n} = \lim_{n\to \infty}{b_n} = \lim_{n\to \infty}{c_n} = 1$[/quote]

I believe that it would be true if it was $ (\lim a_n)^3+(\lim b_n)^3+(\lim c_n)^3=3 \lim a_n \lim b_n \lim c_n$.But from our hypothesis  we cannot say that this equality holds since we dont know wether $ a_n,b_n,c_n$ have limits or not.

[quote="Svejk"][quote="Yuriy Solovyov"]
\[ 3 = \lim_{n\to \infty}\left(a_{n}^3 + b_{n}^3 + c_{n}^3\right)\geq\lim_{n\to \infty}{3\sqrt [3]{a_{n}^3b_{n}^3c_{n}^3}} = 3\lim_{n\to \infty}{a_{n}b_{n}c_{n} = 3
}\]
Equality hold only if $ \lim_{n\to \infty}{a_n} = \lim_{n\to \infty}{b_n} = \lim_{n\to \infty}{c_n} = 1$[/quote]

I believe that it would be true if it was $ (\lim a_n)^3 + (\lim b_n)^3 + (\lim c_n)^3 = 3 \lim a_n \lim b_n \lim c_n$.But from our hypothesis  we cannot say that this equality holds since we dont know wether $ a_n,b_n,c_n$ have limits or not.[/quote]

Manifestly $ a_n,b_n,c_n$ are bounded. Let $ A$ a limit-point of $ a_n$ , so there is  a sub-sequence $ a_{p(n)}$ convergent to $ A$.
Of course there is a convergent sub-sequence $ b_{p(q(n))}$ and a convergent sub-sequence  $ c_{p(q(r(n)))}$ .
So $ a_{p(q(r(n)))},b_{p(q(r(n)))},c_{p(q(r(n)))}$ are convergent , and with Yuriy Solovyov method you prove that $ A = 1$ , so $ a_n$ is convergent to $ 1$ , .. an so on for $ b_n,c_n$ ..
 :cool:

Yes [b]Diogene[/b] , I also thought about this , but I got stuck .Why are $ a_n,b_n,c_n$ convergent to $ 1$ ? Just because some sub-sequence of $ a_n,b_n,c_n$  are convergent to $ 1$ it means that the sequence is convergent ? I dont think this is true .Can you develop your idea ? Maybe I m missing something  :maybe: .Thanks.

If every convergent subsequence of a bounded sequence converges to $ L$, then the entire sequence converges to $ L$. There are many ways to prove it. Two simplest ones are 1) recalling that $ \liminf$ and $ \limsup$ are both limits of subsequences and 2) assuming the contrary and choosing first an infinite subsequence all whose members differ from $ L$ by $ \varepsilon$ and then choosing a converging subsequence of that subsequence.


geometry
trigonometry
ratio
geometry proposed

[i]I'd like to see a synthetic solution to this problem , created- proposed by Virgil Nicula:[/i]

Let a trapez $ABCD ( AB // CD)$ and the midpoint $M$  of $AB$ .Segment $MD$  meets segment $AC$  at point $N$ .Let $P$  the foot of the perpendicular of point $N$ on line $BC$  . Prove that
                       $\angle MPN = \angle DPN$

                           [u]Babis[/u]

  [i](posted in preolympiad section too , but ...forgotten)[/i]

[quote="stergiu"][i]I'd like to see a synthetic solution to this problem , created- proposed by Virgil Nicula:[/i]

Let a trapez $ABCD ( AB // CD)$ and the midpoint $M$  of $AB$ .Segment $MD$  meets segment $AC$  at point $N$ .Let $P$  the foot of the perpendicular of point $N$ on line $BC$  . Prove that
                       $\angle MPN = \angle DPN$

                           [u]Babis[/u]

  [i](posted in preolympiad section too , but ...forgotteten)[/i][/quote]

Don't think my solution is synthethic, sorry. I have this tendency to build castles on top of geometry problems,

1)Prolong $BC$ and $MN$. Let $H$ be the point where they intersect.
2) Then $\frac{MN}{ND} = \frac{AM}{DC} = \frac{MB}{DC} = \frac{MH}{HD}$
So we're left with point $P$ and line $MN$ such that $\frac{MN}{ND}= \frac{MH}{HD}$ and $\angle NPH = 90$, we want to prove that $PN$ bisects $\angle DPM$. These problems with bisectors (for example, to prove the fact that if $PN$ is bisector then $\frac{DP}{PM}= \frac{DN}{NM}$), is good to attack them by drawing parallels to the sides of the triangles.
3) Draw parallel to $PM$ by $D$. Let $R$ the point of intersection of this parallel with $PN$. Also draw $HR$. Then
4) $\frac{PM}{DR} = \frac{MN}{ND} = \frac{HM}{HD}$. And what this is telling you is that if you prolong $PM$ to the point $S$ where it touches line $HR$ then it will be $MS=PM$ so $HM$ is median of $PHS$ so $D$ is the midpoint of $RQ$.
5) As $\angle RPQ = 90$ it is $PD = PR$, hence $\angle DPR = \angle DRP = \angle RPS$ hence $PN$ is a bisector.

There must be certainly a nicer solution, probably even without trig, hope you'll find it  :)

Daniel[/quote]

[quote="dhernandez"][quote="mecrazywong"]Let $DM\cap CB=E$. Note that $\frac{MN}{ND}=\frac{AM}{DC}=\frac{BM}{DC}=\frac{EM}{ED}$. Hence N,E are harmonic conjugates wrt M,D. Since P lies on the circle with diameter NE, that is, the Apollonian Circle for the points D,M and ratio $r=\frac{DN}{NM}$. Therefore the result follows.[/quote]

Yes, that's the idea. It's the same my solution, only that without speaking of harmonic conjugates or Apollonian circles. One might not know all that stuff.

Daniel[/quote]
You successfully cheat me :roll:
Well, when I first read your solution I strongly wonder how someone can work out these stuffs without knowing harmonic conjugates :D Anyway, nice solution.

Yes, it is and the my idea ! Choose another particular harmonical divisions and you can compose another problems.


There is a grid under the tension $ U\equal{}20volt$ and there are 2 resistences $ R_1\equal{}10ohm$ and $ R_2\equal{}20ohm$ such $ R_1$ disjuction with $ R_2$.
 Find $ I\equal{}?$

Perhaps I'm not familiar with Ohm's Law to the extent I thought, but are you asking that if we have (a circuit) with potential difference $ 20\mbox{V}$ and two resistors of $ 10\Omega$ and $ 20\Omega$, what is the current?  And does disjunction mean in series or in parallel?  If they are in series, $ V=IR\rightarrow 20\mbox{V}=10\Omega+20\Omega I\rightarrow I=\frac{2}{3}\mbox{A}$.  If they are in parallel, $ V=IR\rightarrow 20\mbox{V}=(\frac{20}{3}\Omega)I\rightarrow I=3\mbox{A}$.  I hope this answers your question.

Now try [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=166292]these[/url] problems.


I wrote this thread some days ago but didn't get the answer i needed:

------------------------------------------------------------------------------------------------------------

OK, i have quite big problems with this chapter - posibility. You can expect me to stop by several times in the near future since this surely isn't easy for me.

First two different tasks:

1) A person has 10 pair of shoes in a bag. She picks 3 shoes. What is the posibility that there will be at least 1 pair of shoes?

2) In a country every 3 person has a car. What is the posibility that in a gathering of 12 people are:
a. max. 4 people has a car.
b. exactly 4 people has a car.
c. at least 4 people has a car.

Plz set up the equation - i don't have any problems putting the numbers into the calculator. Thx in advance for any help.

(for my own help to remember what tasks we are talking about in this topic - 1024+1025)

------------------------------------------------------------------------------------------------------------

Can someone give me the result using the K(n,r) and !... I believe some of you have done possibility calculations with these means and symbols.

Okay, there we go.

1. (Edit: messed up here in first place) We first pick a shoe at random. Two cases:
* The probability to pick a matching shoe is $\frac{1}{19}$, by Laplace.
** Assume we don't pick a matching shoe, the probability of that happening is $\frac{18}{19}$. Now, we have two shoes in our hands, and two shoes are matching in the box with eightteen shoes remaining. Therefore, the probability to pick a shoe in this case is $\frac{2}{18}$. The probability of this case happenning is therefore $\frac{18\cdot2}{19\cdot18} = \frac{2}{19}$.

In total that's $P = \frac{3}{19}$.

2. Here you should keep in mind that the probability that a person has a car is $\frac{1}{3}$ and for not having one is $\frac{2}{3}$.

(a) Maximum four people have a car, so at least eight people don't. The other four people (that complete the group of twelve) may 'choose' if they have one or not. So:
$P = \underbrace{\frac{2}{3}\cdot\frac{2}{3}\cdots\frac{2}{3}}_{\textrm{8 people mustn't have a car}}\cdot\underbrace{1\cdot1\cdot1\cdot}_{\textrm{4 people may 'choose'}} = \left(\frac{2}{3}\right)^{8}$

(b) Four people must have a car, 8 mustn't, so (in the same way as in (a))
$P = \left(\frac{2}{3}\right)^{8}\cdot\left(\frac{1}{3}\right)^{4}$

(c) And then, as we calculated the probability for eight persons not having a car and the others free to 'choose', we find here:
$P = \left(\frac{1}{3}\right)^{4}$

I hope I made it all clear while writing this  :)

P.S., I like "Non progredi est regredi"  :D

Thx, but since the topic in my math book wants me to use the K(n,r) and ! i would really like to have you (or someone else) explain to me how to do the two tasks using these symbols (as i did mention at the beginning). If no one here can help me at what part of the forum should i try asking?

You could ask it in Combinatorics section, but I wouldn't do it. In fact, I used it in the first problem, but it is so obvious I don't mention it. E.g. I say "Therefore, the probability to pick a shoe in this case is $\frac{2}{18}$".
Normally, that should be $P = \frac{\sharp A}{\sharp U} = \frac{2}{\binom{18}{1}} = \frac{2}{\frac{18!}{17!1!}} = \frac{2}{18}$, because there are two cases that lead to "a good result" (thats #A) and we have in total $\binom{18}{1}$ possibilities to pick a sock... Ridiculous  :P

And the second one is just probablitity, no combination-working... :huh:

[quote="Sporally"]
1) A person has 10 pair of shoes in a bag. She picks 3 shoes. What is the posibility that there will be at least 1 pair of shoes?

2) In a country every 3 person has a car. What is the posibility that in a gathering of 12 people are:
a. max. 4 people has a car.
b. exactly 4 people has a car.
c. at least 4 people has a car.
[/quote]

1) P(at least one pair of shoes) = 1 - P(no pairs of shoes) = $1 - \frac{20 \cdot 18 \cdot 16}{20 \cdot 19 \cdot 18} = 1 - \frac{16}{19}=\frac{3}{19}$.

This should give you an idea how to solve #2... (think about it)

(I don't see how any combinations are involved...) $C(n,r)$ is the number of ways to pick $r$ distinct objects from $n$ choices. ;)

I did it like this using combinatorics and got the same answer:

(C(10, 1)*C(18, 1))/C(20, 3)

Oh no, now i think we are misunderstanding each other. In 2c i found out it was a bit more than a chance of 1%. From these 12 ppl you don't pick 4 and see if those 4 have a car, coz then i would be around 1%, but if you take 4 ppl from the gathering of the 12 and you didn't get your 4 cars you are welcome to see if the 5 person has one. And you can continue until you are through all of the 12 ppl in the gathering. So can it really be true there is still only 1% chance at least 4 of the 12 has a car? I mean, it would be very common that there would be 4 cars in a gathering of 12, and the chance that there are 5 should then be quite big still - or more than 5. I am wrong or did we misunderstand each other?

EDIT: I think i really need a closer explanation on why you place the number in the order and why you pick the numbers you do in this equation:
(C(10, 1)*C(18, 1))/C(20, 3)
I would first of have used something like:
(C(1, 19)*C(18, 2))/C(20, 3)


Gauss
integration
calculus
analytic geometry
real analysis
real analysis unsolved

what is a simple way to calculate $ \int_{\minus{}\infty}^{\plus{}\infty}e^{\minus{}h^2x^2}dx$
?

[quote="Clausius"]what is a simple way to calculate $ \int_{ \minus{} \infty}^{ \plus{} \infty}e^{ \minus{} h^2x^2}dx$
?[/quote]

let u=h*x

$ \int_{ \minus{} \infty}^{ \plus{} \infty}e^{ \minus{} u^2}du/h$

Which is just $ \sqrt {\pi}/h$. If you don't know why, see http://en.wikipedia.org/wiki/Gaussian_integral .

Put it into Mathematica. :wink:

What you can do is use some multivariable calculus and change this into a double integral:
Let $ I\equal{}\int_{\minus{}\infty}^{\infty}e^{\minus{}h^{2}x^{2}}\ dx$.
Then we may define
$ I^{2}\equal{}\iint_{D}e^{\minus{}h^{2}x^{2}\minus{}h^{2}y^{2}}\ dx$ where $ D$ is our region of integration.
Now convert into polar coordinates to obtain
$ I^{2}\equal{}\int_{0}^{2\pi}\int_{0}^{\infty}re^{\minus{}h^{2}r^{2}}\ dr\ d\theta$.
Now you evaluate it:
$ I^{2}\equal{}\frac{\pi}{h^{2}}\Rightarrow I\equal{}\frac{\sqrt{\pi}}{h}$.


Let $a,b,c,d$be positive numbers and assume the equation:$ax^{4}-ax^{3}+bx^{2}-cx+d=0$ has 4 roots in $\left(0;\frac{1}{2}\right)$. Prove that: $21a+164c\geq 80b+320d$

I have posted this problem month ago. But have't received any reply? I think the problem may be wrong? Have any another ideas?


AMC
AIME
USA(J)MO
USAMO
USAJMO
AMC 10
AMC 10 A

If I am taking both the AMC 10 and the AMC 12, what's the formula again for calculating your USAMO index (if the new proposals even have an USAMO index)?

Is it something like AMC 10 + AIME score, or AMC 10 + a number times the AIME score? Because I didn't really see a mention towards the AIME score in considering your USAMO placement (entry) in the proposals.

Thanks!

USAMO qualification is simple now: the AMC 10 does not exist. Your USAMO index is your AMC 12 score plus 10 times your AIME score.

If you don't make the USAMO, you will be considered for the USAJMO, new this year. That's based on AMC 10 plus 10 times AIME.

Actually, if you take the AMC 10, and get at least an 11 on the AIME, you qualify for USAMO. This is one category many students don't fall under.

If you can get a 11 on the AIME, you would not consider the AMC 10 :P .

[quote="AIME15USAMO"]This is one category many students don't fall under.[/quote]
There are some schools which only offer the "best fit" for you. Like if you're under 8th grade, you take the AMC 8, under 10th grade takes AMC 10, etc.

Also, some students may plan to take the AMC 10 the A date, then the AMC 12 the B date, and then take the AMC 10 and not be able to make the B date, so they would have to make USAMO through getting an 11+ on the AIME.

what will the combined aime and amc 10 score will be to make it to UJAMO? most likely?

Well, the USAMO index last year was 215.0 from the AMC 10, and I want to say that the index will be lower, especially given that more people will qualify for the USAJMO than AMC 10-takers for the 2009 USAMO.

I want to say 200-210, but I'm probably failing at the metagaming.

If it was 200-210, that would be nice  :lol: Like a 140 AMC 10 and a 6-7 AIME


[quote="anthony19891019"]Can we use the ABC method in the mathematical competition without proving it?[/quote]

If you are worried about whether a theorem can be used or not, you can always incoporate the proof into your solution!

[quote="quangpbc"]In a any contest , you can use Cauchy-Schwarz or AM-GM only . If you want use a different method you must proof it  :) . And in any contest in your country or international , some inequalities not too hard to use some strong method example SOS-hungkhtn, ABC- ANHCUONG ( http://mathnfriend.org ),...

Some hard inequalities on national contest :

IRAN-1996, VietnamTST 2005 ..., you can use some method, obviously to attach proofs[/quote]

I was under the impression that Holder or some other inequalities on Mildorf's packet could be used too  :maybe:

Yeah - Mildorf's 'Standard Dozen' can all be used for inequalities of a high standard. For easier problems - Muirhead Schur and Jensen maybe Holder might be frowned upon

In Vietnam, all inequalitiy menthods except AM-GM, Cauchy-Schwarz and Jensen can not use  :D


Can somebody help me withh this question, I feel lost.

Let S be the two demensional subspace of $R^3$ spanned by

$x_1 = \left(\begin{array}{cc}1\\0\\2\end{array}\right) \mbox{ } x_2 = \left(\begin{array}{cc}0\\1\\-2\end{array}\right)$

a) Find a basis for $S^\perp$

b) Give a geometric discription of $S$ and $S^\perp$

c) Determin the projection matrix $P$ that projects vectors in $R^3$ onto $S^\perp$

I am pretty sure i found an answer for a)

For a), there's the elementary vector calculus solution: find the cross product of these two vectors.

More in the linear algebra spirit: we want to find $v$ such that $x_1^Tv=0$ and $x_2^Tv=0.$  That means we're looking for the null space of the matrix whose rows are $x_1^T$ and $x_2^T.$ That matrix is

$\begin{bmatrix}1&0&2\\0&1&-2\end{bmatrix}.$

Since this matrix is already in reduced row echelon form, we can simply read off this null space. A basis for the answer would be

$v=\begin{bmatrix}-2\\2\\1\end{bmatrix}.$

For b), I assume that all that is being looked for is the statment that $S$ is a plane and $S^{\perp}$ is a line normal to that plane.

Now, for c): we have an orthonormal basis for $\mathbb{R}^3$ of $\{e_1,e_2,u\}$ where $u=\frac13v$ is a unit vector in the direction of $S^{\perp}$ and $e_1,e_2$ are orthonormal vectors in $S$ which we won't bother to calculate. For a given vector $x,$ the vector $(u\cdot x)u$ is the projection of $x$ into $S^{\perp}.$ If we simply subtract that from $x,$ we will get the projection of $x$ into $S.$ Proceding with that idea, and making things into matrix notation when we can:

$Px=x-(u\cdot x)u=x-u(u^Tx)=x-(uu^T)x=(I-uu^T)x.$

So let $P=I-uu^T=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} -\frac19\begin{bmatrix}-2\\2\\1\end{bmatrix}\begin{bmatrix} -2&2&1 \end{bmatrix}$

$=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} -\frac19\begin{bmatrix}4&-4&-2\\-4&4&2\\-2&2&1\end{bmatrix}= \frac19\begin{bmatrix}5&4&2\\4&5&-2\\2&-2&8\end{bmatrix}.$

sorry, i sort of understand but can explain to me how you got $u = \frac 13v$

Thank you much for help.

I wanted $u$ to be a unit vector. That is, I wanted $||u||=\sqrt{u^Tu}=1.$ If we're given an arbitrary nonzero vector $v$ and want a unit vector in the same direction as $v,$ that standard way of obtaining that is to let $u=\frac v{||v||}.$

In this particular case, $||v||=\sqrt{(-2)^2+2^2+1^2}=3.$

Thank you much, I understand now why you do that.


For natural numbers $n,a,b$ with $(a,b)=1$, prove that
$\frac{1}{a}+\frac{1}{a+b}+...+\frac{1}{a+nb}$ is not a natural number.

The problem was posted by grodij about a year ago:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26137]www.artofproblemsolving.com/Forum/viewtopic.php?t=26137[/url]
and wasn't solved at that time. I think it is a very good problem. Anyone interested in it?

We will prove a "more general" statement: For any natural numbers $a$, $b$ and $n$
\[ \frac 1 a+\frac 1{a+b}+\cdots+\frac 1{a+nb}  \]
is not an integer number.

FIRST OBSERVATION:
If $a>n$ or $b\geq 2(1+\frac 1 2+\cdots+\frac 1 n)$ then the above sum cannot
be integer because it is less than $1$ or (if $a=1$) it is between $1$ and $2$.
So, for every $n$ there are only a finite number of cases to check.

GENERAL TECHNIQUE OF PROOF:
For every $n$, $a$ and $b$ verifying $a\leq n$ and $b<2(1+\frac 1 2+\cdots+\frac 1 n)$ we will
find a prime $p$ such that $p$ divides the denominator of the irreducible fraction representing the
number $\frac 1 a+\frac 1{a+b}+\cdots+\frac 1{a+nb}$.

CASE $n=1$)
There is only one case to check: $a=1$ and $b=1$. But then
\[ \frac 1 1+\frac 1 2=\frac 3 2  \]
and the prime $2$ divides the denominator of the irreducible fraction $\frac 3 2$.

CASE $n=2$)
As $3>b$, then exactly one of the three numbers $a$, $a+b$, $a+2b$ is a multiple of three, so
the prime $3$ divides the denominator of the irreducible fraction representing the number
\[ \frac 1 a+\frac 1{a+b}+\frac 1{a+2b}  \]

CASE $n=3$)
Now $b\leq 3$. If $b=3$ we have the three sums.
First: $\frac 1 1+\frac 1 4+\frac 1 7+\frac 1{10}$, for which the alleged prime is $7$ (coming from
the summand $\frac 1 7$).
Second: $\frac 1 2+\frac 1 5+\frac 1 8+\frac 1{11}$, for which the prime is $11$ (from $\frac 1{11}$).
Third: $\frac 1 3+\frac 1 6+\frac 1 9+\frac 1{12}$, for which the prime is $3$ (because there is
only one summand, $\frac 1 9$ whose denominator is multiple of $3^2$).
If $b=1$ or $b=2$ then at least one of the denominators $a,a+b,a+2b,a+3b$ is multiple of $3$,
so the prime $3$ divides the denominator of the irreducible fraction, unless there are two of the
denominators multiple of three (there cannot be three), such that summing them dissapears the
factor $3$. To be more specific, the two possible subcases are:
First: between the summands there appear $\frac 1{1\times 3}+\frac 1{2\times 3}$. In this case
the sum is $\frac 1 3+\frac 1 4+\frac 1 5+\frac 1 6$ and the prime is $5$.
Second: between the summands there appear $\frac 1{2\times 3}+\frac 1{4\times 3}$. This case
is automatically discarded because $\frac 1 6+\frac 1 8+\frac 1{10}+\frac 1{12}$ has $a=6>3$.

CASE $n=4$)
As $5>b$, exactly one of the numbers $a,a+b,a+2b,a+3b,a+4b$ is a multiple of $5$, so the
prime $5$ is which we wanted.

CASES $n=2,4,6,10,12,16,18,22,\dots$)
The cases $n=2$ and $n=4$ follow the same pattern that applies for every $n$ such that $n+1$
is an odd prime: $n+1>b$, and exactly one of the numbers $a,a+b,a+2b,\dots,a+nb$ is a multiple
of the prime $n+1$, so the prime $n+1$ divides the denominator of the irreducible fraction.

CASE $n=5$)
Now $a+nb\leq 5+5\times 4=25$ where the equality is reached for the sum $\frac 1 5+\frac 1 9+\frac 1{13}+\frac 1{17}+\frac 1{21}+\frac 1{25}$, but now the prime we wanted is $5$, because there is only one denominator multiple of $5^2$.
In any other case, there is at least one denominator multiple of $5$ and at most two of them, and
$5$ divides the denominator of the irreducible fraction unless there appear the summands
$\frac 1{2\times 5}+\frac 1{3\times 5}$ (easily seen imposible) or $\frac 1{1\times 5}+\frac 1{4\times 5}$, that corresponds to the case $\frac 1 5+\frac 1 8+\frac 1{11}+\frac 1{14}+\frac 1{17}+\frac 1{20}$ but in this case we can take the primes $11$ or $17$.

CASE $n=6$)
SOLVED BEFORE ($n+1=7$ is an odd prime).

CASE $n=7$)
Reasoning as in the preceeding cases, the prime searched for is $7$, unless there appear the summands $\frac 1{1\times 7}+\frac 1{6\times 7}$, which is the case of the sum $\frac 1 7+\frac 1{12}+\frac 1{17}+\frac 1{22}+\frac 1{27}+\frac 1{32}+\frac 1{37}+\frac 1{42}$, but now we can take the prime $37$ from the summand $\frac 1{37}$.

CASE $n=8$)
By the same kind of reasons as above, the prime $7$ always divides the denominator of the irreducible fraction unless the summands $\frac 1{1\times 7}+\frac 1{6\times 7}$ appear, which is the case of the sums $\frac 1 2+\frac 1 7+\frac 1{12}+\frac 1{17}+\frac 1{22}+\frac 1{27}+\frac 1{32}+\frac 1{37}+\frac 1{42}$ and $\frac 1 7+\frac 1{12}+\frac 1{17}+\frac 1{22}+\frac 1{27}+\frac 1{32}+\frac 1{37}+\frac 1{42}+\frac 1{47}$ but both sums contain the summand $\frac 1{37}$ from which we can take the prime $37$.

CASE $n=9$)
We can always take the prime $7$ except for $\frac 1 8+\frac 1{11}+\frac 1{14}+\frac 1{17}+\frac 1{20}+\frac 1{23}+\frac 1{26}+\frac 1{29}+\frac 1{32}+\frac 1{35}$, when we take the prime $29$, $\frac 1 2+\frac 1 7+\frac 1{12}+\frac 1{17}+\frac 1{22}+\frac 1{27}+\frac 1{32}+\frac 1{37}+\frac 1{42}+\frac 1{47}$, when we take the prime $37$, and $\frac 1 7+\frac 1{12}+\frac 1{17}+\frac 1{22}+\frac 1{27}+\frac 1{32}+\frac 1{37}+\frac 1{42}+\frac 1{47}+\frac 1{52}$, when we take again the prime $37$.

CASE $n=10$)
SOLVED BEFORE ($n+1=11$ is an odd prime).

CASE $n=11$)
We can always take the prime $11$.

CASE $n=12$)
SOLVED BEFORE ($n+1=13$ is an odd prime).

CASES $n=13,14,15$)
We can always take the prime $13$.

CASE $n=16$)
SOLVED BEFORE ($n+1=17$ is an odd prime).

STOP!!
For the cases to come, we will use the following result.
FUNDAMENTAL RESULT:
Given a positive integer $n$, suppose we find a prime $p$ such that
\[ \frac n 2<p\leq n  \]
\[ p\geq 2(1+\frac 1 2+\cdots+\frac 1 n)  \]
\[ \frac{p+1}2p\geq n+2n(1+\frac 1 2+\cdots+\frac 1n)  \]
then for any couple of positive integers $a$, $b$ such that $a\leq n$ and $b<2(1+\frac 12+\cdots+\frac 1n)$ the denominator of the irreducible fraction representing the rational number
\[ \frac 1a+\frac 1{a+b}+\frac 1{a+2b}+\cdots+\frac 1{a+nb}  \]
is a multiple of $p$.

PROOF OF THE FUNDAMENTAL RESULT:
In the conditions above, at least one of the denominators $a,a+b,a+2b,\dots,a+nb$, and at most two of them, are multiples of p. So we always have the prime $p$ dividing de denominator of the
irreducible fraction, unless there were to multiples of $p$ such that summing them cancels out the factor $p$, but this is possible only if there appear between the summand the terms
\[ \frac 1{p}+\frac 1{p^2-p}  \]
or
\[ \frac 1{2p}+\frac 1{p^2-2p}  \]
or
\[ \frac 1{3p}+\frac 1{p^2-3p}  \]
...
...
or
\[ \frac 1{\frac{p-1}2p}+\frac 1{\frac{p+1}2p}  \]
what is impossible because $\frac{p+1}2p>a+nb$.

                                                                                Q.E.D.

RESUMING THE PROOF
===============

CASE $n=17$)
Take the prime $p=17$ and apply the FUNDAMENTAL RESULT.

CASE $n=18$)
SOLVED BEFORE ($n+1=19$ is an odd prime).

CASES $n=19,20,21$)
Take the prime $p=19$ and apply the FUNDAMENTAL RESULT.

CASE $n=22$)
SOLVED BEFORE ($n+1=23$ is an odd prime).

CASES $n=23,24,25,26,27$)
Take the prime $p=23$ and apply the FUNDAMENTAL RESULT.

CASE $n=28$)
SOLVED BEFORE ($n+1=29$ is an odd prime).

CASE $n=29$)
Take the prime $p=29$ and apply the FUNDAMENTAL RESULT.

CASE $n=30$)
SOLVED BEFORE ($n+1=31$ is an odd prime).

CASES $n=31,32,33,34,35$)
Take the prime $p=31$ and apply the FUNDAMENTAL RESULT.

CASE $n=36$)
SOLVED BEFORE ($n+1=37$ is an odd prime).

CASES $n=37,38,39$)
Take the prime $p=37$ and apply the FUNDAMENTAL RESULT.

CASE $n=40$)
SOLVED BEFORE ($n+1=41$ is an odd prime).

CASE $n=41$)
Take the prime $p=41$ and apply the FUNDAMENTAL RESULT.

CASE $n=42$)
SOLVED BEFORE ($n+1=43$ is an odd prime).

CASES $n=43,44,45$)
Take the prime $p=43$ and apply the FUNDAMENTAL RESULT.

CASE $n=46$)
SOLVED BEFORE ($n+1=47$ is an odd prime).

CASES $n=47,48,49,50,51$)
Take the prime $p=47$ and apply the FUNDAMENTAL RESULT.

CASE $n=52$)
SOLVED BEFORE ($n+1=53$ is an odd prime).

CASES $n=53,54,55,56,57$)
Take the prime $p=53$ and apply the FUNDAMENTAL RESULT.

CASE $n=58$)
SOLVED BEFORE ($n+1=59$ is an odd prime).

CASE $n=59$)
Take the prime $p=59$ and apply the FUNDAMENTAL RESULT.

CASE $n=60$)
SOLVED BEFORE ($n+1=61$ is an odd prime).

CASES $n=61,62,63,64,65$)
Take the prime $p=61$ and apply the FUNDAMENTAL RESULT.

CASE $n=66$)
SOLVED BEFORE ($n+1=67$ is an odd prime).

CASES $n=67,68,69$)
Take the prime $p=67$ and apply the FUNDAMENTAL RESULT.

CASE $n=70$)
SOLVED BEFORE ($n+1=71$ is an odd prime).

CASE $n=71$)
Take the prime $p=71$ and apply the FUNDAMENTAL RESULT.

CASE $n=72$)
SOLVED BEFORE ($n+1=73$ is an odd prime).

CASES $n=73,74,75,76,77$)
Take the prime $p=73$ and apply the FUNDAMENTAL RESULT.

CASE $n=78$)
SOLVED BEFORE ($n+1=79$ is an odd prime).

CASES $n=79,80,81$)
Take the prime $p=79$ and apply the FUNDAMENTAL RESULT.

CASE $n=82$)
SOLVED BEFORE ($n+1=83$ is an odd prime).

CASES $n=83,84,85,86,87$)
Take the prime $p=83$ and apply the FUNDAMENTAL RESULT.

CASE $n=88$)
SOLVED BEFORE ($n+1=89$ is an odd prime).

CASES $n=89,90,91,92,93,94,95$)
Take the prime $p=89$ and apply the FUNDAMENTAL RESULT.

CASE $n=96$)
SOLVED BEFORE ($n+1=97$ is an odd prime).

CASES $n=97,98,99$)
Take the prime $p=97$ and apply the FUNDAMENTAL RESULT.

CASE $n=100$)
SOLVED BEFORE ($n+1=101$ is an odd prime).

CASE $n=101$)
Take the prime $p=101$ and apply the FUNDAMENTAL RESULT.

CASE $n=102$)
SOLVED BEFORE ($n+1=103$ is an odd prime).

CASES $n=103,104,105$)
Take the prime $p=103$ and apply the FUNDAMENTAL RESULT.

CASE $n=106$)
SOLVED BEFORE ($n+1=107$ is an odd prime).

CASE $n=107$)
Take the prime $p=107$ and apply the FUNDAMENTAL RESULT.

CASE $n=108$)
SOLVED BEFORE ($n+1=109$ is an odd prime).

CASES $n=109,110,111$)
Take the prime $p=109$ and apply the FUNDAMENTAL RESULT.

CASE $n=112$)
SOLVED BEFORE ($n+1=113$ is an odd prime).

CASES $n=113,114$)
Take the prime $113$ and apply the FUNDAMENTAL RESULT.

LAST CASE $n\geq 115$)
Take a prime $p$ verifying $\frac n2<p\leq n$ (that exists by Bertrand's Postulate) and apply the FUNDAMENTAL RESULT.

                                                                                Q.E.D.


I decided to start a topic where everyone could share their tricks in doing ordinary math problems. I'm just getting started so if everyone could contribute their "secret" it would be great. Thanx! :lol:

My Trick of the Trade:

[hide][hide][hide]Wow, very desperate for tricks, eh? :) Tricks of the trade can only be learned through doing more problems. Even if you know a formula, you most likely won't remember when to use it even if you memorize it. This is the case when you don't understand the formula, and thus it is important to prove every formula you aren't familiar with once so that you may become familiar with it. [/hide][/hide][/hide]

What whitehorseking said, without all the spoilers.   :D

Yeah, there are quite a few tricks of the trade, but there are too many you can't really learn them well by reading them one or two times. You have to do some problems.